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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 12, pp. 1--23.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/12\hfil A fractional elliptic problem]
{Existence of solutions for a fractional elliptic problem with
 critical Sobolev-Hardy nonlinearities in $\mathbb{R}^N$}

\author[L. Jin, S. Fang \hfil EJDE-2018/12\hfilneg]
{Lingyu Jin, Shaomei Fang}

\address{Lingyu Jin \newline
Department of Mathematics,
South China Agricultural University,
Guangzhou 510642,  China}
\email{jinlingyu300@126.com}

\address{Shaomei Fang \newline
Department of Mathematics,
South China Agricultural  University,
Guangzhou 510642,  China}
\email{fangshaomeidz90@126.com}

\dedicatory{Communicated by Raffaella Servadei}

\thanks{Submitted April 8, 2017. Published January 10, 2018.}
\subjclass[2010]{35J10, 35J20, 35J60}
\keywords{Fractional Laplacian; compactness; positive solution;
\hfill\break\indent  unbounded domain; Sobolev-Hardy nonlinearity}

\begin{abstract}
 In this article, we study the fractional elliptic equation with critical
 Sobolev-Hardy nonlinearity
 \begin{gather*}
 (-\Delta)^{\alpha} u+a(x) u=\frac{|u|^{2^*_{s}-2}u}{|x|^s}+k(x)|u|^{q-2}u,\\
 u\in H^\alpha(\mathbb{R}^N),
 \end{gather*}
 where $2<q< 2^*$, $0<\alpha<1$, $N>4\alpha$, $0<s<2\alpha$,
 $2^*_{s}=2(N-s)/(N-2\alpha)$ is the critical Sobolev-Hardy exponent,
 $2^*=2N/(N-2\alpha)$ is the critical Sobolev exponent,
 $a(x),k(x)\in C(\mathbb{R}^N)$. Through a compactness
 analysis of the functional associated, we
 obtain the existence of positive solutions under
 certain assumptions on $a(x),k(x)$.
\end{abstract}

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

\section{Introduction}

We consider the nonlinear elliptic equation
 \begin{equation}\label{e1.1}
 \begin{gathered}
 (-\Delta)^{\alpha} u+a(x) u=\frac{|u|^{2^*_{s}-2}u}{|x|^s}+k(x)|u|^{q-2}u,
 \quad x\in \mathbb{R}^N,\\
 u \in H^\alpha(\mathbb{R}^N),
 \end{gathered}
 \end{equation}
where $2<q <2^*$, $0<\alpha<1$, $0<s<2\alpha$, $N>4\alpha$,
 $2^*_{s}=2(N-s)/(N-2\alpha)$ is the critical Sobolev-Hardy exponent,
 $2^*=2N/(N-2\alpha)$ is the critical Sobolev exponent,
 $a(x),k(x)\in C(\mathbb{R}^N)$.

Recently the fractional Laplacian and more general nonlocal operators of
elliptic type have been widely studied, both for their interesting
theoretical structure and concrete applications in many fields such as
optimization, finance, phase transitions, stratified materials,
anomalous diffusion and so on (see \cite{BV,DMP,FB,FQ,NPV,SV,SV1,SZY}).
In particular, many results have been obtained for elliptic equations
with critical nonlinearity related to \eqref{e1.1}.
Dipierro et al.\ \cite{DMP} considered the critical problem with Hardy-Leray
potential
 \begin{equation}\label{e1.1j}
 \begin{gathered}
 (-\Delta)^{\alpha} u-\gamma \frac{u}{|x|^{2\alpha}}=|u|^{2^*-2}u, \quad
 x\in \mathbb{R}^N,\\
 u\in \dot{H}^\alpha(\mathbb{R}^N),
 \end{gathered}
 \end{equation}
where $\dot{H}^\alpha (\mathbb{R}^N)$ is defined in \eqref{e1.5jj}.
 They proved existence, certain qualitative properties and asymptotic behavior
of positive solutions to \eqref{e1.1j}. Ghoussoub and Shakerian in \cite{GS}
investigated the double critical problem in $\mathbb{R}^N$,
 \begin{equation}\label{e1.1*}
 \begin{gathered}
 (-\Delta)^{\alpha} u-\gamma \frac{u}{|x|^{2\alpha}}
=\frac{|u|^{2^*_{s}-2}u}{|x|^s}+|u|^{2^*-2}u, \quad x\in \mathbb{R}^N,\\
 u>0,\quad u\in \dot{H}^\alpha(\mathbb{R}^N),
 \end{gathered}
 \end{equation}
with $\gamma>0$.
 There through the non-compactness analysis of the Palais-Smale sequence of
\eqref{e1.1*}, they obtained the existence of the solutions.
 Also Yang etc. in \cite{YY}, \cite{WY} consider a class of critical problems
with a Hardy term for the fractional Laplacian in a bounded domain.
For the two gathered of the spectral fractional Laplacian and of the regional
fractional Laplacian, they obtained the existence of solutions respectively.
In addition, the authors in \cite{JD} established a concentration-compactness
result for a fractional Schr\"{o}dinger equation with the subcritical
nonlinearity $f(x,u)$. Motivated by \cite{DMP,GS,JD,YY,WY} we consider the
existence of positive solutions for problem \eqref{e1.1} in $\mathbb{R}^N$.
The main interest for this type of problems, in addition to the nonlocal
fractional Laplacian is the presence of the singular potential $1/|x|^s$
related to the fractional Sobolev-Hardy's inequality. We recall the Sobolev-Hardy
inequality
 \begin{equation}\label{e1.4j}
 \Big(\int_{\mathbb{R}^N}\frac{|u(x)|^{2^*_s}}{|x|^s}dx\Big)^{2/2^*_s}
\leq c\int_{\mathbb{R}^N}|(-\Delta)^{\alpha/2}u(x)|^2dx,\quad
 \forall  u\in \dot{H}^\alpha(\mathbb{R}^N),
 \end{equation}
where $c$ is a positive constant.
 The Sobolev embedding $\dot{H}^\alpha(\mathbb{R}^N)
\hookrightarrow L^{2^*_s}(|x|^{-s}, \mathbb{R}^N)$ is not compact,
even locally, in any neighborhood of zero. As it is well known, the loss of
the compactness of the embeddings is one of the main difficulties for elliptic
problems with critical nonlinearities. Thus our problem has two factors,
 one is the critical Sobolev-Hardy term, the other is the unbounded domain.
In \cite{DMP} and \cite{GS}, the authors can consider the solutions of critical
problems in the homogeneous fractional Sobolev space $\dot{H}^\alpha(\mathbb{R}^N)$,
while we must deal with \eqref{e1.1} in the nonhomogeneous fractional Sobolev space
$H^\alpha(\mathbb{R}^N)$ given the presence of low sub-critical terms in \eqref{e1.1}.
This is why the methods in \cite{DMP} and \cite{GS} can not be used directly
to \eqref{e1.1}. As far as we know, the existence results for global problems
for the fractional Lapalacian with a mixture of critical Sobolev-Hardy terms
and subcritical terms are relatively new. To overcome the difficulties caused
by the lack of compactness, we carry out a non-compactness analysis which can
distinctly express all the parts which cause non-compactness.
As a result, we are able to obtain the existence of nontrival solutions of the
elliptic problem with the critical nonlinear term on an unbounded domain by
getting rid of these noncompact factors.
 To be more specific, for the Palais-Smale sequences of the variational functional
 corresponding to \eqref{e1.1} we first establish a complete noncompact expression
which includes all the blowing up
 bubbles caused by the critical Sobolev-Hardy nonlinearity and by the unbounded
 domain. Then we derive the existence of positive solutions for \eqref{e1.1}.
Our methods are based on some techniques of 
\cite{Y1,DJP,JD,L1,PP,Sm,S2,Y,ZC}.

\subsection*{Notation and assumptions}
 Denote $c$ and $C$ as
 arbitrary constants which may change from line to line. Let $B(x,r)$
denote a ball centered at $x$ with radius $r$ and
$B(x,r)^C=\mathbb{R}^N\setminus B(x,r)$.

 Let $N\geq 1$, $u\in L^2(\mathbb{R}^N)$, let the Fourier transform of $u$ be
 $$
\widehat{u}(\xi)=\frac{1}{(2\pi)^{N/2}}\int_{\mathbb{R}^N}e^{-i\xi\cdot x}u(x)dx.
$$
 We define the operator $(-\Delta)^\alpha u$ by the Fourier transform
$$
\widehat{(-\Delta)^{\alpha} u}(\xi)=|\xi|^{2\alpha}\hat{u}(\xi),\quad
 \forall u\in C^\infty _0(\mathbb{R}^N).
$$
 Let $\dot{H}^\alpha(\mathbb{R}^N)$ be the homogeneous fractional Sobolev
space as the completion of $C^\infty _0(\mathbb{R}^N)$ under the norm
 \begin{equation}\label{e1.5jj}
 \|u\|_{\dot{H}^\alpha(\mathbb{R}^N)}=\||\xi|^\alpha\widehat{u}\|_{L^2(\mathbb{R}^N)},
 \end{equation}
and denote by $H^\alpha(\mathbb{R}^N)$ the usual nonhomogeneous fractional
 Sobolev space with the norm
\begin{equation}\label{e1.6}
 \|u\|_{H^\alpha(\mathbb{R}^N)}=\|u\|_{L^2(\mathbb{R}^N)}
+\||\xi|^\alpha\widehat{u}\|_{L^2(\mathbb{R}^N)}.
 \end{equation}
 For $0<\alpha<1$, a direct calculation (see e.g.
\cite[Proposition 4.4]{NPV},  or \cite[Proposition 1.2]{DMP}, gives
$$
c_{N,s}\int_{\mathbb{R}^N}\int_{ \mathbb{R}^N}
\frac{|u(x)-u(y)|^2}{|x-y|^{N+2\alpha}}\,dx\,dy
=\int_{\mathbb{R}^N}|(-\Delta )^{\alpha/2}u(x)|^2dx
=\|u\|_{\dot{H}^\alpha(\mathbb{R}^N)}^2,
$$
where $c_{N,s}=2^{2s-1}\pi^{-\frac{N}{2}}\frac{\Gamma(\frac{N+2s}{2})}{|\Gamma(-s)|}$.


Let $u^+=\max \{u,0\}$, $u^-=u^+-u$. From the proof of \eqref{e2.10} in \cite{PAJ},
it follows
\begin{equation}
\|u^+\|_{\dot {H}^\alpha}\leq \|u\|_{\dot {H}^\alpha}.
\end{equation}
We call $u\not\equiv 0$ in $\mathbb{R}^N$ if the measure of the set
$\{x\in \mathbb{R}^N|u(x)\not =0\}
$ is positive.

Recall the definition of Morrey space. A measurable function
$u:\mathbb{R}^N \to \mathbb{R}$ belongs to the Morrey space with $p\in [1,\infty)$
and $\nu \in (0,N]$, if and only if
\[
\|u\|^p_{L^{p,\nu}(\mathbb{R}^N)}
=\sup_{r>0,\bar x\in \mathbb{R}^N}r^{\nu-N}\int_{B(\bar x,r)}|u(x)|^pdx<\infty.
\]
By H\"older inequality, we can verify (refer to \cite{NPV})
 \begin{equation}
 L^{2^*}(\mathbb{R}^N)\hookrightarrow L^{p,\nu}(\mathbb{R}^N),
\quad\text{for } 1\leq p<2^*,
 \end{equation}
and
 \begin{equation}
L^{p,\frac{(N-2\alpha)p}{2}}(\mathbb{R}^N)\hookrightarrow L^{p_1,
\frac{(N-2\alpha)p_1}{2}}(\mathbb{R}^N), \quad \text{for } 1<p_1<p<2^*.
 \end{equation}
 Moreover, we have $ L^{p,\nu}(\mathbb{R}^N)\hookrightarrow
L^{1,\frac{\nu}{p}}(\mathbb{R}^N)$.

