\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 148, pp. 1--11.\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 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/148\hfil Remarks on the gradient]
{Remarks on the gradient of an \\ infinity-harmonic function}

\author[T. Bhattacharya\hfil EJDE-2007/148\hfilneg]
{Tilak Bhattacharya}  

\address{Tilak Bhattacharya \newline
Department of Mathematics\\
Western Kentucky University\\
Bowling Green, KY 42101, USA}
\email{tilak.bhattacharya@wku.edu}


\thanks{Submitted August 10, 2007. Published November 5, 2007.}
\subjclass[2000]{35J60, 35J70}
\keywords{Infinity-harmonic; gradient; maximum principle}

\begin{abstract}
 In this work we (i) prove a maximum principle for the
 modulus of the gradient of infinity-harmonic functions,
 (ii) prove some local properties of the modulus, and
 (iii) prove that if the modulus is constant on the boundary
 of a planar disc then it is constant inside.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

In this work we discuss some local properties of the modulus of
the gradient of the gradient of an infinity-harmonic function.
Differentiability remains an open problem, except in the planar
case \cite{s}; however, a quantity, which would be the modulus
should differentiability hold, does exist. Our effort in this note
is to prove a maximum principle for the modulus and record some
local properties of an infinity-harmonic function at points where
the modulus is a maximum. In particular, we prove that if the
modulus is constant on the boundary of a planar disc then it is
constant inside.

We start with some notations. Let $\Omega\subset \mathbb{R}^n$, $n\ge 2$,
will denote a bounded domain, the origin $o$ will be assumed to
lie in $\Omega$. Let $B_r(x)$, $x\in\mathbb{R}^n$, be the ball of radius $r$
with center $x$. Let $\bar{A}$ denote the closure of a set $A$ and
$A^c=\mathbb{R}^n\setminus A$. An upper semicontinuous function $u$,
defined in $\Omega$, is infinity-subharmonic in $\Omega$ if it solves
\begin{equation} \label{e1}
\Delta_{\infty}u(x)=\sum_{i,j=1}^nD_iu(x)D_ju(x)D_{ij}u(x)\ge
0,\quad x\in\Omega,
\end{equation}
in the viscosity sense. A lower
semicontinuous function $u$ is infinity-superharmonic in $\Omega$ if
$\Delta_{\infty}u(x)\le 0$, $x\in \Omega$, in the viscosity sense.
Moreover, $u$ is infinity-harmonic in $\Omega$ if it is both
infinity-subharmonic and infinity-superharmonic in $\Omega$. Our work
exploits the cone comparison property satisfied by $u$, see
\cite{ceg}. Also see \cite{acj,tb1,tb2,tdm,cgw} in this
connection. For $x\in \Omega$ and $B_r(x)\Subset\Omega$, for
$0\le t\le r$, we define
$M_x(t)=\sup_{B_t(x)}u$, $m_x(t)=\inf_{B_t(x)}u$. For
infinity-subharmonic functions, $M_x(t)=\sup_{\partial B_t(x)}u$, and
for infinity-superharmonic functions, $m_x(t)=\inf_{\partial B_t(x)}u$.
The existence of the following limits is well known \cite[Lemma 2.7]{ceg},
\begin{equation} \label{e2}
\begin{gathered}
\lim_{t\downarrow  0}\frac{M_x(t)-u(x)}{t}=\Lambda^+(x),\quad
\mbox{when $u$ is infinity-subharmonic,}\\
\lim_{t\downarrow 0}\frac{u(x)-m_x(t)}{t}=\Lambda^-(x),\quad
\mbox{when $u$ is infinity-superharmonic}.
\end{gathered}
\end{equation}
Moreover, if $u$ is infinity-harmonic then
$\Lambda^+(x)=\Lambda^-(x)=\Lambda(x)$, and if also differentiable at $x$, then
$\Lambda(x)=|Du(x)|$. See \cite{acj,tb2,ceg,ce}. We now state the two main
results of this work.

\begin{theorem}[Maximum Principle] \label{thm1}
 Let $\Omega\subset \mathbb{R}^n$ and
$\Omega_1\Subset \Omega$. Recall the statements in \eqref{e2}. (i) If
$u$ is infinity-subharmonic in $\Omega$, then $\sup_{x\in
\bar{\Omega}_1}\Lambda^+(x)=\sup_{x\in \partial \Omega_1}\Lambda^+(x)$, and (ii) if
$u$ is infinity-superharmonic in $\Omega$, then $\sup_{x\in
\bar{\Omega}_1}\Lambda^-(x)=\sup_{x\in \partial \Omega_1}\Lambda^-(x)$. In particular,
if $u$ is infinity-harmonic then $\sup_{x\in
\bar{\Omega}_1}\Lambda(x)=\sup_{x\in \partial \Omega_1}\Lambda(x)$.
\end{theorem}

The anonymous referee pointed out this general version of Theorem \ref{thm1}. 
An older version of this theorem was stated only for
infinity-harmonic functions.
A proof will be presented in Section 2.
The main idea of the proof is to
exploit the result about increasing slope estimate in \cite[Lemma 3.3]{ceg}. 
In \cite{tb2}, these have been referred to as
Hopf derivatives in the case of infinity-harmonic functions. The
properties of the latter will be
used to prove the second main result of this work.


\begin{theorem} \label{thm2}
Let $\Omega\subset \mathbb{R}^2$ and $B_r(x)\Subset \Omega$. Let $u$ be
infinity-harmonic in $\Omega$. Suppose that for some $L>0$ and for every $y\in
\partial B_r(x)$, $\Lambda(y)=|Du(y)|=L$, then
\begin{itemize}
\item[(i)] for any $w\in B_r(x)$, $|Du(w)|=L$, and
\item[(ii)] given any point $z\in B_r(x)$ there is a straight segment
$T$, with its end points on $\partial B_r(x)$ and containing $z$,
such that $u$ is linear on $T$. Also, if $e_T$ is a unit vector
parallel to $T$ then for any $\xi$ on $T$ either $Du(\xi)=Le_T$,
or $Du(\xi)=-Le_T$. In addition, if $T_1$ and $T_2$ are any two
such segments then either $T_1$ coincides with $T_2$ or they are
distinct.
\end{itemize}
\end{theorem}

At this time it is unclear whether or not this holds in $\mathbb{R}^n$
with $n\ge 3$. Theorem \ref{thm2} does not hold in general and the
convexity of the domain seems to play a role in the proof this
result. Consider the example $u(x,y)=x^{4/3}-y^{4/3}$ on $\mathbb{R}^2$,
where a point is described as $(x,y)$. Then
$\Lambda(x,y)=|Du(x,y)|=4/3\sqrt{x^{2/3}+y^{2/3}}$, and consider the
regions $D_c$ of the type bounded by $x^{2/3}+y^{2/3}=c>0$. While
$|Du(x,y)|$ is constant on $\partial D_c$,
$|Du(o)|=0$ and $|Du(x,y)|<4c^2/3$, $(x,y)\in D_c$.

We have divided our work as follows. Section 2 presents a proof of
Theorem \ref{thm1}. In Section 3, we study the behaviour of an
infinity-harmonic function $u$ near points of maximum of $\Lambda(x)$.
In Section 4,
we prove the rigidity result in Theorem \ref{thm2}.

