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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 130, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/130\hfil
Extreme values of infinity-harmonic functions]
{Some properties of the extreme values of
infinity-harmonic functions}

\author[T. Bhattacharya\hfil EJDE-2009/130\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 4, 2009. Published October 9, 2009.}
\subjclass[2000]{35J60, 35J70}
\keywords{local behavior, extreme values}

\begin{abstract}
 We study local behavior of infinity-harmonic functions,
 in particular, the extreme values of such functions on
 a ball. We show that the extreme values obey certain
 relationships, and use this to derive an interior growth
 estimate.
\end{abstract}

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


\section{Introduction}

In this work we study some properties of infinity-harmonic
functions. The questions of local behavior and regularity form the
main motivation for our present work, and our hope is that the
results in this work will provide some insight into these matters.

We introduce notation for our discussion. We take
$\Omega\subset\mathbb{R}^n$, $n\ge 2$ to stand for a domain. Our
results, being local in nature, would apply even when $\Omega$ is
unbounded. We will often use $x,y,z$ to denote points on
$\mathbb{R}^n$, and $o$ will stand for the origin. We will
occasionally write a point as $x=(x_1,x_2,\dots ,x_n)$. A ball of
radius $r$ and center $x$ will be denoted by $B_r(x)$.

We define $u:\Omega\to \mathbb{R}$, to be infinity-harmonic in $\Omega$ if it
satisfies, in the sense of viscosity,
\begin{equation} \label{e1}
\Delta_{\infty}u=\sum_{i,j=1}^n \frac{\partial u}{\partial
x_i}\frac{\partial u}{\partial x_j}\frac{\partial^2 u}{\partial
x_i\partial x_j}=0\quad \text{in } \Omega.
\end{equation}
For motivation and a
detailed discussion of various properties of such solutions, see
\cite{acj,tb2,tdm, c,ceg,pssw}. An important aspect of these
functions is that they are completely characterized by the
``cone comparison property", a fact first observed in \cite{ceg}.
Like all other results, our work too exploits this property to
achieve its ends. We refer to \cite{acj,c,ceg} for a detailed
discussion of this issue.

Our motivation for this work is related to the question of local
regularity. Despite great recent progress the matter of local
regularity remains somewhat unresolved.
It is well-known that $u$ is locally Lipschitz continuous
\cite{tb2,c,ceg,j}, and it has been shown that $u$ is $C^{1,\alpha}$
when $n=2$, see \cite{es,s}. However, differentiability is
still open when $n\ge 3$. The work \cite{es} proves deep
results regarding this matter and states a conjecture,
whose proof would lead to $C^{1,\alpha}$ regularity of such functions.
Our goal in this work is to derive new properties of the extreme
values of such functions which we hope may lead to some
additional insight into this problem.

In what follows, we will limit our discussion to a ball
$B_r(x)$ in $\Omega$. Let $\overline{A}$ denote the closure
of a set $A$. Define $M_x(r)=\sup_{\bar{B}_r(x)}u $ and
$m_x(r)=\inf_{\bar{B}_r(x)}u$. We will drop the subscript if $x=o$.
It is well-known that $u$ satisfies the strong maximum principle,
and these values are attained on the boundary of the ball $B_r(x)$
\cite{tb5,ceg}. It is also well-known that $M_x(r)-u(x)$ and
$u(x)-m_x(r)$ are both convex in $r$, and the following limit exists:
\begin{equation} \label{e2}
 \lim_{r\downarrow 0}\frac{M_x(r)-u(x)}{r}
=\lim_{r\downarrow 0}\frac{u(x)-m_x(r)}{r}:=\Lambda(x).
\end{equation}
Moreover, the limit $\Lambda(x)$ would equal $|Du(x)|$, if $u$ is
differentiable at $x$, see \cite{acj,tb2,c,ce,ceg}.
In addition, it was shown in \cite{ce} that if $\omega\in S^{n-1}$
is such that $u(x+r\omega)=M_x(r)$ and the set of $\omega$'s has a
unique limit point as $r\downarrow 0$, then $u$ is differentiable
at $x$. It is also known that if $B_r(x)\subset\subset\Omega$,
then $u$ is differentiable at $x+r\omega$ \cite{tb2}.
Moreover, the limit $\Lambda(x)$ is upper semi-continuous \cite{acj,tb2,c,ceg}
and satisfies a maximum principle, \cite{tb1,ceg}.
An important question in this context is whether or not
infinity-harmonic functions, defined on all of $\mathbb{R}^n$
with bounded $\Lambda$, are affine. This was proven in $n=2$ in \cite{s},
as a consequence of differentiability. Some of the facts mentioned
above will play a role in this work.

