For all $({\beta},h)$, $h>0$, and all $0<\eta<1$, if $h\leq 2\eta{\beta}^{2}q$

$\left(1-{\textstyle{3\over
4(1-\eta^{2})}}\right){\textstyle{{\beta}\sqrt{q}\over h}}\leq
r({\beta},h)\sqrt{{\textstyle{\pi\over
2}}}e^{\frac{1}{2}\left(\frac{h}{{\beta}\sqrt{q}}\right)^{2}}\leq\left(1+{\textstyle{\eta\over
1-\eta^{2}}}\right){\textstyle{2{\beta}\sqrt{q}\over h}},$ (6.5)
${\textstyle{1\over
4}}\sqrt{{\textstyle{1\over[{\beta}\sqrt{q}(1+\eta)]^{2}+1}}}\leq(1-q)\sqrt{{\textstyle{\pi\over
2}}}e^{\frac{1}{2}\left(\frac{h}{{\beta}\sqrt{q}}\right)^{2}}\leq{\textstyle{2\over{\beta}\sqrt{q}(1-\eta^{2})}}.$
(6.6)

The next lemma is given for the sake of completeness and stated without proof.
For $\eta>0$ set

$f({\alpha}_{1},{\alpha}_{2})={\alpha}_{1}+{\alpha}_{2}\sqrt{{\textstyle{2\over\pi}}}{\textstyle{1\over{\beta}\sqrt{q}}}e^{-\frac{1}{2}(2\eta{\beta}\sqrt{q})^{2}},$
(6.7)

######  Lemma 6.3.

For all $({\beta},h)$, $h>0$, and all $\eta>0$, if $h\geq
2(1+\eta){\beta}^{2}q$,

$f(2-3/[4(1+\eta)],1/\eta)\leq
r({\beta},h)e^{2\left(h-{\beta}^{2}q\right)}\leq f(2,3/[4(1+\eta)]),$ (6.8)
$f(1,1/(4\eta))\leq(1-q)e^{2\left(h-{\beta}^{2}q\right)}\leq f(4,2/(1+\eta)).$
(6.9)

######  Remark.

Under the assumptions of Lemma 6.2 and Lemma 6.3, respectively, $r({\beta},h)$
and $1-q$ have a common leading exponential decay. The conditions on
$h/{\beta}$ and $h$ entering the definition of ${\mathcal{D}}^{(3)}$ serve to
control this decay from above. In Lemma 6.3, the pre-factors modulating the
exponential decay are not sharp enough to tell which of $\bar{\varrho}$ or $q$
is larger (this is due to the rough bounds of Lemma 6.13). On the contrary, in
Lemma 6.2, these pre-factors guarantee that $\bar{\varrho}<q$ if
${\beta}^{2}q/h$ is large enough, a fact already clear from (6.1).

######  Remark.

Recall that in the physics literature the magnetic field is the quantity
$h^{\prime}=h/{\beta}$. Under the assumptions of Lemma 6.2, namely if
$h^{\prime}\leq 2\eta{\beta}q$, it follows from (6.6) that when the field
$h^{\prime}$ is large, $q$ is close to one and
${\beta}^{2}(1-q)\sim{\beta}\exp\bigl{(}-\frac{1}{2}h^{\prime 2}\bigr{)}$.
Comparing to formula (23) of [27], we see that up to a constant pre-factor,
${\beta}^{2}(1-q)$ has the same behaviour as
${\beta}^{2}E\left[\cosh^{-4}({\beta}\sqrt{q}Z+{\beta}h^{\prime})\right]$ for
large fields. Thus, under these assumptions on $({\beta},h^{\prime})$, the
condition ${\beta}^{2}(1-q)<1$ is analogous to the AT-condition (1.9). This
sheds light on the domain ${\mathcal{D}}^{(3)}$.

We now turn to the proof of Theorem 6.1. It hinges on two key propositions.
Set

$\begin{split}\psi_{{\beta},h}(\rho)\,=\,&\sqrt{\rho}h\left(1-\sqrt{{\textstyle{\rho\over
q}}}\right)+\frac{{\beta}^{2}}{2}\left(1-q\right)\left(q-\rho\right)\\\
\,+\,&E\left[\log\cosh\left(\sqrt{{\textstyle{\rho\over
q}}}\left({\beta}\sqrt{q}Z+h\right)\right)-\log\cosh({\beta}\sqrt{q}Z+h)\right].\end{split}$
(6.10)

######  Proposition 6.4.

Let $0<\varrho\leq q$ be given.There exists a subset
$\widetilde{\Omega}_{N}\subseteq{\Omega}$ with
${\mathbb{P}}\bigl{(}\widetilde{\Omega}_{N}\bigr{)}\geq 1-e^{-N}$ such that on
$\widetilde{\Omega}_{N}$,

$\sup_{m\in{\mathcal{B}}_{N}(\varrho)}F_{N,{\beta},h}^{HT}(m)\leq
SK({\beta},h)+\sup_{0\leq\rho<\varrho}\psi_{{\beta},h}(\rho)+{\mathcal{O}}\Bigl{(}\sqrt{{\textstyle{\log
N\over N}}}\Bigr{)}.$ (6.11)

(See (6.55) for the precise form of the error term.)

######  Proposition 6.5.

For all $({\beta},h)$ in ${\mathcal{D}}^{(3)}$ and for $\bar{\varrho}$ defined
in (6.2), $\sup_{0\leq\rho<\bar{\varrho}}\psi_{{\beta},h}(\rho)<0$.

###### Proof of Theorem 6.1.

Given Proposition 6.5 and taking $\rho=\bar{\varrho}$ in Proposition 6.4, the
theorem follows from Borel-Cantelli lemma. ∎

In the rest of this section, we first prove Proposition 6.4 and Proposition
6.5, with Lemma 6.2 being proved at the very end, as well as Proposition 1.10,
and Theorems 1.4 and 1.8 from Section 1.

###### Proof of Proposition 6.4.

By (2.5) and (2.2), we can write

$F_{N,{\beta},h}^{HT}(m)=f(m)+\frac{{\beta}^{2}}{4}(1-q)^{2},$ (6.12)

where

$f(m)=\frac{1}{N}\left\\{\frac{{\beta}}{2}(m,{\textstyle{J\over\sqrt{N}}}m)+h({\bf{1}},m)-\sum_{i=1}^{N}I(m_{i})\right\\}+\frac{{\beta}^{2}}{2}\left(1-q\right)\left(q-q_{\text{EA}}(m)\right).$
(6.13)

Our task is thus to bound $\sup_{m\in{\mathcal{B}}_{N}(\varrho)}f(m)$. To this
end, we first replace this quantity by its expectation using a classical
Gaussian concentration inequality, and then apply Gaussian comparison
techniques to linearise the quadratic form. This is the content of the next
two lemmata.

######  Lemma 6.6.

${\mathbb{P}}\left(\sup_{m\in{\mathcal{B}}_{N}(\varrho)}f(m)\geq{\mathbb{E}}\sup_{m\in{\mathcal{B}}_{N}(\varrho)}f(m)+\frac{{\beta}\varrho}{\sqrt{N}}\right)\leq
e^{-N}.$ (6.14)

Let ${\bf{Z}}=(Z_{i})_{1\leq i\leq N}$ be a standard Gaussian random vector in
${\mathbb{R}}^{N}$ and, denoting by ${\bf{E}}$ the expectation with respect to
${\bf{Z}}$, set

$\bar{f}(m)=\frac{1}{N}\left\\{{\beta}({\bf{Z}},m)\frac{\|m\|_{2}^{1}}{\sqrt{N}}+h({\bf{1}},m)-\sum_{i=1}^{N}I(m_{i})\right\\}+\frac{{\beta}^{2}}{2}\left(1-q\right)\left(q-q_{\text{EA}}(m)\right).$
(6.15)

######  Lemma 6.7.

$\displaystyle\quad{\mathbb{E}}\sup_{m\in{\mathcal{B}}_{N}(\varrho)}f(m)\leq{\bf{E}}\sup_{m\in{\mathcal{B}}_{N}(\varrho)}\bar{f}(m).$

###### Proof of Lemma 6.6.

Given $m\in{\mathcal{B}}_{N}(\varrho)$, let
$f(\cdot,m):{\mathbb{R}}^{N(N-1)/2}\rightarrow{\mathbb{R}}$ be the function
that assigns to $x=(x_{ij})_{1\leq i<j\leq N}\in{\mathbb{R}}^{N(N-1)/2}$ the
value

$f(x,m)=\frac{1}{N}\left\\{{\beta}\sum_{1\leq i<j\leq
N}x_{ij}\frac{m_{i}m_{j}}{\sqrt{N}}+h({\bf{1}},m)-\sum_{i=1}^{N}I(m_{i})\right\\}+\frac{{\beta}^{2}}{2}\left(1-q\right)\left(q-q_{\text{EA}}(m)\right).$
(6.16)

By Cauchy-Schwarz’s inequality,

$\displaystyle f(x,m)-f(y,m)$ $\displaystyle\leq$
$\displaystyle\frac{{\beta}q_{\text{EA}}(m)}{\sqrt{2}N}\|x-y\|_{2}\leq\frac{{\beta}\varrho}{\sqrt{2}N}\|x-y\|_{2}.$
(6.17)

Thus, $\sup_{m\in{\mathcal{B}}_{N}(\varrho)}f(x,m)$ is Lipschitz with constant
$L\equiv{{\beta}\varrho}/{\sqrt{2}N}$ and by Tsirelson-Ibragimov-Sudakov
concentration inequality (see [11], Theorem 5.6), for all $t>0$

${\mathbb{P}}\left(\sup_{m\in{\mathcal{B}}_{N}(\varrho)}f(m)-{\mathbb{E}}\sup_{m\in{\mathcal{B}}_{N}(\varrho)}f(m)\geq
t\right)\leq e^{-\frac{1}{2}\left(\frac{t}{L}\right)^{2}}.$ (6.18)

Choosing $t={\beta}\varrho/\sqrt{N}$ then yields the claim of Lemma 6.6. ∎

###### Proof of Lemma 6.7.

This is a straightforward application of Sudakov-Fernique Gaussian comparison
inequality (see [3], Theorem 2.2.3). Since

${\mathbb{E}}f(m)={\bf{E}}\bar{f}(m)=\frac{1}{N}\left\\{h({\bf{1}},m)-\sum_{1\leq
i\leq
N}I(m_{i})\right\\}+\frac{{\beta}^{2}}{2}\left(1-q\right)\left(q-q_{\text{EA}}(m)\right),$
(6.19)

we only have to check that for all $m\in{\mathcal{B}}_{N}(\varrho)$ and all
$\bar{m}\in{\mathcal{B}}_{N}(\varrho)$,

${\mathbb{E}}\left[f(m)-f(\bar{m})\right]^{2}\leq{\bf{E}}\left[\bar{f}(m)-\bar{f}(\bar{m})\right]^{2}.$
(6.20)

Setting $\varphi(m)=\frac{1}{2}(m,{\textstyle{J\over\sqrt{N}}}m)$ and
$\bar{\varphi}(m)=(g,m)\frac{\|m\|_{2}^{1}}{\sqrt{N}}$, (6.20) is equivalent
to

$\Delta\varphi\equiv{\mathbb{E}}\left[\varphi(m)-\varphi(\bar{m})\right]^{2}\leq{\bf{E}}\left[\bar{\varphi}(m)-\bar{\varphi}(\bar{m})\right]^{2}\equiv\Delta\bar{\varphi}.$
(6.21)

