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$C^2$-regularity, operators of mixed order, critical, $L^\infty$-boundedness, H\"older estimate for the gradient.
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\title[Regularity for mixed operators]{ On some regularity properties \\ of mixed local and nonlocal \\
elliptic equations}


\author[X. Su]{Xifeng Su}
\address{School of Mathematical Sciences, Laboratory of Mathematics and Complex Systems (Ministry of Education)\\
Beijing Normal University,
No. 19, XinJieKouWai St., HaiDian District, Beijing 100875, P. R. China}
\email{xfsu@bnu.edu.cn, billy3492@gmail.com}

\author[E. Valdinoci]{Enrico Valdinoci}
\address{Department of Mathematics and Statistics, University of Western Australia, 35 Stirling Highway, WA 6009 Crawley, Australia}
\email{enrico.valdinoci@uwa.edu.au}
	
\author[Y. Wei]{Yuanhong Wei}
\address{School of Mathematics, Jilin University, No. 2699, Qianjin St., Changchun 130012, P. R.  China}
\email{weiyuanhong@jlu.edu.cn}

\author[J. Zhang]{Jiwen Zhang}
\address{School of Mathematical Sciences, Beijing Normal University, No. 19, XinJieKouWai St., HaiDian District, Beijing 100875, P. R. China}
\email{jwzhang628@mail.bnu.edu.cn}

\subjclass[2010]{35B65, 35R11, 35J67.}
 \keywords{$C^2$-regularity, operators of mixed order, critical, $L^\infty$-boundedness, H\"older estimate for the gradient.}

\thanks{Both X. Su and Y. Wei are supported by the National Natural Science Foundation of China (Grant No. 11971060, 11871242). Y. Wei is supported by the Natural Science Foundation of Jilin Province (Grant No. 20200201248JC), and the Scientific Research Project of the Education Department of Jilin Province (Grant No. JJKH20220964KJ).
E. Valdinoci is supported by the Australian Laureate Fellowship (Grant No. FL190100081).}

\begin{document}
\maketitle
\begin{abstract}
This article is concerned with ``up to $C^{2, \alpha}$-regularity results'' about a mixed local-nonlocal nonlinear elliptic equation which is driven by the superposition of Laplacian and fractional Laplacian operators. 

First of all, an estimate on the $L^\infty$ norm of weak solutions is established for more general cases than the ones present in the literature, including here critical nonlinearities.

 We then prove the interior $C^{1,\alpha}$-regularity and the $C^{1,\alpha}$-regularity up to the boundary of weak solutions, which extends previous results by the authors [X. Su, E. Valdinoci, Y. Wei and J. Zhang, Math. Z. (2022)], where the nonlinearities considered were of subcritical type. 

In addition, we establish the interior $C^{2,\alpha}$-regularity of solutions for all $s\in(0,1)$ and the $C^{2,\alpha}$-regularity up to the boundary for all $s\in(0,\frac{1}{2})$, with sharp regularity exponents.

For further perusal, we also include a strong maximum principle and some properties about the principal eigenvalue.
\end{abstract}

\tableofcontents

\section{Introduction}
The present paper is concerned with some regularity results about the following mixed local and nonlocal elliptic equation
\begin{equation}\label{Maineq}
\left\{%
\begin{array}{ll}
    - \Delta u +  (-\Delta)^s u = g(x,u) & \hbox{in $\Omega$,} \\
    u=0 &  \hbox{in $\mathbb{R}^n\backslash\Omega$,} \\
\end{array}%
\right.
\end{equation}
where $s\in (0,1)$ and~$\Omega$ is a bounded open set in  $\mathbb{R}^n$. 

As customary, the fractional Laplacian $ (-\Delta )^{s} $ is defined as
\begin{equation}
(-\Delta )^{s} u(x)=c_{n,s}\,{\text P.V.}\int_{\mathbb{R}^{n}} \frac{u(x)-u(y)}{|x-y|^{n+2s}}\, dy,
\end{equation}
 in which $ c_{n,s}>0 $ is a suitable normalization constant, whose explicit value does not play a role here, and~${\text P.V.}$ means that the integral is taken in the Cauchy principal value sense.

%Nonlocal problems of fractional Laplacian type
%arise in both pure mathematical research and concrete applications,
%such as the thin obstacle problem \cite{CSS08}, minimal
%surfaces \cite{CRS10,CV11}, phase transitions \cite{SV09}.
%  and fractional quantum mechanics
%\cite{Laskin00}. 
%Elliptic equations with fractional Laplacian are also
%studied by many other authors, see
%\cite{CR09,CT10,T11,MRS11,DS10,DPV12} and references therein.  \cite{Silvestre07}

The superposition operator
%the composition of the pseudo-differential elliptic operators
\begin{equation}\label{OPL}
\mathcal{L}=-\Delta  + (-\Delta)^s,
\end{equation}
is of nonlocal type, due to the presence of the fractional Laplacian, hence
equation~\eqref{Maineq} is  a nonlocal elliptic equation in a bounded domain $\Omega$ with homogeneous
Dirichlet exterior data.

We also recall that
the fractional Laplacian is the generator of a L\'{e}vy flight, while the Laplacian is the generator of a Brownian motion. Based on this, equation~\eqref{Maineq} can naturally model the combined effect of Brownian motions and L\'{e}vy flights. As observed in~\cite{DV21}, these operators describe a biological species whose individuals diffuse either by a random walk or by a jump process. The analysis of different types of mixed operators motivated by biological questions has also been
carried out in \cite{MPV13, PV18}.

Elliptic equations driven by the above mixed operator have recently received a great attention from
different points of view, including viscosity solution theory \cite{JK05, BI08, BJK10, BCCI12}, existence and non-existence theory~\cite{AC21, BDVV22a, SVWZ}, eigenvalue problems~\cite{DPFR19, DPLV22},
shape optimization results~\cite{BDVV23},
symmetry properties of the solutions~\cite{BVDV21}, etc. 

In contrast with 
the cases in which the pure classical Laplacian and the pure fractional Laplacian
are separately taken into account, this kind of problems combines the classical setting and the features typical of nonlocal operators in a framework that is not scale-invariant. Hence,
 different scales will lead to different features of the mixed operator from the case of single classical Laplacian or single fractional Laplacian, thereby making either one of them prevail, or both somewhat coexist in the end.

%Among all the topics of elliptic equations, the regularity of weak solution is a significant issue. There are considerable attentions on the regularity of weak solution for nonlinear elliptic problems using various methods. In \cite{CSS08}, Caffarelli, Salsa and Silvestre consider the $C^{1,s}$ regularity estimates for the free boundary of the obstacle problem involving the fractional Laplacian. The essential idea is  to replace  the problem  with an equivalent degenerate problem in one more dimension, which is first  proposed by Caffarelli  and Silvestre \cite{CS07}. In \cite{CT10}, Cabr\'e and Tan study the elliptic equation driven by the operator $A_{1/2}$, which plays the role of the square root of the Laplacian, in a bounded domain with zero Dirichlet boundary conditions
%\begin{equation*}
%\left\{%
%\begin{array}{ll}
%    -A_{1/2} u = u^p& \hbox{in $\Omega$,} \\
%    u=0 &  \hbox{on $\partial\Omega$,} \\
%    u>0 &\hbox{in $\Omega$.}
%\end{array}%
%\right.
%\end{equation*}
%They also treat the nonlocal problem through a local one in one more dimension by a Dirichlet to Neumann map. Assuming that the nonlinearity is subcritical, namely $1<p<\frac{n+1}{n-1}$, they establish the $L^{\infty}$ estimate as well as $C^{2,\alpha}$ regularity of positive weak solution. The method of proof is based on the blow-up technique, which is utilized by Gidas and Spruck  \cite{GS81} for the Laplacian case. 
{
The purpose of this article concentrates on establishing up to $C^{2,\alpha}$-regularity theory for mixed local-nonlocal elliptic equations, both in the interior and up to the boundary.

Our results are divided according to the following aspects:}.
\begin{itemize}
\item First of all, we extend the regularity result of \cite{SVWZ22} to some critical problems. More precisely, we prove the boundedness in $L^\infty$ of any weak solution of nonlinear problem~\eqref{Maineq}  with the critical Sobolev embedding exponent
\begin{equation}
	2^*=\begin{cases}
			\frac{2n}{n-2} &{\mbox{if }}n>2,\\
			\infty &{\mbox{if }}n\leqslant2.
		\end{cases}
	\end{equation}
%\footnote{ \begin{equation}
%	2^*=\begin{cases}
%			\frac{2n}{n-2} &n>2,\\
%			\infty &n\leqslant2,
%		\end{cases}\text{is the critical Sobolev embedding exponent for Laplacian operator $-\Delta$.}
%	\end{equation}

{\item We then  deduce a $C^{1, \alpha}$-regularity result up to the boundary. This will actually be obtained from the $W^{2,p}$-regularity of weak solutions  of problem \eqref{Maineq}. Moreover, we discuss the interior  $C^{1,\alpha}$-regularity for all $\alpha\in(0,1)$. 
%For this,  we use a truncation technique  and regard the fractional Laplacian as a ``perturbation" to avoid the trouble caused by the different scales.
	
	
\item Next we  show the interior  $C^{2,\alpha}$-regularity for all $s\in(0,1)$. We develop a bespoke argument to get over a difficulty caused by the nonlinear terms. Our argument combines a truncation technique with the above  $C^{1, \alpha}$ estimates. We notice when $g$ depends only on $x$, the interior  $C^{2,\alpha}$-regularity of problem \eqref{Maineq}  is studied in \cite{BDVV22b}.
 Our result is thereby a nonlinear version of~\cite{BDVV22b}. 
However, even in the case of linear equations, our method is  different from \cite{BDVV22b} since the working space that we choose here is different.}
\item Finally, we show  the  $C^{2,\alpha}$-regularity up to the boundary for \eqref{Maineq} when~$s\in (0,\frac{1}{2})$ and discuss the case when ~$s\in (\frac{1}{2}, 1)$. 
A restriction on the fractional exponent for this type of regularity is somewhat unavoidable,
as shown by one-dimensional examples in which one picks a solution of~\eqref{Maineq} in~$\Omega:=(0,1)$ which behaves, near~$0^+$, as~$ax+O(x^2)$, for some constant~$a$.

Also, the nonlinear term brings some difficulties for the bootstrap argument, in view of technical regularity issues related to the composition of functions with low and high smoothness. \item As a byproduct of our main results, we also provide in the appendix some auxiliary results in a formulation which is ``ready for use'' in forthcoming papers, such as a strong maximum principle and a characterization of the principal eigenvalue.
\end{itemize}

Now we start presenting our main results. First off, in the strategy that we proposed above, the $L^\infty$-regularity theory goes as follows.
\begin{Theorem}[$L^\infty$-regularity]\label{th: regularity}
	Let $ u$  be   a weak solution\footnote{The precise definition of weak solution will be recalled in Definition~\ref{definition of weak solution}.} of \eqref{Maineq}.
	Assume that there exist $ c>0 $ and $q\in [2,2^*]$ such that
	\begin{equation}\label{eq: H 1}
	|g(x,t)|\leqslant c(1+|t|^{q-1}) \quad \text{ for a.e. } x\in  \Omega ,  t\in \mathbb{R}.
	\end{equation}Then, $ u\in L^{\infty}(\Omega)$. 
	
	More precisely, there exists a constant $C_0= C_0(c,n, q, \Omega)>0$ independent of $u$, such that
	\begin{equation*}
	\|u\|_{L^\infty(\Omega)}\leqslant C_0\left(1+\int_{\Omega}|u|^{2^*\beta_1}\,dx\right)^{\frac{1}{2^*(\beta_1-1)}},
	\end{equation*}where~$\beta_1:=\frac{2^*+1}{2}$.
\end{Theorem}

	\begin{remark}{\rm
In~\cite{SVWZ22}, we established
		the $L^\infty$ boundedness of   weak solution,  provided that the exponent of the  nonlinear term is controlled by
				\begin{equation}
				2^*_s
				=\begin{cases}
					\frac{2n}{n-2s} &{\mbox{if }}n>2s,\\
					\infty &{\mbox{if }}n\leqslant2s.
				\end{cases}
		\end{equation} Here $2^*_s$ is the critical Sobolev embedding exponent for the space $H^s(\mathbb{R}^n)$ and characterizes the critical case for the operator $(-\Delta)^s$.
		
			Obviously, $2^*\geqslant2^*_s$ (and the inequality is strict when $n>2s$). This suggests that, in several cases of interest,  the critical exponent is actually $2^*$ rather than $2^*_s$, due to the interpolation inequality, and correspondingly
Theorem~\ref{th: regularity} can be seen as a sharpening of~\cite[Theorem~1.1]{SVWZ22}.}
	\end{remark}

Based on the $L^\infty$-regularity theory, one can employ the $W^{2,p}$ theory of weak solutions  to deduce a global $ C^{1,\alpha} $-regularity  as follows.

\begin{Theorem}[$C^{1,\alpha}$-regularity up to the boundary]\label{th:C^{1,alpha}}
Let $ u$  be   a weak solution of~\eqref{Maineq}.
	Assume that $g$ satisfies \eqref{eq: H 1}  and 
	$ \partial\Omega $ is of class  $ C^{1,1} $. 
	
	Then, $ u\in C^{1,\alpha}(\overline\Omega)$ for any $\alpha\in(0,\min\left\{1,2-2s\right\})$.
%	\begin{itemize}
%		\item [(1)] If $s\in (0,\frac{1}{2}]$, for any $\alpha \in (0,1)$, we have that~$u\in C^{1,\alpha}(\overline\Omega)$;
%		
%		\item [(2)] If $s\in (\frac{1}{2},1)$, for any $\alpha\in (0,2-2s)$, we have
%		that~$ u\in C^{1,\alpha}(\overline\Omega)$.
%	\end{itemize}
%	

More precisely, for~$s$ and~$\alpha$ as above, it holds that
\begin{equation}
	\|u\|_{C^{1,\alpha}(\overline{\Omega})}\leqslant C \left(1+\|u\|_{L^\infty(\Omega)}^{2^*}\right),
\end{equation}
where the positive constant $C$ depends only on $n,c,\alpha,\Omega,s$.
\end{Theorem}
We remark that Theorems~\ref{th: regularity} and~\ref{th:C^{1,alpha}} can be seen as a critical exponent version
of the $L^\infty$-regularity and of the $ C^{1,\alpha} $-regularity in \cite{SVWZ22}, respectively. 

In light of Theorem~\refeq{th:C^{1,alpha}}, one obtains an interior $ C^{2,\alpha} $-regularity result for the problem  \eqref{Maineq} provided that $\partial\Omega$ is $C^{1,1}$: we present this result in Theorem~\ref{th C^2,alpha interior} below for the sake of completeness, however
this regularity assumption on the domain
can be removed, as showcased in the forthcoming Theorem~\ref{th C^2,alpha interior without boundary condition}.

{\begin{Theorem}[Interior $ C^{2,\alpha} $-regularity]\label{th C^2,alpha interior}
		Let $ u$  be   a weak solution of \eqref{Maineq}. Assume that  $ \partial\Omega $ is  of class  $ C^{1,1} $ and 
			that {$ g\in C^\alpha_{\rm loc}(\Omega\times\mathbb{R}) $ } satisfies \eqref{eq: H 1} for any given $\alpha\in(0,1)$. 			
					Then $u\in C_{\rm loc}^{2,\alpha}(\Omega)$.
					
					More specifically, if~$W$ is an open domain with $\overline{W}\subset\Omega$,  and denote
					\[\rho:=\text{ dist}(W,\partial\Omega),\quad \text{  and } \quad  W_{\frac{\rho}{2}}:=\left\{x\in \Omega: \text{ dist}(x, W)<\frac{\rho}{2}\right\}. \]
					Then, 
				\begin{equation}\label{hjnsdc2kmdP}
							\|u\|_{C^{2,\alpha}(\overline{W})}\leqslant C \left(\|g\|_{L^\infty(\overline W_{{\rho}/{2}}\times\mathbb{R})}+\|g\|_{C^\alpha(\overline W_{{\rho}/{2}}\times\mathbb{R})}\|u\|_{C^1(\overline\Omega)}+ \|u\|_{C^{{1,1-s}}(\overline\Omega)}\right)
						\end{equation}
					for some constant $ C>0 $ depending only on $ n,s,\Omega,\alpha,W$.	
			\end{Theorem}
	}

The above results utilize the external condition in \eqref{Maineq}
	and the regularity of the domain.  We now present an alternative interior regularity theory for weak solutions of the equation without specifying any external condition and without any regularity assumption on the boundary of the domain.
Namely, by combining a suitable truncation method and a covering argument, we obtain the interior $ C^{1,\alpha} $-regularity and the interior $ C^{2,\alpha} $-regularity, as formulated in the following two results: 
	
		\begin{Theorem}[Interior $ C^{1,\alpha} $-regularity]\label{th C^1,alpha interior without boundary condition}
	Let $ g(x,t)\in L^\infty_{\rm loc}(\Omega\times\mathbb{R}) $, $ u \in X^1$ 
	be  a bounded  weak solution of 
	\begin{equation}\label{1.5}
		- \Delta u +  (-\Delta)^s u = g(x,u) \qquad  \hbox{in $\Omega$},
	\end{equation}
where $X^1$ will be defined in \eqref{X_1}.

	Assume that $V$ is an open  domain with $\overline{V}\subset\Omega$.

Then, $u\in C^{1,\alpha}_{\rm loc}(\Omega)$ for any $\alpha\in (0,1)$.	

More specifically, 
	\begin{equation}\label{final inequality }
		\|u\|_{C^{1,\alpha}(\overline V)}\leqslant C\left(\|g\|_{L^\infty( V_{\frac{3\rho}{4}} \times\mathbb{R}^n)}+ \|u\|_{L^\infty(\mathbb{R}^n)}\right)
	\end{equation}
	where the constant $ C>0 $ depends on $ n,s,\alpha,V $.
\end{Theorem}








%%We observe that the $C^{2,\alpha}$-regularity results are  helpful to find multiplesolutions of \eqref{Maineq} (see e.g. \cite{SVWZ}) and establish the following .
%
%
%
%
%We remark that Theorem~\ref{th C^2,alpha interior} uses the assumption that the domain is $C^{1,1} $ in order to have a finite
%right hand side in~\eqref{hjnsdc2kmdP}, due to Theorem~\ref{th:C^{1,alpha}}.
%In several cases, this assumption is suboptimal for local estimates, which can often be obtained without additional assumptions about boundary smoothness. However, the combined presence of the nonlocal term $(-\Delta)^su$ and of the nonlinearity~$g(\cdot,u(\cdot))$ in \eqref{Maineq} make it convenient to leverage the global regularity result to conclude the interior estimates. 

%{To remove any smoothness assumption on the domain,In light of Theorem~\ref{th C^1,alpha interior without boundary condition}, we  provide the following result:

\begin{Theorem}[Interior $ C^{2,\alpha} $-regularity]\label{th C^2,alpha interior without boundary condition}
				Let $ g(x,t)\in C^{\alpha}_{\rm loc}(\Omega\times\mathbb{R}) $ for any  $\alpha\in(0,1)$,  $ u\in X^1$ 
	 be  a  bounded  weak solution of
	 \begin{equation}\label{1.6}
- \Delta u +  (-\Delta)^s u = g(x,u) \qquad  \hbox{in $\Omega$}.
	 \end{equation}
 Assume that $V$ is an open  domain with $\overline{V}\subset\Omega$. 
 
 Then, $u\in C^{2,\alpha}(\overline V)$.
 
 Furthermore,
	\begin{equation}\label{final inequality 2 }
	\|u\|_{C^{2,\alpha}(\overline V)}\leqslant C\left(\|u\|_{L^\infty(\mathbb{R}^n)}+\|g\|_{C^{\alpha}(\overline V_{\frac{7\rho}{8}} \times\mathbb{R}^n)}\right) \left(1+\|g\|_{C^{\alpha}(\overline V_{\frac{7\rho}{8}} \times\mathbb{R}^n)}\right),
	\end{equation}
	where the constant $ C>0 $ depends on $ n,s,\alpha,V $. 
\end{Theorem}

%{We remark that the assumption (a) means  $g(x,t)$ depends linearly on $t$, or  $g(x,t)$  is independent of $t$.}

%{\color{red}Note that, based on Theorem~\ref{th C^1,alpha interior without boundary condition}, one can establish the $C^\alpha$-regularity in $\mathbb{R}^n$ of solution $u$ provided that the $\Omega$ is Lipschitz  domain.
%\begin{Theorem}
%	Let $\Omega$ be a bounded Lipschitz domain,  $ u\in X_0^1$  be   a weak solution of \eqref{Maineq}.
%	Assume that $g\in L^\infty(\Omega\times\mathbb{R})$ . Then we have
%	that~$ u\in C^{\alpha}(\mathbb{R}^n)$ for any $\alpha\in(0,1)$.
%\end{Theorem}}



Now we also consider the global $ C^{2,\alpha} $-regularity up to the boundary:

\begin{Theorem}[$ C^{2,\alpha} $-regularity up to the boundary]\label{theorem C^2 alpha global}
	Suppose that $ \partial\Omega $ is of class  $ C^{2,\alpha} $. Let $ s\in(0,\frac{1}{2}) $ and $ \alpha\in(0,1) $ be such that
		\begin{equation}\label{alpha if s in (0,1/2)}
		\alpha+2s\leqslant 1.
		\end{equation}
	If  $ u$ is a weak solution of  \eqref{Maineq} and $ g\in C^\alpha(\overline\Omega,\mathbb{R}) $ satisfies \eqref{eq: H 1}, then $u\in C^{2,\alpha}(\overline\Omega).$ 	
	
	Moreover, we have that
\begin{equation}
	\|u\|_{ C^{2,\alpha}(\overline\Omega)}\leqslant C \left(\|g\|_{C^\alpha(\overline{\Omega}\times\mathbb{R})}\|u\|_{C^1(\overline\Omega)}+\|g\|_{L^\infty({\Omega}\times\mathbb{R})}+\|u\|_{L^\infty(\Omega)}\right)
\end{equation}
for some positive constant $C$ depending only on $n,s,\alpha,\Omega$.
\end{Theorem}

\begin{remark}{\rm
In \cite[Theorem 1.6]{BDVV22b}, a related~$ C^{2,\alpha} $-regularity result is proved for linear equation. This result relies on the smoothness of the linear term and is not directly applicable to the nonlinear equation \eqref{Maineq}. 

In our context, for the interior $ C^{2,\alpha} $-regularity, we will deal with several technical issues coming from the nonlinearity $ g(x,u(x))$ to guarantee that one can carry out the iterative process.
For the $ C^{2,\alpha} $-regularity up to the boundary, we also treat the mixed operator with $s\in (0,\frac{1}{2})$ as a perturbation of the classical Laplacian and apply the classical results of local problems to obtain the main result.
}\end{remark}

The rest of this paper is organized as follows.
In Section~\ref{sec:preliminary}, we first introduce several preliminaries and define the working space.
Then, the $L^{\infty}$-boundedness of weak solutions
is established in Section~\ref{sec:regularity}, 
while Section~\ref{sec: C^alpha-regularity} is devoted to the~$C^{1,\alpha}$-regularity up to the boundary and  the interior ~$C^{1,\alpha}$-regularity. 

Next, the interior $C^{2,\alpha}$-regularity and the $C^{2,\alpha}$-regularity up to the boundary are shown in Section~ \ref{sec: C^2- regularity}. 

In the appendix we collect some simple consequences of the main results, such as a
strong maximum principle and a variational characterization for the principal eigenvalue.
These results are not stated in their maximal generality, but rather in a form which is ready to use
in nonlinear analysis problems (see~\cite{SVWZ}).

\section{Preliminaries and additional results}\setcounter{equation}{0}\label{sec:preliminary}
Let $ s\in(0,1)$ be given and $ \Omega\subset \mathbb{R}^{n}$ be an open bounded set.  
We introduce our working space, which is slightly different from  e.g.~\cite{BDVV22b,SVWZ22}. 

Let us first define the semi-norm 
\[
[u]_s:= \bigg(\int\!\!\!\int\nolimits_{\mathbb{R}^{2n}}\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\, dxdy\bigg)^{\frac{1}{2}}.
\]

In terms of the mixed  operator $\mathcal{L}$, a ``natural" function space to consider is the following space
\begin{equation}\label{X_1}
	X^1:=
	\left\{\begin{array}{ll}
		u:\mathbb{R}^{n}\rightarrow \mathbb{R} \text{ is Lebesgue measurable:} \vspace{0.2cm} \\ ~ u|_{\Omega} \in H^{1}(\Omega), ~~ [u]_s \text{ is finite}\\
	\end{array}
	\right\}
\end{equation}
equipped with the norm
\begin{equation*}
	\bigg(\|u\|_{H^{1}(\Omega)}^{2}+[u]_s^2 \bigg)^{\frac{1}{2}} .
\end{equation*}
Moreover, in order to study the solvability of Dirichlet problem \eqref{Maineq}, an appropriate space to employ is the following space\footnote{We prefer to use the function space $H_0^1(\Omega)$ rather than $H^1(\Omega)$, or~$H^1(\mathbb{R}^n)$ with zero data outside~$\Omega$,
to avoid trace theorems which may impose additional regularity assumptions on~$\Omega$. Notice indeed
that no regularity on~$\Omega$ is assumed in Theorem~\ref{th: regularity}.
See however the forthcoming Lemma~\ref{lemma equivalence of norms} addressing the equivalence of the norms
involved in each of these choices.} of functions
\begin{equation}\label{working space}
	X^1_0:=
	\left\{\begin{array}{ll}
	u:\mathbb{R}^{n}\rightarrow \mathbb{R} \text{ is Lebesgue measurable:} \vspace{0.2cm} \\ ~ u|_{\Omega} \in H^{1}_{0}(\Omega), ~~ [u]_s \text{ is finite}, ~ ~ u=0 \text{ a.e. in  } \mathbb{R}^n\backslash \Omega \\
	\end{array}
	\right\}
\end{equation}
equipped with the norm
\begin{equation*}
\bigg(\|u\|_{H^{1}_0(\Omega)}^{2}+[u]_s^2 \bigg)^{\frac{1}{2}} .
\end{equation*}
Due to the Poincar\'e inequality, the above norm in $ X_{0}^1 $ is also equivalent to
\begin{equation}
\|u\|_{X_0^1} :=\bigg(\|\nabla u\|^{2}_{2}+[u]_s^2\bigg)^{\frac{1}{2}}.  \label{Xnorm}
\end{equation}
Obviously, $X^1_{0} $ is a Hilbert space equipped with the inner product
\begin{equation*}
\left\langle u,v \right\rangle_{X_{0}^1}=\int_{\Omega} \nabla u\cdot  \nabla v\, dx+\int\!\!\!\int_{\mathbb{R}^{2n}}\frac{(u(x)-u(y))(v(x)-v(y))}{|x-y|^{n+2s}}\, dxdy, \quad \forall\ u,v \in X_{0}^1.
\end{equation*}
The following is the definition of weak solutions of \eqref{Maineq} in $X_{0}^1$: 
\begin{Definition}[Weak solution]\label{definition of weak solution}
	We say that $ u\in X_{0}^1 $ is a weak solution of the mixed elliptic equation \eqref{Maineq} , if $ u $ satisfies
	\begin{equation}
	\begin{split} \label{weaksolution}
	\displaystyle\int_{\Omega}\left\langle \nabla u,\nabla \varphi \right\rangle +& \displaystyle\int\!\!\!\int\nolimits_{\mathbb{R}^{2n}}\frac{(u(x)-u(y))(\varphi(x)-\varphi(y))}{|x-y|^{n+2s}}\, dxdy\\
	&=\int_{\Omega}g(x,u(x))\varphi(x)\, dx, \quad \text{  for any }  \varphi \in X_{0}^1.
	\end{split}
	\end{equation}
\end{Definition}




We investigate some key facts of $ X_0^1 $ in the following lemmata on equivalence of norms,  embedding and density results.

\begin{Lemma}\label{lemma equivalence of norms}
In the space $ X_{0}^1 $, we have the following equivalence of norms:
\[
||\cdot||_{X_0^1}\sim ||\cdot||_{H^1_0(\Omega)}. 
\]
\end{Lemma}
\begin{proof} 
	%Let $ u\in H_0^1(\Omega). $ Thanks to  the regularity assumptions on the domain $ \Omega $, we can extend $ u $ to a function $ \hat{u}:\mathbb{R}^n\rightarrow \mathbb{R} $ such that $ \hat{u}\in H^1(\mathbb{R}^n) $ and $ \|\hat{u}\|_{H^1(\mathbb{R}^n)}\leqslant C\|u\|_{H^1_0(\Omega)} $ for a suitable constant $ C $ (see e.g.\cite[Theorem 7.25]{GTbook}). 
	Let $ u\in H_0^1(\Omega) $ and $ u=0 $ a.e. in $ \mathbb{R}^n\backslash\Omega $.
	Using the change of variable $ z=y-x $, one has 
	\begin{equation}
	\begin{split}\label{H1 -X}
	&\int_{\Omega}\int_{\mathbb{R}^{n}\backslash\Omega}\frac{|u(x)|^2}{|x-y|^{n+2s}}\, dydx\\
	=&\int_{\Omega}\int_{\{x+z\notin\Omega\}\cap\{|z|\leqslant 1\}}\frac{|u(x)|^2}{|z|^{n+2s}}\, dzdx
	+\int_{\Omega}\int_{\{x+z\notin\Omega\}\cap\{|z|>1\}}\frac{|u(x)|^2}{|z|^{n+2s}}\, dzdx\\
	=&:A_1+A_2.
	\end{split}
	\end{equation}
	
	We observe that when~$x+z$ lies in the complement of~$\Omega$ we have
\begin{eqnarray*}&&
|u(x)|^2=|u(x+z)-u(x)|^2=\left| \int_0^1 \nabla u(x+tz)\cdot z\,dt\right|^2\le \int_0^1 |\nabla u(x+tz)|^2 |z|^2\,dt
\end{eqnarray*}
and therefore, making use of the change of variable~$y:=x+tz$ and the fact that~$u$ vanishes outside~$\Omega$,
we find that 
	\begin{equation}
	\begin{split}\label{A1}
	A_1\leqslant C_1(n,s)\|u\|_{H_0^1(\Omega)}^2.
	\end{split}
	\end{equation}
	
	Moreover,
	\begin{equation}\label{A2}
	A_2\leqslant C_2(n,s)\|u\|_{L^2(\Omega)}^2.
	\end{equation}
	Also, it follows from \cite{DPV12} that 
	\begin{equation}\label{A3}
	\int\!\!\!\int\nolimits_{\Omega\times\Omega}\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\, dxdy\leqslant C_3(n,s)\|u\|_{H_0^1(\Omega)}^2.
	\end{equation}
	Combining \eqref{A1}, \eqref{A2} and \eqref{A3}, we deduce that 
	\begin{equation*}
	[u]_s^2\leqslant C(n,s)\|u\|_{H_0^1(\Omega)}^2,
	\end{equation*}
	for some suitable constant $ C(n,s). $ 
	
	As a consequence of this, we obtain the equivalence between both norms $||\cdot||_{X_0^1}$ and $||\cdot||_{H^1_0}$.
\end{proof}

\begin{Lemma}\label{lemma: embedding continuous}
	The embedding $ X_{0}^1\hookrightarrow L^{2^*}(\Omega) $ is continuous.
\end{Lemma}
\begin{proof}
	Let $ u\in X_0^1$, invoking the Sobolev inequality, and we have 
	$$ \|u\|_{L^{2^*}(\Omega)}\leqslant C\|u\|_{H^1_0(\Omega)}\leqslant C\|u\|_{X_0^1} $$
for some suitable positive constant $ C$.
\end{proof}
\begin{Lemma}\label{lemma densely embedding}
	Let $\mathscr{C}_0^\infty:=\left\{u\in C(\mathbb{R}^n): u=0 \text{ in }\mathbb{R}^n\backslash \Omega, \;u|_{\Omega}\in C_0^\infty(\Omega) \right\} $ and denote by $\overline{\mathscr{C}_0^\infty}$ the closure of $ \mathscr{C}_0^\infty $ with respect to $X_0^1$-norm. Then, we have that~$ \overline{\mathscr{C}_0^\infty} = X_0^1 $.
\end{Lemma}
\begin{proof}
	It  is not difficult to see that  $ \overline{\mathscr{C}_0^\infty}\subset X^1_0 . $
	Besides, according to Lemma~\ref{lemma equivalence of norms}  and the fact $ C_0^\infty(\Omega) $ is dense in $ H_0^1(\Omega) $, one can deduce that $ X^1_0\subset\overline{\mathscr{C}_0^\infty}. $
\end{proof}
	
\section{$L^{\infty}$-regularity}\label{sec:regularity}\setcounter{equation}{0}

In this section, we show that the weak solutions $u\in X_0^1$ are bounded
in $ L^{\infty}(\Omega) $. To prove this result, we will use the
Moser iteration method.

