\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 111, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/111\hfil On the first eigenvalue]
{On the first eigenvalue of the Steklov eigenvalue problem for the
infinity Laplacian}

\author[A. L\^e\hfil EJDE-2006/111\hfilneg]
{An L\^e}

\address{ An L\^e \newline
 Department of Mathematics and Statistics,
 Utah State University,
 Logan, Utah 84322 USA}
\email{anle@cc.usu.edu}

\date{}
\thanks{Submitted August 4, 2006. Published September 18, 2006.}
\subjclass[2000]{35J50, 35J55, 35J60, 35J65, 35P30}
\keywords{Nonlinear elliptic equations; eigenvalue problems;
p-Laplacian; \hfill\break\indent
nonlinear boundary condition; Steklov problem; viscosity solutions}

\begin{abstract}
 Let $\Lambda_p^p$ be the best Sobolev embedding constant of
 $W^{1,p}(\Omega )\hookrightarrow L^p(\partial\Omega)$,
 where $\Omega$ is a smooth bounded domain
 in $\mathbb{R}^N$. We prove that as $p \to \infty$ the sequence $\Lambda_p$
 converges to a constant independent of the shape and the volume
 of $\Omega$, namely 1. Moreover, for any sequence of eigenfunctions
 $u_p$ (associated with $\Lambda_p$), normalized by
 $\| u_p \|_{L^\infty(\partial\Omega)}=1$,
 there is a subsequence converging to a limit function $u_\infty$ which
 satisfies, in the viscosity sense, an $\infty$-Laplacian equation
 with a boundary condition.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

\section{Introduction} \label{intro-sec}

Let $\Omega$ be a bounded domain in $\mathbb{R}^N$ with smooth boundary. The
minimum $\Lambda_p^p$ of the Rayleigh quotient, among all nonzero
functions in the Sobolev space $W^{1,p}(\Omega)$,
\[
\frac{\int_\Omega  (|\nabla u |^p + |u|^p) dx}{\int_{\partial\Omega} |u|^p dx}
\]
is the first eigenvalue of the problem
\begin{equation} \label{p-Steklov}
     \begin{gathered}
      -\Delta_p u + |u|^{p-2}u = 0,\quad  \mbox{in } \Omega, \\
          |\nabla u|^{p-2} \frac{\partial u}{\partial\nu }= \lambda |u|^{p-2} u, \quad
          \mbox{on } \partial \Omega.
 \end{gathered}
\end{equation}
where $\Delta_p u = \mathop{\rm div}(|\nabla u|^{p-2}\nabla u)$ is the
$p$-Laplacian and $\partial / \partial \nu$ is the outer normal derivative
along ${\partial\Omega}$. The eigenvalue problem \eqref{p-Steklov} is understood
in the weak sense, i.e, $(u, \lambda) \in W^{1,p}(\Omega) \times \mathbb{R}^+$ is an
eigenpair if
\[
\int_\Omega (| \nabla u|^{p-2} \nabla u \nabla v + |u|^{p-2} u v) dx
= \lambda \int_{\partial\Omega} |u|^{p-2} uv ds,\quad \forall v\in W^{1,p}(\Omega).
\]
The first eigenvalue $\Lambda_p^p=\lambda_1$ is the best constant
of the compact embedding $W^{1,p}(\Omega) \hookrightarrow L^p({\partial\Omega})$ and it satisfies
\[
\Lambda_p \| u \|_{L^p(\partial\Omega) } \leq \| u \|_{W^{1,p}(\Omega)},\quad \forall u \in W^{1,p}(\Omega).
\]
In this paper we are interested in finding the limit as
 $p \to \infty$ of $\Lambda_p$. Alternatively, we want to look at
the limit of the minimum, as $p \to \infty$, of the ratio
\[
\frac{ \left( \int_\Omega  (|\nabla u |^p + |u|^p) dx \right)^{1/p} }{  \left( \int_{\partial\Omega} |u|^p ds \right)^{1/p} }.
\]
It is easy to see that for any positive numbers $a$ and $b$,
\[
\lim_{p \to \infty} (a^p + b^p)^{1/p} = \max\{a,b\}.
\]
Thus we anticipate that
\[
\lim_{p\to \infty} \frac{ \left( \int_\Omega (| \nabla u |^p + |u|^p) dx \right)^{1/p} }
{  \left( \int_{\partial\Omega} |u|^p dx \right)^{1/p} }
= \frac{ \max\{ \|\nabla u\|_{ L^\infty(\Omega)}, \| u \|_{ L^\infty(\Omega)} \} }
{\|u \|_{L^\infty(\partial\Omega)}}.
\]
However, the minimization problem, with minimum value equal 1,
\begin{equation} \label{iy-cha}
\inf_{u \in W^{1,\infty}(\Omega) \setminus \{0\}}
 \frac{ \max\{ \|\nabla u\|_{ L^\infty(\Omega)}, \| u \|_{ L^\infty(\Omega)} \} }
{\|u \|_{L^\infty(\partial\Omega)}},
\end{equation}
has too many solutions. In fact, given a minimizer, we can modify it on any
ball inside the domain to obtain another one. The correct Euler-Lagrange
 equation turns out to be
\begin{equation} \label{iy-eq}
     \begin{gathered}
      \max \{u - |\nabla u|, -\Delta_\infty u \}= 0, \quad\mbox{ in }  \Omega,\\
   \min \{|\nabla u| - u, \frac{ \partial u}{\partial \nu} \} =0,
 \quad \mbox{ on } \partial \Omega,
                \end{gathered}
\end{equation}
where the operator
\[
\Delta_\infty u = \sum_{i,j=1}^N \frac{\partial u}{\partial x_j}
\frac{\partial^2 u}{\partial x_j \partial x_i} \frac{\partial u}{\partial x_i},
\]
is called the $\infty$-Laplacian.

