# Existence and Uniqueness of Permutation-Invariant Optimizers for Parisi
Formula

Victor Issa  0009-0009-1304-046X Department of Mathematics, ENS de Lyon, Lyon,
France<EMAIL_ADDRESS>
###### Abstract.

It has recently been shown in [4] that, upon constraining the system to stay
in a balanced state, the Parisi formula for the mean-field Potts model can be
written as an optimization problem over permutation-invariant functional order
parameters.

In this paper, we focus on permutation-invariant mean-field spin glass models.
After introducing a correction term in the definition of the free energy and
without constraining the system, we show that the limit free energy can be
written as an optimization problem over permutation-invariant functional order
parameters. We also show that for some models this optimization problem admits
a unique optimizer. In the case of Ising spins, the correction term can be
easily removed, and those results transfer to the uncorrected limit free
energy.

We also derive an upper bound for the limit free energy of some nonconvex
permutation-invariant models. This upper bound is expressed as a variational
formula and is related to the solution of some Hamilton-Jacobi equation. We
show that if no first order phase transition occurs, then this upper bound is
equal to the lower bound derived in [17]. We expect that this hypothesis holds
at least in the high temperature regime.

Our method relies on the fact that the free energy of any convex mean-field
spin glass model can be interpreted as the strong solution of some Hamilton-
Jacobi equation.

Keywords and phrases: multi-species spin glass, Hamilton-Jacobi equations,
free energy, Parisi formula.

MSC 2020: 35D40, 60K35, 82B44, 82D30.

###### Contents

  1. 1 Introduction
    1. 1.1 Preamble
    2. 1.2 Main results
    3. 1.3 Motivations
    4. 1.4 Organization of the paper
    5. 1.5 Acknowledgement
  2. 2 The enriched free energy
  3. 3 Hamilton-Jacobi equations on closed convex cones
  4. 4 Permutation-invariant objects
    1. 4.1 Permutation-invariant matrices
    2. 4.2 Permutation-invariant paths
    3. 4.3 Piecewise linear approximations of paths
  5. 5 The free energy and Hamilton-Jacobi equations
    1. 5.1 The equation for models with matrix valued paths
    2. 5.2 The equation for models with vector valued paths
  6. 6 Uniqueness of the optimizer for diagonal models
  7. 7 Upper-bound for nonconvex models
    1. 7.1 An inequality for permutation-invariant covariance functions
    2. 7.2 Positivity principle for multiple species
    3. 7.3 The interpolation
    4. 7.4 Connections with the Hamilton-Jacobi approach

##  1\. Introduction

###  1.1. Preamble

Let $D>1$ be an integer, that we will keep fixed throughout the paper. We
study Gaussian processes $(H_{N}(\sigma))_{\sigma\in\mathbb{R}^{D\times N}}$,
whose covariance is of the form

(1.1)
$\mathbb{E}[H_{N}(\sigma)H_{N}(\tau)]=N\xi\left(\frac{\sigma\tau^{*}}{N}\right).$

Here $\sigma\tau^{*}=(\sigma_{d}\cdot\tau_{d^{\prime}})_{1\leqslant
d,d^{\prime}\leqslant D}$, $x\cdot y$ is the standard scalar product on
$\mathbb{R}^{N}$ and $\xi:\mathbb{R}^{D\times D}\to\mathbb{R}$ is a function
given by an absolutely convergent power series. Unless stated otherwise, we
will always assume that $\xi$ is convex on the set of positive semi-definite
matrices. We give ourselves for each $N$ a reference probability measure
$P_{N}$ on $\mathbb{R}^{D\times N}$ of the form $P_{N}=P_{1}^{\otimes N}$
where $P_{1}$ is a compactly supported probability measure on
$\mathbb{R}^{D}$. For every $t\geqslant 0$, One can associate a random
probability measure on $\mathbb{R}^{D\times N}$ to the process $H_{N}$ called
the Gibbs measure and denoted by $\langle\cdot\rangle$. It is defined by

$\langle h(\sigma)\rangle=\frac{\int
h(\sigma)\exp\left(\sqrt{2t}H_{N}(\sigma)-Nt\xi\left(\frac{\sigma\sigma^{*}}{N}\right)\right)\mathrm{d}P_{N}(\sigma)}{\int\exp\left(\sqrt{2t}H_{N}(\sigma)-Nt\xi\left(\frac{\sigma\sigma^{*}}{N}\right)\right)\mathrm{d}P_{N}(\sigma)}.$

An important step, in understanding the Gibbs measure associated to the family
of processes $(H_{N})_{N}$ is the computation of the large $N$ limit of the
free energy

(1.2)
$\overline{F}_{N}(t)=-\frac{1}{N}\mathbb{E}\log\int\exp\left(\sqrt{2t}H_{N}(\sigma)-Nt\xi\left(\frac{\sigma\sigma^{*}}{N}\right)\right)\mathrm{d}P_{N}(\sigma).$

For the models of interest here, a variational formula for the limiting value
of $\overline{F}_{N}(t)$ is known, this is the celebrated Parisi formula. The
Parisi formula was first conjectured in [24] using a sophisticated non-
rigorous argument now referred to as the replica method. The convergence of
the free energy as $N\to+\infty$ was rigorously established in [14] in the
case of the so-called Sherrington-Kirkpatrick model which corresponds to
$D=1$, $\xi(x)=x^{2}$ and $P_{1}=\text{Unif}(\\{-1,1\\})$. The Parisi formula
for the Sherrington-Kirkpatrick model was then proven in [13, 26]. This was
extended to the case $D=1$, $P_{1}=\text{Unif}(\\{-1,1\\})$ and
$\xi(x)=\sum_{p\geqslant 1}a_{p}x^{p}$ with $a_{p}\geqslant 0$ in [20]. Some
models with $D>1$ such as multispecies models, the Potts model, and a general
class of models with vector spins were treated in [21, 22, 23], under the
assumption that $\xi$ is convex on $\mathbb{R}^{D\times D}$. Finally, the case
$D>1$ was treated in general in [7] assuming only that that $\xi$ is convex on
the set of positive semi-definite matrices. The following version of the
Parisi formula is [7, Corollary 8.2].

######  Theorem 1.1 ([7]).

If $\xi$ is convex on $S^{D}_{+}$, then for every $t>0$,

(1.3)
$\lim_{N\to+\infty}\overline{F}_{N}(t)=\sup_{\mathfrak{q}}\left\\{\psi(\mathfrak{q})-t\int_{0}^{1}\xi^{*}\left(\frac{\mathfrak{q}(u)}{t}\right)\mathrm{d}u\right\\}.$

Here $S^{D}_{+}$ denotes the set of positive semi-definite symmetric matrices
in $\mathbb{R}^{D\times D}$. We equip $\mathbb{R}^{D\times D}$ with the order
$A\leqslant B$ if and only $B-A\in S^{D}_{+}$. The supremum in (1.3) is taken
over the set of bounded functions in

(1.4) $\mathcal{Q}(S^{D}_{+})=\\{\mathfrak{q}:[0,1)\to
S^{D}_{+}\big{|}\;\text{$\mathfrak{q}$ is càdlàg and nondecreasing}\\}.$

The function $\psi$ is the cascade transform of the measure $P_{1}$ and
$\xi^{*}$ is the convex conjugate of $\xi$ with respect to the cone
$S^{D}_{+}$. We postpone the precise definition of these objects to (2.5) and
(3.5) respectively.

Let $\mathscr{S}_{D}$ denote the group of permutations of $\\{1,\dots,D\\}$,
we say that $\xi$ is permutation-invariant and $P_{1}$ is permutation-
invariant when for every permutation $s\in\mathscr{S}_{D}$,

(1.5) $\forall R\in\mathbb{R}^{D\times
D},\;\xi(R)=\xi\left((R_{s(d),s(d^{\prime})})_{1\leqslant
d,d^{\prime}\leqslant D}\right),$ (1.6)
$\forall\chi\in\mathcal{C}_{b}(\mathbb{R}^{D}),\;\int\chi(x_{1},\dots,x_{D})\mathrm{d}P_{1}(x)=\int\chi(x_{s(1)},\dots,x_{s(D)})\mathrm{d}P_{1}(x).$

We will show that when $\xi$ and $P_{1}$ are permutation-invariant, the
supremum in (1.3) can be taken over the set of permutation-invariant paths in
$\mathcal{Q}(S^{D}_{+})$. This result can be interpreted as an absence of
breaking of permutation-invariance by the system. This persistence of
permutation-invariance was predicted to happen for the Potts model (see (1.7)
below) in [11]. It has been rigorously proven that the Potts model does not
break its permutation invariance, but only after constraining the system to
stay in a balanced state [4] or by introducing a correction term of the form
$-Nt\xi\left(\frac{\sigma\sigma^{*}}{N}\right)$ [5]. In (1.2) the correction
term of [5] is also present, but our result is general enough to cover models
where $P_{1}$ is supported on $\\{-1,1\\}^{D}$ and $\xi$ only depends on the
diagonal coefficients of its argument, for those models the correction term is
constant and our main results transfer to the uncorrected free energy.

###  1.2. Main results

Let $\mathcal{Q}(\mathbb{R}_{+})$ denote the set of càdlàg and nondecreasing
functions $[0,1)\to\mathbb{R}_{+}$. Let $U$ denote a uniform random variable
in $[0,1)$, given $p_{1},p_{2}\in\mathcal{Q}(\mathbb{R}_{+})$, we let
$(p_{1},p_{2})^{\perp}:[0,1)\to S^{D}_{+}$ be defined by

$(p_{1},p_{2})^{\perp}=p_{1}\left(\text{id}_{D}-\frac{\mathbf{1}_{D}}{D}\right)+p_{2}\frac{\mathbf{1}_{D}}{D},$

where $\text{id}_{D}$ denotes the $D\times D$ identity matrix and
$\mathbf{1}_{D}$ the $D\times D$ matrix whose coefficients are all equal to
$1$. As will be proven in Section 4, every permutation-invariant path in
$\mathcal{Q}(S^{D}_{+})$ is of the form $(p_{1},p_{2})^{\perp}$.

Heuristically, given a maximizing path $\mathfrak{q}$ in (1.3), the law of the
matrix $\mathfrak{q}(U)$ is the limiting distribution of
$\frac{\sigma\tau^{*}}{N}$ where $\sigma$ and $\tau$ are two independent
random variables with law $\langle\cdot\rangle$. For some specific models, the
distribution of the overlap matrix has some additional properties, and those
additional properties allow us to write the limit free energy as an
optimization over a smaller set of paths. For example, consider the Potts
model which corresponds to the family of processes

(1.7)
$H^{\text{Potts}}_{N}(\sigma)=\frac{1}{\sqrt{N}}\sum_{i,j=1}^{N}J_{ij}\sigma_{i}\cdot\sigma_{j}.$

Here $(J_{ij})_{i,j\geqslant 1}$ denotes a family of independent standard
Gaussian random variables. The process $H_{N}^{\text{Potts}}$ satisfies (1.1)
with $\xi(R)=\sum_{d,d^{\prime}=1}^{D}R^{2}_{dd^{\prime}}$. Often, the Potts
model is considered with reference measure
$P_{1}=\text{Unif}\\{e_{1},\dots,e_{D}\\}$, where $(e_{1},\dots,e_{D})$ denote
the canonical basis of $\mathbb{R}^{D}$. With this assumption, the terms of
the form $\sigma_{i}\cdot\sigma_{j}$ appearing in (1.7) only take the value
$0$ or $1$. In this case, if we sample two independent random variables
$\sigma,\tau\in\mathbb{R}^{D\times N}$ with law $\langle\cdot\rangle$, their
overlap matrix $R=\frac{\sigma\tau^{*}}{N}$ satisfies
$\sum_{d,d^{\prime}=1}^{D}R_{dd^{\prime}}=1$ almost surely and for every
$s\in\mathscr{S}_{D}$, the matrices $R$ and
$(R_{s(d)s(d^{\prime})})_{1\leqslant d,d^{\prime}\leqslant D}$ are equal in
law under $\mathbb{E}\langle\cdot\rangle$. This means that if $\mathfrak{q}$
is a maximizing path for the Parisi formula of the Potts model, we should
expect that

$\mathfrak{q}=p\left(\text{id}_{D}-\frac{1_{D}}{D}\right)+\frac{\mathbf{1}_{D}}{D^{2}}=\left(p,\frac{1}{D}\right)^{\perp}$

for some $p\in\mathcal{Q}(\mathbb{R}_{+})$. This observation on the set of
optimal paths in (1.3) for the Potts model was leveraged in [5, 4]. In [4],
the authors show that when the system is constrained to stay in a balanced
state, the limit free energy of the Potts model can be written a supremum over
$\mathcal{Q}(\mathbb{R}_{+})$. In [5], the author does not constrain the
system but introduces a correction term of the form
$-Nt\xi\left(\frac{\sigma\sigma^{*}}{N}\right)$ like in (1.2) and obtains
results similar to [4].

In this paper, we show that similar results can be obtained in different
settings. We will focus on permutation-invariant models, that is models with
$\xi$ and $P_{1}$ satisfying (1.5) and $\eqref{e. P1 permutation invariance}$.
We will show that for those models, the following variational formula holds.

######  Theorem 1.2.

Assume that $\xi$ is convex on $S^{D}_{+}$ and permutation-invariant, assume
that $P_{1}$ is permutation-invariant. Then, for every $t\geqslant 0$,

(1.8)
$\begin{split}\lim_{N\to+\infty}\overline{F}_{N}(t)=\sup_{(p_{1},p_{2})}&\inf_{(r_{1},r_{2})}\Biggl{\\{}\psi((p_{1},p_{2})^{\perp})\\\
-\langle p_{1},r_{1}\rangle_{L^{2}}&-\langle
p_{2},r_{2}\rangle_{L^{2}}+t\int_{0}^{1}\xi\left(\left(\frac{p_{1}(u)}{D-1},p_{2}(u)\right)^{\perp}\right)\mathrm{d}u\Biggl{\\}}.\end{split}$

Where the supremum and the infimum are taken over
$(\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty})^{2}$.

Similarly to the Potts model, at the heuristic level (1.8) follows from the
fact that given two independent random variables $\sigma$ and $\tau$ under
$\langle\cdot\rangle$, for every $s\in\mathscr{S}_{D}$, the matrices
$\frac{\sigma\tau^{*}}{N}$ and
$\left(\frac{\sigma_{s(d)}\cdot\tau_{s(d^{\prime})}}{N}\right)_{1\leqslant
d,d^{\prime}\leqslant D}$ are equal in law under
$\mathbb{E}\langle\cdot\rangle$. This means that every maximizing path should
be of the form $\mathfrak{q}=(p_{1},p_{2})^{\perp}$ where
$p_{1},p_{2}\in\mathcal{Q}(\mathbb{R}_{+})$ and the limit free energy should
be a supremum over $\mathcal{Q}(\mathbb{R}_{+})^{2}$.

We will also consider permutation-invariant models, where the interaction
function $\xi$ is further assumed to only depend on
$\frac{\sigma_{1}\cdot\tau_{1}}{N},\dots,\frac{\sigma_{D}\cdot\tau_{D}}{N}$.
In this case, we identify the map $R\mapsto\xi(R)$ defined on
$\mathbb{R}^{D\times D}$ with the map $x\mapsto\xi(\text{diag}(x))$ defined on
$\mathbb{R}^{D}$. Since $\xi$ only depends on the diagonal coefficients of its
argument, we only need to keep track of the diagonal of the overlap matrix and
the limit free energy can be written as a supremum over
$\mathcal{Q}(\mathbb{R}_{+})^{D}$. Performing the same heuristic computation
as above, we then expect that under our permutation invariance assumptions,
the limit free energy can in fact be written as a supremum over
$\mathcal{Q}(\mathbb{R}_{+})$. We will show that this is indeed the case.

Given a maximizing path $r$ of the Parisi formula, the law of the random
variable $r(U)$ is called a Parisi measure. When $D=1$, it is known that there
exists a unique Parisi measure [1]. In this case, the proof relies on a strict
concavity property of the Parisi functional. However, when $D>1$, for
technical reasons this strict concavity property does not carry over well to
$\mathcal{Q}(S^{D}_{+})$ or $\mathcal{Q}(\mathbb{R}_{+})^{D}$. But, in (1.9)
below, the Parisi formula is written as a maximization over
$\mathcal{Q}(\mathbb{R}_{+})$, the set of $1$-dimensional paths. Thanks to
this, we can proceed as in [1] to prove uniqueness of Parisi measures.

######  Theorem 1.3.

Assume that $\xi$ is convex on $S^{D}_{+}$, permutation-invariant and only
depends on the diagonal coefficients of its argument, assume that $P_{1}$ is
permutation-invariant. Then, for every $t>0$,

(1.9)
$\lim_{N\to+\infty}\overline{F}_{N}(t)=\sup_{p\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\left\\{\psi(p\text{id}_{D})-t\int_{0}^{1}\xi^{*}\left(\frac{p(u)\text{id}_{D}}{t}\right)\mathrm{d}u\right\\}.$

In addition, the supremum in (1.9) is reached at a unique
$p^{*}\in\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty}$.

Let us now comment on the inclusion of the term
$-Nt\xi\left(\frac{\sigma\sigma^{*}}{N}\right)$ in the definition of the free
energy (1.2). Consider again the Potts model, that is
$\xi(R)=\sum_{d,d^{\prime}=1}^{D}R^{2}_{dd^{\prime}}$ and
$P_{1}=\text{Unif}\\{e_{1},\dots,e_{D}\\}$. For any
$\sigma\in\\{e_{1},\dots,e_{D}\\}^{N}$, we have
$\frac{\sigma\sigma^{*}}{N}=\text{diag}(\alpha_{1},\dots,\alpha_{D})$ where

$\alpha_{d}=\frac{1}{N}\\#\\{i\leqslant N,\sigma_{i}=e_{d}\\}.$

So, $\xi(\sigma\sigma^{*}/N)=\sum_{d=1}^{D}\alpha^{2}_{d}$ with
$\sum_{d=1}^{D}\alpha_{d}=1$ and $\alpha_{d}\geqslant 0$. This means that for
the Potts model, the value of the correction term
$-Nt\xi\left(\frac{\sigma\sigma^{*}}{N}\right)$ is minimal on configurations
of the form $\sigma=(e_{d},\dots,e_{d})$ and maximal on configurations
satisfying $\alpha_{d}=\frac{1}{D}$. In particular, at least in the case of
the Potts model, even though we are not constraining the system to stay in a
balanced state, the correction term favors configurations
$\sigma\in\mathbb{R}^{D\times N}$ which are balanced. Thankfully, this is not
always the case and for some models the correction term
$-Nt\xi\left(\frac{\sigma\sigma^{*}}{N}\right)$ can be removed. For example,
if we assume that $\xi$ only depends on the diagonal coefficients of its
argument and $P_{1}$ is supported on $\\{-1,1\\}^{D}$, then the correction
term is constant. In this case, the variational formula (1.9) can be rewritten
as a variational formula for the uncorrected free energy.

Let $\alpha\geqslant 1$, for $\sigma\in\mathbb{R}^{2\times N}$ define

$H_{N}^{\text{BP+SK}}(\sigma)=\frac{1}{\sqrt{N}}\sum_{i,j=1}^{N}J^{11}_{ij}\sigma_{1i}\sigma_{1j}+\frac{1}{\sqrt{N}}\sum_{i,j=1}^{N}J^{22}_{ij}\sigma_{2i}\sigma_{2j}+\frac{1}{\sqrt{N}}\sum_{i,j=1}^{N}J^{12}_{ij}\sigma_{1i}\sigma_{2j},$

where $J^{11}=(J^{11}_{ij})_{i,j\geqslant 1}$ and
$J^{22}=(J^{22}_{ij})_{i,j\geqslant 1}$ are independent families of
independent centered Gaussian random variables with variance $\alpha/2$ and
$(J^{12}_{ij})_{i,j\geqslant 1}$ is a family of independent Gaussian random
variables with variance $1$ and independent of $J^{11}$ and $J^{22}$. The
process $H_{N}^{\text{BP+SK}}$ satisfies (1.1) with,

$\xi^{BP+SK}(R)=\frac{\alpha}{2}R_{11}^{2}+\frac{\alpha}{2}R_{22}^{2}+R_{11}R_{22}.$

Choosing $P_{1}=\text{Unif}(\\{-1,1\\}^{2})$ as the reference probability
measure, Theorem 1.3 can be applied to $H_{N}^{\text{BP+SK}}$ to discover that
the system does not break its permutation invariance. One can wonder what
happens when $0\leqslant\alpha<1$, in this case $\xi^{BP+SK}$ is nonconvex on
$S^{2}_{+}$ and there is no known generalization of Theorem 1.1. In Section 7,
we will discuss some results that are applicable in this case.

When $\xi$ is not assumed to be convex on $S^{D}_{+}$, to the best of our
knowledge, there is no proof of the fact that $\overline{F}_{N}(t)$ converges
as $N\to+\infty$. Building upon an interpolation argument developed in [2,
Section 2.1], we will show that for permutation-invariant models where $\xi$
only depends on the diagonal coefficients of its argument and is nonconvex on
$S^{D}_{+}$, the right-hand side in (1.9) is an upper bound on the $\limsup$
of the free energy.

######  Theorem 1.4.

Assume that $\xi$ is permutation-invariant and only depends on the diagonal
coefficients of its argument, assume that $P_{1}$ is permutation-invariant.
Then, even when $\xi$ is nonconvex, we have for every $t>0$,

$\limsup_{N\to+\infty}\overline{F}_{N}(t)\leqslant\sup_{p\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\left\\{\psi(p\text{id}_{D})-t\int_{0}^{1}\Xi^{*}\left(\frac{p(u)\text{id}_{D}}{t}\right)\mathrm{d}u\right\\},$

where $\Xi(x)=\frac{1}{D}\sum_{d=1}^{D}\xi(x_{d},\dots,x_{d})$.

