\documentclass[twoside]{article}
\usepackage{amsfonts, amssymb} % used for R in Real numbers
\pagestyle{myheadings}
\markboth{ A remark on infinity harmonic functions }
{ Michael G. Crandall \& L. C. Evans }
\begin{document}
\setcounter{page}{123}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
USA-Chile Workshop on Nonlinear Analysis, \newline
Electron. J. Diff. Eqns., Conf. 06, 2001, pp. 123--129.\newline
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)}
\vspace{\bigskipamount} \\
%
A remark on infinity harmonic functions
%
\thanks{ {\em Mathematics Subject Classifications:} 35D10, 35J60, 35J70.
\hfil\break\indent
{\em Key words:} infnity Laplacian, degenerate elliptic, regularity,
fully nonlinear
\hfil\break\indent
\copyright 2001 Southwest Texas State University.
\hfil\break\indent Published January 8, 2001.
\hfil\break\indent
L.C.E. Supported by NSF Grant DMS-942342 } }
\date{}
\author{ Michael G. Crandall \& L. C. Evans }
\maketitle
\begin{abstract}
A real-valued function $u$ is said to be {\it infinity harmonic} if it
solves the nonlinear degenerate elliptic equation
$-\sum_{i,j=1}^nu_{x_1}u_{x_j}u_{x_ix_j}=0$
in the viscosity sense. This
is equivalent to the requirement that $u$ enjoys comparison with cones,
an elementary notion explained below. Perhaps the primary open problem
concerning infinity harmonic functions is to determine whether or not
they are continuously differentiable. Results in this note reduce the
problem of whether or not a function $u$ which enjoys comparison with
cones has a derivative at a point $x_0$ in its domain to determining
whether or not maximum points of $u$ relative to spheres centered
at $x_0$ have a limiting direction as the radius shrinks to zero.
\end{abstract}
\newcommand{\iprod}[1]{\langle #1\rangle}
\section{Introduction}
Let $U\subset {\mathbb R}^n$ be an open set and $u:U\rightarrow
{\mathbb R}$. Then $u$
is said to be {\it infinity harmonic} in $U$ if it is continuous and
satisfies
\begin{equation}
-\Delta_\infty u=-\sum_{i,j=1}^nu_{x_1}u_{x_j}u_{x_ix_j}=0
\quad\mbox{in } U\label{eqa}
\end{equation}
in the sense of viscosity solutions of nonlinear possibly degenerate
elliptic equations. We refer to Crandall, Evans and Gariepy \cite{ceg}
for terminology not defined here; however, the theory
of viscosity solutions is not needed to read this paper. Indeed, this
note employs only an elementary property which is shown
to be equivalent to infinite harmonicity in \cite{ceg}. One says that
$u$ enjoys comparison with cones from above in $U$ if for every open set
$V\Subset U$ (i.e., $V$ is an open subset of $U$ whose closure is
a compact subset of $U$),
$a,b\in{\mathbb R}$ and $y\in{\mathbb R}^n$ such that
\[
u(x)\le a+b|x-y|\quad\mbox{for }x\in \partial(V\setminus\{y\})
\]
one then has $u(x)\le a+b|x-y|$ for $x\in V$. Similarly, one defines
comparison with cones from below and then defines ``$u$ enjoys comparison
with cones" as the conjunction of the two one-sided notions.
In \cite{ceg} it is shown that $-\Delta_\infty
u=0$ in $U$ in the viscosity sense if and only if $u$ enjoys comparison
with cones in $U$.
