\documentclass[reqno]{amsart} \usepackage{amssymb} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 148, pp. 1--11.\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 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/148\hfil Remarks on the gradient] {Remarks on the gradient of an \\ infinity-harmonic function} \author[T. Bhattacharya\hfil EJDE-2007/148\hfilneg] {Tilak Bhattacharya} \address{Tilak Bhattacharya \newline Department of Mathematics\\ Western Kentucky University\\ Bowling Green, KY 42101, USA} \email{tilak.bhattacharya@wku.edu} \thanks{Submitted August 10, 2007. Published November 5, 2007.} \subjclass[2000]{35J60, 35J70} \keywords{Infinity-harmonic; gradient; maximum principle} \begin{abstract} In this work we (i) prove a maximum principle for the modulus of the gradient of infinity-harmonic functions, (ii) prove some local properties of the modulus, and (iii) prove that if the modulus is constant on the boundary of a planar disc then it is constant inside. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \section{Introduction} In this work we discuss some local properties of the modulus of the gradient of the gradient of an infinity-harmonic function. Differentiability remains an open problem, except in the planar case \cite{s}; however, a quantity, which would be the modulus should differentiability hold, does exist. Our effort in this note is to prove a maximum principle for the modulus and record some local properties of an infinity-harmonic function at points where the modulus is a maximum. In particular, we prove that if the modulus is constant on the boundary of a planar disc then it is constant inside. We start with some notations. Let $\Omega\subset \mathbb{R}^n$, $n\ge 2$, will denote a bounded domain, the origin $o$ will be assumed to lie in $\Omega$. Let $B_r(x)$, $x\in\mathbb{R}^n$, be the ball of radius $r$ with center $x$. Let $\bar{A}$ denote the closure of a set $A$ and $A^c=\mathbb{R}^n\setminus A$. An upper semicontinuous function $u$, defined in $\Omega$, is infinity-subharmonic in $\Omega$ if it solves $$\label{e1} \Delta_{\infty}u(x)=\sum_{i,j=1}^nD_iu(x)D_ju(x)D_{ij}u(x)\ge 0,\quad x\in\Omega,$$ in the viscosity sense. A lower semicontinuous function $u$ is infinity-superharmonic in $\Omega$ if $\Delta_{\infty}u(x)\le 0$, $x\in \Omega$, in the viscosity sense. Moreover, $u$ is infinity-harmonic in $\Omega$ if it is both infinity-subharmonic and infinity-superharmonic in $\Omega$. Our work exploits the cone comparison property satisfied by $u$, see \cite{ceg}. Also see \cite{acj,tb1,tb2,tdm,cgw} in this connection. For $x\in \Omega$ and $B_r(x)\Subset\Omega$, for $0\le t\le r$, we define $M_x(t)=\sup_{B_t(x)}u$, $m_x(t)=\inf_{B_t(x)}u$. For infinity-subharmonic functions, $M_x(t)=\sup_{\partial B_t(x)}u$, and for infinity-superharmonic functions, $m_x(t)=\inf_{\partial B_t(x)}u$. The existence of the following limits is well known \cite[Lemma 2.7]{ceg}, $$\label{e2} \begin{gathered} \lim_{t\downarrow 