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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 149, 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 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/149\hfil  Infinity harmonic functions]
{Extending infinity harmonic functions \\ by rotation}

\author[G. Gripenberg \hfil EJDE-2015/149\hfilneg]
{Gustaf Gripenberg}

\address{Gustaf Gripenberg \newline
Department of Mathematics and Systems Analysis\\
Aalto University,
P.O. Box 11100, FI-00076 Aalto, Finland}
\email{gustaf.gripenberg@aalto.fi}
\urladdr{math.aalto.fi/$\sim$ggripenb}

\thanks{Submitted May 21, 2015. Published June 10, 2015.}
\subjclass[2010]{35J60, 35J70}
\keywords{Infinity harmonic; extension; viscosity solution}

\begin{abstract}
 If $u(\mathbf{x}, y)$ is an infinity harmonic
 function, i.e., a viscosity solution to the equation
 $-\Delta_\infty u=0$ in  $\Omega \subset \mathbb{R}^{m+1}$ then the function
 $v(\mathbf{x}, \mathbf{z})= u(\mathbf{x}, \|\mathbf{z}\|)$ is infinity harmonic 
 in the set $\{(\mathbf{x}, \mathbf{z}): 
 (\mathbf{x}, \|\mathbf{z}\|)\in \Omega\}$ 
 (provided $u(\mathbf{x},-y)=u(\mathbf{x},y)$).
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks


\newcommand{\abs}[1]{|#1|}
\newcommand{\norm}[1]{\|#1\|}
\newcommand{\bigpar}[1]{\bigl( #1 \bigr)}
\newcommand{\set}[1]{\{#1\}}
\newcommand{\inprod}[2]{\langle #1, #2 \rangle}

\section{Introduction and statement of results}

A function $u:\Omega \to \mathbb{R}$, where $\Omega$ is an open subset of
$\mathbb{R}^d$,  is said to be infinity harmonic if $u$ is a viscosity solution
to the equations
 \begin{equation*}
  -\Delta_\infty u = - \sum_{i,j=1}^d u_{x_i}u_{x_j}u_{x_ix_j}=0,
  \end{equation*}
 in $\Omega$. In order for $u$ to be a viscosity solution it has to be
both a subsolution and a supersolution (that is, infinity subharmonic
and infinity superharmonic, respectively) and the requirement for $u$
to be a subsolution to $-\Delta_\infty u=0$ is that $u$ is upper
semicontinuous and if $\varphi$ is twice continuously differentiable
in a neighbourhood of a point $\mathbf{x}_1\in \Omega$,
$u(\mathbf{x}_1)= \varphi(\mathbf{x}_1)$ and
$u(\mathbf{x})\leq \varphi(\mathbf{x})$
when $\abs{\mathbf{x}-\mathbf{x}_1}<\delta$ for some $\delta >0$, then
$- \sum_{i,j=1}^d \varphi_{x_i}\varphi_{x_j}\varphi_{x_ix_j}\leq 0$ at
the point $\mathbf{x}_1$. In the requirements for $u$ to be a
supersolution the inequalities $\leq$ are reversed and $u$ is required
to be lower semicontinuous, so that $u$ is a supersolution if and only
if $-u$ is a subsolution (because $\Delta_\infty(-\varphi)= -
\Delta_\infty \varphi$ if $\varphi$ is twice continuously
differentiable).

This equation arises when one wants to find a Lipschitz continuous
function $u$ in $\Omega$ satisfying given boundary values on $\partial
\Omega$ and one requires that this function in addition is an
absolutely minimizing extension in the sense that if $\Omega_0$ is an
open bounded subset of $\Omega$ and $u=v$ on $\partial \Omega_0$,
where $v$ is a continuous function in the closure of $\Omega_0$, then
the Lipschitz constant of $u$ in $\Omega_0$ is not larger than the one
of $v$, see e.g. \cite{Aronsson1967} and \cite{Jensen1993}. In
addition infinity harmonic functions and their generalizations appear
in several other contexts, see e.g.\ \cite{Barron2008}, in particular
the value of a ``Tug-of-war'' game is an infinity harmonic function,
see \cite{Peres09}.

