\pdfoutput=1\relax\pdfpagewidth=8.26in\pdfpageheight=11.69in\pdfcompresslevel=9 \documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2002(2002), No. 101, pp. 1--22.\newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \thanks{\copyright 2002 Southwest Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE--2002/101\hfil $\infty$-harmonic functions] {On the properties of $\infty$-harmonic functions and an application to capacitary convex rings} \author[Tilak Bhattacharya\hfil EJDE--2002/101\hfilneg] {Tilak Bhattacharya} \address{ Tilak Bhattacharya \hfill\break Mathematics Department\\ Bishop's University \hfill\break Lennoxville, Quebec J1M 1Z7, Canada} \email{tbhattac@ubishops.ca} \date{} \thanks{Submitted August 17, 2002. Published November 28, 2002.} \subjclass[2000]{35J70, 26A16} \keywords{Viscosity solutions, boundary Harnack inequality, $\infty$-Laplacian, \hfill\break\indent capacitary functions, convex rings} \begin{abstract} We study positive $\infty$-harmonic functions in bounded domains. We use the theory of viscosity solutions in this work. We prove a boundary Harnack inequality and a comparison result for such functions near a flat portion of the boundary where they vanish. We also study $\infty$-capacitary functions on convex rings. We show that the gradient satisfies a global maximum principle, it is nonvanishing outside a set of measure zero and the level sets are star-shaped. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{remark}[theorem]{Remark} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \section{Introduction} This is a continuation of the work in [4] and, while we derive a chain of results for $\infty$-harmonic functions, our primary effort in this work will be to prove two sets of results. The first would be for nonnegative $\infty$-harmonic functions, which vanish on a flat portion of the boundary of the set in which they are defined, and the second will be for $\infty$-capacitary functions in convex rings. More precisely, the first result discusses the behaviour of nonnegative $\infty$-harmonic functions near flat boundaries, on which they vanish, and we prove that any two such $\infty$-harmonic functions vanish at the same rate. In the second set of results, we show the non-vanishing of the gradient of $\infty$-capacitary functions on convex rings and the star-shapedness of the level sets of such functions. Clearly, the results are quite different in nature, however, the techniques used have a lot in common. A more detailed discussion follows in Section 2. We now comment on the approach used in this work. Our work utilizes the notion of a viscosity solution in this context and relies on techniques developed in [3,4,7,11,12,13,17,19-22,25,30]. While our results are motivated by the results in [15,18,27,30-35], which are about the weak solutions of the analogous problems with the $p$-Laplacian, for finite $p$, we do not work with approximating weak solutions as has been done in [5,16,18,23,29,30,31,33,34]. The idea in these works was to take the limit as $p\rightarrow \infty$ to capture properties and estimates for the $\infty$-harmonic functions. Instead our approach is closer to the works in [3,4,12,13,21,25,30]. Our intention is to use the framework of viscosity solutions to provide simpler and direct proofs. In this context we also refer the reader to the works in [3,4,13,22,24]. There is some overlap between our current work and [13]. This latter work contains, at times, finer and more detailed versions of some of the results proven here. \section{Notation and statements of the main results} We now introduce the notations we will be using in this work. These will be employed faithfully throughout this work with perhaps minor modifications for local use. By $\Omega$, we will always mean a bounded domain in $\mathbb{R}^n$, $n\ge 2$, and $\bar{A}$ will stand for the closure of a set $A$ in $\mathbb{R}^n$. The letter $O=(0,0,\ldots,0)$ will always stand for the origin in $\mathbb{R}^n$; for a point $x=(x_1,x_2,\ldots,x_n)\in \mathbb{R}^n$, define $\xi=\xi(x)=(x_1,x_2,\ldots,x_{n-1})$ and $x_n(x)=x_n$, then $x=(\xi(x),x_n)$. Also $|\xi(x)|=\sqrt{x_1^2+x_2^2+\ldots+x_{n-1}^2}$. We will sometimes use the notation $y=(0,a)$ to mean $y_1=y_2=\cdots=y_{n-1}=0$ and $y_n=a$. In this context, we will often think of $\mathbb{R}^n=\mathbb{R}^{n-1}\times \mathbb{R}$. We will be working with cylinders in $\mathbb{R}^n$: $A_r(P)=\{x: |\xi(x-P)|0$, then $A_{\lambda r}(P)=\{y\in \mathbb{R}^{n-1}:|y-\xi(P)|<\lambda r\}\times (P_n,P_n +2\lambda r)$ is a $\lambda$ scaling of $A_r(P)$ with the bottom faces situated at the same height. Also $F_r(P)=\{x:|\xi(x-P)|P_n\}$ and $B^-_r(P)$ defined analogously, denote the half-balls. If $A$ and $B$ are two points, with $A\ne B$, then $AB$ stands for the straight segment joining $A$ to $B$. The $\infty$-Laplacian operator $\Delta_{\infty}$ is defined as $\displaystyle{\Delta_{\infty}u=\sum_{i,j=1}^n D_i u D_ju D_{ij}u,}$ where $D_iu=\partial u/\partial x_i$ and $D_{ij}u=\partial^2 u/\partial x_i\partial x_j.