\section{Merging retained patterns}\label{sec:merging}

In this section, we prove the inequality that decides the capacity for binary patterns. By Theorem~\ref{thm:reduction}, the free energy at fixed overlap is a supremum over states correlated with finitely many retained patterns, and we show that such a state is never better than a Gaussian-row state with the single overlap $V$. The inequality is Jensen's inequality on the sphere of overlap vectors. Its field form is due to \cite[Lemma~2.4, Proposition~5.2]{ABCHIM}, and at a fixed slice it takes two lines from the bound \eqref{eq:path-linear-bound}, at the path that is optimal at $V$. What this section adds is the strict inequality with a gap uniform in $r$, the identity of the well sets, and the zero-temperature theorem.

\subsection{Notation}
Let $r\ge1$, $d_r=2^{r-1}$, $\eta\in\{-1,1\}^{r-1}$, and let $\Eeta$ be the uniform average. For $\tv\in[-1,1]^{d_r}$ let $m=\Eeta t_\eta$ and $\ov=\Eeta[\eta t_\eta]\in\R^{r-1}$. We write $x_\eta:=\ov\cdot\eta$, $R_\eta:=m+x_\eta$, and $V=V(\tv)=(m^2+|\ov|^2)^{1/2}$. Since $1,\eta_2,\dots,\eta_r$ are orthonormal in $L^2(\eta)$,
\begin{equation}\label{eq:orthonormal}
  \Eeta R_\eta^2=V^2,\qquad \Eeta[R_\eta t_\eta]=V^2,\qquad V^2\le\Eeta t_\eta^2\le1.
\end{equation}
Recall that $\Lam'(m)=\beta m-h(m)$ and $h(m)\ge\atanh m$ by \eqref{eq:h-bounds}, so $\Lam'\to-\infty$ as $m\to1$, and $\Lam(1)<\max_{[m,1]}\Lam$ for every $m<1$.

For a law $\nu$ with bounded support and quantile path $a$, we will use $\Psi_a$, also written $\Psi_\nu$, $\resp_a=\Psi_a'$ and $LC_a(s)=\inf_u\{\Psi_a(u)-us\}=-g_s(a)$ from Section~\ref{sec:convex}.

\subsection{Jensen on the sphere of overlap vectors}

\begin{lemma}\label{lem:jensen-sphere}
Let $\Psi:\R\to\R$ be even and $C^1$, with $\resp:=\Psi'$ nondecreasing, concave on $[0,\infty)$, and $\resp(0)=0$. Let $LC(s):=\inf_u\{\Psi(u)-us\}$. Then for all $r$ and $\tv\in[-1,1]^{d_r}$,
\begin{equation}\label{eq:jensen-basic}
  \Eeta LC(t_\eta)\le LC(V(\tv)).
\end{equation}
More precisely, if $V>0$, then for every $\lambda\ge0$,
\begin{equation}\label{eq:jensen-lambda}
  \Eeta LC(t_\eta)\le\Eeta\Psi(\lambda R_\eta/V)-\lambda V\le\Psi(\lambda)-\lambda V.
\end{equation}
\end{lemma}

\begin{proof}
Since $\resp$ is concave with $\resp(0)=0$, the function $\resp(u)/u$ is nonincreasing on $(0,\infty)$. Hence $G(q):=\Psi(\sqrt q)$ has $G'(q)=\resp(\sqrt q)/(2\sqrt q)$ nonincreasing, so $G$ is concave on $[0,\infty)$. If $V>0$, we use the trial $u_\eta:=\lambda R_\eta/V$ in the definition of $LC(t_\eta)$ and \eqref{eq:orthonormal} to obtain
\[
  \Eeta LC(t_\eta)\le\Eeta\Psi(\lambda R_\eta/V)-\frac\lambda V\Eeta[R_\eta t_\eta]=\Eeta G(\lambda^2R_\eta^2/V^2)-\lambda V\le G(\lambda^2)-\lambda V=\Psi(\lambda)-\lambda V.
\]
Since $\Psi$ is even, the infimum defining $LC(V)$, $V\ge0$, may be restricted to $u\ge0$, because $\Psi(-u)+uV\ge\Psi(u)-uV$. Taking the infimum over $\lambda\ge0$ gives \eqref{eq:jensen-basic}. If $V=0$, then $m=0$, and concavity of $LC$ gives $\Eeta LC(t_\eta)\le LC(m)=LC(0)$.
\end{proof}

