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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 161, pp. 1--21.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/161\hfil Regularity and  symmetry of positive solutions]
{Regularity and  symmetry of positive solutions to nonlinear
 integral systems}

\author[W. Yao, X. Chen, J. Yan \hfil EJDE-2011/161\hfilneg]
{Wanghe Yao, Xiaoli Chen, Jianfu Yang} 

\address{Wanghe Yao \newline
Department  of Mathematics,
Jiangxi Normal University\\
Nanchang, Jiangxi 330022, China}
\email{yaowanghe198610@sina.com}

\address{Xiaoli Chen \newline
Department of Mathematics,
Jiangxi Normal University\\
Nanchang, Jiangxi 330022, China}
\email{littleli\_chen@163.com}

\address{Jianfu Yang \newline
Department of Mathematics,
Jiangxi Normal University\\
Nanchang, Jiangxi 330022, China}
\email{jfyang\_2000@yahoo.com}

\thanks{Submitted July 9, 2011. Published December 7, 2011.}
\subjclass[2000]{35J25, 47G30, 35B45, 35J70}
\keywords{$L^\infty$ bound; H\"{o}lder continuous;
radial symmetry; fractional Laplacian}

\begin{abstract}
 In this article, we consider the regularity and symmetry of
 positive solutions to the nonlinear integral system
 \[
  u(x)=\int_{\mathbb{R}^n}K_{\alpha}(x-y)\frac{v(y)^q}{|y|^\beta}\,dy,
 \quad
  v(x)=\int_{\mathbb{R}^n}K_{\alpha}(x-y)\frac{u(y)^p}{|y|^\beta}\,dy
 \]
 for $x\in \mathbb{R}^n$, where $K_\alpha(x)$ is the kernel of
 the operator $(- \Delta)^{\alpha}+ id$ of order
 $\alpha$, with $0\leq \beta<2\alpha<n$,
 $1<p$, $q<(n-\beta)/\beta$ and
 \[
 \frac{1}{p+1}+\frac{1}{q+1}>\frac{n-2\alpha+\beta}{n}.
 \]
 We show that positive solution pairs
 $(u,v)\in L^{p+1}(\mathbb{R}^n)\times L^{q+1}(\mathbb{R}^n)$
 are locally H\"older continuous in $\mathbb{R}^N\setminus\{0\}$,
 radially symmetric about the origin, and strictly decreasing.
\end{abstract}

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

\section{Introduction}

In this article, we consider the regularity and symmetry of positive
solutions to the nonlinear integral system
 \begin{equation}\label{eq:1.1}
  u(x)=\int_{\mathbb{R}^n}K_{\alpha}(x-y)\frac{v(y)^q}{|y|^\beta}\,dy,
 \quad
  v(x)=\int_{\mathbb{R}^n}K_{\alpha}(x-y)\frac{u(y)^p}{|y|^\beta}\,dy
  \end{equation}
for $x\in \mathbb{R}^n$, where $K_\alpha(x)$ is the kernel of the
operator $(-\Delta)^\alpha+id$, $0<\alpha<1$,
$0\leq \beta<2\alpha<n$, $1<p$, $q<\frac{n-\beta}{\beta}$ and
\begin{equation}\label{eq:1.2}
\frac{1}{p+1}+\frac{1}{q+1}>\frac{n-2\alpha+\beta}{n}.
\end{equation}
It can be shown that problem \eqref{eq:1.1} is actually equivalent
to the indefinite fractional elliptic systems
\begin{equation}\label{eq:1.3}
{(- \Delta)}^{\alpha}u+ u=\frac{v^q}{|y|^\beta}, \quad
{(- \Delta)}^{\alpha}v + v=\frac{u^p}{|y|^\beta}, \quad\text{in }
 \mathbb{R}^n.
\end{equation}
If $p=q$ and $\beta = 0$, problem \eqref{eq:1.3} is of
particular interest in fractional quantum mechanics in the study of
particles on stochastic fields modelled by L\'evy processes.
A path integral over the L\'evy flights paths and a fractional
Schr\"odinger equation of fractional quantum mechanics are
formulated by Laskin \cite{L}, see also \cite{L1}.
It was shown in \cite{FQT} that in the case $p=q$ and $\beta = 0$,
problem \eqref{eq:1.3} has at least a positive classical solution,
which is radially symmetric and decays at infinity.

On the other hand, the problem
\begin{equation}\label{eq:1.4}
{(- \Delta)}^{\alpha/2}u=v^q, \quad
{(- \Delta)}^{\alpha/2}v=u^p, \quad \text{in }\mathbb{R}^n
\end{equation}
and its generalization have recently been extensively investigated
in \cite{CJL,CL,CL2,CL3,CLO,CLO1} etc. Such a problem is equivalent
to the integral system
 \begin{equation}\label{eq:1.5}
  u(x)=\int_{\mathbb{R}^n}\frac{v(y)^q}{|x-y|^{n-\alpha}}\,dy,\quad
  v(x)=\int_{\mathbb{R}^n}\frac{u(y)^p}{|x-y|^{n-\alpha}}\,dy, \quad
\text{in }\mathbb{R}^n.
  \end{equation}
Solutions $(u,v)$ of \eqref{eq:1.5} are critical points of
the functional associated with the well-known Hardy-Littlewood-Sobolev
inequality, which is precisely stated as follows.

 \begin{proposition}\label{prop:1.1}
Let $0<\lambda<n$ and let $1<p,q<\infty$ such that
$\frac{1}{p}+\frac{1}{q}+\frac{\lambda}{n}=2$. Then
\[\big|\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\frac{f(x)g(y)}{
|x-y|^\lambda}\,dx\,dy\big|\leq C_{q,\lambda,n}\|f\|_{p}\|g\|_{q}\] for
$f\in L^p(\mathbb{R}^n)$ and $g\in L^q(\mathbb{R}^n)$.
\end{proposition}

Regularity and symmetry as well as classification of solutions of
\eqref{eq:1.5} and its generalization have been widely considered,
see \cite{CL} and references therein. The Hardy-Littlewood-Sobolev
inequality plays a key role in the study of these properties.
Meanwhile, the moving plane method and the regularity
lifting method for integral equations have been developed, see also \cite{CL} and
references therein. Furthermore, the double weighted
Hardy-Littlewood-Sobolev inequality, was introduced in \cite{SW},
which is stated as follows.

\begin{proposition}\label{prop:2.2}
Let $0<\lambda<n$, $1<p<\infty,\tau<\frac{n}{p'},\beta<\frac
nq,\tau+\beta\geq0$, $\frac1q + \frac 1{q'} = 1$ and $\frac
1{p}+\frac{1}{q'}+\frac{\lambda+\tau+\beta}{n}=2$. If $p\leq
q<\infty$ and $f\in L^p(\mathbb{R}^n),g\in L^{q'}(\mathbb{R}^n)$,
Then, there exists a constant $C$ independent of $f$ and $g$ such
that the following inequality holds
\begin{equation}\label{eq:1.6}
\big|\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\frac{f(x)g(y)}{|x|^\tau
|x-y|^\lambda|y|^\beta}dx\,dy\big|\leq C\|f\|_{p}\|g\|_{q'}.
\end{equation}
\end{proposition}

Critical points of the functional associated with inequality
\eqref{eq:1.6} will yield solutions of \eqref{eq:1.4} if
$\tau =\beta= 0$.
Essentially, problem \eqref{eq:1.4} is related to the Riesz
potentials $\mathcal{I}_\alpha(f) = (-\Delta)^{-\alpha}$,
$0<\alpha<\frac{n}{2}$, which is defined by
\[
\mathcal{I}_\alpha(f)(x) = \frac 1{C(\alpha)}\int_{\mathbb{R}^n}
\frac{f(y)}{|x-y|^{n-2\alpha}}\,dy
\]
for some $C(\alpha)>0$. It is known that
\[
\|\mathcal{I}_{\alpha}f\|_{q}\leq C\|f\|_{p},\quad\text{where}\quad
\frac 1q=\frac 1p-\frac{2\alpha}n.
\]
In \cite{CY}, the authors studied the regularity and radial symmetry
of solutions of
\begin{equation}\label{eq:1.6a}
  u(x)=\int_{\mathbb{R}^n}G_{\alpha}(x-y)\frac{v(y)^q}{|y|^\beta}\,dy,\quad
  v(x)=\int_{\mathbb{R}^n}G_{\alpha}(x-y)\frac{u(y)^p}{|y|^\beta}\,dy,
  \end{equation}
where $G_\alpha$ is the Bessel kernel; that is,
\[
G_\alpha(x)=\frac{(\sqrt{2\pi})^{-n}}{\Gamma(\frac{n}{2})}
 \int_{0}^\infty e^{-s}e^{-|x|^2/(4s)}s^{(\alpha-n)/2}
\frac{ds}{s},
\]
which is associated with the operator
$(-\Delta + id)^{\frac \alpha 2}$. While problem \eqref{eq:1.1}
is connected with the kernel
$K_\alpha$ of the operator $(-\Delta )^\alpha+id$, such an operator
enjoys different features, for instant, it is not clear whether the
ground state solution of
\[
{(- \Delta)}^{\alpha}u+ u= u^p,  \quad \text{in }\mathbb{R}^n
\]
is exponentially decaying at infinity, see \cite{FQT} for further
properties of the operator $(-\Delta )^\alpha+id$ and results for
 one equation case. We will consider in this paper the regularity and radial symmetry of positive solutions of \eqref{eq:1.1},
 which involves in Hardy type weights. To this purpose, we first establish the following Hardy-Littlewood-Sobolev inequality for
 the potential $K_\alpha$ with double weights.

\begin{theorem}\label{thm:1.1}
Let $0<\alpha<1$, $1<p,q<\frac n{2\alpha}$, $\tau,\beta\ge 0$.
In addition $n(1-\frac 1p-\frac 1q+\frac{2\alpha}n)>\beta+\tau>n(1-\frac 1p-\frac
1q)$. Then, there exists a constant $C$, independent of $f\in
L^p(\mathbb{R}^n)$ and $g\in L^q(\mathbb{R}^n)$, such that the
following inequality holds
\begin{equation}\label{eq:1.8}
\big|\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\frac{f(x)K_\alpha(x-y)h(y)}{|x|^\tau|y|^\beta}dx\,dy\big|\leq
C\|f\|_{p}\|h\|_{q}.\end{equation}
Furthermore, let
$$
Th(x)=\int_{\mathbb{R}^n}\frac{K_\alpha(x-y)h(y)}{|x|^\tau
|y|^\beta}dy,
$$
then
\[
\|Th\|_{{p'}}=\sup_{\|f\|_{p}=1}|\langle Th,f\rangle|\leq
C\|h\|_{q},
\]
where $\frac 1p+\frac 1{p'}=1,\,\,1+\frac 1{p'}\ge\frac
1q+\frac{n-2\alpha+\beta+\tau}{n}$ and $h\in L^q(\mathbb{R}^n)$.
\end{theorem}

Next, we use Theorem \ref{thm:1.1} to investigate properties of
positive solutions of \eqref{eq:1.1}.
In the following, we always assume $1<p,\ q<\frac{n-\beta}{\beta}$
and that \eqref{eq:1.2} holds.  We have the following result.

\begin{theorem}\label{thm:1.2}
If $(u,v)\in L^{p+1}(\mathbb{R}^n)\times L^{q+1}(\mathbb{R}^n) $ is
a solution pair of \eqref{eq:1.1}, then
$(u,v)\in L^\infty(\mathbb{R}^n)\times L^\infty(\mathbb{R}^n) $.
\end{theorem}


Results in Theorem \ref{thm:1.2} hold also for sign-changing
solutions of \eqref{eq:1.1}. In the proof of Theorem \ref{thm:1.2},
we first lift the integrability of a suitable cut-off function of
the solution by the regularity lifting method to some $L^{q_0}$,
and then we show that the solution is actually in $L^\infty$.
From Theorem \ref{thm:1.2}, one may expect the solution
to be smooth. Our next result asserts that the solution is
locally H\"{o}lder continuous. Precisely, let
$\gamma = 1-\frac \beta n$,
under the conditions in Theorem \ref{thm:1.2}, we have the following
result.

\begin{theorem}\label{thm:1.3}
$ u,v\in C^{0,\gamma}_{\rm loc}(\mathbb{R}^n\backslash\{0\})$.
\end{theorem}

Furthermore, we show that the solution is radially symmetric by
the moving plane method.

\begin{theorem}\label{thm:1.4}
Both $u$ and $v$ are  radially symmetric and
strictly decreasing about the origin.
\end{theorem}

In Section 2, we establish the weighted Hardy-Littlewood-Sobolev
inequality related to the kernel $K_\alpha$. Then, using the
inequality, we prove Theorem \ref{thm:1.2} in section 3.
In section 4, we prove Theorem \ref{thm:1.3} by Theorem
\ref{thm:1.1} and the regularity lifting method.
Theorem \ref{thm:1.4} is shown in section 5 by the moving
plane method.