We refer to \cite{ceg} for orientation and
references beyond the following remarks. The equation
(\ref{eqa}) arises as the Euler equation for the problem of minimizing
$\|Du\|$ in the $L^\infty$ norm subject to Dirichlet conditions. That
is, (\ref{eqa}) guarantees that
if $V\Subset U$ and $u=v$ on $\partial V$, then
$\|Du\|_{L^\infty (V)}\le\|Dv\|_{L^\infty (V)}$. We call this
property (AML) for ``absolutely minimizing Lipschitz". This aspect of
(\ref{eqa}) was introduced and studied by Aronsson \cite{arona}, \cite{aronb},
\cite{aronc} in the case of smooth functions. However, the nonsmooth
function
$u(x_1,x_2)=x_1^{4/ 3}-x_2^{4/3}$ solves (\ref{eqa}) (equivalently, enjoys
comparison with cones) in ${\mathbb R}^2$. This function has
H\"older continuous first partial derivatives but is not twice
differentiable along the lines $x_1=0, x_2=0$. This example, again due to
Aronsson, sets limits on the reliability of results derived
for (\ref{eqa}) via pde arguments valid for smooth functions. For
example, Aronsson showed that classical solutions of
(\ref{eqa}) in two variables are either constant or have a nowhere
vanishing gradient. However, the example is not constant and the
gradient vanishes at the origin. Classical solutions of
(\ref{eqa}) also have the property that
$|D(u(X(t))|$ is constant when $dX(t)/dt=Du(X(t))$ (gradient flow). In
fact, this property is easily seen to be equivalent to (\ref{eqa}) for
smooth functions. However $|Du|$ is not constant on gradient flow lines
in the example.
R. Jensen \cite{jen} used approximation of (\ref{eqa}) by the $p$-Laplace
equation to prove the equivalence of (\ref{eqa}) in the viscosity sense
and the extremal property (AML). He proved as well that the Dirichlet
problem for (\ref{eqa}) was uniquely solvable. Jensen used approximation
by the $p$-Laplacian in his remarkable analysis. A direct elementary
proof that functions enjoying comparison with cones have the property
(AML) was given in \cite{ceg}.
It remains a mystery as to whether or not solutions of (\ref{eqa}) are
everywhere differentiable. Our main results imply that $u$ can
only fail to be differentiable via complex phenomena --- roughly
speaking, simple bad behaviors that do not scale to planes are ruled
out.
More precisely, we prove that if $u$ enjoys comparison with cones in
$U$, $x_0\in U$,
$r_j\downarrow0$ and
\begin{equation}
v(x)=\lim_{j\rightarrow \infty}{u(r_jx+x_0)-u(x_0)\over r_j}\label{eqf}
\end{equation}
holds locally uniformly in ${\mathbb R}^n$, then $v$ is linear.
As a consequence of the above and arguments in the proof, we establish
an equivalent condition to differentiability of $u$ at $x_0$. Suppose
$x_0$ is not a point at which
$Du(x_0)$ exists and $Du(x_0)=0$. Then $u$ is differentiable at
$x_0$ if and only if the following condition holds: if
$|z_r-x_0|=r$ and
$u(z_r)=\max_{\{|z-x_0|=r\}}u(z)$ for small $r>0$, then
\begin{equation}
\lim_{r\downarrow 0}{z_r-x_0\over r}\quad\mbox{exists.}
\label{eqc}
\end{equation}
We reduce the problem of showing $u$ is differentiable
at $x_0$ to verifying (\ref{eqc}). However, we have not yet determined
if (\ref{eqc}) is always satisfied.
\section{The Proofs}
We employ the notation
\[
S_r(x_0)=\{z\in{\mathbb R}^n;\ |z-x_0|=r\}
\]
for the sphere of radius $r$ centered at the $x_0$.
Let $u$ enjoy comparison with cones in the open subset $U$ of
${\mathbb R}^n$. We
note that only this simple property is needed in all that
follows. The equivalence with the satisfaction of the
equation is not needed, nor are devices more conveniently dealt with in
terms of the equation (changes of variables, etc.).
