# Fine boundary regularity for fully nonlinear mixed local-nonlocal problems

Mitesh Modasiya and Abhrojyoti Sen  Department of Mathematics, Indian
Institute of Science Education and Research, Dr. Homi Bhabha Road, Pune
411008, India. Email<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We consider Dirichlet problems for fully nonlinear mixed local-nonlocal non-
translation invariant operators. For a bounded $C^{2}$ domain
$\Omega\subset{\mathds{R}^{d}},$ let $u\in C({\mathds{R}^{d}})$ be a viscosity
solution of such Dirichlet problem. We obtain global Lipschitz regularity and
fine boundary regularity for $u$ by constructing appropriate sub and
supersolutions coupled with a weak version of Harnack inequality. We apply
these results to obtain Hölder regularity of $Du$ up to the boundary.

###### Key words and phrases:

Operators of mixed order, viscosity solution, fine boundary regularity, fully
nonlinear integro-PDEs, Harnack inequality, gradient estimate

###### 2020 Mathematics Subject Classification:

Primary: 35D40, 47G20, 35J60, 35B65

##  1\. Introduction and main results

In this article, for a bounded $C^{2}$ domain $\Omega\subset{\mathds{R}^{d}}$
we establish the boundary regularity of the solution $u$ to the in-equations

$\begin{split}{\mathcal{L}}u+C_{0}|Du|&\geq-K\;\quad\text{in}\;\Omega,\\\
{\mathcal{L}}u-C_{0}|Du|&\leq K\quad\quad\text{in}\;\Omega,\\\
u&=0\;\quad\quad\text{in}\;\Omega^{c},\end{split}$ (1.1)

where $C_{0},K\geq 0$ and ${\mathcal{L}}$ is a fully nonlinear integro-
differential operator of the form

$\displaystyle{\mathcal{L}}u(x):={\mathcal{L}}[x,u]=\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}\left\\{\operatorname*{Tr}a_{\theta\nu}(x)D^{2}u(x)+{\mathcal{I}}_{\theta\nu}[x,u]\right\\},$
(1.2)

for some index sets $\Theta,\Gamma$. The coefficient
$a_{\theta\nu}:\Omega\to\mathds{R}^{d\times d}$ is a matrix valued function
and ${\mathcal{I}}_{\theta\nu}$ is a nonlocal operator defined as

$\displaystyle{\mathcal{I}}_{\theta\nu}u(x):={\mathcal{I}}_{\theta\nu}[x,u]=\int_{{\mathds{R}^{d}}}(u(x+y)-u(x)-\mathds{1}_{B_{1}}(y)Du(x)\cdot
y)N_{\theta\nu}(x,y)\;\mathrm{d}{y}.$ (1.3)

The above in-equations (1.1) are motivated by Hamilton-Jacobi equations of the
form

$\displaystyle
Iu(x):=\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}\left\\{L_{\theta\nu}u(x)+f_{\theta\nu}(x)\right\\}=0,$

where

$\displaystyle
L_{\theta\nu}u(x)=\operatorname*{Tr}a_{\theta\nu}(x)D^{2}u(x)+{\mathcal{I}}_{\theta\nu}[x,u]+b_{\theta\nu}(x)\cdot
Du(x),$ (1.4)

$b_{\theta\nu}(\cdot)$ and $f_{\theta\nu}(\cdot)$ are bounded functions on
$\Omega.$ These linear operators (1.4) are extended generator for a wide class
of $d$-dimensional Feller processes (more precisely, jump diffusions) and the
nonlinear operator $Iu(\cdot)$ has its connection to the stochastic control
problems and differential games (see [13, 12] and the references therein). The
first term in (1.4) represents the diffusion, the second term represents the
jump part of a Feller process, and the third represents the drift. We refer to
[15, 14, 1] and the references therein for more on the connections between the
operators of the form (1.4) and stochastic differential equations. For a
precise application of these type of operators in finance and biological
models, we refer to [24, 23, 20] and the references therein.

We set the following assumptions on the coefficient $a_{\theta\nu}(\cdot)$ and
the kernel $N_{\theta\nu}(x,y)$, throughout this article.

######  Assumption 1.1.

  * (a)

$a_{\theta\nu}(\cdot)$ are uniformly continuous and bounded in $\bar{\Omega}$,
uniformly in $\theta,\nu$ for $\theta\in\Theta,\nu\in\Gamma.$ Furthermore,
$a_{\theta\nu}(\cdot)$ satisfies the uniform ellipticity condition
$\lambda\bm{I}\leq a_{\theta\nu}(\cdot)\leq\Lambda\bm{I}$ for some
$0<\lambda\leq\Lambda$ where $\bm{I}$ denotes the $d\times d$ identity matrix.

  * (b)

For each $\theta\in\Theta,\nu\in\Gamma$,
$N_{\theta\nu}:\Omega\times{\mathds{R}^{d}}$ is a measurable function and for
some $\alpha\in(0,2)$ there exists a kernel $k$ that is measurable in
${\mathds{R}^{d}}\setminus\\{0\\}$ such that for any
$\theta\in\Theta,\nu\in\Gamma,x\in\Omega,$ we have

$0\leq N_{\theta\nu}(x,y)\leq k(y)$

and

$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}<+\infty,$

where we denote $p\wedge q:=\min\\{p,q\\}$ for $p,q\in\mathds{R}.$

Let us comment briefly on Assumption 1.1. The uniform continuity of
$a_{\theta\nu}(\cdot)$ is required for the stability of viscosity sub or
supersolutions under appropriate limits and useful in Lemma A.1 which is a key
step for proving interior $C^{1,\gamma}$ regularity (cf. Lemma 2.1). The
Assumption 1.1(b) includes a large class of kernels. We mention some of them
below.

######  Example 1.1.

Consider the following kernels $N_{\theta\nu}(x,y):$

  * (i)

$N_{\theta\nu}(x,y)=\frac{1}{|y|^{d+\sigma}}$ for $\sigma\in(0,2).$ Clearly we
can take $k(y)=\frac{1}{|y|^{d+\sigma}}$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}$ is finite for
$\alpha\in[1+\sigma/2,2).$

  * (ii)

$N_{\theta\nu}(x,y)=\sum_{i=1}^{\infty}\frac{a_{i}}{|y|^{d+\sigma_{i}}}$ for
$\sigma_{i}\in(0,2)$, $\sigma_{0}=\sup_{i}\sigma_{i}<2$ and
$\sum_{i=1}^{\infty}a_{i}=1.$ Similarly taking $N_{\theta\nu}(x,y)=k(y)$ we
can see $\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)<+\infty$ for
$\alpha\in[1+\sigma_{0}/2,2)$.

  * (iii)

$N_{\theta\nu}(x,y)=\begin{cases}\frac{(1-\log|y|)^{\beta}}{|y|^{d+\sigma}}\,\,\,\,\text{for}\,\,\,\,0<|y|\leq
1\\\
\frac{(1+\log|y|)^{-\beta}}{|y|^{d+\sigma}}\,\,\,\,\text{for}\,\,\,\,|y|\geq
1,\end{cases}$
where $\sigma\in(0,2).$
(a) For $2(2-\sigma)>\beta\geq 0,$ taking $N_{\theta\nu}(x,y)=k(y)$ we have
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}<+\infty$ for
$\alpha\in[1+\frac{\sigma}{2}+\frac{\beta}{4},2).$
(b) For $-\sigma<\beta<0,$ taking $N_{\theta\nu}(x,y)=k(y)$ we have
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}<+\infty$ for
$\alpha\in[1+\frac{\sigma}{2},2).$
Proof of (a):

$\displaystyle\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}=\int_{|y|\leq
1}\frac{|y|^{\alpha}(1-\log|y|)^{\beta}}{|y|^{d+\sigma}}\mathrm{d}{y}+\int_{|y|>1}\frac{(1+\log|y|)^{-\beta}}{|y|^{d+\sigma}}\mathrm{d}{y}:=I_{1}+I_{2}.$

Using $(1-\log|y|)\leq\frac{1}{\sqrt{|y|}}+1$ and the convexity of
$\xi(t)=t^{p}$ for $p\geq 1$ we get

$(1-\log|y|)^{\beta}\leq C\left(\frac{1}{|y|^{\beta/2}}+1\right).$

Therefore

$\displaystyle I_{1}\leq\int_{|y|\leq
1}\frac{C\mathrm{d}{y}}{|y|^{\beta/2+d+\sigma-\alpha}}+\int_{|y|\leq
1}\frac{C\mathrm{d}{y}}{|y|^{d+\sigma-\alpha}}<+\infty\,\,\,\,\text{for}\,\,\,\,\alpha\in[1+\sigma/2+\beta/4,2),$

and

$\displaystyle
I_{2}\leq\int_{|y|>1}\frac{\mathrm{d}{y}}{|y|^{d+\sigma}}<+\infty.$

Proof of (b): Since $\beta<0$ in this case, we have $(1-\log|y|)^{\beta}\leq
1$ and $I_{1}<+\infty$ for $\alpha\in[1+\frac{\sigma}{2},2).$ To estimate
$I_{2},$ observe $(1+\log|y|)^{-\beta}\leq(1+|y|)^{-\beta}$ and

$\displaystyle I_{2}\leq
C\int_{|y|>1}\frac{(1+|y|^{-\beta})}{|y|^{d+\sigma}}\mathrm{d}{y}<+\infty\,\,\,\,\text{since}\,\,\,\,\sigma>-\beta.$

  * (iv)

$N_{\theta\nu}(x,y)=\frac{\Psi(1/|y|^{2})}{|y|^{d+\sigma(x,y)}},$ where
$\sigma:{\mathds{R}^{d}}\times{\mathds{R}^{d}}\to\mathds{R}$ satisfying

$0<\sigma^{-}:=\inf_{(x,y)\in{\mathds{R}^{d}}\times{\mathds{R}^{d}}}\sigma(x,y)\leq\sup_{(x,y)\in{\mathds{R}^{d}}\times{\mathds{R}^{d}}}\sigma(x,y):=\sigma^{+}<2.$

and $\Psi$ is a Bernstein function (for several examples of such functions,
see [50]) vanishing at zero. Furthermore, $\Psi$ is non-decreasing, concave
and satisfies a weak upper scaling property i.e, there exists $\mu\geq 0$ and
$c\in(0,1]$ such that

$\Psi(\lambda x)\leq c\lambda^{\mu}\Psi(x)\,\,\,\,\text{for}\,\,\,\,x\geq
s_{0}>0,\lambda\geq 1.$

For $\mu<2(2-\sigma^{+})$, we can take

$\displaystyle
k(y)=\begin{cases}\frac{\Psi(1)}{|y|^{d+2\mu+\sigma^{+}}},\,\,\,\,&\text{if}\,\,\,\,0<|y|\leq
1,\\\
\frac{\Psi(1)}{|y|^{d+\sigma^{-}}},\,\,\,\,&\text{if}\,\,\,\,|y|>1\end{cases}$

and $\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}<+\infty$
for $\alpha\in[1+\mu+\sigma^{+}/2,2).$

The main purpose of this article is to establish a global Lipschitz regularity
and boundary regularity of the solutions satisfying (1.1) under the Assumption
1.1. On the topic of regularity theory for linear elliptic equations, Hölder
estimate plays a key role and it can be obtained by using Harnack inequality.
The pioneering contributions are by DeGiorgi-Nash-Moser [29, 42, 45] who
proved $C^{\alpha}$ regularity for solutions to the second order elliptic
equations in divergence form with measurable coefficients under the assumption
of uniform ellipticity. For equations of non-divergence form, the
corresponding regularity theory was established by Krylov and Safonov [41]. We
refer [16] for a comprehensive overview on the regularity theory for fully
nonlinear elliptic equations. In [40], Krylov studied the boundary regularity
for local second order elliptic equations in non-divergence form with bounded
measurable coefficients. He obtained the Hölder regularity of
$\frac{u}{\delta}$ up to the boundary where $\delta$ denotes the distance
function, i.e, $\delta(x)=\operatorname*{dist}(x,\Omega^{c})$.

Turning our attention towards the nonlocal equations, first Hölder estimates
and Harnack inequalities for $s$-harmonic functions are proved by Bass and
Kassmann [3, 5, 4], however their approach was purely probabilistic. In the
realm of analytic setup, Silvestre [52] proved Hölder continuity of $u$
satisfying (1.3) with some structural assumptions on the operator and kernel
related to the assumptions of Bass and Kassmann. Analogous to the local case
[40], in the nonlocal setting, for a bounded domain
$\Omega\subset{\mathds{R}^{d}}$ with $C^{1,1}$ boundary the first result
concerning boundary regularity of $u$ solving the Dirichlet problem for
$(-\Delta)^{s}$ with bounded right hand side is obtained by Ros-Oton and Serra
[46] where they established a Hölder regularity of $u/\delta^{s}$ up to the
boundary. This result is proved by using a method of Krylov (see [33]). The
idea is to obtain a bound for $u$ with respect to a constant multiple of
$\delta^{s}$ and this controls the oscillation of $u/\delta^{s}$ near the
boundary $\partial\Omega.$ The Hölder regularity of $u/\delta^{s},$ (i) for
more general nonlocal linear operators with $C^{1,\alpha}$ domain is
established in [48], (ii) for smooth domain with smooth right hand side is
established in [30, 31], (iii) for kernel with variable order see [34] and
(iv) for Dirichlet problem for fractional $p$-Laplacian, see [32].

In a seminal paper, Caffarelli and Silvestre [17] studied the regularity
theory for fully nonlinear integro-differential equations of the form :
$\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}{\mathcal{I}}[x,u]$ where
${\mathcal{I}}[x,u]$ is given by (1.3). By obtaining a nonlocal ABP estimate,
they established the Hölder regularity and Harnack inequality when
$N_{\theta\nu}(y)$ ($N_{\theta\nu}(y)$ denotes the $x$-independent form of
$N_{\theta\nu}(x,y)$) is positive, symmetric and comparable with the kernel of
the fractional Laplacian. From a large amount of literature that extend the
work of Caffarelli and Silvestre [17], we mention [36] where the authors
considered integro-PDEs with regularly varying kernel, [9, 19, 37] where
regularity results are obtained for symmetric and non-symmetric stable-like
operators and [35] for kernels with variable order. Also a recent paper [38]
studies Hölder regularity and a scale invariant Harnack inequality under some
weak scaling condition on the kernel. Boundary regularity results for fully
nonlinear integro-differential equations are obtained by Ros-Oton and Serra in
[47]. They considered a restricted class of kernels ${\mathscr{L}}_{*}$ where
$N_{\theta\nu}(x,y)$ is $x$-independent and of the following form

$N_{\theta\nu}(y):=\frac{\mu(y/|y|)}{|y|^{d+2s}}\,\,\,\,\,\text{with}\,\,\,\,\mu\in
L^{\infty}(S^{d-1}),$

satisfying $\mu(\theta)=\mu(-\theta)$ and $\lambda\leq\mu\leq\Lambda$ where
$0<\lambda\leq\Lambda$ are the ellipticity constants. An interesting feature
of ${\mathscr{L}}_{*}$ is

$L(x_{d})_{+}^{s}=0\,\,\,\,\text{in}\,\,\,\,\\{x_{d}>0\\}\,\,\,\,\text{for
all}\,\,\,\,L\in{\mathscr{L}}_{*}$

which is useful to construct barriers in their case. Note that our operators
do not enjoy such property for having different orders. Furthermore, with
Assumption 1.1 the nonlocal part (1.3) is not scale invariant in our case,
that is one may not find any $0\leq\beta\leq 2$ such that
${\mathcal{I}}_{\theta\nu}[x,u(r\cdot)]=r^{\beta}{\mathcal{I}}_{\theta\nu}[rx,u(\cdot)]$
for any $0<r<1.$

Recently, the mathematical study of mixed local-nonlocal integro-differential
equations have been received a considerable attention, for instance see [10,
6, 7, 8, 26, 2]. The regularity results and Harnack inequality for mixed
fractional $p$-Laplace equations are recently obtained in [27, 28]. The
interior $C^{\alpha}$ regularity theory for HJBI-type integro-PDEs has been
studied by Mou [43]. He obtained Hölder regularity for viscosity solutions
under uniform ellipticity condition and a slightly weaker condition on kernels
in compared to the Assumption 1.1 (b), that is
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(y)\mathrm{d}{y}<+\infty.$ More
recently global Lipschitz regularity (compare it with Biagi et. al. [7]) and
fine boundary regularity have been obtained for linear mixed local-nonlocal
operators in [11]. Since the nonlocal operator applied on the distance
function becomes singular near the boundary for certain range of order of the
kernel, one of the main challenges was to construct appropriate sub and
supersolutions and prove an oscillation lemma following [46]. To do such
analysis, along with several careful estimates, the authors borrowed a Harnack
inequality from [25]. Note that for fully nonlinear mixed operators of the
form (1.2) no such Harnack inequality is available in the literature.

In this current contribution, we continue the study started in [11] to obtain
the boundary regularity for fully nonlinear integro-differential problems of
the form (1.1). Below, we present our first result that is the Lipschitz
regularity of $u$ satisfying (1.1) up to the boundary. Note that (1.5) can be
achieved under some weaker assumptions on the domain and kernel. For this
result, we only assume $\partial\Omega$ to be $C^{1,1}$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(y)\mathrm{d}{y}<+\infty.$

######  Theorem 1.1.

Let $\Omega$ be a bounded $C^{1,1}$ domain in ${\mathds{R}^{d}}$ and $u$ be a
continuous function which solves the in-equations (1.1) in viscosity sense.
Then $u$ is in $C^{0,1}({\mathds{R}^{d}})$ and there exists a constant $C$,
depending only on $d,\Omega,\lambda,\Lambda,C_{0},$
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(y)\mathrm{d}{y}$, such that

$\|u\|_{C^{0,1}({\mathds{R}^{d}})}\leq CK.$ (1.5)

To prove Theorem 1.1, the first step is to show that the distance function
$\delta(x)=\operatorname*{dist}(x,\Omega^{c})$ can be used as a barrier to $u$
in $\Omega.$ Once this is done, we can complete the proof by considering
different cases depending on the distance between any two points in $\Omega$
or their distance from $\partial\Omega$ and combining $|u|\leq C\delta$ with
an interior $C^{1,\gamma}$-estimate for scaled operators (cf. Lemma 2.1).

Next we show the fine boundary regularity, that is the Hölder regularity of
$u/\delta$ up to the boundary.

######  Theorem 1.2.

Suppose that Assumption 1.1 holds. Let $\Omega$ be a bounded $C^{2}$ domain
and $u$ be a viscosity solution to the in-equations (1.1). Then there exists
$\upkappa\in(0,\hat{\alpha})$ such that

$\lVert u/\delta\rVert_{C^{\upkappa}(\overline{\Omega})}\leq C_{1}K,$ (1.6)

for some constant $C_{1}$, where $\upkappa,C_{1}$ depend on
$d,\Omega,C_{0},\Lambda,\lambda,\alpha$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}$. Here
$\hat{\alpha}$ is given by

$\displaystyle\hat{\alpha}=\begin{cases}1\,\,\,\,&if\,\,\,\,\alpha\in(0,1]\\\
\frac{2-\alpha}{2}\,\,\,\,&if\,\,\,\,\alpha\in(1,2).\end{cases}$

To prove Theorem 1.2, following [46] we prove an oscillation lemma (cf.
Proposition 4.1). For this, first we need to construct sub and supersolutions
carefully since ${\mathcal{I}}_{\theta\nu}\delta$ becomes singular near the
boundary $\partial\Omega$ for $\alpha\in(1,2).$ Then we shall use a “weak
version” of Harnack inequality (cf. Theorem 4.1). This weak version of Harnack
inequality is new and needed to be developed due to the unavailability of
classical Harnack inequality. Also, we must point out that one needs to bypass
the use of comparison principle [10, Theorem 5.1] in such analysis, since the
mentioned theorem is for translation invariant linear operators. For non-
translation invariant operators, such comparison principle is unavailable, see
Remark 2.1 for details.

Now applying (1.6), we prove the Hölder regularity of $Du$ up to the boundary.

######  Theorem 1.3.

Suppose that Assumption 1.1 holds and $\Omega$ be a bounded $C^{2}$ domain.
Then for any viscosity solution $u$ to the in-equations (1.1) we have

$\displaystyle||Du||_{C^{\eta}(\overline{\Omega})}\leq CK,$

for some $\eta\in(0,1)$ and $C,$ depending only on
$d,\Omega,C_{0},\Lambda,\lambda,\alpha$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}$.

The interior $C^{1,\eta}$-regularity for fully nonlinear integro-differential
equations is studied in [17] by introducing a new ellipticity class where the
kernels are $C^{1}$ away from the origin. Kriventsov [39] extended this result
without the additional assumption on kernels (sometimes referred as rough
kernels). Also see [49] for its parabolic version. For HJBI-type integro-PDEs,
interior $C^{1,\eta}$-regularity is established by Mou and Zhang [44] and for
mixed local nonlocal fractional $p$-Laplacian, see [22]. The
$C^{1,\eta}$-regularity up to the boundary for linear mixed local-nonlocal
operators is recently obtained in [11].

The rest of the article is organized as follows. In Section 2, we introduce
the necessary preliminaries and collect all the auxiliary results which will
be used throughout the article. In Section 3 we prove Theorem 1.1. Theorem 1.2
is proved in Section 4. In Section 5 we prove Theorem 1.3. Lastly, in Appendix
A, following an approximation and scaling argument, we give a proof of
$C^{1,\gamma}$ regularity for a scaled operator i.e, Lemma 2.1.

##  2\. Notation and preliminary results

This section sets the notation which we use throughout the paper and collects
the necessary results.

###  2.1. Notations and Definitions

We start by setting the notations. We use ${\mathscr{B}}_{r}(x)$ to denote an
open ball of radius $r>0$ centred at a point $x\in{\mathds{R}^{d}}$ and for
$x=0,$ we denote ${\mathscr{B}}_{r}:={\mathscr{B}}_{r}(0).$ For any subset
$U\subseteq{\mathds{R}^{d}}$ and for $\alpha\in(0,1),$ we denote
$C^{\alpha}(U)$ as the space of all bounded, $\alpha$-Hölder continuous
functions equipped with the norm

$||f||_{C^{\alpha}(U)}:=\sup_{x\in U}|f(x)|+\sup_{x,y\in
U}\frac{|f(x)-f(y)|}{|x-y|^{\alpha}}.$

Note that for $\alpha=1,$ $C^{0,1}(U)$ denotes the space of all Lipschitz
continuous functions on $U.$ The space of all bounded functions with bounded
$\alpha$-Hölder continuous derivatives is denoted by $C^{1,\alpha}(U)$ with
the norm

$||f||_{C^{1,\alpha}(U)}:=\sup_{x\in U}|f(x)|+||Df||_{C^{\alpha}(U)}.$

We use
$USC({\mathds{R}^{d}}),LSC({\mathds{R}^{d}}),C({\mathds{R}^{d}}),C_{b}({\mathds{R}^{d}}),M^{d}$
to denote the space of upper semicontinuous, lower semicontinuous, continuous
functions, bounded continuous functions on ${\mathds{R}^{d}}$ and $d\times d$
symmetric matrices respectively.

