# Nonlinear and degenerate discounted approximation in discrete weak KAM
theory

Panrui Ni  Maxime Zavidovique

###### Abstract

In this paper, we introduce a discrete version of the nonlinear implicit Lax-
Oleinik operator as studied for instance in [21]. We consider the associated
vanishing discount problem with a non-degenerate condition and prove
convergence of solutions as the discount factor goes to $0$. We also discuss
the uniqueness of the discounted solution. This paper can be thought as the
discrete version of [5], and a generalization of [6] and [26, Chapter 3]. The
convergence result is a selection principle for fixed points of a family of
nonlinear operators.

Keywords. Mather measures, Weak KAM theory, Discretization.

††Panrui Ni: Sorbonne Université, Université de Paris Cité, CNRS, Institut de
Mathématiques de Jussieu-Paris Rive Gauche, Paris 75005, France; e-mail:
<EMAIL_ADDRESS>
Maxime Zavidovique: Sorbonne Université, Université de Paris Cité, CNRS,
Institut de Mathématiques de Jussieu-Paris Rive Gauche, Paris 75005, France;
e-mail<EMAIL_ADDRESS>_Mathematics Subject Classification (2010):_
37J50, 37M15.

_À la mémoire de Nicolas Bergeron
un collègue en or
un mathématicien sensationnel
une incarnation de générosité._

## Introduction

###  0.1 Brief history of the problem

The discounted approximation appeared for Hamilton-Jacobi equations in Lions
Papanicolaou and Varadhan’s celebrated preprint [17]. The goal is to
solve333In this introduction, solutions are intended in the viscosity sense
but no knowledge of this notion is required to understand the paper. an
equation on the torus of the form

$H(x,d_{x}u)=c_{0},\quad x\in\mathbb{T}^{N}$ (0.1)

where $u:\mathbb{T}^{N}\to\mathbb{R}$ and $c_{0}\in\mathbb{R}$ are the unknown
and the Hamiltonian $H:\mathbb{T}^{N}\times\mathbb{R}^{N}\to\mathbb{R}$
satisfies some coercivity condition. The problem of the previous equation is
that if it admits solutions, they are not unique (the set of solutions is
invariant by addition of constants). To deal with this problem, the authors
approximate the equation as follows: given $\lambda>0$, they solve the
discounted equation

$\lambda u_{\lambda}(x)+H(x,d_{x}u_{\lambda})=0,$

and establish that the solution $u_{\lambda}$ is unique, that the family
$(u_{\lambda})_{\lambda\in(0,1)}$ is equicontinuous, that $\lambda
u_{\lambda}$ converges to a constant ($-c_{0}$) as $\lambda\to 0^{+}$ and that
$(u_{\lambda}+c_{0}/\lambda)_{\lambda\in(0,1)}$ is bounded. Therefore, taking
a converging subsequence $u_{\lambda_{n}}+c_{0}/\lambda_{n}\to u_{0}$ provides
a solution $u_{0}$ to (0.1).

The first convergence result for the whole family
$(u_{\lambda}+c_{0}/\lambda)_{\lambda\in(0,1)}$ was established in [16] for a
particular case followed by [7] for a result in full generality under Tonelli
type hypotheses on $H$. A discrete version of this result was published in [6]
around the same time. The necessity of convexity of Hamiltonian in the
convergence result is given by a counterexample in [28].

This convergence phenomenon was followed by many generalizations (see [10, 15]
for non-compact cases, see [13, 14, 18, 27] for second order cases, see [9,
11, 12] for weakly coupled systems, and see [2] for mean field games). Amongst
the ones that are of interest to us here, let us cite also the papers the
papers [3, 4, 22, 20] that prove similar results for equations of the form
$G\big{(}x,\lambda u_{\lambda}(x),d_{x}u_{\lambda}(x)\big{)}=0$ where
$G(x,u,p)$ verifies Tonelli type hypotheses in the variables $(x,p)$ and is
increasing in $u$. This is the nonlinear version of the problem, the results
in [7] corresponding to the particular case $G(x,u,p)=u+H(x,p)$. The
degenerate aspect was studied in [25] for Hamiltonians of the form
$G(x,u,p)=\alpha(x)u+H(x,p)$ where $\alpha$ is a continuous nonnegative
function that verifies some non degeneracy condition but is allowed to vanish
on large portions of $\mathbb{T}^{N}$ (a discrete version of this results is
presented in [26]). Both those settings were merged in a nonlinear degenerate
setting in [5] where general Hamiltonians $G(x,u,p)$ are considered, verifying
Tonelli type hypotheses in the variables $(x,p)$ and being non-decreasing in
$u$. The nondegeneracy hypothesis consists in prescribing that $G$ is
increasing in some regions. When the equation is not non-decreasing in $u$,
the asymptotic behavior is not clear yet. One can refer to [8, 23] for the
approximation process when $\lambda\to 0^{-}$, and see [19] for the asymptotic
behavior of a particular non-monotone case.

###  0.2 The discretization

The philosophy of the discrete problem stems from Lax-Oleinik type formulas.
In the previously mentioned results, if $u_{\lambda}$ solves
$G\big{(}x,\lambda u_{\lambda}(x),d_{x}u_{\lambda}(x)\big{)}=0$ then the
solution $u_{\lambda}$ verifies

$\forall x\in\mathbb{T}^{N},\ \ \forall t>0,\quad
u_{\lambda}(x)=\inf_{\gamma}u_{\lambda}\big{(}\gamma(-t)\big{)}+\int_{-t}^{0}L_{G}\big{(}\gamma(s),u_{\lambda}\big{(}\gamma(s)\big{)},\dot{\gamma}(s)\big{)}ds,$

where $L_{G}:\mathbb{T}^{N}\times\mathbb{R}\times\mathbb{R}^{N}$ is a function
related to $G$ that is non-increasing in $u$, and the infimum is taken amongst
absolutely continuous curves $\gamma:\mathbb{T}^{N}\to\mathbb{R}$ such that
$\gamma(0)=x$. The idea is then to fix $t>0$ (small in spirit) and to consider
an approximation of the integral by a function that may depend on
$\gamma(0)=x$ and $\gamma(-t)$ but also possibly on the values
$u_{\lambda}(x)$ and $u_{\lambda}\big{(}\gamma(-t)\big{)}$. When the
approximation function depends linearly on
$u_{\lambda}\big{(}\gamma(-t)\big{)}$, and does not depend on
$u_{\lambda}(x)$, the whole discrete system then reduces to the case
considered in [6] and [26, Chapter 3]. One can refer to the example given in
Section 3 below.

###  0.3 Setting and statement of results

One advantage of the discretization is that it is non longer necessary to have
a differentiable structure. We then work on $(X,d)$ a compact metric space and
consider a continuous function that is $C^{1}$ with respect to the last two
variables444Our results actually require less regularity., $\ell:X\times
X\times\mathbb{R}\times\mathbb{R}$ such that

  * (1)

there is a constant $\kappa_{u}>0$ such that for all $(x,y,u,v)$,
$0\geqslant\partial_{u}\ell(x,y,u,v)\geqslant-\kappa_{u}$,

  * (2)

there is a constant $\kappa_{v}>0$ such that for all $(x,y,u,v)$,
$0\geqslant\partial_{u}\ell(x,y,u,v)\geqslant-\kappa_{v}$,

  * (3)

$\int_{X\times
X}\big{(}\partial_{u}\ell(z,x,0,0)+\partial_{v}\ell(z,x,0,0)\big{)}d\mu(z,x)<0$
for all Mather measures $\mu$ of $\ell(\cdot,\cdot,0,0)$.

The notion of Mather measure will be detailed later in the paper. Let us
already stress that it may happen (quite often actually) that there is only
one Mather measure. Therefore if this is the case, this last nondegeneracy
condition only requires that $\partial_{u}\ell(z,x,0,0)$ or
$\partial_{v}\ell(z,x,0,0)$ is negative somewhere on the support of this
Mather measure.

Let us denote $c_{0}\in\mathbb{R}$ the critical constant of the function
$\ell(\cdot,\cdot,0,0)$ (its precise definition is given later). Given this
cost function (or discrete Lagrangian) we introduce an implicit Lax-Oleinik
operator:

######  Proposition 0.1.

There is a $\lambda_{0}>0$ such that for $0<\lambda<\lambda_{0}$, if
$\varphi\in C^{0}(X,\mathbb{R})$ there is a unique $T_{\lambda}\varphi\in
C^{0}(X,\mathbb{R})$ such that for all $x\in X$,

$T_{\lambda}\varphi(x)=\min_{z\in
X}\big{\\{}\varphi(z)+\ell\big{(}z,x,\lambda\varphi(z),\lambda
T_{\lambda}\varphi(x)\big{)}\big{\\}}+c_{0}.$

The implicit Lax-Oleinik semigroup was studied in [21] in the continuous
setting. It corresponds to the viscosity solutions of Hamilton-Jacobi
equations depending on the unknown function. It is also meaningful in the
optimal control theory of systems with a non-holonomic constraint, see [1].
Thus, our discrete semigroup here can be thought of as an approximation of the
cost function of a class of optimal control systems.

We then solve the discounted equation:

######  Theorem 1.

For $\lambda\in(0,\lambda_{0})$ the operator $T_{\lambda}$ has a fixed point
$u_{\lambda}$. Moreover if we set $S_{\lambda}$ the set of fixed points of
$T_{\lambda}$, then the family $(S_{\lambda})_{\lambda\in(0,\lambda_{0})}$ is
made of equicontinuous and equibounded functions.

Finally we prove the convergence of solutions of the discounted equations:

######  Theorem 2.

The family $(S_{\lambda})_{\lambda\in(0,\lambda_{0})}$ converges to a
singleton as $\lambda\to 0$ in the sense that there exists
$u_{0}:X\to\mathbb{R}$ such that for any choice $u_{\lambda}\in S_{\lambda}$
for $\lambda\in(0,\lambda_{0})$ the (uniform) convergence $u_{\lambda}\to
u_{0}$ holds as $\lambda\to 0^{+}$.

In establishing those results we also prove two characterizations for the
limit $u_{0}$. We also address the issue of uniqueness of fixed points
$u_{\lambda}\in S_{\lambda}$ under quite natural assumptions. Actually, in the
simplified setting presented above, we prove that $S_{\lambda}$ is a singleton
for $\lambda$ small enough.

###  0.4 Organization of the paper

  * •

In the first Section 1 we recall some needed facts on discrete weak KAM
solutions (corresponding to $\lambda=0$).

  * •

In the following Section 2 we introduce a general theory of implicit Lax-
Oleinik operators.

  * •

Then in Section 3 we define and study solutions to the discounted equations.

  * •

Finally in Section 4 we prove the convergence as $\lambda\to 0$ of solutions
to the discounted equations.

  * •

The last section adresses the uniqueness issue.