It follows from Lemma 2.5 of \cite{ceg} that $u$ is
locally Lipschitz continuous. The functions
\[
L^+_r(y)=\max_{z\in S_r(y)}{u(z)-u(y)\over r}
\quad\mbox{and}\quad L^-_r(y)=\min_{z\in S_r(y)}{u(z)-u(y)\over r}
\]
are defined for $y\in U$ and $r<\mathop{\rm dist}(y,\partial U)$
as are the limits
\begin{equation}
L^+ (y)=\lim_{r\downarrow 0}L^+_r(y)
\quad\mbox{and}\quad L^- (y)=\lim_{r\downarrow 0}L^-_r(y).\label{eqvlim}
\end{equation}
The latter quantities are well defined by Lemma 2.4 of \cite{ceg}, which
establishes that $L^+_r(y)$ (respectively, $L^-_r(y)$) is nonnegative
(respectively, nonpositive) and nondecreasing (respectively,
nonincreasing) in $r$. Moreover, by Lemma 2.7 of \cite{ceg}
\begin{equation}
L^+(y)=-L^-(y).\label{eqequal}
\end{equation}
We will show that if $0\in U$, $r_j\downarrow 0$ and $v\in
C({\mathbb R}^n)$ is
the locally uniform limit
\begin{equation}
v(x)=\lim_{j\rightarrow \infty}{u(r_jx)-u(0)\over r_j}\label{eqlin}
\end{equation}
then $v$ is necessarily linear. Note that as $u$ is locally
Lipschitz continuous, any sequence $R_k\downarrow 0$ has a
subsequence $r_j\downarrow
0$ for which the limit (\ref{eqlin}) is defined on ${\mathbb R}^n$ and the
convergence is uniform on any bounded subset of ${\mathbb R}^n$.
Define
\[
L_0=L^+(0)=-L^-(0).
\]
This number is computed from the original
function $u$. Let $v$ be given by (\ref{eqlin}). It is obvious that
locally uniform limits of functions enjoying comparison with cones
likewise enjoy comparison with cones, so $v$ enjoys comparison with
cones. Hereafter the quantities
\begin{equation}
L^+_r(y)=\max_{z\in S_r(y)}{v(z)-v(y)\over r}
\quad\mbox{and}\quad L^-_r(y)=\min_{z\in S_r(y)}{v(z)-v(y)\over r}\label{eqvvv}
\end{equation}
are computed from $v$. Correspondingly, $L^+(y), L^-(y)$ are now defined
from $v$ and not $u$.
To see that $v$ is linear, we will prove that
\begin{equation}
-L^-_r(y), \ L^+_r(y)\le L_0\quad\mbox{for $y\in{\mathbb R}^n$, 0$<
r$,}\label{eqexact}
\end{equation}
and
\begin{equation}
-L^-(0)=L_0= L^+(0). \label{eqexacta}
\end{equation}
Let us first show why (\ref{eqexact}), (\ref{eqexacta}) imply that $v$
is linear. First (\ref{eqexact}), (\ref{eqexacta}) and $L^+(y)\le
L^+_r(y)$ together with Lemma 2.7 (ii) of \cite{ceg} imply that
\begin{equation}
L_0 \mbox{ is the least Lipschitz constant for } v.\label{eqlip}
\end{equation}
Next, let $z_r^{\pm}\in
S_r(0)$ be such that
\begin{equation}
L^+_r(0)= {v(z^+_r)-v(0)\over r}={v(z^+_r) \over r}, \quad L^-_r(0)=
{v(z^-_r)-v(0)\over r}={v(z^-_r)\over r}.\label{eqvvr}
\end{equation}
Since $L_0\le L^+_r(0)$ and $L_0\le -L^-_r(0)$ because of
(\ref{eqexacta}) and monotonicity, it follows from (\ref{eqexact}) and
(\ref{eqvvr}) that
\begin{equation}
L^+_r(0)= -L^-_r(0)= L_0= {v(z^+_r)-v(z^-_r)\over 2r
}.\label{eqmore}
\end{equation}
The function
\[
g(t)= v(z^-_r+t(z^+_r-z^-_r))-v(z^-_r)
\]
is Lipschitz with constant $L_0 |z^+_r-z^-_r|$ by (\ref{eqlip}) and
satisfies
\[
|g(1)-g(0)|=|g(1)|=|v(z^+_r)-v(z^-_r) |=2rL_0
\]
by (\ref{eqmore}). Hence $2rL_0\le L_0 |z^+_r-z^-_r|$ and this implies
that
\begin{equation}
z^+_r=-z^-_r\label{eqlina}
\end{equation}
since both points lie in $S_r(0)$. Moreover, $|g(1)-g(0)|$
is then the Lipschitz constant for $g$, which forces $g$ to be linear for
$0\le t\le 1$. We conclude that
\[
v(z^-_r+t(z^+_r-z^-_r))-v(z^-_r)=L_02rt\quad\mbox{for } 0\le t\le 1.