\section{Proof of main results}

We first state results we will use in the
proof of Theorem \ref{thm1}. Recall the statements in \eqref{e2}.
Let $\Omega\subset \mathbb{R}^n$ and $B_r(x)\Subset\Omega$. Let (a) $p_t\in
\partial B_t(x)$, $t\le r$, denote a point of maximum of
$u(p)$ on $B_t(x)$, when $u$ is infinity-subharmonic in $\Omega$, and (b)
$q_t\in \partial B_t(x)$ denote a point of minimum of $u$ on $B_t(x)$,
when $u$ is infinity-superharmonic in $\Omega$.
For part (i) of the
theorem we will use the following.
\begin{equation} \label{e3}
\Lambda^+(x)\le \frac{M_x(t)-u(x)}{t}\le \inf_{p_t}\Lambda^+(p_t),\quad
\mbox{and $\Lambda^+$ is upper-semicontinuous.}
\end{equation}
While for part (ii) we use
\begin{equation} \label{e4}
\Lambda^-(x)\le \frac{u(x)-m_x(t)}{t}\le \inf_{q_t}\Lambda^-(q_t),\quad
\mbox{and $\Lambda^-(x)$ is upper-semicontinuous.}
\end{equation}
See \cite[Lemma 3.3]{ceg}. We also point out that minor modifications
of the arguments in \cite{tb1,tb2} will also yield \eqref{e3} and \eqref{e4}. We now
prove Theorem \ref{thm1} by employing the above repeatedly.

\begin{proof}[Proof of Theorem \ref{thm1}]
We first prove part (i).
Let $L=\sup_{x\in\bar{\Omega}_1}\Lambda^+(x)$, we assume that
$L>0$. Since $\Lambda^+$ is upper semi-continuous, for some $y\in
\bar{\Omega}_1$ we have $\Lambda(y)=L$. If $y\in \partial \Omega_1$ then we are
done. Assume then that $y\in \Omega_1$. We will show that in this
case there is a point $\bar{y}\in \partial \Omega_1$ with $\Lambda(\bar{y})=L$.
Set $y_1=y$ and let $d_1=\mathop{\rm dist}(y_1,\Omega_1^c)$. Clearly,
$B_{d_1}(y_1)\subset \Omega_1$; by \eqref{e3}, $\Lambda^+(p)\ge
(u(p)-u(y_1))/d_1\ge\Lambda^+(y_1)=L$, for any $p\in \partial
B_{d_1}(y_1)$ with $u(p)=M_{y_1}(d_1)$. Thus $\Lambda^+(p)=L$ and
$u(p)=u(y_1)+Ld_1$. If $p\in \partial\Omega_1$ then we are done, otherwise
set $y_2=p$. As already noted $u(y_2)=u(y_1)+Ld_1$. Let
$d_2=\mathop{\rm dist}(y_2,\Omega_1^c)$, and any $p\in
\partial B_{d_2}(y_2)$ be such that $u(p)=M_{d_2}(y_2)$. Again by \eqref{e3},
$\Lambda^+(p)\ge(u(p)-u(y_2))/d_2\ge \Lambda^+(y_2)=L$. Thus $\Lambda^+(p)=L$
and $u(p)= u(y_1)+L(d_1+d_2)$. If $p\in \partial \Omega_1$ then we are
done.
Suppose now that we have obtained sequences
$\{y_i\}_{i=1}^k$, $\{d_i\}_{i=1}^k$ such that
\begin{itemize}
\item[(a)] $d_i=\mathop{\rm dist}(y_i,\Omega_1^c)$, $y_i\in \partial B_{d_{i-1}}(y_{i-1})$ and
$y_i\notin\partial\Omega_1$, $2\le i<k$,
\item[(b)]
$u(y_i)=M_{y_{i-1}}(d_{i-1})=u(y_1)+L\sum_{j=1}^{i-1}d_j$, $2\le
i\le k$, and
\item[(c)]
$\Lambda^+(y_i)=\Lambda^+(y_j)=L$, for all $i,j=1,2,\dots ,k$.
\end{itemize}
Suppose that
$y_k\notin \partial \Omega_1$. For any $p\in\partial B_{d_k}(y_k)$, with
$u(p)=M_{d_k}(y_k)$, \eqref{e3} implies $\Lambda^+(p)\ge (u(p)-u(y_k))/d_k\ge
\Lambda^+(y_k)=L$. Thus $\Lambda^+(p)=L$, $u(p)=u(y_k)+Ld_k$. If $p\in \partial
\Omega_1$ then we are done otherwise set $y_{k+1}=p$ and note that
$u(y_{k+1})=u(y_1)+L\sum_{i=1}^{k}d_i$ and $\Lambda^+(y_{k+1})=L$. By
the maximum principle, for every $k$, $u(y_1)+L\sum_{i=1}^{k}d_i=
u(y_{k+1})\le \sup_{\Omega_1}u<\infty$. Thus
$\sum_{i=1}^{\infty}d_i<\infty$ and $d_i\to 0$. Moreover,
for $i<j$, $|y_i-y_j|\le
\sum_{l=i}^{j-1}|y_l-y_{l+1}|=\sum_{l=i}^{j-1}d_l$ is small if $i$
is large. Thus for some $\bar{y}\in\partial\Omega_1$, $y_i\to
\bar{y}$ and $\Lambda^+(\bar{y})\ge \limsup_{k\to
\infty}\Lambda^+(y_k)=L$. Part (ii) may now be proved analogously by
using points of minima and \eqref{e4}. The conclusion follows.
\end{proof}

\begin{remark} \label{rmk1} \rm
In the case of infinity-harmonic functions, we
can show using Lemma \ref{lem1} (see Section 2) that the points
$y_1,y_2,\dots $. all lie on a straight segment terminating at
$\bar{y}$. We also mention in passing that if $\Lambda^s(r)=\sup_{x\in
\partial B_r(o)}\Lambda(x)$ then the upper-semicontinuity of $\Lambda$ and
Theorem \ref{thm1} implies that $\Lambda^s(r)$ is right continuous.
\end{remark}

\section{Comments on the function $\Lambda$}

For the remainder of this work $u$ will denote an infinity-harmonic
function.
Our effort in this section will be to describe the behaviour
of $u$ near points of maximum of $\Lambda$.
We recall some previously defined notations for ease of presentation.
Let $B_r(x)\Subset \Omega$; a point $p_t\in \partial B_t(x)$, $t\le
r$, will denote a point of maximum of $u$ on $B_t(x)$. The
direction $(p_t-x)/t$ will be denoted by $\omega_t$. The quantities
$M_x(t)$ and $m_x(t)$ continue to denote the maximum and the minimum of
$u$ on $B_t(x)$. Note that $M_x(t)$ and $-m_x(t)$ are convex in $t$. We
will drop the subscript $x$ when $x=o$.
Next we summarize the properties of the Hopf-derivaties which will be
used repeatedly in the
rest of this work, see \cite[Theorems 1 and 2]{tb2}. We work in $B_r(o)$.
\begin{itemize}

\item[(i)] $\frac{M(t)-u(o)}{t}$ decreases to $\Lambda(o)$
 as $t\downarrow 0$,
\item[(ii)] for $0\le s\le \tau\le t\le r$,
we have $\Lambda(o)\le\sup_{p_s\in \partial B_s(o)}\Lambda(p_s)\le \Lambda(p_{\tau})\le
\inf_{p_t\in \partial B_t(o)}\Lambda(p_t)\le \Lambda(p_r)$, and
\begin{equation} \label{e5}
\lim_{t\downarrow 0} \sup_{p_t\in\partial B_t(o)}\Lambda(p_t)=\Lambda(o),
\quad t\le r,
\end{equation}

\item[(iii)] $\Lambda(o)\le \frac{M(r)-u(t\omega_r)}{r-t}\le \Lambda(p_r)$,
and $\frac{M(r)-u(t\omega_r)}{r-t}$ increases to $\Lambda(p_r)$ as $t\uparrow r$.