To begin with, we construct an auxiliary function $\varphi$ and provide several fundamental properties of $\varphi$, which are useful for the iterative procedure.
\begin{Lemma}\label{lemma  auxiliary function}
	 Given $ \beta> 1 $ and $ T>0 $, we define
	 \begin{equation*}
	 \varphi(t)=
	 \begin{cases}
	 -\beta T^{\beta-1}(t+T)+T^\beta, &\text{  if }t\leqslant -T\\
	 |t|^\beta, &\text{  if } -T<t<T,\\
	 \beta T^{\beta-1}(t-T)+T^\beta, &\text{  if }t\geqslant T.
	 \end{cases}
	 \end{equation*}
	 Let $ u\in X_0^1 $, then  $ \varphi(u)\in X_0^1 $.
\end{Lemma}
 One can refer to \cite{SVWZ22} for the details of the proof. Note that the difference in the present paper is that we deal with $2^*$ rather than $2_s^*$.

\begin{Lemma}\label{lemma u in L beta1}
	 Under the assumptions of Theorem~\ref{th: regularity}, let $\beta_1$ is such that $ 2\beta_1-1=2^* $.
	 Then, $ u\in L^{2^*\beta_1}(\Omega)$.
\end{Lemma}
\begin{proof}
	Thanks to the convexity of~$\varphi$, one may deduce that
	\begin{equation}
	(-\Delta)^s\varphi(u)\leqslant \varphi'(u)(-\Delta)^su \label{4.2}
	\end{equation}
	in the sense of distributions.
	
	Moreover, we recall that
	\begin{equation*}
	\|u\|^2_{L^{2^*_s}(\Omega)}\leqslant S\iint_{\mathbb{R}^{2n}}\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\, dxdy,
	\end{equation*}
	where $ S$ is a constant depending only on $n$ and $s$.
	
	Hence, by \eqref{4.2}, we have that
	\begin{equation}
	\begin{split}
	\|\varphi(u)\|^2_{L^{2^*_s}(\Omega)}&\leqslant S \iint_{\mathbb{R}^{2n}}\frac{|\varphi(u(x))-\varphi(u(y))|^2}{|x-y|^{n+2s}}\, dxdy\\&
	=2S\int_{\mathbb{R}^{n}}\varphi(u)(-\Delta)^s\varphi(u)\, dx
	 \leqslant 2S\int_{\mathbb{R}^{n}}\varphi(u)\, \varphi'(u) \,(-\Delta)^su\, dx.
	\end{split}\label{3.2BIS}
	\end{equation}
	We multiply the first equation of \eqref{Maineq} by  $ \varphi(u)\varphi'(u) $ and  integrate over $ \mathbb{R}^n $. In this way, we obtain that
	\begin{equation}\label{integration of the main equation}
	-\int_{\mathbb{R}^{n}}(\varphi(u)\varphi'(u))\Delta u\,dx+\int_{\mathbb{R}^{n}}\varphi(u)\,\varphi'(u)\,(-\Delta)^su\, dx=\int_{\mathbb{R}^{n}}\varphi(u)\varphi'(u)g(x,u)\, dx.
	\end{equation}
	The first term on the left side of \eqref{integration of the main equation} can be rewritten as
	\begin{equation}\label{rewriting of the first term}
	\begin{split}
	&-\int_{\Omega} (\varphi(u)\varphi'(u))\Delta u\, dx= \int_{\Omega}\nabla u\cdot \nabla(\varphi(u)\varphi'(u))\, dx\\
	=&\int_{\Omega}|\nabla u|^2|\varphi'(u)|^2\, dx+\int_{\Omega}|\nabla u|^2\varphi(u)\,\varphi''(u)\,  dx.
	\end{split}
	\end{equation}
	Equations  \eqref{3.2BIS}, \eqref{integration of the main equation}  and \eqref{rewriting of the first term} lead to
	\begin{align}
	\int_{\Omega}|\nabla u|^2|\varphi'(u)|^2\, dx
	&\leqslant\int_{\mathbb{R}^n}\varphi(u)\,\varphi'(u)\,g(x,u)\, dx. \label{4.5}
	\end{align}
Hence, recalling~\eqref{eq: H 1} and utilizing the Sobolev inequality, we obtain
\begin{align*}
\|\varphi(u)\|^2_{L^{2^*}(\Omega)}&\leqslant C \|\nabla (\varphi(u))\|_{L^2(\Omega)}^2= C\int_{\Omega}|\varphi'(u) \nabla u|^2\,dx\\
& \leqslant C\int_{\mathbb{R}^n}\varphi(u)\varphi'(u)g(x,u)\, dx
\leqslant C \int_{\mathbb{R}^n}\varphi(u)|\varphi'(u)|(1+|u|^{2^*-1})\, dx\\
& =C\bigg(\int_{\mathbb{R}^n}\varphi(u)|\varphi'(u)|\,dx +\int_{\mathbb{R}^n}\varphi(u)|\varphi'(u)||u|^{2^*-1}\, dx \bigg),
\end{align*}
where the constant $ C $ depends only on $ n,\Omega $ and $ c $.
	
	Employing the estimates \begin{eqnarray*}&& \varphi(u)\leqslant |u|^\beta ,\qquad |\varphi'(u)|\leqslant \beta |u|^{\beta-1} \qquad{\mbox{and}}\qquad |u\varphi'(u)|\leqslant \beta \varphi(u) ,\end{eqnarray*} we see that
	\begin{equation}\label{main estimate}
	\bigg(\int_{\Omega}(\varphi(u))^{2^*}\,dx\bigg)^{2/2^*}\leqslant C\beta\bigg(\int_{\Omega}|u|^{2\beta-1}\,dx +\int_{\Omega}(\varphi(u))^2 |u|^{2^*-2}\, dx \bigg).
	\end{equation}
	Notice that $ C $ is a positive constant that does not depend on $ \beta $ and  that the last integral on the right-hand-side of the inequality \eqref{main estimate} is well defined for every~$ T >0 $. 
	
	Furthermore, for any given $ R>0 $, we apply the H\"older inequality  and obtain that
	\begin{equation}
	\begin{split}
	&\int_{\Omega}(\varphi(u))^2 |u|^{2^*-2}\, dx=\int_{\left\{|u|\leqslant R\right\}}(\varphi(u))^2 |u|^{2^*-2}\, dx +\int_{\left\{|u|>R\right\}} (\varphi(u))^2 |u|^{2^*-2}\, dx \\
	\leqslant& \int_{\left\{|u|\leqslant R\right\}}\frac{(\varphi(u))^2}{|u|} R^{2^*-1}\, dx+\bigg(\int_{\Omega}(\varphi(u))^{2^*}\, dx\bigg)^{2/2^*}\bigg(\int_{\left\{|u|>R\right\}}|u|^{2^*}\, dx\bigg)^{\frac{2^*-2}{2^*}}.
	\end{split}\label{estimate of phi u}
	\end{equation}
	By the Monotone Convergence Theorem, one can choose $R>0$ large enough such that
	\begin{equation}\label{eq:C beta}
	\bigg(\int_{\left\{|u|>R\right\}}|u|^{2^*}\,dx\bigg)^{\frac{2^*-2}{2^*}}\leqslant \frac{1}{2C\beta_1},
	\end{equation}
	where $ C $ is the constant in \eqref{main estimate}. 
	
	Therefore, one can reabsorb the last term in \eqref{estimate of phi u} into the left hand side of \eqref{main estimate} to get
	\begin{equation*}
	\bigg(\int_{\Omega}(\varphi(u))^{2^*}\,dx\bigg)^{2/2^*}\leqslant 2C\beta_1\bigg(\int_{\Omega}|u|^{2^*}\, dx+ R^{2^*-1}\int_{\Omega}\frac{(\varphi(u))^2}{|u|}\, dx \bigg).
	\end{equation*}
	Now, using that $\varphi(u)\leqslant |u|^{\beta_1}$, we conclude that the terms in the right hand side of the above inequality is bounded independently of $T$.
	
	Sending $ T\rightarrow +\infty $, we obtain that
	\begin{equation}\label{eq:u in L beta1}
	\bigg(\int_{\Omega}|u|^{2^*\beta_1}\,dx\bigg)^{2/2^*}\leqslant 2C\beta_1\bigg(\int_{\Omega}|u|^{2^*}\, dx+ R^{2^*-1}\int_{\Omega}|u|^{2^*}\, dx \bigg)<+\infty,
	\end{equation}as desired.
\end{proof}

Now we complete the proof of the $L^\infty$-regularity result:

 \begin{proof}[Proof of Theorem~\ref{th: regularity}]
 	For $ m\geqslant 1, $ we define $ \beta_{m+1} $ such that
 	\begin{equation*}
 	2\beta_{m+1}+2^*-2=2^*\beta_{m}.
 	\end{equation*}
 	As in \cite{SVWZ22}, we infer that
 	\begin{equation*}
 	\bigg(1+\int_{\Omega}|u|^{2^*\beta_{m+1}}\,dx\bigg)^{\frac{1}{2^*(\beta_{m+1}-1)}}\leqslant (C\beta_{m+1})^{\frac{1}{2(\beta_{m+1}-1)}}\bigg(1+\int_{\Omega} |u|^{2^*\beta_{m}}\, dx\bigg)^{\frac{1}{2^*(\beta_{m}-1)}}.
 	\end{equation*}
 	We now define $ C_{m+1}:=C\beta_{m+1} $ and
 	\begin{equation*}
 	A_m:=\bigg(1+\int_{\Omega} |u|^{2^*\beta_{m}}\, dx\bigg)^{\frac{1}{2^*(\beta_{m}-1)}}.
 	\end{equation*}
 	In particular, note that $A_1= \bigg(1+\int_{\Omega} |u|^{2^*\beta_{1}}\, dx\bigg)^{\frac{1}{2^*(\beta_{1}-1)}}$ is bounded by \eqref{eq:u in L beta1}.
 	
 It is not difficult to verify that there exists a constant $ C_0>0 $ independent of $m$, such that
 	\begin{equation}\label{DRI}
 	A_{m+1}\leqslant \prod_{k=2}^{m+1}C_k^{\frac{1}{2(\beta_{k}-1)}}A_1\leqslant C_0 A_1.
 	\end{equation}
 	We stress that once~\eqref{DRI} is established,
 	then, by the H\"older inequality, we conclude that $ u\in L^p(\Omega)$, for every~$ p\in[1,+\infty)$. Furthermore,
 	a limit argument implies that
 	\begin{equation*}
 	\|u\|_{L^\infty(\Omega)}\leqslant C_0 {A_1}<+\infty,
 	\end{equation*}
 	 where $ C_0 $ depends only on $ n,q,\Omega,c $, which finishes the proof of Theorem~\ref{th: regularity}.
 \end{proof}

\section{$ C^{1,\alpha}$-regularity }\label{sec: C^alpha-regularity}\setcounter{equation}{0}

In this section, we focus on the $ C^{1,\alpha}$-regularity up to the boundary  and on the interior $ C^{1,\alpha}$-regularity for operator $\mathcal{L}$.

\subsection{$ C^{1,\alpha}$-regularity up to the boundary }\label{sec: C^1,alpha-regularity}\setcounter{equation}{0}

The $C^{1,\alpha}$-regularity up to the boundary result for $\mathcal{L}$ is now an immediate corollary of  the following  $W^{2,p} $-regularity theorem.

\begin{Theorem}\label{theorem w 2p}
	Let $ \Omega $ be a $ C^{1,1} $ domain in $ \mathbb{R}^n $ and $ s\in (0,1)$. If $ f\in L^p(\Omega) $ 
	where  $p$ satisfies \begin{equation}
		\begin{cases}\label{eq definition of p}
			1<p<\infty ,\qquad &{\mbox{ if }}s\in(0,\frac{1}{2}] ,     \\
			n<p<\frac{n}{2s-1},& {\mbox{ if }}s\in(\frac{1}{2},1).
		\end{cases}
	\end{equation}
Then,
the problem
	\begin{equation}\label{eq:W{2,p} main equation}
	-\Delta u+(-\Delta)^su=f, \qquad \text{in } \Omega
	\end{equation}
	has a unique solution $ u\in W^{2,p}(\Omega)\cap W^{1,p}_0(\Omega)$.
	
	Furthermore,
	\begin{equation}\label{eq:W 2p estimate}
	\|u\|_{W^{2,p}}\leqslant C\bigg(\|u\|_{L^p}+\|f\|_{L^p}\bigg),
	\end{equation}
	where the positive constant $ C$ depends only on $ \Omega, n, s, p $.
\end{Theorem}

Theorem~\ref{theorem w 2p} can be obtained by  following an argument given in \cite{SVWZ22}.
For the reader's convenience we sketch this argument here, by pointing out the following steps:
\begin{itemize}
	\item[Step 1.] Obtain $ L^p $-estimates on the operator $ (-\Delta)^s$;
	\item[Step~2.] Apply the fixed point theorem  to deduce that the problem
	\begin{equation*}
	-\Delta u+\lambda u=f-(-\Delta)^su, \qquad \text{in } \Omega
	\end{equation*}
	has a unique solution $ u\in W^{2,p}(\Omega)\cap W^{1,p}_0(\Omega) $ for $ \lambda>0$ large enough;
	\item[Step~3.] Employ the maximum principle  in  \cite[Theorem 1.2]{BDVV22b} and the
	bootstrap method to conclude.
\end{itemize}

With this, we can now complete the proof of the $C^{1,\alpha}$-regularity theory by proceeding as follows.

\begin{proof}[Proof of Theorem~\ref{th:C^{1,alpha}}]
Let $ u\in X_0^1 $ be the weak solution of \eqref{Maineq}. Theorem~\ref{th: regularity} gives that the map~$x\mapsto g(x,u(x))$
belongs to~$ L^\infty(\Omega)$.

Moreover, due to Theorem~\ref{theorem w 2p}, we  deduce that  there exists a unique solution $ v\in W^{2,p}(\Omega)\cap W^{1,p}_0(\Omega) $ for every $p$ satisfying \eqref{eq definition of p}. By combining this with the Sobolev inequality, we obtain that
 \begin{itemize}
	\item [(1)] if $s\in (0,\frac{1}{2}]$, then for any $\alpha \in (0,1)$,  $v\in C^{1,\alpha}(\overline\Omega)$;
	
	\item [(2)] if $s\in (\frac{1}{2},1)$, then for any $\alpha\in (0,2-2s)$,  $v\in C^{1,\alpha}(\overline\Omega)$.
\end{itemize}

Finally, the Lax-Milgram Theorem yields that $ u=v $. In particular, according to  \eqref{eq:W 2p estimate}, \eqref{eq: H 1} and the Sobolev inequality, we obtain that
\begin{equation}
	\begin{split}
		\|u\|_{C^{1,\alpha}(\overline{\Omega})}&\leqslant C\left(\|u\|_{L^\infty(\Omega)}+\|g(\cdot,u(\cdot))\|_{L^\infty(\Omega)}\right)
		\leqslant C \left(\|u\|_{L^\infty(\Omega)}+c(1+\|u\|_{L^\infty(\Omega)}^{2^*-1})\right),
	\end{split}
\end{equation}
where the positive constant $C$ depends only on $n,\alpha,\Omega,s$.
\end{proof}

\subsection{Interior  $ C^{1,\alpha} $-regularity}\label{sec: C^1-interior regularity}\setcounter{equation}{0}
The  interior $ C^{1,\alpha} $-regularity for the mixed operator $ \mathcal{L}$ will be derived by a mollifier technique and a cutoff argument. We split the proof of Theorem~\ref{th C^1,alpha interior without boundary condition} into three steps.

\noindent{Step 1.} Regularize the solution $u$ by the standard mollifier to get some properties about $u_\epsilon $. See subsection~\ref{sec: Mollifier argument 1 }.

\noindent{Step 2.} Use the cut off argument for $u_\epsilon$ and employing \cite[Proposition 2.18]{MR4560756} to get the interior $C^{1,\alpha}$-norm  of $u_\epsilon$ in $B_R$ for any $R>0$ small enough. See subsection~\ref{sub A priori C{2,alpha}-estimate 1}.
% in some domain by the $C^{1,\alpha}$-norm of $u_\epsilon$ in larger domain and and $C^{\alpha}$-norm  of $u_\epsilon$ in $\mathbb{R}^n$

\noindent{Step 3.}  Apply  Proposition~\ref{pro final estimate 1} and  Arzel\`{a}-Ascoli Theorem  to get the inclusion. See subsection~\ref{sec:Proof of Theorem1.5 1}.

We now introduce some notations.
\begin{itemize}
\item [(a)] Let $V$ be an open set  with $ V\Subset\Omega $, and denote
\begin{equation}\label{definition of rho 1}
	\rho:=\text{ dist}(V,\partial\Omega),\quad \text{  and } \quad  V_{\delta}:=\left\{x\in \Omega: \text{ dist}(x, V)<\delta\right\}.
\end{equation}
\item [(b)] For any given $x_0\in V_{\frac{\rho}{4}}$ where $V_{\frac{\rho}{4}} $ is given in \eqref{definition of rho 1}, denote
\[0<R<\min\left(\frac{1}{2},\frac{\rho}{10}\right),\quad B_{R}(x_0):=\left\{x\in \Omega: |x-x_0|<R \right\}. \]
\item [(c)] Denote the interior norm  as follows:
\begin{equation}
	[u]_{\alpha;B_{R}(x_0)}=\sup\limits_{x,y\in B_{R}(x_0)}\frac{|u(x)-u(y)|}{|x-y|^\alpha},\qquad 0<\alpha<1;
\end{equation}
\begin{equation}
	|u|^\prime_{k;B_{R}(x_0)}=\sum\limits_{j=0}\limits^{k}R^j\|D^j u\|_{L^\infty(B_{R}(x_0))};
\end{equation}
\begin{equation}
	|u|^\prime_{k,\alpha;B_{R}(x_0)}=|u|^\prime_{k;B_{R}(x_0)}+R^{k+\alpha}[D^ku]_{\alpha;B_{R}(x_0)}.
\end{equation}
\end{itemize}
\subsubsection{{A Mollifier technique}}\label{sec: Mollifier argument 1 }		

Let $u\in X^1$ solve the equation \eqref{1.5} and $\eta_\epsilon$ is the standard mollifier. For every $x\in\mathbb{R}^n$, we denote 
\[ u_\epsilon(x):=(\eta_\epsilon\ast u)(x)= \int_{\Omega} \eta_\epsilon(x-y)u(y)\, dy= \int_{|y|\leqslant \epsilon} \eta_\epsilon(y)u(x-y)\, dy\quad \forall 0<\epsilon<R. \]

Recall that if $u$ solves the following equation in the weak sense
\begin{equation}
-\Delta u+(-\Delta)^s u= g(x,u) \qquad \text{  in } \Omega,
\end{equation} 
then, for every $\epsilon\in(0,R)$ small enough, 
\begin{equation}\label{eq u epsilon1}
-\Delta(\eta_\epsilon\ast u)+(-\Delta)^s (\eta_\epsilon \ast u)= \eta_\epsilon \ast g(x,u) \qquad \text{  in } V_{\frac{3\rho}{4}}.
\end{equation}
We may assume that 
$ g_\epsilon:=\eta_\epsilon\ast g(x,u) $. Then, $u_\epsilon$ and $g_\epsilon$ satisfy the following equation 
\begin{equation}
-\Delta u_\epsilon+(-\Delta)^s u_\epsilon= g_\epsilon \qquad \text{  in } V_{\frac{3\rho}{4}}.
\end{equation}
Moreover, from the following two Propositions~\ref{pro u epsilon1} and \ref{pro g epsilon1}, we can see that 
\[ u_\epsilon \in C^{2}(\overline V_{\frac{3\rho}{4}})\cap L^\infty(\mathbb{R}^n) \quad\text{and}\quad g_\epsilon\in L^\infty (B_{R}(x_0) )\qquad \forall 0<\epsilon<R.  \]

\begin{Proposition}\label{pro u epsilon1}
Assume that $u$ is a bounded solution of \eqref{1.5},  then $u_\epsilon\in L^\infty(\mathbb{R}^n)$. In particular,
\begin{equation}\label{eq:u epsilon C11}
	\|u_\epsilon\|_{L^\infty(\mathbb{R}^n)}\leqslant \|u\|_{L^\infty(\mathbb{R}^n)} \qquad \forall 0<\epsilon<R.
\end{equation}
\end{Proposition}
\begin{proof}
$ \forall x\in\mathbb{R}^n$, one has
\begin{equation}
	|u_\epsilon(x)|\leqslant \int_{\Omega}\eta_\epsilon(y)|u(x_1-y)|\, dy
	\leqslant \|u\|_{L^\infty(\mathbb{R}^n)}.
\end{equation}
Consequently,  one can deduce that $u_\epsilon\in L^\infty(\mathbb{R}^n)$ and \eqref{eq:u epsilon C11}.
\end{proof}
\begin{Proposition}\label{pro g epsilon1}
Let $g\in L^\infty_{\rm loc}(\Omega\times\mathbb{R}^n)$,  for any  $\epsilon\in (0,R)$, 
then  \begin{equation}\label{g_epsilon1}
	\|g_\epsilon(\cdot,u(\cdot))\|_{L^\infty( B_{R}(x_0))}\leqslant \|g\|_{L^\infty(\overline B_{2R}(x_0) \times\mathbb{R}^n)}.
\end{equation}
\end{Proposition}
\begin{proof}
For every $x\in B_{R}(x_0)$, one has
\begin{equation}
	|g_\epsilon(x,u(x))|=\int_{|y|\leqslant \epsilon} |\eta_\epsilon(y)g(x-y,u(x-y))|\, dy\leqslant \|g\|_{L^\infty(\overline B_{2R}(x_0) \times\mathbb{R}^n)}.\qedhere
\end{equation}
\end{proof}

\subsubsection{{A priori $C^{1,\alpha}$-estimate}}\label{sub A priori C{2,alpha}-estimate 1}

We now use the cut off argument for $u_\epsilon$ to estimate the $C^{1,\alpha}$-norm  of $u_\epsilon$. 

Consider a cut off function $ \phi\in C^\infty_0(\mathbb{R}^n,\mathbb{R}) $ satisfying 
\begin{equation}\label{cutoff function 21}
\begin{cases}
	\phi \equiv 1 \text{  on } B_{\frac{3}{2}}(0)\\
	\text{supp}(\phi)\subset B_{2}(0);\\
	0\leqslant \phi\leqslant 1 \text{  on } \mathbb{R}^n.
\end{cases}
\end{equation}
Denote 
\begin{equation}\label{R}
	\phi^R(x):=\phi\left(\frac{x-x_0}{R}\right).
\end{equation}
Thus, one has that
\[ B_{4R}(x_0)\subset V_{\frac{3\rho}{4}}\subset \Omega, \quad \text{supp}(\phi^R)\subset B_{2R}(x_0). \]

\begin{Lemma}\label{lemma v epsilon1}
We assume that  $ g(\cdot,u(\cdot))\in L^\infty(\Omega) $.
Suppose that  $ u\in C^{2}(\overline V_{\frac{3\rho}{4}})\cap L^\infty(\mathbb{R}^n)$  solves the following equation 
\begin{equation}\label{u is a solution in V1}
	-\Delta u+(-\Delta)^s u= g \qquad \text{  in } V_{\frac{3\rho}{4}}.
\end{equation}


Then, there exists~$ \psi\in L^{\infty}( B_{R}(x_0),\mathbb{R}) $ such that $ v:=\phi^R u $ satisfies 
\begin{equation}\label{eq: new equation21}
	-\Delta v+(-\Delta)^s v= \psi \qquad \text{  in } V_{\frac{3\rho}{4}}.
\end{equation}
In addition, 
\begin{equation}\label{estimate psi 21}
	R^2\|\psi\|_{L^\infty(B_{R}(x_0))}\leqslant  C_{\rho,n,s}\left(\| g(\cdot,u(\cdot))\|_{L^\infty(B_{R}(x_0))}+\|u\|_{L^\infty(\mathbb{R}^n)}\right)
\end{equation}
for some positive constant $C_{n,s,\rho}$.

\end{Lemma}

\begin{proof}
From  the definition of the function $ v $, we know that
\begin{equation}\label{eq: function space of v21}
	v\in C^{2}_0( B_{2R}(x_0))\subset C^{2}_0(\mathbb{R}^n). 
\end{equation}
Furthermore, to check~\eqref{eq: new equation21}, one can calculate that  
\begin{equation}\label{eq: new equation:PSI2 11}
	\psi= g(x,u)+\Delta (u(1-\phi^R))-(-\Delta)^s (u(1-\phi^R))\quad  \text{  in } V_{\frac{3\rho}{4}}.
\end{equation} 
We also notice that $\phi^R=1$ in $B_{3R/2}(x_0)$ and that  \begin{equation}\label{eq: new equation:PSI21}
	\psi= g(x,u)-(-\Delta)^s (u(1-\phi^R))\quad  \text{  in } B_{3R/2}(x_0).
\end{equation}

We  now obtain suitable $ L^\infty $- estimates on the function $  (-\Delta)^s (u(1-\phi^R)) $.
Observing that for $x\in B_{R}(x_0)$, we have that
\begin{equation}
	(-\Delta)^s (u(1-\phi^R))(x)=\int_{\mathbb{R}^n\backslash B_{3R/2}(x_0)} \frac{-u(1-\phi^R)(y)}{|x-y|^{n+2s}}\, dy.
\end{equation}
{F}rom this and the fact that \begin{equation}\label{fact}
	|x-y|\geqslant |y-x_0-R|\geqslant |y-x_0|-R\geqslant \frac{1}{5}(R+|y-x_0|)\quad \forall y\in \mathbb{R}^n\backslash B_{3R/2}(x_0), x\in B_{R}(x_0),
\end{equation}
one obtains that
\begin{equation}
	\begin{split}\label{step 2 11}
		\|(-\Delta)^s (u(1-\phi^R))\|_{L^\infty(B_{R}(x_0))}\leqslant& 5^{n+2s}\int_{ \mathbb{R}^n\backslash B_{3R/2}(x_0)}\frac{|u(y)|}{(R+|y-x_0|)^{n+2s}}\, dy\\
		\leqslant& C_{n,s}\|u\|_{L^\infty(\mathbb{R}^n)}R^{-2s}.
	\end{split}
\end{equation}
Since $ g(\cdot,u(\cdot))\in L^\infty(B_{R}(x_0)) $, we have 
\begin{equation}
	R^2\|\psi\|_{L^\infty(B_{R}(x_0))}\leqslant  C_{\rho,n,s}\left(\| g(\cdot,u(\cdot))\|_{L^\infty(B_{R}(x_0))}+\|u\|_{L^\infty(\mathbb{R}^n)}\right)
\end{equation}
for some positive constant $C_{n,s,\rho}$.
\end{proof}
According to \eqref{eq u epsilon1}, we know that 		for every $\epsilon\in(0,R)$ small enough,  $u_\epsilon\in C^{2}(\overline V_{\frac{3\rho}{4}})$ solves the following equation 
\begin{equation}
-\Delta u_\epsilon+(-\Delta)^s u_\epsilon= g_\epsilon \qquad \text{  in } V_{\frac{3\rho}{4}},
\end{equation}
where $g_\epsilon$ satisfies the estimate \eqref{g_epsilon1}.  