The purpose of this note is to extend the observation that both the
function $u(x,y)= \abs{x}^{4/3}-\abs{y}^{4/3}$ and its
extension $v(\mathbf{x},\mathbf y) = \norm{x}^{4/3}-\norm{y}^{4/3}$
 are infinity harmonic in $\mathbb{R}^2$ and
$\mathbb{R}^{m+n}$, respectively, where $\norm\cdot$ is the Euclidean
norm. More precisely, we show that if $u(\mathbf{x},y)$ is infinity
harmonic, then so is the function 
$v(\mathbf{x}, \mathbf{z})=u(\mathbf{x}, \norm{\mathbf{z}})$;
 that is, we can extend an infinity harmonic
function to a higher dimensional space by rotation. Here we formulate
this result using coordinates, writing vectors in $\mathbb{R}^{m+1}$ and
$\mathbb{R}^{m+n}$ in the form $(\mathbf{x}, y)$ and $(\mathbf{x},\mathbf{z})$,
respectively, where $\mathbf{x}\in \mathbb{R}^m$, $y\in \mathbb{R}$, and $\mathbf{z}\in
\mathbb{R}^n$ but note that the property of being infinity harmonic does not
depend on the coordinate system, that is, if $u$ is infinity harmonic
in $\Omega$ then $u\circ T$ is infinity harmonic in $T^{-1}\Omega$
when $T$ is an isometry.


Observe that the extension property studied here does not hold for
standard harmonic functions, as for example $\log(\abs{x}^2+
\abs{y}^2)$ is harmonic in $\mathbb{R}^2\setminus \{\mathbf 0\}$ but the
function $\log(\abs{x}^2+ \norm{\mathbf{z}}^2)$ is not harmonic in
$\mathbb{R}\times \mathbb{R}^2 \setminus \{\mathbf 0\}$.



\begin{theorem}\label{T:mainThm} Assume that $m, n \geq 1$ and that
\begin{itemize}

   \item[(i)]   $\Omega_{m,1}\subseteq \mathbb{R}^{m+1}$ is open
    and if $(\mathbf{x},y)\in \Omega_{m,1}$ with $\mathbf{x}\in \mathbb{R}^m$
      and $y< 0$, then
    $(\mathbf{x}, -y)\in \Omega_{m,1}$;

  \item[(ii)] $u:\Omega_{m,1}\to \mathbb{R}$ is infinity harmonic
    (subharmonic, superharmonic) in $\Omega_{m,1}$ and   if
    $(\mathbf{x},y)\in \Omega_{m,1}$ with $\mathbf{x}\in \mathbb{R}^m$
      and $y< 0$, then $u(\mathbf{x},-y)= u(\mathbf{x}, y)$.
  \end{itemize}
   Then $v$ is infinity harmonic  (subharmonic, superharmonic) 
in $\Omega_{m,n}$ where
\begin{itemize}

 \item[(a)]  $\Omega_{m,n}=\set{(\mathbf{x}, \mathbf{z})
   : \mathbf{x}\in \mathbb{R}^m,\; \mathbf{z}\in \mathbb{R}^n,\; (\mathbf{x},
   \norm{\mathbf{z}})\in \Omega_{m,1} }$;

 \item[(b)]  $v(\mathbf{x}, \mathbf{z})= u(\mathbf{x},
\norm{\mathbf{z}})$, $(\mathbf{x}, \mathbf{z})\in
     \Omega_{m,n}$.
  \end{itemize}
\end{theorem}

The main property of infinity harmonic functions that we need in order
to prove this extension property is ``comparison with cones'' which is
formulated in the form we need here in the following theorem. This
well known result and its corollary are stated for subsolutions, but
the corresponding results with inequlities reversed and $\sup$
replaced by $\inf$ hold for supersolutions as well.


