\section{Certified digits at zero temperature}\label{sec:digits}

\newcommand{\dk}{\mathsf{k}}
\newcommand{\dP}{\mathcal{P}}
\newcommand{\dI}{\mathcal{I}}
\newcommand{\dLo}{\mathrm{Lo}}
\newcommand{\deup}{e_{\mathrm{up}}}
\newcommand{\delow}{e_{\mathrm{low}}}
\newcommand{\dE}{\mathsf{E}}
\newcommand{\dA}{\mathsf{A}}
\newcommand{\dB}{\mathsf{B}}
\newcommand{\dKK}{\mathsf{K}}
\newcommand{\dtail}{\mathsf{t}}
\newcommand{\dwfl}{w_{\mathrm{fl}}}
\newcommand{\dQfl}{Q_{\mathrm{fl}}}

In this section, we prove $0.138198129\le\alc(0)\le0.13820106$, with no well at any overlap above the upper end, no hole below the lower end, the same statements for binary patterns, and the same bracket for every limit point of $\alc(T)$ as $T\to0$.

We write $\alo=138198129/10^{9}$ and $\ahi=13820106/10^{8}$, so the bracket has width $2.931\cdot10^{-6}$ and rounds to $0.13820$.

The capacity is the sign of $e'$, by Theorem~\ref{thm:t0-capacity}. One Ising evaluation of the recursion, in ball arithmetic, gives an upper bound on $L(\alpha,m)$ for all $\alpha$ through a trial, and a lower bound for all $\alpha$ through a tangent of the sphere problem. Thus $e'>0$ at a certified pair of overlaps at the lower end, and $e'<0$ on $[0.89,1)$ at the upper end. A separate argument, the slope of the response at the optimizer, excludes wells on $(0,0.89)$.

Appendix~\ref{app:cert} states the ball evaluation of the recursion and what was computed. We will bound the slice energy from below in Subsection~\ref{ssec:sk} (Theorem~\ref{thm:recursion-bound}) and give the upper and lower curves for $L$ and the strong concavity that controls $L'$ in Subsections~\ref{ssec:curves} and~\ref{ssec:sc}. In Subsections~\ref{ssec:bracket}, \ref{ssec:holes}, \ref{ssec:low} and~\ref{ssec:tzero}, we then prove the bracket on a window, exclude holes, remove the window, and pass to $T\to0$.

\subsection{Setting}\label{ssec:dig-notation}
We work in the unit scale of Corollary~\ref{cor:t0-unit} and drop its tildes, so $\kappa,C,\delta,\lambda$ are the variables of \eqref{eq:t0-unit}, and $\Psi_\kappa$ is the recursion at $\alpha=1$: $-\partial_w\Psi_\kappa=\frac12(\partial_x^2\Psi_\kappa+\kappa(\partial_x\Psi_\kappa)^2)$, $\Psi_\kappa(1,x)=|x|$. We will use Lemmas~\ref{lem:flow}, \ref{lem:flow-compare}, \ref{lem:control} and~\ref{lem:response-t0} at $\alpha=1$. On a level $[w,w')$ where $\kappa=c$, we have $\Psi_\kappa(w,\cdot)=\mathcal T_{c,\sqrt{w'-w}}\Psi_\kappa(w',\cdot)$, where $\mathcal T_{c,b}f(x)=\frac1c\log\E\exp(cf(x+bz))$ and $\mathcal T_{0,b}f(x)=\E f(x+bz)$; here and below, $z$ denotes a standard Gaussian. By Corollary~\ref{cor:t0-unit}, we have
\begin{equation}\label{eq:dig-L}
  L(\alpha,m)=\min_{\kappa,C,\lambda}\Bigl\{\frac C2+\frac\alpha2\int_0^1\frac{\dd w}{\delta_{\kappa,C}(w)}+\Psi_\kappa(0,\lambda)-\lambda m\Bigr\},
\end{equation}
with $h=-\partial_mL$ the optimal $\lambda$, and $\partial_\alpha L=\frac12\int_0^1\dd w/\delta^*>0$. Moreover, $L$ is jointly concave, $L\ge\sqrt\alpha$ and $L(\alpha,1)=\sqrt\alpha$. Recall that $e=m^2/2+L^2/2$ and $e'=m-Lh$ (Theorem~\ref{thm:t0-shape}).

We will use two bounds throughout:
\begin{equation}\label{eq:dig-chevet-young}
  \sqrt\alpha+P^*(m)\le L(\alpha,m)\le\sqrt\alpha+\wf(m).
\end{equation}
The upper bound is Lemma~\ref{lem:t0-bracket}, and the lower one is Corollary~\ref{cor:t0-unit}(c), with the slice energy
\begin{equation}\label{eq:dig-P}
  P^*(m)=\inf_{\kappa,\lambda}\dP_m(\kappa,\lambda),\qquad \dP_m(\kappa,\lambda):=\Psi_\kappa(0,\lambda)-\lambda m-\frac12\int_0^1w\kappa(w)\dd w.
\end{equation}
By the $L^1$ contraction, Lemma~\ref{lem:flow-compare}(ii), the infimum may be taken over step functions. The function $P^*$ is even, concave and nonincreasing on $[0,1]$, and $P^*(1)=0$.

For a step function $\kappa_0$ and real $\lambda$, let $u=\partial_x\Psi_{\kappa_0}$ and let $X$ solve $\dd X_w=\kappa_0(w)u(w,X_w)\dd w+\dd B_w$, $X_0=\lambda$. We then set $Q(w)=Q_{\kappa_0,\lambda}(w)=\E u(w,X_w)^2$ and $u_0=u(0,\lambda)$. By Lemma~\ref{lem:control}, $Q$ is nondecreasing, $Q(0)=u_0^2$, $Q(1-)=1$, and for $\kappa\in\mathcal U$ and real $\lambda'$,
\begin{equation}\label{eq:dig-tangent}
  \Psi_\kappa(0,\lambda')\ge\Psi_{\kappa_0}(0,\lambda)+u_0(\lambda'-\lambda)+\frac12\int_0^1(\kappa-\kappa_0)Q\,\dd w.
\end{equation}
The well sets are those of Definition~\ref{def:t0-wells}. A ball $[c\pm r]$ is the set of reals within $r$ of $c$.

All certified numbers below are obtained from the evaluation of the recursion described in Subsection~\ref{ssec:balls}. For a step path with rational data, this evaluation produces balls containing $\Psi_\kappa(0,\lambda)$, $\partial_x\Psi_\kappa(0,\lambda)$, the values of $Q$ at the level boundaries, and the integrals of $1-Q$ over the levels, through \eqref{eq:dig-functionals}.

\subsection{A lower bound for the slice energy}\label{ssec:sk}
A trial path gives an upper bound on $P^*(m)$ at once, by \eqref{eq:dig-P}, and the lower bound comes from convexity of $\dP_m$. More precisely, the tangent \eqref{eq:dig-tangent} at the trial bounds every competitor from below up to the term $\frac12\int\kappa(Q-w)$, which involves the competitor's mass. We will control that mass by the value itself through a short a priori bound, and the inequality then closes.

\begin{lemma}\label{lem:sk-apriori}
Let $n\ge1$ and $|m|<1$. The function $\dP_m$ has a minimizer $(\kappa,\lambda)$ over real $\lambda$ and step functions with at most $n$ steps, and at a minimizer $\int_0^1w\kappa(w)\dd w\le\dP_m(\kappa,\lambda)$.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:flow-compare}(iv), $\Psi_\kappa(w,x)\ge|x|+\frac12\int_w^1\kappa$, so $\dP_m(\kappa,\lambda)\ge\frac12\int(1-w)\kappa+|\lambda|(1-|m|)$, which bounds $\lambda$ and $\kappa$ on every $[0,1-\epsilon]$. A minimizing sequence with levels squeezed to $w=1$ loses nothing in the limit. Indeed, $\Psi_\kappa(w,x)-\frac12\int_w^1w'\kappa(w')\dd w'\ge|x|\ge\E|x+B_{1-w}|-\sqrt{2(1-w)/\pi}$, and monotonicity of the recursion in the terminal data lets us drop those levels at cost $\sqrt{2(1-w)/\pi}$. Thus the infimum is attained on a path with at most $n$ active steps.