Next we give the definition of the Palais-Smale sequence.
 Let $X$ be a Banach space, $\Phi\in C^1(X,\mathbb{R})$, $c\in \mathbb{R}$,
we call $\{u_n\} \subset X$ is a Palais-Smale sequence of $\Phi$ if
 \begin{equation}
 \Phi(u_n)\to c,\; \Phi'(u_n)\to 0\quad \text{as }n\to \infty.
 \end{equation}
In this article we assume that:
\begin{itemize}
\item[(H1)] $ a(x)\in C(\mathbb{R}^N)$, $ k(x)\in C(\mathbb{R}^N)$;

\item[(H2)]
\begin{gather*}
\lim_{|x|\to\infty} a(x)=\bar a>0,\quad
\lim_{|x|\to\infty} k(x)= \bar k>0,\\
\inf_{x\in\mathbb{R}^N}a(x)={\hat a}>0,\quad
\inf_{x\in\mathbb{R}^N}k(x)={\hat k}>0.
\end{gather*}
\end{itemize}
In this article, we assume that $a(x),k(x)$ always satisfy (H1) and (H2).
The energy functional associated with \eqref{e1.1} is for all
$ u\in H^{\alpha}(\mathbb{R}^N)$,
\begin{align*}
I(u)&=\frac{1}{2}\int_{\mathbb{R}^N}\Big (|(-\Delta)^{\alpha/2}u(x)|^2
 +a(x)|u(x)|^2\Big )\,dx \\
&\quad-\frac{1}{2^*_{s}}\int_{\mathbb{R}^N}\frac{(u^+(x))^{2^*_{s}}}{|x|^s}dx
-\frac{1}{q}\int_{\mathbb{R}^N}k(x)(u^+(x))^q \,dx.
\end{align*}
Finally we present some problems
associated to \eqref{e1.1} as follows.

 The limit equation of \eqref{e1.1} involving subcritical terms is
\begin{equation}\label{e2.1}
\begin{gathered}
(-\Delta)^{\alpha} u+\bar a u=\bar{k}|u|^{q-2}u,\\
u\in H^{\alpha}(\mathbb{R}^N),
\end{gathered}
\end{equation}
and its corresponding variational functional is
\begin{align*}
I^\infty(u)&=\frac{1}{2}\int_{\mathbb{R}^N}\Big (|(-\Delta)^{\alpha/2}u(x)|^2+\bar a
|u(x)|^2\Big)dx\\
&\quad -\frac{1}{q}\int_{\mathbb{R}^N}\bar k
(u^+(x))^qdx, \quad
 u\in H^{\alpha}(\mathbb{R}^N).
\end{align*}
The limit equation of \eqref{e1.1}
involving the Sobolev-Hardy critical nonlinear term is
\begin{equation}\label{e1.4}
\begin{gathered}
(-\Delta)^{\alpha} u=\frac{|u|^{2^*_{s}-2}u}{|x|^s},\\
u\in \dot{H}^\alpha(\mathbb{R}^N),
\end{gathered}
\end{equation}
and the corresponding variational functional is
$$
I_s(u)=\frac{1}{2}\int_{\mathbb{R}^N} |(-\Delta)^{\alpha/2}u(x)|^2dx
-\frac{1}{2^*_{s}}\int_{\mathbb{R}^N}\frac{(u^+(x))^{2^*_{s}}}{|x|^s}dx,
\quad u\in \dot{H}^\alpha(\mathbb{R}^N).
$$

In \cite{Y1} Chen and Yang proved that
all the positive solutions of \eqref{e1.4} are of the form
 \begin{equation}\label{e1.5j}
U^\varepsilon(x):=\varepsilon^{\frac{2\alpha-N}{2}} U
(x/\varepsilon),
 \end{equation}
 and $U(x)$ satisfies
\begin{equation}\label{e1.5}
 \frac{C_1}{1+|x|^{N-2\alpha}}\leq U(x)\leq \frac{C_2}{1+|x|^{N-2\alpha}},
 \end{equation}
where $C_2>C_1>0$ are constants.
These solutions are also minimizers for the quotient
 $$
 S_{\alpha,s}=\inf_{u\in
 \dot{H}^\alpha(\mathbb{R}^N) \backslash\{0\}}
\frac{\int_{\mathbb{R}^N}|(-\Delta)^{\alpha/2}
 u(x)|^2dx}{\big(\int_{\mathbb{R}^N}\frac{|u(x)|^{2^*_{s}}}{|x|^s}dx\big)^{2/2^*_{s}}},
$$
which is associated with the fractional Sobolev-Hardy inequality \eqref{e1.4j}.
 Define
\begin{gather}\label{e1.10j}
D_0=\int_{\mathbb{R}^N}\Big(\frac{1}{2}|(-\Delta)^{\alpha/2}
U(x)|^2-\frac{1}{2^*_{s}}\frac{|U(x)|^{2^*_{s}}}{|x|^s}\Big)dx
=\frac{2\alpha-s}{2(N-s)}S_{\alpha,s}^{\frac{N-s}{2\alpha-s}},
\\
\begin{aligned}
\mathcal{N}=\Big\{& u\in H^{\alpha}(\mathbb{R}^N)\setminus
\{0\}: \int_{\mathbb{R}^N}\Big(|(-\Delta)^{\alpha/2}u(x)|^2+\bar
a|u(x)|^2 \\
& -\bar k(u^+(x))^q\Big)\,dx=0\Big\},
\end{aligned} \\
J^\infty=\inf_{u\in \mathcal{N}}I^\infty (u).
\end{gather}
It is known that $\mathcal{N} \neq\emptyset$ since problem \eqref{e2.1} has
at least one positive solution if $N> 2\alpha$ (see \cite{S})
for $1<q<2^*$.

The main result of our paper is as follows.

\begin{theorem}\label{thm1.1}
Suppose $a(x),\,k(x)$ satisfy {\rm (H1)} and {\rm (H2)},
$ 2<q<2^*$, $0<\alpha<1$, $N>4\alpha$, $0<s<2\alpha$.
Assume that $\{u_n\}$ is a positive
Palais-Smale sequence of I at level $d\geq 0$, then there exist
two sequences
$\{R^i_n\}\subset\mathbb{R}^+\,(1\leq i\leq l_1$) and
$\{y^j_n\}\subset\mathbb{R}^N\,(1\leq j\leq
l_2), \, u\in H^{\alpha}(\mathbb{R}^N)$, and
$ u_j\in H^{\alpha}(\mathbb{R}^N) \,(1\leq j\leq l_2), (l_1,l_2\in \mathbb{N})$
such that up to a subsequence:
\begin{gather}
d=I(u)+l_1D_0+\sum^{l_2}_{j=1}I^\infty(u_j); \nonumber\\
\label{b1}
\|u_n-u-\sum^{l_1}_{i=1}U^{R^i_n}
-\sum^{l_2}_{j=1}u_j(x-y^j_n)\|_{H^{\alpha}(\mathbb{R}^N)}=o(1)\quad
\text{as }n\to\infty
\end{gather}
where $u$ and $u_j\,(1\leq j\leq l_2)$ satisfy
\begin{gather*}
I' (u)=0,\quad {I^{\infty} }'(u_j)=0,\\
R^i_n\to 0\; (1\leq i\leq l_1),\quad
|y^j_n|\to \infty\;(1\leq j\leq l_2)\quad \text{as }n\to\infty.
\end{gather*}
In particular, if $u\not\equiv 0$, then $u$ is a weakly solution of \eqref{e1.1}.
Note that the corresponding sum in \eqref{b1} will be treated as zero if
$l_i=0$ $(i=1,2)$.
\end{theorem}

\begin{remark} \label{rmk1} \rm
(1) Similar to \cite[Corollary 3.3]{Sm}, one can show that any Palais-Smale
sequence for $I$ at a level  which is not of the form $m_1D_0+m_2 J^\infty$,
 $m_1,m_2\in\mathbb{N}\bigcup \{0\}$, gives rise to a non-trivial weak
 solution of equation \eqref{e1.1}.

(2) In our non-compactness analysis, we prove
 that the blowing up positive Palais-Smale sequences can bear exactly
 two kinds of bubbles. Up to harmless constants, they are either
 of the form
 $$
U^{R_n}(x), \quad |R_n|\to 0\quad \text{as } n\to  \infty,
$$
or
$$
u(x-y_n)\in H^{\alpha}(\mathbb{R}^N),\quad |y_n|\to\infty,\quad \text{as }
n\to  \infty,
$$
where $u$ is the solution of \eqref{e2.1}. For any Palais-Smale sequence
${u_n} $ for $I$, ruling out the above two bubbles yields the existence of a
 non-trivial weak solution of equation \eqref{e1.1}.

(3) Because of the lower order terms $a(x)u$ and $k(x)|u|^{q-2}u$
in \eqref{e1.1}, we must deal with $u\in H^\alpha(\mathbb{R}^N)$ to ensure that
the functional $I(u)$ is well defined. In fact, if $u\in H^\alpha(\mathbb{R}^N)$,
by the Sobolev inequality, $u\in L^2(\mathbb{R}^N)$ and
$ u\in L^q(\mathbb{R}^N)$ for $2<q<2^*$. Noting that $\|u\|_{L^2}$ and
$\|u\|_{L^q}$ only satisfy the translation invariance and
 $\int_{\mathbb{R}^N}\frac{(u^+(x))^{2^*_s}}{|x|^s}dx$ only satisfies the
scaling invariance, then there exists a new limit equation \eqref{e2.1}
which causes some new structures for the Palais-Smale sequence of \eqref{e1.1}.
\end{remark}

Using the compactness results and the Mountain Pass Theorem \cite{BN}
 we prove the following existence result.

\begin{theorem}\label{thm1.2}
Assume that $2<q<2^*$, $0<\alpha<1$, $0<s<2\alpha$, $N> 4\alpha$.
If $a(x),k(x)$ satisfy {\rm (H1), (H2)} and
\begin{equation}
\bar a\geq a(x), \quad k(x)\geq \bar k>0,\quad k(x)\not \equiv \bar k.
\end{equation}
Then \eqref{e1.1} has a nontrivial solution $u\in H^{\alpha}(\mathbb{R}^N)$ which
satisfies
$$
I(u)< \min\big\{\frac{2\alpha-s}{2(N-s)}S_{\alpha,s}^{\frac{N-s}{2\alpha-s}},
J^\infty\big\}.
$$
\end{theorem}

This paper is organized as follows. In Section 2, we prove Theorem \ref{thm1.1} by
carefully analyzing the features of a positive Palais-Smale
sequence for $I$. Theorem \ref{thm1.2} is proved in Section 3 by
applying Theorem \ref{thm1.1} and the Mountain Pass Theorem.
Finally we put some preliminaries in the last section as an appendix.

