# Multiplicity and concentration of solutions for a fractional $p$-Kirchhoff
type equation

Wenjing Chen111 E-mail address<EMAIL_ADDRESS>(W. Chen),
<EMAIL_ADDRESS>(H. Pan) and Huayu Pan
School of Mathematics and Statistics, Southwest University, Chongqing, 400715,
P.R. China

###### Abstract

This paper is concerned with the following fractional $p$-Kirchhoff equation

$\displaystyle\varepsilon^{sp}M\left({\varepsilon^{sp-N}}\iint_{\mathbb{R}^{2N}}\frac{{{{\left|{u(x)-u(y)}\right|}^{p}}}}{{{{\left|{x-y}\right|}^{N+sp}}}}dxdy\right)(-\Delta)_{p}^{s}u+V(x){u^{p-1}}={u^{p_{s}^{*}-1}}+f(u),\
\ u>0,\ \mbox{in}\ {\mathbb{R}^{N}},$

where $\varepsilon>0$ is a parameter, $M(t)=a+bt^{\theta-1}$ with $a>0$,
$b>0$, $\theta>1$, $(-\Delta)_{p}^{s}$ denotes the fractional $p$-Laplacian
operator with $0<s<1$ and $1<p<\infty$, $N>sp$, $\theta p<p_{s}^{*}$ with
$p_{s}^{*}=\frac{Np}{N-sp}$ is the fractional critical Sobolev exponent, $f$
is a superlinear continuous function with subcritical growth and $V$ is a
positive continuous potential. Using penalization method and Ljusternik-
Schnirelmann theory, we study the existence, multiplicity and concentration of
nontrivial solutions for $\varepsilon>0$ small enough.

_Keywords:_ Fractional $p$-Kirchhoff equation; Critical growth; Penalization
method; Ljusternik-Schnirelmann category.

_2020 Mathematics Subject Classification:_ 35J50, 35J91, 35R03.

##  1 Introduction

In this article, we are interested in the following fractional $p$-Kirchhoff
equation

$\varepsilon^{sp}M\left({\varepsilon^{sp-N}}\iint_{\mathbb{R}^{2N}}\frac{{{{\left|{u(x)-u(y)}\right|}^{p}}}}{{{{\left|{x-y}\right|}^{N+sp}}}}dxdy\right)(-\Delta)_{p}^{s}u+V(x){u^{p-1}}={u^{p_{s}^{*}-1}}+f(u),\
\ u>0,\ \mbox{in}\ {\mathbb{R}^{N}},$ (1.1)

where $\varepsilon>0$ is a parameter, $M(t)=a+bt^{\theta-1}$ with $a>0$,
$b>0$, $\theta>1$, and $0<s<1$, $1<p<\infty$, $N>sp$, $\theta p<p_{s}^{*}$,
$p_{s}^{*}=\frac{Np}{N-sp}$ is the fractional critical Sobolev exponent,
$(-\Delta)_{p}^{s}$ denotes the fractional $p$-Laplacian which (up to
normalization factors) may be defined for every function
$u\in\mathcal{C}_{0}^{\infty}(\mathbb{R}^{N})$ as

$(-\Delta)_{p}^{s}=2\mathop{\lim}\limits_{r\to
0}\int_{\mathbb{R}^{N}\setminus{B_{r}(x)}}\frac{{|u(x)-u(y){|^{p-2}}(u(x)-u(y))}}{{|x-y{|^{N+sp}}}}dy,\
\ for\ \ x\in\mathbb{R}^{N}.$

When $p=2$, $a=1$ and $b=0$, (1.1) is related to the following fractional
Schrödinger equation

$\varepsilon^{2s}(-\Delta)^{s}u+V(x)u=h(x,u)\ \ \ \mbox{in}\ \
\mathbb{R}^{N},$ (1.2)

which has been proposed by Laskin [29] as a result of expending the Feynman
path integral from the Brownian like to the Lévy like quantum mechanical
paths. In the past few years, equation (1.2) has been widely considered, and
the multiplicity of solutions to (1.2) has been established by using
variational methods under different and suitable assumptions on $V$ and $h$,
we refer to [1, 2, 3, 4, 21, 27, 38, 16] and references therein.

In the case $s=\varepsilon=1$, $p=2$, $\theta=2$ and $N=3$, equation (1.1)
becomes the following equation

$-\Big{(}a+b\int_{\mathbb{R}^{3}}|\nabla u|^{2}dx\Big{)}\Delta u+V(x)u=h(x,u)\
\ \ \mbox{in}\ \ \mathbb{R}^{3},$ (1.3)

which is related to the stationary analogue of the equation

$\rho
u_{tt}-\left(\frac{\rho_{0}}{h}+\frac{E}{2L}\int_{0}^{L}|u_{x}|^{2}dx\right)u_{xx}=0,$
(1.4)

introduced by Kirchhoff [28] as an extension of the classical D’Alembert’s
wave equation for describing the transversal oscillations of a stretched
string. The Kirchhoff’s model takes into account the charges in length of the
string produced by transverse vibrations. Here $u=u(x,t)$ is the transverse
string displacement at the space coordinate $x$ and time $t$, $L$ is the
length of the string, $h$ is the area of the cross section, E is Young’s
modulus of the material, $\rho$ is the mass density, and $P_{0}$ is the
initial tension; see [10, 34]. In the classical framework, He and Zou [25]
proved the multiplicity and concentration behavior of solutions for small
$\varepsilon>0$ to the following perturbed Kirchhoff equation.

