Proof. Consider

$$ \begin{align*}u_{\tau }(x) :=(1+\tau)u(x)-\tau\sup_{\bar{\Omega } } u(x) , \quad 0<{\tau }<{\eta}.\end{align*} $$

Since $u_{\tau }(x)\leq u(x)$ , $f(x,t,p)\ge 0$ is nondecreasing in t, and $\iota \longmapsto f(x,t,\iota p)$ is nonincreasing in $[1,1+\eta )$ , we have

(2-3) $$ \begin{align} \Delta_\infty^hu_{\tau }(x)\ge(1+\tau)^h f(x,u(x),Du(x))\ge(1+\tau)^h f(x,u_{\tau}(x),Du_{\tau}(x)) \end{align} $$

in the viscosity sense. That is, $u_{\tau }(x)$ is a viscosity subsolution of $\Delta _\infty ^hu(x)= (1+\tau )^hf(x,u(x),Du(x))$ in $\Omega $ .

We argue by contradiction that there exists $ M_{0}>0$ such that

$$ \begin{align*}M_0:=\sup_{\overline{\Omega}}(u_{\tau}-v)(x)=u_{\tau}(x_0)-v(x_0),\quad x_0\in\overline{\Omega}.\end{align*} $$

For each $\epsilon>0$ small enough, define

$$ \begin{align*} \mathfrak{m}_{\epsilon}(x,y):= u_{\tau}(x)-v(y)-\dfrac{1}{4\epsilon}|x-y|^4 \end{align*} $$

and

$$ \begin{align*}M_{\epsilon}:=\sup_{\overline{\Omega}\times\overline{\Omega}}\mathfrak{m} _{\epsilon}(x,y)<+\infty.\end{align*} $$

Let $(x_{\epsilon },y_{\epsilon })\in \overline {\Omega }\times \overline {\Omega }$ be the point where $\mathfrak {m} _{\epsilon }(x,y)$ attains its maximum.

By [Reference Crandall, Ishii and Lions18, Proposition 3.7], we have

$$ \begin{align*}\lim_{\epsilon\to0^+}\dfrac{1}{4\epsilon}|x_{\epsilon}-y_{\epsilon}|^4=0\quad {\text{and}} \quad \lim_{\epsilon\to 0^+}M_{\epsilon}=M_0.\end{align*} $$

Obviously,

$$ \begin{align*}\lim_{\epsilon\to 0^+}x_{\epsilon}=\lim_{\epsilon\to 0^+}y_{\epsilon}= {x_{0}}.\end{align*} $$

Due to $M_0>0\geq \sup \limits _{\partial \Omega }(u_{\tau }-v)(x)$ , we conclude that $x_{\epsilon }\in \widetilde {\Omega }\Subset \Omega .$

Applying [Reference Crandall, Ishii and Lions18, Theorem 3.2], there exist $X_\epsilon $ and $ Y_\epsilon \in \mathcal {S}_{n\times n}$ such that

(2-4) $$ \begin{align} (p_\epsilon,X_\epsilon )\in\overline{J}_{\Omega}^{2,+}u_{\tau}(x_{\epsilon}),\quad(p_\epsilon,Y_\epsilon)\in\overline{J}_{\Omega}^{2,-}v(y_{\epsilon}), \end{align} $$

and

(2-5) $$ \begin{align} \begin{pmatrix} X_\epsilon & 0 \\ 0 & -Y_\epsilon \end{pmatrix} \leq \dfrac{1}{\epsilon}(|z_\epsilon|^2+2\,|z_\epsilon|^4) \begin{pmatrix} \mathrm{Id}_{n\times n} & -\mathrm{Id}_{n\times n} \\ -\mathrm{Id}_{n\times n} & \mathrm{Id}_{n\times n} \end{pmatrix}+16\begin{pmatrix} z_\epsilon \otimes & z_\epsilon & -z_\epsilon \otimes & z_\epsilon \\ -z_\epsilon \otimes & z_\epsilon & z_\epsilon \otimes & z_\epsilon \end{pmatrix}, \end{align} $$

where $p_\epsilon =({|z_\epsilon |^2z_\epsilon })/({\epsilon })$ , $z_\epsilon =x_{\epsilon }-y_{\epsilon }$ , and $\mathrm {Id}_{n\times n}$ is the $n\times n$ identity matrix. By (2-5), we have $X_\epsilon \leq Y_\epsilon $ .

