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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 118, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/118\hfil Positive ground state solutions]
{Positive ground state solutions to Schr\"{o}dinger-Poisson systems
with a negative non-local term}

\author[Y.-P. Gao, S.-L. Yu, C.-L. Tang \hfil EJDE-2015/118\hfilneg]
{Yan-Ping Gao, Sheng-Long Yu, Chun-Lei Tang}

\address{Yan-Ping Gao \newline
School of Mathematics and Statistics,
Southwest University,
Chongqing 400715, China}
\email{gao0807@swu.edu.cn}

\address{Sheng-Long Yu \newline
School of Mathematics and Statistics,
Southwest University,
Chongqing 400715, China}
\email{ysl345827434@163.com}

\address{Chun-Lei Tang (corresponding author)\newline
School of Mathematics and Statistics,
Southwest University,
Chongqing 400715, China}
\email{tangcl@swu.edu.cn, Phone +86 23 68253135, Fax +86 23 68253135}

\thanks{Submitted January 20, 2015. Published April 30, 2015.}
\subjclass[2010]{35J47, 35J50, 35J99}
\keywords{Schr\"{o}dinger-Poisson system; ground state solution; 
\hfill\break\indent variational methods}

\begin{abstract}
 In this article, we study the  Schr\"{o}dinger-Poisson system
 \begin{gather*}
 -\Delta u+u-\lambda K(x)\phi(x)u=a(x)|u|^{p-1}u, \quad x\in\mathbb{R}^3, \\
 -\Delta\phi=K(x)u^{2},\quad x\in\mathbb{R}^3,
 \end{gather*}
 with $p\in(1,5)$. Assume that $a:\mathbb{R}^3\to \mathbb{R^{+}}$ and
 $K:\mathbb{R}^3\to \mathbb{R^{+}}$ are nonnegative functions  and satisfy
 suitable assumptions, but not requiring any symmetry property on them,
 we prove the existence of a positive ground state solution resolved
 by the variational methods.
\end{abstract}

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


\section{Introduction and main results}

In this article we study the  Schr\"{o}dinger-Poisson system
\begin{equation}
\begin{gathered}
-\Delta u+V(x)u+\lambda K(x)\phi(x)u=f(x,u), \quad x\in\mathbb{R}^3, \\
-\Delta\phi=K(x)u^{2},\quad x\in\mathbb{R}^3,
\end{gathered}\label{e1.1}
\end{equation}
where $V(x)=1$, $\lambda<0$, $f(x,s)=a(x)s^{p}$ and $a(x), K(x)$
satisfying some suitable assumptions, we will prove problem \eqref{e1.1}
 exists a positive ground state solution.

Similar problems have been widely investigated and it is well known they
have a strong physical meaning because they appear in quantum mechanics models
(see e.g. \cite{C5}) and in semiconductor theory
 \cite{C3,C4,C15,C16}. Variational methods and critical point theory
are always powerful tools in studying nonlinear differential equations.
In recent years, system \eqref{e1.1} has been studied widely via modern
variational methods under the various hypotheses, see
\cite{A,AD,C15,C17,DR} and the references therein.
Many researches have been devoted to the study of problem \eqref{e1.1},
but they mainly concern either the autonomous case or, in the non-autonomous
case, the search of the so-called semi-classical states. We refer the reader
interested in a detailed bibliography to the survey paper \cite{A}.
All these works deal with systems like \eqref{e1.1} with $\lambda>0$ and
the nonlinearity $f(x,s)=s^{p}$ with $p$ subcritical.

To the best of our knowledge, there are only a few article on the existence
of ground state solutions to \eqref{e1.1} with the negative coefficient
of the non-local term. Recently, in \cite{GV}, the author
obtained a ground state solution. In \cite{GV1}, the author considered the
nonlinearity $f(x,s)=a(x)s^{2}$ and obtained a ground state solution.
In this article, we consider the nonlinearity $f(x,s)=a(x)s^{p}$ for
 following Schr\"{o}dinger-Poisson system
\begin{equation}
\begin{gathered}
-\Delta u+u-\lambda K(x)\phi(x)u=a(x)|u|^{p-1}u, \quad x\in\mathbb{R}^3, \\
-\Delta\phi=K(x)u^{2},\quad x\in\mathbb{R}^3.
\end{gathered} \label{eSP} %\tag{$SP$}
\end{equation}
It is worth noticing that there are few works concerning on this case up to now.

As we shall see in Section 2, problem \eqref{eSP} can be easily transformed
in a nonlinear Schr\"{o}dinger equation with a non-local term.
 Briefly, the Poisson equation is solved by using the Lax-Milgram theorem,
then, for all $u \in H^{1}(\mathbb\mathbb{R}^{3})$, a unique
 $\phi_{u}\in D^{1,2}(\mathbb{R}^{3})$ is obtained, such that
$-\Delta \phi=K(x)u^{2}$ and that, inserted into the first equation, gives
\begin{align}\label{e1}
-\Delta u+u-\lambda K(x)\phi_{u}(x)u^{2}=a(x)|u|^{p-1}u, \quad x\in\mathbb{R}^3.
\end{align}
This problem is variational and its solutions are the critical points of the
functional defined in $H^{1}(\mathbb{R}^{3})$ by
\begin{align}\label{e1.3}
I(u)=\frac{1}{2}\|u\|^{2}-\frac{\lambda}{4}
\int_{\mathbb{R}^{3}} K(x)\phi_{u}(x) u^{2}dx
-\frac{1}{p+1}\int_{\mathbb{R}^{3}} a(x)|u|^{p+1}dx.
\end{align}

