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\markboth{\hfil An elementary proof of the Harnack inequality \hfil
EJDE--2001/44}
{EJDE--2001/44\hfil Tilak Bhattacharya \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc Electronic Journal of Differential Equations},
Vol. {\bf 2001}(2001), No. 44, pp. 1--8. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu (login: ftp)}
\vspace{\bigskipamount} \\
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An elementary proof of the Harnack inequality for
non-negative infinity-superharmonic functions
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\thanks{ {\em Mathematics Subject Classifications:} 35J70, 26A16.
\hfil\break\indent
{\em Key words:}
Viscosity solutions, Harnack inequality, infinite harmonic operator,
\hfil\break\indent distance function.
\hfil\break\indent
\copyright 2001 Southwest Texas State University. \hfil\break\indent
Submitted January 15, 2001. Revised May 17, 2001. Published June 14, 2001.} }
\date{}
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\author{ Tilak Bhattacharya }
\maketitle
\begin{abstract}
We present an elementary proof of the Harnack inequality for non-negative
viscosity supersolutions of $\Delta_{\infty}u=0$. This was originally proven
by Lindqvist and Manfredi using sequences of solutions of the $p$-Laplacian.
We work directly with the $\Delta_{\infty}$ operator using the distance
function as a test function. We also provide simple proofs of the Liouville
property, Hopf boundary point lemma and Lipschitz continuity.
\end{abstract}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
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\section {Introduction}
Our effort in this note will be to provide an
elementary proof of the Harnack inequality for nonnegative
$\infty$-superharmonic functions.
The $\infty$-harmonic operator, in a domain $\Omega\subset
\mathbb{R}^{n},\;n\ge 1$, is defined as
\begin{equation}
\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}}.
\end{equation}
A function $u=u(x_{1},x_{2},\ldots,x_{n})$ is said to be $\infty$-harmonic
if $u$ is a solution of $\Delta_{\infty}u=0$.
In this work, by a solution $u$ we will
mean a viscosity solution. For definitions and background for such equations
see \cite{a1,c1,e1,j1,j2,j3,j4,l1,l2,l3,r1}
and the references therein, in particular we mention the
remarkable work of Jensen \cite{j1}. These references also discuss the
relevance and
the importance of the notion of viscosity solutions in the context of such
nonlinear equations.
We may also define $\infty$-superharmonicity:
$u$ is said to be $\infty$-superharmonic if
it is a viscosity supersolution of
(1.1) i.e., $u$ is
lower semicontinuous and satisfies
\begin{eqnarray*}
-\Delta_{\infty}u
\ge0,
\end{eqnarray*}
in the viscosity sense. For our work we will take $u\ge 0$.
Let $\Omega \subset\mathbb{R}^{n}$, be a bounded
domain and $\partial \Omega$ be its boundary; also let $B_{r}(P)$ denote the
open ball in $\mathbb{R}^n$, of radius $r$ and center P.
The main result of this work is
\begin{theorem}
Let $u\ge 0$ be a viscosity supersolution of (1.1) in $\Omega$.
Also let $P\in\;\Omega$, $0n$. We thank Peter Lindqvist and
Juan Manfredi for having read an earlier version of this manuscript and for
their comments. We also thank the referee for comments that have clarified
the
presentation greatly.
For more information on the Harnack inequality, in
this context, see \cite{e1,l2}.
\section {Preliminary results and the proof of the main theorem}
We start with the following rather elementary result. It will set
the stage for showing that the function $d$ acts as a barrier for the
$\infty$-harmonic operator.
\begin{lemma} %{\bf Lemma 1.}
Let $B_t(0)$ be the open ball in $\mathbb{R}^{n}$ centered at
$0$ and radius $t$ and let $d(x)=
\mathop{\rm dist}(x,\partial B_t(0))=t-|x|$, $\forall \;x\in
\mathbb{R}^{n}$.
Then for $x\ne 0$ and $|x|\ne t$,
\begin{equation}
\Delta_{\infty} d^{\alpha}(x)=\alpha^{3}(\alpha-1)d(x)^{3\alpha-4}.
\end{equation}
\end{lemma}
\paragraph{Proof.} We observe that
\begin{eqnarray*}
D_id^{\alpha}&=&\alpha
d^{\alpha-1}\left(\frac{-x_i}{|x|}\right)\quad \mbox{and}\\
D_{ij}(d^{\alpha})&=&\alpha d^{\alpha-1}\left(
\frac{x_ix_j}{|x|^3}-\frac{\delta_{ij}}{|x|}\right)
+\alpha(\alpha-1)d^{\alpha-2}
\frac{x_ix_j}{|x|^2}.
