\section{Zero temperature}\label{sec:zero}
\newcommand{\zSph}{\operatorname{Sph}}
\newcommand{\zU}{\mathcal{U}}
\newcommand{\zV}{\mathcal{V}}
\newcommand{\zB}{\mathcal{B}}
\newcommand{\zS}{\Psi^{\mathrm{S}}}
\newcommand{\zT}{\mathcal{T}}
\newcommand{\zP}{\mathcal{P}}
\newcommand{\zN}{\mathcal{N}}
\newcommand{\zd}{\mathsf{d}}
\newcommand{\zr}{\mathsf{r}}
\newcommand{\zI}{\mathcal{I}}
\newcommand{\zIS}{\mathcal{I}^{\mathrm{S}}}
\newcommand{\zK}{\mathcal{K}}
\newcommand{\zA}{\mathcal{A}}
\newcommand{\zF}{\mathbb{F}}

In this section, we show that the energy at fixed overlap $e(m)=m^2/2+L(m)^2/2$ exists and that $L$ is the minimum of an explicit convex functional with a unique minimizer and is jointly concave and $C^1$ in $(\alpha,m)$. We define the capacity at $T=0$ and characterize it by a one-dimensional condition, namely the sign of $e'$ in the overlap window, and Section~\ref{sec:digits} will turn this into digits. The function space and the zero-temperature Parisi recursion, used here through step functions, are those of Auffinger and Chen \cite{AC17}.

\subsection{Setting and existence of the energy}

Throughout, $G$ is an $M\times N$ matrix with i.i.d.\ $N(0,1)$ entries, $M/N\to\alpha>0$, and $z$ is a standard Gaussian. Recall that $I(m)=\frac{1+m}2\log(1+m)+\frac{1-m}2\log(1-m)$, and note that $2^{-N}|\Slice(k)|\le e^{-NI(m_k)}$ exactly. Let $\lambda_0(m):=\Nc^{-1}((1+m)/2)$ and
\[
\wf(m):=2\Nd(\lambda_0(m))=\inf_\lambda\{\E|z+\lambda|-\lambda m\}.
\]
The function $\wf$ is concave on $[-1,1]$, with $\wf'=-\lambda_0$.

For nonempty $A\subset\Cube$, let $L_A(G):=\frac1N\max_{\sigma\in A}|G\sigma|$ and
\[
L_N(k):=\E L_{\Slice(k)}(G),\qquad e_N(k):=\frac{m_k^2}2+\frac12\E L_{\Slice(k)}(G)^2.
\]
In this section, $e_N$ denotes this energy for Gaussian rows, not the energy \eqref{eq:eN} for binary patterns. With a binary pattern $\xi^1$ gauged to all ones, $e_N(k)=\frac1{2N^2}\E\max_{m_1(\sigma)=m_k}[(\xi^1\cdot\sigma)^2+|G\sigma|^2]$.

We write $\Phi_\beta,\Lam_\beta$ for $\Phi,\Lam$ when $\beta$ varies, and we will use three facts from Section~\ref{sec:gaussian}. First, by Theorem~\ref{thm:gaussian-slice}, $\sup_k|\frac1N\E\log Y_{N,k}-\Phi_\beta(m_k)|\to0$, and $\Phi_\beta+I$ is Lipschitz. Second, let $\mu_M$ be the uniform measure on the sphere of radius $\sqrt M$ in $\R^M$ and
\[
Z^{\mathrm{mix}}_N(\gamma;k):=2^{-N}\sum_{\sigma\in\Slice(k)}\int\mu_M(\dd\tau)\exp\big(\sqrt{\gamma/N}\,\tau\cdot G\sigma\big).
\]
This is $Z_{N,k}(s)$ of \eqref{eq:gr-radial-id} at $\gamma=\beta s^2$, so by \eqref{eq:gr-mixed-limit}, for each $\gamma_0$, $\sup_{\gamma\le\gamma_0}\sup_k|\frac1N\E\log Z^{\mathrm{mix}}_N(\gamma;k)-\Phimix(\gamma;m_k)|\to0$. Third, by \eqref{eq:phimix} and \eqref{eq:radial},
\begin{equation}\label{eq:t0-phimix}
\begin{aligned}
\Phimix(\gamma;m)&=\inf_{x,y\in\Qone}\Big\{\frac{\gamma\alpha}2\Big(1-\int_0^1xy\Big)+J_m(x)+\alpha\psiSs(y)\Big\},\\
\Phi_\beta(m)&=\sup_{v>0}\{\rad(v)+\Phimix(\beta v;m)\}.
\end{aligned}
\end{equation}
Let $\zSph(\theta^2):=b-1-\frac12\log b$ with $b:=\frac12(1+\sqrt{1+4\theta^2})$. By Laplace's method with exponent $\sup_u\{\theta u+\frac12\log(1-u^2)\}$, this is the limit of $\frac1M\log\int\mu_M(\dd\tau)\exp(\theta\sqrt M\tau_1)$. Since $b\in[\theta+\frac12,\theta+1]$,
\begin{equation}\label{eq:t0-sph}
\theta-\frac12-\frac12\log(1+\theta)\le\zSph(\theta^2)\le\theta.
\end{equation}

\begin{theorem}\label{thm:t0-existence}
There are continuous functions $L,e$ on $[-1,1]$, with $e=m^2/2+L^2/2$ and $L\ge\sqrt\alpha$, such that the following hold.
\begin{enumerate}[label=(\alph*)]
\item $\sup_k|L_N(k)-L(m_k)|\to0$ and $\sup_k|e_N(k)-e(m_k)|\to0$.
\item For all $\beta>0$ and $m\in[-1,1]$, $\beta L(m)^2/2-\log2\le\Phi_\beta(m)\le\beta L(m)^2/2-I(m)$. So $e(m)-\log2/\beta\le\Lam_\beta(m)/\beta\le e(m)-I(m)/\beta$, and $\sup_m|\Lam_\beta/\beta-e|\le\log2/\beta$.
\item For all $\gamma>0$ and $m$,
\[
-\log2+\alpha\zSph\big(\gamma L(m)^2/\alpha\big)\le\Phimix(\gamma;m)\le-I(m)+\alpha\zSph\big(\gamma L(m)^2/\alpha\big).
\]
So, with $\mathsf{b}(\gamma):=\log2+\frac\alpha2+\frac\alpha2\log\big(1+\sqrt\gamma(1+1/\sqrt\alpha)\big)$,
\begin{equation}\label{eq:t0-mix-sandwich}
-\mathsf{b}(\gamma)\le\Phimix(\gamma;m)-\sqrt{\gamma\alpha}\,L(m)\le-I(m)\le0.
\end{equation}
\item $\P(|L_A(G)-\E L_A(G)|>t)\le2e^{-Nt^2/2}$ for every nonempty $A$, and $0\le\frac12\E L_A^2-\frac12(\E L_A)^2\le\frac1{2N}$. So $L_{\Slice(k)}(G)\to L(m_k)$ almost surely, uniformly in $k$.
\item (Types.) For a partition $(I_\eta)$ of $[N]$ with $|I_\eta|/N\to1/d_r$ and $\tv=(t_\eta)\in[-1,1]^{d_r}$, the limit $L^{(r)}(\tv):=\lim\frac1N\E\max|G\sigma|$ over the type slice with magnetizations $\tv$ exists, is continuous in $\tv$, and is uniform in the slices. Moreover $\frac\beta2L^{(r)2}-\log2\le\Phi^{(r)}(\tv)\le\frac\beta2L^{(r)2}-\Eeta I(t_\eta)$, and (c) holds with $I$ replaced by $\Eeta I(t_\eta)$.
\end{enumerate}
In particular, $\lim_{\gamma\to\infty}\Phimix(\gamma;m)/\sqrt\gamma=\sqrt\alpha L(m)$, uniformly in $m$, with error at most $\mathsf{b}(\gamma)/\sqrt\gamma$.
\end{theorem}

\begin{proof}
The partition function on a set is close to its largest term, up to the entropy $\log2$, which gives existence by a Cauchy argument in $\beta$.

We start with (d). The function $L_A$ is $N^{-1/2}$-Lipschitz in $G$ for the Hilbert--Schmidt norm: $|L_A(G)-L_A(G')|\le\frac1N\max_A|(G-G')\sigma|\le\|G-G'\|_{\mathrm{HS}}/\sqrt N$, so Gaussian concentration gives the tail, and the Poincar\'e inequality gives $\Var L_A\le1/N$.

For (a) and (b), keeping the largest term or bounding all terms by it gives, pointwise in $G$,
\begin{equation}\label{eq:t0-two-sided}
\frac\beta2L_A(G)^2-\log2\le\frac1N\log2^{-N}\sum_{\sigma\in A}\exp\Big(\frac\beta{2N}|G\sigma|^2\Big)\le\frac\beta2L_A(G)^2+\frac1N\log(2^{-N}|A|).
\end{equation}
With $A=\Slice(k)$, taking expectations gives $\frac\beta2\E L_{\Slice(k)}^2-\log2\le\frac1N\E\log Y_{N,k}\le\frac\beta2\E L_{\Slice(k)}^2-I(m_k)$, so $\limsup_N\sup_k|\frac12\E L_{\Slice(k)}^2-\Phi_\beta(m_k)/\beta|\le\log2/\beta$. Comparing two values of $\beta$ gives $\limsup_N\sup_k|\Phi_\beta(m_k)/\beta-\Phi_{\beta'}(m_k)/\beta'|\le\log2\,(1/\beta+1/\beta')$. As $\Phi_\beta+I$ is Lipschitz, $\Phi_\beta$ is continuous, and since $\{m_k\}$ becomes dense, we get $\sup_m|\Phi_\beta/\beta-\Phi_{\beta'}/\beta'|\le\log2\,(1/\beta+1/\beta')$. Hence $\Phi_\beta/\beta$ converges uniformly to a continuous $\mathcal{G}_0$, and $\sup_k|\frac12\E L_{\Slice(k)}^2-\mathcal{G}_0(m_k)|\to0$. Passing to the limit along $m_k\to m$ gives $\beta\mathcal{G}_0(m)-\log2\le\Phi_\beta(m)\le\beta\mathcal{G}_0(m)-I(m)$. For fixed $\sigma$, $\E|G\sigma|/N=\E|g|/\sqrt N\to\sqrt\alpha$ with $g$ standard in $\R^M$, so $\mathcal{G}_0\ge\alpha/2>0$. We set $L:=\sqrt{2\mathcal{G}_0}$, and with (d) we get $L_N(k)\to L(m_k)$ uniformly. This proves (a) and (b).