In our research, we deal with the case in which $p\in{(1, 5)}$,
moreover we always assume that $a(x)$ and $K(x)$ satisfy:
\begin{itemize}
\item[(A1)] There exists a constant $c>0$, such that $a(x)>c$ for all
$x \in\mathbb{R}^{3}$ and $a(x)-c\in L^\frac{6}{5-p}(\mathbb{R}^3)$;

\item[(K1)] $K\in L^2(\mathbb{R}^3)$.
\end{itemize}
Our main result reads as follows.

\begin{theorem} \label{thm1.1}
Suppose $a, K:\mathbb{R}^3\to\mathbb{R}^+$, $\lambda>0$ and $p\in(1,5)$.
Let {\rm (A1), (K1)} hold.
Then problem \eqref{eSP} has a positive ground state solution.
\end{theorem}

\begin{remark} \label{rmk1.1} \rm
To the best of our knowledge, there are only two articles \cite{GV,GV1}
on the existence  of ground state solutions to  \eqref{e1.1} with the negative
coefficient of the non-local.
In \cite{GV}, the author discusses the negative coefficient of the non-local
term under symmetry assumption, but we get the positive ground state solution
without any symmetry assumption. Compared with the \cite{GV1}, we do not need
conditions
$$
\lim_{|x|\to+\infty}a(x)=a_\infty  \quad\text{and}\quad
\lim_{|x|\to+\infty}K(x)=K_\infty.
$$
\end{remark}

The remainder of this paper is organized as follows. In Section 2, notation and preliminaries are presented. In Section 3, we give the proof of Theorem 1.1.

\section{Notation and preliminaries}

Hereafter we use the following notation:

$H^1(\mathbb{R}^3)$ is the usual Sobolev space endowed with the standard scalar
product and norm
$$
(u,v)=\int_{\mathbb{R}^{3}}(\nabla u\cdot\nabla v+uv)dx; \quad
\|u\|^2=\int_{\mathbb{R}^{3}}(|\nabla u|^2+u^2)dx.
$$

$D^{1,2}(\mathbb{R}^3)$ is the completion of $C^{\infty}_0(\mathbb{R}^3)$
with respect to the norm
$$
\|u\|_{D^{1,2}}=\Big(\int_{\mathbb{R}^{3}}|\nabla u|^2dx\Big)^{1/2}.
$$

$L^p(\Omega)$, $1\leq p \leq +\infty$, $\Omega\subseteq\mathbb{R}^3$,
denotes a Lebesgue space, the norm in $L^p(\Omega)$ is denoted by
$\|u\|_{L^p(\Omega)}=|u|_{p,\Omega}$ when $\Omega$ is a proper subset
of $\mathbb{R}^3$,
by $\|u\|_{L^p(\Omega)}=|\cdot|_p$ when $\Omega=\mathbb{R}^3$.

$L^\infty(\Omega)$ is the space of measurable functions in $\Omega $; that is,
$$\operatorname{ess\,sup}_{x\in\Omega}|u(x)|
=\inf\{C>0:|u(x)|\leq C\text{ a. e. in }\Omega\}<+\infty.
$$

  For any $\rho >0$ and for any $z\in\mathbb{R}^3$, $B_\rho(z)$ denotes the ball
 of radius $\rho$ centered at $z$, and $|B_\rho(z)|$ denotes its Lebesgue measure.
 $C,C_0,C_{1},C_{2}$ are various positive constants which can change from
line to line.

 From the embeddings, $H^1(\mathbb{R}^3) \hookrightarrow L^6(\mathbb{R}^3)$ and
$D^{1,2}(\mathbb{R}^3) \hookrightarrow L^6(\mathbb{R}^3)$, we obtain the inequalities
\begin{gather*}
|u|_{6}\leq C_{1}\|u\| \quad \forall u\in H^1(\mathbb{R}^3)\backslash\{0\},\\
|u|_{6}\leq C_{2}\|u\| \quad \forall u\in D^{1,2}(\mathbb{R}^3)\backslash\{0\}.
\end{gather*}