\end{itemize}
Moreover,
\begin{itemize}
\item[(i)] $ u$ is differentiable at any $p_t\in \partial B_t(o)$
and $Du(p_t)=\Lambda(p_t)\omega_t$, $t\le r$,
\item[(ii)]
\begin{equation} \label{e6}
M'(t-)\le \inf_{p_t\in \partial B_t(o)}\Lambda(p_t)\le \sup_{p_t\in
\partial B_t(o)}\Lambda(p_t)\le M'(t+),
\end{equation}
\item[(iii)] There exists $p_t\in \partial B_t(o)$
such that $\Lambda(p_t)=M'(t+)$.

\end{itemize}
Analogous statements also hold for $q_t$ and $m(t)$. Moreover,
for any pair of sequences $r_k\downarrow 0,\omega_{r_k}\in S^{n-1}$,
with $\omega_{r_k}\to \omega$ (by compactness such pairs do exist,
also see \cite{ce}), we have
\begin{equation} \label{e7}
\begin{gathered}
\lim_{r_k\downarrow
0}\frac{u(r_k\omega_{r_k})-u(o)}{r_k}=\lim_{r_k\downarrow
0}\frac{u(r_k\omega)-u(o)}{r_k}=\Lambda(o),\\
\lim_{r_k\downarrow 0}\frac{u(\theta
r_k\eta)-u(o)}{r_k}=\theta\Lambda(o)\langle\omega,\eta\rangle,\quad \forall\;\eta\in
S^{n-1},
\end{gathered}
\end{equation}
for any fixed $\theta>0$. Note that
$M(r_k)=u(r_k\omega_{r_k})$. The above statements also apply to
points of minima. In particular, if $\nu_t=q_t/t$ where
$u(q_t)=m(t)$, then $\nu_{r_k}\to -\omega$. If $\omega$  is the
only limit point of $\omega_t$ as $t\downarrow 0$, then $u$ is
differentiable at $o$, see \cite{ce}. Also, if $\omega\in S^{n-1}$ is
such that \eqref{e7}(ii) holds for any sequence then $\omega$ is a gradient
direction and $u$ is differentiable at $o$. We now prove the following result.

\begin{lemma} \label{lem1}
Let $u\ne 0$ be infinity-harmonic in $\Omega$ and
$B_r(o)\Subset\Omega$.
\begin{itemize}
\item[(a)] If $\Lambda(o)=(M(r)-u(o))/r$,
then $u$ is differentiable at $o$ and
$Du(o)=\Lambda(o)\omega$, for some $\omega\in S^{n-1}$. Moreover, for $0\le t\le r$,
$M(t)=u(t\omega)=u(o)+t\Lambda(o)$, and for very $t>0$ there is exactly one
point $p_{t}\in\partial B_{t}(o)$ such that $u(p_t)=M(t)$.

\item[(b)] If $p_r\in \partial B_r(o)$ is such that $\Lambda(p_r)=\Lambda(o)$ then the
same conclusion holds for $u$ with $\omega=p_r/r$, and
$M(t)=u(t\omega)=u(o)+t\Lambda(o)$, $0\le t\le r$.
\end{itemize}
Furthermore, if $x$ is any point on the segment $op_r$ then
$Du(x)=\Lambda(o)\omega$.
\end{lemma}

\begin{proof}
 We prove part (a). Recall that $M(t)$ is convex in
$t$, thus by 5(i) and the first part of \eqref{e5}(iii),
\begin{equation} \label{e8}
\Lambda(o)\le
\frac{M(t)-u(o)}{t}\le \frac{M(r)-u(o)}{r}=\Lambda(o),\quad 0\le t\le r.
\end{equation}
Thus $M(t)=u(o)+t\Lambda(o)$, and
$u(t\omega)\le u(o)+t\Lambda(o)$, for all $0\le t\le r$.
For $0<t<r$, let $p_{t}\in
\partial B_o(t)$ be any point of maximum of $u$, set $\omega_{t}=p_{t}/t$.
Since $M'(t)=\Lambda(o)$, using \eqref{e6}(ii) and \eqref{e8}, we have that
$\Lambda(p_t)=\Lambda(o)$. For a fixed $t\le r$, an application of
\eqref{e5}(iii) to the ball $B_t(o)$ results in
$$
\Lambda(o)\le \frac{M(t)-u(s\omega_t)}{t-s}\le \Lambda(p_t)=\Lambda(o),\quad
\forall\;0\le s<t.
$$
Thus $u(s\omega_t)=u(o)+s\Lambda(o)$, $0\le s\le t$, and this holds for
every $0\le t< r$. Clearly, for a fixed $0<t<r$,
$(u(s\omega_t)-u(o))/s\to \Lambda(o)$ as $s\downarrow 0$. By the
comments following \eqref{e7}, $u$ is differentiable at $o$ and the
gradient direction is $\omega_t$. This is true of any $\omega_t$ and any
$0<t<r$. Clearly, $\omega=\omega_t$, $0<t\le r$, is unique. Moreover,
$M(t)=u(t\omega)=u(o)+t\Lambda(o)$, $0\le t\le r$. By the second part of
\eqref{e5}(iii) and \eqref{e6}(iii), $\Lambda(r\omega)=\Lambda(o)=M'(r-)$, for any
$p_r\in B_r(o)$. It is also clear that $r\omega$ is a point of
maximum of $u$ on $\partial B_r(o)$.

 Now suppose that $\omega_1\in
S^{n-1}$ is such that $u(r\omega_1)=M(r)$. Using the special nature
of $M(t)$, we see from the second part of \eqref{e5}(iii) that
$\Lambda(r\omega_1)=\Lambda(o)$. Using \eqref{e5}(i) and arguing as above we see
that $\omega_1$ is another gradient direction at $o$. Thus by \eqref{e7},
$\omega_1=\omega$. Also note that by \eqref{e6}(iii), $M'(r)=\Lambda(o)$.
This also proves part (b). To show the last statement let $0\le
s\le r$ be such that $x=s\omega$. Then $B_{\rho}(x)\subset
B_{\rho+s}(o)$, $u(x)=u(o)+s\Lambda(o)$ and
$M_x(\rho)=\sup_{B_{\rho}(x)}u=M(\rho+s)=u(x)+\rho \Lambda(o)$, $\rho\le r-s$.
The rest now follows from the comments following \eqref{e7}, see
\cite{ce}.
\end{proof}

\begin{remark} \label{rmk2} \rm
An analogous version of Lemma \ref{lem1} holds for the case of minima.
\end{remark}

In the rest of this work we will have occasion to use a version of
Rolle's property.
We refer the reader to the Appendix for a proof in the case $n\ge 3$.

