\section{Proof of Theorem~\ref{thm:rapid}}\label{app:rapid}

Throughout this appendix $P$, $k$, $b$, $d$, $\eps_j$, $\gamma_j$ and $\beta_j$ are
as in \eqref{eq:blockparams}. The explicit formula $\eps_j=\frac1{16}e^{-db^j}$
enters only through two exponent computations, in Stages~2 and~3 below. Apart
from these, we use only that
\begin{equation}\label{eq:smalltail}
 0<\eps_{j+1}\le\tfrac14\,\eps_j\le\tfrac14\cdot10^{-12}\qquad(j\ge1),
\end{equation}
which holds because $\eps_1\le2^{-452}$ and $\eps_{j+1}/\eps_j\le2^{-64}$. In
Section~\ref{subsec:A-stage4} we also use that $\eps_1<\frac1{4k}$
(Section~\ref{sec:blocks}): the $\beta_j$ and $\gamma_j$ are distinct, the $\gamma_j$ lie
within $\eps_1$ of $\frac12$, and $\beta_j\le\beta_k=\frac12-\frac1{4k}$.

\emph{Plan.} Part~(e) is a statement about finitely many lattice combs; it is
proved in Section~\ref{subsec:A-finite} by a triangularity argument, which is
used again twice. Parts~(a)--(c) concern the infinite carrier $L_\gamma$. In Zak
coordinates (Section~\ref{subsec:A-zak}) the support condition on $T$ and the
support condition on $\F T$ become triangularity conditions, called
\emph{flags}, on two arrays of moments $\mathbf A$ and $\mathbf B$, and the
Fourier transform becomes an explicit bounded operator $E$ with
$\mathbf B=E\mathbf A$, the \emph{gauge} (Section~\ref{subsec:A-moments}). The
proof then has four stages.
\begin{itemize}[leftmargin=2em]
\item \emph{Stage~1, uniqueness} (Section~\ref{subsec:A-stage1}): the flags
leave at most one free parameter, so $\dim\M_{L_\gamma}\le1$.
\item \emph{Stage~2, exclusion} (Section~\ref{subsec:A-stage2}): for elements of
Hermite order at most $P$, moments with weights that shrink as fast as the
clusters can be used, and in these weights the flags leave no free parameter.
This proves~(b).
\item \emph{Stage~3, construction} (Section~\ref{subsec:A-stage3}): the free
parameter of Stage~1 is realized by a limit of finite approximations whose
Hermite norms of order $18P+2$ are uniformly bounded. This proves~(a) and~(c).
\item \emph{Stage~4, head transfer} (Section~\ref{subsec:A-stage4}): inserting the
finite head $\beta_1,\dots,\beta_k$ costs two Hermite orders. This proves~(d).
\end{itemize}
The same arguments are carried out in the formal development (see the table in
Appendix~\ref{app:lean}); the account below is self-contained.
Table~\ref{tab:appA} lists the objects that the reader has to keep track of.

\begin{table}[!ht]
\centering\small
\begin{tabular}{>{\raggedright\arraybackslash}p{0.25\textwidth}>{\raggedright\arraybackslash}p{0.47\textwidth}>{\raggedright\arraybackslash}p{0.18\textwidth}}
\hline
Object & Meaning & Introduced in\\
\hline
$V$, $gV$ & Zak transforms of $T$ and $\F T$, cut off near the clusters at the
half-integers & Lemma~\ref{lem:zak}\\
$\phi_{i,e}$; $M(U)$; $\mathbb H$ & Newton test functions vanishing at the first $i$
cluster points; the weighted moment array of $U$; the Hilbert space of arrays &
\eqref{eq:newton}, \eqref{eq:moments}\\
$\mathbf A$, $\mathbf B$ & the moment arrays $M(V)$ and $M(gV)$ of $T$ and $\F T$ &
Lemma~\ref{lem:flags}\\
flags & triangularity of $\mathbf A$ and $\mathbf B$, which encodes the support
conditions & \eqref{eq:flags}, \eqref{eq:diagrel}\\
$E$ & the gauge, $\mathbf B=E\mathbf A$, which encodes the Fourier transform &
Lemma~\ref{lem:gauge}\\
$\mathcal H_A,\mathcal H_B,\mathcal D,\mathcal C,\mathcal G$; $P_\bullet$, $Q$ &
subspaces of $\mathbb H$ given by sets of index pairs, and the projections onto them &
before \eqref{eq:kernelop}\\
$\Sigma_A$, $\Sigma_B$ & graph maps, which encode the diagonal relations of the flags &
before \eqref{eq:kernelop}\\
$\mathcal T$; $a=(u,c)$ & the kernel operator, and its unknown: $u$ free, $c$
eliminated & \eqref{eq:kernelop}\\
$\mathcal K$ & the Schur complement, acting on $u$ & Stage~1\\
\hline
\end{tabular}
\caption{The objects of Appendix~\ref{app:rapid}.}\label{tab:appA}
\end{table}

\subsection{Two facts about the Hermite scale}\label{subsec:A-hermite}
We use the norm equivalence of Section~\ref{sec:prelim}: for integers $p\ge0$,
$\norm\varphi_p$ is equivalent to $\sum_{a+b\le2p}\norm{x^a\varphi^{(b)}}_{L^2}$.
Two consequences are used repeatedly. First, if $m\in C^\infty(\R)$ is
$1$-periodic, then multiplication by $m$ is bounded on $H_p$ and, by duality, on
$H_{-p}$, for every integer $p\ge0$; since $\F$ is unitary on $H_{\pm p}$, the same
holds for $\F^{-1}m\F$. Second, for $a\in[-1,1]$ and bounded coefficients
$(c_n)$,
\begin{equation}\label{eq:latticesampling}
 \Bigl|\sum_{n\in\Z}c_n\varphi(n+a)\Bigr|\le C\sup_n|c_n|\,\norm\varphi_1 .
\end{equation}
Indeed, by the Sobolev inequality on unit intervals,
$\sum_n(1+n^2)|\varphi(n+a)|^2\le C\int(1+x^2)(|\varphi|^2+|\varphi'|^2)\,dx\le
C\norm\varphi_1^2$, and it remains to apply the Cauchy--Schwarz inequality.

\subsection{Finitely many phases, and part~(e)}\label{subsec:A-finite}
For $a,b\in\R$ let $w_{a,b}=\sum_{n\in\Z}e^{2\pi\ii nb}\delta_{n+a}$. By
\eqref{eq:latticesampling}, $w_{a,b}\in H_{-1}$, and by Poisson summation
\begin{equation}\label{eq:Fwab}
 \F w_{a,b}=\sum_{l\in\Z}e^{-2\pi\ii a(l+b)}\delta_{l+b}=e^{-2\pi\ii ab}\,w_{b,-a}.
\end{equation}
Thus $w_{a,b}$ is a crystalline measure with support $a+\Z$ and spectrum $b+\Z$.
We write $S(\{x\})$ for the mass at $x$ of a locally atomic distribution $S$.

\begin{lemma}\label{lem:finitephase}
Let $A,B\subset\R$ be finite sets, each consisting of numbers that are pairwise
incongruent modulo $1$, and let $T\in\Sc'$ be locally atomic on $A+\Z$ with
$\F T$ locally atomic on $B+\Z$. Then $T=\sum_{a\in A,\,b\in B}c_{a,b}w_{a,b}$ for
unique coefficients $c_{a,b}\in\C$.
\end{lemma}

\begin{proof}
The trigonometric polynomial $p(\xi)=\prod_{b\in B}(e^{2\pi\ii\xi}-e^{2\pi\ii b})
=\sum_{l=0}^mp_le^{2\pi\ii l\xi}$, $m=\#B$, vanishes on $B+\Z$, so $p\,\F T=0$;
since $\F^{-1}(e^{2\pi\ii l\xi}\F T)=\tau_lT$, this says $\sum_lp_l\tau_lT=0$. For
$a\in A$ let $t_a(n)=T(\{n+a\})$. The mass of $\tau_lT$ at $n+a$ is $t_a(n+l)$,
so $\sum_lp_lt_a(n+l)=0$ for all $n$. This is a linear recurrence whose
characteristic polynomial $\prod_{b\in B}(w-e^{2\pi\ii b})$ has $m$ distinct roots,
so $t_a(n)=\sum_bc_{a,b}e^{2\pi\ii nb}$ for unique $c_{a,b}$. The distributions $T$
and $\sum c_{a,b}w_{a,b}$ are locally atomic on $A+\Z$ with the same masses, so
they coincide.
\end{proof}

\begin{lemma}[Finite critical caps]\label{lem:finitecap}
Let $r\ge1$, let $\alpha_1,\dots,\alpha_r\in(0,\frac12)$ be distinct and
$\alpha'_1,\dots,\alpha'_r\in(0,\frac12)$ be distinct, and let
$T=\sum c_{a,b}w_{a,b}$, the sum over $a\in\{\pm\alpha_j\}$ and
$b\in\{\pm\alpha'_j\}$. Suppose that $T$ has no mass at the points $n\pm\alpha_j$
with $|n|<j$, and $\F T$ has no mass at the points $n\pm\alpha'_j$ with $|n|<j$
\textup{(}$1\le j\le r$\textup{)}. Then $T=0$.
\end{lemma}

\begin{proof}
Let $\Xi=\{\pm\alpha_j\}$ and $\Xi'=\{\pm\alpha'_j\}$, and let $U$ be the measure
$\sum_{a,b}c_{a,b}\,\delta_a\otimes\delta_b$ on the finite grid $\Xi\times\Xi'$. Put
$g(x,y)=e^{-2\pi\ii xy}$. By the definition of $w_{a,b}$ and by \eqref{eq:Fwab},
for all functions $\psi$ on $\Xi$, $\psi'$ on $\Xi'$, and all $n\in\Z$,
\begin{equation}\label{eq:gridextract}
 \sum_{a\in\Xi}\psi(a)\,T(\{n+a\})=U\bigl(\psi(x)e^{2\pi\ii ny}\bigr),\qquad
 \sum_{b\in\Xi'}\psi'(b)\,\F T(\{n+b\})=(gU)\bigl(e^{-2\pi\ii nx}\psi'(y)\bigr).
\end{equation}

\emph{Moments.} Let $s(t)=\sin2\pi t$, $z(t)=1-\cos2\pi t$, $z_j=z(\alpha_j)$,
$D_i=\prod_{l\le i}(z-z_l)$, and $\phi_{i,0}=D_i$, $\phi_{i,1}=sD_i$ for
$0\le i<r$; put $\kappa_i=s(\alpha_{i+1})>0$. For a measure $u$ on $\Xi$ let
$m(u)\in\C^r\otimes\C^2$ have the entries $m(u)_{i,e}=\kappa_i^{-e}u(\phi_{i,e})$.
This map is bijective: the $z_j$ are distinct, so the restrictions to $\Xi$ of the
$D_i$, $i<r$, span the even functions on $\Xi$, and since $s$ is odd and does
not vanish on $\Xi$, the $sD_i$ span the odd ones. Define $m'$ on $\Xi'$ in the
same way, and put $\mathbf A=(m\otimes m')(U)$ and $\mathbf B=(m\otimes m')(gU)$,
with entries $\mathbf A_{ie,jf}$ and $\mathbf B_{ie,jf}$.