\section{Non-compactness analysis}

In this section, we prove Theorem \ref{thm1.1} by using the Concentration-Compactness
Principle and a delicate analysis of the Palais-Smale sequences of
$I$.
Firstly we give the following Lemmas.

\begin{lemma}\label{l6}
 Let $0<\alpha <N/2,\, 0<s<2\alpha,\, r>0 $,
$\{u_n\}\subset {\dot{H}}^\alpha(\mathbb{R}^N)$ be a bounded sequence such that
 \begin{equation}\label{l6.1}
 \inf_{n\in \mathbb{N}}\int_{B(0,r)}\frac{\bigl(u^+_n(x)\bigl)^{2^*_{s}}}{|x|^s}dx\geq c>0.
 \end{equation}
Then, up to subsequence, there exist two sequences $\{r_n\}\subset \mathbb{R}^+$
and $\{x_n\} \subset B(0,2r) $
 such that
 \begin{equation}\label{l6.2}
 \bar u_n \rightharpoonup w\not\equiv 0 \quad \text{in }
 \dot{H}^\alpha(\mathbb{R}^N),
\end{equation}
 where
\begin{equation}\label{e2.3j}
 \bar u_n(x)=\begin{cases}
 r_n^{\frac{N-2\alpha}{2}}u_n(r_n x)
&\text{when $x_n/r_n$ is bounded, }\\
 r_n^{\frac{N-2\alpha}{2}}u_n(r_n x+x_n)
&\text{when } |x_n/r_n|\to \infty.
 \end{cases}
 \end{equation}
\end{lemma}

\begin{proof}
Let $0\leq \eta (x)\leq 1$, $\eta(x)\in C^\infty_0(\mathbb{R}^N),
\eta(x)\equiv 1$  on $B(0,r)$, $\eta(x)\equiv 0$ on $B(0,2r)^C$.
 From \cite[Lemma 5.3]{NPV}, it follows that
 \begin{equation}\label{e2.20j}
 \|\eta u_n\|_{\dot{H}^\alpha(\mathbb{R}^N)}
\leq C \| u_n\|_{\dot{H}^\alpha(\mathbb{R}^N)}.
 \end{equation}
By \cite[Theorem 1.2]{Y1},
 \begin{equation}\label{e2.4jj}
\Big(\int_{\mathbb{R}^N}\frac{|\eta(x)u_n(x)|^{2^*_{s}}}{|x|^s}dx
\Big)^{1/2^*_s}
\leq C \|\eta u_n\|^\theta_{\dot{H}^\alpha(\mathbb{R}^N)}
\|\eta u_n\|^{1-\theta}_{L^{2,N-2\alpha}(\mathbb{R}^N)},
 \end{equation}
where $\max\{\frac{N-2\alpha}{N-s},\frac{2\alpha-s}{N-s}\}\leq \theta<1$.
 From \eqref{e2.20j} and \eqref{e2.4jj}, it follows
 \begin{equation}
 \begin{aligned}
 c&\leq \Big(\int_{B(0,r)}\frac{\bigl(u^+_n(x))^{2^*_{s}}}{|x|^s}dx\Big)^{1/2^*_s}
\leq \Big(\int_{\mathbb{R}^N}\frac{|\eta(x)u_n(x)|^{2^*_{s}}}{|x|^s}dx
 \Big)^{1/2^*_s} \\
&\leq C \|u_n\|^\theta_{\dot{H}^\alpha(\mathbb{R}^N)}
\|\eta u_n\|^{1-\theta}_{L^{2,N-2\alpha}(\mathbb{R}^N)}.
 \end{aligned}
 \end{equation}
Then there exists a constant $c>0$ such that
 \begin{equation}\label{e2.22j}
 \|\eta u_n\|_{L^{2,N-2\alpha}(\mathbb{R}^N)}^2
=\sup_{\bar x\in \mathbb{R}^N,\, R\in \mathbb{R}^+}R^{-2\alpha}\int_{B(\bar x,R)}
|\eta(x)u_n(x)|^2dx\geq c>0.
 \end{equation}

From \eqref{e2.22j}, we may find $r_n>0$ and $x_n \in B(0,2r)$ such that for
$n$ large enough,
 \begin{equation}\label{e2.23j}
 r_n^{-2\alpha}\int_{B(x_n,r_n)}|\eta(x)u_n(x)|^2dx
\geq \|\eta u_n\|_{L^{2,N-2\alpha}(\mathbb{R}^N)}^2-\frac{c}{2n}\geq c/2>0.
 \end{equation}
Denote \begin{equation}\label{e2.3}
 \bar u_n(x)=\begin{cases}
 r_n^{\frac{N-2\alpha}{2}}u_n(r_n x)
&\text{when }\frac{x_n}{r_n} \text{ is bounded, }\\
 r_n^{\frac{N-2\alpha}{2}}u_n(r_n x+x_n)
&\text {when } |\frac{x_n}{r_n}|\to \infty.
 \end{cases}
\end{equation}
 Since $\{u_n\}$ is bounded in $\dot{H}^\alpha(\mathbb{R}^N)$, from the scaling
and translation invariance of $\dot{H}^\alpha(\mathbb{R}^N)$, we have
 $\{\bar u_n\}$ is bounded in $\dot{H}^\alpha(\mathbb{R}^N)$;
 therefore, up to a subsequence (still denoted by $\bar u_n$),
 $$
\bar u_n \rightharpoonup w \text{ in } \dot{H}^\alpha(\mathbb{R}^N),
\quad\text{and}\quad
\bar u_n \to w \text{ in } L^2_{\mathrm{loc}}(\mathbb{R}^N),
\quad \text{as } n\to \infty.
$$

If $x_n/r_n$ is bounded, there exist a $\tilde{R}>1$ such that
$B(\frac{x_n}{r_n},1)\subset B(0,\tilde{R})$, then
 \begin{equation}\label{e2.9}
 c/2<\int_{B(\frac{x_n}{r_n},1)}|\bar u_n(x)\eta(r_nx)|^2dx\leq \int_{B(0,\tilde{R})}|\bar u_n(x)|^2dx\to \int_{B(0,\tilde{R})}|w(x)|^2dx.
 \end{equation}
 If $|\frac{x_n}{r_n}|\to \infty$, then
 \begin{equation}\label{e2.10j}
\begin{aligned}
 c/2&<\int_{B(0,1)}|\bar u_n(x)\eta(r_nx+x_n  )|^2dx
\leq \int_{B(0,\tilde{R})}|\bar u_n(x)|^2dx\\
& \to \int_{B(0,\tilde{R})}|w(x)|^2dx
\end{aligned}
 \end{equation}
where $\tilde{R}>1$.
 Obviously we have $w\not \equiv 0$.
 From \eqref{e2.9} and \eqref{e2.10j}, Lemma \ref{l6} is complete.
\end{proof}

\begin{lemma}\label{lem3.7}
Assume $N>4\alpha,0<s<2\alpha,2<q< 2^*, 0<\alpha<1$.
Let $\{v_n\}\subset H^{\alpha}(\mathbb{R}^N)$ be a Palais-Smale sequence of $I$ at
 level $d_1$ and
$v_n\rightharpoonup 0$ in $H^{\alpha}(\mathbb{R}^N)$  as
$n\to \infty$. If there exists a sequence $\{r_n\}\subset
 \mathbb{R}^+$,  with $r_n\to 0$ as $n\to \infty$ such
 that $\bar v_n(x):=r_n^{\frac{N-2\alpha}{2}}v_n (r_n x)$ converges weakly
 in $\dot{H}^\alpha(\mathbb{R}^N)$ and almost everywhere to some $v_0\in
 \dot{H}^\alpha(\mathbb{R}^N)$ as $n\to \infty$ with $v_0\not\equiv 0$,
then $v_0$ solves  \eqref{e1.4} and the sequence
 $z_n(x):=v_n(x)-v_0(\frac{x}{r_n})r_n^{\frac{2\alpha-N}{2}}$ is a
 Palais-Smale sequence of $I$ at level $d_1-I_s(v_0)$.
 \end{lemma}

\begin{proof}
First, we prove that $v_0$ solves \eqref{e1.4} and
 $I(z_n)=I(v_n)-I_s(v_0)$. Fix a ball $B(0,r)$ and a test
 function $\phi\in C^\infty_0(B(0,r))$.
 Since
 \begin{equation}\label{e2.2j}
\bar v_n\rightharpoonup v_0 \text{ in } \dot{H}^\alpha(\mathbb{R}^N),
 \end{equation}
applying Lemma \ref{jl33}, it implies
 \begin{equation}\label{j62.21}
\begin{split}
&\langle\phi, I'_s( v_0)\rangle+o(1)
=\langle\phi, I'_s(\bar v_n)\rangle\\
&=c_{N,s}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(\bar v_n(x)
 -\bar v_n(y))(\phi(x)-\phi(y))}{|x-y|^{N+2\alpha}} \,dx\,dy \\
&-\int_{\mathbb{R}^N}\frac{\bigl(\bar v^+_n(x)\bigl)^{{2^*_{s}}-1}\phi(x)}{|x|^s}
\,dx\\
&=c_{N,s}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(\bar v_n(x)-\bar v_n(y))(\phi(x)-\phi(y))}{|x-y|^{N+2\alpha}} \,dx\,dy \\
&\quad -\int_{\mathbb{R}^N}\frac{\bigl(\bar v^+_n(x)\bigl)^{{2^*_{s}}-1}
 \phi(x)}{|x|^s}\,dx
 +r_n^{2\alpha}\int_{\mathbb{R}^N} a({r_n x})\phi(x)\bar v_n(x)\,dx \\
&\quad -r_n^{N-\frac{N-2\alpha}{2}q}\int_{\mathbb{R}^N}k(r_n x)\phi(x)
  (\bar v^+_n(x))^{q-1}dx+o(1)\\
&=c_{N,s}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(v_n(x)-v_n(y))(\phi_n(x)
 -\phi_n(y))}{|x-y|^{N+2\alpha}} \,dx\,dy \\
&\quad -\int_{\mathbb{R}^N}\frac{(v^+_n(x))^{{2^*_{s}}-1}\phi_n(x)}{|x|^s}\,dx
  +\int_{\mathbb{R}^N} a(x)\phi_n(x) v_n(x)dx \\
&\quad -\int_{\mathbb{R}^N} k(x)\phi_n(x)(v^+_n(x))^{q-1}\,dx+o(1) \\
&=o(1)\quad \text{as }n\to  \infty,
 \end{split}
 \end{equation}
where $\phi_n=r_n^{-\frac{N-2\alpha}{2}}\phi(\frac{x}{r_n})$.
 The last equality in \eqref{j62.21} holds since
\begin{gather*}
\int_{\mathbb{R}^N}|\phi_n(x)|^2dx=r_n^{2\alpha}
\int_{\mathbb{R}^N}|\phi(x)|^2dx=o(1), \\
\|\phi\|_{\dot{H}^\alpha(\mathbb{R}^N)}
=\|\phi_n\|_{\dot{H}^{\alpha}(\mathbb{R}^N)}
=\|\phi_n\|_{{H}^{\alpha}(\mathbb{R}^N)}+o(1),\quad
\text{as  }n\to \infty.
\end{gather*}
 Thus $v_0$ is a nontrival critical point of $I_s$.
By Lemma \ref{l4.6}, \eqref{e1.5}
and the fact $N>4\alpha$, it follows
 \begin{equation}\label{e2.10}
 \int_{\mathbb{R}^N}|v_0(x)|^pdx
\leq c \int_{\mathbb{R}^N}\frac{1}{(1+|x|^{N-2\alpha})^p}dx\leq c,\quad
 \forall p\geq 2,
 \end{equation}
 which implies that $v_0\in L^2(\mathbb{R}^N)$.
 Let
 $$
z_n(x)=v_n(x)-r_n^{\frac{2\alpha-N}{2}}v_0(\frac{x}{r_n})\in
 H^{\alpha}(\mathbb{R}^N).
$$
Obviously $z_n\rightharpoonup 0 \text{
 in }H^{\alpha}(\mathbb{R}^N)\text{ as }n\to \infty$. Now we prove
 that $\{z_n\}$ is a Palais-Smale sequence of $I$ at level $d_1-I_s
 (v_0)$.
 From \eqref{e2.10}, $v_0\in L^p(\mathbb{R}^N)$ for all $p\in [2,2^*)$.
Then it follows that
\begin{equation}\label{x}
 \int_{\mathbb{R}^N}|v_0(\frac{x}{r_n})r_n^{\frac{2\alpha-N}{2}}|^pdx
=r_n^{N-p\frac{(N-2\alpha)}{2}}\|v_0\|^p_{L^p(\mathbb{R}^N)}\to 0
 \end{equation}
 as $n\to\infty$  for all $2\leq p< {2^*}$.
 By the Br\'{e}zis-Lieb Lemma and the weak convergence, similar to
Lemma \ref{q}, we can prove that
\begin{gather*}
I(z_n)=I(v_n)-I_s(v_0), \\
 \langle I'(z_n),\phi\rangle=o(1)
\end{gather*}
 as $n\to \infty$.
This completes the proof.
 \end{proof}


\begin{proof}[Proof of Theorem \ref{thm1.1}]
By Lemma \ref{3.5} in the appendix, we can assume that $\{u_n\}$ is
bounded. Up to a subsequence,  let $n\to \infty$, and assume that
\begin{gather}
\label{t1.11} u_n\rightharpoonup u \quad\text{in } H^{\alpha}(\mathbb{R}^N),\\
\label{t1.12} u_n\to u \quad\text{in } L^p_{\mathrm{loc}}(\mathbb{R}^N)
\text{ for }2\leq p<{2^*},\\
\label{t1.13} u_n\to u \quad \text{ a.e. in } \mathbb{R}^N.
\end{gather}
Denote $v_n(x)=u_n(x)-u(x)$, then $\{v_n\}$ is
a Palais-Smale sequence of $I$ and 
\begin{gather}
\label{t2.11} v_n\rightharpoonup 0 \quad \text{in } H^{\alpha}(\mathbb{R}^N),\\
\label{t2.12} v_n\to 0 \quad \text{in } L^p_{\mathrm{loc}}(\mathbb{R}^N)
\text{ for }2\leq p<{2^*},\\
\label{t2.13}v_n\to 0 \quad \text{ a.e. in } \mathbb{R}^N.
\end{gather}