\section{Hardy-Littlewood-Sobolev inequality for the Bessel potential}

In this section, we establish a weighted Hardy-Littlewood-Sobolev
inequality for the potential $K_\alpha$. Let $\alpha \in(0,1)$,
the kernel $K_\alpha$ associated with the operator
$(-\Delta)^\alpha+id$ is defined as
\[
K_\alpha(x) = \mathcal{F}^{-1}\big(\frac 1{1+|\xi|^\alpha}\big),
\]
where $ \mathcal{F}^{-1}$ is the inverse Fourier transformation. It
is known from \cite{FQT} that the kernel $K_\alpha$ is radially
symmetric, non-negative and non-increasing in $r = |x|$.
Furthermore, for appropriate constants $C_1$ and $C_2$, the kernel
$K_\alpha$ satisfies
\begin{equation} \label{eq:2.1}
K_\alpha(x)\leq\begin{cases}
 C_1|x|^{-n+2\alpha}&\text{when } |x|\leq 1,\\
 C_2|x|^{-n-2\alpha}&\text{when } |x|\ge 1.
\end{cases}
\end{equation}

\begin{proof}[Proof of Theorem \ref{thm:1.1}]
By \eqref{eq:2.1}, we have
\begin{equation} \label{eq:2.2}
\begin{split}
&\big|\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}
 \frac{f(x)K_\alpha(x-y)h(y)}{|x|^\tau|y|^\beta}\,dy\,dx\big|\\
&=\big|\int_{\mathbb{R}^n}\int_{\{y:|x-y|\ge1\}}
 \frac{f(x)K_\alpha(x-y)h(y)}{|x|^\tau|y|^\beta}\,dy\,dx\\
&+\int_{\mathbb{R}^n}\int_{\{y:|x-y|\le1\}}
 \frac{f(x)K_\alpha(x-y)h(y)}{|x|^\tau|y|^\beta}\,dy\,dx\big|\\
&\leq C\int_{\mathbb{R}^n}\int_{\{y:|x-y|\ge1\}}
 \frac{|f(x)||h(y)|}{|x|^\tau|x-y|^{n+2\alpha}|y|^\beta}\,dy\,dx\\
&\quad +C\int_{\mathbb{R}^n}\int_{\{y:|x-y|\le1\}}
 \frac{|f(x)||h(y)|}{|x|^\tau|x-y|^{n-2\alpha}|y|^\beta}\,dy\,dx
:=C(I+J).
\end{split}
\end{equation}
Firstly, we estimate $I$. We write
\begin{equation}\label{eq:2.3}
\begin{split}
I&=\int_{\{x:|x|\le\frac{1}{2}\}}\int_{\{y:|x-y|\ge1\}}
 \frac{|f(x)||h(y)|}{|x|^\tau|x-y|^{n+2\alpha}|y|^\beta}\,dy\,dx\\
&\quad+\int_{\{x:|x|\ge\frac{1}{2}\}}\int_{\{y:|x-y|\ge1\}}
 \frac{|f(x)||h(y)|}{|x|^\tau|x-y|^{n+2\alpha}|y|^\beta}\,dy\,dx
:=I_1+I_2.
\end{split}
\end{equation}
Notice that if $|x|\le\frac{1}{2},|y-x|\ge1$,
then $|y|\ge\frac{1}{2}$. While the
function $|x|^{-n-2\alpha}$ is in $L^r(R^n\backslash B_1(0))$
for all $\frac n{n+2\alpha}<r<\infty$, we have by
Young's inequality that
\[
\|(|\cdot|^{-n-2\alpha}\chi_{\{|\cdot|\geq1\}})*g\|_{s}
\leq C\|g\|_{q}\Big(\int_{\{x:|x|\geq1\}}
\frac{1}{|x|^{(n+2\alpha)r}}dx\Big)^{1/r},
\]
where $1+\frac{1}{s}=\frac{1}{r}+\frac 1q$.
Choosing in Young's inequality that $r=n/(n-2\alpha)$, then
$s=nq/(n-2\alpha q)$, we obtain
\begin{align*}
I_1
&\leq \int_{\{x:|x|\le\frac{1}{2}\}}\frac{|f(x)|}{|x|^{\tau+\beta}}\,dx
 \Big(\int_{\{y:|y-x|\ge1\}}\frac{|h(y)|}{|x-y|^{n+2\alpha}}dy\Big)\,dx\\
&\leq \|f\|_{p}\||\cdot|^{-n-2\alpha}\chi_{\{|\cdot|\geq1\}}*|h|
 \|_{\frac{nq}{n-2\alpha q}}
 \Big(\int_{\{x:|x|\leq \frac{1}{2}\}}|x|^{-\frac{\tau+\beta}
 {1-\frac 1p-\frac 1q+\frac{2\alpha}{n}}}\,dx\Big)
 ^{1-\frac 1p-\frac 1q+\frac{2\alpha}{n}}\\
&\leq C\|f\|_{p}\|h\|_{q}\Big(\int_{\{x:|x|\geq1\}}
\frac{1}{|x|^{(n+2\alpha)r}}\,dx\Big)^{1/r}\\
&\quad\times \Big(\int_{\{x:|x|\leq 1\}}|x|^{-\frac{\tau+\beta}{1-\frac 1p-\frac
1q+\frac{2\alpha}{n}}}\,dx\Big)^{1-\frac 1p-\frac
1q+\frac{2\alpha}{n}}.
\end{align*}
Since $n-{\frac{\tau+\beta}{1-\frac 1p-\frac 1q+\frac{2\alpha}{n}}}>0$
if and only if $\beta+\tau< n(1-\frac 1p-\frac 1q+\frac{2\alpha}{n})$,
it yields
\[
I_1\leq C\|f\|_{p}\|h\|_{q}.
\]
We decompose  $I_2$ as follows.
\begin{equation} \label{eq:2.4}
\begin{split}
I_2&= \int_{\{x:|x|\ge\frac{1}{2}\}}\int_{\{y:|x-y|\ge1,|y|\ge
\frac{1}{2}\}}\frac{|f(x)||h(y)|}{|x|^\tau|x-y|^{n+2\alpha}|y|^\beta}
 \,dy\,dx\\
&\quad+\int_{\{x:|x|\ge\frac{1}{2}\}}\int_{\{y:|x-y|\ge1,|y|
 \le\frac{1}{2}\}}\frac{|f(x)||h(y)|}{|x|^\tau|x-y|^{n+2\alpha}
 |y|^\beta}\,dy\,dx\\
&:= I_2^1+I_2^2.
\end{split}
\end{equation}
Furthermore,
\begin{equation} \label{eq:2.5}
\begin{split}
 I_2^1&= \int_{\{x:|x|\ge\frac{1}{2}\}}\int_{\{y:|x-y|\ge1,|y|\ge
\frac{1}{2},|y|\geq|x|\}}\frac{|f(x)||h(y)|}{|x|^\tau|x-y|^{n+2\alpha}
 |y|^\beta}\,dy\,dx\\
&\quad+\int_{\{x:|x|\ge\frac{1}{2}\}}\int_{\{y:|x-y|\ge1,|y|
 \ge\frac{1}{2},|y|\leq|x|\}}\frac{|f(x)||h(y)|}{|x|^\tau|x-y|
 ^{n+2\alpha}|y|^\beta}\,dy\,dx\\
&:= I_2^{11}+I_2^{12}.
\end{split}
\end{equation}
Now we estimate $I_2^{11}$ and $I_2^{12}$, respectively. We deduce
by Young's inequality that
\begin{align*}
I_2^{11}
&\leq \int_{\{x:|x|\ge\frac{1}{2}\}}\frac{|f(x)|}{|x|^{\tau+\beta}}
 \Big(\int_{\{y:|y|\ge\frac{1}{2},|x-y|\geq1\}}
 \frac{|h(y)|}{|x-y|^{n+2\alpha}}\,dy\Big)\,dx\\
&\leq \|f\|_{p}\|(|\cdot|^{-n-2\alpha}\chi_{\{|\cdot|\geq1\}})*|h|\|_{q}
 \Big(\int_{\{x:|x|\ge 1\}}|x|^{-(\tau+\beta)\frac{pq}{pq-p-q}}
 \,dx\Big)^{1-\frac 1p-\frac 1{q}}\\
&\leq C\|f\|_{p}\|h\|_{q}\Big(\int_{\{x:|x|\ge 1\}}|x|
 ^{-(\tau+\beta)\frac{pq}{pq-p-q}}\,dx\Big)^{1-\frac 1p-\frac1{q}}.
\end{align*}
Since $\tau+\beta> n(1-\frac 1p-\frac 1{q})$, we have
$n-(\tau+\beta)\frac{pq}{pq-p-q}<0$. Hence,
\[
I_2^{11}\leq C\|f\|_{p}\|h\|_{q}.
\]
By the Fubini theorem, we find
\begin{align*}
I_2^{12}&\leq \int_{\{x:|x|\ge\frac{1}{2}\}}
 \int_{\{y:|y|\ge\frac{1}{2},|x-y|\geq1\}}
 \frac{|f(x)||h(y)|}{|x-y|^{n+2\alpha}|y|^{\tau+\beta}}|\,dy\,dx\\
&\leq \int_{\{y:|y|\ge\frac{1}{2}\}}\frac{|h(y)|}{|y|^{\tau+\beta}}
 \Big(\int_{\{x:|x-y|\geq1\}}\frac{|f(x)|}{|x-y|^{n+2\alpha}}dx
 \Big)\,dy\\
&\leq \|h\|_{q}\|(|\cdot|^{-n-2\alpha}\chi_{\{|\cdot|\geq1\}})
 *|f|\|_{p}\Big(\int_{\{y:|y|\ge
\frac{1}{2}\}}|y|^{-(\tau+\beta)\frac{pq}{pq-p-q}}\,dy\Big)
 ^{1-\frac 1p-\frac 1{q}}\\
&\leq C\|h\|_{q}\|f\|_{p}\Big(\int_{\{y:|y|\ge
\frac{1}{2}\}}|y|^{-(\tau+\beta)\frac{pq}{pq-p-q}}\,dx\Big)
 ^{1-\frac 1p-\frac 1{q}}.
\end{align*}
In the same way,
$n-(\tau+\beta)\frac{pq}{pq-p-q}<0$ if and only if $\tau+\beta>
n(1-\frac 1p-\frac 1{q})$, and then
\[
I_2^{12}\leq C\|f\|_{p}\|h\|_{q}.
\]
Using the Fubini theorem and Young's inequality, we obtain
\begin{align*}
 I_2^2
&\leq \int_{\{y:|y|\le\frac{1}{2}\}}\frac{|h(y)|}{|y|^{\beta+\tau}}
 \int_{\{x:|x|\ge\frac{1}{2},|x-y|\geq1\}}\frac{|f(x)|}
 {|x-y|^{n+2\alpha}}\,dx\,dy\\
&\leq \|h\|_{q}\|(|\cdot|^{-n-2\alpha}\chi_{\{|\cdot|\geq1\}})
 *|f|\|_{\frac{np}{n-2\alpha p}}
 \Big(\int_{\{y:|y|\leq \frac{1}{2}\}}|y|^{-\frac{\tau+\beta}
 {1-\frac 1p-\frac 1q+\frac{2\alpha}{n}}}\,dy\Big)^{1-\frac
 1p-\frac1{q}+\frac{2\alpha}{n}}\\
&\leq C\|h\|_{q}\|f\|_{p}\Big(\int_{\{x:|x|\geq1\}}
 \frac{1}{|x|^{\frac{(n+2\alpha)n}{n-2\alpha}}}\,dx\Big)
 ^{\frac{n-2\alpha}{n}}\\
&\quad\times \Big(\int_{\{y:|y|\leq \frac{1}{2}\}}|y|^{-\frac{\tau+\beta}
 {1-\frac 1p-\frac 1q+\frac{2\alpha}{n}}}\,dx\Big)^{1-\frac
 1p-\frac1{q}+\frac{2\alpha}{n}}.
\end{align*}
Since $n-\frac{\tau+\beta}{1-\frac 1p-\frac 1q+\frac{2\alpha}{n}}>0$
if and only if $\beta+\tau< n(1-\frac 1p-\frac 1q+\frac{2\alpha}{n})$,
it follows that
\[
I_2^{2}\leq C\|f\|_{p}\|h\|_{q}.
\]

