A Remark on the Spherical Bipartite Spin Glass

Abstract

Auffinger and Chen (J Stat Phys 157:40–59, 2014) proved a variational formula for the free energy of the spherical bipartite spin glass in terms of a global minimum over the overlaps. We show that a different optimisation procedure leads to a saddle point, similar to the one achieved for models on the vertices of the hypercube.

1 Introduction

Let $\sigma _{N}(dx)$ denote the uniform probability measure on$S^N:=\{x\in \mathbb {R}^N\,:\,\Vert x\Vert ^2_2=N\}$, where $\Vert x\Vert _2$ is the Euclidean norm. For $x:=(x_1,\ldots x_{N_1})\in \mathbb {R}^{N_{1}}$ and $y:=(y_1,\ldots ,y_{N_2})\in \mathbb {R}^{N_{2}}$ the bipartite spin glass is defined by the energy function

$$\begin{aligned} H_{N_1,N_2}(x,y;\xi ):=-\frac{1}{\sqrt{N}}\sum _{j=1}^{N_2}\sum _{i=1}^{N_1} \xi _{ij}x_iy_j\,. \end{aligned}$$

(1)

Here $\{\xi _{ij}\}_{i\in [N_1],j\in [N_{2}]}$ are ${\mathcal {N}}(0,1)$ i.i.d. quenched r.vs. and we set $N:=N_1+N_2$. The object of interest of this note is the free energy

$$\begin{aligned} A_{N_1,N_2}(\beta ,\xi ):=\frac{1}{N}\log \int \sigma _{N_1}(dx)\sigma _{N_2}(dy)\exp (-\beta H_{N_1,N_2}(x,y;\xi )-b_1(x,1)-b_2(y,1))\, \end{aligned}$$

(2)

in the limit in which $N_1,N_2\rightarrow \infty$ with $N_1/N\rightarrow \alpha \in (0,1)$. Here $\beta \;\geqslant \;0$ is the inverse temperature, $b_1,b_2\in \mathbb {R}$ are external fields and $(\cdot ,\cdot )$ denotes the Euclidean inner product. By concentration of Lipschitz functions of Gaussian random variables one reduces to study the average free energy $A_{N_1,N_2}(\beta ):=E[A_{N_1,N_2}(\beta ,\xi )]$, whose limit we denote by $A(\alpha ,\beta )$.

Auffinger and Chen proved in [1] the following variational formula for $A(\alpha ,\beta )$ for $\beta$ small enough

$$\begin{aligned} A(\alpha ,\beta )= & {} \min _{q_1,q_2\in [0,1]^2} P(q_1,q_2) \end{aligned}$$

(3)

$$\begin{aligned} P(q_1,q_2)= & {} \frac{\beta ^2\alpha (1-\alpha )}{2}(1-q_1q_2)+\frac{\alpha }{2}\left( b_1^2(1-q_1)+\frac{q_1}{1-q_1}+\log (1-q_1)\right) \nonumber \\&\quad +\frac{1-\alpha }{2}\left( b_2^2(1-q_2)+\frac{q_2}{1-q_2}+\log (1-q_2)\right) \, \end{aligned}$$

(4)

(the normalisation in (1) leads to different constants w.r.t. [1]). The above formula was successively proved to hold in the whole range of $\beta \;\geqslant \;0$ in [2, 9]. Yet these proofs are indirect, as in both cases one obtains a formula for the free energy and then verifies a posteriori (analytically for [2] and numerically [14] for [9]) that it coincides with (3). We just mention that the results in [1] have been recently extended in [10, 11] for the complexity and in [5, 6] for the free energy.

The convex variational principle found by Auffinger and Chen appears to be in contrast with the $\min \max$ characterisation given in [4, 7] for models on the vertices of the hypercube (see also [3] for the Hopfield model). The aim of this note is to show that the Auffinger and Chen formula can be equivalently expressed in terms of a $\min \max$.