From hereon we assume that $o\in \Omega$ and work in a ball
$B_R(o)\subset \Omega$. Additionally, we always take $u(o)=0$.
To make our presentation clearer, we redefine
$m(r)=-\inf_{\bar{B}_r(o)}u$, and unless otherwise mentioned,
from hereon we will always take this to be the definition of $m(r)$.
We now state the first of our main results.

\begin{theorem} \label{thm1}
Let $u$ be infinity-harmonic in $B_R(o)$. Suppose that $u(o)=0$;
define for $r\in[0,R)$, $m(r)=-\inf_{\bar{B}_r(o)}u$ and
$M(r)=\sup_{\bar{B}_r(o)}u$. The following two inequalities then hold:
\begin{itemize}
\item[(a)] $m(r)\ge \Big( \sqrt{rM'(r-)}-\sqrt{rM'(r-)-M(r)}\Big)^2$
for $r\in[0,R)$;
\item[(b)] $M(r)\ge \Big( \sqrt{rm'(r-)}-\sqrt{rm'(r-)-m(r)}\Big)^2$ 
for $r\in[0,R)$. 
\end{itemize}
Moreover, if $u$ is infinity-harmonic in
$\mathbb{R}^n$ and equality holds in both the inequalities for
every $r>0$, then $u$ is affine.
\end{theorem}

A proof appears in Section 2.  An easy consequence of
Theorem \ref{thm1} are
the inequalities,
$$
m(r)M'(r-)\ge \frac{M(r)^2}{4r} \quad\mbox{and}\quad
M(r)m'(r-)\ge \frac{m(r)^2}{4r},
$$
see Remark \ref{rmk1}. The inequalities in Theorem \ref{thm1} place no 
restrictions
on how small, for instance, $m(r)$ could be. As a matter of fact
we construct an example in Lemma \ref{lem1} in Section 2 which supports
this observation.
Using the inequalities in Theorem \ref{thm1}, we also prove the following
growth rate in Theorem \ref{thm2}.

\begin{theorem} \label{thm2}
Suppose that $u$ is infinity-harmonic in $B_1(o)$ and $u(o)=0$.
Define for $r\in[0,1)$ $m(r)=-\inf_{\bar{B}_r(o)}u$ and
$M(r)=\inf_{\bar{B}_1(o)}u$.

\noindent (i) Suppose that $ M(r)\le r$ for every $r\in[0,1)$.
Then either $m(r)\le r$ for every $r\in[0,1)$, or there
is an $a\in(0,1)$ such that $m(a)>a$ and
$$
m(r)\ge r(1+k\log(r/a)),\;\forall\;a< r< 1,
$$
where $k=(m(a)-a)^2/4a^2$,

\noindent (ii) Analogously, if $m(r)\le r$ for every $r\in[0,1)$,
then either $M(r)\le r$ for every $r\in[0,1)$, or $M(a)>a$ for
some $a\in(0,1)$ and
$$
M(r)\ge r(1+k\log(r/a)),\;\forall\;a< r< 1,
$$
where $k=(M(a)-a)^2/4a^2$.
\end{theorem}

We provide a proof in Section 2. It is to be noted that the hypothesis
$M(r)\le r$, or for that matter $m(r)\le r$, is not restrictive.
One can scale $u$ to obtain this inequality, see Remark \ref{rmk2}
for a general version.

We now bring up a related matter.  If $u$ is infinity-harmonic
in $B_1(o)$, by convexity, $M'(r+)$ and $M'(r-)$ exist everywhere
and $M'(r)$ exists for almost every $r$. As pointed out
before (see \cite{tb2}), if $p\in \partial B_r(o)$ for $r<1$, is a point
of maximum then $u$ is differentiable at $p$ and 
$$
M'(r-)\le |Du(p)|\le M'(r+).
$$
Moreover, as Lemma \ref{lem2} shows there are points $p$ on $\partial B_r(o)$
where the values $M'(r+)$ and $M'(r-)$ are attained by $|Du(p)|$.
Our question is: does $M'(r)$ exist at every $r\in[0,1)$,
or equivalently, is $M'(r-)=M'(r+)$ ? Clearly, this equality holds
at $r=0$. We have been unable to settle the matter, however, the
example in Lemma \ref{lem4} shows that these may disagree on the boundary
$\partial B_1(o)$ in the event $u\in C(\bar{B}_1(o))$.
In particular, we show that $M'(r)$ exists for every $r\in[0,1)$
but there are points $p$ on $\partial B_1(o)$ such that $u(p)=M(1)$,
and the one sided gradient $\Lambda(p)>M'(1-)$. As our calculations will
show this difference can be made arbitrarily large.

We have divided our work as follows. Proofs of
Theorems \ref{thm1} and \ref{thm2},
and Lemma \ref{lem1} appear in Section 2. Section 3 contains proofs
of Lemmas  \ref{lem2}, \ref{lem3} and \ref{lem4}. In this work, 
all sets are subsets
of $\mathbb{R}^n,\;n\ge 2$, unless otherwise mentioned.