\begin{theorem}\label{T:InfLapCompConeA}
  Assume that $d\geq 1$,  $\Omega\subset \mathbb{R}^d$ is open,  and $w:\Omega
  \to \mathbb{R}$. Then
  $w$ is a subsolution to the equation $-\Delta_\infty w =0$ in
  $\Omega$ if and only if $w$ is locally bounded and
  \begin{equation}\label{Eq:compWithConesA}  
w(\mathbf{x}) \leq
    w(\mathbf{x}_0) + \sup_{\norm{\boldsymbol \xi-\mathbf{x}_0}=r} 
\frac{w(\boldsymbol \xi)-w(\mathbf{x}_0)}r \norm{\mathbf{x}-\mathbf{x}_0}, \quad
  \norm{\mathbf{x}-\mathbf{x}_0} \leq r,
  \end{equation}
 when $\set{\boldsymbol \xi: \norm{\boldsymbol\xi-\mathbf{x}_0}\leq r} \subset \Omega$.
\end{theorem}

This theorem  is proved in \cite{CrandallEvansGariepy01} but for
completeness and since it is
there not formulated in the form above we give a self-contained proof
below. Observe that the ``only if'' part is a consequence of the
comparison principle which holds for infinity harmonic functions, see
e.g.\  \cite{ArmstrongSmart2010}, and of the fact that a cone function
$\mathbf{x} \mapsto a + b \norm{\mathbf{x} -\mathbf{x}_0}$ is infinity
harmonic in $\mathbb{R}^d\setminus \{\mathbf{x}_0\}$.

 A consequence of Theorem \ref{T:InfLapCompConeA} is the well known strong
 maximum principle, which we will need as well.

\begin{corollary}\label{T:strongMaxPrinciple}  
Assume that $d\geq 1$,  $\Omega\subset \mathbb{R}^d$ is open and connected,
and $w:\Omega   \to \mathbb{R}$ is a subsolution to $-\Delta_\infty w =0$ in
  $\Omega$. Then either $w(\mathbf{x}) < \sup_{\boldsymbol \xi\in
    \Omega} w(\boldsymbol \xi)$ for all $\mathbf{x}\in \Omega$ or $w$
  is a constant in $\Omega$.
 \end{corollary}

\section{Proofs}

\begin{proof}[Proof of Theorem \ref{T:mainThm}]
Suppose first that   $u$
  is infinity subharmonic in $\Omega_{m,1}$. Then $u$ and hence $v$ is
  upper semicontinuous so that $v$ is locally bounded. 
Let  $(\mathbf{x}_0, \mathbf{z}_0)\in
  \Omega_{m,n}$ and $r>0$ be  such that $\set{(\boldsymbol \xi,
    \boldsymbol \zeta)\in \mathbb{R}^{m+n} : \norm{(\boldsymbol \xi,
    \boldsymbol \zeta)- (\mathbf{x}_0,
      \mathbf{z}_0)} \leq r} \subset \Omega_{m,n}$. If we can show that
  \begin{equation}\label{Eq:compWithConesB}
  v(\mathbf{x},\mathbf{z}) \leq v(\mathbf{x}_0,\mathbf{z}_0)+
  \max_{\norm{(\boldsymbol \xi,\boldsymbol \zeta)- (\mathbf{x}_0,\mathbf{z}_0)}=r} 
\frac {v(\boldsymbol \xi,\boldsymbol \zeta)- v(\mathbf{x}_0,\mathbf{z}_0)}r 
\norm{(\mathbf{x},\mathbf{z}) -(\mathbf{x}_0,\mathbf{z}_0)},
    \end{equation}
 when  $\norm{(\mathbf{x},\mathbf{z}) -(\mathbf{x}_0,\mathbf{z}_0)}\leq r$
  then we have shown that $v$ satisfies comparison with cones from
  above and it follows from Theorem \ref{T:InfLapCompConeA} that $v$
  is a subsolution to
  $-\Delta_\infty v=0$ in $\Omega_{m,n}$.  If $u$ is infinity superharmonic the
  same argument can be applied to $-u$ and $-v$ and  we can conclude
  that $v$ is infinity superharmonic. Thus the claims of
  the theorem follow provided we can show that inequality
  \eqref{Eq:compWithConesB} holds under the assumption that $u$ is
  infinity subharmonic.

