\documentclass[reqno]{amsart} \usepackage{epic,eepic} \usepackage{hyperref} \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 $\{(u,v)\in\mathbb{R}^2:0L_f$. 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 $$\label{McSh1} u(x)=\sup_{y\in\mathbb{R}}[f(y)-L\sqrt{\delta^2+(x-y)^2}]= \max_{|y-x|\leq D\delta}[f(y)-L\sqrt{\delta^2+(x-y)^2}],$$ 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 $$f(x)+L_f |x-y|-L\sqrt{\delta^2+(x-y)^2}D\delta.$$ \end{proof} Let $y(x)$ be one of the points in $\{|y-x|\leq D\delta\}$, where the maximum in (\ref{McSh1}) is achieved, $$\label{achieved} u(x)=f(y(x))-L\sqrt{\delta^2+(x-y(x))^2}.$$ \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 $$\label{scdder} g''_x(y)>L'_f,\quad \text{for all } |y-x|\leq D\delta.$$ 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 $$\label{maincond} \delta < \frac{L}{L'_f(1+D^2)^{3/2}},$$ 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 $$\label{firstder} f'(y(x))=g_x'(y(x))=\frac{L(y(x)-x)}{\sqrt{\delta^2+(y(x)-x)^2}}$$ (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 $$\label{MFEq} y(x)-x=\frac{\delta f'(y(x))}{\sqrt{L^2-(f'(y(x)))^2}}.$$ Taking $Y(x):=y(x)-x$ we can rewrite this as $$\label{Banach} Y(x)=\frac{\delta f'(Y(x)+x)}{\sqrt{L^2-(f'(Y(x)+x))^2}}=\delta\Phi(f'(Y(x)+x)),$$ 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 $$\label{MFEq1} x(y)=y-\frac{\delta f'(y)}{\sqrt{L^2-(f'(y))^2}},$$ 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