It is well known and easy to show that problem \eqref{eSP} can be reduced to
a single equation with a non-local term. Actually, considering for all
$u\in H^1(\mathbb{R}^3)$, the linear functional $L_u$ defined in
$D^{1,2}(\mathbb{R}^3)$ by
$$
L_u(v)=\int_{\mathbb{R}^3}K(x)u^2v\,dx,
$$
the H\"{o}lder  and Sobolev inequalities imply
\begin{equation}
L_u(v)\leq |K|_2|u^2|_3|v|_6
=|K|_2|u|^{2}_{6}|v|_6
\leq C_{2} |K|_2\cdot|u|^{2}_{6}\|v\|_{D^{1,2}}.\label{Luv}
\end{equation}
Hence, from the Lax-Milgram theorem, for every $u\in H^{1}(\mathbb{R}^3)$,
the Poisson equation $-\Delta\phi=K(x)u^2$ exists a unique
$\phi_u\in D^{1,2}(\mathbb{R}^3)$ such that
\begin{equation} \label{2.2}
\int_{\mathbb{R}^3}K(x)u^2v\,dx
=\int_{\mathbb{R}^3}\nabla\phi_u\cdot\nabla v\,dx,
\end{equation}
for any $v\in D^{1,2}(\mathbb{R}^3)$. Using integration by parts, we get
$$
\int_{\mathbb{R}^3}\nabla\phi_u\cdot\nabla v\,dx
=-\int_{\mathbb{R}^3}v\Delta\phi_udx,
$$
therefore,
$$
-\Delta\phi_u=K(x)u^2,
 $$
in a weak sense and the representation formula
\begin{align}
\phi_u=\int_{\mathbb{R}^3}\frac{K(y)}{|x-y|}u^2(y)dy
=\frac{1}{|x|}\ast Ku^2 \label{Pukxx}
\end{align}
holds. Moreover, $\phi_u>0$ when $u\neq0$, because $K$ does,
and by \eqref{Luv}, \eqref{2.2} and the Sobolev inequality, the relations
\begin{gather}\label{puD}
\|\phi_u\|_{D^{1,2}}\leq C_{2} C^{2}_{1} \cdot|K|_2\|u\|^2, \quad
|\phi_u|_6\leq C_{2}\|\phi_u\|_{D^{1,2}}, \\
\label{y}
\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}\frac{K(x)K(y)}{|x-y|}u^2(x)u^2(y)dxdy
=\int_{\mathbb{R}^3}K(x)\phi_uu^{2}dx
\leq C_{2}^{2} C_{1}^{4}\cdot |K|_{2}^{2}\|u\|^4
\end{gather}
hold. Substituting $\phi_u$ in problem \eqref{eSP}, we are led to \eqref{e1},
whose solutions can be obtained by looking for critical points of the functional
$I:H^{1}(\mathbb{R}^3)\to \mathbb{R}$ where $I$ is defined in \eqref{e1.3}.
Indeed, \eqref{puD} and \eqref{y} imply that $I$ is a well-defined $C^{2}$
functional, and that
\begin{equation}
\langle I'(u),v\rangle=\int_{\mathbb{R}^3}\Big(\nabla u\cdot\nabla v+uv
-\lambda K(x)\phi_uuv-a(x)|u|^{p-1}uv\Big)dx.
\end{equation}
Hence, if $u\in H^1(\mathbb{R}^3)$ is a critical point of $I$,
then the pair $(u,\phi_u)$, with $\phi_u$ as in \eqref{Pukxx}, is a
 solution of  \eqref{eSP}.

Let us define the operator $\Phi$:$H^1(\mathbb{R}^3)\to D^{1,2}(\mathbb{R}^3)$ as
$$
\Phi[u]=\phi_u.
$$
In the next lemma we summarize some properties of $\Phi$, useful for the
study our problem.

\begin{lemma}[\cite{GC}] \label{lem2.1} 
\begin{itemize}
\item[(1)] $\Phi$ is continuous;

\item[(2)] $\Phi$ maps bounded sets into bounded sets;

\item[(3)] if $u_n \rightharpoonup u$ in $H^1(\mathbb{R}^3)$ then 
$\Phi[u_n] \rightharpoonup \Phi[u]$ in $D^{1,2}(\mathbb{R}^3)$;

\item[(4)] $\Phi[tu]=t^2\Phi[u]$ for all $t\in\mathbb{R}$.
\end{itemize}
\end{lemma}


\begin{lemma}[\cite{P}] \label{lem2.2}
 Suppose $r>0$, $2<q<2^*(=6)$. If $\{u_n\}\subset H^1(\mathbb{R}^3)$ 
is bounded and
$$
\sup_{y\in\mathbb{R}^3}\int_{B(y,r)}|u_n|^qdx\to0, \quad\text{as }n\to+\infty,
$$
then $u_n \to 0$ in $L^q(\mathbb{R}^3)$ for $2<q<2^*$.
\end{lemma}

\section{Proof of main results}

First wee give some properties of the  nonlinear Schr\"{o}dinger equation
\begin{equation}
-\Delta u+u=c|u|^{p-1}u, \label{NSE}
\end{equation}
that has been broadly studied in \cite{P,MKK}.
We set
\begin{align}\label{NuH}
\mathcal{N}_\infty:=\{u\in H^1(\mathbb{R}^3)\backslash\{0\}:\|u\|^2
=c|u|_{p+1}^{p+1}\}.
\end{align}
Then for any $u\in\mathcal{N}_\infty$, we have
\begin{equation}
I_\infty(u)=\frac{1}{2}\|u\|^2-\frac{c}{p+1}\int_{\mathbb{R}^3}|u|^{p+1}dx\\
=\Big(\frac{1}{2}-\frac{1}{p+1}\Big)\|u\|^2, \label{hou}
\end{equation}
and
$m_\infty:=\inf\{I_\infty(u):u\in\mathcal{N}_\infty\}$.