$-\Big{(}a\varepsilon^{2}+b\varepsilon\int_{\mathbb{R}^{3}}|\nabla
u|^{2}dx\Big{)}\Delta u+V(x)u=g(u)\ \ \ \mbox{in}\ \ \mathbb{R}^{3},$ (1.5)

under the following global condition on $V$ due to Rabinowitz [36]

$\mathop{\inf}\limits_{x\in\mathbb{R}^{3}}V(x)<\mathop{\lim\inf}\limits_{|x|\to\infty}V(x):=V_{\infty}\leq\infty,$

and $g$ is a subcritical nonlinearity. Subsequently, Wang et al. [39]
investigated the multiplicity and concentration phenomenon for (1.5) in the
presence of the critical term. Under local conditions on the potential $V$, by
the generalized Nehari manifold method and Ljusternik-Schnirelmann theory,
Figueiredo and Junior [22] obtained the multiplicity result for a subcritical
Kirchhoff equation. The existence and concentration of positive solutions for
(1.5) with critical growth has been considered in [24].

Moreover, a great attention has been focused on the study of the fractional
$p$-Laplacian problem. In particular, Pucci et al. [35] obtained the
multiplicity result for nonhomogeneous fractional Kirchhoff-Schrödinger
equation. Assuming that $V$ satisfies a Bartsch-Wang type condition, Fiscella
and Pucci [23] considered stationary fractional Kirchhoff $p$-Laplacian
equations involving a Hardy potential and different critical nonlinearities.
The existence result for a class of quasilinear Kirchhoff system involving the
$p$-Laplacian was proved by Xiang et al. [32]. Recently, Ambrosio and Isernia
[5] proved the multiplicity result and the number of positive solutions is
related with the topology of the set, where the potential $V$ attains its
minimum. Then, by using some appropriate variational arguments, they obtained
the existence result when $f(u)=u^{q-1}+\gamma u^{r-1}$ in (1.1) with $N=3$,
where $\gamma>0$ is sufficiently small, and the powers $q$ and $r$ satisfy
$2p<q<p_{s}^{*}\leq r$.

Motivated by the above works and the fact that several interesting questions
arise from the search of nontrivial non negative solutions, our purpose is to
study the existence, multiplicity and concentration of positive solutions to
(1.1) under suitable assumptions on the potential $V$ and the nonlinearity
$f$.

In order to state our main result, we will assume that
$V:\mathbb{R}^{N}\rightarrow\mathbb{R}$ satisfies the following assumptions
introduced by del Pino and Felmer [19]:

$(V_{1})$ $V\in{\mathcal{C}}({\mathbb{R}^{N}},\mathbb{R})$ and
$V_{1}:=\mathop{\inf}\limits_{x\in{\mathbb{R}^{N}}}V(x)>0$;

$(V_{2})$ there is a bounded open set $\Lambda\subset\mathbb{R}^{N}$ such that

$\mathop{\min}\limits_{\partial\Lambda}V(x)>\mathop{\inf}\limits_{\Lambda}V=:V_{0}>0,$

and the set $M:=\left\\{{x\in\Lambda:V(x)={V_{0}}}\right\\}\neq\emptyset$.

Moreover, we give some assumptions on $f$ as follows. Since we are looking for
positive solutions, we may assume that $f(t)=0$ for $t<0$. Let
$f\in{\mathcal{C}^{1}}({\mathbb{R}^{+}},{\mathbb{R}})$ be such that

$(F_{1})$ $f(t)=o(t^{\theta p-1})$, as $t\rightarrow 0^{+}$;

$(F_{2})$ there exist $q,\sigma\in(\theta p,p_{s}^{*})$ and $\lambda>0$ such
that

$f(t)\geq\lambda t^{q-1},\ \forall t>0,\ \mbox{and}\
\mathop{\lim}\limits_{t\rightarrow\infty}\frac{f(t)}{t^{\sigma-1}}=0,$

where $\lambda$ satisfies

  * •

$\lambda>0$ if either $N\geq sp^{2}$, or $sp<N<sp^{2}$ and
$p_{s}^{*}-\frac{p}{p-1}<q<p_{s}^{*}$,

  * •

$\lambda$ is sufficiently large if $sp<N<sp^{2}$ and $\theta p<q\leq
p_{s}^{*}-\frac{p}{p-1}$;

$(F_{3})$ there exists $\mu\in(\theta p,p_{s}^{*})$ such that $0<\mu F(t)\leq
f(t)t$ for all $t\geq 0$;

$(F_{4})$ the map $t\mapsto\frac{f(t)}{t^{\theta p-1}}$ is increasing in
$(0,\infty)$.

To describe our main result, we recall the following definition of Ljusternik-
Schnirelmann category.