Working out the expectations in the left and right-hand side of (6.21) gives

$\displaystyle\Delta\varphi$ $\displaystyle=$
$\displaystyle\frac{1}{2N}\left[\left(\|m\|_{2}^{2}\right)^{2}-2(m,\bar{m})+\left(\|\bar{m}\|_{2}^{2}\right)^{2}\right]-\frac{1}{2N}\left[\textstyle\sum_{i=1}^{N}\left(m_{i}^{2}-\bar{m}_{i}^{2}\right)^{2}\right],$
(6.22) $\displaystyle\Delta\bar{\varphi}$ $\displaystyle=$
$\displaystyle\frac{1}{N}\left[\left(\|m\|_{2}^{2}\right)^{2}-2(m,\bar{m})\|m\|_{2}^{1}\|\bar{m}\|_{2}^{1}+\left(\|\bar{m}\|_{2}^{2}\right)^{2}\right].$
(6.23)

Since

$(m,\bar{m})^{2}-2(m,\bar{m})\|m\|_{2}^{1}\|\bar{m}\|_{2}^{1}+\|m\|_{2}^{2}\|\bar{m}\|_{2}^{2}=\left\\{(m,\bar{m})-\|m\|_{2}^{1}\|\bar{m}\|_{2}^{1}\right\\}^{2}\geq
0$ (6.24)

then

$-2(m,\bar{m})^{2}\leq
2\left\\{-2(m,\bar{m})\|m\|_{2}^{1}\|\bar{m}\|_{2}^{1}+\|m\|_{2}^{2}\|\bar{m}\|_{2}^{2}\right\\}.$
(6.25)

Inserting (6.25) in the first term on the right-hand side of (6.22) and
dropping the second,

$\displaystyle\Delta\varphi$ $\displaystyle\leq$
$\displaystyle\frac{1}{2N}\left[\left(\|m\|_{2}^{2}\right)^{2}+\left(\|\bar{m}\|_{2}^{2}\right)^{2}-4(m,\bar{m})\|m\|_{2}^{1}\|\bar{m}\|_{2}^{1}+2\|m\|_{2}^{2}\|\bar{m}\|_{2}^{2}\right].$
(6.26)

Now, using that $(a+b)^{2}\leq 2(a^{2}+b^{2})$ with $a=\|m\|_{2}^{2}$ and
$b=\|\bar{m}\|_{2}^{2}$ and recalling (6.23)

$\Delta\varphi\leq\frac{1}{2N}\left\\{2\left[\left(\|m\|_{2}^{2}\right)^{2}+\left(\|\bar{m}\|_{2}^{2}\right)^{2}\right]-4(m,\bar{m})\|m\|_{2}^{1}\|\bar{m}\|_{2}^{1}\right\\}=\Delta\bar{\varphi}.$
(6.27)

This proves (6.21), and hence (6.20). The proof of Lemma 6.7 is complete. ∎

We now return to the proof of Proposition 6.4. Combining Lemma 6.6 and Lemma
6.7, there exists $\widetilde{\Omega}_{N}\subseteq{\Omega}$ with
${\mathbb{P}}\bigl{(}\widetilde{\Omega}_{N}\bigr{)}\geq 1-e^{-N}$ such that on
$\widetilde{\Omega}_{N}$,

$\sup_{m\in{\mathcal{B}}_{N}(\varrho)}f(m)\leq{\bf{E}}\sup_{m\in{\mathcal{B}}_{N}(\varrho)}\bar{f}(m)+\frac{{\beta}\varrho}{\sqrt{N}}.$
(6.28)

By (6.3)

$\sup_{m\in{\mathcal{B}}_{N}(\varrho)}\bar{f}(m)=\sup_{0\leq\rho<\varrho}\sup_{m\in[-1,1]^{N}:q_{\text{EA}}(m)=\rho}\bar{f}(m).$
(6.29)

Going back to the definition (6.15) of $\bar{f}$ and using the fact (seen in
the proof of Proposition 2.1) that the functions ${\mathcal{I}}_{N}$ and
${\mathcal{I}}_{N}^{*}$ defined in (2.8) form a pair of Legendre-Fenchel
conjugates, we have

$\begin{split}&\sup_{m\in[-1,1]^{N}:q_{\text{EA}}(m)=\rho}\bar{f}(m)\leq\sup_{m:q_{\text{EA}}(m)=\rho}\bar{f}(m)\\\
&=\sup_{m:q_{\text{EA}}(m)=\rho}\frac{1}{N}\left\\{({\beta}\sqrt{\rho}{\bf{Z}}+h{\bf{1}},m)-\sum_{i=1}^{N}I(m_{i})\right\\}\\\
&=\sup_{m:q_{\text{EA}}(m)=\rho}\frac{1}{N}\left\\{({\beta}\sqrt{\rho}{\bf{Z}}+h{\bf{1}},m)-\sup_{y\in{\mathbb{R}}^{N}}\left\\{(m,y)-\sum_{i=1}^{N}I^{*}(y_{i})\right\\}\right\\}\\\
&=\inf_{y\in{\mathbb{R}}^{N}}\left\\{\sqrt{\rho}\frac{\|{\beta}\sqrt{\rho}{\bf{Z}}+h{\bf{1}}-y\|_{2}^{1}}{\sqrt{N}}+\frac{1}{N}\sum_{i=1}^{N}I^{*}(y_{i})\right\\}.\end{split}$
(6.30)

Finding the above infimum explicitly is beyond our reach. Instead, we make an
arbitrary (and hopefully judicious) choice of $y$ by taking

$y=\sqrt{\frac{\rho}{q}}\left({\beta}\sqrt{q}{\bf{Z}}+h{\bf{1}}\right).$
(6.31)

With this choice, it follows from (6.28), (6.29) and (6.30) that on
$\widetilde{\Omega}_{N}$

$\begin{split}\sup_{m\in{\mathcal{B}}_{N}(\varrho)}f(m)\leq\,&{\bf{E}}\sup_{0\leq\rho<\varrho}\Biggl{\\{}\sqrt{\rho}h\left(1-\sqrt{{\textstyle{\rho\over
q}}}\right)+\frac{{\beta}^{2}}{2}\left(1-q\right)\left(q-\rho\right)\\\
&+\frac{1}{N}\sum_{i=1}^{N}\log\cosh\left(\sqrt{{\textstyle{\rho\over
q}}}\left({\beta}\sqrt{q}Z_{i}+h\right)\right)\Biggr{\\}}+\log
2+\frac{{\beta}\varrho}{\sqrt{N}}.\end{split}$ (6.32)

To complete the proof of Proposition 6.4, it remains to replace the random
function within braces in (6.32) by its expectation. Let
${\bf{Z}}=(Z_{i})_{1\leq i\leq N}$ be as in (6.15) and set

$g(\rho,{\bf{Z}})\equiv\frac{1}{N}\sum_{i=1}^{N}\log\cosh\left(\sqrt{{\textstyle{\rho\over
q}}}\left({\beta}\sqrt{q}Z_{i}+h\right)\right).$ (6.33)

######  Lemma 6.8.

For all $k>1$

${\mathbb{P}}\left(\sup_{0\leq\rho<\varrho}\left[g(\rho,{\bf{Z}})-{\bf{E}}g(\rho,{\bf{Z}})\right]\geq
2{\beta}\sqrt{\frac{kq\log N}{N}}(1+N^{-k})\right)\leq N^{-k}.$ (6.34)

Moreover,

${\bf{E}}\left\\{\sup_{0\leq\rho<\varrho}\left[g(\rho,{\bf{Z}})-{\bf{E}}g(\rho,{\bf{Z}})\right]^{2}\right\\}\leq({\beta}^{2}q+h^{2})^{2}.$
(6.35)

###### Proof of Lemma 6.8.

To control the probability of the supremum in (6.34), we introduce the
discrete set $\left\\{\rho_{j}\equiv\varrho
jN^{-k},j=0,1,\dots,N^{k}-1\right\\}\subset[0,\varrho]$ where $k>1$ is to be
chosen later. Using this set, we define the sequence of functions

$g^{(j)}(\rho,{\bf{Z}})=g(\rho,{\bf{Z}})-g(\rho_{j},{\bf{Z}}),\quad 0\leq
j<N^{k}.$ (6.36)

The supremum in (6.34) can then be rewritten as

$\begin{split}&\sup_{0\leq\rho<\varrho}\bigl{\\{}g(\rho,{\bf{Z}})-{\bf{E}}g(\rho,{\bf{Z}})\bigr{\\}}\\\
=\,&\sup_{0\leq
j<N^{k}}\biggl{\\{}\sup_{\rho_{j}\leq\rho<\rho_{j+1}}\Bigl{[}g^{(j)}(\rho,{\bf{Z}})-{\bf{E}}g^{(j)}(\rho,{\bf{Z}})\Bigr{]}+\bigl{[}g(\rho_{j},{\bf{Z}})-{\bf{E}}g(\rho_{j},{\bf{Z}})\bigr{]}\biggr{\\}}.\end{split}$
(6.37)

To deal with the first term in braces, note that since

$|\log\cosh(x)-\log\cosh(y)|=\left|\int_{x}^{y}\tanh(t)dt\right|\leq|x-y|,$
(6.38)

then, for each $0\leq j<N^{k}$ and all $\rho_{j}\leq\rho<\rho_{j+1}$

$\begin{split}g^{(j)}(\rho,{\bf{Z}})&\leq\left|\sqrt{{\textstyle{\rho_{j+1}\over
q}}}-\sqrt{{\textstyle{\rho_{j}\over
q}}}\right|\frac{1}{N}\sum_{i=1}^{N}\left|{\beta}\sqrt{q}Z_{i}+h\right|\leq\sqrt{\frac{\varrho}{qN^{k}}}\frac{1}{N}\sum_{i=1}^{N}\left|{\beta}\sqrt{q}Z_{i}+h\right|\\\
&\equiv\bar{g}({\bf{Z}}).\end{split}$ (6.39)

Also note that

$\displaystyle\left|{\bf{E}}g^{(j)}(\rho,{\bf{Z}})\right|\leq{\bf{E}}\bar{g}({\bf{Z}})\leq\sqrt{\frac{\varrho}{qN^{k}}}\frac{1}{N}{\textstyle\left[{\bf{E}}\left(\sum_{i=1}^{N}\left|{\beta}\sqrt{q}Z_{i}+h\right|\right)^{2}\right]^{1/2}}=t_{k}$
(6.40)

where $t_{k}\equiv
t_{k}(\varrho,{\beta},h)=\sqrt{\frac{\varrho({\beta}^{2}q+h^{2})}{qN^{k}}}$.
Thus, by (6.39) and (6.40)

$\displaystyle\sup_{\rho_{j}\leq\rho<\rho_{j+1}}\Bigl{[}g^{(j)}(\rho,{\bf{Z}})-{\bf{E}}g^{(j)}(\rho,{\bf{Z}})\Bigr{]}\leq\bar{g}({\bf{Z}})-{\bf{E}}\bar{g}({\bf{Z}})+2t_{k}.$
(6.41)

Next, observe that for each fixed $\rho_{j}$, $0\leq j<N^{k}$, the function
$\bar{g}(x)+g(\rho_{j},x)$ viewed as a function of $x=(x_{i})_{1\leq i\leq
N}\in{\mathbb{R}}^{N}$ obeys