In addition, we show that if no first order phase transition occurs, then the
upper bound of Theorem 1.4 matches the lower bound derived in [17, 15]. The no
first order phase transition hypothesis is formally written in
(H${}_{\xi,P_{1}}$) below, note that this hypothesis is true when $\xi$ is
convex. We refer to Theorem 7.10 below for a formal version of Theorem 1.5.

######  Theorem 1.5.

Assume that the hypotheses of Theorem 1.4 hold and further assume that no
first order phases transition occurs, then for every $t>0$,
$\overline{F}_{N}(t)$ converges as $N\to+\infty$ and

$\lim_{N\to+\infty}\overline{F}_{N}(t)=\sup_{p\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\left\\{\psi(p\text{id}_{D})-t\int_{0}^{1}\Xi^{*}\left(\frac{p(u)\text{id}_{D}}{t}\right)\mathrm{d}u\right\\},$

where $\Xi(x)=\frac{1}{D}\sum_{d=1}^{D}\xi(x_{d},\dots,x_{d})$.

As explained in Remark 7.12 below, it is plausible that there exists $t_{c}>0$
such that no first order phase transition occurs on $[0,t_{c})$. In this case,
the argument used to prove Theorem 1.5 allows to identify the limit free
energy on $[0,t_{c}]$.

###  1.3. Motivations

Most of the proof we derive below rely on the following result, which we state
informally.

######  Theorem 1.6 ([16, 7]).

Assume that $\xi$ is convex on $S^{D}_{+}$, then for every $t\geqslant 0$,

$\lim_{N\to+\infty}\overline{F}_{N}(t)=f(t,0),$

where $f:\mathbb{R}_{+}\times\mathcal{Q}(S^{D}_{+})\to\mathbb{R}$ is the
unique solution of

(1.10) $\begin{cases}\partial_{t}f-\int\xi(\nabla f)=0\text{ on
}\mathbb{R}_{+}\times\mathcal{Q}(S^{D}_{+})\\\ f(0,\cdot)=\psi\text{ on
}\mathcal{Q}(S^{D}_{+}).\end{cases}$

In Section 2 we will give a precise definition of $f$ and $\psi$, we will also
explain the precise meaning of (1.10). In the context of Theorem 1.2 and
Theorem 1.3, this approach trough partial differential equations does not seem
to possess any clear advantage when compared with the methods developed in
[5]. The true power of Theorem 1.6 is revealed when dealing with nonconvex
models. When $\xi$ is nonconvex, the Parisi formula completely breaks down. To
the best of our knowledge, until recently, it seems that there was no clear
conjecture on what the limit of the free energy should be in this case. In
[16, Conjecture 2.6], it is proposed that results such as Theorem 1.6 should
generalize to nonconvex models. Further development in [17, 15] have led to
the following lower bound for the free energy of nonconvex models.

######  Theorem 1.7 ([17, 15]).

For every $t\geqslant 0$,

$\liminf_{N\to+\infty}\overline{F}_{N}(t)\geqslant f(t,0),$

where $f:\mathbb{R}_{+}\times\mathcal{Q}(S^{D}_{+})\to\mathbb{R}$ is the
unique solution of

(1.11) $\begin{cases}\partial_{t}f-\int\xi(\nabla f)=0\text{ on
}\mathbb{R}_{+}\times\mathcal{Q}(S^{D}_{+})\\\ f(0,\cdot)=\psi\text{ on
}\mathcal{Q}(S^{D}_{+}).\end{cases}$

To identify the limit free energy of nonconvex models, a possible approach is
to prove an upper bound for $\limsup_{N\to+\infty}\overline{F}_{N}(t)$ and
compare it with $f(t,0)$. In Section 7, we will show that

$f(t,0)\leqslant\liminf_{N\to+\infty}\overline{F}_{N}(t)\leqslant\limsup_{N\to+\infty}\overline{F}_{N}(t)\leqslant
g(t,0),$

where $g$ is the solution of an equation similar to (1.10) (see (7.11) below).
We are able to phrase the upper bound this way thanks to the fact that the
proofs of Theorem 1.2 and Theorem 1.3 are formalized in the language of
partial differential equations. In particular, we will show that under some
unproven regularity assumption on $f$, $f(t,0)$ and $g(t,0)$ are equal. We
refer to Section 7.4 for a more precise discussion.

###  1.4. Organization of the paper

We only consider permutation-invariant models. That is, models with $\xi$
permutation-invariant and $P_{N}=P_{1}^{\otimes N}$ with $P_{1}$ permutation-
invariant and compactly supported. Without loss of generality, we will assume
that $P_{1}$ is supported inside the unit ball of $\mathbb{R}^{D}$. Excluding
Section 7, $\xi$ is always assumed to be convex on $S^{D}_{+}$.

In Section 2 we will give a precise definition of the functions $f$ and $\psi$
appearing in Theorem 1.6, we will also explain the precise meaning of (1.10).
In Section 3 we will define an appropriate notion of solution for (1.10)
following [10]. In particular, in Section 2 we introduce a variant of the free
energy $\overline{F}_{N}(t)$ which will depend on $t\in\mathbb{R}_{+}$ and an
extra parameter $\mathfrak{q}\in\mathcal{Q}(S^{D}_{+})$. This enriched version
of the free energy will be denoted $\overline{F}_{N}(t,\mathfrak{q})$ and
satisfies $\overline{F}_{N}(t,0)=\overline{F}_{N}(t)$. It is through the
introduction of this extra parameter that it is possible to obtain a partial
differential equation for the limiting objects. In section 4 we will introduce
some basic notations on permutation-invariant matrices and paths that will be
useful for the later sections. We make the elementary but important remark
that the set of permutation-invariant matrices is isomorphic to
$\mathbb{R}^{2}$ and the set of permutation-invariant vectors is isomorphic to
$\mathbb{R}$. The bulk of our analysis is in Section 5, in this section we
start by showing that the enriched free energy is permutation-invariant,
namely for every permutation $s\in\mathscr{S}_{D}$,

$\overline{F}_{N}(t,(\mathfrak{q}_{dd^{\prime}})_{1\leqslant d,d\leqslant
D})=\overline{F}_{N}(t,(\mathfrak{q}_{s(d)s(d^{\prime})})_{1\leqslant
d,d\leqslant D}).$

Thanks to this property and (1.10), we can write a partial differential
equation satisfied by
$(t,\mathfrak{q})\mapsto\lim_{N\to+\infty}\overline{F}_{N}(t,\mathfrak{q})$ on
the set of permutation-invariant paths. Since the set of permutation-invariant
paths in $\mathcal{Q}(S^{D}_{+})$ is isomorphic to
$\mathcal{Q}(\mathbb{R}^{2}_{+})$, we will deduce Theorem 1.2. Moreover, when
$\xi$ is further assumed to only depend on the diagonal coefficient of its
argument, it is even possible to write a partial differential equation on the
set of permutation-invariant paths in $\mathcal{Q}(\mathbb{R}_{+})^{D}$ which
is isomorphic to $\mathcal{Q}(\mathbb{R}_{+})$, this will yield the
variational formula (1.9) of Theorem 1.3. Using a strict concavity property of
the Parisi functional, in Section 6 we will deduce the existence and the
uniqueness of an optimizer in the variational formula (1.9). Finally, in
Section 7 we assume that $\xi$ is permutation-invariant and depends only on
the diagonal coefficients of its argument, but we do not assume that $\xi$ is
convex. We will show that under those hypotheses if $\xi$ admits an absolutely
convergent power series and satisfies (1.1), then

$\forall
x\in\mathbb{R}^{D}_{+},\;\xi(x)\leqslant\frac{1}{D}\sum_{d=1}^{D}\xi(x_{d},\dots,x_{d}).$

With this inequality, we will prove Theorem 1.4 via a simple interpolation
argument. To conclude, we will explore how this upper bound is related to the
lower bound for the free energy of nonconvex models derived in [17, 15] and we
will prove a rigorous version of Theorem 1.5, namely Theorem 7.10.

###  1.5. Acknowledgement

I warmly thank Jean-Christophe Mourrat for countless fascinating discussions
on spin glasses and many helpful comments and insights during this project.

##  2\. The enriched free energy

The method we propose relies on the fact that the limit free energy
$f(t)=\lim\overline{F}_{N}(t)$ solves a partial differential equation. At this
stage, $f$ only depends on one parameter, in order to obtain a partial
differential equation, we are going to introduce an extra parameter. The
nature of this parameter will not be the same for models that satisfy (1.1)
and models that satisfy (1.1) under the additional assumption that $\xi$ only
depends on the diagonal coefficients of its argument. We will provide the
construction for the first class of models and then explain the additional
adjustments that can be made when $\xi$ only depends on the diagonal
coefficients of its argument.

Here $S^{D}_{+}$ denotes the set of positive semi-definite symmetric matrices
in $\mathbb{R}^{D\times D}$ and $S^{D}_{++}$ the set of positive symmetric
matrices in $\mathbb{R}^{D\times D}$. We equip $R^{D\times D}$ with the order
$A\leqslant B$ if and only $B-A\in S^{D}_{+}$. The extra parameter we
introduce in the definition of the free energy belongs to the space

(2.1) $\mathcal{Q}(S^{D}_{+})=\\{\mathfrak{q}:[0,1)\to
S^{D}_{+}\big{|}\;\text{$\mathfrak{q}$ is càdlàg and nondecreasing}\\}.$

In general paths in $\mathcal{Q}(S^{D}_{+})$ are not defined at $u=1$, but if
$\mathfrak{q}\in\mathcal{Q}(S^{D}_{+})\cap L^{\infty}$ we can set

$\mathfrak{q}(1)=\lim_{u\uparrow 1}\mathfrak{q}(u)\in S^{D}_{+}.$

We let $\mathfrak{R}$ be a Poisson Dirichlet cascade, briefly $\mathfrak{R}$
is a specific random probability measure on the unit sphere of a separable
Hilbert space $(\mathfrak{H},\wedge)$, such that given two independent random
variables $\alpha,\alpha^{\prime}\in\mathfrak{H}$ sampled from $\mathfrak{R}$,
the random variable $\alpha\wedge\alpha^{\prime}$ is uniformly distributed in
$[0,1]$ (see [20, Section 2] and [7, Section 4]). We let
$\langle\cdot\rangle_{\mathfrak{R}}$ denote the expectation with respect to
$\mathfrak{R}^{\otimes\mathbb{N}}$ and $\mathfrak{U}$ the support of
$\mathfrak{R}$. By construction of $\mathfrak{R}$, almost surely
$\mathfrak{U}$ is an ultrametric set, that is almost surely for every
$\alpha^{1},\alpha^{2},\alpha^{3}\in\mathfrak{U}$,

$\alpha^{1}\wedge\alpha^{3}\geqslant\min\left(\alpha^{1}\wedge\alpha^{2},\alpha^{2}\wedge\alpha^{3}\right).$

Let $\mathfrak{q}\in\mathcal{Q}(S^{D}_{+})\cap L^{\infty}$, according to [7,
Propsoition 4.1] $\mathfrak{R}$-almost surely, there exists a
$\mathbb{R}^{D}$-valued centered Gaussian process
$(w^{\mathfrak{q}}(\alpha))_{\alpha\in\mathfrak{U}}$ such that

$\mathbb{E}\left[w^{\mathfrak{q}}(\alpha)\left(w^{\mathfrak{q}}(\alpha^{\prime})\right)^{*}\right]=\mathfrak{q}(\alpha\wedge\alpha^{\prime}).$

Let $N\geqslant 1$, and consider
$(W_{N}^{\mathfrak{q}}(\alpha))_{\alpha\in\mathfrak{U}}$ be a
$(\mathbb{R}^{D})^{N}$-valued random process whose coordinates are independent
and have the same law as $(w^{\mathfrak{q}}(\alpha))_{\alpha\in\mathfrak{U}}$.
We set

(2.2)
$F_{N}(t,\mathfrak{q})=-\frac{1}{N}\log\iint\exp(H_{N}^{t,\mathfrak{q}}(\sigma,\alpha))dP_{N}(\sigma)d\mathfrak{R}(\alpha),$

where

$H_{N}^{t,\mathfrak{q}}(\sigma,\alpha)=\sqrt{2t}H_{N}(\sigma)-tN\xi\left(\frac{\sigma\sigma^{*}}{N}\right)+\sqrt{2}W_{N}^{\mathfrak{q}}(\alpha)\cdot\sigma-\mathfrak{q}(1)\cdot\sigma\sigma^{*}.$

We let $\overline{F}_{N}(t,\mathfrak{q})=\mathbb{E}F_{N}(t,\mathfrak{q})$, the
function
$\overline{F}_{N}:\mathbb{R}_{+}\times\left(\mathcal{Q}(S^{D}_{+})\cap
L^{\infty}\right)\to\mathbb{R}$ is Lipschitz, more precisely according to [7,
Proposition 5.1], we have

(2.3)
$|\overline{F}_{N}(t,\mathfrak{q})-\overline{F}_{N}(t^{\prime},\mathfrak{q}^{\prime})|\leqslant|\mathfrak{q}-\mathfrak{q}^{\prime}|_{L^{1}}+|t-t^{\prime}|\sup_{|a|\leqslant
1}|\xi(a)|.$

Here, $|a|$ denotes the Frobenius norm of $a\in\mathbb{R}^{D\times D}$, that
is the norm associated to the scalar product

$a\cdot b=\sum_{d,d^{\prime}=1}^{D}a_{dd^{\prime}}b_{dd^{\prime}}.$

Using (2.3), we can then extend $\overline{F}_{N}$ by continuity to
$\mathbb{R}_{+}\times\left(\mathcal{Q}(S^{D}_{+})\cap L^{1}\right)$. Note that
if $0$ denotes the null path, we have
$\overline{F}_{N}(t,0)=\overline{F}_{N}(t)$. So the function we have defined
is indeed an extension of the free energy.

Let us now introduce the notion of differentiability that we will use to
formulate the partial differential equation (1.10). Given
$\mathfrak{q}\in\mathcal{Q}(S^{D}_{+})\cap L^{2}$, we let

$\text{Adm}(\mathfrak{q})=\\{\kappa\in L^{2}\big{|}\;\exists r>0,\;\forall
t\in[0,r],\;\mathfrak{q}+t\kappa\in\mathcal{Q}(S^{D}_{+})\cap L^{2}\\}.$

The set $\text{Adm}(\mathfrak{q})$ is the set of admissible directions at
$\mathfrak{q}$. Let $g:\mathcal{Q}(S^{D}_{+})\cap L^{2}\to\mathbb{R}$, we say
that $g$ is Gateaux differentiable at
$\mathfrak{q}\in\mathcal{Q}(S^{D}_{+})\cap L^{2}$ when for every
$\kappa\in\text{Adm}(\mathfrak{q})$, the following limit exists

$g^{\prime}(\mathfrak{q};\kappa)=\lim_{t\downarrow
0}\frac{g(\mathfrak{q}+t\kappa)-g(\mathfrak{q})}{t},$

and there exists a unique $d\in L^{2}$ such that for every
$\kappa\in\text{Adm}(\mathfrak{q})$,

$g^{\prime}(\mathfrak{q};\kappa)=\langle d,\kappa\rangle_{L^{2}}.$

In this case, the vector $d\in L^{2}$ is denoted $\nabla g(\mathfrak{q})$ and
we call it the Gateaux derivative of $g$ at $\mathfrak{q}$. Note that $\nabla
g(\mathfrak{q})\in L^{2}$, so we can make sense of expressions of the form
$\int_{0}^{1}\xi(\nabla g(\mathfrak{q})(u))\mathrm{d}u$, provided that the
integral converges. Such expressions will be simply abbreviated by
$\int\xi(\nabla g(\mathfrak{q}))$ in what follows.

We define $\mathcal{Q}_{\uparrow}(S^{D}_{+})$ as the set of paths
$\mathfrak{q}\in\mathcal{Q}(S^{D}_{+})\cap L^{2}$ such that
$\mathfrak{q}(0)=0$, and there exists a constant $c>0$ such that
$u\mapsto\mathfrak{q}(u)-cu\text{id}_{D}$ is nondecreasing and for all $u<v$,

$\text{Ellipt}(\mathfrak{q}(v)-\mathfrak{q}(u))\leqslant\frac{1}{c}.$

Here, for $m\in S^{D}_{++}$, $\text{Ellipt}(m)$ denotes the ratio of the
biggest and smallest eigenvalue of $m$. When $\xi$ is convex, which will
always be the case except in Section 7, it can be shown that
$\overline{F}_{N}$ converges pointwise to some Lipschitz function
$f:\mathbb{R}_{+}\times\left(\mathcal{Q}(S^{D}_{+})\cap
L^{1}\right)\to\mathbb{R}$. Furthermore, according to [7, Propositions 7.2 &
8.6], $f:\mathbb{R}_{+}\times(\mathcal{Q}(S^{D}_{+})\cap L^{2})\to\mathbb{R}$
is Gateaux differentiable on
$(0,+\infty)\times(\mathcal{Q}_{\uparrow}(S^{D}_{+})\cap L^{\infty})$ and
satisfies

(2.4) $\begin{cases}\partial_{t}f-\int\xi(\nabla f)=0&\text{ on
}(0,+\infty)\times(\mathcal{Q}_{\uparrow}(S^{D}_{+})\cap L^{\infty})\\\
f(0,\cdot)=\psi&\text{ on }\mathcal{Q}_{\uparrow}(S^{D}_{+})\cap
L^{\infty},\end{cases}$

where we have defined $\psi=\lim_{N\to+\infty}\overline{F}_{N}(0,\cdot)$. In
fact, since we have assumed that $P_{N}=P_{1}^{\otimes N}$ we simply have

(2.5) $\psi=\overline{F}_{1}(0,\cdot).$

The fact that $f$ is a solution of (2.4) at every point where it is
differentiable is a consequence of the cavity computations conducted in [7]
(see [7, Propostion 7.2]), this result is robust and holds true even when
$\xi$ is nonconvex. The real miracle is the fact that $f$ is differentiable at
every (nice) points, this is proven in [7, Propostion 8.6] and relies strongly
on the fact that $\xi$ is convex on $S^{D}_{+}$. We stress that in general,
nonlinear partial differential equations such as (2.4) do not always have a
differentiable solution. Therefore, to study (2.4) and other similar
equations, we will need to appeal to the notion of viscosity solution. this
notion will be introduced precisely in Section 3. Also note, that according to
[7, Proposition 3.6], we have $\nabla
f(t,\mathfrak{q})\in\mathcal{Q}(S^{D}_{+})\cap L^{2}$. This means that the
behavior of (2.4) is only governed by the restriction of $\xi$ to $S^{D}_{+}$,
this is why we can assume only that $\xi$ is convex on $S^{D}_{+}$.

In this framework, the Parisi formula can be seen as a consequence of a
variational representation for the viscosity solution of (2.4). According to
[9, Theorem 4.6 (2)], for
$(t,\mathfrak{q})\in\mathbb{R}_{+}\times\mathcal{Q}(S^{D}_{+})\cap L^{2}$, we
have

$f(t,\mathfrak{q})=\sup_{\mathfrak{p}\in\mathcal{Q}(S^{D}_{+})\cap
L^{\infty}}\inf_{\mathfrak{r}\in\mathcal{Q}(S^{D}_{+})\cap
L^{\infty}}\left\\{\psi(\mathfrak{q}+\mathfrak{p})-\langle\mathfrak{p},\mathfrak{r}\rangle_{L^{2}}+t\int_{0}^{1}\xi(\mathfrak{r}(u))\mathrm{d}u\right\\}.$

At $\mathfrak{q}=0$ this yields,
$\lim_{N\to+\infty}\overline{F}_{N}(t)=\sup_{\mathfrak{p}\in\mathcal{Q}(S^{D}_{+})}\mathcal{P}_{t,\xi,P_{1}}(\mathfrak{p})$,
where

(2.6)
$\mathcal{P}_{t,\xi,P_{1}}(\mathfrak{p})=\psi(\mathfrak{p})-\sup_{\mathfrak{r}\in\mathcal{Q}(S^{D}_{+})\cap
L^{\infty}}\left\\{\langle\mathfrak{p},\mathfrak{r}\rangle_{L^{2}}-t\int_{0}^{1}\xi(\mathfrak{r}(u))\mathrm{d}u\right\\}.$

As explained in the proof of [7, Proposition 8.1], one can recover the “usual”
Parisi functional from $\mathcal{P}_{t,\xi,P_{1}}$ by plugging in paths of the
form $\nabla\xi\circ\mathfrak{p}$. If we define
$\theta(a)=a\cdot\nabla\xi(a)-\xi(a)$, we have

$\mathcal{P}_{t,\xi,P_{1}}(t\nabla\xi\circ\mathfrak{p})=\psi(t\nabla\xi\circ\mathfrak{p})-t\int_{0}^{1}\theta(\mathfrak{p}(u))\mathrm{d}u.$

Before moving to the next section, we explain the adjustments that can be made
to obtain a simpler Hamilton-Jacobi equation when $\xi$ is assumed to only
depend on the diagonal coefficients of its argument. Henceforth, we will refer
to models with this additional property as diagonal models. For those models,
we identify the function $A\mapsto\xi(A)$ defined on $\mathbb{R}^{D\times D}$
and the function $x\mapsto\xi(\text{diag}(x))$ defined on $\mathbb{R}^{D}$.
Recall that heuristically, the paths $\mathfrak{q}$ are to be understood as
encoding the limiting distribution of the overlap matrix,
$\frac{\sigma\tau^{*}}{N}$. When the model is diagonal, we do not need to keep
track of the full overlap matrix and encoding the distribution of the diagonal
coefficients is enough. To this end, let us introduce another space of paths,

$\mathcal{Q}(\mathbb{R}^{D}_{+})=\\{q:[0,1)\to\mathbb{R}_{+}^{D}\big{|}\;\text{$q$
is càdlàg nondecreasing}\\}.$

Here, by $q$ is nondecreasing we mean that for every $u\leqslant v$,
$q(v)-q(u)\in\mathbb{R}^{D}_{+}$. Note that $\mathcal{Q}(\mathbb{R}^{D}_{+})$
is isomorphic to the subset of $\mathcal{Q}(S^{D}_{+})$ composed of the paths
that are valued in the set of diagonal matrices. If
$q\in\mathcal{Q}(\mathbb{R}^{D}_{+})$ we denote by $\text{diag}(q)$ the
associated diagonal matrix valued path. We then define
$\mathcal{Q}_{\uparrow}(\mathbb{R}^{D}_{+})$ as the set of paths
$q\in\mathcal{Q}(\mathbb{R}^{D}_{+})$ such that the path $\text{diag}(q)$
belongs to $\mathcal{Q}_{\uparrow}(S^{D}_{+})$. In order to not get mixed up
with the previous definition of enriched free energy, we will denote by
$\overline{F}^{\text{diag}}_{N}$ the restriction of $\overline{F}_{N}$ to
$\mathcal{Q}(\mathbb{R}^{D}_{+})$. That is, for every
$(t,q)\in\mathbb{R}_{+}\times(\mathcal{Q}(\mathbb{R}^{D}_{+})\cap L^{1})$,

$\overline{F}^{\text{diag}}_{N}(t,q)=\overline{F}_{N}(t,\text{diag}(q)).$

As previously, the sequence $(\overline{F}^{\text{diag}}_{N}(t,q))_{N}$
converges to $f(t,\text{diag}(q))$. We let
$f^{\text{diag}}(t,q)=f(t,\text{diag}(q))$, the function $f^{\text{diag}}$ is
Gateaux differentiable on
$(0,+\infty)\times\left(\mathcal{Q}_{\uparrow}(\mathbb{R}^{D}_{+})\cap
L^{\infty}\right)$ and solves

(2.7) $\begin{cases}\partial_{t}f^{\text{diag}}-\int\xi(\nabla
f^{\text{diag}})=0&\text{ on
}(0,+\infty)\times(\mathcal{Q}_{\uparrow}(\mathbb{R}^{D}_{+})\cap
L^{\infty})\\\ f^{\text{diag}}(0,\cdot)=\psi^{\text{diag}}&\text{ on
}\mathcal{Q}_{\uparrow}(\mathbb{R}^{D}_{+})\cap L^{\infty},\end{cases}$

where we have defined $\psi^{\text{diag}}(q)=\psi(\text{diag}(q))$. This can
be checked by differentiating $f^{\text{diag}}$ and using (2.4). The point of
this derivation is that we can now express the limit free energy of diagonal
models as the value at $(t,0)$ of the solution of a Hamilton-Jacobi equation
on $\mathcal{Q}(\mathbb{R}^{D}_{+})$ rather than $\mathcal{Q}(S^{D}_{+})$.
This yields a different variational representation for the limit free energy.