##  1 Classical discrete weak KAM theory

We briefly recall classical results that will be used in the rest of the
paper. References are, amongst many others, [26, 24, 6]. Let $(X,d)$ be a
compact metric space and $\ell_{0}:X\times X\to\mathbb{R}$ a continuous
function sometimes called cost function. The discrete Lax-Oleinik semigroup is

######  Definition 1.1.

The discrete Lax-Oleinik semigroup is the operator
$T_{0}:C^{0}(X,\mathbb{R})\to C^{0}(X,\mathbb{R})$ which to $f:X\to\mathbb{R}$
associates

$T_{0}f:x\mapsto T_{0}f(x)=\min_{y\in X}f(y)+\ell_{0}(y,x).$

It can be checked that $T_{0}$ is non decreasing, $1$-Lipschitz for the sup-
norm and commutes with addition of constants. This allows to prove the
discrete weak KAM theorem:

######  Theorem 3.

There exists a unique constant $c_{0}$ such that there is a function
$u:X\to\mathbb{R}$ verifying $u=T_{0}u+c_{0}$.

The constant $c_{0}$ is called the critical constant of $\ell_{0}$. A function
$u$ verifying $u=T_{0}u+c_{0}$ is called a weak KAM solution. Weak KAM
solutions are not unique, for instance, if $K\in\mathbb{R}$ then $u+K$ is also
a weak KAM solution. Note also that the critical constant for the cost
function $\tilde{\ell}_{0}=\ell_{0}+c_{0}$ is $\tilde{c}_{0}=0$.

By definition of $T_{0}$, a weak KAM solution $u$ verifies
$u(y)-u(x)\leqslant\ell_{0}(x,y)+c_{0}$ for all $x,y\in X$. This motivates the
definition

######  Definition 1.2.

A function $v:X\to\mathbb{R}$ is called a subsolution if it verifies

$\forall(x,y)\in X\times X,\quad v(y)-v(x)\leqslant\ell_{0}(x,y)+c_{0}$

This is equivalent to $T_{0}v+c_{0}\geqslant v$.

An easy but important fact is that

######  Proposition 1.1.

The set of subsolutions is closed (under uniform convergence but also under
pointwise convergence) and it is convex.

An important tool we will use is that of Mather measure. In all the paper, all
measures are Borel measures even if not explicitly stated.

######  Definition 1.3.

A Borel measure $\mu$ on $X\times X$ is closed if its marginals coincide:
$\pi_{1*}\mu=\pi_{2*}\mu$, where $\pi_{1}(x,y)=x$ and $\pi_{2}(x,y)=y$.

If $v:X\to\mathbb{R}$ is a continuous subsolution, integrating the family of
inequalities $v(y)-v(x)\leqslant\ell_{0}(x,y)+c_{0}$ against a closed
probability measure $\mu$, we discover that $\int_{X\times
X}\ell_{0}(x,y)d\mu(x,y)\geqslant-c_{0}$. This leads to the notion of Mather
measures (or equivalently of minimizing measures):

######  Definition 1.4.

A Mather measure (or minimizing measure) $\mu$ is a probability measure on
$X\times X$ that is closed and verifies

$\int_{X\times X}\ell_{0}(x,y)d\mu(x,y)=-c_{0}.$

We will denote by $\mathfrak{M}_{0}$ the set of Mather measures.

Finally we will need an important function associated to $\ell_{0}$ called
Peierls’ barrier. If $n>0$, let

$\forall(x,y)\in X\times X,\quad
h_{n}(x,y)=\min_{\begin{subarray}{c}(x_{0},\cdots,x_{n})\in X^{n+1}\\\
x_{0}=x,x_{n}=y\end{subarray}}\sum_{k=0}^{n-1}\ell_{0}(x_{k},x_{k+1}).$

######  Definition 1.5.

Peierls’ barrier is the function $h:X\times X\to\mathbb{R}$ defined by

$\forall(x,y)\in X\times X,\quad
h(x,y)=\liminf_{n\to+\infty}h_{n}(x,y)+nc_{0}.$

Here are some key properties of Peierls’ barrier

######  Proposition 1.2.

  1. 1.

The function $h$ is finite valued and continuous on $X\times X$.

  2. 2.

For all $x\in X$, the function $h(x,\cdot)$ is a weak KAM solution and the
function $-h(\cdot,x)$ is a subsolution.

  3. 3.

If $v:X\to\mathbb{R}$ is any subsolution, then

$\forall(x,y)\in X\times X,\quad v(y)-v(x)\leqslant h(x,y).$

A crucial set in weak KAM theory is the projected Aubry set:

######  Definition 1.6.

The projected Aubry set is $\mathfrak{A}=\\{x\in X,\ \ h(x,x)=0\\}$.

This set is proven to be non-empty.

We end this section with a comparison principle:

######  Proposition 1.3.

Let $u:X\to\mathbb{R}$ be a weak KAM solution and $v:X\to\mathbb{R}$ be a
subsolution. Assume that $u_{|\mathfrak{A}}\geq v_{|\mathfrak{A}}$ then
$u\geqslant v$ on $X$.

##  2 Discrete version of implicit semigroup

Assume $(X,d)$ is a compact metric space where $d:X\times X\to\mathbb{R}$ is
the distance function. In the paper, $c:X\times
X\times\mathbb{R}^{2}\to\mathbb{R}$ is a continuous function. Hypotheses that
will be needed are the following

  * (Lu)

for each $(z,x,v)\in X\times X\times\mathbb{R}$, $u\mapsto c(z,x,u,v)$ is
$\kappa_{u}$-Lipschitz continuous and $\kappa_{u}\leq 1$.

  * (Lv)

for each $(z,x,u)\in X\times X\times\mathbb{R}$, $v\mapsto c(z,x,u,v)$ is
$\kappa_{v}$-Lipschitz continuous and $\kappa_{v}<1$.

######  Proposition 2.1.

Assume $c$ verifies hypothesis (Lv). For each continuous function
$\varphi:X\to\mathbb{R}$, there is a unique continuous function
$T\varphi:X\to\mathbb{R}$ satisfying

$T\varphi(x)=\min_{z\in X}c\big{(}z,x,\varphi(z),T\varphi(x)\big{)}.$

The operator
$T:(C^{0}(X,\mathbb{R}),\|\cdot\|_{\infty})\to(C^{0}(X,\mathbb{R}),\|\cdot\|_{\infty})$
is continuous and compact.

Moreover, if the family of functions $x\mapsto c(z,x,u,v)$ is locally equi-
Lipschitz continuous, $T\varphi(x)$ is Lipschitz continuous.

###### Proof.

We first prove that $T\varphi(x)$ exists. For a continuous function
$f:X\to\mathbb{R}$, define

$\mathcal{A}f(x)=\min_{z\in X}c\big{(}z,x,\varphi(z),f(x)\big{)}.$

By the continuity of $c$ and $f$ and compactness of $X$, we see that
$\mathcal{A}$ is an operator from $C^{0}(X,\mathbb{R})$ to itself. Indeed,
$\mathcal{A}f$ is an infimum of equicontinuous functions. We are going to find
a fixed point of $\mathcal{A}$. We take two continuous functions $f$ and $g$
on $X$. By compactness of $X$, let $z$ be a minimal point realizing the
minimum in the definition of $\mathcal{A}g(x)$, then we have

$\mathcal{A}f(x)-\mathcal{A}g(x)\leq
c\big{(}z,x,\varphi(z),f(x)\big{)}-c\big{(}z,x,\varphi(z),g(x)\big{)}\leq\kappa_{v}\|f-g\|_{\infty}.$

Exchanging the role of $f$ and $g$, we get that $\mathcal{A}$ is a contraction
in $(C^{0}(X,\mathbb{R}),\|\cdot\|_{\infty})$, since $\kappa_{v}<1$. By the
Banach fixed point theorem, there is a unique fixed point of $\mathcal{A}$,
which is $T\varphi(x)$.

Then we prove the operator $T$ is compact. Let $r>0$ and $\varphi\in
C^{0}(X,\mathbb{R})$ such that $\|\varphi\|_{\infty}<r$. Consider the sequence
$(f_{n})_{n\in\mathbb{N}}$ with $f_{0}=0$ and

$f_{n+1}(x)=\min_{z\in X}c\big{(}z,x,\varphi(z),f_{n}(x)\big{)}.$

Since $\mathcal{A}$ is a contraction, the sequence $f_{n}$ converges to
$T\varphi$ uniformly. We also have

$\displaystyle\|f_{n+1}\|_{\infty}$
$\displaystyle\leq\sum_{i=0}^{n}\|f_{i+1}-f_{i}\|_{\infty}\leq\sum_{i=0}^{n}(\kappa_{v})^{i}\|f_{1}\|_{\infty}$
$\displaystyle\leq\frac{1}{1-\kappa_{v}}\|f_{1}\|_{\infty}\leq\frac{1}{1-\kappa_{v}}\max_{\begin{subarray}{c}z\in
X,x\in X\\\ |u|\leq r\end{subarray}}|c(z,x,u,0)|=:R_{0}.$

Define $R:=\max\\{R_{0},r\\}$. Let $\omega$ be a modulus of continuity of $c$
restricted to the compact set $X\times X\times[-R,R]^{2}$. For each $x,y\in
X$, let $z_{y}$ be a minimal point in the definition of $f_{1}(y)$, we have

$f_{1}(x)-f_{1}(y)\leq
c(z_{y},x,\varphi(z_{y}),0)-c(z_{y},y,\varphi(z_{y}),0)\leq\omega\big{(}d(x,y)\big{)}.$

Exchanging $x$ and $y$, we have

$|f_{1}(x)-f_{1}(y)|\leq\omega\big{(}d(x,y)\big{)}.$

More generally, if $f\in C^{0}(X,\mathbb{R})$ is such that $\|f\|_{\infty}<R$,
and if $\omega_{f}$ is a modulus of continuity of $f$, then for $x,y\in X$,

$\mathcal{A}f(x)-\mathcal{A}f(y)\leqslant
c\big{(}z_{y},x,\varphi(z^{\prime}_{y}),f(x)\big{)}-c\big{(}z^{\prime}_{y},y,\varphi(z_{y}),f(y)\big{)}\\\
=c\big{(}z_{y},x,\varphi(z^{\prime}_{y}),f(x)\big{)}-c\big{(}z_{y},x,\varphi(z^{\prime}_{y}),f(y)\big{)}\\\
+c\big{(}z_{y},x,\varphi(z^{\prime}_{y}),f(y)\big{)}-c\big{(}z^{\prime}_{y},y,\varphi(z_{y}),f(y)\big{)}\\\
\leqslant\kappa_{v}\omega_{f}(x,y)+\omega(x,y).$

It follows, by exchanging the roles of $x$ and $y$, that
$\kappa_{v}\omega_{f}+\omega$ is a modulus of continuity of $\mathcal{A}f$.
Applying to the sequence $(f_{n})_{n}$ we obtain by induction that $f_{n+1}$
has $(1+\kappa_{v}+\cdots+\kappa_{v}^{n})\omega$ as modulus of continuity.
Hence the whole sequence is equicontinuous with modulus
$\frac{\omega}{1-\kappa_{v}}$ and so is $T\varphi$. As
$\|T\varphi\|_{\infty}\leqslant R_{0}$ and $R_{0}$ only depends on $r$, this
proves that $T$ is compact by the Arzela-Ascoli theorem. Next we prove that
$T$ is continuous. Let $(\varphi_{n})_{n}$ be a sequence converging to
$\varphi$. By the previous point, the sequence $(T\varphi_{n})_{n}$ is
precompact. Let $(k_{n})_{n}$ be an extraction such that
$(T\varphi_{k_{n}})_{n}$ uniformly converges to a function $\psi$. Then if
$x\in X$ we can pass to the limit in the relations

$T\varphi_{k_{n}}(x)=\min_{z\in
X}c\big{(}z,x,\varphi_{k_{n}}(z),T\varphi_{k_{n}}(x)\big{)}$

to obtain

$\psi(x)=\min_{z\in X}c\big{(}z,x,\varphi(z),\psi(x)\big{)}$

and by uniqueness, $\psi=T\varphi$. This proves that $T$ is continuous.