Now let us introduce the scaled operators. For $0<s\leq 1,$ we define scaled
version of (1.2) as following.

$\displaystyle{\mathcal{L}}^{s}[x,u]=\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}\left\\{\operatorname*{Tr}a_{\theta\nu}(sx)D^{2}u(x)+{\mathcal{I}}^{s}_{\theta\nu}[x,u]\right\\},$

where

$\displaystyle{\mathcal{I}}^{s}_{\theta\nu}[x,u]=\int_{\mathbb{R}^{d}}(u(x+y)-u(x)-\mathds{1}_{{\mathscr{B}}_{\frac{1}{s}}}(y)\nabla
u(x)\cdot y)s^{d+2}N_{\theta\nu}(sx,sy)\mathrm{d}{y}.$

Next we define extremal Pucci operators for second order term and the nonlocal
term.

$\displaystyle{\mathcal{P}}^{+}u(x)$
$\displaystyle\coloneqq\sup\left\\{\operatorname*{Tr}(AD^{2}u(x)),A\in
M^{d},\lambda\bm{I}\leq A\leq\Lambda\bm{I}\right\\},$
$\displaystyle{\mathcal{P}}^{-}u(x)$
$\displaystyle\coloneqq\inf\left\\{\operatorname*{Tr}(AD^{2}u(x)),A\in
M^{d},\lambda\bm{I}\leq A\leq\Lambda\bm{I}\right\\},$

and

$\displaystyle{\mathcal{P}}^{+}_{k,s}u(x)$
$\displaystyle\coloneqq\int_{{\mathds{R}^{d}}}(u(x+y)-u(x)-\mathds{1}_{{\mathscr{B}}_{\frac{1}{s}}}(y)\nabla
u(x)\cdot y)^{+}s^{d+2}k(sy)\mathrm{d}{y},$
$\displaystyle{\mathcal{P}}^{-}_{k,s}u(x)$
$\displaystyle\coloneqq-\int_{{\mathds{R}^{d}}}(u(x+y)-u(x)-\mathds{1}_{{\mathscr{B}}_{\frac{1}{s}}}(y)\nabla
u(x)\cdot y)^{-}s^{d+2}k(sy)\mathrm{d}{y}.$

Denote ${\mathcal{P}}^{+}_{k,1}={\mathcal{P}}^{+}_{k}$ and
${\mathcal{P}}^{-}_{k,1}={\mathcal{P}}^{-}_{k}.$

We recall the definition of viscosity sub and supersolution. First of all, we
say a function $\varphi$ touches from above (below) at $x$ if, for a small
$r>0,$

$\varphi(x)=u(x)\,\,\,\,\text{and}\,\,\,\,u(y)\leq(\geq)\varphi(y)\,\,\,\,\text{for
all}\,\,\,\,y\in{\mathscr{B}}_{r}(x).$

######  Definition 2.1.

A function $u\in USC({\mathds{R}^{d}})\cap L^{\infty}({\mathds{R}^{d}})$
(resp. $u\in LSC({\mathds{R}^{d}})\cap L^{\infty}({\mathds{R}^{d}})$) is said
to be a viscosity subsolution (resp. supersolution) to (1.1) if whenever
$\varphi$ touches $u$ from above (resp. below) for some bounded test function
$\varphi\in C^{2}({\mathscr{B}}_{r}(x))\cap C({\mathds{R}^{d}}),$ then

$\displaystyle
v=\begin{cases}\varphi\,\,\,\,&\text{in}\,\,\,{\mathscr{B}}_{r}(x)\\\
u\,\,\,\,&\text{in}\,\,\,{\mathscr{B}}^{c}_{r}(x)\end{cases}$

satisfies ${\mathcal{L}}v(x)+C_{0}|Dv(x)|\geq-K$ (resp.
${\mathcal{L}}v(x)-C_{0}|Dv(x)|\leq K$).

###  2.2. Auxiliary lemmas

We collect some preliminary results here. The first result is the interior
$C^{1,\gamma}$ regularity for the scaled operator ${\mathcal{L}}^{s}.$

######  Lemma 2.1.

Let $0<s\leq 1$ and $u\in L^{\infty}({\mathds{R}^{d}})\cap
C({\mathds{R}^{d}})$ solves the in-equations

$\displaystyle{\mathcal{L}}^{s}[x,u]+C_{0}s|Du(x)|\geq-K$ $\displaystyle
in\,\,\,{\mathscr{B}}_{2},$ (2.1)
$\displaystyle{\mathcal{L}}^{s}[x,u]-C_{0}s|Du(x)|\leq K$ $\displaystyle
in\,\,\,{\mathscr{B}}_{2},$

in the viscosity sense. Then there exist constants $0<\gamma<1$ and $C>0$
independent of $s$, such that

$\displaystyle||u||_{C^{1,\gamma}({\mathscr{B}}_{1})}\leq
C\Big{(}||u||_{L^{\infty}({\mathds{R}^{d}})}+K\Big{)},$

where $\gamma$ and $C$ depend only on $d,\lambda,\Lambda,C_{0}$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(y)\mathrm{d}{y}$.

###### Proof.

The proof essentially uses the approximation arguments for nonlocal equations
[18] and we postpone it to Appendix A. ∎

Now we present a maximum principle type result similar to [10, Theorem 5.2].
We report the proof here for reader’s convenience.

######  Lemma 2.2.

Let u be a bounded function on ${\mathds{R}^{d}}$ which is in
$USC(\overline{\Omega})$ and satisfies
${\mathcal{P}}^{+}u+{\mathcal{P}}_{k}^{+}u+C_{0}|Du|\geq 0$ in $\Omega$. Then
we have $\sup_{\Omega}u\leq\sup_{\Omega^{c}}u.$

###### Proof.

From [43, Lemma 5.5] we can find a non-negative function
$\chi\in{C}^{2}(\bar{\Omega})\cap{C}_{b}({\mathds{R}^{d}})$ satisfying

${\mathcal{P}}^{+}\chi+{\mathcal{P}}_{k}^{+}\chi+C_{0}|D\chi|\leq-1\quad\text{in}\;\Omega.$

Note that, since $\chi\in C^{2}(\bar{\Omega})$, the above inequality holds in
the classical sense. For $\varepsilon>0$, we let $\phi_{M}$ to be

$\phi_{M}(x)=M+\varepsilon\chi.$

Then
${\mathcal{P}}^{+}\phi_{M}(x_{0})+{\mathcal{P}}^{+}_{k}\phi_{M}+C_{0}|D\phi_{M}|\leq-\varepsilon$
in $\Omega$.

Let $M_{0}$ be the smallest value of $M$ for which $\phi_{M}\geq u$ in
${\mathds{R}^{d}}$. We show that $M_{0}\leq\sup_{\Omega^{c}}u$. Suppose, to
the contrary, that $M_{0}>\sup_{\Omega^{c}}u$. Then there must be a point
$x_{0}\in\Omega$ for which $u(x_{0})=\phi_{M_{0}}(x_{0}).$ Otherwise using the
upper semicontinuity of $u,$ we get a $M_{1}<M_{0}$ such that
$\phi_{M_{1}}\geq u$ in ${\mathds{R}^{d}},$ which contradicts the minimality
of $M_{0}$. Now $\phi_{M_{0}}$ would touch $u$ from above at $x_{0}$ and thus,
by the definition of the viscosity subsolution, we would have that
${\mathcal{P}}^{+}\phi_{M_{0}}(x_{0})+{\mathcal{P}}^{+}_{k}\phi_{M_{0}}+C_{0}|D\phi_{M_{0}}|\geq
0$. This leads to a contradiction. Therefore, $M_{0}\leq\sup_{\Omega^{c}}u$
which implies that for every $x\in{\mathds{R}^{d}}$

$u\leq\phi_{M_{0}}\leq
M_{0}+\varepsilon\,\sup_{{\mathds{R}^{d}}}\chi\leq\sup_{\Omega^{c}}u+\varepsilon\,\sup_{{\mathds{R}^{d}}}\chi.$

The result follows by taking $\varepsilon\rightarrow 0$.

######  Remark 2.1.

Although we have the above maximum principle, one can not simply compare two
viscosity sub and supersolutions for the operator (1.2). More precisely, if
$u,v$ are bounded functions and $u\in USC({\mathds{R}^{d}}),v\in
LSC({\mathds{R}^{d}})$ satisfy

${\mathcal{L}}u+C|Du|\geq f\,\,\,\,\text{and}\,\,\,\,{\mathcal{L}}v+C|Dv|\leq
g\,\,\,\,\text{in}\,\,\,\,\Omega$

in viscosity sense for two continuous functions $f$ and $g,$ and for some
$C\geq 0,$ then ${\mathcal{L}}(u-v)+C|D(u-v)|\geq f-g$ may not always holds
true in $\Omega$. However, if one of them is $C^{2},$ then we have

${\mathcal{P}}^{+}(u-v)+{\mathcal{P}}^{+}_{k}(u-v)+C|D(u-v)|\geq
f-g\,\,\,\,\text{in}\,\,\,\Omega.$

Indeed, without loss of generality, let us assume $v\in C^{2}(\Omega)$ and
$\varphi$ be a $C^{2}$ test function that touches $u-v$ at $x\in\Omega$ from
above then clearly $\varphi+v$ touches $u$ at $x$ from above. By definition of
viscosity subsolution we have
${\mathcal{L}}(\varphi+v)(x)+C|D(\varphi+v)(x)|\geq f(x),$ which implies

${\mathcal{P}}^{+}\varphi(x)+{\mathcal{P}}^{+}_{k}\varphi(x)+{\mathcal{L}}v(x)+C|D\varphi(x)|+C|Dv(x)|\geq
f(x)$

and hence we obtain

${\mathcal{P}}^{+}\varphi(x)+{\mathcal{P}}^{+}_{k}\varphi(x)+C|D\varphi(x)|\geq
f(x)-g(x).$

∎

##  3\. Global Lipschitz regularity

In this section we establish the Lipschitz regularity of the solution $u$ up
to the boundary. We start by showing that the distance function $\delta(x)$ is
a barrier to $u.$

######  Lemma 3.1.

Let $\Omega$ be a bounded $C^{1,1}$ domain in ${\mathds{R}^{d}}$ and $u$ be a
continuous function which solves (1.1) in the viscosity sense. Then there
exists a constant $C$ which depends only on
$d,\lambda,\Lambda,C_{0},\operatorname*{diam}(\Omega)$, radius of exterior
sphere and $\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(y)\mathrm{d}{y}$, such
that

$|u(x)|\leq CK\delta(x)\quad\text{for all}\;x\in\Omega.$ (3.1)

###### Proof.

First we show that

$|u(x)|\leq\kappa\,K\quad x\in{\mathds{R}^{d}},$ (3.2)

for some constant $\kappa$. From [43, Lemma 5.5], there exists a non-negative
function $\chi\in C^{2}(\bar{\Omega})\cap C_{b}({\mathds{R}^{d}})$, with
$\inf_{{\mathds{R}^{d}}}\chi>0$, satisfying

${\mathcal{P}}^{+}\chi+{\mathcal{P}}^{+}_{k}\chi+C_{0}|D\chi|\leq-1\quad\text{in}\;\,\,\Omega.$

We define $\psi=K\chi$ which gives that $\inf_{{\mathds{R}^{d}}}\psi\geq 0$
and

${\mathcal{P}}^{+}\psi+{\mathcal{P}}^{+}_{k}\psi+C_{0}|D\psi|\leq-K\quad\text{in}\;\,\,\Omega.$

Then by using Remark 2.1, we get

${\mathcal{P}}^{+}(u-\psi)+{\mathcal{P}}^{+}_{k}(u-\psi)+C_{0}|D(u-\psi)|\geq
0.$

Now applying Lemma 2.2 on $u-\psi$ we obtain

$\sup_{\Omega}(u-\psi)\leq\sup_{\Omega^{c}}(u-\psi)\leq 0.$

Note that in the second inequality above we used $u=0$ in $\Omega^{c}.$ This
proves that $u\leq\psi$ in ${\mathds{R}^{d}}$. Similar calculation using $-u$
will also give us $-u\leq\psi$ in ${\mathds{R}^{d}}$. Thus

$|u|\leq\sup_{{\mathds{R}^{d}}}|\chi|\,K\quad\text{in}\;{\mathds{R}^{d}},$

which gives (3.2).

Now we shall prove (3.1). Since $\partial\Omega$ is $C^{1,1}$, $\Omega$
satisfies a uniform exterior sphere condition from outside. Let $r_{\circ}$ be
a radius satisfying uniform exterior condition. From [43, Lemma 5.4] there
exists a bounded, Lipschitz continuous function $\varphi$, Lipschitz constant
being $r_{\circ}^{-1}$, satisfying

$\displaystyle\varphi$
$\displaystyle=0\quad\text{in}\quad\bar{{\mathscr{B}}}_{r_{\circ}},$
$\displaystyle\varphi$
$\displaystyle>0\quad\text{in}\quad\bar{{\mathscr{B}}}_{r_{\circ}}^{c},$
$\displaystyle\varphi$
$\displaystyle\geq\varepsilon\quad\text{in}\quad{\mathscr{B}}_{(1+\delta)r_{\circ}}^{c},$
$\displaystyle{\mathcal{P}}^{+}\varphi+{\mathcal{P}}^{+}_{k}\varphi+C_{0}|D\varphi|$
$\displaystyle\leq-1\quad\text{in}\quad{\mathscr{B}}_{(1+\delta)r_{\circ}}\setminus\bar{{\mathscr{B}}}_{r_{\circ}},$

for some constants $\varepsilon,\delta$, dependent on
$C_{0},d,\lambda,\Lambda,d$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2}){k}(y)\mathrm{d}{y}$. Furthermore,
$\varphi$ is $C^{2}$ in
${\mathscr{B}}_{(1+\delta)r_{\circ}}\setminus\bar{{\mathscr{B}}}_{r_{\circ}}$.
For any point $y\in\partial\Omega$, we can find another point $z\in\Omega^{c}$
such that $\overline{{\mathscr{B}}}_{r_{\circ}}(z)\subset\Omega^{c}$ touches
$\partial\Omega$ at $y$. Let $w(x)=\varepsilon^{-1}\kappa K\varphi(x-z)$. Also
${\mathcal{P}}^{+}(w)+{\mathcal{P}}^{+}_{k}(w)+C_{0}|Dw|\leq-K$. Then by using
Remark 2.1 we have

${\mathcal{P}}^{+}(u-w)+{\mathcal{P}}^{+}_{k}(u-w)+C_{0}|D(u-w)|\geq
0\quad\text{in}\;{\mathscr{B}}_{(1+\delta)r_{\circ}}(z)\cap\Omega.$

Since, by (3.2) $u-w\leq 0$ in
$({\mathscr{B}}_{(1+\delta)r_{\circ}}(z)\cap\Omega)^{c}$, applying Lemma 2.2
on $u-w,$ it follows that $u(x)\leq w(x)$ in ${\mathds{R}^{d}}$. Repeating a
similar calculation for $-u$, we can conclude that $|u(x)|\leq w(x)$ in
${\mathds{R}^{d}}$. Since this relation holds for any $y\in\partial\Omega,$
taking $x\in\Omega$ with $\operatorname*{dist}(x,\partial\Omega)<r_{\circ}$,
one can find $y\in\partial\Omega$ satisfying
$\operatorname*{dist}(x,\partial\Omega)=\lvert x-y\rvert<r_{\circ}.$ Then
using the previous estimate we would obtain

$|u(x)|\leq\varepsilon^{-1}\kappa K\varphi(x-z)\leq\varepsilon^{-1}\kappa
K(\varphi(x-z)-\varphi(y-z))\leq\varepsilon^{-1}\kappa
K\,r_{\circ}^{-1}\operatorname*{dist}(x,\partial\Omega),$

which gives us (3.1). ∎

Now we are ready to prove that $u\in C^{0,1}({\mathds{R}^{d}})$.

###### Proof of Theorem 1.1.

Let $x_{0}\in\Omega$ and $s\in(0,1]$ be such that
$2s=\operatorname*{dist}(x_{0},\partial\Omega)\wedge 1$. Without loss of any
generality, we assume $x_{0}=0$. Define $v(x)=u(sx)$ in ${\mathds{R}^{d}}$.
Using Lemma 3.1 we already have $|u(x)|\leq C_{1}K\delta(x)$, from that one
can deduce

$|v(x)|\leq C_{1}\,Ks(1+\lvert x\rvert)\quad\quad\text{for
all}\,\,\;x\in{\mathds{R}^{d}},$ (3.3)

for some constant $C_{1}$ independent of $s$. We recall the scaled operator

${\mathcal{I}}_{\theta\nu}^{s}[x,f]:=\int_{{\mathds{R}^{d}}}(f(x+y)-f(x)-\mathds{1}_{{\mathscr{B}}_{\frac{1}{s}}}(y)\nabla
f(x)\cdot y)s^{d+2}N_{\theta\nu}(sx,sy)\mathrm{d}{y}.$

To compute ${\mathcal{L}}^{s}[x,v]+C_{0}s|Dv(x)|$ in ${\mathscr{B}}_{2}$,
first we observe that $D^{2}v(x)=s^{2}D^{2}u(sx)$ and $Dv(x)=sDu(sx)$. Also

$\displaystyle{\mathcal{I}}_{\theta\nu}^{s}[x,v]$
$\displaystyle=s^{2}\int_{{\mathds{R}^{d}}}(v(x+y)-v(x)-\mathds{1}_{{\mathscr{B}}_{\frac{1}{s}}}(y)\nabla
v(x)\cdot y)N_{\theta\nu}(sx,sy)s^{d}\mathrm{d}{y}$
$\displaystyle=s^{2}\int_{{\mathds{R}^{d}}}(u(sx+sy)-u(sx)-\mathds{1}_{{\mathscr{B}}_{1}}(sy)\nabla
u(sx)\cdot
sy)N_{\theta\nu}(sx,sy)s^{d}\mathrm{d}{y}=s^{2}{\mathcal{I}}_{\theta\nu}[sx,u].$

Thus, it follows from (1.1) that

$\begin{split}{\mathcal{L}}^{s}[x,v]+C_{0}s|Dv(x)|&\geq-
Ks^{2}\quad\text{in}\quad{\mathscr{B}}_{2},\\\
{\mathcal{L}}^{s}[x,v]-C_{0}s|Dv(x)|&\leq
Ks^{2}\quad\text{in}\quad{\mathscr{B}}_{2}.\end{split}$ (3.4)

Now consider a smooth cut-off function $\varphi,0\leq\varphi\leq 1$,
satisfying

$\varphi=\begin{cases}1\quad\text{in}\;{\mathscr{B}}_{3/2},\\\
0\quad\text{in}\;{\mathscr{B}}^{c}_{2}.\end{cases}$

Let $w=\varphi v$. Clearly, $((\varphi-1)v)(y)=0$ for all
$y\in{\mathscr{B}}_{3/2}$, which gives $D((\varphi-1)v)=0$ and
$D^{2}((\varphi-1)v)=0$ in $x\in{\mathscr{B}}_{3/2}$. Since
$w=v+(\varphi-1)v$, from (3.4) we obtain

$\begin{split}{\mathcal{L}}^{s}[x,w]+C_{0}s|Dw(x)|&\geq-
Ks^{2}-|\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}{\mathcal{I}}_{\theta\nu}^{s}[x,(\varphi-1)v)]|\quad\text{in}\quad{\mathscr{B}}_{1},\\\
{\mathcal{L}}^{s}[x,w]-C_{0}s|Dw(x)|&\leq
Ks^{2}+|\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}{\mathcal{I}}_{\theta\nu}^{s}[x,(\varphi-1)v)]|\quad\text{in}\quad{\mathscr{B}}_{1}.\end{split}$
(3.5)

Again, since $(\varphi-1)v=0$ in ${\mathscr{B}}_{3/2}$, we have in
${\mathscr{B}}_{1}$ that

$\displaystyle|{\mathcal{I}}_{\theta\nu}^{s}[x,(\varphi-1)v]|$
$\displaystyle=\Big{|}\int_{|y|\geq
1/2}((\varphi-1)v)(x+y)-((\varphi-1)v)(x))s^{d+2}N_{\theta\nu}(sx,sy)\mathrm{d}{y}\Big{|}$
$\displaystyle\leq\int_{|y|\geq
1/2}|v(x+y)|s^{d+2}N_{\theta\nu}(sx,sy)\mathrm{d}{y}+|v(x)|\int_{|y|\geq
1/2}s^{d+2}N_{\theta\nu}(sx,sy)\mathrm{d}{y}$ $\displaystyle:=I_{1}+I_{2}.$

Since $x\in{\mathscr{B}}_{1},$ using $s^{d+2}N_{\theta\nu}(sx,sy)\leq
s^{d+2}k(sy)$ and (3.3) we can have the following estimate,

$\displaystyle I_{2}\leq
2C_{1}Ks\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})\mathrm{d}{y}.$

Now write

$\displaystyle I_{1}$ $\displaystyle=\int_{1/2\leq|y|\leq
1/s}|v(x+y)|s^{d+2}N_{\theta\nu}(sx,sy)\mathrm{d}{y}+\int_{|y|\geq
1/s}|v(x+y)|s^{d+2}N_{\theta\nu}(sx,sy)\mathrm{d}{y}$
$\displaystyle=I_{s,1}+I_{s,2}\,.$

Let us first estimate $I_{s,1}.$ Since $x\in{\mathscr{B}}_{1}$ and
$|y|\geq\frac{1}{2}$ we have $1+|x+y|\leq 5|y|.$ By using this estimate and
(3.3) we obtain

$\displaystyle I_{s,1}$ $\displaystyle=s^{d+2}\int_{\frac{1}{2}\leq\lvert
y\rvert\leq\frac{1}{s}}|v(x+y)|N_{\theta\nu}(sx,sy)\mathrm{d}{y}$
$\displaystyle\leq 5C_{1}K\int_{\frac{1}{2}\leq\lvert
y\rvert\leq\frac{1}{s}}|sy|s^{d+2}k(sy)\mathrm{d}{y}\leq
5C_{1}Ks\int_{\frac{s}{2}\leq\lvert z\rvert\leq 1}|sz|k(z)\mathrm{d}{z}$
$\displaystyle\leq C_{2}s\int_{\frac{s}{2}\leq\lvert z\rvert\leq
1}|z|^{2}k(z)\mathrm{d}{z}\leq
C_{2}s\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(z)\mathrm{d}{z}\leq C_{3}s,$

for some constants $C_{3}.$ For $I_{s,2}$, a change of variable and (3.2)
gives

$\displaystyle I_{s,2}\leq\kappa s^{2}K\int_{s\lvert
y\rvert>1}s^{d}k(ry)\mathrm{d}{y}$ $\displaystyle=\kappa s^{2}K\int_{\lvert
y\rvert>1}k(y)\mathrm{d}{y}$ $\displaystyle\leq\kappa
s^{2}K\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(y)\mathrm{d}{y}\leq
C_{4}s^{2}K$

for some constant $C_{4}$. Therefore, putting the estimates of $I_{1}$ and
$I_{2}$ in (3.5) we obtain

$\begin{split}{\mathcal{L}}^{s}[x,w]+C_{0}s|Dw(x)|&\geq-
C_{5}Ks\quad\text{in}\quad{\mathscr{B}}_{1},\\\
{\mathcal{L}}^{s}[x,w]-C_{0}s|Dw(x)|&\geq
C_{5}Ks\quad\text{in}\quad{\mathscr{B}}_{1},\end{split}$ (3.6)

for some constant $C_{5}$. Now applying Lemma 2.1, from (3.6) we have

$\lVert v\rVert_{C^{1}({\mathscr{B}}_{\frac{1}{2}})}\leq C_{6}\Big{(}\lVert
v\rVert_{L^{\infty}({\mathscr{B}}_{2})}+sK\Big{)}$ (3.7)

for some constant $C_{6}$. From (3.3) and (3.7) we then obtain

$\sup_{y\in{\mathscr{B}}_{s/2}(x),y\neq x}\frac{|u(x)-u(y)|}{|x-y|}\leq
C_{7}K,$ (3.8)

for some constant $C_{7}$.