\begin{remark} \label{rmk3} \rm
Suppose that $B_r(o)\subset \Omega$; let $p_r\in
B_r(o)$ be any point such that $u(p_r)=M(r)$. Set $\omega_r=p_r/r$;
we claim that for $0\le \rho<r$, $\Lambda(\rho\omega)\le \Lambda(p_r)$. To see
this note that $B_{(r-\rho)}(\rho \omega)\subset B_r(o)$ and $M_{\rho
\omega}(r-\rho)=M(r)$. Thus using the first part of \eqref{e5}(iii) in
$B_{(r-\rho)}(\rho \omega)$, we see that $\Lambda(\rho\omega)\le
(M(r)-u(\rho\omega))/(r-\rho)\le \Lambda(p_r)$. Moreover, we claim that there
is a sequence of points $x_k$ on the line segment $op_r$ such that
$x_k\to p_r$ and $\Lambda(x_k)\uparrow \Lambda(p_r)$. To see this
note that for $0\le s\le t<r$, \eqref{e5}(iii) implies
\begin{equation} \label{e9}
\sup(\Lambda(o),\Lambda(sw))\le \frac{M(r)-u(s\omega)}{r-s}
\le\frac{M(r)-u(t\omega)}{r-t}\le \Lambda(p_r).
\end{equation}
An application of Rolle's property to $u(p_r)-u(s\omega)$ and
$u(p_r)-u(t\omega)$ in \eqref{e9} implies there are
$\theta_s\in(s,r)$, $\theta_t\in(t,r)$ and $\omega_{\theta_s}$, $\omega_{\theta_t}\in
S^{n-1}$ such that $\Lambda(o)\le
\Lambda(\theta_s\omega)\langle\omega_{\theta_s},\omega\rangle \le
\Lambda(\theta_t\omega)\langle \omega_{\theta_t},\omega\rangle\le \Lambda(p_r)$. Also
from \eqref{e9}, we see that $\Lambda(o)\le\Lambda(s\omega)\le \Lambda(\theta_s\omega)\le
\Lambda(p_r)$. We iterate the latter using \eqref{e9} as follows. Starting
with $s\ge 0$ and setting $s_1=s$, $s_2=\theta_{s_1}$ and
$s_{k+1}=\theta_{s_k}$, $k=2,\dots$, we employ
\begin{equation} \label{e*}
\Lambda(s_k\omega)\le \frac{M(r)-u(s_k\omega)}{r-s_k}=\Lambda(s_{k+1}\omega)\langle
\omega,\omega_{s_{k+1}}\rangle\le\Lambda(s_{k+1}\omega)\le \Lambda(p_r),
\end{equation}
to see that (i) $s_k\uparrow r$,
(ii) $\Lambda(s_1\omega)\le\Lambda(s_1\omega)\le \Lambda(s_2\omega)\le\dots.\le \Lambda(s_k\omega)\le
\dots\le \Lambda(p)$. To see (i) suppose that $s_k\uparrow s<r$, then \eqref{e5}(i) and the second inequality
in \eqref{e*}
would then imply that $\Lambda(s)\le [M(r)-u(s\omega)]/(r-s)\le \Lambda(s)$. 
Lemma \ref{lem1} would then
hold and for $s<t<r$, $\Lambda(t)=\Lambda(s)=\Lambda(p_r)$. We may now select $s_k\uparrow r$.
Finally, employing the definition of $s_k$ and the second part of \eqref{e5}(iii)
in \eqref{e*}, we obtain that
$\Lambda(s_k\omega)\uparrow \Lambda(p)$ as $s_k\uparrow r$ and
$\omega_{s_k}\to \omega$.
\end{remark}

Next we discuss the nature of $u$ near points of maximum of $\Lambda$.
We recall \eqref{e7} and the discussion just following it. Let
$B_r(o)\Subset\Omega$; for $0<t\le r$, again $p_t$ will denote
any point of maximum of $u$ on $\partial B_t(o)$ and $q_t$ any
point of minimum. Once again set $\omega_t=p_t/t$ and $\nu_t=q_t/t$.
We restate \eqref{e7} for ease of presentation. If $t_k\downarrow 0$ with
$\omega_{t_k}\to \omega$ then $\nu_{t_k}\to \nu=-\omega$
and
\begin{equation} \label{e10}
\begin{aligned}
\lim_{t_k\downarrow 0}\frac{u(t_k\omega)-u(o)}{t_k}
&=\lim_{t_k\downarrow
0}\frac{u(t_k\omega_{t_k})-u(o)}{t_k}\\
&=-\lim_{t_k\downarrow 0}\frac{u(t_k\nu_{t_k})-u(o)}{t_k}\\
&= -\lim_{t_k\downarrow 0}\frac{u(t_k\nu)-u(o)}{t_k}=\Lambda(o).
\end{aligned}
\end{equation}
Also $M(t_k)$ and $m(t_k)$ occur near $t_k\omega$
and $-t_k\omega$ when $t_k$ is small. We refer to $\omega$, $\nu$ as limit
directions.

\begin{lemma} \label{lem2}
Let $u$ be infinity-harmonic in $\Omega$ and
$B_r(o)\Subset \Omega$. Also set 
$\Lambda^s=\sup_{x\in \bar{B}_r(o)}\Lambda(x)>0$, let $y\in \partial B_r(o)$ be such that
$\Lambda(y)=\Lambda^s$. Let $H_y$ denote the $n-1$ dimensional plane
tangential to $\partial B_r(o)$ at
$y$. Then only one of the following happens.

\noindent Case(a): There is a straight segment $xy$ with $x\in\partial B_r(o)$
such that $u$ is a linear function on $xy$. More precisely, for
every $0\le t\le |x-y|$, either (i) $u(y+te)=u(y)+t\Lambda^s$, or (ii)
$u(y+te)=u(y)-t\Lambda^s$, where $e=(x-y)/|x-y|$. Moreover, $u$ is
differentiable on the segment $xy$, and if $z\in xy$ then in
(i) $Du(z)=\Lambda^s e$, and in (ii) $Du(z)=-\Lambda^s e$.

\noindent Case (b): For every $s>0$, all the points of extrema of $u$ on
$\partial B_s(y)$ lie outside $\bar{B}_r(o)$. In particular all limit
directions $\omega$, $\nu$ (see comment following \eqref{e10}) lie in $H_y$.
Moreover, if $\omega$ is a limit direction, $s_k\downarrow 0$ the
corresponding sequence, $\eta\in S^{n-1}$ and $y_k\in\partial B_r(o)$ is
the point nearest to $y+s_k\eta$ then $\lim_{s_k\downarrow
0}(u(y_k)-u(y))/s_k=\Lambda^s\langle\omega,\eta\rangle$.
\end{lemma}