\]
The net result of this analysis is that for each $r>0$ there is a unique
point $z_r^+\in S_r(0)$ such that $v$ is linear on the line segment
joining $-z_r^+$ and $z_r^+$ and achieves its Lipschitz constant on these
segments. The uniqueness follows from the result (\ref{eqlina}). Then all
the points $z_r^+$ lie on a common line and thus there is a line through
the origin on which
$v$ is linear and attains its Lipschitz constant. This forces $v$ to be
linear. Indeed, we may change coordinates to assume that the line through
the origin is along the first coordinate axis $(x_1,0,\cdots,0)$,
$-\infty< x_1< \infty$. That is, we assume that
$v(x_1,0,\ldots,0)-L_0x_1\equiv0$. Write
$(x_1,y)$ for a general point in
${\mathbb R}^n$; ``$y$" contains the last $n-1$ coordinates. Consider
$w(x_1,y)=v(x_1,y)/L_0$, which has 1 as a Lipschitz constant and
satisfies $w(x_1,0)\equiv x_1$. Then we have
\begin{eqnarray}
|w(x_1,y)-w(s,0)|^2 & = & |w(x_1,y)-x_1+x_1-w(s,0)|^2 \label{eqsa}
\\ & = & |w(x_1,y)-x_1+x_1-s|^2 \nonumber\\
& = &
|w(x_1,y)-x_1|^2+2(x_1-s)(w(x_1,y)-x_1)+|x_1-s|^2\nonumber\\
\nonumber&\le&|x_1-s|^2+|y|^2.
\end{eqnarray}
Here the last inequality is due to 1 being a
Lipschitz constant for $w$:
\[
|w(x_1,y)-w(s,0)|^2\le |(x_1,y)-(s,0)|^2.
\]
We conclude that $2(x_1-s)(w(x_1,y)-x_1)\le |y|^2$ where $s$ is free. This
can only be if $w(x_1,y)-x_1\equiv0$.
Let $\iprod{\cdot,\cdot}$ be the Euclidean inner-product. For later use, let
us observe that if $v(x)=\iprod{p,x}$ then $|p|$ is determined by
(\ref{eqlip}):
\begin{equation}
|p|=L_0=\lim_{r\downarrow0}\max_{|z|=r}{u(z)-u(0)\over r}.\label{eqnorm}
\end{equation}
\smallskip
\noindent{\bf Remark:}\ The proof just given is essentially a special
case of the proof of Lemma 4.2 of \cite{ceg}. There is is shown that if
$w$ has 1 as a Lipschitz constant then
$x_1\rightarrow w(x_1,y)-x_1$
is nonincreasing and its limits as $x_1\rightarrow\pm\infty$ are independent of
$y$. In our case, when $y=0$ these limits are 0. In view of the
monotonicity, we then have $w(x_1,y)-x_1\equiv 0$, again establishing the
linearity.
\smallskip
It remains to prove (\ref{eqexact}), (\ref{eqexacta}). We treat the case
of the superscript ``$^+$". First fix
$y$ and consider
$z\in S_r(y)$ such that
\[
L^+_r(y)= {v(z)-v(y)\over r}=\lim_{j\rightarrow \infty}{u(r_j z)-u(r_j y)\over
r_jr}.
\]
We have $r_jz\in S_{r_jr}(r_jy)$ and then
\begin{equation}
{u(r_j z)-u(r_j y)\over r_j r}\le L^+_{u,r_jr}(r_jy)\le
L^+_{u,R}(r_jy)\label{eqb}
\end{equation}
for $ r_jr0$ is assumed throughout what
follows.