If $x_{\epsilon }=y_{\epsilon }$ for any $\epsilon>0,$ then we have $p_{\epsilon }=0$ . Thus, the continuous extension $\overline {F}$ equals zero. By (2-3),

$$ \begin{align*} (1+\tau)^hf(y_{\epsilon},v(y_{\epsilon})+M_{\epsilon},0)\leq0. \end{align*} $$

Letting $\epsilon \to 0^+$ ,

$$ \begin{align*} (1+\tau)^hf(x_0,{v}({x_0})+M_0,0)\leq 0, \end{align*} $$

which is impossible due to $M_0>0$ and ${v(x)}\geq 0$ in $\Omega $ .

If there exists a subsequence such that $x_{\epsilon }\ne y_{\epsilon }$ , then $p_{\epsilon }\ne 0.$ We have

(2-6) $$ \begin{align} (1+\tau)^hf(x_{\epsilon},u_{\tau}(x_{\epsilon}),p_{\epsilon}) &=(1+\tau)^h f\bigg(x_{\epsilon},v(y_{\epsilon})+\dfrac{1}{4\epsilon}|x_{\epsilon}-y_{\epsilon}|^4+M_{\epsilon},p_{\epsilon}\bigg) \nonumber\\& \leq |p_{\epsilon}|^{h-3}\cdot X_\epsilon(p_{\epsilon})\cdot ({p_{\epsilon}})\nonumber\\& \leq |p_{\epsilon}|^{h-3}\cdot Y_\epsilon(p_{\epsilon})\cdot (p_{\epsilon})\nonumber\\& \leq f(y_{\epsilon},v(y_{\epsilon}),p_{\epsilon}) \end{align} $$

by (2-4) and (2-5). Since

$$ \begin{align*} u_{\tau}(y_{\epsilon})-v(y_{\epsilon})-\dfrac{1}{4\epsilon}|y_{\epsilon}-y_{\epsilon}|^4\leq u_{\tau}(x_{\epsilon})-v(y_{\epsilon})-\dfrac{1}{4\epsilon}|x_{\epsilon}-y_{\epsilon}|^4, \end{align*} $$

one has

$$ \begin{align*} \dfrac{1}{4\epsilon}|x_{\epsilon}-y_{\epsilon}|^4\leq u_{\tau}(x_{\epsilon})-u_{\tau}(y_{\epsilon}). \end{align*} $$

By the Lipschitz continuity of $u_{\tau }(x)$ , we see that

$$ \begin{align*}\dfrac{1}{4\epsilon}|x_{\epsilon}-y_{\epsilon}|^4\leq u_{\tau}(x_{\epsilon})-u_{\tau}(y_{\epsilon})\leq L|x_{\epsilon}-y_{\epsilon}|, \end{align*} $$

where L is the Lipschitz constant of $u_{\tau }(x)$ . Therefore,

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

Without loss of generality, we can assume

$$ \begin{align*}p_{\epsilon}\to \widetilde{p} \in \overline{B}_{4L},\quad \epsilon\to 0^+,\end{align*} $$

where ${B_{4L}}$ denotes the ball of radius $4L$ centered at the origin. By (2-6),

$$ \begin{align*} (1+\tau)^hf(x_0,{v}({x_0})+M_0,\widetilde{p} )\leq f(x_0,{v}({x_0}),\widetilde{p} ). \end{align*} $$

Since $f(x,t,p)$ is nondecreasing with respect to t and $f(x,t,p)>0$ for each $(x,t,p)\in \Omega \times ( 0,+\infty )\times \mathbb {R} ^{n}$ ,

$$ \begin{align*} 0<(1+\tau)^hf(x_0,{v}({x_0})+M_0,\widetilde{p} )\leq f(x_0,{v}({x_0}),\widetilde{p} )\leq f(x_0,{v}({x_0})+M_0,\widetilde{p}), \end{align*} $$

which contradicts to $f(x_0,{v}({x_0})+M_0,\widetilde {p})>0$ . Thus, we have proven that $u_{\tau }(x)\leq v(x) $ in $ \Omega .$ Letting $\tau \to 0^+$ , we have $u(x)\leq v(x) $ in $ \Omega .$

Proof of Theorem 1.1.