It is well known that \eqref{NSE} has at least a ground state solution
 which we denote $w_\infty$.
By using \eqref{NuH} and \eqref{hou}, we know that
$$
m_\infty=I_\infty(w_\infty)=\Big(\frac{1}{2}-\frac{1}{p+1}\Big)\|w_\infty\|^2 
 $$
and
\begin{equation} \label{wwp}
\|w_\infty\|^2=c\int_{\mathbb{R}^3}|w_\infty|^{p+1}dx.
\end{equation}
For  \eqref{eSP}, it is not difficult to verify that the functional $I$
is bounded either from below or from above.
So it is convenient to consider $I$ restricted to a natural constraint,
the Nehari manifold, that contains all the nonzero critical points of $I$
and on which $I$ turns out to be bounded from below. We set
$$
\mathcal{N}:=\{u\in H^1(\mathbb{R}^3)\backslash\{0\}:G(u)=0\}$$
where
\begin{equation}
G(u)=\langle I'(u),u\rangle
=\|u\|^2-\lambda\int_{\mathbb{R}^3}K(x)\phi_uu^{2}dx
 -\int_{\mathbb{R}^3}a(x)|u|^{p+1}dx.
\label{Gu}
\end{equation}
The following lemma  states the main properties
of $\mathcal{N}$.

\begin{lemma} \label{lem3.1}
 $I$ is bounded from below on $\mathcal{N}$ by a positive constant.
\end{lemma}

\begin{proof} 
Let  $u\in\mathcal{N}$, from {\rm (A1)} and  H\"{o}lder's inequality, we have
\begin{equation} \label{Ituu}
\begin{aligned} 
0&=\|u\|^2-\lambda\int_{\mathbb{R}^3}K(x)\phi_uu^{2}dx
 -\int_{\mathbb{R}^3}a(x)|u|^{p+1}dx\\
&\geq\|u\|^2-C\|u\|^4-C_0\|u\|^{p+1}
\end{aligned}
\end{equation}
from which we have
\begin{align}
\|u\|\geq C_1>0, \quad \forall u\in\mathcal{N}
\end{align}
 Using this inequality, $\lambda>0$, $K>0$, $a>0$, when $1<p<3$, we obtain
\begin{equation} \label{Iu12}
\begin{aligned}
I(u)
&=\Big(\frac{1}{2}-\frac{1}{p+1}\Big)\|u\|^2
+\Big(\frac{1}{p+1}-\frac{1}{4}\Big)\lambda \int_{\mathbb{R}^3}K(x)\phi_uu^{2}dx\\
&\geq \Big(\frac{1}{2}-\frac{1}{p+1}\Big)\|u\|^2 \\
&\geq \Big(\frac{1}{2}-\frac{1}{p+1}\Big)C_1^{2}
>0,
\end{aligned}
\end{equation}
when $3\leq p<5$,
\begin{equation} \label{Iu13}
\begin{aligned}
I(u)&=\frac{1}{4}\|u\|^2+\Big(\frac{1}{4}-\frac{1}{p+1}\Big)
\int_{\mathbb{R}^3}a(x)|u|^{p+1}dx \\
&\geq \frac{1}{4}\|u\|^2 \\
&\geq \frac{1}{4}C_1^{2}
>0.
\end{aligned}
\end{equation}
Setting
$m:=\inf\{I(u):u\in\mathcal{N}\}$,
as a consequence of Lemma 3.1, $m$ turns out to be a positive number.
Then we obtain a sequence $\{u_n\}\subset\mathcal{N}$, such that
\begin{align}\label{Iunm}
\lim_{n\to\infty}I(u_n)=m.
\end{align}
\end{proof}

Now we give the proof of our main result.

\begin{proof}[Proof of Theorem 1.1]
 First, we prove that
\begin{equation}
m<m_\infty. \label{mm}
\end{equation}
We know that $w_\infty \in\mathcal{N}_\infty$ and $I_\infty(w_\infty)=m_\infty$.
We claim that there exists $t_0>0$ such that $t_0w_\infty \in\mathcal{N}$.
Indeed, from \eqref{Gu}, for all $t\geq0$ one has
$$
G(tw_\infty)=t^{2}\|w_\infty\|^2-\lambda t^{4}\int_{\mathbb{R}^3}K(x)
\phi_{w_\infty}w_\infty^{2}dx-t^{p+1}\int_{\mathbb{R}^3}a(x)|w_\infty|^{p+1}dx,
$$
then $G(0)=0$ and $G(tw_\infty)\to-\infty$ as $t\to+\infty$. Moreover,
$$
G_{t}'(tw_\infty)
=t\Big(2\|w_\infty\|^2-4\lambda t^{2}\int_{\mathbb{R}^3}K(x)
 \phi_{w_\infty}w_\infty^{2}dx
 -(p+1)t^{p-1}\int_{\mathbb{R}^3}a(x)|w_\infty|^{p+1}dx\Big),
$$
then there exists $t_{\rm max}>0$ such that $G_{t}'(tw_\infty)>0$
for all $0<t<t_{\rm max}$ and $G_{t}'(tw_\infty)<0$ for all
$t>t_{\rm max}$.
Then $G(tw_\infty)$ is increasing for all $0<t<t_{\rm max}$ and
$G(tw_\infty)$ decreasing for all $t>t_{\rm max}$.
Thus there exists $t_0>0$ such that $G(t_0w_\infty)=0$.
That is, $t_0w_\infty \in\mathcal{N}$. Our claim is true.
It follows that
\begin{equation} \label{m}
\begin{aligned}
m&\leq I(t_0w_\infty)         \\
&=\frac{t_0^2}{2}\|w_\infty\|^2-\frac{t_0^4}{4}
 \lambda\int_{\mathbb{R}^3}K(x)\phi_{w_\infty}(x)w_\infty ^2dx
 -\frac{t_0^{p+1}}{p+1}\int_{\mathbb{R}^3}a(x)|w_\infty|^{p+1}dx     \\
&< \frac{t_0^2}{2}\|w_\infty\|^2-\frac{t_0^{p+1}}{p+1}\int_{\mathbb{R}^3}c|w_\infty|^{p+1}dx   \\
&\leq \Big(\frac{1}{2}-\frac{1}{p+1}\Big)\|w_\infty\|^2   \\
&=I_\infty(w_\infty)
=m_\infty.
\end{aligned}
\end{equation}
We assume that $\{u_n\}$ is what  obtained in \eqref{Iunm}.
 From \eqref{Pukxx}, we can get $\{|u_n|\}$ is also a minimize sequence.
Setting $u_n(x)\geq0$ in $\mathbb{R}^3$ a.e. by \eqref{Iu12} and \eqref{Iu13}, we have if $p\in{(1,3)}$, then
$$
I(u_n)\geq\Big(\frac{1}{2}-\frac{1}{p+1}\Big)\|u_n\|^2,
$$
and if $p\in{[3,5)}$, then
$$
I(u_n)\geq\frac{1}{4}\|u_n\|^2.
$$
In both cases, being $I(u_n)$ is bounded, $\{u_n\}$ is also bounded.

