Proof. Let $d(x_0):=\mathrm {dist}(x_0,\partial \Omega )$ for each $x_0\in \Omega $ . Set

(2-2) $$ \begin{align} k(x_0)= \frac{2(M-m)}{d(x_0)}+1+|\alpha|\mathrm{diam}(\Omega), \end{align} $$

where $M:=\max \nolimits _\Omega u$ and $m:=\min \nolimits _\Omega u$ . For all $y\in B_{{d(x_0)}/{3}}(x_0)$ , we consider the function

$$ \begin{align*}\psi(x):= u(y)+k|x-y|-\frac{|\alpha|}{2}|x-y|^2,\end{align*} $$

where $k:= k(x_0)$ is defined in (2-2). Note that $\psi \in C^{\infty }(\mathbb {R}^{n}-\{y\})$ . For $x\neq y$ ,

$$ \begin{align*} \Delta_{\infty}^{h}\psi(x)=-|\alpha|(k-|\alpha||x-y|)^{h-1}. \end{align*} $$

Since $k\geq 1+|\alpha |\mathrm {diam}(\Omega )$ , we see that $\Delta _{\infty }^{h}\psi \leq \alpha $ in $\Omega \setminus \{y\}$ . Note that $d(y)\geq {2d(x_0)}/{3}$ for any $y\in B_{{d(x_0)}/{3}}(x_0)$ . For any $x\in \partial B_{d(y)}(y)$ ,

$$ \begin{align*} \begin{aligned} \psi(x)&=u(y)+kd(y)-\frac{|\alpha|}{2}d^{2}(y)\\ &\geq m+\frac{d(x_0)}{2}\left(k-\frac{|\alpha|}{2}d(y)\right)\\ &\geq m+\frac{d(x_0)}{2}\left(k-\frac{|\alpha|}{2}\mathrm{diam}(\Omega)\right)\geq M\geq u(x), \end{aligned} \end{align*} $$

where we use (2-2). Therefore, $u\leq \psi $ on $\partial (B_{d(y)}(y)\setminus \{y\})$ . Since $\Delta _{\infty }^{h}\psi \leq \alpha $ and $\Delta _{\infty }^{h}u\geq \alpha $ in $B_{d(y)}(y)\setminus \{y\}$ , by the comparison principle in [Reference Li and Liu25], we have $u\leq \psi $ in $B_{d(y)}(y)$ . Thus, for any $y\in B_{{d(x_0)}/{3}}(x_0)$ and any $z\in B_{d(y)}(y)$ ,

(2-3) $$ \begin{align} u(z)\leq u(y)+k|z-y|-\frac{|\alpha|}{2}|z-y|^{2}. \end{align} $$

For any $p\in B_{{d(x_0)}/{3}}(x_0)$ , one has $B_{{d(x_0)}/{3}}(x_0)\subseteq B_{d(p)}(p)$ . According to (2-3), for any $x, y\in B_{{d(x_0)}/{3}}(x_0),$

$$ \begin{align*} u(y)\leq u(x)+k|x-y|-\frac{|\alpha|}{2}|x-y|^{2} \end{align*} $$

and

$$ \begin{align*} u(x)\leq u(y)+k|x-y|-\frac{|\alpha|}{2}|x-y|^{2}. \end{align*} $$

That is,

$$ \begin{align*} |u(x)-u(y)|\leq \bigg(k-\frac{|\alpha|}{2}|x-y|\bigg)|x-y|\leq k|x-y|\quad \text{for all } x, y\in B_{{d(x_0)}/{3}}(x_0). \end{align*} $$

Thus, for a given $x_{0}\in \Omega $ ,

$$ \begin{align*}|u(x)-u(y)|\leq C|x-y| \quad \text{for all } x,y\in B_{{d(x_0)}/{3}}(x_0),\end{align*} $$

where C depends on $x_{0}, \mathrm {diam}(\Omega ), |\alpha |$ and $\|u\|_{L^{\infty }(\Omega )}$ .

Proof of Theorem 1.1.

Define

$$ \begin{align*}v_{\varepsilon}=v(1+\varepsilon)-\varepsilon\inf\limits_{\overline{\Omega}}v,\quad 0<\varepsilon<\rho-1.\end{align*} $$

Since $F(x,t,p)$ is negative, nondecreasing in t and $\tau \rightarrow F(x,t,\tau p)$ is nondecreasing in $[1,\rho )$ ,

$$ \begin{align*} \Delta_\infty^h v_{\varepsilon}&=(1+\varepsilon)^{h}\Delta_\infty^h v\\ &\leq(1+\varepsilon)^{h}F(x,v,Dv)\\ &\leq(1+\varepsilon)^{h}F(x,v_{\varepsilon},Dv_{\varepsilon})\\ &< F(x,v_{\varepsilon},Dv_{\varepsilon}) \end{align*} $$

in the viscosity sense. That is, $v_{\varepsilon }$ is a viscosity supersolution of (1-6).

Next we want to show $u\leq v_{\varepsilon }$ in $\Omega $ . Suppose in contrast that $u>v_{\varepsilon }$ at some point $x_{0}\in \Omega $ and

$$ \begin{align*} M=\sup\limits_{\Omega}(u-v_{\varepsilon})=u(x_{0})-v_{\varepsilon}(x_{0})>0. \end{align*} $$

According to [Reference Crandall, Ishii and Lions18], we double the variables

$$ \begin{align*} w_{j}(x,y):=u(x)-v_{\varepsilon}(y)-\frac{j}{4}|x-y|^4, \quad(x,y)\in \Omega\times\Omega, \;j=1,2,\ldots. \end{align*} $$

Suppose that $w_{j}$ attains its maximum at $(x_{j},y_{j})\in \overline {\Omega }\times \overline {\Omega }$ . According to [Reference Crandall, Ishii and Lions18, Proposition 3.7],

$$ \begin{align*} \lim\limits_{j\rightarrow \infty}{M_{j}}=\lim\limits_{j\rightarrow\infty}\bigg(u(x_{j})- v_{\varepsilon}(y_{j})-\frac{j|x_{j}-y_{j}|^4}{4}\bigg)=M \end{align*} $$

and

$$ \begin{align*} \lim\limits_{j\rightarrow\infty}\frac{j|x_{j}-y_{j}|^4}{4}=0. \end{align*} $$

It is clear that $x_{j}\rightarrow x_{0}, y_{j}\rightarrow x_{0}$ as $j\rightarrow \infty $ . Since $ M>0\geq \sup \nolimits _{\partial \Omega }(u-v_{\varepsilon })$ , there exists an open set $\Omega _{0}$ such that $x_{0},x_{j}$ and $y_{j}\in \Omega _{0}\subseteq \Omega $ as $j\rightarrow \infty $ .

Let

$$ \begin{align*} \varphi(x)=\frac{j|x-y_{j}|^4}{4} , \quad\psi(y)=-\frac{j|x_{j}-y|^4}{4}. \end{align*} $$

It is obvious that the function $u-\varphi $ has a local maximum at $x_{j}$ and $v_{\varepsilon }-\psi $ has a local minimum at $y_{j}$ .

We consider the following two cases: either $x_{j}= y_{j}$ or $x_{j}\neq y_{j}$ for j large enough.

Case 1: If $x_{j}= y_{j}$ , we have $D\psi (y_{j})=0 $ and $D^{2}\psi (y_{j})=0$ . Since $v_{\varepsilon }$ is a viscosity supersolution,

$$ \begin{align*}F(y_{j},\psi(y_{j}),D\psi(y_{j}))=F(y_{j},v_{\varepsilon}(y_{j}),D\psi (y_{j}))\geq0,\end{align*} $$

which contradicts $F<0$ .