The uniqueness follows by the comparison principle. The existence reduces to finding a viscosity subsolution and a supersolution based on Lemma 2.4. Let

$$ \begin{align*} L_1:=\inf_{x\in \partial\Omega}\phi(x). \end{align*} $$

We can choose constants $K_1$ and $d_1$ such that

$$ \begin{align*} K_1^h\geq \max\bigg\{\sup_{(x,p)\in\Omega\times\mathbb{R} ^{n} } f(x,L_1,p),1\bigg\}\quad \text {and} \quad d_1\leq \frac{L_1}{K_1}-\frac{1}{h+1}h^{(h+1)/h} ({\text{diam}}(\Omega))^{(h+1)/h}. \end{align*} $$

Construct

$$ \begin{align*} \underline {u}(x):= K_1\bigg(\frac{1}{h+1}h^{(h+1)/h}| x-z |^{(h+1)/h} +d_1\bigg), \quad z\in\partial\Omega. \end{align*} $$

We can demonstrate that $\underline {u}(x)\in C(\overline \Omega ), \, \Delta _{\infty }^h\underline {u}(x)=K_1^h\geq \sup _{(x,p)\in \Omega \times \mathbb {R} ^{n} } f(x,L_1,p)$ and $\underline {u}(x)\leq L_1$ by the choice of $K_1$ and $d_1.$ Since $f(x,t,p)$ is nondecreasing with respect to t, we have

$$ \begin{align*} \begin{cases} \Delta_\infty^h\underline{u}(x)\geq f(x,\underline{u}(x),D\underline {u}(x)) \quad & {\text{in}} \ \Omega,\\ \underline{u}(x)\leq\phi(x) \quad & {\text{on}}\ \partial \Omega. \end{cases} \end{align*} $$

Hence, $\underline {u}(x)$ is a viscosity subsolution of (1-1) in $\Omega $ . Let

$$ \begin{align*} \overline{u}(x):=\sup_{x\in \partial\Omega}\phi(x)+1. \end{align*} $$

It is obvious that $\overline {u}(x)$ satisfies

$$ \begin{align*} \begin{cases} \Delta_\infty^h\overline{u}(x)\leq 0 \quad & {\text{in}} \ \Omega,\\ \overline{u}(x)\geq\phi(x) \quad & {\text{on}}\ \partial \Omega. \end{cases} \end{align*} $$

That is, $\overline {u}(x)$ is a viscosity supersolution of (1-1) in $\Omega $ . By Lemma 2.4, we obtain the existence of a viscosity solution to (1-1).

Proof. We argue by contradiction assuming that there exist $\mu _*>0$ and sequences $\{\kappa _i\}$ , $\{\xi _i\}$ ( $i\in \mathbb {N}$ ) with $\kappa _i=o(1),$

$$ \begin{align*}\xi_i(x_0)=0,\quad0\leq\xi_i(x)\leq1,\quad x\in B_1(x_0)\cap\Omega,\end{align*} $$

and

$$ \begin{align*} \Delta_\infty^h\xi_i(x)={ | D\xi_i(x) | }^{h-3}\Delta_{\infty}\xi_i(x)=\kappa^{h+1}_iG(x,\xi_i(x),D\xi_i(x)),\quad x\in B_1(x_0)\cap\Omega \end{align*} $$

such that

$$ \begin{align*} \sup_{B_{({1}/{2})}(x_0)\cap\Omega}\xi_i(x)>\mu_*. \end{align*} $$

Noting that $\{\xi _i\}$ is uniformly bounded and satisfies

$$ \begin{align*} \Delta_\infty^h\xi_i(x)=\kappa^{h+1}_iG(x,\xi_i(x),D\xi_i(x))\geq 0,\quad x\in B_1(x_0)\cap\Omega \end{align*} $$

in the viscosity sense, Lemma 2.3 demonstrates that $\{\xi _i\}$ is a family of locally Lipschitz continuous functions in $\Omega .$ By the Arzela–Ascoli compact theorem, up to a subsequence, $\xi _i(x)$ converges locally uniformly to a function $\xi _{\infty }(x)$ in $ \overline {B_{2/3} (x_0)}\cap \Omega .$ Moreover, by the stability of viscosity solutions, $\xi _{\infty }(x)$ satisfies that $\xi _{\infty }(x_0)=0, 0\leq \xi _{\infty }(x)\leq 1$ , and

$$ \begin{align*}{ | D\xi_{\infty}(x) | }^{h-3}\Delta_{\infty}\xi_{\infty}(x)=0,\quad x\in \overline{B_{{2}/{3}} (x_0)}\cap\Omega \end{align*} $$

in the viscosity sense. By the maximum principle of infinity harmonic functions, we conclude that $\xi _{\infty }(x)\equiv 0$ . This is impossible for i large enough.