On the other hand, since $\{u_n\}$ is bounded in $H^1(\mathbb{R}^3)$, 
there exists $\overline{u}\in H^1(\mathbb{R}^3)$ such that, up to a subsequence,
\begin{gather}
u_n\rightharpoonup\overline{u}, \quad\text{in }H^1(\mathbb{R}^3);\label{unu2}\\
u_n\to\overline{u}, \quad\text{in }L^{p+1}_{\rm loc}(\mathbb{R}^3);\label{ll}\\
u_n(x)\to\overline{u}(x), \quad\text{ a.e. in }\mathbb{R}^3.
\end{gather}
Setting
$$
z_{n}^{1}(x)=u_n(x)-\overline{u}(x).
$$
Obviously, $z_{n}^{1}\rightharpoonup0$ in $H^1(\mathbb{R}^3)$, but not strongly. 
A direct computation gives
\begin{equation}
\|u_n\|^2=\|z_{n}^{1}+\overline{u}\|^2
=\|z_{n}^{1}\|^2+\|\overline{u}\|^2+o(1).\label{un2z}
\end{equation}
According to the Brezis-Lieb Lemma  \cite{HE}, we deduce
\begin{align}\label{unp1p1}
|u_n|_{p+1}^{p+1}=|\overline{u}|_{p+1}^{p+1}+|z_{n}^{1}|_{p+1}^{p+1}+o(1).
\end{align}
Then, we claim that, for any $h\in H^1(\mathbb{R}^3)$, we have
\begin{align}\label{Runp}
\int_{\mathbb{R}^3}|u_n|^{p-1}u_nh\,dx\to\int_{\mathbb{R}^3}
|\overline{u}|^{p-1}\overline{u}h\,dx.
\end{align}
For every $h\in C^{\infty}_0(\mathbb{R}^3)$, there exists a bounded open 
subset $\Omega\subset\mathbb{R}^3$, such that $\operatorname{supp}h\subset\Omega$, 
where $\operatorname{supp}h=\overline{\{x\in\mathbb{R}^3:h(x)\neq0\}}$. 
From \eqref{ll}, we have
\begin{align*}
&\big|\int_{\mathbb{R}^3}|u_n|^{p-1}u_nh\,dx
-|\overline{u}|^{p-1}\overline{u}h\,dx\big|\\
&<\int_{\mathbb{R}^3}\big||u_n|^{p-1}u_nh-|\overline{u}|^{p-1}\overline{u}h\big|dx\\
&\leq \int_{\mathbb{R}^3}p(|u_n|^{p-1}+|\overline{u}|^{p-1})|u_n-\overline{u}||h|dx\\
&=\int_{\mathbb{R}^3}p|u_n|^{p-1}|u_n-\overline{u}||h|dx
 +\int_{\mathbb{R}^3}p|\overline{u}|^{p-1}|u_n-\overline{u}||h|dx\\
&<p|u_n|_{p+1}|u_n-\overline{u}|_{p+1,\Omega}|h|_{p+1}
 +p|\overline{u}|_{p+1}|u_n-\overline{u}|_{p+1,\Omega}|h|_{p+1}
<\varepsilon
\end{align*}
which proves \eqref{Runp}. Let us show that
\begin{gather}\label{k1}
\int_{\mathbb{R}^3}K(x)\phi_{u_n}u^{2}_{n}dx
=\int_{\mathbb{R}^3}K(x)\phi_{\overline{u}}\overline{u}^2dx+o(1), \\
\label{k2}
\int_{\mathbb{R}^3}K(x)\phi_{u_n}u_nh\,dx
=\int_{\mathbb{R}^3}K(x)\phi_{\overline{u}}\overline{u}h\,dx+o(1).
\end{gather}
First let us observe that, in view of the Sobolev embedding theorem, 
\eqref{unu2} and (3) of Lemma 2.1, $u_{n}\rightharpoonup \bar{u}$ 
in $H^{1}(\mathbb{R}^3)$ implies
\begin{gather}
u_n\rightharpoonup\overline{u}, \quad \text{in }L^6(\mathbb{R}^3);\label{a}\\
u^{2}_{n}\to\overline{u}^2, \quad \text{in }L^{3}_{\rm loc}(\mathbb{R}^3);\label{b}\\
\phi_{u_n}\rightharpoonup\phi_{\overline{u}}, 
 \quad\text{in }D^{1,2}(\mathbb{R}^3);\label{c}\\
\phi_{u_n}\to\phi_{\overline{u}}, \quad\text{in }L^{6}_{\rm loc}(\mathbb{R}^3).
\label{d}
\end{gather}
Furthermore, considering \eqref{b} and \eqref{d} respectively, we can assert 
that for any choice of $\varepsilon>0$ and $\rho>0$, the relations
\begin{gather}
|u^{2}_{n}-\overline{u}^2|_{3,B_\rho(0)}<\varepsilon,\label{u2nu}\\
|\phi_{u_n}-\phi_{\overline{u}}|_{6,B_\rho(0)}<\varepsilon \label{punp}
\end{gather}
hold for large $n$.