Case 2: If $x_{j}\neq y_{j}$ , we use jets and the maximum principle for semicontinuous functions [Reference Crandall, Ishii and Lions18, Theorem 3.2]. There exist $n\times n$ symmetric matrices $X_{j}$ and $Y_{j}$ such that $X_{j}\leq Y_{j}$ and

$$ \begin{align*} (p_{j},X_j)\in\overline{\mathcal{J}}^{2,+}u(x_{j}),\; \quad (p_{j},Y_j)\in\overline{\mathcal{J}}^{2,-}v_{\varepsilon}(y_{j}),\end{align*} $$

where $p_{j}=j|x_{j}-y_{j}|^{2}(x_{j}-y_{j})$ . By the definitions of the viscosity subsolution and supersolution, since $\Delta _\infty ^h u\geq F(x,u,Du)$ and $\Delta _\infty ^h v_{\varepsilon }\leq (1+\varepsilon )^{h} F(x,v_{\varepsilon },Dv_{\varepsilon })$ in the viscosity sense,

(2-4) $$ \begin{align} 0&\leq|p_{j}|^{h-3}\langle X_{j}p_{j},p_{j}\rangle -F(x_{j},u(x_{j}),p_{j})\nonumber\\ &\leq|p_{j}|^{h-3}\langle Y_{j}p_{j},p_{j} \rangle -F(x_{j},u(x_{j}),p_{j})\nonumber\\ &\leq (1+\varepsilon)^{h}F(y_{j},v_{\varepsilon}(y_{j}),p_{j}) -F(x_{j},u(x_{j}),p_{j}). \end{align} $$

Since $w_{j}$ attains its maximum at $(x_{j},y_{j})\in \overline {\Omega }\times \overline {\Omega }$ ,

(2-5) $$ \begin{align} u(x)-v_{\varepsilon}(y)-\frac{j}{4}|x-y|^4\leq u(x_{j})- v_{\varepsilon}(y_{j})-\frac{j|x_{j}-y_{j}|^4}{4}\quad \text{for all } x,y\in\overline{\Omega}. \end{align} $$

Since $v_{\varepsilon }$ is a viscosity supersolution, we see that $v_\varepsilon $ is locally Lipschitz continuous by Lemma 2.5. Taking $x=y=x_j$ in (2-5),

$$ \begin{align*} \frac{j|x_{j}-y_{j}|^4}{4}\leq v_{\varepsilon}(x_{j})-v_{\varepsilon}(y_{j})\leq L|x_{j}-y_{j}|, \end{align*} $$

where L is the Lipschitz constant of $v_\varepsilon $ . Thus,

$$ \begin{align*} \frac{j|x_{j}-y_{j}|^3}{4}\leq L. \end{align*} $$

Upon taking a subsequence if necessary, we can assume that $p_j\rightarrow p$ . Taking the limit in (2-4),

$$ \begin{align*} (1+\varepsilon)^{h}F(x_{0},v_{\varepsilon}(x_{0}),p) -F(x_{0},u(x_{0}),p)\geq 0. \end{align*} $$

Thus,

(2-6) $$ \begin{align} F(x_{0},v_{\varepsilon}(x_{0}),p)>(1+\varepsilon)^{h}F(x_{0},v_{\varepsilon}(x_{0}),p) \geq F(x_{0},u(x_{0}),p). \end{align} $$

Since $F(x,t,p)$ is nondecreasing in t and $u(x_{0})>v_{\varepsilon }(x_{0})$ ,

$$ \begin{align*}F(x_{0},u(x_{0}),p)\geq F(x_{0},v_{\varepsilon}(x_{0}),p),\end{align*} $$

which contradicts (2-6).

Thus, we have $u\leq v_{\varepsilon }$ in $\Omega $ . Letting $\varepsilon \rightarrow 0$ , we obtain $u\leq v$ in $\Omega $ .

Proof of Theorem 1.2.

Let

$$ \begin{align*} \mathbb{T}:=\{\alpha\in C(\overline{\Omega}):\Delta_\infty^h\alpha \geq F(x,\alpha,D\alpha)\; \mathrm{in}\; \Omega, \alpha\leq \varphi\; \mathrm{on}\; \partial\Omega\; \mathrm{and}\; \alpha\leq u^{*}\; \mathrm{in}\; \Omega\}. \end{align*} $$

Here $\mathbb {T}$ is nonempty since $u_{*}\in \mathbb {T}.$

Let

(3-1) $$ \begin{align} u(x):=\sup\limits_{\alpha\in\mathbb{T}}\alpha(x),\quad x\in\overline{\Omega}. \end{align} $$

Setting $v_{*}:=\inf \nolimits _{\Omega }u_{*}$ and $v^{*}:=\sup \nolimits _{\Omega }u^{*},$

$$ \begin{align*} v_{*}\leq u_{*}\leq u\leq u^{*}\leq v^{*}\quad \mathrm{in}\; \overline{\Omega}. \end{align*} $$

Step 1. We show that u is a viscosity subsolution. Let $\psi \in C^{2}(\Omega )$ and $u-\psi $ have a local maximum at $x_{0}\in \Omega .$ Then there exists some small ball $B_{\rho }(x_{0})\subseteq \Omega $ such that $u(x)-\psi (x)\leq u(x_{0})-\psi (x_{0})$ for any $x\in B_{\rho }(x_{0}).$ We want to show

$$ \begin{align*} \Delta_{\infty}^{h}\psi(x_{0})\geq F(x_{0},u(x_{0}),D\psi(x_{0})). \end{align*} $$

According to (3-1), $u(x_{0})=\sup \nolimits _{\alpha \in \mathbb {T}}\alpha (x_{0})$ . We fix $0<\delta <\rho ^{2(h+1)}$ and pick a sequence $\{\alpha _{k}\}$ in $\mathbb {T}$ such that $u(x_{0})-\alpha _{k}(x_{0})<{\delta }/{k}$ for each positive integer k. Clearly,

(3-2) $$ \begin{align} \alpha_{k}(x)-\psi(x)\leq u(x)-\psi(x)\leq u(x_{0})-\psi(x_{0}) \leq \alpha_{k}(x_{0})-\psi(x_{0})+\frac{\delta}{k} ,\quad x\in B_{\rho}(x_{0}). \end{align} $$

That is,

$$ \begin{align*} \alpha_{k}(x)-\psi(x)-\frac{\delta}{k}\leq \alpha_{k}(x_{0})-\psi(x_{0}),\quad x\in B_{\rho}(x_{0}). \end{align*} $$

Therefore,

$$ \begin{align*} \alpha_{k}(x)-[\psi(x)+|x-x_{0}|^{2(h+1)}]<\alpha_{k}(x)-\psi(x)-\frac{\delta}{k} \leq \alpha_{k}(x_{0})-\psi(x_{0}) \end{align*} $$

for all $x\in B_{\rho }(x_{0})\backslash \overline {B}_{({\delta }/{k})^{1/[2(h+1)]}}(x_{0}).$ This implies that $\alpha _{k}(x)-[\psi (x)+ |x-x_{0}|^{2(h+1)}]$ attains its maximum at some point $x_{k} \in \overline {B}_{({\delta }/{k})^{1/[2(h+1)]}}(x_{0})$ . In particular,

(3-3) $$ \begin{align} \alpha_{k}(x_{0})-\psi(x_{0})\leq \alpha_{k}(x_{k})-[\psi(x_{k})+|x_{k}-x_{0}|^{2(h+1)}]. \end{align} $$

Let $\psi _{0}(x):=\psi (x)+|x-x_{0}|^{2(h+1)}.$ By a direct calculation,

(3-4) $$ \begin{align} \Delta_{\infty}^{h}\psi_{0}(x_{k}) =\Delta_{\infty}^{h}\psi(x_{k})+O((\delta/k)^{{h}/({h+1})}) \geq F(x_{k},\alpha_{k}(x_{k}),D\psi_{0}(x_{k})), \end{align} $$

where we use $\alpha _{k}\in \mathbb {T}.$ Combining (3-2) and (3-3),

$$ \begin{align*} \alpha_{k}(x_{0})-\psi(x_{0})\leq \alpha_{k}(x_{k})-[\psi(x_{k})+|x_{k}-x_{0}|^{2(h+1)}] \leq u(x_{0})-\psi(x_{0})-|x_{k}-x_{0}|^{2(h+1)}, \end{align*} $$

which implies $\lim \nolimits _{k\rightarrow \infty }\alpha _{k}(x_{k})=u(x_{0})$ . Letting $k\rightarrow \infty $ in (3-4),

$$ \begin{align*} \Delta_{\infty}^{h}\psi(x_{0})\geq F(x_{0},u(x_{0}),D\psi(x_{0})). \end{align*} $$

Therefore, u is a viscosity subsolution.

Step 2. We want to show $u\in C(\overline {\Omega })$ and $u=\varphi $ on $\partial \Omega $ . Take two constants $C_{*}<0<C^{*}$ such that

$$ \begin{align*} C_{*}\leq \inf\{F(x,t,p):(x,t,p)\in\Omega\times[v_{*},v^{*}]\times\mathbb{R}^{n}\} \end{align*} $$

and

$$ \begin{align*} C^{*}\geq \sup\{F(x,t,p):(x,t,p)\in\Omega\times[v_{*},v^{*}]\times\mathbb{R}^{n}\}. \end{align*} $$

Since u is a viscosity subsolution of $\Delta _\infty ^h u=F(x,u,Du)$ and $F(x,u,Du)$ is bounded, u is locally Lipschitz in $\Omega $ by Lemma 2.5. Now we proceed to show that u is continuous on $\partial \Omega $ and $u=\varphi $ on $\partial \Omega $ . By [Reference Li and Liu26], there exist $\alpha _{*},\beta ^{*}\in C(\overline {\Omega })$ such that

$$ \begin{align*} \Delta_\infty^h\alpha_{*}=C^{*}\; \mathrm{in}\; \Omega\quad \mathrm{and}\quad \alpha_{*}=\varphi\; \mathrm{on}\; \partial\Omega, \end{align*} $$

$$ \begin{align*} \Delta_\infty^h\beta^{*}=C_{*}\; \mathrm{in}\; \Omega\quad \mathrm{and}\quad \beta^{*}=\varphi\; \mathrm{on}\; \partial\Omega. \end{align*} $$