Proof of Theorem 1.2.

Set $\tau :=\min \{1,\Arrowvert u\Arrowvert _{L^\infty (\Omega )}^{-1}\}$ , $\rho :=\overline \kappa \tau ^{-({h-\gamma })/({h+1})}$ and

$$ \begin{align*} w_1(x):=\tau u(\rho x+x_0),\quad x\in B_1\cap\Omega, \end{align*} $$

where $\overline \kappa $ is to be determined later. Since $u(\cdot )$ is a viscosity solution of (1-1) in $\Omega $ ,

$$ \begin{align*} \Delta_\infty^hw_1(x)=\overline\kappa^{h+1}\tau^{\gamma}f(\rho x+x_0,\tau^{-1}w_1(x),\tau^{-1}\rho^{-1} Dw_1(x))=:\overline\kappa^{h+1}G_1(x,w_1(x),Dw_1(x)) \end{align*} $$

in the viscosity sense. Moreover, by the selection of $\tau ,$ we find $w_1(0)=0$ , ${0\leq w_1(x)\leq 1}$ , and

$$ \begin{align*} \begin{aligned} 0\leq G_1(x,w_1(x),Dw_1(x))& \leq f(\rho x+x_0,\tau^{-1}w_1(x),\tau^{-1}\rho^{-1} Dw_1(x))\\ & \leq \sup_{(x,t,p)\in\Omega\times[0, ||u||_{L^\infty(\Omega)}]\times\mathbb{R} ^{n} }f(x,t,p). \end{aligned} \end{align*} $$

If we take $\mu =2^{-({h+1})/({h-\gamma })}$ and get the corresponding constant $\kappa _{\mu ,n}>0$ from Lemma 3.1, then we can verify that $w_1(x)$ satisfies the conditions of Lemma 3.1 by choosing $\overline \kappa \leq \min \{1,\kappa _{\mu ,n}, \tau ^{({h-\gamma })/({h+1})}\}.$ This means

$$ \begin{align*} \sup_{B_{({1}/{2})}\cap\Omega}w_1(x)\leq2^{-({h+1})/({h-\gamma})}. \end{align*} $$

Let

$$ \begin{align*} w_2(x):=2^{({h+1})/({h-\gamma})}w_1(2^{-1}x),\quad x\in B_1\cap\Omega. \end{align*} $$

By direct computations, we observe that

$$ \begin{align*} \begin{aligned} &\Delta_\infty^h w_2(x)\\ &\quad =\kappa^{h+1}_{a}2^{{(h+1)\gamma}/{(h-\gamma)}}\tau^{\gamma}f(2^{-1}\rho x+x_0, 2^{-(h+1)/(h-\gamma)}\tau^{-1}w_2(x),\tau^{-1}\rho^{-1}2^{-({2h+1-\gamma})/({h-\gamma})}Dw_2(x))\\ &\quad =:\kappa_{a}^{h+1}G_2(x,w_2(x),Dw_2(x)) \end{aligned} \end{align*} $$

in the viscosity sense. Since $ \tau \leq 1$ , $w_2(0)=0$ , $0\leq w_2(x)\leq 1$ , and

$$ \begin{align*} \begin{aligned} & 0\leq G_2(x,w_2(x),Dw_2(x)) \\ &\quad =2^{{(h+1)\gamma}/({h-\gamma})}\tau^{\gamma}f(2^{-1}\rho x+x_0, 2^{-(h+1)/(h-\gamma)}\tau^{-1}w_2(x),\tau^{-1}\rho^{-1}2^{-{(2h+1-\gamma)}/{(h-\gamma)}}Dw_2(x))\\ & \quad\leq\Lambda(2^{-1}\rho x+x_0)\tau^{\gamma}f(2^{-1}\rho x+x_0,\tau^{-1}w_2(x),\tau^{-1}\rho^{-1}2^{-({2h+1-\gamma})/({h-\gamma})}Dw_2(x))\\ & \quad\leq \|\Lambda(x) \|_{L^\infty(\Omega)}\sup_{(x,t,p)\in\Omega\times{[0,\|u\|_{L^\infty(\Omega)}]}\times \mathbb{R} ^{n} }f(x,t,p),\\ \end{aligned} \end{align*} $$

applying Lemma 3.1 again,

$$ \begin{align*} \sup_{B_{({1}/{2})}\cap\Omega}w_2(x)\leq2^{-({h+1})/({h-\gamma})}. \end{align*} $$