0}\frac{M_x(t)-u(x)}{t}=\Lambda^+(x),\quad \mbox{when u is infinity-subharmonic,}\\ \lim_{t\downarrow 0}\frac{u(x)-m_x(t)}{t}=\Lambda^-(x),\quad \mbox{when u is infinity-superharmonic}. \end{gathered}$$ Moreover, if $u$ is infinity-harmonic then $\Lambda^+(x)=\Lambda^-(x)=\Lambda(x)$, and if also differentiable at $x$, then $\Lambda(x)=|Du(x)|$. See \cite{acj,tb2,ceg,ce}. We now state the two main results of this work. \begin{theorem}[Maximum Principle] \label{thm1} Let $\Omega\subset \mathbb{R}^n$ and $\Omega_1\Subset \Omega$. Recall the statements in \eqref{e2}. (i) If $u$ is infinity-subharmonic in $\Omega$, then $\sup_{x\in \bar{\Omega}_1}\Lambda^+(x)=\sup_{x\in \partial \Omega_1}\Lambda^+(x)$, and (ii) if $u$ is infinity-superharmonic in $\Omega$, then $\sup_{x\in \bar{\Omega}_1}\Lambda^-(x)=\sup_{x\in \partial \Omega_1}\Lambda^-(x)$. In particular, if $u$ is infinity-harmonic then $\sup_{x\in \bar{\Omega}_1}\Lambda(x)=\sup_{x\in \partial \Omega_1}\Lambda(x)$. \end{theorem} The anonymous referee pointed out this general version of Theorem \ref{thm1}. An older version of this theorem was stated only for infinity-harmonic functions. A proof will be presented in Section 2. The main idea of the proof is to exploit the result about increasing slope estimate in \cite[Lemma 3.3]{ceg}. In \cite{tb2}, these have been referred to as Hopf derivatives in the case of infinity-harmonic functions. The properties of the latter will be used to prove the second main result of this work. \begin{theorem} \label{thm2} Let $\Omega\subset \mathbb{R}^2$ and $B_r(x)\Subset \Omega$. Let $u$ be infinity-harmonic in $\Omega$. Suppose that for some $L>0$ and for every $y\in \partial B_r(x)$, $\Lambda(y)=|Du(y)|=L$, then \begin{itemize} \item[(i)] for any $w\in B_r(x)$, $|Du(w)|=L$, and \item[(ii)] given any point $z\in B_r(x)$ there is a straight segment $T$, with its end points on $\partial B_r(x)$ and containing $z$, such that $u$ is linear on $T$. Also, if $e_T$ is a unit vector parallel to $T$ then for any $\xi$ on $T$ either $Du(\xi)=Le_T$, or $Du(\xi)=-Le_T$. In addition, if $T_1$ and $T_2$ are any two such segments then either $T_1$ coincides with $T_2$ or they are distinct. \end{itemize} \end{theorem} At this time it is unclear whether or not this holds in $\mathbb{R}^n$ with $n\ge 3$. Theorem \ref{thm2} does not hold in general and the convexity of the domain seems to play a role in the proof this result. Consider the example $u(x,y)=x^{4/3}-y^{4/3}$ on $\mathbb{R}^2$, where a point is described as $(x,y)$. Then $\Lambda(x,y)=|Du(x,y)|=4/3\sqrt{x^{2/3}+y^{2/3}}$, and consider the regions $D_c$ of the type bounded by $x^{2/3}+y^{2/3}=c>0$. While $|Du(x,y)|$ is constant on $\partial D_c$, $|Du(o)|=0$ and $|Du(x,y)|<4c^2/3$, $(x,y)\in D_c$. We have divided our work as follows. Section 2 presents a proof of Theorem \ref{thm1}. In Section 3, we study the behaviour of an infinity-harmonic function $u$ near points of maximum of $\Lambda(x)$. In Section 