##  3\. Hamilton-Jacobi equations on closed convex cones

As explained above, Hamilton-Jacobi equations on closed convex cones, such as
(2.4) and (2.7), will play an important role in this paper. Here, we summarize
some known results on those equations. We will focus on the theory of
Hamilton-Jacobi equations on closed convex cones in finite dimensional Hilbert
spaces. But, most of the result exposed below remain valid in infinite
dimensional Hilbert spaces [9]. We will use both settings in the following
sections.

Let $(\mathcal{H},\langle\cdot,\cdot\rangle_{\mathcal{H}})$ be a finite
dimensional Hilbert space. Let $\mathcal{C}\subseteq\mathcal{H}$, we say that
$\mathcal{C}$ is a closed convex cone when $\mathcal{C}$ is a closed set and
for every $s,t\geqslant 0$, $x,y\in\mathcal{C}$, we have

$sx+ty\in\mathcal{C}.$

In what follows, we fix a nonempty closed convex cone
$\mathcal{C}\subseteq\mathcal{H}$. We will assume that the interior of
$\mathcal{C}$ is nonempty and that the only vector belonging to $\mathcal{C}$
and $-\mathcal{C}$ simultaneously is $0$. When dealing with infinite
dimensional Hilbert spaces like in [9], $\mathcal{C}$ is allowed to have empty
interior, but for simplicity we do not consider this possibility in this
section.

Let $\psi:\mathcal{C}\to\mathbb{R}$ be a Lipschitz function and
$H:\mathcal{H}\to\mathbb{R}$ be such that $H\big{|}_{\mathcal{C}}$ is locally
Lipschitz. We are interested in equations of the form

(3.1) $\begin{cases}\partial_{t}v-H(\nabla v)=0\text{ on
}(0,+\infty)\times\mathring{\mathcal{C}}\\\ v(0,\cdot)=\psi\text{ on
}\mathcal{C}.\end{cases}$

When $\mathcal{C}=\mathbb{R}^{d}$, there is already a rich theory for
equations like (3.1) [12, Section 10]. One of the main accomplishments of this
theory is the introduction of a suitable notion of solution. Equation (3.1)
may not have any differentiable solution [12, Section 3.3 Example 6]. But, by
introducing the notion of viscosity solutions (see Definition 3.1 below), it
is possible to guarantee existence and uniqueness of solutions under some mild
conditions, provided that we allow non-differentiable functions to solve
(3.1).

Most of the theory of viscosity solutions on $\mathbb{R}^{d}$ can be adapted
for Hamilton-Jacobi equations on closed convex cones [10]. In principle, when
$\mathbb{R}^{d}$ is replaced by $\mathcal{C}$, some boundary conditions should
be enforced to guarantee that (3.1) admits a unique solution. But this
requirement can be bypassed provided that some monotony conditions on $\psi$
and $H$ hold.

Given $\mathcal{D}\subseteq\mathcal{C}$, a function
$\phi:(0,+\infty)\times\mathcal{D}\to\mathbb{R}$ is said to be differentiable
at $(t,x)\in\mathcal{(}0,+\infty)\times\mathcal{D}$ when there exists a unique
$(a,p)\in\mathbb{R}\times\mathcal{H}$ such that the following holds as
$(s,y)\in(0,+\infty)\times\mathcal{D}$ tends to $(t,x)$,

$\phi(s,y)=\phi(t,x)+(s-t)a+\langle
x-y,p\rangle_{\mathcal{H}}+O\left(|s-t|^{2}+|x-y|^{2}_{\mathcal{H}}\right).$

In this case, $(a,p)$ is denoted $(\partial_{t}\phi(t,x),\nabla\phi(t,x))$.
When $\phi$ is differentiable at every point in $(0,+\infty)\times\mathcal{D}$
and the function $(t,x)\mapsto(\partial_{t}\phi(t,x),\nabla\phi(t,x))$ is
continuous, we say that $\phi$ is smooth. Note that the notions of
differentiability and smoothness defined here make sense even when $x$ does
not belong to the interior of $\mathcal{C}$ and does not requires
$\mathcal{D}$ to be an open set.

######  Definition 3.1 (Viscosity solutions).

  1. (1)

An upper semi-continuous function
$v:\mathbb{R}_{+}\times\mathcal{C}\to\mathbb{R}$ is a viscosity subsolution of
(3.1) when for every $(t,x)\in(0,+\infty)\times\mathring{\mathcal{C}}$ and
every smooth function
$\phi:(0,+\infty)\times\mathring{\mathcal{C}}\to\mathbb{R}$ such that $v-\phi$
has a local maximum at $(t,x)$, we have

$\partial_{t}\phi(t,x)-H(\nabla\phi(t,x))\leqslant 0.$

  2. (2)

A lower semi-continuous function
$v:\mathbb{R}_{+}\times\mathcal{C}\to\mathbb{R}$ is a viscosity supersolution
of (3.1) when for every $(t,x)\in(0,+\infty)\times\mathring{\mathcal{C}}$ and
every smooth function
$\phi:\mathbb{R}_{+}\times\mathring{\mathcal{C}}\to\mathbb{R}$ such that
$v-\phi$ has a local minimum at $(t,x)$, we have

$\partial_{t}\phi(t,x)-H(\nabla\phi(t,x))\geqslant 0.$

  3. (3)

A continuous function $v:\mathbb{R}_{+}\times\mathcal{C}\to\mathbb{R}$ is a
viscosity solution when it is both a viscosity subsolution and a viscosity
supersolution.

We define the dual cone $\mathcal{C}^{*}$ of the cone $\mathcal{C}$ by
setting,

(3.2) $\mathcal{C}^{*}=\\{x\in\mathcal{H}\big{|}\;\forall
y\in\mathcal{C},\;\langle x,y\rangle_{\mathcal{H}}\geqslant 0\\}.$

The dual cone $\mathcal{C}^{*}$ is a closed convex cone and satisfies
$(\mathcal{C}^{*})^{*}=\mathcal{C}$. Given
$\mathcal{A},\mathcal{B}\subseteq\mathcal{H}$, we say that
$g:\mathcal{A}\to\mathbb{R}$ is $\mathcal{B}$-nondecreasing when for every
$x,x^{\prime}\in\mathcal{A}$, if $x-x^{\prime}\in\mathcal{B}$ then
$g(x)\geqslant g(x^{\prime})$. Note that, if $h:\mathcal{C}\to\mathbb{R}$ is
differentiable, then $h$ is $\mathcal{C}^{*}$-nondecreasing if and only if
$\nabla h(\mathcal{C})\subseteq\mathcal{C}$. We define
$\mathcal{V}(\mathcal{C})$ to be the set of functions
$v:\mathbb{R}_{+}\times\mathcal{C}\to\mathbb{R}$, such that for every
$t\geqslant 0$, $v(t,\cdot)$ is $\mathcal{C}^{*}$-nondecreasing and such that
the following estimates hold

(3.3) $\sup_{\begin{subarray}{c}t>0\\\
x\in\mathcal{C}\end{subarray}}\frac{|v(0,x)-v(t,x)|}{t}<+\infty\text{ and
}\sup_{t>0}\|v(t,\cdot)\|_{\text{Lip}}<+\infty.$

Here, $\|v(t,\cdot)\|_{\text{Lip}}$ denotes the optimal Lipschitz constant of
$v(t,\cdot)$. The following theorem is extracted from [10, Theorem 1.2].

######  Theorem 3.2 ([10]).

Let $\psi:\mathcal{C}\to\mathbb{R}$ be a Lipschitz and
$\mathcal{C}^{*}$-nondecreasing function and let $H:\mathcal{H}\to\mathbb{R}$
be such that $H\big{|}_{\mathcal{C}}$ is $\mathcal{C}^{*}$-nondecreasing and
locally Lipschitz. Then, the Hamilton-Jacobi equation

(3.4) $\begin{cases}\partial_{t}v-H(\nabla v)=0\text{ on
}(0,+\infty)\times\mathring{\mathcal{C}}\\\ v(0,\cdot)=\psi\text{ on
}\mathcal{C},\end{cases}$

admits a unique viscosity solution in $\mathcal{V}(\mathcal{C})$ that will be
denoted $v$.

Theorem 3.2 ensures that (3.1) is well posed as long as $\psi$ and $H$ are
$\mathcal{C}^{*}$-nondecreasing. When $H$ is convex and some additional
requirements are put on the cone $\mathcal{C}$ it is possible to obtain an
explicit representation for the unique viscosity solution. We refer to this
representation as the Hopf-Lax representation of the viscosity solution.

Given $g:\mathcal{C}\to\mathbb{R}\cup\\{+\infty\\}$, we define the (monotone)
convex conjugate of $g$ over $\mathcal{C}$ by

(3.5) $g^{*}(y)=\sup_{x\in\mathcal{C}}\left\\{\langle
x,y\rangle_{\mathcal{H}}-g(x)\right\\}.$

When $\mathcal{C}=\mathbb{R}^{d}$, the functions that satisfy $g^{**}=g$ are
exactly the lower semicontinuous convex functions [25, Section 12]. If we let

$g^{**}(x)=\sup_{y\in\mathcal{C}^{*}}\left\\{\langle
x,y\rangle_{\mathcal{H}}-g(y)\right\\}$

denote the convex conjugate of $g^{*}$ over $\mathcal{C}^{*}$, we have that
$g^{**}$ is lower semi-continuous, convex and $\mathcal{C}$-nondecreasing. So,
the functions satisfying $g=g^{**}$ must be lower semicontinuous, convex and
$\mathcal{C}$-nondecreasing.

######  Definition 3.3.

We say that a closed convex cone $\mathcal{C}$ has the Fenchel-Moreau property
when for every $g:\mathcal{C}\to\mathbb{R}\cup\\{+\infty\\}$ not identically
equal to $+\infty$, we have $g^{**}=g$ if and only if $g$ is lower
semicontinuous, convex and $\mathcal{C}$-nondecreasing.

######  Theorem 3.4 ([10]).

Assume that $\mathcal{C}$ has the Fenchel-Moreau property. Let
$\psi:\mathcal{C}\to\mathbb{R}$ be a Lipschitz and
$\mathcal{C}^{*}$-nondecreasing function and let $H:\mathcal{H}\to\mathbb{R}$
be such that $H\big{|}_{\mathcal{C}}$ is $\mathcal{C}^{*}$-nondecreasing,
locally Lipschitz and _convex_. Then, the unique viscosity solution $v$ of
(3.4) admits the Hopf-Lax representation, that is for every
$(t,x)\in\mathbb{R}_{+}\times\mathcal{C}$,

$v(t,x)=\sup_{y\in\mathcal{C}}\inf_{z\in\mathcal{C}}\left\\{\psi(x+y)-\langle
y,z\rangle_{\mathcal{H}}+tH(z)\right\\}.$

Most of the cones that we will encounter in this document have the Fenchel-
Moreau property. Examples of cones with the Fenchel-Moreau property include
$\mathbb{R}^{D}_{+}$ and $S^{D}_{+}$. We refer to [8] for an in-depth study of
Fenchel-Moreau cones. Note that the cones $\mathcal{Q}(S^{D}_{+})$,
$\mathcal{Q}(\mathbb{R}^{D}_{+})$ and $\mathcal{Q}^{j}(S^{D}_{+})$,
$\mathcal{Q}^{j}(\mathbb{R}^{D}_{+})$ defined below are all Fenchel-Moreau
cones, for those cones the Hopf-Lax representation will therefore be
available. Note that when $\psi$ is assumed to be convex and $H$ is possibly
nonconvex, another variational representation for the viscosity solution is
available [10, Proposition 6.3]. Finally, we point out that, as proven in [10,
Theorem 1.2 (1)], the comparison principle also remains valid for (3.1). Note
that crucially, to use the comparison principle, the condition to be a
viscosity solution needs only to be checked on the interior of $\mathcal{C}$.
This means that in practice when dealing with solutions of (3.1), the boundary
points of $\mathcal{C}$ do not need to be considered.

######  Theorem 3.5 ([10]).

Let $u,v\in\mathcal{V}(\mathcal{C})$, assume that $u$ is a viscosity
subsolution and $v$ is a viscosity supersolution. Then,

$\sup_{(t,x)\in\mathbb{R}_{+}\times\mathcal{C}}\left\\{u(t,x)-v(t,x)\right\\}=\sup_{x\in\mathcal{C}}\left\\{u(0,x)-v(0,x)\right\\}.$

We conclude this section by using the comparison principle to prove a
stability result for approximate strong solutions of (3.1) on
$\mathring{\mathcal{C}}$.

######  Proposition 3.6.

Let $\psi:\mathcal{C}\to\mathbb{R}$ be a Lipschitz and
$\mathcal{C}^{*}$-nondecreasing function, and let $H:\mathcal{H}\to\mathbb{R}$
be such that $H\big{|}_{\mathcal{C}}$ is $\mathcal{C}^{*}$-nondecreasing and
locally Lipschitz. Let $g\in\mathcal{V}(\mathcal{C})$, assume that
$g(0,\cdot)=\psi$, that $g$ is differentiable on
$(0,+\infty)\times\mathring{\mathcal{C}}$ and that there exists a constant
$c>0$ such that

(3.6) $\sup_{t>0,x\in\mathring{\mathcal{C}}}|\partial_{t}g(t,x)-H(\nabla
g(t,x))|\leqslant c.$

Then, for every $t\geqslant 0$,
$\sup_{x\in\mathcal{C}}|v(t,x)-g(t,x)|\leqslant ct$. Where $v$ is the unique
viscosity solution of (3.4).

###### Proof.

Define $\widetilde{g}(t,x)=g(t,x)+ct$, we have
$\tilde{g}\in\mathcal{V}(\mathcal{C})$ and $\tilde{g}(0,\cdot)=\psi$. In
addition, $\tilde{g}$ is differentiable on
$(0,+\infty)\times\mathring{\mathcal{C}}$ and

$\partial_{t}\tilde{g}-H(\nabla\tilde{g})\geqslant 0\text{ on
}(0,+\infty)\times\mathring{\mathcal{C}}.$

Therefore $\tilde{g}$ is supersolution of (3.4) on $\mathring{\mathcal{C}}$.
According to the comparison principle [10, Theorem 1.2 (1)], we have
$v-\tilde{g}\leqslant 0$ on $\mathbb{R}_{+}\times\mathcal{C}$. So for every
$x\in\mathcal{C}$,

$v(t,\cdot)-g(t,\cdot)\leqslant ct.$

Similarly, we show that $(t,x)\mapsto g(t,x)-ct$ is a subsolution and use the
comparison principle to deduce $g(t,\cdot)-v(t,\cdot)\leqslant ct$ on
$\mathbb{R}_{+}\times\mathcal{C}$. Combining those two bounds, we obtain the
desired result. ∎

##  4\. Permutation-invariant objects

In this section, we introduce basic notations and results for matrix valued
paths. Those results will be useful in later sections for the analysis of
Hamilton-Jacobi equations.

###  4.1. Permutation-invariant matrices

Let $m\in\mathbb{R}^{D\times D}$, for every $s\in\mathscr{S}_{D}$ we define
$m^{s}=(m_{s(d),s(d^{\prime})})_{1\leqslant d,d^{\prime}\leqslant D}$. When
for every $s\in\mathscr{S}_{D}$, $m^{s}=m$ we say that $m$ is permutation-
invariant. An in depth study of permutation-invariant matrices is conducted in
[18], for the sake of completeness we extract the following result. Recall
that $\text{id}_{D}$ denotes the $D\times D$ identity matrix and
$\mathbf{1}_{D}$ denotes the $D\times D$ matrix whose coefficients are all
equal to $1$.

######  Proposition 4.1 ([18]).

Let $m\in\mathbb{R}^{D\times D}$ be permutation-invariant. We have

(4.1) $m=\begin{pmatrix}a&t&\cdots&\cdots&t\\\ t&\ddots&\ddots&&\vdots\\\
\vdots&\ddots&\ddots&\ddots&\vdots\\\ \vdots&&\ddots&\ddots&t\\\
t&\cdots&\cdots&t&a\\\
\end{pmatrix}=a\text{id}_{D}+t\left(\mathbf{1}_{D}-\text{id}_{D}\right),$

where $a=m_{11}$ and $t=m_{12}$. The matrix $m$ admits two eigenvalues
$\lambda_{1}=a-t$ and $\lambda_{2}=a-t+Dt$ (or $1$ eigenvalue when $t=0$) and
can be expressed in terms of its eigenvalues via

(4.2)
$\begin{split}m&=\frac{1}{D}\begin{pmatrix}\lambda_{2}+(D-1)\lambda_{1}&\lambda_{2}-\lambda_{1}&\cdots&\cdots&\lambda_{2}-\lambda_{1}\\\
\lambda_{2}-\lambda_{1}&\ddots&\ddots&&\vdots\\\
\vdots&\ddots&\ddots&\ddots&\vdots\\\
\vdots&&\ddots&\ddots&\lambda_{2}-\lambda_{1}\\\
\lambda_{2}-\lambda_{1}&\cdots&\cdots&\lambda_{2}-\lambda_{1}&\lambda_{2}+(D-1)\lambda_{1}\\\
\end{pmatrix}\\\
&=\lambda_{1}\left(\text{id}_{D}-\frac{\mathbf{1}_{D}}{D}\right)+\lambda_{2}\frac{\mathbf{1}_{D}}{D}.\end{split}$

###### Proof.

Let $m\in\mathbb{R}^{D\times D}$ be permutation-invariant, and let $a=m_{11}$
and $t=m_{12}$. Let $d,d^{\prime}\in\\{1,\dots,D\\}$, if $d\neq d^{\prime}$
there exists $s\in\mathscr{S}_{D}$, such that $(s(d),s(d^{\prime}))=(1,2)$,
since $m$ is permutation-invariant, this yields $m_{d,d^{\prime}}=t$.
Otherwise, $d=d^{\prime}$ and there exists $s\in\mathscr{S}_{D}$, such that
$(s(d),s(d^{\prime}))=(1,1)$, from which it follows that $m_{d,d}=a$. This
proves that $m=aI_{D}+t\left(\mathbf{1}_{D}-\text{id}_{D}\right)$. Moreover,
the characteristic polynomial of the matrix $\mathbf{1}_{D}$ is
$\chi_{\mathbf{1}_{D}}(X)=(X-D)X^{D-1}$. The eigenvalues of $m$ can be deduced
from the fact that if $t\neq 0$,
$\chi_{m}(X)=t^{D}\chi_{\mathbf{1}_{D}}\left(\frac{X-a+t}{t}\right)$. ∎

In what follows, for every $\lambda_{1},\lambda_{2}\in\mathbb{R}$, we will
denote by $m(\lambda_{1},\lambda_{2})$ the matrix
$\lambda_{1}\left(\text{id}_{D}-\frac{\mathbf{1}_{D}}{D}\right)+\lambda_{2}\frac{\mathbf{1}_{D}}{D}$.
According to Proposition 4.1, $m(\lambda_{1},\lambda_{2})$ is the only
$D\times D$ permutation-invariant matrix whose eigenvalues are $\lambda_{1}$
and $\lambda_{2}$. Recall that $\mathbb{R}^{D\times D}$ is equipped with the
Frobenius inner product