It is clear that for each $p>1$ the Sobolev embedding constant $\Lambda_p$
depends on the shape and the volume of domain $\Omega$. We show that when
passing to the limit, $\Lambda_p$ converges to a constant independent of the domain $\Omega$, namely 1. We can also choose a sequence of first eigenfunctions $u_p$ such that the sequence converges uniformly in $C^{\alpha}(\overline{\Omega} )$ to a function $u_\infty$ that satisfies (\ref{iy-eq}) in the viscosity sense. In this case we say $(u_\infty,1)$ is an eigenpair of (\ref{iy-eq}).
Our main result is:

\begin{theorem} \label{thm-1}
For the first eigenvalue of \eqref{p-Steklov} we have
\[
\lim_{p\to \infty} \Lambda^{1/p}_p=1.
\]
For each $p>1$, let $u_p$ positive eigenfunction be a positive eigenfunction
 associated with $\Lambda_p^p$ such that $\| u_p\|_{L^\infty(\partial\Omega)}=1$.
Then there exists a sequence $p_i \to \infty$ such that $u_{p_i} \to u_\infty$ in $C^{\alpha}(\overline{\Omega} )$. The limit $u_\infty$ is a solution of (\ref{iy-eq})
in the viscosity sense.
\end{theorem}

To complete the introduction let us mention some recent work on the subject.
In \cite{juutinen:hep05,juutinen:ep99} the authors study eigenvalue problem
for the $\infty$-Laplacian with Dirichlet boundary condition.
 In \cite{azorero:sep06}, Steklov eigenvalues for the $\infty$-Laplacian
are studied when one considers the limit as $p\to \infty$ of the
eigenvalue problem
\begin{equation} \label{p-har}
 \begin{gathered}
      -\Delta_p u = 0,\\
   |\nabla u|^{p-2} \frac{\partial u}{\partial\nu }=
 \lambda |u|^{p-2} u, \quad \mbox{ on } \partial \Omega.
 \end{gathered}
\end{equation}

It is known that the structure of the spectrum of (\ref{p-har})
is the same as that of \eqref{p-Steklov}, see \cite{bonder:erp01}
and \cite{torne:spi05} . However, the first eigenvalue of (\ref{p-har})
is 1 for any $p>1$ with corresponding constant eigenfunctions in $W^{1,p}(\Omega)$;
thus, theorem \ref{thm-1} is trivial for problem (\ref{p-har}).
Some arguments and technicalities used here are adapted
from  \cite{azorero:sep06,juutinen:hep05,juutinen:ep99}.

\section{\bf Main Results} \label{def-sec}

We first recall the definition of viscosity solutions. Let
\begin{gather*}
F :  \mathbb{R}^N \times \mathbb{R} \times \mathbb{R}^N \times \mathcal{S}(N) \to \mathbb{R},\\
B :  \mathbb{R}^N \times \mathbb{R} \times \mathbb{R}^N \to \mathbb{R},
\end{gather*}
where $\mathcal{S}(N)$ denotes the set of $N \times N$ symmetric matrices.