$ This operator is elliptic but highly degenerate. In this work, we study viscosity solutions of solutions of $$\Delta_{\infty}u=0,\quad \mbox{in } \Omega. \label{*}$$ We provide a definition in this context. We say that $u$ is a viscosity subsolution (or $\infty$-subharmonic) of above equation, in $\Omega$, if $u$ is upper-semicontinuous in $\Omega$, and whenever $x_0\in \Omega$ and $\phi\in C^2(\Omega)$ are such that $\phi(x_0)=u(x_0),\quad\mbox{and}\quad\phi(x)< u(x),\quad\mbox{for}\;x\ne x_0,$ then $\Delta_{\infty}\phi(x_0)\ge 0$. Analogously, we may define a viscosity supersolution (or $\infty$-superharmonic) of \eqref{*}, by requiring that $u$ be lower-semicontinuous and $\Delta \phi(x_0)\le 0$, whenever $u(x)-\phi(x)$ has a local minimum at $x_0$. We say $u$ is a viscosity solution (or $\infty$-harmonic) of \eqref{*} if it is both a subsolution and a supersolution. It is well known that, if $u$ is $\infty$-harmonic then $u$ is locally Lipschitz continuous in $\Omega$ [4,13,19,21,29]. We must point out a key property, we will often exploit here, is that if a function $u$ has cone comparison property then it is $\infty$-harmonic, a fact proven in [13]. Also see [4]. We now state the first result. \begin{theorem}[Comparison] \label{thm1} Let $u_i(x)>0,\;i=1,2$, be $\infty$-harmonic in $A_8(O)$. If $u_1$ and $u_2$ vanish continuously on $F_8(O)$, then there exist positive constants $M_1$, $M_2$ and $M_3$ independent of $u_i$, such that for $x\in A_1(O)$, \begin{itemize} \item[(i)] $u_i(x)\le M_1u_i(z)$, $i=1,2,$ and \item[(ii)] $\displaystyle M_2\frac{u_1(z)}{u_2(z)}\le \frac{u_1(x)}{u_2(x)} \le M_3\frac{u_1(z)}{u_2(z)}$, where $z=(0,2)$. \end{itemize} \end{theorem} One may think of part (i) as a boundary Harnack inequality and plays an important role in the proof of part (ii). This type of comparison result, near a flat portion of the boundary, is well known in the theory of both divergence and nondivergence type elliptic partial differential equations. We refer the reader to the works in [2,6,8,9,15,32] and the references therein. However, the work in [32], done in connection with a Fatou theorem, is the earliest work which proves a result of this type for $p$-harmonic functions, for $11$. A version of the Harnack inequality (see Lemma \ref{lm2} and [29]), part (a) of Lemma \ref{lm4} and the comparison principle permits us to apply the device in Theorem 1.1 [9]. This leads to a proof of part (i) and implies that the solutions are well behaved near $F_2$; away from $F_2$, the solution can be controlled by the Harnack inequality. Putting these together yields the result. The second set of results are concerned with $\infty$-capacitary functions. We now introduce notations for this set up. Let $C_1$ and $C_2$, with $C_2\subset C_1$, be bounded domains in $\mathbb{R}^n,\;n\ge 2$. Let $\Gamma=\Gamma(C_1,\;C_2)=C_1\backslash\bar{C_2}$, denote the annular domain. We take $C_1$ and $C_2$ to be convex $C^2$ domains and we will also assume that the origin $O$ lies in $C_2$. We will refer to $\Gamma$ as a convex ring and $\partial \Gamma=\partial C_1\cup\partial C_2$. If $Q\in \partial C_2$, then the line $L=L(Q)$ will often denote the straight ray normal to $\partial C_2$, at $Q$, directed towards $\partial C_1$. If $\nu=\nu(Q)$ is the unit outer normal to $\partial C_2$ at $Q$ (relative to $C_2$), then the hyperplane $\langle x-Q,\nu(Q)\rangle=0$ will be denoted by $T_Q$. Since $C_2$ is convex, it lies on one side of $T_Q$ and $L\perp T_Q$ at $Q$. We may also define analogously the hyperplane $T_P$ at a point $P\in \partial C_1$. The hyperplane $T_Q$ generates two disjoint half-spaces $$H^+_Q=\{x\in \mathbb{R}^n: \langle x-Q,\nu(Q)\rangle<0\}\quad\mbox{and}\quad H^-_Q=\{x\in \mathbb{R}^n: \langle x-Q,\nu(Q)\rangle>0\}.$$ Clearly $H^+_Q\supset C_2$. For $P \in \partial C_1$, we will again take $H^+_P$ to be the half-space that contains $C_1$. We will be studying the problem $$\Delta_{\infty} u=0,\quad\mbox{in }\Gamma,\quad u\in C(\bar{\Gamma}) \quad\mbox{with}\quad u|_{\partial C_1}=1\quad \mbox{and} \quad u|_{\partial C_2}=0.$$ We again interpret this in the viscosity sense; see [11]. We call $u$ an $\infty$-capacitary function. Invoking the Harnack inequality [4,29], we see that $0\le u\le 1$. As a matter of fact if $P$ is a point of an interior minimum of $u$ then $u-u(P)\ge 0$ in $\Gamma$ and since $u-u(P)>0$ somewhere in $\Gamma$, being connected this would mean $u-u(P)>0$ everywhere. This contradiction implies that $u$ has no interior minimum (nor maximum for that matter) and so $00$, a.e. in $\Omega$. It is not known yet whether $u$ is better than Lipschitz in regularity and hence we are unable to assert the existence of $|Du|$ everwhere. See [12,25] for a discussion regarding this issue. The results in Theorem \ref{thm2} were proven in [33,34] for the $\infty$-Laplacian, by utilizing the approximating procedure involving the $p$-Laplacian, for finite $p$; also see [18,31]. The works [33,34] also deal with star-shaped regions and contain interesting results. However, for convex rings, the result for the $p$-Laplacian, for finite $p$, was originally done in [27]. In this context also see [18,31]. Our approach will be to work directly with the viscosity solutions as discussed before. Our proof utilizes scaling and estimates near the boundaries, proven by employing auxilliary functions as in [27]. While a great many of the comparison type results used in this work may be worked in fairly elementary fashion as in [4,13], the