Now we prove the Lipschitz continuity of $T\varphi$ under the additional
Lipschitz assumption of $c$ with respect to $x$. Let $\kappa^{R}_{x}$ be the
Lipschitz constant of $x\mapsto c(z,x,u,v)$ for $|u|$ and $|v|$ bounded by
$R>0$. Applying the previous method, we obtain that if $f_{n}$ is
$\kappa_{n}$-Lipschitz, then $f_{n+1}$ is $\kappa_{n+1}$-Lipschitz with
$\kappa_{n+1}=\kappa^{R}_{x}+\kappa_{v}\kappa_{n}$. Therefore, $(f_{n})_{n}$
is equi-Lipschitz continuous with constant
$\frac{\kappa^{R}_{x}}{1-\kappa_{v}}$, and uniformly converges to $T\varphi$.
Then $T\varphi$ is Lipschitz continuous. ∎

Now we assume

  * (M)

$u\mapsto c(z,x,u,v)$ is non-decreasing and $v\mapsto c(z,x,u,v)$ is non-
increasing.

######  Proposition 2.2.

(Order preserving). If $f\leq g$, then $Tf\leq Tg$.

###### Proof.

We argue by contradiction. Assume there is $x\in X$ such that $Tf(x)>Tg(x)$.
Let $z_{g}$ be a point realizing the definition of $Tg(x)$, then we have

$Tf(x)\leq c\big{(}z_{g},x,f(z_{g}),Tf(x)\big{)}\leq
c\big{(}z_{g},x,g(z_{g}),Tg(x)\big{)}=Tg(x),$

which leads to a contradiction. ∎

To end this section, assume now that the three hypotheses (Lu), (Lv) and (M)
are satisfied:

######  Proposition 2.3.

(Non-expensiveness). For each $f,g$, we have $\|Tf-
Tg\|_{\infty}\leq\|f-g\|_{\infty}$.

###### Proof.

We are going to prove $Tf(x)-\|f-g\|_{\infty}-Tg(x)\leq 0$ for each $x\in X$.
We argue by contradiction. Assume there is $x\in X$ such that
$Tf(x)-\|f-g\|_{\infty}-Tg(x)>0$. Let $z_{g}$ be a point realizing the
definition of $Tg(x)$. Then we have

$\displaystyle Tf(x)-\|f-g\|_{\infty}-Tg(x)$ $\displaystyle\leq
c\big{(}z_{g},x,f(z_{g}),Tf(x)\big{)}-\|f-g\|_{\infty}-c\big{(}z_{g},x,g(z_{g}),Tg(x)\big{)}$
$\displaystyle\leq
c\big{(}z_{g},x,g(z_{g}),Tf(x)\big{)}-c\big{(}z_{g},x,g(z_{g}),Tg(x)\big{)}\leq
0,$

which leads to a contradiction. For the last inequality, we use (Lu).
Exchanging $f$ and $g$, and then the proof is complete. ∎

##  3 Discounted solutions

Now we consider the discounted problem. Let $\ell:X\times
X\times\mathbb{R}^{2}\to\mathbb{R}$ be continuous and satisfy

  * (l1)

$\ell(z,x,u,v)$ is $\kappa_{u}$-Lipschitz in $u$ and $\kappa_{v}$-Lipschitz in
$v$.

  * (l2)

$\ell(z,x,u,v)$ is non-increasing in $u$ and $v$.

  * (l3)

$\partial_{u}\ell(z,x,0,0)$ and $\partial_{v}\ell(z,x,0,0)$ exist, and

$|\ell(z,x,v,u)-\ell(z,x,0,0)-\partial_{u}\ell(z,x,0,0)u-\partial_{v}\ell(z,x,0,0)v|\leq\frac{|u|+|v|}{2}\eta(\frac{|u|+|v|}{2}),$

where $\eta$ is a modulus of continuity.

  * (l4)

$\int_{X\times
X}\big{(}\partial_{u}\ell(z,x,0,0)+\partial_{v}\ell(z,x,0,0)\big{)}d\mu(z,x)<0$
for all Mather measures $\mu$ of $\ell(\cdot,\cdot,0,0)$.

We assume the critical value of $(x,z)\mapsto\ell(z,x,0,0)$ equals zero555If
this is not the case, all our results apply to the function
$\tilde{\ell}=\ell-c_{0}$ where $c_{0}$ is the critical constant of
$(x,z)\mapsto\ell(z,x,0,0)$..

Note that under hypothesis (l4), both functions
$(z,x)\mapsto\partial_{u}\ell(z,x,0,0)$ and
$(z,x)\mapsto\partial_{v}\ell(z,x,0,0)$ are continuous as uniform limits of
continuous functions.

Let $\lambda_{0}>0$ such that $\lambda_{0}\max(\kappa_{u},\kappa_{v})<1$. For
$\lambda_{0}>\lambda>0$, $u+\ell(z,x,\lambda u,\lambda v)$ satisfies the basic
assumptions (Lu), (Lv) and (M) for $c(z,x,u,v)$ in Section 2. Define for
$\varphi\in C^{0}(X,\mathbb{R})$

$T_{\lambda}\varphi(x)=\min_{z\in
X}\big{\\{}\varphi(z)+\ell\big{(}z,x,\lambda\varphi(z),\lambda
T_{\lambda}\varphi(x)\big{)}\big{\\}}.$

This is well defined by Proposition 2.1.

Example: If $\ell(z,x,u,v)$ is of the form $\ell_{0}(z,x)-\alpha(z)u$, where
$\alpha:X\to\mathbb{R}$ is a non-negative function, we have

$T_{\lambda}\varphi(x)=\min_{z\in
X}\big{\\{}\big{(}1-\lambda\alpha(z)\big{)}\varphi(z)+\ell_{0}(z,x)\big{\\}}=T_{0}\big{(}(1-\lambda\alpha)\varphi\big{)}(x),$

which is the degenerate vanishing discount problem as treated in [26].

######  Proposition 3.1.

For $\lambda<\lambda_{0}$, the operator $T_{\lambda}$ admits at least one
fixed point $u_{\lambda}$.

Moreover, the family of all such fixed points
$(u_{\lambda})_{\lambda\in(0,\lambda_{0})}$ is uniformly bounded and equi-
continuous.

###### Proof.

Step 1. We first prove the existence of $u_{\lambda}$ for
$\lambda<\lambda_{0}$. By the discrete weak KAM theorem, $T_{0}$ has a fixed
point $u$ (recall the critical constant is $0$). Since

$T_{0}(u+k)=T_{0}u+k=u+k,\quad\forall k\in\mathbb{R},$

we can choose $\bar{u}\geq 0$ with $T_{0}\bar{u}=\bar{u}$. We prove that
$T_{\lambda}\bar{u}\leq\bar{u}$. Assume there is a point $x\in X$ such that
$T_{\lambda}\bar{u}(x)>\bar{u}(x)$. Let $z$ be a point realizing the minimum
in the definition of $T_{0}\bar{u}(x)$, we have

$T_{\lambda}\bar{u}(x)\leq\bar{u}(z)+\ell\big{(}z,x,\lambda\bar{u}(z),\lambda
T_{\lambda}\bar{u}(x)\big{)}\leq\bar{u}(z)+\ell(z,x,0,0)=T_{0}\bar{u}(x)=\bar{u}(x),$

which leads to a contradiction. By Proposition 2.2, we have

$\bar{u}\geq T_{\lambda}\bar{u}\geq T_{\lambda}\circ
T_{\lambda}\bar{u}\geq\dots.$

Similarly, let $\underline{u}$ be a negative weak KAM solution, we have

$\underline{u}\leq T_{\lambda}\underline{u}\leq T_{\lambda}\circ
T_{\lambda}\underline{u}\leq\dots.$

Since $\underline{u}\leq\bar{u}$, we have for all $n\geqslant 0$,

$\underline{u}\leq T^{n}_{\lambda}\underline{u}\leq
T^{n}_{\lambda}\bar{u}\leq\bar{u}.$

Now we show that $T^{n}_{\lambda}\underline{u}$ is equi-continuous for all
$n\geq 1$ and $\lambda<\lambda_{0}$. Let $\omega$ be a modulus of continuity
of $\ell$ restricted to $X\times X\times[-M,M]^{2}$ where
$M>\lambda_{0}\max(\|\bar{u}\|_{\infty},\|\underline{u}\|_{\infty})$. By
symmetry of the roles of $x$ and $y$, assume without loss of generality
$T^{n}_{\lambda}\underline{u}(x)\geq T^{n}_{\lambda}\underline{u}(y)$, let $z$
be a minimal point in the definition of
$T_{\lambda}(T^{n-1}_{\lambda}\underline{u})(y)$, then by (l1) we have

$\displaystyle|T^{n}_{\lambda}\underline{u}(x)-T^{n}_{\lambda}\underline{u}(y)|$
$\displaystyle=T^{n}_{\lambda}\underline{u}(x)-T^{n}_{\lambda}\underline{u}(y)$
$\displaystyle\leq\ell\big{(}z,x,\lambda
T^{n-1}_{\lambda}\underline{u}(z),\lambda
T^{n}_{\lambda}\underline{u}(x)\big{)}-\ell\big{(}z,y,\lambda
T^{n-1}_{\lambda}\underline{u}(z),\lambda
T^{n}_{\lambda}\underline{u}(y)\big{)}$
$\displaystyle\leq\ell\big{(}z,x,\lambda
T^{n-1}_{\lambda}\underline{u}(z),\lambda
T^{n}_{\lambda}\underline{u}(y)\big{)}-\ell\big{(}z,y,\lambda
T^{n-1}_{\lambda}\underline{u}(z),\lambda
T^{n}_{\lambda}\underline{u}(y)\big{)}$
$\displaystyle\leq\omega\big{(}d(x,y)\big{)}.$

We finally get the equi-continuity of $T^{n}_{\lambda}\underline{u}$. Then
$T^{n}_{\lambda}\underline{u}$ uniformly converges to a function
$\tilde{u}_{\lambda}$. We have

$\displaystyle\|T_{\lambda}\tilde{u}_{\lambda}-\tilde{u}_{\lambda}\|_{\infty}$
$\displaystyle\leq\|T_{\lambda}\tilde{u}_{\lambda}-T^{n}_{\lambda}\underline{u}\|_{\infty}+\|T^{n}_{\lambda}\underline{u}-\tilde{u}_{\lambda}\|_{\infty}$
$\displaystyle\leq\|\tilde{u}_{\lambda}-T^{n-1}_{\lambda}\underline{u}\|_{\infty}+\|T^{n}_{\lambda}\underline{u}-\tilde{u}_{\lambda}\|_{\infty}\to
0.$

Then $\tilde{u}_{\lambda}$ is a fixed point of $T_{\lambda}$, and
$\underline{u}\leq\tilde{u}_{\lambda}\leq\bar{u}$.

