\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 284, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/284\hfil Regularity lifting result]
{Regularity lifting result for an integral system involving
Riesz potentials}

\author[Y. Li, D. Xu \hfil EJDE-2017/284\hfilneg]
{Yayun Li, Deyun Xu}

\address{Yayun Li \newline
Institute of Mathematics,
School of Mathematical Sciences,
Nanjing Normal University,
Nanjing, 210023, China}
\email{liyayun.njnu@qq.com}

\address{Deyun Xu \newline
Institute of Mathematics,
School of Mathematical Sciences,
Nanjing Normal University,
Nanjing, 210023, China}
\email{954816700@qq.com}

\thanks{Submitted  May 16, 2017. Published November 14, 2017.}
\subjclass[2010]{35J10, 35Q55, 45E10, 45G05}
\keywords{Riesz potential; integral system; regularity lifting lemma;
\hfill\break\indent Hartree equation; Hardy-Littlewood-Sobolev inequality}

\begin{abstract}
 In this article, we study the integral system involving the Riesz potentials
 \begin{gather*}
 u(x)=\sqrt{p} \int_{\mathbb{R}^n}\frac{u^{p-1}(y)v(y)dy}{|x-y|^{n-\alpha}},
 \quad u>0 \text{ in } \mathbb{R}^n,\\
 v(x)=\sqrt{p} \int_{\mathbb{R}^n}\frac{u^p(y)dy}{|x-y|^{n-\alpha}}
 \quad v>0 \text{ in } \mathbb{R}^n,
 \end{gather*}
 where $n \geq 1$, $0<\alpha<n$ and $p>1$. Such a system is related
 to the study of a static Hartree equation and the Hardy-Littlewood-Sobolev
 inequality. We investigate the regularity of positive solutions
 and prove that some integrable solutions belong to $C^1(\mathbb{R}^n)$. An
 essential regularity lifting lemma comes into play, which was
 established by Chen, Li and Ma \cite{ChLM}.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction} 

Recently, many authors have studied the stationary Hartree type equation
\begin{equation} \label{PDE}
(-\Delta)^{\alpha/2} u=pu^{p-1}(|x|^{\alpha-n}*u^p), \quad u>0 \text{ in }
 \mathbb{R}^n,
\end{equation}
where $n \geq 1$, $\alpha \in (0,n)$ and $p>1$.

When $\alpha=2$, \eqref{PDE} is a simplified model of the
Maxwell-Schr\"odinger system (cf. \cite{Amst,CDSS,HW} 
and references therein). It is also 
\cite[Example 3.2.8]{Caz}. 
A more general form is the Choquard type
equation in the papers \cite{Lei,MZ}. Paper \cite{FQT}
studied the existence and the regularity results of positive
solutions of the static Schr\"odinger equation with the fractional
Laplacian. Another interesting work related to \eqref{PDE} are
paper \cite{JL-Z} and the references therein. Equation \eqref{PDE}
is also helpful in understanding the blowing up or the global
existence and scattering of the solutions of the dynamic Hartree
equation (cf. \cite{LMZ}), which arises in the study of boson
stars and other physical phenomena, and also appears as a
continuous-limit model for mesoscopic molecular structures in
chemistry. Such an equation also arises in the Hartree-Fock theory
of the nonlinear Schr\"odinger equations (cf. \cite{LS}). More
related mathematical and physical background can be found in
\cite{GV,KS,Na}.


Since \eqref{PDE} has a convolution term,
it seems difficult to investigate the existence directly.
Write
$$
v(x)=\sqrt{p}\int_{\mathbb{R}^n}\frac{u^p(y)dy}{|x-y|^{n-\alpha}}.
$$
Then $v>0$ in $\mathbb{R}^n$. As in \cite{SIAM,LM,MZ},
we introduce an integral system
\begin{equation} \label{IE}
\begin{gathered}
 u(x)=\sqrt{p} \int_{\mathbb{R}^n}\frac{u^{p-1}(y)v(y)dy}{|x-y|^{n-\alpha}}, 
\quad u>0 \text{ in } \mathbb{R}^n,\\
 v(x)=\sqrt{p} \int_{\mathbb{R}^n}\frac{u^p(y)dy}{|x-y|^{n-\alpha}}, 
\quad v>0 \text{ in } \mathbb{R}^n.
 \end{gathered}
\end{equation}
According to the results in \cite{CL}, we can also see that
the equivalence between \eqref{PDE} and \eqref{IE} if omitting constants.

