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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 72, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/72\hfil Boundary differentiability]
{Boundary differentiability for inhomogeneous infinity Laplace equations}

\author[G. Hong \hfil EJDE-2014/72\hfilneg]
{Guanghao Hong} 

\address{Guanghao Hong \newline
School of Mathematics and Statistics, Xi'an Jiaotong University,
 Xi'an, 710049, China}
\email{ghhongmath@gmail.com}

\thanks{Submitted December 17, 2013. Published March 16, 2014.}
\subjclass[2000]{35J25, 35J70, 49N60}
\keywords{Boundary regularity; infinity Laplacian; comparison principle;
\hfill\break\indent  monotonicity}

\begin{abstract}
 We study the boundary regularity of the solutions to inhomogeneous
 infinity Laplace equations. We prove that if $u\in C(\bar{\Omega})$
 is a viscosity solution to
 $\Delta_{\infty}u:=\sum_{i,j=1}^n u_{x_i}u_{x_j}u_{x_ix_j}=f$
 with $f\in C(\Omega)\cap L^{\infty}(\Omega)$ and for
 $x_0\in \partial\Omega$  both $\partial\Omega$ and $g:=u|_{\partial\Omega}$
 are differentiable at $x_0$, then $u$ is differentiable at $x_0$.
\end{abstract}

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

\section{Introduction}

Infinity Laplace equation $\Delta_{\infty}u=0$ arose as the Euler
equation of $L^{\infty}$ variational problem of $|\nabla u|$, or equivalently, 
absolutely minimizing Lipschitz extension (AML) problem. 
This problem was initially studied by Aronsson \cite{a1} at the classical 
solutions level from 1960's. In 1993,  Jensen \cite{j1} proved that a 
function $u(x)\in C(\Omega)$ is an AML: 
$$
\text{for any}\ V\subset\subset \Omega,\ Lip(u,V)=Lip(u,\partial V)
$$ 
if and only if $u(x)$ is a viscosity solution to $\Delta_{\infty}u=0$.
Moreover, for any bounded domain $\Omega\subset \mathbb{R}^n$ and 
$g\in C(\partial\Omega)$, the Dirichlet problem:
\begin{equation} \label{e1}
\Delta_{\infty}u=0 \text{ in } \Omega,\quad u=g \text{ on } \partial\Omega
\end{equation}
has an unique viscosity solution. Such an solution is called an 
infinity harmonic function.

In 2001, Crandall, Evans and Gariepy \cite{c2} proved that a function 
$u(x)\in C(\Omega)$ is an infinity harmonic function if and only if 
$u$ satisfies the following \textit{comparison with cone property}:
 for any $V\subset\subset\Omega$ and  and $c(x)=a+b|x-x_0|$,
\begin{gather*}
u(x)\leq c(x)   \text{ on } \partial\{V\backslash \{x_0\}\}
 \Rightarrow u(x) \leq c(x)\ \text{in } V,
\\
u(x)\geq c(x) \text{ on } \partial\{V\backslash \{x_0\}\}
 \Rightarrow u(x)\geq c(x)\ \text{in } V.
\end{gather*}
This comparison property turns out to be a very useful tool in the 
study of many aspects of this equation. Especially, it implies the 
following conclusions as a direct result \cite{c2}.

\begin{lemma} \label{lem1}
Let $u(x)\in C(\Omega)$ satisfy comparison with cone property, 
$x_0\in\Omega$, $0<r<\operatorname{dist}(x_0,\partial\Omega)$. Then
\begin{itemize}
\item[(1)]the slope functions
$$
S^+_r(x_0):=\max_{x\in \partial B(x_0,r)}\frac{u(x)-u(x_0)}{r},\quad
S^-_r(x_0):=\max_{x\in \partial B(x_0,r)}\frac{u(x_0)-u(x)}{r}
$$ 
are non-negative and non-decreasing as a function of $r$ for fixed $x_0$. 
So the limits $S^{\pm}(x_0):=\lim_{r\to 0}S^{\pm}_r(x_0)$ exist.

\item[(2)] $S^+(x_0)=S^-(x_0):=S(x_0)$.