In general, if
$u$ is differentiable at $0$ and
$p=Du(0)\not=0$ we have
\begin{equation}
u(x)=u(0)+\iprod{p,x}+o(|x|)\quad\mbox{as }x\rightarrow 0;\label{eqz}
\end{equation}
If $z_r$ is a maximum point of $u$ on the sphere $|x|=r$ then $u(x)\le
u(z_r)$ for $|x|=r$ and so, by (\ref{eqz}),
\[
\iprod{p,x}\le\iprod{p,z_r}+o(r)\quad\mbox{for }|x|=r
\]
and elementary considerations show that this implies
$z_r/r\rightarrow p/|p|$ as
$r\downarrow 0$. For the converse, if $r_j\downarrow 0$ and
\begin{equation}
v(x)=\lim_{j\rightarrow\infty}{u(r_jx)-u(0)\over r_j}\label{eqlim}
\end{equation}
holds locally uniformly, it was proved above $v$ that is linear.
According to (\ref{eqnorm}) and current assumptions $v$ is not constant.
Say $v(x)=\iprod{p,x}$. If we show that $p$ is independent of the sequence
$r_j\downarrow 0$ the result follows. Indeed, then a standard compactness
argument shows that
\[
v(x)=\iprod{p,x}=\lim_{r\downarrow 0}{u(rx)-u(0)\over r}
\mbox{ holds locally uniformly.}
\]
That is, $Du(0)$ exists and $Du(0)=p$.
We already know by (\ref{eqnorm}) that $|p|$ is unique.
Let $z_{r}$ be any maximum point for $u$ on the sphere of radius
$r$ centered at the origin and assume that then
\begin{equation}
\lim_{r\downarrow0}{z_r\over r}= \omega.\label{eqlima}
\end{equation}
Of course, we show $p=|p|\omega$ to complete the argument. Now
$x=z_{r_j}/r_j$ is a maximum point for the function on the
right of (\ref{eqlim}) relative to the sphere
$|x|=1$. Using (\ref{eqlima}) to pass to the limit in the relation
\[
{u(r_jx)-u(0)\over r_j}\le {u(z_{r_j})-u(0)\over
r_j}={u\left(r_j{z_{r_j}\over r_j}\right)-u(0)\over r_j}
\quad\mbox{for }|x|=1
\]
we find $v(x)\le v(\omega)$ for $|x|=1$. As $v(x)=\iprod{p,x}$ we
conclude that
$p=|p|\omega$. This completes the argument.
\begin{thebibliography}{0}
\bibitem{arona} G. Aronsson. {\it Extension of functions satisfying
Lipschitz conditions}, Arkiv f\"ur Mate. {\bf 6} (1967), 551--561.
\bibitem{aronb} G. Aronsson, {\it On the partial differential
equation
$u_x^2u_{xx} + 2u_xu_yu_{xy} + u_y^2u_{yy} = 0$}, Arkiv f\"ur Mate.
{\bf 7} (1968), 395--425
\bibitem{aronc} G. Aronsson, {\it On certain singular solutions
of the
partial differential equation $u_x^2u_{xx} + 2u_xu_yu_{xy} +
u_y^2u_{yy} = 0$}, Manuscripta Math. {\bf 47} (1984),
133--151.
\bibitem{ceg} M. G. Crandall, L. C. Evans and R. Gariepy, {\it
Optimal Lipschitz Extensions and the Infinity Laplacian}, to appear in
Calc.\ Var.\ Partial Differential Equations (preprint temporarily
available at www.math.ucsb.edu/\~ \ crandall).
\bibitem{jen} R. Jensen, {\it Uniqueness of Lipschitz extensions
minimizing the sup-norm of the gradient}, Arch. Rat. Mech. Anal., {\bf 123}
(1993), 51-74.
\end{thebibliography}
\noindent{\sc Michael G. Crandall}\\
Department of Mathematics \\
University of California, Santa Barbara\\
e-mail: crandall@math.ucsb.edu \smallskip
\noindent{\sc Lawrence Craig Evans}\\
Department of Mathematics \\
University of California, Berkeley \\
e-mail: evans@math.berkeley.edu
\end{document}