$m\cdot n=\sum_{d,d^{\prime}=1}^{D}m_{dd^{\prime}}n_{dd^{\prime}}.$

In particular, we have $m(\lambda)\cdot
m(\mu)=(D-1)\lambda_{1}\mu_{1}+\lambda_{2}\mu_{2}$. As per (3.5), we let
$\xi^{*}$ denote the convex conjugate of $\xi$ with respect to the cone
$S^{D}_{+}$, that is

$\xi^{*}(m)=\sup_{n\in S^{D}_{+}}\left\\{m\cdot n-\xi(n)\right\\}.$

For every $\lambda_{1},\lambda_{2}\in\mathbb{R}$, we define

(4.3)
$\xi_{\perp}(\lambda_{1},\lambda_{2})=\xi\left(m\left(\frac{\lambda_{1}}{D-1},\lambda_{2}\right)\right).$

We equip $\mathbb{R}^{D}$ with the standard inner product $x\cdot
y=\sum_{d=1}^{D}x_{d}y_{d}$, and let $\xi^{*}_{\perp}$ denote the convex
conjugate of $\xi_{\perp}$ with respect to $\mathbb{R}_{+}^{2}$,

$\xi_{\perp}^{*}(\lambda)=\sup_{\mu\in\mathbb{R}^{2}_{+}}\left\\{\lambda\cdot\mu-\xi_{\perp}(\mu)\right\\}.$

######  Proposition 4.2.

For every $\lambda_{1},\lambda_{2}\in\mathbb{R}$, we have

$\xi_{\perp}^{*}(\lambda_{1},\lambda_{2})=\xi^{*}\left(m\left(\lambda_{1},\lambda_{2}\right)\right).$

###### Proof.

Clearly,
$\xi^{*}(m(\lambda_{1},\lambda_{2}))\geqslant(\xi_{\perp})^{*}(\lambda_{1},\lambda_{2})$.
Conversely, given $n\in S^{D}_{+}$, we define
$n_{0}=\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}n^{s}$. The matrix $n_{0}$ is
permutation-invariant and positive semi-definite, by Proposition 4.1, there
exists $\mu_{1},\mu_{2}\in\mathbb{R}_{+}^{2}$ such that
$n_{0}=m\left(\mu_{1}/(D-1),\mu_{2}\right)$. Furthermore, since $\xi$ is
convex and permutation-invariant, we have

$\xi(n_{0})\leqslant\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}\xi(n^{s})=\xi(n).$

Therefore,

$\displaystyle m\left(\lambda_{1},\lambda_{2}\right)\cdot n-\xi(n)$
$\displaystyle=m\left(\lambda_{1},\lambda_{2}\right)\cdot n_{0}-\xi(n)$
$\displaystyle=\lambda\cdot\mu-\xi(n)$
$\displaystyle\leqslant\lambda\cdot\mu-\xi\left(m\left(\frac{\mu_{1}}{D-1},\mu_{2}\right)\right)$
$\displaystyle\leqslant(\xi_{\perp})^{*}(\lambda_{1},\lambda_{2}).$

Taking the supremum over $n\in S^{D}_{+}$, we obtain the desired inequality. ∎

###  4.2. Permutation-invariant paths

Let $V=S^{D}$ or $V=\mathbb{R}^{D}$, we equip $V$ with any norm $|\cdot|$. For
$p\in[1,+\infty)$, we denote by $L^{p}([0,1),V)$ or simply $L^{p}$ the set of
functions $h:[0,1)\to V$ such that $|h|^{p}$ is integrable. As usual,
functions in $L^{p}$ are to be understood as equivalence classes of functions
modulo equality almost everywhere. We define the $p$-norm on $L^{p}$, by

$|h|_{L^{p}}=\left(\int_{0}^{1}|h(u)|^{p}\mathrm{d}u\right)^{1/p}.$

The set $L^{\infty}$ is the set of essentially bounded functions, and we equip
it with the sup norm

$|h|_{L^{\infty}}=\text{ess-sup}_{u\in[0,1)}|h(u)|.$

When $V$ is equipped with a scalar product $x\cdot y$ and $|\cdot|$ is the
associated Euclidean norm, the norm $|\cdot|_{L^{2}}$ is Hilbertian and comes
from the scalar product

$\langle h,k\rangle_{L^{2}}=\int_{0}^{1}h(u)\cdot k(u)\mathrm{d}u.$

At fixed $p\in[1,+\infty]$, the different norms $|\cdot|_{L^{p}}$ obtained by
changing the norm $|\cdot|$ on $V$ are all equivalent. Therefore, the
statement $h\in L^{p}$ does not depend on the particular norm $|\cdot|$ we
choose, but the precise value of $|h|_{L^{p}}$ does. Given
$q\in\mathcal{Q}(\mathbb{R}^{2}_{+})$, we let
$q^{\perp}\in\mathcal{Q}(S^{D}_{+})$ be the path defined by
$q^{\perp}(u)=m(q(u))$. For $\mathfrak{q}\in\mathcal{Q}(S^{D}_{+})$ and
$s\in\mathscr{S}_{D}$, we define $\mathfrak{q}^{s}\in\mathcal{Q}(S^{D}_{+})$
by
$\mathfrak{q}^{s}(u)=(\mathfrak{q}(u))^{s}=(\mathfrak{q}_{s(d),s(d^{\prime})}(u))_{1\leqslant
d,d^{\prime}\leqslant D}$. As previously, we equip $S^{D}$ with the Frobenius
inner product and $\mathbb{R}^{2}$ with the standard inner product.

######  Proposition 4.3.

Let $\mathfrak{q}\in\mathcal{Q}(S^{D}_{+})$, assume that for every
$s\in\mathscr{S}_{D}$, we have $\mathfrak{q}^{s}=\mathfrak{q}$, then there
exists $q=(q_{1},q_{2})\in\mathcal{Q}(\mathbb{R}^{2}_{+})$ such that
$\mathfrak{q}=q^{\perp}$. In addition, for every $p\in[1,+\infty]$,
$\mathfrak{q}\in L^{p}$ if and only if $q\in L^{p}$. Moreover, when
$\mathfrak{q}\in L^{2}$, we have for every
$q^{\prime}=(q^{\prime}_{1},q^{\prime}_{2})\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{2}$,

$\langle\mathfrak{q},(q^{\prime})^{\perp}\rangle_{L^{2}}=\langle
q_{1},(D-1)q^{\prime}_{1}\rangle_{L^{2}}+\langle
q_{2},q^{\prime}_{2}\rangle_{L^{2}}.$

###### Proof.

Let $u\in[0,1)$, for every $s\in\mathscr{S}_{D}$, the matrix
$\mathfrak{q}(u)\in S^{D}_{+}$ satisfies
$(\mathfrak{q}(u))^{s}=\mathfrak{q}(u)$. According to Proposition 4.1, there
exists $q(u)\in\mathbb{R}^{2}_{+}$ such that $\mathfrak{q}(u)=m(q(u))$. If
$u\leqslant v$, we have

$m(q(v)-q(u))=\mathfrak{q}(v)-\mathfrak{q}(u)\in S^{D}_{+},$

so $q_{1}(v)-q_{1}(u)\geqslant 0$ and $q_{2}(v)-q_{2}(u)\geqslant 0$.

In addition, the map
$m(\lambda_{1},\lambda_{2})\mapsto(\lambda_{1},\lambda_{2})$ is continuous
according to Proposition 4.1. This justifies that
$q\in\mathcal{Q}(\mathbb{R}^{2}_{+})$ and by definition
$q^{\perp}=\mathfrak{q}$. Let
$|x|_{p}=\left(\frac{1}{D}\sum_{d=1}^{D}|x_{d}|^{p}\right)^{1/p}$ denote the
normalized $p$-norm on $\mathbb{R}^{D}$ and $|x|_{\infty}=\sup_{1\leqslant
d\leqslant D}|x_{d}|$. We equip $S^{D}$ with the norm

$N_{2}(m)=\sup_{|x|_{2}=1}|mx|_{2}.$

When $m\in S^{D}_{+}$, $N_{2}(m)$ is equal to the biggest eigenvalue of $m$.
Therefore, for every $\lambda_{1},\lambda_{2}\in\mathbb{R}_{+}$,

$N_{2}(m(\lambda_{1},\lambda_{2}))=\max\\{\lambda_{1},\lambda_{2}\\}.$

Thus, for every $u\in[0,1)$, $N_{2}(\mathfrak{q}(u))=|q(u)|_{\infty}$, in
particular $\mathfrak{q}\in L^{p}$ if and only if $q\in L^{p}$.

Finally, assume that $\mathfrak{q}\in L^{2}$, then by the previous argument
$q\in L^{2}$. Recall that we have $m(\lambda)\cdot
m(\mu)=(D-1)\lambda_{1}\mu_{1}+\lambda_{2}\mu_{2}$, for every
$q^{\prime}\in\mathbb{R}^{2}_{+}$, we have

$\displaystyle\langle\mathfrak{q},(q^{\prime})^{\perp}\rangle_{L^{2}}$
$\displaystyle=\int_{0}^{1}m(q(u))\cdot m(q^{\prime}(u))\mathrm{d}u$
$\displaystyle=\int_{0}^{1}(D-1)q_{1}(u)q^{\prime}_{1}(u)+q_{2}(u)q^{\prime}_{2}(u)\mathrm{d}u$
$\displaystyle=\langle q_{1},(D-1)q^{\prime}_{1}\rangle_{L^{2}}+\langle
q_{2},q^{\prime}_{2}\rangle_{L^{2}}.$

∎

###  4.3. Piecewise linear approximations of paths

Given a finite dimensional vector space $V$ and a cone $\mathcal{C}\subseteq
V$. For every $j\geqslant 1$, we define

(4.4) $\mathcal{Q}^{j}(\mathcal{C})=\\{x\in\mathcal{C}^{j}\big{|}\;\forall
i\in\\{1,\dots,j\\},\;x_{i}-x_{i-1}\in\mathcal{C}\\}.$

As previously, we only use $(V,\mathcal{C})=(S^{D},S^{D}_{+})$ or
$(V,\mathcal{C})=(\mathbb{R}^{D},\mathbb{R}^{D}_{+})$. We will always adopt
the convention $x_{0}=0$. Given $x\in\mathcal{Q}^{j}(\mathbb{R}^{D}_{+})$, we
define $\Lambda^{j}x\in\mathcal{Q}(\mathbb{R}^{D}_{+})$ to be the path that
linearly interpolates between the values
$(0,0),(\frac{1}{j},x_{1}),(\frac{2}{j},x_{2}),\dots,(1,x_{j})$. More
precisely, for every $u\in[0,1)$,

(4.5)
$\Lambda^{j}x(u)=\sum_{i=1}^{j}\mathbf{1}_{\left[\frac{i-1}{j},\frac{i}{j}\right)}(u)\left(x_{i-1}+j\left(u-\frac{i-1}{j}\right)(x_{i}-x_{i-1})\right).$

We will write $\mathbf{1}_{i}$ instead of
$\mathbf{1}_{\left[\frac{i-1}{j},\frac{i}{j}\right)}$ in what follows, with
the understanding that $\mathbf{1}_{j+1}=0$. We also define

$\widetilde{\mathbf{1}}_{i}(u)=j\left(u-\frac{i-1}{j}\right)\mathbf{1}_{i}(u)+j\left(\frac{i+1}{j}-u\right)\mathbf{1}_{i+1}(u).$

Note that we have
$\Lambda^{j}x=\sum_{i=1}^{j}x_{i}\widetilde{\mathbf{1}}_{i}$.

We equip $\mathbb{R}^{2}$ with the normalized $\ell^{1}$-norm,
$|v|=\frac{1}{2}(|v_{1}|+|v_{2}|)$ and the standard normalized scalar product
$v\cdot w=\frac{1}{2}(v_{1}w_{1}+v_{2}w_{2})$. For every $p\in[1,+\infty)$, we
equip $(\mathbb{R}^{2})^{j}$ with the norm $|\cdot|_{p}$, defined by

(4.6)
$|x|_{p}=\left(\frac{1}{j}\sum_{i=1}^{j}|x_{i}|^{p}\right)^{\frac{1}{p}}.$

We also equip $(\mathbb{R}^{2})^{j}$ with the normalized scalar product,

$\langle x,y\rangle_{j}=\frac{1}{j}\sum_{i=1}^{j}x_{i}\cdot y_{i}.$

Given a path, $p\in\mathcal{Q}(\mathbb{R}_{+})\cap L^{1}$ we set

$\Lambda_{j}p(u)=(\langle
p(u),j\widetilde{\mathbf{1}}_{i}\rangle_{L^{2}})_{1\leqslant i\leqslant j},$

this defines a path $\Lambda_{j}p\in\mathcal{Q}^{j}(\mathbb{R}_{+})$. For
every $q\in\mathcal{Q}(\mathbb{R}_{+}^{D})$, if we write
$q(u)=(q_{d}(u))_{1\leqslant d\leqslant D}$, then
$q_{d}\in\mathcal{Q}(\mathbb{R}_{+})$ and we define

$\Lambda_{j}q(u)=(\Lambda_{j}q_{1}(u),\dots,\Lambda_{j}q_{D}(u)).$

The linear maps
$\Lambda^{j}:\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})\to\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{1}$ and $\Lambda_{j}:\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{1}\to\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})$ form an adjoint pair in the
following sense, for every $p\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{1}$ and
$x\in\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})$, we have

$\langle\Lambda^{j}x,p\rangle_{L^{2}}=\langle x,\Lambda_{j}p\rangle_{j}.$

######  Proposition 4.4.

There exists a constant $c>0$ such that for every $j\geqslant 1$ and every
$q=(q_{1},q_{2})\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{\infty}$, we have

$|q-\Lambda^{j}\Lambda_{j}q|_{L^{1}}\leqslant\frac{c|q|_{L^{\infty}}}{j}.$

###### Proof.

Let $p\in\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty}$, to lighten notations, we
let $p_{i}=p\left(\frac{i}{j}\right)$ and define
$\overline{p}=(p_{i})_{1\leqslant i\leqslant
j}\in\mathcal{Q}^{j}(\mathbb{R}_{+})$. For every $i\in\\{1,\dots,j\\}$, we
have

$\frac{p_{i}+p_{i-1}}{2}\leqslant\langle
p,j\widetilde{\mathbf{1}}_{i}\rangle_{L^{2}}\leqslant\frac{p_{i}+p_{i+1}}{2}.$

Therefore
$|jp\cdot\tilde{\mathbf{1}_{i}}-p_{i}|\leqslant\max\left\\{\frac{p_{i+1}-p_{i}}{2},\frac{p_{i}-p_{i-1}}{2}\right\\}\leqslant\frac{p_{i+1}-p_{i-1}}{2}$.
Summing over $i$, we obtain
$|\Lambda_{j}p-\overline{p}|_{1}\leqslant\frac{|p|_{L^{\infty}}}{j}$. In
addition, given $x\in\mathcal{(}\mathbb{R}_{+})^{j}$, we have

$|\Lambda^{j}(x)|_{L^{1}}\leqslant\sum_{i=1}^{j}|x_{i}||\widetilde{\mathbf{1}_{i}}|_{L^{1}}=|x|_{1}.$

Therefore,

$|\Lambda^{j}\Lambda_{j}p-\Lambda^{j}(\overline{p})|_{L^{1}}\leqslant|\Lambda_{j}p-\overline{p}|_{1}\leqslant\frac{|p|_{L^{\infty}}}{j}.$

Observe, that for $u\in\left[\frac{i-1}{j},\frac{i}{j}\right)$,
$|\Lambda^{j}\overline{p}(u)-p(u)|\leqslant p_{i}-p_{i-1}$, thus

$|\Lambda^{j}(\overline{p})-p|_{L^{1}}\leqslant\frac{|p|_{L^{\infty}}}{j}.$

Combining the previous two displays, we obtain

$|\Lambda^{j}\Lambda_{j}p-p|_{1}\leqslant|\Lambda^{j}\Lambda_{j}p-\Lambda^{j}(\overline{p})|_{1}+|\Lambda^{j}(\overline{p})-p|_{1}\leqslant\frac{|p|_{L^{\infty}}}{j}+\frac{|p|_{L^{\infty}}}{j}=\frac{2|p|_{L^{\infty}}}{j}.$

Finally, given $q\in\mathcal{Q}(\mathbb{R}_{+}^{2})\cap L^{\infty}$ if we
write $q(u)=(q_{1}(u),q_{2}(u))$, then
$q_{1},q_{2}\in\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty}$. Applying the bound
in the previous display to $q_{1}$ and $q_{2}$, we obtain

$\displaystyle|\Lambda^{j}\Lambda_{j}q-q|_{L^{1}}$
$\displaystyle=\frac{1}{2}\left(|\Lambda^{j}\Lambda_{j}q_{1}-q_{1}|_{L^{1}}+|\Lambda^{j}\Lambda_{j}q_{2}-q_{2}|_{L^{1}}\right)$
$\displaystyle\leqslant\frac{|q_{1}|_{L^{\infty}}+|q_{2}|_{L^{\infty}}}{j}$
$\displaystyle\leqslant\frac{2|q|_{L^{\infty}}}{j}.$

∎

Recall that given a closed convex cone $\mathcal{C}$ we have defined its dual
cone $\mathcal{C}^{*}$ in (3.2).

######  Proposition 4.5.

For every $x\in(\mathbb{R}^{2}_{+})^{j}$, we have

$|\Lambda^{j}(x)|_{L^{1}}\leqslant|x|_{1}.$

In addition, if $x\in(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))^{*}$, then
$(\Lambda^{j}(x))^{\perp}\in(\mathcal{Q}(S^{D}_{+})\cap L^{2})^{*}$.

###### Proof.

Let $x=(x_{1},x_{2})\in(\mathbb{R}^{2}_{+})^{j}$, we have
$|\widetilde{\mathbf{1}_{i}}|_{L^{1}}=\frac{1}{j}$. So,

$|\Lambda^{j}x_{1}|_{L^{1}}=\left|\sum_{i=1}^{j}x_{1i}\widetilde{\mathbf{1}_{i}}\right|_{L^{1}}\leqslant\sum_{i=1}^{j}x_{1i}\left|\widetilde{\mathbf{1}_{i}}\right|_{L^{1}}=\frac{1}{j}\sum_{i=1}^{j}|x_{1i}|.$

It follows that,

$|\Lambda^{j}x|_{L^{1}}=\frac{|\Lambda^{j}x_{1}|_{L^{1}}+|\Lambda^{j}x_{2}|_{L^{1}}}{2}\leqslant\frac{|x_{1}|_{1}+|x_{2}|_{1}}{2}=|x|_{1}.$

This proves the first part of the proposition, let us now further assume that
$x\in(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))^{*}$. Let
$\mathfrak{q}\in\mathcal{Q}(S^{D}_{+})\cap L^{2}$, the path
$\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}\mathfrak{q}^{s}$ is permutation-
invariant. According to Proposition 4.3, there exists
$q\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{2}$ such that
$\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}\mathfrak{q}^{s}=q^{\perp}$. We have

$\displaystyle\langle\mathfrak{q},(\Lambda^{j}x)^{\perp}\rangle_{L^{2}}$
$\displaystyle=\left\langle\mathfrak{q},\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}(\Lambda^{j}x)^{\perp,s^{-1}}\right\rangle_{L^{2}}$
$\displaystyle=\left\langle\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}\mathfrak{q}^{s},(\Lambda^{j}x)^{\perp}\right\rangle_{L^{2}}$
$\displaystyle=\left\langle
q^{\perp},(\Lambda^{j}x)^{\perp}\right\rangle_{L^{2}}$
$\displaystyle=\left\langle(D-1)q_{1},\Lambda^{j}x_{1}\right\rangle_{L^{2}}+\left\langle
q_{2},\Lambda^{j}x_{2}\right\rangle_{L^{2}}$
$\displaystyle=\left\langle(D-1)\Lambda_{j}q_{1},x_{1}\right\rangle_{j}+\left\langle\Lambda_{j}q_{2},x_{2}\right\rangle_{j}.$

We have
$((D-1)\Lambda_{j}q_{1},\Lambda_{j}q_{2})\in\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})$,
so by definition of $(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))^{*}$, the last line
in the previous display is $\geqslant 0$. This justifies
$(\Lambda^{j}(x))^{\perp}\in(\mathcal{Q}(S^{D}_{+})\cap L^{2})^{*}$. ∎

Recall the definition of $\xi_{\perp}$ from (4.3). Also recall that, given
$r\in\mathcal{Q}$ we use the notation $\int h(r)$ as a shorthand for
$\int_{0}^{1}h(r(u))\mathrm{d}u$.