Consider the cut off function $\phi^R$ given by \eqref{R}, let $v_\epsilon:=\phi^R u_\epsilon$, then 	applying Lemma~\refeq{lemma v epsilon1}, one has there exists~$ \psi_\epsilon\in L^{\infty}(B_{R}(x_0),\mathbb{R}) $ such that $ v_\epsilon $ satisfies  
\begin{equation}\label{eq: v epsilon equation1}
-\Delta v_\epsilon+(-\Delta)^s v_\epsilon= \psi_\epsilon \qquad \text{  in } V_{\frac{3\rho}{4}},
\end{equation}
where 
\begin{equation}
\psi_\epsilon= g_\epsilon(x,u)+\Delta (u_\epsilon(1-\phi^R))-(-\Delta)^s (u_\epsilon(1-\phi^R))\quad  \text{  in } V_{\frac{3\rho}{4}}.
\end{equation} 
Also, $\phi^R=1$ in $ B_{3R/2}(x_0)$ and 
\begin{equation}\label{eq: new equation:PSI epsilon1}
\psi_\epsilon= 	g_\epsilon(x,u)-(-\Delta)^s (u_\epsilon(1-\phi^R))\quad  \text{  in } B_{3R/2}(x_0).
\end{equation} 
In particular, employing \eqref{estimate psi 21}, one finds that
\begin{equation}
\begin{split}\label{eq:psi epsilon1}
	R^2\|\psi_\epsilon\|_{L^\infty(B_{R}(x_0))}&\leqslant  C_{\rho,n,s} \left(\| g_\epsilon(\cdot,u(\cdot))\|_{L^\infty(B_{R}(x_0))}+\|u_\epsilon\|_{L^\infty(\mathbb{R}^n)}\right),
\end{split}
\end{equation}
where  $ C_{\rho,n,s} $ is a positive constant.

\smallskip
 We notice that,  
for $v_\epsilon\in C^\infty_0( B_{2R}(x_0))$, one has 
\begin{equation}
\begin{split}\label{aa}
	R^2\|(-\Delta)^s v_\epsilon\|_{L^\infty(B_{R}(x_0))}
	&\leqslant C_{n,s,\rho}\left(R^{1+\beta}[Dv_\epsilon]_{0,\beta;B_{2R}(x_0)}+\|v_\epsilon\|_{L^\infty(B_{2R}(x_0))}\right)\\
	&= C_{n,s,\rho}\bigg(\|\phi^R u_\epsilon\|_{L^\infty(B_{2R}(x_0))}+R^{1+\beta}[\phi^R D u_\epsilon+u_\epsilon D\phi^R]_{0,\beta;B_{2R}(x_0)}	\bigg)\\
	&\leqslant C_{n,s,\rho}
	|u_\epsilon|^\prime_{1,\beta;B_{2R}(x_0)},
\end{split}
\end{equation}
for every $\beta\in(\max\left\{0,2s-1\right\},1). $  

As a result, for any given $\alpha\in(\max\left\{0,2s-1\right\},1)$,  taking $\beta=\frac{\max\left\{0,2s-1\right\}+\alpha}{2} $,  invoking \cite[Proposition 2.18]{MR4560756}, and utilizing \eqref{aa} and interpolation inequality, one has   for all~$\delta>0$
there exists $ C_{\delta}>0  $ such that  
\begin{equation}\label{C 2 of v 21}
\begin{split}
	|v_\epsilon|^\prime_{{1,\alpha}; B_{R/2}(x_0)}&\leqslant C_{n,\alpha}\left(\|v_\epsilon\|_{L^\infty(B_{R}(x_0))}+R^2\left(\|\psi_\epsilon\|_{L^\infty(B_{R}(x_0))}+\|(-\Delta)^s v_\epsilon\|_{L^\infty(B_{R}(x_0))}\right)\right)\\
	&\leqslant C_{n,\alpha,\rho,\alpha}\bigg(\|v_\epsilon\|_{L^\infty(B_{R}(x_0))}+R^2\|\psi_\epsilon\|_{L^\infty(B_{R}(x_0))}+|u_\epsilon|^\prime_{1,\beta;B_{2R}(x_0)}\bigg)\\
	&\leqslant C_{n,\alpha,\rho,\alpha}\bigg(\|v_\epsilon\|_{L^\infty(B_{R}(x_0))}+R^2\|\psi_\epsilon\|_{L^\infty(B_{R}(x_0))}\\
	&\qquad\qquad\qquad +\delta |u_\epsilon|^\prime_{1,\alpha;B_{2R}(x_0)} +C_\delta \|u_\epsilon\|_{L^\infty(B_{2R}(x_0))}\bigg). 
\end{split}
\end{equation}
Therefore, for every $x_0\in V_{\frac{\rho}{4}}$, applying Propositions~\ref{pro u epsilon1} and ~\ref{pro g epsilon1}, we know that $ \forall \delta>0$, there exists $C_\delta$ such that 
{\begin{equation}\label{final equation1}
	\begin{split}
		&\quad\|u_\epsilon\|^\prime_{{1,\alpha}; B_{R/2}(x_0)}=|v_\epsilon|^\prime_{{1,\alpha}; B_{R/2}(x_0)}\\
		&\leqslant C_{n,s, \alpha,\rho}\left(R^2\|\psi_\epsilon\|_{L^\infty(B_{R}(x_0))}+\delta |u_\epsilon|^\prime_{{1,\alpha}; B_{2R}(x_0)}+C_\delta \|u_\epsilon\|_{L^\infty(B_{2R}(x_0))}\right)\\
		&\leqslant C_{n,s, \alpha,\rho}\bigg(\|g\|_{L^\infty(V_{3\rho/4} \times\mathbb{R}^n)}+C_\delta\|u\|_{L^\infty(\mathbb{R}^n)} +\delta |u_\epsilon|^\prime_{{1,\alpha}; B_{2R}(x_0)}\bigg),
	\end{split}
\end{equation}
for every $R\in\left(0,\frac{\rho}{10}\right)$, $\epsilon\in (0,R)$,  and for some 
constant $ C_{n,s, \alpha,\rho}>0 $.

\subsubsection{Interior $C^{1,\alpha}$-regularity }\label{sec:Proof of Theorem1.5 1}

Based on \eqref{final equation1}, we can obtain the following Proposition~\ref{pro final estimate 1}  by adapting \cite[Lemma 8]{MR3331523}, which can guarantee that the $C^{1,\alpha}$-norm of $u_\epsilon$ is uniformly bounded in $\epsilon$ in some ball.



\begin{Proposition}\label{pro final estimate 1}
	
	Let $y\in \mathbb{R}^n, d>0$, and $u\in C^{1,\alpha}(B_{d}(y))$. Suppose that  for any $\delta > 0 $, there
	exists $\Lambda_\delta> 0$ such that, for any $x\in B_{d}(y)$ and any $r\in(0, d-|x-y|]$, we have
	\begin{equation}\label{eq:assumption u1}
		|u|^\prime_{1,\alpha;B_{r/8}(x)}\leqslant \Lambda_\delta+ \delta |u|^\prime_{1,\alpha;B_{r/2}(x)}.
	\end{equation}
	
	Then, there exist constants $\delta_0, C > 0$, depending only on $n,\alpha,d$, such that
	\begin{equation}
		\|u\|_{C^{1,\alpha}(B_{d/8}(y))}\leqslant C\Lambda_{\delta_0}.
	\end{equation}
\end{Proposition}
\begin{proof}
	We notice that, since $r<d$, one has
	\[ |u|^\prime_{1,\alpha;B_{r/8}(x)}\leqslant  |u|^\prime_{1,\alpha;B_{d}(y)},\]
	which implies that 
	\[ M:=\sup\limits_{x\in B_{d}(y), r\in(0, d-|x-y|]} |u|^\prime_{1,\alpha;B_{r/8}(x)}<+\infty.\]
	
	We now use a covering argument: pick $\lambda\in(0,\frac{1}{2}]$, to be chosen later, and	fixed any $x\in B_d(y)$ and $ r\in(0, d-|x-y|]$, we cover $B_{r/8}(x)$ with finitely many balls $\left\{B_{\lambda r/8}(x_k)\right\}_{k=1}^N$ with $x_k\in B_{r/8}(x) $,  for some $N $ depending only on $\lambda,n$.
	
	Furthermore, we observe that $B_{r/2}(x_k)\subset B_{d}(y)$. Indeed,
	\[ |x_k-y|+\frac{r}{2}\leqslant |x_k-x|+|x-y|+\frac{r}{2}<r+|x-y|<d. \]
	Consequently, since $\lambda\leqslant\frac{1}{2}$, we use  \eqref{eq:assumption u} (with $x = x_k $ and $ r$ scaled to $\lambda r$) to get 
	\begin{equation}\label{eq lambda1}
		|u|^\prime_{1,\alpha;B_{\lambda r/8}(x_k)}\leqslant \Lambda_\delta+ \delta |u|^\prime_{1,\alpha;B_{\lambda r/2}(x_k)}.
	\end{equation}
	Then, using \cite[Lemma 7]{MR3331523} with $\rho:=r/8$ and $\lambda=1/8$, recalling  the definition of $M$, and employing \eqref{eq lambda1}, one has 
	\begin{equation}
		\begin{split}
			|u|^\prime_{1,\alpha;B_{r/8}(x)}&\leqslant C\sum\limits_{k=1}\limits^{N}|u|^\prime_{1,\alpha;B_{\lambda r/8}(x_k)}
			\leqslant C_\lambda N \Lambda_{\delta}+ C_\lambda \delta \sum\limits_{k=1}\limits^{N}|u|^\prime_{1,\alpha;B_{\lambda r/2}(x_k)}\\
			&\leqslant C N \Lambda_{\delta}+ C\delta \sum\limits_{k=1}\limits^{N}|u|^\prime_{1,\alpha;B_{ r/16}(x_k)}
			\leqslant CN \Lambda_{\delta}+ C N\delta M.
		\end{split}
	\end{equation}
	It follows from the definition of $M$ that 
	\[ M\leqslant  C N \Lambda_{\delta}+ C N\delta M,  \]
	by choosing $\delta_0=1/(CN)$, one has 
	\[ M\leqslant  2C N \Lambda_{\delta_0}.\]
	Thus we have proved that
	\[ |u|^\prime_{1,\alpha;B_{r/8}(x)}\leqslant  2C N \Lambda_{\delta_0}.\qquad \forall x\in B_d(y),  r\in(0, d-|x-y|]. \]
	Setting $x=y, r=d$, we have 
	\[ |u|^\prime_{1,\alpha;B_{d/8}(y)}\leqslant  2 N \Lambda_{\delta_0}, \]
	which implies the desired result that 
	\[ \|u\|_{C^{1,\alpha}(\overline B_{d/8}(y))}\leqslant  2\max\left\{d^{-1-\alpha},1\right\}C N \Lambda_{\delta_0}.\qedhere\]
\end{proof}


\begin{proof}[Proof of Theorem~\ref{th C^1,alpha interior without boundary condition}]
	First of all, one can choose $d=\rho/5 $ in Proposition~\ref{pro final estimate 1}, so that 
	\begin{equation}\label{d}
		B_{d}(y)\subset V_{\rho/4} \quad \text{and}\quad  d/4<\frac{\rho}{10}, \quad \forall y\in V.
	\end{equation}
	As a consequence, applying  Proposition~\ref{pro final estimate 1}, utilizing \eqref{final equation1},
	we know that  
	\begin{equation}
		\|u_\epsilon\|_{C^{1,\alpha}(\overline B_{\rho/40}(y))}\leqslant C\left(\|g\|_{L^\infty( V_{\frac{3\rho}{4}} \times\mathbb{R}^n)}+C_{\delta_0} \|u\|_{L^\infty(\mathbb{R}^n)}\right),
	\end{equation}
	for every $y\in V$, where the constant $ C>0 $ depends on $ n,s,\alpha,\rho $.
	
	
	
	Applying the Arzel\`{a}-Ascoli Theorem, we obtain that  $ u\in C^{1,\alpha}(\overline B_{\rho/40}(y)). $ More specifically, one has that
	\begin{equation}\label{final inequality}
		\|u\|_{C^{1,\alpha}(\overline B_{\rho/40}(y))}\leqslant C\left(\|g\|_{L^\infty( V_{\frac{3\rho}{4}} \times\mathbb{R}^n)}+ \|u\|_{L^\infty(\mathbb{R}^n)}\right),
	\end{equation}
	for every $y\in V$ and $\alpha\in(\max\left\{0,2s-1\right\},1)$, and for	some constant $ C>0 $ depending on $ n,s,\alpha,\rho $. 
\end{proof}

Using the cover argument again, there exist  finitely many balls $\left\{B_{\rho/40}(y_k)\right\}_{k=1}^N$ with $y_k\in V $,  for some $N$ depending on $n,\rho,V$, such that 
\[ V\subset \bigcup\limits_{k=1}\limits^NB_{\rho/40}(y_k). \]
Thus, we can obtain
\begin{equation}\label{final inequality 1}
	\|u\|_{C^{1,\alpha}(\overline V)}\leqslant C\left(\|g\|_{L^\infty( V_{\frac{3\rho}{4}} \times\mathbb{R}^n)}+ \|u\|_{L^\infty(\mathbb{R}^n)}\right),
\end{equation}
for every $\alpha\in(0,1)$,	where the constant $ C>0 $ depends on $ n,s,\alpha,\rho $.


\section{ $ C^{2,\alpha} $-regularity}\label{sec: C^2- regularity}\setcounter{equation}{0}
In this section, we derive the interior $ C^{2,\alpha} $-regularity and the $ C^{2,\alpha} $-regularity  up to the boundary for the mixed operator $ \mathcal{L}$.

\subsection{Interior $ C^{2,\alpha} $-regularity with smooth boundary}\label{sec: interior regularity}
 The aim of this section is to establish an interior $ C^{2,\alpha} $-regularity theory  for the problem \eqref{Maineq}. 
 To this end, we should overcome 
the additional difficulty caused by the nonlinearity $ g(x,u(x))$.
 %which has to be taken into account in the iteration process.
 More precisely, since the nonlinearity $g$ depends on unknown function $u$,  both the H\"{o}lder continuity of nonlinearity $ g $ and the regularity of $u$ impact the regularity of $g(x,u(x))$ as a function of x.
 

% once we  consider the H\"{o}lder continuity of nonlinearity $ g $, the regularity of the solution $u$ has to be taken into account.


From the proof of Theorem~\ref{th:C^{1,alpha}}  for the $C^{1,\alpha}$-regularity obtained in Section~\ref{sec: C^alpha-regularity}, one cannot choose the H\"{o}lder exponent $\alpha$ arbitrarily in $(0,1)$. This restriction 
could at a first glance prevent us from deriving  the interior $ C^{2,\alpha} $-regularity for any $\alpha\in(0,1)$. Therefore, to overcome this difficulty, we  make a detour round such restriction and carry out the proof of Theorem~\ref{th C^2,alpha interior} via the following two steps:
%	Nevertheless, it is worth noting that the restriction of the H\"{o}lder continuous with exponent $\alpha$ of  the  gradient of solution makes it impossible to derive the  $ C^{2,\gamma} $-interior regularity for any $\gamma\in(0,1).$  In terms of this inconvenience, we carry out the proof of Theorem~\ref{th C^2,alpha interior} in two steps:
\begin{itemize}
	\item [(a)] Establish a ``weak version" of Theorem~\ref{th C^2,alpha interior} in subsection~\ref{sec. weak version};
	\item  [(b)] Employ an interpolation inequality  and a truncation argument   to obtain Theorem~\ref{th C^2,alpha interior} in subsection~\ref{sec;Proof of the interior regularity}.	
\end{itemize}


\subsubsection{\textbf{``Weak version" of the regularity result}}\label{sec. weak version}
Now we provide a  ``weak version" of Theorem~\ref{th C^2,alpha interior}, in which
the regularity exponent obtained in the case $s\in (\frac{1}{2},1)$ is suboptimal (we will then utilize this result as a ``pivot''
to obtain Theorem~\ref{th C^2,alpha interior} in its full generality):

\begin{Theorem}\label{th weak C^2,alpha interior}
	Let us assume, in addition to the hypothesis of Theorem~\ref{th:C^{1,alpha}}, that $ g\in C^\gamma_{\rm loc}(\Omega\times\mathbb{R}) $ with $\gamma\in (0,1)$.
	 Then,  we have $u\in C_{\rm loc}^{2,\alpha}(\Omega)$ for any $\alpha<\min\left\{\gamma, \min\left\{1,2-2s\right\}\right\}$.
\end{Theorem}

We remark that  Theorem~\ref{th weak C^2,alpha interior} above provides an interior $C^{2,\alpha_0}$-estimate of the solution $u$ for some $\alpha_0\in(0,2-2s)$, which allows us to invoke an interpolation inequality (see e.g. \eqref{eq:C 1 alpha}) to gain the interior $C^{1,\alpha}$-estimate of $u$ for each $\alpha\in(0,1)$. Moreover, by applying the same arguments as in the proof of Theorem~\ref{th weak C^2,alpha interior} (such as a truncation argument and a mollifier technique), one can finally complete  the proof of Theorem~\ref{th C^2,alpha interior} in Subsection~\ref{sec;Proof of the interior regularity}.  
\smallskip

Let us first give the following $ C^\alpha $-estimates of a truncation function.
\begin{Lemma}\label{lemma v}
	Under the assumptions of Theorem~\ref{th:C^{1,alpha}},  let $V$ be an open set  with $ V\Subset\Omega $, and denote
		\[\rho:=\text{ dist}(V,\partial\Omega),\quad \text{  and } \quad  V_{\delta}:=\left\{x\in \Omega: \text{ dist}(x, V)<\delta\right\}. \]
		Consider a cut off function $ \phi\in C^\infty_0(\mathbb{R}^n,\mathbb{R}) $ satisfying 
		\begin{equation}\label{cutoff function}
			\begin{cases}
				\phi \equiv 1 \text{  on } V_{\frac{\rho}{4}},\\
				\text{supp}(\phi)\subset V_{\frac{\rho}{2}}\subset \Omega,\\
				0\leqslant \phi\leqslant 1 \text{  on } \mathbb{R}^n.
				\end{cases}
			\end{equation}
	
	Then, there exists~$ \psi\in C^{\alpha}(\overline V_{\frac{7\rho}{8}},\mathbb{R}) $ such that $ v:=\phi u $ is a solution of 
	\begin{equation}\label{eq: new equation}
		\begin{cases}
			-\Delta v+(-\Delta)^s v= \psi \qquad &\text{  in } V_{\frac{7\rho}{8}},\\
			v=0&\text{  on } \mathbb{R}^n\backslash V_{\frac{7\rho}{8}}.
			\end{cases}
		\end{equation}
	In particular, 
	\begin{equation}
		\|\psi\|_{C^\alpha(\overline V_{\frac{7\rho}{8}})}\leqslant C \left(1+\|g\|_{C^\gamma(\overline V_{\frac{7\rho}{8}}\times\mathbb{R})}\right)\left(1+\|u\|_{C^{1,\alpha}(\overline\Omega)}\right), \quad   \forall \alpha\in(0, \min\left\{\gamma, \min\left\{1,2-2s\right\}\right\}),
		\end{equation}
where  $ C $ is a positive constant depending only on $ n,s,\Omega,V $.
	
	
\end{Lemma}
\begin{proof}
	From Theorem~\ref{th:C^{1,alpha}} and the definition of the function $ v $, we know that
		\begin{equation}\label{eq: function space of v}
			v\in C^{1,\alpha}_0(\overline V_{\frac{\rho}{2}})\subset C^{1,\alpha}_0(\mathbb{R}^n). 
			\end{equation}
	Furthermore, to check~\eqref{eq: new equation}, one can calculate that  
	\begin{equation}\label{eq: new equation:PSI}\begin{split}& \psi= \phi g(x,u)+u(-\Delta)\phi+u(-\Delta)^s\phi+ 2 \nabla u\nabla \phi \\&\qquad\qquad+ \int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\phi(x)-\phi(x+z))}{|z|^{n+2s}}\, dz.\end{split}\end{equation} 
	Indeed, 
	\[  -\Delta v=\phi(-\Delta)u+ u(-\Delta)\phi+2 \nabla u\nabla \phi,  \]
	and 
	\[  (-\Delta)^s v=\phi(-\Delta)^su+ u(-\Delta)^s\phi+\int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\phi(x)-\phi(x+z))}{|z|^{n+2s}}\, dz . \]
	Recalling the fact that 
	\[ -\Delta u+(-\Delta)^s u= g(x,u) \qquad \text{  in } \Omega, \] one infers that 
	\begin{equation}
		\begin{split}
			-\Delta v+(-\Delta)^s v&=\phi g(x,u)+u(-\Delta)\phi+u(-\Delta)^s\phi+ 2 \nabla u\nabla \phi\\
			&\qquad + \int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\phi(x)-\phi(x+z))}{|z|^{n+2s}}\, dz,
			\end{split}
		\end{equation}
	which is~\eqref{eq: new equation:PSI}, as desired.
	
\noindent\textbf{Step 1. $ C^\alpha $- estimates on the function $ \phi g(x,u) $.}
For each $ i=\left\{1,\dots,n\right\} $, since $ g\in C^\gamma_{\rm loc}({\Omega}\times\mathbb{R}) $ and $ u\in C^{1,\alpha}(\overline{\Omega})  $, for $x_1,x_2\in V_{\frac{7\rho}{8}}$ with $x_1\neq x_2$, we have that 

\begin{equation}
	\begin{split}
			\frac{|g(x_1,u(x_1))-g(x_2,u(x_2))|}{|x_1-x_2|^\alpha}&\leqslant \frac{\|g\|_{C^\gamma(\overline V_{{7\rho}/{8}}\times\mathbb{R})}|(x_1-x_2,u(x_1)-u(x_2))|^\gamma}{|x_1-x_2|^\alpha}\\
			&\leqslant \frac{\|g\|_{C^\gamma(\overline V_{{7\rho}/{8}}\times\mathbb{R})}\left(|x_1-x_2|+|u(x_1)-u(x_2)|\right)^\gamma}{|x_1-x_2|^\alpha}\\
			&\leqslant \frac{\|g\|_{C^\gamma(\overline V_{{7\rho}/{8}}\times\mathbb{R})}\left(|x_1-x_2|+\|u\|_{C^1(\overline\Omega)}|x_1-x_2|\right)^\gamma}{|x_1-x_2|^\alpha}\\
			&\leqslant \big(\text{diam}(\Omega)\big)^{\gamma-\alpha} \|g\|_{C^\gamma(\overline V_{{7\rho}/{8}}\times\mathbb{R})}\left(1+\|u\|_{C^1(\overline\Omega)}\right).
		\end{split}
	\end{equation}
	Consequently,
	\[ [g(\cdot,u(\cdot))]_{C^\alpha(\overline V_{{7\rho}/{8}})}\leqslant \big(\text{diam}(\Omega)\big)^{\gamma-\alpha} \|g\|_{C^\gamma(\overline V_{{7\rho}/{8}}\times\mathbb{R})}\left(1+\|u\|_{C^1(\overline\Omega)}\right). \]	 
	
	Thus, for all~$ \alpha\in (0,\gamma]$, we have 
	\begin{equation}
	\begin{split}
		\|\phi g(\cdot,u(\cdot))\|_{C^\alpha(\overline V_{{7\rho}/{8}})}&\leqslant [\phi]_{C^\alpha(\overline V_{{7\rho}/{8}})}\|g(\cdot,u(\cdot))\|_{L^\infty(V_{{7\rho}/{8}})}+[g(\cdot,u(\cdot))]_{C^\alpha(\overline V_{{7\rho}/{8}})}\|\phi\|_{L^\infty( V_{{7\rho}/{8}})}\\
		&\qquad +\|\phi g(\cdot,u(\cdot))\|_{L^\infty(V_{{7\rho}/{8}})}\\
		&\leqslant C_1 \|g\|_{C^\gamma(\overline V_{{7\rho}/{8}}\times\mathbb{R})}\left(1+\|u\|_{C^1(\overline\Omega)}\right) 
		\end{split}
	\end{equation}
	for some suitable constant $ C_1>0. $	


	
	
\noindent\textbf{Step~2. $ C^\alpha $- estimates on the function $ u(-\Delta)\phi+u(-\Delta)^s\phi+ 2 \nabla u\nabla \phi $.} For every $ x\in V_{\frac{7\rho}{8}} $, we have that
	\begin{equation}
	\begin{split}
		\left|u(-\Delta)\phi+u(-\Delta)^s\phi+ 2 \nabla u\nabla \phi\right|\leqslant C_2\|u\|_{C^1(\overline\Omega)},
		\end{split}
	\end{equation}
	and
	\begin{equation}
	\begin{split}
		&[u(-\Delta)\phi+u(-\Delta)^s\phi+ 2 \nabla u\nabla \phi ]_{C^\alpha(\overline V_{{7\rho}/{8}})}\\
		\leqslant & [u]_{C^\alpha(\overline V_{{7\rho}/{8}})}\|(-\Delta)\phi+(-\Delta)^s\phi\|_{L^\infty(V_{{7\rho}/{8}})}+[(-\Delta)\phi+(-\Delta)^s\phi]_{C^\alpha(\overline V_{{7\rho}/{8}})}\|u\|_{L^\infty(V_{{7\rho}/{8}})}\\
		&\qquad +2\left([\nabla u]_{C^\alpha(\overline V_{{7\rho}/{8}})}\|\nabla \phi\|_{L^\infty(V_{{7\rho}/{8}})}+[\nabla \phi]_{C^\alpha(\overline V_{{7\rho}/{8}})}\|\nabla u\|_{L^\infty(V_{{7\rho}/{8}})}\right)\\
		\leqslant &C_3\|u\|_{C^{1,\alpha}(\overline\Omega)},
		\end{split}
	\end{equation}
	for some suitable constant $ C_2,C_3>0 $.
	
\noindent\textbf{Step~3. $ C^\alpha $- estimates on the function $ h(x):= \int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\phi(x)-\phi(x+z))}{|z|^{n+2s}}\, dz $.}
	Let us first estimate the $ L^\infty $-norm of $h $. To this end, for every $ x\in V_{\frac{7\rho}{8}}, $ since $ u=0 $ on $ \mathbb{R}^n\backslash\Omega $, one has that
	\begin{equation}
	\begin{split}
		|h(x)|&=\left|\int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\phi(x)-\phi(x+z))}{|z|^{n+2s}}\, dz\right|\\
		&\leqslant \int_{\left\{|z|<1\right\}} \frac{[u]_{C^{1}(\overline\Omega)}[\phi]_{C^{1}(\overline\Omega)}}{|z|^{n+2s-2}}\, dz+\int_{\left\{|z|\geqslant 1\right\}}\frac{4\|u\|_{L^\infty(\Omega)}\|\phi\|_{L^\infty(\Omega)}}{|z|^{n+2s}}\, dz
		\leqslant C_4 \|u\|_{C^{1}(\overline\Omega)}.
		\end{split}
	\end{equation}
	
	Now we estimate the $ C^\alpha $-norm of $ h$. For this, given $ x_1,x_2\in V_{\frac{7\rho}{8}} $ with $ x_1\neq x_2 $, let $ r:=|x_1-x_2| $.
	We distinguish two cases.
	