 Note that we may, without loss of generality, assume that 
$(\mathbf{x},y)\in \Omega_{m,1}$ if and only if $(\mathbf{x},-y)\in \Omega_{m,1}$
 and $u(\mathbf{x},y)=u(\mathbf{x},-y)$, because we can, if needed,
 extend $u$ to $\set{(\mathbf{x},y)\in \mathbb{R}^{m+1}: (\mathbf{x},-y)\in
   \Omega_{m,1}}$ by $u(\mathbf{x}, y)= u(\mathbf{x},-y)$ as the the
 property of being a subsolution is a local one and the function
 $(\mathbf{x},y)\mapsto (\mathbf{x},-y)$ is an isometry.  Suppose
 that  $ \norm{(\boldsymbol \xi,\mu)- (\mathbf{x}_0,\norm{\mathbf{z}_0})}\leq r$. 
We want  to show that  $(\boldsymbol \xi,\mu)\in  \Omega_{m,1}$ and
 because we may assume that $\mu\geq 0$ we can  take $\boldsymbol
 \zeta= \frac {\mu}{\norm{\mathbf{z}_0}}\mathbf{z}_0$ if
 $\mathbf{z}_0\neq \mathbf 0$ and otherwise take $\boldsymbol \zeta$ to be an
 arbitrary vector in $\mathbb{R}^n$ so that $\norm
 {\boldsymbol\zeta}=\mu$. Then
 $\abs{\mu-\norm{\mathbf{z}_0}}=\norm{\boldsymbol \zeta-\mathbf{z}_0}$ so that
 $\norm{(\boldsymbol \xi,\boldsymbol \zeta)-(\mathbf{x}_0,\mathbf{z}_0}\leq r$ 
which by the defintion of $\Omega_{m,n}$, our choice of
 $r$ and by the fact that $\norm{\boldsymbol \zeta}=\mu$ implies that
 $(\boldsymbol \xi,\mu)\in \Omega_{m,1}$.

Suppose that $\boldsymbol \xi_0$ and $\mu_0$ are such that $\norm
{(\boldsymbol \xi_0,\mu_0)-(\mathbf{x}_0,\norm{\mathbf{z}_0})}=r$ and
  \begin{equation}\label{Eq:maxAtPoint}
  \max_{\norm{(\boldsymbol \xi,\mu)-(\mathbf{x}_0,\norm{\mathbf{z}_0})}=r}\bigpar{u(\boldsymbol \xi,\mu)- u(\mathbf{x}_0,\norm{\mathbf{z}_0})} = u(\boldsymbol \xi_0,\mu_0)-u(\mathbf{x}_0,\norm{\mathbf{z}_0}).
  \end{equation}
 Since $u$ is upper semicontinuous, such a point $(\boldsymbol
 \xi_0,\mu_0)$ exists.

  If $\mu_0<0$ and $\norm{(\boldsymbol \xi_0,-\mu_0)-(\mathbf{x}_0,
\norm{\mathbf{z}_0})}<r$ then, since $u(\boldsymbol
  \xi_0,-\mu_0)=u(\boldsymbol \xi_0,\mu_0)$, the maximum value is
  obtained in an interior point in the ball with center $(\mathbf{x}_0,
  \norm{\mathbf{z}_0}) $ and radius $r$ which by the strong maximum
  principle implies that $u$ is a constant in this ball. In this case
  we may take $\mu_0$ so that $\mu_0\geq 0$. If on the other hand
  $\mu_0<0$ and $\norm{(\boldsymbol \xi_0,-\mu_0)-(\mathbf{x}_0,
\norm{\mathbf{z}_0})}=r$, then it follows again from the
  assumption that $u(\boldsymbol \xi_0,-\mu_0)=u(\boldsymbol
  \xi_0,\mu_0)$ that $\mu_0$ can be replaced by $-\mu_0$ so that we
  may without loss of generality assume that $\mu_0\geq 0$.

