# On the solutions to $p$-Poisson equation with Robin boundary conditions when
$p$ goes to $+\infty$

Vincenzo Amato, Alba Lia Masiello, Carlo Nitsch, Cristina Trombetti

###### Abstract

We study the behaviour, when $p\to+\infty$, of the first $p$-Laplacian
eigenvalues with Robin boundary conditions and the limit of the associated
eigenfunctions. We prove that the limit of the eigenfunctions is a viscosity
solution to an eigenvalue problem for the so-called $\infty$-Laplacian.

Moreover, in the second part of the paper, we focus our attention on the
$p$-Poisson equation when the datum $f$ belongs to $L^{\infty}(\Omega)$ and we
study the behaviour of solutions when $p\to\infty$.

MSC 2020: 35J92, 35J94, 35P15.
Keywords: $p-$Laplacian, Robin boundary conditions, eigenvalues problem,
infinity Laplacian.

##  1 Introduction

Let $\beta$ be a positive parameter and let $\Omega$ be a bounded and open set
of $\operatorname{\mathbb{R}}^{n}$, $n\geq 2$, with Lipschitz boundary.

We study the $\infty$-Laplacian eigenvalue problem with Robin boundary
conditions

$\begin{cases}\min\Set{{\left|\nabla u\right|}-\Lambda
u,-\Delta_{\infty}u}=0&\text{ in }\Omega,\\\ -\min\Set{{\left|\nabla
u\right|}-\beta u,-\displaystyle{\frac{\partial u}{\partial\nu}}}=0&\text{ on
}\partial\Omega,\end{cases}$ (1)

where $\Delta_{\infty}$, the so-called $\infty$-Laplacian, is defined by

$\Delta_{\infty}u=\left\langle D^{2}u\cdot\nabla u,\nabla u\right\rangle.$

We refer to problem (1) as the $\infty$-Laplacian eigenvalue problem because
it can be seen as limit, in some sense, of the $p$-Laplacian eigenvalues
problem

$\begin{cases}-\Delta_{p}u=\Lambda_{p}{\left|u\right|}^{p-2}u&\text{ in
}\Omega\\\ {\left|\nabla u\right|}^{p-2}\displaystyle{\frac{\partial
u}{\partial\nu}}+\beta^{p}{\left|u\right|}^{p-2}u=0&\text{ on
}\partial\Omega.\end{cases}$ (2)

A function $u_{p}\in W^{1,p}(\Omega)$ is a weak solution to (2) if it
satisfies

$\int_{\Omega}{\left|\nabla u_{p}\right|}^{p-2}\nabla
u_{p}\nabla\varphi\,dx+\beta^{p}\int_{\partial\Omega}{\left|u_{p}\right|}^{p-2}u_{p}\varphi\,d\mathcal{H}^{n-1}=\Lambda_{p}\int_{\Omega}{\left|u_{p}\right|}^{p-2}u_{p}\varphi\,dx,\quad\forall\varphi\in
W^{1,p}(\Omega),$

where $\beta$ and $\Lambda_{p}$ are both positive.

It is well known that the first eigenvalue of the $p$-Laplacian is the minimum
of the following Rayleigh quotient

$\Lambda_{p}=\inf_{\begin{subarray}{c}w\in W^{1,p}(\Omega)\\\
{\left\|w\right\|}_{L^{p}(\Omega)}=1\end{subarray}}\left\\{\int_{\Omega}{\left|\nabla
w\right|}^{p}\,dx+\beta^{p}\int_{\partial\Omega}{\left|w\right|}^{p}\,d\mathcal{H}^{n-1}\right\\}.$
(3)

By classical arguments, one can show that the infimum in (3) is achieved and
in what follows we will denote by $u_{p}\in W^{1,p}(\Omega)$ the eigenfunction
corresponding to the first eigenvalue $\Lambda_{p}$.

In this paper, we firstly prove that

$\lim_{p\rightarrow+\infty}(\Lambda_{p})^{1/p}=\Lambda_{\infty}=:\inf_{\begin{subarray}{c}w\in
W^{1,\infty}(\Omega)\\\
{\left\|w\right\|}_{L^{\infty}(\Omega)}=1\end{subarray}}\max\left\\{{\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)},\beta{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}\right\\},$
(4)

and we give a geometric characterization of this quantity, precisely:

$\Lambda_{\infty}=\frac{1}{1/\beta+R_{\Omega}},$ (5)

where $R_{\Omega}$ denotes the inradius of $\Omega$, i.e. the radius of the l
argest ball contained in $\Omega$. Thereafter, we prove that
$\Lambda_{\infty}$ is the first eigenvalue of the infinite Laplacian, in the
sense that equation (1) admits non-trivial solutions only if
$\Lambda\geq\Lambda_{\infty}$.

Similar results, in the case of Dirichlet and Neumann boundary conditions were
obtained in [JLM99, JL05, BDM89, EKNT15, RS16].

More specifically, in [JLM99, JL05], Juutinen, Lindqvist and Manfredi have
studied the Dirichlet case as $p\rightarrow+\infty$. They provided a complete
characterization of the limiting solutions in terms of geometric quantities.
Indeed, the first eigenvalue of the $p$-Laplace operator
$\\{\lambda_{p}^{D}\\}$ happens to satisfy

$\lim_{p\to\infty}\left(\lambda_{p}^{D}\right)^{{1}/{p}}=\lambda^{D}_{\infty}:=\frac{1}{R_{\Omega}}.$

The related eigenfunctions $v^{D}_{p}$ also converge (up to a subsequence) to
some Lipschitz function $v^{D}_{\infty}$. Most important, the Authors show
that there exists a natural viscosity formulation of the eigenvalue problem
for the $\infty$-Laplacian, for which $\lambda^{D}_{\infty}$ and
$v^{D}_{\infty}$ turn out to be the first eigenvalue and first eigenfunction,
respectively.

The Neumann case seems to be more subtle. It was investigated in [EKNT15,
RS16] and similarly to the Dirichlet case, the Authors established that the
first non-trivial eigenvalues of the $p$-Laplacian $\\{\lambda^{N}_{p}\\}$
satisfy

$\lim_{p\to\infty}\left(\lambda^{N}_{p}\right)^{1/p}=\lambda_{\infty}^{N}:=\frac{2}{\operatorname{\text{diam}}(\Omega)},$

where $\operatorname{\text{diam}}(\Omega)$ is the intrinsic diameter of
$\Omega$, i.e. the supremum of the geodetic distance between two points of
$\Omega$.

However, while both first eigenvalues and first eigenfunctions converge (as
$p\to\infty$) and are solutions to some appropriate eigenvalue problem for the
$\infty$-Laplacian, in [EKNT15], the Authors are able to prove that they
actually converge to the _first_ eigenvalue e _first_ eigenfunction only if
the domain $\Omega$ is convex. Whether or not the same holds true in the
general case, it is still an open problem.

In the second part of the paper, we focus our attention on the study of the
limit of the $p$-Poisson equation with Robin boundary conditions:

$\begin{cases}-\Delta_{p}v=f&\text{ in }\Omega\\\ {\left|\nabla
v\right|}^{p-2}\displaystyle{\frac{\partial
v}{\partial\nu}}+\beta^{p}{\left|v\right|}^{p-2}v=0&\text{ on
}\partial\Omega,\end{cases}$ (6)

when $f\in L^{\infty}(\Omega)$ is a non-negative function.

We prove that there exists (up to a subsequence) a limiting solution
$v_{\infty}$ as $p\to\infty$ and we establish conditions on $f$ which are
equivalent to the uniqueness of $v_{\infty}$.

The $\infty$-Poisson problem for Dirichlet boundary conditions was already
studied in [BDM89] by Bhattacharya, DiBenedetto and Manfredi, while, to the
best of our knowledge, similar results have not been addressed in the case of
Neumann boundary conditions.

##  2 Notations and Preliminaries

Throughout this article, $|\cdot|$ will denote the Euclidean norm in
$\mathbb{R}^{n}$, and $\mathcal{H}^{k}(\cdot)$, for $k\in[0,n)$, will denote
the $k-$dimensional Hausdorff measure in $\mathbb{R}^{n}$.

We denote by $d(x,\partial\Omega)$ the distance function from the boundary,
defined as

$d(x,\partial\Omega)=\inf_{y\in\partial\Omega}{\left|x-y\right|},$ (7)

for an exhaustive discussion about this function and its properties see
[GT01]. Moreover, we recall that the inradius $R_{\Omega}$ of $\Omega$ is

$R_{\Omega}=\sup_{x\in\Omega}\inf_{y\in\partial\Omega}|x-y|={\left\|d(\cdot,\partial\Omega)\right\|}_{L^{\infty}(\Omega)}.$
(8)

The following lemma makes us understand why (4) can be seen as a limit problem
of (3).

######  Lemma 2.1.

Given $f,g\in W^{1,\infty}(\Omega)$, then

$\lim_{p\to\infty}\left(\int_{\Omega}{\left|f\right|}^{p}+\int_{\Omega}{\left|g\right|}^{p}\right)^{1/p}=\max\left\\{{\left\|f\right\|}_{\infty},{\left\|g\right\|}_{\infty}\right\\}.$

###### Proof.

The proof of this lemma can be found in [RS15]. ∎

###  2.1 Viscosity solutions

Before going on, we recall the definition of viscosity solutions to a boundary
value problem, see [CIL92] for more details.