\noindent\textbf{Case 1.} If $ r\geqslant1  $, then 
	\begin{equation}\label{case 1 of h}
	\frac{|h(x_1)-h(x_2)|}{|x_1-x_2|^\alpha}\leqslant 2	\|h\|_{L^\infty(V_{\frac{7\rho}{8}})}\leqslant 2C_4 \|u\|_{C^{1}(\overline\Omega)}.
	\end{equation}
	
\noindent\textbf{Case 2.}  If $ r< 1 $, then 
	\begin{equation}
	\begin{split}
		\frac{|h(x_1)-h(x_2)|}{|x_1-x_2|^\alpha}&=\left|\int_{\mathbb{R}^n}\frac{(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))(\phi(x_2)-\phi(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
		& +\left|\int_{\mathbb{R}^n}\frac{(u(x_1)-u(x_1+z))(\phi(x_1)-\phi(x_1+z)-\phi(x_2)+\phi(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
		&:= I_1+I_2.
		\end{split}
	\end{equation}
	
	
	
	For $ I_1 $,  one sees that 
	\begin{equation}
	\begin{split}&\label{I_1 1}
		\quad	\left|\int_{|z|<1}\frac{(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))(\phi(x_2)-\phi(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
		&\qquad\leqslant\int_{|z|<1}\frac{|(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))|\|\nabla \phi\|_{L^\infty(\Omega)}}{|z|^{n+2s-1}r^\alpha}\, dz\\
		&\qquad\leqslant \|\nabla \phi\|_{L^\infty(\Omega)}\int_{|z|<r}\frac{|(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))|}{|z|^{n+2s-1+\alpha}}\, dz\\
		&\qquad\quad\qquad + \|\nabla \phi\|_{L^\infty(\Omega)}\int_{r<|z|< 1}\frac{2[u]_{C^{1}(\overline{\Omega})}r^\alpha|z|^{1-\alpha}}{|z|^{n+2s-1}r^\alpha}\, dz\\
		&\qquad\leqslant 2 \|\nabla \phi\|_{L^\infty(\Omega)}\int_{|z|< 1}\frac{2[u]_{C^{1}(\overline{\Omega})}}{|z|^{n+2s-2+\alpha}}\, dz\leqslant C_5 [u]_{C^{1}(\overline{\Omega})}, \quad\forall \alpha\in (0, \min\left\{1,2-2s\right\}),
		\end{split}
	\end{equation}
	and
	\begin{equation}
	\begin{split}\label{I_1 2}
		&	\left|\int_{|z|\geqslant 1}\frac{(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))(\phi(x_2)-\phi(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
		&\qquad\qquad\leqslant2\| \phi\|_{L^\infty(\Omega)}\int_{|z|\geqslant 1}\frac{|(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))| }{|z|^{n+2s}r^\alpha}\, dz\\&\qquad\qquad
		\leqslant 
2\| \phi\|_{L^\infty(\Omega)}\int_{|z|\geqslant 1}\frac{2[u]_{C^{\alpha}(\overline{\Omega})}r^\alpha}{|z|^{n+2s}r^\alpha}\, dz\\
		&\qquad\qquad\leqslant 2 \| \phi\|_{L^\infty(\Omega)}\int_{|z|\geqslant 1}\frac{2[u]_{C^{\alpha}(\overline{\Omega})}}{|z|^{n+2s}}\, dz\leqslant C_6 [u]_{C^{\alpha}(\overline{\Omega})}, \quad\forall \alpha\in (0, \min\left\{1,2-2s\right\} ).
		\end{split}
	\end{equation}
	Combining \eqref{I_1 1} with \eqref{I_1 2}, we deduce that 
	\begin{equation}\label{I_1}
	I_1\leqslant \left(C_5 [u]_{C^{1}(\overline{\Omega})}+C_6 [u]_{C^{\alpha}(\overline{\Omega})}\right), \qquad \forall \alpha\in (0,\min\left\{1,2-2s\right\}),
	\end{equation} 
	for some suitable constants $ C_5,C_6>0. $
	
{	For $ I_2, $ using similar arguments to the estimate of $ I_1 $,  one has
	\begin{equation}
			\begin{split}\label{I_2 1}
			& \left|\int_{\mathbb{R}^n}\frac{(u(x_1)-u(x_1+z))(\phi(x_1)-\phi(x_1+z)-\phi(x_2)+\phi(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
			&\qquad\leqslant\int_{|z|<1}\frac{|(\phi(x_1)-\phi(x_1+z)-\phi(x_2)+\phi(x_2+z))|\|\nabla u\|_{L^\infty(\Omega)}}{|z|^{n+2s-1}r^\alpha}\, dz\\
			&\qquad\leqslant \|\nabla u\|_{L^\infty(\Omega)}\int_{|z|<r}\frac{|\phi(x_1)-\phi(x_1+z)-\phi(x_2)+\phi(x_2+z)|}{|z|^{n+2s-1+\alpha}}\, dz\\
			&\quad \qquad\qquad+ \|\nabla u\|_{L^\infty(\Omega)}\int_{r<|z|< 1}\frac{2[\phi]_{C^{1}(\overline{\Omega})}r^\alpha|z|^{1-\alpha}}{|z|^{n+2s-1}r^\alpha}\, dz\\
			&\qquad\leqslant  2 \|\nabla u\|_{L^\infty(\Omega)}\int_{|z|< 1}\frac{2[\phi]_{C^{1}(\overline{\Omega})}}{|z|^{n+2s-2+\alpha}}\, dz\leqslant C_7 [u]_{C^{1}(\overline{\Omega})}, \quad\forall \alpha\in (0,\min\left\{1,2-2s\right\}),
				\end{split}
		\end{equation}
	and 
	\begin{equation}
		\begin{split}\label{I_2 2}
			&\left|\int_{|z|\geqslant 1}\frac{(u(x_1)-u(x_1+z))(\phi(x_1)-\phi(x_1+z)-\phi(x_2)+\phi(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\\
			&\qquad\leqslant 2\| u\|_{L^\infty(\Omega)}\int_{|z|\geqslant 1}\frac{|\phi(x_1)-\phi(x_1+z)-\phi(x_2)+\phi(x_2+z)| }{|z|^{n+2s}r^\alpha}\, dz\\&\qquad
			\leqslant 2\|u\|_{L^\infty(\Omega)}\int_{|z|\geqslant 1}\frac{2[\phi]_{C^{\alpha}(\overline{\Omega})}r^\alpha}{|z|^{n+2s}r^\alpha}\, dz\\&\qquad
			\leqslant 2 \|u \|_{L^\infty(\Omega)}\int_{|z|\geqslant 1}\frac{2[\phi]_{C^{\alpha}(\overline{\Omega})}}{|z|^{n+2s}}\, dz\leqslant C_8\|u\|_{L^\infty(\Omega)}, \quad\forall \alpha\in (0,\min\left\{1,2-2s\right\}).
			\end{split}
		\end{equation}
	From \eqref{I_2 1} and  \eqref{I_2 2} one concludes that 
	\begin{equation}\label{I_2}
		I_2\leqslant  C_7 [u]_{C^{1}(\overline{\Omega})}+C_8\|u\|_{L^\infty(\Omega)}, \qquad \forall \alpha\in (0,\min\left\{1,2-2s\right\}),
		\end{equation} 
	for some suitable constants $ C_7,C_8>0. $}
	
	Combining \eqref{I_1} and \eqref{I_2}, and recalling \eqref{case 1 of h}, we conclude that 
	\begin{equation}
	\frac{|h(x_1)-h(x_2)|}{|x_1-x_2|^\alpha}\leqslant C_8 \|u\|_{C^{1,\alpha}(\overline{\Omega})}, \qquad \forall \alpha\in(0,\min\left\{1,2-2s\right\}).
	\end{equation}	
	As a consequence,  we derive that 
	\begin{equation}
	\|h\|_{C^\alpha(\overline V_{{7\rho}/{8}})}\leqslant C\|u\|_{C^{1,\alpha}(\overline{\Omega})}, \qquad \forall \alpha\in(0,\min\left\{1,2-2s\right\}).
	\end{equation}
		In light of \textbf{Step 1}--\textbf{Step~3}, we obtain that $ \psi\in C^\alpha(\overline V_{\frac{7\rho}{8}}) $ and 
	\begin{equation}\label{estimate psi}
	\|\psi\|_{C^\alpha(\overline V_{{7\rho}/{8}})}\leqslant C\left(\|g\|_{L^\infty(\overline V_{{7\rho}/{8}}\times\mathbb{R})}+\|g\|_{C^\gamma(\overline V_{{7\rho}/{8}}\times\mathbb{R})}\|u\|_{C^1(\overline\Omega)}+ \|u\|_{C^{1,\alpha}(\overline\Omega)}\right),
	\end{equation}
for any $\alpha\in(0,\min\left\{\gamma,\min\left\{1,2-2s\right\}\right\})$ and	for some suitable constant $ C>0 $ depending only on $ n,s,\Omega,V $.	
\end{proof}



We next use a mollifier to regularize $ v $ and obtain the $ C^{2,\alpha} $-estimates on the convolution between the mollifier and $ v, $ then we will pass to the limit. This will allow us to complete the proof of
Theorem~\ref{th weak C^2,alpha interior}. The details are as follows.

\begin{proof}[\textbf{Proof of Theorem~\ref{th weak C^2,alpha interior}}]
	We notice that the function $ v:=\phi u $ satisfies the following  equation
	\begin{equation}\label{v solution}
	\begin{cases}
		-\Delta v+(-\Delta)^s v= \psi \qquad &\text{  in } V_{\frac{7\rho}{8}},\\
		v=0&\text{  on } \mathbb{R}^n\backslash V_{\frac{7\rho}{8}},
		\end{cases}
	\end{equation}
	where $ \psi \in C^\alpha(\overline V_{\frac{7\rho}{8}}) $ and {$ v \in C^{1,\alpha}_0(\overline{ V_{\frac{\rho}{2}}}). $
		
		Recall that if $ v $ solves \eqref{v solution} in the weak sense and $ \eta_\epsilon $ is a standard mollifier then
		\begin{equation}
			-\Delta(\eta_\epsilon\ast v)+(-\Delta)^s (\eta_\epsilon \ast v)= \eta_\epsilon \ast\psi \qquad \text{  in } V_{\frac{3\rho}{4}}, 
			\end{equation}
		for $ 0<\epsilon<\frac{\rho}{10} $ small enough, and supp$ (\eta_\epsilon\ast v)\subset V_{\frac{3\rho}{4}} $. 
		
		Indeed, $ v=0 $ outside $ V_{\frac{\rho}{2}} $, and 
		\[ (\eta_\epsilon\ast v)(x)=\int_{|z|<1} \eta(z)v(x-\epsilon z)\, dz=0, \quad \forall x\in\mathbb{R}^n\backslash V_{\frac{3\rho}{4}}.  \] 
		
		
		Hence, we can regularize $ v $, obtain the desired estimates, then pass to the limit. In this way, we may assume that 
		$ v_\epsilon:=\eta_\epsilon\ast v \in C_0^\infty(V_{\frac{3\rho}{4}}) $  and $ \psi_\epsilon:=\eta_\epsilon\ast \psi \in C^\infty(\overline V_{\frac{3\rho}{4}}) $, then $ v_\epsilon $ and $ \psi_\epsilon $ satisfy the following equation
		\begin{equation}\label{new equation}
			\begin{cases}
				-\Delta v_\epsilon+(-\Delta)^sv_\epsilon= \psi_\epsilon \qquad &\text{  in } V_{\frac{3\rho}{4}}, \\
				v_\epsilon=0 & \text{  on } \mathbb{R}^n\backslash V_{\frac{3\rho}{4}}.
				\end{cases}
			\end{equation}}
	
	According to the global estimate for the Dirichlet problem \eqref{new equation} 	 and the fact that,
	for $v_\epsilon\in C^\infty_0( V_{\frac{3\rho}{4}})$, 
	there exists $ C_{1/2}>0  $ such that  
		\begin{equation}\label{eq:v epsilon}
		\|(-\Delta)^s v_\epsilon \|_{C^\alpha(\overline V_{\frac{3\rho}{4}})} \leqslant \frac{1}{2} \| v_\epsilon\|_{C^{2,\alpha}(\overline V_{\frac{3\rho}{4}})}+C_{1/2}\| v_\epsilon\|_{L^\infty({V_{\frac{3\rho}{4}}})}, 
		\end{equation}
	we have 
	\begin{equation}\label{C 2 of v}
	\begin{split}
		\|v_\epsilon\|_{C^{2,\alpha}(\overline V_{\frac{3\rho}{4}})}&\leqslant C_{n,\alpha,V_{\frac{3\rho}{4}}}\left(\|v_\epsilon\|_{L^\infty(V_{\frac{3\rho}{4}})}+\|\psi_\epsilon\|_{C^\alpha(\overline V_{\frac{3\rho}{4}})}+\|(-\Delta)^s v_\epsilon\|_{C^\alpha(\overline V_{\frac{3\rho}{4}})}\right)\\
		&\leqslant C(n,\alpha,V_{\frac{3\rho}{4}})\left(\|v_\epsilon\|_{L^\infty(V_{\frac{3\rho}{4}})}+\|\psi_\epsilon\|_{C^\alpha(\overline V_{\frac{3\rho}{4}})}\right), 
		\end{split}
	\end{equation}
where in the last inequality we have used the fact that if~$ w\in C^\infty_0(\mathbb{R}^n)  $, then for all~$\tau>0 $ there exists $ C_\tau>0  $ such that  
		\begin{equation}
			\|(-\Delta)^s w\|_{C^\alpha(\overline V_{\frac{3\rho}{4}})} \leqslant \tau \| w\|_{C^{2,\alpha}(\overline V_{\frac{3\rho}{4}})}+C_\tau\| w\|_{L^\infty({V_{\frac{3\rho}{4}}})}.
		\end{equation}	
	
Furthermore, exploiting \cite[Lemma 4.8]{SVWZ22}, there exists $ \omega\in C^\infty(\overline V_{\frac{3\rho}{4}})\cap W^{1,\infty}(\mathbb{R}^n) $ such that for every $ s\in (0,1) $ 
	\begin{equation}
	-\Delta \omega+(-\Delta)^s\omega \geqslant 1, \quad \text{in } V_{\frac{3\rho}{4}},  \quad \omega>0 \text { in } \overline V_{\frac{3\rho}{4}}, \quad  0\leqslant\omega\leqslant C_0 \text { in } \mathbb{R}^n,
	\end{equation}
	for some constant $ C_0 $ depending only on $ n,s,V_{\frac{3\rho}{4}} $. 
	
	As a result, we find that 
	\begin{equation}
	-\Delta \left(\omega-\frac{v_\epsilon}{\|\psi_\epsilon\|_{L^\infty(\mathcal{O})}}\right)+(-\Delta)^s\left(\omega-\frac{v_\epsilon}{\|\psi_\epsilon\|_{L^\infty(\mathcal{O})}}\right)\geqslant 0 \qquad \text{  in } V_{\frac{3\rho}{4}}.\end{equation}
Applying the maximum principle (see \cite[Theorem 1.2]{BDVV22b}), we obtain 
	\begin{equation}
	\|v_\epsilon\|_{L^\infty(V_{\frac{3\rho}{4}})}\leqslant C_0\|\psi_\epsilon\|_{L^\infty(V_{\frac{3\rho}{4}})}\qquad  \text{  in } V_{\frac{3\rho}{4}}.
	\end{equation}
	As a consequence of this and \eqref{C 2 of v}, we have 
	\begin{equation*}
	\|v_\epsilon\|_{C^{2,\alpha}(\overline V_{\frac{3\rho}{4}})}\leqslant C_{n,s, \alpha,V_{\frac{3\rho}{4}}}\left(\|\psi_\epsilon\|_{L^\infty(V_{\frac{3\rho}{4}})}+\|\psi_\epsilon\|_{C^\alpha(\overline V_{\frac{3\rho}{4}})}\right), \quad \forall 0<\epsilon<\frac{\rho}{10}.
	\end{equation*}
	Therefore,  
	\begin{equation}
	\|\eta_\epsilon\ast v\|_{C^{2,\alpha}(\overline V_{\frac{3\rho}{4}})}\leqslant C_{n,s, \alpha,V_{\frac{3\rho}{4}}}\|\eta_\epsilon\ast \psi\|_{C^\alpha(\overline V_{\frac{3\rho}{4}})}\leqslant C \| \psi\|_{C^\alpha(\overline V_{\frac{7\rho}{8}})},
	\end{equation}
	for some suitable constant $ C>0 $ independent  of $ v,\epsilon. $ 
	
As a result, utilizing the  Arzel\`{a}-Ascoli Theorem, one finds that~$ v\in C^{2,\alpha}(\overline V_{\frac{3\rho}{4}}) $, and consequently  that $ u\in C^{2,\alpha}(\overline V_{\frac{\rho}{4}}). $ 

Moreover, it follows from \eqref{estimate psi} that 
	\begin{equation}\label{eq: C 2,alpha estimate of u }
		\|u\|_{C^{2,\alpha}(\overline V_{\frac{\rho}{4}})}\leqslant C \| \psi\|_{C^\alpha(\overline V_{\frac{7\rho}{8}})}\leqslant C\left(\|g\|_{L^\infty(\overline V_{{7\rho}/{8}}\times\mathbb{R})}+\|g\|_{C^\gamma(\overline V_{{7\rho}/{8}}\times\mathbb{R})}\|u\|_{C^1(\overline\Omega)}+ \|u\|_{C^{1,\alpha}(\overline\Omega)}\right),
	\end{equation}
for any $ \alpha\in(0,\min\left\{\gamma,\min\left\{1,2-2s\right\}\right\})$ and for some  positive constant $ C$ depending only on $ n,s,\Omega,V,\alpha,\gamma $.
\end{proof}

\subsubsection{\textbf{Proof of the interior  $ C^{2,\alpha} $-regularity}}\label{sec;Proof of the interior regularity}


In light of the assumptions of Theorem~\refeq{th C^2,alpha interior}, let us denote
\begin{equation}\label{definition of beta}
	\beta:=\min\left\{1-s,\frac{\alpha}{2}\right\}. 
\end{equation}
Without loss of generality, we
suppose that  $ V $ is a $ C^{2,\beta} $ domain in $ \Omega $ with $ \overline{V}\subset \Omega $. Exploiting the interpolation inequality (see e.g. \cite[Lemma 6.35]{GTbook}), for each $ \alpha\in(0,1) $,  one can  deduce that 
\begin{equation}\label{eq:C 1 alpha beta}
	\|u\|_{C^{1,\alpha}(\overline V_{\frac{\rho}{4}})}\leqslant \|u\|_{C^{2,\beta}(\overline V_{\frac{\rho}{4}})}+C_{V,s,\beta}\|u\|_{L^\infty(V_{\frac{\rho}{4}})}, 
\end{equation}
for some suitable constant $ C_{V,s,\beta}.$  

By means of Theorem~\ref{th weak C^2,alpha interior},  combining \eqref{definition of beta} with \eqref{eq: C 2,alpha estimate of u }, one has 
\begin{equation}
	\begin{split}\label{eq:C 1 alpha}
		\|u\|_{C^{1,\alpha}(\overline V_{\frac{\rho}{4}})}&\leqslant C\left(\|g\|_{L^\infty(\overline V_{{7\rho}/{8}}\times\mathbb{R})}+\|g\|_{C^\alpha(\overline V_{{7\rho}/{8}}\times\mathbb{R})}\|u\|_{C^1(\overline\Omega)}+ \|u\|_{C^{1,\beta}(\overline\Omega)}\right)+C_{V,s,\beta}\|u\|_{L^\infty(V_{\frac{\rho}{4}})}\\
		&\leqslant C\left(\|g\|_{L^\infty(\overline V_{{7\rho}/{8}}\times\mathbb{R})}+\|g\|_{C^\alpha(\overline V_{{7\rho}/{8}}\times\mathbb{R})}\|u\|_{C^1(\overline\Omega)}+ \|u\|_{C^{{1,1-s}}(\overline\Omega)}\right)
	\end{split}
\end{equation}
for some  positive constant $ C$ depending only on $ n,s,\Omega,V,\alpha$.


 Based on the $C^{1,\alpha}_{\rm loc} $-estimate of the
solution $u$, using a truncation argument on the solution $u$ on the set $ V $ and adapting the techniques of Theorem~\ref{th weak C^2,alpha interior}, we will now complete the proof of Theorem~\ref{th C^2,alpha interior}. The details go as follows.

We first provide the $ C^\alpha $-estimate of a truncation function.
 \begin{Lemma}\label{lemma v 2}
	Under the assumptions of Theorem~\ref{th:C^{1,alpha}},  let $W$ be an open set  with $ W\Subset V\Subset\Omega $, and denote
	\[\delta:=\text{ dist}(W,\partial V), \]
	Select a cut off function $ \Phi\in C^\infty_0(\mathbb{R}^n,\mathbb{R}) $ satisfying 
	\begin{equation}\label{cut off function}
		\begin{cases}
			\Phi \equiv 1 \text{  on W}, \\
			\text{supp}(\Phi)\subset W_{\frac{\delta}{2}}\subset V,\\
			0\leqslant \Phi\leqslant 1 \text{  on } \mathbb{R}^n.
			\end{cases}
		\end{equation}
	
	Then, there exists~$ \Psi\in C^{\alpha}(\overline{\Omega},\mathbb{R}) $ such that $ w:=\Phi u $ is a solution of 
	\begin{equation}\label{eq: new equation 2}
		\begin{cases}
			-\Delta w+(-\Delta)^s w= \Psi \qquad &\text{  in } V,\\
			w=0&\text{  on } \mathbb{R}^n\backslash V.
			\end{cases}
		\end{equation}
	In particular, 
	\begin{equation}
		\|\Psi\|_{C^\alpha(\overline V)}\leqslant C \left(1+\|g\|_{C^\alpha(\overline V_{\frac{7\rho}{8}}\times\mathbb{R})}\right)^2\left(1+\|u\|_{C^{1,\alpha}(\overline\Omega)}\right)
		\quad \alpha\in(0,1),
		\end{equation}
	and   $ C $ is a positive constant depending only on $ n,s,V,\Phi,\Omega,\alpha $.
\end{Lemma}

{\begin{proof} From \eqref{eq:C 1 alpha} and the definition of the function $ w $, we know that
		\begin{equation}
			\begin{split}
					\|w\|_{C^{1,\alpha}(\overline W_{\frac{\delta}{2}})}&= \|\Phi u\|_{C^{1,\alpha}(\overline W_{\frac{\delta}{2}})}\leqslant\|\Phi u\|_{C^{1,\alpha}(\overline V)}\leqslant C_{W,V}\|u\|_{C^{1,\alpha}(\overline V)}\\
					&\leqslant C_{W,V,n,s,\Omega,\alpha}  \left(1+\|g\|_{C^\alpha(\overline V_{\frac{7\rho}{8}}\times\mathbb{R})}\right)\left(1+\|u\|_{C^{1,1-s}(\overline\Omega)}\right),
				\end{split}
		\end{equation}
		where $C_{W,V,n,s,\Omega,\alpha}$ is a suitable  positive constant.
		
		Moreover, observing that supp$(w)\subset W_{\frac{\delta}{2}} $, one has that
			\begin{equation}\label{eq: function space of w}
				w\in C^{1,\alpha}_0(\overline W_{\frac{\delta}{2}})\subset C^{1,\alpha}_0(\mathbb{R}^n).
			\end{equation}
			
				Furthermore, to check~\eqref{eq: new equation 2}, one can calculate that  
				\begin{equation}\label{eq: new equation:PSI 2}\begin{split}& \Psi= \Phi g(x,u)+u(-\Delta)\Phi+u(-\Delta)^s\Phi+ 2 \nabla u\nabla \Phi \\&\qquad\qquad+ \int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\Phi(x)-\Phi(x+z))}{|z|^{n+2s}}\, dz.\end{split}\end{equation} 
				Indeed, 
				\[  -\Delta w=\Phi(-\Delta)u+ u(-\Delta)\Phi+2 \nabla u\nabla \Phi,  \]
				and 
				\[  (-\Delta)^s w=\Phi(-\Delta)^su+ u(-\Delta)^s\Phi+\int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\Phi(x)-\Phi(x+z))}{|z|^{n+2s}}\, dz . \]
				Recalling the fact that 
				\[ -\Delta u+(-\Delta)^s u= g(x,u) \qquad \text{  in } \Omega, \] we see that 
				\begin{equation}
					\begin{split}
						-\Delta w+(-\Delta)^s w&=\Phi g(x,u)+u(-\Delta)\Phi+u(-\Delta)^s\Phi+ 2 \nabla u\nabla \Phi\\
						&\qquad + \int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\Phi(x)-\Phi(x+z))}{|z|^{n+2s}}\, dz,
						\end{split}
					\end{equation}
				which is~\eqref{eq: new equation:PSI 2}, as desired.
				
				Here below, in analogy with the proof of Lemma~\ref{lemma v}, we state \textbf{Step~1} in a straightforward way. Instead, \textbf{Step~2} and \textbf{Step~3}  need suitable modifications (specifically, we will make use of~\eqref{eq:C 1 alpha} to show \eqref{step2} below).
			
				\noindent\textbf{Step 1. $ C^\alpha $- estimate on the function $ \Phi g(x,u) $}, that is	
			%	Applying Theorem~\ref{th: regularity}, for every $ x\in \Omega $, one has
			%	\begin{equation}
				%	\begin{split}
					%	|g(x,u(x))|\leqslant c\left(2+\|u\|^{2^*}_{L^\infty(\Omega)}\right).
					%	\end{split}
				%	\end{equation}
			%	Moreover,  for each $ i=\left\{1,\dots,n\right\} $, since $ g\in C^1(\overline{\Omega}\times\mathbb{R}) $ and $ u\in C^{1}(\overline{\Omega})  $, \begin{equation}
				%	D_i(g(x,u(x)))=(\partial_ig)(x,u(x))+(\partial_ug)(x,u(x))D_iu(x),
				%	\end{equation}
			%	so that 
			%	\begin{equation}
				%	|D(g(x,u(x)))|=\|g\|_{C^1(\overline{\Omega}\times\mathbb{R})}\left(1+\|u\|_{C^1(\overline\Omega)}\right)\leqslant\|g\|_{C^1(\overline{\Omega}\times\mathbb{R})}\left(1+\|u\|_{C^1(\overline\Omega)}\right). 
				%	\end{equation}
			%	Consequently,
			%	\[ [g(\cdot,u(\cdot))]_{C^\gamma(\overline\Omega)}\leqslant \text{diam}(\Omega)^{1-\gamma}\|g\|_{C^1(\overline{\Omega}\times\mathbb{R})}\left(1+\|u\|_{C^1(\overline\Omega)}\right). \]	
			%	
			%	Moreover, we obtain 
				\begin{equation}
					\begin{split}
						\|\Phi g(\cdot,u(\cdot))\|_{C^\alpha(\overline V)} %\leqslant [\Phi]_{C^\gamma(\overline\Omega)}\|g(\cdot,u(\cdot))\|_{L^\infty(\Omega)}+[g(\cdot,u(\cdot))]_{C^\gamma(\overline\Omega)}\|\Phi\|_{L^\infty(\Omega)}\\
						\leqslant C_1 \|g\|_{C^\alpha(\overline V\times\mathbb{R})}\left(1+\|u\|_{C^1(\overline\Omega)}\right)
						\end{split}
					\end{equation}
				for some suitable constant $ C_1>0. $	
				
				\noindent\textbf{Step~2. $ C^\alpha $- estimate on the function $ u(-\Delta)\Phi+u(-\Delta)^s\Phi+ 2 \nabla u\nabla \Phi $.} For every $ x\in V$, 
				\begin{equation}
					\begin{split}
						\left|u(-\Delta)\Phi+u(-\Delta)^s\Phi+ 2 \nabla u\nabla \Phi\right|\leqslant C_2\|u\|_{C^1(\overline\Omega)},
						\end{split}
					\end{equation}
				and
				\begin{equation}\label{step2}
					\begin{split}
						&[u(-\Delta)\Phi+u(-\Delta)^s\Phi+ 2 \nabla u\nabla \Phi ]_{C^\alpha(\overline V)}\\
						\leqslant & [u]_{C^\alpha(\overline V)}\|(-\Delta)\Phi+(-\Delta)^s\Phi\|_{L^\infty(V)}+[(-\Delta)\Phi+(-\Delta)^s\Phi]_{C^\alpha(\overline V)}\|u\|_{L^\infty(V)}\\
						&\qquad +2\left([\nabla u]_{C^\alpha(\overline V)}\|\nabla \Phi\|_{L^\infty(V)}+[\nabla \Phi]_{C^\alpha(\overline V)}\|\nabla u\|_{L^\infty(V)}\right)\\
						\leqslant &C_3\left(\|u\|_{C^{1}(\overline\Omega)}+\|u\|_{C^{1,\alpha}(\overline V)}\right),
						\end{split}
					\end{equation}
				for some suitable constant $ C_2,C_3>0 $.
				