$\begin{split}|(\bar{g}(x)+g(\rho_{j},x))-(\bar{g}(y)+g(\rho_{j},y))|&\leq\left|\bar{g}(x)-\bar{g}(y)\right|+\left|g(\rho_{j},x)-g(\rho_{j},y)\right|\\\
&\leq\frac{{\beta}}{N^{k}}\sqrt{\frac{q}{N}}\|x-y\|_{2}+{\beta}\sqrt{\frac{q}{N}}\|x-y\|_{2}\end{split}$
(6.42)

where we used in turn (6.38) to bound
$\left|g(\rho_{j},x)-g(\rho_{j},y)\right|$ and Cauchy-Schwarz’s inequality.
Finally, combining (6.37), (6.41) and the above Lipschitz property, (6.34)
follows from Tsirelson-Ibragimov-Sudakov inequality (6.18) with
$L\equiv{\beta}\sqrt{\frac{q}{N}}\left(1+\frac{1}{N^{k}}\right)$ and
$t=2t_{k}+2t^{\prime}_{k}$, where

$t^{\prime}_{k}\equiv{\beta}\sqrt{kqN^{-1}\log N}(1+N^{-k}).$ (6.43)

For later use we denote by ${\Omega}_{k,N}$ the event

${\Omega}_{k,N}=\left\\{{\omega}\in{\Omega}\,:\,\sup_{0\leq\rho<\varrho}\left[g(\rho,{\bf{Z({\omega})}})-{\bf{E}}g(\rho,{\bf{Z({\omega})}})\right]\geq
2t^{\prime}_{k}\right\\}.$ (6.44)

We now turn to (6.35). From (6.33) and the bound $\log\cosh(x)\leq x^{2}/2$,
$x\in{\mathbb{R}}$, we get, for all $0\leq\rho\leq\varrho\leq q$

$g(\rho,{\bf{Z}})\leq\frac{1}{2N}\frac{\rho}{q}\sum_{i=1}^{N}\left({\beta}\sqrt{q}Z_{i}+h\right)^{2}\leq\frac{1}{2N}\sum_{i=1}^{N}\left({\beta}\sqrt{q}Z_{i}+h\right)^{2}\equiv\hat{g}({\bf{Z}}).$
(6.45)

This and the fact that $g(\rho,{\bf{Z}})\geq 0$ yield

$\displaystyle\sup_{0\leq\rho<\varrho}\left[g(\rho,{\bf{Z}})-{\bf{E}}g(\rho,{\bf{Z}})\right]^{2}\leq\hat{g}^{2}({\bf{Z}})+\left({\bf{E}}\hat{g}({\bf{Z}})\right)^{2}.$
(6.46)

Now

$\begin{split}\left({\bf{E}}\hat{g}({\bf{Z}})\right)^{2}&=\frac{1}{4}({\beta}^{2}q+h^{2})^{2},\\\
{\bf{E}}\hat{g}^{2}({\bf{Z}})&=\frac{1}{4}({\beta}^{2}q+h^{2})^{2}+\frac{1}{2N}{\beta}^{2}q({\beta}^{2}q+2h^{2})\leq({\beta}^{2}q+h^{2})^{2}.\end{split}$
(6.47)

Taking the expectation of both sides of (6.46) and inserting the bounds (6.47)
gives (6.35). The proof of Lemma 6.8 is complete. ∎

Set

$\phi(\rho)=\sqrt{\rho}h\left(1-\sqrt{{\textstyle{\rho\over
q}}}\right)+\frac{{\beta}^{2}}{2}\left(1-q\right)\left(q-\rho\right)+E\log\cosh\left(\sqrt{{\textstyle{\rho\over
q}}}\left({\beta}\sqrt{q}Z+h\right)\right).$ (6.48)

Then, on $\widetilde{\Omega}_{N}$ (see the line above (6.28)), (6.32) can be
rewritten as

$\sup_{m\in{\mathcal{B}}_{N}(\varrho)}f(m)\leq\sup_{0\leq\rho<\varrho}\phi(\rho)+\frac{{\beta}\varrho}{\sqrt{N}}+\log
2+{\mathcal{E}}$ (6.49)

where, recalling the definition (6.33) of $g(\rho,{\bf{Z}})$

${\mathcal{E}}\equiv{\bf{E}}\left\\{\sup_{0\leq\rho<\varrho}\left[g(\rho,{\bf{Z}})-{\bf{E}}g(\rho,{\bf{Z}})\right]\right\\}.$
(6.50)

Using the notation (6.44), we decompose (6.50) into
${\mathcal{E}}={\mathcal{E}}^{(1)}+{\mathcal{E}}^{(2)}$ where

$\displaystyle{\mathcal{E}}^{(1)}$ $\displaystyle=$
$\displaystyle{\bf{E}}\left\\{\sup_{0\leq\rho<\varrho}\left[g(\rho,{\bf{Z}})-{\bf{E}}g(\rho,{\bf{Z}})\right]\mathbbm{1}_{\left\\{{\Omega}^{c}_{k,N}\right\\}}\right\\},$
(6.51) $\displaystyle{\mathcal{E}}^{(2)}$ $\displaystyle=$
$\displaystyle{\bf{E}}\left\\{\sup_{0\leq\rho<\varrho}\left[g(\rho,{\bf{Z}})-{\bf{E}}g(\rho,{\bf{Z}})\right]\mathbbm{1}_{\left\\{{\Omega}_{k,N}\right\\}}\right\\}.$
(6.52)

Clearly ${\mathcal{E}}^{(1)}\leq 2t^{\prime}_{k}$ (see (6.43)). To bound
${\mathcal{E}}^{(2)}_{{\beta},h}$ we use successively Cauchy-Schwarz’s
inequality and Lemma 6.8 to write

$\begin{split}{\mathcal{E}}^{(2)}\leq&\sqrt{{\bf{E}}\left\\{\sup_{0\leq\rho<\varrho}\left[g(\rho,{\bf{Z}})-{\bf{E}}g(\rho,{\bf{Z}})\right]^{2}\right\\}}\sqrt{{\mathbb{P}}\left({\Omega}_{k,N}\right)}\leq({\beta}^{2}q+h^{2})N^{-k/2}.\end{split}$
(6.53)

Collecting our bounds we obtain, taking, e.g., $k=2$,

${\mathcal{E}}\leq 2{\beta}\sqrt{{\textstyle{\log N\over
N}}}+({\beta}^{2}q+h^{2})N^{-1}.$ (6.54)

Plugging in (6.49), we have on $\widetilde{\Omega}_{N}$

$\sup_{m\in{\mathcal{B}}_{N}(\varrho)}f(m)\leq\sup_{0\leq\rho<\varrho}\phi(\rho)+\log
2+2{\beta}\sqrt{{\textstyle{\log N\over N}}}+{\beta}\varrho
N^{-1/2}+({\beta}^{2}q+h^{2})N^{-1}.$ (6.55)

We are ready to complete the proof of Proposition 6.4. By (6.12),

$\sup_{m\in{\mathcal{B}}_{N}(\varrho)}F_{N,{\beta},h}^{HT}(m)=\sup_{m\in{\mathcal{B}}_{N}(\varrho)}f(m)+\frac{{\beta}^{2}}{4}(1-q)^{2}$
(6.56)

while by (1.7), (6.10) and (6.48)

$\sup_{0\leq\rho<\varrho}\phi(\rho)+\frac{{\beta}^{2}}{4}(1-q)^{2}+\log
2=SK({\beta},h)+\sup_{0\leq\rho<\varrho}\psi_{{\beta},h}(\rho).$ (6.57)

Proposition 6.4 now follows from (6.55), (6.56) and (6.57). ∎

We now turn to the proof of Proposition 6.5.

###### Proof of Proposition 6.5.

Let us write

$\displaystyle\psi_{{\beta},h}(\rho)$ $\displaystyle=$
$\displaystyle\sqrt{\rho}h\left(1-\sqrt{{\textstyle{\rho\over
q}}}\right)+\frac{{\beta}^{2}}{2}\left(1-q\right)\left(q-\rho\right)+\Delta_{{\beta},h}(\rho),$
(6.58) $\displaystyle\Delta_{{\beta},h}(\rho)$ $\displaystyle\equiv$
$\displaystyle E\left[\log\cosh\left(\sqrt{{\textstyle{\rho\over
q}}}\left({\beta}\sqrt{q}Z+h\right)\right)-\log\cosh({\beta}\sqrt{q}Z+h)\right].$
(6.59)

The next lemma collects properties of $\Delta_{{\beta},h}(\rho)$ and provides
two bounds that will be useful in different range of $\rho$ (and for $h$ large
enough).

######  Lemma 6.9.

For all $0\leq\rho<q$, $\Delta_{{\beta},h}(\rho)<0$,
$\Delta_{{\beta},h}(\rho)\uparrow 0$ as $\rho\uparrow q$ and we have

$\displaystyle\Delta_{{\beta},h}(\rho)\leq\frac{\rho}{2q}E\left({\beta}\sqrt{q}Z+h\right)^{2}-E|{\beta}\sqrt{q}Z+h|+\log
2,$ (6.60)
$\displaystyle\begin{split}\Delta_{{\beta},h}(\rho)\leq&-\left(1-\sqrt{{\textstyle{\rho\over
q}}}\right)\Bigl{(}E|{\beta}\sqrt{q}Z+h|-E2|{\beta}\sqrt{q}Z+h|e^{-2|{\beta}\sqrt{q}Z+h|}\Bigr{)}\quad\\\
&+\frac{q}{4\rho}\left(1-\sqrt{{\textstyle{\rho\over
q}}}\right)^{2}.\end{split}$ (6.61)

###### Proof of Lemma 6.9.

We deduce from the identity

$\textstyle\log\cosh\bigl{(}\sqrt{\rho/q}x\bigr{)}-\log\cosh(x)=-\int_{\sqrt{\rho/q}|x|}^{|x|}\tanh(y)dy,\quad
x\in{\mathbb{R}},$ (6.62)

that $\Delta_{{\beta},h}(\rho)$ is strictly negative for $0<\rho<q$, and
increases to $0$ as $\rho$ increases to $q$. To prove (6.61), consider the
function $f(x)=\log\cosh({\alpha}x)$, $0<{\alpha}<\infty$. By Taylor’s theorem
to second order (with remainder in Lagrange form), $f$ is approximated at
$x_{0}$ by

$f(x)=f(x_{0})+(x-x_{0}){\alpha}\tanh({\alpha}x_{0})+\frac{1}{2}(x-x_{0})^{2}\left({\textstyle{{\alpha}\over\cosh({\alpha}\xi)}}\right)^{2}$
(6.63)

for some $\xi$ between $x$ and $x_{0}$. Since
$\left({\textstyle{{\alpha}\xi\over\cosh({\alpha}\xi)}}\right)^{2}\leq{\textstyle{1\over
2}}$, this implies that for all $0<x\leq x_{0}$

$\displaystyle f(x)$ $\displaystyle\leq$ $\displaystyle
f(x_{0})+(x-x_{0}){\alpha}\tanh({\alpha}x_{0})+\frac{1}{4x^{2}}(x-x_{0})^{2}.$
(6.64)

We now use (6.64) with $x=\sqrt{\rho/q}$, $x_{0}=1$ and
${\alpha}=|{\beta}\sqrt{q}Z+h|$ to bound $\Delta_{{\beta},h}(\rho)$. To do
this, we first introduce a truncation threshold $L>0$, split
$\Delta_{{\beta},h}(\rho)$ into two terms according to whether
$|{\beta}\sqrt{q}Z+h|<L$ or $|{\beta}\sqrt{q}Z+h|\geq L$, apply (6.64) to the
first term, show that the second decays to zero exponentially fast in $L$
(using e.g. the bounds $\log\cosh(y)+\log 2=|y|+\log\left(1+e^{-2|y|}\right)$,
$0\leq\log\left(1+e^{-2|y|}\right)\leq e^{-2|y|}$ ). We skip the simple but
lengthy details of the proof. Doing so and passing to the limit
$L\uparrow\infty$, we get

$\Delta_{{\beta},h}(\rho)\leq-\left(1-\sqrt{{\textstyle{\rho\over
q}}}\right)E|{\beta}\sqrt{q}Z+h|\tanh(|{\beta}\sqrt{q}Z+h|)+{\textstyle{q\over
4\rho}}\left(1-\sqrt{{\textstyle{\rho\over q}}}\right)^{2}.$ (6.65)

Observing that $|y|\tanh(|y|)=|y|\left(1-2{\textstyle{e^{-2|y|}\over
1+e^{-2|y|}}}\right)\geq|y|\left(1-2e^{-2|y|}\right)$ finally gives (6.61).