Now we can complete the proof. Note that if $|x-y|\geq\frac{1}{8}$, then

$\frac{|u(x)-u(y)|}{|x-y|}\leq 2\kappa K,$

by (3.2). So we consider $|x-y|<\frac{1}{8}$. If $|x-y|\geq
8^{-1}(\delta(x)\vee\delta(y))$, then using Lemma 3.1 we get

$\frac{|u(x)-u(y)|}{|x-y|}\leq
4CK(\delta(x)+\delta(y))(\delta(x)\vee\delta(y))^{-1}\leq 8CK.$

Now let $|x-y|<8^{-1}\min\\{\delta(x)\vee\delta(y),1\\}$. Then either
$y\in{\mathscr{B}}_{\frac{\delta(x)\wedge 1}{8}}(x)$ or
$x\in{\mathscr{B}}_{\frac{\delta(y)\wedge 1}{8}}(y)$. Without loss of
generality, we suppose $y\in{\mathscr{B}}_{\frac{\delta(x)\wedge 1}{8}}(x)$.
From (3.8) we get

$\frac{|u(x)-u(y)|}{|x-y|}\leq C_{7}K.$

This completes the proof. ∎

##  4\. Fine boundary regularity

Aim of this section is to prove Theorem 1.2. Since $u$ is Lipschitz, (1.1) can
be written as

$|{\mathcal{L}}u|\leq
CK\,\,\,\,\text{in}\,\,\,\Omega,\,\,\,\,\,\,\text{and}\,\,\,\,u=0\,\,\,\,\text{in}\,\,\,\Omega^{c}.$

We start by constructing subsolutions which will be useful later on to prove
oscillation lemma.

######  Lemma 4.1.

There exists a constant $\tilde{\kappa}$, which depends only on
$d,\lambda,\Lambda,\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(y)\mathrm{d}{y}$,
such that for any $r\in(0,1]$, we have a bounded radial function $\phi_{r}$
satisfying

$\begin{cases}{\mathcal{P}}^{-}\phi_{r}+{\mathcal{P}}^{-}_{k}\phi_{r}\geq
0&\text{in}\;\;{\mathscr{B}}_{4r}\setminus\bar{{\mathscr{B}}}_{r},\\\
0\leq\phi_{r}\leq\tilde{\kappa}r&\text{in}\;\;{\mathscr{B}}_{r},\\\
\phi_{r}\geq\frac{1}{\tilde{\kappa}}(4r-\lvert
x\rvert)&\text{in}\;\;{\mathscr{B}}_{4r}\setminus{\mathscr{B}}_{r},\\\
\phi_{r}\leq
0&\text{in}\;\;{\mathds{R}^{d}}\setminus{\mathscr{B}}_{4r}.\end{cases}$

Moreover, $\phi_{r}\in
C^{2}({\mathscr{B}}_{4r}\setminus\bar{{\mathscr{B}}}_{r})$.

###### Proof.

We use the same subsolution constructed in [11] and show that it is indeed a
subsolution with respect to minimal Pucci operators. Fix $r\in(0,1]$ and
define $v_{r}(x)=\mathrm{e}^{-\eta q(x)}-\mathrm{e}^{-\eta(4r)^{2}}$, where
$q(x)=|x|^{2}\wedge 2(4r)^{2}$ and $\eta>0$. Clearly, $1\geq v_{r}(0)\geq
v_{r}(x)$ for all $x\in{\mathds{R}^{d}}$. Thus using the fact that
$1-e^{-\xi}\leq\xi$ for all $\xi\geq 0$ we have

$v_{r}(x)\leq 1-\mathrm{e}^{-\eta(4r)^{2}}\leq\eta(4r)^{2},$ (4.1)

Again for $x\in{\mathscr{B}}_{4r}\setminus{\mathscr{B}}_{r}$, we have that

$\displaystyle
v_{r}(x)=\mathrm{e}^{-\eta(4r)^{2}}(\mathrm{e}^{\eta((4r)^{2}-q(x))}-1)$
$\displaystyle\geq\eta\mathrm{e}^{-\eta(4r)^{2}}((4r)^{2}-|x|^{2})$
$\displaystyle=\eta\mathrm{e}^{-\eta(4r)^{2}}(4r+|x|)(4r-|x|)\geq 5\eta
r\mathrm{e}^{-\eta(4r)^{2}}(4r-|x|).$ (4.2)

Fix $x\in{\mathscr{B}}_{4r}\setminus\bar{{\mathscr{B}}}_{r}$. We start by
estimating the local minimal Pucci operator ${\mathcal{P}}^{-}$ of $v$. Using
rotational symmetry we may always assume $x=(l,0,\cdots,0)$ Then

$\partial_{i}v_{r}(x)=-2\eta\mathrm{e}^{-\eta|x|^{2}}x_{i}=\begin{cases}-2\eta\mathrm{e}^{-\eta|x|^{2}}l&i=1,\\\
0&i\neq 1\end{cases}$

and

$\displaystyle\partial_{ij}v_{r}(x)$
$\displaystyle=\begin{cases}4\eta^{2}x_{i}^{2}\mathrm{e}^{-\eta|x|^{2}}-2\eta\mathrm{e}^{-\eta|x|^{2}}&i=j,\\\
4\eta^{2}x_{i}x_{j}\mathrm{e}^{-\eta|x|^{2}}&i\neq j.\end{cases}$
$\displaystyle=\begin{cases}4\eta^{2}l^{2}\mathrm{e}^{-\eta|x|^{2}}-2\eta\mathrm{e}^{-\eta|x|^{2}}&i=j=1,\\\
-2\eta\mathrm{e}^{-\eta|x|^{2}}&i=j\neq 1,\\\ 0&i\neq j.\end{cases}$

By the above calculation, for any
$x\in{\mathscr{B}}_{4r}\setminus\bar{{\mathscr{B}}}_{r},$ choosing
$\eta>\frac{1}{r^{2}}$ we have

$\displaystyle{\mathcal{P}}^{-}v_{r}(x)$ $\displaystyle=\lambda
4\eta^{2}l^{2}\mathrm{e}^{-\eta|x|^{2}}-\lambda
2\eta\mathrm{e}^{-\eta|x|^{2}}-\Lambda(d-1)2\eta\mathrm{e}^{-\eta|x|^{2}}$
$\displaystyle\geq\lambda 4\eta^{2}l^{2}\mathrm{e}^{-\eta|x|^{2}}-d\Lambda
2\eta\mathrm{e}^{-\eta|x|^{2}}.$

Now to determine nonlocal minimal Pucci operator, using the convexity of
exponential map we get,

$\displaystyle e^{-\eta|x+y|^{2}}-e^{-\eta|x|^{2}}+2\eta\mathds{1}_{\\{|y|\leq
1\\}}y\cdot xe^{-\eta|x|^{2}}$ $\displaystyle\qquad\geq-\eta
e^{-\eta|x|^{2}}\left(|x+y|^{2}-|x|^{2}-2\mathds{1}_{\\{|y|\leq 1\\}}y\cdot
x\right).$

Since
${\mathcal{P}}^{-}_{k}v_{r}={\mathcal{P}}^{-}_{k}(v_{r}+e^{-\eta(4r)^{2}})$
and using above inequality we obtain

$\displaystyle{\mathcal{P}}^{-}_{k}(e^{-\eta q(\cdot)})(x)$
$\displaystyle=-\int_{{\mathds{R}^{d}}}\left(e^{-\eta q(x+y)}-e^{-\eta
q(x)}-\mathds{1}_{{\mathscr{B}}_{1}}(y)\nabla e^{-\eta q(x)}\cdot
y\right)^{-}k(y)dy$ $\displaystyle\geq-\eta e^{-\eta|x|^{2}}\int_{|y|\leq
r}\Big{(}|x+y|^{2}-|x|^{2}-2y\cdot x\Big{)}k(y)\mathrm{d}{y}$
$\displaystyle-\int_{r<|y|\leq
1}\left|e^{-\eta(|x|^{2}+2(4r)^{2})}-e^{-\eta|x|^{2}}+2\eta y\cdot
xe^{-\eta|x|^{2}}\right|k(y)\mathrm{d}{y}$
$\displaystyle-\int_{|y|>1}\left|e^{-\eta(|x|^{2}+2(4r)^{2})}-e^{-\eta|x|^{2}}\right|k(y)\mathrm{d}{y}$
$\displaystyle\geq-\eta e^{-\eta|x|^{2}}\left[\int_{\lvert y\rvert<r}\lvert
y\rvert^{2}k(y)\mathrm{d}{y}+\int_{r<|y|\leq
1}\left(\left|\frac{1-e^{-2\eta(4r)^{2}}}{\eta}\right|+2|x\cdot
y|\right)k(y)\mathrm{d}{y}\right]$ $\displaystyle-\eta
e^{-\eta|x|^{2}}\int_{|y|>1}\left|\frac{1-e^{-2\eta(4r)^{2}}}{\eta}\right|k(y)\mathrm{d}{y}$
$\displaystyle\geq-\eta e^{-\eta|x|^{2}}\left[\int_{\lvert y\rvert<r}\lvert
y\rvert^{2}k(y)\mathrm{d}{y}+\int_{r<\lvert y\rvert\leq 1}4^{3}\lvert
y\rvert^{2}k(y)\mathrm{d}{y}+\int_{\lvert
y\rvert>1}2(4r)^{2}k(y)\mathrm{d}{y}\right]$ $\displaystyle\geq-\eta
e^{-\eta|x|^{2}}4^{3}\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(y)\mathrm{d}{y},$

where in the second line we used $|x+y|^{2}\wedge
2(4r)^{2}\leq|x|^{2}+2(4r)^{2}$. Combining the above estimates we see that,
for $x\in{\mathscr{B}}_{4r}\setminus\bar{{\mathscr{B}}}_{r}$,

$\displaystyle P^{-}v_{r}(x)+P_{k}^{-}v_{r}(x)$
$\displaystyle\geq\eta\mathrm{e}^{-\eta\lvert
x\rvert^{2}}\Bigl{[}4\eta\lambda|x|^{2}-2d\Lambda-4^{3}\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(y)\mathrm{d}{y}\Bigr{]}$
$\displaystyle\geq\eta\mathrm{e}^{-\eta\lvert x\rvert^{2}}\Bigl{[}4\eta\lambda
r^{2}-2d\Lambda-4^{3}\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(y)\mathrm{d}{y}\Bigr{]}.$

Thus, finally letting $\eta=\frac{1}{\lambda
r^{2}}(2d\Lambda+4^{3}\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(y)\mathrm{d}{y})$,
we obtain

${\mathcal{P}}^{-}v_{r}+{\mathcal{P}}^{-}v_{r}>0\quad\text{in}\;{\mathscr{B}}_{4r}\setminus\bar{{\mathscr{B}}}_{r}.$

Note that the final choice of $\eta$ is admissible since $\frac{1}{\lambda
r^{2}}(2d\Lambda+4^{3}\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(y)\mathrm{d}{y})>\frac{1}{r^{2}}.$
Now set $\phi_{r}=rv_{r}$ and the result follows from (4.1)-(4). ∎

Next we prove a weak version of Harnack inequality.

######  Theorem 4.1.

Let $s\in(0,1],$ $\alpha^{\prime}=1\wedge(2-\alpha)$ and $u$ be a continuous
non-negative function satisfying

$\mathcal{P}^{-}u+\mathcal{P}^{-}_{k,s}u\leq
C_{0}s^{1+\alpha^{\prime}},\quad\mathcal{P}^{+}u+\mathcal{P}^{+}_{k,s}u\geq-
C_{0}s^{1+\alpha^{\prime}}\quad\text{in}\;{\mathscr{B}}_{2}.$

Furthermore if $\sup_{{\mathds{R}^{d}}}|u|\leq M_{0}$ and $|u(x)|\leq
M_{0}s(1+|x|)$ for all $x\in{\mathds{R}^{d}},$ then

$u(x)\leq C(u(0)+(M_{0}\vee C_{0})s^{1+\alpha^{\prime}})$

for every $x\in{\mathscr{B}}_{\frac{1}{2}}$ and for some constant $C$ which
only depends on
$\lambda,\Lambda,d,\int_{\mathbb{R}^{d}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}.$

###### Proof.

Dividing by $u(0)+(M_{0}\vee C_{0})s^{1+\alpha^{\prime}},$ it can be easily
seen that $\sup_{{\mathds{R}^{d}}}|u|\leq s^{-(1+\alpha^{\prime})}$ and
$|u(x)|\leq s^{-\alpha^{\prime}}(1+|x|)$ for all $x\in{\mathds{R}^{d}}$ and
$u$ satisfies

$\displaystyle\mathcal{P}^{-}u+\mathcal{P}^{-}_{k,s}u\leq 1,$
$\displaystyle\mathcal{P}^{+}u+\mathcal{P}^{+}_{k,s}u\geq-1.$

Fix $\varepsilon>0$ from [43, Corollary 3.14] and let
$\gamma=\frac{d}{\varepsilon}$. Let

$t_{0}\coloneqq\min\left\\{t\;:\;u(x)\leq h_{t}(x)\coloneqq t(1-\lvert
x\rvert)^{-\gamma}\;\;\text{for all}\;x\in{\mathscr{B}}_{1}\right\\}.$

Clearly this set is nonempty since $u(0)\leq 1$, thus $t_{0}$ exist. Let
$x_{0}\in{\mathscr{B}}_{1}$ be such that $u(x_{0})=h_{t_{0}}(x_{0})$. Let
$\eta=1-|x_{0}|$ be the distance of $x_{0}$ from $\partial{\mathscr{B}}_{1}$.
For $r=\frac{\eta}{2}$ and $x\in{\mathscr{B}}_{r}(x_{0}),$ we can write

${\mathscr{B}}_{r}(x_{0})=\left\\{u(x)\leq\frac{u(x_{0})}{2}\right\\}\cup\left\\{u(x)>\frac{u(x_{0})}{2}\right\\}:=A+\tilde{A}.$

The goal is to estimate $|{\mathscr{B}}_{r}(x_{0})|$ in terms of $|A|$ and
$|\tilde{A}|.$ Proceeding this way, we show that $t_{0}<C$ for some universal
$C$ which, in turn, implies that $u(x)<C(1-\lvert x\rvert)^{-\gamma}$. This
would prove our result. Next, Using [43, Corollary 3.14] we obtain

$|\tilde{A}\cap{\mathscr{B}}_{1}|\leq
C\bigg{\lvert}\frac{2}{u(x_{0})}\bigg{\rvert}^{\varepsilon}\leq
Ct_{0}^{-\varepsilon}\eta^{d}\,,$

whereas $|{\mathscr{B}}_{r}|=\omega_{d}(\eta/2)^{d}$. In particular,

$\big{\lvert}\tilde{A}\cap{\mathscr{B}}_{r}(x_{0})\big{\rvert}\leq
Ct_{0}^{-\varepsilon}|{\mathscr{B}}_{r}|.$ (4.3)

So if $t_{0}$ is large, $\tilde{A}$ can cover only a small portion of
${\mathscr{B}}_{r}(x_{0})$. We shall show that for some $\delta>0$,
independent of $t_{0}$ we have

$|A\cap{\mathscr{B}}_{r}(x_{0})|\leq(1-\delta)|{\mathscr{B}}_{r}|,$

which will provide an upper bound on $t_{0}$ completing the proof. We start by
estimating $|A\cap{\mathscr{B}}_{\uptheta r}(x_{0})|$ for $\uptheta>0$ small.
For every $x\in{\mathscr{B}}_{\uptheta r}(x_{0})$ we have

$u(x)\leq\,h_{t_{0}}(x)\,\leq\,t_{0}\left(\frac{2\eta-\uptheta\eta}{2}\right)^{-\gamma}\,\leq\,u(x_{0})\left(1-\frac{\uptheta}{2}\right)^{-\gamma},$

with $\left(1-\frac{\uptheta}{2}\right)$ close to 1. Define

$v(x)\coloneqq\left(1-\frac{\uptheta}{2}\right)^{-\gamma}u(x_{0})-u(x).$

Then we get $v\geq 0$ in ${\mathscr{B}}_{\uptheta r}(x_{0})$ and also
$\mathcal{P}^{-}v+\mathcal{P}^{-}_{k,s}v\leq 1$ as
$\mathcal{P}^{+}u+\mathcal{P}^{+}_{k,s}u\geq-1.$

We would like to apply [43, Corollary 3.14] to $v,$ but $v$ need not be non-
negative in the whole of ${\mathds{R}^{d}}$. Thus we consider the positive
part of $v,$ i.e, $w=v^{+}$ and find an upper bound of
$\mathcal{P}^{-}w+\mathcal{P}^{-}_{k,s}w$. Since $v^{-}$ is $C^{2}$ in
${\mathscr{B}}_{\frac{\uptheta r}{4}}(x_{0}),$ we have

$\displaystyle\mathcal{P}^{-}w(x)+\mathcal{P}^{-}_{k,s}w(x)\leq[\mathcal{P}^{-}v(x)+\mathcal{P}^{-}_{k,s}v(x)]+[\mathcal{P}^{+}v^{-}(x)+\mathcal{P}^{+}_{k,s}v^{-}(x)]\leq
1+\mathcal{P}^{+}v^{-}(x)+\mathcal{P}^{+}_{k,s}v^{-}(x).$ (4.4)

Also, using $v^{-}(x)=Dv^{-}(x)=D^{2}v^{-}(x)=0$ for all
$x\in{\mathscr{B}}_{\frac{\uptheta r}{4}}(x_{0}),$ we get

$\displaystyle\mathcal{P}^{+}v^{-}(x)+\mathcal{P}^{+}_{k,s}v^{-}(x)=\int_{{\mathds{R}^{d}}\cap\\{v(x+y)\leq
0\\}}v^{-}(x+y)s^{d+2}k(sy)\mathrm{d}{y}.$ (4.5)

Now plugging (4.5) into (4.4), for all $x\in{\mathscr{B}}_{\frac{\uptheta
r}{4}}(x_{0})$ we obtain

$\displaystyle\mathcal{P}^{-}w(x)+\mathcal{P}^{-}_{k,s}w(x)\leq
1+\int_{{\mathds{R}^{d}}\setminus{\mathscr{B}}_{\frac{\uptheta
r}{2}}(x-x_{0})}\left(u(x+y)-\left(1-\frac{\uptheta}{2}\right)^{-\gamma}u(x_{0})\right)^{+}s^{d+2}k(sy)\mathrm{d}{y}$
$\displaystyle\leq
1+\int_{{\mathds{R}^{d}}\setminus{\mathscr{B}}_{\frac{\uptheta
r}{2}}(x-x_{0})}\left|u(x+y)\right|s^{d+2}k(sy)\mathrm{d}{y}+\int_{{\mathds{R}^{d}}\setminus{\mathscr{B}}_{\frac{\uptheta
r}{2}}(x-x_{0})}\left|\left(1-\frac{\uptheta}{2}\right)^{-\gamma}u(x_{0})\right|s^{d+2}k(sy)\mathrm{d}{y}$
$\displaystyle\leq
1+\int_{{\mathds{R}^{d}}\setminus{\mathscr{B}}_{\frac{\uptheta
r}{4}}}\left|u(x+y)\right|s^{d+2}k(sy)\mathrm{d}{y}+\int_{{\mathds{R}^{d}}\setminus{\mathscr{B}}_{\frac{\uptheta
r}{4}}}\left|\left(1-\frac{\uptheta}{2}\right)^{-\gamma}u(x_{0})\right|s^{d+2}k(sy)\mathrm{d}{y}:=1+I_{1}+I_{2}.$

Estimate of $I_{1}$: Let us write

$\displaystyle I_{1}$ $\displaystyle=\int_{\frac{\uptheta
r}{4}\leq|y|\leq\frac{1}{s}}\left|u(x+y)\right|s^{d+2}k(sy)\mathrm{d}{y}+\int_{|y|\geq\frac{1}{s}}\left|u(x+y)\right|s^{d+2}k(sy)\mathrm{d}{y}:=I_{11}+I_{12}.$

Simply using change of variable and $\sup_{{\mathds{R}^{d}}}|u|\leq
s^{-(1+\alpha^{\prime})},$ we obtain

$I_{12}\leq\int_{|z|\geq 1}k(z)\mathrm{d}{z}.$

Now we estimate $I_{11}$ using $|u(x)|\leq s^{-\alpha^{\prime}}(1+|x|)$ for
all $x\in{\mathds{R}^{d}}$.

