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\hypersetup{pdftitle={The spinless modular bootstrap at large central charge},pdfauthor={Claude (Anthropic)}}

\title{\bfseries The spinless modular bootstrap\\ at large central charge}
\author{Claude (Anthropic)}
\date{November 2026}

\begin{document}
\maketitle
\thispagestyle{empty}

\begin{abstract}
The spinless modular bootstrap bounds the dimension $\Delta_1$ of the lightest primary other than the vacuum in a
two-dimensional conformal field theory, using only modular $S$ invariance of the torus partition function at zero
angular potential.
For unitary, compact theories with central charge $c>1$ and no conserved currents beyond the Virasoro algebra, these
bounds read $\Delta_1\le c/\Lambda+o(c)$ at large $c$.
The best slope $\Lambda$ was known only to lie between $8.503$ and $12$, where the bound would reach the BTZ black hole
threshold.
We show that it is exactly $\Lambda_V=12\pi^2/(\pi^2+4\,\mathrm{arcsinh}^2 1)\approx9.1266$.
No linear functional in a class that contains all finite combinations of derivatives improves on $c/\Lambda_V$ by more
than $O(c^{1/3}\log^{2/3}c)$, and an explicit functional gives $\Delta_1<c/\Lambda_V+0.483\,c^{1/3}+O(1)$ for all
sufficiently large $c$.
The value has an electrostatic origin: as a function of a complexified dimension, $c^{-1}\log\lvert g\rvert$ of a
functional $g$ is a two-dimensional potential with positive charges at the zeros of $g$.
Positivity forces $g$ to fall as fast as the Cardy density of states grows, and the maximum principle bounds the
potential by an explicit harmonic function, which fixes $\Lambda_V$.
The field of this potential is the spectral curve of a mean-field theory of the optimal functionals, and its
charge density is the distribution that the zeros of any functionals attaining $\Lambda_V$ must follow.
We also conjecture the coefficient of the $c^{1/3}$ correction to the optimal bound, supported by numerics up to
$c=2048$.

\end{abstract}

{\small\tableofcontents}
\bigskip\hrule\bigskip

\section{Introduction}
\label{sec:1}
\suppressfloats[t]%

\subsection{The question}
\label{sec:1.1}

Modular invariance of the torus partition function constrains the spectrum of every two-dimensional conformal field
theory (CFT), and the simplest question one can put to it is how heavy the lightest state can be. Let $\Delta_1$ be the
scaling dimension of the lightest primary of the Virasoro algebra, other than the vacuum, in a unitary compact CFT of
central charge $c>1$. Modular invariance bounds $\Delta_1$ from above by a function of $c$ alone~\cite{Hellerman}.

At large $c$ this is a question about quantum gravity in three-dimensional anti-de~Sitter space. In that language the
BTZ black holes~\cite{BTZ} begin at $\Delta=(c-1)/12$, the \emph{BTZ threshold}. A theory of pure
gravity~\cite{Witten3d}, in the sense of~\cite{MaloneyWitten,KellerMaloney}, would have no primary below the black holes other than the vacuum, so that
$\Delta_1\approx c/12$. The known upper bounds on $\Delta_1$ sit well above this value, and one would like to know how
close to the BTZ threshold the consistency of the theory forces its lightest state to be.

The \emph{spinless modular bootstrap} is the simplest rigorous tool for this question. One sets the angular potential
to zero, so that the partition function depends on the inverse temperature $\beta$ alone, primaries are counted by
their dimension whatever their spin, and of the modular group only $S\colon\beta\to1/\beta$ remains. To the resulting
identity one applies a linear functional, for instance a finite combination of derivatives with respect to $\beta$ at
the self-dual point $\beta=1$. If the functional has the right signs, the identity cannot hold when all primaries are
heavier than some value, and this value bounds $\Delta_1$ (\cref{sec:2.2}). At large $c$ every bound obtained in this
way has the form
\begin{equation}\label{eq:1-slope}
  \Delta_1\le\frac{c}{\Lambda}+o(c) ,
\end{equation}
with a number $\Lambda$, the \emph{slope}, that depends on how the functionals are chosen; a larger slope is a stronger
bound, and $\Lambda=12$ would reach the BTZ threshold. The question of this paper is: what is the best slope that this
method can \emph{ever} give, whatever functionals are used and whatever the computing budget?

\subsection{What was known}
\label{sec:1.2}

\Cref{tab:1} is the ladder of slopes obtained so far.
Hellerman~\cite{Hellerman} showed that every unitary compact CFT with $c>1$ and no conserved currents beyond the
Virasoro algebra has $\Delta_1\le c/6+O(1)$, with the
functional of lowest order that gives a bound at all (\cref{sec:2.2}): this is $\Lambda=6$.
Friedan and Keller~\cite{FK} made two observations. First, no functional can exclude a value of $\Delta_1$ below the
BTZ threshold, and so $\Lambda\le12$; they obtain this by an argument of Friedan, Konechny and
Schmidt-Colinet~\cite{FKS}: under $S$ a character below the threshold becomes a continuum of characters above it, with
positive density. In \cref{sec:2.4} the argument is given in terms of the vacuum: its image under $S$ is a continuum of
primaries supported above the BTZ threshold, and the vacuum together with this continuum satisfies the modular
constraint by itself.
Second, they derived heuristically, to leading order in $1/c$ and with numerics, that
with any fixed number of derivatives the slope stays at $6$ as $c\to\infty$: to improve on Hellerman the number of
derivatives must grow with $c$. This statement has been restated without proof, for example by Collier, Lin and
Yin~\cite{CLY}; the analogous statement for the $U(1)^c$ problem, described below, was proved by Cohn, Dong and
Gon\c{c}alves~\cite{CDG}; to our knowledge the Virasoro statement has not been proved.

\begin{table}[t]
\centering\small
\setlength{\tabcolsep}{4pt}
\begin{tabular}{@{}>{\raggedright\arraybackslash}p{0.36\linewidth}ll>{\raggedright\arraybackslash}p{0.2\linewidth}@{}}
\toprule
 & $\Lambda$, Virasoro & $\Lambda$, $U(1)^c$ & status (Virasoro)\\
\midrule
functional of lowest order, every $c>1$ & $6$~\cite{Hellerman} & --- & proved\\
any fixed order, $c\to\infty$ & $6$~\cite{FK} & --- & heuristic, numerical\\
analytic functionals, $c\to\infty$ & $8.503$~\cite{HMR} & $8.856$~\cite{HMR} & proved\\
earlier numerical extrapolations & $9.08$~\cite{AHT} & $9.869$~\cite{ACHLT} & numerical\\
\addlinespace
our extrapolation, $24\le c\le2048$ & $9.1263\pm0.002$ (\cref{sec:5}) & --- & numerical\\
certified by an explicit functional for all large $c$ & $\LV=9.1266$ (\cref{thm:5}) & $\pi^2$~\cite{OpenAI}$^\dagger$
  & proved, computer-assisted\\
no functional in the class $\Abar$ beyond & $\LV$ (\cref{thm:4}) & $\pi^2$~\cite{OpenAI}$^\dagger$ & proved\\
no functional of any kind beyond & $12$~\cite{FKS,FK} & $4\pi$~\cite{Edwin} & proved (\cref{sec:2.4})\\
\bottomrule
\end{tabular}
\caption{The ladder of slopes $\Lambda$ in $\Delta_1\le c/\Lambda+o(c)$ from the spinless modular bootstrap, for the
Virasoro problem and for its twin with $U(1)^c$ characters (\cref{sec:2.3}).
Entries that point to a section or a theorem are results of this paper; the last column is the status of the Virasoro
entry, and for our results it is the tag of \cref{sec:1.3}, where the class $\Abar$ is introduced. $^\dagger$The status
of~\cite{OpenAI} is described in the text; its bound holds for every integrable eigenfunction, and the last row of the $U(1)^c$
column gives the limit that is proved independently of it.}
\label{tab:1}
\end{table}

With the number of derivatives growing, the optimal functional is found by linear programming. Collier, Lin and
Yin~\cite{CLY} saw the slope $\dd\Delta_1/\dd c$ of the resulting bound decrease with $c$ toward a value between
$\frac19$ and $\frac1{12}$.
Afkhami-Jeddi, Hartman and Tajdini~\cite{AHT} reached $c=1800$. Their best bound that involves no extrapolation is
$\Delta_1\le c/8.956$ at $c=1800$; their extrapolation to $c=\infty$, from a fit with a $\log c$ term, is $\Lambda=9.08$.
They state the estimate as $\Delta_1\lesssim c/9.1$ and remark that the extrapolation ``appears to be reliable, but strictly
speaking it is impossible to rule out a stronger bound as $c\to\infty$''.
Hartman, Maz\'a\v{c} and Rastelli~\cite{HMR} adapted to this problem the \emph{analytic functionals} of the four-point bootstrap
on a line (see below), that is, functionals given by a
closed-form integral kernel in $\beta$ rather than by finitely many derivatives at $\beta=1$, and obtained the slope
$8.503$. They also brought out the structure that we use throughout: a functional is an eigenfunction of a Fourier
transform, as already noted for Virasoro in~\cite[App.~C]{FK}, and when the Virasoro characters are replaced by those of $U(1)^c$ the problem becomes the $-1$ eigenfunction form of the linear program of
Cohn and Elkies~\cite{CohnElkies} for sphere packings in dimension $2c$ (\cref{sec:2.3}).

This twin problem is better understood. Afkhami-Jeddi, Cohn, Hartman, de~Laat and Tajdini~\cite{ACHLT} found the slope
$9.869$ numerically and conjectured that it is $\pi^2$; the analytic functionals of~\cite{HMR} give $8.856$. The slope
$\pi^2$ has since been proved in~\cite[Ch.~1]{OpenAI}; we have read the statements and the structure of that argument
but have not independently verified it; nothing in this paper depends on it logically.

For the Virasoro problem the state of the art before this work was $8.503\le\Lambda\le12$ for the best slope, with an
ambiguous extrapolation: a fit with a $c^{1/3}$ term in place of $\log c$ describes the same data equally well and
extrapolates to a different slope (\cref{sec:5.2}).

\paragraph{Related work.}
The method of linear functionals was introduced for four-point functions in~\cite{RRTV}. Applied to the torus partition
function, it also bounds the dimensions of the next primaries~\cite{QuallsShapere}, the entropy and the number of
marginal operators~\cite{HellermanSC}, the spectrum of sigma models on Calabi--Yau manifolds~\cite{KellerOoguri} and
the dimension of the lightest charged state~\cite{BDFK,DFX}; the entropy of a conformal boundary is bounded in the same
way~\cite{FKS}. Constraints that depend on spin are studied in~\cite{CLY,BaeLeeSong1,AMS,BOSW1}; in particular the
twist gap is at most $(c-1)/12$~\cite{CLY,AMS}. Integrality of the degeneracies is imposed
in~\cite{KaidiPerlmutter,FitzpatrickLi}. Modular invariance at genus two constrains the three-point coefficients as
well~\cite{CCY,KMZ}. At high energies invariance under $S$ gives the Cardy growth of the density of
states~\cite{Cardy86}; Tauberian theorems make this rigorous and bound the number of operators in a window of
dimensions~\cite{MZ,GangulyPal,MukhametzhanovPal}, and at large $c$ with a sparse light spectrum the Cardy growth
holds for all $\Delta\ge c/6$~\cite{HKS}.
The analytic functionals of~\cite{HMR} are those constructed earlier for the four-point function of a conformal field
theory on a line~\cite{Mazac,MazacPaulos1,MazacPaulos2}; functionals of this kind have been extended to higher
dimensions~\cite{Paulos1,MRZ,CHMRS} and used as a basis for numerics~\cite{PaulosZan}. It is shown in~\cite{HMR} that
in the twin problem at $c=4$ and $c=12$ they are the parts with $\hat f=-f$ of the functions with which
Viazovska~\cite{Viazovska} and Cohn, Kumar, Miller, Radchenko and Viazovska~\cite{CKMRV} proved that the $E_8$ and Leech lattices are the densest sphere packings
in dimensions $8$ and $24$, and that the complete basis of~\cite{MazacPaulos2} underlies the Fourier interpolation
formulas of~\cite{CKMRV2}, which followed the formula of Radchenko and Viazovska in one
dimension~\cite{RadchenkoViazovska}.
The linear program of Cohn and Elkies is always at least as strong~\cite{CohnZhao} as the bound of Kabatiansky and
Levenshtein~\cite{KL78}, which at large dimension is the slope $9.797$ in the present variables~\cite{ACHLT}. In
the other direction, feasible points of the dual program built from modular forms limit what the linear program can
prove: it is not sharp in dimension~$6$~\cite{dCIDV}, and it stays far from the densest known packings in several
dimensions from $12$ to $32$~\cite{CohnTriantafillou}. At large dimension the linear program cannot prove a density below
$2^{-(0.7786\ldots+o(1))d}$~\cite{TorquatoStillinger} (see also~\cite{ScardicchioEtAl}). For functions with $\hat f=-f$ the twin problem is the $-1$
eigenfunction uncertainty principle of Cohn and Gon\c{c}alves~\cite{CohnGoncalves}, the counterpart of the principle
of Bourgain, Clozel and Kahane~\cite{BCK,GOS}: in dimension $d$ the last sign change lies at a radius larger than
$\sqrt{d/2\pi e}$~\cite{CohnGoncalves}, and for $d\ge5$ at a radius of at least $\sqrt{d/4\pi}$~\cite{Edwin}, which with $d=2c$ and
$\Delta=r^2/2$ is the entry $4\pi$ of \cref{tab:1}. Both bounds hold for every integrable eigenfunction, and the bound
of~\cite[Ch.~1]{OpenAI} that gives $\pi^2$ is stated in the same generality. Extremal functions exist
and vanish at infinitely many radii beyond the last sign change~\cite{CohnGoncalves}.
Free bosons on a torus, the Narain compactifications, have the current algebra of the twin problem; their gap is
bounded with spin at small $c$ in~\cite{ACHT}, and their average over moduli space is computed
in~\cite{ACHT,MaloneyWittenNarain}. For conformal field theories in higher dimensions on a spatial torus, a modular
transformation gives a Cardy formula~\cite{Shaghoulian1,Shaghoulian2}, with an extended range of validity at large $N$
under conditions of sparseness~\cite{BelinEtAl}.

\subsection{Results}
\label{sec:1.3}

In one sentence: the best slope is $\LV=9.1266\ldots$. No functional in a class that contains all those of finite order
does better (\cref{thm:3,thm:4}), and an explicit functional does as well for all sufficiently large $c$
(\cref{thm:5}, computer-assisted).

\paragraph{The words of the problem.}
The terms are defined in \cref{sec:2}. We write $c'=c-1$, $a=c'/12$ and
$t=\Delta-a$ (\cref{sec:2.1}), so that the \emph{BTZ threshold} is $t=0$ and the vacuum is at $t=-a$. An \emph{admissible
spectrum} is one that obeys our hypotheses on the CFT: unitary, compact, $c>1$, invariant under $S$, no conserved
currents beyond the Virasoro algebra (\cref{def:2-admissible}). A functional is a function $g(t)$, its value on a
primary of dimension $t+a$. It is \emph{feasible at $T$} if $g\ge0$ on $[T,\infty)$ and its value $\gvac$ on the vacuum is
strictly positive, and \emph{weakly feasible} if $\gvac\ge0$. Applied to the partition function term by term, a
functional feasible at $T$ shows that $\Delta_1<T+a$ (\cref{sec:2.2}); this is immediate for finitely many derivatives
and is \cref{thm:2} in general. The last sign change $t_1(g)$ of $g$ is the \emph{gap bound} that $g$ certifies
(\cref{def:2-feasible}). The numerically optimal functionals are negative on the \emph{lobe interval} $[0,t_1)$, up to a
sliver next to the threshold, and have a \emph{band} of double zeros beyond $t_1$ (\cref{sec:2.5}, \cref{fig:2}). The
\emph{Cardy weight} $V(t)$, which grows like $e^{4\pi\sqrt{at}}$, is $1/2\pi$ times the density of primaries in the
$S$-image of the vacuum, and $M[g]=\int_0^\infty\abs g\,V\,\dd t$ is the \emph{Cardy-weighted mass} of $g$
(\cref{sec:2.4}). In a plot of $\frac1{c'}\log\abs g$ against $x=\sqrt{3t/c'}$ the \emph{Cardy line} is the straight
line along which a functional decays exactly as fast as this density grows (\cref{sec:2.5}). Each result carries a status
tag (legend at the end of this subsection).

\smallskip\noindent\textbf{\Cref{thm:1} (the slope)} \tagpar{\provedCA}.
Let $\Abar$ be the class of \cref{def:2-Abar}: the continuous integrable functions $g$ that the Fourier transform of
\cref{sec:2.3} maps to $-g$ and that have finite Cardy-weighted mass $M[g]$. It contains every functional of finite
order (\cref{lem:2.1}). Let $\Dbest(c)$ be the best bound on $\Delta_1$ that feasible functionals in $\Abar$ certify
(\cref{sec:2.6}). Then
\begin{equation}\label{eq:1-thm1}
  \lim_{c\to\infty}\frac{c}{\Dbest(c)}=\LV=\frac{12\pi^2}{\pi^2+4\arcsinh^21}=9.12663\ldots\,.
\end{equation}
Equivalently, the best gap bound, measured from the BTZ threshold, is $\xis\,(c-1)\,(1+o(1))$, where
$1/\LV=\frac1{12}+\xis$ and $\xis=(\xs)^2/3=0.0262361$, with $\xs=\arcsinh(1)/\pi$ the solution of $\sinh\pi x=1$. The
hyperbolic sine comes from electrostatics on a strip of unit width in the plane of complex $x=\sqrt{3t/c'}$:
$\sinh\pi x=1$ is where the potential that bounds $\frac1{c'}\log\abs g$ for every functional in $\Abar$ first touches
the Cardy line (\cref{sec:1.4}). On the word ``best'': $\Dbest(c)$ is
at most the limit, as the order $P\to\infty$, of the order-$P$ optimal gap bounds that numerical work computes; equality
is conjectural (\cref{sec:2.6}).

\smallskip\noindent\textbf{\Cref{thm:2} (certification)} \tagpar{\proved}.
For $c\ge40$, every $g\in\Abar$ that is feasible at some $T>0$ gives $\Delta_1<T+a$ for every admissible spectrum,
although $g$ need not be of finite order. Hence $\Dbest$ is a bound on theories, not only the value of a linear program
(\cref{sec:2.6,sec:8}).

\smallskip\noindent\textbf{\Cref{thm:3,thm:4} (optimality)} \tagpar{\proved}.
For every $g\in\Abar$, whatever its signs, the part of the Cardy-weighted mass $M[g]$ that lies below
$t=(\xis-\epsilon)\,c'$ is a fraction of $M[g]$ exponentially small in $c$, for every $0<\epsilon<\xis$ and
$c\ge40$ (\cref{thm:3}). But a weakly feasible functional must carry at least half of its mass below $t_1$. Hence no
weakly feasible functional in $\Abar$, other than zero, has $t_1<\xis c'-0.44\,c'^{1/3}\log^{2/3}c'$ once $c\ge\copt$
(\cref{thm:4}; both in \cref{sec:6}).

\smallskip\noindent\textbf{\Cref{thm:5} (attainment) and \cref{cor:6}} \tagpar{\provedCA}.
An explicit functional in $\Abar$, built from a product over prescribed zeros (\cref{sec:7.2}), is feasible at
$\xis c'+0.483\,c'^{1/3}+O(1)$ for $c\ge\catt$ (\cref{thm:5}, \cref{sec:7}). Hence every admissible CFT with $c\ge\catt$
has $\Delta_1<c/\LV+0.483\,c^{1/3}+O(1)$ (\cref{cor:6}, \cref{sec:8}).

\smallskip\noindent\textbf{\Cref{thm:7} (structure)} \tagpar{\proved}.
Any sequence of weakly feasible functionals in $\Abar$ that attains $\LV$ has its zeros equidistributed according to one
explicit law $\rhoU$ (\cref{sec:9}); the subscript refers to the extremal potential $U$ of \cref{sec:6}, not to $U(1)$.

\smallskip\noindent\textbf{\Cref{conj:1} (fine structure)} \tagpar{\conjtag}.
$\Dbest(c)=c/\LV+0.27666\,c^{1/3}+O(1)$, with no $\log c$ term at order $c^{1/3}$ or above, and the anatomy of the optimal
functional (\cref{sec:4.6}). Numerics for $24\le c\le2048$ support it and extrapolate to
$\Lmeas=9.1263\pm0.002$ (\cref{sec:5}).

\paragraph{What the result does and does not say.}
The window
\begin{equation}\label{eq:1-window}
  \frac{c-1}{12}<\Delta_1<\frac{c-1}{12}+\xis\,(c-1) ,
\end{equation}
which is the lower $31\%$, in units of $c/12$, of the range $c/12\le\Delta\le c/6$, is out of reach of spinless
$S$-invariance used through functionals in $\Abar$, whatever the computing budget. Nothing is claimed about whether
conformal field theories live there, nor about constraints that depend on spin, about integrality of the degeneracies,
or about pure gravity. Four caveats apply. The central charges $\copt$ and $\catt$ are not computed, and at every $c$
accessible to numerics the numerical bounds remain the best available. The attainment proof is computer-assisted, in
the sense of the legend. The constant $0.483$ is not the best one. Finally,
\cref{thm:3,thm:4,thm:7} are statements about the method, not about which conformal field theories exist.

\paragraph{Legend of the status tags.}
A tag on a statement or on a section applies to all of it.
{\small
\begin{description}[style=multiline,leftmargin=0.25\linewidth,font=\normalfont,itemsep=1pt,parsep=0pt,topsep=4pt]
\item[\tagpar{\proved}] Proved in this paper from classical facts that are cited to textbooks.
\item[\tagpar{\provedCA}] Proved in this paper, modulo finitely many explicitly stated inequalities, independent of $c$,
  for elementary functions of one variable and, in one case, integrals of such functions, that are certified in interval
  arithmetic: rigorous floating-point arithmetic with outward-rounded error bounds.
\item[\tagpar{\conjtag}] A precise statement, supported by heuristics and numerics, not proved.
\item[\tagpar{\heuristic}] A derivation with named uncontrolled steps.
\item[\tagpar{\numerical}] The result of a computation, with a stated error estimate.
\item[\tagpar{\orientation}] Explanatory material that no proof uses.
\item[\cert] A mark, not a tag, on an individual $c$-independent numerical inequality: it is one of the certified
  inequalities, and it has a row in \cref{tab:4} (\cref{app:G}).
\end{description}}

\subsection{The ideas in one page}
\label{sec:1.4}

\paragraph{(i) The problem (\cref{sec:2}).}
In the variable $t=\abs{\br}^2/2$ a functional is a radial function in a plane that the Fourier transform maps to minus
itself. Because of this property its value on the vacuum, which sits at negative $t$, equals minus its integral over
$t\ge0$ against the Cardy density: the vacuum acts on a functional as a continuum of heavy states. If $g$ is feasible, its
negative part lies below $t_1$ and must outweigh in this integral everything beyond, so it carries at least half of the
Cardy-weighted mass, \eqref{eq:2-halfmass}. The weight grows like $e^{4\pi\sqrt{at}}$, and a function that is its
own Fourier transform up to a sign resists falling that fast: this tension is the whole problem.

\paragraph{(ii) The heuristic picture (\cref{sec:3,sec:4}).}
At large $c$ the optimal functional has the form of a WKB wave function, without oscillations on the lobe interval and
oscillating on the band, where it has double zeros with some density. The Fourier transform acts on such wave functions
by a quarter-turn of phase space. That the functional is mapped to minus itself, and that on the band it runs along the
Cardy line, together fix the density of zeros through a self-consistent equation of mean-field type; for $U(1)^c$ it is
literally the Curie--Weiss equation of a ferromagnet (\cref{res:H1}). The band begins at the critical point of this
equation, where the density vanishes, and its position gives the slope. For Virasoro the solution is the curve
\begin{equation}\label{eq:1-curve}
  \sinh\pi b=i\,\sinh\pi p
\end{equation}
in the plane of $b=\sqrt{3t/c'}$ and its conjugate momentum $p$ (\cref{res:H2}); its critical point is $\sinh\pi b=1$,
that is, $b=\xs=\arcsinh(1)/\pi$ and $\Lambda=\LV$. The derivation rests on named hypotheses, one of which, a selection
rule, we cannot derive.

\paragraph{(iii) Optimality (\cref{sec:6}).}
As a function of $b$, every functional in $\Abar$ continues analytically to the strip $\abs{\Im b}\le\half$, whose edges
pass through the vacuum, $b=\pm\frac i2$ (\cref{fig:3}). There $\frac1{c'}\log\abs g$ is a two-dimensional electrostatic
potential with positive point charges at the zeros of $g$, in the sign convention, used throughout, in which the
potential dips to $-\infty$ at a positive charge. Such a potential has no maximum inside a domain (the maximum
principle); it is \emph{subharmonic}. Measure $g$ in units of $M[g]$. On the edges the potential is at most $o(1)$, because there $g$ is,
as at the vacuum in (i), an integral of itself against a kernel with exactly Cardy growth; on the real axis beyond any
point $s$ it lies, in the mean, on or below the Cardy line, because $M[g]$ is finite. The maximum
principle then bounds it inside by an explicit harmonic function on the strip slit along $\abs x\ge s$. For $s=\xs$ this
function, the \emph{extremal potential} $U$, has the $p(b)$ of \eqref{eq:1-curve}, up to a constant factor, as its
complex electric field; it lies strictly below the Cardy line for $\abs x<\xs$ and first touches it at $\xs$. Since
$\abs g\,V$ is the exponential of $c'$ times the height of the potential above the Cardy line, a functional has
exponentially little mass below any $x<\xs$, and by (i) its last sign change cannot lie there. Signs enter
only in this last step, and no computer is used. The strategy is that of the proof for $U(1)^c$
in~\cite[Ch.~1]{OpenAI}; its status is described in \cref{sec:1.2,sec:6.1}.

\paragraph{(iv) Attainment (\cref{sec:7,sec:8}).}
The potential $U$ is produced by a definite density of charge $\rhoU$ on the slits, and this is a blueprint for a
functional: we place zeros at the quantiles of this density. A zero enters through a factor whose logarithm is the
potential of a point charge together with its images in the edges of the strip; the factor has modulus one on the
edges. The product $h_0$ of these factors and of a fixed function without zeros is not an eigenfunction of the Fourier
transform, but $g=\half(h_0-\hat h_0)$ is, and $\hat h_0$ is computable at large $c$ by steepest descent along an edge of
the strip, where the size of $h_0$ is known exactly and the zeros enter only through a phase. On the band $h_0\ge0$, and
$g\ge0$ reduces to an inequality between two numbers that do not depend on $c$: the relative amplitude of $\hat h_0$,
about $\half$, and the smallest value of $h_0$ on the band in the same units, which exceeds $\half$ once the zeros have
been moved slightly off the real axis in conjugate pairs. This inequality, and a similar one for $\gvac$, is what the
computer certifies. The resulting functional is not of finite order; \cref{thm:2}, proved by a convergence argument,
makes it a bound on conformal field theories. A second, quite different functional that attains the slope, the imaginary
part of a rotation of phase space by $45^\circ$ applied to one real function, is described in \cref{sec:alt}; it
needs no computer and gives no rate.

\smallskip
The curve \eqref{eq:1-curve} thus appears four times: as a heuristic spectral curve in (ii), as the
field of the extremal potential in (iii), and as the blueprint for the zeros in (iv). The fourth time it is a
necessity (\cref{sec:9}): limits of such potentials exist along subsequences, and a sequence of functionals that
reaches $\LV$ must converge to $U$, charges included.

\PartHeading{I}{The problem}\section{The spinless modular bootstrap}
\label{sec:2}

This section states the problem and fixes the language of the paper. It ends with the class of functionals that we
consider and the statements of the two main theorems (\cref{sec:2.6}).

\subsection{Partition function and crossing equation}
\label{sec:2.1}

The idea is to keep, of modular invariance, only the one equation that involves scaling dimensions alone, and to bring
it to a form in which every primary other than the vacuum contributes a single exponential with a positive
coefficient.

We consider a two-dimensional conformal field theory that is unitary and compact, with central charge $c>1$ in both
chiralities, a unique normalizable vacuum and a discrete spectrum. We assume that it has no conserved currents beyond
the Virasoro algebra, so that every primary other than the vacuum has weights $h,\bar h>0$ and a Virasoro module
without null states. Its torus partition function is invariant under $\tau\to-1/\tau$. We restrict to zero angular
potential, $\tau_{\rm torus}=i\beta$ with $\beta>0$ the inverse temperature. A state is then weighted by
$e^{-2\pi\beta(\Delta-c/12)}$ with $\Delta=h+\bar h$: the spin $h-\bar h$ drops out, primaries of all spins are counted by
their dimension $\Delta$ alone---hence \emph{spinless}---and of the modular group only $S\colon\beta\to1/\beta$ survives.

Let $q=e^{-2\pi\beta}$ and let $\eta$ be the Dedekind function. A primary with $h,\bar h>0$ contributes
$q^{\Delta-(c-1)/12}/\abs{\eta(i\beta)}^2$ to the partition function $Z(\beta)$; the vacuum, whose descendants
$L_{-1}\lvert0\rangle$ and $\bar L_{-1}\lvert0\rangle$ are null, contributes
$q^{-(c-1)/12}(1-q)^2/\abs{\eta(i\beta)}^2$. The exponents carry $c-1$ rather than $c$ because each factor $1/\eta$ brings
its own $q^{-1/24}$. Since $\abs{\eta(i/\beta)}^2=\beta\,\abs{\eta(i\beta)}^2$, multiplying $Z$ by
$\abs{\eta(i\beta)}^2$ strips off all descendants at the price of one power of $\beta$ in the transformation law. We set
\begin{equation}\label{eq:2-vars}
  c':=c-1,\qquad a:=\frac{c'}{12},\qquad t:=\Delta-a ,
\end{equation}
and define the \emph{reduced partition function}
\begin{equation}\label{eq:2-Zhat}
  \Zhat(\beta):=\abs{\eta(i\beta)}^2Z(\beta)
  =e^{2\pi\beta a}\bigl(1-e^{-2\pi\beta}\bigr)^2+\sum_{n\ge1}d_n\,e^{-2\pi\beta t_n},
\end{equation}
where $d_n>0$ is the number of primaries with $\Delta=t_n+a$. Modular invariance becomes the \emph{crossing equation}
\begin{equation}\label{eq:2-crossing}
  \Zhat(\beta)=\beta^{-1}\,\Zhat(1/\beta)\qquad\text{for all }\beta>0 .
\end{equation}
The reduced partition function is the right object because \eqref{eq:2-crossing} relates sums of pure exponentials in
which the unknowns $d_n$ enter linearly and are positive; nothing else about the theory will be used. The central charge
enters only through $a$. The vacuum sits at $t_0:=-a$, and it is the only term that is not a single positive
exponential: its null states produce the factor $(1-e^{-2\pi\beta})^2=1-2e^{-2\pi\beta}+e^{-4\pi\beta}$. The point
$t=0$, that is $\Delta=(c-1)/12$, is the \emph{BTZ threshold}.\footnote{The classical BTZ value is $\Delta=c/12$; the
threshold $(c-1)/12$ includes the one-loop shift $c\to c-1$.}

The hypothesis on currents is what makes every $d_n$ multiply a single exponential. A conserved current $J$ of weights
$(s,0)$ has the null descendant $\bar L_{-1}J=0$; its module contributes
$d_J\,(e^{-2\pi\beta t_J}-e^{-2\pi\beta(t_J+1)})$ to $\Zhat$ and enters every sum rule below as $d_J[g(t_J)-g(t_J+1)]$,
which has no definite sign.

\begin{definition}\label{def:2-admissible}
An \emph{admissible spectrum} at central charge $c$ is a countable family $\{(t_n,d_n)\}_{n\ge1}$ with $d_n>0$ and
$t_n>-a$ (unitarity, $\Delta_n>0$) that is locally finite (every bounded interval contains finitely many $t_n$) and
such that $\Zhat(\beta)$ in \eqref{eq:2-Zhat} is finite and satisfies \eqref{eq:2-crossing} for every $\beta>0$.
Primaries with $t_n<0$ are called \emph{light}; there are finitely many of them. An \emph{admissible CFT} is one whose primaries form an
admissible spectrum; its \emph{gap} is $\Delta_1:=a+\min_nt_n$.
\end{definition}

\begin{remark}[\orientation]\label{rem:2-liouville}
In Liouville notation, $c=1+6Q_L^2$ and $h=Q_L^2/4+P_L^2$, so that $h-c'/24=P_L^2$ and a primary with momenta
$(P_L,\bar P_L)$ has $t=P_L^2+\bar P_L^2$. For a scalar, the variable $b:=\sqrt{3t/c'}$ that we use at large $c$
(\cref{sec:2.5}) is $b=P_L/Q_L$, and the vacuum, $h=\bar h=0$, sits at $b=\pm i/2$. Thus $b$ is a rescaled momentum; it
is not the Liouville coupling.
\end{remark}

\subsection{Linear functionals, bounds, and duality}
\label{sec:2.2}

The idea is that a linear functional turns the crossing equation into a single linear relation, a \emph{sum rule},
among the values of one function $g$ at the dimensions of the primaries. If $g$ is non-negative at every dimension that
an assumed gap allows, and positive on the vacuum, the relation cannot hold, and the assumed gap is excluded.

Let $\omega$ be a linear functional acting on functions of $\beta$, for instance a finite combination of derivatives at
a point (the general class is fixed in \cref{sec:2.6}). To $\omega$ we associate the function
\begin{equation}\label{eq:2-g}
  g(t):=\omega\bigl[\Xib(t)\bigr],\qquad \Xib(t):=e^{-2\pi\beta t}-\beta^{-1}e^{-2\pi t/\beta},
\end{equation}
where $\Xib$ is the \emph{crossing kernel}: \eqref{eq:2-crossing} says that the vacuum and the primaries, weighted by
$d_n$, sum to zero when inserted in $\Xib$. Applying $\omega$ term by term therefore gives the sum rule
\begin{equation}\label{eq:2-sumrule}
  \gvac+\sum_{n\ge1}d_n\,g(t_n)=0,\qquad \gvac:=g(-a)-2g(-a+1)+g(-a+2).
\end{equation}
The \emph{vacuum value} $\gvac$ is what $\omega$ returns on the vacuum term of \eqref{eq:2-Zhat}; it is a second
difference because the null-state factor spreads the vacuum over the three exponentials at $t=-a$, $-a+1$, $-a+2$. For a
functional of finite order the exchange of $\omega$ with the sum over primaries is termwise differentiation of a
convergent series of exponentials; for the general class it is the content of \cref{thm:2}. From now on we identify a
functional with its function $g$.

\begin{definition}\label{def:2-feasible}
Let $T\in\RR$. A functional $g$ is \emph{feasible at $T$} if $g(t)\ge0$ for all $t\ge T$ and $\gvac>0$. It is
\emph{weakly feasible} at $T$ if instead $\gvac\ge0$. The number $t_1(g):=\inf\{T:g\ge0\text{ on }[T,\infty)\}$ is the
last sign change of $g$; we call it the \emph{gap bound certified by} $g$. Without a named $T$, (weakly) feasible means
(weakly) feasible at some $T$.
\end{definition}

If $g$ is feasible at $T$ and all primaries had $t_n\ge T$, the left side of \eqref{eq:2-sumrule} would be at least
$\gvac>0$. Hence some primary has $t_n<T$, that is, $\Delta_1<T+a$. A sign condition gives a bound because the unknown
degeneracies enter the sum rule only through their positivity.

Minimizing $T$ over pairs $(g,T)$ is a linear program: $g$ enters linearly, and there is one inequality,
$g(t)\ge0$, for each $t\ge T$, together with $\gvac\ge0$. The dual variables, one non-negative multiplier for each
inequality, form a spectrum. Concretely, the dual problem asks for points $t_n\ge T$ and weights $d_n>0$ that satisfy
the sum rule \eqref{eq:2-sumrule} for every $g$ in the search space: by the argument above such a spectrum excludes
every functional feasible at $T$, and at the optimal $T$ the best functional and this obstruction meet. By complementary slackness---a constraint that is not tight
carries zero dual weight---$d_n>0$ only where $g(t_n)=0$. At the optimum, therefore, $\gvac=0$ and $g$ vanishes on an
\emph{extremal spectrum} $\{t_n\}$ with positive weights $d_n$~\cite{EFM}: it has a simple zero at $t_1$, where it changes sign, and
double zeros beyond, because $g\ge0$ there. An extremal functional is weakly feasible only: it gives an
extremality statement, not a bound; bounds come from feasible functionals at $T$ slightly above $t_1$.

In practice one uses derivatives at the self-dual point~\cite{Hellerman,FK,CLY,AHT}, $\omega=\sum_{k=1}^{2P-1}\omega_k\,\partial_\beta^k\big|_{\beta=1}$.
Each derivative of $\Xib$ at $\beta=1$ is a polynomial in $t$ times $e^{-2\pi t}$; for instance
$\partial_\beta\Xib\big|_{\beta=1}=(1-4\pi t)e^{-2\pi t}=L_1(4\pi t)e^{-2\pi t}$, with $L_k$ the Laguerre polynomials.
In general
\begin{equation}\label{eq:2-deriv}
  g(t)=\pol_P(t)\,e^{-2\pi t},\qquad \pol_P\in\operatorname{span}\bigl\{L_{2k-1}(4\pi t):1\le k\le P\bigr\},
\end{equation}
for all real $t$, and we call $P$ the \emph{order} of the functional; why only odd Laguerre polynomials occur is
explained in \cref{sec:2.3}. We write $t_1^{(P)}(c)$ for the infimum of $t_1(g)$ over feasible functionals of order $P$
(with the value $+\infty$ if there are none), and $\Delta_1^{(P)}(c):=a+t_1^{(P)}(c)$ for the corresponding bound on the gap; both are non-increasing in $P$, with
limits $t_1^{(\infty)}(c)$ and $\Delta_1^{(\infty)}(c)$.

As an example, at order $P=1$ the only candidate, $g=(4\pi t-1)e^{-2\pi t}$, has $\gvac<0$ and gives no bound. At order
$P=2$, $\pol_2$ is a cubic, a combination of $L_1(4\pi t)$ and $L_3(4\pi t)$;
one of its two coefficients is a normalization and the other is fixed by $\gvac=0$. The resulting cubic is positive
beyond its largest root $t_1$, nearby combinations with $\gvac>0$ are feasible at $T$ arbitrarily close to $t_1$, and
\begin{equation}\label{eq:2-hellerman}
  t_1=a+\frac2\pi+\frac{2}{e^{2\pi}-1}-\epsilon_c,\quad 0<\epsilon_c=O(1/c),
  \qquad\text{so that}\qquad \Delta_1<\frac c6+0.4737 .
\end{equation}
This is Hellerman's bound~\cite{Hellerman}, valid for every $c>1$; its slope is
$c/\Delta_1\to6$. Two exact optima serve as anchors for conventions and numerics: at $c=4$ and $c=12$ the optimal gap
bound is exactly $\Delta_1\le1$ and $\Delta_1\le2$, and it is saturated by a solution of the crossing equation
\eqref{eq:2-crossing} with positive $d_n$ and integer dimensions~\cite{HMR}.

\subsection{Functionals are Fourier eigenfunctions}
\label{sec:2.3}

The idea is that the crossing kernel is a Gaussian minus its own Fourier transform, provided $t$ is read as
$\abs\br^2/2$ for a point $\br$ of a plane. Every functional is then a radial function on the plane that is odd under the
Fourier transform, and the linear program becomes a problem about such functions in which $\beta$ no longer
appears~\cite{FK,HMR}.

Let $\br\in\RR^2$, $t=\abs\br^2/2$, and take the Fourier transform in the convention
\begin{equation}\label{eq:2-fourier}
  \hat f(\mathbf y)=\int_{\RR^2}f(\br)\,e^{-2\pi i\,\br\cdot\mathbf y}\,\dd^2\br,\qquad\text{so that}\qquad
  e^{-\pi\beta\abs\br^2}\;\longmapsto\;\beta^{-1}e^{-\pi\abs{\mathbf y}^2/\beta}.
\end{equation}
The power $\beta^{-1}$ is $\beta^{-d/2}$ with $d=2$, and it is the power of $\beta$ in the crossing equation
\eqref{eq:2-crossing}, which came from the two Dedekind functions; this is what fixes the dimension of the plane.
In the variable $t$ the Gaussian is $e^{-2\pi\beta t}$ and its transform is $\beta^{-1}e^{-2\pi s/\beta}$ with
$s=\abs{\mathbf y}^2/2$. The crossing kernel is thus a function minus its transform, and since the transform is an
involution on radial functions, $\hat\Xi_\beta=-\Xib$. The functional $\omega$ acts on $\beta$ and commutes with the
transform, so
\begin{equation}\label{eq:2-eigen}
  \hat g=-g,\qquad\text{where}\qquad \hat g(s)=2\pi\int_0^\infty g(t)\,J_0\bigl(4\pi\sqrt{st}\bigr)\,\dd t
\end{equation}
is the planar transform of a radial function written in the variable $t$, as we always write it.

The Laguerre functions
\begin{equation}\label{eq:2-psik}
  \psi_k(t):=L_k(4\pi t)\,e^{-2\pi t},\qquad \hat\psi_k=(-1)^k\psi_k,\qquad k=0,1,2,\dots,
\end{equation}
are the rotation-invariant eigenstates of the two-dimensional harmonic oscillator, $\psi_k$ carrying $2k$ quanta. The
Fourier transform is the evolution of the oscillator through a quarter period, a quarter-turn
$(\br,\mathbf p)\mapsto(\mathbf p,-\br)$ of its phase space, which multiplies a state of $N$ quanta by $(-i)^N$; this
is the eigenvalue in \eqref{eq:2-psik} (for a computation see \cref{sec:8.2}). The condition $\hat g=-g$ selects the
odd $\psi_k$, and these are exactly the derivative functionals \eqref{eq:2-deriv}. The quarter-turn returns in
\cref{sec:4.1} as the symmetry of the spectral curve.

\begin{remark}[\orientation]\label{rem:2-skernel}
In the notation of \cref{rem:2-liouville}, $\br=\sqrt2\,(P_L,\bar P_L)$, and the planar Fourier kernel
$e^{2\pi i\,\br\cdot\br'}\,\dd^2\br'$ is the product of the left and right modular $S$-kernels
$\sqrt2\,e^{4\pi iP_LP_L'}\,\dd P_L'$ on non-degenerate characters: one direction for each chirality, hence dimension $2$ for
every $c$.
\end{remark}

\paragraph{The twin problem.} For a theory with $U(1)^c$ current algebra the same steps, with characters
$q^{\Delta}/\abs{\eta(i\beta)}^{2c}$ and $\Delta$ the dimension of a $U(1)^c$ primary, give
$\Zhat(\beta)=1+\sum_nd_ne^{-2\pi\beta\Delta_n}=\beta^{-c}\Zhat(1/\beta)$. The power $\beta^{-c}$ is that of a Gaussian in
$\RR^{2c}$: functionals are radial functions $f(\Delta)$, $\Delta=\abs{\mathbf x}^2/2$, $\mathbf x\in\RR^{2c}$, with
$\hat f=-f$; the vacuum sits at the origin, and $f$ proves $\Delta_1<T$ if $f(0)>0$ and $f\ge0$ on $[T,\infty)$. This is
the $-1$ eigenfunction form of the linear program of Cohn and Elkies for sphere packing in $d=2c$ dimensions~\cite{CohnElkies};
that the two have the same optimum in every dimension is their Conjecture~7.2, of which only the direction shown below is proved in
general. The dictionary is that
of~\cite{HMR}:
\begin{ideabox}[title={Theorem (Cohn--Elkies)}]
Let $h\colon\RR^d\to\RR$ be a radial Schwartz function, not identically zero, and $r>0$. If $h(\mathbf x)\le0$ for
$\abs{\mathbf x}\ge r$, $\hat h(\mathbf y)\ge0$ for all $\mathbf y$, and $h(0)=\hat h(0)>0$, then every sphere packing in
$\RR^d$ has density at most the volume of a ball of radius $r/2$.
\end{ideabox}
\noindent For such an $h$ the function $f=\hat h-h$ is radial, has $\hat f=-f$ and $f(0)=0$, and is non-negative for
$\abs{\mathbf x}\ge r$: it is a weakly feasible functional of the $U(1)^c$ problem at $T=r^2/2$. The two problems differ
in one respect. For $U(1)^c$ the central charge is the dimension of the space; for Virasoro the space is a plane for
every $c$, and $c$ enters only through the position $-a$ of the vacuum and its null-state factor. \Cref{sec:3} treats the
twin problem as a warm-up.

\subsection{The vacuum is a Cardy continuum}
\label{sec:2.4}

The vacuum enters the sum rule through values of $g$ at negative $t$, where no sign condition is imposed. The idea is
that modular invariance converts them into an integral over $t\ge0$: the $S$-transform of the vacuum is a continuum of
non-degenerate characters with the Cardy density, and acting with $g$ on the vacuum is integrating $-g$ against that
density. Feasibility becomes a contest between the negative and the positive part of $g$ under a growing weight.

The continuum comes from the Laplace transform of a Bessel function. For $\alpha\ge0$ and $\beta>0$,
\begin{equation}\label{eq:2-Kalpha}
  \beta^{-1}e^{2\pi\alpha/\beta}=2\pi\int_0^\infty K_\alpha(t)\,e^{-2\pi\beta t}\,\dd t,\qquad
  K_\alpha(t):=I_0\bigl(4\pi\sqrt{\alpha t}\bigr),
\end{equation}
so that the $S$-image of the vacuum term of \eqref{eq:2-Zhat} is
\begin{equation}\label{eq:2-rhoC}
  \beta^{-1}e^{2\pi a/\beta}\bigl(1-e^{-2\pi/\beta}\bigr)^2=\int_0^\infty\rhoC(t)\,e^{-2\pi\beta t}\,\dd t,\quad
  \rhoC:=2\pi V,\quad V:=K_a-2K_{a-1}+K_{a-2}.
\end{equation}
This is a continuum of non-degenerate characters supported on $t\ge0$, that is, above the BTZ threshold. Since
$I_0(z)\simeq e^z/\sqrt{2\pi z}$, the \emph{Cardy kernel} $K_\alpha$ grows like $e^{4\pi\sqrt{\alpha t}}$, and $\rhoC$ is the
Cardy density of states~\cite{Cardy86} with all its corrections. For $a\ge2$ the \emph{Cardy weight} $V$ has a power series with
non-negative coefficients, because $\alpha\mapsto\alpha^k$ is convex:
\begin{equation}\label{eq:2-Vseries}
  V(t)=\sum_{k\ge2}\frac{(4\pi^2t)^k}{(k!)^2}\bigl[a^k-2(a-1)^k+(a-2)^k\bigr]\;\ge\;8\pi^4t^2 .
\end{equation}
Thus $V$ is positive and increasing on $(0,\infty)$; further properties are collected in \cref{lem:A.2}.

Now let $g$ be a functional. The equation $\hat g=-g$ reads
$g(s)=-2\pi\int_0^\infty g(t)J_0(4\pi\sqrt{st})\,\dd t$ for $s\ge0$. For $g$ of the form \eqref{eq:2-deriv} both sides are entire
functions of $s$, because $J_0(4\pi\sqrt{st})$ is a power series in $st$; the equation therefore persists at
$s=-\alpha<0$, where $J_0(4\pi\sqrt{-\alpha t})=I_0(4\pi\sqrt{\alpha t})$. This is the \emph{eigen-identity}
\begin{equation}\label{eq:2-eigenid}
  g(-\alpha)=-2\pi\int_0^\infty g(t)\,K_\alpha(t)\,\dd t,\qquad 0\le\alpha\le a .
\end{equation}
The values of $g$ below the BTZ threshold are Bessel-weighted integrals of $g$ above it. The general statement, for the
class of \cref{sec:2.6} and for complex $\alpha$ in a parabolic region, is \cref{lem:A.1}. Taking the second difference
in \eqref{eq:2-sumrule},
\begin{equation}\label{eq:2-gvac-cardy}
  \gvac=-\int_0^\infty g(t)\,\rhoC(t)\,\dd t :
\end{equation}
\emph{acting on the vacuum is integrating $-g$ against the Cardy density.}

Let $g_\pm:=\max(\pm g,0)$. If $g$ is weakly feasible at $t_1=t_1(g)$, then $g_-$ is supported on $[0,t_1]$, and
\eqref{eq:2-gvac-cardy} with $\gvac\ge0$ says $\int g_-V\ge\int g_+V$. Adding $\int g_-V$ to both sides gives the
\emph{half-mass condition}
\begin{equation}\label{eq:2-halfmass}
  \int_0^{t_1}g_-(t)\,V(t)\,\dd t\;\ge\;\half\,M[g],\qquad M[g]:=\int_0^\infty\abs{g(t)}\,V(t)\,\dd t ,
\end{equation}
where $M[g]$ is the \emph{Cardy-weighted mass} of $g$. The negative part of $g$ on $[0,t_1)$ must outweigh everything
beyond $t_1$, although the weight grows like $e^{4\pi\sqrt{at}}$. This is possible only if $\abs g$ falls at least as fast
as the Cardy weight grows, and a function that is its own Fourier transform up to a sign resists falling fast. This
tension is the whole problem.

A first consequence is due to Friedan and Keller~\cite{FK}, who attribute the argument to Friedan, Konechny and
Schmidt-Colinet~\cite{FKS}. By \eqref{eq:2-rhoC} the vacuum together with the continuum
$\rhoC(t)\,\dd t$ solves the crossing equation by itself, with nothing below $t=0$. Accordingly, if $g\ge0$ on
$[0,\infty)$ then \eqref{eq:2-gvac-cardy} gives $\gvac\le0$, with equality only for $g\equiv0$: no nonzero functional
is even weakly feasible at $T\le0$, no bound reaches below the BTZ threshold, and the slope of any bound
$\Delta_1\le c/\Lambda+o(c)$ obeys $\Lambda\le12$.

\subsection{The large-\texorpdfstring{$c$}{c} regime}
\label{sec:2.5}

\Cref{fig:2} is a schematic picture of a numerically optimal functional at moderate $c$. This subsection introduces the
variables in which its features do not depend on $c$, and then lists the features.

\begin{figure}[t]
\centering
\figfile[0.78\linewidth]{fig2_anatomy}
\caption{Anatomy of an optimal functional at moderate $c$: a schematic, drawn for $c=48$ from the leading-order
formulas of \cref{sec:4.6}, not from data. Plotted is $\frac1{c'}\log\abs g$, normalized to $g(t_0)=1$, from the vacuum down to the BTZ threshold along $b=iy$,
then along the physical axis $b=x$ (top axis: $t$).
The curve is solid where $g>0$ and dashed where $g<0$; zeros appear as downward cusps. The functional is negative on the
lobe interval $[0,t_1)$, except for a positive sliver between the BTZ threshold and the interior zero
$\tint\approx1/4\pi$; it has a simple zero at $t_1$ and a band of double zeros beyond, on which the local upper envelope
of $\abs g$ falls like $e^{-c'L(x)}$: it runs along the Cardy line (dashed straight line). At $c=48$ the corrections to the limit
of large $c$ are still large: $x_1\approx0.39$ lies well above its limit $\xs=0.2805$, and part of the lobe lies above
the Cardy line, whereas the limiting profile stays below it up to $\xs$ (\cref{fig:7}(b)).}
\label{fig:2}
\end{figure}

All limits are taken in the order $P\to\infty$ at fixed $c$ first, then $c\to\infty$. The order matters: at fixed $P$ the
slope of the optimal bound falls back to Hellerman's $6$ as $c\to\infty$. This was derived heuristically, to leading
order in $1/c$ and with numerics, by Friedan and Keller~\cite{FK}. In the limit so defined the
natural variable is
\begin{equation}\label{eq:2-scaling}
  b:=\sqrt{3t/c'}=x+iy,\qquad t=\frac{c'b^2}{3},\qquad G(b):=g\bigl(c'b^2/3\bigr),\qquad x_1(g):=\sqrt{3\,t_1(g)/c'} ,
\end{equation}
with dimensions $t\ge0$ on the physical axis $b=x\ge0$. The reason for the square root is the Cardy growth: with
\begin{equation}\label{eq:2-L}
  L(x):=\frac{2\pi}3\abs x,\qquad e^{4\pi\sqrt{at}}=e^{c'L(x)},\qquad
  \lim_{c\to\infty}\frac1{c'}\log V\bigl(c'x^2/3\bigr)=L(x)\quad(x>0),
\end{equation}
the exponent of the Cardy weight is linear in $x$ and proportional to $c'$. In a plot of $\frac1{c'}\log\abs G$ against $x$
we call the straight line $-L(x)$ \emph{the Cardy line}; a functional that decays exactly as fast as the Cardy weight
grows runs parallel to it. The vacuum, $t=-a$, sits at $b=\pm\frac i2$, off the physical axis. \Cref{fig:3} shows the
$b$-plane and the strip $\abs{\Im b}\le\half$, on which the kernel of the eigen-identity of \cref{sec:2.4}, continued to
complex $t$, is bounded by $K_a$ and so grows no faster than the Cardy weight (\cref{lem:A.1}); it becomes
the stage in \cref{sec:4.2} and \cref{sec:6}. Finally, a family of functionals, one for each $c$, that are feasible at
$T=t_1(g)+o(c)$ with $x_1(g)\to x_1$ certifies $\Delta_1\le c/\Lambda+o(c)$ with
\begin{equation}\label{eq:2-slope}
  \frac1\Lambda=\frac1{12}+\xi_1,\qquad \xi_1=\frac{x_1^2}3 .
\end{equation}

\begin{figure}[t]
\centering
\figfile[\linewidth]{fig3_strip}
\caption{The $t$-plane and the $b$-plane, $t=c'b^2/3$, $b=x+iy$. (a)~The parabolic region of the $t$-plane whose vertex
is the vacuum, $t_0=-a$, and whose focus is the BTZ threshold, $t=0$. (b)~Its image, the strip
$\strip\colon\abs y<\half$ with edges $\Epm\colon y=\pm\half$. The vacuum is at $b=\pm\frac i2$; the dashed segment is the
range $-a<t<0$ below the BTZ threshold, and the physical axis is $x\ge0$. The band becomes the slits $\abs x\ge x_1$ of
the real axis, drawn here for $x_1=\xs$ of \cref{thm:1}. Diamonds mark the saddle points $b_0=\pm v_0(x)+\frac i2$ on the
edge $E_+$ that govern a functional at a point $x$ of the band (squares); see \cref{sec:4.2,sec:7.3}.}
\label{fig:3}
\end{figure}

In these variables the optimal functionals found numerically have the following anatomy (\cref{fig:2}). On the
\emph{lobe interval} $[0,t_1)$ the functional is negative; this negative \emph{lobe} carries the half mass in
\eqref{eq:2-halfmass}. The one exception is a positive sliver next to the BTZ threshold: $g>0$ on $(0,\tint)$ with
$\tint\approx1/4\pi$, which is explained in \cref{sec:4.6}. At $c=12$ this zero is already visible in the numerics: it is the
``additional single zero near $\Delta\approx1$'' of~\cite[Figure~1]{AHT} and the single root at $\Delta=1$ of~\cite[\S2.5]{HMR}.
At $t_1$ there is a simple zero. Beyond it lies the
\emph{band} of double zeros, the extremal spectrum of \cref{sec:2.2}, and on the band $\abs g$ follows $e^{-c'L(x)}$:
the functional runs along the Cardy line. These are the features to be explained.

\subsection{The class \texorpdfstring{$\Abar$}{A-bar}, the best gap bound, and the main theorems}
\label{sec:2.6}

To speak of all functionals, of any order, one needs a class that contains all of them and in which the vacuum value
still makes sense. We define it directly, by the three properties that the previous subsections have singled out.
\emph{Standing assumption:} in all rigorous statements $c\ge40$, so that $a\ge3.25$. It is used through the Bessel
inequalities of \cref{app:A}, and in \cref{prop:7.2}, where it keeps the poles of the comb of \cref{sec:7} out of the
closed strip.

\begin{definition}\label{def:2-Abar}
$\Abar$ is the set of real continuous functions $g$ on $[0,\infty)$ such that
\textup{(i)}~$g(\abs\br^2/2)\in L^1(\RR^2)$, that is, $\int_0^\infty\abs g\,\dd t<\infty$;
\textup{(ii)}~$\hat g=-g$, with the transform \eqref{eq:2-eigen}; and
\textup{(iii)}~$M[g]<\infty$.
For $g\in\Abar$ the values below the BTZ threshold are \emph{defined} by the eigen-identity \eqref{eq:2-eigenid},
$g(-\alpha):=-2\pi\int_0^\infty gK_\alpha\,\dd t$ for $0<\alpha\le a$, so that $\gvac=-\int_0^\infty g\,\rhoC\,\dd t$;
feasibility is then as in \cref{def:2-feasible}.
\end{definition}

Conditions (i) and (ii) say that $g$ is a functional in the sense of \cref{sec:2.3}, and (iii) is exactly the absolute
convergence of the vacuum action \eqref{eq:2-gvac-cardy}; the integrals that define $g(-\alpha)$ converge absolutely by
(i), (iii) and \cref{lem:A.2}. Nothing else is required: no smoothness, no rate of decay, no analytic continuation.

\begin{lemma}[\proved]\label{lem:2.1}
$\Abar$ contains the function $g$ of every functional $\omega$ that is a real distribution of finite order with compact
support in $\beta\in(0,\infty)$, with its values at $t<0$ given by \eqref{eq:2-g}. In particular it contains all
derivative functionals \eqref{eq:2-deriv}, for every order $P$.
\end{lemma}
\begin{proof}
Let $\omega$ have order $N$ and support in $(\beta_{\min},\beta_{\max})$, $0<\beta_{\min}<\beta_{\max}$. Then
$g=\omega[\Xib]$ is an entire function of $t$ with
$\abs{g(t)}\le C(1+\abs t)^Ne^{-2\pi\min(\beta_{\min},1/\beta_{\max})\Re t}$ for $\Re t\ge0$, and likewise for its
derivatives, so $g(\abs\br^2/2)$ is a radial Schwartz function; $\hat g=-g$ because $\hat\Xi_\beta=-\Xib$. Since
$V\le2K_a\le2e^{4\pi\sqrt{at}}$ grows more slowly than any exponential, $M[g]<\infty$. Finally \eqref{eq:2-Kalpha} gives
$-2\pi\int_0^\infty\Xib K_\alpha\,\dd t=\Xib(-\alpha)$ for every $\alpha\ge0$, and applying $\omega$ shows that the values
defined by the eigen-identity agree with the entire function.
\end{proof}

The \emph{best gap bound} is
\begin{equation}\label{eq:2-TV}
  \TV(c):=\inf\bigl\{T:\text{some }g\in\Abar\text{ is feasible at }T\bigr\},\qquad \Dbest:=\TV+a ,
\end{equation}
with the strict inequality $\gvac>0$ of \cref{def:2-feasible}, and with the value $+\infty$ if no $g\in\Abar$ is feasible
at any $T$; by \cref{sec:2.4}, $\TV\ge0$, and by \cref{thm:5}, $\TV(c)<\infty$ for all sufficiently large $c$. The subscript $V$ in
$\TV$ and $\LV$ stands for Virasoro, as opposed to the $U(1)^c$ problem; the Cardy weight $V(t)$ never carries a
subscript.

\begin{theorem}[The slope; \provedCA]\label{thm:1}
Let $\xs:=\arcsinh(1)/\pi=0.2805499$, $\xis:=(\xs)^2/3=0.0262361$ and
$\LV:=1/(\frac1{12}+\xis)=12\pi^2/(\pi^2+4\arcsinh^21)=9.1266340$. Then
\begin{equation}\label{eq:2-thm1}
  \lim_{c\to\infty}\frac{\TV(c)}{c'}=\xis ,\qquad\text{or equivalently}\qquad
  \lim_{c\to\infty}\frac{c}{\Dbest(c)}=\LV .
\end{equation}
\end{theorem}

The constants $\xs$, $\xis$ and $\LV$ are defined here for the whole paper; Part~II explains where they come from. The
theorem has two halves. The lower bound on $\TV$, the \emph{optimality half}, says that no nonzero functional in $\Abar$, even a
weakly feasible one, does better; it is \cref{thm:4}, proved in \cref{sec:6} by hand, with no computer. The upper
bound, the \emph{attainment half}, is \cref{thm:5}, proved in \cref{sec:7}: an explicit functional in $\Abar$ that is
feasible at $T=\xis c'+O(c'^{1/3})$. Its proof uses finitely many inequalities between elementary functions of one
variable, independent of $c$, that are certified in interval arithmetic and listed in \cref{app:G}; this is the only
input from a computer and the reason for the tag.

\begin{proof}[Proof of \cref{thm:1} from \cref{thm:4,thm:5}]
Let $c\ge\copt$, and let $g\in\Abar$ be feasible at $T$. Then $g\not\equiv0$ because $\gvac>0$, $g$ is weakly feasible,
and $T\ge t_1(g)$. So \eqref{eq:6-rate} gives $T\ge\xis c'-0.44\,c'^{1/3}\log^{2/3}c'$, and therefore
$\TV(c)\ge\xis c'-0.44\,c'^{1/3}\log^{2/3}c'$. For $c\ge\catt$ the function $g$ of \cref{thm:5} is feasible at
$\tstar+1$, so $\TV(c)\le\tstar+1=\xis c'+O(c'^{1/3})$ by \eqref{eq:7-tstar}. Dividing by $c'$ gives the first limit
in \eqref{eq:2-thm1}. Since $\Dbest=\TV+a$ and $a=c'/12$, it is equivalent to $\Dbest(c)/c'\to\xis+\frac1{12}=1/\LV$; since
$c/c'\to1$, this is the second limit.
\end{proof}

\begin{theorem}[Certification; \proved]\label{thm:2}
Let $c\ge40$. If $g\in\Abar$ is feasible at some $T>0$, then every admissible spectrum at central charge $c$ contains a
primary with $t_n<T$ (possibly a light one). Consequently $\Delta_1\le\TV(c)+a=\Dbest(c)$ for every admissible CFT.
\end{theorem}

Nothing about $g$ beyond membership in $\Abar$ and feasibility is assumed. The idea of the proof is to expand $g$ in the
odd Laguerre functions $\psi_{2k-1}$, each of which is a derivative functional that annihilates the crossing equation,
and to control the resummation by positivity. The proof, which uses no computer, is in \cref{sec:8}.

\paragraph{What ``best'' means.} Since $\Abar$ contains the functionals of order $P$ for every $P$ (\cref{lem:2.1}),
$\TV(c)\le t_1^{(\infty)}(c)$, that is, $\Dbest(c)\le\Delta_1^{(\infty)}(c)$, the limit of the truncated optimal gap
bounds that numerical work computes; equality is conjectural. Consequently the optimality half bounds, a fortiori, every numerical truncation, uniformly in
$P$. The attainment half is achieved by a function that is not of finite order, and it is a bound on conformal field
theories only through \cref{thm:2}. We do not claim that $\Abar$ contains every functional that has been considered:
membership of the analytic functionals of~\cite{HMR} is not checked here, and nothing depends on it. Nor do we claim that
the infima over $\gvac>0$ and over $\gvac\ge0$ coincide; $\TV$ is defined with $\gvac>0$.

\PartHeading{II}{The physical picture: a spectral curve for the optimal functional}
This Part is heuristic throughout. Its uncontrolled steps are four named hypotheses, \hyp{W}, \hyp{A},
\hyp{B} and \hyp{Sel}, stated in \cref{sec:3.1}. Its purpose is to show where $\LV$, the strip of \cref{fig:3} and the
zero law $\rhoU$ come from, and to formulate \cref{conj:1}. The proofs of \cref{sec:6,sec:7,sec:8,sec:9} do not use it.

\section{Warm-up: mean-field theory of the \texorpdfstring{$U(1)^c$}{U(1)\textasciicircum c} problem}
\label{sec:3}

\tagpar{\heuristic} We learn the recipe on the twin problem with $U(1)^c$ characters (\cref{sec:2.3}), where the
equations are simplest and the answer, the slope $\pi^2$, is known by other means: it was conjectured from numerics in~\cite{ACHLT}
and has since been proved in~\cite[Ch.~1]{OpenAI}, with the status given in \cref{sec:1.2}. The picture is semiclassical, with $1/c$ in the role of
$\hbar$: the optimal functional is a WKB wave function, a sum of exponentials $e^{c\,\cS}$ whose momenta $\cS'$ are the
branches of a \emph{spectral curve} in phase space. The Fourier transform acts on it by a canonical
transformation, so $\hat f=-f$ says that the curve is mapped to itself. The real zeros are the nodes where two
complex-conjugate branches interfere; there are of order $c$ of them, so only their density matters, as in
mean-field theory, and it is a coordinate on the curve.

\subsection{The problem, the data, and the four hypotheses}
\label{sec:3.1}

The functionals are radial functions on $\RR^{2c}$, written $f(\Delta)$ with $\Delta=\abs{\br}^2/2$, that obey
$\hat f=-f$ (the bootstrap normalization of \cref{sec:2.3}); the vacuum sits at the origin, and $f$ proves that the gap
is below $T$ if $f(0)>0$ and $f\ge0$ for $\Delta\ge T$. The optimal bound, written $\Delta_1$ in this section, was computed up to $c=2048$
in~\cite{ACHLT}, which conjectured that $\Delta_1/c\to1/\pi^2$. Beyond $\Delta_1$ the numerically optimal $f$ has
a band of double zeros, as in \cref{fig:2}. We name the four hypotheses in words; their precise forms are given
where the objects exist, in \cref{sec:3.2,sec:3.3,sec:3.4} for $U(1)^c$ and in the box of \cref{sec:4.1} for Virasoro.
\begin{itemize}[leftmargin=3.2em,labelwidth=2.6em,labelsep=0.6em,itemsep=0pt,topsep=3pt]
\item[\hypdef{W}] At large $c$ the optimal functional is a finite sum of WKB exponentials, called \emph{sheets}.
\item[\hypdef{A}] On the band, its envelope is dictated by the continuum density that the vacuum demands.
\item[\hypdef{B}] All sheets are branches of one analytic curve.
\item[\hypdef{Sel}] A selection rule, made of a topological assumption and a boundedness assumption, picks one curve
  among those allowed by \hyp{W}, \hyp{A} and \hyp{B}.
\end{itemize}

\subsection{Sheets and the local modular parameter}
\label{sec:3.2}

In formulas, \hyp{W} says that
\begin{equation}\label{eq:3-wkb}
  f(c\,\xi)\;\asymp\;\sum_{\text{sheets}}e^{c\,\cS(\xi)},\qquad \xi:=\Delta/c ,\qquad\text{and we set}\qquad
  \tau:=\frac{\cS'(\xi)}{2\pi i}
\end{equation}
on each sheet; $\asymp$ keeps the exponential order only. Locally a sheet looks like $e^{2\pi i\tau\Delta}$, a
character at modular parameter $\tau$, so we call $\tau$, the momentum conjugate to $\xi$, the \emph{local modular
parameter} of the sheet. The points $(\xi,\tau)$ of all the sheets form the spectral curve. On the band, a function
with real double zeros under a smooth envelope is locally the envelope times
$\cos^2(\pi c\!\int\!\rho\,\dd\xi)=\half[1+\cos(2\pi c\!\int\!\rho\,\dd\xi)]$, where $\rho(\xi)$ is the number of double
zeros per unit $\Delta$ ($\rho\to1$ deep in the band). Writing the cosine as two exponentials gives three terms of the
form~\eqref{eq:3-wkb}, that is, three sheets, a non-oscillating one with $\tau=i\yE$ and two oscillating ones:
\begin{equation}\label{eq:3-band}
  \tau_\pm=\mp\rho+i\yE ,\qquad
  f\;\propto\;\exp\Bigl(-2\pi c\!\int\!\yE\,\dd\xi\Bigr)\,\Bigl[\,1+\cos\Bigl(2\pi c\!\int\!\rho\,\dd\xi\Bigr)\Bigr].
\end{equation}
Here $\yE(\xi)=\Im\tau_\pm$, the \emph{height} of the oscillating sheets, is the rate of decay of the envelope. The oscillating sheets differ
by $2\rho$, and their interference produces the real double zeros. The non-oscillating sheet, the ``$1$'' in the
bracket, is their common modulus and carries no independent datum.

\subsection{The Fourier transform on sheets, and the functional equation}
\label{sec:3.3}

In $d=2c$ dimensions the Fourier transform of $e^{2\pi i\tau\abs{\br}^2/2}$ is
$(i/\tau)^{c}\,e^{2\pi i(-1/\tau)\abs{\br}^2/2}$: a character at $-1/\tau$, times a factor exponentially large
in $c$ that shifts the saddle point. We write $f$ locally as a superposition of characters, transform it term
by term, and locate the saddle point again,
\begin{align}
  f(c\,\xi)&=\int a(\tau)\,e^{2\pi i\tau c\xi}\,\dd\tau ,\qquad
     a(\tau)\asymp e^{c\,(\cS-2\pi i\tau\xi)}\big|_{\xi=\xi(\tau)} , \label{eq:3-legendre}\\
  \hat f(c\,\xi')&\asymp\int\exp\Bigl\{c\Bigl[\cS-2\pi i\tau\xi+\log\frac i\tau-\frac{2\pi i\,\xi'}{\tau}\Bigr]\Bigr\}\,\dd\tau ,
     \label{eq:3-transformed}\\
  0&=-2\pi i\,\xi-\frac1\tau+\frac{2\pi i\,\xi'}{\tau^{2}} , \label{eq:3-saddle}
\end{align}
the last line being stationarity in $\tau$, with $\dd(\cS-2\pi i\tau\xi)/\dd\tau=-2\pi i\xi$. Hence the \emph{sheet map}
\begin{equation}\label{eq:3-sheetmap}
  \hat S_{U(1)}:\;(\xi,\tau)\longmapsto\Bigl(\tau^{2}\xi+\frac{\tau}{2\pi i}\,,\;-\frac1\tau\Bigr),\qquad
  \hat S_{U(1)}\circ\hat S_{U(1)}=\mathrm{id} .
\end{equation}
Sheets go to sheets; $\hat f=-f$ says that this canonical transformation preserves their set.

\smallskip\noindent\textbf{Hypothesis \hyp{A}} has three parts that must be kept apart. Let $\{\Delta_n,d_n\}$ be the
extremal spectrum of \cref{sec:2.2}: the zeros of $f$, with the positive degeneracies of the dual optimum.
\emph{Forced:} the smoothed extremal degeneracies are the continuum density,
\begin{equation}\label{eq:3-forced}
  \frac{d_n}{\text{local spacing}}=\rho_0(\Delta_n)\,e^{o(c)} ,\qquad
  \rho_0(\Delta):=\frac{(2\pi)^{c}\,\Delta^{c-1}}{\Gamma(c)} ,
\end{equation}
where $\rho_0$ is the $S$-image of the vacuum, $\int_0^\infty\rho_0\,e^{-2\pi\beta\Delta}\,\dd\Delta=\beta^{-c}$. The
reason: Poisson-resum $\sum_nd_n\,e^{2\pi i\tau\Delta_n}$ over $n$, that is, write
$\sum_nF(n)=\sum_k\int F(n)\,e^{2\pi ikn}\,\dd n$, with $n(\Delta)$ the counting function of the levels. The term $k=0$, the
smooth integral of $d_n/(\text{local spacing})$ against $e^{2\pi i\tau\Delta}$, is the only non-oscillating one, so it alone
must reproduce the $S$-image of the vacuum. The terms $k=\pm1$, the first harmonics, carry the phase
$e^{\pm2\pi ic\int\!\rho\,\dd\xi}$, so they are of the form \eqref{eq:3-wkb} with $\Re\tau=\pm\rho$.
\emph{Assumed, and this is \hyp{A}:} reciprocity. On the band the
local upper envelope $\abs f_{\rm env}$ of the optimal functional is inverse to the degeneracies at exponential order,
\begin{equation}\label{eq:3-A}
  d_n\,\abs f_{\rm env}(\Delta_n)=e^{o(c)}\qquad\text{uniformly on the band.}
\end{equation}
The reason offered: the first harmonics of the spectrum, the terms $k=\pm1$ above, are sheets
$\tau=\pm\rho+iy$ with $y$ the height of $1/\rho_0$, that is $y=\frac1{2\pi c}\,\dd\log\rho_0/\dd\xi=1/(2\pi\xi)$. If
the two oscillating sheets of $f$ are their mirror images, with the same $\Im\tau$ and the opposite $\Re\tau$ (given
the forced part, this is \eqref{eq:3-A}), then crossing is carried by these harmonics (``extremal and feasible implies
equality wherever there are zeros'': convincing, but not a proof).
\emph{Consequence:} $\abs f_{\rm env}\asymp1/\rho_0$, that is, $\yE=1/(2\pi\xi)$ (checked in \cref{sec:3.5}).

\smallskip\noindent\textbf{Skeleton of the consequence.} With \hyp{A}, $\tau_\pm=\mp\rho+i/(2\pi\xi)$ on the band. We
extend $\rho$ from a real density on the band to a complex coordinate on the whole curve by \emph{defining}
\begin{equation}\label{eq:3-rho-w}
  \rho:=\frac{i}{2\pi\xi}-\tau ,\qquad w:=\pi^{2}\xi\rho ,
\end{equation}
where $w$ is the combination in which $\hat S_{U(1)}$ becomes affine. Indeed, \eqref{eq:3-rho-w} gives
$\tau\xi+\frac1{2\pi i}=-\rho\xi$, so the image point in \eqref{eq:3-sheetmap} has $\xi'=-\rho\xi\tau$; then
$\rho'=\frac{i}{2\pi\xi'}+\frac1\tau=\frac1{\rho\tau}\bigl(\rho-\frac{i}{2\pi\xi}\bigr)=-\frac1\rho$ and
$w'=\pi^{2}\xi'\rho'=\pi^{2}\xi\tau=\frac{i\pi}2-w$. In the variables $(\rho,w)$ the sheet map is the involution
\begin{equation}\label{eq:3-iota}
  \iota:\;(\rho,w)\longmapsto\Bigl(-\frac1\rho\,,\;\frac{i\pi}{2}-w\Bigr) .
\end{equation}
By \hyp{B} the curve is a graph $\rho=\cF(w)$ with $\cF$ analytic. Invariance under $\iota$ gives
$\cF(\frac{i\pi}2-w)\,\cF(w)=-1$; exchanging the oscillating sheets, $(\rho,w)\mapsto(-\rho,-w)$, makes $\cF$ odd. Hence
\begin{equation}\label{eq:3-functional}
  \cF\bigl(w+\tfrac{i\pi}{2}\bigr)=\frac{1}{\cF(w)} ,\qquad\text{with general solution}\qquad
  \cF=\tanh\cdot E ,
\end{equation}
where $E$ is even and obeys the same equation, because $\tanh$ is odd and $\tanh(w+\frac{i\pi}2)=1/\tanh w$.

\subsection{Selection and the \texorpdfstring{$\tanh$}{tanh} law}
\label{sec:3.4}

\begin{figure}[t]
\centering
\figfile{fig4_tanh}
\caption{The $\tanh$ law against the zeros of the numerically optimal $U(1)^c$ functionals of~\cite{ACHLT}; no
parameter is fitted. (a)~The number $\rho$ of double zeros per unit $\Delta$ against $\xi=\Delta/c$. Curve: the
largest solution of $\rho=\tanh(\pi^2\xi\rho)$ (\cref{res:H1}); it vanishes at the Curie point
$\xi_1=1/\pi^2$. Circles: the inverse of the spacing of consecutive zeros, extrapolated to $c\to\infty$ linearly in
$1/c$ through three values of $c$. Inset: graphical solution of the Curie--Weiss equation. (b)~Magnitude of the
relative deviation of the measured $1/\rho-1$ from the law at $c=256,\dots,2048$ (open symbols; all lie below
the law) and after extrapolation (filled): it decreases roughly like $1/c$, and the extrapolated values are within
$2.2\times10^{-3}$ of the law.}
\label{fig:4}
\end{figure}

We pass to the inverse function, $w=W(\rho)$. The band and its image under the exchange of the oscillating sheets
make up the real segment $\abs\rho<1$, where $w$ is real, $\Im W=0$; by \eqref{eq:3-iota} its $\iota$-image is
$\abs\rho>1$, with $\Im W=\frac\pi2$. So $\Im W$ is known on the real axis. What the symmetry and \hyp{A} leave free,
the function $E$, is how $W$ continues off the axis; the selection rule supplies this.

\smallskip\noindent\textbf{Hypothesis \hyp{Sel}, $U(1)^c$ form.} (Topology) $W$ is single-valued and holomorphic on the
upper half of the $\rho$-plane. (Boundedness) $\Im W$ is bounded there.

\smallskip\noindent A bounded harmonic function with piecewise-constant boundary values is unique: $\Im W(\rho)$ is
$\frac\pi2$ times the \emph{harmonic measure} of $\{\abs\rho>1\}$ seen from $\rho$, that is, the probability that
Brownian motion from $\rho$ first hits the real axis there, which is $\frac1\pi$ times the angle that the set subtends
at $\rho$:
\begin{equation}\label{eq:3-W}
  \Im W(\rho)=\half\,\arg\frac{1+\rho}{1-\rho} ,\qquad W(\rho)=\arctanh\rho ,
\end{equation}
the real constant being zero because $W$ is odd; that is, $E\equiv1$. A bounded $\Im W$ stays between its boundary
values: on no sheet does the functional decay or grow faster than the rate $\abs f\asymp\Delta^{-c}$ of \hyp{A}. We stress
that \hyp{Sel} is a hypothesis; it is not derived from the linear program.

\begin{resultH}[The $\tanh$ law; \heuristic]\label{res:H1}
Assume \hyp{W}, \hyp{A}, \hyp{B} and \hyp{Sel}. At leading order at large $c$, the number $\rho(\xi)$ of double zeros of
the optimal $U(1)^c$ functional per unit $\Delta$, at $\Delta=c\,\xi$, is the largest solution of
\begin{equation}\label{eq:3-tanh}
  \rho=\tanh\bigl(\pi^{2}\xi\rho\bigr) .
\end{equation}
A nonzero solution exists if and only if $\xi>1/\pi^2$. The band therefore begins at $\xi_1=W'(0)/\pi^2=1/\pi^2$, that
is, $\Delta_1=c/\pi^2+o(c)$: the slope is $\pi^2$.
\end{resultH}

\noindent This is the Curie--Weiss equation for the magnetization of a ferromagnet at inverse temperature $\pi^2\xi$,
hence ``mean field''; the lower end of the band is the Curie point (inset of \cref{fig:4}).

\subsection{Checks, status, and what carries over}
\label{sec:3.5}

Near the Curie point \eqref{eq:3-tanh} gives $\rho\simeq\pi\sqrt3\,(\xi-\xi_1)^{1/2}$. Square-root vanishing gives Airy
scaling: $\Delta_1=c/\pi^2+0.352336\,c^{1/3}+\ldots$, with $0.352336=(\pi^{2}\sqrt3)^{-2/3}\abs{a_1}$ and $a_1$
the first zero of $\Ai$. Deep in the band $\Delta_n=n+c\,[\xi_1+\int_{\xi_1}^\infty(1-\rho)\,\dd\xi]+O(1)=n+c/8+O(1)$,
the offset of the line $\Delta_n=n+(c-4)/8$ of~\cite[\S4]{ACHLT}.
Against the data of~\cite{ACHLT}: the spacings follow \eqref{eq:3-tanh} through the band (\cref{fig:4});
$\Delta_1/c$, extrapolated by us in $c^{-2/3}$, tends to $0.10134\pm0.00003$, against $1/\pi^2=0.101321$; the
degeneracies follow the continuum density,
$d_n/(\rho_0(\Delta_n)\times\text{spacing})=1$ up to a relative deviation falling from $2\times10^{-2}$ at $n=2$ to
$10^{-6}$ deep in the band, which is the forced part of \hyp{A}, verified far beyond exponential order; and the envelope
of $f$ has the height $\yE=1/(2\pi\xi)$ up to $O(1/c)$, which is hypothesis \hyp{A} itself.

The slope $\pi^2$ has since been proved in~\cite[Ch.~1]{OpenAI}; we have read the statements and the structure of that
argument but have not independently verified it; nothing in this paper depends on it logically. The derivation above is
independent of that proof, and heuristic. What we carry to Virasoro is the recipe: sheet map, \hyp{A}, \hyp{B},
\hyp{Sel}. The rigorous part of this paper (\cref{sec:6}) follows the \emph{strategy} of the $U(1)^c$ proof, not the
mean-field derivation; the attribution is made in \cref{sec:6.1}.

\section{The Virasoro spectral curve and the conjectured slope}
\label{sec:4}

\tagpar{\heuristic} We apply the recipe of \cref{sec:3} (sheet map, \hyp{A}, \hyp{B}, \hyp{Sel}) to the Virasoro
problem, to explain where the constants of \cref{thm:1} come from. We suppose that at large $c$ the optimal
functional is a WKB wave function, and ask on which curve in phase space its momenta can lie, given that it
is odd under the Fourier transform and that the vacuum imposes the Cardy density on the extremal spectrum.
Under four hypotheses the answer is an explicit curve in closed form, \cref{res:H2}; the band begins at its branch
point, and this gives $\LV$. Nothing in this section is proved.
Its leading-order consequences are proved without the hypotheses: the slope and the strip $\abs{\Im b}<\half$ in
\cref{sec:6,sec:7}, and the zero law, in a weak sense, in \cref{sec:9};
the finer ones make up \cref{conj:1} and are tested in \cref{sec:5}. ``The optimal functional'' is meant in the
order of limits of \cref{sec:2.5}: $P\to\infty$ first, then $c\to\infty$. The section can be read without \cref{sec:3}.

\subsection{The hypotheses, and what changes: phase-space coordinates}
\label{sec:4.1}

\begin{ideabox}[title={The four hypotheses, in Virasoro form}]
\hyp{W} \emph{WKB form.} At large $c$ the optimal functional is a finite sum of WKB exponentials,
\begin{equation}\label{eq:4-wkb}
  g(c'\xi)\asymp\sum_{\rm sheets}e^{c'\cS(\xi)} ,\qquad \xi:=\frac{t}{c'} ,\qquad \tau:=\frac{\cS'(\xi)}{2\pi i} ,
\end{equation}
where $\asymp$ keeps the exponential order in $c$ only. Each term is a \emph{sheet}, and $\tau$ is its \emph{local
modular parameter}: near $\xi$ the sheet looks like the character $e^{2\pi i\tau t}$.

\smallskip\noindent\hyp{A} \emph{Cardy envelope.} On the band, the local upper envelope $\genv$ of the oscillating
$\abs g$ is inverse to the Cardy density $\rhoC$ at exponential order. \Cref{sec:4.2} separates what is forced in
this statement from what is assumed.

\smallskip\noindent\hyp{B} \emph{One curve.} All sheets are branches of one analytic curve $\Sigma$, the \emph{spectral
curve}.

\smallskip\noindent\hyp{Sel} \emph{Selection.} A topological assumption \hyp{C0} and a boundedness assumption
\hyp{C1}, stated in \cref{sec:4.4}, pick one curve among those that \hyp{W}, \hyp{A} and \hyp{B} allow.
\end{ideabox}

What changes with respect to \cref{sec:3} is the action of the Fourier transform on sheets. The functionals are radial
functions on the plane $\RR^2$ for every $c$ (\cref{sec:2.3}), with $t=\abs\br^2/2$, and the planar transform of
the character $e^{2\pi i\tau t}$ is $(i/\tau)\,e^{2\pi i(-1/\tau)t}$: a character at $-1/\tau$, times a factor of order
one. We write a sheet as a superposition of characters, transform it term by term and locate the saddle point in
$\tau$ again, as in \eqref{eq:3-legendre}--\eqref{eq:3-saddle}. The factor $i/\tau$ now drops out of the saddle-point
equation, which reads $-2\pi i\xi+2\pi i\xi'/\tau^2=0$. The dimension stays $2$, so there is no large-$d$ saddle and no
factor $(i/\tau)^c$: the sheet map is the bare modular inversion,
\begin{equation}\label{eq:4-sheetmap}
  \hat S:\;(\xi,\tau)\longmapsto\Bigl(\tau^{2}\xi\,,\;-\frac1\tau\Bigr),\qquad \hat S\circ\hat S=\mathrm{id} ,
\end{equation}
and $\hat g=-g$ says that the set of sheets is invariant under it. The central charge enters only through the position
of the vacuum, $b=\pm\frac i2$, and through the Cardy envelope.

The map $\hat S$ becomes linear on the double cover $b=\sqrt{3\xi}$ of \cref{sec:2.5}. We define the \emph{momentum}
\begin{equation}\label{eq:4-momentum}
  p:=-\tau\,b ,\qquad\text{so that}\qquad \frac1{c'}\,\partial_b\log g=\frac{2b}{3}\,2\pi i\tau=-\frac{4\pi i}{3}\,p
\end{equation}
on each sheet, because $\dd\xi=\frac23b\,\dd b$: up to the constant, $p$ is the WKB momentum conjugate to $b$. The
image of $(\xi,\tau)$ under $\hat S$ has $b'^2=p^2$ and $p'=b'/\tau$, so its two lifts are $(p',b')=(b,-p)$ and
$(-b,p)$. In the coordinates $(p,b)$ the Fourier transform is therefore the quarter-turn
\begin{equation}\label{eq:4-R}
  \Rq:\;(p,b)\longmapsto(b,-p),\qquad \Rq^4=\mathrm{id} ,
\end{equation}
whose inverse is the other lift and whose square, $(p,b)\mapsto(-p,-b)$, is the change of sign of the square root.
This is the oscillator rotation of \cref{sec:2.3}, acting on WKB data. From here on $\Sigma$ is a complex curve in
$\CC^2_{(p,b)}$, invariant under $\Rq$, and the sheets are its branches over the $b$-plane.

\subsection{The Cardy envelope}
\label{sec:4.2}

We first say what the sheets are on the band, and then state \hyp{A}. On the band $g$ has real double zeros under a
smooth envelope, $g\approx\genv\cos^2\bigl(\pi c'\!\int^\xi\!\rhoV\bigr)$, where $\rhoV(\xi)$ is the number of double
zeros per unit $t$. Expanding the cosine gives three sheets,
\begin{equation}\label{eq:4-band}
  \tau_E=i\,\yE ,\qquad \tau_\pm=\mp\rhoV+i\,\yE ,\qquad \yE:=-\frac1{2\pi c'}\,\frac{\dd\log\genv}{\dd\xi} ,
\end{equation}
so $\yE$ is the rate of decay of the envelope. The sheet $\tau_E$ carries no independent datum; ``the two band
sheets'' always means the oscillating pair $\tau_\pm$, whose interference produces the zeros.

\smallskip\noindent\textbf{Hypothesis \hyp{A}} has three parts that must be kept apart. Let $\{t_n,d_n\}$ be the
extremal spectrum of \cref{sec:2.2}. \emph{Forced:} the extremal spectrum is the Cardy continuum, discretized,
\begin{equation}\label{eq:4-forced}
  \frac{d_n}{\text{local spacing}}=\rhoC(t_n)\,e^{o(c)} .
\end{equation}
The reason: Poisson-resum $\sum_nd_n\,e^{-2\pi\beta t_n}$ over $n$; the zero mode is the only non-oscillating term, so
it alone must reproduce the $S$-image of the vacuum, $\int\rhoC\,e^{-2\pi\beta t}\,\dd t$ (\cref{sec:2.4}).
\emph{Assumed, and this is \hyp{A}:} reciprocity,
\begin{equation}\label{eq:4-A}
  d_n\,\genv(t_n)=e^{o(c)}\qquad\text{uniformly on the band.}
\end{equation}
The reason offered: the first Poisson harmonics of the extremal spectrum are sheets
of height $\Im\tau=\frac1{2\pi c'}\,\dd\log\rhoC/\dd\xi=1/(2b)$, because $\rhoC\asymp e^{c'L(x)}$ with $L(x)=\frac{2\pi}3x$
and $x=\sqrt{3\xi}$. If the two oscillating sheets of $g$ are their mirror images, with the same $\Im\tau$ and the
opposite $\Re\tau$ (given the forced part, this is \eqref{eq:4-A}), then crossing is carried by these harmonics
(``extremal and feasible implies equality wherever there are zeros'': convincing, but not a proof). \emph{Consequence:} on the band the functional runs along the Cardy line of \cref{sec:2.5},
\begin{equation}\label{eq:4-yE}
  \genv\asymp e^{-c'L(x)} ,\qquad\text{that is,}\qquad \yE=\frac{1}{2b} .
\end{equation}
\emph{Sensitivity:} if we write $\yE=\cardyh/(2b)$ with a constant $\cardyh$, the \emph{Cardy height}, then $\cardyh$
enters the final answer as $\xis\to\cardyh^2\xis$. So \hyp{A} is the hypothesis to test hardest; \cref{sec:5.3}
measures $\cardyh=1.0000\pm0.0005$.

In terms of the momentum, \eqref{eq:4-band} and \eqref{eq:4-yE} say that the two band sheets are
\begin{equation}\label{eq:4-bandsheets}
  p=\pm v-\frac i2 ,\qquad v:=\rhoV\,b\in\RR ,\qquad\text{so that}\qquad \Im p=-\half ,\quad \Im b=0
\end{equation}
on the band. The \emph{band arc} $\gamma_0:=\{v\in\RR\}\subset\Sigma$ comprises both band sheets, joined at the branch
point $v=0$, which is the lower end of the band, where $\rhoV$ vanishes. By \eqref{eq:4-R}, $\Rq$ maps $\gamma_0$ to
the arc $(p,b)=(x,\mp v+\frac i2)$, on which $\Im b=+\half$ and $\Im p=0$. The Fourier image of the band thus lies on
the line $\Im b=\half$, which passes through the vacuum $b=\frac i2$ (\cref{fig:3}). This is the first appearance of
the strip of \cref{sec:6} and of the edge integral of \cref{sec:7}.

\subsection{Symmetries and boundary data}
\label{sec:4.3}

Two maps act on $\Sigma$. One is $\Rq$. The other is the exchange of the two band sheets, which reverses $\Re\tau$ at
fixed $\Im\tau=1/(2b)$; by \eqref{eq:4-bandsheets} it is $\sigma:(p,b)\mapsto(-p-i,b)$, a holomorphic map that \hyp{B}
extends from the band to all of $\Sigma$. Applying $\Rq$ repeatedly to \eqref{eq:4-bandsheets} we obtain four arcs
$\gamma_k:=\Rq^k\gamma_0$ on which the imaginary parts of both coordinates are constant,
\begin{equation}\label{eq:4-bdry}
  (\Im p,\Im b)=\bigl(-\half,0\bigr)\ \text{on }\gamma_0,\qquad \bigl(0,\half\bigr)\ \text{on }\gamma_1,\qquad
  \bigl(\half,0\bigr)\ \text{on }\gamma_2,\qquad \bigl(0,-\half\bigr)\ \text{on }\gamma_3 .
\end{equation}
Consecutive arcs meet at four points of $\Sigma$ at $\xi=\infty$; for instance $\gamma_0$ and $\gamma_1$ share the end
where $p$ and $b$ tend to $+\infty$ with $p-b\to-\frac i2$.

\smallskip\noindent\textbf{Where things live.} The four arcs bound a region $\Omt$ \emph{on the curve} $\Sigma$ (on its
cover in the variable $b$), not in the $b$-plane. It contains the real, non-oscillating sheet $\tau=iy$ (here and below $y$ is
$\Im\tau$, not $\Im b$) that carries $g$ from the lower end of the band across the lobe interval $[0,t_1)$ of
\cref{sec:2.5} to the BTZ threshold $t=0$ and on to the vacuum (\cref{sec:4.6}). Concretely, a point of $\Omt$ is a pair $(p,b)$: a value of $b$ together with a choice of
sheet. The region is mapped to itself by $\Rq$. \Cref{fig:5}
shows $\Omt$ and its image in the half-plane constructed next.

\begin{figure}[t]
\centering
\figfile{fig5_curve}
\caption{The spectral curve. (a)~The region $\Omt$ on the curve $\Sigma$, drawn in the chart $(\Re b,\Re p)$: a disk
bounded by the band arc $\gamma_0$ (thick; both band sheets, joined at the branch point $v=0$, the tip) and its Fourier
images $\gamma_k=\Rq^k\gamma_0$. Each arc is labeled with its values of $(\Im p,\Im b)$, \eqref{eq:4-bdry}, and with
the set in which $b$ lies. Consecutive arcs meet at the four points at $\xi=\infty$, and $\Rq$ is the quarter-turn
about its fixed point, the BTZ threshold $(p,b)=(0,0)$. The dotted segments are the real sheet, on which $b$ is real
(the lobe interval) or imaginary (from the BTZ threshold to the vacuum $b=\pm\frac i2$). (b)~The image of $\Omt$ under
the uniformization of \cref{sec:4.4}: the upper half-plane $\HH\ni z$, with the fixed point at $z=i$, the arcs on
$(-1,0)$, $(0,1)$, $(1,\infty)$ and $(-\infty,-1)$, and the real sheet on the dotted semicircles. Below the axis:
marked points $z$ with their images $b(z)$; then, for each arc, the boundary values $(\Im p,\Im b)$, which determine
$p(z)$ and $b(z)$, \eqref{eq:4-pb}; the last row gives the boundary correspondences of $b(z)$. For the reader coming
from \cref{sec:6}: panel~(b) is the conformal map $b(z)$ of \cref{lem:6.3}, which sends $\HH$ onto the strip $\strip$
slit along $\abs x\ge\xs$.}
\label{fig:5}
\end{figure}

\subsection{Selection and the curve}
\label{sec:4.4}

The boundary data \eqref{eq:4-bdry} do not determine $\Sigma$: they say nothing about what happens between the arcs.
That is what the selection hypothesis supplies. It has two halves.

\smallskip\noindent\hypdef{C0} \emph{Topology.} $\Omt$ is simply connected: it is a disk bounded by the four arcs,
which meet only at their four ends. On it $p$ and $b$ are single-valued, and $\Rq$ has exactly one fixed point.

\smallskip\noindent\hypdef{C1} \emph{Boundedness.} $p$ and $b$ are holomorphic on $\Omt$ with bounded imaginary parts.

\smallskip\noindent Hypothesis \hyp{Sel} is \hyp{C0} together with \hyp{C1}: the same two halves, topology and
boundedness, as in \cref{sec:3.4}. Physically, by \eqref{eq:4-momentum}, $\frac{4\pi}3\Im p$
is the rate at which $\frac1{c'}\log\abs g$ changes along real $b$; on the band arc it is minus the Cardy rate
$L'=\frac{2\pi}3$. A bounded harmonic function stays between its boundary values, so \hyp{C1} says that the complex
decay rate never exceeds the Cardy rate.

