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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 98, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/98\hfil Optimization problems]
{Optimization problems involving the \\ fractional Laplacian}

\author[C. Qiu, Y. Huang, Y. Zhou \hfil EJDE-2016/98\hfilneg]
{Chong Qiu, Yisheng Huang, Yuying Zhou}

\address{Chong Qiu \newline
Department of Mathematics,
Taizhou University,
Taizhou 225300, China}
\email{qchsuda@163.com}

\address{Yisheng Huang \newline
Department of Mathematics,
Soochow University,
Suzhou 215006, China}
\email{yishengh@suda.edu.cn}

\address{Yuying Zhou (corresponding author) \newline
Department of Mathematics,
Soochow University,
Suzhou 215006, China}
\email{yuyingz@suda.edu.cn}

\thanks{Submitted January 17, 2016. Published April 19, 2016.}
\subjclass[2010]{49K30, 35A15, 47J30}
\keywords{Optimization; fractional Laplacian; rearrangements}

\begin{abstract}
 In this article we study rearrangement optimization problems
 related to  boundary-value problems involving the fractional Laplacian.
 We establish the existence and uniqueness of a solution
 under suitable assumptions.
\end{abstract}

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


\section{Introduction}

A rearrangement optimization problem is referred to an optimization problem 
in which the admissible set consists of functions that are rearrangements 
of a prescribed function. The theory of rearrangement optimizations has 
been established by Burton \cite{G.R87,G.R89}.
  Since then, this topic has been widely
studied by many authors in different aspects, see for example
\cite{C.S.C, F.B.G06, F.B09,F.B11, B.R07, M.M10,  NE03, QHZ15}.
Burton \cite{G.R89}  proved that  both the  minimization and maximization
problems for the boundary value problem involving the Laplacian  have solutions.
However,  the results obtained in  \cite{G.R89} can not be directly applied 
to the optimization problems for the boundary value problem involving 
$p$-Laplacian$ (1<p<\infty)$.
So by using a new approach, Cuccu et al \cite{F.B.G06} proved  that the 
 minimization problem has a solution for $ 1<p<\infty$. 
But their approach is not efficient for the  maximization problem. 
 Marras \cite{M.M10} obtained the solvability of the maximization for $ 1<p<\infty$ 
 by using another method. While  Cuccu et al \cite{F.B11} obtained a result 
of uniqueness for a class of $p$-Laplace equations under non-standard assumptions. 
 Recently, Qiu et al \cite{QHZ15} considered a rearrangement optimization problem 
related to the  quasilinear elliptic boundary value problem, where under 
suitable assumptions, it is  shown that both  the minimization and maximization 
problems  are solvable, which extends the results in \cite{G.R89,F.B.G06,M.M10}.


 It is worth to note  that most of the rearrangement optimization problems 
considered in the above papers are related to boundary value problems 
involving the Laplacian or $p$-Laplacian. In recent years, fractional and 
nonlocal operators of elliptic type have been attracted a lot of interests
since these operators appear in concrete applications in  fields 
such as minimal surface \cite{CV11},
thin obstacle problem \cite{CS08}, anomalous diffusion \cite{AT05},  
phase transition \cite{SV09},  hemivariational inequality \cite{T13,XHZ2015}, 
shape optimization \cite{AD13}, optimal transportation theory \cite{LJA15} 
and so on.  Interesting studies involving nonlocal fractional problems 
by variational methods can be found in \cite{GVR16,GD15} and  the
references therein.

  In this article, we study  several rearrangement optimization problems 
related to a class of boundary value problems involving the fractional Laplacian. 
We need to overcome the difficulties coming from both the fractional
Laplacian  and the rearrangement optimization problem. As our knowledge, 
this kind of problems has not  been considered in literature.

Let $\Omega$ be a smooth bounded
 domain of $\mathbb{R}^N$ $(N\geq 2)$  and let $k\in L^{q}(\Omega)$ 
with  $1\leq q \leq \infty$.  We recall that a
rearrangement of $k$ is an element of the set $\mathscr{R}(k)$  of all 
 measurable functions $g$ on $\Omega$ satisfying
$$
\operatorname{meas}(\{x\in\Omega:g(x)\geq
a\})=\operatorname{meas}(\{x\in\Omega:k(x)\geq
a\}),\quad \forall a\in\mathbb{R}.
$$
It is easy to prove that if $g\in\mathscr{R}(k)$, then 
$g\in L^q(\Omega)$ and $\|g\|_{L^{q}}=\|k\|_{L^{q}}$ (cf.  \cite[Lemma
2.1]{G.R89}).

Let $h(x,t):\Omega\times\mathbb{R}\to\mathbb{R}$ be
 a  Carath\'{e}odory function and $f\in L^{{\infty}}(\Omega)$.
 Under suitable assumptions we can show that the  boundary value problem
\begin{equation}
\begin{gathered}
{-L_\theta^s}  u+h(x,u)=f(x)\quad \text{in }\Omega,\\
u=0 \quad\text{in }\mathbb{R}^N\backslash\Omega
\end{gathered} \label{Qs}
\end{equation}
has a unique solution $u_f\in H^s(\Omega)$
 (cf. Proposition \ref{prop2}), where $L_\theta^s $ is the fractional Laplace
type operator   defined as
\begin{equation}\label{e4}
L_\theta^s u(x)=\int_{\mathbb{R}^n} \frac {u(x+y)+u(x-y)-2u(x)}{|y|^{N+2s}}
\theta(y)dy,\ \ x\in\mathbb{R}^N
\end{equation}
with $0<s<1$ and $\theta:\mathbb{R}^N\to (0,+\infty)$.