For each $k\in \mathbb {N},\ k\geq 2$ , we construct a series of iterative functions

$$ \begin{align*} w_k(x):=2^{({h+1})/({h-\gamma})}w_{k-1}(2^{-1}x), \quad x\in B_1\cap\Omega. \end{align*} $$

Similarly, we can verify that $w_k(x)$ satisfies the conditions of Lemma 3.1 and then

$$ \begin{align*} \sup_{B_{({1}/{2})}\cap\Omega}w_k(x)\leq2^{-({h+1})/({h-\gamma})},\quad k=2,3,\ldots. \end{align*} $$

By the iterative relationship of $\{w_k\},$ we obtain

(3-1) $$ \begin{align} \sup_{B_{{2}^{-k}}\cap\Omega}w_1(x)\leq2^{-({(h+1)k})/({h-\gamma})},\quad k=1,2,3,\ldots. \end{align} $$

For any $0<r\leq ({\rho }/{2})$ , we can select some $k\in \mathbb {N}$ such that

$$ \begin{align*} \frac{r}{\rho} \in [2^{-(k+1)},2^{-k}]. \end{align*} $$

Therefore, by the construction of $w_1(x)$ and (3-1),

$$ \begin{align*} \begin{aligned} \sup_{B_r(x_0)\cap\Omega}u(x) &\leq\sup_{B_{\rho2^{-k}}(x_0)\cap\Omega}u(x) \leq\tau^{-1}\sup_{B_{{2}^{-k}}\cap\Omega}w_1(x) \leq\tau^{-1}2^{({h+1})/({h-\gamma})}2^{-{(h+1)(k+1)}/{(h-\gamma)}}\\ &\leq\tau^{-1}(2\rho^{-1})^{({h+1})/({h-\gamma})}r^{(h+1)/(h-\gamma)} :=Cr^{({h+1})/({h-\gamma})}. \end{aligned} \end{align*} $$

This completes the proof of Theorem 1.2.

Proof of Theorem 1.3.

By continuity, we only need to prove (1-2) for the point $x_0\in {\{u>0\}}\cap {{\Omega }^{\prime }} .$

Consider a function

$$ \begin{align*}\varphi(x):= c\,|x-x_0|^{{({h+1})/({h-\gamma})} },\quad x\in B_r(x_0),\end{align*} $$

where the constant $c>0$ is to be determined later. For $x\ne x_0,$

$$ \begin{align*} D\varphi(x)=c\, \bigg( {\frac{h+1}{h-\gamma}} \bigg) \, |x-x_0|^{({1-h+2\gamma})/({h-\gamma})}{(x-x_0)}, \end{align*} $$

$$ \begin{align*} \begin{aligned} D^2\varphi(x)& =c\,\bigg({\frac{h+1}{h-\gamma}}\bigg) \bigg(\frac{1+\gamma}{h-\gamma}\bigg)\,|x-x_0|^{({1-h+2\gamma})/({h-\gamma})} {\frac{(x-x_0)\otimes (x-x_0)}{|x-x_0|^2}}\\ & \quad +c\, \bigg({\frac{h+1}{h-\gamma}}\bigg) \, |x-x_0|^{({1-h+2\gamma})/({h-\gamma})}\bigg(\mathrm{Id}_{n\times n}-\frac{(x-x_0)\otimes (x-x_0)}{|x-x_0|^2}\bigg). \end{aligned} \end{align*} $$