(c). Let $\mathcal{E}_M(r):=\int\mu_M(\dd\tau)e^{r\tau_1}$, which is even, log-convex, and nondecreasing on $[0,\infty)$. For fixed $\sigma$, $\int\mu_M(\dd\tau)\exp(\sqrt{\gamma/N}\,\tau\cdot G\sigma)=\mathcal{E}_M(\sqrt{\gamma/N}|G\sigma|)$, so with $A=\Slice(k)$ and $r_*:=\sqrt{\gamma/N}\,NL_A(G)$,
\[
2^{-N}\mathcal{E}_M(r_*)\le Z^{\mathrm{mix}}_N(\gamma;k)\le2^{-N}|A|\mathcal{E}_M(r_*).
\]
The function $\Pi_M(\theta):=\frac1M\log\mathcal{E}_M(\sqrt M\theta)$ is convex, $|\Pi_M'|\le1$ because $\mu_M$ lives on the sphere of radius $\sqrt M$, and $\Pi_M\to\zSph(\theta^2)$ locally uniformly. Here $r_*=\sqrt M\theta_*$ with $\theta_*:=\sqrt{\gamma N/M}\,L_A(G)$, and $\frac1N\log\mathcal{E}_M(r_*)=\frac MN\Pi_M(\theta_*)$ is Lipschitz in $L_A$ with constant $\sqrt{\gamma M/N}$, so by (d) its expectation tends to $\alpha\zSph(\gamma L(m)^2/\alpha)$ along $m_k\to m$. We take logarithms in the previous display, divide by $N$, and take expectations and limits. Then \eqref{eq:t0-mix-sandwich} follows from \eqref{eq:t0-sph} with $\theta=\sqrt{\gamma/\alpha}\,L$, using $\sqrt\alpha\le L\le1+\sqrt\alpha$, where the upper bound holds because $L_A(G)\le\|G\|_{\mathrm{op}}/\sqrt N$ and $\E\|G\|_{\mathrm{op}}\le\sqrt N+\sqrt M$.

(e). The same proof works for types, with $\Phi^{(r)}$ in place of $\Phi$ and the type slices in place of $\Slice(k)$.
\end{proof}

\subsection{The zero-temperature recursion}

Let $\zU$ be the set of nondecreasing, right-continuous $\kappa:[0,1)\to[0,\infty)$ with $\int_0^1\kappa<\infty$, which is the space of \cite{AC17}, and let $\zV:=\zU\times[0,\infty)$. For $(\kappa,C)\in\zV$,
\[
\delta_\kappa(w):=\int_w^1\kappa,\qquad\delta(w):=\delta_{\kappa,C}(w):=C+\delta_\kappa(w),\qquad\zS(\kappa,C):=C+\int_0^1\frac{\dd w}{\delta(w)}\in(0,\infty].
\]
We fix $\alpha$. For $c,b\ge0$ and $f$ 1-Lipschitz, let $\zT_{c,b}f(x):=\frac1c\log\E\exp(cf(x+bz))$ for $c>0$, and $\zT_{0,b}f(x):=\E f(x+bz)$.

\begin{lemma}\label{lem:flow}
\begin{enumerate}[label=(\alph*)]
\item $\zT_{c,b}$ is monotone and $\zT_{c,b}(f+K)=\zT_{c,b}f+K$, so $\sup|\zT_{c,b}f-\zT_{c,b}g|\le\sup|f-g|$.
\item $\zT_{c,b}$ preserves the 1-Lipschitz property, convexity (by H\"older) and evenness, and $\zT_{c,b}\zT_{c,b'}=\zT_{c,\sqrt{b^2+b'^2}}$.
\item $\zT_{c,b}(sx+K)=sx+K+cs^2b^2/2$.
\item For $f$ 1-Lipschitz and $c'\ge c\ge0$, $0\le\zT_{c',b}f-\zT_{c,b}f\le(c'-c)b^2/2$.
\end{enumerate}
\end{lemma}

\begin{proof}
Parts (a)--(c) are straightforward. For (d), let $c>0$ and $\vartheta:=cf(x+b\,\cdot)$, a function of $z$. Then $\partial_c\zT_{c,b}f=c^{-2}\operatorname{Ent}(e^\vartheta)/\E e^\vartheta\ge0$. The Gaussian log-Sobolev inequality gives $\operatorname{Ent}(e^\vartheta)\le\frac12\E[|\partial_z\vartheta|^2e^\vartheta]\le\frac12c^2b^2\E e^\vartheta$, so $0\le\partial_c\zT_{c,b}f\le b^2/2$. Finally, we let $c\downarrow0$, using continuity.
\end{proof}

For a step function $\kappa$ on $[w,w')$, with partition $w=w_0<\dots<w_n=w'$, value $c_i$ on $[w_i,w_{i+1})$ and $b_i^2:=\alpha(w_{i+1}-w_i)$, let
\[
\zP_\kappa^{w,w'}:=\zT_{c_0,b_0}\cdots\zT_{c_{n-1},b_{n-1}},
\]
which does not depend on the partition, by Lemma~\ref{lem:flow}(b).

\begin{lemma}\label{lem:flow-compare}
For $f,g$ 1-Lipschitz and step functions $\kappa,\kappa'$:
\begin{enumerate}[label=(\roman*)]
\item $\zP_\kappa^{w,w'}$ is monotone, commutes with constants, preserves the 1-Lipschitz property, convexity and evenness, and $\sup|\zP_\kappa^{w,w'}f-\zP_\kappa^{w,w'}g|\le\sup|f-g|$.
\item ($L^1$ contraction.) $\sup_x|\zP_\kappa^{w,w'}f-\zP_{\kappa'}^{w,w'}f|\le\frac\alpha2\int_w^{w'}|\kappa-\kappa'|$. If $\kappa\le\kappa'$ then $\zP_\kappa^{w,w'}f\le\zP_{\kappa'}^{w,w'}f$.
\item For $s=\pm1$, $\zP_\kappa^{w,w'}(sx+K)=sx+K+\frac\alpha2\int_w^{w'}\kappa$.
\item With $\bar z:=\sqrt{\alpha(w'-w)}\,z$,
\[
\max\Big\{|x|+\frac\alpha2\int_w^{w'}\kappa,\ \E|x+\bar z|\Big\}\le\zP_\kappa^{w,w'}|\cdot|(x)\le\E|x+\bar z|+\frac\alpha2\int_w^{w'}\kappa\le|x|+\sqrt{\frac{2\alpha(w'-w)}\pi}+\frac\alpha2\int_w^{w'}\kappa.
\]
\end{enumerate}
\end{lemma}

\begin{proof}
Parts (i) and (iii) follow from Lemma~\ref{lem:flow}, and (ii) follows on a common partition by induction, with Lemma~\ref{lem:flow}(a),(d). For the lower bound in (iv), we write $|x|=\max_{s=\pm1}sx$ and use monotonicity and (iii); the second term follows from $\kappa\ge0$ and (ii). For the upper bound, we induct from the top: if $\mathsf{u}(x):=\E|x+\sqrt{v'}z|+K$ is the bound at a level, then by monotonicity and Lemma~\ref{lem:flow}(d) with $c=0$, $\zT_{c,b}\mathsf{u}\le\zT_{0,b}\mathsf{u}+cb^2/2=\E|x+\sqrt{v'+b^2}\,z|+K+cb^2/2$.
\end{proof}

By (ii), $\zP_\kappa^{w,w'}$ extends to every $\kappa\in\zU$ (truncate to $\kappa\wedge n$ and take step functions below it), and (i)--(iv) still hold; compare \cite[Corollary~2]{AC17}. For $\kappa\in\zU$ and $f$ 1-Lipschitz, let
\[
\Psi^f_\kappa(w,x):=\zP_\kappa^{w,1}f(x),\quad\Psi_\kappa:=\Psi_\kappa^{|\cdot|},\quad W_m^{[f]}(\kappa):=\inf_\lambda\{\Psi_\kappa^f(0,\lambda)-\lambda m\},\quad W_m:=W_m^{[|\cdot|]},
\]
and $W_m^{[0,w_0]}(\kappa):=\inf_\lambda\{\zP_\kappa^{0,w_0}|\cdot|(\lambda)-\lambda m\}$. This is the zero-temperature Parisi recursion, in the level variable $w$, with steps $\frac1c\log\E\exp(c\,\cdot)$ and terminal function $|x|$. For step $\kappa$ and $w<1$, $\Psi_\kappa$ is smooth on each slab, its $x$-derivatives are continuous across slabs and bounded for $w\le w_0<1$, and by Cole--Hopf
\begin{equation}\label{eq:t0-pde}
-\partial_w\Psi=\frac\alpha2\big(\partial_x^2\Psi+\kappa(\partial_x\Psi)^2\big).
\end{equation}
By Lemma~\ref{lem:flow-compare}(iv), for every $\kappa\in\zU$, $|m|\le1$ and $w_0<1$,
\begin{gather}
W_m(\kappa)\ge\frac\alpha2\delta_\kappa(0),\label{eq:t0-W-mass}\\
\frac\alpha2\delta_\kappa(w_0)+W_m^{[0,w_0]}(\kappa)\le W_m(\kappa)\le\frac\alpha2\delta_\kappa(w_0)+W_m^{[0,w_0]}(\kappa)+\sqrt{\frac{2\alpha(1-w_0)}\pi}.\label{eq:t0-W-cut}
\end{gather}