One disadvantage of the spherical prior is that the associated moment generating function

$$\begin{aligned} \Gamma _N(h):=\frac{1}{N}\log \int \sigma _N(dx)e^{(h,x)}\,,\quad h\in \mathbb {R}^N\,, \end{aligned}$$

(5)

is not easy to compute. If h is random with i.i.d. ${\mathcal {N}}(b,q)$ components it is convenient to set

$$\begin{aligned} \Gamma (b,q):=\lim _NE \Gamma _N(h)\,. \end{aligned}$$

(6)

The so-called Crisanti–Sommers variational characterisation of it as $N\rightarrow \infty$ reads as follows.

Lemma 1

Let $b\in \mathbb {R}$, $q>0$, $h\in \mathbb {R}^N$ with i.i.d ${\mathcal {N}}(b,\sqrt{q})$ components. Then

$$\begin{aligned} \Gamma (b,q)= \frac{1}{2}\min _{r\in [0,1)}\left( (b^2+q)(1-r)+\frac{r}{1-r}+\log (1-r)\right) \, \end{aligned}$$

(7)

At the end of this note we give a simple proof of this statement, based on the method of [8, 9]. We first get a variational characterisation of the moment generating function of a Gaussian distribution (whose variance is Legendre conjugate to q) and then use concentration of measure.

A direct computation shows that the minimum of (7) is attained for

$$\begin{aligned} \frac{r}{(1-r)^2}=q+b^2\,. \end{aligned}$$

(8)

A standard replica symmetric interpolation gives that for any $q_1,q_2\in [0,1]$

$$\begin{aligned} A _{N_1,N_2}(\beta )\!\!\!&\,=&\,\!\!\!\frac{\beta ^2\alpha (1-\alpha )}{2}(1-q_1)(1-q_2)+(1-\alpha )\Gamma (b_2,\beta ^2\alpha q_1)\nonumber \\&\quad +\alpha \Gamma (b_1,\beta ^2(1-\alpha ) q_2)+{\text {Error}}_N(q_1,q_2)\,. \end{aligned}$$

(9)

The last summand is an error term whose specific form is not important here. What matters is that by [1, Lemma 1] there is a choice of $(q_1,q_2)$ (see below) for which this remainder goes to zero as $N\rightarrow \infty$ if $\beta$ is small enough. Combining (7) and (8) we can rewrite the first line of (9) as

$$\begin{aligned} {{\,\mathrm{RS}\,}}(q_1,q_2):= & {} \frac{\beta ^2\alpha (1-\alpha )}{2}(1-q_1)(1-q_2)\nonumber \\&\quad +\frac{\beta ^2\alpha (1-\alpha )}{2}\left( \left( q_2+\frac{b_1^2}{\beta ^2(1-\alpha )}\right) (1-r_1) +\left( q_1+\frac{b_2^2}{\beta ^2\alpha }\right) (1-r_2)\right) \nonumber \\&\quad +\frac{\alpha }{2}\frac{r_1}{1-r_1}+\frac{\alpha }{2}\log (1-r_1)+\frac{1-\alpha }{2}\frac{r_2}{1-r_2} +\frac{1-\alpha }{2}\log (1-r_2)\,, \end{aligned}$$

(10)

under the condition

$$\begin{aligned} \frac{r_1}{(1-r_1)^2}=\beta ^2(1-\alpha ) q_2+b_1^2\,,\quad \frac{r_2}{(1-r_2)^2}=\beta ^2\alpha q_1+b_2^2\,. \end{aligned}$$

(11)

Here we used that there is a sequence $o_N\rightarrow 0$ uniformly in $q_1,q_2,\beta ,\alpha$ such that

$$\begin{aligned} \frac{\beta ^2\alpha (1-\alpha )}{2}(1-q_1)(1-q_2)+(1-\alpha )\Gamma _N(\beta ^2\alpha q_1)+\alpha \Gamma _N(\beta ^2(1-\alpha ) q_2)={{\,\mathrm{RS}\,}}(q_1,q_2)+o_N\,. \end{aligned}$$

(12)

Indeed (12) follows easily once we use Lemma 1 for the limit of the functions $\Gamma _N$ and we note that (11) are the critical point equations related to the minimisation of (7).