Let the minimizer have active levels $[s_j,s_{j+1})$ with values $c_j$, and let $Q=Q_{\kappa,\lambda}$. Then $\partial_\lambda\dP_m=u_0-m$, $\partial_{c_j}\dP_m=\frac12\int_{s_j}^{s_{j+1}}(Q-w)$, and $\partial_{s_j}\dP_m=\frac12(c_{j-1}-c_j)(Q(s_j)-s_j)$, so $u_0=m$, $\int_{s_j}^{s_{j+1}}(Q-w)=0$ on positive active levels, and $Q(s_j)=s_j$ at internal positive jumps. An initial zero level satisfies only a one-sided condition, but it contributes nothing when multiplied by $\kappa$.

Let $u_w=u(w,X_w)$, $\mathsf p_w=\partial_x^2\Psi_\kappa(w,X_w)\ge0$, $\mathsf a(w)=\E\mathsf p_w$. Differentiating the recursion in $x$ and It\^o's formula give $\dd u_w=\mathsf p_w\dd B_w$ and $\dd\mathsf p_w=-\kappa\mathsf p_w^2\dd w+{}$martingale, so $Q'=\E\mathsf p_w^2$ and $\mathsf a'=-\kappa Q'$. The equality case of Lemma~\ref{lem:control}(a) gives $\Psi_\kappa(0,\lambda)=\E|X_1|-\frac12\int\kappa Q$. By Lemma~\ref{lem:control}(b), $\E|X_1|=\E[\sgn(X_1)X_1]=\lambda m+\int\kappa Q+\E[\sgn(X_1)B_1]$, where $\E[\sgn(X_1)B_1]=\int_0^1\mathsf a$ by the It\^o isometry. Thus $\dP_m(\kappa,\lambda)=\frac12\int\kappa(Q-w)+\int_0^1\mathsf a=\int_0^1\mathsf a$. Integrating by parts, we obtain $\int_0^t\mathsf a=t\mathsf a(t)+\int_0^tw\kappa Q'\ge\int_0^tw\kappa Q'$. On each active level $\int wQ'=[wQ]-\int Q=(s_{j+1}^2-s_j^2)/2=\int w\,\dd w$, by the stationarity conditions and $Q(1-)=1$. Letting $t\to1$ completes the proof.
\end{proof}

\begin{theorem}\label{thm:recursion-bound}
Let $\kappa$ be a step function in $\mathcal U$, $\lambda$ real, $|m|<1$, $Q=Q_{\kappa,\lambda}$, and let
\[
  \Gamma_{\mathrm{id}}(s)=\int_s^1(Q(w)-w)\dd w,\qquad D=\min\Bigl(0,\inf_{s\in[0,1)}\frac{2\Gamma_{\mathrm{id}}(s)}{1-s^2}\Bigr),
\]
\[
  \dP_{\mathrm{tan}}=\dP_m(\kappa,\lambda)-\frac12\int_0^1\kappa(Q-w)\dd w,\qquad\Xi=|\lambda|+\frac{\dP_m(\kappa,\lambda)}{1-|m|}.
\]
Then
\begin{equation}\label{eq:dig-T}
  P^*(m)\ge\frac{\dP_{\mathrm{tan}}-|u_0-m|\Xi}{1+|D|/2}.
\end{equation}
\end{theorem}

\begin{proof}
Let $(\kappa_n,\lambda_n)$ minimize over at most $n$ steps, with $n$ at least the number of steps of $\kappa$, and let $\dP_n$ be its value. Then $\dP_n\le\dP_m(\kappa,\lambda)$ and $|\lambda_n|(1-|m|)\le\dP_n$, so $|\lambda_n-\lambda|\le\Xi$. By Fubini, with the Stieltjes measure including the atom $\kappa_n(0)\delta_0$,
\[
  \int\kappa_n(Q-w)=\int_{[0,1)}\Gamma_{\mathrm{id}}\,\dd\kappa_n,\qquad\int w\kappa_n=\int_{[0,1)}\frac{1-s^2}2\dd\kappa_n(s).
\]
Thus $\int\kappa_n(Q-w)\ge D\int w\kappa_n\ge D\dP_n$, by Lemma~\ref{lem:sk-apriori} and $D\le0$. Subtracting $\frac12\int w\kappa_n$ from the tangent \eqref{eq:dig-tangent} gives $\dP_n\ge\dP_{\mathrm{tan}}+\frac12D\dP_n-|u_0-m|\Xi$, and we conclude by letting $n\to\infty$ and using \eqref{eq:dig-P}.
\end{proof}

At an exactly stationary trial, $\dP_{\mathrm{tan}}=\dP_m(\kappa,\lambda)$, while for a rational trial, the difference is first order in the distance to the stationary one and is included in the ball evaluation.

The quantity $D$ is an infimum over a continuum, which we will reduce to finitely many values of $\Gamma_{\mathrm{id}}$ in the next lemma. These come from \eqref{eq:dig-functionals} with no quadrature in $w$: for levels cut at $s_0<\dots<s_M=1$, we have $\Gamma_{\mathrm{id}}(s_i)=\sum_{j\ge i}[b_j^2-\int_{s_j}^{s_{j+1}}(1-Q)-(s_{j+1}^2-s_j^2)/2]$.

\begin{lemma}\label{lem:sk-continuum}
Let $s_i<s_{i+1}$ be consecutive grid points and $t_i=1-s_i$. On $[s_i,s_{i+1}]$,
\[
  \frac{\Gamma_{\mathrm{id}}(s)}{1-s}\ge\min\Bigl(\frac{\Gamma_{\mathrm{id}}(s_i)}{t_i},\frac{\Gamma_{\mathrm{id}}(s_{i+1})}{t_{i+1}}\Bigr)-\frac12(\sqrt{t_i}-\sqrt{t_{i+1}})^2.
\]
Let $\ell_i$ be the right side, replaced by $0$ if it is positive. Then $2\Gamma_{\mathrm{id}}(s)/(1-s^2)\ge2\ell_i/(1+s_i)$ on the cell. Moreover:
\begin{enumerate}[label=\textup{(\alph*)}]
\item if $Q(s_i)\ge s_{i+1}$, then $\Gamma_{\mathrm{id}}\ge\Gamma_{\mathrm{id}}(s_{i+1})$ on the cell, and if $Q(s_{i+1})\le s_i$, then $\Gamma_{\mathrm{id}}\ge\Gamma_{\mathrm{id}}(s_i)$ on the cell;
\item on the last cell $[s_{M-1},1)$, $2\Gamma_{\mathrm{id}}(s)/(1-s^2)\ge Q(s_{M-1})-1$;
\item below the first point, $\Gamma_{\mathrm{id}}(s)\ge\Gamma_{\mathrm{id}}(s_0)-(s_0-u_0^2)_+^2/2$.
\end{enumerate}
\end{lemma}

\begin{proof}
As $\Gamma_{\mathrm{id}}(s)-s^2/2$ has derivative $-Q$, it is concave, and $\Gamma_{\mathrm{id}}(s)\ge\mathrm{chord}(s)-(s-s_i)(s_{i+1}-s)/2$. With $t=1-s$, the chord divided by $t$ is monotone in $t$, and $(t_i-t)(t-t_{i+1})/t\le(\sqrt{t_i}-\sqrt{t_{i+1}})^2$. Multiplying by $2/(1+s)\le2/(1+s_i)$ keeps a lower bound only for a nonpositive bound, which is why $\ell_i$ is capped at $0$. For (a), the sign of $\Gamma_{\mathrm{id}}'=s-Q(s)$ is fixed on the cell; for (b), we have $\Gamma_{\mathrm{id}}(s)\ge Q(s_{M-1})(1-s)-(1-s^2)/2$; and for (c), we use $Q\ge u_0^2$.
\end{proof}

Uniform spacing in $\sqrt{1-s}$ makes the slack in Lemma~\ref{lem:sk-continuum} constant. We take the largest available lower bound on each cell and evaluate \eqref{eq:dig-T} in balls. In this way, \eqref{eq:dig-T} with Lemma~\ref{lem:sk-continuum} gives a certified lower bound on $P^*(m)$ at a rational $m$, which we denote by $P(m)$, and we call such a rational $m$ an SK node.