Since $\Delta _\infty ^h u^{*}\leq C^{*}$ and $\alpha _{*}\leq u^{*}$ on $\partial \Omega $ , we have $\alpha _{*}\leq u^{*}$ in $\Omega $ by the comparison principle (see Remark 2.7). Similarly, we have $u_{*}\leq \beta ^{*}$ in $\Omega $ . Let

$$ \begin{align*} \hat{\alpha}:=\max\{\alpha_{*},u_{*}\}\quad \mathrm{and} \quad \hat{\beta}:=\min\{\beta^{*},u^{*}\}. \end{align*} $$

Then $\hat {\alpha }\in C(\overline {\Omega }), \hat {\beta }\in C(\overline {\Omega })$ , and

$$ \begin{align*} v_{*}\leq u_{*}\leq \hat{\alpha},\,\hat{\beta}\leq u^{*}\leq v^{*}\quad \mathrm{in}\; \Omega \text{ and } \hat{\alpha}=\hat{\beta}=\varphi\; \mathrm{on}\; \partial\Omega. \end{align*} $$

Note that $\alpha _{*}$ and $\beta ^{*}$ are a viscosity subsolution and a viscosity supersolution of (1-8), respectively. By Remark 2.4, we conclude that $\hat {\alpha }$ and $\hat {\beta }$ are a viscosity subsolution and a viscosity supersolution of (1-8), respectively. Thus, $\hat {\alpha }\in \mathbb {T}$ and $\hat {\alpha }\leq u$ in $\overline {\Omega }$ . For any $z\in \mathbb {T}$ , we have $z\leq u^{*}$ in $\overline {\Omega }$ . Additionally, by the comparison principle, we also have $z\leq \beta ^{*}$ in $\overline {\Omega }$ . Therefore, $z\leq \hat {\beta }$ in $\overline {\Omega }$ . This implies $u \leq \hat {\beta }$ in $\overline {\Omega }$ . Then we have $\hat {\alpha }\leq u \leq \hat {\beta }$ in $\overline {\Omega }$ .

Since $\hat {\alpha }, \hat {\beta }\in C(\overline {\Omega })$ and $\hat {\alpha }=\hat {\beta }=\varphi $ on $\partial \Omega $ , it follows that $u\in C(\overline {\Omega })$ and $u=\varphi $ on $\partial \Omega $ .

Step 3. We show that u is a viscosity supersolution. Suppose that it does not hold. Then there exists a point $x_{0}\in \Omega $ and a function $\psi \in C^{2}(\Omega )$ such that $u-\psi $ has a local minimum at $x_{0},$ but

(3-5) $$ \begin{align} \Delta_{\infty}^{h}\psi(x_{0})>F(x_{0},u(x_{0}),D\psi(x_{0})). \end{align} $$

Suppose that $u(x_{0})=u^{*}(x_{0})$ . Then for any x near $x_{0}$ ,

$$ \begin{align*} u^{*}(x)-\psi(x)\geq u(x)-\psi(x)\geq u(x_{0})-\psi(x_{0})=u^{*}(x_{0})-\psi(x_{0}). \end{align*} $$

Since $u^{*}$ is a viscosity supersolution,

$$ \begin{align*} \Delta_{\infty}^{h}\psi(x_{0})\leq F(x_{0},u^{*}(x_{0}),D\psi(x_{0})) =F(x_{0},u(x_{0}),D\psi(x_{0})), \end{align*} $$

which is contrary to (3-5).

Now suppose that $u(x_{0})<u^{*}(x_{0})$ . Let $d(x_{0}):=\mathrm {dist} (x_{0},\partial \Omega )$ . For any $x\in B_{\rho }(x_{0})$ ,

$$ \begin{align*} u(x)-\psi(x)\geq u(x_{0})-\psi(x_{0}). \end{align*} $$

Define $\phi (x):=\psi (x)+(u(x_{0})-\psi (x_{0})).$ For any $x\in B_{\rho }(x_{0})$ ,

$$ \begin{align*} \Delta_{\infty}^{h}\phi(x_{0}) =\Delta_{\infty}^{h}\psi(x_{0})>F(x_{0},\phi(x_{0}),D\phi(x_{0})). \end{align*} $$

Since $F(x,t,p)$ is continuous, one can take $0<\varepsilon _{0}<\min \{1,\rho ,({d(x_{0})}/{2})^{4(h+1)}\}$ small enough such that

(3-6) $$ \begin{align} \Delta_{\infty}^{h}\phi(x_{0})>F(x_{0},\phi(x_{0})+\varepsilon,D\phi(x_{0}))\; \text{ for all } 0<\varepsilon\leq\varepsilon_{0}. \end{align} $$

For $0<\varepsilon \leq \varepsilon _{0}$ , define $\phi _{\varepsilon }(x):=\phi (x)-\sqrt {\varepsilon }|x-x_{0}|^{2(h+1)}+\varepsilon $ . By a direct computation,

$$ \begin{align*} \Delta_{\infty}^{h}\phi_{\varepsilon}(x) =\Delta_{\infty}^{h}\phi(x)+O(\sqrt{\varepsilon}|x-x_{0}|^{2h}),\quad \mathrm{as}\; x\rightarrow x_{0}. \end{align*} $$

Thus, by (3-6),

$$ \begin{align*} \Delta_{\infty}^{h}\phi_{\varepsilon}(x_{0}) =\Delta_{\infty}^{h}\phi(x_{0})>F(x_{0},\phi(x_{0})+\varepsilon,D\phi(x_{0})) =F(x_{0},\phi_{\varepsilon}(x_{0}),D\phi_{\varepsilon}(x_{0})). \end{align*} $$

We claim that there exists a small $\varepsilon _{1}$ , with $0<\varepsilon _{1}<\varepsilon _{0}$ , such that

$$ \begin{align*} \Delta_{\infty}^{h}\phi_{\varepsilon_{1}}(x)>F(x,\phi_{\varepsilon_{1}}(x),D\phi_{\varepsilon_{1}}(x))\quad \text{ for all } x\in B_{\varepsilon_{1}^{1/[4(h+1)]}}(x_{0}). \end{align*} $$

Suppose that our claim does not hold. Then for each small $\varepsilon>0$ , there exists an $x_{\varepsilon }\in B_{\varepsilon ^{1/[4(h+1)]}}(x_{0})$ such that $\Delta _{\infty }^{h}\phi _{\varepsilon }(x_{\varepsilon }) \leq F(x_{\varepsilon },\phi _{\varepsilon }(x_{\varepsilon }),D\phi _{\varepsilon }(x_{\varepsilon }))$ . Since $x_{\varepsilon }\rightarrow x_{0}$ as $\varepsilon \rightarrow 0$ , we observe that

$$ \begin{align*} \lim\limits_{\varepsilon\rightarrow 0} \Delta_{\infty}^{h}\phi_{\varepsilon}(x_{\varepsilon}) =\Delta_{\infty}^{h}\phi(x_{0})\quad \mathrm{and} \quad \lim\limits_{\varepsilon\rightarrow 0} F(x_{\varepsilon},\phi_{\varepsilon}(x_{\varepsilon}),D\phi_{\varepsilon}(x_{\varepsilon})) =F(x_{0},\phi(x_{0}),D\phi(x_{0})). \end{align*} $$

We conclude that $\Delta _{\infty }^{h}\phi (x_{0})\leq F(x_{0},\phi (x_{0}),D\phi (x_{0})),$ which is a contradiction and the claim is proved.