The function $\Psi_a$ satisfies these hypotheses by Lemma~\ref{lem:response}(i), and so does $\Psi_\kappa(0,\cdot)$ by Lemma~\ref{lem:response-t0}.

\begin{theorem}\label{thm:merging}
For $\alpha,\beta,\gamma>0$, $r\ge1$ and $\tv\in[-1,1]^{d_r}$,
\begin{equation}\label{eq:merging}
  \Phimix_r(\gamma;\tv)\le\Phimix(\gamma;V(\tv)),\qquad\Phi^{(r)}(\tv)\le\Phi(V(\tv)),\qquad\Fr(\tv)\le\Lam(V(\tv)).
\end{equation}
Hence for $m\in[0,1]$,
\begin{equation}\label{eq:envelope}
  \Lam(m)\le\LamR(m)\le\max_{V\in[m,1]}\Lam(V),\qquad\LamR(0)=\max_{[0,1]}\Lam.
\end{equation}
\end{theorem}

\begin{proof}
We bound the multi-type problem by a linear functional of the path, and choose the path that is optimal at the merged overlap $V$, not at $m$.

First let $V:=V(\tv)<1$ and $\gamma>0$. We put $\tau=\gamma\alpha/2$ and $a:=\tau y$, where $y$ denotes the optimal path at the slice $V$. By \eqref{eq:path-linear-bound}, used for $r$ types and then for one type with equality, and Lemma~\ref{lem:jensen-sphere}, we have
\[
  \Phimix_r(\gamma;\tv)\le\tau+\alpha\psiSs(y)+\Eeta LC_a(t_\eta)\le\tau+\alpha\psiSs(y)+LC_a(V)=\Phimix(\gamma;V).
\]
At $\gamma=\beta v$ we add $\rad(v)$, take the supremum over $v$ using \eqref{eq:radial}, and add $\beta V^2/2$. If $V=1$, we apply the above to $(1-\varepsilon)\tv$, whose $V$ is $1-\varepsilon$, and let $\varepsilon\to0$ by continuity, using \eqref{eq:mix-lipschitz} for the first inequality.

It remains to prove \eqref{eq:envelope}. The bounds $V(\tv)\ge|m|$ and $V(\tv)\le1$ give the upper bound, and the case $r=1$ gives the lower bound. At $m=0$, for $s\in[0,1]$, the choice $r=2$, $\tv=(s,-s)$ has $m=0$ and $\ov=s$. Since $g_{-s}=g_s$, by Proposition~\ref{prop:parisi-facts}(b) for $s<1$ and by \eqref{eq:gr-g-values} for $s=1$, \eqref{eq:gr-y-form} gives $\Phimix_2(\gamma;\tv)=\Phimix(\gamma;s)$, so $\Phi^{(2)}(s,-s)=\Phi(s)$ by \eqref{eq:radial}, and $\mathcal F_2(\tv)=\Lam(s)$. This completes the proof.
\end{proof}

\begin{remark}\label{rem:abchim-link}
In \cite{ABCHIM}, Lemma~2.4 is the concavity in the square of the field, and Proposition~5.2, (5.21), compares $k$ retained fields with one at fixed radius. Thus \eqref{eq:merging} at a fixed slice also follows from (5.21), their Proposition~5.1, (5.3), and convexity in $m$. The upper bound in $\LamR(0)=\max\Lam$ is their Theorem~1.4 and Corollary~1.5 with the gauge, and the same holds at $T=0$ as $\beta\to\infty$.
\end{remark}

\begin{corollary}\label{cor:r-one-envelope}
Let $E:=\{m\in[0,1]:\Lam(m)=\max_{[m,1]}\Lam\}$. On $E$, $\LamR=\Lam$. If $\alpha\notin\Well_T$, then $E=[0,1]$, so $\LamR=\Lam$ on $[0,1]$ and $\LamR$ is nonincreasing there.
\end{corollary}

\begin{proof}
This is \eqref{eq:envelope}. If $\alpha\notin\Well_T$, then $\Lam$ is nonincreasing on $(0,1)$, and it is continuous.
\end{proof}

\subsection{The strict form}
The bound \eqref{eq:envelope} alone does not give $\Well_T(m_s)\subset\WellR_T(m_s)$, since the value $\LamR(m_1)$ could rise up to $\max_{[m_1,1]}\Lam$ and fill the well. We will need strictness at the barrier, with a gap that does not disappear as $r\to\infty$. The gap comes from the difference between Jensen's inequality for the pair $\pm\eta$ and for the mean, and from the fact that a Rademacher sum cannot concentrate near zero.

