\documentclass[twoside]{article} \usepackage{amssymb} % font used for R in Real numbers \pagestyle{myheadings} \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} \\ % An elementary proof of the Harnack inequality for non-negative infinity-superharmonic functions % \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{} % \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} \renewcommand{\theequation}{\thesection.\arabic{equation}} \catcode@=11 \@addtoreset{equation}{section} \catcode@=12 \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 $$\Delta_{\infty}u=\sum_{i,j=1}^{n}\frac{\partial u}{\partial x_{i}}\frac{\partial u}{\partial x_{j}}\frac{\partial^{2}u}{\partial x_{i}\partial x_{j}}.$$ 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 $|x|\ne t$, $$\Delta_{\infty} d^{\alpha}(x)=\alpha^{3}(\alpha-1)d(x)^{3\alpha-4}.$$ \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 $z$. 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