Since $\phi (x_{0})=u(x_{0})<u^{*}(x_{0})$ , we can suppose that $\varepsilon _{1}$ is small enough and for all $x\in B_{\varepsilon _{1}^{1/[4(h+1)]}}(x_{0})$ , $\phi (x)\leq \phi _{\varepsilon _{1}}(x) \leq u^{*}(x)$ . Since $u(x_{0})<u(x_{0})+\varepsilon _{1}=\phi (x_{0})+\varepsilon _{1} =\phi _{\varepsilon _{1}}(x_{0})$ , there exists $0<s_{1}<\varepsilon _{1}^{1/[4(h+1)]}$ such that $u(x)<\phi _{\varepsilon _{1}}(x)$ for all $x\in B_{s_{1}}(x_{0})$ . We note that

$$ \begin{align*} u(x)\geq \phi(x)\quad \text{for all } x\in B_{\rho}(x_{0}), \end{align*} $$

and

$$ \begin{align*} &u(x)-\phi_{\varepsilon_{1}}(x)=u(x)-\phi(x) +\sqrt{\varepsilon_{1}}|x-x_{0}|^{2(h+1)}-\varepsilon_{1}>0\\ &\quad\text{for all } x\in B_{\rho}(x_{0})\backslash \overline{B}_{\varepsilon_{1}^{1/[4(h+1)]}} (x_{0}). \end{align*} $$

To sum up,

(3-7) $$ \begin{align} \begin{cases} (1)\;\Delta_\infty^h \phi_{\varepsilon_{1}}(x)>F(x,\phi_{\varepsilon_{1}}(x),D\phi_{\varepsilon_{1}}(x)) \quad &\text{for all } x\in B_{\varepsilon_{1}^{1/[4(h+1)]}}(x_{0}), \\ (2)\;\phi_{\varepsilon_{1}}(x)<u^{*}(x)\quad &\text{for all } x\in B_{\varepsilon_{1}^{1/[4(h+1)]}}(x_{0}),\\ (3)\;u(x)<\phi_{\varepsilon_{1}}(x) \quad &\text{for all } x\in B_{s_{1}}(x_{0}),\\ (4)\;u(x)>\phi_{\varepsilon_{1}}(x) \quad &\text{for all } x\in B_{\rho}(x_{0})\backslash \overline{B}_{\varepsilon_{1}^{1/[4(h+1)]}}(x_{0}). \end{cases} \end{align} $$

Now define

$$ \begin{align*} \hat{u}(x)=\begin{cases} u(x)\quad &\mathrm{if}\; x\in\; \Omega\backslash \overline{B}_{\varepsilon_{1}^{1/[4(h+1)]}}(x_{0}), \\ \sup\{\phi_{\varepsilon_{1}}(x),u(x)\}\quad &\mathrm{if}\; x\in B_{\varepsilon_{1}^{1/[4(h+1)]}}(x_{0}). \end{cases} \end{align*} $$

It is obvious that $\hat {u}\in C(\overline {\Omega })$ and $u_{*}\leq u\leq \hat {u}\leq u^{*}$ in $\Omega $ . Next we want to show that $\hat {u}\in \mathbb {T}$ . Take $\hat {\psi }\in C^{2}(\Omega )$ such that $\hat {u}-\hat {\psi }$ has a local maximum at some point $y\in \Omega $ , that is, $\hat {u}(x)-\hat {\psi }(x)\leq \hat {u}(y)-\hat {\psi }(y)$ for x in some ball $B_{\delta }(y)$ . It is clear that $\hat {u}(y)=u(y)$ or $\hat {u}(y)=\phi _{\varepsilon _{1}}(y)$ . If $\hat {u}(y)=u(y)$ , we note that $u\leq \hat {u}$ in $\Omega $ . For all $x\in B_{\delta }(y)$ ,

$$ \begin{align*} u(x)-\hat{\psi}(x)\leq \hat{u}(x)-\hat{\psi}(x)\leq \hat{u}(y)-\hat{\psi}(y)=u(y)-\hat{\psi}(y). \end{align*} $$

Thus, $u-\hat {\psi }$ has a local maximum at y. Since u is a viscosity subsolution, we have $\Delta _{\infty }^{h}\hat {\psi }(y)\geq F(y,u(y),D\hat {\psi }(y))=F(y,\hat {u}(y),D\hat {\psi }(y))$ . If $\hat {u}(y)=\phi _{\varepsilon _{1}}(y)$ , we can see that $u(y)<\phi _{\varepsilon _{1}}(y)$ . According to (3-7)(4), we have $y\in B_{\varepsilon _{1}^{1/[4(h+1)]}}(x_{0})$ . Then note that $\phi _{\varepsilon _{1}}\leq \hat {u}$ and for any $x\in B_{\varepsilon _{1}^{1/[4(h+1)]}}(x_{0})\bigcap B_{\delta }(y)$ ,

$$ \begin{align*} \phi_{\varepsilon_{1}}(x)-\hat{\psi}(x)\leq \hat{u}(x)-\hat{\psi}(x)\leq \hat{u}(y)-\hat{\psi}(y) =\phi_{\varepsilon_{1}}(y)-\hat{\psi}(y). \end{align*} $$

Thus, $\phi _{\varepsilon _{1}}-\hat {\psi }$ has a local maximum at y. This implies that $\Delta _{\infty }^{h}\phi _{\varepsilon _{1}}(y)\leq \Delta _{\infty }^{h}\hat {\psi }(y)$ . Together with (3-7)(1), one has $\Delta _{\infty }^{h}\hat {\psi }(y)\geq F(y,\phi _{\varepsilon _{1}}(y),D\phi _{\varepsilon _{1}}(y)) =F(y,\hat {u}(y),D\hat {\psi }(y))$ . In any case, we show that $\hat {u}$ is a viscosity subsolution, that is, $\hat {u}\in \mathbb {T}$ . By (3-7)(3), we see that $\hat {u}=\phi _{\varepsilon _{1}}>u$ in $B_{s_{1}}(x_{0})$ , which contradicts the definition of u. Therefore, u is a viscosity supersolution.

We have completed the proof that u is a viscosity solution to Problem (1-8) in $\Omega $ .

Proof. Define the cut-off function $\hat {F}(x,t,p):\Omega \times \mathbb {R}\times \mathbb {R}^{n}\rightarrow \mathbb {R}$ by

(4-2) $$ \begin{align} \hat{F}(x,t,p)=\begin{cases} F(x,t,p)\quad &\mathrm{for} \ t\geq \underline{u}(x), \\ F(x,\underline{u}(x),p)\quad &\mathrm{for} \ t<\underline{u}(x). \end{cases} \end{align} $$

Since $\underline {u}>0$ in $\Omega $ , we have $\underset {\Omega \times I\times \mathbb {R}^{n}}\sup |\hat {F}(x,t,p)|<\infty $ for any compact interval $I\subseteq \mathbb {R}$ . Let $\{\Omega _{j}\}$ be a sequence of sub-domains of $\Omega $ such that

$$ \begin{align*} \Omega_{j}\Subset\Omega_{j+1}\quad \text{and} \quad\Omega=\bigcup_{k=1}^{\infty}\Omega_{k}, \quad (j=1,2,\ldots). \end{align*} $$

For each positive integer j, we consider the following Dirichlet problem:

(4-3) $$ \begin{align} \begin{cases} \Delta_\infty^h u=\hat{F}(x,u,Du) \quad &\mathrm{in}\; \Omega_{j}, \\ u=\underline{u} \quad &\mathrm{on} \; \partial\Omega_{j}. \end{cases} \end{align} $$

Since $\overline {u}$ is a viscosity supersolution of (4-1) and $0<\underline {u}\leq \overline {u}$ in $\Omega $ ,

$$ \begin{align*}\Delta_\infty^h \overline{u}\leq F(x,\overline{u},D\overline{u})=\hat{F}(x,\overline{u},D\overline{u}) \quad \mathrm{in}\; \Omega_{j}\quad \mathrm{and} \quad\overline{u}\geq \underline{u}\quad \mathrm{on}\; \partial\Omega_{j}\end{align*} $$

in the viscosity sense, where we use (4-2). That is, $\overline {u}$ is a viscosity supersolution of (4-3) in $\Omega _{j}$ for each positive integer j. Similarly, $\underline {u}$ is a viscosity subsolution of (4-3) in $\Omega _{j}$ . Since $\underline {u}\leq \overline {u}$ in $\Omega $ , according to Theorem 1.2, we can find a viscosity solution $\tilde {u}_{j}\in C(\overline {\Omega }_{j})$ of (4-3) such that $\underline {u}\leq \tilde {u}_{j}\leq \overline {u}$ in $\overline {\Omega }_{j}$ . We extend $u_{j}$ to $\overline {\Omega }$ by defining

$$ \begin{align*} u_{j}=\begin{cases} \tilde{u}_{j} \quad &\mathrm{in}\; \Omega_{j}, \\ \underline{u} \quad &\mathrm{on} \; \overline{\Omega}\setminus\Omega_{j}. \end{cases} \end{align*} $$

Then we obtain a sequence $\{u_{j}\}$ in $C(\overline {\Omega })$ with $\underline {u}\leq u_{j}\leq \overline {u}$ in $\overline {\Omega }$ for all j. In particular, $\{u_{j}\}$ is uniformly bounded in $\overline {\Omega }$ . We also note that $\{u_{j}\}$ is locally Lipschitz continuous in $\Omega $ from Lemma 2.5. Thus, $\{u_{j}\}$ is equicontinuous. Therefore, there exists a subsequence of $\{u_{j}\}$ that converges locally uniformly to some $u\in C(\Omega )$ . Since $\underline {u}\leq u\leq \overline {u}$ in $\overline {\Omega }$ and $\underline {u}=u=\overline {u}$ on $\partial \Omega $ , we have $u\in C(\overline {\Omega })$ .

By (4-2) and $u\geq \underline {u}> 0$ in $\Omega $ , we have $\hat {F}(x,u,Du)=F(x,u,Du)$ in $\Omega $ . If we want to show that u is a viscosity solution of $\Delta _\infty ^h u=F(x,u,Du)$ , we only need to show that u is a viscosity solution of $\Delta _\infty ^h u=\hat {F}(x,u,Du)$ in $\Omega $ . First, we show $\Delta _\infty ^h u\leq \hat {F}(x,u,Du)$ in the viscosity sense.

