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\begin{document}
{\noindent\small {\sc Electronic Journal of Differential Equations}\newline
Vol. 1993(1993), No. 03, pp. 1-9. Published October 12, 1993.\newline
ISSN 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp (login: ftp) 147.26.103.110 or 129.120.3.113
}\thanks{\copyright 1993 Southwest Texas State University and
University of North Texas.}
\vspace{1.5cm}
\title[\hfilneg EJDE--1993/03\hfil Absolutely minimizing Lipschitz
extensions]{Estimates for Smooth Absolutely Minimizing
Lipschitz Extensions}
\author[Lawrence C. Evans\hfil EJDE--1993/03\hfilneg]{Lawrence C. Evans}
\date{}
\thanks{Submitted on July 20, 1993.}
\thanks{Supported in part by NSF grant DMS-9203440.}
\subjclass{35J70, 26A16 }
\keywords{Lipschitz extensions, Harnack inequalities}
\begin{abstract}
I present some elementary maximum principle arguments,
establishing interior gradient bounds and Harnack inequalities for
both $u$ and $|Du|$, where $u$ is a smooth solution of the degenerate
elliptic PDE $\D_{\i}u=0$. These calculations in particular extend
to higher dimensions G.~Aronsson's assertion \cite{aronb} that a
nonconstant, smooth solution can have no interior critical point.
\end{abstract}
\maketitle
\section{Introduction}
G. Aronsson initiated in \cite{arona}, \cite{aronb} investigation of
highly degenerate elliptic boundary value problem:
\begin{equation}
\label{eqn1}
u_{x_i}u_{x_j}u_{x_ix_j} = 0 \mbox{ in } U
\end{equation}
\begin{equation}
\label{eqn2}
u = g \mbox{ on } \p U,
\end{equation}
where $U$ is a bounded, connected, open subset of ${\Bbb R}^n$,
$g: \p U \to {\Bbb R}$ is a given Lipschitz function,
and $u: {\bar U} \to {\Bbb R}$ is the unknown.
The PDE (\ref{eqn1}) arises naturally if we consider ``optimal''
Lipschitz extensions of $g$ into $U$.
A function $g$ defined on $\p U$ has in general many extensions into
${\bar U}$ which preserve its Lipschitz constant. Aronsson
proposed trying to find a ``best'' Lipschitz extension $u$,
characterized by the property that for each subdomain $V \subset U$,
the Lipschitz constant of $u$ within $V$ equals the Lipschitz
constant
of $u$ restricted to $\p V$. More precisely, and following Jensen
\cite{jen}, let us say $u \in W^{1,\i}(U)$ is an {\em absolutely
minimizing Lipschitz extension} of $g$ into $U$ provided (\ref{eqn2})
holds and also
\begin{equation}
\label{eqn3}
\|Du\|_{L^{\i}(V)} \le \|D{\tilde u}\|_{L^{\i}(V)}
\end{equation}
for each open set $V \subset U$ and each ${\tilde u} \in W^{1,\i}(V)$
such that
\begin{equation}
\label{eqn4}
u - {\tilde u} \in W_0^{1,\i}(V).
\end{equation}
See Jensen \cite{jen} for more discussion, in particular concerning
the equality
\[
\|Du\|_{L^{\i}(V)} = \sup \Sb x,y \in {\bar V} \\ x \ne y \endSb
\left\{ \frac {|u(x)-u(y)|}{d_V(x,y)} \right\} ,
\]
$d_V(x,y)$ denoting the distance from $x$ to $y$ within $V$.
