\section{A strictly convex problem and the capacity for Gaussian rows} \label{sec:convex} \newcommand{\Dylow}{\hat y_{\mathrm{low}}} In this section, we show that the Gaussian-row problem of Section~2, at fixed overlap and with the radius eliminated in closed form, is a strictly convex minimization in the law of the overlap path: \begin{equation*} \Phi(m)=\min_{x\in\Qone}\bigl\{J_m(x)+\alpha\Sbeta(x)\bigr\},\qquad \Sbeta(x)=\frac\beta2\int_0^1\frac{\mathrm{d}w}{D_x(w)}. \end{equation*} We then deduce that the optimizer is unique at every $(\alpha,T,m)$ and that the free energy $\Lam$ is $C^1$ with $\Lam'(m)=\beta m-h(m)$, where $h$ is the multiplier at the optimizer. This characterizes the capacity $\alc(T)$ of Gaussian rows at every $T>0$, including the replica symmetry breaking region, through a generalized Mattis equation. Strict convexity and uniqueness at fixed radius and field, with the spherical law as variable, are proved in \cite[Proposition~1.2]{ABCHIM}, while our form keeps the Ising law and removes the radius. \subsection{Paths, the Parisi equation and the quoted facts} Let $\Qone$ be the set of nondecreasing right-continuous paths $[0,1)\to[0,1]$, and $\Qinf$ the set of bounded nondecreasing right-continuous paths with values in $\R_+$. A path $x\in\Qone$ and its law $\Law_x$ on $[0,1]$, the law of $x(U)$ with $U$ uniform, determine each other. Since $L^1$ convergence of paths in $\Qone$ is weak convergence of laws, $\Qone$ is compact and convex in the variable $\Law$. Throughout, convexity in the law refers to mixtures of laws, not to the pointwise sum of paths. We write $\bar x:=\int_0^1x$, $\hat y:=y(1):=\lim_{\theta\to1}y(\theta)$ and $\check y:=y(0)$, and $z$ denotes a standard Gaussian variable. For a law $\Law$ on $[0,1]$, the law of a path $y$, we set \begin{equation*} \delta_{\Law}(w):=\E\bigl[1-y(U)\vee w\bigr]=\int_w^1\Law([0,w'])\dd w'. \end{equation*} It is affine and injective in $\Law$, with $\delta_\Law(w)\le1-w$ and $\delta_\Law(w)=1-w$ above $\hat y$. We write $\delta_x:=\delta_{\Law_x}$ and \begin{equation*} D_x(w):=1-\beta\delta_x(w), \end{equation*} which is increasing in $w$, with $D_x(1)=1$ and $D_x(0)=1-\beta+\beta\bar x$. For $\gamma>0$ we put $\tau:=\gamma\alpha/2$ and $c:=\gamma/2$, so $\tau=\alpha c$. At radius $v$, that is at $\gamma=\beta v$, we write $\tau_v:=\alpha\beta v/2$ and $c_v:=\beta v/2$. Recall that $\rad(v)=\frac\alpha2(1+\log v-v)$. By \eqref{eq:phimix} and \eqref{eq:radial}, \begin{equation}\label{eq:convex-H} \Phimix(\gamma;m)=\min_{x,y\in\Qone}H(x,y),\qquad H(x,y):=\tau\Bigl(1-\int xy\Bigr)+J_m(x)+\alpha\psiSs(y), \end{equation} and $\Phi(m)=\sup_{v>0}\{\rad(v)+\Phimix(\beta v;m)\}$, with the supremum attained at a unique $v^*$. Recall also that $\Lam(m)=\beta m^2/2+\Phi(m)$. For types $\eta\in\{-1,1\}^{r-1}$ with uniform weights and $\tv=(t_\eta)\in[-1,1]^{d_r}$, Theorem~\ref{thm:gaussian-slice} gives $\Phi^{(r)}(\tv)=\sup_{v>0}\{\rad(v)+\Phimix_r(\beta v;\tv)\}$, with the supremum attained at a unique $v$. For one type, this gives back $\Phi$ and $\Phimix$. On the Ising side, let $\nu$ be a law on $\R_+$ with bounded support, $a\in\Qinf$ its path, and $S\ge\sup\supp\nu$ a horizon. Recall that the function $\phi_\nu$ solves the Parisi equation \eqref{eq:parisi-pde}, whose space variable we now call $u$, because $x$ is a path: \begin{equation*} -\partial_s\phi_\nu=\partial_u^2\phi_\nu+\nu([0,s])(\partial_u\phi_\nu)^2\quad\text{on }[0,S]\times\R,\qquad \phi_\nu(S,\cdot)=\log\cosh. \end{equation*} In this notation, $\resp_\nu(u)=\partial_u\phi_\nu(0,u)$. Along $\dd X_s=2\nu([0,s])\partial_u\phi_\nu(s,X_s)\dd s+\sqrt2\dd B_s$, we put $\mathsf m_s:=\partial_u\phi_\nu(s,X_s)$. Recall that $g_m(a)=g_m(\nu)$, with $g_m$ as in \cite[(1.7)]{BM}, $J_m(x)=\sup_{a\in\Qinf}\{\int xa-g_m(a)\}$ \cite[(9.1)]{BM}, and $I(m)=\frac{1+m}2\log(1+m)+\frac{1-m}2\log(1-m)$. Above the support of $\nu$, we have $\nu([0,s])=1$ and $\phi_\nu(s,u)=\log\cosh u+S-s$. Hence $g_m$ does not depend on the horizon $S\ge\sup\supp\nu$, nor does $\mathsf m_s$ on $[0,S]$, and we continue $\mathsf m_s$ to all $s\ge0$ accordingly, setting $\mathsf m_s=\tanh X_s$ above the support. The potential of \cite[(6.19)]{BM} is stopped at the horizon; it agrees with the continued one on $[0,S]$, and we always take $S$ above the supports of all laws in play. \begin{proposition}\label{prop:parisi-facts} Let $|m|<1$, and let $\nu$ be a law on $\R_+$ with bounded support and path $a$. \begin{enumerate}[label=(\alph*)] \item \cite[(1.7), (6.14), (6.15)]{BM} One has $g_m(\nu)=S+\sup_{u\in\R}\{um-\phi_\nu(0,u)\}$. The maximizer $\lambda(\nu,m)$ is unique, and $\resp_\nu(\lambda(\nu,m))=m$. \item \cite[Lemma~6.7, (6.25)--(6.29)]{BM} $\partial_mg_m(\nu)=\lambda(\nu,m)$. The function $m\mapsto g_m(\nu)-I(m)$ is even, convex and $2\int s\dd\nu(s)$-Lipschitz. \item \cite[Corollary~6.8]{BM}, \cite[(2.9)]{ABCHIM} $\nu\mapsto g_m(\nu)$ is concave. Put $z_\nu(s):=\E[\mathsf m_s^2]\in[m^2,1)$, with $z_\nu$ nondecreasing, increasing to $1$, by the continuation above. \item \cite[Proposition~6.10, (6.45), (6.46)]{BM} $g_m$ is convex in the path and $1$-Lipschitz for $L^1$, with supporting path $z_a(\theta):=z_\nu(a(\theta))$: \begin{equation}\label{eq:path-support} g_m(a')\ge g_m(a)+\int(a'-a)z_a. \end{equation} \item \cite[Lemmas~9.1 and 9.5]{BM} $J_m(z_a)=\int z_aa-g_m(a)$, and $J_m$ is convex in the law. $J_m(x)=-I(m)$ for constant $x\le m^2$. \item \cite[Lemma~6.3]{BM} If $\nu_n\to\nu$ in $W_1$ with supports in a common interval $[0,S]$, then $\phi_{\nu_n}\to\phi_\nu$ and $\partial_u\phi_{\nu_n}\to\partial_u\phi_\nu$ uniformly. Also $|g_m(\nu)-g_m(\nu')|\le W_1(\nu,\nu')$, which is the Lipschitz bound of \textup{(d)}, since $W_1(\nu,\nu')$ is the $L^1$ distance of the paths. \end{enumerate} At $|m|=1$ one has $g_{\pm1}(a)=\int a+\log2$ \cite[(6.22)]{BM}, $z_a\equiv1$ and $J_{\pm1}\equiv-\log2$ on $\Qone$, by Lemma~\ref{lem:ising-conjugate-constant}. When the slice varies we write $z_{\nu,m}$, $z_{a,m}$. \end{proposition} On the sphere side, we use the functional of Section~2, changing its notation in this section for convenience: its path is $\pi$, written $\vartheta$ there, and $\mathcal D$, $K$, $\zeta_\pi$, $\mathcal R_y$ below are $\mathsf T$, $\mathsf k$, $\mathsf z_\vartheta$, $\mathsf r_y$ there. For $\pi\in\Qinf$ let $d_{\pi,\ell}(p):=\ell+2\int_0^1(p-\pi(\theta))_+\dd\theta$, $K(\pi):=\pi(1)-\int\pi$ and $K_\pi(p):=\int_0^1(\pi(1)-\pi(\theta)\vee p)\dd\theta$. As in \cite[(2.23)--(2.25)]{CM}, \begin{equation*} \psiS(\pi)=-\inf_{b>2K(\pi)}\mathcal D(\pi,b),\qquad\mathcal D(\pi,b):=\int_0^{\pi(1)}\frac{\mathrm{d}p}{b-2K_\pi(p)}+\frac12(b-1-\log b)-\pi(1). \end{equation*} By \cite[Lemma~2.4]{CM}, there is a unique $\ell_0(\pi)>0$ with $2\int_0^\infty d_{\pi,\ell_0}^{-2}\dd p=1$, and the minimizer is $b_0:=\ell_0+2K(\pi)=d_{\pi,\ell_0}(\pi(1))$ \cite[(2.35)]{CM}. We put $\mathsf y_\pi(p):=1-2\int_p^\infty d_{\pi,\ell_0(\pi)}^{-2}\dd p'$ and $\zeta_\pi:=\mathsf y_\pi\circ\pi$ \cite[(2.37)]{CM}, so $\zeta_\pi\in\Qone$ and $\zeta_\pi(1)=1-1/b_0<1$. The functional $\psiS$ is convex and $1$-Lipschitz in the path, with supporting path $\zeta_\pi$ \cite[Proposition~2.8, (2.52), Lemma~2.9]{CM}: \begin{equation}\label{eq:sphere-support} \psiS(\pi')\ge\psiS(\pi)+\int(\pi'-\pi)\zeta_\pi. \end{equation} Recall that its conjugate is $\psiSs(y)=\sup_{\pi\in\Qinf}\{\int y\pi-\psiS(\pi)\}$ for $y\in\Qone$. \subsection{Reduced forms and the radius-free formula} Eliminating one of the two paths in \eqref{eq:convex-H} leaves a problem in the other. \begin{proposition}\label{prop:reduced-forms} For $|m|\le1$ and $\gamma>0$, \begin{align} \Phimix(\gamma;m)&=\tau+\min_{y\in\Qone}\bigl\{\alpha\psiSs(y)-g_m(\tau y)\bigr\}\label{eq:reduced-y}\\ &=\tau+\min_{x\in\Qone}\bigl\{J_m(x)-\alpha\psiS(cx)\bigr\}.\label{eq:reduced-x} \end{align} For types, \begin{equation}\label{eq:reduced-types} \Phimix_r(\gamma;\tv)=\tau+\min_{y\in\Qone}\Bigl\{\alpha\psiSs(y)-\Eeta g_{t_\eta}(\tau y)\Bigr\}. \end{equation} \end{proposition} \begin{proof} For $a:=\tau y$ we have $J_m(x)-\int xa\ge-g_m(a)$ by the definition of $J_m$, with equality at $x=z_a\in\Qone$ by Proposition~\ref{prop:parisi-facts}(e), so minimizing over $x$ in \eqref{eq:convex-H} gives \eqref{eq:reduced-y}. Similarly, for $\pi:=cx$ we have $\alpha[\psiSs(y)-\int y\pi]\ge-\alpha\psiS(\pi)$, with equality at $y=\zeta_\pi\in\Qone$ by \eqref{eq:sphere-support}, and since $\tau=\alpha c$, minimizing over $y$ gives \eqref{eq:reduced-x}. For types, we minimize over each $x_\eta$ separately in the first step. \end{proof} \begin{proposition}\label{prop:concavity-m} For fixed $\gamma$, the function $m\mapsto\Phimix(\gamma;m)+I(m)$ is concave and even on $[-1,1]$. So $m\mapsto\Phimix(\gamma;m)$ is concave. \end{proposition} \begin{proof} In \eqref{eq:reduced-y}, $-g_m(a)+I(m)$ is concave in $m$ by Proposition~\ref{prop:parisi-facts}(b), and an infimum of concave functions is concave. \end{proof} In the next lemma, we write the spherical terms in the law, the variable in which they are convex. We set \begin{equation*} \mathcal R_\Law(w):=\frac12\int_0^w\frac{\mathrm{d}w'}{\delta_\Law(w')^2}, \end{equation*} and $\mathcal R_y:=\mathcal R_{\Law_y}$. \begin{lemma}\label{lem:sphere-law} \begin{enumerate}[label=(\roman*)] \item For every $y\in\Qone$, with values in $[0,+\infty]$, \begin{equation}\label{eq:sphere-law} \psiSs(y)=\mathcal C(\Law_y):=\frac12\int_0^1\Bigl[\frac1{\delta_{\Law_y}(w)}-\frac1{1-w}\Bigr]\dd w. \end{equation} So $\psiSs$ is a strictly convex function of the law. \item $\psiS$ is strictly concave in the law of its