In addition, \eqref{IE} is analogous to the system
\begin{equation}
\begin{gathered}
 u(x)=  \int_{\mathbb{R}^n}\frac{v^q(y)dy}{|x-y|^{n-\alpha}}, \quad
u,v>0 \text{ in }\mathbb{R}^n,\\
 v(x)= \int_{\mathbb{R}^n}\frac{u^p(y)dy}{|x-y|^{n-\alpha}}, \quad
 p,q>0.
 \end{gathered}\label{1.3s} 
 \end{equation}
It is the Euler-Lagrange equations which the extremal functions of
the following Hardy-Littlewood-Sobolev inequality satisfies
$$
\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}
\frac{f(x)g(y)}{|x|^{\alpha}|x-y|^{\lambda}|y|^{\beta}}\,dx\,dy
 \leq C_{\alpha,\beta,s,\lambda,n }\|f\|_r \|g\|_s,
$$
where $1 < s,r < \infty$, $0 < \lambda < n$, 
$\lambda \leq \overline{\lambda} =\lambda+\alpha+\beta \leq n$, 
$\frac{1}{r}+\frac{1}{s} +\frac{\overline{\lambda}}{n} =2$, 
$\frac{\alpha}{n}< 1- \frac{1}{r}<\frac{\lambda +\alpha}{n}$, 
$\frac{\beta}{n}< 1- \frac{1}{s}<\frac{\lambda +\beta}{n}$. Some
classical work can be found in \cite{CDM,CLBook,CLO,Lieb} and many other papers.


The main conclusions of this paper are stated as follows, which
are proved in section 2.

\begin{theorem} \label{thm1.1}
Let $n \geq 1$ and $0<\alpha<n$.
If $1<p \leq \frac{n}{n-\alpha}$, \eqref{IE} does not
have any positive solution.
\end{theorem}


\begin{theorem} \label{thm1.2}
Assume $u$ is a positive solution of \eqref{IE} and $1<\alpha<n$.
If $u \in L^{\frac{n(p-1)}{\alpha}}(\mathbb{R}^n)$,
then $u \in C^1(\mathbb{R}^n)$.
\end{theorem}

To prove Theorem \ref{thm1.2}, we need a regularity
lifting lemma in \cite{CLBook} which was established by Chen, 
Li and Ma \cite{ChLM}. This powerful technique was successfully
applied to obtain the Lipschitz continuity of positive solutions 
of integral systems involving the Riesz potential, 
Bessel potential and the Wolff potential (cf. \cite{Lei,ChLM,Zhang}). In particular,
those regularity properties of \eqref{1.3s} are helpful to understand well the shape
of the extremal functions of the Hardy-Littlewood-Sobolev inequality.

Let $V$ be a function space equipped with two norms $\|\cdot\|_X$ and $\|\cdot\|_Y$.
Define
$$
X=\{v \in V: \|v\|_X<\infty\}, \quad Y=\{v \in V: \|v\|_Y<\infty\}.
$$
Assume that spaces $X$ and $Y$ are complete under the corresponding norms
and the convergence in $X$ or in $Y$ implies the convergence in $V$.

From \cite[Theorem 3.3.5 and Remark 3.3.5]{CLBook}, we have the following
regularity lifting lemma.

\begin{lemma} \label{lem2.2}
Let $X=L^{\infty}(\mathbb{R}^n) \times L^{\infty}(\mathbb{R}^n)$ and 
$Y=C^{0,1}(\mathbb{R}^n) \times C^{0,1}(\mathbb{R}^n)$
with the norms
$$
\|(f,g)\|_X= \|f\|_\infty+\|g\|_\infty, \quad\text{and}\quad 
\|(f,g)\|_Y= \|f\|_{0,1}+\|g\|_{0,1}.
$$
Define their closed subset
\begin{gather*}
X_1=\{(f,g) \in X;\|f\|_\infty+\|g\|_\infty \leq C(\|u\|_{\infty}+\|v\|_\infty) \}, \\
Y_1=\{(f,g) \in Y;\|f\|_\infty+\|g\|_\infty \leq C(\|u\|_{\infty}+\|v\|_\infty) \}.
\end{gather*}
Assume
\begin{itemize}
\item[(i)] $T$ is a contraction map from $X_1 \to X$;

\item[(ii)] $T$ is a shrinking map from $Y_1 \to Y$;

\item[(iii)] $(F,G) \in X_1 \cap Y_1$;

