\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{mathrsfs} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 109, pp. 1--17.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/109\hfil Normalized $F$-infinity Laplacian] {Existence of solutions to a normalized $F$-infinity Laplacian equation} \author[H. Wang, Y. He \hfil EJDE-2014/109\hfilneg] {Hua Wang, Yijun He} \address{Hua Wang \newline School of Mathematical Sciences, Shanxi University, Taiyuan 030006, China} \email{197wang@163.com} \address{Yijun He (Corresponding author)\newline School of Mathematical Sciences, Shanxi University, Taiyuan 030006, China} \email{heyijun@sxu.edu.cn} \thanks{Submitted March 18, 2014. Published April 16, 2014.} \subjclass[2000]{35D40, 35J60, 35J70} \keywords{Inhomogeneous equation; normalized $F$-infinity Laplacian; \hfill\break\indent viscosity solution} \begin{abstract} In this article, for a continuous function $F$ that is twice differentiable at a point $x_0$, we define the normalized $F$-infinity Laplacian $\Delta_{F; \infty}^N$ which is a generalization of the usual normalized infinity Laplacian. Then for a bounded domain $\Omega\subset\mathbb{R}^n$, $f\in C(\Omega)$ with $\inf_\Omega f(x)>0$ and $g\in C(\partial\Omega)$, we obtain existence and uniqueness of viscosity solutions to the Dirichlet boundary-value problem \begin{gather*} \Delta_{F; \infty}^N u=f, \quad \text{in }\Omega,\\ u=g, \quad \text{on }\partial\Omega. \end{gather*} \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Let $F: \mathbb{R}^n\to[0, +\infty)$ be a function which satisfies the following conditions: \begin{itemize} \item[(a)] $F\in C^2(\mathbb{R}^n\setminus\{0\})$, $F(0)=0$, $F(p)>0$, for any $p\in\mathbb{R}^n\setminus\{0\}$; \item[(b)] $F$ is positively homogeneous of degree 1: $F(tp)=tF(p)$, for any $t>0$ and $p\in\mathbb{R}^n$; \item[(c)] $\operatorname{Hess}(F^2)$ is positive definite in $\mathbb{R}^n\setminus\{0\}$. \end{itemize} Let $\Omega$ be a bounded domain in $\mathbb{R}^n$. For a $C^2(\Omega)$ function $u$, we define the $F$-infinity Laplacian $\Delta_{F; \infty}$ and the normalized $F$-infinity Laplacian $\Delta_{F; \infty}^N$ by \begin{gather} \Delta_{F; \infty}u=F^2(Du)\sum_{i, j=1}^n \frac{\partial^2u}{\partial x_i\partial x_j} \frac{\partial F}{\partial p_i}(Du)\frac{\partial F}{\partial p_j}(Du), \\ \Delta_{F; \infty}^Nu=\sum_{i, j=1}^n \frac{\partial^2u}{\partial x_i\partial x_j} \frac{\partial F}{\partial p_i}(Du)\frac{\partial F}{\partial p_j}(Du) \end{gather} respectively. Clearly when $F(p)=p$, they are the usual infinity Laplacian and the normalized infinity Laplacian, respectively. The operator $\Delta_{F; \infty}$ is a kind of Aronsson operator. A general Aronsson operator $\mathscr{A}_H$ is defined by $$ \mathscr{A}_Hu(x)=\langle D_x H(Du(x),u(x),x), H_p(Du(x), u(x),x)\rangle $$ for a function $H: \mathbb{R}^n\times\mathbb{R}\times\Omega\to\mathbb{R}$, where $H_p$ denotes the gradient of $H(p,s,x)$ with respect to the first variable and $D_x H(Du(x),u(x),x)$ is the gradient of the map $x \mapsto H(Du(x),u(x),x)$. Clearly, $\Delta_{F; \infty}$ is the Aronsson operator $\mathscr{A}_H$ for $H(p, s, x)=\frac12F^2(p)$. The Aronsson equation $\mathscr{A}_H=0$ was proposed by Aronsson in 1960's \cite{Aronsson1,Aronsson1.1,Aronsson2}, which is the Euler-Lagrange equation associated with the variational problem for $L^{\infty}$-functional $$ \mathscr{F}(u,\Omega)=\underset{x \in \Omega}{\operatorname{ess\,sup}} H (Du(x), u(x), x ), \quad u \in W^{1, \infty}(\Omega). $$ In recent years, there have been many studies of properties of the Aronsson equation, especially of the infinity Laplace equation $\Delta_\infty u=0$ which is corresponding to the special case $H(p)=\frac12|p|^2$, see \cite{Aronsson3,ArGrJu,BB,Bhattacharya,JS,LW0,LW2,SaWaYu,Wang,WaYu}, etc. Uniqueness of the viscosity solution of the homogeneous infinity Laplacian equation was established by Jensen in \cite{Jensen}. Later, Barles and Busca gave a second proof of the uniqueness of the infinity harmonic function in \cite{BB}, their proof is quite different from Jensen's work and applies to many degenerate elliptic equations without zeroth-order term. But, largely due to the degeneracy of Aronsson operator, even the basic existence and uniqueness questions have been proven difficult. Several approaches were developed to overcome this difficulty, including the notion of viscosity solutions \cite{cran-ishii-lions} and the method of comparison with cones \cite{BEJ,cran-evans-gar,EY,generalcomparisonGWY}. In \cite{WH}, the authors studied the existence of viscosity solutions for the Dirichlet problem of the inhomogeneous equation $F^{-h}(Du)\Delta_{F; \infty}u=f$, where $0\le h<2$. The special case $F(p)=p$ was studied in \cite{LW0} and \cite{LY}. The existence and uniqueness of the viscosity solutions of the Dirichlet problem $\Delta_\infty^N u=f$ were established by Peres, Schramm, Sheffield and Wilson in \cite{PSSW} using differential game theory and later reproved by Lu and Wang in \cite{LW} using the theory of partial differential equations. In this paper, we study the existence of viscosity solutions for the Dirichlet problem of the inhomogeneous normalized $F$-infinity Laplacian equation. In this paper, $\Omega$ is