\section{Proofs of Theorems \ref{thm1} and \ref{thm2}}

We first state a few properties of infinity-harmonic functions
that could be thought of as "monotonicity" along radial
segments, see \cite{tb4,c}. For completeness, we provide short
proofs of these, also see Exercise 16 in \cite{c}.
The proofs will use the comparison principle \cite{acj,bb,c,cgw,j}.

Let $v$ be a positive infinity-harmonic function in a domain $\Omega$
and $B_{\rho}(o)\subset  \Omega$. Take $y\in B_{\rho}(o)$ and let
$d>0$ be chosen to be any value that does not exceed the distance
of $y$ from $\partial \Omega$.
By comparing $v(x)$ to the infinity-harmonic function $v(y)(1-|x-y|/d)$,
we see
$$
v(x)\ge v(y)\Big(1-\frac{|x-y|}{d}\big),\quad \text{for }x\in B_d(y),
$$
In this discussion, we refer to the above inequality as
{\em cone comparison in $B_{d}(y)$}.
Taking $d=\rho-|y|$, a rearrangement leads to the inequality
$$
\frac{v(y)-v(x)}{|x-y|}\le \frac{v(x)}{d-|x-y|}.
$$
It is clear by using \eqref{e2} that $\Lambda(y)\le v(y)/d$,
for every $y\in B_{\rho}(o)$. Moreover, if $\omega\in S^{n-1}$,
by selecting the points $x$ and $y$ on the radial ray, in
$B_{\rho}(o)$ and along $\omega$, it is clear that
$ v(\theta \omega)/(\rho-\theta)$ is an increasing function of
$\theta$ in $[0,\rho)$.

Next we show that $v(\theta\omega)(\rho-\theta)$ is decreasing.
To see this, take $x=\theta\omega$ with $\theta>0$, small,
so that $B_{\rho}(x)\subset \Omega$ (one may need the assumption
$\bar{B}_{\rho}(o)\subset \Omega$, but this is not restrictive).
 Using cone comparison in $B_{\rho}(x)$, we have
$v(x)(\rho-|x|)\le v(o)\rho$. Taking $y=\bar{\theta}\omega$
with $\bar{\theta}>\theta$ and $\bar{\theta}$ close to $\theta$,
 cone comparison in $B_{\rho-|x|}(y)$, leads to
$v(x)(\rho-|x|)\ge v(y)(\rho-|y|)$. Proceeding in this way
we obtain our claim. An alternative is to use the Harnack
inequality \cite{tb4}. Collecting our conclusions,
for any $\omega\in S^{n-1}$ and $\theta\in[0,\rho)$,
\begin{equation} \label{e3}
\parbox{9cm}{
(i) $v(\theta\omega)/(\rho-\theta)$ is increasing in $\theta$,
\\
(ii)  $v(\theta\omega)(\rho-\theta)$ is decreasing in $\theta$, and
\\
(iii) $\Lambda(y)\le v(y)/(\rho-|y|)$, for all
$y\in B_{\rho}(o)$.}
\end{equation}
For applications, we will often use these properties with
$v=M-u$ or $v=u-l$, where $M$ and $l$ are any numbers with
$M\ge \sup_{\bar{B}_{\rho}(o)}u$ and $l\le \inf_{\bar{B}_{\rho}(o)}u$.
We also note another property. For $r\le \rho$, if
$\gamma,\bar{\gamma}\in S^{n-1}$ are such that
$v(r\gamma)=\sup_{\bar{B}_r(o)}v$ and
$v(r\bar{\gamma})=\inf_{\bar{B}_r(o)}v$, then $u$ is differentiable
at $r\gamma$ and $r\bar{\gamma}$ and
\begin{equation} \label{e4}
Du(r\gamma)=\alpha \gamma,\quad
Du(\bar{\gamma}r)=\beta\bar{\gamma}.
\end{equation}
for some $M'(r-)\le \alpha\le M'(r+)$ and $m'(r-)\le\beta\le m'(r+)$.
For a proof see \cite{tb2}, also see Lemma \ref{lem2} for a related statement.