Since $u$ is infinity subharmonic we can apply Theorem
 \ref{T:InfLapCompConeA} and  using \eqref{Eq:maxAtPoint} we obtain
 \begin{equation}\label{Eq:mainInEqA}
\begin{aligned}
u(\mathbf{x}, \norm{\mathbf{z}})
&\leq   u(\mathbf{x}_0,\norm{\mathbf{z}_0}) 
+ \max_{\norm{(\boldsymbol \xi,\mu)-(\mathbf{x}_0,\norm{\mathbf{z}_0})}=r}
\frac {u(\boldsymbol \xi,\mu)- u(\mathbf{x}_0,\norm{\mathbf{z}_0})}r \\
&\quad\times \norm{(\mathbf{x}, \norm{\mathbf{z}}) 
- (\mathbf{x}_0,\norm{\mathbf{z}_0})}  \\ 
&= u(\mathbf{x}_0,\norm{\mathbf{z}_0})+ \frac { u(\boldsymbol
    \xi_0,\mu_0)-u(\mathbf{x}_0,\norm{\mathbf{z}_0})} r
  \norm{(\mathbf{x}, \norm{\mathbf{z}}) - (\mathbf{x}_0,\norm{\mathbf{z}_0})}.
  \end{aligned}
\end{equation}
By  the definition of $v$, the triangle
inequality, and by the fact that $u(\boldsymbol
    \xi_0,\mu_0)-u(\mathbf{x}_0,\norm{\mathbf{z}_0})\geq 0$ (use
    Corollary  \ref{T:strongMaxPrinciple})  we conclude from
    inequality \eqref{Eq:mainInEqA} that
  \begin{equation}\label{Eq:mainInEqB}
     v(\mathbf{x},\mathbf{z})\leq v(\mathbf{x}_0,\mathbf{z}_0) +  \frac {
       u(\boldsymbol
    \xi_0,\mu_0)-u(\mathbf{x}_0,\norm{\mathbf{z}_0})} r
\norm{(\mathbf{x}, \mathbf{z}) - (\mathbf{x}_0,\mathbf{z}_0)}.
  \end{equation}