always assumed to be a bounded open subset of $\mathbb{R}^n$, $f\in C(\Omega)$ with $\inf_\Omega f (x)>0$ or $\sup_\Omega f (x)<0$ and $g\in C(\partial\Omega)$, we concentrate on the Dirichlet problem \begin{equation}\label{eq1.4} \begin{gathered} \Delta_{F; \infty}^Nu=f, \quad \text{in }\Omega,\\ u=g, \quad \text{on }\partial\Omega. \end{gathered} \end{equation} We find the ``radial'' solution to \begin{equation}\label{eq1.5} \Delta_{F; \infty}^Nu=f, \end{equation} where $f=2a$ is a constant. Additionally, we obtain the existence and uniqueness of solutions to the Dirichlet problem in the viscosity sense. When $F(p)=\frac12|p|^2$, these reduce to the cases discussed in \cite{PSSW} and \cite{LW}. We employ the classical Perron's method to get the result of existence. The rest of this paper is organized as follows. In Section 2, we give the notations, definitions related to $\Delta_{F; \infty}^Nu$. In Section 3, we give the ``radial'' solution of the equation $\Delta_{F; \infty}^Nu=1$, and the properties of this solution. In Section 4, we prove our main existence result by Perron's method. \section{Preliminaries} In this paper, $\Omega$ will always be a bounded open subset of $\mathbb{R}^n$. We denote the set of continuous functions on a set $V\subset\mathbb{R}^n$ by $C(V)$. If $V$ is a subset of $\mathbb{R}^n$, $\partial V$ denotes its boundary and $\overline{V}$ its closure. The notation $V\subset\subset\Omega$ means that $V$ is an open subset of $\Omega$ whose closure $\overline{V}$ is a compact subset of $\Omega$. $o(\epsilon)$ means that $\lim_{\epsilon\to0}\frac{o(\epsilon)}{\epsilon}=0$. $\langle\cdot, \cdot\rangle$ denotes the usual Euclidean inner product. $|\cdot|$ denotes the Euclidean norm. $\mathcal S_{n\times n}$ denotes the set of all $n\times n$ symmetric matrices with real entries. $u\in \mathrm{USC}(\Omega)$ denotes the set of all upper semi-continuous functions and $u\in \mathrm{LSC}(\Omega)$ denotes the set of all lower semi-continuous functions. $u\prec_{x_0}\phi$ means $u-\phi$ has a local maximum at $x_0$. On the other hand, $u\succ_{x_0}\phi$ means $u-\phi$ has a local minimum at $x_0$. Almost always in this paper, $u\prec_{x_0}\phi$ (resp. $u\succ_{x_0}\phi$) is understood as $u(x)\le\phi(x)$ (resp. $u(x)\ge\phi(x)$) for all $x\in\Omega$ in interest and $u(x_0)=\phi(x_0)$, as subtracting a constant from $\phi$ does not cause any problem in the standard viscosity solution argument applied in the paper. We define $F^*: \mathbb{R}^n\to[0, \infty)$ to be \begin{equation}\label{eq2.2} F^*(x)=\sup_{\xi\ne0}\frac{\langle x, \xi\rangle}{F(\xi)}, \quad \text{for any $x\in\mathbb{R}^n$}, \end{equation} then $F^*$ has same properties (a), (b), (c) as $F$. Let \[ \alpha={\inf_{\xi\ne0}}\frac{|\xi|}{F(\xi)}, \quad \beta={\sup_{\xi\ne0}}\frac{|\xi|}{F(\xi)}, \] then, by \eqref{eq2.2} and the conditions (a), (b) on $F$, we have $0<\alpha\leq\beta$ and \begin{equation}\label{eq2.3} \alpha|x|\leq F^*(x)\leq\beta|x|, \text{for any $x\in\mathbb{R}^n$}. \end{equation} From \eqref{eq2.3}, we easily get \begin{equation}\label{eq2.4} F^*(-x)\le\frac{\beta}{\alpha}F^*(x),\quad\text{for any $x\in\mathbb{R}^n$}. \end{equation} \begin{definition} \rm For $y\in\mathbb{R}^n$ and $r>0$, we define $B_r^{+}(y)$ by $B_r^+(y)=\{x\in\mathbb{R}^n: F^*(x-y)0$ with $\overline{B_r^+(x_0)\cup B_r^-(x_0)}\subset\Omega$, we define $g(r)=\max_{F^*(x-x_0)=r}u(x)$ and $h(r)=\min_{F^*(x_0-x)=r}u(x)$. In addition, $x_r^+$ denotes any point with $F^*(x_r^+-x_0)=r$ such that $u(x_r^+)=g(r)$, while $x_r^-$ denotes any point with $F^*(x_0-x_r^-)=r$ such that $u(x_r^-)=h(r)$. If $x_0\in\Omega$ and $u\in C(\Omega)$ such that $u$ is twice differentiable at $x_0$, we define the set of maximum directions of $u$ at $x_0$ to be the set $$ E^+(x_0)=\{e=\lim_k\frac{x_{r_k}^+-x_0}{r_k} \text{ for some sequence $r_k\downarrow0$}\} $$ and the set of minimum directions of $u$ at $x_0$ to be the set $$ E^-(x_0)=\{e=\lim_k\frac{x_{r_k}^--x_0}{r_k} \text{ for some sequence $r_k\downarrow0$}\}. $$ \begin{definition} \rm If $u\in C(\Omega)$ is twice differentiable at $x_0$, we define the upper $F$-infinity Laplacian of $u$ at $x_0$ to be $\Delta_{F; \infty}^+u(x_0)=\langle D^2u(x_0)e, e\rangle$, where $e$ is any maximum direction of $u$ at $x_0$ . Similarly, the lower $F$-infinity Laplacian of $u$ at $x_0$ is defined to be $\Delta_{F; \infty}^-u(x_0)=\langle D^2u(x_0)e, e\rangle$, where $e$ is any minimum direction of $u$ at $x_0$ . \end{definition} \begin{remark} \rm From Proposition \ref{prop2.4} which will be proved below, the definition of $\Delta_{F; \infty}^+u(x_0)$ (resp. $\Delta_{F; \infty}^-u(x_0)$) is independent of the choice of maximum (resp. minimum) direction of $u$ at $x_0$. \end{remark} \begin{lemma}[{\cite[page 7]{BCS}}]\label{lm1} For any $y\in\mathbb{R}^n\setminus\{0\}$ and $w\in\mathbb{R}^n$, we have \begin{equation} \label{1.0} w\cdot DF(y)\le F(w), \end{equation} and equality holds if and only of $w=\alpha y$ for some $\alpha\ge 0$. \end{lemma} \begin{proposition}\label{prop2.4} Suppose $u\in C(\Omega)$ is twice differentiable at $x_0$. (1) If $Du(x_0)\ne 0$, then $$ \Delta_{F; \infty}^+u(x_0)=\Delta_{F; \infty}^-u(x_0) =\langle D^2u(x_0)DF(Du(x_0)), DF(Du(x_0))\rangle. $$ (2) If $Du(x_0)=0$, then $$ \Delta_{F; \infty}^+u(x_0)=\max\{\langle D^2u(x_0)e, e\rangle: F^*(e)=1\}, $$ $$\Delta_{F; \infty}^-u(x_0)=\min\{ \langle D^2u(x_0)e, e\rangle : F^*(e) = 1\}. $$ \end{proposition} \begin{proof} (1) There exists a positive-valued function $\rho$ with $\rho(r)\to 0$ as $r\downarrow 0$, defined for all small positive numbers $r$, such that \begin{equation} \label{1.1} |u(x)-u(x_0)-Du(x_0)\cdot(x-x_0)|\le \rho(r)r \end{equation} for all $x$ with $F^*(x-x_0)=r$. Take $\tilde{x}_r^+=x_0+rDF(Du(x_0))$. Then \begin{align*} &u(x_0)+Du(x_0)\cdot(x_r^+-x_0)-\rho(r)r\\ &\le u(x_r^+)\le u(x_0)+Du(x_0)\cdot(\tilde{x}_r^+-x_0)+\rho(r)r. \end{align*} The second inequality is due to the choice of $\tilde{x}_r^+$ and Lemma \ref{lm1}. So, $Du(x_0)\cdot (x_r^+-\tilde{x}_r^+)\le 2\rho(r)r$. On the other hand, the chain of inequalities \begin{align*} &u(x_0)+Du(x_0)\cdot(\tilde{x}_r^+-x_0)-\rho(r)r\\ &\le u(\tilde{x}_r^+) \le u(x_r^+) \le u(x_0)+Du(x_0)\cdot(x_r^+-x_0)+\rho(r)r \end{align*} implies $Du(x_0)\cdot (x_r^+-\tilde{x}_r^+)\ge -2\rho(r)r$. So \begin{equation}\label{1.2} |Du(x_0)\cdot \frac{x_r^+-x_0}{r}-F(Du(x_0))|=|Du(x_0)\cdot \frac{x_r^+-\tilde{x}_r^+}{r}|\le 2\rho(r). \end{equation} Thus, \begin{equation} \label{1.3} \lim_{r\downarrow0}Du(x_0)\cdot \frac{x_r^+-x_0}{r}=F(Du(x_0)). \end{equation} Then, for any $r_k\downarrow0$ such that $\lim_k\frac{x_{r_k}^+-x_0}{r_k}=DF(y_0)$ exists, we must have $Du(x_0)\cdot DF(y_0)=F(Du(x_0))$. So, by Lemma \ref{lm1}, $DF(y_0)=DF(Du(x_0))$. Thus, for any $e\in E^+(x_0)$, $e=DF(Du(x_0))$ holds. Similarly, $E^-(x_0)=\{-DF(Du(x_0))\}$. Therefore, $$ \Delta_{F; \infty}^+u(x_0)=\Delta_{F; \infty}^-u(x_0) =\langle D^2u(x_0)DF(Du(x_0)), DF(Du(x_0))\rangle. $$ (2) If $Du(x_0)=0$, then there exists a positive-valued function $\rho$ with $\rho(r)\to 0$ as $r\downarrow 0$, defined for all small positive numbers $r$, such that \begin{equation} |u(x)-u(x_0)-\langle D^2u(x_0)(x-x_0), x-x_0\rangle|\le \rho(r)r^2 \end{equation} for all $x$ with $F^*(x-x_0)=r$. Let $\lambda^+=\max\{\langle D^2u(x_0)e, e\rangle:F^*(e)=1\}$ and $e^+\in S^+_1(0)$ be such that $\lambda^+=\langle D^2u(x_0)e^+, e^+\rangle$. Take $\tilde{x}_r^+=x_0+re^+$. Then \begin{align*} & u(x_0)+\langle D^2u(x_0)(x_r^+-x_0), x_r^+-x_0\rangle-\rho(r)r^2\\ &\le u(x_r^+)\\ &\le u(x_0)+\langle D^2u(x_0)(\tilde{x}_r^+-x_0), \tilde{x}_r^+-x_0\rangle+\rho(r)r^2. \end{align*} So, $$ \langle D^2u(x_0)(x_r^+-x_0), x_r^+-x_0\rangle -\langle D^2u(x_0)(\tilde{x}_r^+-x_0), \tilde{x}_r^+-x_0\rangle\le2\rho(r)r^2. $$ On the other hand, the chain of inequalities \begin{align*} & u(x_0)+\langle D^2u(x_0)(\tilde{x}_r^+-x_0), \tilde{x}_r^+-x_0\rangle-\rho(r)r^2\\ &\le u(\tilde{x}^+_r)\le u(x_r^+)\\ &\le u(x_0)+\langle D^2u(x_0)(x_r^+-x_0), x_r^+-x_0\rangle+\rho(r)r^2 \end{align*} implies $$ \langle D^2u(x_0)(x_r^+-x_0), x_r^+-x_0\rangle -\langle D^2u(x_0)(\tilde{x}_r^+-x_0), \tilde{x}_r^+-x_0\rangle\ge-2\rho(r)r^2. $$ So \begin{equation} |\langle D^2u(x_0)(\frac{x_r^+-x_0}{r}), \frac{x_r^+-x_0}{r}\rangle -\lambda^+|\le2\rho(r). \end{equation} Then, take any $r_k\downarrow0$ such that $\lim_k\frac{x_{r_k}^+-x_0}{r_k}=e\in E^+(x_0)$, we see $\Delta_{F; \infty}^+u(x_0)=\lambda^+$. Similarly, we have $\Delta_{F; \infty}^-u(x_0) =\min\{ \langle D^2u(x_0)e, e\rangle : F^*(e) = 1\}$. \end{proof} We are then concerned with the viscosity solutions of \eqref{eq1.5} given in the following definition. \begin{definition} \rm $u: \Omega\to\mathbb{R}$ is called a viscosity subsolution of the partial differential equation $\Delta_{F; \infty}^Nu(x)=f(x)$ in $\Omega$, if for any $x_0\in\Omega$ and any test function $\phi\in C^2(\Omega)$ with $u\prec_{x_0}\phi$, there holds $$ \Delta_{F; \infty}^+\phi(x_0)\ge f(x_0). $$ In this case, we say $\Delta_{F; \infty}^Nu\ge f$ in the viscosity sense. Similarly, $u: \Omega\to\mathbb{R}$ is called a viscosity supersolution of the partial differential equation $\Delta_{F; \infty}^Nu(x)=f(x)$ in $\Omega$, if for any $x_0\in\Omega$ and any test function $\phi\in C^2(\Omega)$ with $u\succ_{x_0}\phi$, there holds $$ \Delta_{F; \infty}^-\phi(x_0)\le f(x_0). $$ In this case, we say $\Delta_{F; \infty}^Nu\le f$ in the viscosity sense. A viscosity solution of the partial differential equation $\Delta_{F; \infty}^Nu(x)=f(x)$ in $\Omega$ is both a viscosity subsolution and viscosity supersolution of the equation. \end{definition} Furthermore, viscosity solutions of the Dirichlet problem \eqref{eq1.4} are defined as follows. \begin{definition} \rm A function $u: \Omega\to\mathbb{R}$ is called a viscosity subsolution (resp., supersolution) of \eqref{eq1.4} if $u$ is a viscosity subsolution (resp., supersolution) in $\Omega$ of \eqref{eq1.5} and $u\le g$ (resp., $u\ge g$) on $\partial\Omega$. Furthermore, $u: \Omega\to\mathbb{R}$ is a viscosity solution of \eqref{eq1.4} if it is both a viscosity subsolution and a viscosity supersolution of \eqref{eq1.4}. \end{definition} We will need the concepts of superjets and subjets in our approach. \begin{definition} \rm Suppose $u\in C(\Omega)$. The second-order superjet of $u$ at $x_0$ is defined to be the set $$ J_\Omega^{2,+}u(x_0)=\{(D\phi(x_0), D^2\phi(x_0)):\phi \text{ is $C^2$ and } u\prec_{x_0}\phi\}, $$ whose closure is defined to be \begin{align*} \bar{J}_\Omega^{2, +}u(x_0) =\Big\{&(p, X)\in\mathbb{R}^n\times\mathcal{S}_{n\times n}: \exists (x_n, p_n, X_n)\in\Omega\times\mathbb{R}^n\times\mathcal{S}_{n\times n}\\ \text{ such that}\\ & (p_n, X_n)\in J_\Omega^{2,+}u(x_n) \text{ and } (x_n, u(x_n), p_n, X_n)\to (x_0, u(x_0), p, X)\Big\}. \end{align*} The second-order subjet of $u$ at $x_0$ is defined to be the set $$ J_\Omega^{2,-}u(x_0)=\{(D\phi(x_0), D^2\phi(x_0)):\phi\ \text{is}\ C^2 \text{ and } u\succ_{x_0}\phi\}, $$ whose closure is defined to be \begin{align*} \bar{J}_\Omega^{2, -}u(x_0) =\Big\{&(p, X)\in\mathbb{R}^n\times\mathcal{S}_{n\times n}: \exists (x_n, p_n, X_n)\in\Omega\times\mathbb{R}^n\times\mathcal{S}_{n\times n} \text{ such that }\\ & (p_n, X_n)\in J_\Omega^{2,-}u(x_n) \text{ and } (x_n, u(x_n), p_n, X_n)\to (x_0, u(x_0), p, X)\}. \end{align*} \end{definition} \begin{lemma}[\cite{CS}] \label{lm2.4} (i) \begin{gather}\label{2.5} F^*(DF(p))=1\text{ for $p\in\mathbb{R}^n\setminus\{0\}$}, \\ \label{2.6} F(DF^*(x))=1\text{ for $x\in\mathbb{R}^n\setminus\{0\}$}; \end{gather} (ii) the map $FDF: \mathbb{R}^n\to\mathbb{R}^n$ is invertible and \begin{equation}\label{2.7} FDF=(F^*DF^*)^{-1}. \end{equation} Here, and in what follows, $FDF$ and $F^*DF^*$ are continued by 0 at 0. \end{lemma} \begin{remark} \rm We note we only assume $F$ to be positively homogenous of degree 1, not homogenous of degree 1, so $F(-x)\ne F(x)$ in general, thus $F^*(-x)\ne F^*(x)$ in general either. \end{remark} \begin{lemma}\label{lm3.1} (1) $I$ is an index set, $f\in C(\Omega)$, for any $\lambda\in I$, $\Delta_{F; \infty}^Nu_\lambda\ge f$ in $\Omega$ in the viscosity sense, $u(x)=\sup_{x\in\Omega}u_\lambda(x)<\infty$, then $\Delta_{F; \infty}^Nu\ge f$ in $\Omega$ in the viscosity sense. (2) $I$ is an index set, $f\in C(\Omega)$, for any $\lambda\in I$, $\Delta_{F; \infty}^Nu_\lambda\le f$ in $\Omega$ in the viscosity sense, $u(x)=\inf_{x\in\Omega}u_\lambda(x)>-\infty$, then $\Delta_{F; \infty}^Nu\le f$ in $\Omega$ in the viscosity sense. \end{lemma} \begin{proof} Because the proof of (2) is similar to that of (1), we only present the proof of (1). Suppose $\Delta_{F; \infty}^Nu\ge f$ in the viscosity sense is not true in $\Omega$. Then there exists a point $x_0\in\Omega$ and a test function $\phi\in C^2(\Omega)$ such that $u\prec_{x_0}\phi$ and $\Delta_{F; \infty}^+\phi(x_0)0$, then $u-\phi_\delta$ has a strict maximum at point $x_0$; i.e., $u(x_0)=\phi_\delta(x_0)$, $u(x)<\phi_\delta(x), x\ne x_0$, and we have $$ \Delta_{F; \infty}^+\phi_\delta(x_0)=\Delta_{F; \infty}^+\phi(x_0)+O(\delta)0$ is taken small enough. So we can assume that the original test function $\phi$ satisfies $$ \phi(x)\ge u(x)+\delta|x-x_0|^2 $$ for some $\delta>0$. We claim that $\Delta_{F; \infty}^+\phi(x)u(x_0)-\epsilon$. Let $\hat{\phi}(x)=\phi(x)-\epsilon$. Then $\hat{\phi}(x_0)u(x)\ge u_\lambda(x) $$ on $\partial B_r(x_0)$. So there exists $x_*\in B_r(x_0)$ such that $u_\lambda-\hat{\phi}$ has maximum at $x_*$. As $\Delta_{F; \infty}^+u_\lambda\ge f$ in $\Omega$ in the viscosity sense and $u_\lambda\prec_{x_*}\hat{\phi}$, we have $$ \Delta_{F; \infty}^+\hat{\phi}(x_*)\ge f(x_*), $$ which is a contradiction with the claim we just have derived, $$ \Delta_{F; \infty}^+\hat{\phi}(x)=\Delta_{F; \infty}^+\phi(x)0\}$ is a nonempty domain, in this domain, we calculate: \begin{gather}\label{3.1} \frac{\partial u}{\partial x_i}=[2aF^*(x)+B]\frac{\partial F^*}{\partial x_i},\\ \label{3.2} \frac{\partial^2 u}{\partial x_i\partial x_j} =2a\frac{\partial F^*}{\partial x_i} \cdot\frac{\partial F^*}{\partial x_j} +[2aF^*(x)+B]\frac{\partial^2F^*}{\partial x_i\partial x_j}. \end{gather} As $F$ is positively homogeneous of degree 1, $\frac{\partial F}{\partial p_i}$ is positively homogeneous of degree 0. So by \eqref{2.6} and \eqref{2.7}, we have \begin{equation}\label{3.3} \frac{\partial F}{\partial p_i}(DF^*(x))=\frac{x_i}{F^*(x)}. \end{equation} Thus, by \eqref{2.6}, \eqref{3.1} and \eqref{3.3}, we obtain \begin{gather}\label{3.4} F(Du(x))=2aF^*(x)+B, \\ \label{3.5} \frac{\partial F}{\partial p_i}(Du(x))=\frac{x_i}{F^*(x)}. \end{gather} Since $F^*$ is of class $C^2(\mathbb{R}^n\setminus\{0\})$ and positively homogeneous of degree 1, we have \begin{equation}\label{3.6} \sum_{i=1}^n\frac{\partial F^*}{\partial x_i}x_i=F^*(x), \quad\sum_{i=1}^n\frac{\partial^2 F^*}{\partial x_i\partial x_j}x_i=0, \quad\text{for all }x\ne0. \end{equation} Using \eqref{3.2}, \eqref{3.4}, \eqref{3.5} and \eqref{3.6}, through direct calculation, we obtain $$ \Delta_{F; \infty}^Nu=\sum_{i, j=1}^n\frac{\partial^2u}{\partial x_i\partial x_j} \cdot \frac{\partial F}{\partial p_i}(Du(x))\cdot\frac{\partial F}{\partial p_j} (Du(x))=2a. $$ Thus, we proved that $u(x)=a[F^*(x)]^2+BF^*(x)+C$ is a solution of the equation \begin{equation}\label{eq3.1} \Delta_{F; \infty}^Nu=2a \end{equation} in the domain $\{x\in\mathbb{R}^n\setminus\{0\}: 2aF^*(x)+B>0\}$. Since \eqref{eq3.1} is invariant by translation, $$ \Psi_{x_0, BC}(x)=a[F^*(x-x_0)]^2+BF^*(x-x_0)+C $$ is its $C^2$ solution in $$ D^+(x_0, B):=\{x\in\mathbb{R}^n\setminus\{x_0\}: 2aF^*(x-x_0)+B>0\}. $$ In particular, we have the following lemma. \begin{lemma} $\Psi_{x_0, BC}(x)$ is a viscosity solution of \eqref{eq3.1} in $D^+(x_0, B)$. \end{lemma} \begin{proof} The fact that a