######  Definition 1.1.

If $Y$ is a given closed set of a topological space $X$, we denote by
$cat_{Y}X$ the Ljusternik-Schnirelmann category of $Y$ in $X$, that is the
least number of closed and contractible sets in $X$ which cover $Y$.

For more properties about the Ljusternik-Schnirelmann category, we refer to
[9, 40].

The main result can be stated as follows.

######  Theorem 1.2.

Suppose that $(V_{1})$-$(V_{2})$ and $(F_{1})$-$(F_{4})$ hold. Then, for any
$\delta>0$ such that

${M_{\delta}}=\left\\{{x\in\mathbb{R}^{N}:dist\left({x,M}\right)\leq\delta}\right\\}\subset\Lambda,$

there exists ${\varepsilon_{\delta}}>0$ such that, for any
$\varepsilon\in\left({0,{\varepsilon_{\delta}}}\right)$, problem (1.1) has at
least $ca{t_{{M_{\delta}}}}(M)$ positive solutions. Moreover, if
$u_{\varepsilon}$ denotes one of these solutions and
$x_{\varepsilon}\in\mathbb{R}^{N}$ is a global maximum point of
${u_{\varepsilon}}$, then we have

$\mathop{\lim}\limits_{\varepsilon\to 0}V({x_{\varepsilon}})={V_{0}}$

and there exists $C>0$ such that

$0<{u_{\varepsilon}(x)}\leq\frac{{C{\varepsilon^{N+sp}}}}{{{\varepsilon^{N+sp}}+{{\left|{x-{x_{\varepsilon}}}\right|}^{N+sp}}}},\
\ \forall x\in\mathbb{R}^{N}.$

######  Remark 1.3.

We obtain the result of Theorem 1.2 for $1<\theta<\frac{N}{N-sp}$ and all
$a,b>0$. This expands results of [14] with $\alpha=0$:
$1<\theta<\frac{N}{N-sp}$ and $a=0$, $b>0$. Moreover, we discuss both the case
$sp<N<sp^{2}$ and $\theta p<q\leq p_{s}^{*}-\frac{p}{p-1}$, which is more
complicated than that of [14, 15].

######  Remark 1.4.

We give an example of function $f$ satisfying conditions $(F_{1})$-$(F_{4})$,
$f(t)=\sum\limits_{i=1}^{k}\alpha_{i}t^{q_{i}-1}$ with $\alpha_{i}\geq 0$ not
all null and $\theta p<q_{i}<p_{s}^{*}$ for all $i\in\\{1,\ldots,k\\}$.

Under the local assumption $(V_{2})$, Theorem 1.2 will be proved by the
penalization methods developed by del Pino and Felmer [19]. Comparing our
result with the result of the classical Kirchhoff problem (1.5), there are
some new difficulties:

(i) The key point is to overcome the compactness for the associated Lagrange-
Euler functional, namely, the Palais-Smale ((PS) for short) condition. Since
the nonlinearity term in equation (1.1) contains the critical Sobolev term,
the functional does not satisfy the Palais-Smale condition in all range, we
will use a fractional version of the concentration compactness principle to
show that the energy functional satisfies the local $(PS)_{c}$ condition for
$c$ less than a suitable compactness threshold when the perturbation term $f$
satisfies condition $(F_{2})$.

On the other hand, the compactness threshold is usually related to the best
fractional critical Sobolev constant $S$, namely

$\displaystyle S:=\inf\limits_{u\in
D^{s,p}(\mathbb{R}^{N})\setminus\\{0\\}}\frac{\iint_{\mathbb{R}^{2N}}\frac{{{{\left|{u(x)-u(y)}\right|}^{p}}}}{{{{\left|{x-y}\right|}^{N+sp}}}}dxdy}{\left(\int_{\mathbb{R}^{N}}|u|^{p_{s}^{\ast}}\right)^{p/p_{s}^{\ast}}}.$
(1.6)

For the critical fractional case with $p\neq 2$, the main difficulty is the
lack of an explicit formula for minimizers of $S$ which is very often a key
tool to handle the estimates leading to the compactness range of the
functional. It was conjectured that, up to a multiplicative constant, all
minimizers are of the form $U((x-x_{0})/\varepsilon),$ with

$U(x)=(1+|x|^{\frac{p}{p-1}})^{-\frac{N-ps}{p}},\quad x\in\mathbb{R}^{N}.$

This conjecture was proved in [13] for $p=2$, but for $p\neq 2$, it is not
even known if these functions are minimizers of $S$. As in [14, 15], we can
overcome this difficulty by the optimal asymptotic behavior of minimizers,
which was obtained in [11].

(ii) The Lagrange-Euler functional of problem (1.1) is related to the presence
of the Kirchhoff term $[u]_{s,p}^{p}(-\Delta)_{p}^{s}$ and the lack of
compactness caused by the unboundedness of the domain.

