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 $0__0$, 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 $0 0$ 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 __