$\displaystyle I_{11}$ $\displaystyle\leq\int_{\frac{\uptheta
r}{4}\leq|y|\leq\frac{1}{s}}\left(1+|x+y|\right)s^{d+2-\alpha^{\prime}}k(sy)\mathrm{d}{y}$
$\displaystyle\leq\frac{5}{4}\int_{\frac{\uptheta
r}{4}\leq|y|\leq\frac{1}{s}}s^{d+2-\alpha^{\prime}}k(sy)\mathrm{d}{y}+\int_{\frac{\uptheta
r}{4}\leq|y|\leq\frac{1}{s}}s^{d+2-\alpha^{\prime}}|y|k(sy)\mathrm{d}{y}\,.$

We consider two cases. First consider the case $\alpha^{\prime}=1$ so
$\alpha\leq 1$. This implies

$I_{11}\leq\frac{5}{4}\int_{\frac{\uptheta rs}{4}\leq|z|\leq
1}sk(z)\mathrm{d}{z}+\int_{\frac{\uptheta rs}{4}\leq|z|\leq
1}|z|k(z)\mathrm{d}{z}\leq 6(\uptheta
r)^{-1}\int_{{\mathds{R}^{d}}}(1\wedge|z|^{\alpha})k(z)\mathrm{d}{z}.$

Now consider the case $\alpha^{\prime}=2-\alpha,$ and hence $\alpha>1.$ In
this case

$\displaystyle I_{11}$ $\displaystyle\leq\frac{5}{4}\int_{\frac{\uptheta
r}{4}\leq|y|\leq\frac{1}{s}}s^{\alpha}s^{d}k(sy)\mathrm{d}{y}+\int_{\frac{\uptheta
r}{4}\leq|y|\leq\frac{1}{s}}s^{\alpha-1}|sy|s^{d}k(sy)\mathrm{d}{y}$
$\displaystyle=5\cdot 4^{\alpha-1}(\uptheta r)^{-\alpha}\int_{\frac{\uptheta
rs}{4}\leq|z|\leq 1}\left(\frac{\uptheta
r}{4}s\right)^{\alpha}k(z)\mathrm{d}{z}+\left(\frac{\uptheta
r}{4}\right)^{1-\alpha}\int_{\frac{\uptheta rs}{4}\leq|z|\leq
1}\left(\frac{\uptheta r}{4}s\right)^{\alpha-1}|z|k(z)\mathrm{d}{z}$
$\displaystyle\leq C(\uptheta
r)^{-2}\int_{{\mathds{R}^{d}}}(1\wedge|z|^{\alpha})k(z)\mathrm{d}{z}\,.$

Combining the estimates of $I_{11}$ and $I_{12},$ we get

$\displaystyle I_{1}\leq C(\uptheta
r)^{-2}\int_{{\mathds{R}^{d}}}(1\wedge|z|^{\alpha})k(z)\mathrm{d}{z}.$

Estimate of $I_{2}$: If $\alpha^{\prime}=1,$ then $\alpha\leq 1$ and using
$|u(x_{0})|\leq s^{-\alpha^{\prime}}(1+|x_{0}|)$ we have

$\displaystyle
I_{2}:=\int_{{\mathds{R}^{d}}\setminus{\mathscr{B}}_{\frac{\uptheta
r}{4}}}\left|\left(1-\frac{\uptheta}{2}\right)^{-\gamma}u(x_{0})\right|s^{d+2}k(sy)\mathrm{d}{y}\leq
C\int_{{\mathds{R}^{d}}\setminus{\mathscr{B}}_{\frac{\uptheta
r}{4}}}s^{d+2-\alpha^{\prime}}k(sy)\mathrm{d}{y}$
$\displaystyle=C\int_{{\mathds{R}^{d}}\setminus{\mathscr{B}}_{\frac{\uptheta
rs}{4}}}sk(z)\mathrm{d}{z}\leq C\left[\int_{\frac{\uptheta rs}{4}\leq|z|\leq
1}sk(z)\mathrm{d}{z}+\int_{|z|\geq 1}sk(z)\mathrm{d}{z}\right]$
$\displaystyle\leq C\left[\frac{4}{\uptheta r}\int_{\frac{\uptheta
rs}{4}\leq|z|\leq
1}|z|^{\alpha}k(z)\mathrm{d}{z}+\int_{|z|>1}k(z)\mathrm{d}{z}\right]\leq
C(\uptheta
r)^{-1}\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(z)\mathrm{d}{z}.$

If $\alpha^{\prime}=2-\alpha$ then $\alpha>1.$ In that case, using similar
calculation as above we have

$\displaystyle
I_{2}:=\int_{{\mathds{R}^{d}}\setminus{\mathscr{B}}_{\frac{\uptheta
r}{4}}}\left|\left(1-\frac{\uptheta}{2}\right)^{-\gamma}u(x_{0})\right|s^{d+2}k(sy)\mathrm{d}{y}\leq
C\int_{{\mathds{R}^{d}}\setminus{\mathscr{B}}_{\frac{\uptheta
r}{4}}}s^{d+\alpha}k(sy)\mathrm{d}{y}$
$\displaystyle=C\int_{{\mathds{R}^{d}}\setminus{\mathscr{B}}_{\frac{\uptheta
rs}{4}}}s^{\alpha}k(z)\mathrm{d}{z}\leq C(\uptheta
r)^{-\alpha}\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(z)\mathrm{d}{z}.$

Since $\alpha\in(0,2),$ combining the above estimates we obtain

$\mathcal{P}^{-}w+\mathcal{P}^{-}_{k,s}w\leq\frac{C}{(\uptheta
r)^{2}}\,\,\,\,\textnormal{in}\,\,\,\,{\mathscr{B}}_{\frac{\uptheta
r}{4}}(x_{0})\,.$

Now using [43, Corollary 3.14] for $w$ we get

$\displaystyle|A\cap{\mathscr{B}}_{\frac{\uptheta
r}{8}}(x_{0})|=\bigg{\lvert}\bigg{\\{}w\geq
u(x_{0})((1-\nicefrac{{\uptheta}}{{2}})^{-\gamma}-\nicefrac{{1}}{{2}})\bigg{\\}}\cap{\mathscr{B}}_{\frac{\uptheta
r}{8}}(x_{0})\bigg{\rvert}$ $\displaystyle\qquad\leq C(\uptheta
r)^{d}\left(\inf_{{\mathscr{B}}_{\frac{\uptheta r}{8}}(x_{0})}w+\frac{\uptheta
r}{8}\cdot\frac{C}{(\uptheta
r)^{2}}\right)^{\varepsilon}\cdot\left[u(x_{0})((1-\nicefrac{{\uptheta}}{{2}})^{-\gamma}-\nicefrac{{1}}{{2}})\right]^{-\varepsilon}$
$\displaystyle\quad\quad\leq C(\uptheta
r)^{d}\Big{[}\left((1-\frac{\uptheta}{2})^{-\gamma}-\frac{1}{2}\right)+\frac{C}{8}(\uptheta
r)^{-1}t_{0}^{-1}(2r)^{d}\Big{]}^{\varepsilon}$ $\displaystyle\quad\quad\leq
C(\uptheta
r)^{d}\left(\left((1-\nicefrac{{\uptheta}}{{2}})^{-\gamma}-1\right)^{\varepsilon}+C_{0}(\uptheta
r)^{-\varepsilon}t_{0}^{-\varepsilon}r^{d\varepsilon}\right)\,.$

Now let us choose $\uptheta>0$ small enough (independent of $t_{0}$) to
satisfy

$C(\uptheta
r)^{d}\left((1-\nicefrac{{\uptheta}}{{2}})^{-\gamma}-1\right)^{\varepsilon}\leq\;\frac{1}{4}|{\mathscr{B}}_{\frac{\uptheta
r}{8}}(x_{0})|\,.$

With this choice of $\uptheta$ if $t_{0}$ becomes large, then we also have

$C(\uptheta
r)^{d}\uptheta^{-\varepsilon}r^{(n-1)\varepsilon}t_{0}^{-\varepsilon}\leq\frac{1}{4}|{\mathscr{B}}_{\frac{\uptheta
r}{8}}(x_{0})|\,,$

and hence

$|A\cap{\mathscr{B}}_{\frac{\uptheta
r}{8}}(x_{0})|\leq\frac{1}{2}|{\mathscr{B}}_{\frac{\uptheta r}{8}}(x_{0})|\,.$

This estimate of course implies that

$|\tilde{A}\cap{\mathscr{B}}_{\frac{\uptheta r}{8}}(x_{0})|\geq
C_{2}|{\mathscr{B}}_{r}|,$

but this is contradicting (4.3). Therefore $t_{0}$ cannot be large and this
completes the proof. ∎

######  Corollary 4.1.

Let $u$ satisfies the conditions of Theorem 4.1, then the following holds.

$\sup_{{\mathscr{B}}_{\frac{1}{4}}}u\leq
C\left(\inf_{{\mathscr{B}}_{\frac{1}{4}}}u+(M_{0}\vee
C_{0})s^{1+\alpha^{\prime}}\right).$

###### Proof.

Take any point $x_{0}\in\overline{{\mathscr{B}}_{\frac{1}{4}}}$ such that
$u(x_{0})=\inf_{{\mathscr{B}}_{\frac{1}{4}}}u(x).$ Clearly
${\mathscr{B}}_{\frac{1}{4}}\subset{\mathscr{B}}_{\frac{1}{2}}(x_{0}).$
Defining $\tilde{u}(x):=u(x+x_{0})$ and applying Theorem 4.1 on $\tilde{u}$ we
find

$\tilde{u}(x)\leq C\left(\tilde{u}(0)+(M_{0}\vee
C_{0})s^{1+\alpha^{\prime}}\right)\,\,\,\text{in}\,\,\,\,{\mathscr{B}}_{\frac{1}{2}}.$

This implies

$\sup_{{\mathscr{B}}_{\frac{1}{4}}}u(x)\leq\sup_{{\mathscr{B}}_{\frac{1}{2}}(x_{0})}u(x)\leq
C\left(\inf_{{\mathscr{B}}_{\frac{1}{4}}}u(x)+(M_{0}\vee
C_{0})s^{1+\alpha^{\prime}}\right).$

This proves the claim. ∎

Now we will give some auxiliary lemmas which will be used to construct
appropriate supersolutions that are crucial to prove the oscillation estimate.

######  Lemma 4.2.

Let $\Omega$ be a bounded $C^{2}$ domain in ${\mathds{R}^{d}}$, then for any
$0<\upepsilon<1,$ we have the following estimate

$\left|{\mathcal{I}}_{\theta\nu}(\delta^{1+\upepsilon})\right|\leq
C\left(1+\mathds{1}_{(1,2)}(\alpha)\delta^{1-\alpha}\right)\,\,\,\,\text{in}\,\,\,\,\Omega,$
(4.6)

where $C>0$ depends only on $d,\Omega$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}.$

###### Proof.

Since $\delta\in C^{0,1}({\mathds{R}^{d}})\cap C^{2}(\bar{\Omega})$ [21,
Theorem 5.4.3], using the Lipschtiz continuity of $\delta^{1+\upepsilon}$ near
the origin and boundedness away from the origin we can easily obtain the
estimate (4.6) for $\alpha\in(0,1].$ Next consider the case $\alpha\in(1,2).$
For any $x\in\Omega$ we have

$\displaystyle\left|{\mathcal{I}}_{\theta\nu}(\delta^{1+\upepsilon})(x)\right|\leq$
$\displaystyle\int_{{\mathds{R}^{d}}}\left|\delta^{1+\upepsilon}(x+y)-\delta^{1+\upepsilon}(x)-\mathds{1}_{{\mathscr{B}}_{1}}(y)y\cdot\nabla\delta^{1+\upepsilon}(x)\right|k(y)\mathrm{d}{y}$
$\displaystyle=\int_{|y|<\frac{\delta(x)}{2}}+\int_{\frac{\delta(x)}{2}\leq|y|\leq
1}+\int_{|y|>1}:=I_{1}+I_{2}+I_{3}\,.$

Since $|y|\leq\frac{\delta(x)}{2}$ and $\delta(x)<1,$ we have the following
estimate on $I_{1}.$

$\displaystyle\left|\delta^{1+\upepsilon}(x+y)-\delta^{1+\upepsilon}(x)-\mathds{1}_{{\mathscr{B}}_{1}}(y)y\cdot\nabla\delta^{1+\upepsilon}(x)\right|$
$\displaystyle\leq||\delta^{1+\upepsilon}||_{C^{2}({\mathscr{B}}_{\frac{\delta(x)}{2}}(x))}|y|^{2}$
$\displaystyle\leq
4C\frac{||\delta||_{C^{2}(\bar{\Omega})}}{\delta(x)^{1-\upepsilon}}|y|^{2}$
$\displaystyle\leq
4C\frac{||\delta||_{C^{2}(\bar{\Omega})}\delta(x)^{2-\alpha}}{\delta(x)^{1-\upepsilon}}|y|^{\alpha}.$

This implies

$\displaystyle I_{1}\leq
4C||\delta||_{C^{2}(\bar{\Omega})}\delta(x)^{1+\upepsilon-\alpha}\int_{{\mathds{R}^{d}}}|y|^{\alpha}k(y)\mathrm{d}{y}\leq
4C_{0}C||\delta||_{C^{2}(\bar{\Omega})}\delta(x)^{1+\upepsilon-\alpha}.$ (4.7)

Again for $I_{2}$ we have

$\displaystyle I_{2}\leq C\int_{\frac{\delta(x)}{2}\leq|y|\leq
1}|y|k(y)\mathrm{d}{y}$
$\displaystyle\leq\left(\frac{C\delta(x)}{2}\right)^{1-\alpha}\int_{\frac{\delta(x)}{2}\leq|y|\leq
1}|y|^{\alpha}k(y)\mathrm{d}{y}$
$\displaystyle\leq\left(\frac{C\delta(x)}{2}\right)^{1-\alpha}\int_{{\mathds{R}^{d}}}\left(1\wedge|y|^{\alpha}\right)k(y)\mathrm{d}{y}.$

Finally,

$\displaystyle
I_{3}=\int_{|y|>1}|\delta^{1+\upepsilon}(x+y)-\delta^{1+\upepsilon}(x)|k(y)\mathrm{d}{y}\leq
2(\operatorname*{diam}\Omega)^{1+\upepsilon}\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}.$
(4.8)

Combining (4.7)-(4.8) we obtain (4.6). ∎

Next we obtain an estimate on minimal Pucci operator ${\mathcal{P}}^{-}$
applied on $\delta^{1+\upepsilon}.$

######  Lemma 4.3.

Let $\Omega$ be a bounded $C^{2}$ domain in ${\mathds{R}^{d}}$, then for any
$0<\upepsilon<1,$ we have the following estimate

$\displaystyle\mathcal{P}^{-}\left(\delta^{1+\upepsilon}\right)\geq
C_{1}\cdot\upepsilon\delta^{\upepsilon-1}-C_{2}\,\,\,\,\text{in}\,\,\,\,\Omega,$

where $C_{1},C_{2}$ depends only on $d,\Omega,\lambda,\Lambda$.

###### Proof.

Since $\partial\Omega$ is $C^{2},$ we have $\delta^{1+\upepsilon}\in
C^{2}(\Omega)$ and for any $x\in\Omega$

$\displaystyle\frac{\partial^{2}}{\partial x_{i}\partial
x_{j}}\delta^{1+\upepsilon}(x)=(1+\upepsilon)\Big{[}\delta^{\upepsilon}(x)\frac{\partial^{2}}{\partial
x_{i}\partial
x_{j}}\delta(x)+\upepsilon\delta^{\upepsilon-1}(x)\frac{\partial\delta(x)}{\partial
x_{i}}\cdot\frac{\partial\delta(x)}{\partial x_{j}}\Big{]}:=A+B$

where $A,B$ are two $d\times d$ matrices given by

$A:=(a_{i,j})_{1\leq i,j\leq
d}=(1+\upepsilon)\delta^{\upepsilon}(x)\frac{\partial^{2}}{\partial
x_{i}\partial x_{j}}\delta(x)$

and

$B:=(b_{i,j})_{1\leq i,j\leq
d}=(1+\upepsilon)\upepsilon\delta^{\upepsilon-1}(x)\frac{\partial\delta(x)}{\partial
x_{i}}\cdot\frac{\partial\delta(x)}{\partial x_{j}}.$

Note that $B$ is a positive definite matrix and $||A||\leq
d^{2}(1+\upepsilon)(\operatorname*{diam}\Omega)^{\upepsilon}||\delta||_{C^{2}(\bar{\Omega})}.$
Therefore we have

$\displaystyle\mathcal{P}^{-}(\delta^{1+\upepsilon}(x))=\mathcal{P}^{-}(A+B)$
$\displaystyle\geq\mathcal{P}^{-}(B)+\mathcal{P}^{-}(A)$
$\displaystyle\geq\mathcal{P}^{-}(B)-d^{2}\Lambda(1+\upepsilon)(\operatorname*{diam}\Omega)^{\upepsilon}||\delta||_{C^{2}(\bar{\Omega})}$
$\displaystyle\geq\upepsilon(1+\epsilon)\delta^{\upepsilon-1}(x)\lambda|D\delta(x)|^{2}-d^{2}\Lambda(1+\upepsilon)(\operatorname*{diam}\Omega)^{\upepsilon}||\delta||_{C^{2}(\bar{\Omega})}$
$\displaystyle\geq C_{1}\cdot\upepsilon\delta^{\upepsilon-1}(x)-C_{2}.$

∎

Next we obtain an estimate on ${\mathcal{L}}\delta$ in $\Omega.$

######  Lemma 4.4.

Let $\Omega$ be a bounded $C^{2}$ domain in ${\mathds{R}^{d}}$. Then we have
the following estimate

$\displaystyle|{\mathcal{L}}\delta|\leq
C(1+\mathds{1}_{(1,2)}\delta^{1-\alpha})\,\,\,in\,\,\,\Omega,$ (4.9)

where constant $C$ depends only on $d,\Omega,\lambda,\Lambda$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}.$

###### Proof.

First of all, for all $x\in\Omega$ we have

$\displaystyle|{\mathcal{L}}\delta(x)|\leq\sup_{\theta,\nu}|\operatorname*{Tr}(a_{\theta\nu}(x)D^{2}\delta(x))|+\sup_{\theta,\nu}|{\mathcal{I}}_{\theta\nu}\delta(x)|\leq\kappa+\sup_{\theta,\nu}|{\mathcal{I}}_{\theta\nu}\delta(x)|,$
(4.10)

for some constant $\kappa$, depending on $\Omega$ and uniform bound of
$a_{\theta\nu}$. For $\alpha\in(0,1],$ (4.9) follows from the same arguments
of Lemma 4.2. For $\alpha\in(1,2),$ it is enough to obtain the estimate (4.9)
for all $x\in\Omega$ such that $\delta(x)<1$. We follow the similar
calculation as in Lemma 4.2 and get

$\displaystyle|{\mathcal{I}}_{\theta\nu}\delta(x)|$
$\displaystyle\leq\int_{{\mathds{R}^{d}}}|\delta(x+y)-\delta(x)-\mathds{1}_{B_{1}}(y)y\cdot\nabla\delta(x)|k(y)\mathrm{d}{y}$
$\displaystyle=\int_{|y|\leq\frac{\delta(x)}{2}}+\int_{\frac{\delta(x)}{2}<|y|<1}+\int_{|y|>1}$

and

$\displaystyle|I_{\theta\nu}\delta(x)|\leq\kappa_{1}\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}\delta^{1-\alpha}(x)$

for some constant $\kappa_{1}$. Inserting these estimates in (4.10) we obtain

$|{\mathcal{L}}\delta(x)|\leq\kappa_{2}\delta^{1-\alpha}(x)$

for some constant $\kappa_{2}$ and (4.9) follows. ∎

Let us now define the sets that we use for our oscillation estimates. We
borrow the notations of [46].

######  Definition 4.1.

Let $\kappa\in(0,\frac{1}{16})$ be a fixed small constant and let
$\kappa^{\prime}=1/2+2\kappa$. Given a point $x_{0}\in\partial\Omega$ and
$R>0$, we define

$\mathscr{D}_{R}=\mathscr{D}_{R}(x_{0})={\mathscr{B}}_{R}(x_{0})\cap\Omega,$

and

$\mathscr{D}^{+}_{\kappa^{\prime}R}=\mathscr{D}^{+}_{\kappa^{\prime}R}(x_{0})={\mathscr{B}}_{\kappa^{\prime}R}(x_{0})\cap\left\\{x\in\Omega:(x-x_{0})\cdot{\rm
n}(x_{0})\geq 2\kappa R\right\\},$

where ${\rm n}(x_{0})$ is the unit inward normal at $x_{0}$. For any bounded
$C^{1,1}$-domain, we know that there exists $\rho>0$, depending on $\Omega$,
such that the following inclusions hold for each $x_{0}\in\partial\Omega$ and
$R\leq\rho$:

${\mathscr{B}}_{\kappa R}(y)\subset\mathscr{D}_{R}(x_{0})\qquad\text{ for
all}\;y\in\mathscr{D}^{+}_{\kappa^{\prime}R}(x_{0}),$ (4.11)

and

${\mathscr{B}}_{4\kappa R}(y^{\ast}+4\kappa R{\rm
n}(y^{\ast}))\subset\mathscr{D}_{R}(x_{0}),\quad\text{and}\quad{\mathscr{B}}_{\kappa
R}(y^{\ast}+4\kappa R{\rm
n}(y^{\ast}))\subset\mathscr{D}^{+}_{\kappa^{\prime}R}(x_{0})$ (4.12)

for all $y\in\mathscr{D}_{R/2}$, where $y^{\ast}\in\partial\Omega$ is the
unique boundary point satisfying
$|y-y^{\ast}|=\operatorname*{dist}(y,\partial\Omega)$. Note that, since
$R\leq\rho$, $y\in\mathscr{D}_{R/2}$ is close enough to $\partial\Omega$ and
hence the point $y^{\ast}+4\kappa R\,{\rm n}(y^{\ast})$ belongs to the line
joining $y$ and $y^{\ast}$.

######  Remark 4.1.

In the remaining part of this section, we fix $\rho>0$ to be a small constant
depending only on $\Omega$, so that (4.11)-(4.12) hold whenever $R\leq\rho$
and $x_{0}\in\partial\Omega$. Also, every point on $\partial\Omega$ can be
touched from both inside and outside $\Omega$ by balls of radius $\rho$. We
also fix $\upsigma>0$ small enough so that for $0<r\leq\rho$ and
$x_{0}\in\partial\Omega$ we have

${\mathscr{B}}_{\eta
r}(x_{0})\cap\Omega\subset{\mathscr{B}}_{(1+\sigma)r}(z)\setminus\bar{{\mathscr{B}}}_{r}(z)\quad\text{for}\quad\eta=\sigma/8,\,\sigma\in(0,\upgamma),$

for any $x^{\prime}\in\partial\Omega\cap{\mathscr{B}}_{\eta r}(x_{0})$, where
${\mathscr{B}}_{r}(z)$ is a ball contained in
${\mathds{R}^{d}}\setminus\Omega$ that touches $\partial\Omega$ at point
$x^{\prime}$.