\end{eqnarray*}
Thus we see that
\begin{eqnarray*}
\Delta_{\infty} d^{\alpha}&=& (\alpha d^{\alpha-1})^2 \frac{x_ix_j}{|x|^2}
\left[
\alpha(\alpha-1)d^{\alpha-2}
\frac{x_ix_j}{|x|^2}
+ \alpha d^{\alpha-1}\left
(\frac{x_ix_j}{|x|^3} - \frac{\delta_{ij}}{|x|}\right)\right]\\
&=&\left(\alpha d^{\alpha-1}\right)^{2}\left[ \alpha (\alpha-1)
d^{\alpha-2}+ \alpha
d^{\alpha-1}\left(\frac{1}{|x|}-\frac{1}{|x|}\right)\right]\\
&=& \alpha^{3}(\alpha-1) d^{3\alpha-4}.\quad\diamondsuit
\end{eqnarray*}
We now prove the main estimate mentioned in the introduction. It appears
as part (i) of Lemma 2.
In what follows, we will take $u$
to be
lower semicontinuous. Note that the estimates are stated in terms of
distances of points from the boundary of a certain ball they lie in.
The basic observation is that the function $ku(x)-d(x)$ attains its infimum
at the center of the ball which is being used for defining $d$. Here $k$
is a suitable scaling constant. This fact leads to the estimate pointed
out in Section 1.
\begin{lemma} %{\bf Lemma 2.}
Let $P\;\in\Omega$, $r\le \mathop{\rm dist}(P,\partial\Omega)$ and
$B_{r}=B_r(P)$
be the open ball of radius $r$ and center $P$. Set
$d(x)=r-|x-P|=\mathop{\rm dist}(x,\partial B_{r})$ for all
$x\;\in\Omega$. Let $u(x)\ge 0$ solve
$-\Delta_{\infty}u \ge 0$ in the viscosity
sense. Assume $u(P)>0$ and if $k>0$ is such that $d(P)=ku(P)=r$,
then $\forall\; x\in B_{r}$,
\begin{enumerate}
\item[(i)] $u(x)\ge u(P) \frac{d(x)}{d(P)}$;
\item[(ii)] $u(x)-u(P)\geq -|x-P|/k$ or $u(x)+\frac{|x-P|}{k}\ge u(P)$.
\end{enumerate}\end{lemma}
\paragraph{Proof.}
We will scale the functions $u$ and $d$
as follows. We define $u_c(x)=cu(x)/r$ and $v(x)=d(x)/r$ where $01$ and
we look at $w_{\alpha}(x)=u_{c}(x)-v^{\alpha}(x)$. Clearly,
$$w_{\alpha}(P)=u_{c}(P)-1<0\;\mbox{and}\;w_{\alpha}(x)\ge
0\quad\mbox{on }\partial B_{r}.
$$
Choose $\alpha$ sufficiently close to 1 so that the point of infimum
of $w_{\alpha}$, denoted by $x_{c,\alpha}$, is different from $P$ and
$$w_{\alpha}(x_{c,\alpha})0,
$$
since $\alpha>1$, which results in a contradiction. Thus the infimum
of $w$ occurs at $P$. Hence, $
u_{c}(x)-v(x)\ge u_{c}(P)-1$, i. e.,
$$
\frac{cu(x)}{r}-\frac{d(x)}{r}\ge
\frac{cu(P)}{r}-1,\;\forall\;x\in\;B_{r}\;\mbox{and}\;\forall\;c0$, is open. Suppose $y$ is a limit point of $S$.
There are two possiblities: either
$y$ is in $\partial \Omega$ or it is in the interior of $\Omega$. If the
latter
happens then there is a $B_{\delta}(y)$ that lies completely in
$\Omega$. Clearly there is a point $z\in\;S$ that lies in
$B_{\delta/4}(y)$. Now
$u(z)>0$ and $B_{\delta/2}(z)\subset B_{\delta}(y)$ with
$y\in\;B_{\delta/2}(z)$. Part (i) of Lemma 2 implies $u(y)\ge u(z)/2>0$.
Thus
$S$ is both open and closed and $\Omega$ being connected we have
$S=\Omega$. \medskip
We now present the proof of the main result.
\paragraph{Proof of the main Theorem.}
By Remark 2, $u>0$ in $\Omega$. Let Q be the point of infimum of $u$ on
$B_{r/2}(P)$. By Remark 1,
$u(Q)\ge u(P)/2$. Let $x$ be in $B_{r/2}(P)$. Let $R$ be the midpoint
of the
segment joining $x$ to $P$. Let $l=|x-P|$, then by applying Remark 1
to the $B_{l}(x)$, we see that
$u(R)\ge u(x)/2$. Clearly, $l\le r/2$ and $|R-P|\le r/4$. Finally,
by applying part (i) of Lemma 2 to the ball $B_{r/2}(R)$ ( now P lies
in this ball with distance from P to the boundary of this ball is
at least $r/2-r/4=r/4$), we get $u(P)\ge u(R)/2$. Putting these
inequalities together we obtain
$$u(Q)\ge u(P)/2\ge u(R)/4\ge u(x)/8,\quad\forall\;x\in\;B_{r/2}(x).$$
\section{Appendix}
We now present the proofs of the Hopf boundary lemma, the Liouville
property and the local Lipschitz regularity. They follow from the
basic estimates proved in Lemma 2.