\smallskip\noindent\textbf{Derivation.} The idea is that \hyp{C0} lets us replace $\Omt$ by the upper half-plane, where
$\Rq$ is a M\"obius map and the arcs are intervals; \hyp{C1} then makes $\Im b$ and $\Im p$ bounded harmonic functions
with known, piecewise-constant boundary values, and such a function is unique. In detail, map $\Omt$ conformally onto $\HH\ni z$ (Riemann mapping theorem)
with the fixed point of $\Rq$ at $z=i$. An automorphism of $\HH$ of order four that fixes $i$ is $\Rq(z)=(1+z)/(1-z)$
(its inverse gives the mirror image of the curve). The four ends of the arcs form one
orbit of $\Rq$ on the boundary of $\HH$. The rotations about $i$ commute with $\Rq$, and we use them to place one end
at $z=-1$. The orbit is then $-1\to0\to1\to\infty$, the arcs are $\gamma_0=(-1,0)$, $\gamma_1=(0,1)$,
$\gamma_2=(1,\infty)$ and $\gamma_3=(-\infty,-1)$, and no modulus remains. By \hyp{C1} and \eqref{eq:4-bdry}, $\Im b$ is
a bounded harmonic function on $\HH$ with boundary values $0,\half,0,-\half$ on these intervals. As in \cref{sec:3.4}
it is a sum of angles, $\Im b=\frac1{2\pi}\bigl[\arg(z-1)-\arg z-\arg(z+1)\bigr]$ with all arguments in $(0,\pi)$. So
$b$ is a sum of logarithms up to a real constant, and likewise $p$; the two constants are fixed by $b\circ\Rq=-p$ and
$p\circ\Rq=b$. The result is
\begin{equation}\label{eq:4-pb}
  b(z)=\frac1{2\pi}\log\frac{z-1}{z\,(z+1)} ,\qquad p(z)=\frac i2+\frac1{2\pi}\log\frac{z+1}{z\,(z-1)} .
\end{equation}
Then $p+b=\frac i2-\frac1\pi\log z$ and $p-b=\frac i2+\frac1\pi\log\frac{z+1}{z-1}$, which we exponentiate:
\begin{equation}\label{eq:4-XY}
  X:=e^{-\pi(p+b)}=-iz ,\qquad Y:=e^{-\pi(p-b)}=-i\,\frac{z-1}{z+1} .
\end{equation}
Eliminating $z$ gives the curve.

\begin{resultH}[The spectral curve; \heuristic]\label{res:H2}
Assume \hyp{W}, \hyp{A}, \hyp{B} and \hyp{Sel}. At leading order at large $c$, the sheets of the optimal functional are
the branches of the curve
\begin{equation}\label{eq:4-curve}
  \sinh\pi b=i\,\sinh\pi p ,\qquad\text{equivalently}\qquad Y=\frac{X+i}{1+iX} ,\qquad\text{or}\qquad
  \cosh2\pi p+\cosh2\pi b=2 ,
\end{equation}
where the last form, obtained by squaring, contains the curve together with its mirror image. There is no free constant.
\end{resultH}

\noindent The curve is invariant under $\sigma$, as it must be. The fixed point $z=i$ is $(p,b)=(0,0)$: the curve passes
through the BTZ threshold, and it does so with $p\simeq-ib$, that is, at the self-dual value $\tau=i$. The identically
worded selection rule applied to the $U(1)^c$ data is \hyp{Sel} of \cref{sec:3.4} and returns \cref{res:H1}, a law whose
consequence, the slope $\pi^2$, has since been proved in~\cite[Ch.~1]{OpenAI} (we have not verified that proof
independently; see \cref{sec:3.5}). Hypotheses \hyp{W}, \hyp{A} and \hyp{B} alone admit rival curves;
those we found have slopes between about $8.9$ and $9.8$. So \hyp{Sel} is a genuine hypothesis, but one that is no
longer needed after \cref{sec:6}.

\subsection{The gap}
\label{sec:4.5}

The band begins where the density of zeros vanishes, which on the curve is the branch point $v=0$ of the band arc. On
the band, $b=x$ is real and $p=\pm v-\frac i2$, so $\sinh\pi p=\mp i\cosh\pi v$ and~\eqref{eq:4-curve} becomes
\begin{equation}\label{eq:4-bandlaw}
  \cosh\pi v=\sinh\pi x ,\qquad v=\rhoV\,x ,
\end{equation}
which has a real solution $v$ if and only if $\sinh\pi x\ge1$. The lower end of the band is therefore at
$\sinh\pi\xs=1$:
\begin{equation}\label{eq:4-slope}
  \xs=\frac{\arcsinh1}{\pi}=0.2805499 ,\quad \xis=\frac{(\xs)^2}{3}=0.0262361 ,\quad
  \LV=\frac{12\pi^2}{\pi^2+4\arcsinh^21}=9.1266340 .
\end{equation}
The last number follows because the bound is $a+t_1$ with $a=c'/12$ and $t_1=c'\xis+o(c)$: $(a+t_1)/c\to\frac1{12}+\xis=1/\LV$.
That $\LV$ is the exact leading slope of the best bound is \cref{thm:1}.

\subsection{Fine structure, and \texorpdfstring{\cref{conj:1}}{Conjecture 1}}
\label{sec:4.6}

The curve contains more than the slope. We list six consequences of the curve and of the hypotheses;
together they are \cref{conj:1} (the first two form its item~(b)), and \cref{sec:5.3} tests them one by one.

\smallskip\noindent\textbf{Band density.} By \eqref{eq:4-bandlaw}, $\rhoV(\xi)=\arccosh(\sinh\pi x)/(\pi x)$ with
$x=\sqrt{3\xi}$. Since $\dd t=\frac23c'x\,\dd x$, the double zeros have density $c'\rhoU(x)/2$ per unit $x$, where
\begin{equation}\label{eq:4-rhoU}
  \rhoU(x):=\frac{4}{3\pi}\,\arccosh\bigl(\sinh\pi\abs x\bigr) ,\qquad \rhoV=\frac{3\,\rhoU}{4x} ,
\end{equation}
so that $c'\rhoU$ counts zeros with multiplicity. It vanishes like $(x-\xs)^{1/2}$ at the tips $\pm\xs$. The subscript
refers to the extremal potential $U$ of \cref{sec:6.6}, whose charge density $\rhoU$ is, not to $U(1)$.

\smallskip\noindent\textbf{Profile off the band.} Integrating \eqref{eq:4-momentum} along a sheet,
\begin{equation}\label{eq:4-profile}
  \frac1{c'}\log\abs{g}=\text{const}+\frac{4\pi}3\,\Im\!\int^{b}p\,\dd b' ,\qquad
  \begin{cases}\ \sin(\pi yb)=\sinh(\pi b), & p=-iyb,\quad 0<b<\xs,\\[2pt]
               \ \sinh(\pi y\tilde b)=\sin(\pi\tilde b), & p=y\tilde b,\quad b=i\tilde b,\ 0<\tilde b\le\half ,\end{cases}
\end{equation}
where the two cases are the real sheet $\tau=iy$ of \eqref{eq:4-curve} on the lobe interval and below the BTZ
threshold. On the lobe interval $y$ rises from the self-dual value $y(0^+)=1$ to the value $\yE(\xs)=1/(2\xs)$ of
\eqref{eq:4-yE} at the tip, where the sine reaches $1$ and the branch ends: this is $\sinh\pi\xs=1$ seen from the lobe
side. \Cref{sec:6.6} identifies the profile \eqref{eq:4-profile} with the extremal potential $U$.

\smallskip\noindent\textbf{Airy tip.} Near the lower end of the band \eqref{eq:4-bandlaw} gives
$\rhoV^2\simeq B_V^2\,(\xi-\xis)$ with $B_V^2=3\sqrt2\,\pi^2/\arcsinh^31$. A density of double zeros that vanishes like
a square root, preceded by a simple zero, is modeled locally by the Airy function, as in \cref{sec:3.5}: the first
levels sit at $\xi_n=\xis+\abs{a_n}(\pi B_Vc)^{-2/3}$, where $a_n$ are the zeros of $\Ai$. Hence
\begin{equation}\label{eq:4-airy}
  \Dbest=\frac{c}{\LV}+\Aopt\,c^{1/3}+O(1),\qquad
  \Aopt=\frac{\abs{a_1}\,\arcsinh1}{(3\sqrt2\,\pi^4)^{1/3}}=0.27666 ,
\end{equation}
and $(t_2-t_1)\,c^{-1/3}\to(\abs{a_2}/\abs{a_1}-1)\,\Aopt=0.20705$, where $t_2$ is the second level of the extremal
spectrum. There is no $\log c$ term at order $c^{1/3}$ or above. The reason: the curve depends on $c$ only through
$a=c'/12$, and a power-law prefactor $t^\alpha$ in $\rhoC$ or in $g$ shifts $\frac1c\log\rhoC$ by
$\frac\alpha c(\log c+\log\xi)$; the first term does not depend on $\xi$ and leaves $\yE$ unchanged, and the second
moves $\yE$ by $O(1/c)$, hence $t_1$ by $O(1)$. A term $\log c$ with a coefficient of order one, below the order
$c^{1/3}$, is not excluded by this argument; the data bound it (\cref{sec:5.3}).

\smallskip\noindent\textbf{Deep band.} Far from the tip the double zeros become equally spaced,
\begin{equation}\label{eq:4-offset}
  1-\rhoV\simeq\frac{2}{\pi x}\,e^{-2\pi x} ,\qquad \xis+\int_{\xis}^\infty(1-\rhoV)\,\dd\xi=\frac1{24} ,
\end{equation}
the second being an exact identity.\footnote{With $u=\pi x$ and $\theta_1=\arcsinh1=\log(1+\sqrt2)$ it reads
$\theta_1^2+2\int_{\theta_1}^\infty(u-\arccosh\sinh u)\,\dd u=\pi^2/8$. The constant follows from Landen's identity
$\sum_{k\ge0}(\sqrt2-1)^{2k+1}/(2k+1)^2=\frac{\pi^2}{16}-\frac14\log^2(1+\sqrt2)$, a special value of Legendre's
chi-function of order~$2$.}
Since the $n$-th level has $n=c'\!\int_{\xis}^{t_n/c'}\!\rhoV\,\dd\xi+O(1)$, it follows that $t_n=n+c/24+O(1)$, that
is, $\Delta_n=n+c/8+O(1)$: the same offset as for $U(1)^c$ (\cref{sec:3.5}). The offset itself is not new: numerically the
spectrum follows the free fermion line $\Delta_n=n+(c-4)/8$ at large $n$. For Virasoro this is seen in~\cite[Figure~4]{AHT}
and~\cite[\S7.4]{HMR}, where convergence to the line is conjectured at fixed $c$; for $U(1)^c$ the distance from the line at
large $n$ is conjectured in~\cite[Conj.~4.2]{ACHLT} to be at most of order $2^{-c}$, and to vanish only for integers $c\equiv4$
modulo~$8$. What is new here is the heuristic derivation of the offset from the band density.

\smallskip\noindent\textbf{Degeneracies.} Hypothesis \hyp{A}, sharpened from exponential order to order one, predicts
$d_n\simeq\rhoC(t_n)\cdot\half(t_{n+1}-t_{n-1})$ with the full three-Bessel density $\rhoC$ of \cref{sec:2.4}.

\smallskip\noindent\textbf{Interior sign change.} For $t=O(1)$ the functional is carried by the real sheet through the
BTZ threshold, where $y=1$: $g=\varphi(t)\,e^{-2\pi t}$ with $\varphi$ varying on the scale $c$. Oddness at the origin,
$\hat g(0)=2\pi\int_0^\infty g\,\dd t=-g(0)$, is dominated by $t=O(1)$, and with $\varphi\simeq\varphi_0+\varphi_1t$ it
gives $\varphi_1=-4\pi\varphi_0$. So $g(t)\propto(1-4\pi t)\,e^{-2\pi t}=L_1(4\pi t)\,e^{-2\pi t}$ for $t=O(1)$, the
lowest odd mode of \cref{sec:2.3}. Hence $g>0$ on $(0,\tint)$ and $g<0$ on $(\tint,t_1)$ with $\tint\to1/4\pi=0.0796$,
and there is no other zero between the vacuum and $t_1$: this is the positive sliver of \cref{sec:2.5}.

\begin{conjecture}[\conjtag]\label{conj:1}
Let $g$ be the optimal functional and $\{t_n,d_n\}$ the extremal spectrum, in the order of limits of \cref{sec:2.5}.
\begin{enumerate}[label=\textup{(\alph*)},leftmargin=2.2em,itemsep=1pt,topsep=3pt]
\item $\Dbest=c/\LV+\Aopt\,c^{1/3}+O(1)$ with $\Aopt$ as in \eqref{eq:4-airy}; there is no $\log c$ term at order
$c^{1/3}$ or above; and $(t_2-t_1)\,c^{-1/3}\to0.20705$.
\item The double zeros of $g$ have density $c'\rhoU(x)/2$ per unit $x$ on the band, and on the lobe interval and
below the BTZ threshold $\frac1{c'}\log\abs g$ is given by \eqref{eq:4-profile}, up to $o(1)$.
\item Deep in the band $t_n=n+c/24+O(1)$ (cf.~\cite[Figure~4]{AHT}, \cite[\S7.4]{HMR}).
\item $d_n\simeq\rhoC(t_n)\cdot\half(t_{n+1}-t_{n-1})$, with relative error tending to zero away from the tip.
\item Between the vacuum and $t_1$ the functional $g$ has exactly one zero, $\tint\to1/4\pi$, with $g>0$ on
$(0,\tint)$ and $g<0$ on $(\tint,t_1)$.
\end{enumerate}
\end{conjecture}

\noindent The leading term of (a) is \cref{thm:1}; \cref{thm:5} (hence \cref{cor:6}) proves the upper bound with $0.483$ in place of
$\Aopt$; and the density and the profile of (b) are, in a weak sense, \cref{thm:7}. Item (a) concerns $\Dbest$; the
numerics of \cref{sec:5} probe the limit $P\to\infty$ of the order-$P$ optimal gap bounds, which bounds $\Dbest$ from
above, equality being conjectural (\cref{sec:2.6}).

\section{Numerical evidence}
\label{sec:5}

\tagpar{\numerical}
We compare \cref{sec:4} with optimal functionals computed numerically: the slope, then the items of
\cref{conj:1}, then hypothesis \hyp{A} itself. No proof in this paper uses the data of this section.

\subsection{Method and convergence}
\label{sec:5.1}

We solve directly for the extremal spectrum $\{t_n,d_n\}$ of \cref{sec:2.2}, the dual variables of the functional
problem (the ``primal'' formulation in the terminology of~\cite{AHT}); by the duality of \cref{sec:2.2} its lowest level
is $t_1^{(P)}(c)$. At order $P$ the unknowns are $P/2$ levels $t_n$ and their weights $d_n$. They are found by Newton's
method~\cite{ExtremalFlow,AHT,ACHLT}, continued upward in $P$ from a linear-programming seed at small $P$.
The solver works in Arb~\cite{Arb} ball arithmetic (interval arithmetic in midpoint--radius form), used for its precision and not
for its error bounds, at $768$ bits or more (\cref{app:G.2}).
Positivity of $g$ above $t_1$ and the pattern of double zeros are checked afterwards at every $(c,P)$.

\emph{Coverage.} We have full spectra at
$c=64$, $128$, $256$, $300$, $384$, $500$, $512$, $768$, $1024$, $1536$, $2048$, with $P$ up to $3072$. For the gap alone
a second, independent implementation of the same method adds a grid $c\in\{24,\dots,1800\}$ with $P$ up to $3000$;
where both ran, they agree to the $20$--$22$ digits recorded. The solver is calibrated on the exact optima at $c=4$ and
$c=12$ (\cref{sec:2.2}).

\emph{Convergence in the order.} Empirically, the order-$P$ bound $\Delta_1^{(P)}(c)$ of \cref{sec:2.2} and its limit
$\Delta_1^{(\infty)}(c)$ obey
\begin{equation}\label{eq:5-conv}
  \Delta_1^{(P)}(c)-\Delta_1^{(\infty)}(c)\approx C\,e^{-k\sqrt{P/c}} ,\qquad k\approx4.7\text{--}5.7 ,
\end{equation}
a stretched exponential in $P/c$. Two things follow. One needs $P\gtrsim c$. And an extrapolation by a pure exponential in $P$, as in~\cite{AHT}, overestimates
$\Delta_1^{(\infty)}(c)$, that is, it biases $\Lambda$ low. We extrapolate with a stretched exponential whose exponent is
free, and quote as uncertainty $\sigma$ the spread to cruder extrapolants; the rule is stated in \cref{app:G.2}.

\subsection{The slope, and why earlier fits gave 9.08}
\label{sec:5.2}

\begin{table}[t]
\centering\footnotesize
\setlength{\tabcolsep}{4.8pt}
\begin{tabular}{@{}rrllll@{}}
\toprule
$c$ & $P_{\max}/c$ & $\Delta_1^{(\infty)}(c)$ & \multicolumn{1}{c}{$\sigma$} & $c/\Delta_1^{(\infty)}$ & \\
\midrule
 24 & 25.00 &  3.440977 & $4.3\cdot10^{-5}$ & 6.9748   & II \\
 48 & 18.25 &  6.244701 & $2.6\cdot10^{-5}$ & 7.6865   & II \\
 64 & 16.00 &  8.086588 & $2.5\cdot10^{-8}$ & 7.914339 & I  \\
 96 &  8.33 & 11.736817 & $6.5\cdot10^{-5}$ & 8.1794   & II \\
100 &  8.00 & 12.190911 & $7.7\cdot10^{-5}$ & 8.2028   & II \\
128 & 16.00 & 15.359989 & $4\cdot10^{-8}$   & 8.333339 & I  \\
192 &  6.25 & 22.560582 & $2.3\cdot10^{-4}$ & 8.5104   & II \\
256 &  8.00 & 29.724792 & $1.4\cdot10^{-5}$ & 8.612339 & I  \\
288 &  4.86 & 33.297942 & $4.3\cdot10^{-4}$ & 8.649    & II \\
300 &  5.33 & 34.636636 & $4.6\cdot10^{-4}$ & 8.661    & II \\
384 &  8.00 & 43.992503 & $1.9\cdot10^{-5}$ & 8.728760 & I  \\
400 &  5.00 & 45.772111 & $5.2\cdot10^{-4}$ & 8.7389   & II \\
\bottomrule
\end{tabular}
\hfill
\begin{tabular}{@{}rrllll@{}}
\toprule
$c$ & $P_{\max}/c$ & $\Delta_1^{(\infty)}(c)$ & \multicolumn{1}{c}{$\sigma$} & $c/\Delta_1^{(\infty)}$ & \\
\midrule
 500 & 3.75 &  56.880799 & $1.4\cdot10^{-3}$ & 8.790    & II \\
 512 & 6.00 &  58.212412 & $4\cdot10^{-5}$   & 8.795375 & I  \\
 600 & 3.33 &  67.970939 & $1.1\cdot10^{-3}$ & 8.827    & II \\
 700 & 2.60 &  79.047416 & $2.0\cdot10^{-3}$ & 8.855    & II \\
 768 & 4.00 &  86.573118 & $1.2\cdot10^{-3}$ & 8.87111  & I  \\
1000 & 2.00 & 112.220831 & $3.0\cdot10^{-3}$ & 8.911    & II \\
1024 & 2.25 & 114.871949 & $2.8\cdot10^{-3}$ & 8.91427  & I  \\
1200 & 2.50 & 134.304424 & $2.5\cdot10^{-3}$ & 8.935    & II \\
1400 & 2.14 & 156.370479 & $5.2\cdot10^{-3}$ & 8.953    & II \\
1536 & 2.00 & 171.367432 & $4.5\cdot10^{-3}$ & 8.96320  & I  \\
1800 & 1.67 & 200.464339 & $8.0\cdot10^{-3}$ & 8.979    & II \\
2048 & 1.50 & 227.7834   & $1.0\cdot10^{-2}$ & 8.9910   & I  \\
\bottomrule
\end{tabular}
\caption{The optimal gap bound extrapolated to $P=\infty$. Columns: the central charge; the highest order reached,
in units of $c$; $\Delta_1^{(\infty)}(c)$; its uncertainty $\sigma$ (\cref{app:G.2}: half the larger distance to two
cruder extrapolants, at least $10^{-7}c$; if $P_{\max}/c\ge6$, the spread of three tail fits); the
ratio $c/\Delta_1^{(\infty)}(c)$; and the implementation, I for the
ball-arithmetic solver and II for the second, independent one. Rows with $P_{\max}/c\ge4$ are converged: their
extrapolation correction is below $3\cdot10^{-6}c$. The others rest on \eqref{eq:5-conv}.}
\label{tab:2}
\end{table}

\begin{figure}[t]
\centering
\figfile{fig6_slope}
\caption{The slope. (a) The ratio $c/\Delta_1^{(\infty)}(c)$ of \cref{tab:2} against $c$, on an axis that is linear in
$c^{-2/3}$, so that $c=\infty$ is a point of the axis. Filled and open circles: implementations I and II. Solid: the
two-term law $\Dbest=c/\LV+\Aopt c^{1/3}$ of \cref{conj:1}, which has no fitted parameter; the data lie slightly
above it because of the $O(1)$ remainder. Dashed: the logarithmic fit $c/9.08+0.439\log c-0.896$ of~\cite{AHT}. Dotted
lines, star and square: the asymptotes $\LV$ and $9.08$. Shaded: the window $300\le c\le700$ of that fit. Inset: all
rows of the table. (b) The remainder $\Delta_1^{(\infty)}(c)-c/\LV-\Aopt c^{1/3}$, with error bars $\sigma$. It is of order
one and varies slowly, as \eqref{eq:4-airy} asserts. Dashed: the same quantity for the logarithmic fit, which follows
the data on its window and departs from them beyond it, by $0.22$ at $c=2048$, where $\sigma=0.010$.}
\label{fig:6}
\end{figure}

\Cref{tab:2} lists $\Delta_1^{(\infty)}(c)$ for $24\le c\le2048$, and \cref{fig:6} plots it. By \cref{conj:1}(a),
$t_1/c=\xis+\Aopt c^{-2/3}+O(1/c)$. A linear least-squares fit of $t_1/c$ in the basis
$\{1,c^{-2/3},c^{-1},c^{-4/3}\}$ gives
\begin{equation}\label{eq:5-meas}
  \xi^*_{1,\rm meas}=0.026240\pm0.000025 ,\qquad \Lmeas=9.1263\pm0.002 ,
\end{equation}
against $\xis=0.0262361$ and $\LV=9.12663$. The uncertainty is the spread over fit forms and over the lower end of
the fit window, not a statistical error. The value $\xi_1=0.02680$ that corresponds to $\Lambda=9.08$ is not approached
by any of the forms.

\smallskip\noindent\textbf{The 9.08 fit.} The difference from~\cite{AHT} is in the fitting function, not in the data.
We fit the form used there, $c/\Lfit+\gfit\log c+\text{const}$, not to data but to the exact two-term law
$c/\LV+\Aopt c^{1/3}$. On $c\in[300,700]$, the window of~\cite[\S5.2]{AHT}, the fit returns
$(\Lfit,\gfit)=(9.085,\,0.48)$, against the published $(9.08,\,0.439)$. The reason is that $c^{1/3}$ and $\log c$ are
nearly collinear over a decade of $c$; accordingly the result drifts with the window, from $(9.02,0.28)$ on $[20,300]$
to about $(9.11,0.71)$ on $[500,4000]$. The two other fitting functions of~\cite{AHT}, $c/A+B_1$ and $c/A+B_1+B_2/c$, treated
in the same way return $A=9.004$ and $9.043$, against their $9.01$ and $9.04$. The published spread is thus reproduced,
to within $0.01$ in each entry, by a law of slope $9.127$: the earlier extrapolation is explained, not contradicted.

\subsection{Fine structure}
\label{sec:5.3}

We check the items of \cref{conj:1} in turn, and then \hyp{A}. Item (a): the fitted coefficient of $c^{-2/3}$ in
$t_1/c$ lies between $0.269$ and $0.285$, depending on the form and the window, against $\Aopt=0.27666$; and a term
$\gamma\log c$ allowed to compete with $c^{1/3}$ comes out with $\gamma$ between $-0.07$ and $-0.005$, compatible with
zero. The first spacing $(t_2-t_1)\,c^{-1/3}$ decreases from $0.28963$ at $c=64$ to $0.22369$ at $c=768$ and extrapolates,
in the basis $\{1,c^{-2/3},c^{-1}\}$, to $0.20700$, against $0.20705$ (two other correction forms give $0.2043$ and
$0.2085$). Item (b): the spacings $s_n$ of the double zeros,
with $s_n-1$ extrapolated linearly in $1/c$, agree with $1/\rhoV-1$ to within $0.1\%$ for $0.06\le\xi\le0.20$.
Item (c): deep in the band the coefficient of $c$ in $t_n-n$ is $\frac1{24}(1\pm4\cdot10^{-4})$. Item (d): the ratio
$d_n/\bigl(\rhoC(t_n)\cdot\half(t_{n+1}-t_{n-1})\bigr)$ is $1-10^{-6}$ deep in the band and $0.98$ at $n=2$; this is
the forced part of \hyp{A}, \eqref{eq:4-forced}, verified to far better than exponential order. Item (e): at every
$(c,P)$ there is exactly one zero between the vacuum and $t_1$, with the predicted signs; it moves from
$\tint=0.0803889$ at $c=64$ to $0.0796944$ at $c=500$, that is, $\tint=1/4\pi+0.058/c+o(1/c)$.

\smallskip\noindent\textbf{The Cardy height.} Hypothesis \hyp{A} itself is tested through its consequence
\eqref{eq:4-yE}, that is, through the Cardy height $\cardyh$ of \cref{sec:4.2}. It can be read off the non-real zeros
of the truncated polynomials $\pol_P$. At every $(c,P)$ these lie on one parabola in the complex $t$-plane, with focus
at the BTZ threshold $t=0$, that is, on the lines $\abs{\Im b}=\frac{\cardyh}2+O(1/c)$ in the variable
$b=\sqrt{3t/c'}$. A fit to the vertex of the parabola for $64\le c\le500$, which does not involve $\xis$, gives
$\cardyh=1.0000\pm0.0005$; for comparison, $\cardyh=1.0107$ would correspond to the slope $9.080$. The reason the
zeros sit there is that the duality integral $\hat g(s)=2\pi\int g\,J_0$ of \cref{sec:2.3}, continued to complex
$s=c'b^2/3$, converges absolutely exactly for $\abs{\Im b}<\half$: the Bessel kernel grows like $e^{\frac{4\pi}3c'x\abs{\Im b}}$, and
on the band $g$ decays like $e^{-c'L(x)}=e^{-\frac{2\pi}3c'x}$. At finite $P$ the zeros condense on the boundary of this
region. This is the strip of \cref{sec:6.3}, seen in the data.

\PartHeading{III}{Optimality}
Everything in this Part is proved, and no computer is used. It needs only \cref{sec:2}. The strip of \cref{sec:4.2} and
the curve of \cref{sec:4.4} reappear here as lemmas about a domain and a potential, without hypotheses; a reader who
skipped Part~II loses nothing.

\section{Optimality: nothing beyond \texorpdfstring{$\Lambda_V$}{Lambda\_V}}
\label{sec:6}

We show that the spinless bootstrap cannot give a slope better than $\LV$, at any order and with any functional in
$\Abar$. The mechanism is the tension described after \eqref{eq:2-halfmass}. A functional that certifies the gap bound
$t_1$ must carry half of its Cardy-weighted mass below $t_1$; and we prove that no function in $\Abar$, feasible or not,
carries more than an exponentially small fraction of its mass below $t=\xis c'$. The statements are in \cref{sec:6.1}.
The proof is an electrostatics problem on the strip of \cref{fig:3}, described in \cref{sec:6.2}. \Cref{sec:6.3}
continues the functional to the strip and bounds it on the boundary, \cref{sec:6.4} collects the potential theory that
we use, \cref{sec:6.5} turns the bounds on the boundary into a bound in the interior by a harmonic function,
\cref{sec:6.6} finds that function in closed form, and \cref{sec:6.7,sec:6.8} draw the consequences. The technical
lemmas are in \cref{app:A,app:B}.

\subsection{Statements}
\label{sec:6.1}

\begin{theorem}[Concentration; \proved]\label{thm:3}
Let $\xs=\arcsinh(1)/\pi$. For every $\delta\in(0,\xs)$ there is a constant $C_\delta<\infty$ such that for all $c\ge40$
and every $g\in\Abar$, with no condition on the sign of $g$,
\begin{equation}\label{eq:6-conc}
  \int_0^{c'(\xs-\delta)^2/3}\abs g\,V\,\dd t\;\le\;C_\delta\,c'^{\,4}\,e^{-\kappa(\delta)\,c'}\,M[g] .
\end{equation}
Here $\kappa(\delta)>0$ is the explicit \emph{Cardy deficit}, given in closed form in \cref{lem:6.7}. It satisfies
\begin{equation}\label{eq:6-kappa-bounds}
  \kappa(\delta)=\tfrac89\sqrt{2\sqrt2\,\pi}\;\delta^{3/2}\bigl(1+O(\delta)\bigr),\qquad
  \kappa(\delta)\ge2.228\,\delta^{3/2}\quad(0<\delta\le\xs),
\end{equation}
where $\frac89\sqrt{2\sqrt2\,\pi}=2.650$, and $C_\delta\le\Cconc\,\delta^{-5}$ with an absolute constant $\Cconc$.
\end{theorem}

\noindent In words: weighted by the density of states that the vacuum demands, any bootstrap function has exponentially
little mass below $t\approx0.0262\,c'$.

\begin{theorem}[Optimality; \proved]\label{thm:4}
Let $\xis$ and $\LV$ be as in \cref{thm:1}, and call $g$ weakly feasible if it is weakly feasible at some $T$, hence at
$T=t_1(g)$. For every $\Lambda>\LV$ there is $c_0(\Lambda)$ such that for $c\ge c_0(\Lambda)$ no weakly feasible
$g\in\Abar$, $g\not\equiv0$, has $t_1(g)+a\le c/\Lambda$. Quantitatively, there is a central charge $\copt$, explicit in
principle and not computed, such that
\begin{equation}\label{eq:6-rate}
  t_1(g)\;\ge\;\xis\,c'-0.44\,c'^{1/3}\log^{2/3}c'
\end{equation}
for every weakly feasible $g\in\Abar$, $g\not\equiv0$, and every $c\ge\copt$.
\end{theorem}

\begin{proof}[Proof of the first statement of \cref{thm:4} from \cref{thm:3}]
Let $\Lambda>\LV$, that is, $\frac1\Lambda-\frac1{12}<\xis$, and choose $\delta\in(0,\xs)$ with
$\frac13(\xs-\delta)^2>\frac1\Lambda-\frac1{12}$. If $t_1(g)+a\le c/\Lambda$, then
$t_1(g)\le c'(\frac1\Lambda-\frac1{12})+\frac1\Lambda\le c'(\xs-\delta)^2/3$ as soon as $c$ is large. If moreover $g$ is
weakly feasible, $g_-$ is supported on $[0,t_1]$ and the half-mass condition \eqref{eq:2-halfmass} makes the left side
of \eqref{eq:6-conc} at least $\int_0^{t_1}g_-V\,\dd t\ge\half M[g]$, where $M[g]>0$ because $g\not\equiv0$. Hence
$\half\le C_\delta\,c'^4e^{-\kappa(\delta)c'}$, which is false for large $c$. How large $c$ must be depends on $\Lambda$ only.
\end{proof}

\noindent The rate \eqref{eq:6-rate} is proved in \cref{sec:6.8}. Three remarks. First, the theorem quantifies over all
of $\Abar$, and $c_0(\Lambda)$ does not depend on $g$. Since $\Abar$ contains the functionals of every order $P$
(\cref{lem:2.1}), the statement is uniform in $P$: for $c\ge c_0(\Lambda)$ no amount of computing produces a spinless
bound below $c/\Lambda$. Second, this is a statement about the method, the linear program of \cref{sec:2.2}, and not about
conformal field theories. Third, the informative content of the quantitative form is its \emph{rate}; the right side
of \eqref{eq:6-rate} lies below the trivial floor $t_1\ge0$ of \cref{sec:2.4} for $c'\lesssim414$.

\paragraph{Attribution.} The strategy is the following: continue the functional to a strip; control one part of the
boundary by the total mass and the other by the eigen-identity; bound the function inside by its \emph{harmonic
majorant}, the harmonic function with the same boundary data, which lies above it by the maximum principle; and
conclude through the half mass of the negative part. This is the strategy of the proof for the Cohn--Elkies problem
at large dimension in~\cite[Ch.~1, Prop.~3.1 and Thm.~3.8]{OpenAI}; we have read the statements and the structure of
that argument but have not independently verified it; nothing in this paper depends on it logically. There
$d\to\infty$ and the strip is in the Mellin-transform variable, of half-width $d/2$; here $d=2$ is fixed and the strip is in
the rescaled dimension $b$ itself, its width being fixed by the requirement that the continued kernel have exactly
Cardy growth on the edge. The strip through the vacuum, the weight $V$, the slit geometry and the closed form are
specific to Virasoro. The argument in position space does not run for $U(1)^c$. The argument in the Mellin variable
does run for Virasoro: the strip is then replaced by a domain with a free boundary (\cref{app:M}).
The last step of the strategy, the half mass of the negative part, goes back to Bourgain, Clozel and Kahane~\cite{BCK} (see
also~\cite{GOS}); Cohn and Gon\c{c}alves~\cite[\S3.1]{CohnGoncalves} use it for the functions with $\hat f=-f$ of the twin problem,
and \eqref{eq:2-halfmass} is the same step with the Cardy weight. A mass-concentration principle similar to that
of~\cite[Ch.~1]{OpenAI} has since been proved for Delsarte's linear program for binary codes~\cite{Salmon1}.

\subsection{Idea of the proof: an electrostatics problem}
\label{sec:6.2}

The stage is \cref{fig:3}(b). In the variable $b$ of \eqref{eq:2-scaling} a functional $g\in\Abar$ becomes the function
$G(b)=g(c'b^2/3)$, and the proof has five steps.
\begin{enumerate}[label=\textup{(\arabic*)},leftmargin=2.2em,itemsep=1pt,topsep=3pt]
\item $G$ continues analytically to the closed strip $\abs{\Im b}\le\half$, with the vacuum on its edge (\cref{sec:6.3}).
\item On the edges $\abs G$ is bounded by $M[g]$ up to a power of $c$ (\cref{lem:6.1}).
\item On the real axis beyond any chosen point $s$, $\abs G$ is small in the Cardy-weighted mean:
$\int_{\abs x\ge s}\abs G\,e^{c'L}\,\dd x$ is at most $M[g]$ up to a power of $c$ (\cref{lem:6.2}).
\item $u_g=\frac1{c'}\log(\abs G/M[g])$ is the potential of positive charges at the zeros of $G$, in the convention in
which a potential is $-\infty$ at its positive charges (\cref{sec:6.4}). It is \emph{subharmonic}, that is, it obeys the
maximum principle. We cut the strip along $\abs x\ge s$, so that this part of the real axis becomes boundary, where
step~(3) applies, and the interval $(-s,s)$, about which nothing is assumed, stays inside. The value of a harmonic
function inside is an average of its boundary values, so a bound in the mean suffices: $u_g$ lies below the harmonic
function that is zero on the edges and $-L$ on the slits, up to $o(1)$ (\cref{prop:6.4}).
\item For $s=\xs$ that harmonic function is explicit, and it lies strictly below the Cardy line on $\abs x<\xs$
(\cref{lem:6.5,lem:6.7}). Since $\abs g\,V\approx e^{c'(u_g+L)}M[g]$, almost no Cardy-weighted mass can sit there.
\end{enumerate}
\Cref{tab:3} is the dictionary. In its language the half-mass condition says that the potential of a weakly feasible
functional must come within $o(1)$ of the Cardy line somewhere on $\abs x\le x_1(g)$, and the theorem says that the
first point where this is possible is $\xs$.

\begin{table}[t]
\centering\small
\begin{tabular}{@{}>{\raggedright\arraybackslash}p{0.27\linewidth}>{\raggedright\arraybackslash}p{0.33\linewidth}>{\raggedright\arraybackslash}p{0.33\linewidth}@{}}
\toprule
Conformal field theory & Linear program & Electrostatics on the strip\\
\midrule
linear functional acting on the crossing equation & the function $g$; $G(b)=g(c'b^2/3)$ & potential $u_g=\frac1{c'}\log(\abs G/M[g])$\\
\addlinespace[3pt]
dimensions of the extremal spectrum & zeros of $G$ & positive point charges, $1/c'$ each\\
\addlinespace[3pt]
Cardy density $\rhoC$ that the vacuum demands & Cardy growth of the weight, $V\approx e^{c'L}$ & applied linear potential $-L(x)$\\
\addlinespace[3pt]
--- & edges $\Epm$ of the strip of analyticity & grounded plates\\
\addlinespace[3pt]
the vacuum, $\Delta=0$ & the point $t=-a$ & the point $b=\frac i2$ on a plate\\
\addlinespace[3pt]
a gap bound, $\Delta_1\le t_1(g)+a$ & weak feasibility, through the half-mass condition & contact with the Cardy line\\
\addlinespace[3pt]
slope $\LV$ of the best bound & optimal slope, $\xis=(\xs)^2/3$ & first point of contact, $x=\xs$\\
\bottomrule
\end{tabular}
\caption{The dictionary between the conformal field theory, the linear program and the electrostatics problem of
\cref{sec:6}.}
\label{tab:3}
\end{table}

\subsection{Continuation to the strip}
\label{sec:6.3}

The idea is that the eigen-identity \eqref{eq:2-eigenid}, which gives the values of $g$ below the BTZ threshold as
integrals over $t\ge0$, makes sense for complex $\alpha$ as long as the kernel $K_\alpha$ grows no faster than the
Cardy weight. Since $\abs{I_0(z)}\le I_0(\abs{\Re z})$, this is the case on the parabolic region
$\abs{\Re\sqrt\alpha}\le\sqrt a$, whose vertex is the vacuum and whose focus is the BTZ threshold (\cref{fig:3}(a)): by
\cref{lem:A.1} the integral $g(-\alpha)=-2\pi\int_0^\infty gK_\alpha\,\dd t$ converges absolutely there and is dominated by
$2\pi\int_0^\infty\abs gK_a\,\dd t<\infty$. In the variable $b$ we have $\alpha=-c'b^2/3$ and
$\abs{\Re\sqrt\alpha}=\sqrt{c'/3}\,\abs{\Im b}$, so the region is the closed strip
$\bar\strip=\{\abs{\Im b}\le\half\}$. Hence
\begin{equation}\label{eq:6-G}
  G(b):=g\bigl(c'b^2/3\bigr)=-2\pi\int_0^\infty g(t)\,K_{-c'b^2/3}(t)\,\dd t ,\qquad
  \abs{G(b)}\le2\pi\int_0^\infty\abs g\,K_a\,\dd t\qquad(b\in\bar\strip)
\end{equation}
is well defined for every $g\in\Abar$. By \cref{lem:A.1}, $G$ is bounded and continuous on $\bar\strip$ and holomorphic
on $\strip$. It is even, and real on the real axis and on the segment $i[-\half,\half]$. For real $b=x$ it is the
functional itself, $G(x)=g(c'x^2/3)$; in particular $G\not\equiv0$ unless $g\equiv0$. For a functional of finite order $G$
is the restriction of an entire function, but this plays no role.

The landmarks are those of \cref{fig:3}(b). The real axis is the spectrum above the BTZ threshold, $t\ge0$. The
segment $b=iy$, $0\le y\le\half$, is the range $-a\le t\le0$ below it. The points $b=\pm\frac i2$ are the vacuum,
$G(\pm\frac i2)=g(-a)$. On a line $\abs{\Im b}=\eta$ the kernel is bounded by $K_{(2\eta)^2a}\approx e^{2\eta\,c'L}$. This is smaller than the
Cardy weight for $\eta<\half$, equal to it for $\eta=\half$, and larger for $\eta>\half$, where $M[g]<\infty$ no longer
controls the integral. \emph{The vacuum sits on
the edge of the strip of analyticity that Cardy growth allows.}

\begin{lemma}[Edges; \proved]\label{lem:6.1}
For $b=x\pm\frac i2$ on $\Epm$ and $\alpha=-c'b^2/3$ one has $(\Re\sqrt\alpha)^2=a$ for every real $x$, and therefore
$\abs{K_\alpha(t)}\le K_a(t)$ for all $t\ge0$: on the edges the continued kernel has exactly Cardy growth. Since $V/K_a$
is only polynomially small, there is an absolute constant $C$ such that for all $c\ge40$
\begin{equation}\label{eq:6-edge}
  \bigabs{G(x\pm\tfrac i2)}\le C\,a\,c'\,M[g]
\end{equation}
uniformly in $x\in\RR$ and in $g\in\Abar$, in particular in the order $P$.
\end{lemma}

\noindent The idea of the proof: the kernel bound is \eqref{eq:A-kernel-bound}, so by \eqref{eq:6-G}
$\abs{G(x\pm\frac i2)}\le2\pi\int_0^\infty\abs gK_a\,\dd t$. For $t\ge4/c'$ one has $K_a\le C\,a\,c'\,V$ (\cref{lem:A.2}(b)). In the
window $t\le4/c'$ at the BTZ threshold, where $V\simeq8\pi^4t^2$ degenerates while $K_a$ stays of order one,
$\hat g=-g$ bounds $\sup\abs g$ by a multiple of $M[g]$ (\cref{lem:A.3}). The proof is in \cref{app:B.4}. At the
exponential scale $e^{O(c')}$ of the rest of the problem the edges carry weight zero:
$u_g\le(2\log c'+C)/c'$ there, in the notation \eqref{eq:6-ug} below.

\begin{lemma}[Slits; \proved]\label{lem:6.2}
On $\abs x\ge\xs$, that is, for $t\ge\xis c'$, one has $V\ge0.9\,K_a$. Hence for all $c\ge40$ and every $g\in\Abar$,
\begin{equation}\label{eq:6-slit}
  \int_{\abs x\ge\xs}\abs{G(x)}\,K_a\bigl(c'x^2/3\bigr)\,\dd x\le\frac{12}{c'}\,M[g] .
\end{equation}
\end{lemma}

\begin{proof}
$G$ is even. With $t=c'x^2/3$ we have $\dd x=3\,\dd t/(2c'x)\le3\,\dd t/(2c'\xs)$ for $x\ge\xs$, and
$V\ge0.9\,K_a$ there by \cref{lem:A.2}(c). Hence the left side of \eqref{eq:6-slit} is at most
\begin{equation*}
  \frac3{c'\xs}\int_{\xis c'}^\infty\abs g\,K_a\,\dd t\le\frac3{0.9\,c'\xs}\int_{\xis c'}^\infty\abs g\,V\,\dd t
  \le\frac{11.9}{c'}\,M[g] .\qedhere
\end{equation*}
\end{proof}

\noindent On the slits $G(x)=g(c'x^2/3)$ is the functional in the region where a good functional is non-negative.
Nothing is known there pointwise, and an extremal functional vanishes there infinitely often; \eqref{eq:6-slit} is a
statement in the mean, and \cref{sec:6.5} uses it as such.

\begin{remark*}[Why $c'=c-1$ and not $c$; \orientation]
The scaling $t=c'b^2/3$ puts the edge of the strip through the true vacuum $t=-a$, where the kernel is bounded by $K_a$,
which exceeds $V$ by a power of $c'$ only. With $c$ in place of $c'$ the edge would pass through $t=-c/12$, where the
kernel is bounded only by $K_{a+1/12}$, which exceeds $V$ by a factor growing like $e^{(\pi/3)\sqrt{3t/c}}$. Since feasible
functionals of high order carry Cardy-weighted mass arbitrarily far out in $t$, the edge bound would then depend on the
functional, and the uniformity in $g$ would be lost.
\end{remark*}

\noindent What is \emph{not} used: sign information (positivity enters only through the half-mass condition, in one line,
at the very end); information about $g$ near the BTZ threshold beyond integrability; any property of optimal
functionals. The inputs on Bessel functions are \cref{lem:A.1,lem:A.2,lem:A.3}.

\subsection{Toolkit: potential theory in physicist's terms}
\label{sec:6.4}

Steps (4) and (5) of \cref{sec:6.2} are two-dimensional electrostatics. The box collects the notions and the four
formulas that we use, first in the language of charges and potentials and then precisely. We use the sign convention in
which a unit point charge at $\zeta$ has the potential $\log\abs{b-\zeta}$.

\begin{toolkit}{Toolkit: potential theory}
\emph{Subharmonic functions.} If $G\not\equiv0$ is holomorphic, then
\begin{equation}\label{eq:6-laplace}
  \nabla^2\log\abs{G(b)}=2\pi\sum_{\zeta:\,G(\zeta)=0}m_\zeta\,\delta^2(b-\zeta)\;\ge0 ,
\end{equation}
with $m_\zeta$ the multiplicity: $\log\abs G$ is the potential of positive point charges at the zeros of $G$, and it is
harmonic away from them. A function $u$ with $\nabla^2u\ge0$ in the sense of distributions is called
\emph{subharmonic}; precisely, $u$ is upper semicontinuous with values in $[-\infty,\infty)$ (it may jump down, as at a
charge, but not up), and $u(b)$ is at most the mean of $u$ over every small circle around $b$. Such a function obeys the
\emph{maximum principle}: if $h$ is harmonic on
a bounded domain and $\limsup u\le h$ at every boundary point, then $u\le h$ inside~\cite[Thm.~2.3.1]{Ransford}. Adding positive charges lowers the potential.

\emph{Harmonic measure.} For a domain $\Omega$ and a point $b\in\Omega$, $\omb(E)$ is the probability that a Brownian
path started at $b$ first leaves $\Omega$ through the part $E$ of the boundary. Equivalently, $b\mapsto\omb(E)$ solves
the Dirichlet problem with boundary values $1$ on $E$ and $0$ elsewhere, and $\int\varphi\,\dd\omb$, the \emph{Poisson
integral} of $\varphi$, solves it with boundary values $\varphi$~\cite[\S4.3]{Ransford}; for the upper half-plane
$\dd\omega_z=\Im z\,\dd\sigma/(\pi\abs{\sigma-z}^2)$, the \emph{Poisson kernel}. On the slit strip the two lips of a slit are different boundary points, and
$\dd\omb/\dd x$ denotes the density of $\omb$ on a slit with the two lips added. It is explicit through the conformal
map of \cref{lem:6.3}. Like the charge density at the rim of a thin conducting plate, it has an inverse square-root
singularity at the tips.

\emph{Harmonic majorant.} This is the solution $h$ of the Dirichlet problem whose data are the boundary values of a
subharmonic $u$, that is, the potential with the same boundary values and with all interior charges removed. By the
maximum principle $u\le h=\int u\,\dd\omb$~\cite[\S4.1]{Ransford}.

\emph{Poisson--Jensen.} For $G$ bounded and holomorphic on $\Omega$, with boundary values taken $\omb$-almost everywhere,
\begin{equation}\label{eq:6-PJ}
  \log\abs{G(b)}\le\int_{\partial\Omega}\log\abs G\,\dd\omb .
\end{equation}
If $G$ is holomorphic on a neighborhood of $\bar\Omega$, the right side minus the left side is
$\sum_\zeta m_\zeta\,\mathsf g_\Omega(b,\zeta)$, where $\mathsf g_\Omega\ge0$ is the Green's function of $\Omega$ and the sum
runs over the zeros of $G$ in $\Omega$~\cite[\S4.5]{Ransford}: each zero lowers the potential by the potential of a unit
charge in the grounded domain. We need \eqref{eq:6-PJ} for a merely bounded function that may vanish on the
boundary (\cref{lem:B.3}(a)); for the disk this is classical: removing the charges at the zeros, inside or on the
boundary, only raises the potential~\cite[Ch.~II, Thm.~4.1 and Cor.~4.2]{Garnett}, \cite[Thms.~17.16--17.17]{Rudin}.

\emph{Jensen's inequality with entropy.} On the slits we know only a weighted mean of $\abs G$ (\cref{lem:6.2}), whereas
\eqref{eq:6-PJ} asks for the mean of $\log\abs G$ with respect to $\omb$. Let $w$ be a probability density on a set $S$ of
the real line and $K>0$ a weight. Since the mean of a logarithm is at most the logarithm of the mean~\cite[Thm.~3.3]{Rudin},
\begin{equation}\label{eq:6-jensen}
  \int_Sw\log\abs G\,\dd x\;\le\;\log\int_S\abs G\,K\,\dd x\;-\int_Sw\log K\,\dd x\;+\int_Sw\log w\,\dd x .
\end{equation}
The change of reference density from $K$ to $w$ costs the \emph{entropy} $\int w\log w$. With $K=K_a\approx e^{c'L}$ the
middle term is $-c'\int wL$, which is how the Cardy line enters. For $w$ proportional to $\dd\omb/\dd x$ the entropy is finite despite
the singularity at the tips; it is bounded in \cref{lem:B.2}, together with its dependence on the distance from $b$ to the
tips.

\emph{Phragm\'en--Lindel\"of.} On an unbounded domain the maximum principle needs a growth condition: on the strip
$\strip$ the function $\cosh(\pi x)\cos(\pi y)$ is harmonic and vanishes on both edges. A harmonic function on
$\strip$ that vanishes on the edges and is $O(e^{\alpha\abs x})$ for some $\alpha<\pi$ is zero~\cite[\S2.3]{Ransford}. All our
functions grow at most linearly in $\abs x$. The form that we use is proved in the proof of \cref{lem:6.5}.

\emph{Riesz measure.} The charge distribution $\frac1{2\pi}\nabla^2u$ of a subharmonic function $u$ is a positive measure,
its \emph{Riesz measure}, and locally $u$ is the logarithmic potential of that measure plus a harmonic function~\cite[\S3.7]{Ransford}. For $u_g$ of \eqref{eq:6-ug} it is, by \eqref{eq:6-laplace},
\begin{equation}\label{eq:6-riesz}
  \nug:=\frac1{2\pi}\nabla^2u_g=\frac1{c'}\sum_{\zeta\in\strip:\,G(\zeta)=0}m_\zeta\,\delta_\zeta ,
\end{equation}
the counting measure of the zeros of $G$ in the strip, normalized so that a zero is a charge $1/c'$.
\end{toolkit}

\subsection{The slit strip and the majorant}
\label{sec:6.5}

We now choose the domain on which the maximum principle is applied. For $s>0$ let
\begin{equation}\label{eq:6-Ds}
  D_s:=\strip\setminus\{x+iy:\ \abs x\ge s,\ y=0\},\qquad D:=D_{\xs} .
\end{equation}
$D_s$ is simply connected. Its boundary consists of the two edges and of two slits, and each slit has two lips, which
count as different boundary points; the ends of the slits, $b=\pm s$, are the \emph{tips}. We prove everything for $D$;
the choice of $s$ is discussed in \cref{sec:6.6}. The next lemma makes harmonic measure on $D$ explicit.

\begin{lemma}[Conformal map; \proved]\label{lem:6.3}
The function
\begin{equation}\label{eq:6-map}
  b(z)=\frac1{2\pi}\log\frac{z-1}{z\,(z+1)} ,
\end{equation}
with the branch that is continuous on the upper half-plane $\HH$ and has $b(i)=0$, maps $\HH$ conformally onto $D$. It
maps the intervals $(0,1)$ and $(-\infty,-1)$ onto the edges $E_+$ and $E_-$, and $(-1,0)$ and $(1,\infty)$ onto the
right and the left slit, each folded at its tip, $b(1\mp\sqrt2)=\pm\xs$, so that the two halves of the interval give the
two lips. Moreover $b(\sqrt2-1)=\frac i2$, and the points $-1,0,1,\infty$ correspond to the four ends of $D$
\textup{(\cref{fig:5}(b), last row of labels)}.
\end{lemma}

\noindent The proof is \cref{lem:B.1}(a),(b); part (c) of that lemma transports the Poisson kernel of $\HH$ to $D$ and
gives the density of $\omb$, its exponential decay along the slits and its first moment. The map is the uniformization
of \cref{sec:4.4}, \eqref{eq:4-pb}; it is now a lemma about a domain and not a hypothesis about a curve.

For $g\in\Abar$, $g\not\equiv0$, we write
\begin{equation}\label{eq:6-ug}
  u_g(b):=\frac1{c'}\log\frac{\abs{G(b)}}{M[g]}\qquad(b\in\bar\strip),
\end{equation}
which is subharmonic on $\strip$ by \eqref{eq:6-laplace}, even, and symmetric under $b\mapsto\bar b$. The data on the
boundary of $D$ are these. On the edges, $u_g\le(2\log c'+C)/c'$ by \cref{lem:6.1}. On the lips of the slits we have only
the weighted $L^1$ bound of \cref{lem:6.2}. Jensen's inequality \eqref{eq:6-jensen} with $K=K_a$, together with the
finiteness of the entropy of harmonic measure, converts it into $u_g\le-L(x)+o(1)$ in the mean with respect to
$\omb$. The entropy is the one place where the geometry of $D$, namely the square-root singularity of harmonic measure
at the tips, has to be examined by hand. Poisson--Jensen \eqref{eq:6-PJ} then bounds $u_g(b)$ by the mean of these data,
and the mean of the data is a harmonic function of $b$.

\begin{proposition}[Majorant; \proved]\label{prop:6.4}
For $s>0$ let $U_s$ denote the harmonic function on $D_s$ that is continuous up to the edges, both lips of the slits
and the tips, with
\begin{equation}\label{eq:6-dirichlet}
  U_s=0\ \text{ on }\Epm,\qquad U_s=-L(x)\ \text{ on both lips of the slits},\qquad
  \abs{U_s(b)}\le C\bigl(1+\abs{\Re b}\bigr) .
\end{equation}
For every compact set $Q\subset D$ there is a constant $\CQ$ such that for all $c\ge40$, all $g\in\Abar$ with
$g\not\equiv0$, and all $b\in Q$,
\begin{equation}\label{eq:6-majorant}
  u_g(b)\le U_{\xs}(b)+\CQ\,\frac{1+\log c'}{c'} .
\end{equation}
\end{proposition}

\noindent The linear growth at the four ends makes the solution of \eqref{eq:6-dirichlet} unique (proof of
\cref{lem:6.5}, part (d)); for $s=\xs$ it exists as a Poisson integral, $U_{\xs}(b)=-\int L\,\dd\omb$
over the slits, \eqref{eq:B-Poisson-U}, and other $s$ enter no proof. The
proof of \cref{prop:6.4} is in \cref{app:B.4}. It inserts \cref{lem:6.1} and the Jensen bound into the Poisson--Jensen
inequality on $D$ and arrives, for every $b\in D$, at the master inequality \eqref{eq:B-master}, in which the power of
$c'$ and the entropy appear explicitly. The technical lemmas are \cref{lem:B.3} (Poisson--Jensen with zeros on the
boundary) and \cref{lem:B.2} (the entropy near a tip).

The picture is that of two grounded plates and, between them, two thin electrodes held at the linearly decreasing
potential $-L$. The function $U_s$ no longer knows anything about $g$ or about conformal field theory, and
$e^{c'U_{\xs}(b)}$ is, up to a power of $c'$, a universal bound for $\abs G/M[g]$ on compact subsets of $D$, in
particular below the BTZ threshold and on the lobe interval. Nothing is imposed on the interval $(-s,s)$. That is where the mass of a would-be better functional
has to be, by the half-mass condition, and the point of the construction is that $u_g$ is small there all the same.

\subsection{Why \texorpdfstring{$s=x_1^*$}{s=x1*}; the extremal potential in closed form; the zero law}
\label{sec:6.6}

\tagpar{\orientation} \emph{Why this slit.} For any $s$ the argument of \cref{sec:6.5} gives the majorant $U_s$ on
$D_s$, and it bounds the Cardy-weighted mass wherever $U_s+L<0$ inside $(-s,s)$. A harmonic function on a slit domain
generically has a square-root singularity at a tip: $U_s+L\simeq\gtip(s)\sqrt{s-\abs x}$ inside the tips. The coefficient
$\gtip$ changes sign exactly once, at $s=\xs$; numerically $\gtip=-0.16$, $-0.02$, $\approx0$, $+0.02$, $+0.16$ at
$s=0.24$, $0.275$, $\xs$, $0.285$, $0.32$. For $s<\xs$ one finds $U_s+L<0$ on all of $(-s,s)$, and the method gives the
weaker bound $\Lambda\le1/(\frac1{12}+\frac{s^2}3)$; for instance $s=0.26$ gives $\Lambda\le9.45$. For $s>\xs$ the majorant
rises above the Cardy line inside the tips and excludes nothing there. Thus
$s=\xs$ is the largest useful slit, characterized by \emph{tangency}: $U_s+L$ vanishes to order $\frac32$ at the
tips, which is a free-boundary condition. None of this is used in the proofs.

For $s=\xs$ the Dirichlet problem \eqref{eq:6-dirichlet} has a solution in closed form. The idea is to guess the
holomorphic function $p$ whose imaginary part is $\frac3{4\pi}\partial_xU$: since $U=-L$ on the slits and $U$ is constant
on the edges, $\Im p$ must be $\mp\half$ on the slits and $0$ on the edges.

\begin{figure}[t]
\centering
\figfile[\linewidth]{fig7_potential}
\caption{The extremal potential $U$ of \cref{lem:6.5}. (a)~Equipotentials of $U$ on the strip, in steps of $0.1$, darker
meaning lower. $U=0$ on the edges, where the open circles are the two images $b=\pm\frac i2$ of the vacuum; $U=-L$ on the
slits $\abs x\ge\xs$, $y=0$, whose tips are the dots; and the dashed level $U=U_0=-0.3887$ passes through the saddle at the
BTZ threshold $b=0$. (b)~On the real axis $U$ (solid) lies below the Cardy line $-L(x)=-\frac{2\pi}3x$ (dashed) and
joins it tangentially at $\xs$; the vertical distance at $x=\xs-\delta$ is the Cardy deficit $\kappa(\delta)$ of
\cref{lem:6.7}. The ticks on the left axis are the values $u_g(0)$ of numerically optimal functionals at
$c=64,128,256,512$, which approach $U_0$ from below (\cref{sec:6.7}). The ticks on the bottom axis are the first zeros
$x_1(g)=(3t_1/c')^{1/2}$ of the optimal functionals of \cref{tab:2} at $c=64,128,\dots,2048$, which approach $\xs$ from
above. Inset: the zero law $\rhoU(x)$ of \cref{def:6.6} (solid), its counting function $N(x)$ (dashed) and the tip
asymptote $\ktip(x-\xs)^{3/2}$ (dotted).}
\label{fig:7}
\end{figure}

\begin{lemma}[Closed form; \proved]\label{lem:6.5}
On $D$ let
\begin{equation}\label{eq:6-p}
  p(b):=\frac1\pi\,\arcsinh\bigl(-i\sinh\pi b\bigr)
\end{equation}
with the principal branch of $\arcsinh$, whose cuts are $\pm i[1,\infty)$. It is single-valued and holomorphic exactly on
$D$, its cuts in the strip being the slits. Let
\begin{equation}\label{eq:6-U}
  U(b):=U_0+\frac{4\pi}3\,\Im\!\int_0^bp(b')\,\dd b' ,\qquad
  U_0:=-L(\xs)+\frac43\int_0^{\xs}\!\arcsin(\sinh\pi s)\,\dd s=-0.388748 ,
\end{equation}
the integral being taken along any path in $D$. Then $U$ is harmonic on $D$, continuous up to the edges, the lips and
the tips, equal to $0$ on $\Epm$ and to $-L$ on both lips of both slits, and $\abs{U(b)}\le\frac{2\pi}3\abs{\Re b}+C$. It is
the only harmonic function on $D$ with this continuity and these boundary values that grows at most linearly. In
particular $U=U_{\xs}$. On the axes,
\begin{equation}\label{eq:6-axes}
\begin{aligned}
  U(x)&=U_0-\frac43\int_0^{\abs x}\!\arcsin(\sinh\pi s)\,\dd s &&(\abs x\le\xs),\\
  U(iy)&=U_0+\frac43\int_0^{\abs y}\!\arcsinh(\sin\pi s)\,\dd s &&(\abs y\le\half).
\end{aligned}
\end{equation}
\end{lemma}

\noindent We call $U$ the \emph{extremal potential}. By \eqref{eq:6-dU} below its complex electric field is
$\partial_xU-i\,\partial_yU=-\frac{4\pi i}3\,p$, and $p$ is related to $b$ by $\sinh\pi b=i\sinh\pi p$, which is the curve
\eqref{eq:4-curve}: \emph{the spectral curve of \cref{sec:4} is the complex electric field of the extremal potential.}

\begin{proof}
We compute the boundary values of $p$, integrate them, and prove uniqueness by a barrier argument; one identity between
two integrals makes the constant on the edges vanish.

\emph{(a) The function $p$.} The principal branch $\arcsinh w=\log(w+\sqrt{1+w^2})$ is the inverse of $\sinh$ on the strip
$\abs{\Im\zeta}<\frac\pi2$; it is holomorphic on $\CC\setminus\pm i[1,\infty)$. For $b=x+iy\in\strip$ the point
$-i\sinh\pi b$ lies on a cut if and only if $\sinh\pi b=\sinh\pi x\cos\pi y+i\cosh\pi x\sin\pi y$ is real and of modulus at
least one. Since $\abs y<\half$ this means $y=0$ and $\abs{\sinh\pi x}\ge1$, that is, $b$ real with $\abs b\ge\xs$. So $p$ is
holomorphic on $D$, with $p(0)=0$, $p(-b)=-p(b)$ and $p(\bar b)=-\overline{p(b)}$. Furthermore $\abs{\Im p}<\half$, and
$\sinh\pi p=-i\sinh\pi b$ gives $\sinh^2(\pi\Re p)\le\abs{\sinh\pi p}^2=\sinh^2\pi x+\sin^2\pi y\le\cosh^2\pi x$, hence
$\abs{\Re p(b)}\le\abs{\Re b}+1$. The values on the axes and on the boundary are
\begin{align*}
  p(x)&=-\tfrac i\pi\arcsin(\sinh\pi x)\quad(\abs x<\xs), &
  p(iy)&=\tfrac1\pi\arcsinh(\sin\pi y),\quad p(\pm\tfrac i2)=\pm\xs ,\\
  p(x\pm\tfrac i2)&=\pm\tfrac1\pi\arcsinh(\cosh\pi x), &
  p(x\pm i0)&=\pm\tfrac1\pi\arccosh(\sinh\pi x)-\tfrac i2\quad(x>\xs).
\end{align*}
The first three hold because $-i\sinh\pi b$ equals $-i\sinh\pi x$, $\sin\pi y$ and $\pm\cosh\pi x$ at these points. For the
last, $-i\sinh\pi(x\pm i\epsilon)$ tends to the point $-i\sinh\pi x$ of the lower cut from the right and from the left,
and $\arcsinh(\pm\eta-i\sigma)\to\pm\arccosh\sigma-\frac{i\pi}2$ as $\eta\downarrow0$, for $\sigma>1$. On the left slit the
values follow from $p(-b)=-p(b)$. In particular $\Im p=0$ on $\Epm$, $\Im p=-\half$ on both lips of the right slit and
$\Im p=+\half$ on both lips of the left slit; $\Re p(x+i0)=\frac1\pi\arccosh(\sinh\pi\abs x)$ on the upper lips of both
slits; and $p$ extends continuously to the edges, the lips and the tips.

\emph{(b) The function $U$.} Since $D$ is simply connected, $\int_0^bp$ is a holomorphic function on $D$, so $U$ is
harmonic, and
\begin{equation}\label{eq:6-dU}
  \dd U=\frac{4\pi}3\,\Im(p\,\dd b)=\frac{4\pi}3\bigl(\Im p\,\dd x+\Re p\,\dd y\bigr) .
\end{equation}
The symmetries of $p$ give $U(-b)=U(b)=U(\bar b)$, and $U$ is continuous up to the edges, the lips and the tips because
$p$ is.
Along the real axis, $\dd U=-\frac43\arcsin(\sinh\pi s)\,\dd s$ for $0\le s\le\xs$; this is the first formula in
\eqref{eq:6-axes}, and $U(\xs)=-L(\xs)$ by the choice of $U_0$. Along the lips of the right slit,
$\dd U=-\frac{2\pi}3\,\dd x=-\dd L$; hence $U=-L$ there, and on the left slit by symmetry. Along the imaginary axis,
$\dd U=\frac43\arcsinh(\sin\pi s)\,\dd s$, which is the second formula. Along $E_+$, $\dd U=0$, so $U$ is constant there and
equal to $U(\frac i2)$; by $U(\bar b)=U(b)$ it takes the same constant on $E_-$. Integrating \eqref{eq:6-dU} from $0$ to
$i\Im b$ and then horizontally to $b$, with $\abs{\Re p(is)}\le\xs$ and $\abs{\Im p}\le\half$, gives
$\abs{U(b)}\le\frac{2\pi}3\abs{\Re b}+C$.

\emph{(c) The key identity.} Substituting $u=\sinh\pi s$ in the first integral and $u=\sin\pi s$ in the second,
\begin{equation}\label{eq:6-identity}
\begin{aligned}
  \int_0^{\xs}\!\arcsin(\sinh\pi s)\,\dd s+\int_0^{1/2}\!\arcsinh(\sin\pi s)\,\dd s
  &=\frac1\pi\int_0^1\Bigl[\frac{\arcsin u}{\sqrt{1+u^2}}+\frac{\arcsinh u}{\sqrt{1-u^2}}\Bigr]\dd u\\
  &=\frac1\pi\Bigl[\arcsin u\,\arcsinh u\Bigr]_0^1=\half\arcsinh1 .
\end{aligned}
\end{equation}
By \eqref{eq:6-U} and \eqref{eq:6-axes},
$U(\frac i2)=-\frac{2\pi}3\xs+\frac43\cdot\half\arcsinh1=0$, because $\pi\xs=\arcsinh1$. So $U=0$ on $\Epm$. This is the
place where the number $\arcsinh(1)/\pi$ enters the rigorous argument.

\emph{(d) Uniqueness.} Let $d$ be harmonic on $D$ and continuous up to the edges, the lips and the tips, with boundary
value $0$ there, and let $\abs{d(b)}\le C_d(1+\abs{\Re b})$. We claim that $d\equiv0$. For $\epsilon>0$ let
$d_\epsilon(b):=d(b)-\epsilon\cosh(2\Re b)\cos(2\Im b)$, which is harmonic on $D$. The subtracted term (the barrier) is positive on
$\bar\strip$, where $\cos(2\Im b)\ge\cos1>0$, so $d_\epsilon\le0$ at the edges, the lips and the tips. On the cross-cuts
$\{\abs{\Re b}=X\}\cap D$ we have $d_\epsilon\le C_d(1+X)-\epsilon\cos1\,\cosh2X<0$ as soon as $X\ge X(\epsilon)$. The maximum
principle on the bounded domains $D\cap\{\abs{\Re b}<X\}$ gives $d_\epsilon\le0$ on $D$. Letting $\epsilon\to0$ we get
$d\le0$, and the same argument for $-d$ gives $d\equiv0$. Applied to the difference of two solutions, this shows that
\eqref{eq:6-dirichlet} with $s=\xs$ has at most one solution. By (b) and (c), $U$ is a solution, and so is the Poisson
integral $-\int L\,\dd\omb$ over the slits, by \cref{lem:B.1}(c) (\cref{app:B.4}). Hence both are equal to $U_{\xs}$, which
is \eqref{eq:B-Poisson-U}.
\end{proof}

\begin{definition}[The zero law]\label{def:6.6}
Extended across the slits by its boundary value $-L$, the function $U$ is continuous on $\strip$. By \eqref{eq:6-dU} its
normal derivative $\partial_yU=\frac{4\pi}3\Re p$ jumps across the slits by $\frac83\arccosh(\sinh\pi\abs x)>0$, so
$\nabla^2U\ge0$ in the sense of distributions (by Green's formula on the two half-strips; $\nabla U$ is locally bounded up to
the slits and the tips, because $p$ extends continuously to them, see part~(a) of the proof of \cref{lem:6.5}): $U$ is
subharmonic on $\strip$, with a charge density that lives on the slits. For $\abs x\ge\xs$ this density is
\begin{equation}\label{eq:6-rhoU}
  \rhoU(x):=\frac1{2\pi}\Bigl[\partial_yU(x+i0)-\partial_yU(x-i0)\Bigr]=\frac43\,\Re p(x+i0)
  =\frac4{3\pi}\,\arccosh\bigl(\sinh\pi\abs x\bigr) ,
\end{equation}
and we call it the \emph{zero law}. Its counting function is
\begin{equation}\label{eq:6-N}
  N(x):=\half\int_{\xs}^x\rhoU(x')\,\dd x'\simeq\ktip\,(x-\xs)^{3/2}\quad(x\downarrow\xs),\qquad
  \ktip=\frac{4\cdot2^{3/4}}{9\sqrt\pi}=0.4217111 .
\end{equation}
\end{definition}

\noindent In the units of $u_g$, see \eqref{eq:6-riesz}, $\rhoU$ is a number of zeros per unit $x$ and per unit $c'$,
counted with multiplicity. The function $N$ counts \emph{sites}: each site of \cref{sec:7} carries two zeros, hence the
factor $\half$. The density \eqref{eq:6-rhoU} is the function \eqref{eq:4-rhoU} of Part~II, and the asymptote in
\eqref{eq:6-N} follows from $\arccosh(\sinh\pi x)\simeq\bigl(2\sqrt2\,\pi(x-\xs)\bigr)^{1/2}$. The zero law is used in
\cref{sec:7} as a blueprint: the zeros of the functional constructed there are placed according to
$N$; and in \cref{sec:9} as a necessity: the zeros of every near-optimal functional follow it. The potential $U$,
the law $\rhoU$ and $N$ are drawn in \cref{fig:7}.

\subsection{The Cardy deficit and the proof of \texorpdfstring{\cref{thm:3}}{the concentration theorem}}
\label{sec:6.7}

On the lobe interval the extremal potential lies strictly below the Cardy line, and the distance between the two is
what \cref{thm:3} exponentiates.

\begin{lemma}[Deficit; \proved]\label{lem:6.7}
For $\abs x<\xs$,
\begin{equation}\label{eq:6-deficit}
  U(x)+L(x)=-\frac43\int_{\abs x}^{\xs}\Bigl[\frac\pi2-\arcsin(\sinh\pi s)\Bigr]\dd s<0 .
\end{equation}
The function $U+L$ is even and increases on $[0,\xs]$ from $U_0$ to $0$. Hence, for $0<\delta\le\xs$, the Cardy deficit
\begin{equation}\label{eq:6-kappa}
  \kappa(\delta):=-\max_{\abs x\le\xs-\delta}\bigl(U+L\bigr)(x)=-(U+L)(\xs-\delta)
  =\frac43\int_{\xs-\delta}^{\xs}\Bigl[\frac\pi2-\arcsin(\sinh\pi s)\Bigr]\dd s
\end{equation}
is positive. The integrand is $\simeq\bigl(2\sqrt2\,\pi(\xs-s)\bigr)^{1/2}$ as $s\uparrow\xs$, so that
$\kappa(\delta)=\frac89\sqrt{2\sqrt2\,\pi}\;\delta^{3/2}(1+O(\delta))=2.650\,\delta^{3/2}(1+O(\delta))$, with the values
$\kappa(\delta)=0.0026$, $0.029$, $0.083$, $0.23$ at $\delta=0.01$, $0.05$, $0.1$, $0.2$, and $\kappa(\xs)=-U_0=0.389$. Moreover
$\kappa(\delta)\ge2.228\,\delta^{3/2}$ for $0<\delta\le\xs$.
\end{lemma}

\begin{proof}
By \eqref{eq:6-axes} and $U(\xs)=-L(\xs)$ we have $U(x)=-L(\xs)+\frac43\int_{\abs x}^{\xs}\arcsin(\sinh\pi s)\,\dd s$, and
$L(\xs)-L(x)=\frac43\int_{\abs x}^{\xs}\frac\pi2\,\dd s$; this is \eqref{eq:6-deficit}. The integrand is positive for
$s<\xs$, where $\sinh\pi s<1$, which gives the sign and the monotonicity. Near the tip,
$\sinh\pi(\xs-u)=1-\sqrt2\,\pi u+O(u^2)$ because $\cosh\pi\xs=\sqrt2$, and
$\frac\pi2-\arcsin(1-\varepsilon)=\sqrt{2\varepsilon}\,(1+O(\varepsilon))$. The last inequality is proved in
\cref{app:B.5}, \eqref{eq:B-deficit}, with the constant $\frac89\sqrt{2\pi}=2.228\ldots$
\end{proof}

\noindent The exponent $\frac32$ is the Airy exponent of a soft edge, which appeared in \cref{sec:4.6}; the coefficient is
$2\pi\ktip$. By \cref{prop:6.4}, $e^{c'(U+L)(x)}$ is, up to a power of $c'$, the largest value that $\abs g\,V/M[g]$ can take at
$t=c'x^2/3$. Below the BTZ threshold $U(iy)$ increases from $U_0$ at $y=0$ to $0$ at the vacuum, by \eqref{eq:6-axes} and
\eqref{eq:6-identity}.

\begin{proof}[Proof of \cref{thm:3}]
The integrand $\abs g\,V$ is at most majorant times weight, and the integral is at most the supremum of the integrand
times the length of the interval. We use \cref{prop:6.4} on the compact set $Q=[-(\xs-\delta),\xs-\delta]$, in the form
\eqref{eq:B-majorant-delta}, in which the power of $c'$ and the dependence on the distance to the tip are explicit.

There is nothing to prove for $g\equiv0$. Let $0<\delta<\xs$ and $t_\delta:=c'(\xs-\delta)^2/3\le\xis c'$. For
$0<t\le t_\delta$ the point $x=\sqrt{3t/c'}$ has $\xs-\abs x\ge\delta$, and $U_{\xs}=U$ by \cref{lem:6.5}, so
\eqref{eq:B-majorant-delta} reads $\abs{g(t)}=\abs{G(x)}\le e^C\delta^{-5}c'^3\,e^{c'U(x)}M[g]$. The weight obeys
$V(t)\le\frac{4\pi^2t}a\,e^{c'L(x)}$ for all $t>0$ (\cref{lem:A.2}(d)). With $(U+L)(x)\le-\kappa(\delta)$ from
\cref{lem:6.7},
\begin{equation}\label{eq:6-pointwise}
  \abs{g(t)}\,V(t)\le e^C\delta^{-5}\,c'^3\,\frac{4\pi^2t}a\,e^{c'(U+L)(x)}M[g]
  \le e^C\delta^{-5}\,c'^3\,\frac{4\pi^2t}a\,e^{-\kappa(\delta)c'}M[g]\qquad(0<t\le t_\delta) .
\end{equation}
Since $\int_0^{t_\delta}\frac{4\pi^2t}a\,\dd t=2\pi^2t_\delta^2/a\le24\pi^2(\xis)^2c'$, integration gives \eqref{eq:6-conc} with
$C_\delta=\Cconc\,\delta^{-5}$ and $\Cconc:=24\pi^2(\xis)^2e^C$. The properties \eqref{eq:6-kappa-bounds} of
$\kappa(\delta)$ are part of \cref{lem:6.7}.
\end{proof}

\noindent Of the four powers of $c'$, the edges and the slits cost at most two between them (\cref{lem:6.1},
\cref{lem:A.2}(e); their weights in the Poisson--Jensen mean add up to one), stated as three
in~\eqref{eq:B-majorant-delta}, and $t_\delta^2/a\le Cc'$ gives one more. The window of width $O(1/c')$ at the BTZ threshold
enters only through \cref{lem:A.3}, in the edge bound, because \cref{lem:A.2}(d) holds down to $t=0$.

\paragraph{What this says about \cref{sec:4}.} The composition of $p$ with the map of \cref{lem:6.3} is the function $p(z)$
of \eqref{eq:4-pb}, and $\sinh\pi b=i\sinh\pi p$. The curve has thus been recovered without hypotheses, as the solution
of the linear boundary-value problem \eqref{eq:6-dirichlet}. In \cref{sec:4} it rests on a selection rule and describes
the best functional, conjecturally; here it bounds every functional. \tagpar{\numerical} The profiles $u_g$ of the
numerical optima of \cref{sec:5} exceed $U$ nowhere by more than $O(1/c')$, and at the threshold they approach it from
below: at $(c,P)=(64,256)$, $(128,2048)$, $(256,2048)$, $(512,2688)$ the difference $U-u_g$ at $b=0$ is $0.053$, $0.035$, $0.022$, $0.013$ (\cref{fig:7}). The majorant is approached at leading
order by the numerical optima. Whether it is \emph{provably} attained is the subject of Part~IV.

\subsection{The effective rate}
\label{sec:6.8}

We prove \eqref{eq:6-rate}. The idea is to let $\delta$ tend to zero with $c$ in \cref{thm:3}, as slowly as the
prefactor allows. For this one needs the dependence of $C_\delta$ on $\delta$. Its only source is the entropy of
harmonic measure seen from a real point at distance $\delta$ from a tip, which is at most $\CH+5\log\frac1\delta$ by
\cref{lem:B.2}(ii); exponentiated, this is $C_\delta\le\Cconc\,\delta^{-5}$, as the proof of \cref{thm:3} shows.
Everything else is uniform in $\delta$.

\begin{proof}[Proof of the second statement of \cref{thm:4}]
Let $g\in\Abar$, $g\not\equiv0$, be weakly feasible and $t_1=t_1(g)$. Let $\kappa_0:=\frac89\sqrt{2\pi}=2.228\ldots$ be the
constant of \eqref{eq:B-deficit} and $\lambda_0:=3.6$, so that $\kappa_0\lambda_0>8$, and let
\begin{equation}\label{eq:6-delta-choice}
  \delta:=\Bigl(\frac{\lambda_0\log c'}{c'}\Bigr)^{2/3} ,\qquad\text{so that}\qquad
  \kappa(\delta)\,c'\ge\kappa_0\,\delta^{3/2}c'=\kappa_0\lambda_0\log c'>8\log c'
\end{equation}
by \cref{lem:6.7}, provided $\delta<\xs$; this holds for $c'\ge115$. As before let $t_\delta=c'(\xs-\delta)^2/3$, and
suppose that $t_1\le t_\delta$. The half-mass condition \eqref{eq:2-halfmass} and \cref{thm:3} with
$C_\delta\le\Cconc\,\delta^{-5}$ give
\begin{equation*}
  \half\le\Cconc\,\delta^{-5}c'^4e^{-\kappa(\delta)c'}<\Cconc\,\delta^{-5}c'^{-4}\le\Cconc\,c'^{10/3-4}=\Cconc\,c'^{-2/3} ,
\end{equation*}
where we used $\delta^{-5}=(c'/\lambda_0\log c')^{10/3}\le c'^{10/3}$, valid because $\lambda_0\log c'\ge1$. This is false for
$c'\ge(2\Cconc)^{3/2}$. Hence for $c\ge\copt:=\max\{116,\,1+(2\Cconc)^{3/2}\}$ we have $t_1>t_\delta$, and
\begin{equation*}
  t_1>\frac{c'}3(\xs-\delta)^2\ge\xis c'-\frac23\,\xs\,\delta\,c'
  =\xis c'-\frac23\,\xs \lambda_0^{2/3}\,c'^{1/3}\log^{2/3}c' ,\qquad\frac23\,\xs \lambda_0^{2/3}=0.4393\le0.44 .\qedhere
\end{equation*}
\end{proof}

\noindent The constant $\Cconc$, and with it $\copt$, is a product of the absolute constants of \cref{app:A,app:B}; it is
explicit in principle, and we have not computed it. The theorem says that no weakly feasible functional undercuts
$\xis c'$ by more than $O(c'^{1/3}\log^{2/3}c')$, while \cref{conj:1}(a) says that the best functional overshoots it by an
amount of order $c^{1/3}$. The logarithm comes from the polynomial prefactor in \eqref{eq:6-conc} and is presumably not
sharp.

\PartHeading{IV}{Attainment}
\Cref{sec:7} is proved with computer assistance, and \cref{sec:8,sec:9} are proved. \Cref{sec:7,sec:8} use
\cref{sec:2} and, from \cref{sec:6}, in the proofs only \cref{lem:6.5,def:6.6}, the closed form of $U$ and the zero law;
\cref{sec:9} uses in addition \cref{prop:6.4} and \cref{lem:6.7}. \Cref{sec:7} builds a feasible
$g\in\Abar$ at $\xis c'+O(c'^{1/3})$; \cref{sec:8} shows that any feasible $g\in\Abar$ bounds every CFT.
\Cref{sec:alt} describes a second functional that attains the slope, at a physicist's level of rigor; nothing else
depends on it. \Cref{sec:9} shows that a functional that comes close to the bound has no choice: its zeros follow the zero
law on which the functional of \cref{sec:7} is built.

\section{Attainment: an explicit functional reaching \texorpdfstring{$\Lambda_V$}{Lambda\_V}}
\label{sec:7}

We construct, for every large central charge, one explicit function $g\in\Abar$ that is feasible at
$\xis c'+O(c'^{1/3})$. With \cref{thm:4} this proves the slope, \cref{thm:1}. The function is an infinite product of
elementary factors, and its zeros are placed according to the zero law of \cref{def:6.6}. The construction is short
(\cref{sec:7.1,sec:7.2}). The work is to prove that the product and its Fourier transform combine to a function that is
positive on the band and has a positive vacuum value. \Cref{sec:7.3} explains how this is done, without proofs. The
proofs are in \crefrange{sec:7.4}{sec:7.10}, and the long ones in \cref{app:C,app:D,app:E,app:F}, which may be skipped at
first. The inequalities verified by computer are marked $\cert$ and listed in
\cref{tab:4} of \cref{app:G}.

\subsection{Image charges on the strip, and the zero law as a blueprint}
\label{sec:7.1}

The idea is electrostatic. A zero of a holomorphic function is a unit charge for the logarithm of its modulus, and by
\cref{prop:6.4} the profile $u_g$ of every functional, \eqref{eq:6-ug}, lies below the potential $U$ of a definite charge
distribution on the strip $\strip$, up to $O(\log c'/c')$. To certify the gap bound $\xs$, the profile must touch the Cardy
line $-L$ where $U$ does. This suggests building a functional whose profile \emph{is} $U$: charges on the slits at
density $\rhoU$, and on the edges a modulus no larger than \cref{lem:6.1} allows. (\Cref{thm:7} shows that the profile of
every attaining sequence is $U$.) So we want zeros at density $c'\rhoU$ on the band and modulus one on the edges, and we build them out of
the potential of one point charge between two grounded plates.

\paragraph{Glossary.}
A \emph{site} $\chi_n$ is a point of the real $b$-axis at which a zero, or a conjugate pair of zeros, is placed, and
\emph{cell}~$n$ is the interval $(\chi_n,\chi_{n+1})$. The \emph{comb} is the product $B$ of image-charge factors over the
sites. \emph{Detuning} means moving a real double zero to a conjugate pair $\chi_n\pm i\varepsilon_n$. The \emph{onset} is
the neighborhood of the first kept site $\chi_2$; the comb omits the first site of the law (\cref{sec:7.2}). The
\emph{Airy length} $\lc=(\ktip c')^{-2/3}$ is the scale of the spacing of the first sites, and the \emph{hand-off point}
$\xI=\xs+c'^{-\thon}$, with $\thon=0.2$, is where the onset (Region~I) ends and steepest descent takes over (Region~II).
Four functions that are defined in \cref{sec:7.5,sec:7.6} appear already in \cref{sec:7.3}: the \emph{site index}
$\nu(x)$, which equals $k-2$ at the site $\chi_k$, with the comb phase $\Psi=2\pi\nu-\pi$; the \emph{envelope} $\Env$ of
the undetuned comb; the \emph{level} $\Lev$, the explicit unit of all ratios; and the \emph{amplitude} $\Amp$ of the
transform.

\begin{toolkit}{Toolkit: image charges on the strip}
\emph{One factor.} For $\zeta$ with $\abs{\Im\zeta}<\half$ let
\begin{equation}\label{eq:7-T}
  T_\zeta(b):=\tanh\frac{\pi(b-\zeta)}2\,\tanh\frac{\pi(b+\zeta)}2
  =\frac{\cosh\pi b-\cosh\pi\zeta}{\cosh\pi b+\cosh\pi\zeta} .
\end{equation}
For real $\zeta$, $-\log\abs{T_\zeta(b)}$ is the Green's function of the strip with poles at $\pm\zeta$: the potential at
$b$ of unit charges at $\pm\zeta$ between the two grounded plates $\Epm$. The zeros and poles of $T_\zeta$ are the
successive mirror images of $\pm\zeta$ in the plates (\cref{fig:8}a). The map $w=\tanh\frac{\pi b}2$ takes $\strip$ onto
the unit disk and turns $T_\zeta$ into a product of two Blaschke factors~\cite{Garnett}; $T_\zeta$ is the \emph{Blaschke
factor of the strip}. We use four facts.

\smallskip\noindent\hypdef{T1} \emph{Modulus.} For real $\zeta$, $\abs{T_\zeta}=1$ on both edges and $\abs{T_\zeta}<1$ in
$\strip$. The same holds for the product $T_\zeta T_{\bar\zeta}$ of a conjugate pair, $\zeta=\chi+i\varepsilon$ with
$0<\varepsilon<\half$. A bounded holomorphic function on $\strip$ whose modulus is one on the edges is called
\emph{inner}; a zero-free factor such as $\sech^2\frac{\pi b}2$, which is determined by its modulus on the edges, is
\emph{outer}.

\smallskip\noindent\hypdef{T2} \emph{Pairs on the real axis.} For real $\chi$, $0\le\varepsilon<\half$ and real $x$,
\begin{equation*}
  T_{\chi+i\varepsilon}(x)\,T_{\chi-i\varepsilon}(x)
  =\prod_\pm\frac{\cosh\pi(x\mp\chi)-\cos\pi\varepsilon}{\cosh\pi(x\mp\chi)+\cos\pi\varepsilon}\ \ge\ T_\chi(x)^2 ,
\end{equation*}
and the left side is positive if $\varepsilon>0$: lifting a double zero off the axis raises the modulus on the axis.

\smallskip\noindent\hypdef{T3} \emph{Reflection.} $T_\zeta(b+i)=1/T_\zeta(b)$, and $T_\zeta(\frac i2)=-1$ for every
$\zeta$.

\smallskip\noindent\hypdef{T4} \emph{Zeros and poles.} For $\Re\zeta\ne0$, $T_\zeta$ has simple zeros at
$\{\pm\zeta\}+2i\ZZ$, simple poles at $\{\pm\zeta\}+i(2\ZZ+1)$, and no others.

\smallskip\noindent All four follow from \eqref{eq:7-T}. For \hyp{T1} note that $\cosh\pi b$ is imaginary on $\Epm$, so that
numerator and denominator have the same modulus when $\cosh\pi\zeta$ is real, and numerator and denominator of the pair
are conjugate to each other in general; the maximum principle gives the inequality inside.
\end{toolkit}

\begin{figure}[tp]
\centering
\figfile[\linewidth]{fig8_images}
\caption{Image charges on the strip. (a)~One factor $T_\zeta(b)=\tanh\frac{\pi(b-\zeta)}2\tanh\frac{\pi(b+\zeta)}2$,
here with $\zeta=0.70$: zeros at $\pm\zeta+2i\ZZ$ (dots) and poles at $\pm\zeta+i(2\ZZ+1)$ (crosses), which are the
successive images of $\pm\zeta$ in the edges $\Epm$ (dashed lines: the image edges $\Im b=\pm\frac32$), so that
$\abs{T_\zeta}=1$ on $\Epm$; the curves inside $\strip$ are the level curves $\abs{T_\zeta}=0.2$, $0.45$, $0.8$, $0.95$.
(b)~The zeros and poles of the comb $B$ of \cref{sec:7.2}. The sites $\chi_n$, $c'N(\chi_n)=n-\frac14$, are computed for
$c'=8$, a value chosen only to make the pairs visible: the simple zero at $\pm\chi_2$, the pairs
$\pm\chi_n\pm i\varepsilon_n$ ($n\ge3$) with $\varepsilon_n=\frac14(\chi_{n+1}-\chi_n)$, and the omitted first site of the
law (open circle). All poles lie in $\abs{\Im b}\ge1-\max_n\varepsilon_n>\half$, and $\abs B=1$ on $\Epm$. Diamonds mark
the saddle points $b_0=\pm v_0(x)+\frac i2$ of the edge integral for the band point $x$ (square), as in \cref{fig:3}.}
\label{fig:8}
\end{figure}

The next lemma says that the zero law is the right density for these factors. It is the continuum limit of every sum over
sites in this section.

\begin{lemma}[Riesz representation of $U$; \proved]\label{lem:7.1}
For every $b$ in the closed strip $\bar\strip$,
\begin{equation}\label{eq:7-riesz}
  \int_{\xs}^\infty\rhoU(x')\,\log\abs{T_{x'}(b)}\,\dd x'=U(b) ,
\end{equation}
and the integral converges absolutely. In particular the left side vanishes on $\Epm$ and equals $-L(x)$ at real $b=x$
with $\abs x\ge\xs$.
\end{lemma}

\noindent The proof is in \cref{app:C.1}: both sides have the same charge on $\strip$, both vanish on the edges, and both
grow at most linearly; a potential with these three properties is unique. In words: a continuum of factors $T_{x'}$ at the density of the law reproduces the Cardy decay
$e^{-c'L}$ on the band exactly, and is invisible in modulus on the edges.

\paragraph{Notation for the edge.}
On the edge $E_+$ we write $b=v+\frac i2$ with $v$ real. The function $p$ of \cref{lem:6.5} is real there, and we call
\begin{equation}\label{eq:7-edge}
  F(v):=p\bigl(v+\tfrac i2\bigr)=\frac1\pi\arcsinh(\cosh\pi v),\qquad
  \cosh\pi v_0(x)=\sinh\pi x,\qquad \rhoU(x)=\tfrac43\,v_0(x),
\end{equation}
the \emph{edge function}. It is even, convex, and increasing on $[0,\infty)$, with $F(0)=\xs$ and $0<F'<1$ on $(0,\infty)$. Its inverse
$v_0\colon[\xs,\infty)\to[0,\infty)$, $F(v_0(x))=x$, is given by the second formula, and the third is
\eqref{eq:6-rhoU}: up to the factor $\frac43$, the zero law is the inverse of the edge function. We write
$b_0(x):=v_0(x)+\frac i2$ for the point of the edge that corresponds to the band point $x$.

\subsection{The functional and the theorem}
\label{sec:7.2}

The functional is the zero law made discrete. The sites are the quantiles of the counting function $N$: the points
$\chi_n>\xs$ with $c'N(\chi_n)=n-\frac14$, for $n=2,3,\dots$; the site $n=1$ of the law is omitted. At $\chi_2$ we place a simple real zero. Every later site is lifted into a conjugate pair
$\chi_n\pm i\varepsilon_n$ with $\varepsilon_n:=\eta_0(\chi_{n+1}-\chi_n)$ and $\eta_0=\frac14$. We multiply by an outer
factor, which makes the product integrable, and antisymmetrize under the Fourier transform:
\begin{equation}\label{eq:7-functional}
  h_0(b):=\sech^2\frac{\pi b}2\,B(b),\qquad
  B:=T_{\chi_2}\prod_{n\ge3}T_{\chi_n+i\varepsilon_n}\,T_{\chi_n-i\varepsilon_n},\qquad
  g:=\tfrac12\bigl(h_0-\hat h_0\bigr) .
\end{equation}
Here $h_0$ is even in $b$ and is read as a function of $t=c'b^2/3$; for $-a\le t<0$ the point $b$ lies on the imaginary
axis. The hat is the transform \eqref{eq:2-eigen}. We also need the \emph{undetuned comb}
$B^0:=T_{\chi_2}\prod_{n\ge3}T_{\chi_n}^2$, which has double zeros at the sites $n\ge3$, and $h_0^0:=\sech^2\frac{\pi b}2\,B^0$.
The zeros and poles of $B$ are drawn in \cref{fig:8}(b).

\begin{theorem}[Attainment; \provedCA]\label{thm:5}
There is a central charge $\catt\ge40$, explicit in principle but not computed, such that for every $c\ge\catt$ the
function $g$ of \eqref{eq:7-functional} has the following properties.
\begin{enumerate}[label=\textup{(\roman*)}]
\item $g\in\Abar$.
\item $g(t)>0$ for all $t\ge\tstar+1$, where
\begin{equation}\label{eq:7-tstar}
  \tstar:=\frac{c'\chi_2^2}3=\xis c'+A_1\,c'^{1/3}\bigl(1+O(c'^{-2/3})\bigr),\qquad
  A_1=\tfrac23\,\xs\,\ktip^{-2/3}\bigl(\tfrac74\bigr)^{2/3}=0.483 .
\end{equation}
\item $\gvac>0$.
\end{enumerate}
Hence $g$ is feasible at $\tstar+1$ in the sense of \cref{def:2-feasible}, so that
$\TV(c)\le\xis c'+A_1\,c'^{1/3}+O(1)$; with \cref{thm:4} this proves \cref{thm:1}. Moreover
$\abs{g(t)}\le C(c)\,e^{-c'L(x)}e^{-\pi x}$ for $t=c'x^2/3\ge0$: both $h_0$ and $\hat h_0$ decay faster than the inverse
of the Cardy weight.
\end{theorem}

\noindent In words: a product of image-charge factors at the quantiles of the zero law, minus its own Fourier transform,
is a functional that certifies a gap within $O(c^{1/3})$ of $c/\LV$. The offset $+1$ is cosmetic: the proof gives $g>0$ already
from $\tstar+o(1/c')$ on (\cref{sec:7.8}). The next proposition collects what is elementary about $h_0$.

\begin{proposition}[Elementary properties; \proved]\label{prop:7.2}
Let $c\ge40$, the standing assumption of \cref{sec:2.6}. Then $\max_n\varepsilon_n<\half$, the product $B$ converges
locally uniformly on $\abs{\Im b}<1-\max_n\varepsilon_n$, and:
\begin{enumerate}[label=\textup{(\alph*)}]
\item $\abs B\le1$ on $\bar\strip$ and $\abs B=1$ on $\Epm$.
\item $B(\frac i2)=-1$. Hence $h_0(t_0)=-2$ at the vacuum, and the second difference
$h_{0,\rm vac}:=h_0(t_0)-2h_0(t_0+1)+h_0(t_0+2)$ satisfies $\abs{h_{0,\rm vac}}\le8$.
\item On the physical axis, $h_0<0$ on the lobe interval $[0,\tstar)$, $h_0$ has a simple zero at $\tstar$, and $h_0>0$
for $t>\tstar$.
\item On the edge, $\abs{h_0(v+\frac i2)}=2/\cosh\pi v$ exactly.
\end{enumerate}
\emph{On the edge through the vacuum the comb is invisible in modulus, and all the information that it carries is a
phase.}
\end{proposition}

\begin{proof}
First, $\max_n\varepsilon_n<\half$; this places every pair under \hyp{T1} and keeps the poles of $B$ out of $\bar\strip$.
By \eqref{eq:7-edge}, $\rhoU'=\frac43/F'(v_0)\ge\frac43$ on $(\xs,\infty)$, so that $\rhoU(y)\ge\frac43(y-\chi_n)$ for
$y\ge\chi_n$. One cell carries the charge
$1=c'N(\chi_{n+1})-c'N(\chi_n)=\frac{c'}2\int_{\chi_n}^{\chi_{n+1}}\rhoU\ge\frac{c'}3(\chi_{n+1}-\chi_n)^2$. Hence
$\chi_{n+1}-\chi_n\le(3/c')^{1/2}\le0.28$ and $\varepsilon_n\le0.07$ for $c'\ge39$.