\subsection{Upper and lower curves for $L$}\label{ssec:curves}
We fix a rational $m\in(0,1)$ and a rational step trial $(\kappa_0,\lambda_{\mathrm t})$ with values $0<c_1\le\dots\le c_k$ on $[\hat w_j,\hat w_{j+1})$, $\hat w_j=1-t_j$, $t_1>\dots>t_k>t_{k+1}:=0$, and $\kappa_0=0$ below $\hat w_1$. Let $Q_0=Q_{\kappa_0,\lambda_{\mathrm t}}$, $u_0=\partial_x\Psi_{\kappa_0}(0,\lambda_{\mathrm t})$, and
\[
  \mathsf W=\Psi_{\kappa_0}(0,\lambda_{\mathrm t})-\lambda_{\mathrm t}m,\qquad\mathsf I=\mathsf W-\frac12\int_0^1\kappa_0Q_0.
\]
By \eqref{eq:dig-L}, for every $C'>0$ and every $\alpha$,
\begin{equation}\label{eq:dig-U}
  L(\alpha,m)\le U(\alpha,m):=\frac{C'}2+\frac\alpha2\dI(C')+\mathsf W,
\end{equation}
where, with $\delta_j=\delta_{\kappa_0,C'}(\hat w_j)$ and $\delta_{k+1}=C'$,
\[
  \dI(C')=\int_0^1\frac{\dd w}{\delta_{\kappa_0,C'}(w)}=\frac{1-t_1}{\delta_1}+\sum_{j=1}^k\frac1{c_j}\log\frac{\delta_j}{\delta_{j+1}}.
\]
Thus one Ising evaluation gives $U$ for all $\alpha$, affine in $\alpha$ at fixed $C'$.

For the lower curve, we take one tangent of the Ising part at the trial and one of the convex function $\delta\mapsto1/\delta$ for the sphere part. The second is a function $\vartheta$ on $[0,1)$, admissible as long as a one-dimensional inequality holds, and no bound on the mass of the unknown minimizer is needed. The choice $\vartheta\equiv1$ corresponds to the lower bound in \eqref{eq:dig-chevet-young}.

\begin{lemma}\label{lem:young-lower}
Let $\vartheta>0$ be measurable on $[0,1)$, $R_\vartheta(s)=\int_0^s\vartheta^2$, with $R_\vartheta(1)\le1$ and $\Gamma_\vartheta(s):=\int_s^1(Q_0-R_\vartheta)\ge0$ on $[0,1)$. Then for every $\alpha>0$,
\begin{equation}\label{eq:dig-Y}
  L(\alpha,m)\ge\sqrt\alpha\int_0^1\vartheta+\mathsf I-|u_0-m|\Xi(\alpha),\qquad\Xi(\alpha):=|\lambda_{\mathrm t}|+\frac{\sqrt\alpha+1}{1-|m|}.
\end{equation}
\end{lemma}

\begin{proof}
We take the minimizer $(\kappa,C,\lambda)$ of \eqref{eq:dig-L}. By \eqref{eq:dig-tangent},
\[
  \Psi_\kappa(0,\lambda)-\lambda m\ge\mathsf W+(u_0-m)(\lambda-\lambda_{\mathrm t})+\frac12\int(\kappa-\kappa_0)Q_0.
\]
Also $|\lambda|(1-|m|)\le\Psi_\kappa(0,\lambda)-\lambda m\le L\le\sqrt\alpha+\wf(m)\le\sqrt\alpha+1$, since $\Psi_\kappa(0,x)\ge|x|$, the sphere part is nonnegative, and by \eqref{eq:dig-chevet-young}. Thus the second term is at least $-|u_0-m|\Xi(\alpha)$. For the sphere part, we use $1/\delta\ge2\theta-\delta\theta^2$ for any $\theta>0$, and with $\theta=\vartheta/\sqrt\alpha$ this reads $(\alpha/2)/\delta\ge\sqrt\alpha\vartheta-\delta\vartheta^2/2$. By Fubini $\int_0^1\delta\vartheta^2=CR_\vartheta(1)+\int\kappa R_\vartheta$, so
\[
  \frac C2+\frac\alpha2\int\frac{\dd w}\delta+\frac12\int\kappa Q_0\ge\sqrt\alpha\int\vartheta+\frac C2(1-R_\vartheta(1))+\frac12\int\kappa(Q_0-R_\vartheta).
\]
Finally $\int\kappa(Q_0-R_\vartheta)=\int_{[0,1)}\Gamma_\vartheta\,\dd\kappa\ge0$ by Fubini with the atom of $\dd\kappa$ at $0$; the integrals converge absolutely since $|Q_0-R_\vartheta|\le1$ and $\kappa\in L^1$. Collecting the above bounds proves the lemma.
\end{proof}

The numbers $S_\vartheta:=\int\vartheta$, $\mathsf I$, $\mathsf W$, $u_0$ do not depend on $\alpha$, so we write
\begin{equation}\label{eq:dig-Lo}
  \dLo(\alpha,m):=\sqrt\alpha S_\vartheta+\mathsf I-|u_0-m|\Xi(\alpha)=S'\sqrt\alpha+I',
\end{equation}
with $S'=S_\vartheta-|u_0-m|/(1-m)$ and $I'=\mathsf I-|u_0-m|(|\lambda_{\mathrm t}|+1/(1-m))$. The curve is affine in $\sqrt\alpha$. Below, $\vartheta$ is rational and piecewise constant, proposed by a floating-point solver of the sphere problem; any $\vartheta$ is allowed, since the next proposition rechecks it.

Let $s_0<\dots<s_{M-1}<1$ be rational nodes with all $\hat w_j$ among them, and cut the levels at the nodes. We take $\vartheta$ constant on $[0,s_0]$, on each $[s_i,s_{i+1}]$ with value $\vartheta_i$, and on $[s_{M-1},1)$. The recursion gives balls for $Q_i=Q_0(s_i)$, for the cell integrals of $Q_0$, and for $\mathsf W$, $\mathsf I$, $u_0$. Then $R_i=R_\vartheta(s_i)$ are exact rationals, and $\Gamma_i=\Gamma_\vartheta(s_i)$ are balls, through $\Gamma_i=\Gamma_{i+1}+\int_{s_i}^{s_{i+1}}Q_0-(s_{i+1}-s_i)(R_i+R_{i+1})/2$.

\begin{proposition}\label{prop:young-checks}
Suppose the following hold.
\begin{enumerate}[label=\textup{(\roman*)}]
\item On each cell $[s_i,s_{i+1}]$, with $\mathsf D_i=s_{i+1}-s_i$, the largest of the applicable lower bounds below is nonnegative: $\Gamma_{i+1}$ if $Q_i\ge R_{i+1}$; $\Gamma_i$ if $Q_{i+1}\le R_i$; and $\min(\Gamma_i,\Gamma_{i+1})-(\vartheta_i^2-k_i)_+\mathsf D_i^2/8$, where $k_i$ is any lower bound of $Q_0'$ on the cell.
\item $Q_{M-1}\ge R_\vartheta(1)$.
\item $u_0^2\ge R_0$ and $\Gamma_0\ge0$.
\end{enumerate}
Then $\vartheta$ satisfies the hypotheses of Lemma~\ref{lem:young-lower}.
\end{proposition}

\begin{proof}
As $\Gamma_\vartheta$ is $C^1$ with $\Gamma_\vartheta'=R_\vartheta-Q_0$, and $Q_0,R_\vartheta$ are nondecreasing, $\Gamma_\vartheta$ is nonincreasing on the cell if $Q_i\ge R_{i+1}$, and nondecreasing if $Q_{i+1}\le R_i$. Always $\Gamma_\vartheta''=\vartheta_i^2-Q_0'\le\bar k:=(\vartheta_i^2-k_i)_+$, so $x\mapsto\Gamma_\vartheta(x)+\bar k(x-s_i)(s_{i+1}-x)/2$ is concave. This gives (i). On $[s_{M-1},1)$, it follows from (ii) that $Q_0\ge R_\vartheta$, so $\Gamma_\vartheta\ge\Gamma_\vartheta(1)=0$; on $[0,s_0]$, $Q_0\ge u_0^2\ge R_0$, so $\Gamma_\vartheta$ is nonincreasing there. Finally $R_\vartheta(1)\le Q_{M-1}\le1$.
\end{proof}