In particular,  if $h\equiv0$,  then  the boundary value problem \eqref{Qs} becomes
\begin{equation}
\begin{gathered}
{-L_\theta^s}  u=f(x)\quad \text{in }\Omega,\\
u=0 \quad \text{in }\mathbb{R}^N\backslash\Omega,
\end{gathered} \label{Ps}
\end{equation}
and it obviously has a unique solution $u_f$ (cf. Proposition \ref{prop1}),
thus  it deduces that (cf.  Remark \ref{rmk1}),
\begin{equation}\label{e3}
\begin{aligned}
&\sup_{v\in H^s(\Omega)}\Big(\int_{\Omega}2fvd x-\int_{\mathbb{R}^N}
\int_{\mathbb{R}^N}\frac{(v(x)-v(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\Big)\\
&=\int_{\Omega}2fu_{f}\,dx-\int_{\mathbb{R}^N}
 \int_{\mathbb{R}^N}\frac{(u_f(x)-u_f(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\\
&=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u_f(x)-u_f(y))^2}{| x-y|^{N+2s}}
\theta(x-y)\,dx\,dy=\int_{\Omega} fu_{f}\,dx.
\end{aligned}
\end{equation}
Therefore, we can define the functional $\Phi$ on $ L^{\infty}(\Omega)$ as
\begin{equation}\label{e00}
\Phi(f)=\int_{\Omega} fu_{f}\,dx
\end{equation}
and we are able to consider the following two optimization problems:
\begin{itemize}
 \item[(Opt1)]  Find $ f_1\in\mathscr{R}(f_0)$ such that
$ \Phi(f_1)=\sup_{f\in\mathscr{R}(f_0)}\Phi(f)$,
 \item[(Opt2)] Find $ f_2\in\mathscr{R}(f_0)$ such that
$ \Phi(f_2)=\inf_{f\in\mathscr{R}(f_0)}\Phi(f)$,
\end{itemize}
where $f_0\in L^{\infty}(\Omega)$ is a given function and $\mathscr{R}(f_0)$
is the set of all rearrangement of $f_0$.

If $h\not\equiv 0$,  then it is difficult {to} consider the above optimization 
problems related to the boundary value problem \eqref{Qs} since
the functional $\Psi$ on $ L^{\infty}(\Omega)$, corresponding to the 
optimization problems, defined by
\begin{equation} \label{e0}
\begin{aligned}
\Psi(f)&=\frac{1}{2}\Big(\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(u_f(x)-u_f(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\Big)\\
&\quad +\int_{\Omega} H(x,u_f)\,d x-\int_{\Omega} fu_f\,d x
\end{aligned}
\end{equation}
is hard to be reduced to a simple formula like \eqref{e00}.  However, 
in the case of $h\not\equiv0$, we can consider the following minimization  
optimization problem related to \eqref{Qs}:
 \begin{itemize}
 \item [(Opt*)] Find $ f_*\in\mathscr{R}(f_0)$ such that 
$ \Psi(f_*)=\inf_{f\in\mathscr{R}(f_0)}\Psi(f)$,
\end{itemize}
where $f_0\in L^{\infty}(\Omega)$ is a given function and $\mathscr{R}(f_0)$
is the set of all rearrangement of $f_0$.


This article is organized as follows. In Section \ref{s2}, we give
some preliminaries. Section \ref{s4}
is devoted to discussing   the minimization problem (Opt*) in detail. 
After establishing  the uniqueness result of solutions for the  
problem \eqref{Qs}, we show that the minimization problem  (Opt*) is solvable. 
In Section \ref{s3}, we prove that both the maximization and minimization 
optimization problems (Opt1) and (Opt2) are solvable.
To our best of knowledge, the results of this paper are new and nontrivial.


\section{Preliminaries}\label{s2}

Given $0<s<1$, we define the fractional Sobolev space 
\[
 H^s(\Omega)=\big\{u\in L^{2}(\mathbb{R}^N):u\equiv 0\text{ in }
\mathbb{R}^N\backslash\Omega, \int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\,dx\,dy<\infty\big\}
\]
 with the inner product 
\[
\langle u,v\rangle=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u(x)-u(y))(v(x)-v(y))}{| x-y|^{N+2s}}\,dx\,dy
\quad \forall u,v\in H^s(\Omega).
\]
Then the norm of $u$ is 
\[ 
\|u\|=\Big(\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\,dx\,dy\Big)^{1/2}.
\] %\label{e02}

   Throughout this article, we denote by $\|u\|$ and $\|u\|_p$ the usual norm
in spaces $H^s(\Omega)$ and $L^p (\Omega)$ $(1\leq p\leq\infty)$, respectively. 
As usual,  ``$\to $'' and
``$\rightharpoonup $'' denote the strong and weak convergence.

 We now list some lemmas which are useful in the proof of our main results.

 \begin{lemma}[{\cite[Lemma 8]{RE12}}] \label{l1}
 $H^s(\Omega)\hookrightarrow L^{r}(\mathbb{R}^N)$, 
for $1\leq r\leq\frac{2N}{N-2s}$, and the embedding is compact if 
 $1\leq r<\frac{2N}{N-2s}$.
 \end{lemma}

 \begin{lemma}[{\cite[Lemma 2.2]{G.R89}}] \label{l7}
 Assume that $ 1\leq r<\infty$ and for $ f\in L^{r}(\Omega)$ denote 
by $\overline{\mathscr{R}(f)}$ the weak closure of $ \mathscr{R}(f)$ 
in $ L^{r}(\Omega)$. Then $\overline{\mathscr{R}(f)}$ is convex and weakly 
compact in $ L^{r}(\Omega)$.
 \end{lemma}

 \begin{lemma}[{\cite[Lemma 2.9]{G.R89} or \cite[Lemma 2.1]{F.B09})}] 
\label{l2}
  Let $ f,g:\Omega\to\mathbb{R}$ be measurable functions and
 suppose that for each $t\in\mathbb{R}$, the  level set of $g$ at $t$, 
i.e., $\{x\in\Omega:g(x)=t\}$,  has zero measure.
 Then there exists an increasing (decreasing) function $ \varphi$ such that
 $\varphi\circ g$ is a rearrangement of $f$ where $ \varphi\circ g$ denotes 
a composite function defined by
 $$
  (\varphi\circ g)(x)=\varphi(g(x)), \forall x\in\Omega.
 $$
 \end{lemma}

 \begin{lemma}[{\cite[Lemma 2.4]{G.R89} or \cite[Lemma 2.2]{F.B09}}] \label{l3}
For any $ 1\leq r<\infty$ define $r'= \frac{r}{r-1}$ if $r>1$ and 
$r'=\infty$ if $r=1$.
 Let $ f\in L^{r}(\Omega)$ and $ g\in L^{r'}(\Omega)$. Suppose that there 
exists an increasing (decreasing) function
 $\varphi:\mathbb{R}\to\mathbb{R}$ such that 
$\varphi\circ g\in\mathscr{R}(f)$.  Then $ \varphi\circ g$ is the unique 
maximizer (minimizer) of the linear functional $\int_{\Omega} hg \,dx$,
 relative to $ h\in\overline{\mathscr{R}(f)}$.
 \end{lemma}

\begin{lemma}[{\cite[Lemma 2.3]{B.J09}}] \label{l4}
Suppose that $ f\in L^{r}(\Omega)$ and $ g\in L^{r'}(\Omega)$. Then there 
exists $ \hat{f}\in\mathscr{R}(f)$ which maximizes (minimizes) the linear 
functional $\int_{\Omega} hgd x$,
relative to $ h\in\overline{\mathscr{R}(f)}$.
\end{lemma}