\item[(iv)] $T(\cdot,\cdot)+(F,G)$ is a map from $X_1 \cap Y_1$ to itself.
\end{itemize}
If $(u,v) \in X$ is a pair of solutions of the operator equation
$(f,g)=T(f,g)+(F,G)$, then $(u,v) \in Y$.
\end{lemma}

\section{Proof of main results}


\begin{theorem} \label{thm2.1}
If $1<p\leq n/(n-\alpha)$, then there is no positive solution of \eqref{IE}.
\end{theorem}

\begin{proof}
If $u,v$ are positive solutions, we can deduce a contradiction by
the ideas in \cite{CDM}.
Clearly,
\begin{equation} \label{jiang}
u(x) \geq c\int_{B_R(0)}\frac{u^{p-1}(y)v(y)dy}{|x-y|^{n-\alpha}}
\geq \frac{c}{(R+|x|)^{n-\alpha}}\int_{B_R(0)}u^{p-1}(y)v(y)dy.
\end{equation}
Therefore,
\begin{equation} \label{iin}
\begin{aligned}
 \int_{B_R(0)}u^p(x)dx
&\geq c \int_{B_R(0)}\frac{dx}{(R+|x|)^{p(n-\alpha)}}
(\int_{B_R(0)}u^{p-1}(y)v(y)dy)^p\\
&\geq  \frac{c}{R^{p(n-\alpha)-n}}(\int_{B_R(0)}u^{p-1}(y)v(y)dy)^p.
\end{aligned}
\end{equation}
Here $c$ is independent of $R$. Similarly, from
\begin{equation} \label{quaf}
v(x) \geq \frac{c}{(R+|x|)^{n-\alpha}}\int_{B_R(0)}u^p(y)dy,
\end{equation}
and \eqref{jiang}, \eqref{iin}, we  deduce
\begin{align*}
 \int_{B_R(0)}u^{p-1}(x)v(x)dx
&\geq  \int_{B_R(0)}\frac{cu^{p-1}(x)dx}{(R+|x|)^{n-\alpha}}
 \int_{B_R(0)}u^p(y)dy\\
&\geq  \frac{c}{R^{2[p(n-\alpha)-n]}}(\int_{B_R(0)}u^{p-1}(y)v(y)dy)^p,
\end{align*}
which implies
\begin{equation} \label{xiannv}
\int_{B_R(0)}u^{p-1}(x)v(x)dx \leq CR^{2[p(n-\alpha)-n]/(p-1)}.
\end{equation}

If $1<p<n/(n-\alpha)$, then \eqref{xiannv} with $R \to \infty$ leads to 
$\|u^{p-1}v\|_{L^1(\mathbb{R}^n)}=0$.
This contradicts with $u^{p-1}v>0$.

If $p=n/(n-\alpha)$, then \eqref{xiannv} implies $u^{p-1}v \in L^1(\mathbb{R}^n)$ 
if we let $R \to \infty$.
Multiplying \eqref{quaf} by $u^{p-1}$ and integrating on $A_R:=B_R(0)
\setminus B_{R/2}(0)$, we still have
$$
\int_{A_R}u^{p-1}(x)v(x)dx
\geq c(\int_{B_R(0)}u^{p-1}(y)v(y)dy)^p.
$$
Letting $R \to \infty$ and noting $u^{p-1}v \in L^1(\mathbb{R}^n)$,
we obtain $\|u^{p-1}v\|_{L^1(\mathbb{R}^n)}=0$. It is also impossible.
\end{proof}

Note that Theorem \ref{thm2.1} implies 
\begin{equation} \label{base}
p>\frac{n}{n-\alpha}
\end{equation}
which is the necessary condition of the existence of positive solutions
for \eqref{IE}.


\begin{theorem} \label{thm3.2}
Assume $u$ is a positive solution of \eqref{IE} with
$\alpha \in (1,n)$. If $u \in L^{\frac{n(p-1)}{\alpha}}(\mathbb{R}^n)$,
then $u \in C^1(\mathbb{R}^n)$.
\end{theorem}

\begin{proof}
\textbf{Step 1.} By \cite[Lemmas 2.3 and 2.4]{XL}, we know that $u,v$ are bounded.
\smallskip