				\noindent\textbf{Step~3. $ C^\alpha $- estimate on the function $ H(x):= \int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\Phi(x)-\Phi(x+z))}{|z|^{n+2s}}\, dz $}.
			%	Let us first estimate the $ L^\infty $-norm of $H $. To this end, for every $ x\in \Omega, $ since $ u=0 $ on $ \mathbb{R}^n\backslash\Omega $, one has that
			%	\begin{equation}
				%	\begin{split}
					%	|H(x)|&=\left|\int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\Phi(x)-\Phi(x+z))}{|z|^{n+2s}}\, dz\right|\\
					%	&\leqslant \int_{\left\{|z|<1\right\}} \frac{[u]_{C^{1}(\overline\Omega)}[\Phi]_{C^{1}(\overline\Omega)}}{|z|^{n+2s-2}}\, dz+\int_{\left\{|z|\geqslant 1\right\}}\frac{4\|u\|_{L^\infty(\Omega)}\|\Phi\|_{L^\infty(\Omega)}}{|z|^{n+2s}}\, dz\\
					%	&\leqslant C_4 \|u\|_{C^{1}(\overline\Omega)}.
					%	\end{split}
				%	\end{equation}
			%	
			%	Now we estimate the $ C^\gamma $-norm of $ H$. For this, given $ x_1,x_2\in \Omega $ with $ x_1\neq x_2 $, let $ r:=|x_1-x_2| $.
			%	We distinguish two cases.
			%	
			%	\noindent\textbf{Case 1.} If $ r\geqslant 1 $, then 
			%	\begin{equation}\label{case 1 of H}
				%	\frac{|H(x_1)-H(x_2)|}{|x_1-x_2|^\gamma}\leqslant 2	\|H\|_{L^\infty(\Omega)}\leqslant 2C_4 \|u\|_{C^{1}(\overline\Omega)}.
				%	\end{equation}
			%	
			%	\noindent\textbf{Case 2.}  If $ r< 1 $, then 
			%	\begin{equation}
				%	\begin{split}
					%	\frac{|H(x_1)-H(x_2)|}{|x_1-x_2|^\gamma}&=\left|\int_{\mathbb{R}^n}\frac{(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))(\Phi(x_2)-\Phi(x_2+z))}{|z|^{n+2s}r^\gamma}\, dz\right|\\
					%	& +\left|\int_{\mathbb{R}^n}\frac{(u(x_1)-u(x_1+z))(\Phi(x_1)-\Phi(x_1+z)-\Phi(x_2)+\Phi(x_2+z))}{|z|^{n+2s}r^\gamma}\, dz\right|\\
					%	&:= \Theta_1+\Theta_2.
					%	\end{split}
				%	\end{equation}
			%	
			%	
			%	
			%	For $ \Theta_1 $,  one has 
			%	\begin{equation}
				%	\begin{split}\label{Theta 1}
					%	&	\left|\int_{|z|<1}\frac{(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))(\Phi(x_2)-\Phi(x_2+z))}{|z|^{n+2s}r^\gamma}\, dz\right|\\
					%	\leqslant&\int_{|z|<1}\frac{|(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))|\|\nabla \Phi\|_{L^\infty(\Omega)}}{|z|^{n+2s-1}r^\gamma}\, dz\\
					%	\leqslant & \|\nabla \Phi\|_{L^\infty(\Omega)}\int_{|z|<r}\frac{|(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))|}{|z|^{n+2s-1+\gamma}}\, dz\\
					%	&\quad + \|\nabla \Phi\|_{L^\infty(\Omega)}\int_{r<|z|< 1}\frac{2[u]_{C^{1}(\overline{\Omega})}r^\gamma|z|^{1-\gamma}}{|z|^{n+2s-1}r^\gamma}\, dz\\
					%	\leqslant & 2 \|\nabla \Phi\|_{L^\infty(\Omega)}\int_{|z|< 1}\frac{2[u]_{C^{1}(\overline{\Omega})}}{|z|^{n+2s-2+\gamma}}\, dz\leqslant C_5 [u]_{C^{1}(\overline{\Omega})},
					%	\end{split}
				%	\end{equation}
			%	and
			%	\begin{equation}
				%	\begin{split}\label{Theta 2}
					%	&	\left|\int_{|z|\geqslant 1}\frac{(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))(\Phi(x_2)-\Phi(x_2+z))}{|z|^{n+2s}r^\gamma}\, dz\right|\\
					%	\leqslant& 2\| \Phi\|_{L^\infty(\Omega)}\int_{|z|\geqslant 1}\frac{|(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))| }{|z|^{n+2s}r^\gamma}\, dz\\
					%	\leqslant 
					%	&2\| \Phi\|_{L^\infty(\Omega)}\int_{|z|\geqslant 1}\frac{2[u]_{C^{\gamma}(\overline{\Omega})}r^\gamma}{|z|^{n+2s}r^\gamma}\, dz\\
					%	\leqslant & 2 \| \Phi\|_{L^\infty(\Omega)}\int_{|z|\geqslant 1}\frac{2[u]_{C^{\gamma}(\overline{\Omega})}}{|z|^{n+2s}}\, dz\leqslant C_6 [u]_{C^{\gamma}(\overline{\Omega})}.
					%	\end{split}
				%	\end{equation}
			%	Combining \eqref{Theta 1} with \eqref{Theta 2}, we deduce that 
			%	\begin{equation}\label{Theta_1}
				%	I_1\leqslant \left(C_5 [u]_{C^{1}(\overline{\Omega})}+C_6 [u]_{C^{\gamma}(\overline{\Omega})}\right), \qquad \forall \gamma\in (0,1),
				%	\end{equation} 
			%	for some suitable constants $ C_5,C_6>0. $
			%	
			%	{	For $ \Theta_2, $ using similar arguments to the estimate of $ \Theta_1 $,  one has
				%		\begin{equation}
					%		\begin{split}\label{Theta_2 1}
						%		&\left|\int_{\mathbb{R}^n}\frac{(u(x_1)-u(x_1+z))(\Phi(x_1)-\Phi(x_1+z)-\Phi(x_2)+\Phi(x_2+z))}{|z|^{n+2s}r^\gamma}\, dz\right|\\
						%		\leqslant&\int_{|z|<1}\frac{|(\Phi(x_1)-\Phi(x_1+z)-\Phi(x_2)+\Phi(x_2+z))|\|\nabla u\|_{L^\infty(\Omega)}}{|z|^{n+2s-1}r^\gamma}\, dz\\
						%		\leqslant & \|\nabla u\|_{L^\infty(\Omega)}\int_{|z|<r}\frac{|\Phi(x_1)-\Phi(x_1+z)-\Phi(x_2)+\Phi(x_2+z)|}{|z|^{n+2s-1+\gamma}}\, dz\\
						%		&\quad + \|\nabla u\|_{L^\infty(\Omega)}\int_{r<|z|< 1}\frac{2[\Phi]_{C^{1}(\overline{\Omega})}r^\gamma|z|^{1-\gamma}}{|z|^{n+2s-1}r^\gamma}\, dz\\
						%		\leqslant & 2 \|\nabla u\|_{L^\infty(\Omega)}\int_{|z|< 1}\frac{2[\Phi]_{C^{1}(\overline{\Omega})}}{|z|^{n+2s-2+\gamma}}\, dz\leqslant C_7 [u]_{C^{1}(\overline{\Omega})},
						%		\end{split}
					%		\end{equation}
				%		and 
				%		\begin{equation}
					%		\begin{split}\label{Theta_2 2}
						%		&\left|\int_{|z|\geqslant 1}\frac{(u(x_1)-u(x_1+z))(\Phi(x_1)-\Phi(x_1+z)-\Phi(x_2)+\Phi(x_2+z))}{|z|^{n+2s}r^\gamma}\, dz\right|\\\
						%		\leqslant& 2\| u\|_{L^\infty(\Omega)}\int_{|z|\geqslant 1}\frac{|\Phi(x_1)-\Phi(x_1+z)-\Phi(x_2)+\Phi(x_2+z)| }{|z|^{n+2s}r^\gamma}\, dz\\
						%		\leqslant 
						%		&2\|u\|_{L^\infty(\Omega)}\int_{|z|\geqslant 1}\frac{2[\Phi]_{C^{\gamma}(\overline{\Omega})}r^\gamma}{|z|^{n+2s}r^\gamma}\, dz\\
						%		\leqslant & 2 \|u \|_{L^\infty(\Omega)}\int_{|z|\geqslant 1}\frac{2[\Phi]_{C^{\gamma}(\overline{\Omega})}}{|z|^{n+2s}}\, dz\leqslant C_8\|u\|_{L^\infty(\Omega)}.
						%		\end{split}
					%		\end{equation}
				%		From \eqref{Theta_2 1} and  \eqref{Theta_2 2}, one derive that 
				%		\begin{equation}\label{Theta_2}
					%		I_2\leqslant  C_7 [u]_{C^{1}(\overline{\Omega})}+C_8\|u\|_{L^\infty(\Omega)}, \qquad \forall \gamma\in (0,1),
					%		\end{equation} 
				%		for some suitable constants $ C_7,C_8>0. $}
			%	
			%	Combining \eqref{Theta_1} and \eqref{Theta_2}, and recalling \eqref{case 1 of h}, we can obtain that 
			%	\begin{equation}
				%	\frac{|H(x_1)-H(x_2)|}{|x_1-x_2|^\gamma}\leqslant C_8 \|u\|_{C^{1}(\overline{\Omega})}, \qquad \forall \gamma\in(0,1).
				%	\end{equation}	
			%	As a consequence,  we derive that 
			%	\begin{equation}
				%	\|H\|_{C^\alpha(\overline V)}\leqslant C\|u\|_{C^{1}(\overline{\Omega})}, \qquad \forall \alpha\in(0,1).
				%	\end{equation}
			Let us first estimate the $ L^\infty $-norm of $H $. 
			
			For every $ x\in V, $ since $ u=0 $ on $ \mathbb{R}^n\backslash\Omega $, one has that
			\begin{equation}
					|h(x)|=\left|\int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\Phi(x)-\Phi(x+z))}{|z|^{n+2s}}\, dz\right|\leqslant C_4 \|u\|_{C^{1}(\overline\Omega)}.
			\end{equation}
			
			Now we estimate the $ C^\alpha $-norm of $ H$. For this, given $ x_1,x_2\in V $ with $ x_1\neq x_2 $, let $ r:=|x_1-x_2| $.
			We distinguish two cases.
			
			\noindent\textbf{Case 1.} If $ r\geqslant \frac{\rho}{10}  $, then 
			\begin{equation}\label{case 1 of H}
				\frac{|h(x_1)-h(x_2)|}{|x_1-x_2|^\alpha}\leqslant 20	\|h\|_{L^\infty(V)}\leqslant 20C_4\rho^{-\alpha} \|u\|_{C^{1}(\overline\Omega)}.
			\end{equation}
			
			\noindent\textbf{Case 2.}  If $ r< \frac{\rho}{10} $, then 
			\begin{equation}
				\begin{split}
						\frac{|h(x_1)-h(x_2)|}{|x_1-x_2|^\alpha}&=\left|\int_{\mathbb{R}^n}\frac{(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))(\Phi(x_2)-\Phi(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
						& +\left|\int_{\mathbb{R}^n}\frac{(u(x_1)-u(x_1+z))(\Phi(x_1)-\Phi(x_1+z)-\Phi(x_2)+\Phi(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
						&:= I_1+I_2.
					\end{split}
			\end{equation}
			
			For $ I_1 $, if $|z|<\frac{\rho}{10}$, then $x_i\pm z\in V_{\frac{\rho}{4}}$,  one sees that 
			\begin{equation}
				\begin{split}\label{I_1 1 H}
						&	\left|\int_{|z|<\frac{\rho}{10}}\frac{(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))(\Phi(x_2)-\Phi(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
						\leqslant &\int_{|z|<\frac{\rho}{10}}\frac{|(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))|\|\nabla \Phi\|_{L^\infty(\Omega)}}{|z|^{n+2s-1}r^\alpha}\, dz\\
						\leqslant & \|\nabla \Phi\|_{L^\infty(\Omega)}\int_{|z|< \frac{\rho}{10}}\frac{\int_{0}^{1}|Du(t(x_1+z)+(1-t)x_1)-Du(t(x_2+z)+(1-t)x_2)|\, dt}{|z|^{n+2s-2}r^\alpha}\, dz\\
						\leqslant & \|\nabla \Phi\|_{L^\infty(\Omega)}\int_{|z|< \frac{\rho}{10}}\frac{2[u]_{C^{1,\alpha}(\overline V_{\rho/4})}r^\alpha}{|z|^{n+2s-2}r^\alpha}\, dz\leqslant C_5\rho^{2-2s} [u]_{C^{1,\alpha}(\overline V_{\rho/4})},
					\end{split}
			\end{equation}
			and
			\begin{equation}
				\begin{split}\label{I_1 2 H}
						&	\left|\int_{|z|\geqslant \frac{\rho}{10}}\frac{(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))(\Phi(x_2)-\Phi(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
					&\qquad\leqslant  2\| \Phi\|_{L^\infty(\Omega)}\int_{|z|\geqslant \frac{\rho}{10}}\frac{|(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))| }{|z|^{n+2s}r^\alpha}\, dz\\
						&\qquad\leqslant 2\| \Phi\|_{L^\infty(\Omega)}\int_{|z|\geqslant \frac{\rho}{10}}\frac{2[u]_{C^{\alpha}(\overline{\Omega})}r^\alpha}{|z|^{n+2s}r^\alpha}\, dz  \leqslant C_6 \rho^{-2s}[u]_{C^{\alpha}(\overline{\Omega})}.
					\end{split}
			\end{equation}
			Combining \eqref{I_1 1 H} with \eqref{I_1 2 H}, we deduce that 
			\begin{equation}\label{I_1 H}
				I_1\leqslant \left(C_5\rho^{2-2s} [u]_{C^{1,\alpha}(\overline V_{\rho/4})}+C_6 \rho^{-2s}[u]_{C^{\alpha}(\overline{\Omega})}\right)
			\end{equation} 
			for some suitable constants $ C_5,C_6>0. $
			
			{	For $ I_2, $ using similar arguments to the estimate of $ I_1 $,  one has
					\begin{equation}
							\begin{split}\label{I_2 1 H}
									&\left|\int_{\mathbb{R}^n}\frac{(u(x_1)-u(x_1+z))(\Phi(x_1)-\Phi(x_1+z)-\Phi(x_2)+\Phi(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
									\leqslant&\int_{|z|<1}\frac{|(\Phi(x_1)-\Phi(x_1+z)-\Phi(x_2)+\Phi(x_2+z))|\|\nabla u\|_{L^\infty(\Omega)}}{|z|^{n+2s-1}r^\alpha}\, dz\\
									\leqslant & 2\|\nabla u\|_{L^\infty(\Omega)}\int_{|z|< 1}\frac{2[\Phi]_{C^{1,\alpha}(\mathbb{R}^n)}r^\alpha|z|^{1}}{|z|^{n+2s-1}r^\alpha}\, dz\leqslant C _7[u]_{C^{1}(\overline{\Omega})}, 
								\end{split}
						\end{equation}
					and 
					\begin{equation}
							\begin{split}\label{I_2 2 H}
									&\left|\int_{|z|\geqslant 1}\frac{(u(x_1)-u(x_1+z))(\Phi(x_1)-\Phi(x_1+z)-\Phi(x_2)+\phi(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
									\leqslant 
									&2\|u\|_{L^\infty(\Omega)}\int_{|z|\geqslant 1}\frac{2[\Phi]_{C^{\alpha}(\overline{\Omega})}r^\alpha}{|z|^{n+2s}r^\alpha}\, dz\leqslant C_8\|u\|_{L^\infty(\Omega)}.
								\end{split}
						\end{equation}
					From \eqref{I_2 1 H} and  \eqref{I_2 2 H} one concludes that 
					\begin{equation}\label{I_2 H}
							I_2\leqslant  C_7 [u]_{C^{1}(\overline{\Omega})}+C_8\|u\|_{L^\infty(\Omega)},
						\end{equation} 
					for some suitable constants $ C_7,C_8>0. $}
			
			Combining \eqref{I_1 H} and \eqref{I_2 H}, and recalling \eqref{case 1 of H}, we conclude that 
			\begin{equation}
				\frac{|h(x_1)-h(x_2)|}{|x_1-x_2|^\alpha}\leqslant C_8 \left(\|u\|_{C^{1}(\overline{\Omega})}+[u]_{C^{1,\alpha}(\overline V_{\rho/4})}\right).
			\end{equation}	
			As a consequence,  we derive that 
			\begin{equation}
				\|h\|_{C^\alpha(\overline V)}\leqslant C\left(\|u\|_{C^{1}(\overline{\Omega})}+[u]_{C^{1,\alpha}(\overline V_{\rho/4})}\right).
			\end{equation}
			
			In light of \textbf{Step 1}--\textbf{Step~3}, and combining~\eqref{eq: C 2,alpha estimate of u }
			with~\eqref{eq:C 1 alpha}, we obtain that $ \Psi\in C^\alpha(\overline V) $ and 
				\begin{equation}
						\begin{split}\label{eq:Psi C alpha }
							\|\Psi\|_{C^\alpha(\overline V)}&\leqslant C \left(\|g\|_{L^\infty(\overline V\times\mathbb{R})}+\|g\|_{C^\alpha(\overline V\times\mathbb{R})}\|u\|_{C^1(\overline\Omega)}+\|u\|_{C^{1}(\overline{\Omega})}+[u]_{C^{1,\alpha}(\overline V_{\rho/4})}\right)\\
							&\leqslant C \left(\|g\|_{L^\infty(\overline V\times\mathbb{R})}+\|g\|_{C^\alpha(\overline V\times\mathbb{R})}\|u\|_{C^1(\overline\Omega)}+\|u\|_{C^{1}(\overline{\Omega})}+\|u\|_{C^{2,\beta}(\overline V_{\frac{\rho}{4}})}+\|u\|_{L^\infty(V_{\frac{\rho}{4}})}\right)\\
							&\leqslant C \left(\|g\|_{L^\infty(\overline V_{{7\rho}/{8}}\times\mathbb{R})}+\|g\|_{C^\alpha(\overline V_{{7\rho}/{8}}\times\mathbb{R})}\|u\|_{C^1(\overline\Omega)}+ \|u\|_{C^{{1,1-s}}(\overline\Omega)}\right),
							 \quad \forall \alpha\in(0,1),
							\end{split}
					\end{equation}
				for some constant $ C>0 $ depending only on $ n,s,\Omega,V,W,\alpha $.
			\end{proof}
			
			The gist of this argument consists of employing a mollifier to regularize $ w $, establishing the $ C^{2,\alpha} $-estimates on the mollification, and then passing to the limit.
			
			\begin{proof}[\textbf{Proof of Theorem~\ref{th C^2,alpha interior}}]
				We observe that the function $ w:=\Phi u $ satisfies the following  equation
				\begin{equation}\label{w solution}
					\begin{cases}
						-\Delta w+(-\Delta)^s w= \Psi \qquad &\text{  in } V,\\
						w=0&\text{  on } \mathbb{R}^n\backslash V,
						\end{cases}
					\end{equation}
				where $ \Psi \in C^\alpha(\overline V) $ and {$ w \in C^{1,\alpha}_0(\overline{ W}_{\frac{\delta}{2}}). $ 		
				
				Hence, by the mollifier argument, we can consider the functions 
				%		we can employ the standard mollifier $ \eta_\epsilon $  to regularize $ w $, that is
						\[ w_\epsilon:=\eta_\epsilon\ast w ,\quad   \Psi_\epsilon:=\eta_\epsilon\ast \Psi, \qquad \forall 0<\epsilon<\frac{\delta}{10}.\] In this way, we have that $ w_\epsilon\in C_0^\infty(W_{\frac{3\delta}{4}}) $, that$ \Psi_\epsilon\in C^\infty(\overline W_{\frac{3\delta}{4}}) $ and that
						\begin{equation}\label{new equation 2}
								\begin{cases}
										-\Delta w_\epsilon+(-\Delta)^s w_\epsilon= \Psi_\epsilon \qquad &\text{  in } W_{\frac{3\delta}{4}}, \\
										w_\epsilon=0 & \text{  on } \mathbb{R}^n\backslash W_{\frac{3\delta}{4}}.
									\end{cases}
						\end{equation}}
				We readily demonstrate the $C^{2,\alpha}$-estimate of $ w_\epsilon$, referring to  \eqref{C 2 of v}, finding that
				\begin{equation}\label{C 2 of w}
						\begin{split}
								\|w_\epsilon\|_{C^{2,\alpha}(\overline W_{\frac{3\delta}{4}})}&\leqslant C_{n,\alpha,W_{\frac{3\delta}{4}}}\left(\|w_\epsilon\|_{L^\infty(W_{\frac{3\delta}{4}})}+\|\Psi_\epsilon\|_{C^\alpha(\overline W_{\frac{3\delta}{4}})}+\|(-\Delta)^s w_\epsilon\|_{C^\alpha(\overline W_{\frac{3\delta}{4}})}\right)\\
								&\leqslant C(n,\alpha,W_{\frac{3\delta}{4}})\left(\|w_\epsilon\|_{L^\infty(W_{\frac{3\delta}{4}})}+\|\Psi_\epsilon\|_{C^\alpha(\overline W_{\frac{3\delta}{4}})}\right).
							\end{split}
					\end{equation}	
			
				
				
			%		Recall that if $ w $ solves \eqref{v solution} in the weak sense and $ \eta_\epsilon $ is the standard mollifier then
			%		\begin{equation}
				%		-\Delta(\eta_\epsilon\ast w)+(-\Delta)^s (\eta_\epsilon \ast w)= \eta_\epsilon \ast\psi \qquad \text{  in } W_{\frac{3\delta}{4}}, 
				%		\end{equation}
			%		for $ 0<\epsilon<\frac{\delta}{10} $ small enough, and supp$ (\eta_\epsilon\ast w)\subset W_{\frac{3\delta}{4}} $. Indeed, $ w=0 $ outside $ W_{\frac{\delta}{2}} $, and 
			%		\[ (\eta_\epsilon\ast v)(x)=\int_{|z|<1} \eta(z)v(x-\epsilon z)\, dz=0, \quad \forall x\in\mathbb{R}^n\backslash W_{\frac{3\delta}{4}}.  \] 
			%		
			%	According to the global estimate for the Dirichlet problem \eqref{new equation 2} and the fact that 
			%%	let $ \omega\in C^\infty_0(\mathbb{R}^n)  $, 
			%	there exists $ C_{1/2}>0  $ such that  
			%		\begin{equation}
				%		\|(-\Delta)^s w_\epsilon \|_{C^\alpha(\overline{W_{\frac{3\delta}{4}}})} \leqslant \frac{1}{2} \| w_\epsilon\|_{C^{2,\alpha}(\overline{W_{\frac{3\delta}{4}}})}+C_{1/2}\| w_\epsilon\|_{L^\infty({W_{\frac{3\delta}{4}}})}, 
				%		\end{equation}
			
			%	{where in the last inequality we have used the fact that:\newline
				%	 let $ \omega\in C^\infty_0(\mathbb{R}^n)  $, then $ \forall \tau>0 $, there exists $ C_\tau>0  $ such that  
				%		\begin{equation}
					%		\|(-\Delta)^s \omega\|_{C^\gamma(\overline{W_{\frac{3\delta}{4}}})} \leqslant \tau \| \omega\|_{C^{2,\gamma}(\overline{W_{\frac{3\delta}{4}}})}+C_\tau\| \omega\|_{L^\infty({W_{\frac{3\delta}{4}}})}.
					%		\end{equation}
						
				
				
				% there exists $ \omega\in C^\infty(\overline{V_{\frac{3\rho}{4}}})\cap W^{1,\infty}(\mathbb{R}^n) $ such that for every $ s\in (0,1) $ 
				%	\begin{equation}
					%	-\Delta \omega+(-\Delta)^s\omega \geqslant 1, \quad \text{in } V_{\frac{3\rho}{4}},  \quad \omega>0 \text { in } \overline{V_{\frac{3\rho}{4}}}, \quad  0\leqslant\omega\leqslant C_0 \text { in } \mathbb{R}^n,
					%	\end{equation}
				%	for some constant $ C_0 $ depending on $ n,s,V_{\frac{3\rho}{4}} $. As a result, we get 
				%	\begin{equation}
					%	-\Delta \left(\omega-\frac{v_\epsilon}{\|\psi_\epsilon\|_{L^\infty(\mathcal{O})}}\right)+(-\Delta)^s\left(\omega-\frac{v_\epsilon}{\|\psi_\epsilon\|_{L^\infty(\mathcal{O})}}\right)\geqslant 0 \qquad \text{  in } V_{\frac{3\rho}{4}}.\end{equation}
				 
				
				% that  
				%\begin{equation}
				%	\|w_\epsilon\|_{L^\infty(W_{\frac{3\delta}{4}})}\leqslant \mu\|\Psi_\epsilon\|_{L^\infty(W_{\frac{3\delta}{4}})}\qquad  \text{  in } W_{\frac{3\delta}{4}},
				%\end{equation}
				%for some constant $\mu $ depending on $ n,s, W_{\frac{3\delta}{4}} $.
				%	
				
				As a consequence of this, and combining 
				 ~\cite[Lemma 4.8]{SVWZ22} with the maximum principle (see \cite[Theorem 1.2]{BDVV22b}), we conclude that 
					\begin{equation*}
						\|w_\epsilon\|_{C^{2,\alpha}(\overline W_{\frac{3\delta}{4}})}\leqslant C_{n,s, \alpha,W_{\frac{3\delta}{4}}}\left(\|\Psi_\epsilon\|_{L^\infty(W_{\frac{3\delta}{4}})}+\|\Psi_\epsilon\|_{C^\alpha(\overline W_{\frac{3\delta}{4}})}\right), \quad \forall 0<\epsilon<\frac{\delta}{10}.
						\end{equation*}
					Therefore,  
					\begin{equation}\label{eq:mollifer}
						\|\eta_\epsilon\ast w\|_{C^{2,\alpha}(\overline W_{\frac{3\delta}{4}})}\leqslant C_{n,s, \alpha,W_{\frac{3\delta}{4}}}\|\eta_\epsilon\ast \Psi\|_{C^\alpha(\overline W_{\frac{3\delta}{4}})}\leqslant C \| \Psi\|_{C^\alpha(\overline V)},
						\end{equation}
					for some  constant $ C>0 $ independent  of $ w,\epsilon. $ 
					
					As a result, applying the Arzel\`{a}-Ascoli Theorem and exploiting the definition of  $ w$, we obtain that  $ u\in C^{2,\alpha}(\overline{W}). $ Moreover, it follows from \eqref{eq:mollifer} and \eqref{eq:Psi C alpha } that
					\begin{equation}
							\|u\|_{C^{2,\alpha}(\overline{W})}\leqslant  C \| \Psi\|_{C^\alpha(\overline V)}\leqslant C \left(\|g\|_{L^\infty(\overline V_{{7\rho}/{8}}\times\mathbb{R})}+\|g\|_{C^\alpha(\overline V_{{7\rho}/{8}}\times\mathbb{R})}\|u\|_{C^1(\overline\Omega)}+ \|u\|_{C^{{1,1-s}}(\overline\Omega)}\right)
						\end{equation}
					for some constant $ C>0 $ depending only on $ n,s,\Omega,V,\alpha,W$.	
				\end{proof}
				
				
\subsection{General interior  $ C^{2,\alpha} $-regularity}\label{sec: C^2-interior regularity}\setcounter{equation}{0}
				
Now we discuss how to remove the $C^{1,1}$ smoothness assumption of the domain. For this, 
				we 
				apply a suitable truncation argument again of
				the solution $ u $ with an interior $ C^{1,\alpha} $-regularity argument to have a version of $C^{2,\alpha}$ regularity result stated in Theorem~\ref{th C^2,alpha interior without boundary condition}.
				
				
				We  split the proof of Theorem~\ref{th C^2,alpha interior without boundary condition} into two steps:
				
				\begin{itemize}
					\item [(a)]  Employ a mollifier technique and a cut off argument  to get the interior $C^{2,\alpha}$-norm of convolution  $u_\epsilon$ in subsection~\ref{sub A priori C{2,alpha}-estimate}.
					\item [(b)]Apply  Proposition~\ref{pro final estimate} and  Arzel\`{a}-Ascoli Theorem  to get the inclusion in subsection~\ref{sec:Proof of Theorem1.5}.
				\end{itemize}
				
%			To begin with, we need to introduce some notations.
%				
%				
%				\begin{itemize}
%					\item [(a)] Let $V$ be an open set  with $ V\Subset\Omega $, and denote
%					\begin{equation}\label{definition of rho}
%							\rho:=\text{ dist}(V,\partial\Omega),\quad \text{  and } \quad  V_{\delta}:=\left\{x\in \Omega: \text{ dist}(x, V)<\delta\right\}.
%						\end{equation}
%				\item [(b)] For any given $x_0\in V_{\frac{\rho}{4}}$ where $V_{\frac{\rho}{4}} $ is given in \eqref{definition of rho}, denote
%				\[0<R<\min\left(\frac{1}{2},\frac{\rho}{10}\right),\quad B_{R}(x_0):=\left\{x\in \Omega: |x-x_0|<R \right\}. \]
%				\item [(c)] Denote the interior norm  as follows:
%				\begin{equation}
%					[u]_{\alpha;B_{R}(x_0)}=\sup\limits_{x,y\in B_{R}(x_0)}\frac{|u(x)-u(y)|}{|x-y|^\alpha},\qquad 0<\alpha<1;
%				\end{equation}
%				\begin{equation}
%					|u|^\prime_{k;B_{R}(x_0)}=\sum\limits_{j=0}\limits^{k}R^j\|D^j u\|_{L^\infty(B_{R}(x_0))};
%				\end{equation}
%				\begin{equation}
%					|u|^\prime_{k,\alpha;B_{R}(x_0)}=|u|^\prime_{k;B_{R}(x_0)}+R^{k+\alpha}[D^ku]_{\alpha;B_{R}(x_0)}.
%				\end{equation}
%				\end{itemize}
%				
				
				
			\subsubsection{{A mollifier technique and a truncation argument}}\label{sub A priori C{2,alpha}-estimate}
			
			Let $u\in X^1$ solve the equation \eqref{1.6}, then the convolutions
			$u_\epsilon$ and $g_\epsilon$ satisfy the following equation 
			\begin{equation}\label{u epsilon}
				-\Delta u_\epsilon+(-\Delta)^s u_\epsilon= g_\epsilon \qquad \text{  in } V_{\frac{3\rho}{4}}.
			\end{equation}
			%and $\eta_\epsilon$ is the standard mollifier. For every $x\in\mathbb{R}^n$, we denote 
			%\[ u_\epsilon(x):=(\eta_\epsilon\ast u)(x)= \int_{|y|\leqslant \epsilon} \eta_\epsilon(y)u(x-y)\, dy\quad \forall 0<\epsilon<R. \]
			
			
			
			
			
			%Recall that if $u$ solves the following equation in the weak sense, 
			%\begin{equation}
			%	-\Delta u+(-\Delta)^s u= g(x,u) \qquad \text{  in } \Omega
			%\end{equation} 
			%Then, for every $0<\epsilon<R$ small enough, 
			%\begin{equation}\label{eq u epsilon}
			%	-\Delta(\eta_\epsilon\ast u)+(-\Delta)^s (\eta_\epsilon \ast u)= \eta_\epsilon \ast g(x,u) \qquad \text{  in } V_{\frac{3\rho}{4}}.
			%\end{equation}
			%We may assume that 
			%$ g_\epsilon:=\eta_\epsilon\ast g(x,u) $, then
			
			Moreover, from  Proposition~\ref{pro u epsilon1} and the following Proposition~\ref{pro g epsilon}, we can see that 
			\[ u_\epsilon \in C^{2,\alpha}(\overline V_{\frac{3\rho}{4}})\cap L^\infty(\mathbb{R}^n) \quad\text{and}\quad g_\epsilon\in C^{\alpha}(\overline B_{R}(x_0) )\qquad \forall 0<\epsilon<R.  \]
			