On the other hand,  $u_n$ being bounded in $H^1(\mathbb{R}^3)$, 
$\phi_{u_n}$ is bounded in $D^{1,2}(\mathbb{R}^3)$ and 
in $L^{6}(\mathbb{R}^3)$, because of (2) of Lemma 2.1 and the continuity 
of the Sobolev embedding of $D^{1,2}(\mathbb{R}^3)$ in $L^{6}(\mathbb{R}^3)$. 
Moreover $K\in L^2(\mathbb{R}^3)$, for any $\varepsilon>0$, there exists 
$\overline{\rho}=\overline{\rho}(\varepsilon)$ such that
\begin{align}\label{k2B0}
|K|_{2,\mathbb{R}^3\backslash B_\rho(0)}<\varepsilon, \quad 
\forall\rho\geq\overline{\rho}.
\end{align}
Hence, by \eqref{u2nu} and \eqref{k2B0},  for large $n$, we deduce that
\begin{align*}
&\Big|\int_{\mathbb{R}^3}K(x)\phi_{u_n}u^{2}_{n}dx
-\int_{\mathbb{R}^3}K(x)\phi_{\overline{u}}\overline{u}^2dx\Big|\\
&\leq \Big|\int_{\mathbb{R}^3}K(x)\phi_{u_n}(u^{2}_{n}-\overline{u}^2)dx
 +\int_{\mathbb{R}^3}K(x)(\phi_{u_n}-\phi_{\overline{u}})\overline{u}^2dx\Big|\\
&\leq \Big|\int_{\mathbb{R}^3}K(x)\phi_{u_n}(u^{2}_{n}-\overline{u}^2)dx\Big|
 +\Big|\int_{\mathbb{R}^3}K(x)(\phi_{u_n}-\phi_{\overline{u}})\overline{u}^2dx\Big|\\
&\leq \Big|\int_{\mathbb{R}^3\backslash B_\rho(0)}K(x)\phi_{u_n}(u^{2}_{n}
 -\overline{u}^2)dx\Big|
 +\Big|\int_{B_\rho(0)}K(x)\phi_{u_n}(u^{2}_{n}-\overline{u}^2)dx\Big|\\
&\quad +\Big|\int_{\mathbb{R}^3\backslash B_\rho(0)}K(x)(\phi_{u_n}
 -\phi_{\overline{u}})\overline{u}^2dx\Big|
+\Big|\int_{B_\rho(0)}K(x)(\phi_{u_n}-\phi_{\overline{u}})\overline{u}^2dx\Big|\\
&\leq |K|_{2,\mathbb{R}^3\backslash B_\rho(0)}
 \Big(|\phi_{u_n}|_{6,\mathbb{R}^3\backslash B_\rho(0)}|u^{2}_{n}
 -\overline{u}^2|_{3,\mathbb{R}^3\backslash B_\rho(0)} \\
&\quad  +|\phi_{u_n}-\phi_{\overline{u}}|_{6,\mathbb{R}^3\backslash B_\rho(0)}
 |\overline{u}^2|_{3,\mathbb{R}^3\backslash B_\rho(0)}\Big)
+|K|_{2,B_\rho(0)}|\phi_{u_n}|_{6,B_\rho(0)}|u^{2}_{n}
 -\overline{u}^2|_{3,B_\rho(0)}\\
&\quad +|K|_{2,B_\rho(0)}|\phi_{u_n}
 -\phi_{\overline{u}}|_{6,B_\rho(0)}|\overline{u}^2|_{3,B_\rho(0)}\\
&\leq C\varepsilon
\end{align*}
which proves \eqref{k1}.

