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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2002(2002), No. 101, pp. 1--22.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or 
http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2002 Southwest Texas State University.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2002/101\hfil   $\infty$-harmonic functions]
{On the properties of $\infty$-harmonic functions
 and an application to capacitary convex rings}

\author[Tilak Bhattacharya\hfil EJDE--2002/101\hfilneg]
{Tilak Bhattacharya}

\address{ Tilak Bhattacharya \hfill\break
Mathematics Department\\
Bishop's University \hfill\break
Lennoxville, Quebec J1M 1Z7, Canada}
\email{tbhattac@ubishops.ca}

\date{}
\thanks{Submitted August 17, 2002. Published November 28, 2002.}
\subjclass[2000]{35J70, 26A16}
\keywords{Viscosity solutions, boundary Harnack inequality,
  $\infty$-Laplacian, \hfill\break\indent
  capacitary functions, convex rings}

\begin{abstract}
  We study positive $\infty$-harmonic functions in bounded domains. 
  We use the theory of viscosity solutions in this work. We  prove a 
  boundary Harnack inequality and a comparison result for such
  functions near a flat portion of the boundary where they vanish. 
  We also study $\infty$-capacitary functions on convex rings. 
  We show that the gradient satisfies a global maximum  principle, 
  it is nonvanishing outside a set of measure zero and the level sets
  are star-shaped.
\end{abstract}

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


\section{Introduction}

 This is a continuation of the work in [4] and,
while we derive a chain of results for $\infty$-harmonic functions, our
primary effort in this work will be to prove two
sets of results. The first would be for nonnegative $\infty$-harmonic
functions, which vanish on a flat portion of the boundary of the set in
which they are defined, and the
second will be for $\infty$-capacitary functions in convex
rings. More precisely, the first result discusses the behaviour of
nonnegative $\infty$-harmonic functions near flat boundaries, on which
they vanish, and we prove that any two such $\infty$-harmonic
functions vanish at the same rate. In the second set of results, we
show the non-vanishing of the gradient of $\infty$-capacitary
functions on convex rings and the star-shapedness of the level sets of
such functions. Clearly, the results are quite different in nature,
however, the techniques used have a lot in common. A more detailed
discussion follows in Section 2. We now comment on the approach used
in this work. Our work utilizes the notion of a viscosity solution in
this context and relies on techniques developed
in [3,4,7,11,12,13,17,19-22,25,30].
While our results are motivated by the results in [15,18,27,30-35], which are
about the weak solutions of the analogous problems with the
$p$-Laplacian, for finite $p$, we do not work with approximating weak
solutions as has been done in [5,16,18,23,29,30,31,33,34]. The idea in these
works was to take the limit as
$p\rightarrow \infty$ to capture properties and estimates for the
$\infty$-harmonic functions.
Instead our approach is closer to the works in
[3,4,12,13,21,25,30].
Our intention is to use the framework of
viscosity solutions to provide simpler and direct proofs. In this
context we also refer the reader to the works in
[3,4,13,22,24]. There is some overlap between our current work and
[13]. This latter work contains, at times, finer and more detailed
versions of some of the results proven here.

\section{Notation and statements of the main results}

We now introduce the notations we will be using in this work.
These will be employed faithfully
throughout this work with perhaps minor modifications for local use.
By $\Omega$, we
will always mean a bounded domain in $\mathbb{R}^n$, $n\ge 2$, and $\bar{A}$
will stand for the closure of a set $A$ in $\mathbb{R}^n$.
The letter $O=(0,0,\ldots,0)$ will always
stand for the origin in
$\mathbb{R}^n$; for a point $x=(x_1,x_2,\ldots,x_n)\in \mathbb{R}^n$, define
$\xi=\xi(x)=(x_1,x_2,\ldots,x_{n-1})$ and $x_n(x)=x_n$, then
$x=(\xi(x),x_n)$.
Also
$|\xi(x)|=\sqrt{x_1^2+x_2^2+\ldots+x_{n-1}^2}$. We will sometimes use
the notation $y=(0,a)$ to mean $y_1=y_2=\cdots=y_{n-1}=0$ and
$y_n=a$. In this context, we will often think of
$\mathbb{R}^n=\mathbb{R}^{n-1}\times \mathbb{R}$.
We will be working with cylinders in $\mathbb{R}^n$: $A_r(P)=\{x: |\xi(x-P)|<r,\;P_n<x_n<P_n+2r\}
=\{y\in \mathbb{R}^{n-1}:|y-\xi(P)|<r\}\times (P_n,P_n+2r)$ is the cylinder with
axis parallel to the $x_n$-axis, radius $r$, length $2r$ with $P$
being the
center of the bottom face. Let $\lambda>0$, then
$A_{\lambda r}(P)=\{y\in \mathbb{R}^{n-1}:|y-\xi(P)|<\lambda r\}\times
(P_n,P_n +2\lambda r)$ is a $\lambda$ scaling of $A_r(P)$ with the bottom
faces situated at the same height.
Also $F_r(P)=\{x:|\xi(x-P)|<r,\;x_n=P_n\}=\{y\in
\mathbb{R}^{n-1}:|y-\xi(P)|<r\}\times \{P_n\}$ will denote the bottom face.
We also describe a
cylinder using the point half way on its axis. Let
$K_r(P)=\{x:|\xi(x-P)|<r,\;P_n-r<x_n<P_n+r\}=\{y\in \mathbb{R}^{n-1}:|y-\xi(P)|
<r\}\times (P_n-r,P_n+r)$; then $K_r(P)$ and
$K_{\lambda r}(P)=\{y\in \mathbb{R}^{n-1}:|y-\xi(P)|<\lambda r\}\times
(P_n-\lambda r,P_n+\lambda r)$ are concentric cylinders with center $P$. By $B_r(P)$,
we will always mean the ball of radius $r$, centered at $P$. For ease
of notation, we take $A_r=A_r(O)$, $K_r=K_r(O)$ and
$B_r=B_r(O)$. Their use will be clear from the context.
The sets
$B^+_r(P)=\{x\in B_r(P): x_n>P_n\}$ and $B^-_r(P)$ defined
analogously, denote the half-balls. If $A$ and $B$ are two points, with $A\ne B$, then $AB$
stands for the straight segment joining $A$ to $B$.

The
$\infty$-Laplacian operator $\Delta_{\infty}$ is defined as
$\displaystyle{\Delta_{\infty}u=\sum_{i,j=1}^n D_i u D_ju D_{ij}u,}$
where $D_iu=\partial u/\partial x_i$ and
$D_{ij}u=\partial^2 u/\partial x_i\partial x_j.$
This operator is elliptic but highly degenerate. In this work, we
study viscosity solutions of
solutions of
\begin{equation}
\Delta_{\infty}u=0,\quad \mbox{in } \Omega. \label{*}
\end{equation}
We provide a definition in this context. We say that $u$
is a viscosity subsolution (or $\infty$-subharmonic) of above equation,
 in $\Omega$, if
$u$ is upper-semicontinuous in $\Omega$, and whenever $x_0\in \Omega$
and $\phi\in C^2(\Omega)$ are such that
\[
\phi(x_0)=u(x_0),\quad\mbox{and}\quad\phi(x)< u(x),\quad\mbox{for}\;x\ne x_0,
\]
then $\Delta_{\infty}\phi(x_0)\ge 0$. Analogously, we may
define a viscosity supersolution (or $\infty$-superharmonic) of \eqref{*}, by
requiring that $u$ be lower-semicontinuous and $\Delta \phi(x_0)\le 0$, whenever $u(x)-\phi(x)$ has a
local minimum at $x_0$. We say $u$ is a viscosity solution (or
$\infty$-harmonic) of \eqref{*} if it is both a subsolution and a supersolution.
It is well
known that, if $u$ is $\infty$-harmonic then $u$ is locally
Lipschitz continuous in $\Omega$ [4,13,19,21,29]. We must point out
a key
property, we will often exploit here, is that if a function $u$ has cone
comparison property then it is $\infty$-harmonic, a fact proven
in [13]. Also see [4].
We now state the first result.

\begin{theorem}[Comparison] \label{thm1}
Let $u_i(x)>0,\;i=1,2$, be $\infty$-harmonic
  in $A_8(O)$. If $u_1$ and $u_2$ vanish continuously on $F_8(O)$,
  then there exist positive constants $M_1$, $M_2$ and $M_3$ independent of
  $u_i$, such that for $x\in A_1(O)$,
\begin{itemize}
\item[(i)] $u_i(x)\le M_1u_i(z)$, $i=1,2,$ and
\item[(ii)] $\displaystyle M_2\frac{u_1(z)}{u_2(z)}\le 
\frac{u_1(x)}{u_2(x)}
\le M_3\frac{u_1(z)}{u_2(z)}$,
where $z=(0,2)$.
\end{itemize}
\end{theorem}

One may think of part (i) as a boundary Harnack inequality and 
plays
an important role in the proof of part (ii).
This type of comparison result, near a flat portion of the boundary, 
is
well known in the theory of both divergence and nondivergence type
elliptic partial differential equations. We refer the reader to the
works in [2,6,8,9,15,32] and the references therein. However, the 
work
in [32], done in connection with a Fatou theorem, is the earliest 
work
which proves a result of this type for $p$-harmonic
functions, for $1<p<3+2/(n-2)$. Among other
things, the work used associated kernel functions and is based on 
the earlier
fundamental works [6], for nondivergence
type equations, and [9]. The result for the entire range of
$1<p<\infty$ was later proven in
[15], by carrying out a detailed refinement of the works in [6] and
[33]. These works are quite nontrivial in nature. To the best of our
knowledge, no attempt has been made yet to study the behaviour of the
constants $M_2$ and $M_3$ as $p\rightarrow \infty$, in the context of
the $p$-Laplacian. For the case
$p=\infty$, however, we do not utilize the notion of a kernel as employed
in these aforementioned works. But it needs to be mentioned that while we use ideas from
[4,13] and work directly with \eqref{*}, the work in [9] continues to
be very useful, even in this context. Our approach is as follows. We
first prove that the oscillation
$\nu(r)=\mathop{\rm osc}_{K_r}u$ satisfies
$\nu(2r)\ge C\nu(r)$, for some $C>1$. A version of the Harnack
inequality (see Lemma \ref{lm2} and [29]), part (a) of Lemma \ref{lm4} and the
comparison principle permits us to apply the device in Theorem 1.1
[9]. This leads to a proof of part (i) and implies that the solutions are well behaved near
$F_2$; away from $F_2$, the solution can be controlled by the
Harnack inequality. Putting these together yields the result.