argument. \item For $\pi\in\Qinf$ and $y:=\zeta_\pi$ one has $\mathcal R_y(\mathsf y_\pi(p))=p$ for all $p\ge0$. In particular $\mathcal R_y(y)=\pi$. \end{enumerate} \end{lemma} \begin{proof} (i) For $\hat y<1$, Lemma~\ref{lem:sphere-conjugate} reads $\psiSs(y)=\frac12[\int_0^{\hat y}\mathrm{d}w/\delta_y(w)+\log(1-\hat y)]$. Moreover, $\int_0^{\hat y}\mathrm{d}w/(1-w)=-\log(1-\hat y)$, and the integrand of \eqref{eq:sphere-law} vanishes above $\hat y$. For $\hat y=1$, we use $y_\varepsilon:=(1-\varepsilon)y$: path convexity and $\psiSs(0)=0$ give $\psiSs(y_\varepsilon)\le(1-\varepsilon)\psiSs(y)$, and lower semicontinuity gives $\psiSs(y_\varepsilon)\to\psiSs(y)$. Since $\delta_{y_\varepsilon}$ decreases to $\delta_y$, we obtain \eqref{eq:sphere-law} by monotone convergence. Strictness holds because $\Law\mapsto\delta_\Law$ is affine and injective and $d\mapsto1/d$ is strictly convex. (ii) Let $\pi\in\Qinf$. The function $d_{\pi,\ell}(p)=\ell+2\int(p-p')_+\dd\Law_\pi(p')$ is affine in $(\Law_\pi,\ell)$, and with $\ell=b-2K(\pi)>0$ we have $b-2K_\pi(p)=d_{\pi,\ell}(p)$ on $[0,\pi(1)]$ and $d_{\pi,\ell}(p)=b+2(p-\pi(1))$ above, by \cite[Lemma~2.4]{CM}. Thus for any $r_0\ge\pi(1)$, \begin{equation}\label{eq:sphere-affine} \mathcal D(\pi,b)=\mathcal E(\Law_\pi,\ell):=\int_0^{r_0}\frac{\mathrm{d}p}{d_{\pi,\ell}(p)}-\frac12\log d_{\pi,\ell}(r_0)+\frac12d_{\pi,\ell}(r_0)-r_0-\frac12. \end{equation} For $r_0$ above the supports of the laws in play, each term is convex in $(\Law_\pi,\ell)$, and the first strictly so. Hence $\inf_\ell\mathcal E(\Law_\pi,\ell)$, which is attained by \cite[Lemma~2.4]{CM}, is strictly convex in $\Law_\pi$, and $\psiS(\pi)=-\inf_\ell\mathcal E(\Law_\pi,\ell)$. (iii) We put $\ell:=\ell_0(\pi)$ and $d:=d_{\pi,\ell}$. The map $\mathsf y_\pi$ is an increasing bijection $[0,\infty)\to[0,1)$ with $\mathsf y_\pi'=2/d^2$, and $\{y\le\mathsf y_\pi(p)\}=\{\pi\le p\}$. Since $d'(p)=2\P(\pi(U)\le p)$ and $\delta_y'(w)=-\P(y(U)\le w)$ almost everywhere, the functions $p\mapsto\delta_y(\mathsf y_\pi(p))$ and $p\mapsto1/d(p)$ have the same derivative $-2\P(\pi(U)\le p)/d(p)^2$, and both tend to $0$ at infinity. It follows that $\delta_y(\mathsf y_\pi(p))=1/d(p)$. The substitution $w'=\mathsf y_\pi(p')$ gives $\mathcal R_y(\mathsf y_\pi(p))=\frac12\int_0^p\mathsf y_\pi'(p')d(p')^2\dd p'=p$. \end{proof} \begin{theorem}\label{thm:radius-free} Let $\alpha,\beta>0$ and $|m|<1$. For $x\in\Qone$ let \begin{equation}\label{eq:Sbeta} \Sbeta(x):=\frac\beta2\int_0^1\frac{\mathrm{d}w}{D_x(w)}\ \text{ if }D_x(0)>0,\qquad\Sbeta(x):=+\infty\ \text{ otherwise.} \end{equation} Then \begin{equation}\label{eq:radius-free} \Phi(m)=\min_{x\in\Qone}\bigl\{J_m(x)+\alpha\Sbeta(x)\bigr\}. \end{equation} The objective is strictly convex in the law of $x$, so the optimizer $x^*$ is unique. The optimal radius is \begin{equation}\label{eq:radius-formula} v^*=1+\beta R_{x^*}(1),\qquad R_x(w):=\int_0^w\frac{\mathrm{d}w'}{D_x(w')^2}. \end{equation} \end{theorem} The proof is based on the fact that at fixed radius the objective is convex in the law of $x$, and at fixed law concave in the radius, so that a minimax theorem exchanges the two. The supremum over the radius is then explicit, because the radius and the spherical multiplier enter through one affine term once the multiplier is scaled by $v$. \begin{proof} \emph{Step 1: minimax.} By \eqref{eq:radial} and \eqref{eq:reduced-x}, $\Phi(m)=\sup_v\min_\Law f(v,\Law)$ with \begin{equation*} f(v,\Law):=\rad(v)+\alpha c_v+J_m(\Law)-\alpha\psiS(c_vx_\Law), \end{equation*} where $x_\Law$ is the path with law $\Law$. For fixed $v$, $f$ is convex by Proposition~\ref{prop:parisi-facts}(e) and Lemma~\ref{lem:sphere-law}(ii), and lower semicontinuous in $\Law$ on the compact convex set of laws on $[0,1]$, while for fixed $\Law$ with $J_m(\Law)<\infty$, $f$ is concave and continuous in $v$, since $\rad$ is concave, $c_v$ is affine and $\psiS$ is convex in the path. As $f$ takes the value $+\infty$ where $J_m=+\infty$, we apply Sion's theorem \cite{Sion58} on $v\in[1/n,n]$ and $\Law$ in the convex set $\{J_m<\infty\}$, where $f$ is real, and let $n\to\infty$, using lower semicontinuity on the compact space of laws. This gives $\Phi(m)=\min_\Law\sup_vf(v,\Law)$, with attainment. \emph{Step 2: the supremum over $v$.