######  Definition 2.1.

We consider the following boundary value problem

$\begin{cases}F(x,u,\nabla u,D^{2}u)=0&\text{ in }\Omega,\\\ B(x,u,\nabla
u)=0&\text{ on }\partial\Omega,\end{cases}$ (9)

where
$F:\operatorname{\mathbb{R}}^{n}\times\operatorname{\mathbb{R}}\times\operatorname{\mathbb{R}}^{n}\times\operatorname{\mathbb{R}}^{n\times
n}\to\operatorname{\mathbb{R}}$ and
$B:\operatorname{\mathbb{R}}^{n}\times\operatorname{\mathbb{R}}\times\operatorname{\mathbb{R}}^{n}\to\operatorname{\mathbb{R}}$
are two continuous functions.

Viscosity supersolution

A lower semi-continuous function $u$ is a viscosity supersolution to (9) if,
whenever we fix $x_{0}\in\overline{\Omega}$, for every $\phi\in
C^{2}(\overline{\Omega})$ such that $u(x_{0})=\phi(x_{0})$ and $x_{0}$ is a
strict minimum in $\Omega$ for $u-\phi$, then

  * •

if $x_{0}\in\Omega$, the following holds

$F\left(x_{0},\phi(x_{0}),\nabla\phi(x_{0}),D^{2}\phi(x_{0})\right)\geq 0$

  * •

if $x_{0}\in\partial\Omega$, the following holds

$\max\Set{F\left(x_{0},\phi(x_{0}),\nabla\phi(x_{0}),D^{2}\phi(x_{0})\right),B\left(x_{0},\phi(x_{0}),\nabla\phi(x_{0})\right)}\geq
0$

Viscosity subsolution

An upper semi-continuous function $u$ is a viscosity subsolution to (9) if,
whenever we fix $x_{0}\in\overline{\Omega}$, for every $\phi\in
C^{2}(\overline{\Omega})$ such that $u(x_{0})=\phi(x_{0})$ and $x_{0}$ is a
strict maximum in $\Omega$ for $u-\phi$, then

  * •

if $x_{0}\in\Omega$, the following holds

$F\left(x_{0},\phi(x_{0}),\nabla\phi(x_{0}),D^{2}\phi(x_{0})\right)\leq 0$

  * •

if $x_{0}\in\partial\Omega$, the following holds

$\min\Set{F\left(x_{0},\phi(x_{0}),\nabla\phi(x_{0}),D^{2}\phi(x_{0})\right),B\left(x_{0},\phi(x_{0}),\nabla\phi(x_{0})\right)}\leq
0$

Viscosity solution

A continuous function $u$ is a viscosity solution to (9) if it is both a super
and subsolution.

######  Remark 2.1.

The condition $u-\phi$ has a strict maximum or minimum can be relaxed: it is
sufficient to ask that $u-\phi$ has a local maximum or minimum in a ball
$B_{R}(x_{0})$ for some positive $R$.

##  3 The $\infty$-eigenvalue problem

Let us start this section observing that Lemma 2.1 brings to the following
estimate

$\limsup_{p\to\infty}\Lambda_{p}^{1/p}\leq\Lambda_{\infty}.$ (10)

We can say something more

######  Lemma 3.1.

Let $\Set{\Lambda_{p}}_{p>1}$ be the sequence of the first eigenvalues of the
$p$-Laplacian operator with Robin boundary condition. Then,

$\lim_{p\to\infty}\left(\Lambda_{p}\right)^{\frac{1}{p}}=\Lambda_{\infty},$
(11)

where $\Lambda_{\infty}$ is defined in (4).

Moreover, if $\Set{u_{p}}_{p>1}$ is the sequence of eigenfunctions associated
to $\\{\Lambda_{p}\\}_{p>1}$, then there exists a function $u_{\infty}\in
W^{1,\infty}(\Omega)$ such that, up to a subsequence,

$\displaystyle u_{p}\to u_{\infty}$ $\displaystyle\text{ uniformly in
}\,\Omega$ $\displaystyle\nabla u_{p}\to\nabla u_{\infty}$
$\displaystyle\text{ weakly in }\,L^{q}(\Omega),\forall q.$

###### Proof.

As a consequence of (10), the sequence $\Set{u_{p}}_{p>1}$ of eigenfunctions
associated to $\Lambda_{p}$ is uniformly bounded in $W^{1,q}(\Omega)$: indeed,
if $q<p$, by Hölder inequality,

$\displaystyle{\left\|\nabla u_{p}\right\|}_{L^{q}(\Omega)}\leq{\left\|\nabla
u_{p}\right\|}_{L^{p}(\Omega)}{\left|\Omega\right|}^{\frac{1}{q}-\frac{1}{p}}\leq\Lambda_{p}^{1/p}{\left|\Omega\right|}^{\frac{1}{q}-\frac{1}{p}}\leq
C,$ (12)
$\displaystyle{\left\|u_{p}\right\|}_{L^{q}(\Omega)}\leq{\left\|u_{p}\right\|}_{L^{p}(\Omega)}{\left|\Omega\right|}^{\frac{1}{q}-\frac{1}{p}}\leq{\left|\Omega\right|}^{\frac{1}{q}-\frac{1}{p}}\leq
C,$ (13)

where the constant $C$ is independent of $p$.

By a classical argument of diagonalization, see for instance [BDM89], we can
extract a subsequence $u_{p_{j}}$ such that

$\displaystyle u_{p_{j}}\to u_{\infty}\,\text{
uniformly}\implies\|u_{p_{j}}\|_{L^{p_{j}}}\to{\left\|u_{\infty}\right\|}_{L^{\infty}},$
$\displaystyle\nabla u_{p_{j}}\to\nabla u_{\infty}\,\text{ weakly in
}\,L^{q}(\Omega),\,\forall q>1.$

Moreover, from (12) and (13), the following inequality holds

$\displaystyle\frac{{\left\|\nabla
u_{\infty}\right\|}_{L^{q}(\Omega)}}{{\left\|u_{\infty}\right\|}_{L^{q}(\Omega)}}$
$\displaystyle\leq\liminf_{p\to\infty}\frac{{\left\|\nabla
u_{p}\right\|}_{L^{q}(\Omega)}}{{\left\|u_{p}\right\|}_{L^{q}(\Omega)}}\leq\liminf_{p\to\infty}\frac{{\left\|\nabla
u_{p}\right\|}_{L^{p}(\Omega)}}{{\left\|u_{p}\right\|}_{L^{q}(\Omega)}}{\left|\Omega\right|}^{\frac{1}{q}-\frac{1}{p}}$
$\displaystyle\leq\frac{{\left|\Omega\right|}^{\frac{1}{q}}}{{\left\|u_{\infty}\right\|}_{L^{q}(\Omega)}}\liminf_{p\to\infty}\left(\Lambda_{p}\right)^{\frac{1}{p}}.$

Letting $q\to\infty$ we obtain

${\left\|\nabla
u_{\infty}\right\|}_{L^{\infty}(\Omega)}\leq\liminf_{p\to\infty}\left(\Lambda_{p}\right)^{\frac{1}{p}}$

Similarly

$\beta{\left\|u_{p}\right\|}_{L^{q}(\partial\Omega)}\leq\beta{\left\|\
u_{p}\right\|}_{L^{p}(\partial\Omega)}{\left|\partial\Omega\right|}^{\frac{1}{q}-\frac{1}{p}}\leq\Lambda_{p}^{1/p}{\left|\partial\Omega\right|}^{\frac{1}{q}-\frac{1}{p}}\leq
C,$

gives us

$\beta{\left\|u_{\infty}\right\|}_{L^{\infty}(\partial\Omega)}\leq\liminf_{p\to\infty}\left(\Lambda_{p}\right)^{\frac{1}{p}},$

hence

$\Lambda_{\infty}\leq\liminf_{p\to\infty}\Lambda_{p}^{1/p}.$

∎

Now we want to show that the limit $u_{\infty}$ solves (1) in viscosity sense,
but before we need the following proposition

######  Proposition 3.2.

A continuous weak solution $u$ to (2) is a viscosity solution to (2).

###### Proof.

The proof is similar to the one in [JLM99, EKNT15] for the $p$-Laplacian with
other boundary conditions.

We only write explicitly the proof that $u_{\infty}$ satisfies the boundary
conditions in the viscosity sense.

Let $u$ be a continuous weak solution to (2), let $x_{0}\in\partial\Omega$ and
let us consider a function $\phi$ such that $\phi(x_{0})=u(x_{0})$ and such
that $u-\phi$ has a strict minimum at $x_{0}$. Then

$\begin{multlined}\max\Big{\\{}-{\left|\nabla\phi(x_{0})\right|}^{p-2}\Delta\phi(x_{0})-(p-2){\left|\nabla\phi(x_{0})\right|}^{p-4}\Delta_{\infty}\phi(x_{0})-\Lambda_{p}{\left|\phi(x_{0})\right|}^{p-2}\phi(x_{0}),\\\
{\left|\nabla\phi(x_{0})\right|}^{p-2}\frac{\partial\phi(x_{0})}{\partial\nu}+\beta^{p}{\left|\phi(x_{0})\right|}^{p-2}\phi(x_{0})\Big{\\}}\geq
0.\end{multlined}\max\Big{\\{}-{\left|\nabla\phi(x_{0})\right|}^{p-2}\Delta\phi(x_{0})-(p-2){\left|\nabla\phi(x_{0})\right|}^{p-4}\Delta_{\infty}\phi(x_{0})-\Lambda_{p}{\left|\phi(x_{0})\right|}^{p-2}\phi(x_{0}),\\\
{\left|\nabla\phi(x_{0})\right|}^{p-2}\frac{\partial\phi(x_{0})}{\partial\nu}+\beta^{p}{\left|\phi(x_{0})\right|}^{p-2}\phi(x_{0})\Big{\\}}\geq
0.$ (14)

Assume by contradiction that both terms are negative. If we choose $r$
sufficiently small, in $\overline{\Omega}\cap B_{r}(x_{0})$, we have