We can verify

$$ \begin{align*} \begin{aligned} &\Delta_\infty^h\varphi(x)-f(x,\varphi(x),D\varphi(x)) =c^h{\bigg(\frac{h+1}{h-\gamma}\bigg)}^h\bigg(\frac{1+\gamma}{h-\gamma}\bigg)\,|x-x_0|^{({(h+1)\gamma})/({h-\gamma})}-f(x,\varphi(x),D\varphi(x))\\ &\quad \leq -\bigg[c^{\gamma}\lambda(x) f(x,1,D\varphi(x)) -c^h\bigg(\frac{h+1}{h-\gamma}\bigg)^h\bigg(\frac{1+\gamma}{h-\gamma}\bigg)\bigg]\,|x-x_0|^{{(h+1)\gamma}/{h-\gamma}}\\ &\quad \leq -\bigg[\inf_{x\in\Omega}\lambda(x) \inf_{(x,p)\in\Omega\times\mathbb{R} ^{n} } f(x,1,p)-c^{h-\gamma}\bigg(\frac{h+1}{h-\gamma}\bigg)^h\bigg(\frac{1+\gamma}{h-\gamma}\bigg)\bigg]c^{\gamma}\,|x-x_0|^{({(h+1)\gamma})/({h-\gamma})}\\ &\quad \leq 0, \end{aligned} \end{align*} $$

where we have chosen

$$ \begin{align*}0<c<\Big[\Big({\inf _{x\in \Omega }\lambda (x) \inf _{(x,p)\in \Omega \times \mathbb {R} ^{n}} f(x,1,p)(h-\gamma )^{h+1}}\Big)/({(h+1)^h(1+\gamma )})\Big]^{{1}/({h-\gamma })}. \end{align*} $$

For $x=x_0,$ we can also derive $\Delta _\infty ^h\varphi (x)-f(x,\varphi (x),D\varphi (x))\leq 0.$ Hence, $\varphi (x)$ is a viscosity supersolution of $\Delta _{\infty }^hu(x)=f(x,u(x),Du(x))$ in $B_r(x_0).$

Now, we illustrate that there must exist a point $\bar {x}\in \partial B_r(x_0)$ such that $u(\bar {x})> \varphi (\bar {x}).$ In fact, if $u(x)\leq \varphi (x)$ on $\partial B_r(x_0)$ , we can conclude $u(x)\leq \varphi (x)$ in $B_r(x_0)$ by the comparison principle. This contradicts $u(x_0)>0= \varphi (x_0)$ . Therefore,

$$ \begin{align*} \begin{aligned} \sup_{B_r(x_0)}u(x)\geq\sup_{\partial B_r(x_0)}u(x)\geq \varphi(\bar{x})\geq cr^{({h+1})/({h-\gamma})}. \end{aligned} \\[-41pt] \end{align*} $$

Proof. For any $r>0$ and $x\in \partial \{u>0\}$ , we can select some $\xi \in \overline {B_r(x)}$ such that

$$ \begin{align*} u(\xi)=\sup_{B_r(x)}u(x). \end{align*} $$

According to Theorem 1.3, $u(\xi )$ satisfies the following nondegeneracy estimate:

(4-1) $$ \begin{align} u(\xi)\geq cr^{({h+1})/({h-\gamma})}. \end{align} $$

By Theorem 1.2,

(4-2) $$ \begin{align} u(\xi)\leq C[d(\xi)]^{({h+1})/({h-\gamma})}, \end{align} $$

where $d(\xi ):={\mathrm {dist}}(\xi ,\partial \{u>0\})>0.$ By (4-1) and (4-2),

$$ \begin{align*} \bigg(\dfrac{c}{C}\bigg)^{({h-\gamma})/{(h+1)}}\leq \frac{d(\xi)}{r}. \end{align*} $$

Selecting $\sigma =\tfrac 12({c}/{C})^{({h-\gamma })/({h+1})},$

$$ \begin{align*} B_{2\sigma r}(\xi)\subset\{u>0\}. \end{align*} $$

For $\overline {\eta }\in (0,1),$ choosing a point $z=\overline {\eta }\xi +(1-\overline {\eta })x $ such that $|\xi -z|=\sigma r$ ,

$$ \begin{align*} B_{\sigma r}(z)\subset B_{2\sigma r}(\xi)\cap B_r(x)\subset B_r(x)\setminus\partial\{u>0\}, \end{align*} $$

which means that $\partial \{u>0\}$ is porous with porosity $\sigma .$

Using Remark 4.2, we can immediately demonstrate that the Lebesgue measure of $\partial \{u>0\}$ is zero.