We thank the referee for his/her suggestions that have simplified
the following proof.

\begin{proof}[Proof of Theorem \ref{thm1}]
  We prove inequality (a) of the theorem, as inequality (b) follows
by symmetry. First observe that
$(r-t)(M(t)+m(r))$ is non-increasing in $t$. This may be argued by
using 3(ii) with $v=u+m(r)$. Alternatively, one could use
\cite[Lemma 4.6]{c} or \eqref{e3}(iii) to first derive the following
differential inequality
$$
M'(t+)\le \frac{M(t)+m(r)}{r-t},
$$
and then conclude the same. In any case, since $u(o)=0$,
$$
M(t)\le \Big(\frac{t}{r-t}\Big)m(r),\quad 0\le t<r.
$$
Combining this with the convexity property $M(t)\ge M(r)-(r-t)M'(r-)$,
we have
\begin{equation} \label{e5}
\Big(\frac{r-t}{t}\Big) \Big(M(r)-(r-t)M'(r-)\Big)\le m(r),\quad 0<t\le r.
\end{equation}
Our idea is to select a value of $t$ that optimizes the above
inequality in $0<t\le r$. To this end define
$$
f(t)=\Big(\frac{r-t}{t}\Big)\Big(M(r)-(r-t)M'(r-)\Big)\quad 0<t\le r,
$$
and observe that $f(r)=0,\;f(t)\ge 0$, for $t$ near $r$, and 
$\lim_{t\downarrow 0}f(t)< 0$ unless $M(r)=rM'(r-)$.
It is easy to show that $f$ is optimized  when
$$
t_0=\Big(\frac{r[rM'(r-)-M(r)]}{M'(r-)}\Big)^{1/2}.
$$
As $rM'(r-)\ge M(r)$ by convexity, we have $0\le t_0\le r$.
Inserting the value of $t_0$ in \eqref{e5} the first inequality of the
theorem follows,
\begin{equation} \label{e6}
m(r)\ge\left (\sqrt{rM'(r-)}-\sqrt{rM'(r-)-M(r)}\right)^2.
\end{equation}
We discuss briefly the cases when $t_0=0$ and $t_0=r$. If $t_0=0$
then $rM'(r-)=M(r)$, and convexity leads to  $M(t)=tM(r)/r$.
If $t_0=r$, then $M(r)=0$ leading to $u= 0$
in $B_r(o)$. Also see  \cite[Lemma 3.1]{tb1}.

Let us now assume that $u$ is infinity-harmonic in $\mathbb{R}^n$
and equality holds every where, in the two inequalities of the
theorem. We use an equivalent form of \eqref{e6},
\begin{equation} \label{e7}
M(r)^2\le m(r)\left(\sqrt{rM'(r-)}+\sqrt{rM'(r-)-M(r)}\right)^2.
\end{equation}
With equality in place in \eqref{e7} and replacing $rM'(r-)$
by the smaller quantity $M(r)$ and disregarding the
second term on the right hand side, we deduce that $M(r)\ge m(r)$.
Analogously, by exploiting inequality (b) of the theorem,
$m(r)\ge M(r)$, which  leads to $M(r)=m(r)$. Next, we use this
in \eqref{e6} and \eqref{e7}, with equality in place, and compare
the two right hand sides to conclude that $rM'(r-)=M(r)$.
This leads to $M(r)=m(r)=kr$, for some $k>0$. Now applying
the ``tight on a line" result in Section 7.2 in \cite{c},
it follows that $u$ is affine.
\end{proof}

\begin{remark} \label{rmk1} \rm
We discuss briefly the inequalities in Theorem \ref{thm1}. From \eqref{e7},
it follows easily that $m(r)M'(r-)\ge M(r)^2/4r$,
and, similarly, $M(r)m'(r-)\ge m(r)^2/4r$.
Next in \eqref{e7}, we set $x=rM'(r-)$ and $y=M(r)$, and
use an expansion for $\sqrt{x-y}$.
Working with
\[
\sqrt{x-y}\le\sqrt{x}-\frac{y}{2x^{1/2}}-\frac{y^2}{8x^{3/2}}
-\frac{y^3}{16x^{5/2}},\quad 0<y\le x,
\]
and, for instance, using the first two terms and convexity,
$$
rM'(r-)\ge \frac{M(r)^{3/2}}{2m(r)^{1/2}}+\frac{M(r)}{4}.
$$
An analogue, from the second inequality in Theorem \ref{thm1}, can be
easily worked out.
\end{remark}

Next we construct an example to show that the inequalities
of Theorem \ref{thm1} place no restrictions on the lower bounds of
 $m(r)$ and $M(r)$.  First we discuss some preliminaries.

Our construction will involve the Aronsson singular example, and
we recall below some of its properties that will be used
in Lemma \ref{lem1}. For a more detailed discussion, see \cite{a},
\cite[Lemmas 2.5, 2.6, 2.9]{tb3}, and (b) in part II of the appendix in
\cite{tb4}. We use $x=(x_1,\dots ,x_n)=(\bar{x},x_n)$ to denote a
point in $\mathbb{R}^n$. Let $e_n$ be the unit vector along the
positive $x_n$-axis and $\theta=\cos^{-1}(\langle x,e_n\rangle/|x|)$,
the angle made with the $x_n$-axis. The Aronsson example is given
by $v(x)=\psi(\theta)/|x|^{1/3}$, where $v(x)$ is infinity-harmonic
in $\mathbb{R}^n\setminus\{o\}$. We list only what we need. For
$x\ne o$,
\begin{equation} \label{e8}
\parbox{9cm}{
(i) $\psi(\theta)>0$ if $x_n>0$, and $\psi(\pm\pi/2)=v(\bar{x},0)=0$;\\
(ii) $\sup_{\partial B_r(o)}v=\psi(0)/r^{1/3}$;\\
(iii) $\psi(\theta)=\psi(-\theta)$, and $\psi(\theta)$ is decreasing in
$[0,\pi/2]$.}
\end{equation}