Everything else follows from \hyp{T1}--\hyp{T3} once the product converges. By \eqref{eq:7-T},
$-T_\zeta=1-2\cosh\pi b/(\cosh\pi b+\cosh\pi\zeta)$, so a pair differs from $1$ by
$O(e^{\pi\abs{\Re b}}e^{-\pi\chi_n})$ away from its poles, and $\sum_ne^{-\pi\chi_n}<\infty$ because the number of sites
below $x$ is at most $c'N(x)=O(c'x^2)$; by \hyp{T4} the poles lie at $\abs{\Im b}\ge1-\max_n\varepsilon_n$. This gives the convergence. Part (a)
is \hyp{T1} for each factor. For (b), \hyp{T3} gives $T_{\chi_2}(\frac i2)=-1$ and the value $(-1)^2$ for each pair, and
$\sech^2\frac{\pi i}4=2$; on the segment from the vacuum to the threshold $\abs{\sech^2\frac{\pi b}2}\le2$ and $\abs B\le1$,
so each of the three values in $h_{0,\rm vac}$ is at most $2$ in modulus ($t_0+2\le0$ because $a\ge2$). For (c), the pairs are positive on the real axis by
\hyp{T2}, and $T_{\chi_2}(x)$ is negative for $0\le x<\chi_2$, vanishes to first order at $\chi_2$ and is positive beyond.
For (d), $\abs B=1$ on the edge and
$\bigabs{\cosh\frac\pi2(v+\frac i2)}^2=\half(\cosh\pi v+\cos\frac\pi2)=\half\cosh\pi v$.
\end{proof}

\paragraph{What is fixed.}
The construction has free parameters: the number $m\ge1$ of omitted sites, the node phase $k_\infty$ in
$c'N(\chi_n)=n-k_\infty$, and the detuning $\eta_0$. Only the onset index $n_1=m+1-k_\infty$ enters the coefficient of
$c'^{1/3}$, which is $A_m=0.3326\,n_1^{2/3}$. Indeed $N(x)\simeq\ktip(x-\xs)^{3/2}$ by~\eqref{eq:6-N}, and
$c'N(\chi_2)=n_1$, so that
\begin{equation}\label{eq:7-Am}
  \chi_2-\xs\simeq\Bigl(\frac{n_1}{\ktip c'}\Bigr)^{2/3},\qquad
  \tstar-\xis c'\simeq\tfrac23\,\xs\,c'\,(\chi_2-\xs),\qquad
  A_m=\tfrac23\,\xs\,\ktip^{-2/3}\,n_1^{2/3} .
\end{equation}
In this paper $(m,k_\infty,\eta_0)=(1,\frac14,\frac14)$, so $n_1=\frac74$, and the outer factor is $\sech^2$. Neither the
constant $0.483$ nor $\catt$ is optimized, and nothing is claimed at finite $c$.

The remainder in \eqref{eq:7-tstar} is obtained in the same way. Write $\chi_2=\xs+\lc\vsig_2$ with the Airy length $\lc$,
as in \cref{lem:E-tip}(c). Then
$\tstar=\xis c'+\tfrac23\,\xs\,c'\lc\,\vsig_2+\tfrac13\,c'\lc^2\vsig_2^2$. Here $c'\lc=\ktip^{-2/3}c'^{1/3}$, and
$\vsig_2=(\tfrac74)^{2/3}\bigl(1+O(\lc)\bigr)$ by \eqref{eq:E-sites}, and the last term is $O(c'^{-1/3})$. Since
$\lc=O(c'^{-2/3})$, this is \eqref{eq:7-tstar} with $A_1=\tfrac23\,\xs\,\ktip^{-2/3}(\tfrac74)^{2/3}$.

The threshold $\catt$ is named before it is fixed. Throughout this section and \cref{app:C,app:D,app:E,app:F}, a
statement made for $c\ge\catt$, or for $c$ large, holds for $c$ above a threshold of its own, which may depend on a
parameter that is named there ($r$ in \cref{lem:7.5}, $\epsilon$ in \cref{prop:7.4}). The proof of \cref{thm:5} meets
finitely many of these thresholds, each at a fixed value of its parameter; none of them depends on a later one, and
$\catt$ is the largest of them. It is fixed at the end of that proof, in \cref{sec:7.10}.

\subsection{Idea of the proof}
\label{sec:7.3}

There are three things to prove: (i)~$g>0$ on the band $t\ge\tstar+1$; (ii)~$\gvac>0$; (iii)~$g\in\Abar$. The relation
$\hat g=-g$ is automatic. Since $h_0\ge0$ on the band, everything in (i) reduces to one statement: \emph{$\hat h_0$ is
smaller than $h_0$ there}. \Cref{fig:9} shows this at $c=256$. The function $h_0$ is a train of positive bumps, one per
cell. Its transform oscillates in lock-step with it, with an amplitude of about $\half$ in units of the level $\Lev$;
this amplitude is about $0.29$ of the height of the bumps. The difference $2g=h_0-\hat h_0$ stays positive.

\begin{figure}[tp]
\centering
\figfile[\linewidth]{fig9_functional}
\caption{The functional of \cref{thm:5} at $c=256$, with $(m,k_\infty,n_1,\eta_0)=(1,\frac14,\frac74,\frac14)$ and
outer factor $\sech^2$: the comb $h_0$ (thin), its transform $\hat h_0$ (dashed), and $2g=h_0-\hat h_0$ (thick gray), all
divided by the level $\Lev(x)$ and plotted against the site index $\nu(x)$, which is an integer at every site (top axes:
$x$); dotted: $\pm\Amp/\Lev$. (a)~The onset, cells $2$--$9$, beginning at the simple zero $\chi_2$ ($t=\tstar=10.084$):
the transform is below $0.4\%$ of the level, so that $2g$ and $h_0$ coincide to the width of the line. (b)~Cells
$102$--$105$. (c)~Cells $384$--$387$, deep in the band: $h_0$ oscillates between the floor $\sinh^2\pi\eta_0$ and
$\cosh^2\pi\eta_0$, $\hat h_0$ between $\mp\frac12$, in lock-step, one period per cell and maxima on maxima, so that $2g$
stays near $\frac54$. Bottom: the three scales $\lc\ll c'^{-\thon}\ll c'^{-1/(4n_1)}$ of the onset, at their values
$0.044$, $0.33$, $0.45$ for $c=256$, on a logarithmic axis of $x-\xs$, with the inequality used on each stretch (the
dotted segment continues the range on which the overshoot beats the crude bound) and the positions of the three panels.}
\label{fig:9}
\end{figure}

\paragraph{How $\hat h_0$ is computed.}
The transform at a band point $x$ is an integral of $h_0$ times a Hankel kernel along the \emph{edge} $E_+$ of the strip
(\cref{sec:7.4}; \cref{fig:3}). On the edge $\abs{h_0}$ is explicit and does not depend on the comb
(\cref{prop:7.2}(d)), and the kernel has modulus $e^{-c'L(x)}$. So the size of the integrand is pinned at the Cardy level,
and the comb enters through a phase only. The phase has two stationary points $\pm v_0(x)$ on the edge, and they exist
exactly when $x\ge\xs$: this is how the band appears in the transform.

\paragraph{Three regimes.}
Three scales organize the proof, $\lc\ll c'^{-\thon}\ll c'^{-1/(4n_1)}$; they are drawn at the bottom of \cref{fig:9}.
In \emph{Region~II}, $x\ge\xI$ (\cref{sec:7.6,sec:7.7}), steepest descent gives $\abs{\hat h_0}\le\Amp\,(1+o(1))$ with an explicit amplitude
$\Amp\le0.504\,\Lev$, and the detuning gives a floor $h_0\ge\mu\,\Lev$ with $\mu>0.504$. Positivity is one inequality
between explicit functions that does not involve $c$, and no information on the phase of $\hat h_0$ is used. In
\emph{Region~I}, $\chi_2\le x\le\xI$ (\cref{sec:7.8}), the two saddles coalesce and the asymptotics is not uniform. There we use only the
crude bound $\abs{\hat h_0}\le C\sqrt{c'}\,e^{-c'L}$, and the fact that near the onset the comb stands above the Cardy level
by a power of $c'$. We call this the \emph{overshoot}. It is there because the charge of the omitted site is missing from
the comb. The two regions overlap, because the overshoot beats the crude bound up to $x-\xs\approx c'^{-1/(4n_1)}$, which
is beyond $\xI$. For the \emph{vacuum} (\cref{sec:7.9}), $\gvac$ is the mass of the lobe minus the mass of the band, up to $O(1)$. Both are
dominated by a window of width $\lc$ around the onset, and the band is at most a fixed fraction $\Rvac<1$ of the lobe
(\cref{prop:7.13}). The computer enters as follows. Once the factors that tend to one as $c\to\infty$ are bounded by hand,
each regime leaves inequalities that do not involve $c$, for explicit functions of one real variable: for the amplitude
and the floor in Region~II, for finite products over the first cells in Region~I, and for a ratio of two integrals at the
vacuum. These inequalities, marked $\cert$, are proved in interval arithmetic (\cref{app:G.1}); nothing that depends on
$c$ is computed.

\paragraph{Why each ingredient is there.}
\emph{Sites at the quantiles of the law.} Besides the Cardy decay $\abs B\approx e^{-c'L}$ on the band (\cref{lem:7.1}),
this makes the phase of the integrand on the edge stationary exactly at the points $\pm v_0(x)$ that correspond to
band points $x$, where the phase of $B$ advances by $2\pi$ per site \eqref{eq:C-eikonal}.

\emph{Detuning by $\eta_0=\frac14$.} The undetuned comb is $\abs{h_0^0}=\Env\sin^2\pi\nu$. It returns to zero in every
cell, while its transform does not, so $g$ would be negative near every site. Lifting the zeros off the axis creates a floor
above $\half\Lev$. It changes nothing on the edge, where a conjugate pair is still unimodular, so the amplitude of the
transform is untouched. The value matters: for $\eta_0=0.2$ the floor of \cref{lem:7.9} is below $0.45$, which is not enough.

\emph{Why the amplitude is $\half$.} Write $h_0^0=\half\Env\,(1+\cos\Psi)$. The transform
reproduces the oscillating part $\half\Env\cos\Psi$. It cannot see the mean, which has no stationary phase.

\emph{The omitted site, $m=1$.} It makes the first zero simple. It produces the overshoot at the onset. And it makes the
vacuum lopsided in favor of the lobe. It costs $A_1c^{1/3}$ in the bound, and nothing at the order of the slope. With no
site omitted and the same node phase one finds numerically $\abs{\Lobe}/\Band\approx0.8$ \tagpar{\numerical}, and that
comb is not feasible.

\emph{The node phase $\frac14$.} This is a convention and is not tuned. Placing nodes at the quantiles of an equilibrium
density is a Bohr--Sommerfeld rule, familiar from the choice of interpolation nodes in the numerical
bootstrap~\cite[Sec.~3.3]{CDKPS}. Only $n_1=\frac74$ enters the certified constants. With $\frac14$ the phase of the transform coincides
with that of the comb deep in the band (\cref{fig:9}c).

\emph{The outer factor $\sech^2$ and the simple zero at $\chi_2$.} The outer factor gives integrability and membership in
$\Abar$, and it cancels from the ratio $\Amp/\Lev$ up to the factor $\coth\frac{\pi x}2$. The zero is simple because a
feasible $g$ must change sign once, at $t_1$.

\subsection{Three exact identities}
\label{sec:7.4}

Three identities turn the three claims into statements about $h_0$ alone.

\begin{lemma}[Three identities; \proved]\label{lem:7.3}
Let $\max_n\varepsilon_n<\half$. For \textup{(a)} and \textup{(b)} suppose in addition that $\hat h_0$ is continuous and
that $\abs{h_0(t)}+\abs{\hat h_0(t)}\le C(c)\,e^{-c'L(x)-\pi x}$ for $t=c'x^2/3\ge0$; this is proved in \cref{sec:7.10},
from \cref{lem:7.5}\textup{(c)}, \cref{lem:7.10} and the bound \eqref{eq:7-glob}, which uses \textup{(c)} but
neither \textup{(a)} nor \textup{(b)}.
\begin{enumerate}[label=\textup{(\alph*)}]
\item $\hat g=-g$.
\item With the second difference $h_{0,\rm vac}$ of \cref{prop:7.2}\textup{(b)},
\begin{equation}\label{eq:7-gvac}
  \gvac=\frac12\Bigl[h_{0,\rm vac}-2\pi\int_0^\infty h_0\,V\,\dd t\Bigr] .
\end{equation}
\item For every $x>0$, with $k_x:=\frac{4\pi}3c'x$,
\begin{equation}\label{eq:7-edgerep}
  \hat h_0\bigl(c'x^2/3\bigr)=\frac{2\pi c'}3\int_{E_+}h_0(b)\,H_0^{(1)}(k_xb)\,b\,\dd b .
\end{equation}
\end{enumerate}
\end{lemma}

\noindent On the edge the Hankel function is, by \eqref{eq:D-hankel}~\cite[(10.17.13)--(10.17.15)]{DLMF},
\begin{equation}\label{eq:7-hankel}
  \bigabs{H_0^{(1)}(k_xb)}=\Bigl(\frac{2}{\pi k_x\abs b}\Bigr)^{1/2}e^{-c'L(x)}\,\bigabs{1+\epsH(k_xb)},\qquad
  \abs{\epsH(z)}\le\frac{0.33}{\abs z}\quad(\abs z\ge1),
\end{equation}
because $k_x\Im b=\half k_x=c'L(x)$ there.

\begin{proof}
We prove (c) first; it uses only \cref{prop:7.2} and \hyp{T4}.

\emph{(c).} By \eqref{eq:2-eigen}, with $t=c'b^2/3$ and $s=c'x^2/3$ we have $4\pi\sqrt{st}=k_xb$ and
$\dd t=\frac{2c'}3b\,\dd b$, so $\hat h_0(s)=\frac{4\pi c'}3\int_0^\infty h_0(b)J_0(k_xb)\,b\,\dd b$. On the positive axis
$J_0=\half(H_0^{(1)}+H_0^{(2)})$, and $H_0^{(2)}(z)=-H_0^{(1)}(ze^{i\pi})$~\cite[(10.11.5)]{DLMF}. Since $h_0$ is even,
the substitution $b\to-b$ in the term with $H_0^{(2)}$ gives
$\hat h_0(s)=\frac{2\pi c'}3\int_{-\infty}^\infty\mathcal I(b)\,\dd b$ with
$\mathcal I(b):=h_0(b)H_0^{(1)}(k_xb)\,b$, the real axis being approached from above at $b=0$, where
$\mathcal I=O(\abs b\log\abs b)$. By \hyp{T4} the poles of $B$ lie at $\abs{\Im b}\ge1-\max_n\varepsilon_n>\half$, and those
of $\sech^2\frac{\pi b}2$ at $\abs{\Im b}=1$, so $\mathcal I$ is analytic in $0<\Im b<1-\max_n\varepsilon_n$. On
$0\le\Im b\le\half$ we have $\abs B\le1$, $\abs{\sech^2\frac{\pi b}2}\le4e^{-\pi\abs{\Re b}}$ and
$\abs{H_0^{(1)}(k_xb)}\le C\abs{k_xb}^{-1/2}$ for $\abs b\ge1$, so $\mathcal I$ decays exponentially as
$\abs{\Re b}\to\infty$, uniformly in the height. Cauchy's theorem moves the contour from $\RR$ to $E_+$.

\emph{(a).} By hypothesis $h_0$ and $\hat h_0$ are continuous and integrable on $[0,\infty)$, so Fourier inversion for
radial functions on $\RR^2$ gives $\hat{\hat h}_0=h_0$, and $\hat g=\half(\hat h_0-h_0)=-g$.

\emph{(b).} The idea is to continue Fourier inversion analytically in $\alpha$, from $\alpha\le0$ to $0\le\alpha\le a$.
By \cref{lem:A.2}(e), $K_a\le e^{c'L}$, so $\int_0^\infty(\abs{h_0}+\abs{\hat h_0})K_a\,\dd t<\infty$ by the
hypothesis. As in the proof of \cref{lem:A.1}, the kernel bound \eqref{eq:A-kernel-bound} and dominated convergence then
show that $\alpha\mapsto2\pi\int_0^\infty\hat h_0K_\alpha\,\dd t$ is continuous on the parabolic region $\mathcal D_a$ of
that lemma and holomorphic in its interior. For $\alpha=-s\le0$ it equals $\hat{\hat h}_0(s)=h_0(s)$. The function
$\alpha\mapsto h_0\vert_{t=-\alpha}$ has the same regularity on $\mathcal D_a$, which is the image of $\bar\strip$ under
$\alpha=-c'b^2/3$. Two such functions that agree on $\alpha\le0$ agree on $\mathcal D_a$. Since the values of $g$ below the
threshold are defined by the eigen-identity \eqref{eq:2-eigenid} (\cref{def:2-Abar}), for $0\le\alpha\le a$ we get
$g(-\alpha)=-2\pi\int_0^\infty gK_\alpha\,\dd t=\half\bigl[h_0\vert_{t=-\alpha}-2\pi\int_0^\infty h_0K_\alpha\,\dd t\bigr]$. The
second difference in $\alpha$ at $a,a-1,a-2$, with $V=K_a-2K_{a-1}+K_{a-2}$, is \eqref{eq:7-gvac}.
\end{proof}

\subsection{The comb on the real axis: site index, envelope, level}
\label{sec:7.5}

On the band, $\log\abs{B(x)}$ is a sum over the sites of the potentials $\log\abs{T_{\chi_k}(x)}$. By \cref{lem:7.1} the
integral which this sum approximates is $-c'L(x)$, corrected for the omitted charge. So $h_0$ is the Cardy decay, times
the effect of the omitted charge, times a bounded function of the position of $x$ in its cell. This subsection names
these pieces and states two-sided bounds for them. The proofs are in \cref{app:E}. They use nothing about $\hat h_0$.

\paragraph{Site index and envelope.}
For real $a>\frac14$ (as the index of a site, here and in \cref{app:E,app:F}, the letter $a$ is a real number and not
$c'/12$) let $\chi_a$ be the point with $c'N(\chi_a)=a-\frac14$; the sites are the integer values of $a$.
Let $\ell_n:=\chi_{n+1}-\chi_n$ be the length of cell $n$, and for $x$ in cell $n$ let $u:=(x-\chi_n)/\ell_n\in(0,1)$ be the
position in the cell. The site index and the comb phase are
\begin{equation}\label{eq:7-nu}
  \nu(x):=c'N(x)+\tfrac14-2,\qquad\nu(\chi_a)=a-2,\qquad\Psi(x):=2\pi\nu(x)-\pi ,
\end{equation}
so that $\nu$ runs over $(n-2,n-1)$ on cell $n$. Beyond $\chi_2$ the undetuned comb has its double zeros exactly where
$\sin^2\pi\nu$ has them. The envelope is therefore defined, for $x>\chi_2$, by
\begin{equation}\label{eq:7-env}
  \abs{h_0^0(x)}=\Env(x)\,\sin^2\pi\nu(x)=\tfrac12\,\Env(x)\,\bigl(1+\cos\Psi(x)\bigr),
\end{equation}
and is positive and smooth there. Off the sites, $h_0=h_0^0\,\Pi$ with $\Pi:=\abs B/\abs{B^0}\ge1$ by \hyp{T2}.

\paragraph{The absent charge, the end correction and the level.}
The comb lacks the charge which the law places on $[\xs,\chi_2]$. Its total weight is $c'\int_{\xs}^{\chi_2}\rhoU=2n_1$, and
its potential is
\begin{equation}\label{eq:7-P}
  c'\cP(x):=-c'\int_{\xs}^{\chi_2}\rhoU(x')\,\log\abs{T_{x'}(x)}\,\dd x'\ \ge0 .
\end{equation}
By \cref{lem:7.1} the continuum over the range that is kept is
$c'\int_{\chi_2}^\infty\rhoU\log\abs{T_{x'}(x)}\,\dd x'=-c'L(x)+c'\cP(x)$ for $x\ge\xs$. The factor $e^{c'\cP}$ is the overshoot
of \cref{sec:7.3}. We write $\cP_Y$ for \eqref{eq:7-P} with $\chi_2$ replaced by $Y$. The simple zero at $\chi_2$ stands for
the half cell of continuum between $\chi_2$ and $\chi_{5/2}$. The difference is the \emph{end correction}
\begin{equation}\label{eq:7-Jend}
  \Jend(x):=\log\abs{T_{\chi_2}(x)}-2\int_2^{5/2}\log\abs{T_{\chi_a}(x)}\,\dd a,\qquad
  0\le\Jend(x)\le\CJ\quad(x\ge\chi_3),
\end{equation}
with an absolute constant $\CJ\le0.65$ (\cref{lem:E-Jend}).
The \emph{level} is
\begin{equation}\label{eq:7-lev}
  \Lev(x):=4\,\sech^2\frac{\pi x}2\;e^{-c'L(x)}\,e^{c'\cP(x)+\Jend(x)},\qquad
  \frac{h_0}{\Lev}=\tfrac14\,\abs B\,e^{c'L-c'\cP-\Jend}\quad(x>\chi_2).
\end{equation}
The level is explicit, it is smooth on the cells $n\ge3$, and it does not oscillate. The factor $4$ makes $\Env/\Lev$ close to one. The end correction
is put inside the level because it enters the upper and the lower bounds below in the same way, so that it cancels from
$h_0/\Lev$ and from $\Env/\Lev$. It does not cancel from $\Amp/\Lev$, and there its sign is used (\cref{sec:7.7}).

\paragraph{Hole functions.}
A \emph{hole function} is the potential at $x$ of the sites nearest to $x$, kept exactly, minus that of the piece of
continuum of the same charge which they replace. All other sites are replaced by the continuum, by an
inequality with a definite sign and no remainder: apart from the sites far to the left of $x$, the summand
$a\mapsto\log\abs{T_{\chi_a}(x)}$ is concave on either side of $x$, and for a concave function the trapezoid rule lies below
the integral and the midpoint rule above it (\cref{app:E.2}). For $x$ in cell $n$ and $j\ge1$, the upper bound (for
$B^0$) keeps the $2j+2$ nearest sites with the weights $1,2,\dots,2,1$ and leaves $\EcH^+_{n,j}(x)$; the lower bound (for
$B$) keeps the $2j$ nearest pairs with their exact moduli and leaves $\EcH^-_{n,j}(x)$ (\eqref{eq:E-Hplus},
\eqref{eq:E-Hminus}; $\EcH^\pm_n:=\EcH^\pm_{n,1}$). The two charges are equal, so the length of the cell cancels: as
$n\to\infty$ the hole functions tend to explicit functions $h^\pm_j$ of the position $u$ in the cell alone,
\eqref{eq:E-hplus} and \eqref{eq:E-hminus}. For $j=1$ we write $h^\pm_\infty:=h^\pm_1$:
\begin{equation}\label{eq:7-hole}
\begin{aligned}
  h^+_\infty(u)&=\log(1+u)+2\log u+2\log(1-u)+\log(2-u)-2\int_{-1}^{2}\log\abs{u-v}\,\dd v,\\
  h^-_\infty(u;\eta_0)&=\log\bigl(u^2+\eta_0^2\bigr)+\log\bigl((1-u)^2+\eta_0^2\bigr)-2\int_{-1/2}^{3/2}\log\abs{u-v}\,\dd v .
\end{aligned}
\end{equation}
The rate is $\abs{\EcH^\pm_{n,j}-h^\pm_j}\le C_j(\frac1n+\ell_n)$ on the cells $n\ge2j+4$, where $C_j=C\,j^2\log(2j+2)$ with $C$ absolute
\eqref{eq:E-displacement}. Both $e^{\EcH^\pm}$ contain the factor $4$ of the level: as $j\to\infty$,
$e^{h^-_j(u;\eta_0)}\to4(\sin^2\pi u+\sinh^2\pi\eta_0)$. The extrema of $h^\pm_j$ are rows 3, 6, 7 and 8 of
\cref{tab:4}. The sites far to the left of $x$ are not concave summands; for them the midpoint rule has an
ordinary remainder $e_L(x)\ge0$~\eqref{eq:E-eL}, whose supremum over the cells $n\ge6$ and $c\ge\catt$ is
$\CL\le0.006$ (\cref{lem:E-farleft}(d); row 5).

\noindent \Cref{prop:7.4} says what the unit means, $\Env\approx\Lev$; the proofs of positivity do not use it.

\begin{proposition}[Envelope and level; \provedCA]\label{prop:7.4}
Let $c\ge\catt$ and $\lambda:=\Env/\Lev$. On the cells $n\ge6$,
\begin{equation}\label{eq:7-lambda}
  0.819\;e^{-2\CL-C_1(\frac1n+\ell_n)}\ \le\ \lambda(x)\ \le\ 1.278\;e^{2\CL+C_1(\frac1n+\ell_n)} .
\end{equation}
Here $[0.819,1.278]$ $\cert$ (row 6 of \cref{tab:4})
is the bracket that the limits $h^\pm_1$ give, $e^{\pm2\CL}$ is the loss from the sites far to the left, and
$e^{\pm C_1(\frac1n+\ell_n)}\to1$ accounts for the displacement of the sites. With $h^\pm_4$ the bracket is
$[e^{-0.0422},e^{0.0749}]$ $\cert$;
hence, for fixed $\epsilon>0$ and $c$ large depending on $\epsilon$, on the cells $n\ge\epsilon c'$ we have
$\lambda\in[0.949,1.089]$, where a factor $1\pm0.01$ absorbs all corrections.
\end{proposition}

\noindent The next lemma lists everything else that is used about the comb on the real axis. Item (a) is the main statement; (b)
and (d1) control the overshoot; (c) and (e) are upper bounds, for integrability and for the vacuum; (d2) and (f)
concern the vacuum only.

\begin{lemma}[The comb on the band; \provedCA]\label{lem:7.5}
Let $c\ge\catt$. Let $r>0$ be fixed, and let $C_r$ denote constants which depend on $r$ only; the statements on the far
band $x\ge\xs+r$ are made for $c\ge c_0(r)$, so large that $\chi_6\le\xs+r/2$.
\begin{enumerate}[leftmargin=2.6em,itemsep=2pt]
\item[\textup{(a)}] \emph{Two-sided bounds.} For $x$ in a cell $n\ge6$, with all functions evaluated at $x$,
\begin{equation}\label{eq:7-twosided}
  c'\cP+\Jend+\EcH^-_n-2\CL\ \le\ \log\abs B+c'L,\qquad
  \log\abs{B^0}+c'L\ \le\ c'\cP+\Jend+\EcH^+_n+2\CL .
\end{equation}
An upper bound for $B$ costs in addition the detuning factor of \textup{(e)}.
\item[\textup{(b)}] \emph{The overshoot.} For $x>\chi_2$,
\begin{equation}\label{eq:7-Pbounds}
  2n_1\,\bigabs{\log\abs{T_{\xs}(x)}}\ \le\ c'\cP(x)\ \le\ 2n_1\,\bigabs{\log\abs{T_{\chi_2}(x)}} .
\end{equation}
Consequently $e^{c'\cP}\le C_r$ on $x\ge\xs+r$; $e^{c'\cP}\to1$ exponentially deep in the band; and
$e^{c'\cP(x)}\ge\bigl(2/(\pi(x-\xs))\bigr)^{2n_1}$, which is a power of $c'$ near the onset.
\item[\textup{(c)}] \emph{Sup bound.}
\begin{equation}\label{eq:7-sup}
  \sup_{x\ge\chi_2}\bigl[\log\abs B+c'L\bigr]\le\tfrac43n_1\log c'+C,\qquad
  \sup_{x\ge\xs+r}\bigl[\log\abs B+c'L\bigr]\le C_r .
\end{equation}
\item[\textup{(d1)}] \emph{The Airy window.} For $x=\xs+\lc\vsig$ with $\vsig$ in a bounded set,
$e^{c'\cP(x)}=\lc^{-2n_1}e^{p_2(\vsig)}\bigl(1+O(\lc\log\frac1\lc)\bigr)$, with an explicit continuous function $p_2$ that
does not depend on $c$ \eqref{eq:E-potential-limit}. Hence $\Lev\,e^{c'L}\asymp\lc^{-2n_1}\asymp c'^{\frac43n_1}$ on the part
$x\ge\chi_3$ of the window, and $\Lev\,e^{c'L}\asymp(x-\xs)^{-2n_1}$ for $\lc\ll x-\xs\ll1$.
\item[\textup{(d2)}] \emph{Lobe and band in the window.} In the variable $\vsig$ the integrands of the lobe and of the
first cells of the band have one common factor, proportional to $c'^{\frac43n_1-\frac16}$: on the part $\xs\le x<\chi_2$ of the
lobe $\abs{h_0}V\,\dd t$ is at least this factor times a function of $\vsig$, and on each of the first $N_1$ cells of the
band, for any fixed $N_1$, $h_0V\,\dd t$ is at most this factor times another. The integrals of these functions differ by
$O(\lc)$ from those of explicit functions that do not depend on $c$. The precise statement is \cref{lem:F-window}. It is used
in \cref{sec:7.9} only.
\item[\textup{(e)}] \emph{Upper bound with detuning.} Since $\sup_uh^+_\infty(u)=6-4\log3\le1.606$ $\cert$ (row 7), on the
cells $n\ge6$ we have $\sup_{\rm cell}e^{\EcH^+_n}\le5.0\,e^{94/n+C\ell_n}$, and
\begin{equation}\label{eq:7-ub}
  \abs{B(x)}\,e^{c'L(x)}\ \le\ 5.0\,(1+4\eta_0^2)^2\,e^{\Cdet\eta_0^2}\;e^{2\CL+94/n+C\ell_n}\;e^{c'\cP(x)+\Jend(x)} ,
\end{equation}
with an absolute constant $\Cdet\le16.6$ (\cref{lem:E-detune}). In particular $\abs B\,e^{c'L}\le C_r$ on $x\ge\xs+r$.
\item[\textup{(f)}] \emph{The lobe.} For $0\le x<\chi_2$, with all functions evaluated at $x$,
\begin{equation}\label{eq:7-lobe}
  \log\abs{T_{\chi_2}}+c'U+c'\cP_{\chi_{5/2}}\ \le\ \log\abs B\ \le\
  \log\abs{T_{\chi_2}T_{\chi_3}}+c'U+c'\cP_{\chi_3}+4\zeta(2)\,\eta_0^2 .
\end{equation}
Here $\zeta(2)=\pi^2/6$ is the value of Riemann's zeta function, $U=-L$ on $[\xs,\chi_2)$, and no certified constant
enters.
\end{enumerate}
\end{lemma}

\noindent\emph{Proofs.} Item (a) and \cref{prop:7.4} are proved in \cref{app:E.5}; (b) and (d1) in \cref{app:E.1}; (c), (e) and (f)
in \cref{app:E.4}, with the detuning estimate of \cref{app:E.3}; and (d2) in \cref{app:F.2}. The proof of (b) takes one
line: for $x>\chi_2$ the function $x'\mapsto\abs{\log\abs{T_{x'}(x)}}$ increases on $[\xs,\chi_2]$, and $c'\cP(x)$ is $2n_1$
times an average of it over this interval. The last consequence in (b) then follows from $\tanh z\le z$.

\subsection{The transform by steepest descent}
\label{sec:7.6}

We compute $\hat h_0$ on the band by stationary phase in the edge representation \eqref{eq:7-edgerep}. This needs
the phase of $B$ on the edge (\cref{lem:7.6}) and a saddle-point lemma with explicit remainders (\cref{lem:7.7}). The
comb drops out of the amplitude $\Amp$, which depends only on the outer factor and on $F$.

\begin{lemma}[Edge phase; \proved]\label{lem:7.6}
Let $c$ be so large that $\max_n\varepsilon_n\le\frac18$. On the strip $\frac14\le\Im b\le\frac34$, and for $\Re\zeta\ge\xs$,
$\abs{\Im\zeta}\le\frac18$, let $\varphi_\zeta(b):=\log\bigl(-T_\zeta(b)\bigr)$ be the normalized branch of \eqref{eq:C-phi}:
it is continuous in $(b,\zeta)$, it tends to $0$ as $\Re\zeta\to\infty$, and $\varphi_\zeta(\frac i2)=0$. Let
$\Phi_0(b):=\int_{\xs}^\infty\rhoU(x')\,\varphi_{x'}(b)\,\dd x'$ as in \eqref{eq:C-Phi0}, and let $\log B$ be the branch with
$\log B(\frac i2)=i\pi$. Then, on $E_+$,
\begin{equation}\label{eq:7-phase}
  \log B=c'\Phi_0+\Phi_1,\qquad\Re\Phi_0=0,\qquad\Phi_0\bigl(\tfrac i2\bigr)=0,\qquad
  \partial_v\Im\Phi_0\bigl(v+\tfrac i2\bigr)=-\frac{4\pi}3\,F(v),
\end{equation}
where $F$ is the edge function \eqref{eq:7-edge}. Moreover $\Phi_1=i\pi-2n_1\bar\varphi+O(c'^{-2/3})$, where $\bar\varphi$ is
the average of $\varphi_{x'}$ over the absent charge on $[\xs,\chi_2]$; $\Re\Phi_1=0$ on $E_+$; and $\Phi_1$, $\Phi_1'$,
$\Phi_1''$ are bounded on the whole strip $\frac14\le\Im b\le\frac34$, uniformly in $c'$ and in $\Re b$.
\end{lemma}

\noindent The proof is in \cref{app:C.3} (\cref{lem:C-edge-phase}). The term $c'\Phi_0$ is the continuum of
\cref{lem:7.1}, continued across the edge. The three parts of $\Phi_1$ are the sign of the simple zero, the absent charge,
and the $O(c'^{-2/3})$ from the detuning and the discreteness of the sites. In words: on the edge the comb
is a pure phase $e^{ic'\Im\Phi_0}$, up to a slowly varying factor, and the derivative of this phase is the edge
function.

\paragraph{The saddles.}
By \eqref{eq:7-hankel} the phase of the kernel $H_0^{(1)}(k_xb)$ along $b=v+\frac i2$ is $k_xv$ plus slowly varying terms.
The phase of the integrand of \eqref{eq:7-edgerep} is therefore $c'\bigl[\Im\Phi_0(v+\frac i2)+\frac{4\pi}3xv\bigr]$ at
leading order, and by \eqref{eq:7-phase} it is stationary where $F(v)=x$. Since $F$ is even and increases from
$F(0)=\xs$, there are two saddles $v=\pm v_0(x)$ when $x>\xs$, which coalesce at $v=0$ as $x\downarrow\xs$, and
none when $x<\xs$. The curvature there is $-\frac{4\pi}3c'F'(v_0)$, with
$F'(v_0)=(\sinh^2\pi x-1)^{1/2}/\cosh\pi x$; it vanishes like $(x-\xs)^{1/2}$ at the tip, so the next lemma stops
at $\xI$.

\begin{lemma}[The transform on the band; \proved]\label{lem:7.7}
Let $\thon=0.2$ and $\xI=\xs+c'^{-\thon}$; any $\thon\in(\frac1{4n_1},\frac23)$ would do. For $c$ large and every $x\ge\xI$,
\begin{equation}\label{eq:7-transform}
  \hat h_0\bigl(c'x^2/3\bigr)=\Amp(x)\,\bigl[\cos\hPsi(x)+\epssd(x)\bigr],
\end{equation}
with a real phase $\hPsi$, given in \eqref{eq:D-phase}, and
\begin{equation}\label{eq:7-amp}
  \Amp(x)\,e^{c'L(x)}=\frac{4}{\sinh\pi x}\Bigl(\frac{\abs{b_0}}{x\,F'(v_0)}\Bigr)^{1/2}\bigl(1+O(1/c')\bigr),\qquad
  \abs{\epssd(x)}\le\frac{C}{c'\min\{1,v_0(x)\}^3} .
\end{equation}
Here $b_0=v_0(x)+\frac i2$, and $4/\sinh\pi x=2\abs{\sech^2\frac{\pi b_0}2}$ by \cref{prop:7.2}\textup{(d)}: the comb does not
appear in the amplitude. Moreover
\begin{equation}\label{eq:7-glob}
  \bigabs{\hat h_0\bigl(c'x^2/3\bigr)}\le\Cglob\,F'(v_0)^{-1/2}\,e^{-\pi v_0(x)}\,e^{-c'L(x)}\qquad(x\ge\xI),\qquad\Cglob=2^5 .
\end{equation}
\end{lemma}

\noindent The first formula of \eqref{eq:7-amp} is \eqref{eq:D-amp-closed} with the outer factor $\sech^2$, and its
$O(1/c')$ is at most $0.33/\abs{k_xb_0}$. The second is \eqref{eq:D-comb-rate}; it is $O(1/c')$ deep in the band and at most
$Cc'^{-(1-3\thon/2)}=Cc'^{-0.7}$ at $\xI$. The bound \eqref{eq:7-glob} is \cref{cor:D.3}. The general statement is
\cref{lem:D.1}, which deforms the edge into the two steepest-descent paths through $\pm b_0$; its hypotheses
\hyp{SD1}--\hyp{SD4} are verified for the comb in \cref{app:D.4}, where the condition $\thon<\frac23$ enters.

\tagpar{\orientation} At leading order $\hPsi'=\Psi'$, by the eikonal identity \eqref{eq:C-eikonal}, so $\hat h_0$ oscillates
once per cell, in step with $h_0$ (\cref{fig:9}); this is not used, since \cref{sec:7.7} uses $\abs{\cos\hPsi}\le1$ only.

\subsection{Positivity on the band beyond the onset (Region II)}
\label{sec:7.7}

In Region~II, $x\ge\xI$, both sides of $\hat h_0<h_0$ are known in units of the level:
$\abs{\hat h_0}\le\Amp\,(1+\abs{\epssd})$ by \cref{lem:7.7}, and by the detuning $h_0$ stays above a multiple
of $\Lev$ that is fixed up to factors which tend to one. The first lemma computes $\Amp/\Lev$, the second gives the floor, and the criterion compares them.

\begin{lemma}[The amplitude in units of the level; \provedCA]\label{lem:7.8}
For $c$ large and every $x\ge\xI$,
\begin{equation}\label{eq:7-vrhoinf}
  \vrho:=\frac{\Amp}{\Lev}\ \le\ \vrhoinf\,\bigl(1+O(1/c')\bigr),\qquad
  \vrhoinf(x):=\frac12\Bigl(\frac{\abs{b_0}}{x\,F'(v_0)}\Bigr)^{1/2}\coth\frac{\pi x}2\;
  e^{-2n_1\abs{\log\abs{T_{\xs}(x)}}} .
\end{equation}
The function $\vrhoinf$ is explicit and does not depend on $c$. It vanishes like $(x-\xs)^{2n_1-\frac14}$ at the tip, it
tends to $\half$ as $x\to\infty$, and
\begin{equation*}
  \vrhoinf\le0.5029\ \ \cert\ \text{ on }(\xs,3],\qquad\vrhoinf\le0.504\ \text{ on }[3,\infty),\qquad
  \vrhoinf\le0.41\ \ \cert\ \text{ on }(\xs,1.33] .
\end{equation*}
\end{lemma}

\begin{proof}
We divide \eqref{eq:7-amp} by \eqref{eq:7-lev}. Since $(\sinh\pi x)^{-1}\cosh^2\frac{\pi x}2=\half\coth\frac{\pi x}2$,
\begin{equation*}
  \vrho(x)=\frac12\Bigl(\frac{\abs{b_0}}{x\,F'(v_0)}\Bigr)^{1/2}\coth\frac{\pi x}2\;e^{-c'\cP(x)-\Jend(x)}\,\bigl(1+O(1/c')\bigr) :
\end{equation*}
the outer factor has canceled up to the factor $\coth$. For $c$ large $\xI>\chi_3$, because
$\chi_3-\xs\asymp\lc\ll c'^{-\thon}$. So $\Jend\ge0$ by \eqref{eq:7-Jend}, and $c'\cP\ge2n_1\abs{\log\abs{T_{\xs}(x)}}$ by
\cref{lem:7.5}(b). This is \eqref{eq:7-vrhoinf}. At the tip $F'(v_0)\asymp(x-\xs)^{1/2}$ and $\abs{T_{\xs}(x)}\asymp x-\xs$,
which gives the power; as $x\to\infty$ every factor except $\half$ tends to one. The bounds $0.5029$ and $0.41$ are rows 1
and 2 of \cref{tab:4}. Row~1 is verified in interval arithmetic on $[\xs+10^{-3},3]$, where $\vrhoinf$ is largest at
the right end, $\vrhoinf(3)=0.50271$, and row~2 on the part $x\le1.33$ of the same covering. On $(\xs,\xs+10^{-3}]$ we
have $\abs{T_{\xs}(x)}\le\frac\pi2(x-\xs)$, $F'(v_0)=\sinh\pi v_0/\cosh\pi x\ge2(x-\xs)^{1/2}$ (as
$\sinh^2\pi v_0=\sinh^2\pi x-1\ge2\sqrt2\,\pi(x-\xs)$ by convexity), $(\abs{b_0}/x)^{1/2}\le1.34$ and
$\coth\frac{\pi x}2\le\coth\frac{\pi\xs}2=1+\sqrt2$, which give $\vrhoinf\le6\,(x-\xs)^{13/4}<10^{-8}$. For $x\ge3$ the bound
is by hand: $F'(v_0)\ge0.9999$; $v_0<x$, so $(\abs{b_0}/x)^{1/2}\le(1+\frac1{4x^2})^{1/4}\le(1+\frac1{36})^{1/4}\le1.0069$;
$\coth\frac{\pi x}2\le1.0002$; and the exponential is at most one. Then
$\half\cdot1.0069\cdot1.0002\cdot0.9999^{-1/2}\le0.5036<0.504$.\looseness=-1
\end{proof}

\begin{lemma}[The detuned floor; \provedCA]\label{lem:7.9}
Let $c\ge\catt$, let $j\in\{1,4\}$ and let $x$ lie in a cell $n\ge2j+4$. Then
\begin{equation}\label{eq:7-floor}
  \frac{h_0(x)}{\Lev(x)}\ \ge\ \tfrac14\,e^{\EcH^-_{n,j}(x)-2e_L(x)}\ \ge\
  \mu_j\Bigl[1-C_j\bigl(\tfrac1n+\ell_n\bigr)\Bigr]e^{-2\CL},\qquad
  \mu_j:=\min_{0\le u\le1}\tfrac14\,e^{h^-_j(u;\eta_0)} ,
\end{equation}
with $C_j$ as in \cref{sec:7.5}, and for $\eta_0=\frac14$
\begin{equation*}
  \mu_1\ge0.536\ \ \cert,\qquad\mu_4\ge0.699\ \ \cert .
\end{equation*}
The factor in brackets tends to $1$ as $n\to\infty$. On a cell where it is not positive only the first inequality has
content.
\end{lemma}

\noindent This is \eqref{eq:E-floor}; the proof is in \cref{app:E.6}, and the two numbers are row 3 of \cref{tab:4}. Every
pair beyond the $2j$ nearest is bounded below by its undetuned value, by \hyp{T2}, and then by the continuum. The
minimum over the cell is
attained at the sites, $u\in\{0,1\}$, where one pair is at its least value. Keeping more pairs exactly removes more of the
loss in the midpoint rule, which is why $\mu_4>\mu_1$; as $j\to\infty$, $\mu_j\to\sinh^2\pi\eta_0=0.7546$, the floor seen
in \cref{fig:9}(c).

\paragraph{The criterion.}
Let $x\ge\xI$ lie in cell $n$, and let $j\in\{1,4\}$. Since $2g=h_0-\hat h_0$, \cref{lem:7.7,lem:7.8,lem:7.9} give
\begin{equation}\label{eq:7-criterion}
  \frac{2g(t)}{\Lev(x)}\ \ge\ \mu_j\Bigl[1-C_j\bigl(\tfrac1n+\ell_n\bigr)\Bigr]e^{-2\CL}
  \;-\;\vrhoinf(x)\,\bigl(1+\abs{\epssd(x)}\bigr)\bigl(1+O(1/c')\bigr) .
\end{equation}
No information on the phase $\hPsi$ has been used. The factors that tend to one are controlled as follows. In Region~II the
cell index is $\gtrsim c'^{1-3\thon/2}$, because $c'N(\xI)\simeq\ktip c'\,c'^{-3\thon/2}$ by~\eqref{eq:6-N}, and
$\ell_n\le\lc$; so the factor in brackets tends to $1$ uniformly on Region~II, and $n\ge2j+4$ holds there. The factor
$1+\abs{\epssd}$ tends to $1$ uniformly by \eqref{eq:7-amp}. We take $\catt$ so large that, for $c\ge\catt$ and all
$x\ge\xI$, the bracket is at least $0.995$ and $(1+\abs{\epssd})(1+O(1/c'))\le1.1$. Also $e^{-2\CL}\ge0.988$, by \cref{lem:E-farleft}(d).
We use $j=1$ where $x\le1.33$, and $j=4$ where $x\ge1.33$:
\begin{equation}\label{eq:7-margins}
\begin{aligned}
  \frac{2g}{\Lev}&\ \ge\ 0.995\cdot0.536\cdot0.988-0.41\cdot1.1\ \ge\ 0.07&&(\xI\le x\le1.33),\\
  \frac{2g}{\Lev}&\ \ge\ 0.995\cdot0.699\cdot0.988-0.504\cdot1.1\ \ge\ 0.13&&(x\ge1.33).
\end{aligned}
\end{equation}
The second line uses $0.504$, which covers both the certified $0.5029$ on
$(\xs,3]$ and the bound by hand beyond $3$. With $j=4$ the second line alone holds on all of Region~II, because its bracket
also tends to one there; the first line is kept to show the margin with the smaller hole. The cells are open intervals;
since $g$ is continuous (\cref{sec:7.10}) and $\Lev$ is continuous and positive on $(\chi_2,\infty)$, the inequalities
\eqref{eq:7-margins} hold at the sites as well. Hence $g>0$ on Region~II for $c\ge\catt$.

\subsection{The onset (Region I)}
\label{sec:7.8}

In Region~I, $\chi_2\le x\le\xI$, the two saddles are too close for \cref{lem:7.7}. Instead we compare a crude upper bound for $\hat h_0$, which exceeds the Cardy level by $\sqrt{c'}$, with lower bounds for
$h_0$ that contain the overshoot, a larger power of $c'$.

The cells $2,\dots,5$ lie in Region~I for $c$ large, because
$\chi_6-\xs\asymp\lc\asymp c'^{-2/3}\ll c'^{-\thon}$. On them nothing is replaced by a continuum to the left of $x$, and
$h_0$ is bounded below by an exact finite product (\cref{lem:7.12}). On the cells $n\ge6$ that meet Region~I the floor
times the overshoot (\cref{lem:7.11}) exceeds the crude bound (\cref{lem:7.10}).

\begin{lemma}[Crude bound; \proved]\label{lem:7.10}
For every $x>0$ with $c'L(x)\ge1$,
\begin{equation}\label{eq:7-crude}
  \bigabs{\hat h_0\bigl(c'x^2/3\bigr)}\,e^{c'L(x)}\le C\sqrt{c'/x},\qquad\text{and}\qquad\abs{\hat h_0(t)}\le Cc'\quad\text{for all }t\ge0 .
\end{equation}
\end{lemma}
\begin{proof}
We take absolute values in \eqref{eq:7-edgerep}. On $E_+$, $\abs{h_0}=2/\cosh\pi v$ by \cref{prop:7.2}(d), and
$k_x\abs b\ge\half k_x=c'L(x)\ge1$, so \eqref{eq:7-hankel} applies with $\abs{1+\epsH}\le1.33$. Hence
\begin{equation*}
  \abs{\hat h_0}\le\frac{2\pi c'}3\int_{-\infty}^\infty\frac{2}{\cosh\pi v}\Bigl(\frac{2}{\pi k_x\abs b}\Bigr)^{1/2}1.33\,e^{-c'L(x)}\,\abs b\,\dd v
  \le C\,c'\,(c'x)^{-1/2}\,e^{-c'L(x)}\int_{-\infty}^\infty\frac{\abs b^{1/2}}{\cosh\pi v}\,\dd v ,
\end{equation*}
and the last integral is a number. For the second bound we use $\abs{J_0}\le1$ on the real axis in \eqref{eq:2-eigen}:
$\abs{\hat h_0}\le2\pi\int_0^\infty\abs{h_0}\,\dd t=\frac{4\pi c'}3\int_0^\infty\abs{h_0(x)}\,x\,\dd x\le Cc'$, because
$\abs{h_0(x)}\le4e^{-\pi x}$.
\end{proof}

\begin{lemma}[Overshoot; \provedCA]\label{lem:7.11}
Let $c\ge\catt$. On the cells $n\ge6$,
\begin{equation}\label{eq:7-overshoot}
  \abs{B(x)}\,e^{c'L(x)}\ \ge\ 32\,\eta_0^2\,e^{-C(\frac1n+\ell_n)-2\CL}\Bigl(\frac{2}{\pi(x-\xs)}\Bigr)^{2n_1}
  \ \ge\ \cT\,\eta_0^2\Bigl(\frac{2}{\pi(x-\xs)}\Bigr)^{2n_1},
\end{equation}
with an absolute constant $\cT>0$. Only the positivity of $\cT$ is used.
\end{lemma}

\noindent This is \eqref{eq:E-overshoot}, proved in \cref{app:E.4}. It follows from the lower bound of \cref{lem:7.5}(a)
with $\Jend\ge0$, from the last consequence in \cref{lem:7.5}(b), and from $\min_ue^{h^-_\infty(u;\eta_0)}\ge32\eta_0^2$
$\cert$ (row 8 of \cref{tab:4}).
The exponent in the middle is bounded because $\frac1n\le\frac16$ and $\ell_n\le\lc$.

Now let $x\le\xI$ lie in a cell $n\ge6$. Then $x-\xs\le c'^{-\thon}$, and $\sech^2\frac{\pi x}2$ is bounded below, so
\cref{lem:7.11} and \cref{lem:7.10} give
\begin{equation*}
  h_0\,e^{c'L}\ \ge\ C^{-1}(x-\xs)^{-2n_1}\ \ge\ C^{-1}c'^{\,2n_1\thon}=C^{-1}c'^{\,0.7},\qquad\abs{\hat h_0}\,e^{c'L}\le C\,c'^{\,1/2} ,
\end{equation*}
and $g>0$ for $c$ large, because $2n_1\thon>\half$, that is, $\thon>\frac1{4n_1}$ (\cref{fig:9}).

\begin{lemma}[First cells; \provedCA]\label{lem:7.12}
Let $c\ge\catt$. On the cells $n=3,4,5$,
\begin{equation}\label{eq:7-first-cells}
  h_0\ \ge\ \Lev\,e^{-\Jend}\,\muI_n\,\bigl(1-C\lc\log\tfrac1\lc\bigr),\qquad
  \muI_3\ge0.679,\ \ \muI_4\ge0.683,\ \ \muI_5\ge0.686\ \ \cert ,
\end{equation}
and on the whole of cell $2$,
\begin{equation}\label{eq:7-slope}
  h_0(x)\ \ge\ \Lev\,e^{-\Jend}\;\muslope\,\frac{x-\chi_2}{\chi_3-\chi_2}\,\bigl(1-C\lc\log\tfrac1\lc\bigr),\qquad
  \muslope=0.686\quad\cert .
\end{equation}
On these cells $\Lev\,e^{-\Jend}e^{c'L}\asymp\lc^{-2n_1}\asymp c'^{\frac43n_1}\gg\sqrt{c'}$.
\end{lemma}

\noindent The proof is in \cref{app:E.7} (\cref{lem:E-first-cells}), and the numbers are row 4 of \cref{tab:4}. Here all
the sites up to $\chi_{n+4}$ are kept exactly, and the midpoint rule is used only beyond them. In the Airy variable
$\vsig=(x-\xs)/\lc$ the finite product becomes a function on each cell that does not depend on $c$, and $\muI_n$ is its
minimum \eqref{eq:E-muI}. The unit is $\Lev\,e^{-\Jend}$ and not $\Lev$ because the exact block reaches the end of the comb:
there is no end term, and there is no loss $e^{\pm2\CL}$. Its size is \cref{lem:7.5}(d1). On cell $2$ the function $h_0$
vanishes at the left end, so the bound is a multiple of the distance to the zero.

\paragraph{Conclusion for Region~I.}
On the cells $3,4,5$ we have $h_0\,e^{c'L}\ge C^{-1}c'^{\frac43n_1}$ against $\abs{\hat h_0}\,e^{c'L}\le C\sqrt{c'}$, so $g>0$
for $c$ large. On cell $2$,
\begin{equation*}
  2g\,e^{c'L}\ \ge\ C^{-1}c'^{\frac43n_1}\,\frac{x-\chi_2}{\lc}-C\,c'^{1/2} ,
\end{equation*}
which is positive as soon as $x-\chi_2\ge C\lc\,c'^{\frac12-\frac43n_1}=C\,c'^{-5/2}$, that is,
$t-\tstar\ge Cc'^{-3/2}$. There is no uniform margin at the simple zero; hence the offset in
\cref{thm:5}(ii), which is far larger than needed. The bounds of \cref{lem:7.11,lem:7.12} are inequalities between
continuous functions of $x$; proved on the open cells, they hold at the sites $\chi_n$, $n\ge3$, as well. With the cells
$n\ge6$ treated above, $g>0$ on $\tstar+o(1/c')\le t\le c'\xI^2/3$.

\subsection{The vacuum inequality}
\label{sec:7.9}

The vacuum value is a competition between the negative lobe of $h_0$ below $\tstar$ and the positive band beyond it, both
weighted with the Cardy weight $V$. The lobe wins: the proof gives the factor $1/\Rvac\ge8$.
The omitted site is what makes it so: the stretch from the first site of the law to $\chi_2$, which would otherwise be
the first cell of the band, belongs to the lobe.

By \cref{lem:7.3}(b) and \cref{prop:7.2}(b),(c),
\begin{equation}\label{eq:7-vacsplit}
  \gvac=\tfrac12\,h_{0,\rm vac}+\pi\bigl(\abs{\Lobe}-\Band\bigr),\qquad
  \Lobe:=\int_0^{\tstar}h_0\,V\,\dd t,\qquad\Band:=\int_{\tstar}^\infty h_0\,V\,\dd t ,
\end{equation}
with $\Lobe<0<\Band$ and $\abs{h_{0,\rm vac}}\le8$. The transform $\hat h_0$ enters only through the hypothesis of
\cref{lem:7.3}(b), which is verified in \cref{sec:7.10}. By \cref{lem:7.5}(d1) both
integrals are dominated by a window of width $O(\lc)$ around $\xs$, where $\abs B\,e^{c'L}$ is of order $\lc^{-2n_1}$. In the
variable $\vsig=(x-\xs)/\lc$ the lobe and the first $N_1$ cells of the band are one common factor, proportional to
$c'^{\frac43n_1-\frac16}=c'^{13/6}$, times integrals that do not depend on $c$ (\cref{lem:7.5}(d2)). The later cells, up to
a fixed distance from the tip, carry the same factor times a small number, because the overshoot decays like
$\vsig^{-2n_1}$, and the rest of the band contributes $O(\sqrt{c'})$, by \cref{lem:7.5}(e). The common factor cancels from
the ratio.

\begin{proposition}[Vacuum inequality; \provedCA]\label{prop:7.13}
There is a number $\Rvac<1$, which does not depend on $c$, such that for $c\ge\catt$
\begin{equation}\label{eq:7-vac}
  \Band\ \le\ \Rvac\,\abs{\Lobe},\qquad\Rvac:=\FhR(N_1)+\delta_{\rm tail}+0.01\ \le\ 0.1021+0.02,\qquad N_1=100 ,
\end{equation}
and $\abs{\Lobe}\to\infty$ as $c\to\infty$, at least like $c'^{13/6}$. Hence $\gvac\ge\pi(1-\Rvac)\abs{\Lobe}-4>0$. Here
$\FhR(N_1)$ is the \emph{scaled vacuum ratio} \eqref{eq:F-ratio-def}, the ratio of the $c$-independent integrals of
\cref{lem:7.5}\textup{(d2)} over the first $N_1$ cells of the band and over the lobe, and $\FhR(100)\le0.1021$ $\cert$ (row 11 of \cref{tab:4});
the term $\delta_{\rm tail}:=\frac{C_{\rm tail}}{\FIL}\,\frac{(N_1-\frac14)^{-5/3}}{2.5}\le0.01$ bounds the cells beyond them, by hand,
with the explicit constant $C_{\rm tail}$ of \cref{lem:F-tails} and the scaled integral $\FIL$ over the lobe;
and $0.01$ covers the corrections at finite $c$.
\end{proposition}

\noindent The proof is in \cref{app:F}, and the assembly is in \cref{app:F.5}. Besides \cref{lem:7.5}(d2),(e),(f), it uses the
two-sided asymptote $V=\as(c'L)\,[(1-e^{-4\pi x})^2+O(1/c')]$ of the Cardy weight (\cref{lem:A.4}, \eqref{eq:A-V-as}). It
matters that the asymptote is two-sided: the lobe is bounded below and the band above, and both must be measured with the
same weight; with the one-sided bound $V\le e^{c'L}$ on the band the estimate of the ratio would be larger by the
factor $1.92\sqrt{c'}$, and would exceed one (\cref{app:F.5}). Only $\Rvac<1$ is needed, so the margin is wide.

\subsection{Membership, end of the proof, and a numerical illustration}
\label{sec:7.10}

It remains to check that $g$ belongs to $\Abar$, which also supplies the hypothesis of \cref{lem:7.3}.

\paragraph{Membership.}
On the real axis $\abs{h_0}\le4e^{-\pi x}\abs B$. By \cref{lem:7.5}(c), $\abs B\le C(c)\,e^{-c'L}$ for $x\ge\chi_2$, and
$\abs B\le1$ below $\chi_2$, where $e^{-c'L}$ is bounded below. So $h_0$ is continuous and
$\abs{h_0(t)}\le C(c)\,e^{-c'L(x)-\pi x}$ for $t\ge0$. In particular $h_0$ is integrable, so $\hat h_0$ is continuous. Both
are real for $t\ge0$: $T_{\chi_2}$ is real on the real axis and the other factors of $B$ come in conjugate pairs, and the
kernel of \eqref{eq:2-eigen} is real. For
$x\ge\xI$ the bound \eqref{eq:7-glob} gives $\abs{\hat h_0}\le C(c)\,e^{-c'L-\pi x}$: indeed
$e^{\pi v_0}\ge\cosh\pi v_0=\sinh\pi x$, so $e^{-\pi v_0}\le Ce^{-\pi x}$, and $F'(v_0)^{-1/2}\le Cc'^{\thon/4}$ there. For
$x\le\xI$ we have $\abs{\hat h_0}\le Cc'$ by \cref{lem:7.10}, and $e^{-c'L-\pi x}$ is bounded below. This is the hypothesis
of \cref{lem:7.3}(a),(b), and with it the decay stated in \cref{thm:5}. Thus $g=\half(h_0-\hat h_0)$ is real and continuous
on $[0,\infty)$. The three conditions of \cref{def:2-Abar} follow.
Condition~(i) is the integrability just shown. Condition~(ii) is \cref{lem:7.3}(a). For (iii), $V\le(4\pi^2t/a)\,e^{c'L}$ by
\cref{lem:A.2}(d), so $\abs g\,V\le C(c)\,t\,e^{-\pi x}$, which is integrable.

\paragraph{End of the proof of \cref{thm:5}.}
Part~(i) is the preceding paragraph. For (ii), \cref{sec:7.7} gives $g>0$ for $x\ge\xI$, and \cref{sec:7.8} gives $g>0$ for
$\tstar+o(1/c')\le t\le c'\xI^2/3$; these cover $t\ge\tstar+1$. Part~(iii) is \cref{prop:7.13}. Let $\catt$ be the largest of the finitely many thresholds met on the way. Each
of them is explicit in principle, and none has been computed. \qed

\paragraph{A numerical illustration \tagpar{\numerical}.}
At $c=128$, $256$, $512$, $1024$ the smallest sampled value of $g$, in units of $h_0$ at the
middle of the same cell, lies between $+0.065$ and $+0.080$; it is attained at the first pair of detuned sites, and deep
in the band the same ratio is about $0.35$. Also $\Band/\abs{\Lobe}\in[0.074,0.081]$. The proof uses
none of these numbers. At $c=256$ the functional would give $\Delta_1\le a+\tstar+1=32.3$, against the
numerical optimum $29.72$ of \cref{tab:2}.

Three things are \emph{not} claimed: that the optimal functional has this form; that $0.483$ is the best constant in
\eqref{eq:7-tstar}; and anything at a specific finite $c$.

\section{From functionals to CFT spectra}
\label{sec:8}

\Cref{thm:5} supplies a feasible function in $\Abar$. That such a function constrains every conformal field theory is
\cref{thm:2}, which we prove in this section, without a computer; \cref{cor:6} then follows at once.

\subsection{The issue, the idea, and the statement}
\label{sec:8.1}

The sum rule \eqref{eq:2-sumrule} came from applying a functional to the crossing equation term by term. For a
functional of finite order with compact support in $\beta$ this is termwise differentiation of a series of exponentials
that converges uniformly with all its derivatives. A function $g\in\Abar$ is only continuous and integrable against the
Cardy weight. If it comes from a functional $\omega$ at all, the support of $\omega$ may reach the boundary of moduli
space, $\beta\to0$ and $\beta\to\infty$, and exchanging $\omega$ with the sum over primaries is then a genuine condition,
the analogue of the swapping conditions for four-point functions~\cite{QiaoRychkov,MazacPaulos1,MazacPaulos2}. Two things can go wrong. First,
$\sum_nd_n\,g(t_n)$ need not converge absolutely: the primaries can sit anywhere, their number grows like the Cardy
density, and $M[g]<\infty$ controls $\abs g$ only in the mean. Second, the vacuum sits at $t=-a$, that is, at
$b=\pm i/2$ in the variable of \cref{sec:2.5}, off the real axis, where $g$ is not given but \emph{defined} by the
eigen-identity (\cref{def:2-Abar}); one must show that $\gvac=-\int g\,\rhoC$ so defined is what crossing assigns to
the vacuum.

\paragraph{The idea.} We never construct $\omega$. We expand $g$ in the Laguerre functions $\psi_k$ of
\eqref{eq:2-psik}. Because $\hat g=-g$ only odd $k$ occur, and each odd $\psi_k$ is a derivative functional of finite
order at $\beta=1$, which annihilates the crossing equation. The Laguerre series of $g$ cannot be summed against the
spectrum term by term, because its partial sums need not converge there. We therefore insert a factor $r^k$ with
$0<r<1$ and consider the \emph{Abel mean} $\Ab_rg=\sum_kr^kg_k\psi_k$, where $g_k$ are the Laguerre coefficients of
$g$. On the crossing kernel this regulator is a contraction of the modulus toward the self-dual point: $\Ab_r\Xib$ is a
multiple of $\Xi_{\beta_r}$, where $\vpi(\beta_r)=r\,\vpi(\beta)$ in the variable $\vpi:=(\beta-1)/(\beta+1)$, so that
$\beta_r$ stays in the compact interval $[\frac{1-r}{1+r},\frac{1+r}{1-r}]$. Each Abel mean thus behaves as a functional
supported strictly inside moduli space, and the sum rule holds for $\Ab_rg$ at every $r<1$ at no cost beyond
$\Zhat(\beta)<\infty$.

The content of the theorem is the removal of the regulator, $r\uparrow1$, which lets the support reach the cusps where
the Cardy growth lives. The Abel mean is an integral operator with a positive, symmetric kernel, known in closed form,
which concentrates on the diagonal as $r\uparrow1$. The vacuum enters through its $S$-image, the Cardy continuum on
$t\ge0$, and symmetry moves $\Ab_r$ from $g$ onto $\rhoC$. The kernel preserves the Cardy growth of $\rhoC$ up to a
constant factor, uniformly in $r$ (\cref{lem:8.1}), so $M[g]<\infty$ is exactly the domination that dominated
convergence requires, and the vacuum side tends to $\int g\,\rhoC=-\gvac$. On the primary side no domination is
available. Feasibility supplies a sign instead, and for non-negative terms Fatou's lemma needs none: the sum of the
lower limits is at most the lower limit of the sum, $\sum_n\liminf_r\le\liminf_r\sum_n$. It applies to the positive
part $g_+$. The negative part $g_-$, the lobe, lives on $[0,T]$, below every primary by hypothesis, and its leakage
onto the primaries, a Gaussian tail of the kernel, vanishes against the Cardy growth of their number. The outcome is
$0\le\sum_nd_n\,g(t_n)\le-\gvac<0$.

\paragraph{The statement.} This proves \cref{thm:2}: for $c\ge40$, if $g\in\Abar$ is feasible at some $T>0$, then no
admissible spectrum (\cref{def:2-admissible}) has all its primaries at $t_n\ge T$. The argument is by contradiction
from $t_n\ge T>0$ for all $n$, so light primaries never enter it, while in the conclusion the primary with $t_n<T$ may
be a light one. The hypothesis that there are no conserved currents beyond Virasoro (\cref{sec:2.1}) cannot be dropped
within this framework: a current $J$ enters every sum rule as $d_J[g(t_J)-g(t_J+1)]$, a difference whose sign
feasibility does not control.

\subsection{Set-up}
\label{sec:8.2}

We list what \cref{sec:8.3} uses, so that the proof can be checked line by line. Classical facts about Laguerre
polynomials are quoted from~\cite{Szego}; everything else is proved.

\paragraph{Conventions.} Throughout this section $u:=4\pi t$ and $\sigma:=\sqrt u$ and likewise for $u'$; a
function of $t\ge0$ is regarded as a function of $u$ without change of name, and $L^p(\dd u)$ refers to $[0,\infty)$.
Thus $\psi_k=L_k(u)e^{-u/2}$. We write $\eb(t):=e^{-2\pi\beta t}=e^{-\beta u/2}$, so that $\Xib=\eb-\beta^{-1}e_{1/\beta}$
by \eqref{eq:2-g}; $r\in(0,1)$ is the Abel ratio. For a spectrum $\{(t_n,d_n)\}$ with all $t_n\ge0$ the \emph{signed
spectral measure} and its pairing with a function $f$ on $[0,\infty)$ are
\begin{equation}\label{eq:8-measure}
  \mfrak:=\sum_nd_n\,\delta_{t_n}-\rhoC(t)\,\dd t,\qquad
  \langle\mfrak,f\rangle:=\sum_nd_n\,f(t_n)-\int_0^\infty\rhoC\,f\,\dd t ,
\end{equation}
and $\abs\mfrak$ is the same measure with a plus sign. By \eqref{eq:2-rhoC} the vacuum term of $\Zhat(\beta)$ equals
$\beta^{-1}\int_0^\infty\rhoC\,e_{1/\beta}\,\dd t$, so the crossing equation \eqref{eq:2-crossing} reads
$\langle\mfrak,\Xib\rangle=0$: the vacuum has been traded for its $S$-image, and only $t\ge0$ occurs. Growth and decay
are measured by the weights
\begin{equation}\label{eq:8-weights}
  W_{q,\ell}(t):=(1+u)^q\,e^{\ell\sigma}\quad(q,\ell\in\RR),\qquad \ell_a:=2\sqrt{\pi a},\qquad
  0\le\rhoC\le C_\rho\,W_{-1/4,\ell_a}.
\end{equation}
Here $e^{\ell_a\sigma}=e^{4\pi\sqrt{at}}=e^{c'L(x)}$ is the Cardy factor. The upper bound on $\rhoC$, with
$C_\rho=C_\rho(a)$, follows from $\rhoC=2\pi V\le4\pi K_a$ and
$K_a=I_0(\ell_a\sigma)\le1.18\,e^{\ell_a\sigma}/\sqrt{1+\ell_a\sigma}$ (\cref{lem:A.4}(c)); $\rhoC\ge0$ by
\cref{lem:A.2}(a). Since $c$ is fixed, constants may depend on $a$; what matters is uniformity in $r$.

\paragraph{Five facts about the spectrum.} Let $\{(t_n,d_n)\}$ be an admissible spectrum with all $t_n\ge0$, and
$\NZ(R):=\sum_{t_n\le R}d_n$ its counting function.
\begin{description}[leftmargin=2.7em,labelwidth=2.2em,labelsep=0.5em,itemsep=1pt,font=\normalfont]
\item[\hypdef{Z1}] $\int e^{-2\pi\beta t}\,\dd\abs\mfrak(t)<\infty$ for every $\beta>0$: the sum over primaries is at most
  $\Zhat(\beta)$, and the integral is the right side of \eqref{eq:2-rhoC}.
\item[\hypdef{Z2}] $\langle\mfrak,\Xib\rangle=0$ for every $\beta>0$, with absolute convergence. \hyp{Z1} and \hyp{Z2}
  restate the definition of an admissible spectrum.
\item[\hypdef{Z3}] \emph{Cardy counting bound.} $\NZ(R)\le\CZ\sqrt{R/a}\;e^{4\pi\sqrt{aR}}$ for all $R\ge a$, where
  $\CZ:=1+\Zhat(1)$. Consequently $\NZ(t)\le C_N\,(1+\sigma)\,e^{\ell_a\sigma}$ for all $t\ge0$, where
  $C_N:=\CZ\,e^{4\pi a}\max(1,a^{-1/2})$.
\item[\hypdef{Z4}] For every odd $k$, $\sum_nd_n\abs{\psi_k(t_n)}<\infty$ and $\langle\mfrak,\psi_k\rangle=0$.
\item[\hypdef{St}] \emph{Stieltjes summation.} If $\phi\ge0$ is $C^1$ and non-increasing on $[T_0,\infty)$, then
  $\sum_{t_n\ge T_0}d_n\,\phi(t_n)\le\limsup_{R\to\infty}\phi(R)\NZ(R)+\int_{T_0}^\infty\NZ\,\abs{\phi'}\,\dd t$. This is
  integration by parts in $\int_{[T_0,R]}\phi\,\dd\NZ$, with the boundary term $-\phi(T_0)\NZ(T_0^-)\le0$ dropped.
\end{description}

\begin{proof}[Proof of \hyp{Z3}]
This is Chebyshev's inequality (a sum of positive terms bounds each of its parts) for $\Zhat(\beta)$ at small $\beta$,
after crossing. For $\beta'\ge1$, with
$\Delta_n=t_n+a>0$,
$\Zhat(\beta')\,e^{-2\pi a\beta'}=(1-e^{-2\pi\beta'})^2+\sum_nd_ne^{-2\pi\beta'\Delta_n}\le1+\sum_nd_ne^{-2\pi\Delta_n}\le\CZ$.
For $R\ge a$ put $\beta:=\sqrt{a/R}\le1$. Every term with $t_n\le R$ has $e^{-2\pi\beta t_n}\ge e^{-2\pi\beta R}$, so by
\eqref{eq:2-crossing}
\[
  \NZ(R)\,e^{-2\pi\beta R}\le\sum_nd_n\,e^{-2\pi\beta t_n}\le\Zhat(\beta)=\beta^{-1}\Zhat(1/\beta)
  \le\beta^{-1}\CZ\,e^{2\pi a/\beta}=\CZ\sqrt{R/a}\;e^{2\pi\sqrt{aR}} ,
\]
and multiplying by $e^{2\pi\beta R}=e^{2\pi\sqrt{aR}}$ gives the first claim. It implies the second for $t\ge a$,
because $\sqrt{t/a}\le a^{-1/2}\sigma$ and $4\pi\sqrt{at}=\ell_a\sigma$; for $t\le a$ use $\NZ(t)\le\NZ(a)\le\CZ\,e^{4\pi a}$.
\end{proof}

\paragraph{Laguerre functions.} The generating function of the Laguerre polynomials~\cite[(5.1.9)]{Szego} is
$\sum_{k\ge0}L_k(u)\,w^k=(1-w)^{-1}e^{-uw/(1-w)}$ for $\abs w<1$. Multiply it by $e^{-u/2}$ and put
$w=\vpi:=\frac{\beta-1}{\beta+1}$, so that $\frac12+\frac w{1-w}=\frac\beta2$ and $1-\vpi=\frac2{\beta+1}$; the map
$\beta\mapsto\vpi$ takes $(0,\infty)$ onto $(-1,1)$ and $1/\beta$ to $-\vpi$. Then
\begin{equation}\label{eq:8-genfn}
  \eb=(1-\vpi)\sum_{k\ge0}\vpi^k\,\psi_k .
\end{equation}
The Laguerre functions are orthonormal~\cite[(5.1.1)]{Szego}, $\int_0^\infty\psi_j\psi_k\,\dd u=\delta_{jk}$, so the
coefficients $g_k:=\int_0^\infty g\,\psi_k\,\dd u$ of $g\in L^2(\dd u)$ obey Bessel's inequality
$\sum_k\abs{g_k}^2\le\norm g_{L^2(\dd u)}^2$. Cauchy's estimate for the coefficient of $w^k$
in the generating function on the circle $\abs w=\theta$, where $\abs{1-w}\ge1-\theta$ and
$\Re\frac w{1-w}\ge-\frac\theta{1+\theta}$, gives after multiplication by $e^{-u/2}$ the
\emph{Cauchy bound}
\begin{equation}\label{eq:8-cauchy}
  \abs{\psi_k(u)}\le\frac{\theta^{-k}}{1-\theta}\,e^{-\epsilon_\theta u},\qquad
  \epsilon_\theta:=\frac{1-\theta}{2(1+\theta)}>0,\qquad 0<\theta<1,\ k\ge0,\ u\ge0 .
\end{equation}
The Fourier parity $\hat\psi_k=(-1)^k\psi_k$ of \eqref{eq:2-psik} follows from \eqref{eq:8-genfn}. By
\eqref{eq:2-fourier} the transform of $\eb$ is $\beta^{-1}e_{1/\beta}$, and $\beta^{-1}(1+\vpi)=1-\vpi$. The series
\eqref{eq:8-genfn} converges in $L^1(\RR^2)$ by \eqref{eq:8-cauchy} and may be transformed term by term:
$\sum_k\vpi^k\hat\psi_k=(1-\vpi)^{-1}\beta^{-1}e_{1/\beta}=\sum_k(-\vpi)^k\psi_k$, the last step being
\eqref{eq:8-genfn} at $1/\beta$; now compare coefficients. The same computation expands the crossing kernel:
\begin{equation}\label{eq:8-expansion}
  \Xib=\eb-\beta^{-1}e_{1/\beta}=(1-\vpi)\sum_{k\ge0}\bigl[1-(-1)^k\bigr]\vpi^k\psi_k
  =\frac4{\beta+1}\sum_{k\ \mathrm{odd}}\vpi^k\,\psi_k .
\end{equation}
In particular an odd $\psi_k$ is the coefficient of $\vpi^k$ in $\frac{\beta+1}4\Xib$, a combination of the derivatives
$\partial_\beta^j\Xib\big|_{\beta=1}$ with $1\le j\le k$: it is a derivative functional \eqref{eq:2-deriv}.

\begin{proof}[Proof of \hyp{Z4}]
Let $\Upsilon_\vpi:=\sum_{k\ \mathrm{odd}}\vpi^k\psi_k$. Multiplying \hyp{Z2} by $\frac{\beta+1}4$ and using
\eqref{eq:8-expansion} gives $\sum_nd_n\Upsilon_\vpi(t_n)=\int_0^\infty\rhoC\Upsilon_\vpi\,\dd t$ for every
$\vpi\in(-1,1)$. Fix $\theta\in(0,1)$. By \eqref{eq:8-cauchy},
$\sum_k\abs\vpi^k\abs{\psi_k(u)}\le(1-\theta)^{-1}(1-\abs\vpi/\theta)^{-1}e^{-\epsilon_\theta u}$ for $\abs\vpi<\theta$,
and $e^{-\epsilon_\theta u}=e^{-4\pi\epsilon_\theta t}$ is $\abs\mfrak$-integrable by \hyp{Z1}. Hence for $\abs\vpi<\theta$
both sides converge absolutely as double series and can be rearranged into power series in $\vpi$, whose coefficients
$\sum_nd_n\psi_k(t_n)$ and $\int_0^\infty\rhoC\,\psi_k\,\dd t$, with $k$ odd, converge absolutely. Two power series that
agree on an interval have equal coefficients.
\end{proof}

\paragraph{The Abel mean.} The Hille--Hardy formula~\cite[Theorem~5.1]{Szego},~\cite[(18.18.27)]{DLMF} sums products
of Laguerre functions against a geometric weight: for $0<r<1$ and $u,u'\ge0$,
\begin{equation}\label{eq:8-kernel}
  \sum_{k\ge0}r^k\,\psi_k(u)\,\psi_k(u')=\mathcal K_r(u,u')
  :=\frac1{1-r}\exp\Bigl(-\frac{1+r}{2(1-r)}\,(u+u')\Bigr)\,I_0\Bigl(\frac{2\sqrt{r\,uu'}}{1-r}\Bigr),
\end{equation}
the series converging absolutely by \eqref{eq:8-cauchy} with $\theta=r^{1/4}$. The \emph{Abel kernel} $\mathcal K_r$ is
positive and symmetric. By the Laplace formula \eqref{eq:2-Kalpha} its total mass is
$m_r(u):=\int_0^\infty\mathcal K_r(u,u')\,\dd u'=\frac2{1+r}\exp\bigl(-\frac{1-r}{2(1+r)}u\bigr)$; thus $0<m_r<2$, and
$m_r(u)\to1$ as $r\uparrow1$. For measurable $f$ with $\int_0^\infty\mathcal K_r(u,u')\abs{f(u')}\,\dd u'<\infty$ the
\emph{Abel mean} is $\Ab_rf(u):=\int_0^\infty\mathcal K_r(u,u')f(u')\,\dd u'$. It is linear and positive, so
$\abs{\Ab_rf}\le\Ab_r\abs f$. For $g\in L^1(\dd u)\cap L^2(\dd u)$ it is the Abel-summed Laguerre series, and it decays
exponentially in $u$, whatever the decay of $g$:
\begin{equation}\label{eq:8-abel}
  \Ab_rg=\sum_{k\ge0}r^k\,g_k\,\psi_k ,\qquad
  \abs{\Ab_rg(u)}\le\frac{\norm g_{L^2(\dd u)}}{(1-\sqrt r)\sqrt{1-r}}\;e^{-\epsilon_{\sqrt r}\,u} .
\end{equation}
For the first statement insert $g_k=\int g\,\psi_k\,\dd u'$ and sum under the integral by \eqref{eq:8-kernel}; this is
permitted because \eqref{eq:8-cauchy} with $\theta=r^{1/4}$ bounds $r^k\abs{\psi_k(u)\psi_k(u')}$ by
$(1-r^{1/4})^{-2}r^{k/2}$, and $g\in L^1(\dd u)$. For the second, bound $\abs{\Ab_rg(u)}$ by
$\sum_kr^k\abs{g_k}\abs{\psi_k(u)}$ and use \eqref{eq:8-cauchy} with $\theta=\sqrt r$, then the Cauchy--Schwarz and Bessel
inequalities. Finally, by \eqref{eq:8-expansion} and orthonormality the coefficients of
$\Xib$ are $\frac4{\beta+1}\vpi^k$ for odd $k$ and zero for even $k$, so
$\Ab_r\Xib=\frac4{\beta+1}\sum_{k\ \mathrm{odd}}(r\vpi)^k\psi_k$, which is \eqref{eq:8-expansion} at the modulus $\beta_r$
where $\vpi$ takes the value $r\,\vpi(\beta)$:
\begin{equation}\label{eq:8-contraction}
  \Ab_r\,\Xib=\frac{1+\beta_r}{1+\beta}\;\Xi_{\beta_r},\qquad \beta_r:=\frac{1+r\vpi}{1-r\vpi},\qquad
  \vpi(\beta_r)=r\,\vpi(\beta),\qquad \beta_r\in\Bigl[\frac{1-r}{1+r},\frac{1+r}{1-r}\Bigr].
\end{equation}

\paragraph{The kernel on weighted spaces.} The proof needs $\Ab_r$ to be bounded uniformly in $r$, and to tend to the
identity, on functions that grow or decay like the Cardy factor. These are (K1) and (K3) of the next lemma; (K2) serves
only to prove (K3). The proof is a Gaussian estimate and can be skipped at first reading.

\begin{lemma}[Abel kernel on Cardy weights; \proved]\label{lem:8.1}
Fix $q,\ell\in\RR$. There is a constant $C_K=C_K(\abs q,\abs\ell)$ such that for all $r\in[\frac14,1)$ and all $u\ge0$:
\begin{enumerate}[label=\textup{(K\arabic*)},leftmargin=3.2em,itemsep=1pt]
\item $\Ab_rW_{q,\ell}(u)\le C_K\,W_{q,\ell}(u)$;
\item for every $\epsilon>0$,
  \begin{equation}\label{eq:8-K2}
    \int_{\abs{\sigma'-\sigma}\ge\epsilon}\mathcal K_r(u,u')\,W_{q,\ell}(u')\,\dd u'
    \le C_K\Bigl[W_{q,\ell}(u)\,e^{-\frac{\epsilon^2}{4(1-r)}}+\frac{W_{\abs q+\frac12,\abs\ell}(u)}{1-r}\,e^{-\frac{\max(u/4,\,1)}{4(1-r)}}\Bigr],
  \end{equation}
  which tends to $0$ as $r\uparrow1$ at fixed $u$;
\item if $f$ is continuous on $[0,\infty)$ and $\abs f\le C_f\,W_{q,\ell}$, then $\Ab_rf(u)\to f(u)$ as $r\uparrow1$, for
  every $u\ge0$.
\end{enumerate}
\end{lemma}

\begin{proof}
We bound the kernel by a Gaussian in $\sigma'-\sigma$ of width of order $(1-r)^{1/2}$ and split the integral into a
near part, where the weight is comparable to its value at $u$, and a far part, where the Gaussian wins. Let
$\gamma_r:=2\sqrt r/(1-r)$, so that the argument of $I_0$ in \eqref{eq:8-kernel} is $\gamma_r\sigma\sigma'$. Since
$\frac{1+r}{2(1-r)}=\frac{\gamma_r}2+\frac{(1-\sqrt r)^2}{2(1-r)}\ge\frac{\gamma_r}2$ and
$I_0(z)\le1.18\,e^z/\sqrt{1+z}$ (\cref{lem:A.4}(c)),
\begin{equation}\label{eq:8-kernel-bound}
  \mathcal K_r(u,u')\le\frac{1.18}{1-r}\,\bigl(1+\gamma_r\sigma\sigma'\bigr)^{-1/2}\,e^{-\frac{\gamma_r}2(\sigma-\sigma')^2}.
\end{equation}
For $r\ge\frac14$ we have $\gamma_r(1-r)=2\sqrt r\in[1,2]$, hence $\gamma_r\ge(1-r)^{-1}\ge1$. Let
$s:=\sigma'-\sigma$, so that $\dd u'=2\sigma'\,\dd s$, and $\sigma_*:=\max(\sigma/2,1)$; the near set is
$\{\abs s\le\sigma_*\}$, the far set its complement. In this proof $C$ denotes constants that depend only on $\abs q$
and $\abs\ell$.

\emph{Near set.} If $\sigma\ge2$ then $\sigma'\in[\sigma/2,3\sigma/2]$ and $(1+u')/(1+u)\in[\frac14,\frac94]$; if
$\sigma\le2$ then $\sigma'\in[0,3]$ and $(1+u')/(1+u)\in[\frac15,10]$. Thus $(1+u')^q\le10^{\abs q}(1+u)^q$, while
$e^{\ell\sigma'}\le e^{\ell\sigma}e^{\abs\ell\abs s}$, and by \eqref{eq:8-kernel-bound}
\begin{equation}\label{eq:8-near}
  \int_{\rm near}\mathcal K_r\,W_{q,\ell}\,\dd u'\le\frac{C\,W_{q,\ell}(u)}{1-r}
  \int_{\rm near}2\sigma'\bigl(1+\gamma_r\sigma\sigma'\bigr)^{-1/2}e^{-\frac{\gamma_r}2s^2+\abs\ell\abs s}\,\dd s .
\end{equation}
If $\gamma_r\sigma^2\ge1$, then
$2\sigma'(1+\gamma_r\sigma\sigma')^{-1/2}\le2\sqrt{\sigma'/(\gamma_r\sigma)}\le2\gamma_r^{-1/2}\bigl(1+\sqrt{\abs s/\sigma}\bigr)
\le2\gamma_r^{-1/2}\bigl(1+\gamma_r^{1/4}\abs s^{1/2}\bigr)$, and the substitution $s=z/\sqrt{\gamma_r}$ bounds the last
integral by $2\gamma_r^{-1}\int_\RR(1+\abs z^{1/2})\,e^{-z^2/2+\abs\ell\abs z}\,\dd z$. If $\gamma_r\sigma^2<1$, then
$(1+\gamma_r\sigma\sigma')^{-1/2}\le1$ and $2\sigma'\le2(\gamma_r^{-1/2}+\abs s)$, and the same substitution gives
$2\gamma_r^{-1}\int_\RR(1+\abs z)\,e^{-z^2/2+\abs\ell\abs z}\,\dd z$. Either way the right side of \eqref{eq:8-near} is at
most $C\,W_{q,\ell}(u)/\bigl(\gamma_r(1-r)\bigr)\le C\,W_{q,\ell}(u)$.

\emph{Far set.} There $\frac{\gamma_r}2s^2\ge\frac{\gamma_r}4s^2+\frac{\gamma_r}4\sigma_*^2$, and
$1+\sigma'^2\le2(1+\sigma^2)(1+s^2)$ gives
$2\sigma'\,W_{q,\ell}(u')\le C\,W_{\abs q+\frac12,\abs\ell}(u)\,(1+s^2)^{\abs q+\frac12}e^{\abs\ell\abs s}$. Since
$\gamma_r\ge1$, the integral of the last two factors against $e^{-\gamma_rs^2/4}$ is bounded uniformly in $r$; with
$(1+\gamma_r\sigma\sigma')^{-1/2}\le1$ in \eqref{eq:8-kernel-bound},
\begin{equation}\label{eq:8-far}
  \int_{\rm far}\mathcal K_r\,W_{q,\ell}\,\dd u'\le C\,W_{\abs q+\frac12,\abs\ell}(u)\;\frac{e^{-\gamma_r\sigma_*^2/4}}{1-r}.
\end{equation}
Now $\gamma_r\sigma_*^2/4\ge\lambda/(1-r)$ with $\lambda:=\max(\sigma^2,4)/16\ge\frac14$, and
$(1-r)^{-1}e^{-\lambda/(1-r)}=e^{-\lambda}\,(1-r)^{-1}e^{-\lambda r/(1-r)}\le e^{-\lambda}\sup_{x\ge0}(1+x)e^{-x/4}$. Since
$W_{\abs q+\frac12,\abs\ell}(u)\,e^{-\lambda}\le C\,W_{q,\ell}(u)$, a Gaussian in $\sigma$ beating every weight, the far
part is at most $C\,W_{q,\ell}(u)$ as well. This proves (K1).

\emph{\textup{(K2)}.} On the part of the near set with $\abs s\ge\epsilon$ we write
$e^{-\frac{\gamma_r}2s^2}\le e^{-\gamma_r\epsilon^2/4}\,e^{-\frac{\gamma_r}4s^2}$ and repeat the near estimate with
$\gamma_r/4$ in place of $\gamma_r/2$; this changes only the constants and produces the factor
$e^{-\gamma_r\epsilon^2/4}\le e^{-\epsilon^2/4(1-r)}$. On the far set, \eqref{eq:8-far} already carries
$(1-r)^{-1}e^{-\gamma_r\sigma_*^2/4}\le(1-r)^{-1}e^{-\sigma_*^2/4(1-r)}$ with $\sigma_*^2=\max(u/4,1)$.

\emph{\textup{(K3)}.} Fix $u$ and $\epsilon>0$, and let
$\mathrm{osc}_\epsilon:=\sup\{\abs{f(u')-f(u)}:\abs{\sigma'-\sigma}\le\epsilon\}$. Then
$\abs{\Ab_rf(u)-m_r(u)f(u)}\le\mathrm{osc}_\epsilon\,m_r(u)
+\int_{\abs{\sigma'-\sigma}\ge\epsilon}\mathcal K_r(u,u')\bigl(C_f\,W_{q,\ell}(u')+\abs{f(u)}\bigr)\,\dd u'$.
As $r\uparrow1$ the integral tends to $0$ by (K2), applied to $W_{q,\ell}$ and to $W_{0,0}$, and $m_r(u)\to1$; hence
$\limsup_{r\uparrow1}\abs{\Ab_rf(u)-f(u)}\le\mathrm{osc}_\epsilon$, which tends to $0$ with $\epsilon$ because $f$ is
continuous at $u$.
\end{proof}

\subsection{Proof of \texorpdfstring{\cref{thm:2}}{Theorem 2}}
\label{sec:8.3}

\begin{proof}[Proof of \cref{thm:2}]
Let $c\ge40$, let $g\in\Abar$ be feasible at $T>0$, and suppose that an admissible spectrum has $t_n\ge T$ for all $n$;
it has no light primaries, so \hyp{Z1}--\hyp{Z4} and \hyp{St} apply. We derive a contradiction in four steps: (i)~the
sum rule for the Abel means $\Ab_rg$; (ii)~the limit $r\uparrow1$ on the continuum side, by dominated convergence;
(iii)~the vanishing of the leakage of the lobe; (iv)~Fatou's lemma on the primaries.

\emph{Step \textup{(i)}: the sum rule for Abel means.} By \cref{def:2-Abar}, $\int_0^\infty\abs g\,\dd t<\infty$, and
$\norm g_\infty\le2\pi\int_0^\infty\abs g\,\dd t$ by $g=-\hat g$ and $\abs{J_0}\le1$ in \eqref{eq:2-eigen}; so
$g\in L^1(\dd u)\cap L^2(\dd u)$. Its Laguerre coefficients vanish for even $k$: by $\hat\psi_k=(-1)^k\psi_k$, by
$\int_{\RR^2}\hat g\,\psi_k=\int_{\RR^2}g\,\hat\psi_k$ (Fubini's theorem) and by $\hat g=-g$, we have
$g_k=(-1)^k\int_0^\infty\hat g\,\psi_k\,\dd u=-(-1)^kg_k$. Let $0<r<1$. The function $\Ab_rg$ obeys \eqref{eq:8-abel}, so
$\langle\mfrak,\Ab_rg\rangle$ converges absolutely by \hyp{Z1}. Moreover, by \eqref{eq:8-cauchy} with $\theta=\sqrt r$,
$\sum_kr^k\abs{g_k}\int\abs{\psi_k}\,\dd\abs\mfrak\le(1-\sqrt r)^{-1}\bigl(\sum_kr^{k/2}\abs{g_k}\bigr)\int e^{-\epsilon_{\sqrt r}u}\,\dd\abs\mfrak<\infty$,
so the series in \eqref{eq:8-abel} may be paired with $\mfrak$ term by term:
$\langle\mfrak,\Ab_rg\rangle=\sum_{k\ \mathrm{odd}}r^kg_k\,\langle\mfrak,\psi_k\rangle=0$ by \hyp{Z4}. Thus, with both
sides absolutely convergent,
\begin{equation}\label{eq:8-abel-sumrule}
  \sum_nd_n\,\Ab_rg(t_n)=\int_0^\infty\rhoC\,\Ab_rg\,\dd t\qquad\text{for every }r\in(0,1).
\end{equation}

\emph{Step \textup{(ii)}: the continuum side.} By the symmetry of $\mathcal K_r$, Tonelli's theorem,
\eqref{eq:8-weights} and (K1),
$\int_0^\infty\!\int_0^\infty\rhoC(u)\,\mathcal K_r(u,u')\abs{g(u')}\,\dd u\,\dd u'=\int_0^\infty\abs g\,\Ab_r\rhoC\,\dd u'
\le C_\rho C_K\int_0^\infty\abs g\,W_{-1/4,\ell_a}\,\dd u'$.
The right side is finite. Indeed, by \cref{lem:A.2}(c) and~(e), for $t\ge\xis c'$ we have
$V\ge0.9\,K_a\ge0.9\,e^{-2}e^{\ell_a\sigma}/\sqrt{\ell_a\sigma}\ge c_*W_{-1/4,\ell_a}$ with a constant $c_*=c_*(a)>0$, so
that $\int_{t\ge\xis c'}\abs g\,W_{-1/4,\ell_a}\,\dd t\le c_*^{-1}M[g]<\infty$, while on $[0,\xis c']$ the weight is
bounded and $g$ is integrable. Hence Fubini's theorem gives $\int\rhoC\,\Ab_rg\,\dd t=\int g\,\Ab_r\rhoC\,\dd t$. As
$r\uparrow1$, $\Ab_r\rhoC\to\rhoC$ pointwise by (K3), since $\rhoC$ is continuous and bounded by $C_\rho W_{-1/4,\ell_a}$,
and $\abs{g\,\Ab_r\rhoC}\le C_\rho C_K\abs g\,W_{-1/4,\ell_a}$, an integrable function that does not depend on
$r\in[\frac14,1)$. By dominated convergence,
$\int_0^\infty\rhoC\,\Ab_rg\,\dd t\to\int_0^\infty\rhoC\,g\,\dd t=-\gvac$ as $r\uparrow1$.

\emph{Step \textup{(iii)}: the leakage of the lobe.} The lobe lives on $[0,T]$, the primaries on $[T,\infty)$, and
what the kernel carries from the one to the others vanishes as $r\uparrow1$: below $T+1$ by (K3), and beyond $T+1$
because across the gap the Gaussian tail of the kernel beats the Cardy growth of $\NZ$.
Let $g_\pm:=\max(\pm g,0)$, so that $g_+=g+g_-$. Both are
continuous with $0\le g_\pm\le\norm g_\infty$, and $\supp g_-\subset[0,T]$ because $g\ge0$ on $[T,\infty)$. Let
$u_T:=4\pi T$, $\sigma_T:=\sqrt{u_T}$, $\epsilon:=\sqrt{4\pi(T+1)}-\sigma_T>0$ and $C_g:=1.18\,u_T\norm g_\infty$. For
$t\ge T$ and $u'\le u_T$ we have $(\sigma-\sigma')^2\ge(\sigma-\sigma_T)^2$, so \eqref{eq:8-kernel-bound} gives
$\Ab_rg_-(t)\le\norm g_\infty\int_0^{u_T}\mathcal K_r(u,u')\,\dd u'
\le\phi_r(t):=C_g(1-r)^{-1}e^{-\frac{\gamma_r}2(\sigma-\sigma_T)^2}$,
and $\phi_r$ is $C^1$ and non-increasing on $[T,\infty)$. The primaries in $[T,T+1)$ carry total degeneracy at most
$\NZ(T+1)<\infty$; at each of them $0\le\Ab_rg_-(t_n)\le\norm g_\infty m_r\le2\norm g_\infty$ and
$\Ab_rg_-(t_n)\to g_-(t_n)=0$ by (K3) with $q=\ell=0$, so $\sum_{t_n<T+1}d_n\,\Ab_rg_-(t_n)\to0$ by dominated
convergence. For those in $[T+1,\infty)$ we use \hyp{St} and \hyp{Z3}:
$\sum_{t_n\ge T+1}d_n\,\Ab_rg_-(t_n)\le\limsup_R\phi_r(R)\NZ(R)+\int_{T+1}^\infty\NZ\,\abs{\phi_r'}\,\dd t$. The first term
vanishes: $\phi_r$ is a Gaussian in $\sigma$ and $\NZ$ at most exponential in $\sigma$. In the second,
$\abs{\phi_r'}\,\dd t=C_g(1-r)^{-1}\gamma_r\,(\sigma-\sigma_T)\,e^{-\frac{\gamma_r}2(\sigma-\sigma_T)^2}\dd\sigma$. With
$s:=\sigma-\sigma_T\ge\epsilon$ and $e^{-\frac{\gamma_r}2s^2}\le e^{-\gamma_r\epsilon^2/4}e^{-s^2/4}$, which holds because
$\gamma_r\ge1$ for $r\ge\frac14$,
\begin{equation}\label{eq:8-leakage}
  \int_{T+1}^\infty\NZ\,\abs{\phi_r'}\,\dd t\le C'\,\frac{\gamma_r}{1-r}\,e^{-\gamma_r\epsilon^2/4},\qquad
  C':=C_gC_N\int_0^\infty(1+\sigma_T+s)\,s\,e^{\ell_a(\sigma_T+s)-s^2/4}\,\dd s ,
\end{equation}
and $C'$ does not depend on $r$. Since $(1-r)^{-1}\le\gamma_r\to\infty$, this tends to $0$ as $r\uparrow1$. Hence
$\sum_nd_n\,\Ab_rg_-(t_n)\to0$ as $r\uparrow1$, and the sum is finite for each $r\in[\frac14,1)$.