######  Proposition 4.6.

Let $q\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{1}$, for every $j\geqslant 1$,
we have

$\int\xi_{\perp}(\Lambda^{j}\Lambda_{j}q)\leqslant\int\xi_{\perp}(q).$

###### Proof.

We let $\rho_{1},\dots,\rho_{j}\in\mathbb{R}^{2}_{+}$ denote the coordinates
of $\Lambda_{j}q$. By convexity of $\xi$, we have

$\displaystyle\int\xi_{\perp}(\Lambda^{j}\Lambda_{j}q)$
$\displaystyle=\int_{0}^{1}\xi_{\perp}(\Lambda^{j}\Lambda_{j}q(u))\mathrm{d}u$
$\displaystyle=\sum_{i=1}^{j}\int_{\left[\frac{i-1}{j},\frac{i}{j}\right)}\xi_{\perp}\left(\rho_{i-1}+j\left(u-\frac{i-1}{j}\right)(\rho_{i}-\rho_{i-1})\right)\mathrm{d}u$
$\displaystyle=\sum_{i=1}^{j}\int_{\left[\frac{i-1}{j},\frac{i}{j}\right)}\xi_{\perp}\left(j\left(\frac{i}{j}-u\right)\rho_{i-1}+j\left(u-\frac{i-1}{j}\right)\rho_{i}\right)\mathrm{d}u$
$\displaystyle\leqslant\sum_{i=1}^{j}\int_{\left[\frac{i-1}{j},\frac{i}{j}\right)}j\left(\frac{i}{j}-u\right)\xi_{\perp}\left(\rho_{i-1}\right)+j\left(u-\frac{i-1}{j}\right)\xi_{\perp}\left(\rho_{i}\right)\mathrm{d}u$
$\displaystyle=\sum_{i=1}^{j}\frac{\xi_{\perp}\left(\rho_{i-1}\right)+\xi_{\perp}\left(\rho_{i}\right)}{2j}.$

Using Jensen’s inequality, it follows that
$\xi_{\perp}\left(\rho_{i}\right)\leqslant\langle\xi_{\perp}(q),j\tilde{\mathbf{1}_{i}}\rangle_{L^{2}}$.
Finally, since $\sum_{i=1}^{j}\widetilde{\mathbf{1}}_{i}=1$, we obtain

$\int\xi_{\perp}(\Lambda^{j}\Lambda_{j}q)\leqslant\int\xi_{\perp}(q).$

∎

##  5\. The free energy and Hamilton-Jacobi equations

We recall that
$f(t,\mathfrak{q})=\lim_{N\to+\infty}\overline{F}_{N}(t,\mathfrak{q})$. In
this section, we use the fact that $f$ solves (2.4) and the permutation
invariance of the model to show that $\lim\overline{F}_{N}(t,0)$ can be
expressed as the value at $(t,0)$ of the solution of a reduced Hamilton-Jacobi
equation.

###  5.1. The equation for models with matrix valued paths

Given $q\in\mathcal{Q}(\mathbb{R}^{2}_{+})$, recall that we have defined
$q^{\perp}(u)=m(q(u))$. The goal of this section is to show that the functions
defined by $f^{\perp}(t,q)=f(t,q^{\perp})$ is the viscosity solution of some
Hamilton-Jacobi on $\mathcal{Q}(\mathbb{R}_{+}^{2})$. Since we know that $f$
is Gateaux differentiable and solves $\eqref{e. HJ smooth}$, this will
basically only amount to computing the Gateaux derivatives of $f^{\perp}$.

######  Proposition 5.1.

For every $t\geqslant 0$, $\mathfrak{q}\in\mathcal{Q}(S^{D}_{+})\cap L^{2}$
and $s\in\mathscr{S}_{D}$, we have

(5.1) $f(t,\mathfrak{q}^{s})=f(t,\mathfrak{q}).$

###### Proof.

Let $\mathfrak{q}\in\mathcal{Q}(S^{D}_{+})$ and $t\geqslant 0$, recall from
(2.2) that we have

$F_{N}(t,\mathfrak{q})=-\frac{1}{N}\mathbb{E}\log\iint\exp(H_{N}^{t,\mathfrak{q}}(\sigma,\alpha))dP_{N}(\sigma)d\mathfrak{R}(\alpha),$

where

$H_{N}^{t,\mathfrak{q}}(\sigma,\alpha)=\sqrt{2t}H_{N}(\sigma)-tN\xi\left(\frac{\sigma\sigma^{*}}{N}\right)+\sqrt{2}W_{N}^{\mathfrak{q}}(\alpha)\cdot\sigma-\mathfrak{q}(1)\cdot\sigma\sigma^{*}.$

The process
$(H_{N}^{t,\mathfrak{q}}(\sigma,\alpha))_{\sigma\in\mathbb{R}^{D},\alpha\in\mathfrak{U}}$
is a Gaussian process with the following mean and covariance,

$\displaystyle\mathbb{E}H_{N}^{t,\mathfrak{q}}(\sigma,\alpha)$
$\displaystyle=-tN\xi\left(\frac{\sigma\sigma^{*}}{N}\right)-\mathfrak{q}(1)\cdot\sigma\sigma^{*},$
$\displaystyle\text{Cov}\left(H_{N}^{t,\mathfrak{q}}(\sigma,\alpha),H_{N}^{t,\mathfrak{q}}(\tau,\beta)\right)$
$\displaystyle=2N\left(t\xi\left(\frac{\sigma\tau^{*}}{N}\right)+\frac{1}{N}\sum_{i=1}^{N}\sigma_{i}\cdot\mathfrak{q}(\alpha\wedge\beta)\tau_{i}\right).$

Let us show that for every $s\in\mathscr{S}_{D}$, we have

$(H_{N}^{t,\mathfrak{q}^{s^{-1}}}(\sigma,\alpha))_{\sigma\in\mathbb{R}^{D},\alpha\in\mathfrak{U}}\overset{(d)}{=}(H_{N}^{t,\mathfrak{q}}(\sigma^{s},\alpha))_{\sigma\in\mathbb{R}^{D},\alpha\in\mathfrak{U}}.$

To proceed, we compute the covariance and the mean of those two Gaussian
processes and discover that they are equal. Let $\sigma,\tau\in\mathbb{R}^{D}$
and $\alpha,\beta\in\mathfrak{U}$, since $\xi$ is permutation-invariant, we
have
$\xi\left(\frac{(\sigma^{s})(\tau^{s})^{*}}{N}\right)=\xi\left(\frac{\sigma\tau^{*}}{N}\right)$.
Therefore,

$\displaystyle\mathbb{E}H_{N}^{t,\mathfrak{q}^{s^{-1}}}(\sigma,\alpha)H_{N}^{t,\mathfrak{q}^{s^{-1}}}(\tau,\beta)$
$\displaystyle=2N\left(t\xi\left(\frac{\sigma\tau^{*}}{N}\right)+\frac{1}{N}\sum_{i=1}^{N}\sigma_{i}\cdot\mathfrak{q}^{s^{-1}}(\alpha\wedge\beta)\tau_{i}\right)$
$\displaystyle=2N\left(t\xi\left(\frac{\sigma\tau^{*}}{N}\right)+\frac{1}{N}\sum_{i=1}^{N}\sigma^{s}_{i}\cdot\mathfrak{q}(\alpha\wedge\beta)\tau^{s}_{i}\right)$
$\displaystyle=2N\left(t\xi\left(\frac{(\sigma^{s})(\tau^{s})^{*}}{N}\right)+\frac{1}{N}\sum_{i=1}^{N}\sigma^{s}_{i}\cdot\mathfrak{q}(\alpha\wedge\beta)\tau^{s}_{i}\right)$
$\displaystyle=\mathbb{E}H_{N}^{t,\mathfrak{q}}(\sigma^{s},\alpha)H_{N}^{t,\mathfrak{q}}(\tau^{s},\beta).$

In addition,

$\displaystyle\mathbb{E}H_{N}^{t,\mathfrak{q}^{s^{-1}}}(\sigma,\alpha)$
$\displaystyle=-Nt\xi\left(\frac{\sigma\sigma^{*}}{N}\right)-\mathfrak{q}^{s^{-1}}(1)\cdot\sigma\sigma^{*}$
$\displaystyle=-Nt\xi\left(\left(\frac{\sigma\sigma^{*}}{N}\right)^{s}\right)-\mathfrak{q}(1)\cdot\left(\sigma\sigma^{*}\right)^{s}$
$\displaystyle=-Nt\xi\left(\frac{\sigma^{s}(\sigma^{s})^{*}}{N}\right)-\mathfrak{q}(1)\cdot\sigma^{s}(\sigma^{s})^{*}$
$\displaystyle=\mathbb{E}H_{N}^{t,\mathfrak{q}}(\sigma^{s},\alpha).$

The desired equality in law follows. We now show that
$f(t,\mathfrak{q}^{s})=f(t,\mathfrak{q})$. According to the previous result
and the permutation invariance of $P_{N}$, we have

$\displaystyle F_{N}(t,\mathfrak{q}^{s})$
$\displaystyle=-\frac{1}{N}\mathbb{E}\log\iint\exp(H_{N}^{t,\mathfrak{q}^{s}}(\sigma,\alpha))dP_{N}(\sigma)d\mathfrak{R}(\alpha)$
$\displaystyle=-\frac{1}{N}\mathbb{E}\log\iint\exp(H_{N}^{t,\mathfrak{q}}(\sigma^{s^{-1}},\alpha))dP_{N}(\sigma)d\mathfrak{R}(\alpha)$
$\displaystyle=-\frac{1}{N}\mathbb{E}\log\iint\exp(H_{N}^{t,\mathfrak{q}}(\sigma,\alpha))dP_{N}(\sigma)d\mathfrak{R}(\alpha)$
$\displaystyle=F_{N}(t,\mathfrak{q}).$

Letting $N\to+\infty$, we obtain the desired result.

∎

######  Proposition 5.2.

For every
$(t,q)\in(0,+\infty)\times(\mathcal{Q}_{\uparrow}(\mathbb{R}_{+}^{2})\cap
L^{\infty})$, $f(t,\cdot)$ is Gateaux differentiable at $q^{\perp}$ and the
path $\nabla f(t,q^{\perp})\in\mathcal{Q}(S^{D}_{+})$ is permutation-invariant
that is, for every $u\in[0,1)$ and $s\in\mathscr{S}_{D}$,

$(\nabla f(t,q^{\perp})(u))^{s}=\nabla f(t,q^{\perp})(u).$

###### Proof.

According to [7, Proposition 8.6], $f(t,\cdot)$ is Gateaux differentiable at
$q^{\perp}$. Let $\kappa\in L^{2}([0,1),S^{D})$ such that for $\varepsilon>0$
small enough $q^{\perp}+\varepsilon\kappa\in\mathcal{Q}(S^{D}_{+})$. According
to Proposition 5.1, we have for every $s\in\mathscr{S}_{D}$,

$\frac{f(t,q^{\perp}+\varepsilon\kappa^{s})-f(t,q^{\perp})}{\varepsilon}=\frac{f(t,q^{\perp}+\varepsilon\kappa)-f(t,q^{\perp})}{\varepsilon}.$

Letting $\varepsilon\to 0$, we obtain that $\langle\nabla
f(t,q^{\perp}),\kappa\rangle_{L^{2}}=\langle\nabla
f(t,q^{\perp}),\kappa^{s}\rangle_{L^{2}}$. So,

$\langle\nabla f(t,q^{\perp}),\kappa\rangle_{L^{2}}=\langle(\nabla
f(t,q^{\perp}))^{s^{-1}},\kappa\rangle_{L^{2}}.$

This means that $(\nabla f(t,q^{\perp}))^{s^{-1}}=\nabla f(t,q^{\perp})$. ∎

For $q\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{2}$, we define
$f^{\perp}(t,q)=f(t,q^{\perp})$, $\psi^{\perp}(q)=\psi(q^{\perp})$. Recall the
definition of $\xi_{\perp}$ in (4.3).

######  Proposition 5.3.

The function $f^{\perp}$ is Gateaux differentiable at every
$(t,q)\in(0,+\infty)\times(\mathcal{Q}_{\uparrow}(\mathbb{R}^{2}_{+})\cap
L^{\infty})$, and it satisfies

(5.2) $\begin{cases}\partial_{t}f^{\perp}-\int\xi_{\perp}(\nabla
f^{\perp})=0\text{ on
}(0,+\infty)\times(\mathcal{Q}_{\uparrow}(\mathbb{R}^{2}_{+})\cap
L^{\infty})\\\ f^{\perp}(0,\cdot)=\psi^{\perp}.\end{cases}$

###### Proof.

According to [7, Propositions 7.2 & 8.6], $f$ is Gateaux differentiable at
every
$(t,\mathfrak{q})\in(0,+\infty)\times(\mathcal{Q}_{\uparrow}(S^{D}_{+})\cap
L^{\infty})$ and we have

$\partial_{t}f(t,\mathfrak{q})=\int\xi(\nabla f(t,\mathfrak{q})).$

Let $t>0$ and
$q=(q_{1},q_{2})\in\mathcal{Q}_{\uparrow}(\mathbb{R}^{2}_{+})\cap L^{\infty}$
and let $\kappa=(\kappa_{1},\kappa_{2})\in L^{2}([0,1),\mathbb{R}^{2})$ such
that for $\varepsilon>0$ small enough
$q+\varepsilon\kappa\in\mathcal{Q}_{\uparrow}(\mathbb{R}^{2}_{+})$. Passing to
the limit as $\varepsilon\to 0$ in

$\frac{f^{\perp}(t,q+\varepsilon\kappa)-f^{\perp}(t,q)}{\varepsilon}=\frac{f(t,q^{\perp}+\varepsilon\kappa^{\perp})-f(t,q^{\perp})}{\varepsilon},$

yields $\langle\nabla f^{\perp}(t,q),\kappa\rangle_{L^{2}}=\langle\nabla
f(t,q^{\perp}),\kappa^{\perp}\rangle_{L^{2}}$. According to Proposition 5.2,
the path $\nabla f(t,q^{\perp})$ is permutation-invariant. According to
Proposition 4.3, there exists
$r=(r_{1},r_{2})\in\mathcal{Q}(\mathbb{R}^{2}_{+})$ such that, $\nabla
f(t,q^{\perp})=r^{\perp}$ and we have

$\langle(\nabla f^{\perp}(t,q))_{1},\kappa_{1}\rangle_{L^{2}}+\langle(\nabla
f^{\perp}(t,q))_{2},\kappa_{2}\rangle_{L^{2}}=\langle(D-1)r_{1},\kappa_{1}\rangle_{L^{2}}+\langle
r_{2},\kappa_{2}\rangle_{L^{2}}.$

So, $\nabla f^{\perp}(t,q)=\left((D-1)r_{1},r_{2}\right)$. In particular, for
every $u\in[0,1)$,

$\xi_{\perp}(\nabla f^{\perp}(t,q))=\xi(r^{\perp})=\xi(\nabla
f(t,q^{\perp})).$

Finally, we have

$\displaystyle\partial_{t}f^{\perp}(t,q)=\partial_{t}f(t,q^{\perp})=\int\xi(\nabla
f(t,q^{\perp}))=\int\xi_{\perp}(\nabla f^{\perp}(t,q)).$

∎

If (5.2) was written on $\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{2}$ rather
than $\mathcal{Q}_{\uparrow}(\mathbb{R}^{2}_{+})\cap L^{\infty}$, then we
could immediately conclude that $f^{\perp}$ is a solution in the viscosity
sense and (1.8) would directly follow from the Hopf-Lax representation. This
is indeed how are going to argue, but to do so we need to show that we can
neglect boundary points. That is, paths in
$\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{2}$ that do not belong to
$\mathcal{Q}_{\uparrow}(\mathbb{R}^{2}_{+})\cap L^{\infty}$. This can be done
using the content of Section 3. To connect the two settings, we will need to
consider finite dimensional approximations of (5.2). Recall the definition of
the lift and projection maps $\Lambda^{j}$ and $\Lambda_{j}$ from Section 4.3.
Given $(t,x)\in\mathbb{R}_{+}\times\mathcal{Q}^{j}(\mathbb{R}_{+}^{2})$, we
define $f^{\perp,j}(t,x)=f^{\perp}(t,\Lambda^{j}x)$ and
$H_{\perp}^{j}(x)=\int\xi_{\perp}(\Lambda^{j}x)$.

######  Proposition 5.4.

There exists a constant $c>0$ such that the following holds. For every
$j\geqslant 1$, the function $f^{\perp,j}$ is differentiable on
$(0,+\infty)\times\mathring{\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})}$ and for
every
$(t,x)\in(0,+\infty)\times\mathring{\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})}$,

$|\partial_{t}f^{\perp,j}(t,x)-H_{\perp}^{j}(\nabla
f^{\perp,j}(t,x))|\leqslant\frac{c}{j}.$

Furthermore, $f^{\perp,j}\in\mathcal{V}(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))$.

###### Proof.

_Step 1._ We show that
$f^{\perp,j}\in\mathcal{V}(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))$.

According to [7, Proposition 5.1],
$f:\mathbb{R}_{+}\times(\mathcal{Q}(S^{D}_{+})\cap L^{1})\to\mathbb{R}$ is
Lipschitz. More precisely, we have for every
$(t,\mathfrak{q}),(t^{\prime},\mathfrak{q}^{\prime})\in\mathbb{R}_{+}\times(\mathcal{Q}(S^{D}_{+})\cap
L^{1})$,

$|f(t,\mathfrak{q})-f(t^{\prime},\mathfrak{q}^{\prime})|\leqslant|\mathfrak{q}-\mathfrak{q}^{\prime}|_{L^{1}}+|t-t^{\prime}|\sup_{|a|\leqslant
1}|\xi(a)|.$

Let $|\cdot|$ denote the Frobenius norm on $S^{D}$. There exists a constant
$c>0$ depending only on $|\cdot|$ such that the following holds. For every
$x\in\mathcal{Q}^{j}(\mathbb{R}_{+}^{2})$, we have

$\displaystyle|(\Lambda^{j}x)^{\perp}|_{L^{1}}$
$\displaystyle=\int_{0}^{1}|m(\Lambda^{j}x(u))|\mathrm{d}u$
$\displaystyle=\int_{0}^{1}|\Lambda^{j}x_{1}(u)|\left|\text{id}_{D}-\frac{\mathbf{1}_{D}}{D}\right|+|\Lambda^{j}x_{2}(u)|\left|\frac{\mathbf{1}_{D}}{D}\right|\mathrm{d}u$
$\displaystyle\leqslant
c\int_{0}^{1}\frac{|\Lambda^{j}x_{1}(u)|+|\Lambda^{j}x_{2}(u)|}{2}\mathrm{d}u$
$\displaystyle=c|\Lambda^{j}x|_{L^{1}}$ $\displaystyle\leqslant c|x|_{1},$

where the last line is a consequence of Proposition 4.5. Since
$f^{\perp,j}(t,x)=f(t,(\Lambda^{j}x)^{\perp})$, it follows that
$f^{\perp,j}:\mathbb{R}_{+}\times\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})\to\mathbb{R}$
is Lipschitz. In particular,

$\sup_{t>0,x\in\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})}\frac{|f^{\perp,j}(t,x)-f^{\perp,j}(0,x)|}{t}<+\infty\text{
and }\sup_{t>0}\|f^{\perp,j}(t,\cdot)\|_{\text{Lip}}<+\infty.$

Furthermore, according to [7, Proposition 3.6], for every $t\geqslant 0$,
$f(t,\cdot)$ is $(\mathcal{Q}(S^{D}_{+})\cap L^{2})^{*}$-nondecreasing. Let
$x,y\in\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})$, such that
$y-x\in(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))^{*}$, we have
$(\Lambda^{j}x)^{\perp},(\Lambda^{j}y)^{\perp}\in\mathcal{Q}(S^{D}_{+})\cap
L^{\infty}$ and according to Proposition 4.5,
$(\Lambda^{j}(y-x))^{\perp}\in(\mathcal{Q}(S^{D}_{+})\cap L^{2})^{*}$.
Therefore,

$f^{\perp,j}(t,y)-f^{\perp,j}(t,x)=f(t,(\Lambda^{j}y)^{\perp})-f(t,(\Lambda^{j}x)^{\perp})\geqslant
0.$

Thus, $f^{\perp,j}(t,\cdot)$ is
$(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))^{*}$-nondecreasing and we have proven
that $f^{\perp,j}\in\mathcal{V}(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))$. This
concludes Step 1.

_Step 2._ We show that there exists $c>0$ such that for every $j\geqslant 1$
and every
$(t,x)\in(0,+\infty)\times\mathring{\mathcal{Q}^{j}(\mathbb{R}_{+}^{2})}$,
$f^{\perp,j}$ is differentiable at $(t,x)$ and

$|\partial_{t}f^{j}(t,x)-H_{\perp}^{j}(\nabla
f^{j}(t,x))|\leqslant\frac{c}{j}.$

For every $x\in\mathring{\mathcal{Q}^{j}(\mathbb{R}_{+}^{2})}$,
$\Lambda^{j}x\in\mathcal{Q}_{\uparrow}(\mathbb{R}^{2}_{+})\cap L^{\infty}$.
Using Proposition 5.3, we deduce that $f^{\perp,j}$ is differentiable on
$(0,+\infty)\times\mathring{\mathcal{Q}^{j}(\mathbb{R}_{+}^{2})}$ and for
every
$(t,x)\in(0,+\infty)\times\mathring{\mathcal{Q}^{j}(\mathbb{R}_{+}^{2})}$ we
have $\nabla f^{\perp,j}(t,x)=\Lambda_{j}(\nabla f^{\perp}(t,\Lambda^{j}x))$.
In addition, we have

$\displaystyle\partial_{t}f^{\perp,j}(t,x)$
$\displaystyle=\partial_{t}f^{\perp}(t,\Lambda^{j}x)$
$\displaystyle=\int\xi_{\perp}(\nabla f^{\perp}(t,\Lambda^{j}x))$
$\displaystyle=H_{\perp}^{j}(\nabla
f^{\perp,j}(t,x))+\int\left(\xi_{\perp}(\nabla
f^{\perp}(t,\Lambda^{j}x))-\xi_{\perp}(\Lambda^{j}\Lambda_{j}\nabla
f^{\perp}(t,\Lambda^{j}x))\right).$

Since $f^{\perp}$ is $1$-Lipschitz with respect to $L^{1}$-norm, we have
$|\nabla f^{\perp}(t,\Lambda^{j}x))|_{L^{\infty}}\leqslant 1$. Now, using Step
1 and Proposition 4.4 we discover that

$\displaystyle\left|\partial_{t}f^{\perp,j}(t,x)-H_{\perp}^{j}(\nabla
f^{\perp,j}(t,x))\right|$ $\displaystyle=|\xi_{\perp}(\nabla
f^{\perp}(t,\Lambda^{j}x))-\xi_{\perp}(\Lambda^{j}\Lambda_{j}\nabla
f^{\perp}(t,\Lambda^{j}x))|_{1}$ $\displaystyle\leqslant\ell_{\perp}|\nabla
f^{\perp}(t,\Lambda^{j}x)-\Lambda^{j}\Lambda_{j}\nabla
f^{\perp}(t,\Lambda^{j}x)|_{L^{1}}$ $\displaystyle\leqslant\frac{c}{j},$

where $\ell_{\perp}=\sup_{|a|\leqslant 1}|\xi_{\perp}(a)|$ and $c\geqslant 0$
is some constant depending on $\xi$ and $|\cdot|$. ∎

######  Theorem 5.5.

The function $f^{\perp}$ is the unique viscosity solution of

(5.3) $\begin{cases}\partial_{t}u^{\perp}-\int\xi_{\perp}(\nabla
u^{\perp})=0\text{ on }(0,+\infty)\times(\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{2})\\\ u^{\perp}(0,\cdot)=\psi^{\perp}.\end{cases}$

In addition, for every
$(t,q)\in(0,+\infty)\times(\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{2})$,
$f^{\perp}$ admits the Hopf-Lax representation at $(t,q)$,

(5.4) $f^{\perp}(t,q)=\sup_{p\in\mathcal{Q}(\mathbb{R}_{+}^{2})\cap
L^{\infty}}\inf_{r\in\mathcal{Q}(\mathbb{R}_{+}^{2})\cap
L^{\infty}}\left\\{\psi^{\perp}(p+q)-\langle
p,r\rangle_{L^{2}}+t\int_{0}^{1}\xi_{\perp}\left(r\right)\right\\}.$

######  Remark 5.6.

Taking $q=0$ in (5.4), we obtain Theorem 1.2.