Eq. (6.60) follows from the classical bounds, valid for all
$x\in{\mathbb{R}}$,

$|x|-\log 2\leq|x|+\log\left(1+e^{-2|x|}\right)-\log 2=\log\cosh(x)\leq
x^{2}/2.$ (6.66)

The proof of Lemma 6.9 is complete. ∎

The next lemma is needed to estimate the expectations appearing in Lemma 6.9.
Define

${\rm{erfc}}(z)=2\int_{z}^{\infty}e^{-\frac{x^{2}}{2}}\frac{dx}{\sqrt{2\pi}}.$
(6.67)

######  Lemma 6.10.

Let $Z$ be standard Gaussian random variable. For all $a>0$ and $b\geq 0$

$\displaystyle E|aZ+b|$ $\displaystyle=$ $\displaystyle
a\sqrt{{\textstyle{2\over\pi}}}e^{-\frac{1}{2}\left(\frac{b}{a}\right)^{2}}+b\left[1-{\rm{erfc}}\left({\textstyle{b\over
a}}\right)\right],$ (6.68) $\displaystyle Ee^{-|aZ+b|}$ $\displaystyle=$
$\displaystyle t^{-}_{a,b}+t^{+}_{a,b},$ (6.69) $\displaystyle
E|aZ+b|e^{-|aZ+b|}$ $\displaystyle=$ $\displaystyle
b\left(t^{-}_{a,b}-t^{+}_{a,b}\right)-a^{2}\left(t^{-}_{a,b}+t^{+}_{a,b}\right)+a\sqrt{{\textstyle{2\over\pi}}}e^{-\frac{1}{2}\left(\frac{b}{a}\right)^{2}},\,$
(6.70)

where

$\displaystyle t^{-}_{a,b}$ $\displaystyle=$ $\displaystyle
e^{{\textstyle{a^{2}\over 2}}-b}\left[1-{\textstyle{1\over
2}}{\rm{erfc}}\left({\textstyle{b\over
a}}-a\right)\right]1_{\left\\{-\frac{b}{a}+a\leq
0\right\\}}+{\textstyle{1\over
2}}e^{\frac{a^{2}}{2}-b}{\rm{erfc}}\left(-{\textstyle{b\over
a}}+a\right)1_{\left\\{-\frac{b}{a}+a\geq 0\right\\}},$ $\displaystyle
t^{+}_{a,b}$ $\displaystyle=$ $\displaystyle{\textstyle{1\over
2}}e^{\frac{a^{2}}{2}+b}{\rm{erfc}}\left({\textstyle{b\over a}}+a\right).$

###### Proof of Lemma 6.10.

Eq. (6.68) and (6.69) are straightforward. Eq. (6.70) relies on Gaussian
integration by parts. ∎

It is well known (see[1], inequalities 7.1.13) that ${\rm{erfc}}(z)$ obeys the
bounds

$C^{-}(z)=\frac{2}{z+\sqrt{z^{2}+4}}\leq\sqrt{\frac{\pi}{2}}e^{\frac{z^{2}}{2}}{\rm{erfc}}\bigl{(}z\bigr{)}\leq
C^{+}(z)=\frac{2}{z+\sqrt{z^{2}+\frac{8}{\pi}}}\leq\frac{1}{z}.\quad$ (6.71)

######  Corollary 6.11.

For all $({\beta},h)$, $E|{\beta}\sqrt{q}Z+h|\geq h$.

###### Proof.

Use (6.68) and the rightmost upper bound of (6.71). ∎

We now proceed in three steps, using the two bounds of Lemma 6.9 in turn.

Step 1: Using Corollary 6.11 in (6.60) together with our assumptions on
$({\beta},h)$

$\Delta_{{\beta},h}(\rho)\leq\frac{\rho}{2q}\left({\beta}^{2}q+h^{2}\right)-h+\log
2\leq\frac{5}{8}\frac{\rho}{q}h^{2}-h+\log 2,$ (6.72)

and inserting in (6.10), we get

$\psi_{{\beta},h}(\rho)\leq\Upsilon_{{\beta},h}(\rho)\equiv
h\left\\{\sqrt{\rho}+\frac{q{\beta}^{2}}{2h}\left(1-q\right)+\frac{5}{8}\frac{\rho}{q}h-1+\frac{\log
2}{h}\right\\}.$ (6.73)

The right-hand-side of (6.73) is a quadratic fonction of $\sqrt{\rho}$. One
checks that $\Upsilon_{{\beta},h}(\rho)=0$ has a single strictly positive
root, $\rho^{+}>0$, and that $\Upsilon_{{\beta},h}(\rho)<0$ in the interval
$[0,\rho^{+})$. One also checks that for all $({\beta},h)$ such that
${\beta}^{2}(1-q)\leq 1$ and $h\geq 4$,

$\rho^{+}\geq\rho^{(1)}\equiv\frac{q}{h}\left({4}/{5}\right)^{3}(4/3)$ (6.74)

and so, on that domain, $\psi_{{\beta},h}(\rho)<0$ for all

$0\leq\rho<\rho^{(1)}.$ (6.75)

Step 2: We now assume that $q>\rho\geq\rho^{(1)}$. Inserting (6.61) in (6.58),
observing that

$\sqrt{\rho}\left(1-\sqrt{{\textstyle{\rho\over
q}}}\right)\leq\sqrt{{\textstyle{\rho\over
q}}}\left(1-\sqrt{{\textstyle{\rho\over
q}}}\right),\quad\frac{q}{4h\rho}\leq\frac{3}{8},\quad\frac{1}{2}\left(q-\rho\right)\leq
q\left(1-\sqrt{{\textstyle{\rho\over q}}}\right),$ (6.76)

and using again Corollary 6.11, we get

$\displaystyle\psi_{{\beta},h}(\rho)$ $\displaystyle\leq$ $\displaystyle
h\left(1-\sqrt{{\textstyle{\rho\over
q}}}\right)\left\\{-\left(1-{\textstyle{3\over
8}}\right)\left(1-\sqrt{{\textstyle{\rho\over
q}}}\right)+r({\beta},h)\right\\}$ (6.77)

where $r({\beta},h)\equiv
h^{-1}\left[{\beta}^{2}q\left(1-q\right)+E2|{\beta}\sqrt{q}Z+h|e^{-2|{\beta}\sqrt{q}Z+h|}\right].$
Thus $\psi_{{\beta},h}(\rho)<0$ if

$\sqrt{\frac{\rho}{q}}<1-\frac{8}{5}r({\beta},h).$ (6.78)

The next lemma provides bounds on $r({\beta},h)$. We postpone its proof to the
end of the section.

######  Lemma 6.12.

For all $({\beta},h)$, $h>0$

$r({\beta},h)\leq
r^{*}({\beta},h)\equiv\begin{cases}\left(3+{\textstyle{4\over\sqrt{2\pi}}}\frac{{\beta}\sqrt{q}}{h}\right)e^{-h}&\text{if
$h>2{\beta}^{2}q$,}\\\
\left(1+{\textstyle{6\over\sqrt{2\pi}}}\frac{{\beta}\sqrt{q}}{h}\right)e^{-\frac{1}{2}\left(\frac{h}{{\beta}\sqrt{q}}\right)^{2}}&\text{if
$h\leq 2{\beta}^{2}q$.}\end{cases}$ (6.79)

One checks, using Lemma 6.12, that for all $({\beta},h)$ such that $h\geq 4$
and $h/{\beta}\geq 2$, $0<1-\frac{8}{5}r^{*}({\beta},h)\leq 1$ and
$\rho^{(1)}<q\left(1-\frac{8}{5}r^{*}({\beta},h)\right)^{2}$. Hence, on that
domain, $\psi_{{\beta},h}(\rho)<0$ for all

$\rho^{(1)}\leq\rho<\rho^{(2)}\equiv
q\left(1-\frac{8}{5}r({\beta},h)\right)^{2}.$ (6.80)

Step 3: In this last step we repeat step 2 assuming this time that
$q>\rho\geq\rho^{(2)}$. Doing this we get that $\psi_{{\beta},h}(\rho)<0$ if

$\sqrt{\frac{\rho}{q}}<\sqrt{\frac{\rho^{(3)}}{q}}\equiv
1-\left(1-\left[4h\left(1-{\textstyle{8\over
5}}r({\beta},h)\right)^{2}\right]^{-1}\right)^{-1}r({\beta},h).$ (6.81)

One checks, using Lemma 6.12, that for $({\beta},h)$ as in Step 2,
$4\left(1-{\textstyle{8\over 5}}r({\beta},h)\right)^{2}>1$. Hence

$\rho^{(3)}>q\left[1-\left(1+(h-1)^{-1}\right)r({\beta},h)\right]^{2}\geq
q\left[1-(4/3)r({\beta},h)\right]^{2}>\rho^{(2)}.$ (6.82)

To go from (6.81) to (6.82) we checked, again by Lemma 6.12, that under our
assumptions on $({\beta},h)$ the right-hand side of (6.81) and all the
quantities in square brackets in (6.82) are positive.

If $\left[1-\left(1+(h-1)^{-1}\right)r({\beta},h)\right]^{2}<q$, the lower
bound on $\rho^{(3)}$ obtained in (6.82) can be improved. Indeed, for all
$\rho<q$ we can write, instead of (6.76)

$\frac{q}{4h\rho}\left(1-\sqrt{{\textstyle{\rho\over
q}}}\right)\leq\frac{3}{8}\left(1-\sqrt{\rho}\right),\quad\frac{1}{2}\left(q-\rho\right)\leq
q\left(1-\sqrt{{\textstyle{\rho\over q}}}\right).$ (6.83)

Eq. (6.77) then becomes

$\displaystyle\psi_{{\beta},h}(\rho)$ $\displaystyle\leq$ $\displaystyle
h\left(1-\sqrt{{\textstyle{\rho\over
q}}}\right)\left\\{-\left(1-{\textstyle{3\over
8}}\right)\left(1-\sqrt{\rho}\right)+r({\beta},h)\right\\}.$ (6.84)

Setting $q$ to one in the definitions of $\rho^{(2)}$ and $\rho^{(3)}$ and
calling $\hat{\rho}^{(2)}$ and $\hat{\rho}^{(3)}$ the resulting quantities,
the conclusions of Step 2 and Step 3 above hold unchanged for
$\hat{\rho}^{(2)}$ and $\hat{\rho}^{(3)}$ whenever $\hat{\rho}^{(3)}\leq q$.