Proof. By (5-2), taking $R\gg 1$ sufficiently large,

(5-3) $$ \begin{align} \sup_{\partial B_R(x_0)}\frac{u(x)}{|x-x_0|^{({h+1})/({h-\gamma})}} <\theta \bigg({\frac{b(h-\gamma)^{h+1} }{(\gamma+1)(h+1)^h} }\bigg)^{{1}/({h-\gamma})} \end{align} $$

for some $0<\theta <1.$ Choosing $a=\sup _{\partial B_{R}(x_0) } u(x)$ in (5-1), we get that

$$ \begin{align*} v(x)=\rho(b,\gamma)\bigg(|x-x_0|-R+ \bigg(\frac{\sup_{\partial B_{R}(x_0) } u}{\rho(b,\gamma)}\bigg)^{({h-\gamma})/({h+1})}\bigg)^{({h+1})/({h-\gamma})}_+ \end{align*} $$

is the unique viscosity solution of (5-1) in $B_{R}(x_0)$ .

The selection of b implies that $u(x)$ is a viscosity subsolution of (5-1) in $B_{R}(x_0)$ . Since $u(x)\leq \sup _{\partial B_{R}(x_0) } u(x)$ on $\partial B_{R}{(x_{0})},$ by the comparison principle, we conclude $u(x)\leq v(x) $ in $ B_{R}{(x_{0})}$ . Combining (5-3),

$$ \begin{align*} u(x)\leq\rho(b,\gamma) (|x-x_0|-(1-\theta^{({h-\gamma})/({h+1})})R)^{({h+1})/({h-\gamma})}_+. \end{align*} $$

Letting $R \to +\infty $ , we observe $u(x)\equiv 0$ , which completes the proof.

Proof. For $k\in \mathbb {N}$ and $0<r\ll 1$ , let

$$ \begin{align*} u_k(x):=k^{-({h+1})/({h-\gamma})}u(x_0+kx),\quad x\in B_r. \end{align*} $$

By computations, it is easy to check

$$ \begin{align*} \begin{aligned} &\Delta_\infty^hu_k(x)\\ &\quad={|Du_k(x)|}^{h-3}\Delta_{\infty}u_k(x) =k^{-{(h+1)\gamma}/({h-\gamma})}{|Du(x_0+kx)|}^{h-3}\Delta_{\infty}u(x_0+kx)\\ &\quad =k^{-{(h+1)\gamma}/({h-\gamma})}f(x_0+kx,k^{({h+1})/({h-\gamma})}u_k(x), k^{({1+\gamma})/({h-\gamma})} Du_k(x))=:\hat{G}(x,u_k(x),Du_k(x)), \end{aligned} \end{align*} $$

and $\hat {G}$ satisfies Conditions (f $_{b}$ ) and (f $_{\Lambda }$ ).

We are going to prove

$$ \begin{align*} \Arrowvert u_k\Arrowvert_{L^{\infty}(\overline{B_r})}\rightarrow 0^+ \quad{\text{as}}\ k\rightarrow+\infty. \end{align*} $$

Let $x_k$ be the maximum point of $u_k(x)$ in $\overline {B_r}.$ Now, we consider two cases of $|kx_k|$ . If $|kx_k|$ remains bounded as $k\rightarrow +\infty ,$ then

$$ \begin{align*} u_k(x_k)\rightarrow0^+ \quad{\text{as}}\ k\rightarrow+\infty. \end{align*} $$

If $|kx_k|$ goes to infinity as $k\rightarrow +\infty $ , we can use (5-4) to get

$$ \begin{align*} u_k(x_k)\rightarrow0^+ \quad{\text{as}}\ k\rightarrow+\infty. \end{align*} $$

Applying Theorem 1.2,

(5-5) $$ \begin{align} u_k(x)\leq C_k\,|x|^{({h+1})/({h-\gamma})}, \end{align} $$

where $C_k>0$ and $C_k\rightarrow 0^+$ for $k\rightarrow +\infty .$

We complete the rest of the proof by contradiction. Suppose that there exists some point $y\in \mathbb {R}^n$ such that $u(x_0+y)>0.$ Choosing k large enough so that $| y| \leq kr$ and

(5-6) $$ \begin{align} \sup_{B_r}\frac{u_k(x)}{|x|^{({h+1})/({h-\gamma})}}\leq \frac{u(x_0+y)}{20\,|y|^{({h+1})/({h-\gamma})}}, \end{align} $$

where (5-6) follows from (5-5), we can show

$$ \begin{align*} \frac{u(x_0+y)}{|y|^{({h+1})/({h-\gamma})}}\leq\sup_{B_{kr}}\frac{u(x_0+x)}{|x|^{({h+1})/({h-\gamma})}}\leq\sup_{B_r}\frac{u_k(x)}{|x|^{({h+1})/({h-\gamma})}}\leq \frac{u(x_0+y)}{20|y|^{({h+1})/({h-\gamma})}}. \end{align*} $$

That is, $u(x_0+y)=0,$ which is a contradiction.