All comparisons in Proposition~\ref{prop:young-checks} are made in balls. Since $Q_0$ is steep near the top of a step path, a good lower bound $k_i$ matters, and the next lemma provides one.

\begin{lemma}\label{lem:Q-slope}
For a step path, let $\mathsf p=\partial_x^2\Psi_{\kappa_0}$ and $\bar{\mathsf p}(w)=\sup_x\mathsf p(w,x)=\mathsf p(w,0)$. Then $Q_0'=\E\mathsf p(w,X_w)^2$ is continuous across levels, and on each level $Q_0''\ge-2\kappa_0\bar{\mathsf p}Q_0'$. Hence, on consecutive cells $i-1,i$ with widths $\mathsf D_{i-1},\mathsf D_i$, values $c_{i-1},c_i$, and upper bounds $\bar{\mathsf p}_{i-1},\bar{\mathsf p}_i$ of $\bar{\mathsf p}$, the number
\[
  k_i:=\frac{Q_i-Q_{i-1}}{\mathsf D_{i-1}}\exp\bigl(-2c_{i-1}\bar{\mathsf p}_{i-1}\mathsf D_{i-1}-2c_i\bar{\mathsf p}_i\mathsf D_i\bigr)
\]
is a lower bound of $Q_0'$ on cell $i$. On a level with constant $c>0$, with $\ell$ the distance from $w$ to the top of the level and $\bar{\mathsf p}_+$ the bound at the top,
\[
  \bar{\mathsf p}(w)\le\frac{c+\sqrt{c^2+4/\ell}}2,\qquad\bar{\mathsf p}(w)\le\bar{\mathsf p}_++c\,\bar{\mathsf p}_+^2\ell\,\dk(c\sqrt\ell)^2,\qquad\dk(\eta):=\frac{\eta+\sqrt{\eta^2+4}}2.
\]
\end{lemma}

\begin{proof}
From $-\partial_w\mathsf p=\frac12\partial_x^2\mathsf p+\kappa_0(\mathsf p^2+u\partial_x\mathsf p)$ and It\^o, $\dd\mathsf p_w=-\kappa_0\mathsf p_w^2\dd w+\partial_x\mathsf p(w,X_w)\dd B_w$. Hence $Q_0''=\E[(\partial_x\mathsf p)^2-2\kappa_0\mathsf p^3]\ge-2\kappa_0\bar{\mathsf p}Q_0'$, because $0\le\mathsf p(w,x)\le\mathsf p(w,0)$ by Lemma~\ref{lem:response-t0}(b). Thus $\log Q_0'$ has derivative at least $-2\kappa_0\bar{\mathsf p}$, and the mean slope of $Q_0$ on cell $i-1$ is attained at some point of that cell.

For the first bound, let $f=\Psi_{\kappa_0}$ at the top of the level and $\mathsf x(x)=\frac1c\log\E\exp(cf(x+\sqrt\ell z))$. Completing the square gives $\mathsf x''(0)=(\E_{\mathrm t}z^2-1)/(c\ell)$, with $\E_{\mathrm t}$ the tilted law at $x=0$. Since $f$ is even and $1$-Lipschitz, $\exp(\eta|z|-cf(\sqrt\ell z))$ is nondecreasing in $|z|$, with $\eta=c\sqrt\ell$. Hence $\E_{\mathrm t}z^2\le\E_\eta z^2$ under the density proportional to $\exp(\eta|z|-z^2/2)$. Integration by parts gives $\E_\eta z^2=\eta\E_\eta|z|+1\le\eta\sqrt{\E_\eta z^2}+1$, so $\E_{\mathrm t}z^2\le\dk(\eta)^2$ and $\bar{\mathsf p}\le\dk(\eta)/\sqrt\ell$. For the second, we use $\mathsf x''=\E_{\mathrm t}f''+c\Var_{\mathrm t}f'$ and $|f'(y)|\le\bar{\mathsf p}_+|y|$.
\end{proof}

On a cell we take the smaller of the first bound at the lowest $\ell$ in the cell and the second at the largest, and we use $k=0$ in the first cell and no $k$ in the last.

\subsection{Strong concavity}\label{ssec:sc}
To bound $e'$, we need a bound on $L'$. Concavity of $L$ gives $L'$ from a chord, with an error of the order of the square root of $U-\dLo$, which is too large on a mesh of reasonable size. The next lemma shows that $L$ is strongly concave near overlap one, with a constant independent of $\alpha$. The idea is that $h'=1/\resp_{\kappa^*}'$, up to the regularity needed, and that $\resp_{\kappa^*}'$ is small at large fields because the optimal path is flat below a point $w_1$ close to $1$.

Recall from Lemma~\ref{lem:t0-kappa-nonzero}(c), (d), in unit scale, that with $w_1=w_1(m)$ as there, we have $w_1\ge\dwfl(m)$, $\dwfl$ is nondecreasing in $m$, and $h(m)\ge\sqrt{w_1}\lambda_0(m)$. Thus a lower bound of $\dwfl(y)$ is a lower bound of $w_1(m')$ for all $m'\ge y$. Let $\mathsf b_u(T)=\frac12[\Nd(u-T)+\Nd(u+T)]$ for $T\le u-1$, and $\mathsf b_u(T)=\frac12[\Nd(0)+\Nd(u)]$ for $T>u-1$.

\begin{lemma}\label{lem:strong-concavity}
Let $y\in(0,1)$, $\underline w\in(0,1)$ with $\underline w\le w_1(m')$ for all $m'\ge y$, and $u\ge1$ with $u\le h(y)$. Then for $y\le m_1<m_2<1$,
\[
  h(m_2)-h(m_1)\ge\frac{m_2-m_1}{\dKK(u,\underline w)},\qquad\dKK(u,\underline w):=\frac2{\sqrt{\underline w}}\E\mathsf b_u\Bigl(\sqrt{\frac{1-\underline w}{\underline w}}|z|\Bigr).
\]
So $L+\mu m^2/2$ is concave on $[y,1)$, with $\mu=1/\dKK(u,\underline w)$.
\end{lemma}

\begin{proof}
Let $(\kappa_i,C_i,h_i)$ be the minimizers at $m_i$, $\Psi_i^0=\Psi_{\kappa_i}(0,\cdot)$, $\resp_i=\resp_{\kappa_i}$. Testing $(\kappa_1,C_1,h_2)$ at $m_2$ and $(\kappa_2,C_2,h_1)$ at $m_1$ in \eqref{eq:dig-L} and adding gives $\Psi_1^0(h_1)+\Psi_2^0(h_2)\le\Psi_1^0(h_2)+\Psi_2^0(h_1)$, that is $\int_{h_1}^{h_2}\resp_2\le\int_{h_1}^{h_2}\resp_1$. Let $\dKK$ bound $\resp_i'$ on $[h_1,h_2]$. As $\resp_1(h_1)=m_1$, $\resp_2(h_2)=m_2$, we have $\resp_1(x)\le m_1+\dKK(x-h_1)$ and $\resp_2(x)\ge m_2-\dKK(h_2-x)$, so $(m_2-m_1)\mathsf D\le\dKK\mathsf D^2$ with $\mathsf D=h_2-h_1$. Here $\mathsf D>0$. Indeed $h$ is nondecreasing, and if $h_1=h_2=:h$, the two tests are equalities, so both $(\kappa_i,C_i)$ minimize $C/2+(\alpha/2)\int\dd w/\delta+\Psi_\kappa(0,h)$, which is strictly convex in $\delta$. Then $\kappa_1=\kappa_2$, and $m_1=\resp_1(h)=\resp_2(h)=m_2$, a contradiction.