As noted by Aronsson, any smooth absolutely minimizing Lipschitz
extension solves the PDE (\ref{eqn1}) within $U$, and Jensen provides
a
somewhat different proof. The best insight for this equation is had
by considering instead of (\ref{eqn1}), (\ref{eqn2}) the
corresponding
boundary-value problem for the $p$-Laplacian:
\begin{equation}
\label{eqn5}
\mbox{div}(|Du_p|^{p-2}Du_p) = 0 \mbox{ in } U,
\end{equation}
\begin{equation}
\label{eqn6}
u_p = g \mbox{ on } \p U,
\end{equation}
when $n < p < \i$. This is the Euler--Lagrange equation for the
variational problem of minimizing the energy $\|D{\tilde
u}\|_{L^{p}(U)}$ among
all ${\tilde u} \in W^{1,p}(U)$ with ${\tilde u} = g$ on $\p U$. In
particular
\begin{equation}
\label{eqn7}
\|Du_p\|_{L^p(V)} \le \|D{\tilde u}\|_{L^p(V)}
\end{equation}
for each open $V \subset U$ and each ${\tilde u}$ such that $u -
{\tilde u} \in W_0^{1,p}(V)$. Assuming $u_p$ is smooth and $|Du_p|
\ne 0$, we may rewrite (\ref{eqn5}) to read
\begin{equation}
\label{eqn8}
\frac {1}{(p-2)} \D u_p + \frac {u_{p,x_i}u_{p,x_j}}{|Du_p|^2}
u_{p,x_ix_j} = 0.
\end{equation}
Suppose also we knew that as $p \to \i$, the function $u_p$ converge
in some sufficiently strong sense to a limit $u$. Formally passing
to limits in (\ref{eqn7}), we would expect $u$ to be an absolutely
miminizing Lipschitz extension, and passing to limits in (\ref{eqn8})
we expect as well $u$ to solve the PDE (\ref{eqn1}).
R. Jensen in his important paper \cite{jen} has made these insights
rigorous. In addition, he has proved that (a) any absolutely
minimizing Lipschitz extension is a weak solution of (\ref{eqn1}),
and
(b) any weak solution is unique. (Here ``weak solution'' means a
solution in the so-called viscosity sense, cf. Crandall--Ishii--Lions
\cite{cran}).
In view of the construction of absolutely minimizing Lipschitz
extensions as limits of solution of the $p$-Laplacian, it seems
reasonable to define, at least at points where $|Du| \ne 0$,
the nonlinear operator
\begin{equation}
\label{eqn9}
\D_{\i} u = \frac {u_{x_i}u_{x_j}}{|Du|^2} u_{x_ix_j}
\end{equation}
as the ``$\i$-Laplacian''.
This paper is a small contribution to the further study of smooth
solutions of the highly degenerate elliptic PDE $\D_{\i}u = 0$, and,
equivalently, of smooth absolutely minimizing Lipschitz extensions.
I
provide some elementary maximum principle arguments establishing
interior sup-norm bounds on both $|D(\log u)|$ (if $u > 0$) and
$|D(\log |Du|)|$. These imply in particular Harnack inequalities for
$u$ and $|Du|$. One consequence is that if $u$ is not constant, then
$|Du|$ can never vanish. This is an extension to dimensions $n \ge
3$
of a corresponding assertion of Aronsson in $n = 2$; cf. also Fuglede
\cite{fug}.
I should point out explicitly however that in general the $\D_{\i}u =
0$ does not admit smooth solutions, and consequently the calculations
presented here, although I think interesting, have limited
applicability in practice. For instance, Aronsson in \cite{aronc}
has
constructed a $C^1$ nonconstant weak solution, which does indeed
possess an interior critical point. This example shows that it is
not
merely a question of finding some reasonable approximation structure
to
which to modify the computation from \S 3.
My calculations estimating $|D(\log |Du|)|$ are somewhat reminiscent
of standard computations for minimal surfaces. This suggests
comparison of the PDE $\D_{\i} u = 0$ with the ``dual'' equation
$\D_1u = 0$, where
\[
\D_1u = \left(\d_{ij} - \frac {u_{x_i}u_{x_j}}{|Du|^2} \right)
u_{x_ix_j} = |Du| \mbox{ div}\left( \frac {Du}{|Du|} \right).
\]
Note that the operator $\D_1$ is degenerate, but only in the one
direction normal to each level set. By contrast, the operator
$\D_{\i}$ is nondegenerate only in this direction. Observe also
$\D_1u = 0$ is
a ``geometric'' equation, since it says that the level sets of $u$
have zero mean curvature (at least in regions where $u$ is smooth and
$|Du| > 0$). This fact suggests that the PDE $\D_{\i}u = 0$ is
somehow strongly ``nongeometric'', or rather that all its geometric
information concerns not the level sets of $u$, but rather the curves
normal to level sets. The concluding remark in \S 3 makes this
comment a bit more precise.