 The second set of results are concerned with $\infty$-capacitary
functions. We now introduce notations for this set up.
Let $C_1$ and $C_2$, with $C_2\subset C_1$, be bounded
domains in $\mathbb{R}^n,\;n\ge 2$. Let
$\Gamma=\Gamma(C_1,\;C_2)=C_1\backslash\bar{C_2}$, denote the annular
domain. We take $C_1$ and $C_2$ to be convex $C^2$ domains and we will also
assume that the origin $O$ lies in $C_2$. We will refer to $\Gamma$ as a convex
ring and $\partial \Gamma=\partial C_1\cup\partial C_2$. If $Q\in \partial C_2$, then the line $L=L(Q)$ will often denote the
straight ray normal to $\partial C_2$, at $Q$, directed towards
$\partial C_1$. If $\nu=\nu(Q)$ is the unit outer normal to $\partial C_2$ at
$Q$ (relative to $C_2$), then the hyperplane $\langle x-Q,\nu(Q)\rangle=0$ will be denoted
by $T_Q$. Since $C_2$ is convex, it lies on one side of $T_Q$ and
$L\perp T_Q$ at $Q$. We may also define analogously the hyperplane $T_P$ at a
point $P\in \partial C_1$. The hyperplane $T_Q$ generates two disjoint
half-spaces
$$
H^+_Q=\{x\in \mathbb{R}^n: \langle x-Q,\nu(Q)\rangle<0\}\quad\mbox{and}\quad
H^-_Q=\{x\in \mathbb{R}^n: \langle x-Q,\nu(Q)\rangle>0\}.
$$
Clearly $H^+_Q\supset C_2$. For $P \in \partial C_1$, we will again take
$H^+_P$ to be the half-space that contains
$C_1$. We will be studying the problem
$$
\Delta_{\infty} u=0,\quad\mbox{in }\Gamma,\quad u\in C(\bar{\Gamma})
\quad\mbox{with}\quad u|_{\partial C_1}=1\quad \mbox{and}
\quad u|_{\partial C_2}=0.
$$
We again interpret this in the viscosity sense; see [11]. We call $u$ an $\infty$-capacitary function. Invoking the
Harnack inequality [4,29], we see that $0\le u\le 1$. As a matter of fact if $P$ is a
point of an interior minimum of $u$ then $u-u(P)\ge 0$ in $\Gamma$ and
since $u-u(P)>0$ somewhere in $\Gamma$, being connected this would
mean $u-u(P)>0$ everywhere. This contradiction implies that $u$ has no
interior minimum (nor maximum for that matter) and so $0<u<1$. We
will derive better bounds for $u$. By $\Gamma_t$, we
mean the set $\{x\in \Gamma:u(x)<t\}$. This part of our work has
been motivated by the prior works in [18,27,31,33,34].

\begin{theorem} \label{thm2}
Let $u$ be an $\infty$-capacitary function in a
  bounded convex ring $\Gamma\subset \mathbb{R}^n$. Then\\
(A) the level
  sets $\{x\in\Gamma :u(x)=t\},\;0<t<1,$ are star-shaped and
  satisfy a cone condition;\\
(B) there exists a positive number $\lambda,$ depending only on the
geometry of the domain, such that for any $x\in \Gamma$, there is a direction
  $\vec{e}=\vec{e(x)}$, such that
\[
|u(x+t\vec{e})-u(x)|\ge \lambda t,\quad \forall\; 0<t<t_0=t_0(x).
\]
\end{theorem}

It clearly follows that $|Du(x)|\ge \lambda>0$, a.e. in
$\Omega$. It is not known yet whether $u$ is better than Lipschitz in regularity and hence we are
unable to assert the existence of $|Du|$ everwhere. See [12,25] for a
discussion regarding this issue. The results in
Theorem \ref{thm2} were proven in [33,34] for the $\infty$-Laplacian, by
utilizing the approximating procedure involving the $p$-Laplacian, for
finite $p$; also see [18,31]. The works [33,34] also deal with star-shaped regions and contain
interesting results. However, for convex rings,
the result for the $p$-Laplacian, for finite $p$, was
originally done in [27]. In this context also see [18,31].
Our approach will be to work directly with the viscosity solutions as
discussed before. Our proof utilizes scaling and estimates near the
boundaries, proven by employing auxilliary functions as in [27]. While a great
many of the comparison type results
used in this work may be worked in fairly elementary fashion as in
[4,13], the comparison principle employed to compare $u$ to its scaled
version requires the application of a stronger result. More general
versions of a comparison
principle for such functions, originally proven in [19], may be
found in [3,7,21]. Also see [17,20,23] for related works.
Our approach also utilizes a property of
nonnegative $\infty$-harmonic functions first alluded to in [4, see
Remark 6] which follows from cone comparsion. See \eqref{1} and part (a) of Lemma \ref{lm4} in Section 3. This is used in the proof of the existence of normal
derivatives of $u$ at the boundaries and also in the proof of a
general bound for the gradients. We must point out that at this time we do
not have a proof of the convexity of level sets uitilizing the
viscosity framework. This fact was proven in [27] for the
$p$-Laplacian, for finite $p$, and also holds for $p=\infty$ and appears
in [33,34]. We make some remarks about this issue in Section 6.

 We have divided our work as follows. Section 3 contains
preliminary results needed for our work and Section 4 contains the proof of
Theorem \ref{thm1}.
Section 5 contains results applicable to the context of convex rings
and the proof of Theorem \ref{thm2} appears in Section 6. Appendix contains (i) the
proof of the fact that odd reflections of $\infty$-harmonic functions
are also $\infty$-harmonic, and (ii) the proof of Theorem 1.1 in
[9].

 We thank Michael Crandall for showing
us a short proof of a sharper version of the Harnack inequality
(see Lemma \ref{lm2}) and also for showing us some elegant proofs of
results related to those in [4]. We also thank Juan
Manfredi for several discussions in connection with this work and also for
pointing out the work in [31,32]. We are also indebted to the referee
whose comments have greatly improved the presentation of this work and
also for pointing out the reference [18].
This research was partially supported by a grant from NSERC.

\section{Preliminary results}

In this section we will state and prove a
sequence of results which lead to the proofs of Theorems 2.1 and 2.2. To
achieve our end we will require somewhat more refined versions of the
Hopf principle and the Harnack inequality. The proofs rely on the comparison
principle and some auxilliary functions. A general version of the
comparison principle is proven in [3; also see 7,19,21], however
simpler arguments, such as those used in [4,13], will also suffice in
many instances.

 We will first recall Remark 6 in [4]. Also see Lemme \ref{lm4}.
Let $u>0$ be an $\infty$-harmonic function in a domain $\Omega$ and
$B_r(O)\subset\subset\Omega$. We set
$d(x)=\mathop{\rm dist}(x,\partial B_r(O))=r-|x|,\;x\in B_r(O)$.
Part (i) of Lemma 2 [4], then states
\[
\frac{u(x)}{d(x)}\ge \frac{u(O)}{d(O)}=\frac{u(O)}{r},
\quad \forall x\in B_r(O).
\]
Utilizing this, we showed, in Remark 6 [4], that if $\vec{e}$ is a unit
vector, $x=s\vec{e}$ and
  $y=t\vec{e}$, where $0<s$, $t<r$ (clearly, $x$, $y\in B_r(O)$, $d(x)=r-s$
and $d(y)=r-t$),   then
\begin{equation}
\frac{u(x)}{d(x)}=\frac{u(x)}{r-s}\le
\frac{u(y)}{r-t}=\frac{u(y)}{d(y)},\quad \forall \;0<s<t<r. \label{1}
\end{equation}
In other words, $u(x)/(r-|x|)$ is monotonic along radial lines through
$O$ and is increasing as $x\rightarrow \partial B_r(O)$ along $\vec{e}$. One notes that this may prove
useful especially when $u(r\vec{e})=0$, i. e., $u$ vanishes at some
boundary point.
This observation leads to viscosity proofs of some well-known
results and proves important in our work. We prove \eqref{1} in Lemma \ref{lm4}.
We first start with a more general version of the Hopf boundary
point lemma; in this context also see [4,31,33,34].

\begin{lemma}[Hopf boundary point lemma] \label{lm1}
Let $\Omega $ be a $C^2$
domain and $u\in C(\bar{\Omega})$ be $\infty$-harmonic in
$\Omega$. Suppose $S\in \partial \Omega$ is such that there is a ball
$B_r(P)\subset\Omega$ with $\partial B_r(P)\cap \partial\Omega \ni
S$. Let $x\in B_r(P)$, $I$ be the fixed straight line segment containing
$x$ and $S$, $\vec{a}=(P-S)/r$ and $\vec{b}=(x-S)/|x-S|$. If
$u(S)=\mbox{max}_{x\in B_r(P)}u(x)$, then
\[
\limsup_{x\rightarrow s,\;x\in I}\frac{u(x)-u(S)}{|x-S|}\le\langle
  \vec{a},\vec{b}\rangle\frac{u(P)-u(S)}{r}<0.
\]
\end{lemma}

\noindent {\bf Proof:} We use the result in [4].  By comparison it follows
that $\forall x\in B_r(P)$,
\begin{equation}
u(x)-u(S)\le (u(P)-u(S))\Big(1-\frac{|x-P|}{r}\Big). \label{2}
\end{equation}
Writing $x-P=(x-S)-(P-S)$, $\epsilon=|x-S|/r$, we see that
\begin{equation} \label{3}
\begin{aligned}
1-\frac{|x-P|}{r}=&1-|\vec{a}-\epsilon
\vec{b}|=1-\{1-(2\langle\vec{a},\vec{b}\rangle-\epsilon)\epsilon\}^{1/2}\\
=&\frac{(2\langle\vec{a},\vec{b}\rangle
-\epsilon)\epsilon}{1+\{1-(2\langle\vec{a},\vec{b}\rangle-\epsilon)
\epsilon\}^{1/2}}\ge 0. \end{aligned}
\end{equation}
Letting $\epsilon\rightarrow 0$, \eqref{2} and \eqref{3} yield
$$
\limsup_{x\rightarrow s,\;x\in I}\frac{u(x)-u(S)}{|x-S|}\le\langle
  \vec{a},\vec{b}\rangle\frac{u(P)-u(S)}{r}<0.
$$
\mbox{}\hfill$\square$

 We now present a proof of a sharper version of the Harnack inequality
for nonnegative $\infty$-harmonic functions. This version was first
proven in [29; also see 21] using approximating $p$-harmonic
functions. The proof
below uses the notion of viscosity solutions and uses the estimates
proven in [4]. Note that it uses none of the differentiation theory
utilized in [29]. This proof was pointed out to us by Michael Crandall.

\begin{lemma}[The Harnack inequality] \label{lm2}
Let $u>0$ be $\infty$-harmonic
  in $\Omega$, and $\delta>0$ be such that the set $\Omega_{\delta}=\{x\in\Omega:
  \;\mathop{\rm dist}(x,\partial \Omega)\ge \delta\}\ne \emptyset$. Suppose $A$
  and $B$ are points, in $\Omega_{\delta}$, such that the segment
  $AB\subset \Omega_{\delta}$. Then
\[
u(B)\ge e^{-\frac{|A-B|}{\delta}}u(A).
\]
\end{lemma}

\noindent {\bf Proof:} We note that, by employing the comparison principle,
if $y\in \Omega_{\delta}$ then
\begin{equation} \label{4}
u(y)\Big(1-\frac{|x-y|}{\delta}\Big)\le u(x),\quad \forall x\in
B_{\delta}(y).
\end{equation}
Let $x_0,x_1,x_2,\ldots,x_m$ be points on the segment $AB$ such
that $x_0=A,x_m=B$ and $|x_i-x_{i+1}|=|A-B|/m$, $\forall i=0,1,\ldots,\;m-1$.
Choose $m$ large so that $|A-B|/m\le \delta/2$. Since $x_{i+1}\in
B_{\delta}(x_i)$, applying \eqref{4}, we find
that
$$ u(x_{i+1})\ge u(x_i) \Big(1-\frac{|A-B|}{m\delta}\Big),\quad
\forall i=0,1,\ldots,\;m-1.$$
Thus
\begin{equation} \label{5}
u(B)\ge u(A) \quad(1-\frac{|A-B|}{m\delta}\quad)^m .
\end{equation}
The lemma follows by letting $m\rightarrow \infty$ in \eqref{5}. \qed

\begin{remark} \label{rmk1} \rm
 It is clear that above estimate can be extended very
easily to $\infty$-superharmonic functions and to
polygonal paths joining two points in $\Omega_{\delta}$.
\end{remark}

 We now prove a result about the oscillation of $u$ which will prove
important in our proof of
Theorem \ref{thm1}. Calling $w(r)=\max_{B_r(O)}|u(x)-u(y)|=\mathop{\rm osc}_{B_r(O)}u$,
we show that $w(r)$ is convex and in particular $w(2r)\ge 2w(r)$. This
fact together with Theorem 1.1 in [9] will lead to a proof of
Theorem \ref{thm1}. Note in Lemma \ref{lm3}, we do not assume that $u>0$.