Combining (6.75), (6.80), (6.82) and the above observation we proved that
$\psi_{{\beta},h}(\rho)<0$ for all $0\leq\rho\leq\bar{\varrho}$ and
$({\beta},h)\in{\mathcal{D}}^{(3)}$ with $\bar{\varrho}$ and
${\mathcal{D}}^{(3)}$ defined in (6.2) and (1.33), respectively. The proof of
Proposition 6.5 is complete. ∎

We now prove Lemma 6.12 and Lemma 6.2 together. The proof uses the next

######  Lemma 6.13.

$Ee^{-2|{\beta}\sqrt{q}Z+h|}\leq 1-q\leq 4Ee^{-2|{\beta}\sqrt{q}Z+h|}$.

###### Proof of Lemma 6.13.

By (1.6), $1-q=E\cosh^{-2}({\beta}\sqrt{q}Z+h)$ . The claim of the lemma then
follows from the bounds $e^{-2|x|}\leq\cosh^{-2}(x)\leq 4e^{-2|x|}$,
$x\in{\mathbb{R}}$. ∎

###### Proof of Lemma 6.12 and 6.2.

Recall (6.1) and set $a=2{\beta}\sqrt{q}$ and $b=2h$. By (6.70)

$\displaystyle r({\beta},h)$ $\displaystyle=$
$\displaystyle\frac{a^{2}}{4h}\left(1-q\right)+\frac{1}{h}\Bigl{[}(b-a^{2})\left(t^{-}_{a,b}+t^{+}_{a,b}\right)-2bt^{+}_{a,b}+a\sqrt{{\textstyle{2\over\pi}}}e^{-\frac{1}{2}\left(\frac{b}{a}\right)^{2}}\Bigr{]}$
(6.85) $\displaystyle\leq$ $\displaystyle
2Ee^{-|aZ+b|}+\frac{4a}{b\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{b}{a}\right)^{2}}$
(6.86)

where the last inequality follows from Lemma 6.13 and (6.69). We next use
(6.69) and (6.71) to bound $Ee^{-|aZ+b|}$, distinguishing two cases. Assume
first that $-\frac{b}{a}+a\leq 0$. Then

$\displaystyle r({\beta},h)$ (6.87) $\displaystyle\leq$ $\displaystyle
2\left\\{e^{{\textstyle{a^{2}\over 2}}-b}\left[1-{\textstyle{1\over
2}}{\rm{erfc}}\left({\textstyle{b\over a}}-a\right)\right]+{\textstyle{1\over
2}}e^{\frac{a^{2}}{2}+b}{\rm{erfc}}\left({\textstyle{b\over
a}}+a\right)\right\\}+{\textstyle{4a\over
b\sqrt{2\pi}}}e^{-\frac{1}{2}\left(\frac{b}{a}\right)^{2}}$
$\displaystyle\leq$
$\displaystyle\Bigl{[}2+\sqrt{{\textstyle{2\over\pi}}}C^{+}\left({\textstyle{b\over
a}}+a\right)+{\textstyle{4a\over b\sqrt{2\pi}}}\Bigr{]}e^{-{\textstyle{b\over
2}}}$ (6.88)

where we used (6.71) in the last inequality together with the following two
facts: ${\textstyle{a^{2}\over 2}}-b\leq-\frac{b}{2}$ (which follows from the
assumption that $-\frac{b}{a}+a\leq 0$) and ${\textstyle{a^{2}\over
2}}-b\geq-\frac{1}{2}\left(\frac{b}{a}\right)^{2}$. The first inequality of
(6.79) then follows from (6.88) and the bound $\sqrt{{2}/{\pi}}C^{+}(z)\leq 1$
$\forall z\geq 0$. Assume now that $-\frac{b}{a}+a>0$.

$\displaystyle r({\beta},h)$ $\displaystyle\leq$ $\displaystyle
2\left[{\textstyle{1\over
2}}e^{\frac{a^{2}}{2}-b}{\rm{erfc}}\left(-{\textstyle{b\over
a}}+a\right)+{\textstyle{1\over
2}}e^{\frac{a^{2}}{2}+b}{\rm{erfc}}\left({\textstyle{b\over
a}}+a\right)\right]+{\textstyle{4a\over
b\sqrt{2\pi}}}e^{-\frac{1}{2}\left(\frac{b}{a}\right)^{2}}$ (6.89)
$\displaystyle\leq$
$\displaystyle\Bigl{\\{}\sqrt{{\textstyle{2\over\pi}}}\left[C^{+}\left(-{\textstyle{b\over
a}}+a\right)+C^{+}\left({\textstyle{b\over
a}}+a\right)\right]+{\textstyle{4a\over
b\sqrt{2\pi}}}\Bigr{\\}}e^{-\frac{1}{2}\left(\frac{b}{a}\right)^{2}}$ (6.90)
$\displaystyle\leq$
$\displaystyle\Bigl{[}1+\sqrt{{\textstyle{2\over\pi}}}\left({\textstyle{b\over
a}}+a\right)^{-1}+\sqrt{{\textstyle{2\over\pi}}}{\textstyle{2a\over
b}}\Bigr{]}e^{-\frac{1}{2}\left(\frac{b}{a}\right)^{2}}$ (6.91)

where we used that $\sqrt{{2}/{\pi}}C^{+}(z)\leq 1$ and $C^{+}(z)\leq z^{-1}$
to bound, respectively, the first and second occurence of this function in
(6.90). Since $\left({\textstyle{b\over
a}}+a\right)^{-1}\leq{\textstyle{a\over b}}$, the second inequality of (6.79)
follows. The upper bound on $r({\beta},h)$ of Lemma 6.2 is proved in the same
way but we now bound both occurrences of $C^{+}(z)$ in (6.90) by $C^{+}(z)\leq
z^{-1}$, namely,

$C^{+}\left(-{\textstyle{b\over a}}+a\right)+C^{+}\left({\textstyle{b\over
a}}+a\right)\leq{\textstyle{2a\over
a^{2}-\left(\frac{b}{a}\right)^{2}}}\leq{\textstyle{2\over
a(1-\eta^{2})}}\leq{\textstyle{2\eta\over 1-\eta^{2}}}{\textstyle{a\over b}}$
(6.92)

where the last two inequalities follow from the assumption that $b\leq\eta
a^{2}$. To prove the associated lower bound we go back to (6.85)
(equivalently, to (6.1)). By (6.70), the observation that
$t^{-}_{a,b}-t^{+}_{a,b}\geq 0$ and the lower bound on $(1-q)$ of Lemma 6.13,
we have

$\displaystyle r({\beta},h)$ $\displaystyle\geq$ $\displaystyle
h^{-1}\Bigl{[}-{\textstyle{3\over
4}}a^{2}Ee^{-|aZ+b|}+a\sqrt{{\textstyle{2\over\pi}}}e^{-\frac{1}{2}\left(\frac{b}{a}\right)^{2}}\Bigr{]}.$
(6.93)

The lower bound of (6.5) now follows from the upper bound on $Ee^{-|aZ+b|}$
established in the proof of the upper bound of (6.5) using (6.92). However,
here, we do not use the last inequality of (6.92) but only but the one before
last.

We now turn to (6.6). In view of the upper bound of Lemma 6.13, $1-q$ is
bounded above by twice the first term in the second line of (6.86). Based on
this observation, the proof of the upper bound on $1-q$ is a by-product of the
proof of the upper bound on $r({\beta},h)$ (note that here again, we do not
use the last inequality in (6.92) but the one before last). To prove the
associated lower bound we write, combining the lower bound of Lemma 6.13,
(6.69), the lower bound (6.71) and the assumption that $b\leq\eta a^{2}$,

$\displaystyle 1-q\geq{\textstyle{1\over
2}}e^{\frac{a^{2}}{2}+b}{\rm{erfc}}\left({\textstyle{b\over
a}}+a\right)\geq{\textstyle{1\over
2}}\sqrt{{\textstyle{2\over\pi}}}C^{-}\left({\textstyle{b\over
a}}+a\right)e^{-\frac{1}{2}\left(\frac{b}{a}\right)^{2}}\geq{\textstyle{e^{-\frac{1}{2}\left(\frac{b}{a}\right)^{2}}\over\sqrt{2\pi}\sqrt{a^{2}(1+\eta)^{2}+4}}}.$
(6.94)

Lemma 6.12 and Lemma 6.2 are proved. ∎

###### Proof of Proposition 1.10.

Let us establish that
$\widetilde{\mathcal{D}}^{(2)}\subset{\mathcal{D}}^{(2)}_{\bar{\varrho}({\beta},h)}\cap{\mathcal{D}}^{(3)}\cap{\mathcal{D}}^{(4)}$.
The condition $3\leq{h}/{{\beta}}\leq{\beta}q/10$ implies that
${h}/{{\beta}}\geq 3$ and ${\beta}\geq 30/q$ so that
$h=({h}/{{\beta}}){\beta}\geq 90/q\geq 90$. It also implies that
$10h\leq{\beta}^{2}q$. Thus, the assumptions of Lemma 6.2 are satisfied with
$\eta=1/20$. This key lemma is used throughout the proof. It first guarantees
that $\sqrt{3/4}\leq q\leq 1$. Indeed, using that ${\beta}\sqrt{q}\geq
10h/({\beta}\sqrt{q})$ and setting $f(x)=x^{-1}e^{-x^{2}/2}$, we deduce from
the upper bound of (6.6) that

$1-q\leq{\textstyle{2\over
10(1-\eta^{2})}}\sqrt{{\textstyle{2\over\pi}}}f(h/({\beta}\sqrt{q}))\leq{\textstyle{2\over
10(1-\eta^{2})}}\sqrt{{\textstyle{2\over\pi}}}f(3)\leq 1-\sqrt{3/4},$ (6.95)

where we used that $f$ is strictly decreasing on ${\mathbb{R}}^{+}$ and that
$h/({\beta}\sqrt{q})\geq h/{\beta}\geq 3$. Next, it guarantees that
${\beta}^{2}(1-q)<1$. Indeed, using again the upper bound of (6.6) and the
fact just established that $q\geq\sqrt{3/4}$, we get

${\beta}^{2}(1-q)\leq{\textstyle{4\over 1-\eta^{2}}}\sqrt{{\textstyle{2\over
3\pi}}}{\beta}e^{-\frac{1}{2}(h/{\beta})^{2}}\leq
12{\beta}e^{-\frac{1}{9}(h/{\beta})^{2}}<1,$ (6.96)

where the last inequality is the first condition in the definition of
$\widetilde{\mathcal{D}}^{(2)}$. So far, we have established that
$\widetilde{\mathcal{D}}^{(2)}\subseteq{\mathcal{D}}^{(3)}$. Let us now check
that $\widetilde{\mathcal{D}}^{(2)}\subseteq{\mathcal{D}}^{(4)}$. Comparing
the prefactors of the lower bound of (6.5) and of the upper bound of (6.6), we
see that $1-\left(1+{\textstyle{1\over h-1}}\right)r({\beta},h)<q$ on
$\widetilde{\mathcal{D}}^{(2)}$. Thus, by (6.2),
$\bar{\varrho}({\beta},h)=\left[1-\left(1+{\textstyle{1\over
h-1}}\right)r({\beta},h)\right]^{2}$. By the lower bound of (6.5) this implies
that $\bar{\varrho}({\beta},h)>\sqrt{3/4}$ if (for $f$ defined as in (6.95))

$f(h/({\beta}\sqrt{q}))\leq\left(1-(3/4)^{1/4}\right)/\left(2\left(1+{\textstyle{1\over
h-1}}\right)\left(1+{\textstyle{\eta\over
1-\eta^{2}}}\right)\sqrt{{2}/{\pi}}\right),$ (6.97)

which is satisfied on $\widetilde{\mathcal{D}}^{(2)}$. Thus,
$\widetilde{\mathcal{D}}^{(2)}\subseteq{\mathcal{D}}^{(4)}$. Using Lemma 6.2
once more, one proves that ${\beta}\vartheta(\bar{\varrho}({\beta},h))<1$ if
the first condition in the definition of $\widetilde{\mathcal{D}}^{(2)}$ is
satisfied. Hence,
$\widetilde{\mathcal{D}}^{(2)}\subset{\mathcal{D}}^{(2)}_{\bar{\varrho}({\beta},h)}$.
We skip the elementary details.