\begin{proof}
 Assume that Case (b) is not true. There is a ball
$B_{\delta}(y)$ and a point $p\in \partial B_{\delta}(y)\cap \bar{B}_r(o)$
such that $u(p)=M_{y}(\delta)$. Our assumption of a point of maximum
of $u$ on $\partial B_{\delta}(y)$, lying in $\bar{B}_r(o)$, is not
restrictive and the arguments we use will apply equally to a
minimum. By \eqref{e5}(iii) or even \eqref{e3}, $\Lambda(p)\ge \Lambda(y)$ implying that
$\Lambda(p)=\Lambda^s$. Set $\omega=(p-y)/\delta$; by Lemma \ref{lem1}, we see that (i)
$u(y+t\omega)=u(y)+t\Lambda^s$, $0\le t\le \delta$, (ii) $u$ is
differentiable everywhere on the segment $yp$ with $Du(z)=\Lambda^s
\omega$, for any $z$ on $yp$, and (iii) $p$ is the only point of
maximum on $\partial B_{\delta}(y)$. If $p\in \partial B_r(o)$, then $x=p$
and the lemma holds. Assume that $p\in B_r(o)$; set
$y_1=y$, $y_2=p$, $\omega_1=\omega$ and $d_1=\delta$. Note that $\omega_1$
points into $B_r(o)$. We repeat the argument at $y_2$ as follows.
Set $d_2=r-|y_2|$ and $y_3\in \partial B_{d_2}(y_2)$ be a point of
maximum. By 5(iii), $\Lambda(y_3)\ge \Lambda(y_2)=\Lambda^s$ implying
$\Lambda(y_3)=\Lambda^s$; set $\omega_2=(y_3-y_2)/d_2$. Again by Lemma 
\ref{lem1}, $u$
is differentiable on $y_2y_3$ with
$u(y_2+t\omega_2)=u(y_2)+t\Lambda^s=u(y_1)+(t+d_1)\Lambda^s$, $0\le t\le d_2$.
By the uniqueness of gradient direction at $y_2$,
$\omega_2=\omega_1=\omega$ and $y_1y_3$ is a straight segment. If $y_3\in
\partial B_r(o)$ the process stops. Otherwise assume that we have a
sequence of points $\{y_i\}_{i=1}^k\in B_r(o)$, with
$\omega_i=(y_{i+1}-y_i)/d_i=\omega$, $i=1,2,\dots,k-1$; i.e., $y_1y_k$ a
straight segment parallel to $\omega$, and
$u(y_1+t\omega)=u(y_1)+t\Lambda^s$, $0\le t\le \sum_{i=1}^k d_i$.
Moreover, $u$ is differentiable at every point $z$ on $y_1y_k$ and
$Du(z)=\Lambda^s\omega$. Now let $d_{k+1}=r-|y_k|$ and $y_{k+1}\in \partial
B_{d_{k+1}}(y_k)$ such that $u(y_{k+1})=M_{y_k}(d_{k+1})$. Set
$\omega_{k+1}=(y_{k+1}-y_k)/d_k$; then by Lemma \ref{lem1},
$\Lambda(y_{k+1})=\Lambda(y_k)=\Lambda^s$, $y_{k}y_{k+1}$ is a straight
segment and $u$ is differentiable on $y_{k}y_{k+1}$. Thus
$\omega_{k+1}=\omega$, i.e., $y_1y_{k+1}$ is a straight segment parallel
to $\omega$. Moreover, on $y_1y_{k+1}$,
$u(y_1+t\omega)=u(y_1)+t\Lambda^s$, $0\le t\le \sum_{i=1}^{k+1}d_i$, and
$Du(z)= \Lambda^s \omega$, for any $z$ on $y_1y_{k+1}$. Either
$y_{k+1}\in \partial B_r(o)$ in which case the process stops or we
continue. By the maximum principle,
$u(y_1)+\Lambda^s\sum_{i=1}^{k}d_i\le M_o(r)<\infty$, for all $k\ge
1$. Thus $d_i\to 0$, $y_k\to x$ where $x\in \partial
B_r(o)$. Thus we obtain a straight segment $xy$ where $x\in\partial
B_r(o)$ and the
conclusions of Case (a) hold with $e=\omega$.

Now assume that Case (b) holds. We suppose that for every $s>0$,
the points of extrema of $u$ on $\partial B_s(o)$ lie outside
$\bar{B}_r(o)$. Given any sequence $s_k\downarrow 0$ and
$\omega_{s_k}\to \omega$ we also have $\nu_{s_k}\to
-\omega$, see remarks preceding \eqref{e10}. Thus any limit direction $\omega$
lies in $H_y$. Let $\omega$ be a limit direction and $s_k\downarrow
0$ be such that $(u(y+s_k\omega)-u(y))/s_k\to \Lambda(y)$. For
$k=1,2,\dots$, let $y_k\in \partial B_R(o)\cap \partial B_{s_k}(y)$ be the point
nearest to $y+s_k\omega$. Thus $y_k=y+s_k\zeta_k$, where $\zeta_k\in
S^{n-1}$. Since the sphere is $C^2$ at $y$, $|y_k-(y+s_k\omega)|/s_k
\to 0$ and $\langle \omega,\zeta_k\rangle \to 1$.
Thus we have that, near $y$,
\begin{align*}
\big|\frac{u(y_k)-u(y)}{s_k}-\Lambda^s \big|
&\le \big|\frac{u(y_k)-u(y+s_k\omega)}{s_k}\big|+
\big|\frac{u(y+s_k\omega)-u(y)}{s_k}-\Lambda^s\big|\\
&\le C|\zeta_k-\omega |+\big|\frac{u(y+s_k\omega)-u(y)}{s_k}-\Lambda^s\big|,
\end{align*}
where $C>0$ is the local Lipschitz constant. Clearly, the conclusion
holds when $\eta=\omega$ by letting $s_k\to 0$. The
statement for general $\eta$ may now be derived
by using \eqref{e7}.
\end{proof}

\begin{remark} \label{rmk4} \rm
 In Case (a) of Lemma \ref{lem2}, if $z$ is any point in the interior of the segment
$xy$ and $B_s(z)\subset B_r(o)$, then $u$ has exactly one point of maximum
and one point of minimum on $\partial B_s(z)$. Both these lie on $xy$.
One may show this by applying \eqref{e5}(iii) or \eqref{e4}.
Lemma \ref{lem2} also holds if a limit direction $\omega$ or $-\omega$,
at $y$, points into $\bar{B}_r(o)$. One can find a small $\delta>0$ such that
$M_y(\delta)$ occurs near $\delta\omega$ (analogous for a minimum) and hence
lies in $\bar{B}_r(o)$. See discussion at the beginning of this section.
Using Lemma \ref{lem1}, one may show that $xy$ is parallel to $\omega$.
\end{remark}

\begin{remark} \label{rmk5} \rm
 By \eqref{e6}(iii) there is at least one point $p_r\in
\partial B_r(o)$, where $\Lambda(p_r)=M'(r+)$. Thus $\Lambda^s\ge
M'(r+)$. The existence of a straight line segment on which
$u$ is linear need not imply that $u$ is affine. Take
$u(x)=|x|$, $x\ne 0$. Also see capacitary rings \cite{tb1}.
\end{remark}

\begin{remark} \label{rmk6}\rm
 If $y\in \partial B_r(o)$ is a point of extrema of $u$
and $\Lambda(y)=\Lambda^s$, then by \eqref{e6}(i) $Du(y)=\pm \Lambda^s \omega$, where
$\omega=y/r$. Clearly, case (a) of Lemma \ref{lem2} applies and $u$ is linear
and differentiable on $xy$, where $x=-y$. For $0\le t\le r$,
either $u(x+t\omega)=u(x)+t\Lambda^s$ or $u(x+t\omega)=u(x)-t\Lambda^s$. Since
$\Lambda(y)=\Lambda(o)=\Lambda^s$, by Lemma \ref{lem1} and Remark \ref{rmk2}, for $0\le t\le r$,
we have $M'(t)=-m'(t)=\Lambda^s$; we also have
$|M(t)-m(t)|\le 2t\Lambda^s$. Assume that $u(y)=M(r)$; linearity
implies that for any $0\le t\le r$, $u(o)=M(t)-t\Lambda^s$,
$m(t)=u(-t\omega)=M(t)-2t\Lambda^s$, and in particular,
$u(x)=m(r)=M(r)-2r\Lambda^s$. Employing Lemma \ref{lem1}, we see that
$t\omega$, $-t\omega$ are the only points of extrema on $\partial B_t(o)$,
$t\omega$ being the maximum and $-t\omega$ being the minimum. Thus for
every $0<t\le r$, $m(t)<u(x)<M(t)$, $x\in\partial B_t(o)\setminus\{\pm
t\omega\}$.
\end{remark}