4, we prove the rigidity result in Theorem \ref{thm2}. \section{Proof of main results} We first state results we will use in the proof of Theorem \ref{thm1}. Recall the statements in \eqref{e2}. Let $\Omega\subset \mathbb{R}^n$ and $B_r(x)\Subset\Omega$. Let (a) $p_t\in \partial B_t(x)$, $t\le r$, denote a point of maximum of $u(p)$ on $B_t(x)$, when $u$ is infinity-subharmonic in $\Omega$, and (b) $q_t\in \partial B_t(x)$ denote a point of minimum of $u$ on $B_t(x)$, when $u$ is infinity-superharmonic in $\Omega$. For part (i) of the theorem we will use the following. $$\label{e3} \Lambda^+(x)\le \frac{M_x(t)-u(x)}{t}\le \inf_{p_t}\Lambda^+(p_t),\quad \mbox{and \Lambda^+ is upper-semicontinuous.}$$ While for part (ii) we use $$\label{e4} \Lambda^-(x)\le \frac{u(x)-m_x(t)}{t}\le \inf_{q_t}\Lambda^-(q_t),\quad \mbox{and \Lambda^-(x) is upper-semicontinuous.}$$ See \cite[Lemma 3.3]{ceg}. We also point out that minor modifications of the arguments in \cite{tb1,tb2} will also yield \eqref{e3} and \eqref{e4}. We now prove Theorem \ref{thm1} by employing the above repeatedly. \begin{proof}[Proof of Theorem \ref{thm1}] We first prove part (i). Let $L=\sup_{x\in\bar{\Omega}_1}\Lambda^+(x)$, we assume that $L>0$. Since $\Lambda^+$ is upper semi-continuous, for some $y\in \bar{\Omega}_1$ we have $\Lambda(y)=L$. If $y\in \partial \Omega_1$ then we are done. Assume then that $y\in \Omega_1$. We will show that in this case there is a point $\bar{y}\in \partial \Omega_1$ with $\Lambda(\bar{y})=L$. Set $y_1=y$ and let $d_1=\mathop{\rm dist}(y_1,\Omega_1^c)$. Clearly, $B_{d_1}(y_1)\subset \Omega_1$; by \eqref{e3}, $\Lambda^+(p)\ge (u(p)-u(y_1))/d_1\ge\Lambda^+(y_1)=L$, for any $p\in \partial B_{d_1}(y_1)$ with $u(p)=M_{y_1}(d_1)$. Thus $\Lambda^+(p)=L$ and $u(p)=u(y_1)+Ld_1$. If $p\in \partial\Omega_1$ then we are done, otherwise set $y_2=p$. As already noted $u(y_2)=u(y_1)+Ld_1$. Let $d_2=\mathop{\rm dist}(y_2,\Omega_1^c)$, and any $p\in \partial B_{d_2}(y_2)$ be such that $u(p)=M_{d_2}(y_2)$. Again by \eqref{e3}, $\Lambda^+(p)\ge(u(p)-u(y_2))/d_2\ge \Lambda^+(y_2)=L$. Thus $\Lambda^+(p)=L$ and $u(p)= u(y_1)+L(d_1+d_2)$. If $p\in \partial \Omega_1$ then we are done. Suppose now that we have obtained sequences $\{y_i\}_{i=1}^k$, $\{d_i\}_{i=1}^k$ such that \begin{itemize} \item[(a)] $d_i=\mathop{\rm dist}(y_i,\Omega_1^c)$, $y_i\in \partial B_{d_{i-1}}(y_{i-1})$ and $y_i\notin\partial\Omega_1$, $2\le i0$. Note that $M(r_k)=u(r_k\omega_{r_k})$. The above statements also apply to points of minima. In particular, if $\nu_t=q_t/t$ where $u(q_t)=m(t)$, then $\nu_{r_k}\to -\omega$. If $\omega$ is the only limit point of $\omega_t$ as $t\downarrow 0$, then $u$ is differentiable at $o$, see \cite{ce}. Also, if $\omega\in S^{n-1}$ is such that \eqref{e7}(ii) holds for any sequence then $\omega$ is a gradient direction and $u$ is differentiable at $o$. We now prove the following result. \begin{lemma} \label{lem1} Let $u\ne 0$ be infinity-harmonic in $\Omega$ and $B_r(o)\Subset\Omega$. \begin{itemize} \item[(a)] If $\Lambda(o)=(M(r)-u(o))/r$, then $u$ is differentiable at $o$ and $Du(o)=\Lambda(o)\omega$, for some $\omega\in S^{n-1}$. Moreover, for $0\le t\le r$, $M(t)=u(t\omega)=u(o)+t\Lambda(o)$, and for very $t>0$ there is exactly one point $p_{t}\in\partial B_{t}(o)$ such that $u(p_t)=M(t)$. \item[(b)] If $p_r\in \partial B_r(o)$ is such that $\Lambda(p_r)=\Lambda(o)$ then the same conclusion holds for $u$ with $\omega=p_r/r$, and $M(t)=u(t\omega)=u(o)+t\Lambda(o)$, $0\le t\le r$. \end{itemize} Furthermore, if $x$ is any point on the segment $op_r$ then $Du(x)=\Lambda(o)\omega$. \end{lemma} \begin{proof} We prove part (a). Recall that $M(t)$ is convex in $t$, thus by 5(i) and the first part of \eqref{e5}(iii), $$\label{e8} \Lambda(o)\le \frac{M(t)-u(o)}{t}\le \frac{M(r)-u(o)}{r}=\Lambda(o),\quad 0\le t\le r.$$ Thus $M(t)=u(o)+t\Lambda(o)$, and $u(t\omega)\le u(o)+t\Lambda(o)$, for all $0\le t\le r$. For $00$, let $y\in \partial B_r(o)$ be such that $\Lambda(y)=\Lambda^s$. Let $H_y$ denote the $n-1$ dimensional plane tangential to $\partial B_r(o)$ at $y$. Then only one of the following happens. \noindent Case(a): There is a straight segment $xy$ with $x\in\partial B_r(o)$ such that $u$ is a linear function on $xy$. More precisely, for every $0\le t\le |x-y|$, either (i) $u(y+te)=u(y)+t\Lambda^s$, or (ii) $u(y+te)=u(y)-t\Lambda^s$, where $e=(x-y)/|x-y|$. Moreover, $u$ is differentiable on the segment $xy$, and if $z\in xy$ then in (i) $Du(z)=\Lambda^s e$, and in (ii) $Du(z)=-\Lambda^s e$. \noindent Case (b): For every $s>0$, all the points of extrema of $u$ on $\partial B_s(y)$ lie outside $\bar{B}_r(o)$. In particular all limit directions $\omega$, $\nu$ (see comment following \eqref{e10}) lie in $H_y$. Moreover, if $\omega$ is a limit direction, $s_k\downarrow 0$ the corresponding sequence, $\eta\in S^{n-1}$ and $y_k\in\partial B_r(o)$ is the point nearest to $y+s_k\eta$ then $\lim_{s_k\downarrow 0}(u(y_k)-u(y))/s_k=\Lambda^s\langle\omega,\eta\rangle$. \end{lemma} \begin{proof} Assume that Case (b) is not true. There is a ball $B_{\delta}(y)$ and a point $p\in \partial B_{\delta}(y)\cap \bar{B}_r(o)$ such that $u(p)=M_{y}(\delta)$. Our assumption of a point of maximum of $u$ on $\partial B_{\delta}(y)$, lying in $\bar{B}_r(o)$, is not restrictive and the arguments we use will apply equally to a minimum. By \eqref{e5}(iii) or even \eqref{e3}, $\Lambda(p)\ge \Lambda(y)$ implying that $\Lambda(p)=\Lambda^s$. Set $\omega=(p-y)/\delta$; by Lemma \ref{lem1}, we see that (i) $u(y+t\omega)=u(y)+t\Lambda^s$, $0\le t\le \delta$, (ii) $u$ is differentiable everywhere on the segment $yp$ with $Du(z)=\Lambda^s \omega$, for any $z$ on $yp$, and (iii) $p$ is the only point of maximum on $\partial B_{\delta}(y)$. If $p\in \partial B_r(o)$, then $x=p$ and the lemma holds. Assume that $p\in B_r(o)$; set $y_1=y$, $y_2=p$, $\omega_1=\omega$ and $d_1=\delta$. Note that $\omega_1$ points into $B_r(o)$. We repeat the