We now bound $\resp_i'$. Let $\kappa=\kappa_i$, $w_1=w_1(m_i)$, $\bar r_1^2=(1-w_1)/w_1$. By Lemma~\ref{lem:t0-kappa-nonzero}(e) in unit scale, with $\mathsf f=\partial_x\Psi_\kappa(w_1,\cdot)$ and the probability measure $\varsigma=\frac12\dd\mathsf f$,
\[
  \resp_\kappa(x)=\int\Bigl(2\Nc\Bigl(\frac{x-t}{\sqrt{w_1}}\Bigr)-1\Bigr)\varsigma(\dd t),\qquad\resp_\kappa'(x)=\frac2{\sqrt{w_1}}\int\frac12\bigl[\Nd(x'-t')+\Nd(x'+t')\bigr]\varsigma(\dd t),
\]
where $x'=x/\sqrt{w_1}$, $t'=t/\sqrt{w_1}$, using the symmetry of $\varsigma$. By Lemma~\ref{lem:response-t0}(b), $\varsigma(|t|>T)=1-\mathsf f(T)\le\P(\sqrt{1-w_1}|z|>T)$, so $|t'|$ is stochastically below $\bar r_1|z|$. Also $x\ge h_1\ge h(y)\ge u$ and $w_1\le1$, so $x'\ge u$. Let $\mathsf v(x',t')=\frac12[\Nd(x'-t')+\Nd(x'+t')]$, $x'\ge u$, $t'\ge0$. Since $x\Nd(x)$ decreases on $x\ge1$, $\partial_{t'}\mathsf v\ge0$ when $t'\le x'-1$, and $\partial_{x'}\mathsf v\le0$ when $x'\ge t'$. Thus for $T\le u-1$, $\sup_{x'\ge u,t'\le T}\mathsf v=\mathsf v(u,T)$, and always $\mathsf v\le\frac12[\Nd(0)+\Nd(u)]$, because $x'+t'\ge u$. The function $\mathsf b_u$ is nondecreasing in $T$, where at $T=u-1$ we use $\Nd(1)+\Nd(2u-1)\le\Nd(0)+\Nd(u)$. It follows that $\resp_\kappa'(x)\le(2/\sqrt{w_1})\E\mathsf b_u(\bar r_1|z|)\le\dKK(u,\underline w)$, where in the last step we used that $w_1\ge\underline w$ decreases both $2/\sqrt{w_1}$ and $\bar r_1$.
\end{proof}

Of the two sources for $u$, the universal one, $u=\sqrt{\dwfl(y)}\lambda_0(y)$, holds at any $\alpha$. The other holds at a given $\alpha$ and comes from a chord: with $\mu_{\mathrm u}$ the universal $\mu$ on $[y',1)$,
\begin{equation}\label{eq:dig-u-chord}
  h(y)\ge\frac{L(y')-L(y)}{y-y'}+\frac{\mu_{\mathrm u}(y-y')}2,\qquad y'<y.
\end{equation}
This follows by integrating $h(t)\le h(y)-\mu_{\mathrm u}(y-t)$ over $[y',y]$.

In balls, we bound $\dwfl$ from below by lower Riemann sums of $\dQfl$, with $\lambda_0$ replaced by a rational below it within $2\cdot10^{-9}$ and the iterates rounded down to multiples of $10^{-7}$, bound $\dKK$ from above by a Riemann sum in $|z|$ on $|z|\le z_{\max}$, with $\sqrt{(1-\underline w)/\underline w}\,z_{\max}\le u-1$, and by the second branch of $\mathsf b_u$ on the tail, and we check $u\ge1$.

Let $\tilde\mu\ge0$ be a step function such that $h(t_2)-h(t_1)\ge\int_{t_1}^{t_2}\tilde\mu$ for $a\le t_1<t_2<1$, and set $\bar\mu(a,b)=(b-a)^{-1}\int_a^b(t-a)\tilde\mu(t)\dd t$. Integrating over $[a,b]$ gives, for $a<b\le m<1$,
\begin{equation}\label{eq:dig-slope}
  L'(m)\le\frac{L(b)-L(a)}{b-a}-\bar\mu(a,b)-\int_b^m\tilde\mu.
\end{equation}

\subsection{The bracket on a window}\label{ssec:bracket}
The main result of this subsection is the following.

\begin{theorem}\label{thm:bracket-window}
Consider Gaussian rows.
\begin{enumerate}[label=\textup{(\alph*)}]
\item At $\alpha=\alo$, $e(0.9668708)>e(0.9666236)$. So $\alo\in\Well_0(m_s)$ for every $m_s\le0.9666236$.
\item For every $\alpha\ge\ahi$, $e'<0$ on $[0.89,1)$.
\item For every $m_s\in[0.89,0.9666236]$, $\alo\le\alc(0;m_s)\le\ahi$.
\end{enumerate}
\end{theorem}

Part (c) follows from (a), (b) and the two forms of $\Well_0(m_s)$. Part (a) is one evaluation at two overlaps, while (b) needs a mesh. A trial node is a rational $y$ with a rational trial and a $\vartheta$ that passes Proposition~\ref{prop:young-checks}; it provides $U(\alpha,y)$ and $\dLo(\alpha,y)$ for all $\alpha$. An SK node provides, by \eqref{eq:dig-chevet-young}, two bounds on $L$ that differ by $\wf(y)-P(y)$, independently of $\alpha$. Trial nodes are tight and serve near the threshold, while SK nodes are cruder and serve elsewhere.

\begin{proof}[Proof of \textup{(a)}]
We set $\deup=m^2/2+U^2/2$ and $\delow=m^2/2+\dLo^2/2$. If $\dLo>0$, then $e\ge\delow$ by Lemma~\ref{lem:young-lower}, and $e\le\deup$ by \eqref{eq:dig-U}. At $\alpha=\alo$, with $m_1=0.9666236$, $m_2=0.9668708$, the computation gives
\begin{align*}
  U(\alo,m_1)&=[0.4547205791577915\pm7.78\cdot10^{-17}],\\
  \deup(m_1)&=[0.5705659945932787\pm7.26\cdot10^{-17}],\\
  \dLo(\alo,m_2)&=[0.4541947217816\pm3.86\cdot10^{-14}],\\
  \delow(m_2)&=[0.5705659945935\pm4.63\cdot10^{-14}],
\end{align*}
and $\delow(m_2)-\deup(m_1)=[2\cdot10^{-13}\pm2.50\cdot10^{-14}]>0$.
\end{proof}

\begin{proof}[Proof of \textup{(b)}]
On a mesh cell, \eqref{eq:dig-slope} bounds $L'$ by a chord through two nodes, corrected by strong concavity, and the lower curve bounds $L$. Their product bounds $e'=m+LL'$ by a convex function of the position in the cell, so the endpoints suffice. Since the chord numerator is convex in $\alpha$, the endpoints of a load interval suffice too, and SK nodes need no upper limit on $\alpha$.

For the slopes, let $\tilde\mu$ be a step function on the mesh with value $\mu_i$ on $[y_i,y_{i+1})$, where $\mu_i=1/\dKK(u,\underline w)$ of Lemma~\ref{lem:strong-concavity}, $\underline w$ is a rational lower bound of $\dwfl$ at the point of the lattice $\frac1{2000}\mathbb Z$ below $y_i$, and $u$ is a rational lower bound of $h(y_i)$. For SK nodes $u$ is the universal one, and for trial nodes on a load interval $u$ is the larger of the universal one and the chord value \eqref{eq:dig-u-chord} with $\dLo(\alpha,y')$ and $U(\alpha,y)$. Since the numerator $\dLo(\alpha,y')-U(\alpha,y)$ is concave in $\alpha$, the endpoints of the interval give a lower bound.