Secondly, we estimate $J$. There holds
\begin{equation}\label{eq:2.6}
\begin{split}
J&=\int_{\{x:|x|\ge2\}}\int_{\{y:|x-y|\le1\}}
 \frac{|f(x)||h(y)|}{|x|^\tau|x-y|^{n-2\alpha}|y|^\beta}\,dy\,dx\\
&\quad+\int_{\{x:|x|\le2\}}\int_{\{y:|x-y|\le1\}}
 \frac{|f(x)|h(y)|}{|x|^\tau|x-y|^{n-2\alpha}|y|^\beta}\,dy\,dx
:=J_1+J_2.
\end{split}
\end{equation}
Note that if $|x|\ge2,|y-x|\le1$, then
$|y|\ge|x|-|x-y|\ge1\ge|x-y|$ and $|x|>|x-y|$. By Young's
inequality,
\begin{equation}\label{eq:2.7}
\begin{split}
J_1
&=\int_{\{x:|x|\ge2\}}f(x)\int_{\{y:|x-y|\le1\}}
 \frac{|h(y)|}{|x-y|^{n-2\alpha+\beta+\tau}}\,dy\,dx\\
&\le\|f\|_p\|h*(\frac{\chi_{\{|\cdot|\le1\}}}{|\cdot|
 ^{n-2\alpha+\beta+\tau}})\|_{p'}\\
&\le\|f\|_p\|h\|_q\Big(\int_{\{y:|y|\le1\}}
 \frac{1}{|y|^{(n-2\alpha+\beta+\tau)l}}\,dy\Big)^{1/l},
\end{split}
\end{equation}
where $\frac{1}{p'}=\frac{1}{q}+\frac{1}{l}-1$, that is
$\frac1l=2-\frac 1p-\frac 1q$. Thus, $n-(n-2\alpha+\beta+\tau)l\ge0$
if and only if  $\beta+\tau<n(1-\frac
1p-\frac1q+\frac{2\alpha}{n})$, it follows that
\[
J_1\leq C\|f\|_p\|h\|_q.
\]

To estimate $J_2$, we decompose  it as follows.
\begin{equation}\label{eq:2.8}
\begin{split}
J_2
&=\int_{\{x:|x|\le2\}}\int_{\{y:|x-y|\le1,|y|\ge
|x|\}}\frac{|f(x)||h(y)|}{|x|^\tau|x-y|^{n-2\alpha}|y|^\beta}\,dy\,dx\\
&\quad+\int_{\{x:|x|\le2\}}\int_{\{y:|x-y|\le1,|y|\le|x|\}}
 \frac{|f(x)||h(y)|}{|x|^\tau|x-y|^{n-2\alpha}|y|^\beta}\,dy\,dx
:=J_2^1+J_2^2.
\end{split}
\end{equation}
Now we estimate $J_2^1$ and $J_2^2$ respectively.
\begin{equation}\label{eq:2.9}
\begin{split}
J_2^1
&\le\int_{\{x:|x|\le2\}}\frac{|f(x)|}{|x|^{\tau+\beta}}
 \Big(\int_{\{y:|x-y|\le1\}}\frac{|h(y)|}{|x-y|^{n-2\alpha}}\,dy\Big)
 \,dx\\
&\le\int_{\{x:|x|\le2\}}\frac{|f(x)|}{|x|^{\tau+\beta}}I_{\alpha}(|h|)(x)\,dx\\
&\le\|f\|_p\|I_{\alpha}(|h|)\|_{\frac{nq}{n-2\alpha
q}}\Big(\int_{\{x:|x|\le2\}}\frac{1}{|x|
 ^{\frac{\tau+\beta}{1-1/p-1/q+2\alpha/n}}}\,dx\Big)
 ^{1-1/p-1/q+2\alpha/n}\\
&\leq C\|f\|_p\|h\|_q\Big(\int_{\{x:|x|\le2\}}\frac{1}{|x|
 ^{\frac{\tau+\beta}{1-1/p-1/q+2\alpha/n}}}\,dx
 \Big)^{1-1/p-1/q+2\alpha/n}.\\
\end{split}
\end{equation}
Since $n-\frac{\tau+\beta}{1-1/p-1/q+2\alpha/n}>0$ if
and only if $\tau+\beta<n(1-1/p-1/q+2\alpha/n)$, it yields
\[
J_2^1\leq C\|f\|_p\|h\|_q.
\]

In the same way,
\begin{equation}\label{eq:2.10}
\begin{split}
J_2^2&\le\int_{\{x:|x|\le2\}}\int_{\{y:|y|\le|x|\}}
 \frac{|f(x)||h(y)|}{|x|^\tau|x-y|^{n-2\alpha}|y|^\beta}\,dy\,dx\\
&\le\int_{\{y:|y|\le2\}}\frac{|h(y)|}{|y|^{\tau+\beta}}
 \Big(\int_{\{x:|x|\ge|y|\}}\frac{|f(x)|}{|x-y|^{n-2\alpha}}\,dx\Big)\,dy\\
&\le\int_{\{y:|y|\le2\}}\frac{|h(y)|}{|y|^{\tau+\beta}}
 I_{\alpha}(|f|)(y)\,dy\\
&\le\|h\|_q\|I_{\alpha}(|f|)\|_{\frac{np}{n-2\alpha
p}}\Big(\int_{\{y:|y|\le2\}}\frac{1}{|y|
 ^{\frac{\tau+\beta}{1-1/p-1/q+2\alpha/n}}}\Big)^{1-1/p-1/q+2\alpha/n}\\
&\leq C\|f\|_p\|h\|_q\Big(\int_{\{y:|y|\le2\}}\frac{1}{|y|
 ^{\frac{\tau+\beta}{1-1/p-1/q+2\alpha/n}}}\Big)^{1-1/p-1/q+2\alpha/n}.
\end{split}
\end{equation}
The inequality $\tau+\beta<n(1-1/p-1/q+2\alpha/n)$ implies
\[
J_2^2\leq C\|f\|_p\|h\|_q.
\]
The proof is complete.
\end{proof}

\section{$L^\infty$-bound of solutions}




In this section, we show that any solution of \eqref{eq:1.1}
in $L^{p+1}(\mathbb{R}^n)\times
L^{q+1}(\mathbb{R}^n)$ actually belongs  to
$L^\infty(\mathbb{R}^n)\times L^\infty(\mathbb{R}^n)$.
To this purpose, we will use the
regularity lifting method developed in \cite{CL}, which we will
state as follows.

Let $\mathbf{Z}$ be a given vector space, $\|\cdot\|_{\mathbf{X}}$
and $\|\cdot\|_{\mathbf{Y}}$ be two norms on $\mathbf{Z}$. Define a
new norm $\|\cdot\|_{\mathbf{Z}}$ by
\[\|\cdot\|_{\mathbf{Z}}=\sqrt[p]{\|\cdot\|_{\mathbf{X}}^p+\|\cdot\|_{\mathbf{Y}}^p}.\]
Suppose that $\mathbf{Z}$ is complete with respect to the norm
$\|\cdot\|_{\mathbf{Z}}$. Let $\mathbf{X}$ and $\mathbf{Y}$ be the
completion under $\|\cdot\|_{\mathbf{X}}$ and
$\|\cdot\|_{\mathbf{Y}}$, respectively. Here one can choose $p$ such that $1\leq
p\le\infty$. According to what one needs, it is easy to see that
$\mathbf{Z}=\mathbf{X}\cap\mathbf{Y}$. The following regularity
lifting theorem was obtained in \cite{CL}.

\begin{lemma}[Regularity Lifting I] \label{lem:3.1}
 Let $T$ be a contracting map from $\mathbf{X}$ into itself and from
 $\mathbf{Y}$  into itself. Assume that $f\in\mathbf{X}$ and that
 there exists a function $g\in \mathbf{Z}$ such that $f=Tf+g$, then
 $f$ also belongs to $\mathbf{Z}$.
\end{lemma}

\begin{proof}[Proof of Theorem \ref{thm:1.2}]
 Let $(u,v)\in L^{p+1}(\mathbb{R}^n)\times L^{q+1}(\mathbb{R}^n)$
be a pair of solution to integral systems \eqref{eq:1.1}.
We first show by Lemma \ref{lem:3.1} that $(u_\xi,v_\xi)$,
a cut-off function of $(u,v)$
defined below, belongs to $L^{\tilde p}(\mathbb{R}^n)\times
L^{\tilde q}(\mathbb{R}^n)$ for $\tilde p>\frac{np}{2\alpha-\beta},
\,\tilde q> \frac{nq}{2\alpha-\beta}$ and
$\frac{1}{\tilde{p}}-\frac{1}{\tilde{q}}=\frac{1}{p+1}-\frac{1}{q+1}$,
 then we prove that $(u,v)\in L^{\infty}(\mathbb{R}^n)\times
L^{\infty}(\mathbb{R}^n)$.