(iii) The non-Hilbert structure of the fractional Sobolev space
$W^{s,p}(\mathbb{R}^{N})$ when $p\neq 2$ makes our study rather tough, we have
to develop some suitable arguments which take care of the nonlocal character
of the leading operator $(-\Delta)_{p}^{s}$, see Lemmas 3.7 and 3.8. Moreover,
we borrow ideas of Moser iteration argument [33] with Hölder continuity result
established for $(-\Delta)_{p}^{s}$, which implies that the solution of the
modified problem is also a solution of the original one.

The paper is organized as follows. In Section 2, we first present some
preliminaries, and we introduce the modified problem in Section 3. In Section
3, we will prove the limiting problem has a positive ground state solution. In
Section 4, we show the modified problem has multiple solutions. Section 5 is
devoted to prove Theorem 1.2.

##  2 Preliminaries

In this section, we outline the variational framework for studying problem
(1.1) and list some preliminary lemmas which will be used later. In the
sequel, we denote by $\|\cdot\|_{p}$ the usual norm of the space
$L^{p}(\mathbb{R}^{N})$ and denote by $|\cdot|_{\infty}$ the usual norm of the
space $L^{\infty}(\mathbb{R}^{N})$, the letters $C_{i}(i=1,2,\ldots)$ and $C$
denote some positive constants. We denote by $B_{r}(x)$ the ball centered at
$x\in\mathbb{R}^{N}$ with radius $r>0$. For $s\in(0,1)$, we define
$D^{s,p}(\mathbb{R}^{N})$ as the closure of
$\mathcal{C}_{c}^{\infty}(\mathbb{R}^{N})$ with respect to

$\left[u\right]_{s,p}^{p}:=\iint_{\mathbb{R}^{2N}}\frac{{{{\left|{u(x)-u(y)}\right|}^{p}}}}{{{{\left|{x-y}\right|}^{N+sp}}}}dxdy.$

Consider the fractional Sobolev space

${W^{s,p}}(\mathbb{R}^{N})=\left\\{{u\in{L^{p}}(\mathbb{R}^{N}):{{\left[u\right]}_{s,p}}<+\infty}\right\\}.$

Then ${W^{s,p}}(\mathbb{R}^{N})$ is a Banach Space with respect to the norm

$\|u\|_{s,p}^{p}:={\left[u\right]}_{s,p}^{p}+\|u\|_{p}^{p}.$

######  Lemma 2.1.

[20] Let $s\in(0,1)$ and $p\in(1,\infty)$ be such that $sp<N$. Then there
exists a constant $C_{*}:=C_{*}(s,p)>0$ such that, for any
$u\in{D^{s,p}}(\mathbb{R}^{N})$, we have

$\left|u\right|_{p_{s}^{*}}^{p}\leq C_{*}{\left[u\right]}_{s,p}^{p}.$

Moreover, ${W^{s,p}}(\mathbb{R}^{N})$ is continuously embedded in
$L^{q}(\mathbb{R}^{N})$ for any $q\in\left[p,p_{s}^{*}\right]$, and compactly
in $L_{loc}^{q}(\mathbb{R}^{N})$ for any $q\in\left[1,p_{s}^{*}\right)$.

######  Lemma 2.2.

[6] Let $s\in(0,1)$ and $p\in(1,\infty)$ be such that $sp<N$, and
$r\in[p,p_{s}^{*})$. If $\\{u_{n}\\}_{n\in\mathbb{N}}$ is a bounded sequence
in $W^{s,p}(\mathbb{R}^{N})$ and

$\mathop{\lim}\limits_{n\to\infty}\mathop{\sup}\limits_{y\in\mathbb{R}^{N}}\int_{{B_{R}}(y)}{|{u_{n}}{|^{r}}dx}=0,$

where $R>0$, then $u_{n}\rightarrow 0$ in $L^{\sigma}(\mathbb{R}^{N})$ for all
$\sigma\in\left(p,p_{s}^{*}\right)$.

######  Lemma 2.3.

[6] Let $u\in W^{s,p}(\mathbb{R}^{N})$ and
$\phi\in\mathcal{C}_{c}^{\infty}(\mathbb{R}^{N})$ be such that $0\leq\phi\leq
1$, $\phi=1$ in $B_{1}$ and $\phi=0$ in $B_{2}^{c}$. Set
$\phi_{\varsigma}(x)=\phi(x/\varsigma)$. Then,

${\left[{u{\phi_{\varsigma}}-u}\right]_{s,p}}\to 0\ \ \text{and}\ \
|u\phi_{\varsigma}-u|_{p}\to 0,$

as $\varsigma\to 0$.

Next, we introduce the well-known Simon inequality.