\begin{lemma}[Convexity of oscillation] \label{lm3}
Let $\Omega\subset \mathbb{R}^n$ be a
  domain and $u$ be $\infty$-harmonic in $\Omega$. Let
  $B_R(O)\subset\subset\Omega$, be the ball of radius $R$, centered at
  $O$. Suppose that $M(r)=\sup _{B_r(O)}u(x)$, and
  $m(r)=\inf _{B_r(O)}u(x)$. Then for $0\le r\le R$, \\
(i) $w(r)=\mathop{\rm osc}_{B_r(O)}u(x)$ is convex  and \\
(ii) $\displaystyle{\frac{w(r)}{r}=\frac{M(r)-m(r)}{r}\downarrow}$
as $r\downarrow 0$.
\end{lemma}

\noindent {\bf Proof:} Let $0<\delta <R$; set
\begin{align*}
&W_{\delta}(x)=\frac{M(R)-M(\delta)}{R-\delta}|x|+\frac{RM(\delta)-\delta
  M(R)}{R-\delta},\\
&w_{\delta}(x)=\frac{m(R)-m(\delta)}{R-\delta}|x|+\frac{Rm(\delta)-\delta
  m(R)}{R-\delta},
\end{align*}
for all $x\in B_R(O)\backslash B_{\delta}(O)$. It is easily checked that
$W_{\delta}(x)\ge u(x)$ and $w_{\delta}(x)\le u(x)$, when $|x|=R$ and
$|x|=\delta$. Since $W_{\delta}$ and $w_{\delta}$ are both
$\infty$-harmonic, it follows that $w_{\delta}\le u(x)\le W_{\delta}(x),\;
\delta \le |x|\le R$. Set $r=|x|$, it then follows that
$$ M(r)\le \frac{M(R)-M(\delta)}{R-\delta}r+\frac{RM(\delta)-\delta
  M(R)}{R-\delta}=\frac{r-\delta}{R-\delta}M(R)+\frac{R-r}{R-\delta}M(\delta),$$
\noindent and
$$ m(r)\ge \frac{m(R)-m(\delta)}{R-\delta}r+\frac{Rm(\delta)-\delta
  m(R)}{R-\delta}=\frac{r-\delta}{R-\delta}m(R)+\frac{R-r}{R-\delta}m(\delta).$$
It is clear that $M(r)$ is convex and increasing while $m(r)$ is
concave and decreasing. It is clear from above that $M(r)-m(r)$ is
convex. Letting $\delta\rightarrow 0$ and noting that $M(0)=m(0)=u(0)$, we find that $(M(r)-m(r))/r \le
(M(R)-m(R))/R$. This proves (i) and (ii). \qed

\begin{remark} \label{rmk2} \rm
From Lemma \ref{lm3} (ii), it follows that for $t>1$, $w(tr)\ge
t w(r)$ and in particular, $w(2r)\ge 2w(r)$. Now let
$K_r=K_r(O)=\{x:|\xi(x)|< r,\;|x_n|<r\}=\{x:|\xi(x)|< r,\}\times(-r,r)$
be the cylinder of length $2r$, radius $r$ and centered at $O$;
let $\nu(r)=\mathop{\rm osc}_{K_r}u$. Then $K_{2r}\supset B_{2r}\supset
B_{\sqrt{2}r}\supset K_r$ (see Section 2). Thus
\begin{equation} \label{6}
\nu(2 r)\ge w(2r)\ge \sqrt{2}
w(\sqrt{2}r)\ge \sqrt{2}\nu(r).
\end{equation}
\end{remark}

We now work a proof of Lemma \ref{lm4}; this will imply \eqref{1}. We introduce
some additional notation. Let $u$ be $\infty$-harmonic in
$\Omega$, $B_r(z)\subset\subset\Omega$, $M=\sup _{x\in
  B_r(z)}u(x)$ and $m=\mbox{inf}_{x\in B_r(z)}u(x)$.
By the maximum principle, these are attained on the
boundary $\partial B_r(z)$. Let $P_M,\;P_m\in \partial B_r(z)$ be such that
$u(P_M)=M$ and $u(P_m)=m$. We define the following difference quotients on
the segments $zP_m$ and $zP_M$;
$$
\forall x\in
zP_m,\;\;D_1(x,P_m)=\frac{u(x)-m}{|x-P_m|}\quad\mbox{and}\quad
\forall x\in zP_M,\;D_2(x,P_M)=\frac{M-u(x)}{|x-P_M|}.
$$
Note that $D_1(x,P_m)\ge 0$ and $D_2(x,P_M)\ge 0$; also define
$$
D(m)=\lim_{x\rightarrow
  P_m}D_1(x,P_m),\;x\in zP_m\quad\mbox{and}\quad
D(M)=\lim_{x\rightarrow
  P_M}D_2(x,P_M),\; x\in zP_M.
$$
whenever these exist. For any $x\in B_r(z),$ set
$d(x)=\mathop{\rm dist}(x,\partial B_r(z))=r-|x-z|$, and for
 $x\ne z$, take $\vec{e}=\frac{z-x}{|x-z|}$ and
 $y=x-(z-r\vec{e})$. Note that $(z-r\vec{e})\in \partial B_r(z)$ and
 $|y|=r-|x-z|=d(x)$. We will also take $u(y)$ to stand for the value $u(x)$.

\begin{lemma}[Monotonicity] \label{lm4}
 Let $u$ be  $\infty$-harmonic in $\Omega$ and $B_r(z)\subset\subset\Omega$. \\
\\
(a) If $u>0$ in $\Omega$ and $\vec{\eta}$ is such that
  $|\vec{\eta}|=1$, then for $0\le a\le r$,
$$
\frac{u(z+a\vec{\eta})}{r-a}=\frac{u(z+a\vec{\eta})}{r-|a\vec{\eta}|}\ge
  \frac{u(z+b\vec{\eta})}{r-|b\vec{\eta}|}=\frac{u(z+b\vec{\eta})}{r-b},\;\;\forall \;0\le b\le a\le
  r.
$$
In other words, $\forall\;x\in B_r(z)$, $u(x)/d(x)$, is increasing as
$x\rightarrow\partial B_r(z)$ along a radial line. We also note that
by taking $b=0$,
$u(z+a\vec{\eta})/(r-a)\ge u(z)/r,\;\forall 0\le a\le r$.
\\
 (b) Under the assumptions in (a), $x\ne z$, $\vec{e}$ and $y$ as
defined above, we have
$$u(ty)\le
  tu(y),\quad\mbox{whenever}\quad t\ge 1\quad\mbox{and}\quad 0<t|y|\le r.
$$
Furthermore, if $Du(x)$ exists then
$\displaystyle{\langle Du(x),\vec{e}\rangle\le\frac{u(x)}{d(x)}}$.
\\
(c) Moreover, if $u>0$ and $|Du|(z)$ exists then
$$
(i)\;\;|Du|(z)\le \frac{u(z)}{\mathop{\rm dist}(z,\partial B_r(z))};
\quad\mbox{and}\quad (ii)\;\;\;|Du|(z)\le
\frac{u(z)}{\mathop{\rm dist}(z,\partial \Omega)}.
$$
(d) Regardless of the sign of $u$, $D(M)$ and $D(m)$ exist on $B_r(z)$ and
$|Du|(z)\le \min(D(m),\;D(M))$.
\end{lemma}

\noindent {\bf Proof:} The proof follows by an application of part (i) Lemma
2 in [4]; see \eqref{1}.

\noindent{\it Part (a):} For $0\le b\le a\le r$, set $x=z+a\vec{\eta}$ and
$v=z+b\vec{\eta}$; then $d(x)=\mathop{\rm dist}(x,\partial B_r(z))=r-a$,
$d(v)=\mathop{\rm dist}(v,\partial B_r(z))=r-b$ and $d(v)\ge d(x)$. Clearly, $x$
lies in the ball $B_{d(v)}(v)$; applying Lemma 2 in [4] (also see
discussion preceding \eqref{1}), we see that $u(x)\ge u(v)
d(x)/d(v)$. This proves part (a). The rest of the assertions are
consequences of this fact.

\noindent{\it Part (b):} We reinterpret part (a). One notes that for $x\in
B_r(z)$, the ray $z-s\vec{e}$, $s\ge 0$, cuts $\partial B_r(z)$ at
$z-r\vec{e}$. Thus by part (a) we find
that $u(x)/(r-|x-z|)=u(y)/|y|$ increases as $x\rightarrow z-r\vec{e}$,
i.e., as $|y|=d(x)\downarrow 0$. If $t>1$ is such that $0<t|y|\le r$, then
$$ \frac{u(ty)}{|ty|}\le \frac{u(y)}{|y|}\;\;\Rightarrow\;\;u(ty)\le tu(y).
$$
If $x_1,\; x_2\in B_r(z)$ such that
$x_i=z-s_i\vec{e},\;i=1,2$, for some $\vec{e}\in \mathbb{R}^n$ with $|\vec{e}|=1$, $0\le s_2< s_1\le r$, and
$\theta=s_1-s_2$, then $x_2=x_1+\theta\vec{e}$, $d(x_i)=r-s_i$ and
$d(x_1)<d(x_2)$. Using part (a), i.e., $u(x_1)/d(x_1)\ge u(x_2)/d(x_2)$,
\begin{equation} \label{7}
\frac{u(x_1+\theta\vec{e})-u(x_1)}{\theta}
=\frac{u(x_2)-u(x_1)}{|x_1-x_2|}\le
\begin{cases}
\frac{u(x_1)}{d(x_1)}=\frac{u(x_1)}{r-|x_1-z|},\\[2pt]
\frac{u(x_2)}{d(x_2)}=\frac{u(x_2)}{r-|x_2-z|}.
\end{cases}
\end{equation}
Note that $s_2=|x_2-z|< |x_1-z|=s_1$. Suppose $Du(x_1)$ exists;
letting $x_2\rightarrow x_1$ in \eqref{7} yields the
conclusion. Recall that the
solution $u$ is Lipschitz continuous hence such directional
derivatives exist a. e. on these rays.

\noindent{\it Part (c):} We use part \eqref{7} with $x_2=z$, $\vec{e}$ any unit
vector and $x_1=z+s\vec{e},\;s>0$. Then
$$
\frac{u(z)-u(z+s(\vec{e}))}{s}\le \frac{u(z)}{r}\;\;\Rightarrow\;
\langle Du(z), -\vec{e}\rangle \le\frac{u(z)}{r},\quad \forall
\vec{e}\in \mathbb{R}^n,
$$
if $Du(z)$ exists. If $Du(z)\ne 0$, we may take
$\vec{e}=-Du(z)/|Du(z)|$ and part (i) follows.
To prove (ii), let $R=\mathop{\rm dist}(z,\partial
\Omega)$, then the ball $B_R(z)\subset \Omega$, and (i)
continues to hold by considering an increasing sequence of balls.

\noindent {\it Part(d):} All the results discussed above require that
$u>0$. Now we drop this requirement. With
$m$ and $M$ as above, part (c) holds for $u-m$ and
$M-u$, i. e.,
$$ |Du(z)|\le \min \Big(\frac{M-u(z)}{r},\;\frac{u(z)-m}{r}\Big).
$$
Applying part (a) to $M-u$, it follows that $(M-u(z))/r\le (M-u(x))/|x-P_M|\le
D(M),\forall x\in zP_M$, if $D(M)$ exists. To see this, note
\[
\forall x\in OP_M,\;
0<D_2(x,P_M)=\frac{M-u(x)}{|x-P_M|}=\frac{u(P_M)-u(x)}{|x-P_M|}\;\uparrow\;\mbox{as}\;x\rightarrow P_M.
\]
The case of the minimum follows analogously.
Clearly the directional derivatives of $u$ at $P_M$ along $OP_M$, and
at $P_m$ along $OP_m$, either exist or blow up. Since $P_m$ and $P_M$
are points in the interior of $\Omega$, the local Lipschitz regularity
of $u$ thus implies that $D_1(x,P_m)$ and $D_2(x,P_M)$ are all
uniformly bounded. Hence the normal derivatives $D(m)$ and $D(M)$ exist and are strictly
positive and finite. Thus
$$|Du(z)| \le \min (D(m),D(M)).
$$

\section{Proof of Theorem \ref{thm1}}

We follow the outline of Theorem 1.1 in [9] (see Appendix) and use Lemmas
3.2, 3.4 and 3.6. We point out that part (i) of Theorem \ref{thm1} (also see
\eqref{10}) turns out to be crucial for the proof of part (ii). One may
think of (i) as a boundary Harnack inequality.
For notations, see Section 2. Recall the expression
$x=(\xi(x),x_n)$, for all $x\in\mathbb{R}^n$ and $A_r=\{x:\;|\xi(x)|<r
,\;\mbox{and}\;0<x_n<2r\}$. We proceed as follows.
Assume that
$u_i>0,\;i=1,2$; by scaling if necessary, we may take $u_i$'s to be
$\infty$-harmonic in $A_{8}$ and vanishing
continuously on the face $F_8=\{x:|\xi(x)|<8,\;x_n=0\}\subset
\{x_n=0\}$. We suppress the subscript and work with a general $u$ that
satisfies the requirements of the theorem. The constants $M$, $C$ are
positive constants, that are independent of $u$, but may depend on the
geometry. We will often write $x=(\xi(x),x_n)$ and set $z=(0,2)$. We
achieve our proof in five steps.