\item[(3)] $S(x)$ is upper-semicontinuous, i.e., 
$\limsup_{y\to x} S(y)\leq S(x)$ for all $x\in\Omega$.
\end{itemize}
\end{lemma}

The lemma implies locally Lipschitz continuity of $u$ immediately. 
Crandall and Evans \cite{c1} applied this lemma to prove that at any interior 
point $x_0$, a blow-up limit 
$$
v(x)=\lim_{r_j\to 0}\frac{u(x_0+r_jx)-u(x_0)}{r_j}
$$ 
of an infinity harmonic function $u$ must be a linear function, i.e., 
$v(x)=a\cdot x$ for some $a\in \mathbb{R}^n$ with $|a|=S(x_0)$. 
The sketch of their proof is the following. 
Firstly, (3) of Lemma \ref{lem1} implies $Lip(v, \mathbb{R}^n)\leq S(x_0)$. 
Secondly, for any $R>0$ fixed, for every $j$ there exists a maximal 
direction $e_j\in \mathbb{R}^n$ with $|e_j|=1$ such that 
$u(x_0+Rr_j e_j)=\max_{x\in \partial B_{Rr_j}(x_0)}u(x)$. 
The sequence $\{e_j\}$ must have an accumulating point say $e^+$, 
then $v(Re^+)=Re^+$. For all $R$, we will have the same $e^+$. 
By considering the minimum directions we will get an $e^-$ and moreover 
$e^-=-e^+$. So $v$ is tight on the line $te^+$, $t\in(-\infty, \infty)$. 
Finally, a Lipschitz function on $\mathbb{R}^n$ that is tight on a line
 must be linear. However, this result does not imply the differentiability 
of $u$ in general since for different sequences $r_j$ one may get different 
linear functions $v$ although they must have same slope $S(x_0)$. 
Ten years later, by using much deeper pde techniques Evans and 
Smart \cite{ev1} proved that the blow-up limits are unique and accomplished the 
proof of interior differentiability. The continuously differentiability 
is still left open as the most prominent problem in this field although 
in 2 dimension $C^1$ and $C^{1,\alpha}$ regularity was achieved by
 Savin \cite{s1} and Evans-Savin \cite{ev1} respectively.

Boundary regularity for infinity harmonic function was initially studied
 by Wang and Yu \cite{w1}. They proved the following result.

\begin{theorem} \label{thm1}
For $n\geq 2$, let $\Omega\subset R^n$ be a bounded domain with 
$\partial\Omega\in C^1$ and $g\in C^1(R^n)$. Assume that $u\in C(\bar{\Omega})$ 
is the viscosity solution of the infinity Laplace equation \eqref{e1}. 
Then $u$ is differentiable on the boundary, i.e., for any 
$x_0\in \partial\Omega$, there exists $Du(x_0)\in R^n$ such that 
$$
u(x)=u(x_0)+Du(x_0)\cdot (x-x_0)+o(|x-x_0|),\quad  \forall x\in \bar{\Omega}.
$$
\end{theorem}

The boundary differentiability is much easier than interior differentiability. 
They defined the slope functions near and on the boundary by 
$$
S^+_r(x)=\sup_{y\in \partial(B(x,r)\cap\Omega)\backslash\{x\}}
\frac{u(y)-u(x)}{|y-x|}\ \text{and}\ S^-_r(x)
=\sup_{y\in \partial(B(x,r)\cap\Omega)\backslash\{x\}}\frac{u(x)-u(y)}{|y-x|}
$$ 
for $x\in \bar{\Omega}$ and $r>0$ small. $S^{\pm}_r(x)$ are still monotone 
and have limits $S^{\pm}(x)$. But $S^+(x)\neq S^-(x)$ in general if 
$x\in \partial \Omega$. Denote $S(x):=\max\{S^+(x), S^-(x)\}$. 
$S(x)$ is upper-semicontinuous $\forall x\in\bar{\Omega}$ with the 
assumption that both $\partial\Omega$ and $g$ are $C^1$. 
They applied a similar argument as in \cite{c1} and proved that any blow-up 
limit of $u$ at a boundary point $x_0$ is a linear function 
$v(x)=e\cdot x$ with $|e|=S(x_0)$ on the half space $\mathbb{R}^n_+=\{x_n>0\}$. 
But this time it is very easy to prove the uniqueness of blow-up limits 
since the tangential part of $e$ is already given by the boundary data. 
So $e=(\sqrt{S(x_0)^2-|D_Tg(x_0)|^2}, D_Tg(x_0))$ or 
$e=(-\sqrt{S(x_0)^2-|D_Tg(x_0)|^2}, D_Tg(x_0))$. The former happens when 
$S(x_0)=S^+(x_0)$ and the latter happens when $S(x_0)=S^-(x_0)$.

