Some properties of the phase diagram for mixed p-spin glasses

Abstract

In this paper we study the Parisi variational problem for mixed p-spin glasses with Ising spins. Our starting point is a characterization of Parisi measures whose origin lies in the first order optimality conditions for the Parisi functional, which is known to be strictly convex. Using this characterization, we study the phase diagram in the temperature-external field plane. We begin by deriving self-consistency conditions for Parisi measures that generalize those of de Almeida and Thouless to all levels of Replica Symmetry Breaking (RSB) and all models. As a consequence, we conjecture that for all models the Replica Symmetric phase is the region determined by the natural analogue of the de Almeida–Thouless condition. We show that for all models, the complement of this region is in the RSB phase. Furthermore, we show that the conjectured phase boundary is exactly the phase boundary in the plane less a bounded set. In the case of the Sherrington–Kirkpatrick model, we extend this last result to show that this bounded set does not contain the critical point at zero external field.

Appendix

The appendix is organized as follows. In “Well-posedness of the Parisi PDE” we state some preliminary facts about the Parisi PDE. In Sect. “The Auffinger–Chen SDE” we state some useful formulas regarding the Auffinger-Chen SDE. In Sect. “A change of measure formula” we state a useful change of variables through Girsanov’s theorem. In Sect. “Asymptotic spectral theory for certain operators” we record some important spectral estimates to be used in Sect. 7. In Sect. “Proof of Fact 2” we bound a certain integral whose sign is of interest. In Sect. “The elementary argument for RS” we give an elementary argument for RS at sufficiently high temperature and external field, which when combined with the main theorems proves boundedness of the exceptional set for our arguments. We end in Sect. “Is the AT line a line?” with a discussion regarding topological properties of the level sets of $\alpha$.

1.1 Well-posedness of the Parisi PDE

The following three propositions are taken from the authors’ paper [12]. We call a continuous function $u:\left[ 0,1\right] \times \mathbb {R}\rightarrow \mathbb {R}$ with essentially bounded weak derivative $u_{x}$ a weak solution of the Parisi PDE (1.2) if it satisfies

$$\begin{aligned} 0=\int _{0}^{1}\int _{\mathbb {R}}-u\phi _{t}+\frac{\xi ''\left( t\right) }{2}\left( u\phi _{xx}+\mu \left[ 0,t\right] u_{x}^{2}\phi \right) \, dxdt+\int _{\mathbb {R}}\phi \left( 1,x\right) \log \cosh x\, dx \end{aligned}$$

for every $\phi \in C_{c}^{\infty }\left( (0,1]\times \mathbb {R}\right) .$

Proposition 8.1

Let $\mu \in \Pr \left[ 0,1\right]$. There exists a unique weak solution u to the Parisi PDE. The weak solution u to (1.2) has higher regularity:

• $\partial _{x}^{j}u\in C_{b}\left( \left[ 0,1\right] \times \mathbb {R}\right)$ for $j\ge 1$

• $\partial _{t}\partial _{x}^{j}u\in L^{\infty }\left( \left[ 0,1\right] \times \mathbb {R}\right)$ for $j\ge 0$.

For all $j\ge 1$, the derivative $\partial _{x}^{j}u$ is a weak solution to

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( \partial _{x}^{j}u\right) _{t}+\frac{\xi ''\left( t\right) }{2}\left( \left( \partial _{x}^{j}u\right) _{xx}+\mu \left[ 0,t\right] \partial _{x}^{j}u_{x}^{2}\right) =0 &{} \left( t,x\right) \in \left( 0,1\right) \times \mathbb {R}\\ \partial _{x}^{j}u\left( 1,x\right) =\frac{d^{j}}{dx^{j}}\log \cosh x &{} x\in \mathbb {R}\end{array}\right. }. \end{aligned}$$

Proposition 8.2

Let $\mu ,\tilde{\mu }\in \Pr [0,1]$ and $u,\tilde{u}$ be the corresponding solutions to the Parisi PDE. Then

$$\begin{aligned} ||u-\tilde{u}||_{\infty } \le \xi ''\left( 1\right) d(\mu ,\tilde{\mu }) \qquad \text { and }\qquad ||u_x-\tilde{u}_x||_{\infty } \le \exp \left( \xi '\left( 1\right) \right) \xi ''\left( 1\right) d(\mu ,\tilde{\mu }). \end{aligned}$$

Proposition 8.3

The solution u to the Parisi PDE satisfies $|u_{x}|<1$ and $0<u_{xx}\le 1$.

