\section{Gaussian rows at fixed magnetization}\label{sec:gaussian}

\newcommand{\bu}{\mathsf{u}}            % normalized measure on the sphere

In this section, we identify the limit of the free energy of Gaussian rows on a magnetization slice, uniformly in the slice. We will use it in Section~\ref{sec:reduction}, where the free energy at fixed overlap for binary patterns is reduced to it, and in Section~\ref{sec:convex}, where the capacity is read off its formula. A radial identity turns the rows into an Ising slice coupled to a sphere. At fixed radius, this mixed model is solved in \cite[Theorem~3.3, Proposition~4.1]{ABCHIM}, where the Hamilton--Jacobi method of \cite{BM} is extended to it, and a Laplace principle in the radius then yields the formula for $\Phi$.

\subsection{The formula}\label{subsec:gr-formula}

Let $\mathcal{Q}$ be the set of right-continuous nondecreasing maps $[0,1)\to\R_+$, which we call paths. For a path $p$, we set $p(1):=\lim_{r\to1}p(r)$. As before, $\Qone$ denotes the set of paths bounded by $1$, and $\Qinf$ the set of bounded paths. We write $\Law_y$ for the law of a path $y$ under Lebesgue measure on $[0,1)$.

We recall that, for a law $\nu$ on $\R_+$ with bounded support, $\phi_\nu$ solves \eqref{eq:parisi-pde} and $g_m(\nu)=S+\sup_u\{um-\phi_\nu(0,u)\}$ \cite[(1.7)--(1.8)]{BM}; for $a\in\Qinf$, $g_m(a)$ is $g_m$ at the law of $a$. By \cite[Lemma~2.1]{ABCHIM}, $g_m$ is convex and $1$-Lipschitz for $L^1$ on paths, $g_m(0)=I(m):=\frac{1+m}2\log(1+m)+\frac{1-m}2\log(1-m)$, and
\begin{equation}\label{eq:gr-g-lip}
  \bigl|g_m(a)-g_{m'}(a)-I(m)+I(m')\bigr|\le2|a|_{L^1}|m-m'|.
\end{equation}
At a constant path $c\ge0$ and at $m=\pm1$, we have
\begin{equation}\label{eq:gr-g-values}
  g_m(c)=c+\sup_u\bigl\{um-\E\log\cosh(u+\sqrt{2c}\,z)\bigr\},\qquad
  g_{\pm1}(a)=\log2+\int_0^1a,
\end{equation}
with $z$ standard normal; the last value is \cite[(6.22)]{BM}. The conjugate is $J_m(x):=\sup_{a\in\Qinf}\{\int_0^1xa-g_m(a)\}$ for $x\in\Qone$ \cite[(9.1)]{BM}, and by \cite[(4.5)]{ABCHIM}, for every $a\in\Qinf$,
\begin{equation}\label{eq:gr-g-max}
  g_m(a)=\max_{x\in\Qone}\Bigl\{\int_0^1xa-J_m(x)\Bigr\}.
\end{equation}

For the sphere, we put, as in \cite[(2.10)--(2.11)]{ABCHIM}, $\mathsf{k}(\vartheta):=\vartheta(1)-\int_0^1\vartheta$ and $\mathsf{k}_\vartheta(r):=\int_0^1(\vartheta(1)-\vartheta(u)\vee r)\dd u$ for $\vartheta\in\Qinf$, and
\begin{equation}\label{eq:gr-psiS}
  \mathsf{T}(\vartheta,b):=\int_0^{\vartheta(1)}\frac{\dd r}{b-2\mathsf{k}_\vartheta(r)}+\frac12(b-1-\log b)-\vartheta(1),\qquad
  \psiS(\vartheta):=-\inf_{b>2\mathsf{k}(\vartheta)}\mathsf{T}(\vartheta,b).
\end{equation}
Its conjugate is $\psiSs(y):=\sup_{\vartheta\in\Qinf}\{\int_0^1y\vartheta-\psiS(\vartheta)\}$ for $y\in\Qone$. Recall from \eqref{eq:sbeta-def} that $\delta_y(w)=\int_0^1(1-y(u)\vee w)\dd u$.