   As above  we choose  $\boldsymbol \zeta_0= \frac
{\mu_0}{\norm{\mathbf{z}_0}}\mathbf{z}_0$ if $\mathbf{z}_0\neq \mathbf 0$
and otherwise we take $\boldsymbol \zeta_0$ to be
an arbitrary vector in $\mathbb{R}^n$ such that $\norm{\boldsymbol
  \zeta_0}=\mu_0$ so that  $\norm{\boldsymbol
  \zeta_0}= \mu_0$ as we have $\mu_0\geq 0$. Hence it  follows from the
definition of $v$ that
 \begin{equation}\label{Eq:equalDiff}
   u(\boldsymbol \xi_0,\mu_0)-u(\mathbf{x}_0,\norm{\mathbf{z}_0}) =
  v(\boldsymbol \xi_0,\boldsymbol \zeta_0)- v(\mathbf{x}_0,\mathbf{z}_0).
  \end{equation}
  Our choice of  $\boldsymbol \zeta_0$ implies in addition that
  $\norm{\boldsymbol \zeta_0-\mathbf{z}_0}=\abs{\mu_0-\norm{\mathbf{z}_0}}$ 
from which it follows that
  \begin{equation*}
  \norm{(\boldsymbol \xi_0,\boldsymbol \zeta_0)- (\mathbf{x}_0,\mathbf{z}_0)} = \norm
{(\boldsymbol \xi_0,\mu_0)-(\mathbf{x}_0,\norm{\mathbf{z}_0})}=r.
  \end{equation*}
 This result combined with \eqref{Eq:equalDiff} shows that
  \begin{equation*}
  \frac {
       u(\boldsymbol
    \xi_0,\mu_0)-u(\mathbf{x}_0,\norm{\mathbf{z}_0})} r \leq
\max_{\norm{(\boldsymbol \xi,\boldsymbol \zeta)-(\mathbf{x}_0,\mathbf{z}_0)}=r} \frac{v(\boldsymbol \xi,\boldsymbol \zeta)- v(\mathbf{x}_0,\mathbf{z}_0)}r.
  \end{equation*}
 This inequality combined with \eqref{Eq:mainInEqB} shows that inequality
   \eqref{Eq:compWithConesB} holds and the proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{T:InfLapCompConeA}]
 Assume first that $w$ is a subsolution to
  $-\Delta_\infty w =0$ in $\Omega$. By definition $w$ is upper
  semicontinuous and hence locally bounded. Suppose  $\set{\boldsymbol
    \xi : \norm{\boldsymbol \xi-\mathbf{x}_0}\leq r} \subset \Omega$
  for some $\mathbf{x}_0\in \mathbb{R}^d$
  and  $r> 0$. Let $\alpha\in (0,1) $,
  choose $\beta$  to be
  \begin{equation*}
  \beta = \frac 1{r^\alpha}\max_{\norm{\boldsymbol \xi-\mathbf{x}_0}=r}(w(\boldsymbol \xi)-w(\mathbf{x}_0)),
  \end{equation*}
 and define
  \begin{equation*}
  \psi(\mathbf{x}) = w(\mathbf{x}_0) + \beta \norm{\mathbf{x}-\mathbf{x}_0}^\alpha, 
\quad   \mathbf{x}\in \mathbb{R}^d.
  \end{equation*}
 By definition, $w(\mathbf{x})\leq \psi(\mathbf{x})$ when $\mathbf{x}=
 \mathbf{x}_0$ and when $\norm{\mathbf{x}-\mathbf{x}_0}=r$ so that if we
 have $\max_{\norm{x-x_0}\leq r}\bigpar{w(\mathbf{x})-\psi(\mathbf{x})}>0$ 
and define $\varphi(\mathbf{x})= \psi(\mathbf{x})+\max_{\norm{x-x_0}\leq r}
\bigpar{w(\mathbf{x})-\psi(\mathbf{x})}$,
 then there is a point $\mathbf{x}_1$ such that 
$ 0<\norm{\mathbf{x}_1-\mathbf{x}_0} < r$, $\varphi(\mathbf{x}_1)= w(\mathbf{x}_1)$ 
and  $w(\mathbf{x}) \leq \varphi(\mathbf{x})$ when 
$\norm{\mathbf{x}-\mathbf{x}_1} < \min\{\norm{\mathbf{x}_1-\mathbf{x}_0},
 r-\norm{\mathbf{x}_1-\mathbf{x}_0}\}$. But since $w$ is a subsolution it 
follows from  the definition of a subsolution that $-\Delta_{\infty}
 \varphi(\mathbf{x}_1)\leq 0$ which is a contradiction since a
 calculation shows that $-\Delta_\infty \varphi(\mathbf{x}) =
 -\beta^3\alpha^3(\alpha-1)\norm{\mathbf{x}-\mathbf{x}_0}^{3\alpha-4}>0$
 when $\mathbf{x}\neq \mathbf{x}_0$. Thus we know that $w(x)\leq
 \psi(x)$ when $\norm{\mathbf{x}-\mathbf{x}_0}\leq r$ and taking the
 limit $\alpha \uparrow 1$ we get inequality
 \eqref{Eq:compWithConesA}.