 Then by Lemma \ref{q} we know that
\begin{gather}
\label{e4.1}I(v_n)=I(u_n)-I(u)+o(1),\text{ as }n\to \infty,\\
\label{e4.2}I'(v_n)=o(1),\text{ as }n\to \infty,\\
\label{e4.3}\|v_n\|_{H^{\alpha}(\mathbb{R}^N)}=\|u_n\|_{H^{\alpha}(\mathbb{R}^N)}-\|u\|_{H^{\alpha}(\mathbb{R}^N)}+o(1),\text{
as }n\to \infty.
\end{gather}

Without loss of generality, we  assume that
$$
\|v_n\|^2_{H^{\alpha}(\mathbb{R}^N)}\to l>0\quad \text{as }n\to\infty.
$$
In fact if $l=0$, Theorem \ref{thm1.1} is proved for $l_1=0,l_2=0$.
\smallskip


\noindent\textbf{Step 1.} Getting rid of the blowing up bubbles caused by
the Sobolev-Hardy term.
Suppose there exists $0<\delta<\infty$ such that
 \begin{equation}\label{e4.6}
\inf_{n\in \mathbb{N}}\int_{|x|<R}\frac{\big(v_n^+(x)\big)^{2^*_s}}{|x|^s}\,dx
\geq \delta>0,\quad \text{for some } 0<R<\infty.
\end{equation}

It follows from Lemma \ref{l6} that there exist two sequences
$\{r_n\}\subset \mathbb{R}^+$ and $\{x_n\}\subset B(0, 2R)$,
such that
\begin{equation}\label{e2.25}
\bar v_n(x)\rightharpoonup v_0\not\equiv 0\quad \text{in }
\dot{H}^{\alpha}(\mathbb{R}^N),
\end{equation}
where
\begin{equation}\label{e2.3jj}
 \bar v_n(x)=\begin{cases}
 r_n^{\frac{N-2\alpha}{2}}v_n(r_n x) &\text{when }
 \frac{x_n}{r_n} \text{ is bounded, }\\
 r_n^{\frac{N-2\alpha}{2}}v_n(r_n x+x_n) &\text {when } |\frac{x_n}{r_n}|\to \infty.
 \end{cases}
 \end{equation}

Now we claim that $r_n\to 0$  as $n\to \infty$.
In fact there exists a $R_1>0$ such that
\begin{equation}\label{jt3.8}
 \int_{B(0,R_1)}|v_0(x)|^pdx=\delta_1>0, \quad \text{for }2\leq p<2^*.
\end{equation}
From the Sobolev compact embedding, \eqref{t1.12}, \eqref{e2.25} and \eqref{jt3.8},
we have that for all $r>0$,
\begin{gather*}
v_n\to 0\quad \text{in $L^p(B(0,r))$ for all }2\leq p<2^*,\\
 \bar v_n\to v_0 \quad \text{in $L^p(B(0,r))$  for all }2\leq p<2^*,
\end{gather*}
\begin{equation}\label{t1.18}
\begin{aligned}
0&\neq\|v_0\|_{L^2(B(0,R_1))}^2+o(1)\\
&=\int_{B(0,R_1)}|\bar v_n(x)|^2dx\\
&=\begin{cases}
r_n^{-2\alpha}\int_{B(0,r_nR_1)}|v_n(x)|^2dx,
 &\text{if }\frac{x_n}{r_n} \text{ is bounded, }\\
r_n^{-2\alpha}\int_{B(x_n,r_nR_1)}|v_n(x)|^2dx
&\text{if } |\frac{x_n}{r_n}|\to \infty.
\end{cases}
\end{aligned}
\end{equation}

If $r_n\to r_0>0$, then
\begin{equation}
\begin{gathered}
r_n^{-2\alpha}\int_{B(0,r_nR_1)}|v_n(x)|^2dx
\leq c r_0^{-2\alpha}\|v_n\|^2_{L^2(B(0,cR_1))}\to 0;\\
r_n^{-2\alpha}\int_{B(x_n,r_nR_1)}|v_n(x)|^2dx
\leq c r_0^{-2\alpha}\|v_n\|^2_{L^2(B(0,cR_1+4R))}\to 0.
\end{gathered}
\end{equation}
If $r_n\to \infty$, then
\begin{equation}
\begin{gathered}
 r_n^{-2\alpha}\int_{B(0,r_nR_1)}|v_n(x)|^2dx
\leq r_n^{-2\alpha}\|v_n\|^2_{H^\alpha(\mathbb{R}^N)}\to 0,\\
r_n^{-2\alpha}\int_{B(x_n,r_nR_1)}|v_n(x)|^2dx
\leq r_n^{-2\alpha}\|v_n\|^2_{H^\alpha(\mathbb{R}^N)}\to 0.
\end{gathered}
\end{equation}
 A contradiction to \eqref{t1.18}. Thus we have $r_n\to 0$.

 Next we claim that $x_n/r_n$ is bounded.
Indeed, if on the contrary, $|\frac{x_n}{r_n}|\to \infty$, fix a ball $B(0,r)$
and a test function $\phi\in C^\infty_0(B(0,r))$, then
 \begin{equation}
 \begin{aligned}
\int_{\mathbb{R}^N}\frac{ (\bar v^+_n(x))^{{2^*_{s}}-1}\phi(x)}
{|x+\frac{x_n}{r_n}|^s} \,dx
&=\int_{B(0,r)}\frac{ (\bar v^+_n(x))^{{2^*_{s}}-1}\phi(x)}
 {|x+\frac{x_n}{r_n}|^s}\,dx\\
&\leq \frac{c}{|\frac{x_n}{r_n}|}\int_{B(0,r)}
 {(\bar v_n^+(x))^{{2^*_{s}}-1}\phi(x)}\,dx\to 0,
 \end{aligned}
\end{equation}
similar to \eqref{j62.21}, it follows that
\begin{equation}
(-\Delta)^\alpha u=0,\quad x\in \mathbb{R}^N
\end{equation}
which implies that $\|v_0\|_{\dot{H}^\alpha(\mathbb{R}^N)}=0$.
By the Sobolev inequality and the H\"older inequality it follows
 \begin{equation}
 \|v_0\|_{L^p(B(0,R_1))}\leq c\|v_0\|_{L^{2^*}(B(0,R_1))}
\leq c\|v_0\|_{L^{2^*}(\mathbb{R}^N)}
\leq C\|v_0\|_{\dot{H}^\alpha(\mathbb{R}^N)}=0
 \end{equation}
 for $2\leq p<2^*$.
This contradicts \eqref{jt3.8}. So we can deduce that $x_n/r_n$
is bounded and $\bar v_n(x)=r_n^{\frac{N-2\alpha}{2}}v_n(r_n x)$.

Define $z_n(x) =v_n(x)-v_0(\frac{x}{r_n})r_n^{\frac{2\alpha-N}{2}}$,
then $ z_n \rightharpoonup 0 $ in $H^\alpha (\mathbb{R}^N)$.
It follows from Lemma  \ref{lem3.7} that $\{z_n\}$ is a Palais-Smale sequence of $I$
 satisfying
 \begin{equation}  \label{e4.12}
I(z_n)=I(v_n)-I_s(v_0)+o(1), \quad \text{as } n\to\infty.
 \end{equation}
Since $v_0$ satisfies \eqref{e1.4}, from Lemma \ref{l4.6}, \eqref{e1.5j}
and \eqref{e1.10j} there exists $\varepsilon_1>0$ such that
 \begin{equation}\label{e2.25j}
 v_0(x)=\varepsilon_1^{\frac{2\alpha-N}{2}}U(\frac{x}{\varepsilon_1}),\quad
 I_s(v_0)=D_0.
 \end{equation}
Let $R_n^1=r_n\varepsilon_1$, from \eqref{e2.25j}, it follows
 \begin{equation}\label{e2.26j}
r_n^{\frac{2\alpha-N}{2}} v_0(\frac{x}{r_n})
={(R^1_n)}^{\frac{2\alpha-N}{2}}U(\frac{x}{R_n^1})=U^{R^1_n}(x),
 \end{equation}
with $R^1_n\to 0$. Then from \eqref{e4.1} it follows that
 \begin{equation}
 \begin{gathered}
 z_n(x)=v_n(x)-U^{R^1_n}(x)=u_n(x)-u(x)-U^{R^1_n}(x),\\
I(z_n)=I(v_n)-D_0+o(1)=I(u_n)-I(u)-D_0+o(1)
 \end{gathered}
\end{equation}
with $R^1_n\to 0$.
From Lemma \ref{l4.8}, letting $a=v_n,b=U^{R^1_n}$, it follows
 \begin{equation}\label{e3.18}
 \begin{aligned}
\int_{|x|<R}\frac{\big(z_n^+(x)\big)^{2^*_s}}{|x|^s}dx
&=\int_{\tilde B(0,R)}\frac{\big(z_n(x)\big)^{2^*_s}}{|x|^s}dx\\
&\leq \int_{\tilde B(0,R)}\frac{\big(v_n(x)\big)^{2^*_s}
 -(U^{R^1_n}(x))^{2^*_s}}{|x|^s}dx\\
&=\int_{\tilde B(0,R)}\frac{\big(v_n^+(x)\big)^{2^*_s}}{|x|^s}dx-C\\
&\leq  \int_{|x|<R}\frac{\big(v_n^+(x)\big)^{2^*_s}}{|x|^s}dx-C
 \end{aligned}
 \end{equation}
where $\tilde{B}(0,R)=\{x|z_n(x)\geq 0\}\cap B(0,R)$.