\subsection{The optimally controlled process and the supporting hyperplane}

The recursion has a stochastic control representation, compare \cite[Theorem~2]{AC17}, whose value is the supremum of functions convex in $(\kappa,\lambda)$; this gives the convexity of $W_m$ in $\kappa$ and a supporting hyperplane with slope the overlap function $Q$.

\begin{lemma}\label{lem:control}
Let $\kappa_\circ$ be a step function, $\lambda_\circ$ real, and $u(w,x):=\partial_x\Psi_{\kappa_\circ}(w,x)$, so $|u|\le1$. Let $X$ solve $\dd X_w=\alpha\kappa_\circ(w)u(w,X_w)\dd w+\sqrt\alpha\dd B_w$, $X_0=\lambda_\circ$, and put $u_w:=u(w,X_w)$, $Q(w):=Q_{\kappa_\circ,\lambda_\circ}(w):=\E u_w^2$.
\begin{enumerate}[label=(\alph*)]
\item (Verification.) For $\kappa\in\zU$, $\lambda$ real and any bounded adapted control $\vartheta$,
\begin{equation}\label{eq:t0-verification}
\Psi_\kappa(0,\lambda)\ge\E\Big|\lambda+\alpha\int_0^1\kappa\vartheta_w\dd w+\sqrt\alpha B_1\Big|-\frac\alpha2\int_0^1\kappa\,\E\vartheta_w^2\dd w,
\end{equation}
with equality when $\kappa=\kappa_\circ$, $\lambda=\lambda_\circ$, $\vartheta=u_w$. Consequently $(\kappa,\lambda)\mapsto\Psi_\kappa(0,\lambda)$ is convex on $\zU\times\R$.
\item $u_w$ is a bounded martingale on $[0,1)$ with $u_{1-}=\sgn(X_1)$ almost surely and $u_0=\partial_x\Psi_{\kappa_\circ}(0,\lambda_\circ)$. So $Q$ is nondecreasing, $Q(0)=u_0^2$, $Q(1-)=1$, and $\E[\sgn(X_1)u_w]=Q(w)$.
\item (Supporting hyperplane.) For $\kappa\in\zU$ and $\lambda$ real,
\begin{equation}\label{eq:t0-support}
\Psi_\kappa(0,\lambda)\ge\Psi_{\kappa_\circ}(0,\lambda_\circ)+u_0(\lambda-\lambda_\circ)+\frac\alpha2\int_0^1(\kappa-\kappa_\circ)Q\dd w.
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
(a). Let $\kappa$ be a step function and $w\le w_0<1$, let $X^\vartheta$ denote the process inside \eqref{eq:t0-verification}, and put $\mathsf{p}:=\partial_x\Psi_\kappa(w,X^\vartheta_w)$. By It\^o and \eqref{eq:t0-pde},
\[
\dd\Psi_\kappa(w,X^\vartheta_w)=\frac\alpha2\kappa(2\vartheta_w\mathsf{p}-\mathsf{p}^2)\dd w+\text{martingale}\le\frac\alpha2\kappa\vartheta_w^2\dd w+\text{martingale},
\]
with equality when $\vartheta_w=\mathsf{p}$. We take expectations and let $w_0\to1$, using $\sup_x|\Psi_\kappa(w_0,x)-|x||\to0$ from Lemma~\ref{lem:flow-compare}(iv). The case of general $\kappa$ follows from Lemma~\ref{lem:flow-compare}(ii). For step functions, the right side of \eqref{eq:t0-verification} is convex in $(\kappa,\lambda)$ for each control, and $\Psi_\kappa(0,\lambda)$ is the supremum over controls; convexity is preserved in the limit.

(b). Differentiating \eqref{eq:t0-pde} gives $-\partial_wu=\frac\alpha2\partial_x^2u+\alpha\kappa_\circ u\partial_xu$, so by It\^o $\dd u_w=\sqrt\alpha\,\partial_xu(w,X_w)\dd B_w$. As $w\to1$, $u(w,x)\to\sgn(x)$ locally uniformly on $x\ne0$, by the closed form of $\zT_{c,b}|\cdot|$, and $\P(X_1=0)=0$ by Girsanov.

(c). By \eqref{eq:t0-verification} with $\vartheta=u_w$, we have $\Psi_\kappa(0,\lambda)\ge\E|X^\kappa_1|-\frac\alpha2\int\kappa Q$, where $X^\kappa_1=X_1+(\lambda-\lambda_\circ)+\alpha\int(\kappa-\kappa_\circ)u_w\dd w$. Convexity of $|\cdot|$ at $X_1$ gives
\[
\begin{aligned}
\E|X^\kappa_1|&\ge\E|X_1|+\E[\sgn(X_1)](\lambda-\lambda_\circ)+\alpha\int(\kappa-\kappa_\circ)\E[\sgn(X_1)u_w]\dd w\\
&=\E|X_1|+u_0(\lambda-\lambda_\circ)+\alpha\int(\kappa-\kappa_\circ)Q.
\end{aligned}
\]
By the equality case of (a), $\E|X_1|=\Psi_{\kappa_\circ}(0,\lambda_\circ)+\frac\alpha2\int\kappa_\circ Q$, and summing gives (c).
\end{proof}

For $\kappa\in\zU$ let $\resp_\kappa:=\partial_x\Psi_\kappa(0,\cdot)$.

\begin{lemma}\label{lem:response-t0}
Let $\kappa\in\zU$.
\begin{enumerate}[label=(\alph*)]
\item $\Psi_\kappa(0,x)>|x|+\frac\alpha2\delta_\kappa(0)$ for all $x$.
\item For $w<1$, $\partial_x\Psi_\kappa(w,\cdot)$ exists, is odd, nondecreasing, Lipschitz, and concave on $[0,\infty)$. For $x\ge0$, $\partial_x\Psi_\kappa(w,x)\ge2\Nc(x/\sqrt{\alpha(1-w)})-1$.
\item $\resp_\kappa$ is a strictly increasing bijection $\R\to(-1,1)$. So for $|m|<1$ the infimum in $W_m(\kappa)$ is attained only at $\lambda(\kappa,m):=\resp_\kappa^{-1}(m)$, and $m\mapsto W_m(\kappa)$ is concave and $C^1$ on $(-1,1)$ with derivative $-\lambda(\kappa,m)$ and slope tending to $\mp\infty$ at $\pm1$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a). Let $x\ge0$. Taking $\vartheta=1$ in \eqref{eq:t0-verification}, with $\bar\delta:=\delta_\kappa(0)$ and $Y:=x+\alpha\bar\delta+\sqrt\alpha z$, we get $\Psi_\kappa(0,x)\ge\E|Y|-\frac\alpha2\bar\delta$, and $\E|Y|=\E Y+2\E Y_->x+\alpha\bar\delta$. The case $x<0$ follows by evenness.

(b). We first take step $\kappa$, with terminal function $f_n(x):=n^{-1}\log\cosh(nx)$, and let $u:=\partial_x\Psi^{f_n}_\kappa$, which is odd. Then $\mathsf{s}:=\partial_x^2u$ solves the linear equation $-\partial_w\mathsf{s}=\frac\alpha2\partial_x^2\mathsf{s}+\alpha\kappa(u\partial_x\mathsf{s}+3(\partial_xu)\mathsf{s})$, with $\mathsf{s}(w,0)=0$ and terminal data $f_n'''\le0$ on $x>0$, so $\mathsf{s}\le0$ on $x\ge0$ by the maximum principle on the half line. We next prove the lower bound. Let $u^{\mathrm{h}}$ be the heat flow of $\tanh(nx)$. Then $\mathsf{v}:=u-u^{\mathrm{h}}$ solves $-\partial_w\mathsf{v}=\frac\alpha2\partial_x^2\mathsf{v}+\alpha\kappa u\partial_xu$, whose source is nonnegative on $x>0$ because $u,\partial_xu\ge0$ there, and $\mathsf{v}(w,0)=0$, $\mathsf{v}(1,\cdot)=0$, so $\mathsf{v}\ge0$ on $x\ge0$. We then let $n\to\infty$, using that $|f_n-|x||\le\log2/n$ and that derivatives of convex functions converge where the limit is differentiable.

For the Lipschitz bound, let $w\le w_0<1$, so $\kappa\le K:=\kappa(w_0)$ on $[0,w_0]$. Then $u$ solves the linear equation $-\partial_wu=\frac\alpha2\partial_x^2u+\mathsf{c}\,\partial_xu$ with $|\mathsf{c}|\le\alpha K$. Duhamel from level $w_0$, the bound on the $x$-derivative of the heat kernel and a singular Gronwall inequality give $\sup_x|\partial_xu(w,x)|\le\mathrm{const}(\alpha,K,w_0)(w_0-w)^{-1/2}$, independent of the step approximation. Hence the same bound holds for $\kappa\in\zU$, and (b) passes to the limit.