Assume next that $w$ is locally bounded and \eqref{Eq:compWithConesA}
holds when $\set{\boldsymbol \xi: \norm{\boldsymbol\xi-\mathbf{x}_0}\leq r} 
\subset \Omega$. Then if follows from
\eqref{Eq:compWithConesA} that $w$ is upper semicontinuous. If $w$ is
not a subsolution to $-\Delta_\infty w= 0$ in $\Omega$, then there is
a point $\mathbf{x}_1\in \Omega$ and a function $\varphi$ which is
twice continuously differentiable in the set $\set{\mathbf{x}:
  \norm{\boldsymbol\xi -\mathbf{x}_1}<\delta}\subset \Omega$ where
$\delta >0$, such that $w(\mathbf{x}) \leq \varphi(\mathbf{x})$ when
$\norm{\mathbf{x}-\mathbf{x}_1}< \delta$, $w(\mathbf{x}_1)=\varphi(\mathbf{x}_1)$, 
and $-\Delta_\infty \varphi(\mathbf{x}_1) > 0$.  The Taylor expansion of $\varphi$ 
is
  \begin{equation}\label{Eq:TaylorA}
  \varphi(\mathbf{x})= \varphi(\mathbf{x}_1)
+ \inprod{\mathbf p}{\mathbf{x}-\mathbf{x}_1} 
+ \tfrac 12 \inprod{A(\mathbf{x}-\mathbf{x}_1)}{\mathbf{x}-\mathbf{x}_1} + o(\norm{\mathbf{x}-\mathbf{x}_1}^2),
  \end{equation}
where $\mathbf p = D\varphi(\mathbf{x}_1)$, $A = D^2\varphi(\mathbf{x}_1)$ and
$\inprod{\cdot}{\cdot}$ is the inner product in $\mathbb{R}^d$.
 Note that  since $-\Delta_\infty \varphi(\mathbf{x}_1) > 0$ we have  
$\inprod{A\mathbf   p}{\mathbf p}<0$ and therefore also 
 $\mathbf p \neq \mathbf 0$.   We define $\mathbf p_0= \frac
1{\norm{\mathbf p}} \mathbf p$ and introduce new coordinates 
$(t,\mathbf y)$ by writing  $\mathbf{x}= \mathbf{x}_1 + t\mathbf p_0 + \mathbf y$
 where $\inprod {\mathbf p}{\mathbf y}=0$. Then  we have
\begin{equation} \label{Eq:secondTermExpansion}
\begin{aligned}
  \inprod{A(t\mathbf p_0+\mathbf y)}{t\mathbf p_0+\mathbf y}
  \leq  t^2\inprod{A\mathbf p_0}{\mathbf p_0} +
  2t\norm{A\mathbf p_0}\norm{\mathbf y} +  \norm{A}\norm{\mathbf y}^2\\ \leq -2c_1 t^2+ \tfrac 12 c_2 \norm{\mathbf y}^2,
  \end{aligned}
\end{equation}
  where $c_1$ and $c_2$ are positive constants,
(in fact we can choose  $c_1= -\frac 13\inprod{A\mathbf p_0}{\mathbf p_0}$ and
  $c_2=2(\frac{\norm{A\mathbf p_0}^2}{c_1}+ \norm{A})$).

Since $w(\mathbf{x})\leq \varphi(\mathbf{x})$ when 
$\norm{\mathbf{x}-\mathbf{x}_1}<\delta$ and
$w(\mathbf{x}_1)= \varphi(\mathbf{x}_1)$ we can by \eqref{Eq:TaylorA} and
\eqref{Eq:secondTermExpansion} choose  $r$ so
that
\begin{equation}\label{Eq:TaylorB}
  w(\mathbf{x}_1+t\mathbf p_0+\mathbf y)\leq w(\mathbf{x}_1) 
+ \norm{\mathbf p} t -c_1 t^2+ c_2
  \norm{\mathbf y}^2 , \quad 
\norm{t\mathbf p_0+ \mathbf y} \leq \frac 32r,\quad 
\inprod{\mathbf p}{\mathbf y}=0,
  \end{equation}
and
 \begin{equation}\label{Eq:rChoice}
 0<r < \min \big\{\frac {2\delta}3,\frac{\norm{\mathbf p}}{c_1+2c_2}\big\}.
  \end{equation}