Analogously, by \eqref{punp} and \eqref{k2B0}, for large $n$, we infer that
\[
\Big|\int_{\mathbb{R}^3}K(x)\phi_{u_n}u_nh\,dx
-\int_{\mathbb{R}^3}K(x)\phi_{\overline{u}}\overline{u}h\,dx\Big|
\leq\varepsilon
\]
which proves \eqref{k2}.
Therefore, by \eqref{un2z}, \eqref{unp1p1} and \eqref{k1} respectively, 
we obtain
\begin{equation}
\begin{aligned}
I(u_n)
&=\frac{1}{2}\|u_n\|^2-\frac{\lambda}{4}\int_{\mathbb{R}^3}K(x)\phi_{u_n}u^{2}_{n}dx
 -\frac{1}{p+1}\int_{\mathbb{R}^3}a(x)|u_n|^{p+1}dx\\
&=\frac{1}{2}\|z_{n}^{1}\|^2+\frac{1}{2}\|\overline{u}\|^2
-\frac{\lambda}{4}\int_{\mathbb{R}^3}K(x)\phi_{\overline{u}}\overline{u}^2dx
-\frac{1}{p+1}\int_{\mathbb{R}^3}a(x)|\overline{u}|^{p+1}dx\\
&\quad -\frac{c}{p+1}\int_{\mathbb{R}^3}|z_{n}^{1}|^{p+1}dx+o(1)\\
&=I(\overline{u})+I_\infty(z_{n}^{1})+o(1). 
\end{aligned} \label{IuIz}
\end{equation}
By \eqref{Runp} and \eqref{k2} for any $h\in C^{\infty}_0(\mathbb{R}^3)$,
\begin{equation}
\begin{aligned}
\langle I'(u_n),h\rangle
&=\int_{\mathbb{R}^3}(\nabla u_n\cdot\nabla h+u_nh
 -\lambda K(x)\phi_{u_n}u_nh-a(x)|u_n|^{p-1}u_nh)dx\\
&=\int_{\mathbb{R}^3}(\nabla \overline{u}\cdot\nabla h
 +\overline{u}h-\lambda K(x)\phi_{\overline{u}}\overline{u}h
 -a(x)|\overline{u}|^{p-1}\overline{u}h)dx+o(1)\\
&=\langle I'(\overline{u}),h\rangle+o(1).
\end{aligned} \label{zz}
\end{equation}
We now claim that
\begin{equation}\label{nIuu}
\nabla I(u_n)\to 0, \quad\text{in } H^1(\mathbb{R}^3).
\end{equation}
By Lagrange's multiplier theorem, we know that there exists
$\lambda_n\in\mathbb{R}$ such that
\begin{equation}  \label{o1nI}
o(1)=\nabla I|_\mathcal{N}(u_n)=\nabla I(u_n)-\lambda_n\nabla G(u_n).
\end{equation}
So, taking the scalar product with $u_n$, we obtain
\[
o(1)=(\nabla I(u_n),u_n)-\lambda_n(\nabla G(u_n),u_n).
\]
$G$ turns out to be a $C^1$ functional. Using \eqref{Ituu} and
$\lambda>0, K>0, a>0$, when $1<p\leq3$, we deduce
\begin{equation} \label{GuuI}
\begin{aligned}
\langle G'(u),u\rangle
&= 2\|u\|^2-4\lambda\int_{\mathbb{R}^3}K(x)\phi_uu^{2}dx-(p+1)\int_{\mathbb{R}^3}a(x)|u|^{p+1}dx\\
&= (1-p)\|u\|^2+\lambda(p-3)\int_{\mathbb{R}^3}k(x)\phi_uu^{2}dx\\
&\leq (1-p)\|u\|^2\\
&\leq -(p-1)C_1
<0,
\end{aligned}
\end{equation}
when $3<p<5$,
\begin{equation} \label{GuuI2}
\begin{aligned}
\langle G'(u),u\rangle
&= 2\|u\|^2-4\lambda\int_{\mathbb{R}^3}K(x)\phi_uu^{2}dx
 -(p+1)\int_{\mathbb{R}^3}a(x)|u|^{p+1}dx\\
&= -2\|u\|^2+(3-p)\int_{\mathbb{R}^3}a(x)|u|^{p+1}dx\\
&\leq -2\|u\|^2\\
&\leq -2C_2 <0.
\end{aligned}
\end{equation}
Since $u_n\in\mathcal{N}$, we have $(\nabla I(u_n),u_n)=0$; by inequalities
\eqref{GuuI} and \eqref{GuuI2}, we have $(\nabla G(u_n),u_n)<C<0$.
Thus $\lambda_n\to0$ for  $n\to+\infty$.
Moreover, by the boundedness of $\{u_n\}$,
$\nabla G(u_n)$ is bounded and this implies $\lambda_n\nabla G(u_n)\to0$,
so \eqref{o1nI} follows from \eqref{nIuu}. By \eqref{zz} and \eqref{nIuu},
we have $\langle I'(\overline{u}),h\rangle=0$, so
$\overline{u}$ is a solution of  problem \eqref{eSP}. By \eqref{k1}, we have
\begin{align*}
\langle I'(u_n),u_n\rangle
&= \|u_n\|^2-\lambda\int_{\mathbb{R}^3}K(x)\phi_{u_n}u_{n}^{2}dx
 -\int_{\mathbb{R}^3}a(x)|u_n|^{p+1})dx\\
&= \|\overline{u}\|^2+\|z_{n}^{1}\|^2-\lambda\int_{\mathbb{R}^3}
 K(x)\phi_{\overline{u}}\overline{u}^2dx
 -\int_{\mathbb{R}^3}a(x)|\overline{u}|^{p+1}dx\\
&\quad -c\int_{\mathbb{R}^3}|z_{n}^{1}|^{p+1}dx+o(1)\\
&= \langle I'(\overline{u}),\overline{u}\rangle
 +\langle I'_\infty(z_{n}^{1}),z_{n}^{1}\rangle+o(1),
\end{align*}
which implies that
\begin{equation}
o(1)=\langle I'_\infty(z_{n}^{1}),z_{n}^{1}\rangle
=\|z_{n}^{1}\|^2-c|z_{n}^{1}|^{p+1}_{p+1}.\label{1I}
\end{equation}
Setting
$$
\delta:=\limsup_{n\to+\infty}
\Big(\sup_{y\in\mathbb{R}^3}\int_{B_1(y)}|z_{n}^{1}|^{p+1}dx\Big).
$$
We claim $\delta=0$. By \cite[Lemma 1.21]{C17}, one has
\begin{align}\label{haokan}
z_{n}^{1}\to0, \quad\text{in } L^{p+1}(\mathbb{R}^3).
\end{align}
From \eqref{1I} and \eqref{haokan}, we obtain
\begin{align*}
o(1)&= \langle I'_\infty(z_{n}^{1}),z_{n}^{1}\rangle\\
&= \|z_{n}^{1}\|^2-c|z_{n}^{1}|^{p+1}_{p+1}\\
&= \|z_{n}^{1}\|^2+o(1)\\
&= \|u_n-\overline{u}\|^2+o(1),
\end{align*}
so $u_n\to\overline{u}$ in $H^1(\mathbb{R}^3)$.
Let $u=\overline{u}$, so $I(u)=m$, $I'(u)=0$ and $u(x)>0$ a.e.
in $\mathbb{R}^3$.