$-{\left|\nabla\phi(x)\right|}^{p-2}\Delta\phi(x)-(p-2){\left|\nabla\phi(x)\right|}^{p-4}\Delta_{\infty}\phi(x)-\Lambda_{p}{\left|\phi(x)\right|}^{p-2}\phi(x)<0$

and, in $\partial\Omega\cap B_{r}(x_{0})$,

${\left|\nabla\psi(x)\right|}^{p-2}\frac{\partial\psi(x)}{\partial\nu}+\beta^{p}{\left|\psi(x)\right|}^{p-2}\psi(x)<0,\quad\text{where
}\psi=\phi+\frac{m}{2}.$

Then

$\begin{multlined}\int_{\Set{\psi>u}\cap
B_{r}(x_{0})}{\left|\nabla\psi\right|}^{p-2}\nabla\psi\nabla(\psi-u)\,dx\\\
<\Lambda_{p}\int_{\Set{\psi>u}\cap
B_{r}(x_{0})}{\left|\phi\right|}^{p-2}\phi(\psi-u)\,dx-\beta^{p}\int_{\partial\Omega\cap
B_{r}(x_{0})\cap\Set{\psi>u}}{\left|\psi\right|}^{p-2}\psi(\psi-u)\,d\mathcal{H}^{n-1},\end{multlined}\int_{\Set{\psi>u}\cap
B_{r}(x_{0})}{\left|\nabla\psi\right|}^{p-2}\nabla\psi\nabla(\psi-u)\,dx\\\
<\Lambda_{p}\int_{\Set{\psi>u}\cap
B_{r}(x_{0})}{\left|\phi\right|}^{p-2}\phi(\psi-u)\,dx-\beta^{p}\int_{\partial\Omega\cap
B_{r}(x_{0})\cap\Set{\psi>u}}{\left|\psi\right|}^{p-2}\psi(\psi-u)\,d\mathcal{H}^{n-1},$

using the definition of weak solution, we have

$\displaystyle C(N,p)\int_{\Set{\psi>u}\cap B_{r}(x_{0})}$
$\displaystyle{\left|\nabla\psi-\nabla u\right|}^{p}\,dx$
$\displaystyle\leq\int_{\Set{\psi>u}\cap
B_{r}(x_{0})}\left\langle{\left|\nabla\psi\right|}^{p-2}\nabla\psi-{\left|\nabla
u\right|}^{p-2}\nabla u,\nabla(\psi-u)\right\rangle\,dx$
$\displaystyle<\Lambda_{p}\int_{\Set{\psi>u}\cap
B_{r}(x_{0})}\left({\left|\phi\right|}^{p-2}\phi-{\left|u\right|}^{p-2}u\right)(\psi-u)\,dx$
$\displaystyle-\beta^{p}\int_{\partial\Omega\cap
B_{r}(x_{0})\cap\Set{\psi>u}}\left({\left|\psi\right|}^{p-2}\psi-{\left|u\right|}^{p-2}u\right)(\psi-u)\,d\mathcal{H}^{n-1}<0$

which gives a contradiction. ∎

Now we can prove the following

######  Theorem 3.3.

Let $u_{\infty}$ be the function given in Theorem 3.1. Then $u_{\infty}$ is a
viscosity solution to

$\begin{cases}\min\Set{{\left|\nabla
u\right|}-\Lambda_{\infty}u,-\Delta_{\infty}u}=0&\text{ in }\Omega,\\\
-\min\Set{{\left|\nabla u\right|}-\beta u,-\displaystyle{\frac{\partial
u}{\partial\nu}}}=0&\text{ on }\partial\Omega.\end{cases}$ (15)

###### Proof.

We divide the proof in two steps.

Step 1 $u_{\infty}$ is a viscosity supersolution.
Let $x_{0}\in\Omega$ and let $\phi\in C^{2}(\Omega)$ be such that
$u_{\infty}-\phi$ has a strict minimum in $x_{0}$. We want to show

$\min\Set{{\left|\nabla\phi(x_{0})\right|}-\Lambda_{\infty}\phi(x_{0}),-\Delta_{\infty}\phi(x_{0})}\geq
0$

Notice that $u_{p}-\phi$ has a minimum in $x_{p}$ and $x_{p}\to x_{0}$. If we
set $\phi_{p}(x)=\phi(x)+c_{p}$ with $c_{p}=u_{p}(x_{p})-\phi(x_{p})\to 0$
when $p$ goes to infinity, we have that $u_{p}(x_{p})=\phi_{p}(x_{p})$ and
$u_{p}-\phi_{p}$ has a minimum in $x_{p}$, so Proposition 3.2 implies

$-{\left|\nabla\phi_{p}(x_{p})\right|}^{p-2}\Delta\phi_{p}(x_{p})-(p-2){\left|\nabla\phi_{p}(x_{p})\right|}^{p-4}\Delta_{\infty}\phi(x_{p})-\Lambda_{p}{\left|\phi_{p}(x_{p})\right|}^{p-2}\phi_{p}(x_{p})\geq
0.$ (16)

Now dividing by $(p-2){\left|\nabla\phi_{p}(x_{p})\right|}^{p-4}$, we obtain

$-\Delta_{\infty}\phi_{p}(x_{p})-\frac{{\left|\nabla\phi_{p}(x_{p})\right|}^{2}\Delta\phi_{p}(x_{p})}{p-2}\geq\frac{{\left|\nabla\phi_{p}(x_{p})\right|}^{4}}{(p-2)\phi_{p}(x_{p})}\left(\frac{\Lambda_{p}^{1/p}\phi_{p}(x_{p})}{{\left|\nabla\phi_{p}(x_{p})\right|}}\right)^{p}$
(17)

This gives us
${\left|\nabla\phi(x_{0})\right|}-\Lambda_{\infty}\phi(x_{0})\geq 0$ since,
otherwise, the right-hand side of (17) would go to infinity, in contradiction
with the fact that $\phi\in C^{2}(\Omega)$. Moreover
$-\Delta_{\infty}\phi(x_{0})\geq 0$, just taking the limit.

Then,
$\min\Set{{\left|\nabla\phi(x_{0})\right|}-\Lambda_{\infty}\phi(x_{0}),-\Delta_{\infty}\phi(x_{0})}\geq
0$ and $u_{\infty}$ is a viscosity supersolution.

Let us fix $x_{0}\in\partial\Omega$, $\phi\in C^{2}(\overline{\Omega})$ such
that $u-\phi$ has a strict minimum in $x_{0}$, our aim is to prove that

$\max\Set{\min\Set{{\left|\nabla\phi(x_{0})\right|}-\Lambda_{\infty}\phi(x_{0}),-\Delta_{\infty}\phi(x_{0})},-\min\Set{{\left|\nabla\phi(x_{0})\right|}-\beta\phi(x_{0}),-\frac{\partial\phi}{\partial\nu}(x_{0})}}\geq
0$

If for infinitely many $x_{p}\in\Omega$ (16) holds true, then we get

$\min\Set{{\left|\nabla\phi(x_{0})\right|}-\Lambda_{\infty}\phi(x_{0}),-\Delta_{\infty}\phi(x_{0})}\geq
0.$

If for infinitely many $p$, $x_{p}\in\partial\Omega$ the following holds true

${\left|\nabla\phi_{p}(x_{p})\right|}^{p-2}\frac{\partial\phi_{p}(x_{p})}{\partial\nu}+\beta^{p}{\left|\phi_{p}(x_{p})\right|}^{p-2}\phi_{p}(x_{p})\geq
0,$

then

${\left|\nabla\phi_{p}(x_{p})\right|}^{p-2}\left(-\frac{\partial\phi_{p}(x_{p})}{\partial\nu}\right)\leq\beta^{p}{\left|\phi_{p}(x_{p})\right|}^{p-2}\phi_{p}(x_{p}).$

Only two cases can occur:

  * •

$\displaystyle{-\frac{\partial\phi}{\partial\nu}(x_{0})\leq 0}$;

  * •

$\displaystyle{-\frac{\partial\phi}{\partial\nu}(x_{0})>0}$, then letting $p$
to infinity in the following

$\left({\left|\nabla\phi_{p}(x_{p})\right|}^{p-2}\left(-\frac{\partial\phi_{p}(x_{p})}{\partial\nu}\right)\right)^{1/p}\leq\left(\beta^{p}{\left|\phi_{p}(x_{p})\right|}^{p-2}\phi_{p}(x_{p})\right)^{1/p}$

we get $\displaystyle{{\left|\nabla\phi(x_{0})\right|}\leq\beta\phi(x_{0})}$.

That is

$-\min\Set{{\left|\nabla\phi(x_{0})\right|}-\beta\phi(x_{0}),-\frac{\partial\phi}{\partial\nu}(x_{0})}\geq
0.$

Step 2 $u_{\infty}$ is a viscosity subsolution.

Let us fix $x_{0}\in\Omega$, $\phi\in C^{2}(\Omega)$ such that
$u_{\infty}-\phi$ has a strict maximum. We want to prove that

$\min\left\\{{\left|\nabla\phi(x_{0})\right|}-\Lambda_{\infty}\phi(x_{0}),-\Delta_{\infty}\phi(x_{0})\right\\}\leq
0,$

so it is enough to prove that only one of the two terms in the bracket is non
positive.