######  Lemma 2.4.

[37] There exist constants $c_{p},C_{p}>0$ such that for any
$a,b\in{\mathbb{R}}$, it holds

$\displaystyle|a-b|^{p}\leq\left\\{\begin{array}[]{ll}{c_{p}}\left({|a{|^{p-2}}a-|b{|^{p-2}}b}\right)\left(a-b\right)&\mbox{if}\
p\geq 2\\\
{C_{p}}{\left[{\left({|a{|^{p-2}}a-|b{|^{p-2}}b}\right)(a-b)}\right]^{\frac{p}{2}}}{\left({|a{|^{p}}+|b{|^{p}}}\right)^{\frac{{(2-p)}}{2}}}&\mbox{if}\
1<p<2.\end{array}\right.$

In what follows, we provide some useful estimates, which will be needed to
overcome the difficulty coming from the critical nonlinear term. Let $S$ be
the best fractional Sobolev embedding constant defined by (1.6). As shown in
[11], there exists a radially symmetric nonnegative decreasing minimizer
$U=U(r)$ for $S$ such that $(-\Delta)_{p}^{s}U={U^{p_{s}^{*}-1}}$ in
$\mathbb{R}^{N}$, and
$[U]_{s,p}^{p}=|U|_{p_{s}^{*}}^{p_{s}^{*}}=S^{\frac{N}{{sp}}}$. Moreover,
$U\in L^{\infty}(\mathbb{R}^{N})\cap\mathcal{C}^{0}(\mathbb{R}^{N})$,

$\mathop{\lim}\limits_{|x|\to\infty}|x{|^{\frac{{N-sp}}{{p-1}}}}U(x)={U_{\infty}}\in\mathbb{R}\setminus\\{0\\}$

and verifies the following decay estimates.

######  Lemma 2.5.

[11] There exist constants $c_{1},c_{2}>0$ and $\kappa>1$ such that for all
$r\geq 1$,

$\frac{{{c_{1}}}}{{{r^{\frac{{N-sp}}{{p-1}}}}}}\leq
U(r)\leq\frac{{{c_{2}}}}{{{r^{\frac{{N-sp}}{{p-1}}}}}}$ (2.2)

and

$\frac{{U(\kappa r)}}{{U(r)}}\leq\frac{1}{2}.$

For any $\epsilon>0$, we consider the following family of minimizers for $S$
given by

${U_{\epsilon}}(x):=\frac{1}{{{\epsilon^{\frac{{N-sp}}{p}}}}}U\left(\frac{{|x|}}{\epsilon}\right).$

For $\epsilon,\delta>0$, set

${m_{\epsilon,\delta}}:=\frac{{{U_{\epsilon}}(\delta)}}{{{U_{\epsilon}}(\delta)-{U_{\epsilon}}(\kappa\delta)}},$

and define

$\displaystyle{g_{\epsilon,\delta}}(t):=\left\\{\begin{array}[]{ll}0&\text{if}\
{0\leq t\leq{U_{\epsilon}}(\kappa\delta)},\\\
m_{\epsilon,\delta}^{p}(t-{U_{\epsilon}}(\kappa\delta)&\text{if}\
{{U_{\epsilon}}(\kappa\delta)\leq t\leq{U_{\epsilon}}(\delta)},\\\
t+{U_{\epsilon}}(\delta)(m_{\epsilon,\delta}^{p-1}-1)&\text{if}\
{t\geq{U_{\epsilon}}(\delta)},\end{array}\right.$

and

$\displaystyle{G_{\epsilon,\delta}}(t):=\int_{0}^{t}{{{({{g^{\prime}}_{\epsilon,\delta}}(\tau))}^{\frac{1}{p}}}}d\tau=\left\\{\begin{array}[]{ll}0&\text{if}\
{0\leq t\leq{U_{\epsilon}}(\kappa\delta)},\\\
{m_{\epsilon,\delta}}(t-{U_{\epsilon}}(\kappa\delta))&\text{if}\
{{U_{\epsilon}}(\kappa\delta)\leq t\leq{U_{\epsilon}}(\delta)},\\\
t&\text{if}\ {t\geq{U_{\epsilon}}(\delta)}.\end{array}\right.$

Let us observe that ${g_{\epsilon,\delta}}$ and ${G_{\epsilon,\delta}}$ are
nondecreasing and absolutely continuous functions. Now, we consider the
radially symmetric nonincreasing function

${u_{\epsilon,\delta}}(r)={G_{\epsilon,\delta}}({U_{\epsilon}}(r)),$ (2.5)

which, in view of the definition of ${G_{\epsilon,\delta}}$, satisfies

$\displaystyle{u_{\epsilon,\delta}}(r)=\left\\{\begin{array}[]{ll}{U_{\epsilon}}(r)&\text{if}\
{r\leq\delta}\\\ 0&\text{if}\ {r\geq\kappa\delta}.\end{array}\right.$

######  Lemma 2.6.

[31] There exists a constant $C=C(N,p,s)>0$ such that for any
$\epsilon<\frac{\delta}{2}$ the following estimates hold

$[{u_{\varepsilon,\delta}}]_{s,p}^{p}\leq
S^{\frac{N}{{sp}}}+C\left({\left(\frac{\epsilon}{\delta}\right)^{\frac{{N-sp}}{{p-1}}}}\right),$
$\|{u_{\varepsilon,\delta}}\|_{p_{s}^{*}}^{p_{s}^{*}}\geq
S^{\frac{N}{{sp}}}-C\left({\left(\frac{\epsilon}{\delta}\right)^{\frac{N}{{p-1}}}}\right).$

######  Lemma 2.7.

[8, 15] There exists a constant $C=C(N,p,s)>0$ such that for any
$\epsilon<\frac{\delta}{2}$