\noindent {\bf Step 1.}
We first show that
\begin{equation} \label{8}
\frac{u(x)}{u(z)}\ge \frac{x_n}{4},\quad \forall x\in A_1.
\end{equation}
For $x\in A_1$, write $x=(\xi(x),x_n)$ and set $P=(\xi(x), 2)$; then
$P$ lies in the hyperplane $x_n=2$ and
$x\in B_{2}(P)\subset A_8$. Applying Lemma \ref{lm4} (a), in $B_2(P)$, we see
$$ \frac{u(x)}{x_n}\ge \frac{u(P)}{2}.
$$
Again $P\in B_2(z)$, and applying once more Lemma \ref{lm4} (a) to $B_2(z)$ and noting
that $d(P)=\mathop{\rm dist}(P,\partial B_2(z))\ge 1$, we have
$$ u(P)\ge\frac{u(P)}{d(P)}\ge \frac{u(z)}{2}.
$$
Combining these observations yields \eqref{8}.

\noindent {\bf Step 2.} We now make a few remarks which again follow from
Lemmas 3.2 and 3.4. For $x\in A_2$ with
$0<x_n< 3/2$, an application of Lemma \ref{lm2}, to the
points $(\xi(x),x_n)$ and $(\xi(x),2x_n)$, implies
$$
u(\xi(x),x_n)\le e^{(|2x_n-x_n|/x_n)}u(\xi(x),2x_n)=e^1 u(\xi(x),2x_n).
$$
Now for $x\in A_2$ with $1<x_n<3$, we see that dist$(x,F_2)\ge
1$ and $|z-x|\le \sqrt{5}$. For these $x$'s, Lemma \ref{lm2} again implies
$u(\xi(x),x_n)\le e^{\sqrt{5}}u(z)$. To recap, with $M=e^{\sqrt{5}}$,
\begin{equation} \label{9}
u(x)=u(\xi(x),x_n)\le \begin{cases}
Mu(\xi(x),2x_n): & |\xi(x)|\le 2,\;0<x_n<3/2,\\[2pt]
Mu(z): & |\xi(x)|\le 2,\;1<x_n<3.
\end{cases}
\end{equation}
We also recall, with the notations of Remark \ref{rmk2}, that if $K_r$ and
$K_{r/2}$ are concentric cylinders and $\nu(r)=\mathop{\rm osc}_{K_r}u$, then
$$\nu(r)\ge \sqrt{2}\nu(r/2).$$
This together with \eqref{8}, the fact that odd reflection of
$u$ about $x_n=0$ continues to be $\infty$-harmonic (see Appendix)
and Theorem 1.1 [9] implies that there is a universal
constant $C$ such that
\begin{equation} \label{10}
\sup _{A_1}u(x)\le Cu(z)\;\Rightarrow\;\sup _{A_1}u_i(x)\le Cu_i(z),\;i=1,2.
\end{equation}
This achieves the proof of part (i) of Theorem \ref{thm1}.

\noindent{\bf Step 3.} Our next goal is to prove that for some universal
constant $C$,
\begin{equation} \label{11}
u(x)\le C u(z) x_n,\quad \forall x\in A_1.
\end{equation}
We first describe a proof of \eqref{11} when $\xi(x)=0$ and $0<x_n\le 2$, i.e.,
when $x$ is on the segment $Oz$. It is clear that $A_1\subset
B^+_{\sqrt{5}}(O)\subset A_8$. Let
$M_O=\sup _{B^+_{\sqrt{5}}(O)}u>0$. By the maximum principle, $M_O=\sup _{S}u$, where $S=\partial
B^+_{\sqrt{5}}(O)$, since
$u=0$ on $F_8$. By comparison, $u(x)\le M_O|x|/\sqrt{5},\;\;\forall
x\in B^+_{\sqrt{5}}(O)\supset
A_1$. In particular,
\begin{equation}
u(x) \le \frac{M_O x_n}{\sqrt{5}},\quad\forall x=(0,x_n),\; 0< x_n \le 2.
\end{equation} \label{12}
Our next task now will be to estimate $M_O$ in terms of $u(z)$.
We do this as follows. If the maximum $M_O$ occurs near $F_8$, then it
can be controlled first by $u$ at a point away from $F_8$ by an application
of \eqref{10}. This in turn can be estimated by $u(z)$ by the
Harnack inequality. Note that a direct application of the Harnack
inequality is not possible since the constants degenerate near
$F_8$. If the maximum occurs away from $F_8$, then the Harnack
inequality suffices to achieve our end.
We set $T=F_8\cap \partial
B_{\sqrt{5}}(O)$. For $P\in T$, $x_n(P)=0$, $|\xi(P)|=\sqrt{5}$ and
the cylinder $A_4(P)\subset A_8$. Thus by \eqref{10} and
scaling
\begin{equation} \label{13}
u(x)\le C_1 u(\bar{P}),\;\;\forall x\in A_{1/2}(P),
\end{equation}
where $\bar{P}=(\xi(P),1)$ and $C_1$ is the constant in \eqref{10}.
Clearly, \eqref{13} holds in $I=\cup_{P\in T} A_{1/2}(P)$. Let $E=\cup_{P\in T}\{\bar{P}\}=\{x:
|\xi(x)|=\sqrt{5},\;x_n=1\}$. Observe
that dist$(\bar{P},
F_8)=\mathop{\rm dist}(\bar{P},\partial A_8)=1,\;\forall \bar{P}\in E$. Employing
Lemma \ref{lm2}, \eqref{13} and
recalling that $z=(0,2)$, we have
\begin{equation} \label{14}
u(\bar{P})\le u(z)e^{|\bar{P}-z|}\le u(z)
e^{\sqrt{6}}\;\;\Rightarrow\;\;u(x)\le C_2 u(z),\;\;\forall
x\in I.
\end{equation}
Clearly, this also holds on $\partial
B^+_{\sqrt{5}}(O)\cap I$. If the maximum $M_O$ occurs in $I$, \eqref{14}
applies. Now for $x\in \partial
B^+_{\sqrt{5}}(O)\setminus I$, dist$(x,\partial A_8)=\mathop{\rm dist}(x,F_8)\ge
1$ and $|x-z|\le \sqrt{6}$; we may apply Lemma \ref{lm2}, as done above, to conclude that
\[
u(x)\le e^{\sqrt{6}}u(z),\;\;\forall x \in\partial
B_{\sqrt{5}}(O)\setminus I.
\]
This together with \eqref{14} implies that $M_O\le C_3u(z)$, where $C_3$ is again a
universal constant. The inequality in \eqref{12} now implies for some
appropriate constant $C$,
\begin{equation} \label{15}
u(x)\le C u(z) x_n,\quad \forall x=(0,x_n),\;\;0<x_n\le 2.
\end{equation}
\noindent {\bf Step 4.} We now show that \eqref{11} holds in all of
$A_1$. Let $x\in A_1$, with $\xi(x)\ne 0$; define $L=(\xi(x),0)$ and
$\bar{L}=(\xi(x),2)$. Then $L$ and $\bar{L}$ belong to $\partial
A_1$; the cylinders $A_1(L)$, $A_4(L)$ all lie in
$A_8(O)$ and the half-ball $B^+_{\sqrt{5}}(L)\subset
A_8(O)$. Furthermore, $\forall N\in F_8\cap
\partial B_{\sqrt{5}}(L),\;A_4(N)\subset A_8(O)$. Now working with $A_{1/2}(N)$, we may now
apply Step 3 to conclude that for all $\bar{L}\in \partial A_1$, with
$x_n(\bar{L})=2$, \eqref{15} holds i.e.,
\[
u(x)\le Cu(\bar{L})x_n,\;\;\forall x\in A_1,\;\;\mbox{with}\;\;
\xi(x)=\xi(\bar{L}),\;0<x_n\le 2.
\]
All that remains now is to relate $u(\bar{L})$ to $u(z)$ and this is
achieved by Lemma \ref{lm2}. Clearly, $u(\bar{L})\le e^{1}u(z)$ and hence
we see that
\begin{equation} \label{16}
u(x)\le C u(z) x_n,\quad x\in A_1.
\end{equation}

\noindent {\bf Step 5.} We combine \eqref{8} and \eqref{16} to deduce that for some
universal $C>0$ and $\bar{C}>0$
\[
\bar{C}\frac{u_1(z)}{u_2(z)}\le \frac{u_1(x)}{u_2(x)}
\le C\frac{u_1(z)}{u_2(z)},\;\;\forall x\in
A_1.
\]
This finishes the proof of part (ii) of Theorem \ref{thm1}.

\section{$\infty$-Capacitary functions in convex rings}

 Our effort, in Sections 5 and 6, is to prove Theorem \ref{thm2} (see
Section 2 for notation). The main ideas used here are similar to
those in Section 3. Our strategy will be to prove bounds for $u$, show strict
monotonicity by using scaling, a global maximum principle for $|Du|$
and make remarks about a global lower bound.

 We start with bounds for $u$ in $\Gamma$. Our approach is to use
appropriate barrier functions and comparison and while these will suffice
for our purposes, an approach based on Lemma \ref{lm4} can also be worked
out. We make comments along this direction later in this work. The
function $u\in C(\bar{\Gamma})$, from hereon, is an
$\infty$-capacitary function with $u|_{\partial C_1}=1,\;u|_{\partial C_2}=0$,
and, as observed in Section 2, $0<u<1$ in $\Gamma$. We take the origin
$O$ to lie in $C_2$. We also remind the reader that, for $Q\in
\partial C_2$, $L=L(Q)$ will be the straight ray
  originating from $Q$, normal to $\partial C_2$, directed toward
  $\partial C_1$.

\begin{lemma}[Lower bound] \label{lm5}
Let $Q\in \partial C_2$ and $L$ be as
  described above. Let $P=P(Q)=L\cap \partial C_1$ and $R=|P-Q|$, then
\[
u(x)\ge 1-\frac{|x-P|}{|Q-P|}=1-\frac{|x-P|}{R}>0,\quad \forall x\in
  B_R(P)\cap \Gamma.
\]
\end{lemma}

\noindent{\bf Proof:} Set $w=w_{P(Q)}(x)=1-\left(|x-P|/R \right)$; then $w$
is (i) $\infty$-harmonic in $B_R(P)\backslash\{P\}$, (ii) $w=0$ on $\partial B_R(P)$
and (iii) $0<w(x)<1$ in $B_R(P)\backslash \{P \}$ with $w(P)=1$. We compare $w$ to $u$
in $B_R(P)\cap \Gamma$ and conclude that $u\ge w$ on $\partial (B_R(P)\cap
\Gamma)
=(\partial B_R(P)\cap \Gamma) \cup (B_R(P)\cap \partial
\Gamma)$.
Both being $\infty$-harmonic in $B_R(P)\cap \Gamma$, comparison
yields $u\ge w$ in $B_R(P)\cap \Gamma$ and
$$u(x)\ge w_{P(Q)}= 1- \frac{|x-P|}{R}>0,\;\;x\in B_R(P)\cap
\Gamma.$$
Note that the function $\tilde{w}(x)=\sup _{Q\in \partial
  C_2}w_{P(Q)}(x)$ is $\infty$-subharmonic and $u(x)\ge\tilde{w}(x)$.\qed