argument at $y_2$ as follows. Set $d_2=r-|y_2|$ and $y_3\in \partial B_{d_2}(y_2)$ be a point of maximum. By 5(iii), $\Lambda(y_3)\ge \Lambda(y_2)=\Lambda^s$ implying $\Lambda(y_3)=\Lambda^s$; set $\omega_2=(y_3-y_2)/d_2$. Again by Lemma \ref{lem1}, $u$ is differentiable on $y_2y_3$ with $u(y_2+t\omega_2)=u(y_2)+t\Lambda^s=u(y_1)+(t+d_1)\Lambda^s$, $0\le t\le d_2$. By the uniqueness of gradient direction at $y_2$, $\omega_2=\omega_1=\omega$ and $y_1y_3$ is a straight segment. If $y_3\in \partial B_r(o)$ the process stops. Otherwise assume that we have a sequence of points $\{y_i\}_{i=1}^k\in B_r(o)$, with $\omega_i=(y_{i+1}-y_i)/d_i=\omega$, $i=1,2,\dots,k-1$; i.e., $y_1y_k$ a straight segment parallel to $\omega$, and $u(y_1+t\omega)=u(y_1)+t\Lambda^s$, $0\le t\le \sum_{i=1}^k d_i$. Moreover, $u$ is differentiable at every point $z$ on $y_1y_k$ and $Du(z)=\Lambda^s\omega$. Now let $d_{k+1}=r-|y_k|$ and $y_{k+1}\in \partial B_{d_{k+1}}(y_k)$ such that $u(y_{k+1})=M_{y_k}(d_{k+1})$. Set $\omega_{k+1}=(y_{k+1}-y_k)/d_k$; then by Lemma \ref{lem1}, $\Lambda(y_{k+1})=\Lambda(y_k)=\Lambda^s$, $y_{k}y_{k+1}$ is a straight segment and $u$ is differentiable on $y_{k}y_{k+1}$. Thus $\omega_{k+1}=\omega$, i.e., $y_1y_{k+1}$ is a straight segment parallel to $\omega$. Moreover, on $y_1y_{k+1}$, $u(y_1+t\omega)=u(y_1)+t\Lambda^s$, $0\le t\le \sum_{i=1}^{k+1}d_i$, and $Du(z)= \Lambda^s \omega$, for any $z$ on $y_1y_{k+1}$. Either $y_{k+1}\in \partial B_r(o)$ in which case the process stops or we continue. By the maximum principle, $u(y_1)+\Lambda^s\sum_{i=1}^{k}d_i\le M_o(r)<\infty$, for all $k\ge 1$. Thus $d_i\to 0$, $y_k\to x$ where $x\in \partial B_r(o)$. Thus we obtain a straight segment $xy$ where $x\in\partial B_r(o)$ and the conclusions of Case (a) hold with $e=\omega$. Now assume that Case (b) holds. We suppose that for every $s>0$, the points of extrema of $u$ on $\partial B_s(o)$ lie outside $\bar{B}_r(o)$. Given any sequence $s_k\downarrow 0$ and $\omega_{s_k}\to \omega$ we also have $\nu_{s_k}\to -\omega$, see remarks preceding \eqref{e10}. Thus any limit direction $\omega$ lies in $H_y$. Let $\omega$ be a limit direction and $s_k\downarrow 0$ be such that $(u(y+s_k\omega)-u(y))/s_k\to \Lambda(y)$. For $k=1,2,\dots$, let $y_k\in \partial B_R(o)\cap \partial B_{s_k}(y)$ be the point nearest to $y+s_k\omega$. Thus $y_k=y+s_k\zeta_k$, where $\zeta_k\in S^{n-1}$. Since the sphere is $C^2$ at $y$, $|y_k-(y+s_k\omega)|/s_k \to 0$ and $\langle \omega,\zeta_k\rangle \to 1$. Thus we have that, near $y$, \begin{align*} \big|\frac{u(y_k)-u(y)}{s_k}-\Lambda^s \big| &\le \big|\frac{u(y_k)-u(y+s_k\omega)}{s_k}\big|+ \big|\frac{u(y+s_k\omega)-u(y)}{s_k}-\Lambda^s\big|\\ &\le C|\zeta_k-\omega |+\big|\frac{u(y+s_k\omega)-u(y)}{s_k}-\Lambda^s\big|, \end{align*} where $C>0$ is the local Lipschitz constant. Clearly, the conclusion holds when $\eta=\omega$ by letting $s_k\to 0$. The statement for general $\eta$ may now be derived by using \eqref{e7}. \end{proof} \begin{remark} \label{rmk4} \rm In Case (a) of Lemma \ref{lem2}, if $z$ is any point in the interior of the segment $xy$ and $B_s(z)\subset B_r(o)$, then $u$ has exactly one point of maximum and one point of minimum on $\partial B_s(z)$. Both these lie on $xy$. One may show this by applying \eqref{e5}(iii) or \eqref{e4}. Lemma \ref{lem2} also holds if a limit direction $\omega$ or $-\omega$, at $y$, points into $\bar{B}_r(o)$. One can find a small $\delta>0$ such that $M_y(\delta)$ occurs near $\delta\omega$ (analogous for a minimum) and hence lies in $\bar{B}_r(o)$. See discussion at the beginning of this section. Using Lemma \ref{lem1}, one may show that $xy$ is parallel to $\omega$. \end{remark} \begin{remark} \label{rmk5} \rm By \eqref{e6}(iii) there is at least one point $p_r\in \partial B_r(o)$, where $\Lambda(p_r)=M'(r+)$. Thus $\Lambda^s\ge M'(r+)$. The existence of a straight line segment on which $u$ is linear need not imply that $u$ is affine. Take $u(x)=|x|$, $x\ne 0$. Also see capacitary rings \cite{tb1}. \end{remark} \begin{remark} \label{rmk6}\rm If $y\in \partial B_r(o)$ is a point of extrema of $u$ and $\Lambda(y)=\Lambda^s$, then by \eqref{e6}(i) $Du(y)=\pm \Lambda^s \omega$, where $\omega=y/r$. Clearly, case (a) of Lemma \ref{lem2} applies and $u$ is linear and differentiable on $xy$, where $x=-y$. For $0\le t\le r$, either $u(x+t\omega)=u(x)+t\Lambda^s$ or $u(x+t\omega)=u(x)-t\Lambda^s$. Since $\Lambda(y)=\Lambda(o)=\Lambda^s$, by Lemma \ref{lem1} and Remark \ref{rmk2}, for $0\le t\le r$, we have $M'(t)=-m'(t)=\Lambda^s$; we also have $|M(t)-m(t)|\le 2t\Lambda^s$. Assume that $u(y)=M(r)$; linearity implies that for any $0\le t\le r$, $u(o)=M(t)-t\Lambda^s$, $m(t)=u(-t\omega)=M(t)-2t\Lambda^s$, and in particular, $u(x)=m(r)=M(r)-2r\Lambda^s$. Employing Lemma \ref{lem1}, we see that $t\omega$, $-t\omega$ are the only points of extrema on $\partial B_t(o)$, $t\omega$ being the maximum and $-t\omega$ being the minimum. Thus for every $0u(q)$, we see that $u(z)=u(x,y)$, $z\in I-$, is increasing in $y$. Recalling \eqref{e11}(i) and (ii), we see that for $a,\;b\in I-$, $a\ne b$, $u(a)-u(b)=\langle Du(d),e_d\rangle l(a,b)=\langle Du(c),a-b\rangle\ne 0$. Let $\omega_d$ denote the gradient direction of $u$ at $d$. Noting that $|Du(d)|=L\ge |Du(c)|$ and $l(a,b)>|a-b|$, it follows that $\langle \omega_d,e_d\rangle\ne 0,\pm 1$. This implies that $\omega_d$ does not lie in the tangent space of $\partial D$ at $d$ nor is it parallel to segment $od$. Case(a) of Lemma \ref{lem2} now applies and we have a straight segment originating from $d$ and terminating at $\bar{d}\in \partial D$ such that $u$ is linear on the segment $d\bar{d}$, and $|Du(z)|=L$, $z\in d\bar{d}$, and if $\zeta=(d-\bar{d})/|d-\bar{d}|$ then $Du(z)=\pm L \zeta$. From here on $T$ will denote a segment of the type $d\bar{d}$, as described in Step 1. Let $z_T=(x_T,y_T)$ and $\bar{z}_T=(\bar{x}_T,\bar{y}_T)$ denote the two end points that lie on the unit circle $\partial