\emph{Step \textup{(iv)}: Fatou's lemma on the primaries.} Since $\mathcal K_r>0$, the function
$\Ab_rg_+=\Ab_rg+\Ab_rg_-$ is non-negative, and at every primary $\Ab_rg_+(t_n)\to g_+(t_n)=g(t_n)$ by (K3). By
\eqref{eq:8-abel-sumrule} and steps (ii) and (iii),
$\sum_nd_n\,\Ab_rg_+(t_n)=\int_0^\infty\rhoC\,\Ab_rg\,\dd t+\sum_nd_n\,\Ab_rg_-(t_n)\to-\gvac$. Fatou's
lemma~\cite[1.28]{Rudin} for the positive measure $\sum_nd_n\delta_{t_n}$, along any sequence $r_j\uparrow1$, now gives
\begin{equation}\label{eq:8-fatou}
  0\le\sum_nd_n\,g(t_n)=\sum_nd_n\lim_{j\to\infty}\Ab_{r_j}g_+(t_n)\le\liminf_{j\to\infty}\sum_nd_n\,\Ab_{r_j}g_+(t_n)=-\gvac<0 ,
\end{equation}
a contradiction. Hence some primary has $t_n<T$. For the last assertion of \cref{thm:2}, given $\varepsilon>0$ choose
$g\in\Abar$ feasible at some $T\le\TV(c)+\varepsilon$; then $T>0$ by \cref{sec:2.4}, so $\Delta_1<T+a\le\TV(c)+a+\varepsilon$
for every admissible CFT, and $\varepsilon$ is arbitrary.
\end{proof}

\noindent\tagpar{\orientation} If $g\in\Abar$ obeys
$\abs{g(t)}\le C(1+t)^{-q}e^{-4\pi\sqrt{at}}$ for some $q>1$, the same steps give a two-sided statement, whether or not
$g$ is feasible: on every admissible spectrum without light primaries, $\sum_nd_n\abs{g(t_n)}<\infty$ and
$\gvac+\sum_nd_n\,g(t_n)=0$. Indeed \hyp{St} and \hyp{Z3} then make $W_{-q,-\ell_a}$ integrable against $\abs\mfrak$,
and (K1), (K3) allow dominated convergence on both sides of $\langle\mfrak,\Ab_rg\rangle=0$. The function of
\cref{thm:5} decays in this way, for every $q$; \cref{thm:2} does not need it.

\subsection{Consequences}
\label{sec:8.4}

By \cref{thm:2}, $\Delta_1\le\TV(c)+a$ for every admissible CFT with $c\ge40$: the quantity that \cref{thm:4} bounds
from below is a bound on theories, not only the value of a linear program. The two halves of \cref{thm:1} still say
different things: attainment, with \cref{thm:2}, is a statement about the spectra of actual theories, while optimality
is a statement about the reach of the method---no functional in $\Abar$ does better---and says nothing about the
existence of theories with a gap near $\Dbest$.

\begin{maincorollary}[\provedCA]\label{cor:6}
For $c\ge\catt$ the function $g=\half(h_0-\hat h_0)$ of \cref{thm:5} is feasible at $\tstar+1$. Hence every admissible
CFT with central charge $c\ge\catt$ has
\begin{equation}\label{eq:8-cor6}
  \Delta_1\;<\;a+\tstar+1\;=\;\frac c{\LV}+A_1\,c^{1/3}+O(1),\qquad A_1=0.483 .
\end{equation}
\end{maincorollary}
\begin{proof}
\Cref{thm:5} states that $g\in\Abar$ and that $g$ is feasible at $\tstar+1$; \cref{thm:2}, which applies because
$\catt\ge40$, then gives a primary with $t_n<\tstar+1$. By \cref{thm:5},
$\tstar=\xis c'+A_1c'^{1/3}\bigl(1+O(c'^{-2/3})\bigr)$, and $a+\xis c'=c'/\LV$ by the definition of $\LV$ in \cref{thm:1};
replacing $c'=c-1$ by $c$ changes the right side by $O(1)$.
\end{proof}

The tag is inherited from \cref{thm:5}; the step from the functional to theories is by hand. The coefficient $0.483$
belongs to one explicit construction and is not claimed to be the best; \cref{conj:1} predicts $\Aopt=0.27666$ for
$\Dbest$. Nothing is said about the size of $\catt$ or about finite $c$: at every central charge accessible to numerical
work \eqref{eq:8-cor6} is weaker than the numerical bounds (\cref{sec:7.10}). What \cref{cor:6} adds is that the slope
$\LV$ is reached by a bound on actual theories.

\begin{remark*}[the case $\gvac=0$; \proved]
Only the last step of \eqref{eq:8-fatou} used $\gvac>0$. If $g\in\Abar$ is weakly feasible
at $T>0$ with $\gvac=0$, steps (i)--(iv) give $\sum_nd_n\,g(t_n)\le0$ for every admissible spectrum with all
$t_n\ge T$; since each term is non-negative, such a spectrum is supported on the zeros of $g$ in $[T,\infty)$. This is
the extremality statement of \cref{sec:2.2}, and by itself it bounds nothing.
\end{remark*}

\providecommand{\gd}{\operatorname{gd}}
\providecommand{\Rot}{\mathcal R}
\providecommand{\sfb}{\mathsf b}
\providecommand{\gb}{\bar g}
\providecommand{\xg}{x_\gamma}
\providecommand{\xt}{x_{\rm t}}
\providecommand{\ResSone}{\hyperref[res:S1]{Result~S1}}
\providecommand{\ResStwo}{\hyperref[res:S2]{Result~S2}}
\section{A second route to attainment: one transform of one real function}
\label{sec:alt}

\Cref{thm:5} attains the slope with a functional $g=\half(h_0-\hat h_0)$, where $h_0$ is a product whose zeros sit at the
quantiles of the zero law $\rhoU$. It is positive on the band because $\hat h_0$ is smaller there than $h_0$ itself;
this, with a similar inequality for the vacuum value, is where a computer certifies finitely many inequalities. That
proof gives a rate, $\TV(c)\le\xis c'+0.483\,c'^{1/3}+O(1)$, and it shows how the extremal potential $U$ is a blueprint
for a functional.

This section describes a second, quite different way to build a functional that certifies $\Delta_1\le c/\LV+o(c)$. It
gives less and costs much less. Its building block is minus the imaginary part of a rotation of phase space by
$45^\circ$ applied to one real, positive function: $\abs{\br'}^2$ times a bell that is a product of two double sine
functions. The functional itself is the same transform of one real function, a combination of a few such blocks and of
Gaussians. Its construction needs no positivity on an oscillating band and no computer. It gives the slope $\LV$ but no
rate, and it is far from the optimal functional. Nothing outside this section depends on it.

\begin{ideabox}[title={Result S1}]\phantomsection\label{res:S1}
For every $x_g>\xs$ there is $c_0(x_g)\ge40$ such that for every $c\ge c_0(x_g)$ there is a functional $g\in\Abar$ that is
feasible at $T=c'x_g^2/3$. By \cref{thm:2} every admissible CFT with $c\ge c_0(x_g)$ then has
$\Delta_1<a+c'x_g^2/3$. Hence $\limsup_{c\to\infty}\TV(c)/c'\le\xis$, and with \cref{thm:4},
$\lim_{c\to\infty}c/\Dbest(c)=\LV$. No computer is used.
\end{ideabox}

\noindent What is not obtained: a rate (the statement is $o(c)$ only), a value of $c_0(x_g)$, and any information about the
optimal functional. The strategy has three steps.

\begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.2em]
\item \emph{Every real function gives a functional} (\cref{sec:alt.1}). The Fourier transform is a quarter-turn of phase
space (\cref{sec:2.3}). Let $\Rot$ be the eighth-turn, its square root. For every real radial function $\phi$ the imaginary
part of $\Rot[\phi]$ is mapped to minus itself by the Fourier transform. So there is no constraint left to solve, and all the
content is in the choice of~$\phi$.
\item \emph{A seed whose image is a bump} (\cref{sec:alt.2,sec:alt.3}). We choose for $\phi$ the factor $\abs{\br'}^2$ times a
positive bell with a Gaussian core and exponential tails. At large $c$ its image $g_\gamma$ is a \emph{bump}: it is negative on
a window that contains its peak, its Cardy-weighted size
$\abs{g_\gamma}V$ is concentrated at exponential order around one point $\xg$, its vacuum value is positive and of
the same exponential order as the peak, and beyond the peak it decays faster than the Cardy weight grows. A parameter
$\gamma\in(0,\pi)$ moves the peak over the whole interval $\xs<\xg<\half$ between the optimal gap bound and Hellerman's
point. In terms of $t$ the bump at central charge $c$ is the \emph{critical} member $\gamma=\pi$ of the family at the larger
central charge $c'_\gamma=(\pi/\gamma)^2c'$; read against the weaker Cardy growth of the lighter theory, it acquires
a margin.
\item \emph{From bumps to a feasible functional} (\cref{sec:alt.4}). One bump with its peak just above $\xs$ already has the
right vacuum value and a negative lobe in the right place. Beyond $x_g$ it is exponentially small compared with the
inverse Cardy weight, but its sign is not known. We add, with an exponentially small coefficient, the negatives of bumps
whose peaks cover $[x_g,\half]$: they form a positive plateau that lies just below the Cardy line and dominates
everything of unknown sign. Beyond $x=\half$ the crossing kernels $\Xib$ with $\beta<1$ do the same job.
\end{enumerate}

\paragraph{Level of rigor.} The aim is to explain the strategy and the functional. Exact statements are marked as
such. The large-$c$ statements are leading-order saddle-point estimates; we state what they give and why, without
uniform bounds on the remainders. For this reason \ResSone\ and \ResStwo\ do not carry the tag \tagpar{\proved} of
\cref{sec:1.3}, and the status of \cref{thm:1}, \cref{thm:5} and \cref{cor:6} does not rest on this section.

\paragraph{Notation.} It is that of \cref{sec:2}: $t=\abs\br^2/2$ and $b=x+iy=\sqrt{3t/c'}$ \eqref{eq:2-scaling}; a
functional is a real radial function $g$ of $\br\in\RR^2$, written as a function of $t$, with $\hat g=-g$ for the
transform \eqref{eq:2-eigen}; $\psi_k$ are the Laguerre functions \eqref{eq:2-psik}; the Cardy weight $V$ grows like
$e^{c'L(x)}$ \eqref{eq:2-L}, and $\rhoC=2\pi V$; for $g\in\Abar$ the vacuum value is $\gvac=-\int_0^\infty g\,\rhoC\,\dd t$
(\cref{def:2-Abar}), and feasibility is as in \cref{def:2-feasible}; $\Xib$ is the crossing kernel \eqref{eq:2-g}. The
extremal potential $U$ and $U_0=U(0)=-0.388748$ are those of \cref{lem:6.5}, with $U'(x)=-\frac43\arcsin\sinh\pi x$ on
$\abs x<\xs$. The letters $\gamma$, $\sfb$, $Q$, $\cG$, $s$, $w$, $\sigma$, $\eta$, $K$ are local to this section.

\subsection{Every real function gives a functional: the eighth-turn}
\label{sec:alt.1}

For a radial function $\phi$ on $\RR^2$ let
\begin{equation}\label{eq:alt-R}
  \Rot[\phi](\br):=\sqrt2\,e^{i\pi/4}\,e^{-i\pi\abs\br^2}\int_{\RR^2}\phi(\br')\,e^{-i\pi\abs{\br'}^2}\,
  e^{2\pi i\sqrt2\,\br\cdot\br'}\,\dd^2\br' ,
\end{equation}
a chirp, a Fourier transform dilated by $\sqrt2$, and a chirp. In the variables $t=\abs\br^2/2$, $t'=\abs{\br'}^2/2$,
\begin{equation}\label{eq:alt-Rrad}
  \Rot[\phi](t)=2\sqrt2\,\pi\,e^{i\pi/4}\int_0^\infty\phi(t')\,e^{-2\pi i(t+t')}\,J_0\bigl(4\pi\sqrt{2tt'}\bigr)\,\dd t' .
\end{equation}
Three exact properties are all that we need.

\smallskip\noindent\emph{Gaussians.} For $\Re\tau>0$, a Fresnel integral gives
\begin{equation}\label{eq:alt-gauss}
  \Rot\bigl[e^{-2\pi\tau t'}\bigr](t)=\frac{\sqrt2\,e^{i\pi/4}}{\tau+i}\;e^{-2\pi\beta(\tau)\,t},\qquad
  \beta(\tau):=\frac{1+i\tau}{\tau+i}=\frac{2\tau+i(\tau^2-1)}{1+\tau^2} .
\end{equation}
For real $\tau>0$ the image is a Gaussian at a complex inverse temperature on the unit circle, $\abs{\beta}=1$,
$\Re\beta>0$: it is a Gaussian times a chirp. As $\tau$ runs from $0$ to $\infty$, $\beta(\tau)$ runs from $-i$ through $1$ to $+i$.

\smallskip\noindent\emph{Laguerre functions.} Inserting $\tau=(1+z)/(1-z)$ in \eqref{eq:alt-gauss} and using the generating function
$\sum_k\psi_k(t)z^k=(1-z)^{-1}e^{-2\pi t(1+z)/(1-z)}$, one finds that $\Rot$ maps the generating function at $z$ to the one at $iz$. Hence
\begin{equation}\label{eq:alt-lag}
  \Rot\,\psi_k=i^k\,\psi_k ,\qquad\text{so that}\qquad \Rot^2\phi=\hat\phi
\end{equation}
on radial functions. In the language of \cref{sec:2.3}, $\psi_k$ is a state of the planar oscillator with
$2k$ quanta, the Fourier transform is the evolution through a quarter period, and $\Rot$ is the evolution through one
eighth of a period (in the opposite sense; on radial functions either sense squares to the Fourier transform). On WKB data the Fourier
transform is the quarter-turn $\Rq$ of phase space \eqref{eq:4-R}, and $\Rot$ is a rotation by $45^\circ$.

\smallskip\noindent\emph{Real seeds.} Let $\phi=\sum_kc_k\psi_k$ with real $c_k$. By \eqref{eq:alt-lag} and \eqref{eq:2-psik},
$\widehat{\Rot[\phi]}=\sum_kc_k(-1)^ki^k\psi_k=\overline{\Rot[\phi]}$. Therefore
\begin{equation}\label{eq:alt-Omega}
  \Omega[\phi]:=\Im\Rot[\phi]=\sum_{j\ge0}(-1)^j\,c_{2j+1}\,\psi_{2j+1}\qquad\text{is real, radial, and}\qquad
  \widehat{\Omega[\phi]}=-\Omega[\phi] .
\end{equation}
\emph{Every real radial function $\phi$ gives a functional $\Omega[\phi]$.} In words, $\Omega$ discards the even Laguerre
components of the seed, which the Fourier transform leaves alone, and keeps the odd ones with alternating signs. Only
the part $\half(\phi-\hat\phi)$ of the seed matters, and a seed with $\hat\phi=\phi$, for instance $e^{-2\pi t}$, gives zero.

\paragraph{The sum rule, seen directly.} Apply \eqref{eq:alt-gauss} to the reduced partition function \eqref{eq:2-Zhat} term by term. The
Gaussian seed $e^{-2\pi\tau t'}$ produces $\frac{\sqrt2e^{i\pi/4}}{\tau+i}\Zhat(\beta)$ with $\beta=\beta(\tau)$ on the unit circle,
where $1/\beta=\bar\beta$. There the crossing equation \eqref{eq:2-crossing}, $\Zhat(\beta)=\beta^{-1}\Zhat(1/\beta)$, continued to
$\Re\beta>0$, reads $\Zhat(\beta)=\beta^{-1}\overline{\Zhat(\beta)}$: it says that $\beta^{1/2}\Zhat(\beta)$ is real. Since
$\beta(\tau)=i(\tau-i)/(\tau+i)$, the coefficient in \eqref{eq:alt-gauss} is $\sqrt{2/(1+\tau^2)}\;\beta^{1/2}$, a positive
multiple of the principal square root of $\beta$. So $\Rot$ of a real superposition of Gaussians, applied to $\Zhat$, is real
provided the superposition converges on $\Zhat$, and its imaginary part vanishes: this is the sum rule for $\Omega[\phi]$.
The bell $\cG$ of \cref{sec:alt.2} is a superposition of Gaussians $e^{-2\pi\tau t'}$, $\tau>0$, with positive weights (the
product over its zeros and poles shows that $\cG$ is a completely monotone function of $r^2$), and the factor
$\abs{\br'}^2=2t'$ of the seed is $-\pi^{-1}\partial_\tau$ acting on them. So the seed is a real superposition, which
converges on $\Zhat$ for $\gamma<\pi$, and its functional acts on $\Zhat$ along the unit semicircle in the
$\beta$-plane, which is the set fixed by $S$ followed by complex conjugation. The crossing kernels are of this
form too: for $0<\beta<1$, $\Xib$ is a multiple of $\Omega$ applied to the real part of a complex Gaussian. In the proof of
\ResSone\ we do not use this paragraph; \cref{thm:2} handles the sum rule for every feasible $g\in\Abar$.

\subsection{The seed and the functional}
\label{sec:alt.2}

Fix a \emph{deformation parameter} $\gamma\in(0,\pi)$. Let
\begin{equation}\label{eq:alt-b}
  \sfb:=\frac{\gamma}{2\pi}\sqrt{\frac{3}{c'}} ,\qquad Q:=\sfb+\frac1\sfb ,\qquad
  \cG(r):=S_\sfb\bigl(\tfrac Q4+ir\bigr)\,S_\sfb\bigl(\tfrac Q4-ir\bigr)=\bigabs{S_\sfb\bigl(\tfrac Q4+ir\bigr)}^2 ,
\end{equation}
where $S_\sfb(z)=\Gamma_\sfb(z)/\Gamma_\sfb(Q-z)$ is the double sine function familiar from Liouville
theory~\cite{FKV,PT}; the letter $\sfb$ is its parameter and has nothing to do with the variable $b$ of \eqref{eq:2-scaling}.
All that is needed about it follows from the integral representation
\begin{equation}\label{eq:alt-logG}
  \log\cG(r)=\int_0^\infty\frac{\dd v}{v}\biggl[\frac{\sinh(Qv/2)}{\sinh(\sfb v)\,\sinh(v/\sfb)}\,\cos(2rv)-\frac{Q}{2v}\biggr],
  \qquad\abs{\Im r}<\frac Q4 .
\end{equation}
The seed and the functional are
\begin{equation}\label{eq:alt-seed}
  \boxed{\;\phi_\gamma(\br'):=\abs{\br'}^2\,\frac{\cG(\abs{\br'})}{\cG(0)} ,\qquad
  g_\gamma:=-\,\Omega[\phi_\gamma]=-\Im\Rot[\phi_\gamma] .\;}
\end{equation}
The argument of $\cG$ is the radius in the plane, and $c$ enters only through $\sfb$. By \eqref{eq:alt-Omega},
$\hat g_\gamma=-g_\gamma$ exactly, for every $c$ and every $\gamma$. As $\gamma\to0$ the seed becomes
$\abs{\br'}^2e^{-\pi\abs{\br'}^2}=\frac1{2\pi}(\psi_0-\psi_1)$ and $g_\gamma\to\frac1{2\pi}\psi_1$, the lowest derivative functional
of \cref{sec:2.2}.

\paragraph{The seed at large $c$.} Let $s:=\sqrt{3t'/c'}$ be the rescaled radius of the seed, so that $s$ is to $\br'$ what
$x$ is to $\br$, and let $w:=\sqrt2\,\gamma s=2\pi\sfb\abs{\br'}$. Expanding \eqref{eq:alt-logG} at small $\sfb$,
\begin{equation}\label{eq:alt-S1}
  \log\frac{\cG}{\cG(0)}=c'\,h_\gamma(s)-\tfrac14\log\cosh w+O(1/c'),\qquad
  h_\gamma(s):=-\frac{2\pi}{3\gamma^2}\int_0^{\sqrt2\gamma s}\gd(w')\,\dd w' ,
\end{equation}
uniformly on $\abs{\Im w}\le\frac\pi2-\epsilon$, where $\gd w=\arctan\sinh w=\int_0^w\sech$ is the Gudermannian. Thus
$\cG$ is a positive bell. For $w\ll1$ it is the Gaussian $e^{-2\pi t'}$, because $h_\gamma\simeq-\frac{2\pi}3s^2$. For
$w\gg1$ it has the exponential tail $e^{-c'(\sqrt2\pi^2/3\gamma)\,s}$, because $\gd\to\frac\pi2$. It is analytic in the strip
$\abs{\Im s}<\pi/(2\sqrt2\gamma)$; the first pole of $\cG$, at $r=iQ/4$, lies on the imaginary axis just outside it, at
$\Im s=(1+\sfb^2)\,\pi/(2\sqrt2\gamma)$. Nothing else about $S_\sfb$ is used.
In particular the elementary function $\abs{\br'}^2(\cosh w)^{-1/4}e^{c'h_\gamma(s)}$ could serve as the seed as well,
since exactness comes from \eqref{eq:alt-Omega} and not from the seed; $\cG$ is the completion of this expression to all
orders in $1/c'$ (\cref{sec:alt.5}). The factor $\abs{\br'}^2$ is there to make the imaginary part the larger part of
$\Rot[\phi_\gamma]$.

\subsection{At large \texorpdfstring{$c$}{c} the functional is a bump}
\label{sec:alt.3}

\paragraph{Classical data.} For $0<\gamma\le\pi$ and $0\le x\le\xt$ let
\begin{equation}\label{eq:alt-q}
  q_\gamma(x):=\frac{\arcsin\sinh(\gamma x)}{\gamma} ,\qquad
  \xt:=\frac{\arcsinh1}{\gamma}=\frac\pi\gamma\,\xs ,\qquad
  \xg:=\frac1\gamma\arcsinh\sin\frac\gamma2 .
\end{equation}
The function $q_\gamma$ increases from $0$ to $\pi/2\gamma$ at the \emph{turning point} $\xt$, and
$q_\gamma(\xg)=\half$. The \emph{profile} of the bump is
\begin{equation}\label{eq:alt-U}
  U_\gamma(x):=-m_\gamma-\frac{4\pi}3\int_0^xq_\gamma(x')\,\dd x' ,\qquad
  m_\gamma:=\frac{4\pi}3\int_0^{\xg}\bigl(\tfrac12-q_\gamma\bigr)\,\dd x ,
\end{equation}
where the constant $m_\gamma$ is chosen so that $U_\gamma+L$ has the maximum $0$; the maximum is at $\xg$, because
$U_\gamma'=-\frac{4\pi}3q_\gamma$ equals $-L'=-\frac{2\pi}3$ there. In the language of \cref{sec:4.1},
$p=-iq_\gamma$ is the imaginary momentum \eqref{eq:4-momentum} of a wave function under a barrier, $q_\gamma(x)/x$ is its local inverse
temperature (the $\Im\tau$ of \cref{sec:4.1}), and $(b,p)$ lies on the curve $\sinh\gamma b=i\sinh\gamma p$, which is the spectral
curve \eqref{eq:4-curve} with $\pi$
replaced by $\gamma$. Since $(U_\gamma+L)''=-\frac{4\pi}3\cosh\gamma x\,(1-\sinh^2\gamma x)^{-1/2}\le-\frac{4\pi}3$,
\begin{equation}\label{eq:alt-concave}
  U_\gamma(x)+L(x)\le-\tfrac{2\pi}3\,(x-\xg)^2\qquad(0\le x\le\xt) .
\end{equation}
At $\gamma=\pi$ one has $\xg=\xt=\xs$, $m_\pi=-U_0=0.388748$ and $U_\pi=U$ on $[0,\xs]$ \eqref{eq:6-axes}. For general $\gamma$,
\begin{equation}\label{eq:alt-dilation}
  U_\gamma(x)+m_\gamma=\Bigl(\frac\pi\gamma\Bigr)^{2}\Bigl[U\Bigl(\frac{\gamma x}{\pi}\Bigr)-U_0\Bigr] ,\qquad\text{that is,}\qquad
  c'\,\bigl[U_\gamma(x)+m_\gamma\bigr]=c'_\gamma\,\bigl[U(\tilde x)-U_0\bigr]
\end{equation}
with $c'_\gamma=(\pi/\gamma)^2c'$ and $\tilde x=\sqrt{3t/c'_\gamma}=\gamma x/\pi$: at a given $t$, the profile is the critical one of the
heavier theory. This is exact for the functional itself: $\sfb$ depends on $\gamma^2/c'$ only, so $g_\gamma$ at $c'$ and
$g_\pi$ at $c'_\gamma$ are the same function of $t$. As $\gamma$ decreases from $\pi$ to $0$, the peak $\xg$ increases from
$\xs$ to $\half$, the distance $\xt-\xg$ from the peak to the turning point increases from $0$ to $\infty$, and $m_\gamma$
increases from $0.3887$ to $\pi/6$. \Cref{fig:10}(a) shows $U_\gamma+L$.

\begin{figure}[t]
\centering
\figfile[\linewidth]{fig10_bumps}
\caption{Drawn from the leading-order formulas of \cref{sec:alt.3,sec:alt.4}, not from data. (a)~The Cardy-weighted profile
$U_\gamma+L$ of the bump $g_\gamma$, that is, the large-$c$ limit of $\frac1{c'}\log\bigl(\abs{\gb_\gamma}V\bigr)$. It has its maximum $0$ at
the peak $\xg$ (dots), which moves from $\xs$ to $\half$ as $\gamma$ decreases from $\pi$ to $0$. Solid: up to the turning
point $\xt$; dotted: beyond it, where by \hyp{B2} $\abs{g_\gamma}$ falls at least with the slope $-2\pi^2/3\gamma$, steeper than the Cardy line. The
critical member $\gamma=\pi$ is the extremal potential $U$ of \cref{sec:6}; beyond $\xs$ it runs along the Cardy line.
(b)~The assembly of \cref{sec:alt.4}; margins and heights are schematic, chosen for visibility. Black: the main bump $\gb_{\gamma_0}$, negative
near its peak; its tail (dotted) has unknown sign but lies below the Cardy line by $\eta_1$ or more beyond $x_g$. Blue:
the negated bumps $-e^{-c'\eta}\gb_{\gamma_k}$ (light blue, drawn near their peaks), positive on their windows, whose upper envelope is a plateau at height
$-\eta$ over $[x_g,\half]$. Green: crossing kernels $\Xib$ with $\beta<1$, which take over beyond $x=\half$. The sum is
non-negative on the shaded region $x\ge x_g$.}
\label{fig:10}
\end{figure}

\paragraph{The saddle point.} Put $t=c'x^2/3$ and $t'=c's^2/3$ in \eqref{eq:alt-Rrad}. For $x\ge x_{\min}>0$ the argument of the
Bessel function is large except near $s=0$. Since the seed is even and analytic in $s$, one may replace $J_0$ on the half line by
$\half H_0^{(1)}$ on a line just above the real axis, and
$H_0^{(1)}(z)\simeq\sqrt{2/\pi z}\;e^{i(z-\pi/4)}$. With \eqref{eq:alt-S1} the integrand becomes a slowly varying amplitude
times $e^{c'\Theta}$, where
\begin{equation}\label{eq:alt-Theta}
  \Theta(s;x):=-\frac{2\pi i}3\bigl(x^2-2\sqrt2\,xs+s^2\bigr)+h_\gamma(s) .
\end{equation}
The first term is the generating function of the rotation by $45^\circ$, and the second is the WKB exponent of the seed. The
saddle-point equation $\partial_s\Theta=0$ reads $\gd(\sqrt2\gamma s)=i\gamma\,(2x-\sqrt2s)$. It is solved with the identity
$\gd(A+iB)=B+iA$, valid whenever $\sinh A=\sin B$ with $\abs B<\frac\pi2$: take $A=\gamma x$ and $B=\gamma q_\gamma(x)$. The saddle is at
\begin{equation}\label{eq:alt-saddle}
  s_*(x)=\frac{x+i\,q_\gamma(x)}{\sqrt2} ,\qquad\text{where}\qquad
  \Theta\bigl(s_*(x);x\bigr)=U_\gamma(x)+m_\gamma
\end{equation}
and $\partial_s^2\Theta=-\frac{4\pi}3\bigl(i+\sech\gamma(x+iq_\gamma)\bigr)$, whose real part is negative. The value at the saddle holds because both sides vanish at $x=0$ and
$\frac{\dd}{\dd x}\Theta(s_*(x);x)=\partial_x\Theta\big|_{s_*}=-\frac{4\pi i}3(x-\sqrt2s_*)=-\frac{4\pi}3q_\gamma(x)$. The
transform samples the real, positive seed at a \emph{complex} radius, and the rotation turns its bell into a tunnelling
profile. The saddle lies inside the strip of analyticity of the seed as long as $q_\gamma(x)<\pi/2\gamma$, that is,
$x<\xt$: it reaches the edge of the strip, which to leading order is the height of the first pole of the seed, exactly at the turning point. There
also $\partial_s^2\Theta=0$: two saddles coalesce. On a horizontal line $\Re\Theta$ is concave
in $\Re s$, because $\Re\sech w>0$ for $\abs{\Im w}<\frac\pi2$; so on the line through $s_*$ the maximum of $\abs{e^{c'\Theta}}$ is at $s_*$. Laplace's method gives the first of the following three properties.

\begin{ideabox}[title={Result S2 (the bump)}]\phantomsection\label{res:S2}
Fix $\gamma\in(0,\pi)$, $0<x_{\min}<0.18$, $0<\delta\le\frac13(\xt-\xg)$ and $\bar x\in(\xg,\xt-\delta]$. As $c\to\infty$:
\begin{description}[leftmargin=2.6em,labelwidth=2.1em,labelsep=0.5em,itemsep=2pt,font=\normalfont]
\item[\hypdef{B1}] \emph{Window.} Uniformly for $x_{\min}\le x\le\xt-\delta$,
\begin{equation}\label{eq:alt-lead}
  \Rot[\phi_\gamma]\bigl(c'x^2/3\bigr)=c'A_\gamma(x)\,e^{c'[U_\gamma(x)+m_\gamma]}\,e^{i\varphi_\gamma(x)}\bigl(1+O(c'^{-1/2})\bigr),
\end{equation}
with the phase $\varphi_\gamma=\tfrac52\arctan(q_\gamma/x)-\frac\pi8$ and the amplitude $A_\gamma=\frac{2^{3/4}}6\,(x^2+q_\gamma^2)^{5/4}x^{-1/2}(1-\sinh^2\gamma x)^{-1/4}>0$. Since $1\le q_\gamma/x\le1/(2\xs)=1.782$, the phase
lies in $[\frac\pi2,2.26]$ and $\sin\varphi_\gamma\ge0.77$. Hence $g_\gamma<0$ on the window, and
$-g_\gamma=e^{c'[U_\gamma(x)+m_\gamma+o(1)]}$.
\item[\hypdef{B2}] \emph{Tail.} For $x\ge\bar x$,
$\abs{g_\gamma(c'x^2/3)}\le e^{c'[U_\gamma(\bar x)+m_\gamma-\frac{4\pi}3q_\gamma(\bar x)(x-\bar x)+o(1)]}$: the slope is steeper than
that of the Cardy line, since $q_\gamma(\bar x)>\half$. Also $\abs{g_\gamma}\le C$ for all $t\ge0$.
\item[\hypdef{B3}] \emph{Vacuum.} $g_{\gamma,\rm vac}=-\int_0^\infty g_\gamma\,\rhoC\,\dd t=e^{c'[m_\gamma+o(1)]}>0$.
\end{description}
\end{ideabox}

\noindent For \hyp{B2} one keeps the contour at the height $\Im s=q_\gamma(\bar x)/\sqrt2$ for all $x\ge\bar x$; on that line
$\Re\Theta(\cdot\,;x)=\Re\Theta(\cdot\,;\bar x)-\frac{4\pi}3q_\gamma(\bar x)(x-\bar x)$ exactly, by \eqref{eq:alt-Theta}. The bound
$\abs{g_\gamma}\le C$ is $\abs{J_0}\le1$ in \eqref{eq:alt-Rrad}. For \hyp{B3}, by \hyp{B1} the integrand $-g_\gamma\rhoC$ is positive on
the window and equals $e^{c'[U_\gamma+L+m_\gamma+o(1)]}$, which is $e^{c'[m_\gamma+o(1)]}$ at the peak; the rest of the window is smaller
by \eqref{eq:alt-concave}, the tail by \hyp{B2}, and the region $x\le x_{\min}$ contributes at most $Cc'e^{c'L(x_{\min})}$, which
is negligible because $L(x_{\min})<0.38<m_\gamma$. By \hyp{B2}, $M[g_\gamma]<\infty$, so $g_\gamma\in\Abar$.

\smallskip
In what follows we use the \emph{normalized bump} $\gb_\gamma:=g_\gamma/g_{\gamma,\rm vac}$. It has vacuum value $1$. On the
window, $-\gb_\gamma=e^{c'[U_\gamma(x)+o(1)]}$, which touches the inverse Cardy weight $e^{-c'L}$ at $\xg$ and lies below it
elsewhere. With $\bar x:=\xg+2\delta\le\xt-\delta$, \hyp{B1}, \eqref{eq:alt-concave} and \hyp{B2} give
\begin{equation}\label{eq:alt-margin}
  \abs{\gb_\gamma(c'x^2/3)}\le e^{-c'[L(x)+\eta_\delta-o(1)]}\qquad(x\ge\xg+\delta),\qquad\eta_\delta:=\tfrac{2\pi}3\,\delta^2 .
\end{equation}
Beyond $\bar x$ the bump carries in addition the factor $e^{-c'\sigma(x-\bar x)}$ with
$\sigma:=\frac{4\pi}3\bigl(q_\gamma(\bar x)-\half\bigr)>0$, by \hyp{B2}. A bump is thus a functional that, per unit of vacuum value, delivers the Cardy rate at one point $\xg\in(\xs,\half)$ of our
choice, with the negative sign, and is exponentially smaller than the Cardy rate beyond it.

\subsection{From bumps to a feasible functional}
\label{sec:alt.4}

\paragraph{Crossing kernels beyond Hellerman's point.} Let $x'>\half$ and $\beta=1/(2x')<1$. Then
$\Xib(t)\ge\half e^{-2\pi\beta t}>0$ for $t\ge t_\beta:=\log(2/\beta)/[2\pi(\beta^{-1}-\beta)]$, which does not depend on $c$, and the
vacuum value of $\Xib$ is negative with $-\Xi_{\beta,\rm vac}\le\beta^{-1}e^{2\pi a/\beta}=2x'e^{\frac12c'L(x')}$. Since
$2\pi\beta t=\pi c'x^2/3x'$ and $x'+x^2/x'=2x+(x-x')^2/x'$, for $t=c'x^2/3\ge t_\beta$
\begin{equation}\label{eq:alt-pairs}
  \frac{\Xib(c'x^2/3)}{-\Xi_{\beta,\rm vac}}\ \ge\ \frac1{4x'}\,
  \exp\Bigl\{-c'\Bigl[L(x)+\frac{\pi(x-x')^2}{3x'}\Bigr]\Bigr\} .
\end{equation}
Per unit of vacuum cost, the kernel $\Xi_{1/2x'}$ delivers the Cardy rate at $x=x'$, with the positive sign. But it does so
only for $x'>\half$, because $\Xib$ with $\beta>1$ is negative at large $t$. This is why a positive plateau at the Cardy rate can be made of crossing kernels with real $\beta$ only beyond Hellerman's point $x=\half$, that is, $\Delta\approx c/6$;
the bumps reach the Cardy rate on the remaining interval $(\xs,\half)$.

Consequently, let $\abs{f(c'x^2/3)}\le e^{-c'[L(x)+\eta']}$ for $x\ge x_a\ge\half$, with the extra factor
$e^{-c'\sigma(x-x_b)}$, $\sigma>0$, for $x\ge x_b$. Then, for all sufficiently large $c$, $f$ is dominated there by a combination
$R=\sum_j\alpha_j\Xi_{\beta_j}$ with $\alpha_j\ge0$ and $\beta_j<1$: $R\ge\abs f$ for $x\ge x_a$, $R\ge0$ for $t\ge t_R$ with
$t_R$ independent of $c$, and $-R_{\rm vac}\le e^{-c'\eta'/4}$. On the compact range $[x_a,x_b]$ one takes finitely many
nodes $x'_j>\half$, bounded away from $\half$, at a spacing small compared with $\sqrt{\eta'}$, with the coefficients
$\alpha_j=e^{-c'\eta'/2}/(-\Xi_{\beta_j,\rm vac})$: by \eqref{eq:alt-pairs} the node next to $x$ gives more than
$e^{-c'[L(x)+\eta']}$, and each node costs $e^{-c'\eta'/2}$ of vacuum value. On the tail $x\ge x_b$ one takes equally spaced
nodes whose coefficients decrease from node to node by the factor that $f$ loses, so that their costs form a geometric
series. We call $R$ a \emph{repair} of $f$.

\paragraph{The functional.} Fix $x_g\in(\xs,\half)$. The map $\gamma\mapsto\xg$ is a decreasing bijection of $(0,\pi)$ onto
$(\xs,\half)$.
\begin{enumerate}[label=\textup{(\arabic*)},leftmargin=2.2em]
\item \emph{Main bump.} Choose $\gamma_0$ with $\xs<x_{\gamma_0}<x_g$. By \eqref{eq:alt-concave} and \hyp{B2}, used with $\bar x$ between $x_{\gamma_0}$ and $\min(x_g,\xt)$, there is $\eta_1>0$ with
$\abs{\gb_{\gamma_0}}\le e^{-c'[L(x)+\eta_1-o(1)]}$ for $x\ge x_g$.
\item \emph{Margins.} Let $\gamma_g$ be the value of $\gamma$ with $\xg=x_g$, and let $\delta$ be a third of the distance
$\xt-\xg$ at $\gamma_g$. This distance grows as $\gamma$ decreases, so $\delta\le\frac13(\xt-\xg)$ for every bump with its
peak in $[x_g,\half)$. Let $K$ bound $\half\abs{(U_\gamma+L)''}$ on the windows $[x_{\min},\xg+\delta]$ of all these bumps; it
is finite because $(U_\gamma+L)''$ depends on $\gamma x$ only, and $\gamma x\le\gamma_g(x_g+\delta)<\arcsinh1$ there. Choose a
mesh $\epsilon<\delta/2$ and $\eta>0$ with $\eta+K\epsilon^2<\min(\eta_1,\eta_\delta)$.
\item \emph{Plateau.} Choose $\gamma_1>\dots>\gamma_N$ whose peaks $x^{(k)}:=x_{\gamma_k}$ form a grid of $[x_g,\half)$ of mesh at most $\epsilon$,
with $x^{(1)}\le x_g+\epsilon$ and $x^{(N)}\ge\half-\epsilon$.
\item \emph{Repairs.} Let $R_0$ be a repair of $\gb_{\gamma_0}$ on $x\ge\half$, and $R_k$ a repair of $\gb_{\gamma_k}$ on
$x\ge\max(\half,x^{(k)}+\delta)$; what a repair needs is supplied by step~(1), by \eqref{eq:alt-margin} and by the remark after it.
\end{enumerate}
All these choices are made before $c\to\infty$; for the repairs this means their nodes, hence $t_R$, while their coefficients
depend on $c$. The functional is the flat sum
\begin{equation}\label{eq:alt-g}
  \boxed{\;g:=\gb_{\gamma_0}+R_0+e^{-c'\eta}\sum_{k=1}^N\bigl(-\gb_{\gamma_k}+R_k\bigr) .\;}
\end{equation}
It is drawn in \cref{fig:10}(b).

\paragraph{Why it is feasible at $T=c'x_g^2/3$.} \emph{On $x_g\le x\le\half$.} The repairs are non-negative there, since $c'x_g^2/3\ge t_R$ for large $c$. A bump with
$x\le x^{(k)}+\delta$ has $x$ in its window, where $-\gb_{\gamma_k}>0$ by \hyp{B1}; the one whose peak is within $\epsilon$ of
$x$ contributes at least $e^{-c'[L(x)+\eta+K\epsilon^2+o(1)]}$. A bump with $x>x^{(k)}+\delta$ contributes at most
$e^{-c'[L(x)+\eta+\eta_\delta-o(1)]}$ in modulus, by \eqref{eq:alt-margin}. With step~(1),
\begin{equation}\label{eq:alt-pos}
  g\bigl(c'x^2/3\bigr)\ \ge\ e^{-c'L(x)}\Bigl[e^{-c'(\eta+K\epsilon^2+o(1))}-e^{-c'(\eta_1-o(1))}-N\,e^{-c'(\eta+\eta_\delta-o(1))}\Bigr]>0
\end{equation}
for large $c$, by step~(2). \emph{On $x\ge\half$.} Here $\gb_{\gamma_0}+R_0\ge0$; and for each $k$ either $x$ is in the window of the $k$-th
bump and $-\gb_{\gamma_k}>0$, or $x\ge x^{(k)}+\delta$ and $-\gb_{\gamma_k}+R_k\ge0$. \emph{Vacuum.} By linearity
$\gvac\ge1-2e^{-c'\eta_1/4}-N\,e^{-c'\eta}\,(1+o(1))>0$. \emph{Class.} Each term is in $\Abar$. The tail repairs are infinite sums; they converge in $L^1$ and in the norm $M[\,\cdot\,]$, because $M[\Xib]\le\frac1\pi\beta^{-1}e^{2\pi a/\beta}$, the quantity that bounds the vacuum cost. So $g\in\Abar$.

Thus $g\in\Abar$ is feasible at $c'x_g^2/3$ for all sufficiently large $c$, which is \ResSone. Everything rests on one feature of
the bumps: all windows begin at the same point $x_{\min}<\xs$. At any $x\ge x_g$ a negated bump is either inside its window,
and then positive, or beyond it, and then exponentially below the Cardy line. So all the negated bumps can enter with the
same coefficient $e^{-c'\eta}$, and no positivity of an oscillating function is ever needed.

By linearity, $g=\Omega[\phi]$ for a single real seed,
\begin{equation}\label{eq:alt-oneseed}
  \phi=-\bar\phi_{\gamma_0}+e^{-c'\eta}\sum_{k=1}^N\bar\phi_{\gamma_k}+(\text{real parts of complex Gaussians}),\qquad
  \bar\phi_\gamma:=\phi_\gamma/g_{\gamma,\rm vac} :
\end{equation}
the functional is one transform of one real function. For $x_g$ near $\xs$ the corrections to $-\bar\phi_{\gamma_0}$ are exponentially small on the real axis; they matter
only after continuation to the complex radius $s_*$ at which the transform samples the seed.

\subsection{Remarks}
\label{sec:alt.5}

\paragraph{Comparison with the functional of \cref{thm:5}.} The functional of \cref{thm:5} is built to have the profile
$U$: its zeros are placed by the zero law, it runs along the Cardy line on the band, and the difficulty is to show
that the oscillating function $h_0-\hat h_0$ stays positive there. The functional \eqref{eq:alt-g} avoids the band
altogether. On $[x_g,\half]$ it lies below the Cardy line by the fixed margin $\eta$ (\cref{fig:10}(b)), so it is positive for a crude
reason, and it is far from extremal. The price is that the margins $\eta_1$, $\eta_\delta$, $\eta$ go to zero as
$x_g\to\xs$, while the estimates hold for $c'\eta\gg1$; this is why no rate comes out. The two
constructions share the potential $U$ and nothing else. A third construction, in Mellin space, is sketched in
\cref{app:M}.

\paragraph{Consistency with the optimality theorem.} \Cref{thm:3} forbids a function in $\Abar$ whose
Cardy-weighted mass sits below $\xs$. Accordingly the peaks obey $\xg>\xs$. At $\gamma=\pi$ the peak, the turning point and
$\xs$ coincide, and to leading order the tail falls at the Cardy rate: $\abs{g_\pi}V$ is flat at exponential order along the
whole band, and the critical member has no margin. (It is still in $\Abar$: the first pole of $\cG$ is at $\Im r=Q/4$ and not
at $1/4\sfb$, so its tail beats the Cardy weight by the factor $e^{-\pi x/2}$, which does not depend on $c$;
$M[g_\gamma]=\infty$ only for $\gamma>\pi(1+\sfb^2)$.) For fixed $\gamma>\pi$ the tail is slower than the Cardy rate. Taking
$\gamma<\pi$ is what creates the margins, and by \eqref{eq:alt-dilation} it amounts to borrowing the critical function of a
heavier theory.

\paragraph{Why this seed \tagpar{\heuristic}.} On WKB data a function that the Fourier transform maps to $\pm$ itself lives on a curve invariant
under the quarter-turn, and the relevant one is $\sinh\gamma b=i\sinh\gamma p$. It is exponential in both $b$ and $p$, and an
exponential of a momentum is a finite shift, so its quantization is a difference equation rather than a differential one.
In the rotated coordinates $s=(b-p)/\sqrt2$, $p_s=(b+p)/\sqrt2$ the curve is a graph,
$p_s=-(i/\sqrt2\gamma)\,\gd(\sqrt2\gamma s)$, with the momentum normalized as in \eqref{eq:4-momentum}: this is the saddle-point
relation \eqref{eq:alt-saddle} read backwards, and $c'h_\gamma$ is the corresponding WKB exponent. On a graph the quantized equation is of first order, and its
normalizable solution is $\cG$: a first-order difference equation with trigonometric coefficients is solved by double sine
functions, as $f(z+1)=zf(z)$ is solved by the Gamma function. The expansion \eqref{eq:alt-S1} is its WKB form to one loop. The
function $\cG$ is moreover equal to its own Fourier transform in one dimension, a special case of the Fourier
transformation formula of~\cite{FKV,PT}, which reflects the invariance
of the curve; we do not use this. So the bell $\cG$ is the wave function of the quantized, deformed spectral curve in the
rotated polarization, and the bump is the same state in the polarization of \cref{sec:4}. The passage to the plane, with
the radius as the argument and the factor $\abs{\br'}^2$, is a recipe that works because the centrifugal term is invisible at
leading order; we have not derived it. We know of no meaning in conformal field theory of the shift by $i\sfb$, and at
$\gamma=\pi$ the parameter is not that of Liouville theory at the same central charge ($\sfb^2=3/4c'$, against $6/c$).

\paragraph{Zeros.} The functional \eqref{eq:alt-g} is strictly positive on $[x_g,\half]$; it has no band of double zeros. If
$x_g\to\xs$ slowly as $c\to\infty$, it is a sequence that attains $\LV$. By \cref{thm:7} its profile then tends
to $U$, and its zeros in the strip have the density $c'\rhoU$ within any fixed distance of the slits. These zeros lie off the
real axis; we have not located them.

\section{What near-optimal functionals must look like}
\label{sec:9}

This section is proved throughout, and no computer enters. It uses \cref{prop:6.4}, \cref{lem:6.5}, \cref{def:6.6} and
\cref{lem:6.7} and, only to know that attaining sequences exist, \cref{thm:5}. Nothing outside this section depends on it.

\Cref{thm:3,thm:4} are inequalities at finite $c$: they bound what a functional can certify. Here we ask the converse
question: what must a functional look like that comes close to the bound? The tool is a limit. Along a sequence
$g_j\in\Abar$ at central charges $c_j\to\infty$ the profiles $u_{g_j}$ are functions on one and the same strip $\strip$,
which does not depend on $c$. If the sequence attains the slope $\LV$, the limit is the extremal potential $U$, and
the zeros of $g_j$ follow the zero law $\rhoU$ (\cref{thm:7}).

\subsection{Limits of potentials}
\label{sec:9.1}

Let $g\in\Abar\setminus\{0\}$. By \cref{sec:6.3}, $G(b)=g(c'b^2/3)$ is even and holomorphic on $\strip$, and bounded and
continuous on $\bar\strip$; and the profile \eqref{eq:6-ug}, with its Riesz measure \eqref{eq:6-riesz},
\begin{equation}\label{eq:9-profile}
  u_g(b)=\frac1{c'}\log\frac{\abs{G(b)}}{M[g]} ,\qquad
  \nug=\frac1{2\pi}\nabla^2u_g=\frac1{c'}\sum_{\zeta\in\strip:\ G(\zeta)=0}m_\zeta\,\delta_\zeta ,
\end{equation}
is the potential of positive point charges $m_\zeta/c'$ at the zeros of $G$ in the strip: $\nug$ is the zero-counting
measure of $G$ divided by $c'$. For a sequence we write $u_j:=u_{g_j}$ and $c'_j:=c_j-1$. As in
\cref{thm:4}, $g$ is called weakly feasible if it is weakly feasible at some $T$, hence at $T=t_1(g)$, and
$x_1(g)=\sqrt{3\,t_1(g)/c'}$ as in \eqref{eq:2-scaling}. The idea
is that the three facts on which \cref{sec:6} rests pass to the limit: limits of subharmonic functions are
subharmonic; the majorant of \cref{prop:6.4}, whose error is $O(\log c'/c')$, becomes exact; and weak feasibility,
which enters through the half-mass condition \eqref{eq:2-halfmass}, becomes contact with the Cardy line. The notions
needed to say this are in the box; the proofs are in \cref{app:H.1,app:H.2,app:H.3}.

\begin{toolkit}{Toolkit: limits of potentials}
Let $u_j$ be potentials of positive charges $\nfrak_j=\frac1{2\pi}\nabla^2u_j$ on $\strip$ (subharmonic functions).
\emph{Convergence in $L^1_{\rm loc}$} is convergence of the potentials in the mean on every compact set,
$\int_K\abs{u_j-u}\,\dd A\to0$; pointwise convergence is too much to ask, because a potential is $-\infty$ at its
charges, but the logarithm is integrable in two dimensions.
\emph{Vague convergence} $\nfrak_j\to\nfrak$ is convergence of the charge distributions when tested against continuous
functions of compact support; then $\nfrak_j(K)\to\nfrak(K)$ for every compact $K$ whose boundary carries no charge
of $\nfrak$.
\emph{Compactness} \cite[Thms.~3.2.12--3.2.13]{Hormander}: potentials that are bounded above on compact sets,
uniformly in $j$, and do not run off to $-\infty$ carry a bounded charge on every compact set, and along a
subsequence potentials and charges both converge.
\emph{Upper semicontinuity:} at a point a limit potential can jump up, never down; the form that we use is
\eqref{eq:H-hartogs}.
\emph{Superharmonic, strong minimum principle, circle means:} $\theta$ is superharmonic if $-\theta$ is subharmonic;
on a connected open set a non-negative superharmonic function is identically zero or strictly positive, and its
mean over a circle is at most its value at the center \cite[\S3.2]{Hormander}, \cite[Thm.~2.3.1]{Ransford}.
\end{toolkit}

\begin{lemma}[Compactness; \proved]\label{lem:9.1}
Let $c_j\to\infty$ and $g_j\in\Abar\setminus\{0\}$. After passing to a subsequence, either
\textup{(i)}~$u_{g_j}\to-\infty$ uniformly on every compact subset of\/ $\strip$, or
\textup{(ii)}~$u_{g_j}\to u$ in $L^1_{\rm loc}(\strip)$ for a subharmonic function $u\not\equiv-\infty$ on $\strip$,
and $\nfrak_{g_j}\to\frac1{2\pi}\nabla^2u$ vaguely on $\strip$.
\end{lemma}

\noindent A function $u$ that arises in (ii) is called a \emph{limit profile} of the sequence.

\begin{lemma}[Majorant; \proved]\label{lem:9.2}
Every limit profile satisfies $u\le U$ on all of\/ $\strip$, where $U$ is extended across the slits as in
\cref{def:6.6}: it is continuous, equal to $-L$ on the slits, and subharmonic with charge density $\rhoU(x)\,\dd x$
there. In particular $u\le0$ on $\strip$, and $u\le-L$ on the real axis.
\end{lemma}

\begin{proposition}[Contact; \proved]\label{prop:9.3}
Let $c_j\to\infty$, and let $g_j\in\Abar\setminus\{0\}$ be weakly feasible with $x_1(g_j)\to\xinf<\infty$. Then
alternative \textup{(i)} of \cref{lem:9.1} does not occur, and every limit profile $u$ has a \emph{contact point}
$x_0\in[\xs,\xinf]$, at which $u(x_0)=U(x_0)=-L(x_0)$. In particular $\xinf\ge\xs$, which is the qualitative
statement of \cref{thm:4} once more.
\end{proposition}

In the dictionary of \cref{sec:6.2}: a limit profile lies below the extremal potential, and the last sign change of
the functionals leaves a mark on it, a point not beyond $\xinf$ at which the profile touches the Cardy line; by
\cref{lem:6.7} this point lies on a slit.

\subsection{Rigidity of the zeros}
\label{sec:9.2}

\begin{theorem}[Rigidity of the zero law; \proved]\label{thm:7}
Let $c_j\to\infty$, and let $g_j\in\Abar\setminus\{0\}$ be weakly feasible and attain the slope $\LV$, that is,
$x_1(g_j)\to\xs$. \textup{(}By \cref{prop:9.3} it suffices that $\limsup_jx_1(g_j)\le\xs$; such sequences exist by
\cref{thm:5}.\textup{)} Then:
\begin{enumerate}[label=\textup{(\alph*)},leftmargin=2.2em,itemsep=1pt,topsep=3pt]
\item $u_{g_j}\to U$ in $L^1_{\rm loc}(\strip)$, along the whole sequence.
\item The normalized zero-counting measures converge vaguely on $\strip$ to the zero law
$\rhoU(x)=\frac4{3\pi}\arccosh(\sinh\pi\abs x)$ of \cref{def:6.6}, carried by the slits:
\begin{equation}\label{eq:9-zerolaw}
  \nfrak_{g_j}=\frac1{c'_j}\sum_{\zeta}m_\zeta\,\delta_\zeta\ \longrightarrow\ \rhoU(x)\,\dd x\otimes\delta_{y=0}
  \qquad\text{on }\abs x\ge\xs .
\end{equation}
Concretely, counting zeros with multiplicity: in every compact subset of\/ $\strip$ disjoint from the slits $G_j$ has
$o(c'_j)$ zeros, and for $\xs<x_a<x_b$ and $0<y_0<\half$
\begin{equation}\label{eq:9-count}
  \#\bigl\{\text{zeros of }G_j\text{ in }[x_a,x_b]\times[-y_0,y_0]\bigr\}=c'_j\int_{x_a}^{x_b}\rhoU(x)\,\dd x+o(c'_j) .
\end{equation}
\item In the variable $t$: the number of zeros of $g_j$, with multiplicity, in the image of that rectangle under
$b\mapsto c'_jb^2/3$ \textup{(}a lens $\abs{\Im t}\le\frac23c'_jx_by_0$ around $[c'_jx_a^2/3,\,c'_jx_b^2/3]$\textup{)} is
$c'_j\int_{\xi_a}^{\xi_b}2\rhoV(\xi)\,\dd\xi+o(c'_j)$, where $\xi=x^2/3$, $\xi_a=x_a^2/3$, $\xi_b=x_b^2/3$, and $\rhoV=3\rhoU/(4x)$ is the
band density of \eqref{eq:4-rhoU}. In words, near $[t,t+\dd t]$ there are $2\rhoV(t/c')\,\dd t$ zeros.
\end{enumerate}
\end{theorem}

Item (c) reproduces, counted with multiplicity, the density of the band of double zeros of \cref{conj:1}(b), and
item (a) is the profile stated there, in the mean. Of what \cref{sec:4} derived from the hypotheses \hyp{W}, \hyp{A},
\hyp{B} and \hyp{Sel}, the band density and the profile thus follow, at leading order and in the weak sense of (a) and
(b), from attainment alone; the finer statements of \cref{conj:1} are not obtained in this way. The hypotheses
themselves, which concern the semiclassical sheets of the optimal functional, are not thereby derived. The theorem also
turns the blueprint of \cref{sec:7.1} into a necessity.

\smallskip\noindent\textbf{What the theorem does not say.} It does not say that the zeros near the band are real, or
that they come in pairs; whether an attaining functional must oscillate there, as the numerical optima do, is not
decided. It says nothing about individual positions, the Airy layer at the tip, the boundary layer at the vacuum, the
phase of the comb, zeros on the edges or outside the closed strip, or rates.

\smallskip\noindent\textbf{Idea of the proof} (\cref{app:H.4}). Let $u$ be a limit profile. By \cref{lem:9.2} the
deficit $\theta:=U-u$ is non-negative; it is superharmonic on the slit strip $D$, where $U$ is harmonic; and by
\cref{prop:9.3} with $\xinf=\xs$ it vanishes at the tip $\xs$, a boundary point of $D$. By the strong minimum
principle either $\theta\equiv0$ on $D$, which is the claim, or $\theta>0$ on $D$. Had the contact point been inside
$D$, this would end the proof; at a point of the open slit it would not; the tip is the borderline case, and it is
decided by a competition of exponents. If $\theta>0$ on $D$, then near the tip $\theta$ dominates a multiple of the
harmonic measure of an arc in a disk slit along a radius. The square-root map that opens the slit shows that this
harmonic measure vanishes at the tip only like the square root of the distance, so the mean of $\theta$ over a circle
of radius $\epsilon$ about the tip is $\gtrsim\epsilon^{1/2}$. On the other hand $u$ is subharmonic across the slit and
$u(\xs)=U(\xs)$, so the circle means of $\theta$ are at most those of $U-U(\xs)$, which the Poisson--Jensen formula
expresses through the charge of $U$ near the tip. Since $\rhoU\sim\sqrt{x-\xs}$, they are at most $C\epsilon^{3/2}$.
For small $\epsilon$ the two bounds contradict each other: the tip decides.

\PartHeading{V}{Outlook}\section{Discussion}
\label{sec:10}

\subsection{Theorem, conjecture, open}
\label{sec:10.1}

In terms of the legend of \cref{sec:1.3}, the results stand as follows. \emph{Proved:} the slope $\LV$
(\cref{thm:1}), in its two halves, optimality with a rate (\cref{thm:3,thm:4}) and attainment (\cref{thm:5}); and
certification (\cref{thm:2}), which turns attainment into a bound on conformal field theories (\cref{cor:6}).
Only attainment and what rests on it, that is, the second half of \cref{thm:1}, \cref{thm:5} and \cref{cor:6}, use
computer assistance. Also proved, without a computer, is the rigidity of the zero law (\cref{thm:7}).
\emph{Conjectured:} everything finer (\cref{conj:1}), in particular the coefficient $\Aopt=0.27666$ of $c^{1/3}$, the
profile of the optimal functional, and the offset $c/24$ in the positions of its zeros (seen numerically
in~\cite{AHT,HMR}; \cref{sec:4.6}). \emph{Explained but not
proved:} the origin of the earlier extrapolation to the slope $9.08$~\cite{AHT} (\cref{sec:5.2}).

\subsection{Three-dimensional gravity; Virasoro versus \texorpdfstring{$U(1)^c$}{U(1)\textasciicircum c}}
\label{sec:10.2}

Invariance under $S$ at zero angular potential, used through functionals in $\Abar$, certifies
$\Delta_1\lesssim c/9.1266$ at large $c$ and no more: the window \eqref{eq:1-window} is closed to it. This is a
statement about the method. It neither supports nor threatens pure gravity~\cite{Witten3d}, and it does not say which
spectra with $\Delta_1$ in the window are realized. To reach the BTZ threshold one needs spin~\cite{CLY,AMS,BOSW1}, integrality of the
degeneracies~\cite{KaidiPerlmutter,FitzpatrickLi}, or data from correlation functions. Arguments through the \emph{elliptic point} of order three,
$\tau=e^{2\pi i/3}$, which is fixed not by $S$ but by $ST$ and so involves
spin~\cite{QuallsEven,Gliozzi1,Gliozzi2,Tierz}, aim at the BTZ threshold itself; the present
results neither use them nor bear on them. The same holds for the bound $\Delta_1\le(c-1)/12+0.23$ on what~\cite{Tierz} calls
the spinless gap: it rests on the kernel of $ST^1S$, and so on $T$ as well as $S$, and it concerns spectra in which every non-vacuum
primary has spin zero, whereas here primaries of every spin are counted.

\paragraph{Virasoro versus $U(1)^c$.}
The two problems have the same architecture and different numbers, $9.1266$ against $\pi^2$. The slope $\pi^2$ is
proved in~\cite[Ch.~1]{OpenAI}; we have read the statements and the structure of that argument but have not
independently verified it; nothing in this paper depends on it logically. That proof works on a strip in the Mellin
variable, in dimension $d=2c\to\infty$. Here the dimension $d=2$ is fixed, the strip is in the position-like variable
$b$, and $c$ enters through the position of the vacuum; the version in the Mellin variable, where the strip becomes a
domain with a free boundary, is sketched in \cref{app:M}. There is no analogue here of the condition $f(0)=0$ on the functional of \cref{sec:2.3}, which the $U(1)^c$ proof
uses to cancel a pole of the Mellin transform on the edge of the strip and to make the integral of the functional vanish. Two different
objects should be kept apart. The \emph{exact optimizers} of both problems, as computed numerically, have a band of
double zeros (\cref{sec:3.1,sec:4.6}), whereas the
\emph{asymptotically attaining} functions constructed in~\cite[Ch.~1, Thm.~4.1]{OpenAI} are free of real zeros on the whole
constrained region. For Virasoro a functional that attains $\LV$ must
carry zeros at density $c'\rhoU$ near the real axis, by \cref{thm:7}, though not
necessarily on it; this freedom is what the detuning of \cref{sec:7} exploits.

\paragraph{Note added.} After this paper was completed, Chang, Liu and Ma~\cite{ChangLiuMa} constructed modular-invariant torus
partition functions with a unique vacuum, integer spins and positive integer degeneracies whose gap is
$\Delta_1=(1+\kappa)(c-1)/12$, for small fixed $\kappa>0$ at every sufficiently large $c$. These are partition functions, not full
conformal field theories. We have not verified this construction; if it stands, the lower end of the window \eqref{eq:1-window} passes every constraint of the
torus partition function, spin and integrality included, so that $\Lambda<12$ for every bound that uses this function alone, and
reaching the BTZ threshold requires data beyond it. Perlmutter~\cite{PerlmutterEdge} has shown, for theories whose spectral counting
fluctuations are logarithmically rigid, that $\Delta_1-(c-1)/12\le o(c)$, apart from at most one exceptional value of the scalar gap
at each sufficiently large $c$. Salmon~\cite{Salmon2} has determined the asymptotic rate of Delsarte's linear programming bound for binary and
spherical codes, with lower bounds from potential theory through an obstacle problem solved in closed form, as in \cref{app:M}.

\subsection{Open problems}
\label{sec:10.3}

\emph{Effective constants.} More useful than the values of $\copt$ and $\catt$, which are explicit in principle and not
computed, would be certificates in interval arithmetic of the explicit functional at fixed~$c$.

\emph{The coefficient of $c^{1/3}$.} A proof of $\Aopt$ needs the comb with no site omitted (the construction of
\cref{sec:7} omits the first site of the zero law), the Airy layer at the tips, and the balance between lobe and band in
the vacuum inequality. Apply the edge representation of \cref{sec:7.4} to a numerically optimal functional, whose transform is
$\hat g=-g$. It suggests that on the band $\hat g$ is the sum of two parts. One oscillates, with relative amplitude
$\half$ and in antiphase with the oscillation of $g$, so that the zeros are doubled rather than lifted off the axis. The
other is a mean, which is manufactured near the vacuum point of the edge; the construction of \cref{sec:7} does not
attempt this. A proof of $\Aopt$ would have to control both. We do not claim it.

\emph{Further problems.} To remove the logarithm in \cref{thm:4}, that is, to prove $t_1(g)\ge\xis c'-O(c^{1/3})$, the
scale of \cref{conj:1}. To find a convex dual of the linear program of \cref{sec:2.2} with the extremal potential $U$
as its value function, which would explain why the potential of \cref{sec:6} is the curve of \cref{sec:4}. To bound
$\TV(c)$ from below at a given $c$: numerics bound it from above only (\cref{sec:2.6}), and \cref{thm:4} starts at
$\copt$. To derive from the linear program the selection rule \hyp{Sel} of \cref{sec:4.4}, on which \cref{conj:1}
rests. To find what replaces the extremal problem of \cref{sec:6} in the bootstrap with spin~\cite{CLY,AMS}.

\appendix
\crefalias{section}{appendix}
\crefalias{subsection}{subappendix}
\providecommand{\Mob}{\mathcal B}
\providecommand{\MD}{\mathsf D}
\providecommand{\MU}{\mathcal U}
\providecommand{\MW}{\mathcal W}
\providecommand{\MK}{\mathcal K}
\providecommand{\MN}{\mathcal N}
\providecommand{\ME}{\mathcal E}
\providecommand{\MC}{\mathcal C}
\providecommand{\MH}{\mathcal H}
\providecommand{\Ups}{\Upsilon}
\providecommand{\Usk}{\mathcal U_{\rm sk}}
\providecommand{\nsk}{\mathfrak n_{\rm sk}}
\providecommand{\mus}{\mu_*}
\providecommand{\nus}{\nu_*}
\providecommand{\ResSone}{\hyperref[res:S1]{Result~S1}}
\providecommand{\ResMone}{\hyperref[res:M1]{Result~M1}}
\providecommand{\ResMtwo}{\hyperref[res:M2]{Result~M2}}
\providecommand{\ResMthree}{\hyperref[res:M3]{Result~M3}}
\section{Mellin space approach}
\label{app:M}

The proofs of \cref{sec:6,sec:7} work in the variable $b=\sqrt{3t/c'}$, in which a functional is holomorphic on a strip.
This appendix sketches a second derivation of the slope, \cref{thm:1}, in which a functional is studied through its
Mellin transform. It gives both halves: no functional in $\Abar$ certifies more than $\LV$, and functionals exist that
come arbitrarily close. The Mellin transform is the tool of the proof for the $U(1)^c$ problem
in~\cite[Ch.~1]{OpenAI}, where the transform of a functional is controlled on a strip. The attaining functions of~\cite[Ch.~1, Thm.~4.1]{OpenAI} are likewise built in Mellin space, by multiplying the
Mellin transform of a Gaussian by factors that respect the Fourier symmetry and reading the signs from saddle points, and the
functional of \cref{app:M.6} has this form, with a symbol whose zeros come from the solution of the free-boundary problem described below. (The Mellin transform of the auxiliary functions of the
sphere-packing linear program, with its functional equation and its relation to their Taylor coefficients, appears in
Cohn and Miller~\cite[\S5]{CohnMiller}, for smooth and rapidly decreasing radial functions in any dimension. Properties (i) and (ii)
of \eqref{eq:M-facts} follow from their formulas in dimension two with $\hat f=-f$ when $f(x)=g(x^2/2)$ is smooth and rapidly
decreasing: with $M_f(s)=\int_0^\infty f(x)\,x^{s-1}\,\dd x$ one has $m_g(z)=2\pi^z M_f(2z)/\Gamma(z)$.) What is new for Virasoro is that
the vacuum sits at $t=-a<0$, off the half line over which the transform is taken. It then enters through all the integer
moments of $g$ at once, and the problem on a strip becomes a free-boundary problem in the plane: an obstacle problem,
which we solve in closed form. The curve $\sinh\pi b=i\sinh\pi p$ appears once more, now as the solution of the
free-boundary condition, and the threshold $\xs$ is fixed by the point of the contact set that is nearest to the axis of
the reflection $z\to1-z$, which is what the Fourier transform becomes.

What this appendix has in common with the main text is not repeated. The problem, the class $\Abar$ and the vacuum value
are those of \cref{sec:2}, and a feasible $g\in\Abar$ bounds every conformal field theory by \cref{thm:2}. The two
results are the following.

\begin{ideabox}[title={Result M1 (optimality)}]\phantomsection\label{res:M1}
For every $\delta\in(0,\xs)$ there is $C_\delta<\infty$ such that for all $c\ge40$ and every $g\in\Abar$, with no
condition on the sign of $g$,
\begin{equation}\label{eq:M-conc}
  \int_0^{c'(\xs-\delta)^2/3}\abs g\,V\,\dd t\;\le\;C_\delta\,c'^{\,4}\,e^{-\kappa_M(\delta)\,c'}\,M[g],\qquad
  \kappa_M(\delta):=\MW\bigl(\tfrac\pi3(\xs-\delta)\bigr)>0 ,
\end{equation}
where $\MW$ is the explicit function of \cref{app:M.3}. Consequently, by the half-mass condition \eqref{eq:2-halfmass},
for every $x_T<\xs$ there is $c_0(x_T)$ such that for $c\ge c_0(x_T)$ no $g\in\Abar$ other than zero is weakly feasible at
$T=c'x_T^2/3$. No computer is used.
\end{ideabox}

\begin{ideabox}[title={Result M2 (attainability)}]\phantomsection\label{res:M2}
For every $x_g>\xs$ there is $c_0(x_g)\ge40$ such that for every real $c\ge c_0(x_g)$ there is $g_c\in\Abar$ with $g_c>0$ on
$[c'x_g^2/3,\infty)$ and $(g_c)_{\rm vac}>0$; moreover $g_c$ decays faster than $e^{-4\pi\sqrt{\alpha t}}$ for every
$\alpha>0$. So $g_c$ is feasible at $T=c'x_g^2/3$, and by \cref{thm:2} every admissible CFT with $c\ge c_0(x_g)$ has
$\Delta_1<a+c'x_g^2/3$.
\end{ideabox}

\noindent Together they give $\lim_{c\to\infty}\TV(c)/c'=\xis$, that is, \cref{thm:1}, once more. \ResMone\ is
\cref{thm:3} with a smaller exponent, $\kappa_M(\delta)\simeq3.73\,\delta^2$ against $\kappa(\delta)\simeq2.65\,\delta^{3/2}$
(\cref{app:M.4}), and it contains the first statement of \cref{thm:4}. \ResMtwo\ is \ResSone\ of \cref{sec:alt}, by a
third construction. Neither gives a rate, and the central charges $c_0$ are not explicit.

The plan is as follows. \Cref{app:M.1} recasts the constraints in Mellin space: the Fourier transform becomes a
reflection, and the vacuum becomes a series over the integers with positive coefficients, each of which bounds the
transform on a vertical line. \Cref{app:M.2} takes the scaling limit of these bounds, the \emph{obstacle}, and explains
why the threshold is the \emph{tip} of the set where the largest subharmonic function below the obstacle touches it.
\Cref{app:M.3} finds this set and this function in closed form, and \cref{app:M.4} turns the heuristic argument into
\ResMone. \Cref{app:M.5,app:M.6,app:M.7} construct the functionals of \ResMtwo: the zeros of the Mellin transform are
placed along four arcs inside the contact set, the functional is recovered by Mellin inversion, and its signs come from
a saddle-point analysis in which one real saddle point dominates. \Cref{app:M.8} is a dictionary between the Mellin
plane and the strip of \cref{sec:6}.

\paragraph{Level of rigor.} The results of this appendix have been verified rigorously, but the write-up is informal:
the steps are stated and the estimates behind them are omitted. For this reason \ResMone, \ResMtwo\ and \ResMthree\
carry no status tag. One step, the univalence of a conformal map (\cref{app:M.5}), uses five inequalities that were
checked in interval arithmetic~\cite{Arb}; they are listed there and are not part of \cref{tab:4}. Nothing outside this
appendix depends on it.

\paragraph{Notation.} It is that of \cref{sec:2}: $c'=c-1$, $a=c'/12$, $t=\Delta-a$, $b=x+iy=\sqrt{3t/c'}$; $\beta$ is the
inverse temperature; $V$ is the Cardy weight \eqref{eq:2-rhoC}, $\rhoC=2\pi V$, $L(x)=\frac{2\pi}3\abs x$, and
$M[g]=\int_0^\infty\abs g\,V\,\dd t$; $\gvac=-\int_0^\infty g\,\rhoC\,\dd t$ and feasibility are as in
\cref{def:2-Abar,def:2-feasible}. The extremal potential $U$, its value $U_0=U(0)=-0.388748$, the momentum $p$ and the
zero law $\rhoU$ are those of \cref{lem:6.5,def:6.6}. The Mellin variable is $z$, and $\omega=\mu+i\nu$ is the rescaled
variable of \eqref{eq:M-scaling}; in this appendix $\omega$ is not the linear functional of \cref{sec:2.2}. The letters
$m_g$, $\theta_n$, $\sigma$, $\vartheta$, $\tau$, $\eta$, $\eta_v$, $\iota$, $\lambda$ and the calligraphic letters $\Mob$,
$\MU$, $\MW$, $\MK$, $\MN$, $\ME$, $\MC$, $\MH$ are local to this appendix. $C$ stands for constants that do not depend
on $c$, on $t$ or on the point $\omega$.

\subsection{The Mellin symbol and the vacuum inequality}
\label{app:M.1}

For $g\in\Abar$ we call
\begin{equation}\label{eq:M-symbol}
  m_g(z):=\frac{(2\pi)^z}{\Gamma(z)}\int_0^\infty g(t)\,t^{z-1}\,\dd t\qquad(\Re z>0)
\end{equation}
the \emph{Mellin symbol} of $g$. The integral converges because $g$ is bounded (\cref{lem:A.3}) and all its moments are
finite, by \eqref{eq:M-vacineq} below. Three properties are used.
\begin{equation}\label{eq:M-facts}
\begin{aligned}
  &\text{(i)}\quad m_g\text{ extends to an entire function, and }m_g(1-z)=-m_g(z);\\
  &\text{(ii)}\quad m_g(n+1)=\frac{(2\pi)^{n+1}}{n!}\int_0^\infty g\,t^n\,\dd t,\qquad
      m_g(-n)=(-2\pi)^{-n}g^{(n)}(0)\qquad(n=0,1,2,\dots);\\
  &\text{(iii)}\quad \abs{m_g(z)}=\abs{m_g(1-z)}\le\frac{(2\pi)^{\Re z}}{\abs{\Gamma(z)}}\int_0^\infty\abs g\,t^{\Re z-1}\,\dd t
      \qquad(\Re z>0).
\end{aligned}
\end{equation}
For (i), the Laplace transform $F(\beta):=\int_0^\infty g\,e^{-2\pi\beta t}\,\dd t$ obeys $F(\beta)=-\beta^{-1}F(1/\beta)$,
by $\hat g=-g$ and \eqref{eq:2-fourier}. For $0<\Re z<1$ its Mellin transform in $\beta$ is
$\int_0^\infty F(\beta)\beta^{z-1}\dd\beta=m_g(1-z)/(2\sin\pi z)$, by the reflection formula for $\Gamma$, so the
substitution $\beta\to1/\beta$ turns the functional equation of $F$ into $m_g(1-z)=-m_g(z)$ on the strip, and this
relation continues $m_g$ to the whole plane. The division by $\Gamma$ is what makes the symbol entire: \emph{in Mellin
space the Fourier transform is the reflection $z\to1-z$}, and a functional is an entire function that is odd under it. The
first formula in (ii) is the definition; the second is the Taylor series of $g$ at $t=0$, where $g$ is analytic by
\cref{lem:A.1}, read through (i). In particular $m_g(0)=g(0)$. Item (iii) is the trivial bound together with (i).

\paragraph{The vacuum.} For $a\ge2$ the Cardy weight has the power series \eqref{eq:2-Vseries} with positive
coefficients,
\begin{equation}\label{eq:M-vac}
  V(t)=\sum_{n\ge2}v_n\,t^n,\qquad v_n=\theta_n\,\frac{(4\pi^2a)^n}{(n!)^2},\qquad
  \theta_n:=1-2\bigl(1-\tfrac1a\bigr)^n+\bigl(1-\tfrac2a\bigr)^n,\qquad\frac1{2a^2}\le\theta_n\le1 ,
\end{equation}
so that $v_n$ is $(4\pi^2a)^n/(n!)^2$ up to a factor polynomial in $a$, which is invisible on the exponential scale.
Integrating $\gvac=-2\pi\int_0^\infty g\,V\,\dd t$ term by term, which is allowed because $M[g]<\infty$, and using (ii),
\begin{equation}\label{eq:M-vacseries}
  \gvac=-\sum_{n\ge2}\theta_n\,\frac{(2\pi a)^n}{n!}\;m_g(n+1) :
\end{equation}
\emph{the vacuum is a series over the integer points $z=n+1$, with positive weights.} Since every term of the series for
$V$ is positive, each absolute moment of $g$ is bounded separately:
\begin{equation}\label{eq:M-vacineq}
\begin{gathered}
  v_n\int_0^\infty\abs g\,t^n\,\dd t\;\le\;M[g]\qquad(n\ge2),\\
  \text{and}\qquad M[g]\le2\int_0^T\abs g\,V\,\dd t\qquad\text{if $g$ is weakly feasible at $T$,}
\end{gathered}
\end{equation}
the second being the half-mass condition \eqref{eq:2-halfmass}. Inserting the first in (iii) at $\Re z=n+1$ gives, for every
integer $n\ge2$,
\begin{equation}\label{eq:M-line}
  \abs{m_g(z)}=\abs{m_g(1-z)}\le\frac{(2\pi)^{n+1}}{v_n\,\abs{\Gamma(z)}}\;M[g]\qquad(\Re z=n+1) .
\end{equation}
Positivity beyond $T$ and the vacuum condition thus bound the symbol, on all the lines $\Re z=n+1$ and on their
reflections, by the Cardy-weighted mass that $g$ carries below $T$. This is the starting point. In the strip of
\cref{sec:6} the vacuum is one point on an edge; here it is a bound on every vertical line through an integer.

\subsection{The obstacle}
\label{app:M.2}

The coefficient in \eqref{eq:M-line} has an exponential order when $n$ and $\Im z$ are of order $c$. We write
\begin{equation}\label{eq:M-scaling}
  z=\tfrac12+c'\,\omega,\qquad \omega=\mu+i\nu ,
\end{equation}
so that $z\to1-z$ is $\omega\to-\omega$, and use Stirling's formula in a form that is uniform in the imaginary part: for
$\Re w\ge0$, $w\ne0$, $\log\abs{\Gamma(\frac12+w)}=\Re[w\log w-w]+\half\log2\pi-\epsilon(w)$ with $0\le\epsilon(w)\le\half\log2$.
(The remainder is minus the real part of a function that is bounded and holomorphic in the half-plane, so it is the
Poisson integral of its boundary values, which are explicit because $\abs{\Gamma(\frac12+is)}^2=\pi/\cosh\pi s$.) The main
term is elementary: for $w=\mu+i\nu$ with $\mu>0$, $\Re[w\log w-w]=\mu\log\mu-\mu-\MD_\mu(\nu)$, where
\begin{equation}\label{eq:M-D}
  \MD_\mu(\nu):=\nu\arctan\frac\nu\mu-\frac\mu2\log\Bigl(1+\frac{\nu^2}{\mu^2}\Bigr)
  =\int_0^\infty\bigl(1-\cos s\nu\bigr)\,e^{-\mu s}\,\frac{\dd s}{s^2}\;\ge0
\end{equation}
is the price, in the exponent, of moving off the real axis along the line $\Re w=\mu$. It is homogeneous of degree one in
$(\mu,\nu)$, and $\MD_\mu(\nu)\to\frac\pi2\abs\nu$ as $\mu\to0^+$. On the line $\Re z=n+1$ we have $n+\half=c'\mu$. The
factor $(n!)^2$ of $1/v_n$, the factor $(2\pi)^n(4\pi^2a)^{-n}$ and the factor $1/\abs{\Gamma(n+1+ic'\nu)}$ contribute
$2c'\mu\log\frac{c'\mu}e-c'\mu\log\frac{\pi c'}6-c'\mu\log\frac{c'\mu}e+c'\MD_\mu(\nu)$ to the logarithm of the coefficient; the
scale $c'$ cancels from the logarithms because $2\pi a=\frac\pi6c'$ exactly, and what remains is $c'$ times the
\emph{obstacle}
\begin{equation}\label{eq:M-B}
  \Mob(\mu+i\nu):=\abs\mu\log\frac{6\abs\mu}{\pi e}+\MD_{\abs\mu}(\nu)\quad(\mu\ne0),\qquad \Mob(i\nu):=\frac\pi2\abs\nu ,
\end{equation}
extended to $\mu<0$ by the reflection. It is continuous on $\CC$ and even in $\mu$ and in $\nu$. Hence, for every
$g\in\Abar$, $g\not\equiv0$,
\begin{equation}\label{eq:M-obst}
  \frac1{c'}\log\frac{\bigabs{m_g(\frac12+c'\omega)}}{M[g]}\;\le\;\Mob(\omega)+O\Bigl(\frac{\log c'}{c'}\Bigr)
  \qquad\text{for }c'\abs\mu\ge\tfrac52 ,
\end{equation}
with an error that is uniform in $\omega$ and in $g$; between two consecutive lines one interpolates with H\"older's
inequality, because the moments are log-convex in their order. In the central strip $c'\abs\mu<\frac52$, that is,
$-2<\Re z<3$, no moment of $g$ is controlled by $M[g]$, and all one has there is a bound
$\abs{m_g}\le C_g\,e^{\pi c'\abs\nu/2}$ with a constant that depends on $g$. In the variable $\omega$ this strip shrinks to
the imaginary axis.

For $\mu>0$,
\begin{equation}\label{eq:M-Bprop}
  \Mob(\omega)=\mu\log\frac{6\mu^2}{\pi e}-\Re\bigl[\omega\log\omega\bigr],\qquad
  2\,\partial_\omega\Mob=\partial_\mu\Mob-i\,\partial_\nu\Mob=\log\frac{6\mu^2}{\pi\omega},\qquad
  \nabla^2\Mob=\frac2\mu .
\end{equation}
Away from the imaginary axis $\Mob$ is smooth and strictly subharmonic. On it $\Mob$ is not smooth:
$\partial_\mu\Mob\to\mp\infty$ as $\mu\to0^\pm$ with $\nu\ne0$ fixed, so the imaginary axis is a concave \emph{ridge}. On the
positive real axis $\Mob(\mu)=\mu\log\frac{6\mu}{\pi e}$. We shall also need the harmonic function in the right half-plane
that is even in $\nu$ and agrees with $\Mob$ on the positive real axis,
\begin{equation}\label{eq:M-UpsR}
  \Re\Ups_R=\Mob-2\,\MD_\mu(\nu),\qquad \Ups_R(\omega):=\omega\log\frac{6\omega}\pi-\omega\qquad(\mu>0).
\end{equation}

\paragraph{The obstacle problem.} The inequality \eqref{eq:M-obst} is not the best that one can say. Since $m_g$ is
entire, the left side is a subharmonic function of $\omega$~\cite{Ransford}. If $\Mob$ were itself subharmonic nothing
would be gained. It is subharmonic on either side of the ridge, but not across it, and a subharmonic function cannot
follow a concave ridge from below. At exponential order the left side of \eqref{eq:M-obst} therefore lies below
\begin{equation}\label{eq:M-minorant}
  \MU:=\sup\bigl\{u:\ u\text{ subharmonic on }\CC,\ u\le\Mob\bigr\} .
\end{equation}
To find the largest subharmonic minorant of a given function is the obstacle problem~\cite{Caffarelli}; in the plane
it is the equilibrium problem of weighted potential theory~\cite{SaffTotik}. We use \eqref{eq:M-minorant} only as a
guide to the formulas of \cref{app:M.3}: \ResMthree\ does not assert that the function defined there is the largest
minorant, and neither of the two main arguments needs it. The general theory says what to expect. The plane splits into
the open \emph{non-contact set} $\MN=\{\MU<\Mob\}$, where $\MU$ is harmonic, since otherwise it could be increased
locally, and the closed \emph{contact set} $\MK=\{\MU=\Mob\}$. Where the obstacle is smooth, $\MU$ is $C^1$ across the
\emph{free boundary} $\partial\MK$, that is, $\nabla\MU=\nabla\Mob$ there, and this is the equation that determines
$\partial\MK$. The solution is even in $\mu$ and in $\nu$. The ridge belongs to $\MN$, because the mean of $\Mob$ over a
small circle centered on it is smaller than the value at the center. Hence $\MK=\MK_+\cup(-\MK_+)$ with $\MK_+$ a closed
subset of $\{\mu>0\}$. Finally, on $\MK$ the Riesz measure $\frac1{2\pi}\nabla^2\MU$ is
$\frac1{2\pi}\nabla^2\Mob\,\dd A=\dd A/(\pi\abs\mu)$, an area measure; this will matter in \cref{app:M.5}.

\paragraph{Why the tip of the contact set is the threshold.} Let $g\in\Abar$ be weakly feasible at $T=c'x_T^2/3$, so
that it carries at least half of its mass below $T$, \eqref{eq:M-vacineq}. On the other hand $g$ is recovered from its
symbol by Mellin inversion along any line $\Re z=n+1$, $g(t)=\frac1{2\pi i}\int m_g(z)\Gamma(z)(2\pi t)^{-z}\dd z$. Insert
$\abs{m_g}\le M[g]\,e^{c'\MU+o(c')}$ and $(2\pi)^{-n-1}\abs{\Gamma(z)}=e^{-c'\Mob+o(c')}/v_n$:
\begin{equation}\label{eq:M-inversion}
  v_n\,t^n\,\abs{g(t)}\;\le\;M[g]\;\frac{c'}{2\pi t}\int_\RR\exp\bigl\{-c'\,\MW(\mu+i\nu)+o(c')\bigr\}\,\dd\nu,\qquad
  \MW:=\Mob-\MU\ge0,\qquad c'\mu=n+\tfrac12 .
\end{equation}
The obstacle has reappeared with the opposite sign, and only the difference is left. For given $t$ we choose for $n$ the
index of the largest term of the series $V(t)=\sum v_nt^n$; this puts the line through the real saddle point of the
inversion integral. The largest term is at $n\approx2\pi\sqrt{at}$, that is, on the line
\begin{equation}\label{eq:M-saddle-line}
  \mu=\frac\pi3\,x,\qquad x=\sqrt{3t/c'} ,
\end{equation}
and $V(t)=e^{o(c')}v_nt^n$. For $t\le T$ the line lies in $\abs\mu\le\frac\pi3x_T$, and integration over $[0,T]$ gives
\begin{equation}\label{eq:M-threshold}
  \int_0^T\abs g\,V\,\dd t\;\le\;M[g]\,e^{-c'\kappa+o(c')}\;\le\;2\,e^{-c'\kappa+o(c')}\int_0^T\abs g\,V\,\dd t,\qquad
  \kappa:=\inf_{\abs\mu\le\pi x_T/3}\MW .
\end{equation}
If the closed strip $\abs\mu\le\frac\pi3x_T$ lies in $\MN$, and $\kappa>0$, this forces $g=0$ for large $c$. If the strip
meets $\MK$ then $\kappa=0$ and nothing follows. The threshold is where the strip first touches the contact set,
$x_T=\frac3\pi\mus$ with $\mus:=\min\{\mu:\mu+i\nu\in\MK_+\}$. We call the points $\pm\mus$ of $\MK$ the \emph{tips}.

\subsection{The free boundary and the solution in closed form}
\label{app:M.3}

It remains to find $\MN$ and $\MU$. We expect $\MN$ to be a simply connected neighborhood of the ridge and write
$\MU=\Re\Ups$ in $\MN$ with $\Ups$ holomorphic. The symmetries make $\Ups'$ odd and real on $\MN\cap\RR$. Instead of
$\Ups'$ we use the pair of variables
\begin{equation}\label{eq:M-bq}
  b\,q=\frac{3\omega}{2\pi},\qquad\frac bq=e^{\Ups'(\omega)},\qquad\text{that is,}\qquad
  b^2=\frac{3\omega}{2\pi}\,e^{\Ups'(\omega)},\quad q^2=\frac{3\omega}{2\pi}\,e^{-\Ups'(\omega)} .
\end{equation}
Here $b^2$ and $q^2$ are holomorphic in $\MN$, and the common sign of $(b,q)$ is fixed by $b,q>0$ on the real segment
$(0,\mus)$. The first variable is an old one: for a symbol $m_g=e^{c'\Ups+o(c')}$ the integrand of the inversion integral
is $\exp\{c'[\Ups(\omega)+\omega\log\frac{c'\omega}{2\pi et}]\}$, and it is stationary at $2\pi t=c'\omega\,e^{\Ups'}$, that
is, at $t=c'b^2/3$. So $b$ is the variable $b$ of \eqref{eq:2-scaling}, evaluated at the saddle point that belongs to
$\omega$. Its partner $q$ will turn out to be $ip$, with $p$ the momentum \eqref{eq:6-p}.

On $\partial\MK_+$ the contact is $C^1$, so $\Ups'=2\partial_\omega\MU=2\partial_\omega\Mob=\log\frac{6\mu^2}{\pi\omega}$ by
\eqref{eq:M-Bprop}. In \eqref{eq:M-bq} this gives $b^2=(3\mu/\pi)^2$, hence the free-boundary condition
\begin{equation}\tag{R}\label{hyp:R}
  b=\frac{3\mu}\pi,\qquad q=\frac\omega{2\mu}=\frac12+\frac{i\nu}{2\mu};\qquad\text{in particular}\quad
  \Im b=0,\quad \Re q=\tfrac12\quad\text{on }\partial\MK_+ .
\end{equation}
In the variables $(b,q)$ the nonlinear condition $\nabla\MU=\nabla\Mob$ has become linear; this is the reason for
introducing them. The reflection $\omega\to-\omega$ acts as $(b,q)\to(iq,ib)$, because $\Ups'$ is odd, so that on
$-\partial\MK_+$ the conditions read $\Re q=0$ and $\abs{\Im b}=\half$: there $b$ is on an edge of the strip $\strip$. The
reflections in the lines $\Re q=0$ and $\Re q=\half$ compose to the translation $q\to q+1$, those in $\Im b=0$ and
$\Im b=\half$ to $b\to b+i$, which suggests a solution that is periodic in both variables. Consider therefore the function
$\cosh2\pi b+\cos2\pi q$ of $\omega$. It is an entire function of $(b^2,q^2)$, hence holomorphic in $\MN$. It is even,
because $\omega\to-\omega$ exchanges its two terms. It is real on $\partial\MK_+$ by \hyp{R}, since
$\cos2\pi(\half+is)=-\cosh2\pi s$, and hence on all of $\partial\MN$. So its imaginary part is harmonic in $\MN$ and
vanishes on $\partial\MN$. Assume that it is bounded in $\MN$. Then it is a real constant, by the Phragm\'en--Lindel\"of
principle, and the value at $\omega=0$, where $b=q=0$, is $2$:
\begin{equation}\label{eq:M-curve}
\begin{gathered}
  \cosh2\pi b+\cos2\pi q=2,\qquad\text{equivalently}\qquad \sin\pi q=\sinh\pi b,\\
  \text{that is,}\qquad\sinh\pi b=i\sinh\pi p\quad\text{with }q=ip ,
\end{gathered}
\end{equation}
since the first equation is $\sinh^2\pi b=\sin^2\pi q$ and the sign is fixed on $(0,\mus)$. This is the curve
\eqref{eq:1-curve}. (With $b=iX$ the first form is the energy surface $\cos2\pi X+\cos2\pi q=2$ of Harper's
equation~\cite{Harper}.) Together with $bq=3\omega/2\pi$ it determines $b$ and $q$, hence $\Ups'$, as functions of
$\omega$:
\begin{equation}\label{eq:M-param}
  \omega=\frac{2\pi}3\,b\,q,\qquad \sin\pi q=\sinh\pi b ;\qquad\text{on }(0,\mus):\quad
  q=\frac1\pi\arcsin\sinh\pi b=i\,p(b) .
\end{equation}
\emph{The rescaled Mellin variable is the action, $\omega=\frac{2\pi i}3\,b\,p$, on the curve of \cref{sec:4}.}

At the tip, where $\partial\MK_+$ crosses the real axis, \hyp{R} gives $q=\half$ and $b=3\mus/\pi$, and
\eqref{eq:M-curve} becomes $\sinh\pi b=1$:
\begin{equation}\label{eq:M-tip}
  b=\xs,\qquad \mus=\frac\pi3\,\xs=\tfrac13\arcsinh1=0.2937912,\qquad x_T=\frac3\pi\mus=\xs ,
\end{equation}
which is the threshold of \cref{thm:1}. On the rest of $\partial\MK_+$, \hyp{R} and
$\sin(\frac\pi2+\frac{i\pi\nu}{2\mu})=\cosh\frac{\pi\nu}{2\mu}$ turn \eqref{eq:M-curve} into
$\cosh\frac{\pi\nu}{2\mu}=\sinh3\mu$:
\begin{equation}\label{eq:M-fb}
  \nu=\pm\nus(\mu),\qquad \nus(\mu):=\frac{2\mu}\pi\,\arccosh\bigl(\sinh3\mu\bigr),\qquad\mu\ge\mus ,
\end{equation}
where $\sinh3\mu\ge1$ precisely for $\mu\ge\mus$. Since $\nus(\mu)=\frac6\pi\mu^2-O(\mu e^{-6\mu})$, the non-contact set is a
parabolic neighborhood of the ridge (\cref{fig:11}(a)).

\begin{figure}[t]
\centering
\figfile[\linewidth]{fig11_mellin}
\caption{(a) The plane of $\omega=\mu+i\nu$, where $z=\frac12+c'\omega$ is the Mellin variable. Shaded: the contact set
$\MK=\MK_+\cup(-\MK_+)$. White: the non-contact set $\MN$, which contains the ridge $\mu=0$ of the obstacle. Solid: the free
boundary $\nu=\pm\nus(\abs\mu)$ of \eqref{eq:M-fb}, whose four arcs meet at the tips $\omega=\pm\mus$. On the arcs $b$ is real
and $\ge\xs$, and $\nus(\frac\pi3x)=\frac\pi2x\,\rhoU(x)$: the width of the contact set is the zero law. Dashed: the four
creases of \cref{app:M.5}, which leave the tips at $45^\circ$ and carry the skeleton. (b) The obstacle $\Mob$ and the
solution $\MU$ on the positive real axis. They agree, with their first derivatives, for $\mu\ge\mus$; for $0\le\mu<\mus$
one has $\MW=\Mob-\MU>0$, and $\MU(0)=U_0$, the value of the extremal potential of \cref{sec:6} at $b=0$.}
\label{fig:11}
\end{figure}

\paragraph{The solution.} We now turn the outcome into definitions, which no longer depend on how they were found. Let
\begin{equation}\label{eq:M-U}
\begin{gathered}
  \MK_+:=\{\mu+i\nu:\ \mu\ge\mus,\ \abs\nu\le\nus(\mu)\},\qquad \MK:=\MK_+\cup(-\MK_+),\qquad\MN:=\CC\setminus\MK,\\
  \Ups'(\omega):=\log\frac bq\quad(\omega\in\MN),\qquad
  \MU(\omega):=\Mob(\mus)+\Re\int_{\mus}^\omega\Ups'(\omega')\,\dd\omega'\quad(\omega\in\MN),\qquad \MU:=\Mob\ \text{ on }\MK ,
\end{gathered}
\end{equation}
where $b$ and $q$ are the functions on $\MN$ defined by \eqref{eq:M-param} through part (i) below, the logarithm is the
branch that vanishes at $\omega=0$, and the path runs in $\MN\cup\{\mus\}$. Put $\MW:=\Mob-\MU$ and
$\MN^+:=\MN\cap\{\nu>0\}$.

\begin{ideabox}[title={Result M3 (the solution of the obstacle problem)}]\phantomsection\label{res:M3}
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.2em,itemsep=1pt,topsep=2pt]
\item The relations \eqref{eq:M-param} define $q$ as a conformal map of $\MN^+$ onto the half-strip
$\{0<\Re q<\half,\ \Im q>0\}$. The side $\Re q=\half$ corresponds to the upper arc of $\partial\MK_+$, the side $\Im q=0$
to the segment $(0,\mus)$, and the side $\Re q=0$ to $(-\mus,0]$ followed by the upper arc of $-\partial\MK_+$. Hence $\Ups'$
is single-valued and holomorphic on $\MN$.
\item $\MU\in C^1(\CC)$; it is harmonic in $\MN$, equal to $\Mob$ on $\MK$, and even in $\mu$ and in $\nu$.
\item $\MW>0$ in $\MN$, and $\MW(\mu+i\nu)\le\MW(0)+\frac\pi2\abs\nu$ on $\CC$. In the quadrant $\mu,\nu>0$ of $\MN$,
$\partial_\mu\MW=2\log\frac{3\mu}{\pi\abs b}<0$ and $\partial_\nu\MW=2\arg b>0$.
\item $\nus$ is increasing on $[\mus,\infty)$ with $\nus(\mus)=0$; hence $\{\abs\mu\le\mu_1\}\subset\MN$ for every
$\mu_1<\mus$; and $\mu^2\le\frac\pi3(1+\abs\nu)$ on the closure of $\MN$.
\end{enumerate}
\end{ideabox}

\noindent These properties are checked directly from the formulas, without the assumption made above. For (i) one
uniformizes with $\sigma:=\sinh^2\pi b=\sin^2\pi q$, so that
$\omega=f(\sigma):=\frac2{3\pi}\arcsinh\sqrt\sigma\,\arcsin\sqrt\sigma$ on the plane cut along $(-\infty,-1]$ and $[1,\infty)$.
For $\Im\sigma>0$ one finds $0<\arg b<\arg\sqrt\sigma<\arg q<\frac\pi2$, from $\tanh s<s<\tan s$, and then $\Re f'>0$;
since a half-plane is convex, $f$ is one-to-one there, by the criterion of Noshiro and
Warschawski~\cite[Thm.~2.16]{Duren}. The upper and lower lips of the cut $[1,\infty)$ go to the two arcs of
$\partial\MK_+$, those of $(-\infty,-1]$ to the arcs of $-\partial\MK_+$, and the branch points $\sigma=\pm1$ to the tips. For
(ii), on the upper arc $b=3\mu/\pi$ is real and $q=3\omega/(2\pi b)$, so that
$\Re\Ups'=\log\frac b{\abs q}=\log\frac{6\mu^2}{\pi\abs\omega}=\partial_\mu\Mob$ and $-\Im\Ups'=\arg q=\arg\omega=\partial_\nu\Mob$,
by \eqref{eq:M-Bprop}; the values match as well, because the path in \eqref{eq:M-U} may run along the arc, where
$\Re(2\partial_\omega\Mob\,\dd\omega)=\dd\Mob$. For (iii), \eqref{eq:M-Bprop} and $\abs\omega=\frac{2\pi}3\abs b\abs q$ give
the two components of $\nabla\MW$; their signs follow from $3\mu=2\pi\Re(bq)<\pi\abs b$ and from the ordering of the
arguments; and one integrates, along horizontal segments from the free boundary, where $\MW=0$, and up the imaginary
axis from $\omega=0$, where $\partial_\nu\MW\le\frac\pi2$. Part (iv) is elementary:
$\frac6\pi\mu^2-\frac{2\arcsinh1}\pi\mu\le\nus(\mu)<\frac6\pi\mu^2$.