\emph{Flags.} We claim that
\begin{equation}\label{eq:finiteflags}
 \mathbf A_{ie,jf}=0\ \text{ if } i\ge j+f,\qquad
 \mathbf B_{ie,jf}=0\ \text{ if } j\ge i+e .
\end{equation}
The function $\phi'_{j,f}$ is a trigonometric polynomial with frequencies in
$[-j-f,j+f]$. By \eqref{eq:gridextract},
$U(\phi_{i,e}(x)e^{2\pi\ii ny})=\sum_a\phi_{i,e}(a)T(\{n+a\})$. For $|n|\le j+f$
the mass $T(\{n\pm\alpha_l\})$ vanishes unless $l\le|n|\le j+f$, while
$\phi_{i,e}(\pm\alpha_l)=0$ for $l\le i$; hence all terms vanish if $i\ge j+f$.
The second assertion follows in the same way from the second identity in
\eqref{eq:gridextract}.

\emph{The gauge.} Let $X=m\circ x\circ m^{-1}$ be the matrix of multiplication by
$x$ in the coordinates $m$, and define $X'$ likewise. The function
$x\phi_{i,e}$ vanishes at $\pm\alpha_l$ for $l\le i$, and the functions on $\Xi$
with this property are spanned by the $\phi_{i',e'}$ with $i'\ge i$; hence
$X_{(i,e),(i',e')}=0$ for $i'<i$. Evaluating at $\pm\alpha_{i+1}$, where the
$\phi_{i',e'}$ with $i'>i$ vanish, and using $\phi_{i,0}(\pm\alpha_{i+1})=D_i(\alpha_{i+1})\ne0$
and $\phi_{i,1}(\pm\alpha_{i+1})=\pm\kappa_iD_i(\alpha_{i+1})$, we find that
$x\phi_{i,0}-(\alpha_{i+1}/\kappa_i)\phi_{i,1}$ and
$x\phi_{i,1}-\alpha_{i+1}\kappa_i\phi_{i,0}$ lie in the span of the $\phi_{i',e'}$ with
$i'>i$. Thus the diagonal block of $X$ at level $i$ is $\alpha_{i+1}Q$, with
$Q=\bigl(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\bigr)$. On the finite grid,
multiplication by $x^ny^n$ is represented by $X^n\otimes X'^n$, so
$\mathbf B=E\mathbf A$ with $E=\exp(-2\pi\ii X\otimes X')$. Consequently the block
$(E\mathbf A)_{(i,j)}=((E\mathbf A)_{ie,jf})_{e,f}$ involves only the blocks
$\mathbf A_{(i',j')}$ with $i'\ge i$ and $j'\ge j$, and $E$ acts from $(i,j)$ to
$(i,j)$ by
\[
 \exp(-2\pi\ii\alpha_{i+1}\alpha'_{j+1}Q\otimes Q)=\cos\theta_{ij}\,I-\ii\sin\theta_{ij}\,Q\otimes Q,
 \qquad\theta_{ij}=2\pi\alpha_{i+1}\alpha'_{j+1}\in(0,\tfrac\pi2).
\]

\emph{Conclusion.} Suppose $\mathbf A\ne0$, and choose a nonzero block
$\mathbf A_{(i,j)}$ with $i+j$ maximal. Then $i\le j$ by \eqref{eq:finiteflags},
and $(E\mathbf A)_{(i,j)}=(\cos\theta_{ij}I-\ii\sin\theta_{ij}Q\otimes Q)\mathbf A_{(i,j)}$,
since all other contributing blocks have a larger value of $i+j$. If $i<j$, all
four entries of $\mathbf B_{(i,j)}$ vanish by \eqref{eq:finiteflags}, and the
$4\times4$ matrix is invertible, so $\mathbf A_{(i,j)}=0$. If $i=j$, the entries
of $\mathbf A_{(i,i)}$ with $f=0$ vanish, and the entries of $\mathbf B_{(i,i)}$
with $e=0$ vanish; in terms of the remaining entries this reads
$-\ii\sin\theta_{ii}\,\mathbf A_{i1,i1}=0$ and $\cos\theta_{ii}\,\mathbf A_{i0,i1}=0$,
so again $\mathbf A_{(i,i)}=0$. This contradiction shows that $\mathbf A=0$; hence
$U=0$ and $T=0$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:rapid}(e)]
An element $w\in W_k$ is a combination of the $w_{a,b}$ with
$a,b\in\{\pm\beta_j:j\le k\}$, and the hypothesis of~(e) is that of
Lemma~\ref{lem:finitecap} with $r=k$ and $\alpha_j=\alpha'_j=\beta_j$.
\end{proof}

\subsection{Zak coordinates}\label{subsec:A-zak}
For $\Phi\in\Sc(\R^2)$ let $\widehat\Phi^{(2)}(x,n)=\int\Phi(x,y)e^{-2\pi\ii ny}\,dy$
and
\[
 \Zak^t\Phi(t)=\sum_{n\in\Z}(-1)^n\,\widehat\Phi^{(2)}\bigl(t-n-\tfrac12,n\bigr).
\]
This is a continuous linear map $\Sc(\R^2)\to\Sc(\R)$, and for $T\in\Sc'(\R)$ we
define $\Zak T\in\Sc'(\R^2)$ by $\Zak T(\Phi)=T(\Zak^t\Phi)$. For $T\in\Sc(\R)$,
$\Zak T(x,y)=\sum_n(-1)^nT(x+n+\frac12)e^{-2\pi\ii ny}$; thus $\Zak$ is the Zak
transform, recentred at the half-integers.

For a sequence $\eta=(\eta_j)_{j\ge1}$ with $0\le\eta_j\le10^{-12}$ and $\eta_j\to0$ we put
$\Xi_\eta=\{0\}\cup\{\pm\eta_j:j\ge1\}$. We fix an even function
$\chi\in C_c^\infty(-\frac14,\frac14)$ equal to $1$ on $[-10^{-6},10^{-6}]$ and,
for a distribution $S$ and $c\in\R$, we call $S_c(\psi)=S((\chi\psi)(\cdot-c))$
the \emph{cluster} of $S$ at $c$. If $S$ is locally atomic and its only atoms in
$c+\supp\chi$ lie in $c+\Xi_\eta$, then $S_c=\sum_{x\in\Xi_\eta}S(\{c+x\})\delta_x$.

\begin{lemma}\label{lem:zak}
Let $\eta,\eta'$ be as above, let $T\in\Sc'$ be supported in
$\frac12+\Xi_\eta+\Z$ with $\F T$ supported in $\frac12+\Xi_{\eta'}+\Z$, and put
\[
 V=e^{-\pi\ii y}\,\chi(x)\chi(y)\,\Zak T,\qquad g(x,y)=e^{-2\pi\ii xy}.
\]
Then $V$ is supported in $\Xi_\eta\times\Xi_{\eta'}$, and for all
$\psi\in C^\infty(\R)$ and $n\in\Z$,
\begin{align}
 V\bigl(\psi(x)\,e^{2\pi\ii(n+\frac12)y}\bigr)&=(-1)^n\,T_{n+\frac12}(\psi),\label{eq:zakphys}\\
 (gV)\bigl(e^{-2\pi\ii(n+\frac12)x}\,\psi(y)\bigr)&=\ii(-1)^n\,(\F T)_{n+\frac12}(\psi).\label{eq:zakfour}
\end{align}
Moreover, $T(\varphi)=V(G_\varphi)$ for $\varphi\in\Sc$, where
$G_\varphi(x,y)=\sum_n(-1)^n\varphi(x+n+\frac12)e^{2\pi\ii(n+\frac12)y}$. If
$T\in H_{-p}$ for an integer $p\ge0$, then
\begin{equation}\label{eq:zakorder}
 |V(q)|\le C_p\norm T_{-p}\norm q_{C^{2p}}\qquad(q\in C^\infty(\R^2)),
\end{equation}
where the $C^{2p}$ norm is taken on $\supp\chi\times\supp\chi$.
\end{lemma}

\begin{proof}
For $T\in\Sc$, Poisson summation applied to
$t\mapsto T(x+\frac12+t)e^{-2\pi\ii t(y+\frac12)}$ gives
\begin{equation}\label{eq:zakfourier}
 \Zak T(x,y)=e^{2\pi\ii(x+\frac12)(y+\frac12)}\,\Zak(\F T)(y,-x),
\end{equation}
and since both sides depend continuously on $T\in\Sc'$, \eqref{eq:zakfourier}
holds on $\Sc'$. Also $\Zak T(x,y+1)=\Zak T(x,y)$.

If $\Phi\in C_c^\infty(\R^2)$ vanishes on $N\times\R$, where $N$ is a $1$-periodic
open neighbourhood of $\Xi_\eta+\Z$, then $\Zak^t\Phi$ vanishes on $\frac12+N$,
which is a neighbourhood of $\supp T$; hence $\Zak T$ is supported in
$(\Xi_\eta+\Z)\times\R$. Applying this to $\F T$ and using \eqref{eq:zakfourier},
we see that $\Zak T$ is also supported in $\R\times(\Xi_{\eta'}+\Z)$. This gives the
support of $V$.

Let $\varrho\in C_c^\infty(-\frac34,\frac34)$ with $\varrho=1$ on $[-\frac14,\frac14]$
and $\sum_k\varrho(y+k)=1$. Then $\int\varrho(y)e^{-2\pi\ii my}\,dy=\delta_{m0}$ for
$m\in\Z$, so for $\psi\in C_c^\infty(\R)$ we have
$\Zak^t(\psi\otimes\varrho e^{2\pi\ii n\cdot})=(-1)^n\psi(\cdot-n-\frac12)$, that is,
$\Zak T(\psi\otimes\varrho e^{2\pi\ii n\cdot})=(-1)^nT(\psi(\cdot-n-\frac12))$. Since
$\chi=\varrho$ near $\Xi_{\eta'}+\Z$ and $\Zak T$ is supported in
$\R\times(\Xi_{\eta'}+\Z)$, we may replace $\varrho$ by $\chi$; with $\chi\psi$ in
place of $\psi$ this is \eqref{eq:zakphys}. For \eqref{eq:zakfour},
\eqref{eq:zakfourier} gives $gV=\ii\,e^{\pi\ii x}\chi(x)\chi(y)\,\Zak(\F T)(y,-x)$.
After the change of variables $(u,v)=(y,-x)$, and since $\chi$ is even, the left
side of \eqref{eq:zakfour} becomes
$\ii\,\Zak(\F T)\bigl((\chi\psi)(u)\,\chi(v)e^{2\pi\ii nv}\bigr)$, and the previous
argument, applied to $\F T$ (whose Fourier transform $\check T$ is supported in
$\frac12+\Xi_\eta+\Z$, because $\Xi_\eta$ is symmetric and $-\frac12\equiv\frac12$ modulo
$1$), gives \eqref{eq:zakfour}.