If still there exists a $\bar \delta>0$ such that
\[
\int_{|x|<R}\frac{\big(z_n^+(x)\big)^{2^*_s}}{|x|^s}dx\geq \bar \delta
 >0,
\]
then we repeat the previous argument.
 From \eqref{e3.18} and the fact
\[
\int_{|x|<R}\frac{\big(v_n^+(x)\big)^{2^*_s}}{|x|^s}dx\leq \|v
 _n\|_{H^\alpha}^{2^*_s}\leq c,
\]
we deduce that the iteration must stop after  finite times.
 That is to see, there exist a positive constant $l_1$ and a new Palais-Smale
sequence of $I$,
 (without loss of generality) denoted by $\{v_n\}$, such that as $ n\to \infty$,
 \begin{equation}\label{e2.30j}
 d=I(v_n)+I(u)+l_1D_0,\quad
 v_n(x)=u_n(x)-u(x)-\sum^{l_1}_{i=1}U^{R^i_n}(x),
 \end{equation}
 with $R^i_n\to 0$,
\begin{gather}\label{e4.15}
\int_{|x|<R}\frac{\big(v_n^+(x)\big)^{2^*_s}}{|x|^s}\,dx=o(1)\quad
\text{for any } 0<R<\infty, \\
\label{e4.15jj}v_n\rightharpoonup 0\quad \text{in } H^{\alpha}(\mathbb{R}^N).
 \end{gather}
\smallskip

\noindent\textbf{Step 2.} Getting rid of the blowing up bubbles caused
by unbounded domains.
Suppose there exists $0<\delta <\infty$ such that
\begin{equation}\label{e4.5}
\Big(\int_{\mathbb{R}^N}(v_n^+(x))^{q}dx\Big)^{2/q}\geq \delta >0,\quad
\text{for } 2<q<2^*.
\end{equation}
By the interpolation inequality, it follows that
\[
\|v_n\|_{L^q}\leq \|v_n\|_{L^2}^\lambda \|v_n\|^{1-\lambda}_{L^{2^*}},
\quad \text{for } 2<q<2^*
\]
where $0<\lambda<1$. Thus there exist $\tilde{\delta}>0$ such that
\[
\|v_n\|_{L^2}^2\geq \tilde{\delta}>0.
\]
 By Lemma \ref{lem4.1}, there exists a subsequence still denoted by $\{v_n\}$,
such that one of the following two gathered occurs.

(i) Vanish occurs: for all $0< R<\infty$,
\[
\sup_{y\in\mathbb{R}^N}\int_{B(y,R)}|v_n(x)|^2dx\to 0\quad
\text{as }n\to \infty.
\]
By Lemma \ref{3.2}, \eqref{e4.10} and Sobolev inequality, it follows
 $$
\int_{\mathbb{R}^N}(v_n^+(x))^{q}dx\to 0\quad \text{as  }n\to \infty,\;
\forall\,2< q<2^*,
$$
which contradicts  \eqref{e4.5}.

(ii) Nonvanish occurs:
 there exist $\beta >0$, $0<\bar R<\infty$, $\{y_n\}\subset\mathbb{R}^N$,
such that
 \begin{equation}\label{e4.4}
 \liminf_{n\to \infty}\int_{y_n+B_{\bar R}} |v_n(x)|^2dx\geq \beta >0.
 \end{equation}

 We claim that $|y_n|\to\infty$ as  $n\to\infty$. Otherwise, if there exists
 a constant $M>0$ such that $|y_n|\leq M$, then we can choose a $R_2>0$
large enough such that
 \begin{equation}\label{e2.30}
 \int_{y_n+B_{\bar R}}
 |v_n(x)|^2dx\leq \|v_n\|_{L^2 (B(0,R_2))}^2\to 0 \text{ as } n \to \infty.
 \end{equation}
 which contradicts \eqref{e4.4}.

To proceed, we first construct the Palais-Smale sequences of $I^\infty$.
Denote $\bar  v_n(x)=v_n(x+y_n)$.
 Since $\|\bar v_n\|_{H^{\alpha}(\mathbb{R}^N)}
=\|v_n\|_{H^{\alpha}(\mathbb{R}^N)}\leq c$, without
 loss of generality, we assume that  as $n\to  \infty$,
 \begin{equation}\label{e2.46j}
 \begin{gathered}
\bar v_n\rightharpoonup v_0 \quad \text{in }H^{\alpha}(\mathbb{R}^N),\\
\bar v_n\to v_0\quad  \text{in } L^p_{\mathrm{loc}}(\mathbb{R}^N),\text{ for any }
1< p < {2^*}.
 \end{gathered}
 \end{equation}
By \eqref{e4.15}, we have that for all $\phi\in C^\infty_0(\mathbb{R}^N)$
 as $ n\to \infty$,
 \begin{equation}\label{e2.43j}
 \begin{aligned}
&\int_{\mathbb{R}^N}\frac{
 (\bar v^+_n(x))^{2^*_{s}-1} \phi(x)}{|x+y_n|^s}dx\\
&=\int_{\mathbb{R}^N}\frac{ ( v^+_n(x))^{2^*_{s}-1}\phi_n(x)}{|x|^s}dx\\
&=\int_{|x|>r}\frac{ ( v^+_n(x))^{2^*_{s}-1}\phi_n(x)}{|x|^s}dx+o(1) \\
&\leq\frac{1}{r^s}\Big(\int_{\mathbb{R}^N}|v_n(x)|^{2^*}dx\Big)^{\frac{2^*_s-1}{2^*}}
\Big(\int_{\mathbb{R}^N}|\phi_n(x)|^{q_1}dx\Big)^{1/q_1}+o(1),
 \end{aligned}
 \end{equation}
where $\phi_n=\phi(x-y_n)$ and $q_1=\frac{2^*}{2^*+1-2^*_s}$. Obviously
 \begin{equation}\label{e2.44}
 \int_{\mathbb{R}^N}|\phi_n(x)|^{q_1}dx
= \int_{\mathbb{R}^N}|\phi(x)|^{q_1}dx
\leq c, \quad \int_{\mathbb{R}^N}|v_n(x)|^{2^*}dx\leq c.
 \end{equation}
Let $r\to \infty$, from \eqref{e2.43j} and \eqref{e2.44}, we have
 \begin{equation}\label{e4.16}
\int_{\mathbb{R}^N}  \frac{ (\bar v^+_n(x))^{2^*_{s}-1} \phi(x)}{|x+y_n|^s}dx
=o(1)\quad \text{as }  n\to \infty.
 \end{equation}
Similarly we have
\begin{equation}\label{e4.17}
\int_{\mathbb{R}^N}\frac{(\bar v^+_n(x))^{2^*_{s}}}{|x+y_n|^s}dx=o(1)
\quad \text{as }  n\to \infty.
 \end{equation}
Since $v_n\rightharpoonup 0$ in
$H^{\alpha}(\mathbb{R}^N)$ and $\lim_{n\to\infty}a(x+y_n)=\bar a$,
 we have as $n\to\infty$,
\begin{align*}
o(1)&= \int_{\mathbb{R}^N}a(x)v_n(x)\phi_n(x)\,dx \\
&=\int_{\mathbb{R}^N}\bar a \bar v_n(x)\phi(x)\,dx
+\int_{\mathbb{R}^N}[a(x+y_n)-\bar a] \bar v_n(x)\phi(x)\,dx
\end{align*}
 and
$$
|\int_{\mathbb{R}^N}[a(x+y_n)-\bar a] \bar v_n(x)\phi(x)\,dx|
\leq c(\int_{\mathbb{R}^N}|a(x+y_n)-\bar  a
 |^2\phi(x)^2dx)^{1/2}=o(1);
$$
that is,
 \begin{equation}\label{s3}
\int_{\mathbb{R}^N}\bar a \bar v_n(x)\phi(x)\,dx=o(1)
=\int_{\mathbb{R}^N}a(x)v_n(x)\phi_n(x)
\,dx\quad \text{as } n\to \infty.
\end{equation}
Similarly we have
 \begin{equation}\label{s4}
\int_{\mathbb{R}^N} k(x) (v_n^+(x))^{q-1} \phi_n(x)\,dx
=\int_{\mathbb{R}^N}\bar k (\bar
v^+_n(x))^{q-1}\phi(x)\,dx=o(1)
\end{equation}
as $n\to \infty$.
Recall that $v_n$ is a Palais-Smale
 sequence of $I$, by \eqref{e2.46j} and \eqref{e4.16}-\eqref{s4} we have
 \begin{equation}
o(1)=\langle I'(v_n),\phi_n\rangle=\langle {I^\infty}' (\bar
 v_n),\phi\rangle+o(1)=\langle {I^\infty}' (
 v_0),\phi\rangle+o(1),
\end{equation}
 as $n\to \infty$.
 This shows  that  $v_0$ is a weak solution of
\eqref{e2.1}.

 We claim that $v_0\not\equiv 0$.
From \eqref{e4.5}, we may assume that there exists a sequence $\{y_n\}$
satisfying \eqref{e4.4} and
\begin{equation}\label{z3}
\int_{B(y_n,R)}(v^+_n(x))^qdx=b+o(1)>0,\quad \text{as }n\to \infty,
\end{equation}
where $b>0$ is a constant.
If $v_0\equiv 0$, we have $$\int_{B(0,R)}(\bar v^+_n(x))^qdx=\int_{B(y_n,R)}
(v^+_n(x))^qdx=o(1)\text{ as } n\to \infty\text{ for }0<R<\infty$$ which
contradicts \eqref{z3}.

 Denote $z_n(x)= v_n(x)-v_0(x-y_n)$. Since
\begin{align*}
 I(v_n)&=\frac{1}{2}\int_{\mathbb{R}^N}\Big (|(-\Delta)^{\alpha/2}v_n(x)|^2
 +a(x)|v_n(x)|^2 \Big )dx \\
&\quad -\frac{1}{{2^*_{s}}}\int_{\mathbb{R}^N}\frac{(v_n^+(x))^{{2^*_{s}}}}{|x|^s}dx
 -\frac{1}{q}\int_{\mathbb{R}^N}k(x)(v_n^+(x))^{q} \,dx\\
&=\frac{1}{2}\int_{\mathbb{R}^N} \Big (|(-\Delta)^{\alpha/2}\bar v_n(x)|^2
 +a(x+y_n)|\bar v_n(x)|^2 \Big )dx \\
&\quad -\frac{1}{{2^*_{s}}}\int_{\mathbb{R}^N}\frac{(\bar
 v_n^+(x))^{{2^*_{s}}}}{|x+y_n|^s}dx
 -\frac{1}{q}\int_{\mathbb{R}^N}k(x+y_n)(\bar v_n^+(x))^q\,dx\\
&=\frac{1}{2}\int_{\mathbb{R}^N}\Big (|(-\Delta)^{\alpha/2}
 \bar v_n(x)|^2+\bar a|\bar v_n(x)|^2 \Big )dx\\
&\quad -\frac{1}{q}\int_{\mathbb{R}^N}\bar k(\bar v_n^+(x))^q\,dx+o(1),
\end{align*}
where the last equality is a result of \eqref{e4.17}, therefore, as $n\to \infty$,
\begin{gather}\label{e4.20}
\|z_n\|_{H^{\alpha}(\mathbb{R}^N)}=\|\bar
v_n\|_{H^{\alpha}(\mathbb{R}^N)}-\|v_0\|_{H^{\alpha}(\mathbb{R}^N)}+o(1), \\
\label{e4.21} I(z_n)=I^\infty(\bar v_n)-I^\infty
(v_0)+o(1)=I(v_n)-I^\infty(v_0)+o(1).
\end{gather}
Hence
$z_n\rightharpoonup0$ in $H^{\alpha}(\mathbb{R}^N)\text{ as } n\to
\infty$,
 and $z_n$ is a Palais-Smale sequence of
$I$. From \eqref{e4.10} in Lemma \ref{l4.5}, it follows
$ \|v_0^-\|_{H^\alpha}=0$, that is $v_0\geq 0$ a.e. in $\mathbb{R}^N$.
Then by Brezis-Lieb Lemma and \eqref{e4.10}, there exists a constant $c>0$ such that
\begin{equation}\label{e2.56j}
\begin{aligned}
\int_{\mathbb{R}^N}(z_n^+(x))^qdx
&= \int_{\mathbb{R}^N}(v_n^+(x))^qdx-\int_{\mathbb{R}^N}(v_0^+(x))^qdx+o(1) \\
&\leq \int_{\mathbb{R}^N}(v_n^+(x))^qdx-c
\end{aligned}
\end{equation}
where the last inequality follows from the fact $v_0\not\equiv 0$.
 If $\|z_n\|_{L^q(\mathbb{R}^N)}\to \delta_2>0$ as
$n\to\infty$, from \eqref{e2.56j}
and the boundedness of $\|v_n\|_{L^q}$, then one can repeat Step 2 for
finite times ($l_2$ times).
Thus from \eqref{e2.30j} and Step 2, we obtain a new Palais-Smale sequence
of $I$, without loss of generality still denoted by $v_n$, such that
\begin{gather}\label{e2.45}
d=I(u)+I(v_n)+l_1D_0+\sum^{l_2}_{j=1}I^\infty(u_j)+o(1), \\
\label{e2.46}
v_n(x)=u_n(x)-u(x)-\sum^{l_1}_{i=1}U^{R^i_n}(x)
-\sum^{l_2}_{j=1}u_j(x-y_n^j),\quad \text{with } R^i_n\to 0, \\
\label{e2.57}
\|v_n^+\|_{L^q(\mathbb{R}^N)}\to 0,\quad
 \int_{\mathbb{R}^N}\frac{\big(v_n^+(x)\big)^{2^*_s}}{|x|^s}dx\to 0
\end{gather}
as $n\to\infty$. Then from the fact $<I'(v_n),v_n>=o(1)$, it follows that
\begin{equation}\label{e2.58}
\begin{aligned}
\|v_n\|^2_{H^\alpha(\mathbb{R}^N)}
&\leq c\int_{\mathbb{R}^N}(| \big(-\Delta )^{\alpha/2}v_n(x)|^2
+a(x)|v_n(x)|^2\big)dx\\
&=c\bigl(\int_{\mathbb{R}^N}k(x)(v_n^+(x))^{q}dx
+ \int_{\mathbb{R}^N}\frac{\big(v_n^+(x)\big)^{2^*_s}}{|x|^s}dx\big) \to 0
\end{aligned}
\end{equation}
as $n\to \infty$. From \eqref{e2.57} and \eqref{e2.58}, it gives
\begin{equation}\label{e2.59}
I(v_n)=o(1).
\end{equation}
From \eqref{e2.45}-\eqref{e2.59}, the proof is complete.
\end{proof}