1.2 The Auffinger–Chen SDE

Recall the Auffinger–Chen SDE from (1.5),

$$\begin{aligned} dX_{t}&=\xi ''\left( t\right) \mu [0,t]u_x\left( t,X_{t}\right) \, dt +\sqrt{\xi ''\left( t\right) }\, dW_{t}\\ X_{0}&=h \end{aligned}$$

which has infinitesimal generator

$$\begin{aligned} \mathcal {L}_{t,\mu }=\frac{\xi ''(t)}{2}({\varDelta }+ 2\mu [0,t] u_x(t,x)\partial _x ). \end{aligned}$$

Note this has coefficients which are uniformly bounded in time and Lipschitz in space by Sect. “Well-posedness of the Parisi PDE” of the appendix. We now summarize some basic properties of the SDE which will be used in the subsequent. Their proofs are standard applications of Itô’s lemma (see [19]) so they are omitted.

Lemma 8.4

We have

$$\begin{aligned} \mathbb {E}_{h}\left[ u_{x}^{2}\left( s,X_{s}\right) \right]&= \int _{0}^{s}\xi ''\left( t\right) \mathbb {E}_{h}\left[ u_{xx}^{2}\left( t,X_{t}\right) \right] \, dt+u_{x}^{2}\left( 0,h\right) \\ \mathbb {E}_{h}\left[ u_{xx}^{2}\left( s,X_{s}\right) \right]&= \int _{0}^{s}\xi ''\left( t\right) \mathbb {E}_{h}\left[ u_{xxx}^{2}\left( t,X_{t}\right) -2\mu \left[ 0,t\right] u_{xx}^{3}\left( t,X_{t}\right) \right] \, dt\\&\quad +u_{xx}^{2}\left( 0,h\right) \\ \frac{d}{ds}\mathbb {E}_{h}\left[ u_{x}^{2}\left( s,X_{s}\right) \right]&=\xi ''\left( s\right) \mathbb {E}_{h}\left[ u_{xx}^{2}\left( s,X_{s}\right) \right] \\ \frac{d}{ds}^{+}\mathbb {E}_{h}\left[ u_{xx}^{2}\left( s,X_{s}\right) \right]&=\xi ''\left( s\right) \mathbb {E}_{h}\left[ u_{xxx}^{2}\left( s,X_{s}\right) -2\mu \left[ 0,s\right] u_{xx}^{3}\left( s,X_{s}\right) \right] \\ \frac{d}{ds}^{-}\mathbb {E}_{h}\left[ u_{xx}^{2}\left( s,X_{s}\right) \right]&=\xi ''\left( s\right) \mathbb {E}_{h}\left[ u_{xxx}^{2}\left( s,X_{s}\right) -2\mu [0,s)u_{xx}^{3}\left( s,X_{s}\right) \right] . \end{aligned}$$

1.3 A change of measure formula

Lemma 8.5

Fix a measurable space $(\Omega ,\mathcal {F})$. Let Q be a probability measure such that $X_t$ solves (1.5). Then, there is a unique probability measure P with

$$\begin{aligned} R(t)= \frac{dQ}{dP}= \exp \left[ \int _0^{t} \mu [0,s]\, du\left( s,X_s\right) \right] . \end{aligned}$$

Moreover, $X_t$ is distributed like $Y_t$ with respect to P, where $Y_t$ solves $dY_t = \sqrt{\xi ''(t)}dW_t$ and $W_t$ is a standard Brownian motion with respect to P.

