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\usepackage{epic,eepic}
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\AtBeginDocument{{\noindent\small 
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 18, pp. 1--5.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.} 
\vspace{9mm}}

\begin{document} 

\title[\hfilneg EJDE-2005/18\hfil $\infty$-harmonic function, which is not $C^2$]
{Example of an $\infty$-harmonic function which is not $C^2$ on a dense subset} 

\author[H. Mikayelyan\hfil EJDE-2005/18\hfilneg]
{Hayk Mikayelyan} 

\address{Hayk Mikayelyan \hfill\break
Max-Planck-Institut f\"ur Mathematik in den Naturwissenschaften \\
Inselstrasse 22 \\
04103 Leipzig, Germany}
\email{hayk@mis.mpg.de}

\date{}
\thanks{Submitted November 24, 2004. Published February 5, 2005.}
\subjclass[2000]{35B65, 35J70, 26B05}
\keywords{Infinity-Laplacian}

\begin{abstract}
 We show that for certain boundary values,
 McShane-Whitney's minimal-extension-like function
 is $\infty$-harmonic near the boundary and is not
 $C^2$ on a dense subset.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{lemma}{Lemma}[section]
\newtheorem{prop}[lemma]{Proposition}
\newtheorem{cor}[lemma]{Corollary}
\newtheorem{rem}[lemma]{Remark}

\section{Results}

Let us consider the strip  $\{(u,v)\in\mathbb{R}^2:0<v<\delta\}$, which is
going to be the domain for a function constructed in this article.
Take a function $f\in C^{1,1}(\mathbb{R})$ and let
$L_f:=\|f'\|_\infty$ and $L'_f:=\mathop{\rm Lip}(f')$.
Let us consider an analogue of the minimal extension of McShane and Whitney,
\begin{equation}
\label{McSh}
u(x,d):=\sup_{y\in\mathbb{R}}[f(y)-L|(x,d)-(y,0)|],
\end{equation}
where $0<d<\delta$ and $L>L_f$. Note that to obtain the classical minimal
extension of McShane and Whitney we have to take $L=L_f$.

 For the rest of this article we fix the function $f$ and the constants 
$L>L_f$, $\delta>0$. We will find conditions on $\delta>0$,
which make our statements true. The real number $x$ will be
associated with
the point $(x,\delta)\in\Gamma_\delta:=\{(u,v)\in\mathbb{R}^2:v=\delta\}$,
and the real number $y$ with the point $(y,0)\in\Gamma_0$.
In the sequel the values of $u$ on the line $\Gamma_\delta$ will be
of our interest and we write $u(x)$ for $u(x,\delta)$ (see Figure \ref{nkar1}).

\begin{prop} \label{prop1}
The function $u$ defined above satisfies
\begin{equation}
\label{McSh1}
u(x)=\sup_{y\in\mathbb{R}}[f(y)-L\sqrt{\delta^2+(x-y)^2}]=
\max_{|y-x|\leq D\delta}[f(y)-L\sqrt{\delta^2+(x-y)^2}],
\end{equation}
where $D:=\frac{2LL_f}{L^2-L^2_f}$.
\end{prop}

\begin{proof}
 From the definition of $u$ we have $f(x)-L\delta\leq u(x)$
so it is sufficient to show that if $|x-y|>D\delta$ then
$$
f(y)-L\sqrt{\delta^2+(x-y)^2}< f(x)-L\delta.
$$
On the other hand, from the bound of $f'$ we have
$$
f(y)-L\sqrt{\delta^2+(x-y)^2}\leq f(x)+L_f |x-y|-L\sqrt{\delta^2+(x-y)^2}.
$$
Thus we note that all values of $y$ for which
$$
f(x)+L_f |x-y|-L\sqrt{\delta^2+(x-y)^2}<f(x)-L\delta
$$
can be ignored in taking supremum in the definition of $u$.
We write
$$
L_f |x-y|+L\delta<L\sqrt{\delta^2+(x-y)^2}
$$
and arrive at
$$
L_f^2|x-y|^2+2LL_f\delta|x-y|+L^2\delta^2<L^2\delta^2+L^2|x-y|^2\,.
$$
Therefore,
$$
2LL_f\delta<(L^2-L_f^2)|x-y| \quad \Longleftrightarrow   \quad  |x-y|>D\delta.
$$
\end{proof}