\section{Interior gradient bounds, Harnack inequality for $u$}
In this section we present a very simple proof of interior gradient
bounds and a Harnack inequality for $u$.
\begin{theorem}
Let $u$ be a $C^2$ solution of
\begin{equation}
\label{eqn10}
\D_{\i}u = 0 \mbox{ in } U.
\end{equation}
{\rm (i)} There exists a constant $C$ such that
\begin{equation}
\label{eqn11}
|Du(x_0)| \le C\|u\|_{L^{\i}(U)} \mbox{dist }(x_0,\p U)^{-1}
\end{equation}
for each point $x_0 \in U$.
{\rm (ii)} Suppose also $u \ge 0$. Then for each connected open set
$V \subset\subset U$, there exists a constant $C = C(V)$ such that
\begin{equation}
\label{eqn12}
\sup_V u \le C \inf_V u\,.
\end{equation}
{\rm (iii)} In particular, if $U$ is connected and $u > 0$ at some
point in $U$, then $u > 0$ everywhere in $U$. In this case, we have
the estimate
\[
\sup_V \left(\frac {|Du|}{u}\right) \le C\,,
\]
$C$ depending only on $n$ and $\mbox{dist}\,(V,\p U)$.
\end{theorem}
\medskip
\noindent
{\bf Proof}. 1. Write $v = |Du|$ and suppose for the moment we know
$v \ne 0$ in $U$. Then (\ref{eqn10}) implies
\begin{equation}
\label{eqn13}
u_{x_i}v_{x_i} = 0 \mbox{ in } U.
\end{equation}
Define
\[
w = \zeta \Phi(u,v),
\]
where $\zeta \in C_c^{\i}(U)$ and $\Phi \in C^2({\Bbb R}^2)$ are
smooth nonnegative functions, to be selected below. If $w$ attains
its maximum over ${\bar U}$ at a point $x_1 \in U$, we have
\[
\zeta\Phi_u u_{x_i} + \zeta\Phi_v v_{x_i} = -\zeta_{x_i}\Phi \quad (1
\le i \le n)
\]
at $x_1$. Multiply by
\[
\nu^i = u_{x_i}/|Du| = u_{x_i}/v \quad (1 \le i \le n)
\]
and sum on $i$, recalling (\ref{eqn13}) to deduce
\[
\zeta\Phi_uv = -(D\zeta\cdot\nu)\Phi
\]
at $x_1$. Consequently,
\begin{equation}
\label{eqn14}
\zeta|\Phi_u|v \le |D\zeta|\Phi
\end{equation}
at $x_1$.
2. Now take
\[
\Phi(u,v) = e^{\l u}v \quad (\l > 0).
\]
Then (\ref{eqn14}) reads
\begin{equation}
\label{eqn15}
\l\zeta(x_1)v(x_1) \le |D\zeta(x_1)|.
\end{equation}
Fix $x_0 \in U$ and take $\zeta \in C_c^{\i}(U)$ such that
$\zeta(x_0)
= 1$, $|D\zeta| \le 2\,\mbox{dist}\,(x_0,\p U)^{-1}$.
As $w = \zeta\Phi$ attains its maximum at $x_1$, we see that
\begin{eqnarray*}
|Du(x_0)| &= &v(x_0) \\
&\le &\frac {2}{\lambda} e^{2\lambda\|u\|_{L^{\i}}}
\mbox{dist}\,(x_0,\p U)^{-1}.
\end{eqnarray*}
Set
\[
\l = \|u\|_{L^{\i}}^{-1}
\]
to prove estimate (\ref{eqn11}).
3. Next, assume $u \ge 0$, fix $\d > 0$, and take
\[
\Phi(u,v) = \frac {v}{u+\d} .
\]
Then (\ref{eqn14}) implies
\[
\zeta \frac {v^2}{(u+\d)^2} \le |D\zeta| \frac {v}{u+\d}
\]
at $x_1$. Consequently
\begin{equation}
\label{eqn16}
w(x_1) = \frac {\zeta(x_1)v(x_1)}{u(x_1)+\d} \le |D\zeta(x_1)|\,.