For $\varphi\in\Sc$ we have $T(\varphi)=\sum_nT(\chi(\cdot-n-\frac12)\varphi)$,
because $\sum_n\chi(\cdot-n-\frac12)=1$ near $\supp T$, and by \eqref{eq:zakphys}
the $n$-th term is $(-1)^nV\bigl(\varphi(x+n+\frac12)e^{2\pi\ii(n+\frac12)y}\bigr)$.
Summing gives $T(\varphi)=V(G_\varphi)$, since the series defining $G_\varphi$
converges in $C^\infty(\R^2)$.

Finally, $V(q)=T(\Zak^t\Phi)$ with $\Phi=e^{-\pi\ii y}\chi(x)\chi(y)q$. Since $\Phi$
is supported in $(-\frac14,\frac14)^2$, we have
$\Zak^t\Phi(t)=(-1)^n\widehat\Phi^{(2)}(t-n-\frac12,n)$ for $|t-n-\frac12|<\frac12$. For
$a+b\le2p$ we expand $(x+n+\frac12)^a=\sum_{l\le a}\binom aln^l(x+\frac12)^{a-l}$ and
use $n^l\widehat\Phi^{(2)}(x,n)=(2\pi\ii)^{-l}\widehat{\partial_y^l\Phi}{}^{(2)}(x,n)$
together with Parseval's identity in $y$; this gives
$\norm{t^a\partial^b\Zak^t\Phi}_{L^2}\le C\sum_{l\le a}\norm{\partial_y^l\partial_x^b\Phi}_{L^2}
\le C\norm q_{C^{2p}}$, and \eqref{eq:zakorder} follows from the norm equivalence.
\end{proof}

\subsection{Newton moments and the gauge}\label{subsec:A-moments}
Let $\eta$ be a nonincreasing sequence with $0\le\eta_j\le10^{-12}$ and $\eta_j\to0$;
zeros are allowed. Put
\begin{equation}\label{eq:newton}
\begin{gathered}
 z(t)=1-\cos2\pi t,\qquad a_j=z(\eta_j),\qquad D_i=\prod_{l\le i}(z-a_l),\\
 \phi_{i,0}=\cos(\pi t)\,D_i,\qquad\phi_{i,1}=\sin(\pi t)\,D_i\qquad(i\ge0).
\end{gathered}
\end{equation}
The function $\phi_{i,e}$ vanishes at $\pm\eta_l$ for $l\le i$. It is a
trigonometric polynomial with frequencies in $\{\pm\frac12,\dots,\pm(i+\frac12)\}$,
and its coefficients at the extreme frequencies $i+\frac12$ and $-(i+\frac12)$ are
$h_i$ and $h_i$ if $e=0$, and $-\ii h_i$ and $\ii h_i$ if $e=1$, where
$h_i=(-1)^i2^{-i-1}$. Since $D_i$ is a polynomial of degree exactly $i$ in
$\cos2\pi t$, the $\phi_{i,e}$ span the trigonometric polynomials with
half-integer frequencies.

Given two such sequences, $\eta$ for the variable $x$ and $\eta'$ for $y$, positive
weights $(\omega_i)$ and $(\omega'_j)$, and $\kappa>0$, we define, for a
distribution $U$ on $\R^2$ with compact support in $(-10^{-3},10^{-3})^2$, the
\emph{moment array}
\begin{equation}\label{eq:moments}
 M(U)_{ie,jf}=\frac{U\bigl(\phi_{i,e}(x)\,\phi'_{j,f}(y)\bigr)}{\omega_i\,\omega'_j\,\kappa^{e+f}}
 \qquad(i,j\ge0,\ e,f\in\{0,1\}),
\end{equation}
where $\phi'$ is built from $\eta'$. Write $I=\NN_0\times\{0,1\}$ and
$\mathbb H=\ell^2(I)\otimes\ell^2(I)=\ell^2(I\times I)$.

\begin{lemma}\label{lem:faithful}
If $U$ is a distribution with compact support in $(-\frac14,\frac14)^2$ such that
$U(\phi_{i,e}\otimes\phi'_{j,f})=0$ for all $i,j,e,f$, then $U=0$.
\end{lemma}

\begin{proof}
The products $\phi_{i,e}(x)\phi'_{j,f}(y)$ span the trigonometric polynomials in
$(x,y)$ with half-integer frequencies, so $e^{\pi\ii(x+y)}U$ annihilates every
$e^{2\pi\ii(mx+m'y)}$, $m,m'\in\Z$. Its periodization to $\R^2/\Z^2$ therefore
vanishes, and since its support lies inside a single period cell,
$e^{\pi\ii(x+y)}U=0$.
\end{proof}

\begin{lemma}[Flags]\label{lem:flags}
Let $\eps,\delta$ be nonincreasing sequences with values in $[0,10^{-12}]$ tending to
$0$, put
$\alpha_j=\frac12-\eps_j$ and $\alpha'_j=\frac12-\delta_j$, and let
$T\in\Ncal(L_{\alpha'},L_\alpha)$, where
$L_\alpha=\{n\pm\alpha_j:j\ge1,\ |n|\ge j\}$ as in Section~\ref{sec:blocks} (if
$\eps_j=0$, the corresponding points are half-integers). Let $V$ be as in
Lemma~\ref{lem:zak} with $\eta=\eps$ and $\eta'=\delta$, and let
$\mathbf A=M(V)$ and $\mathbf B=M(gV)$, built with $\eps$ in $x$ and $\delta$ in
$y$, for arbitrary weights and $\kappa$. Then
\begin{equation}\label{eq:flags}
 \mathbf A_{ie,jf}=0\ \ (i>j),\qquad\mathbf B_{ie,jf}=0\ \ (j>i),
\end{equation}
and, with $t_i=\tan(\pi\eps_{i+1})$ and $t'_i=\tan(\pi\delta_{i+1})$,
\begin{equation}\label{eq:diagrel}
 \mathbf A_{i1,i0}=\ii t_i\mathbf A_{i0,i1},\quad
 \mathbf A_{i1,i1}=-\frac{\ii t_i}{\kappa^2}\mathbf A_{i0,i0},\quad
 \mathbf B_{i0,i1}=-\ii t'_i\mathbf B_{i1,i0},\quad
 \mathbf B_{i1,i1}=\frac{\ii t'_i}{\kappa^2}\mathbf B_{i0,i0}.
\end{equation}
\end{lemma}

\begin{proof}
Since $n+\alpha_l=n+\frac12-\eps_l$ and $(n+1)-\alpha_l=n+\frac12+\eps_l$, the
cluster of $T$ at $n+\frac12$ is $\sum_{x\in\Xi_n}T(\{n+\frac12+x\})\delta_x$, where
\begin{equation}\label{eq:clusterset}
 \Xi_n=\{-\eps_l:\ l\le|n|\}\cup\{\eps_l:\ l\le|n+1|\};
\end{equation}
the clusters of $\F T$ are described in the same way, with $\delta$ in place of
$\eps$. Expanding $\phi'_{j,f}$ in the exponentials $e^{2\pi\ii(n+\frac12)y}$,
$-j-1\le n\le j$, and using \eqref{eq:zakphys}, we see that
$V(\phi_{i,e}\otimes\phi'_{j,f})$ is a combination of the numbers
$T_{n+\frac12}(\phi_{i,e})$ with $-j-1\le n\le j$. For these $n$ the set $\Xi_n$
contains only points $\pm\eps_l$ with $l\le j+1$, at which $\phi_{i,e}$ vanishes if
$i\ge j+1$. This proves the first part of \eqref{eq:flags}; the second follows in
the same way from \eqref{eq:zakfour}, by expanding $\phi_{i,e}(x)$ in the
exponentials $e^{-2\pi\ii(n+\frac12)x}$, $-i-1\le n\le i$.

Now let $i=j$. Then $\phi_{i,e}$ vanishes at every point of the sets $\Xi_n$ with
$-i-1\le n\le i$ except at $\eps_{i+1}\in\Xi_i$ and at $-\eps_{i+1}\in\Xi_{-i-1}$, and
these two clusters are extracted by the extreme frequencies $\pm(i+\frac12)$ of
$\phi'_{i,f}$. Put $X_+=(-1)^iT(\{i+\frac12+\eps_{i+1}\})$,
$X_-=(-1)^{i+1}T(\{-i-\frac12-\eps_{i+1}\})$, $c=\cos\pi\eps_{i+1}$,
$s=\sin\pi\eps_{i+1}$ and $D=D_i(\eps_{i+1})$. By \eqref{eq:zakphys} and the values of
the extreme coefficients,
\begin{gather*}
 V(\phi_{i,0}\otimes\phi'_{i,0})=h_icD(X_++X_-),\qquad
 V(\phi_{i,0}\otimes\phi'_{i,1})=-\ii h_icD(X_+-X_-),\\
 V(\phi_{i,1}\otimes\phi'_{i,0})=h_isD(X_+-X_-),\qquad
 V(\phi_{i,1}\otimes\phi'_{i,1})=-\ii h_isD(X_++X_-),
\end{gather*}
and the first two relations of \eqref{eq:diagrel} follow after normalization. For
$\mathbf B$ the roles of the variables are exchanged. By \eqref{eq:zakfour}, the
frequency $i+\frac12$ of $\phi_{i,e}(x)$ extracts the cluster of $\F T$ at
$-i-\frac12$, whose only point not annihilated by $\phi'_{i,f}$ is $-\delta_{i+1}$,
and the frequency $-(i+\frac12)$ extracts the cluster at $i+\frac12$, with the point
$\delta_{i+1}$. With $Y_+=\ii(-1)^i\,\F T(\{i+\frac12+\delta_{i+1}\})$ and
$Y_-=\ii(-1)^{i+1}\,\F T(\{-i-\frac12-\delta_{i+1}\})$, the same computation gives
$(gV)(\phi_{i,1}\otimes\phi'_{i,0})=\ii h_ic'D'(Y_+-Y_-)$,
$(gV)(\phi_{i,0}\otimes\phi'_{i,1})=h_is'D'(Y_+-Y_-)$,
$(gV)(\phi_{i,0}\otimes\phi'_{i,0})=h_ic'D'(Y_++Y_-)$ and
$(gV)(\phi_{i,1}\otimes\phi'_{i,1})=\ii h_is'D'(Y_++Y_-)$, where $c',s',D'$ are formed
with $\delta_{i+1}$. This gives the last two relations.
\end{proof}