Before we prove an upper bound for $u$, we note the following. Being a $C^2$ domain, $C_1$ satisfies an interior ball
condition at every point on $\partial C_1$. For $\eta>0$, let $C_{1,\eta}=\{x\in \Gamma
: \mathop{\rm dist}(x,\partial C_1)\le \eta \}$.
Since $\partial C_1$ is
$C^2$, for every $A\in
\partial C_1$, there is a $\delta_A>0$ and an $H_A\in \Gamma$ with the property that
$B_{\delta_A}(H_A)\subset \Gamma
\;\mbox{and}\;B_{\delta_A}(H_A)\cap \partial
C_1\ni A$. We take $\delta_A$ to be the largest such number, and if
$\delta_0=\inf_A\{\delta_A\}$, then $\delta_0>0$. Let $l=\mathop{\rm dist}(\partial C_1,\partial C_2)/2$ and $\delta=$ min$( \delta_0,l)$. This choice is made for
technical reasons. For notational ease, define
$\delta(x)=\mathop{\rm dist}(x, \partial C_1)$. By $\Delta$, we denote the
diameter of $C_1$. We should point out that while Lemma \ref{lm6},
as stated below, provides a bound only for points near $\partial C_1$, its
derivation requires the calculation of upper bounds in the rest of $\Gamma$.

\begin{lemma}[Upper bound] \label{lm6}
Let $\delta,\;\delta(x),\;\Delta,\;l$ and $C_{1,\delta}$
be as described above. If $x\in C_{1,\delta}$, i.e., $\delta(x)\le
\delta$, then
\[
u(x)\le 1-\frac{ e^{-(\Delta/\delta)}}{2\delta}\delta(x).
\]
\end{lemma}

\noindent {\bf Proof:} We do this in three steps. \\
{\bf Step 1 (Upper bound near $\partial C_2$)}
For every $Q\in \partial C_2$, the ball $B_{2l}(Q)\subset C_1$. Fix
$Q$ and set $v=v(x)=|x-Q|/(2l)$. Observe that
(i) $v$ is $\infty$-harmonic in $B_{2l}(Q)\cap \Gamma$,
(ii) $v=1$ on $\partial
B_{2l}(Q)\cap \Gamma$ and $0\le v \le 1$ on $\partial C_2 \cap
B_{2l}(Q)$. Since $u\le v$ on $\partial(B_{2l}(Q)\cap \Gamma)$,
comparison implies
\begin{equation} \label{17}
u(x)\le \frac{|x-Q|}{2l}\le 1,\;\; x\in B_{2l}(Q)\cap\Gamma.
\end{equation}
Clearly $u\le 1/2$ in $B_{l}(Q)$, and so defining $E_l=
\cup_{Q\in \partial C_2}(B_{l}(Q)\cap
\Gamma)=\{x\in\Gamma:\;\mathop{\rm dist}(x,\partial C_2<l)\}$,
\eqref{17} yields
\begin{equation} \label{18}
u(x)\le 1/2,\quad\forall x\in \bar{E}_l.
\end{equation}

{\bf Step 2 (Upper bound away from $\partial C_2$ and $\partial  C_1$)}
Let $Q_1\in (\partial E_l\setminus \partial C_2)$ and
$P_1\in (\partial C_{1,\delta}\setminus\partial C_1)$;
hence by \eqref{18}
\begin{equation} \label{19}
u(Q_1)\le 1/2.
\end{equation}
Now $Q_1$ and $P_1$ are in $\Gamma_{\delta}=\{x\in \Gamma:\;
\mathop{\rm dist}(x,\partial \Gamma)\ge \delta\}$. We now apply Lemma \ref{lm2} to
the function $1-u(x)\ge 0$ in $\Gamma$, along the segment $P_1Q_1$, to
conclude that, $\forall x\in P_1Q_1$,
\[
1-u(x)\ge(1-u(Q_1)) e^{-(|x-Q_1|/\delta)}.
\]
Noting \eqref{19}, it follows that
$$
u(x)\le 1-\frac{e^{-(|Q_1-x|/\delta)}}{2},\quad \forall x\in P_1Q_1.
$$
Taking $x=P_1$, and observing that $|P_1-Q_1|\le \Delta$, we see
\begin{equation} \label{20}
u(P_1)\le 1-\frac{e^{-(|Q_1-P_1|/\delta)}}{2}\le 1
-\frac{e^{-(\Delta/\delta)}}{2}
\;\;\Rightarrow\;1-u(P_1)\ge\frac{e^{-(\Delta/\delta)}}{2}.
\end{equation}

{\bf Step 3 (Bound near $\partial C_1$)}
To get our target estimate for every $x\in C_{1,\delta}$, we proceed
as follows. For a fixed $x\in C_{1,\delta}$, let $H\in \partial C_1$ be such
that $|x-H|=\delta(x)$. Then the straight line $R$ containing
the segment $xH$ is perpendicular to $\partial C_1$ at $H$. Set
$P_1=R\cap (\partial C_{1,\delta}\setminus\partial C_1)$ and
$Q_1\in \bar{E}_l$ be such that dist$(P_1,E_l)=|Q_1-P_1|$.
We now employ Lemma \ref{lm1} as follows.
Observe that $B_{\delta}(P_1)\subset \Gamma$,
$\partial B_{\delta}(P_1)\cap \partial C_1 \ni H$, $u(H)=1=\max_{x\in
B_{\delta}(P_1)}u(x)$ and $\delta(x)=\delta-|x-P_1|$. Then \eqref{2} implies
\begin{eqnarray}
u(x)-1&\le&
(u(P_1)-1)\left(1-\frac{|x-P_1|}{\delta}\right)=(u(P_1)-1)
\frac{\delta(x)}{\delta}\nonumber\\
\Rightarrow\;u(x)&\le&
1-(1-u(P_1))\frac{\delta(x)}{\delta},\;\;\;\forall x\in P_1H.
\label{21}
\end{eqnarray}
Clearly our estimate in \eqref{19} now holds for $u(Q_1)$ and
$|P_1-Q_1|\le \Delta$; an
application of Lemma \ref{lm2}, \eqref{20} and \eqref{21} implies
\begin{equation} \label{22}
u(x)\le 1-\Big(\frac{e^{-(\Delta/\delta)}}{2\delta}\Big)\delta(x),
\quad \forall x\in C_{1,\delta}.
\end{equation}
\mbox{}\hfill$\square$

\begin{remark} \label{rmk3} \rm
The boundaries of $\Gamma$ being $C^2$, the distance functions
$d_i(x)=\mathop{\rm dist}(x,\partial C_i),\;i=1,2,$ is $C^2$ in a neighborhood of
$\partial C_i$. Note that $d_2(x)$ is $C^2,\;\forall x\in
\mathbb{R}^n\backslash C_2$. Thus
$\Delta_{\infty}d_2=0,\;\forall x\in\Gamma$, while
$\Delta_{\infty}d_1=0$ only for $x$ near $\partial C_1$. One may also
use these functions as barriers.
\end{remark}

 Our next step is to prove strict monotonicity of $u$ along rays emanating
from points in $C_2$. This lemma relies on the comparison principle
proven in [3,7,19,21]. A stronger result is proven in Theorem \ref{thm2} but this
weaker result will prove adequate for what is to follow. We use
scaling as was done in [27]. We introduce the following notations. For
$t>0$, define $C_i^t=tC_i=\{tx,\;\forall x\in C_i\},\;\mbox{and}\;\partial
C_i^t=t\partial C_i=\{tx:x\in\partial C_i\},\;i=1,2$.
If $0<t<1$ then $C^t_i\subset C_i.$ Also set
$u_t(y)=u(y/t)$. We will take $t$ to be close to 1 and assume that
$C_2\subset C^t_1\subset C_1$.

\begin{lemma}[Strict monotonicity] \label{lm7}
 Let $T$ denote a straight ray
  originating from $O\in C_2$
and $Q=T\cap\partial C_2$. Then $u$ is strictly increasing along $T\cap
\Gamma$, in the direction of $\partial C_1$.
\end{lemma}

\noindent {\bf Proof:} We employ scaling. Let $P_0$ and $P_1$ be on $T\cap
\Gamma$ with $0<|P_0|<|P_1|$. Set $t=|P_0|/|P_1|<1$, $y=tx$
and $u_t(y)=u(x)=u(y/t)$. Clearly, $tP_1=P_0$,
$u_t(P_0)=u_t(tP_1)=u(P_1)$ and $u_t$ is $\infty$-harmonic in
$\Gamma^t=\Gamma(C_1^t,C_2^t)$. Notice that $0<u_t<1$ in $\Gamma^t$,
$u_t=0$ on $\partial C_2^t$, $u_t=1$ on $\partial C_1^t$ and
$\partial(\Gamma\cap
\Gamma^t)=\partial C_2\cup \partial C_1^t$. Now
\[
\Delta_{\infty}u=\Delta_{\infty}u_t=0, \;\;\mbox{in}\;\; \Gamma\cap
\Gamma^t,\;\;
u<u_t=1\;\;\mbox{on}\;\partial
C_1^t\;\;\;\mbox{and}\;\;u=0<u_t,\;\mbox{on}\;\partial C_2.
\]
Applying the comparison principle, we obtain
\[
u_t(y)\ge u(y),\;\forall y\in\Gamma^t\cap \Gamma,\;\;\Rightarrow\;\;
u(P_1)=u_t(tP_1)=u_t(P_0)\ge u(P_0).
\]
Since $\Delta_{\infty}$ is translation and rotation invariant, we may move $O$
around in $C_2$ and conclude that $u$ is nondecreasing along the cone
of rays passing through $Q$ (call it $K_Q$) and in particular if $P\in \Gamma$
then there is a cone $K_P$ (with apex at $P$) of rays through $P$, opening towards $\partial
C_1$, along which $u$ is nondecreasing. To see this just join $O$ to
$P$ and move $O$ in $C_2$. Fix $Q\in\partial C_2$; we now show that $u$ is
strictly increasing along $T\cap \Gamma$. Let $P_0\;\mbox{and}\;P_1\in
T$ as before and $K_{P_0}$ be the cone as
described above. Clearly,
$u(x)\ge u(P_0)$ in $K_{P_0}$. Recall that $u(P_0)<1$, and
if $S \in T\subset K_{P_0}$, close enough to $\partial C_1$, then
Lemma \ref{lm5} ensures that $u(S)>u(P_0)$.
To see this, let $L\ni S$ be a straight line such that $L\perp \partial
C_2$ and take $P=L\cap \partial C_1$.
Thus $u(S)-u(P_0)>0$ and $u(x)-u(P_0)\ge 0$ in $K_{P_0}$. The
Harnack inequality
implies that $u(P_1)>u(P_0)$.
Clearly $u$ is strictly
increasing along rays in a strict sub-cone of $K_{P_0}$. \qed

 One of our goals is to prove the strict positivity of $|Du|$,
whenever it exists, in $\Gamma$. To do this, we will need to derive
bounds on $u$, at points near $\partial \Gamma$, which in turn
require estimates of distances.

\begin{lemma}[Distance estimate] \label{lm8}
Let $0<t<1$, $t$ close to 1, be
  fixed. For $i=1,2$, let $\partial C_i^t=t\partial C_i$
  be the $t$ scaling of $\partial C_i$, and $l_i=\mathop{\rm dist}(O,\partial
  C_i)$. Then
\[
\mathop{\rm dist}(\partial C_i, \partial C_i^t)\ge l_i(1-t),\;i=1,2.
\]
\end{lemma}