\end{equation}
Given any ball $B \subset\subset U$, select $\zeta$ so that $\zeta
\equiv 1$ on $B$, $|D\zeta| \le C \mbox{ dist}\,(B,\p U)^{-1}$.
As $w = \zeta \Phi$ attains its maximum at $x_1$, we conclude
from (\ref{eqn16}) that
\[
\sup_B \left( \frac {v}{u+\d} \right) \le C,
\]
in constant $C$ depending only on $\mbox{dist}\,(B,\p U)$. But $v =
|Du|$, and so
\begin{equation}
\label{eqn17}
\|D(\log (u+\d))\|_{L^{\i}(B)} \le C.
\end{equation}
Now take any pair of points $x_1,x_2 \in B$. Let $P$ denote the path
\[
\{tx_2 + (1-t)x_1 \mid 0 \le t \le 1\}.
\]
Then
\begin{eqnarray*}
\log(u(x_2) + \d) - \log(u(x_1) + \d) &= &\int_0^1 \frac {d}{dt}
[\log(u(tx_2 + (1-t)x_1) + \d)]\,dt \\
&= &\int_0^1 D(\log(u(\cdot)+\d))\cdot (x_2-x_1)\,dt \\
&\le &C \mbox{ diam}(B),
\end{eqnarray*}
according to (\ref{eqn17}). Consequently
\[
u(x_2) + \d \le (u(x_1)+\d)e^{C \mbox{ diam }(B)}.
\]
Letting $\d \to 0$, we deduce
\[
u(x_2) \le Cu(x_1)
\]
for some constant $C$ and each $x_1,x_2 \in B$. If $V \subset\subset
U$ is connected, we cover $V$ with balls and iteratively apply the
foregoing result to each $B$, finally to deduce
\[
u(x_2) \le Cu(x_1)
\]
for each pair of points $x_1,x_2 \in V$, the constant $C$ depending
only on $V$.
4. Finally, we remove the restriction $v = |Du| > 0$. For this
define
\[
{\tilde u}({\tilde x}) = u(x) + \e x_{n+1},
\]
where ${\tilde x} = (x_1,\dots,x_{n+1}) = (x,x_{n+1})$, $\e > 0$.
Then ${\tilde u}$ is a $C^2$ solution of
\[
{\tilde u}_{x_i}{\tilde u}_{x_j}{\tilde u}_{x_ix_j} = 0 \mbox{ in }
{\tilde U},
\]
${\tilde U} = U \times {\Bbb R}$, and $|D{\tilde u}| \ge \e > 0$. Apply
the calculations in steps 1-3 to ${\tilde u}$ in place of $u$. \qed
\medskip
\noindent
{\bf Remark}. It is somewhat surprising in light of the extremely
strong degeneracy of the nonlinear operator $\D_{\i}$ that (smooth)
solutions verify the interior gradient bound (\ref{eqn11}) and the
Harnack inequality (\ref{eqn12}). Such estimates are usually the
hallmarks of averaging effects resulting from uniform ellipticity.
It is therefore perhaps worth noting that solutions do not in general
satisfy the strong maximum principle. For example, let $u(x) = |x|$
and ${\tilde u}(x) = x_n$, and take $U$ to be the open ball of radius
one, centered at the point $(0,\dots,2)$. Then $u,{\tilde u}$ are
$C^{\i}$ solutions in $U$, $u \ge {\tilde u}$ in $\p U$, but $u =
{\tilde u}$ in $U$ along the line $x' = 0$, $x' =
(x_1,\dots,x_{n-1})$. This example shows also the Harnack inequality
(\ref{eqn12}) is false if we ``tilt'' coordinates: it is {\em not}
true that
\[
\sup_V (u-L) \le C \inf_V (u-L)
\]
for each linear function $L$. \qed
\section{Harnack inequality for $|Du|$}
Our goal next is to establish a Harnack inequality for $v = |Du|$,
and in particular to show a smooth, nonconstant $u$ cannot have any
critical point.
\begin{theorem}
Let $u$ be a $C^4$ solution of
\begin{equation}
\label{eqn18}
\D_{\i}u = 0 \mbox{ in } U.
\end{equation}
(i) Then for each smooth, connected $V \subset\subset U$ there
exists a
constant $C = C(V)$ such that
\begin{equation}
\label{eqn19}
\sup_V |Du| \le C \inf_V |Du|.