Next we show a property of $u$ in the situation when Case (a) of
Lemma \ref{lem2} holds. For $z\in \mathbb{R}^n$ and $e\in S^{n-1}$, let
$\gamma(z,e)$ be the interior of the cone that has vertex $z$,
aperture $\pi/3$  and opens along $e$.

\begin{lemma} \label{lem3}
Let $y\in\partial B_r(o)$ be such that $\Lambda(y)=\Lambda^s$. Assume Case (a)
of Lemma \ref{lem2} holds, that is, there is a segment $xy$ in
$\bar{B}_r(o)$, with $x\in \partial B_r(o)$, such that $u$ is linear and
differentiable on $xy$. Assume that $u(y+te)=u(y)+t\Lambda^s$, $0\le
t\le d$, where $d=|x-y|$ and $e=(x-y)/d$. Let $y_t=y+te$, $0\le
t<d$, then (i) $u(z)\ge u(y_t)$, $z\in \gamma(y_t,e)\cap B_r(o)\cap
B_{d-t}(y_t)$, and (ii) $u(z)\le u(y_t)$, $z\in \gamma(y_t,-e)\cap
B_r(o)\cap B_{t}(y_t)$. The case when $u(y+te)=u(y)-t\Lambda^s$, $0\le
t\le d$, is analogous.
\end{lemma}

\begin{proof} Let $0\le \varepsilon\le d-t$, set $y_{t+\varepsilon}=y_t+\varepsilon e$. Now
select $z\in B_r(o)$ such that $|z-y_t|=\varepsilon$ and
set $e_{\varepsilon}=(z-y_{t+\varepsilon})/|z-y_{t+\varepsilon}|$.
Let $\theta$ be the angle between segments $zy_t$ and $xy_t$. By
the Rolle's property, for some
point $a$ on the straight segment $zy_t$ and limit direction $\omega$, we
 have
$u(z)-u(y_{t+\varepsilon})=u(z)-u(y_t)-\varepsilon\Lambda^s=2\varepsilon\Lambda(a)\langle\omega,e_{\varepsilon}\rangle
\sin (\theta/2)$. Thus
$u(z)-u(y_t)=2\varepsilon\left(\Lambda^s+\Lambda(a)\langle\omega,e_{\varepsilon}\rangle\sin
(\theta/2)\right)\ge \varepsilon\Lambda^s\left(1-2\sin(\theta/2)\right)$. It follows
that $u(z)\ge u(y_t)$, if $\theta\le \pi/3$. We now take
$y_{t-\varepsilon}=y_t-\varepsilon e$, $z\in B_r(o)$ with $|z-y_t|=\varepsilon$ and
$\bar{e}_{\varepsilon}=(z-y_{t-\varepsilon})/|z-y_{t-\varepsilon}|$. With $\theta$ as defined
before, argue similarly to see that for some $\bar{a}$ on
$zy_{t-\varepsilon}$ and a limit direction $\bar{\omega}$, $u(z)-u(y_{t-\varepsilon})
=u(z)-u(y_t)+\varepsilon \Lambda^s=2\varepsilon\Lambda(\bar{a})\langle
\bar{\omega},\bar{e}_{\varepsilon}\rangle\sin[(\pi-\theta)/2]$. Thus
$u(z)-u(y_t)\le\varepsilon \Lambda^s(-1+2\sin[(\pi-\theta)/2])$. If $\theta\ge
2\pi/3$ then $u(z)\le u(y_t)$.
\end{proof}


\begin{remark} \label{rmk7}\rm
 Let $B_r(o)$, $x$, $y$ and $e$ and  be as in
Lemma \ref{lem3}. Set $2l=|x-y|$ and consider the triangle $\bigtriangleup
oyx$. The angles $\angle oyx=\angle oxy\le \pi/3$ if and only if
$l\ge r/2$. Let $l\ge r/2$ and $y_t=y+te$ be such that $\angle
oy_tx=\pi/3$ then $t=l-\sqrt{(r^2-l^2)/3}$. Since $o$ lies in the
cone $\gamma(y_t,e)$, Lemma \ref{lem3} implies
$$
u(y)+\Lambda^s[l-\sqrt{(r^2-l^2)/3}]\le u(o)\le
u(x)-\Lambda^s[l-\sqrt{(r^2-l^2)/3}].
$$
Also $u(o)-r\Lambda^s\le u(y)\le
u(x)\le u(o)+r\Lambda^s$. If $l\uparrow r$, we
 have
$u(y)\to u(o)-r\Lambda^s(=m(r))$ and $u(x)\to
u(o)+r\Lambda^s(=M(r))$. See Remark \ref{rmk6}.
\end{remark}

\section{Proof of Theorem \ref{thm2}}

Let $D\subset\mathbb{R}^2$ be the unit disc centered at $o$. We will
often describe a point $z\in \mathbb{R}^2$ as $z=(x,y)$. Also set $e_1$
and $e_2$ to be the unit vectors along the positive $x$-axis and
the positive $y$-axis. Let $u$ be infinity-harmonic in a domain
$\Omega\subset \mathbb{R}^2$ and $D\Subset\Omega$. Recall that $u$ is
$C^1$ \cite{s}, and the use of this fact simplifies our
presentation. However, a proof can be worked out without using
this fact. Without any loss of generality, assume that $u(o)=0$.
Let $M=\sup_D u$ and $m=\inf_D u$. Also let $p,\;q\in\partial D$ be such
that $u(p)=M$ and $u(q)=m$. By Theorem \ref{thm1}, $L=\sup_{x\in
\bar{D}}|Du(x)|$. By Remark \ref{rmk6}, $p$ and $q$ are antipodal points
and we may take both of them on the $y$-axis with $p=(0,1)$ and
$q=(0,-1)$. Also $u(0,t)=m+(t+1)L=M-(1-t)L$, $-1\le t\le 1$.
Moreover, for $-1\le t\le 1$, and $Du(0,t)=Le_2$. Let
$H+=\{z\in\mathbb{R}^2:\;x(z)\ge 0\}$ denote the right half disc and
$H-=\{z\in\mathbb{R}^2:\;x(z)\le 0\}$ the left half-disc. Let the right
semi-circle be denoted by $I+=\partial D\cap H+$ and the left
semi-circle by $I-=\partial D\cap H-$. We will work in $H-$ and the
analysis is analogous in $H+$. Let $a,\;b\in I-$ with $a\ne b$. We
will denote the circular arc on $\partial D$, with end points $a$ and
$b$, by $\widehat{ab}$, and use $\bar{ab}$ for the straight
segment with end points $a$ and $b$. Also $l(a,b)$ will denote the
arc length of $\widehat{ab}$.