The main observation of this note is that (10) under (11) is optimised as a $\min \max$.

Proposition 1

Assume $b_1^2+b_2^2>0$. The function ${{\,\mathrm{RS}\,}}(q_1,q_2)$ has a unique stationary point $(\bar{q}_1, \bar{q}_2)$. It solves

$$\begin{aligned} \frac{q_2}{(1-q_2)^2}=\beta ^2\alpha q_1+b_1^2\,,\qquad \frac{q_1}{(1-q_1)^2}=\beta ^2(1-\alpha ) q_2+b_2^2\,. \end{aligned}$$

(13)

Moreover

$$\begin{aligned} {{\,\mathrm{RS}\,}}(\bar{q}_1,\bar{q}_2)=\min _{q_2\in [0,1]}\max _{q_1\in [0,1]} {{\,\mathrm{RS}\,}}(q_1,q_2)\,. \end{aligned}$$

(14)

If $b_1=b_2=0$ and

$$\begin{aligned} \beta ^4\alpha (1-\alpha )<1\, \end{aligned}$$

(15)

the origin is the unique solution of (13) and

$$\begin{aligned} {{\,\mathrm{RS}\,}}(0,0)=\min _{q_2\in [0,1]}\max _{q_1\in [0,1]} {{\,\mathrm{RS}\,}}(q_1,q_2)\,. \end{aligned}$$

(16)

If $b_1=b_2=0$ and

$$\begin{aligned} \beta ^4\alpha (1-\alpha )>1\, \end{aligned}$$

(17)

there is a unique $(\bar{q}_1, \bar{q}_2)\ne (0,0)$ which solves (13) and such that (14) holds. Moreover

$$\begin{aligned} {{\,\mathrm{RS}\,}}(0,0)=\max _{q_2\in [0,1]}\max _{q_1\in [0,1]} {{\,\mathrm{RS}\,}}(q_1,q_2)\,. \end{aligned}$$

(18)

The crucial point of [1, Lemma 1] (for us) is that from the Latala argument [13, Sect. 1.4] it follows that the overlaps self-average as $N\rightarrow \infty$ at a point $(\tilde{q}_1, \tilde{q}_2)$ uniquely given by

$$\begin{aligned} \frac{\tilde{q}_1}{(1-\tilde{q}_1)^2}=\beta ^2(1-\alpha ) \tilde{q}_2+b_1^2\,,\quad \frac{\tilde{q}_2}{(1-\tilde{q}_2)^2}=\beta ^2\alpha \tilde{q}_1+b_2^2\,, \end{aligned}$$

(19)

which (see [12, Lemma 7]) are indeed asymptotically equivalent to

$$\begin{aligned} q_{1,N}:= & {} \frac{1}{N}E\left[ \frac{\int \sigma _{N_1}(dy)\sigma _{N_1}(dy') (y,y') e^{\beta \sqrt{q_2}(y+y',h)}}{\left( \int \sigma _{N_1}(dy) e^{\beta \sqrt{q_2}(y,h)}\right) ^2}\right] \,,\end{aligned}$$

(20)

$$\begin{aligned} q_{2,N}:= & {} \frac{1}{N} E\left[ \frac{\int \sigma _{N_2}(dx)\sigma _{N_2}(dx') (x,x') e^{\beta \sqrt{q_1}(x+x',h)}}{\left( \int \sigma _{N_2}(dx) e^{\beta \sqrt{q_1}(x,h)}\right) ^2}\right] \,, \end{aligned}$$

(21)

naturally arising from the replica symmetric interpolation (here h is random with i.i.d. $\mathcal N(0,1)$ entries). Comparing (11) and (19) readily implies that we can plug $(r_1,r_2)=(q_1,q_2)$ into (10) and obtain the convex function $P(q_1,q_2)$ of [1, Theorem 1], optimised by (19).