\begin{lemma}[The gauge]\label{lem:gauge}
In the setting of \eqref{eq:moments}, suppose that $\omega_{i+1}=\lambda_{i+1}\omega_i$
with $0<\lambda_i\le10^{-3}$, and let $S$ be the backward shift on $\ell^2(\NN_0)$,
$(Su)_i=u_{i+1}$. Put
\[
 J=\operatorname{diag}(a_{i+1})+\operatorname{diag}(\lambda_{i+1})\,S,\qquad
 \mathsf s=\begin{pmatrix}0&\kappa(2-J)\\ J/\kappa&0\end{pmatrix}
\]
on $\ell^2(I)=\ell^2(\NN_0)\oplus\ell^2(\NN_0)$ \textup{(}parities $0$ and $1$\textup{)},
and define $J'$ and $\mathsf s'$ from $\eta'$ and $\lambda'$ in the same way. Assume
that $\norm{\mathsf s},\norm{\mathsf s'}\le\frac12$, and put
$X=\frac1{2\pi}\arcsin\mathsf s$, $X'=\frac1{2\pi}\arcsin\mathsf s'$ and
$E=\exp(-2\pi\ii X\otimes X')$. Let $K=[-10^{-2},10^{-2}]^2$, and let $U$ be a
distribution with compact support in $(-10^{-3},10^{-3})^2$ such that
$M(fU)\in\mathbb H$ and $\norm{M(fU)}_{\mathbb H}\le C\norm f_{C^N(K)}$ for all
$f\in C^\infty(\R^2)$, for some $C$ and $N$. Then $M(gU)=E\,M(U)$.
\end{lemma}

\begin{proof}
The identities
\[
 zD_i=D_{i+1}+a_{i+1}D_i,\qquad\sin(2\pi t)\cos(\pi t)=(2-z)\sin(\pi t),\qquad
 \sin(2\pi t)\sin(\pi t)=z\cos(\pi t)
\]
show that, entry by entry,
\[
 M(\sin(2\pi x)\,U)=(\mathsf s\otimes I)M(U),\qquad\text{hence}\qquad
 M(p(\sin2\pi x)\,U)=(p(\mathsf s)\otimes I)M(U)
\]
for every polynomial $p$; the entries on the right are finite sums. On $[-10^{-2},10^{-2}]$ we have
$2\pi x=\arcsin(\sin2\pi x)$, and the Taylor series of $\arcsin$ converges there in
every $C^N$ norm, since $|\sin2\pi x|<\frac1{10}$; it also converges in operator norm
at $\mathsf s$. By the assumed continuity of $f\mapsto M(fU)$ in $C^N(K)$,
$M(xU)=(X\otimes I)M(U)$; likewise $M(yU)=(I\otimes X')M(U)$, and the same holds
with $fU$ in place of $U$. The exponential series of $g$ converges in every
$C^N(K)$ norm, and that of $E$ in operator norm, so
\[
 M(gU)=\sum_n\frac{(-2\pi\ii)^n}{n!}(X\otimes X')^nM(U)=E\,M(U).
\]
\end{proof}

\emph{The kernel equation.} We use the following subspaces of $\mathbb H$,
described by the index pairs $((i,e),(j,f))$ they contain: $\mathcal H_A$, the
pairs with $i<j$ and the diagonal pairs ($i=j$) of parity $ef\in\{00,01\}$;
$\mathcal H_B$, the pairs with $i>j$ and the diagonal pairs of parity
$ef\in\{00,10\}$; $\mathcal D$ and $\mathcal G$, the diagonal pairs of parity $00$
and $11$; and $\mathcal C=\mathcal H_A\ominus\mathcal D$. Let
$P_A,P_B,P_{\mathcal D},P_{\mathcal C},P_{\mathcal G}$ be the orthogonal projections,
and $Q=I-P_B$, so that $Q\mathbb H=\mathcal C\oplus\mathcal G$. Given sequences
$(t_i)$ and $(t'_i)$, define the \emph{graph maps} $\Sigma_A$ and $\Sigma_B$ by
\[
 \Sigma_A:\ e_{i0,i0}\mapsto-\frac{\ii t_i}{\kappa^2}e_{i1,i1},\quad e_{i0,i1}\mapsto\ii t_ie_{i1,i0};
 \qquad
 \Sigma_B:\ e_{i0,i0}\mapsto\frac{\ii t'_i}{\kappa^2}e_{i1,i1},\quad e_{i1,i0}\mapsto-\ii t'_ie_{i0,i1},
\]
and zero on all other basis vectors. Then \eqref{eq:flags} and \eqref{eq:diagrel}
say exactly that $\mathbf A=(I+\Sigma_A)P_A\mathbf A$ and
$\mathbf B=(I+\Sigma_B)P_B\mathbf B$. If moreover $\mathbf B=E\mathbf A$, then,
applying $Q$ and using $QP_B=0$ and $Q\Sigma_B=\Sigma_B$, we get
$QE\mathbf A=\Sigma_BP_BE\mathbf A$, that is,
\begin{equation}\label{eq:kernelop}
 \mathcal Ta=0,\qquad
 \mathcal T=(Q-\Sigma_BP_B)\,E\,(I+\Sigma_A)\big|_{\mathcal H_A},\qquad a=P_A\mathbf A .
\end{equation}
We write $a=(u,c)$ with $u=P_{\mathcal D}a$, identified with the sequence
$(u_i)=(\mathbf A_{i0,i0})\in\ell^2(\NN_0)$, and $c=P_{\mathcal C}a$. The operator
$\mathcal T$ maps $\mathcal D\oplus\mathcal C$ to $\mathcal C\oplus\mathcal G$, and
$Q|_{\mathcal H_A}$ is the map $(u,c)\mapsto(c,0)$.

\emph{How the kernel equation is used.} The diagonal part $u\in\mathcal D$ of the
unknown is free, and its remaining part $c\in\mathcal C$ is eliminated. Since
$\mathcal T$ is a small perturbation of $Q|_{\mathcal H_A}$, the equation splits into
two components:
\[
\begin{array}{lll}
 \mathcal C\text{-component:} & \mathcal T_{\mathcal{CD}}u+\mathcal T_{\mathcal{CC}}c=0,
 \quad \mathcal T_{\mathcal{CC}}\approx I & \Longrightarrow\ c\text{ is determined by }u;\\[2pt]
 \mathcal G\text{-component:} & \mathcal T_{\mathcal{GD}}u+\mathcal T_{\mathcal{GC}}c=0
 & \Longrightarrow\ \text{one equation for }u\in\ell^2(\NN_0).
\end{array}
\]
The dimension of the space of solutions is therefore decided by the
$\mathcal G$-component alone. In Stage~1, with the weights $\omega_i=\vartheta^i$, it
becomes $\mathcal Ku=0$, where $\mathcal K$ is a perturbation of a multiple of the
backward shift, whose kernel is one-dimensional. In Stage~2, with weights that
shrink as fast as the clusters, its leading part is a positive diagonal operator,
and it forces $u=0$.

\subsection{Stage 1: at most one dimension}\label{subsec:A-stage1}
Let $\eps$ be a nonincreasing sequence with values in $[0,10^{-12}]$ tending to $0$, let
$\gamma^\eps_j=\frac12-\eps_j$, and consider
$L_{\gamma^\eps}=\{n\pm\gamma^\eps_j:\ j\ge1,\ |n|\ge j\}$. In this subsection
$\vartheta=2^{-12}$, and we use \eqref{eq:moments} with $\eta=\eta'=\eps$,
$\omega_i=\omega'_i=\vartheta^i$ and $\kappa=\vartheta^{1/2}$; thus
$\lambda_i=\vartheta$ in Lemma~\ref{lem:gauge}, and $t_i=t'_i=\tan(\pi\eps_{i+1})\le\vartheta^3$
and $a_i\le\vartheta^6$. For $T\in\M_{L_{\gamma^\eps}}$ we let $V$ be as in
Lemma~\ref{lem:zak}, $\mathbf A=M(V)$ and $\mathbf B=M(gV)$.

\begin{proposition}\label{prop:stage1}
In this setting $\mathbf A,\mathbf B\in\mathbb H$ and $\mathbf B=E\mathbf A$, and there
is a vector $\bar u=\bar u(\eps)\in\ell^2(\NN_0)$ with $\bar u_0=1$, depending only
on $\eps$, such that $u=(\mathbf A_{i0,i0})_{i\ge0}$ satisfies $u=u_0\,\bar u$ for
every $T\in\M_{L_{\gamma^\eps}}$. If $u_0=0$, then $T=0$; in particular
$\dim\M_{L_{\gamma^\eps}}\le1$. Finally $\norm{\bar u}\le\frac{11}{10}$ and
$\norm{\mathbf A}\le2|u_0|$.
\end{proposition}

\begin{proof}
\emph{Membership.} Let $\chi_0\in C_c^\infty(-10^{-3},10^{-3})$ equal $1$ near
$[-10^{-12},10^{-12}]$. For $|t|\le10^{-3}$ we have $z(t)\le2\pi^2t^2<\vartheta/4$, so
$|z(t)-a_l|<\vartheta/2$, and the Leibniz rule gives
$\norm{\chi_0\phi_{i,e}}_{C^N}\le C_N(1+i)^N(\vartheta/2)^i$. Since $V$ has compact
support, it has finite order $N$, and
$|M(fV)_{ie,jf}|\le C\norm f_{C^N}(1+i)^N(1+j)^N2^{-i-j}\kappa^{-2}$. Hence
$M(fV)\in\mathbb H$, continuously in $f\in C^N$, and Lemma~\ref{lem:gauge} gives
$\mathbf B=E\mathbf A$ (the hypothesis $\norm{\mathsf s}\le\frac12$ is verified
below). By Lemma~\ref{lem:flags} and \eqref{eq:kernelop}, $a=P_A\mathbf A$
satisfies $\mathcal Ta=0$, and $\mathbf A=(I+\Sigma_A)a$.

\emph{Estimates.} We have $\norm J\le\vartheta+\vartheta^6<2\vartheta$ and, since
$\vartheta/\kappa=\kappa$, $\norm{\mathsf s}\le\kappa(2+2\vartheta)<3\kappa$. As the
Taylor coefficients of $\arcsin$ are nonnegative,
$\norm X\le\frac1{2\pi}\arcsin(3\kappa)<\frac\kappa2$, so that
$Z_0=-2\pi\ii X\otimes X$ satisfies $\norm{Z_0}<\frac\pi2\kappa^2<2\vartheta$. Hence
\[
 \norm{E-I}<3\vartheta,\qquad\norm{E-I-Z_0}<4\vartheta^2,\qquad
 \norm{\Sigma_A},\norm{\Sigma_B}\le\vartheta^2 ,
\]
and, writing $\mathcal T-Q=Q(E-I)(I+\Sigma_A)+Q\Sigma_A-\Sigma_BP_BE(I+\Sigma_A)$ on
$\mathcal H_A$,
\begin{equation}\label{eq:TminusQ}
 \norm{\mathcal T-Q|_{\mathcal H_A}}\le3\vartheta(1+\vartheta^2)+\vartheta^2
 +\vartheta^2(1+3\vartheta)(1+\vartheta^2)<4\vartheta .
\end{equation}