$\displaystyle\|{u_{\epsilon,\delta}}\|_{p}^{p}\leq\left\\{\begin{array}[]{ll}C\epsilon^{sp}&\text{if}\
{N>sp^{2}},\\\
C\epsilon^{sp}\log\left(\frac{\delta}{\epsilon}\right)&\text{if}\
{N=sp^{2}},\\\
C{\epsilon^{\frac{{N-sp}}{{p-1}}}}{\delta^{\frac{{s{p^{2}}-N}}{{p-1}}}}-C{\epsilon^{sp}}&\text{if}\
{N<sp^{2}},\end{array}\right.$

and

$\displaystyle\|{u_{\epsilon,\delta}}\|_{q}^{q}\leq\left\\{\begin{array}[]{ll}C{\epsilon^{N-\frac{{(N-sp)}}{p}q}}&\text{if}\
{q>\frac{{N(p-1)}}{{N-sp}}},\\\
C{\epsilon^{N-\frac{{(N-sp)}}{p}q}}|\log\epsilon|&\text{if}\
{q=\frac{{N(p-1)}}{{N-sp}}},\\\
C{\epsilon^{\frac{{(N-sp)q}}{{p(p-1)}}}}&\text{if}\
{q<\frac{{N(p-1)}}{{N-sp}}}.\end{array}\right.$

##  3 The modified problem

Taking the change of variable $x\mapsto\varepsilon x$, equation (1.1) becomes
the following equation

$\displaystyle(a+b\left[u\right]_{s,p}^{(\theta-1)p})(-\Delta)_{p}^{s}u+V(\varepsilon
x){u^{p-1}}={u^{p_{s}^{*}-1}}+f(u),\ u>0,\ \mbox{in}\ {\mathbb{R}^{N}}.$ (3.1)

Now, we introduce a penalized function in the spirit of [19], which will be
fundamental to obtain our main result. Let $K>2$ and $a>0$ be such that
$a^{p_{s}^{*}-1}+f(a)=\frac{V_{0}}{K}a^{p-1}$, and we consider the function

$\displaystyle\tilde{f}(t):=\left\\{\begin{array}[]{ll}{({t^{+}})^{p_{s}^{*}-1}}+f(t)&\mbox{if}\
t\leq a,\\\ \frac{{{V_{0}}}}{K}{t^{p-1}}&\mbox{if}\ t>a.\end{array}\right.$

If $\chi_{\Lambda}$ denotes the characteristic function of the set $\Lambda$,
we introduce the penalized nonlinearity
$g:\mathbb{R}^{N}\times\mathbb{R}\rightarrow\mathbb{R}$ by setting

$g(x,t)=\chi_{\Lambda}(x)\left(t^{p_{s}^{*}-1}+f(t)\right)+\left(1-\chi_{\Lambda}(x)\right)\tilde{f}(t).$
(3.3)

From conditions $(F_{1})$-$(F_{4})$, it is easy to check that $g$ satisfies
the following properties.

$(G_{1})$ $\mathop{\lim}\limits_{t\to{0^{+}}}\frac{{g(x,t)}}{{{t^{\theta
p-1}}}}=0$ uniformly in $x\in\mathbb{R}^{N}$;

$(G_{2})$ $g(x,t)\leq t^{p_{s}^{*}-1}+f(t)$ for all $x\in\mathbb{R}^{N},t>0$;

$(G_{3})$ $(i)$ $0<\mu G(x,t):=\mu\int_{0}^{t}{g(x,\tau)}d\tau\leq g(x,t)t$
for all $x\in\Lambda$ and $t>0$,

$(ii)$ $0\leq pG(x,t)<g(x,t)t\leq\frac{V_{1}}{K}a^{p}$ for all
$x\in\mathbb{R}^{N}\setminus\Lambda$ and $t>0$;

$(G_{4})$ the map $t\mapsto\frac{g(x,t)}{t^{\theta p-1}}$ is increasing for
all $x\in\Lambda$ and $t>0$, and the map $t\mapsto\frac{g(x,t)}{t^{\theta
p-1}}$ is increasing for all $x\in\mathbb{R}^{N}\setminus\Lambda$ and
$t\in(0,a)$.
Then, we consider the following modified problem

$\displaystyle(a+b\left[u\right]_{s,p}^{(\theta-1)p})(-\Delta)_{p}^{s}u+V(\varepsilon
x){u^{p-1}}=g(\varepsilon x,u),\ \ u>0,\ \ \mbox{in}\ {\mathbb{R}^{N}}.$ (3.4)

Let us note that if $u$ is a solution of (1.1) such that

$|u(x)|\leq a\ \ \text{for}\ \text{any}\
x\in\mathbb{R}^{N}\setminus\Lambda_{\varepsilon},$

where $\Lambda_{\varepsilon}:=\left\\{x\in\mathbb{R}^{N}:\varepsilon
x\in\Lambda\right\\}$, then $u$ is also a solution of (3.1). Therefore, in
order to study weak solutions of (3.4), we seek the critical points of the
following functional