Step 2. We prove for $\lambda<\lambda_{0}$, all such fixed points
$u_{\lambda}$ are uniformly bounded, more precisely, $\underline{u}\leq
u_{\lambda}\leq\bar{u}$. . We prove that $u_{\lambda}\leq\bar{u}$, the lower
bound of $u_{\lambda}$ is similar. Assume there is $x_{0}\in X$ such that

$u_{\lambda}(x_{0})-\bar{u}(x_{0})=\max_{x\in
X}\big{(}u_{\lambda}(x)-\bar{u}(x)\big{)}>0.$

Let $(x_{-k})_{k\in\mathbb{N}}$ be a sequence obtained inductively such that
for all $k\geqslant 0$, $x_{-k-1}$ is a point realizing the minimum in the
definition of $T_{0}\bar{u}(x_{-k})$. It follows that for all $k\geqslant 0$,
equatlity $\bar{u}(x_{-k})-\bar{u}(x_{-k-1})=\ell(x_{-k-1},x_{-k},0,0)$ holds.
We first show that if $\lambda<\lambda_{0}$, we have
$u_{\lambda}(x_{-k})>\bar{u}(x_{-k})$ for all $k\geq 0$. Assume
$u_{\lambda}(x_{-1})\leq\bar{u}(x_{-1})$. By (l1) and (l2), we have

$\displaystyle u_{\lambda}(x_{0})-u_{\lambda}(x_{-1})$
$\displaystyle\leq\ell\big{(}x_{-1},x_{0},\lambda u_{\lambda}(x_{-1}),\lambda
u_{\lambda}(x_{0})\big{)}$ $\displaystyle\leq\ell\big{(}x_{-1},x_{0},\lambda
u_{\lambda}(x_{-1}),\lambda\bar{u}(x_{0})\big{)}$
$\displaystyle\leq\ell\big{(}x_{-1},x_{0},\lambda\bar{u}(x_{-1}),\lambda\bar{u}(x_{0})\big{)}+\lambda\kappa_{u}(\bar{u}-u_{\lambda})(x_{-1})$
$\displaystyle\leq\ell(x_{-1},x_{0},0,0)+\lambda\kappa_{u}(\bar{u}-u_{\lambda})(x_{-1})$
$\displaystyle=\bar{u}(x_{0})-\bar{u}(x_{-1})+\lambda\kappa_{u}(\bar{u}-u_{\lambda})(x_{-1}),$

which implies that

$(1-\lambda\kappa_{u})(u_{\lambda}-\bar{u})(x_{-1})\geq
u_{\lambda}(x_{0})-\bar{u}(x_{0})>0,$

which leads to a contradiction as $0<1-\lambda\kappa_{u}<1$. Then
$u_{\lambda}(x_{-1})>\bar{u}(x_{-1})$.

Note that by (l2), we have

$\displaystyle\bar{u}(x_{0})-\bar{u}(x_{-1})$
$\displaystyle=\ell(x_{-1},x_{0},0,0)$
$\displaystyle\geq\ell\big{(}x_{-1},x_{0},\lambda\bar{u}(x_{-1}),\lambda\bar{u}(x_{0})\big{)}$
$\displaystyle\geq\ell\big{(}x_{-1},x_{0},\lambda u_{\lambda}(x_{-1}),\lambda
u_{\lambda}(x_{0})\big{)}$ $\displaystyle\geq
u_{\lambda}(x_{0})-u_{\lambda}(x_{-1}),$

which implies $u_{\lambda}(x_{-1})-\bar{u}(x_{-1})\geq
u_{\lambda}(x_{0})-\bar{u}(x_{0})$. By the definition of $x_{0}$, this must be
an equality, that is,
$u_{\lambda}(x_{-1})-\bar{u}(x_{-1})=u_{\lambda}(x_{0})-\bar{u}(x_{0})$.
Moreover, as all previous inequalities are equalities, we obtain that
$\ell(x_{-1},x_{0},0,0)=\ell\big{(}x_{-1},x_{0},\lambda\bar{u}(x_{-1}),\lambda\bar{u}(x_{0})\big{)}$.

By induction, the same proof then shows that

$u_{\lambda}(x_{-k})-\bar{u}(x_{-k})=\max_{x\in
X}\big{(}u_{\lambda}(x)-\bar{u}(x)\big{)}>0,\quad\forall k\geq 0,$

and that
$\ell(x_{-k-1},x_{-k},0,0)=\ell\big{(}x_{-k-1},x_{-k},\lambda\bar{u}(x_{-k-1}),\lambda\bar{u}(x_{-k})\big{)}$
for all $k\geqslant 0$.

Define the probability measure on $X\times X$, for $N>0$,

$\mu_{N}:=N^{-1}\sum_{k=-N}^{-1}\delta_{(x_{k},x_{k+1})}.$

By weak compactness of measures on $X\times X$ let $N_{n}\to+\infty$ be an
extraction and $\mu$ a probability measure on $X\times X$ such that
$\mu_{N_{n}}\to\mu$, as $n\to+\infty$.

Let $f\in C^{0}(X,\mathbb{R})$. Since

$\int_{X\times
X}\big{(}f(x)-f(y)\big{)}d\mu_{N}=\frac{f(x_{0})-f(x_{-N})}{N}\leq\frac{2\|f\|_{\infty}}{N}\to
0,$

the measure $\mu$ is closed. We also have

$\int_{X\times
X}\ell(z,x,0,0)d\mu_{N}=\frac{\bar{u}(x_{0})-\bar{u}(x_{-N})}{N}\to 0.$

Thus, $\mu$ is a Mather measure. Since $\mu_{N_{n}}\to\mu$, for each
$(z,x)\in\textrm{supp}(\mu)$, there is a sequence
$(z_{n},x_{n})\in\textrm{supp}(\mu_{N_{n}})$ with $(z_{n},x_{n})\to(z,x)$. We
have known that $u_{\lambda}(z_{n})-\bar{u}(z_{n})$ equals a constant $M>0$.
Therefore, $u_{\lambda}(z)-\bar{u}(z)=M>0$. Similarly, we have
$u_{\lambda}(x)-\bar{u}(x)=M>0$.

By the same argument, from
$\ell(z_{n},x_{n},0,0)=\ell\big{(}z_{n},x_{n},\lambda\bar{u}(z_{n}),\lambda\bar{u}(x_{n})\big{)}$
we obtain
$\ell(z,x,0,0)=\ell\big{(}z,x,\lambda\bar{u}(z),\lambda\bar{u}(x)\big{)}$.

Since $u_{\lambda}>\bar{u}\geq 0$ on supp$(\mu)$, by (l2), we have

$\ell(z,x,r,s)=\ell\big{(}z,x,\lambda u_{\lambda}(z),\lambda
u_{\lambda}(x)\big{)},$

for all $(z,x)\in\textrm{supp}(\mu)$ and $r\in[0,\lambda u_{\lambda}(z)]$,
$s\in[0,\lambda u_{\lambda}(x)]$. Then

$\partial_{u,v}\ell(z,x,0,0)=0,\quad\forall(z,x)\in\textrm{supp}(\mu),$

which contradicts (l4).

Step 3. We finally prove the equi-continuity of $u_{\lambda}$. Let $x,y\in X$.
Assume without loss of generality that $u_{\lambda}(x)\geq u_{\lambda}(y)$,
let $z$ the a minimal point of $u_{\lambda}(y)$, then by (l1) we have

$\displaystyle u_{\lambda}(x)-u_{\lambda}(y)$
$\displaystyle\leq\ell\big{(}z,x,\lambda u_{\lambda}(z),\lambda
u_{\lambda}(x)\big{)}-\ell\big{(}z,y,\lambda u_{\lambda}(z),\lambda
u_{\lambda}(y)\big{)}$ $\displaystyle\leq\ell\big{(}z,x,\lambda
u_{\lambda}(z),\lambda u_{\lambda}(y)\big{)}-\ell\big{(}z,y,\lambda
u_{\lambda}(z),\lambda u_{\lambda}(y)\big{)}$
$\displaystyle\leq\omega\big{(}d(x,y)\big{)}.$

∎

##  4 Vanishing discount convergence

Let $\lambda\to 0$, by Proposition 3.1, there is a sequence $\lambda_{n}\to 0$
such that $u_{\lambda_{n}}$ uniformly converges. Let $u_{*}$ be a limit
function of the family $(u_{\lambda})_{\lambda\in(0,\lambda_{0})}$. The
vanishing discount problem concerns the uniqueness of $u_{*}$.

Notation: Let us define $\ell_{0}:X\times X\to\mathbb{R}$ the function defined
by $(z,x)\mapsto\ell(z,x,0,0)$.

Let $\mathcal{S}_{0}$ be the set of subsolutions $w$ of $\ell_{0}$ that
satisfy

$\int_{X\times
X}\big{(}\partial_{u}\ell(z,x,0,0)w(z)+\partial_{v}\ell(z,x,0,0)w(x)\big{)}d\mu(z,x)\geq
0,$ (4.1)

for all Mather measures $\mu$ of $\ell_{0}$. The set $\mathcal{S}_{0}$ is non-
empty since negative weak KAM solutions fullfill (4.1).

######  Theorem 4.

Let $\lambda\to 0$, $u_{\lambda}$ uniformly converges to

$u_{0}:=\sup_{w\in\mathcal{S}_{0}}w,$

where the supremum is taken pointwise. The function $u_{0}$ is therefore a
fixed point of $T_{0}$.

We also establish an alternative formula for the limit $u_{0}$:

######  Theorem 5.

The following holds for all $x\in X$:

$u_{0}(x)=\min_{\mu\in\mathfrak{M}_{0}}\frac{\int_{X\times
X}\Big{(}\partial_{u}\ell(z,y,0,0)h(z,x)+\partial_{v}\ell(z,y,0,0)h(y,x)\Big{)}d\mu(z,y)}{\int_{X\times
X}\Lambda(z,y)d\mu(z,y)},$

where $\mathfrak{M}_{0}$ denotes the set of Mather measures of $\ell_{0}$,

$\Lambda(z,y):=\partial_{u}\ell(z,y,0,0)+\partial_{v}\ell(z,y,0,0),$

and $h(z,x)$ is Peierls’ barrier of $\ell_{0}$.

######  Remark 4.1.

  1. 1.

When $\ell$ satisfies that $\partial_{v}\ell(\cdot,\cdot,0,0)$ is constant,
the previous equality reduces to

$u_{0}(x)=\min_{\mu\in\mathfrak{M}_{0}}\bigg{(}\int_{X\times
X}\Lambda(z,y)d\mu(z,y)\bigg{)}^{-1}\int_{X\times
X}\Lambda(z,y)h(z,x)d\mu(z,y).$

  2. 2.