Next, for trial nodes on a load interval $[\alpha_{\mathrm l},\alpha_{\mathrm h}]$, we fix $C'$ at each node for the interval. Then $U$ is affine in $\alpha$, and $\dLo=S'\sqrt\alpha+I'$ with $S'>0$ is concave in $\alpha$, so $U-\dLo$ is convex in $\alpha$. For nodes $y_j<y_i$ put
\[
  \bar{\mathsf s}=\max_{\alpha\in\{\alpha_{\mathrm l},\alpha_{\mathrm h}\}}\frac{U(\alpha,y_i)-\dLo(\alpha,y_j)}{y_i-y_j}-\bar\mu(y_j,y_i).
\]
By \eqref{eq:dig-slope}, for $\alpha$ in the interval and $m=y_i+x$ with $0\le x\le\nu:=y_{i+1}-y_i$, we have $L'\le\bar{\mathsf s}-\mu_ix$. Also $L(\alpha,m)\ge L(\alpha_{\mathrm l},m)\ge\ell(x)$ by $\partial_\alpha L>0$ and concavity, where $\ell$ is the chord of $\dLo(\alpha_{\mathrm l},\cdot)$ on the cell. If $\bar{\mathsf s}<0$, then
\[
  e'(m)\le q(x):=y_i+x+\ell(x)(\bar{\mathsf s}-\mu_ix).
\]
When $\ell$ also has negative slope, $q''=-2\mu_i\ell'\ge0$, so $q$ is convex and $q(0),q(\nu)<0$ suffices. The inequalities $\bar{\mathsf s}<0$, $\ell'<0$, $q(0)<0$, $q(\nu)<0$ are checked in balls, at an anchor $j$ chosen in floating point.

For SK nodes and $\alpha\ge\alpha_{\mathrm l}$, we obtain from \eqref{eq:dig-chevet-young} that $L(\alpha,b)-L(\alpha,a)\le\wf(b)-P(a)$, independent of $\alpha$. Thus, for nodes $y_j<y_i$ and $\bar{\mathsf s}=(\wf(y_i)-P(y_j))/(y_i-y_j)-\bar\mu(y_j,y_i)$, we have $L'\le\bar{\mathsf s}-\mu_ix$, and $L(\alpha,m)\ge\sqrt{\alpha_{\mathrm l}}+P^*(m)\ge\ell(x):=\sqrt{\alpha_{\mathrm l}}+$ chord of $P$ on the cell, by concavity of $P^*$. The same $q$ applies, with $\ell'<0$ checked, and one check at $\alpha_{\mathrm l}$ covers all larger $\alpha$. On the final cell $[m_{\mathrm{top}},1)$, concavity with $L(\alpha,m_{\mathrm{top}})\ge\sqrt\alpha$ and $L(\alpha,1)=\sqrt\alpha$ gives $L\ge\sqrt\alpha$, and $L'\le\bar{\mathsf s}<0$, so $e'\le1+\sqrt{\alpha_{\mathrm l}}\bar{\mathsf s}$.

The certificate uses $164$ trial nodes on $[0.931,0.985]$ and $48$ SK nodes on $[0.88,0.9975]$, and the load range $[\ahi,164868419/10^9]$ is split into $35$ intervals with exactly matching ends. With these nodes and intervals, we find that all the checks above pass, which proves (b). Table~\ref{tab:dig-upper} lists the largest upper end of the ball for $q$ on each part, rounded to nearest. The worst cell at $\ahi$ is $[0.9667225,0.9667349]$, with anchor $0.9666856$.
\end{proof}

\begin{table}[ht]
\caption{Coverage of $\{(\alpha,m):\alpha\ge\ahi,\ m\in[0.89,1)\}$ in the proof of Theorem~\ref{thm:bracket-window}(b), and the largest value of $q$ found. The final cell $[0.9975,1)$ has $q\le-0.1093$ at $\ahi$.}\label{tab:dig-upper}
\centering
\begin{tabular}{lllr}
\toprule
nodes & load & overlap & worst $q$\\
\midrule
trial & $[\ahi,\,164868419/10^9]$ & $[0.9322,0.985]$ & $-2.041\cdot10^{-8}$\\
SK & $[\ahi,\infty)$ & $[0.89,0.9325]$ & $-6.506\cdot10^{-3}$\\
SK & $[\ahi,\infty)$ & $[0.985,1)$ & $-1.784\cdot10^{-2}$\\
SK & $[164868419/10^9,\infty)$ & $[0.89,1)$ & $-6.221\cdot10^{-2}$\\
\bottomrule
\end{tabular}
\end{table}

\subsection{No holes below the lower end}\label{ssec:holes}

\begin{theorem}\label{thm:certified-wells}
Every $\alpha\in(0,\alo]$ lies in $\Well_0(m_s)\cap\WellR_0(m_s)$ for all $m_s\in(0,0.9616]$. Explicitly, for Gaussian rows, with $\lambda_b=2.0709518$ and $m_b=2\Nc(\lambda_b)-1=[0.961636699353\pm1.20\cdot10^{-14}]$:
\begin{enumerate}[label=\textup{(\alph*)}]
\item $e(0.974)>e(m_b)$ for every $\alpha\in(0,0.13755521]$;
\item $e(0.9668835)>e(0.9666236)$ for every $\alpha\in[0.13755521,\alo]$.
\end{enumerate}
\end{theorem}

\begin{proof}
Both left overlaps are at least $0.9616$, so (a) and (b) give the membership in $\Well_0(m_s)$. For binary patterns, $\WellR_0(m_s)=\Well_0(m_s)$ by Theorem~\ref{thm:merging-t0}(c).

For (a), let $m^*=0.974$, let $P$ be a lower bound of $P^*(m^*)$, and write $w_b=2\Nd(\lambda_b)=\wf(m_b)$. By \eqref{eq:dig-chevet-young},
\[
  e(m^*)-e(m_b)\ge\frac{m^{*2}-m_b^2}2+\frac{(\sqrt\alpha+P)^2-(\sqrt\alpha+w_b)^2}2=\frac{m^{*2}-m_b^2}2-\frac{w_b-P}2\bigl(2\sqrt\alpha+P+w_b\bigr).
\]
For $w_b>P$ this is positive exactly when
\begin{equation}\label{eq:dig-alphaD}
  \sqrt\alpha<\sqrt{\alpha_{\mathrm D}}:=\frac12\Bigl[\frac{m^{*2}-m_b^2}{w_b-P}-P-w_b\Bigr],
\end{equation}
which is monotone in $\alpha$, so no grid in $\alpha$ is needed. Theorem~\ref{thm:recursion-bound} with a $12$-step trial, cut into $215$ levels, gives $\dP_m=[0.066934867211846\pm5.15\cdot10^{-16}]$, $\dP_{\mathrm{tan}}=[0.066934867141065\pm4.66\cdot10^{-16}]$, $D=-1.634\cdot10^{-7}$, $|u_0-m^*|\le2.6\cdot10^{-10}$, and
\[
  P^*(0.974)\ge P=[0.06693486043888495\pm1.53\cdot10^{-18}],\quad\text{so }P^*(0.974)\ge0.066934860438.
\]
With $w_b=[0.09346090751068000\pm2.78\cdot10^{-18}]$, the right side of \eqref{eq:dig-alphaD} gives
\[
  \alpha_{\mathrm D}\in[0.1375552194487222\pm9.37\cdot10^{-18}],
\]
which exceeds $0.13755521$. This proves (a).

For (b), we make the argument of Theorem~\ref{thm:bracket-window}(a) uniform in $\alpha$. With trial nodes $m_1=0.9666236$, $m_2=0.9668835$ and $C'=167090722627/549755813888$, and $x=\sqrt\alpha$, we have $U(\alpha,m_1)=\mathsf c_0+\mathsf c_1x^2$, where $\mathsf c_0=C'/2+\mathsf W(m_1)$ and $\mathsf c_1=\dI(C')/2$, and $\dLo(\alpha,m_2)=S'x+I'$, with $S'>0$ and $I'>0$ in balls, so $\dLo>0$ and
\[
  f(x):=\delow(m_2)-\deup(m_1)=\frac{m_2^2-m_1^2}2+\frac12\bigl[(S'x+I')^2-(\mathsf c_0+\mathsf c_1x^2)^2\bigr]
\]
has $f''(x)=S'^2-2\mathsf c_1(\mathsf c_0+\mathsf c_1x^2)-4\mathsf c_1^2x^2$, which decreases in $x>0$. In balls, we find that $f\ge2.103\cdot10^{-8}$ at the lower end of $[0.13755521,\alo]$, $f\ge9.521\cdot10^{-14}$ at the upper end, and $f''\le-1.216$ at the lower end, hence throughout. Thus $f$ is concave in $x$ and positive on the interval, and this completes the proof.
\end{proof}

\subsection{No well below the window}\label{ssec:low}
Theorem~\ref{thm:bracket-window}(b) says nothing below $0.89$, where the constants of Lemma~\ref{lem:strong-concavity} fail, because they need $h\ge1$ and a large $\dwfl$. A cruder fact works at every overlap, namely that the response of the optimal path has slope less than one, by an explicit margin, at its own field. Thus $h$ grows at least like $\int_0^m(1+s^2/\wf(s)^2)^{1/2}\dd s$, which, combined with $L\ge\sqrt\alpha+P^*(m)$, makes $e'=m-Lh$ negative, down to $m=0$.