\section{Proof of Theorem \ref{thm1.2}}
To this end we use the mountain pass theorem \cite{BN} and
Theorem \ref{thm1.1}.

\begin{proof}[Proof of Theorem \ref{thm1.2}]
From
\begin{align*}
I (tu)
&=\frac{t^2}{2}\Big[\int_{\mathbb{R}^N}|(-\Delta)^{\alpha/2}
u(x)|^2dx+\int_{\mathbb{R}^N}a(x)|u(x)|^2\,dx\Big]\\
&\quad -\frac{|t|^{{2^*_{s}}}}{{2^*_{s}}}
\int_{\mathbb{R}^N}\frac{(u^+(x))^{{2^*_{s}}}}{|x|^s}dx
-\frac{|t|^q}{q}\int_{\mathbb{R}^N}k(x)(u^+(x))^q\,dx,
\end{align*}
we deduce that for a fixed $u\not\equiv 0$ in $H^{\alpha}(\mathbb{R}^N)$,
$I(tu)\to -\infty$ if $t\to \infty$. Since
 $$
\int_{\mathbb{R}^N}(u^+(x))^q\,dx\leq
C\|u\|^q_{H^{\alpha}({\mathbb{R}^N})},\quad\text{and}\quad
\int_{\mathbb{R}^N}\frac{(u^+(x))^{{2^*_{s}}}}{|x|^s}\,dx\leq
C\|u\|^{{2^*_{s}}}_{H^{\alpha}({\mathbb{R}^N})},
 $$
 we have
$$
I(u)\geq
c\|u\|_{H^{\alpha}(\mathbb{R}^N)}^2-C(\|u\|_{H^{\alpha}(\mathbb{R}^N)}^q
+\|u\|_{H^{\alpha}(\mathbb{R}^N)}^{{2^*_{s}}}).
$$
Hence, there exists $r_0>0$ small such that
$I(u)\Big|_{\partial B(0,r_0)}\geq\rho>0$ for $q,\,{2^*_{s}}>2$.

As a consequence, $I(u)$ satisfies the geometry structure of
Mountain-Pass Theorem. Now, we define
$$
c^*=:\inf_{\gamma \in \Gamma} \sup_{t\in [0,1]}I(\gamma (t)),
$$
where $ \Gamma=\{\gamma \in
C([0,1],H^{\alpha}(\mathbb{R}^N)):\gamma(0)=0,
\gamma(1)=\psi_0\in H^{\alpha}(\mathbb{R}^N)\}$
with $I(t\psi_0)\leq 0$ for all $t\geq 1$.

To complete the proof of Theorem \ref{thm1.2}, we need to verify that
$I(u)$ satisfies the local Palais-Smale conditions. According to
By \ref{rmk1},  we only need to verify that
\begin{equation}\label{e4.28}
c^*<\min\{\frac{2\alpha-s}{2(N-s)}S_{\alpha,s}^{\frac{N-s}{2\alpha-s}},
J^\infty\}.
\end{equation}
Set
\[v_\varepsilon(x)=\frac{U_\varepsilon(x)}{(\int_{\mathbb{R}^N}
\frac{|U_\varepsilon(x)|^{{2^*_{s}}}}{|x|^s}dx)^{1/{2^*_{s}}}}.
\]
We  claim that
\begin{equation}\label{e4.29}
\max_{t>0}I(tv_\varepsilon)<\frac{2\alpha-s}{2(N-s)}
S_{\alpha,s}^{\frac{N-s}{2\alpha-s}}.
\end{equation}
In fact, from \eqref{e1.5} it is easy to calculate the following estimates
\begin{gather}
\|v_\varepsilon\|_{\dot{H}^\alpha(\mathbb{R}^N)}^2=S_{\alpha,s}, \label{jt3.19} \\
\int_{\mathbb{R}^N}(v_\varepsilon(x))^2dx\leq c\varepsilon^{2\alpha-N}
\int_{\mathbb{R}^N}\frac{1}{(1+|\frac{x}{\varepsilon}|^2)^{N-2\alpha}}dx\leq
O(\varepsilon^{2\alpha}), \quad \text{for }N> 4\alpha,  \label{jt3.20} \\
\int_{\mathbb{R}^N}(v_\varepsilon(x))^qdx= O(\varepsilon^{\frac{(2\alpha-N)q}{2}+N}).\label{jt3.21}
\end{gather}
 Since $2^*>q>2$, we  have
 \begin{equation}\label{e3.6}
 O(\varepsilon^{2\alpha})=o(\varepsilon^{\frac{(2\alpha-N)q}{2}+N}).
 \end{equation}
 Denote by $t_\varepsilon$ the attaining point of $\max_{t>0}I(tv_\varepsilon)$,
similar to the proof of \cite[Lemma 3.5]{DG} we can
prove that $t_\varepsilon$ is uniformly bounded.
In fact, we consider the function
\begin{equation}
\begin{split}
h(t)&=I \left(tv_\varepsilon\right)\\&=\frac{t^2}{2}
 (\|(-\Delta)^{\alpha/2}v_\varepsilon\|^2_{L^2(\mathbb{R}^N)}
 +\int_{\mathbb{R}^N}a(x) (v_\varepsilon(x))^2\,dx) \\
&\quad -\frac{t^{{2^*_s}}}{{2^*_s}}
 \int_{\mathbb{R}^N} \frac{(v_\varepsilon(x))^{{2^*_s}}}{|x|^s}dx
 -\frac{t^{{q}}}{{q}} \int_{\mathbb{R}^N}(k(x)v_\varepsilon(x))^qdx \\
& \geq \frac{ct^2}{2}\|v_\varepsilon\|^2_{H^\alpha(\mathbb{R}^N)}
 -\frac{Ct^{{2^*_s}}}{{2^*_s}}\|v_\varepsilon\|_{H^\alpha(\mathbb{R}^N)}^{2^*_s}
 -\frac{Ct^{{q}}}{{q}}\|v_\varepsilon\|_{H^\alpha(\mathbb{R}^N)}^q.
\end{split}
\end{equation}
Since $\lim_{t\to +\infty} h(t)=-\infty$ and $h(t)>0$ when $t$ is
closed to $0$, it follows that $\max_{t>0}h(t)$ is attained for $t_\varepsilon>0$.
From the fact $\int_{\mathbb{R}^N}\frac{(v_\varepsilon(x))^{{2^*_s}}}{|x|^s}dx=1$,
we have
\begin{equation}\label{e3.11j}
\begin{aligned}
h'(t_\varepsilon)
&=t_\varepsilon(\|(-\Delta)^{\alpha/2}v_\varepsilon\|^2_{L^2(\mathbb{R}^N)}
 +\int_{\mathbb{R}^N}a(x) (v_\varepsilon(x))^2\,dx) \\
&\quad -{t_\varepsilon^{{2^*_s-1}}} -t_\varepsilon^{q-1}\int_{\mathbb{R}^N}k(x)(v_\varepsilon(x))^qdx=0.
\end{aligned}
\end{equation}
Since $k(x)>0$, from \eqref{jt3.19} and \eqref{jt3.20} for $\varepsilon$ sufficiently small,
we have
\begin{equation}\label{e3.12j}
t_\varepsilon^{{2^*_s-2}}\leq {\|(-\Delta)^\alpha v_\varepsilon\|^2_{L^2(\mathbb{R}^N)}+\int_{\mathbb{R}^N} a(x)(v_\varepsilon(x))^{2}\,dx}<2S_{\alpha,s}.\end{equation}
Then \begin{equation}\label{e3.10}
\begin{aligned}
&\|(-\Delta)^{\alpha/2}v_\varepsilon\|^2_{L^2(\mathbb{R}^N)}
 +\int_{\mathbb{R}^N}a(x) (v_\varepsilon(x))^2\,dx\\
 &= t_\varepsilon^{2^*_s-2}+t_\varepsilon^{q-2}\int_{\mathbb{R}^N}k(x)(v_\varepsilon(x))^qdx\\
&\leq t_\varepsilon^{2^*_s-2}+(2S_{\alpha,s})^{\frac{q-2}{2^*_s-2}}
\int_{\mathbb{R}^N}k(x)(v_\varepsilon(x))^qdx.
\end{aligned}
\end{equation}
Choosing $\varepsilon>0$ small enough, by \eqref{jt3.19}-\eqref{jt3.21}, there
exists a constant $\mu>0$ such that $t_\varepsilon>\mu>0$.
 Combining this with \eqref{e3.12j}, it implies that $t_\varepsilon$ is bounded
for $\varepsilon>0$ small enough.
Then, for $\varepsilon>0$ small,
\begin{align*}
 \max_{t>0}I(tv_\varepsilon)
&=I (t_\varepsilon v_\varepsilon)\\
&\leq  \max_{t>0}\Big\{\frac{t^2}{2}\int_{\mathbb{R}^N}|(-\Delta)^{\alpha/2}
v_\varepsilon(x)|^2dx-\frac{t^{{2^*_{s}}}}{{2^*_{s}}}\int_{\mathbb{R}^N}
\frac{(v_\varepsilon(x))^{{2^*_{s}}}}{|x|^s}dx\Big\}\\
&\quad -O(\varepsilon^{\frac{(2\alpha-N)q}{2}+N}) + O(\varepsilon^{2\alpha}), \\
&<\frac{2\alpha-s}{2(N-s)}S_{\alpha,s}^{\frac{N-s}{2\alpha-s}}\quad
\text{(by \eqref{e3.6})}.
\end{align*}
This  proves \eqref{e4.29}.
By the definition of
$c^*$, we have $c^*<\frac{2\alpha-s}{2(N-s)}S_{\alpha,s}^{\frac{N-s}{2\alpha-s}}$.