On the other hand, without using the Latala method one might still optimise (10) as a function of four variables, ignoring (11). Taking derivatives first in $q_1,q_2$, the critical point equations (24), (2425) below select exactly $(q_1,q_2)=(r_1,r_2)$. This procedure is however unjustified a priori and this particular application of Latala’s method legitimises the exchange in the order of the optimisation of the q and the r variables for small $\beta$, which a posteriori can be extended to all $\beta$ [2, 9].

We stress that by itself the Latala method is not variational, it only gives the self-consistent equations for the critical points. It is the Crisanti–Sommers formula (7) which makes it implicitly variational. Such a variational representation is not necessary in other cases of interest, for instance for the bipartite SK model (namely Hamiltonian (1) with $\pm 1$ spins), for which one simply has the $\log \cosh$. Indeed in this case a direct use of the Latala method yields the validity of the $\min \max$ formula of [4] for $\beta$ and $|b_1|,|b_2|$ small enough. The proof is essentially an exercise after [13, Proposition 1.4.8] and [1, Formula (9)] and will not be reproduced here in details. The replica symmetric sum-rule for the free energy (analogue of formula (9)) reads as

$$\begin{aligned} A _{N_1,N_2}(\beta )= & {} \frac{\beta ^2\alpha (1-\alpha )}{2}(1-q_1)(1-q_2)+(1-\alpha )E\log \cosh (b_2+\beta \sqrt{\alpha q_1}g)\nonumber \\&\quad +\,\alpha E\log \cosh (b_1+\beta \sqrt{(1-\alpha ) q_2}g)\nonumber \\&\quad +\,{\text {Error}}_N(q_1,q_2)\,, \end{aligned}$$

(22)

(here $g\sim {\mathcal N}(0,1)$) and the error term can be shown by the Latala method to vanish for small $\beta ,|b_1|, |b_2|$, if $(q_1,q_2)=(\bar{q}_1, \bar{q}_2)$ are given by

$$\begin{aligned} \bar{q}_1=E[\tanh (b_1+\beta \sqrt{(1-\alpha )\bar{q}_2} g)]\,,\qquad \bar{q}_2=E[\tanh (b_2+\beta \sqrt{\alpha \bar{q}_1} g)]\,. \end{aligned}$$

(23)

Therefore the free energy equals the first two lines on the r.h.s. of (22) evaluated in $(q_1,q_2)=(\bar{q}_1, \bar{q}_2)$, which is the value attained at the $\min \max$, as shown in [4, 7].

2 Proofs

Proof of Proposition 1

Assume first $b_1^2+b^2_2>0$. We differentiate (10) and by (11) we get

$$\begin{aligned} \partial _{q_1} {{\,\mathrm{RS}\,}}= & {} \frac{\beta ^2\alpha (1-\alpha )}{2}(q_2-r_2(q_1))\, \end{aligned}$$

(24)

$$\begin{aligned} \partial _{q_2} {{\,\mathrm{RS}\,}}= & {} \frac{\beta ^2\alpha (1-\alpha )}{2}(q_1-r_1(q_2))\,. \end{aligned}$$

(25)

The functions $r_1,r_2$ write explicitly as

$$\begin{aligned} r_1(q_2)= & {} \frac{\sqrt{1+4(\beta ^2(1-\alpha )q_2+b_1^2)}-1}{\sqrt{1+4(\beta ^2(1-\alpha )q_2+b_1^2)}+1} \end{aligned}$$

(26)

$$\begin{aligned} r_2(q_1)= & {} \frac{\sqrt{1+4(\beta ^2\alpha q_1+b_2^2)}-1}{\sqrt{1+4(\beta ^2\alpha q_1+b_2^2)}+1}\,. \end{aligned}$$