It remains to check that $\widetilde{\mathcal{D}}^{(2)}$ is contained in the
AT-region. Using that $\cosh^{-4}(x)\leq 16e^{-4|x|}$ and proceeding as in the
proof of the upper bound of (6.6), we readily get that if $h\leq
4\eta{\beta}^{2}q$, choosing $\eta=1/40$, the AT-region contains the region
$4{\beta}e^{-\frac{1}{2}(h/{\beta})^{2}}<1$, which itself contains
$12{\beta}e^{-\frac{1}{9}(h/{\beta})^{2}}<1$. The proof of the proposition is
complete. ∎

###### Proofs of Theorem 1.8 and Theorem 1.4.

Let ${\mathcal{D}}$ be given by (1.36). Combining Theorem 1.6, Theorem 1.7 and
(1.24) from Theorem 1.3 proves Theorem 1.8 and, as a consequence, Theorem 1.4.
The claim that with ${\mathbb{P}}$-probability one, for all large enough $N$,
$F_{N,{\beta},h}^{HT}$ has a unique global maximum over $[-1,1]^{N}$, follows
from Lemma 5.13 and the fact that its global maximum is achieved in
${\mathcal{B}}^{c}_{N,{\epsilon}}(\varrho)$. ∎

##  7\. An integral representation formula

In this section, an integral representation of the partition function (1.3)
derived from the Hubbard-Stratonovitch transformation [33, 46] is used to
prove Theorem 1.5. This transformation has proved to be a useful tool for
identifying the free energy functionals of mean-field models, from the early
work of Kac [35] on the Curie-Weiss model to the more recent analysis of the
Hopfield model [15, 12, 13, 14]. Here, it allows us to identify the function
$F_{N,{\beta},h}^{HT}$ in (1.15) as the free energy functional of the SK model
at high temperature

Consider the matrix $A\equiv A_{N}$ defined in (2.1). Since $M$ is real
symmetric, there exists an orthogonal matrix ${\mathcal{O}}$ and a diagonal
matrix
${\Lambda}=\textrm{diag}({\lambda}_{1},{\lambda}_{2},\dots,{\lambda}_{N})$
such that $A={\mathcal{O}}^{t}{\Lambda}{\mathcal{O}}$. Let $\sqrt{A}$ denote
the matrix $\sqrt{A}\equiv{\mathcal{O}}^{t}\sqrt{{\Lambda}}{\mathcal{O}}$
where
$\sqrt{{\Lambda}}\equiv\textrm{diag}(\sqrt{{\lambda}_{1}},\sqrt{{\lambda}_{2}},\dots,\sqrt{{\lambda}_{N}})$
with the convention that if ${\alpha}$ is a real number,

$\sqrt{{\alpha}}=\zeta\sqrt{|{\alpha}|}\quad\text{where}\quad\zeta=\begin{cases}i&\mbox{if}\,\,{\alpha}<0,\\\
1&\mbox{if}\,\,{\alpha}\geq 0,\end{cases}$ (7.1)

where $i$ is the unit imaginary number. Thus $\sqrt{A}$ is a complex symmetric
matrix that satisfies $\sqrt{A}\sqrt{A}=A$. Let
${\mathcal{F}}_{N,{\beta},h}:{\mathbb{C}}^{N}\mapsto{\mathbb{C}}$ be the
function

${\mathcal{F}}_{N,{\beta},h}(x)=-\frac{1}{2}\sum_{j=1}^{N}x_{j}^{2}+\sum_{j=1}^{N}\log\cosh\left(\sqrt{{\beta}}(\sqrt{A}x)_{j}+h\right)+N\log
2$ (7.2)

where the log function is defined in the principal branch (it is therefore
continuous in any open set in the complex plane from which the negative axes
and zero have been removed).

######  Lemma 7.1.

$Z_{N,{\beta},h}=e^{\frac{1}{2}N{\beta}^{2}(1-q)}\int_{{\mathbb{R}}^{N}}\left(\prod_{j=1}^{N}\frac{dx_{j}}{\sqrt{2\pi}}\right)\,e^{{\mathcal{F}}_{N,{\beta},h}(x)}.$
(7.3)

To evaluate such an integral, one usually starts by looking for a critical
point that maximises the real part of ${\mathcal{F}}_{N,{\beta},h}$ and, if
this point is unique, one tries to deform the integration path so that it
passes through this point in the direction of steepest descent (equivalently,
a direction of constant phase). On such a contour, in the vicinity of the
critical point, the integral should resemble a Laplace integral. Although our
attempts to compute this integral failed, part of this programme can be
carried out as we now explain.

Our first lemma links the critical points of ${\mathcal{F}}_{N,{\beta},h}$ to
the solutions of the TAP equations. Recall the definitions of
$\Psi_{N,{\beta},{\bf{h}}}$ and $\Phi_{N,{\beta},{\bf{h}}}$ from (2.2).
Observe that for all $z\in{\mathbb{R}}^{N}$,

${\mathcal{F}}_{N,{\beta},h}\bigl{(}\sqrt{{\beta}}\sqrt{A}z\bigr{)}=\Phi_{N,{\beta},{\bf{h}}}(z).$
(7.4)

######  Lemma 7.2.

If $z\in{\mathbb{R}}^{N}$ is a solution of the TAP equations (1.10) satisfying
$q_{\text{EA}}(z)=q$, then $x=\sqrt{{\beta}}\sqrt{A}z\in{\mathbb{C}}^{N}$ is a
critical point of ${\mathcal{F}}_{N,{\beta},h}$, and for each such pair of
critical points (7.4) holds.

Points of the form $x=\sqrt{{\beta}}\sqrt{A}z$, $z\in{\mathbb{R}}^{N}$, are in
general complex. Thus the prospective maximiser of
$\Re({\mathcal{F}}_{N,{\beta},h})$ is in ${\mathbb{C}}^{N}$ while the
integration contour in (7.3) is the collection of the $N$ real axes. The next
lemma state that we can shift the contour so that it passes through a given
point in ${\mathbb{C}}^{N}$.

######  Lemma 7.3.

The following holds with ${\mathbb{P}}$-probability one for sufficiently
enough $N$. For all $x^{*}=a+ib$, $a,b\in{\mathbb{R}}^{N}$ such that
$\|b\|_{\infty}<\infty$

$Z_{N,{\beta},h}=e^{\frac{1}{2}N{\beta}^{2}(1-q)}\int_{{\mathbb{R}}^{N}}\left(\prod_{j=1}^{N}\frac{dx_{j}}{\sqrt{2\pi}}\right)\,e^{{\mathcal{F}}_{N,{\beta},h}(x+x^{*})}.$
(7.5)

We next must choose the shift in Lemma 7.3. Given $z\in{\mathbb{R}}^{N}$, set
$x^{*}(z)=\sqrt{{\beta}}\sqrt{A}z$ and let
${\mathcal{R}}_{N,{\beta},{\bf{h}}}:{\mathbb{R}}^{N}\mapsto{\mathbb{R}}$ be
the function

${\mathcal{R}}_{N,{\beta},{\bf{h}}}(z)=\frac{1}{N}\log\left(\int_{{\mathbb{R}}^{N}}\left(\textstyle{\prod_{j=1}^{N}\frac{dx_{j}}{\sqrt{2\pi}}}\right)\,e^{{\mathcal{F}}_{N,{\beta},h}(x+x^{*}(z))-{\mathcal{F}}_{N,{\beta},h}(x^{*}(z))}\right).$
(7.6)

######  Proposition 7.4.

With ${\mathbb{P}}$-probability one, for all $N$ large enough, the following
holds:

(i) For any $z\in{\mathbb{R}}^{N}$ such that $\|z\|_{\infty}<\infty$,

$\frac{1}{N}\log
Z_{N,{\beta},h}=\Phi_{N,{\beta},{\bf{h}}}(z)+\frac{1}{2}{\beta}^{2}(1-q)+{\mathcal{R}}_{N,{\beta},{\bf{h}}}(z).$
(7.7)

(ii) If $\Psi_{N,{\beta},{\bf{h}}}$ attains its global maximum uniquely at a
point $z^{*}$ lying in $(-1,1)^{N}$, then

$\Phi_{N,{\beta},{\bf{h}}}(z^{*})=\sup_{z\in[-1,1]^{N}}\Psi_{N,{\beta},{\bf{h}}}(z).$
(7.8)

Moreover, $\Psi_{N,{\beta},{\bf{h}}}$ cannot be replaced with
$\Phi_{N,{\beta},{\bf{h}}}$ in (7.8).

######  Remark.

One can prove that under the assumptions and with the notation of Proposition
7.4, (ii), the supremum of
$\Re\bigl{(}{\mathcal{F}}_{N,{\beta},h}(x+x^{*}(z^{*}))\bigr{)}$ over the
integration contour ${\mathbb{R}}^{N}$ is attained uniquely at $x=0$. Hence,
on the integration contour, the exponent in (7.6) has a unique saddle point at
$x=0$ and its real part is strictly negative away from this point. (To limit
the length of this paper, we refrain from giving the proof of this result.)
While it would be unwise to draw too close a parallel with the classical
setting (where the dimension of ${\mathcal{F}}$ does not diverge with the
asymptotic parameter), we note that the above properties would typically put
us in a position to apply Laplace method.

######  Remark.

More generally, Lemma 7.2 holds for $z\in{\mathbb{C}}^{N}$. In this case the
TAP equations become a system of equations in ${\mathbb{C}}^{N}$ and the
function $\Phi_{N,{\beta},{\bf{h}}}$ also takes values in ${\mathbb{C}}^{N}$.
While we could not rule out the existence of such critical points, they are of
no interest to us since a complex $\Phi_{N,{\beta},{\bf{h}}}$ in (7.7) would
not lead to a meaningful representation of $Z_{N}$.

We first prove Theorem 1.5, assuming the above results. Then we successively
prove Lemma 7.1, Lemma 7.2, Proposition 7.4 and Lemma 7.3 in this order.