\end{equation}
(ii) In particular, if $U$ is connected and $|Du| > 0$ at some point
in $U$, then $|Du| > 0$ everywhere in $U$. In this case
\[
\sup_V \left( \frac {|D|Du||}{|Du|} \right) \le C,
\]
$C$ depending only on $n$ and $\mbox{dist}\,(V,\p U)$.
\end{theorem}
\medskip
\noindent
{\bf Proof}. 1. Assume first $v = |Du| > 0$ everywhere in $U$. As
above we write
\begin{equation}
\label{eqn20}
\nu^i = u_{x_i}/|Du| = u_{x_i}/v \quad (1 \le i \le n),
\end{equation}
and also write
\begin{equation}
\label{eqn21}
h_{ij} = \nu^i\nu^j,\ g_{ij} = \d_{ij} - \nu^i\nu^j \quad (1 \le i,j
\le n).
\end{equation}
Notice
\begin{equation}
\label{eqn22}
\nu_{x_j}^i = \frac {1}{v} g_{ik}u_{x_kx_j} \quad (1 \le i,j \le n),
\end{equation}
\begin{equation}
\label{eqn23}
v_{x_i} = \nu^j u_{x_jx_i} \quad (1 \le i \le n).
\end{equation}
Observe further that the PDE (\ref{eqn18}) says
\begin{equation}
\label{eqn24}
\nu^iv_{x_i} = 0.
\end{equation}
2. We derive a PDE $v$ satisfies. We first differentiate
(\ref{eqn24}) with respect to $x_j$ and then utilize (\ref{eqn22}) to
find
\[
\nu^iv_{x_ix_j} = -\frac {1}{v} g_{ik}u_{x_kx_j}v_{x_i}.
\]
Consequently
\begin{eqnarray*}
h_{ij}v_{x_ix_j} &= &- \frac {\nu^j}{v} g_{ik}u_{x_kx_j}v_{x_i} \\
&= &- \frac {g_{ik}}{v} v_{x_k}v_{x_j} \mbox{ by (23)} \\
&= &- \frac {|Du|^2}{v} \mbox{ by (24).}
\end{eqnarray*}
Therefore
\begin{equation}
\label{eqn25}
-h_{ij}v_{x_ix_j} = |A|^2v,
\end{equation}
where we have written
\begin{equation}
\label{eqn26}
|A|^2 = \frac {|Dv|^2}{v^2} .
\end{equation}
3. Next we compute a differential inequality satisfied by $z =
|A|^2$. For this, first write $w = \log v$. Then equation
(\ref{eqn25}) becomes, in light of (\ref{eqn24}),
\begin{equation}
\label{eqn27}
-h_{ij}w_{x_ix_j} = |A|^2 = z.
\end{equation}
Now $z = |Dw|^2$, and so
\[
z_{x_ix_j} = 2w_{x_kx_i}w_{x_kx_j} + 2w_{x_k}w_{x_kx_ix_j} \quad (1
\le i,j \le n).
\]
Hence
\begin{equation}
\label{eqn28}
-h_{ij}z_{x_ix_j} = -2\nu^i\nu^jw_{x_kx_i}w_{x_kx_j} +
2w_{x_k}(-h_{ij}w_{x_kx_ix_j}).
\end{equation}
We differentiate (\ref{eqn27}) with respect to $x_k$ and substitute
above, thereby deducing
\begin{equation}
\label{eqn29}
-h_{ij}z_{x_ix_j} = -2\nu^i\nu^jw_{x_kx_i}w_{x_kx_j} +
2w_{x_k}z_{x_k}
4 w_{x_k}\nu^j\nu^i_{x_k}w_{x_ix_j}\,. \end{equation}
Now (\ref{eqn24}) implies
\begin{equation}
\label{eqn30}
\nu^jw_{x_j} = 0,
\end{equation}
and so
\[
\nu^jw_{x_ix_j} = -\nu^j_{x_i}w_{x_j} \quad (1 \le i \le n).
\]
We can therefore rewrite the last term in (\ref{eqn29}) as
\begin{equation}
\label{eqn31}
4w_{x_k}\nu^j\nu^i_{x_k}w_{x_ix_j} =
-4w_{x_k}w_{x_j}\nu^j_{x_i}\nu^i_{x_k}.