Proof

We apply Girsanov’s theorem (see [19, Lemma 6.4.1]) directly. In particular, in the notation of the reference, if let

$$\begin{aligned} c(t) = \mu [0,t]u_x(t,Y_t), \end{aligned}$$

$a(t)=\xi ''(t)$, and $b=0$, we see that the Cameron–Martin–Girsanov exponential is of the form

$$\begin{aligned} R(t)=\exp \left[ \int _0^{t}\mu [0,t]u_x(t,Y_t)dY_t-\frac{1}{2}\int _0^{t} \xi ''(t)\mu [0,t]^2u_x(t,Y_t)^2dt\right] . \end{aligned}$$

Since u solves the Parisi PDE, we see that its Itô differential with respect to $dY_t$ is

$$\begin{aligned} du(t,Y_t)=-\frac{\xi ''}{2}\mu u_x^2 dt + u_x dY_t. \end{aligned}$$

The result then follows by rearrangement. $\square$

Lemma 8.6

We get the integration by parts formula:

$$\begin{aligned} \int _{0}^{t}\nu [0,s]du(t,Y_{t})=\int _{0}^{t}u(t,Y_{t})-u(s,Y_{s})d\nu (s) \end{aligned}$$

for $\nu$ a probability measure on [0, 1].

Corollary 8.7

We have

$$\begin{aligned} R(t)=e^{\int _{0}^{t}u(t,Y_{t})-u(q,Y_{q})d\mu (q)}. \end{aligned}$$

In particular if $\mu =\delta _q$, we have

$$\begin{aligned} R(t)=\frac{\text {sech}(Y_q)}{\text {sech}(Y_t)}e^{- \frac{1}{2}(\xi '(t)-\xi '(q))},\quad t\ge q. \end{aligned}$$

For the reader more familiar with the work of [24, 25], we would like to demonstrate that this Girsanov argument also allows one to translate between that work, the work of Auffinger and Chen in [3, 4], and the authors in [12]. For example, in the notation of [25, Chap. 13] the function g defined in (4.1), $g'$, and the main family of integrals studied in Sect. 7 can also be written as follows. Let $z,z'$ be standard Gaussians, let $Y=h+\xi '(q_*)^{1/2}z$, $Y'=Y+(\xi (t)-\xi (q_*))^{1/2}z'$ and let $\mathbb {E}$ and $\mathbb {E}'$ denote the expectations with respect to z and $z'$ respectively. Then

$$\begin{aligned} \begin{aligned} g(y)&= \mathbb {E}\frac{\mathbb {E}'\left( \tanh ^2(Y')\cosh (Y'))\right) }{\mathbb {E}_{z'}\cosh \left( Y'\right) }\\ g'(y)&= \xi ''(t)\mathbb {E}\frac{\mathbb {E}'\text {sech}^4(Y')\cosh (Y')}{\mathbb {E}'\cosh (Y')}\\ \mathbb {E}4\text {sech}^4(X_t)-6\text {sech}^6(X_t)&=4\mathbb {E}\frac{\mathbb {E}'\text {sech}^3(Y')}{\mathbb {E}'\cosh (Y')}-6\mathbb {E}\frac{\mathbb {E}'\text {sech}^5(Y')}{\mathbb {E}'\cosh (Y')} \end{aligned} \end{aligned}$$

(8.1)

In particular, the reader will observe that, judiciously applied, this Girsanov argument can be seen to relate the representation for these functions obtained through the dynamic programming principle and the Cole–Hopf formula.

1.4 Asymptotic spectral theory for certain operators

In this section we prove Lemma 7.3. We begin with some preliminary estimates. Then the lemma is proved at the end of this section. The definitions from Sect. 7.1 will be used throughout this section.

We observe the following fact from calculus that will be used repeatedly in the subsequent.