Suppose that $\psi \in C^{2}(\Omega )$ and $u-\psi $ has a local minimum at some $x_{0}\in \Omega $ . Then,

$$ \begin{align*}u(x)-\psi(x)\geq u(x_{0})-\psi(x_{0}),\end{align*} $$

where $x\in B_{r}(x_{0})\Subset \Omega _{l}$ for some small $r>0$ and some positive integer l. Fix $\varepsilon>0$ small enough and let $x_{j}\in \overline {B}_{r}(x_{0})$ be a point of minimum of

$$ \begin{align*}u_{j}(x)-\bigg(\psi(x)-\frac{\varepsilon}{2}|x-x_{0}|^{2}\bigg),\quad j\geq l.\end{align*} $$

Then,

(4-4) $$ \begin{align} u_{j}(x_{j})-\bigg(\psi(x_{j})-\frac{\varepsilon}{2}|x_{j}-x_{0}|^{2}\bigg) \leq u_{j}(x_{0})-\psi(x_{0}). \end{align} $$

Since $x_{j}\in \overline {B}_{r}(x_{0})$ , we assume that $x_{j}\rightarrow \hat {x}$ for some $\hat {x}\in \overline {B}_{r}(x_{0})$ . Taking the limit in (4-4) as $j\rightarrow \infty $ ,

$$ \begin{align*}u(\hat{x})-\bigg(\psi(\hat{x})-\frac{\varepsilon}{2}|\hat{x}-x_{0}|^{2}\bigg) \leq u(x_{0})-\psi(x_{0}).\end{align*} $$

Therefore,

$$ \begin{align*}\frac{\varepsilon}{2}|\hat{x}-x_{0}|^{2}\leq u(x_{0})-\psi(x_{0})- (u(\hat{x})-\psi(\hat{x}))\leq 0.\end{align*} $$

This yields $\hat {x}=x_{0}$ . Since $u_{j}$ is a viscosity solution of (4-3) and $x_{j}$ is a point of minimum of $u_{j}(x)-(\psi (x)-{\varepsilon }/{2}|x-x_{0}|^{2})$ in $B_{r}(x_{0})$ , we take $\phi _{\varepsilon }(x):=\psi (x)-{\varepsilon }/{2}|x-x_{0}|^{2}$ as a test function. Then,

$$ \begin{align*} \Delta_\infty^h \phi_{\varepsilon}(x_{j})\leq\hat{F}(x_{j},u_{j}(x_{j}),D\phi_{\varepsilon}(x_{j})). \end{align*} $$

Recalling that $u_{j}\rightarrow u$ uniformly in $B_{r}(x_{0})$ and taking the limit as $j\rightarrow \infty $ ,

$$ \begin{align*}\Delta_\infty^h \phi_{\varepsilon}(x_{0})\leq\hat{F}(x_{0},u(x_{0}),D\phi_{\varepsilon}(x_{0})).\end{align*} $$

Letting $\varepsilon \rightarrow 0$ ,

$$ \begin{align*}\Delta_\infty^h \psi(x_{0})\leq\hat{F}(x_{0},u(x_{0}),D\psi(x_{0})),\end{align*} $$

that is, u is a viscosity supersolution of (4-3). Thus, u is a viscosity supersolution of (4-1).

Similarly, one can prove that u satisfies $\Delta _\infty ^h u\geq \hat {F }(x,u,Du)$ in the viscosity sense. We leave it to the reader.

Proof. Define a continuous function $G:\mathbb {R}^{+}\times \mathbb {R}^{+}\rightarrow \mathbb {R}^{+}$

$$ \begin{align*} G(t,\tau) := \begin{cases} t ^{h}\underset{t\leq s\leq \tau }\sup g(s) s^{-h} & 0<t\leq \tau , \\ t ^{h} g(\tau) \tau^{-h} & 0<\tau \leq t. \end{cases} \end{align*} $$

It is easy to see that $G(t, \tau ) t^{-h}$ is nonincreasing in t. Let

(4-7) $$ \begin{align} \Gamma(t)=\frac{2}{t} \int_{t / 2}^t \frac{G(s, t)}{s^h} \,ds,\quad t>0. \end{align} $$

Noting that $\Gamma $ is a $C^1$ function,

$$ \begin{align*} \begin{aligned} \Gamma'(t) &=-\frac{2}{t^2} \int_{t/2}^t \frac{G(s, t)}{s^h}\,ds +\frac{2}{t}\bigg(\frac{G(t,t)}{t^h}-\frac{1}{2} \frac{G(t / 2, t)}{(t / 2)^h}\bigg) \\ & \leq-\frac{1}{t} \frac{G(t, t)}{t^h}+\frac{2}{t} \frac{G(t, t)}{t^h} -\frac{1}{t} \frac{G(t / 2, t)}{(t / 2)^h} \\ &=\frac{1}{t}\bigg( \frac{G(t, t)}{t^h}- \frac{G(t / 2, t)}{(t / 2)^h} \bigg)\\ & \leq 0. \end{aligned} \end{align*} $$

Hence, $\Gamma $ is a nonincreasing function in $\mathbb {R}^{+}$ . It is obvious that

(4-8) $$ \begin{align} \frac{g(t)}{t^{h}} =\frac{G(t,t)}{t^{h}}\leq\Gamma(t) \leq \frac{G(\frac{t}{2},t)}{(\frac{t}{2})^{h}} =\sup_{({t}/{2})\leq s \leq t}\frac{g(s)}{s^{h}}\leq\sup_{({t}/{2})\leq s < \infty}\frac{g(s)}{s^{h}}. \end{align} $$

Therefore,

$$ \begin{align*} \lim_{t\rightarrow\infty}\Gamma(t)=g_{\infty}. \end{align*} $$

Now we suppose that $0<g_{\infty }<\infty $ and, in this case, we define

(4-9) $$ \begin{align} \lambda_{\Omega}^{\ast}:=\frac{1}{g_{\infty}\|w_{\Omega}\|_{\infty}^{h}}. \end{align} $$

For $0<\lambda <\lambda _{\Omega }^{\ast }$ ,

$$ \begin{align*} \lim_{t\rightarrow\infty}\Gamma(t) =\frac{1}{\lambda_{\Omega}^{\ast}\|w_{\Omega}\|_{\infty}^{h}} <\frac{1}{\lambda\|w_{\Omega}\|_{\infty}^{h}}. \end{align*} $$

Consider the function

$$ \begin{align*} \Psi(t):=\int_{0}^{t}\frac{1}{\sqrt[h]{\Gamma(s)}} ds ,\quad t>0. \end{align*} $$

We have

$$ \begin{align*} \lim_{t\rightarrow\infty}\frac{\Psi(t)}{t}=\lim_{t\rightarrow\infty}\Psi'(t) =\lim_{t\rightarrow\infty}\frac{1}{\sqrt[h]{\Gamma(t)}}>\sqrt[h]{\lambda} \|w_{\Omega}\|_{\infty}. \end{align*} $$

Thus, we can find $\theta \geq 1$ large enough such that

(4-10) $$ \begin{align} \Psi(\theta)\geq\theta \sqrt[h]{\lambda}\|w_{\Omega}\|_{\infty}. \end{align} $$

In the other case, we suppose $g_{\infty }=\infty $ . One can verify that

$$ \begin{align*} \alpha_{\ast}:=\sup_{t\geq1}\frac{\Psi(t)}{t}<\infty. \end{align*} $$

Set

(4-11) $$ \begin{align} \lambda_{\Omega}^{\ast}:=\bigg(\frac{\alpha_{\ast}}{\|w_{\Omega}\|_{\infty}}\bigg)^{h}. \end{align} $$

Then for $0<\lambda <\lambda _{\Omega }^{\ast },$

$$ \begin{align*} \sqrt[h]{\lambda}\|w_{\Omega}\|_{\infty}<\sqrt[h]{\lambda_{\Omega}^{\ast}}\|w_{\Omega}\|_{\infty} =\alpha_{\ast}. \end{align*} $$

Therefore, there exists $\gamma _{\infty }\geq 1$ , depending on $\lambda $ , such that

(4-12) $$ \begin{align} \sqrt[h]{\lambda}\|w_{\Omega}\|_{\infty}<\frac{\Psi(\gamma_{\infty})}{\gamma_{\infty}}. \end{align} $$

Hence, by (4-10) and (4-12), there exists $\theta \geq 1$ such that

(4-13) $$ \begin{align} \Psi(\theta)>\theta\sqrt[h]{\lambda}\|w_{\Omega}\|_{\infty}. \end{align} $$

Let $\Phi $ be the inverse of $\Psi $ , that is, $\Phi $ satisfies

$$ \begin{align*} \int_{0}^{\Phi(t)}\frac{1}{\sqrt[h]{\Gamma(s)}}\,ds =t, \quad t\geq0. \end{align*} $$

Define

(4-14) $$ \begin{align} v_{\Omega,\lambda}(x):=\Phi(\theta\sqrt[h]{\lambda} w_{\Omega}(x)),\quad x\in\Omega. \end{align} $$

It is clear that $v_{\Omega ,\lambda }>0$ in $\Omega $ and $v_{\Omega ,\lambda }=0$ on $\partial \Omega $ .