We fix $\alpha>0$ and $m\in(0,1)$ and use the notation of Lemma~\ref{lem:t0-kappa-nonzero} in unit scale: $M^*=\delta^*(0)/\sqrt\alpha$, $w_1$ as there, and $A_0(m)=\resp_{\kappa^*}'(h(m))$.

\begin{lemma}\label{lem:slope}
For all $\alpha>0$, $m\in(0,1)$,
\begin{equation}\label{eq:dig-slope-A}
  A_0(m)^2\le\frac1{M^{*2}}-\frac{m^2}{w_1},\qquad A_0(m)\le\frac{\wf(m)}{\sqrt{w_1}},\qquad\text{hence}\quad A_0(m)\le\frac{\wf(m)}{\sqrt{m^2+\wf(m)^2}}<1.
\end{equation}
\end{lemma}

\begin{proof}
On $[0,w_1)$ the path vanishes, so $u_w$ and $\mathsf p_w=\partial_x^2\Psi_{\kappa^*}(w,X_w)$ are martingales for the heat flow, with $Q'=\E\mathsf p_w^2$ and $\E\mathsf p_w=A_0$. By Cauchy--Schwarz $Q'\ge A_0^2$, so $Q(w_1)\ge m^2+A_0^2w_1$. By Lemma~\ref{lem:t0-kappa-nonzero}(b), $Q(w_1)=w_1/M^{*2}$, which gives the first inequality, and the second is in \eqref{eq:t0-top-mass}.

For the third, $M^*\ge1$ by Lemma~\ref{lem:t0-kappa-nonzero}(a), so $A_0^2\le1-m^2/w_1$ and $A_0^2\le\wf^2/w_1$. The first bound increases in $w_1$, the second decreases, and they cross at $w_1=m^2+\wf^2$, where both equal $\wf^2/(m^2+\wf^2)$.
\end{proof}

\begin{lemma}\label{lem:h-growth}
$h$ is nondecreasing, $h(0)=0$, and for $0\le a<b<1$,
\begin{equation}\label{eq:dig-h-growth}
  h(b)-h(a)\ge\int_a^b\mu_0(s)\dd s,\qquad\mu_0(s):=\sqrt{1+\frac{s^2}{\wf(s)^2}}.
\end{equation}
So $h(m)\ge\mathsf H(m):=\int_0^m\mu_0>m$ on $(0,1)$. $\mathsf H$ is convex with $\mathsf H(0)=0$. Neither $\mu_0$ nor $\mathsf H$ depends on $\alpha$.
\end{lemma}

\begin{proof}
By Theorem~\ref{thm:t0-shape}(b), $h=-L'$ is continuous and nondecreasing, and $h(0)=0$. Thus $h$ is differentiable at almost every $m_0$, and at such a point $L$ has a second-order Taylor expansion. We fix such $m_0\in(0,1)$, with minimizer $(\kappa_0,C_0)$. For all $m$,
\[
  L(\alpha,m)\le\mathsf l(m):=\frac{C_0}2+\frac\alpha2\int\frac{\dd w}{\delta_{\kappa_0,C_0}}+\inf_\lambda\{\Psi_{\kappa_0}(0,\lambda)-\lambda m\},
\]
with equality at $m_0$. Below $w_1$ the flow is the heat flow, so $\resp_{\kappa_0}$ is real analytic, and $\resp_{\kappa_0}'>0$, since $\resp_{\kappa_0}$ is concave and strictly increasing on $[0,\infty)$ (Lemma~\ref{lem:response-t0}(b),(c)). It follows that $\mathsf l$ is $C^2$ near $m_0$, with $\mathsf l'=-\resp_{\kappa_0}^{-1}$ and $\mathsf l''(m_0)=-1/A_0(m_0)$. Since $\mathsf l-L\ge0$ has a minimum at $m_0$, we have $L''(m_0)\le\mathsf l''(m_0)$, that is $h'(m_0)\ge1/A_0(m_0)\ge\mu_0(m_0)$ by \eqref{eq:dig-slope-A}. For a nondecreasing function $h(b)-h(a)\ge\int_a^bh'$. Finally $\mu_0$ is continuous and nondecreasing, because $(s/\wf(s))'=(\wf+s\lambda_0)/\wf^2>0$.
\end{proof}

\begin{corollary}\label{cor:alpha-one}
For $\alpha\ge1$, $e'<0$ on $(0,1)$. So $\Well_0\subset(0,1)$ and $\alc(0)\le1$.
\end{corollary}

\begin{proof}
Since $L\ge\sqrt\alpha$, we have $e'=m-Lh\le m-\sqrt\alpha\mathsf H(m)<0$.
\end{proof}

\begin{theorem}\label{thm:no-low-well}
For every $\alpha\ge\ahi$ and every $m\in(0,1)$, $e'(m)<0$. Consequently:
\begin{enumerate}[label=\textup{(\alph*)}]
\item $\Well_0\subset(0,\ahi)$, $\alo\le\alc(0;m_s)\le\ahi$ for all $m_s\in(0,0.9666236]$, and $\alo\le\alc(0)\le\ahi$.
\item The same holds for binary patterns, with $\WellR_0$ and $\alcR$. For $\alpha\ge\ahi$, $\eR=e$ on $[0,1]$, and $\eR$ is strictly decreasing on $(0,1)$.
\item For $\alpha\ge\ahi$ the slice-minimum profile has no barrier of extensive depth at any overlap, in the sense of Theorem~\ref{thm:t0-capacity}\textup{(b)}.
\end{enumerate}
\end{theorem}

\begin{proof}
On $[0.89,1)$ this is Theorem~\ref{thm:bracket-window}(b), and it remains to treat smaller overlaps. Recall that $P(m)$ denotes the certified lower bound of $P^*(m)$ at SK nodes. We will use three facts, which hold for all $\alpha\ge\alpha_{\mathrm l}:=\ahi$:
\begin{enumerate}[label=\textup{(\roman*)}]
\item $L(\alpha,m)\ge\sqrt{\alpha_{\mathrm l}}+P^*(m)$, and $P^*$ is nonincreasing on $[0,1]$ and at least the chord of $P$ on a cell;
\item $L(\alpha,b)-L(\alpha,a)\le\wf(b)-P^*(a)$ for $a<b$, independent of $\alpha$, by \eqref{eq:dig-chevet-young};
\item \eqref{eq:dig-h-growth}, independent of $\alpha$.
\end{enumerate}

\emph{Part A, $m\in(0,0.65]$.} On a cell $[a,b]$,
\[
  \frac{e'(m)}m=1-\frac{Lh}m\le1-(\sqrt{\alpha_{\mathrm l}}+P(b))\frac{\mathsf H(a)}a,
\]
because $\mathsf H(m)/m$ is nondecreasing. On the first cell, with $a=0$, we use $\mathsf H(m)\ge m$. It is enough that the right side is negative in balls.