(27)

We easily see that $r_1,r_2$ are increasing from $r_1(0),r_2(0)>0$ (obviously computable by the formulas above) to 1 and concave. Moreover we record for later use that if $b_1=b_2=0$ we have

$$\begin{aligned} \frac{d}{dq_2}r_1(q_2)\Big |_{q_2=0}=\beta ^2(1-\alpha )\,,\quad \frac{d}{dq_1}r_2(q_1)\Big |_{q_1=0}=\beta ^2\alpha \,. \end{aligned}$$

(28)

Now we take the derivative w.r.t. $q_1$ and note that the r.h.s. of (24) is decreasing as a function of $q_1$, thus $\partial ^2_{q_1} {{\,\mathrm{RS}\,}}<0$. Therefore by the implicit function theorem there is a unique function $q_1$ such that $q_2=r_2(q_1)$. As a function of $q_2$, $q_1$ is non-negative, increasing and convex and it is $q_1(r_2(0))=0$. We set

$$\begin{aligned} {{\,\mathrm{RS}\,}}_1(q_2):=\max _{q_1}{{\,\mathrm{RS}\,}}(q_1,q_2)={{\,\mathrm{RS}\,}}(q_1(q_2),q_2)\, \end{aligned}$$

(29)

and compute

$$\begin{aligned} \partial _{q_2}{{\,\mathrm{RS}\,}}_1(q_2)=\frac{\beta ^2\alpha (1-\alpha )}{2}\left( q_1(q_2)-r_1(q_2)\right) \,. \end{aligned}$$

(30)

By the properties of the functions $q_1$ and $r_1$ it is clear that there is a unique intersection point $\bar{q}_2$; moreover $q_1\;\leqslant \;r_1$ for $q_2\;\leqslant \;\bar{q}_2$ and otherwise $q_1\;\geqslant \;r_1$. Therefore $\partial _{q_2}{{\,\mathrm{RS}\,}}_1(q_2)$ is increasing in a neighbourhood of $\bar{q}_2$ which allows us to conclude $\partial ^2_{q_2}{{\,\mathrm{RS}\,}}_1>0$. This finishes the proof if $b_1^2+b_2^2>0$.

If $b_1=b_2=0$ the origin is always a stationary point. It is unique if

$$\begin{aligned} \left[ \frac{d}{dq_1}r_2(q_1)\big |_{q_1=0}\right] ^{-1} =\frac{d}{dq_2}q_1(q_2)\big |_{q_2=0}>\frac{d}{dq_2}r_1(q_2)\big |_{q_2=0}\,, \end{aligned}$$

(31)

which, bearing in mind (28), amounts to ask (15).

Since $r_2$ is increasing around the origin, we have $\partial ^2_{q_1}{{\,\mathrm{RS}\,}}<0$ and by the implicit function theorem we define locally a function $q_1(q_2)$ increasing and positive, vanishing at the origin. We set

$$\begin{aligned} {{\,\mathrm{RS}\,}}_1(q_2):=\max _{q_1}{{\,\mathrm{RS}\,}}(q_1,q_2)={{\,\mathrm{RS}\,}}(q_1(q_2),q_2)\, \end{aligned}$$

(32)

and compute

$$\begin{aligned} \partial _{q_2}{{\,\mathrm{RS}\,}}_1(q_2)=\frac{\beta ^2\alpha (1-\alpha )}{2}\left( q_1(q_2)-r_1(q_2)\right) \,. \end{aligned}$$

(33)

By (31) we have $\partial ^2_{q_2}{{\,\mathrm{RS}\,}}_1\big |_{q_2=0}>0\,$, whence we obtain (16).