Symmetrically, when $\ell$ satisfies that $\partial_{u}\ell(\cdot,\cdot,0,0)$
is constant, then

$u_{0}(x)=\min_{\mu\in\mathfrak{M}_{0}}\bigg{(}\int_{X\times
X}\Lambda(y,z)d\mu(y,z)\bigg{)}^{-1}\int_{X\times
X}\Lambda(z,y)h(y,x)d\mu(z,y).$

######  Proposition 4.1.

For each Mather measure $\mu$ of $\ell_{0}$, we have

$\int_{X}\big{(}\partial_{u}\ell(z,x,0,0)u_{*}(z)+\partial_{v}\ell(z,x,0,0)u_{*}(x)\big{)}d\mu(z,x)\geq
0.$

###### Proof.

Let $\mu$ be a Mather measure, Since $T_{\lambda}u_{\lambda}=u_{\lambda}$,
recalling that $\int\ell_{0}d\mu=0$, we have

$\displaystyle\int_{X\times
X}\big{(}u_{\lambda}(x)-u_{\lambda}(z)\big{)}d\mu(z,x)$
$\displaystyle\leq\int_{X\times X}\ell\big{(}z,x,\lambda
u_{\lambda}(z),\lambda u_{\lambda}(x)\big{)}d\mu(z,x)$
$\displaystyle\leq\int_{X\times
X}\big{(}\ell(z,x,0,0)+\lambda\partial_{u}\ell(z,x,0,0)u_{\lambda}(z)+\lambda\partial_{v}\ell(z,x,0,0)u_{\lambda}(x)\big{)}d\mu(z,x)+\lambda\varepsilon(\lambda)$
$\displaystyle=\int_{X\times
X}\big{(}\lambda\partial_{u}\ell(z,x,0,0)u_{\lambda}(z)+\lambda\partial_{v}\ell(z,x,0,0)u_{\lambda}(x)\big{)}d\mu(z,x)+\lambda\varepsilon(\lambda),$

where
$\varepsilon(\lambda)=\|u_{\lambda}\|_{\infty}\eta(\lambda\|u_{\lambda}\|_{\infty})$.
Since $\mu$ is closed, we have

$\int_{X\times X}\big{(}u_{\lambda}(x)-u_{\lambda}(z)\big{)}d\mu(z,x)=0.$

Therefore, we have

$\int_{X\times
X}\big{(}\partial_{u}\ell(z,x,0,0)u_{\lambda}(z)+\partial_{v}\ell(z,x,0,0)u_{\lambda}(x)\big{)}d\mu(z,x)\geq-\varepsilon(\lambda).$

Letting $\lambda\to 0$ along the sequence $(\lambda_{n})_{n}$, we then get the
result. ∎

######  Remark 4.2.

As observed in the proof of Proposition 3.1, since $u_{\lambda}$ is a fixed
point of $T_{\lambda}$, for each $x\in X$, there is a sequence
$(x_{n})_{-n\in\mathbb{N}}$ with $x_{0}=x$, such that

$\forall n\leqslant 0,\quad
u_{\lambda}(x_{n})=u_{\lambda}(x_{n-1})+\ell\big{(}x_{n-1},x_{n},\lambda
u_{\lambda}(x_{n-1}),\lambda u_{\lambda}(x_{n})\big{)}.$

Here we note that the sequence $(x_{n})_{-n\in\mathbb{N}}$ depends on $x$ and
$\lambda$.

######  Lemma 4.1.

For $-n\in\mathbb{N}_{+}$, we define

$\beta_{n}=\frac{\prod_{i=n}^{-1}\big{(}1+\lambda\partial_{u}\ell(x_{i},x_{i+1},0,0)\big{)}}{\prod_{i=n}^{0}\big{(}1-\lambda\partial_{v}\ell(x_{i-1},x_{i},0,0)\big{)}},\quad\beta_{0}=\frac{1}{1-\lambda\partial_{v}\ell(x_{-1},x,0,0)}.$

Since $(x_{n})_{-n\in\mathbb{N}}$ depends on $x$ and $\lambda$, the sequence
$(\beta_{n})_{-n\in\mathbb{N}}$ also depends on $x$ and $\lambda$. For each
integer $N>0$, we have

$\displaystyle u_{\lambda}(x)$
$\displaystyle=\sum_{n=-N+1}^{0}\beta_{n}\bigg{(}\ell(x_{n-1},x_{n},0,0)+\theta_{k}(\lambda)\bigg{)}$
(4.2)
$\displaystyle\quad+\big{(}1+\lambda\partial_{u}\ell(x_{-N},x_{-N+1},0,0)\big{)}\beta_{-N+1}u_{\lambda}(x_{-N}),$

where $|\theta_{k}(\lambda)|\leq\lambda\varepsilon(\lambda)$.

###### Proof.

Since $u_{\lambda}$ is a fixed point of $T_{\lambda}$, we have

$\displaystyle u_{\lambda}(x)$
$\displaystyle=u_{\lambda}(x_{-1})+\ell\big{(}x_{-1},x,\lambda
u_{\lambda}(x_{-1}),\lambda u_{\lambda}(x)\big{)}$
$\displaystyle=u_{\lambda}(x_{-1})+\ell(x_{-1},x,0,0)+\partial_{u}\ell(x_{-1},x,0,0)\lambda
u_{\lambda}(x_{-1})+\partial_{v}\ell(x_{-1},x,0,0)\lambda
u_{\lambda}(x)+\theta_{0}(\lambda),$

where $|\theta_{0}(\lambda)|\leq\lambda\varepsilon(\lambda)$, which implies

$\displaystyle u_{\lambda}(x)$
$\displaystyle=\frac{1+\lambda\partial_{u}\ell(x_{-1},x,0,0)}{1-\lambda\partial_{v}\ell(x_{-1},x,0,0)}u_{\lambda}(x_{-1})+\frac{1}{1-\lambda\partial_{v}\ell(x_{-1},x,0,0)}\big{(}\ell(x_{-1},x,0)+\theta_{0}(\lambda)\big{)}$
$\displaystyle=\big{(}1+\lambda\partial_{u}\ell(x_{-1},x,0,0)\big{)}\beta_{0}u_{\lambda}(x_{-1})+\beta_{0}\big{(}\ell(x_{-1},x,0)+\theta_{0}(\lambda)\big{)}.$

We also have

$u_{\lambda}(x_{-1})=u_{\lambda}(x_{-2})+\ell\big{(}x_{-2},x_{-1},\lambda
u_{\lambda}(x_{-2}),\lambda u_{\lambda}(x_{-1})\big{)},$

which implies

$\displaystyle\big{(}1+\lambda\partial_{u}\ell(x_{-1},x,0,0)\big{)}\beta_{0}u_{\lambda}(x_{-1})$
$\displaystyle=\big{(}1+\lambda\partial_{u}\ell(x_{-2},x_{-1},0,0)\big{)}\beta_{-1}u_{\lambda}(x_{-2})+\beta_{-1}\big{(}\ell(x_{-2},x_{-1},0,0)+\theta_{-1}(\lambda)\big{)},$

where $|\theta_{-1}(\lambda)|\leq\lambda\varepsilon(\lambda)$. Letting this
procedure go on, and adding all equalities up, we get (4.2). ∎

######  Proposition 4.2.

There is $r>0$ such that for each $\lambda\in(0,r)$, there is $K>0$,
independent of $\lambda$ and $x$, such that $\lambda\sum_{k\leq
0}\beta_{k}\leq K$.

###### Proof.

We argue by contradiction. Assume there is a sequence
$(\lambda_{n})_{n\in\mathbb{N}}\to 0$ and $(x_{n})_{n\in\mathbb{N}}\in
X^{\mathbb{N}}$, and for all $n$, a minimal sequence
$(x^{n}_{k})_{-k\in\mathbb{N}}$ associated to $u_{\lambda_{n}}(x_{n})$ and an
integer $N_{n}>0$ such that

$\lambda_{n}\sum_{k=-N_{n}}^{-1}\beta^{n}_{k+1}\to+\infty.$

Here $(\beta^{n}_{k})_{-k\in\mathbb{N}}$ is the sequence associated to
$(x^{n}_{k})_{-k\in\mathbb{N}}$ as defined in Lemma 4.1, which depends on
$x_{n}$ and $\lambda_{n}$. As for $k$ fixed, $\beta^{n}_{k}\to 1$ as
$n\to+\infty$, we have $N_{n}\to+\infty$.

Define the probability measure

$\mu_{n}:=C_{n}^{-1}\sum_{k=-N_{n}}^{-1}\beta_{k+1}^{n}\delta_{(x^{n}_{k},x^{n}_{k+1})},$

where $C_{n}=\sum_{k=-N_{n}}^{-1}\beta^{n}_{k+1}$. Up to an extraction, we
assume $\mu_{n}\to\mu$.

$\mu$ is closed: Let $f\in C^{0}(X,\mathbb{R})$, we have

$\displaystyle\bigg{|}\int_{X\times
X}\big{(}f(x)-f(z)\big{)}d\mu_{n}(z,x)\bigg{|}$
$\displaystyle=C_{n}^{-1}\bigg{|}\sum_{k=-N_{n}}^{-1}\beta_{k+1}^{n}\big{(}f(x^{n}_{k+1})-f(x^{n}_{k})\big{)}\bigg{|}$
$\displaystyle=C_{n}^{-1}\bigg{|}\sum_{k=-N_{n}}^{-1}(\beta_{k}^{n}-\beta_{k+1}^{n})f(x^{n}_{k})-\beta^{n}_{-N_{n}}f(x^{n}_{-N_{n}})+\beta^{n}_{0}f(x_{n})\bigg{|}$
$\displaystyle\leq
C_{n}^{-1}\bigg{(}\sum_{k=-N_{n}}^{-1}(\beta_{k+1}^{n}-\beta_{k}^{n})\|f\|_{\infty}+2\|f\|_{\infty}\bigg{)}\leq
4C^{-1}_{n}\|f\|_{\infty},$

where we use the fact $\beta^{n}_{k}\leq\beta^{n}_{k+1}$. Since
$C_{n}\to+\infty$, $\mu$ is closed.

$\mu$ is minimizing: By definition we have

$\displaystyle\bigg{|}\int_{X\times X}\ell(z,x,0,0)d\mu_{n}(z,x)\bigg{|}$
$\displaystyle=C_{n}^{-1}\bigg{|}\sum_{k=-N_{n}}^{-1}\beta_{k+1}^{n}\ell(x^{n}_{k},x^{n}_{k+1},0,0)\bigg{|}$
$\displaystyle=C_{n}^{-1}\bigg{|}u_{\lambda_{n}}(x_{n})-\sum_{k=-N_{n}}^{-1}\beta^{n}_{k+1}\theta^{n}_{k+1}(\lambda_{n})-\big{(}1+\lambda\partial_{u}\ell(x^{n}_{-N},x^{n}_{-N+1},0,0)\big{)}\beta^{n}_{-N+1}u_{\lambda}(x^{n}_{-N})\bigg{|}$
$\displaystyle\leq
2C^{-1}_{n}\|u_{\lambda_{n}}\|_{\infty}+\lambda_{n}\omega(\lambda_{n})\to 0.$

Now using

$\frac{1+x}{1-y}=1+\frac{x+y}{1-y}\leq\exp\bigg{\\{}\frac{x+y}{1-y}\bigg{\\}}\leq\exp\\{x+y\\},\quad\textrm{for}\
\ y\leq 0,$

we get

$\displaystyle-\int_{X\times X}\Lambda(z,x)d\mu_{n}(z,x)$
$\displaystyle=-C_{n}^{-1}\sum_{k=-N_{n}}^{-1}\beta^{n}_{k+1}\Lambda(x^{n}_{k},x^{n}_{k+1})$
$\displaystyle\leq-
C_{n}^{-1}\sum_{k=-N_{n}}^{-1}\exp\\{\lambda_{n}\sum_{i=k+1}^{-1}\Lambda(x^{n}_{i},x^{n}_{i+1})\\}\Lambda(x^{n}_{k},x^{n}_{k+1})$
$\displaystyle\leq-
C_{n}^{-1}\exp\\{\|\Lambda\|_{\infty}\\}\sum_{k=-N_{n}}^{-1}\exp\\{\lambda_{n}\sum_{i=k}^{-1}\Lambda(x^{n}_{i},x^{n}_{i+1})\\}\Lambda(x^{n}_{k},x^{n}_{k+1})$
$\displaystyle\leq
C_{n}^{-1}\exp\\{\|\Lambda\|_{\infty}\\}\int_{0}^{+\infty}e^{-\lambda_{n}x}dx=\frac{\exp\\{\|\Lambda\|_{\infty}\\}}{\lambda_{n}C_{n}}\to
0,$

which contradicts (l4). ∎

In the following, let $(x_{n})_{-n\in\mathbb{N}}$ and
$(\beta_{n})_{-n\in\mathbb{N}}$ be the sequences defined in Remark 4.2 and
Lemma 4.1 respectively, associated to the pair $x_{0}\in X$ and
$\lambda\in(0,r)$.