} We fix $x$ with law $\Law$ and let $\Delta_x(w):=\int_0^1(w-x(\theta))_+\dd\theta$, so $\Delta_x(1)=1-\bar x$ and $\delta_x(w)=1-\bar x-\Delta_x(w)$. The substitution $p=cw$ in \eqref{eq:sphere-affine} with $r_0=c$ gives, for $\ell>0$ and $b=\ell+2K(cx)$, \begin{equation*} \mathcal D(cx,b)=c\int_0^1\frac{\mathrm{d}w}{\ell+2c\Delta_x(w)}-\frac12\log\bigl(\ell+2c(1-\bar x)\bigr)+\frac\ell2-c\bar x-\frac12. \end{equation*} With $\ell=v\vartheta$ and $2c=\beta v$, where $\vartheta>0$, the two terms in $\log v$ cancel, and \begin{gather*} \frac12(1+\log v-v)+c+\mathcal D(cx,b)=\Gamma_x(\vartheta)+\frac v2\bigl(\vartheta-D_x(0)\bigr),\\ \Gamma_x(\vartheta):=\frac\beta2\int_0^1\frac{\mathrm{d}w}{\vartheta+\beta\Delta_x(w)}-\frac12\log\bigl(\vartheta+\beta(1-\bar x)\bigr). \end{gather*} Hence, for each fixed $x\in\Qone$ and every $v>0$, \begin{equation}\label{eq:k-inf} \mathsf k_x(v):=\rad(v)+\alpha c_v-\alpha\psiS(c_vx)=\alpha\inf_{\vartheta>0}\Bigl\{\Gamma_x(\vartheta)+\frac v2\bigl(\vartheta-D_x(0)\bigr)\Bigr\}. \end{equation} The function $\Gamma_x$ is convex and strictly decreasing, and $\Gamma_x(0+)=+\infty$ since $\Delta_x(w)\le w$. Also, $\sup_vf(v,\Law)-J_m(\Law)=\sup_v\mathsf k_x(v)$. If $D_x(0)>0$, the choice $\vartheta=D_x(0)$ shows that $\sup_v\inf_\vartheta\{\cdots\}\le\Gamma_x(D_x(0))$, and $v_x:=-2\Gamma_x'(D_x(0))$ makes $D_x(0)$ the minimizer in $\vartheta$, so the value equals $\Gamma_x(D_x(0))$. Since $D_x(0)+\beta\Delta_x(w)=D_x(w)$ and $D_x(0)+\beta(1-\bar x)=1$, this value is $\Sbeta(x)$, and $v_x=1+\beta\int_0^1\mathrm{d}w/D_x(w)^2$. If $D_x(0)<0$, then for $v\ge1$, $\Gamma_x(\vartheta)+v\vartheta/2\ge-C_x$ with $C_x$ finite, so the value is at least $-C_x+\frac v2|D_x(0)|\to\infty$. Finally, if $D_x(0)=0$, then for each $n$ we pick $\vartheta_n>0$ with $\Gamma_x\ge n$ on $(0,\vartheta_n)$. Since $\Gamma_x(\vartheta)+v\vartheta/2\ge-C_x+(v-1)\vartheta_n/2$ on $\vartheta\ge\vartheta_n$, the value tends to infinity with $v$. Thus $\sup_vf(v,\Law)=J_m(\Law)+\alpha\Sbeta(x)$, which with Step 1 gives \eqref{eq:radius-free}. At the optimum, $(v^*,\Law_{x^*})$ is a saddle point of $f$, so $v^*$ is the maximizer $v_{x^*}$ of $f(\cdot,\Law_{x^*})$, which is \eqref{eq:radius-formula}. \emph{Step 3: convexity and attainment.} The function $D_x$ is affine in $\Law$ and positive where $\Sbeta<\infty$, $d\mapsto1/d$ is strictly convex, $\Law\mapsto\delta_\Law$ is injective, and $J_m$ is convex in $\Law$. The objective is lower semicontinuous, being a supremum over $v$ of lower semicontinuous functions, and finite at the constant $x=q$ with $q\in(\max(m^2,1-1/\beta),1)$, since $J_m(q)<\infty$ for $|m|<1$ by Lemma~\ref{lem:ising-conjugate-constant}. This completes the proof. \end{proof} \subsection{The optimal paths} We fix $(\gamma,m)$ with $|m|<1$. For a law $\Law$ on $[0,1]$, let $\nu_\Law$ be the law of $\tau y$, where $y$ has law $\Law$, and put $\mathcal G(\Law):=\alpha\mathcal C(\Law)-g_m(\nu_\Law)$. \begin{theorem}\label{thm:fixed-radius-dual} $\mathcal G$ is strictly convex and lower semicontinuous, and it has a unique minimizer $\Law^*$ on the laws on $[0,1]$, with $\hat y^*<1$. Its value is $\Phimix(\gamma;m)-\tau$. If $(x,y)$ minimizes $H$, then $y=\zeta_{cx}$. \end{theorem} \begin{proof} Strict convexity follows from \eqref{eq:sphere-law} and Proposition~\ref{prop:parisi-facts}(c). Compactness and semicontinuity give a minimizer, and \eqref{eq:reduced-y} gives the value. By Proposition~\ref{prop:reduced-forms}, there is a minimizer $(x,y)$ of $H$, and then $y$ minimizes $\mathcal G$. Also, $y$ minimizes $y'\mapsto\alpha\psiSs(y')-\tau\int xy'$, which has the minimizer $\zeta_{cx}$ by \eqref{eq:sphere-support}. Hence $(x,\zeta_{cx})$ minimizes $H$ too, and $\zeta_{cx}$ minimizes $\mathcal G$, so $y=\zeta_{cx}$ by strict convexity. In particular, $\hat y=1-1/b_0<1$. \end{proof} \begin{lemma}\label{lem:conjugate-variations} Let $x\in\Qone$ with $D_x(0)>0$, and put $v:=1+\beta R_x(1)$ and $y:=\beta R_x(x)/v$. Then $y=\zeta_{c_vx}$. \end{lemma} \begin{proof} We put $\ell:=vD_x(0)$ and $\pi:=c_vx$. Then $d_{\pi,\ell}(c_vq)=vD_x(q)$ for all $q\ge0$, with $D_x(q)=1+\beta(q-1)$ above $1$. Thus $2\int_0^\infty d_{\pi,\ell}^{-2}\dd p=\frac\beta v(R_x(1)+\frac1\beta)=1$, which says that $\ell$ is the multiplier $\ell_0(\pi)$, and \begin{equation*} \mathsf y_\pi(c_vq)=1-\frac\beta v\int_q^\infty\frac{\mathrm{d}q'}{D_x(q')^2}=1-\frac1v-\frac\beta v\bigl(R_x(1)-R_x(q)\bigr)=\frac{\beta R_x(q)}v, \end{equation*} where we used $v-1=\beta R_x(1)$ in the last step. Hence $\zeta_\pi=\beta R_x(x)/v=y$. \end{proof} \subsection{Uniqueness and the derivative} We will obtain the derivative of $\Phi$ in $m$ from an envelope argument with the unique optimizer, and control its size by the slope of the magnetization response, which has an explicit formula. Let $\nu$ be a law on $\R_+$ with bounded support, and write $A_\nu:=\resp_\nu'$. \begin{lemma}\label{lem:response} \begin{enumerate}[label=(\roman*)] \item $\resp_\nu$ is odd and strictly increasing, $\resp_\nu(\pm\infty)=\pm1$, and $\resp_\nu$ is concave on $[0,\infty)$. So $A_\nu$ is even and nonincreasing in $|u|$. \item If $m=\resp_\nu(u)$ then $A_\nu(u)=1-\int z_{\nu,m}\dd\nu$. So $\partial_m\lambda(\nu,m)=1/(1-\bar x(\nu,m))$, where $\bar x(\nu,\tilde m):=\int z_{a,\tilde m}$ is the mean