classical solution is a viscosity solution follows easily from the definition of a viscosity solution. \end{proof} \begin{remark} \rm Similarly, let \begin{gather*} \Phi_{x_0, BC}(x)=-a[F^*(x_0-x)]^2+BF^*(x_0-x)+C,\\ D^-(x_0, B)=\{x\in\mathbb{R}^n\setminus\{x_0\}: 2aF^*(x_0-x)+B>0\}, \end{gather*} then $\Phi_{x_0, BC}(x)$ is a viscosity solution of equation \begin{equation}\label{eq3.2} \Delta_{F; \infty}^Nu=-2a \end{equation} in $D^-(x_0, B)$. \end{remark} For simplicity, taking $a=1/2$. Letting $B=0$, $\Psi_{x_0}(x)=\frac{1}{2}[F^*(x-x_0)]^2+C$ and $D(x_0)=D^+(x_0, B)=\mathbb{R}^n\setminus\{x_0\}$. \section{A strict comparison principle} \begin{theorem} \label{thm4.1} For $j=1, 2$, suppose $u_j\in C(\overline{\Omega})$ and $$ \Delta_{F; \infty}^Nu_1\le f_1,\quad \Delta_{F; \infty}^Nu_2\ge f_2 $$ in $\Omega$, where $f_1 0$, let $u_\delta = u_2-\delta$. Then $u_\delta < u_1$ on $\partial\Omega$ and $\Delta_{F; \infty}^Nu_\delta\ge f_2$ in $\Omega$. If we can show that $u_\delta < u_1$ in $\Omega$ for every small $\delta > 0$, then it follows that $u_2\le u_1$ in $\Omega$. So we may additionally assume $u_2 < u_1$ on $\partial\Omega$ in the following proof. We apply the sup- and inf-convolution technique here. Take any \[ A\ge\max\{\|u_1\|_{L^\infty(\Omega)}, \|u_2\|_{L^\infty(\Omega)}\}. \] For any sufficiently small real number $\epsilon>0$, we take $\delta=3\sqrt{A\epsilon}$ and $\Omega_\delta=\{x\in\Omega:\operatorname{dist}(x, \partial\Omega)>\delta\}$. We define, on $\mathbb{R}^n$, \begin{gather} \label{2.8} u_{1, \epsilon}(x)=\inf_{y\in\Omega}(u_1(y)+\frac{1}{2\epsilon}|x-y|^2),\\ \label{2.9} u_{2}^\epsilon(x)=\sup_{y\in\Omega}(u_2(y)-\frac{1}{2\epsilon}|x-y|^2) \end{gather} For any $y\in\Omega$ such that $|y-x|\ge 2\sqrt{A\epsilon}$, $u_1(y)+\frac{1}{2\epsilon}|x-y|^2\ge u_1(x)$ holds. So, in $\Omega_\delta$, \begin{equation} \label{2.10} u_{1, \epsilon}(x)=\inf_{y\in\Omega, |x-y| \le 2\sqrt{A\epsilon}}(u_1(y)+\frac{1}{2\epsilon}|x-y|^2) =\inf_{|z|\le 2\sqrt{A\epsilon}}(u_1(x+z)+\frac{1}{2\epsilon}|z|^2), \end{equation} as $x+z\in\Omega$ for any $x\in\Omega_\delta$ and $|z|\le 2\sqrt{A\epsilon}$. Similarly, for $x\in\Omega_\delta$, \begin{equation} \label{2.11} u_{2}^\epsilon(x)=\sup_{y\in\Omega, |x-y|\le 2\sqrt{A\epsilon}}(u_2(y) -\frac{1}{2\epsilon}|x-y|^2)=\sup_{|z|\le 2\sqrt{A\epsilon}}(u_2(x+z) -\frac{1}{2\epsilon}|z|^2), \end{equation} Let \begin{gather} \label{2.12} f_1^\epsilon(x)=\sup_{x+z\in\Omega, |z|\le 2\sqrt{A\epsilon}}f_1(x+z) =\sup_{|z|\le 2\sqrt{A\epsilon}}f_1(x+z),\\ \label{2.13} f_{2, \epsilon}(x)=\inf_{x+z\in\Omega, |z|\le 2\sqrt{A\epsilon}}f_2(x+z) =\inf_{|z|\le 2\sqrt{A\epsilon}}f_2(x+z), \end{gather} for $x\in\Omega_\delta$. Clearly, $f_1^\epsilon$ is upper-semicontinuous. It is continuous due to the equicontinuity of the one parameter family of the functions $x\mapsto f_1(x+z)$ in any compact subset of $\Omega$. $f_{2, \epsilon}$ is continuous for a similar reason. We notice that, for every $z$ with $|z|\le 2\sqrt{A\epsilon}$ and $x\in\Omega_\delta$, \begin{gather} \label{2.14} \Delta_{F; \infty}^N(u_1(x+z)+\frac{1}{2\epsilon}|z|^2) \le f_1(x+z)\le f_1^\epsilon(x),\\ \label{2.15} \Delta_{F; \infty}^N(u_2(x+z)-\frac{1}{2\epsilon}|z|^2)\ge f_2(x+z) \ge f_{2, \epsilon}(x). \end{gather} Lemma \ref{lm3.1} implies that $\Delta_{F; \infty}^Nu_{1, \epsilon}\le f_1^\epsilon$ and $\Delta_{F; \infty}^Nu_2^\epsilon\ge f_{2, \epsilon}$ in $\Omega_\delta$ in the viscosity sense. By \cite[Proposition 6.4]{ArGrJu}, we have the following result. \begin{proposition} \label{prop4.2} $-u_{1, \epsilon}$ and $u_2^\epsilon$ are semi-convex in $\mathbb{R}^n$. $u_{1, \epsilon}\le u_1$ and $u_2^\epsilon\ge u_2$ in $\Omega$. $u_{1, \epsilon}$ and $u_2^\epsilon$ converge locally uniformly to $u_1$ and $u_2$ in $\Omega$, as $\epsilon\to 0$. $u_{1, \epsilon}$ and $u_2^\epsilon$ are both differentiable at the maximum points of $u_2^\epsilon-u_{1, \epsilon}$. \end{proposition} As a result, if we take the value of $\epsilon$ smaller if necessary, then $u_{1, \epsilon}>u_2^\epsilon$ on $\partial\Omega_\delta$, $\Delta_{F; \infty}^Nu_{1, \epsilon}\le f_1^\epsilon$ and $\Delta_{F; \infty}^Nu_2^\epsilon\ge f_{2, \epsilon}$ in $\Omega_\delta$, and $f_1^\epsilon0$ and $\delta=3\sqrt{A\epsilon}$, then $u_2\le u_1$ in $\Omega$ holds. So we may without loss of generality assume that $-u_1$ and $u_2$ are semi-convex in $\mathbb{R}^n$. Suppose $u_1(x_0)0$ such that for any $h\in\mathbb{R}^n$ with $|h|<\delta$, we have $u_1(x_0)f_1(x)$, for all $x\in\Omega_\delta$. For any small positive number $\epsilon$ and $h\in\mathbb{R}^n$ with $|h|<\delta$, we define \begin{equation} \label{3.9} w_{\epsilon, h}(x, y)=u_2(x+h)-u_1(y)-\frac{1}{2\epsilon}|x-y|^2, \end{equation} for all $(x, y)\in\overline{\Omega}_\delta\times\overline{\Omega}_\delta$. Let \begin{gather}\label{3.10} M_0=\max_\Omega(u_2-u_1), \\ \label{3.11} M_h=\max_{\overline{\Omega}_\delta}(u_2(\cdot+h)-u_1(\cdot)), \\ \label{3.12} M_{\epsilon, h}=\max_{\overline{\Omega}_\delta \times\overline{\Omega}_\delta}w_{\epsilon, h}=u_2(x_{\epsilon, h}) -u_1(y_{\epsilon, h})-\frac{1}{2\epsilon}|x_{\epsilon, h}-y_{\epsilon, h}|^2 \end{gather} for some $(x_{\epsilon, h}, y_{\epsilon, h})\in\overline{\Omega}_\delta \times\overline{\Omega}_\delta$. Our