If (17) holds, then

$$\begin{aligned} \frac{d}{dq_2}q_1(q_2)\big |_{q_2=0}<\frac{d}{dq_2}r_1(q_2)\big |_{q_2=0}\,, \end{aligned}$$

(34)

which proceeding as before leads to (18).

However also in the case $b_1=b_2=0$ we can repeat all the steps done in the case $b_1^2+b_2^2>0$, showing the existence of a point $(\bar{q}_1, \bar{q}_2)$ in which a $\min \max$ of ${{\,\mathrm{RS}\,}}$ is attained. If (31) (i.e. (15)) holds then it must be $(\bar{q}_1, \bar{q}_2)=(0,0)$. If (17) holds, then (34) enforces

$$\begin{aligned} q_1(q_2)-r_1(q_2)\;\leqslant \;0 \end{aligned}$$

in a neighbourhood of the origin (as $q_1(0)=r_1(0)=0$), which implies that the critical point $(\bar{q}_1, \bar{q}_2)$ must fall elsewhere. $\square$

Proof of Lemma 1

We will prove that for all $u\in \sqrt{q}S^N$

$$\begin{aligned} \Gamma ^{(\sigma )} (q):=\lim _N \Gamma _N (u)=\frac{1}{2}\min _{r\in [0,1)}\left( q(1-r)+\frac{r}{1-r}+\log (1-r)\right) \,. \end{aligned}$$

(35)

We show first that (35) implies the assertion. Let h be a random vector with i.i.d. $\mathcal N(0,q)$ entries. (As customary we write $X\simeq Y$ if there are constants $c,C>0$ such that $cY\;\leqslant \;X\;\leqslant \;CY$). The classical estimates

$$\begin{aligned} \Gamma _N(h)\;\leqslant \;\frac{\Vert h\Vert _2}{\sqrt{N}}\,,\quad P\left( \left| \frac{\Vert h\Vert _2}{\sqrt{N}}-\sqrt{q}\right| \;\geqslant \;t\right) \simeq e^{-\frac{t^2N}{2}}\, \end{aligned}$$

(36)

permit us to write for all $t>0$ (small)

$$\begin{aligned} |E[\Gamma _N]-\Gamma ^{(\sigma )} (q)|&\;\leqslant \;|E[\Gamma _N1_{\left\{ \left| \frac{\Vert h\Vert }{\sqrt{N}}-\sqrt{q}\right|< t\right\} }]-\Gamma ^{(\sigma )} (q)|+\left| E\left[ \frac{\Vert h\Vert _2}{\sqrt{N}}1_{\left\{ \left| \frac{\Vert h\Vert }{\sqrt{N}}-\sqrt{q}\right| \;\geqslant \;t\right\} }\right] \right| \nonumber \\&\quad \simeq \left| \Gamma _N(u^*)P\left( \left| \frac{\Vert h\Vert _2}{\sqrt{N}}-\sqrt{q}\right| < t\right) -\Gamma ^{(\sigma )} (q)\right| +o(t)+e^{-t^2N/2}\,\nonumber \\&\quad \simeq \left| \Gamma _N(u^*)-\Gamma ^{(\sigma )} (q)\right| +o(t)+e^{-t^2N/2}\,, \end{aligned}$$

(37)

for some $u^*\in \sqrt{q}S^{N}$ and $o(t)\rightarrow 0$ as $t\rightarrow 0$. Since $t>0$ is arbitrary we obtain

$$\begin{aligned} |E[\Gamma _N]-\Gamma ^{(\sigma )} (q)|\;\leqslant \;\left| \Gamma _N(u^*)-\Gamma ^{(\sigma )} (q)\right| \,. \end{aligned}$$