of the supporting path at slice $\tilde m$ for the fixed law $\nu$. \item $\nu\mapsto A_\nu$ is continuous for $W_1$, uniformly on compacts, on sets of $\nu$ with a common bound on the support. \end{enumerate} \end{lemma} \begin{proof} It suffices to treat finite-step $\nu$: the general case follows from Proposition~\ref{prop:parisi-facts}(f), since concavity and monotonicity pass to uniform limits, and (ii), (iii) will follow from the bound on $\partial_u^3\phi_\nu$ below. We write $\phi=\phi_\nu$ and $\phi^{(k)}:=\partial_u^k\phi$. On each piece, $\nu([0,s])=:\varsigma$ is constant. Evenness of $\phi$ in $u$ follows from the terminal datum and uniqueness. Differentiating the equation, we obtain \begin{gather*} -\partial_s\phi^{(1)}=\partial_u^2\phi^{(1)}+2\varsigma\phi^{(1)}\phi^{(2)},\qquad-\partial_s\phi^{(2)}=\partial_u^2\phi^{(2)}+2\varsigma\bigl(\phi^{(1)}\partial_u\phi^{(2)}+(\phi^{(2)})^2\bigr),\\ -\partial_s\phi^{(3)}=\partial_u^2\phi^{(3)}+2\varsigma\phi^{(1)}\partial_u\phi^{(3)}+6\varsigma\phi^{(2)}\phi^{(3)}. \end{gather*} Thus $\phi^{(3)}$ is a bounded solution of a linear backward parabolic equation with bounded coefficients on $u>0$, with $\phi^{(3)}(s,0)=0$ by oddness. At the top of the piece, $\phi^{(3)}\le0$ on $u\ge0$. Indeed, this holds initially, as $\tanh''=-2\tanh\,\mathrm{sech}^2\le0$, and then by induction over pieces, since at a breakpoint the function is continuous and only $\varsigma$ changes. The maximum principle then gives $\phi^{(3)}\le0$ on $u\ge0$, which is the concavity in (i). Oddness is the evenness of $\phi$, and strict monotonicity is $\phi^{(2)}>0$, which we will obtain from the representation below. The limits at $\pm\infty$ hold because $\resp_\nu$ takes values in $(-1,1)$ and hits every $m\in(-1,1)$, by Proposition~\ref{prop:parisi-facts}(a). (ii) Let $\mathsf p_s:=\phi^{(2)}(s,X_s)$. With $\mathcal A:=\partial_s+\partial_u^2+2\varsigma\phi^{(1)}\partial_u$, the generator along $X$, we have $\mathcal A\phi^{(1)}=0$ and $\mathcal A\phi^{(2)}=-2\varsigma(\phi^{(2)})^2$. Hence $\dd\mathsf p_s=-2\varsigma\mathsf p_s^2\dd s+$ martingale, and $\dd\,\E\mathsf m_s^2=2\E\mathsf p_s^2\dd s$ as $\dd\mathsf m_s=\sqrt2\mathsf p_s\dd B_s$. Since $\mathsf p_{S}=\mathrm{sech}^2X_{S}=1-\mathsf m_{S}^2$, with $S$ above the support, taking expectations gives \begin{equation*} A_\nu(u)=\mathsf p_0=1-z_\nu(S)+\int_0^{S}\nu([0,s])\dd z_\nu(s)=1-z_\nu(S)+\int\bigl(z_\nu(S)-z_\nu(s)\bigr)\dd\nu(s)=1-\int z_\nu\dd\nu. \end{equation*} The same computation started at $(s,u)$ gives $\phi^{(2)}(s,u)\in(0,1]$, and the formula for $\partial_m\lambda$ follows as $\lambda(\nu,\cdot)$ is the inverse of $\resp_\nu$. (iii) By the maximum principle applied to its linear equation, $\phi^{(3)}$ is bounded on $[0,S]\times\R$ by $C\mathrm e^{6S}$, uniformly over $\nu$ with support in $[0,S]$. With Proposition~\ref{prop:parisi-facts}(f) and Arzel\`a--Ascoli this gives (iii). \end{proof} \begin{theorem}\label{thm:uniqueness-derivative} Let $\alpha,\beta>0$ and $|m|<1$. \begin{enumerate}[label=(\alph*)] \item \textup{(Uniqueness.)} At fixed radius, the minimizer $(x,y)$ of $H$ is unique. At the optimal radius, the triple $(x^*,y^*,v^*)$ is unique, with \begin{equation}\label{eq:optimal-triple} y^*=\beta R_{x^*}(x^*)/v^*,\qquad a^*:=\frac{\alpha\beta^2}2R_{x^*}(x^*)=\tau_{v^*}y^*,\qquad x^*=z_{a^*,m}. \end{equation} In all cases $\supp\Law_{x^*}\subset[m^2,1)$. \item \textup{(Continuity.)} The map $(m,\alpha,\beta)\mapsto(\Law_{x^*},\Law_{y^*},v^*)$ is continuous on $(-1,1)\times(0,\infty)^2$, and the optimal law of $y$ at fixed radius is continuous in $(\gamma,m)$. \item \textup{(Differentiability.)} $\Phimix(\gamma;\cdot)$ and $\Phi$ are $C^1$ on $(-1,1)$, with \begin{equation}\label{eq:derivative} \Phi'(m)=-h(m),\qquad h(m):=\lambda(\nu^*(m),m),\qquad\Lam'(m)=\beta m-h(m), \end{equation} where $\nu^*(m)$ is the law of $a^*$ at slice $m$. The function $h$ is odd and continuous, and for $m\in[0,1)$, \begin{equation}\label{eq:h-bounds} \max\Bigl\{\atanh m,\frac m{1-\bar x(\nu^*(m),0)}\Bigr\}\le h(m)\le\min\Bigl\{\atanh m+2\int s\dd\nu^*(m)(s),\frac m{1-\bar x(\nu^*(m),m)}\Bigr\}, \end{equation} with $\bar x(\nu^*(m),m)=\int x^*$. So $\Lam'(m)\to-\infty$ as $m\to1$. \item \textup{($m=0$.)