Fact 3

We have that

$$\begin{aligned} 0\le \sqrt{x+1}-1\le \frac{1}{2}x,\quad x\ge 0. \end{aligned}$$

We will also use the following bound frequently: for $(\beta ,h,\tau )\in T_{\tilde{\alpha },\tau }$,

$$\begin{aligned} \bigg |{\frac{a-1}{2}}\bigg |&=\bigg |{\frac{\sigma (t)-\sigma (q_{*})}{2\sigma (q_{*})}\bigg |} =\frac{\tau }{2}\bigg |{\frac{\sigma (1)-\sigma (q_{*})}{\sigma (q_{*})}\bigg |}\nonumber \\&\le \frac{\tau }{2}\frac{\sigma '(1)}{\sigma (q_*)\sigma '(q_*)}\sigma '(q_{*})|1-q_{*}| \le \frac{1}{\beta ^{2}}C_{1}(\tilde{\alpha },\beta ,q_*;\tau ). \end{aligned}$$

(8.2)

The notation $||\cdot ||$ will refer to the $\ell _2$-norm throughout.

1.4.1 Estimates on the eigenvectors

Our goal will be to show

Lemma 8.8

For $(\beta ,h,\tau )\in T_{\tilde{\alpha },\tau }$, we have that

$$\begin{aligned} |\left\langle w_1,v_{2}\right\rangle |=|\left\langle w_2,v_{1}\right\rangle |\le \frac{C_{1}(\tilde{\alpha },\beta ,q_*;\tau )}{\sqrt{2}\beta ^{2}}\left( 1+\frac{C_{1}(\tilde{\alpha },\beta ,q_*;\tau )}{2\beta ^{2}}\right) . \end{aligned}$$

Proof

Since $||\tilde{v}_{1}||\ge 1$,

$$\begin{aligned} |\left\langle w_2,v_{1}\right\rangle |&\le \frac{1}{\sqrt{2}}|\left\langle \tilde{w_2},\tilde{v}_{1}\right\rangle | =\frac{1}{\sqrt{2}}|1+\frac{1}{2}\left( -\left( a-1\right) -\sqrt{\left( a-1\right) ^{2}+4}\right) |\\&=\frac{1}{\sqrt{2}}|(1-\sqrt{1+\frac{\left( a-1\right) ^{2}}{4}})-\frac{a-1}{2}|\\&\le \frac{1}{\sqrt{2}}\left( |\frac{a-1}{2}|+|1-\sqrt{1+\frac{\left( a-1\right) ^{2}}{4}}|\right) \!\le \!\frac{1}{\sqrt{2}}\left( |\frac{a-1}{2}|+\frac{1}{2}|\frac{a-1}{2}|^{2}\right) \\&= \frac{1}{\sqrt{2}}|\frac{a-1}{2}|\left( 1+\frac{1}{2}|\frac{a-1}{2}|\right) . \end{aligned}$$

By (8.2),

$$\begin{aligned} |\left\langle w_2,v_{1}\right\rangle | \le \frac{1}{\sqrt{2}}\frac{1}{\beta ^{2}}C_{1}\left( 1+\frac{1}{2\beta ^{2}}C_{1}\right) . \end{aligned}$$

$\square$

1.4.2 Estimates on the eigenvalues

We will prove

Lemma 8.9

For $(\beta ,h,\tau )\in T_{\tilde{\alpha },\tau }$, we have that

$$\begin{aligned} |\lambda _{1}-\nu |\le \frac{1}{\beta }C_2(\tilde{\alpha },\beta ,q_*;\tau )\qquad \text {and}\qquad |\frac{1}{\lambda _{2}^{1/2}}|&\le \frac{1}{\beta \sqrt{2\sigma (q_{*})}}. \end{aligned}$$

Proof

The second estimate follows from the fact that since $a\ge 1$, so that

$$\begin{aligned} \tilde{\lambda }_{2}&=\frac{1}{2}\left( 2+\left( a-1\right) +\sqrt{\left( a-1\right) ^{2}+4}\right) \ge \frac{1}{2}\left( 2+\sqrt{4}\right) =2. \end{aligned}$$