In the following lemma, using Lemma 4.2 and Lemma 4.3 we construct
supersolutions. We denote
$\Omega_{\rho}:=\left\\{x\in\Omega|\operatorname*{dist}(x,\Omega^{c})<\rho\right\\}$.

######  Lemma 4.5.

Let $\Omega$ be a bounded $C^{2}$ domain in ${\mathds{R}^{d}}$ and
$\alpha\in(1,2),$ then there exist $\rho_{1}>0$ and a $C^{2}$ function
$\phi_{1}$ satisfying

$\begin{cases}{\mathcal{P}}^{+}\phi_{1}(x)+{\mathcal{P}}^{+}_{k}\phi_{1}(x)\leq-C\delta^{-\frac{\alpha}{2}}(x)&\quad\text{in
}\quad\Omega_{\rho_{1}},\\\ C^{-1}\delta(x)\leq\phi_{1}(x)\leq
C\delta(x)&\quad\text{in}\quad\Omega,\\\
\phi_{1}(x)=0&\quad\text{in}\quad{\mathds{R}^{d}}\setminus\Omega,\end{cases}$

where the constants $\rho_{1}$ and $C$ depend only on
$d,\alpha,\Omega,\lambda,\Lambda$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}$.

###### Proof.

Let $\upepsilon=\frac{2-\alpha}{2}$ and
$c=\frac{1}{(\operatorname*{diam}\Omega)^{2}}$, and define

$\phi_{1}(x)=\delta(x)-c\delta^{1+\upepsilon}(x).$

Since both $\delta$ and $\delta^{1+\upepsilon}$ are in $C^{2}(\Omega)$, we
have
${\mathcal{P}}^{+}\phi_{1}(x)\leq{\mathcal{P}}^{+}\delta(x)-c{\mathcal{P}}^{-}\delta^{1+\upepsilon}(x).$
Then by Lemma 4.3 and
$\sup_{\theta\nu}|\operatorname*{Tr}(a_{\theta\nu}(x)D^{2}\delta(x))|\leq\tilde{C}$,
we get for all $x\in\Omega_{\rho}$

$\displaystyle{\mathcal{P}}^{+}\phi_{1}(x)$
$\displaystyle\leq{\mathcal{P}}^{+}\delta(x)-c{\mathcal{P}}^{-}\delta^{1+\epsilon}(x)\leq
C-c(C_{1}\cdot\upepsilon\delta^{\upepsilon-1}(x)).$

Similarly for all $x\in\Omega_{\rho}$, using Lemma 4.2 and Lemma 4.4 we get

${\mathcal{P}}^{+}_{k}\phi_{1}(x)\leq|{\mathcal{P}}^{+}_{k}\delta(x)|+c|{\mathcal{P}}^{-}_{k}\delta^{1+\upepsilon}(x)|\leq
C_{2}\delta^{1-\alpha}(x).$

Combining the above inequalities we have

$\displaystyle{\mathcal{P}}^{+}\phi_{1}(x)+{\mathcal{P}}_{k}^{+}\phi_{1}(x)$
$\displaystyle\leq
C-cC_{1}\upepsilon\delta^{\upepsilon-1}(x)+C_{2}\delta^{1-\alpha}(x)$
$\displaystyle\leq-\delta^{\upepsilon-1}(x)\left(\frac{C_{1}(2-\alpha)}{2(\operatorname*{diam}\Omega)^{2}}-C\delta^{\frac{\alpha}{2}}(x)-C_{2}\delta^{\frac{2-\alpha}{2}}(x)\right),$

for all $x\in\Omega_{\rho}$. Now choose $0<\rho_{1}\leq\rho<1$ such that

$\left(\frac{C_{1}(2-\alpha)}{2(\operatorname*{diam}\Omega)^{2}}-C\rho_{1}^{\frac{\alpha}{2}}-C_{2}\rho_{1}^{\frac{2-\alpha}{2}}\right)\geq\frac{C_{1}(2-\alpha)}{4(\operatorname*{diam}\Omega)^{2}}.$

Thus for all $x\in\Omega_{\rho_{1}},$ we have

${\mathcal{P}}^{+}\phi_{1}(x)+{\mathcal{P}}_{k}^{+}\phi_{1}(x)\leq-\frac{C_{1}(2-\alpha)}{4(\operatorname*{diam}\Omega)^{2}}\delta^{-\frac{\alpha}{2}}(x).$

Finally the construction of $\phi_{1}$ immediately gives us that

$C^{-1}\delta(x)\leq\phi_{1}(x)\leq C\delta(x)\quad\text{in}\quad\Omega,$

and $\phi_{1}=0$ in $\Omega^{c}.$ This completes the proof of the lemma. ∎

As mentioned earlier, the key step of proving Theorem 1.2 is to obtain the
oscillation lemma Proposition 4.1. For this we next prove two preparatory
lemmas. In the first lemma we obtain a lower bound of
$\inf_{\mathscr{D}_{\frac{R}{2}}}\frac{u}{\delta}$ whereas the second lemma
controls $\sup_{\mathscr{D}^{+}_{\kappa^{\prime}R}}\frac{u}{\delta}$ by using
that lower bound.

######  Lemma 4.6.

Let $\alpha\in(0,2)$ and $\Omega$ be a bounded $C^{2}$ domain in
${\mathds{R}^{d}}$. Also, let $u$ be such that $u\geq 0$ in
${\mathds{R}^{d}}$, and $\lvert{\mathcal{L}}u\rvert\leq
C_{2}(1+\mathds{1}_{(1,2)}(\alpha)\delta^{1-\alpha})$ in $\mathscr{D}_{R}$,
for some constant $C_{2}$. If $\hat{\alpha}$ is given by

$\displaystyle\hat{\alpha}=\begin{cases}1\,\,\,\,&if\,\,\,\,\alpha\in(0,1],\\\
\frac{2-\alpha}{2}\,\,\,\,&if\,\,\,\,\alpha\in(1,2),\end{cases}$

then there exists a positive constant $C$ depending only on
$d,\Omega,\Lambda,\lambda,\alpha,\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}$,
such that

$\inf_{\mathscr{D}^{+}_{\kappa^{\prime}R}}\frac{u}{\delta}\leq
C\left(\inf_{\mathscr{D}_{\frac{R}{2}}}\frac{u}{\delta}+C_{2}R^{\hat{\alpha}}\right)$
(4.13)

for all $R\leq\rho_{0}$, where the constant $\rho_{0}$ depends only on
$d,\Omega,\lambda,\Lambda,\alpha$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}$.

###### Proof.

Suppose $R\leq\eta\rho$, where $\rho$ is given by Remark 4.1 and $\eta\leq 1$
be some constant that will be chosen later. Define
$m=\inf_{\mathscr{D}^{+}_{\kappa^{{}^{\prime}}R}}u/\delta\geq 0$. Let us first
observe that by (4.11) we have,

$u\geq m\delta\geq m\,(\kappa
R)\quad\text{in}\;\;\mathscr{D}^{+}_{\kappa^{\prime}R}.$ (4.14)

Moreover by (4.12), for any $y\in\mathscr{D}_{R/2},$ we have either
$y\in\mathscr{D}^{+}_{\kappa^{\prime}R}$ or $\delta(y)<4\kappa R.$ If
$y\in\mathscr{D}^{+}_{\kappa^{\prime}R},$ then by the definition of $m$ we get
$m\leq u(y)/\delta(y).$

Next we consider $\delta(y)<4\kappa R$. Let $y^{\ast}$ be the nearest point to
$y$ on $\partial\Omega,$ i.e,
$\operatorname*{dist}(y,\partial\Omega)=|y-y^{*}|$ and define
$\tilde{y}=y^{\ast}+4\kappa R\,{\rm n}(y^{\ast})$. Again by (4.12), we have

${\mathscr{B}}_{4\kappa
R}(\tilde{y})\subset\mathscr{D}_{R}\,\,\,\,\text{and}\,\,\,\,{\mathscr{B}}_{\kappa
R}(\tilde{y})\subset\mathscr{D}^{+}_{\kappa^{\prime}R}.$

Denoting $r=\kappa R$ and using the subsolution constructed in Lemma 4.1,
define $\tilde{\phi}_{r}(x):=\frac{1}{\tilde{\kappa}}\phi_{r}(x-\tilde{y}).$
We will consider two cases.
Case 1: $\alpha\in(0,1]$. Take $r^{\prime}=\frac{R}{\eta}$. Since
$r^{\prime}\leq\rho$, points of $\partial\Omega$ can be touched by exterior
ball of radius $r^{\prime}$. In particular, for $y^{\ast}\in\partial\Omega$,
we can find a point $z\in\Omega^{c}$ such that
$\bar{{\mathscr{B}}}_{r^{\prime}}(z)\subset\Omega^{c}$ touches
$\partial\Omega$ at $y^{\ast}$. Now from [43, Lemma 5.4] there exists a
bounded, Lipschitz continuous function $\varphi_{r^{\prime}}$, with Lipschitz
constant $\frac{1}{r^{\prime}}$, that satisfies

$\begin{cases}\varphi_{r^{\prime}}=0,&\quad\text{in}\quad\bar{{\mathscr{B}}}_{r^{\prime}},\\\
\varphi_{r^{\prime}}>0,&\quad\text{in}\quad\bar{{\mathscr{B}}}_{r^{\prime}}^{c},\\\
{\mathcal{P}}^{+}\varphi_{r^{\prime}}+{\mathcal{P}}^{+}_{k}\varphi_{r^{\prime}}\leq-\frac{1}{(r^{\prime})^{2}},&\quad\text{in}\quad{\mathscr{B}}_{(1+\sigma)r^{\prime}}\setminus\bar{{\mathscr{B}}}_{r^{\prime}},\end{cases}$

for some constant $\sigma$, independent of $r^{\prime}$. Without any loss of
any generality we may assume $\sigma\leq\upgamma$ (see Remark 4.1). Then
setting $\eta=\frac{\sigma}{8}$ and using Remark 4.1, we have

$\mathscr{D}_{R}\subset{\mathscr{B}}_{(1+\sigma)r^{\prime}}(z)\setminus\overline{{\mathscr{B}}}_{r^{\prime}}(z)$

and by (4.12) we have

${\mathscr{B}}_{4r}(\tilde{y})\setminus\overline{{\mathscr{B}}}_{r}(\tilde{y})\subset\mathscr{D}_{R}\subset{\mathscr{B}}_{(1+\sigma)r^{\prime}}(z)\setminus\overline{{\mathscr{B}}}_{r^{\prime}}(z).$

We show that
$v(x)=m\tilde{\phi}_{r}(x)-C_{2}(r^{\prime})^{2}\varphi_{r^{\prime}}(x-z)$ is
an appropriate subsolution. Since both $\tilde{\phi}_{r}$ and
$\varphi_{r^{\prime}}$ are $C^{2}$ functions in
${\mathscr{B}}_{4r}(\tilde{y})\setminus\bar{{\mathscr{B}}}_{r}(\tilde{y}),$ we
conclude that $v$ is $C^{2}$ function in
${\mathscr{B}}_{4r}(\tilde{y})\setminus\bar{{\mathscr{B}}}_{r}(\tilde{y}).$
For
$x\in{\mathscr{B}}_{4r}(\tilde{y})\setminus\bar{{\mathscr{B}}}_{r}(\tilde{y}),$

${\mathcal{P}}^{-}v(x)+{\mathcal{P}}^{-}_{k}v(x)\geq
m\left[{\mathcal{P}}^{-}\tilde{\phi}_{r}(x)+{\mathcal{P}}^{-}_{k}\tilde{\phi}_{r}(x)\right]-C_{2}(r^{\prime})^{2}\left[{\mathcal{P}}^{+}\varphi_{r^{\prime}}(x-z)+{\mathcal{P}}^{+}_{k}\varphi_{r^{\prime}}(x-z)\right]\geq
C_{2}.$

Therefore by Remark 2.1 we have

${\mathcal{P}}^{+}(v-u)+{\mathcal{P}}^{+}_{k}(v-u)\geq
0\,\,\,\,\text{in}\,\,\,\,{\mathscr{B}}_{4r}(\tilde{y})\setminus\bar{{\mathscr{B}}}_{r}(\tilde{y}).$

Furthermore, using (4.14) and $u\geq 0$ in ${\mathds{R}^{d}}$ we obtain
$u(x)\geq m\tilde{\phi}_{r}(x)-C_{2}(r^{\prime})^{2}\varphi_{r^{\prime}}(x-z)$
in
$\left({\mathscr{B}}_{4r}(\tilde{y})\setminus\bar{{\mathscr{B}}}_{r}(\tilde{y})\right)^{c}.$
Hence an application of maximum principle (cf. Lemma 2.2) gives $u\geq v$ in
${\mathds{R}^{d}}.$ Now for $y\in\mathscr{D}_{R/2}$, using the Lipschitz
continuity of $\varphi_{r^{\prime}}$ we get

$\displaystyle m\tilde{\phi}_{r}(y)\leq
u(y)+C_{2}(r^{\prime})^{2}\left[\varphi_{r^{\prime}}(y-z)-\varphi_{r^{\prime}}(y^{*}-z)\right]\leq
u(y)+C_{2}{r^{\prime}}\cdot\delta(y)$

and as $y$ lies on the line segment joining $y^{*}$ to $\tilde{y}$ we get

$\frac{u(y)}{\delta(y)}+C_{2}r^{\prime}\geq\frac{m}{(\tilde{\kappa})^{2}}.$

This gives

$\inf_{\mathscr{D}^{+}_{\kappa^{\prime}R}}\frac{u}{\delta}\leq
C\left(\inf_{\mathscr{D}_{R/2}}\frac{u}{\delta}+C_{2}\frac{R}{\eta}\right)$

and finally choosing $\rho_{0}=\eta\rho$ we have (4.13).

Case 2: $\alpha\in(1,2)$. Let $\rho_{1}$ as in Lemma 4.5 and consider
$R\leq\rho_{1}<1$. Here we aim to construct an appropriate subsolution using
$\tilde{\phi}_{r}(x)$ and supersolution constructed in Lemma 4.5. Since
$\delta(x)\leq 1$ in $\mathscr{D}_{R},$ we have
$\lvert{\mathcal{L}}u(x)\rvert\leq C_{2}(1+\delta^{1-\alpha}(x))\leq
2C_{2}\delta^{1-\alpha}(x)$ in $\mathscr{D}_{R}$. Also by Lemma 4.5, we have a
bounded function $\phi_{1}$ which is $C^{2}$ in
$\Omega_{\rho_{1}}\supset\mathscr{D}_{R}$ and satisfies

$\displaystyle{\mathcal{P}}^{+}\phi_{1}(x)+{\mathcal{P}}^{+}_{k}\phi_{1}(x)\leq-C\delta^{-\frac{\alpha}{2}}(x)=-C\frac{1}{\delta^{\frac{2-\alpha}{2}}(x)}\delta^{1-\alpha}(x)\leq\frac{-C}{R^{\hat{\alpha}}}\delta^{1-\alpha}(x),$

for all $x\in\mathscr{D}_{R}$. Now we define the subsolutions as

$v(x)=m\tilde{\phi}_{r}(x)-\upmu\,R^{\hat{\alpha}}\phi_{1}(x),$

where the constant $\upmu$ is chosen suitably so that
${\mathcal{P}}^{-}v(x)+{\mathcal{P}}^{-}_{k}v(x)\geq
2C_{2}\delta^{1-\alpha}(x)$ in
${\mathscr{B}}_{4r}(\tilde{y})\setminus\bar{{\mathscr{B}}}_{r}(\tilde{y})$
(i.e. $\upmu=\frac{2C_{2}}{C}$). Also $u\geq v$ in
$({\mathscr{B}}_{4r}(\tilde{y})\setminus\bar{{\mathscr{B}}}_{r}(\tilde{y}))^{c}$.
Using the same calculation as previous case for $v-u$ and maximum principle
Lemma 2.2 we derive that $u\geq v$ in ${\mathds{R}^{d}}$. Again, repeating the
arguments of Case 1 we get

$\inf_{\mathscr{D}^{+}_{\kappa^{\prime}R}}\frac{u}{\delta}\leq
C\left(\inf_{\mathscr{D}_{\frac{R}{2}}}\frac{u}{\delta}+2C_{2}R^{\hat{\alpha}}\right)\,.$

Choosing $\rho_{0}=\eta\rho\wedge\rho_{1}$ completes the proof. ∎

######  Lemma 4.7.

Let $\alpha^{\prime}=1\wedge(2-\alpha)$ and $\Omega$ be a bounded $C^{2}$
domain in ${\mathds{R}^{d}}$. Also, let $u$ be a bounded continuous function
such that $u\geq 0$ and $u\leq M_{0}\delta(x)$ in ${\mathds{R}^{d}}$, and
$\lvert{\mathcal{L}}u\rvert\leq
C_{2}(1+\mathds{1}_{(1,2)}(\alpha)\delta^{1-\alpha})$ in $\mathscr{D}_{R}$,
for some constant $C_{2}$. Then, there exists a positive constant $C$,
depending only on $d,\lambda,\Lambda,\Omega$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha}){k}(y)\mathrm{d}{y}$, such that

$\sup_{\mathscr{D}^{+}_{\kappa^{\prime}R}}\frac{u}{\delta}\leq
C\left(\inf_{\mathscr{D}^{+}_{\kappa^{\prime}R}}\frac{u}{\delta}+(M_{0}\vee
C_{2})R^{\alpha^{\prime}}\right)$ (4.15)

for all $R\leq\rho$, where constant $\rho$ is given by Remark 4.1.

###### Proof.

We will use the weak Harnack inequality proved in Theorem 4.1 to show (4.15).
Let $R\leq\rho$. Then for each $y\in\mathscr{D}^{+}_{\kappa^{\prime}R}$, we
have ${\mathscr{B}}_{\kappa R}(y)\subset\mathscr{D}_{R}$. Hence we have
$\lvert{\mathcal{L}}u\rvert\leq
C_{2}(1+\mathds{1}_{(1,2)}(\alpha)\delta^{1-\alpha}(x))$ in
${\mathscr{B}}_{\kappa R}(y)$. Without loss of generality, we may assume
$y=0$. Let $s=\kappa R$ and define $v(x)=u(sx)$ for all
$x\in{\mathds{R}^{d}}$. Then, it can be easily seen that

$s^{2}{\mathcal{L}}[sx,u]={\mathcal{L}}^{s}[x,v]\coloneqq\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}\left\\{\operatorname*{Tr}a_{\theta\nu}(sx)D^{2}v(x)+{\mathcal{I}}^{s}_{\theta\nu}[x,v]\right\\}\qquad\text{for
all }\;x\in{\mathscr{B}}_{2}.$

This gives

$\displaystyle\left|{\mathcal{L}}^{s}[x,v]\right|$ $\displaystyle\leq
C_{2}s^{2}(1+\mathds{1}_{(1,2)}(\alpha)\delta^{1-\alpha}(sx))$
$\displaystyle\leq
C_{2}\left(s^{2}+\mathds{1}_{(1,2)}(\alpha)s^{2}\left(\kappa
R\right)^{1-\alpha}\right)$ $\displaystyle\leq C_{2}s^{1+\alpha^{\prime}}\,,$

in ${\mathscr{B}}_{2}$ where $\alpha^{\prime}=1\wedge(2-\alpha)$. In second
line, we used that for each $x\in{\mathscr{B}}_{\kappa R},$ $|sx|<\kappa R$
and hence $\delta(sx)>\frac{\kappa R}{2}=\frac{s}{2}$. From $u\leq
M_{0}\delta(x)$ we have $v(y)\leq M_{0}\operatorname*{diam}\Omega$ and
$v(y)\leq M_{0}s(1+|y|)$ in whole ${\mathds{R}^{d}}$. Hence by Corollary 4.1,
we obtain

$\sup_{{\mathscr{B}}_{\frac{1}{4}}}v\leq
C\left(\inf_{{\mathscr{B}}_{\frac{1}{4}}}v+(M_{0}\vee
C_{2})s^{1+\alpha^{\prime}}\right),$

where constant $C$ does not depend on $s,M_{0},C_{2}$. This of course, implies

$\sup_{{\mathscr{B}}_{\frac{\kappa R}{64}}(y)}u\leq
C\left(\inf_{{\mathscr{B}}_{\frac{\kappa R}{64}}(y)}u+(M_{0}\vee
C_{2})R^{1+\alpha^{\prime}}\right),$

for all $y\in\mathscr{D}^{+}_{\kappa^{\prime}R}$. Now cover
$\mathscr{D}^{+}_{\kappa^{\prime}R}$ by a finite number of balls
${\mathscr{B}}_{\kappa R/64}(y_{i})$, independent of $R$, to obtain

$\sup_{\mathscr{D}^{+}_{\kappa^{\prime}R}}u\leq
C\left(\inf_{\mathscr{D}^{+}_{\kappa^{\prime}R}}u+(M_{0}\vee
C_{2})R^{1+\alpha^{\prime}}\right).$

Then (4.15) follows since $\kappa R/2\leq\delta\leq 3\kappa R/2$ in
$\mathscr{D}^{+}_{\kappa^{\prime}R}$. ∎

Now we are ready to prove the oscillation lemma.

######  Proposition 4.1.

Let $u$ be a bounded continuous function such that $|{\mathcal{L}}u|\leq K$ in
$\Omega$, for some constant $K$, and $u=0$ in $\Omega^{c}$. Given any
$x_{0}\in\partial\Omega$, let $\mathscr{D}_{R}$ be as in the Definition 4.1.
Then for some $\uptau\in(0,\hat{\alpha})$ there exists $C$, dependent on
$\Omega,d,\lambda,\Lambda,\alpha$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}$ but not on
$x_{0}$, such that

$\sup_{\mathscr{D}_{R}}\frac{u}{\delta}-\inf_{\mathscr{D}_{R}}\frac{u}{\delta}\leq
CKR^{\uptau}$ (4.16)

for all $R\leq\rho_{0}$, where $\rho_{0}>0$ is a constant depending only on
$\Omega,d,\lambda,\Lambda,\alpha$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}$.