For instance, assume that $-\Delta_{\infty}\phi(x_{0})>0$, we can argue as in
(16), but now, all the inequality involving the second order differential
operator are reversed and we get

$\Lambda_{p}\phi_{p}^{p-1}(x_{p})\geq(p-2){\left|\nabla\phi_{p}(x_{p})\right|}^{p-4}\left[-\frac{{\left|\nabla\phi_{p}(x_{p})\right|}^{2}\Delta\phi_{p}(x_{p})}{p-2}-\Delta_{\infty}\phi_{p}(x_{p})\right].$

As $-\Delta_{\infty}\phi(x_{0})>0$, the term in the big parenthesis is non-
negative, we can erase everything to the power $1/p$, obtaining

$\Lambda_{\infty}\phi(x_{0})\geq{\left|\nabla\phi(x_{0})\right|},$

which shows that $u_{\infty}$ is a viscosity subsolution to (15).

Similar arguments to step 1 give us the boundary conditions for viscosity
subsolution. ∎

We are also able to give a geometric characterization of $\Lambda_{\infty}$.

######  Lemma 3.4.

Let $\Lambda_{\infty}$ be the quantity defined in (4) and let $R_{\Omega}$ be
the inradius of $\Omega$. Then

$\Lambda_{\infty}=\min_{x_{0}\in\Omega}\frac{1}{\frac{1}{\beta}+d(x_{0},\partial\Omega)}=\frac{1}{\frac{1}{\beta}+R_{\Omega}}.$

###### Proof.

The function $\frac{1}{\beta}+d(x,\partial\Omega)\in W^{1,\infty}(\Omega)$,
moreover

${\left\|\nabla\left({1}/{\beta}+d(x,\partial\Omega)\right)\right\|}_{L^{\infty}(\Omega)}=1\quad\text{
and
}\quad\beta{\left\|{1}/{\beta}+d(x,\partial\Omega)\right\|}_{L^{\infty}(\partial\Omega)}=1.$

Then

$\Lambda_{\infty}\leq\min_{x_{0}\in\Omega}\frac{1}{\frac{1}{\beta}+d(x_{0},\partial\Omega)}.$

In order to prove the reverse inequality, we take $w\in W^{1,\infty}(\Omega)$
such that ${\left\|w\right\|}_{L^{\infty}(\Omega)}=1$.

The following facts can occur

Case 1

If $\beta{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}\leq{\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)}$, then

$\max\left\\{{\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)},\beta{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}\right\\}={\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)}.$

We choose $x\in\Omega$ and $y$ equal to the point on the boundary such that
${\left|x-y\right|}=d(x,\partial\Omega)$. So, we have

$\displaystyle{\left|w(x)\right|}$
$\displaystyle\leq{\left|w(x)-w(y)\right|}+{\left|w(y)\right|}$
$\displaystyle\leq{\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)}{\left|x-y\right|}+{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}$
$\displaystyle\leq{\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)}d(x,\partial\Omega)+\frac{1}{\beta}{\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)}$ $\displaystyle={\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)}\left(\frac{1}{\beta}+d(x,\partial\Omega)\right)$
$\displaystyle\leq{\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)}{\left\|{1}/{\beta}+d(x,\partial\Omega)\right\|}_{L^{\infty}(\Omega)},$

and then

$\frac{{\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)}}{{\left\|w\right\|}_{L^{\infty}(\Omega)}}\geq\frac{1}{{\left\|{1}/{\beta}+d(x,\partial\Omega)\right\|}_{L^{\infty}(\Omega)}}$

Case 2

If $\beta{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}>{\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)}$, then

$\max\left\\{{\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)},\beta{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}\right\\}=\beta{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}.$

With the same choice of $x$ and $y$, we have

$\displaystyle{\left|w(x)\right|}$
$\displaystyle\leq{\left|w(x)-w(y)\right|}+{\left|w(y)\right|}$
$\displaystyle\leq{\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)}{\left|x-y\right|}+{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}$
$\displaystyle\leq\beta{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}d(x,\partial\Omega)+{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}$
$\displaystyle=\beta{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}\left(d(x,\partial\Omega)+\frac{1}{\beta}\right)$
$\displaystyle\leq\beta{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}{\left\|{1}/{\beta}+d(x,\partial\Omega)\right\|}_{L^{\infty}(\Omega)}.$

Hence,

$\frac{\beta{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}}{{\left\|w\right\|}_{L^{\infty}(\Omega)}}\geq\frac{1}{{\left\|{1}/{\beta}+d(x,\partial\Omega)\right\|}_{L^{\infty}(\Omega)}}.$

Finally we get $\forall w\in
W^{1,\infty}(\Omega):{\left\|w\right\|}_{L^{\infty}(\Omega)}=1,$

$\max\left\\{{\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)},\beta{\left\|w\right\|}_{L^{\infty}(\partial\Omega)}\right\\}\geq\frac{1}{{\left\|{1}/{\beta}+d(x,\partial\Omega)\right\|}_{L^{\infty}(\Omega)}}$

and then the desired inequality

$\Lambda_{\infty}\geq\min_{x_{0}\in\Omega}\frac{1}{\frac{1}{\beta}+d(x_{0},\partial\Omega)}.$

∎

######  Theorem 3.5.

Let $\Lambda_{\infty}$ be the quantity defined in (4). Then

$\Lambda_{\infty}(\Omega)\geq\Lambda_{\infty}(\Omega^{\sharp}),$

where $\Omega^{\sharp}$ is the ball centered at the origin with the same
measure of $\Omega$.

###### Proof.

The Faber-Krahn inequality for the first eigenvalue of the $p$-Laplacian with
Robin boundary condition (for instance see [BD10]) states that

$\Lambda_{p}(\Omega)\geq\Lambda_{p}(\Omega^{\sharp}).$

Letting $p$ go to infinity, we have

$\Lambda_{\infty}(\Omega)\geq\Lambda_{\infty}(\Omega^{\sharp}).$

This can follow also from the geometric characterization in Lemma 3.4

$\Lambda_{\infty}=\frac{1}{\frac{1}{\beta}+R_{\Omega}}$

as the ball maximizes the inradius among sets of given volume. ∎

######  Remark 3.1.

One can easily prove that the function $\frac{1}{\beta}+d(x,\partial\Omega)$
is an eigenfunction if the domain $\Omega=B_{R}(x_{0})$. This is not true if
$\Omega$ is a square: see for instance [JLM99].

###  3.1 The first Robin $\infty$-eigenvalue

Now we want to show that $\Lambda_{\infty}$ is the first eigenvalue of (1),
that is the smallest $\Lambda$ such that

$\begin{cases}\min\Set{{\left|\nabla u\right|}-\Lambda
u,-\Delta_{\infty}u}=0&\text{ in }\Omega,\\\ -\min\Set{{\left|\nabla
u\right|}-\beta u,-\displaystyle{\frac{\partial u}{\partial\nu}}}=0&\text{ on
}\partial\Omega\end{cases}$

admits a non-trivial solution.

######  Theorem 3.6.

Let $\Omega$ be a bounded and open set of class $C^{2}$ in
$\operatorname{\mathbb{R}}^{n}$. If for some $\Lambda$, problem (1) admits a
non-trivial eigenfunction $u$, then $\Lambda\geq\Lambda_{\infty}$.

###### Proof.

Let $\Lambda$ be an eigenvalue to (1), let $u$ be a corresponding
eigenfunction. We normalize it in this way

$\max_{x\in\Omega}u(x)=\frac{1}{\Lambda}.$

Then $u$ is viscosity subsolution to

$\min\Set{{\left|\nabla u\right|}-1,-\Delta_{\infty}u}=0\text{ in }\Omega.$

For every $\varepsilon>0$ and $\gamma>0$, let us consider the function

$g_{\varepsilon,\gamma}=\frac{1}{\beta}+(1+\varepsilon)d(x,\partial\Omega)-\gamma
d(x,\partial\Omega)^{2}.$

It is well known (see [GT01]) that in a small tubular neighbourhood
$\Gamma_{\mu}$ of $\partial\Omega$, the boundary $\partial\Omega$ and the
distance function $d(x,\partial\Omega)$ share the same regularity: so both
$d(x,\partial\Omega)$ and $g_{\varepsilon,\gamma}$ are $C^{2}(\Gamma_{\mu})$.

Moreover, by a direct computation, one can check that if

$\gamma<\frac{\varepsilon}{2R_{\Omega}},$

then $g_{\varepsilon,\gamma}$ is a viscosity supersolution to

$\min\Set{{\left|\nabla
g_{\varepsilon,\gamma}\right|}-1,-\Delta_{\infty}g_{\varepsilon,\gamma}}=0\text{
in }\Omega.$

Hence, Theorem $2.1$ in [Jen93] ensures that

$m_{\varepsilon}=\inf_{x\in\Omega}(g_{\varepsilon,\gamma}(x)-u(x))=\inf_{x\in\partial\Omega}(g_{\varepsilon,\gamma}(x)-u(x)).$

Assume by contradiction that $m_{\varepsilon}<-\frac{\varepsilon}{\beta}$, and
set $v=g_{\varepsilon,\gamma}-m_{\varepsilon}$. We observe that $v\geq u$ in
$\Omega$ and $v(x_{0})=u(x_{0})$, where $x_{0}$ is the point which achieves
the minimum on the boundary, so we can use it as test function in the
definition of viscosity subsolution for $u$.