Next we want to show that $v_{\Omega ,\lambda }$ is a viscosity supersolution of Problem (1-1). Let $x_{0}\in \Omega $ and $\varphi \in C^{2}(\Omega )$ such that $v_{\Omega ,\lambda }-\varphi $ has a local minimum at $x_{0}\in \Omega $ . Without loss of generality, we can suppose that $x_{0}$ is a global minimum point of $v_{\Omega ,\lambda }-\varphi $ in $\Omega $ and $v_{\Omega ,\lambda }(x_{0})=\varphi (x_{0})$ . Setting

$$ \begin{align*} \eta(x):=\frac{1}{\theta\sqrt[h]{\lambda}}\Psi(\varphi)\in C^{2}(\Omega), \end{align*} $$

we have $w_{\Omega }(x_{0})=\eta (x_{0})$ and $w_{\Omega }-\eta $ has a minimum at $x_{0}$ . Since $w_{\Omega }$ is a viscosity supersolution of Problem (1-10),

$$ \begin{align*} \Delta_{\infty}^{h}\eta(x_{0})\leq -b(x_{0}). \end{align*} $$

Since $\Psi '(\varphi )>0$ and $\Psi "(\varphi )\geq 0$ , one can easily check that

$$ \begin{align*} \begin{aligned} \Delta_{\infty}^{h}\eta &=\bigg(\frac{1}{\theta\sqrt[h]{\lambda}}\bigg)^{h} (\Psi"(\varphi)[\Psi'(\varphi)]^{h-1}|D\varphi|^{h+1} +[\Psi'(\varphi)]^{h}\Delta_{\infty}^{h}\varphi) \\ &\geq\bigg(\frac{1}{\theta\sqrt[h]{\lambda}}\bigg)^{h}[\Psi'(\varphi)]^{h}\Delta_{\infty}^{h}\varphi\\ &=\bigg(\frac{1}{\theta\sqrt[h]{\lambda}}\bigg)^{h} \frac{1}{\Gamma(\varphi)}\Delta_{\infty}^{h}\varphi. \end{aligned} \end{align*} $$

Then we have the following inequalities at $x_{0}$

(4-15) $$ \begin{align} \Delta_{\infty}^{h}\varphi &\leq(\theta\sqrt[h]{\lambda})^{h}\Gamma(\varphi) \Delta_{\infty}^{h}\eta\nonumber\\ &\leq-b(\theta\sqrt[h]{\lambda})^{h}\Gamma(\varphi)\nonumber\\ &\leq-b(\theta\sqrt[h]{\lambda})^{h}\bigg(\frac{g(\varphi)}{\varphi^{h}}\bigg)\nonumber\\ &=-\lambda b\bigg(\frac{\theta}{\varphi}\bigg)^{h}g(\varphi)\nonumber\\ &\leq-\lambda bg(\varphi)\nonumber\\ &\leq\lambda f(x_{0},\varphi,D\varphi), \end{align} $$

where we use (4-8) and $\varphi (x_0)=v_{\Omega ,\lambda }(x_0)=\Phi (\theta \sqrt [h]{\lambda }w_{\Omega }(x_0)) \leq \Phi (\theta \sqrt [h]{\lambda }\|w_{\Omega }\|_{\infty })\leq \theta $ . This shows that $v_{\Omega ,\lambda }$ is a viscosity supersolution of Problem (1-1).

Finally, since $\theta \geq 1$ , it follows from (4-15) that $\Delta _{\infty }^{h}\varphi \leq -\lambda b(x)\Gamma (\varphi )$ . Therefore,

$$ \begin{align*} \Delta_{\infty}^{h}v_{\Omega,\lambda}\leq-\lambda b(x)\Gamma(v_{\Omega,\lambda}).\\[-37pt] \end{align*} $$

Proof. Fix $\lambda $ such that $\lambda>\Lambda _{1}(\Omega )k_{0}^{-1}$ . First, if the constant $0<k_{0}<\infty $ , where $k_{0}$ is defined in (1-9), then there exists a $\delta :=\delta (\lambda )$ with $0<\delta <t_{0}$ such that

$$ \begin{align*} k(t)\geq\bigg(\frac{\Lambda_{1}(\Omega)k_{0}^{-1}}{\lambda}\bigg)k_{0}t^{h} =\frac{\Lambda_{1}(\Omega)}{\lambda}t^{h}\quad\text{for all } 0<t\leq\delta. \end{align*} $$

However, if $k_{0}=\infty $ , then for any $\lambda>0 $ , there exists $0<\delta <t_{0}$ such that

$$ \begin{align*} k(t)\geq\frac{\Lambda_{1}(\Omega)}{\lambda}t^{h}\quad\text{for all } 0<t\leq\delta. \end{align*} $$

In any case, let $\phi _{\Omega ,\lambda }$ be a positive eigenfunction of Problem (1-11) corresponding to the principal eigenvalue $\Lambda _{1}(\Omega )$ such that $0<\phi _{\Omega ,\lambda }\leq \min \{1,\delta (\lambda )\}$ in $\Omega $ . Then under the assumption on $\lambda $ and condition (F-1),

$$ \begin{align*} \Delta_{\infty}^{h}\phi_{\Omega,\lambda}=-\Lambda_{1}(\Omega) a(x)\phi_{\Omega,\lambda}^{h} \geq-\lambda a(x)k(\phi_{\Omega,\lambda})\geq\lambda f(x,\phi_{\Omega,\lambda},D\phi_{\Omega,\lambda}) \end{align*} $$

in $\Omega $ . This shows that $\phi _{\Omega ,\lambda }$ is a viscosity subsolution of Problem (1-1).

Proof of Theorem 1.3.

When $\lambda>\Lambda _{1}(\Omega )k_{0}^{-1}$ , according to Lemma 4.5, we have a viscosity subsolution $\phi _{\Omega ,\lambda }$ of Problem (1-1) and $\phi _{\Omega ,\lambda }\leq 1$ in $\Omega $ . Then,

$$ \begin{align*} \begin{aligned} \Delta_{\infty}^{h}\phi_{\Omega,\lambda}&\geq \lambda f(x,\phi_{\Omega,\lambda},D\phi_{\Omega,\lambda})\\ &\geq -\lambda b(x)g(\phi_{\Omega,\lambda})\\ &\geq -\lambda b(x)\phi_{\Omega,\lambda}^{h}\Gamma(\phi_{\Omega,\lambda})\\ &\geq -\lambda b(x)\Gamma(\phi_{\Omega,\lambda}) \end{aligned} \end{align*} $$

in the viscosity sense, where $\Gamma $ is the function defined in (4-7).

When $0<\lambda <\lambda _{\Omega }^{\ast }$ , according to Lemma 4.3, we know that $v_{\Omega ,\lambda }$ is a viscosity supersolution of Problem (1-1) and satisfies (4-6).

Now we claim that $\phi _{\Omega ,\lambda }\leq v_{\Omega ,\lambda }$ in $\Omega $ . Assume that $\phi _{\Omega ,\lambda }\leq v_{\Omega ,\lambda }$ is not valid. Let

$$ \begin{align*} \mathbb{D}:=\{x\in\Omega:\phi_{\Omega,\lambda}(x)>v_{\Omega,\lambda}(x)\}. \end{align*} $$

Since $\Gamma $ is nonincreasing,

$$ \begin{align*} \Delta_{\infty}^{h}\phi_{\Omega,\lambda}\geq -\lambda b(x)\Gamma(\phi_{\Omega,\lambda}) \geq -\lambda b(x)\Gamma(v_{\Omega,\lambda})\quad \mathrm{in} \; \mathbb{D}. \end{align*} $$

Then in $\mathbb {D}$ ,

$$ \begin{align*} \Delta_{\infty}^{h}\phi_{\Omega,\lambda} \geq -\lambda b(x)\Gamma(v_{\Omega,\lambda})\quad \text{and}\quad \Delta_{\infty}^{h}v_{\Omega,\lambda} \leq -\lambda b(x)\Gamma(v_{\Omega,\lambda}). \end{align*} $$

According to the comparison principle (Theorem 1.1), we see that $\phi _{\Omega ,\lambda }\leq v_{\Omega ,\lambda }$ in $\mathbb {D}$ , which is a contradiction.