			
			%{\begin{Proposition}\label{pro u epsilon}
					%		Assume that $u$ is a bounded solution of \eqref{1.5},  then $u_\epsilon\in L^\infty(\mathbb{R}^n)$. In particular,
					%		\begin{equation}\label{eq:u epsilon C1}
						%			\|u_\epsilon\|_{L^\infty(\mathbb{R}^n)}\leqslant \|u\|_{L^\infty(\mathbb{R}^n)} \qquad \forall 0<\epsilon<R.
						%		\end{equation}
					%\end{Proposition}}
					%\begin{proof}
					%	$ \forall x\in\mathbb{R}^n$, one has
					%	\begin{equation}
						%		|u_\epsilon(x)|\leqslant \int_{\Omega}\eta_\epsilon(y)|u(x_1-y)|\, dy
						%		\leqslant \|u\|_{L^\infty(\mathbb{R}^n)}.
						%	\end{equation}
					%	Consequently,  one can deduce that $u_\epsilon\in L^\infty(\mathbb{R}^n)$ and \eqref{eq:u epsilon C1}.
					%\end{proof}
					%
					%	\begin{Proposition}\label{pro u epsilon}
						%		Assume that $u$ is a bounded weak solution of \eqref{1.5} satisfying \eqref{*},  then 
						%		\begin{equation}\label{eq:u epsilon C1}
							%			\int_{ \mathbb{R}^n}\frac{|u_\epsilon(x)|}{(1+|x|)^{n+2s}}\, dx\leqslant 2\kappa, \qquad \forall 0<\epsilon<R.
							%		\end{equation}
						%\end{Proposition}}
						%\begin{proof}
						%	Recalling  the definition of $u_\epsilon$ and $R$, utilizing the assumption \eqref{*}, one has
						%	\begin{equation}
							%		\begin{split}
								%			\int_{ \mathbb{R}^n}\frac{|u_\epsilon(x)|}{(1+|x|)^{n+2s}}\, dx&\leqslant \int_{|z|\leqslant\epsilon}\eta_\epsilon(z)\int_{ \mathbb{R}^n}\frac{|u(x-z)|}{(1+|x|)^{n+2s}}\, dx\, dz\\
								%			& \leqslant \int_{|z|\leqslant \epsilon}\eta_\epsilon(z)\int_{ \mathbb{R}^n}\frac{|u(x)|}{(1+|x+z|)^{n+2s}}\, dx\, dz\\
								%			& \leqslant \int_{|z|\leqslant \epsilon}\eta_\epsilon(z)\int_{ \mathbb{R}^n}\frac{|u(x)|}{(1+|x|-R|)^{n+2s}}\, dx\, dz\\
								%			& \leqslant \int_{|z|\leqslant \epsilon}\eta_\epsilon(z)\int_{ \mathbb{R}^n}\frac{|u(x)|}{(1/2+|x|)^{n+2s}}\, dx\, dz\\
								%				& \leqslant \int_{ \mathbb{R}^n}\frac{|u(x)|}{(1/2)^{n+2s}+|x|^{n+2s}}\, dx\\
								%				&=2^{n+2s} \int_{ \mathbb{R}^n}\frac{|u(x)|}{1+2^{n+2s}|x|^{n+2s}}\, dx\leqslant 2^{n+2s}\kappa
								%		\end{split}
							%	\end{equation}
						%	Consequently,  one can obtain the desired result \eqref{eq:u epsilon C1}.
						%\end{proof}
						%
						
\begin{Proposition}\label{pro g epsilon}
							Let $g\in C^{\alpha}_{\rm loc}(\Omega\times\mathbb{R}^n)$ for any  $\alpha\in (0,1)$ and  $\epsilon\in (0,R)$, 
							If $u\in C^1_{\rm loc}(\Omega)$,  then  
							\begin{equation}\label{g_epsilon *}
								\|g_\epsilon(\cdot,u(\cdot))\|_{C^{\alpha}(\overline B_{R}(x_0))}\leqslant \|g\|_{C^{\alpha}(\overline B_{2R}(x_0) \times\mathbb{R}^n)}\left(1+\|Du\|_{L^\infty(B_{2R}(x_0))}\right).
							\end{equation}
						\end{Proposition}
						
						\begin{proof}
							For every $x\in B_{R}(x_0)$, one has
							\begin{equation}
								|g_\epsilon(x)|=\left|\int_{|y|\leqslant \epsilon} \eta_\epsilon(y)g(x-y,u(x-y))\, dy\right|\leqslant \|g(\cdot,u(\cdot))\|_{L^\infty(B_{2R}(x_0))},
							\end{equation}
							and, for all~$ x_1,x_2\in B_{R}(x_0)$ with $x_1\neq x_2$, one has that 
							%	\begin{equation}
								%		\begin{split}
									%			&{g_\epsilon(x_1)-g_\epsilon(x_2)}= \int_{|y|\leqslant\epsilon} \eta_\epsilon(y) g(x_1-y,u(x_1-y))-g(x_2-y,u(x_2-y))\, dy\\
									%			=&\int_{|y|\leqslant\epsilon} \eta_\epsilon(y)\bigg(\int_{0}^{1} \partial_xg(tx_1+(1-t)x_2-y,tu(x_1-y)+(1-t)u(x_2-y))(x_1-x_2)\\
									%			&+ \partial_tg(tx_1+(1-t)x_2-y,tu(x_1-y)+(1-t)u(x_2-y))(u(x_1-y)-u(x_2-y))\, dt\bigg)\, dy.
									%		\end{split}
								%	\end{equation}
							% So that, as for the assumption (a), one has 
							% \begin{equation}
								% 	\begin{split}
									% 	{|g_\epsilon(x_1)-g_\epsilon(x_2)}|&\leqslant \|g\|_{C^{0,1}(\overline B_{2R}(x_0)\times\mathbb{R})}|x_1-x_2|+|\kappa|[u_\epsilon]_{0,\alpha; B_{R}(x_0)}|x_1-x_2|^\alpha.
									% 	\end{split}
								% \end{equation}
							%	As a result, one has 
							%	\begin{equation}
								%		\|g_\epsilon\|_{C^{\alpha}(\overline B_{R}(x_0))}\leqslant \|g\|_{C^{0,1}(\overline B_{2R}(x_0) \times\mathbb{R}^n)}+ |\kappa|[u_\epsilon]_{0,\alpha; B_{R}(x_0)}.
								%	\end{equation}
							%As for the assumption (b), one has 
\begin{equation}
								\|g_\epsilon(\cdot,u(\cdot))\|_{C^{\alpha}(\overline B_{R}(x_0))}\leqslant \|g\|_{C^{\alpha}(\overline B_{2R}(x_0) \times\mathbb{R}^n)}\left(1+\|Du\|_{L^\infty( B_{2R}(x_0))}\right).\qedhere
							\end{equation}
						\end{proof}
						
						
						
						We now use the cut off argument again for $u_\epsilon$ to get the $C^{2,\alpha}$-norm  of $u_\epsilon$. 
						
						%Consider a cut off function $ \phi^R(x) $ defined in \eqref{R}. 
						%satisfying 
						%\begin{equation}\label{cutoff function 2}
						%	\begin{cases}
							%		\phi \equiv 1 \text{  on } B_1(0)\\
							%		\text{supp}(\phi)\subset B_{\frac{3}{2}}(0);\\
							%		0\leqslant \phi\leqslant 1 \text{  on } \mathbb{R}^n.
							%	\end{cases}
						%\end{equation} Denote 
						%\[ \quad :=\phi\left(\frac{x-x_0}{R}\right). \]
						%Thus, one has 
						%\[ B_{4R}(x_0)\subset V_{\frac{3\rho}{4}}\subset \Omega, \quad \text{supp}(\phi^R)\subset B_{\frac{3R}{2}}(x_0). \]
						
						\begin{Lemma}\label{lemma v epsilon}
							For any  $\alpha\in(0,1)$,	we assume that  $ g(\cdot,u(\cdot))\in C^\alpha_{\rm loc}(\Omega) $.
							Suppose that  $ u\in C^{2,\alpha}(\overline V_{\frac{3\rho}{4}})\cap L^\infty(\mathbb{R}^n)$  solves the following equation 
							\begin{equation}\label{u is a solution in V}
								-\Delta u+(-\Delta)^s u= g \qquad \text{  in } V_{\frac{3\rho}{4}}.
							\end{equation}
							
							
							Then, there exists~$ \psi\in C^{\alpha}( B_{R}(x_0),\mathbb{R}) $ such that $ v:=\phi^R u $ satisfies 
							\begin{equation}\label{eq: new equation2}
								-\Delta v+(-\Delta)^s v= \psi \qquad \text{  in } V_{\frac{3\rho}{4}},
							\end{equation}
							where $ \phi^R(x) $ is defined in \eqref{R}. In particular, 
							\begin{equation}\label{estimate psi 2}
								R^2|\psi|^\prime_{0,\alpha;B_{R}(x_0)}\leqslant  C_{\rho,n,s}\left(R^2| g(\cdot,u(\cdot))|^\prime_{0,\alpha;B_{R}(x_0)}+\|u\|_{L^\infty(\mathbb{R}^n)}\right)
							\end{equation}
							for some positive constant $C_{n,s,\rho}$.
							
						\end{Lemma}
						
						\begin{proof}
							From  the definition of the function $ v $, we know that
							\begin{equation}\label{eq: function space of v2}
								v\in C^{2,\alpha}_0( B_{2R}(x_0))\subset C^{2,\alpha}_0(\mathbb{R}^n). 
							\end{equation}
							%		Furthermore, to check~\eqref{eq: new equation2}, one can calculate that  
							%	\begin{equation}\label{eq: new equation:PSI2 1}
								%			\psi= g(x,u)+\Delta (u(1-\phi^R))-(-\Delta)^s (u(1-\phi^R))\quad  \text{  in } V_{\frac{3\rho}{4}}.
								%		\end{equation} 
							We also notice that $\phi^R=1$ in $B_{3R/2}(x_0)$ and that  \begin{equation}\label{eq: new equation:PSI2}
								\psi= g(x,u)-(-\Delta)^s (u(1-\phi^R))\quad  \text{  in } B_{3R/2}(x_0).
							\end{equation}
							\noindent\textbf{Step 1. $ C^\alpha $- estimates on the function $  g(x,u) $.}
							Since $ g(\cdot,u(\cdot))\in C^\alpha(\overline B_{R}(x_0)) $, we have 
							\begin{equation}
								\begin{split}
									R^2| g(\cdot,u(\cdot))|^\prime_{0,\alpha;B_{R}(x_0)}&=	R^2\| g(\cdot,u(\cdot))\|_{L^\infty(B_{R}(x_0))}+R^{2+\alpha}[g(\cdot,u(\cdot))]_{0,\alpha; B_{R}(x_0)}. 
								\end{split}
							\end{equation}
							
							\noindent\textbf{Step 2. $ C^\alpha $- estimates on the function $  (-\Delta)^s (u(1-\phi^R)) $.}
							
							Observing that for $x\in B_{R}(x_0)$, we have 
							\begin{equation}
								(-\Delta)^s (u(1-\phi^R))(x)=\int_{\mathbb{R}^n\backslash B_{3R/2}(x_0)} \frac{-u(1-\phi^R)(y)}{|x-y|^{n+2s}}\, dy.
							\end{equation}
							From the above expression and the inequality \eqref{fact},
							% the following fact that \[ |x-y|\geqslant |y-x_0-R|\geqslant |y-x_0|-R\geqslant \frac{1}{5}(R+|y-x_0|)\quad \forall y\in \mathbb{R}^n\backslash B_{3R/2}(x_0), x\in B_{R}(x_0),  \] 
							one obtains that
							\begin{equation}
								\begin{split}\label{step 2 1}
									\|(-\Delta)^s (u(1-\phi^R))\|_{L^\infty(B_{R}(x_0))}\leqslant& 5^{n+2s}\int_{ \mathbb{R}^n\backslash B_{3R/2}(x_0)}\frac{|u(y)|}{(R+|y-x_0|)^{n+2s}}\, dy\\
									\leqslant& C_{n,s}\|u\|_{L^\infty(\mathbb{R}^n)}R^{-2s}.
								\end{split}
							\end{equation}
							%			\begin{equation}
								%				\begin{split}\label{step 2 1}
									%					\|(-\Delta)^s (u(1-\phi^R))\|_{L^\infty(B_{R}(x_0))}&\leqslant 5^{n+2s}\|(R+|y|)^{-n-2s} u(y)\|_{L^1(\mathbb{R}^n)}\\
									%					&\leqslant C_{n,s} \int_{ \mathbb{R}^n}\frac{|u(y)|}{(R+|y|)^{n+2s}}\, dy\\
									%					&\leqslant C_{n,s} R^{-2s}\int_{ \mathbb{R}^n}\frac{|u(Ry)|}{(1+|y|)^{n+2s}}\, dy\\
									%					&\leqslant C_{n,s} R^{-2s}\int_{ \mathbb{R}^n}\frac{|u(Ry)|}{1+|Ry|^{n+2s}}\, dy\\
									%				\end{split}
								%			\end{equation}
							For every $x_1,x_2\in B_{R}(x_0)$ with $x_1\neq x_2$, 
							\begin{equation}
								\begin{split}
									&|(-\Delta)^s (u(1-\phi^R))(x_1)-(-\Delta)^s (u(1-\phi^R))(x_2)|\\
									\leqslant & \int_{\mathbb{R}^n\backslash B_{3R/2}(x_0)} |-u(1-\phi^R)(y)| \left|\frac{1}{|x_1-y|^{n+2s}}-\frac{1}{|x_2-y|^{n+2s}}\right|\, dy\\
									\leqslant &\int_{|x_1-x_2|>\frac{|x_1-y|}{2}\cap \mathbb{R}^n\backslash B_{3R/2}(x_0)}|u(y)|\left|\frac{1}{|x_1-y|^{n+2s}}-\frac{1}{|x_2-y|^{n+2s}}\right|\, dy\\
									&\quad +\int_{|x_1-x_2|\leqslant\frac{|x_1-y|}{2}\cap \mathbb{R}^n\backslash B_{3R/2}(x_0)}|u(y)|\left|\frac{1}{|x_1-y|^{n+2s}}-\frac{1}{|x_2-y|^{n+2s}}\right|\, dy\\
									:=& I_1+I_2.
								\end{split}
							\end{equation}
							For $I_1$, we see that
							\begin{equation}
								\begin{split}\label{I1}
									 I_1 &\leqslant 2\int_{|x_1-x_2|>\frac{|x_1-y|}{2}\cap \mathbb{R}^n\backslash B_{3R/2}(x_0)}\frac{|u(y)||x_1-x_2|^\alpha}{|x_1-y|^{n+2s+\alpha}}\, dy\\
									&\leqslant 2\int_{ \mathbb{R}^n\backslash B_{3R/2}(x_0)}\frac{|u(y)||x_1-x_2|^\alpha}{|x_1-y|^{n+2s+\alpha}}\, dy\\
									&\leqslant  C_{n,s}|x_1-x_2|^\alpha \int_{ \mathbb{R}^n}\frac{|u(y)|}{(R+|y-x_0|)^{n+2s+\alpha}}\, dy\\
									&\leqslant C_{n,s}\|u\|_{L^\infty(\mathbb{R}^n)}|x_1-x_2|^\alpha R^{-2s-\alpha}.
								\end{split}
							\end{equation}
							We also observe that, for every $t\in [0,1]$,
							\[ |tx_1+(1-t)x_2-y|\geqslant |x_1-y|-|x_1-x_2|\geqslant \frac{|x_1-y|}{2}, \quad \text{if} \quad |x_1-x_2|\leqslant\frac{|x_1-y|}{2}, \] so that,
							\begin{equation}
								\begin{split}\label{I2}
									I_2&\leqslant \int_{|x_1-x_2|\leqslant\frac{|x_1-y|}{2}\cap \mathbb{R}^n\backslash B_{3R/2}(x_0)}|u(y)|\left|\int_{0}^{1}\frac{|x_1-x_2|}{|tx_1+(1-t)x_2-y|^{n+2s+1}}\, dt\right|\, dy\\
									&\leqslant C_{n,s}\int_{\mathbb{R}^n\backslash B_{3R/2}(x_0)}|u(y)|\left|\int_{0}^{1}\frac{|x_1-x_2|^\alpha |x_1-y|^{1-\alpha} }{|x_1-y|^{n+2s+1}}\, dt\right|\, dy\\
									&\leqslant C_{n,s}\int_{\mathbb{R}^n\backslash B_{3R/2}(x_0)}|u(y)|\frac{|x_1-x_2|^\alpha  }{|x_1-y|^{n+2s+\alpha}}\, dy\\
									&\leqslant  C_{n,s}|x_1-x_2|^\alpha \int_{ \mathbb{R}^n}\frac{|u(y)|}{(R+|y-x_0|)^{n+2s+\alpha}}\, dy\\
									&\leqslant C_{n,s}\|u\|_{L^\infty(\mathbb{R}^n)}|x_1-x_2|^\alpha R^{-2s-\alpha}.
								\end{split}
							\end{equation}
							Combining \eqref{I1} and \eqref{I2}, one has 
							\begin{equation}
								[(-\Delta)^s (u(1-\phi^R))]_{0,\alpha;B_{R}(x_0)}\leqslant C_{n,s}\|u\|_{L^\infty(\mathbb{R}^n)}R^{-2s-\alpha}.
							\end{equation}
							As a consequence of this and \eqref{step 2 1}, we have that
							\begin{equation}
								\begin{split}
									&	R^2| (-\Delta)^s (u(1-\phi^R))|^\prime_{0,\alpha;B_{R}(x_0)}\\
									=&	R^2\| (-\Delta)^s (u(1-\phi^R))\|_{L^\infty(B_{R}(x_0))}+R^{2+\alpha}[(-\Delta)^s (u(1-\phi^R))]_{\alpha; B_{R}(x_0)}\\
									\leqslant& C_{n,s}\|u\|_{L^\infty(\mathbb{R}^n)}R^{2-2s}.
								\end{split}
							\end{equation}	
							In light of \textbf{Step 1} and \textbf{Step 2}, we obtain
							\begin{equation}
								\begin{split}
									R^2|\psi|^\prime_{0,\alpha;B_{R}(x_0)}&=	R^2\| \psi\|_{L^\infty(B_{R}(x_0))}+R^{2+\alpha}[\psi]_{\alpha; B_{R}(x_0)}\\
									&\leqslant C_{\rho,n,s} \left(R^2| g(\cdot,u(\cdot))|^\prime_{0,\alpha;B_{R}(x_0)}+\|u\|_{L^\infty(\mathbb{R}^n)}\right)
								\end{split}
							\end{equation}
							for some positive constant $C_{n,s,\rho}$.
						\end{proof}
						
						%\begin{Lemma}\label{lemma v epsilon}
						%	For any  $\alpha\in(0,1)$,	we assume that  $ g\in C^\alpha_{\rm loc}(\Omega) $.
						%	Suppose that  $ u\in C^{2,\alpha}(\overline V_{\frac{3\rho}{4}})\cap C^\alpha(\mathbb{R}^n)$  solves the following equation 
						%	\begin{equation}\label{u is a solution in V}
							%		-\Delta u+(-\Delta)^s u= g \qquad \text{  in } V_{\frac{3\rho}{4}}.
							%	\end{equation}
						%
						%	
						%	Then, there exists~$ \psi\in C^{\alpha}( B_{2R}(x_0),\mathbb{R}) $ such that $ v:=\phi^R u $ satisties 
						%	\begin{equation}\label{eq: new equation2}
							%			-\Delta v+(-\Delta)^s v= \psi \qquad \text{  in } V_{\frac{3\rho}{4}}.
							%	\end{equation}
						%	In particular, 
						%	\begin{equation}\label{estimate psi 2}
							%		R^2|\psi|^\prime_{0,\alpha;B_{2R}(x_0)}\leqslant C\left(|u|^\prime_{1,\alpha;B_{3R}(x_0)}+\|u\|_{C^{\alpha}(\mathbb{R}^n)}+\|g\|_{C^\alpha(\overline B_{2R}(x_0))}\right)
							%\end{equation}
							%for any $\alpha\in(0,1)$ and	for some suitable constant $ C>0 $ depending only on $ n,s,\rho,\alpha$.	
							%	
							%\end{Lemma}
							%
							%
							%\begin{proof}
							%	From  the definition of the function $ v $, we know that
							%	\begin{equation}\label{eq: function space of v2}
								%		v\in C^{2,\alpha}_0( B_{2R}(x_0))\subset C^{2,\alpha}_0(\mathbb{R}^n). 
								%	\end{equation}
							%	Furthermore, to check~\eqref{eq: new equation2}, one can calculate that  
							%	\begin{equation}\label{eq: new equation:PSI2}
								%		\begin{split}
									%			& \psi= \phi^R g+u(-\Delta)\phi^R+u(-\Delta)^s\phi^R+ 2 \nabla u\nabla \phi^R \\&\qquad\qquad+ \int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\phi^R(x)-\phi^R(x+z))}{|z|^{n+2s}}\, dz.\end{split}\end{equation} 
							%	Indeed, 
							%	\[  -\Delta v=\phi^R(-\Delta)u+ u(-\Delta)\phi^R+2 \nabla u\nabla \phi^R,  \]
							%	and 
							%	\[  (-\Delta)^s v=\phi^R(-\Delta)^su+ u(-\Delta)^s\phi^R+\int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\phi^R(x)-\phi^R(x+z))}{|z|^{n+2s}}\, dz . \]
							%	Recalling \eqref{u is a solution in V}, one infers that 
							%	\begin{equation}
								%		\begin{split}
									%			-\Delta v+(-\Delta)^s v&=\phi^R g(x,u)+u(-\Delta)\phi^R+u(-\Delta)^s\phi^R+ 2 \nabla u\nabla \phi^R\\
									%			&\qquad + \int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\phi^R(x)-\phi^R(x+z))}{|z|^{n+2s}}\, dz,
									%		\end{split}
								%	\end{equation}
							%	which is~\eqref{eq: new equation:PSI2}, as desired.
							%	
							%	\noindent\textbf{Step 1. $ C^\alpha $- estimates on the function $ \phi g $.}
							%	Since $ g\in C^\alpha(\overline B_{2R}(x_0)) $ and $ u\in C^{2,\alpha}(\overline B_{2R}(x_0))  $, we have 
							%	\[ \|\phi^R g\|_{L^\infty(B_{2R}(x_0))}\leqslant\|g\|_{L^\infty(B_{2R}(x_0))}  \]
							%	and 
							%	\begin{equation}
								%		\begin{split}
									%			[\phi^R g]_{\alpha; B_{2R}(x_0)}&\leqslant [\phi^R]_{\alpha; B_{2R}(x_0)}\|g\|_{L^\infty(B_{2R}(x_0))}+[g]_{\alpha; B_{2R}(x_0)}\|\phi^R\|_{L^\infty(B_{2R}(x_0))}\\
									%			&\leqslant C_1R^{-\alpha}[\phi]_{\alpha; B_{2}(0)}\|g\|_{C^\alpha(\overline B_{2R}(x_0))}.
									%		\end{split}
								%	\end{equation}
							%	for some suitable constant $ C_1>0. $	
							%	Thus, 
							%	\begin{equation}
								%	R^2|\phi^R g|^\prime_{0,\alpha;B_{2R}(x_0)}=	R^2\|\phi^R g\|_{L^\infty(B_{2R}(x_0))}+R^{2+\alpha}[\phi^R g]_{\alpha; B_{2R}(x_0)}\leqslant C_1R^2\|g\|_{C^\alpha(\overline B_{2R}(x_0))}. 
								%	\end{equation}
							%	
							%	
							%	
							%	\noindent\textbf{Step~2. $ C^\alpha $- estimates on the function $ u(-\Delta)\phi^R+u(-\Delta)^s\phi^R+ 2 \nabla u\nabla \phi^R $.} For every $ x\in B_{2R}(x_0)$, we have that
							%	\begin{equation}
								%		\begin{split}
									%			&\left|u(-\Delta)\phi^R+u(-\Delta)^s\phi^R+ 2 \nabla u\nabla \phi^R\right|\\
									%			\leqslant& C_2\left((R^{-2}+R^{-2s})\|u\|_{L^\infty(B_{2R}(x_0))}+R^{-1}\|Du\|_{L^\infty( B_{2R}(x_0))}\right),
									%		\end{split}
								%	\end{equation}
							%	and
							%	\begin{equation}
								%		\begin{split}
									%			&[u(-\Delta)\phi^R+u(-\Delta)^s\phi^R+ 2 \nabla u\nabla \phi^R ]_{\alpha; B_{2R}(x_0)}\\
									%			\leqslant & [u]_{\alpha; B_{2R}(x_0)}\|(-\Delta)\phi^R+(-\Delta)^s\phi^R\|_{L^\infty( B_{2R}(x_0))} +[(-\Delta)\phi^R+(-\Delta)^s\phi^R]_{\alpha; B_{2R}(x_0)}\|u\|_{L^\infty( B_{2R}(x_0))}\\
									%			&\quad +2\left([\nabla u]_{\alpha; B_{2R}(x_0)}\|\nabla \phi^R\|_{L^\infty( B_{2R}(x_0))}+[\nabla \phi^R]_{\alpha; B_{2R}(x_0)}\|\nabla u\|_{L^\infty( B_{2R}(x_0))}\right)\\
									%			\leqslant &C_3\bigg((R^{-2}+R^{-2s})R^{1-\alpha}\|Du\|_{L^{\infty}( B_{2R}(x_0))}+R^{-1}[ Du]_{\alpha; B_{2R}(x_0)}\\
									%			&\quad +R^{-1-\alpha}\|Du\|_{L^{\infty}( B_{2R}(x_0))}+(R^{-2-\alpha}+R^{-2s-\alpha})\|u\|_{L^\infty( B_{2R}(x_0))}\bigg),
									%		\end{split}
								%	\end{equation}
							%	for some suitable constant $ C_2,C_3>0 $.
							%	
							%	Consequently, for every $0<R<\frac{\rho}{10}<$diam$(\Omega)$, we have
							%	\begin{equation}
								%		\begin{split}
									%			&	R^2|u(-\Delta)\phi^R+u(-\Delta)^s\phi^R+ 2 \nabla u\nabla \phi^R|^\prime_{0,\alpha;B_{2R}(x_0)}\\
									%		\leqslant 	& 	R^2\|u(-\Delta)\phi^R+u(-\Delta)^s\phi^R+ 2 \nabla u\nabla \phi^R\|_{L^\infty(B_{2R}(x_0))}\\
									%		&\qquad +R^{2+\alpha}[u(-\Delta)\phi^R+u(-\Delta)^s\phi^R+ 2 \nabla u\nabla \phi^R]_{\alpha; B_{2R}(x_0)}\\
									%		\leqslant  & C_2\left(\|u\|_{L^\infty(B_{2R}(x_0))}+R\|Du\|_{L^\infty( B_{2R}(x_0))}+R^{1+\alpha}[ Du]_{\alpha; B_{2R}(x_0)}\right)
									%		\end{split}
								%	\end{equation}
							%
							%
							%	\noindent\textbf{Step~3. $ C^\alpha $- estimates on the function $ \xi(x):= \int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\phi^R(x)-\phi^R(x+z))}{|z|^{n+2s}}\, dz $.}
							%	\textbf{1. $ L^\infty $-norm of $h $.}
							%	 For every $ x\in   B_{2R}(x_0), $ if $|z|<R$ , then one has $x\pm z\in B_{3R}(x_0) $. Thus,  one has that
							%	\begin{equation}
								%		\begin{split}
									%			|\xi(x)|&=\left|\int_{\mathbb{R}^n}\frac{(u(x)-u(x+z))(\phi^R(x)-\phi^R(x+z))}{|z|^{n+2s}}\, dz\right|\\
									%			&\leqslant \left|\int_{|z|<R}\frac{(u(x)-u(x+z))(\phi^R(x)-\phi^R(x+z))}{|z|^{n+2s}}\, dz\right|\\
									%			&\quad +\left|\int_{|z|\geqslant R}\frac{(u(x)-u(x+z))(\phi^R(x)-\phi^R(x+z))}{|z|^{n+2s}}\, dz\right|\\
									%			&\leqslant C_4\left(R^{1-2s}\|Du\|_{L^\infty(B_{3R}(x_0))}+R^{-2s}\|u\|_{L^\infty(\mathbb{R}^n)}\right).
									%		\end{split}
								%	\end{equation}
							%	
							%	\textbf{2.  $ C^\alpha $-norm of $ h$.} For this, given $ x_1,x_2\in B_{2R}(x_0)$ with $ x_1\neq x_2 $, let $ r:=|x_1-x_2| $.
							%	We distinguish two cases.
							%	
							%	\noindent{Case 1.} If $ r\geqslant R $, then 
							%	\begin{equation}\label{case 1 of h 2}
								%		\frac{|\xi(x_1)-\xi(x_2)|}{|x_1-x_2|^\alpha}\leqslant 	\|\xi\|_{L^\infty(\Omega)}R^{-\alpha}\leqslant  C_4\left(R^{1-2s-\alpha}\|Du\|_{L^\infty(B_{3R}(x_0))}+R^{-2s-\alpha}\|u\|_{L^\infty(\mathbb{R}^n)}\right).
								%	\end{equation}
							%	
							%	\noindent{Case 2.}  If $ r< R $, then 
							%	\begin{equation}
								%		\begin{split}
									%			\frac{|\xi(x_1)-\xi(x_2)|}{|x_1-x_2|^\alpha}&\leqslant \left|\int_{\mathbb{R}^n}\frac{(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))(\phi^R(x_2)-\phi^R(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
									%			& +\left|\int_{\mathbb{R}^n}\frac{(u(x_1)-u(x_1+z))(\phi^R(x_1)-\phi^R(x_1+z)-\phi^R(x_2)+\phi^R(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
									%			&:= I_1+I_2.
									%		\end{split}
								%	\end{equation}
							%	Furthermore, 	for $ I_1 $,  one sees that 
							%\begin{equation}
							%		\begin{split}\label{I_1 1 1}
								%			&	\left|\int_{|z|<R}\frac{(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))(\phi^R(x_2)-\phi^R(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
								%			&\qquad\leqslant R^{-1}\int_{|z|<R}\frac{|(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))|\|\nabla \phi\|_{L^\infty(\mathbb{R}^n)}}{|z|^{n+2s-1}r^\alpha}\, dz\\
								%			&\qquad\leqslant  R^{-1}\|\nabla \phi\|_{L^\infty(\mathbb{R}^n)}\int_{|z|<R}\frac{2[Du]_{\alpha; B_{3R}(x_0)}r^\alpha|z|^{1}}{|z|^{n+2s-1}r^\alpha}\, dz\\&\qquad\leqslant C_5 R^{1-2s} [Du]_{\alpha; B_{3R}(x_0)}, 
								%		\end{split}
							%	\end{equation}
						%	and
						%	\begin{equation}
							%		\begin{split}\label{I_1 2 1}
								%			&	\left|\int_{|z|\geqslant R}\frac{(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))(\phi^R(x_2)-\phi^R(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
								%	&\qquad		\leqslant 2\| \phi\|_{L^\infty(\mathbb{R}^n)}\int_{|z|\geqslant R}\frac{|(u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z))| }{|z|^{n+2s}r^\alpha}\, dz\\
								%			&\qquad\leqslant 
								%			2\| \phi\|_{L^\infty(\mathbb{R}^n)}\int_{|z|\geqslant R}\frac{2[u]_{C^{\alpha}(\mathbb{R}^n)}r^\alpha}{|z|^{n+2s}r^\alpha}\, dz\\&\qquad\leqslant C_6 [u]_{C^{\alpha}(\mathbb{R}^n)}R^{-2s}.
								%		\end{split}
							%	\end{equation}
						%	Combining \eqref{I_1 1 1} with \eqref{I_1 2 1}, we deduce that 
						%	\begin{equation}\label{I_12}
							%		I_1\leqslant  C_5 R^{1-2s} [Du]_{\alpha; B_{3R}(x_0)}+C_6R^{-2s} [u]_{C^{\alpha}(\mathbb{R}^n)},
							%	\end{equation} 
						%	for some suitable constants $ C_5,C_6>0. $
						%	
						%	{	For $ I_2, $ using similar arguments to the estimate of $ I_1 $,  one has
							%		\begin{equation}
								%			\begin{split}\label{I_21}
									%				&\left|\int_{|z|<R}\frac{(u(x_1)-u(x_1+z))(\phi^R(x_1)-\phi^R(x_1+z)-\phi^R(x_2)+\phi^R(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\
									%&\qquad				\leqslant\int_{|z|<R}\frac{|(\phi^R(x_1)-\phi^R(x_1+z)-\phi^R(x_2)+\phi^R(x_2+z))|\|\nabla u\|_{L^\infty(B_{3R}(x_0))}}{|z|^{n+2s-1}r^\alpha}\, dz\\
									%			&\qquad	\leqslant  \|\nabla u\|_{L^\infty(B_{3R}(x_0))}\int_{|z|<R}\frac{2R^{-1-\alpha}[\phi]_{C^{1,\alpha}(\mathbb{R}^n)}r^\alpha|z|^{1}}{|z|^{n+2s-1}r^\alpha}\, dz\\&\qquad\leqslant C_7 R^{1-\alpha-2s} \|Du\|_{L^{\infty}( B_{3R}(x_0))},
									%			\end{split}
								%		\end{equation}
							%		and 
							%		\begin{equation}
								%			\begin{split}\label{I_22 }
									%				&\left|\int_{|z|\geqslant R}\frac{(u(x_1)-u(x_1+z))(\phi^R(x_1)-\phi^R(x_1+z)-\phi^R(x_2)+\phi^R(x_2+z))}{|z|^{n+2s}r^\alpha}\, dz\right|\\ &\qquad
									%				\leqslant 2\| u\|_{L^\infty(\mathbb{R}^n)}\int_{|z|\geqslant R}\frac{|\phi^R(x_1)-\phi^R(x_1+z)-\phi^R(x_2)+\phi^R(x_2+z)| }{|z|^{n+2s}r^\alpha}\, dz\\ &\qquad
									%				\leqslant 2\|u\|_{L^\infty(\mathbb{R}^n)}\int_{|z|\geqslant R}\frac{2R^{-\alpha}[\phi]_{C^{\alpha}(\mathbb{R}^n)}r^\alpha}{|z|^{n+2s}r^\alpha}\, dz\\&\qquad\leqslant C_8R^{-\alpha-2s}\|u\|_{L^\infty(\mathbb{R}^n)}.
									%			\end{split}
								%		\end{equation}
							%		From \eqref{I_21} and  \eqref{I_22 } one concludes that 
							%		\begin{equation}\label{I_23}
								%			I_2\leqslant  C_7 R^{1-\alpha-2s} \|Du\|_{L^{\infty}( B_{3R}(x_0))}+C_8R^{-\alpha-2s}\|u\|_{L^\infty(\mathbb{R}^n)},
								%		\end{equation} 
							%		for some suitable constants $ C_7,C_8>0. $}
						%	
						%	Combining \eqref{I_12} and \eqref{I_23}, and recalling \eqref{case 1 of h 2}, we conclude that 
						%	\begin{equation}
							%		\begin{split}
								%			[\xi ]_{\alpha; B_{2R}(x_0)}	&\leqslant C_9\Big(R^{1-2s-\alpha}\|Du\|_{L^\infty(B_{3R}(x_0))}+R^{-2s-\alpha}\|u\|_{L^\infty(\mathbb{R}^n)}\\
								%			&\quad + R^{1-2s} [Du]_{\alpha; B_{3R}(x_0)}+R^{-2s} [u]_{C^{\alpha}(\mathbb{R}^n)}\Big).
								%		\end{split}
							%	\end{equation}	
						%	As a consequence,  for any $R\in\left(0,\frac{\rho}{10}\right)$, we find that 
						%	\begin{equation}
							%		\begin{split}
								%				R^2|\xi|^\prime_{0,\alpha;B_{2R}(x_0)}
								%			\leqslant 	& 	R^2\|\xi\|_{L^\infty(B_{2R}(x_0))} +R^{2+\alpha}[\xi]_{\alpha; B_{2R}(x_0)}\\
								%			\leqslant  & C_{10}\left(R\|Du\|_{L^\infty(B_{3R}(x_0))}+R^{1+\alpha} [Du]_{\alpha; B_{3R}(x_0)}+\|u\|_{L^\infty(\mathbb{R}^n)}+[u]_{C^{\alpha}(\mathbb{R}^n)}\right).
								%		\end{split}
							%	\end{equation}
						%
						%	In light of \textbf{Step 1}--\textbf{Step~3}, we obtain that $ \psi\in C^\alpha (B_{2R}(x_0))  $ and 
						%	\begin{equation}
							%		\begin{split}
								%				R^2|\psi|^\prime_{0,\alpha;B_{2R}(x_0)}&\leqslant C \bigg(R\|Du\|_{L^\infty(B_{3R}(x_0))}+R^{1+\alpha} [Du]_{\alpha; B_{3R}(x_0)}+\|u\|_{L^\infty(B_{3R}(x_0))}\\
								%				&\qquad +\|u\|_{L^\infty(\mathbb{R}^n)}+[u]_{C^{\alpha}(\mathbb{R}^n)}+\|g\|_{C^\alpha(\overline B_{2R}(x_0))}\bigg)\\
								%				&\leqslant C\left(|u|^\prime_{1,\alpha;B_{3R}(x_0)}+[u]_{C^{\alpha}(\mathbb{R}^n)}+\|g\|_{C^\alpha(\overline B_{2R}(x_0))}+\|u\|_{L^\infty(\mathbb{R}^n)}\right)
								%		\end{split}
							%	\end{equation}
						%	for any $\alpha\in(0,1)$ and	for some suitable constant $ C>0 $ depending only on $ n,s,\rho,\alpha$.	
						%\end{proof}	
						