\noindent\textbf{Step 1.} Let $a,\;b\in I-$ with $a\ne b$. Then
\begin{equation} \label{e11}
\parbox{10cm}{ (i) there is a point point $c\in D$, on the straight segment
$ab$ such that $u(a)-u(b)=\langle Du(c),a-b\rangle$, and
(ii) there is a point $d\in \partial D$, on $\widehat{ab}$,
and a vector $e_d\in S^1$, with $e_d$ tangential to $\partial D$
(perpendicular to the segment $od$) at $d$, such that $u(a)-u(b)=\langle
Du(d),e_d\rangle l(a,b)$.
}
\end{equation}
 In \eqref{e11}(ii), if
$u(a)=u(b)$ then $\langle Du(d),e_d\rangle=0$, implying
$Du(d)\perp e_d$ and parallel to $od$. Noting that $Du(d)=L$, by
Case (a) of Lemma \ref{lem2}, we have a straight segment originating at
$d$, along $od$ and lying in $D$, on which $u$ is linear. Since
this segment terminates on $\partial D$, it passes through $o$, and
differentiability of $u$ at $o$ implies that $\omega_d=Du(d)/L=e_2$.
Thus either $a=b=p$ or $a=b=q$. Also see Remark \ref{rmk6} and the remarks
preceding Step 1. Clearly, $u(a)\ne u(b)$ if $a,\;b\in I-$ and
$a\ne b$. Since $u(p)>u(q)$, we see that $u(z)=u(x,y)$, $z\in I-$,
is increasing in $y$. Recalling \eqref{e11}(i) and (ii), we see that for
$a,\;b\in I-$, $a\ne b$, $u(a)-u(b)=\langle Du(d),e_d\rangle
l(a,b)=\langle Du(c),a-b\rangle\ne 0$. Let $\omega_d$ denote the
gradient direction of $u$ at $d$. Noting that $|Du(d)|=L\ge
|Du(c)|$ and $l(a,b)>|a-b|$, it follows that $\langle
\omega_d,e_d\rangle\ne 0,\pm 1$. This implies that $\omega_d$ does not
lie in the tangent space of $\partial D$ at $d$ nor is it parallel to
segment $od$. Case(a) of Lemma \ref{lem2} now applies and we have a
straight segment originating from $d$ and terminating at
$\bar{d}\in \partial D$ such that $u$ is linear on the segment
$d\bar{d}$, and $|Du(z)|=L$, $z\in d\bar{d}$, and if
$\zeta=(d-\bar{d})/|d-\bar{d}|$ then $Du(z)=\pm L
\zeta$.

 From here on $T$ will denote a segment of the type $d\bar{d}$, as
described in Step 1. Let $z_T=(x_T,y_T)$ and
$\bar{z}_T=(\bar{x}_T,\bar{y}_T)$ denote the two end points that
lie on the unit circle $\partial D$. We set $z_T$ to be the higher end
point and $\bar{z}_T$ will denote the lower end point, i.e.,
$y_T\ge \bar{y}_T$. Also set $e_T$ to be the unit vector parallel
to $T$ and pointing towards $z_T$. By the comments in Step 1,
$u(z_T)\ge u(\bar{z}_T)$, $u$ is linear on $T$ and $Du(x)=L e_T$
for any $x$ on $T$. Also let $\lambda(T)$ denote the length of $T$.
>From now on we will call such segments $T$, as described in Step 1, as
segments of type $S$.


\noindent\textbf{Step 2.}
By taking arbitrary points $a,\;b\in I-$, $a\ne b$ in \eqref{e11}(ii), we
see that the points $d$, on the arc $\widehat{ab}$ form a dense set in
the unit circle $\partial D$.
By Step 1, we obtain infinitely many such segments $T$ of type $S$.
By the uniqueness of gradient directions any two such segments
intersect if and only if they are identical. By the discussion preceding
Step 1, $pq$ is one such segment. It also
follows then that segments $T$ of type $S$ either lie
completely in $H+$ or in $H-$. Suppose that
$T_1$ and $T_2$ are two such non-overlapping segments
in $H-$ then one lies to the "left" of the other. More precisely,
if $y_{T_1}>y_{T_2}$, then
\begin{equation} \label{e12}
\bar{y}_{T_1}<\bar{y}_{T_2},\quad
\lambda(T_1)>\lambda(T_2),\quad
\mathop{\rm dist}(o,T_1)<\mathop{\rm dist}(o,T_2).
\end{equation}
An analogous property holds in $H+$.

\noindent\textbf{Step 3.}
 For $k=1,2,3,\dots$ let $T_k$ be a segment of type
$S$ in $H-$ such that $y_{T_k}\uparrow 1$. Since the end points
$z_{T_k}$ and $\bar{z}_{T_k}$ lie on the unit circle,
$z_{T_k}\to p$ and $x_{T_k}\uparrow 0$. Moreover by Step 2
and \eqref{e12}, $\bar{y}_{T_k}\downarrow y_{\infty}\ge -1$ and
$\bar{x}_{T_k}\to x_{\infty}$. Set
$e_{\infty}=(-x_{\infty},1-y_{\infty})/\sqrt{x_{\infty}^2+(1-y_{\infty})^2}$,
clearly, $e_{T_k}\to e_{\infty}$. Thus the segments $T_k$
tend to the segment $T_{\infty}$ with end points
$z_{T_{\infty}}=(0,1)$ and
$\bar{z}_{T_{\infty}}=(x_{\infty},y_{\infty})$. Also by Step 1,
for every $k$ and any $0\le t\le
\lambda(T_k)$, $u(z_{T_k}-te_{T_k})=u(z_{T_k})-t L$, and
$Du(z_{T_k}-te_{T_k})=Le_{T_k}$. Since $u$ is $C^1$ we see that
for any $0\le t\le\lambda(T_{\infty})$, $u(p-t
e_{\infty})=M-tL$, $Du(p-t e_{\infty})=Le_{\infty}$, and
$T_{\infty}$ is of type $S$. By the comments preceding Step 1,
$Du(p)=Le_2=Le_{\infty}$, and $(x_0,y_0)=q$. Thus the segments
$T_k$ move right to the segment $pq$. As noted in Step 1, since
the set of end points $z_T$ and $\bar{z}_T$, of segments $T$ of
type $S$, are dense in $\partial D$, it is clear now that we can always
find segments $T$ arbitrarily close to the segment $pq$ and lying
in $H-$.