\begin{lemma} \label{lem1}
Let  $\varepsilon>0$ and $B_1(o)\subset\mathbb{R}^n$, $n\ge 2$. There
exists a function $u_\varepsilon\in C(\bar{B}_1(o))$, infinity-harmonic in
$B_1(o)$ with $u_{\varepsilon}(o)=0$ and $\sup_{\bar{B}_1(o)}u_{\varepsilon}=1$,
such that $\inf_{\bar{B}_1(o)}u_{\varepsilon}\ge -\varepsilon$.
\end{lemma}

\begin{proof} To construct our example $u_{\varepsilon}$ on $B_1(o)$,
we employ a translate of the Aronsson singular function $v$ and
use the properties stated in \eqref{e8}. Let $\delta>0$ and let
$p_{\delta}=(0,0,0,\dots ,-(1+\delta))$. Define
$$
v_{\delta}(x)=\frac{\psi(\theta)}{|x-p_{\delta}|^{1/3}}
=\frac{\psi(\theta)}{|x+(1+\delta)e_n|^{1/3}},\quad
\theta=\theta(x)=\frac{\langle x-p_{\delta},\;e_n\rangle}{|x-p_{\delta}|}.
$$
We scale $\psi(0)=1$. By \eqref{e8}(i) $v>0$ on $B_1(o)$, and by
\eqref{e8}(ii) and (iii),
$\sup_{\partial B_{\delta}(p_{\delta})}v_{\delta}=\sup_{B_1(o)}v=1/\delta^{1/3}$.
Next for $x\in B_1(o)$, define
\begin{align*}
w_{\delta}(x)
&=\frac{\delta^{1/3}(1+\delta)^{1/3}(v_{\delta}-v(o))}{(1+\delta)^{1/3}
-\delta^{1/3}}\\
&=\frac{\delta^{1/3}(1+\delta)^{1/3}}{(1+\delta)^{1/3}-\delta^{1/3}}
\Big(\frac{\psi(\theta)}{|x+(1+\delta)e_n|^{1/3}}-\frac{1}{(1+\delta)^{1/3}}\Big).
\end{align*}
Then $w_{\delta}$ is infinity-harmonic in $x_n>-(1+\delta)$ and clearly so
in $B_1(o)$. Also $\sup_{B_1(o)}w_{\delta}=1,\;w_{\delta}(o)=0$ and
$w_{\delta}\in C(\bar{B}_1(o))$. Noting that $\psi(\theta)>0$
when $x\in B_1(o)$
(see \eqref{e8}(i)), we see that
$$
0\ge \inf_{\bar{B}_1(o)}w_{\delta}\ge
\frac{-\delta^{1/3}}{(1+\delta)^{1/3}-\delta^{1/3}}.
$$
Choosing $\delta$ small enough, we obtain our desired
infinity-harmonic function $u_{\varepsilon}$.
\end{proof}

Next we present a proof of the growth estimate in Theorem \ref{thm2}.
We utilize the inequalities proven in Theorem \ref{thm1}.

\begin{proof}[Proof of Theorem \ref{thm2}]
 We will only prove part (i), part (ii) will follow analogously.
To achieve our goal we use inequality (b) of Theorem \ref{thm1}.
Using the hypothesis, we note
\begin{equation} \label{e9}
\Big(\sqrt{rm'(r-)}-\sqrt{rm'(r-)-m(r)}\Big)^2\le r,\quad
0\le r<1.
\end{equation}
Let us assume that $m(a)> a$, for some $0<a<1$.
By the convexity of $m(r)$,
$$
m'(r-)\ge m'(a-)> 1,\quad
m(r)>r,\quad \forall\;a<r<1.
$$
Observing that $2r m'(r-)\ge r+m(r)$, squaring \eqref{e9},
rearranging terms and squaring again we obtain for $r>a$,
\begin{equation} \label{e10}
m'(r-)\ge \frac{(m(r)+r)^2}{4r^2}.
\end{equation}
Setting $w=m(r)/r$ in the second inequality in \eqref{e10},
we obtain the differential inequality $4rw'\ge (w-1)^2$.
An integration from $c$ to $r$, for any $a\le c< 1$ yields
$$
\frac{c}{m(c)-c}-\frac{r}{m(r)-r}\ge \frac{\log(r/c)}{4},\quad
c\le r<1.
$$
Noting that $m(r)-r\ge m(c)-c$, a further rearrangement yields
\[
\frac{m(r)}{r}\ge\frac{m(c)}{c}+\Big(\frac{m(c)-c}{c}\Big)^2
\frac{\log(r/c)}{4},\quad c\le r<1.
\]
Selecting $k=(m(a)-a)^2/4a^2$ we have
$$
m(r)\ge \frac{r}{a} m(a)+k r\log(r/a)\ge r+k r\log(r/a),\quad
a\le r<1.
$$
The theorem follows.
\end{proof}