As in the proof of \cite[Lemma 2.1]{TWY12} we have the following result.

\begin{lemma}\label{l8}
  Given $u\in H^{1/2}(\Omega)$, there exists a unique extension 
$v \in H^{1}(\mathbb{R}_{+}^{N+1})$ of $u$ such that
\begin{gather*}
-\Delta v(x,y)=0,\quad \text{for } x\in\mathbb{R}^N,y>0,\\
v(x,0)=u(x),\quad \text{for } x\in\mathbb{R}^N.
\end{gather*}
Moreover,
\[
-\partial_{y}v(x,0)=(-\Delta)^{1/2}u(x)
\]
in the sense that
\[
-\int_{\mathbb{R}^N\times\{0\}}\frac{\partial v}{\partial y}\varphi dx
=\int_{\mathbb{R}^N}\varphi(-\Delta)^{1/2}u dx
\]
for every $\varphi\in H^{1/2}(\mathbb{R}^N)$.
\end{lemma}



\section{Minimization related to \eqref{Qs}}  \label{s4}

Recall that the energy functional $I:H^s(\Omega)\to \mathbb{R}$ corresponding
to \eqref{Qs} is 
\begin{equation}\label{e1}
   I(u)=\frac{1}{2}\int_{\mathbb{R}^N}
\int_{\mathbb{R}^N}\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy
+\int_{\Omega} H(x,u)d x-\int_{\Omega} fud x,
\end{equation}
where $H(x,u)=\int_0^{u}h(x,t)d t$.

 We use the following hypotheses on the functions 
$h:\Omega\times\mathbb{R}\to\mathbb{R}$ and  $\theta:\mathbb{R}^N\to (0,+\infty)$:
\begin{itemize}
\item [(H1)] $h(x,t)$ is Carath\'{e}odory and is non-decreasing with respect 
to the second variable for almost all $x\in\Omega$.

\item [(H2)] There exist $a(x), b(x)\in L^{\infty}(\Omega)$, $0<l<1$, such 
that $|h(x,t)|\leq a(x)+b(x)|t|^l$, for all  $t\in\mathbb{R}$, a.e. $x\in\Omega$.

\item[(H3)]
   $\theta(x)=\theta(-x)$ for any $x\in\mathbb{R}^N\backslash \{0\}$;
  
\item[(H4)] $\theta\in L^{\infty}(\mathbb{R}^N)$ and there exists 
$\theta_0 \in \mathbb{R}_+$ such that $\theta(x)\geq\theta_0$,
for any $x\in\mathbb{R}^N$.
\end{itemize}

\begin{proposition}\label{prop2}
 Suppose that {\rm(H1)--(H4)} hold. Then
\eqref{Qs} has a unique solution $u_f\in H^s(\Omega)$ and 
$ I(u_f)=\inf_{v\in H^s(\Omega)}I(v)$.
 \end{proposition}
 
\begin{proof}
   First, we show that  the problem \eqref{Qs} has a  solution.
Let $C$ be a  positive constant.
From (H2), (H4),  and the Sobolev embedding inequality it follows that
   \begin{gather}\label{e5}
    \big|\int_{\Omega} H(x,u)d x\big|\leq 
\int_{\Omega} \big|\int_0^{u}h(x,v)\, dv \big| \,d x
\leq C\|u\|+C\|u\|^{l+1}, \\
\label{e115}
   \big|\int_{\Omega} fu\,dx\big| \leq \|f\|_{\infty}\|u\|_{1}\leq C\|u\|, \\
 \label{e2}
\theta_0\|u\|^{2}\leq\int_{\mathbb{R}^N}
\int_{\mathbb{R}^N}\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy,
\quad \forall u\in H^s(\Omega).
\end{gather}
 Hence we deduce from \eqref{e1}, \eqref{e5},
   \eqref{e115} and  \eqref{e2} that
\begin{equation}\nonumber
     I(u)\geq \frac{\theta_0}{2}\|u\|^{2}-C\|u\|^{l+1}-C\|u\|
\to\infty
   \end{equation}
as $\|u\|\to\infty$, which shows that the functional $ I$ is
coercive.
We will prove that the functional $ I$ is weakly lower
semi-continuous.  To do this, let $ v_{n}\rightharpoonup v$
in $ H^s(\Omega)$ as $n\to\infty$, noting both the embeddings
$H^s(\Omega)\hookrightarrow L^{1}(\Omega)$ and 
$H^s(\Omega)\hookrightarrow L^{l+1}(\Omega)$ are compact,
 we see that $ v_{n}\to v$ in 
$L^{l+1}(\Omega)$ and $ L^{1}(\Omega)$ as $n\to\infty$. Therefore,
$H(x,v_n)\to H(x,v)$ in $L^{1}(\Omega)$ as $n\to\infty$ by
the continuity of the operator $u\mapsto H(x,u)$ from
$L^{l+1}(\Omega)$ to $ L^{1}(\Omega)$, which implies that
\begin{equation}\label{e110}
  \int_{\Omega} H(x,v_n)d x\to\int_{\Omega} H(x,v)\,dx
\end{equation}
as $n\to\infty$.  Then we have
\begin{align*}
&\liminf_{n\to\infty}I(v_n)\\
&= \liminf_{n\to\infty}\int_{\Omega} \Big(\frac{1}{2}
\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(v_n(x)-v_n(y))^2}
{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy+H(x,v_{n})-fv_{n}\Big)\,d x\\
&\geq \frac{1}{2}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(v(x)-v(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy
 +\int_{\Omega} H(x,v)d x-\limsup_{n\to\infty}\int_{\Omega} fv_nd x\\
&\geq I(v)-\limsup_{n\to\infty}\|f\|_{\infty}\|v_n-v\|_{1}\\
&= I(v).
\end{align*}
Thus the functional $ I$ is weakly lower semi-continuous 
(which we will denote by w.l.s.c for short).
So that the functional $I$ has a minimizer 
$ u_f\in H^s(\Omega)$ with $I(u_f)=\inf_{v\in H^s(\Omega)} I(v)$.  
 By assumptions (H1), (H2), and using a standard argument 
  \cite[Lemma 2.16]{MW96}, we can easily
show that $ I\in C^{1}(H^s(\Omega),\mathbb{R})$, therefore $u_f$
is a critical point of $I$, i.e.,
\begin{equation}\label{e67}
\begin{aligned}
I'(u_{f})v
&=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u_{f}(x)-u_{f}(y))(v(x)-v(y))}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\\
&\quad +\int_{\Omega} (h(x,u_{f})v-fv)d x
=0,\quad \forall v\in H^s(\Omega).
\end{aligned}
\end{equation}
  By the definition of the fractional Laplace type operator  
$L_\theta^s $ (see \eqref{e4}), we have
  \begin{equation}\label{e6}
  \begin{aligned}
  &\int_{\mathbb{R}^N}-L_\theta^s u(x)v(x)dx\\
  &= -\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac {u(x+y)+u(x-y)-2u(x)}{|y|^{N+2s}}
  \theta(y)v(x)\,dy\,dx\\
  &= -\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} 
 \frac {u(x+y)-u(x)}{|y|^{N+2s}} \theta(y)v(x)\,dy\,dx\\
&\quad -\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac {u(x-y)-u(x)}{|y|^{N+2s}} 
 \theta(y)v(x)\,dy\,dx.
\end{aligned}
\end{equation}
Let $z=x+y$, $t=x-y$. By using  (H3) we obtain
\begin{equation}\label{e13}
\begin{aligned}
&-\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac {u(x+y)-u(x)}{|y|^{N+2s}} 
 \theta(y)v(x)\,dy\,dx \\
&=\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac {u(x)-u(z)}{|x-z|^{N+2s}} 
 \theta(x-z)v(x)\,dz\,dx,\\
&-\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac {u(x-y)-u(x)}{|y|^{N+2s}}
 \theta(y)v(x)\,dy\,dx \\
&=\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac {u(x)-u(t)}{|x-t|^{N+2s}}
  \theta(x-t)v(x)\,dt\,dx.
\end{aligned}
\end{equation}