###### Proof.

For the proof we follow a standard method, similar to [46], with the help of
Lemmas 4.4, 4.7, and 4.6. Fix $x_{0}\in\partial\Omega$ and consider
$\rho_{0}>0$ to be chosen later. With no loss of generality, we assume
$x_{0}=0$. In view of (3.2), we only consider the case $K>0$. By considering
$u/K$ instead of $u$, we may assume that $K=1$, that is, $|{\mathcal{L}}u|\leq
1$ in $\Omega$. From Theorem 1.1 we note that
$||u||_{C^{0,1}({\mathds{R}^{d}})}\leq C_{1}$. Below, we consider two cases.

Case 1: For $\alpha\in(0,1]$, ${\mathcal{I}}_{\theta\nu}u$ is classically
defined and $|{\mathcal{I}}_{\theta\nu}u|\leq\tilde{C}$ in $\Omega$ for all
$\theta$ and $\nu.$ Consequently, one can combine the nonlocal term on the rhs
and only deal with local nonlinear operator
$\tilde{\mathcal{L}}[x,u]:=\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}\left\\{\operatorname*{Tr}a_{\theta\nu}(x)D^{2}u(x)\right\\}.$
In this case the proof is simpler and can be done following the same method as
for the local case. However, the method we use below would also work with an
appropriate modification.

Case 2: Now we deal with the case $\alpha\in(1,2)$. We show that there exists
${\mathscr{K}}>0,$ $\rho_{1}\in(0,\rho_{0})$ and $\uptau\in(0,1)$, dependent
only on $\Omega,d,\lambda,\Lambda,\alpha$ and
$\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\mathrm{d}{y}$, and monotone
sequences $\\{M_{k}\\}$ and $\\{m_{k}\\}$ such that, for all $k\geq 0,$

$\displaystyle M_{k}-m_{k}=\frac{1}{4^{k\uptau}},\,\,\,\,\,-1\leq m_{k}\leq
m_{k+1}<M_{k+1}\leq M_{k}\leq 1,$ (4.17)

and

$\displaystyle m_{k}\leq{\mathscr{K}}^{-1}\frac{u}{\delta}\leq
M_{k}\quad\text{in}\quad\mathscr{D}_{R_{k}},\quad\text{where}\quad
R_{k}=\frac{\rho_{1}}{4^{k}}.$ (4.18)

Note that (4.18) is equivalent to the following

$\displaystyle m_{k}\delta\leq{\mathscr{K}}^{-1}u\leq
M_{k}\delta,\quad\text{in}\quad{\mathscr{B}}_{R_{k}},\quad\text{where}\quad
R_{k}=\frac{\rho_{1}}{4^{k}}.$ (4.19)

Next we construct monotone sequences $\\{M_{k}\\}$ and $\\{m_{k}\\}$ by
induction.

The existence of $M_{0}$ and $m_{0}$ such that (4.17) and (4.19) hold for
$k=0$ is guaranteed by Lemma 3.1. Assume that we have the sequences up to
$M_{k}$ and $m_{k}$. We want to show the existence of $M_{k+1}$ and $m_{k+1}$
such that (4.17)-(4.19) hold. We set

$\displaystyle u_{k}=\frac{1}{{\mathscr{K}}}u-m_{k}\delta.$

Note that to apply Lemma 4.7 we need $u_{k}$ to be nonnegative in
${\mathds{R}^{d}}$. Therefore we shall work with $u^{+}_{k}$, the positive
part of $u_{k}$. Let $u_{k}=u^{+}_{k}-u^{-}_{k}$ and by the induction
hypothesis,

$\displaystyle u^{+}_{k}=u_{k}\quad\text{and}\quad
u^{-}_{k}=0\quad\text{in}\quad{\mathscr{B}}_{R_{k}}.$ (4.20)

We need to find a lower bound on $u_{k}$. Since $u_{k}\geq 0$ in
${\mathscr{B}}_{R_{k}}$ and $u_{k}$ is Lipschitz in ${\mathds{R}^{d}},$ we get
for $x\in{\mathscr{B}}^{c}_{R_{k}}$ that

$\displaystyle u_{k}(x)=u_{k}(R_{k}x_{\rm u})+u_{k}(x)-u_{k}(R_{k}x_{\rm
u})\geq-C_{L}|x-R_{k}x_{\rm u}|,$ (4.21)

where $z_{\rm u}=\frac{1}{|z|}z$ for $z\neq 0$ and $C_{L}$ denotes a Lipschitz
constant of $u_{k}$ which can be chosen independent of $k$. Using Lemma 3.1 we
also have $|u_{k}|\leq{\mathscr{K}}^{-1}+\operatorname*{diam}(\Omega)=C_{1}$
for all $x\in{\mathds{R}^{d}}$. Thus using (4.20) and (4.21) we calculate
${\mathcal{L}}[x,u^{-}_{k}]$ in $\mathscr{D}_{\frac{R_{k}}{2}}$. Let
$x\in\mathscr{D}_{R_{k}/2}$. By (4.20), $D^{2}u^{-}_{k}(x)=0$. Then

$\displaystyle 0\leq{\mathcal{I}}_{\theta\nu}[x,u^{-}_{k}]$
$\displaystyle=\int_{x+y\not\in{\mathscr{B}}_{R_{k}}}u^{-}_{k}(x+y)N_{\theta\nu}(x,y)\mathrm{d}{y}$
$\displaystyle\leq\int_{\left\\{|y|\geq\frac{R_{k}}{2},x+y\neq
0\right\\}}u^{-}_{k}(x+y)k(y)\mathrm{d}{y}$ $\displaystyle\leq
C_{L}\int_{\left\\{\frac{R_{k}}{2}\leq|y|\leq 1,\;x+y\neq
0\right\\}}\Big{|}(x+y)-R_{k}(x+y)_{\rm
u}\Big{|}k(y)\mathrm{d}{y}+C_{1}\int_{|y|\geq 1}k(y)\mathrm{d}{y}$
$\displaystyle\leq C_{L}\int_{\frac{R_{k}}{2}\leq|y|\leq
1}\left(|x|+R_{k}\right)k(y)\,\mathrm{d}{y}+C_{L}\int_{\frac{R_{k}}{2}\leq|y|\leq
1}|y|k(y)\,\mathrm{d}{y}+C_{1}\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\,\mathrm{d}{y}$
$\displaystyle\leq\kappa_{3}\left[\int_{{\mathds{R}^{d}}}(1\wedge|y|^{\alpha})k(y)\,\mathrm{d}{y}\right]\left(R_{k}^{1-\alpha}+1\right)$
$\displaystyle\leq\kappa_{4}R_{k}^{1-\alpha},$ (4.22)

for some constants $\kappa_{3},\kappa_{4}$, independent of $k$.

Now we write $u^{+}_{k}={\mathscr{K}}^{-1}u-m_{k}\delta+u^{-}_{k}$. Since
$\delta$ is $C^{2}$ and $u^{-}_{k}=0$ in $\mathscr{D}_{\frac{R_{k}}{2}},$
first note that

$\displaystyle{\mathcal{L}}u^{+}_{k}$
$\displaystyle\leq{\mathscr{K}}^{-1}-({\mathcal{P}}^{-}+{\mathcal{P}}^{-}_{k})(m_{k}\delta)+({\mathcal{P}}^{+}+{\mathcal{P}}^{+}_{k})(u_{k}^{-}),$
$\displaystyle{\mathcal{L}}u^{+}_{k}$
$\displaystyle\geq-{\mathscr{K}}^{-1}-({\mathcal{P}}^{+}+{\mathcal{P}}^{+}_{k})(m_{k}\delta)+({\mathcal{P}}^{-}+{\mathcal{P}}^{-}_{k})(u_{k}^{-}).$

Using Lemma 4.4 and (4) in the above estimate we have

$\displaystyle|{\mathcal{L}}u^{+}_{k}|\leq{\mathscr{K}}^{-1}+m_{k}C\delta^{1-\alpha}+\kappa_{4}(R_{k})^{1-\alpha}\,\,\,\text{in}\,\,\,\mathscr{D}_{\frac{R_{k}}{2}}.$
(4.23)

Since $\rho_{1}\geq R_{k}\geq\delta$ in $\mathscr{D}_{R_{k}}$, for $\alpha>1,$
we have $R^{1-\alpha}_{k}\leq\delta^{1-\alpha}$, and hence, from (4.23), we
have

$|{\mathcal{L}}u^{+}_{k}|\leq\Big{[}{\mathscr{K}}^{-1}[(\rho_{1})]^{\alpha-1}+C+\kappa_{4}\Big{]}\delta^{1-\alpha}(x):=\kappa_{5}\delta^{1-\alpha}(x)\quad\text{in}\quad\mathscr{D}_{R_{k}/2}.$

Now we are in a position to apply Lemmas 4.6 and 4.7. Recalling that

$u^{+}_{k}=u_{k}={\mathscr{K}}^{-1}u-m_{k}\delta\quad\text{in}\quad\mathscr{D}_{R_{k}},$

and using Lemma 3.1 we also have
$|u^{+}_{k}|\leq|u_{k}|\leq({\mathscr{K}}^{-1}+1)\delta(x)=C_{1}\delta(x)$ for
all $x\in{\mathds{R}^{d}}$. We get from Lemmas 4.6 and 4.7 that

$\displaystyle\sup_{\mathscr{D}^{+}_{\kappa^{\prime}R_{k}/2}}\Big{(}{\mathscr{K}}^{-1}\frac{u}{\delta}-m_{k}\Big{)}$
$\displaystyle\leq
C\Big{(}\inf_{\mathscr{D}^{+}_{\kappa^{\prime}R_{k}/2}}\Big{(}{\mathscr{K}}^{-1}\frac{u}{\delta}-m_{k}\Big{)}+(\kappa_{5}\vee
C_{1})R^{\hat{\alpha}}_{k}\Big{)}$ (4.24) $\displaystyle\leq
C\Big{(}\inf_{\mathscr{D}_{R_{k}/4}}\Big{(}{\mathscr{K}}^{-1}\frac{u}{\delta}-m_{k}\Big{)}+(\kappa_{5}\vee
C_{1})R^{\hat{\alpha}}_{k}\Big{)}.$

Repeating a similar argument for the function
$\tilde{u}_{k}=M_{k}\delta-{\mathscr{K}}^{-1}u$, we find

$\displaystyle\sup_{\mathscr{D}^{+}_{\kappa^{\prime}R_{k}/2}}\Big{(}M_{k}-{\mathscr{K}}^{-1}\frac{u}{\delta}\Big{)}\leq
C\Big{(}\inf_{\mathscr{D}_{R_{k}/4}}\Big{(}M_{k}-{\mathscr{K}}^{-1}\frac{u}{\delta}\Big{)}+(\kappa_{5}\vee
C_{1})R^{\hat{\alpha}}_{k}\Big{)}.$ (4.25)

Combining (4.24) and (4.25) we obtain

$\displaystyle M_{k}-m_{k}$ $\displaystyle\leq
C\Big{(}\inf_{\mathscr{D}^{+}_{R_{k}/4}}\Big{(}M_{k}-{\mathscr{K}}^{-1}\frac{u}{\delta}\Big{)}+\inf_{\mathscr{D}^{+}_{R_{k}/4}}\Big{(}{\mathscr{K}}^{-1}\frac{u}{\delta}-m_{k}\Big{)}+(\kappa_{5}\vee
C_{1})R^{\hat{\alpha}}_{k}\Big{)}$
$\displaystyle=C\Big{(}\inf_{\mathscr{D}_{R_{k+1}}}{\mathscr{K}}^{-1}\frac{u}{\delta}-\sup_{\mathscr{D}_{R_{k+1}}}{\mathscr{K}}^{-1}\frac{u}{\delta}+M_{k}-m_{k}+(\kappa_{5}\vee
C_{1})R^{\hat{\alpha}}_{k}\Big{)}.$ (4.26)

Putting $M_{k}-m_{k}=\frac{1}{4^{\uptau k}}$ in (4), we have

$\displaystyle\sup_{\mathscr{D}_{R_{k+1}}}{\mathscr{K}}^{-1}\frac{u}{\delta}-\inf_{\mathscr{D}_{R_{k+1}}}{\mathscr{K}}^{-1}\frac{u}{\delta}$
$\displaystyle\leq\Big{(}\frac{C-1}{C}\frac{1}{4^{\uptau k}}+(\kappa_{5}\vee
C_{1})R^{\hat{\alpha}}_{k}\Big{)}$ $\displaystyle=\frac{1}{4^{\uptau
k}}\Big{(}\frac{C-1}{C}+(\kappa_{5}\vee C_{1})R^{\hat{\alpha}}_{k}4^{\uptau
k}\Big{)}.$ (4.27)

Since $R_{k}=\frac{\rho_{1}}{4^{k}}$ for $\rho_{1}\in(0,\rho_{0})$, we can
choose $\rho_{0}$ and $\uptau$ small so that

$\Big{(}\frac{C-1}{C}+(\kappa_{5}\vee C_{1})R^{\hat{\alpha}}_{k}4^{\uptau
k}\Big{)}\leq\frac{1}{4^{\uptau}}.$

Putting in (4) we obtain

$\sup_{\mathscr{D}_{R_{k+1}}}{\mathscr{K}}^{-1}\frac{u}{\delta}-\inf_{\mathscr{D}_{R_{k+1}}}{\mathscr{K}}^{-1}\frac{u}{\delta}\leq\frac{1}{4^{\uptau(k+1)}}.$

Thus we find $m_{k+1}$ and $M_{k+1}$ such that (4.17) and (4.18) hold. It is
easy to prove (4.16) from (4.17)-(4.18). ∎

Next we establish Hölder regularity of $u/\delta$ up to the boundary, that is
Theorem 1.2.

###### Proof of Theorem 1.2.

Replacing $u$ by $\frac{u}{CK}$ we may assume that $|{\mathcal{L}}u|\leq 1$ in
$\Omega$. Let $v=u/\delta$. From Lemma 3.1 we then have

$\lVert v\rVert_{L^{\infty}(\Omega)}\leq C,$

for some constant $C$ and from Theorem 1.1 we have

$\lVert u\rVert_{C^{0,1}({\mathds{R}^{d}})}\leq C.$ (4.28)

Also from Proposition 4.1 for each $x_{0}\in\partial\Omega$ and for all $r>0$
we have

$\sup_{\mathscr{D}_{r}(x_{0})}v-\inf_{\mathscr{D}_{r}(x_{0})}v\leq
Cr^{\uptau}.$ (4.29)

where $\mathscr{D}_{r}(x_{0})={\mathscr{B}}_{r}(x_{0})\cap\Omega$ as before.
To complete the proof we shall show that

$\sup_{x,y\in\Omega,x\neq y}\frac{|v(x)-v(y)|}{|x-y|^{\upkappa}}\leq C,$
(4.30)

for some $\upkappa>0$. Let $r=|x-y|$ and there exists
$x_{0},y_{0}\in\partial\Omega$ such that $\delta(x)=|x-x_{0}|$ and
$\delta(y)=|y-y_{0}|.$ If $r\geq\frac{1}{8},$ then

$\displaystyle\frac{|v(x)-v(y)|}{|x-y|^{\upkappa}}\leq 2\cdot
8^{\upkappa}||v||_{L^{\infty}(\Omega)}.$

If $r<\frac{1}{8}$ and $r\geq\frac{1}{8}(\delta(x)\vee\delta(y))^{p}$ for some
$p>2$ then clearly $y\in{\mathscr{B}}_{\kappa r^{1/p}}(x_{0})$ for some
$\kappa>0.$ Now using (4.29) we obtain

$\displaystyle|v(x)-v(y)|\leq\sup_{\mathscr{D}_{\kappa
r^{1/p}}(x_{0})}v-\inf_{\mathscr{D}_{\kappa r^{1/p}}(x_{0})}v\leq
C_{\kappa}r^{\uptau/p}.$

If $r<\frac{1}{8}$ and $r<\frac{1}{8}(\delta(x)\vee\delta(y))^{p},$ then
$r<\frac{1}{8}(\delta(x)\vee\delta(y))$ and this implies
$y\in{\mathscr{B}}_{\frac{1}{8}(\delta(x)\vee\delta(y))}(x)$ or
$x\in{\mathscr{B}}_{\frac{1}{8}(\delta(x)\vee\delta(y))}(y).$ Without loss of
any generality assume $\delta(x)\geq\delta(y)$ and
$y\in{\mathscr{B}}_{\frac{\delta(x)}{8}}(x).$ Using (4.28) and the Lipschitz
continuity of $\delta,$ we get

$\displaystyle|v(x)-v(y)|=\left|\frac{u(x)}{\delta(x)}-\frac{u(y)}{\delta(y)}\right|\leq\frac{M(K,\operatorname*{diam}\Omega)r}{\delta(x)\cdot\delta(y)}.$

Also we have $(8r)^{1/p}\delta(y)<\delta(x)\cdot\delta(y).$ This implies

$\displaystyle|v(x)-v(y)|\leq\frac{M(K,\operatorname*{diam}\Omega)r}{\delta(x)\cdot\delta(y)}<\frac{M(K,\operatorname*{diam}\Omega)}{8^{1/p}}\cdot\frac{r^{1-1/p}}{\delta(y)}.$

Now if $r<\frac{1}{8}(\delta(y))^{p}$ then one obtains

$\displaystyle|v(x)-v(y)|<\frac{M(K,\operatorname*{diam}\Omega)}{8^{1/p}}\cdot\frac{r^{1-1/p}}{\delta(y)}\leq
Cr^{1-2/p}.$

On the other hand, if $r\geq\frac{1}{8}(\delta(y))^{p},$ since
$\delta(y)>\frac{1}{64}\delta(x)$ we have
$r\geq\frac{1}{8}\cdot\left(\frac{1}{64}\right)^{p}(\delta(x))^{p}$ and this
case can be treated as previous. Therefore choosing
$\upkappa=(1-\frac{2}{p})\wedge\frac{\uptau}{p}$ we conclude (4.30). This
completes the proof. ∎

##  5\. Global Hölder regularity of the gradient

In this section we prove the Hölder regularity of $Du$ up to the boundary.
First, let us recall

${\mathcal{L}}[x,u]=\sup_{\theta\in\Theta}\inf_{\nu\in\gamma}\left\\{\operatorname*{Tr}a_{\theta\nu}(x)D^{2}u(x)+{\mathcal{I}}_{\theta\nu}[x,u]\right\\}.$

We denote $v=\frac{u}{\delta}.$ Following [11], next we obtain the in-
equations satisfied by $v.$

######  Lemma 5.1.

Let $\Omega$ be bounded $C^{2}$ domain in ${\mathds{R}^{d}}$. If
$|{\mathcal{L}}u|\leq K$ in $\Omega$ and $u=0$ in $\Omega^{c},$ then we have

$\displaystyle{\mathcal{L}}v+2K_{0}d^{2}\frac{|D\delta|}{\delta}|Dv|\geq\frac{1}{\delta}\Big{[}-K-|v|(P^{+}+P^{+}_{k})\delta-\sup_{\theta,\nu}Z_{\theta\nu}[v,\delta]\Big{]},$
(5.1)
$\displaystyle{\mathcal{L}}v-2K_{0}d^{2}\frac{|D\delta|}{\delta}|Dv|\leq\frac{1}{\delta}\Big{[}K-|v|(P^{-}+P^{-}_{k})\delta-\inf_{\theta,\nu}Z_{\theta\nu}[v,\delta]\Big{]}$

for some $K_{0},$ where

$Z_{\theta\nu}[v,\delta](x)=\int_{{\mathds{R}^{d}}}(v(y)-v(x))(\delta(y)-\delta(x))N_{\theta\nu}(x,y-x)dy.$

###### Proof.

First note that, since $u\in C^{1}(\Omega)$ by Lemma 2.1, we have $v\in
C^{1}(\Omega)$. Therefore, $Z_{\theta\nu}[v,\delta]$ is continuous in
$\Omega$. Consider a test function $\psi\in C^{2}(\Omega)$ that touches $v$
from above at $x\in\Omega$. Define

$\psi_{r}(z)=\left\\{\begin{array}[]{lll}\psi(z)&\text{in}\;{\mathscr{B}}_{r}(x),\\\
v(z)&\text{in}\;{\mathscr{B}}^{c}_{r}(x).\end{array}\right.$

By our assertion, we have $\psi_{r}\geq v$ for all $r$ small. To verify the
first inequality in (5.1) we must show that

$\displaystyle{\mathcal{L}}[x,\psi_{r}]+2k_{0}d^{2}\frac{|D\delta(x)|}{\delta(x)}\cdot|D\psi_{r}(x)|\geq\frac{1}{\delta(x)}[-K-|v(x)|({\mathcal{P}}^{+}+{\mathcal{P}}^{+}_{k})\delta(x)-\sup_{\theta,\nu}Z_{\theta\nu}[v,\delta](x)],$
(5.2)

for some $r$ small. We define

$\tilde{\psi}_{r}(z)=\left\\{\begin{array}[]{lll}\delta(z)\psi(z)&\text{in}\;{\mathscr{B}}_{r}(x),\\\
u(z)&\text{in}\;{\mathscr{B}}^{c}_{r}(x).\end{array}\right.$

Then, $\tilde{\psi}_{r}\geq u$ for all $r$ small. Since $|{\mathcal{L}}u|\leq
K$ and $\delta\psi_{r}=\tilde{\psi}_{r}$, we obtain at a point $x$

$\displaystyle-K\leq$ $\displaystyle{\mathcal{L}}[x,\tilde{\psi}_{r}]$
$\displaystyle=$
$\displaystyle\sup_{\theta\in\Theta}\inf_{\nu\in\gamma}\biggl{[}\delta(x)\left(\operatorname*{Tr}a_{\theta\nu}(x)D^{2}\psi_{r}(x)+{\mathcal{I}}_{\theta\nu}\psi_{r}(x)\right)+\psi_{r}(x)\left(\operatorname*{Tr}a_{\theta\nu}(x)D^{2}\delta(x)+{\mathcal{I}}_{\theta\nu}\delta(x)\right)$
$\displaystyle+\operatorname*{Tr}\big{[}\left(a_{\theta\nu}(x)+a^{T}_{\theta\nu}(x)\right)\cdot\left(D\delta(x)\otimes
D\psi_{r}(x)\right)\big{]}+Z_{\theta\nu}[\psi_{r},\delta](x)\biggr{]}$
$\displaystyle\leq\delta(x){\mathcal{L}}[x,\psi_{r}]+\sup_{\theta,\nu}\biggl{[}|\psi_{r}(x)|\left(\operatorname*{Tr}a_{\theta\nu}(x)D^{2}\delta(x)+{\mathcal{I}}_{\theta\nu}\delta(x)\right)$
$\displaystyle+\operatorname*{Tr}\left[\left(a_{\theta\nu}(x)+a^{T}_{\theta\nu}(x)\right)\cdot\left(D\delta(x)\otimes
D\psi_{r}(x)\right)\right]+Z_{\theta\nu}[\psi_{r},\delta](x)\biggr{]}$
$\displaystyle\leq\delta(x){\mathcal{L}}[x,\psi_{r}]+|v(x)|\left({\mathcal{P}}^{+}+{\mathcal{P}}^{+}_{k}\right)\delta(x)+2K_{0}d^{2}|D\delta(x)|\cdot|D\psi_{r}(x)|+\sup_{\theta,\nu}Z_{\theta\nu}[\psi_{r},\delta](x),$

for all $r$ small and some constant $K_{0}$, where $D\delta(x)\otimes
D\psi_{r}(x):=\left(\frac{\partial\delta}{\partial
x_{i}}\cdot\frac{\partial\psi_{r}}{\partial x_{j}}\right)_{i,j}.$ Rearranging
the terms we have

$-K-|v(x)|\left({\mathcal{P}}^{+}+{\mathcal{P}}^{+}_{k}\right)\delta(x)-\sup_{\theta,\nu}Z_{\theta\nu}[\psi_{r},\delta](x)\leq\delta(x){\mathcal{L}}[x,\psi_{r}]+2K_{0}d^{2}|D\delta(x)|\cdot|D\psi_{r}(x)|.$
(5.3)

Let $r_{1}\leq r$. Since $\psi_{r}$ is decreasing with $r$, we get from (5.3)
that

$\displaystyle\delta(x){\mathcal{L}}[x,\psi_{r}]+2K_{0}d^{2}|D{\delta(x)}|\cdot|D\psi_{r}(x)|$
$\displaystyle\geq\delta(x){\mathcal{L}}[x,\psi_{r_{1}}]+2K_{0}d^{2}|D{\delta(x)}|\cdot|D\psi_{r_{1}}(x)|$
$\displaystyle\geq\lim_{r_{1}\to
0}\left[-K-\left|v(x)\right|\left({\mathcal{P}}^{+}+{\mathcal{P}}^{+}_{k}\right)\delta(x)-\sup_{\theta,\nu}Z_{\theta\nu}[\psi_{r_{1}},\delta](x)\right]$
$\displaystyle=\left[-K-\left|v(x)\right|\left({\mathcal{P}}^{+}+{\mathcal{P}}^{+}_{k}\right)\delta(x)-\sup_{\theta,\nu}Z_{\theta\nu}[v,\delta](x)\right],$

by dominated convergence theorem. This gives (5.2). Similarly we can verify
the second inequality of (5.1). ∎

Next we obtain a the following estimate on $v,$ away from the boundary. Denote
$\Omega^{\sigma}=\\{x\in\Omega\;:\;\operatorname*{dist}(x,\Omega^{c})\geq\sigma\\}$.