Furthermore, we see that $0<\phi _{\Omega ,\lambda }\leq v_{\Omega ,\lambda }$ in $\Omega $ and $\phi _{\Omega ,\lambda }(x)=v_{\Omega ,\lambda }=0$ on $\partial \Omega $ . By Corollary 4.2, we conclude that Problem (1-1) admits a viscosity solution $u\in C(\overline {\Omega })$ such that $\phi _{\Omega ,\lambda }\leq u\leq v_{\Omega ,\lambda }$ in $\overline {\Omega }$ .

Proof. Let $B_{r}:=B(O,r)$ be the ball of radius r centred at the origin O. By [Reference Li and Liu26], we note that the problem

(5-2) $$ \begin{align} \begin{cases} \Delta_\infty^h w=-b(x) \quad &\mathrm{{in}}\; B_r, \\ w>0 \quad &\mathrm{in}\; B_r,\\ w=0 \quad &\mathrm{on} \; \partial B_r \end{cases} \end{align} $$

admits a viscosity solution $w_r$ . Taking $2a=\inf \nolimits _{\mathbb {R}^{n}}(-b(x))$ in Lemma 5.1, it is obvious that $v_{x_{0},BC}(x)$ is a viscosity supersolution of the equation $\Delta _\infty ^h w=-b(x)$ in $B_r$ . Then,

$$ \begin{align*} 0< w_r\leq v_{x_{0},BC}(x)\quad \mathrm{in}\; B_r. \end{align*} $$

This implies that the sequence $\{w_{r}\}$ is uniformly bounded in $\mathbb {R}^{n}$ . According to Lemma 2.5, we note that the sequence $\{w_{r}\}$ is locally uniformly Lipschitz. Therefore, $\{w_{r}\}$ is equicontinuous. Then we can obtain a subsequence that converges locally uniformly to some w and $0<w\leq v_{x_{0},BC}(x)$ in $\mathbb {R}^{n}$ .

Then we want to show that w is a viscosity solution of $\Delta _\infty ^h w=-b(x)$ in $\mathbb {R}^{n}$ . We can use the similar argument to the one in the proof of Theorem 4.1. First, we show $\Delta _\infty ^h w\leq -b(x)$ in the viscosity sense.

Let $\varphi \in C^{2}(\mathbb {R}^{n})$ and suppose $w-\varphi $ has a local minimum at some $x_{0}\in \mathbb {R}^{n}$ , that is,

$$ \begin{align*}w(x)-\varphi(x)\geq w(x_{0})-\varphi(x_{0}),\quad x\in B_{\delta}(x_{0})\subset B_s(x_0)\end{align*} $$

for some small $\delta>0$ and some positive integer s. We take $\varepsilon>0$ small enough and let $x_{r}\in \overline {B}_{\delta }(x_{0})$ be a point of minimum of

$$ \begin{align*}w_{r}(x)-\bigg(\varphi(x)-\frac{\varepsilon}{2}|x-x_{0}|^{2}\bigg),\quad r\geq s.\end{align*} $$

Then,

(5-3) $$ \begin{align} w_{r}(x_{r})-\bigg(\varphi(x_{r})-\frac{\varepsilon}{2}|x_{r}-x_{0}|^{2}\bigg) \leq w_{r}(x_{0})-\varphi(x_{0}). \end{align} $$

We can assume that $x_{r}\rightarrow \hat {x}$ for some $\hat {x}\in \overline {B}_{r}(x_{0})$ . As $r\rightarrow \infty $ , it follows from (5-3) that

$$ \begin{align*}w(\hat{x})-\bigg(\varphi(\hat{x})-\frac{\varepsilon}{2}|\hat{x}-x_{0}|^{2}\bigg) \leq w(x_{0})-\varphi(x_{0}).\end{align*} $$

That is,

$$ \begin{align*}\frac{\varepsilon}{2}|\hat{x}-x_{0}|^{2}\leq w(x_{0})-\varphi(x_{0})- (w(\hat{x})-\varphi(\hat{x}))\leq 0.\end{align*} $$

This implies $\hat {x}=x_{0}$ . Since $w_{r}$ is a viscosity solution of (5-2) and $x_{r}$ is a point of minimum of $w_{r}(x)-(\varphi (x)- \frac {\varepsilon }{2}|x-x_{0}|^{2})$ in $B_{r}(x_{0})$ , we take a test function $\phi _{\varepsilon }(x):=\varphi (x)- \frac {\varepsilon }{2}|x-x_{0}|^{2}$ . Then,

$$ \begin{align*} \Delta_\infty^h \phi_{\varepsilon}(x_{r})\leq -b(x_r). \end{align*} $$

Taking the limit as $r\rightarrow \infty $ ,

$$ \begin{align*}\Delta_\infty^h \phi_{\varepsilon}(x_{0})\leq-b(x_0).\end{align*} $$

Letting $\varepsilon \rightarrow 0$ ,

$$ \begin{align*}\Delta_\infty^h \varphi(x_{0})\leq-b(x_0),\end{align*} $$

that is, w is a viscosity supersolution of $\Delta _\infty ^h w=-b(x)$ in $\mathbb {R}^{n}$ . Similarly, one can prove that w is a viscosity subsolution of $\Delta _\infty ^h w=-b(x)$ in $\mathbb {R}^{n}$ . Clearly, $w(x)\rightarrow 0$ as $|x|\rightarrow \infty $ . Therefore, w is a viscosity solution of (5-1).

Proof of Theorem 1.4.

Let $\lambda ^{*}:=\lambda _{\mathbb {R}^{n}}^{*}$ be the positive constant in Theorem 1.3 corresponding to $\Omega =\mathbb {R}^{n}$ and $\Lambda _{1}(B(O,1))k_{0}^{-1}<\lambda <\lambda ^{*}$ . For each positive integer k, let $B_{k}:=B(O,k)$ be the ball of radius k centred at the origin O. Since $\Lambda _{1}(B(O,k))\leq \Lambda _{1}(B(O,1))$ and $\lambda ^{*}\leq \lambda _{B(O,k)}^{*}$ , we have $\Lambda _{1}(B(O,k))k_{0}^{-1}<\lambda <\lambda _{B(O,k)}^{*}$ . Since f satisfies Conditions (F-1), (F-2) and b satisfies (B-w) in $\Omega =B(O,k)$ , by Theorem 1.3, we obtain a viscosity solution $u_{k}:=u_{B_{k},\lambda }\in C(\overline {B}_{k}) (k=1,2,\ldots )$ of the singular boundary value problem

(5-4) $$ \begin{align} \begin{cases} \Delta_\infty^h u=\lambda f(x,u,Du) \quad &\mathrm{in}\; B_{k}, \\ u(x)=0 \quad &\mathrm{on} \; \partial B_{k}. \end{cases} \end{align} $$

Letting $v_{\lambda }$ be a viscosity supersolution of Problem (1-12) given by the method in Lemma 4.3 in $\Omega =\mathbb {R}^{n},$

(5-5) $$ \begin{align} 0<u_{k}\leq v_{\lambda}\quad \mathrm{in}\; B_{k}. \end{align} $$

Equation (5-5) shows that the sequence $\{u_{k}\}$ is uniformly bounded in $\mathbb {R}^{n}$ . According to Lemma 2.5, we see that the sequence $\{u_{k}\}$ is locally uniformly Lipschitz. Thus, $\{u_{k}\}$ is equicontinuous. Then we can obtain a subsequence that converges locally uniformly to some $u_{\lambda }\in C(\mathbb {R}^{n})$ and $0\leq u_{\lambda }\leq v_{\lambda }$ in $\mathbb {R}^{n}$ .

Next, for a given positive integer l, let $\varphi _{l}:=\varphi _{B_{l}}$ be an eigenfunction of Problem (1-11) corresponding to the principal eigenvalue $\Lambda _{1}(B(O,l))$ on the ball $B_{l}:=B(O,l)$ . For any viscosity solution $u_{k}$ of (5-4), we want to show that

(5-6) $$ \begin{align} u_{k}\geq \varphi_{l} \quad \mathrm{in}\;B(O,l)\text{ for all } k>l. \end{align} $$

For each $(x,t,p)\in \mathbb {R}^{n}\times \mathbb {R}^{+}\times \mathbb {R}^{n}$ , let

$$ \begin{align*} \hat{f}(x,t,p)=\frac{f(x,e^{t},e^{t}p)}{a(x)(e^{t})^{h}}. \end{align*} $$

For convenience, we set $\Lambda _{1}:=\Lambda _{1}(B(O,1))$ and $\Lambda _{l}:=\Lambda _{1}(B(O,l))$ . Note that

$$ \begin{align*} \lambda>\Lambda_{1}k_{0}^{-1}, \end{align*} $$

where $k_{0}$ is defined in (1-9) and $k_{0}^{-1}$ is interpreted to be zero if $k_{0}=\infty $ . Fix $\tau $ such that