For any sufficient large positive real number $\xi$, define
\begin{equation}\label{eq:3.1}
u_\xi(x)= \begin{cases}
u(x),& \text{if } |u(x)|\ge \xi \text{ or }  |x|>\xi,\\
u_\xi(x)=0,& \text{otherwise}.
 \end{cases}
\end{equation}
Similarly, we define $v_\xi$. Let
\[
T_1^\xi g(x)=\int_{\mathbb{R}^n}\frac{K_\alpha(x-y)
|v_\xi|^{q-1}g(y)}{|y|^{\beta}}\,dy,\quad
T_2^\xi f(x)=\int_{\mathbb{R}^n}\frac{K_\alpha(x-y)
|u_\xi|^{p-1}f(y)}{|y|^{\beta}}\,dy
\]
and \[
T_\xi(f,g)=(T_1^\xi g, T_2^\xi f).
\]
Let $\tilde u_\xi(x)=u(x)- u_\xi(x)$, and
$E_\xi^u=\{x\in \mathbb{R}^n:|u(x)|\ge \xi  {\rm\ or\ }|x|>\xi\}$.
Similarly, we define $\tilde v_\xi$ and $E_\xi^v$.
By \eqref{eq:1.1}, we have
\begin{equation} \label{eq:3.2}
\begin{split}
 u(x)
&= \int_{\mathbb{R}^n}\frac{K_\alpha(x-y)|v(y)|^{q-1}v(y)}
 {|y|^{\beta}}\,dy\\
&= \int_{E_\xi^v}\frac{K_\alpha(x-y)|v(y)|^{q-1}v(y)}{|y|^{\beta}}\,dy
 +\int_{\mathbb{R}^n\backslash E_\xi^v}\frac{K_\alpha(x-y)|v(y)|^{q-1}
 v(y)}{|y|^{\beta}}\,dy\\
&= \int_{\mathbb{R}^n}\frac{K_\alpha(x-y)|v_\xi(y)|^{q-1}
 v_\xi(y)}{|y|^{\beta}}\,dy+\int_{\mathbb{R}^n}\frac{K_\alpha(x-y)|
 \tilde v_\xi(y)|^{q-1}\tilde v_\xi(y)}{|y|^{\beta}}\,dy.
\end{split}
\end{equation}
Moreover,
\begin{equation}\label{eq:3.3}
u_\xi(x)=\int_{\mathbb{R}^n}\frac{K_\alpha(x-y)|
v_\xi(y)|^{q-1}v_\xi(y)}{|y|^{\beta}}\,dy+M_1(x),
\end{equation}
where
\[
M_1(x)=\int_{\mathbb{R}^n}\frac{K_\alpha(x-y)|
 \tilde v_\xi(y)|^{q-1}\tilde v_\xi(y)}{|y|^{\beta}}\,dy
 -\tilde u_\xi(x).
\]
Similarly,
\begin{equation}\label{eq:3.4}
v_\xi(x)=\int_{\mathbb{R}^n}\frac{K_\alpha(x-y)|
u_\xi(y)|^{p-1}u_\xi(y)}{|y|^{\beta}}\,dy+M_2(x),
\end{equation}
where
\[
M_2(x)=\int_{\mathbb{R}^n}\frac{K_\alpha(x-y)|
 \tilde u_\xi(y)|^{p-1}\tilde u_\xi(y)}{|y|^{\beta}}\,dy
 -\tilde v_\xi(x).
\]
It yields
\[
(u_\xi, v_\xi)=T_\xi( u_\xi, v_\xi)+(M_1(x),M_2(x)),
\]
where $T_\xi(u_\xi, v_\xi)=(T_1^\xi  v_\xi,T_2^\xi u_\xi)$.

We claim that $M_1(x),M_2(x)\in
L^\infty(\mathbb{R}^n)\cap L^s(\mathbb{R}^n)$ for $s>1$. Obviously,
$\tilde u_\xi,\tilde v_\xi\in L^\infty(\mathbb{R}^n)\cap L^s(\mathbb{R}^n)$. So
it suffices to show that $H_1,H_2\in L^\infty(\mathbb{R}^n)\cap
L^s(\mathbb{R}^n)$, where
\begin{gather*}
H_1(x)=\int_{\mathbb{R}^n}\frac{K_\alpha(x-y)|
 \tilde v_\xi(y)|^{q-1}\tilde v_\xi(y)}{|y|^{\beta}}\,dy,\\
H_2(x)=\int_{\mathbb{R}^n}\frac{K_\alpha(x-y)|
 \tilde u_\xi(y)|^{p-1}\tilde u_\xi(y)}{|y|^{\beta}}\,dy.
\end{gather*}

Now, we estimate $H_1$, the estimation for $H_2$ can be obtained
in the same way. By the definition of $\tilde v_\xi(x)$, for $x\in
\mathbb{R}^n$, we obtain
\begin{align*}  %\label{eq:3.5}
|H_1(x)|
&\leq C \int_{\{y:|y|\leq \xi\}}\frac{K_\alpha(x-y) }{|y|^{\beta}}\,dy\\
&\leq C\int_{\{y:|y|\leq \xi,|x-y|\ge 1\}}
 \frac{K_\alpha(x-y) }{|y|^{\beta}}\,dy
 + C\int_{\{y:|y|\leq \xi,|x-y|\leq 1\}}\frac{K_\alpha(x-y) }{|y|^{\beta}}\,dy\\
&\leq C\int_{\{y:|y|\leq \xi,|x-y|\ge 1\}}
 \frac{1}{|x-y|^{n+2\alpha}|y|^{\beta}}\,dy\\
&\quad + C \int_{\{y:|y|\leq \xi,|x-y|\leq 1\}}
 \frac{1}{|y|^{\beta}|x-y|^{n-2\alpha}}\,dy\\
&= A(x)+B(x),
\end{align*}
where $C>0$ depends on $\xi$. Since $0\leq \beta<2\alpha<n$,
\[
A(x)\leq C\int_{\{y:|y|\leq \xi\}}\frac{1}{|y|^{\beta}}\,dy\leq C.
\]
Similarly,
\begin{align*} %\label{eq:3.6}
&B(x)\\
&\leq  \Big(\int_{\{y:|y|\leq \xi,|x-y|\leq 1,|x-y|\ge |y|\}}
+\int_{\{y:|y|\leq \xi,|x-y|\leq 1,|x-y|\leq |y|\}}\Big)
 \frac{C}{|y|^{\beta}|x-y|^{n-2\alpha}}\,dy\\
&\leq \int_{\{y:|y|\leq \xi\}}\frac{C}{|y|^{n-2\alpha+\beta}}dy
 +\int_{\{y:|x-y|\leq 1\}}\frac{C}{|x-y|^{n-2\alpha+\beta}}\,dy\leq C.
\end{align*}
As a result, $H_1\in L^\infty(\mathbb{R}^n)$. On the other hand, by
Theorem \ref{thm:1.1},  for $r>\frac{ns}{2\alpha s+n-s\beta}$, we
have
  \[
\|H_1\|_{L^s(\mathbb{R}^n)}\leq \|\tilde v_\xi^q\|_{L^r(B_\xi(0))}
\leq C;
\]
that is, $H_1\in L^s(\mathbb{R}^n)$. The claim follows.

Next, we show that $T_\xi(f,g)$ is a  contraction map from
$L^{\tilde{p}}(\mathbb{R}^n)\times L^{\tilde{q}}(\mathbb{R}^n)$
into $L^{\tilde{p}}(\mathbb{R}^n)\times
L^{\tilde{q}}(\mathbb{R}^n)$ for $\tilde{q},\tilde{p}>1$ satisfying
\begin{equation}\label{eq:3.7}
\frac{1}{\tilde{p}}-\frac{1}{\tilde{q}}=\frac{1}{p+1}-\frac{1}{q+1}.
\end{equation}
We may verify by the fact $p,q>1$, \eqref{eq:1.2} and \eqref{eq:3.7}
that
\[
\tilde{q}>\frac{n\tilde{p}}{n+(2\alpha-\beta)\tilde{p}},\quad
\tilde{p}>\frac{n\tilde{q}}{n+(2\alpha-\beta)\tilde{q}}.
\]
Choosing $d_1$ such that
\begin{equation}\label{eq:3.8}
\frac{1}{d_1}=\frac{1}{\tilde{q}}+\frac{q-1}{q+1},
\end{equation}
we  verify by \eqref{eq:1.2} that
\begin{equation}\label{eq:3.9}
\tilde{q}>d_1>\frac{n\tilde{p}}{n+(2\alpha-\beta)\tilde{p}}.
\end{equation}
By Theorem \ref{thm:1.1}, we find
 \begin{equation}\label{eq:3.10}
 \|T_1^\xi  g\|_{{\tilde{p}}}\leq C\||v_\xi|^{q-1}g\|_{d_1}.
 \end{equation}
This and H\"{o}lder's inequality yield
\begin{equation}\label{eq:3.11}
\|T_1^\xi
 g\|_{{\tilde{p}}}\leq C\|v_\xi\|_{q+1}^{q-1}\|g\|_{{\tilde{q}}}.
 \end{equation}
In the same way, choosing $\frac{1}{d_2}=\frac{1}{\tilde{p}}+\frac{p-1}{p+1}$,
 we obtain
\begin{equation}\label{eq:3.12}
 \|T_2^\xi f\|_{{\tilde{q}}}\leq C\||u_\xi|^{p-1}f\|_{d_2}\leq C\|u_\xi\|_{p+1}^{p-1}\|f\|_{{\tilde{p}}}.\end{equation}
Since $(u,v)\in L^{p+1}(\mathbb{R}^n)\times L^{q+1}(\mathbb{R}^n)$, one can choose  $\xi$ sufficiently
large so that
\begin{equation}\label{eq:3.13}
\|T_1^\xi g\|_{{\tilde{p}}}\leq \frac{1}{2}\|g\|_{{\tilde{q}}},\quad \|T_2^\xi
 f\|_{{\tilde{q}}}\leq \frac{1}{2}\|f\|_{{\tilde{p}}}.
 \end{equation}
Therefore,
\begin{equation} \label{eq:3.14}
\begin{split}
  \|T_\xi(f,g)\|_{{\tilde{p}}\times {\tilde{q}}}
 &=  \|(T_1^\xi g,T_2^\xi f)\|_{{\tilde{p}}\times {\tilde{q}}}
=\|T_1^\xi g\|_{{\tilde{p}}}+\|T_2^\xi f\|_{{\tilde{q}}}\\
&\leq \frac{1}{2}\|g\|_{{\tilde{q}}}+\frac{1}{2}\|f\|_{{\tilde{p}}}=\frac{1}{2}\|(f,g)\|_{{\tilde{p}}\times
{\tilde{q}}}.
\end{split}
\end{equation}
In other words, $T_\xi(f,g)$ is a contraction map from
$L^{\tilde{p}}(\mathbb{R}^n)\times L^{\tilde{q}}(\mathbb{R}^n)$ into
itself for
$\tilde{p},\tilde{q}>1,\frac{1}{\tilde{p}}-\frac{1}{\tilde{q}}=\frac{1}{p+1}-\frac{1}{q+1}$.
In particular, for $\tilde p = p+1$ and $\tilde q = q +1$, we see
that $T_\xi(f,g)$ is also a contraction map from
$L^{p+1}(\mathbb{R}^n)\times L^{q+1}(\mathbb{R}^n)$ into itself.
Choosing  $\tilde{p},\tilde{q}$ large enough such that
$\tilde{p}>\frac{np}{2\alpha-\beta},\tilde{q}>\frac{nq}{2\alpha-\beta},\frac{1}{\tilde{p}}-\frac{1}{\tilde{q}}=\frac{1}{p+1}-\frac{1}{q+1}$,
by Lemma \ref{lem:3.1}, we conclude that $(u_\xi,v_\xi)\in
(L^{\tilde{p}}(\mathbb{R}^n)\times L^{\tilde{q}}(\mathbb{R}^n))\cap
(L^{p+1}(\mathbb{R}^n)\times L^{q+1}(\mathbb{R}^n))$.