Now we prove the first estimate. By the triangle inequality,

$$\begin{aligned} |\lambda _{1}-\nu |&\le \beta ^{2}\sigma (q_{*})|\tilde{\lambda }_{1}-\frac{a-1}{2}|+|\beta ^{2}\sigma (q_{*})\frac{a-1}{2}-\frac{3}{4}\tilde{\alpha }\tau |=(i)+(ii). \end{aligned}$$

Since

$$\begin{aligned} \tilde{\lambda }_{1}=\frac{1}{2}\left( 2+\left( a-1\right) -\sqrt{\left( a-1\right) ^{2}+4}\right) =\frac{\left( a-1\right) }{2}+\left( 1-\sqrt{\frac{\left( a-1\right) ^{2}}{4}+1}\right) , \end{aligned}$$

we see that by Fact 3,

$$\begin{aligned} |\tilde{\lambda _1} - \frac{a-1}{2} | = |1-\sqrt{1+\frac{(a-1)^2}{4}}| \le \frac{1}{2}|\frac{a-1}{2}|^2. \end{aligned}$$

Hence by (8.2) and the fact that $\sigma$ is non-decreasing,

$$\begin{aligned} (i) \le \beta ^{2}\sigma (q_{*})\frac{1}{2}|\frac{a-1}{2}|^2 \le \frac{\sigma (1)}{2\beta ^{2}}C_{1}(\tilde{\alpha },\beta ,q_*;\tau )^2. \end{aligned}$$

Now to study (ii). For some $c(q_*)\in (q_*,1)$ we have that

$$\begin{aligned} (ii)&=\frac{\tau }{2}|\beta ^{2}(\sigma (1)-\sigma (q_{*}))-\frac{3}{2}\tilde{\alpha }| =\frac{\tau }{2}|\beta ^{2}\sigma '(c(q_{*}))(1-q_{*})-\frac{3}{2}\tilde{\alpha }|\\&\le \frac{\tau }{2}|\sigma '(c(q_{*}))-\sigma '(q_{*})|\beta ^{2}(1-q_{*})+\frac{\tau }{2}|\xi ''(q_{*})\left( 1-q_{*}\right) -\frac{3}{2}\tilde{\alpha }| \! \, = \,\!(iii)+(iv). \end{aligned}$$

We already know a bound on (iv) by Corollary 6.5, so it remains to bound (iii). Since $\sigma ''$ is non-decreasing,

$$\begin{aligned} |\sigma '(c(q_{*}))-\sigma '(q_{*})|\le \sigma ''(1)(1-q_{*})\le \frac{\sigma ''(1)}{\sigma '(q_*)}\frac{C_{0}(\tilde{\alpha },\beta ,q_*)}{\beta ^{2}}, \end{aligned}$$

where we have used Corollary 6.5 in the last inequality. Thus,

$$\begin{aligned} (i)+(ii)\le \frac{1}{2\beta ^{2}}\sigma (1)C_{1}^2 + \frac{1}{\beta }\frac{\tau }{2}{\varLambda }_0\frac{\sigma '(1)}{(\sigma (q_*))^{3/2}}+\frac{1}{\beta ^{2}}\frac{\tau }{2}\frac{\sigma ''(1)}{\sigma '(q_*)}C_{0} = \frac{C_2}{\beta }. \end{aligned}$$

$\square$

Lemma 8.10

For all $(\beta ,h,\tau )\in T_{\tilde{\alpha },\tau }$, we have that

$$\begin{aligned} |\tilde{\lambda }_{2}-2|\le \frac{C_1(\tilde{\alpha },\beta ,q_*;\tau )}{\beta ^2} \left( 1+\frac{C_1(\tilde{\alpha },\beta ,q_*;\tau )}{2\beta ^2}\right) . \end{aligned}$$