						According to \eqref{u epsilon}, we know that 		for every $0<\epsilon<R$ small enough,  $u_\epsilon\in C^{2,\alpha}(\overline V_{\frac{3\rho}{4}})$ solves the following equation 
						\begin{equation}
							-\Delta u_\epsilon+(-\Delta)^s u_\epsilon= g_\epsilon \qquad \text{  in } V_{\frac{3\rho}{4}},
						\end{equation}
						where $g_\epsilon$ satisfies the estimate \eqref{g_epsilon *}.  
						
						Consider the cut off function $\phi^R$ given by \eqref{cutoff function 21}, let $v_\epsilon:=\phi^R u_\epsilon$, then 	applying Lemma~\refeq{lemma v epsilon}, one has there exists~$ \psi_\epsilon\in C^{\alpha}(B_{R}(x_0),\mathbb{R}) $ such that $ v_\epsilon $ satisfies  
						\begin{equation}\label{eq: v epsilon equation}
							-\Delta v_\epsilon+(-\Delta)^s v_\epsilon= \psi_\epsilon \qquad \text{  in } V_{\frac{3\rho}{4}},
						\end{equation}
						%where 
						%	\begin{equation}
							%	\psi_\epsilon= g_\epsilon(x,u)+\Delta (u_\epsilon(1-\phi^R))-(-\Delta)^s (u_\epsilon(1-\phi^R))\quad  \text{  in } V_{\frac{3\rho}{4}},
							%\end{equation} 
							%and 
							where
							\begin{equation}\label{eq: new equation:PSI epsilon}
								\psi_\epsilon= 	g_\epsilon(x,u)-(-\Delta)^s (u_\epsilon(1-\phi^R))\quad  \text{  in } B_{3R/2}(x_0).
							\end{equation} 
							In particular, employing \eqref{estimate psi 2}, one finds that
							\begin{equation}
								\begin{split}\label{eq:psi epsilon}
									R^2|\psi_\epsilon|^\prime_{0,\alpha;B_{R}(x_0)}&\leqslant  C_{\rho,n,s} \left(R^2| g_\epsilon(\cdot,u(\cdot))|^\prime_{0,\alpha;B_{R}(x_0)}+\|u_\epsilon\|_{L^\infty(\mathbb{R}^n)}\right),
								\end{split}
							\end{equation}
							where  $ C_{\rho,n,s} $ is a positive constant.
							
							\smallskip
							{Observing that 
								for $v_\epsilon\in C^\infty_0( B_{2R}(x_0))$, for all~$\delta>0$,
								there exists $ C_{\delta}>0  $ such that  
								\begin{equation}
									\begin{split}\label{bb}
										R^2|(-\Delta)^s v_\epsilon|^\prime_{0,\alpha;B_{R}(x_0)}&= R^2\|(-\Delta)^s v_\epsilon\|_{L^\infty(B_{R}(x_0))} +R^{2+\alpha}[(-\Delta)^s v_\epsilon]_{\alpha; B_{R}(x_0)}\\
										&\leqslant C_{n,s,\rho}\bigg(R^2\|D^2 v_\epsilon\|_{L^\infty(B_{2R}(x_0))}+\|v_\epsilon\|_{L^\infty(B_{2R}(x_0))}\\
										&\qquad+R^{2+\alpha_0}[D^2 v_\epsilon]_{\alpha_0;B_{2R}(x_0)}+R\|D v_\epsilon\|_{L^\infty(B_{2R}(x_0))}
										\bigg)\\
										&= C_{n,s,\rho}\bigg(R^2\|u_\epsilon D^2\phi^R +2Du_\epsilon D\phi^R+\phi^RD^2 u_\epsilon\|_{L^\infty(B_{2R}(x_0))}\\
										&\qquad+\|\phi^Ru_\epsilon\|_{L^\infty(B_{2R}(x_0))}+R\|\phi^R D u_\epsilon+u_\epsilon D\phi^R\|_{L^\infty(B_{2R}(x_0))}
										\\
										&\qquad +R^{2+\alpha_0}[u_\epsilon D^2\phi^R +2Du_\epsilon D\phi^R+\phi^RD^2 u_\epsilon]_{\alpha_0;B_{2R}(x_0)}	\bigg)\\
										&\leqslant C_{n,s,\rho}
										|u_\epsilon|^\prime_{2,\alpha_0;B_{2R}(x_0)}
										\leqslant\delta |u_\epsilon|^\prime_{2,\alpha;B_{2R}(x_0)} +C_\delta \|u_\epsilon\|_{L^\infty(B_{2R}(x_0))},
									\end{split}
								\end{equation}
								where 
								\begin{equation}
									\alpha_0=\begin{cases}
										0\qquad & \alpha<2-2s\\
										\alpha-(1-s) &\alpha \geqslant 2-2s.
									\end{cases}
								\end{equation}
								Thus, combining \cite[Theorem 4.6]{GTbook} 	} and \eqref{bb}, one has for all~$\delta>0$
							there exists $ C_{\delta}>0  $ such that  
							\begin{equation}\label{C 2 of v 2}
								\begin{split}
									|v_\epsilon|^\prime_{2,\alpha;B_{R/2}(x_0)}&\leqslant C_{n,\alpha}\left(\|v_\epsilon\|_{L^\infty(B_{R}(x_0))}+R^2\left(|\psi_\epsilon|^\prime_{0,\alpha;B_{R}(x_0)}+|(-\Delta)^s v_\epsilon|^\prime_{0,\alpha;B_{R}(x_0)}\right)\right)\\
									&\leqslant C_{n,\alpha,\rho,\alpha}\bigg(\|v_\epsilon\|_{L^\infty(B_{R}(x_0))}+R^2|\psi_\epsilon|^\prime_{0,\alpha;B_{R}(x_0)}\\
									&\qquad\qquad\qquad +\delta |u_\epsilon|^\prime_{2,\alpha;B_{2R}(x_0)}+C_\delta \|u_\epsilon\|_{L^\infty(B_{2R}(x_0))}\bigg). 
								\end{split}
							\end{equation}
							
							%{F}rom the definition of $v_\epsilon$, one has
							%\begin{equation}
							%	\|v_\epsilon\|_{L^\infty(B_{R}(x_0))}=\|\phi^R u_\epsilon\|_{L^\infty(B_{R}(x_0))}= \| u_\epsilon\|_{L^\infty(B_{R}(x_0))}\leqslant \| u\|_{L^\infty(B_{2R}(x_0))}.
							%\end{equation}
							%As a consequence of this and \eqref{C 2 of v 2}, for all~$R\in\left(0,\frac{\rho}{10}\right)$ and~$\epsilon\in\left(0,R\right)$, we have 
							%\begin{equation*}
							%	|v_\epsilon|^\prime_{2,\alpha;B_{R/2}(x_0)}\leqslant C_{n,s, \alpha,\rho}\left(R^2|\psi_\epsilon|^\prime_{0,\alpha;B_{R}(x_0)}+\delta |u_\epsilon|^\prime_{2,\alpha;B_{2R}(x_0)}+C_\delta \|u\|_{L^\infty(B_{2R}(x_0))}\right).
							%\end{equation*}
							Therefore, for every $x_0\in V_{\frac{\rho}{4}}$, recalling the definition of $v_\epsilon$ and Proposition~\ref{pro u epsilon1}, and applying  Proposition~\ref{pro g epsilon}, we know that $ \forall \delta>0$, there exists $C_\delta$ such that 
							%{\begin{equation}\label{final equation}
									%	\begin{split}
										%		 |u_\epsilon|^\prime_{2,\alpha;B_{R/2}(x_0)}&=|v_\epsilon|^\prime_{2,\alpha;B_{R/2}(x_0)}\\&\leqslant C_{n,s, \alpha,\rho}\left(R^2|\psi_\epsilon|^\prime_{0,\alpha;B_{R}(x_0)}+\delta |u_\epsilon|^\prime_{2,\alpha;B_{2R}(x_0)}+C_\delta \|u\|_{L^\infty(B_{2R}(x_0))}\right)\\
										%		&\leqslant C\bigg(\|g\|_{C^{0,1}(\overline B_{2R}(x_0) \times\mathbb{R}^n)}+|\kappa| |u_\epsilon|^\prime_{0,\alpha;B_{R}(x_0)}+\|u\|_{L^\infty(\mathbb{R}^n)}\\
										%		&\qquad +\delta |u_\epsilon|^\prime_{2,\alpha;B_{2R}(x_0)}+C_\delta \|u\|_{L^\infty(B_{2R}(x_0))}\bigg)\\
										%		&\leqslant C\bigg(\|g\|_{C^{0,1}(\overline V_{\frac{7\rho}{8}} \times\mathbb{R}^n)}+C_\delta \|u\|_{L^\infty(\mathbb{R}^n)}+\delta |u_\epsilon|^\prime_{2,\alpha;B_{2R}(x_0)}\bigg),
										%	\end{split}
									%\end{equation}
									%and 
									\begin{equation}\label{final equation 2}
										\begin{split}
											&|u_\epsilon|^\prime_{2,\alpha;B_{R/2}(x_0)}=|v_\epsilon|^\prime_{2,\alpha;B_{R/2}(x_0)}\\  \leqslant& \ C_{n,s, \alpha,\rho}\left(R^2|\psi_\epsilon|^\prime_{0,\alpha;B_{R}(x_0)}+\delta |u_\epsilon|^\prime_{2,\alpha;B_{2R}(x_0)}+C_\delta \|u\|_{L^\infty(B_{2R}(x_0))}\right)\\
											\leqslant &\ C\bigg(\|g\|_{C^{\alpha}(\overline B_{2R}(x_0) \times\mathbb{R}^n)}\left(1+\|Du\|_{L^\infty(B_{2R}(x_0))}\right)+\|u\|_{L^\infty(\mathbb{R}^n)}\\
											&\qquad +\delta |u_\epsilon|^\prime_{2,\alpha;B_{2R}(x_0)}+C_\delta \|u\|_{L^\infty(B_{2R}(x_0))}\bigg)\\
											\leqslant &\ C\bigg(\|g\|_{C^{\alpha}(\overline V_{\frac{3\rho}{4}} \times\mathbb{R}^n)}\left(1+\|Du\|_{L^\infty( V_{\frac{3\rho}{4}})}\right)+C_\delta \|u\|_{L^\infty(\mathbb{R}^n)}+\delta |u_\epsilon|^\prime_{2,\alpha;B_{2R}(x_0)}\bigg),
										\end{split}
									\end{equation} for every $R\in\left(0,\frac{\rho}{10}\right)$ and $\epsilon\in (0,R)$,
									where the constant $ C>0 $ depends only on $ n,s,\alpha,\rho $.
								}
								
								\subsubsection{Interior $C^{2,\alpha}$-regularity}\label{sec:Proof of Theorem1.5}
								
								
								In light of the estimate \eqref{final equation 2}, we can obtain the following Proposition~\ref{pro final estimate}  by adapting Proposition~\ref{pro final estimate 1}, which can guarantee the $C^{2,\alpha}$-norm of $u_\epsilon$ is uniformly bounded in $\epsilon$ in some ball.
								
								
								
								\begin{Proposition}\label{pro final estimate}
									
									Let $y\in \mathbb{R}^n, d>0$, and $u\in C^{2,\alpha}(B_{d}(y))$. Suppose that  for any $\delta > 0 $, there
									exists $\Lambda_\delta> 0$ such that, for any $x\in B_{d}(y)$ and any $r\in(0, d-|x-y|]$, we have
									\begin{equation}\label{eq:assumption u}
										|u|^\prime_{2,\alpha;B_{r/8}(x)}\leqslant \Lambda_\delta+ \delta |u|^\prime_{2,\alpha;B_{r/2}(x)}.
									\end{equation}
									
									Then, there exist constants $\delta_0, C > 0$, depending only on $n,\alpha,d$, such that
									\begin{equation}
										\|u\|_{C^{2,\alpha}(B_{d/8}(y))}\leqslant C\Lambda_{\delta_0}.
									\end{equation}
								\end{Proposition}
								%\begin{proof}
								%	We notice that, since $r<d$, one has
								%	\[ |u|^\prime_{2,\alpha;B_{r/8}(x)}\leqslant  |u|^\prime_{2,\alpha;B_{d}(y)},\]
								%	which implies that 
								%	\[ M:=\sup\limits_{x\in B_{d}(y), r\in(0, d-|x-y|]} |u|^\prime_{2,\alpha;B_{r/8}(x)}<+\infty.\]
								%	
								%	We now use a covering argument: pick $\lambda\in(0,\frac{1}{2}]$, to be chosen later, and	fixed any $x\in B_d(y)$ and $ r\in(0, d-|x-y|]$, we cover $B_{r/8}(x)$ with finitely many balls $\left\{B_{\lambda r/8}(x_k)\right\}_{k=1}^N$ with $x_k\in B_{r/8}(x) $,  for some $N $ depending only on $\lambda,n$.
								%	
								%	Furthermore, we observe that $B_{r/2}(x_k)\subset B_{d}(y)$. Indeed,
								%	\[ |x_k-y|+\frac{r}{2}\leqslant |x_k-x|+|x-y|+\frac{r}{2}<r+|x-y|<d. \]
								%Consequently, since $\lambda\leqslant\frac{1}{2}$, we use  \eqref{eq:assumption u} (with $x = x_k $ and $ r$ scaled to $\lambda r$) to get 
								%\begin{equation}\label{eq lambda}
								%	|u|^\prime_{2,\alpha;B_{\lambda r/8}(x_k)}\leqslant \Lambda_\delta+ \delta |u|^\prime_{2,\alpha;B_{\lambda r/2}(x_k)}.
								%\end{equation}
								%	Then, using \cite[Lemma 7]{MR3331523} with $\rho:=r/8$ and $\lambda=1/8$, recalling  the definition of $M$, and employing \eqref{eq lambda}, one has 
								%	\begin{equation}
									%		\begin{split}
										%				|u|^\prime_{2,\alpha;B_{r/8}(x)}&\leqslant C\sum\limits_{k=1}\limits^{N}|u|^\prime_{2,\alpha;B_{\lambda r/8}(x_k)}\\
										%				&\leqslant C_\lambda N \Lambda_{\delta}+ C_\lambda \delta \sum\limits_{k=1}\limits^{N}|u|^\prime_{2,\alpha;B_{\lambda r/2}(x_k)}\\
										%				&\leqslant C N \Lambda_{\delta}+ C\delta \sum\limits_{k=1}\limits^{N}|u|^\prime_{2,\alpha;B_{ r/16}(x_k)}\\
										%				&\leqslant CN \Lambda_{\delta}+ C N\delta M.
										%		\end{split}
									%	\end{equation}
								%	It follows from the definition of $M$ that 
								%	\[ M\leqslant  C N \Lambda_{\delta}+ C N\delta M,  \]
								%	by choosing $\delta_0=1/(CN)$, one has 
								%	\[ M\leqslant  2C N \Lambda_{\delta_0}.\]
								%	Thus we have proved that
								%	\[ |u|^\prime_{2,\alpha;B_{r/8}(x)}\leqslant  2C N \Lambda_{\delta_0}.\qquad \forall x\in B_d(y),  r\in(0, d-|x-y|]. \]
								%	Setting $x=y, r=d$, we have 
								%	\[ |u|^\prime_{2,\alpha;B_{d/8}(y)}\leqslant  2 N \Lambda_{\delta_0}, \]
								%	which implies the desired result that 
								%	\[ \|u\|_{C^{2,\alpha}(\overline B_{d/8}(y))}\leqslant  2\max\left\{d^{-2-\alpha},1\right\}C N \Lambda_{\delta_0}.\qedhere\]
								%\end{proof}
								
								
								\begin{proof}[Proof of Theorem~\ref{th C^2,alpha interior without boundary condition}]
									First of all, one can choose $d=\rho/5 $ in Proposition~\ref{pro final estimate}, so that 
									\begin{equation}\label{d}
										B_{d}(y)\subset V_{\rho/4} \quad \text{and}\quad  d/4<\frac{\rho}{10}, \quad \forall y\in V.
									\end{equation}
									As a consequence, applying  Proposition~\ref{pro final estimate}, and utilizing   \eqref{final equation 2},
									we know that  
									%\begin{equation}
									%	\|u_\epsilon\|_{C^{2,\alpha}(\overline B_{\rho/40}(y))}\leqslant C\left(\|g\|_{C^{0,1}(\overline V_{\frac{7\rho}{8}} \times\mathbb{R}^n)}+C_{\delta_0} \|u\|_{L^\infty(\mathbb{R}^n)}\right),
									%\end{equation}
									%and 
									\begin{equation}
										\|u_\epsilon\|_{C^{2,\alpha}(\overline B_{\rho/40}(y))}\leqslant C\left(\|g\|_{C^{\alpha}(\overline V_{\frac{3\rho}{4}} \times\mathbb{R}^n)}\left(1+\|Du\|_{L^\infty(V_{\frac{3\rho}{4}})}\right)+C_{\delta_0} \|u\|_{L^\infty(\mathbb{R}^n)}\right)
									\end{equation}
									for every $y\in V$, where the constant $ C>0 $ depends on $ n,s,\alpha,\rho $.
									
									
									
									Applying the Arzel\`{a}-Ascoli Theorem, we obtain that  $ u\in C^{2,\alpha}(\overline B_{\rho/40}(y)). $ In particular,
									%	since $u\equiv 0$ in $\mathbb{R}^n\backslash\Omega$, 
									% for the case (a) in Theorem~\ref{th C^2,alpha interior without boundary condition}, one has
									%\begin{equation}\label{final inequality}
									%	\|u\|_{C^{2,\alpha}(\overline B_{\rho/40}(y))}\leqslant C\left(\|g\|_{C^{0,1}(\overline V_{\frac{7\rho}{8}} \times\mathbb{R}^n)}+ \|u\|_{L^\infty(\mathbb{R}^n)}\right),
									%\end{equation}
									%and for the case (b) in Theorem~\ref{th C^2,alpha interior without boundary condition},
									one has 
									\begin{equation}\label{final inequality 2}
										\|u\|_{C^{2,\alpha}(\overline B_{\rho/40}(y))}\leqslant C\left(\|g\|_{C^{\alpha}(\overline V_{\frac{3\rho}{4}} \times\mathbb{R}^n)}\left(1+\|Du\|_{L^\infty(V_{\frac{3\rho}{4}})}\right)+ \|u\|_{L^\infty(\mathbb{R}^n)}\right),
									\end{equation}
									for every $y\in V$, and for	some constant $ C>0 $ depending on $ n,s,\alpha,\rho $. 
									
									
									Using the cover argument again, 
									%there exist  finitely many balls $\left\{B_{\rho/40}(y_k)\right\}_{k=1}^N$ with $y_k\in V $,  for some $N$ depending on $n,\rho,V$, such that 
									%\[ V\subset \bigcup\limits_{k=1}\limits^NB_{\rho/40}(y_k), \]
									and employing Theorem~\ref{th C^1,alpha interior without boundary condition}, we can obtain the desired result that 
									% in the case (a)
									%\begin{equation}\label{final inequality 1}
									%	\|u\|_{C^{2,\alpha}(\overline V)}\leqslant C\left(\|g\|_{C^{0,1}(\overline V_{\frac{7\rho}{8}} \times\mathbb{R}^n)}+ \|u\|_{L^\infty(\mathbb{R}^n)}\right),
									%\end{equation}
									%and in the case (b)
									\begin{equation}\label{final inequality2}
										\begin{split}
											\|u\|_{C^{2,\alpha}(\overline V)}&\leqslant C\left(\|g\|_{C^{\alpha}(\overline V_{\frac{3\rho}{4}} \times\mathbb{R}^n)}\left(1+\|Du\|_{L^\infty(V_{\frac{3\rho}{4}}))}\right)+ \|u\|_{L^\infty(\mathbb{R}^n)}\right)\\
											&\leqslant C\left(\|g\|_{C^{\alpha}(\overline V_{\frac{3\rho}{4}} \times\mathbb{R}^n)}\left(1+	\|g\|_{L^\infty( V_{\frac{7\rho}{8}} \times\mathbb{R}^n)}+ \|u\|_{L^\infty(\mathbb{R}^n)}\right)+ \|u\|_{L^\infty(\mathbb{R}^n)}\right)\\
											&\leqslant C\left(\|u\|_{L^\infty(\mathbb{R}^n)}+\|g\|_{C^{\alpha}(\overline V_{\frac{7\rho}{8}} \times\mathbb{R}^n)}\right) \left(1+\|g\|_{C^{\alpha}(\overline V_{\frac{7\rho}{8}} \times\mathbb{R}^n)}\right)
										\end{split}
									\end{equation}
									where the constant $ C>0 $ depends on $ n,s,\alpha,\rho $.
								\end{proof}
								
								











\subsection{$ C^{2,\alpha} $-regularity up to the boundary}\label{sec: C^2alpha-global regularity}\setcounter{equation}{0}
The main ideas used here come from {\rm \cite[Theorem B.1]{FABEJK102}.
 We split the proof of Theorem~\ref{theorem C^2 alpha global} into the following two subsections:
\begin{itemize}
	\item[Section \ref{subsubsection 4.21}:]Estimate $ \|(-\Delta)^{s} u\|_{C^\alpha(\overline{\Omega})} $.
	\item[Section \ref{subsubsection 4.22}:] Apply the method of continuity to complete the proof of Theorem~\ref{theorem C^2 alpha global}.
\end{itemize}



	
	
\subsubsection{H\"{o}lder estimate of the fractional Laplacian term}\label{subsubsection 4.21}


Let $ \alpha\in(0,1) $, we consider the function space ${\mathcal{B}}_\alpha(\Omega)  $ defined as follows:
\begin{equation}\label{definition of B}
 {\mathcal{B}}_\alpha(\Omega):=\left\{u\in C(\mathbb{R}^n):u\equiv 0 \text{ in } \mathbb{R}^n\backslash \Omega \text{  and } u|_{\Omega}\in C^{2,\alpha}(\overline\Omega) \right\}.
\end{equation}
\begin{Lemma}\label{lemma C^alpha estimate }
	Let $ \Omega $ be a $ C^{2,\alpha} $ domain in $ \mathbb{R}^n $, $ u\in {\mathcal{B}}_\alpha(\Omega)$,  $ s\in(0,\frac{1}{2}) $ and $ \alpha+2s<1 $.
	
	Then, $ (-\Delta)^{s} u\in C^\alpha(\overline{\Omega})  $ and
		\begin{equation}\label{eq: C^alpha norm estimate of (0,1/2)}
		\|(-\Delta)^{s} u\|_{C^\alpha(\overline{\Omega})}\leqslant C(\|u\|_{C^{\alpha+2s}(\overline{\Omega})}+\|u\|_{C^{2}(\overline{\Omega})}),
		\end{equation}	
		where $ C $ depends only on $ n,s,\alpha,\Omega$.
	
\end{Lemma}

We split the proof of Lemma~\ref{lemma C^alpha estimate } into two steps:
\begin{itemize}
	\item[{Step 1.}] Estimate $ [(-\Delta)^{s} u]_{C^\alpha(\overline{\Omega})} $ for every $ s\in(0,1) $ (see Lemma~\ref{lemma: C^alpha estimate of fraction laplacian }).
	\item[Step~2.] Estimate the $ L^\infty $-norm of $ (-\Delta)^{s} u $ for every $ s\in(0,1) $ (see Lemma~\ref{lemma: boundedness}).
\end{itemize}


\begin{Lemma}\label{lemma: C^alpha estimate of fraction laplacian }
	Suppose that $ \Omega $ is a $ C^{2,\alpha} $ domain in $ \mathbb{R}^n$, $ u\in {\mathcal{B}}_\alpha(\Omega) $, $ s\in(0,\frac{1}{2}) $ and $ \alpha+2s\leqslant 1 $.
	