Finally, we show that $u,v\in L^\infty(\mathbb{R}^n)$. Since
$u(x)=u_\xi(x)+ \tilde u_\xi(x),v(x)=v_\xi(x)+ \tilde v_\xi(x)$ and
$\tilde u_\xi, \tilde v_\xi\in L^\infty(\mathbb{R}^n)$, we only need
to verify $u_\xi,v_\xi\in L^\infty(\mathbb{R}^n)$. By
\eqref{eq:3.3},\eqref{eq:3.4} and $M_1,M_2\in
L^\infty(\mathbb{R}^n)$, it is sufficient to verify that $I_1,
I_2\in L^\infty(\mathbb{R}^n)$, where
\[
I_1(x)=\int_{R^n}\frac{K_\alpha(x-y)|v_\xi|^{q-1}
 v_\xi(y)}{|y|^{\beta}}\,dy,\quad
I_2(x)=\int_{R^n}\frac{K_\alpha(x-y)|u_\xi|^{p-1}
 u_\xi(y)}{|y|^{\beta}}\,dy.
\]
There holds
\begin{equation} \label{eq:3.15}
\begin{split}
  |I_1(x)|&\leq
\int_{\{y:|y|\leq \xi\}}\frac{K_\alpha(x-y)|
 v_\xi|^q}{|y|^{\beta}}\,dy+\int_{\{y:|y|\ge \xi\}}
 \frac{K_\alpha(x-y)|v_\xi|^q}{|y|^{\beta}}\,dy\\
&:= J(x)+G(x).
\end{split}
\end{equation}
If $x\in \mathbb{R}^n\backslash B_{2\xi}(0)$,
$y\in B_{\xi}(0)$, then
$|x-y|>|x|-|y|>\xi>|y|$. Thus,
\begin{equation} \label{eq:3.16}
\begin{split}
  J(x)
&\leq  C\int_{\{y:|y|\leq \xi\}}
 \frac{|v_\xi|^q}{|x-y|^{n+2\alpha}|y|^{\beta}}\,dy
 \leq C\int_{\{y:|y|\leq \xi\}}\frac{|v_\xi|^q}{|y|^{\beta}}\,dy\\
&\leq  \Big(\int_{\{y:|y|\leq
\xi\}}|v_\xi|^{q+1}\,dy\Big)^{q/(q+1)} \Big(\int_{\{y:|y|\leq
\xi\}}\frac{1}{|y|^{(q+1)\beta}}\,dy\Big)^{1/(q+1)}\leq C
\end{split}
\end{equation}
since $q<(n-\beta)/\beta$. If $x\in B_{2\xi}(0)$, we have
\begin{align*} %\label{eq:3.17}
J(x)
&\leq  \int_{\{y:|y|\leq \xi,|x-y|\ge 1\}}\frac{K_\alpha(x-y)|v_\xi|^q}{|y|^{\beta}}\,dy
+\int_{\{y:|y|\leq \xi,|x-y|\leq 1\}}\frac{K_\alpha(x-y)|v_\xi|^q}{|y|^{\beta}}dy\\
&\leq C\int_{\{y:|y|\leq \xi,|x-y|\ge 1\}}
 \frac{|v_\xi|^q}{|x-y|^{n+2\alpha}|y|^{\beta}}\,dy\\
&\quad +C\int_{\{y:|y|\leq \xi,|x-y|\leq 1\}}
\frac{|v_\xi|^q}{|x-y|^{n-2\alpha}|y|^{\beta}}\,dy
:=CJ_1(x)+CJ_2(x).
\end{align*}
Now we estimate $J_1(x),J_2(x)$ respectively. By H\"{o}lder's
inequality, we have
\begin{equation} \label{eq:3.18}
\begin{split}
J_1(x)
&\leq C\Big(\int_{\{y:|y|\leq
\xi\}}|v_\xi|^{q+1}\,dy\Big)^{q/(q+1)} \Big(\int_{\{y:|y|\leq
\xi\}}\frac{1}{|y|^{(q+1)\beta}}\,dy\Big)^{1/(q+1)}\\
&\leq C
\end{split}
\end{equation}
and because of $\tilde{q}>nq/(2\alpha-\beta)$, we deduce
\begin{equation} \label{eq:3.19}
\begin{split}
&J_2(x)\\
&\leq  \int_{\{y:|y|\leq \xi,|x-y|\leq 1,|x-y|\ge
|y|\}}\frac{|v_\xi|^q}{|x-y|^{n-2\alpha}|y|^{\beta}}\,dy\\
&\quad +\int_{\{y:|y|\leq \xi,|x-y|\leq 1,|x-y|\le|y|\}}\frac{|v_\xi|^q}{|x-y|^{n-2\alpha}|y|^{\beta}}\,dy\\
&\leq  \int_{\{y:|y|\leq \xi\}}\frac{|v_\xi|^q}{|y|^{n-2\alpha+\beta}}\,dy+\int_{\{y:|x-y|\leq 1,|y|\leq \xi\}}\frac{|v_\xi|^q}{|x-y|^{n-2\alpha+\beta}}\,dy\\
&\leq \Big(\int_{\{y:|y|\leq
\xi\}}|v_\xi|^{\tilde{q}}\,dy\Big)^{q/\tilde{q}}
\Big(\int_{\{y:|y|\leq
\xi\}}\frac{1}{|y|^{\frac{\tilde{q}}{\tilde{q}-q}(n-2\alpha+\beta)}}\,dy
\Big)^{(\tilde{q}-q)/\tilde{q}}\\
&\quad + \Big(\int_{\{y:|y|\leq
\xi\}}|v_\xi|^{\tilde{q}}\,dy\Big)^{q/\tilde{q}}
\Big(\int_{\{y:|x-y|\le1\}}\frac{1}{|x-y|^{\frac{\tilde{q}}{\tilde{q}-q}(n-2\alpha+\beta)}}\,dy
\Big)^{(\tilde{q}-q)/\tilde{q}}\\
&\leq  C.
\end{split}
\end{equation}
Inequalities \eqref{eq:3.16},\eqref{eq:3.18} and \eqref{eq:3.19}
imply that $J\in L^\infty(\mathbb{R}^n)$. Now we estimate $G(x)$.
For any $x\in \mathbb{R}^n$,
\begin{align*} % \label{eq:3.20}
G(x)
&\leq  \int_{\{y:|y|\ge \xi,|x-y|\ge
1\}}\frac{K_\alpha(x-y)|v_\xi|^q}{|y|^{\beta}}\,dy
+\int_{\{y:|y|\ge \xi,|x-y|\leq 1\}}\frac{K_\alpha(x-y)|v_\xi|^q}{|y|^{\beta}}\,dy\\
&\leq  C\int_{\{y:|y|\ge \xi,|x-y|\ge
1\}}\frac{|v_\xi|^q}{|x-y|^{n+2\alpha}|y|^{\beta}}\,dy\\
&\quad +C\int_{\{y:|y|\ge \xi,|x-y|\leq 1\}}
 \frac{|v_\xi|^q}{|x-y|^{n-2\alpha}|y|^{\beta}}\,dy\\
&\leq C\int_{\{y:|y|\ge \xi,|x-y|\ge 1\}}
 \frac{|v_\xi|^q}{|x-y|^{n+2\alpha}|y|^{\beta}}\,dy
 +C\int_{\{y:|x-y|\leq 1\}}\frac{|v_\xi|^q}{|x-y|^{n-2\alpha+\beta}}\,dy\\
&:= CG_1(x)+CG_2(x).
\end{align*}
By H\"{o}lder's inequality,
\[
G_2(x)\le\Big(\int_{\mathbb{R}^n}|v_\xi|^{\tilde{q}}dy\Big)^{q/\tilde{q}}
\Big(\int_{\{y:|x-y|\le1\}}\frac{1}{|x-y|^{\frac{\tilde{q}}{\tilde{q}-q}(n-2\alpha+\beta)}}dy
\Big)^{(\tilde{q}-q)/\tilde{q}}\leq C.
\]

 Now we estimate
$G_1(x)$. Since
$\tilde{q}>\frac{nq}{2\alpha-\beta}>\frac{nq}{n-\beta}$, we can
choose an $r$ such that
$1<r<\frac{n\tilde{q}}{n\tilde{q}-nq-\tilde{q}\beta}$. Hence,
H\"older's inequality implies that
\begin{equation}\label{eq:3.21}
\begin{split}
G_1(x)
&\leq  \Big(\int_{\{y:|y|\ge \xi\}}|v_\xi|^{\tilde{q}}dy
 \Big)^{q/\tilde{q}}
\Big(\int_{\{|x-y|\ge 1\}} \frac{1}{y:|x-y|^{(n+2\alpha)r}}dy
 \Big)^{1/r}\\
&\quad\times \Big(\int_{\{y:|y|\ge
\xi\}}\frac{1}{|y|^{\frac{\beta}{1-q/{\tilde{q}}-1/r}}}dy
\Big)^{1-q/{\tilde{q}}-1/{r}}\leq C.
\end{split}
\end{equation}
Consequently, both $J$ and $G$ belong to $L^\infty\mathbb{(R}^n)$,
so is $I_1$. Similarly, we have $I_2\in L^\infty(\mathbb{R}^n)$.
Therefore, $u,v\in L^\infty(\mathbb{R}^n)$. The proof of Theorem
\ref{thm:1.1} is completed.
\end{proof}


\section{Regularity  of solutions to integral systems}

In this section, we show that the solution of \eqref{eq:1.1}
is H\"older continuous.
We recall the regularity lifting theorem II given in \cite{CL}.
Let $V$ be a Hausdorff topological vector space.
Suppose there are two extended norms defined on $V$,
\[
\|\cdot\|_{X},\|\cdot\|_{Y}: V\to [0,\infty].
\]
Let
\[
X:=\{v\in V: \|v\|_{X}<\infty\},\quad Y:=\{v\in V:
\|v\|_{Y}<\infty\}.
\]
We also assume that $X$ is complete and that
the topology in $V$ is weaker than the topology of $X$ and the weak
topology of $Y$, which means that the convergence in $X$ or weak
convergence in $Y$ will imply convergence in $V$.
The pair of spaces $(X,Y)$ described as above is called an
$XY-$pair, if whenever the sequence $\{u_n\}\subset X$ with
$u_n\to u$ in $X$ and $\|u_n\|_{Y}<C$ will imply $u\in Y$.

 From \cite[Remark 3.3.5]{CL}, we know that if $X=L^p(U)$
for $1\leq p\le\infty, Y=C^{0,\gamma}(U)$ for $0<\gamma\le1$,
and $V$ is the space of distributions, where $U$ can be any subset
of $\mathbb{R}^n$ or $\mathbb{R}^n$ itself, then $(X,Y)$ is
 an $XY-$pair.

\begin{lemma}[Regularity Lifting II]\label{lem:4.1}
Suppose that Banach spaces $X,Y$ are an $XY$-pair, both contained
in some larger topological space $V$ satisfying properties described
above. Let $\mathfrak{X}$ and $\mathfrak{Y}$ be closed subsets of $X$
and $Y$ respectively. Suppose that $T:\mathfrak{X}\to X$ is
a contraction:
\[
\|Tf-Tg\|_{X}\leq \eta\|f-g\|_{X},\quad
\forall f,g\in \mathfrak{X}\text{ and  for  some }
0<\eta<1;
\]
and
$T:\mathfrak{Y}\to Y$ is shrinking:
\[
\|Tg\|_{Y}\leq \theta\|g\|_{Y},\quad \forall g\in \mathfrak{Y}
\text{ and  for  some } 0<\theta<1;
\]
Define
\[
Sf=Tf+F\quad\text{for  some } F\in \mathfrak{X}\cap\mathfrak{Y}.
\]
Moreover, assume that
$S:\mathfrak{X}\cap\mathfrak{Y}\to \mathfrak{X}\cap\mathfrak{Y}$.
Then there exists a unique solution $u$ of the equation
\[
u=Tf+F\quad\text{in }\mathfrak{X},
\]
and more importantly, $u\in Y$.
\end{lemma}