Obviously,
\begin{equation}\label{e15}
\begin{aligned}
&\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac {u(x)-u(z)}{|x-z|^{N+2s}}
  \theta(x-z)v(x)\,dz\,dx\\
&= \int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac {u(x)-u(t)}{|x-t|^{N+2s}}
  \theta(x-t)v(x)\,dt\,dx\\
&= \int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac {u(x)-u(y)}{|x-y|^{N+2s}}
  \theta(x-y)v(x)\,dy\,dx\\
&= \int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac {u(y)-u(x)}{|x-y|^{N+2s}}
  \theta(x-y)v(y)\,dy\,dx.
\end{aligned}
\end{equation}
It follows from \eqref{e6}, \eqref{e13} and \eqref{e15}  that
\begin{align*}
&\int_{\mathbb{R}^N}-L_\theta^s u(x)v(x)dx\\
&= \int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac {u(x)-u(y)}{|x-y|^{N+2s}} 
 \theta(x-y)v(x)\,dy\,dx
 +\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \frac {u(y)-u(x)}{|x-y|^{N+2s}}
  \theta(x-y)v(y)\,dy\,dx\\
&= \int_{\mathbb{R}^n}\int_{\mathbb{R}^n} 
 \frac {(u(x)-u(y))(v(x)-v(y))}{|x-y|^{N+2s}} \theta(x-y)\,dy\,dx.
\end{align*}
This, combined with  \eqref{e67}  yields  
$$
  \int_{\mathbb{R}^N}-L_\theta^s u_f(x)v(x)dx+\int_{\Omega}(h(x,u_f(x))-f(x))v(x)dx=0,
\quad \forall v\in H^s(\Omega).
  $$
Thus, $u_f$ is a solution of  problem \eqref{Qs}.

Next, we show that $u_f$ is the unique solution of  \eqref{Qs}.
Assume that $ w\in H^s(\Omega)$ is another solution of  \eqref{Qs} and 
$ u_{f}\neq w$. Then 
   \begin{equation}\label{e104}
\|u_{f}-w\|>0.
   \end{equation}
Since $ h(x,\cdot)$ is non-decreasing,
\begin{equation}\label{e109}
\int_{\Omega} (h(x,u_{f})-h(x,w))(u_{f}-w)d x\geq0.
\end{equation}
From \eqref{e67}  we obtain that for every $v\in H^s(\Omega)$,
\begin{gather} \label{e7}
\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(u_{f}(x)-u_{f}(y))(v(x)-v(y))}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy
+\int_{\Omega} h(x,u_{f})vdx
=\int_{\Omega} fv\,dx,\\
\label{e8}
\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(w(x)-w(y))(v(x)-v(y))}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy
 +\int_{\Omega} h(x,w)vdx=\int_{\Omega} fv\,dx.
\end{gather}
From these two equalities, we obtain that for every $v\in H^s(\Omega)$,
\begin{equation}\nonumber
\begin{aligned}
&\int_{\Omega} (h(x,u_{f})-h(x,w))vdx\\
&= \int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(w(x)-u_{f}(x)-w(y)
+u_{f}(y))(v(x)-v(y))}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy.
\end{aligned}
\end{equation}
Let $ v=u_{f}-w$ we have
\begin{equation}\label{e14}
\begin{aligned} 
&\int_{\Omega} (h(x,u_{f})-h(x,w))(u_{f}-w)dx\\
&=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{-(v(x)-v(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\\
&\leq -\theta_0\|v\|^2 <0,
\end{aligned}
\end{equation}
 the last inequality above comes from \eqref{e104}.  So,  \eqref{e14} contradicts
\eqref{e109}. Therefore, we have proved that $u_f$ is the unique
solution of  \eqref{Qs}.
\end{proof}

 Let $u_f$ be the unique solution of  \eqref{Qs}. Recall that $\Psi(f)$ 
is defined by \eqref{e0}. considering the  optimization problem (Opt*),
we have the following result.

 \begin{theorem}\label{thm3}
Suppose that {\rm (H1--(H4)} hold. Then for a fixed nonnegative function 
$f_0\in L^{\infty}(\Omega)$ there exists $ f_*\in\mathscr{R}(f_0)$ which 
solves the minimization optimization problem {\rm (Opt*)}, i.e.,
 $$
 \Psi(f_*)=\inf_{f\in\mathscr{R}(f_0)}\Psi(f).
 $$
 \end{theorem}

 \begin{proof}
   Let $ A=\inf_{f\in\mathscr{R}(f_0)}\Psi(f)$, then $ A$ is well-defined.
Indeed, for each $f\in\mathscr{R}(f_0)$, we have
  \begin{equation}\label{e12}
  \begin{aligned}
    \Psi(f)
    &=\frac{1}{2}\Big(\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u_f(x)-u_f(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\Big) \\
&\quad +\int_{\Omega} H(x,u_f)d x-\int_{\Omega} fu_f\,d x\\
&\geq \frac{\theta_0}{2}\|u_{f}\|^{2}-C(\|f\|_{\infty}\|u_{f}\|+\|u_{f}\|^{l+1}).
    \end{aligned}
   \end{equation}
Noting that $ \|f\|_{\infty}=\|f_0\|_{\infty}$ and $l+1<2$, we deduce
that $A$ must be finite.