\begin{lemma}\label{lem:strict-analytic}
Let $\Psi:\R\to\R$ be even and $C^1$, with $\resp:=\Psi'$ real analytic, strictly increasing, bounded, and concave on $[0,\infty)$. Then $\resp(u)/u$ is strictly decreasing on $(0,\infty)$, and $q\mapsto\Psi(\sqrt q)$ is strictly concave on $[0,\infty)$.
\end{lemma}

\begin{proof}
By oddness, $\resp(0)=0$, so $\resp(u)/u$ is nonincreasing by concavity. If it were constant, equal to $c$, on an interval in $(0,\infty)$, then $\resp=cu$ on $\R$ by analyticity, and boundedness gives $c=0$, contrary to strict monotonicity. The last claim follows as in the proof of Lemma~\ref{lem:jensen-sphere}, since $\frac{\dd}{\dd q}\Psi(\sqrt q)=\resp(\sqrt q)/(2\sqrt q)$.
\end{proof}

\begin{theorem}\label{thm:merging-strict}
Let $0<m_b<1$ and $L_*:=\max_{[m_b,1]}\Lam>\Lam(m_b)$. Then $\LamR(m_b)<L_*$.
\end{theorem}

\begin{proof}
Fix $r$, $\tv$ with $\Eeta t_\eta=m_b$ and $V:=V(\tv)\in(0,1)$. Let $v$ be the maximizer of \eqref{eq:radial} at $\tv$, let $y$ be the optimal path at the slice $V$, and put $\gamma=\beta v$ and $a:=\tau_vy$. We write $\nu$ for its law, and we set $\lambda:=\lambda(\nu,V)=\resp_\nu^{-1}(V)>0$ and $G:=\Psi_\nu(\sqrt\cdot)$. By \eqref{eq:path-linear-bound} and \eqref{eq:jensen-lambda} at this $\lambda$, we have
\[
  \Fr(\tv)\le\Lam(V)-g(\tv),\qquad g(\tv):=G(\lambda^2)-\Eeta G(\lambda^2R_\eta^2/V^2).
\]
We pair $\eta$ with $-\eta$: since $R_{\pm\eta}=m_b\pm x_\eta$, we set
\[
  \Delta(x):=G\Bigl(\frac{\lambda^2(m_b^2+x^2)}{V^2}\Bigr)-\frac12\Bigl[G\Bigl(\frac{\lambda^2(m_b+x)^2}{V^2}\Bigr)+G\Bigl(\frac{\lambda^2(m_b-x)^2}{V^2}\Bigr)\Bigr]\ge0.
\]
Then $\Eeta G(\lambda^2R_\eta^2/V^2)=\Eeta G(\lambda^2(m_b^2+x_\eta^2)/V^2)-\Eeta\Delta(x_\eta)$, and $\Eeta G(\lambda^2(m_b^2+x_\eta^2)/V^2)\le G(\lambda^2)$ by Jensen, since $\Eeta x_\eta^2=|\ov|^2$. Thus $g(\tv)\ge\Eeta\Delta(x_\eta)$.

Suppose $\LamR(m_b)\ge L_*$. By \eqref{eq:envelope} equality holds, and there are $(r_n,\tv_n)$ with $\mathcal F_{r_n}(\tv_n)\to L_*$. A pair with $V(\tv)=1$ has $\Fr(\tv)\le\Lam(1)<L_*$, so $V_n\in[m_b,1)$ for $n$ large and the bound above applies. Since $\Lam(V_n)\le L_*$, along a subsequence $V_n\to V_\infty$ with $\Lam(V_\infty)=L_*$, and $g(\tv_n)\to0$. Now $\Lam(m_b)<L_*$ and $\Lam(1)<L_*$ give $V_\infty\in(m_b,1)$, hence $|\ov_n|^2\to\varsigma^2:=V_\infty^2-m_b^2>0$. Theorem~\ref{thm:gaussian-slice}(c) gives $v_n\le v_1$, so the laws $\nu_n$ have support in $[0,S]$ with $S:=\alpha\beta v_1/2$, and along a subsequence $v_n\to v_\infty$. By Theorem~\ref{thm:uniqueness-derivative}(b), the optimal path at $(\beta v_n,V_n)$ converges to the one at $(\beta v_\infty,V_\infty)$, so $\nu_n\to\nu_\infty$ in $W_1$, where $\nu_\infty$ is the law of $\tau_{v_\infty}y_\infty$ with $y_\infty$ optimal at $(\beta v_\infty,V_\infty)$. Moreover, $g_{m}-I(m)$ is even, convex and $2\int s\dd\nu_n$-Lipschitz in $m$ by Proposition~\ref{prop:parisi-facts}, so $\atanh V_n\le\lambda_n\le\atanh V_n+2S$. Thus $\lambda_n\to\lambda_\infty>0$ along a subsequence, with $\resp_{\nu_\infty}(\lambda_\infty)=V_\infty$.