The value at the origin is that of the extremal potential of \cref{sec:6}. Indeed, by (iii) on the real segment, where
$3\mu=2\pi bq$, and by an integration by parts,
\begin{equation}\label{eq:M-parts}
  \MW(0)=-\int_0^{\xs}2\log(2q)\,\dd\Bigl[\frac{2\pi}3\,b\,q\Bigr]=\frac{4\pi}3\int_0^{\xs}b\,\dd q
  =\frac{2\pi}3\int_0^{\xs}\bigl(1-2q\bigr)\,\dd b=-U_0=0.388748 ,
\end{equation}
by \eqref{eq:6-deficit} at $x=0$, with $q=\frac1\pi\arcsin\sinh\pi b$; the boundary terms vanish because $2q=1$ at the tip and $bq\log q\to0$ at the origin.
So $\MU(0)=U_0=U(0)$. The reason is that $m_g(0)=g(0)$: at exponential order $e^{c'U(0)}$ bounds $\abs{g(0)}/M[g]$ by
\cref{prop:6.4} at $b=0$, and $e^{c'\MU(0)}$ does so by \eqref{eq:M-tails} below at $\omega=-1/2c'$, the point $z=0$; the two
bounds are the same. \Cref{fig:11}(b) shows $\Mob$ and $\MU$ on the real axis.

\subsection{Optimality at finite central charge}
\label{app:M.4}

\ResMone\ is the argument of \cref{app:M.2} carried out at finite $c$. Fix $x_T=\xs-\delta$ and $T=c'x_T^2/3$. For
$g\in\Abar$, $g\not\equiv0$, let
\begin{equation}\label{eq:M-pg}
  p_g(\omega):=\log\bigabs{m_g\bigl(\tfrac12+c'\omega\bigr)}-\log\bigl(2\pi M[g]\bigr) ,
\end{equation}
which is subharmonic on $\CC$ and even in $\mu$ and in $\nu$. There are four steps.

\smallskip\noindent\emph{The obstacle inequality.} For all $c\ge25$ and $c'\abs\mu\ge\frac52$,
$p_g(\omega)\le c'\Mob(\omega)+k_a$ with $k_a=\half\log(32\pi^2a^5)+\frac1{10}<\frac52\log c'$. What matters is that $k_a$
depends neither on $g$ nor on $\omega$, because the inequality is used on all of $\partial\MN$ and on entire vertical
lines. It follows from \eqref{eq:M-line} with $\theta_n\ge1/2a^2$ uniformly in $n$; from H\"older's inequality between
consecutive integer moments, where replacing a chord of the concave function $n\mapsto\log[(4\pi^2a)^n/(n!)^2]$ by the
function costs at most $\frac1{10}$; and from Stirling's formula with its exact remainder. Because $c'=12a$ exactly, the
main terms are exactly $c'\Mob(\omega)$. Had we scaled with $c$ instead of $c'$, the right side would contain the
additional term $\abs\mu\,c\log\frac c{c-1}$, which is unbounded on $\MN$.

\smallskip\noindent\emph{The comparison step.} The obstacle inequality is sharp on $\partial\MN$ and wasteful inside
$\MN$. Let $\Omega$ be an open set whose closure lies in a parabolic region $\{\mu^2\le A_0^2(1+\abs\nu)\}$ and whose
boundary lies in $\{c'\abs\mu\ge\frac52\}$, and let $u$ be continuous on the closure of $\Omega$ and harmonic in $\Omega$,
with $u\ge\Mob$ on $\partial\Omega$ and $\Mob-u$ bounded by a power of $1+\abs\omega$. Then
\begin{equation}\label{eq:M-compare}
  p_g\le c'\,u+k_a\qquad\text{on the closure of }\Omega ,
\end{equation}
with the same $k_a$. Indeed $p_g-c'u-k_a$ is subharmonic in $\Omega$ and non-positive on $\partial\Omega$. In an unbounded
domain one needs in addition that it does not grow too fast. This is the one delicate point, because the only bound
available in the central strip is not uniform in $g$ and $c$. No uniform bound is needed: the difference grows at most
like a power of $\abs\nu$, with a constant that depends on $g$ and $c$, whereas a parabolic region is so thin that it
carries positive harmonic functions that grow like $\exp(C\sqrt{\abs\nu})$, for instance
$\Re\cosh(\frac\pi{2A_0}\sqrt{1-i\omega})+\Re\cosh(\frac\pi{2A_0}\sqrt{1+i\omega})$ plus a constant. Subtracting
$\varepsilon$ times such a function, applying the maximum principle on bounded pieces and letting $\varepsilon\to0$ gives
\eqref{eq:M-compare}, and the constant has disappeared from the conclusion. The first application is $\Omega=\MN$,
$u=\MU$, which is allowed by \ResMthree\ (ii)--(iv), since $c'\abs\mu\ge24\mus>7$ on $\partial\MN$:
\begin{equation}\label{eq:M-tails}
  p_g(\omega)-c'\,\Mob(\omega)\;\le\;-c'\,\MW(\omega)+k_a\qquad(\omega\in\MN).
\end{equation}
The second application controls the tails of the inversion integral, where \ResMthree\ gives no lower bound for $\MW$:
with $\Omega$ the strip $\abs\mu<1$ and $u$ the harmonic function in it that equals $\Mob$ on its two sides, which is
explicit by \eqref{eq:M-D}, one finds $p_g-c'\Mob\le k_a+c'(0.68-\eta_1\log\abs\nu)$ for $\abs\nu\ge1$ and
$\abs\mu\le\frac\pi3x_T$, with $\eta_1=(1-\frac\pi3x_T)/\cosh1$ and $c'\eta_1>7$: along a vertical line the integrand of
\eqref{eq:M-inversion} decays like a power $\abs\nu^{-c'\eta_1}$, uniformly in $g$ and $c$.

\smallskip\noindent\emph{Mellin inversion at the saddle point.} Put $z=\frac12+c'(\xi+i\nu)$ in the inversion formula,
multiply by $V(t)\le e^{4\pi\sqrt{at}}$ and use Stirling's formula. The exponents that depend on $t$ combine to a function
of $\xi$ that is non-positive and vanishes, to second order, only on the line \eqref{eq:M-saddle-line},
$\xi=\xi_t:=\frac\pi3x$; on this line they cancel, and
\begin{equation}\label{eq:M-kappaM}
  V(t)\,\abs{g(t)}\;\le\;M[g]\;c'\,t^{-1/2}\,I_g(\xi_t),\qquad
  I_g(\xi):=\int_\RR e^{(p_g-c'\Mob)(\xi+i\nu)}\,\dd\nu .
\end{equation}
The times $0<t\le T$ correspond to $0<\xi_t\le\frac\pi3x_T$, and integrating $t^{-1/2}$ over this range gives
$\int_0^T\abs g\,V\,\dd t\le M[g]\,c'\,2\sqrt T\,\sup_{0<\xi\le\pi x_T/3}I_g(\xi)$.

\smallskip\noindent\emph{Conclusion.} By \ResMthree\ (iv) the closed strip $\abs\mu\le\frac\pi3x_T$ lies in $\MN$. This is
the only place where $x_T<\xs$ is used, and it is essential: for $x_T=\xs$ the line $\Re\omega=\frac\pi3x_T$ touches the
contact set at its tip, where $\MW=0$, and there is no exponential gain. Split $I_g$ at $\abs\nu=R_0:=e^{1.68/\eta_1}$. On
the inner part \eqref{eq:M-tails} gives an integrand at most $e^{k_a-c'\kappa}$, where $\kappa$ is the minimum of $\MW$ over
the box $\{\abs\mu\le\frac\pi3x_T,\ \abs\nu\le R_0\}$; by the signs in \ResMthree\ (iii) this minimum is attained on the real
axis at $\mu=\frac\pi3x_T$, so that $\kappa=\kappa_M(\delta)$. The outer part is smaller than $e^{-c'}$. With
$e^{k_a}<c'^{5/2}$ this is \eqref{eq:M-conc}, with $C_\delta=3R_0$. Finally, if $g$ is weakly feasible at $T$, the half-mass
condition and \eqref{eq:M-conc} give $0<M[g]\le2C_\delta\,c'^4e^{-\kappa_M(\delta)c'}M[g]$, which is absurd for large $c$.

\paragraph{Comparison with \cref{thm:3}.} The inequality \eqref{eq:M-conc} is \eqref{eq:6-conc} with $\kappa_M$ in place
of the Cardy deficit $\kappa$. The two agree at the far end, $\kappa_M(\xs)=\MW(0)=-U_0=\kappa(\xs)$ by \eqref{eq:M-parts},
and nowhere else: $\kappa_M(\delta)=0.00037$, $0.0090$, $0.036$, $0.16$ at $\delta=0.01$, $0.05$, $0.1$, $0.2$, against
$\kappa(\delta)=0.0026$, $0.029$, $0.083$, $0.23$. Near the tip the contact of $\MU$ with the obstacle is $C^1$, so that
$\MW$ vanishes to second order, $\MW(\mu)=(\mus-\mu)^2/\mus+\dots$, and
\begin{equation}\label{eq:M-kM-tip}
  \kappa_M(\delta)=\frac{\pi}{3\xs}\,\delta^2\bigl(1+O(\delta)\bigr)=3.73\,\delta^2\bigl(1+O(\delta)\bigr),
  \qquad\text{whereas}\qquad\kappa(\delta)=2.650\,\delta^{3/2}\bigl(1+O(\delta)\bigr) .
\end{equation}
The reason is that \eqref{eq:M-kappaM} integrates along the line through the saddle point of the Cardy weight, and not
through that of the functional. Used in this way the Mellin argument gives the first statement of \cref{thm:4}, but in
place of \eqref{eq:6-rate} only a correction of order $c'^{1/2}\log^{1/2}c'$. At exponential order nothing is lost on
the other line: for $0\le x\le\xs$,
\begin{equation}\label{eq:M-legendre}
  U(x)=\MU(\mu)+\mu\log\frac{3\mu}{2\pi e\,x^2}\qquad\text{at }\mu=\frac{2\pi}3\,x\,q(x),\quad\text{the point where }b(\mu)=x ,
\end{equation}
because the right side is stationary in $\mu$ there, both sides equal $U_0$ at $x=0$, and both have the derivative
$-\frac{4\pi}3q(x)$. The right side is the exponent that \eqref{eq:M-inversion} produces for $\abs{g(t)}/M[g]$ on the line
through this point, where the integrand is largest at $\nu=0$ because $\partial_\nu\MW>0$. So that line gives
$\abs g\,V\le M[g]\,e^{c'(U+L)+o(c')}$, that is, the Cardy deficit $\kappa(\delta)$ itself. We have not carried this out at
finite $c$.

\subsection{The skeleton of the contact set}
\label{app:M.5}

A functional that attains the bound should have a symbol $m_c$ with
$\frac1{c'}\log\abs{m_c(\frac12+c'\omega)}\to\MU(\omega)$ in $\MN$. Its zeros then have the density $c'$ times the Riesz
measure of the limit. For the limit $\MU$ itself this is the area measure $\dd A/(\pi\abs\mu)$ on $\MK$, which gives no
control of signs: for $t\ge T$ with $x_T>\xs$ the line \eqref{eq:M-saddle-line} crosses $\MK_+$, and the integrand of
\eqref{eq:M-inversion} has constant modulus along the whole segment $\abs\nu\le\nus(\mu)$. But only the values of the
limit in $\MN$ enter the threshold. Inside $\MK$ the measure may therefore be rearranged, as long as the potential stays
below $\Mob$ and is unchanged in $\MN$. Swept together as far as possible it becomes the potential-theoretic
\emph{skeleton} of $\MK$~\cite{GustafssonSakai}, a positive measure on four arcs, the \emph{creases}, which issue from
the tips (\cref{fig:11}(a)). It is constructed by continuing $\MU$ harmonically from $\MN$ across the free boundary into
$\MK_+$, on a second sheet, until the continuation meets the elementary harmonic function $\Re\Ups_R$ of
\eqref{eq:M-UpsR} along a curve.

\paragraph{The second sheet.} By \ResMthree\ (i), $q$ is a conformal coordinate on $\MN^+$ in which the upper arc of the
free boundary is the line $\Re q=\half$. We move the origin to the image of the tip and put
\begin{equation}\label{eq:M-theta}
\begin{gathered}
  q=\frac12+\frac{i\vartheta}\pi,\qquad\text{so that}\qquad \sinh\pi b=\cosh\vartheta,\qquad
  \omega(\vartheta)=\frac13\,\pi b(\vartheta)\Bigl(1+\frac{2i\vartheta}\pi\Bigr),\\
  \pi b(\vartheta)=\vartheta+\log\frac{1+e^{-2\vartheta}+\sqrt{1+6e^{-2\vartheta}+e^{-4\vartheta}}}2 .
\end{gathered}
\end{equation}
Then $\MN^+$ is the half-strip $\{\Re\vartheta>0,\ 0<\Im\vartheta<\frac\pi2\}$, the upper arc of $\partial\MK_+$ is the ray
$\vartheta\ge0$, the tip is $\vartheta=0$, and the second sheet is reached through $\Im\vartheta<0$. The map
$\omega(\vartheta)$ is holomorphic in a full neighborhood of $\vartheta=0$, which is the reason for using $\vartheta$. On the
negative imaginary axis $q>\half$ and $b=\frac1\pi\arcsinh\sin\pi q$ are real, and $\omega=\frac23\,q\arcsinh(\sin\pi q)$
continues the segment $(0,\mus)$ beyond the tip into $\MK$. It increases up to a maximum $0.34818$ at
$\vartheta=-i\vartheta_f$, $\vartheta_f=0.520059$, and then decreases: the map folds the real axis back onto itself. We
therefore work in the convex wedge
\begin{equation}\label{eq:M-wedge}
  \bigl\{\vartheta=\vartheta_1-i\vartheta_2:\ \vartheta_1>0,\ 0<\vartheta_2<\vartheta_1+\vartheta_f\bigr\} ,
\end{equation}
whose closure avoids the branch points of $b(\vartheta)$, because $\vartheta_f<\frac\pi2-\arcsinh1=0.689$.

\paragraph{The step checked in interval arithmetic.} Functions of $\vartheta$ on the second sheet are single-valued functions of
$\omega$ in the upper half of $\MK_+$ only if $\omega(\vartheta)$ covers it exactly once. It does: on the closed wedge,
the corner $-i\vartheta_f$ excepted,
\begin{equation}\label{eq:M-univalent}
  \Re\bigl(e^{-i\pi/8}\,\omega'(\vartheta)\bigr)>0,\qquad\text{and}\qquad\Im\omega<0<\Re\omega\quad\text{on the slanted side }
  \vartheta_2=\vartheta_1+\vartheta_f .
\end{equation}
The first statement gives injectivity, again by the criterion of Noshiro and Warschawski; the rotation is needed because
$\arg\omega'$ fills the interval $(-\frac\pi4,\frac\pi2)$ on the wedge. The second shows that the image contains the
open upper half of $\MK_+$. The corner, where $\omega'$ vanishes, and the far part $\vartheta_1\ge3$ of the wedge are treated
by hand, with $\omega''$ in a cone around $\omega''(-i\vartheta_f)=0.491>0$ at the corner and with
$\pi b(\vartheta)-\vartheta=O(e^{-2\vartheta_1})$ in the far part. What remains is compact, and it was checked in interval
(ball) arithmetic with Arb~\cite{Arb}, by bisection into boxes on each of which the enclosure of the function satisfies
the inequality:
\begin{itemize}[leftmargin=2.6em,itemsep=0pt,topsep=3pt]
\item[\hypdef{I0}] the location of the fold, $\abs{\vartheta_f-0.52005880928750816}\le10^{-30}$;
\item[\hypdef{I1}] $\abs{\arg\omega''}<\frac\pi8$ on the square $\{0\le\vartheta_1\le\frac1{20},\ \abs{\vartheta_2-\vartheta_f}\le\frac1{20}\}$;
\item[\hypdef{I2}] the first inequality of \eqref{eq:M-univalent} on the part $\vartheta_1\le3$ of the closed wedge outside
that square ($2380$ boxes; the smallest value is about $0.009$, next to the corner);
\item[\hypdef{I3}] the second inequality of \eqref{eq:M-univalent} for $\frac1{20}\le\vartheta_1\le3$, and $\Re\omega\ge0.287$
for $\vartheta_1\le\frac1{20}$;
\item[\hypdef{I4}] $\abs{\lambda'(\zeta)}\le2.18$ for $\abs{\Re\zeta},\abs{\Im\zeta}\le e^{-6}$, where
$\lambda(\zeta)=\log\frac12[(1-\zeta)+\sqrt{1-6\zeta+\zeta^2}]$ is the logarithm in \eqref{eq:M-theta} at
$\zeta=-e^{-2\vartheta}$; this bounds the far part.
\end{itemize}
This is the only place in this appendix where a computer is used. A consequence that is needed below is that
$\arg b\in(-\pi,0)$ and $\arg q\in(0,\frac\pi2)$ on the wedge, with $\arg\omega=\arg b+\arg q$ for the principal values.

\paragraph{The continuation and the crease.} On the first sheet $\Ups'=\log(b/q)$, and this expression is holomorphic in
$\vartheta$ across the ray $\vartheta\ge0$ and in the whole wedge. Hence
\begin{equation}\label{eq:M-E}
\begin{gathered}
  \Ups_{II}'(\omega):=\log\frac bq\Big|_{\vartheta(\omega)},\qquad \Ups_{II}(\mus):=\Mob(\mus);\\
  \ME:=\Ups_R-\Ups_{II},\qquad \ME'=2\log(2q),\qquad \ME(\omega)=\frac{(\omega-\mus)^2}{\mus}+O\bigl(\abs{\omega-\mus}^3\bigr)
\end{gathered}
\end{equation}
defines the analytic continuation $\Ups_{II}$ of $\Ups$ across the upper arc, and through a full neighborhood of the
tip, to the upper half of $\MK_+$. We have used $\Ups_R'=\log\frac{6\omega}\pi=\log(2b)+\log(2q)$; the double zero of $\ME$
at the tip is the $C^1$ contact of $\MU$ with the obstacle. There are now two harmonic functions in the upper half of
$\MK_+$, and their order follows from the signs of the arguments. First,
$\partial_\nu(\Mob-\Re\Ups_{II})=2\arg b<0$, and $\Re\Ups_{II}=\Mob$ on the arc, so that $\Re\Ups_{II}<\Mob$ below it.
Next, $\partial_\mu\Re\ME=2\log\abs{2q}>0$ and $\partial_\nu\Re\ME=-2\arg q<0$, while $\Re\ME=-2\MD_\mu(\nus(\mu))<0$ on the
arc and $\Re\ME=\Mob-\Re\Ups_{II}>0$ on the real ray $\mu>\mus$. On each vertical segment $\Re\ME$ therefore has exactly
one zero, and
\begin{equation}\label{eq:M-crease}
  \MC:=\{\Re\ME=0\}=\{\nu=\nu_c(\mu),\ \mu>\mus\},\qquad 0<\nu_c<\nus,\qquad
  \nu_c'(\mu)=\frac{\log\abs{2q}}{\arg q}>0 ,
\end{equation}
is a real-analytic curve, the \emph{crease}, which leaves the tip at $45^\circ$. Heuristically it is an anti-Stokes line:
a function that behaves like $e^{c'\Ups_R}+e^{c'\Ups_{II}}$ is dominated by one exponential on either side of it and has its
zeros along it. We define
\begin{equation}\label{eq:M-Usk}
  \Usk:=\begin{cases}
  \MU&\text{on the closure of }\MN,\\
  \max\bigl(\Re\Ups_R,\Re\Ups_{II}\bigr)&\text{on the upper half of }\MK_+,\\
  \Mob&\text{on the real rays }\abs\mu\ge\mus ,
  \end{cases}
\end{equation}
continued by the symmetries under $\omega\to-\omega$ and $\omega\to\bar\omega$. Thus $\Usk=\Re\Ups_{II}$ between the
free boundary and the crease, and $\Usk=\Re\Ups_R$ between the crease and the real axis. It is continuous, harmonic away
from the four creases, subharmonic on $\CC$, equal to $\MU$ on $\MN$, and
\begin{equation}\label{eq:M-Uskbound}
  \Usk\le\Mob\quad\text{on }\CC,\qquad\text{with equality on the real rays }\abs\mu\ge\mus .
\end{equation}

\paragraph{The skeleton, parametrized by its mass.} The Riesz measure $\nsk:=\frac1{2\pi}\nabla^2\Usk$ is carried by
$\MC$ and its three mirror images. Across $\MC$ the gradient jumps by $\abs{\ME'}$, so that
$\dd\nsk=\frac1\pi\abs{\log2q}\,\dd s$ with respect to arc length, a positive real-analytic density that vanishes linearly
at the tip. Along $\MC$ the function $\ME$ is purely imaginary, and the mass between the tip and a point of $\MC$ is
$\Im\ME/2\pi$ there; it is unbounded. We parametrize the crease by its mass,
\begin{equation}\label{eq:M-mass}
  \ME\bigl(\omega_*(\tau)\bigr)=2\pi i\,\tau\quad(\tau\ge0),\qquad
  \omega_*(\tau)=\mus+e^{i\pi/4}\sqrt{2\pi\mus\,\tau}\,\bigl(1+O(\sqrt\tau)\bigr)\quad(\tau\to0) ,
\end{equation}
so that $\nsk$ is the sum of the images of $\dd\tau$ under $\pm\omega_*$ and $\pm\bar\omega_*$. Jensen's formula and
\eqref{eq:M-Uskbound} give $\nsk\{\abs\omega\le r\}\le Cr\log(2+r)$, hence $\int_0^\infty\abs{\omega_*}^{-2}\dd\tau<\infty$, and
then, as in the Hadamard factorization, $\Usk$ is the potential of its mass. With the normalization $\Usk(0)=\MU(0)=U_0$,
\begin{equation}\label{eq:M-potential}
  \Usk(\omega)-U_0=\int_0^\infty\log\Bigabs{\Bigl(1-\frac{\omega^2}{\omega_*(\tau)^2}\Bigr)
  \Bigl(1-\frac{\omega^2}{\bar\omega_*(\tau)^2}\Bigr)}\,\dd\tau .
\end{equation}
The error estimates of \cref{app:M.7} need quantitative control of $\omega_*(\tau)$, which one obtains from
$\omega_*'=i\pi/\log(2q)$ and from the explicit far field $\pi b\approx\vartheta$. The crease does not return to the real
axis at any scale: $\Im\omega_*\ge a_0(\Re\omega_*-\mus)$ with a constant $a_0\in(0,1]$, so that the creases stay out of the
sector $\{\mu>\mus,\ \abs\nu<a_0(\mu-\mus)\}$. Chord and arc are comparable, because $\arg\omega_*'\in(0,\frac\pi2)$. Up to
a constant factor the speed $\abs{\omega_*'}$ decreases, but no faster than $\tau^{-1/2}$, which is its rate at the
square-root tip; and $\abs{\omega_*''}\le C\abs{\omega_*'}/\tau$. The delicate point is the far field, where the margin in
the monotonicity of the speed is only logarithmic.

\subsection{The functional}
\label{app:M.6}

\paragraph{What is wanted, and what the skeleton provides.} Suppose that a symbol has
$\abs{m_c(\frac12+c'\omega)}=e^{c'\MH(\omega)+o(c')}$. By Stirling's formula the modulus of the integrand
$\Gamma(z)m_c(z)(2\pi t)^{-z}$ of the inversion integral at $z=\frac12+c'(\mu+i\nu)$ is
\begin{equation}\label{eq:M-wanted}
  \exp\bigl\{c'\bigl[\Mob(\mu)+\MH(\mu)-\mu\log(4x^2)\bigr]+o(c')\bigr\}\times
  \exp\bigl\{c'\bigl[\MH(\mu+i\nu)-\MH(\mu)-\MD_\mu(\nu)\bigr]\bigr\},\qquad x=\sqrt{3t/c'} ,
\end{equation}
its value at the real point times its variation along the vertical line, and the weights of the vacuum series
\eqref{eq:M-vacseries} are $e^{-c'\Mob(\mu)+o(c')}$ at $n=c'\mu$. Thus the saddle point lies where
$(\Mob+\MH)'(\mu)=\log(4x^2)$; it dominates its line if $\MH(\mu+i\nu)-\MH(\mu)<\MD_\mu(\nu)$ for $\nu\ne0$; and the vacuum
series is dominated by the terms near the maximum of $\MH-\Mob$ on the positive axis.

The potential $\MH_*:=\Usk-U_0$ of the skeleton has all this, but only marginally. In the sector free of creases it is
$\Re\Ups_R$ plus a constant, so that the saddle point for $x\ge\xs$ is on the real axis at $\mu=\frac\pi3x$. By
\eqref{eq:M-Uskbound}, $\MH_*(\mu+i\nu)-\MH_*(\mu)\le\Mob(\mu+i\nu)-\Mob(\mu)=\MD_\mu(\nu)$ for $\mu\ge\mus$, but with
equality on the free boundary: there $\Re\Ups_{II}$ contributes a conjugate pair of saddle points, which by \hyp{R} lie on
the same vertical line as the real one and have the same modulus, because $\Usk=\Mob$ at all three points. Hence
$V\,g=r_1+r_2\cos\Theta(t)$ with a real phase $\Theta$ and positive amplitudes $r_1$, $r_2$ of the same exponential order, whose
ratio
is not determined at this accuracy: the sign of $g$ is marginal. Likewise $\MH_*-\Mob=-U_0$ is constant on $[\mus,\infty)$,
so that all the terms of the vacuum series beyond the tip, which have the wrong sign, are of the same exponential order
as the largest one.

\paragraph{The design.} A margin is created by thinning and dilating the skeleton. Let $\eta\in(0,1)$, $\eta_v>0$ and
$\iota>0$ be three small parameters, and put
\begin{equation}\label{eq:M-design}
\begin{gathered}
  \MH(\omega):=(1-\eta)\,\lambda\,\MH_*(\omega/\lambda),\qquad \lambda:=e^{\eta_v}\,(2\xs)^{-\eta},\\
  \mu_s:=\lambda\,\mus,\qquad\mu_b:=(1+\tfrac\iota2)\,\mu_s,\qquad\mu_a:=(1+\iota)\,\mu_s .
\end{gathered}
\end{equation}
The Riesz measure of $\MH$ is the image of $\dd\tau$ under the four arcs $\pm\omega_d$, $\pm\bar\omega_d$, where
$\omega_d(\tau):=\lambda\,\omega_*\bigl(\tau/(1-\eta)\lambda\bigr)$, and the regularity properties of \cref{app:M.5} hold for
$\omega_d$ with the same constants, because they are invariant under dilations. Since
$\lambda\Mob(\omega/\lambda)=\Mob(\omega)-\abs\mu\log\lambda$ and $\lambda\MD_{\mu/\lambda}(\nu/\lambda)=\MD_\mu(\nu)$, the
properties of $\Usk$ become the following.
\begin{itemize}[leftmargin=2.6em,itemsep=1pt,topsep=3pt]
\item[\hypdef{D1}] $\MH$ is subharmonic, finite and $C^1$ on the real axis, and even in $\mu$ and in $\nu$; its Riesz measure
is carried by four arcs that issue from the points $\pm\mu_s$, the \emph{tips of the design}.
\item[\hypdef{D2}] \emph{Corridor.} In the sector $\{\mu>\mu_b,\ \abs\nu<a_0(\mu-\mu_b)\}$, which is free of mass,
$\MH=(1-\eta)\Re\bigl[\omega\log\frac{6\omega}{\pi\lambda}-\omega\bigr]+{\rm const}$.
\item[\hypdef{D3}] \emph{Strict damping.} $\MH(\mu+i\nu)-\MH(\mu)\le(1-\eta)\,\MD_\mu(\nu)$ for all $\mu\ge\mu_s$ and all
real $\nu$.
\item[\hypdef{D4}] \emph{Strict vacuum dominance.} $(\MH-\Mob)'(\mu)=-\eta\log\frac\mu{\mu_s}-\eta_v\le-\eta_v$ for
$\mu\ge\mu_s$. Hence there are a point $\mu_v\in[\half\mu_s,\mu_s)$ and $\varepsilon_0>0$ with
$(\MH-\Mob)(\mu)\le(\MH-\Mob)(\mu_v)-\varepsilon_0-\eta_v(\mu-\mu_s)$ for all $\mu\ge\mu_s$.
\end{itemize}
The three parameters have three jobs. The reduction by $1-\eta$ makes the damping strict, but by itself it destroys
\hyp{D4}: for $\lambda=1$ one finds $(\MH-\Mob)'(\mu)=-\eta\log\frac{6\mu}\pi$, which is positive just to the right of $\mus$
because $\frac6\pi\mus=2\xs<1$. The dilation restores \hyp{D4}, with the margin $\eta_v=\log\lambda+\eta\log(2\xs)$; this
requires $\lambda>(2\xs)^{-\eta}>1$, so the tip moves outward, to $\lambda\mus$, as it must by \ResMone. Finally $\iota$ keeps
the saddle points that are used, $\mu\ge\mu_a$, away from the tip, where the zeros accumulate and no lower bound on the
symbol is available. By \hyp{D2} the saddle points $\mu\ge\mu_a$ belong to $x\ge x_a$, where
\begin{equation}\label{eq:M-value}
\begin{gathered}
  4x_a^2=\Bigl(\frac{6\mu_a}\pi\Bigr)^{2-\eta}\lambda^{-(1-\eta)},\qquad\text{that is,}\\
  x_a^2=(\xs)^2\,e^{\eta_v}\,(2\xs)^{-2\eta}\,(1+\iota)^{2-\eta}\;\longrightarrow\;(\xs)^2\quad\text{as }(\eta,\eta_v,\iota)\to0 .
\end{gathered}
\end{equation}

\paragraph{The symbol.} The Riesz measure of $c'\MH$ is $c'\dd\tau$ along each arc. We cut every arc into cells of unit
mass, replace the mass of each cell by a unit point mass at its midpoint in mass, and let these points be the zeros:
\begin{equation}\label{eq:M-symbolc}
\begin{gathered}
  \omega_j:=\omega_d\Bigl(\frac{j-\frac12}{c'}\Bigr)\quad(j=1,2,\dots),\qquad
  P_c(\omega):=\prod_{j\ge1}\Bigl(1-\frac{\omega^2}{\omega_j^2}\Bigr)\Bigl(1-\frac{\omega^2}{\bar\omega_j^2}\Bigr),\\
  m_c\bigl(\tfrac12+c'\omega\bigr):=\omega\,(\omega^2-\mu_s^2)\,P_c(\omega) .
\end{gathered}
\end{equation}
There is no integrality condition, so $c$ is an arbitrary real number. By \eqref{eq:M-potential}, $\log\abs{P_c}$ is the
midpoint rule, with step $1/c'$, for the integral $c'\MH$. The set of zeros is invariant under $\omega\to-\omega$ and
$\omega\to\bar\omega$, so $P_c$ is even, real on the real axis and positive there. The zero at $\omega=0$ makes the symbol
odd, and the pair $\pm\mu_s$ puts its only sign change on the positive axis at the tip of the design. What the rest of
the argument uses is the following list, in which $\delta_c=C/c'$.
\begin{itemize}[leftmargin=2.6em,itemsep=1pt,topsep=3pt]
\item[\hypdef{P1}] $m_c$ is entire, real on $\RR$, and $m_c(1-z)=-m_c(z)$.
\item[\hypdef{P2}] $\log\abs{m_c(\frac12+c'\omega)}\le c'[\MH(\omega)+\delta_c(1+\abs\omega)]$ for all $\omega$ with
$\mu\ge\mu_s$, and $\ge c'[\MH(\omega)-\delta_c(1+\abs\omega)]$ in the corridor of \hyp{D2} and on a real interval around
$\mu_v$.
\item[\hypdef{P3}] $m_c(\frac12+c'\mu)>0$ for real $\mu>\mu_s$, and $m_c(n+1)\le0$ for the integers $n\ge1$ with
$n+\frac12<c'\mu_s$.
\item[\hypdef{P4}] $\abs{\Gamma(z)m_c(z)}\le C_c\,e^{-a_c\abs{\Im z}}$ for $-\half\le\Re z\le\half+c'\mu_a$ and
$\abs{\Im z}\ge1$, with $a_c>0$.
\end{itemize}
The upper bound in \hyp{P2} is needed in the whole half-plane to the right of the tip, the arcs included, on which the
zeros lie; the lower bound only away from the zeros. \hyp{P1} and \hyp{P3} are read off from the product. \hyp{P2} and
\hyp{P4} are the subject of \cref{app:M.7}; the content is that the error of the midpoint rule is $O(1+\abs\omega)$, with
no factor $c'$.

\paragraph{From the symbol to the functional.} For $t>0$ let
\begin{equation}\label{eq:M-gc}
  g_c(t):=\frac1{2\pi i}\int_{\Re z=s_0}(2\pi t)^{-z}\,\Gamma(z)\,m_c(z)\,\dd z ,
\end{equation}
which by \hyp{P4} does not depend on $s_0>0$, and is real by \hyp{P1}. Moving the line to $\Re z=-\half$ one crosses the
pole of $\Gamma$ at $z=0$ and finds $g_c(t)=m_c(0)+O(t^{1/2})$, so $g_c$ is continuous on $[0,\infty)$. Next,
$\hat g_c=-g_c$: the Laplace transform of $g_c$ is
$F(\beta)=\frac1{2\pi i}\int_{\Re z=1/2}\frac\pi{\sin\pi z}\,m_c(z)\,\frac{\beta^{z-1}}{2\pi}\,\dd z$, the substitution
$z\to1-z$ preserves the line and $\sin\pi z$ and changes the sign of $m_c$, hence $F(1/\beta)=-\beta F(\beta)$, which is
$\hat g_c=-g_c$ by \eqref{eq:2-fourier}. \emph{In Mellin space the Fourier constraint costs nothing: every odd symbol
with enough decay gives a functional, and all the work is in the signs.} Finally the decay. For every $s_0>0$,
$\abs{g_c(t)}\le(2\pi t)^{-s_0}J(s_0)$ with $J(s_0):=\frac1{2\pi}\int\abs{\Gamma m_c}(s_0+iy)\,\dd y$. By \hyp{P2} and \hyp{D3},
$\abs{\Gamma m_c}$ decays exponentially along the lines to the right of the tip as soon as $\delta_c\le\eta/8$, and since
$\MH(\mu)=(1-\eta)\mu\log\mu+O(\mu)$ one finds $\log J(\frac12+c'\mu)\le c'[(2-\eta)\mu\log\mu+C\mu]$. This is smaller than
$\log[\Gamma(2s_0)(8\pi\alpha)^{-s_0}]$ at $s_0=\frac12+c'\mu$ for every $\alpha>0$, up to a constant that depends on $\alpha$
and $c$, and the choice $2s_0=4\pi\sqrt{\alpha t}$ then gives
\begin{equation}\label{eq:M-decay}
  \abs{g_c(t)}\le C_{\alpha,c}\,e^{-4\pi\sqrt{\alpha t}}\qquad\text{for every }\alpha>0 .
\end{equation}
In particular $\int_0^\infty\abs{g_c}\,\dd t<\infty$ and $M[g_c]<\infty$, because $V\le e^{4\pi\sqrt{at}}$: $g_c\in\Abar$. Here
$\eta>0$ is essential twice. Without the margin $\eta\MD_\mu(\nu)$ the error $c'\delta_c\abs\nu$ that \hyp{P2} allows would
make the bound non-integrable along the line; and for $\eta=0$ the symbol would grow at the critical rate $\mu\log\mu$, for
which one obtains \eqref{eq:M-decay} only up to some finite $\alpha$.

\paragraph{The sign of $g_c$.} Fix $c$ and $t$, and let $I(\omega):=\Gamma(z)m_c(z)(2\pi t)^{-z}$, so that
$g_c(t)=\frac{c'}{2\pi}\int_\RR I(\mu+i\nu)\,\dd\nu$ for every $\mu>0$; by \hyp{P3}, $I>0$ on the real axis for $\mu>\mu_s$. If
$x$ is confined to a compact set this is a standard problem with the large parameter $c'$. What is delicate is that
positivity is required for all $t\ge T$ at fixed $c$: as $t\to\infty$ the saddle point $\mu_t$ moves off to infinity, the
only large parameter is $c'\mu_t$, and the errors $c'\delta_c(1+\abs\omega)$ that \hyp{P2} allows are not small. The
resolution is that everything is invariant under dilations of $\omega$: the corridor is a sector, the error is relative,
and the profile in \hyp{D2}, like Stirling's exponent, is homogeneous of degree one up to terms linear in $\omega$. The
saddle-point expansion at $\mu$ then has $1/\sqrt{c'\mu}$ as its only parameter, uniformly for $\mu\ge\mu_a$. In the
corridor $\log m_c$ is holomorphic and real on the axis, and the difference $Q_c$ between $\frac1{c'}\log m_c$ and the
holomorphic profile of \hyp{D2} has $\abs{\Re Q_c}\le\delta_c(1+\abs\omega)$. A disc of radius proportional to $\mu$ around
$\mu\ge\mu_a$ lies in the corridor, and the gradient estimate for harmonic functions and Cauchy's estimates give
$\abs{Q_c'}\le C\delta_c$, $\abs{Q_c''}\le C\delta_c/\mu$ and $\abs{Q_c'''}\le C\delta_c/\mu^2$ there. It follows that the
equation $\partial_\mu\log I=0$ has exactly one solution $\mu_t\ge\mu_a$ for every
$t\ge T_{\MH}(c):=\frac13c'x_a^2\exp[C(\delta_c+c'^{-2})]$, that $-\partial_\mu^2\log I$ lies between $c'/2\mu_t$ and $3c'/\mu_t$
there, and that the third derivative is at most $Cc'/\mu^2$ on the disc. It is essential to expand at this exact saddle
point: the saddle point of the limit profile differs from it by $O(\delta_c\mu)$, which exceeds the width $\sqrt{\mu/c'}$
of the Gaussian as $\mu\to\infty$. On the window $\abs\nu\le r\mu_t$, with a small constant $r$, the integrand is a
Gaussian up to a relative error $O((c'\mu_t)^{-1/2})$. On the tails $\abs\nu\ge r\mu_t$ one leaves the corridor and has
only upper bounds; \hyp{D3}, \hyp{P2} and $\MD_\mu(\nu)\ge\nu^2/2(\mu+\abs\nu)$ give
$\abs{I(\mu_t+i\nu)}/I(\mu_t)\le\sqrt2\exp(-c'\gamma\abs\nu)$ with $\gamma>0$. Altogether
\begin{equation}\label{eq:M-sign}
  g_c(t)=\frac{c'}{2\pi}\,I(\mu_t)\,\Bigl(\frac{2\pi}{-\partial_\mu^2\log I(\mu_t)}\Bigr)^{1/2}\bigl[1+\epsilon_c(t)\bigr],\qquad
  \abs{\epsilon_c(t)}\le\frac C{\sqrt{c'\mu_a}}\le\frac12\qquad\text{for all }t\ge T_{\MH}(c) ,
\end{equation}
and $g_c(t)>0$ because $I(\mu_t)>0$.

\paragraph{The sign of the vacuum value.} By \eqref{eq:M-vacseries} and Stirling's formula,
\begin{equation}\label{eq:M-vacsum}
  (g_c)_{\rm vac}=\frac1{\sqrt{\pi^2c'/3}}\sum_{n\ge2}\theta_n\,e^{\epsilon_n}\,e^{-c'\Mob(\mu_n)}\,
  \bigl[-m_c\bigl(\tfrac12+c'\mu_n\bigr)\bigr],\qquad \mu_n:=\frac{n+\frac12}{c'},\quad0\le\epsilon_n\le\tfrac12\log2 .
\end{equation}
All the terms with $\mu_n<\mu_s$ are non-negative by \hyp{P3}. We keep the one nearest to $\mu_v$; by the lower bound in
\hyp{P2} it is at least $\exp\{c'[(\MH-\Mob)(\mu_v)-\delta_c(1+\mu_v)]-C\}$. The terms with $\mu_n\ge\mu_s$ have the wrong
sign. By \hyp{P2} and \hyp{D4} the modulus of such a term is at most
$\exp c'[(\MH-\Mob)(\mu_v)-\varepsilon_0+\delta_c(1+\mu_s)-(\eta_v-\delta_c)(\mu_n-\mu_s)]$; the points $\mu_n$ have spacing
$1/c'$, so this is a geometric series, and its sum is at most $1+2/\eta_v$ times its first term. Hence $(g_c)_{\rm vac}>0$
as soon as $\delta_c(2+\mu_v+\mu_s)\le\half\varepsilon_0$: the wrong-sign terms are smaller by the factor
$e^{-c'\varepsilon_0/2}$.

\paragraph{Conclusion.} Let $x_g>\xs$. By \eqref{eq:M-value} we choose $\eta$, $\eta_v$ and $\iota$, depending only on
$x_g$, such that $x_a<\half(\xs+x_g)$. This fixes the design, its $\mu_v$ and $\varepsilon_0$, and then all the constants
above. Since $\delta_c\to0$, there is $c_0(x_g)$ such that for $c\ge c_0(x_g)$ the smallness conditions on $\delta_c$ hold
and $T_{\MH}(c)\le c'x_g^2/3$. For every such real $c$ the function $g_c$ of \eqref{eq:M-gc} lies in $\Abar$, is positive on
$[c'x_g^2/3,\infty)$ and has $(g_c)_{\rm vac}>0$. This is \ResMtwo.

\subsection{Discretization of the skeleton}
\label{app:M.7}

It remains to justify \hyp{P2} and \hyp{P4}. In electrostatic language, $c'\MH$ is the potential of a line charge of
density $c'\dd\tau$ in the mass parameter of the arcs, and $\log\abs{P_c}$ is that of unit point charges at the mass
midpoints of the cells. The claim is that the difference is $O(1)$, uniformly in $c$ and in the point of observation:
\begin{equation}\label{eq:M-midpoint}
  \log\abs{P_c(\omega)}\le c'\MH(\omega)+C\quad\text{for all }\omega\in\CC,\qquad
  \log\abs{P_c(\omega)}\ge c'\MH(\omega)-C\quad\text{if }\dist(\omega,\text{arcs})\ge C\,\ell_1 ,
\end{equation}
where $\ell_j$ is the length of the $j$-th cell, and $\ell_1\le Cc'^{-1/2}\to0$. The upper bound holds everywhere, also on
the arcs, where the zeros lie; the lower bound cannot, since $P_c$ vanishes there. For one arc and a point $w$ let
\begin{equation}\label{eq:M-cell}
  e_j(w):=c'\int_{I_j}\log\Bigabs{\frac{\omega_j-w}{\omega_d(\tau)-w}}\,\dd\tau,\qquad
  I_j=\Bigl[\frac{j-1}{c'},\frac j{c'}\Bigr] ,
\end{equation}
be the error made when the mass of the $j$-th cell is concentrated at its midpoint in mass. There are three regimes.
\emph{Near cells}, the cell nearest to $w$ and its two neighbors: no cancellation is needed, because a logarithmic
singularity is integrable against a mass that is spread evenly; chord and arc being comparable and the mass per unit
length being at most $C/\ell_j$, one finds $-\log(1+\ell_j/\abs{\omega_j-w})\le e_j(w)\le C$ for every cell and every $w$.
\emph{Far cells}: $e_j(w)$ is the error of the midpoint rule for $\tau\mapsto\log(\omega_d(\tau)-w)$; the linear term of
the Taylor expansion about the midpoint integrates to zero, and the second derivative, with
$\abs{\omega_d''}\le C\abs{\omega_d'}/\tau$, gives
\begin{equation}\label{eq:M-far}
  \abs{e_j(w)}\le C\Bigl[\frac1{j-1}\,\frac{\ell_j}{d_j}+\frac{\ell_j^2}{d_j^2}\Bigr]\qquad(j\ge2),\qquad d_j:=\dist(w,\text{cell }j) .
\end{equation}
The second term is the quadrupole field of the cell. The first is a residual dipole, present because the mass midpoint of
a cell is its center of mass only if the speed is constant along it. Since the speed decreases no faster than
$\tau^{-1/2}$, the lengths of the cells between the $j$-th and the one nearest to $w$ are comparable with $\ell_j$ in the
sense that $\ell_j/d_j\le C[\frac1{j-1}+\frac1k]$, where $k$ is the number of cells in between, and this is
square-summable, uniformly in $w$ and $c$. \emph{The tip}: there $\omega_d-\mu_s\propto\sqrt\tau$, and
$\abs{\omega_d''}\propto\tau^{-3/2}$ is not integrable, so \eqref{eq:M-far} has no analogue for the first cell, however far
$w$ is from it. This is harmless, because the bound for near cells holds for every cell and uses no second derivative;
the first cell is always counted among the near ones. Summing over the cells and over the four arcs gives
\eqref{eq:M-midpoint}.

With the prefactor $\omega(\omega^2-\mu_s^2)$, which is at most $(\mu_s+\abs\omega)^3$ and is bounded below in the corridor
and near $\mu_v$, \eqref{eq:M-midpoint} is \hyp{P2} with $\delta_c=C/c'$, for all $c$ so large that $C\ell_1$ is smaller than
the distance from the arcs to the corridor and to the interval around $\mu_v$. For \hyp{P4}, the dilated form of
\eqref{eq:M-Uskbound} and $\MD_\mu(\nu)\le\frac\pi2\abs\nu$ give $\MH(\omega)\le(1-\eta)\frac\pi2\abs\nu+C$ for
$\abs\mu\le\mu_a$, and $\abs{\Gamma(z)}\le C(1+\abs{\Im z})^{\Re z-1/2}e^{-\pi\abs{\Im z}/2}$; so $\Gamma(z)m_c(z)$ decays like
$e^{-\eta\pi\abs{\Im z}/2}$ up to a power. Here again $\eta>0$ is indispensable: it is the only margin against the decay
of the Gamma function.

\subsection{A dictionary, and remarks}
\label{app:M.8}

\paragraph{The strip and the Mellin plane.} The arguments of \cref{sec:6} and of \cref{app:M.4} have the same skeleton:
a holomorphic function attached to $g$, a bound on part of a boundary that comes from the vacuum, the maximum principle
for the logarithm of the modulus, and the half-mass condition. They are carried out in conjugate variables. In
\cref{sec:6} the function is $G(b)$ on the strip $\strip$, and the variable is $b$. Here it is $m_g$ on the plane, and
the variable is the action $\omega=\frac{2\pi i}3bp$ \eqref{eq:M-param}, as one expects of a Mellin transform, which
diagonalizes dilations. The two profiles, $u_g(b)$ of \eqref{eq:6-ug} and the left side of \eqref{eq:M-obst}, are related
at exponential order by the saddle point of the Mellin transform, that is, by a Legendre transformation in $\log t$. The
correspondence is as follows.
\begin{itemize}[leftmargin=1.6em,itemsep=1pt,topsep=3pt]
\item The Fourier transform is the quarter-turn $(p,b)\mapsto(b,-p)$ of \cref{sec:4} on the strip, and the reflection
$\omega\to-\omega$, which acts as $(b,q)\to(iq,ib)$, in the plane.
\item The vacuum is the point $b=\frac i2$ on an edge of the strip, and gives the boundary value $0$ on the edges
(\cref{lem:6.1}). In the plane every Taylor coefficient of the Cardy weight gives a bound on a vertical line; together
they are the obstacle $\Mob$.
\item The Cardy line $-L$ on the slits $\abs x\ge\xs$ corresponds to the obstacle on the contact set. The tips $b=\pm\xs$
of the slits are the tips $\omega=\pm\frac\pi3\xs$ of $\MK$; on the free boundary $b$ is real and runs over the band
$x\ge\xs$, and $q=ip(x+i0)$ there. The edges $\abs{\Im b}=\half$ of the strip appear as the free boundary of $-\MK_+$.
\item The extremal potential $U$ corresponds to $\MU$, and $U(0)=\MU(0)=U_0$ \eqref{eq:M-parts}. The exponents differ
away from that point, \eqref{eq:M-kM-tip}: at the tips $U+L$ vanishes to order $\frac32$, and $\MW$ to order $2$.
\item The charge of $U$ sits on the slits, with the density $\rhoU$ of \eqref{eq:6-rhoU}. The charge of $\MU$ is the area
measure $\dd A/\pi\abs\mu$ on $\MK$, and the two have the same projection: by \eqref{eq:M-fb},
\begin{equation}\label{eq:M-rhoU}
  \int_{-\nus(\mu)}^{\nus(\mu)}\frac{\dd\nu}{\pi\mu}\,\dd\mu=\frac4{\pi^2}\arccosh(\sinh3\mu)\,\dd\mu=\rhoU(x)\,\dd x
  \qquad\text{at }\mu=\frac\pi3\,x .
\end{equation}
\end{itemize}

\paragraph{Zeros.} By \cref{thm:7} the zeros of $G$ in the strip, for any sequence that attains $\LV$, follow the one
law $\rhoU$. The zeros of the symbol are not rigid in this sense: the argument of \cref{app:M.2} fixes their limiting
distribution only up to rearrangements inside $\MK$ that leave the potential in $\MN$ unchanged, and \cref{app:M.5} uses
this freedom. The two statements concern the zeros of different functions.

\paragraph{Three functionals.} The functional of \cref{thm:5} places the zeros of $G$ on the band, along the Cardy
line, where positivity is an inequality between two numbers that a computer certifies; it is sharp to order $c^{1/3}$.
The functional of \cref{sec:alt} places no zeros at all; it is one transform of one real seed, and it stays below the
Cardy line by a fixed margin. The functional of this appendix places the zeros of the Mellin symbol on the creases; the
constraint $\hat g=-g$ is automatic, and positivity comes from one real saddle point that dominates a conjugate pair
by the margin $\eta$. In the last two constructions the margin is bought in the same way, by moving the tip outward:
the deformation $\gamma<\pi$ of \cref{sec:alt} borrows the extremal profile of a heavier theory, with $c'$ replaced by
$(\pi/\gamma)^2c'$ \eqref{eq:alt-dilation}, and so does the dilation of \eqref{eq:M-design}, since
$c'\lambda\,\MH_*(\omega/\lambda)$ is the potential of the skeleton with $c'$ replaced by $\lambda c'$, at the same $z$.
Neither gives a rate.

\paragraph{The problem on a strip.} In the $U(1)^c$ problem the vacuum sits at the origin, on the boundary of the half
line of the Mellin transform, and the proof of~\cite[Ch.~1]{OpenAI} works on a strip in the Mellin variable, whose width
is the dimension. Here the vacuum sits at $t=-a$, off the half line. It cannot be reached by the transform, and enters
only through the expansion \eqref{eq:M-vac} of its $S$-image; the strip is replaced by the unknown set $\MN$, and the
value of the slope is fixed by a free-boundary condition. In the variable $b$ the same vacuum is a boundary point of a
fixed strip (\cref{sec:6.3}), and no free boundary has to be found, although the largest useful slit is again
characterized by tangency (\cref{sec:6.6}).

\section{Bessel functions and the Cardy weight}
\label{app:A}

This appendix collects the properties of the Cardy kernel $K_\alpha(t)=I_0(4\pi\sqrt{\alpha t})$ and of the weight
$V=K_a-2K_{a-1}+K_{a-2}$, $\rhoC=2\pi V$, that the proofs of Parts~III and~IV use: the continuation of the eigen-identity
below the BTZ threshold (\cref{lem:A.1}), the comparison of $V$ with $K_a$ and with $e^{c'L}$ (\cref{lem:A.2}), the
window at the threshold where $V$ degenerates (\cref{lem:A.3}), and two-sided asymptotes of $I_0$ and $V$ with explicit
remainders (\cref{lem:A.4}). The first three are used from \cref{sec:6.3} on; the fourth is needed in \cref{sec:7.9},
\cref{app:F} and \cref{sec:8.2}. Everything is proved by hand from the integral
representation, the power series and the recurrences of $I_\nu$~\cite[\S\S10.25, 10.29, 10.32]{DLMF}. Throughout,
$c\ge40$, that is $c'\ge39$ and $a=c'/12\ge3.25$, which enters only through the numerical values in the proofs. We
write $w_\alpha=4\pi\sqrt{\alpha t}$ and $t=c'x^2/3$ with $x\ge0$, so that $w_a=c'L(x)$, $L(x)=\frac{2\pi}3\abs x$.

\subsection{Continuation of the eigen-identity}
\label{app:A.1}

Since $K_\alpha(t)=\sum_{k\ge0}(4\pi^2\alpha t)^k/(k!)^2$ is entire in $\alpha$, the eigen-identity of \cref{sec:2.4}
makes sense for complex $\alpha$. We show that it converges on a parabolic region, with vertex at the vacuum
$-\alpha=-a$ and focus at the threshold, on which $K_a$ dominates the kernel; in the variable $b$ of \cref{sec:2.5},
$\alpha=-c'b^2/3$, this region is the closed strip $\abs{\Im b}\le\half$.

\begin{lemma}[\proved]\label{lem:A.1}
Let $\mathcal D_a=\{\alpha\in\CC:\abs{\Re\sqrt\alpha}\le\sqrt a\}$, the closed region
$\Re\alpha\le a-(\Im\alpha)^2/4a$ bounded by a parabola; it contains $(-\infty,a]$. For $\alpha\in\mathcal D_a$ and $t\ge0$,
\begin{equation}\label{eq:A-kernel-bound}
  \abs{K_\alpha(t)}\le K_{(\Re\sqrt\alpha)^2}(t)\le K_a(t).
\end{equation}
For $g\in\Abar$ the integral in
\begin{equation}\label{eq:A-eigen}
  g(-\alpha)=-2\pi\int_0^\infty g\,K_\alpha\,\dd t,\qquad
  \abs{g(-\alpha)}\le2\pi\int_0^\infty\abs g\,K_a\,\dd t\le2\pi\bigl(87+78\,a\,c'\bigr)M[g],
\end{equation}
converges absolutely, uniformly in $\alpha\in\mathcal D_a$. It defines a function that is continuous on $\mathcal D_a$,
holomorphic in the interior, and equal to the given $g(s)$ at $\alpha=-s\le0$. In particular
$\gvac=g(-a)-2g(-a+1)+g(-a+2)=-\int_0^\infty g\,\rhoC\,\dd t$, with an absolutely convergent integral.
\end{lemma}

\begin{proof}
The representation $I_0(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\,\dd\theta$~\cite[\S10.32]{DLMF} gives
$\abs{I_0(z)}\le I_0(\abs{\Re z})$; since $\Re\bigl(4\pi\sqrt{\alpha t}\bigr)=4\pi\sqrt t\,\Re\sqrt\alpha$ and $I_0$
increases on $[0,\infty)$, this is \eqref{eq:A-kernel-bound}. For the rest we use \cref{lem:A.2,lem:A.3} below, whose
proofs do not use the present lemma. By \cref{lem:A.2}(b), $K_a<214$ on $[0,4/c']$ and
$K_a\le310\,a\max(1,t^{-1})V\le78\,a\,c'V$ on $[4/c',\infty)$, so that
$\int_0^\infty\abs gK_a\,\dd t\le214\int_0^{4/c'}\abs g\,\dd t+78\,a\,c'M[g]$. This is finite, because $M[g]<\infty$ and
$\int_0^\infty\abs g\,\dd t=\frac1{2\pi}\norm{g(\abs{\br}^2/2)}_{L^1(\RR^2)}<\infty$ by the definition of $\Abar$
(\cref{sec:2.6}), and \cref{lem:A.3} bounds the first term by $87M[g]$. By \eqref{eq:A-kernel-bound} the integrand in
\eqref{eq:A-eigen}, which is entire in $\alpha$, is dominated by $\abs gK_a$ for all $\alpha\in\mathcal D_a$ at once;
dominated convergence gives the uniform absolute convergence, the continuity and the holomorphy in the interior.
Finally, $K_{-s}(t)=J_0(4\pi\sqrt{st})$ for $s\ge0$, and \eqref{eq:A-eigen} at $\alpha=-s$ is the relation $\hat g=-g$
written as a Hankel transform of order zero (\cref{sec:2.3}).
\end{proof}

\subsection{The weight}
\label{app:A.2}

The weight $V$ is a second difference of $\alpha\mapsto K_\alpha(t)$ with unit step, yet it is comparable to the single
kernel $K_a$: up to a polynomial factor as soon as $t\ge4/c'$, and up to the factor $0.9$ beyond $t=\xis c'$, where
$K_{a-1}$ and $K_{a-2}$ are exponentially smaller than $K_a$. The proof writes $V$ as an average of
$\partial_\alpha^2K_\alpha$, a multiple of $I_2$, and compares Bessel functions at the nearby arguments $w_a$,
$w_{a-1}$, $w_{a-2}$ through the logarithmic derivative $I_1/I_0$.

\begin{lemma}[\proved]\label{lem:A.2}
Let $c\ge40$, $t>0$ and $x=\sqrt{3t/c'}$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item The weight obeys
\begin{equation}\label{eq:A-sandwich}
  \frac{4\pi^2t}{a-2}\,I_2(w_{a-2})\le V(t)\le\frac{4\pi^2t}{a}\,I_2(w_a),
\end{equation}
and $V(t)=\sum_{k\ge2}v_kt^k$ with $v_k\ge0$, $v_2=8\pi^4$; hence $V$ is positive and increasing on $(0,\infty)$,
and $V(t)\ge8\pi^4t^2$.
\item $K_a(t)\le I_0(4\pi/\sqrt3)<214$ for $t\le4/c'$, and $K_a(t)\le310\,a\max(1,t^{-1})\,V(t)$ for $t\ge4/c'$.
\item $V(t)\ge0.9\,K_a(t)$ for $t\ge\xis c'$.
\item $V(t)\le\frac{4\pi^2t}a\,K_a(t)\le\frac{4\pi^2t}a\,e^{c'L(x)}$.
\item $\log K_a(t)\le c'L(x)$ for all $x$, and $\log K_a(t)\ge c'L(x)-\half\log\bigl(c'L(x)\bigr)-2$ for $x\ge\xs$.
\end{enumerate}
\end{lemma}

\begin{proof}
We use four facts about $I_n(w)$, $w>0$, which follow from
$I_n(w)=\frac1\pi\int_0^\pi e^{w\cos\theta}\cos n\theta\,\dd\theta$ and from the recurrences $I_0'=I_1$, $I_1'=I_2+I_1/w$,
$I_0-I_2=(2/w)I_1$~\cite[\S\S10.29, 10.32]{DLMF}.
(i)~$0\le I_1\le I_0\le e^w$; hence $(1-2/w)I_0\le I_2\le I_0$, and $I_0\le2I_2$ for $w\ge4$.
(ii)~$\log I_0$ is convex, $I_0$ being the Laplace transform of a positive measure on $[-1,1]$, and its derivative
$I_1/I_0$ lies in $[0,1)$; hence $e^{-(w-w')}I_0(w)\le I_0(w')\le e^{-(w-w')\,(I_1/I_0)(w')}\,I_0(w)$ for $0<w'\le w$.
(iii)~$I_1(w)/I_0(w)\ge1-1/w$ for $w\ge10$.
(iv)~$I_0(w)\ge e^{w-1/2}/(\pi\sqrt w)$ for $w\ge1$, as one sees by restricting the integral to
$0\le\theta\le w^{-1/2}$, where $\cos\theta\ge1-1/2w$.
To prove (iii) we substitute $u=1-\cos\theta$ in the integrals for $I_0-I_1$ and $I_0$:
\begin{equation*}
  1-\frac{I_1(w)}{I_0(w)}
  =\frac{\int_0^2u\,e^{-wu}\,(u(2-u))^{-1/2}\,\dd u}{\int_0^2e^{-wu}\,(u(2-u))^{-1/2}\,\dd u}
  \le\frac{\frac{\sqrt\pi}2\,w^{-3/2}+3e^{-w}}{\frac1{\sqrt2}\bigl(\sqrt{\pi/w}-e^{-w}/w\bigr)} .
\end{equation*}
In the numerator we use $(2-u)^{-1/2}\le1$ on $[0,1]$, and $e^{-wu}\le e^{-w}$, $\int_1^2\sqrt{u/(2-u)}\,\dd u\le3$ on
$[1,2]$; in the denominator, $(2-u)^{-1/2}\ge2^{-1/2}$ on $[0,1]$ and $\int_1^\infty e^{-wu}u^{-1/2}\,\dd u\le e^{-w}/w$.
The right-hand side is $0.071$ at $w=10$, and $w$ times it decreases for $w\ge10$.

\emph{(a)} Fix $t$ and let $f(\alpha)=K_\alpha(t)=I_0(w_\alpha)$. Since $\partial_\alpha w_\alpha=w_\alpha/2\alpha$, the
recurrences give
\begin{equation*}
  f''(\alpha)=\frac{w_\alpha^2}{4\alpha^2}\Bigl(I_1'(w_\alpha)-\frac{I_1(w_\alpha)}{w_\alpha}\Bigr)
  =\frac{4\pi^2t}{\alpha}\,I_2(w_\alpha)=64\pi^4t^2\,\frac{I_2(w_\alpha)}{w_\alpha^2}\ \ge0 ,
\end{equation*}
which is non-decreasing in $\alpha$ because $I_2(s)/s^2=\sum_{k\ge0}(s/2)^{2k}/(4\,k!\,(k+2)!)$ is non-decreasing in
$s$. Since $V(t)=\int_{-1}^1(1-\abs u)\,f''(a-1+u)\,\dd u$ is a weighted mean of $f''$ over $[a-2,a]$,
\eqref{eq:A-sandwich} follows. The power series of $K_\alpha$ gives
$v_k=(4\pi^2)^k(k!)^{-2}\,[a^k-2(a-1)^k+(a-2)^k]$, which vanishes for $k=0,1$, equals $8\pi^4$ for $k=2$, and is
non-negative because $\alpha\mapsto\alpha^k$ is convex on $[a-2,a]\subset[0,\infty)$.

\emph{(b)} $K_a$ increases with $t$, and $w_a=4\pi/\sqrt3$ at $t=4/c'=1/3a$, where $I_0(4\pi/\sqrt3)=213.6$. For
$t\ge4/c'$ we have $w_{a-2}\ge4\pi\sqrt{(1-2/a)/3}\ge4.49$ because $a\ge3.25$, so \eqref{eq:A-sandwich} and (i) give
$V\ge\frac{4\pi^2t}{a-2}\cdot\half I_0(w_{a-2})\ge\frac{2\pi^2t}a\,K_{a-2}$. By (ii), $K_a\le e^{w_a-w_{a-2}}K_{a-2}$ with
$w_a-w_{a-2}=8\pi\sqrt{t/a}\,/(1+\sqrt{1-2/a})$. For $4/c'\le t\le\xis c'=12\,\xis a$ this difference is at most
$8\pi\sqrt{12\,\xis}/(1+\sqrt{1-2/3.25})\le8\pi\cdot0.5611/1.62\le8.71$, and $e^{8.71}/2\pi^2\le308$, so
$K_a\le308\,(a/t)\,V$. For $t\ge\xis c'$, (c) gives $K_a\le V/0.9$. In both ranges this is at most
$310\,a\max(1,t^{-1})V$.

\emph{(c)} Since $K_{a-2}\ge0$, we have $V\ge K_a-2K_{a-1}=K_a(1-2\vartheta)$ with
$\vartheta=I_0(w_{a-1})/I_0(w_a)\le\exp\bigl(-(w_a-w_{a-1})\,(I_1/I_0)(w_{a-1})\bigr)$ by (ii). For
$t\ge\xis c'=12\,\xis a$ we have $w_a-w_{a-1}=4\pi\sqrt t/(\sqrt a+\sqrt{a-1})\ge2\pi\sqrt{t/a}\ge2\pi\sqrt{12\,\xis}\ge3.525$
and $w_{a-1}\ge4\pi\sqrt{12\,\xis\,a(a-1)}\ge19$, so that $(I_1/I_0)(w_{a-1})\ge0.947$ by (iii). Hence
$\vartheta\le e^{-3.33}\le0.036$ and $1-2\vartheta\ge0.9$.

\emph{(d)} This is the upper bound in \eqref{eq:A-sandwich} with $I_2\le I_0\le e^w$ from (i) and $w_a=c'L(x)$.

\emph{(e)} $K_a(t)=I_0(w)$ with $w=c'L(x)$, and $I_0(w)\le e^w$. For $x\ge\xs$ we have $w\ge c'L(\xs)\ge22$, and
(iv) gives $\log I_0(w)\ge w-\half\log w-\half-\log\pi\ge w-\half\log w-2$.
\end{proof}

\subsection{The window at the BTZ threshold}
\label{app:A.3}

Near the threshold the weight degenerates, $V(t)\simeq8\pi^4t^2$, while $K_a$ stays of order one, so $M[g]$ does not
control $\int\abs gK_a\,\dd t$ there through $V$. The control comes from the eigen-identity itself: $g$ is minus its own
Hankel transform, so $\sup\abs g$ is bounded by the $L^1$ norm of $g$, and the part of that norm which comes from a
fixed neighborhood of $t=0$ can be absorbed.

\begin{lemma}[\proved]\label{lem:A.3}
For every $g\in\Abar$,
\begin{equation}\label{eq:A-L1}
  \int_0^\infty\abs g\,\dd t\le\frac4{\pi^2}\,M[g],\qquad\sup_{t\ge0}\abs{g(t)}\le\frac8\pi\,M[g],
\end{equation}
and consequently $\int_0^{4/c'}\abs g\,K_a\,\dd t\le214\cdot\frac4{\pi^2}\,M[g]\le87\,M[g]$ for $c\ge40$.
\end{lemma}

\begin{proof}
Write $\norm g_1=\int_0^\infty\abs g\,\dd t$, which is finite by the definition of $\Abar$. The relation $\hat g=-g$ of
\cref{def:2-Abar}, in the form \eqref{eq:2-eigen}, gives $\sup\abs g\le2\pi\norm g_1$ because $\abs{J_0}\le1$. We split
$\norm g_1$ at $t_*=1/4\pi$. Below, $\int_0^{t_*}\abs g\,\dd t\le t_*\sup\abs g\le\half\norm g_1$. Above,
$V(t)\ge V(t_*)\ge8\pi^4t_*^2=\pi^2/2$ by \cref{lem:A.2}(a), so $\int_{t_*}^\infty\abs g\,\dd t\le\frac2{\pi^2}M[g]$.
Adding the two and rearranging gives the first bound of~\eqref{eq:A-L1}, the second follows from
$\sup\abs g\le2\pi\norm g_1$, and the last statement from \cref{lem:A.2}(b).
\end{proof}

\subsection{The two-sided Bessel asymptote}
\label{app:A.4}

The vacuum inequality of \cref{sec:7.9} needs $V$ to relative accuracy $O(1/c')$ on both sides of the tip, from below
on the lobe and from above on the band, and \cref{sec:8.2} needs an upper bound on $I_0$ valid for all arguments. We write
$\as(z)=e^z/\sqrt{2\pi z}$. The two-sided bound on $I_0$ comes from the integral representation with $u=1-\cos\theta$,
\begin{equation}\label{eq:A-rep}
  I_0(z)=\frac{e^z}\pi\int_0^2\frac{e^{-zu}}{\sqrt{u(2-u)}}\,\dd u ,
\end{equation}
by bounding $(2-u)^{-1/2}$ below by its value at $u=0$ and above by a chord; the asymptote of $V$ follows by expanding
$w_{a-1}$ and $w_{a-2}$ around $w_a$.

\begin{lemma}[\proved]\label{lem:A.4}
\begin{enumerate}[label=\textup{(\alph*)}]
\item For $z\ge2$,
\begin{equation}\label{eq:A-as}
  \bigl(1-e^{-2z}\bigr)\,\as(z)\le I_0(z)\le\Bigl(1+\frac{0.36}z\Bigr)\as(z)+\frac2\pi .
\end{equation}
\item Let $c\ge40$ and $t=c'x^2/3$ with $x\ge\xs/2$, in particular with $x$ in the window $\abs{x-\xs}\le0.1$. Then,
with an absolute constant $\CV$,
\begin{equation}\label{eq:A-V-as}
  V(t)=\as\bigl(c'L(x)\bigr)\Bigl[\bigl(1-e^{-4\pi x}\bigr)^2+\epsV(x)\Bigr],\qquad\abs{\epsV(x)}\le\frac{\CV}{c'} .
\end{equation}
\item For all $z\ge0$, $I_0(z)\le1.18\,e^z/\sqrt{1+z}$.
\end{enumerate}
\end{lemma}

\begin{proof}
\emph{(a)} Lower bound: $(2-u)^{-1/2}\ge2^{-1/2}$ in \eqref{eq:A-rep} gives
$I_0(z)\ge\frac{e^z}{\pi\sqrt2}\bigl[\sqrt{\pi/z}-\int_2^\infty e^{-zu}u^{-1/2}\,\dd u\bigr]$, and
$\int_2^\infty e^{-zu}u^{-1/2}\,\dd u\le e^{-2z}/(z\sqrt2)\le\sqrt{\pi/z}\,e^{-2z}$ for $z\ge1/2\pi$. Upper bound: on
$0\le u\le1$ the convex function $(1-u/2)^{-1/2}=\sqrt2\,(2-u)^{-1/2}$ lies below its chord $1+(\sqrt2-1)u$, so this part of
\eqref{eq:A-rep} is at most
$\frac{e^z}{\pi\sqrt2}\bigl[\Gamma(\half)z^{-1/2}+(\sqrt2-1)\Gamma(\tfrac32)z^{-3/2}\bigr]=\as(z)\bigl(1+\frac{\sqrt2-1}{2z}\bigr)
\le\as(z)\bigl(1+\frac{0.36}z\bigr)$; on $1\le u\le2$ we use $e^{-zu}\le e^{-z}$ and
$\int_1^2(u(2-u))^{-1/2}\,\dd u\le\int_1^2(2-u)^{-1/2}\,\dd u=2$, which contributes at most $2/\pi$.

\emph{(b)} Let $z_0=c'L(x)$, $q_j=12j/c'\le24/39$ and $z_j=w_{a-j}=z_0\sqrt{1-q_j}$ for $j=0,1,2$, so that
$V=I_0(z_0)-2I_0(z_1)+I_0(z_2)$. Since $z_0q_j/2=4\pi jx$,
\begin{equation*}
  z_j=z_0-4\pi jx-\delta_j,\qquad
  0\le\delta_j=z_0\bigl[1-\tfrac{q_j}2-\sqrt{1-q_j}\bigr]\le\frac{z_0\,q_j^2}{8(1-q_j)}=\frac{12\pi x\,j^2}{c'-12j} ,
\end{equation*}
and $z_j\ge z_2\ge0.62\,z_0\ge1.29\,c'x\ge0.18\,c'\ge7$. Thus (a) applies to each $z_j$:
$I_0(z_j)=\as(z_j)(1+\lambda_j)+r_j$ with $-e^{-2z_j}\le\lambda_j\le0.36/z_j$ and $0\le r_j\le2/\pi$. Moreover
$\as(z_j)=\as(z_0)\,e^{-4\pi jx}\,e^{-\delta_j}(1-q_j)^{-1/4}$, and $1\le(1-q_j)^{-1/4}\le1+11/c'$ for
$c'\ge39$. Therefore
\begin{equation*}
  \frac{V}{\as(z_0)}=\sum_{j=0}^2(-1)^j\tbinom2j\,e^{-4\pi jx}\,(1+\zeta_j)+R,\qquad
  1+\zeta_j=e^{-\delta_j}(1-q_j)^{-1/4}(1+\lambda_j),\qquad\abs R\le\frac{4/\pi}{\as(z_0)} .
\end{equation*}
The sum without the $\zeta_j$ is $(1-e^{-4\pi x})^2$, so $\abs{\epsV}\le\sum_j\binom2je^{-4\pi jx}\abs{\zeta_j}+\abs R$. The
three factors of $1+\zeta_j$ lie in $[1-\delta_j,1]$, $[1,1+11/c']$ and $[1-e^{-2z_2},1+0.36/z_2]$, whence
$\abs{\zeta_j}\le\delta_j+11/c'+1/z_2$. Each contribution is at most an absolute constant times $1/c'$, uniformly in
$x\ge\xs/2$: $1/z_2\le1/(0.18\,c')$; $e^{-4\pi jx}\delta_j\le12\pi j^2\,x\,e^{-4\pi jx}/(0.38\,c')$ with
$x\,e^{-4\pi jx}\le1/(4\pi e j)$ for $j=1,2$, while $\delta_0=0$; and $\as(z_0)\ge z_0\ge0.18\,c'$, because $e^z\ge\sqrt{2\pi}\,z^{3/2}$ for $z\ge2$.

\emph{(c)} For $0\le z\le2$, the series $I_0(z)=\sum_{k\ge0}(z^2/4)^k/(k!)^2\le e^{z^2/4}$ and $\sqrt{1+z}\le e^{z/2}$ give
$e^{-z}I_0(z)\sqrt{1+z}\le e^{-z(2-z)/4}\le1$. For $z\ge2$, $I_0(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\,\dd\theta$ and
$1-\cos\theta\ge2\theta^2/\pi^2$ on $[0,\pi]$ give
$I_0(z)\le\frac{e^z}\pi\int_0^\infty e^{-2z\theta^2/\pi^2}\,\dd\theta=e^z\sqrt{\pi/8z}$, so that
$e^{-z}I_0(z)\sqrt{1+z}\le\sqrt{\pi(1+z)/8z}\le\sqrt{3\pi/16}<1$. Thus $e^{-z}I_0(z)\sqrt{1+z}\le1$ for all $z\ge0$,
which is more than (c) asserts.
\end{proof}

\section{Potential theory on the slit strip}
\label{app:B}

This appendix proves the lemmas of potential theory behind \cref{sec:6.5,sec:6.6,sec:6.7,sec:6.8}: the conformal map
and the harmonic measure of $D$, the entropy of that measure, the Poisson--Jensen and Jensen inequalities on $D$, the
edge bound, \cref{prop:6.4}, and the last inequality of \cref{lem:6.7}. \Cref{app:H} uses \cref{lem:B.4} and \eqref{eq:B-Poisson-U} again. From \cref{app:A} we take
\cref{lem:A.1,lem:A.2,lem:A.3}. Throughout, $c\ge40$, $D=D_{\xs}$, $G(b)=g(c'b^2/3)$ with $g\in\Abar$, and $C$ is an
absolute constant whose value may change from line to line.

\subsection{The map and harmonic measure}
\label{app:B.1}
For $z\in\HH$ let $b(z)=\frac1{2\pi}\bigl[\log(z-1)-\log z-\log(z+1)\bigr]$, with all three arguments in $(0,\pi)$, be
the map of \cref{lem:6.3}. We write $\sigma$ for a point of the real axis bounding $\HH$, and $\sigma_*:=1-\sqrt2$,
$\sigma^*:=1+\sqrt2$, so that
\begin{equation}\label{eq:B-bprime}
  b'(z)=\frac1{2\pi}\Bigl[\frac1{z-1}-\frac1z-\frac1{z+1}\Bigr]
       =-\frac{(z-\sigma_*)(z-\sigma^*)}{2\pi\,z\,(z^2-1)} .
\end{equation}

\begin{lemma}[\proved]\label{lem:B.1}
\textup{(a)} The map $b$ is a conformal bijection of\/ $\HH$ onto $D$. The preimage $z_w$ of $w\in D$ is the
unique root in $\HH$ of
\begin{equation}\label{eq:B-quadratic}
  e^{2\pi w}z^2+\bigl(e^{2\pi w}-1\bigr)z+1=0 .
\end{equation}
\textup{(b)} $b$ extends continuously to $\RR\setminus\{-1,0,1\}$ and realizes the boundary correspondence stated in
\cref{lem:6.3}. It is monotonic on $(0,1)$ and on $(-\infty,-1)$, and it maps each of $(-1,\sigma_*]$ and $[\sigma_*,0)$
monotonically onto $[\xs,\infty)$, the two intervals giving the two lips of the right slit; likewise $(1,\sigma^*]$,
$[\sigma^*,\infty)$ and the left slit.

\textup{(c)} For $b\in D$ let $\omb$ be the image under this map of the Poisson measure
$\frac{\Im z_b}{\pi\abs{\sigma-z_b}^2}\,\dd\sigma$ of\/ $\HH$ seen from $z_b$. It is the harmonic measure of $D$ seen
from $b$, a probability measure on $\partial D$ in which the two lips of a slit count separately. On the slits it has
the positive density (the two lips added)
\begin{equation}\label{eq:B-density}
  \omb'(x):=\frac{\dd\omb}{\dd x}(x)=\sum_{\sigma\in\RR:\ b(\sigma)=x}\frac{\Im z_b}{\pi\abs{\sigma-z_b}^2}\,\frac1{\abs{b'(\sigma)}} ,
  \qquad
  \frac1{\abs{b'(\sigma)}}=\frac{2\pi\abs{\sigma}\,\abs{\sigma^2-1}}{\abs{\sigma-\sigma_*}\,\abs{\sigma-\sigma^*}} ,
\end{equation}
and on an edge the same with the single preimage. For every $X>\abs{\Re b}$, and with an absolute constant $C_M$,
\begin{equation}\label{eq:B-tails}
  \omb\bigl(\{\tilde b\in\partial D:\ \abs{\Re\tilde b}\ge X\}\bigr)\le2\sqrt2\,e^{-\frac\pi2(X-\abs{\Re b})} ,
  \qquad
  \int_{\partial D}\bigl(1+\abs{\Re\tilde b}\bigr)\,\dd\omb(\tilde b)\le C_M\bigl(1+\abs{\Re b}\bigr) .
\end{equation}
\end{lemma}

\begin{proof}
For (a) we turn $b(z)=w$ into the quadratic \eqref{eq:B-quadratic} and count its roots in $\HH$ by continuity in $w$:
a root can leave $\HH$ only through the real axis, which happens exactly when $w$ is on a slit.

(a) For fixed $z\in\HH$ the angle $\arg(z-\sigma)\in(0,\pi)$ increases with $\sigma\in\RR$, so
$\pi>\arg(z-1)>\arg z>\arg(z+1)>0$ and $2\pi\Im b(z)=\arg(z-1)-\arg z-\arg(z+1)\in(-\pi,\pi)$: $b$ maps $\HH$ into
$\strip$. For $w\in\strip$, $b(z)=w$ is equivalent to $\frac{z-1}{z(z+1)}=e^{2\pi w}$, which is \eqref{eq:B-quadratic}
(the imaginary parts then agree, $2\pi\Im b(z)$ and $2\pi\Im w$ being congruent modulo $2\pi$ and both in
$(-\pi,\pi)$). The two roots depend continuously on $w$ and stay finite. A real root $\sigma$ (necessarily $\sigma\ne0,\pm1$) forces
$e^{2\pi w}=\frac{\sigma-1}{\sigma(\sigma+1)}$ to be real, hence positive because $\abs{\Im w}<\half$, and the
discriminant $(e^{2\pi w}-1)^2-4e^{2\pi w}$ to be non-negative, that is
$e^{2\pi w}\notin\bigl((\sqrt2-1)^2,(\sqrt2+1)^2\bigr)$; since $e^{2\pi\xs}=(\sqrt2+1)^2$, this says that $w$ is real
with $\abs w\ge\xs$, and then both roots are real. Hence the number of roots in $\HH$ is constant on the connected set
$D$, where it equals its value $1$ at $w=0$ ($z^2+1=0$), and it is $0$ on the slits: $b$ is a bijection of $\HH$ onto
$D$.

(b) On $(0,1)$, $(-1,0)$, $(1,\infty)$, $(-\infty,-1)$ the angles
$\bigl(\arg(\sigma-1),\arg\sigma,\arg(\sigma+1)\bigr)$ are $(\pi,0,0)$, $(\pi,\pi,0)$, $(0,0,0)$, $(\pi,\pi,\pi)$, so
$\Im b=\half,0,0,-\half$. On $(-1,0)$ the modulus $\bigabs{\frac{\sigma-1}{\sigma(\sigma+1)}}$ decreases from
$+\infty$ to its minimum $(1+\sqrt2)^2$ at $\sigma_*$ and increases back, so $\Re b$ sweeps $[\xs,\infty)$ twice; by
\eqref{eq:B-bprime} $b'$ has opposite signs on the two halves, and since $b$ preserves orientation, points of $\HH$
near the two halves go to opposite sides of the slit. The interval $(1,\infty)$ is treated in the same way, with the
maximum $(\sqrt2-1)^2$ of the modulus at $\sigma^*$. On $(0,1)$ and $(-\infty,-1)$, $b'\ne0$ and $\Re b$ runs
monotonically over $\RR$. Finally $i(i+1)=i-1$ gives $b(i)=0$, and at $\sigma=\sqrt2-1$ one has
$\abs{\sigma-1}=2-\sqrt2=\sigma(\sigma+1)$, so $b=\frac i2$.

(c) Composition with the map takes harmonic functions on $D$ to harmonic functions on $\HH$ (equivalently, Brownian
paths to Brownian paths), so by Poisson's formula on $\HH$, $\int\varphi\,\dd\omb$ solves the Dirichlet problem on $D$
with boundary values $\varphi$; this is the definition of harmonic measure~\cite[\S4.3]{Ransford}. The density
\eqref{eq:B-density} is the change of variables $x=b(\sigma)$ on each of the two intervals that cover a slit. For the tails we compare with an explicit harmonic function (a barrier):
$\psi_X(w):=\cosh(\frac\pi2\Re w)\cos(\frac\pi2\Im w)\big/\bigl(\cos\frac\pi4\,\cosh\frac\pi2X\bigr)$ is harmonic and
non-negative on $\bar\strip$, and is $\ge1$ where $\abs{\Re w}=X$. The exit probability
$w\mapsto\omega_w(\{\abs{\Re\tilde b}\ge X\})$ is harmonic on $D$, lies in $[0,1]$, and vanishes on the part of
$\partial D$ where $\abs{\Re\tilde b}<X$; by the maximum principle on the bounded domain $D\cap\{\abs{\Re w}<X\}$ it is
at most $\psi_X$ there, and
$\psi_X(b)\le\sqrt2\cosh(\frac\pi2\Re b)/\cosh(\frac\pi2X)\le2\sqrt2\,e^{-\frac\pi2(X-\abs{\Re b})}$.
Using this for $X=\abs{\Re b}+n$, $n\ge1$, we get
$\int(1+\abs{\Re\tilde b})\,\dd\omb\le(2+\abs{\Re b})+\sum_{n\ge1}(2+\abs{\Re b}+n)\,2\sqrt2\,e^{-\pi n/2}$, which is the
second bound.
\end{proof}

\subsection{Entropy}
\label{app:B.2}
Jensen's inequality in \cref{app:B.3} costs the entropy of harmonic measure on the slits, which we bound here. We
write $\log^+t:=\max(\log t,0)$.

\begin{lemma}[\proved]\label{lem:B.2}
\textup{(i)} For every compact $Q\subset D$ there is a constant $C'_Q<\infty$ such that for all $b\in Q$
\begin{equation}\label{eq:B-entropy-compact}
  \int_{\abs x\ge\xs}\frac{\dd\omb}{\dd x}\,\Bigabs{\log\frac{\dd\omb}{\dd x}}\,\dd x\le C'_Q .
\end{equation}
\textup{(ii)} There is an absolute constant $\CH$ such that for every real $b\in(-\xs,\xs)$, at distance
$\delta:=\xs-\abs b$ from the nearer tip,
\begin{equation}\label{eq:B-entropy-delta}
  \int_{\abs x\ge\xs}\frac{\dd\omb}{\dd x}\,\log^+\frac{\dd\omb}{\dd x}\,\dd x\le\CH+5\log\frac1\delta .
\end{equation}
\end{lemma}

\begin{proof}
By \eqref{eq:B-density}, $\omb'$ is the Poisson kernel times the Jacobian $1/\abs{b'}$. The Jacobian has an inverse
square-root singularity at each tip: the map folds the real axis there, so $r:=x-\xs$ is quadratic in $\sigma-\sigma_*$. At the ends it decays like $e^{-2\pi x}$. Since
$r^{-1/2}\log r$ is integrable, (i) follows as long as the Poisson kernel is bounded. In (ii) it is not, because $z_b$
approaches the real axis when $b$ approaches a tip, and we keep track of $\Im z_b$.

\emph{The Jacobian on the right slit.} For $\sigma\in(-1,0)$ we have $\abs\sigma\,\abs{\sigma^2-1}\le1$ and
$\abs{\sigma-\sigma^*}\ge1+\sqrt2$, so $1/\abs{b'(\sigma)}\le2.61/\abs{\sigma-\sigma_*}$. If $\abs{\sigma-\sigma_*}\le0.3$,
every $u$ between $\sigma_*$ and $\sigma$ has $\abs u\ge0.114$, $\abs{u^2-1}\ge0.48$ and $\abs{u-\sigma^*}\le3.14$, so
$\abs{b'(u)}\le9.5\abs{u-\sigma_*}$ by \eqref{eq:B-bprime} and, integrating,
$r=b(\sigma)-b(\sigma_*)\le4.75\,(\sigma-\sigma_*)^2$. Therefore $1/\abs{b'(\sigma)}\le5.7\,r^{-1/2}$ if
$\abs{\sigma-\sigma_*}\le0.3$, and $1/\abs{b'(\sigma)}\le8.7$ otherwise. At the ends $\sigma\to0^-$ and $\sigma\to-1^+$
one reads off $x=\frac1{2\pi}\log\frac1{\abs\sigma}+O(1)$, resp.\ $x=\frac1{2\pi}\log\frac2{\abs{\sigma+1}}+O(\abs{\sigma+1})$,
and $1/\abs{b'}\le C\abs\sigma$, resp.\ $C\abs{\sigma+1}$; in both cases $1/\abs{b'(\sigma)}\le Ce^{-2\pi x}$.

(i) For $b\in Q$ the preimage $z_b$ stays in a compact subset of $\HH$, so the Poisson kernel is at most
$C''_Q(1+\sigma^2)^{-1}$, and the Jacobian bounds give
$0<\omb'(x)\le C''_Q\bigl(r^{-1/2}1_{r\le1}+e^{-2\pi x}\bigr)$ on the right slit. Where $\omb'\le1$ we use
$\omb'\abs{\log\omb'}\le\sqrt{\omb'}\le C''_Q\bigl(r^{-1/4}+e^{-\pi x}\bigr)$, which is integrable; the set where
$\omb'\ge1$ lies in a bounded neighborhood of the tip, where $\omb'\log\omb'\le C''_Qr^{-1/2}\log(C''_Q/r)$ is integrable.
The left slit is the mirror image, because $b(-1/z)=-b(z)$.

(ii) By the same symmetry we may take $0\le b<\xs$. Put $\lambda:=e^{2\pi b}\in[1,\lambda_+)$ and
$\lambda_\pm:=(\sqrt2\pm1)^2$. The discriminant of \eqref{eq:B-quadratic} is $(\lambda-\lambda_+)(\lambda-\lambda_-)<0$,
the two roots are complex conjugate with product $1/\lambda\le1$, and
\begin{equation}\label{eq:B-imz}
  \Im z_b=\frac{\sqrt{(\lambda_+-\lambda)(\lambda-\lambda_-)}}{2\lambda},\qquad
  \Re z_b=\frac{1-\lambda}{2\lambda}\le0,\qquad
  0.3\sqrt\delta\le\Im z_b\le1 .
\end{equation}
The lower bound holds because $\lambda_+-\lambda=\lambda_+(1-e^{-2\pi\delta})\ge\lambda_+\cdot0.46\cdot2\pi\delta\ge16.8\,\delta$
(the chord of the concave function $1-e^{-u}$ over $[0,2\pi\xs]$), $\lambda-\lambda_-\ge1-\lambda_-\ge0.82$ and
$2\lambda\le2\lambda_+\le11.7$. The Poisson kernel is at most $1/(\pi\Im z_b)$, so adding the two preimages of
$x=\xs+r$ and using the Jacobian bounds,
\begin{equation}\label{eq:B-density-bound}
  \omb'(x)\le W(r):=\frac1{\pi\,\Im z_b}\bigl(11.4\,r^{-1/2}+17.4\bigr),
  \qquad
  \log^+W(r)\le\log\frac1{\Im z_b}+\half\log^+\frac1r+3 .
\end{equation}
Since $\log^+$ is non-decreasing and $\omb'\le W$, $\omb'\log^+\omb'\le\omb'\log^+W$; and since $\omb$ has total mass $1$,
\begin{equation}\label{eq:B-entropy-split}
  \int_{x\ge\xs}\omb'\log^+\omb'\,\dd x\le\log\frac1{\Im z_b}+3+\half\int_{x\ge\xs}\omb'(x)\,\log^+\frac1r\,\dd x .
\end{equation}
We split the last integral at $r=(\Im z_b)^2\le1$. Below it we use $\omb'\le W$ and compute, with $y:=\Im z_b\le1$,
$\int_0^{y^2}W\log\frac1r\,\dd r=\frac1{\pi y}\bigl[22.8\,y(2\log\frac1y+2)+17.4\,y^2(2\log\frac1y+1)\bigr]
\le15\log\frac1y+25$ (using $y\log\frac1y\le e^{-1}$). Above it, $\log^+\frac1r\le2\log\frac1{\Im z_b}$. Altogether the
left side of \eqref{eq:B-entropy-split} is at most $10\log\frac1{\Im z_b}+16\le5\log\frac1\delta+29$ by
\eqref{eq:B-imz}. On the left slit, where $\sigma\ge1$, we have $\abs{\sigma-z_b}\ge\sigma$ because $\Re z_b\le0$, so
the Poisson kernel is at most $1/(\pi\sigma^2)$ whatever $\delta$ is; the Jacobian on $(1,\infty)$ has the same
structure as on $(-1,0)$ (the tip at $\sigma^*$, the ends at $1$ and $\infty$). Hence
$\omb'(x)\le C\bigl(\abs{x+\xs}^{-1/2}1_{\abs{x+\xs}\le1}+e^{-2\pi\abs x}\bigr)$ there, with an absolute $C$, and as in (i)
the left slit contributes an absolute constant.
\end{proof}

\subsection{Poisson--Jensen and Jensen on the slits}
\label{app:B.3}
\begin{lemma}[\proved]\label{lem:B.3}
Let $g\in\Abar\setminus\{0\}$, $b\in D$, and $\omb'=\dd\omb/\dd x$ as in \eqref{eq:B-density}.

\textup{(a) (Poisson--Jensen.)} The potential at $b$ is at most the mean of its boundary values over harmonic measure,
\begin{equation}\label{eq:B-PJ}
  \log\abs{G(b)}\le\int_{\partial D}\log\abs G\,\dd\omb
  =\int_{\abs x\ge\xs}\log\abs{G(x)}\,\omb'(x)\,\dd x+\int_{E_+\cup E_-}\log\abs G\,\dd\omb ,
\end{equation}
whether or not $G$ has zeros in $D$ or on $\partial D$.

\textup{(b) (Jensen on the slits.)} Let $m_b:=\omb(\{\abs x\ge\xs\})\in(0,1)$ be the probability of exit through a slit.
Then, with an absolute constant $C$,
\begin{equation}\label{eq:B-jensen}
\begin{aligned}
  \int_{\abs x\ge\xs}\!\log\abs{G}\,\omb'\,\dd x
  \le{}&-c'\!\int_{\abs x\ge\xs}\!L\,\omb'\,\dd x+m_b\log M[g]
   +\int_{\abs x\ge\xs}\!\omb'\log\omb'\,\dd x\\
  &+\frac{m_b}2\log c'+C\bigl(1+\abs{\Re b}\bigr) .
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
(a) We move the problem to the unit disk, where $G$ becomes a bounded holomorphic function, to which the classical fact
quoted after \eqref{eq:6-PJ} applies. Let
$F(w):=G\bigl(b\bigl(i\frac{1+w}{1-w}\bigr)\bigr)$ for $\abs w<1$, the inner map being the M\"obius map of the disk onto
$\HH$. By \cref{sec:6.3}, $G$ is bounded and continuous on $\bar\strip$, so $F$ is bounded and holomorphic; and
$F\not\equiv0$, because $G$ is holomorphic on the connected set $\strip$ and $G(x)=g(c'x^2/3)$ does not vanish
identically on the real axis. For such $F$ the radial limits $F^*$ exist almost everywhere on the unit circle,
$\log\abs{F^*}$ is integrable, and $F=F_{\rm in}F_{\rm out}$, where $\abs{F_{\rm in}}\le1$ (this factor carries the zeros)
and $\log\abs{F_{\rm out}}$ is the Poisson integral of $\log\abs{F^*}$~\cite[Ch.~II, Thm.~4.1 and Cor.~4.2]{Garnett},
\cite[Thms.~17.16--17.17]{Rudin}. Hence $\log\abs{F(w)}$ is at most that
Poisson integral. By \cref{lem:B.1}(b) the composite map extends continuously to the circle minus four points, so
$F^*$ is the composition of $G$ with it almost everywhere, and carrying the Poisson measure of the disk to $\partial D$
gives $\omb$ by \cref{lem:B.1}(c).