###### Proof of Theorem 1.5.

Taking $z=m^{(k)}$ in item (i) of Proposition 7.4, it follows from (7.7) that
on a set of full measure, for all large enough $N$

$\frac{1}{N}\log
Z_{N,{\beta},h}=\frac{1}{N}\Phi_{N,{\beta},h{\bf{1}}}\left(m^{(k)}\right)+\frac{1}{2}{\beta}^{2}(1-q)+{\mathcal{R}}_{N,{\beta},h{\bf{1}}}\left(m^{(k)}\right).$
(7.9)

We know from Section 3 that $m^{(k)}$ is a near solution of the system of
specialised TAP equations (1.21) and that the functions
$\Phi_{N,{\beta},h{\bf{1}}}$ and $\Psi_{N,{\beta},h{\bf{1}}}$ can be modified
according to the strategy of Section 2.2 (see (2.9)) to make $m^{(k)}$ an
exact critical point for which the duality formula holds (see Lemma 2.3). We
will not repeat the details of this argument which we have used many times
before (see, e.g., (4.3)-(4.5) in the proof Lemma 4.1 and the proof of Theorem
1.6 after the proof of Lemma 5.13). Proceeding in this way, we obtain that for
all $({\beta},h)$ in the AT-region,

$\lim_{k\rightarrow\infty}\lim_{N\rightarrow\infty}\left|\frac{1}{N}\Phi_{N,{\beta},h{\bf{1}}}\left(m^{(k)}\right)-\frac{1}{N}\Psi_{N,{\beta},h{\bf{1}}}\left(m^{(k)}\right)\right|=0\quad{\mathbb{P}}-\text{a.s.}$
(7.10)

We also know by Theorem 1.8 that for all $({\beta},h)$ in the AT-region
intersected with the region ${\mathcal{D}}$ defined by (1.36),
$F_{N,{\beta},h}^{HT}\left(m^{(k)}\right)$ converges ${\mathbb{P}}$-almost
surely to the global maximum of $F_{N,{\beta},h}^{HT}$. Hence, by (2.5), the
same holds true for the function $\frac{1}{N}\Psi_{N,{\beta},h{\bf{1}}}$, and
so,

$\lim_{k\rightarrow\infty}\lim_{N\rightarrow\infty}\left|\frac{1}{N}\Psi_{N,{\beta},h{\bf{1}}}\left(m^{(k)}\right)-\sup_{z\in[-1,1]^{N}}\frac{1}{N}\Psi_{N,{\beta},h{\bf{1}}}(z)\right|=0\quad{\mathbb{P}}-\text{a.s.}$
(7.11)

Thus, using (7.10), (7.11) and (2.5), for all $({\beta},h)$ in the
intersection of the AT-region and the region ${\mathcal{D}}$, (7.9) can be
written as

$\frac{1}{N}\log
Z_{N,{\beta},h}=\left\\{\sup_{x\in[-1,1]^{N}}F_{N,{\beta},h}^{HT}(x)\right\\}+\frac{{\beta}^{2}}{4}\left(1-q\right)^{2}+{\mathcal{R}}_{N,{\beta},h{\bf{1}}}\left(m^{(k)}\right)+r_{k,N}({\beta},h)$
(7.12)

where $r_{k,N}({\beta},h)$ satisfies
$\lim_{k\rightarrow\infty}\lim_{N\rightarrow\infty}r_{k,N}({\beta},h)=0$
${\mathbb{P}}$-almost surely.

As already mentioned, we have not been able to work out the term
${\mathcal{R}}_{N,{\beta},h{\bf{1}}}\left(m^{(k)}\right)$ by direct methods.
With additional information, however, we can identify its limits. More
precisely, we know from (1.5) that for all $({\beta},h)$ in the high-
temperature region of Definition 1.1

$\lim_{N\rightarrow\infty}\frac{1}{N}\log
Z_{N,{\beta},h}=SK({\beta},h)\quad{\mathbb{P}}\textstyle{-a.s.}.$ (7.13)

On the other hand, by Lemma 4.2, for all $({\beta},h)$ in the AT-region,

$\lim_{k\rightarrow\infty}\lim_{N\rightarrow\infty}\frac{1}{N}\Phi_{N,{\beta},h{\bf{1}}}\left(m^{(k)}\right)=SK({\beta},h)-\frac{{\beta}^{2}}{4}\left(1-q^{2}\right)\quad{\mathbb{P}}\textstyle{-a.s.}$
(7.14)

Since convergence in (7.13) and (7.14) holds ${\mathbb{P}}$-almost surely, it
follows from (7.9) that in the high-temperature region

$\lim_{k\rightarrow\infty}\lim_{N\rightarrow\infty}{\mathcal{R}}_{N,{\beta},h{\bf{1}}}\left(m^{(k)}\right)=-\frac{{\beta}^{2}}{4}\left(1-q\right)^{2}\quad{\mathbb{P}}\textstyle{-a.s.}$
(7.15)

It should be remembered here that, although it is generally accepted in the
physics literature that the high-temperature region of the SK model coincides
with the AT-region, from a rigorous point of view it is only known that the
former is a subregion of the latter (see the discussion below (1.9)). Thus,
combining (7.15) and (7.12) proves (1.27) in the intersection of
${\mathcal{D}}$ and the high-temperature region.

To see see that (1.27) holds in ${\mathbb{P}}$-probability when replacing the
assumption of almost sure convergence in (7.11) by convergence in
${\mathbb{P}}$-probability, simply recall that almost sure convergence implies
convergence in probability, and that if two sequences converge in probability,
then so does their sum. ∎

###### Proof of Lemma 7.1.

By the definition and notation (1.1) and (1.3) of Section 1.1,

$Z_{N,{\beta},h}=\sum_{{\sigma}\in\Sigma_{N}}e^{\frac{1}{2}N{\beta}^{2}(1-q)}e^{\frac{{\beta}}{2}\left({\sigma},A{\sigma}\right)+h(1,{\sigma})}.$
(7.16)

Next, using the identity $A={\mathcal{O}}^{t}{\Lambda}{\mathcal{O}}$ to
express the quadratic form in (7.16) yields

$e^{\frac{{\beta}}{2}\left({\sigma},A{\sigma}\right)}=\prod_{j=1}^{N}e^{\frac{{\beta}{\lambda}_{j}}{2}({\mathcal{O}}{\sigma})_{j}^{2}}.$
(7.17)

Eq. (7.17) can be further expressed applying the Hubbard-Stratonovich
transformation which, for any pair of real numbers ${\alpha}$ and $y$, is
defined through the identity

$e^{{\alpha}\frac{y^{2}}{2}}=\int_{-\infty}^{+\infty}\frac{dx}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}-\sqrt{{\alpha}}xy},$
(7.18)

where $\sqrt{{\alpha}}$ is as in (7.1). To check (7.18), simply note that for
${\alpha}<0$, it is the Fourier transform of a Gaussian density while for
${\alpha}\geq 0$, it is its two-sided Laplace transform. Eq. (7.17) can thus
be rewritten as

$e^{\frac{{\beta}}{2}\left({\sigma},A{\sigma}\right)}=\prod_{j=1}^{N}\int_{-\infty}^{+\infty}\frac{dx_{j}}{\sqrt{2\pi}}e^{-\frac{1}{2}x_{j}^{2}-\sqrt{{\lambda}_{j}}({\mathcal{O}}{\sigma})_{j}x_{j}}.$
(7.19)

Next observe that

$\textstyle\sum_{j=1}^{N}\sqrt{{\lambda}_{j}}({\mathcal{O}}{\sigma})_{j}x_{j}=\sum_{j^{\prime}=1}^{N}{\sigma}_{j^{\prime}}({\mathcal{O}}^{t}\sqrt{{\Lambda}}x)_{j^{\prime}}=\sum_{j^{\prime}=1}^{N}{\sigma}_{j^{\prime}}(\sqrt{A}{\mathcal{O}}^{t}x)_{j^{\prime}}.$
(7.20)

By this and the change of variable $x\mapsto{\mathcal{O}}x$, (7.19) becomes

$e^{\frac{{\beta}}{2}\left({\sigma},A{\sigma}\right)}=\int_{-\infty}^{+\infty}\frac{dx_{1}}{\sqrt{2\pi}}\dots\int_{-\infty}^{+\infty}\frac{dx_{N}}{\sqrt{2\pi}}e^{-\frac{1}{2}\sum_{j=1}^{N}x_{j}^{2}-\sum_{j=1}^{N}{\sigma}_{j}(\sqrt{A}x)_{j}}.$
(7.21)

Inserting (7.21) in (7.16), the summation in ${\sigma}$ is easily carried out
and we get

$Z_{N,{\beta},h}=e^{\frac{1}{2}N{\beta}^{2}(1-q)}\int_{{\mathbb{R}}^{N}}\left(\prod_{j=1}^{N}\frac{dx_{j}}{\sqrt{2\pi}}\right)\,\Xi_{N,{\beta},h}(x),$
(7.22)

where $\Xi_{N,{\beta},h}:{\mathbb{R}}^{N}\rightarrow{\mathbb{C}}$ is defined
by

$\Xi_{N,{\beta},h}(x)=2^{N}e^{-\frac{1}{2}\sum_{j=1}^{N}x_{j}^{2}}\prod_{j=1}^{N}\cosh\left(\sqrt{{\beta}}(\sqrt{A}x)_{j}+h\right).$
(7.23)

We can view the function $\Xi_{N,{\beta},h}$ as the restriction to the
hyperplane ${\mathbb{R}}^{N}$ of a holomorphic function defined on the whole
of ${\mathbb{C}}^{N}$. Eq. (7.3) then follows from (7.23) and the identity
$z=e^{\log z}$, $z\neq 0$, where the log function is defined in the principal
branch. ∎

###### Proof of Lemma 7.2.

The claim of the lemma follows in a straightforward way by differentiation of
${\mathcal{F}}_{N,{\beta},h}$. ∎

###### Proof of Proposition 7.4.

Choosing $x^{*}=x^{*}(z)$ in Lemma 7.3, item (i) follows from Lemma 7.1, the
identity (7.4) and the definition (7.6). As seen in the proof of Proposition
2.1, the assumption in item (ii) that $\Psi_{N,{\beta},{\bf{h}}}$ attains its
global maximum at a point $z^{*}$ in
$\rm{int}(\rm{dom\,}\Psi_{N,{\beta},{\bf{h}}})=(-1,1)^{N}$ guarantees that
$z^{*}$ is a critical point of $\Psi_{N,{\beta},{\bf{h}}}(z)$. By Proposition
2.1, $z^{*}$ is also a critical point of $\Phi_{N,{\beta},{\bf{h}}}$ and by
(2.4), $\Phi_{N,{\beta},{\bf{h}}}(z^{*})=\Psi_{N,{\beta},{\bf{h}}}(z^{*})$.
This proves (7.8). As explained in the remark at the end of Section 2.1, below
(2.5), $\Phi_{N,{\beta},{\bf{h}}}$ is unbounded. Thus,
$\Psi_{N,{\beta},{\bf{h}}}$ cannot be replaced with
$\Phi_{N,{\beta},{\bf{h}}}$ in the right-hand side of (7.8). ∎

###### Proof of Lemma 7.3.

The probabilistic part of the statement of the lemma serves to guarantee, as
Theorem 4.3 permits, that

$\max_{j}|{\lambda}_{j}|\leq 2(1+o(1))+{\beta}|1-q|\equiv c_{q}.$ (7.24)

From now on, we place ourselves on the set of ${\mathbb{P}}$-probability one
for which Theorem 4.3 is obtained, and assume that $N$ is sufficiently large
for (7.24) to be satisfied.