comparison principle employed to compare $u$ to its scaled version requires the application of a stronger result. More general versions of a comparison principle for such functions, originally proven in [19], may be found in [3,7,21]. Also see [17,20,23] for related works. Our approach also utilizes a property of nonnegative $\infty$-harmonic functions first alluded to in [4, see Remark 6] which follows from cone comparsion. See \eqref{1} and part (a) of Lemma \ref{lm4} in Section 3. This is used in the proof of the existence of normal derivatives of $u$ at the boundaries and also in the proof of a general bound for the gradients. We must point out that at this time we do not have a proof of the convexity of level sets uitilizing the viscosity framework. This fact was proven in [27] for the $p$-Laplacian, for finite $p$, and also holds for $p=\infty$ and appears in [33,34]. We make some remarks about this issue in Section 6. We have divided our work as follows. Section 3 contains preliminary results needed for our work and Section 4 contains the proof of Theorem \ref{thm1}. Section 5 contains results applicable to the context of convex rings and the proof of Theorem \ref{thm2} appears in Section 6. Appendix contains (i) the proof of the fact that odd reflections of $\infty$-harmonic functions are also $\infty$-harmonic, and (ii) the proof of Theorem 1.1 in [9]. We thank Michael Crandall for showing us a short proof of a sharper version of the Harnack inequality (see Lemma \ref{lm2}) and also for showing us some elegant proofs of results related to those in [4]. We also thank Juan Manfredi for several discussions in connection with this work and also for pointing out the work in [31,32]. We are also indebted to the referee whose comments have greatly improved the presentation of this work and also for pointing out the reference [18]. This research was partially supported by a grant from NSERC. \section{Preliminary results} In this section we will state and prove a sequence of results which lead to the proofs of Theorems 2.1 and 2.2. To achieve our end we will require somewhat more refined versions of the Hopf principle and the Harnack inequality. The proofs rely on the comparison principle and some auxilliary functions. A general version of the comparison principle is proven in [3; also see 7,19,21], however simpler arguments, such as those used in [4,13], will also suffice in many instances. We will first recall Remark 6 in [4]. Also see Lemme \ref{lm4}. Let $u>0$ be an $\infty$-harmonic function in a domain $\Omega$ and $B_r(O)\subset\subset\Omega$. We set $d(x)=\mathop{\rm dist}(x,\partial B_r(O))=r-|x|,\;x\in B_r(O)$. Part (i) of Lemma 2 [4], then states $\frac{u(x)}{d(x)}\ge \frac{u(O)}{d(O)}=\frac{u(O)}{r}, \quad \forall x\in B_r(O).$ Utilizing this, we showed, in Remark 6 [4], that if $\vec{e}$ is a unit vector, $x=s\vec{e}$ and $y=t\vec{e}$, where $00$ be $\infty$-harmonic in $\Omega$, and $\delta>0$ be such that the set $\Omega_{\delta}=\{x\in\Omega: \;\mathop{\rm dist}(x,\partial \Omega)\ge \delta\}\ne \emptyset$. Suppose $A$ and $B$ are points, in $\Omega_{\delta}$, such that the segment $AB\subset \Omega_{\delta}$. Then $u(B)\ge e^{-\frac{|A-B|}{\delta}}u(A).$ \end{lemma} \noindent {\bf Proof:} We note that, by employing the comparison principle, if $y\in \Omega_{\delta}$ then $$\label{4} u(y)\Big(1-\frac{|x-y|}{\delta}\Big)\le u(x),\quad \forall x\in B_{\delta}(y).$$ Let $x_0,x_1,x_2,\ldots,x_m$ be points on the segment $AB$ such that $x_0=A,x_m=B$ and $|x_i-x_{i+1}|=|A-B|/m$, $\forall i=0,1,\ldots,\;m-1$. Choose $m$ large so that $|A-B|/m\le \delta/2$. Since $x_{i+1}\in B_{\delta}(x_i)$, applying \eqref{4}, we find that $$u(x_{i+1})\ge u(x_i) \Big(1-\frac{|A-B|}{m\delta}\Big),\quad \forall i=0,1,\ldots,\;m-1.$$ Thus $$\label{5} u(B)\ge u(A) \quad(1-\frac{|A-B|}{m\delta}\quad)^m .$$ The lemma follows by letting $m\rightarrow \infty$ in \eqref{5}. \qed \begin{remark} \label{rmk1} \rm It is clear that above estimate can be extended very easily to $\infty$-superharmonic functions and to polygonal paths joining two points in $\Omega_{\delta}$. \end{remark} We now prove a result about the oscillation of $u$ which will prove important in our proof of Theorem \ref{thm1}. Calling $w(r)=\max_{B_r(O)}|u(x)-u(y)|=\mathop{\rm osc}_{B_r(O)}u$, we show that $w(r)$ is convex and in particular $w(2r)\ge 2w(r)$. This fact together with Theorem 1.1 in [9] will lead to a proof of Theorem \ref{thm1}. Note in Lemma \ref{lm3}, we do not assume that $u>0$. \begin{lemma}[Convexity of oscillation] \label{lm3} Let $\Omega\subset \mathbb{R}^n$ be a domain and $u$ be $\infty$-harmonic in $\Omega$. Let $B_R(O)\subset\subset\Omega$, be the ball of radius $R$, centered at $O$. Suppose that $M(r)=\sup _{B_r(O)}u(x)$, and $m(r)=\inf _{B_r(O)}u(x)$. Then for $0\le r\le R$, \\ (i) $w(r)=\mathop{\rm osc}_{B_r(O)}u(x)$ is convex and \\ (ii) $\displaystyle{\frac{w(r)}{r}=\frac{M(r)-m(r)}{r}\downarrow}$ as $r\downarrow 0$. \end{lemma} \noindent {\bf Proof:} Let $0<\delta 1$, $w(tr)\ge t w(r)$ and in particular, $w(2r)\ge 2w(r)$. Now let $K_r=K_r(O)=\{x:|\xi(x)|< r,\;|x_n|0$ in $\Omega$ and $\vec{\eta}$ is such that $|\vec{\eta}|=1$, then for $0\le a\le r$, $$\frac{u(z+a\vec{\eta})}{r-a}=\frac{u(z+a\vec{\eta})}{r-|a\vec{\eta}|}\ge \frac{u(z+b\vec{\eta})}{r-|b\vec{\eta}|}=\frac{u(z+b\vec{\eta})}{r-b},\;\;\forall \;0\le b\le a\le r.