######  Lemma 5.2.

Let $\Omega$ be bounded $C^{2}$ domain in ${\mathds{R}^{d}}$. If
$|{\mathcal{L}}u|\leq K$ in $\Omega$ and $u=0$ in $\Omega^{c},$ then for some
constant $C$ it holds that

$\lVert Dv\rVert_{L^{\infty}(\Omega^{\sigma})}\leq
CK\sigma^{\upkappa-1}\quad\text{for all}\;\sigma\in(0,1).$ (5.4)

Furthermore, there exists $\eta\in(0,1)$ such that for any
$x\in\Omega^{\sigma}$ and $0<|x-y|\leq\sigma/8$ we have

$\frac{|Dv(y)-Dv(x)|}{|x-y|^{\eta}}\leq CK\sigma^{\upkappa-1-\eta},$

for all $\sigma\in(0,1)$.

###### Proof.

Using Lemma 5.1 we have

$\displaystyle{\mathcal{L}}v+2K_{0}d^{2}\frac{|D\delta|}{\delta}|Dv|\geq\frac{1}{\delta}\Big{[}-K-|v|({\mathcal{P}}^{+}+{\mathcal{P}}^{+}_{k})\delta-\sup_{\theta,\nu}Z_{\theta\nu}[v,\delta]\Big{]},$
(5.5)
$\displaystyle{\mathcal{L}}v-2K_{0}d^{2}\frac{|D\delta|}{\delta}|Dv|\leq\frac{1}{\delta}\Big{[}K-|v|({\mathcal{P}}^{-}+{\mathcal{P}}^{-}_{k})\delta-\inf_{\theta,\nu}Z_{\theta\nu}[v,\delta]\Big{]}$

in $\Omega$. Fix a point $x_{0}\in\Omega^{\sigma}$ and define

$w(x)=v(x)-v(x_{0}).$

From (5.5) we then obtain

$\displaystyle{\mathcal{L}}w+2K_{0}d^{2}\frac{|D\delta|}{\delta}|Dw|\geq\Big{[}-\frac{1}{\delta}K-\ell_{1}\Big{]},$
(5.6)
$\displaystyle{\mathcal{L}}w-2K_{0}d^{2}\frac{|D\delta|}{\delta}|Dw|\leq\Big{[}\frac{1}{\delta}K+\ell_{2}\Big{]}$

in $\Omega$, where

$\ell_{1}(x)=\frac{1}{\delta(x)}\left[|w(x)|({\mathcal{P}}^{+}+{\mathcal{P}}^{+}_{k})\delta(x)+\sup_{\theta,\nu}Z_{\theta\nu}[w,\delta](x)+|v(x_{0})|({\mathcal{P}}^{+}+{\mathcal{P}}^{+}_{k})\delta(x)\right]$

And

$\ell_{2}(x)=\frac{1}{\delta(x)}\left[|w(x)|({\mathcal{P}}^{-}+{\mathcal{P}}^{-}_{k})\delta(x)-\inf_{\theta,\nu}Z_{\theta\nu}[w,\delta](x)-|v(x_{0})|({\mathcal{P}}^{-}+{\mathcal{P}}^{-}_{k})\delta(x)\right].$

We set $r=\frac{\sigma}{2}$ and claim that

$\lVert\ell_{i}\rVert_{L^{\infty}(B_{r}(x_{0}))}\leq\kappa_{1}\sigma^{\upkappa-2},\quad\text{for
all}\;\sigma\in(0,1)\;\text{and}\;i=1,2,$ (5.7)

for some constant $\kappa_{1}$. Let us denote by

$\xi_{1}^{\pm}=\frac{|w(x)|({\mathcal{P}}^{\pm}+{\mathcal{P}}^{\pm}_{k})\delta}{\delta},\quad\xi_{2}=\frac{1}{\delta}\sup_{\theta,\nu}Z_{\theta\nu}[w,\delta],\quad\xi_{3}^{\pm}=\frac{|v(x_{0})|({\mathcal{P}}^{\pm}+{\mathcal{P}}^{\pm}_{k})\delta}{\delta},\quad\xi_{4}=\frac{1}{\delta}\inf_{\theta,\nu}Z_{\theta\nu}[v,\delta].$

Recall that $\upkappa\in(0,\hat{\alpha}).$ Since

$\lVert{\mathcal{P}}^{\pm}\delta\rVert_{L^{\infty}(\Omega)}<\infty\quad\text{and}\quad\lVert{\mathcal{P}}^{\pm}_{k}\delta\rVert_{L^{\infty}(\Omega_{\sigma})}\lesssim\left(1+\mathds{1}_{(1,2)}(\alpha)\delta^{1-\alpha}\right)$

(cf Lemma 4.4 ), and

$\lVert v\rVert_{L^{\infty}}({\mathds{R}^{d}})<\infty,\quad\lVert
w\rVert_{L^{\infty}(B_{r}(x_{0}))}\lesssim r^{\upkappa},$

it follows that

$\lVert\xi_{3}^{\pm}\rVert_{L^{\infty}(B_{r}(x_{0}))}\lesssim\left.\begin{cases}\frac{1}{\sigma}&\quad\text{if}\;\alpha\in(0,1],\\\
\frac{1}{\sigma^{\alpha}}&\quad\text{if}\;\alpha\in(1,2)\end{cases}\right\\}\lesssim\sigma^{\upkappa-2},$

and

$\lVert\xi_{1}^{\pm}\rVert_{L^{\infty}(B_{r}(x_{0}))}\lesssim\left.\begin{cases}\frac{\sigma^{\upkappa}}{\delta^{2}}&\quad\text{if}\;\alpha\in(0,1],\\\
\frac{\sigma^{\upkappa}}{\delta^{\alpha}}&\quad\text{if}\;\alpha\in(1,2)\end{cases}\right\\}\lesssim\sigma^{\upkappa-2}.$

Next we estimate $\xi_{2}$ and $\xi_{4}$. Let $x\in{\mathscr{B}}_{r}(x_{0})$.
Denote by $\hat{r}=\delta(x)/4$. Note that

$\delta(x)\geq\delta(x_{0})-|x-x_{0}|\geq 2r-r=r\Rightarrow\hat{r}\geq r/4.$

Since $u\in C^{1}(\Omega)$ by Lemma 2.1 and $|u|\leq C\delta$ in
${\mathds{R}^{d}}$ by Lemma 3.1. Thus we have

$|Dv|\leq\left|\frac{Du}{\delta}\right|+\left|\frac{uD\delta}{\delta^{2}}\right|\lesssim\frac{1}{\delta(x)}\quad\text{in}\;{\mathscr{B}}_{\hat{r}}(x).$
(5.8)

Now we calculate

$\displaystyle|Z_{\theta\nu}[w,\delta](x)|\leq\int_{{\mathds{R}^{d}}}|\delta(x)-\delta(y)||v(x)-v(y)|k(y-x)\mathrm{d}{y}$
$\displaystyle=\int_{{\mathscr{B}}_{\hat{r}}(x)}+\int_{{\mathscr{B}}_{1}(x)\setminus{\mathscr{B}}_{\hat{r}}(x)}+\int_{{\mathscr{B}}^{c}_{1}(x)}$
$\displaystyle=I_{1}+I_{2}+I_{3}.$

To estimate $I_{1},$ first we consider $\alpha\leq 1.$ Since $\delta$ is
Lipschitz continuous and $v$ bounded on ${\mathds{R}^{d}},$ $I_{1}$ can be
written as

$\displaystyle I_{1}$
$\displaystyle=\int_{{\mathscr{B}}_{\hat{r}}(x)}\frac{|\delta(x)-\delta(y)|}{|x-y|}|v(x)-v(y)|\cdot|x-y|k(y-x)\mathrm{d}{y}$
$\displaystyle\lesssim\int_{{\mathscr{B}}_{\hat{r}}(x)}|x-y|^{\alpha}k(y-x)\mathrm{d}{y}\leq\int_{{\mathds{R}^{d}}}(1\wedge|z|^{\alpha})k(z)\mathrm{d}{z}.$

For $\alpha\in(1,2),$ using the Lipschitz continuity of $\delta$ and (5.8) we
get

$\displaystyle I_{1}$
$\displaystyle=\int_{{\mathscr{B}}_{\hat{r}}(x)}\frac{|\delta(x)-\delta(y)|}{|x-y|}\cdot\frac{|v(x)-v(y)|}{|x-y|}\cdot|x-y|^{\alpha}|x-y|^{2-\alpha}k(y-x)\mathrm{d}{y}$
$\displaystyle\lesssim\frac{\hat{r}^{2-\alpha}}{\delta(x)}\int_{{\mathscr{B}}_{\hat{r}}(x)}|x-y|^{\alpha}k(y-x)\mathrm{d}{y}\lesssim\delta(x)^{1-\alpha}\int_{{\mathds{R}^{d}}}(1\wedge|z|^{\alpha})k(z)\mathrm{d}{z}\lesssim\sigma^{\upkappa-1}.$

Bounds on $I_{2}$ can be computed as follows: for $\alpha\leq 1,$ we write

$\displaystyle
I_{2}=\int_{{\mathscr{B}}_{1}(x)\setminus{\mathscr{B}}_{\hat{r}}(x)}|\delta(x)-\delta(y)||v(x)-v(y)|k(y-x)\mathrm{d}{y}$
$\displaystyle\lesssim\int_{{\mathscr{B}}_{1}(x)\setminus{\mathscr{B}}_{\hat{r}}(x)}|x-y|^{\alpha}k(y-x)\mathrm{d}{y}$
$\displaystyle\lesssim\int_{{\mathds{R}^{d}}}(1\wedge|z|^{\alpha})k(z)\mathrm{d}{z}.$

In the second line of the above inequality we used

$|\delta(x)-\delta(y)|\lesssim|x-y|\,\,\,\text{and}\,\,\,\,||v||_{L^{\infty}({\mathds{R}^{d}})}<\infty.$

For $\alpha\in(1,2)$ we can compute $I_{2}$ as

$\displaystyle\int_{{\mathscr{B}}_{1}(x)\setminus{\mathscr{B}}_{\hat{r}}(x)}|\delta(x)-\delta(y)||v(x)-v(y)|k(y-x)\mathrm{d}{y}$
$\displaystyle\lesssim\int_{{\mathscr{B}}_{1}(x)\setminus{\mathscr{B}}_{\hat{r}}(x)}|x-y|^{1-\alpha}\cdot|x-y|^{\alpha}k(y-x)\mathrm{d}{y}$
$\displaystyle\lesssim\delta(x)^{1-\alpha}\int_{{\mathds{R}^{d}}}(1\wedge|z|^{\alpha})k(z)\mathrm{d}{z}\lesssim\sigma^{\upkappa-1}.$

Moreover, since $\delta$ and $v$ are bounded in ${\mathds{R}^{d}}$, we get
$I_{3}\leq\kappa_{3}.$ Combining the above estimates we obtain

$\lVert\xi_{i}\rVert_{L^{\infty}{\mathscr{B}}_{r}(x_{0})}\lesssim\sigma^{\upkappa-2}\,\,\,\text{for}\,\,\,i=2,4.$

Thus the claim (5.7) is established.

Let us now define $\zeta(z)=w(\frac{r}{2}z+x_{0})$. Letting
$b(z)=\frac{D\delta(\frac{r}{2}z+x_{0})}{2\delta(\frac{r}{2}z+x_{0})}$ it
follows from (5.6) that

$\displaystyle\tilde{{\mathcal{L}}}^{r}\zeta+K_{0}d^{2}rb(z)\cdot|D\zeta|\geq-\frac{r^{2}}{4}\left[\frac{1}{\delta}K+l_{1}\right]\left(\frac{r}{2}z+x_{0}\right)$
(5.9) $\displaystyle\tilde{{\mathcal{L}}}^{r}\zeta-
K_{0}d^{2}rb(z)\cdot|D\zeta|\leq\frac{r^{2}}{4}\left[\frac{1}{\delta}K+l_{2}\right]\left(\frac{r}{2}z+x_{0}\right)$

in ${\mathscr{B}}_{2}(0),$ where

$\tilde{{\mathcal{L}}}^{r}[x,u]:=\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}\left\\{\operatorname*{Tr}\left(a_{\theta\nu}\left(\frac{r}{2}x+x_{0}\right)D^{2}u(x)\right)+\tilde{{\mathcal{I}}}^{r}_{\theta\nu}[x,u]\right\\}$

and $\tilde{{\mathcal{I}}}^{r}_{\theta\nu}$ is given by

$\tilde{{\mathcal{I}}}^{r}_{\theta\nu}[x,f]=\int_{{\mathds{R}^{d}}}\left(f(x+y)-f(x)-\mathds{1}_{B_{\frac{1}{r}}(y)}\nabla
f(x)\cdot
y\right)\left(\frac{r}{2}\right)^{d+2}N_{\theta\nu}\left(\frac{r}{2}x+x_{0},ry\right)\mathrm{d}{y}.$

Consider a cut-off function $\varphi$ satisfying $\varphi=1$ in
${\mathscr{B}}_{3/2}$ and $\varphi=0$ in ${\mathscr{B}}^{c}_{2}$. Defining
$\tilde{\zeta}=\zeta\varphi$ we get from (5.9) that

$\displaystyle\tilde{{\mathcal{L}}}^{r}[z,\tilde{\zeta}]+K_{0}d^{2}rb(z).|D\tilde{\zeta}(z)|\geq-\frac{r^{2}}{4}\left[\frac{K}{\delta}+|l_{1}|\right]\left(\frac{r}{2}z+x_{0}\right)-\left|\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}\tilde{{\mathcal{I}}}^{r}_{\theta\nu}[z,(\varphi-1)\zeta]\right|$
$\displaystyle\tilde{{\mathcal{L}}}^{r}[z,\tilde{\zeta}]-K_{0}d^{2}rb(z).|D\tilde{\zeta}(z)|\leq\frac{r^{2}}{4}\left[\frac{K}{\delta}+|l_{1}|\right]\left(\frac{r}{2}z+x_{0}\right)-\left|\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}\tilde{{\mathcal{I}}}^{r}_{\theta\nu}[z,(\varphi-1)\zeta]\right|$

in ${\mathscr{B}}_{1}$. Since

$\lVert
rb\rVert_{L^{\infty}({\mathscr{B}}_{1}(0))}\leq\kappa_{3}\quad\text{for
all}\;\sigma\in(0,1),$

applying Lemma 2.1 we obtain, for some $\eta\in(0,1)$,

$\lVert
D\zeta\rVert_{C^{\eta}({\mathscr{B}}_{1/2}(0))}\leq\kappa_{6}\left(\lVert\tilde{\zeta}\rVert_{L^{\infty}({\mathds{R}^{d}})}+\kappa_{4}\sigma+\kappa_{5}\sigma^{\upkappa}\right),$
(5.10)

for some constant $\kappa_{6}$ independent of $\sigma\in(0,1),$ where we used

$\displaystyle\left|\tilde{{\mathcal{I}}}^{r}_{\theta\nu}[z,(\varphi-1)\zeta]\right|\lesssim\sigma\,\,\,\,\,(\text{cf.
the proof of
\lx@cref{creftype~refnum}{T1.1}})\,\,\,\,\text{and}\,\,\,\,|l_{1}|(\frac{r}{2}\cdot+x_{0})\lesssim\sigma^{\upkappa-2}.$

Since $v$ is in $C^{\upkappa}({\mathds{R}^{d}})$, it follows that

$\lVert\tilde{\zeta}\rVert_{L^{\infty}({\mathds{R}^{d}})}=\lVert\tilde{\zeta}\rVert_{L^{\infty}({\mathscr{B}}_{2})}\leq\lVert\zeta\rVert_{L^{\infty}({\mathscr{B}}_{2})}\lesssim
r^{\upkappa}.$

Putting these estimates in (5.10) and calculating the gradient at $z=0$ we
obtain

$|Dv(x_{0})|\lesssim\sigma^{\upkappa-1},$

for all $\sigma\in(0,1)$. This proves the Hölder estimate (5.4).

For the second part, compute the Hölder ratio with $D\zeta(0)-D\zeta(z)$ where
$z=\frac{2}{r}(y-x_{0})$ for $|x_{0}-y|\leq\sigma/8$. This completes the
proof.

∎

Now we can complete the proof of Theorem 1.3. If $u$ is solution of the in-
equation (1.1) then using Theorem 1.1 we have $|{\mathcal{L}}u|\leq CK$. Now
the proof can be obtained by following the same lines as in [11, Theorem 1.3].
We present it here for the sake of completeness.

###### Proof of Theorem 1.3.

Since $u=v\delta$ it follows that

$Du=vD\delta+\delta Dv.$

Since $\delta\in C^{2}(\bar{\Omega})$, it follows from Theorem 1.2 that
$vD\delta\in C^{\upkappa}(\bar{\Omega})$. Thus, we only need to concentrate on
$\vartheta=\delta Dv$. Consider $\eta$ from Lemma 5.2 and with no loss of
generality, we may fix $\eta\in(0,\upkappa)$.

For $|x-y|\geq\frac{1}{8}(\delta(x)\vee\delta(y))$ it follows from Eq. 5.4
that

$\frac{|\vartheta(x)-\vartheta(y)|}{|x-y|^{\eta}}\leq
CK(\delta^{\upkappa}(x)+\delta^{\upkappa}(y))(\delta(x)\vee\delta(y))^{-\eta}\leq
2CK.$

So consider the case $|x-y|<\frac{1}{8}(\delta(x)\vee\delta(y))$. Without loss
of generality, we may assume that $|x-y|<\frac{1}{8}\delta(x)$. Then

$\frac{9}{8}\delta(x)\geq|x-y|+\delta(x)\geq\delta(y)\geq\delta(x)-|x-y|\geq\frac{7}{8}\delta(x).$

By Lemma 5.2, it follows

$\displaystyle\frac{|\vartheta(x)-\vartheta(y)|}{|x-y|^{\eta}}$
$\displaystyle\leq|Dv(x)|\frac{|\delta(x)-\delta(y)|}{|x-y|^{\eta}}+\delta(y)\frac{|Dv(x)-Dv(y)|}{|x-y|^{\eta}}$
$\displaystyle\lesssim\delta(x)^{\upkappa-1}(\delta(x))^{1-\eta}+\delta(y)[\delta(x)]^{\upkappa-1-\eta}$
$\displaystyle\leq CK.$

This completes the proof. ∎

##  Appendix A Appendix

In this section we aim to present a proof of Lemma 2.1. For this purpose, we
first introduce the scaled operator. Let $x_{0}\in\Omega$ and $r>0,$ we define
the doubly scaled operator as

$\displaystyle{\mathcal{L}}^{r,s}(x_{0})[x,u]=\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}\left\\{\operatorname*{Tr}a_{\theta\nu}(sr(x-x_{0})+sx_{0})D^{2}u(x)+{\mathcal{I}}^{r,s}_{\theta\nu}(x_{0})[x,u]\right\\}$
(A.1)

where

$\displaystyle{\mathcal{I}}^{r,s}_{\theta\nu}(x_{0})[x,u]=\int_{{\mathds{R}^{d}}}(u(x+y)-u(x)-\mathds{1}_{{\mathscr{B}}_{\frac{1}{sr}}}(y)\nabla
u(x)\cdot
y)r^{d+2}(s^{d+2}N_{\theta\nu}(rs(x-x_{0})+sx_{0},sry)\mathrm{d}{y}.$

Further, we define

${\mathcal{L}}^{0,s}(x_{0})[x,u]:=\displaystyle{\sup_{\theta\in\Theta}}\inf_{\nu\in\Gamma}\left\\{\operatorname*{Tr}a_{\theta\nu}(sx_{0})D^{2}u(x)\right\\}.$
(A.2)

Now we give the definition of weak convergence of operators.