######  Proposition 4.3.

For each $\lambda\in(0,r)$ and $x_{0}\in X$, we have

$u_{\lambda}(x_{0})=\sum_{n\leq
0}\beta_{n}\ell(x_{n-1},x_{n},0,0)+\Omega(\lambda),$ (4.3)

where

$\lim_{\lambda\to 0}\Omega(\lambda)\to 0.$

###### Proof.

Since $\lambda\sum_{n\leq 0}\beta_{n}\leq K$, we have $\beta_{-N+1}\to 0$ and

$\bigg{|}\sum_{n=-N+1}^{0}\beta_{n}\theta_{n}(\lambda)\bigg{|}\leq\sum_{n\leq
0}\beta_{n}\lambda\varepsilon(\lambda)\leq K\varepsilon(\lambda)\to 0.$

By (4.2) we get (4.3), where $\Omega(\lambda):=\sum_{n\leq
0}\beta_{n}\theta_{n}(\lambda)$. ∎

######  Definition 4.1.

We define the following probability measures on $X\times X$

$\mu^{1}_{\lambda}=C_{\lambda}^{-1}\sum_{k\leq-1}\frac{\beta_{k+1}}{1-\lambda\partial_{v}\ell(x_{k-1},x_{k},0,0)}\delta_{(x_{k},x_{k+1})},$

and

$\mu^{2}_{\lambda}=C_{\lambda}^{-1}\sum_{k\leq-1}\frac{\beta_{k+1}}{1-\lambda\partial_{v}\ell(x_{k-1},x_{k},0,0)}\delta_{(x_{k-1},x_{k})},$

where

$C_{\lambda}:=\sum_{k\leq-1}\frac{\beta_{k+1}}{1-\lambda\partial_{v}\ell(x_{k-1},x_{k},0,0)}.$

Note that as $x_{0}$ will be fixed in what follows, we only specify explicitly
the dependance of those measures in $\lambda$ but they also depend on $x_{0}$.

Since the functions $(z,x)\mapsto\partial_{v}\ell(z,x,0,0)$ and
$(z,x)\mapsto\partial_{u}\ell(z,x,0,0)$ are bounded respectively by
$\kappa_{v}$ and $\kappa_{u}$, it is easily observed that
$C_{\lambda}\to+\infty$ (each term of the sum converges to $1$ as $\lambda\to
0$). Moreover $\lambda C_{\lambda}\leq\lambda\sum_{k\leq 0}\beta_{k}\leq K$.

######  Proposition 4.4.

For each subsolution $w$ of $\ell_{0}$ and $\lambda\in(0,r)$, we have

$\displaystyle u_{\lambda}(x)$ $\displaystyle\geq\beta_{0}w(x)$
$\displaystyle\quad+\lambda C_{\lambda}\int_{X\times
X}\partial_{u}\ell(z,x,0,0)w(z)d\mu^{1}_{\lambda}(z,x)$
$\displaystyle\quad+\lambda C_{\lambda}\int_{X\times
X}\partial_{v}\ell(z,x,0,0)w(x)d\mu^{2}_{\lambda}(z,x)+\Omega(\lambda).$

###### Proof.

By (4.3) we have

$\displaystyle u_{\lambda}(x)$ $\displaystyle\geq\sum_{n\leq
0}\beta_{n}\big{(}w(x_{n})-w(x_{n-1})\big{)}+\Omega(\lambda)$
$\displaystyle=\beta_{0}w(x)+\lambda\sum_{k\leq-1}\frac{\beta_{k+1}}{1-\lambda\partial_{v}\ell(x_{k-1},x_{k},0,0)}\partial_{u}\ell(x_{k},x_{k+1},0,0)w(x_{k})$
$\displaystyle\quad+\lambda\sum_{k\leq-1}\frac{\beta_{k+1}}{1-\lambda\partial_{v}\ell(x_{k-1},x_{k},0,0)}\partial_{v}\ell(x_{k-1},x_{k},0,0)w(x_{k})+\Omega(\lambda)$
$\displaystyle\geq\beta_{0}w(x)+\lambda C_{\lambda}\int_{X\times
X}\partial_{u}\ell(z,x,0,0)w(z)d\mu^{1}_{\lambda}$ $\displaystyle\quad+\lambda
C_{\lambda}\int_{X\times
X}\partial_{v}\ell(z,x,0,0)w(x)d\mu^{2}_{\lambda}(z,x)+\Omega(\lambda).$

∎

######  Lemma 4.2.

The limits of $\mu^{1}_{\lambda}$ and $\mu^{2}_{\lambda}$ coincide in the
weak* topology as $\lambda\to 0$. That is, if there is a sequence
$\lambda_{n}\to 0$ such that $\mu^{1}_{\lambda_{n}}\to\mu$, then
$\mu^{2}_{\lambda_{n}}\to\mu$.

###### Proof.

For all $f\in C^{0}(X\times X,\mathbb{R})$, we have

$\displaystyle\bigg{|}\int_{X\times
X}f(z,x)d(\mu^{1}_{\lambda}-\mu^{2}_{\lambda})\bigg{|}$
$\displaystyle=C_{\lambda}^{-1}\bigg{|}\sum_{k\leq-1}\frac{\beta_{k+1}}{1-\lambda\partial_{v}\ell(x_{k-1},x_{k},0,0)}\big{(}f(x_{k},x_{k+1})-f(x_{k-1},x_{k})\big{)}\bigg{|}$
$\displaystyle=C_{\lambda}^{-1}\bigg{|}\frac{\beta_{0}}{1-\lambda\partial_{v}\ell(x_{-2},x_{-1},0,0)}f(x_{-1},x_{0})+\sum_{k\leq-2}\frac{\beta_{k+1}}{1-\lambda\partial_{v}\ell(x_{k-1},x_{k},0,0)}f(x_{k},x_{k+1})$
$\displaystyle\quad-\sum_{k\leq-2}\frac{\beta_{k+1}}{1+\lambda\partial_{u}\ell(x_{k+1},x_{k+2},0,0)}f(x_{k},x_{k+1})\bigg{|}$
$\displaystyle\leq C_{\lambda}^{-1}\|f\|_{\infty}$ $\displaystyle\quad+\lambda
C_{\lambda}^{-1}\bigg{|}\sum_{k\leq-2}\frac{\partial_{u}\ell(x_{k+1},x_{k+2},0,0)+\partial_{v}\ell(x_{k-1},x_{k},0,0)}{\big{(}1-\lambda\partial_{v}\ell(x_{k-1},x_{k},0,0)\big{)}\big{(}1+\lambda\partial_{u}\ell(x_{k+1},x_{k+2},0,0)\big{)}}\beta_{k+1}f(x_{k},x_{k+1})\bigg{|}$
$\displaystyle\leq C_{\lambda}^{-1}\|f\|_{\infty}+\lambda
C_{\lambda}^{-1}\frac{\kappa_{u}+\kappa_{v}}{1-\lambda\kappa_{u}}\sum_{k\leq-1}\beta_{k+1}\|f\|_{\infty}$
$\displaystyle\leq\left(1+\frac{\kappa_{u}+\kappa_{v}}{1-\lambda\kappa_{u}}K\right)C_{\lambda}^{-1}\|f\|_{\infty}\to
0.$

Indeed, recall that $C_{\lambda}\to+\infty$ as $\lambda\to 0$. ∎

######  Proposition 4.5.

Any limit $\mu$ of $d\mu^{1,2}_{\lambda_{n}}$ as $\lambda_{n}\to 0$ is a
Mather measure of $\ell(z,x,0,0)$.

###### Proof.

We first prove that $\mu$ is closed. Let $f\in C^{0}(X,\mathbb{R})$, then

$\displaystyle\int_{X\times X}\big{(}f(x)-f(z)\big{)}d\mu^{1}_{\lambda}(z,x)$
$\displaystyle=C_{\lambda}^{-1}\sum_{k\leq-1}\frac{\beta_{k+1}}{1-\lambda\partial_{v}\ell(x_{k-1},x_{k},0,0)}\big{(}f(x_{k+1})-f(x_{k})\big{)}$
$\displaystyle=C_{\lambda}^{-1}\frac{\beta_{0}}{1-\lambda\partial_{v}\ell(x_{-2},x_{-1},0,0)}f(x)$
$\displaystyle\quad+\lambda
C_{\lambda}^{-1}\sum_{k\leq-1}\beta_{k+1}\frac{\partial_{u}\ell(x_{k-2},x_{k-1},0,0)+\partial_{v}\ell(x_{k},x_{k+1},0,0)}{\big{(}1-\lambda\partial_{v}\ell(x_{k-2},x_{k-1},0,0)\big{)}\big{(}1-\lambda\partial_{v}\ell(x_{k-1},x_{k},0,0)\big{)}}f(x_{k}).$

Since $-\kappa_{v}<\partial_{v}\ell(z,x,0,0)\leq 0$ and
$-\kappa_{v}\leq\partial_{u}\ell(z,x,0,0)\leq 0$, we have

$\bigg{|}\int_{X\times
X}\big{(}f(x)-f(z)\big{)}d\mu^{1}_{\lambda}(z,x)\bigg{|}\\\ \leq
C_{\lambda}^{-1}\|f\|_{\infty}(1+(\kappa_{u}+\kappa_{v})\lambda\sum_{k\leq
0}\beta_{k})\\\
\leq(1+(\kappa_{u}+\kappa_{v})K)C_{\lambda}^{-1}\|f\|_{\infty}\to 0.$