$$ In other words, $\forall\;x\in B_r(z)$, $u(x)/d(x)$, is increasing as $x\rightarrow\partial B_r(z)$ along a radial line. We also note that by taking $b=0$, $u(z+a\vec{\eta})/(r-a)\ge u(z)/r,\;\forall 0\le a\le r$. \\ (b) Under the assumptions in (a), $x\ne z$, $\vec{e}$ and $y$ as defined above, we have $$u(ty)\le tu(y),\quad\mbox{whenever}\quad t\ge 1\quad\mbox{and}\quad 00 and |Du|(z) exists then$$ (i)\;\;|Du|(z)\le \frac{u(z)}{\mathop{\rm dist}(z,\partial B_r(z))}; \quad\mbox{and}\quad (ii)\;\;\;|Du|(z)\le \frac{u(z)}{\mathop{\rm dist}(z,\partial \Omega)}. $$(d) Regardless of the sign of u, D(M) and D(m) exist on B_r(z) and |Du|(z)\le \min(D(m),\;D(M)). \end{lemma} \noindent {\bf Proof:} The proof follows by an application of part (i) Lemma 2 in [4]; see \eqref{1}. \noindent{\it Part (a):} For 0\le b\le a\le r, set x=z+a\vec{\eta} and v=z+b\vec{\eta}; then d(x)=\mathop{\rm dist}(x,\partial B_r(z))=r-a, d(v)=\mathop{\rm dist}(v,\partial B_r(z))=r-b and d(v)\ge d(x). Clearly, x lies in the ball B_{d(v)}(v); applying Lemma 2 in [4] (also see discussion preceding \eqref{1}), we see that u(x)\ge u(v) d(x)/d(v). This proves part (a). The rest of the assertions are consequences of this fact. \noindent{\it Part (b):} We reinterpret part (a). One notes that for x\in B_r(z), the ray z-s\vec{e}, s\ge 0, cuts \partial B_r(z) at z-r\vec{e}. Thus by part (a) we find that u(x)/(r-|x-z|)=u(y)/|y| increases as x\rightarrow z-r\vec{e}, i.e., as |y|=d(x)\downarrow 0. If t>1 is such that 00. Then$$ \frac{u(z)-u(z+s(\vec{e}))}{s}\le \frac{u(z)}{r}\;\;\Rightarrow\; \langle Du(z), -\vec{e}\rangle \le\frac{u(z)}{r},\quad \forall \vec{e}\in \mathbb{R}^n, $$if Du(z) exists. If Du(z)\ne 0, we may take \vec{e}=-Du(z)/|Du(z)| and part (i) follows. To prove (ii), let R=\mathop{\rm dist}(z,\partial \Omega), then the ball B_R(z)\subset \Omega, and (i) continues to hold by considering an increasing sequence of balls. \noindent {\it Part(d):} All the results discussed above require that u>0. Now we drop this requirement. With m and M as above, part (c) holds for u-m and M-u, i. e.,$$ |Du(z)|\le \min \Big(\frac{M-u(z)}{r},\;\frac{u(z)-m}{r}\Big). $$Applying part (a) to M-u, it follows that (M-u(z))/r\le (M-u(x))/|x-P_M|\le D(M),\forall x\in zP_M, if D(M) exists. To see this, note $\forall x\in OP_M,\; 00,\;i=1,2; by scaling if necessary, we may take u_i's to be \infty-harmonic in A_{8} and vanishing continuously on the face F_8=\{x:|\xi(x)|<8,\;x_n=0\}\subset \{x_n=0\}. We suppress the subscript and work with a general u that satisfies the requirements of the theorem. The constants M, C are positive constants, that are independent of u, but may depend on the geometry. We will often write x=(\xi(x),x_n) and set z=(0,2). We achieve our proof in five steps. \noindent {\bf Step 1.} We first show that $$\label{8} \frac{u(x)}{u(z)}\ge \frac{x_n}{4},\quad \forall x\in A_1.$$ For x\in A_1, write x=(\xi(x),x_n) and set P=(\xi(x), 2); then P lies in the hyperplane x_n=2 and x\in B_{2}(P)\subset A_8. Applying Lemma \ref{lm4} (a), in B_2(P), we see \frac{u(x)}{x_n}\ge \frac{u(P)}{2}. Again P\in B_2(z), and applying once more Lemma \ref{lm4} (a) to B_2(z) and noting that d(P)=\mathop{\rm dist}(P,\partial B_2(z))\ge 1, we have u(P)\ge\frac{u(P)}{d(P)}\ge \frac{u(z)}{2}. Combining these observations yields \eqref{8}. \noindent {\bf Step 2.} We now make a few remarks which again follow from Lemmas 3.2 and 3.4. For x\in A_2 with 00. By the maximum principle, M_O=\sup _{S}u, where S=\partial B^+_{\sqrt{5}}(O), since u=0 on F_8. By comparison, u(x)\le M_O|x|/\sqrt{5},\;\;\forall x\in B^+_{\sqrt{5}}(O)\supset A_1. In particular, $$u(x) \le \frac{M_O x_n}{\sqrt{5}},\quad\forall x=(0,x_n),\; 0< x_n \le 2.$$ \label{12} Our next task now will be to estimate M_O in terms of u(z). We do this as follows. If the maximum M_O occurs near F_8, then it can be controlled first by u at a point away from F_8 by an application of \eqref{10}. This in turn can be estimated by u(z) by the Harnack inequality. Note that a direct application of the Harnack inequality is not possible since the constants degenerate near F_8. If the maximum occurs away from F_8, then the Harnack inequality suffices to achieve our end. We set T=F_8\cap \partial B_{\sqrt{5}}(O). For P\in T, x_n(P)=0, |\xi(P)|=\sqrt{5} and the cylinder A_4(P)\subset A_8. Thus by \eqref{10} and scaling $$\label{13} u(x)\le C_1 u(\bar{P}),\;\;\forall x\in A_{1/2}(P),$$ where \bar{P}=(\xi(P),1) and C_1 is the constant in \eqref{10}. Clearly, \eqref{13} holds in I=\cup_{P\in T} A_{1/2}(P). Let E=\cup_{P\in T}\{\bar{P}\}=\{x: |\xi(x)|=\sqrt{5},\;x_n=1\}. Observe that dist(\bar{P}, F_8)=\mathop{\rm dist}(\bar{P},\partial A_8)=1,\;\forall \bar{P}\in E. Employing Lemma \ref{lm2}, \eqref{13} and recalling that z=(0,2), we have $$\label{14} u(\bar{P})\le u(z)e^{|\bar{P}-z|}\le u(z) e^{\sqrt{6}}\;\;\Rightarrow\;\;u(x)\le C_2 u(z),\;\;\forall x\in I.$$ Clearly, this also holds on \partial B^+_{\sqrt{5}}(O)\cap I. If the maximum M_O occurs in I, \eqref{14} applies. Now for x\in \partial B^+_{\sqrt{5}}(O)\setminus I, dist(x,\partial A_8)=\mathop{\rm dist}(x,F_8)\ge 1 and |x-z|\le \sqrt{6}; we may apply Lemma \ref{lm2}, as done above, to conclude that \[ u(x)\le e^{\sqrt{6}}u(z),\;\;\forall x \in\partial B_{\sqrt{5}}(O)\setminus I.