It remains to show (35). Given $\varepsilon >0$ we introduce the spherical shell

$$\begin{aligned} S^{N,\varepsilon }:=S^N+\sqrt{\frac{ \varepsilon }{N}} S^N \end{aligned}$$

and the measure $\sigma _N^{(\varepsilon )}$ as the uniform probability on it. For any $\theta >0$ we have

$$\begin{aligned} \int \sigma ^{(\varepsilon )}_{N}(dx) e^{(u,x)}&\;\leqslant \;e^{\frac{\theta (N+\varepsilon )}{2}}\int \sigma ^{(\varepsilon )}_{N}(dx) e^{-\frac{\theta }{2}\Vert x\Vert _2^2+(u,x)}\nonumber \\&\;\leqslant \;e^{\frac{\theta (N+\varepsilon )}{2}}\frac{\sqrt{2\pi }^N}{\theta ^{\frac{N}{2}}|S^{N,\varepsilon }|}\int e^{-\frac{\theta }{2}\Vert x\Vert _2^2+(u,x)}\frac{dx}{\sqrt{2\pi }^N}\nonumber \\&\quad =e^{\frac{\theta (N+\varepsilon )}{2}+\frac{q N}{2\theta }}\frac{\sqrt{2\pi }^N}{\theta ^{\frac{N}{2}}|S^{N,\varepsilon }|}\,. \end{aligned}$$

(38)

Therefore for $C>0$ large enough

$$\begin{aligned} \frac{1}{N}\log \int \sigma ^{(\varepsilon )}_{N}(dx) e^{(u,x)}\;\leqslant \;\frac{\theta }{2}+\frac{q}{2\theta }-\frac{1}{2}(\log \theta +1)+C\theta \frac{\varepsilon }{N}\,. \end{aligned}$$

(39)

Since this inequality holds for all $\theta >0$ and $\varepsilon >0$ we have

$$\begin{aligned} \limsup _N \Gamma _N (u)\;\leqslant \;\inf _{\theta >0}\left( \frac{q}{2\theta }+\frac{\theta -1}{2}-\frac{1}{2}\log \theta \right) \,. \end{aligned}$$

(40)

We set for brevity

$$\begin{aligned} \Gamma _1(\theta ):=\frac{q}{2\theta }+\frac{\theta -1}{2}-\frac{1}{2}\log \theta \, \end{aligned}$$

and notice that $\Gamma _1$ is uniformly convex in all the intervals $(0,\theta _0)$ for finite $\theta _0>0$.

For the reverse bound, again we let $\theta >0$ and write

$$\begin{aligned} \int \sigma ^{(\varepsilon )}_{N}(dx) e^{(u,x)}\;\geqslant \;e^{\frac{\theta }{2} N}\int _{\mathbb {R}^N} \frac{dx}{|S^{N,\varepsilon }|} e^{-\frac{\theta }{2}\Vert x\Vert _2^2+(u,x)}-e^{\frac{\theta }{2} N}\int _{(S^{N,\varepsilon })^c} \frac{dx}{|S^{N,\varepsilon }|} e^{-\frac{\theta }{2}\Vert x\Vert _2^2+(u,x)}\,. \end{aligned}$$

(41)

The first summand on the r.h.s. can be written as before

$$\begin{aligned} e^{\frac{\theta }{2} N}\int _{\mathbb {R}^N} \frac{dx}{|S^{N,\varepsilon }|} e^{-\frac{\theta }{2}\Vert x\Vert _2^2+(u,x)}=e^{\frac{\theta N}{2}+\frac{q N}{2\theta }}\frac{\sqrt{2\pi }^N}{\theta ^{\frac{N}{2}}|S^{N,\varepsilon }|}\,. \end{aligned}$$

(42)

For the second summand we introduce $\eta \in (0,\frac{\theta }{2})$ and bound

$$\begin{aligned} e^{\frac{\theta }{2} N}\int _{\Vert x\Vert ^2\;\leqslant \;N-\varepsilon } \frac{dx}{|S^{N,\varepsilon }|} e^{-\frac{\theta }{2}\Vert x\Vert _2^2+(u,x)}&\;\leqslant \;&e^{\frac{\theta }{2} N+(N-\varepsilon )\frac{\eta }{2}+\frac{q N}{2(\theta +\eta )}} \frac{\sqrt{2\pi }^N}{\theta ^{\frac{N}{2}}|S^{N,\varepsilon }|} \end{aligned}$$