assumption implies $M_h>0$ for all $h$ with $0\le |h|<\delta$, and clearly $\lim_{h\to 0}M_h=M_0$. As the semi-convex functions $u_2(\cdot+h)$ and $-u_1$ are locally Lipschitz continuous, the function $M_h$ is Lipschitz continuous in $h\in\mathbb{R}^n$ with $|h|<\delta$, if $\delta$ is taken smaller. By \cite[Lemma 3.1]{cran-ishii-lions}, we know that \begin{gather}\label{3.13} \lim_{\epsilon\downarrow0}M_{\epsilon, h}=M_h, \\ \label{3.14} \lim_{\epsilon\downarrow0}\frac{1}{2\epsilon}|x_{\epsilon, h}-y_{\epsilon, h}|^2=0,\\ \label{3.15} \lim_{\epsilon\downarrow0}(u_2(x_{\epsilon, h}+h)-u_1(y_{\epsilon, h}))=M_h. \end{gather} As a result of the second equality, $\lim_{\epsilon\downarrow0}|x_{\epsilon, h}-y_{\epsilon, h}|=0$. As $M_h>0\ge\max_{\partial\Omega_\delta}(u_2(\cdot+h)-u_1(\cdot))$, we know $x_{\epsilon, h}, y_{\epsilon, h}\in \Omega_1$ for some $\Omega_1\subset\subset\Omega_\delta$ and all small $\epsilon>0$. Then \cite[Theorem 3.2]{cran-ishii-lions} implies that there exist $X=X_{\epsilon, h}, Y=Y_{\epsilon, h}\in\mathcal{S}_{n\times n}$ such that $(\frac{x_{\epsilon, h}-y_{\epsilon, h}}{\epsilon}, X) \in \bar{J}_\Omega^{2, +}u_2(x_\epsilon+h)$, $(\frac{x_{\epsilon, h}-y_{\epsilon, h}}{\epsilon}, Y)\in\bar{J}_\Omega^{2, -} u_1(y_\epsilon)$ and \begin{equation} \label{3.16} -\frac{3}{\epsilon}\begin{pmatrix} I & 0 \\ 0 & I \\ \end{pmatrix}\le\begin{pmatrix} X & 0 \\ 0 & -Y \\ \end{pmatrix}\le\frac{3}{\epsilon}\begin{pmatrix} I & -I \\ -I & I \\ \end{pmatrix}. \end{equation} In particular, $X\le Y$. Again, we solve the problem via a dichotomy. Case 1. Suppose that $\exists h$ with $|h|<\delta$, and $\epsilon_k\to0$ such that $x_{\epsilon_k, h}\ne y_{\epsilon_k, h}$. Then it is easy to see that \begin{align*} % 4.17 f_2(x_{\epsilon_k, h}) &\le \langle X(DF(\frac{x_{\epsilon_k, h}-y_{\epsilon_k, h}}{\epsilon_k})), DF(\frac{x_{\epsilon_k, h}-y_{\epsilon_k, h}}{\epsilon_k})\rangle \\ &\le \langle Y(DF(\frac{x_{\epsilon_k, h}-y_{\epsilon_k, h}}{\epsilon_k})), DF(\frac{x_{\epsilon_k, h}-y_{\epsilon_k, h}}{\epsilon_k})\rangle \\ &\le f_1(y_{\epsilon_k, h}). \end{align*} For a subsequence of $\{\epsilon_k\}$, $x_{\epsilon_k, h}\to x_h$ and $y_{\epsilon_k, h}\to y_h$. As $\lim_{\epsilon\downarrow0}|x_{\epsilon_k, h}-y_{\epsilon_k, h}|=0$, we know that $x_h=y_h$, which leads to a contradiction with the assumption $f_1(x_h)0$. Then $M_{\epsilon, h}=u_2(x_{\epsilon, h}+h)-u_1(y_{\epsilon, h})=M_h$. We simply write $x_{\epsilon, h}=y_{\epsilon, h}=x_h$. The semi-convexity of $u_2(\cdot+h)$ and $-u_1(\cdot)$ implies that the two functions are differentiable at the maximum point $x_h$ of their sum. The definition of $x_h$ shows that \begin{equation} \label{3.20} u_2(x_h+h)-u_1(x_h)\ge u_2(y+h)-u_1(x_h)-\frac{1}{2\epsilon}|x_h-y|^2, \end{equation} which in turn implies \begin{equation} \label{3.21} u_2(x_h+h)\ge u_2(y+h)-\frac{1}{2\epsilon}|x_h-y|^2, \end{equation} for small $\epsilon>0$. So $Du_2(x_h+h)=Du_1(x_h)=0$. For small $h, k\in\mathbb{R}^n$, \begin{align*} M_h &= u_2(x_h+h)-u_1(x_h)\ge u_2(x_k+h)-u_1(x_k)\\ &= M_k+u_2(x_k+h)-u_2(x_k+k)\ge M_k-o(|h-k|), \end{align*} as $Du_2(x_k+k)=0$. So $DM_h=0\ a. e.$ as $M_h$ is Lipschitz continuous, which implies $M_h=M_0$ for all small $h\in\mathbb{R}^n$. At $x_0$, either $f_1(x_0) < 0$ or $f_2(x_0) > 0$ holds due to the fact $f_1 < f_2$. Without loss of generality, we assume that $f_2(x_0) > 0$. The proof for the case $f_1(x_0) < 0$ is parallel. So $u_2$ is $\infty$-subharmonic in a neighborhood of $x_0$. For any $h$ with $|h|<\delta$, \begin{equation} \label{3.22} u_2(x_0+h)-u_1(x_0)\le u_2(x_h+h)-u_1(x_h)=u_2(x_0)-u_1(x_0). \end{equation} So $u_2(x_0)$ is a local maximum of $u_2$. As $\Delta_{F; \infty} u_2\ge 0$, the maximum principle for infinity harmonic functions implies that $u_2$ is constant near $x_0$. So we have \begin{equation} \Delta_{F; \infty}^N u_2(x_0)=\max\{\langle D^2u_2(x_0)e, e\rangle: F^*(e)=1\}=0 0$, we take \begin{equation} u_\delta(x)=(1+\delta)u(x)-\delta\|u\|_{L^\infty(\partial\Omega)}. \end{equation} Then $u_\delta\le u\le v$ on $\partial\Omega$, and it is easily checked by the standard viscosity solution theory that \begin{equation} \Delta_{F; \infty}^N u_\delta(x)=(1+\delta)\Delta_{F; \infty}^N u(x) \ge(1+\delta)f(x)>f(x)\ge\Delta_{F; \infty}^N v(x) \end{equation} in $\Omega$ in the viscosity sense. Applying the preceding strict comparison theorem to $v$ and $u_\delta$, we have $u_\delta\le v$ in $\Omega$ for any small $\delta > 0$. Sending $\delta$ to 0, we have $u\le v$ in $\Omega$ as desired. \end{proof} \section{Existence theorem} In this section, we prove existence of \eqref{eq1.4} by Perron's method. Firstly we prove some lemmas. \begin{lemma}\label{lm4.1} Let $U$ be bounded, $u\in\mathrm{USC}(\overline{U})$ and $\Delta_{F; \infty}u\ge0$ in $U$. If $x_0\in\mathbb{R}^n$, $a\in\mathbb{R}$, $b\ge0$ and \begin{equation}\label{4.1} u(x)\le C(x)=a+bF^*(x-x_0) \quad\text{for $x\in\partial(U\setminus\{x_0\})$}, \end{equation} then \begin{equation}\label{4.2} u(x)\le C(x) \quad\text{for $x\in U$}. \end{equation} \end{lemma} \begin{proof} Firstly we assume $b>0$. Assume that $u(\hat{x})-C(\hat{x})>0$ at some point $\hat{x}\in U \setminus\{x_0\}$. Choose $R$ so large that $F^*(x-x_0)\le R$ on $\partial U$ and put $w=a+bF^*(x-x_0)+\epsilon(R^2-[F^*(x-x_0)]^2)$. Then $u\le w$ on $\partial(U\setminus\{x_0\})$, whereas $u(\hat x)-w(\hat x) > 0$ if $\epsilon$ is sufficiently small. We may assume that $\hat x$ is the maximum of $u-w$ on $U\setminus\{x_0\}$. Through direct calculation, we have \begin{gather}\label{4.3} \frac{\partial w}{\partial x_i}=[b-2\epsilon F^*(x-x_0)] \frac{\partial F^*}{\partial x_i}(x-x_0), \\ \label{4.4} \frac{\partial^2 w}{\partial x_i\partial x_j} =[b-2\epsilon F^*(x-x_0)]\frac{\partial^2F^*}{\partial x_i\partial x_j} (x-x_0)-2\epsilon\frac{\partial F^*}{\partial x_i}(x-x_0) \cdot\frac{\partial F^*}{\partial x_j}(x-x_0). \end{gather} Since $b>0$, we have $b-2\epsilon F^*(\hat x-x_0)>0$, if we choose $\epsilon$ sufficiently small. So the 1-positively homogeneous of $F$, \eqref{2.6}, \eqref{2.7} and \eqref{4.3} imply \begin{equation}\label{4.5} F(Dw)(\hat x)=b-2\epsilon F^*(\hat x-x_0),\quad DF(Dw)(\hat x) =\frac{\hat x-x_0}{F^*(\hat x-x_0)}. \end{equation} Using \eqref{3.6}, \eqref{4.4} and \eqref{4.5}, we obtain $\Delta_{F; \infty}w (\hat x)= -2\epsilon(b-2\epsilon F^*(\hat x-x_0))^2$, and this is strictly negative. This contradicts the assumption $\Delta_{F; \infty}u\ge0$. If $b=0$, we substitute $b$ by $\delta>0$ in \eqref{4.1} and let $\delta\to 0$. \end{proof} \begin{lemma}\label{lm4.2} Let $U$ be bounded, $u\in\mathrm{USC}(\overline{U})$ and $\Delta_{F; \infty}u\ge0$ in $U$. Then the function defined for $y\in U$ and $r<\alpha d(y, \partial U)$ by \begin{equation}\label{eq4.3} L_r^+(y):=\inf\{k\ge0: u(z)\le u(y)+kr, \forall z\in S_r^+(y)\} \end{equation} is nondecreasing in $r$. \end{lemma} \begin{proof} $L_r^+(y)$ is the smallest nonnegative constant for which $$ u(x)\le u(y)+L_r^+(y)F^*(x-y) $$ holds for $F^*(x-y)=r$. Lemma \ref{lm4.1} then implies the inequality holds for $F^*(x-y)\le r$. Thus $(u(x)-u(y))/F^*(x-y)\le L_r^+(y)$ for $F^*(x-y)\le r$. This implies that $L_r^+(y)$ is nondecreasing as a function of $r$ for fixed $y$. \end{proof} \begin{lemma}\label{lm4.21} Let $U$ be bounded, $u\in\mathrm{USC}(\overline{U})$ and $\Delta_{F; \infty}u\ge0$ in $U$. Then $u$ is locally Lipschitz continuous. \end{lemma} \begin{proof} Firstly we show $u$ is bounded below on compact subsets of $U$. Let $x\in U$, $00$, then $L_r^+(y)=\max_{z\in S_r^+(y)}\frac{u(z)-u(y)}{r}$. From \eqref{eq2.3} and Lemma \ref{lm4.2}, we have \begin{equation} \begin{aligned} u(x) &\le u(y)+\max_{z\in S_r^+(y)}\frac{u(z)-u(y)}{r}F^*(x-y) \\ &\le u(y)+\max_{z\in S_r^+(y)}\frac{u(z)-u(y)}{r}\beta|x-y|. \end{aligned}\label{eq4.7} \end{equation} Since $|x-y|2r/\alpha$. Interchanging $x$ and $y$ in \eqref{eq4.7} and putting the resulting relations together yields \begin{equation} |u(x)-u(y)|\le\beta\max(L_r^+(y), L_r^+(x))|x-y|, \end{equation} for $|x-y|\le r/\beta$ and $2r/\alpha<\max(\operatorname{dist} (x, \partial U), \operatorname{dist}(y, \partial U))$. We conclude that $u$ is locally Lipschitz continuous. \end{proof} Now we are ready to prove the existence of a viscosity solution of the Dirichlet boundary problem \eqref{eq1.4} by constructing a solution as the infimum of a family of admissible supersolutions. \begin{theorem} Suppose $\Omega$ is a bounded open subset of $\mathbb{R}^n$, $f\in C(\Omega)$, $\inf_\Omega f (x)>0$ or $\sup_\Omega f (x)<0$, and $g\in C(\partial\Omega)$. Then there exists a unique $u\in C(\overline{\Omega})$ such that $u=g$ on $\partial\Omega$ and $\Delta_{F; \infty}^Nu(x)=f(x)$ in $\Omega$ in the viscosity sense. \end{theorem} \begin{proof} Let $\tilde\Omega=\{x\in\mathbb{R}^n:-x\in\Omega\}$, then $u\in C(\overline{\Omega})$ satisfies $\Delta_{F; \infty}^Nu(x)=f(x), x\in\Omega$ and $u(x)=g(x), x\in\partial\Omega$ in the viscosity sense if and only if $w(x)=-u(-x)\in C(\tilde\Omega)$ satisfies the Dirichlet boundary problem \begin{equation} %5.10 \begin{gathered} \Delta_{F; \infty}^Nw(x)=-f(-x), \quad x \text{in } \tilde\Omega,\\ w(x)=-g(-x), \quad x \text{ on }\partial \tilde\Omega,\\ \end{gathered} \end{equation} in the viscosity sense. Thus, it is sufficient to consider the case $\inf_\Omega f (x)>0$ only, since $-\sup_{x\in\Omega}f (x)=\inf_{x\in\tilde\Omega} \{-f (-x)\}$. In the following, we assume $\inf_\Omega f (x)>0$. We define the admissible sets $S$ and $T$ to be \begin{gather*} S=\{v\in \mathrm{C}(\overline{\Omega}): \Delta_{F; \infty}^Nv\le f \text{ and $v\ge g$ on $\partial\Omega$}\},\\ T=\{w\in \mathrm{C}(\overline{\Omega}): \Delta_{F; \infty}^Nw\ge f \text{ and $w\le g$ on $\partial\Omega$}\}, \end{gather*} where $\Delta_{F; \infty}^Nv\le f$ and $\Delta_{F; \infty}^Nw\ge f$ are satisfied in the viscosity sense. Firstly, we show $S$ and $T$ are nonempty. The constant function $$ \Phi(x)=\|g\|_{L^\infty(\partial\Omega)}+1,\quad x\in\overline{\Omega} $$ is clearly an element of the set $S$. So the admissible set $S$ is nonempty. For any fixed point $z\in\partial\Omega$, take $\Psi(x)=\frac{a}{2}[F^*(x-z)]^2-C$, where $a>\|f\|_{L^\infty(\Omega)}$ and $C>0$ sufficiently large such that $\Psi\le g$ on $\partial\Omega$. Because $\Delta_{F; \infty}^N\psi=a>\|f\|_{L^\infty(\Omega)}\ge f$ in $\Omega$, $\Psi\in T$. That is $T$ is nonempty. Take \begin{gather*} u(x)=\inf_{v\in S}v(x),\quad x\in\overline{\Omega},\\ \bar{u}(x)=\sup_{w\in T}w(x), \quad