\begin{remark} \label{rmk2}\rm
 Firstly, we note that if $M(1)=1$, by convexity $M(r)\le r$.
We now state Theorem \ref{thm2} for the general case.  
If $v$ is infinity-harmonic in $B_R(o)$ and $\sup_{B_R(o)}v<\infty$,
then we scale $v$ and define $u(y)=(v(x)-v(o))/(\sup_{B_R(o)}v-v(o))$,
where $y=x/R$.
Then $\sup_{B_1(o)}u=1$ and $u$ satisfies the conditions of
Theorem \ref{thm2}.
Thus either
$$
\inf_{B_r(o)}v\ge v(o)-\frac{r}{R}(\sup_{B_R(o)}u-u(o)),\quad
\forall\;r\in[0,R),
$$
 or for some $0<t<R$,
\[
\inf_{B_r(o)}v\le v(o)-\frac{r}{R}\left(1+k\log\frac{r}{t}\right)
(\sup_{B_R(o)}v-v(o)),\quad \forall\;r\in[t,R),
\]
where
\[
k=\frac{R^2}{4t^2}\Big(\frac{v(o)-\inf_{B_t(o)}v
 -\frac{t}{R}(\sup_{B_R(o)}v-v(o))}{\sup_{B_R(o)}v-v(o)}\Big)^2.
\]
\end{remark}

\section{Results about $M'(r)$}

 Our effort in this section is to construct an example in connection
with the question raised, in Section 1, about $M'(r)$.
To this end, we start with Lemma \ref{lem2}. In this connection,
recall the definition of $\Lambda$ in \eqref{e2} and the
result in \eqref{e4}.

\begin{lemma} \label{lem2}
Let $u$ be infinity-harmonic in $B_R(o)$. For $0<r<R$,
let $p\in \partial B_r(o)$ be a point of maximum of $u$
on $B_r(o)$,  then $u$ is differentiable at $p$ and
$$
\Lambda(o)\le M'(r-)\le |Du(p)|\le M'(r+).
$$
 Moreover, there are points $p^+$ and $p^-$ on $\partial B_r(o)$
such that
$u(p^+)=u(p^-)=M(r)$, with $|Du(p^+)|=M'(r+)$ and $|Du(p^-)|=M'(r-)$.
 A similar result holds for $m'(r-)$ and $m'(r+)$.
 \end{lemma}

\begin{proof}
 For the inequality in the lemma, see \cite[Remark 2]{tb2}.
Fix $r<R$ and let $p^+\in \partial B_r(o)$ denote a limit
point of a sequence of points of maximum $p\in \partial B_{\rho}(o)$
with $\rho>r$ and $\rho\downarrow r$. Clearly for each
$p\in\partial B_{\rho}(o)$, $|Du(p)|\ge M'(\rho-)$.
Using the convexity of $M(r)$ and the upper semicontinuity
of $\Lambda$ \cite{tb2, ceg}, we see that $|Du(p^+)|\ge M'(r+)$.
 Hence equality follows.

Now let $p^-$ be a limit point of a sequence of such points
$p\in \partial B_{\rho}(o)$ with $\rho<r$ and $\rho\uparrow r$.
Applying \eqref{e3}(i) to $M(\rho)-u$ in $B_{\rho}(o)$,
noting \eqref{e4} and the inequality in the lemma, we see that
for a fixed $0<t<1$,
$$
\frac{M(\rho)-u(tp)}{\rho(1-t)}\le |Du(p)|\le M'(\rho+).
$$
Letting $\rho\uparrow r$ and selecting a subsequence if needed,
$$
\frac{M(r)-u(tp^-)}{r(1-t)}\le M'(r-).
$$
Letting $t\uparrow 1$, using \eqref{e4} and the inequality in
this lemma, we obtain  $|Du(p^-)|=M'(r-)$.
\end{proof}

We introduce additional notations before stating Lemma \ref{lem3}.
For $\alpha\in (0,\pi)$, let $C_{\alpha}\subset \partial B_1(o)$
denote the spherical cap, of aperture $2\alpha$, centered on
the negative $x_n$-axis. We state an estimate which is based
 on the Aronsson singular example \cite{pssw}.