Proof

By Fact 3 and (8.2),

$$\begin{aligned} |\tilde{\lambda }_{2}-2|&= \bigg |{1+\left( \frac{a-1}{2}\right) \!+\!\sqrt{1+\left( \frac{a-1}{2}\right) ^{2}}-2}\bigg |\!\le \! \bigg |{\frac{a-1}{2}} \bigg |\!+\! \bigg |{\sqrt{1+\left( \frac{a-1}{2}\right) ^2} \!-\!1\bigg |}\\&\le \bigg |{\frac{a-1}{2}}\bigg |\left( 1 + \frac{1}{2}\left( \frac{a-1}{2}\right) \right) \le \frac{C_1}{\beta ^2} \left( 1+\frac{C_1}{2\beta ^2}\right) . \end{aligned}$$

$\square$

1.4.3 Proof of Lemma 7.3

Proof of Lemma 7.3

Assembling the estimates in Lemmas 8.8–8.10, we have for all $(\beta ,h,\tau )\in T_{\tilde{\alpha },\tau }$ that

$$\begin{aligned} |\left\langle w_1,v_{2}\right\rangle |&=|\left\langle w_2,v_{1}\right\rangle |\le \frac{C_{1}(\tilde{\alpha },\beta ,q_*;\tau )}{\sqrt{2}\beta ^{2}}\left( 1+\frac{C_{1}(\tilde{\alpha },\beta ,q_*;\tau )}{2\beta ^{2}}\right) \\ |\lambda _{1}-\nu |&\le \frac{1}{\beta }C_2(\tilde{\alpha },\beta ,q_*;\tau )\\ |\frac{1}{\lambda _{2}^{1/2}}|&\le \frac{1}{\beta \sqrt{2\sigma (q_{*})}}\\ |\tilde{\lambda }_{2}-2|&\le \frac{C_1(\tilde{\alpha },\beta ,q_*;\tau )}{\beta ^2} \left( 1+\frac{C_1(\tilde{\alpha },\beta ,q_*;\tau )}{2\beta ^2}\right) . \end{aligned}$$

Note that each $C_i(a,b,\tilde{q};\theta )$ is non-decreasing in a and $\theta$ and non-increasing in b and $\tilde{q}$.

By definition of $q_0$ from (7.1), for $i=1,2$ we have that

$$\begin{aligned} C_i(\tilde{\alpha },\beta ,q_*;\tau ) \le C_i(1,\beta _0,q_0(\xi _0,\beta _0,h_0); 1). \end{aligned}$$

This implies the result. $\square$

1.5 Proof of Fact 2

Lemma 8.11

We have that

$$\begin{aligned} \left\langle {\varPsi }\right\rangle \left( x\right) =\int \left( 4\text {sech}^{3}\left( \frac{x+y}{\sqrt{2}}\right) -6\text {sech}^{5}\left( \frac{x+y}{\sqrt{2}}\right) \right) \text {sech}\left( \frac{y-x}{\sqrt{2}}\right) \, dy<0 \end{aligned}$$

for all $x\in \mathbb {R}$.

Proof

A change of variables shows it is enough to prove that

$$\begin{aligned} g\left( x\right) =\frac{1}{\sqrt{2}}\left\langle {\varPsi }\right\rangle \left( \sqrt{2}x\right) =\int \left( 4\text {sech}^{3}y-6\text {sech}^{5}y\right) \text {sech}\left( y-2x\right) \, dy<0 \end{aligned}$$

for all $x\in \mathbb {R}$. Note that

$$\begin{aligned} \text {sech}\left( y-2x\right) =\frac{1}{\cosh \left( -2x\right) \cosh \left( y\right) +\sinh \left( -2x\right) \sinh \left( y\right) }=\frac{1}{a\cosh y+b\sinh y} \end{aligned}$$

where $a\left( x\right) =\cosh \left( -2x\right)$, $b\left( x\right) =\sinh \left( -2x\right)$. Thus

$$\begin{aligned} g\left( x\right)&=\int _{-\infty }^{\infty }\frac{4\text {sech}^{3}y-6\text {sech}^{5}y}{a\cosh y+b\sinh y}\, dy =\int _{-\infty }^{\infty }\frac{\left( 4\text {sech}^{2}y-6\text {sech}^{4}y\right) \text {sech}^{2}y}{a+b\tanh y}\, dy\\&=-\frac{2}{\sinh ^{5}\left( 2x\right) }\left( -3\sinh \left( 4x\right) +4x\cosh \left( 4x\right) +8x\right) . \end{aligned}$$