Let $\{f_i\}\subset\mathscr{R}(f_0)$ be such that $\Psi(f_i)\to A$
as $i\to\infty$ and we denote $u_i=u_{f_i}$.  It follows from
\eqref{e12} that $\{u_i\}$ is bounded in $H^s(\Omega)$, then
 it has a subsequence (still denoted $\{u_i\}$) which converges weakly
to $ u\in H^s(\Omega)$ and strongly  to $ u$ in $ L^{2}(\Omega)$.
 On the other hand, since $
\|f_i\|_{L_{{\infty}}}\equiv\|f_0\|_{L_{{\infty}}}$, 
$\{f_i\}$ is bounded in $ L^{2}(\Omega)$, it must contain a subsequence
(still denoted $\{f_i\}$) converging weakly to some
$\bar{f}\in  \overline{\mathscr{R}(f_0)}$, the weak closure of
 $\mathscr{R}(f_0)$ in $  L^{2}(\Omega)$.
  Then
 $$
 \big|\int_{\Omega} (f_i-\bar{f})ud x\big|\to 0\quad \text{as } i\to\infty,
 $$
since $u\in L^{2}(\Omega)$.
It follows from the H\"{o}lder inequality that
\begin{equation}\label{e72}
\begin{aligned}
\big|\int_{\Omega} (f_iu_i-\bar{f}u)d x\big|
& \leq\big|\int_{\Omega} f_i(u_i-u)d x\big|
 +\big|\int_{\Omega} (f_i-\bar{f})ud x\big|\\
&\leq\|f_i\|_{2}\|u_i-u\|_{2}+\big|\int_{\Omega} (f_i-\bar{f})u\,dx\big|\to 0
\end{aligned}
\end{equation}
as $i\to\infty$.   By \eqref{e110}, \eqref{e72} and the weak lower
semi-continuity of the norm in $ H^s(\Omega)$, we
obtain 
 \begin{equation}\label{e0016}
 \begin{aligned}
A&=\lim_{i\to\infty}\Psi(f_i)\\
&\geq\frac{1}{2}(\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy)\\
&\quad +\int_{\Omega} H(x,u)d x-\int_{\Omega} \bar{f}ud x.
\end{aligned}
\end{equation}
   From Lemma \ref{l4} we infer the existence of $\hat{f}\in\mathscr{R}(f_0)$ 
which maximizes the linear functional $\int_{\Omega} hud x$,
relative to $ h\in\overline{\mathscr{R}(f_0)}$.  As a consequence,
\[ %\label{e017}
\int_{\Omega} \bar{f}ud x\leq\int_{\Omega} \hat{f}ud x.
\]
Combining this with \eqref{e0016}, we obtain
\begin{equation}\label{e74}
A\geq \frac{1}{2}(\int_{\mathbb{R}^N}
\int_{\mathbb{R}^N}\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy)
+\int_{\Omega} H(x,u)d x-\int_{\Omega} \hat{f}u\,d x.
\end{equation}
By Proposition \ref{prop2},
\begin{equation}\label{e75}
\begin{aligned}
\Psi(\hat{f})&=\inf_{v\in H^s(\Omega)}
\Big(\frac{1}{2}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(v(x)-v(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy
+\int_{\Omega} (H(x,v)-\hat{f}v)d x\Big)\\
&\leq \frac{1}{2}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy
 +\int_{\Omega}(H(x,u)-\hat{f}u) dx.
\end{aligned}
\end{equation}
It  follows from \eqref{e74} and \eqref{e75} that $ \Psi(\hat{f})\leq A$.

On the other hand, recall that $A=\inf_{f\in\mathscr{R}(f_0)}\Psi(f)$ and 
$ \hat{f}\in\mathscr{R}(f_0)$,
we must have $ A\leq \Psi(\hat{f})$. So that $ A=\Psi(\hat{f})$.  We
complete the proof by letting $ f_{*}= \hat{f}$.
 \end{proof}

\section{Maximization and minimization related to \eqref{Ps}} \label{s3}

In this section, we consider two optimization problems  (Opt1) and  (Opt2) 
related to \eqref{Ps}.
The energy functional of \eqref{Ps} is
\[
J(u)=\frac{1}{2}(\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy)
-\int_{\Omega} f(x)u(x)d x.
\]
It is easy to see $ J\in C^{1}(H^s(\Omega),\mathbb{R})$ and 
$$
J'(u)v=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u(x)-u(y))
(v(x)-v(y))}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy-\int_{\Omega} f(x)v(x)d x.
$$
 $u\in H^s(\Omega)$ is a solution of the problem \eqref{Ps} 
if and only if $J'(u)v=0,\forall v\in H^s(\Omega)$.

By proposition \ref{prop2}, we have the following result.

\begin{proposition}\label{prop1}
Assume that $\theta $ satisfies {\rm (H3)} and {\rm (H4)}. Then for each
 $f$ in $L^{\infty}(\Omega)$,   \eqref{Ps} has a unique solution $u_f\in
H^s(\Omega)$, and $ J(u_f)=\inf_{v\in H^s(\Omega)}J(v)$.
\end{proposition}

\begin{remark}\label{rmk1} \rm
  Since $J'(u_f)u_f=0$,
  \begin{equation}\label{e16}
\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u_f(x)-u_f(y))^2}{| x-y|^{N+2s}}
\theta(x-y)\,dx\,dy=\int_{\Omega} f(x)u_f(x)\,d x.
\end{equation}
  Recall that $\Phi(f)=\int_{\Omega}fu_f d x$, we have
  \begin{equation}\nonumber
  \begin{aligned}
  &\Phi(f)\\
  &= -2\Big(\frac{1}{2}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(u_f(x)-u_f(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy
 -\int_{\Omega} f(x)u_f(x)d x\Big)\\
  &= -2\inf_{v\in H^s(\Omega)}\Big(\frac{1}{2}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(v(x)-v(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy-\int_{\Omega} f(x)v(x)d x\Big)\\
  &= -2J(u_f).
  \end{aligned}
  \end{equation}
\end{remark}