(c). The function $\resp_\kappa$ is odd, concave on $[0,\infty)$, and bounded by $1$. If it were not strictly increasing on $[0,\infty)$, it would be constant on some $[x_0,\infty)$, so $\Psi_\kappa(0,x)=a_\infty x+b_\infty$ there. By Lemma~\ref{lem:flow-compare}(iv), $a_\infty=1$ and $b_\infty=\frac\alpha2\delta_\kappa(0)$, which contradicts (a). The limit at infinity is $1$, again by Lemma~\ref{lem:flow-compare}(iv). The rest follows by Legendre duality.
\end{proof}

\begin{lemma}\label{lem:t0-derivative}
For $\kappa\in\zU$ and $\lambda$ real, the equation of Lemma~\ref{lem:control} has a unique strong solution on $[0,1)$, and $Q_{\kappa,\lambda}(w):=\E u_w^2$ is nondecreasing with values in $[\resp_\kappa(\lambda)^2,1]$. If $\kappa_n\to\kappa$ in $L^1$ within $\zU$ and $\lambda_n\to\lambda$, then $Q_{\kappa_n,\lambda_n}\to Q_{\kappa,\lambda}$ almost everywhere on $[0,1)$. The inequality \eqref{eq:t0-support} holds for every $\kappa_\circ\in\zU$. For $|m|<1$ put $Q_{\kappa,m}:=Q_{\kappa,\lambda(\kappa,m)}$. Then $W_m$ is convex on $\zU$, and for $\kappa,\kappa'\in\zU$,
\begin{equation}\label{eq:t0-dW}
W_m(\kappa')\ge W_m(\kappa)+\frac\alpha2\int(\kappa'-\kappa)Q_{\kappa,m},\qquad\lim_{t\downarrow0}\frac{W_m(\kappa+t(\kappa'-\kappa))-W_m(\kappa)}t=\frac\alpha2\int(\kappa'-\kappa)Q_{\kappa,m}.
\end{equation}
\end{lemma}

\begin{proof}
The drift is Lipschitz on $w\le w_0$ by Lemma~\ref{lem:response-t0}(b). On $[0,w_0]$ the $\kappa_n$ are uniformly bounded, $\kappa_n\to\kappa$ almost everywhere and $\Psi_{\kappa_n}\to\Psi_\kappa$ uniformly, and by Lemma~\ref{lem:response-t0}(b) the derivatives converge locally uniformly, with a uniform Lipschitz constant. By SDE stability, we have $X^n_w\to X_w$ in $L^2$, so $Q_n\to Q$ on $[0,w_0]$. The inequality \eqref{eq:t0-support} for step approximations of $\kappa_\circ$ passes to the limit, since $Q_n$ is bounded and $\kappa-\kappa_\circ^n$ converges in $L^1$. The convexity of $W_m$ follows by partial minimization of a jointly convex function. The first part of \eqref{eq:t0-dW} is \eqref{eq:t0-support} with $u_0=m$. For the derivative, the inequality $\ge$ follows from the first part. For $\le$, we apply the first part at $\kappa_t:=\kappa+t(\kappa'-\kappa)$ with $\kappa'$ replaced by $\kappa$, and use $Q_{\kappa_t,m}\to Q_{\kappa,m}$ almost everywhere, which holds as $\lambda(\kappa_t,m)\to\lambda(\kappa,m)$ by Lemma~\ref{lem:response-t0}(c) and uniform convergence; here $Q$ is bounded and $\kappa'-\kappa\in L^1$.
\end{proof}

\subsection{The variational formula}

The main result of this subsection is the following.

\begin{theorem}\label{thm:t0-formula}
For all $\alpha>0$ and $m\in[-1,1]$,
\begin{equation}\label{eq:t0-formula}
\sqrt\alpha L(m)=\lim_{\gamma\to\infty}\frac{\Phimix(\gamma;m)}{\sqrt\gamma}=\min_{(\kappa,C)\in\zV}\zB_m(\kappa,C),\qquad\zB_m(\kappa,C):=\frac\alpha2\zS(\kappa,C)+W_m(\kappa).
\end{equation}
The minimum is attained at a unique $(\kappa^*,C^*)=(\kappa^*_m,C^*_m)$. Bounded step functions with $C>0$ are enough for the infimum. For types, in the notation of Theorem~\ref{thm:t0-existence}(e), the same proof gives for $\tv\in[-1,1]^{d_r}$
\begin{equation}\label{eq:t0-formula-types}
\sqrt\alpha L^{(r)}(\tv)=\min_{(\kappa,C)\in\zV}\Big\{\frac\alpha2\zS(\kappa,C)+\Eeta W_{t_\eta}(\kappa)\Big\}.
\end{equation}
\end{theorem}

We prove the theorem by taking the limit $\gamma\to\infty$ in \eqref{eq:t0-phimix}. We will eliminate the Ising path $x$ exactly by conjugacy, and the two remaining terms then become functions of the rescaled path $\kappa=\sqrt\gamma F$, with $F$ the distribution function of $y$. The upper bound will come from scaled trials, and the lower bound from compactness, in which the mass of $\kappa$ that moves to the top level $w=1$ lands in $C$.

\begin{lemma}\label{lem:t0-exact}
Let $\zF$ be the set of $F:[0,1]\to[0,1]$ nondecreasing, right-continuous, with $F(1)=1$. Such an $F$ is the distribution function of a path $y\in\Qone$ with $\hat y:=y(1-)$, and $\delta_F(w):=\int_w^1F=\int_0^1(1-y(u)\vee w)\dd u\le1-w$. For $\gamma>0$ and $m\in(-1,1)$,
\[
\Phimix(\gamma;m)=\inf\{\zA_\gamma(F;m):F\in\zF,\hat y<1\},
\]
where, with $\kappa:=\sqrt\gamma F$ and $\ell_\gamma(x):=\gamma^{-1/2}\log\cosh(\sqrt\gamma x)$,
\begin{equation}\label{eq:t0-exact}
\frac{\zA_\gamma(F;m)}{\sqrt\gamma}=\frac\alpha2\int_0^1\Big[\frac1{\delta_\kappa(w)}-\frac1{\sqrt\gamma(1-w)}\Big]\dd w+W_m^{[\ell_\gamma]}(\kappa).
\end{equation}
The integrand is nonnegative. The right side is a term independent of $m$ plus an infimum over $\lambda$ of functions affine in $m$, so $\Phimix(\gamma;\cdot)$ is concave for every $\gamma$. Also $|x|-\log2/\sqrt\gamma\le\ell_\gamma(x)\le|x|$.
\end{lemma}

\begin{proof}
We first eliminate $x$. By \eqref{eq:gr-y-form} with one type,
\[
\Phimix(\gamma;m)=\inf_{y\in\Qone}\Big\{\frac{\gamma\alpha}2-g_m\Big(\frac{\gamma\alpha}2y\Big)+\alpha\psiSs(y)\Big\},
\]
and by Theorem~\ref{thm:gaussian-slice}(b), the path $y$ may be restricted to $\hat y<1$.

For the sphere term, by Lemma~\ref{lem:sphere-conjugate}, we have $\psiSs(y)=\frac12[\int_0^{\hat y}\dd w/\delta_F(w)+\log(1-\hat y)]=\frac12\int_0^1[1/\delta_F(w)-1/(1-w)]\dd w$, because $\delta_F(w)=1-w$ on $[\hat y,1]$ and $\int_0^{\hat y}\dd w/(1-w)=-\log(1-\hat y)$; also $\delta_F=\delta_\kappa/\sqrt\gamma$.

For the Ising term, we take in the definition of $g_m$ in \cite[(1.7), (1.8)]{BM} the horizon $S:=\gamma\alpha/2$ and $\nu([0,v])=F(2v/(\gamma\alpha))$, so that $-\partial_v\phi_\nu=\partial_x^2\phi_\nu+F(2v/(\gamma\alpha))(\partial_x\phi_\nu)^2$, $\phi_\nu(S,\cdot)=\log\cosh$. For step $F$ with values $f_i$ on levels $w_i$, and $v_i=Sw_i$, we have $\phi_\nu(v_i,x)=\frac1{f_i}\log\E\exp(f_i\phi_\nu(v_{i+1},x+\sqrt{2(v_{i+1}-v_i)}z))$. Writing $\phi_\nu(v,x)=\sqrt\gamma\hat\phi(2v/(\gamma\alpha),x/\sqrt\gamma)$, $c_i:=\sqrt\gamma f_i$, $b_i^2:=\alpha(w_{i+1}-w_i)$, we get $\hat\phi(w_i,\cdot)=\zT_{c_i,b_i}\hat\phi(w_{i+1},\cdot)$ and $\hat\phi(1,\cdot)=\ell_\gamma$, so $\phi_\nu(0,\sqrt\gamma\lambda)=\sqrt\gamma\Psi_\kappa^{\ell_\gamma}(0,\lambda)$ and
\[
g_m(a)=\sup_\lambda\{S-\phi_\nu(0,\sqrt\gamma\lambda)+\sqrt\gamma\lambda m\}=\frac{\gamma\alpha}2-\sqrt\gamma W_m^{[\ell_\gamma]}(\kappa).
\]
The term $\gamma\alpha/2$ then cancels. Since both sides are $L^1$-Lipschitz in $F$, by \cite[(6.43)]{BM}, $|y_F-y_{F'}|_{L^1}=|F-F'|_{L^1}$ and Lemma~\ref{lem:flow-compare}(ii), this covers general $F$.
\end{proof}

\begin{proof}[Proof of the upper bound in \eqref{eq:t0-formula}]
We fix $m\in(-1,1)$, $C>0$ and a bounded $\kappa\le K$. For $\sqrt\gamma>\max(K,C)$, let $\hat y_\gamma:=1-C/\sqrt\gamma$, $\kappa_\gamma:=\kappa$ on $[0,\hat y_\gamma)$, $\kappa_\gamma:=\sqrt\gamma$ on $[\hat y_\gamma,1)$, $F_\gamma:=\kappa_\gamma/\sqrt\gamma$.

For the sphere term, $\delta_{\kappa_\gamma}(w)=C+\int_w^{\hat y_\gamma}\kappa$ for $w<\hat y_\gamma$, so the first term of \eqref{eq:t0-exact} is $\frac\alpha2[\int_0^{\hat y_\gamma}\dd w/(C+\int_w^{\hat y_\gamma}\kappa)+\gamma^{-1/2}\log(C/\sqrt\gamma)]$, which tends to $\frac\alpha2\int_0^1\dd w/\delta(w)$.

For the Ising term, $\ell_\gamma\le|\cdot|$, and by Lemma~\ref{lem:flow}, bounding the exponential of a maximum by the sum, $\zT_{c,b}|\cdot|(x)\le|x|+\log2/c+cb^2/2$. With $c=\sqrt\gamma$, $b^2=\alpha C/\sqrt\gamma$ we get $cb^2/2=\alpha C/2$. By Lemma~\ref{lem:flow-compare}(iv), $|x|\le\Psi_\kappa(\hat y_\gamma,x)$, so $\Psi_{\kappa_\gamma}^{\ell_\gamma}(0,\lambda)\le\Psi_\kappa(0,\lambda)+\alpha C/2+\log2/\sqrt\gamma$ for all $\lambda$, and $W_m^{[\ell_\gamma]}(\kappa_\gamma)\le W_m(\kappa)+\alpha C/2+\log2/\sqrt\gamma$. Using Lemma~\ref{lem:t0-exact} and Theorem~\ref{thm:t0-existence}(c), we obtain $\sqrt\alpha L(m)\le\zB_m(\kappa,C)$.