Thus to show negativity of g it suffices to show positivity of

$$\begin{aligned} R\left( x\right) =\frac{-3\sinh \left( 4x\right) +4x\cosh \left( 4x\right) +8x}{\sinh ^{5}\left( 2x\right) } \end{aligned}$$

for all x. Note the denominator is negative for negative x and positive for positive x, so it suffices to show that the same is true for

$$\begin{aligned} N\left( x\right) =-3\sinh \left( 4x\right) +4x\cosh \left( 4x\right) +8x, \end{aligned}$$

and to check that $R\left( 0\right) >0$. Since $\frac{d^{j}}{dx^{j}}N\left( 0\right) =0$ for $j=0,\dots ,4$ and

$$\begin{aligned} \frac{d^{5}}{dx^{5}}N\left( x\right) =2048\cosh \left( 4x\right) \left( 1+2x\tanh \left( 4x\right) \right) >0 \end{aligned}$$

the result follows. $\square$

1.6 The elementary argument for RS

In this section we prove the proposition:

Proposition 8.12

For all models $\xi _0$ and $\beta _0>0$, there is an $h_0\left( \beta _0,\xi _0\right)$ such that

$$\begin{aligned}{}[0,\beta _0]\times [h_0,\infty ) \subset RS. \end{aligned}$$

Recall the definitions of $q_*$ and $\alpha$ from (1.8). We will need the following preliminary result.

Lemma 8.13

Fix a model $\xi _0$ and $\beta _0>0$. Then for all $\epsilon >0$, there is an $h_{0}\left( \epsilon ,\beta _0,\xi _0\right)$ such that

$$\begin{aligned} \alpha \le \epsilon \end{aligned}$$

for all $\beta \le \beta _{0}$ and $h\ge h_{0}$.

Proof

Note that since $\text {sech}^{4}\le 1$ and since $\xi '_0$ and $\xi ''_0$ are non-decreasing, we have that

$$\begin{aligned}&\beta ^{2}\xi _0''(q_*)\mathbb {E}\text {sech}^{4}\left( \beta \sqrt{\xi _0'(q_{*})}Z+h\right) \\&\quad \le \beta _0^{2}\xi _0''(1)\left( P\left( |Z|\ge \frac{\delta }{\beta \sqrt{\xi '_0(q_{*})}}\right) +\text {sech}^{4}(h-\delta )\right) \\&\quad \le \beta _0^{2}\xi _0''(1)\left( 2e^{-\frac{\delta ^{2}}{2\beta _{0}^{2}\xi _0'(1)}}+\text {sech}^{4}(h-\delta )\right) \end{aligned}$$

for $0<\delta <h$. Taking $\delta =h/2$ and $h\rightarrow \infty$ proves the result.

Proof of Proposition 8.12

By Lemma 4.3 we may assume that $\beta \ge \beta _{*}=\frac{1}{\sqrt{\xi ''_0(1)}}$. Then as in Lemma 6.2, we observe that

$$\begin{aligned} q_*&\ge q_0\\ 1-q_*&\le \frac{\sqrt{\alpha }}{\beta _{*}\sqrt{\xi _0''(q_0)}} \end{aligned}$$

where $q_0(h) = \frac{1}{2}\tanh ^2(h)$. Recall by Lemma 4.2, that if $(\beta ,h)\in AT$ it suffices to prove that $g'\le 0$ on $[q_*,1]$ to conclude RS. We observe as in the proof of Theorem 1.9 that

$$\begin{aligned} g'(y)\le \frac{\xi ''_0(1)}{\xi ''(q_*)}(\alpha -1)+\frac{\xi '''_0(1)}{\xi ''_0(q_*)}(1-q_*)+ C\beta _0^4\left( \xi ''_0(1)\right) ^2(1-q_*). \end{aligned}$$

Using Lemma 8.13 and the estimates on $q_*$ given above, we may take $h\rightarrow \infty$ to conclude the result. $\square$

1.7 Is the AT line a line?

In this section we briefly discuss some questions regarding the nature of the quantities and sets defined in (1.8).