###### Proof.

According to [9, Theorem 4.6], (5.3) has a unique Lipschitz viscosity solution
$u^{\perp}$ and it is given by the variational formula (5.4). Let us show that
$f^{\perp}$ and $u^{\perp}$ coincide on
$[0,+\infty)\times(\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{2})$.

Let $j\geqslant 1$, define Let $\psi^{\perp,j}=f^{\perp,j}(0,\cdot)$.
According to Proposition 5.4, $\psi^{\perp,j}$ is
$(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))^{*}$-nondecreasing and Lipschitz. In
addition, $H_{\perp}^{j}\big{|}_{\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})}$ is
$(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))^{*}$-nondecreasing and locally
Lipschitz. According to Theorem 3.2, the following Hamilton-Jacobi equation is
well posed

$\begin{cases}\partial_{t}u^{\perp,j}-H_{\perp}^{j}(\nabla
u^{\perp,j})=0\text{ on
}(0,+\infty)\times\mathring{\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})}\\\
u^{\perp,j}(0,\cdot)=\psi^{\perp,j}.\end{cases}$

We let $u^{\perp,j}\in\mathcal{V}(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))$ be its
unique viscosity solution.

_Step 1._ We show that, for every
$(t,q)\in\mathbb{R}_{+}\times(\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty})$,

$f^{\perp}(t,q)=\lim_{j\to+\infty}u^{\perp,j}(t,\Lambda_{j}q).$

According to Proposition 3.6 and Proposition 5.4, there exists $c>0$ such that
for every $j\geqslant 1$ and
$(t,x)\in\mathbb{R}_{+}\times\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})$, we have

(5.5) $|f^{\perp,j}(t,x)-u^{\perp,j}(t,x)|\leqslant\frac{ct}{j}.$

Let $q\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{\infty}$, according to
Proposition 4.4 as $j\to+\infty$ we have $\Lambda^{j}\Lambda_{j}q\to q$ in
$L^{1}$. By Lipschitz continuity of $f^{\perp}$ we have

$f^{\perp}(t,q)=\lim_{j\to+\infty}f^{\perp,j}(t,\Lambda_{j}q)=\lim_{j\to+\infty}u^{\perp,j}(t,\Lambda_{j}q).$

_Step 2._ We show that, for every
$(t,q)\in\mathbb{R}_{+}\times(\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty})$,

$\lim_{j\to+\infty}u^{\perp,j}(t,\Lambda_{j}q)=u^{\perp}(t,q).$

According to [10, Theorem 1.2 (2) (d)], $u^{\perp,j}$ admits the Hopf-Lax
representation. That is, for every
$(t,x)\in\mathbb{R}_{+}\times\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})$,

$u^{\perp,j}(t,x)=\sup_{y\in\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})}\inf_{z\in\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})}\left\\{\psi^{\perp,j}(x+y)-\langle
y,z\rangle_{j}+t\int\xi_{\perp}(\Lambda^{j}z)\right\\}.$

Similarly, from [9, Theorem 1.1], we know that for every
$(t,q)\in\mathbb{R}_{+}\times(\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{2})$ we
have

$u^{\perp}(t,q)=\sup_{p\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}}\inf_{r\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}}\left\\{\psi^{\perp}(q+p)-\langle
p,r\rangle_{L^{2}}+t\int\xi_{\perp}(r)\right\\}.$

_Step 2.1_ We show that,
$\lim_{j\to+\infty}u^{\perp,j}(t,\Lambda_{j}q)\leqslant u^{\perp}(t,q)$.

Observe that
$\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})=\\{\Lambda_{j}p,p\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}\\}$. So,

$\begin{split}u^{\perp,j}(t,\Lambda_{j}q)=\sup_{p\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}}\inf_{r\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}}\biggl{\\{}\psi^{\perp}(\Lambda^{j}\Lambda_{j}q+\Lambda^{j}\Lambda_{j}p)\\\
-\langle\Lambda_{j}p,\Lambda_{j}r\rangle_{j}+t\int\xi_{\perp}(\Lambda^{j}\Lambda_{j}r)\biggl{\\}}.\end{split}$

Since,
$\langle\Lambda_{j}p,\Lambda_{j}r\rangle_{j}=\langle\Lambda^{j}\Lambda_{j}p,r\rangle_{L^{2}}$
and $\\{\Lambda^{j}\Lambda_{j}p,p\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}\\}\subseteq\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{\infty}$. We have

$\begin{split}u^{\perp,j}(t,\Lambda_{j}q)\leqslant\sup_{p\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}}\inf_{r\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}}\biggl{\\{}\psi^{\perp}(\Lambda^{j}\Lambda_{j}q+p)\\\ -\langle
p,r\rangle_{L^{2}}+t\int\xi_{\perp}(\Lambda^{j}\Lambda_{j}r)\biggl{\\}}.\end{split}$

Finally, according to Proposition 4.6, we have
$\int\xi_{\perp}(\Lambda^{j}\Lambda_{j}r)\leqslant\int\xi_{\perp}(r)$, so

$u^{\perp,j}(t,\Lambda_{j}q)\leqslant\sup_{p\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}}\inf_{r\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}}\left\\{\psi^{\perp}(\Lambda^{j}\Lambda_{j}q+p)-\langle
p,r\rangle_{L^{2}}+t\int\xi_{\perp}(r)\right\\}.$

Using the Lipschitz continuity of $\psi^{\perp}$, we discover that

$\begin{split}u^{\perp,j}(t,\Lambda_{j}q)\leqslant&|\Lambda^{j}\Lambda_{j}q-q|_{1}\\\
&+\sup_{p\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}}\inf_{r\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}}\left\\{\psi^{\perp}(q+p)-\langle
p,r\rangle_{L^{2}}+t\int\xi_{\perp}(r)\right\\}.\end{split}$

Using Proposition 4.4, we obtain
$\lim_{j\to+\infty}u^{\perp,j}(t,\Lambda_{j}q)\leqslant u^{\perp}(t,q)$.

_Step 2.2_ We show that,
$\lim_{j\to+\infty}u^{\perp,j}(t,\Lambda_{j}q)\geqslant u^{\perp}(t,q)$.

For every $\varepsilon>0$, there exists
$p\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{\infty}$, such that

$\displaystyle
u^{\perp}(t,q)\leqslant\varepsilon+\psi^{\perp}(p+q)+\inf_{r\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}}\left\\{-\langle p,r\rangle_{L^{2}}+t\int\xi_{\perp}(r)\right\\}.$

We have
$\\{\Lambda^{j}z,z\in\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})\\}\subseteq\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}$. So,

$\displaystyle
u^{\perp}(t,q)\leqslant\varepsilon+\psi^{\perp}(p+q)+\inf_{z\in\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})}\left\\{-\langle
p,\Lambda^{j}z\rangle_{L^{2}}+t\int\xi_{\perp}(\Lambda^{j}z)\right\\}.$

Using the Lipschitz continuity of $\psi^{\perp}$, we have
$\psi^{\perp}(p+q)\leqslant\psi^{\perp,j}(\Lambda_{j}p+\Lambda_{j}q)+|\Lambda^{j}\Lambda_{j}(p+q)-(p+q)|_{L^{1}}$.
And we have $\langle
p,\Lambda^{j}z\rangle_{L^{2}}=\langle\Lambda_{j}p,z\rangle_{j}$. So,

$\displaystyle u^{\perp}(t,q)\leqslant\varepsilon$
$\displaystyle+|\Lambda^{j}\Lambda_{j}(p+q)-(p+q)|_{L^{1}}+\psi^{\perp,j}(\Lambda_{j}p+\Lambda_{j}q)$
$\displaystyle+\inf_{z\in\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})}\left\\{-\langle\Lambda_{j}p,z\rangle_{j}+t\int\xi_{\perp}(\Lambda^{j}z)\right\\}.$

Since, $\\{\Lambda_{j}p,p\in\mathcal{Q}(\mathbb{R}^{2}_{+})\cap
L^{\infty}\\}\subseteq\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})$, obtain

$\displaystyle
u^{\perp}(t,q)\leqslant\varepsilon+|\Lambda^{j}\Lambda_{j}(p+q)-(p+q)|_{L^{1}}+u^{\perp,j}(t,\Lambda_{j}q).$

Using Proposition 4.4 and letting $j\to+\infty$, we obtain

$u^{\perp}(t,q)\leqslant\varepsilon+\lim_{j\to+\infty}u^{\perp,j}(t,\Lambda_{j}q),$

since $\varepsilon>0$ is arbitrary this concludes Step 2.

_Step 3._ Conclusion.

From Step 1 and Step 2, we deduce that $f^{\perp}$ and $u^{\perp}$ coincide on
$\mathbb{R}_{+}\times(\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{\infty})$. Since
$\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{\infty}$ is dense in
$\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{2}$ with respect to $L^{1}$
convergence and $f^{\perp}$ and $u^{\perp}$ are both Lipschitz continuous with
respect to $|\cdot|_{L^{1}}$, we conclude that $f^{\perp}$ and $u^{\perp}$
coincide on $\mathbb{R}_{+}\times(\mathcal{Q}(\mathbb{R}^{2}_{+})\cap L^{2})$.
∎

###  5.2. The equation for models with vector valued paths

In this section, we repeat the analysis of Section 5.1 under the additional
assumption that the interaction function $\xi$ only depends on the diagonal
coefficients of the overlap matrix. In this case, we can use the fact the
value of the limit free energy is encoded by a Hamilton-Jacobi equation on
$\mathcal{Q}(\mathbb{R}^{D}_{+})$, as pointed out in (2.7). In this setting,
permutation-invariant paths $q\in\mathcal{Q}(\mathbb{R}^{D}_{+})$ are of the
form $q=(p,\dots,p)$ with $p\in\mathcal{Q}(\mathbb{R}_{+})$. Leveraging
symmetries of the problem, this will allow us to show that the limit free
energy is in fact encoded by a Hamilton-Jacobi equation on
$\mathcal{Q}(\mathbb{R}_{+})$. Since the proofs are very similar, we will not
write them out in as many details as in the previous section.

For $p\in\mathcal{Q}(\mathbb{R}_{+})$, we let
$\mathfrak{q}=\text{diag}(p,\dots,p)\in\mathcal{Q}(S^{D}_{+})$ denote the path
such that $\mathfrak{q}(u)$ is the diagonal matrix with diagonal coefficients
$(p(u),\dots,p(u))$. We define $f^{\dagger}(t,p)=f(t,\text{diag}(p,\dots,p))$,
$\psi^{\dagger}(p)=\psi(\text{diag}(p,\dots,p))$. Recall that here, we
identify the function $A\mapsto\xi(A)$ defined on $\mathbb{R}^{D\times D}$ and
the function $x\mapsto\xi(\text{diag}(x))$ defined on $\mathbb{R}^{D}$. With
this in mind, we set $\xi_{\dagger}(\lambda)=\xi(\lambda,\dots,\lambda)$.

######  Proposition 5.7.

The function $f^{\dagger}$ is Gateaux differentiable at every
$(t,p)\in(0,+\infty)\times\left(\mathcal{Q}_{\uparrow}(\mathbb{R}_{+})\cap
L^{\infty}\right)$ and satisfies

$\begin{cases}\partial_{t}f^{\dagger}-\int\xi_{\dagger}\left(\frac{\nabla
f^{\dagger}}{D}\right)=0\text{ on
}(0,+\infty)\times\left(\mathcal{Q}_{\uparrow}(\mathbb{R}_{+})\cap
L^{\infty}\right)\\\ f^{\dagger}(0,\cdot)=\psi^{\dagger}\text{ on
}\mathcal{Q}_{\uparrow}(\mathbb{R}_{+})\cap L^{\infty}.\end{cases}$

###### Proof.

We reproduce the proof of Proposition 5.3, replacing $f$ by $f^{\text{diag}}$,
the function defined by $f^{\text{diag}}(t,q)=f(t,\text{diag}(q))$ for
$q\in\mathcal{Q}(\mathbb{R}^{D}_{+})$. Let
$(t,p)\in\mathbb{R}_{+}\times\left(\mathcal{Q}_{\uparrow}(\mathbb{R}_{+})\cap
L^{\infty}\right)$ according to Proposition 5.2, $f$ is Gateaux differentiable
at $(t,\text{diag}(p,\dots,p))$ and its gradient is a permutation-invariant
path. Therefore, $f^{\text{diag}}$ is Gateaux differentiable at
$(t,(p,\dots,p))$ and its gradient $\nabla
f^{\text{diag}}(t,(p,\dots,p))\in\mathcal{Q}(\mathbb{R}^{D}_{+})$ satisfies
$\nabla f^{\text{diag}}(t,(p,\dots,p))=(r,\dots,r)$ for some
$r\in\mathcal{Q}(\mathbb{R}_{+})$. It follows that $f^{\dagger}$ is Gateaux
differentiable at $(t,p)$ and its Gateaux derivative satisfies $\nabla
f^{\dagger}(t,p)=Dr$. According to (2.7), we have

$\displaystyle\partial_{t}f^{\dagger}(t,p)=\partial_{t}f^{\text{diag}}(t,p)=\int\xi(\nabla
f^{\text{diag}}(t,(p,\dots,p)))=\int\xi_{\dagger}\left(\frac{\nabla
f^{\dagger}(t,p)}{D}\right).$

∎

Recall from (4.4) that

$\mathcal{Q}^{j}(\mathbb{R}_{+})=\\{x\in\mathbb{R}_{+}^{j}\big{|}\;x_{1}\leqslant\dots\leqslant
x_{j}\\}.$

Also recall that given $x\in\mathcal{Q}^{j}(\mathbb{R}_{+})$, $\Lambda^{j}x$
denote the path which linearly interpolates between the values
$(0,0),(\frac{1}{j},x_{1}),\dots,(1,x_{j})$ and that given
$p\in\mathcal{Q}(\mathbb{R}_{+})$ the vector
$\Lambda_{j}p\in\mathcal{Q}^{j}(\mathbb{R}_{+})$ is defined so that
$\Lambda_{j}$ and $\Lambda^{j}$ form an adjoint pair. As previously, we define
$f^{\dagger,j}(t,x)=f^{\dagger}(t,\Lambda^{j}x)$ and
$H_{\dagger}^{j}(x)=\int_{0}^{1}\xi_{\dagger}(\Lambda^{j}x)$.

######  Proposition 5.8.

There exists a constant $c>0$ such that the following holds. For every
$j\geqslant 1$, the function $f^{\dagger,j}$ is differentiable on
$(0,+\infty)\times\mathring{\mathcal{Q}^{j}(\mathbb{R}_{+})}$ and for every
$(t,x)\in(0,+\infty)\times\mathring{\mathcal{Q}^{j}(\mathbb{R}_{+})}$,

$\left|\partial_{t}f^{\dagger,j}(t,x)-H_{\dagger}^{j}\left(\frac{\nabla
f^{\dagger,j}(t,x)}{D}\right)\right|\leqslant\frac{c}{j}.$

Furthermore, $f^{\dagger,j}\in\mathcal{V}(\mathcal{Q}^{j}(\mathbb{R}_{+}))$.

###### Proof.

The proof will follow from Proposition 5.4 and the fact that
$f^{\dagger}(t,p)=f^{\perp}(t,(p,p))$.

_Step 1._ We show that
$f^{\dagger,j}\in\mathcal{V}(\mathcal{Q}^{j}(\mathbb{R}_{+}))$.

According to Proposition 5.4,
$f^{\perp,j}\in\mathcal{V}(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))$. In
particular,

$\displaystyle\sup_{\begin{subarray}{c}t>0\\\
x\in\mathcal{Q}^{j}(\mathbb{R}_{+})\end{subarray}}\frac{|f^{\dagger,j}(t,x)-f^{\dagger,j}(0,x)|}{t}$
$\displaystyle=\sup_{\begin{subarray}{c}t>0\\\
x\in\mathcal{Q}^{j}(\mathbb{R}_{+})\end{subarray}}\frac{|f^{\perp,j}(t,(x,x))-f^{\perp,j}(0,(x,x))|}{t}$
$\displaystyle\leqslant\sup_{\begin{subarray}{c}t>0\\\
y\in\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})\end{subarray}}\frac{|f^{\perp,j}(t,y)-f^{\perp,j}(0,y)|}{t}$
$\displaystyle<+\infty.$

Similarly,

$\displaystyle\sup_{t>0}\|f^{\dagger,j}(t,\cdot)\|_{\text{Lip}}$
$\displaystyle=\sup_{t>0}\|x\mapsto f^{\perp,j}(t,(x,x))\|_{\text{Lip}}$
$\displaystyle\leqslant\sup_{t>0}\|f^{\perp,j}(t,\cdot)\|_{\text{Lip}}$
$\displaystyle<+\infty.$

Furthermore, given $x,x^{\prime}\in\mathcal{Q}^{j}(\mathbb{R}_{+})$, it is
clear that if $x^{\prime}-x\in(\mathcal{Q}^{j}(\mathbb{R}_{+}))^{*}$, then
$(x^{\prime},x^{\prime})-(x,x)\in(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))^{*}$.
So,

$f^{\dagger,j}(t,x^{\prime})-f^{\dagger,j}(t,x)=f^{\perp,j}(t,(x^{\prime},x^{\prime}))-f^{\perp,j}(t,(x,x))\geqslant
0.$

Thus, $f^{\dagger,j}(t,\cdot)$ is
$(\mathcal{Q}^{j}(\mathbb{R}_{+}))^{*}$-nondecreasing. We have proven that
$f^{\dagger,j}\in\mathcal{V}(\mathcal{Q}^{j}(\mathbb{R}^{2}_{+}))$, this
concludes Step 1.

_Step 2._ We show that there exists $c>0$ such that for every $j\geqslant 1$
and every
$(t,x)\in(0,+\infty)\times\mathring{\mathcal{Q}^{j}(\mathbb{R}_{+})}$,
$f^{\dagger,j}$ is differentiable at $(t,x)$ and

$\left|\partial_{t}f^{\dagger,j}(t,x)-H_{\dagger}^{j}\left(\frac{\nabla
f^{\dagger,j}(t,x)}{D}\right)\right|\leqslant\frac{c}{j}.$

For every $x\in\mathring{\mathcal{Q}^{j}(\mathbb{R}_{+})}$,
$(x,x)\in\mathring{\mathcal{Q}^{j}(\mathbb{R}^{2}_{+})}$. Using Proposition
5.4, we deduce that $f^{\dagger,j}$ is differentiable on
$(0,+\infty)\times\mathring{\mathcal{Q}^{j}(\mathbb{R}_{+})}$ and for every
$(t,x)\in(0,+\infty)\times\mathring{\mathcal{Q}^{j}(\mathbb{R}_{+})}$, we have

$\partial_{t}f^{\dagger,j}(t,x)-H^{j}_{\dagger}\left(\frac{\nabla
f^{\dagger,j}(t,x)}{D}\right)=\partial_{t}f^{\perp,j}(t,(x,x))-H^{j}_{\perp}(\nabla
f^{\perp,j}(t,(x,x))).$

The result then follows from Proposition 5.4. ∎

######  Theorem 5.9.

The function $f^{\dagger}$ is the unique viscosity solution of,

(5.6)
$\begin{cases}\partial_{t}u^{\dagger}-\int\xi_{\dagger}\left(\frac{\nabla
u^{\dagger}}{D}\right)=0\text{ on
}(0,+\infty)\times\left(\mathcal{Q}(\mathbb{R}_{+})\cap L^{2}\right)\\\
u^{\dagger}(0,\cdot)=\psi^{\dagger}.\end{cases}$

In addition, for every
$(t,r)\in[0,+\infty)\times(\mathcal{Q}(\mathbb{R}_{+})\cap L^{2})$,
$f^{\dagger}$ admits the Hopf-Lax representation,

(5.7) $f^{\dagger}(t,r)=\sup_{p\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\left\\{\psi^{\dagger}(p)-t\int_{0}^{1}\xi^{*}\left(\frac{p-r}{t},\dots,\frac{p-r}{t}\right)\right\\}.$

######  Remark 5.10.

Taking $r=0$ in (5.7), we obtain the variational formula (1.9) of Theorem 1.3.

######  Remark 5.11.

Here, since (5.6) is written over the set of $1$-dimensional paths, a
different version of the Hopf-Lax representation is available [9, Proposition
A.3]. This allows us to write the viscosity solution as a “sup” rather than a
“sup inf”.