It is not natural to put $C^1$ assumption on the boundary conditions in 
order to prove merely differentiability of a solution. In a recent work 
\cite{h1} we improved Wang-Yu's Theorem to the following sharp version.

 \begin{theorem} \label{thm2}
 Let $\Omega\subset R^n$ be a domain and $u\in C(\bar{\Omega})$ 
is an infinity harmonic function in $\Omega$. Assume that for 
$x_0\in \partial\Omega$, $\partial \Omega$ and $g:=u|_{\partial \Omega}$ 
are differentiable at $x_0$. Then $u$ is differentiable at $x_0$.
 \end{theorem}

Under this weaker assumption, it is not true that $S(x)$ is
 upper-semicontinuous at $x_0$. However we managed to show that
 $\limsup_{x\to x_0} S(x)\leq S(x_0)$ if $x\to x_0$ in a non-tangentially way. 
This is enough to imply $Lip(v, \mathbb{R}^n_+)\leq S(x_0)$.

The inhomogeneous infinity Laplace Equation $\Delta_{\infty}u=f$
was studied by Lu and Wang \cite{l2}. They proved existence and uniqueness 
of a viscosity solution of the Dirichlet problem
\begin{equation} \label{e2}
\Delta_{\infty}u=f \text{ in } \Omega,\quad u=g \text{ on } \partial\Omega
\end{equation}
under the conditions that $\Omega\subset \mathbb{R}^n$ is bounded, 
$f\in C(\Omega)$ with $\inf_{\Omega}f>0$ or $\sup_{\Omega}f<0$ and 
$g\in C(\partial\Omega)$. They also proved some comparison principles and 
stability results.  Lindgren \cite{l1} investigated the interior regularity of 
viscosity solutions of \eqref{e2}. He proved that the blow-ups are linear 
if $f\in C(\Omega)\cap L^{\infty}(\Omega)$ and $u$ is differentiable if
 $f\in C^1(\Omega)\cap L^{\infty}(\Omega)$. For inhomogeneous equation \eqref{e2}, 
the slope functions $S^{\pm}_r(x)$ is not monotone anymore, but so is 
$S^{\pm}_r(x)+r$ \cite[Corollary 1]{l1}. Hence the limits 
$S^{\pm}(x):=\lim_{r\to 0}S^{\pm}_r(x)$ still exist and the arguments 
in \cite{c1} and \cite{ev1} work.

In this paper, we combine the techniques used in \cite{h1,l1,w1} 
to prove boundary differentiability for inhomogeneous infinity Laplace equation.

 \begin{theorem} \label{thm3}
 Let $\Omega\subset R^n$ and $u\in C(\bar{\Omega})$ is a viscosity solution 
of the inhomogeneous infinity Laplace equation \eqref{e2}. 
Assume that $f\in C(\Omega)\cap L^{\infty}(\Omega)$ and for 
$x_0\in \partial\Omega$, both $\partial \Omega$ and $g$ are differentiable 
at $x_0$. Then $u$ is differentiable at $x_0$.
 \end{theorem}

\section{Proof of Theorem \ref{thm3}}

Without lost of generality, we may assume that $x_0=0$ and the tangential
 plane of $\partial\Omega$ at $0$ is $\{x=(x',x_n)\in R^n:x_n=0\}$. 
Denote $B(x,r):=\{y\in R^n: |y-x|<r\}$ for $x\in R^n$, $B(r):=B(0,r)$, 
$\hat{B}(x',r):=\{y'\in R^{n-1}: |y'-x'|<r\}$ for $x'\in R^{n-1}$ and 
$\hat{B}(r):=\hat{B}(0,r)$. We assume for some $0<r_0<1$, 
$$
\Omega \cap B(r_0)= \{x\in B(r_0):x_n>f(x')\},
$$ 
where $f\in C(\hat{B}(r_0))$ is differentiable at $0$ with $f(0)=Df(0)=0$. 
Denote $\hat{g}(x')=g(x',f(x'))$ for $x'\in\hat{B}(r_0)$, then 
$\hat{g}(x')\in C(\hat{B}(r_0))$ is differentiable at $0$.