} $m=0$ is a strict local maximum of $\Lam$ at every $T>0$: with $\bar x_0$ the mean of $x^*$ at slice $m=0$, one has $1-\beta+\beta\bar x_0>0$ and \begin{equation}\label{eq:slope-at-zero} \limsup_{m\to0+}\frac{\Lam'(m)}m\le\beta-\frac1{1-\bar x_0}=-\frac{1-\beta+\beta\bar x_0}{1-\bar x_0}<0. \end{equation} \end{enumerate} \end{theorem} \begin{proof} (a) At fixed radius, we use Theorem~\ref{thm:fixed-radius-dual} for $y$, and for $x$ the objective in \eqref{eq:reduced-x} is convex in $\Law_x$ by Proposition~\ref{prop:parisi-facts}(e), and strictly so by Lemma~\ref{lem:sphere-law}(ii), since $\tau,c$ are fixed and $cx$ has an injective law. For the triple at the optimal radius, Theorem~\ref{thm:radius-free} gives the unique $x^*$ with $v^*=1+\beta R_{x^*}(1)$, and by Lemma~\ref{lem:conjugate-variations}, we have $\zeta_{c_{v^*}x^*}=\beta R_{x^*}(x^*)/v^*$, which is $y^*$ by Theorem~\ref{thm:fixed-radius-dual}. Finally, $(z_{\tau_{v^*}y^*},y^*)$ minimizes $H$ at $v^*$ by Proposition~\ref{prop:reduced-forms}, so $x^*=z_{a^*,m}$ by uniqueness of $x$, and the values of $x^*$ lie in $[m^2,1)$ by Proposition~\ref{prop:parisi-facts}(c). (b) By Proposition~\ref{prop:parisi-facts}(b),(f), and since $\nu_\Law$ has support in $[0,\tau]$, $g_m(\nu_\Law)$ is jointly continuous in $(m,\tau,\Law)$ on $(-1,1)\times\R_+\times\{\text{laws on }[0,1]\}$. The functional $\mathcal C$ is lower semicontinuous and independent of the parameters, so $\mathcal G$ $\Gamma$-converges when the parameters converge. As the domain is compact and limits of minimizers are minimizers, uniqueness gives continuity of the optimal law of $y$. The function $v\mapsto\rad(v)+\Phimix(\beta v;m)$ is strictly concave in $v$ and jointly continuous in all parameters, with a unique maximizer by Theorem~\ref{thm:gaussian-slice}. We pick $a0$ since $\Sbeta(x^*)<\infty$ with $\alpha>0$, so $\beta<1/(1-\bar x_0)$. As $m\to0+$, $\nu^*(m)\to\nu^*(0)$ by (b), and $\bar x(\nu^*(m),0)\to\bar x(\nu^*(0),0)$ by Lemma~\ref{lem:response}(iii),(ii), and this limit equals $\bar x_0$ by \eqref{eq:optimal-triple}. Then \eqref{eq:h-bounds} gives $h(m)\ge m/(1-\bar x(\nu^*(m),0))$, which is \eqref{eq:slope-at-zero}. Since $\Lam$ is even and $\Lam'<0$ on a right neighborhood of $0$, $m=0$ is a strict local maximum. \end{proof} \subsection{The potential of a path} In Section~6, we will compare a configuration of several retained patterns with one pattern at larger overlap through a bound linear in a path. For $a\in\Qinf$ with law $\nu$, we put \begin{equation}\label{eq:LC} LC_a(m):=-g_m(a)=\inf_u\{\Psi_a(u)-um\},\qquad\Psi_a(u):=\phi_\nu(0,u)-S,\qquad\resp_a:=\Psi_a'=\resp_\nu. \end{equation} Proposition~\ref{prop:parisi-facts} shows that $LC_a$ is concave and $-\partial_mLC_a(m)=\lambda(\nu,m)$, and Lemma~\ref{lem:response} that it is $C^2$ on $(-1,1)$, with $\partial_m^2LC_a(m)=-1/A_\nu(\lambda(\nu,m))$. By \eqref{eq:reduced-types}, for every $y\in\Qone$ and $\gamma>0$, \begin{equation}\label{eq:path-linear-bound} \Phimix_r(\gamma;\tv)\le\tau+\alpha\psiSs(y)+\Eeta LC_{\tau y}(t_\eta), \end{equation} with equality when $r=1$ and $y$ is the optimal path at the slice. \subsection{The capacity for Gaussian rows} We now define the capacity of Gaussian rows with one binary pattern, at every $T>0$, by a condition on the multiplier $h$ of Theorem~\ref{thm:uniqueness-derivative}. \begin{definition}\label{def:capacity-gaussian} For $\alpha>0$ and $T=1/\beta>0$, let $h(m)=h(m;\alpha,\beta)$ and \begin{equation}\label{eq:capacity-def} \mathsf h(\alpha,\beta):=\sup_{m\in(0,1)}\bigl(\beta m-h(m;\alpha,\beta)\bigr)\ge0,\quad\Well_T:=\{\alpha>0:\mathsf h(\alpha,1/T)>0\},\quad\alc(T):=\sup\Well_T, \end{equation} with the supremum of the empty set equal to $0$. A \emph{well} at $(\alpha,T)$ means: there exist $00:\exists m\in(0,1),\ \beta m>\lambda(\nu^*(m),m)\bigr\}, \end{equation*} where $\nu^*(m)$ is the law of $a^*=\frac{\alpha\beta^2}2R_{x^*}(x^*)$ and $x^*$ is the unique minimizer of the strictly convex functional $x\mapsto J_m(x)+\alpha\Sbeta(x)$. By the mean value theorem and \eqref{eq:derivative}, $\Well_T$ is the union over $m_s$ of the sets \eqref{eq:well-def}, and $\Well_T(m_s)=\{\alpha>0:\exists m\in[m_s,1),\ \beta m>h(m;\alpha,\beta)\}$. \begin{remark} By Theorem~\ref{thm:uniqueness-derivative}(d), the state $m=0$ is always a strict local maximum for $\alpha>0$. Hence a point with $\Lam'>0$ always lies on the inner wall of a true double well, however small $m$ is, and no window near $m=1$ is needed. \end{remark} \begin{theorem}\label{thm:capacity-gaussian} For every $T>0$ the following hold. \begin{enumerate}[label=(\alph*)] \item $\alpha\in\Well_T$ if and only if $\Lam$ has a well at $(\alpha,T)$, if and only if there is $m\in(0,1)$ with $\resp_{a^*(m)}(\beta m)>m$, where $a^*(m)$ is the optimal path of Theorem~\ref{thm:uniqueness-derivative} at slice $m$. So the stationary points $m>0$ of $\Lam$ are exactly the solutions of the generalized Mattis equation \begin{equation}\label{eq:mattis} m=\resp_{a^*(m)}(\beta m). \end{equation} \item $\Well_T$ is open in $(0,\infty)$. \item For $T\ge1$, $\Well_T$ is empty and $\alc(T)=0$. For $T<1$, $\Well_T$ contains $(0,\alpha_0(T))$ for some $\alpha_0(T)>0$, so $\alc(T)>0$. \item At a positive boundary point $\alpha_b>0$ of $\Well_T$, in particular at $\alc(T)$ when $T<1$ by Corollary~\ref{cor:load-bound}, one