Next we verify that
\begin{equation}\label{e4.30}
c^*<J^\infty.
\end{equation}
Let $\{u_0\}$ be the minimizer of
$J^\infty,\,I^\infty(u_0)=J^\infty$ and
\[
\int_{\mathbb{R}^N}\Big(|(-\Delta)^{\alpha/2} u_0(x)|^2+\bar a
|u_0(x)|^2 \Big)dx
=\int_{\mathbb{R}^N}\bar k(u_0^+(x))^q\,dx.
\]
Let
\begin{gather*}
\begin{aligned}
g(t)&=I^\infty(tu_0) \\
&=\frac{1}{2}t^2\int_{\mathbb{R}^N}\Big(|(-\Delta)^{\alpha/2} u_0(x)|^2+\bar a
|u_0(x)|^2 \Big)dx
-\frac{t^q}{q}\int_{\mathbb{R}^N}\bar k(u_0^+(x))^q\,dx,
\end{aligned}\\
g'(t)=t\int_{\mathbb{R}^N}\Big(|(-\Delta)^{\alpha/2} u_0(x)|^2+\bar a
|u_0(x)|^2 \Big)dx
-t^{q-1}\int_{\mathbb{R}^N}\bar k(u_0^+(x))^q\,dx.
\end{gather*}
Thus $g'(t)\geq0$ if $t\in(0,1)$; $g'(t)\leq 0$ if $t\geq 1$. Then
\begin{equation}\label{e4.31}
g(1)=I^\infty(u_0)=\max_{l}I^\infty (u),
\end{equation}
 where $l=\{tu_0,t\geq 0\}$ for a fixed $u_0$.

Since there exists a $t_0>0$ such that $ \sup_{t\geq
0}I(tu_0)=I(t_0u_0)$, from \eqref{e4.31} and the assumptions on
$a(x)$  and $k(x)$, we have
$$
J^\infty=I^\infty (u_0)\geq I^\infty (t_0u_0)>I(t_0 u_0)
=\sup_{t\geq 0}I(t u_0).
$$
This  proves \eqref{e4.30}. By \eqref{e4.29} and \eqref{e4.30} we have
\eqref{e4.28}. Then the proof is completed.
\end{proof}

\section{Appendix}

In this section, we give some lemmas and detailed proofs for the convenience
of the reader.

\begin{lemma}[{\cite[Lemma 2.1]{ZC}}] \label{lem4.1}
 Let $\{\rho_n\}_{n\geq 1}$ be a sequence in $L^1(\mathbb{R}^N)$ satisfying
 \begin{equation}\label{e2.4}
\rho_n\geq 0\text{ on } \mathbb{R}^N,\quad
\lim_{n\to\infty}\int_{\mathbb{R}^N}\rho_n(x)dx=\lambda>0,
 \end{equation}
where $\lambda>0$ is fixed. Then there exists a
 subsequence $\{\rho_{n_k}\}$ satisfying one of the following two
 possibilities:
\begin{itemize}
\item[(1)] (Vanishing):
 \begin{equation}\label{e2.5}
 \lim_{k\to\infty} \sup_{y\in\mathbb{R}^N}\int_{B(y,R)}\rho_{n_k}(x)
\,dx=0,\quad \text{for all } R<+\infty.
 \end{equation}

\item[(ii)] (Nonvanishing): there exist $\alpha>0$, $R<+\infty$ and
 $\{y_k\}\subset \mathbb{R}^N$ such that
$$
\liminf_{k\to  +\infty}\int_{y_k+B_R}\rho_{n_k}(x)dx\geq \alpha >0.
$$
\end{itemize}
\end{lemma}

\begin{lemma}[{\cite[Lemma 2.2]{PAJ}}] \label{3.2}
 If $\{u_n\}$ is bounded in $H^\alpha(\mathbb{R}^N) $ and for some $R>0$, we have
 $$
\sup_{y\in\mathbb{R}^N}\int_{B(y,R)}|u_n(x)|^2\,dx\to 0
 \quad \text{as }n\to \infty.
$$
 Then $u_n\to 0$ in $L^q (\mathbb{R}^N)$, for $2<q<\frac{2N}{N-2\alpha}$.
\end{lemma}

\begin{lemma}\label{jl33}
 Suppose that $0<s<2\alpha$ and $N>2\alpha$. Then there exists $C>0$ such
that for any $u\in \dot{H}^\alpha(\mathbb{R}^N)$,
 \begin{equation}\label{j2.3}
 \Big(\int_{\mathbb{R}^N}\frac{|u(x)|^{p}}{|x|^s}dx\Big)^{2/p}
\leq C\|u\|^2_{\dot{H}^\alpha(\mathbb{R}^N)},
 \end{equation}
 i.e.,
$\dot{H}^\alpha(\mathbb{R}^N)\hookrightarrow L^{2^*_{s}}_{}(\mathbb{R}^N,|x|^{-s})$
 is continuous. In addition, the inclusion
 \[
\dot{H}^\alpha(\mathbb{R}^N)\hookrightarrow L_{\mathrm{loc}}^p
(\mathbb{R}^N,|x|^{-s}),
\]
 is compact if $2\leq p<2^*_{s}$.
\end{lemma}

\begin{proof}
The proof of \eqref{j2.3} is similar to that of \cite[Lemma 3.1]{Y}.
Now we prove the compact impeding if $2\leq p<2^*_{s}$.
 Let $\{u_n\}$ be a bounded sequence in $\dot{H}^\alpha(\mathbb{R}^N)$,
then up to a subsequence (still denoted by $\{u_n\}$),
 \[
u_n \rightharpoonup u\text{ in } \dot{H}^\alpha(\mathbb{R}^N).
\]
 Denote $v_n(x)=u_n(x)-u(x)$, then for any $B(0,r)$,
 \[
v_n \rightharpoonup 0 \text{ in } \dot{H}^\alpha(\mathbb{R}^N),\quad
 v_n \to 0 \text{ in } L^q(B(0,r)),\quad
 2\leq q < 2^*=\frac{2N}{N-2\alpha}.
\]
Fixing $r>0$,
 since $(p-\frac{s}{\alpha})(\frac{2\alpha}{2\alpha-s})<2^*$, it follows that
 \begin{equation}\label{l2.41}
 \begin{split}
&\int_{B(0,r)}\frac{|v_n(x)|^{p}}{|x|^s}dx\\
&=\int_{B(0,r)}\frac{|v_n(x)|^{s/\alpha}}{|x|^s}|v_n(x)|^{p-s/\alpha}dx\\
&\leq \Big(\int_{B(0,r)}（\frac{|v_n(x)|^{2}}{|x|^{2\alpha}}dx
 \Big )^{\frac{s}{2\alpha}}
\Big (\int_{B(0,r)}|v_n(x)|^{(p-\frac{s}{\alpha})(\frac{2\alpha}{2\alpha-s})}dx
\Big)^{\frac{2\alpha-s}{2\alpha}} \\
&\leq c\|(-\Delta )^{\alpha/2}v_n(x)\|_{L^2(\mathbb{R}^N)}^{\frac{s}{2\alpha}}
\Big (\int_{B(0,r)}|v_n(x)|^{(p-\frac{s}{\alpha})(\frac{2\alpha}{2\alpha-s})}
dx\Big)^{\frac{2\alpha-s}{2\alpha}}\to 0.
 \end{split}
 \end{equation}
Then we have
 \[
u_n\to u \text{ in } L_{\mathrm{loc}}^{p}(\mathbb{R}^N,|x|^{-s}),
\]
which completes the proof.
\end{proof}

\begin{lemma}\label{3.5}
 Let $\{u_n\}$ be a Palais-Smale sequence of $I$ at level
 $d\in\mathbb{R}$. Then $d\geq 0$ and $\{u_n\}$ $\subset$ $H^{\alpha}
 (\mathbb{R}^N)$ is bounded. Moreover,
every Palais-Smale sequence for $I$ at a level zero converges
 strongly to zero.
\end{lemma}

\begin{proof}
Since $a(x)\geq 0$, $\bar a>0$ and
 $\inf_{x\in \mathbb{R}^N} a(x)=\hat{a}>0$, we have
 $$
\|u_n\|_{\dot{H}^\alpha (\mathbb{R}^N)}^2+\int_{\mathbb{R}^N}
 a(x)|u_n(x)|^2dx\geq
 c\|u_n\|_{H^{\alpha}(\mathbb{R}^N)}^2,
$$
and hence for $q\leq 2^*_{s}$,
 \begin{equation}\label{l3.1}
 \begin{split}
 d+1+o(\|u_n\|) 
&\geq  I(u_n)-\frac{1}{q}\langle I'(u_n), u_n\rangle\\
&= (\frac{1}{2}-\frac{1}{q})\int_{\mathbb{R}^N}
\big(|(-\Delta)^{\alpha/2}u_n(x)|^2+a(x)|u_n(x)|^2\big)dx \\
&\quad +(\frac{1}{q}-\frac{1}{{2^*_{s}}})\int_{\mathbb{R}^N}
 \frac{(u_n^+(x))^{{2^*_{s}}}}{|x|^s}\,dx\\
&\geq  C\|u_n\|_{H^{\alpha}(\mathbb{R}^N)}^2 .
 \end{split}
 \end{equation}
For $2^*_{s}<q<2^*$,
 \begin{equation}\label{l3.2}
 \begin{split}
d+1+o(\|u_n\|) \
&\geq I(u_n)-\frac{1}{2^*_{s}}\langle I'(u_n), u_n\rangle\\
&=(\frac{1}{2}-\frac{1}{2^*_{s}})\int_{\mathbb{R}^N}\Big(|(-\Delta)^{\alpha/2}
 u_n(x)|^2+a(x)|u_n(x)|^2\Big)dx \\
 &\quad +(\frac{1}{{2^*_{s}}}-\frac{1}{q})\int_{\mathbb{R}^N}k(x)(u_n^+(x))^{q}\,dx\\
 &\geq C\|u_n\|_{H^{\alpha}(\mathbb{R}^N)}^2.
 \end{split}
 \end{equation}
It follows from \eqref{l3.1} and \eqref{l3.2} that $\{u_n\}$ is bounded in
$H^{\alpha}(\mathbb{R}^N)$ for $2<q<2^*$. Since
 $$
d=\lim_{n\to  \infty}I(u_n)-\max\{\frac{1}{q},\frac{1}{2^*_{s}}\}
\langle I'(u_n),u_n\rangle\geq
 C\limsup_{n\to \infty}\|u_n\|_{H^{\alpha}(\mathbb{R}^N)}^2,
$$
we have  $d\geq 0$. Suppose now that $d=0$, we obtain from the above inequality
 that
 $$
\lim_{n\to \infty}\|u_n\|_{H^{\alpha}(\mathbb{R}^N)} =0.
$$
\end{proof}

\begin{lemma}\label{l4.5}
Let $\{u_n\}$ be a Palais-Smale sequence of $I$ at level
$d\in\mathbb{R}$ and $u^+_n=\max \{u_n,0\}$.
 Then $\{u_n^+\}$ is also a Palais-Smale sequence of $I$ at level $d$.
\end{lemma}

\begin{proof}
From the definition of $I$ we have that as $n\to \infty$,
\begin{align*}
I(u_n)&=\frac{1}{2}\int_{\mathbb{R}^N}\Big (|(-\Delta)^{\alpha/2}u_n(x)|^2
 +a(x)|u_n(x)|^2\Big )\,dx
-\frac{1}{2^*_{s}}\int_{\mathbb{R}^N}\frac{(u_n^+(x))^{2^*_{s}}}{|x|^s}dx \\
&\quad -\frac{1}{q}\int_{\mathbb{R}^N}k(x)(u_n^+(x))^q \,dx\to d,
\end{align*}
and
 \begin{equation}
 \begin{aligned}
 &\langle I'(u_n),\phi\rangle\\
&=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u_n(x)-u_n(y))(\phi(x)
 -\phi(y))}{|x-y|^{N+2\alpha}}\,dx\,dy
 +\int_{\mathbb{R}^N}a(x)u_n(x)\phi(x)\,dx\\
&\quad -\int_{\mathbb{R}^N}\frac{(u^+_n(x))^{2^*_s-1}\phi(x)}{|x|^s}dx
 -\int_{\mathbb{R}^N}k(x)(u_n^+(x))^{q-1}\phi(x)dx\to 0,
 \end{aligned}
 \end{equation}
for all $\phi \in H^\alpha(\mathbb{R}^N)$.