Assuming $\gamma<\frac{\varepsilon}{2R_{\Omega}}$, we obtain

$\displaystyle\nabla v(x)=\left[1+\varepsilon-2\gamma
d(x,\partial\Omega)\right]\nabla d(x,\partial\Omega)$
$\displaystyle{\left|\nabla v(x_{0})\right|}=1+\varepsilon-2\gamma
d(x_{0},\partial\Omega)>1$ $\displaystyle-\frac{\partial
v}{\partial\nu}(x_{0})=-\left[1+\varepsilon-2\gamma
d(x_{0},\partial\Omega)\right]\nabla d(x_{0},\partial\Omega)\cdot\nu>0$
$\displaystyle-\Delta_{\infty}v(x_{0})=2\gamma\left[1+\varepsilon-2\gamma
d(x_{0},\partial\Omega)\right]^{2}{\left|\nabla
d(x_{0},\partial\Omega)\right|}^{4}>0.$

The fact that $m_{\varepsilon}<-\frac{\varepsilon}{\beta}$ implies

${\left|\nabla v(x_{0})\right|}-\beta v(x_{0})=\varepsilon+\beta
m_{\varepsilon}<0.$

Therefore

$-\min\left\\{{\left|\nabla v\right|}-\beta v,-\frac{\partial
v}{\partial\nu}\right\\}>0\quad\text{ and }\quad\min\left\\{{\left|\nabla
v\right|}-1,-\Delta_{\infty}v\right\\}>0$

against the fact that

$\min\left\\{\min\left\\{{\left|\nabla
v\right|}-1,-\Delta_{\infty}v\right\\},-\min\left\\{{\left|\nabla
v\right|}-\beta v,-\frac{\partial v}{\partial\nu}\right\\}\right\\}\leq 0.$

So we have

$g_{\varepsilon,\gamma}(x)-u(x)\geq
m_{\varepsilon}\geq-\frac{\varepsilon}{\beta},$

letting $\varepsilon$ and $\gamma$ go to zero, it follows

$\frac{1}{\beta}+d(x,\partial\Omega)\geq u(x)\qquad\forall x\in\Omega.$

Hence

$\frac{1}{\Lambda_{\infty}}=\max_{x\in\Omega}\left(\frac{1}{\beta}+d(x,\partial\Omega)\right)\geq\max_{x\in\Omega}u(x)=\frac{1}{\Lambda},$

which concludes the proof. ∎

##  4 The $p$-Poisson equation

Let $f$ be a function in $L^{\infty}(\Omega)$ and let $\beta>0$. We consider
the following $p$-Poisson equation with Robin boundary conditions

$\begin{cases}-\Delta_{p}v=f&\text{ in }\Omega\\\ {\left|\nabla
v\right|}^{p-2}\displaystyle{\frac{\partial
v}{\partial\nu}}+\beta^{p}{\left|v\right|}^{p-2}v=0&\text{ on
}\partial\Omega.\end{cases}$ (18)

A function $v_{p}$ is a weak solution to (18) if it satisfies

$\int_{\Omega}{\left|\nabla v_{p}\right|}^{p-2}\nabla
v_{p}\nabla\varphi\,dx+\beta^{p}\int_{\partial\Omega}{\left|v_{p}\right|}^{p-2}v_{p}\varphi\,d\mathcal{H}^{n-1}=\int_{\Omega}f\varphi\,dx,\quad\forall\varphi\in
W^{1,p}(\Omega).$ (19)

It is well known that the solution to this equation is the unique minimum of
the functional

$J_{p}(\varphi)=\frac{1}{p}\int_{\Omega}{\left|\nabla\varphi\right|}^{p}\,dx+\frac{\beta^{p}}{p}\int_{\partial\Omega}{\left|\varphi\right|}^{p}\,d\mathcal{H}^{n-1}(x)-\int_{\Omega}f\varphi\,dx.$
(20)

Indeed, it is possible to prove the existence and the uniqueness of the
minimum of the functional thanks to the so-called direct method of calculus of
variation, see for instance [Dac08, Giu94, Lin06, AGM21].

If we let _formally_ $p$ go to $\infty$ in (20), we obtain the functional

$\varphi\longrightarrow\min\int_{\Omega}-f\varphi\,dx\quad\varphi\in
W^{1,\infty}(\Omega).$ (21)

The limit procedure imposes two extra constraints to (21), namely

${\left\|\nabla\varphi\right\|}_{\infty}\leq
1,\quad\beta{\left\|\varphi\right\|}_{L^{\infty}(\partial\Omega)}\leq 1.$

The following proposition holds true.

######  Proposition 4.1.

Let $v_{p}$ be the solution to (18). Then there exists a subsequence
$\Set{v_{p_{j}}}_{j}$ such that

$v_{p_{j}}\to v_{\infty}\,\text{ uniformly,}\qquad\nabla v_{p_{j}}\to\nabla
v_{\infty}\,\text{ weakly in }\,L^{m}(\Omega),\,\forall m>1.$

Moreover

${\left\|\nabla v_{\infty}\right\|}_{\infty}\leq
1\qquad\beta{\left\|v_{\infty}\right\|}_{L^{\infty}(\partial\Omega)}\leq 1.$

###### Proof.

Choosing $\varphi=v_{p}$ in (19), we have

$\int_{\Omega}{\left|\nabla
v_{p}\right|}^{p}+\beta^{p}\int_{\partial\Omega}v_{p}^{p}=\int_{\Omega}fv_{p},$

and Young inequality gives

$\int_{\Omega}{\left|\nabla
v_{p}\right|}^{p}+\beta^{p}\int_{\partial\Omega}v_{p}^{p}-\frac{1}{\varepsilon_{p}^{p}p}\int_{\Omega}v_{p}^{p}\leq\frac{\varepsilon_{p}^{p^{\prime}}}{p^{\prime}}\int_{\Omega}f^{p^{\prime}}.$
(22)

Taking into account (3), we get

$\begin{multlined}\left(1-\frac{1}{p\Lambda_{p}\varepsilon_{p}^{p}}\right)\left[\int_{\Omega}{\left|\nabla
v_{p}\right|}^{p}+\beta^{p}\int_{\partial\Omega}v_{p}^{p}\right]\leq\qquad\qquad\qquad\\\
\qquad\leq\int_{\Omega}{\left|\nabla
v_{p}\right|}^{p}+\beta^{p}\int_{\partial\Omega}v_{p}^{p}-\frac{1}{\varepsilon_{p}^{p}p}\int_{\Omega}v_{p}^{p}\leq\frac{\varepsilon_{p}^{p^{\prime}}}{p^{\prime}}\int_{\Omega}f^{p^{\prime}}.\end{multlined}\left(1-\frac{1}{p\Lambda_{p}\varepsilon_{p}^{p}}\right)\left[\int_{\Omega}{\left|\nabla
v_{p}\right|}^{p}+\beta^{p}\int_{\partial\Omega}v_{p}^{p}\right]\leq\qquad\qquad\qquad\\\
\qquad\leq\int_{\Omega}{\left|\nabla
v_{p}\right|}^{p}+\beta^{p}\int_{\partial\Omega}v_{p}^{p}-\frac{1}{\varepsilon_{p}^{p}p}\int_{\Omega}v_{p}^{p}\leq\frac{\varepsilon_{p}^{p^{\prime}}}{p^{\prime}}\int_{\Omega}f^{p^{\prime}}.$

Choosing $\varepsilon_{p}$ such that
$\displaystyle{1-\frac{1}{p\Lambda_{p}\varepsilon_{p}^{p}}=\frac{1}{2}}$, we
have

$\int_{\Omega}{\left|\nabla
v_{p}\right|}^{p}+\beta^{p}\int_{\partial\Omega}v_{p}^{p}\leq
2\frac{\varepsilon_{p}^{p^{\prime}}}{p^{\prime}}\int_{\Omega}f^{p^{\prime}}\leq
C\int_{\Omega}f^{p^{\prime}}$ (23)

where the constant $C$ is independent of $p$, thanks to Lemma 2.1.

Hence

$\displaystyle\left(\int_{\Omega}{\left|\nabla
v_{p}\right|}^{p}\right)^{1/p}\leq\left(C\int_{\Omega}f^{p^{\prime}}\right)^{1/p}\leq\left(C{\left|\Omega\right|}\,{\left\|f\right\|}^{p^{\prime}}_{\infty}\right)^{1/p},$
$\displaystyle\left(\beta^{p}\int_{\partial\Omega}v_{p}^{p}\right)^{1/p}\leq\left(C\int_{\Omega}f^{p^{\prime}}\right)^{1/p}\leq\left(C{\left|\Omega\right|}\,{\left\|f\right\|}^{p^{\prime}}_{\infty}\right)^{1/p}.$

Analogously

$\left(\int_{\Omega}v_{p}^{p}\right)^{1/p}\leq
C\left(\int_{\Omega}f^{p^{\prime}}\right)^{1/p}.$ (24)

If $p>m$, Hölder inequality gives

$\begin{gathered}\left(\int_{\Omega}{\left|\nabla
v_{p}\right|}^{m}\right)^{1/m}\leq\left(\int_{\Omega}{\left|\nabla
v_{p}\right|}^{p}\right)^{1/p}{\left|\Omega\right|}^{1/m-1/p}\leq\left(C{\left\|f\right\|}^{p^{\prime}}_{\infty}\right)^{1/p}{\left|\Omega\right|}^{1/m},\\\
\beta\left(\int_{\partial\Omega}v_{p}^{m}\right)^{1/m}\leq\left(\beta^{p}\int_{\partial\Omega}v_{p}^{p}\right)^{1/p}{\left|\partial\Omega\right|}^{1/m-1/p}\leq\left(C{\left\|f\right\|}^{p^{\prime}}_{\infty}\right)^{1/p}{\left|\partial\Omega\right|}^{1/m},\\\
\left(\int_{\Omega}v_{p}^{m}\right)^{1/m}\leq\left(\int_{\Omega}v_{p}^{p}\right)^{1/p}{\left|\Omega\right|}^{1/m-1/p}\leq
C\left({\left\|f\right\|}_{\infty}^{p^{\prime}}\right)^{1/p}{\left|\Omega\right|}^{1/m},\end{gathered}$
(25)