has $\mathsf h(\alpha_b,1/T)=0$, and there is $m_c\in(0,1)$ with $h(m_c)=\beta m_c$ and $\Lam'\le0$ on $[0,1)$. So at the boundary $\Lam$ has an interior stationary contact: the well disappears at positive overlap, with no collapse to $m=0$ or $m=1$. \end{enumerate} \end{theorem} For a constant path $a\equiv a_0$, we have $\resp_a(u)=\E\tanh(u+\sqrt{2a_0}\,z)$, and with $2a_0=\alpha\beta^2\mathsf r$, equation \eqref{eq:mattis} becomes the Mattis equation $m=\E\tanh(\beta m+\beta\sqrt{\alpha\mathsf r}\,z)$ of \cite{AGS85,AGS87}, which is the replica symmetric case. We will use three facts about $\Lam'$: it is negative just above $0$, it tends to $-\infty$ at $1$, and it depends continuously on the load. Hence a single point with $\Lam'>0$ produces a true double well, and at a boundary load the contact stays inside $(0,1)$. \begin{proof} (a) If there is a well, the mean value theorem gives $m$ with $\Lam'(m)>0$. Conversely, let $\Lam'(m_\circ)>0$ with $m_\circ\in(0,1)$. Since $\Lam$ is strictly decreasing near $0+$ by Theorem~\ref{thm:uniqueness-derivative}(d) and $\Lam'(m_\circ)>0$, the minimum of $\Lam$ on $[0,m_\circ]$ is attained at an interior point. By \eqref{eq:h-bounds}, $\Lam'\to-\infty$ at $1$, so the maximum of $\Lam$ on $[m_\circ,1)$ is attained at some $m^*\in(m_\circ,1)$, with $\Lam(m^*)>\Lam(m_\circ)$. We let $m_b$ be a minimizer of $\Lam$ on $[0,m^*]$; it is interior and a local minimum, and $\Lam(m_b)\le\Lam(m_\circ)<\Lam(m^*)$. It remains to treat the last form. Here $\resp_{a^*(m)}$ is strictly increasing with $\resp_{a^*(m)}(h(m))=m$, so $\beta m>h(m)$ if and only if $\resp_{a^*(m)}(\beta m)>m$, and likewise with equalities. This gives \eqref{eq:mattis}, by \eqref{eq:derivative}. (b) The map $(m,\alpha)\mapsto h$ is continuous by Theorem~\ref{thm:uniqueness-derivative}(b),(c) and Proposition~\ref{prop:parisi-facts}(f), so a point with $\beta m-h(m)>0$ persists under small changes of $\alpha$. (c) For $T\ge1$, $h(m)\ge\atanh m>m\ge\beta m$ for $m\in(0,1)$, by \eqref{eq:h-bounds}. If $T<1$, let $m_{\mathrm{CW}}$ be the positive maximizer of $m\mapsto\beta m^2/2-I(m)$, so $\beta(1-m_{\mathrm{CW}}^2)<1$, as the Curie--Weiss root is stable, and pick $m_\circ\in(0,m_{\mathrm{CW}})$ with $D_{m_\circ^2}(0)=1-\beta(1-m_\circ^2)>0$. We have $\Phi(m)\ge-I(m)$, because $J_m\ge-g_m(0)=-I(m)$ and $\Sbeta>0$. On the other hand, by \eqref{eq:radius-free} with $x=m_\circ^2$ and $J_{m_\circ}(m_\circ^2)=-I(m_\circ)$, we have $\Phi(m_\circ)\le-I(m_\circ)+\alpha\Sbeta(m_\circ^2)$. Hence for $\alpha$ below $[\beta m_{\mathrm{CW}}^2/2-I(m_{\mathrm{CW}})-\beta m_\circ^2/2+I(m_\circ)]/\Sbeta(m_\circ^2)$, we get $\Lam(m_\circ)<\beta m_{\mathrm{CW}}^2/2-I(m_{\mathrm{CW}})\le\Lam(m_{\mathrm{CW}})$, and the mean value theorem gives a point with $\Lam'>0$ in $(m_\circ,m_{\mathrm{CW}})$. (d) By (b), $\alpha_b\notin\Well_T$, so $\mathsf h(\alpha_b,1/T)=0$, that is, $\Lam'\le0$ on $[0,1)$ at $\alpha_b$. We take $\alpha_n\to\alpha_b$ in $\Well_T$, with $m_n$ such that $\beta m_n-h(m_n;\alpha_n)>0$. By \eqref{eq:h-bounds}, $m_n$ stays below $1-\varepsilon$, since $h\ge\atanh m>\beta$ for $m$ near $1$. Also, $m_n$ stays away from $0$. Indeed, as in the proof of Theorem~\ref{thm:uniqueness-derivative}(d), $\bar x(\nu^*(m;\alpha),0)$ tends to $\bar x_0(\alpha_b)$ as $(m,\alpha)\to(0,\alpha_b)$, so \eqref{eq:h-bounds} and \eqref{eq:slope-at-zero} give $h(m;\alpha)\ge(\beta+\varepsilon)m$ near $(0,\alpha_b)$. Hence $(m_n)$ has limit points in $(0,1)$, and every limit point $m_c$ has $\beta m_c-h(m_c;\alpha_b)\ge0$, and $\le0$ since $\mathsf h=0$ at $\alpha_b$. \end{proof} It remains to bound the load. The next lemma is an estimate at every fixed radius: the optimal spherical path has a top value $\hat y$ bounded below explicitly, which bounds the slope of the response from above. Using $v^*=1+\beta R_{x^*}(1)\ge1+\beta$, which holds because $00$ and $m\in(0,1)$, and let $y$ be the optimal path of \eqref{eq:reduced-y} at slice $m$, $a:=\tau y$, and $\nu$ its law. Let $\mathsf f(s):=\E\tanh^2(\sqrt s\,z)$, which is increasing in $s\ge0$, and \begin{equation*} \Dylow(\gamma,\alpha):=\inf\Bigl\{w\in(0,1):\frac w{(1-w)^2}\ge\gamma\mathsf f(\gamma\alpha w)\Bigr\}. \end{equation*} \begin{lemma}\label{lem:finite-gamma} For all $\alpha,\gamma>0$ and $m\in(0,1)$: \begin{enumerate}[label=(\alph*)] \item $z_\nu(\tau w)=\frac2\gamma\mathcal R_y(w)$ on the closure of the range of $y$. In particular $\check y>0$, $z_\nu(\tau\check y)=\check y/(\gamma\delta_y(0)^2)$ and $z_\nu(\tau\hat y)=\frac2\gamma\mathcal R_y(\hat y)$. \item $\alpha\gamma A_\nu(\lambda(\nu,m))^2\le\dfrac1{\gamma\delta_y(0)^2}-\dfrac{m^2}{\check y}\le\dfrac1{\hat y}$. \item $\hat y/(1-\hat y)^2\ge\gamma z_\nu(\tau\hat y)\ge\gamma\mathsf f(\gamma\alpha\hat y)$. So $\hat