\begin{proof}[Proof of Theorem \ref{thm:1.3}]
Since $(u,v)$ is a solution pair for \eqref{eq:1.1}, we have
\begin{align*}
u(x)&= \int_{\mathbb{R}^n}K_\alpha(|x-y|)\frac{v^q(y)}{|y|^{\beta}}\,dy\\
&= -\int_{\mathbb{R}^n}\int^\infty_{|x-y|}K_\alpha'(s)
 \frac{v^q(y)}{|y|^{\beta}}\,ds\,dy \\
&= -\int_0^\infty\int_{B_s(x)}K_\alpha'(s)\frac{v^q(y)}{|y|^{\beta}}
\,ds\,dy
\end{align*}
and
\[
v(x)=
-\int_0^\infty\int_{B_s(x)}K_\alpha'(s)
\frac{u^p(y)}{|y|^{\beta}}\,ds\,dy.
\]
For any $\Omega\subset\subset\mathbb{R}^n\setminus\{0\}$, denote
$d= \operatorname{dist}(0,\Omega)>0$. Let $X = L^\infty(\Omega)$
and $Y = C^{0,\gamma}(\Omega)$. By Theorem \ref{thm:1.2}, $u,v\in
L^\infty(\mathbb{R}^n)$, we  define
\begin{gather*}
\mathfrak{X}=\{w\in X\big|
\|w\|_{L^\infty}\leq 2\|u\|_{L^\infty}+2\|v\|_{L^\infty}\},\\
\mathfrak{Y}=\{w\in Y\big|
\|w\|_{L^\infty}\leq 2\|u\|_{L^\infty}+2\|v\|_{L^\infty}\}.
\end{gather*}
For every $\varepsilon>0$ such that $0<\varepsilon<\frac d2$,
we define
\begin{gather*}
T_\varepsilon^q \hat{v}(x) = -\int_0^\varepsilon
\int_{B_s(x)}K_\alpha'(s)\frac{\hat{v}^q(y)}{|y|^{\beta}}\,ds\,dy,\\
T_\varepsilon^p \hat{u}(x) = -\int_0^\varepsilon
\int_{B_s(x)}K_\alpha'(s)\frac{\hat{u}^p(y)}{|y|^{\beta}}\,ds\,dy,
\quad
T_\varepsilon (\hat{u}, \hat{v}) = (T_\varepsilon^q \hat{v},
T_\varepsilon^p \hat{u}).
\end{gather*}
Furthermore, we define
\begin{gather*}
F^q v(x) = -\int_\varepsilon^\infty
\int_{B_s(x)}K_\alpha'(s)\frac{v^q(y)}{|y|^{\beta}}\,ds\,dy,\\
F^p u(x) = -\int_\varepsilon^\infty
\int_{B_s(x)}K_\alpha'(s)\frac{u^p(y)}{|y|^{\beta}}\,ds\,dy,\quad
F = (F^p v, F^q u).
\end{gather*}
Obviously, a solution $(u,v)$ of \eqref{eq:1.1} is a solution
of the equation
\[
(\hat{u}, \hat{v}) = T_\varepsilon(\hat{u},\hat{v}) + F.
\]
Write $S_\varepsilon (\hat{u},\hat{v}) =
T_\varepsilon(\hat{u},\hat{v}) + F$. We will show for
$\varepsilon>0$ small that $T_\varepsilon$ is a contracting operator
from $\mathfrak{X}\times \mathfrak{X}$ to $X\times X$, and also is a
shrinking operator from $\mathfrak{Y}\times\mathfrak{Y}$ to $Y\times
Y$. Furthermore,
$F\in(\mathfrak{X}\times\mathfrak{X})\cap(\mathfrak{Y}\times\mathfrak{Y})$,
and
$S_\varepsilon:(\mathfrak{X}\times\mathfrak{X})\cap(\mathfrak{Y}\times\mathfrak{Y})\to
(\mathfrak{X}\times\mathfrak{X})\cap(\mathfrak{Y}\times\mathfrak{Y})$.
This then will yields $(u,v)\in Y\times Y$ by Lemma \ref{lem:4.1}.

We first show that $T_\varepsilon$ is a contracting operator from
$\mathfrak{X}\times \mathfrak{X}$ to $X\times X$. For any $f,g\in
\mathfrak{X}$, we denote here and below that
 $f^p =f_+^p$. By the mean value theorem, we have
\begin{align*}
\big|T_\varepsilon^q f(x)-T_\varepsilon^q g(x)\big|
&\leq \int^\varepsilon_0\int_{B_s(x)} |f^q(y) - g^q(y)|
 \frac{|K'_\alpha(s)|}{|y|^{\beta}}\,ds\,dy\\
&\leq C\max\{\|f\|_{L^\infty}^{q-1}, \|g\|_{L^\infty}^{q-1}\}
 \|f - g\|_{L^\infty}\int^\varepsilon_0 s^{n-\beta}|K_\alpha'(s)|\,ds.
\end{align*}
By \eqref{eq:2.1}, we obtain
\[
\int^\varepsilon_0 s^{n-\beta}|K_\alpha'(s)|\,ds \leq
O(\varepsilon^{2\alpha-\beta})
\]
as $\varepsilon\to 0$. Hence, for $\varepsilon>0$ small,
\begin{equation}\label{eq:4.1}
\big|T_\varepsilon^q f(x)-T_\varepsilon^q g(x)\big| \leq
C\max\{\|f\|_{L^\infty}^{q-1}, \|g\|_{L^\infty}^{q-1}\}\|f -
g\|_{L^\infty}\varepsilon^{2\alpha-\beta}.
\end{equation}
Choosing $\varepsilon>0$ small so that
$C\max\{\|f\|_{L^\infty}^{q-1},
\|g\|_{L^\infty}^{q-1}\}\varepsilon^{2\alpha-\beta}\leq 1/4$,
 we see that $T_\varepsilon^q$ is a contracting operator from $\mathfrak{X}$ to $X$. Similarly,
 $T_\varepsilon^p$ is also a  contracting operator from $\mathfrak{X}$ to $X$. Therefore, $T_\varepsilon$ is a contracting operator
from $\mathfrak{X}\times \mathfrak{X}$ to $X\times X$.

Next, we verify that $T_\varepsilon$ is a shrinking operator from $\mathfrak{Y}\times \mathfrak{Y}$ to $Y\times Y$. We only show it for $T_\varepsilon^q$, it can be done in the same way for $T_\varepsilon^p$. Assume $f\in \mathfrak{Y}$. Then for any $x,z\in \Omega$, we have
\begin{align*}
|T_\varepsilon^q f(x)-T_\varepsilon^q f(z)|
&=\big|\int^\varepsilon_0\Big\{\int_{B_s(x)} \frac{f^q(y)}{|y|^{\beta}}\,dy - \int_{B_s(z)}\frac{f^q(y)}{|y|^{\beta}}\,dy\Big\}K'_\alpha(s)\,ds\big|\\
&= \big|\int^\varepsilon_0\int_{B_s(x)}
 \big[\frac{f^q(y)}{|y|^{\beta}}
 - \frac{f^q(y+z -x)}{|y+z-x|^{\beta}}\big]K'_\alpha(s)\,dy\,ds\big|\\
&\leq \big|\int^\varepsilon_0\int_{B_s(x)} \Big[f^q(y)
\big(\frac{1}{|y|^{\beta}} - \frac{1}{|y+z-x|^{\beta}}\big)\\
&\quad + \Big(\frac{f^q(y) - f^q(y+z -x)} {|y+z-x|^{\beta}}\Big)\Big]
K'_\alpha(s)\,dy\,ds\big|.
\end{align*}
For $y\in B_s(x)$, $0< s < \varepsilon$, we have
$|y|\geq |x| -s\geq d-\frac d2 = \frac d2>0$ and
$|y +z -x|\geq |z|- |y-x|\geq d - s\geq \frac d2$.
So  both $\frac{1}{|y|^{\beta}}$ and
$\frac{1}{|y+z-x|^{\beta}}$ are regular in $B_s(x)$ for
$0< s < \varepsilon$. In particular, there exists $C>0$ such that
$|\frac{1}{|y|^{\beta}} - \frac{1}{|y+z-x|^{\beta}}|\leq C|x-z|$.
Hence,
\begin{align*}
&\big|\int^\varepsilon_0\int_{B_s(x)} f^q(y)\big(\frac{1}{|y|^{\beta}} - \frac{1}{|y+z-x|^{\beta}}\big)K'_\alpha(s)\,dy\,ds\big|\\
&\leq C\|f\|^q_{L^\infty} |x-z|\big|\int^\varepsilon_0 s^nK'_\alpha(s)\,ds\big|\\
&\leq C\|f\|^q_{L^\infty} |x-z|\varepsilon^{2\alpha}\\
&\leq C\|f\|^{q-1}_{L^\infty} \|f\|_{C^{0,\gamma}}
|x-z|\varepsilon^{2\alpha}.
\end{align*}

If $|x-z|\leq 1$, $|x-z|\leq|x-z|^\gamma$; if $|x-z|> 1$,
$|x-z|\leq(diam \Omega)^{1-\gamma}|x-z|^\gamma$. Therefore,
\begin{align*}
&\big|\int^\varepsilon_0\int_{B_s(x)} f^q(y)\big(\frac{1}{|y|^{\beta}} - \frac{1}{|y+z-x|^{\beta}}\big)K'_\alpha(s)\,dy\,ds\big|\\
&\leq C\|f\|^{q-1}_{L^\infty} \|f\|_{C^{0,\gamma}}
|x-z|^\gamma\varepsilon^{2\alpha}.
\end{align*}
On the other hand, by the mean value theorem,
\begin{align*}
&\big|\int^\varepsilon_0\int_{B_s(x)}
 \frac{f^q(y) - f^q(y+z -x)} {|y+z-x|^{\beta}}K'_\alpha(s)\,dy\,ds\big|\\
&\leq \big|\int^\varepsilon_0\int_{B_s(x)}|w(\xi)|^{q-1}
 \frac{|f(y) - f(y+z -x)|} {|y+z-x|^{\beta}}K'_\alpha(s)\,dy\,ds\big|\\
&\leq  C\|f\|^{q-1}_{L^\infty}\|f\|_{C^{0,\gamma}} |x-z|^{\gamma}
 \big|\int^\varepsilon_0 s^n K'_\alpha(s)\,ds\big|\\
&\leq C\|f\|^{q-1}_{L^\infty}\|f\|_{C^{0,\gamma}}
|x-z|^{\gamma}\varepsilon^{2\alpha},
\end{align*}
where $w$ is valued between $f(y)$ and $f(y+z -x)$. Consequently,
\[
|T_\varepsilon^q f(x)-T_\varepsilon^q f(z)| \leq
C\|f\|_{C^{0,\gamma}} |x-z|^{\gamma}\varepsilon^{2\alpha}.
\]
Choosing $\varepsilon>0$ sufficiently small, we obtain
\[
\sup_{x\not=z}\frac{\big|T_\varepsilon^q f(x)-T_\varepsilon^q
f(z)\big|}{|x-z|}\leq \frac 14\|f\|_{C^{0,\gamma}}.
\]
We may derive in the same way as \eqref{eq:4.1} that
\[
\big|T_\varepsilon^q f(x)\big| \leq
C\|f\|_{L^\infty}\varepsilon^{2\alpha-\beta}
\leq\frac14\|f\|_{C^{0,\gamma}}.
\]
Therefore, for any $f\in \mathfrak{Y}$,
\[
\|T_\varepsilon^q f(x)\|_{C^{0,\gamma}}
\leq\frac12\|f\|_{C^{0,\gamma}};
\]
that is, $T_\varepsilon^q$ is a shrinking operator from
$\mathfrak{Y}$ to $Y$.