$ This together with \eqref{14} implies that M_O\le C_3u(z), where C_3 is again a universal constant. The inequality in \eqref{12} now implies for some appropriate constant C, \label{15} u(x)\le C u(z) x_n,\quad \forall x=(0,x_n),\;\;00 and \bar{C}>0 $\bar{C}\frac{u_1(z)}{u_2(z)}\le \frac{u_1(x)}{u_2(x)} \le C\frac{u_1(z)}{u_2(z)},\;\;\forall x\in A_1.$ This finishes the proof of part (ii) of Theorem \ref{thm1}. \section{\infty-Capacitary functions in convex rings} Our effort, in Sections 5 and 6, is to prove Theorem \ref{thm2} (see Section 2 for notation). The main ideas used here are similar to those in Section 3. Our strategy will be to prove bounds for u, show strict monotonicity by using scaling, a global maximum principle for |Du| and make remarks about a global lower bound. We start with bounds for u in \Gamma. Our approach is to use appropriate barrier functions and comparison and while these will suffice for our purposes, an approach based on Lemma \ref{lm4} can also be worked out. We make comments along this direction later in this work. The function u\in C(\bar{\Gamma}), from hereon, is an \infty-capacitary function with u|_{\partial C_1}=1,\;u|_{\partial C_2}=0, and, as observed in Section 2, 00,\quad \forall x\in B_R(P)\cap \Gamma. \] \end{lemma} \noindent{\bf Proof:} Set w=w_{P(Q)}(x)=1-\left(|x-P|/R \right); then w is (i) \infty-harmonic in B_R(P)\backslash\{P\}, (ii) w=0 on \partial B_R(P) and (iii) 00,\;\;x\in B_R(P)\cap \Gamma.$$ Note that the function $\tilde{w}(x)=\sup _{Q\in \partial C_2}w_{P(Q)}(x)$ is $\infty$-subharmonic and $u(x)\ge\tilde{w}(x)$.\qed Before we prove an upper bound for $u$, we note the following. Being a $C^2$ domain, $C_1$ satisfies an interior ball condition at every point on $\partial C_1$. For $\eta>0$, let $C_{1,\eta}=\{x\in \Gamma : \mathop{\rm dist}(x,\partial C_1)\le \eta \}$. Since $\partial C_1$ is $C^2$, for every $A\in \partial C_1$, there is a $\delta_A>0$ and an $H_A\in \Gamma$ with the property that $B_{\delta_A}(H_A)\subset \Gamma \;\mbox{and}\;B_{\delta_A}(H_A)\cap \partial C_1\ni A$. We take $\delta_A$ to be the largest such number, and if $\delta_0=\inf_A\{\delta_A\}$, then $\delta_0>0$. Let $l=\mathop{\rm dist}(\partial C_1,\partial C_2)/2$ and $\delta=$ min$( \delta_0,l)$. This choice is made for technical reasons. For notational ease, define $\delta(x)=\mathop{\rm dist}(x, \partial C_1)$. By $\Delta$, we denote the diameter of $C_1$. We should point out that while Lemma \ref{lm6}, as stated below, provides a bound only for points near $\partial C_1$, its derivation requires the calculation of upper bounds in the rest of $\Gamma$. \begin{lemma}[Upper bound] \label{lm6} Let $\delta,\;\delta(x),\;\Delta,\;l$ and $C_{1,\delta}$ be as described above. If $x\in C_{1,\delta}$, i.e., $\delta(x)\le \delta$, then $u(x)\le 1-\frac{ e^{-(\Delta/\delta)}}{2\delta}\delta(x).$ \end{lemma} \noindent {\bf Proof:} We do this in three steps. \\ {\bf Step 1 (Upper bound near $\partial C_2$)} For every $Q\in \partial C_2$, the ball $B_{2l}(Q)\subset C_1$. Fix $Q$ and set $v=v(x)=|x-Q|/(2l)$. Observe that (i) $v$ is $\infty$-harmonic in $B_{2l}(Q)\cap \Gamma$, (ii) $v=1$ on $\partial B_{2l}(Q)\cap \Gamma$ and $0\le v \le 1$ on $\partial C_2 \cap B_{2l}(Q)$. Since $u\le v$ on $\partial(B_{2l}(Q)\cap \Gamma)$, comparison implies $$\label{17} u(x)\le \frac{|x-Q|}{2l}\le 1,\;\; x\in B_{2l}(Q)\cap\Gamma.$$ Clearly $u\le 1/2$ in $B_{l}(Q)$, and so defining $E_l= \cup_{Q\in \partial C_2}(B_{l}(Q)\cap \Gamma)=\{x\in\Gamma:\;\mathop{\rm dist}(x,\partial C_20$, define $C_i^t=tC_i=\{tx,\;\forall x\in C_i\},\;\mbox{and}\;\partial C_i^t=t\partial C_i=\{tx:x\in\partial C_i\},\;i=1,2$. If $0u(P_0)$. To see this, let $L\ni S$ be a straight line such that $L\perp \partial C_2$ and take $P=L\cap \partial C_1$. Thus $u(S)-u(P_0)>0$ and $u(x)-u(P_0)\ge 0$ in $K_{P_0}$. The Harnack inequality implies that $u(P_1)>u(P_0)$. Clearly $u$ is strictly increasing along rays in a strict sub-cone of $K_{P_0}$. \qed One of our goals is to prove the strict positivity of $|Du|$, whenever it exists, in $\Gamma$. To do this, we will need to derive bounds on $u$, at points near $\partial \Gamma$, which in turn require estimates of distances. \begin{lemma}[Distance estimate] \label{lm8} Let $00$, $C_{i,\eta}=\{x\in \Gamma : \mathop{\rm dist}(x,\partial C_i)\le \eta \},\;i=1,2.$ For $Q\in\partial C_2$, let $L=L(Q)$ be the line $\perp$ to $\partial C_2$, $P=P(Q)=L\cap\partial C_1$. Set $\eta_0=\delta/10$, where $\delta$ is the number in Lemma \ref{lm6}, and $\Delta=$diameter$(C_1)$. For every such $Q$, let $\bar{Q}=\bar{Q}(Q,t)\in\partial C_2^t$ be such that $|Q-\bar{Q}|=\mathop{\rm dist}(Q,\partial C^t_2)$. We select $t$ such that sup$_{Q\in \partial C_2}|Q-\bar{Q}|\le \eta_0$. Also take $u_t$ as in Lemma \ref{lm7}. \begin{lemma}[Bounds on $u$ and $u_t$] \label{lm9} Let $t\in (0,1)$ be such that $\partial C_i^t\subset C_{i,{\eta_0}},\;i=1,2.