\noindent{\bf Proof:}
Let $A_1\in \partial C_1$ and $A_2\in \partial C_1^t$ be such
that dist$(\partial C_1^t,\partial C_1)=|A_1-A_2|$. Since $\partial
C_1^t,\;\partial C_1$
are boundaries of $C^2$ convex sets, it follows that the supporting
hyperplanes $T_1$ (at $A_1$) and $T_2$ (at $A_2$) are parallel and the
segment $A_1A_2$ is orthogonal to both $T_1$ and $T_2$. We show that
$T_2$ is obtained form $T_1$ by the $t$ scaling. Let $R$ be the line
containing $O$ and $A_1$; this intersects $\partial C_1^t$ at
$B$. Clearly $|B|=t|A_1|$ and the supporting hyperplane $T$ at $B$ is
parallel to $T_1$ since $T$ is the $t$ scaling of $T_1$. Then
$T\parallel T_2$ and by convexity $T_2=T$, which in turn implies that
$T_2$ is the $t$ scaling of $T_1$. If $C_1$ is strictly convex this
would also imply $B=A_2$.
Let $L$ be the straight line containing $O$ and perpendicular to $T_1$ and
$T_2$. Let the intersection of $L$ with $\partial C_1^t$ be $C$, with
$T_2$ be $D$ and with $T_1$ be $E$. Then
$|E-D|=|A_1-A_2|=\mathop{\rm dist}(\partial C_1^t,\partial C_1)$. Clearly
$|D|=t|E|$ and $tl_1\le |D|$. The latter follows
since $D$ lies on the supporting hyperplane $T_2$ and $D\not\in C_1^t$. Thus
$|E-D|=(1-t)|D|/t\ge (1-t)l_1$. A similar argument now proves the
statement when $i=2$.\qed

 Define, for $\eta>0$, $C_{i,\eta}=\{x\in \Gamma
: \mathop{\rm dist}(x,\partial C_i)\le \eta \},\;i=1,2.$ For $Q\in\partial
C_2$, let $L=L(Q)$ be the line $\perp$ to $\partial
C_2$, $P=P(Q)=L\cap\partial
C_1$. Set $\eta_0=\delta/10$,
where $\delta$ is the number in Lemma \ref{lm6}, and $\Delta=$diameter$(C_1)$.
For every such $Q$, let $\bar{Q}=\bar{Q}(Q,t)\in\partial C_2^t$ be such
that $|Q-\bar{Q}|=\mathop{\rm dist}(Q,\partial C^t_2)$. We select $t$ such
that sup$_{Q\in \partial C_2}|Q-\bar{Q}|\le \eta_0$. Also take $u_t$ as in
Lemma \ref{lm7}.

\begin{lemma}[Bounds on $u$ and $u_t$] \label{lm9}
Let $t\in (0,1)$ be such that
  $\partial C_i^t\subset C_{i,{\eta_0}},\;i=1,2.$ Then
  there exist positive constants $\eta_1$ and $\eta_2$, depending only
  on the geometry of $\Gamma$, such that
\begin{itemize}
\item[(i)] $u(x)\le 1- \eta_1(1-t)$, for all $x\in \partial C_1^t$
and
\item[(ii)] $u_t(x)\ge \eta_2(1-t)$, for all $x\in \partial C_2$.
\end{itemize}
\end{lemma}

\noindent{\bf Proof:} To prove (i), we use Lemma \ref{lm5}. Let $Q\in \partial C_2$ and select $\bar{Q}\in \partial
C_2^t$ closest to $Q$. The line $L$ containing $\bar{Q}$ and $Q$ is orthogonal
to $\partial C_2^t$. Let $P=L\cap \partial C_1$, $P_t=L\cap\partial
C_1^t$; set $R=|Q-P|$ and $R_t=|\bar{Q}-P_t|$. Then $R_t= R+|\bar{Q}-Q|-|P-P_t|\le
R+\eta_0\le \Delta+\eta_0$. Applying Lemma \ref{lm5} to $u_t(Q)$, we see
$$u_t(y)\ge 1-(|Q-P_t|/R_t)=(R_t-|Q-P_t|)/R_t=|Q-\bar{Q}|/R_t.$$
Since $Q$
lies on $\partial C_2$, by Lemma \ref{lm8}, $|Q-\bar{Q}|\ge (1-t)l_2$, and we
have
$$
u_t(y)\ge \frac{(1-t)l_2}{\Delta+\eta_0}\;\;\mbox{on}\;\; \partial C_2.
$$
To prove (ii) we use
Lemma \ref{lm6}. For $x\in \partial C_1^t$, $\delta(x)\ge \mathop{\rm dist}
(\partial C_1, \partial C_1^t)\ge l_1(1-t)$.
We use \eqref{22} to conclude
\[
u(x)\le 1- \frac{C(\delta,\eta_0)\delta(x)}{2\delta},
\]
and this in turn yields,
$u(x)+ C(\delta) (1-t)l_1\le 1$ on $\partial C_1^t$.\qed

 We now begin our study of the boundary behaviour of
$\infty$-capacitary functions in convex rings. We will utilize the
observation in \eqref{1} in Section 2 and Lemma \ref{lm4}.
We recall and introduce some notations.
For $Q\in
  \partial C_2$, let $\nu(Q)$ denote the unit outer normal to
  $C_2$, and for $A\in \partial C_1$, let $\nu(A)$ stand for the unit
  outer normal to $C_1$. For $Q\in \partial C_2$, let $L=L(Q)\ni Q$ be the
  straight line with $L\perp \partial C_2$. Call $P=P(Q)=L\cap\partial
  C_1$; for $x\in
  L\cap \Gamma$, define $d(x)=d(x,Q)=|x-Q|$. For $A\in \partial C_1$,
  let $B_r(H)=B_r(H,A)\subset\Gamma$ be the interior ball at $A$,
  centered at $H$, and for $x$ on the segment formed by $HA$, set
  $\delta(x)=\delta(x,A)=|x-A|$. Note that $HA$ is directed
  along $\nu(A)$. The following notation is set up for directional
  derivatives of $u$ along $\nu(Q)$ and along $\nu(A)$. For
  $Z\in \partial \Gamma$, set
\[
\Delta(x,Z)=\Delta(x,\nu(Z))=\frac{du(x+\theta\nu(Z))}{d\theta}|_{\theta=0},\;\; x\in\Gamma;
\]
and when $x=Z$, we write $\Delta(Z)=\Delta(Z,Z)=\Delta(Z,\nu(Z))$, where
\[
\Delta(Z)=\begin{cases}
\lim_{\theta\rightarrow0^+}\frac{u(Z+\theta\nu(Z))-u(Z)}
{\theta}:& Z\in \partial C_2,\\[2pt]
\lim_{\theta\rightarrow 0^-}\frac{u(Z+\theta\nu(Z))-u(Z)}{\theta}:& Z\in
\partial C_1,\end{cases}
\]
whenever they exist. We make an
observation before we start. Suppose $A\in\partial C_1$ and $T_A$ is
the supporting hyperplane at $A$. Let $J\in \partial C_2$ be such that
the supporting hyperplane $T_J\parallel T_A$, i.e.,
$\nu(J)=\nu(A)$. This is possible since $C_1$ and $C_2$ are both
$C^2$. Recall the definitions of the hyperplanes $H^+_A$ and $H^-_J$;
see Section 2. Set
$G=H^{-}_{J}\cap H^{+}_{A}$ and define
$w_A(x)=1+\langle x-A, \nu(A)\rangle/R$, where
$R=R(A)=\mathop{\rm dist}(T_A,T_J)$. Note that convexity of $C_1$ implies $\langle
x-A, \nu(A)\rangle\le 0,\;\forall x\in H^+_A$; it
is easily seen that $w_A|_{T_J}=0$, $w_A|_{T_A}=1$ and
$\Delta_{\infty}w_A=0$ in $G$. Now $u\ge w_A=0$ on $T_J\cap\Gamma$ and
$w_A\le u=1$ on $\partial C_1$. Comparison in $G\cap \Gamma$ yields that
\begin{equation} \label{23}
u(x)\ge w_A(x)=1+\frac{\langle x-A, \nu(A)\rangle}{R},\quad x\in G\cap \Gamma.
\end{equation}
Thus, for $x\in G\cap \Gamma$ with $(x-A)\parallel \nu(A)$, i.e.,
$x=A-t\nu(A)$, for some $t>0$, we have
\begin{equation} \label{24}
1-u(x)\le \frac{\delta(x)}{R}\;\Rightarrow\;
\frac{1-u(x)}{\delta(x)}\le \frac{1}{R}.
\end{equation}

\begin{theorem}[A global maximum principle for $|Du|$] \label{thm3}
Let $u$ be the
  $\infty$-capacitary function in $\Gamma$; for $Q\in\partial C_2$, let
$d(x)$, $\delta(x)$, $L=L(Q)$ and $P=P(Q)$ be as described above. Then
the following are true.

\noindent (a) The normal derivatives of $u$ exist on $\partial\Gamma$,
i.e., $\forall A\in \partial C_1$ and $\forall Q\in \partial C_2$,
$$\Delta(A)>0,\;\;\Delta(Q)>0,\;\;\mbox{and}\;\;\max(\sup_Q\Delta(Q),\;
\sup_A\Delta(A))<\infty.
$$
\noindent(b) Let $x\in
  \Gamma$ and $Q\in\partial C_2$ be such that $|x-Q|=\mathop{\rm dist}(x,\partial
  C_2)$. If $x_1,\;x_2\in L(Q)$
  are such that $d(x_1)\le d(x_2)$,
  then
$$
0<\frac{u(x_2)-u(x_1)}{|x_1-x_2|}\le \frac{u(x_2)-u(Q)}{d(x_2)}\le
  \frac{u(x_1)-u(Q)}{d(x_1)}\le \Delta(Q).
$$
In particular,
$$ u(y)<u(ty)\le tu(y),\;\;\forall t\;\;\mbox{with}\;\;1\le t\le|Q-P|/|x-Q|,
$$
where $y=x-Q$ and $u(y)$ stands for the value $u(x)$. Moreover, if the
directional derivative of $u$ along $L$ exists at $x$,
then $0<\Delta(x,Q)\le\Delta(Q)$.

\noindent (c) Suppose $A\in \partial C_1$ and $B_r(H)$ is the interior ball at $A$.
Let $x_1,\;x_2\in HA$, with $x_1\ne x_2$, and
  $\delta(x_1)\le \delta(x_2)$. Then
$$
\frac{u(x_1)-u(x_2)}{|x_2-x_1|}\le
  \frac{u(A)-u(x_1)}{\delta(x_1)}\le
  \frac{u(A)-u(x_2)}{\delta(x_2)}\le
\Delta(A).
$$
In particular, if the directional derivative
  of $u$ along $HA$ exists at $x$, then $0<\Delta(x,A)\le\Delta(A)$.