Consider the boundary-value problem
\begin{equation} \label{bvp}
\begin{gathered}
      F(x, u, \nabla u, D^2 u) = 0,\\
        B(x,u,\nabla u)= 0, \quad\mbox{on } \partial \Omega.
\end{gathered}
\end{equation}

\begin{definition} \label{vis-def} \rm
\begin{itemize}

\item[(i)] An upper semicontinuous function $u$ is a
\emph{viscosity subsolution} of (\ref{bvp}) if for every $\phi \in C^2(\overline{\Omega} )$
such that $u - \phi$ has a strict maximum at the point $x_0 \in \overline{\Omega} $
with $u(x_0) = \phi(x_0)$, we have
\begin{gather*}
\min\{F(x_0,\phi(x_0), \nabla \phi(x_0), D^2 \phi(x_0)), B(x_0, \phi(x_0),
\nabla \phi(x_0) ) \}  \leq  0, \quad x_0 \in {\partial\Omega},\\
F(x_0, \phi(x_0),\nabla \phi(x_0), D^2 \phi(x_0)) \leq  0,
\quad x_0 \in \Omega.
\end{gather*}

\item[(ii)] A lower semicontinuous function $u$ is a
\emph{viscosity supersolution} of (\ref{bvp}) if for every
$\phi \in C^2(\overline{\Omega} )$ such that
$u - \phi$ has a strict minimum at the point $x_0 \in \overline{\Omega} $
with $u(x_0) = \phi(x_0)$, we have
\begin{gather*}
\max\{F(x_0,\phi(x_0), \nabla \phi(x_0), D^2 \phi(x_0)), B(x_0,
\phi(x_0), \nabla \phi(x_0) ) \}  \geq  0, \quad x_0 \in {\partial\Omega},\\
F(x_0, \phi(x_0),\nabla \phi(x_0), D^2 \phi(x_0))  \geq  0, \quad
 x_0 \in \Omega.
\end{gather*}

\item[(iii)] $u$ is a \emph{viscosity solution} of (\ref{bvp})
if it is both a supersolution and a subsolution.
\end{itemize}
\end{definition}

In (i) and (ii) the extrema at $x_0$ need not be strict.
We refer to \cite{crandall:ugv92} for the theory of viscosity
solutions in general and \cite{barles:fnn93} for viscosity solutions
with general boundary conditions.

If $u$ is a smooth eigenfunction of \eqref{p-Steklov} then by
differentiation we get
\begin{equation} \label{p-vis-eq}
\begin{gathered}
      - |\nabla u|^{p-2} \Delta u - (p-2)|\nabla u|^{p-4} \Delta_\infty u + |u|^{p-2}u = 0,
\quad \mbox{in } \Omega,\\
          |\nabla u|^{p-2} \frac{\partial u}{\partial\nu } - \lambda |u|^{p-2} u = 0,
\quad \mbox{on } \partial \Omega.
\end{gathered}
\end{equation}
In this case,
\begin{gather}
F(x,z,X,S)  =  -|X|^{p-2}\mbox{trace}(S)
 - (p-2)|X|^{p-4}\langle S\cdot X, X \rangle + |z|^{p-2}z, \label{F-form} \\
B(x,z,X)  = |X|^{p-2} \langle X, \nu (x) \rangle - \lambda |z|^{p-2}z.
\end{gather}

It is known that eigenfunctions of \eqref{p-Steklov} are in
$C^{1,\alpha}(\overline{\Omega} )$, see \cite{an:epp06,lieberman:brs88} and references therein.
 Thus it makes sense to talk about viscosity solutions.
The following lemma tells us that an eigenfunction is a viscosity solution.

\begin{lemma} \label{lem2.2}
A weak solution $u$ of \eqref{p-Steklov} is a viscosity solution of
\eqref{p-vis-eq}.
\end{lemma}