Let $y(x)$ be one of the points in $\{|y-x|\leq D\delta\}$, where the
maximum in (\ref{McSh1}) is achieved,
\begin{equation}
\label{achieved}
u(x)=f(y(x))-L\sqrt{\delta^2+(x-y(x))^2}.
\end{equation}

\begin{figure}[th]
\label{nkar1}
\begin{center}
\setlength{\unitlength}{0.9mm}
\begin{picture}(140,110)(0,0)
\drawline(10,10)(100,10)
\drawline(30,40)(120,40)
\drawline(20,-2)(20,85)
\drawline(40,30)(40,110)
\drawline(13.3,0)(45.4,48)
\put(19.16,85){$\uparrow$}
\put(39.16,110){$\uparrow$}
\put(99.8,9.12){$\rightarrow$}
\put(119.8,39.12){$\rightarrow$}
\dashline{0.8}(54,10)(54,16)(74,68)(74,40)
\qbezier(54,94)(75,42.5)(94,94)
\qbezier(25,60)(50,72)(74,68)
\qbezier(74,68)(92,65)(110,70)
\qbezier(110,70)(125,73)(140,65)
\qbezier(2,9)(22,26)(54,16)
\qbezier(54,16)(65,13)(75,18)
\qbezier(75,18)(95,28)(120,14)
\put(42,108){$f(y), g_x(y)$}
\put(95,90){$g_{x_0}(y)$}
\put(22,83){$u(x)$}
\put(120,73){$f(y)$}
\put(120,42){$\Gamma_0$}
\put(42,36){$0$}
\put(71,36){$y(x_0)$}
\put(120,36){$y$}
\put(105,21){$u(x)$}
\put(100,12){$\Gamma_\delta$}
\put(22,6){$\delta$}
\put(52,6){$x_0$}
\put(100,6){$x$}
\end{picture}
\end{center}
\caption{Touched by hyperbola}
\label{cone}
\end{figure}


\begin{lemma}
If $\delta>0$ is small enough
then for every $x\in\Gamma_\delta$ the point $y(x)$ is unique
and $y(x):\mathbb{R}\to\mathbb{R}$ is a bijective Lipschitz map.
\end{lemma}

\begin{proof}
For each $x\in\Gamma_\delta$ consider the function
$g_x(y):=u(x)+L\sqrt{\delta^2+(x-y)^2}$ defined on $\Gamma_0$ (see Figure \ref{nkar1}).
The graph of $g_x$ is a hyperbola and the graph of any other function
$g_{x'}$ can be obtained by a translation.
Obviously $f(y)\leq g_x(y)$ on $\Gamma_0$ and $g_x(y(x))=f(y(x))$.
If at every point $y\in\Gamma_0$ the graph of $f$ can
be touched from above by some hyperbola $g_x(y)$
then we will get the surjectivity of $y(x)$.
To obtain this result, the following will be sufficient
\begin{equation}
\label{scdder}
g''_x(y)>L'_f,\quad \text{for all } |y-x|\leq D\delta.
\end{equation}
For a fixed $y_0\in\Gamma_0$, we can find a hyperbola
$h_{x_0}(y)=C+L\sqrt{\delta^2+(x_0-y)^2}$ such that
$h_{x_0}(y_0)=f(y_0)$ and $h'_{x_0}(y_0)=f'(y_0)$;
then obviously $f(y)\leq h_{x_0}(y)$ for
$|y-x_0|\leq D\delta$ (see (\ref{scdder}))
and for $|y-x_0|>D\delta$ (see Proposition \ref{prop1}).
In other words, $h_{x_0}(y)=g_{x_0}(y)$.
So (\ref{scdder}) gives us
\begin{equation} \label{maincond}
\delta < \frac{L}{L'_f(1+D^2)^{3/2}},
\end{equation}
where $D$ is defined in Proposition \ref{prop1}.