Taking $\phi=-u_n^-=\min \{u_n,0\}$, from
 \begin{equation}\label{e4.8}
u_n(x)=u_n^+(x)-u_n^-(x), \quad u_n^+(x)u_n^-(x)=0,
 \end{equation}
we have
 \begin{equation}\label{e4.9}
 \begin{aligned}
\langle I'(u_n),-u_n^-\rangle 
&=-\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(u_n(x)-u_n(y))(u_n^-(x)-u_n^-(y))}{|x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad -\int_{\mathbb{R}^N}a(x)u_n(x)u_n^-(x)\,dx
 +\int_{\mathbb{R}^N}\frac{(u^+_n(x))^{2^*_s-1}u_n^-(x)}{|x|^s}dx \\
&\quad +\int_{\mathbb{R}^N}k(x)(u_n^+(x))^{q-1}u_n^-(x)dx\\
&=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{(u_n^-(x)-u_n^-(y))^2}{|x-y|^{N+2\alpha}}\,dx\,dy \\
&\quad + \int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{u_n^+(x)u_n^-(y)
 +u_n^+(y)u_n^-(x)}{|x-y|^{N+2\alpha}}\,dx\,dy \\
&\quad +\int_{\mathbb{R}^N}a(x)(u_n^-(x))^2\,dx
 \to 0\,.
 \end{aligned}
 \end{equation}
From \eqref{e4.9},  $u^+_n(x)\geq 0$, $u^-_n(x)\geq 0$ and $a(x)>0$,
it follows that
 \begin{equation}\label{e4.10}
 \|u_n^-\|_{H^\alpha}\to 0,
 \end{equation}
 and
 \begin{equation}\label{e4.11}
 \int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{2(u^+_n(x)-u^+_n(y))(u^-_n(x)
-u^-_n(y))}{|x-y|^{N+2\alpha}}\,dx\,dy\to 0.
 \end{equation}
Then  from \eqref{e4.8} and \eqref{e4.10}-\eqref{e4.11}, we have
 \begin{equation}\label{e4.12j}
 \begin{aligned}
& \int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u_n(x)-u_n(y))^2}{|x-y|^{N+2\alpha}}
\,dx\,dy\\
&=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\Big((u^+_n(x)-u^+_n(y))^2+(u^-_n(x)-u^-_n(y))^2 \\
&\quad -2(u^+_n(x)-u^+_n(y))
(u^-_n(x)-u^-_n(y))\Big)\big/
|x-y|^{N+2\alpha} \,dx\,dy\\
&=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u_n^+(x)-u^+_n(y))^2}
{|x-y|^{N+2\alpha}}\,dx\,dy+o(1).
 \end{aligned}
 \end{equation}
That is
 \begin{equation}\label{e4.13}
 \|u_n\|_{\dot{H}^\alpha}=\|u_n^+\|_{\dot{H}^\alpha}+o(1).
 \end{equation}
Therefore
 \begin{gather*}
\lim_{n\to \infty}I(u_n^+)=\lim_{n\to \infty}I(u_n)=d,\\
I'(u_n^+,\phi)=I'(u_n,\phi)\to 0
\end{gather*}
as $n\to \infty$. This complete the proof.
\end{proof}

\begin{lemma}\label{l4.6}
All nontrivial critical points of $I_s$ are  positive solutions of \eqref{e1.4}.
\end{lemma}

\begin{proof}
Let $u\not\equiv 0$ and $ u\in H^\alpha(\mathbb{R}^N)$ be a nontrivial
critical point of $I_s$.
 First, arguing as in the proof of Lemma \ref{l4.5}
(similar to \eqref{e4.9} and \eqref{e4.10}), we can obtain that
$\|u^-\|_{H^\alpha}=0$ which gives that
 \begin{equation}\label{e4.14}
 u\geq 0 \quad \text{a.e. in } \mathbb{R}^N.
 \end{equation}
Then for any $x_0\in \mathbb{R}^N$,
 \begin{equation}\label{e4.15j}
\begin{gathered}
 (-\Delta)^\alpha u=\frac{|u|^{2^*_s-2}u}{|x|^s}\geq 0,\quad
\text{a.e. in } B(x_0,1), \\
\int_{\mathbb{R}^N}\frac{|u(x)|}{1+|x|^{N+2\alpha}}dx\leq c\|u\|_{L^2}\leq c\,.
\end{gathered}
 \end{equation}
From \cite[Proposition 2.2.6]{SL}, we have
$u$ is lower semicontinuous in $B(x_0,1)$. Combining this with \eqref{e4.14},
 it follows $u(x_0)\geq 0$. Then $u(x)\geq 0$ pointwise in $\mathbb{R}^N$.

Next we claim that $u>0$ in $\mathbb{R}^N$. Otherwise there exist
$x_1\in \mathbb{R}^N$ such that $u(x_1)=0$.
Then  $u$ is lower semicontinuous in $\overline {B(x_1,1/2)}$.
 From \cite[Proposition 2.2.8]{SL}, it follows $u\equiv 0$  in
$\mathbb{R}^N$. This contradicts the assumption $u$ is nontrivial.
\end{proof}

Let $\{u_n\}$ be a Palais-Smale sequence at level $d$. Up to a subsequence, we
assume that
$$
u_n\rightharpoonup u \quad \text{in } H^{\alpha}(\mathbb{R}^N)\text{ as
}n\to \infty.
$$
Obviously, we have $I'(u)=0$.
Let $v_n(x)=u_n(x)-u(x)$, from Lemma \ref{jl33} as $n\to \infty$,
\begin{gather}
v_n\rightharpoonup 0\quad \text{in }H^{\alpha}(\mathbb{R}^N),  \label{jl2.7}\\
v_n\to 0\quad \text{in }L_{\mathrm{loc}}^{p}(\mathbb{R}^N, |x|^{-s})
\text{ for all } 2\leq p<2^*_{s}, \\
v_n\to 0\quad \text{in }L^{q}_{\mathrm{loc}}(\mathbb{R}^N)
 \text{ for all } 2< q<2^*,\\
v_n\to 0,\quad  \text{a.e. in } \mathbb{R}^N.\label{jl4.19}
\end{gather}
As a consequence, we have the following Lemma.

\begin{lemma}\label{q}
 $\{v_n\}$ is a Palais-Smale sequence for $I$ at level
 $d_0=d- I(u)$.
\end{lemma}

\begin{proof}
 For $\phi(x)\in C^\infty_0(\mathbb{R}^N)$, there exists a $B(0,r)$ such that
$\mathrm{supp}\phi \subset B(0,r)$. Then  as $n\to \infty$,
 \begin{equation}\label{l5.16}
 \big|\int_{\mathbb{R}^N}k(x)(v_n^+(x))^{q-1} \phi(x)\,dx\big|
\leq c \big|\int_{B(0,r)}(v_n^+(x))^{q-1} \phi(x)dx\big|=o(1),
\end{equation}
and from Lemma \ref{jl33},
 \begin{equation}\label{l5.17}
\begin{aligned}
 \big|\int_{\mathbb{R}^N }\frac{(v_n^+(x))^{2^*_{s}-1} \phi(x)}{|x|^s}dx\big|
&\leq \big|\int_{|x|\leq r }\frac{(v_n^+(x))^{2^*_{s}-1} \phi(x)}{|x|^s}dx\big|\\
&\leq c \int_{|x|\leq r }\frac{(v_n^+(x))^{2^*_{s}-1}}{|x|^s}dx=o(1).
\end{aligned}
 \end{equation}
By \eqref{jl2.7}, \eqref{l5.16} and \eqref{l5.17}, we have
$\langle \phi, I'(v_n)\rangle=o(1)\text{ as }n\to  \infty$.
Then similar to \eqref{e4.10}, we have
\begin{equation} \label{e4.22j}
\|v_n^-\|_{\dot{H}^\alpha}\to 0,\|u^-\|_{\dot{H}^\alpha}=0.
 \end{equation}
By Sobolev inequality, \eqref{e4.10} and \eqref{e4.22j} it follows that
 \[
\|u_n\|_{L^q}=\|u_n^+\|_{L^q}+o(1),\,\|v_n\|_{L^q}
=\|v_n^+\|_{L^q}+o(1), \|u\|_{L^q}=\|u^+\|_{L^q}.
\]
Then by the Br\'{e}zis-Lieb Lemma in \cite{BN} as
 $n\to\infty$, we have
 \begin{equation}\label{e4.22}
\int_{\mathbb{R}^N}( v_n^+(x))^{q}dx
=\int_{\mathbb{R}^N} (u_n^+(x))^{q}dx-\int_{\mathbb{R}^N}(u^+(x))^{q}dx+o(1)
\end{equation}
for all $2\leq q\leq {2^*_{s}}$.
 Similarly
 \begin{gather}\label{e4.23}
\int_{\mathbb{R}^N }\frac{\big(v_n^+(x)\big)^{2^*_s}}{|x|^s}dx
=\int_{\mathbb{R}^N} \frac{\bigl(u^+_n(x)\bigl)^{2^*_{s}}}{|x|^s}dx
-\int_{\mathbb{R}^N} \frac{(u^+(x))^{2^*(s)}}{|x|^s}dx+o(1), \\
\label{e4.24}
 \begin{split}
&\int_{\mathbb{R}^N}\int_{ \mathbb{R}^N}
 \frac{|u_{n}(x)-u_{n}(y)|^2}{|x-y|^{N+2\alpha}}\,dx\,dy\\
&= \int_{\mathbb{R}^N}\int_{ \mathbb{R}^N}
 \frac{|(v_{n}(x)+u(x))-(v_{n}(y)+u(y))|^2}{|x-y|^{N+2\alpha}}\,dx\,dy\\
&=\int_{\mathbb{R}^N}\int_{ \mathbb{R}^N}
\Big(|v_{n}(x)-v_{n}(y)|^2+|u(x)-u(y)|^2 \\
&\quad +2(v_n(x)-v_n(y))(u(x)-u(y))\Big)
\big/  |x-y|^{N+2\alpha} \,dx\,dy\\
&=\int_{\mathbb{R}^N}\int_{ \mathbb{R}^N}
\frac{|v_{n}(x)-v_{n}(y)|^2}{|x-y|^{N+2\alpha}}\,dx\,dy
+\int_{\mathbb{R}^N}\int_{ \mathbb{R}^N}
\frac{|u(x)-u(y)|^2}{|x-y|^{N+2\alpha}}\,dx\,dy+o(1).
 \end{split}
 \end{gather}
 Then from \eqref{e4.22}-\eqref{e4.24}, it follows that
 $I(v_n)=I(u_n)-I(u)+o(1)=d-I(u)+o(1)$.
\end{proof}

\begin{lemma}\label{l4.8}
Assume $t\geq b> 0$ and $q>1$, then
$ t^q-(t-b)^q\geq b^q$.
\end{lemma}

\begin{proof}
 Let $f(t)= t^q-(t-b)^q$, it follows
 \[
f'(t)=qt^{q-1}-q(t-b)^{q-1}>0\quad \text{for } t\geq b> 0,\,q>1.
\]
 Then $f(t)= t^q-(t-b)^q\geq f(b)=b^q$.
\end{proof}

\subsection*{Acknowledgements}
 This research was supported by the Natural Science Foundation of China
 (11101160, 11271141), and by the China Scholarship Council (201508440330).
We would like to thank the anonymous referee for the carefully reading this
 paper, and for the  useful comments and suggestions.


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\end{document}