D$. We set $z_T$ to be the higher end point and $\bar{z}_T$ will denote the lower end point, i.e., $y_T\ge \bar{y}_T$. Also set $e_T$ to be the unit vector parallel to $T$ and pointing towards $z_T$. By the comments in Step 1, $u(z_T)\ge u(\bar{z}_T)$, $u$ is linear on $T$ and $Du(x)=L e_T$ for any $x$ on $T$. Also let $\lambda(T)$ denote the length of $T$. >From now on we will call such segments $T$, as described in Step 1, as segments of type $S$. \noindent\textbf{Step 2.} By taking arbitrary points $a,\;b\in I-$, $a\ne b$ in \eqref{e11}(ii), we see that the points $d$, on the arc $\widehat{ab}$ form a dense set in the unit circle $\partial D$. By Step 1, we obtain infinitely many such segments $T$ of type $S$. By the uniqueness of gradient directions any two such segments intersect if and only if they are identical. By the discussion preceding Step 1, $pq$ is one such segment. It also follows then that segments $T$ of type $S$ either lie completely in $H+$ or in $H-$. Suppose that $T_1$ and $T_2$ are two such non-overlapping segments in $H-$ then one lies to the "left" of the other. More precisely, if $y_{T_1}>y_{T_2}$, then $$\label{e12} \bar{y}_{T_1}<\bar{y}_{T_2},\quad \lambda(T_1)>\lambda(T_2),\quad \mathop{\rm dist}(o,T_1)<\mathop{\rm dist}(o,T_2).$$ An analogous property holds in $H+$. \noindent\textbf{Step 3.} For $k=1,2,3,\dots$ let $T_k$ be a segment of type $S$ in $H-$ such that $y_{T_k}\uparrow 1$. Since the end points $z_{T_k}$ and $\bar{z}_{T_k}$ lie on the unit circle, $z_{T_k}\to p$ and $x_{T_k}\uparrow 0$. Moreover by Step 2 and \eqref{e12}, $\bar{y}_{T_k}\downarrow y_{\infty}\ge -1$ and $\bar{x}_{T_k}\to x_{\infty}$. Set $e_{\infty}=(-x_{\infty},1-y_{\infty})/\sqrt{x_{\infty}^2+(1-y_{\infty})^2}$, clearly, $e_{T_k}\to e_{\infty}$. Thus the segments $T_k$ tend to the segment $T_{\infty}$ with end points $z_{T_{\infty}}=(0,1)$ and $\bar{z}_{T_{\infty}}=(x_{\infty},y_{\infty})$. Also by Step 1, for every $k$ and any $0\le t\le \lambda(T_k)$, $u(z_{T_k}-te_{T_k})=u(z_{T_k})-t L$, and $Du(z_{T_k}-te_{T_k})=Le_{T_k}$. Since $u$ is $C^1$ we see that for any $0\le t\le\lambda(T_{\infty})$, $u(p-t e_{\infty})=M-tL$, $Du(p-t e_{\infty})=Le_{\infty}$, and $T_{\infty}$ is of type $S$. By the comments preceding Step 1, $Du(p)=Le_2=Le_{\infty}$, and $(x_0,y_0)=q$. Thus the segments $T_k$ move right to the segment $pq$. As noted in Step 1, since the set of end points $z_T$ and $\bar{z}_T$, of segments $T$ of type $S$, are dense in $\partial D$, it is clear now that we can always find segments $T$ arbitrarily close to the segment $pq$ and lying in $H-$. \noindent\textbf{Step 4.} Suppose now that there is an $a\in D$ such that $|Du(a)|\bar{y}_0$ and $\inf_k\mathop{\rm dist}(o, T_k)>\mathop{\rm dist}(o,z_0\bar{z}_0)$. Let $\bar{y}_1=\inf_k \bar{y}_{T_k}$ and $\bar{z}_1=(\bar{x}_1,\bar{y}_1)\in I-$. It follows easily that the segment $z_0\bar{z}_1$ is type $S$. Let $\bar{e}=(z_0-\bar{z}_1)/|z_0-\bar{z}_1|$, then $e\ne \bar{e}$ since $\bar{z}_0\ne \bar{z}_1$. It now follows that on the segment $z_0\bar{z}_1$, $$u(z_0-t\bar{e})=u(z_0)-tL,\quad 0\le t\le |z_0-\bar{z}_1|,\quad Du(z_0-t\bar{e})=L\bar{e}.