\begin{theorem}\label{t1}
Assume that $\theta $ satisfies {\rm (H3)} and {\rm (H4)}. 
Then for each non-negative function $f_0\in L^{\infty}(\Omega)$,  
there exists $ f_1\in\mathscr{R}(f_0)$ which solves the maximization optimization 
problem {\rm (Opt1)}, i.e.,
\[
{\Phi(f_1)=}\int_{\Omega} f_1u_{1}\,dx=\sup_{f\in\mathscr{R}(f_0)}
\int_{\Omega} fu_{f}\,dx{
=\sup_{f\in\mathscr{R}(f_0)}\Phi(f)}
\]
where $u_{1}=u_{f_1}$. Moreover,
   if $s=\frac{1}{2}$ then there exists an increasing function $\phi$ such that
$f_1=\phi(u_1)$
almost everywhere in $\Omega$.
\end{theorem}

\begin{proof}
Let
\[
M=\sup_{f\in\mathscr{R}(f_0)}\int_{\Omega} fu_{f}\,dx.
\]
We first show that $ M$ is finite. Let $ f\in\mathscr{R}(f_0)$. 
From  (H4), {\eqref{e16}}  and  H\"{o}lder's inequality we
find
\begin{equation}\label{e9}
\begin{aligned}
\theta_0\|u_{f}\|^{2}
&\leq\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u_f(x)-u_f(y))^2}{| x-y|^{N+2s}}
 \theta(x-y)\,dx\,dy \\
&=\int_{\Omega} fu_{f}\,dx
\leq \|f\|_{\infty}\|u_{f}\|_{1}.
\end{aligned}
\end{equation}
Since $\|f\|_{\infty}=\|f_0\|_{\infty}$, it follows from \eqref{e9}
and Lemma \ref{l1} that $ M$ is finite.

Let $\{f_i\}$ be a maximizing sequence and let $ u_i=u_{f_i}$. 
From \eqref{e9} it is clear that $\{u_i\}$ is bounded in $  H^{s}(\Omega)$,
hence it has a subsequence (still denoted $\{u_i\}$) that converges weakly 
to $ u\in H^{s}(\Omega)$. We also infer that $\{u_i\}$ converges strongly to $ u$ 
in $ L^{2}(\mathbb{R}^N)$. On the other hand, since $\{f_i\}$ is bounded 
in $ L^{\infty}(\Omega)$, it must contain a subsequence(still denoted $\{f_i\}$)
 converging weakly to $\bar{f}\in L^{2}(\Omega)$. Note that 
$ \bar{f}\in\overline{\mathscr{R}(f_0)}$, 
the weak closure of $\mathscr{R}(f_0)$ in $  L^{2}(\Omega)$.
Thus, using the weak lower semi-continuity of the $ H^{s}(\Omega)$ 
norm and \eqref{e5} we obtain
\begin{equation}\label{e10}
\begin{aligned}
M&=\lim_{i\to\infty}\int_{\Omega} f_iu_i\,dx\\
&=\lim_{i\to\infty}(\int_{\Omega}2f_iu_i\,dx
 -\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u_i(x)-u_i(y))^2}{| x-y|^{N+2s}}
 \theta(x-y)\,dx\,dy)\\
&\leq\int_{\Omega}2\bar{f}ud x-\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy.
\end{aligned}
\end{equation}
Note that from Lemma \ref{l4} we infer the existence of 
$\hat{f}\in\mathscr{R}(f_0)$ that maximizes the linear functional 
$\int_{\Omega} hu\,dx$,
relative to $ h\in\overline{\mathscr{R}(f_0)}$. As a consequence we obtain
\begin{equation}\label{e11}
\int_{\Omega} \bar{f}u\,d x\leq\int_{\Omega} \hat{f}u\,d x.
\end{equation}
Applying \eqref{e16}, \eqref{e10} and \eqref{e11} we find
\begin{align*}
  M
&\leq\int_{\Omega} 2\hat{f}ud x-\int_{\mathbb{R}^N}
\int_{\mathbb{R}^N}\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\\
  &\leq\int_{\Omega}2\hat{f}\hat{u}\,dx
-\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(\hat{u}(x)
-\hat{u}(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\\
  &=\int_{\Omega}\hat{f}\hat{u}\,dx\leq M
\end{align*}
where $\hat{u}=u_{\hat{f}}$. Thus let $f_1= \hat{f}$ we complete the 
first part of the proof.

We then show that $f_1$ is also a maximizer of the functional 
$\int_{\Omega} hu_1d x$,
relative to $ h\in\overline{\mathscr{R}(f_0)}$.
In fact, we notice that for each $g\in\mathscr{R}(f_0)$,
\begin{align*}
  \int_{\Omega} f_1u_1dx{\geq}& \int_{\Omega} gu_gdx\\
 &\geq \int_{\Omega}2gu_1d x-\int_{\mathbb{R}^N}
 \int_{\mathbb{R}^N}\frac{(u_1(x)-u_1(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\\
 &= \int_{\Omega}2gu_1d x-\int_{\Omega} f_1u_1dx,
 \end{align*}
 which implies that
\begin{equation}\label{e113}
 \int_{\Omega} f_1u_1 dx\geq \int_{\Omega} gu_1 dx,\quad \forall g\in\mathscr{R}(f_0).
 \end{equation}
 If $g\in\overline{\mathscr{R}(f_0)}$  then  we may choose a sequence 
$\{g_n\}\subset\mathscr{R}(f_0)$ such that $\{g_n\}$ converge weakly to $g$ 
in $L^{2}(\Omega)$.
 By \eqref{e113}, we obtain
 $$
 \int_{\Omega} f_1u_1\, dx\geq \int_{\Omega} g_nu_{1} dx\to \int_{\Omega} gu_1\,dx
 $$
 as $n\to\infty$.
So that
\begin{equation}\label{e114}
\int_{\Omega} f_1u_1 \,dx\geq \int_{\Omega} gu_1 \,dx,\quad
\forall g\in\overline{\mathscr{R}(f_0)}
 \end{equation}
and our claim is valid.