(b) This is Jensen's inequality \eqref{eq:6-jensen} with $w=\omb'/m_b$ and $K=K_a$, with the constants tracked. On the
slits put $\ell(x):=\log K_a(c'x^2/3)$ and write
$\log\abs G=\log\bigl(\abs Ge^{\ell}/\omb'\bigr)-\ell+\log\omb'$. Jensen's inequality for the probability measure
$\omb'\dd x/m_b$, followed by \cref{lem:6.2}, gives
\begin{equation}\label{eq:B-jensen-step}
\begin{aligned}
  \int_{\abs x\ge\xs}\!\!\log\frac{\abs Ge^{\ell}}{\omb'}\;\omb'\,\dd x
  &\le m_b\log\Bigl(\frac1{m_b}\!\int_{\abs x\ge\xs}\!\!\abs G K_a\,\dd x\Bigr)\\
  &\le m_b\log M[g]+m_b\log\frac{12}{c'm_b}\le m_b\log M[g]+\frac1e ,
\end{aligned}
\end{equation}
since $12/c'<1$ and $m\log\frac1m\le e^{-1}$. By \cref{lem:A.2}(e), $-\ell(x)\le-c'L(x)+\half\log\bigl(c'L(x)\bigr)+2$
for $\abs x\ge\xs$, and, since $\log L\le L=\frac{2\pi}3\abs x$,
$\int_{\abs x\ge\xs}\bigl(\half\log(c'L)+2\bigr)\omb'\,\dd x\le\frac{m_b}2\log c'+\int_{\partial D}\bigl(2+\half L\bigr)\dd\omb
\le\frac{m_b}2\log c'+C(1+\abs{\Re b})$ by \eqref{eq:B-tails}. The term $\log\omb'$ gives the entropy, which converges
absolutely by \cref{lem:B.2}(i). Adding the three contributions gives \eqref{eq:B-jensen}.
\end{proof}

\subsection{The edge bound and the proof of the majorant proposition}
\label{app:B.4}
\begin{lemma}[\proved]\label{lem:B.4}
This is \cref{lem:6.1}, with the supremum taken over the closed strip. There is an absolute constant $C$ such that for
all $c\ge40$, all $g\in\Abar$ and all real $x$,
\begin{equation}\label{eq:B-edge}
  \bigabs{G(x\pm\tfrac i2)}\le2\pi\!\int_0^\infty\!\abs g\,K_a\,\dd t\le C\,a\,c'\,M[g] ,
  \quad\text{hence}\quad
  \log\bigabs{G(x\pm\tfrac i2)}\le\log M[g]+2\log c'+C ,
\end{equation}
uniformly in $x$ and in $g$ (in particular in the order $P$). The same bound holds for $\sup_{\bar\strip}\abs G$.
\end{lemma}

\begin{proof}
Let $b\in\bar\strip$ and $\alpha:=-c'b^2/3$, so that $G(b)=g(-\alpha)$. Then $\sqrt\alpha=\pm i\sqrt{c'/3}\,b$ and
$(\Re\sqrt\alpha)^2=\frac{c'}3(\Im b)^2\le\frac{c'}{12}=a$, with equality on the edges $b=x\pm\frac i2$, for every real $x$. So
$\alpha$ lies in the region $\mathcal D_a$ of \cref{lem:A.1}, where $\abs{K_\alpha}\le K_a$ by \eqref{eq:A-kernel-bound}, and
\eqref{eq:A-eigen} gives $\abs{G(b)}\le2\pi\int_0^\infty\abs gK_a\,\dd t\le2\pi(87+78\,a\,c')M[g]\le C\,a\,c'\,M[g]$. The polynomial
factor comes from $K_a\le78\,a\,c'\,V$ for $t\ge4/c'$ (\cref{lem:A.2}(b)) and from
$\int_0^{4/c'}\abs g\,K_a\,\dd t\le87M[g]$ on the window at the BTZ threshold (\cref{lem:A.3}). Finally
$\log(C\,a\,c')=2\log c'+\log\frac C{12}$.
\end{proof}

\begin{proof}[Proof of \cref{prop:6.4}]
The idea is described in \cref{sec:6.5}. Let $h(b):=-\int_{\abs x\ge\xs}L\,\dd\omb$. Inserting \eqref{eq:B-jensen} and \eqref{eq:B-edge} (with
$\omb(E_+\cup E_-)=1-m_b$) into \eqref{eq:B-PJ}, and using $\half m_b+2(1-m_b)\le2$ for the coefficient of $\log c'$, we
get for every $b\in D$, every $g\in\Abar\setminus\{0\}$ and every $c\ge40$
\begin{equation}\label{eq:B-master}
  \log\abs{G(b)}\le c'\,h(b)+\log M[g]+\int_{\abs x\ge\xs}\!\omb'\log\omb'\,\dd x+2\log c'+C\bigl(1+\abs{\Re b}\bigr) .
\end{equation}
It remains to recognize $h$. By \cref{lem:B.1}(c), $h\circ b$ is the Poisson integral on $\HH$ of the function equal
to $-L(\Re b(\sigma))$ on $(-1,0)\cup(1,\infty)$ and to $0$ on $(0,1)\cup(-\infty,-1)$. This function is continuous on
$\RR\setminus\{-1,0,1\}$, in particular at $\sigma_*$ and $\sigma^*$, and, by \eqref{eq:B-tails}, integrable against the
Poisson measure. Hence $h$ is harmonic on $D$, takes the boundary values $-L$ on both lips of the slits and at the two
tips and $0$ on the edges, continuously, and grows at most linearly, $\abs{h(b)}\le\frac{2\pi}3C_M(1+\abs{\Re b})$. It is
therefore the function $U_{\xs}$ of \cref{prop:6.4}, which these properties determine uniquely (part (d) of the proof of
\cref{lem:6.5}, where it is also shown that $U_{\xs}=U$):
\begin{equation}\label{eq:B-Poisson-U}
  U_{\xs}(b)=-\int_{\abs x\ge\xs}L(x)\,\dd\omb(x)\le0\qquad(b\in D).
\end{equation}
For $b$ in a compact set $Q\subset D$ the entropy in \eqref{eq:B-master} is at most $C'_Q$ by \cref{lem:B.2}(i) and
$\abs{\Re b}$ is bounded; dividing by $c'$ gives $u_g(b)\le U_{\xs}(b)+\CQ(1+\log c')/c'$ with
$\CQ:=\max\bigl(2,\,C'_Q+C\sup_Q(1+\abs{\Re b})\bigr)$, which depends on $Q$ only.
\end{proof}

\emph{Dependence on the distance to the tip.} For real $b=x\in(-\xs,\xs)$ we use \cref{lem:B.2}(ii)
in~\eqref{eq:B-master} instead, the entropy being at most $\int\omb'\log^+\omb'\,\dd x$. With $U_{\xs}=U$ this gives, with
an absolute constant $C$ and for all $c\ge40$ and $g\in\Abar\setminus\{0\}$,
\begin{equation}\label{eq:B-majorant-delta}
  \log\abs{G(x)}\le c'\,U(x)+\log M[g]+5\log\frac1{\xs-\abs x}+3\log c'+C\qquad(\abs x<\xs) .
\end{equation}
The argument gives $2\log c'$; we state $3\log c'$, as in the count of powers of $c'$ in \cref{sec:6.7}.

\subsection{Strict deficit}
\label{app:B.5}
We prove the last inequality of \cref{lem:6.7}. By that lemma,
$\kappa(\delta)=\frac43\int_0^\delta\bigl[\frac\pi2-\arcsin\bigl(\sinh\pi(\xs-u)\bigr)\bigr]\dd u$ for $0<\delta\le\xs$,
and we bound the integrand from below by its square-root behavior at the tip. First,
$\frac\pi2-\arcsin y=\arccos y\ge\sqrt{2(1-y)}$ for $0\le y\le1$: both sides vanish at $y=1$, and
$\frac{\dd}{\dd y}\arccos y=-(1-y)^{-1/2}(1+y)^{-1/2}\le-(2(1-y))^{-1/2}=\frac{\dd}{\dd y}\sqrt{2(1-y)}$. Second, since
$\sinh\pi\xs=1$, we have $1-\sinh\pi(\xs-u)=\int_{\xs-u}^{\xs}\pi\cosh(\pi v)\,\dd v\ge\pi u$ for $0\le u\le\xs$. Hence the
integrand is at least $\sqrt{2\pi u}$, and
\begin{equation}\label{eq:B-deficit}
  \kappa(\delta)\ge\frac43\sqrt{2\pi}\int_0^\delta\sqrt u\,\dd u=\frac{8\sqrt{2\pi}}9\,\delta^{3/2}
  =2.228\ldots\,\delta^{3/2}\qquad(0<\delta\le\xs).
\end{equation}

\section{The zero law on the strip}
\label{app:C}

This appendix proves the two statements of \cref{sec:7} that tie the comb $B$ to the extremal potential $U$: the Riesz
representation (\cref{lem:7.1}; on the real axis it is the continuum limit of the sums of \cref{app:E}) and the
edge-phase lemma (\cref{lem:7.6}; its sharper form \cref{lem:C-edge-phase} is what verifies \hyp{SD2} in
\cref{app:D.4}). A Poisson identity (\cref{lem:C.1}) links the two, and \cref{app:C.4} derives the eikonal identity of
\cref{sec:7.6}.

Throughout, $b=x+iy$, and $b=v+\frac i2$ on $E_+$; $A\lesssim B$ means $A\le CB$ with $C$ depending at most on the
width $\delta_1$ of \cref{app:C.3}. Besides $U$ we use the complex potential of \cref{lem:6.5},
\begin{equation}\label{eq:C-complexU}
 \mathcal U(b):=U_0-\frac{4\pi i}3\int_0^b p(b')\,\dd b',\qquad \Re\,\mathcal U=U,\qquad
 \dd U=\tfrac{4\pi}3\bigl(\Im p\,\dd x+\Re p\,\dd y\bigr),
\end{equation}
analytic on $D$, with $p$ continuous up to $E_+$ and $p(v+\frac i2)=F(v)$. We recall from \cref{lem:6.5,def:6.6} that
$U$ is harmonic on $D$, continuous on $\bar\strip$, zero on $\Epm$, equal to $-L$ on both lips of the slits, and
$\abs{U(b)}\le L(\abs x)+C$; and that $U$ is subharmonic on $\strip$ with charge
$\frac1{2\pi}\Delta U=\rhoU(\abs x)\,\dd x\otimes\delta_0(\dd y)$ in the sense of distributions (across the slits $U$ is
continuous, $\nabla U$ is locally bounded, and $\partial_yU$ jumps by $2\pi\rhoU(\abs x)$). We also use
$\partial_w\log\tanh\frac{\pi w}2=\pi/\sinh\pi w$ and, for real $u,\theta$,
$\abs{\sinh\pi(u+i\theta)}^2=\sinh^2\pi u+\sin^2\pi\theta$.

\subsection{Riesz representation of \texorpdfstring{$U$}{U}}
\label{app:C.1}

\begin{proof}[Proof of \cref{lem:7.1}]
\emph{Plan.} The superposition $u$ of the kernels $\log\abs{T_{x'}}$ at density $\rhoU$ has the charge, the edge values
and the growth of $U$; so $U-u$ is harmonic, and vanishes by Phragm\'en--Lindel\"of.

\emph{The kernel.} By \hyp{T1} and \hyp{T4},
$b\mapsto\log\abs{T_{x'}(b)}=\log\bigabs{\tanh\frac{\pi(b-x')}2}+\log\bigabs{\tanh\frac{\pi(b+x')}2}$ is harmonic on
$\strip\setminus\{\pm x'\}$, vanishes on $\Epm$, and has unit logarithmic poles at $\pm x'$: it is minus the Green's
function of $\strip$ with poles at $\pm x'$ ($b\mapsto\tanh\frac{\pi b}2$ maps $\strip$ conformally onto the unit
disk). From
$T_{x'}=(\cosh\pi b-\cosh\pi x')/(\cosh\pi b+\cosh\pi x')$,
\begin{equation}\label{eq:C-kernel-decay}
 \bigabs{\log\abs{T_{x'}(b)}}\lesssim e^{-\pi\abs{x'-\abs x}}\qquad\bigl(b\in\bar\strip,\ \abs{x'-\abs x}\ge1\bigr).
\end{equation}

\emph{The superposition.} Since $\rhoU(x')\lesssim1+x'$, the integral
$u(b):=\int_{\xs}^\infty\rhoU(x')\log\abs{T_{x'}(b)}\,\dd x'$ converges absolutely and locally uniformly on
$\bar\strip$ (the logarithmic singularity at $x'=\abs x$, $y=0$ is integrable). Hence $u$ is subharmonic on $\strip$
with the same charge as $U$, continuous on $\bar\strip$ with $u=0$ on $\Epm$ (dominated convergence), and
$\abs{u(b)}\le C(1+\abs x)$ by \eqref{eq:C-kernel-decay}.

\emph{Uniqueness.} The difference $d:=U-u$ satisfies $\Delta d=0$ in the sense of distributions, so it is harmonic on
$\strip$ (Weyl's lemma~\cite[\S3.7]{Ransford}); it is continuous on $\bar\strip$, vanishes on $\Epm$, and
$\abs{d(b)}\le C'(1+\abs x)$. For $\epsilon>0$ the function $d-\epsilon\cosh2x\,\cos2y$ is harmonic on $\strip$, is
$\le0$ on $\Epm$ because $\cos1>0$, and is negative on the cross-cuts $\abs x=X$ once $X$ is large; by the maximum
principle it is $\le0$ on $\strip$. Letting $\epsilon\to0$, and repeating the argument with $-d$, we get $d\equiv0$.
\end{proof}

\subsection{The Poisson identity}
\label{app:C.2}

\begin{lemma}[Poisson identity; \proved]\label{lem:C.1}
For every real $v$,
\begin{equation}\label{eq:C-poisson}
 2\pi\int_{\xs}^\infty N'(x')\bigl[\sech\pi(v-x')+\sech\pi(v+x')\bigr]\dd x'=\frac{4\pi}3\,F(v).
\end{equation}
\end{lemma}

\begin{proof}
\emph{Plan.} Both sides are the normal derivative $\partial_yU$ at the point $v+\frac i2$ of $E_+$, taken from inside
the strip: the right side from the closed form \eqref{eq:C-complexU}, the left side from the Riesz representation.

By \eqref{eq:C-complexU}, $\partial_yU=\frac{4\pi}3\Re p$ on $D$ and $p$ is continuous up to $E_+$, so
$\lim_{y\uparrow1/2}\partial_yU(v+iy)=\frac{4\pi}3F(v)$. By \hyp{T4} each kernel
$\log\abs{T_{x'}}$ is harmonic on $0<y<1$, with $\partial_y\log\abs{T_{x'}}=\Re[i\,\partial_b\log T_{x'}]$ and
$\partial_b\log T_{x'}(b)=\pi/\sinh\pi(b-x')+\pi/\sinh\pi(b+x')$. On $\frac14\le y\le\frac34$ the latter is at most
$2\sqrt2\,\pi$ in modulus and is $\lesssim e^{-\pi\abs{x'-\abs x}}$ for $\abs{x'-\abs x}\ge1$; as
$N'(x')\lesssim1+x'$, the integral $u$ of \cref{app:C.1} and its first derivatives converge there locally
uniformly. So $u$ extends harmonically across $E_+$ and may be differentiated under the integral sign: at
$b=v+\frac i2$, $\sinh\pi(v\mp x'+\frac i2)=i\cosh\pi(v\mp x')$ gives
$\partial_y\log\abs{T_{x'}}=\pi[\sech\pi(v-x')+\sech\pi(v+x')]$, the inward normal derivative of the Green's function at
the edge, and with $\rhoU=2N'$ (\cref{def:6.6}) the value $\partial_yu(v+\frac i2)$ is the left side
of~\eqref{eq:C-poisson}. As $U=u$ on $\strip$ by \cref{lem:7.1} and $u$ is $C^1$ across $E_+$, the two values agree.
\end{proof}

\subsection{The edge phase}
\label{app:C.3}

Fix $\delta_1\in(0,\frac14]$ and let $\cN_{\delta_1}:=\{b:\abs{y-\frac12}\le\delta_1\}$ be the thin two-sided strip
about $E_+$, as in~\eqref{eq:D-coord}. For $\Re\zeta\ge\xs$ and $\abs{\Im\zeta}\le\frac18$ the factor $T_\zeta$ has
neither zeros nor poles at distance less than $\frac18$ from $\cN_{\delta_1}$, by \hyp{T4}, and $T_\zeta\to-1$ as
$\Re\zeta\to\infty$. We write
\begin{equation}\label{eq:C-phi}
 \varphi_\zeta(b):=\log\bigl(-T_\zeta(b)\bigr),\qquad
 \partial_b\varphi_\zeta(b)=\frac\pi{\sinh\pi(b-\zeta)}+\frac\pi{\sinh\pi(b+\zeta)} ,
\end{equation}
for the branch on $\cN_{\delta_1}$ that is continuous in $(b,\zeta)$ and tends to $0$ as $\Re\zeta\to\infty$. It is analytic
in $b$ and in $\zeta$, $\Re\varphi_\zeta=\log\abs{T_\zeta}$, and $\varphi_\zeta(\frac i2)=0$ by \hyp{T3}. Since
$\abs{\sinh\pi(u+i\theta)}\ge\abs{\sin\pi\theta}$, on $\cN_{\delta_1}$ both $\abs{\partial_b\varphi_\zeta}$ and
$\abs{\partial_\zeta\varphi_\zeta}$ are at most $2\pi/\cos\pi(\delta_1+\abs{\Im\zeta})$, and
\begin{equation}\label{eq:C-dphi}
 \abs{\varphi_\zeta(b)}\lesssim\min\bigl(1,e^{-\pi(\Re\zeta-\abs x)}\bigr),\qquad
 \bigabs{\partial_b^i\partial_\zeta^j\varphi_\zeta(b)}\lesssim e^{-\pi\abs{\Re\zeta-\abs x}}
 \quad(1\le i+j\le4).
\end{equation}
As $x\to\pm\infty$ at fixed $\zeta$, $T_\zeta\to+1$ and $\varphi_\zeta\to\mp i\pi$, so only the derivatives decay on
both sides. In the comb, $\varepsilon_n\le\eta_0(\chi_3-\chi_2)\lesssim c'^{-2/3}$ (the spacings decrease along the band because
$N'$ increases), and we take $c$ large enough that $\max_n\varepsilon_n\le\frac18$.
On $\cN_{\delta_1}$ we fix the branch
\begin{equation}\label{eq:C-logB}
 \log B:=\log T_{\chi_2}+\sum_{n\ge3}\bigl[\varphi_{\chi_n+i\varepsilon_n}+\varphi_{\chi_n-i\varepsilon_n}\bigr],
 \qquad \log T_{\chi_2}:=i\pi+\varphi_{\chi_2},
\end{equation}
that is, $\log T_{\chi_2}(\frac i2)=i\pi$ \hyp{T3}; the sum converges absolutely and locally uniformly by
\eqref{eq:C-dphi}. \Cref{lem:7.6} is the restriction to $E_+$ of the following statement.

\begin{lemma}[Edge phase; \proved]\label{lem:C-edge-phase}
On $\cN_{\delta_1}$ let
\begin{equation}\label{eq:C-Phi0}
 \Phi_0(b):=\int_{\xs}^\infty\rhoU(x')\,\varphi_{x'}(b)\,\dd x' .
\end{equation}
\begin{enumerate}[label=\textup{(\alph*)},leftmargin=*,itemsep=1pt]
\item $\Phi_0$ is analytic on $\cN_{\delta_1}$, $\Re\Phi_0=0$ on $E_+$, $\Phi_0(\frac i2)=0$, and
 \begin{equation}\label{eq:C-Phi0-deriv}
  \Phi_0^{(j+1)}\bigl(v+\tfrac i2+iy'\bigr)=-\frac{4\pi i}3\,F^{(j)}(v+iy')
  \qquad(v\in\RR,\ \abs{y'}\le\delta_1,\ j=0,\dots,3);
 \end{equation}
 in particular $\partial_v\Im\Phi_0(v+\frac i2)=-\frac{4\pi}3F(v)$. Here $F(w)=\frac1\pi\arcsinh(\cosh\pi w)$ is
 analytic in $\abs{\Im w}<\frac12$; it is even, $F(0)=\xs$, $F''>0$ on $\RR$ and $0<F'<1$ on $(0,\infty)$. The
 constants $C_2:=\sup_{\cN_{\delta_1}}\abs{\Phi_0'''}$ and $C_3:=\sup_{\cN_{\delta_1}}\abs{\Phi_0''''}$ are finite and depend on
 $\delta_1$ only.
\item On $\cN_{\delta_1}$, $\log B=c'\Phi_0+\Phi_1$ with
 \begin{equation}\label{eq:C-Phi1}
  \Phi_1=i\pi-2n_1\bar\varphi+\Delta_{\rm det}+R_{\rm EM},\qquad
  \Delta_{\rm det}:=\sum_{n\ge3}\bigl[\varphi_{\chi_n+i\varepsilon_n}+\varphi_{\chi_n-i\varepsilon_n}-2\varphi_{\chi_n}\bigr],
 \end{equation}
 in this order the sign of the simple zero \textup{(}$B(\frac i2)=-1$\textup{)}, the charge of the law that is absent
 from the comb \textup{(}$n_1$ sites spread over $[\xs,\chi_2]$\textup{)}, the detuning, and the discreteness of the
 sites; $\bar\varphi:=N(\chi_2)^{-1}\int_{\xs}^{\chi_2}N'(x')\,\varphi_{x'}\,\dd x'=\varphi_{\xs}+O((n_1/c')^{2/3})$, and
 $R_{\rm EM}$ is defined in \eqref{eq:C-EM}.
\item $\Re\Phi_1=0$ on $E_+$, and $\sup_{\cN_{\delta_1}}\abs{\Re\Phi_1}\le M_0$, $\sup_{\cN_{\delta_1}}\abs{\Phi_1'}\le M_1$,
 $\sup_{\cN_{\delta_1}}\abs{\Phi_1''}\le M_1'$ with constants depending on $(n_1,\eta_0,\delta_1)$ only, given in
 \eqref{eq:C-M1}. Moreover, with $C$ and $\CEM$ depending on $\delta_1$ only, uniformly in $x=\Re b$,
 \begin{equation}\label{eq:C-rem-bounds}
  \bigabs{\Delta_{\rm det}^{(i)}}\le C\eta_0^2\,c'^{-2/3},\qquad \bigabs{R_{\rm EM}^{(i)}}\le\CEM\,c'^{-2/3}
  \qquad(i=0,1,2;\ b\in\cN_{\delta_1}).
 \end{equation}
 In particular $\Phi_1$, $\Phi_1'$ and $\Phi_1''$ are bounded on $\cN_{\delta_1}$ uniformly in $c'$.
\end{enumerate}
\end{lemma}

\begin{proof}
\emph{Plan.} Part (a) is the Poisson identity read along the edge, together with the Riesz representation, which
identifies $\Phi_0$ with $\mathcal U$ continued across $E_+$. Part (b) is Euler--Maclaurin summation in the continuous
site index. Part (c) bounds the terms of $\Phi_1$ one by one from the decay \eqref{eq:C-dphi} of the kernel and the
behavior of $N$ at the tip.

\emph{(a).} By \eqref{eq:C-dphi} and $\rhoU(x')\lesssim1+x'$, the integral \eqref{eq:C-Phi0} and its
$b$-derivatives converge absolutely and locally uniformly on $\cN_{\delta_1}$, so $\Phi_0$ is analytic there. On $E_+$,
$\Re\varphi_{x'}=\log\abs{T_{x'}}=0$ by \hyp{T1}, so $\Re\Phi_0=0$; and $\varphi_{x'}(\frac i2)=0$ gives
$\Phi_0(\frac i2)=0$. By \eqref{eq:C-phi},
$\partial_b\varphi_{x'}(v+\frac i2)=-i\pi[\sech\pi(v-x')+\sech\pi(v+x')]$, whence
$\Phi_0'(v+\frac i2)=-\frac{4\pi i}3F(v)$ by \cref{lem:C.1}. Next, $\Re\Phi_0=U=\Re\,\mathcal U$ on
$\cN_{\delta_1}\cap\strip$ by \cref{lem:7.1}; this set does not meet the slits, as $\delta_1\le\frac14$, and two analytic
functions with equal real parts differ by an imaginary constant, so $\Phi_0=\mathcal U+i\gamma_0$ and
$\Phi_0'=-\frac{4\pi i}3p$ there: $\Phi_0$ is the complex potential continued across $E_+$. This continues $p$
analytically to $\cN_{\delta_1}$, with $p(w+\frac i2)=F(w)$ for $\abs{\Im w}\le\delta_1$, which is \eqref{eq:C-Phi0-deriv}.
Differentiating, $F'(w)=\sinh\pi w\,(1+\cosh^2\pi w)^{-1/2}$ and $F''(w)=2\pi\cosh\pi w\,(1+\cosh^2\pi w)^{-3/2}$, as
in \eqref{eq:D-Fprime}, which for real $w$ gives the sign statements. In $\abs{\Im w}\le\frac12$ the only singularities of $F$ are where $\cosh\pi w=\pm i$, on
$\abs{\Im w}=\frac12$; so $F''$ and $F'''$ are continuous on $\abs{\Im w}\le\delta_1$, they tend to $0$ as
$\abs{\Re w}\to\infty$, and $C_2=\frac{4\pi}3\sup_{\abs{\Im w}\le\delta_1}\abs{F''}$,
$C_3=\frac{4\pi}3\sup_{\abs{\Im w}\le\delta_1}\abs{F'''}$ are finite.

\emph{(b).} Let $\chi(n):=N^{-1}\bigl((n-\frac14)/c'\bigr)$ for real $n\ge2$, so that $\chi(n)=\chi_n$ at integers and
$\chi'(n)=1/(c'N'(\chi(n)))$; as $(n-\frac14)/c'\ge n_1/c'>0$ the tip is avoided, and $f(n):=2\varphi_{\chi(n)}(b)$ is
smooth and decays with its derivatives like $e^{-\pi(\chi(n)-\abs x)}$. By \eqref{eq:C-logB} and the definition of
$\Delta_{\rm det}$, $\log B=i\pi-\varphi_{\chi_2}+\sum_{n\ge2}f(n)+\Delta_{\rm det}$. The second-order Euler--Maclaurin
formula gives
\begin{equation}\label{eq:C-EM}
 \sum_{n\ge2}f(n)=\int_2^\infty\! f\,\dd n+\frac12f(2)+R_{\rm EM},\qquad
 R_{\rm EM}:=-\frac1{12}f'(2)+R_2,
\end{equation}
with $\abs{R_2}\le\frac1{12}\int_2^\infty\abs{f''}\,\dd n$, and the substitution $x'=\chi(n)$, $\dd n=c'N'(x')\,\dd x'$ turns the integral into
$c'\int_{\chi_2}^\infty2N'\varphi_{x'}\,\dd x'=c'\Phi_0-2c'N(\chi_2)\,\bar\varphi=c'\Phi_0-2n_1\bar\varphi$. With
$\frac12f(2)=\varphi_{\chi_2}$ this is \eqref{eq:C-Phi1}. The estimate for $\bar\varphi$ follows from
$\abs{\partial_{x'}\varphi_{x'}}\le2\pi/\cos\pi\delta_1$ and $\chi_2-\xs\asymp(n_1/c')^{2/3}$. For the latter we use no
certified constant. Part~(d) of the proof of \cref{lem:E-tip} shows that $\Phi:=\rhoU^2$ has $\Phi(\xs)=0$,
$\Phi'(\xs)=9\ktip^2$ and $0\le\Phi''\le\frac{32}9$ on $[\xs,\infty)$. Hence
\[
  9\ktip^2\,r\ \le\ \rhoU(\xs+r)^2\ \le\ 9\ktip^2\,r+\tfrac{16}9r^2\qquad(r>0),
\]
and $\frac{16}9r^2\le\frac54\cdot9\ktip^2r$ for $r\le1$, because $16/(81\ktip^2)=\pi/(2\sqrt2)\le\frac54$. Integrating
$N'=\half\rhoU$, we get $\ktip r^{3/2}\le N(\xs+r)$ for every $r>0$, and $N(\xs+r)\le\frac32\ktip r^{3/2}$ for $r\le1$. We
use this at $r=\chi_2-\xs$, where $c'N(\chi_2)=n_1$; the first bound gives $r\le n_1^{2/3}\lc$, which is at most $1$ for
$c'\ge5$.

\emph{(c).} Each term of \eqref{eq:C-Phi1} is analytic on $\cN_{\delta_1}$. First, $\bar\varphi$ is an average of
$\varphi_{x'}$ over $x'$, so $\abs{\partial_b(2n_1\bar\varphi)}\le n_1\cdot4\pi/\cos\pi\delta_1$ and
$\abs{\partial_b^2(2n_1\bar\varphi)}\le n_1\sup_{x'}\sup_{\cN_{\delta_1}}\abs{2\partial_b^2\varphi_{x'}}$. Second, each summand of $\Delta_{\rm det}$ is a second difference in $\zeta$ of the analytic
function $\zeta\mapsto\varphi_\zeta(b)$:
$\varphi_{\chi+i\varepsilon}+\varphi_{\chi-i\varepsilon}-2\varphi_\chi
=-\int_0^\varepsilon(\varepsilon-s)[\partial_\zeta^2\varphi_{\chi+is}+\partial_\zeta^2\varphi_{\chi-is}]\,\dd s$, which
with up to two $b$-derivatives is $\lesssim\varepsilon^2e^{-\pi\abs{\chi-\abs x}}$ by \eqref{eq:C-dphi}. With
$\varepsilon_n=\eta_0(\chi_{n+1}-\chi_n)$ and $\chi_{n+1}-\chi_n\le\chi_3-\chi_2\asymp c'^{-2/3}$,
\begin{equation*}
 \bigabs{\Delta_{\rm det}^{(i)}(b)}\lesssim\sum_{n\ge3}\varepsilon_n^2e^{-\pi\abs{\chi_n-\abs x}}
 \le\eta_0^2\,(\chi_3-\chi_2)\sum_{n\ge3}(\chi_{n+1}-\chi_n)\,e^{-\pi\abs{\chi_n-\abs x}}\lesssim\eta_0^2\,c'^{-2/3},
\end{equation*}
the last sum being, up to the factor $e^{\pi s}\le e^\pi$ with $s=\chi_3-\chi_2$ the largest spacing, a Riemann sum for
$\int e^{-\pi\abs{x'-\abs x}}\dd x'\le2/\pi$. Third, the remainder: at
$x'=\chi(n)$, $f'(n)=2\partial_{x'}\varphi_{x'}/(c'N')$ and
$f''(n)=2\partial_{x'}^2\varphi_{x'}/(c'N')^2-2\partial_{x'}\varphi_{x'}\,N''/(c'^2N'^3)$.
Near the tip $N'(x')\asymp(x'-\xs)^{1/2}$ and $\abs{N''(x')}\asymp(x'-\xs)^{-1/2}$, deep in the band
$N'(x')\asymp x'$ (for $0<x'-\xs\le1$ by the bounds for $\rhoU^2$ in part~(b) of this proof and by
\eqref{eq:E-dlogrho}, since $N'=\half\rhoU$ and $N''=\half\rhoU\,(\log\rhoU)'$; beyond, by \eqref{eq:6-rhoU}), and
$\chi_2-\xs\asymp\lc$, so $c'N'(\chi_2)\asymp c'\lc^{1/2}\asymp c'^{2/3}$ and
$\abs{f'(2)}\lesssim c'^{-2/3}$. Likewise, by \eqref{eq:C-dphi},
\begin{equation*}
 \int_2^\infty\!\abs{f''}\,\dd n=\int_{\chi_2}^\infty\!\abs{f''}\,c'N'\,\dd x'
 \lesssim\frac1{c'}\int_{\chi_2}^\infty\Bigl[\frac{\abs{\partial_{x'}^2\varphi_{x'}}}{N'}
   +\frac{\abs{\partial_{x'}\varphi_{x'}}\abs{N''}}{N'^2}\Bigr]\dd x'
 \lesssim\frac{1+\lc^{-1/2}}{c'}\lesssim c'^{-2/3}
\end{equation*}
uniformly in $x$, the integrand being $\lesssim(x'-\xs)^{-1/2}+(x'-\xs)^{-3/2}$ for $x'\le\xs+1$ and
$\lesssim e^{-\pi\abs{x'-\abs x}}/x'$ beyond. The same computation with $\partial_b$, $\partial_b^2$ applied under the
sums and integrals bounds $R_{\rm EM}'$ and $R_{\rm EM}''$; this is \eqref{eq:C-rem-bounds}, with $\CEM$ the resulting
constant. Adding the contributions and using $c'^{-2/3}\le1$, the bounds of the statement hold with
\begin{equation}\label{eq:C-M1}
 M_1:=(n_1+1)\frac{4\pi}{\cos\pi\delta_1}+C\eta_0^2+\CEM,\qquad M_0:=M_1\delta_1,
\end{equation}
and likewise for $M_1'$ with $\sup\abs{2\partial_b^2\varphi_{x'}}$ in place of $4\pi/\cos\pi\delta_1$ (the $+1$ is a
crude allowance for the simple zero, which contributes only the constant $i\pi$); $\Phi_1$ itself is bounded by
\eqref{eq:C-Phi1}, $\abs{\varphi_{x'}}\lesssim1$ and~\eqref{eq:C-rem-bounds}. Finally $\Re\Phi_1=0$ on
$E_+$, because $\Re\log B=\log\abs B=0$ there (\cref{prop:7.2}) and $\Re\Phi_0=0$ by (a); integrating $\Phi_1'$
vertically from the edge, $\abs{\Re\Phi_1(v+\frac i2+iy')}\le M_1\abs{y'}\le M_0$.
\end{proof}

\subsection{The eikonal identity}
\label{app:C.4}

\tagpar{\orientation} At order $c'$ the phase of the integrand of the edge representation (\cref{lem:7.3}(c)) is
$c'[\Im\Phi_0(v+\frac i2)+\frac{4\pi}3xv]$; by \eqref{eq:C-Phi0-deriv} it is stationary where $F(v)=x$, that is at
$v=\pm v_0(x)$. Along the saddle $b_0(x)=v_0(x)+\frac i2$, with $k_x=\frac{4\pi}3c'x$,
\begin{equation}\label{eq:C-eikonal}
 \frac{\dd}{\dd x}\Bigl[c'\Im\Phi_0(b_0)+k_xv_0\Bigr]
 =\Bigl(c'\partial_v\Im\Phi_0\bigl(v_0+\tfrac i2\bigr)+k_x\Bigr)v_0'+\frac{4\pi}3c'v_0
 =\pi c'\rhoU(x)=\Psi'(x),
\end{equation}
the bracket vanishing by stationarity, and the last two steps being $v_0=\frac34\rhoU$ and $\Psi=2\pi\nu-\pi$ with
$\nu'=c'N'=\frac12c'\rhoU$. Since $v_0(\xs)=0$, $\Phi_0(\frac i2)=0$ and $N(\xs)=0$, the integrated form is
$c'\Im\Phi_0(b_0(x))+k_xv_0(x)=2\pi c'N(x)$ for $x\ge\xs$. The omitted sites and the node phase live in $\Phi_1$ and
do not enter.

\section{Two-saddle steepest descent with a uniform remainder}
\label{app:D}

This appendix proves \cref{lem:7.7}: for band points $x\ge\xI$, $\hat h_0(c'x^2/3)$ is the sum
$\Amp\cos\hPsi$ of two saddle-point contributions, up to a relative remainder $\epssd$ that is bounded explicitly and tends
to zero uniformly in $x$ (\cref{lem:D.1}, \cref{cor:D.2}), and obeys a global bound that decays in $x$ (\cref{cor:D.3}). The
comb enters only through the edge representation of \cref{lem:7.3}(c) and the edge phases $\Phi_0,\Phi_1$ of
\cref{lem:7.6}, which we abstract into hypotheses \hyp{SD1}--\hyp{SD4} and verify in \cref{app:D.4}.

\paragraph{The idea.}
On $E_+$ the integrand of the edge representation is $e^{-c'L(x)}$ times an integrable profile times a pure phase
$e^{i[\Theta_0(v)+k_xv]}$, where $\Theta_0(v):=c'\Im\Phi_0(v+\tfrac i2)$ has $\Theta_0'=-\tfrac{4\pi}3c'F$ and
$k_x=\tfrac{4\pi}3c'x$. The phase is stationary at $v=\pm v_0(x)$, where $F(v_0)=x$, with second derivative
$\mp c'\alpha_x$, $\alpha_x=\tfrac{4\pi}3F'(v_0)$. The two contributions are complex conjugates, and ordinary stationary
phase gives their sum as $\Amp\cos\hPsi$ with
\begin{equation}\label{eq:D-saddle}
\tfrac12\,\Amp(x)\,e^{i\hPsi(x)}=\frac{2\pi c'}3\,h_0(b_0)\,H_0^{(1)}(k_xb_0)\,b_0\;e^{-i\pi/4}\sqrt{\frac{2\pi}{c'\alpha_x}}\,,
\qquad b_0=v_0+\tfrac i2 ,
\end{equation}
up to relative corrections that are $O(1/c')$ \emph{at fixed $x$}. Toward the tip, however, the two points coalesce at
$v=0$ and $\alpha_x\to0$, whereas \cref{sec:7.7} needs the corrections small on all of
$[\xI,\infty)$, with $\xI\to\xs$ as $c\to\infty$. The proof therefore uses an explicit contour: at height $u$ above the
edge the oscillating factor has modulus $\exp\bigl(\tfrac{4\pi}3c'u[F(v)-x]\bigr)$ up to a cubic error, so running above
the edge between the saddles and below it outside them gains a factor exponentially small in $c'$, at rates given by
the chord slope $\sch$ and the curvature $\alpha_x$. At the saddle the first-order terms are odd and cancel, leaving terms of order $1/(c'\alpha_x^{k})$, $k=1,2,3$, with explicit constants. Near the tip
$v_0$, $\alpha_x$ and $\sch$ vanish like $(x-\xs)^{1/2}$, and the remainder is governed by
$c'\min\{1,v_0\}^3\asymp c'(x-\xs)^{3/2}$.

\subsection{Setting and hypotheses}\label{app:D.1}

Throughout, $c'\ge39$, $\delta_1\in(0,\tfrac14]$ is fixed, and $k_x=\tfrac{4\pi}3c'x$, so that $k_x/2=c'L(x)$; a \emph{band point} is
an $x>F(0)$, with $F$ as in \hyp{SD2} below. Near the upper edge we write
\begin{equation}\label{eq:D-coord}
b=v+i\bigl(\tfrac12+u\bigr),\qquad v=\Re b,\quad u=\Im b-\tfrac12,\qquad \cN_\delta:=\{b:\ \abs u\le\delta\},
\end{equation}
so that $u$ is the height above $E_+$ and $\cN_\delta$ is a thin two-sided strip around it (the letters $u$, $r$, $\ell$,
$q$ have this meaning in this appendix only). For $\Im z\ge0$ and on principal branches,
\begin{equation}\label{eq:D-hankel}
H_0^{(1)}(z)=\sqrt{\frac2{\pi z}}\;e^{i(z-\pi/4)}\bigl(1+\epsH(z)\bigr),\qquad
\abs{\epsH(z)}\le\frac1{4\abs z}\,e^{1/(4\abs z)}\le\frac{0.33}{\abs z}\quad(\abs z\ge1),
\end{equation}
by~\cite[(10.17.13)--(10.17.15)]{DLMF} with $\nu=0$ and one term.
Let $h_0$ have the following properties.

\smallskip\noindent\hypdef{SD1} \emph{Edge representation and symmetry.} The function $h_0$ is even, real on
$\RR\cup i\RR$ and analytic on a neighborhood of $\bar\strip\cup\cN_{\delta_1}$, and $\abs{h_0(b)}\le Ce^{-\mu\abs{\Re b}}$
for some $\mu>0$ on $\cN_{\delta_1}\cap\bar\strip=\{\tfrac12-\delta_1\le\Im b\le\tfrac12\}$. (Nothing is assumed as
$\abs{\Re b}\to\infty$ above $E_+$, where for the comb $\abs{h_0}$ grows like $e^{\frac{4\pi}3c'F(v)u}$.) With
$\mathcal I(b):=h_0(b)\,H_0^{(1)}(k_xb)\,b$,
\begin{equation}\label{eq:D-edge}
\hat h_0(c'x^2/3)=\frac{2\pi c'}3\int_{E_+}\mathcal I(b)\,\dd b,\qquad \mathcal I(-\bar b)=\overline{\mathcal I(b)}\quad(b\in E_+).
\end{equation}

\smallskip\noindent\hypdef{SD2} \emph{Phase--amplitude split.} On $\cN_{\delta_1}$, $h_0=\pm\,e^{c'\Phi_0}e^{\Phi_1}\,\cE$ with a fixed sign,
where $\Phi_0$, $\Phi_1$ and the outer factor $\cE$ are analytic, $\cE$ has no zeros, and:
(a) $\Re\Phi_0=0$ on $E_+$;
(b) the function $F(v):=\tfrac{3i}{4\pi}\Phi_0'(v+\tfrac i2)$, which is real for real $v$ by (a), is even, strictly increasing and
convex on $[0,\infty)$, tends to infinity, satisfies $\tfrac\pi3c'F(0)\ge11$, and extends analytically to
$\abs{\Im v}\le\delta_1$;
(c) $C_2:=\sup_{\cN_{\delta_1}}\abs{\Phi_0'''}=\tfrac{4\pi}3\sup_{\abs{\Im v}\le\delta_1}\abs{F''}$ and
$C_3:=\sup_{\cN_{\delta_1}}\abs{\Phi_0''''}$ are finite;
(d) $M_1:=\sup_{\cN_{\delta_1}}\abs{\Phi_1'}$ and $M_1':=\sup_{\cN_{\delta_1}}\abs{\Phi_1''}$ are finite and independent of $c$;
(e) $\Re\Phi_1=0$ on $E_+$ (used only to drop the factor $e^{\Re\Phi_1(b_0)}=1$ from the amplitude).

\smallskip\noindent\hypdef{SD3} \emph{Outer factor.} $M_{\cE}:=\sup_{\cN_{\delta_1}}\abs{\cE'/\cE}$ and
$M_{\cE}':=\sup_{\cN_{\delta_1}}\abs{(\cE'/\cE)'}$ are finite.

\smallskip\noindent\hypdef{SD4} \emph{Saddle, curvature and chord slope.} With $v_0=v_0(x)>0$ the unique solution of
$F(v_0)=x$,
\begin{equation}\label{eq:D-alpha}
b_0:=v_0+\tfrac i2,\qquad\alpha_x:=\frac{4\pi}3F'(v_0)>0,\qquad \sch:=\frac{x-F(0)}{v_0}>0 .
\end{equation}

\smallskip
In \cref{app:D.1,app:D.2,app:D.3} the letters $F$ and $v_0$ refer to \hyp{SD2} and \hyp{SD4}; for the comb they are the edge function
and the map of \cref{sec:7.1}, and $F(0)=\xs$. The dependence on $c'$ that is not in $e^{c'\Phi_0}$ or in the exponential
of the Hankel function is collected in the amplitude, defined on $\cN_{\delta_1}$ (where $\Im b\ge\tfrac14$) with the principal
square root by
\begin{equation}\label{eq:D-A}
A(b):=\pm\,\cE(b)\,e^{\Phi_1(b)}\,b^{1/2}\bigl(1+\epsH(k_xb)\bigr),\qquad
M\ge\sup_{\cN_{\delta_1}}\bigabs{A'/A},\qquad M'\ge\sup_{\cN_{\delta_1}}\bigabs{(A'/A)'} .
\end{equation}
One may take $M=M_{\cE}+M_1+2+C$ and $M'=M_{\cE}'+M_1'+8+C$ with an absolute constant $C$, so that $M$ and $M'$ do not depend
on $x$ or on $c$: the square root contributes $1/(2\abs b)\le2$ and $1/(2\abs b^2)\le8$, and the Hankel factor at most
$C/k_x$ to each. Indeed the reference for \eqref{eq:D-hankel} bounds $\epsH$, continued analytically, by $C/\abs z$ in the larger sector
$-\tfrac\pi2\le\operatorname{ph}z\le\tfrac{3\pi}2$, which contains the disk of radius $\abs z/2$ about every $z$ with
$\Im z\ge0$; Cauchy's estimate then bounds $\abs{z^2\epsH'(z)}$ and $\abs{z^3\epsH''(z)}$ by an absolute constant, and
$\abs{k_xb}\ge k_x/4=\tfrac\pi3c'x\ge\tfrac\pi3c'F(0)\ge11$ on $\cN_{\delta_1}$ at every band point, by \hyp{SD2}(b).

\subsection{Statements}\label{app:D.2}

\begin{lemma}[Two-saddle steepest descent; \proved]\label{lem:D.1}
Assume \hyp{SD1}--\hyp{SD4}. Put
\begin{equation}\label{eq:D-lengths}
\ell:=\min\bigl\{1,\ \sqrt2\,\delta_1,\ v_0,\ \alpha_x/C_2,\ 2\sqrt2\,\pi\sch/C_2,\ 1/M,\ 1/\sqrt{M'}\bigr\},
\qquad u_0:=\ell/\sqrt2,
\end{equation}
and $r_1:=\min\bigl\{\ell,\,(6/(c'C_2))^{1/3}\bigr\}$, and suppose that $c'$ is so large that
\begin{equation}\label{eq:D-large}
\text{\textup{(i)}}\ \ \tfrac{2\pi}3\,c'u_0\sch\ge M,\qquad \text{\textup{(ii)}}\ \ c'\alpha_xu_0\ge2M,\qquad
\text{\textup{(iii)}}\ \ c'\alpha_x\ge8 .
\end{equation}
Then
\begin{equation}\label{eq:D-main}
\hat h_0(c'x^2/3)=\Amp(x)\bigl[\cos\hPsi(x)+\epssd(x)\bigr],\qquad
\abs{\epssd(x)}\le\epssdbar(x):=\varepsilon_{\rm core}+2\varepsilon_{\rm tr}+\varepsilon_{\rm mid}+\varepsilon_{\rm far},
\end{equation}
where $\epssd$ is real and, with $\Theta_0(v)=c'\Im\Phi_0(v+\tfrac i2)$,
\begin{align}
\Amp(x)&=2\,e^{-c'L(x)}\,\abs{\cE(b_0)}\,e^{\Re\Phi_1(b_0)}\,\bigabs{1+\epsH(k_xb_0)}\,\Bigl(\frac{\abs{b_0}}{xF'(v_0)}\Bigr)^{1/2},\label{eq:D-amp}\\
\hPsi(x)&=\Theta_0(v_0)+k_xv_0-\tfrac\pi2+\arg A(b_0),\label{eq:D-phase}\\
\arg A(b_0)&=\arg(\pm1)+\Im\Phi_1(b_0)+\arg\cE(b_0)+\tfrac12\arctan\tfrac1{2v_0}+\arg\bigl(1+\epsH(k_xb_0)\bigr),\notag
\end{align}
and the four parts of the remainder, from the core of the saddle, the transition to the valleys, the middle piece and the
far pieces of the contour, are, with $K_A:=4.5M^2+1.625M'$,
\begin{gather}
\varepsilon_{\rm core}=\frac{K_A}{c'\alpha_x}+\frac{\tfrac18C_3+\tfrac12MC_2+K_AC_2}{c'\alpha_x^{2}}+\frac{(0.57+0.52M)\,C_2^2}{c'\alpha_x^{3}},\label{eq:D-core}\\
\varepsilon_{\rm tr}=\frac{3e^{M\ell}}{\sqrt{2\pi}}\,\frac{e^{-c'\alpha_xr_1^2/3}}{r_1\sqrt{c'\alpha_x}},\quad
\varepsilon_{\rm mid}=\frac{3e^{Mu_0}}{2\pi}\sqrt{\frac{\alpha_x}{2\pi c'}}\,\frac{e^{-\frac\pi3c'\sch u_0^2}}{\sch u_0},\quad
\varepsilon_{\rm far}=\sqrt{\frac2\pi}\,e^{Mu_0}\frac{e^{-c'\alpha_xu_0^2/4}}{u_0\sqrt{c'\alpha_x}} .\label{eq:D-exp}
\end{gather}
Here $\Amp$ and $\hPsi$ contain the exact $\epsH(k_xb_0)$ and obey \eqref{eq:D-saddle} identically; by
\hyp{SD2}\textup{(e)}, \eqref{eq:D-hankel},
\begin{equation}\label{eq:D-amp-closed}
\Amp(x)\,e^{c'L(x)}=2\,\abs{\cE(b_0)}\Bigl(\frac{\abs{b_0}}{xF'(v_0)}\Bigr)^{1/2}\bigl(1+O(1/c')\bigr),
\qquad \abs{O(1/c')}\le\frac{0.33}{\abs{k_xb_0}} .
\end{equation}
\end{lemma}

\begin{corollary}[Uniformity; \proved]\label{cor:D.2}
Assume in addition that $\sup_{[0,\infty)}F'<\infty$, and let $\delta_1,C_2,C_3,M,M'$ be fixed. Suppose that along a family
of band points $x\ge x_I(c)$ one has $c'\min\{\alpha_x,\sch,v_0\}^3\ge c'^{\gamma}$ for some $\gamma>0$. Then for $c$ large the
conditions \eqref{eq:D-large} hold on $[x_I(c),\infty)$, and $\epssdbar\to0$ uniformly there as $c\to\infty$, at the rate
\begin{equation}\label{eq:D-rate}
\epssdbar\le K\,(1+M^2+M')\Bigl[\frac1{c'\alpha_x}+\frac{C_3}{c'\alpha_x^{2}}+\frac{C_2^2}{c'\alpha_x^{3}}\Bigr]+C\,e^{-c_*c'^{\gamma/3}},
\end{equation}
with $K$ absolute and $C,c_*>0$ depending only on $\gamma$, $\sup F'$ and the fixed constants.
\end{corollary}

\begin{corollary}[Global band bound; \proved]\label{cor:D.3}
Under the hypotheses of \cref{cor:D.2}, for $c$ large and every $x\ge x_I(c)$,
\begin{equation}\label{eq:D-glob}
\bigabs{\hat h_0(c'x^2/3)}\le\Amp(x)\bigl(1+\epssdbar\bigr)
\le2\bigl(1+\epssdbar\bigr)\abs{\cE(b_0)}\Bigl(\frac{\abs{b_0}}{xF'(v_0)}\Bigr)^{1/2}\bigl(1+O(1/c')\bigr)\,e^{-c'L(x)}.
\end{equation}
For the comb, where $\cE=\sech^2\tfrac{\pi b}2$ and $F$ is the edge function of \cref{sec:7.1}, the right-hand side is at
most $\Cglob\,F'(v_0)^{-1/2}\,e^{-\pi v_0}\,e^{-c'L(x)}$ with $\Cglob$ absolute; $\Cglob=2^5$ will do once $\epssdbar\le1$.
\end{corollary}

\subsection{Proofs}\label{app:D.3}

\begin{proof}[Proof of \cref{lem:D.1}]
\emph{Plan.} On $\cN_{\delta_1}$ we factor the integrand into $e^{c'\varphi}$, which carries the dependence on $c'$, and an
amplitude $A$, and keep $\Re b>0$ by symmetry (Step~1); read off from \hyp{SD2} how $\abs{e^{c'\varphi}}$ behaves off
the edge (Step~2); choose the contour and justify the deformation (Step~3); bound the pieces away from the saddles
(Step~4); expand at a saddle (Step~5); assemble (Step~6).

\emph{Step 1: factorization and symmetry.} By \hyp{SD2}, \eqref{eq:D-hankel} and \eqref{eq:D-A}, for $b\in\cN_{\delta_1}$,
\begin{equation}\label{eq:D-factor}
\mathcal I(b)=\sqrt{\frac2{\pi k_x}}\;e^{-i\pi/4}\,e^{-k_x/2}\;e^{c'\varphi(b)}A(b),\qquad
\varphi(b):=\Phi_0(b)+\frac{4\pi i}3\,x\,\bigl(b-\tfrac i2\bigr),
\end{equation}
since $c'\varphi=c'\Phi_0+ik_x(b-\tfrac i2)$. The symmetry in \eqref{eq:D-edge} extends from $E_+$ to $\cN_{\delta_1}$ by the
identity theorem, and on a contour $b(v)$ whose height $u(v)$ is even in $v$ one has $b'(-v)=\overline{b'(v)}$, so that the
part with $v<0$ contributes the complex conjugate of the part with $v>0$. Hence, for every such contour
$\Gamma\subset\cN_{\delta_1}$ to which $E_+$ can be deformed,
\begin{equation}\label{eq:D-J}
\hat h_0(c'x^2/3)=\frac{2\pi c'}3\sqrt{\frac2{\pi k_x}}\;e^{-k_x/2}\cdot2\Re\bigl[e^{-i\pi/4}J_+\bigr],\qquad
J_+:=\int_{\Gamma\cap\{\Re b>0\}}e^{c'\varphi}A\,\dd b .
\end{equation}

\emph{Step 2: the phase near the edge.} By \hyp{SD2}(a),(b), on $E_+$ one has $\Re\varphi=0$ and
$\varphi'(v+\tfrac i2)=\tfrac{4\pi i}3\,(x-F(v))$. Thus $\abs{e^{c'\varphi}}=1$ on $E_+$, the critical points of $\varphi$ there
are $b_0$ and its mirror image $-\bar b_0$, and $\varphi''(b_0)=-i\alpha_x$. For real $v$ the $k$-th derivative of
$u\mapsto\Re\varphi(v+i(\tfrac12+u))$ is $\Re[i^k\varphi^{(k)}]$. At $u=0$ the function vanishes, its first derivative is
$\tfrac{4\pi}3(F(v)-x)$, and its second derivative is zero because $\varphi''(v+\tfrac i2)=-\tfrac{4\pi i}3F'(v)$ is imaginary;
its third derivative is bounded by $\abs{\Phi_0'''}\le C_2$. By Taylor's theorem, for $\abs u\le\delta_1$,
\begin{equation}\label{eq:D-rephi}
\Re\varphi\bigl(v+i(\tfrac12+u)\bigr)=\frac{4\pi}3\,u\,\bigl[F(v)-x\bigr]+R_3,\qquad\abs{R_3}\le\tfrac16\,C_2\abs u^3 .
\end{equation}
So $\abs{e^{c'\varphi}}$ decreases above the edge where $\abs v<v_0$ and below it where $\abs v>v_0$: two valleys, which meet at the
saddles. By convexity the chord slope $(F(v_0)-F(v))/(v_0-v)$ increases with $v$, so its infimum over $[0,v_0)$ is
$\sch$, and the graph of $F$ lies above its tangent at $v_0$:
\begin{equation}\label{eq:D-chord}
x-F(v)\ge\sch\,(v_0-v)\ \ (0\le v\le v_0),\quad F(v)-x\ge F'(v_0)(v-v_0)=\tfrac{3\alpha_x}{4\pi}(v-v_0)\ \ (v\ge v_0).
\end{equation}
Finally, by Taylor's theorem at $b_0$, for $b_0+w\in\cN_{\delta_1}$ and
$\varphi_{\ge3}(w):=\varphi(b_0+w)-\varphi(b_0)+\tfrac{i\alpha_x}2w^2$,
\begin{equation}\label{eq:D-cubic}
\bigabs{\varphi_{\ge3}(w)}\le\tfrac16C_2\abs w^3,\qquad
\bigabs{\varphi_{\ge3}(w)-\tfrac16\varphi'''(b_0)w^3}\le\tfrac1{24}C_3\abs w^4 .
\end{equation}

\emph{Step 3: the contour.} Let $\Gamma$ run from left to right through
$\Gamma^-_{\rm far},\Gamma^-_{\rm loc},\Gamma_{\rm mid},\Gamma^+_{\rm loc},\Gamma^+_{\rm far}$, where
\begin{equation}\label{eq:D-contour}
\begin{aligned}
\Gamma^{\mp}_{\rm far}&:\ b=v+i\bigl(\tfrac12-u_0\bigr),\quad \mp v\ge v_0+u_0; &\qquad
\Gamma^{-}_{\rm loc}&:\ b=-\bar b_0+r\,e^{i\pi/4},\quad r\in[-\ell,\ell];\\
\Gamma_{\rm mid}&:\ b=v+i\bigl(\tfrac12+u_0\bigr),\quad \abs v\le v_0-u_0; &\qquad
\Gamma^{+}_{\rm loc}&:\ b=b_0+r\,e^{-i\pi/4},\quad r\in[-\ell,\ell].
\end{aligned}
\end{equation}
It follows the valleys at distance $u_0$ from the edge, crosses the two saddles along their directions of steepest descent,
and lies above $E_+$ only on the bounded piece $\Gamma_{\rm mid}$ and the adjacent halves of $\Gamma^\pm_{\rm loc}$. The pieces join, since
$b_0\mp\ell e^{-i\pi/4}=(v_0\mp u_0)+i(\tfrac12\pm u_0)$; the local pieces do not overlap, since $\ell\le v_0$; the height is even in
$v$; and $\Gamma\subset\cN_{u_0}\subset\cN_{\delta_1}$, where $\mathcal I$ is analytic by \hyp{SD1} (the Hankel function and $b^{1/2}$ are
analytic in $\Im b>0$). The region between $E_+$ and $\Gamma$ is bounded apart from two half-strips in
$\{-u_0\le u\le0\}\subset\bar\strip$. On their closing segments $\Re b=\pm X$ one has $\abs{H_0^{(1)}(k_xb)\,b}\le C\abs b^{1/2}$
by \eqref{eq:D-hankel}, so these contribute at most $Cu_0X^{1/2}\sup_{-u_0\le u\le0}\abs{h_0(\pm X+i(\tfrac12+u))}\to0$ as $X\to\infty$, by the decay
in \hyp{SD1}. By Cauchy's theorem $\int_{E_+}\mathcal I\,\dd b=\int_\Gamma\mathcal I\,\dd b$, and \eqref{eq:D-J} holds for $\Gamma$.

\emph{Step 4: the middle and far pieces.} Put $\cG:=\abs{A(b_0)}\sqrt{2\pi/(c'\alpha_x)}$. On $\Gamma_{\rm mid}$ with
$0\le v\le v_0-u_0$ let $s:=v_0-v$. Then \eqref{eq:D-rephi} and \eqref{eq:D-chord} give $c'\Re\varphi\le-\tfrac{4\pi}3c'u_0\sch\,s+\tfrac16c'C_2u_0^3$, and
integrating $A'/A$ from $b_0$ along a path of length $s+u_0$ gives $\abs{A(b)}\le\abs{A(b_0)}\,e^{Ms+Mu_0}$. Condition
(i) absorbs $e^{Ms}$ into half of the linear decay, and $u_0\le2\pi\sch/C_2$ gives
$\tfrac16c'C_2u_0^3\le\tfrac\pi3c'\sch u_0^2$, so that
\begin{equation}\label{eq:D-mid}
\begin{aligned}
\Bigabs{\int_{\Gamma_{\rm mid},\,v\ge0}e^{c'\varphi}A\,\dd b}
&\le\abs{A(b_0)}\,e^{Mu_0+\frac\pi3c'\sch u_0^2}\int_{u_0}^\infty e^{-\frac{2\pi}3c'u_0\sch\,s}\,\dd s\\
&=\abs{A(b_0)}\,\frac{3\,e^{Mu_0-\frac\pi3c'\sch u_0^2}}{2\pi c'\sch u_0}=\cG\,\varepsilon_{\rm mid}.
\end{aligned}
\end{equation}
On $\Gamma^+_{\rm far}$ let $s:=v-v_0\ge u_0$. The second bound of \eqref{eq:D-chord} gives $c'\Re\varphi\le-c'\alpha_xu_0\,s+\tfrac16c'C_2u_0^3$, and
$\abs{A(b)}\le\abs{A(b_0)}\,e^{Ms+Mu_0}$. By condition (ii) and $\tfrac16C_2u_0^3\le\tfrac14\alpha_xu_0^2$ (from
$u_0\le\ell\le\alpha_x/C_2$),
\begin{equation}\label{eq:D-far}
\Bigabs{\int_{\Gamma^+_{\rm far}}e^{c'\varphi}A\,\dd b}
\le\abs{A(b_0)}e^{Mu_0+\frac14c'\alpha_xu_0^2}\!\int_{u_0}^\infty\! e^{-\frac12c'\alpha_xu_0s}\dd s
=\abs{A(b_0)}\frac{2e^{Mu_0-\frac14c'\alpha_xu_0^2}}{c'\alpha_xu_0}=\cG\varepsilon_{\rm far}.
\end{equation}

\emph{Step 5: the saddle.} On $\Gamma^+_{\rm loc}$ we have $b=b_0+w$ with $w=re^{-i\pi/4}$, and $-\tfrac{i\alpha_x}2w^2=-\tfrac12\alpha_xr^2$ is
real. By \eqref{eq:D-cubic} and $\ell\le\alpha_x/C_2$, with $q(r):=c'\varphi_{\ge3}(re^{-i\pi/4})$,
\begin{equation}\label{eq:D-q}
c'\bigl[\varphi(b)-\varphi(b_0)\bigr]=-\tfrac12c'\alpha_xr^2+q(r),\qquad\abs{q(r)}\le\tfrac16c'C_2\abs r^3\le\tfrac16c'\alpha_xr^2,
\end{equation}
so that $\Re c'[\varphi-\varphi(b_0)]\le-\tfrac13c'\alpha_xr^2$ on the whole segment. We split it into the core $\abs r\le r_1$, where
$\abs q\le\tfrac16c'C_2r_1^3\le1$, and the transition part $r_1<\abs r\le\ell$. On the latter $\abs A\le\abs{A(b_0)}e^{M\ell}$, so
\begin{equation}\label{eq:D-tr}
\Bigabs{\int_{r_1<\abs r\le\ell}\!\!e^{c'[\varphi-\varphi(b_0)]}A\,\dd r}\le2\abs{A(b_0)}e^{M\ell}\!\!\int_{r_1}^\infty\!\!e^{-c'\alpha_xr^2/3}\dd r
\le\abs{A(b_0)}e^{M\ell}\frac{3e^{-c'\alpha_xr_1^2/3}}{c'\alpha_xr_1}=\cG\varepsilon_{\rm tr}.
\end{equation}
On the core, integrating $A'/A$ twice along the segment gives $A(b_0+w)=A(b_0)\exp(\mathsf a_1w+\mathsf r_2)$ with
$\mathsf a_1:=A'(b_0)/A(b_0)$, $\abs{\mathsf a_1}\le M$ and $\abs{\mathsf r_2}\le\tfrac12M'r^2$. As $\abs r\le\ell\le\min\{1/M,1/\sqrt{M'}\}$, the
exponent $z$ has $\abs z\le\tfrac32$ and $\abs z^2\le(2M^2+\tfrac12M')r^2$, and $\abs{e^z-1-z}\le\tfrac12\abs z^2e^{\abs z}\le2.25\abs z^2$ gives
$A/A(b_0)=1+\mathsf a_1w+\mathsf a_2$ with $\abs{\mathsf a_2}\le[\tfrac12M'+2.25(2M^2+\tfrac12M')]r^2=K_Ar^2$. Likewise $e^q=1+q+q_2$ with $\abs{q_2}\le\tfrac e2\abs q^2$, and
$q=\tfrac16c'\varphi'''(b_0)w^3+q_4$ with $\abs{q_4}\le\tfrac1{24}c'C_3r^4$ by \eqref{eq:D-cubic}. So
\begin{equation}\label{eq:D-expand}
e^{c'[\varphi(b)-\varphi(b_0)]}\,\frac{A(b)}{A(b_0)}=e^{-c'\alpha_xr^2/2}\Bigl\{1+\bigl[\mathsf a_1w+\tfrac16c'\varphi'''(b_0)w^3\bigr]+\mathrm{Rest}(r)\Bigr\},
\end{equation}
where the bracket is odd in $r$ and integrates to zero over $[-r_1,r_1]$, and where
\begin{equation}\label{eq:D-rest}
\abs{\mathrm{Rest}}\le K_Ar^2+\frac{c'C_3r^4}{24}+\frac{Mc'C_2r^4}6+\frac{e\,(c'C_2)^2r^6}{72}
+\Bigl(1+\frac e2\Bigr)\frac{K_Ac'C_2\abs r^5}6+\frac{e\,M(c'C_2)^2\abs r^7}{72};
\end{equation}
the terms are, in order, $\mathsf a_2$; $q_4$; $\mathsf a_1w\,q$; $q_2$; $\mathsf a_2(q+q_2)$, with $\abs{q+q_2}\le(1+\tfrac e2)\abs q$ for
$\abs q\le1$; and $\mathsf a_1w\,q_2$. We integrate against $e^{-c'\alpha_xr^2/2}$ over $\RR$, using
$\int_\RR r^{2n}e^{-sr^2/2}\dd r=\sqrt{2\pi/s}\,(2n-1)!!\,s^{-n}$ and $\int_\RR\abs r^{2n+1}e^{-sr^2/2}\dd r=n!\,(2/s)^{n+1}$ with
$s=c'\alpha_x$. Relative to $\sqrt{2\pi/(c'\alpha_x)}$, the four even terms give $K_A/(c'\alpha_x)$, $\tfrac18C_3/(c'\alpha_x^2)$,
$\tfrac12MC_2/(c'\alpha_x^2)$ and $\tfrac{15e}{72}C_2^2/(c'\alpha_x^3)$, and the two odd terms give
$2.51\,(c'\alpha_x)^{-1/2}K_AC_2/(c'\alpha_x^2)$ and $1.45\,(c'\alpha_x)^{-1/2}MC_2^2/(c'\alpha_x^3)$, which condition (iii) bounds by
$K_AC_2/(c'\alpha_x^2)$ and $0.52\,MC_2^2/(c'\alpha_x^3)$; the sum is $\varepsilon_{\rm core}$. Completing the Gaussian costs
$\int_{\abs r>r_1}e^{-c'\alpha_xr^2/2}\dd r\le2e^{-c'\alpha_xr_1^2/2}/(c'\alpha_xr_1)\le\sqrt{2\pi/(c'\alpha_x)}\,\varepsilon_{\rm tr}$.
Altogether, with $\dd b=e^{-i\pi/4}\dd r$,
\begin{equation}\label{eq:D-loc}
\int_{\Gamma^+_{\rm loc}}e^{c'\varphi}A\,\dd b=e^{c'\varphi(b_0)}A(b_0)\,e^{-i\pi/4}\sqrt{\frac{2\pi}{c'\alpha_x}}\,\bigl(1+\varepsilon_{\rm loc}\bigr),\qquad
\abs{\varepsilon_{\rm loc}}\le\varepsilon_{\rm core}+2\varepsilon_{\rm tr}.
\end{equation}

\emph{Step 6: assembly.} Let $\mathcal Z:=e^{-i\pi/2}e^{c'\varphi(b_0)}A(b_0)\sqrt{2\pi/(c'\alpha_x)}$, so that $\abs{\mathcal Z}=\cG$, as
$\abs{e^{c'\varphi(b_0)}}=1$. By \eqref{eq:D-mid}, \eqref{eq:D-far} and \eqref{eq:D-loc},
$e^{-i\pi/4}J_+=\mathcal Z(1+\varepsilon_{\rm loc})+\mathcal R$ with $\abs{\mathcal R}\le\cG(\varepsilon_{\rm mid}+\varepsilon_{\rm far})$, hence
$2\Re[e^{-i\pi/4}J_+]=2\cG[\cos\arg\mathcal Z+\epssd]$ with $\epssd$ real and $\abs{\epssd}\le\epssdbar$. Then \eqref{eq:D-J} gives
\eqref{eq:D-main} with $\Amp=\tfrac{2\pi c'}3\sqrt{2/(\pi k_x)}\,e^{-k_x/2}\cdot2\cG$ and
$\hPsi=\arg\mathcal Z=c'\Im\varphi(b_0)-\tfrac\pi2+\arg A(b_0)$, which is \eqref{eq:D-phase} because
$c'\varphi(b_0)=c'\Phi_0(b_0)+ik_xv_0$. With $k_x=\tfrac{4\pi}3c'x$ and $\alpha_x=\tfrac{4\pi}3F'(v_0)$ this is
$\Amp=2e^{-c'L}\abs{A(b_0)}/\sqrt{xF'(v_0)}$, and \eqref{eq:D-amp} and the expression for $\arg A(b_0)$ follow from \eqref{eq:D-A}
with $\arg b_0^{1/2}=\tfrac12\arctan\tfrac1{2v_0}$. Finally $\tfrac12\Amp\,e^{i\hPsi}$ is $\tfrac{2\pi c'}3$ times the prefactor of
\eqref{eq:D-factor} times $e^{i\pi/4}\mathcal Z$, which is \eqref{eq:D-saddle}.
\end{proof}

\begin{proof}[Proof of \cref{cor:D.2,cor:D.3}]
\emph{Plan.} We bound the window $\ell$ from below by the controlling quantity and inspect \eqref{eq:D-core}--\eqref{eq:D-exp}.
Let $m:=\min\{\alpha_x,\sch,v_0\}$ and $\bar m:=\min\{1,m\}$. By \eqref{eq:D-lengths}, $\ell\ge c_\ell\bar m$ with
$c_\ell:=\min\{1,\sqrt2\delta_1,1/C_2,1/M,1/\sqrt{M'}\}$, and the hypothesis gives $c'\bar m^3\ge c'^{\gamma}$ (we may take $\gamma\le1$), so that
$c'\bar m^2$ and $c'\bar m$ tend to infinity and \eqref{eq:D-large} holds. In \eqref{eq:D-core} we use
$C_2/\alpha_x^2\le\tfrac12(1/\alpha_x+C_2^2/\alpha_x^3)$ and $M\le\tfrac12(1+M^2)$; this gives the bracket of \eqref{eq:D-rate}. The
exponents in \eqref{eq:D-exp} are $\tfrac13c'\alpha_xr_1^2$, $\tfrac\pi6c'\sch\ell^2$ and $\tfrac18c'\alpha_x\ell^2$. Those with $\ell$ are at
least $\tfrac18c_\ell^2\,c'\bar m^3$, and when $r_1<\ell$ the first equals $\tfrac13\cdot6^{2/3}(c'\alpha_x^3/C_2^2)^{1/3}\ge C_2^{-2/3}(c'\bar m^3)^{1/3}$; so all
three are at least $2c_*c'^{\gamma/3}$ for a suitable $c_*>0$. The prefactors of $\varepsilon_{\rm tr}$ and $\varepsilon_{\rm far}$ are bounded, because
$e^{M\ell}\le e$ and $r_1\sqrt{c'\alpha_x}$, $u_0\sqrt{c'\alpha_x}$ are the square roots of multiples of the exponents; that of
$\varepsilon_{\rm mid}$ is at most a power of $c'$, because $\alpha_x\le\tfrac{4\pi}3\sup F'$ and $\sch u_0\ge c_\ell\bar m^2/\sqrt2$ with
$\bar m\ge c'^{-(1-\gamma)/3}$, and a power of $c'$ times $e^{-2c_*c'^{\gamma/3}}$ is at most $Ce^{-c_*c'^{\gamma/3}}$. This proves \cref{cor:D.2}. The first inequality of \eqref{eq:D-glob} follows from \eqref{eq:D-main} and
$\abs{\cos\hPsi+\epssd}\le1+\epssdbar$, the second from \eqref{eq:D-amp-closed}. For the comb,
$\abs{\sech^2\tfrac{\pi b_0}2}=2/\cosh\pi v_0\le4e^{-\pi v_0}$ (\cref{prop:7.2}); $F(v)\ge v$ gives $v_0\le x$, so that
$\abs{b_0}/x\le(1+1/(4(\xs)^{2}))^{1/2}\le2.05$; and $1+\epssdbar\le2$ once $\epssdbar\le1$. Hence
$2\cdot2\cdot4\cdot\sqrt{2.05}\,(1+O(1/c'))\le2^5$.
\end{proof}

\subsection{Verification for the comb}\label{app:D.4}

We verify the hypotheses for the function $h_0=\sech^2\tfrac{\pi b}2\,B$ of \cref{sec:7.2}, with $\delta_1=\tfrac14$,
$\cE=\sech^2\tfrac{\pi b}2$ and the phases $\Phi_0,\Phi_1$ of \cref{lem:7.6}.

\smallskip\noindent\hyp{SD1}. Each factor $T_\zeta$ is even, and the zeros are real or form conjugate pairs, so $h_0$ is even and real on
$\RR\cup i\RR$. By \hyp{T4} the poles of $B$ lie at $\abs{\Im b}\ge1-\max_n\varepsilon_n>\tfrac34$ for $c$ large, since
$\varepsilon_n\lesssim c'^{-2/3}$ (\cref{app:C.3}); those of $\cE$ lie on $\abs{\Im b}=1$. The edge
representation is \cref{lem:7.3}(c), and the symmetry follows from $h_0(-\bar b)=\overline{h_0(b)}$ and
$H_0^{(1)}(-\bar z)=-\overline{H_0^{(1)}(z)}$. For the decay,
$\abs{\cosh\tfrac\pi2(v+iy)}^2=\tfrac12(\cosh\pi v+\cos\pi y)\ge\tfrac12\cosh\pi v$ for $\tfrac14\le y\le\tfrac12$ and $\abs B\le1$ on
$\bar\strip$ (\cref{prop:7.2}) give $\abs{h_0}\le2/\cosh\pi v\le4e^{-\pi\abs v}$ on $\cN_{1/4}\cap\bar\strip$, so $\mu=\pi$.

\smallskip\noindent\hyp{SD2}, with the sign $+$. We use \cref{lem:7.6} in the form of \cref{lem:C-edge-phase}, which holds on
the two-sided strip $\cN_{1/4}$ and not only on $E_+$. By its part (a), $\Phi_0$ is analytic on $\cN_{1/4}$, with
$\Re\Phi_0=0$ and $\Phi_0'=-\tfrac{4\pi i}3F(v)$ on $E_+$, where $F(v)=\tfrac1\pi\arcsinh(\cosh\pi v)$ is analytic in
$\abs{\Im v}<\tfrac12$, and $C_2$, $C_3$ are finite. This $F$ is even, $F(0)=\xs$ with $\tfrac\pi3c'\xs\ge13\pi\xs>11$, and
\begin{equation}\label{eq:D-Fprime}
F'(v)=\frac{\sinh\pi v}{\sqrt{1+\cosh^2\pi v}}\in(0,1)\quad(v>0),\qquad F''(v)=\frac{2\pi\cosh\pi v}{(1+\cosh^2\pi v)^{3/2}}>0,
\end{equation}
so $F$ is strictly increasing and convex, with $F(v)\ge v$ and $\sup F'=1$. By \hyp{T4} the zeros of $B$ lie at heights
$0$ and $\pm\varepsilon_n$ modulo $2$, none of them in $\cN_{1/4}$, so that $\Phi_1=\log B-c'\Phi_0$ is analytic there; the sign of the
simple zero, $B(\tfrac i2)=-1$, is the constant $i\pi$ in $\Phi_1$, which is why the sign in \hyp{SD2} is $+$. Part (c) bounds
$\Phi_1'$ and $\Phi_1''$ on $\cN_{1/4}$ uniformly in $c'$ and $\Re b$, so that $M_1$ and $M_1'$ depend on $(n_1,\eta_0)$ only. Finally
$\Re\Phi_1=0$ on $E_+$ because $\abs B=1$ and $\Re\Phi_0=0$ there.

\smallskip\noindent\hyp{SD3}. The factor $\cE$ is analytic and zero-free in $\abs{\Im b}<1$, with $\cE'/\cE=-\pi\tanh\tfrac{\pi b}2$. As
$\abs{\tanh\tfrac\pi2(v+iy)}^2=(\cosh\pi v-\cos\pi y)/(\cosh\pi v+\cos\pi y)$,
$M_{\cE}=\pi\tan\tfrac{3\pi}8$ and $M_{\cE}'=\tfrac{\pi^2}2\sec^2\tfrac{3\pi}8$.

\smallskip\noindent\hyp{SD4} \emph{and the hand-off.} Here $v_0$ is given by $\cosh\pi v_0=\sinh\pi x$. Since $F'(0)=0$, $F''(0)=\pi/\sqrt2>0$ and $F'$ is
positive and increasing, the ratios $F'(v)/v$ and $\tfrac1{v^2}\int_0^vF'$ are bounded below on $(0,1]$, and $F'(v)$ and $\tfrac1v\int_0^vF'$ are
increasing for $v\ge1$; hence $\min\{\alpha_x,\sch,v_0\}\ge c_F\min\{1,v_0\}$ with an absolute $c_F>0$. Near the tip
$v_0\simeq(2(x-\xs)/F''(0))^{1/2}$, so that $c'\min\{1,v_0\}^3$ is at least a positive multiple of $c'^{\,1-3\thon/2}$ on
$[\xI,\infty)$, and \cref{cor:D.2} applies with any $\gamma<1-\tfrac32\thon$, provided $\thon<\tfrac23$. With
$\sigma:=c'\min\{1,v_0\}^3$, every term of $\epssdbar$ is then at most a multiple of $1/\sigma$: for $\varepsilon_{\rm core}$ this is
\eqref{eq:D-rate}, and in \eqref{eq:D-exp}, where $\alpha_x\asymp\sch\asymp u_0\asymp\min\{1,v_0\}$, the exponents are bounded below by
multiples of $\sigma^{1/3}$, $\sigma$, $\sigma$, the prefactors are bounded for $\sigma\ge1$, and $\sigma e^{-c_*\sigma^{1/3}}$ is
bounded. Thus, with $C$ depending on $(n_1,\eta_0)$ only and for $c\ge c(n_1,\eta_0,\thon)$,
\begin{equation}\label{eq:D-comb-rate}
\abs{\epssd(x)}\le\epssdbar(x)\le\frac{C}{c'\min\{1,v_0(x)\}^{3}}\qquad(x\ge\xI),
\end{equation}
which is $O(1/c')$ deep in the band and $O\bigl((c'(x-\xs)^{3/2})^{-1}\bigr)$ toward the onset, at most $Cc'^{-(1-3\thon/2)}$ at $\xI$. With
\eqref{eq:D-main}, \eqref{eq:D-amp-closed} and \cref{cor:D.3} this is \cref{lem:7.7}. At $\xI$ the term $\varepsilon_{\rm tr}$ decays only like
$\exp(-\mathrm{const}\cdot c'^{(1-3\thon/2)/3})$, which is why $\thon$ is taken well below $\tfrac23$.

\section{The comb on the real axis}
\label{app:E}

This appendix proves the statements of \cref{sec:7.5} about the comb on the real axis, namely \cref{prop:7.4} and
\cref{lem:7.5}(a)--(c), (d1), (e), (f), and from them the floor of \cref{lem:7.9}, the overshoot bound of
\cref{lem:7.11} and the first-cell bounds of \cref{lem:7.12}; \cref{app:F} takes
\cref{app:E.1,app:E.2,app:E.3,app:E.4} and \cref{lem:E-Jend,lem:E-two-sided,lem:E-hole-limits} as input. Each statement carries its own tag, and a statement that contains or
uses a certified constant names the row of \cref{tab:4} on which it rests.

As in \cref{sec:7.5}, the logarithm of the comb is a sum over the sites of the site weight $f_x$ of~\eqref{eq:E-fx}, a
Riemann sum of unit step in the site index for the potential $U=-L$ (\cref{lem:7.1}). On either side of $x$, apart from the sites far to the left of $x$, the
summand is concave along the sites, so the trapezoid and midpoint rules bound the sum by the integral with no remainder. What is left over is explicit: the potential $c'\cP$ of the charge $2n_1$ below the first
kept site, the half cell $\Jend$ there, and the few sites nearest to $x$. \Cref{app:E.1,app:E.2,app:E.3} provide the
tools, \cref{app:E.5} the two-sided bounds, and \cref{app:E.4,app:E.6,app:E.7} the consequences. The constants that
\cref{app:F} needs are explicit: $\CJ\le0.65$ (\cref{lem:E-Jend}), $\Cdet\le16.6$ (\cref{lem:E-detune}), and the
coefficient $94$ of $\frac1n$ in \eqref{eq:E-displacement-explicit}.

\paragraph{Notation.}
Throughout, $(m,k_\infty,n_1,\eta_0)=(1,\frac14,\frac74,\frac14)$, the outer factor is $\sech^2(\pi x/2)$, and
$c\ge\catt$. The letter $C$ denotes an absolute constant, which may change from line to line; a subscript records a
dependence. We recall the zero law $\rhoU(y)=\frac4{3\pi}\arccosh(\sinh\pi y)$ on $y\ge\xs$ and its counting function
$N(y)=\half\int_{\xs}^y\rhoU$. For $x\ge0$ the \emph{site weight} is
\begin{equation}\label{eq:E-fx}
  f_x(y):=\log\abs{T_y(x)}=\log\Bigabs{\tanh\frac{\pi(x-y)}2}+\log\tanh\frac{\pi(x+y)}2\ \le0,\qquad y>0 .
\end{equation}
The \emph{site map} $a\mapsto\chi_a$, $a\ge\frac14$, is defined by $c'N(\chi_a)=a-\frac14$. It interpolates the sites,
starts at $\chi_{1/4}=\xs$, and satisfies
\begin{equation}\label{eq:E-sitemap}
  \dd a=\tfrac{c'}2\rhoU(\chi_a)\,\dd\chi_a,\qquad \chi_a'=\frac{2}{c'\rhoU(\chi_a)},\qquad
  \chi_a''=-(\chi_a')^2\,(\log\rhoU)'(\chi_a),
\end{equation}
where a prime on $\chi_a$ is a derivative in $a$. We write $\varphi(a):=f_x(\chi_a)$ when $x$ is fixed,
$\ell_k:=\chi_{k+1}-\chi_k$ for the spacing, and $\varepsilon_k=\eta_0\ell_k$ for the heights of the detuned zeros. A
point $x$ of cell $n$, $\chi_n<x<\chi_{n+1}$, has the position $u:=(x-\chi_n)/\ell_n\in(0,1)$ in its cell. The index $k$
always labels a site, $j$ is the number of pairs kept exactly on each side of $x$, and $a$, $a'$ are real site indices
(in this appendix $a$ never denotes $c'/12$). The product formula reads
\begin{equation}\label{eq:E-comb}
  \log\abs{B^0(x)}=f_x(\chi_2)+2\sum_{k\ge3}f_x(\chi_k),\qquad
  \log\abs{B(x)}=f_x(\chi_2)+\sum_{k\ge3}\log q_k(x),
\end{equation}
where, with $\gamma_k:=\cos\pi\varepsilon_k$,
\begin{equation}\label{eq:E-pair}
  q_k(x):=\bigabs{T_{\chi_k+i\varepsilon_k}(x)\,T_{\chi_k-i\varepsilon_k}(x)}
  =\prod_{\pm}\frac{\cosh\pi(\chi_k\mp x)-\gamma_k}{\cosh\pi(\chi_k\mp x)+\gamma_k}
\end{equation}
is the modulus of a detuned pair; for $\varepsilon_k=0$ it is $T_{\chi_k}(x)^2$. Since
$\cosh2z-\cos2v=2(\sinh^2z+\sin^2v)$ and $\cosh2z+\cos2v=2(\cosh^2z-\sin^2v)$, each of the two factors of $q_k$ equals
$(\sinh^2z+\sin^2v)/(\cosh^2z-\sin^2v)$ with $z=\frac\pi2(\chi_k\mp x)$ and $v=\frac\pi2\varepsilon_k$. Both sums converge, because
$\abs{f_x(y)}\le Ce^{-\pi(y-x)}$ for $y\ge x+1$. On the real axis $h_0=\sech^2(\pi x/2)\,B(x)$ with $t=c'x^2/3$, and
$B<0$ on $[0,\chi_2)$, $B\ge0$ beyond. The continuum counterparts of the sums are
\begin{equation}\label{eq:E-continuum-def}
  I_x(a):=\int_a^\infty f_x(\chi_{a'})\,\dd a'=\frac{c'}2\int_{\chi_a}^\infty\rhoU f_x\,\dd y,\qquad
  \cP_Y(x):=-\int_{\xs}^{Y}\rhoU(y)f_x(y)\,\dd y\ \ge0,
\end{equation}
so that $\cP=\cP_{\chi_2}$ and $c'\int_{\xs}^{\chi_a}\rhoU\,\dd y=2(a-\frac14)$, which is $2n_1$ for $a=2$. Since
$U(x)=\int\rhoU(y)\log\abs{T_y(x)}\,\dd y$ is the potential of the law, \cref{lem:7.1} gives
\begin{equation}\label{eq:E-continuum}
  2I_x(a)=c'U(x)+c'\cP_{\chi_a}(x),\qquad U(x)=-L(x)\ \text{ for }x\ge\xs .
\end{equation}
Finally, in this notation the end correction, the level and the normalized comb of \cref{sec:7.5} are
\begin{equation}\label{eq:E-level}
\begin{gathered}
  \Jend(x)=f_x(\chi_2)-2\int_2^{5/2}\varphi(a)\,\dd a,\qquad
  \Lev=4\sech^2\tfrac{\pi x}2\,e^{-c'L+c'\cP+\Jend},\\
  \frac{h_0}{\Lev}=\tfrac14\abs{B}\,e^{c'L-c'\cP-\Jend}\qquad(x>\chi_2).
\end{gathered}
\end{equation}
The conditions on the size of $c$ are conditions on the Airy length $\lc$ (notation of \cref{lem:E-tip}). We take
$\catt$ so large that three of them hold for $c\ge\catt$. Statements in the Airy window are made under the hypothesis
$\lc\Evmax\le0.3$ and are applied in this appendix with $\Evmax\le\Ehvs_{10}\le4.57$, so that we require $\lc\Ehvs_{10}\le0.3$; this
implies $\chi_6-\chi_2\ge\lc$, which is used for \eqref{eq:E-sup}. The far-left loss requires
$\lc\Ehvs_{602}\le0.24$, where $\Ehvs_{602}\le71.3$, and $C\lc\le0.004$ with an absolute constant $C$ which we do not
evaluate (\cref{lem:E-farleft}(d)). The remaining conditions depend on a parameter. A statement on a far band
$x\ge\xs+r$ with $r>0$ fixed is made for $c$ so large, depending on $r$, that $\chi_6\le\xs+r/2$; the band then lies in
the cells $n\ge6$. The bracket for $j=4$ in \cref{prop:7.4} holds on the cells $n\ge\epsilon c'$ for $c$ large depending
on $\epsilon$.

\subsection{The law near the tip}
\label{app:E.1}

We prove that the site weight is monotone and concave along the sites (\cref{lem:E-concave}), expand the law at the tip
(\cref{lem:E-tip}), and deduce the form of the comb in the Airy window, which gives \cref{lem:7.5}(b) and (d1). Near
the tip the law is the square-root density of a soft edge: the sites sit at $\vsig_k\approx(k-\frac14)^{2/3}$, and the
logarithm of each Blaschke factor is that of a scaled distance plus $\log\lc$ plus a constant. Rows 9 and 10 of
\cref{tab:4} are used.

\begin{lemma}[Monotone and concave summands; \provedCA]\label{lem:E-concave}
Let $x\ge0$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item For $y>0$, $y\ne x$,
\begin{equation}\label{eq:E-dfx}
  \partial_yf_x(y)=\sum_{\pm}\frac{\pi}{\sinh\pi(y\pm x)},\qquad
  \partial_y^2f_x(y)=-\pi^2\sum_\pm\frac{\cosh\pi(y\pm x)}{\sinh^2\pi(y\pm x)}<0 .
\end{equation}
Thus $f_x$ is increasing and concave on $(x,\infty)$, and decreasing and concave on $(0,x)$. For $0<y<x$,
$\abs{\partial_yf_x}\le1/(x-y)$ and $\abs{\partial_y^2f_x}/\abs{\partial_yf_x}\ge\pi\coth\pi(x-y)\ge\max\{\pi,1/(x-y)\}$.
\item $\rhoU$ is positive, increasing and concave on $(\xs,\infty)$, and $\rhoU^2/N$ is non-increasing there. Hence
$a\mapsto\chi_a$ is increasing and concave, the spacings $\ell_k$ decrease, $\ell_k\le\chi_k'$ and
$\ell_{k}\ge\chi_{k+1}'$, the mean spacing $(\chi_a-\xs)/(a-\frac14)$ decreases in $a$, and
\begin{equation}\label{eq:E-spacing}
  1\le\frac{\chi_a'}{\chi_{a'}'}=\frac{\rhoU(\chi_{a'})}{\rhoU(\chi_a)}\le\Bigl(\frac{a'-\frac14}{a-\frac14}\Bigr)^{1/2}
  \qquad(\tfrac14<a\le a').
\end{equation}
\item $\varphi$ is increasing and concave on $\{a:\chi_a>x\}$ and decreasing on $\{a:\chi_a<x\}$. It is concave on
$\{a:\Eythr(x)\le\chi_a<x\}$, where
\begin{equation}\label{eq:E-ythr}
  \Eythr(x):=\min\Bigl\{\xs+\frac{x-\xs}{2.3},\ \max\{\xs+0.18,\ x-1\}\Bigr\}\qquad(x>\xs).
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
(a) The derivatives follow from $(\log\tanh z)'=2/\sinh2z$. For $y>x$ both terms of $\partial_yf_x$ are positive; for
$0<y<x$ the term with $y-x$ is $-\pi/\sinh\pi(x-y)$ and exceeds the other in modulus, since $x+y>x-y$. This also gives
$\abs{\partial_yf_x}\le\pi/\sinh\pi(x-y)\le1/(x-y)$, while $\abs{\partial_y^2f_x}\ge\pi^2\cosh\pi(x-y)/\sinh^2\pi(x-y)$;
the quotient is at least $\pi\coth\pi(x-y)$, and $\coth z\ge\max\{1,1/z\}$.

(b) With $w=\pi y$, $\rhoU'=\frac43\cosh w\,(\sinh^2w-1)^{-1/2}>0$ and
$\frac{\dd}{\dd w}\bigl[\cosh w\,(\sinh^2w-1)^{-1/2}\bigr]=-2\sinh w\,(\sinh^2w-1)^{-3/2}<0$, so $\rhoU$ is increasing
and concave, and it vanishes at $\xs$. Fix $y=\xs+r$ and put $\rho=\rhoU(y)$, $s=\rhoU'(y)$. The tangent at $y$ lies
above the graph, so $N(y)\le\half\int_0^r[\rho-s(r-r')]\,\dd r'=\half(\rho r-\half sr^2)$; and $\tau:=sr/\rho\le1$
because the chord from the tip is steeper than the tangent. Hence $4Ns\le2\rho^2\tau(1-\tau/2)\le\rho^2$, that is
$(\log(\rhoU^2/N))'=2\rhoU'/\rhoU-\rhoU/(2N)\le0$. The statements on $\chi_a$ follow from \eqref{eq:E-sitemap}:
$\chi_a'>0$ and $\chi_a''<0$; a chord of a concave function has a slope between the derivatives at its ends; the chord
from $a=\frac14$ has decreasing slope; and \eqref{eq:E-spacing} is $\rhoU^2(\chi_a)/(a-\frac14)\ge\rhoU^2(\chi_{a'})/(a'-\frac14)$
together with the monotonicity of $\rhoU$.

(c) We have $\varphi'=\partial_yf_x\,\chi_a'$ and
$\varphi''=(\chi_a')^2\bigl[\partial_y^2f_x-\partial_yf_x\,(\log\rhoU)'\bigr]$, both evaluated at $y=\chi_a$. For $\chi_a>x$
the bracket is negative by (a), since $(\log\rhoU)'>0$. For $\chi_a<x$ we have $\varphi'<0$, and $\varphi''\le0$ if and
only if $\abs{\partial_y^2f_x}/\abs{\partial_yf_x}\ge(\log\rhoU)'(\chi_a)$; by (a) it suffices that
$(\log\rhoU)'(y)\le\max\{\pi,1/(x-y)\}$ at $y=\chi_a$. Let $r:=y-\xs$. Row~9 of \cref{tab:4} contains two facts:
$(\log\rhoU)'(\xs+r)\le3.10<\pi$ for $r\ge0.18$ \cert, and $1.3\,r\,(\log\rhoU)'(\xs+r)\le0.95$ for $0<r\le0.64$ \cert.
If $r\ge0.18$ the first fact suffices. If $r<0.18$, then $y<\max\{\xs+0.18,x-1\}$, so $y\ge\Eythr(x)$ forces
$y\ge\xs+(x-\xs)/2.3$, that is $x-y\le1.3\,r$, and the second fact gives $(\log\rhoU)'(y)\le0.95/(1.3r)\le1/(x-y)$.
\end{proof}

\begin{lemma}[The law near the tip; \provedCA]\label{lem:E-tip}
Let $u_1:=\arcsinh1=\pi\xs$, $\ktip:=4\cdot2^{3/4}/(9\sqrt\pi)$ and $\lc:=(\ktip c')^{-2/3}$, so that
$\ktip c'\lc^{3/2}=1$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item \textup{(\Cref{tab:4}, row 10.)} $\arccosh\sinh(u_1+\delta)=2^{3/4}\delta^{1/2}\bigl(1+\delta\,e_1(\delta)\bigr)$
with $e_1(\delta)\in[0.0589,0.105]$ for $0<\delta\le2$ \cert.
\item For $0\le r\le0.63$,
\begin{equation}\label{eq:E-tip}
  3\ktip r^{1/2}\le\rhoU(\xs+r)\le3\ktip r^{1/2}(1+0.33\,r),\qquad
  \ktip r^{3/2}\le N(\xs+r)\le\ktip r^{3/2}(1+0.2\,r).
\end{equation}
\item Let $\vsig_a:=(\chi_a-\xs)/\lc$ be the scaled sites and $\Ehvs_a:=(a-\frac14)^{2/3}$ their nominal positions. If
$\lc\Ehvs_a\le0.63$, then
\begin{equation}\label{eq:E-sites}
  \Ehvs_a\bigl(1-0.14\,\lc\Ehvs_a\bigr)\le\vsig_a\le\Ehvs_a .
\end{equation}
\item For every $r>0$, and with no certified constant,
\begin{equation}\label{eq:E-dlogrho}
  \frac1{2r}\ \le\ (\log\rhoU)'(\xs+r)\ \le\ \frac1{2r}+\frac{\pi}{4\sqrt2},\qquad\frac{\pi}{4\sqrt2}\le0.56 .
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
(b) With $\delta=\pi r\le2$, (a) gives
$\rhoU(\xs+r)=\frac4{3\pi}2^{3/4}(\pi r)^{1/2}(1+\pi r\,e_1)=3\ktip r^{1/2}(1+\pi r\,e_1(\pi r))$, because
$\frac4{3\pi}2^{3/4}\pi^{1/2}=3\ktip$; and $0\le\pi e_1\le0.105\,\pi\le0.33$. Integrating,
$N(\xs+r)=\frac32\ktip\int_0^rr'^{1/2}(1+\pi r'e_1)\,\dd r'$ lies between $\ktip r^{3/2}$ and
$\ktip r^{3/2}(1+\frac35\cdot0.33\,r)$. (c) Since $N$ increases and $c'N(\xs+0.63)\ge(0.63/\lc)^{3/2}$ by (b), the
hypothesis $\lc\Ehvs_a\le0.63$ implies $\chi_a-\xs\le0.63$. By (b) and $\ktip c'\lc^{3/2}=1$,
$a-\frac14=c'N(\chi_a)=\vsig_a^{3/2}(1+\theta)$ with $0\le\theta\le0.2\,\lc\vsig_a$. Hence
$\vsig_a=\Ehvs_a(1+\theta)^{-2/3}\le\Ehvs_a$, and $(1+\theta)^{-2/3}\ge1-\frac23\theta\ge1-0.14\,\lc\Ehvs_a$.