x\in\overline{\Omega}. \end{gather*} By Theorem \ref{comparison}, we have $w\le v$, $\forall v\in S$, for all $w\in T$. Since $\Phi=\|g\|_{L^\infty(\partial\Omega)}+1\in S$ and $\Psi\in T$, we obtain $u(x)\ge \Psi(x)>-\infty$ and $\bar{u}(x)\le\Phi(x)<\infty$. Thus, by Lemma \ref{lm3.1}, $u$ is a viscosity supersolution of \eqref{eq1.4} in $\Omega$, $\bar u$ is a viscosity subsolution of \eqref{eq1.4} in $\Omega$, and the inequality $\bar u\le g\le u$ holds on $\partial\Omega$. As the infimum of a family of upper semi-continuous functions, $u$ is upper semi-continuous on $\overline{\Omega}$. We have $\Delta_{F; \infty}^Nu\ge f$ in $\Omega$ in the viscosity sense. Suppose not, there exists a $C^2$ function $\phi$ and a point $x_0$ such that $u\prec_{x_0}\phi$, but $\Delta_{F; \infty}^+\phi(x_0)0$, we define \begin{equation} \phi_\epsilon(x)=\phi(x_0)+\langle D\phi(x_0), x-x_0\rangle +\frac12\langle D^2\phi(x_0)(x-x_0), x-x_0\rangle+\epsilon|x-x_0|^2. \end{equation} Clearly, $u\prec_{x_0}\phi\prec_{x_0}\phi_\epsilon$, and $\Delta_{F; \infty}^+\phi_\epsilon(x)u$ for all $x$ near but other than $x_0$ and $\phi_\epsilon(x_0)=u(x_0)$. We define $\hat{\phi}(x)=\phi_\epsilon(x)-\delta$ for a small positive number $\delta$. Then $\hat{\phi}(x) 0$, there is a neighborhood $B_r^+ (z)$ of $z$ such that $|g(x) - g(z)| < \epsilon$ for all $x\in B_r^+(z)\cap\partial\Omega$. Take a large number $C > 0$ such that $Cr > 2\|g\|_{L^\infty(\partial\Omega)}$. We define \begin{equation} v(x) = g(z) + \epsilon + CF^*(x - z) \end{equation} for $x\in\Omega$. For $x\in\partial\Omega$ and $F^*(x-z) < r$, $v(x) \ge g(z) + \epsilon \ge g(x)$; while for $x \in\partial\Omega$ and $F^*(x - z) \ge r$, $v(x) \ge g(z) + \epsilon + Cr >g(z) + \epsilon + 2\|g\|_{L^\infty(\partial\Omega)} \ge g(x)$, that is $v \ge g$ on $\partial\Omega$. In addition, through direct calculation we have $\Delta_{F; \infty}^Nv=0$ in $\Omega$ and since $\inf_\Omega f (x) > 0$, $\Delta_{F; \infty}^Nv=0\le f(x) $ in $\Omega$. So $v\in S$ and $v(z) = g(z) + \epsilon$. Thus $g(z) \le u(z) \le v(z) = g(z) + \epsilon$, for arbitrary $\epsilon > 0$. Letting $\epsilon\to0^+$, we have $u(z) = g(z)$ for any $z\in \partial\Omega$. Indeed, as $\Delta_{F; \infty}^+u(x) = f (x) \ge 0$, $\Delta_{F; \infty}u\ge0$, so by Lemma \ref{lm4.21} $u$ is locally Lipschitz continuous in $\Omega$. Therefore $u$ is continuous in $\Omega$. The following is to prove $u \in C(\overline{\Omega})$. By Lemma \ref{lm3.1}, $\bar{u}$ verifies $\Delta_{F; \infty}^N\bar u(x)\ge f(x)$ in the viscosity sense. Clearly, $\bar u$ is lower semi-continuous in $\overline\Omega$ as the supremum of a family of lower semi-continuous functions and $\bar u \le g$ on $\partial\Omega$. We now show $\bar u \ge g$ on $\partial\Omega$. Fix a point $z\in\partial\Omega$ and a positive number $\epsilon$. Since $g$ is continuous on $\partial\Omega$, there exists a positive number $r$ such that $|g(x) -g(z)| < \epsilon$, for all $x\in \Omega\cap B_r^- (z)$. As $\Omega$ is a bounded domain, the values of $F^*(z-x)$ are bounded above and bounded below from zero for all $x\in\Omega\setminus B_r^- (z)$. We take a large number $A$ such that $A > \sup_{x\in\Omega} F^*(z-x)$ and a large number $C\ge\|f\|_{L^\infty(\Omega)}$ such that $$ C[A^2-(A-r)^2] \ge 2\|g\|_{L^\infty(\partial\Omega)}. $$ We define $$ w(x)=g(z)-\epsilon- C[A^2-(A-F^*(z-x))^2], \quad x\in\overline{\Omega} $$ with $A, C$ as chosen. For $x\in\Omega$, $$ Dw(x)=2C(A-F^*(z-x))DF^*(z-x)\neq0, $$ and \begin{align*} \Delta_{F; \infty}^Nw(x) &= \langle D^2w(x)DF(Dw(x)), DF(Dw(x))\rangle\\ &= 2C\ge\|f\|_{L^\infty(\Omega)}\ge f(x). \end{align*} That is, $w$ is a viscosity subsolution of $\Delta_{F; \infty}^Nu(x)=f(x)$ for all $x\in\Omega$. On $\partial\Omega\cap B_r^-(z)$, $w(x)\le g(z)-\epsilon\le g(x)$; while on $\partial\Omega\setminus B_r^-(z)$, \begin{align*} w(x) &\le g(z)-\epsilon-C[A^2-(A-F^*(z-x))^2]\\ &\le g(z)-\epsilon-2\|g\|_{L^\infty(\partial\Omega)}\\ &\le -\|g\|_{L^\infty(\partial\Omega)}\le g(x). \end{align*} That is to say $w \le g$ on $\partial\Omega$. So the function $w$ defined above is in the family $T$. Thus, from the definition of $\bar{u}$, we obtain $\bar{u}\ge w$. Since $w(z) = g(z) -\epsilon$, we have $\bar{u}(z)\ge g(z)-\epsilon$ for any $\epsilon > 0$, which implies that $\bar{u}(z) \ge g(z)$ for any $z \in \partial\Omega$. As the supremum of a family of lower semi-continuous functions on $\overline\Omega$, $\bar u$ is lower semi-continuous on $\overline\Omega$. Therefore $$ g(z) \le \bar{u}(z) \le \liminf_{x\in\Omega\to z}\bar{u}(x), \quad \forall z \in\partial\Omega. $$ The comparison principle (Theorem \ref{comparison}) implies $v \le w$ on $\Omega$ for any $w \in S$ and $v \in T$ . In particular, $\bar{u}\le u$ in $\Omega$. So $$ g(z) \le \liminf_{x\in\Omega\to z}\bar{u}(x) \le \liminf_{x\in\Omega\to z}u(x), \quad \forall z \in \partial\Omega. $$ On the other hand, the upper semi-continuity of $u$ on $\overline\Omega$ implies that $$ \limsup_{x\in\Omega\to z}u(x) \le u(z) = g(z), \forall z \in\partial\Omega. $$ So $\lim_{x\in\Omega\to z}u(x) = g(z), \forall z \in\partial\Omega$. This shows that $u\in C(\overline\Omega)$. The uniqueness follows from \cite[Theorem 1.4]{LW2}. 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