For general $(\kappa,C)$ let $\kappa_K:=\kappa\wedge K$, $C_K:=C+\int(\kappa-K)_+$. Then $C_K+\delta_{\kappa_K}\ge C+\delta_\kappa$, so $\int\dd w/\delta$ does not increase; $W_m(\kappa_K)\le W_m(\kappa)$; and $C_K\to C$. Hence $\zB_m(\kappa_K,C_K)\le\zB_m(\kappa,C)+\frac\alpha2\int(\kappa-K)_+$, and bounded $\kappa$ with $C>0$ suffices (if $C_K=0$, then $\kappa$ is bounded and $C=0$, and $\zB_m=+\infty$). Step functions suffice by Lemma~\ref{lem:flow-compare}(ii) and continuity of $\zS$ at $C>0$.
\end{proof}

\begin{lemma}\label{lem:t0-compact}
Let $(\kappa_n,C_n)\in\zV$ with $\delta_n:=C_n+\delta_{\kappa_n}$ and $\delta_n(0)\le\bar\delta$, and let $f_n$ be 1-Lipschitz with $|x|-\epsilon_n\le f_n$, $\epsilon_n\to0$. Along a subsequence $\delta_n(0)\to\delta_\infty$, $\kappa_n\to\kappa$ almost everywhere, $\kappa\in\zU$, and with $C:=\delta_\infty-\int\kappa\ge0$:
\begin{enumerate}[label=(\alph*)]
\item $\delta_n(w)\to C+\delta_\kappa(w)$ for all $w\in[0,1)$;
\item for $\lambda_n\to\lambda$, $\liminf\{\frac\alpha2C_n+\Psi_{\kappa_n}^{f_n}(0,\lambda_n)\}\ge\frac\alpha2C+\Psi_\kappa(0,\lambda)$;
\item for $m\in[-1,1]$, $\liminf\{\frac\alpha2C_n+W_m^{[f_n]}(\kappa_n)\}\ge\frac\alpha2C+W_m(\kappa)$.
\end{enumerate}
\end{lemma}

\begin{proof}
Since $\kappa_n(w)(1-w)\le\delta_{\kappa_n}(w)\le\bar\delta$, the $\kappa_n$ are uniformly bounded and nondecreasing on $[0,w_0]$. We obtain the almost-everywhere limit from Helly's theorem and a diagonal argument, and Fatou gives $\int\kappa\le\delta_\infty$. For $w<1$, $\int_0^w\kappa_n\to\int_0^w\kappa$, which is (a). To prove (b) and (c), we fix $w_0<1$. By Lemma~\ref{lem:flow-compare}(iii),(i), $\Psi_{\kappa_n}^{f_n}(w_0,x)\ge|x|+\frac\alpha2\delta_{\kappa_n}(w_0)-\epsilon_n$. Hence, for all $\lambda$,
\[
\frac\alpha2C_n+\Psi_{\kappa_n}^{f_n}(0,\lambda)-\lambda m\ge\frac\alpha2\delta_n(w_0)-\epsilon_n+\zP_{\kappa_n}^{0,w_0}|\cdot|(\lambda)-\lambda m.
\]
By Lemma~\ref{lem:flow-compare}(ii), $\zP_{\kappa_n}^{0,w_0}|\cdot|\to\zP_\kappa^{0,w_0}|\cdot|$ uniformly, with error $\frac\alpha2\int_0^{w_0}|\kappa_n-\kappa|\to0$. Thus the liminf of the left side, also after the infimum over $\lambda$, is at least $\frac\alpha2(C+\delta_\kappa(w_0))+\zP_\kappa^{0,w_0}|\cdot|(\lambda)-\lambda m$. By \eqref{eq:t0-W-cut} and Lemma~\ref{lem:flow-compare}(iv) this is at least $\frac\alpha2C+\Psi_\kappa(0,\lambda)-\lambda m-\sqrt{2\alpha(1-w_0)/\pi}$. Letting $w_0\to1$ gives (b) and (c).
\end{proof}

The mass that $\kappa_n$ carries to $w=1$ lands in $C$, which is why $C$ belongs to the space: $\zU$ alone is not closed for this convergence, while $\zU\times[0,\infty)$ with the total mass is.

\begin{proof}[End of the proof of Theorem~\ref{thm:t0-formula}]
For the lower bound, let $m\in(-1,1)$ and take $F_\gamma$ with $\zA_\gamma(F_\gamma;m)\le\Phimix(\gamma;m)+1$. We put $\kappa_\gamma:=\sqrt\gamma F_\gamma$ and let $\gamma_n\to\infty$. By \eqref{eq:t0-W-mass} for $\ell_\gamma$, $W_m^{[\ell_\gamma]}(\kappa_\gamma)\ge\frac\alpha2\delta_{\kappa_\gamma}(0)-\log2/\sqrt\gamma$. The sphere term in \eqref{eq:t0-exact} is nonnegative, so with \eqref{eq:t0-mix-sandwich} we get $\delta_{\kappa_\gamma}(0)\le2L/\sqrt\alpha+o(1)$, which is bounded. Applying Lemma~\ref{lem:t0-compact} with $C_n=0$ and $f_n=\ell_{\gamma_n}$, we obtain a limit $(\kappa,C)$. The integrand in \eqref{eq:t0-exact} is nonnegative and for $w<1$ tends to $1/(C+\delta_\kappa(w))$, or $+\infty$ if the limit is $0$, so Fatou gives $\liminf\ge\int_0^1\dd w/\delta(w)$. Lemma~\ref{lem:t0-compact}(c) handles the Ising term, and summing gives $\sqrt\alpha L(m)\ge\frac\alpha2\int\dd w/\delta+\frac\alpha2C+W_m(\kappa)=\zB_m(\kappa,C)$, which is the lower bound.

For existence, the same steps apply to a minimizing sequence of $\zB_m$. Indeed, its masses are bounded by \eqref{eq:t0-W-mass}, and Lemma~\ref{lem:t0-compact} with Fatou for $1/\delta$ gives lower semicontinuity under the convergence of that lemma.

For uniqueness, note that $\zS$ is convex in $(\kappa,C)$, since $\delta$ is linear in $(\kappa,C)$ and $t\mapsto1/t$ is strictly convex, and $W_m$ is convex by Lemma~\ref{lem:t0-derivative} (at $m=\pm1$, see \eqref{eq:t0-m1} below). Therefore, two minimizers would have $\delta_1=\delta_2$ almost everywhere, hence everywhere on $[0,1)$ by continuity, so $\kappa_1=\kappa_2$ and $C_1=C_2$.

We next treat $m=\pm1$. Here $L(\pm1)=\sqrt\alpha$, as $\Slice(N)$ is a single point. By \eqref{eq:t0-W-mass} and $\lambda\to\infty$ in Lemma~\ref{lem:flow-compare}(iv),
\begin{equation}\label{eq:t0-m1}
W_1(\kappa)=\frac\alpha2\delta_\kappa(0),\qquad\zB_1(\kappa,C)=\frac\alpha2\Big[\delta(0)+\int_0^1\frac{\dd w}{\delta(w)}\Big]\ge\frac\alpha2\Big(\delta(0)+\frac1{\delta(0)}\Big)\ge\alpha,
\end{equation}
with equality only at $\kappa=0$, $C=1$. This is the formula at $m=1$, and $m=-1$ follows by evenness.

For types, the elimination of the Ising paths runs type by type and gives $\Eeta g_{t_\eta}$ in place of $g_m$. The rest of the proof is unchanged.
\end{proof}

\begin{lemma}\label{lem:t0-bracket}
For $m\in[0,1]$,
\begin{equation}\label{eq:t0-bracket}
\sqrt\alpha\,m+\wf(m)\le L(m)\le\sqrt\alpha+\wf(m).
\end{equation}
\end{lemma}

\begin{proof}
For the upper bound, we take $\kappa=0$, $C=1$ in \eqref{eq:t0-formula}, so $\zB_m(0,1)=\alpha+W_m(0)$, where $W_m(0)=\inf_\lambda\{\E|\lambda+\sqrt\alpha z|-\lambda m\}=\sqrt\alpha\,\wf(m)$. For the lower bound, let $G_i$ be the columns of $G$, $G_\Sigma:=\sum_iG_i$, and $\mathsf{e}:=G_\Sigma/|G_\Sigma|$. Given $G_\Sigma$, the variables $h_i:=(G_i-G_\Sigma/N)\cdot\mathsf{e}$ are centered Gaussian with variance $1-1/N$, exchangeable, with sum $0$, and independent of $G_\Sigma$. If $\sigma$ flips a set $F$ of size $pN$, $p=(1-m)/2$, then $|G\sigma|\ge G\sigma\cdot\mathsf{e}=|G_\Sigma|(1-2p)-2\sum_Fh_i$. With $F$ the $pN$ smallest $h_i$, we get $-\frac2N\sum_Fh_i\to2\Nd(\Nc^{-1}(p))=\wf(m)$, and $|G_\Sigma|/N\to\sqrt\alpha$.
\end{proof}

\subsection{Optimality conditions}

We now state the optimality conditions for the minimizer.