\noindent (d) Finally, we have
$$
||Du||_{L^{\infty}(\Gamma)}\le \max\Big(\sup _{Q\in \partial C_2}\Delta(Q),
\;\sup _{P\in \partial C_1}\Delta(P)\Big).
$$
\end{theorem}

\noindent{\bf Proof.}
 {\it Part (a):} Let $A\in \partial C_1$ and $Q\in \partial C_2$.
 We first note that $u(Q)=0$ and $u(A)=1$. Since $C_2$ is
$C^2$ and
convex, we may find an outer ball $B_r(S)\subset\Gamma$ at $Q$; note
that $(S-Q)/|S-Q|=\nu(Q)$. Recall that $u>0$ in $B_r(S)$ and so an
application of part (a) of Lemma \ref{lm4}, yields that for $x\in SQ$,
\begin{equation} \label{25}
0<\frac{u(x)}{d(x)}=\frac{u(x)-u(Q)}{|x-Q|}\uparrow\;\;\mbox{as}\;\;d(x)\downarrow
0,\;\mbox{i.e.,}\;
x\rightarrow Q.
\end{equation}
Recalling Lemma \ref{lm6}, in particular \eqref{17}, we know that
$u(x)\le |x-Q|/d$ for an appropriate $D=D(Q)$. Thus
$$0<\frac{u(S)}{r}\le\frac{u(x)}{d(x)}=\frac{u(x)-u(Q)}{|x-Q|}
\le \frac{1}{D},\;\;\forall
x=Q+\theta \nu(Q),\;0<\theta\le \min (r,D).
$$
Letting $\theta \rightarrow 0^+ $ yields the result for $\Delta(Q)$.
To see the result for $\Delta(A)$, note that $1-u(x)>0$ in $\Gamma$
and $1-u(A)=0$. Let $B_r(H)\subset\Gamma$ be
the interior ball at $A$. Then $(A-H)/|A-H|=\nu(A)$. An application of
Lemma \ref{lm4} (a) to $1-u(x)$ in $B_r(H)$, and \eqref{24} implies that, for $x\in HA$,
\begin{equation} \label{26}
\begin{gathered}
\frac{1-u(x)}{\delta(x)}\;\uparrow \;\mbox{as}\;x\rightarrow
A,\;\;\mbox{and}\\
0< \frac{1-u(H)}{r}\le \frac{1-u(x)}{\delta(x)}
=\frac{u(A)-u(x)}{|x-A|}\le \frac{1}{R(A)}.
\end{gathered}
\end{equation}
The result for $\Delta(A)$ now follows. Note that $\Delta(Q)\le
1/D(Q)$ and $\Delta(A)\le 1/R(A)$. An inspection of \eqref{17} and \eqref{24} shows
that the supremum of each of these quantities is also finite.

\noindent {\it Part (b):} Let $Q$, $P$ and $L$ be as above; set $r=|P-Q|$ and
$w(x)=1-|x-P|/r$ in $B_r(P)$. From Lemma \ref{lm5},
$$
u(x)\ge w(x)=1-\frac{|x-P|}{r}=\frac{|x-Q|}{r}=\frac{d(x)}{r},
\;\;\;x\in L\cap B_r(P)\cap \Gamma.
$$
Now let $x_1,\;x_2\in L\cap \Gamma$ such that $d(x_1)\le
d(x_2)$. Then the ball $B=B_{d(x_2)}(x_2)\subset B_r(P)$ and $B\ni
x_1$. If we fix $x_2$ then
$$
v(z)=u(x_2)\left(1-\frac{|z-x_2|}{d(x_2)}\right)\le u(z),\;\forall z\in
B\cap\Gamma.
$$
To see this, note that (i) $u(z)\ge v(z)=0$
on $\partial B\cap \Gamma$, (ii) $u(x_2)=v(x_2)$, and (iii) $u(z)=1>
u(x_2)\ge v(z)$ on $B\cap\partial C_1$; now apply
comparison. For $z$ on the segment $x_1x_2$, $d(z)=d(x_2)-|z-x_2|$;
taking $z=x_1$ yields that $u(x_1)/d(x_1)\ge u(x_2)/d(x_2)$.
Thus for all $x\in L$, $u(x)/d(x)\uparrow$ as
$d(x)\downarrow 0$, i.e., as $x\rightarrow Q$.
This implies the assertion about scaling. Also
$$\frac{u(x_1)}{d(x_1)}\ge \frac{u(x_2)}{d(x_2)}\ge\frac{u(P)}{R}
=\frac{1}{R}\;\Rightarrow\;
0<\frac{u(x_2)-u(x_1)}{|x_2-x_1|}\le \frac{u(x_2)}{d(x_2)}
\le \frac{u(x_1)}{d(x_1)}.
$$
The positivity follows from Lemma \ref{lm7}. By considering $x$ close to $Q$,
applying \eqref{25} and part (a), we obtain the complete
assertion in part (b).

\noindent {\it Part (c):} We work with $v(x)=1-u(x)$ and use \eqref{26} much the way we
did in part (b).

\noindent {\it Part (d):} Let $x\in \Gamma$ and $\mu(x)=\mathop{\rm dist}(x,\partial
\Gamma)=$min$(\mathop{\rm dist}(x,\partial C_1)$, dist$(x,\partial C_2))$. Thus
ball $B_{\mu(x)}(x)$ either touches $\partial C_1$ or $\partial C_2$ or
both. In the first case, calling the point of tangency as $P$,
$u(P)=1=\sup _{\Gamma}u$. By Lemma \ref{lm4} (d) and part (c) above,
$$|Du|(x)\le D(M)=\Delta(P),\;\mbox{where}\; M=1.
$$
An analogous situation arises if the second case happens; calling the
point of tangency to be $Q$, we see that from part (b)
$$|Du|(x)\le D(m)=\Delta(Q),\;\mbox{where}\; m=0.$$
Clearly, the statement follows from part (a). \qed

\begin{remark} \label{rmk4} \rm
The assertion in Theorem \ref{thm3} (b) yields some type of
concavity of $u$ along $L$. Note also that if we take $Q=O$ then, along $L$,
$u(P)=u(|P|x/|x|)\le u(x)|P|/|x|$, implying thereby $u(x)\ge
|x|/|P|$. This fact has been derived in Lemma \ref{lm5}.
The inequalities in Theorem \ref{thm3} (a), clearly imply
\begin{eqnarray*}
u(x_2)&\le& u(x_1)+\Delta(Q)|x_2-x_1|\;\;\Rightarrow\;\;u(x)\le
\Delta(Q)|x-Q|,\;(\mbox{take}\;x_1=Q),\;\mbox{and}\\
u(x)&\ge& 1-\Delta(Q)|x-P|,\;(\mbox{take}\;x_2=P),\;\forall x\in QP.
\end{eqnarray*}
The latter is similar to Lemma \ref{lm6}.
\end{remark}

We now prove a global lower bound for $u$ in $\Gamma$. This consists in
taking the supremum (call it $w$) of affine functions that lie below
$u$; see \eqref{23} and \eqref{24}. It turns out that this lower bound is a solution in many special
cases. It is not clear whether this is actually a solution in more
general situations. We adopt the notations used in \eqref{23} and \eqref{24}; also
see Section 2. Let $P\in \partial C_1$,
$R(P)$ and $H_P^{\pm}$, as before. Suppose $T_P$ is the supporting
hyperplane at $P$ and $Q\in \partial C_2$ is such that the supporting
hyperplane $T_Q$, at $Q$, is parallel to $T_P$. Let $H_Q^{\pm}$ be
corresponding the half-spaces. Set $G=H^+_P\cap H_Q^-$ and
$\nu(P)$
is the unit outer normal to $\partial C_1$ at $P$.

\begin{lemma}[Universal lower bound for u] \label{lm10}
Let $P\in \partial C_1$,
$Q\in \partial C_2$, $G$ and $\nu(P)$ as described above. Set
\[
w_P(x)=1+\frac{\langle x-P,\nu(P)\rangle}{\delta(Q)},\;\;\forall x\in
  \Gamma\;\;\mbox{and}\;\;w(x)=\sup _{P\in\partial C_1}w_P(x).
\]
Then
$w|_{\partial C_1}=1,\;\; w|_{\partial
  C_2}=0,\;w$ is $\infty$-subharmonic and $u(x)\ge w(x),\;x\in\Gamma.$
\end{lemma}

\noindent {\bf Proof:} From \eqref{23}, we see that $w_P(x)\le u(x)$.
Clearly then $w(x)\le u(x)$ and it is well known
that $w(x)$ is $\infty$-subharmonic. Note that
$w(x)=w_P(x)=1-|x-P|/|P-Q|$ along the segment $PQ$. It is also to be noted
that in case $\Gamma$ is a spherical annulus or more generally if the
geometry of $\Gamma$ is such that
$C_1=\cup_{x\in C_2}B_{r}(x)=\{x\in \mathbb{R}^n:\;\mathop{\rm dist}(x,C_2)<r\}$,  for
some $r>0$, then $u(x)=w(x)$. As a matter of fact
$u(x)=\mathop{\rm dist}(x,\partial C_2)/r$.\qed

\begin{remark} \label{rmk5} \rm
Let $P\in \partial C_1$ and $Q\in \partial C_2$ be such that
$|P-Q|=\mathop{\rm dist}(\partial C_1,\partial C_2)=\delta.$ Clearly the
smallest ball in $\Gamma$, that touches both $\partial C_1$ and $\partial
C_2$, has radius $\delta/2$. From Lemma \ref{lm5} and \eqref{17} ($2l=\delta$),
it follows that $u(x)\le |x-Q|/\delta$ and
$u(x)\ge 1-(|x-P|/\delta)$. Note that the segment $PQ$ is
orthogonal to both $\partial C_1$ and $\partial C_2$. It then follows
that $u$ is linear on $PQ$ and $u(x)=|x-Q|/\delta,\;\forall x\in
PQ$.
\end{remark}

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

\noindent{\bf (a) Proof of part A of Theorem \ref{thm2}: Star-shapedness of level
  sets $\{u=t\}$ and cone condition.}\\
For $0<t<1$, let $\Gamma_t=\{x\in \Gamma: u(x)<t\}$, then
$\partial \Gamma_t=\partial C_2\cup\{u(x)=t\}$. This follows from
Lemma \ref{lm7}, since any $x\in \Gamma$, with $u(x)=t$, may be approached by points
where $u(x)<t$. This is seen by considering the straight line
containing $x$ and a point in $C_2$. Also at $x$, there are two cones
with the apex at $x$ such that $u>t$ in one and $u<t$ in the
other. Thus
$\{x\in\Gamma:u(x)=t\}$ satisfies an interior and an exterior cone
condition.
The shape of the cones depend on the
geometry of $\Gamma$. This also implies the set $\{u(x)=t\}$ is locally
Lipschitz. It is also clear that $\Gamma_t$ is star-shaped with
respect to any point in $C_2$.

\noindent{\bf (b) Proof of part B of Theorem \ref{thm2}: Strict
  positivity of the difference quotient and $|Du|$.}\\
We employ the idea of Lemma \ref{lm7} again. The selection of $O\in C_2$
will influence the lower bound $\lambda$, but it will stay
positive. We recall the notations and the scaling in Lemma \ref{lm7}. We
consider $P_0$ and $P_1$, in $\Gamma$, such that
$|P_0|=|P_0-O|<|P_1-O|=|P_1|$ and $t=|P_0|/|P_1|<1$.
We select $t$ close to $1$.
Note that $u_t(y)$ is
$\infty$-harmonic in $\Gamma^t=\Gamma(C_1^t, C_2^t)$ and
$${C_i^t\subset C_i,\;i=1,2,\;\;u_t|_{\partial C_1^t}=1 \;\;\mbox{and}\;\;
u_t|_{\partial C_2^t}=0.}
$$
From Lemma \ref{lm9} there exists a positive $\eta$, depending only on
the geometry, such that
$$
u|_{\partial C_1^t}+\eta(1-t)\le 1=u_t|_{\partial C_1^t}\;\;\mbox{and}\;\;
u|_{\partial C_2}=0\le \eta (1-t)+ u|_{\partial C_2}\le u_t|_{\partial C_2}.
$$
Thus $u_t\ge u+\eta(1-t)$ on $\partial(\Gamma^t\cap\Gamma)$.
Thus by comparison $u_t(y)\ge u(y)+\eta (1-t)$ in
$\Gamma^t\cap\Gamma$. Recalling that $P_0=tP_1$ and $u_t(y)=u(y/t)$,
it is seen that
$$
\frac{u_t(P_0)-u(P_0)}{|P_1-P_0|}=\frac{u(P_1)-u(P_0)}{|P_1-P_0|}\ge
\frac{\eta (1-t)}{(1-t)|P_1|}=\frac{\eta}{|P_1|}\ge \frac{\eta}{\Delta}>0,
$$
where $\Delta$ is the diameter of $C_1$. The result follows.