(43)

$$\begin{aligned} e^{\frac{\theta }{2} N}\int _{\Vert x\Vert ^2\;\geqslant \;N+\varepsilon } \frac{dx}{|S^{N,\varepsilon }|} e^{-\frac{\theta }{2}\Vert x\Vert _2^2+(u,x)}&\;\leqslant \;&e^{\frac{\theta }{2} N-(N+\varepsilon )\frac{\eta }{2}+\frac{q N}{2(\theta -\eta )}}\frac{\sqrt{2\pi }^N}{\theta ^{\frac{N}{2}}|S^{N,\varepsilon }|} \,. \end{aligned}$$

(44)

Thus

$$\begin{aligned} \liminf _N\frac{1}{N}\log \int \sigma ^{(\varepsilon )}_{N}(dx) e^{(u,x)}\;\geqslant \;\max (\Gamma _1, \Gamma _2,\Gamma _3) \end{aligned}$$

(45)

with

$$\begin{aligned} \Gamma _2(\eta ,\theta ):= & {} \frac{q}{2(\theta -\eta )}+\frac{\eta (1-\frac{\varepsilon }{N})}{2}+\frac{\theta -1}{2}-\frac{1}{2}\log \theta \,,\\ \Gamma _3(\eta ,\theta ):= & {} \frac{q}{2(\theta +\eta )}-\frac{\eta (1+\frac{\varepsilon }{N})}{2}+\frac{\theta -1}{2}-\frac{1}{2}\log \theta . \end{aligned}$$

Now we define

$$\begin{aligned} \Delta _{12}(\eta ,\theta ):=\Gamma _1(\theta )-\Gamma _2(\eta ,\theta )\,,\quad \Delta _{13}(\eta ,\theta ):=\Gamma _1(\theta )-\Gamma _3(\eta ,\theta )\,, \end{aligned}$$

(46)

and we seek $\bar{\theta }>0$ for which $\Delta _{12},\Delta _{13}\;\geqslant \;0$ for sufficiently small $\eta$. Since $\Delta _{12}(0,\theta )=\Delta _{13}(0,\theta )=0$ it suffices to study

$$\begin{aligned} \frac{d}{d\eta } \Delta _{12}\Big |_{\eta =0}\,,\quad \frac{d}{d\eta } \Delta _{13}\Big |_{\eta =0}\,. \end{aligned}$$

(47)

A direct computation shows

$$\begin{aligned} \frac{d}{d\eta } \Delta _{12}\Big |_{\eta =0}= & {} \frac{\varepsilon }{2N}-\partial _\theta \Gamma _1(\theta )\,,\end{aligned}$$

(48)

$$\begin{aligned} \frac{d}{d\eta } \Delta _{13}\Big |_{\eta =0}= & {} \frac{\varepsilon }{2N}+\partial _\theta \Gamma _1(\theta )\,. \end{aligned}$$

(49)

Combining (47), (48) and (49) we see that plugging $\bar{\theta }=\arg \min \Gamma _1$ into (45) we arrive to

$$\begin{aligned} \lim \inf _N \Gamma _N (u)\;\geqslant \;\min _{\theta >0}\left( \frac{q}{2\theta }+\frac{\theta -1}{2}-\frac{1}{2}\log \theta \right) \,. \end{aligned}$$

(50)

Therefore (40) and (50) give

$$\begin{aligned} \lim _N \Gamma _N (u)= \min _{\theta >0}\left( \frac{q}{2\theta }+\frac{\theta -1}{2}-\frac{1}{2}\log \theta \right) \nonumber \\ \end{aligned}$$

and changing variable $\theta =(1-r)^{-1}$ we obtain (35). $\square$