y\ge\Dylow(\gamma,\alpha)$. The function $\Dylow$ increases in $\gamma$ and in $\alpha$. \item $-\partial_m\Phimix(\gamma;m)\ge m\sqrt{\alpha\gamma\Dylow(\gamma,\alpha)}$. \end{enumerate} \end{lemma} \begin{proof} (a) By Theorem~\ref{thm:fixed-radius-dual} and Proposition~\ref{prop:reduced-forms}, $y=\zeta_{cx}$ and $x=z_a$, and by Lemma~\ref{lem:sphere-law}(iii), $cx=\mathcal R_y(y)$, with $c=\gamma/2$. Hence $z_\nu(\tau y(\theta))=\frac2\gamma\mathcal R_y(y(\theta))$ for all $\theta$, and both sides are continuous in the value of $y$. At $w=0$, $z_\nu(0)=m^2>0=\mathcal R_y(0)$, so $0$ is not in the closure of the range. Also, $\delta_y=\delta_y(0)$ on $[0,\check y]$. (b) On $[0,\tau\check y)$ we have $\nu([0,s])=0$, so there $X_s=\lambda(\nu,m)+\sqrt2B_s$, the function $\partial_u\phi_\nu$ is a bounded heat flow, and $\mathsf p_s:=\partial_u^2\phi_\nu(s,X_s)$ is a martingale with $\E\mathsf p_s=A_\nu(\lambda(\nu,m))$. By Cauchy--Schwarz, $\partial_sz_\nu(s)=2\E\mathsf p_s^2\ge2A_\nu(\lambda(\nu,m))^2$, so, using (a) and $2\tau=\alpha\gamma$, we obtain \begin{equation*} \frac{\check y}{\gamma\delta_y(0)^2}=z_\nu(\tau\check y)\ge m^2+\alpha\gamma A_\nu(\lambda(\nu,m))^2\check y. \end{equation*} For the last inequality of (b), $\delta_y\le\delta_y(0)$ on $[0,\hat y]$ gives $\hat y/(\gamma\delta_y(0)^2)\le\frac2\gamma\mathcal R_y(\hat y)=z_\nu(\tau\hat y)\le1$. (c) As $\delta_y\ge\delta_y(\hat y)=1-\hat y$ on $[0,\hat y]$, we have $\frac2\gamma\mathcal R_y(\hat y)\le\hat y/(\gamma(1-\hat y)^2)$, which with (a) gives the first inequality. Above $\tau\hat y$, the top of the support of $\nu$, $\partial_u\phi_\nu=\tanh$, so $z_\nu(\tau\hat y)=\E\tanh^2X_{\tau\hat y}$. On $[0,\tau\hat y)$ the drift of $X$ has the sign of $X$, by Lemma~\ref{lem:response}(i). Thus, by Tanaka's formula, $|X|$ dominates reflected Brownian motion of variance $2s$ started from $|\lambda(\nu,m)|$, hence the one started from $0$, and as $\tanh^2$ increases in $|\cdot|$, we obtain $z_\nu(\tau\hat y)\ge\mathsf f(2\tau\hat y)=\mathsf f(\gamma\alpha\hat y)$. Since $\hat y\ge\check y>0$, $\hat y$ lies in the set defining $\Dylow$, and monotonicity holds because $\mathsf f$ increases. This completes the proof of (c). (d) The function $\Phimix(\gamma;\cdot)$ is concave and even by Proposition~\ref{prop:concavity-m}, and differentiable by Theorem~\ref{thm:uniqueness-derivative}(c), so $\partial_m\Phimix(\gamma;0)=0$. Let $y^\circ$ be optimal at $m_0\in(0,1)$, and $\nu^\circ$ the law of $\tau y^\circ$. By \eqref{eq:reduced-y}, $\Phimix(\gamma;\cdot)\le\tau+\alpha\psiSs(y^\circ)+LC_{\tau y^\circ}$ on $[-1,1]$, with equality at $m_0$, and $LC_{\tau y^\circ}$ is $C^2$ with second derivative $-1/A_{\nu^\circ}(\lambda(\nu^\circ,\cdot))$. Hence $\partial_m^2\Phimix(\gamma;m_0)\le-1/A_{\nu^\circ}(\lambda(\nu^\circ,m_0))$ at every $m_0$ where $\partial_m\Phimix(\gamma;\cdot)$ is differentiable, and by (b) and (c), $1/A_{\nu^\circ}(\lambda(\nu^\circ,m_0))\ge\sqrt{\alpha\gamma\hat y^\circ}\ge\sqrt{\alpha\gamma\Dylow(\gamma,\alpha)}$. Finally, $-\partial_m\Phimix(\gamma;\cdot)$ is nondecreasing, so it is at least the integral of its almost everywhere derivative, which proves (d). \end{proof} \begin{proposition}\label{prop:low-overlap-criterion} For $m\in(0,1)$, with $v^*(m)$ the optimal radius and $\gamma^*:=\beta v^*(m)$, \begin{equation}\label{eq:low-overlap-criterion} \Lam'(m)\le m\Bigl(\beta-\sqrt{\alpha\beta v^*(m)\Dylow(\gamma^*,\alpha)}\Bigr). \end{equation} So $\Lam'(m)<0$ at every $m>0$ with $\alpha v^*(m)\Dylow(\gamma^*,\alpha)>\beta$. At a contact point $m_c$ of Theorem~\ref{thm:capacity-gaussian}\textup{(d)}, $\alpha v^*(m_c)\Dylow(\gamma^*,\alpha)\le\beta$. \end{proposition} \begin{proof} We have $\Lam'=\beta m-h(m)$, and $h(m)=\lambda(\nu^*(m),m)=-\partial_m\Phimix(\gamma^*;m)$ at this $m$, by the proof of Theorem~\ref{thm:uniqueness-derivative}(c), so Lemma~\ref{lem:finite-gamma}(d) applies at $\gamma^*$. \end{proof} \begin{corollary}\label{cor:load-bound} For every $T>0$, $\Well_T\subset(0,2)$. So $\alc(T)\le2$ uniformly in $T$. \end{corollary} \begin{proof} For $T\ge1$, $\Well_T$ is empty. Let $T<1$. By \eqref{eq:radius-formula}, $v^*\ge1+\beta$, so $\gamma^*\ge\beta(1+\beta)\ge2$ and $\alpha v^*/\beta\ge\alpha(1+T)$. For $\alpha\ge2$, monotonicity gives $\Dylow(\gamma^*,\alpha)\ge\Dylow(\beta(1+\beta),2)$. In ball arithmetic, we have $\Dylow(2,2)>.35$ and $\Dylow(4.44,2)>.5$; the computation will be given in Subsection~\ref{ssec:computed}. If $T\ge.6$, then $\alpha(1+T)\Dylow(\gamma^*,\alpha)>2(1+.6)(.35)>1$, while if $T\le.6$, then $\gamma^*\ge\frac53\cdot\frac83>4.44$, so $\Dylow(\gamma^*,\alpha)>.5$ and $\alpha(1+T)\Dylow(\gamma^*,\alpha)>1$. In both cases Proposition~\ref{prop:low-overlap-criterion} gives $\Lam'<0$ on $(0,1)$, so $\alpha\notin\Well_T$. \end{proof}