\section{Appendix}
In this section we put together results needed in the proofs of the theorems.
We will show that odd reflections of $\infty$-harmonic functions stay
$\infty$-harmonic and also include the proof of Theorem 1.1 [9].

 In the proof of Theorem \ref{thm1}, we required the the
following result which we now prove. Let $F$ be the $n-1$ dimensional
hyperplane given by $x_n=0$. Let us write $x=(\xi, x_n)$, where
$\xi=\xi(x)=(x_1,\ldots,x_{n-1})$;
also take
$$B^+=B^{+}_{R}(O)=\{x\in B_{R}(O):\;x_n>0\},\;B^-=B^-_{R}(O)
=\{x\in B_{R}(O):\;x_n <0\}
$$
and $F_R=F\cap B_{R}(O)$.

\begin{proposition}[Odd reflection of $u$] \label{prop1}
 Let $F$ and $O$ be as above and
  $u$ be $\infty$-harmonic in $B^+$; also assume that $u$ vanishes
  continuously on $F$. Define
$$
v(x)=v(\xi(x),x_n)=\begin{cases}
 u(\xi(x),x_n):& x_n\ge 0\\
 -u(\xi(x),-x_n) :& x_n\le 0.
 \end{cases}
$$
Then $v$ is $\infty$-harmonic in $B_{R}(O)$.
\end{proposition}

\noindent{\bf Proof:}
We will show that $\Delta_{\infty}v=0$ in the viscosity
sense. Let $\psi\in C^2$ and $P\in B_{R}(O)$
be such that $v-\psi$ attains a local minimum at $P$. We show that
$\Delta_{\infty}\psi(P)\le 0$. We will concern ourselves with the
cases when  $P\in B^-$ and when $P\in F_R$.

\noindent {\bf Case A ($\bf{P\in B^-}$):} Since $v(x)-\psi(x)\ge v(P)-
\psi(P)$, it
follows that $u(\xi,-x_n)-\phi(\xi,-x_n)\le
u(\xi(P),-P_n)-\phi(\xi(P),-P_n)$, where
$\phi(\xi,-x_n)=-\psi(\xi,x_n)$. Clearly
$\Delta_{\infty}\phi(\xi(P),-P_n)\ge 0$, since $(\xi(P),-P_n)\in B^+$
is a point of local maximum of $u-\phi$. Clearly then
$\Delta_{\infty}\psi(P)\le 0$.

\noindent{\bf Case B ($\bf{P\in F_R}$):} Note $v(P)=v(\xi,0)=0$. 
Thus $v(x)\ge
\psi(x)-\psi(P)$, which in turn implies
\begin{equation}
v(x)\ge \langle D\psi(P), x-p\rangle + \frac{1}{2}\langle
D^2\psi(P)(x-P),x-P\rangle + o(|x-P|^2),\;\;x\rightarrow P. \label{**}
\end{equation}
We study various situations. Suppose that $x\in F_R$, i.e.,
$x_n=x_n-P_n=0$. Since $v(\xi,0)=0$, \eqref{**} implies
$$
0\ge\sum_{i=1}^{i=n-1}D_i\psi(P)(x-P)_i+
\frac{1}{2}\sum_{i,j=1}^{n-1}D_{ij}\psi(P)(x-P)_i(x-P)_j +
o(|\xi(x-P)|^2),
$$
$x\rightarrow P$.
Now select $x$ such that $(x-P)_i=t$ and $(x-P)_j=0$,
$j=1,\ldots,n-1$, $j\ne i$. Then
$$
0\ge tD_i\psi(P)+\frac{t^2}{2}D_{ii}\psi(P)+ o(t^2),\;\;
t\rightarrow 0.
$$
Since this holds for all $t\in (-\varepsilon, \varepsilon)$ for small
$\varepsilon>0$, it follows that
$D_i\psi(P)=0,\;i=1,\ldots,n-1$. From \eqref{**}, it follows that
$$
v(x)=v(\xi,x_n)\ge
D_n\psi(P)(x-P)_n+\frac{1}{2}\sum_{i,j=1}^{n}D_{ij}\psi(P)(x-P)_i(x-
P)_j
+o(|x-P|^2),
$$
$x\rightarrow P$, and
$\Delta_{\infty}\psi(P)=(D_n\psi(P))^2 D_{nn}\psi(P)$. To prove this
is non-positive, we consider $x$'s such that 
$\xi(x)=\xi(P)\;(\mbox{i.e.},\; x_i=P_i,\;i=1,\ldots,n-1)$ and
$x_n=\pm t$, for small $t$. Let $t>0$, then the above inequality for
$\psi$ yields
\begin{gather*}
v(\xi(P),t)=u(\xi(P),t)\ge t D_n\psi(P)+ 
\frac{t^2}{2}D_{nn}\psi(P)+o(t^2),\\
v(\xi(P),-t)=-u(\xi(P),t)\ge-t D_n\psi(P)+
\frac{t^2}{2}D_{nn}\psi(P)+o(t^2),
\end{gather*}
as $t\rightarrow 0$. Adding the two inequalities, dividing by $t^2$
and letting $t\rightarrow 0$, we obtain that $D_{nn}\psi(P)\le
0$. Thus $\Delta_{\infty}\psi(P)\le 0$. The case of local maximum 
may
be handled analogously.\qed

\noindent {\bf Proof of Theorem 1.1 [9]}
 For easy reference, we now include the proof of Theorem 1.1 in 
[9], as
applied to our situation. This is essentially a repetition of the
proof in [9], nonetheless we provide details.

\begin{theorem}[Boundary Harnack Principle] \label{thm4}
Let
  $A_8=\{x:\;|\xi(x)|<8,\;0<x_n<16\}$,
  $A_1=\{x:\;|\xi(x)|<1,\;0<x_n<2\}$ and $X_0=(0,1)$. Let $u>0$ be
  $\infty$-harmonic in $A_8$. Then there exists a constant $C$,
  independent of $u$ but depending on the geometry, such that
$\sup _{A_{1}}\le C u(X_0)$.
\end{theorem}

\noindent{\bf Proof.} Recall \eqref{9} from the proof of Theorem 
\ref{thm1} and \eqref{6}.
Let us continue to call $u$ the extended function
obtained by the odd reflection about $F_8$. Let us note that 
Lemma \ref{lm3}
continues to apply to this extended function. Clearly then $u$ is
Lipschitz continuous in any sub-cylinder of $A_8$. Our selection of 
$X_0$ is different
from $z$. This means the Harnack
constant $M$ will need modification (see \eqref{9}). For $x$ with 
$1<x_n<3$
and $|\xi(x)|\le 2$, dist$(x,F_8)\ge 1$ and dist$(x,X_0)\le
2\sqrt{2}$. This implies $u(x)\le e^{2\sqrt{2}/1}u(X_0)$ by Lemma
\ref{lm2}. Take $M=e^{2\sqrt{2}}$ in \eqref{9}. The letters $l$, $m$ 
and $k$ denote
positive integers. Rewritten
\begin{equation} \label{27}
u(x)=u(\xi(x),x_n)\le \begin{cases}
Mu(\xi(x),2x_n):& |\xi(x)|\le 2,\;0<x_n<3/2,\\
Mu(X_0):& |\xi(x)|\le 2,\;1<x_n<3.
\end{cases}
\end{equation}
 We argue by contradiction. Suppose that there is a $Y_0\in A_1$
such that $u(Y_0)\ge M^{l+2}u(X_0)$, where $l$ is large and its 
value will
be determined later in the
proof. We now make an observation which will be used repeatedly 
in the proof.
If $x\in \bar{A}_1$ is such that dist$(x,F_8)=x_n\ge 2^{-l}$ then 
$u(x)\le
M^{l+1}u(X_0)$. This follows by an application of \eqref{27}. If 
$x_n\ge
1$ then \eqref{27} implies the result. If $0<x_n<1$ and $s$ is the
smallest integer such that $2^sx_n\ge 1$, then \eqref{27} implies
\begin{equation} \label{28}
u(x)=u(\xi(x),x_n)\le Mu(\xi(x), 2x_n)\le\ldots\le M^su(\xi(x),2^s
x_n)\le M^{s+1}u(X_0).
\end{equation}
Since $l\ge s$, $u(x)\le M^{l+1}u(X_0)$. It follows from \eqref{28} 
that
\[
\mathop{\rm dist}(Y_0,F_8)\le 2^{-l}.
\]
Let $K(r,Z)=K_r(Z)$ be the cylinder of dimension $r$ and center 
$Z$ (see
Section 2) and $\nu(Z,r)=\mathop{\rm osc}_{K(r,Z)}u$. Recall from 
Remark \ref{rmk2} that
$\nu(Z,r)\ge C \nu(Z,r/2)$. We consider $K(r,Y_0)$. Clearly
$\bar{K}(Y_0,2^{-l})\cap F_8\ne\emptyset$ and $\nu(Y_0, 2^{-l})\ge
u(Y_0)$. It follows then
\[
\nu(Y_0, 2^{-l+m})\ge C^m \nu(Y_0, 2^{-l})\ge C^m M^{l+2}u(X_0).
\]
Choose $m$ so that $C^m\ge 2M^2$. Thus
$\nu(Y_0,2^{-l+m})\ge 2 M^{l+4}u(X_0)$. Noting $u$ has been 
extended as
an odd function about $F_8$, there is a $Y_1$ such that
\[
Y_1\in K(Y_0,2^{-l+m})\cap \{x_n>0\}\;\;\mbox{and}\;\;u(Y_1)\ge
M^{l+4}u(X_0).
\]
Thus dist$(Y_1, F_8)\le 2^{-l-2}$ (if not, an argument along the lines
of \eqref{28} will imply $u(Y_0)\le M^{l+3}$) and
$\nu(Y_1,2^{-l-2+m})\ge C^m\nu(Y_1,2^{-l-2})\ge C^m u(Y_1)\ge
2M^{l+6}u(X_0)$. Once again there exists a
\[
Y_2\in K(Y_1,2^{-l-2+m})\cap \{x_n>0\}\quad\mbox{and}\quad 
u(Y_2)\ge
M^{l+6}u(X_0).
\]
Again dist$(Y_2, F_8)\le 2^{-l-4}$ and $\nu(Y_2,2^{-l-4+m})
\ge C^m\nu(Y_2,2^{-l-4})\ge
2M^{l+8}u(X_0)$. We obtain by induction a sequence of points 
$\{Y_k\}$
such that
\begin{equation} \label{29}
\begin{gathered}
\mathop{\rm dist}(Y_k,F_8)\le 2^{-l-2k},\;\;Y_k\in K(Y_{k-1},
2^{-l-2(k-1)+m})\cap \{x_n>0\},\\
u(Y_k)\ge   M^{l+2(k+1)}u(X_0).
\end{gathered}
\end{equation}
Recalling that $K(Y_{k-1}, 2^{-l-2(k-1)+m})$ is a cylinder with center
$Y_{k-1}$ with radius $2^{-l-2(k-1)+m}$ and long axis $2(2^{-l-2(k-
1)+m})$,
\[
|Y_{k}|\le |Y_{k-1}-Y_{k}|+|Y_{k-1}|\le
2^{-l-2(k-1)+m+1}+|Y_{k-1}|.
\]
Thus $|Y_{k}|\le 2^{-l+m+1}\sum_{j=1}^{k}2^{-2(j-1)}+|Y_{0}|\le
2^{-l+m+1}\sum_{j=0}^{k}2^{-2j}+1$. Now choose $l$ so large that
$|Y_{k}|\le 3/2,\;\forall k.$ Thus if $Y\in K(Y_k,2^{-l-2k+m})$ then 
$|Y|\le
3$. Thus each $K(Y_k, 2^{-l-2k+m})$ lies in a fixed sub-cylinder of
$A_8$. Letting $k\rightarrow\infty$ results in a contradiction.

The above proves that for some constant $C>0$, $u(x)\le C 
u(X_0)\le CMu(z)$.
This completes the proof.

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\end{document}