\end{equation}
Next, we return to (\ref{eqn22}) and compute
\begin{eqnarray*}
\nu_{x_i}^j\nu_{x_k}^i &= &\frac {1}{v^2}
g_{jm}u_{x_mx_i}g_{il}u_{x_lx_k} \\
&= &\frac {1}{v} g_{jm}\nu_{x_m}^lu_{x_lx_k} \\
&= &\frac {1}{v} g_{jm}\nu_{x_m}^lg_{lj}u_{x_jx_k} \\
&= &g_{jm}\nu_{x_m}^l\nu_{x_k}^l,
\end{eqnarray*}
the penultimate equality holding since $\nu^l\nu_{x_m}^l = 0$.
Inserting this computation into (\ref{eqn31}) yields
\begin{eqnarray*}
4w_{x_k}\nu^j\nu^i_{x_k}w_{x_ix_j} &= &-4
w_{x_k}w_{x_j}g_{jm}\nu_{x_m}^l\nu_{x_k}^l \\
&= &-4 w_{x_k}w_{x_m}\nu_{x_m}^l\nu_{x_k}^l \mbox{ by (\ref{eqn30})}
\\
&\le &0.
\end{eqnarray*}
Consequently, (\ref{eqn29}) implies
\begin{equation}
\label{eqn32}
-h_{ij}z_{x_ix_j} \le -2\nu^i\nu^jw_{x_kx_i}w_{x_kx_j} +
2w_{x_k}z_{x_k}.
\end{equation}
But (\ref{eqn27}) tells us
\begin{eqnarray*}
z^2 &= &(\nu^i\nu^jw_{x_ix_j})^2 \\
&\le &\nu^iw_{x_kx_i}\nu^jw_{x_kx_j}.
\end{eqnarray*}
Thus from (\ref{eqn32}) we discover the differential inequality
\begin{equation}
\label{eqn33}
-h_{ij}z_{x_ix_j} \le -2z^2 + 2w_{x_k}z_{x_k}.
\end{equation}
4. We intend next to deduce from (\ref{eqn33}) an interior estimate
on $z$. This is possible owing to the $z^2$ term in (\ref{eqn33}).
Indeed, let $\zeta \in C_c^{\i}(U)$, $0 \le \zeta \le 1$, and write
$r
= \zeta^4z$. Then
\begin{equation}
\label{eqn34}
r_{x_i} = \zeta^4z_{x_i} + 4\zeta^3\zeta_{x_i}z \quad (1 \le i \le
n),
\end{equation}
\begin{equation}
\label{eqn35}
r_{x_ix_j} = \zeta^4z_{x_ix_j} + 4\zeta^3(\zeta_{x_j}z_{x_i} +
\zeta_{x_i}z_{x_j}) + (4\zeta^3\zeta_{x_i})_{x_j}z.
\end{equation}
Assume $r$ attains its positive maximum over ${\bar U}$ at a point
$x_0 \in U$.
Then
\[
Dr = 0,\ D^2r \le 0 \mbox{ at } x_0.
\]
Thus at the point $x_0$,
\begin{eqnarray*}
0 &\le &-h_{ij}r_{x_ix_j} = \zeta^4(-h_{ij}z_{x_ix_j})
-8\zeta^3h_{ij}\zeta_{x_i}z_{x_j} + C\zeta^2z \\
&\le &-2 \zeta^4z^2 + 2\zeta^4w_{x_i}z_{x_i} -
8\zeta^3h_{ij}\zeta_{x_i}z_{x_j}+C\zeta^2z\,,
\end{eqnarray*}
according to (\ref{eqn33}). Now, since $Dr = 0$ at $x_0$, we deduce
from (\ref{eqn34}) that
\[
\zeta z_{x_i} = -4\zeta_{x_i}z \quad (1 \le i \le n).
\]
Substituting above, we compute
\begin{eqnarray*}
2\zeta^4z^2 &\le &-8\zeta^3\zeta_{x_i}w_{x_i}z +
32\zeta^2h_{ij}\zeta_{x_i}\zeta_{x_j}z + C\zeta^2z \\
&\le &C\zeta^3|Dw|z + C\zeta^2z \\
&= &C\zeta^3z^{3/2} + C\zeta^2z.