\begin{proposition}\label{prop:t0-kkt}
Let $m\in(-1,1)$. $\zB_m$ is convex on $\zV$, strictly in $\delta$. For $(\kappa,C)\in\zV$ with $\zB_m(\kappa,C)<\infty$ put $Q:=Q_{\kappa,m}$, $R(s):=R_{\kappa,C}(s):=\int_0^s\dd w/\delta(w)^2$, $\Gamma(s):=\int_s^1(Q-R)\dd w$. If $(\kappa,C)$ is the minimizer, then
\begin{enumerate}[label=(K\arabic*)]
\item $R(1-)\le1$, and $C(1-R(1-))=0$;
\item $\Gamma(s)\ge0$ for all $s\in[0,1)$;
\item $\int_{[0,1)}\Gamma\dd\kappa=0$, where $\dd\kappa$ includes the atom $\kappa(0)$ at $s=0$.
\end{enumerate}
On the interior of $\supp\dd\kappa$, $Q=R$. At the minimizer, $\delta(0)\ge1$, by (K1) and $\delta\le\delta(0)$.
\end{proposition}

\begin{proof}
By Fubini, the directional derivative of $\zS$ along $(\kappa'-\kappa,C'-C)$ is $(C'-C)(1-R(1-))-\int(\kappa'-\kappa)R$, so with \eqref{eq:t0-dW} the derivative of $\zB_m$ is $\frac\alpha2[\int(\kappa'-\kappa)(Q-R)+(C'-C)(1-R(1-))]$, and at the minimizer it is nonnegative in every admissible direction. Taking $C'=C+t$, we see by monotone convergence that the derivative in $C$ is $\frac\alpha2(1-R(1-))$, so $R(1-)\le1$. If $C>0$, decreasing $C$ is also allowed, which gives (K1). With $\kappa'=\kappa+\theta\one_{[s,1)}$, we get (K2). Finally, $\kappa'=(1\pm\epsilon)\kappa$ gives $\int\kappa(Q-R)=0$, and $\int\kappa(Q-R)=\int\Gamma\dd\kappa$ by Fubini, the integrand being integrable since $\int\kappa<\infty$ and $|Q-R|\le1+R(1-)$; this is (K3).
\end{proof}

\subsection{Unit scale and joint concavity in $(\alpha,m)$}

Rescaling removes $\alpha$ from the Ising part. More precisely, we have $\Psi_\kappa(w,x)=\sqrt\alpha\tilde\Psi_{\sqrt\alpha\kappa}(w,x/\sqrt\alpha)$, where $\tilde\Psi$ is the recursion at $\alpha=1$: $-\partial_w\tilde\Psi=\frac12(\partial_x^2\tilde\Psi+\tilde\kappa(\partial_x\tilde\Psi)^2)$. Let $\tilde\kappa:=\sqrt\alpha\kappa$, $\tilde C:=\sqrt\alpha C$, $\tilde\delta:=\sqrt\alpha\delta$, and let $W^1_m$ denote $W_m$ at $\alpha=1$. Then $W_m(\kappa)=\sqrt\alpha W^1_m(\tilde\kappa)$, and $Q$ and $R$ are unchanged, with $R'=\alpha/\tilde\delta^2$.

\begin{corollary}\label{cor:t0-unit}
For all $\alpha>0$ and $m\in[-1,1]$,
\begin{equation}\label{eq:t0-unit}
L(\alpha,m)=\min_{(\tilde\kappa,\tilde C)\in\zV}\Big\{\frac{\tilde C}2+\frac\alpha2\int_0^1\frac{\dd w}{\tilde\delta(w)}+W^1_m(\tilde\kappa)\Big\}.
\end{equation}
The functional involves $\alpha$ only through the factor $\alpha/2$. So:
\begin{enumerate}[label=(\alph*)]
\item $L(\alpha,m)=\inf_{\tilde\kappa,\tilde C,\tilde\lambda}\{\frac{\tilde C}2+\frac\alpha2\int_0^1\dd w/\tilde\delta+\tilde\Psi_{\tilde\kappa}(0,\tilde\lambda)-\tilde\lambda m\}$ is an infimum of functions affine in $(\alpha,m)$, and $L$ is jointly concave on $(0,\infty)\times[-1,1]$.
\item $L$ is $C^1$ on $(0,\infty)\times(-1,1)$, with $\partial_mL=-\tilde\lambda^*(\alpha,m)$ and $\partial_\alpha L=\frac12\Upsilon(\alpha,m)$, $\Upsilon:=\int_0^1\dd w/\tilde\delta^*(w)$. If $(\alpha_n,m_n)\to(\alpha,m)$ in this set, the minimizers converge: $\tilde\delta_n\to\tilde\delta^*$ pointwise on $[0,1)$, $\tilde\kappa_n\to\tilde\kappa^*$ almost everywhere, $\tilde\lambda_n\to\tilde\lambda^*$, and $\Upsilon_n\to\Upsilon$.
\item Put
\begin{equation}\label{eq:t0-Pstar}
P^*(m):=\inf_{\tilde\kappa\in\zU}\Big\{W^1_m(\tilde\kappa)-\frac12\int_0^1w\tilde\kappa(w)\dd w\Big\}.
\end{equation}
$P^*$ is even and concave, $P^*\ge0$ and $P^*(1)=0$. So $P^*$ is nonincreasing on $[0,1]$. And $L\ge\sqrt\alpha+P^*(m)$ for all $\alpha$.
\end{enumerate}
By \cite[Theorem~1]{AC17} and a Legendre transform in the field, $P^*$ is the ground state energy of the Sherrington--Kirkpatrick model on the slice. This is why we call it the slice energy. No proof below uses that identification.
\end{corollary}

\begin{proof}
Part (a) follows by writing out $W^1_m$ in \eqref{eq:t0-unit}.

(b). By (a), at each $(\alpha,m)$ with $|m|<1$ the vector $(\Upsilon/2,-\tilde\lambda^*)$ is a supergradient of $L$, where $(\tilde\kappa^*,\tilde C^*)$ is the minimizer and $\tilde\lambda^*:=\lambda(\tilde\kappa^*,m)$ at $\alpha=1$. Let $(\alpha_n,m_n)\to(\alpha,m)$, with minimizing triples $(\tilde\kappa_n,\tilde C_n,\tilde\lambda_n)$. The masses are bounded, since \eqref{eq:t0-W-mass} gives $\frac12\tilde C_n+\frac12\delta_{\tilde\kappa_n}(0)\le L(\alpha_n,m_n)$, and so is $|\tilde\lambda_n|$, since $\tilde\Psi(0,\lambda)-\lambda m\ge|\lambda|(1-|m|)$. By Lemma~\ref{lem:t0-compact}(b) and Fatou, a limit point $(\kappa',C',\lambda')$ has value at most $\lim L(\alpha_n,m_n)=L(\alpha,m)$, so it is the minimizer, by Theorem~\ref{thm:t0-formula} and Lemma~\ref{lem:response-t0}(c), and the whole sequence converges. Also $\int\dd w/\tilde\delta_n^2\le1/\alpha_n$ by (K1), so $1/\tilde\delta_n$ is uniformly integrable and $\Upsilon_n\to\Upsilon$. Hence every limit of supergradients at nearby points is $(\Upsilon/2,-\tilde\lambda^*)$. The superdifferential of a concave function is the closed convex hull of limits of gradients at nearby points \cite[Theorem~25.6]{Rockafellar70}, so it is a singleton, and $L$ is differentiable. A differentiable concave function on an open convex set is $C^1$ \cite[Corollary~25.5.1]{Rockafellar70}.

(c). The function $P^*$ is an infimum of functions affine in $m$, hence concave, and it is even because $\tilde\Psi_{\tilde\kappa}(0,\cdot)$ is even. By Lemma~\ref{lem:flow-compare}(iv), $\tilde\Psi_{\tilde\kappa}(0,\lambda)-\lambda m-\frac12\int w\tilde\kappa\ge|\lambda|(1-|m|)+\frac12\int(1-w)\tilde\kappa\ge0$, so $P^*\ge0$. At $m=1$, we take $\tilde\kappa=0$ and let $\lambda\to\infty$: $\E|\lambda+z|-\lambda\to0$, so $P^*(1)=0$.

It remains to prove the lower bound on $L$. Since $\tilde\delta/2+(\alpha/2)/\tilde\delta\ge\sqrt\alpha$ pointwise and $\frac{\tilde C}2=\frac12\int_0^1(\tilde\delta-\delta_{\tilde\kappa})\dd w$, we get $\frac{\tilde C}2+\frac\alpha2\int\dd w/\tilde\delta\ge\sqrt\alpha-\frac12\int\delta_{\tilde\kappa}$, with $\int_0^1\delta_{\tilde\kappa}=\int w\tilde\kappa$, so \eqref{eq:t0-unit} and \eqref{eq:t0-Pstar} give $L\ge\sqrt\alpha+P^*(m)$. This completes the proof of (c).
\end{proof}

\subsection{Shape of $L$ and $e$}

At fixed load, the previous results give the following.

\begin{theorem}\label{thm:t0-shape}
Fix $\alpha>0$.
\begin{enumerate}[label=(\alph*)]
\item $L$ is concave and even on $[-1,1]$, and $L(m)=\inf_{\kappa,C,\lambda}\alpha^{-1/2}\{\frac\alpha2\zS(\kappa,C)+\Psi_\kappa(0,\lambda)-\lambda m\}$ is an infimum of functions affine in $m$.
\item $L$ is $C^1$ on $(-1,1)$ with $L'(m)=-h(m)$, $h(m):=\lambda(\kappa^*_m,m)/\sqrt\alpha=\tilde\lambda^*(m)$. The function $h$ is continuous and nondecreasing, $h(0)=0$, and $L'(1-)=-\infty$.
\item $e$ is $C^1$ on $(-1,1)$, $e'(m)=m+LL'=m-L(m)h(m)$, $e'(0)=0$, and $e'(m)\to-\infty$ as $m\to1$.
\end{enumerate}
\end{theorem}

\begin{proof}
Part (a) follows from Theorem~\ref{thm:t0-formula} and evenness under $\sigma\mapsto-\sigma$.

For (b), we apply Corollary~\ref{cor:t0-unit}(b), and for $L'(1-)=-\infty$ we use \eqref{eq:t0-bracket}, concavity and $L(1)=\sqrt\alpha$: for $q<1$, $L'(1-)\le(L(1)-L(q))/(1-q)\le\sqrt\alpha-\wf(q)/(1-q)$, and the right side tends to $-\infty$ as $q\to1$.