The anti-concentration step is independent of $r$. Indeed, $x_\eta=\ov\cdot\eta$ is a Rademacher sum, with $\Eeta x^2=|\ov|^2$ and $\Eeta x^4\le3|\ov|^4$, so the Paley--Zygmund inequality gives $\P(x^2\ge|\ov|^2/2)\ge1/12$, and Markov's inequality gives $\P(x^2>24|\ov|^2)\le1/24$. Thus $x_\eta$ lies in the set $\mathcal K_n:=\{|\ov_n|^2/2\le x^2\le24|\ov_n|^2\}$ with weight at least $1/24$, and for $n$ large, $\mathcal K_n$ lies in the compact set $\mathcal K:=\{\varsigma^2/3\le x^2\le25\varsigma^2\}$.

By Proposition~\ref{prop:parisi-facts}, $(\nu,u)\mapsto\Psi_\nu(u)$ is jointly continuous, so $\Delta_n\to\Delta_\infty$ uniformly on $\mathcal K$, where $\Delta_\infty$ denotes $\Delta$ for $\nu_\infty,\lambda_\infty,V_\infty$. Using Lemma~\ref{lem:finite-gamma}(a), we obtain $\check y_\infty>0$, so $\nu_\infty$ puts no mass on $[0,s_0)$, $s_0:=\tau_{v_\infty}\check y_\infty>0$. There the Parisi equation is the heat equation, whence $\phi_{\nu_\infty}(0,u)=\E\phi_{\nu_\infty}(s_0,u+\sqrt{2s_0}z)$ and $\resp_{\nu_\infty}$ is real analytic. With Lemma~\ref{lem:response}(i), Lemma~\ref{lem:strict-analytic} applies, so $G_\infty$ is strictly concave. Hence $\Delta_\infty>0$ on $\mathcal K$, because $m_b>0$ and $x\ne0$ make $(m_b+x)^2\ne(m_b-x)^2$. Thus $\liminf g(\tv_n)\ge\frac1{24}\min_{\mathcal K}\Delta_\infty>0$, which contradicts $g(\tv_n)\to0$.
\end{proof}

\subsection{The capacity for binary patterns}\label{ssec:capacity-pm}

\begin{theorem}\label{thm:capacity-pm}
Let $T>0$ and $m_s\in(0,1)$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item $\WellR_T(m_s)=\Well_T(m_s)$. So $\alcR(T)=\alc(T)$ and
\[
  \alcR(T;m_s)=\alc(T;m_s)=\sup\bigl\{\alpha:\ \exists\, m\in[m_s,1),\ \beta m>\lambda(\nu^*(m;\alpha,\beta),m)\bigr\}.
\]
For $T\in(0,1)$ there is $m_0(T)>0$ with $\alcR(T;m_0)=\alc(T)$.
\item If $\alpha\in\Well_T(m_s)$, there are $m_s\le m_b<m_*<1$ with $\Delta^{\mathrm R}:=\LamR(m_*)-\LamR(m_b)>0$, and for $N$ large, with probability at least $1-e^{-cN}$ for some $c>0$, for all $\mu$ simultaneously, $Z^\mu(k_b)\le\exp(-N\Delta^{\mathrm R}/3)Z^\mu(k_*)$.
\end{enumerate}
\end{theorem}

\begin{proof}
(a) For $\WellR_T(m_s)\subset\Well_T(m_s)$, let $\LamR(m_2)>\LamR(m_1)$ with $m_s\le m_1<m_2<1$. By \eqref{eq:envelope} there is $V_0\in[m_2,1]$ with $\Lam(V_0)\ge\LamR(m_2)>\LamR(m_1)\ge\Lam(m_1)$. If $V_0=1$, we replace it by a point of $(m_1,1)$ with a larger value, as $\Lam'\to-\infty$ at $1$. Thus $\alpha\in\Well_T(m_s)$.