Let us prove $\delta=0$. Actually, if $\delta>0$, there exists sequence 
$\{y_{n}^{1}\}\subset\mathbb{R}^3$, such that
$$
\int_{B_1(y_{n}^{1})}|z_{n}^{1}|^{p+1}dx>\frac{\delta}{2}.
$$
Let us now consider $z_{n}^{1}(\cdot+y_{n}^{1})$. 
We assume that $z_{n}^{1}(\cdot+y_{n}^{1})\rightharpoonup u^1$ in 
$H^1(\mathbb{R}^3)$ and, then, $z_{n}^{1}(x+y_{n}^{1})$ $\to$ $u^1(x)$ a.e. on
 $\mathbb{R}^3$. Since
$$
\int_{B_1(0)}|z_{n}^{1}(x+y_{n}^{1})|^{p+1}dx>\frac{\delta}{2},
$$
from the Rellich theorem it follows that
$$
\int_{B_1(0)}|u^1(x)|^{p+1}dx\geq\frac{\delta}{2}, 
$$
and thus $u^1\neq0$. Finally, let us set
$$
z_{n}^{2}(x)=z_{n}^{1}(x+y_{n}^{1})-u^1(x). 
$$
Then, using \eqref{un2z}, \eqref{unp1p1} and the Brezis-Lieb Lemma, we have
\begin{gather}
\|z_{n}^{2}\|^2=\|z_{n}^{1}\|^2-\|u^1\|^2+o(1),\label{shang} \\
|z_{n}^{2}|_{p+1}^{p+1}=|u_n|_{p+1}^{p+1}
 -|\overline{u}|_{p+1}^{p+1}-|u^1|_{p+1}^{p+1}+o(1). \label{xia}
\end{gather}
These equalities imply
$$
I_\infty(z_{n}^{2})=I_\infty(z_{n}^{1})-I_\infty(u^1)+o(1), 
$$
hence, by using \eqref{IuIz}, we obtain
\begin{equation} \label{un}
\begin{aligned}
I(u_n)&=I(\overline{u})+I_\infty(z_{n}^{1})+o(1)\\
&=I(\overline{u})+I_\infty(u^1)+I_\infty(z_{n}^{2})+o(1).
\end{aligned}
\end{equation}
Using \eqref{1I}, \eqref{shang} and \eqref{xia}, we obtain
\begin{align*}
\langle I'_\infty(z_{n}^{1}),z_{n}^{1}\rangle
&= \|z_{n}^{1}\|^2-c|z_{n}^{1}|_{p+1}^{p+1}\\
&= \|u^1\|^2-c|u^1|_{p+1}^{p+1}+\|z_{n}^{2}\|^2-c|z_{n}^{2}|_{p+1}^{p+1}+o(1) \\
&= \langle I'_\infty(u^1),u^1\rangle
 +\langle I'_\infty(z_{n}^{2}),z_{n}^{2}\rangle+o(1),
\end{align*}
which implies
\[
o(1)=\langle I'_\infty(z_{n}^{2}),z_{n}^{2}\rangle
=\|z_{n}^{2}\|^2-c|z_{n}^{2}|^{p+1}_{p+1}.
\]
Moreover, we obtain
\begin{equation} \label{zn}
I_\infty(z_{n}^{2})
= \frac{1}{2}\|z_{n}^{2}\|^2-\frac{c}{p+1}|z_{n}^{2}|^{p+1}_{p+1}
= \Big(\frac{1}{2}-\frac{1}{p+1}\Big)\|z_{n}^{2}\|^2+o(1).
\end{equation}
Since $z_{n}^{1}\rightharpoonup u^1$ in $H^1(\mathbb{R}^3)$ and
$u^{1}\neq0$, according to \eqref{1I}, one has $u^{1}\in\mathcal{N}_\infty$.
Because of  $\overline{u} \in \mathcal{N}$, from Lemma 3.1,
we obtain $I(\overline{u})>0$. Thus, using \eqref{un} and \eqref{zn}, we obtain
\begin{align*}
m&=\liminf_{n\to\infty}I(u_n)   \\
 &\geq  I(\overline{u})+I_\infty(u^1)+\liminf_{n\to\infty}I_\infty(z_{n}^{2})  \\
 &\geq I_\infty(u^1)
 \geq m_\infty
\end{align*}
which contradicts with \eqref{mm}.
\end{proof}

\subsection*{Acknowledgments}
This research was supported by National Natural Science Foundation of
China (No.11471267).

The authors would like to thank the anonymous referees for their valuable 
suggestions.

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\end{thebibliography}                                                                                                                                                                                                                                                                                                                          \end{document}