Part (c) is straightforward, using $L\ge\sqrt\alpha$.
\end{proof}

\subsection{The energy for binary patterns}

We recall the objects of Section~\ref{sec:reduction}. For $r\ge1$, $d_r=2^{r-1}$, and $\eta$ is uniform on $\{-1,1\}^{r-1}$. For $\tv=(t_\eta)\in[-1,1]^{d_r}$, $\ov(\tv)=(\Eeta\eta_at_\eta)_{a=2,\dots,r}$, $V(\tv)^2=(\Eeta t_\eta)^2+|\ov(\tv)|^2$, $P_r(m)=\{\tv:\Eeta t_\eta=m\}$, $\Theta_r(m)=\max_{P_r(m)}\Fr$, $\LamR=\sup_r\Theta_r$, and $\eR=\lim_{\beta\to\infty}\LamR_\beta/\beta$, uniformly on $[-1,1]$ (Corollary~\ref{cor:energy-limit}). With $L^{(r)}$ from Theorem~\ref{thm:t0-existence}(e), let
\begin{equation}\label{eq:theta-t0}
\Theta^{0}_r(m):=\max_{\tv\in P_r(m)}\Big\{\frac{V(\tv)^2}2+\frac{L^{(r)}(\tv)^2}2\Big\}.
\end{equation}

\begin{lemma}\label{lem:t0-eR}
For all $r$ and $\beta$, $\sup_m|\Theta_r(m)/\beta-\Theta^{0}_r(m)|\le\log2/\beta$. So $\eR=\sup_r\Theta^{0}_r$. Moreover $\eR\ge\Theta^{0}_1=e$.
\end{lemma}

\begin{proof}
By Theorem~\ref{thm:t0-existence}(e), $\frac\beta2L^{(r)}(\tv)^2-\log2\le\Phi^{(r)}(\tv)\le\frac\beta2L^{(r)}(\tv)^2$, with rates independent of $r$, so $|\Fr(\tv)/\beta-\frac12V(\tv)^2-\frac12L^{(r)}(\tv)^2|\le\log2/\beta$ for all $\tv$. We take the maximum on $P_r(m)$, then the supremum over $r$, and let $\beta\to\infty$. For $r=1$ the polytope is a point.
\end{proof}

\subsection{The capacity at zero temperature}

\begin{definition}\label{def:t0-wells}
For $m_s\in(0,1)$,
\begin{align*}
\Well_0(m_s)&:=\{\alpha>0:e_\alpha(m_2)>e_\alpha(m_1)\text{ for some }m_s\le m_1<m_2\le1\}\\
&\phantom{:}=\{\alpha>0:\sup_{m\in(m_s,1)}e_\alpha'(m)>0\},
\end{align*}
and $\alc(0;m_s):=\sup\Well_0(m_s)$, $\Well_0:=\bigcup_{m_s\in(0,1)}\Well_0(m_s)$, $\alc(0):=\sup\Well_0$. For binary patterns, $\WellR_0(m_s)$, $\alcR(0;m_s)$, $\WellR_0$ and $\alcR(0)$ are defined by the first form, with $\eR$ in place of $e$.
\end{definition}

The two forms agree by the mean value theorem and Theorem~\ref{thm:t0-shape}(c), which also gives $e_\alpha'(m)=m-L_\alpha(m)h_\alpha(m)$, where $(L,h)$ come from the convex program \eqref{eq:t0-unit}. In particular, $\alpha\in\Well_0(m_s)$ if and only if $m>L_\alpha(m)h_\alpha(m)$ somewhere in the window. This is a variational characterization of $\alc(0;m_s)$.

For the landscape statement, $H_N$ denotes the energy \eqref{eq:hamiltonian} for binary patterns, and $H_N(\sigma):=-\frac1{2N}[(\xi^1\cdot\sigma)^2+|G\sigma|^2]$ for Gaussian rows with one binary pattern.

\begin{theorem}\label{thm:t0-capacity}
Fix $m_s\in(0,1)$ and write $\zd(\alpha):=\max_{[m_s,1)}e_\alpha'$.
\begin{enumerate}[label=(\alph*)]
\item $\Well_0(m_s)$ is open. The function $\zd$ is continuous, $e_\alpha'\to-\infty$ at $m\to1$ locally uniformly in $\alpha$, and $\Well_0(m_s)=\{\zd>0\}$.
\item (Landscape, for all patterns.) Let $e_\bullet$ be $e$ for Gaussian rows with one binary pattern and $\eR$ for binary patterns, and let $\Well_\bullet(m_s)$ be the corresponding well set. Let $m_1<m_2$ with $\Delta:=e_\bullet(m_2)-e_\bullet(m_1)>0$, and $m_{k_i}\to m_i$. For some $c>0$ and $N$ large, with probability at least $1-e^{-cN}$, for all patterns $\mu$ simultaneously (for Gaussian rows, only $\mu=1$),
\[
\min\{H_N(\sigma):m_\mu(\sigma)=m_{k_1}\}\ge\min\{H_N(\sigma):m_\mu(\sigma)=m_{k_2}\}+\frac\Delta3N.
\]
If $\alpha\notin\Well_\bullet(m_s)$, then for every $\epsilon>0$, with probability tending to $1$, for all $\mu$ (for Gaussian rows, only $\mu=1$) and all $k_1<k_2$ with $m_{k_1}\ge m_s$,
\[
\min\{H_N:m_\mu=m_{k_1}\}\le\min\{H_N:m_\mu=m_{k_2}\}+\epsilon N.
\]
So the profile $k\mapsto\min\{H_N:m_\mu=m_k\}$ has no barrier of extensive depth on the window, and for each $\alpha$, membership in the well set is exactly the existence of an extensive well in this profile.
\item (Binary patterns.) By Lemma~\ref{lem:t0-eR}, $\eR\ge e$, and $\WellR_0(m_s)$ and $\alcR(0;m_s)$ are read from $\eR=\sup_r\Theta^{0}_r$. Section~\ref{sec:merging} identifies the two well sets, Theorem~\ref{thm:merging-t0}.
\item ($T\to0$.) For $T=1/\beta>0$, let $\Well_T(m_s)$ and $\alc(T;m_s)$ be as in \eqref{eq:well-def}, for Gaussian rows. If $\alpha\in\Well_0(m_s)$, with $m_1<m_2$ and $\Delta:=e(m_2)-e(m_1)>0$, then $\alpha\in\Well_T(m_s)$ for $T<1/\beta_0$, $\beta_0:=(\log2-I(m_1))/\Delta$. So
\[
\liminf_{T\to0}\alc(T;m_s)\ge\alc(0;m_s).
\]
Let $v^*(m)$ be the optimal radius in \eqref{eq:t0-phimix}. Then, locally uniformly in $\alpha$, $v^*(m)/\beta\to L(m)^2/\alpha$ uniformly in $m\in[-1,1]$, and $\Lam_\beta'(m)/\beta\to e'(m)$ uniformly on compact subsets of $(-1,1)$. For every compact $A\subset(0,\infty)$ there is $\bar m<1$ such that $\Lam_\beta'<0$ on $[\bar m,1)$ for all $\alpha\in A$ and all large $\beta$.
\end{enumerate}
\end{theorem}

\begin{proof}
(a). Continuity follows from Corollary~\ref{cor:t0-unit}(b) and compactness once $m$ near $1$ is excluded uniformly. By concavity $L'(m)\le L'(\bar m)$ for $m\ge\bar m$, and $L\ge\sqrt\alpha$, so $e_\alpha'(m)\le1+\sqrt\alpha L'(\bar m)$ once $L'(\bar m)<0$. For $q<\bar m<1$, concavity and \eqref{eq:t0-bracket} give
\[
L'(\bar m)\le\frac{L(\bar m)-L(q)}{\bar m-q}\le\frac{\sqrt\alpha(1-q)+\wf(\bar m)-\wf(q)}{\bar m-q}.
\]
Taking $q$ near $1$ and then $\bar m$ near $1$ makes this as negative as we like, locally uniformly in $\alpha$. Openness and $\Well_0(m_s)=\{\zd>0\}$ follow from the second form of the definition.

(b). We fix $\mu$. By the gauge of Section~\ref{sec:reduction}, $\max_{m_\mu(\sigma)=m_k}(-H_N(\sigma))/N=m_k^2/2+\frac12L^\mu(k)^2$, where $L^\mu(k)$ has the law of $L_{\Slice(k)}$ on the Gaussian rows or on the Rademacher matrix of the gauged patterns. Let $\mathsf{q}^\mu(k):=\frac12L^\mu(k)^2$. For every $\tau>0$ there is $c(\tau)>0$ with $\P(|\mathsf{q}^\mu(k)-\E\mathsf{q}^\mu(k)|>\tau)\le e^{-c(\tau)N}$ for $N$ large, uniformly in $k$; for binary patterns this follows from \eqref{eq:conc-L}, and for Gaussian rows from Theorem~\ref{thm:t0-existence}(d), as $\E L_A$ is bounded. By Corollary~\ref{cor:energy-limit}, respectively Theorem~\ref{thm:t0-existence}(a), $\sup_k|m_k^2/2+\E\mathsf{q}^\mu(k)-e_\bullet(m_k)|\to0$. We take $\tau=\Delta/7$ and a union bound over the $M$ patterns and the two slices. On that event the difference of the two minima, divided by $N$, is at least $\Delta-2\tau-o(1)\ge\Delta/3$, by continuity of $e_\bullet$. For the second statement, $e_\bullet$ is nonincreasing on the window, and the union bound runs over $(N+1)M$ events, with $\tau=\epsilon/3$.

Part (c) follows from Lemma~\ref{lem:t0-eR}.

(d). For the first part, by Theorem~\ref{thm:t0-existence}(b), we have $\Lam_\beta(m_2)-\Lam_\beta(m_1)\ge\beta\Delta-\log2+I(m_1)$; the endpoint $m_2=1$ makes no difference, by continuity of $\Lam_\beta$.