Then there exists $v_{p_{j}}$ such that

$\displaystyle v_{p_{j}}\to v_{\infty}\,$ $\displaystyle\text{
uniformly,}\qquad\nabla v_{p_{j}}\to\nabla v_{\infty}\,\text{ weakly in
}\,L^{m}(\Omega),\,\forall m>1,$

moreover

$\displaystyle{\left\|\nabla
v_{\infty}\right\|}_{m}\leq\liminf_{j\to\infty}{\left\|\nabla
v_{p_{j}}\right\|}_{m}\leq\lim_{j\to\infty}\left(C{\left\|f\right\|}^{p_{j}^{\prime}}_{\infty}\right)^{1/{p_{j}}}{\left|\Omega\right|}^{1/m}={\left|\Omega\right|}^{1/m}$
$\displaystyle\beta{\left\|v_{\infty}\right\|}_{L^{m}(\partial\Omega)}=\beta\lim_{j\to\infty}{\left\|v_{p_{j}}\right\|}_{L^{m}(\partial\Omega)}\leq\lim_{j\to\infty}\left(C{\left\|f\right\|}^{p_{j}^{\prime}}_{\infty}\right)^{1/p_{j}}{\left|\partial\Omega\right|}^{1/m}={\left|\partial\Omega\right|}^{1/m}$

and then

${\left\|\nabla v_{\infty}\right\|}_{L^{\infty}(\Omega)}\leq
1\quad\quad\beta{\left\|v_{\infty}\right\|}_{L^{\infty}(\partial\Omega)}\leq
1.$

∎

We are also able to link the so obtained function $v_{\infty}$ with the
functional (21), indeed

######  Theorem 4.2.

The functional

$J_{\infty}(\varphi)=-\int_{\Omega}f\varphi\qquad\varphi\in
W^{1,\infty}(\Omega)$ (26)

admits at least one minimum $\overline{\varphi}$ satisfying
${\left\|\nabla\overline{\varphi}\right\|}_{L^{\infty}(\Omega)}\leq 1$ and
$\beta{\left\|\overline{\varphi}\right\|}_{L^{\infty}(\partial\Omega)}\leq 1$.

Moreover, if $v_{\infty}$ is a limit of a subsequence of $\\{v_{p}\\}$, then
$v_{\infty}$ is a minimizer of (26).

###### Proof.

Let $v_{\infty}$ a limit of a subsequence of $\\{v_{p}\\}$ and let us assume
it is not a minimum of $J_{\infty}$. This implies that $\exists\varphi\in
W^{1,\infty}(\Omega)$ such that

$-\int_{\Omega}f\varphi<-\int_{\Omega}fv_{\infty}.$

We want to show that there exists a function $\phi$ and an exponent $p$, such
that $J_{p}(\phi)<J_{p}(v_{p})$, which contradicts the minimality of $v_{p}$.

First of all, we recall that exists a sequence $v_{p_{i}}\rightharpoonup
v_{\infty}$ in $W^{1,m}(\Omega)\,\,\forall m$. Then

$\int_{\Omega}fv_{p_{i}}\to\int_{\Omega}fv_{\infty}$

and so there exists $\overline{i}$ for which we still have

$-\int_{\Omega}f\varphi<-\int_{\Omega}fv_{p_{i}}\qquad\forall i>\overline{i}.$
(27)

Two cases can occur:

case 1

$\exists i>\overline{i}$

$\int_{\Omega}{\left|\nabla\varphi\right|}^{p_{i}}+\beta^{p_{i}}\int_{\partial\Omega}{\left|\varphi\right|}^{p_{i}}\leq\int_{\Omega}{\left|\nabla
v_{p_{i}}\right|}^{p_{i}}+\beta^{p_{i}}\int_{\partial\Omega}v_{p_{i}}^{p_{i}}$

Then

$\begin{split}J_{p_{i}}(\varphi)&=\frac{1}{p}\int_{\Omega}{\left|\nabla\varphi\right|}^{p_{i}}+\frac{\beta^{p_{i}}}{p}\int_{\partial\Omega}{\left|\varphi\right|}^{p_{i}}-\int_{\Omega}f\varphi\qquad\\\
&<\frac{1}{p}\int_{\Omega}{\left|\nabla
v_{p_{i}}\right|}^{p_{i}}+\frac{\beta^{p_{i}}}{p}\int_{\partial\Omega}v_{p_{i}}^{p_{i}}-\int_{\Omega}fv_{p_{i}}=J_{p_{i}}(v_{p_{i}}),\end{split}$

which is a contradiction.

case 2

$\forall i>\overline{i}$

$\int_{\Omega}{\left|\nabla\varphi\right|}^{p_{i}}+\beta^{p_{i}}\int_{\partial\Omega}{\left|\varphi\right|}^{p_{i}}>\int_{\Omega}{\left|\nabla
v_{p_{i}}\right|}^{p_{i}}+\beta^{p_{i}}\int_{\partial\Omega}v_{p_{i}}^{p_{i}}.$

Considering $\phi=\alpha\varphi$ with $\alpha\in(0,1)$:

$-\int_{\Omega}f\phi=-\alpha\int_{\Omega}f\varphi<-\int_{\Omega}fv_{p_{i}}<0\qquad\forall
i>\overline{i},$

we have

$\int_{\Omega}{\left|\nabla\phi\right|}^{p_{i}}+\beta^{p_{i}}\int_{\partial\Omega}{\left|\phi\right|}^{p_{i}}=\alpha^{p_{i}}\left[\int_{\Omega}{\left|\nabla\varphi\right|}^{p_{i}}+\beta^{p_{i}}\int_{\partial\Omega}{\left|\varphi\right|}^{p_{i}}\right].$

Moreover

$\begin{cases}\displaystyle{M\leq\int_{\Omega}fv_{p_{i}}=\int_{\Omega}{\left|\nabla
v_{p_{i}}\right|}^{p_{i}}+\beta^{p_{i}}\int_{\partial\Omega}v_{p_{i}}^{p_{i}}}\\\
\displaystyle{\int_{\Omega}{\left|\nabla\varphi\right|}^{p_{i}}+\beta^{p_{i}}\int_{\partial\Omega}{\left|\varphi\right|}^{p_{i}}\leq{\left|\Omega\right|}{\left\|\nabla\varphi\right\|}_{L^{\infty}(\Omega)}^{p_{i}}+\beta^{p_{i}}{\left|\partial\Omega\right|}{\left\|\varphi\right\|}_{L^{\infty}(\partial\Omega)}^{p_{i}}\leq{\left|\Omega\right|}+{\left|\partial\Omega\right|}}.\end{cases}$

We now choose $p_{i}$:

$0\xleftarrow{i\to\infty}\alpha^{p_{i}}\leq\frac{M}{{\left|\Omega\right|}+{\left|\partial\Omega\right|}}\leq\frac{\int_{\Omega}{\left|\nabla
v_{p_{i}}\right|}^{p_{i}}+\beta^{p_{i}}\int_{\partial\Omega}v_{p_{i}}^{p_{i}}}{\int_{\Omega}{\left|\nabla\varphi\right|}^{p_{i}}+\beta^{p_{i}}\int_{\partial\Omega}{\left|\varphi\right|}^{p_{i}}}$

obtaining

$\int_{\Omega}{\left|\nabla\phi\right|}^{p_{i}}+\beta^{p_{i}}\int_{\partial\Omega}{\left|\phi\right|}^{p_{i}}\leq\int_{\Omega}{\left|\nabla
v_{p_{i}}\right|}^{p_{i}}+\beta^{p_{i}}\int_{\partial\Omega}v_{p_{i}}^{p_{i}}.\qed$

######  Proposition 4.3.

Let $v_{p}$ be the solution to (18) and let $v_{\infty}$ be any limit of a
subsequence of $\Set{v_{p}}_{p>1}$. Then

$v_{\infty}(x)\leq\frac{1}{\beta}+d(x,\partial\Omega).$ (28)

###### Proof.

We notice that

${\left|v_{\infty}(x)-v_{\infty}(y)\right|}\leq{\left|x-y\right|}$

as we have proven that ${\left\|\nabla v_{\infty}\right\|}_{\infty}\leq 1$.
This holds true for every $x,y$ in $\Omega$. In particular, we can choose $y$
equal to the point on the boundary which realizes
${\left|x-y\right|}=d(x,\partial\Omega)$. So, we have

$v_{\infty}(x)\leq
v_{\infty}(y)+d(x,\partial\Omega)\leq\frac{1}{\beta}+d(x,\partial\Omega),$

as $v_{\infty}$ also satisfies
$\beta{\left\|v_{\infty}\right\|}_{L^{\infty}(\partial\Omega)}\leq 1$. ∎

######  Remark 4.1.

We explicitly observe that the estimate
$\varphi(x)\leq\frac{1}{\beta}+d(x,\partial\Omega)$ holds for every admissible
function $\varphi$.

######  Proposition 4.4.

Assume $f>0$ in $\Omega$. Then the sequence of solution to (18) converges
strongly in $W^{1,m}(\Omega)$, for all $m>1$, to

$\overline{v}_{\infty}(x)=\frac{1}{\beta}+d(x,\partial\Omega).$

###### Proof.

Let $v_{\infty}$ be any limit of a subsequence
$\Set{v_{p_{j}}}\subset\Set{v_{p}}$. Theorem 4.2 gives that $v_{\infty}$ is a
minimum of the functional $J_{\infty}$, in the class $\\{\varphi\in
W^{1,\infty}(\Omega):\,{\left\|\nabla\varphi\right\|}_{\infty}\leq
1,\,\beta{\left\|\varphi\right\|}_{L^{\infty}(\partial\Omega)}\leq 1\\}$.

The function $\frac{1}{\beta}+d(x,\partial\Omega)$ is a competitor and then

$\int_{\Omega}f\left(v_{\infty}-\frac{1}{\beta}-d(x,\partial\Omega)\right)\geq
0.$ (29)

Then (28) gives $v_{\infty}(x)=\frac{1}{\beta}+d(x,\partial\Omega)$.