(5-7) $$ \begin{align} \Lambda_{1}<\tau<k_{0}\lambda. \end{align} $$

Then by the definition of $k_{0}$ , there exists $0<t^{*}:=t^{*}(\lambda )\leq t_{0}$ such that $k(t)\geq (\frac {\tau }{\lambda })t^{h}$ for $0<t<t^{*}$ . Thus,

$$ \begin{align*} f(x,t,p)\leq -a(x)k(t)\leq -\frac{\tau}{\lambda}a(x)t^{h},\quad (x,t,p) \in\mathbb{R}^{n}\times(0,t^{*})\times\mathbb{R}^{n}. \end{align*} $$

Therefore,

(5-8) $$ \begin{align} \hat{f}(x,t,p)\leq -\frac{\tau}{\lambda},\quad (x,t,p)\in\mathbb{R}^{n}\times(-\infty,\ln t^{*})\times\mathbb{R}^{n}. \end{align} $$

Suppose that $\varphi _{l}$ has been normalized so that $0<\varphi _{l}\leq t^{*}$ in $B_{l}$ . Suppose that Inequality (5-6) is not true. Since $u_{k}\geq \varphi _{l}$ on $\partial B_{l}$ , it follows that $\frac {\varphi _{l}}{u_{k}}$ attains its maximum in $B_{l}$ . Since $u_{k}$ and $\varphi _{l}$ are positive in $ B_{l}$ , we see that

$$ \begin{align*} \ln\bigg(\frac{\varphi_{l}}{u_{k}}\bigg)=\ln \varphi_{l}-\ln u_{k} \end{align*} $$

attains its positive maximum at some point $x_{0}\in B_{l}$ . Set

$$ \begin{align*} \alpha(x)=\ln \varphi_{l}(x)\quad \text{and} \quad \beta(x)=\ln u_{k}(x). \end{align*} $$

By direct calculations,

(5-9) $$ \begin{align} \Delta_\infty^h \alpha=-\frac{1}{\varphi_{l}^{h+1}}|D\varphi_{l}|^{h+1}+\frac{1}{\varphi_{l}^{h}}\Delta_\infty^h \varphi_{l} =-|D\alpha|^{h+1}-\Lambda_{l}a, \end{align} $$

(5-10) $$ \begin{align} \Delta_\infty^h \beta=-\frac{1}{u_{k}^{h+1}}|Du_{k}|^{h+1}+\frac{1}{u_{k}^{h}}\Delta_\infty^h u_{k} =-|D\beta|^{h+1}+\frac{\lambda f(x,e^{\beta},e^{\beta}D\beta)}{(e^{\,\beta})^{h}}. \end{align} $$

Therefore,

$$ \begin{align*} \Delta_\infty^h \alpha+|D\alpha|^{h+1}=-a\Lambda_{l} \quad \mathrm{and} \quad \Delta_\infty^h \beta+|D\beta|^{h+1}=\frac{\lambda f(x,e^{\beta},e^{\beta}D\beta)}{(e^{\,\beta})^{h}}. \end{align*} $$

Note that

$$ \begin{align*} \alpha-\beta=\ln\bigg(\frac{\varphi_{l}}{u_{k}}\bigg) \end{align*} $$

has a positive maximum in $\overline {B}_{l}$ , that is, $M:=\max \nolimits _{\overline {B}_{l}}(\alpha -\beta )>0$ . Define

$$ \begin{align*} \Phi_{\varepsilon}(x,y)=\alpha(x)-\beta(y)-\frac{1}{2\varepsilon}|x-y|^{2},\quad (x,y)\in \overline{B}_{l}\times\overline{B}_{l}. \end{align*} $$

Let

$$ \begin{align*} M_{\varepsilon}:=\Phi_{\varepsilon}(x_{\varepsilon},y_{\varepsilon})=\max_{\overline{B}_{l}\times \overline{B}_{l}}\Phi_{\varepsilon}(x,y)\quad \mathrm{for\;some\;}(x_{\varepsilon},y_{\varepsilon}) \in\overline{B}_{l}\times\overline{B}_{l}. \end{align*} $$

Without loss of generality, we can assume that the sequence $\{(x_{\varepsilon },y_{\varepsilon })\}$ converges to $(z,z)\in \overline {B}_{l}\times \overline {B}_{l}$ as $\varepsilon \rightarrow 0$ , where $M=(\alpha -\beta )(z)$ . Since $M>0$ , we must have $z\in B_{l}$ . Then we can suppose that $(x_{\varepsilon },y_{\varepsilon })\in B_{l}$ for sufficiently small $\varepsilon>0$ . According to the maximum principle in [Reference Crandall, Ishii and Lions18, Reference Ishii23], there are $n\times n$ symmetric matrices $X_{\varepsilon }, Y_{\varepsilon }\in \mathbb {S}$ such that

$$ \begin{align*} (\eta_{\varepsilon},X_{\varepsilon})\in\overline{\mathcal{J}}^{2,+}\alpha(x_{\varepsilon}), \quad(\eta_{\varepsilon},Y_{\varepsilon})\in\overline{\mathcal{J}}^{2,-}\beta(y_{\varepsilon}) \end{align*} $$

and

(5-11) $$ \begin{align} -\frac{3}{\varepsilon}\left({\begin{array}{cc}I&0\\0&I\\ \end{array}}\right) \leq \left({\begin{array}{cc}X_{\varepsilon}&0\\0&-Y_{\varepsilon}\\ \end{array}}\right) \leq \frac{3}{\varepsilon}\left({\begin{array}{cc}I&-I\\-I&I\\ \end{array}}\right), \end{align} $$

where $\eta _{\varepsilon }:=\frac {1}{\varepsilon }(x_{\varepsilon }-y_{\varepsilon })$ . Following from Inequality (5-11), we have $X_{\varepsilon }\leq Y_{\varepsilon }$ . Applying the definitions of the viscosity subsolution and viscosity supersolution to (5-9) and (5-10), respectively,

$$ \begin{align*} &-\Lambda_{l}a(x_{\varepsilon})\leq\langle X_{\varepsilon}\eta_{\varepsilon},\eta_{\varepsilon}\rangle |\eta_{\varepsilon}|^{h-3}+|\eta_{\varepsilon}|^{h+1} \quad \mathrm{and} \\ &\qquad\langle Y_{\varepsilon}\eta_{\varepsilon},\eta_{\varepsilon}\rangle |\eta_{\varepsilon}|^{h-3} +|\eta_{\varepsilon}|^{h+1}\leq \frac{\lambda f(y_{\varepsilon},e^{\beta},e^{\beta}\eta_{\varepsilon})} {(e^{\,\beta})^{h}}. \end{align*} $$

Since $X_{\varepsilon }\leq Y_{\varepsilon }$ ,

$$ \begin{align*} \frac{\lambda f(y_{\varepsilon},e^{\beta},e^{\beta}\eta_{\varepsilon})}{(e^{\,\beta})^{h}} \geq -\Lambda_{l}a(x_{\varepsilon}). \end{align*} $$

As $\varepsilon \rightarrow 0$ , we get the inequality

$$ \begin{align*} \frac{\lambda f(z,e^{\beta}(z),e^{\beta}(z)\eta_{\varepsilon})}{(e^{\beta}(z))^{h}}\geq -\Lambda_{l}a(z). \end{align*} $$

Therefore,

$$ \begin{align*} \lambda \hat{f}(z,e^{\beta}(z),e^{\beta}(z)\eta_{\varepsilon}))\geq -\Lambda_{l}\geq-\Lambda_{1}. \end{align*} $$

By the assumption $\varphi _{l}(z)>u_{k}(z)$ and since $\hat {f}(x,t,p)$ is nondecreasing in t, we conclude that

$$ \begin{align*} \lambda \hat{f}(z,e^{\alpha}(z),e^{\alpha}(z)\eta_{\varepsilon}))\geq-\Lambda_{1}. \end{align*} $$

Together with Inequality (5-8),

(5-12) $$ \begin{align} \tau\leq\Lambda_{1}. \end{align} $$

Inequality (5-12) is an obvious contradiction to (5-7). Thus, the claim that $u_{k}\geq \varphi _{l}$ in $B(O,l)$ for all $k>l$ holds.

Then,

$$ \begin{align*}0<\varphi_{l}\leq u_{k} \quad \mathrm{in}\; B(O,l)\text{ for all } k\geq l.\end{align*} $$

As a consequence of this inequality, we have $0<\varphi _{l}\leq u_{\lambda }$ in $B(O,l)$ . Thus,

$$ \begin{align*} 0<u_{\lambda}\leq v_{\lambda} \quad \mathrm{in}\; B(O,l). \end{align*} $$

Since l is arbitrary, we conclude that $0<u_{\lambda }\leq v_{\lambda }$ in $\mathbb {R}^{n}$ . By a similar argument to the one used in the proof of Theorem 4.1, we can show that $u_{\lambda }$ is a viscosity solution of Problem (1-12). Clearly, $u(x)\rightarrow 0$ as $|x|\rightarrow \infty $ .