\begin{proof}
We present the details for the case of supersolutions.
Let $x_0 \in \Omega$ and a function $\phi \in C^2(\Omega)$ such that
$u(x_0)=\phi(x_0)$ and $u (x) > \phi(x)$, for $x \neq x_0$.
We want to show that
\[
 -|\nabla \phi|^{p-2} \Delta \phi(x_0) - (p-2)|\nabla \phi|^{p-4} \Delta_\infty \phi(x_0)
+ |u|^{p-2} u(x_0)  \geq  0.
\]
Suppose that this is not the case, then by continuity there exists a
radius $r >0$ such that, for any $x \in B(x_0,r)$,
\[
 -|\nabla \phi|^{p-2} \Delta \phi(x) - (p-2)|\nabla \phi|^{p-4} \Delta_\infty \phi(x)
+  |u|^{p-2} u(x) < 0.
\]
Set $m = \inf \{u(x) - \phi(x): |x-x_0|=r \} >0$ and let
$\Phi(x) = \phi + \frac{1}{2} m$. The function $\Phi$ satisfies $\Phi < u $
on $\partial B(x_0,r)$, $\Phi(x_0) > u(x_0)$ and
\[
 -\mathop{\rm div} (|\nabla \Phi|^{p-2} \nabla \Phi(x)) +  |u|^{p-2} u(x) < 0.
\]
Multiplying by $(\Phi - u)^+$ extended by zero outside $B(x_0,r)$ we get
\begin{equation} \label{Phi-ineq}
\int_{\{\Phi >u \}} |\nabla \Phi|^{p-2} \nabla \Phi \cdot \nabla (\Phi - u)
+ \int_{\{\Phi >u \}} |u|^{p-2} u (\Phi -u) < 0.
\end{equation}
Since $u$ is a weak solution, we have
\begin{equation} \label{u-eq}
\int_{\{\Phi >u \}} |\nabla u|^{p-2}  \nabla u \cdot \nabla (\Phi - u)
+ \int_{\{\Phi >u \}} |u|^{p-2} u (\Phi -u) = 0.
\end{equation}
Subtracting (\ref{u-eq}) from (\ref{Phi-ineq}) we get
\[
\int_{\{\Phi >u \}} \langle |\nabla \Phi|^{p-2}\nabla \Phi
- |\nabla u|^{p-2}  \nabla u, \nabla (\Phi - u) \rangle < 0.
\]
We obtain a contradiction since the left hand side is bounded below by
\[
C(N,p) \int_{\{\Phi >u \}} |\nabla \Phi - \nabla u|^p,
\]
where $C(N,p)$ is a positive constant depending only on $N$ and $p$.

Let $\lambda$ be the eigenvalue corresponding to $u$ in \eqref{p-Steklov}.
If $x_0 \in {\partial\Omega}$ and $\phi$ is a function in $C^2(\overline{\Omega} )$ such
that $u(x_0)=\phi(x_0)$ and $u(x) > \phi(x)$, for $x \neq x_0$.
We want to prove that either
\begin{gather*}
-|\nabla \phi|^{p-2} \Delta \phi(x_0) - (p-2)|\nabla \phi|^{p-4} \Delta_\infty \phi(x_0)
 + |u|^{p-2} u(x_0)  \geq  0, \\
\mbox{or }\quad  |\nabla \phi|^{p-2} \frac{\partial \phi}{\partial\nu }(x_0)
- \lambda |u|^{p-2} u(x_0)  \geq  0.
\end{gather*}
Suppose that this is not the case. We repeat the previous argument to obtain
\begin{align*}
&\int_{\{\Phi >u \}} |\nabla \Phi|^{p-2} \nabla \Phi \cdot \nabla (\Phi - u)  +  \int_{\{\Phi >u \}} |u|^{p-2} u (\Phi -u) \\
 &  <   \lambda \int_{{\partial\Omega} \cap \{\Phi >u \}} |u|^{p-2} u (\Phi -u),
\end{align*}
and
\begin{align*}
&\int_{\{\Phi >u \}} |\nabla u|^{p-2} \nabla u \cdot \nabla (\Phi - u)
 +  \int_{\{\Phi >u \}} |u|^{p-2} u (\Phi -u) \\
 &  =   \lambda \int_{{\partial\Omega} \cap \{\Phi >u \}} |u|^{p-2} u (\Phi -u),
\end{align*}
which implies
\[
\int_{\{\Phi >u \}} \langle |\nabla \Phi|^{p-2}\nabla \Phi - |\nabla u|^{p-2}  \nabla u, \nabla (\Phi - u) \rangle < 0.
\]
A contradiction is established. This proves that $u$ is a
viscosity supersolution. As mentioned, the proof that
$u$ is a viscosity subsolution is similar.
\end{proof}


We are ready to pass to limit as $p \to \infty$ in the eigenvalue problem.
Using the characterization
\begin{equation} \label{p-cha}
\Lambda_p = \min_{u \neq 0}
\frac{ \left( \int_\Omega (| \nabla u |^p + |u|^p) dx \right)^{1/p} }
{  \left( \int_{\partial\Omega} |u|^p ds \right)^{1/p} },
\end{equation}
one can show the following statement.