\emph{Part B, $m\in[0.65,0.89]$.} For nodes $y_j<y_i$, integrating \eqref{eq:dig-h-growth} gives $L(y_j)-L(y_i)=\int_{y_j}^{y_i}h\le(y_i-y_j)h(y_i)-\int_{y_j}^{y_i}(t-y_j)\mu_0(t)\dd t$, so $-h(y_i)\le\mathsf s_{ij}$, where
\[
  \mathsf s_{ij}:=\frac{\wf(y_i)-P(y_j)}{y_i-y_j}-\frac1{y_i-y_j}\int_{y_j}^{y_i}(t-y_j)\mu_0(t)\dd t.
\]
For $m=y_i+x$, $x\in[0,\nu]$, we have $L'\le\mathsf s_{ij}-\mu_0(y_i)x$ and $L\ge\ell(x):=\sqrt{\alpha_{\mathrm l}}+$ chord of $P$. If $\mathsf s_{ij}<0$, then $e'\le q(x):=y_i+x+\ell(x)(\mathsf s_{ij}-\mu_0(y_i)x)$, which is convex when $\ell'\le0$, so $q(0),q(\nu)<0$ suffices. These are checked in balls, with the anchor $j$ chosen in floating point, $\mu_0$ bounded below on the lattice $\frac1{800}\mathbb Z$ by its value at the left end, and $\wf$ from above.

The certificate uses $43$ SK nodes in $[0.05,0.9]$, in $160$-bit balls, and Part A holds on $25$ cells, up to $0.65$. On $[0.65,0.675]$ its expression is positive, and Part B applies there. Part B holds on $14$ cells, with $\mathsf s_{ij}<0$ and $P$ decreasing in every cell, and $\max q<-0.0284228$ over all cells, which proves the first statement. Table~\ref{tab:dig-low} shows samples.

For (a), the first statement gives $\Well_0\subset(0,\ahi)$, $\alc(0)\le\ahi$, and $\alc(0;m_s)\le\ahi$ for all $m_s\in(0,1)$, and the lower end is Theorem~\ref{thm:bracket-window}(a). Part (b) follows from Theorem~\ref{thm:merging-t0}(a), (c), and $e'<0$, and (c) is Theorem~\ref{thm:t0-capacity}(b).
\end{proof}

\begin{table}[ht]
\caption{Samples from the proof of Theorem~\ref{thm:no-low-well}. Decimals are midpoints of balls, and acceptance uses ball comparisons only. Left: the bound $1-(\sqrt{\ahi}+P(b))\mathsf H(a)/a$ in Part A. Right: Part B, with $y_i$, anchor $y_j$, $\mathsf s_{ij}$, $q(0)$ and $q(\nu)$.}\label{tab:dig-low}
\centering
\begin{tabular}{lr}
\toprule
$[a,b]$ & bound\\
\midrule
$[0,0.05]$ & $-0.13310$\\
$[0.25,0.275]$ & $-0.11322$\\
$[0.5,0.525]$ & $-0.05332$\\
$[0.6,0.625]$ & $-0.01467$\\
$[0.625,0.65]$ & $-0.00336$\\
\bottomrule
\end{tabular}
\qquad
\begin{tabular}{lllll}
\toprule
$y_i$ & $y_j$ & $\mathsf s_{ij}$ & $q(0)$ & $q(\nu)$\\
\midrule
$0.65$ & $0.425$ & $-0.8132$ & $-0.06530$ & $-0.05614$\\
$0.75$ & $0.625$ & $-1.0581$ & $-0.07494$ & $-0.05853$\\
$0.85$ & $0.775$ & $-1.3853$ & $-0.05549$ & $-0.02842$\\
$0.875$ & $0.825$ & $-1.4902$ & $-0.04433$ & $-0.03919$\\
$0.8875$ & $0.85$ & $-1.5476$ & $-0.03736$ & $-0.03475$\\
\bottomrule
\end{tabular}
\end{table}

\subsection{The limit $T\to0$}\label{ssec:tzero}
Theorem~\ref{thm:t0-capacity}(d) compares $\alc(T;m_s)$ with the capacity at $T=0$ on a window. For the unrestricted $\alc(T)$, we must also exclude a contact point that sinks to $m=0$ as $T\to0$, where $e'(0)=0$ and the convergence of $\Lam'/\beta$ to $e'$ says nothing. We will do this with Proposition~\ref{prop:low-overlap-criterion}.

\begin{theorem}\label{thm:t-to-zero}
For $T$ small enough, $\alc(T)\le\ahi$. Also $\liminf_{T\to0}\alc(T)\ge\alo$. So every limit point of $\alc(T)$ as $T\to0$ lies in $[\alo,\ahi]$, and by Theorem~\ref{thm:capacity-pm} the same holds for $\alcR(T)$. The same holds for $\alc(T;m_s)$ and $\alcR(T;m_s)$, for every $m_s\in(0,0.9666236]$.
\end{theorem}

\begin{proof}
The lower bound is Theorem~\ref{thm:t0-capacity}(d) with Theorem~\ref{thm:bracket-window}(c), since $\Well_T$ contains $\Well_T(m_s)$.

For the upper bound, we may take $\alpha\in[\ahi,2]$ by Corollary~\ref{cor:load-bound}. By Theorem~\ref{thm:no-low-well} and continuity of $e'$ in $(\alpha,m)$, we have $e'<0$ uniformly on $[\ahi,2]\times[0.4,\bar m]$, for each $\bar m<1$. Theorem~\ref{thm:t0-capacity}(d) gives $\Lam'/\beta\to e'$ uniformly there, and $\Lam'<0$ on $[\bar m,1)$ for some $\bar m<1$ and $\beta$ large. Thus $\Lam'<0$ on $[0.4,1)$ for $\beta$ large, uniformly on the load interval.

On $(0,0.4]$, let $v^*(m)$ be the optimal radius and $\gamma^*=\beta v^*(m)$. By Theorem~\ref{thm:t0-capacity}(d), $\alpha v^*/\beta\to L(m)^2$ uniformly, and $L(m)\ge L(0.4)\ge\sqrt{\ahi}+P^*(0.4)>1.04$, by $P^*(0.4)\ge0.6747288067$ at the SK node $0.4$. The same limit and $L\ge\sqrt\alpha$ give $\gamma^*=\beta^2v^*/\beta\ge\beta^2/2$ for $\beta$ large. Hence $\Dylow(\gamma^*,\alpha)\ge\Dylow(\beta^2/2,\ahi)$ by Lemma~\ref{lem:finite-gamma}(c), and this tends to $1$. Indeed, since $\mathsf f(s)\ge c\min(s,1)$ for some $c>0$, for each $w_0<1$ the inequality defining $\Dylow$ fails on $(0,w_0]$ once $\gamma$ is large. Therefore $\alpha v^*\Dylow(\gamma^*,\alpha)>\beta$ on $(0,0.4]$ for $\beta$ large, and Proposition~\ref{prop:low-overlap-criterion} gives $\Lam'<0$ there, so no $\alpha\ge\ahi$ lies in $\Well_T$ for $T$ small. This proves the upper bound.

For a window, $\alc(T;m_s)\le\alc(T)$ gives the upper side, and Theorem~\ref{thm:t0-capacity}(d) with Theorem~\ref{thm:no-low-well}(a) gives $\liminf_{T\to0}\alc(T;m_s)\ge\alc(0;m_s)\ge\alo$. Finally, Theorem~\ref{thm:capacity-pm} covers binary patterns.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main-zero}]
Part (a) is Corollary~\ref{cor:energy-limit}, Theorem~\ref{thm:t0-existence}, Corollary~\ref{cor:t0-unit} and Theorem~\ref{thm:t0-shape}(c); (b) is Theorem~\ref{thm:merging-t0}(c) with Theorem~\ref{thm:t0-shape}(c) and Definition~\ref{def:t0-wells}; (c) is Theorem~\ref{thm:no-low-well}(a) with Theorem~\ref{thm:merging-t0}(c); (d) is Theorem~\ref{thm:no-low-well}(b) and Theorem~\ref{thm:certified-wells}; (e) is Theorem~\ref{thm:t0-capacity}(b); and (f) is Theorem~\ref{thm:t-to-zero}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Part (a) is Theorem~\ref{thm:capacity-pm}(a) with Theorem~\ref{thm:main-capacity}(a), (b) is Theorem~\ref{thm:main-capacity}(b), and (c) and (d) are Theorems~\ref{thm:no-low-well} and~\ref{thm:certified-wells}.
\end{proof}