Then we prove $\mu$ is minimizing. By (4.3) we have

$\displaystyle\bigg{|}\int_{X\times X}$
$\displaystyle\ell(z,x,0,0)\big{(}1-\lambda\partial_{v}\ell_{0}(z,x)\big{)}d\mu^{2}_{\lambda}(z,x)-C_{\lambda}^{-1}u_{\lambda}(x_{0})\bigg{|}$
$\displaystyle=C_{\lambda}^{-1}\bigg{|}\sum_{k\leq-1}(\beta_{k+1}-\beta_{k})\ell(x_{k-1},x_{k},0,0)-\beta_{0}\ell_{0}(x_{-1},x)-\Omega(\lambda)\bigg{|}$
$\displaystyle\leq
C_{\lambda}^{-1}\bigg{(}\bigg{|}\lambda\sum_{k\leq-1}\frac{\partial_{u}\ell_{0}(x_{k},x_{k+1},0,0)+\partial_{v}\ell(x_{k-1},x_{k},0,0)}{1-\lambda\partial_{v}\ell(x_{k-1},x_{k},0,0)}\beta_{k+1}\ell_{0}(x_{k-1},x_{k})\bigg{|}$
$\displaystyle\qquad\qquad+\|\ell_{0}\|_{\infty}+K\varepsilon(\lambda)\bigg{)}$
$\displaystyle\leq C_{\lambda}^{-1}\big{(}\lambda\sum_{k\leq
0}(\kappa_{u}+\kappa_{v})\beta_{k}\|\ell_{0}\|_{\infty}+\|\ell_{0}\|_{\infty}+K\varepsilon(\lambda)\big{)}$
$\displaystyle\leq
C_{\lambda}^{-1}\big{(}((\kappa_{u}+\kappa_{v})K+1)\|\ell_{0}\|_{\infty}+K\varepsilon(\lambda)\big{)}\to
0.$

Let $\lambda\to 0$, we get $\int_{X\times
X}\ell_{0}(z,x)d\mu_{\lambda}(z,x)=0$. ∎

We now turn to the proof of the main theorem. Recall that $\mathcal{S}_{0}$ is
the set of subsolutions $w$ of $\ell_{0}$ that satisfy

$\int_{X\times
X}\big{(}\partial_{u}\ell(z,x,0,0)w(z)+\partial_{v}\ell(z,x,0,0)w(x)\big{)}d\mu(z,x)\geq
0,$

for all Mather measures $\mu$ of $\ell_{0}$.

_Proof of Theorem 4._ We first show that $u_{0}$ is well-define, that is,
functions in $\mathcal{S}_{0}$ are uniformly bounded from above. Assume there
is $w\in\mathcal{S}_{0}$ such that $w\geq\delta>0$, then by (4.1) we have

$\displaystyle 0$ $\displaystyle\leq\int_{X\times
X}\big{(}\partial_{u}\ell(z,x,0,0)w(z)+\partial_{v}\ell(z,x,0,0)w(x)\big{)}d\mu(z,x)$
$\displaystyle\leq\delta\int_{X\times
X}\big{(}\partial_{u}\ell(z,x,0,0)+\partial_{v}\ell(z,x,0,0)\big{)}d\mu(z,x),$

which contradicts (l4). Therefore, for all $w\in\mathcal{S}_{0}$, there is
$x_{0}\in X$ such that $w(x_{0})\leq 0$. By the equi-continuity of
subsolutions of $\ell_{0}$, the result follows.

Recall that we consider a decreasing sequence $\lambda_{n}\to 0$ such that
$u_{\lambda_{n}}\to u_{*}$ uniformly.

Since for all $(x,z)\in X\times X$,

$u_{\lambda}(x)-u_{\lambda}(z)\leq\ell\big{(}z,x,\lambda
u_{\lambda}(x),\lambda u_{\lambda}(z)\big{)}.$

Let $\lambda\to 0$ we get

$u_{*}(z)-u_{*}(x)\leq\ell(z,x,0,0),$

which means $u_{*}$ is a subsolution of $\ell_{0}$. By Proposition 4.1,
$u_{*}\in\mathcal{S}_{0}$, which implies $u_{*}\leq u_{0}$.

Let now $x_{0}\in X$ and up to a further extraction, assume that the
associated measures converge. By Lemma 4.2 the limits are the same and
$\lim\limits_{n\to+\infty}\mu_{\lambda_{n}}^{1}=\lim\limits_{n\to+\infty}\mu_{\lambda_{n}}^{2}=\mu$
is a Mather measure by Proposition 4.5. For each $w\in\mathcal{S}_{0}$, by
Proposition 4.4, we have

$\displaystyle u_{*}(x_{0})$
$\displaystyle\geq\lim_{n\to+\infty}\beta_{0}w(x_{0})$
$\displaystyle\quad+\limsup_{n\to+\infty}\lambda_{n}C_{\lambda_{n}}\bigg{(}\int_{X\times
X}\partial_{u}\ell(z,x,0,0)w(z)d\mu^{1}_{\lambda_{n}}+\int_{X\times
X}\partial_{v}\ell(z,x,0,0)w(x)d\mu^{2}_{\lambda_{n}}\bigg{)}$
$\displaystyle=w(x_{0})+\limsup_{n\to+\infty}\lambda_{n}C_{\lambda_{n}}\int_{X\times
X}\big{(}\partial_{u}\ell(z,x,0,0)+\partial_{v}\ell(z,x,0,0)\big{)}w(x)d\mu$
$\displaystyle\geq w(x_{0}),$

where we have used that $w\in\mathcal{S}_{0}$. Therefore,
$u_{*}(x_{0})\geq\sup_{w\in\mathcal{S}_{0}}w(x_{0})=u_{0}(x_{0})$. We finally
get $u_{*}=u_{0}$.

Now we prove that $u_{0}$ is a fixed point of $T_{0}$. We have seen that
$u_{0}=u_{*}$ is a subsolution. Let $x_{0}\in X$. Since $X$ is compact, let
$z_{n}$ be a point realizing the minimum in $u_{\lambda_{n}}(x_{0})$ and up to
extracting, assume $z_{\lambda_{n}}\to z_{*}$. By

$u_{\lambda_{n}}(x_{0})-u_{\lambda_{n}}(z_{\lambda_{n}})=\ell\big{(}z_{\lambda_{n}},x_{0},\lambda_{n}u_{\lambda_{n}}(z_{\lambda_{n}}),\lambda_{n}u_{\lambda_{n}}(x_{0})\big{)},$

we get

$u_{0}(x_{0})-u_{0}(z_{*})=\ell(z_{*},x_{0},0,0).$

Thus, $u_{0}$ is a fixed point of $T_{0}$. ∎

######  Remark 4.3.

As a byproduct of the previous proof, we have also proven that
$u_{0}\in\mathcal{S}_{0}$.

We finish this section by the alternative representation formula of $u_{0}$.

Proof of Theorem 5. Define

$\hat{u}_{0}(x)=\min_{\mu\in\mathfrak{M}_{0}}\frac{\int_{X\times
X}\Big{(}\partial_{u}\ell(z,y,0,0)h(z,x)+\partial_{v}\ell(z,y,0,0)h(y,x)\Big{)}d\mu(z,y)}{\int_{X\times
X}\Lambda(z,y)d\mu(z,y)},$

where $\mathfrak{M}_{0}$ denotes the set of Mather measures of $\ell_{0}$.
Note first that for each $\mu\in\mathfrak{M}_{0}$, the function

$x\mapsto\frac{\int_{X\times
X}\Big{(}\partial_{u}\ell(z,y,0,0)h(z,x)+\partial_{v}\ell(z,y,0,0)h(y,x)\Big{)}d\mu(z,y)}{\int_{X\times
X}\Lambda(z,y)d\mu(z,y)},$

is a subsolution of $\ell_{0}$. Indeed each $-h(z,\cdot)$ is a subsolution
(Proposition 1.2) hence the integral is a barycenter of subsolutions
(Proposition 1.1). Hence $\hat{u}_{0}$ is also a subsolution for $\ell_{0}$ as
an infimum of subsolutions.

proof that $u_{0}\leq\hat{u}_{0}$: Let $x\in X$ and $\mu\in\mathfrak{M}_{0}$.
Integrating the inequalities $u_{0}(x)\leqslant u_{0}(z)+h(z,x)$ recalled in
Proposition 1.2, we find that

$u_{0}(x)\int_{X\times X}\partial_{u}\ell(z,y,0,0)d\mu(z,y)\geq\int_{X\times
X}\partial_{u}\ell(z,y,0,0)\big{(}u_{0}(z)+h(z,x)\big{)}d\mu(z,y),$ (4.4)
$u_{0}(x)\int_{X\times X}\partial_{v}\ell(z,y,0,0)d\mu(z,y)\geq\int_{X\times
X}\partial_{v}\ell(z,y,0,0)\big{(}u_{0}(y)+h(y,x)\big{)}d\mu(z,y).$ (4.5)

Recall that $u_{0}\in\mathcal{S}_{0}$ so that

$\int_{X\times
X}\big{(}\partial_{u}\ell(z,y,0,0)u_{0}(z)+\partial_{v}\ell(z,y,0,0)u_{0}(y)\big{)}d\mu(z,y)\geq
0.$

Therefore, summing (4.4) and (4.5) and using the previous inequality, we
obtain

$u_{0}(x)\int_{X\times X}\Lambda(z,y)d\mu(z,y)\geq\int_{X\times
X}\Big{(}\partial_{u}\ell(z,y,0,0)h(z,x)+\partial_{v}\ell(z,y,0,0)h(y,x)\Big{)}d\mu(z,y).$

Dividing by $\int_{X\times X}\Lambda(z,y)d\mu(z,y)<0$ and taking a minimum
over $\mu\in\mathfrak{M}_{0}$ yields the desired
$u_{0}(x)\leqslant\hat{u}_{0}(x)$.

proof that $u_{0}\geq\hat{u}_{0}$: We first show
$v_{y}(\cdot):=-h(\cdot,y)+\hat{u}_{0}(y)\in\mathcal{S}_{0}$ for all $y\in X$.
Let $\mu\in\mathfrak{M}_{0}$. We get

$\displaystyle\int_{X\times
X}\big{(}\partial_{u}\ell(z,x,0,0)v_{y}(z)+\partial_{v}\ell(z,x,0,0)v_{y}(x)\big{)}d\mu(z,x)$
$\displaystyle\quad=-\int_{X\times
X}\Big{(}h(z,y)\partial_{u}\ell(z,x,0,0)+h(x,y)\partial_{v}\ell(z,x,0,0)\Big{)}d\mu(z,x)$
$\displaystyle\quad\quad+\int_{X\times X}\Lambda(z,x)\hat{u}_{0}(y)d\mu(z,x)$
$\displaystyle\quad=-\int_{X\times
X}\Big{(}h(z,y)\partial_{u}\ell(z,x,0,0)+h(x,y)\partial_{v}\ell(z,x,0,0)\Big{)}d\mu(z,x)$
$\displaystyle+\int_{X\times
X}\Lambda(z,x)d\mu(z,x)\min_{\tilde{\mu}\in\mathfrak{M}_{0}}\frac{\int_{X\times
X}\Big{(}\partial_{u}\ell(z,x,0,0)h(z,y)+\partial_{v}\ell(z,x,0,0)h(x,y)\Big{)}d\tilde{\mu}(z,x)}{\int_{X\times
X}\Lambda(z,x)d\tilde{\mu}(z,x)}$ $\displaystyle{\geq 0}.$

It follows that $v_{y}\leqslant u_{0}$ and evaluating at $y$ yields
$u_{0}(y)\geq-h(y,y)+\hat{u}_{0}(y)$. Let $y\in\mathfrak{A}$, we have
$h(y,y)=0$, and $u_{0}(y)\geq\hat{u}_{0}(y)$. By comparison (Proposition 1.3),
as $u_{0}$ is a solution and $\hat{u}_{0}$ a subsolution, we finally get
$u_{0}\geq\hat{u}_{0}$. ∎

##  5 Uniqueness of $u_{\lambda}$

######  Theorem 6.

The fixed point $u_{\lambda}$ of $T_{\lambda}$ is unique if $\lambda$ is small
and one of the following holds

  * (1)

$\ell(z,x,u,v)$ is concave in $u$ and concave in $v$;

  * (2)

$\partial_{u,v}\ell(z,x,u,v)$ exist and are continuous for $(u,v)$ near
$(0,0)$.