\begin{lemma}[{Sphere conjugate, \cite[Lemma~2.3, (2.19)]{ABCHIM}}]\label{lem:sphere-conjugate}
Let $y\in\Qone$ with $\hat y:=y(1)<1$. Then
\begin{equation}\label{eq:gr-sphere-conj}
  \psiSs(y)=\frac12\Bigl[\int_0^{\hat y}\frac{\dd w}{\delta_y(w)}+\log(1-\hat y)\Bigr]\ge0,
\end{equation}
with equality only at $y=0$. In particular $\psiSs(q_0)=\frac12\bigl(\frac{q_0}{1-q_0}+\log(1-q_0)\bigr)$ for a constant $q_0\in[0,1)$.
\end{lemma}

\begin{lemma}[Ising conjugate at constant paths]\label{lem:ising-conjugate-constant}
Let $m\in[-1,1]$ and $q_0\in[0,1)$. Then $J_m(q_0)=\sup_{c\ge0}\{cq_0-g_m(c)\}<\infty$ if $|m|<1$. If $q_0\le m^2$, then $J_m(q_0)=-I(m)$. If $|m|=1$, then $J_m=-\log2$ on all of $\Qone$. Always $J_m\ge-I(m)$.
\end{lemma}

\begin{proof}
Let $|m|<1$ and $c\ge0$. By \cite[Proposition~6.10]{BM}, the supporting inequality holds at the constant path $c$, with a constant supporting path $\mathsf{q}(c,m):=\E\tanh^2(u_c+\sqrt{2c}\,z)$, where $\E\tanh(u_c+\sqrt{2c}\,z)=m$, and hence $J_m(\mathsf{q}(c,m))=c\,\mathsf{q}(c,m)-g_m(c)$ by \cite[Lemma~9.1]{BM}. The map $c\mapsto g_m(c)$ is convex with derivative $\mathsf{q}(c,m)$ \cite[(6.44)]{BM}. This derivative is continuous, with $\mathsf{q}(0,m)=m^2$ and $\mathsf{q}(c,m)\to1$ as $c\to\infty$, since $\E\operatorname{sech}^2(u+\sqrt{2c}\,z)\le2(2\pi)^{-1/2}(2c)^{-1/2}$ for every $u$. Hence, for $q_0\in[m^2,1)$, there is $c$ with $\mathsf{q}(c,m)=q_0$, and $cq_0-g_m(c)=\sup_{c'}\{c'q_0-g_m(c')\}$. For $q_0<m^2$, \cite[(6.46)]{BM} at the path $0$, whose slope is $m^2$, gives $\int_0^1q_0a-g_m(a)\le\int_0^1(q_0-m^2)a-I(m)\le-I(m)$, with equality at $a=0$. If $|m|=1$, we obtain from \eqref{eq:gr-g-values} that $J_{\pm1}(x)=-\log2+\sup_a\int_0^1(x-1)a=-\log2$. Finally, taking $a=0$ in the definition gives $J_m\ge-g_m(0)=-I(m)$.
\end{proof}