Now, we show that $F^qv(x)$ and $F^pu(x)$ are H\"{o}lder continuous
for $u,v\in \mathfrak{Y}$. We only deal with $F^qv(x)$. For
$F^pu(x)$, it can be shown similarly. We write
\begin{align*}
F^q v(x)
&= -\int_\varepsilon^1 \int_{B_s(x)}K_\alpha'(s)
 \frac{v^q(y)}{|y|^{\beta}}\,ds\,dy
- \int_1^\infty \int_{B_s(x)}K_\alpha'(s)
 \frac{v^q(y)}{|y|^{\beta}}\,ds\,dy\\
&:= F_1(x) + F_2(x).
\end{align*}
For $x, z\in \Omega$, we have
\begin{align*}
|F_1(x) - F_1(z)|
&= \big|\int_\varepsilon^1
\Big(\int_{B_s(x)}\frac{v^q(y)}{|y|^{\beta}}\,dy
 - \int_{B_s(x)}\frac{v^q(y)}{|y|^{\beta}}\,dy\Big)
 K_\alpha'(s)\,ds\big| \\
&\leq \|v\|^q_{L^\infty} \int_\varepsilon^1
\Big(\int_{(B_s(x)\setminus B_s(z))\cup (B_s(z)\setminus
B_s(x))}\frac{1}{|y|^{\beta}}\,dy \Big) |K_\alpha'(s)|\,ds.
\end{align*}
Denote by $A^*$ the symmetric rearrangement of the set $A$,
and $f^*$ the symmetric-decreasing rearrangement of a function $f$.
It is known that for the characteristic function $\chi_A$ of a
set $A$,
$\chi_A^* = \chi_{A^*}$.
Moreover, for nonnegative functions $f$ and $g$, there holds
\[
\int_{\mathbb{R}^n}fg\,dx\leq \int_{\mathbb{R}^n}f^*g^*\,dx.
\]
If $|x-z|\geq 2s$, then
\begin{align*}
\int_{(B_s(x)\setminus B_s(z))\cup (B_s(z)\setminus B_s(x))}
\frac{1}{|y|^{\beta}}\,dy
=&\int_{ B_s(x)}\frac{1}{|y|^{\beta}}\,dy + \int_{B_s(z)}\frac{1}{|y|^{\beta}}\,dy\\
=&\int_{\mathbb{R}^n}\chi_{B_s(x)}\frac{1}{|y|^{\beta}}\,dy + \int_{\mathbb{R}^n}\chi_{B_s(z)}\frac{1}{|y|^{\beta}}\,dy\\
&\leq  \int_{\mathbb{R}^n}\chi_{B_s(0)}\frac{1}{|y|^{\beta}}\,dy + \int_{\mathbb{R}^n}\chi_{B_s(0)}\frac{1}{|y|^{\beta}}\,dy\\
&\leq  Cs^{n-\beta}\\
&\leq  Cs^{n-\beta-1}|x-z|^{\gamma}.\\
\end{align*}
If $|x-z|< 2s$, we have
$$
(B_s(x)\setminus B_s(z))\cup (B_s(z)\setminus B_s(x))
\subset (B_s(x)\cup B_s(z))\setminus B_{s-\frac{|x-z|}2}(\frac{x+z}2).
$$
Let $r = \big(s^n-\big(s-\frac{|x-z|}2\big)^n\big)^{1/n}$.
Noting $0<s<1$ and reasoning in the same way, we obtain
\begin{equation}\label{eq:4.1a}
\begin{split}
\int_{(B_s(x)\setminus B_s(z))\cup (B_s(z)\setminus B_s(x))}
\frac{1}{|y|^{\beta}}\,dy
&\leq  \int_{(B_s(x)\cup (B_s(z))
 \setminus B_{s-\frac{|x-z|}2}(\frac{x+z}2)}\frac{1}{|y|^{\beta}}\,dy \\
&\leq  2\int_{B_r(0)}\frac{1}{|y|^{\beta}}\,dy\\
&\leq  C\Big(s^n-\big(s-\frac{|x-z|}2\big)^n\Big)^{\frac{n-\beta}n}\\
&\leq  C s^{n-1-\beta}|x-z|^{\gamma}.
\end{split}
\end{equation}
As a result,
\[
|F_1(x) - F_1(z)|
\leq  - \|v\|^q_{L^\infty}|x-z|^{1-\frac {\beta}n}
 \int_\varepsilon^1s^{n-\beta-1}K_\alpha'(s)\,ds
\leq  C(\varepsilon) \|v\|^q_{L^\infty}|x-z|^{\gamma};
\]
that is,
\begin{equation}\label{eq:4.2}
\sup_{x\not=z}\frac{|F_1(x) - F_1(z)|}{|x-z|^{\gamma}}
\leq C(\varepsilon).
\end{equation}
Now we estimate the H\"older norm of $F_2$.
For $x, z\in \Omega$, by \eqref{eq:4.1a},
\[
\begin{split}
|F_2(x) - F_2(z)|
&= \big|\int_1^\infty
\Big(\int_{B_s(x)}\frac{v^q(y)}{|y|^{\beta}}\,dy - \int_{B_s(x)}\frac{v^q(y)}{|y|^{\beta}}\,dy\Big) K_\alpha'(s)\,ds\big| \\
&\leq \|v\|^q_{L^\infty} \int_1^\infty\Big(\int_{(B_s(x)\setminus
B_s(z))\cup (B_s(z)\setminus
B_s(x))}\frac{1}{|y|^{\beta}}\,dy \Big) |K_\alpha'(s)|\,ds\\
&\leq - \|v\|^q_{L^\infty}|x-z|^{\gamma}\int_1^\infty
s^{n-\beta-1}K_\alpha'(s)\,ds\\
&\leq - \|v\|^q_{L^\infty}|x-z|^{\gamma}
\int_1^\infty s^{n-\beta}K_\alpha'(s)\,ds.
\end{split}
\]
We may verify that
\begin{align*}
 -\int_1^\infty s^{n-\beta-1}K_\alpha'(s)
&= K_\alpha(1)+(n-\beta)\int_1^\infty s^{n-\beta-1}K_\alpha(s)\,ds\\
&\leq K_\alpha(1)+C(n-\beta)\int_1^\infty s^{n-\beta-1}
 s^{-n-2\alpha}\,ds\\
&\leq K_\alpha(1)+C.
\end{align*}
Thus,
\[
|F_2(x) - F_2(z)|\leq C \|v\|^q_{L^\infty}|x-z|^{\gamma};
\]
that is,
\begin{equation}\label{eq:4.3}
\sup_{x\not=z}\frac{|F_2(x) - F_2(z)|}{|x-z|^{\gamma}}
\leq C.
\end{equation}
Inequalities \eqref{eq:4.2} and \eqref{eq:4.3} yield
\begin{equation}\label{eq:4.4}
\sup_{x\not=z}\frac{|F^qv(x) - F^qv(z)|}{|x-z|^{\gamma}}
\leq C(\varepsilon).
\end{equation}
From the definition of $F_1(x)$ and $F_2(x)$, we have
\begin{equation}\label{eq:4.5}
\begin{split}|F_1(x)|
&\leq \|v\|^q_{L^\infty} \int_\varepsilon^1
\Big(\int_{B_s(x)}\frac{1}{|y|^{\beta}}\,dy \Big) |K_\alpha'(s)|\,ds\\
&\leq C\|v\|^q_{L^\infty} \int_\varepsilon^1 s^{n-\beta}|K_\alpha'(s)|\,ds\\
&\leq C(\varepsilon)\|v\|^q_{L^\infty},
\end{split}
\end{equation}
and
\begin{equation}\label{eq:4.6}
\begin{split}
|F_2(x)|&\leq \|v\|^q_{L^\infty} \int_1^\infty
\Big(\int_{B_s(x)}\frac{1}{|y|^{\beta}}\,dy \Big) |K_\alpha'(s)|\,ds\\
&\leq C\|v\|^q_{L^\infty} \int_1^\infty s^{n-\beta}|K_\alpha'(s)|\,ds\\
&\leq C\|v\|^q_{L^\infty}.
\end{split}
\end{equation}
It follows from \eqref{eq:4.5} and \eqref{eq:4.6} that
\begin{equation}\label{eq:4.7}
|F^qv|_{L^\infty}\leq
C(\varepsilon)\|v\|^q_{L^\infty}.
\end{equation}
Inequalities \eqref{eq:4.5} and \eqref{eq:4.7} imply that $F^qv$ is
 H\"older continuous, and this together with \eqref{eq:4.7} imply
$F^qv\in\mathfrak{X}\cap\mathfrak{Y}$.

Finally, we show that $S_\varepsilon$ maps
 $(\mathfrak{X}\times\mathfrak{X})\cap(\mathfrak{Y}\times\mathfrak{Y})$
 to itself. We need only to verify that if $\|w\|_{L^\infty}\leq
2\|u\|_{L^\infty}+2\|v\|_{L^\infty}$, then
\begin{equation}\label{eq:4.8}
\|T^q_\varepsilon
w\|_{L^\infty}\leq 2\|u\|_{L^\infty}+2\|v\|_{L^\infty}.
\end{equation}
In the same way, we can prove
\begin{equation}\label{eq:4.9}
\|T^p_\varepsilon w\|_{L^\infty}\leq
2\|u\|_{L^\infty}+2\|v\|_{L^\infty}.
\end{equation}
Now, we verify \eqref{eq:4.8}. Indeed,
\begin{align*}
|T^q_\varepsilon w(x)|
&= -\int_0^\varepsilon
\int_{B_s(x)}K_\alpha'(s)\frac{w^q(y)}{|y|^{\beta}}\,ds\,dy\\
&\leq (2\|u\|_{L^\infty}+2\|v\|_{L^\infty})^q\int_0^\varepsilon
\int_{B_s(x)}\frac 1{|y|^{\beta}}|K_\alpha'(s)|\,dyds\\
&\leq C(2\|u\|_{L^\infty}+2\|v\|_{L^\infty})^q\int_0^\varepsilon
s^{n-\beta}|K_\alpha'(s)|ds\\
&\leq C(2\|u\|_{L^\infty}+2\|v\|_{L^\infty})^q
\varepsilon^{2\alpha-\beta}.
\end{align*}
Therefore, choosing $\varepsilon$ sufficiently small, but
independent of $w$, we obtain \eqref{eq:4.8}. This completes
the proof of the theorem.
\end{proof}


\section{Symmetry of solutions}

In this section, we show that positive solutions of \eqref{eq:1.1}
are radially symmetric. For a given real number $\lambda$, we may
define
\[\Sigma_\lambda =\{x=(x_1,x_2,\cdots, x_n)\in\mathbb{R}^n|x_1 \leq
\lambda\},\quad T_\lambda=\{x\in\mathbb{R}^n| x_1=\lambda\}.\]
For $x\in\Sigma_\lambda$, let $x_\lambda=(2\lambda -x_1, x_2,\cdots,
x_n)$, and define \[u_\lambda(x) =u(x_\lambda),\quad v_\lambda(x)
=v(x_\lambda).\]

\begin{lemma}\label{lem:5.1}
For any positive solution $u$ of \eqref{eq:1.1}, we have
\begin{gather}\label{eq-uv-1}
u(x)-u_\lambda(x) =\int_{\Sigma_\lambda}
\Big(K_\alpha(x-y)-K_\alpha(x_\lambda
-y)\Big)\Big(\frac{v^q(y)}{|y|^{\beta}}
-\frac{v_\lambda^q(y)}{|y_\lambda|^{\beta}}\Big)\,dy,
\\
\label{eq-uv-2}
v(x)-v_\lambda(x) =\int_{\Sigma_\lambda}
\Big(K_\alpha(x-y)-K_\alpha(x_\lambda
-y)\Big)\Big(\frac{u^p(y)}{|y|^{\beta}}
-\frac{u_\lambda^p(y)}{|y_\lambda|^{\beta}}\Big)\,dy.
\end{gather}
\end{lemma}