\begin{lemma} \label{limsup-p}
\[
\limsup_{p \to \infty} \Lambda_p \leq 1.
\]
\end{lemma}

\begin{proof}
Let $v(x) \equiv 1$, $x \in \overline{\Omega} $. It follows from (\ref{p-cha}) that
\[
\Lambda_p \leq \frac{ \left( \int_\Omega (| \nabla v |^p + |v|^p) dx \right)^{1/p} }
{  \left( \int_{\partial\Omega} |v|^p ds \right)^{1/p} }
= \frac{|\Omega|^{1/p}}{|{\partial\Omega}|^{1/p}},
\]
where $|\Omega|$ is the Lebesgue measure of $\Omega$ and $|{\partial\Omega}|$ is the boundary measure of ${\partial\Omega}$.
Letting $p \to \infty$ we obtain the inequality.
\end{proof}

We recall from  \cite{an:epp06,martinez:isf02} that the first eigenvalue
$\Lambda_p^p$ is isolated and simple. Any eigenfunction associated
with $\Lambda_p^p$ is either positive or negative in $\Omega$ and any other
eigenfunction (not associated with $\Lambda_p^p$) has to change sign.
We show that $\Lambda_p$ converges to $\Lambda_\infty=1$, the minimum value
of \eqref{iy-cha}. We also construct a minimizer of \eqref{iy-cha}.

\begin{proposition} \label{iy-prop}
Given $u_p$, a positive eigenfunction of \eqref{p-Steklov} associated
with eigenvalue $\Lambda_p^p$, normalized by $\| u_p \|_{L^\infty({\partial\Omega})}=1$.
Then there exists a sequence $p_i \to \infty$ such that
$u_{p_i} \to u_\infty \mbox{ in } C^\alpha(\overline{\Omega} )$,
where the limit $u_{\infty}$ is a minimizer of \eqref{iy-cha} and
\[
\lim_{p \to \infty} \Lambda_p = 1.
\]
\end{proposition}

\begin{proof}
Fix $q>N$. For any $p>q$, one has
\begin{equation} \label{p-q-ineq}
\begin{aligned}
\Big(\int_\Omega |\nabla u_p |^q + |u_p|^q \Big)^{1/q}
& \leq  |\Omega|^{(1/q) -(1/p)}
 \Big[\int_\Omega (|\nabla u_p |^q + |u_p|^q ) ^{p/q} \Big]^{1/p} \\
& \leq  (2|\Omega|)^{(1/q) -(1/p)} \Big(\int_\Omega |\nabla u_p |^p
+ |u_p|^p \Big)^{1/p} \\
&=  (2|\Omega|)^{(1/q) -(1/p)} \Lambda_p \Big(\int_{\partial\Omega}  |u_p|^p \Big)^{1/p} \\
& \leq \Lambda_p (2|\Omega|)^{(1/q) -(1/p)}  |{\partial\Omega}|^{1/p}.
\end{aligned}
\end{equation}
In above expression, the first inequality follows from a H\"{o}lder inequality.
We have used that  $(a+b)^r \leq 2^{r-1}(a^r+b^r)$, $r \geq 1$,
for the second inequality and that $\|u_p\|_{L^\infty({\partial\Omega})}=1$ for the
last inequality.
We obtain from \eqref{p-q-ineq} that $\{u_p \}$ is uniformly bounded
in $W^{1,q}(\Omega)$. Thus there exists a subsequence $\{ u_{p_i} \}$
converging to a function $u_{\infty}$ weakly in $W^{1,q}(\Omega)$.
Since $q>N$, the Sobolev compact embedding
$W^{1,q}(\Omega) \hookrightarrow C^{\alpha}(\overline{\Omega} )$ holds for any
$\alpha \in (0, 1-N/q)$. It follows that $\{ u_{p_i} \}$ converges
 to $u_{\infty}$ uniformly in $C^{\alpha}(\overline{\Omega} )$. Moreover,
as $p_i \to \infty$, \eqref{p-q-ineq} becomes
\begin{equation} \label{iy-q-ineq}
\Big(\int_\Omega |\nabla u_{\infty} |^q + |u_{\infty}|^q \Big)^{1/q}
 \leq \liminf_{p_i \to \infty} \Lambda_{p_i} (2|\Omega|)^{1/q} \leq (2|\Omega|)^{1/q}.
\end{equation}
On the other hand,
\begin{equation} \label{iy-max-ineq}
\max\{ \| \nabla u_{\infty}\|_{L^q(\Omega)}, \| u_{\infty} \|_{L^q(\Omega} \}
 \leq \Big(\int_\Omega |\nabla u_{\infty} |^q + |u_{\infty}|^q \Big)^{1/q}
\end{equation}
Letting $q \to \infty$,  (\ref{iy-q-ineq}) and (\ref{iy-max-ineq}) imply that
\[
\max\{ \| \nabla u_{\infty}\|_{ L^\infty(\Omega)}, \| u_{\infty} \|_{ L^\infty(\Omega)} \} \leq 1.
\]
The uniform convergence of $\{ u_{p_i} \}$ in $C^\alpha(\overline{\Omega} )$ gives
$\| u_{\infty} \|_{L^\infty(\partial\Omega)}=1$. Hence
\[
\| \nabla u_{\infty}\|_{ L^\infty(\Omega)} \leq 1 \mbox{ and } \| u_{\infty} \|_{ L^\infty(\Omega)}
= \| u_{\infty} \|_{ L^\infty(\partial\Omega))} = 1.
\]
Clearly, $u_{\infty}$ is the minimizer of \eqref{iy-cha}. Furthermore,
\[
1 = \| u_{\infty} \|_{ L^\infty(\Omega)}  \leq \lim_{q\to \infty}
\Big(\int_\Omega |\nabla u_{\infty} |^q + |u_{\infty}|^q \Big)^{1/q}
\leq \liminf_{p_i \to \infty} \Lambda_{p_i},
\]
which together with the lemma \ref{limsup-p} gives
$ \lim_{p_i \to \infty} \Lambda_{p_i} = 1 $. Since the limit holds for any
subsequence, we conclude that
$ \lim_{p \to \infty} \Lambda_p = 1$.
The proof is complete.
\end{proof}