\begin{theorem}[Gaussian rows on slices]\label{thm:gaussian-slice}
Let $r\ge1$, $d_r=2^{r-1}$, let $(I_\eta)_{\eta\in\{-1,1\}^{r-1}}$ be a partition of $[N]$ with $|I_\eta|/N\to1/d_r$, let $M/N\to\alpha>0$, let $G$ be an $M\times N$ matrix of i.i.d.\ $N(0,1)$ entries, and let $\beta>0$. For a slice $k=(k_\eta)_\eta$, $0\le k_\eta\le|I_\eta|$, let $\tv_k:=(2k_\eta/|I_\eta|-1)_\eta$ and
\[
  Y_{N,k}:=2^{-N}\sum_{\sigma:k}\exp\Bigl(\frac\beta{2N}|G\sigma|^2\Bigr),
\]
the sum over $\sigma\in\Cube$ with $k_\eta$ plus signs on $I_\eta$ for every $\eta$. For $\tv\in[-1,1]^{d_r}$ and $\gamma\ge0$ let
\begin{equation}\label{eq:phimix}
  \Phimix_r(\gamma;\tv):=\inf_{(x_\eta)_\eta,\,y\in\Qone}\Bigl\{\frac{\gamma\alpha}2\Bigl(1-\Eeta\int_0^1x_\eta y\Bigr)+\Eeta J_{t_\eta}(x_\eta)+\alpha\psiSs(y)\Bigr\},
\end{equation}
and, with $\rad(v)=\frac\alpha2(1+\log v-v)$,
\begin{equation}\label{eq:radial}
  \Phi^{(r)}(\tv):=\sup_{v>0}\bigl\{\rad(v)+\Phimix_r(\beta v;\tv)\bigr\}.
\end{equation}
Then:
\begin{enumerate}[label=\textup{(\alph*)}]
\item $\sup_k\bigl|\frac1N\E\log Y_{N,k}-\Phi^{(r)}(\tv_k)\bigr|\to0$.
\item The infimum in \eqref{eq:phimix} is attained, at some $y\le\gamma/(1+\gamma)$.
\item The supremum in \eqref{eq:radial} is attained at a unique $v^*$, and $v^*$ lies in a compact interval $[v_0,v_1]\subset(0,\infty)$ depending only on $(\alpha,\beta)$.
\item With $\tau=\gamma\alpha/2$, for all $\tv,\tv'$,
\begin{equation}\label{eq:mix-lipschitz}
  \Bigl|\bigl[\Phimix_r(\gamma;\tv)+\Eeta I(t_\eta)\bigr]-\bigl[\Phimix_r(\gamma;\tv')+\Eeta I(t'_\eta)\bigr]\Bigr|\le2\tau\,\Eeta|t_\eta-t'_\eta|,
\end{equation}
and $\Phi^{(r)}+\Eeta I(t_\eta)$ is Lipschitz on $[-1,1]^{d_r}$, with a constant depending only on $(\alpha,\beta)$ for the norm $\Eeta|\cdot|$.
\end{enumerate}
\end{theorem}

With one type, we drop the index:
\begin{gather*}
  \Phimix(\gamma;m)=\inf_{x,y\in\Qone}\Bigl\{\frac{\gamma\alpha}2\Bigl(1-\int_0^1xy\Bigr)+J_m(x)+\alpha\psiSs(y)\Bigr\},\\
  \Phi(m)=\sup_{v>0}\bigl\{\rad(v)+\Phimix(\beta v;m)\bigr\}.
\end{gather*}

\subsection{The mixed model at fixed radius}\label{subsec:gr-mixed}

If $\mathbf{z}\sim N(0,I_M)$, then $\mathbf{z}=s\tau$ with $s:=|\mathbf{z}|/\sqrt M$ and $\tau:=\sqrt M\mathbf{z}/|\mathbf{z}|$ independent. Here $\tau$ is uniform on the sphere of radius $\sqrt M$, with law $\bu_M$, and $s\sim\mathsf{n}_M(\dd s):=c_Ms^{M-1}e^{-Ms^2/2}\dd s$, where $c_M:=2(M/2)^{M/2}/\Gamma(M/2)$ and $\frac1M\log c_M\to\frac12$. Since $\E_{\mathbf{z}}\exp(\sqrt{\beta/N}\,\mathbf{z}\cdot G\sigma)=\exp(\frac\beta{2N}|G\sigma|^2)$, we obtain
\begin{equation}\label{eq:gr-radial-id}
  Y_{N,k}=\int_0^\infty\mathsf{n}_M(\dd s)Z_{N,k}(s),\qquad Z_{N,k}(s):=2^{-N}\sum_{\sigma:k}\int\bu_M(\dd\tau)\exp\bigl(s\sqrt{\beta/N}\,\tau\cdot G\sigma\bigr).
\end{equation}
We set $\mathsf{A}_N(s):=\frac1N\log Z_{N,k}(s)$ and $\alpha_N:=M/N$.