\begin{lemma} \label{lem3}
Let $u\in C(\bar{B}_1(o))$ be infinity-harmonic
and $0\le u\le 1$. Suppose that $0<\alpha<\pi$, and $u(x)=1$
 on $C_{\alpha}$. If $\pi-\alpha$ is small then
$$
u(x)\ge 1-\frac{c(\pi-\alpha)^{1/3}}{|x-(1+k(\alpha))e_n|^{1/3}},\quad
x\in B_1(o),
$$
for some positive constants $c$, independent of $n$ and $\alpha$, and
$k(\alpha)$, where $k(\alpha)\to 0$ as $\alpha\to \pi$.
\end{lemma}

For $0<\alpha<\pi$, let $C_{\alpha}$ be the spherical cap as described
above and $\hat{C}_{\alpha}\subset \partial B_1(o)$ be the complementary
spherical cap of aperture $2\pi-2\alpha$. We write the point $x$
as $x=(\bar{x},x_n)$, and define
$|\bar{x}|=(\sum_{i=1}^{n-1}x_i^2)^{1/2}$. We will study
functions $u\in C(\bar{B}_1(o))$, infinity-harmonic in $B_1(o)$, with
\begin{equation} \label{e11}
u(o)=0,\quad u<1\quad \text{in } B_1(o)\quad\text{and}\quad
u=\phi\quad \text{on }\partial B_1(o),
\end{equation}
where $\phi(x)=\phi(|\bar{x}|,x_n)$ is axially symmetric about
$x_n$ axis, $\phi=1$ on $C_{\alpha}$, and $\phi(x)$ is decreasing
in $x_n$ for $x\in \hat{C}_{\alpha}$.
Such functions $u$ are easily constructed. Our chief interest
is their behavior when $\alpha$ is near $\pi$. As we will see
in Lemma \ref{lem4}, when $r\in[0,1)$, $M'(r)$ exists for $r<1$.
However, this does not extend to the boundary and the disparity
 between the one-sided gradients at points of maximum on $r=1$
can be made quite large by making $\alpha$ close to $\pi$.
Before we state Lemma \ref{lem4}, we mention that if $\omega$ is such
that $u(r\omega)=M(r)$ then the limit
$$
\lim_{t\uparrow r}\frac{M(r)-u(t\omega)}{r-t}=L(r\omega),
$$
exists (may be unbounded) \cite{tb2}.
This will be referred to as the one sided gradient at $r\omega$ and
equals $|Du(r\omega)|$ if $r<1$.

\begin{lemma} \label{lem4}
For $0<\alpha<\pi$, let $C_{\alpha}$, $\hat{C}_{\alpha}$ and $L$ be as
described above. Suppose that $u$ is an infinity-harmonic
function that satisfies \eqref{e11} and $\pi-\alpha$ is sufficiently small.
The following then hold.
\begin{itemize}
\item[(i)] For $r\in[0,1]$, $m(r)=-u(re_n)$, and as $\alpha\to \pi$, $m(1)$ and
$m'(1-)$ become unbounded.

\item[(ii)] For every $0\le r< 1$, $M(r)=u(-re_n)\ge 1-(1-r)|\sec \alpha |$,
$M'(r)$ exists and $1\le M'(1-)\le |\sec\alpha|$.
\end{itemize}
Moreover, if $p$ is any point on the boundary of $C_{\alpha}$, relative
to $\partial B_1(o)$, then the one sided gradient $L(p)\to \infty$
as $\alpha\to \pi$
\end{lemma}

\begin{proof}
  Before proving parts (i) and (ii),  we state some symmetry-related
properties of the the solution $u$, obtained by utilizing
reflections and the comparison principle.  Since $u$
satisfies \eqref{e11}, we note that $m(1)$ is attained
at $x=e_n$. Moreover, by using reflection about any $n-1$ plane,
containing the $x_n$  axis, and the comparison principle
it follows that $u$ is axially symmetric about $x_n$-axis.
For a set $A$, define $-A=\{-x:\;x\in A\}$.

Let $\kappa$ denote a unit vector with $\kappa_n\ge 0$ and
$P_{\kappa}$ be the $n-1$ dimensional plane passing through $o$
and having $\kappa$ as its normal. Define the half-space
$P^+_{\kappa}$ to be the set of all points $x\in \mathbb{R}^n$
with $\langle x, \kappa \rangle\ge 0$.  Define the half-space
$P^-_{\kappa}=-P^+_{\kappa}$. We now prove a monotonicity property
of $u$ to be used later. Now recall that $u=1$ on $C_{\alpha}$ and
$u<1$, and consider $u$ in the half ball $B_1(o)\cap
P^-_{\kappa}$. Using its reflection about $P_{\kappa}$ and
comparing with $u$, in $B_1(o)\cap P^+_{\kappa}$, we claim
\begin{equation} \label{e12}
\text{for all $y\in P_{\kappa}\cap B_1(o)$ and $t>0$ with
$y\pm t\kappa\in B_1(o)$, $u(y-t\kappa)\ge u(y+t\kappa)$.}
\end{equation}
The assertion in \eqref{e12} follows quite easily if the
plane $P_{\kappa}$ does not intersect $\hat{C}_{\alpha}$.
In case it does, we recall that $\phi$ is axially symmetric
and decreasing in $x_n$. These properties allows us to compare
boundary data after reflection and \eqref{e12} follows.