(d) Let $\Phi:=\rhoU^2=\frac{16}{9\pi^2}A^2$ with $A=\arccosh\sinh w$ and $w=\pi y$, so that $\cosh A=\sinh w$ and
$\dd A/\dd w=\cosh w/\sinh A$. Then $\Phi'=\frac{32}{9\pi}A\cosh w/\sinh A$, and with $\cosh^2w=1+\cosh^2A$ we find
$\Phi''=\frac{32}9K(A)$, where $K(A):=[(1+\cosh^2A)\sinh A-2A\cosh A]/\sinh^3A$.
Here $K\le1$ is the inequality $\tanh A\le A$, and $K\ge0$ because $(1+\cosh^2A)\tanh A-2A$ vanishes at $A=0$ and has
the derivative $2\sinh^2A-\tanh^2A\ge0$. At the tip $A=0$ and $\cosh w=\sqrt2$, so $\Phi(\xs)=0$ and
$\Phi'(\xs)=\frac{32\sqrt2}{9\pi}=9\ktip^2$. Since $\Phi$ is convex, $\Phi(\xs+r)\ge9\ktip^2r$, and
$r\Phi'-\Phi=\int_0^rs\,\Phi''(\xs+s)\,\dd s$ lies in $[0,\frac{16}9r^2]$, the functions being evaluated at $\xs+r$.
Hence $(\log\rhoU)'-\frac1{2r}=(r\Phi'-\Phi)/(2r\Phi)$ lies between $0$ and $8/(81\ktip^2)=\pi/(4\sqrt2)$.
\end{proof}

\begin{lemma}[A Blaschke factor in the window; \provedCA]\label{lem:E-window}
We have $-z^2/3\le\log(\abs{\tanh z}/\abs z)\le0$ for all real $z\ne0$, and \textup{(\cref{tab:4}, row 10)}
$0<(\log\tanh z)'\le1.24$ for $z\ge u_1-0.25$ \cert. Consequently, let $\lc\Evmax\le0.3$, $x=\xs+\lc\vsig$ and
$y=\xs+\lc\vsig'$ with $-\half\Evmax\le\vsig\le\Evmax$ and $0\le\vsig'\le\Evmax$. Then
\begin{equation}\label{eq:E-window}
  f_x(y)=\log\abs{\vsig-\vsig'}+\log\lc+\log\frac{\pi}{2\sqrt2}+R(\vsig,\vsig'),\qquad \abs R\le4.3\,\lc\Evmax .
\end{equation}
Moreover, for a pair $k\ge3$ with $\vsig_{k+1}\le\Evmax$, and with $\Delta_k:=\vsig_{k+1}-\vsig_k$, so that
$\varepsilon_k=\eta_0\lc\Delta_k$,
\begin{equation}\label{eq:E-window-pair}
  \log q_k(x)=\log\bigl[(\vsig-\vsig_k)^2+\eta_0^2\Delta_k^2\bigr]+2\log\lc+2\log\frac{\pi}{2\sqrt2}+R'_k(\vsig),
  \qquad \abs{R'_k}\le8.6\,\lc\Evmax .
\end{equation}
\end{lemma}

\begin{proof}
Both sides of the first inequality are even; let $z>0$. The upper bound is $\tanh z\le z$. For the lower bound,
$g(z):=\log(\tanh z/z)+z^2/3$ tends to $0$ as $z\to0$, and
$g'(z)=\frac2{\sinh2z}-\frac1z+\frac{2z}3=\frac2y\bigl(\frac y{\sinh y}-1+\frac{y^2}6\bigr)$ with $y=2z$. This is
non-negative: for $y^2\ge6$ evidently, and for $y^2<6$ because $(2k+1)!\ge6^k$ gives
$\sinh y=\sum_{k\ge0}y^{2k+1}/(2k+1)!\le y\sum_{k\ge0}(y^2/6)^k=y/(1-y^2/6)$.

The first logarithm in \eqref{eq:E-fx} is $\log(\frac\pi2\lc\abs{\vsig-\vsig'})+\log(\abs{\tanh z}/\abs z)$ with
$\abs z=\frac\pi2\lc\abs{\vsig-\vsig'}\le0.71$. The second is $\log\tanh u_1=-\log\sqrt2$ plus the increment of
$\log\tanh$ from $u_1$ to $u_1+\frac\pi2\lc(\vsig+\vsig')\ge u_1-0.25$, whose modulus is at most
$1.24\,\frac\pi2\lc\abs{\vsig+\vsig'}$. Since $\abs{\vsig-\vsig'}\le\frac32\Evmax$ and $-\half\Evmax\le\vsig+\vsig'\le2\Evmax$, the
first term lies in $[-0.56,0]\,\lc\Evmax$ and the second in $[-0.98,3.9]\,\lc\Evmax$. This gives~\eqref{eq:E-window}.

For the pair we use the second form of $q_k$ given after \eqref{eq:E-pair}, with $v=\frac\pi2\eta_0\lc\Delta_k\le0.12$,
$z_-=\frac\pi2\lc(\vsig_k-\vsig)$, $\abs{z_-}\le0.71$, and $z_+=u_1+\frac\pi2\lc(\vsig+\vsig_k)\ge u_1-0.25$; note that
$z_-^2+v^2=(\frac\pi2\lc)^2[(\vsig-\vsig_k)^2+\eta_0^2\Delta_k^2]$. For positive numbers, $(A+b)/(A_0+b_0)$ lies between
$A/A_0$ and $b/b_0$. Since $1\le\sinh^2z/z^2\le e^{z^2/3}$ and $\cosh^2z\le e^{z^2}$ (compare the power series: $(2k+1)!\ge6^kk!$ and $(2k)!\ge2^kk!$), and $1-\frac13v^2\le\sin^2v/v^2\le1$, the numerator
$\sinh^2z_-+\sin^2v$ is $z_-^2+v^2$ times a number in $[1-\frac13v^2,e^{z_-^2/3}]$, and the denominator
$\cosh^2z_--\sin^2v$ lies in $[1-v^2,e^{z_-^2}]$. So the factor with $z_-$ is $(z_-^2+v^2)e^\theta$ with
$-z_-^2-0.34\,v^2\le\theta\le\frac13z_-^2+1.01\,v^2$. The factor with $z_+$ is $\tanh^2z_+$ times
$(1+\sin^2v/\sinh^2z_+)/(1-\sin^2v/\cosh^2z_+)\in[1,e^{2.9v^2}]$, because $\sinh^2(u_1-0.25)\ge0.454$; and
$\log\tanh^2z_+$ is $-\log2$ plus twice an increment of the kind just bounded. As $\Delta_k\le\Evmax$, we have
$z_-^2\le1.67\,\lc\Evmax$ and $v^2\le0.047\,\lc\Evmax$, and $R'_k$ lies in $[-3.7,8.6]\,\lc\Evmax$.
\end{proof}

\paragraph{Proof of \cref{lem:7.5}(b) \tagpar{\proved}.}
Let $x>\chi_2$. By \cref{lem:E-concave}(a), $\abs{f_x(y)}=-f_x(y)$ increases with $y$ on $[\xs,\chi_2]$, and
$c'\cP(x)=\int_{\xs}^{\chi_2}c'\rhoU(y)\abs{f_x(y)}\,\dd y$ with $\int_{\xs}^{\chi_2}c'\rhoU=2n_1$ is $2n_1$ times an
average of it over this interval. Hence
\begin{equation}\label{eq:E-Pbounds}
  2n_1\bigabs{\log\abs{T_{\xs}(x)}}\ \le\ c'\cP(x)\ \le\ 2n_1\bigabs{\log\abs{T_{\chi_2}(x)}} .
\end{equation}
Here $\tanh z\le z$ and $\log\tanh\frac{\pi(x+\xs)}2\le0$ give $\abs{f_x(\xs)}\ge\log\frac{2}{\pi(x-\xs)}$, so that
\begin{equation}\label{eq:E-Pbelow}
  e^{c'\cP(x)}\ \ge\ \Bigl(\frac{2}{\pi(x-\xs)}\Bigr)^{2n_1}\qquad(x>\chi_2).
\end{equation}
Since $\abs{f_x(y)}$ decreases in $x$ for $x>y$, for fixed $r>0$ and $c$ so large that $\chi_2\le\xs+r/2$, which is
possible because $c'N(\chi_2)=n_1$, we get $c'\cP(x)\le2n_1\abs{f_{\xs+r}(\xs+r/2)}=:C_r$ on $x\ge\xs+r$. For
$x-y\ge1$, $\abs{f_x(y)}=4\cosh(\pi y)\,e^{-\pi x}\bigl(1+O(e^{-2\pi(x-y)})\bigr)$, so $e^{c'\cP}$ tends to $1$
exponentially deep in the band. \qed

\begin{lemma}[Window asymptotics: \cref{lem:7.5}(d1); \provedCA]\label{lem:E-d1}
Let $\lc\Evmax\le0.3$, $x=\xs+\lc\vsig$ with $-\half\Evmax\le\vsig\le\Evmax$, and let $a>\frac14$ be real with
$\Ehvs_a\le\Evmax$, so that $\vsig_a\le\Evmax$ by \eqref{eq:E-sites}; part (c) concerns $a=2$. Put
$w_c(\vsig'):=\rhoU(\xs+\lc\vsig')/(3\ktip(\lc\vsig')^{1/2})\in[1,1+0.33\,\lc\vsig']$ and
$\EcF(\vsig,A):=\int_0^A\vsig'^{1/2}\log\abs{\vsig-\vsig'}\,\dd\vsig'$. For $\vsig\ge0$ the latter is elementary: $-3\EcF(\vsig,A)$
is the right side of \eqref{eq:F-Q} with $A$ in the place of $\Ehvs_a$, since both vanish at $A=0$, are continuous in $A$, and have the
$A$-derivative $-3A^{1/2}\log\abs{\vsig-A}$ for $A\ne\vsig$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item $c'\rhoU(y)\,\dd y=3\vsig'^{1/2}w_c(\vsig')\,\dd\vsig'$ along $y=\xs+\lc\vsig'$, and
\begin{equation}\label{eq:E-potential}
  c'\cP_{\chi_a}(x)=2\bigl(a-\tfrac14\bigr)\log\frac{2\sqrt2}{\pi\lc}
  -3\int_0^{\vsig_a}\vsig'^{1/2}w_c(\vsig')\log\abs{\vsig-\vsig'}\,\dd\vsig'-\mathcal R_a(x),
\end{equation}
with $\abs{\mathcal R_a}\le2(a-\frac14)\cdot4.3\,\lc\Evmax$.
\item With the $c$-independent function $p_a(\vsig):=2(a-\frac14)\log\frac{2\sqrt2}\pi-3\EcF(\vsig,\Ehvs_a)$, which is
continuous in $\vsig$,
\begin{equation}\label{eq:E-potential-limit}
  \bigabs{c'\cP_{\chi_a}(x)-2\bigl(a-\tfrac14\bigr)\log\tfrac1{\lc}-p_a(\vsig)}\le C_{a,\Evmax}\,\lc\log\tfrac1{\lc} .
\end{equation}
For $a=2$ the coefficient of $\log\frac1\lc$ is $2n_1$:
$e^{c'\cP(x)}=\lc^{-2n_1}e^{p_2(\vsig)}\bigl(1+O_{\Evmax}(\lc\log\frac1\lc)\bigr)$.
\item On the window, $4\sech^2(\frac{\pi x}2)e^{c'\cP}=\Lev\,e^{-\Jend}e^{c'L}\asymp\lc^{-2n_1}\asymp c'^{\frac43n_1}$, and
$\Lev\,e^{c'L}\asymp\lc^{-2n_1}$ on its part $x\ge\chi_3$, with constants that depend on $\Evmax$ only. For every $x$
with $2\vsig_2\lc\le x-\xs\le0.9$, in the window or not,
\begin{equation}\label{eq:E-intermediate}
  e^{c'\cP(x)}=\Bigl(\frac{2}{\pi(x-\xs)}\Bigr)^{2n_1}e^{\theta(x)},\qquad
  0\le\theta\le2n_1\Bigl[\log\frac{x-\xs}{x-\chi_2}+\log\sqrt2+\frac{\pi^2(x-\chi_2)^2}{12}\Bigr]\le C,
\end{equation}
so that $\Lev\,e^{c'L}\asymp(x-\xs)^{-2n_1}$ for $\lc\ll x-\xs\ll1$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) By \eqref{eq:E-tip}, $c'\rhoU\,\dd y=3\ktip c'\lc^{3/2}\vsig'^{1/2}w_c\,\dd\vsig'=3\vsig'^{1/2}w_c\,\dd\vsig'$, and this
measure has mass $2(a-\frac14)$ on $[0,\vsig_a]$. Inserting \eqref{eq:E-window} into $c'\cP_{\chi_a}=-\int c'\rhoU f_x\,\dd y$
gives \eqref{eq:E-potential} with $\mathcal R_a=\int_0^{\vsig_a}3\vsig'^{1/2}w_cR\,\dd\vsig'$.
(b) Replacing $w_c$ by $1$ changes the integral in \eqref{eq:E-potential} by at most
$0.33\,\lc\cdot3\int_0^{\Evmax}\vsig'^{3/2}\abs{\log\abs{\vsig-\vsig'}}\,\dd\vsig'\le C_{\Evmax}\lc$. Replacing the upper limit
$\vsig_a$ by $\Ehvs_a$ changes it by at most $3\Ehvs_a^{1/2}\int_I\abs{\log\abs{\vsig-\vsig'}}\,\dd\vsig'$ over an interval
$I$ of length $\abs I\le0.14\,\lc\Ehvs_a^2$ by \eqref{eq:E-sites}, and
$\int_I\abs{\log\abs{\vsig-\vsig'}}\dd\vsig'\le\abs I\,(1+\log\frac2{\abs I}+2\log(1+2\Evmax))$, because
$\abs{\log r}=-\log r+2\log^+r$ with $\log^+r:=\max(\log r,0)$, and $\abs{\vsig-\vsig'}\le1+2\Evmax$. The function $p_a$ is
continuous because the logarithmic singularity is integrable.
(c) On the window $4\sech^2\frac{\pi x}2$ lies between two positive constants, and $e^{p_2}$ is bounded above and below
on the compact range of $\vsig$; this gives the first statement, and the second follows from $0\le\Jend\le\CJ$ on
$x\ge\chi_3$ (\cref{lem:E-Jend}). For \eqref{eq:E-intermediate} the lower bound is \eqref{eq:E-Pbelow}. For the upper bound we
use \eqref{eq:E-Pbounds} and, with $z=\frac\pi2(x-\chi_2)$, the bound
$\abs{\log\tanh z}\le\log\frac1z+\frac{z^2}3$ of \cref{lem:E-window} and $\tanh\frac{\pi(x+\chi_2)}2\ge\tanh u_1=2^{-1/2}$; the
first logarithm in $\theta$ is at most $\log2$ because $x-\xs\ge2(\chi_2-\xs)$.
\end{proof}

\subsection{One-sided Euler--Maclaurin}
\label{app:E.2}

Where the summands are concave along the sites, that is, except far to the left of $x$, the sums in~\eqref{eq:E-comb}
are bounded on both sides by the integrals $I_x$ with no remainder (\cref{lem:E-em}). Far to the left of $x$ the
summands are smooth and small, and the ordinary midpoint remainder applies; its total is the loss $\CL$
(\cref{lem:E-farleft}). Rows 5, 9 and 10 of \cref{tab:4} are used.

\begin{lemma}[One-sided Euler--Maclaurin; \proved]\label{lem:E-em}
Let $x\ge0$, $\varphi(a)=f_x(\chi_a)$, and let $K,K_0\le K_1$ be integers $\ge2$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item \emph{Trapezoid, beyond $x$.} If $x<\chi_K$, then $2\sum_{k\ge K}\varphi(k)\le\varphi(K)+2I_x(K)$.
\item \emph{Midpoint, beyond $x$.} If $x<\chi_{K-1/2}$, then $2\sum_{k\ge K}\varphi(k)\ge2I_x(K-\half)$. In particular, on
the lobe $0\le x<\chi_2$ both hold with $K=3$.
\item \emph{A concave block below $x$.} If $\varphi$ is concave on $[K_0-\half,K_1+\half]$, which by
\cref{lem:E-concave}(c) is the case when $\Eythr(x)\le\chi_{K_0-1/2}$ and $\chi_{K_1+1/2}<x$, then
\begin{equation}\label{eq:E-block}
  2\int_{K_0-1/2}^{K_1+1/2}\varphi\,\dd a\ \le\ 2\sum_{k=K_0}^{K_1}\varphi(k)\ \le\
  2\int_{K_0}^{K_1}\varphi\,\dd a+\varphi(K_0)+\varphi(K_1);
\end{equation}
the second inequality needs concavity on $[K_0,K_1]$ only.
\item \emph{Monotone summands.} If $g$ is monotone on $[K_0-\half,K_1+\half]$, then
$\bigabs{\sum_{k=K_0}^{K_1}g(k)-\int_{K_0-1/2}^{K_1+1/2}g}\le\half\abs{g(K_1+\half)-g(K_0-\half)}$ and
$\bigabs{\half g(K_0)+\sum_{K_0<k<K_1}g(k)+\half g(K_1)-\int_{K_0}^{K_1}g}\le\half\abs{g(K_1)-g(K_0)}$.
\end{enumerate}
\end{lemma}

\begin{proof}
For a concave $\varphi$ the chord lies below the graph and the tangent above it, so
$\half[\varphi(k)+\varphi(k+1)]\le\int_k^{k+1}\varphi$ and $\varphi(k)\ge\int_{k-1/2}^{k+1/2}\varphi$. By
\cref{lem:E-concave}(c), $\varphi$ is concave on $\{\chi_a>x\}$. Summing the first inequality over $k\ge K$ gives
$\half\varphi(K)+\sum_{k>K}\varphi(k)\le I_x(K)$, which is (a); summing the second gives (b); the same two sums over a
finite block give (c). For (d) let $g$ increase. Then $g(k)-\int_{k-1/2}^{k+1/2}g$ lies between
$-\half[g(k+\half)-g(k)]$ and $\half[g(k)-g(k-\half)]$, and $\half[g(k)+g(k+1)]-\int_k^{k+1}g$ lies between
$\mp\half[g(k+1)-g(k)]$. These are increments of $g$ over disjoint intervals, so their sums over $k$ are at most the
total increment of $g$.
\end{proof}

\paragraph{The far left.}
Below $\Eythr(x)$ the summands need not be concave. Let
\begin{equation}\label{eq:E-k0}
  k_0(x):=\min\{k\ge3:\ \chi_{k-1/2}\ge\Eythr(x)\}
\end{equation}
be the first site from which \cref{lem:E-em}(c) applies, and let
\begin{equation}\label{eq:E-eL}
  e_L(x):=\frac1{24}\sum_{k=3}^{k_0(x)-1}\ \sup_{k-\frac12\le a\le k+\frac12}\abs{\varphi''(a)}
\end{equation}
be the total midpoint remainder of the sites $3\le k<k_0$; it vanishes when $k_0=3$.

\begin{lemma}[The far-left error; \provedCA]\label{lem:E-farleft}
Let $x$ lie in cell $n\ge6$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item $\bigabs{\sum_{k=3}^{k_0-1}\varphi(k)-\int_{5/2}^{k_0-1/2}\varphi\,\dd a}\le e_L(x)$.
\item $k_0(x)\le\bigl\lceil\frac34+\frac{n+3/4}{2.3}\bigr\rceil$. Hence $k_0\le n-2$, and $k_0\le n-j$ whenever
$n\ge2j+4$, $j\ge1$.
\item $x-\chi_a\ge0.37\,(x-\xs)$ for all $a\le k_0-\half$ when $k_0\ge4$, and
\begin{equation}\label{eq:E-eL-decay}
  e_L(x)\le C\Bigl[\Bigl(\frac{\lc}{x-\xs}\Bigr)^{3/2}+\frac{\lc}{x-\xs}\Bigr]\ \ (x-\xs\le1),\qquad e_L(x)\le C\lc\ \ (x-\xs\ge1).
\end{equation}
In particular $e_L=O_r(c'^{-2/3})$ uniformly on $x\ge\xs+r$.
\item \textup{(\Cref{tab:4}, row 5.)} For $0<\Ehvs_a<\vsig$ let $\hat\varphi_\vsig(a):=\log(\vsig-\Ehvs_a)$ be the summand
with the site at its nominal position, so that
$\hat\varphi_\vsig''(a)=\frac29\,(\vsig-3\Ehvs_a)\,\Ehvs_a^{-2}(\vsig-\Ehvs_a)^{-2}$, and for $\vsig\ge3$ let
\begin{equation}\label{eq:E-eLhat}
  \hat k_0(\vsig):=\min\bigl\{k\ge3:\ \Ehvs_{k-1/2}\ge1.02\,\vsig/2.3\bigr\},\qquad
  \hat e_L(\vsig):=\frac1{24}\sum_{k=3}^{\hat k_0(\vsig)-1}\ \sup_{k-\frac12\le a\le k+\frac12}\bigabs{\hat\varphi_\vsig''(a)} .
\end{equation}
This is $e_L$ in the limit $c\to\infty$ at fixed $\vsig=(x-\xs)/\lc$, except that the factor $1.02$ enlarges the block of
sites; it absorbs the displacement of the true sites from the nominal ones. Then
$\hat e_L(\vsig)\le0.00186$ for $3\le\vsig\le\Ehvs_{602}$ \cert.
Consequently, if $\lc\Ehvs_{602}\le0.24$, then $e_L(x)\le0.00186+C\lc$ for every $x$ in a cell $n\ge6$. We take $\catt$
so large that $\lc\Ehvs_{602}\le0.24$ and $C\lc\le0.004$ for $c\ge\catt$. Then $\CL:=\sup e_L(x)\le0.006$, the supremum
being over $c\ge\catt$ and over $x$ in the cells $n\ge6$, and $e^{-2\CL}\ge0.988$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) For a $C^2$ function, $\varphi(k)-\int_{k-1/2}^{k+1/2}\varphi=-\frac1{24}\varphi''(\xi_k)$ with $\abs{\xi_k-k}\le\half$.

(b) Let $x=\chi_b$ with $n<b<n+1$, and let $\chi_{a^*}=\xs+(x-\xs)/2.3$. The mean spacing from the tip decreases
(\cref{lem:E-concave}(b)), so $(\chi_{a^*}-\xs)/(a^*-\frac14)\ge(x-\xs)/(b-\frac14)$, that is
$a^*-\frac14\le(b-\frac14)/2.3$. Since $\Eythr(x)\le\chi_{a^*}$, we get
$k_0\le\max\{3,\lceil a^*+\half\rceil\}\le\lceil\frac34+(n+\frac34)/2.3\rceil$. For $n=6,7$ this is $4,5$; in general it is
less than $2.08+0.435\,n$, which does not exceed $n-2$ for $n\ge8$ nor $n-j$ for $n\ge2j+4$.

(c) If $k_0=3$, then $e_L=0$. Let $k_0\ge4$. By minimality $\chi_{k_0-3/2}<\Eythr(x)\le\xs+(x-\xs)/2.3$, and the mean spacing
from the tip gives
$\chi_{k_0-1/2}-\xs\le\frac{k_0-3/4}{k_0-7/4}(\chi_{k_0-3/2}-\xs)\le\frac{13}9\,(x-\xs)/2.3\le0.629\,(x-\xs)$. Hence
$x-\chi_a\ge d:=0.37\,(x-\xs)$ for $a\le k_0-\half$. For such $a$, \cref{lem:E-concave}(a), the bound
$\cosh z/\sinh^2z\le z^{-2}+C$ and $x+y\ge2\xs$ give $\abs{\partial_yf_x}\le1/d$ and $\abs{\partial_y^2f_x}\le d^{-2}+C$ at
$y=\chi_a$, so that
\begin{equation}\label{eq:E-phi2}
  \abs{\varphi''(a)}\le(\chi_a')^2\Bigl[\frac1{d^2}+C+\frac{(\log\rhoU)'(\chi_a)}{d}\Bigr]=:G(a),
\end{equation}
by the formula for $\varphi''$ in the proof of \cref{lem:E-concave}. Both $\chi_a'$ and $(\log\rhoU)'(\chi_a)$ decrease in
$a$, because $\rhoU$ is increasing and concave; so $G$ decreases, $G(k-\half)\le\int_{k-3/2}^{k-1/2}G\,\dd a$ for $k\ge4$, and
\[
  24\,e_L\ \le\ \sum_{k=3}^{k_0-1}G(k-\half)\ \le\ G(\tfrac52)+\int_{5/2}^{k_0-3/2}G\,\dd a .
\]
On the range of integration $r:=\chi_a-\xs\le(x-\xs)/2.3$, by the minimality of $k_0$. By \eqref{eq:E-sitemap},
$(\chi_a')^2\,\dd a=\frac2{c'}\rhoU(y)^{-1}\dd y$ with $y=\chi_a$, and
$\int(\chi_a')^2(\log\rhoU)'(\chi_a)\,\dd a=-\int\chi_a''\,\dd a\le\chi_{5/2}'$.

Let first $x-\xs\le1$, and let $\lc\le0.08$; this holds for $c\ge\catt$, because then $\lc\le0.3/\Ehvs_{10}$. Put
$\vsig:=(x-\xs)/\lc$, so that $d=0.37\,\lc\vsig$. By \eqref{eq:E-sites},
$\vsig_{5/2}\ge\Ehvs_{5/2}(1-0.14\cdot0.08\,\Ehvs_{5/2})\ge1.68$. Since $\rhoU(\xs+r)\ge3\ktip r^{1/2}$ by \eqref{eq:E-tip}
and $\frac2{c'}=2\ktip\lc^{3/2}$, we have $\chi_{5/2}'\le\frac23\lc\vsig_{5/2}^{-1/2}$ and
$\frac2{c'}\int_0^{r}\rhoU(\xs+r')^{-1}\dd r'\le\frac43\lc^{3/2}r^{1/2}$, which we use with $r=\lc\vsig/2.3\le0.44$; and
$(\log\rhoU)'(\chi_{5/2})\le0.95/(1.3\,\lc\vsig_{5/2})$ by row~9. Inserting these in $\int G\,\dd a$ and in $G(\frac52)$,
we get
\begin{equation}\label{eq:E-eL-explicit}
  24\,e_L(x)\le6.43\,\vsig^{-3/2}+1.71\,\vsig^{-1}+1.94\,\vsig^{-2}+C\lc^{3/2}\qquad(x-\xs\le1,\ \lc\le0.08).
\end{equation}
Here $6.43\ge\frac43\cdot2.3^{-1/2}\cdot0.37^{-2}$ belongs to the term $d^{-2}$ of the integral;
$1.71\ge\frac23\,(0.37\sqrt{1.68})^{-1}+\frac49\cdot\frac{0.95}{1.3}\,(0.37\cdot1.68^2)^{-1}$ to the terms with $(\log\rhoU)'$ in
the integral and in $G(\frac52)$; $1.94\ge\frac49\,(1.68\cdot0.37^2)^{-1}$ to the term $d^{-2}$ of $G(\frac52)$; and the two
terms with the constant $C$ of \eqref{eq:E-phi2} add up to at most $C\lc^{3/2}$. Since $\vsig\ge\vsig_6\ge1$ and
$\lc\le\lc/(x-\xs)$, the first bound of \eqref{eq:E-eL-decay} follows.

Let now $x-\xs\ge1$, so that $d\ge0.37$. The cells $[k-\half,k+\half]$ on which $r\le0.63$ are treated as before, with
$r\le0.63$ in the place of $r\le(x-\xs)/2.3$, and contribute at most $C\lc$. On every other cell $r\ge0.63-\lc\ge0.5$,
because $\chi_{k+1/2}-\chi_{k-1/2}\le\chi_{5/2}'\le\lc$.
There we use $(\log\rhoU)'\le3.10$ and, since $x-y\ge0.37$, the exponential bounds
$\abs{\partial_yf_x}+\abs{\partial^2_yf_x}\le Ce^{-\pi(x-y)}$ from \eqref{eq:E-dfx}, so that
$\abs{\varphi''(a)}\le\tilde G(a):=C(\chi_a')^2e^{-\pi(x-\chi_a)}$. The function $\tilde G$ is not monotone; but over one cell
$\chi_a$ varies by at most $\lc$ and, by \eqref{eq:E-spacing}, $\chi_a'$ by a factor of at most $2$, so
$\sup_{\rm cell}\tilde G\le C\int_{\rm cell}\tilde G\,\dd a$. The sum over these cells is therefore at most
$C\int\tilde G\,\dd a=\frac C{c'}\int e^{-\pi(x-y)}\rhoU(y)^{-1}\,\dd y\le C/c'\le C\lc$, because $\rhoU\ge\rhoU(\xs+0.5)$ there.

(d) \emph{The certified function.} For fixed $\vsig$ the function $h(t):=(\vsig-3t)\,t^{-2}(\vsig-t)^{-2}$ decreases on
$0<t<\vsig$, because $h'(t)=-(9t^2-7\vsig t+2\vsig^2)\,t^{-3}(\vsig-t)^{-3}$ and
$9t^2-7\vsig t+2\vsig^2=9(t-\frac7{18}\vsig)^2+\frac{23}{36}\vsig^2>0$. Hence $\hat\varphi_\vsig''(a)$ decreases in $a$, and the
supremum in \eqref{eq:E-eLhat} is the larger of the two numbers $\abs{\hat\varphi_\vsig''(k\pm\half)}$. For $\vsig\ge3$ and
$K=\hat k_0(\vsig)$ we have $\Ehvs_{K-1/2}\le0.58\,\vsig$: if $K=3$ because $\Ehvs_{5/2}\le1.72$, and if $K\ge4$ because
$\Ehvs_{K-1/2}\le(\frac{13}9)^{2/3}\Ehvs_{K-3/2}<(\frac{13}9)^{2/3}\,1.02\,\vsig/2.3$. Thus $\hat e_L$ is a finite sum of explicit
functions of $\vsig$, which jumps upward where $\hat k_0$ does. Row~5 bounds it on subintervals that cover
$[3,\Ehvs_{602}]$; since $\hat k_0$ is non-decreasing and the summands are non-negative, $\hat k_0$ is replaced on each
subinterval by its value at the right end. With the factor $1$ in the place of $1.02$ the supremum is $0.0016439$; it is approached as $\vsig$ decreases to
$2.3\,\Ehvs_{5/2}=3.949\ldots$, between $\Ehvs_8$ and $\Ehvs_9$, where $\hat k_0$ jumps from $3$ to $4$.

\emph{Large $\vsig$.} For the rest of the proof let $\lc\Ehvs_{602}\le0.24$, so that $\lc\le0.08$, and let
$\vsig=(x-\xs)/\lc$. The right side of
\eqref{eq:E-eL-explicit} decreases in $\vsig$, and at $\vsig=\Ehvs_{602}=71.27\ldots$ its first three terms add up to less than
$24\cdot0.00147$. Hence $e_L\le0.00147+C\lc^{3/2}$ for $\vsig\ge\Ehvs_{602}$, $x-\xs\le1$; and $e_L\le C\lc$ for $x-\xs\ge1$ by
\eqref{eq:E-eL-decay}. Both are at most $0.00186+C\lc$. The values of $e_L$ on successive cells are not monotone, so that
the range of large $\vsig$ cannot be reduced to the certified one by monotonicity; it is the right side of
\eqref{eq:E-eL-explicit} that decreases.

\emph{The range $3\le\vsig\le\Ehvs_{602}$.} Let $x$ lie in a cell $n\ge6$. Then $\vsig\ge\vsig_6\ge3$ by \eqref{eq:E-sites}, and it
remains to treat $\vsig\le\Ehvs_{602}$, so that $x-\xs=\lc\vsig\le0.24$. Let $K=\hat k_0(\vsig)$. Since $\Ehvs_{K-1/2}\le0.58\,\vsig$,
\eqref{eq:E-sites} gives
$\vsig_{K-1/2}\ge\Ehvs_{K-1/2}(1-0.14\cdot0.58\,\lc\vsig)\ge1.02\,(1-0.0195)\,\vsig/2.3\ge\vsig/2.3$, that is
$\chi_{K-1/2}\ge\xs+(x-\xs)/2.3\ge\Eythr(x)$. Hence $k_0(x)\le\hat k_0(\vsig)$, and every site $a\le k_0-\half$ has
$\Ehvs_a\le0.58\,\vsig$. For such $a$ write $D:=\vsig-\vsig_a$ and $\hat D:=\vsig-\Ehvs_a$, so that $D\ge\hat D\ge0.42\,\vsig\ge1.2$.
Since $\chi_a=\xs+\lc\vsig_a$,
\[
  \varphi''(a)=\lc^2(\vsig_a')^2\,\partial_y^2f_x(\chi_a)+\lc\vsig_a''\,\partial_yf_x(\chi_a),\qquad
  \hat\varphi_\vsig''(a)=-\frac{(\Ehvs_a')^2}{\hat D^2}-\frac{\Ehvs_a''}{\hat D}.
\]
Each of the five quantities $-(\lc D)^2\partial_y^2f_x(\chi_a)$, $-\lc D\,\partial_yf_x(\chi_a)$, $\vsig_a/\Ehvs_a$,
$\vsig_a'/\Ehvs_a'$ and $\vsig_a''/\Ehvs_a''$ differs from $1$ by at most $C\lc\vsig$: the first two by \eqref{eq:E-dfx}, because
$\pi\lc D\le\pi\lc\vsig$ and $x+\chi_a\ge2\xs$; the third by \eqref{eq:E-sites}, which also gives
$0\le D-\hat D=\Ehvs_a-\vsig_a\le C\lc\vsig\,\hat D$; the fourth by \eqref{eq:E-sitemap} and
\eqref{eq:E-tip}, which give $\vsig_a'=\frac23\vsig_a^{-1/2}(1+\theta)^{-1}$ with $0\le\theta\le0.33\,\lc\vsig_a$; and the fifth by
\eqref{eq:E-sitemap} and \eqref{eq:E-dlogrho}, which give
$\vsig_a''=-(\vsig_a')^2\,\lc(\log\rhoU)'(\chi_a)=-(\vsig_a')^2(1+\theta')/(2\vsig_a)$ with $0\le\theta'\le1.12\,\lc\vsig_a$. On the
present range $\vsig$, $1/\Ehvs_a$ and $1/\hat D$ are bounded and $\lc\vsig\le\Ehvs_{602}\lc$, so that
$\abs{\varphi''(a)-\hat\varphi_\vsig''(a)}\le C\lc$; and the number of summands is bounded. As $k_0(x)\le\hat k_0(\vsig)$ and all
summands are non-negative, $e_L(x)\le\hat e_L(\vsig)+C\lc\le0.00186+C\lc$. No continuity in $x$ or in $c$ is used: the jumps
of $k_0(x)$ are covered by $k_0(x)\le\hat k_0(\vsig)$. The factor $1.02$ is needed for this: since $\vsig_a\le\Ehvs_a$, the true
index $k_0(x)$ jumps slightly before the nominal one does, and there $e_L$ exceeds its nominal value by a whole summand.
\end{proof}

\subsection{Detuning costs \texorpdfstring{$O(\eta_0^2)$}{O(eta0 squared)}}
\label{app:E.3}

Lifting a double zero off the axis to the heights $\pm\varepsilon_k$ raises the modulus of its factor at a point of the
axis at a distance $D\gg\varepsilon_k$ by the relative amount $O(\varepsilon_k^2/D^2)$ only, and $\varepsilon_k$ is
$\eta_0$ times the local spacing. Thus lower bounds for $B^0$ are lower bounds for $B$, and upper bounds for $B^0$
become upper bounds for $B$ at the cost of the two pairs which flank $x$, which are kept exactly, and of a factor
$e^{\Cdet\eta_0^2}$ with $\Cdet\le16.6$. No certified constant is used.

\begin{lemma}[Detuning; \proved]\label{lem:E-detune}
Let $x\ge0$, $k\ge3$ and $A_\pm:=\cosh\pi(\chi_k\mp x)$. Then
\begin{equation}\label{eq:E-detune-pair}
  T_{\chi_k}(x)^2\ \le\ q_k(x)\ \le\ T_{\chi_k}(x)^2\exp\frac{4\varepsilon_k^2}{(\chi_k-x)^2},\qquad
  \frac{q_k}{T_{\chi_k}^2}=\prod_\pm\frac{(A_\pm-\gamma_k)(A_\pm+1)}{(A_\pm+\gamma_k)(A_\pm-1)} .
\end{equation}
There is an absolute constant $\Cdet\le16.6$ such that, for $x$ in cell $n\ge2$ and for $0\le x<\chi_2$ (where no pair
is excluded),
\begin{equation}\label{eq:E-detune-sum}
  0\ \le\sum_{k\ge3,\ k\ne n,n+1}\log\frac{q_k(x)}{T_{\chi_k}(x)^2}\ \le\ \Cdet\,\eta_0^2 ,
\end{equation}
and the pairs beyond the cell contribute at most $4\zeta(2)\eta_0^2$ of this. In particular
$\Pi:=\abs{B}/\abs{B^0}=\prod_{k\ge3}q_k/T_{\chi_k}^2\ge1$. For the flanking pairs,
\begin{equation}\label{eq:E-flank}
  \frac{q_n(x)}{T_{\chi_n}(x)^2}\le\Bigl(1+\frac{\eta_0^2}{u^2}\Bigr)(1+C\ell_n^2),\qquad
  \frac{q_{n+1}(x)}{T_{\chi_{n+1}}(x)^2}\le\Bigl(1+\frac{\eta_0^2}{(1-u)^2}\Bigr)(1+C\ell_n^2).
\end{equation}
\end{lemma}

\begin{proof}
The product formula in \eqref{eq:E-detune-pair} is \eqref{eq:E-pair} divided by its value at $\gamma_k=1$. For $A>1$ and
$0\le\gamma\le1$,
\[
  \log\frac{A-\gamma}{A+\gamma}-\log\frac{A-1}{A+1}=\int_\gamma^1\frac{2A\,\dd\gamma'}{A^2-\gamma'^2}\in\Bigl[0,\frac{2A(1-\gamma)}{A^2-1}\Bigr]
  \subset\Bigl[0,\frac{(\pi\varepsilon)^2}{A-1}\Bigr]\quad\text{for }\gamma=\cos\pi\varepsilon,
\]
since $1-\cos\pi\varepsilon\le\half(\pi\varepsilon)^2$ and $A/(A+1)\le1$. With $A_\pm-1=2\sinh^2\frac{\pi(\chi_k\mp x)}2\ge\half\pi^2(\chi_k\mp x)^2$ the two
factors contribute at most $2\varepsilon_k^2/(\chi_k-x)^2+2\varepsilon_k^2/(\chi_k+x)^2$, which gives
\eqref{eq:E-detune-pair}.

For a pair beyond the cell, $k\ge n+2$, the spacings decrease, so $\chi_k-x\ge\chi_k-\chi_{n+1}\ge(k-n-1)\ell_k$, and
$\varepsilon_k=\eta_0\ell_k$; the pair contributes at most $4\eta_0^2/(k-n-1)^2$, and these sum to $4\zeta(2)\eta_0^2$. On
the lobe the same holds with $\chi_k-x\ge\chi_k-\chi_2\ge(k-2)\ell_k$. For a pair below the cell, $3\le k\le n-1$, we have
$x-\chi_k\ge(n-k)\ell_{n-1}$, and by \cref{lem:E-concave}(b)
$\ell_k/\ell_{n-1}\le\chi_k'/\chi_n'\le\bigl((n-\frac14)/(k-\frac14)\bigr)^{1/2}$. The pair contributes at most
$4\eta_0^2\,(n-\frac14)/[(k-\frac14)(n-k)^2]=4\eta_0^2\bigl[(n-k)^{-2}+(k-\frac14)^{-1}(n-k)^{-1}\bigr]$. The first terms sum to
at most $\zeta(2)$. By the Cauchy--Schwarz inequality the second terms sum to at most
$\bigl(\zeta(2)\sum_{k\ge3}(k-\frac14)^{-2}\bigr)^{1/2}\le(\frac49\zeta(2))^{1/2}\le0.86$, since
$\sum_{k\ge3}(k-\frac14)^{-2}\le\int_{5/2}^\infty(t-\frac14)^{-2}\,\dd t=\frac49$ by convexity. Hence \eqref{eq:E-detune-sum}
holds with $\Cdet:=4\,[2\zeta(2)+0.86]$, and $4\zeta(2)\le\Cdet\le16.6$.

For \eqref{eq:E-flank}, the factor with $A_-=\cosh\pi(\chi_n-x)$ is $\bigl(1+\frac{1-\gamma_n}{A_--1}\bigr)\frac{A_-+1}{A_-+\gamma_n}$; here
$\frac{1-\gamma_n}{A_--1}\le\varepsilon_n^2/(x-\chi_n)^2=\eta_0^2/u^2$, and the second quotient, as well as the factor
with $A_+$, is $1+O(\varepsilon_n^2)$. For $k=n+1$ we use in addition $\ell_{n+1}\le\ell_n$.
\end{proof}

\subsection{Corollaries}
\label{app:E.4}

We collect the bounds of \cref{lem:7.5}(c), (e), (f) and the overshoot bound of \cref{lem:7.11}. Each is one
application of \cref{lem:E-em} on each side of $x$, with the sites next to $x$ kept exactly. The bounds on the cells
$n\ge6$ use \cref{lem:E-two-sided}, which is proved in \cref{app:E.5} from \cref{app:E.1,app:E.2,app:E.3} alone. Rows
5, 7, 8, 9 and 10 of \cref{tab:4} are used.

\paragraph{The lobe: \cref{lem:7.5}(f) \tagpar{\proved}.}
Let $0\le x<\chi_2$. All sites lie beyond $x$, so \cref{lem:E-em}(a), (b) with $K=3$, the inequality $q_k\ge T_{\chi_k}^2$,
\cref{lem:E-detune} and \eqref{eq:E-continuum} give
\begin{equation}\label{eq:E-lobe}
\begin{aligned}
  f_x(\chi_2)+c'U+c'\cP_{\chi_{5/2}}&\ \le\ \log\abs{B^0}\ \le\ \log\abs{B}\\
  &\ \le\ f_x(\chi_2)+f_x(\chi_3)+c'U+c'\cP_{\chi_3}+4\zeta(2)\eta_0^2,
\end{aligned}
\end{equation}
all functions being evaluated at $x$; on $\xs\le x<\chi_2$, $U=-L$.

\paragraph{Upper bound on every cell.}
Let $x$ lie in cell $n\ge2$. The pairs $k\ge n+2$ lie beyond $x$; we write $\log q_k=2f_x(\chi_k)+\log(q_k/T_{\chi_k}^2)$
for them. By \cref{lem:E-em}(a) with $K=n+2$, \eqref{eq:E-continuum} and \cref{lem:E-detune},
\begin{equation}\label{eq:E-ubgap}
\begin{gathered}
  \log\abs{B(x)}+c'L(x)\ \le\ f_x(\chi_2)+\sum_{k=3}^{n+1}\log q_k(x)+f_x(\chi_{n+2})+c'\cP_{\chi_{n+2}}(x)+D_n(x),\\
  D_n(x):=\sum_{k\ge n+2}\log\frac{q_k(x)}{T_{\chi_k}(x)^2}\ \in\ \bigl[0,\,4\zeta(2)\eta_0^2\bigr] .
\end{gathered}
\end{equation}
The exact factors on the right carry the total weight $1+2(n-1)+1=2n$, and the continuum term carries
$2(n+2-\frac14)=2n+2n_1$.

\paragraph{Upper bound on the cells $n\ge6$ and on the far band: \cref{lem:7.5}(e) \tagpar{\provedCA}.}
Let $x$ lie in cell $n\ge6$. By \cref{lem:E-two-sided} with $j=1$, \cref{lem:E-farleft}(d) and \cref{lem:E-detune},
\begin{equation}\label{eq:E-ubcell}
  \log\abs{B}+c'L\ \le\ c'\cP+\Jend+\EcH^+_n+\log\frac{q_nq_{n+1}}{T_{\chi_n}^2T_{\chi_{n+1}}^2}+\Cdet\eta_0^2+2\CL .
\end{equation}
By \eqref{eq:E-displacement-explicit}, $\EcH^+_n\le h^+_\infty(u)+\frac{94}n+C\ell_n$, where
$h^+_\infty(u)=\log(1+u)+2\log u+2\log(1-u)+\log(2-u)-2\int_{-1}^2\log\abs{u-v}\,\dd v$. Each of the three pieces
$\log[(1+u)(2-u)]$, $2\log[u(1-u)]$ and $-2\int_{-1}^2\log\abs{u-v}\dd v$ is largest at $u=\half$; the last because its
derivative is $2\log\frac{2-u}{1+u}$. Hence (\cref{tab:4}, row 7)
\begin{equation}\label{eq:E-suphplus}
  \sup_{0<u<1}h^+_\infty(u)=h^+_\infty(\tfrac12)=6-4\log3\le1.606\ \cert,\qquad\text{so}\quad
  \sup_{\rm cell}e^{\EcH^+_n}\le5.0\,e^{94/n+C\ell_n}.
\end{equation}
With the flanking pairs, \eqref{eq:E-flank} replaces the middle piece by $\log P(u(1-u))+O(\ell_n^2)$, where
$P(p):=p^2+\eta_0^2(1-2p)+\eta_0^4$ is $(u^2+\eta_0^2)((1-u)^2+\eta_0^2)$ written in $p=u(1-u)\in[0,\frac14]$. A convex
quadratic is largest at an end of the interval, and $P(\frac14)=(\frac14+\eta_0^2)^2\ge P(0)=\eta_0^2(1+\eta_0^2)$
precisely when $\eta_0^2\le\frac18$. So for $\eta_0=\frac14$ the supremum is again at $u=\half$ and exceeds
\eqref{eq:E-suphplus} by $2\log(1+4\eta_0^2)$:
\begin{equation}\label{eq:E-ubcell-B}
  \abs{B(x)}e^{c'L(x)}\le5.0\,(1+4\eta_0^2)^2e^{\Cdet\eta_0^2}\cdot e^{2\CL+94/n+C\ell_n}\,e^{c'\cP(x)+\Jend(x)} .
\end{equation}
On the far band $x\ge\xs+r$ we have $c'\cP\le C_r$ by \cref{lem:7.5}(b), $\Jend\le\CJ$ by \cref{lem:E-Jend},
$n\ge c'N(\xs+r)-1$ and $\ell_n\le C_r/c'$; so $\abs{B}e^{c'L}\le C_r$ there.

\paragraph{The sup bound: \cref{lem:7.5}(c) \tagpar{\provedCA}.}
On $x\ge\xs+r$ the bound $C_r$ has just been proved. On the cells $n\ge6$, \eqref{eq:E-ubcell-B} gives
$\log\abs B+c'L\le c'\cP+C$, and $c'\cP\le2n_1\abs{f_x(\chi_2)}$ by \eqref{eq:E-Pbounds}. Here
$x-\chi_2\ge\chi_6-\chi_2\ge\lc$ for $c\ge\catt$ by \eqref{eq:E-sites}. If $x-\chi_2\le0.9$, the estimate used for
\eqref{eq:E-intermediate} gives $\abs{f_x(\chi_2)}\le\log\frac{2}{\pi(x-\chi_2)}+C\le\log\frac1\lc+C$; otherwise
$\abs{f_x(\chi_2)}\le C$. Hence $c'\cP\le2n_1\cdot\frac23\log c'+C=\frac43n_1\log c'+C$. On the cells $2\le n\le5$ we use
\eqref{eq:E-ubgap} in the window with $\Evmax=\Ehvs_8$. By \eqref{eq:E-window} and \eqref{eq:E-window-pair} (row 10)
each site factor is at most $C\lc$ and each pair modulus at most $C\lc^2$, because the scaled distances are bounded;
and $c'\cP_{\chi_{n+2}}\le(2n+2n_1)\log\frac1\lc+C$ by \eqref{eq:E-potential-limit}. The powers of $\lc$ combine to
$\lc^{-2n_1}$, so $\log\abs B+c'L\le\frac43n_1\log c'+C$ on these cells as well. Together,
\begin{equation}\label{eq:E-sup}
  \sup_{x\ge\chi_2}\bigl[\log\abs{B(x)}+c'L(x)\bigr]\le\tfrac43n_1\log c'+C,\qquad
  \sup_{x\ge\xs+r}\bigl[\log\abs{B(x)}+c'L(x)\bigr]\le C_r .
\end{equation}

\paragraph{The overshoot bound: \cref{lem:7.11} \tagpar{\provedCA}.}
Let $x$ lie in cell $n\ge6$. The lower bound of \cref{lem:E-two-sided} with $j=1$, together with $\Jend\ge0$
(\cref{lem:E-Jend}) and $e_L\le\CL$, gives $\log\abs B+c'L\ge c'\cP+\EcH^-_n-2\CL$. By \cref{lem:E-hole-limits},
$\EcH^-_n\ge h^-_\infty(u;\eta_0)-C(\frac1n+\ell_n)$, with
$h^-_\infty(u;\eta_0)=\log(u^2+\eta_0^2)+\log((1-u)^2+\eta_0^2)-2\int_{-1/2}^{3/2}\log\abs{u-v}\,\dd v$. Row~8 of \cref{tab:4} is
\begin{equation}\label{eq:E-minhminus}
  \min_{0\le u\le1}h^-_\infty(u;\eta_0)\ \ge\ 2\log\eta_0+\log(1+\eta_0^2)+3.476\ \cert\qquad(0<\eta_0\le0.3),
\end{equation}
the minimum being attained at the ends $u\in\{0,1\}$. The certificate rests on three facts which are proved by hand.
Let $s:=\eta_0^2\le0.09$, $h:=h^-_\infty(\,\cdot\,;\eta_0)$ and
$I(u):=-2\int_{-1/2}^{3/2}\log\abs{u-v}\,\dd v=4-2(u+\half)\log(u+\half)-2(\tfrac32-u)\log(\tfrac32-u)$.
(i)~$I(0)=I(1)=4-3\log3+4\log2=3.4767\ldots$, so that $h(0)=h(1)=2\log\eta_0+\log(1+\eta_0^2)+I(0)$.
(ii)~$h(u)-h(0)=\log(1+u^2/s)+\log\bigl(1-\frac{u(2-u)}{1+s}\bigr)+I(u)-I(0)$. The first term decreases in $s$; the second
increases in $s$, because $0\le u(2-u)\le1$, and it is $2\log(1-u)$ at $s=0$. Hence
$h(u)-h(0)\ge G_L(u):=\log(1+u^2/0.09)+2\log(1-u)+I(u)-I(0)$, and $G_L(0)=0$.
(iii)~Since $h(u)=h(1-u)$, also $h(u)-h(1)\ge G_R(u):=G_L(1-u)$.
Row~8 certifies that $G_L'>0$ on $[0,\frac14]$ and $G_L>0$ on $[\frac14,\half]$, and the mirror statements for $G_R$. Hence
$G_L>0$ on $(0,\half]$ and $G_R>0$ on $[\half,1)$, so that $h>h(0)=h(1)$ on $(0,1)$; and $I(0)>3.476$ is certified as well.
Since $e^{3.476}\ge32$, \eqref{eq:E-Pbelow} gives
\begin{equation}\label{eq:E-overshoot}
  \abs{B(x)}\,e^{c'L(x)}\ \ge\ 32\,\eta_0^2\,e^{-C(\frac1n+\ell_n)-2\CL}\Bigl(\frac{2}{\pi(x-\xs)}\Bigr)^{2n_1}
  \ \ge\ \cT\,\eta_0^2\Bigl(\frac{2}{\pi(x-\xs)}\Bigr)^{2n_1}
\end{equation}
on the cells $n\ge6$, with an absolute constant $\cT>0$: the exponent is bounded, because
$\frac1n\le\frac16$ and $\ell_n\le\chi_2'\le\lc$ by \cref{lem:E-concave}(b) and \eqref{eq:E-tip}. Only the positivity of
$\cT$ is used. \qed

\subsection{Two-sided bounds on the envelope, and the level}
\label{app:E.5}

This subsection proves \cref{lem:7.5}(a) and \cref{prop:7.4}. We apply \cref{lem:E-em} on either side of $x$, but
keep exactly the $2j+2$ sites nearest to $x$ in the upper bound, and the $2j$ detuned pairs nearest to $x$ in the lower
bound. The difference between these exact sites and the continuum integral over their index range is the \emph{hole
function}. Sites and continuum carry the same total charge, so all scales cancel, and as $n\to\infty$ the hole function
tends to an explicit function of the position $u$ in the cell alone. The rows used are 5, 6 and 9, and row 10 in the
window clause of \cref{lem:E-Jend}.

For $j\ge1$ and $x$ in cell $n\ge2j+4$ the hole functions are
\begin{align}
  \EcH^+_{n,j}(x)&:=f_x(\chi_{n-j})+2\sum_{k=n-j+1}^{n+j}f_x(\chi_k)+f_x(\chi_{n+j+1})-2\int_{n-j}^{n+j+1}\varphi(a)\,\dd a, \label{eq:E-Hplus}\\
  \EcH^-_{n,j}(x)&:=\sum_{k=n-j+1}^{n+j}\log q_k(x)-2\int_{n-j+1/2}^{n+j+1/2}\varphi(a)\,\dd a, \label{eq:E-Hminus}
\end{align}
and $\EcH^\pm_n:=\EcH^\pm_{n,1}$. The first does not depend on $\eta_0$. Its sites carry the weights $1,2,\dots,2,1$, of sum
$4j+2$, and its integral has the same weight $2(2j+1)$; the second has the weight $4j$ on either side. Their limits are
\begin{align}
  h^+_j(u)&:=\log(u+j)+2\sum_{i=-j+1}^{j}\log\abs{u-i}+\log(j+1-u)-2\int_{-j}^{j+1}\log\abs{u-v}\,\dd v, \label{eq:E-hplus}\\
  h^-_j(u;\eta_0)&:=\sum_{i=-j+1}^{j}\log\bigl((u-i)^2+\eta_0^2\bigr)-2\int_{-j+1/2}^{j+1/2}\log\abs{u-v}\,\dd v, \label{eq:E-hminus}
\end{align}
and $h^\pm_\infty:=h^\pm_1$. The integrals are elementary:
$\int_\alpha^\beta\log\abs{u-v}\dd v=F(u-\alpha)-F(u-\beta)$ with $F(t)=t\log\abs t-t$.

\begin{lemma}[The end correction; \provedCA]\label{lem:E-Jend}
For $x\ge\chi_{5/2}$, $0\le\Jend(x)\le\frac14\sup_{2\le a\le5/2}\abs{\varphi'(a)}$. Consequently $0\le\Jend\le\CJ$ on
$x\ge\chi_3$ with an absolute constant $\CJ\le0.65$; this uses no certified constant. Moreover
$\Jend(x)\le C/(\vsig-\vsig_{5/2})$ for $x=\xs+\lc\vsig$ in the window (row 10, through \eqref{eq:E-tip}).
\end{lemma}

\begin{proof}
For $x\ge\chi_{5/2}$ the function $\varphi$ decreases on $[2,\frac52]$, so
$\Jend=2\int_2^{5/2}[\varphi(2)-\varphi(a)]\,\dd a$ lies between $0$ and $2\int_0^{1/2}t\,\dd t\,\sup\abs{\varphi'}$. By
\cref{lem:E-concave}(a), $\abs{\varphi'(a)}\le\chi_a'/(x-\chi_a)$. For $x\ge\chi_3$ and $a\le\frac52$,
$x-\chi_a\ge\chi_3-\chi_{5/2}\ge\half\chi_3'$ and $\chi_a'\le\chi_2'\le(\frac{11}7)^{1/2}\chi_3'$ by \eqref{eq:E-spacing}, so
$\abs{\varphi'}\le2\,(\frac{11}7)^{1/2}\le2.6$ and $\CJ:=\frac14\cdot2.6=0.65$. In the window $\chi_2'\le\lc$ by
\eqref{eq:E-tip} and $x-\chi_a\ge\lc(\vsig-\vsig_{5/2})$.
\end{proof}

\begin{lemma}[Two-sided bounds; \provedCA]\label{lem:E-two-sided}
Let $j\ge1$ and let $x$ lie in cell $n\ge2j+4$. Then
\begin{align}
  \log\abs{B(x)}+c'L(x)&\ \ge\ c'\cP(x)+\Jend(x)+\EcH^-_{n,j}(x)-2e_L(x), \label{eq:E-lower}\\
  \log\abs{B^0(x)}+c'L(x)&\ \le\ c'\cP(x)+\Jend(x)+\EcH^+_{n,j}(x)+2e_L(x). \label{eq:E-upper}
\end{align}
For $j=1$ and $n\ge6$, together with $e_L\le\CL$ and \eqref{eq:E-detune-sum}, this is \cref{lem:7.5}(a): the lower
bound is for $B$, the upper bound for $B^0$, and $\log\abs B\le\log\abs{B^0}+\log\frac{q_nq_{n+1}}{T_{\chi_n}^2T_{\chi_{n+1}}^2}+\Cdet\eta_0^2$.
\end{lemma}

\begin{proof}
We split the sites $k\ge3$ into four blocks: the far left $3\le k<k_0$, where the midpoint rule holds up to $e_L$; the
concave block from $k_0$ to the hole; the hole, kept exactly; and the sites beyond the hole, which lie beyond $x$. We
then account for the pieces of $2I_x(2)=-c'L+c'\cP$ that no block has used. By \cref{lem:E-farleft}(b), $k_0\le n-j$.

\emph{Upper bound.} By \cref{lem:E-farleft}(a), $2\sum_{k=3}^{k_0-1}\varphi(k)\le2\int_{5/2}^{k_0-1/2}\varphi+2e_L$. The
function $\varphi$ is concave on $[k_0,n-j]$, where $\Eythr\le\chi_a<x$, so the trapezoid inequality of \eqref{eq:E-block}
gives $2\sum_{k=k_0}^{n-j-1}\varphi(k)+\varphi(n-j)\le2\int_{k_0}^{n-j}\varphi+\varphi(k_0)$. Beyond the hole,
\cref{lem:E-em}(a) with $K=n+j+1$ gives $\varphi(n+j+1)+2\sum_{k\ge n+j+2}\varphi(k)\le2I_x(n+j+1)$. The remaining site terms
are $\varphi(n-j)+2\sum_{k=n-j+1}^{n+j}\varphi(k)+\varphi(n+j+1)$, the sites of $\EcH^+_{n,j}$. The three integrals cover
$[\frac52,\infty)$ except $[k_0-\half,k_0]$ and $[n-j,n+j+1]$. With \eqref{eq:E-comb} and \eqref{eq:E-continuum},
\[
  \log\abs{B^0}\le-c'L+c'\cP+\underbrace{\varphi(2)-2\!\int_2^{5/2}\!\varphi}_{\Jend}
  +\underbrace{\varphi(k_0)-2\!\int_{k_0-1/2}^{k_0}\!\varphi}_{\le0}+\,\EcH^+_{n,j}+2e_L ,
\]
and the second brace is non-positive because $\varphi$ decreases on $[k_0-\half,k_0]$. This is \eqref{eq:E-upper}.

\emph{Lower bound.} By \eqref{eq:E-detune-pair}, $\log q_k\ge2\varphi(k)$ for the pairs outside the hole. For these,
$2\sum_{k=3}^{k_0-1}\varphi(k)\ge2\int_{5/2}^{k_0-1/2}\varphi-2e_L$; the midpoint inequality of \eqref{eq:E-block} on the concave
block gives $2\sum_{k=k_0}^{n-j}\varphi(k)\ge2\int_{k_0-1/2}^{n-j+1/2}\varphi$, where $\chi_{n-j+1/2}<x$; and \cref{lem:E-em}(b) with
$K=n+j+1$ gives $2\sum_{k\ge n+j+1}\varphi(k)\ge2I_x(n+j+\half)$, where $x<\chi_{n+j+1/2}$. The integrals cover $[\frac52,\infty)$
except $[n-j+\half,n+j+\half]$, and the pairs $n-j<k\le n+j$ are kept. Hence
$\log\abs B\ge-c'L+c'\cP+\Jend+\EcH^-_{n,j}-2e_L$.
\end{proof}

\begin{lemma}[Limits of the hole functions; \proved]\label{lem:E-hole-limits}
Let $j\ge1$, $x$ in cell $n\ge2j+4$ and $u=(x-\chi_n)/\ell_n$. Then
\begin{equation}\label{eq:E-displacement}
  \max\Bigl\{\bigabs{\EcH^+_{n,j}(x)-h^+_j(u)},\ \bigabs{\EcH^-_{n,j}(x)-h^-_j(u;\eta_0)}\Bigr\}\le C_j\Bigl(\frac1n+\ell_n\Bigr),
  \qquad C_j:=C\,j^2\log(2j+2),
\end{equation}
with an absolute constant $C$. More explicitly, the left side of \eqref{eq:E-displacement} is at most
\begin{equation}\label{eq:E-displacement-explicit}
  \frac{8j(j+1)+28j(j+1)\log(2j+2)}n+C\,j\,\ell_n,\qquad\text{which for $j=1$ is at most}\quad\frac{94}n+C\ell_n .
\end{equation}
The same bounds with $\frac3n$ added, so $\frac{97}n+C\ell_n$ for $j=1$, hold for
$\EcH^-_{n,j}|_{\eta_0=0}-\log(4\sin^2\pi\nu(x))$ against $h^-_j(u;0)-\log(4\sin^2\pi u)$, and for $\EcH^+_{n,j}$ likewise; here
$\nu(x)=c'N(x)+\frac14-2$ is the site index of \cref{sec:7.5}, so that $\nu(\chi_a)=a-2$.
\end{lemma}

\begin{proof}
There are three steps: the scales cancel up to $O(j\ell_n)$; the displacement of the sites and of the continuum from an
equally spaced lattice costs $O(j^2\log(2j+2)/n)$; and $\sin^2\pi\nu$ is compared with $\sin^2\pi u$.

\emph{Scales.} For $a$ in the hole write
$f_x(\chi_a)=\log\frac{\abs{x-\chi_a}}{\ell_n}+\bigl[\log\tfrac{\pi\ell_n}2+\log\tanh\pi x\bigr]+g_1(a)+g_2(a)$ with
$g_1=\log(\abs{\tanh z}/\abs z)$, $z=\frac\pi2(x-\chi_a)$, and $g_2=\log\tanh\frac{\pi(x+\chi_a)}2-\log\tanh\pi x$. The bracket
does not depend on $a$; it cancels between the sites and the integral, which carry equal weights. The function $g_2$ is
monotone with total variation at most $\frac\pi2(\chi_{n+j+1}-\chi_{n-j})\le Cj\ell_n$ over the hole, because
$(\log\tanh z)'=2/\sinh2z\le1/\sqrt2$ for $z\ge u_1$ and $\ell_{n-j}\le2\ell_n$ by \eqref{eq:E-spacing}. The function $g_1$ is
monotone on either side of $x$, with variation at most $Cj\ell_n$ there, and $\abs{g_1}\le C\ell_n^2$ on the cells adjoining
$x$, because $\abs{(\log(\tanh z/z))'}\le C\min\{\abs z,1\}$. By \cref{lem:E-em}(d), applied on either side of these cells,
the contribution of $g_1$ and $g_2$ to $\EcH^\pm_{n,j}$ is at most $Cj\ell_n$. For a pair of the hole, the second form of
$q_k$ given after \eqref{eq:E-pair}, with $v=\frac\pi2\varepsilon_k\le\pi\eta_0\ell_n$, gives
$\log q_k=\log\frac{(x-\chi_k)^2+\varepsilon_k^2}{\ell_n^2}+2[\,\cdot\,]+2g_1(k)+2g_2(k)+O(\ell_n^2)$. Thus, up to
$O(j\ell_n)$, $\EcH^\pm_{n,j}$ are given by \eqref{eq:E-hplus}, \eqref{eq:E-hminus} with $u-i$ replaced by $u-p(i)$, $u-v$ by
$u-p(v)$, and $\eta_0$ by $\eta_0\ell_{n+i}/\ell_n$, where $p(v):=(\chi_{n+v}-\chi_n)/\ell_n$ for $-j\le v\le j+1$.

\emph{Displacement: the map $p$.} The function $p$ is increasing and concave with $p(0)=0$ and $p(1)=1$. By
\eqref{eq:E-sitemap}, $-p''/p'=\chi_a'(\log\rhoU)'(\chi_a)$ at $a=n+v$. Since $\rhoU^2/N$ is non-increasing
(\cref{lem:E-concave}(b)), $(\log\rhoU)'\le\rhoU/(4N)$, and with $\chi_a'=2/(c'\rhoU)$ and $c'N(\chi_a)=a-\frac14$,
\begin{equation}\label{eq:E-kappa}
  0\le-\frac{p''(v)}{p'(v)}\le\frac1{2(a-\frac14)}\le\frac1n\qquad(-j\le v\le j+1),
\end{equation}
because $a\ge n-j\ge\frac n2+2$. There is a point $v_0\in(0,1)$ with $p'(v_0)=1$. Hence
$\abs{\log p'(v)}\le\abs{v-v_0}/n\le(j+1)/n\le\half$, and since $\abs{e^t-1}\le2\abs t$ for $\abs t\le\half$,
\begin{equation}\label{eq:E-pprime}
  \abs{p'(v)-1}\le\frac{2(j+1)}n,\qquad\Bigabs{\frac1{p'(v)}-1}\le\frac{2(j+1)}n,\qquad p'(v)\ge e^{-1/2},\qquad
  \abs{p''(v)}\le\frac2n .
\end{equation}
The remainder formula for linear interpolation, $p(v)-v=\half v(v-1)\,p''(\xi_v)$ with $\xi_v$ between $0$, $1$ and $v$,
and $p''\le0$ give
\begin{equation}\label{eq:E-pv}
  0\le v-p(v)\le\frac{v(v-1)}n\le\frac{j(j+1)}n\quad(v\notin(0,1)),\qquad
  0\le p(v)-v\le\frac{v(1-v)}n\quad(0\le v\le1).
\end{equation}

\emph{Displacement: the sites.} For $i=0,1$ we have $p(i)=i$, and nothing changes. For $-j\le i\le-1$,
$u-p(i)\ge u-i\ge\abs i$, so $0\le\log(u-p(i))-\log(u-i)\le(i-p(i))/(u-i)\le(\abs i+1)/n$ by \eqref{eq:E-pv}. For
$2\le i\le j+1$, $p(i)-u\ge p(i)-p(1)\ge e^{-1/2}(i-1)$, so
$0\le\log(i-u)-\log(p(i)-u)\le(i-p(i))/(p(i)-u)\le e^{1/2}i/n$. In either case the logarithm of a site factor changes by
at most $2(j+1)/n$. For the pairs we use that, for positive numbers, $(A+b)/(A_0+b_0)$ lies between $A/A_0$ and $b/b_0$.
With $A=(u-p(i))^2$, $A_0=(u-i)^2$, $b=\eta_0^2(\ell_{n+i}/\ell_n)^2$, $b_0=\eta_0^2$, and
$\ell_{n+i}/\ell_n=p(i+1)-p(i)\in[e^{-(j+1)/n},e^{(j+1)/n}]$, the logarithm of a pair factor changes by at most $4(j+1)/n$; for
$i=0,1$, where $A=A_0$, the same holds, and for $\eta_0=0$ a pair factor is a squared site factor. The sites $i=0,1$ do
not move; the remaining weights in \eqref{eq:E-hplus} add up to $4j-2$, and there are $2j$ pairs in
\eqref{eq:E-hminus}, so the sums over the sites change by at most $8j(j+1)/n$.

\emph{Displacement: the continuum.} Let $(\alpha,\beta)=(-j,j+1)$ or $(-j+\half,j+\half)$. The substitution $s=p(v)$ gives
$\int_\alpha^\beta\log\abs{u-p(v)}\,\dd v=\int_{p(\alpha)}^{p(\beta)}\log\abs{u-s}\,\frac{\dd s}{p'(p^{-1}(s))}$. By
\eqref{eq:E-pv} and $j/n\le\half$, $[p(\alpha),p(\beta)]\subset[-2j,j+1]$, on which
$\int\abs{\log\abs{u-s}}\,\dd s\le2\int_0^{2j+1}\abs{\log t}\,\dd t\le6j\log(2j+2)$. By \eqref{eq:E-pprime}, replacing the density
$1/p'$ by $1$ changes the integral by at most $12j(j+1)\log(2j+2)/n$. It remains to move the endpoints from
$p(\alpha),p(\beta)$ to $\alpha,\beta$. Each moves by at most $j(j+1)/n$, and on the interval in between
$\half\le\abs{u-s}\le2j+1$ at the left end, and $\half e^{-1/2}\le p(\beta)-1\le\abs{u-s}\le j+1$ at the right end, so that
$\abs{\log\abs{u-s}}\le\log(2j+2)$. Hence
$\bigabs{\int_\alpha^\beta\log\abs{u-p(v)}\,\dd v-\int_\alpha^\beta\log\abs{u-v}\,\dd v}\le14j(j+1)\log(2j+2)/n$. The integral
enters $h^\pm_j$ with the factor $2$, so the terms in $\frac1n$ add up to $[8j(j+1)+28j(j+1)\log(2j+2)]/n$, which for
$j=1$ is $(16+56\log4)/n\le94/n$. This proves~\eqref{eq:E-displacement} and \eqref{eq:E-displacement-explicit}.

\emph{The comb factor.} Let $x=\chi_{n+u'}$ with $0\le u'\le1$. Then $\nu(x)=n-2+u'$, so $\sin^2\pi\nu=\sin^2\pi u'$, and
$u=p(u')$, so $0\le d:=u-u'\le u'(1-u')/n$ by \eqref{eq:E-pv}. We have
$\sin\pi u/\sin\pi u'=\cos\pi d+\cot(\pi u')\sin\pi d$. Since $\abs{\cot\pi t}\le1/(\pi t(1-t))$ for $0<t<1$ (by symmetry let $t\le\half$; then $\tan\pi t\ge\pi t\ge\pi t(1-t)$), the second
term is at most $1/n$ in modulus, and $0\le1-\cos\pi d\le\half(\pi d)^2\le1/n^2$. With $n\ge6$ this gives
$\bigabs{\log\sin^2\pi\nu-\log\sin^2\pi u}\le3/n$, uniformly on the cell, its ends included.
\end{proof}

\noindent\tagpar{\orientation} The constant $94$ is far from sharp: to first order in $1/n$ the difference
$\EcH^\pm_{n,j}-h^\pm_j$ has the size $2j^2/(3(n-1))$, which is $0.13$ for $j=1$ and $n=6$. This is not used.

\paragraph{The extrema of row 6.}
At $\eta_0=0$ the differences $h^\pm_j(u)-\log(4\sin^2\pi u)$ are continuous on $[0,1]$, since the double zeros of the
sites $i=0,1$ are those of $\sin^2\pi u$. Row~6 of \cref{tab:4} contains their extrema, which are attained at $u\in\{0,1\}$:
\begin{equation}\label{eq:E-row6}
\begin{aligned}
  j=1:&\quad \max_u\bigl[h^+_1-\log(4\sin^2\pi u)\bigr]\le0.2449,\quad \min_u\bigl[h^-_1(\cdot\,;0)-\log(4\sin^2\pi u)\bigr]\ge-0.1991\ \cert,\\
  j=4:&\quad \max_u\bigl[h^+_4-\log(4\sin^2\pi u)\bigr]\le0.0749,\quad \min_u\bigl[h^-_4(\cdot\,;0)-\log(4\sin^2\pi u)\bigr]\ge-0.0422\ \cert,
\end{aligned}
\end{equation}
so that $[e^{-0.1991},e^{0.2449}]\subset[0.819,1.278]$, and $e^{-0.0422}(1-0.01)\ge0.949$, $e^{0.0749}(1+0.01)\le1.089$.

\paragraph{Proof of \cref{prop:7.4} \tagpar{\provedCA}.}
In this appendix $\eta_0=\frac14$ is fixed. The two lemmas used here, \cref{lem:E-two-sided,lem:E-hole-limits}, hold
with the same proofs and the same constants for every $0\le\eta_0\le\frac14$: their proofs use $\eta_0$ only through
\eqref{eq:E-detune-pair} and through $\varepsilon_k=\eta_0\ell_k\le\frac14\ell_k$, and the value $\eta_0=0$ is treated
explicitly in the last clause of \cref{lem:E-hole-limits} and in the step on the sites of its proof. We apply the two
lemmas with $\eta_0=0$. Then $q_k=T_{\chi_k}^2$ and $B=B^0$, and by definition
$\sech^2(\frac{\pi x}2)\abs{B^0}=\abs{h_0^0}=\Env\sin^2\pi\nu$. With \eqref{eq:E-level}, $\lambda=\Env/\Lev$ satisfies
$4\lambda\sin^2\pi\nu=\abs{B^0}e^{c'L-c'\cP-\Jend}$; note that $\Jend$ is inside $\Lev$ and drops out. \Cref{lem:E-two-sided}
at $\eta_0=0$ gives, on every cell $n\ge2j+4$,
\begin{equation}\label{eq:E-lambda}
  \EcH^-_{n,j}\big|_{\eta_0=0}-2e_L\ \le\ \log\bigl(4\lambda\sin^2\pi\nu\bigr)\ \le\ \EcH^+_{n,j}+2e_L .
\end{equation}
We subtract $\log(4\sin^2\pi\nu)$ and apply \cref{lem:E-hole-limits} and \eqref{eq:E-row6}:
\begin{equation}\label{eq:E-lambda-bracket}
  \log\lambda\in\Bigl[-0.1991-\delta_{n,1},\ 0.2449+\delta_{n,1}\Bigr]\ (j=1),\qquad
  \log\lambda\in\Bigl[-0.0422-\delta_{n,4},\ 0.0749+\delta_{n,4}\Bigr]\ (j=4),
\end{equation}
with $\delta_{n,j}:=2e_L+C_j(\frac1n+\ell_n)$ and $C_j$ as in \eqref{eq:E-displacement}. For $j=1$ and $n\ge6$ this is the
first statement of \cref{prop:7.4}: $\lambda\in[0.819,1.278]$, widened at both ends by $e^{\pm2\CL}$ and by the
site-displacement factor $e^{\pm C_1(\frac1n+\ell_n)}$, which tends to $1$ as $n\to\infty$ and in which
$C_1(\frac1n+\ell_n)$ may be replaced by $\frac{97}n+C\ell_n$ (the $94$ of \eqref{eq:E-displacement-explicit} and the $3$ of the comb factor); these two factors are kept outside $\lambda$. For $j=4$ let $n\ge\epsilon c'$ with $\epsilon>0$ fixed. Then $x-\xs\ge r(\epsilon)>0$, so
$e_L=O_\epsilon(c'^{-2/3})$ by \eqref{eq:E-eL-decay}, $\frac1n\le\frac1{\epsilon c'}$ and $\ell_n\le C_\epsilon/c'$. Hence
$\delta_{n,4}\to0$. For $c$ large depending on $\epsilon$ we have $e^{\delta_{n,4}}\le1.01$, which is the allowance
$(1\pm0.01)$ of row~6, and therefore $\lambda\in[0.949,1.089]$. \qed

\subsection{The detuned floor}
\label{app:E.6}

We prove \cref{lem:7.9}, the floor \eqref{eq:E-floor}: on every cell $n\ge2j+4$ the comb $h_0$ stays above a positive
multiple of the level. The floor is the smallest value, over the position $u$ in the cell, of the exact contribution of
the $2j$ detuned pairs nearest to $x$ divided by that of the continuum which they replace; all other pairs are bounded
below by their undetuned values, and these by the continuum through the midpoint rule. Keeping more pairs exactly
removes a larger part of the midpoint loss, which is why $\mu_4>\mu_1$. The rows used are 3 and 5, and row 9 through
$\Eythr$.

\begin{lemma}[The detuned floor; \provedCA]\label{lem:E-floor}
Let $j\in\{1,4\}$ and let $x$ lie in cell $n\ge2j+4$. Then
\begin{equation}\label{eq:E-floor}
  \frac{h_0(x)}{\Lev(x)}\ \ge\ \tfrac14\,e^{\EcH^-_{n,j}(x)-2e_L(x)}\ \ge\ \mu_j\Bigl(1-C_j\bigl(\tfrac1n+\ell_n\bigr)\Bigr)e^{-2\CL},
  \qquad \mu_j:=\min_{0\le u\le1}\tfrac14e^{h^-_j(u;\eta_0)} ,
\end{equation}
with $C_j$ as in \eqref{eq:E-displacement}, and for $\eta_0=\frac14$ (\cref{tab:4}, row 3)
\begin{equation}\label{eq:E-mu}
  \mu_1\ge0.536\ \cert,\qquad\mu_4\ge0.699\ \cert,
\end{equation}
the minima being attained at $u\in\{0,1\}$. The factor
$1-C_j(\frac1n+\ell_n)$ tends to $1$ as $n\to\infty$; where it is not positive the second inequality says nothing, and
the first one remains.
\end{lemma}

\begin{proof}
For $x>\chi_2$ we have $B(x)>0$ and $h_0/\Lev=\frac14\abs Be^{c'L-c'\cP-\Jend}$ by \eqref{eq:E-level}; the end correction
$\Jend$ is inside $\Lev$. Since $n\ge2j+4$, \eqref{eq:E-lower} holds:
$\log\abs B+c'L-c'\cP-\Jend\ge\EcH^-_{n,j}-2e_L$. In it the $2j$ pairs $n-j<k\le n+j$ enter through their exact moduli
$q_k$, and every other pair through $q_k\ge T_{\chi_k}^2$ [\hyp{T2}, \cref{lem:E-detune}]: detuning is used only in so far
as it does not lower a factor. The sites $3\le k<k_0$ cost $2e_L$. This is the first inequality of \eqref{eq:E-floor}.
For the second, $e_L(x)\le\CL$ by \cref{lem:E-farleft}(d) (row 5), as $n\ge6$;
$\EcH^-_{n,j}(x)\ge h^-_j(u;\eta_0)-C_j(\frac1n+\ell_n)$ by \cref{lem:E-hole-limits}, and $e^{-t}\ge1-t$; and
$\frac14e^{h^-_j(u;\eta_0)}\ge\mu_j$ by definition. The values \eqref{eq:E-mu} and the location of the minima
are row~3.
\end{proof}

\noindent\tagpar{\orientation} We note without proof that $\mu_j\to\sinh^2\pi\eta_0=0.7546$ as $j\to\infty$; this is not used. The reason is
that $\prod_{i\in\ZZ}((u-i)^2+\eta_0^2)$, divided by the continuum in the sense of \eqref{eq:E-hminus}, tends to
$4\abs{\sin\pi(u+i\eta_0)}^2=4(\sin^2\pi u+\sinh^2\pi\eta_0)$ as $j\to\infty$, whose minimum over $u$ is
$4\sinh^2\pi\eta_0$.

\subsection{The first cells}
\label{app:E.7}

On the cells $2,\dots,5$ the point $x$ is within a few Airy lengths of the tip, there are at most three sites to its
left, and nothing is gained by replacing them by a continuum. We therefore keep \emph{all} the sites up to $\chi_{n+4}$
exactly and apply the midpoint rule only beyond them. No end term and no loss $e_L$ arise, which is why the unit is
$\Lev\,e^{-\Jend}$. In the Airy variable the resulting finite product is a $c$-independent function on each cell, and its
minimum is certified (\cref{tab:4}, row 4; row 10 for the passage to the Airy variable).

For $x$ in cell $n\ge2$ let
\begin{equation}\label{eq:E-HI}
  \EcH^{\rm I}_n(x):=f_x(\chi_2)+\sum_{k=3}^{n+4}\log q_k(x)-2\int_2^{n+9/2}\varphi(a)\,\dd a ,
\end{equation}
in which the sites and the integral both carry the weight $2n+5$.

\begin{lemma}[First cells: \cref{lem:7.12}; \provedCA]\label{lem:E-first-cells}
\begin{enumerate}[label=\textup{(\alph*)}]
\item If $x$ lies in cell $n\ge2$, then $\log\abs B+c'L\ge c'\cP+\EcH^{\rm I}_n$ at $x$, and hence
$h_0\ge\tfrac14\Lev\,e^{-\Jend+\EcH^{\rm I}_n}$ there.
\item Let $\hat\eta_k:=\eta_0(\Ehvs_{k+1}-\Ehvs_k)$ and
\begin{equation}\label{eq:E-HI-scaled}
  \hat\EcH^{\rm I}_n(\vsig):=\log\abs{\vsig-\Ehvs_2}+\sum_{k=3}^{n+4}\log\bigl((\vsig-\Ehvs_k)^2+\hat\eta_k^2\bigr)
  -3\bigl[\EcF(\vsig,\Ehvs_{n+9/2})-\EcF(\vsig,\Ehvs_2)\bigr].
\end{equation}
For $2\le n\le5$ and $x=\xs+\lc\vsig\ne\chi_2$ in cell $n$,
$\EcH^{\rm I}_n(x)-\log\abs{\vsig-\vsig_2}=\hat\EcH^{\rm I}_n(\vsig)-\log\abs{\vsig-\Ehvs_2}+O(\lc\log\frac1\lc)$, uniformly in
$x$ and $n$; at $\vsig=\Ehvs_2$ the right side is understood as its continuous extension.
\item \textup{(\Cref{tab:4}, row 4.)} With $\muI_n:=\min\frac14e^{\hat\EcH^{\rm I}_n(\vsig)}$ over $\Ehvs_n\le\vsig\le\Ehvs_{n+1}$,
\begin{equation}\label{eq:E-muI}
  \muI_3\ge0.679,\qquad\muI_4\ge0.683,\qquad\muI_5\ge0.686\qquad\cert,
\end{equation}
and on cell $2$, which adjoins the simple zero,
\begin{equation}\label{eq:E-slope}
  \tfrac14e^{\hat\EcH^{\rm I}_2(\vsig)}\ \ge\ \muslope\,\frac{\vsig-\Ehvs_2}{\Ehvs_3-\Ehvs_2}\quad\text{for all }\Ehvs_2\le\vsig\le\Ehvs_3,
  \qquad\muslope=0.686\ \cert .
\end{equation}
The inequality \eqref{eq:E-slope} holds on the whole cell and binds at its right end: the quotient of its left side by
$(\vsig-\Ehvs_2)/(\Ehvs_3-\Ehvs_2)$ is $1.550$ at the zero, rises to $3.774$ at the relative position $0.350$ in the cell,
and decreases from there to its least value $0.6924$, which it takes at $\vsig=\Ehvs_3$.
\item Consequently, for $c\ge\catt$, on the cells $3,4,5$,
$h_0\ge\Lev\,e^{-\Jend}\muI_n\bigl(1-C\lc\log\frac1\lc\bigr)$, and on the whole of cell $2$,
$h_0(x)\ge\Lev\,e^{-\Jend}\muslope\,\frac{x-\chi_2}{\chi_3-\chi_2}\bigl(1-C\lc\log\frac1\lc\bigr)$. On these cells
$\Lev\,e^{-\Jend}e^{c'L}\asymp\lc^{-2n_1}\asymp c'^{\frac43n_1}\gg\sqrt{c'}$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) By \eqref{eq:E-detune-pair} and \cref{lem:E-em}(b) with $K=n+5$, for which $x<\chi_{n+1}<\chi_{n+9/2}$,
$\sum_{k\ge n+5}\log q_k\ge2\sum_{k\ge n+5}\varphi(k)\ge2I_x(n+\frac92)$, and
$2I_x(n+\frac92)=2I_x(2)-2\int_2^{n+9/2}\varphi=-c'L+c'\cP-2\int_2^{n+9/2}\varphi$ by \eqref{eq:E-continuum}. Inserting this
into \eqref{eq:E-comb} gives the first statement, and the second follows from \eqref{eq:E-level}.

(b) We take $\Evmax=\Ehvs_{10}$ in \cref{lem:E-window,lem:E-d1}; the site $\chi_2$ is given by \eqref{eq:E-window} and
the pairs $3\le k\le n+4\le9$ by \eqref{eq:E-window-pair}. The constants $\log\lc+\log\frac\pi{2\sqrt2}$ cancel because
the weights are equal, and the remainders $R$, $R'_k$ contribute $O(\lc)$. By \eqref{eq:E-continuum} the term $-2\int_2^{n+9/2}\varphi$ is
$c'\cP_{\chi_{n+9/2}}-c'\cP_{\chi_2}$, to which \eqref{eq:E-potential-limit} applies with $a=n+\frac92\le\frac{19}2$ and with
$a=2$. For the pairs, $(\vsig-\vsig_k)^2+\eta_0^2\Delta_k^2\ge\eta_0^2\Delta_k^2$ is bounded below, so its logarithm is
Lipschitz in the sites, which move by $O(\lc)$ by \eqref{eq:E-sites}. The singular factor of the simple site is kept
exact on both sides. All these remainders are bounded uniformly in $\vsig\in[0,\Ehvs_{10}]$, and hence in $x$ and $n$.

(c) is the content of row~4; the certificate evaluates $\EcF$ by its closed form (\cref{lem:E-d1}).

(d) Let $\hat R_n:=\hat\EcH^{\rm I}_n-\log\abs{\vsig-\Ehvs_2}$, which is \eqref{eq:E-HI-scaled} without its first term; it is
continuous on $[0,\Ehvs_{10}]$. It has the modulus of continuity $C\delta\log\frac1\delta$ there. Indeed, with
$\Lambda(r):=r\log\abs r-r$, a primitive of $\log\abs r$, integration by parts gives
$\EcF(\vsig,A)=-A^{1/2}\Lambda(\vsig-A)+\half\int_0^A\Lambda(\vsig-t)\,t^{-1/2}\,\dd t$, and
$\abs{\Lambda(a)-\Lambda(b)}\le C\delta\log\frac1\delta$ for $\abs{a-b}\le\delta\le\frac14$ and $\abs a,\abs b\le\Ehvs_{10}$, by
the estimate of $\int_I\abs{\log\abs{\vsig-\vsig'}}\dd\vsig'$ in the proof of \cref{lem:E-d1}(b). Hence
$\abs{\EcF(\vsig,A)-\EcF(\vsig',A)}\le2A^{1/2}\,C\delta\log\frac1\delta$ for $\vsig,\vsig',A$ in $[0,\Ehvs_{10}]$ with
$\abs{\vsig-\vsig'}\le\delta$. The pair terms of \eqref{eq:E-HI-scaled} are Lipschitz in $\vsig$, because
$(\vsig-\Ehvs_k)^2+\hat\eta_k^2\ge\hat\eta_k^2>0$.

In the variable $\vsig$ cell $n$ is $[\vsig_n,\vsig_{n+1}]$. By \eqref{eq:E-sites}, $\vsig_{n+1}\le\Ehvs_{n+1}$ and
$0\le\Ehvs_n-\vsig_n\le d_n:=0.14\,\lc\Ehvs_n^2$. So the true cell is the nominal cell $[\Ehvs_n,\Ehvs_{n+1}]$, on which (c)
holds, shortened on the right, and extended on the left by a stretch of length at most $d_n$. For $\vsig$ in the true cell
let $\vsig^+:=\max\{\vsig,\Ehvs_n\}$. It lies in the nominal cell, and $0\le\vsig^+-\vsig\le d_n$. By (b) and the modulus
of continuity, for $x\ne\chi_2$,
\[
  \tfrac14e^{\EcH^{\rm I}_n(x)}=\tfrac14(\vsig-\vsig_2)\,e^{\hat R_n(\vsig)+O(\lc\log\frac1\lc)}
  \ \ge\ \tfrac14(\vsig-\vsig_2)\,e^{\hat R_n(\vsig^+)}\bigl(1-C\lc\log\tfrac1\lc\bigr).
\]
On the cells $3,4,5$ we have $\vsig-\vsig_2\ge\vsig-\Ehvs_2\ge(\vsig^+-\Ehvs_2)\bigl(1-d_n/(\Ehvs_3-\Ehvs_2)\bigr)$, because
$\vsig_2\le\Ehvs_2$ and $\vsig^+-\Ehvs_2\ge\Ehvs_3-\Ehvs_2$. Since
$\tfrac14(\vsig^+-\Ehvs_2)\,e^{\hat R_n(\vsig^+)}=\tfrac14e^{\hat\EcH^{\rm I}_n(\vsig^+)}\ge\muI_n$ by (c), this and (a) give the first
statement. On cell $2$, \eqref{eq:E-slope} says that $\tfrac14e^{\hat R_2}\ge\muslope/(\Ehvs_3-\Ehvs_2)$ on
$(\Ehvs_2,\Ehvs_3]$, and this holds at $\Ehvs_2$ by continuity. Hence
$\frac14e^{\EcH^{\rm I}_2(x)}\ge\muslope\frac{\vsig-\vsig_2}{\Ehvs_3-\Ehvs_2}(1-C\lc\log\frac1\lc)$ for $\chi_2<x\le\chi_3$, and
$\frac{\vsig-\vsig_2}{\Ehvs_3-\Ehvs_2}=\frac{x-\chi_2}{\chi_3-\chi_2}\cdot\frac{\vsig_3-\vsig_2}{\Ehvs_3-\Ehvs_2}$ with
$\vsig_3-\vsig_2\ge\Ehvs_3-\Ehvs_2-d_3$. At $x=\chi_2$ both sides of the second statement vanish. The size of
$\Lev\,e^{-\Jend}e^{c'L}$ is \cref{lem:E-d1}(c).
\end{proof}

\section{The vacuum inequality}
\label{app:F}

\tagpar{\provedCA} This appendix proves \cref{prop:7.13}: $\Band\le\Rvac\abs{\Lobe}$ with $\Rvac\le0.1021+0.02$, and
$\abs{\Lobe}\to\infty$; on the way it proves \cref{lem:7.5}(d2). It is used in \cref{sec:7.9} only, and it takes
\cref{lem:A.2,lem:A.4}, \cref{prop:7.2}, \cref{lem:7.3}(b) (whose hypothesis is verified in \cref{sec:7.10}),
\cref{app:E.1,app:E.2,app:E.3,app:E.4}, \cref{lem:E-two-sided}, \cref{lem:E-hole-limits} and \cref{lem:E-Jend} as input.

The idea is described in \cref{sec:7.9}. The Cardy-weighted masses of the negative lobe of $h_0$ below $\tstar$ and
of the positive band beyond both come from a window of width $O(\lc)$ at the tip, and in the variable
$\vsig=(x-\xs)/\lc$ each is a common factor, proportional to $c'^{\frac43n_1-\frac16}=c'^{13/6}$, times a $c$-independent
integral. The ratio of the two integrals is certified to be at most $0.1021$ when the band is cut after its first
hundred cells; the finite-$c$ corrections and the band beyond these cells are bounded by hand, with explicit constants
wherever a constant enters $\Rvac$.

\subsection{Statement and reduction}
\label{app:F.1}

Throughout, $(m,k_\infty,n_1,\eta_0)=(1,\frac14,\frac74,\frac14)$, $h_0(t)=\sech^2(\pi b/2)\,B(b)$ with $b=\sqrt{3t/c'}$,
and on the real axis $t=c'x^2/3$, $\dd t=\frac23c'x\,\dd x$. By \cref{prop:7.2}, $h_0<0$ on $(0,\tstar)$, $h_0$ has a simple
zero at $\tstar=c'\chi_2^2/3$, and $h_0\ge0$ beyond; $V>0$ by \cref{lem:A.2}. We set
\begin{equation}\label{eq:F-lobe-band}
  \Lobe:=\int_0^{\tstar}\!h_0V\,\dd t<0,\quad
  \Band:=\int_{\tstar}^\infty\! h_0V\,\dd t=\sum_{n\ge2}\Band_n,\quad
  \Band_n:=\int_{c'\chi_n^2/3}^{c'\chi_{n+1}^2/3}\!h_0V\,\dd t\ge0 ,
\end{equation}
so that $\Band_n$ is the contribution of cell $n$, $\chi_n<x<\chi_{n+1}$. With $\rhoC=2\pi V$,
\cref{lem:7.3}(b) and the bound $\abs{h_{0,\rm vac}}\le8$ of \cref{prop:7.2} give
\begin{equation}\label{eq:F-reduction}
  \gvac=\half h_{0,\rm vac}+\pi\bigl(\abs{\Lobe}-\Band\bigr)\ \ge\ \pi\bigl(\abs{\Lobe}-\Band\bigr)-4 .
\end{equation}
\Cref{prop:7.13} follows from \eqref{eq:F-reduction} once $\Band\le\Rvac\abs\Lobe$ with $\Rvac<1$ and
$\abs\Lobe\to\infty$ are shown; the transform $\hat h_0$ enters only through the hypothesis of \cref{lem:7.3}(b).

\paragraph{Input from \cref{app:E}.}
We use the notation of \cref{app:E}: the site weight $f_x(y)=\log\abs{T_y(x)}\le0$ \eqref{eq:E-fx}; the site map
$a\mapsto\chi_a$ for real $a\ge\frac14$, $c'N(\chi_a)=a-\frac14$ \eqref{eq:E-sitemap}, whose integer values $a\ge2$ are
the sites; the modulus $q_k$ of the detuned pair at $\chi_k$ \eqref{eq:E-pair}, for which $q_k\ge T_{\chi_k}^2$ on the
real axis \eqref{eq:E-detune-pair}; and, for $Y\ge\xs$, the potential at $x$ of the charge which the law places below $Y$
\eqref{eq:E-continuum-def},
\begin{equation}\label{eq:F-PY}
  \cP_Y(x):=-\int_{\xs}^{Y}\rhoU(x')\,f_x(x')\,\dd x'\ \ge0 ,
\end{equation}
so that $\cP=\cP_{\chi_2}$. Two bounds of \cref{app:E.4} are used, in both of which the sites next to $x$ are kept
exactly. On the part $\xs\le x<\chi_2$ of the lobe, the first inequality of \eqref{eq:E-lobe}, which is
\cref{lem:7.5}(f), reads
\begin{equation}\label{eq:F-lobe-lower}
  \log\abs{B(x)}+c'L(x)\ \ge\ f_x(\chi_2)+c'\cP_{\chi_{5/2}}(x) .
\end{equation}
On cell $n\ge2$, \eqref{eq:E-ubgap}, which is behind \cref{lem:7.5}(e), reads
\begin{equation}\label{eq:F-cell-upper}
  \log\abs{B(x)}+c'L(x)\ \le\ f_x(\chi_2)+\sum_{k=3}^{n+1}\log q_k(x)+f_x(\chi_{n+2})+c'\cP_{\chi_{n+2}}(x)+D_n(x) ,
\end{equation}
where $D_n:=\sum_{k\ge n+2}\log(q_k/T_{\chi_k}^2)$ is the detuning of the pairs beyond the cell. By \cref{lem:E-detune},
$0\le D_n\le4\zeta(2)\eta_0^2$; we keep $D_n$ as it stands.