\emph{The leading term.} Write $\mathcal T$ in block form with respect to
$\mathcal D\oplus\mathcal C\to\mathcal C\oplus\mathcal G$. By \eqref{eq:TminusQ},
$\mathcal T_{\mathcal{CC}}$ is invertible with $\norm{\mathcal T_{\mathcal{CC}}^{-1}}<2$,
and $\norm{\mathcal T_{\mathcal{CD}}},\norm{\mathcal T_{\mathcal{GC}}}<4\vartheta$. If
$\mathcal T(u,c)=0$, then $c=-\mathcal T_{\mathcal{CC}}^{-1}\mathcal T_{\mathcal{CD}}u$ and
$\mathcal Ku=0$, where
\[
 \mathcal K=\mathcal T_{\mathcal{GD}}-\mathcal T_{\mathcal{GC}}\mathcal T_{\mathcal{CC}}^{-1}\mathcal T_{\mathcal{CD}}
 :\ \ell^2(\NN_0)\to\ell^2(\NN_0),
\]
$\mathcal D$ and $\mathcal G$ being identified with $\ell^2(\NN_0)$ through the
diagonal index. For $u\in\mathcal D$ we have $(I+\Sigma_A)u=u+\Sigma_Au$ with
$\Sigma_Au\in\mathcal G$, and $P_{\mathcal G}u=0$, so
\[
 \mathcal T_{\mathcal{GD}}u=P_{\mathcal G}Z_0u+P_{\mathcal G}(E-I-Z_0)u+\Sigma_Au
 +P_{\mathcal G}(E-I)\Sigma_Au-P_{\mathcal G}\Sigma_BP_BE(I+\Sigma_A)u ,
\]
where the last four terms have total norm less than $7\vartheta^2\norm u$. In
$P_{\mathcal G}Z_0|_{\mathcal D}$ only the block $X_{10}$ of $X$ from parity $0$ to
parity $1$ enters: $(P_{\mathcal G}Z_0u)_i=-2\pi\ii\sum_l(X_{10})_{il}^2u_l$. Now
$\arcsin\mathsf s=\mathsf s\,h(\mathsf s^2)$ with
$h(w)=\sum_m\binom{2m}m\frac{w^m}{4^m(2m+1)}$, and $\mathsf s^2=\Psi\oplus\Psi$ with
$\Psi=J(2-J)$, so $X_{10}=Y\,h(\Psi)$ with $Y=J/(2\pi\kappa)$. From
$\norm\Psi<5\vartheta$ we get $\norm{h(\Psi)-I}<\vartheta$, and with $\norm Y<\kappa/3$
this gives $\norm{X_{10}-Y}<\kappa\vartheta/3$ and $\norm{X_{10}}<\kappa/2$. For the
entrywise squares $M\circ M$ of bounded matrices,
$\norm{M\circ M-N\circ N}\le\norm{M-N}\norm{M+N}$, so replacing $X_{10}$ by $Y$
costs less than $2\vartheta^2$; and the entrywise square of
$Y=(\vartheta S+\operatorname{diag}(a_{i+1}))/(2\pi\kappa)$ is
$(\vartheta^2S+\operatorname{diag}(a_{i+1}^2))/(4\pi^2\vartheta)$. Altogether
\begin{equation}\label{eq:schur}
 \mathcal K=c_0S+\mathcal E,\qquad c_0=-\frac{\ii\vartheta}{2\pi},\qquad
 \norm{\mathcal E}<3\vartheta^2+7\vartheta^2+2\,(4\vartheta)^2=42\vartheta^2,
\end{equation}
so that $\norm{\mathcal E}/|c_0|<270\vartheta<\frac1{10}$. The leading term of
$\mathcal K$ is thus a multiple of the backward shift.

\emph{The kernel.} Let $S^*$ be the forward shift, so that $SS^*=I$. Then
$\mathcal KS^*=c_0I+\mathcal ES^*$ is invertible. If $\mathcal Ku=0$, write
$u=u_0e_0+S^*Su$; then $u_0\mathcal Ke_0+\mathcal KS^*(Su)=0$, so
$Su=-u_0(\mathcal KS^*)^{-1}\mathcal Ke_0$ and $u=u_0\bar u$ with
\begin{equation}\label{eq:ubar}
 \bar u=e_0-S^*(\mathcal KS^*)^{-1}\mathcal Ke_0 .
\end{equation}
Since $Se_0=0$, $\norm{\mathcal Ke_0}=\norm{\mathcal Ee_0}$, and therefore
$\norm{\bar u-e_0}\le\norm{\mathcal E}/(|c_0|-\norm{\mathcal E})<\frac1{10}$. If
$u_0=0$, then $u=0$, hence $c=0$, $a=0$ and $\mathbf A=0$, so that $V=0$ by
Lemma~\ref{lem:faithful}; by \eqref{eq:zakphys} all clusters of $T$ vanish, and
$T=0$ because $T$ is supported in $\frac12+\Xi_\eps+\Z$. Finally,
$\norm c\le8\vartheta\norm u$, so
$\norm{\mathbf A}\le(1+\vartheta^2)(1+8\vartheta)\frac{11}{10}|u_0|<2|u_0|$.
\end{proof}

\begin{lemma}\label{lem:stage1shift}
In the setting of Proposition~\ref{prop:stage1}:
\begin{enumerate}[label=\textup{(\roman*)}]
\item there is an absolute constant $C_*$ with
$\norm{\bar u(\eps)-e_0}\le C_*\sup_j\eps_j$;
\item for $h\ge0$ let $\eps^{[h]}=(\eps_{h+1},\eps_{h+2},\dots)$; then
$(u_{h+i})_{i\ge0}=u_h\,\bar u(\eps^{[h]})$, and the restriction of $\mathbf A$ to
the index pairs with $\max(i,j)\ge h$ has norm at most $2|u_h|$.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) For the zero sequence, $J=\vartheta S$ annihilates $e_0$, so $X$ annihilates
$e_{(0,0)}$, $E$ fixes $e_{00,00}$, and $\Sigma_A=\Sigma_B=0$; hence
$\mathcal Te_{00,00}=Qe_{00,00}=0$, and $\mathcal K(0)e_0=0$. In general
$\norm{J(\eps)-J(0)}\le2\pi^2\sup_j\eps_j^2$ and
$\norm{\Sigma_A},\norm{\Sigma_B}\le4\sup_j\eps_j/\vartheta$. The operators $X$,
$E$, $\mathcal T$, $\mathcal T_{\mathcal{CC}}^{-1}$ and $\mathcal K$ are obtained from
$J$, $\Sigma_A$ and $\Sigma_B$ by norm-convergent power series, products and
inverses whose arguments stay in fixed balls, so
$\norm{\mathcal K(\eps)-\mathcal K(0)}\le C\sup_j\eps_j$ with an absolute constant $C$.
By \eqref{eq:ubar}, $\norm{\bar u(\eps)-e_0}\le\norm{(\mathcal KS^*)^{-1}}
\norm{(\mathcal K(\eps)-\mathcal K(0))e_0}\le C_*\sup_j\eps_j$.

(ii) The operators $J$ and $\mathsf s$, and hence $X$, $E$, $\Sigma_A$, $\Sigma_B$ and
$\mathcal T$, have the following property: each entry of the image involves only
entries at indices that are at least as large, coordinatewise. Consequently the
subarray $(\mathbf A_{(h+i)e,(h+j)f})_{i,j\ge0}$ satisfies \eqref{eq:flags},
\eqref{eq:diagrel} and $\mathbf B=E\mathbf A$ for the sequence $\eps^{[h]}$, with the
same weights, and its diagonal of parity $00$ is $(u_{h+i})_{i\ge0}$. The proof of
Proposition~\ref{prop:stage1} used only these relations, so
$(u_{h+i})=u_h\bar u(\eps^{[h]})$. For the second assertion, let $U_h$ be the set of
index pairs with $\max(i,j)\ge h$. By the same property, the restriction to $U_h$
of the equation $\mathcal T_{\mathcal{CC}}c=-\mathcal T_{\mathcal{CD}}u$ involves only
$c|_{U_h}$ and $(u_l)_{l\ge h}$, and the compression of $\mathcal T_{\mathcal{CC}}$ to
$U_h$ is within $4\vartheta$ of the identity. Hence
$\norm{c|_{U_h}}\le8\vartheta\norm{(u_l)_{l\ge h}}\le9\vartheta|u_h|$, and
$\norm{\mathbf A|_{U_h}}\le(1+\vartheta^2)(\frac{11}{10}+9\vartheta)|u_h|<2|u_h|$.
\end{proof}

\subsection{Stage 2: exclusion at low Hermite order}\label{subsec:A-stage2}
In this subsection $\rho=10^{-12}$ and $\kappa=\rho^{1/4}$. Let $\eps$ and $\delta$
be nonincreasing sequences of positive numbers tending to $0$, and let the weights
be
\[
 \omega_i=\prod_{l\le i}\lambda_l,\qquad\omega'_j=\prod_{l\le j}\lambda'_l,
 \qquad\text{where}\quad\eps_l\le\lambda_l\le\min\{\rho,64\eps_l\},\quad
 \delta_l\le\lambda'_l\le\min\{\rho,64\delta_l\}.
\]
These weights shrink as fast as the clusters do.

\begin{proposition}[Weighted contraction]\label{prop:contraction}
Let $\mathbf A\in\mathbb H$ satisfy \eqref{eq:flags} and \eqref{eq:diagrel}, with
$t_i=\tan\pi\eps_{i+1}$ and $t'_i=\tan\pi\delta_{i+1}$, and suppose that
$\mathbf B=E\mathbf A$ satisfies them as well, where $E$ is the operator of
Lemma~\ref{lem:gauge} for the present weights. Then $\mathbf A=0$.
\end{proposition}

\begin{proof}
Here $J=D_\lambda J_1$ with $D_\lambda=\operatorname{diag}(\lambda_{i+1})$ and
$J_1=S+\operatorname{diag}(a_{i+1}/\lambda_{i+1})$; since $a_l\le2\pi^2\eps_l^2$ and
$\lambda_l\ge\eps_l$, we have $\norm{J_1}<2$ and $\norm J<2\rho$. Hence
$\norm{\mathsf s}<3\kappa$, the operator $\Psi=J(2-J)$ has norm less than $5\rho$,
and
\[
 X=\frac1{2\pi}\,\mathsf s\,h(\mathsf s^2)
 =\begin{pmatrix}0&\kappa U_1\\ D_\lambda V_1/\kappa&0\end{pmatrix},\qquad
 X^2=\begin{pmatrix}\Upsilon&0\\0&\Upsilon\end{pmatrix},\qquad\Upsilon=D_\lambda W_1,
\]
with $U_1=(2-J)h(\Psi)/2\pi$, $V_1=J_1h(\Psi)/2\pi$ and
$W_1=J_1(2-J)h(\Psi)^2/4\pi^2$, all of norm less than $1$; in particular
$\norm X\le\kappa$ and $\norm\Upsilon\le\rho$. The same holds for $X'$, with
$\lambda'$. Consequently $\norm{E-I}\le e^{2\pi\kappa^2}-1<7\sqrt\rho$, and
$\norm{\Sigma_A},\norm{\Sigma_B}\le\tan(\pi\rho)/\kappa^2<4\sqrt\rho$. As in
\eqref{eq:TminusQ},
\[
 \norm{\mathcal T-Q|_{\mathcal H_A}}\le d_0:=16\sqrt\rho .
\]

The key point is that the diagonal entries of parity $11$ of $(E-I)\mathbf A$
carry the factor $\lambda_{i+1}\lambda'_{i+1}$:
\begin{equation}\label{eq:rowfactor}
 P_{\mathcal G}(E-I)=\kappa^{-2}D_{\lambda\lambda'}Z,\qquad\norm Z<8,
\end{equation}
where $D_{\lambda\lambda'}=\operatorname{diag}(\lambda_{i+1}\lambda'_{i+1})$. Indeed, in
$(X\otimes X')^n$ an output of parity $11$ arises, in the first factor, through
the block $D_\lambda V_1\Upsilon^q/\kappa$ (from parity $0$) if $n=2q+1$, and
through $\Upsilon^q=D_\lambda W_1\Upsilon^{q-1}$ (from parity $1$) if $n=2q\ge2$, and
similarly in the second factor. In both cases the diagonal factors $D_\lambda$ and
$D_{\lambda'}$ appear on the left, and summing the exponential series gives
\eqref{eq:rowfactor} with
$\norm Z\le\sum_{q\ge0}\frac{(2\pi)^{2q+1}\rho^{2q}}{(2q+1)!}
+\kappa^2\sum_{q\ge1}\frac{(2\pi)^{2q}\rho^{2q-2}}{(2q)!}<8$.