Let $E=\{x\in \Omega: f_1(x)=0\}$ and define $S=sup_{x\in E}(u_1(x))$, 
we claim that $u_1(x)\geq S$ on $E^c$ almost everywhere. Arguing by 
contradiction suppose the claim is false. Therefore there exists a 
number $S_1<S$ and a subset $A$ of $E^c$ with $|A|>0$ such that $u_1(x)<S_1$ 
on $A$ almost everywhere. Now let $S_1<S_2<S$. We can find a set $D$ 
of positive measure contained in $E$ such that $u_1(x)>S_2$ on $D$ 
almost everywhere. We can assume $|A|=|D|$. Next, consider a measure-preserving 
map $T:A\to D$. Using $T$ we define a particular rearrangement of $f_1$, 
denoted by $\overline{f}$, as follows:
$$
\overline{f}=\begin{cases}
 f_1(Tx) & x\in A,\\
 f_1(T^{-1}x) & x\in D,\\
 f_1(x) & x\in \Omega\backslash(A\cup D).
\end{cases}
$$
Thus
\begin{align*}
& \int_{\Omega} \overline{f}u_1dx-\int_{\Omega} f_1u_1dx\\
&=\int_{A\bigcup D}\overline{f}u_1dx-\int_{A\cup D}f_1u_1dx\\
&=\int_{A}\overline{f}u_1dx+\int_{A}(u_1\circ T)f_1 dx
 -\int_{A}f_1u_1dx-\int_{A}(u_1\circ T)\overline{f}dx\\
&=\int_{A}(u_1\circ T-u_1)(f_1-\overline{f})dx>(S_2-S_1)\int_{A}f_1dx>0.
\end{align*}
Therefore $\int_{\Omega} \overline{f}u_1dx>\int_{\Omega} f_1u_1dx$,
 which contradicts \eqref{e114}.

Since $s=\frac{1}{2}$, by Lemma \ref{l8} we know that $u_1$ has no level 
set of positive measure on $E^c$, then by Lemma \ref{l2} we infer 
the existence of an increasing function $\phi_1$ such that
$\phi_1(u_1)$ is a rearrangement of $f_1$ on $E^c$. Now we define an 
increasing function $\phi_2$ as
$$
\phi_2(t)=\begin{cases}
 0 &t<S,\\
 \phi_1(t) &t\geq S.
\end{cases}
$$
It is easy to check that $\phi_2(u_1)$ is a rearrangement of $f_1$ 
on $\Omega$. By  \eqref{e114} and Lemma \ref{l3} we infer that 
$f_1=\phi_2(u_1)$, so that we complete the proof.
\end{proof}

\begin{theorem}\label{thm2}
   Suppose that $f_0$ is positive, $\theta(x)\equiv m\geq\theta_0$ and  $s=1/2$. 
Then  there exists $f_2\in\mathscr{R}(f_0)$ which solves the minimization 
optimization problem {\rm (Opt2)}, i.e.,
\[
{ \Phi(f_2)=}\int_{\Omega} f_2u_{2}\,dx
=\inf_{f\in\mathscr{R}(f_0)}\int_{\Omega} fu_{f}\,dx{ 
=\inf_{f\in\mathscr{R}(f_0)}\Phi(f)}
\]
where $u_{2}=u_{f_2}$.
\end{theorem}

We need some lemmas before we give the proof of the above theorem.

\begin{lemma}\label{l5}
The functional $\Phi|_{\overline{\mathscr{R}(f_0)}}$ is strictly convex.
\end{lemma}

\begin{proof}
 Let $g,h\in\overline{\mathscr{R}(f_0)}$ and $ v\in H^s(\Omega)$,
then for all $t\in (0,1)$, we have
\begin{align*} %\label{e49}
 &2\int_{\Omega} (tg+(1-t)h)vd x-m\|v\|^2\\
&= t\Big(\int_{\Omega} 2gvd x-m\|v\|^2\Big)
+(1-t)\Big(\int_{\Omega}2hvd x-m\|v\|^2\Big).
\end{align*}
 By  \eqref{e3} and \eqref{e00}, and taking the superior relative to 
$ v \in H^s(\Omega)$ in both sides of the above equality, we obtain
 \[ %\nonumber
 \Phi(tg+(1-t)h)\leq t\Phi(g)+(1-t)\Phi(h);
 \]
 that is, the convexity of $\Phi$ has been proved.
 Now, suppose that equality holds in the above inequality for some $ t\in(0,1)$.
 Then, denote by $ u_{t}$  the solution of the problem \eqref{Ps} corresponding 
to $ tg+(1-t)h$, we have
%\label{e51}
  \begin{align*}
 &t\Big(\int_{\Omega}2gu_{t}\,dx-m\|u_t\|^2\Big)
+(1-t)\Big(\int_{\Omega}2hu_{t}\,dx  -m\|u_t\|^2\Big)\\
&= t\Big(\int_{\Omega}2gu_{g}\,dx-m\|u_g\|^2\Big)
+(1-t)\Big(\int_{\Omega}2hu_{h}\,dx-m\|u_g\|^2\Big).
 \end{align*}
 It follows that
\begin{gather*}
 \int_{\Omega}2gu_{t}\,dx-m\|u_t\|^2=\int_{\Omega}2gu_{g}\,dx-m\|u_g\|^2,\\
 \int_{\Omega}2hu_{t}\,dx-m\|u_t\|^2=\int_{\Omega}2hu_{h}\,dx-m\|u_h\|^2.
  \end{gather*}
 By the uniqueness of the minimizer of the functional $ J$, we must have 
$ u_{t}=u_{g}=u_{h}$. Moreover, since
\begin{gather*}
 m\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u_{g}(x)-u_{g}(y))(v(x)-v(y))}
 {| x-y|^{N+2s}}\,dx\,dy=\int_{\Omega} g(x)v(x)d x,\quad
   \forall v \in H^s(\Omega),\\
 m\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}
 {| x-y|^{N+2s}}\,dx\,dy=\int_{\Omega} h(x)v(x)d x,\quad 
  \forall v \in H^s(\Omega),
 \end{gather*}
 if $ u_{g}=u_{h}$, we must have $ g(x)=h(x)$ a.e. in $ \Omega$, 
and the strict concavity is proved.
\end{proof}


\begin{lemma}\label{l6}
The functional $\Phi|_{\overline{\mathscr{R}(f_0)}}$ is weakly continuous.
\end{lemma}