######  Definition A.1.

Let $\Omega\subset{\mathds{R}^{d}}$ be open and $0<r<1.$ A sequence of
operators ${\mathcal{L}}^{m}$ is said to converge weakly to ${\mathcal{L}}$ in
$\Omega$, if for any test function $\varphi\in
L^{\infty}({\mathds{R}^{d}})\cap C^{2}({\mathscr{B}}_{r}(x_{0}))$ for some
${\mathscr{B}}_{r}(x_{0})\subset\Omega,$ we have

${\mathcal{L}}^{m}[x,\varphi]\to{\mathcal{L}}[x,\varphi]\,\,\,\,\,\,\text{uniformly}\,\,\,\text{in}\,\,\,{\mathscr{B}}_{\frac{r}{2}}(x_{0})\,\,\,\text{as}\,\,\,m\to\infty.$

The next lemma is a slightly modified version of [44, Lemma 4.1] which can be
proved by similar arguments.

######  Lemma A.1.

For any $x_{0}\in{\mathscr{B}}_{1},$ $r>0$ and $0<s<1,$ Let
${\mathcal{L}}^{r,s}(x_{0})$ and ${\mathcal{L}}^{0,s}(x_{0})$ is given by
(A.1) and (A.2) respectively where the Assumption 1.1 are satisfied by the
corresponding coefficients with $\Omega={\mathscr{B}}_{2}.$ Moreover, for
given $M,\varepsilon>0$ and a modulus of continuity $\rho,$ there exists
$r_{0},\eta>0$ independent of $x_{0}$ and $s$ such that if

  1. (i)

$r<r_{0}$, ${\mathcal{L}}^{0,s}(x_{0})[x,u]=0$ in ${\mathscr{B}}_{1},$

  2. (ii)

$\displaystyle{\mathcal{L}}^{r,s}(x_{0})[x,u]+C_{0}rs|Du(x)|\geq-\eta\,\,\,\,$
$\displaystyle in\,\,\,{\mathscr{B}}_{1}$
$\displaystyle{\mathcal{L}}^{r,s}(x_{0})[x,u]-C_{0}rs|Du(x)|\leq\eta\,\,\,\,$
$\displaystyle in\,\,\,{\mathscr{B}}_{1},$ $\displaystyle u=v\,\,\,$
$\displaystyle in\,\,\,\partial{\mathscr{B}}_{1}.$

  3. (iii)

$|u(x)|+|v(x)|\leq M$ in ${\mathds{R}^{d}}$ and
$|u(x)-u(y)|+|v(x)-v(y)|\leq\rho(|x-y|)$ for all
$x,y\in\overline{{\mathscr{B}}}_{1},$

then we have

$|u-v|\leq\varepsilon$

in ${\mathscr{B}}_{1}.$

It is worth mentioning that in [44], the authors have set a uniform continuity
assumption on the nonlocal kernels $N_{\theta\nu}(x,y)$ ( for the precise
assumption, see Assumption (C) of [44, p. 391] ) which is a standard
assumption to make for the stability property of viscosity solutions. Namely,
if we have a sequence of integro-differential operators ${\mathcal{L}}^{m}$
converging weakly to ${\mathcal{L}}$ in $\Omega$ and a sequence of
subsolutions (or supersolutions) in $\Omega$ converging locally uniformly on
any compact subset of $\Omega$, then the limit is also a subsolution (or
supersolution) with respect to ${\mathcal{L}}$. However in the case of the
operator ${\mathcal{L}}^{r,s}$ defined in (A.1), the nonlocal term
${\mathcal{I}}^{r,s}_{\theta\nu}$ can be treated as a lower order term that
converges to zero as $r\to 0$ without any kind of continuity assumptions on
nonlocal kernels $N_{\theta\nu}$.

Now we give the proof of Lemma 2.1.

###### Proof of Lemma 2.1.

We will closely follow the proof of [44, Theorem 4.1]. Fix any
$x_{0}\in{\mathscr{B}}_{1}$, let ${\mathcal{L}}^{r_{k},s}(x_{0})$ and
${\mathcal{L}}^{0,s}(x_{0})$ is given by (A.1) and (A.2) respectively. Then by
[44, Lemma 3.1] as $r_{k}\to 0,$ we have

${\mathcal{L}}^{r_{k},s}(x_{0})\to{\mathcal{L}}^{0,s}(x_{0}),$

in the sense of Definition A.1. By interior regularity [16, Corollary 5.7],
${\mathcal{L}}^{0,s}(x_{0})$ has $C^{1,\beta}$ estimate for an universal
constant $\beta>0.$ Now without loss of any generality we may assume that
$x_{0}=0$. Also dividing $u$ by $||u||_{L^{\infty}({\mathds{R}^{d}})}+K$ in
(2.1) we may assume that $K=1$ and $||u||_{L^{\infty}({\mathds{R}^{d}})}\leq
1.$
Using the Hölder regularity [44, Lemma 2.1], we have $u\in
C^{\beta}({\mathscr{B}}_{1}).$ Following [18, Theorem 52], we will show that
there exists $\delta,\mu\in(0,\frac{1}{4}),$ independent of $s$ and a sequence
of linear functions $l_{k}(x)=a_{k}+b_{k}x$ such that

$\displaystyle\begin{cases}(i)\,\,\,\displaystyle{\sup_{B_{2\delta\nu^{k}}}}|u-l_{k}|\leq\mu^{k(1+\gamma)}\;,\\\
(ii)\,\,\,\left|a_{k}-a_{k-1}\right|\leq\mu^{(k-1)(1+\gamma)}\;,\\\
(iii)\,\,\,\mu^{k-1}|b_{k}-b_{k-1}|\leq C\mu^{(k-1)(1+\gamma)}\;,\\\
(iv)\,\,\,\left|u-l_{k}\right|\leq\mu^{-k(\gamma^{\prime}-\gamma)}\delta^{-(1+\gamma^{\prime})}|x|^{1+\gamma^{\prime}}\,\,\,\,\text{for}\,\,\,x\in{\mathscr{B}}^{c}_{2\delta\mu^{k}}\;,\end{cases}$
(A.3)

where $0<\gamma<\gamma^{\prime}<\beta$ do not depend on $s$. We plan to
proceed by induction, when $k=0,$ since
$||u||_{L^{\infty}({\mathds{R}^{d}})}\leq 1$, (A.3) holds with
$l_{-1}=l_{0}=0$. Assume (A.3) holds for some $k$ and we shall show (A.3) for
$k+1.$

Let $\xi:\mathbb{R}^{d}\to[0,1]$ be a continuous function such that

$\displaystyle\xi(x)=\begin{cases}1\,\,\,\text{for}\,\,\,x\in{\mathscr{B}}_{3},\\\
0\,\,\,\text{for}\,\,\,x\in{\mathscr{B}}^{c}_{4}.\end{cases}$

Let us define

$\displaystyle w_{k}(x)=\frac{(u-\xi
l_{k})(\delta\mu^{k}x)}{\mu^{k(1+\gamma)}}.$

We claim that there exists a universal constant $C>0$, such that for all $k,$
we have

$\displaystyle{\mathcal{L}}^{r_{k},s}[x,w_{k}]-C_{0}r_{k}s|Dw_{k}(x)|\leq
C\delta^{2}\mu^{k(1-\gamma)}\leq C\delta^{2},$ (A.4)
$\displaystyle{\mathcal{L}}^{r_{k},s}[x,w_{k}]+C_{0}r_{k}s|Dw_{k}(x)|\geq-C\delta^{2}\mu^{k(1-\gamma)}\geq-C\delta^{2},$

in ${\mathscr{B}}_{2}$ in viscosity sense. Let $\phi\in
C^{2}({\mathscr{B}}_{2})\cap C({\mathds{R}^{d}})$ which touches $w_{k}$ from
below at $x^{\prime}$ in ${\mathscr{B}}_{2}.$ Let

$\psi(x):=\mu^{k(1+\gamma)}\phi\left(\frac{x}{\delta\mu^{k}}\right)+\xi
l_{k}(x).$

Then $\psi\in C^{2}({\mathscr{B}}_{2\delta\mu^{k}})\cap C({\mathds{R}^{d}})$
is bounded and touches $u$ from below at $\delta\mu^{k}x^{\prime}.$ Taking
$r_{k}=\delta\mu^{k},$ we have

${\mathcal{I}}^{r_{k},s}_{\theta\nu}[x^{\prime},\phi]=\delta^{2}\mu^{k(1-\gamma)}{\mathcal{I}}^{s}_{\theta\nu}[r_{k}x^{\prime},\psi-\xi
l_{k}].$

Thus we get

$\displaystyle{\mathcal{L}}^{r_{k},s}[x^{\prime},\phi]-C_{0}r_{k}s|D\phi(x^{\prime})|$
$\displaystyle=\delta^{2}\mu^{k(1-\gamma)}\Big{[}\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}\left\\{\operatorname*{Tr}a_{\theta\nu}(sr_{k}x^{\prime})D^{2}\psi(r_{k}x^{\prime})+{\mathcal{I}}^{s}_{\theta\nu}[r_{k}x^{\prime},\psi-\xi
l_{k}]\right\\}-sC_{0}|D\psi(r_{k}x^{\prime})-b_{k}|\Big{]}$
$\displaystyle\leq\delta^{2}\mu^{k(1-\gamma)}\Big{[}{\mathcal{L}}^{s}[r_{k}x^{\prime},\psi]-sC_{0}|D\psi(r_{k}x^{\prime})|+\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}\\{-{\mathcal{I}}^{s}_{\theta\nu}[r_{k}x^{\prime},\xi
l_{k}]\\}+sC_{0}|b_{k}|\Big{]}$ $\displaystyle\leq
C\delta^{2}\mu^{k(1-\gamma)}\leq C\delta^{2}.$

In the second last inequality we use that

${\mathcal{L}}^{s}[x,u]-C_{0}s|Du(x)|\leq 1,$

and $|a_{k}|$, $|b_{k}|$ are uniformly bounded and for all
$x^{\prime}\in{\mathscr{B}}_{2}$,
$\displaystyle{\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}}\\{-{\mathcal{I}}^{s}_{\theta\nu}[r_{k}x^{\prime},\xi
l_{k}]\\}$ is bounded independent of $s$ and $k$ . Thus we have proved

${\mathcal{L}}^{r_{k},s}[x,w_{k}]-C_{0}r_{k}s|Dw_{k}(x)|\leq
C\delta^{2}\,\,\,\text{in}\,\,\,{\mathscr{B}}_{2},$

in viscosity sense. Similarly the other inequality in (A.4) can be proven.

Define
$w^{\prime}_{k}(x):=\max\left\\{\min\left\\{w_{k}(x),1\right\\},-1\right\\}.$
We see that $w^{\prime}_{k}$ is uniformly bounded independent of $k.$ We claim
that in ${\mathscr{B}}_{\frac{3}{2}}$

$\displaystyle{\mathcal{L}}^{r_{k},s}[x,w^{\prime}_{k}]-C_{0}r_{k}s|Dw^{\prime}_{k}(x)|\leq
C\delta^{2}+\omega_{1}(\delta),$ (A.5)
$\displaystyle{\mathcal{L}}^{r_{k},s}[x,w^{\prime}_{k}]+C_{0}r_{k}s|Dw^{\prime}_{k}(x)|\geq-C\delta^{2}-\omega_{1}(\delta)$

Now take any bounded $\phi\in C^{2}({\mathscr{B}}_{2})\cap
C({\mathds{R}^{d}})$ that touches $w^{\prime}_{k}$ from below at $x^{\prime}$
in ${\mathscr{B}}_{3/2}.$ By the definition of $w^{\prime}_{k},$ in
${\mathscr{B}}_{2}$ we have $|w_{k}|=|w^{\prime}_{k}|\leq 1$ and $\phi$
touches $w_{k}$ from below at $x^{\prime}.$ Hence

$\displaystyle\sup_{\theta\in\Theta}\inf_{\nu\in\Gamma}\Big{\\{}\operatorname*{Tr}a_{\theta\nu}(sr_{k}x^{\prime})D^{2}\phi(x^{\prime})$
$\displaystyle+\int_{{\mathscr{B}}_{1/2}}(\phi(x^{\prime}+z)+\phi(x^{\prime})-\mathds{1}_{{\mathscr{B}}_{\frac{1}{rs}}}(z)D\phi(x^{\prime})\cdot
z)(r_{k}s)^{d+2}N_{\theta\nu}(r_{k}sx,sr_{k}z)\mathrm{d}{z}$
$\displaystyle-\int_{{\mathds{R}^{d}}\setminus{\mathscr{B}}_{1/2}}(w_{k}(x^{\prime}+z)-w^{\prime}_{k}(x^{\prime}+z)$
$\displaystyle+w^{\prime}_{k}(x^{\prime}+z)-\phi(x^{\prime})-\mathds{1}_{{\mathscr{B}}_{\frac{1}{rs}}}(z)D\phi(x^{\prime})\cdot
z))(r_{k}s)^{d+2}N_{\theta\nu}(r_{k}sx,sr_{k}z)\mathrm{d}{z}\Big{\\}}$
$\displaystyle-C_{0}r_{k}s|D\phi(x^{\prime})|\leq C\delta^{2}$

Therefore by Definition 2.1 of viscosity supersolution and using the bounds on
the kernel we get the following estimate:

$\displaystyle{\mathcal{L}}^{r_{k},s}[x,w^{\prime}_{k}]-C_{0}r_{k}s|Dw^{\prime}_{k}(x)|$
$\displaystyle\leq\int_{{\mathds{R}^{d}}\setminus{\mathscr{B}}_{1/2}}\left|w_{k}(x^{\prime}+z)-w^{\prime}_{k}(x^{\prime}+z)\right|(r_{k}s)^{d+2}k(r_{k}sz)\mathrm{d}{z}+C\delta^{2}.$

in the viscosity sense. By the inductive assumptions, we have $a_{k}$ and
$b_{k}$ uniformly bounded. Since
$\left|\left|u\right|\right|_{L^{\infty}({\mathds{R}^{d}})}\leq 1$ and $\xi
l_{k}$ is uniformly bounded, $\left|w_{k}\right|\leq C\mu^{-k(1+\gamma)}$ in
${\mathds{R}^{d}}.$ Using $(iv)$ from (A.3) we have

$|w_{k}(x)|=\frac{(u-\xi
l_{k})(r_{k}x)}{\mu^{k(1+\gamma)}}\leq\left(\frac{1}{r_{k}}\right)^{1+\gamma^{\prime}}|r_{k}x|^{1+\gamma^{\prime}}=|x|^{1+\gamma^{\prime}},$

for any $x\in{\mathscr{B}}^{c}_{2}\cap{\mathscr{B}}_{\frac{2}{r_{k}}}.$ Again
for any $x\in{\mathscr{B}}^{c}_{2/r_{k}},$ we find

$|w_{k}(x)|\leq
C\mu^{-k(1+\gamma^{\prime})}\cdot\mu^{-k(\gamma-\gamma^{\prime})}\leq
C\mu^{-k(1+\gamma^{\prime})}\leq
C\frac{\delta^{1+\gamma^{\prime}}}{2}|x|^{1+\gamma^{\prime}}\leq
C|x|^{1+\gamma^{\prime}}.$

Now, since $w^{\prime}_{k}$ is uniformly bounded, we have for
$x\in{\mathscr{B}}^{c}_{2},$

$|w_{k}|+|w^{\prime}_{k}-w_{k}|\leq
C\min\\{|x|^{1+\gamma^{\prime}},\mu^{-k(1+\gamma)}\\}.$ (A.6)

For $x^{\prime}\in{\mathscr{B}}_{3/2},$ using (A.6) we have the following
estimate.

$\displaystyle\int_{{\mathds{R}^{d}}}\left|w_{k}(x^{\prime}+z)-w^{\prime}_{k}(x^{\prime}+z)\right|(r_{k}s)^{d+2}k(r_{k}sz)\mathrm{d}{z}$
$\displaystyle\leq\int_{\\{z:|x^{\prime}+z|\geq
2\\}\cap{\mathscr{B}}_{1/r_{k}}}\left|w_{k}-w^{\prime}_{k}\right|(x^{\prime}+z)(r_{k}s)^{d+2}k(r_{k}sz)\mathrm{d}{z}+\delta^{2}\mu^{k(1-\gamma)}\int_{{\mathscr{B}}^{c}_{\frac{1}{r_{k}}}}\frac{(r_{k}s)^{d+2}k(r_{k}sz)}{(\delta\mu^{-k})^{2}}\mathrm{d}{z}$
$\displaystyle\leq
C\Big{[}\int_{{\mathscr{B}}^{c}_{1/2}\cap{\mathscr{B}}_{\frac{1}{\sqrt{r_{k}}}}}|z|^{2}(r_{k}s)^{d+2}k(r_{k}sz)\mathrm{d}{z}+r^{\frac{(1-\gamma^{\prime})}{2}}_{k}\int_{{\mathscr{B}}^{c}_{\frac{1}{\sqrt{r_{k}}}}\cap{\mathscr{B}}_{\frac{1}{r_{k}}}}|z|^{2}(r_{k}s)^{d+2}k(r_{k}sz)\mathrm{d}{z}$
$\displaystyle+\delta^{2}\mu^{k(1-\gamma)}\int_{{\mathscr{B}}^{c}_{s}}s^{2}k(z)\mathrm{d}{z}\Big{]}$
$\displaystyle\leq
C\Big{[}\int_{{\mathscr{B}}_{\sqrt{r_{k}}}}|y|^{2}k(y)\mathrm{d}{y}+(r^{\frac{(1-\gamma^{\prime})}{2}}_{k}+\delta^{2}\mu^{k(1-\gamma)})\int_{{\mathds{R}^{d}}}(1\wedge|y|^{2})k(y)\mathrm{d}{y}\Big{]}.$

Hence,

$\displaystyle\int_{{\mathds{R}^{d}}}\left|w_{k}(x^{\prime}+z)-w^{\prime}_{k}(x^{\prime}+z)\right|k^{r_{k},s}(z)\mathrm{d}{z}\leq\tilde{C}\left(\int_{{\mathscr{B}}_{\sqrt{\delta}}}|y|^{2}k(y)\mathrm{d}{y}+\delta^{\frac{1-\gamma^{\prime}}{2}}+\delta^{2}\right)=\omega_{1}(\delta)$

where $\omega_{1}(\delta)\to 0$ as $\delta\to 0.$ Therefore we proved
${\mathcal{L}}^{r_{k},s}[x,w^{\prime}_{k}]-C_{0}r_{k}s|Dw^{\prime}_{k}(x)|\leq
C\delta^{2}+\omega_{1}(\delta).$ The other inequality of (A.5) can be proved
in a similar manner.

Since $w^{\prime}_{k}$ satisfies the equation (A.5), by [44, Lemma 2.1] we
have $||w^{\prime}_{k}||_{C^{\beta}(\overline{{\mathscr{B}}_{1}})}\leq M_{1}$
for some $M_{1}$ independent of $k,s.$ Now we consider the a function $h$
which solves

$\displaystyle{\mathcal{L}}^{0,s}(x_{0})[x,h]=0\,\,\,\,$ $\displaystyle
in\,\,\,{\mathscr{B}}_{1}$ $\displaystyle h=w^{\prime}_{k}\,\,\,\,$
$\displaystyle on\,\,\,\partial{\mathscr{B}}_{1}.$

Existence of such $h$ can be seen from [51, Theorem 1]. Moreover, using [51,
Theorem 2] we have $||h||_{C^{\alpha}(\overline{{\mathscr{B}}_{1}})}\leq
M_{2}$ where $\alpha<\frac{\beta}{2}$ and $M_{2}$ is independent of $k,s.$ Now
for any $0<\varepsilon<1,$ let $r_{0}:=r_{0}(\varepsilon)$ and
$\eta:=\eta(\varepsilon)$ as given in Lemma A.1. Also for
$x\in{\mathscr{B}}_{1}$ and $\delta:=\delta(\varepsilon)\leq r_{0},$ we have

$\displaystyle{\mathcal{L}}^{r_{k},s}[x,w^{\prime}_{k}]+C_{0}r_{k}s|Dw^{\prime}_{k}(x)|\geq-\eta,$
$\displaystyle{\mathcal{L}}^{r_{k},s}[x,w^{\prime}_{k}]-C_{0}r_{k}s|Dw^{\prime}_{k}(x)|\leq\eta.$

Therefore by Lemma A.1, we conclude $|w^{\prime}_{k}-h|\leq\varepsilon$ in
${\mathscr{B}}_{1}.$ Again by using [16, Corollary 5.7], we have $h\in
C^{1,\beta}({\mathscr{B}}_{1/2})$ and we can take a linear part $l(x):=a+bx$
of $h$ at the origin. By $C^{1,\beta}$ estimate of
${\mathcal{L}}^{0,s}(x_{0})$ and $|w^{\prime}_{k}|\leq 1$ in
${\mathscr{B}}_{1}$ we obtain that the coefficients of $l,$ i.e, $a,b$ are
bounded independent of $k,s.$ Further for $x\in{\mathscr{B}}_{1/2},$ we have

$|h(x)-l(x)|\leq C_{1}|x|^{1+\beta},$

where $C_{1}$ is independent of $k,s.$ Hence using the previous estimate we
get

$|w^{\prime}_{k}(x)-l(x)|\leq\epsilon+C_{1}|x|^{1+\beta}\,\,\,\,in\,\,{\mathscr{B}}_{1/2}.$

Again using (A.6) and $|w_{k}|\leq 1$ in ${\mathscr{B}}_{2}$ we have

$\displaystyle|w_{k}(x)-l(x)|\leq 1+|a|+|b|\leq
C_{2}\,\,\,in\,\,\,{\mathscr{B}}_{1},$
$\displaystyle|w_{k}(x)-\xi(\delta\mu^{k}x)l(x)|\leq
C|x|^{1+\gamma^{\prime}}+C_{3}|x|\,\,\,in\,\,\,{\mathscr{B}}^{c}_{1}.$

Next defining

$\displaystyle
l_{k+1}(x):=l_{k}(x)+\mu^{k(1+\gamma)}l\left(\delta^{-1}\mu^{-k}x\right),$
$\displaystyle w_{k+1}(x):=\frac{(u-\xi
l_{k+1})(\delta\mu^{k+1}x)}{\mu^{(k+1)(1+\gamma)}},$

and following the proof of [44, Theorem 4.1] we conclude that (A.3) holds for
$k+1.$ This completes the proof. ∎

### Acknowledgement

We thank Anup Biswas for several helpful discussions during the preparation of
this article. Mitesh Modasiya is partially supported by CSIR PhD fellowship
(File no. 09/936(0200)/2018-EMR-I).

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