\end{eqnarray*}
Finally we employ Young's inequality in the form
\[
ab \le \e a^p + C(\e)b^q \quad (a,b > 0,\ \frac {1}{p} + \frac {1}{q}
= 1),
\]
with $p = q = 2$ and $p = \frac {4}{3}$, $q = 4$, to deduce
\[
2\zeta^4z^2 \le C\e\zeta^4z^2 + C(\e).
\]
Fix $\e > 0$ small enough to conclude
\[
\zeta^4z^2(x_0) \le C.
\]
Since $r = \zeta^4z$ attains its maximum over ${\bar U}$ at $x_0$, we
deduce that
\[
\max_{\bar U} \zeta^4z \le C,
\]
the constant $C$ depending only on $n$ and $\zeta$. Given any region
$V \subset\subset U$, we may select $\zeta \equiv 1$ on $V$, thereby
concluding
\begin{equation}
\label{eqn36}
\max_V z \le C(V).
\end{equation}
As $z = |Dw|^2 = |D(\log v)|^2$, we deduce the Harnack inequality
from
(\ref{eqn36}) as in the proof of Theorem~2.1.
5. If it is not true that $v = |Du| > 0$ everywhere in $U$, apply
the
above reasoning to ${\tilde u}(x) = u(x) + \e x_{n+1}$, and send $\e
\to 0$ to deduce $|Du| = 0$ everywhere in $U$. \qed
\medskip
\noindent
{\bf Remark}. The expression $z = |A|^2$ has the following geometric
interpretation (cf. Aronsson \cite{aronb}). Given a smooth solution
$u$ of $\D_{\i}u = 0$, with $|Du| > 0$, set as above $\nu = Du/|Du|$
to denote the field of normals to the level sets of $u$. Consider
then the ODE
\begin{equation}
\label{eqn37}
{\dot x}(s) = \nu(x(s)) \quad (s \in {\Bbb R}),
\end{equation}
whose trajectories are curves in $U$ normal to the level sets. Then
$v = |Du|$ is constant along each such curve, according to
(\ref{eqn13}). The curvature is
\[
\kappa = \left| \frac {d}{ds} \nu(x(s))\right| = |D\nu\cdot\nu|.
\]
But
\[
\nu_{x_j}^i\nu^j = \frac {1}{v} g_{ik}u_{x_kx_j}\nu^j
=\frac {1}{v} g_{ik}v_{x_k}
=\frac {v_{x_i}}{v} = w_{x_i}\,.\]
Thus $\kappa = |A|$. Theorem~3.1 in particular asserts that the
curvatures
of each normal curve are bounded in each region $V \subset\subset U$.
\qed
\begin{thebibliography}{6}
\bibitem{arona} G. Aronsson, {\em Extension of functions satisfying
Lipschitz conditions}, Arkiv f\"ur Mate. {\bf 6} (1967), 551--561.
\bibitem{aronb} G. Aronsson, {\em 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, {\em 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{cran} M. G. Crandall, H. Ishii, and P.-L. Lions, {\em User's
guide to viscosity solutions of second-order partial differential
equations}, Bull. Am. Math. Soc. {\bf 27} (1992), 1--67.
\bibitem{fug} B. Fuglede, {\em A criterion of non-vanishing
differential of a smooth map}, Bull. London Math. Soc. {\bf 14} (1982), 98--102.
\bibitem{jen} R. Jensen, {\em Uniqueness of Lipschitz extensions
minimizing the sup-norm of the gradient}, preprint, 1992.
\end{thebibliography}
{\sc Department of Mathematics,
University of California,
Berkeley, CA 94720}
\noindent{\em E-mail:} evans@@math.berkeley.edu
\section*{Addenda}
{\bf March 9, 1994}.
I should have referenced as well the interesting paper
\noindent T. Bhattacharya, E. DiBenedetto, and J. Manfredi,
{\em Limits as $p \rightarrow \infty$ of $\Delta_pu_p = f$ and
related external problems},
Rend. Sem. Mat. Univers. Politecn. Torino,
Fascicolo Speciale (1989), Nonlinear PDE's, pages 15-68.
\end{document}