The first question along these lines is as follows. Fix a model $\xi _0$.

Question 8.14

For what region in the plane $(\beta ,h)$ is $Q_*$ a singleton?

This question, it turns out, is very difficult to answer. For the SK model, this question has been resolved by Guerra and Latała [10, 25], where they (separately) showed uniqueness everywhere except for the set $h=0,\beta \ge 1$. For models other than SK, it is far more complicated. For example, numerical studies show that in general the solution to this fixed point equation is not unique. These studies suggest that when h is large or when $\beta \ge h-\delta$, the solution is unique. As the reader will see, the condition $\alpha \le 1+\epsilon$ comes up in the analysis of related questions, so one is led to ask if this is exactly the region in which the unicity fails.

Another natural question is regarding the set $\alpha =1$.

Question 8.15

Is the AT line actually a line? That is, is the AT line a (topological or smooth) curve?

This is also a delicate question. A step toward studying this question is the following lemma.

Lemma 8.16

For any model and point $(\beta ,h)$ with $\beta ,h>0$ and $\alpha \le 1$, the map $(\beta ,h)\mapsto (q_*,\alpha )$ is $C^1$.

We note here that the condition $h>0$ is in general necessary as $\alpha$ should be zero on the set $h=0$ when $\xi ''(0)=0$.

Proof

For ease of notation let $f=\xi '_0$. Consider, as usual, the map $F:\mathbb {R}^2\times \mathbb {R}^2 \rightarrow \mathbb {R}^2$, defined by

$$\begin{aligned} F(\beta ,h;q,\alpha )=(\mathbb {E}\tanh ^2(\beta \sqrt{f(q)}z+h)-q,f'(q)\mathbb {E}\text {sech}^4(\beta \sqrt{f(q)}z+h)-\alpha ). \end{aligned}$$

Note that this is $C^1$. We see that the differential in $(q,\alpha )$ is lower triangular

$$\begin{aligned} \left( \begin{matrix} \partial _q F_1&{}\partial _\alpha F_1\\ \partial _q F_2 &{}\partial _\alpha F_2\\ \end{matrix}\right) = \left( \begin{matrix} a &{} 0 \\ b &{} -1 \\ \end{matrix}\right) \end{aligned}$$

so that it suffices to show that $\partial _q F_1$ is non-zero. To see this, note that

$$\begin{aligned} \frac{\partial }{\partial q}\mathbb {E}\tanh ^{2}(\beta \sqrt{f(q)}z+h)-q&=\mathbb {E}2\tanh \text {sech}^{2}(\beta \sqrt{f(q)}z+h)\beta \frac{f'(q)}{2\sqrt{f(q)}}z-1\\&=\beta ^{2}f'(q)\mathbb {E}\left[ 2-\cosh \left( 2X\right) \right] \text {sech}^{4}(X)-1\\&=\alpha -1+\beta ^{2}f'(q)\mathbb {E}\text {sech}^{4}(X)\left( 1-\cosh (2X)\right) <0 \end{aligned}$$

provided $\alpha \le 1+\delta$ for $\delta$ sufficiently small. The second line follows from an integration by parts. Thus by the implicit function theorem the map from $(\beta ,h)\mapsto (q_*,\alpha )$ is $C^1$. $\square$

This does not show that the set $\alpha =1$ is a curve, however it does show that for almost every $\alpha \in [0,1]$, the set $\alpha (\beta ,h)=\alpha$ is a curve. In particular, it shows this for every $\alpha$ that is regular in the sense of Sard. To get the result precisely when $\alpha =1$ is a difficult calculus question.