\noindent\textbf{Step 2.} Moreover, we claim that $u,v \in C^{0,1}(\mathbb{R}^n)$.
We use the regularity lifting lemma (Lemma \ref{lem2.2}) to prove this claim.
Let $X=L^{\infty}(\mathbb{R}^n) \times L^{\infty}(\mathbb{R}^n)$ and 
$Y=C^{0,1}(\mathbb{R}^n) \times C^{0,1}(\mathbb{R}^n)$
with the norms
\begin{gather*}
\|(f,g)\|_X= \|f\|_\infty+\|g\|_\infty\,, \\
\|(f,g)\|_Y= \|f\|_{0,1}+\|g\|_{0,1}.
\end{gather*}
Define their closed subset
\begin{gather*}
X_1=\{(f,g) \in X;\|f\|_\infty+\|g\|_\infty \leq C(\|u\|_{\infty}+\|v\|_\infty) \}, \\
Y_1=\{(f,g) \in Y;\|f\|_\infty+\|g\|_\infty \leq C(\|u\|_{\infty}+\|v\|_\infty) \}.
\end{gather*}
Let $d>0$. Set
\begin{gather*}
T_1(f,g)=\sqrt{p}\int_{B_d(x)} \frac{f^{p-1}(y)g(y)dy}{|x-y|^{n-\alpha}}, \\
T_2(f)=\sqrt{p}\int_{B_d(x)}\frac{f^p(y)dy}{|x-y|^{n-\alpha}}, \\
F(x)=\sqrt{p}\int_{\mathbb{R}^n\setminus B_d(x)} 
\frac{u^{p-1}(y)v(y)dy}{|x-y|^{n-\alpha}}, \\
G(x)=\sqrt{p}\int_{\mathbb{R}^n\setminus B_d(x)}\frac{u^p(y)dy}{|x-y|^{n-\alpha}},
\end{gather*}
and $T(f,g)=(T_1(f,g),T_2(f)).$ Then $(u,v)$ solves the operator equation
$$
(f,g)=T(f,g)+(F,G).
$$
\smallskip

\noindent\textbf{Claim 1.} $T$ is a contracting map from $X_1$ to $X$.
In fact, for two functions $(f_1,g_1),(f_2,g_2)\in X_1$, we deduce that
\begin{align*}
&\|T_1(f_1,g_1)-T_1(f_2,g_2)\|_\infty \\
&\leq C(\| \int_{B_d(x)} \frac{|g_1(f_1^{p-1}-f_2^{p-1})|}{|x-y|^{n-\alpha}}dy
\|_\infty\\
&\quad +\| \int_{B_d(x)} \frac{|(g_1-g_2)f_2^{p-1}|}{|x-y|^{n-\alpha}}dy\|_\infty).
\end{align*}
By the mean value theorem and noting the definition of $X_1$, we obtain
\begin{align*}
& \|T_1(f_1,g_1)-T_1(f_2,g_2)\|_\infty \\
&\leq Cd^\alpha
(\|u\|_\infty+\|v\|_\infty)^{p-1}[\|g_1-g_2\|_\infty+\|f_1-f_2\|_\infty].
\end{align*}
Similarly, we obtain
$$
\|T_2(f_1)-T_2(f_2)\|_\infty \leq Cd^{\alpha}
(\|u\|_\infty+\|v\|_\infty)^{p-1}\|f_1-f_2\|_\infty.
$$
Choose $d$ sufficiently small such that 
$C(\|u\|_\infty+\|v\|_\infty)^{p-1}d^{\alpha}<1$, then $T$ is a
contracting map.
\smallskip

\noindent\textbf{Claim 2.} 
$T$ is a shrinking map from $Y_1$ to $Y$.
In fact, for $(f,g) \in Y_1$ and for any $x_1,x_2\in \mathbb{R}^n$, we have
\begin{equation} \label{4.2}
\begin{aligned}
& |T_1(f,g)(x_1)-T_1(f,g)(x_2)|\\
&\leq C| \int_{B_d(0)} |y|^{\alpha-n}
((gf^{p-1})(y+x_1)-(gf^{p-1})(y+x_2))dy|\\
&\leq C d^\alpha (\|u\|_\infty+\|v\|_\infty)^{p-1}
(\|f\|_{0,1}+\|g\|_{0,1})|x_1-x_2|.
\end{aligned}
\end{equation}
Choosing $d$ sufficiently small, we have
$$
\frac{|T_1(f,g)(x_1)-T_1(f,g)(x_2)|}{|x_1-x_2|} \leq
\frac{1}{3}(\|f\|_{0,1}+\|g\|_{0,1}).
$$
Similarly, we  deduce that
$$
\frac{|T_2(f)(x_1)-T_2(f)(x_2)|}{|x_1-x_2|} 
\leq Cd^{\alpha}(\|u\|_\infty+\|v\|_\infty)^{p-1}\|f\|_{0,1}
\leq \frac{1}{3}\|f\|_{0,1}.
$$
Thus, $T$ is a shrinking map.
\smallskip