Note that also uniqueness of $y(x)$ follows from (\ref{scdder});
assume we have $y(x)$ and ${\tilde y}(x)$, then
$$
L'_f|y(x)-{\tilde y}(x)|<\Big|\int_{y(x)}^{{\tilde y}(x)}g''_x(t)dt\Big|=
|f'(y(x))-f'({\tilde y}(x))|\leq L'_f|y(x)-{\tilde y}(x)|.
$$
We have used here that
\begin{equation}
\label{firstder}
f'(y(x))=g_x'(y(x))=\frac{L(y(x)-x)}{\sqrt{\delta^2+(y(x)-x)^2}}
\end{equation}
(derivatives in $y$ at the point $y(x)$).

The injectivity of the map $y(x)$ follows from differentiability of $f$.
Assume $y_0=y(x)=y({\tilde x})$, so we have
$f(y_0)=g_x(y_0)=g_{{\tilde x}}(y_0)$.
On the other hand, $f(y)\leq \min (g_x(y),g_{{\tilde x}}(y))$; this contradicts
differentiability of $f$ at $y_0$.

The monotonicity of $y(x)$ can be obtained using the same arguments;
if $x<{\tilde x}$ then the `left' hyperbola $g_x(y)$ touches the
graph of $f$ `lefter' than the `right' hyperbola $g_{{\tilde x}}(y)$,
since both hyperbolas are above the graph of $f$.

Now we will prove that $y(x)$ is Lipschitz.
 From (\ref{firstder}) it follows that
\begin{equation}
\label{MFEq}
y(x)-x=\frac{\delta f'(y(x))}{\sqrt{L^2-(f'(y(x)))^2}}.
\end{equation}
Taking $Y(x):=y(x)-x$ we can rewrite this as
\begin{equation}
\label{Banach}
Y(x)=\frac{\delta f'(Y(x)+x)}{\sqrt{L^2-(f'(Y(x)+x))^2}}=\delta\Phi(f'(Y(x)+x)),
\end{equation}
where $\Phi(t)=t/\sqrt{L^2-t^2}$.
For $ \delta < \frac{(L^2-L^2_f)^\frac{3}{2}}{L^2L'_f}$, we
can use Banach's fix point theorem and get that this functional
equation has unique continuous solution. On the other hand, it is
not difficult to check that
$$
\big|\frac{Y(x_2)-Y(x_1)}{x_2-x_1}\big|\leq \frac{\delta C}{1-\delta C},
$$
where $C=\frac{L^2L'_f}{(L^2-L^2_f)^{3/2}}$.
\end{proof}

\begin{cor}
If $\delta$ is as small as in the previous Lemma, then the function
$u$ is $\infty$-harmonic in the strip  between $\Gamma_0$ and $\Gamma_\delta$.
\end{cor}

\begin{proof}
This follows from the fact that if we take the strip with boundary values
$f$ on $\Gamma_0$ and $u$ on $\Gamma_\delta$ then
McShane-Whitney's minimal and maximal
solutions will coincide, obviously with $u$.
\end{proof}


\begin{rem} \rm
We can rewrite (\ref{MFEq}) in the form
\begin{equation} \label{MFEq1}
x(y)=y-\frac{\delta f'(y)}{\sqrt{L^2-(f'(y))^2}},
\end{equation}
where $x(y)$ is the inverse of $y(x)$.
This together with (\ref{achieved}) gives us 
$$
u(x(y))=f(y)-\frac{\delta L^2}{\sqrt{L^2-(f'(y))^2}}.
$$
Using the recent result of O.Savin that $u$ is $C^1$, we conclude
that function $x(y)$ is as regular as $f'$,
so we cannot expect to have better regularity than Lipschitz.
\end{rem}


\begin{lemma}
If $\delta>0$ is as small as above and function $f$ is not twice
differentiable at $y_0$, then the function $u$ is not twice
differentiable at $x_0:=x(y_0)$.
\end{lemma}

\begin{proof}
First note that for all $x$ and $y$, such that $x=x(y)$
we have
$$
u'(x)=f'(y).
$$
This can be checked analytically but actually is a trivial
geometrical fact; the hyperbola 'slides' in the direction of the
growth of $f$ at point $y$, thus the cone which generates this hyperbola
and 'draws' with its peak the graph of $u$ moves in same direction which is
the direction of the growth of $u$ at point $x=x(y)$.