At fixed radius, we use \cite{ABCHIM} with $J=\{-1,1\}^{r-1}$, so that $k=r-1$ in their (4.1), with $n=N+M$ sites, species $I_\eta$ and a spherical species of size $M$, proportions $\lambda_\eta=d_r^{-1}/(1+\alpha)$, $\lambda_{\mathrm S}=\alpha/(1+\alpha)$, and covariance $\xi_{\alpha,r-1}(a)=(1+\alpha)a_{\mathrm S}\sum_\eta a_\eta$. A Gaussian field with this covariance is $H_n(\sigma,\tau)=\sqrt{(1+\alpha)/n}\;\tau\cdot G\sigma$, and their measure \cite[(3.2)]{ABCHIM} is $2^{-|I_\eta|}$ times counting measure on each Ising slice, and $\bu_M$ on the sphere. We let $\gamma=\beta s^2$ and $t_N:=\gamma n/(2N(1+\alpha))$, so that $\sqrt{2t_N}H_n=s\sqrt{\beta/N}\,\tau\cdot G\sigma$ and $nt_N\xi_{\alpha,r-1}(\lambda_n)=\gamma M/2$. Their free energy \cite[(3.6)]{ABCHIM} at $q=0$ is thus
\begin{equation}\label{eq:gr-mixed}
  \E\mathsf{A}_N(s)=\frac{\gamma\alpha_N}2-\frac nN\bar F_{n,k}(t_N,0).
\end{equation}
Since $t_N\to\gamma/2$ and $n/N\to1+\alpha$, \cite[Theorem~3.3 and (3.7)]{ABCHIM} gives $\bar F_{n,k}(t_N,0)-f_{\alpha,r-1}(\gamma/2,\tv_k)\to0$ uniformly in $k$ and in $\gamma$ on bounded sets. Their theorem is stated along a sequence indexed by all $n$, while our $n=N+M$ may repeat or skip integers; this is harmless, since every sequence of $N$ has a subsequence along which $n$ increases strictly, and the missing $n$ can be filled with any partition satisfying their (3.1).

Finally, \cite[(4.7)]{ABCHIM}, with $p_\varepsilon=x_\eta$, $p_{\mathrm S}=y$, $\alpha t=\gamma\alpha/2$, $g^*_m=J_m$ and $\psi^{\mathsf S,*}=\psiSs$, reads
\begin{equation}\label{eq:gr-phimix-hj}
  \Phimix_r(\gamma;\tv)=\frac{\gamma\alpha}2-(1+\alpha)f_{\alpha,r-1}\Bigl(\frac\gamma2,\tv\Bigr).
\end{equation}
Hence, for each $\bar s$,
\begin{equation}\label{eq:gr-mixed-limit}
  \sup_{s\le\bar s}\sup_k\bigl|\E\mathsf{A}_N(s)-\Phimix_r(\beta s^2;\tv_k)\bigr|\to0.
\end{equation}