###### Proof.

We argue by contradiction. Let $u_{\lambda}$ and $v_{\lambda}$ be two fixed
points. Assume

$u_{\lambda}(x_{0})-v_{\lambda}(x_{0})=\max_{x\in
X}\big{(}u_{\lambda}(x)-v_{\lambda}(x)\big{)}>0.$

Let $(x_{-k})_{k\in\mathbb{N}}$ be a minimizing sequence associated to
$v_{\lambda}(x_{0})$ as defined in Remark 4.2.

Step 1. We first show that if $\lambda$ is small enough, we have
$u_{\lambda}(x_{-k})>v_{\lambda}(x_{-k})$ for all $k\geq 0$. Assume
$u_{\lambda}(x_{-1})\leq v_{\lambda}(x_{-1})$. By (l1) and (l2), we have

$\displaystyle u_{\lambda}(x_{0})-u_{\lambda}(x_{-1})$
$\displaystyle\leq\ell\big{(}x_{-1},x,\lambda u_{\lambda}(x_{-1}),\lambda
u_{\lambda}(x_{0})\big{)}$ $\displaystyle\leq\ell\big{(}x_{-1},x_{0},\lambda
u_{\lambda}(x_{-1}),\lambda v_{\lambda}(x_{0})\big{)}$
$\displaystyle\leq\ell\big{(}x_{-1},x_{0},\lambda v_{\lambda}(x_{-1}),\lambda
v_{\lambda}(x_{0})\big{)}+\lambda\kappa_{u}(v_{\lambda}-u_{\lambda})(x_{-1})$
$\displaystyle=v_{\lambda}(x_{0})-v_{\lambda}(x_{-1})+\lambda\kappa_{u}(v_{\lambda}-u_{\lambda})(x_{-1}),$

which implies that

$(1-\lambda\kappa_{u})(u_{\lambda}-v_{\lambda})(x_{-1})\geq
u_{\lambda}(x_{0})-v_{\lambda}(x_{0})>0,$

which leads to a contradiction. Then
$u_{\lambda}(x_{-1})>v_{\lambda}(x_{-1})$. We then go on to find, using (l2),
that

$\displaystyle u_{\lambda}(x_{-1})-u_{\lambda}(x_{-2})$
$\displaystyle\leq\ell\big{(}x_{-2},x,\lambda u_{\lambda}(x_{-2}),\lambda
u_{\lambda}(x_{-1})\big{)}$ $\displaystyle\leq\ell\big{(}x_{-2},x_{-1},\lambda
u_{\lambda}(x_{-2}),\lambda v_{\lambda}(x_{-1})\big{)}$
$\displaystyle\leq\ell\big{(}x_{-2},x_{-1},\lambda v_{\lambda}(x_{-2}),\lambda
v_{\lambda}(x_{-1})\big{)}$
$\displaystyle=v_{\lambda}(x_{-1})-v_{\lambda}(x_{-2}),$

therefore $u_{\lambda}(x_{-2})-v_{\lambda}(x_{-2})\geq
u_{\lambda}(x_{-1})-v_{\lambda}(x_{-1})>0$.

By induction, we have $u_{\lambda}(x_{-k})>v_{\lambda}(x_{-k})$ for all $k\geq
0$.

Step 2. By Step 1 and (l2), we have

$\displaystyle v_{\lambda}(x_{-k+1})-v_{\lambda}(x_{-k})$
$\displaystyle=\ell\big{(}x_{-k},x_{-k+1},\lambda v_{\lambda}(x_{-k}),\lambda
v_{\lambda}(x_{-k+1})\big{)}$ (5.1)
$\displaystyle\geq\ell\big{(}x_{-k},x_{-k+1},\lambda
u_{\lambda}(x_{-k}),\lambda u_{\lambda}(x_{-k+1})\big{)}$ $\displaystyle\geq
u_{\lambda}(x_{-k+1})-u_{\lambda}(x_{-k}),$

which implies

$u_{\lambda}(x_{-k})-v_{\lambda}(x_{-k})\geq\dots\geq
u_{\lambda}(x_{0})-v_{\lambda}(x_{0}).$

By the definition of $x_{0}$, all inequalities above are equalities, that is,

$u_{\lambda}(x_{-k})-v_{\lambda}(x_{-k})=\max_{x\in
X}\big{(}u_{\lambda}(x)-v_{\lambda}(x)\big{)}>0,\quad\forall k\geq 0.$ (5.2)

Step 3. Define the probability measure

$\mu_{N}:=N^{-1}\sum_{k=-N}^{-1}\delta_{(x_{k},x_{k+1})}.$ (5.3)

By compactness of measures, let $(N_{n})_{n}$ be an extraction such that
$\mu_{N_{n}}\to\mu$ as $N\to+\infty$. Since

$\left|\int_{X\times
X}\big{(}f(x)-f(y)\big{)}d\mu_{N}\right|=\left|\frac{f(x_{0})-f(x_{-N})}{N}\right|\leq\frac{2\|f\|_{\infty}}{N}\to
0,$

the measure $\mu$ is closed. We also have

$\int_{X\times X}\ell\big{(}z,x,\lambda v_{\lambda}(z),\lambda
v_{\lambda}(x)\big{)}d\mu_{N}=\frac{v_{\lambda}(x_{0})-v_{\lambda}(x_{-N})}{N}\to
0.$

As all inequalities in (5.1) are equalities, it follows that

$\int_{X\times X}\ell\big{(}z,x,\lambda v_{\lambda}(z),\lambda
v_{\lambda}(x)\big{)}d\mu=\int_{X\times X}\ell\big{(}z,x,\lambda
u_{\lambda}(z),\lambda u_{\lambda}(x)\big{)}d\mu=0.$ (5.4)

Since $\mu_{N_{n}}\to\mu$, for each $(z,x)\in\textrm{supp}(\mu)$, there is a
sequence $(z_{n},x_{n})\in\textrm{supp}(\mu_{N_{n}})$ with
$(z_{n},x_{n})\to(z,x)$. By Step 2, $u_{\lambda}(z_{n})-v_{\lambda}(z_{n})$
equals a positive constant $M$. Therefore,
$u_{\lambda}(z)-v_{\lambda}(z)=M>0$. Similarly, we have
$u_{\lambda}(x)-v_{\lambda}(x)=M>0$. By (l2), we have

$\ell(z,x,r,s)=\ell\big{(}z,x,\lambda u_{\lambda}(z),\lambda
u_{\lambda}(x)\big{)},$

for all $(z,x)\in\textrm{supp}(\mu)$ and $r\in[\lambda v_{\lambda}(z),\lambda
u_{\lambda}(z)]$, $s\in[\lambda v_{\lambda}(x),\lambda u_{\lambda}(x)]$.

Conclusion under hypothesis (1). By (5.2), the intervals $[\lambda
v_{\lambda}(z),\lambda u_{\lambda}(z)]$ and $[\lambda v_{\lambda}(x),\lambda
u_{\lambda}(x)]$ have no-empty interior, by the concavity we have

$\ell\big{(}z,x,\lambda u_{\lambda}(z),\lambda
u_{\lambda}(x)\big{)}=\max_{r,s}\ell(z,x,r,s),\quad\forall(z,x)\in\textrm{supp}(\mu).$

By (l2), we have

$\ell(z,x,r,s)=\ell\big{(}z,x,\lambda u_{\lambda}(z),\lambda
u_{\lambda}(x)\big{)},\quad\forall r\leq\lambda u_{\lambda}(z),\ s\leq\lambda
u_{\lambda}(x).$

Let $u_{0}<0$ and $u_{0}\leq\min_{x\in X}\lambda u_{\lambda}(x)$, we have

$\int_{X\times X}\ell(z,x,u_{0},u_{0})d\mu=\int_{X\times
X}\ell\big{(}z,x,\lambda u_{\lambda}(z),\lambda u_{\lambda}(x)\big{)}d\mu=0.$

By (l2) we also have

$0=\int_{X\times X}\ell(z,x,u_{0},u_{0})d\mu\geq\int_{X\times
X}\ell(z,x,0,0)d\mu.$

Thus, $\mu$ is a Mather measure of $\ell(z,x,0,0)$. By (l2) again we have

$\ell(z,x,r,s)=\ell(z,x,u_{0},u_{0}),\quad\forall(z,x)\in\textrm{supp}(\mu),\
\forall r,s\in[u_{0},0].$

Since $u_{0}<0$, we have

$\partial_{u,v}\ell(z,x,0,0)=0,\quad\forall(z,x)\in\textrm{supp}(\mu),$

which contradicts (l4).

Conclusion under hypothesis (2). Since the set of Mather measures is compact,
there is $\epsilon>0$ such that

$\int_{X\times
X}\Lambda(z,x)d\mu(z,x)<-2\epsilon,\quad\forall\mu\in\mathfrak{M}_{0}.$

We first show that there is $r>0$ and $N_{0}>0$ such that

$N^{-1}\sum_{k=-N}^{-1}\Lambda(x_{k},x_{k+1})<-2\epsilon,\quad\forall\lambda\in(0,r),\
\forall N\geq N_{0}.$

If not, we assume that there is a sequence $\lambda_{n}\to 0$ and
$N_{n}\to+\infty$ such that

$N_{n}^{-1}\sum_{k=-N_{n}}^{-1}\Lambda(x_{k},x_{k+1})\geq-2\epsilon.$

Extracting a subsequence if necessary, let $\mu_{\lambda_{n}}$ be the limit
given by (5.3) and $\mu_{\lambda_{n}}\to\mu$. By (5.4), $\mu$ is a Mather
measure. We then get a contradiction.

Since $\partial_{u,v}\ell$ is continuous for $(u,v)$ near $(0,0)$, for
$\lambda$ small and $N$ large, we have

$N^{-1}\sum_{k=-N}^{-1}\Big{[}\partial_{u}\ell\big{(}x_{k},x_{k+1},\lambda
u_{\lambda}(x_{k}),\lambda
u_{\lambda}(x_{k+1})\big{)}+\partial_{v}\ell\big{(}x_{k},x_{k+1},\lambda
u_{\lambda}(x_{k}),\lambda u_{\lambda}(x_{k+1})\big{)}\Big{]}<-\epsilon.$

Since all inequalities in (5.1) are equalities, we have

$\partial_{u}\ell\big{(}x_{k},x_{k+1},\lambda u_{\lambda}(x_{k}),\lambda
u_{\lambda}(x_{k+1})\big{)}=\partial_{v}\ell\big{(}x_{k},x_{k+1},\lambda
u_{\lambda}(x_{k}),\lambda u_{\lambda}(x_{k+1})\big{)}=0,$

which leads to a contradiction. ∎

## Acknowledgements

The authors are supported by ANR CoSyDy (ANR-CE40-0014).

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