Now we choose $\mathbf{x}_0= \mathbf{x}_1-\frac r2\mathbf p_0$ and we will show that
the inequality in \eqref{Eq:compWithConesA} does not hold if 
$\mathbf{x}=\mathbf{x}_1$.
 Since $\norm{\mathbf{x}_1-\mathbf{x}_0}=\frac r2$  we have to show that
\begin{equation}\label{Eq:contradiction}
  w(\mathbf{x}_1) > \frac 12 w(\mathbf{x}_0) + \frac 12  \max_{\norm{\boldsymbol
      \zeta-\mathbf{x}_0}=r} w(\boldsymbol \zeta).
\end{equation}
 With the aid of the  upper bounds for both terms on the
 right-hand side in this inequality that we get from   \eqref{Eq:TaylorB} 
 we see that  it suffices to show that
  \[
  - \norm{\mathbf p}\frac r2 - \frac {c_1r^2}4 + \max_{(t+\frac
    r2)^2+\norm{\mathbf y}^2=r^2}\Bigl (\norm{\mathbf p} t-c_1t^2
+ c_2\norm{\mathbf y}^2\Bigr)<0.
  \]
 This inequality holds  because
  \begin{align*}
&\max_{(t+\frac
    r2)^2+\norm{\mathbf y}^2=r^2}\Bigl (\norm{\mathbf p} t-c_1t^2
 + c_2\norm{\mathbf{y}}^2\Bigr) \\
&= \max_{-\frac 32 r \leq t \leq \frac r2}\Big(\norm{\mathbf p}t-c_1t^2 
+c_2\Big(r^2-\big(t+\frac r2\big)^2\Big)\Big)  \\
&= \norm{\mathbf p}\frac r2 - c_1\frac{r^2}4,
  \end{align*}
where we used the fact that  assumption \eqref{Eq:rChoice} implies that the 
function to be   maximized is increasing in the interval $[-\frac {3r}2, \frac r2]$.
Thus  we get the desired contradiction \eqref{Eq:contradiction} and the proof 
is completed.
\end{proof}

\begin{proof}[Proof of Corollary \ref{T:strongMaxPrinciple}]
  Let $S=\sup_{\boldsymbol \xi\in     \Omega} w(\boldsymbol \xi)$. Suppose
  there is a point $\mathbf{x} \in \Omega$ such that $w(\mathbf{x})= S$ and
  $w$ is not a constant.
  Then there are, because $\Omega$ is open and connected, points $\mathbf{x}_0$
  and $\mathbf{x}_1\in \Omega$ such that $\set{\boldsymbol \xi:
    \norm{\boldsymbol \xi-\mathbf{x}_0}\leq
    2\norm {\mathbf{x}_1-\mathbf{x}_0}}\subset \Omega$ and
  \begin{equation}\label{Eq:supNotSup}
  w(\mathbf{x}_0)< S= w(\mathbf{x}_1).
  \end{equation}
 Since $S$ is the supremum we have $\sup_{\norm{\boldsymbol \xi
      -\mathbf{x}_0} = 2\norm{\mathbf{x}_1-\mathbf{x}_0}}(w(\boldsymbol \xi)
-w(\mathbf{x}_0))       \leq S- w(\mathbf{x}_0)$. 
But then it follows from
\eqref{Eq:compWithConesA} that
\begin{equation*}
  w(\mathbf{x}_1) \leq w(\mathbf{x}_0)+ \frac 12 (S- w(\mathbf{x}_0))  < S,
  \end{equation*}
which is a contradiction by \eqref{Eq:supNotSup}.
\end{proof}

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\end{document}