With $\tau=\gamma\alpha/2$, minimizing over each $x_\eta$ in \eqref{eq:phimix} by \eqref{eq:gr-g-max} gives the form in $y$,
\begin{equation}\label{eq:gr-y-form}
  \Phimix_r(\gamma;\tv)=\tau+\inf_{y\in\Qone}\Bigl\{\alpha\psiSs(y)-\Eeta g_{t_\eta}(\tau y)\Bigr\}.
\end{equation}
By \cite[Proposition~4.1]{ABCHIM}, the infimum is attained at a path bounded by $a_{\gamma/2}=\gamma/(1+\gamma)<1$, which is Theorem~\ref{thm:gaussian-slice}(b). From \eqref{eq:gr-y-form} and the properties of $g_m$, we read off three facts. At $\gamma=0$, we have $\Phimix_r(0;\tv)=-\Eeta I(t_\eta)$, by Lemma~\ref{lem:sphere-conjugate}. Since $g_m$ is $1$-Lipschitz and $y\le1$, $\Phimix_r$ is Lipschitz in $\gamma$, with constant $\alpha$, and using \eqref{eq:gr-g-lip} with $|\tau y|_{L^1}\le\tau$, we obtain \eqref{eq:mix-lipschitz}. Moreover, by \eqref{eq:phimix}, $\gamma\mapsto\Phimix_r(\gamma;\tv)$ is an infimum of affine functions, and hence concave.

\subsection{The radial Laplace principle}\label{subsec:gr-radial}

We first show that the mixed free energy is monotone in the radius, which allows us to work on a grid in the radius. Since $\log Z_{N,k}(s)$ is a log-Laplace transform, it is convex in $s$, and it is even because $\tau\mapsto-\tau$ preserves $\bu_M$. It follows that $\mathsf{A}_N$ is nondecreasing on $[0,\infty)$, and that $\mathsf{A}_N(s)\ge\mathsf{A}_N(0)=\frac1N\log(2^{-N}\#\text{slice})\ge-\log2$. On the other hand, since $|\tau\cdot G\sigma|\le\|G\|_{\mathrm{op}}\sqrt{MN}$, we have, on the whole probability space,
\begin{equation}\label{eq:gr-growth}
  \mathsf{A}_N(s)\le\mathsf{A}_N(0)+s\mathsf{B}_G,\qquad\mathsf{B}_G:=\sqrt{\beta\alpha_N}\,\|G\|_{\mathrm{op}}/\sqrt N.
\end{equation}

\begin{proof}[Proof of Theorem~\ref{thm:gaussian-slice}\textup{(c), (d)}]
For (c), taking expectations and limits in \eqref{eq:gr-growth}, with \eqref{eq:gr-mixed-limit} and $\E\|G\|_{\mathrm{op}}\le\sqrt N+\sqrt M$, we obtain $\Phimix_r(\beta s^2;\tv)+\Eeta I(t_\eta)\le s\sqrt{\beta\alpha}(1+\sqrt\alpha)$, while monotonicity gives $\Phimix_r(\beta;\tv)+\Eeta I(t_\eta)\ge0$, and $\rad(1)=0$. Hence any maximizer in \eqref{eq:radial} lies in a compact interval $[v_0,v_1]\subset(0,\infty)$ depending only on $(\alpha,\beta)$, and the supremum is attained. Uniqueness holds because $v\mapsto\Phimix_r(\beta v;\tv)$ is concave and $\rad$ is strictly concave.

For (d), \eqref{eq:mix-lipschitz} with $v\le v_1$ gives the Lipschitz bound, with constant $\alpha\beta v_1$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:gaussian-slice}\textup{(a)}]
Let $\mathsf{E}_N:=\{\|G\|_{\mathrm{op}}\le C_0\sqrt N\}$ with $C_0:=2(1+\sqrt\alpha)+1$, so that $\P(\mathsf{E}_N^c)\le e^{-cN}$, and $\mathsf{B}_G\le\mathsf{B}_0$ on $\mathsf{E}_N$ for a constant $\mathsf{B}_0$. We fix $\bar s>1$ with $\bar s^2\ge v_1$ such that, for $N$ large and $s\ge\bar s$,
\begin{equation}\label{eq:gr-tail}
  \log c_M+(M-1)\log s-Ms^2/2+Ns\mathsf{B}_0\le-N(1+s),
\end{equation}
which is possible as $M/N\to\alpha>0$. Consider a grid $s_j=j\varepsilon$, $j=0,\dots,\mathsf{j}$, with $s_{\mathsf{j}}=\bar s$, and let $s_j^*$ be a maximizer of $s\mapsto\rad(s^2)$ on $[s_j,s_{j+1}]$. In what follows, $j$ runs over $0,\dots,\mathsf{j}-1$.