######  Remark 5.12.

Reproducing the proof of Proposition 4.2, we have for every $\lambda\geqslant
0$,

$\xi^{*}(\lambda,\dots,\lambda)=\left(\xi_{\dagger}\left(\frac{\cdot}{D}\right)\right)^{*}(\lambda).$

###### Proof.

According to [9, Theorem 4.6] and [9, Proposition A.3], (5.6) has a unique
Lipschitz viscosity solution $u^{\dagger}$ and it is given by the variational
formula (5.7). Let us show that $f^{\dagger}$ and $u^{\dagger}$ coincide on
$[0,+\infty)\times(\mathcal{Q}(\mathbb{R}_{+})\cap L^{2})$.

Let $j\geqslant 1$, define $\psi^{\dagger,j}=f^{\dagger,j}(0,\cdot)$,
according to Proposition 5.8, $\psi^{\dagger,j}$ is
$(\mathcal{Q}^{j}(\mathbb{R}_{+}))^{*}$-nondecreasing and Lipschitz. In
addition, $H_{\dagger}^{j}\big{|}_{\mathcal{Q}^{j}(\mathbb{R}_{+})}$ is
$(\mathcal{Q}^{j}(\mathbb{R}_{+}))^{*}$-nondecreasing and locally Lipschitz.
Let $u^{\dagger,j}\in\mathcal{V}(\mathcal{Q}^{j}(\mathbb{R}_{+}))$ be the
unique viscosity solution of

$\begin{cases}\partial_{t}u^{\dagger,j}-H_{\dagger}^{j}\left(\frac{\nabla
u^{\dagger,j}}{D}\right)=0\text{ on
}(0,+\infty)\times\mathring{\mathcal{Q}^{j}(\mathbb{R}_{+})}\\\
u^{\dagger,j}(0,\cdot)=\psi^{\dagger,j}.\end{cases}$

_Step 1._ We show that, for every
$(t,p)\in\mathbb{R}_{+}\times(\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty})$,

$f^{\dagger}(t,p)=\lim_{j\to+\infty}u^{\dagger,j}(t,\Lambda_{j}p).$

According to Proposition 3.6 and Proposition 5.8, there exists $c>0$ such that
for every $j\geqslant 1$ and
$(t,x)\in\mathbb{R}_{+}\times\mathcal{Q}^{j}(\mathbb{R}_{+})$, we have

$|f^{\dagger,j}(t,x)-u^{\dagger,j}(t,x)|\leqslant\frac{ct}{j}.$

Let $p\in\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty}$, according to Proposition
4.4, as $j\to+\infty$ we have $\Lambda^{j}\Lambda_{j}p\to p$ in $L^{1}$. By
Lipschitz continuity of $f^{\dagger}$ we have

$f^{\dagger}(t,p)=\lim_{j\to+\infty}f^{\dagger,j}(t,\Lambda_{j}p)=\lim_{j\to+\infty}u^{\dagger,j}(t,\Lambda_{j}p).$

_Step 2._ We show that, for every
$(t,p)\in\mathbb{R}_{+}\times(\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty})$,

$\lim_{j\to+\infty}u^{\dagger,j}(t,\Lambda_{j}p)=u^{\dagger}(t,p).$

According to [10, Theorem 1.2 (2) (d)], $u^{\dagger,j}$ admits the Hopf-Lax
representation. That is, for every
$(t,x)\in\mathbb{R}_{+}\times\mathcal{Q}^{j}(\mathbb{R}_{+})$,

$u^{\dagger,j}(t,x)=\sup_{y\in\mathcal{Q}^{j}(\mathbb{R}_{+})}\inf_{z\in\mathcal{Q}^{j}(\mathbb{R}_{+})}\left\\{\psi^{\dagger,j}(x+y)-\langle
y,z\rangle_{j}+t\int\xi_{\dagger}\left(\frac{\Lambda^{j}z}{D}\right)\right\\}.$

Similarly, from [9, Theorem 4.6], we know that for every
$(t,p)\in\mathbb{R}_{+}\times\mathcal{(}Q(\mathbb{R}_{+})\cap L^{2})$ we have

$u^{\dagger}(t,p)=\sup_{r\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\inf_{s\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\left\\{\psi^{\dagger}(p+r)-\langle
r,s\rangle_{L^{2}}+t\int\xi_{\dagger}\left(\frac{s}{D}\right)\right\\}.$

_Step 2.1_ We show that,
$\lim_{j\to+\infty}u^{\dagger,j}(t,\Lambda_{j}q)\leqslant u^{\dagger}(t,q)$.

Observe that
$\mathcal{Q}^{j}(\mathbb{R}_{+})=\\{\Lambda_{j}p,p\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}\\}$. So,

$\begin{split}u^{\dagger,j}(t,\Lambda_{j}p)=\sup_{r\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\inf_{s\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\biggl{\\{}\psi^{\dagger}(\Lambda^{j}\Lambda_{j}p+\Lambda^{j}\Lambda_{j}r)\\\
-\langle\Lambda_{j}r,\Lambda_{j}s\rangle_{j}+t\int\xi(\Lambda^{j}\Lambda_{j}s)\biggl{\\}}.\end{split}$

Since,
$\langle\Lambda_{j}r,\Lambda_{j}s\rangle_{j}=\langle\Lambda^{j}\Lambda_{j}r,s\rangle_{L^{2}}$
and $\\{\Lambda^{j}\Lambda_{j}r,r\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}\\}\subseteq\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty}$. We have

$\begin{split}u^{\dagger,j}(t,\Lambda_{j}p)\leqslant\sup_{r\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\inf_{s\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\bigg{\\{}\psi^{\dagger}(\Lambda^{j}\Lambda_{j}p+r)\\\ -\langle
r,s\rangle_{L^{2}}+t\int\xi_{\dagger}\left(\frac{\Lambda^{j}\Lambda_{j}s}{D}\right)\bigg{\\}}.\end{split}$

Finally, according to Proposition 4.6, we have
$\int\xi_{\dagger}(\Lambda^{j}\Lambda_{j}s/D)\leqslant\int\xi_{\dagger}(s/D)$,
so

$\begin{split}u^{\dagger,j}(t,\Lambda_{j}p)\leqslant\sup_{p\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\inf_{r\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\bigg{\\{}\psi^{\dagger}(\Lambda^{j}\Lambda_{j}p+r)\\\ -\langle
r,s\rangle_{L^{2}}+t\int\xi_{\dagger}\left(\frac{s}{D}\right)\bigg{\\}}.\end{split}$

Using the Lipschitz continuity of $\psi^{\dagger}$, we discover that

$\begin{split}u^{\dagger,j}(t,\Lambda_{j}p)\leqslant&|\Lambda^{j}\Lambda_{j}p-p|_{1}\\\
&+\sup_{p\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\inf_{r\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\biggl{\\{}\psi^{\dagger}(p+r)-\langle
r,s\rangle_{L^{2}}+t\int\xi_{\dagger}\left(\frac{s}{D}\right)\biggl{\\}}.\end{split}$

Using Proposition 4.4, we finally obtain
$\lim_{j\to+\infty}u^{\dagger,j}(t,\Lambda_{j}p)\leqslant u^{\dagger}(t,p)$.

_Step 2.2_ We show that,
$\lim_{j\to+\infty}u^{\dagger,j}(t,\Lambda_{j}p)\geqslant u^{\dagger}(t,p)$.

For every $\varepsilon>0$, there exists $r\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}$, such that

$\displaystyle
u^{\dagger}(t,p)\leqslant\varepsilon+\psi^{\dagger}(p+r)+\inf_{s\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\left\\{-\langle
r,s\rangle_{L^{2}}+t\int\xi_{\dagger}\left(\frac{s}{D}\right)\right\\}.$

We have
$\\{\Lambda^{j}z,z\in\mathcal{Q}^{j}(\mathbb{R}_{+})\\}\subseteq\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}$. So,

$\displaystyle
u^{\dagger}(t,p)\leqslant\varepsilon+\psi^{\dagger}(p+r)+\inf_{z\in\mathcal{Q}^{j}(\mathbb{R}_{+})}\left\\{-\langle
r,\Lambda^{j}z\rangle_{L^{2}}+t\int\xi_{\dagger}\left(\frac{\Lambda^{j}z}{D}\right)\right\\}.$

Using the Lipschitz continuity of $\psi^{\dagger}$, we have
$\psi^{\dagger}(p+r)\leqslant\psi^{\dagger,j}(\Lambda_{j}p+\Lambda_{j}r)+|\Lambda^{j}\Lambda_{j}(p+r)-(p+r)|_{L^{1}}$.
And we have $\langle
r,\Lambda^{j}z\rangle_{L^{2}}=\langle\Lambda_{j}r,z\rangle_{j}$. So,

$\displaystyle u^{\dagger}(t,p)\leqslant\varepsilon$
$\displaystyle+|\Lambda^{j}\Lambda_{j}(p+r)-(p+r)|_{L^{1}}+\psi^{\dagger,j}(\Lambda_{j}p+\Lambda_{j}r)$
$\displaystyle+\inf_{z\in\mathcal{Q}^{j}(\mathbb{R}_{+})}\left\\{-\langle\Lambda_{j}r,z\rangle_{j}+t\int\xi_{\dagger}\left(\frac{\Lambda^{j}z}{D}\right)\right\\}.$

Since, $\Lambda_{j}r\in\mathcal{Q}^{j}(\mathbb{R}_{+})$, obtain

$\displaystyle
u^{\dagger}(t,p)\leqslant\varepsilon+|\Lambda^{j}\Lambda_{j}(p+r)-(p+r)|_{L^{1}}+u^{\dagger,j}(t,\Lambda_{j}p).$

Using Proposition 4.4 and letting $j\to+\infty$, we obtain

$u^{\dagger}(t,p)\leqslant\varepsilon+\lim_{j\to+\infty}u^{\dagger,j}(t,\Lambda_{j}p),$

since $\varepsilon>0$ is arbitrary this concludes Step 2.

_Step 3._ Conclusion.

From Step 1 and Step 2, we deduce that $f^{\dagger}$ and $u^{\dagger}$
coincide on $\mathbb{R}_{+}\times(\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty})$. Since $\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty}$ is dense in
$\mathcal{Q}(\mathbb{R}_{+})\cap L^{2}$ with respect to $L^{1}$ convergence
and $f^{\dagger}$ and $u^{\dagger}$ are both Lipschitz continuous with respect
to $L^{1}$, we conclude that $f^{\dagger}$ and $u^{\dagger}$ coincide on
$\mathbb{R}_{+}\times(\mathcal{Q}(\mathbb{R}_{+})\cap L^{2})$. ∎

##  6\. Uniqueness of the optimizer for diagonal models

Let $\mathcal{P}^{\uparrow}(\mathbb{R}^{D}_{+})$ denote the set of probability
measures on $\mathbb{R}^{D}_{+}$ of the form $\text{Law}(q(U))$ with
$q\in\mathcal{Q}(\mathbb{R}^{D}_{+})$ and $U$ a uniform random variable on
$[0,1)$. The map $q\mapsto\text{Law}(q(U))$ is an isometric bijection from
$\mathcal{Q}(\mathbb{R}^{D}_{+})$ to
$\mathcal{P}^{\uparrow}(\mathbb{R}^{D}_{+})$. Therefore, we can also think of
the Parisi functional

$q\mapsto\psi(q)-t\int_{0}^{1}\xi^{*}\left(\frac{q}{t}\right),$

as a functional depending on
$\mu\in\mathcal{P}^{\uparrow}(\mathbb{R}^{D}_{+})$ rather than
$q\in\mathcal{Q}(\mathbb{R}^{D}_{+})$. In a sense, replacing
$\mathcal{Q}(\mathbb{R}^{D}_{+})$ by
$\mathcal{P}^{\uparrow}(\mathbb{R}^{D}_{+})$ changes the geometry. Given
$\mu_{0},\mu_{1}\in\mathcal{P}^{\uparrow}(\mathbb{R}^{D}_{+})$ and
$q_{0},q_{1}\in\mathcal{Q}(\mathbb{R}^{D}_{+})$ such that
$\mu_{i}=\text{Law}(q_{i}(U))$, in general, for $\lambda\in(0,1)$ we have

$\lambda\mu_{1}+(1-\lambda)\mu_{0}\neq\text{Law}(\lambda
q_{1}(U)+(1-\lambda)q_{0}(U)).$

When $D=1$, this allows us to reveal a hidden concavity property of the Parisi
functional, according to [1, Theorem 2], the Parisi functional is strictly
concave on
$\mathcal{P}^{\uparrow}(\mathbb{R}_{+})=\mathcal{P}(\mathbb{R}_{+})$. This is
the key property used to establish the uniqueness of Parisi measures for
models with $D=1$ [1, Corollary 1].

In the setting, $D>1$ this approach seems to fail. A difficulty is that the
convexity of the set $\mathcal{P}^{\uparrow}(\mathbb{R}_{+})$ is an exception
rather than the rule. For every $D>1$, the set
$\mathcal{P}^{\uparrow}(\mathbb{R}^{D}_{+})$ is not convex, since for example,
$\frac{\delta_{(0,1)}+\delta_{(1,0)}}{2}\notin\mathcal{P}^{\uparrow}(\mathbb{R}^{2}_{+})$.
This is problematic, as given two possible maximizing probability measures
$\mu_{0}\neq\mu_{1}$ in $\mathcal{P}^{\uparrow}(\mathbb{R}^{D}_{+})$, the
Parisi functional may not even be defined at $\frac{\mu_{0}+\mu_{1}}{2}$.

In [6, Theorem 1.2], it was pointed out that, that the restriction of the
Parisi functional to any “one-dimensional subspace of
$\mathcal{P}^{\uparrow}(\mathbb{R}^{D}_{+})$” is strictly concave. Thanks to
Theorem 1.3, the Parisi formula can be written as a supremum over probability
measures in $\mathcal{P}^{\uparrow}(\mathbb{R}^{D}_{+})$ supported on
$\\{(\lambda,\dots,\lambda),\lambda\geqslant 0\\}$. This space is of course a
“one-dimensional subspace” in the sense of [6, Theorem 1.2], this allows us to
prove uniqueness of Parisi measures.

Let $\mathcal{P}_{1}(\mathbb{R}_{+})$ denote the set of Borel probability
measures on $\mathbb{R}_{+}$ with finite first moment. Given
$\mu\in\mathcal{P}_{1}(\mathbb{R}_{+})$, for every $u\in[0,1)$, we define

$p_{\mu}(u)=\inf\\{x\geqslant 0\big{|}\;\mu([0,x])>u\\}.$

The path $p_{\mu}\in\mathcal{Q}(\mathbb{R}_{+})\cap L^{1}$ is the quantile
function of the probability measure $\mu$ and classically,
$\text{Law}(p_{\mu}(U))=\mu$.

######  Proposition 6.1 ([6]).

The map $\mu\mapsto\psi^{\dagger}(p_{\mu})$ is strictly concave on
$\mathcal{P}_{1}(\mathbb{R}_{+})$.

###### Proof.

First, note that in [6] the Parisi functional depends on paths in,

$\Pi(S^{D}_{+})=\\{\pi:[0,1]\to S^{D}_{+}\big{|}\;\pi\text{ is left-continuous
and nondecreasing}\\}.$

The Parisi functional is continuous with respect to $|\cdot|_{L^{1}}$, so one
can replace paths $p\in\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty}$ with their
left continuous version without affecting the value of the Parisi functional.
Hence, the distinction between $\Pi$ and $\mathcal{Q}$ is simply a matter of
taste. Let $\widetilde{\xi}:\mathbb{R}^{D\times D}\to\mathbb{R}$ be a convex
function satisfying (1.1) for some Gaussian process $\widetilde{H}_{N}$.
Presumably, as a consequence of [6, Theorem 1.2], choosing
$\Psi=s\text{id}_{D}$, we only have that the function
$\mu\mapsto\psi(\nabla\widetilde{\xi}\circ\Psi\circ p_{\mu})$ is strictly
concave on $\mathcal{P}_{1}(\mathbb{R}_{+})$. But, $\psi$ is independent of
the choice of $\widetilde{\xi}$, and with
$\widetilde{\xi}(R)=\frac{1}{2}\sum_{d,d^{\prime}=1}^{D}R^{2}_{dd^{\prime}}$
we obtain that $\mu\mapsto\psi^{\dagger}(p_{\mu})$ is strictly concave on
$\mathcal{P}_{1}(\mathbb{R}_{+})$. ∎

###### Proof of Theorem 1.3.

The variational formula (1.9) follows from Theorem 5.9, we now prove existence
and uniqueness of an optimizer in (1.9). In the first three steps of the
proof, we explain that in (1.9), the supremum can be taken over a subset of
paths bounded in $L^{\infty}$, this yields existence of an optimizer. The last
step is dedicated to showing uniqueness of the optimizer through Proposition
6.1.

Step 1. We show that there exists $c\geqslant 0$, such that for every
$\lambda>\lambda^{\prime}\geqslant c$ we have

$\frac{\xi_{\dagger}^{*}(\lambda)-\xi_{\dagger}^{*}(\lambda^{\prime})}{\lambda-\lambda^{\prime}}\geqslant
1.$

Let $a>0$, there exists $b\geqslant 0$ such that
$\xi_{\dagger}(\lambda)\leqslant b$ on $[0,a]$. Let $I_{[0,a]}$ denote the
convex function taking the value $0$ on $[0,a]$ and $+\infty$ otherwise. We
have $\xi_{\dagger}\leqslant b+I_{[0,a]}$ on $\mathbb{R}_{+}$. Thus, for
$\lambda\geqslant 0$,

$\displaystyle\xi^{*}_{\dagger}(\lambda)$
$\displaystyle=\sup_{\lambda^{\prime}\geqslant
0}\left\\{\lambda\lambda^{\prime}-\xi_{\dagger}(\lambda)\right\\}$
$\displaystyle\geqslant\sup_{\lambda^{\prime}\geqslant
0}\left\\{\lambda\lambda^{\prime}-b-I_{[0,a]}\right\\}$
$\displaystyle=\sup_{\lambda^{\prime}\in[0,a]}\left\\{\lambda\lambda^{\prime}-b\right\\}$
$\displaystyle=a\lambda-b.$

This imposes
$\liminf_{\lambda\to+\infty}\frac{\xi_{\dagger}^{*}(\lambda)}{\lambda}\geqslant
a$. Since $a>0$ is arbitrary, we have

$\liminf_{\lambda\to+\infty}\frac{\xi_{\dagger}^{*}(\lambda)}{\lambda}=+\infty.$

Let $c\geqslant 0$ be so that for every $\lambda\geqslant c$,

$\frac{\xi_{\dagger}^{*}(\lambda)-\xi_{\dagger}^{*}(0)}{\lambda}\geqslant 1.$

Then, since $\xi_{\dagger}^{*}$ is convex, its slope increases and for every
$\lambda>\lambda^{\prime}\geqslant c$, we have

$\frac{\xi_{\dagger}^{*}(\lambda)-\xi_{\dagger}^{*}(\lambda^{\prime})}{\lambda-\lambda^{\prime}}\geqslant\frac{\xi_{\dagger}^{*}(\lambda)-\xi_{\dagger}^{*}(0)}{\lambda}\geqslant
1.$

This concludes Step 1.

Step 2. Let $L^{\infty}_{\leqslant tc}$ denote the ball of center $0$ and
radius $tc$ in $L^{\infty}$. We show that

$\sup_{p\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}}\left\\{\psi^{\dagger}(p)-\int\xi^{*}_{\dagger}\left(\frac{p}{t}\right)\right\\}=\sup_{p\in\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}_{\leqslant
tc}}\left\\{\psi^{\dagger}(p)-\int\xi^{*}_{\dagger}\left(\frac{p}{t}\right)\right\\}.$

Let LHS and RHS denote the left-hand side and right-hand side in the previous
display. It is clear that LHS $\geqslant$ RHS, we only prove the other
inequality. Let $p\in\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty}$, and let
$\widetilde{p}=p\wedge(t\overline{c})$ be the path which coincides with $p$
until $u^{*}=\inf\\{u\in[0,1),p(u)>tc\\}$ and is constant $=tc$ on
$[u^{*},1)$. We have

$\displaystyle\left(\psi^{\dagger}(p)-t\int_{0}^{1}\xi_{\dagger}^{*}\left(\frac{p}{t}\right)\right)-\left(\psi^{\dagger}(\widetilde{p})-t\int_{0}^{1}\xi_{\dagger}^{*}\left(\frac{\widetilde{p}}{t}\right)\right)$
$\displaystyle\leqslant\int_{0}^{1}|p-\widetilde{p}|-t\int_{0}^{1}\left(\xi^{*}_{\dagger}\left(\frac{p}{t}\right)-\xi^{*}_{\dagger}\left(\frac{\widetilde{p}}{t}\right)\right)$
$\displaystyle=\int_{u^{*}}^{1}p-tc-t\left(\xi^{*}_{\dagger}\left(\frac{p}{t}\right)-\xi^{*}_{\dagger}\left(c\right)\right)$
$\displaystyle=\int_{u^{*}}^{1}(p-tc)\left(1-\frac{\xi^{*}_{\dagger}\left(\frac{p}{t}\right)-\xi^{*}_{\dagger}\left(c\right)}{\frac{p}{t}-c}\right)$
$\displaystyle\leqslant 0.$

Since $|\widetilde{p}|_{L^{\infty}}\leqslant tc$, this concludes Step 2.

Step 3. We show that the sup in the variational formula of Theorem 1.3 is
reached at some $p^{*}\in\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty}$.