${J_{\varepsilon}}(u)=\frac{1}{p}\|u\|_{\varepsilon}^{p}+\frac{b}{{\theta
p}}[u]_{{}_{s,p}}^{\theta p}-\int_{\mathbb{R}^{N}}{G(\varepsilon x,u)}dx,$

which is well-defined for all $u:\mathbb{R}^{N}\rightarrow\mathbb{R}$
belonging to the following fractional space

${\mathcal{W}}_{\varepsilon}:=\left\\{u\in{W^{s,p}}(\mathbb{R}^{N}):\int_{\mathbb{R}^{N}}{V(\varepsilon
x)}|u{|^{p}}dx<+\infty\right\\}$

endowed with the norm

$\|u\|_{\varepsilon}^{p}:=a[u]_{s,p}^{p}+\int_{\mathbb{R}^{N}}{V(\varepsilon
x)}|u{|^{p}}dx.$

Standard arguments show that
${J_{\varepsilon}}\in\mathcal{C}^{1}({\mathcal{W}}_{\varepsilon},\mathbb{R})$
and its differential is given by

$\displaystyle\left\langle{{J^{\prime}_{\varepsilon}}(u),v}\right\rangle=$
$\displaystyle(a+b[u]_{s,p}^{(\theta-1)p})\iint\limits_{\mathbb{R}^{2N}}\frac{{|u(x)-u(y){|^{p-2}}(u(x)-u(y))(v(x)-v(y))}}{{|x-y{|^{N+sp}}}}dxdy$
$\displaystyle+\int_{\mathbb{R}^{N}}{V(\varepsilon
x)}|u{|^{p-2}}uvdx-\int_{\mathbb{R}^{N}}g(\varepsilon x,u)vdx$

for any $u,v\in{\mathcal{W}}_{\varepsilon}$. Let us introduce the Nehari
manifold associated with (3.4), that is,

${\mathcal{N}}_{\varepsilon}:=\left\\{u\in{{\mathcal{W}}_{\varepsilon}\setminus\\{0\\}}:\left\langle{{{J^{\prime}}_{\varepsilon}}(u),u}\right\rangle=0\right\\}.$

We first shown that ${J_{\varepsilon}}$ possesses geometric structure.

######  Lemma 3.1.

The functional ${J_{\varepsilon}}$ satisfies the following conditions:

$(i)$ there exist $\alpha,\rho>0$ such that ${J_{\varepsilon}}(u)\geq\alpha$
for any $u\in{\mathcal{W}}_{\varepsilon}$ such that
$\|u\|_{\varepsilon}=\rho$;

$(ii)$ there exists $e\in{\mathcal{W}}_{\varepsilon}$ with
$\|e\|_{\varepsilon}>\rho$ such that ${J_{\varepsilon}}(e)<0$.

###### Proof.

$(i)$ For $\varepsilon>0$ small, by $(G_{1})$ and $(G_{2})$, we have

$|g(\varepsilon x,t)|\leq\varepsilon|t|^{\theta
p-1}+C_{\varepsilon}|t|^{p_{s}^{*}-1},\ \ \text{for}\ \text{all}\ \
x\in\mathbb{R}^{N},\ \ t\in\mathbb{R}.$ (3.5)

Therefore, for any $u\in\mathcal{W}_{\varepsilon}\setminus\\{0\\}$, we get

${J_{\varepsilon}(u)}\geq\frac{{\|u\|_{\varepsilon}^{p}}}{p}-\frac{{{C_{1}}\varepsilon}}{{\theta
p}}\|u\|_{\varepsilon}^{\theta
p}-\frac{{{C_{2}}{C_{\varepsilon}}}}{{p_{s}^{*}}}\|u\|_{\varepsilon}^{p_{s}^{*}}.$

Hence, since $\theta>1$, there exist $\alpha,\rho>0$ such that
${J_{\varepsilon}}(u)\geq\alpha$ with $\|u\|_{\varepsilon}=\rho$.

$(ii)$ For all $t>0$, we infer that

$\displaystyle{J_{\varepsilon}}(u)\leq$
$\displaystyle\frac{t^{p}}{p}\|u\|_{\varepsilon}^{p}+\frac{bt^{\theta
p}}{{\theta p}}[u]_{{}_{s,p}}^{\theta
p}-\int_{\Lambda_{\varepsilon}}{G(\varepsilon x,tu)}dx$ $\displaystyle\leq$
$\displaystyle\frac{t^{p}}{p}\|u\|_{\varepsilon}^{p}+\frac{bt^{\theta
p}}{{\theta p}}[u]_{{}_{s,p}}^{\theta
p}-\frac{t^{p_{s}^{*}}}{p_{s}^{*}}\int_{\Lambda_{\varepsilon}}{|u^{+}|^{p_{s}^{*}}}dx.$
(3.6)

Thanks to $\theta p<p_{s}^{*}$, the conclusion $(ii)$ holds. ∎

######  Lemma 3.2.

Define the function $H(t)=\frac{a}{p}{S}{t^{p}}+\frac{b}{{\theta
p}}S^{\theta}{t^{\theta p}}-\frac{1}{{p_{s}^{*}}}{t^{p_{s}^{*}}}$. Then there
exists a unique ${\hat{T}}>0$ such that