Let us verify that the limit of \eqref{p-vis-eq} as $p \to \infty$
is (\ref{iy-eq}) in the viscosity sense. We obtain from
proposition \ref{iy-prop} that there is a sequence of
positive eigenfuntions $\{u_{p_i}\}$ converging to $u_\infty$
uniformly in $\overline{\Omega} $ as $p_i \to \infty$. Consequently, $u_\infty \geq 0$
in $\overline{\Omega} $.

\begin{lemma} \label{iy-vis}
$u_\infty$ is a viscosity solution of (\ref{iy-eq}), i.e.,
\begin{equation}
 \begin{gathered}
  \max \{u_\infty - |\nabla u_\infty|, -\Delta_\infty u_\infty \}= 0, \quad\mbox{ in }  \Omega,\\
  \min \{|\nabla u_\infty| - u, \frac{ \partial u_\infty}{\partial \nu} \} = 0,\quad
 \mbox{ on } \partial \Omega.
 \end{gathered}
\end{equation}
\end{lemma}

\begin{proof}
First let us check
\begin{equation} \label{interior}
\max \{u_\infty - |\nabla u_\infty|, -\Delta_\infty u_\infty \}= 0 \quad \mbox{in } \Omega.
\end{equation}
Fix $x_0 \in \Omega$ and a function $\phi \in C^2(\Omega)$ such that
$u_\infty(x_0)=\phi(x_0)$ and $u(x) < \phi(x)$, for $x \neq x_0$.
Also fix $R>0$ such that $B(x_0,2R) \subset \Omega$.
For $0 < r < R$ we have
\[
\sup\{ u_\infty(x)  - \phi(x): x \in B(x_0,R) \setminus B(x_0,r) \} < 0.
\]
As $u_{p_i} \to u_\infty$ uniformly in $\overline{B(x_0,R)}$, for $i$
large enough we conclude that
\[
\sup\{ u_{p_i}(x)  - \phi(x): x \in B(x_0,R) \setminus B(x_0,r) \}
< u_{p_i}(x_0)  - \phi(x_0).
\]
Therefore for such indices $i$, $u_{p_i} - \phi$ attains its maximum
at $x_i \in B(x_0,r)$. By letting $r \to 0$ we obtain $x_i \to x_0$
as $i \to \infty$. We relabel and denote by $\{x_i\}$ and $\{p_i\}$
the subsequences $\{x_{i_r}\}$ and $\{p_{i_r}\}$. Since $u_{p_i}$
is a subsolution of \eqref{p-vis-eq} and $x_0 \in \Omega$,
\begin{equation} \label{sub-ineq1}
- |\nabla \phi|^{p_i-2} \Delta \phi(x_i) - (p_i-2)|\nabla \phi|^{p_i-4} \Delta_\infty \phi(x_i) + |u_{p_i}|^{p_i-2}u_{p_i}(x_i) \leq  0.
\end{equation}