We will prove the following easier conclusion first and then apply it to prove 
Theorem \ref{thm3}.

 \begin{proposition} \label{prop1}
Assume that $u$, $f$, $\Omega$ and $g$ satisfy the conditions in 
Theorem \ref{thm3}. 
We assume additionally $\hat{g}(x')\in C^1(\hat{B}(r_0))$. Then $u$ is 
differentiable at $0$.
 \end{proposition}

For $x\in \bar{\Omega}\cap B_{r_0/2}$ and $0<r< r_0/2$, we define 
$$
S^+_r(x)=\sup_{y\in \partial(B(x,r)\cap\Omega)\backslash\{x\}}\frac{u(y)-u(x)}{|y-x|}
$$ 
and 
$$
S^-_r(x)=\sup_{y\in \partial(B(x,r)\cap\Omega)\backslash\{x\}}
\frac{u(x)-u(y)}{|y-x|}.
$$
We make the following two assumptions on the solution $u$ of \eqref{e2} as 
in \cite{l1}:
\begin{itemize}
\item[(A1)] $S^{\pm}_r(x)\geq 1$ for all $x$ and $r$;

\item[(A2)] $-\frac{3}{4}\leq f \leq -\frac{1}{4}$.

\end{itemize}
This is not a restriction since we can define
$$
\tilde{u}(x_1,\dots,x_{n+2})=\frac{u(x_1,\dots,x_n)}{4^{\frac{1}{3}}
\|f\|^{\frac{1}{3}}_{L^{\infty}(\Omega)}}+x_{n+1}
-\frac{3^{\frac{4}{3}}}{2^{\frac{1}{3}}\cdot 4}|x_{n+2}|^{\frac{3}{4}}
$$
to make $\tilde{u}$ satisfy the assumptions. Any regularity result 
(up to $C^{1,\frac{1}{3}}$) on $\tilde{u}$ in general dimension also holds for $u$.

\begin{lemma} \label{lem2}
Under the assumptions and notation above, $S^{\pm}_r(x)+r$ is
 non-decreasing for all $x\in \bar{\Omega}\cap B_{r_0/2}$ and 
$0<r<\frac{r_0}{2}$. So the limit $S^{\pm}(x):=\lim_{r\to 0}S^{\pm}_r(x)+r$ exist.
\end{lemma}

\begin{proof}
Fix a point $x\in \bar{\Omega}\cap B_{r_0/2}$ and $0<r<\frac{r_0}{2}$. 
Define 
$$
\phi(y):=u(x)+S^+_r(x)r^{\frac{r}{S^+_r(x)}}\cdot |y-x|^{1-\frac{r}{S^+_r(x)}}.
$$ 
Direct computation shows that 
$$
\Delta_{\infty}\phi(y)=S^+_r(x)^3r^{\frac{3r}{S^+_r(x)}}
(-\frac{r}{S^+_r(x)})|y-x|^{-\frac{3r}{S^+_r(x)}-1}\leq -S^+_r(x)^2\leq -1<f
$$ 
when $y\in B_r(x)\cap\Omega\backslash \{x\}$. And $\phi(y)\geq u(y)$ 
on $\partial(B(x,r)\cap\Omega)\cup \{x\}$. 
So $\phi(y)\geq u(y)$ in $B(x,r)\cap\Omega$ from the comparison principle 
\cite[Theorem 3]{l2}.

For $0<\rho<r$, let $y\in \partial(B(x,\rho)\cap\Omega)\backslash\{x\}$. 
If $y\in \partial \Omega\cap B(x,\rho)\backslash\{x\}$ then 
$y\in \partial \Omega\cap B(x,r)\backslash\{x\}$, so 
$\frac{u(y)-u(x)}{|y-x|}\leq S^+_r(x)$. If $y\in \partial B(x,\rho) \cap \Omega$, 
then 
$$
\frac{u(y)-u(x)}{|y-x|}\leq \frac{\phi(y)-u(x)}{\rho}
= S^+_r(x)(\frac{r}{\rho})^{\frac{r}{S^+_r(x)}}.
$$
Hence, $S^+_{\rho}(x)\leq S^+_r(x)(\frac{r}{\rho})^{\frac{r}{S^+_r(x)}}$. 
Therefore, 
$$
\liminf_{\rho\to r}\frac{S^+_r(x)-S^+_{\rho}(x)}{r-\rho}
\geq \liminf_{\rho\to r}\frac{S^+_r(x)
(1-(\frac{r}{\rho})^{\frac{r}{S^+_r(x)}})}{r-\rho}=-1.
$$
The same argument applies to $S^-_r(x)$.
\end{proof}

Define $S(x):=max\{S^+(x), S^-(x)\}$. we prove that $S(x)$ is upper-semicontinuous 
at 0 under the conditions of Proposition 1.