Since every subsequence of $\Set{v_{p}}$ has a subsequence converging to
$\frac{1}{\beta}+d(x,\partial\Omega)$ weakly in $W^{1,m}(\Omega)$, the whole
sequence $\Set{v_{p}}$ converges to $\frac{1}{\beta}+d(x,\partial\Omega)$
weakly in $W^{1,m}(\Omega)$, and in particular, in
$C^{\alpha}(\overline{\Omega})$ and its gradient weakly in $L^{m}(\Omega)$.

It remains to prove the strong convergence in $W^{1,m}(\Omega)$.

Clarkson’s inequality implies for $p,q>m$

$\begin{multlined}\int_{\Omega}\frac{{\left|\nabla v_{p}+\nabla
v_{q}\right|}^{m}}{2^{m}}+\int_{\Omega}\frac{{\left|\nabla v_{p}-\nabla
v_{q}\right|}^{m}}{2^{m}}\leq\frac{1}{2}\int_{\Omega}{\left|\nabla
v_{p}\right|}^{m}+\frac{1}{2}\int_{\Omega}{\left|\nabla
v_{q}\right|}^{m}\end{multlined}\int_{\Omega}\frac{{\left|\nabla v_{p}+\nabla
v_{q}\right|}^{m}}{2^{m}}+\int_{\Omega}\frac{{\left|\nabla v_{p}-\nabla
v_{q}\right|}^{m}}{2^{m}}\leq\frac{1}{2}\int_{\Omega}{\left|\nabla
v_{p}\right|}^{m}+\frac{1}{2}\int_{\Omega}{\left|\nabla v_{q}\right|}^{m}$

From (25) we deduce

$\lim_{p\to\infty}\int_{\Omega}{\left|\nabla
v_{p}\right|}^{m}\leq{\left|\Omega\right|},$

then semicontinuity of $L^{m}$-norm gives

$\limsup_{p,q}\int_{\Omega}\frac{{\left|\nabla v_{p}+\nabla
v_{q}\right|}^{m}}{2^{m}}\leq{\left|\Omega\right|}=\int_{\Omega}{\left|\nabla
d(x,\partial\Omega)\right|}^{m}\leq\liminf_{p,q}\int_{\Omega}\frac{{\left|\nabla
v_{p}+\nabla v_{q}\right|}^{m}}{2^{m}}$

and then

$\limsup_{p,q}\int_{\Omega}\frac{{\left|\nabla v_{p}-\nabla
v_{q}\right|}^{m}}{2^{m}}=0.$

∎

######  Remark 4.2.

If $\text{Supp}f\subset\Omega$, then we can deduce that
$v_{\infty}(x)=\frac{1}{\beta}+d(x,\partial\Omega)$ for all
$x\in\text{Supp}f$, while in $\Omega\setminus\text{Supp}f$ inequality (28) can
be strict.

######  Definition 4.1.

We denote by $\mathcal{R}$ the set of discontinuity of the function $\nabla
d(x,\partial\Omega)$. This set consists of points $x\in\Omega$ for which
$d(x,\partial\Omega)$ is achieved by more than one point $y$ on the boundary.

Then it holds true the following

######  Theorem 4.5.

Let $f$ be a non-negative function in $\Omega$, then function
$\displaystyle{\overline{v}_{\infty}(x)=\frac{1}{\beta}+d(x,\partial\Omega)}$
is the unique extremal function of (26) if and only if
$\mathcal{R}\subset\text{Supp}f$.

###### Proof.

Suppose that $\mathcal{R}\subset\text{Supp}f$ and let $w$ be a minimum of
(26)in the class $\\{\varphi\in
W^{1,\infty}(\Omega):\,{\left\|\nabla\varphi\right\|}_{\infty}\leq
1,\,\beta{\left\|\varphi\right\|}_{L^{\infty}(\partial\Omega)}\leq 1\\}$. By
Remark 4.1 we have

$w(x)\leq\frac{1}{\beta}+d(x,\partial\Omega)\qquad\forall x\in\Omega,$

and arguing as in Remark 4.2 we have

$w(x)=\frac{1}{\beta}+d(x,\partial\Omega)\qquad\forall
x\in\overline{\text{supp}f}.$

Assume by contradiction that there exists $x\in(\overline{\text{Supp}f})^{c}$
such that

$w(x)<\frac{1}{\beta}+d(x,\partial\Omega),$

setting $\eta=\nabla d(x,\partial\Omega)$, we choose the smallest $t$ such
that $y=x+t\eta$ belongs to $\partial(\text{Supp}f)$ (Lemma 4.6 will justify
this choice), thus

$\displaystyle{\left\|\nabla
w\right\|}_{L^{\infty}(\Omega)}{\left|y-x\right|}$ $\displaystyle\geq
w(y)-w(x)$ $\displaystyle>d(y,\partial\Omega)-d(x,\partial\Omega)=\nabla
d(\xi,\partial\Omega)\cdot(y-x)={\left|y-x\right|}$

where the last equality follows from Lemma 4.6.

So we have ${\left\|\nabla w\right\|}_{L^{\infty}(\Omega)}>1$ that is a
contradiction.

Assume now that $\displaystyle{w(x)=\frac{1}{\beta}+d(x,\partial\Omega)}$ is
the unique extremal of (26), and assume by contradiction that
$\mathcal{R}\not\subset\operatorname{\text{Supp}}f$. Then, thanks to the
Aronsson theorem (see [Aro67]), we can construct a function $\varphi$
different from $w$ which coincide with $w$ in $\operatorname{\text{Supp}}f$
and such that it is admissible for (21). Then $\varphi$ is a minimum too.
This contradicts the fact that $w$ is the unique minimum of (21), so
$\mathcal{R}\subset\text{Supp}f$.

∎

We have to prove Lemma 4.6 to complete the proof.

######  Lemma 4.6.

Let $x\in\Omega\setminus\mathcal{R}$ and set
$\eta=\nabla(d(x,\partial\Omega))$. Let us consider $y_{t}=x+t\eta$, then
there exists $T$ such that $y_{T}\in\mathcal{R}$ and $y_{t}\notin\mathcal{R}$
for all $t<T$. Moreover,

$\nabla d(x+t\eta,\partial\Omega)=\eta\qquad\forall t\in[0,T).$

###### Proof.

Consider the following Cauchy problem

$\begin{cases}\dot{\gamma}(t)=\nabla d(\gamma(t),\partial\Omega),\\\
\gamma(0)=x\end{cases}$ (30)

in a maximal interval $[0,T)$. We have that

  * •

$\displaystyle{L(\gamma)=\int_{0}^{T}{\left|\dot{\gamma}(t)\right|}\,ds=T}$;

  * •

$\displaystyle{\frac{d}{dt}d(\gamma(t),\partial\Omega)=\nabla
d(\gamma(t),\partial\Omega)\dot{\gamma}(t)=1,}$ then

$T=\int_{0}^{T}\frac{d}{dt}d(\gamma(t),\partial\Omega)\,dt=d(\gamma(T),\partial\Omega)-d(x,\partial\Omega).$

These considerations give us the following:

  * •

$T<\infty$, otherwise $d$ is unbounded, and this is a contradiction as
$\Omega$ is bounded;

  * •

$\gamma(T)\in\mathcal{R}$, otherwise one can extend the solution for $t>T$, in
contradiction with the fact that $[0,T)$ is the maximal interval.

In the end, if $y=\gamma(T)$, we have

$d(y,\partial\Omega)=d(x,\partial\Omega)+T=d(x,\partial\Omega)+L(\gamma),$

$L(\gamma)\geq{\left|y-x\right|}$, and they are equal if and only if $\gamma$
is a segment.

If $L(\gamma)>{\left|y-x\right|}$ then

$d(y,\partial\Omega)=d(x,\partial\Omega)+L(\gamma)>d(x,\partial\Omega)+{\left|y-x\right|}\geq{\left|y-z\right|}$

with $z\in\partial\Omega$ such that $d(x,\partial\Omega)={\left|x-z\right|}$,
and this is a contradiction, because $d(y,\partial\Omega)$ is the infimum.
Then $L(\gamma)={\left|y-x\right|}$ and, remembering the fact
$\dot{\gamma}(t)=\nabla d(\gamma(t),\partial\Omega)$, whose norm is 1, then
$\gamma$ is a segment and

$\nabla d(\gamma(t),\partial\Omega)=\eta\qquad\forall t\in[0,T).$

This conclude the proof. ∎

###  4.1 The limit PDE

We have proved that any limit $v_{\infty}$ of subsequence $\Set{v_{p}}$ is a
minimum of the functional (26) defined in $W^{1,\infty}(\Omega)$. Now we want
to understand if such limits are solutions to a certain PDE, which, in some
sense, is the Euler-Lagrange equation of the functional (26).

######  Proposition 4.7.

Let $f\in L^{\infty}(\Omega)\cap C(\overline{\Omega})$ be a non-negative
function. Then any limit $v_{\infty}$ of a subsequence $\Set{v_{p}}$ satisfies

${\left|\nabla v_{\infty}\right|}\leq 1\quad\qquad\qquad\text{in the viscosity
sense}.$ (31)

###### Proof.

The proof follows the techniques contained in [BDM89, JLM99, Yu07, LW08].