\noindent $\bullet$ Case 1: $\phi(x_0)=u(x_0) > 0$.
Then $u_{p_i}(x_i) > 0$ for large $i$, which implies that
$|\nabla \phi(x_i)| \neq 0$ due to (\ref{sub-ineq1}).
Dividing by $(p_i -2)|\nabla \phi(x_i)|^{p_i-4}$ we get
\begin{equation} \label{sub-ineq2}
-\frac{|\nabla \phi|^2 \Delta \phi(x_i)}{p_i-2} - \Delta_\infty \phi(x_i) \leq
- \Big( \frac{u_{p_i}(x_i)}{|\nabla \phi(x_i)|}\Big)^{p_i-4}
 \frac{u_{p_i}^3(x_i)}{p_i-2}.
\end{equation}
Letting $p_i \to \infty$ we obtain from (\ref{sub-ineq2}) that
\[
\frac{\phi(x_0)}{|\nabla \phi(x_0)|} \leq 1 \mbox{ and} - \Delta_\infty \phi(x_i) \leq 0 .
\]
Therefore,
\begin{equation} \label{sub-ineq3}
\max \{\phi(x_0) - |\nabla \phi(x_0)|, -\Delta_\infty \phi(x_0) \} \leq 0.
\end{equation}

\noindent$\bullet$ Case 2: $\phi(x_0)=u(x_0)=0$.
If $|\nabla \phi(x_0)| =0$, then $\Delta_\infty \phi(x_0) =0$ and
thus (\ref{sub-ineq3}) holds. If $|\nabla \phi(x_0)| \neq 0$,
then  $|\nabla \phi(x_i)| \neq 0$ for $i$ large.
We then obtain (\ref{sub-ineq2}). The right-hand side of (\ref{sub-ineq2})
tends to zero as $p_i \to \infty$, since
\[
\lim_{p_i \to \infty} \Big( \frac{u_{p_i}(x_i)}{|\nabla \phi(x_i)|}\Big)^{p_i-4} =0.
\]
Thus $- \Delta_\infty \phi(x_i) \leq 0$ and  (\ref{sub-ineq3}) holds in this case.
From both cases we conclude that $u_\infty$ is a viscosity subsolution
of (\ref{interior}).

Next we claim that $u_\infty$ is a viscosity supersolution of
(\ref{interior}) in $\Omega$. Fix a point $x_0 \in \Omega$ and a function
$\phi \in C^2(\Omega)$ such that $u_\infty(x_0)=\phi(x_0)$ and
$u_\infty(x) > \phi(x)$, for $x \neq x_0$. We will show that
\begin{equation} \label{super-ineq1}
\max \{\phi(x_0) - |\nabla \phi(x_0)|, -\Delta_\infty \phi(x_0) \} \geq 0.
\end{equation}
If $|\nabla \phi(x_0)| =0$, there is nothing to prove. It suffices to show
that if $|\nabla \phi(x_0)| \neq 0$ and $\phi(x_0) - |\nabla \phi(x_0)| < 0$,
then $-\Delta_\infty \phi(x_0) \geq 0$. We follow the arguments made
in the subsolution case. An analogue of (\ref{sub-ineq2}) is
\begin{equation} \label{super-ineq2}
-\frac{|\nabla \phi|^2 \Delta \phi(x_i)}{p_i-2} - \Delta_\infty \phi(x_i) \geq
- \Big( \frac{u_{p_i}(x_i)}{|\nabla \phi(x_i)|}\Big)^{p_i-4}
\frac{u_{p_i}^3(x_i)}{p_i-2}.
\end{equation}
Since $\phi(x_0) - |\nabla \phi(x_0)| < 0$,
$\frac{\phi(x_0)}{|\nabla \phi(x_0)|} \leq 1$. Letting $p_i \to \infty$
it follows from (\ref{super-ineq2}) that
$-\Delta_\infty \phi(x_0) \geq 0$ as claimed. Therefore $u_\infty$ is a
viscosity solution of (\ref{interior}).