\noindent\textbf{Claim 3.} $(F,G) \in X_1 \cap Y_1$.
First, \eqref{IE} and
the definitions of $F$ and $G$ imply $F \leq u$ and $G \leq v$.
So $(F,G) \in X_1$.

Next, for any $x_1,x_2 \in \mathbb{R}^n$ satisfying 
$|x_1-x_2|:=\delta<d/3$,
we have
\begin{align*}
&|F(x_2)-F(x_1)|/\sqrt{p}\\
&\leq  \int_{\mathbb{R}^n\setminus B_d(x_1)}||x_2-y|^{\alpha-n}
-|x_1-y|^{\alpha-n}|u^{p-1}(y)v(y)dy\\
&\quad + \int_{B_d(x_1)\setminus
B_{d-\delta}(x_1)}|x_2-y|^{\alpha-n}u^{p-1}(y)v(y)dy\\
&:=I_1+I_2.
\end{align*}
Using the mean value theorem and the integrability, we obtain
$$
I_1 \leq C\|u\|_s^{p-1}\|v\|_\infty \Big(\int_d^\infty
r^{n-t(n-\alpha+1)}\frac{dr}{r}\Big)^{1/t}  |x_1-x_2| \leq C\delta,
$$
where $\frac{p-1}{s}+\frac{1}{t}=1$ with
$s=\frac{n+\epsilon}{n-\alpha}$. Here $\epsilon>0$ is 
suitably small such that $n<t(n-\alpha+1)$. On the other hand,
$$
I_2 \leq C\|u\|_\infty^{p-1}\|v\|_\infty
\int_{B_d(x_1)\setminus B_{d-\delta}(x_1)}|x_2-y|^{\alpha-n}dy \leq C\delta.
$$
Combining the estimates of $I_1$ and $I_2$, we see $F \in C^{0,1}(\mathbb{R}^n)$.

Finally, we prove $G \in C^{0,1}(\mathbb{R}^n)$.
Interchanging the order of integration, we obtain
\begin{align*}
G(x)&=\sqrt{p}(n-\alpha)( \int_d^1\frac{\int_{B_t(x)}u^p(y)dy}{t^{n-\alpha}}
\frac{dt}{t}+\int_1^\infty
\frac{\int_{B_t(x)}u^p(y)dy}{t^{n-\alpha}} \frac{dt}{t})\\
&:=\sqrt{p}(n-\alpha)[G_1(x)+G_2(x)].
\end{align*}
For any $x_1,x_2 \in \mathbb{R}^n$ satisfying $|x_1-x_2|:=\delta<1/3$,
 by scaling we obtain
$$
G_2(x_2) \leq \int_1^\infty
\frac{\int_{B_{t+\delta}(x_1)}u^p(y)dy}{t^{n-\alpha}} \frac{dt}{t}
\leq G_2(x_1)(1+\delta)^{n-\alpha+1}.
$$
Therefore,
$|G_2(x_2)-G_2(x_1)| \leq G_2(x_1)[(1+\delta)^{n-\alpha+1}-1] \leq C\delta$.
In addition,
$$
|G_1(x_2)-G_1(x_1)| 
\leq C\int_d^1 \frac{\int_{B_{t+\delta}(x_1)\setminus B_t(x_1)}u^p(y)dy}{t^{n-\alpha}}
\frac{dt}{t}
\leq C\|u\|_\infty^p\delta.
$$
Thus, we deduce $G \in C^{0,1}(\mathbb{R}^n)$.
Hence, $(F,G) \in Y$. Claim 3 is complete.
\smallskip

\noindent\textbf{Claim 4.} $T(\cdot,\cdot)+(F,G)$ is a map from $X_1 \cap Y_1$ 
to itself.
In fact, for $(f,g) \in X_1 \cap Y_1$,
\begin{equation}
\begin{aligned}
\|T(f,g)\|_\infty
&=\|T_1(f,g)\|_\infty+\|T_2(f)\|_\infty\\
&\leq C(\|u\|_\infty+\|v\|_\infty)^p d^\alpha.
\end{aligned}\label{4.3}
\end{equation}
Similar to \eqref{4.2}, we  have
$$
\|T(f,g)\|_{0,1}=\|T_1(f,g)\|_{0,1}+\|T_2(f)\|_{0,1} \leq C.
$$
Thus, $T(f,g) \in X \cap Y$.