Clearly, by the change of variable $x+a\mapsto x$, it suffices to prove (7.5)
for pure imaginary vectors $x^{*}=ib$. Since $\Xi_{N,{\beta},h}(x)$ is
holomorphic and continuous, it follows from Osgood’s lemma that it is
holomorphic in each variable separately. We can thus apply the one variable
Cauchy-Goursat integral formula to each of the variables $x_{j}$,
$j=1,2,\dots,N$, successively.

We begin with the variable $x_{1}$. Let
$\theta_{1}(x_{1}):{\mathbb{C}}\mapsto{\mathbb{C}}$ be the function defined by
integrating $\Xi_{N,{\beta},h}$ over all variables except $x_{1}$, which is
kept fixed

$\theta_{1}(x_{1})=\int_{{\mathbb{R}}^{N-1}}\left(\prod_{j=2}^{N}\frac{dx_{j}}{\sqrt{2\pi}}\right)\,\Xi_{N,{\beta},h}(x).$
(7.25)

According to the Cauchy-Goursat integral formula

$\int_{{\mathcal{C}}_{1}}\frac{dx_{1}}{\sqrt{2\pi}}\theta_{1}(x_{1})=0$ (7.26)

where, writing $x_{1}=u_{1}+iv_{1}$, ${\mathcal{C}}_{1}$ is the rectangular
closed path in the plane $(u_{1},v_{1})$ defined as the boundary of the
rectangle of vertices $A=(-R_{N},0)$, $B=(R_{N},0)$, $C=(R_{N},b_{1})$, and
$D(-R_{N},b_{1})$, oriented counter-clockwise. Here $R_{N}=NR$, $R>0$, and
$b_{1}>0$ so that the rectangle lies in the upper half-plane (the case
$b_{1}<0$, which corresponds to a rectangle in the lower half-plane, is
treated in the same way). The left-hand side of (7.26) naturally decomposes
into four integrals, each of them along the path that follows a given side of
the rectangle

$\int_{{\mathcal{C}}_{1}}\frac{dx_{1}}{\sqrt{2\pi}}\theta_{1}(x_{1})={\mathcal{I}}_{AB}+{\mathcal{I}}_{BC}+{\mathcal{I}}_{CD}+{\mathcal{I}}_{DA}.$
(7.27)

Let us establish that

$\lim_{R\rightarrow\infty}{\mathcal{I}}_{BC}\equiv\int_{R_{N}+i0}^{R_{N}+ib_{1}}\frac{dx_{1}}{\sqrt{2\pi}}\theta_{1}(x_{1})=0.$
(7.28)

By a change of variables

${\mathcal{I}}_{BC}=2^{N}\int_{0}^{b_{1}}\frac{dv_{1}}{\sqrt{2\pi}}e^{-\frac{1}{2}(R_{N}+iv_{1})^{2}}\int_{{\mathbb{R}}^{N-1}}\left(\prod_{j=2}^{N}\frac{du_{j}}{\sqrt{2\pi}}\right)e^{-\frac{1}{2}\sum_{j=2}^{N}u_{j}^{2}}\Theta(u+iv)$
(7.29)

where $u=(R_{N},u_{2},\dots,u_{N})$ and $v=(v_{1},0,\dots,0)$ are vectors in
${\mathbb{R}}^{N}$ and

$\Theta(u+iv)=\prod_{j=1}^{N}\cosh\left(\sqrt{{\beta}}(\sqrt{A}(u+iv))_{j}+h\right).$
(7.30)

Taking the modulus,

$\left|{\mathcal{I}}_{BC}\right|\leq
2^{N}\int_{0}^{b_{1}}\frac{dv_{1}}{\sqrt{2\pi}}e^{-\frac{1}{2}(R^{2}_{N}-v_{1}^{2})}\int_{{\mathbb{R}}^{N-1}}\left(\prod_{j=2}^{N}\frac{du_{j}}{\sqrt{2\pi}}\right)e^{-\frac{1}{2}\sum_{j=2}^{N}u_{j}^{2}}\left|\Theta(u+iv)\right|.$
(7.31)

To bound $\left|\Theta(u+iv)\right|$, first note that

$\left|\cosh(u_{0}+iv_{0})\right|=\left|\cosh(u_{0})\right|\sqrt{\cos^{2}(v_{0})+\tanh^{2}(u_{0})\sin^{2}(v_{0})}\leq\left|\cosh(u_{0})\right|\leq
e^{\left|u_{0}\right|}.$

Now, writing $\sqrt{A}=U+iV$ where
$U={\mathcal{O}}^{t}\Re(\sqrt{{\Lambda}}){\mathcal{O}}$ and
$V={\mathcal{O}}^{t}\Im(\sqrt{{\Lambda}}){\mathcal{O}}$, and using the above
bound

$\left|\Theta(u+iv)\right|\leq
e^{\sqrt{{\beta}}\sum_{j=1}^{N}\left|(Uu)_{j}-(Vv)_{j}\right|+Nh}.$ (7.32)

Observing that $UV=VU=0$,

$\displaystyle\textstyle\sum_{j=1}^{N}\left|(Uu)_{j}-(Vv)_{j}\right|$
$\displaystyle\leq$
$\displaystyle\textstyle\sqrt{\sum_{j=1}^{N}((Uu)_{j}-(Vv)_{j})^{2}}$
$\displaystyle=$ $\displaystyle\textstyle\sqrt{(uU^{2}u)+(vV^{2}v)}$
$\displaystyle\leq$
$\displaystyle\textstyle\sqrt{\max_{j}|{\lambda}_{j}|(\left\|u\right\|_{2}^{2}+\left\|v\right\|_{2}^{2})}\leq\textstyle\sqrt{\max_{j}|{\lambda}_{j}|}(\left\|u\right\|_{1}+\left\|v\right\|_{1})$

where $\max_{j}|{\lambda}_{j}|$ is bounded in (7.24) and, by definition of $u$
and $v$, $\left\|u\right\|_{1}=R_{N}+\sum_{j=2}^{N}|u_{j}|$ and
$\left\|v\right\|_{1}=b_{1}$. Thus, recalling the definition of $c_{q}$ from
(7.24)

$\left|\Theta(u+iv)\right|\leq
e^{\sqrt{{\beta}c_{q}}\left(R_{N}+\sum_{j=2}^{N}|u_{j}|+b_{1}\right)+Nh}.$
(7.34)

Inserting this bound in (7.31),

$\displaystyle\left|{\mathcal{I}}_{BC}\right|$ $\displaystyle\leq$
$\displaystyle
2^{N}\int_{0}^{b_{1}}\frac{dv_{1}}{\sqrt{2\pi}}e^{-\frac{1}{2}(R^{2}_{N}-v_{1}^{2})+\sqrt{{\beta}c_{q}}(R_{N}+b_{1})+Nh}\prod_{j=2}^{N}\int_{{\mathbb{R}}}\frac{du_{j}}{\sqrt{2\pi}}e^{-\frac{1}{2}u_{j}^{2}+\sqrt{{\beta}c_{q}}|u_{j}|}$
(7.35) $\displaystyle\leq$
$\displaystyle\frac{2b_{1}}{\sqrt{2\pi}}e^{-\frac{1}{2}(R^{2}_{N}-b_{1}^{2})+\sqrt{{\beta}c_{q}}(R_{N}+b_{1})+Nh}\left(4e^{\frac{1}{2}{\beta}c_{q}}\right)^{N-1},$

and taking the limit $R\rightarrow\infty$ of both sides of (7.35), we obtain
(7.28). We prove in exactly the same way that

$\lim_{R\rightarrow\infty}{\mathcal{I}}_{DA}\equiv\int_{-R_{N}+ib_{1}}^{-R_{N}+i0}\frac{dx_{1}}{\sqrt{2\pi}}\theta_{1}(x_{1})=0.$
(7.36)

Passing to the limit $R\rightarrow\infty$ in (7.27), it follows from (7.26),
(7.28) and (7.36) that

$\int_{-\infty}^{\infty}\frac{dx_{1}}{\sqrt{2\pi}}\theta_{1}(x_{1})=\int_{-\infty}^{\infty}\frac{dx_{1}}{\sqrt{2\pi}}\theta_{1}(x_{1}+ib_{1}).$
(7.37)

By definition of $\theta_{1}(x_{1})$ (see (7.25)), setting
$b^{(1)}=(b_{1},0,\dots,0)$, this is equivalent to

$\int_{{\mathbb{R}}^{N}}\left(\prod_{j=1}^{N}\frac{dx_{j}}{\sqrt{2\pi}}\right)\,\Xi_{N,{\beta},h}(x)=\int_{{\mathbb{R}}^{N}}\left(\prod_{j=1}^{N}\frac{dx_{j}}{\sqrt{2\pi}}\right)\,\Xi_{N,{\beta},h}(x+ib^{(1)}).$
(7.38)

To deal with the next variable, $x_{2}$, we start from the right-hand side of
(7.38) and, in complete analogy to (7.25), we let
$\theta_{2}(x_{2}):{\mathbb{C}}\rightarrow{\mathbb{C}}$ be the function
defined by integrating $\Xi_{N,{\beta},h}$ over all variables except $x_{2}$,
which is kept fixed

$\theta_{2}(x_{2})=\int_{{\mathbb{R}}^{N-1}}\left(\prod_{1\leq j\leq N:j\neq
2}\frac{dx_{j}}{\sqrt{2\pi}}\right)\,\Xi_{N,{\beta},h}(x+b^{(1)}).$ (7.39)

We then consider the Cauchy-Goursat integral formula

$\int_{{\mathcal{C}}_{2}}\frac{dx_{1}}{\sqrt{2\pi}}\theta_{2}(x_{2})=0$ (7.40)

where ${\mathcal{C}}_{2}$ is defined as ${\mathcal{C}}_{1}$, replacing $b_{1}$
by $b_{2}$, and use it, proceeding exactly as in the proof of (7.28) to prove
that

$\int_{-\infty}^{\infty}\frac{dx_{2}}{\sqrt{2\pi}}\theta_{2}(x_{2})=\int_{-\infty}^{\infty}\frac{dx_{2}}{\sqrt{2\pi}}\theta_{2}(x_{2}+ib_{2}).$
(7.41)

We omit the details of the straightforward adaptation of the bounds
(7.31)-(7.35). Setting $b^{(2)}=(b_{1},b_{2},\dots,0)$, (7.41) is equivalent
to

$\int_{{\mathbb{R}}^{N}}\left(\prod_{j=1}^{N}\frac{dx_{j}}{\sqrt{2\pi}}\right)\,\Xi_{N,{\beta},h}(x)=\int_{{\mathbb{R}}^{N}}\left(\prod_{j=1}^{N}\frac{dx_{j}}{\sqrt{2\pi}}\right)\,\Xi_{N,{\beta},h}(x+ib^{(2)}).$
(7.42)

Iterating this procedure over the variable $x_{j}$, we obtain (7.5). Lemma 7.3
is proven. ∎

We conclude this section with the

###### Proofs of Theorem 1.9 and 1.2.

Theorem 1.9 follows from Theorem 1.5 and Theorem 1.8. Theorem 1.2 follows from
Theorem 1.9. ∎

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