Let $S$ be a great semicircle, centered at $o$, with end points
$re_n$ and $-re_n$.
Applying \eqref{e12} with suitable planes $P_{\kappa}$ one
sees that for $x\in S$, $u(x)$ decreasing in $x_n$.
Thus for every $r\in(0,1)$, $u(re_n)=-m(r)$ and $M(r)=u(-re_n)$.
We have thus proven the first parts of (i) and (ii) but for
the estimates.

To prove the rest of part (i), we note that the function
$$
v(x)=\frac{u(x)+m(1)}{1+m(1)}
$$
satisfies the hypothesis of Lemma \ref{lem2}.
For $\pi-\alpha$ small,
 $v(o)\ge 1-c(\pi-\alpha)^{1/3}$. Since $u(o)=0$,
it is clear that for $\alpha$ close to $\pi$,
$m(1)\ge \hat{c}(\pi-\alpha)^{-1/3}$, for some $\hat{c}>0$.
Next by taking $x=re_n$ and applying the inequality in
Lemma \ref{lem3} to $v$, we see
\[
m(1)-m(r)\ge(1+m(1))\Big(1-\frac{c(\pi-\alpha)^{1/3}}{(1+k(\alpha)-r)^{1/3}}
\Big).
\]
 Using the convexity of $m$ the above yields, for any $0<r<1$,
\[
m'(1-)\ge \frac{m(1)-m(r)}{1-r}\ge \frac{1}{2}
\Big(\frac{1+m(1)}{1-r}\Big),
\]
as $\alpha\to \pi$. The conclusion for $m'(1-)$ follows.

Next  we show that $u\ge 0$ on a large portion of $B_1(o)$.
For $x\in B_1(o),\;x\ne o$, let us denote by $\Pi(x)$ the $n-1$
dimensional plane passing through $x$ with $x/|x|$ as  its
normal vector. By the set $T$, let us denote those points
$x\in B_1(o)$ such that $x/|x|\in C_{\alpha-\pi/2}$. For such
$x$'s $\Pi(x)$ does not intersect $\hat{C}_{\alpha}$.
Recall that $u=1$ on $C_{\alpha-\pi/2}$. Let $x\in T$,
using reflection about $\Pi(x)$ and the comparison principle,
one notes that  for $t>0$, $u(x+tx/|x|)$ is increasing in $t$.
Since $u(o)=0$, we have $u(x)\ge 0$ whenever $x\in T$.

We now prove the rest of the lemma. From the foregoing,
$B_{|\cos \alpha|}(-e_n)\cap B_1(o)\subset T$. If we set
$w(x)=(1-|x+e_n| |\sec \alpha|)$ then by comparison
$u(x)\ge w(x),\;x\in B_{|\cos \alpha|}(-e_n)\cap B_1(o)$, and
$M(r)=u(-re_n)\ge 1-(1-r)|\sec\alpha |$. Working with $1-u$,
applying \eqref{e3} (i) and convexity, the one-sided gradient
$M'(1-)$ exists and
$$
1\le M'(1-)=\lim_{r\uparrow 1}\frac{1-u(-re_n)}{1-r}
\le |\sec\alpha|\to 1\quad \text{as }\alpha\to \pi.
$$
This proves part (ii).

To show the last part, recall from (i) that $m(r)=-u(re_n)$. To
estimate the one-sided gradient $L(p)$, we apply \eqref{e3}(i) to
$1-u$ near $p$, \eqref{e4} and the Harnack inequality
\cite{acj,tb4,lm} to conclude
\[
L(p)\ge \frac{1-u(p-\varepsilon p)}{\varepsilon} \ge
\frac{1+m(1-\varepsilon)}{\varepsilon}\exp
\Big(\frac{-(1-\varepsilon)(\pi-\alpha)}{\varepsilon}\Big).
\]
 We may take $\pi-\alpha=\varepsilon$ and the conclusion now follows by
taking $\varepsilon\to 0$.
\end{proof}

\subsection*{Acknowledgments}
We thank the referee for several comments and suggestions that
have led to a clearer presentation.
This work was partially supported by a Junior Faculty Scholarship
Award funded by Western Kentucky University.


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\end{document}