In addition, \eqref{4.3} implies $\|T(f,g)\|_\infty \leq
\|u\|_\infty+\|v\|_\infty$ as long as $d$ is chosen suitably
small. Thus,
$$
\|T(f,g)+(F,G)\|_{\infty} \leq \|T(f,g)\|_\infty+\|(F,G)\|_\infty
\leq C(\|u\|_\infty+\|v\|_\infty).
$$
Claim 4 is verified.

Since $(u,v)$ solves $(f,g)=T(f,g)+(F,G)$, claims 1-4 lead to 
$u,v \in C^{0,1}(\mathbb{R}^n)$ by Lemma \ref{lem2.2}.
\smallskip

\noindent\textbf{Step 3.} We claim that $u \in C^1(\mathbb{R}^n)$.
We use the classical potential estimation to
verify $u \in C^1(\mathbb{R}^n)$ and $\nabla u$ can be expressed formally as
\begin{equation}\label{xiaonei}
\nabla u(x)=(\alpha-n)\int_{\mathbb{R}^n}u^{p-1}(y)v(y)
\frac{x-y}{|x-y|^{n-\alpha+2}}dy.
\end{equation}
Write
\begin{gather*}
J_1=(\alpha-n) \int_{\mathbb{R}^n\setminus B_d(x)}u^{p-1}(y)v(y)
\frac{x-y}{|x-y|^{n-\alpha+2}}dy\\
J_2= \int_{B_d(x)\setminus B_\varepsilon(x)}
u^{p-1}(y)v(y)\nabla (|x-y|^{\alpha-n})dy.
\end{gather*}
We claim that the improper integral $J_1$ converges uniformly about $x$.
In fact,
\begin{align*}
|J_1| 
&\leq C \int_{\mathbb{R}^n\setminus B_d(x)}\frac{u^{p-1}(y)v(y)dy}{|x-y|^{n-\alpha+1}}
 \\
&\leq C\|u\|_s^{p-1}\|v\|_\infty( \int_d^\infty \rho^{n-(n-\alpha+1)t}
\frac{d\rho}{\rho})^{1/t},
\end{align*}
where $\frac{p-1}{s}+\frac{1}{t}=1$. Let $s=\frac{n+\delta}{n-\alpha}$.
Here $\delta>0$ is sufficiently small such that
$\frac{1}{t}<\frac{n-\alpha+1}{n}$. Thus, from the integrability it follows 
$J_1<\infty$.

Clearly,
\begin{align*}
|J_2| 
&\leq \int_{B_d(x)\setminus B_\varepsilon(x)}
\frac{|u^{p-1}(y)v(y)-u^{p-1}(x)v(x)|}{|x-y|^{n-\alpha+1}}dy\\
&\quad +u^{p-1}(x)v(x)| \int_{B_d(x)\setminus
B_\varepsilon(x)}\nabla (|x-y|^{\alpha-n})dy|\\
&:=J_{21}+J_{22}.
\end{align*}
In view of $u,v \in C^{0,1}(\mathbb{R}^n)$,
\[
J_{21} \leq C(\|u^{p-1}\|_\infty\|v\|_{0,1}+
\|u^{p-2}\|_\infty\|v\|_\infty\|u\|_{0,1})
 \int_{B_d(x)\setminus B_\varepsilon(x)}
\frac{|x-y|dy}{|x-y|^{n-\alpha+1}}<\infty.
\]
On the other hand, integration by parts yields
$$
J_{22} \leq C\|u\|_\infty^{p-1}\|v\|_\infty|
\int_{\partial (B_d(x)\setminus B_\varepsilon(x))}
|x-y|^{\alpha-n}ds|<\infty
$$
as long as $\alpha>1$. Hence, $J_\varepsilon$ is convergent uniformly 
about $x$ when $\varepsilon \to 0$.

Combining the estimates of $J_1$ and $J_2$, we know that
\eqref{xiaonei} makes sense, and hence $u \in C^1(\mathbb{R}^n)$.
\end{proof}

\subsection*{Acknowledgements} 
This research was supported by the NSF (No. 11471164) of China.

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