We next need to check on the boundary using definition \ref{vis-def}.
Fix $x_0 \in {\partial\Omega}$ and a function $\phi \in C^2(\overline{\Omega} )$ such that
$u_\infty(x_0)=\phi(x_0)$ and $u_\infty(x) < \phi(x)$, for $x \neq x_0$.
Using the uniform convergence of $u_{p_i}$ to $u_\infty$ we obtain
that $u_{p_i} - \phi$ attains a maximum at $x_i \in \overline{\Omega} $ with
$x_i \to x_0$.
If (\ref{sub-ineq1}) holds for infinitely many $x_i$,
we use the argument before to obtain (\ref{sub-ineq3}).
Thus we may assume that, for infinitely many $x_i \in {\partial\Omega}$,
\[
|\nabla \phi(x_i)|^{p_i-2} \frac{\partial \phi}{\partial \nu}(x_i)
\leq \Lambda^{p_i}_{p_i} |u_{p_i}|^{p_i -2} u_{p_i}(x_i).
\]
If $|\nabla \phi(x_0)| =0$, then $\frac{\partial \phi}{\partial \nu}(x_0) =0$.
If $|\nabla \phi(x_0)| \neq 0$ we obtain
\[
\frac{\partial \phi}{\partial \nu}(x_i)
\leq  \Big( \frac{\Lambda_{p_i} |u_{p_i}(x_i)|}{|\nabla \phi(x_i)|}\Big)^{p_i -2}
\Lambda^2 _{p_i}u_{p_i}(x_i).
\]
Since $\Lambda_{p_i} \to 1$ as $p_i \to \infty$, we conclude that either
\[
\frac{\phi(x_0)}{|\nabla \phi(x_0)|} \geq 1 \quad\mbox{or}\quad
 \frac{\partial \phi}{\partial \nu}(x_0) \leq 0,
\]
which implies that
$\min \{|\nabla \phi(x_0)| - \phi(x_0), \frac{\partial \phi}{\partial \nu}(x_0) \} \leq 0$.
Therefore, at $x_0$,
\begin{equation} \label{sub}
\min\big\{\max \{\phi - |\nabla \phi|, -\Delta_\infty \phi \},
 \min \{|\nabla \phi| - \phi, \frac{\partial \phi}{\partial \nu} \} \big\} \leq 0.
\end{equation}

Fix $x_0 \in {\partial\Omega}$ and a function $\phi \in C^2(\overline{\Omega} )$ such that
$u_\infty(x_0)=\phi(x_0)$ and $u_\infty(x) > \phi(x)$, for $x \neq x_0$.
Using the uniform convergence of $u_{p_i}$ to $u_\infty$ we obtain
that $u_{p_i} - \phi$ attains a minimum at $x_i \in \overline{\Omega} $ with $x_i \to x_0$.
If
\[
- |\nabla \phi|^{p_i-2} \Delta \phi(x_i) - (p_i-2)|\nabla \phi|^{p_i-4} \Delta_\infty \phi(x_i)
+ |u_{p_i}|^{p_i-2}u_{p_i}(x_i) \geq  0
\]
holds for infinitely many $x_i$,
we use the argument before to obtain (\ref{super-ineq1}).
Thus we may assume that, for infinitely many $x_i \in {\partial\Omega}$,
\[
|\nabla \phi(x_i)|^{p_i-2} \frac{\partial \phi}{\partial \nu}(x_i)
\geq \Lambda^{p_i}_{p_i} |u_{p_i}|^{p_i -2} u_{p_i}(x_i).
\]
If $|\nabla \phi(x_0)| =0$,
\[
\max \{\phi(x_0) - |\nabla \phi|(x_0), -\Delta_\infty \phi(x_0)\} \geq 0
\]
If $|\nabla \phi(x_0)| \neq 0$ we obtain
\[
\frac{\partial \phi}{\partial \nu}(x_i)
\geq  \Big(\frac{\Lambda_{p_i}|u_{p_i}(x_i)|}{|\nabla \phi(x_i)|}\Big)^{p_i -2}
\Lambda^2_{p_i} u_{p_i}(x_i).
\]
We conclude that
\[
\frac{\phi(x_0)}{|\nabla \phi(x_0)|} \geq 1 \quad\mbox{and}\quad
 \frac{\partial \phi}{\partial \nu}(x_0) \geq 0,
\]
which implies that
$\min \{|\nabla \phi(x_0)| - \phi(x_0), \frac{\partial \phi}{\partial \nu}(x_0) \} \geq 0$.
Therefore, at $x_0$,
\begin{equation} \label{super}
\max\big\{\max \{\phi - |\nabla \phi|, -\Delta_\infty \phi \} ,
 \min \{|\nabla \phi| - \phi, \frac{\partial \phi}{\partial \nu} \} \big\} \geq 0.
\end{equation}
Inequalities (\ref{sub}) and (\ref{super}) prove that $u_\infty$ satisfies
in the viscosity sense the boundary condition of (\ref{iy-eq}).
\end{proof}

Theorem \ref{thm-1} follows from proposition \ref{iy-prop}
and lemma \ref{iy-vis}.

\subsection*{Acknowledgments}
The author is grateful to Professor Juan Manfredi for
many interesting discussions  which leads to the current work.
The author also would like to thank Professor Klaus Schmitt
for his careful reading of the manuscript and his valuable comments.
This work was done while the author was visiting the MSRI in Berkeley,
California, in the fall of 2005 and Utah State University in the
spring of 2006.

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\end{document}