Now let $a=(u,c)$ satisfy $\mathcal Ta=0$, as in \eqref{eq:kernelop}. The
$\mathcal C$-component of this equation gives $\norm c\le d_0(1-d_0)^{-1}\norm u$.
For the $\mathcal G$-component, note that $P_{\mathcal G}a=0$,
$P_{\mathcal G}\Sigma_A=-\ii\operatorname{diag}(t_i/\kappa^2)P_{\mathcal D}$,
$P_{\mathcal D}\Sigma_A=0$ and
$P_{\mathcal G}\Sigma_BP_B=\ii\operatorname{diag}(t'_i/\kappa^2)P_{\mathcal D}$. Writing
$E=I+(E-I)$ we obtain the exact identity
\[
 P_{\mathcal G}\mathcal T=-\ii D_0P_{\mathcal D}+\mathcal R,\qquad
 D_0=\operatorname{diag}\Bigl(\frac{t_i+t'_i}{\kappa^2}\Bigr),
\]
\[
 \mathcal R=P_{\mathcal G}(E-I)(I+\Sigma_A)
 -\ii\operatorname{diag}\bigl(t'_i/\kappa^2\bigr)P_{\mathcal D}(E-I)(I+\Sigma_A).
\]
The diagonal operator $D_0$ has positive entries but an unbounded inverse.
However, $t_i\ge\pi\eps_{i+1}\ge\pi\lambda_{i+1}/64$, similarly for $t'_i$, and
$\lambda\lambda'/(\lambda+\lambda')\le\rho/2$, so
\[
 \frac{\lambda_{i+1}\lambda'_{i+1}}{t_i+t'_i}\le\frac{64\rho}{2\pi}<11\rho ,
\]
and by \eqref{eq:rowfactor} the composition $D_0^{-1}\mathcal R$ is bounded, with
\[
 \norm{D_0^{-1}\mathcal R}\le(88\rho+7\sqrt\rho)(1+4\sqrt\rho)<8\sqrt\rho=:e_0 .
\]
The equation $P_{\mathcal G}\mathcal Ta=0$ reads $\ii D_0u=\mathcal Ra$. Dividing
coordinatewise, $u=-\ii D_0^{-1}\mathcal Ra$, so that
$\norm u\le e_0(\norm u+\norm c)\le e_0(1-d_0)^{-1}\norm u$. Since $e_0<1-d_0$, we
get $u=0$, then $c=0$, $a=0$ and $\mathbf A=(I+\Sigma_A)a=0$.
\end{proof}

\begin{lemma}\label{lem:shrinkmember}
Let $\eps$ be as in \eqref{eq:blockparams}, let $m\ge0$ be an integer with
$3m<b/(b-1)$, let $T\in\M_{L_\gamma}$, and let $V$ be as in Lemma~\ref{lem:zak}
with $\eta=\eta'=\eps$. Suppose that $|V(q)|\le C\norm q_{C^m}$ for all
$q\in C^\infty(\R^2)$. Then, for the weights $\lambda=\lambda'=\eps$, $M(fV)\in\mathbb H$
and $\norm{M(fV)}_{\mathbb H}\le C'\norm f_{C^m}$ for all $f\in C^\infty(\R^2)$.
\end{lemma}

\begin{proof}
We first show that $V$ has no transversal derivatives along the isolated lines
$x=\pm\eps_l$. Let $x_0=\pm\eps_l$ and let $\psi\in C_c^\infty$ equal $1$ near $x_0$,
with support meeting $\Xi_\eps$ only at $x_0$. The distribution $\psi(x)V$ has
compact support in the line $\{x_0\}\times\R$, so
$\psi(x)V=\sum_{r\le N}\delta_{x_0}^{(r)}\otimes S_r$ for some compactly supported
distributions $S_r$ on $\R$ \cite[Section~2.3]{Hor03}. By \eqref{eq:zakphys}, the
value of this distribution at $\varphi(x)e^{2\pi\ii(n+\frac12)y}$ depends on $\varphi$
only through $\varphi(x_0)$, because the clusters of $T$ are combinations of point
masses. Hence $S_r(e^{2\pi\ii(n+\frac12)\cdot})=0$ for all $n$ and $r\ge1$, and since
$\supp S_r\subset(-\frac14,\frac14)$, this forces $S_r=0$ for $r\ge1$. Thus
$\psi(x)V=\delta_{x_0}\otimes S_0$, and $D_i(x)\psi(x)V=0$ if $D_i(x_0)=0$. The same
argument, applied to $gV$ with \eqref{eq:zakfour}, handles the lines $y=\pm\eps_l$.
Consequently $D_i(x)D_j(y)V$ is supported in $\Xi^{>i}\times\Xi^{>j}$, where
$\Xi^{>i}=\{0\}\cup\{\pm\eps_l:l>i\}\subset[-\eps_{i+1},\eps_{i+1}]$.

For $i\ge1$ let $\chi_i\in C_c^\infty$ equal $1$ near $[-\eps_{i+1},\eps_{i+1}]$ and
be supported in $|t|<\frac12(\eps_i+\eps_{i+1})$, with
$\norm{\chi_i^{(r)}}_\infty\le C_r(\eps_i-\eps_{i+1})^{-r}$, and let $\chi_0$ be as in
Stage~1. Then
$(fV)(\phi_{i,e}\otimes\phi_{j,f})=V\bigl(f\cdot(\chi_i\phi_{i,e}\otimes\chi_j\phi_{j,f})\bigr)$.
On the support of $\chi_i$ we have $z(t)\le a_i$, so $|z(t)-a_l|\le a_l\le2\pi^2\eps_l^2$
for $l\le i$, and differentiating one of the factors of $D_i$ costs at most a
factor $C\eps_i^{-2}$. Hence
\[
 \norm{\chi_i\phi_{i,e}}_{C^m}\le C_m(1+i)^m(\eps_i-\eps_{i+1})^{-m}\eps_i^{-2m}\,
 20^i\prod_{l\le i}\eps_l^2 ,
\]
and $|M(fV)_{ie,jf}|\le C\norm f_{C^m}\kappa^{-2}\beta_i\beta_j$ with
$\beta_i=(1+i)^m(\eps_i-\eps_{i+1})^{-m}\eps_i^{-2m}20^i\prod_{l\le i}\eps_l$. Since
$\eps_i-\eps_{i+1}\ge\eps_i/2$ and
$\sum_{l\le i}\log\eps_l=-d(b^{i+1}-b)/(b-1)-i\log16$,
\[
 \log\beta_i\le-d\Bigl(\frac b{b-1}-3m\Bigr)b^i+O(1+i),
\]
so that $(\beta_i)\in\ell^2$ when $3m<b/(b-1)$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:rapid}(b)]
Let $1\le p\le P$ and $T\in\M_{L_\gamma}\cap H_{-p}$. By \eqref{eq:zakorder}, $V$
satisfies the hypothesis of Lemma~\ref{lem:shrinkmember} with $m=2p$, and
$3m=6p<6P+1=b/(b-1)$. Hence the moment arrays for the weights
$\lambda=\lambda'=\eps$ lie in $\mathbb H$, continuously in the multiplier, and
Lemma~\ref{lem:gauge} gives $\mathbf B=E\mathbf A$; by Lemma~\ref{lem:flags} the
hypotheses of Proposition~\ref{prop:contraction} hold, with $\delta=\eps$ (note that
$\eps_1\le\rho$). Thus $\mathbf A=0$, so $V=0$ by Lemma~\ref{lem:faithful}, and $T=0$
by \eqref{eq:zakphys}.
\end{proof}

\subsection{Stage 3: existence and admission}\label{subsec:A-stage3}
For $N\ge0$ let $\eps^{(N)}=(\eps_1,\dots,\eps_{N},0,0,\dots)$, and let
$L^{(N)}=L_{\gamma^{\eps^{(N)}}}$ in the notation of Section~\ref{subsec:A-stage1}:
\[
 L^{(N)}=\{n\pm\gamma_j:\ j\le N,\ |n|\ge j\}\cup\{n+\tfrac12:\ n\ge N\text{ or }n\le-N-1\}.
\]
Thus $L^{(N)}$ consists of the points of $L_\gamma$ with phase index at most $N$,
together with the half-integers outside $[-N+\frac12,N-\frac12]$.

\begin{lemma}\label{lem:endpoint}
$\M_{L^{(N)}}$ is spanned by a crystalline measure $T_{N}$ with $\F T_{N}=-\ii T_{N}$
and $\check T_{N}=-T_{N}$, normalized so that its moment array in the weights of
Section~\ref{subsec:A-stage1} has $u_0=\mathbf A_{00,00}=1$.
\end{lemma}

\begin{proof}
For $N=0$ the set $L^{(0)}$ consists of the half-integers, $\M_{L^{(0)}}=\C w_{1/2,1/2}$
by Lemma~\ref{lem:finitephase}, and the assertions below about $w_{1/2,1/2}$ apply
directly; so let $N\ge1$. Let $\mathcal A=\{\pm\gamma_j:j\le N\}\cup\{\frac12\}$, a set of $2N+1$ numbers that
are pairwise incongruent modulo $1$. By Lemma~\ref{lem:finitephase}, every
$T\in\M_{L^{(N)}}$ is a combination $\sum_{a,b\in\mathcal A}c_{a,b}w_{a,b}$, and such
a combination belongs to $\M_{L^{(N)}}$ if and only if neither it nor its Fourier
transform has mass at the $2N^2+2N$ points of $(\mathcal A+\Z)\setminus L^{(N)}$.
These are $4N^2+4N<(2N+1)^2$ linear conditions, so $\M_{L^{(N)}}\ne\{0\}$.