\begin{lemma} \label{lem3}
For any $\epsilon>0$, there exists $r(\epsilon, u)>0$, such that 
$$
\sup_{x\in \bar{\Omega}\cap B(r)}S(x)\leq S(0)+\epsilon.
$$
\end{lemma}

\begin{proof}
For $\epsilon>0$, since $\hat{g}(x')\in C^1(\hat{B}(r_0))$ and 
$|D\hat{g}(0)|\leq S(0)$, there exists $r_1>0$ such that
\begin{equation} \label{e3}
\sup_{x\neq y\in \partial\Omega \cap B(r_1)}
\frac{|u(x)-u(y)|}{|x-y|}\leq \sup_{x\neq y\in \partial\Omega \cap B(r_1)}
\frac{|\hat{g}(x')-\hat{g}(y')|}{|x'-y'|} \leq S(0)+\frac{\epsilon}{3}.
\end{equation}
Since $\lim_{r\to 0}S_r(0)=S(0)$, there exists 
$0<r_2\leq \min (r_1/2,\frac{\epsilon}{3})$, such that 
$$
S_{r_2}(0)\leq S(0)+\frac{\epsilon}{4}.
$$ 
From the continuity of $u$, there exists $0<r_3\ll r_2$, such that
\begin{equation} \label{e4}
\sup_{y\in \partial B(x, r_2)\cap\Omega}\frac{|u(y)-u(x)|}{r_2}
\leq S(0)+\frac{\epsilon}{3}\quad\text{for}\quad x\in\bar{\Omega}\cap B(r_3).
\end{equation}
From \eqref{e3} and \eqref{e4}, we have
\[
S_{r_2}(x)=\sup_{y\in \partial(B(x, r_2)\cap\Omega)
\backslash\{x\}}\frac{|u(y)-u(x)|}{|y-x|}\leq S(0)
+\frac{\epsilon}{3}\quad\text{for}\quad 
x\in\partial\Omega\cap B(r_3).
\]
From Lemma \ref{lem2}, we have
\begin{equation} \label{e5}
\frac{|u(y)-u(x)|}{|y-x|}\leq S_{r_2}(x)+r_2\leq S(0)+\frac{2\epsilon}{3} 
\end{equation}
for $x\in\partial\Omega\cap B(r_3)$ and $y\in\Omega\cap B(r_3)$.
From the continuity of $u$ again, there exists $0<r_4\leq r_3/2$, such that
\begin{equation} \label{e6}
\sup_{y\in\partial B(x,r_3/2)\cap \Omega}\frac{|u(y)-u(x)|}{r_3/2}
\leq S(0)+\frac{2\epsilon}{3} \quad\text{for } x\in\bar{\Omega}\cap B(r_4).
\end{equation}
From \eqref{e5} and \eqref{e6} and Lemma \ref{lem2}, we have 
$$
S(x)\leq S_{r_3/2}(x)+\frac{r_3}{2}=\sup_{y\in\partial(B(x,r_3/2)
\cap\Omega)\backslash\{x\}}\frac{|u(y)-u(x)|}{|y-x|}
+\frac{r_3}{2}\leq S(0)+\epsilon
$$ 
for $x\in\bar{\Omega}\cap B(r_4)$. Finally we choose $r(\epsilon, u)=r_4$.
\end{proof}

The rest of the proof of Proposition 1 is the same as that in \cite{w1}.
We have described the idea in the introduction and refer the readers 
to \cite{w1} for the details.

Now we prove the non-tangentially upper-semicontinuity of $S(x)$ at 0 
under the conditions of Theorem \ref{thm3} and assumptions (A1) and (A2).

\begin{lemma} \label{lem4}
Given any $0<\theta\ll 1$, we have that for all $0<\epsilon<\frac{1}{8}$, 
there exists $r(\epsilon,\theta,u)>0$, such that 
$$
\sup_{x\in\bar{\Omega}\cap B(r)\cap \{x_n\geq \theta|x'|\}}S(x)\leq S(0)+\epsilon.
$$
\end{lemma}

The proof of Lemma \ref{lem4} is essentially same as the proof of 
\cite[Lemma 2]{h1} for the homogeneous equation. Several places 
need minor modification, but this can be easily justified.
 So we omit the proof and refer the readers to \cite{h1}.

With the result in Lemma \ref{lem4} the rest of the proof of Theorem \ref{thm3} 
follows the same way as in the homogeneous equation case.

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\end{document}