Proof of Theorem 1.4.

We argue by contradiction. Assume that there exists a point $y_0\in \Omega $ such that $u(y_0)>0$ and

$$ \begin{align*} r:={\text{dist}}(y_0,\{u=0\})<\tfrac{1}{10}{\text{dist}}(y_0,\partial\Omega). \end{align*} $$

Now, we aim to construct a suitable barrier function as a viscosity subsolution to (1-1). Let

$$ \begin{align*} \Gamma_\xi(x):=\begin{cases} e^{-\xi({r}/{2})^{2}}-e^{-\xi r^{2}} & \text{ in } B_{({r}/{2})}(y_0) , \\ e^{-\xi|x-y_0|^{2}}-e^{-\xi r^{2}} & \text { in } B_{r}(y_0) \backslash B_{({r}/{2})}(y_0) , \\ 0 & \text { in } \Omega \backslash B_{r}(y_0), \end{cases} \end{align*} $$

where $\xi $ is a positive constant to be determined later. Direct computations show that there exists some $\varepsilon _0>0$ such that

$$ \begin{align*} \inf_{B_r(y_0)\backslash B_{({r}/{2})}(y_0)}{|D\Gamma_\xi(x)|}\geq \varepsilon_0. \end{align*} $$

Choosing $0<\beta \ll 1$ and $\xi \gg 1,$ we can verify

$$ \begin{align*} \beta\Gamma_\xi(x)\leq u(x) \quad{\text{on}}\ \partial B_{r}(y_0) \cup \partial B_{({r}/{2})}(y_0) \end{align*} $$

and

$$ \begin{align*} \Delta_\infty^h(\beta\Gamma_\xi(x))-f(x,\beta\Gamma_\xi(x),\beta D\Gamma_\xi(x)) &\geq\beta^h(\Delta_\infty^h\Gamma_\xi(x)-\Lambda(x) f(x,\Gamma_\xi(x), \beta D\Gamma_\xi(x)))\\ & \geq 0 \quad{\text{in}}\ B_{r}(y_0) \backslash B_{({r}/{2})}(y_0). \end{align*} $$

Thus, $\beta \Gamma _\xi (x)$ is a viscosity subsolution to (1-1) in $B_{r}(y_0) \backslash B_{({r}/{2})}(y_0)$ and then

$$ \begin{align*} \beta\Gamma_\xi(x)\leq u(x)\quad {\text{in}}\ B_{r}(y_0) \backslash B_{({r}/{2})}(y_0) \end{align*} $$

by the comparison principle.

Based on Condition (f $_h$ ),

$$ \begin{align*} 0\leq f(x,\delta t,p)\leq \Lambda(x)\delta^{h-1}\delta f(x,t,p)=\Lambda'(x)\delta f(x,t,p), \end{align*} $$

where $\Lambda '(x)=\Lambda (x)\delta ^{h-1}>0, \, t>0$ , and $\delta>0 $ is small enough. By Theorem 1.2,

$$ \begin{align*} \sup_{B_d(z)}u(x)\leq Cd^{({h+1})/({h-1})} \end{align*} $$

for $z\in \partial B_r(y_0)\cap \partial \{u>0\}$ and $0<d\ll ({\beta \varepsilon _0}/{8C})^{({h-1})/{2}}.$ Hence,

$$ \begin{align*} 0<\tfrac{1}{2}\beta\varepsilon_0 d\leq\sup_{B_d(z)}\beta|\Gamma_\xi(x)-\Gamma_\xi(z)|\leq\sup_{B_d(z)}\beta\Gamma_\xi(x)\leq\sup_{B_d(z)}u(x)\leq Cd^{({h+1})/({h-1})}\leq\tfrac{1}{8}\beta\varepsilon_0d, \end{align*} $$

which provides us with a contradiction.