$$ By \eqref{e13}, $Du(z_0)=Le=L\bar{e}$ and we have a contradiction. Thus the theorem holds and $|Du(w)|=L$, for all $w\in D$. \section{Appendix} We now prove a version of the Rolle's property in $\mathbb{R}^n$, $n\ge 3$. \begin{lemma} \label{lem4} Let $u$ be infinity-harmonic in $\Omega\subset \mathbb{R}^n$, $n\ge 3$. Let $x,\;y\in \Omega$ and $\sigma(s)$, $0\le s\le 1$ be a $C^1$ curve that lies completely in $\Omega$ with $\sigma(0)=x$ and $\sigma(1)=y$. Let $l(s)$ denote the arclength of the curve from $\sigma(0)$ to $\sigma(s)$. Then for some $0<\tau<1$, and vector $\omega_{\tau}\in S^{n-1}$, we have $$u(y)-u(x)=\Lambda(\sigma(\tau))l(1)\langle\omega_\tau, \sigma'(\tau)\rangle/|\sigma'(\tau)|.$$ \end{lemma} \begin{proof} The proof utilizes simple calculus ideas and \eqref{e7}(i). Without any loss of generality, take $x=o$, $u(o)=0$, and set $v(s)=u(\sigma(s))-u(y)l(s)/l(1)$, $0\le s\le 1$. Then $v(s)$ is continuous and $v(0)=v(1)=0$. Suppose that $v$ has a positive maximum at some $0<\tau<1$. Thus $u(\sigma(\tau))-u(y)l(\tau)/l(1)\ge u(\sigma(s))-u(y)l(s)/l(1)$, $0\le s\le 1$, and $$\label{e14} u(\sigma(s))-u(\sigma(\tau))\le u(y)(l(s)-l(\tau))/l(1),\quad 0\le s\le 1.$$ Set $z=\sigma(\tau)$ and $e=\sigma'(\tau)/|\sigma'(\tau)|$. By \eqref{e7}(i), there exists a limit direction $\omega_{\tau}\in S^{n-1}$ and $r_k\downarrow 0$ such that $\lim_{r_k\downarrow 0}(u(z+r_k\omega_{\tau})-u(z))/r_k=\Lambda(z)$. Let $z_k=z-r_ke$, $\xi_k=z+r_ke$; denote by $s_k$, the largest value of $s\le \tau$ such that $\sigma(s)\in \partial B_{r_k}(z)$, and by $\bar{s}_k$, the smallest value of $s\ge \tau$ such that $\sigma(\bar{s}_k)\in \partial B_{r_k}(z)$. Since $\sigma$ is $C^1$ and $u$ is locally Lipschitz, the following hold for small $r_k$: $$\label{e15} \begin{gathered} |\sigma'(\tau)|(\tau-s_k),\quad |\sigma'(\tau)|(\bar{s}_k-\tau)\approx r_k,\\\ |\sigma(s)-z|-|\sigma'(\tau)(s-\tau)|=o(|s-\tau|),\\ |\sigma(s_k)-z_k|,\quad |\sigma(\bar{s}_k)-\xi_k|=o(r_k),\\ |u(z_k)-u(\sigma(s_k)|,\quad |u(\xi_k)-u(\sigma(\bar{s}_k)|=o(r_k). \end{gathered}$$ From \eqref{e14}, \begin{gather*} \frac{u(\sigma(s_k))-u(z)}{r_k} \le -\frac{u(y)[l(\tau)-l(s_k)]}{l(1)r_k},\quad \frac{u(\sigma(\bar{s}_k))-u(z)}{r_k}\le \frac{u(y)[l(\bar{s_k})-l(\tau)]}{l(1)r_k}. \end{gather*} Using \eqref{e15} and taking limits in the above stated inequalities, we obtain that \label{e16} \begin{aligned} \lim_{r_k\downarrow 0}\frac{u(\sigma(s_k))-u(z)}{r_k} &=\lim_{r_k\downarrow 0}\frac{u(z_k)-u(z)}{r_k}=-\Lambda(z)\langle \omega_{\tau},e\rangle\\ &\le \lim_{r_k\downarrow 0}-\frac{u(y)(l(\tau)-l(s_k))}{l(1)r_k }=-\frac{u(y)}{l(1)}. \end{aligned} Using $\bar{s}_k$ and $\xi_k$, and taking limits as in \eqref{e16}, we see that $\Lambda(z)\langle\omega_z,e\rangle \le u(y)/l(1)$. 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