For the rest, we show that the optimal radius converges after scaling and that $\Phimix(\gamma;\cdot)/\sqrt\gamma$ converges with its derivative because it is concave. Let $\tau:=v/\beta$. Since $\rad(v)=\frac\alpha2(1+\log v-v)$, \eqref{eq:t0-mix-sandwich} gives, uniformly for $\tau\in(0,\infty)$ in compact sets, in $m$, and in $\alpha$ in a compact subset of $(0,\infty)$,
\[
\frac{\rad(\beta\tau)+\Phimix(\beta^2\tau;m)}\beta\to\sqrt{\tau\alpha}\,L(m)-\frac\alpha2\tau,
\]
which is strictly concave in $\tau$ with maximum at $L^2/\alpha$. As the left side is concave in $\tau$, its maximizer $\tau^*=v^*/\beta$ converges to $L(m)^2/\alpha$ uniformly. Let $m<m'$. Since $\Phi_\beta(m)\ge\Phimix(\beta v^*(m');m)+\rad(v^*(m'))$,
\[
\Phi_\beta(m')-\Phi_\beta(m)\le\Phimix(\beta v^*(m');m')-\Phimix(\beta v^*(m');m)\le(m'-m)\partial_m^+\Phimix(\beta v^*(m');m).
\]
The function $f_\gamma:=\Phimix(\gamma;\cdot)/\sqrt\gamma$ is concave by Lemma~\ref{lem:t0-exact}, and by \eqref{eq:t0-mix-sandwich} it converges to $\sqrt\alpha L$ uniformly, also uniformly on compact $\alpha$ sets. Since $L$ is $C^1$, a chord comparison gives $\partial_m^\pm f_\gamma\to\sqrt\alpha L'$ uniformly on compact subsets of $(-1,1)$. With $\gamma=\beta v^*(m')=\beta^2\tau^*(m')$ we have $\sqrt\gamma/\beta\to L(m')/\sqrt\alpha$, so
\[
\frac{\Lam_\beta(m')-\Lam_\beta(m)}{\beta(m'-m)}\le\frac{m+m'}2+L(m')L'(m)+o(1).
\]
The same argument, with $v^*(m)$ in place of $v^*(m')$ and $\partial_m^-$ at $m'$, gives the lower bound $\frac{m+m'}2+L(m)L'(m')-o(1)$. Since $\Lam_\beta$ is $C^1$ by Theorem~\ref{thm:uniqueness-derivative}, letting $m'\downarrow m$ gives $\Lam_\beta'(m)/\beta\to m+L(m)L'(m)=e'(m)$, with the stated uniformity.

It remains to treat the region near $1$. For $\bar m\le m<m'\le1$, concavity gives $\partial_m^+f_\gamma(m)\le\partial_m^+f_\gamma(\bar m)$, and $\sqrt\gamma/\beta\ge1-o(1)$ since $L\ge\sqrt\alpha$, so once $L'(\bar m)<0$ the quotient is at most $1+\sqrt\alpha L'(\bar m)+o(1)$. As we saw in the proof of (a), $\bar m$ can be chosen with $1+\sqrt\alpha L'(\bar m)\le-1$ for all $\alpha\in A$, which proves the last claim of (d).
\end{proof}

\subsection{Structure of the minimizer}

We show that the support of the minimizer stays away from level $0$. Below its first level the flow is the pure heat flow, which gives closed bounds.

\begin{lemma}\label{lem:t0-kappa-nonzero}
Let $\alpha>0$, $m\in(0,1)$. Let $(\kappa^*,C^*)$ be the minimizer, with $\delta^*$, $M_*=\delta^*(0)$, $Q^*=Q_{\kappa^*,m}$, $R^*$, $\Gamma^*$ as in Proposition~\ref{prop:t0-kkt}, and $\lambda^*=\lambda(\kappa^*,m)$. Let $w_1:=\min(\supp\dd\kappa^*\cup\{1\})$, where $\dd\kappa^*$ includes the atom $\kappa^*(0)$ at $0$. Then:
\begin{enumerate}[label=(\alph*)]
\item $w_1>0$, $\delta^*=M_*$ on $[0,w_1]$, and $M_*\ge1$;
\item $Q^*(w_1)=R^*(w_1)=w_1/M_*^2$, so $w_1=M_*^2Q^*(w_1)\ge Q^*(w_1)\ge m^2$ (at $w_1=1$ read $Q^*(1-)$, $R^*(1-)$);
\item $\sqrt{\alpha w_1}\,\lambda_0(m)\le\lambda^*\le\sqrt\alpha\,\lambda_0(m)$;
\item $w_1\ge w_{\mathrm{fl}}(m)$, where $w_{\mathrm{fl}}(m)$ is the limit of the monotone iteration $w^{(0)}=m^2$, $w^{(n+1)}=Q_{\mathrm{fl}}(w^{(n)})$, and
\[
Q_{\mathrm{fl}}(w):=\E\Big(2\Nc\Big(\sqrt{\frac w{1-w}}\,|\lambda_0(m)+z|\Big)-1\Big)^2.
\]
The function $w_{\mathrm{fl}}$ is nondecreasing in $m$;
\item let $\mathsf{f}:=\partial_x\Psi_{\kappa^*}(w_1,\cdot)$, with $\mathsf f=\sgn$ if $w_1=1$, and let $\varsigma$ be the probability measure $\frac12\dd\mathsf{f}$. Then
\[
\resp_{\kappa^*}(x)=\int\Big(2\Nc\Big(\frac{x-t}{\sqrt{\alpha w_1}}\Big)-1\Big)\varsigma(\dd t),\qquad\resp_{\kappa^*}'\le\frac{\wf(\resp_{\kappa^*})}{\sqrt{\alpha w_1}}.
\]
\end{enumerate}
\end{lemma}

\begin{proof}
(a), (b). If $w_1=1$, then $\kappa^*=0$, $\delta^*\equiv C^*>0$, and (K1) forces $C^*=1$, so $R^*(1-)=1=Q^*(1-)$ and both items hold. Now let $w_1<1$. Since $(Q^*-R^*)(0)=m^2>0$ and $Q^*,R^*$ are continuous, $\Gamma^*$ is strictly decreasing near $0$, and (K2) gives $\Gamma^*(0)>0$. Hence $\Gamma^*>0$ on some $[0,s_0)$, and (K3) excludes support there. By (K1), $R^*(1-)\le1$, while $\delta^*\le M_*$ gives $1/M_*^2\le R^*(1-)$, so $M_*\ge1$. By (K2) and (K3), we have $\Gamma^*=0$ at $w_1$, an interior minimum, so $Q^*(w_1)=R^*(w_1)$. Also $R^*(w_1)=w_1/M_*^2$ as $\delta^*=M_*$ on $[0,w_1]$, and $Q^*\ge m^2$.

(e). Below $w_1$, the flow is the pure heat flow, so $\resp_{\kappa^*}(x)=\E\mathsf{f}(x+\sqrt{\alpha w_1}z)$. By Lemma~\ref{lem:response-t0}(b), $\mathsf f$ is odd and nondecreasing with limits $\pm1$, so $\varsigma$ is a probability measure and the representation follows. Each $\mathsf M_t(x):=2\Nc((x-t)/\sqrt{\alpha w_1})-1$ satisfies $\mathsf M_t'=\wf(\mathsf M_t)/\sqrt{\alpha w_1}$, since $\wf(2\Nc(y)-1)=2\Nd(y)$, and as $\wf$ is concave, Jensen gives the bound on $\resp_{\kappa^*}'$.

(c). For $x\ge0$, $t\mapsto\mathsf M_t(x)+\mathsf M_{-t}(x)$ is largest at $t=0$, and $\varsigma$ is symmetric, so $m=\resp_{\kappa^*}(\lambda^*)\le\mathsf M_0(\lambda^*)=2\Nc(\lambda^*/\sqrt{\alpha w_1})-1$. For the upper bound, $m\ge2\Nc(\lambda^*/\sqrt\alpha)-1$ by Lemma~\ref{lem:response-t0}(b).

Part (d) is trivial if $w_1=1$. Otherwise $X_{w_1}\sim N(\lambda^*,\alpha w_1)$, and by Lemma~\ref{lem:response-t0}(b), $\mathsf{f}(x)^2\ge(2\Nc(|x|/\sqrt{\alpha(1-w_1)})-1)^2$. By (c), and since $\E\mathsf{n}(|a+z|)$ is nondecreasing in $a\ge0$ for nondecreasing $\mathsf{n}$, we get $w_1\ge Q^*(w_1)\ge Q_{\mathrm{fl}}(w_1)$. The function $Q_{\mathrm{fl}}$ is nondecreasing in $w$ and in $m$, so $Q_{\mathrm{fl}}(w)\le Q_{\mathrm{fl}}(w_1)\le w_1$ whenever $w\le w_1$. Since $w^{(0)}=m^2\le w_1$, all iterates are at most $w_1$, and they are monotone because $Q_{\mathrm{fl}}$ is nondecreasing. By induction each $w^{(n)}$ is nondecreasing in $m$.
\end{proof}

In unit scale (Corollary~\ref{cor:t0-unit}), $w_1$ and $Q^*$ are unchanged, $M_*=\tilde\delta^*(0)/\sqrt\alpha$, and the optimal field is $\tilde\lambda^*=\lambda^*/\sqrt\alpha=h(m)$. Hence (c) and (e) read $\sqrt{w_1}\lambda_0(m)\le h(m)\le\lambda_0(m)$, $\resp_{\tilde\kappa^*}(y)=\int(2\Nc((y-t)/\sqrt{w_1})-1)\tilde\varsigma(\dd t)$ with $\tilde\varsigma$ the image of $\varsigma$ under $t\mapsto t/\sqrt\alpha$, and
\begin{equation}\label{eq:t0-top-mass}
\mathsf A_0(m):=\resp_{\tilde\kappa^*}'(h(m))\le\frac{\wf(m)}{\sqrt{w_1}}.
\end{equation}