	Then,
		\begin{equation}\label{eq: C^alpha estimate of (0,1/2)}
		[(-\Delta)^{s} u]_{C^\alpha(\mathbb{R}^n)}\leqslant C[u]_{C^{\alpha+2s}(\overline{\Omega})},
		\end{equation}	
		where $ C $ depends only on $ n,s,\alpha $.
\end{Lemma}
\begin{proof}
	Let us estimate the difference $ (-\Delta)^{s}u(x_1)-(-\Delta)^{s}u(x_2) $ for $ x_1,x_2\in\mathbb{R}^n  $. For this, we note that
	\begin{equation*}
	|(-\Delta)^{s}u(x_1)-(-\Delta)^{s}u(x_2)|\leqslant A_1+A_2,
	\end{equation*}
	where 	\begin{align*}
	A_1&= \left|\int_{B(r)}\frac{u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z)}{|z|^{n+2s}}\,dz\right|,\\
	A_2&= \left|\int_{\mathbb{R}^n\backslash B(r)}\frac{u(x_1)-u(x_1+z)-u(x_2)+u(x_2+z)}{|z|^{n+2s}}\,dz\right|.
	\end{align*}
	\textbf{Case 1.} Let $ \beta:=\alpha+2s\leqslant 1 $.
	For $ A_1 $,  we use that $ \left|u(x_i)-u(x_i+z)\right|\leqslant[u]_{C^\beta(\overline{\Omega})}|z|^\beta $ for $ i=1,2 $. Therefore,
	\begin{equation}
	\begin{split}\label{eq: A_1 (0,1/2)}
	A_1&\leqslant \int_{B(r)}\frac{2[u]_{C^\beta(\overline{\Omega})}|z|^\beta}{|z|^{n+2s}}\,dz
	\leqslant C(n,s,\alpha)[u]_{C^\beta(\overline{\Omega})}r^\alpha.
	\end{split}
	\end{equation}
	For $ A_2 $, we use that $ \left|u(x_1)-u(x_2)\right|\leqslant[u]_{C^\beta(\overline{\Omega})}|x_1-x_2|^\beta $ for $ i=1,2 $ As a consequence,
	\begin{equation}
	\begin{split}\label{eq: A_2 (0,1/2)}
	A_2&\leqslant \int_{\mathbb{R}^n\backslash B(r)}\frac{2[u]_{C^\beta(\overline{\Omega})}|x_1-x_2|^\beta}{|z|^{n+2s}}\,dz
	\leqslant C(n,s)[u]_{C^\beta(\overline{\Omega})}r^{-2s}|x_1-x_2|^\beta.
	\end{split}
	\end{equation}
	Taking $ r=|x_1-x_2| $ and adding $ A_1 $ and $ A_2 $, we obtain
	\begin{equation}\label{eq: C^alpha<C^1 (0,1/2)}
	[(-\Delta)^{s} u]_{C^\alpha(\mathbb{R}^n)}\leqslant C_{n,s,\alpha}[u]_{C^\beta(\overline{\Omega})},
	\end{equation}
	where $ C_{n,s,\alpha}>0 $ is a suitable constant.
\end{proof}




\begin{Lemma}\label{lemma: boundedness}
		Let $ \Omega\subset\mathbb{R}^n  $ be a $ C^{2,\alpha} $ domain and $ u\in {\mathcal{B}}_\alpha(\Omega). $    Then, for every $ s\in(0,1) $ we have that $ (-\Delta)^{s} u\in L^\infty(\mathbb{R}^n) $ and
	\begin{equation}\label{eq:boundedness}
	\|(-\Delta)^{s} u\|_{L^\infty(\mathbb{R}^n)}\leqslant C\|u\|_{C^2(\overline{\Omega})},
	\end{equation}
	where $ C>0 $ depends only on $ n,s,\Omega$.
\end{Lemma}

\begin{proof}
	 We choose $ x_0\in\Omega $ such that
	 \begin{equation*}
	 	d_0=:\text{dist}(x_0,\partial\Omega)=\sup\limits_{x\in{\Omega}}\text{dist}(x,\partial\Omega).
	 \end{equation*}
Then, we have that
	\begin{equation}
		\begin{split}\label{eq:(-Delta)^{s}u(x_0)}
		|(-\Delta)^{s}u(x_0)|&\leqslant\left|\int_{|z|<d_0}\frac{u(x_0+z)-u(x_0)}{|z|^{n+2s}}\,dx\right|+\left|\int_{|z|\geqslant d_0}\frac{u(x_0+z)-u(x_0)}{|z|^{n+2s}}\,dx\right|\\
		&\leqslant \int_{|z|<d_0}\frac{\int_{0}^{1}|\nabla u(x_0+tz)|\,dt}{|z|^{n+2s-1}}\, dx+ 2\|u\|_{L^\infty(\Omega)}\int_{|z|\geqslant d_0}\frac{1}{|z|^{n+2s}}\, dx\\
	 &\leqslant c_{n,s}(d_0^{1-2s}+d_0^{-2s})\|u\|_{C^1(\overline\Omega)},
		\end{split}
	\end{equation}
	where $ c_{n,s}>0 $ is another  constant independent of $ u $.
	
	Combining Lemma~\ref{lemma: C^alpha estimate of fraction laplacian } with \eqref{eq:(-Delta)^{s}u(x_0)},
	for every $ x\in\overline{\Omega} $, one has
	\begin{equation*}
		\begin{split}
		|(-\Delta)^{s}u(x)|&\leqslant[(-\Delta)^{s} u]_{C^\alpha(\mathbb{R}^n)}|x-x_0|^\alpha+|(-\Delta)^{s}u(x_0)|\\
		&\leqslant C_{n,s,\alpha}\text{  diam}(\Omega)^\alpha\|u\|_{C^{\alpha+2s}(\overline{\Omega})}+|(-\Delta)^{s}u(x_0)|\\
		&\leqslant C_{n,s,\alpha}\|u\|_{C^{\alpha+2s}(\overline{\Omega})}+c_{n,s}(d_0^{1-2s}+d_0^{-2s})\|u\|_{C^1(\overline\Omega)}.
		\end{split}
	\end{equation*}
	 As a consequence of this and Lemma~\ref{lemma: C^alpha estimate of fraction laplacian }, we finally obtain
	\begin{equation*}
		\begin{split}
	\|(-\Delta)^{s}u\|_{C^{\alpha}(\overline{\Omega})}&=\|(-\Delta)^{s}u\|_{L^\infty(\overline{\Omega})}+[(-\Delta)^{s} u]_{C^\alpha(\overline{\Omega})}
	\leqslant C_{n,s,\Omega,\alpha}(\|u\|_{C^{\alpha+2s}(\overline{\Omega})}+\|u\|_{C^{1}(\overline{\Omega})}),
		\end{split}
	\end{equation*}
for some suitable $ C_{n,s,\Omega,\alpha}>0 $.
\end{proof}


	\subsubsection{Proof of the $ C^{2,\alpha} $-regularity up to the boundary}\label{subsubsection 4.22}
	
	First of all, we establish the following facts:
	\begin{Lemma}\label{lemma facts relating to method of continutiy}
		\begin{itemize}
			\item[(1)] $ \mathcal{L} u\in C^\alpha(\overline{\Omega}) $ for every $ u\in {\mathcal{B}}_\alpha(\Omega) $;
			\item[(2)] there exists a constant $ C=C(n,s,\alpha)>0 $ such that
			\begin{equation}\label{eq:C^2 estimate of u }
				\|u\|_{C^{2,\alpha}(\overline{\Omega})}\leqslant C\left(\|\mathcal{L}u\|_{C^{\alpha}(\overline{\Omega})}+\sup\limits_{x\in{\Omega}}|u|\right)\qquad \forall u\in {\mathcal{B}}_\alpha(\Omega).
			\end{equation}
		\end{itemize}
			\end{Lemma}
		\begin{proof}
	As regards (1), it follows directly from Lemma~\ref{lemma C^alpha estimate } and the fact that $ \Delta u\in C^\alpha(\overline{\Omega}) $  if $ u\in {\mathcal{B}}_\alpha(\Omega) $.  We now turn to prove \eqref{eq:C^2 estimate of u }.
	
	To this end, we observe that $ \partial\Omega $ is of class $ C^{2,\alpha} $, which allows us to apply \cite[Theorem 6.14]{GTbook}: there exists a constant $ C=C(n,\alpha)>0 $ such that
	\begin{equation}\label{eq:Laplacian estimate of C^2}
		\|u\|_{C^{2,\alpha}(\overline{\Omega})}\leqslant C\left(\|\Delta u\|_{C^{\alpha}(\overline{\Omega})}+\sup\limits_{x\in{\Omega}}|u|\right),
	\end{equation}
	for every $ u\in C^{2,\alpha}(\overline{\Omega}) $ satisfying $ u\equiv 0 $ in $ \mathbb{R}^n\backslash\Omega. $	 
	
	Thus, combining Lemma~\ref{lemma C^alpha estimate } and \eqref{eq:Laplacian estimate of C^2},  for every $ u\in {\mathcal{B}}_\alpha(\Omega) $ one has
	\begin{equation}
		\begin{split}\label{eq:Laplacian estimate of C^2 alpha}
		\|u\|_{C^{2,\alpha}(\overline{\Omega})}&\leqslant C\left(\|\Delta u\|_{C^{\alpha}(\overline{\Omega})}+\sup\limits_{x\in{\Omega}}|u|\right)
		=C\left(\|\mathcal{L}u-(-\Delta)^s u\|_{C^{\alpha}(\overline{\Omega})}+\sup\limits_{x\in{\Omega}}|u|\right)\\
		&\leqslant C\left(\|\mathcal{L}u\|_{C^{\alpha}(\overline{\Omega})}+\|(-\Delta)^s u\|_{C^{\alpha}(\overline{\Omega})}+\sup\limits_{x\in{\Omega}}|u|\right)\\
		&\leqslant \bar{C}\left(\|\mathcal{L}u\|_{C^{\alpha}(\overline{\Omega})}+\| u\|_{C^{\alpha+2s}(\overline{\Omega})}+\| u\|_{C^{1}(\overline{\Omega})}+\sup\limits_{x\in{\Omega}}|u|\right).
		\end{split}
	\end{equation}
		Owing again to the regularity of $ \partial\Omega $, we can utilize the global interpolation inequality (see
		e.g.~\cite[Chapter 6]{GTbook}): there exists a constant $ \theta=\theta(\bar{C})>0 $ such that
		\begin{equation*}
			\| u\|_{C^{\alpha+2s}(\overline{\Omega})}+\| u\|_{C^{1}(\overline{\Omega})}\leqslant \frac{1}{2\bar{C}}\| u\|_{C^{2,\alpha}(\overline{\Omega})}+\theta \sup\limits_{x\in{\Omega}}|u|.
		\end{equation*}
		As a consequence of this and \eqref{eq:Laplacian estimate of C^2 alpha}, we obtain the estimate in~\eqref{eq:C^2 estimate of u }, as desired.
		\end{proof}
	
	
			We now use the method of continuity (See. e.g. \cite[Theorem 5.2]{GTbook}) to complete the proof.
			
			We first notice that $ {\mathcal{B}}_\alpha(\Omega) $ is  endowed with a structure of Banach space by the norm
			\begin{equation*}
				\| u\|_{{\mathcal{B}}_\alpha(\Omega)}:=\| u\|_{C^{2,\alpha}(\overline{\Omega})}\qquad \forall u\in {\mathcal{B}}_\alpha(\Omega).
			\end{equation*}
			Furthermore, for every $ t\in[0,1] $, we set
			\begin{equation*}
			\mathcal{L}_t=(1-t)(-\Delta)+t\mathcal{L}=-\Delta+t(-\Delta)^s.
			\end{equation*}
			Utilizing (1)-(2) in Lemma~\ref{lemma facts relating to method of continutiy}, one concludes that $ \mathcal{L}_t $ maps $ {\mathcal{B}}_\alpha(\Omega) $ to $ C^\alpha(\overline{\Omega}) $ and
			\begin{equation}\label{eq: B norm estimate of u}
				\| u\|_{{\mathcal{B}}_\alpha(\Omega)}:=\| u\|_{C^{2,\alpha}(\overline{\Omega})}\leqslant C\left(\|\mathcal{L}_t u\|_{C^{\alpha}(\overline{\Omega})}+\sup\limits_{x\in{\Omega}}|u|\right) \qquad \forall u\in {\mathcal{B}}_\alpha(\Omega),
			\end{equation}
			where $ C>0 $ is a suitable constant independent of $ u $ and $ t $. 
			
			Furthermore, using~\cite[Theorem 4.7]{BDVV22b}, it is not difficult to see that
			\begin{equation}\label{eq: sup u in Omega}
				\sup\limits_{x\in{\Omega}}|u|=\sup\limits_{x\in\mathbb{R}^n}|u|\leqslant\rho \|\mathcal{L}_t u\|_{L^\infty(\Omega)}\leqslant\rho\|\mathcal{L}_t u\|_{C^{\alpha}(\overline{\Omega})},
			\end{equation}
			where $ \rho $ is an another constant independent of $ u $ and $ t $.
	

Due to \eqref{eq: B norm estimate of u} and \eqref{eq: sup u in Omega}, we can apply the method of continuity in the following settings:
\begin{itemize}
	\item[(i)] $ {\mathcal{B}}_\alpha(\Omega) $ and $ C^\alpha(\Omega) $ are Banach space;
	\item[(ii)] $ \mathcal{L}_0 $ and $ \mathcal{L}_1 $ are linear and bounded from $ {\mathcal{B}}_\alpha(\Omega) $ to $ C^\alpha(\overline\Omega) $;
	\item[(iii)] There exists a constant $ \hat{C}>0 $ such that
	\begin{equation*}
		 \| u\|_{{\mathcal{B}}_\alpha(\Omega)}\leqslant \hat{C}\|\mathcal{L}_t u\|_{C^{\alpha}(\overline{\Omega})}\qquad \forall u\in {\mathcal{B}}_\alpha(\Omega) \text{ and } t\in[0,1].
	\end{equation*}
\end{itemize}
 Since $  \mathcal{L}_0=-\Delta $ is surjective, one can deduce that $ \mathcal{L}_1=\mathcal{L} $ is  also surjective: for every $ f\in C^\alpha(\overline{\Omega}) $ there exists unique $ u\in {\mathcal{B}}_\alpha(\Omega) $ such that
 \begin{equation}
 	\mathcal{L}u=f \qquad \text{ a.e. in  } \Omega.
 \end{equation}
	We  observe that $ u\in C^{2,\alpha}(\overline{\Omega}) $ and $ u\equiv0$ in $ \mathbb{R}^n\backslash \Omega $.
	
\begin{proof}[Proof of Theorem~\ref{theorem C^2 alpha global}]	
	Let $ u\in X_0^1 $ be the weak solution of \eqref{Maineq}. Theorem~\ref{th:C^{1,alpha}} implies that the map~$ x\mapsto g(x,u(x))$ lies in~$ C^\alpha(\overline{\Omega})$. 
	
	Moreover, there exists unique solution  $ v\in {\mathcal{B}}_\alpha(\Omega)\subset X_0^1.  $ In addition, the Lax-Milgram Theorem yields that $u=v$, and in particular, from
	\eqref{eq:C^2 estimate of u }, we deduce that 
	\begin{equation}
		\begin{split}
				\|u\|_{C^{2,\alpha}(\overline{\Omega})}&\leqslant C\left(\|\mathcal{L}u\|_{C^{\alpha}(\overline{\Omega})}+\sup\limits_{x\in{\Omega}}|u|\right)
				\leqslant C\left(\|g(\cdot,u(\cdot))\|_{C^{\alpha}(\overline{\Omega})}+\sup\limits_{x\in{\Omega}}|u|\right)\\
				&\leqslant C \left([g]_{C^\alpha(\overline{\Omega}\times\mathbb{R})}\left(1+\|Du\|_{L^\infty(\Omega)}\right)+\|g\|_{L^\infty(\overline\Omega\times \mathbb{R}^n)}+\|u\|_{L^\infty(\Omega)}\right)\\
				&\leqslant C \left(\|g\|_{C^\alpha(\overline{\Omega}\times\mathbb{R})}\left(1+\|u\|_{C^1(\overline\Omega)}\right)+\|u\|_{L^\infty(\Omega)}\right),
			\end{split}
	\end{equation}
	for some constant $C$ depending only on $n,s,\alpha,\Omega$.
	Hence, the proof of Theorem~\ref{theorem C^2 alpha global} is completed.
\end{proof}
	
	{
	\begin{appendix}
	
	\section{The strong maximum principle}
	
	We collect here some auxiliary results which follow as a simple byproduct of our main theorems.
	These results are not stated in their full generality, but rather in a way which makes them directly utilizable in nonlinear analysis problems (in particular, we will use them in the present form in the article~\cite{SVWZ}).

First, we present a strong maximum principle (see~\cite{BDVV22b} for a weak maximum principle).

	\begin{Theorem}[Strong maximum principle]\label{coro: solution u>0}
	Let 
	$\Omega$ be a bounded $C^{1,1}$ domain, 
	$ u$  be   a weak solution of \eqref{Maineq}.
	Assume that
	 $ g\in C^\alpha_{\rm loc}(\Omega\times\mathbb{R}) $  satisfies \eqref{eq: H 1} for any given  $\alpha\in(0,1)$
		and   $ g\geqslant 0 $.
		
		Then, $u\in C^{2, \alpha}_{\rm loc}(\Omega)\cap C(\mathbb{R}^n)  $ for any $\alpha\in(0,1)$ and furthermore
		$$ u>0 \quad \text{  in  } \Omega. $$
	\end{Theorem}	 
	
	\begin{proof}
		We divide the proof  into the following two steps.
		
		\noindent\textbf{Step 1.}	We first show that $ u\geqslant 0$ in $ \Omega.$	
		Since $ u $ is a nontrivial weak solution of \eqref{Maineq}, utilizing Theorems~\ref{th: regularity} and \ref{th C^2,alpha interior}, one deduces that
		\begin{equation*}
			u\in C^{2,\alpha}_{\rm loc}(\Omega)\cap C(\mathbb{R}^n).
		\end{equation*}
		Moreover, due to $ g(x,u)\geqslant 0 $,
		\begin{equation*}
			-\Delta u+(-\Delta)^su\geqslant 0, \text{ a.e. in } \Omega \text{ and }  u\equiv 0  \text{ in } \mathbb{R}^n\backslash \Omega.
		\end{equation*}
		Suppose by contradiction that there exists a point $ \eta\in\Omega $ such that $ u(\eta)<0 $. Since $ u\in C(\mathbb{R}^n) $ and $ \overline{\Omega} $ is compact,  one can find  $ x_0\in\overline\Omega $ such that
		$$ u(x_0)=\min\limits_{\overline\Omega} u(x)<0. $$
		We notice that $x_0\in \Omega$ because $u$ is continuous and $ u=0 $ in $ \mathbb{R}\backslash\Omega$.  
		
		Thus, $ \Delta u(x_0)\geqslant 0 $ and so
		\begin{equation}\label{eq: contradiction for u }
			\begin{split}
				0\leqslant \big(-\Delta +(-\Delta)^s \big) u(x_0) &=-\Delta u(x_0)+ \int_{\mathbb{R}^{n}}\frac{u(x_0)-u(y)}{|x_0-y|^{n+2s}}\, dy\\
				&\leqslant \int_{\mathbb{R}^{n}}\frac{u(x_0)-u(y)}{|x_0-y|^{n+2s}}\, dy \leqslant 0.
			\end{split}
		\end{equation}Here, the last inequality holds due to the fact that $ u(x_0)\leqslant u(x)$ for all $ x\in\mathbb{R}^n$.
		%	Additionally, we know that $ u(x_0)\leqslant u(x)$ for all $ x\in\mathbb{R}^n$. Comparing this with \eqref{eq: contradiction for u },
		Hence, we obtain
		that $ u\equiv 0 $ throughout $ \mathbb{R}^n $, which
		is a contradiction.
		
		\noindent\textbf{Step 2.} We now show that~$u>0 $ in $ \Omega.$	We argue by contradiction.
		
		If this were not true, there would exist a point $ \xi\in\Omega $ such that $ u(\xi)=0 $, which is an interior minimum point for $ u $ in $ \Omega$.
		Replacing  $ x_0 $ with $ \xi $  in \eqref{eq: contradiction for u }, due to Step~1, one has $ u\equiv 0 $ throughout $ \mathbb{R}^n $, which
		is a contradiction.
\end{proof}

\section{The principal eigenvalue problem}

Now we give a characterization
of the principal eigenvalue and eigenfunction. 

\begin{Theorem}[Variational properties for the principal eigenvalue]\label{th: eigenvalue problem}		
%		The first eigenvalue of \eqref{eq:eigenvalue problem} is defined by
%		\begin{equation}\label{definition of eigenvalue}
%			\lambda_1:=\inf\limits_{u\in X_0^1\backslash\{0\}}\frac{\|u\|}{\int_{\Omega}u^2\,dx}
%		\end{equation}
%		The infimum is achieve. More precisely, 	
%Let $\lambda$ be the first eigenvalue
Consider the eigenvalue problem 	 	\begin{equation}\label{eq:eigenvalue problem}
	 	\begin{cases}
	 		-\Delta u+(-\Delta)^s u=\lambda u\quad &\text{in } \Omega,\\
	 		u=0&\text{in }\mathbb{R}^n\backslash \Omega.
	 	\end{cases}
	 \end{equation}

Then,	
		\begin{itemize}
			\item [(a)] We have 
				\begin{equation*}\label{definition of eigenvalue}
							\lambda_1=\min\left\{\|u\|^2_{X_0^1}\hspace{2mm}| \hspace{2mm} u\in X^1_0,\|u\|_{L^2(\Omega)}=1 \right\}>0.
						\end{equation*}
		\item [(b)] $ \lambda_1 $ is simple.
			\item[(c)]  The first eigenfunction $\varphi_1$ is of $C^{2,\alpha}_{\rm loc}(\Omega)$ class.
				\item[(d)]   $\varphi_1\geqslant 0$.
		\end{itemize}	
				Furthermore, if $\Omega$  is a $C^{1,1}$ domain, then
				\begin{itemize}
				\item[(e)]  $\varphi_1\in C^{2,\alpha}_{\rm loc}(\Omega)\cap C^{1,\alpha}(\overline \Omega)$.
			\item[(f)] $\varphi_1>0$ in $\Omega$ and $\varphi_1=0$ in $\mathbb{R}^n\backslash\Omega$.
				\end{itemize}  
	\end{Theorem}

	\begin{proof}
		As for \textbf{(a)} and \textbf{{(d)}}, we set
		\begin{equation*}
			I(u):=\|u\|^2_{X_0^1}, \qquad  \mathcal{S}= \left\{u\in X_0^1, \|u\|_{L^2(\Omega)}=1  \right\}.
		\end{equation*}
		Let  $ \left\{u_j\right\}\subset \mathcal{S} $ be a minimizing sequence for $ I $, that is 
		\begin{equation*}
			\lim\limits_{j \rightarrow \infty}I{(u_j)}=\mathop{\text{inf }}\limits_{\mathcal{S} } \|u\|^2_{X_0^1}\geqslant 0.
		\end{equation*}
		Then $ \left\{u_j\right\} $ is bounded in $ X_0^1 $. Up to a subsequence, still defined by $ \left\{u_j\right\} $, there exists $ u^*\in X_0^1 $ such that 
		\begin{equation}
			\begin{split}
				u_j\rightharpoonup u^* &\text{ in }   X_0^1;\\
				u_j\rightarrow u^*& \text{ in } L^2(\Omega).
			\end{split}
		\end{equation}
		Since $ \|u_j\|_{L^2(\Omega)}=1 $, we know that  $ \|u^*\|_{L^2(\Omega)}=1  $ and  $ u^*\in \mathcal{S} $.  According to the weak lower semi-continuity of the norm, one has that
		\begin{equation}
			\lim\limits_{j \rightarrow \infty}I{(u_j)}\geqslant I(u^*)\geqslant \mathop{\text{inf }}\limits_{\mathcal{S} } \|u\|^2_{X_0^1},
		\end{equation}
		which  implies  that $ I(u^*)= \mathop{\text{inf }}\limits_{\mathcal{S} } \|u\|^2_{X_0^1} $. 
		
		We claim that for any $ v\in X_0^1, $ 
		\begin{equation}\label{5.3}
			\left\langle u^*, v\right\rangle_{X_0^1}=I(u^*)\int_{\Omega} u^* v\,dx.
		\end{equation}
		In fact,	let $ t\in (-1,1), $ $ u_t=\frac{u^*+tv}{\|u^*+tv\|_2} $, then $ u_t\in \mathcal{S} $ and one has
		\begin{align*}
			I(u_t)&=\frac{\|u^*\|^2_{X_0^1}+2t\left\langle u^*, v\right\rangle_{X_0^1}+t^2 \|v\|^2_{X_0^1}}{1+2t\int_{\Omega}u^* v\, dx+t^2|v|^2_{2}}\\
			&=\frac{\|u^*\|^2_{X_0^1}+2t\left\langle u^*, v\right\rangle_{X_0^1}+t^2 \|v\|^2_{X_0^1}}{1-4t^2(\int_{\Omega}u^* v\, dx)^2+2t^2|v|^2_{2}+t^4|v|^4_{2}}\left(1-
			2t\int_{\Omega}u^* v\, dx+t^2|v|^2_{2}\right)\\
			&\leqslant \frac{\|u^*\|^2_{X_0^1}+2t\left\langle u^*, v\right\rangle_{X_0^1}+t^2 \|v\|^2_{X_0^1}}{(1-t^2|v|^2_2)^2}\left(1-
			2t\int_{\Omega}u^* v\, dx+t^2|v|^2_{2}\right)\\
			&=\frac{1}{(1-t^2|v|^2_2)^2}\left(I(u^*)+2t\big(\left\langle u^*, v\right\rangle_{X_0^1}-I(u^*)\int_{\Omega}u^* v\, dx\big)+P(t^2)\right).
		\end{align*} 
		If there existed $ v\in X_0^1 $ such that 
		\begin{equation*}
			\left\langle u^*, v\right\rangle_{X_0^1}-I(u^*)\int_{\Omega} u^* v\,dx=l>0,
		\end{equation*}  
		then for $ t<0 $ small enough,
		\begin{equation*}
			I(u_t)\leqslant \frac{1}{(1-t^2|v|^2_2)^2}\bigg(I(u^*)+2tl+P(t^2)\bigg)<I(u^*),
		\end{equation*}
		which would contradict  the definition of $ u^* $.  We can use the same argument to verify that the case of $ l<0 $ cannot hold as well, thus deducing that~\eqref{5.3} is valid. 
		
		As a consequence of \eqref{5.3}, we infer that $ \lambda_1 $ is attained at some $ \varphi_1\in X_0^1 $, with $ \|\varphi_1\|_{L^2(\Omega)}=1 $  and $ I(\varphi_1)=\lambda_1 $.
		
		In order to  prove that~$ \varphi_1\geqslant0 $ a.e. in $ \Omega $, we first need to show that if $ \varphi_1 $ is an eigenfunction related to $ \lambda_1 $, 
		then both $ \varphi_1 $ and $ |\varphi_1| $ realize the minimum in the definition of $\lambda_1$.%\eqref{definition of eigenvalue}. 
		
		Indeed, let  $ x_1\in\left\{x\in\mathbb{R}^n,\varphi_1 >0\right\} $ and $ x_2\in\left\{x\in\mathbb{R}^n,\varphi_1 <0\right\} $. Then,
		\begin{equation}\label{eq:eigen}
			\left||\varphi_1(x_1)|-|\varphi_1(x_2)|\right|=|\varphi_1(x_1)+\varphi_1(x_2)|<\varphi_1(x_1)-\varphi_1(x_2)=|\varphi_1(x_1)-\varphi_1(x_2)|.
		\end{equation}
		If both the sets $ \left\{x\in\mathbb{R}^n,\varphi_1 >0\right\} $ and $ \left\{x\in\mathbb{R}^n,\varphi_1 <0\right\} $  have positive measure, exploiting  \eqref{eq:eigen},
		we get $ I(|\varphi_1|)<I(\varphi_1)=\lambda_1 $, whence, since $ |\varphi_1|\in\mathcal{S} $, we have that~$ I(|\varphi_1|)=I(\varphi_1)=\lambda_1 $, which implies that $ \varphi_1\geqslant0  $ or $ \varphi_1\leqslant0 $ a.e. in $ \Omega $. By replacing $ \varphi_1 $ by $ |\varphi_1| $, we may conclude that~$ \varphi_1\geqslant 0. $
		
		\bigskip
		
		Next, we turn to prove \textbf{(b)}.  Suppose by contradiction that there exists another eigenfunction $ \varphi_2\in X_0^1, $ $ \varphi_1\neq \varphi_2 $, corresponding to $ \lambda_1 $. Setting $ \bar{\varphi}_2=\frac{\varphi_2}{\|\varphi_2\|_{L^2(\Omega)}} $ and $ u_1=\varphi_1-\bar{\varphi}_2 $.
		
		We claim that $ u_1=0 $ a.e. in $ \Omega $. In fact, it is easy to see that $ u_1 $ is also the eigenfunction related to $ \lambda_1 $,  so we deduce $ u_1\leqslant0 $ or $ u_1\geqslant 0 $ a.e. in $ \Omega $, that is, $ \varphi_1\leqslant\bar{\varphi}_2  $ or $ \varphi_1\geqslant\bar{\varphi}_2  $ a.e. in $ \Omega $, which implies that $ {\varphi_1}^2\leqslant \bar{\varphi}_2^2 $ or $ \varphi_1^2\geqslant\bar{\varphi}_2^2 $. However, $ \|\varphi_1\|_{L^2(\Omega)}=1=\|\bar{\varphi}_2\|_{L^2(\Omega)} $, we get $ \varphi_1^2=\bar{\varphi}_2^2 $, hence $ u_1=0 $ a.e. in $ \Omega $. Therefore, if $ \varphi\in X_0 $ is an eigenfunction related to $ \lambda_1 $, then $ \varphi= |\varphi|_2\varphi_1 $. 
		
		\bigskip
		
		
		Finally, as for \textbf{(c)}  \textbf{(e)} and \textbf{(f)}, since $\varphi_1$ is a weak solution of \eqref{eq:eigenvalue problem} and satisfies the assumption \eqref{eq: H 1}, applying  Theorems~\ref{th: regularity} and \ref{th:C^{1,alpha}}, one has $\varphi_1\in C^{1,\alpha}(\overline{\Omega})$ for some $\alpha\in(0,1)$. Moreover, it follows from Theorem~\refeq{th C^2,alpha interior}  that $ \varphi_1\in C^{2,\alpha}_{\rm loc}(\Omega)\cap C^{1,\alpha}(\overline{\Omega}) $. 	As a consequence of this and \textbf{(a)}, employing the strong maximum principle, we obtain that~$\varphi_1>0$.
	\end{proof}


	\end{appendix}
	}


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