Let $T=\sum c_{a,b}w_{a,b}\in\M_{L^{(N)}}$ with $c_{1/2,1/2}=0$. The masses of $T$ at
the half-integers $n+\frac12$ are $\sum_{b\ne1/2}c_{1/2,b}e^{2\pi\ii nb}$, a
combination of $2N$ distinct characters which vanishes for the $2N$ consecutive
values $-N\le n\le N-1$; hence $c_{1/2,b}=0$ for all $b$. By \eqref{eq:Fwab} the
masses of $\F T$ at the points $n+\frac12$ are
$\sum_{a\ne1/2}c_{a,1/2}e^{-2\pi\ii a(n+1/2)}$, and the same argument gives
$c_{a,1/2}=0$. Then $T$ satisfies the hypotheses of Lemma~\ref{lem:finitecap} with
$r=N$ and $\alpha_j=\alpha'_j=\gamma_j$, so $T=0$. Thus $T\mapsto c_{1/2,1/2}$ is
injective on $\M_{L^{(N)}}$, and $\dim\M_{L^{(N)}}=1$.

Since $L^{(N)}$ is symmetric, $\M_{L^{(N)}}$ is invariant under $\F$ and under the
reflection $T\mapsto\check T$. By \eqref{eq:Fwab}, $\F w_{1/2,1/2}=-\ii w_{1/2,1/2}$,
and $\check w_{1/2,1/2}=w_{-1/2,-1/2}=-w_{1/2,1/2}$, while neither operation maps any
other $w_{a,b}$, $a,b\in\mathcal A$, to a multiple of $w_{1/2,1/2}$; comparing the
coefficients of $w_{1/2,1/2}$ gives $\F T=-\ii T$ and $\check T=-T$ for
$T\in\M_{L^{(N)}}$. Finally, Proposition~\ref{prop:stage1} applies to $\eps^{(N)}$, so
a nonzero element of $\M_{L^{(N)}}$ has $u_0\ne0$, and we may normalize $u_0=1$.
\end{proof}

\begin{lemma}\label{lem:momentdecay}
Let $V_{N}$ be the distribution of Lemma~\ref{lem:zak} for $T_{N}$, and let the
$\phi_{i,e}$ be built from $\eps^{(N)}$. Then, for all $N$, $i,j\ge0$ and $e,f$,
\[
 |V_{N}(\phi_{i,e}\otimes\phi_{j,f})|\le2\,\vartheta^{i+j}C_*^{m}\prod_{l\le m}\eps_l,
 \qquad m=\max(i,j),
\]
where $C_*$ is the constant of Lemma~\ref{lem:stage1shift}.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:stage1shift}, applied to $\eps^{(N)}$,
$|u_{h+1}|=|u_h|\,|\bar u(\eps^{(N),[h]})_1|\le C_*\eps^{(N)}_{h+1}|u_h|$, so that
$|u_h|\le C_*^h\prod_{l\le h}\eps_l$, and every entry of $\mathbf A$ with
$\max(i,j)=m$ is bounded by $2|u_m|$. Undoing the normalization
\eqref{eq:moments}, with $\kappa\le1$, gives the claim.
\end{proof}

To convert these moment bounds into bounds for $V_{N}(f)$ with $f$ smooth, we
interpolate $f$ on the grid that supports $V_{N}$. We use $C^{r,r}$ norms, which
control the partial derivatives $\partial_x^\alpha\partial_y^{\alpha'}$ with
$\alpha,\alpha'\le r$.

\begin{lemma}\label{lem:divdiff}
Let $t_1>t_2>\dots>t_n\ge0$ with $t_{l+1}\le\frac12t_l$, let $L\ge1$, and let
$G\in C^L([0,t_1])$. Then, for all $s$ and $i$ with $s+i\le n$, the divided
differences of $G$ satisfy
\[
 |G[t_s,\dots,t_{s+i}]|\le\frac{4^i\norm G_{C^L}}{\prod_{l=s}^{s+i-L-1}t_l},
\]
an empty product being $1$.
\end{lemma}

\begin{proof}
For $i\le L$ this is the mean value bound $|G[t_s,\dots,t_{s+i}]|\le\norm{G^{(i)}}_\infty/i!$.
For $i>L$ we use
$G[t_s,\dots,t_{s+i}]=(G[t_s,\dots,t_{s+i-1}]-G[t_{s+1},\dots,t_{s+i}])/(t_s-t_{s+i})$
and $t_s-t_{s+i}\ge\frac12t_s$, and argue by induction on $i$. The second term
contributes at most $2\cdot4^{i-1}\norm G_{C^L}/\prod_{l=s}^{s+i-L-1}t_l$, and so
does the first, because its bound
$2\cdot4^{i-1}\norm G_{C^L}/(t_s\prod_{l=s}^{s+i-L-2}t_l)$ is at most this, as
$t_s\ge t_{s+i-L-1}$.
\end{proof}

\begin{lemma}\label{lem:parity}
Let $L\ge1$ and $c=\frac1{100}$. Every $f\in C^{2L+1,2L+1}([-c,c]^2)$ can be written
on $[-c,c]^2$ as $f(x,y)=\sum_{e,e'\in\{0,1\}}s_e(x)s_{e'}(y)G_{ee'}(z(x),z(y))$,
with $s_0=\cos\pi\cdot$, $s_1=\sin\pi\cdot$ and
$\norm{G_{ee'}}_{C^{L,L}([0,z(c)]^2)}\le C_L\norm f_{C^{2L+1,2L+1}}$.
\end{lemma}

\begin{proof}
In one variable, write $f=f_0+f_1$ with $f_0$ even and $f_1$ odd. Then $f_0/s_0$ is
even and of class $C^{2L+1}$, and $f_1/s_1=(f_1(x)/x)(x/\sin\pi x)$ is even and of
class $C^{2L}$, since $f_1(x)/x=\int_0^1f_1'(\sigma x)\,d\sigma$. An even function
$v\in C^{2L}([-c,c])$ is of the form $v(x)=W(x^2)$ with
$\norm W_{C^L([0,c^2])}\le C\norm v_{C^{2L}}$, as one sees by induction from
$W'(x^2)=v'(x)/2x=\frac12\int_0^1v''(\sigma x)\,d\sigma$. Finally $z(x)=2\sin^2\pi x$ is
a function of $x^2$ with nonvanishing derivative at $0$, so it is a smooth
change of variables. In two variables we apply these operations in each variable
separately; they are linear and commute with differentiation in the other
variable.
\end{proof}

\begin{proposition}\label{prop:uniform}
Let $L\ge1$ be an integer with $b^L>4$. Then $\sup_{N}\norm{T_{N}}_{-(2L+2)}<\infty$.
\end{proposition}

\begin{proof}
Let $f\in C^\infty(\R^2)$, and write $f=\sum s_e(x)s_{e'}(y)G_{ee'}(z(x),z(y))$ as in
Lemma~\ref{lem:parity}. The distribution $V_{N}$ is a finite combination of point
masses at the points of $\Xi_{\eps^{(N)}}\times\Xi_{\eps^{(N)}}$: indeed $T_{N}$ is a
combination of combs $w_{a,b}$, and Poisson summation in $y$ shows that $\Zak w_{a,b}$
is a sum of point masses on $(a-\frac12+\Z)\times(b+\frac12+\Z)$. The
$z$-coordinates are $a_1>\dots>a_{N}>a_{N+1}=0$, and $a_{l+1}\le\frac12a_l$ by
\eqref{eq:smalltail}. On these points $G_{ee'}$ coincides with its tensor Newton
interpolant $\sum_{i,j\le N}N^{ee'}_{ij}D_i(x)D_j(y)$, where $N^{ee'}_{ij}$ is the
divided difference of $G_{ee'}$ at the nodes $a_1,\dots,a_{i+1}$ in the first
variable and $a_1,\dots,a_{j+1}$ in the second; and $s_eD_i=\phi_{i,e}$. Applying
Lemma~\ref{lem:divdiff} in each variable, and Lemma~\ref{lem:parity},
\[
 |N^{ee'}_{ij}|\le\frac{C\,16^{\max(i,j)}\norm f_{C^{2L+1,2L+1}}}{\Pi_{(i-L)_+}\,\Pi_{(j-L)_+}},
 \qquad\Pi_q=\prod_{l\le q}a_l .
\]
Since $a_l<1$, $\Pi_q$ decreases with $q$; with $m=\max(i,j)$ and
Lemma~\ref{lem:momentdecay} we obtain
\[
 |V_{N}(f)|\le\sum_{e,e'}\sum_{i,j}|N^{ee'}_{ij}|\,|V_{N}(\phi_{i,e}\otimes\phi_{j,e'})|
 \le C\norm f_{C^{2L+1,2L+1}}\sum_{m\ge0}(2m+1)(16C_*)^m
 \frac{\prod_{l\le m}\eps_l}{\Pi_{(m-L)_+}^2}.
\]
Since $a_l\ge8\eps_l^2$, the logarithm of the $m$-th term is at most
$\sum_{l\le m}\log\eps_l-4\sum_{l\le m-L}\log\eps_l+O(1+m)
=-\frac{d\,b^{m+1}}{b-1}(1-4b^{-L})+O(1+m)$, and the series converges because
$b^L>4$. Thus $|V_{N}(f)|\le C\norm f_{C^{2L+1,2L+1}([-c,c]^2)}$, uniformly in $N$.

Now let $\varphi\in\Sc$ and $r=2L+1$. By Lemma~\ref{lem:zak},
$T_{N}(\varphi)=V_{N}(G_\varphi)$, and for $\alpha,\alpha'\le r$,
\[
 |\partial_x^\alpha\partial_y^{\alpha'}G_\varphi(x,y)|\le\sum_n\bigl(2\pi(|n|+1)\bigr)^{\alpha'}
 \sup_{|x|\le c}|\varphi^{(\alpha)}(x+n+\tfrac12)|
 \le C\sum_{\sigma\le1}\norm{(1+|x|)^{\alpha'+1}\varphi^{(\alpha+\sigma)}}_{L^2},
\]
by the Sobolev inequality on unit intervals and the Cauchy--Schwarz inequality.
Since $(\alpha'+1)+(\alpha+1)\le2(r+1)$, the norm equivalence bounds the right side
by $C\norm\varphi_{r+1}$. Hence $|T_{N}(\varphi)|\le C\norm\varphi_{2L+2}$ for all $N$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:rapid}(a) and~(c)]
Let $L=9P$. Then $b^L=(1+\frac1{6P})^{9P}\ge(\frac76)^9>4$, by the monotonicity of
$(1+1/x)^x$. By Proposition~\ref{prop:uniform}, a subsequence of $(T_{N})$
converges weakly in $H_{-(2L+2)}$ to some $T$; hence $T_{N}\to T$ and
$\F T_{N}\to\F T$ in $\Sc'$ along the subsequence. Let $I$ be a bounded open
interval. For large $N$, $L^{(N)}\cap I=L_\gamma\cap I$, because the half-integers in
$L^{(N)}$ have modulus at least $N+\frac12$ and the points of $L_\gamma$ in $I$ have
bounded phase index. The distributions on $I$ that are combinations of the point
masses at the finitely many points of $L_\gamma\cap I$ form a finite-dimensional,
hence closed, subspace. Therefore $T$ and $\F T$ are locally atomic on $L_\gamma$,
and $T\in\M_{L_\gamma}$. By Lemma~\ref{lem:zak},
$1=u_0=V_{N}(\cos\pi x\cos\pi y)=T_{N}(\Phi_0)$ with
$\Phi_0=\Zak^t\bigl(e^{-\pi\ii y}\chi(x)\chi(y)\cos\pi x\cos\pi y\bigr)\in\Sc$, so
$T(\Phi_0)=1$ and $T\ne0$; moreover $\F T=-\ii T$ and $\check T=-T$, by
Lemma~\ref{lem:endpoint}. By Proposition~\ref{prop:stage1}, applied to $\eps$
itself, $\dim\M_{L_\gamma}\le1$. Hence $\M_{L_\gamma}=\C T\subset H_{-(18P+2)}$,
which proves~(a) and~(c).
\end{proof}