\begin{proof}
Let $\Sigma_\lambda^c=\{x=(x_1,x_2,\cdots,
x_n)\in\textsc{R}^n|x_1
> \lambda\}$. It follows from \eqref{eq:1.1} that
\begin{align*}
u(x)
&=\int_{\Sigma_\lambda}\frac{K_\alpha(x-y)v^q(y)}{|y|^{\beta}}\,dy
+\int_{\Sigma_\lambda^c}\frac{K_\alpha(x-y)v^q(y)}{|y|^{\beta}}dy\\
&=\int_{\Sigma_\lambda}\frac{K_\alpha(x-y)v^q(y)}{|y|^{\beta}}\,dy
+\int_{\Sigma_\lambda}\frac{K_\alpha(x-y_\lambda)v^q(y_\lambda)}{|y_\lambda|^{\beta}}\,dy\\
&=\int_{\Sigma_\lambda}\frac{K_\alpha(x-y)v^q(y)}{|y|^{\beta}}\,dy
+\int_{\Sigma_\lambda}\frac{K_\alpha(x_\lambda-y)
v_\lambda^q(y)}{|y_\lambda|^{\beta}}\,dy.
\end{align*}
Here we have used the fact that $|x-y_\lambda|=|x_\lambda-y|$ and
the fact that $K_\alpha$ is radially symmetric in $\textsc{R}^n$.
Substituting $x$ by $x_\lambda$ gives
\[
u(x_\lambda)=\int_{\Sigma_\lambda}
 \frac{G_\alpha(x_\lambda-y)v^q(y)}{|y|^{\beta}}\,dy
+\int_{\Sigma_\lambda}
\frac{G_\alpha(x-y)v_\lambda^q(y)}{|y_\lambda|^{\beta}}\,dy.
\]
Hence,
\[
u(x)-u_\lambda(x) =\int_{\Sigma_\lambda}
\Big(K_\alpha(x-y)-K_\alpha(x_\lambda
-y)\Big)\Big(\frac{v^q(y)}{|y|^{\beta}}
-\frac{v_\lambda^q(y)}{|y_\lambda|^{\beta}}\Big)\,dy.
\]
Similarly, we have
\[
v(x)-v_\lambda(x) =\int_{\Sigma_\lambda}
\Big(K_\alpha(x-y)-K_\alpha(x_\lambda
-y)\Big)\Big(\frac{u^p(y)}{|y|^{\beta}}
-\frac{u_\lambda^p(y)}{|y_\lambda|^{\beta}}\Big)\,dy.
\]
This completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:1.4}]
We use the moving plane
method developed for integral equations in \cite{CLO} to prove the
result.
First, we show for sufficiently negative $\lambda$ that
\begin{equation}\label{eq:5.1}
u(x)\leq u(x_\lambda),\quad v(x)\leq v(x_\lambda),\quad \forall x\in
\Sigma_\lambda.
\end{equation}
Set
\begin{gather*}
w_\lambda(x)=u(x)-u(x_\lambda),\quad z_\lambda(x)=v(x)-v(x_\lambda),\\
\Sigma_{\lambda}^{u,-}=\{x\in \Sigma_\lambda|u(x)>u(x_\lambda)\},\quad
\Sigma_{\lambda}^{v,-}=\{x\in \Sigma_\lambda| v(x)>v(x_\lambda)\}.
\end{gather*}
From Lemma \ref{lem:5.1}, we deduce that
\begin{align*}
u(x)-u_\lambda(x)
&= \int_{\Sigma_\lambda\backslash \Sigma_\lambda^{v,-}}
\Big(K_\alpha(x-y)-K_\alpha(x_\lambda
-y)\Big)\Big(\frac{v^q(y)}{|y|^{\beta}}
-\frac{v_\lambda^q(y)}{|y_\lambda|^{\beta}}\Big)\,dy\\
&\quad+\int_{\Sigma_\lambda^{v,-}}
\Big(K_\alpha(x-y)-K_\alpha(x_\lambda
-y)\Big)\Big(\frac{v^q(y)}{|y|^{\beta}}
-\frac{v_\lambda^q(y)}{|y_\lambda|^{\beta}}\Big)\,dy.
\end{align*}
Since $|x-y|<|x_\lambda-y|$ and $|y|>|y_\lambda|$ in
$\Sigma_\lambda$, taking into account that $K_\alpha(x)$
is decreasing as well as that $t^q$ is convex, we obtain
\begin{align*}
 u(x)-u_\lambda(x)&\leq \int_{\Sigma_\lambda^{v,-}}
\Big(K_\alpha(x-y)-K_\alpha(x_\lambda-y)\Big)
\Big(\frac{v^q(y)}{|y|^{\beta}}
-\frac{v_\lambda^q(y)}{|y_\lambda|^{\beta}}\Big)\,dy\\
&\leq \int_{\Sigma_\lambda^{v,-}}
\Big(K_\alpha(x-y)-K_\alpha(x_\lambda
-y)\Big)\Big(\frac{v^q(y)-v_\lambda^q(y)}{|y|^{\beta}}\Big)\,dy\\
&\leq C\int_{\Sigma_\lambda^{v,-}}K_\alpha(x-y)
\frac{v^{q-1}(v-v_\lambda)}{|y|^{\beta}}\,dy.
\end{align*}
We may derive as in the proof of \eqref{eq:3.11} and \eqref{eq:3.12}
for
\[\frac{1}{\tilde{p}}-\frac{1}{\tilde{q}}=\frac{1}{p+1}-\frac{1}{q+1}\]
and
\[\frac{1}{d_1}=\frac{1}{\tilde{q}}+\frac{q-1}{q+1},\quad\frac{1}{d_2}=\frac{1}{\tilde{p}}+\frac{p-1}{p+1}\]
that
\begin{equation}\label{eq:5.2}
\|w_\lambda\|_{L^{\tilde{p}}(\Sigma_\lambda^{u,-})}\leq C\|v^{q-1}z_\lambda\|_{L^{d_1}(\Sigma_\lambda^{v,-})}
\leq C\|v\|_{L^{q+1}(\Sigma_\lambda^{v,-})}^{q-1}
\|z_\lambda\|_{L^{\tilde{q}}(\Sigma_\lambda^{v,-})}
\end{equation}
and
\begin{equation}\label{eq:5.3}
\|z_\lambda\|_{L^{\tilde{q}}(\Sigma_\lambda^{v,-})}\leq
C\|u^{p-1}w_\lambda\|_{L^{d_2}(\Sigma_\lambda^{u,-})} \leq
C\|u\|_{L^{p+1}(\Sigma_\lambda^{u,-})}^{p-1}
\|w_\lambda\|_{L^{\tilde{p}}(\Sigma_\lambda^{u,-})}.
\end{equation}
As a result,
\begin{equation}\label{eq:5.4}
\|w_\lambda\|_{L^{\tilde{p}}(\Sigma_\lambda^{u,-})}\leq
C\|u\|_{L^{p+1}(\Sigma_\lambda^{u,-})}^{p-1}
\|v\|_{L^{q+1}(\Sigma_\lambda^{v,-})}^{q-1}
\|w_\lambda\|_{L^{\tilde{p}}(\Sigma_\lambda^{u,-})}.
\end{equation}
Since $(u,v)\in L^{p+1}(\mathbb{R}^n)\times L^{q+1}(\mathbb{R}^n)$,
for sufficiently negative $\lambda$,
\[
C\|u\|_{L^{p+1}(\Sigma_\lambda^{u,-})}^{p-1}
\|v\|_{L^{q+1}(\Sigma_\lambda^{v,-})}^{q-1}\leq \frac{1}{2}.
\]
 Hence,
 \[
\|w_\lambda\|_{L^{\tilde{p}}(\Sigma_\lambda^{u,-})}
\le\frac{1}{2}\|w_\lambda\|_{L^{\tilde{p}}(\Sigma_\lambda^{u,-})}.
\]
This implies $\Sigma_\lambda^{u,-}$ must be a set of measure zero.
Similarly,
the measure of $\Sigma_\lambda^{v,-}$ is zero. Consequently,
\eqref{eq:5.1} holds.

Next, we increase the value of $\lambda$ continuously; that is,
we move the plane $T_\lambda$ to the right
as long as the inequality \eqref{eq:5.1} holds. We show that by
moving $T_\lambda$ in this way, it will not stop before the
plane hitting the origin. Let
\begin{equation}\label{eq:5.5}
\lambda_0=\sup\{\lambda|u(x)-u_{\lambda}(x)\leq
0,v(x)-v_{\lambda}(x)\leq 0,\forall x\in \Sigma_{\lambda}\}.
\end{equation}
Obviously $\lambda_0\leq 0$, We claim that
\begin{equation}\label{eq:5.6}
\lambda_0=0.
\end{equation}
In fact, if it were not the case, we would show that the  plane
could be moved further to the
right by a small distance, and this would contradict with the
definition of $\lambda_0$. Suppose by the contrary that $\lambda_0<0$,
and that there exist some points $x_0,x_1$ in $\Sigma_{\lambda_0}$
such that $u(x_0)=u_{\lambda_0}(x_0),v(x_1)=v_{\lambda_0}(x_1)$. By
Lemma \ref{lem:5.1} and noting that $x_{\lambda_0}=(x_0)_{\lambda_0}$,
we obtain
\begin{align*}
0&= u(x_0)-u_{\lambda_0}(x_0) \\
&= \int_{\Sigma_{\lambda_0}}
\Big(K_\alpha(x_0-y)-K_\alpha(x_{\lambda_0}
-y)\Big)\Big(\frac{v^q(y)}{|y|^{\beta}}
-\frac{v_{\lambda_0}^q(y)}{|y_{\lambda_0}|^{\beta}}\Big)\,dy.
\end{align*}
Since $|y|>|y_0|$ in  $\Sigma_{\lambda_0}$,
\[
\frac{v^q(y)}{|y|^{\beta}}
<\frac{v_{\lambda_0}^q(y)}{|y_{\lambda_0}|^{\beta}}\quad\text{in }
 \Sigma_{\lambda_0}.
\]
Moreover, $|x_0-y|<|x_{\lambda_0}-y|$ in $\Sigma_{\lambda_0}$, we
infer that
\[
v(x)\equiv v_{\lambda_0}(x)\equiv 0, \quad a.e.\
x\in\Sigma_{\lambda_0}.
\]
This also implies that $v(x)\equiv 0$, which is a contradiction to
the fact that $v$ is positive. So we have
\[
u(x)<u_{\lambda_0}(x),\quad a.e.\ x\in \Sigma_{\lambda_0}.
\]
Similarly,
\[
v(x)<v_{\lambda_0}(x),\quad\text{a.e. } x\in \Sigma_{\lambda_0}.
\]
Since $(u,v)\in L^{p+1}(\mathbb{R}^n)\times L^{q+1}(\mathbb{R}^n)$,
for any $\varepsilon>0 $ there exists $R>0$ such that
\[
\int_{R^n\backslash
B_R(0)}u^{p+1}\,dx<\varepsilon,\quad\int_{R^n\backslash
B_R(0)}v^{q+1}\,dx<\varepsilon.
\]
By  Lusin theorem, for any $\delta>0$, there exists a closed
set $F_\delta$ with $F_\delta\subset B_R(0)\cup \Sigma_{\lambda_0}=E$
and $m(E-F_\delta)<\delta$ such that
$w_{\lambda_0}|F_{\delta},z_{\lambda_0}|F_{\delta}$ is continuous.

As $w_{\lambda_0},z_{\lambda_0}<0$ in the interior of
$\Sigma_{\lambda_0},w_{\lambda_0},z_{\lambda_0}<0$ in  $F_\delta$.
Choosing $\varepsilon_0>0$ sufficiently small so that for any
$\lambda\in [\lambda_0,\lambda_0+\varepsilon_0)$, it holds that
$w_\lambda,z_\lambda<0$ in $F_\delta$. For such a
$\lambda$,
\begin{gather*}
\Sigma_\lambda^{u,-}\subset M^u:=(R^n\backslash
B_R(0))\cup(E\backslash
F_\delta)\cup[(\Sigma_\lambda\backslash\Sigma_{\lambda_0}^{u,-})\cap
B_R(0)],\\
\Sigma_\lambda^{v,-}\subset M^v:=(R^n\backslash
B_R(0))\cup(E\backslash
F_\delta)\cup[(\Sigma_\lambda\backslash\Sigma_{\lambda_0}^{v,-})\cap
B_R(0)].
\end{gather*}
We may choose $\varepsilon,\delta$ and $\varepsilon_0$
small so that
\[
C\|u\|_{L^{p+1}(\Sigma_\lambda^{u,-})}^{p-1}\|v\|_{L^{q+1}
(\Sigma_\lambda^{v,-})}^{q-1}\leq \frac{1}{2}.
\]
Hence,
 \[
\|w_\lambda\|_{L^{\tilde{p}}(\Sigma_\lambda^{u,-})}
\le\frac{1}{2}\|w_\lambda\|_{L^{\tilde{p}}(\Sigma_\lambda^{u,-})},
\]
which implies that $\Sigma_{\lambda}^{u,-}$ must be of measure zero.
Again, it contradicts the definition of $\lambda_0$. Equation
\eqref{eq:5.6} is proved.

On the other hand, we can also move the plane from positive infinite
to  zero by the similar procedure. Hence, $u(x),v(x)$ are symmetric
and monotonic with respect to $x_1=0$. Moreover, since the $x_1$
direction can be chosen arbitrarily, $u(x),v(x)$ are radial
symmetric and strictly monotonic with respect to the origin. Thus we
have completed the proof of Theorem \ref{thm:1.4}.
\end{proof}



\subsection*{Acknowledgments}
 X. Chen was supported by grant 10961015 from the NNSF of China.
J. Yang was supported by grants 10961016 from the NNSF of China,
and 2009GZS0011 from the NSF of Jiangxi.


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\end{document}