For the lower bound, monotonicity gives, for each $j$, $Y_{N,k}\ge\mathsf{n}_M([s_j,s_{j+1}])\exp(N\mathsf{A}_N(s_j))$, and hence $\frac1N\E\log Y_{N,k}\ge\frac1N\log\mathsf{n}_M([s_j,s_{j+1}])+\E\mathsf{A}_N(s_j)$. Since $\frac1N\log\mathsf{n}_M([s_j,s_{j+1}])\to\rad((s_j^*)^2)$, it follows from \eqref{eq:gr-mixed-limit} that
\[
  \liminf_N\inf_k\Bigl[\frac1N\E\log Y_{N,k}-\max_j\bigl\{\Phimix_r(\beta s_j^2;\tv_k)+\rad((s_j^*)^2)\bigr\}\Bigr]\ge0.
\]

For the upper bound, \eqref{eq:gr-growth} and \eqref{eq:gr-tail} give $\int_{\bar s}^\infty\mathsf{n}_M(\dd s)Z_{N,k}(s)\le e^{N\mathsf{A}_N(0)}e^{-N}$ on $\mathsf{E}_N$. Since $\mathsf{n}_M([0,\bar s])\ge\frac12$, the bulk is at least $\frac12e^{N\mathsf{A}_N(0)}$, so the tail is smaller than the bulk. By monotonicity on $[0,\bar s]$, we obtain, on $\mathsf{E}_N$,
\[
  \frac1N\log Y_{N,k}\le\max_j\Bigl\{\frac1N\log\mathsf{n}_M([s_j,s_{j+1}])+\mathsf{A}_N(s_{j+1})\Bigr\}+\frac{\log(2\mathsf{j}+2)}N.
\]
The quantity $\mathsf{A}_N(s)$ is a function of $G$ with Lipschitz constant $s\sqrt{\beta M}/N$, since $\nabla_G\log Z_{N,k}(s)=s\sqrt{\beta/N}\langle\tau\sigma^T\rangle$ has norm at most $s\sqrt{\beta/N}\sqrt{MN}$. Gaussian concentration gives $\E\max_j|\mathsf{A}_N(s_j)-\E\mathsf{A}_N(s_j)|\le C\bar s\sqrt{\beta\alpha_N\log(2\mathsf{j}+2)/N}\to0$, uniformly in $k$. On $\mathsf{E}_N^c$, $-\log2\le\frac1N\log Y_{N,k}\le\beta\|G\|_{\mathrm{op}}^2/(2N)$, and the expectation of the upper bound on $\mathsf{E}_N^c$ is $o(1)$. Hence, uniformly in $k$,
\[
  \limsup_N\sup_k\Bigl[\frac1N\E\log Y_{N,k}-\max_j\bigl\{\Phimix_r(\beta s_{j+1}^2;\tv_k)+\rad((s_j^*)^2)\bigr\}\Bigr]\le0.
\]

Finally, since $\bar s^2\ge v_1$, we have $\sup_{s\le\bar s}\{\Phimix_r(\beta s^2;\tv_k)+\rad(s^2)\}=\Phi^{(r)}(\tv_k)$. Let $s^*$ be the maximizer, and $[s_j,s_{j+1}]$ its cell. Then $\rad((s_j^*)^2)\ge\rad((s^*)^2)$, and by the Lipschitz bound in $\gamma$ the first maximum above is at least $\Phi^{(r)}(\tv_k)-C\varepsilon$. In the second maximum, $s_j^*$ and $s_{j+1}$ lie in one cell, so it is at most $\Phi^{(r)}(\tv_k)+C\varepsilon$. As $C$ does not depend on $k$, letting $\varepsilon\to0$ completes the proof of (a).
\end{proof}