\subsection{Stage 4: the head}\label{subsec:A-stage4}
Recall that $\gamma'=(\beta_1,\dots,\beta_k,\gamma_1,\gamma_2,\dots)$. Let
\[
 p_k(x)=\prod_{j\le k}\bigl(z(x)-z(\beta_j)\bigr)=\sum_{|l|\le k}c_le^{2\pi\ii lx},
\]
a real even trigonometric polynomial with $c_{\pm k}=(-\frac12)^k$, whose zeros are
exactly the points of $\{\pm\beta_j:j\le k\}+\Z$, all of them simple. Put
$A_kT=p_kT$ and $B_kT=\F^{-1}(p_k\F T)=\sum_lc_l\tau_lT$. Since $p_k$ is
$1$-periodic, $A_k$ and $B_k$ commute, and since $p_k$ is even, $\F A_k=B_k\F$
and $\F B_k=A_k\F$. Let $\mathcal R=A_kB_k$, so that $\F\mathcal R=\mathcal R\F$. We
write $\Lambda^{\mathrm{tail}}=\{n\pm\gamma_j:\ |n|\ge k+j\}$, the part of $L_{\gamma'}$
with phases $\gamma_j$.

\begin{lemma}\label{lem:forward}
$\mathcal R$ maps $\M_{L_{\gamma'}}$ injectively into $\M_{L_\gamma}$, and
$\norm{\mathcal RT}_{-p}\le C_p\norm T_{-p}$ for every integer $p\ge0$.
\end{lemma}

\begin{proof}
The bound follows from Section~\ref{subsec:A-hermite}. Let $T\in\M_{L_{\gamma'}}$.
The measure $A_kT$ has no mass on $\{\pm\beta_j\}+\Z$, so it is locally atomic on
$\Lambda^{\mathrm{tail}}$, and $B_k$ translates it by integers $l$ with $|l|\le k$.
Since $|n-l|\ge j$ when $|n|\ge k+j$, $\mathcal RT=B_kA_kT$ is locally atomic on
$L_\gamma$. Since $\F T\in\M_{L_{\gamma'}}$ and $\F\mathcal RT=\mathcal R\F T$, the same
holds for $\F\mathcal RT$. Thus $\mathcal RT\in\M_{L_\gamma}$.

Suppose that $\mathcal RT=0$, and put $H=A_kT$, so that $B_kH=0$. For a phase
$\pm\gamma_j$ let $t(n)$ be the mass of $H$ at $n\pm\gamma_j$. Then
$\sum_lc_lt(n+l)=0$ for all $n$, a recurrence of order $2k$ with nonzero extreme
coefficients, and $t(n)=0$ for $|n|\le k$, because $|n|<k+j$. Hence $t=0$ and
$H=0$, i.e.\ $T$ is supported in $\{\pm\beta_j\}+\Z$. Then
$B_k\F T=\F A_kT=0$, and the same recurrence argument shows that $\F T$ has no
mass at the points $n\pm\gamma_j$, so $\F T$ is also supported in
$\{\pm\beta_j\}+\Z$. By Lemma~\ref{lem:finitephase}, $T\in W_k$, and $T$ satisfies
the hypothesis of Theorem~\ref{thm:rapid}(e), since $L_{\gamma'}$ does not contain
the points $n\pm\beta_j$ with $|n|<j$. Hence $T=0$.
\end{proof}

\begin{lemma}\label{lem:green}
Let $D=\sum_{|l|\le k}c_l\tau_l$, where $\sum_lc_lw^l$ has $2k$ simple zeros on the
unit circle. There are bounded sequences $(g^+_t)_{t\ge k}$ and $(g^-_t)_{t\le-k}$
such that, as formal Laurent series,
\[
 \Bigl(\sum_lc_lw^l\Bigr)\Bigl(\sum_{t\ge k}g^+_tw^t\Bigr)=1=
 \Bigl(\sum_lc_lw^l\Bigr)\Bigl(\sum_{t\le-k}g^-_tw^t\Bigr).
\]
Let $\chi_++\chi_-=1$ be a smooth partition of unity with $\chi_+=0$ on
$(-\infty,-\frac14]$ and $\chi_+=1$ on $[\frac14,\infty)$, and put
\[
 \mathcal QH=\sum_{t\le-k}g^-_t\,\tau_t(\chi_+H)+\sum_{t\ge k}g^+_t\,\tau_t(\chi_-H)
 \qquad(H\in\Sc').
\]
Then $\mathcal Q$ is continuous on $\Sc'$, $D\mathcal Q=I$, $\mathcal Q$ commutes with
multiplication by $1$-periodic smooth functions, and
$\norm{\mathcal QH}_{-(p+1)}\le C_p\norm H_{-p}$ for every integer $p\ge0$.
\end{lemma}

\begin{proof}
Let $q(w)=w^k\sum_lc_lw^l$, a polynomial of degree $2k$ with $q(0)\ne0$ and
simple zeros on the unit circle. By partial fractions, the Taylor coefficients of
$1/q$ at $0$ and its Laurent coefficients at $\infty$ are bounded; after the shift
by $w^k$ they give $g^+$ and $g^-$. The translates $\tau_t(\chi_+H)$, $t\le-k$, are supported in
$[|t|-\frac14,\infty)$, and the $\tau_t(\chi_-H)$, $t\ge k$, in
$(-\infty,\frac14-t]$, so both sums are locally finite, and $D\mathcal QH=\chi_+H+\chi_-H=H$
by the two formal identities. Multiplication by a $1$-periodic function commutes
with integer translations and with multiplication by $\chi_\pm$. The transpose of
$\mathcal Q$ is
$\mathcal Q^t\varphi=\chi_+\sum_{t\le-k}g^-_t\varphi(\cdot-t)+\chi_-\sum_{t\ge k}g^+_t\varphi(\cdot-t)$.
For the first sum, the term $x^a\chi_+^{(\sigma)}(x)\varphi^{(b-\sigma)}(x+|t|)$, with
$a+b\le2p$, is supported where $x\ge-\frac14$; there, with $y=x+|t|$, we have
$1+|x|\le C(1+|y|)$ and $1+|t|\le C(1+|y|)$, so its $L^2$ norm is at most
$C(1+|t|)^{-2}\norm{(1+|y|)^{a+2}\varphi^{(b-\sigma)}}_{L^2}\le C(1+|t|)^{-2}\norm\varphi_{p+1}$
by the norm equivalence. Summing over $t$, and treating the second sum in the
same way, gives $\norm{\mathcal Q^t\varphi}_p\le C\norm\varphi_{p+1}$; this proves the
continuity of $\mathcal Q$ and, by duality, the bound.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:rapid}(d)]
By Lemma~\ref{lem:forward} it remains to show that every $H\in V_p(L_\gamma)$ is
$\mathcal RT$ for some $T\in V_{p+2}(L_{\gamma'})$. Let $\mathcal Q_B$ be the operator of
Lemma~\ref{lem:green} for $D=B_k$, and $\mathcal Q_A=\F^{-1}\mathcal Q_B\F$. Since
$\F A_k\F^{-1}=B_k$, we have $B_k\mathcal Q_B=I$ and $A_k\mathcal Q_A=I$; moreover
$\mathcal Q_A$ commutes with $B_k=\F^{-1}p_k\F$, because $\mathcal Q_B$ commutes with
multiplication by $p_k$. Put $T_0=\mathcal Q_A\mathcal Q_BH\in H_{-(p+2)}$. Then
\[
 A_kT_0=\mathcal Q_BH,\qquad B_kT_0=\mathcal Q_AB_k\mathcal Q_BH=\mathcal Q_AH,\qquad
 \mathcal RT_0=A_k\mathcal Q_AH=H .
\]
Since $L_\gamma$ has no points in $[-\frac12,\frac12]$, $\chi_\pm H$ is the restriction
of $H$ to $\{\pm x>0\}$. An atom of $H$ at $n\pm\gamma_j$ with $n\ge j$ is moved by
$\mathcal Q_B$ to the points $n+|t|\pm\gamma_j$, $t\le-k$, and an atom with $n\le-j$ to
the points $n-t\pm\gamma_j$, $t\ge k$; all of these lie in $\Lambda^{\mathrm{tail}}$.
Thus $\mathcal Q_BH$ is locally atomic on $\Lambda^{\mathrm{tail}}$, and so is
$\F\mathcal Q_AH=\mathcal Q_B\F H$, since $\F H\in\M_{L_\gamma}$.

From $p_kT_0=\mathcal Q_BH$ we read off the support of $T_0$. Away from the zeros of
$p_k$, $T_0=p_k^{-1}\mathcal Q_BH$ is locally atomic on $\Lambda^{\mathrm{tail}}$. Near a
zero $x_0\in\{\pm\beta_j\}+\Z$, the right-hand side vanishes, because
$\Lambda^{\mathrm{tail}}$ lies within $\eps_1$ of the half-integers and
$\beta_j\le\frac12-\frac1{4k}$; since $x_0$ is a simple zero of $p_k$, $T_0$ is a
multiple of $\delta_{x_0}$ near $x_0$. Thus $T_0$ is locally atomic on
$\Lambda^{\mathrm{tail}}\cup(\{\pm\beta_j\}+\Z)$, and the same argument, applied to
$p_k\F T_0=\F B_kT_0=\F\mathcal Q_AH$, shows that so is $\F T_0$. The only points of
this set outside $L_{\gamma'}$ are the points $n\pm\beta_j$ with $|n|<j$, which give
$4k^2$ coordinates on the two sides. By Theorem~\ref{thm:rapid}(e) and
$\dim W_k=4k^2$, the map recording the masses of $w\in W_k$ and of $\F w$ at these
points is bijective, so there is a unique $w\in W_k$ with the same masses there as
$T_0$ and $\F T_0$; it depends linearly on these masses, which are bounded by
$C\norm{T_0}_{-(p+2)}$. Since $A_kw=B_kw=0$, the distribution $T=T_0-w$ belongs to
$\M_{L_{\gamma'}}\cap H_{-(p+2)}$, $\mathcal RT=H$, and
$\norm T_{-(p+2)}\le C\norm H_{-p}$.
\end{proof}