\begin{proof}
  Suppose $\{g_i\}\subset\overline{\mathscr{R}(f_0)}$ and
 $g_i\rightharpoonup g\in\overline{\mathscr{R}(f_0)}$. 
If $u_i$ and $u_g$ are the corresponding solutions of  \eqref{Ps} we have
 \begin{align*}
\Phi(g)+\int_{\Omega} 2(g_i-g)u_{g}\,dx
&= \int_{\Omega}2g_iu_gdx-m\|u_g\|^2\\
&\leq \Phi(g_i)\\
&= \int_{\Omega}2gu_idx-m\|u_i\|^2+\int_{\Omega} 2(g_i-g)u_i\,dx\\
&\leq  \Phi(g)+\int_{\Omega} 2(g_i-g)u_i\,dx.
 \end{align*}
 We have
\[
  \lim_{i\to\infty}\int_{\Omega}(g_i-g)u_gdx=0.
\]
  We only need to prove that
  \begin{equation}\label{e116}
  \lim_{i\to\infty}\int_{\Omega}(g_i-g)u_idx=0.
  \end{equation}
Since $J(u_i)\leq J(0)=0$ and
\[
  J(u_i)\geq \frac{m}{2}\|u_i\|^2-C\|g_i\|\|u_i\|,
\]
it follows that $\{u_i\}$ is bounded in $ H^s(\Omega)$.
  Hence,
  $$
 \big|\int_{\Omega} (g_i-g)u_i\,dx\big|\leq C\|u_i\|\leq
 C.
$$
Now we can choose a subsequence $ \{u_{n_j}\}$ such that
\[
 \lim_{j\to\infty}\big|\int_{\Omega}(g_{n_j}-g)u_{n_j}\,dx\big|
=\limsup_{i\to\infty}\big|\int_{\Omega}(g_i-g)u_i\,dx\big|.
\]
 Noting that $\{u_{n_j}\}$ is also bounded in $ H^s(\Omega)$, going if
necessary to a subsequence, we may assume that
$u_{n_j}\rightharpoonup u$ in $ H^s(\Omega)$  and $ u_{n_j}\to u$ in
$ L^{2}(\Omega)$ as $j\to\infty$.
By  the H\"{o}lder inequality, we obtain
\begin{align*}
 \big|\int_{\Omega}(g_{n_j}-g)u_{n_j}\,dx\big|
&\leq\big|\int_{\Omega} (g_{n_j}-g)(u_{n_j}-u)d x\big|
+\big|\int_{\Omega}(g_{n_j}-g)ud x\big|\\
 &\leq\|g_{n_j}-g\|_{L^{2}}\|u_{n_j}-u\|_{L^{2}}
+\big|\int_{\Omega}(g_{n_j}-g)ud x\big|\to 0
   \end{align*}
 as $j\to\infty$.
 So that
 $$
 0\leq\liminf_{i\to\infty}\big|\int_{\Omega}(g_i-g)u_i\,dx\big|
\leq\limsup_{i\to\infty}\big|\int_{\Omega}(g_i-g)u_i\,dx\big|\leq0,
 $$
 which implies \eqref{e116}, and then $\Phi(g_i)\to\Phi(g)$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
 Denote by $\overline{\mathscr{R}(f_0)}$ the weak closure of
 $ \mathscr{R}(f_0)$ in $ L^{2}(\Omega)$, then $\overline{\mathscr{R}(f_0)}$ 
is convex and weakly compact in $L^{2}(\Omega)$ by  Lemma \ref{l7}. 
By using Lemma \ref{l5} and Lemma \ref{l6}, one can prove easily that 
$\Phi$ has a unique minimizer in $\overline{\mathscr{R}(f_0)}$.
 Let $\underline{g}\in \overline{\mathscr{R}(f_0)}$ be the minimizer of 
$\Phi$. If $g\in\overline{\mathscr{R}(f_0)}$ and $0<t<1$ we have 
$g_t=t\underline{g}+(1-t)g\in\overline{\mathscr{R}(f_0)}$. 
By the minimality condition, we have
 \begin{equation}\label{e42}
 \Phi(\underline{g})\leq \Phi(g_t).
 \end{equation}
 If $u_g$, $u_{\underline{g}}$ and $u_{t}$ are solutions of  \eqref{Ps}
 corresponding to $g$, $\underline{g}$ and $g_t$, respectively, we have
 \begin{equation}\nonumber
u_t=tu_{\underline{g}}+(1-t)u_g.
 \end{equation}
 Therefore, by  \eqref{e3} \eqref{e00} and \eqref{e42}, we have
\begin{align*}
 &\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(u_{\underline{g}}(x)-u_{\underline{g}}(y))^2}{| x-y|^{N+2s}}\,dx\,dy\\
&\leq \int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\Big(\frac{t(u_{\underline{g}}(x)-u_{\underline{g}}(y))+(1-t)(u_{g}(x)-u_{g}(y))}
{| x-y|^{\frac{N+2s}{2}}}\Big)^2\,dx\,dy.
 \end{align*}
 After simplification, we obtain
\begin{align*}
&(1+t)\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(u_{\underline{g}}(x)-u_{\underline{g}}(y))^2}{| x-y|^{N+2s}}\,dx\,dy\\
&\leq  2t\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(u_{\underline{g}}(x)-u_{\underline{g}}(y))(u_{g}(x)-u_{g}(y))}
 {| x-y|^{N+2s}}\,dx\,dy\\
&\quad  + (1-t)\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(u_{g}(x)-u_{g}(y))^2}{| x-y|^{N+2s}}\,dx\,dy.
\end{align*}
Letting $t\to 1$, we find
\[
 \int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u_{\underline{g}}(x)-u_{\underline{g}}(y))^2}{| x-y|^{N+2s}}\,dx\,dy
\leq \int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u_{\underline{g}}(x)-u_{\underline{g}}(y))(u_{g}(x)-u_{g}(y))}
{| x-y|^{N+2s}}\,dx\,dy.
\]
 We can rewrite the latter inequality as
\[
 \int_{\Omega}\underline{g}u_{\underline{g}}dx
\leq \int_{\Omega}gu_{\underline{g}}dx, \quad
\forall g\in \overline{\mathscr{R}(f_0)}.
\]
Since $f_0$ is positive it is easy to deduce that 
$\underline{g}>0$ a.e. $x\in \Omega$. By Lemma \ref{l8} we have that
 each level set of $u_{\underline{g}}$ has zero measure. By Lemmas \ref{l2} 
and \ref{l3}, there exists a decreasing function $\phi$ such that 
$\phi\circ u_{\underline{g}}$ is a rearrangement of $\underline{g}$ and 
the unique minimizer of the linear functional $\int_{\Omega}gu_{\underline{g}}dx$, 
related to $g\in\overline{\mathscr{R}(f_0)}$. 
So that $\underline{g}=\phi\circ u_{\underline{g}}\in \mathscr{R}(f_0)$. 
We complete the proof by letting $f_2=\underline{g}$.
\end{proof}

\subsection*{Acknowledgements}
This work was supported by Natural Science Foundation of China
(11171247, 11371273, 11471235,11501437) and the scientific research
fund project  of Taizhou University (TZXY2015QDXM033).
The authors are  grateful to the anonymous  referee for
the useful comments. 


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\end{document}