For $\Well_T(m_s)\subset\WellR_T(m_s)$, let $\Lam(m_2)>\Lam(m_1)$ with $m_s\le m_1<m_2<1$, and let $m_*$ be a maximizer of $\Lam$ on $[m_1,1]$, so $m_*\in(m_1,1)$. Theorem~\ref{thm:merging-strict} with $m_b=m_1$ gives $\LamR(m_1)<\Lam(m_*)\le\LamR(m_*)$, hence $\alpha\in\WellR_T(m_s)$.

The formula is the second form of the definition (Theorem~\ref{thm:capacity-gaussian}), and taking unions gives $\alcR(T)=\alc(T)$. For $m_0$, let $T<1$. Then $\alc(T)$ is positive by Theorem~\ref{thm:capacity-gaussian} and finite by Corollary~\ref{cor:load-bound}, so Theorem~\ref{thm:capacity-gaussian} gives a contact point $m_c>0$, the limit of points $m_n$ with $\beta m_n>h(m_n;\alpha_n)$, $\alpha_n\to\alc(T)$, $\alpha_n\in\Well_T$. With $m_0:=m_c/2$, for $n$ large $\alpha_n\in\Well_T(m_0)$, hence $\alc(T;m_0)=\alc(T)$. This proves (a).

(b) We apply Theorem~\ref{thm:reduction}(iii) and the concentration bound \eqref{eq:conc} to the two slices $k_b$, $k_*$, for each pattern, with deviation parameter $\Delta^{\mathrm R}/4$, and take a union bound over $M$ patterns.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main-free-energy}]
Part (a) and the statements on \eqref{eq:main-formula} are Theorem~\ref{thm:gaussian-slice} with Theorems~\ref{thm:radius-free} and~\ref{thm:uniqueness-derivative}; part (b) is Theorem~\ref{thm:reduction}(iii); part (c) is Theorem~\ref{thm:merging}, Corollary~\ref{cor:r-one-envelope} and Theorem~\ref{thm:merging-strict}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main-capacity}]
Parts (a), (b), (c) are Theorem~\ref{thm:capacity-gaussian} and Corollary~\ref{cor:load-bound}; part (d) is Theorem~\ref{thm:capacity-pm}(b); part (e) is Corollary~\ref{cor:r-one-envelope}.
\end{proof}

\begin{remark}
Here $\Delta^{\mathrm R}$ comes from compactness and is not explicit. The identity $\LamR=\Lam$ fails at $m=0$ as soon as a state at positive overlap beats $m=0$, and by continuity on an interval next to $0$. The set where the identity holds contains $E$, and the capacity does not depend on it.
\end{remark}

\subsection{Zero temperature}
With the notation of Section~\ref{sec:zero}, let $L^{(r)}(\tv)$ be the limit of $\frac1N\E\max|G\sigma|$ on the type slice. For simplicity of notation, in this subsection we write $\Psi_\kappa:=\Psi_\kappa(0,\cdot)$ and $\resp_\kappa:=\Psi_\kappa'$, and we put $W_s(\kappa)=\inf_\lambda\{\Psi_\kappa(\lambda)-\lambda s\}$. Recall from \eqref{eq:theta-t0} that $\eR=\sup_r\Theta^0_r$.

\begin{theorem}\label{thm:merging-t0}
Fix $\alpha>0$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item For all $r$ and $\tv$: $L^{(r)}(\tv)\le L(V(\tv))$, so $\frac12(m^2+|\ov|^2)+\frac12L^{(r)}(\tv)^2\le e(V(\tv))$, and
\[
  e(m)\le\eR(m)\le\max_{[m,1]}e,\qquad\eR(0)=\max_{[0,1]}e.
\]
On $\{m:e(m)=\max_{[m,1]}e\}$, $\eR=e$. If $\alpha\notin\Well_0$, then $\eR=e$ on $[0,1]$ and $\eR$ is nonincreasing.
\item If $0<m_b<1$ and $e_*:=\max_{[m_b,1]}e>e(m_b)$, then $\eR(m_b)<e_*$.
\item $\WellR_0(m_s)=\Well_0(m_s)$ for all $m_s\in(0,1)$, so $\alcR(0;m_s)=\alc(0;m_s)$. If $\alpha\in\Well_0(m_s)$, Theorem~\ref{thm:t0-capacity}(b) gives the landscape statement for all patterns with $\Delta=\eR(m_*)-\eR(m_b)>0$.
\end{enumerate}
\end{theorem}