\noindent\textbf{Step 4.}
Suppose now that there is an $a\in D$ such that
$|Du(a)|<L$, then there is a disc $D_{\varepsilon}(a)\subset D$ such that
$|Du(w)|<L$, $w\in D_{\varepsilon}(a)$. Since $D_{\varepsilon}(a)$ cannot intersect
the segment $pq$, it lies either in $H+$ or in $H-$. Assume that
$D_{\varepsilon}(a)\subset H-$. Let $\eta_a=a/|a|$ and $w_{\varepsilon}$ be the
point on $\partial D_{\varepsilon}(a)$ nearest to $o$, i.e.,
$w_{\varepsilon}=(|a|-\varepsilon)\eta_a$. By the comment made at the end of Step
3, there are segments $T$ of type $S$ that intersect the segment
$ow_{\varepsilon}$. These lie completely in $H-$. Consider now the set of
such segments $T$ and set $y_0$ to be the infimum of
$y_T$'s($y$-coordinates of the higher end points) of these
segments. Let $z_0=(x_0,y_0)\in I-$. Also by \eqref{e12}, the supremum
$\bar{y}_0$ of the $\bar{y}_T$'s($y$-coordinates of the lower end
points) of these particular segments exists. Clearly,
$\bar{y}_0\le y_0$; set $\bar{z}_0=(\bar{x}_0,\bar{y}_0)\in I-$.
By employing \eqref{e12}, one can easily find a sequence segments $T_k$
of type $S$, that intersect $ow_{\varepsilon}$, such that $T_k$'s tend to
the segment $z_0\bar{z}_0$, i.e., $e_{T_k}\to e$, where
$e=(z_0-\bar{z}_0)/|z_0-\bar{z}_0|$. Morever, since $u$ is $C^1$,
the straight segment $z_0\bar{z}_0$ is of type $S$, it intersects
$ow_{\varepsilon}$ and 
\begin{equation} \label{e13}
u(z_0-te)=u(z_0)-tL,\quad 
0\le t\le |z_0-\bar{z}_0|,\quad 
Du(z_0-te)=Le. 
\end{equation} 
Now let $T_k$
be segments of type $S$ with $z_{T_k}\to z_0$ (this is
possible by the density of $z_T$'s). We choose these to lie to the
left of $z\bar{z}$, i.e., $y_{T_k}\uparrow y_0$ (see above). By
the definition of $z_0$ and our assumption about $D_{\varepsilon}(a)$, the
segments $T_k$ neither intersect $ow_{\varepsilon}$ nor $D_{\varepsilon}(a)$. We
now consider the lower end points $\bar{z}_{T_k}$ of these
$T_k$'s. Since $y_{T_k}\le y_0$, \eqref{e12} implies that
$\inf_k\bar{y}_{T_k}>\bar{y}_0$ and $\inf_k\mathop{\rm dist}(o,
T_k)>\mathop{\rm dist}(o,z_0\bar{z}_0)$. Let $\bar{y}_1=\inf_k \bar{y}_{T_k}$
and $\bar{z}_1=(\bar{x}_1,\bar{y}_1)\in I-$. It follows easily
that the segment $z_0\bar{z}_1$ is type $S$. Let
$\bar{e}=(z_0-\bar{z}_1)/|z_0-\bar{z}_1|$, then $e\ne \bar{e}$
since $\bar{z}_0\ne \bar{z}_1$. It now follows that on the segment
$z_0\bar{z}_1$,
$$
u(z_0-t\bar{e})=u(z_0)-tL,\quad
0\le t\le |z_0-\bar{z}_1|,\quad
Du(z_0-t\bar{e})=L\bar{e}.
$$
By \eqref{e13}, $Du(z_0)=Le=L\bar{e}$ and we have a contradiction. Thus the
theorem holds and $|Du(w)|=L$, for all $w\in D$.

\section{Appendix}

We now prove a version of the Rolle's property in $\mathbb{R}^n$, $n\ge 3$.

\begin{lemma} \label{lem4}
Let $u$ be infinity-harmonic in $\Omega\subset \mathbb{R}^n$, $n\ge 3$.
Let $x,\;y\in \Omega$ and $\sigma(s)$, $0\le s\le 1$ be a
$C^1$ curve that
lies completely in $\Omega$ with $\sigma(0)=x$ and $\sigma(1)=y$.
Let $l(s)$ denote the arclength of the curve from $\sigma(0)$ to
$\sigma(s)$.
Then for some $0<\tau<1$, and vector $\omega_{\tau}\in
S^{n-1}$, we have
$$
u(y)-u(x)=\Lambda(\sigma(\tau))l(1)\langle\omega_\tau,
\sigma'(\tau)\rangle/|\sigma'(\tau)|.
$$
\end{lemma}

\begin{proof}
 The proof utilizes simple calculus ideas and
\eqref{e7}(i). Without any loss of generality, take $x=o$, $u(o)=0$, and
set $v(s)=u(\sigma(s))-u(y)l(s)/l(1)$, $0\le s\le 1$. Then $v(s)$
is continuous and $v(0)=v(1)=0$. Suppose that $v$ has a positive
maximum at some $0<\tau<1$. Thus
$u(\sigma(\tau))-u(y)l(\tau)/l(1)\ge
u(\sigma(s))-u(y)l(s)/l(1)$, $0\le s\le 1$, and
\begin{equation} \label{e14}
u(\sigma(s))-u(\sigma(\tau))\le u(y)(l(s)-l(\tau))/l(1),\quad 
0\le s\le 1.
 \end{equation}
Set $z=\sigma(\tau)$ and  $e=\sigma'(\tau)/|\sigma'(\tau)|$.
By \eqref{e7}(i), there exists a limit direction $\omega_{\tau}\in S^{n-1}$
and $r_k\downarrow 0$ such that $\lim_{r_k\downarrow
0}(u(z+r_k\omega_{\tau})-u(z))/r_k=\Lambda(z)$. Let
$z_k=z-r_ke$, $\xi_k=z+r_ke$; denote by $s_k$, the largest value of
$s\le \tau$ such that $\sigma(s)\in \partial B_{r_k}(z)$, and by
$\bar{s}_k$, the smallest value of $s\ge \tau$ such that
$\sigma(\bar{s}_k)\in \partial B_{r_k}(z)$. Since $\sigma$ is $C^1$ and
$u$ is locally Lipschitz, the following hold for small $r_k$:
 \begin{equation} \label{e15}
\begin{gathered}
|\sigma'(\tau)|(\tau-s_k),\quad
|\sigma'(\tau)|(\bar{s}_k-\tau)\approx r_k,\\\
|\sigma(s)-z|-|\sigma'(\tau)(s-\tau)|=o(|s-\tau|),\\
|\sigma(s_k)-z_k|,\quad |\sigma(\bar{s}_k)-\xi_k|=o(r_k),\\
|u(z_k)-u(\sigma(s_k)|,\quad |u(\xi_k)-u(\sigma(\bar{s}_k)|=o(r_k).
\end{gathered}
\end{equation}
 From \eqref{e14},
\begin{gather*}
\frac{u(\sigma(s_k))-u(z)}{r_k} \le -\frac{u(y)[l(\tau)-l(s_k)]}{l(1)r_k},\quad
\frac{u(\sigma(\bar{s}_k))-u(z)}{r_k}\le
\frac{u(y)[l(\bar{s_k})-l(\tau)]}{l(1)r_k}.
\end{gather*}
Using \eqref{e15} and taking limits in the above stated inequalities,
we obtain that
\begin{equation} \label{e16}
\begin{aligned}
\lim_{r_k\downarrow 0}\frac{u(\sigma(s_k))-u(z)}{r_k}
&=\lim_{r_k\downarrow 0}\frac{u(z_k)-u(z)}{r_k}=-\Lambda(z)\langle
\omega_{\tau},e\rangle\\
&\le \lim_{r_k\downarrow 0}-\frac{u(y)(l(\tau)-l(s_k))}{l(1)r_k
}=-\frac{u(y)}{l(1)}.
\end{aligned}
\end{equation}
 Using $\bar{s}_k$ and $\xi_k$, and
taking limits as in \eqref{e16}, we see that $\Lambda(z)\langle\omega_z,e\rangle
\le u(y)/l(1)$. The conclusion of the lemma holds. The analyses
when $v(s)=0$, for all $s>0$, or when $v(s)$ has a negative minimum
are analogous.
\end{proof}


\subsection*{Acknowledgments}
We thank the anonymous referee for several comments that have helped
simplify and improve this work.

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