$ Then there exist positive constants $\eta_1$ and $\eta_2$, depending only on the geometry of $\Gamma$, such that \begin{itemize} \item[(i)] $u(x)\le 1- \eta_1(1-t)$, for all $x\in \partial C_1^t$ and \item[(ii)] $u_t(x)\ge \eta_2(1-t)$, for all $x\in \partial C_2$. \end{itemize} \end{lemma} \noindent{\bf Proof:} To prove (i), we use Lemma \ref{lm5}. Let $Q\in \partial C_2$ and select $\bar{Q}\in \partial C_2^t$ closest to $Q$. The line $L$ containing $\bar{Q}$ and $Q$ is orthogonal to $\partial C_2^t$. Let $P=L\cap \partial C_1$, $P_t=L\cap\partial C_1^t$; set $R=|Q-P|$ and $R_t=|\bar{Q}-P_t|$. Then $R_t= R+|\bar{Q}-Q|-|P-P_t|\le R+\eta_0\le \Delta+\eta_0$. Applying Lemma \ref{lm5} to $u_t(Q)$, we see $$u_t(y)\ge 1-(|Q-P_t|/R_t)=(R_t-|Q-P_t|)/R_t=|Q-\bar{Q}|/R_t.$$ Since $Q$ lies on $\partial C_2$, by Lemma \ref{lm8}, $|Q-\bar{Q}|\ge (1-t)l_2$, and we have $$u_t(y)\ge \frac{(1-t)l_2}{\Delta+\eta_0}\;\;\mbox{on}\;\; \partial C_2.$$ To prove (ii) we use Lemma \ref{lm6}. For $x\in \partial C_1^t$, $\delta(x)\ge \mathop{\rm dist} (\partial C_1, \partial C_1^t)\ge l_1(1-t)$. We use \eqref{22} to conclude $u(x)\le 1- \frac{C(\delta,\eta_0)\delta(x)}{2\delta},$ and this in turn yields, $u(x)+ C(\delta) (1-t)l_1\le 1$ on $\partial C_1^t$.\qed We now begin our study of the boundary behaviour of $\infty$-capacitary functions in convex rings. We will utilize the observation in \eqref{1} in Section 2 and Lemma \ref{lm4}. We recall and introduce some notations. For $Q\in \partial C_2$, let $\nu(Q)$ denote the unit outer normal to $C_2$, and for $A\in \partial C_1$, let $\nu(A)$ stand for the unit outer normal to $C_1$. For $Q\in \partial C_2$, let $L=L(Q)\ni Q$ be the straight line with $L\perp \partial C_2$. Call $P=P(Q)=L\cap\partial C_1$; for $x\in L\cap \Gamma$, define $d(x)=d(x,Q)=|x-Q|$. For $A\in \partial C_1$, let $B_r(H)=B_r(H,A)\subset\Gamma$ be the interior ball at $A$, centered at $H$, and for $x$ on the segment formed by $HA$, set $\delta(x)=\delta(x,A)=|x-A|$. Note that $HA$ is directed along $\nu(A)$. The following notation is set up for directional derivatives of $u$ along $\nu(Q)$ and along $\nu(A)$. For $Z\in \partial \Gamma$, set $\Delta(x,Z)=\Delta(x,\nu(Z))=\frac{du(x+\theta\nu(Z))}{d\theta}|_{\theta=0},\;\; x\in\Gamma;$ and when $x=Z$, we write $\Delta(Z)=\Delta(Z,Z)=\Delta(Z,\nu(Z))$, where $\Delta(Z)=\begin{cases} \lim_{\theta\rightarrow0^+}\frac{u(Z+\theta\nu(Z))-u(Z)} {\theta}:& Z\in \partial C_2,\\[2pt] \lim_{\theta\rightarrow 0^-}\frac{u(Z+\theta\nu(Z))-u(Z)}{\theta}:& Z\in \partial C_1,\end{cases}$ whenever they exist. We make an observation before we start. Suppose $A\in\partial C_1$ and $T_A$ is the supporting hyperplane at $A$. Let $J\in \partial C_2$ be such that the supporting hyperplane $T_J\parallel T_A$, i.e., $\nu(J)=\nu(A)$. This is possible since $C_1$ and $C_2$ are both $C^2$. Recall the definitions of the hyperplanes $H^+_A$ and $H^-_J$; see Section 2. Set $G=H^{-}_{J}\cap H^{+}_{A}$ and define $w_A(x)=1+\langle x-A, \nu(A)\rangle/R$, where $R=R(A)=\mathop{\rm dist}(T_A,T_J)$. Note that convexity of $C_1$ implies $\langle x-A, \nu(A)\rangle\le 0,\;\forall x\in H^+_A$; it is easily seen that $w_A|_{T_J}=0$, $w_A|_{T_A}=1$ and $\Delta_{\infty}w_A=0$ in $G$. Now $u\ge w_A=0$ on $T_J\cap\Gamma$ and $w_A\le u=1$ on $\partial C_1$. Comparison in $G\cap \Gamma$ yields that $$\label{23} u(x)\ge w_A(x)=1+\frac{\langle x-A, \nu(A)\rangle}{R},\quad x\in G\cap \Gamma.$$ Thus, for $x\in G\cap \Gamma$ with $(x-A)\parallel \nu(A)$, i.e., $x=A-t\nu(A)$, for some $t>0$, we have $$\label{24} 1-u(x)\le \frac{\delta(x)}{R}\;\Rightarrow\; \frac{1-u(x)}{\delta(x)}\le \frac{1}{R}.$$ \begin{theorem}[A global maximum principle for $|Du|$] \label{thm3} Let $u$ be the $\infty$-capacitary function in $\Gamma$; for $Q\in\partial C_2$, let $d(x)$, $\delta(x)$, $L=L(Q)$ and $P=P(Q)$ be as described above. Then the following are true. \noindent (a) The normal derivatives of $u$ exist on $\partial\Gamma$, i.e., $\forall A\in \partial C_1$ and $\forall Q\in \partial C_2$, $$\Delta(A)>0,\;\;\Delta(Q)>0,\;\;\mbox{and}\;\;\max(\sup_Q\Delta(Q),\; \sup_A\Delta(A))<\infty.$$ \noindent(b) Let $x\in \Gamma$ and $Q\in\partial C_2$ be such that $|x-Q|=\mathop{\rm dist}(x,\partial C_2)$. If $x_1,\;x_2\in L(Q)$ are such that $d(x_1)\le d(x_2)$, then $$0<\frac{u(x_2)-u(x_1)}{|x_1-x_2|}\le \frac{u(x_2)-u(Q)}{d(x_2)}\le \frac{u(x_1)-u(Q)}{d(x_1)}\le \Delta(Q).$$ In particular, $$u(y)0 in B_r(S) and so an application of part (a) of Lemma \ref{lm4}, yields that for x\in SQ, $$\label{25} 0<\frac{u(x)}{d(x)}=\frac{u(x)-u(Q)}{|x-Q|}\uparrow\;\;\mbox{as}\;\;d(x)\downarrow 0,\;\mbox{i.e.,}\; x\rightarrow Q.$$ Recalling Lemma \ref{lm6}, in particular \eqref{17}, we know that u(x)\le |x-Q|/d for an appropriate D=D(Q). Thus$$0<\frac{u(S)}{r}\le\frac{u(x)}{d(x)}=\frac{u(x)-u(Q)}{|x-Q|} \le \frac{1}{D},\;\;\forall x=Q+\theta \nu(Q),\;0<\theta\le \min (r,D). $$Letting \theta \rightarrow 0^+  yields the result for \Delta(Q). To see the result for \Delta(A), note that 1-u(x)>0 in \Gamma and 1-u(A)=0. Let B_r(H)\subset\Gamma be the interior ball at A. Then (A-H)/|A-H|=\nu(A). An application of Lemma \ref{lm4} (a) to 1-u(x) in B_r(H), and \eqref{24} implies that, for x\in HA, $$\label{26} \begin{gathered} \frac{1-u(x)}{\delta(x)}\;\uparrow \;\mbox{as}\;x\rightarrow A,\;\;\mbox{and}\\ 0< \frac{1-u(H)}{r}\le \frac{1-u(x)}{\delta(x)} =\frac{u(A)-u(x)}{|x-A|}\le \frac{1}{R(A)}. \end{gathered}$$ The result for \Delta(A) now follows. Note that \Delta(Q)\le 1/D(Q) and \Delta(A)\le 1/R(A). An inspection of \eqref{17} and \eqref{24} shows that the supremum of each of these quantities is also finite. \noindent {\it Part (b):} Let Q, P and L be as above; set r=|P-Q| and w(x)=1-|x-P|/r in B_r(P). From Lemma \ref{lm5},$$ u(x)\ge w(x)=1-\frac{|x-P|}{r}=\frac{|x-Q|}{r}=\frac{d(x)}{r}, \;\;\;x\in L\cap B_r(P)\cap \Gamma. $$Now let x_1,\;x_2\in L\cap \Gamma such that d(x_1)\le d(x_2). Then the ball B=B_{d(x_2)}(x_2)\subset B_r(P) and B\ni x_1. If we fix x_2 then$$ v(z)=u(x_2)\left(1-\frac{|z-x_2|}{d(x_2)}\right)\le u(z),\;\forall z\in B\cap\Gamma. $$To see this, note that (i) u(z)\ge v(z)=0 on \partial B\cap \Gamma, (ii) u(x_2)=v(x_2), and (iii) u(z)=1> u(x_2)\ge v(z) on B\cap\partial C_1; now apply comparison. For z on the segment x_1x_2, d(z)=d(x_2)-|z-x_2|; taking z=x_1 yields that u(x_1)/d(x_1)\ge u(x_2)/d(x_2). Thus for all x\in L, u(x)/d(x)\uparrow as d(x)\downarrow 0, i.e., as x\rightarrow Q. This implies the assertion about scaling. Also$$\frac{u(x_1)}{d(x_1)}\ge \frac{u(x_2)}{d(x_2)}\ge\frac{u(P)}{R} =\frac{1}{R}\;\Rightarrow\; 0<\frac{u(x_2)-u(x_1)}{|x_2-x_1|}\le \frac{u(x_2)}{d(x_2)} \le \frac{u(x_1)}{d(x_1)}. $$The positivity follows from Lemma \ref{lm7}. By considering x close to Q, applying \eqref{25} and part (a), we obtain the complete assertion in part (b). \noindent {\it Part (c):} We work with v(x)=1-u(x) and use \eqref{26} much the way we did in part (b). \noindent {\it Part (d):} Let x\in \Gamma and \mu(x)=\mathop{\rm dist}(x,\partial \Gamma)=min(\mathop{\rm dist}(x,\partial C_1), dist(x,\partial C_2)). Thus ball B_{\mu(x)}(x) either touches \partial C_1 or \partial C_2 or both. In the first case, calling the point of tangency as P, u(P)=1=\sup _{\Gamma}u. By Lemma \ref{lm4} (d) and part (c) above,$$|Du|(x)\le D(M)=\Delta(P),\;\mbox{where}\; M=1. $$An analogous situation arises if the second case happens; calling the point of tangency to be Q, we see that from part (b)$$|Du|(x)\le D(m)=\Delta(Q),\;\mbox{where}\; m=0.$$Clearly, the statement follows from part (a). \qed \begin{remark} \label{rmk4} \rm The assertion in Theorem \ref{thm3} (b) yields some type of concavity of u along L. Note also that if we take Q=O then, along L, u(P)=u(|P|x/|x|)\le u(x)|P|/|x|, implying thereby u(x)\ge |x|/|P|. This fact has been derived in Lemma \ref{lm5}. The inequalities in Theorem \ref{thm3} (a), clearly imply \begin{eqnarray*} u(x_2)&\le& u(x_1)+\Delta(Q)|x_2-x_1|\;\;\Rightarrow\;\;u(x)\le \Delta(Q)|x-Q|,\;(\mbox{take}\;x_1=Q),\;\mbox{and}\\ u(x)&\ge& 1-\Delta(Q)|x-P|,\;(\mbox{take}\;x_2=P),\;\forall x\in QP. \end{eqnarray*} The latter is similar to Lemma \ref{lm6}. \end{remark} We now prove a global lower bound for u in \Gamma. This consists in taking the supremum (call it w) of affine functions that lie below u; see \eqref{23} and \eqref{24}. It turns out that this lower bound is a solution in many special cases. It is not clear whether this is actually a solution in more general situations. We adopt the notations used in \eqref{23} and \eqref{24}; also see Section 2. Let P\in \partial C_1, R(P) and H_P^{\pm}, as before. Suppose T_P is the supporting hyperplane at P and Q\in \partial C_2 is such that the supporting hyperplane T_Q, at Q, is parallel to T_P. Let H_Q^{\pm} be corresponding the half-spaces. Set G=H^+_P\cap H_Q^- and \nu(P) is the unit outer normal to \partial C_1 at P. \begin{lemma}[Universal lower bound for u] \label{lm10} Let P\in \partial C_1, Q\in \partial C_2, G and \nu(P) as described above. Set $w_P(x)=1+\frac{\langle x-P,\nu(P)\rangle}{\delta(Q)},\;\;\forall x\in \Gamma\;\;\mbox{and}\;\;w(x)=\sup _{P\in\partial C_1}w_P(x).$ Then w|_{\partial C_1}=1,\;\; w|_{\partial C_2}=0,\;w is \infty-subharmonic and u(x)\ge w(x),\;x\in\Gamma. \end{lemma} \noindent {\bf Proof:} From \eqref{23}, we see that w_P(x)\le u(x). Clearly then w(x)\le u(x) and it is well known that w(x) is \infty-subharmonic. Note that w(x)=w_P(x)=1-|x-P|/|P-Q| along the segment PQ. It is also to be noted that in case \Gamma is a spherical annulus or more generally if the geometry of \Gamma is such that C_1=\cup_{x\in C_2}B_{r}(x)=\{x\in \mathbb{R}^n:\;\mathop{\rm dist}(x,C_2)0, then u(x)=w(x). As a matter of fact u(x)=\mathop{\rm dist}(x,\partial C_2)/r.\qed \begin{remark} \label{rmk5} \rm Let P\in \partial C_1 and Q\in \partial C_2 be such that |P-Q|=\mathop{\rm dist}(\partial C_1,\partial C_2)=\delta. Clearly the smallest ball in \Gamma, that touches both \partial C_1 and \partial C_2, has radius \delta/2. From Lemma \ref{lm5} and \eqref{17} (2l=\delta), it follows that u(x)\le |x-Q|/\delta and u(x)\ge 1-(|x-P|/\delta). Note that the segment PQ is orthogonal to both \partial C_1 and \partial C_2. It then follows that u is linear on PQ and u(x)=|x-Q|/\delta,\;\forall x\in PQ. \end{remark} \section{Proof of Theorem \ref{thm2}} \noindent{\bf (a) Proof of part A of Theorem \ref{thm2}: Star-shapedness of level sets \{u=t\} and cone condition.