\begin{proof}
(a) By Theorem~\ref{thm:merging}, $\Phi^{(r)}(\tv)\le\Phi(V(\tv))$ at every $\beta$, and by Theorem~\ref{thm:t0-existence}(b), (e), $\frac\beta2L^{(r)}(\tv)^2-\log2\le\Phi^{(r)}(\tv)$ and $\Phi(V)\le\frac\beta2L(V)^2$. We divide by $\beta$ and let $\beta\to\infty$. The bounds on $\eR$ follow in the same way from \eqref{eq:envelope}, since $\Lam_\beta/\beta\to e$ and $\LamR_\beta/\beta\to\eR$ uniformly on $[0,1]$ (Theorem~\ref{thm:t0-existence}(b), Corollary~\ref{cor:energy-limit}). If $\alpha\notin\Well_0$, then $e$ is nonincreasing on $[0,1]$ by Definition~\ref{def:t0-wells} and continuity, so $e(m)=\max_{[m,1]}e$ for all $m$, which proves (a).

(b) We follow the proof of Theorem~\ref{thm:merging-strict}, with these changes. Let $(\kappa_n,C_n)$ be the minimizer at $V_n$, and put $\lambda_n:=\lambda(\kappa_n,V_n)$ and $G_n:=\Psi_{\kappa_n}(\sqrt\cdot)$. Then $L^{(r_n)}(\tv_n)\le L(V_n)-g(\tv_n)/\sqrt\alpha$ and $g(\tv_n)\ge\Eeta\Delta_n(x_\eta)$. If the value tends to $e_*$, then, as before, $V_n<1$ for $n$ large and $V_n\to V_\infty\in(m_b,1)$, because $e'\to-\infty$ at $1$ (Theorem~\ref{thm:t0-shape}(c)) gives $e(1)<e_*$, and $g(\tv_n)\to0$ since $L\ge\sqrt\alpha>0$.

By the proof of Corollary~\ref{cor:t0-unit}(b), $(\kappa_n,C_n)\to(\kappa_\infty,C_\infty)$, the minimizer at $V_\infty$, in the sense used there, and $\lambda_n\to\lambda_\infty>0$. Hence $\kappa_n\to\kappa_\infty$ a.e., with $\int_0^{w_0}|\kappa_n-\kappa_\infty|\to0$ for $w_0<1$. By Lemma~\ref{lem:flow-compare}(iv) with $w=w_0$, $w'=1$, we have $|x|\le\Psi_\kappa(w_0,x)-\frac\alpha2\delta_\kappa(w_0)\le|x|+\sqrt{2\alpha(1-w_0)/\pi}$ for every $\kappa\in\mathcal U$. Since the flow is a sup-norm contraction which commutes with constants, and Lemma~\ref{lem:flow-compare}(ii) compares $\kappa_n$ with $\kappa_\infty$ on $[0,w_0]$, we get $\Psi_{\kappa_n}(x)-\Psi_{\kappa_n}(0)\to\Psi_{\kappa_\infty}(x)-\Psi_{\kappa_\infty}(0)$ uniformly in $x$. As $\Delta_n$ depends only on such differences, $\Delta_n\to\Delta_\infty$ uniformly on $\mathcal K$.

We next prove strictness of $\Delta_\infty$. With $w_1$ as in Lemma~\ref{lem:t0-kappa-nonzero} for the minimizer at $V_\infty$, part (b) there gives $w_1\ge V_\infty^2>0$. Below $w_1$ the flow is pure heat, so $\resp_{\kappa_\infty}$ is real analytic, and Lemma~\ref{lem:strict-analytic} applies by Lemma~\ref{lem:response-t0}. Hence $G_\infty$ is strictly concave and $\Delta_\infty>0$ on $\mathcal K$. The anti-concentration step is unchanged. This completes the proof of (b).

(c) We argue as for Theorem~\ref{thm:capacity-pm}. Since $e'\to-\infty$ at $1$, the value $e(1)$ is exceeded just below $1$, which handles the replacement of $V_0=1$. The landscape statement is Theorem~\ref{thm:t0-capacity}(b) with $\eR$.
\end{proof}