Now assume we have two sequences $y_k\to y_0$ and
${\tilde y}_k\to y_0$ such that
$$
\frac{f'(y_k)-f'(y_0)}{y_k-y_0}\to\underline{f''}(y_0)
 \quad \text{and}\quad
\frac{f'({\tilde y}_k)-f'(y_0)}{{\tilde y}_k-y_0}\to\overline{f''}(y_0)
$$
and $\underline{f''}(y_0)<\overline{f''}(y_0)$.
Let us define appropriate sequences on $\Gamma_\delta$ denoting by
$x_k:=x(y_k)$ and by ${\tilde x}_k:=x({\tilde y}_k)$ and
compute the limits of
$$
\frac{u'(x_k)-u'(x_0)}{x_k-x_0} \quad \text{and}\quad
\frac{u'({\tilde x}_k)-u'(x_0)}{{\tilde x}_k-x_0}.
$$
We have
$$
\frac{u'(x_k)-u'(x_0)}{x_k-x_0}=
\frac{f'(y_k)-f'(y_0)}{y_k-y_0}\frac{y_k-y_0}{x_k-x_0}
$$
the first multiplier converges to $\underline{f''}(y_0)$,
let us compute the limit of
the second one. From (\ref{MFEq1}) we get that
$$
\frac{x_k-x_0}{y_k-y_0}\to 1-\delta \Phi'(f'(y_0))\underline{f''}(y_0),
$$
where $\Phi(t)=t/\sqrt{L^2-t^2}$.
Thus
$$
\frac{u'(x_k)-u'(x_0)}{x_k-x_0}\to
\frac{\underline{f''}(y_0)}{1-\delta \Phi'(f'(y_0))\underline{f''}(y_0)},
$$
and analogously
$$
\frac{u'({\tilde x}_k)-u'(x_0)}{{\tilde x}_k-x_0}\to
\frac{\overline{f''}(y_0)}{1-\delta \Phi'(f'(y_0))\overline{f''}(y_0)}.
$$
To complete the proof we need to use the monotonicity of the function
$$
\frac{t}{1-\delta C t}, \quad -L'_f<t<L'_f,
$$
where $\frac{1}{L}<C<L^2/ (L^2-L^2_f)^{3/2}$.
\end{proof}

Note that if the function $f$ is not $C^2$ at a point $y$ then
$u$ constructed here is not $C^2$ on the whole line connecting $y$ and $x(y)$.
So choosing $f$ to be not twice differentiable
on a dense set we can get a function
$u$ which is not $C^2$ on the collection of corresponding line-segments.
A similar example is the distance function from a convex
set, whose boundary is $C^1$ and not $C^2$ on a dense subset.
Then the distance function is $\infty$-harmonic and
is not $C^2$ on appropriate lines.


\section{Motivation}

Our example $u$ has the property of having constant $|\nabla u|$
on gradient flow curves (lines in our case). It would be interesting to
find a general answer to the question:

\begin{quote}
 What geometry do the gradient flow curves of an
$\infty$-harmonic function $u$ have,
on which $|\nabla u|$ is not constant?
\end{quote}

 From Aronsson's results we know that $u$ is not $C^2$ on such a curve.
This is our motivation for the investigation of $C^2$-differentiability
of $\infty$-harmonic functions.


The author has only one item in the list of references. The
history and the recent developments of the theory of $\infty$-harmonic
functions, as well as a complete reference list could be found in that paper.

\subsection*{Acknowledgement}
The author is grateful to Gunnar Aronsson, Michael Crandall and
Arshak Petrosyan for valuable discussions.


\begin{thebibliography}{0}

\bibitem{ACJ} G. Aronsson, M. Crandall, P. Juutinen 
\textit{A tour of the theory of absolutely minimizing functions}
Bull. Amer. Math. Soc. {\bf 41}  (2004),  no. 4, 439--505.

\end{thebibliography}


\end{document}