The functional
$p\mapsto\psi^{\dagger}(p)-t\int_{0}^{1}\xi^{*}_{\dagger}\left(\frac{p}{t}\right)$
is continuous with respect to the $L^{1}$-norm. So, according to Step 2 it is
enough to prove that the set $\mathcal{Q}(\mathbb{R}_{+})\cap
L^{\infty}_{\leqslant tc}$ is compact with respect to $L^{1}$ convergence. Let
$(p_{n})_{n}$ be a sequence of paths in $\mathcal{Q}(\mathbb{R}_{+})$ which is
bounded by $tc$ in $|\cdot|_{L^{\infty}}$. By a diagonal argument, up to
extraction (not relabelled) we may assume that for every
$u\in\mathbb{Q}\cap[0,1)$, we have $\lim_{n\to+\infty}p_{n}(u)=p(u)$ for some
$p(u)\in\mathbb{R}_{+}$. The map $u\mapsto p(u)$ is nondecreasing and bounded
by $tc$ on $\mathbb{Q}\cap[0,1)$. For every $v\in[0,1)$, the quantity $p(u)$
converges as $u\to v$ in $\mathbb{Q}\cap[v,1)$, we denote by $p(v)$ its
limiting value. The function $p:[0,1)\to\mathbb{R}_{+}$ thus defined is
càdlàg, nondecreasing and bounded by $tc$. Since $p$ is monotone, the set of
discontinuities of $p$ is countable. Let $v\in[0,1)$ be a point at which $p$
is continuous, let $u\in\mathbb{Q}\cap[v,1)$ we have $p_{n}(v)\leqslant
p_{n}(u)$, letting $n\to+\infty$ we obtain
$\limsup_{n\to+\infty}p_{n}(v)\leqslant p(u)$ and since $p$ is continuous at
$v$, letting $u\to v$ yields $\limsup_{n\to+\infty}p_{n}(v)\leqslant p(v)$. By
considering $u\in\mathbb{Q}\cap[0,v]$, we can repeat the same argument to
discover that $\liminf_{n\to+\infty}p_{n}(v)\geqslant p(v)$. In conclusion,
$p_{n}\to p$ pointwise on $[0,1)$ outside a countable set of points. Since
$|p_{n}|_{L^{\infty}}\leqslant tc$, by dominated convergence, it follows that
$p_{n}\to p$ in $L^{1}$.

Step 4. We show that the sup in the variational formula of Theorem 1.3 is
reached at most at one $p^{*}\in\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty}$.

By contradiction, assume that there are two maximizers $p_{0}\neq p_{1}$ of
(1.9) in $\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty}$. Let
$\mu_{i}\in\mathcal{P}(\mathbb{R}_{+})$ be the law of the random variable
$p_{i}(U)$ where $U$ is a uniform random variable in $[0,1)$. Let
$p=p_{\mu}\in\mathcal{Q}(\mathbb{R}_{+})\cap L^{\infty}$ denote the quantile
function of $\mu=\frac{\mu_{0}+\mu_{1}}{2}\in\mathcal{P}(\mathbb{R}_{+})$.
According to Proposition 6.1, we have

$\psi^{\dagger}(p)>\frac{\psi^{\dagger}(p_{0})+\psi^{\dagger}(p_{1})}{2}.$

By definition

$\int\xi_{\dagger}^{*}\left(\frac{p}{t}\right)=\frac{1}{2}\int\xi_{\dagger}^{*}\left(\frac{p_{0}}{t}\right)+\frac{1}{2}\int\xi_{\dagger}^{*}\left(\frac{p_{1}}{t}\right).$

Therefore,

$\psi^{\dagger}(p)-t\int\xi^{*}_{\dagger}\left(\frac{p}{t}\right)>\frac{1}{2}\left(\psi^{\dagger}(p_{0})-t\int\xi^{*}_{\dagger}\left(\frac{p_{0}}{t}\right)\right)+\frac{1}{2}\left(\psi^{\dagger}(p_{1})-t\int\xi^{*}_{\dagger}\left(\frac{p_{1}}{t}\right)\right).$

This is a contradiction since in the previous display the left-hand side is
upper-bounded by $\lim_{N\to+\infty}\overline{F}_{N}(t)$ and the right-hand
side is equal to $\lim_{N\to+\infty}\overline{F}_{N}(t)$.

∎

##  7\. Upper-bound for nonconvex models

In this section, we assume that the interaction function $\xi$ only depends on
the diagonal coefficients of its argument, and we do _not_ assume that $\xi$
is convex. We give a proof of Theorem 1.4. We believe that this Theorem is
related to the approach developed in [17, 15] where the limit free energy of
nonconvex models is lower bounded in terms of the value at $(t,0)$ of the
viscosity solution of some Hamilton-Jacobi equation. Theorem 1.4 will follow
from a classic interpolation argument after observing that $\xi$ satisfies the
following inequality

$\forall
x\in\mathbb{R}^{D}_{+},\;\xi(x)\leqslant\frac{1}{D}\sum_{d=1}^{D}\xi(x_{d},\dots,x_{d}).$

###  7.1. An inequality for permutation-invariant covariance functions

######  Proposition 7.1.

Let $i_{1},\dots,i_{D}\in\mathbb{N}$ and $I=\sum_{d=1}^{D}i_{d}$, for every
$x_{1},\dots,x_{D}\in\mathbb{R}_{+}$, we have

(7.1)
$\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}\prod_{d=1}^{D}x_{s(d)}^{i_{d}}\leqslant\frac{1}{D}\sum_{d=1}^{D}x^{I}_{d}.$

In addition, if $I$ is even then (7.1) holds for
$x_{1},\dots,x_{D}\in\mathbb{R}$.

###### Proof.

If $I=0$ then the inequality is clear, otherwise by the inequality of
arithmetic and geometric means we have

$\prod_{d=1}^{D}x_{s(d)}^{i_{d}}\leqslant\frac{1}{I}\sum_{\delta=1}^{D}i_{\delta}x^{I}_{s(\delta)}.$

Summing over $s\in\mathscr{S}_{D}$, we obtain

$\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}\prod_{d=1}^{D}x_{s(d)}^{i_{d}}\leqslant\frac{1}{I}\sum_{\delta=1}^{D}i_{\delta}\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}x^{I}_{s(\delta)}.$

Observing that
$\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}x^{I}_{s(\delta)}=\frac{1}{D}\sum_{d=1}^{D}x^{I}_{d}$
is independent of $\delta$, we obtain (7.1). Assume now that $I$ is even, and
let $y\in\mathbb{R}^{D}$, define $x_{d}=|y_{d}|$. Let $E$ be the set of
$d\in\\{1,\dots,D\\}$ such that $y_{d}<0$. We have

$\prod_{d=1}^{D}y_{s(d)}^{i_{d}}=(-1)^{\sum_{s(d)\in
E}i_{d}}\prod_{d=1}^{D}x_{s(d)}^{i_{d}}\leqslant\prod_{d=1}^{D}x_{s(d)}^{i_{d}}.$

Summing over $s\in\mathscr{S}_{D}$ and applying (7.1) to $x$, we obtain

$\displaystyle\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}\prod_{d=1}^{D}y_{s(d)}^{i_{d}}$
$\displaystyle\leqslant\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}\prod_{d=1}^{D}x_{s(d)}^{i_{d}}$
$\displaystyle\leqslant\frac{1}{D}\sum_{d=1}^{D}x_{d}^{I}$
$\displaystyle=\frac{1}{D}\sum_{d=1}^{D}y_{d}^{I}.$

Where the last line follows from the fact that $x_{d}=\pm y_{d}$ and $I$ is
even. ∎

Recall that we have defined for every $\lambda\in\mathbb{R}$,
$\xi_{\dagger}(\lambda)=\xi(\lambda,\dots,\lambda)$ and we say that $\xi$ is
permutation-invariant when for every $s\in\mathscr{S}_{D}$ and
$x\in\mathbb{R}^{D}$, $\xi(x^{s})=\xi(x)$. Let $A\in\mathbb{R}^{D\times D}$ be
a $D\times D$ matrix and $p\in\mathbb{N}^{*}$, we denote

$A^{\otimes p}=(A_{i_{1},j_{1}}\times\dots\times A_{i_{p},j_{p}})_{1\leqslant
i_{1},\dots,i_{p},j_{1},\dots,j_{p}\leqslant D}\in\mathbb{R}^{D^{p}\times
D^{p}}$

the $p$-fold tensor product of $A$ with itself. For every
$C,C^{\prime}\in\mathbb{R}^{D^{p}\times D^{p}}$ we define

$C\cdot C^{\prime}=\sum_{1\leqslant
i_{1},\dots,i_{p},j_{1},\dots,j_{p}\leqslant
D}C_{(i_{1},j_{1}),\dots,(i_{p},j_{p})}C^{\prime}_{(i_{1},j_{1}),\dots,(i_{p},j_{p})}.$

######  Proposition 7.2.

Let $\xi:\mathbb{R}^{D}\to\mathbb{R}$, such that $\xi$ is permutation-
invariant and admits an absolutely convergent power series. Assume that for
every, $N\geqslant 1$ there exists a Gaussian process
$(H_{N}(\sigma))_{\sigma\in\mathbb{R}^{D\times N}}$ such that for every
$\sigma,\tau\in\mathbb{R}^{D\times N}$,

$\mathbb{E}\left[H_{N}(\sigma)H_{N}(\tau)\right]=N\xi\left(\frac{\sigma_{1}\cdot\tau_{1}}{N},\dots,\frac{\sigma_{D}\cdot\tau_{D}}{N}\right).$

Then, $\xi_{\dagger}$ is convex on $\mathbb{R}_{+}$ and for every
$x\in\mathbb{R}^{D}_{+}$, we have

(7.2) $\xi(x)\leqslant\frac{1}{D}\sum_{d=1}^{D}\xi_{\dagger}(x_{d}).$

######  Remark 7.3.

For some $\xi$, (7.2) is not satisfied on $\mathbb{R}^{D}$. For example,
assume that $D=2$ and consider $\xi(x_{1},x_{2})=x_{1}x_{2}(x_{1}+x_{2})$, we
have $\xi_{\dagger}(\lambda)=2\lambda^{3}$, and

$\xi(-2,1)=2>-7=\frac{\xi_{\dagger}(-2)+\xi_{\dagger}(1)}{2}.$

###### Proof.

Given $\sigma,\tau\in\mathbb{R}^{D\times N}$, we consider their overlap matrix

$\frac{\sigma\tau^{*}}{N}=\left(\frac{\sigma_{d}\cdot\tau_{d^{\prime}}}{N}\right)_{1\leqslant
d,d^{\prime}\leqslant D},$

the diagonal of the overlap matrix $\frac{\sigma\tau^{*}}{N}$ is the overlap
vector

$R(\sigma,\tau)=\left(\frac{\sigma_{1}\cdot\tau_{1}}{N},\dots,\frac{\sigma_{D}\cdot\tau_{D}}{N}\right).$

For every $A\in\mathbb{R}^{D\times D}$, we define

$\overline{\xi}(A)=\xi(A_{11},\dots,A_{dd}).$

We have
$\overline{\xi}(\frac{\sigma\tau^{*}}{N})=\xi\left(\frac{\sigma_{1}\cdot\tau_{1}}{N},\dots,\frac{\sigma_{D}\cdot\tau_{D}}{N}\right)$.
According to [17, Poposition 6.6], there exists a sequence of matrices
$(C^{(p)})_{p\geqslant 1}$ such that $C^{(p)}\in S^{D^{p}}_{+}$ and, for every
$A\in\mathbb{R}^{D\times D}$,

$\overline{\xi}(A)=\sum_{p\geqslant 1}C^{(p)}\cdot A^{\otimes p}.$

Since $\overline{\xi}(A)$ only depends on the diagonal of $A$, the matrices
$C^{(p)}$ must be diagonal with nonnegative coefficients. In particular, there
exists nonnegative real numbers
$(a_{i_{1},\dots,i_{D}})_{i_{1},\dots,i_{D}\geqslant 0}$, such that for every
$x\in\mathbb{R}^{D}$,

$\xi(x)=\sum_{i_{1},\dots,i_{D}\geqslant
0}a_{i_{1},\dots,i_{D}}\prod_{d=1}^{D}x_{d}^{i_{d}}.$

Therefore, for every $\lambda\in\mathbb{R}$,

$\xi_{\dagger}(\lambda)=\sum_{I\geqslant
0}\lambda^{I}\sum_{\begin{subarray}{c}i_{1},\dots,i_{D}\geqslant 0\\\
\sum_{\delta=1}^{D}i_{\delta}=I\end{subarray}}a_{i_{1},\dots,i_{D}}$

This shows that $\xi_{\dagger}$ has a power series expansion with nonnegative
coefficients, so $\xi_{\dagger}$ is convex on $\mathbb{R}_{+}$. In addition,
since $\xi$ is permutation-invariant, given $x\in\mathbb{R}^{D}_{+}$ we have

$\displaystyle\xi(x)$
$\displaystyle=\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}\xi(x^{s})$
$\displaystyle=\sum_{i_{1},\dots,i_{D}\geqslant
0}a_{i_{1},\dots,i_{D}}\frac{1}{D!}\sum_{s\in\mathscr{S}_{D}}\prod_{d=1}^{D}x_{s(d)}^{i_{d}}.$

According to Proposition 7.1, we thus have

$\displaystyle\xi(x)$ $\displaystyle\leqslant\sum_{i_{1},\dots,i_{D}\geqslant
0}a_{i_{1},\dots,i_{D}}\frac{1}{D}\sum_{d=1}^{D}x_{d}^{\sum_{\delta=1}^{D}i_{\delta}}$
$\displaystyle=\frac{1}{D}\sum_{d=1}^{D}\sum_{I\geqslant
0}x_{d}^{I}\sum_{\begin{subarray}{c}i_{1},\dots,i_{D}\geqslant 0\\\
\sum_{\delta=1}^{D}i_{\delta}=I\end{subarray}}a_{i_{1},\dots,i_{D}}$
$\displaystyle=\frac{1}{D}\sum_{d=1}^{D}\xi_{\dagger}(x_{d}).$

∎

######  Remark 7.4.

As a consequence of the proof of (7.2) we have just given and the second part
of Proposition 7.1, if the power series of $\xi$ only has terms of even
degree, then (7.2) holds for $x\in\mathbb{R}^{D}$.

###  7.2. Positivity principle for multiple species

Roughly speaking, the positivity principle is the statement that if
$\sigma,\tau\in\mathbb{R}^{D\times N}$ are independent random variables drawn
from the Gibbs measure, then almost surely in the limit $N\to+\infty$ the
overlaps $\frac{\sigma_{d}\cdot\tau_{d}}{N}$ are all nonnegative. When $D=1$,
it is well known that the positivity principle is verified as soon as the
Gibbs measure satisfies the so-called Ghirlanda–Guerra identities [20, Theorem
3.4]. When $D>1$ and $P_{N}$ is the uniform probability measure on a product
of $D$ spheres, it was shown that the statement still holds true [3, Section
3.2]. In this section, for the convenience of the reader and the sake of
completeness, we briefly explain how the arguments given to justify [3, Lemma
3.3] can be adapted into a proof of the positivity principle for models with
$D>1$ and $P_{N}=P_{1}^{\otimes N}$. The proofs are unchanged, and we simply
replace the uniform probability measure on a product of $D$ spheres by
$P_{1}^{\otimes N}$.

Given $\sigma,\tau\in\mathbb{R}^{D\times N}$, we let
$R(\sigma,\tau)=(\frac{\sigma_{d}\cdot\tau_{d}}{N})_{1\leqslant d\leqslant
D}$. Given $w\in[0,1]^{D}$, we define

$R^{w}(\sigma,\tau)=\frac{w\cdot
R(\sigma,\tau)}{D}=\frac{1}{D}\sum_{d=1}^{D}w_{d}\frac{\sigma_{d}\cdot\tau_{d}}{N}.$

We let $\mathscr{W}=(w_{q})_{q\geqslant 1}$ denote a countable dense subset of
$[0,1]^{D}$. We assume that $\mathscr{W}$ contains $e_{1}\dots,e_{D}$ and does
not contain the null vector. We fix $p\geqslant 1$, for every $q\geqslant 1$,
we consider the centered Gaussian Process $H_{N,p,q}^{\text{pert}}$ with
covariance

(7.3)
$\mathbb{E}H_{N,p,q}^{\text{pert}}(\sigma)H_{N,p,q}^{\text{pert}}(\tau)=\frac{N}{4^{p+q}}\left(R^{w_{q}}(\sigma,\tau)\right)^{p}.$

Such a process can be explicitly defined by setting

$H_{N,p,q}^{\text{pert}}(\sigma)=\frac{2^{-(p+q)}}{N^{\frac{p-1}{2}}D^{\frac{p}{2}}}\sum_{d_{1},\dots,d_{p}=1}^{D}\sum_{i_{1},\dots,i_{p}=1}^{N}J_{(d_{1},\dots,d_{p}),(i_{1},\dots,i_{p})}\prod_{k=1}^{p}\sqrt{w_{d_{k}}}\sigma_{d_{k},i_{k}}\tau_{d_{k},i_{k}},$

Where the coefficients $J_{(d_{1},\dots,d_{p}),(i_{1},\dots,i_{p})}$ are
independent standard normal random variables. We fix $(u_{p,q})_{p,q\geqslant
1}$ a sequence of numbers in $[1,2]$ and we let

(7.4) $H_{N}^{\text{pert}}(\sigma)=\sum_{p,q\geqslant
1}u_{p,q}H_{N,p,q}^{\text{pert}}(\sigma).$

Soon we are going to choose random coefficients $(u_{p,q})_{p,q\geqslant 1}$
that are independent and uniformly distributed in $[1,2]$ and independent of
every other source of randomness. We denote by $\mathbb{E}_{u}$ the
expectation with respect to this law. We let $c_{N}=N^{-\omega}$ with
$0<\omega<1/2$, and for every nonrandom function $H:\mathbb{R}^{D\times
N}\to\mathbb{R}$, we consider

$\widetilde{H}_{N}(\sigma)=H(\sigma)+c_{N}H_{N}^{\text{pert}}(\sigma).$

First, note that by introducing the perturbation $c_{N}H_{N}^{\text{pert}}$ we
do not change the limiting value of the free energy. Indeed, by applying
Jensen’s inequality twice, we obtain

(7.5) $0\leqslant\left(-\frac{1}{N}\mathbb{E}\log\int
e^{H(\sigma)}\mathrm{d}P_{N}(\sigma)\right)-\left(-\frac{1}{N}\mathbb{E}\log\int
e^{\widetilde{H}(\sigma)}\mathrm{d}P_{N}(\sigma)\right)\leqslant\frac{c_{N}^{2}}{2}.$

We let $\widetilde{G}_{N}$ denote the Gibbs measure associated to
$\widetilde{H}_{N}$, that is

(7.6) $d\widetilde{G}_{N}(\sigma)\propto
e^{\widetilde{H}_{N}(\sigma)}dP_{N}(\sigma).$

We denote by $\mathbb{E}_{H_{N}^{\text{pert}}}$ the expectation with respect
to the randomness of the Gaussian process $H_{N}^{\text{pert}}$. We denote by
$\widetilde{G}^{\otimes k}_{N}(A)$ the probability of the event $A$ under the
probability measure $\widetilde{G}^{\otimes k}_{N}$. As usual, let
$\sigma,\tau$ denote two random variables sampled independently with law
$\widetilde{G}_{N}$. The following lemma is [3, Lemma 3.3].

######  Lemma 7.5 ([3]).

For every $\varepsilon>0$,

$\lim_{N\to+\infty}\sup_{H(\sigma)}\mathbb{E}_{u}\mathbb{E}_{H_{N}^{\text{pert}}}\widetilde{G}_{N}^{\otimes
2}\left(\exists
d,\;\frac{\sigma_{d}\cdot\tau_{d}}{N}\leqslant-\varepsilon\right)=0,$

where the sup is taken over all nonrandom functions $H:\mathbb{R}^{D\times
N}\to\mathbb{R}$ satisfying
$\int\exp|H(\sigma)|\mathrm{d}P_{N}(\sigma)<+\infty$ and $\mathbb{E}_{u}$
denote the expectation with respect to the parameters $(u_{p,q})_{p,q\geqslant
1}$ in (7.4).

###### Proof.

It suffices to show that for every $d\in\\{1,\dots,D\\}$,

$\lim_{N\to+\infty}\sup_{H(\sigma)}\mathbb{E}_{u}\mathbb{E}_{H_{N}^{\text{pert}}}\widetilde{G}_{N}^{\otimes
2}\left(\frac{\sigma_{d}\cdot\tau_{d}}{N}\leqslant-\varepsilon\right)=0.$

Let $(\sigma^{\ell})_{\ell\geqslant 1}$ be independent and identically
distributed random variables with law $\widetilde{G}_{N}$. We let
$R_{\ell,\ell^{\prime}}=R(\sigma^{\ell},\sigma^{\ell^{\prime}})$, and for
every $d\in\\{1,\dots,D\\}$,
$R^{d}_{\ell,\ell^{\prime}}=\frac{\sigma_{d}^{\ell}\cdot\sigma_{d}^{\ell^{\prime}}}{N}$.
The result will follow from the fact that the array
$(R^{d}_{\ell,\ell^{\prime}})_{\ell,\ell^{\prime}\geqslant 1}$ satisfies the
Ghirlanda-Guerra identities in the limit $N\to+\infty$. More precisely, if we
define for any $n\geqslant 1$,
$R(n)=(R_{\ell,\ell^{\prime}})_{1\leqslant\ell,\ell^{\prime}\leqslant n}$ and
for every $d\in\\{1,\dots,D\\}$, for every real valued bound measurable
function $g=g(R(n))$ and every continuous function,
$h:\mathbb{R}\to\mathbb{R}$,

$\begin{split}\Delta^{d}(g,n,h)=\biggl{|}\mathbb{E}\widetilde{G}_{N}\left(gh(R^{d}_{1,{n+1}})\right)-\frac{1}{n}\mathbb{E}\widetilde{G}_{N}\left(gh(R^{d}_{1,2})\right)\\\
-\frac{1}{n}\sum_{l=2}^{n}\mathbb{E}\widetilde{G}_{N}\left(gh(R^{d}_{1,{\ell}})\right)\biggl{|}.\end{split}$

Then, according to [3, Theorem A.3], we have

$\lim_{N\to+\infty}\sup_{H}\mathbb{E}_{u}\Delta^{d}(g,n,h)=0.$

We can then proceed as in [20, Theorem 3.4] to deduce the desired result. ∎

###  7.3. The interpolation