\paragraph{Remark 3 (The Liouville Property).}
If $u\ge 0$ is a viscosity supersolution of (1.1)
defined on all of $\mathbb{R}^{n}$ then it is a constant function. To see
this, we take two distinct points $x\;\mbox{and}\;z$ in $\mathbb{R}^{n}$.
Consider the ball $B_{R}(z)$ with $R>|x-z|$. By part (i) of Lemma 2,
$$u(z)\le u(x)\frac{d(z)}{d(x)},\;\mbox{and}\;d(z)=d(x)+|x-z|=R.$$
Letting $R\rightarrow\infty$ we get $u(z)\le u(x)$. Switching the
roles of $x$ and $z$ we get the reverse inequality.
\paragraph{Remark 4 (The Hopf Boundary Point Lemma)}.
We drop the requirement that $u\ge 0$.
Let $Q\in\partial\Omega$ be such that there is a ball
$B_{r}(P)\subset\Omega$
with
$Q\in\partial\Omega\cap\partial B_{r}$. Assume that
$u(Q)=inf_{\Omega}\;u$ and $u(P)>u(Q)$. We apply part (i) of Lemma 2
to the function $v(x)=u(x)-u(Q)\ge 0$ in the ball $B_{r}(P)$. Then
$$v(x)\ge v(P) \frac{d(x)}{d(P)}= v(P) \frac{d(x)}{r}.$$
Clearly, we obtain
$$ \frac{u(x)-u(Q)}{d(x)}\ge \frac{u(P)-u(Q)}{d(P)}.$$
Implying then
$$ liminf_{x\rightarrow Q}\frac{u(x)-u(Q)}{d(x)}\ge
\frac{u(P)-u(Q)}{d(P)}>0.$$
Also see the work in \cite{r1} in this regard.
\quad\quad$\diamondsuit$ \medskip
The estimates in Lemma 2 also imply local Lipschitz continuity of $u$.
See \cite{l2} in this regard. We first prove this for $u\ge 0$ which are
supersolutions of (1.1). A somewhat modified estimate continues to hold
if the assumption of nonnegativity is dropped. We do this in Remark 5.
\begin{lemma}[Lipschitz Continuity] % Lemma 3.
Let $y$ be in $\Omega$ and
$\delta=\mathop{\rm dist}(y,\partial\Omega)$. Then for all $x$ in
$B_{\delta/4}(y)$, we have
$$ |u(x)-u(y)|\le \frac{4u(y)|x-y|}{\delta}\le
\frac{4M|x-y|}{\mathop{\rm dist}(y,\partial\Omega)},$$
where $M=sup u$.
\end{lemma}
\paragraph{Proof.} We apply part (ii) of Lemma 2. Let $x$ and $y$ be as in
the statement of the lemma. Clearly,
$$
u(x)-u(y)\ge -\frac{|x-y|}{k}=-\frac{u(y)|x-y|}{r}.
$$
The ball $B_{\delta/2}(x)$ lies in $B_{\delta}(y)$ and contains $y$.
Another application of part (ii) of Lemma 2 to $B_{\delta/2}(x)$ implies
$$
u(y)-u(x)\ge -\frac{u(x)|x-y|}{\delta/2}.
$$
Putting together these two inequalities, we obtain
$$
-\frac{u(y)|x-y|}{\delta}\le u(x)-u(y)\le \frac{2u(x)|x-y|}{\delta}.
$$
The conclusion is now obtained by observing that $u(y)\ge u(x)/2$
(apply part (i) of Lemma 2 to $B_{\delta/2}(x)$ with the observation
that $|y-x|\le \delta/4$).\quad$\diamondsuit$
\paragraph{Remark 5.} In order to prove Lemma 3 for a more general $u$,
we proceed as follows. First redefine $\delta=\mathop{\rm dist}(y,\partial
\Omega)/2$. Let $m=inf_{B_{\delta}(y)}\;u$; clearly, $v(x)=u(x)-m\ge 0$ in
$B_{\delta}(y)$ and is a supersolution of (1.1). Going through the
proof of Lemma 4, we find that
\begin{eqnarray*}
|u(x)-u(y)|&=&|v(x)-v(y)|
\le \frac{4v(y)|x-y|}{\delta}\\
&\le& \frac{(u(y)-m)|x-y|}{\delta}\\
&\le& \frac{8 sup|u|\;|x-y|}{\mathop{\rm dist}(y,\partial\Omega)}.
\end{eqnarray*}
Finally, we state a somewhat more precise version of part (i) of Lemma 2.
\paragraph{Remark 6.} Let $B_{r}(z)$ be in $\Omega$. We show that for $x\in
B_{r}(z)$, the function $u(x)/d(x)$ is increasing along radial lines
emanating from $z$. To state this from precisely, let $e$ be a unit
vector in $\mathbb{R}^n$ and $0