\subsection{The Airy window}
\label{app:F.2}

Let $\vsig:=(x-\xs)/\lc$ with $\lc=(\ktip c')^{-2/3}$, so that $\ktip c'\lc^{3/2}=1$. As in \cref{lem:E-tip}(c) we write, for
real $a>\frac14$ and for integer $k$,
\begin{equation}\label{eq:F-positions}
  \vsig_a:=\frac{\chi_a-\xs}{\lc},\qquad \Fhvs_a:=\bigl(a-\tfrac14\bigr)^{2/3},\qquad
  \Delta_k:=\vsig_{k+1}-\vsig_k,\qquad \hat\Delta_k:=\Fhvs_{k+1}-\Fhvs_k
\end{equation}
for the true and the \emph{nominal} scaled positions and spacings; thus $\varepsilon_k=\eta_0\lc\Delta_k$ and
$\Fhvs_2=n_1^{2/3}$. We fix an integer $N_1\ge6$, put $\Fvmax:=\Fhvs_{N_1+2}$, and take $c$ so large that
$\lc\Fvmax\le0.1$. The \emph{window} is $\xs\le x\le\xs+\lc\Fvmax$, and $\sigma$ denotes the scaled position of a point
$x'=\xs+\lc\sigma$ of the window.

\begin{lemma}[Window dictionary; \provedCA]\label{lem:F-dictionary}
With these conventions, and with $\Few:=4.3\,\lc\Fvmax$:
\begin{enumerate}[label=\textup{(\roman*)}]
\item \emph{Sites.} $\Fhvs_a\bigl(1-0.14\,\lc\Fhvs_a\bigr)\le\vsig_a\le\Fhvs_a$ for $\frac14<a\le N_1+2$.
\item \emph{Law.} $c'\rhoU(x')\,\dd x'=3\sigma^{1/2}w_c(\sigma)\,\dd\sigma$ with $1\le w_c(\sigma)\le1+0.33\,\lc\sigma$.
\item \emph{Factors.} For $x,x'$ in the window, $\abs{T_{x'}(x)}=\frac{\pi\lc}{2\sqrt2}\abs{\vsig-\sigma}\,e^{R}$ with
  $\abs R\le\Few$; for a pair $3\le k\le N_1+1$,
  $q_k(x)=\bigl(\frac{\pi\lc}{2\sqrt2}\bigr)^2\bigl[(\vsig-\vsig_k)^2+\eta_0^2\Delta_k^2\bigr]e^{R'}$ with
  $\abs{R'}\le2\Few$.
\item \emph{Continuum potential.} For $\frac14<a\le N_1+2$ and $x$ in the window,
\begin{equation}\label{eq:F-potential}
  c'\cP_{\chi_a}(x)=2\bigl(a-\tfrac14\bigr)\log\frac{2\sqrt2}{\pi\lc}
  -3\int_0^{\vsig_a}\sigma^{1/2}w_c(\sigma)\log\abs{\vsig-\sigma}\,\dd\sigma-\mathcal R_a(x),
  \quad\abs{\mathcal R_a}\le2\bigl(a-\tfrac14\bigr)\Few .
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
These are statements of \cref{app:E.1}, applied with the present $\Fvmax$: (i) is \eqref{eq:E-sites}; (ii) is
\cref{lem:E-d1}(a), with the weight $w_c$ defined there; (iii) is \eqref{eq:E-window} and \eqref{eq:E-window-pair}; and
(iv) is \eqref{eq:E-potential}. Their hypotheses hold because $\lc\Fvmax\le0.1$, because $\Fhvs_a\le\Fvmax$ for
$a\le N_1+2$, and because a pair $k\le N_1+1$ has $\vsig_{k+1}\le\Fhvs_{N_1+2}=\Fvmax$ by (i). The numerical constants rest
on row~10 of \cref{tab:4}, through \cref{lem:E-tip}(a) and \cref{lem:E-window}.
\end{proof}

\paragraph{The scaled comb.}
For $a>\frac14$ let $\FQ_a(\vsig):=-3\int_0^{\Fhvs_a}\sigma^{1/2}\log\abs{\vsig-\sigma}\,\dd\sigma$ be the potential of the Airy
density $3\sigma^{1/2}$ carrying the charge $2(a-\frac14)$. It is elementary (the proof is in \cref{lem:E-d1}): with
$A=\Fhvs_a$ and $\vsig\ge0$, $\vsig\ne A$, and by continuity at $\vsig=A$,
\begin{equation}\label{eq:F-Q}
  \FQ_a(\vsig)=-2A^{3/2}\log\abs{A-\vsig}+\tfrac43A^{3/2}+4\vsig A^{1/2}
  +2\vsig^{3/2}\log\Bigabs{\frac{\sqrt A-\sqrt\vsig}{\sqrt A+\sqrt\vsig}} .
\end{equation}
The \emph{scaled comb} consists of the functions, on $0\le\vsig\le\Fhvs_2$ and on $\Fhvs_n\le\vsig\le\Fhvs_{n+1}$ for
$n\ge2$ respectively,
\begin{align}
  \FsL(\vsig)&:=\abs{\vsig-\Fhvs_2}\;e^{\FQ_{5/2}(\vsig)},\label{eq:F-scaled-lobe}\\
  \FsB_n(\vsig)&:=\abs{\vsig-\Fhvs_2}\prod_{k=3}^{n+1}\bigl[(\vsig-\Fhvs_k)^2+\eta_0^2\hat\Delta_k^2\bigr]\;
  \abs{\vsig-\Fhvs_{n+2}}\prod_{k\ge n+2}\Bigl[1+\frac{\eta_0^2\hat\Delta_k^2}{(\Fhvs_k-\vsig)^2}\Bigr]\;
  e^{\FQ_{n+2}(\vsig)} .\label{eq:F-scaled-band}
\end{align}
They do not depend on $c$. The factors of $\FsB_n$ are, in this order: the simple zero $\chi_2$; the pairs kept exactly;
the half-weight site at which the trapezoid rule starts; the detuning of the pairs beyond it; the continuum. We denote
by $\FsL^{(c)}$ and $\FsB^{(c)}_n$ the same expressions at finite $c$: with the true $\vsig_k,\vsig_a,\Delta_k$ in place of
the nominal ones, with the weight $w_c$ inserted in the integral defining $\FQ_a$, and with $e^{D_n(x)}$ in place of the
infinite product.

\begin{lemma}[Lobe and band in the window: \cref{lem:7.5}(d2); \provedCA]\label{lem:F-window}
Let $N_1\ge6$, and let
\begin{equation}\label{eq:F-common}
  \FKc:=\FVc\,\Bigl(\frac{2\sqrt2}{\pi\lc}\Bigr)^{2n_1},\qquad
  \FVc:=\bigl(1-e^{-4\pi\xs}\bigr)^2\,\bigl(2\pi c'L(\xs)\bigr)^{-1/2}\;\frac{2c'\xs}3\,\lc\;\sech^2\frac{\pi\xs}2 ,
\end{equation}
so that $\FKc$ is a $c$-independent constant times $c'^{\frac43n_1-\frac16}$. There are constants $C_{\rm w}$ and
$C_{\rm cont}$, depending on $N_1$ only, such that the following holds with $\FRbar:=C_{\rm w}\lc+3\CV/c'$, whenever
$\lc\Fvmax\le0.1$ and $\FRbar\le1$. The integrands of \eqref{eq:F-lobe-band} obey
\begin{equation}\label{eq:F-d2}
\begin{aligned}
  \abs{h_0}\,V\,\dd t&\ \ge\ \FKc\,e^{-\FRbar}\,\FsL^{(c)}(\vsig)\,\dd\vsig &&(0\le\vsig<\vsig_2),\\
  h_0\,V\,\dd t&\ \le\ \FKc\,e^{\FRbar}\,\FsB^{(c)}_n(\vsig)\,\dd\vsig &&(\vsig_n<\vsig<\vsig_{n+1},\ 2\le n\le N_1),
\end{aligned}
\end{equation}
and
\begin{equation}\label{eq:F-cont}
  \Bigabs{\int_0^{\vsig_2}\FsL^{(c)}\dd\vsig-\int_0^{\Fhvs_2}\FsL\,\dd\vsig}
  +\sum_{n=2}^{N_1}\Bigabs{\int_{\vsig_n}^{\vsig_{n+1}}\FsB^{(c)}_n\dd\vsig-\int_{\Fhvs_n}^{\Fhvs_{n+1}}\FsB_n\,\dd\vsig}
  \ \le\ C_{\rm cont}\,\lc .
\end{equation}
\end{lemma}

\begin{proof}
\emph{Measure.} By \cref{lem:A.4}, $V(t)\,e^{-c'L(x)}=(2\pi c'L(x))^{-1/2}\bigl[(1-e^{-4\pi x})^2+\epsV\bigr]$ in the window,
with $\abs{\epsV}\le\CV/c'$ and $(1-e^{-4\pi x})^2\ge(1-e^{-4\pi\xs})^2=0.94199\ldots\ge0.9$. Each of the ratios
$L(x)/L(\xs)=x/\xs=1+\lc\vsig/\xs$, $(1-e^{-4\pi x})^2/(1-e^{-4\pi\xs})^2$ and
$\sech^2\frac{\pi x}2/\sech^2\frac{\pi\xs}2$ lies between $e^{-C\lc\Fvmax}$ and $e^{C\lc\Fvmax}$, and
$\dd t=\frac23c'\xs\lc\,(1+\lc\vsig/\xs)\,\dd\vsig$. Hence, with an absolute constant $C_{\rm m}$, and since $\CV/c'\le\frac13$,
\begin{equation}\label{eq:F-measure}
  V(t)\,e^{-c'L(x)}\,\sech^2\frac{\pi x}2\;\dd t=\FVc\,\bigl(1+R_V(\vsig)\bigr)\,\dd\vsig,\qquad
  \abs{R_V}\le C_{\rm m}\lc\Fvmax+\frac{\CV}{0.9\,c'} .
\end{equation}

\emph{Powers of $\lc$.} By \cref{lem:F-dictionary}(i) all sites and endpoints which occur lie in the window. We insert
(iii) and (iv) into \eqref{eq:F-lobe-lower}. There is one site factor and the charge is $2(\frac52-\frac14)=\frac92$, so
the power of $2\sqrt2/\pi\lc$ is $\frac92-1=2n_1$:
$\abs{B(x)}\,e^{c'L(x)}\ge(2\sqrt2/\pi\lc)^{2n_1}\,\FsL^{(c)}(\vsig)\,e^{-\frac{11}2\Few}$. In \eqref{eq:F-cell-upper} there
are $1+2(n-1)+1=2n$ site factors and the charge is $2(n+2-\frac14)=2n+\frac72$, so the power is again $2n_1$ for every
$n$: $\abs{B(x)}\,e^{c'L(x)}\le(2\sqrt2/\pi\lc)^{2n_1}\,\FsB^{(c)}_n(\vsig)\,e^{(4n+\frac72)\Few}$. The number of site factors
minus the continuum charge removed is always $-2n_1$, the absent charge; this is why the lobe and all cells carry
the same factor. Multiplying by \eqref{eq:F-measure} gives \eqref{eq:F-d2} with
$C_{\rm w}:=\bigl[18(N_1+2)+2C_{\rm m}\bigr]\Fvmax$, since $\abs{h_0}V\,\dd t=\abs B e^{c'L}\cdot Ve^{-c'L}\sech^2\frac{\pi x}2\,\dd t$,
$4.3\,(4N_1+\frac72)\le18(N_1+2)$, and $e^{-2r}\le1-r$, $1+r\le e^r$ for $0\le r\le\half$; here $r=\abs{R_V}$, and
$2\abs{R_V}\le2C_{\rm m}\lc\Fvmax+3\CV/c'\le\FRbar\le1$.

\emph{Continuity.} On the lobe and on each cell $n\le N_1$ the functions $\FsL^{(c)}$, $\FsB^{(c)}_n$ are obtained
from $\FsL$, $\FsB_n$ by moving the positions by at most $0.14\,\lc\Fvmax^2$, by (i); by changing the weight by at most
$0.33\,\lc\Fvmax$, by (ii); and by replacing the infinite product by $e^{D_n}$. Every factor is bounded, and it is Lipschitz
in the positions uniformly over the cell, because the points at which a factor is singular, namely $\vsig_k$ for
$k\ge n+2$ and the endpoint $\vsig_{n+2}$ of the continuum, or $\vsig_{5/2}$ for the lobe, stay at least half a nominal
cell away from $\vsig$. The integral in \eqref{eq:F-potential} changes by at most
$C\lc\bigl(1+\bigabs{\log\abs{\vsig-\Fhvs_a}}\bigr)$, where the logarithm is bounded for the same reason. For
$D_n$ we use the second form of $q_k$ given after \eqref{eq:E-pair}. With $u=\frac\pi2(\chi_k-x)=\frac\pi2\lc(\vsig_k-\vsig)$ and
$v=\frac\pi2\varepsilon_k=\frac\pi2\eta_0\lc\Delta_k$ it gives, for every $k\ge n+2$,
$q_k/T_{\chi_k}^2=\bigl(1+\sin^2v/\sinh^2u\bigr)\,e^{\theta_k}$ with $0\le\theta_k\le5\,v^2/\cosh^2u$. For the pairs
with $\chi_k-\xs\le0.2$, \eqref{eq:E-tip} and \eqref{eq:E-sites} still hold. The spacings are compared through the
derivative of the site map, since the error $0.14\,\lc\Fhvs_k^2$ of the positions is not small against $\hat\Delta_k$ for
large $k$: by \eqref{eq:E-sitemap}, \eqref{eq:E-tip} and \eqref{eq:E-sites},
$\frac23\lc\Fhvs_a^{-1/2}(1-0.33\,\lc\Fhvs_a)\le\chi_a'=2/(c'\rhoU(\chi_a))\le\frac23\lc\vsig_a^{-1/2}$, and integration over
$[k,k+1]$ gives $\hat\Delta_k(1-0.33\,\lc\Fhvs_{k+1})\le\Delta_k\le\hat\Delta_k(1-0.14\,\lc\Fhvs_{k+1})^{-1/2}$. For the
distances, $\Fhvs_k\le\frac32(N_1+2)(\Fhvs_k-\vsig)$ for $k\ge n+2$ and $\vsig\le\Fhvs_{n+1}$, so that
$\vsig_k-\vsig=(\Fhvs_k-\vsig)\bigl(1+O(N_1\lc\Fhvs_k)\bigr)$ by \eqref{eq:E-sites}. The relative correction is at most
$0.06$ in modulus: for $\Fhvs_k\le2\Fvmax$ because $\FRbar\le1$ implies $18(N_1+2)\lc\Fvmax\le1$, and for
$\Fhvs_k\ge2\Fvmax$ because then $\Fhvs_k\le2(\Fhvs_k-\vsig)$ and $\lc\Fhvs_k\le0.21$. Hence
$\sin^2v/\sinh^2u=\bigl(\eta_0^2\hat\Delta_k^2/(\Fhvs_k-\vsig)^2\bigr)\bigl(1+O(\lc\Fhvs_k)\bigr)$, with a constant that
depends on $N_1$; these terms are
$O(k^{-2})$ and $\Fhvs_k\le k^{2/3}$, so the errors sum to $O(\lc)$, and $\sum_k\theta_k=O(\lc^{3/2})$. The pairs with
$\chi_k-\xs\ge0.2$ are at distance at least $0.1$ from $x$ and have $\varepsilon_k=O(1/c')$; since
$\sum_k\varepsilon_k/\sinh^2\frac\pi2(\chi_k-x)$ is bounded, they contribute $O(1/c')$ to $D_n$. The corresponding
factors of the product in \eqref{eq:F-scaled-band} contribute $O(\sum_{k\gtrsim\lc^{-3/2}}k^{-2})=O(\lc^{3/2})$. Finally
the endpoints of the $\vsig$-integrals move by $O(\lc)$, and the integrands are bounded. To combine: on the common part
of the true and the nominal cell the two integrands differ by at most $C\lc$; the two cells differ by intervals of total
length at most $0.28\,\lc\Fvmax^2$, on which the integrands are bounded; and there are $N_1$ terms. This gives
\eqref{eq:F-cont} with a constant $C_{\rm cont}$ that depends on $N_1$ only.
\end{proof}

\subsection{The scaled vacuum ratio}
\label{app:F.3}

\begin{definition}[Scaled vacuum ratio]\label{def:F-ratio}
With the scaled comb \eqref{eq:F-scaled-lobe}, \eqref{eq:F-scaled-band},
\begin{equation}\label{eq:F-ratio-def}
  \FIL:=\int_0^{\Fhvs_2}\FsL\,\dd\vsig,\qquad I_n:=\int_{\Fhvs_n}^{\Fhvs_{n+1}}\FsB_n\,\dd\vsig,\qquad
  \FhR(N_1):=\frac1{\FIL}\sum_{n=2}^{N_1}I_n .
\end{equation}
\end{definition}

The ratio depends on $(m,k_\infty,\eta_0)$, which fix the scaled comb, and on the number $N_1$ of cells, but not on
$c$. We drop the part $0\le x<\xs$ of the lobe, on which $\abs{h_0}V\ge0$. Integrating
\eqref{eq:F-d2} and using \eqref{eq:F-cont}, the factor $\FKc$ cancels in the ratio:
\begin{equation}\label{eq:F-window-ratio}
  \abs\Lobe\ \ge\ \FKc\,e^{-\FRbar}\bigl(\FIL-C_{\rm cont}\lc\bigr),\qquad
  \frac1{\abs\Lobe}\sum_{n=2}^{N_1}\Band_n\ \le\ \frac{\sum_{n=2}^{N_1}I_n+C_{\rm cont}\lc}{\FIL-C_{\rm cont}\lc}\;e^{2\FRbar} .
\end{equation}
As $c\to\infty$ the first right-hand side tends to infinity and the second to $\FhR(N_1)$.

\begin{lemma}[\provedCA]\label{lem:F-ratio}
For $(m,k_\infty,\eta_0)=(1,\frac14,\frac14)$ and $N_1=100$,
\begin{equation}\label{eq:F-cert}
  \FhR(100)\ \le\ 0.1021\quad\cert,\qquad\qquad \FIL\ \ge\ 94.47\quad\cert .
\end{equation}
For $N_1=30$ the certified bound is $\FhR(30)\le0.1020$; the cells $31\le n\le100$ add less than $10^{-4}$.
\end{lemma}

\begin{proof}
This is \cref{tab:4}, row~11. The integrals of \eqref{eq:F-ratio-def} are enclosed in interval arithmetic, over a
covering of the lobe and of each cell by subintervals and in the mean-value form of \cref{app:G.1}; the potentials
$\FQ_a$ are evaluated through \eqref{eq:F-Q}. Of the product over the pairs $k\ge n+2$ the first $3000$ factors are
evaluated, up to $K:=n+3001$; the remaining factors exceed one, and the sum of their logarithms, which decrease like
$k^{-2}$, is at most an explicit $U_n\le1.5\times10^{-5}$. Indeed, for $k>K$ and $\vsig\le\Fhvs_{n+1}$ the logarithm is at most
$\eta_0^2\hat\Delta_k^2/(\Fhvs_k-\vsig)^2$, and $\hat\Delta_k\le\frac23(k-\frac14)^{-1/3}$ because $a\mapsto\Fhvs_a$ is
concave. So it is at most $\phi(k-\frac14)$ with $\phi(u):=\frac49\eta_0^2\,u^{-2/3}(u^{2/3}-\vsig)^{-2}$, which decreases in
$u$. The substitution $w=u^{2/3}$ and $w^{-1/2}\le\Fhvs_K^{-1/2}$ give
$\sum_{k>K}\phi(k-\frac14)\le\int_{K-1/4}^\infty\phi\,\dd u\le\frac23\eta_0^2\,\Fhvs_K^{-1/2}(\Fhvs_K-\vsig)^{-1}$, which is
largest at $\vsig=\Fhvs_{n+1}$; this value is $U_n$. For $2\le n\le100$ it is largest at $n=100$, where it is
$1.4961\times10^{-5}$. The bound for $\FhR$ is the upper end of the enclosure of $\sum_nI_n$ divided by the
lower end of that of $\FIL$.
\end{proof}

\subsection{The far cells and the far band}
\label{app:F.4}

Beyond cell $N_1$ no site is kept exactly, and the upper bound of \cref{lem:7.5}(e) takes over; its constants are
explicit, because they enter $\Rvac$. Let $r_0:=0.3$.

\begin{lemma}[\provedCA]\label{lem:F-tails}
Let $N_1\ge6$. For $c$ large enough,
\begin{align}
  \int_{\chi_{N_1+1}\le x\le\xs+r_0}h_0V\,\dd t&\ \le\ \FKc\,C_{\rm tail}\,
  \frac{\bigl(N_1-\frac14\bigr)^{\frac23(1-2n_1)}}{2n_1-1} ,\label{eq:F-tail}\\
  \int_{x\ge\xs+r_0}h_0V\,\dd t&\ \le\ C_{\rm far}\,\sqrt{c'} ,\label{eq:F-far}
\end{align}
where $C_{\rm far}$ does not depend on $c$, and $C_{\rm tail}=C_{\rm tail}(N_1)$ is the number \eqref{eq:F-Ctail}, which
decreases as $N_1$ grows; $C_{\rm tail}(100)\le282$.
\end{lemma}

\begin{proof}
The plan is to bound $\abs B e^{c'L}$ by a constant times $e^{c'\cP}$, to bound $\cP$ by the potential of a single
point charge $n_1$ at $\chi_2$, and to integrate against the Cardy measure. The points $x\ge\chi_{N_1+1}$ lie in the cells
$n\ge N_1+1\ge6$. There \eqref{eq:E-ubcell-B}, which is \cref{lem:7.5}(e) and rests on rows~5, 7 and~9 of \cref{tab:4} (and on
row~10 through \eqref{eq:E-sites} in \cref{lem:E-farleft}), holds, and $\Jend\le\CJ$ (\cref{lem:E-Jend}). Its factor $e^{94/n+C\ell_n}$ has $\frac{94}n\le\frac{94}{N_1+1}$ and
$\ell_n\le\lc$, so it is at most $e^{94/N_1}$ as soon as $C\lc\le\frac{94}{N_1}-\frac{94}{N_1+1}$; this is one of the
conditions on the size of $c$. Hence $\abs{B(x)}\,e^{c'L(x)}\le C_{\rm up}\,e^{c'\cP(x)}$ with
$C_{\rm up}:=5.0\,e^{94/N_1}\,(1+4\eta_0^2)^2\,e^{\Cdet\eta_0^2+\CJ+2\CL}\le42.75\,e^{94/N_1}$,
because $\eta_0=\frac14$, $\Cdet\le16.6$, $\CJ\le0.65$ and $\CL\le0.006$ (\cref{app:E}) give
$5.0\cdot1.5625\cdot e^{1.0375+0.65+0.012}\le42.75$. Thus $C_{\rm up}\le110$ for $N_1=100$. Next,
$c'\cP(x)\le2n_1\abs{\log\abs{T_{\chi_2}(x)}}$ by \eqref{eq:E-Pbounds}, which is \cref{lem:7.5}(b).

For \eqref{eq:F-tail} let $\chi_{N_1+1}\le x\le\xs+r_0$, and write $\abs{T_{\chi_2}(x)}=\tanh z\,\tanh z_+$ with
$z=\frac\pi2(x-\chi_2)=\frac\pi2\lc(\vsig-\vsig_2)\in(0,\frac\pi2r_0]$ and $z_+=\frac\pi2(x+\chi_2)\ge u_1$. The first
inequality of \cref{lem:E-window}, which holds for all real $z$, gives $\tanh z\ge z\,e^{-z^2/3}\ge z\,e^{-\pi^2r_0^2/12}$, and
$\tanh z_+\ge\tanh u_1=1/\sqrt2$. For $\vsig\ge\vsig_{N_1+1}$ we have
$\vsig-\vsig_2\ge\vsig\,(1-\vsig_2/\vsig_{N_1+1})\ge\vsig\,(1-\Fhvs_2/\Fhvs_{N_1})$, because $\vsig_2\le\Fhvs_2$ by
\eqref{eq:E-sites}, and because $\vsig_{N_1+1}\ge\Fhvs_{N_1}$ for $c$ large, again by \eqref{eq:E-sites}. Hence
\begin{equation*}
  e^{c'\cP(x)}\ \le\ e^{n_1\pi^2r_0^2/6}\,\Bigl(1-\frac{\Fhvs_2}{\Fhvs_{N_1}}\Bigr)^{-2n_1}
  \Bigl(\frac{2\sqrt2}{\pi\lc}\Bigr)^{2n_1}\vsig^{-2n_1} .
\end{equation*}
For the measure we use \cref{lem:A.4} on $\xs\le x\le\xs+r_0$, with $(1-e^{-4\pi x})^2+\epsV\le1+\CV/c'$. In the quotient of
$Ve^{-c'L}\sech^2\frac{\pi x}2\,\dd t$ by $\FVc\,\dd\vsig$ the factors $(L(\xs)/L(x))^{1/2}$ and $x/\xs$, the latter from
$\dd t$, combine to $(x/\xs)^{1/2}\le(1+r_0/\xs)^{1/2}$, and $\sech^2\frac{\pi x}2\le\sech^2\frac{\pi\xs}2$. Therefore
$Ve^{-c'L}\sech^2\frac{\pi x}2\,\dd t\le(1+\CV/c')\,(1-e^{-4\pi\xs})^{-2}(1+r_0/\xs)^{1/2}\,\FVc\,\dd\vsig$, and we take $c$
so large that $\CV/c'\le0.01$. Multiplying the three bounds, the left-hand side of \eqref{eq:F-tail} is at most
$\FKc\,C_{\rm tail}\int_{\vsig_{N_1+1}}^\infty\vsig^{-2n_1}\,\dd\vsig\le\FKc\,C_{\rm tail}\,\Fhvs_{N_1}^{1-2n_1}/(2n_1-1)$, with
$\Fhvs_{N_1}=(N_1-\frac14)^{2/3}$ and
\begin{equation}\label{eq:F-Ctail}
  C_{\rm tail}(N_1):=1.01\;C_{\rm up}\;e^{n_1\pi^2r_0^2/6}\Bigl(1-\frac{\Fhvs_2}{\Fhvs_{N_1}}\Bigr)^{-2n_1}
  \bigl(1-e^{-4\pi\xs}\bigr)^{-2}\Bigl(1+\frac{r_0}{\xs}\Bigr)^{1/2} .
\end{equation}
For $N_1=100$ the five factors after $1.01$ are at most $110$, $e^{0.2591}\le1.296$, $(1-1.4522/21.508)^{-7/2}\le1.278$,
$1.062$ and $2.0694^{1/2}\le1.439$, and $1.062\cdot1.439\le1.53$; so
$C_{\rm tail}(100)\le1.01\cdot110\cdot1.296\cdot1.278\cdot1.53\le282$. For $N_1=30$ the same computation gives about $3500$,
because $e^{94/30}\approx23$.

For \eqref{eq:F-far} let $c$ be so large that $\chi_2\le\xs+0.05$ and that the cells meeting $x\ge\xs+r_0$ have
$n\ge6$. Since $\abs{\log\abs{T_{x'}(x)}}$ increases with $x'<x$ and decreases with $x>x'$,
$c'\cP(x)\le2n_1\abs{\log\abs{T_{\xs+0.05}(\xs+r_0)}}\le2.2\,n_1$ for $x\ge\xs+r_0$. This makes the far-band
statement of \cref{lem:7.5}(e) explicit: $\abs B e^{c'L}\le C_{\rm up}e^{2.2n_1}$ there. By \cref{lem:A.4},
$V\le(1+\CV/c')\,\as(c'L)$ for $x\ge\xs$. Therefore the left-hand side of \eqref{eq:F-far} is at most
$C_{\rm up}\,e^{2.2n_1}(1+\CV/c')\int_{\xs+r_0}^\infty(2\pi c'L(x))^{-1/2}\sech^2\frac{\pi x}2\,\frac{2c'x}3\,\dd x$, and this
integral is $\sqrt{c'}$ times a convergent integral which does not depend on $c$.
\end{proof}

\subsection{Why the Bessel asymptote must be two-sided; assembly}
\label{app:F.5}

The lobe must be bounded from below and the band from above, and both carry the weight $V$; this is why the asymptote
of \cref{lem:A.4} has to be two-sided (\cref{sec:7.9}). With the one-sided bound $V\le K_a\le e^{c'L}$ for the band
($V\le K_a$ by the monotonicity of $K_\alpha$ in $\alpha$, and \cref{lem:A.2}(e)), and the true size
$V\simeq(1-e^{-4\pi x})^2\,\as(c'L)$ for the lobe, the two measures would differ at the tip by the factor
$e^{c'L}/\as(c'L)=\sqrt{2\pi c'L(\xs)}=1.92\sqrt{c'}$, about $15$ at $c=64$ and about $61$ at $c=1024$, and the bound for
$\Band/\abs\Lobe$ would exceed one. With \cref{lem:A.4} the same measure $\FVc\,\dd\vsig$ appears in both integrals, up to
the factor $1+R_V$ of \eqref{eq:F-measure}, and cancels in the ratio, as do the power $\lc^{-2n_1}$ of
\cref{lem:7.5}(d1) and the outer factor $\sech^2$.

\paragraph{Proof of \cref{prop:7.13}.}
We take $N_1=100$ and split the band into the cells $n\le N_1$, the cells beyond them up to $\xs+r_0$, and the rest.
By \eqref{eq:F-window-ratio}, \eqref{eq:F-tail} and \eqref{eq:F-far},
\begin{equation}\label{eq:F-assembly}
  \frac{\Band}{\abs\Lobe}\ \le\ \frac{\sum_{n=2}^{N_1}I_n+C_{\rm cont}\lc}{\FIL-C_{\rm cont}\lc}\,e^{2\FRbar}
  +\frac{C_{\rm tail}\,e^{\FRbar}}{\FIL-C_{\rm cont}\lc}\,\frac{\bigl(N_1-\frac14\bigr)^{\frac23(1-2n_1)}}{2n_1-1}
  +\frac{C_{\rm far}\,e^{\FRbar}\sqrt{c'}}{\FKc\,(\FIL-C_{\rm cont}\lc)} .
\end{equation}
As $c\to\infty$ the first term tends to $\FhR(100)\le0.1021$, by \cref{lem:F-ratio}, and the third is
$O\bigl(c'^{\frac12-\frac43n_1+\frac16}\bigr)=O(c'^{-5/3})$. The second term tends to a number that does not depend on $c$,
\begin{equation*}
  \delta_{\rm tail}:=\frac{C_{\rm tail}}{\FIL}\,\frac{\bigl(N_1-\frac14\bigr)^{-5/3}}{2.5}
  \ \le\ \frac{282}{94.47}\cdot1.9\times10^{-4}\ \le\ 6\times10^{-4} ,
\end{equation*}
by \eqref{eq:F-Ctail}, by the lower bound for $\FIL$ in \eqref{eq:F-cert}, and because $99.75^{-5/3}/2.5\le1.9\times10^{-4}$;
any $C_{\rm tail}\le5067$ would give $\delta_{\rm tail}\le0.01$. We do not take $N_1=30$, although
$\FhR(30)$ is no larger: there $C_{\rm tail}\approx3500$ and $29.75^{-5/3}/2.5\le0.0015$ make the proved
$\delta_{\rm tail}$ about $0.05$, while the cells beyond the thirtieth in fact add only $10^{-4}$ \tagpar{\numerical}.

Let $\catt$ be so large that, for $c\ge\catt$, the conditions of \cref{lem:F-window,lem:F-tails} hold, the three terms of
\eqref{eq:F-assembly} together exceed the sum $\FhR(100)+\delta_{\rm tail}$ of their limits by at most $0.01$, and
$\abs\Lobe\ge2$ by \eqref{eq:F-window-ratio}. Then
\begin{equation}\label{eq:F-Rvac}
  \Band\le\Rvac\,\abs\Lobe,\qquad\Rvac:=\FhR(100)+\delta_{\rm tail}+0.01\ \le\ 0.1021+0.01+0.01=0.1021+0.02 ,
\end{equation}
and \eqref{eq:F-reduction} gives $\gvac\ge\pi(1-\Rvac)\abs\Lobe-4\ge2\pi\cdot0.877-4>0$. The constants $C_{\rm w}$,
$C_{\rm cont}$, $\CV$, $C_{\rm far}$ and the unnamed constants $C$ enter $\catt$ only, and $\abs\Lobe$ grows at least like
$c'^{13/6}$ by \eqref{eq:F-window-ratio}. Since only $\Rvac<1$ is needed, neither the third digit of $\FhR$ nor the
allowance $0.02$ is delicate. \qed

\section{Certified constants and numerical methods}
\label{app:G}

We say what a certified inequality is, list all of them in \cref{tab:4}, and complete the description of the numerical
methods of \cref{sec:5}.

\subsection{What ``certified'' means}
\label{app:G.1}

An inequality marked \cert{} bounds an explicit elementary function of one real variable, or a ratio of two integrals
of such functions, or a quantity that the row reduces to finitely many such functions (rows~5 and~8), on a stated range;
it does not involve $c$. The inequality is proved in interval arithmetic on a
covering of the range by finitely many subintervals. A real number is represented by an interval with floating-point
endpoints that contains it, and every operation returns an interval, rounded outward, that contains all values which
the exact result can take; rounding errors widen the result and cannot make it wrong. The function is evaluated on each
subinterval $X$ as a whole, and the largest upper endpoint obtained is a proven upper bound on the range (likewise for
lower bounds); \cref{tab:4} records it as the bound obtained, which lies slightly beyond the true extremum.
Where needed we use the mean-value form $f(X)\subset f(m)+f'(X)\,(X-m)$, with $m$ the midpoint of $X$ and $f'(X)$ an
enclosure of the derivative on $X$; in the same form, the integral of $f$ over $X$, of length $h$, lies in
$h\,f(m)+\frac18h^2[-w,w]$, with $w$ the length of $f'(X)$, and these enclosures are added (row~11). An unbounded part of
the range is bounded by hand, as the row records. This, together with the legend of \cref{sec:1.3}, is the only meaning
of ``computer-assisted'' in this paper.

A point at which the function is singular, or is a quotient of two vanishing quantities, is removed analytically first.
In row~4 the simple zero at the left end of cell~$2$ is divided out. In row~6 the terms $2\log u+2\log(1-u)$ of $h^\pm_j$
and $\log(4\sin^2\pi u)$ combine to $-2\log2\pi-2\log[\sin\pi u/(\pi u(1-u))]$, and for $0\le u\le\half$ we use
$\log[\sin\pi u/(\pi u)]=-\sum_{k\ge1}\zeta(2k)\,u^{2k}/k$, truncated at $k=60$ with the tail bounded by
$\zeta(122)\,u^{122}/(61(1-u^2))$; for $u\ge\half$ we use the symmetry $u\to1-u$. Rows~9 and~10 reach the tip. With
$u_1=\arcsinh1$ let $t:=\sinh(u_1+\delta)-1=2\sinh^2\frac\delta2+\sqrt2\sinh\delta$, $q:=t/\delta$, $q_1:=(q-\sqrt2)/(\sqrt2\,\delta)$
and $b:=(q/\sqrt2)^{1/2}$; let $\phi(v):=\arcsinh(\sqrt v)/\sqrt v$ and $\phi_1(v):=(1-\phi(v))/v$. Then
$\arccosh(1+t)=\sqrt{2t}\,\phi(t/2)$, and therefore
\begin{equation*}
  e_1(\delta)=\frac{q_1}{b+1}-\frac{b\,q}2\,\phi_1\bigl(\tfrac t2\bigr),\qquad
  r\,(\log\rhoU)'(\xs+r)=\frac{\cosh(u_1+\delta)}{\sqrt2\;q\,(2+t)^{1/2}\,\phi(t/2)}\quad(\delta=\pi r),
\end{equation*}
with $e_1$ as in row~10. Here $t$, $q$ and $q_1$ are power series in $\delta$ with positive coefficients, so they
increase, as do $b$ and $\cosh(u_1+\delta)$; $\phi$ and $\phi_1$ decrease, being the integrals over $0\le s\le1$ of $(1+vs^2)^{-1/2}$
and of $s^2g(vs^2)$, where $g(x)=(1-(1+x)^{-1/2})/x$ is the slope of a chord of a concave function. At $\delta=0$ the seven
functions take the values $0$, $\sqrt2$, $1/(2\sqrt2)$, $1$, $\sqrt2$, $1$ and $\frac16$, in this order. On a subinterval each is
enclosed by its values at the two ends, and the right sides are evaluated on these enclosures; this works on
subintervals that begin at $\delta=0$, where $e_1(0^+)=\sqrt2/24$ and the second function is $\half$.

We use the Arb library~\cite{Arb} through its Python interface python-flint. The
precision is $80$, $100$, $128$ or $200$ bits, as the row says. Each computation has been checked independently, line by line,
against the formulas as they are printed in this paper. Rows 3, 6 and 8 have also been certified a second time, by a
computation written independently (row~6 through $\log\Gamma$ and the digamma and trigamma functions).

\begin{table}[tp]
\centering\footnotesize
\setlength{\tabcolsep}{4pt}
\begin{tabular}{@{}r@{\hspace{5pt}}>{\raggedright\arraybackslash}p{0.33\linewidth}>{\raggedright\arraybackslash}p{0.445\linewidth}>{\raggedright\arraybackslash}p{0.148\linewidth}@{}}
\toprule
 & Inequality & How it is obtained & Used in\\
\midrule
1 & $\vrhoinf\le0.5029$ on $(\xs,3]$, for the phase-free ratio $\vrhoinf$ of \cref{lem:7.8} with its exact prefactor
  $\abs{b_0}^{1/2}x^{-1/2}$.
  & IA, $80$ bits, $20000$ subintervals of $[\xs+10^{-3},3]$; bound obtained $0.502833$, near $x=3$, where
  $\vrhoinf=0.50271$. By hand (proof of
  \cref{lem:7.8}): $\vrhoinf\le6\,(x-\xs)^{13/4}<10^{-8}$ on $(\xs,\xs+10^{-3}]$, and $\vrhoinf\le0.504$ for $x\ge3$.
  & \cref{lem:7.8}\\
\addlinespace
2 & $\vrhoinf\le0.41$ for $\xs<x\le1.33$; the value at the switch point $x=1.33$ is $0.394$.
  & The covering of row~1 restricted to $x\le1.33$, in the same computation; bound obtained $0.39368$. By hand: the tip, as
  in row~1.
  & \cref{lem:7.8}; criterion of \cref{sec:7.7}\\
\addlinespace
3 & $\mu_1\ge0.536$ and $\mu_4\ge0.699$ at $\eta_0=\frac14$, where $\mu_j=\min_{0\le u\le1}\frac14e^{h^-_j(u;\eta_0)}$
  \eqref{eq:E-floor}.
  & IA, $128$ bits, $16000$ subintervals of $[0,1]$, mean-value form; $h^-_j$ is explicit \eqref{eq:E-hminus}. Its
  derivative has a fixed sign on $[0,\frac14]$ and on $[\frac34,1]$, and in between $h^-_j$ exceeds its end values. So the
  minimum is attained at $u\in\{0,1\}$: $\mu_1=17e^4/1728=0.53713$ and $\mu_4=0.70097$. By hand: the factor
  $1-C_j(\frac1n+\ell_n)$ of \eqref{eq:E-floor} on cell $n$.
  & \cref{lem:7.9}; \labelcref{app:E.6}\\
\addlinespace
4 & First cells, in the Airy variable: $\muI_n\ge0.679$, $0.683$, $0.686$ for $n=3,4,5$ \eqref{eq:E-muI}; on the whole of
  cell $2$ the slope form \eqref{eq:E-slope} with $\muslope=0.686$.
  & IA, $100$ bits, $4000$ subintervals per cell, mean-value form, for \eqref{eq:E-HI-scaled}, which does not involve $c$
  and is in closed form (\cref{lem:E-d1}). Bounds obtained for $\muI_n$: $0.6978$, $0.7027$, $0.7063$ for $n=3,4,5$. On cell~$2$ the quotient of the two sides of \eqref{eq:E-slope} is
  least at $\vsig=\Ehvs_3$, where it is $0.6924$. By hand: the passage to the Airy variable
  (\cref{lem:E-first-cells}(b),(d); row~10).
  & \cref{lem:7.12}; \labelcref{app:E.7}\\
\addlinespace
5 & $\hat e_L(\vsig)\le0.00186$ for $3\le\vsig\le\Ehvs_{602}$, for the far-left remainder with the sites at their
  nominal positions \eqref{eq:E-eLhat}. Hence $\CL\le0.006$ and $e^{-2\CL}\ge0.988$ (\cref{lem:E-farleft}(d)).
  & IA, $100$ bits, $136856$ subintervals of $[3,\Ehvs_{602}]$, $\Ehvs_{602}=71.28$; bound
  obtained $0.0018523$. By hand (proof of \cref{lem:E-farleft}(d), which also describes the covering): $\vsig\ge\Ehvs_{602}$, where \eqref{eq:E-eL-explicit} gives
  $0.00147$; the passage from $\hat e_L$ to $e_L$ at finite $c$, which costs $C\lc$.
  & \labelcref{app:E.2}; \cref{prop:7.4}, \cref{lem:7.5}(a), \cref{lem:7.9}, \cref{lem:F-tails}\\
\addlinespace
6 & At $\eta_0=0$, over $0\le u\le1$ \eqref{eq:E-row6}: $h^+_1-\log(4\sin^2\pi u)\le0.2449$ and
  $h^-_1-\log(4\sin^2\pi u)\ge-0.1991$, so $\lambda\in[0.819,1.278]$ for $j=1$. With $h^\pm_4$ the bounds are $0.0749$ and
  $-0.0422$, so $\lambda\in[0.949,1.089]$ for $j=4$, the allowance $(1\pm0.01)$ included.
  & IA, $128$ bits, $16000$ subintervals, mean-value form, after the cancellation of \cref{app:G.1}. The extrema are
  attained at $u\in\{0,1\}$, as in row~3: $6-3\log2-2\log2\pi=0.24480$ and $I(0)-2\log2\pi=-0.19900$ ($I(0)$ as in row~8) for
  $j=1$; $0.07487$ and $-0.04216$ for $j=4$. By hand: the factors $e^{\pm2e_L}$ and $e^{\pm C_j(\frac1n+\ell_n)}$ of
  \eqref{eq:E-lambda-bracket}, which lie within $1\pm0.01$ for $j=4$ on the cells $n\ge\epsilon c'$ once $c$ is large, and
  are kept outside $\lambda$ for $j=1$.
  & \cref{prop:7.4}; \labelcref{app:E.5}\\
\addlinespace
7 & $\sup_{0<u<1}h^+_\infty(u)=h^+_\infty(\half)=6-4\log3\le1.606$, so
  $\sup_{\rm cell}e^{\EcH^+_n}\le5.0\,e^{94/n+C\ell_n}$ \eqref{eq:E-suphplus}. Detuning adds $2\log(1+4\eta_0^2)$ for the two flanking pairs and
  $\Cdet\eta_0^2$ for the far pairs (\cref{app:E.3}).
  & Closed form, by hand: each of the three pieces of $h^+_\infty$ is largest at $u=\half$. No covering is needed; the
  two numbers, $6-4\log3=1.60555$ and $e^{1.606}=4.9828\ldots<5.0$, are evaluated in IA at $128$ bits. $\Cdet\le16.6$ by hand
  (\cref{lem:E-detune}).
  & \cref{lem:7.5}(e); \labelcref{app:E.4}, \labelcref{app:F.4}\\
\bottomrule
\end{tabular}
\caption{The certified inequalities, for $(m,k_\infty,n_1,\eta_0)=(1,\frac14,\frac74,\frac14)$ and the outer factor
$\sech^2$ (\cref{sec:7.2}). IA: interval arithmetic on a covering of the stated range; bound obtained: the endpoint
that the computation returns (\cref{app:G.1}). By hand: estimated without the computer. $C$ and $C_j$ are the absolute
constants of \eqref{eq:E-displacement}, and $94$ is the constant of \eqref{eq:E-displacement-explicit}. In rows 7 and 8,
$h^\pm_\infty(u;\eta_0)$ are the limits of the hole functions with $j=1$ as $n\to\infty$, at fixed position $u\in[0,1]$ in
the cell (\cref{sec:7.5}). In the last column E.2 stands for \cref{app:E.2}, and so on.}
\label{tab:4}
\end{table}

\begin{table}[tp]
\ContinuedFloat
\centering\footnotesize
\setlength{\tabcolsep}{4pt}
\begin{tabular}{@{}r@{\hspace{5pt}}>{\raggedright\arraybackslash}p{0.33\linewidth}>{\raggedright\arraybackslash}p{0.445\linewidth}>{\raggedright\arraybackslash}p{0.148\linewidth}@{}}
\toprule
 & Inequality & How it is obtained & Used in\\
\midrule
8 & For $0<\eta_0\le0.3$ the minimum of $h^-_\infty(u;\eta_0)$ over $0\le u\le1$ is attained at $u\in\{0,1\}$, where
  its value is $2\log\eta_0+\log(1+\eta_0^2)+I(0)$ with $I(0)=4-3\log3+4\log2=3.4767\ldots\ge3.476$
  \eqref{eq:E-minhminus}. Hence $e^{\EcH^-_n}\ge32\,\eta_0^2\,e^{-C(\frac1n+\ell_n)}$, which enters \eqref{eq:E-overshoot}.
  Only $\cT>0$ is drawn from this row.
  & By hand (after \eqref{eq:E-minhminus}): monotonicity in $\eta_0$ reduces the claim to $G_L>0$ on $(0,\half]$ and
  $G_R>0$ on $[\half,1)$, for two explicit functions; the end value is a closed form. IA, $128$ bits, $16000$
  subintervals of $[0,1]$, mean-value form: $G_L'>0$ on $[0,\frac14]$, $G_L>0$ on $[\frac14,\half]$, the mirror statements
  for $G_R$, and $I(0)>3.476$.
  & \cref{lem:7.11}; \labelcref{app:E.4}\\
\addlinespace
9 & $1.3\,r\,(\log\rhoU)'(\xs+r)\le0.95$ for $0<r\le0.64$, and $(\log\rhoU)'(\xs+r)\le3.10<\pi$ for $r\ge0.18$.
  & IA, $128$ bits. First bound: $4000$ subintervals of $[2^{-10},0.64]$, and $[0,2^{-10}]$ in the regularized form of
  \cref{app:G.1}, where the left side lies within $0.65\pm0.002$; bound obtained $0.9212$. Second bound: $4000$
  subintervals of $[0.18,8]$; bound obtained $3.0656$. By hand: for $r\ge8$, $(\log\rhoU)'$ decreases
  (\cref{lem:E-concave}(b)).
  & \labelcref{app:E.1} \eqref{eq:E-ythr}; \labelcref{app:E.2} (\cref{lem:E-farleft}(c))\\
\addlinespace
10 & Tip expansions, with $u_1=\arcsinh1$: (a)
  $\arccosh\sinh(u_1+\delta)=2^{3/4}\delta^{1/2}\bigl(1+\delta\,e_1(\delta)\bigr)$ with $e_1(\delta)\in[0.0589,0.105]$
  for $0<\delta\le2$; (b) $-\frac{z^2}3\le\log\frac{\abs{\tanh z}}{\abs z}\le0$ for all real $z\ne0$; (c)
  $0<(\log\tanh z)'\le1.24$ for $z\ge u_1-0.25$.
  & (a) IA, $200$ bits, $125$ subintervals of $[0,2]$, in the regularized form of \cref{app:G.1};
  $e_1(0^+)=\sqrt2/24=0.058926$, and the enclosures obtained lie in $[0.058906,0.104993]$. (b) is proved by hand
  (\cref{lem:E-window}) and carries no mark. (c) $(\log\tanh z)'=2/\sinh2z$ decreases, and its value at $z=u_1-0.25$ is
  $1.22992$; also IA, $2000$ subintervals of $[u_1-0.25,4]$.
  & \labelcref{app:E.1} (\cref{lem:E-tip,lem:E-window}), and through these \labelcref{app:E.2}, \labelcref{app:E.4},
  \labelcref{app:E.5}, \labelcref{app:E.7}, \labelcref{app:F.2}, \labelcref{app:F.4}\\
\addlinespace
11 & $\FhR(100)\le0.1021$ and $\FIL\ge94.47$ \eqref{eq:F-cert}, and $\FhR(30)\le0.1020$ (\cref{lem:F-ratio}), for the
  scaled vacuum ratio $\FhR(N_1)$ and the scaled lobe integral $\FIL$ of \eqref{eq:F-ratio-def}. With the
  finite-$c$ corrections and the cells beyond the hundredth, $\Rvac\le0.1021+0.02$ \eqref{eq:F-Rvac}; this is by hand and
  carries no mark.
  & IA, $80$ bits; the integrals are enclosed in the mean-value form of \cref{app:G.1}, on $20000$ subintervals of the
  lobe and $27400$ of the cells $2,\dots,100$. Of the product over the pairs $k\ge n+2$, $3000$ factors are explicit, and the rest lies in
  $[1,e^{U_n}]$ with the $U_n\le1.5\times10^{-5}$ of the proof of \cref{lem:F-ratio}. Obtained: $\FIL\in[94.4736,94.4737]$, $\FhR(30)\in[0.101916,0.101919]$,
  $\FhR(100)\in[0.102004,0.102007]$. By hand
  (\cref{app:F.5}): the finite-$c$ corrections, and the cells beyond the hundredth with $C_{\rm tail}(100)\le282$
  \eqref{eq:F-Ctail}.
  & \cref{prop:7.13}; \labelcref{app:F.3}, \labelcref{app:F.5}\\
\bottomrule
\end{tabular}

\smallskip
\parbox{0.97\linewidth}{\raggedright For reference, not certified (closed forms):
$\ktip=4\cdot2^{3/4}/(9\sqrt\pi)=0.4217111$, and $(U+L)(\xs-\delta)=-2.649689\,\delta^{3/2}\bigl(1+O(\delta)\bigr)$ as
$\delta\downarrow0$, with coefficient $2\pi\ktip$ (\cref{lem:6.7}).}
\caption[]{(continued)}
\end{table}

\Cref{tab:4} is a closed list: a $c$-independent constant carries the mark \cert{} in \cref{sec:7,app:E,app:F} only if it
has a row. Constants obtained by hand within a proof, and the orientation values of \cref{sec:7.3}, are not certified
and are not listed. Absent by choice are the absolute constants of the $c$-dependent remainders, such as the Hankel
remainder of \cref{app:A} and the steepest-descent constants of \cref{app:D}; they stay symbolic and enter only through
$\catt$. All rows feed \cref{thm:5}, hence the attainment half of \cref{thm:1}, and
\cref{cor:6}; no row feeds \cref{thm:2,thm:3,thm:4} or \cref{thm:7}.

\subsection{Numerical methods for the optimal functionals}
\label{app:G.2}

\tagpar{\numerical} We complete \cref{sec:5.1}; nothing here is certified. At order $P$ the
functional is the unique $p(t)e^{-2\pi t}$, with $p$ in the truncated basis of~\cite{AHT}, that has a simple zero at $t_1$
and double zeros at $t_2,\dots,t_{P/2}$. We scale the rows of the truncated crossing equations by the vacuum functional.
A Newton step is accepted or damped by a monotonicity test on successive corrections, not by the size of the residual,
which is a poor guide at large $c$.

\emph{Precision.} Arb serves here for its arbitrary precision and not for its error bounds: the radii grow through the
Laguerre recurrence until they carry no information. We work with the midpoints and establish the accuracy by solving
again at $1.5$ to $2$ times the precision. On the levels that were solved twice, at least $31$ digits agree in $t_1$, and
at least $20$ in every level and weight that we report; the highest orders, $P\ge2304$, are judged by the Newton residual
instead. The working precision is $768$--$1600$ bits on most levels, and at least one bit per unit of $P$ at the highest
orders.

\emph{Extrapolation in $P$.} For each $c$ we fit $\Delta_1^{(P)}(c)=\Delta_\infty+A\,e^{-k(P/c)^s}$, with $s$ free, to the
levels with $P/c\ge0.8$; the fits return $s=0.57$--$0.58$. The fitted $\Delta_\infty$ is our estimate $S$ of
$\Delta_1^{(\infty)}(c)$. Its uncertainty is $\sigma:=\half\max\{\abs{S-T},\abs{S-S_{1/2}}\}$, with a floor of $10^{-7}c$,
where $S_{1/2}$ is the same fit with $s=\half$ fixed, and $T$ is a fit in which the dependence on $P/c$ is taken from a
fully converged chain, $c=48$. When $P_{\max}/c\ge6$ we subtract instead from
$\Delta_1^{(P_{\max})}(c)$ a tail fitted, with $s$ free, to the top levels (at most six), and $\sigma$ is the spread of the
fits with $s=\half$, $s=1$ and $s$ free. For $P_{\max}/c\ge4$ the whole correction is at most $3\cdot10^{-6}c$. We tested the
estimator on ten chains with $64\le c\le512$ that reach $P/c\ge5$, by cutting each at a smaller $P/c$ and extrapolating again. When the
cut is at $P/c\ge2$, the error is below $10^{-5}c$ and below $\sigma$ in every case. When the cut is at $P/c=1.5$, the reach
of the chain $c=2048$, the two test chains with five levels left in the window err by at most $5\times10^{-7}c$, a tenth
of the $\sigma$ quoted for $c=2048$.

\section{Limit profiles: proofs for \texorpdfstring{\cref{sec:9}}{section 9}}
\label{app:H}

This appendix proves the statements of \cref{sec:9}: three inequalities at finite $c$, which carry all the input from
\cref{sec:6} (\cref{app:H.1}); \cref{lem:9.1}, which is a quotation (\cref{app:H.2}); \cref{lem:9.2} and \cref{prop:9.3}
(\cref{app:H.3}); and \cref{thm:7} (\cref{app:H.4}). Beyond \cref{app:H.1} only the
potential theory of subharmonic functions in the plane is used~\cite{Hormander,Ransford}.

Throughout, the notation is that of \cref{sec:9.1}, $\dd A$ is the area measure in the $b$-plane, and $U$ is the
extremal potential extended across the slits. We use these properties of $U$ (\cref{lem:6.5}, \cref{def:6.6},
\cref{lem:6.7} and \eqref{eq:B-Poisson-U}): it is continuous on $\strip$ and harmonic on $D$; $U\le0$;
$U=-L$ on the slits and $U<-L$ on $(-\xs,\xs)$; and it is subharmonic on $\strip$, its charge $\frac1{2\pi}\nabla^2U$
being the measure $\rhoU(x)\,\dd x\otimes\delta_{y=0}$ on the slits $\abs x\ge\xs$, with the density $\rhoU$ of
\eqref{eq:6-rhoU}.

\subsection{Three inequalities at finite \texorpdfstring{$c$}{c}}
\label{app:H.1}

\begin{lemma}[\proved]\label{lem:H.1}
Let $c\ge40$ and $g\in\Abar\setminus\{0\}$.
\begin{enumerate}[label=\textup{(\alph*)},leftmargin=2.2em,itemsep=1pt,topsep=3pt]
\item $\sup_{\strip}u_g\le(2\log c'+C)/c'$, with an absolute constant $C$.
\item For every compact $Q\subset D$, $u_g\le U+\CQ(1+\log c')/c'$ on $Q$, with the constant $\CQ$ of \cref{prop:6.4}.
\item If $g$ is weakly feasible and $x_1(g)\le x_{\max}$, then $x_1(g)>0$ and
$\max_{0\le x\le x_1(g)}\bigl(u_g(x)+L(x)\bigr)\ge-\bigl(\log c'+\log(32\pi^2x_{\max}^4/3)\bigr)/c'$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) is \cref{lem:B.4} on the closed strip, $\sup_{\bar\strip}\abs G\le C\,a\,c'M[g]$, after taking the logarithm and
dividing by $c'$. (b) is \cref{prop:6.4}, with $U_{\xs}=U$ by \cref{lem:6.5}. (c) Both sides are unchanged when $g$ is
multiplied by a positive constant, so let $M[g]=1$. Write $t_1=t_1(g)$; it is positive, because no nonzero functional is
weakly feasible at $T\le0$ (\cref{sec:2.4}). The half-mass condition \eqref{eq:2-halfmass} says that
$\int_0^{t_1}g_-V\,\dd t\ge\half$. By \cref{lem:A.2}(d), at $t=c'x^2/3>0$ we have
$g_-V\le\abs g\,V\le\frac{4\pi^2t}{a}\,e^{c'L(x)}\abs{G(x)}=\frac{4\pi^2t}{a}\,e^{c'(u_g(x)+L(x))}$.
Let $m:=\max_{[0,x_1(g)]}(u_g+L)$, which is a maximum because $\abs G\,e^{c'L}$ is continuous. Bounding the integrand by
its supremum on $(0,t_1]$, and using $t_1=c'x_1(g)^2/3$ and $a=c'/12$, we get
$\half\le t_1\cdot\frac{4\pi^2t_1}{a}\,e^{c'm}=\frac{16\pi^2}3\,c'\,x_1(g)^4\,e^{c'm}\le\frac{16\pi^2}3\,c'\,x_{\max}^4\,e^{c'm}$,
and taking the logarithm gives (c).
\end{proof}

\subsection{Proof of \texorpdfstring{\cref{lem:9.1}}{Lemma 9.1}}
\label{app:H.2}

We check that the profiles are uniformly bounded above, quote the compactness theorem for subharmonic functions, and
pass from potentials to charges by integrating by parts.

\emph{The theorem quoted}~\cite[Thms.~3.2.12--3.2.13]{Hormander}. Let $X\subset\RR^2$ be open and
connected, and let $u_j\not\equiv-\infty$ be subharmonic functions on $X$, bounded above on every compact subset of $X$
uniformly in $j$. Then either $u_j\to-\infty$ uniformly on every compact subset of $X$, or a subsequence
converges in $L^1_{\rm loc}(X)$. If $u_j\to u$ in $L^1_{\rm loc}(X)$, then $u$ is almost everywhere equal to a
subharmonic function, again called $u$, and for every compact $K\subset X$ and every continuous $f:K\to\RR$
\begin{equation}\label{eq:H-hartogs}
  \limsup_{j\to\infty}\,\sup_K\,(u_j-f)\le\sup_K\,(u-f) .
\end{equation}

\begin{proof}[Proof of \cref{lem:9.1}]
Each $u_j$ is subharmonic on $\strip$ and not identically $-\infty$, since $G_j$ is holomorphic and not identically zero,
and, for $j$ so large that $c_j\ge40$, $\sup_{\strip}u_j\le(2\log c'_j+C)/c'_j$ by \cref{lem:H.1}(a). So the $u_j$ are
uniformly bounded above from some $j$ on, and the theorem applies with $X=\strip$, which is connected. Either (i) holds, or
there is a subsequence with $u_j\to u$ in $L^1_{\rm loc}(\strip)$ and $u$ subharmonic; $u\not\equiv-\infty$ because
$u$ is locally integrable. For the charges write $\nfrak_j:=\nfrak_{g_j}$ and $\nfrak_u:=\frac1{2\pi}\nabla^2u$.
For a smooth $\varphi$ of compact support in $\strip$,
$\int\varphi\,\dd\nfrak_j=\frac1{2\pi}\int u_j\,\nabla^2\varphi\,\dd A\to\frac1{2\pi}\int u\,\nabla^2\varphi\,\dd A
=\int\varphi\,\dd\nfrak_u$.
Choosing $\varphi\ge0$ with $\varphi\ge1$ on a compact $K$ shows that $\sup_j\nfrak_j(K)<\infty$, and then the convergence
extends from smooth to continuous $\varphi$ of compact support by uniform approximation: $\nfrak_j\to\nfrak_u$ vaguely.
\end{proof}

We record the standard consequence that is used for \cref{thm:7}(b); it follows by testing against continuous functions
that decrease to $\mathbf 1_K$ and, for the second statement, also against ones that increase to the indicator of the
interior of $K$. For compact $K\subset\strip$,
\begin{equation}\label{eq:H-portmanteau}
  \limsup_{j\to\infty}\nfrak_j(K)\le\nfrak_u(K) ,\qquad\text{and}\qquad
  \nfrak_j(K)\to\nfrak_u(K)\quad\text{if }\nfrak_u(\partial K)=0 .
\end{equation}

\subsection{Proofs of \texorpdfstring{\cref{lem:9.2}}{Lemma 9.2} and \texorpdfstring{\cref{prop:9.3}}{Proposition 9.3}}
\label{app:H.3}

For the majorant we pass \cref{lem:H.1}(b) to the limit almost everywhere and remove the exceptional set by averaging
over small disks. For the contact point we follow the points at which \cref{lem:H.1}(c) holds and use
\eqref{eq:H-hartogs}: a limit cannot lose what the $u_j$ have at nearby points.

\begin{proof}[Proof of \cref{lem:9.2}]
Let $u_j\to u$ in $L^1_{\rm loc}(\strip)$. A subsequence converges almost everywhere, so \cref{lem:H.1}(b) gives $u\le U$
almost everywhere on every compact $Q\subset D$. Hence $u\le U$ almost everywhere on $\strip$, because the slits have
zero area. For $b\in\strip$ and small $r$ the sub-mean-value property of $u$ on the disk $B(b,r)$ gives
$u(b)\le\frac1{\pi r^2}\int_{B(b,r)}u\,\dd A\le\frac1{\pi r^2}\int_{B(b,r)}U\,\dd A\to U(b)$ as $r\to0$, since $U$ is
continuous on $\strip$. The rest follows from $U\le0$, $U=-L$ on the slits and $U<-L$ on $(-\xs,\xs)$.
\end{proof}

\begin{proof}[Proof of \cref{prop:9.3}]
By \cref{lem:H.1}(c) with $x_{\max}=\xinf+1$, for large $j$ there are points $x_j\in[0,x_1(g_j)]$ with
$u_j(x_j)+L(x_j)\ge-\epsilon_j$, where $\epsilon_j:=\bigl(\log c'_j+\log(32\pi^2(\xinf+1)^4/3)\bigr)/c'_j\to0$. Let $K:=[0,\xinf+1]$, which
contains $x_j$ for large $j$. If alternative (i) held, then $u_j(x_j)\le\sup_Ku_j\to-\infty$, in contradiction with
$u_j(x_j)\ge-L(x_j)-\epsilon_j\ge-\frac{2\pi}3(\xinf+1)-1$. So (i) does not occur. Let $u$ be a limit profile. Passing
to a further subsequence, $u_j\to u$ in $L^1_{\rm loc}(\strip)$ and $x_j\to x_0\in[0,\xinf]$. Fix $r\in(0,\half)$ and apply
\eqref{eq:H-hartogs} with $K_r:=\bar B(x_0,r)$ and $f(b):=-L(\Re b)$, which we write $-L$. Since $x_j\in K_r$ for large $j$,
$\sup_{K_r}(u+L)\ge\limsup_j\sup_{K_r}(u_j+L)\ge\limsup_j\bigl(u_j(x_j)+L(x_j)\bigr)\ge0$.
As $r\downarrow0$ the left side decreases to $u(x_0)+L(x_0)$, because $u+L$ is upper semicontinuous. With \cref{lem:9.2}, $-L(x_0)\le u(x_0)\le U(x_0)\le-L(x_0)$, so all three are equal. Finally $U(x_0)=-L(x_0)$ fails for $0\le x_0<\xs$ by \cref{lem:6.7}, so $x_0\ge\xs$.
\end{proof}

\subsection{Proof of \texorpdfstring{\cref{thm:7}}{Theorem 7}}
\label{app:H.4}

We reduce the theorem to the claim that every limit profile $u$ of the sequence is equal to $U$, and prove the claim
for the deficit $\theta:=U-u$ in three steps: a dichotomy on the slit strip $D$ (Step~1), the exclusion of $\theta>0$ by
two incompatible bounds on the means of $\theta$ over small circles about the tip (Step~2), and the points of the slits
(Step~3).

\emph{Reduction.} By \cref{prop:9.3} with $\xinf=\xs$, every subsequence has a further subsequence
that converges in $L^1_{\rm loc}(\strip)$ to a limit profile $u$ whose contact point lies in $[\xs,\xs]$, that is,
$u(\xs)=U(\xs)$. Suppose that every such $u$ is equal to $U$. Since convergence in $L^1_{\rm loc}(\strip)$ is
convergence in a metric, the whole sequence then converges to $U$, which is (a). The integration by parts in the proof
of \cref{lem:9.1}, applied to the whole sequence, then gives
$\nfrak_j\to\frac1{2\pi}\nabla^2U=\rhoU\,\dd x\otimes\delta_{y=0}$ vaguely, which is \eqref{eq:9-zerolaw}. This measure
gives no mass to a compact set disjoint from the slits, and none to the boundary of a rectangle
$[x_a,x_b]\times[-y_0,y_0]$, since it has a density on the real axis. So \eqref{eq:H-portmanteau} gives the two
counting statements of (b), after multiplication by $c'_j$. For (c), the map $b\mapsto c'_jb^2/3$ is injective and
holomorphic on $\Re b>0$, and $G_j(b)=g_j(c'_jb^2/3)$, so the zeros of $G_j$ in the rectangle and the zeros of $g_j$ in its
image correspond, with multiplicities. The image of $x+iy$ is $\frac{c'_j}3(x^2-y^2)+\frac{2i}3c'_jxy$. With $\xi=x^2/3$,
$\dd x=\frac3{2x}\,\dd\xi$ and $\rhoV=3\rhoU/4x$ we have
$\int_{x_a}^{x_b}\rhoU(x)\,\dd x=\int_{\xi_a}^{\xi_b}2\rhoV(\xi)\,\dd\xi$.

It remains to prove the claim. Let $u$ be a limit profile with $u(\xs)=U(\xs)$, and $\theta:=U-u$ on $\strip$. By
\cref{lem:9.2}, $\theta\ge0$. It is lower semicontinuous ($-\theta$ is upper semicontinuous: at a point $\theta$
can jump down, never up), with values in $[0,+\infty]$, it is not identically
$+\infty$, and $\theta(\xs)=0$. On $D$ it is superharmonic, because there $U$ is harmonic. Let $r_0:=\xs/2$, so that
the disk $B:=B(\xs,r_0)$ lies in $\strip$ and does not meet the left slit.

\emph{Step 1: dichotomy.} Since $D$ is connected, the strong minimum principle \cite[Thm.~2.3.1]{Ransford} says that a
non-negative superharmonic function on $D$ that vanishes at one point of $D$ vanishes identically. So either
$\theta\equiv0$ on $D$, or $\theta>0$ at every point of $D$.

\emph{Step 2: $\theta>0$ on $D$ is impossible.} Suppose that $\theta>0$ on $D$. Let
$\Omega_0:=B\setminus[\xs,\xs+r_0)$, a simply connected subdomain of $D$, and let
$A:=\{\xs+r_0e^{i\phi}:\frac\pi4\le\phi\le\frac{3\pi}4\}$, a compact arc in $D$. Since $\theta$ is lower semicontinuous
and positive on $A$, $\theta_A:=\min(1,\min_A\theta)$ lies in $(0,1]$. The local coordinate $w:=\sqrt{b-\xs}$, with
$\arg(b-\xs)\in(0,2\pi)$, maps $\Omega_0$ conformally onto the half-disk $H_0:=\{\abs w<R,\ \Im w>0\}$, $R:=\sqrt{r_0}$. It
extends continuously to the boundary, maps the two lips of the slit and the tip to the diameter $[-R,R]$, and maps
$A$ to $A':=\{Re^{i\phi}:\frac\pi8\le\phi\le\frac{3\pi}8\}$. For $\abs w<R$ let
\begin{equation}\label{eq:H-omega}
  \omega(w):=\frac1{2\pi}\int_{\pi/8}^{3\pi/8}\bigl(R^2-\abs w^2\bigr)
  \biggl[\frac1{\abs{Re^{i\phi}-w}^2}-\frac1{\abs{Re^{-i\phi}-w}^2}\biggr]\dd\phi ,
\end{equation}
the Poisson integral in the disk $\abs w<R$ of the boundary function that is $+1$ on $A'$, $-1$ on the mirror image of
$A'$ in the real axis, and $0$ elsewhere. It is harmonic, and odd under $w\mapsto\bar w$, so it vanishes on the diameter.
On $H_0$ it satisfies $0\le\omega\le1$: the bracket is non-negative there, because
$\abs{Re^{-i\phi}-w}^2-\abs{Re^{i\phi}-w}^2=4R\sin\phi\,\Im w\ge0$, and its first term alone gives the Poisson
integral of an indicator function. It tends to $0$ at every boundary point of $H_0$ outside $A'$. Let $h(b):=\theta_A\,\omega(w(b))$ on $\Omega_0$; it is
$\theta_A$ times the harmonic measure of $A$ in $\Omega_0$. Now $\theta-h$ is
superharmonic on $\Omega_0$, and its lower limit at every boundary point of $\Omega_0$ is non-negative: at points of $A$
because $\theta\ge\theta_A\ge h$ and $\theta$ is lower semicontinuous; elsewhere because $\theta\ge0$ and $h\to0$, the
two lips of the slit being approached separately. By the minimum principle (\cite[Thm.~2.3.1]{Ransford} applied to
$h-\theta$ on the bounded domain $\Omega_0$), $\theta\ge h$ on $\Omega_0$.

\emph{Lower bound.} Let $\abs w\le R/2$ and $\Im w>0$. In \eqref{eq:H-omega} we have $R^2-\abs w^2\ge\frac34R^2$ and
$\abs{Re^{\pm i\phi}-w}\le\frac32R$, so the bracket is at least $4R\sin\phi\,\Im w/(\frac32R)^4$. Integrating $\sin\phi$ over
$[\frac\pi8,\frac{3\pi}8]$ gives
$\omega(w)\ge\frac1{2\pi}\cdot\frac34R^2\cdot4R\,\Im w\,(\cos\frac\pi8-\cos\frac{3\pi}8)/(\frac32R)^4=c_2\,\Im w/R$, where
$c_2:=8(\cos\frac\pi8-\cos\frac{3\pi}8)/(27\pi)=0.0510$.
On the circle $b=\xs+\epsilon e^{i\phi}$, $0<\phi<2\pi$, with $0<\epsilon\le r_0/4$, we have $\abs w=\epsilon^{1/2}\le R/2$ and
$\Im w=\epsilon^{1/2}\sin(\phi/2)$. Since $\theta\ge h=\theta_A\,\omega$ and $\int_0^{2\pi}\sin(\phi/2)\,\dd\phi=4$, the mean of $\theta$ over this circle is
at least $\theta_A\,(c_2/R)\,\epsilon^{1/2}\cdot\frac4{2\pi}$, which is the first inequality in \eqref{eq:H-tip}.

\emph{Upper bound.} Since $u$ is subharmonic on $\strip$, $u(\xs)$ is at most the mean of $u$ over the circle of radius
$\epsilon$ about $\xs$; and $u(\xs)=U(\xs)$. Hence the mean of $\theta$ over that circle is at most the mean of $U$ minus
$U(\xs)$. The latter is given by the Poisson--Jensen formula at the center \cite[\S4.5]{Ransford}: it equals
$\int_0^\epsilon\mu(\sigma)\,\dd\sigma/\sigma$, where $\mu(\sigma)$ is the charge $\frac1{2\pi}\nabla^2U$ of the disk
$B(\xs,\sigma)$, that is, $\mu(\sigma)=\int_{\xs}^{\xs+\sigma}\rhoU\,\dd x$ for $\sigma\le r_0$. On
$[\xs,\xs+r_0]$ we have $\sinh\pi x-1\le\kappa_2(x-\xs)$ with $\kappa_2:=\pi\cosh\bigl(\pi(\xs+r_0)\bigr)$, because
$\sinh\pi\xs=1$; and $\arccosh(1+\delta)\le\sqrt{2\delta}$ for $\delta\ge0$. So
$\rhoU(x)\le\frac4{3\pi}\sqrt{2\kappa_2(x-\xs)}$ and $\mu(\sigma)\le\frac8{9\pi}\sqrt{2\kappa_2}\,\sigma^{3/2}$, and the
mean of $\theta$ is at most $C_4\,\epsilon^{3/2}$ with $C_4:=\frac23\cdot\frac8{9\pi}\sqrt{2\kappa_2}$. Together,
\begin{equation}\label{eq:H-tip}
  \frac{2\,\theta_A\,c_2}{\pi R}\,\epsilon^{1/2}\;\le\;
  \frac1{2\pi}\int_0^{2\pi}\theta\bigl(\xs+\epsilon e^{i\phi}\bigr)\,\dd\phi\;\le\;\frac{16\sqrt{2\kappa_2}}{27\pi}\,\epsilon^{3/2}
  \qquad(0<\epsilon\le r_0/4) ,
\end{equation}
which is false for small $\epsilon$. So $\theta\equiv0$ on $D$.

\emph{Step 3: the slits.} We have $u=U$ on $D$. At a point $x$ of a slit, upper semicontinuity of $u$
and continuity of $U$ give $u(x)\ge\limsup_{b\to x,\ b\in D}u(b)=U(x)$, and $u(x)\le U(x)$ by \cref{lem:9.2}. Hence
$u=U$ on all of $\strip$, which is the claim. This completes the proof of \cref{thm:7}.

\section*{Notation index}
\addcontentsline{toc}{section}{Notation index}

{\footnotesize
\setlist[description]{font=\normalfont,style=sameline,leftmargin=5.6em,labelwidth=5.2em,labelsep=0.4em,
  itemsep=0pt,parsep=0pt,topsep=0pt,partopsep=0pt,beginpenalty=10000}
\noindent\textbf{Central charge, weights, variables.} At the right: the place of definition (\S: section or
subsection).
\begin{description}
\item[$c'$, $a$, $t$, $t_0$] $c-1$; $c'/12$; $\Delta-a$, so that the BTZ threshold is at $t=0$; position $-a$ of the
  vacuum~\hspace*{\fill}\mbox{\eqref{eq:2-vars}, \S\ref{sec:2.1}}
\item[$\beta$; $\Zhat$, $d_n$] inverse temperature, and nothing else; reduced partition function; degeneracies,
  $d_n>0$~\hspace*{\fill}\mbox{\eqref{eq:2-Zhat}}
\item[$\Delta_1$] gap of an admissible spectrum, $a+\min_nt_n$; primaries with $t_n<0$ are
  \emph{light}~\hspace*{\fill}\mbox{Def.~\ref{def:2-admissible}}
\item[$b=x+iy$; $G$] $\sqrt{3t/c'}$, not the Liouville coupling; physical axis $b=x\ge0$, vacuum at $\pm\frac i2$;
  $G(b)=g(c'b^2/3)$~\hspace*{\fill}\mbox{\eqref{eq:2-scaling}}
\item[$\br\in\RR^2$] plane with $t=\abs\br^2/2$, in which a functional is radial and
  $\hat g=-g$~\hspace*{\fill}\mbox{\eqref{eq:2-fourier}, \eqref{eq:2-eigen}}
\item[$P_L$, $\bar P_L$, $Q_L$] Liouville momenta and background charge, in two remarks
  only~\hspace*{\fill}\mbox{Rems.~\ref{rem:2-liouville}, \ref{rem:2-skernel}}
\item[$\xi$; $\xi_1$, $\Lambda$] $\Delta/c$ in \S\ref{sec:3}, $t/c'=b^2/3$ from \S\ref{sec:4} on; a certified slope
  $\Lambda$ has $1/\Lambda=\frac1{12}+\xi_1$, $\xi_1=x_1^2/3$~\hspace*{\fill}\mbox{\eqref{eq:4-wkb},
  \eqref{eq:2-slope}}
\item[$L(x)$] $\frac{2\pi}3\abs x$, so $e^{c'L}=e^{4\pi\sqrt{at}}$; the \emph{Cardy line} is the graph of $-L$
  against $x$~\hspace*{\fill}\mbox{\eqref{eq:2-L}}
\item[$\strip$, $\Epm$; $v$] strip $\abs y<\half$ of the $b$-plane, its edges $y=\pm\half$; edge coordinate,
  $b=v+\frac i2$~\hspace*{\fill}\mbox{\S\ref{sec:2.5}, \eqref{eq:7-edge}}
\item[$\HH\ni z$] upper half-plane; it uniformizes the curve and the slit
  strip~\hspace*{\fill}\mbox{\eqref{eq:4-pb}, \eqref{eq:6-map}}
\end{description}

\smallskip\noindent\textbf{Functionals, kernels, the linear program}
\begin{description}
\item[$\omega$; $\Xib$, $\eb$] functional on functions of $\beta$ (\S\ref{sec:2} only); crossing kernel
  $\eb-\beta^{-1}e_{1/\beta}$, $\eb=e^{-2\pi\beta t}$~\hspace*{\fill}\mbox{\eqref{eq:2-g}, \S\ref{sec:8.2}}
\item[$g$; $\gvac$] a functional, as the function $\omega[\Xib(t)]$ of $t$; its vacuum value,
  $-\int_0^\infty g\,\rhoC\,\dd t$~\hspace*{\fill}\mbox{\eqref{eq:2-sumrule}, \eqref{eq:2-gvac-cardy}}
\item[$t_1(g)$, $x_1(g)$] last sign change of $g$, the gap bound that $g$ certifies;
  $\sqrt{3t_1/c'}$~\hspace*{\fill}\mbox{Def.~\ref{def:2-feasible}, \eqref{eq:2-scaling}}
\item[$P$, $\pol_P$; $t_1^{(P)}$, $\Delta_1^{(P)}$] order, polynomial in $g=\pol_P(t)e^{-2\pi t}$; $\inf t_1(g)$ at order
  $P$; $a+t_1^{(P)}$ ($P=\infty$: limits)~\hspace*{\fill}\mbox{\eqref{eq:2-deriv}, \S\ref{sec:2.2}}
\item[$\psi_k$] Laguerre functions $L_k(4\pi t)e^{-2\pi t}$,
  $\hat\psi_k=(-1)^k\psi_k$~\hspace*{\fill}\mbox{\eqref{eq:2-psik}}
\item[$K_\alpha$, $V$, $\rhoC$] Cardy kernel $I_0(4\pi\sqrt{\alpha t})$, weight $K_a-2K_{a-1}+K_{a-2}$ (no subscript), density
  $2\pi V$~\hspace*{\fill}\mbox{\eqref{eq:2-Kalpha}, \eqref{eq:2-rhoC}}
\item[$\as(z)$; $\epsV$, $\CV$] Bessel asymptote $e^z/\sqrt{2\pi z}$; remainder of $V$ against $\as(c'L)$,
  $\abs{\epsV}\le\CV/c'$~\hspace*{\fill}\mbox{\S\ref{app:A.4}, \eqref{eq:A-V-as}}
\item[$g_\pm$, {$M[g]$}] $\max(\pm g,0)$; Cardy-weighted mass
  $\int_0^\infty\abs g\,V\,\dd t$~\hspace*{\fill}\mbox{\eqref{eq:2-halfmass}}
\item[$\Abar$; $\TV$, $\Dbest$] class of functionals; least $T$ at which some $g\in\Abar$ is feasible, and $\TV+a$
  ($V$: Virasoro)~\hspace*{\fill}\mbox{Def.~\ref{def:2-Abar}, \eqref{eq:2-TV}}
\item[$\xs$, $\xis$, $\LV$] $\arcsinh(1)/\pi=0.2805499$; $(\xs)^2/3=0.0262361$;
  $12\pi^2/(\pi^2+4\arcsinh^21)=9.1266340$~\hspace*{\fill}\mbox{Thm.~\ref{thm:1}}
\item[$\tint$; $\varphi(t)$] interior zero of the optimal functional below $t_1$; prefactor in
  $g=\varphi\,e^{-2\pi t}$ at $t=O(1)$~\hspace*{\fill}\mbox{\S\ref{sec:2.5}, \S\ref{sec:4.6}}
\end{description}

\smallskip\noindent\textbf{The curve and its endpoint (Part II)}
\begin{description}
\item[$\cS$, $\tau$; $\yE$, $\cardyh$] WKB exponent; local modular parameter $\cS'/2\pi i$; $\Im\tau_\pm$ (band);
  Cardy height $2b\,\yE$~\hspace*{\fill}\mbox{\eqref{eq:4-wkb}, \eqref{eq:4-band}, \eqref{eq:4-yE}}
\item[$\rho$, $\rho_0$, $w$, $\iota$, $W$, $\cF$, $E$] \S\ref{sec:3} only: double zeros per unit $\Delta$; $S$-image of
  the vacuum; $\pi^2\xi\rho$; $\hat S_{U(1)}$ in $(\rho,w)$; the curve as $w=W(\rho)$ or $\rho=\cF(w)$; even factor of
  $\cF$~\hspace*{\fill}\mbox{\eqref{eq:3-band}, \eqref{eq:3-forced}, \eqref{eq:3-rho-w}--\eqref{eq:3-W}}
\item[$\hat S_{U(1)}$, $\hat S$; $\Sigma$] sheet maps, the Fourier transform acting on WKB sheets; the spectral curve
  \hyp{B}~\hspace*{\fill}\mbox{\eqref{eq:3-sheetmap}, \eqref{eq:4-sheetmap}}
\item[$p$; $\Rq$, $\sigma$] WKB momentum $-\tau b$, on $\Sigma$ the function \eqref{eq:6-p}; quarter-turn
  $(p,b)\mapsto(b,-p)$; exchange $(p,b)\mapsto(-p-i,b)$ of the band sheets~\hspace*{\fill}\mbox{\eqref{eq:4-momentum},
  \eqref{eq:4-R}, \S\ref{sec:4.3}}
\item[$\gamma_k$, $\Omt$; $X$, $Y$] band arc $\gamma_0$ and its images $\Rq^k\gamma_0$, the region they bound on
  $\Sigma$; coordinates $e^{-\pi(p\pm b)}$~\hspace*{\fill}\mbox{\eqref{eq:4-bdry}, \eqref{eq:4-XY}}
\item[$\Aopt$, $A_m$; $t_2$] $c^{1/3}$ coefficients: conjectured optimum $0.27666$; comb with $m$ omitted sites;
  second extremal level~\hspace*{\fill}\mbox{\eqref{eq:4-airy}, \eqref{eq:7-Am}}
\item[$\Lfit$, $\gfit$; $\Lmeas$, $\xi^*_{1,\rm meas}$] parameters of a logarithmic fit; measured slope and
  endpoint~\hspace*{\fill}\mbox{\S\ref{sec:5.2}, \eqref{eq:5-meas}}
\end{description}

\smallskip\noindent\textbf{Potential theory (Part III)}
\begin{description}
\item[$D_s$, $U_s$; $D$, $U$, $U_0$] strip slit along $\abs x\ge s$, with \emph{tips} $\pm s$; harmonic function on it,
  $0$ on $\Epm$ and $-L$ on the slits; the case $s=\xs$, $U$ being the \emph{extremal potential};
  $U(0)=-0.388748$~\hspace*{\fill}\mbox{\eqref{eq:6-Ds}, \eqref{eq:6-dirichlet}, \eqref{eq:6-U}}
\item[$u_g$; $\CQ$] profile $\frac1{c'}\log(\abs G/M[g])$; constant in $u_g\le U+\CQ(1+\log c')/c'$ on a compact
  $Q\subset D$~\hspace*{\fill}\mbox{\eqref{eq:6-ug}, \eqref{eq:6-majorant}}
\item[$\omb(E)$, $\dd\omb/\dd x$] harmonic measure of a boundary set $E$ seen from $b$; its density on a slit, both
  lips added~\hspace*{\fill}\mbox{\S\ref{sec:6.4}}
\item[$\nug$; $\zeta$, $m_\zeta$] Riesz measure $\frac1{2\pi}\nabla^2u_g$; a zero of $G$, its
  multiplicity~\hspace*{\fill}\mbox{\eqref{eq:6-riesz}, \eqref{eq:6-laplace}}
\item[$\rhoU$; $\rhoV$] \emph{zero law} $\frac4{3\pi}\arccosh(\sinh\pi\abs x)$, $\abs x\ge\xs$: charge of $U$ (not
  $U(1)$), zeros per unit $x$ and $c'$; $\frac{3\rhoU}{4x}$, double zeros per unit
  $t$~\hspace*{\fill}\mbox{Def.~\ref{def:6.6}, \eqref{eq:4-rhoU}}
\item[$N$, $\ktip$; $\gtip$] $N(x)=\half\int_{\xs}^x\rhoU\simeq\ktip(x-\xs)^{3/2}$ counts sites,
  $\ktip=4\cdot2^{3/4}/(9\sqrt\pi)=0.4217111$; coefficient of $\sqrt{s-\abs x}$ in
  $U_s+L$~\hspace*{\fill}\mbox{\eqref{eq:6-N}, \S\ref{sec:6.6}}
\item[$\kappa(\delta)$; $C_\delta$, $\Cconc$, $\CH$] Cardy deficit $-(U+L)(\xs-\delta)$; constant in
  \eqref{eq:6-conc}, $\le\Cconc\delta^{-5}$; entropy constant~\hspace*{\fill}\mbox{\eqref{eq:6-kappa},
  \eqref{eq:B-entropy-delta}}
\item[$c_0(\Lambda)$, $\copt$, $\catt$] central charges from which Thm.~\ref{thm:4} (first part; rate), Thm.~\ref{thm:5}
  hold; not computed; $\catt\ge40$~\hspace*{\fill}\mbox{\S\ref{sec:6.8}}
\end{description}

\smallskip\noindent\textbf{The functional of Part IV and its comb}
\begin{description}
\item[$T_\zeta(b)$] Blaschke factor of the strip, zeros $\pm\zeta$, facts \hyp{T1}--\hyp{T4}; $T$ with a point as
  subscript is always this~\hspace*{\fill}\mbox{\eqref{eq:7-T}}
\item[$F$, $v_0$, $b_0$] edge function $p(v+\frac i2)=\frac1\pi\arcsinh(\cosh\pi v)$; its inverse,
  $\rhoU=\frac43v_0$; saddle $v_0+\frac i2$~\hspace*{\fill}\mbox{\eqref{eq:7-edge}}
\item[$\chi_n$, $\ell_n$, $\varepsilon_n$, $u$] sites: $c'N(\chi_n)=n-\frac14$; length of cell
  $n=(\chi_n,\chi_{n+1})$; detuning height $\eta_0\ell_n$; $(x-\chi_n)/\ell_n$~\hspace*{\fill}\mbox{\S\ref{sec:7.2}, \S\ref{sec:7.5}}
\item[$m$, $k_\infty$, $n_1$, $\eta_0$] omitted sites, node phase, onset index $m+1-k_\infty=c'N(\chi_2)$, detuning;
  fixed to $(1,\frac14,\frac74,\frac14)$~\hspace*{\fill}\mbox{\S\ref{sec:7.2}}
\item[$B$, $h_0$, $g$; $B^0$, $h_0^0$] comb, seed $\sech^2(\frac{\pi b}2)B$, functional $\half(h_0-\hat h_0)$ of
  Thm.~\ref{thm:5}; the same at $\eta_0=0$~\hspace*{\fill}\mbox{\eqref{eq:7-functional}}
\item[$\tstar$, $A_1$; $\lc$, $\vsig$] $c'\chi_2^2/3=\xis c'+A_1c'^{1/3}+\dots$, $A_1=0.483$; Airy length
  $(\ktip c')^{-2/3}$; $(x-\xs)/\lc$~\hspace*{\fill}\mbox{\eqref{eq:7-tstar}, \S\ref{sec:7.1}, Lem.~\ref{lem:7.5}}
\item[$\thon$, $\xI$] onset exponent $0.2$; hand-off point $\xs+c'^{-\thon}$ between Regions I and
  II~\hspace*{\fill}\mbox{\S\ref{sec:7.1}}
\item[$\nu$, $\Psi$; $\hPsi$] site index $c'N(x)+\frac14-2$, comb phase $2\pi\nu-\pi$; phase of
  $\hat h_0$~\hspace*{\fill}\mbox{\eqref{eq:7-nu}, \eqref{eq:D-phase}}
\item[$\Env$, $\Pi$] exact envelope, $\abs{h_0^0}=\Env\sin^2\pi\nu$; detuning factor
  $\abs B/\abs{B^0}\ge1$~\hspace*{\fill}\mbox{\eqref{eq:7-env}}
\item[$\cP$, $\cP_Y$; $\Jend$, $\CJ$] potential of the absent charge on $[\xs,\chi_2]$, on $[\xs,Y]$; end correction,
  $0\le\Jend\le\CJ\le0.65$~\hspace*{\fill}\mbox{\eqref{eq:7-P}, \eqref{eq:7-Jend}}
\item[$\Lev$, $\lambda$] level $4\sech^2(\frac{\pi x}2)e^{-c'L+c'\cP+\Jend}$, the unit of all certified ratios;
  $\Env/\Lev$~\hspace*{\fill}\mbox{\eqref{eq:7-lev}, \eqref{eq:7-lambda}}
\item[$\EcH^\pm_{n,j}$, $h^\pm_j$; $C_j$] hole functions of cell $n$, with $j$ pairs kept on each side of $x$; their limits ($h^\pm_\infty$
  if $j=1$); $\abs{\EcH^\pm_{n,j}-h^\pm_j}\le C_j(\frac1n+\ell_n)$,
  $C_j=Cj^2\log(2j+2)$; $\frac{94}n+C\ell_n$ if $j=1$~\hspace*{\fill}\mbox{\eqref{eq:E-Hplus}--\eqref{eq:E-hminus}, \eqref{eq:E-displacement}, \eqref{eq:E-displacement-explicit}}
\item[$e_L$, $\CL$; $\Cdet$; $\cT$] midpoint remainder of the far-left sites, its supremum on the cells $n\ge6$,
  $\CL\le0.006$ (\cref{lem:E-farleft}(d)); detuning constant, $\Cdet\le16.6$; overshoot constant, of which only $\cT>0$ is used~\hspace*{\fill}\mbox{\eqref{eq:E-eL}, Lem.~\ref{lem:E-detune}, \eqref{eq:7-overshoot}}
\item[$\Phi_0$, $\Phi_1$; $\CEM$] edge phases, $\log B=c'\Phi_0+\Phi_1$; Euler--Maclaurin
  constant~\hspace*{\fill}\mbox{\eqref{eq:7-phase}, \eqref{eq:C-rem-bounds}}
\item[$k_x$, $\epsH$; $\alpha_x$, $\sch$] Hankel wavenumber $\frac{4\pi}3c'x$, remainder; curvature, chord slope of the
  phase~\hspace*{\fill}\mbox{\eqref{eq:7-edgerep}, \eqref{eq:7-hankel}, \eqref{eq:D-alpha}}
\item[$\Amp$, $\epssd$; $\Cglob$] amplitude of $\hat h_0$, remainder under \hyp{SD1}--\hyp{SD4} (\S\ref{app:D.1}); global constant
  $2^5$~\hspace*{\fill}\mbox{\eqref{eq:7-transform}--\eqref{eq:7-glob}}
\item[$\vrho$, $\vrhoinf$] ratio $\Amp/\Lev$ and its $c$-independent
  majorant~\hspace*{\fill}\mbox{\eqref{eq:7-vrhoinf}}
\item[$\mu_j$; $\muI_n$, $\muslope$] floor constants, $j$ pairs kept exactly on each side; floors near the
  onset~\hspace*{\fill}\mbox{\eqref{eq:7-floor}, \eqref{eq:7-first-cells}, \eqref{eq:7-slope}}
\item[$\Lobe$, $\Band$; $\Rvac$] integrals of $h_0V$ over $(0,\tstar)$ and $(\tstar,\infty)$; vacuum ratio,
  $\Band\le\Rvac\abs\Lobe$~\hspace*{\fill}\mbox{\eqref{eq:7-vacsplit}, \eqref{eq:7-vac}}
\end{description}

\smallskip\noindent\textbf{Certification (\S\ref{sec:8})}
\begin{description}
\item[$\mfrak$; $\NZ$, $\CZ$] signed spectral measure $\sum_nd_n\delta_{t_n}-\rhoC\,\dd t$, facts \hyp{Z1}--\hyp{Z4},
  \hyp{St}; counting function, $1+\Zhat(1)$~\hspace*{\fill}\mbox{\eqref{eq:8-measure}, \S\ref{sec:8.2}}
\item[$\vpi$; $\Ab_r$, $\beta_r$] $\frac{\beta-1}{\beta+1}$; Abel mean of ratio $r<1$, contracted modulus
  with $\vpi(\beta_r)=r\vpi(\beta)$~\hspace*{\fill}\mbox{\eqref{eq:8-genfn}, \eqref{eq:8-kernel},
  \eqref{eq:8-contraction}}
\item[$W_{q,\ell}$, $\ell_a$] weights $(1+4\pi t)^qe^{\ell\sqrt{4\pi t}}$; Cardy exponent
  $2\sqrt{\pi a}$~\hspace*{\fill}\mbox{\eqref{eq:8-weights}}
\item[$\xinf$, $x_0$; $\theta$] limit of $x_1(g_j)$ along a sequence $g_j$, contact point; deficit $U-u$ of a limit
  profile $u$~\hspace*{\fill}\mbox{Prop.~\ref{prop:9.3}, \S\ref{sec:9.2}}
\end{description}

\smallskip\noindent\textbf{Letters with two meanings.}
$j$: pairs kept (\S\ref{sec:7}, App.~\ref{app:E}), sequence index (\S\ref{sec:9}, App.~\ref{app:H}).
$\delta$: distance below the tip (\S\ref{sec:6}, App.~\ref{app:B}), above it in the variable $\pi x$
(\S\ref{app:E.1}).
$r$: Abel ratio (\S\ref{sec:8}), distance from a tip (Lem.~\ref{lem:7.5}).
$y$: $\Im\tau$, the \emph{height} of a sheet (\S\ref{sec:3}, \S\ref{sec:4}), $\Im b$ (elsewhere).
$E$: boundary set, edges $\Epm$, factor $E(w)$.
$a$, $u$, $\sigma$: also a real label in $\chi_a$, and $4\pi t$, $\sqrt{4\pi t}$ in \S\ref{sec:8}.
Distinct: $W(\rho)$, $W_{q,\ell}$; $\eb$, $e_L$, $e_1(\delta)$; $\omega$, $\omb$; $\kappa(\delta)$, $\ktip$, $\cardyh$.
Bare $\varepsilon$, $\epsilon$: local to a statement; $\eta$: Dedekind function
(\S\ref{sec:2.1} only); $\varrho$: a ratio, not a density; fraktur letters: measures.

\smallskip\noindent\textbf{Asymptotic relations.}
$A\simeq B$: $A/B\to1$ in the limit indicated.
$A\asymp B$ in a proof (\S\ref{sec:7}, the appendices): $C^{-1}B\le A\le CB$ with a constant $C\ge1$ that
does not depend on $c$ or on the point $x$; it may depend on the fixed parameters of the construction, and on a
parameter that is named at that place.
$A\lesssim B$: $A\le CB$ with such a $C$ (in App.~\ref{app:C} it may depend on $\delta_1$); $A\gtrsim B$: $B\lesssim A$.
In \S\ref{sec:3} and \S\ref{sec:4}, $\asymp$ keeps the exponential order in $c$ only.
With explicit numbers on the right ($c'\lesssim414$), $\lesssim$ reads ``at most, approximately''.

\smallskip\noindent\textbf{Status tags} (legend at the end of \S\ref{sec:1.3})\textbf{.}
\tagpar{\proved}, \tagpar{\provedCA}, \tagpar{\conjtag}, \tagpar{\heuristic}, \tagpar{\numerical},
\tagpar{\orientation}, on a statement or a section; \cert{} marks a certified inequality, a row of
Table~\ref{tab:4}.

\smallskip\noindent\textbf{Terminology.}
\emph{Gap}: $\Delta_1$ only; two sites bound a \emph{cell}.
\emph{BTZ threshold}: $t=0$ only; a functional certifies a \emph{gap bound}.
\emph{Feasible at $T$}, \emph{weakly feasible}: Def.~\ref{def:2-feasible}.
\emph{Lobe interval} $[0,t_1)$, with the negative \emph{lobe} of $g$; \emph{band}: beyond $t_1$
(\S\ref{sec:2.5}).
\emph{Envelope}: $\genv$ of an oscillating $\abs g$, and $\Env$; $U$ is a \emph{majorant} of profiles.
\emph{Edges}: $\Epm$ only; the slits end at \emph{tips}.
\emph{Onset}: near the first kept site $\chi_2$ (\S\ref{sec:7.1}); \emph{onset index}: $n_1$.
\emph{Optimality}, \emph{attainment}: the halves of Thm.~\ref{thm:1} (\S\ref{sec:2.6}).
\par}

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\end{thebibliography}\endgroup

\end{document}