}\\ For 0t in one and u0,$$ where $\Delta$ is the diameter of $C_1$. The result follows. \section{Appendix} In this section we put together results needed in the proofs of the theorems. We will show that odd reflections of $\infty$-harmonic functions stay $\infty$-harmonic and also include the proof of Theorem 1.1 [9]. In the proof of Theorem \ref{thm1}, we required the the following result which we now prove. Let $F$ be the $n-1$ dimensional hyperplane given by $x_n=0$. Let us write $x=(\xi, x_n)$, where $\xi=\xi(x)=(x_1,\ldots,x_{n-1})$; also take $$B^+=B^{+}_{R}(O)=\{x\in B_{R}(O):\;x_n>0\},\;B^-=B^-_{R}(O) =\{x\in B_{R}(O):\;x_n <0\}$$ and $F_R=F\cap B_{R}(O)$. \begin{proposition}[Odd reflection of $u$] \label{prop1} Let $F$ and $O$ be as above and $u$ be $\infty$-harmonic in $B^+$; also assume that $u$ vanishes continuously on $F$. Define $$v(x)=v(\xi(x),x_n)=\begin{cases} u(\xi(x),x_n):& x_n\ge 0\\ -u(\xi(x),-x_n) :& x_n\le 0. \end{cases}$$ Then $v$ is $\infty$-harmonic in $B_{R}(O)$. \end{proposition} \noindent{\bf Proof:} We will show that $\Delta_{\infty}v=0$ in the viscosity sense. Let $\psi\in C^2$ and $P\in B_{R}(O)$ be such that $v-\psi$ attains a local minimum at $P$. We show that $\Delta_{\infty}\psi(P)\le 0$. We will concern ourselves with the cases when $P\in B^-$ and when $P\in F_R$. \noindent {\bf Case A ($\bf{P\in B^-}$):} Since $v(x)-\psi(x)\ge v(P)- \psi(P)$, it follows that $u(\xi,-x_n)-\phi(\xi,-x_n)\le u(\xi(P),-P_n)-\phi(\xi(P),-P_n)$, where $\phi(\xi,-x_n)=-\psi(\xi,x_n)$. Clearly $\Delta_{\infty}\phi(\xi(P),-P_n)\ge 0$, since $(\xi(P),-P_n)\in B^+$ is a point of local maximum of $u-\phi$. Clearly then $\Delta_{\infty}\psi(P)\le 0$. \noindent{\bf Case B ($\bf{P\in F_R}$):} Note $v(P)=v(\xi,0)=0$. Thus $v(x)\ge \psi(x)-\psi(P)$, which in turn implies $$v(x)\ge \langle D\psi(P), x-p\rangle + \frac{1}{2}\langle D^2\psi(P)(x-P),x-P\rangle + o(|x-P|^2),\;\;x\rightarrow P. \label{**}$$ We study various situations. Suppose that $x\in F_R$, i.e., $x_n=x_n-P_n=0$. Since $v(\xi,0)=0$, \eqref{**} implies $$0\ge\sum_{i=1}^{i=n-1}D_i\psi(P)(x-P)_i+ \frac{1}{2}\sum_{i,j=1}^{n-1}D_{ij}\psi(P)(x-P)_i(x-P)_j + o(|\xi(x-P)|^2),$$ $x\rightarrow P$. Now select $x$ such that $(x-P)_i=t$ and $(x-P)_j=0$, $j=1,\ldots,n-1$, $j\ne i$. Then $$0\ge tD_i\psi(P)+\frac{t^2}{2}D_{ii}\psi(P)+ o(t^2),\;\; t\rightarrow 0.$$ Since this holds for all $t\in (-\varepsilon, \varepsilon)$ for small $\varepsilon>0$, it follows that $D_i\psi(P)=0,\;i=1,\ldots,n-1$. From \eqref{**}, it follows that $$v(x)=v(\xi,x_n)\ge D_n\psi(P)(x-P)_n+\frac{1}{2}\sum_{i,j=1}^{n}D_{ij}\psi(P)(x-P)_i(x- P)_j +o(|x-P|^2),$$ $x\rightarrow P$, and $\Delta_{\infty}\psi(P)=(D_n\psi(P))^2 D_{nn}\psi(P)$. To prove this is non-positive, we consider $x$'s such that $\xi(x)=\xi(P)\;(\mbox{i.e.},\; x_i=P_i,\;i=1,\ldots,n-1)$ and $x_n=\pm t$, for small $t$. Let $t>0$, then the above inequality for $\psi$ yields \begin{gather*} v(\xi(P),t)=u(\xi(P),t)\ge t D_n\psi(P)+ \frac{t^2}{2}D_{nn}\psi(P)+o(t^2),\\ v(\xi(P),-t)=-u(\xi(P),t)\ge-t D_n\psi(P)+ \frac{t^2}{2}D_{nn}\psi(P)+o(t^2), \end{gather*} as $t\rightarrow 0$. Adding the two inequalities, dividing by $t^2$ and letting $t\rightarrow 0$, we obtain that $D_{nn}\psi(P)\le 0$. Thus $\Delta_{\infty}\psi(P)\le 0$. The case of local maximum may be handled analogously.\qed \noindent {\bf Proof of Theorem 1.1 [9]} For easy reference, we now include the proof of Theorem 1.1 in [9], as applied to our situation. This is essentially a repetition of the proof in [9], nonetheless we provide details. \begin{theorem}[Boundary Harnack Principle] \label{thm4} Let $A_8=\{x:\;|\xi(x)|<8,\;00$ be $\infty$-harmonic in $A_8$. Then there exists a constant $C$, independent of $u$ but depending on the geometry, such that $\sup _{A_{1}}\le C u(X_0)$. \end{theorem} \noindent{\bf Proof.} Recall \eqref{9} from the proof of Theorem \ref{thm1} and \eqref{6}. Let us continue to call $u$ the extended function obtained by the odd reflection about $F_8$. Let us note that Lemma \ref{lm3} continues to apply to this extended function. Clearly then $u$ is Lipschitz continuous in any sub-cylinder of $A_8$. Our selection of $X_0$ is different from $z$. This means the Harnack constant $M$ will need modification (see \eqref{9}). For $x$ with $10\}\;\;\mbox{and}\;\;u(Y_1)\ge M^{l+4}u(X_0). \] Thus dist$(Y_1, F_8)\le 2^{-l-2}$(if not, an argument along the lines of \eqref{28} will imply$u(Y_0)\le M^{l+3}$) and$\nu(Y_1,2^{-l-2+m})\ge C^m\nu(Y_1,2^{-l-2})\ge C^m u(Y_1)\ge 2M^{l+6}u(X_0)$. Once again there exists a $Y_2\in K(Y_1,2^{-l-2+m})\cap \{x_n>0\}\quad\mbox{and}\quad u(Y_2)\ge M^{l+6}u(X_0).$ Again dist$(Y_2, F_8)\le 2^{-l-4}$and$\nu(Y_2,2^{-l-4+m}) \ge C^m\nu(Y_2,2^{-l-4})\ge 2M^{l+8}u(X_0)$. We obtain by induction a sequence of points$\{Y_k\}$such that $$\label{29} \begin{gathered} \mathop{\rm dist}(Y_k,F_8)\le 2^{-l-2k},\;\;Y_k\in K(Y_{k-1}, 2^{-l-2(k-1)+m})\cap \{x_n>0\},\\ u(Y_k)\ge M^{l+2(k+1)}u(X_0). \end{gathered}$$ Recalling that$K(Y_{k-1}, 2^{-l-2(k-1)+m})$is a cylinder with center$Y_{k-1}$with radius$2^{-l-2(k-1)+m}$and long axis$2(2^{-l-2(k- 1)+m})$, $|Y_{k}|\le |Y_{k-1}-Y_{k}|+|Y_{k-1}|\le 2^{-l-2(k-1)+m+1}+|Y_{k-1}|.$ Thus$|Y_{k}|\le 2^{-l+m+1}\sum_{j=1}^{k}2^{-2(j-1)}+|Y_{0}|\le 2^{-l+m+1}\sum_{j=0}^{k}2^{-2j}+1$. Now choose$l$so large that$|Y_{k}|\le 3/2,\;\forall k.$Thus if$Y\in K(Y_k,2^{-l-2k+m})$then$|Y|\le 3$. Thus each$K(Y_k, 2^{-l-2k+m})$lies in a fixed sub-cylinder of$A_8$. Letting$k\rightarrow\infty$results in a contradiction. The above proves that for some constant$C>0$,$u(x)\le C u(X_0)\le CMu(z)$. 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