If $x_{0}\in\Omega$, let $\varphi\in C^{2}(\Omega)$ be such that
$v_{\infty}-\varphi$ has a local maximum at $x_{0}$

$(v_{\infty}-\varphi)(x_{0})\geq(v_{\infty}-\varphi)(x),\,\,\forall x\in
B_{R}(x_{0}),$

then

${\left|\nabla\varphi(x_{0})\right|}\leq 1.$ (32)

Indeed, let

$C=\sup_{q>1}\max{\Set{{\left\|v_{q}\right\|}_{L^{\infty}\left(B_{R}(x_{0})\right)};{\left\|\varphi\right\|}_{L^{\infty}\left(B_{R}(x_{0})\right)}}},$

we consider the following sequence

$f_{q}(x)=v_{q}(x)-\varphi(x)-k{\left|x-x_{0}\right|}^{a},\quad
k=\frac{4C}{R^{a}},\quad a>2.$

So $f_{q}(x)\leq-3C$ for $x\in\partial B_{R}(x_{0})$, and
$f_{q}(x_{0})\geq-2C$. Then $f_{q}(x)$ attains its maximum at some point
$x_{q}$ in the interior of $B_{R}(x_{0})$. and it holds

$\nabla
v_{q}(x_{q})=\nabla\varphi(x_{q})+ka(x_{q}-x_{0}){\left|x_{q}-x_{0}\right|}^{a-2}.$

Moreover, the sequence $\Set{x_{q}}$ must converge to $x_{0}$.

Assume by contradiction that ${\left|\nabla\varphi(x_{0})\right|}>1$, so there
exists $\delta\in(0,1)$ such that ${\left|\nabla\varphi(x_{0})\right|}\geq
1+\delta$. Choosing $\overline{q}$ large enough, we have

${\left|\nabla
v_{q}(x_{q})\right|}>1+\frac{\delta}{2}-ka{\left|x_{q}-x_{0}\right|}^{a-1}\geq
1+\frac{\delta}{4},\quad\forall q\geq\overline{q}.$

By Lemma 1.1 of Part III in [BDM89],

${\left|\nabla
v_{q}(x)\right|}\leq\left(\frac{\gamma}{R^{n}}\right)^{\frac{1}{q}}\left(\int_{B_{\frac{R}{2}(x_{0})}}\left(1+{\left|\nabla
v_{q}\right|}\right)^{q}\,dx\right)^{\frac{1}{q}},\qquad\forall x\in
B_{\frac{R}{2}}(x_{0})$ (33)

where $\gamma$ is a constant independent of $q$.

For $q$ sufficiently large, this contradict ${\left|\nabla
v_{q}(x_{q})\right|}>1+\frac{\delta}{4}$. ∎

######  Proposition 4.8.

Let $f\in L^{\infty}(\Omega)\cap C(\Omega)$ be a non-negative function, then a
continuous weak solution to (18) is a viscosity solution.

###### Proof.

The proof follows the same techniques of Proposition 3.2. ∎

######  Theorem 4.9.

Let $f\in L^{\infty}(\Omega)\cap C(\Omega)$ be a non-negative function. Then
any $v_{\infty}$ satisfies

$\displaystyle{\left|\nabla
v_{\infty}\right|}=1\quad\qquad\qquad\,\text{on}\,\Set{f>0}$ in the viscosity
sense (34)
$\displaystyle-\Delta_{\infty}v_{\infty}=0\quad\qquad\qquad\text{on}\,(\overline{\Set{f>0}})^{c}$
in the viscosity sense (35)

###### Proof.

We first prove (34). Let $x_{0}\in\Omega\cap\Set{f>0}$ and let $\varphi\in
C^{2}(\Omega)$ such that $v_{\infty}-\varphi$ has a strict minimum in $x_{0}$.
We want to show

${\left|\nabla\varphi(x_{0})\right|}\geq 1$

Let us denote by $x_{p}$ the minimum of $v_{p}-\varphi$, we have that
$x_{p}\to x_{0}$, then $x_{p}\in B_{R}(x_{0})\subset\Set{f>0}$ for $p$ large
enough. Setting $\varphi_{p}(x)=\varphi(x)+c_{p}$, then
$c_{p}=v_{p}(x_{p})-\varphi(x_{p})\to 0$ when $p$ goes to infinity. We notice
that $v_{p}(x_{p})=\varphi_{p}(x_{p})$ and $v_{p}-\varphi_{p}$ has a minimum
in $x_{p}$, so by Proposition 4.8,

$-{\left|\nabla\varphi_{p}(x_{p})\right|}^{p-2}\Delta\varphi_{p}(x_{p})-(p-2){\left|\nabla\varphi_{p}(x_{p})\right|}^{p-4}\Delta_{\infty}\varphi(x_{p})\geq
f(x_{p})>0.$

Dividing by $(p-2){\left|\nabla\varphi_{p}(x_{p})\right|}^{p-4}$, we obtain

$-\Delta_{\infty}\varphi_{p}(x_{p})-\frac{{\left|\nabla\varphi_{p}(x_{p})\right|}^{2}\Delta\varphi_{p}(x_{p})}{p-2}\geq\frac{f(x_{p})}{(p-2){\left|\nabla\varphi_{p}(x_{p})\right|}^{p-4}}$
(36)

This gives us ${\left|\nabla\varphi(x_{0})\right|}\geq 1$, otherwise the
right-hand side would go to infinity, in contradiction with the fact that
$\varphi\in C^{2}(\Omega)$.

We stress that, if we let $p\to\infty$ in (36), we get
$-\Delta_{\infty}\varphi(x_{0})\geq 0$, so $v_{\infty}$ is a supersolution to
$-\Delta_{\infty}u=0$.

Now, we only have to prove that $v_{\infty}$ is a viscosity solution to
$\Delta_{\infty}\varphi=0$ in $(\overline{\Set{f>0}})^{c}$.

If we fix $x_{0}\in(\overline{\Set{f>0}})^{c}$, and $\varphi\in C^{2}(\Omega)$
such that $v_{\infty}-\varphi$ has a strict maximum in $x_{0}$, we can choose
$R$ such that $B_{R}(x_{0})\subset(\overline{\Set{f>0}})^{c}$. The function
$v_{p}-\varphi$ has a maximum $x_{p}\to x_{0}$, and $x_{p}\in B_{R}(x_{0})$
for $p$ large enough.

The definition of viscosity subsolution implies

$-{\left|\nabla\varphi_{p}(x_{p})\right|}^{p-2}\Delta\varphi_{p}(x_{p})-(p-2){\left|\nabla\varphi_{p}(x_{p})\right|}^{p-4}\Delta_{\infty}\varphi(x_{p})\leq
f(x_{p})=0.$

Without loss of generality, we may assume
${\left|\nabla\varphi(x_{0})\right|}\neq 0$. Dividing both side of the last
equation by $(p-2){\left|\nabla\varphi_{p}(x_{p})\right|}^{p-4}$, we obtain

$-\Delta_{\infty}\varphi_{p}(x_{p})\leq\frac{{\left|\nabla\varphi_{p}(x_{p})\right|}^{2}\Delta\varphi_{p}(x_{p})}{p-2}.$

Letting $p\to\infty$, we get

$-\Delta_{\infty}\varphi(x_{0})\leq 0.$

Analogously if $\varphi\in C^{2}(\Omega)$ is such that $v_{\infty}-\varphi$
has a minimum at $x_{0}$, a symmetric argument shows that
$-\Delta_{\infty}\varphi(x_{0})\geq 0$. ∎

### Some examples

######  Example 4.3.

We consider the case $\Omega=B_{1}(0)$ and $f=1$. The uniqueness of the
solution and the invariance under rotation of the $p$-Laplacian ensure that
the solution $v$ is radially symmetric and

$r^{n-1}\Delta_{p}v_{p}=\frac{d}{dr}\left(r^{n-1}\left\lvert
v_{p}^{\prime}\right\rvert^{p-2}v_{p}^{\prime}\right).$

Setting $\alpha=1/(p-1)$, we have

$v_{p}(x)=-\frac{p-1}{n^{\alpha}p}{\left|x\right|}^{\frac{p}{p-1}}+\frac{1}{(n\beta^{p})^{\alpha}}+\frac{p-1}{n^{\alpha}p}$

Then

$v_{\infty}(x)=-{\left|x\right|}+\frac{1}{\beta}+1=\frac{1}{\beta}+d(x,\partial\Omega)$

######  Example 4.4.

We fix $0<\varepsilon<1$ and we consider

$\left.f=\right.\begin{cases}1&\text{ if }x\in B_{\varepsilon}(0)\\\ 0&\text{
if }x\in B_{1}(0)\setminus B_{\varepsilon}(0).\end{cases}$

In this case, $v_{p}$ is radially symmetric, and

$\left.v_{p}=\right.\begin{cases}\frac{p-1}{n^{\alpha}p}\left(\varepsilon^{\frac{p}{p-1}}-{\left|x\right|}^{\frac{p}{p-1}}\right)+\frac{\varepsilon^{n\alpha}(p-1)}{n^{\alpha}(p-n)}\left(1-\varepsilon^{\frac{p-n}{p-1}}\right)+\frac{\varepsilon^{n\alpha}}{(n\beta^{p})^{\alpha}}&\text{
if }x\in B_{\varepsilon}(0)\\\
\frac{\varepsilon^{n\alpha}(p-1)}{n^{\alpha}(p-n)}\left(1-{\left|x\right|}^{\frac{p-n}{p-1}}\right)+\frac{\varepsilon^{n\alpha}}{(n\beta^{p})^{\alpha}}&\text{
if }x\in B_{1}(0)\setminus B_{\varepsilon}(0)\end{cases}$

Letting $p$ go to infinity, we obtain

$v_{\infty}=\frac{1}{\beta}+d(x,\partial\Omega).$

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Dipartimento di Matematica e Applicazioni “R. Caccioppoli”, Università degli
studi di Napoli Federico II, Via Cintia, Complesso Universitario Monte S.
Angelo, 80126 Napoli, Italy.

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