$H(\hat{T})=\sup\limits_{t\geq 0}H(t).$

###### Proof.

For $t>0$, we have

$H^{\prime}(t)={t^{p-1}}\hat{H}(t),\ \
\hat{H}(t)=a{S}+bS^{\theta}{t^{(\theta-1)p}}-{t^{p_{s}^{*}-p}}.$ (3.7)

Moreover,

$\hat{H}^{\prime}(t)={t^{(\theta-1)p-1}}\left((\theta-1)pbS^{\theta}-(p_{s}^{*}-p){t^{p_{s}^{*}-\theta
p}}\right).$

Thus there exists a unique $\hat{T}>0$ such that $\hat{H}>0$ if
$t\in(0,\hat{T})$ and $\hat{H}<0$ if $t\in(\hat{T},+\infty)$. Thus, $\hat{T}$
is the unique maximum point of $H(t)$. ∎

In view of Lemma 3.1, we can define the minimax level

${c_{\varepsilon}}=\mathop{\inf}\limits_{\gamma\in{\Gamma_{\varepsilon}}}\mathop{\max}\limits_{t\in[0,1]}{J_{\varepsilon}}(\gamma(t)),$

where ${\Gamma_{\varepsilon}}=\left\\{{\gamma\in
C([0,1]),{\mathcal{W}_{\varepsilon}}):\gamma(0)=0\ \ \text{and}\ \
{J_{\varepsilon}}(\gamma(1))<0}\right\\}$. From [40, Theorem 4.2], it is
standard to verify that $c_{\varepsilon}$ can be characterized as follows

${c_{\varepsilon}}=\mathop{\inf}\limits_{u\in{\mathcal{W}_{\varepsilon}}\setminus\\{0\\}}\mathop{\sup}\limits_{t\geq
0}{J_{\varepsilon}}(tu)=\mathop{\inf}\limits_{u\in\mathcal{N}_{\varepsilon}}{J_{\varepsilon}}(u).$
(3.8)

######  Lemma 3.3.

There holds

${c_{\varepsilon}}<{c_{0}}=:H(\hat{T})$

for all $\varepsilon>0$, where $H(\hat{T})$ was given in Lemma 3.2.

###### Proof.

Let $u_{\epsilon,\delta}$ be the function defined in (2.5) such that
$supp(u_{\epsilon,\delta})\subset
B_{\theta\delta}\subset\Lambda_{\varepsilon}$. For simplicity, we take
$\delta=1$ and we set $u_{\epsilon}:=u_{\epsilon,1}$. Then, by $(F_{2})$, we
have

$\displaystyle{J_{\varepsilon}}(tu_{\epsilon})\leq\frac{at^{p}}{{p}}[u_{\epsilon}]_{{}_{s,p}}^{p}+\frac{V_{2}t^{p}}{{p}}\|u_{\epsilon}\|_{p}^{p}+\frac{bt^{\theta
p}}{{\theta p}}[u_{\epsilon}]_{{}_{s,p}}^{\theta p}-\frac{\lambda
t^{q}}{q}\|u_{\epsilon}\|_{q}^{q}-\frac{t^{p_{s^{*}}}}{{p_{s^{*}}}}\|u_{\epsilon}\|_{p_{s^{*}}}^{{p_{s^{*}}}}\rightarrow-\infty\
\ \mbox{as}\ \ t\rightarrow\infty,$

where $V_{2}=|V|_{\infty}$. Therefore, there exists $t_{\epsilon}>0$ such that

${J_{\varepsilon}}(t_{\epsilon}{u_{\epsilon}})=\mathop{\max}\limits_{t\geq
0}{J_{\varepsilon}}(t{u_{\epsilon}}).$

Next, we can show that there exist $t_{1},t_{2}>0$ (independent of
$\epsilon>0$) satisfying $t_{1}\leq t_{\epsilon}\leq t_{2}$. Indeed, since
$\left\langle{{J^{\prime}_{\varepsilon}}({t_{\epsilon}}{u_{\epsilon}}),{u_{\epsilon}}}\right\rangle=0$,
we have

$t_{\epsilon}^{p}{\|u_{\epsilon}\|_{\epsilon}^{p}}+bt_{\epsilon}^{\theta
p}[u_{\epsilon}]_{{}_{s,p}}^{\theta
p}=t_{\epsilon}^{p_{s}^{*}}{\|u_{\epsilon}\|_{p_{s}^{*}}^{p_{s}^{*}}}+\int_{\mathbb{R}^{N}}f(t_{\epsilon}{u_{\epsilon}}){t_{\epsilon}u_{\epsilon}}dx.$

Moreover

$t_{\epsilon}^{p}{\|u_{\epsilon}\|_{\epsilon}^{p}}+bt_{\epsilon}^{\theta
p}[u_{\epsilon}]_{{}_{s,p}}^{\theta
p}\geq{t_{\epsilon}^{p_{s}^{*}}}{\|u\|_{p_{s}^{*}}^{p_{s}^{*}}},$

which implies that $|t_{\epsilon}|\leq t_{2}$. On the other hand, there exists
$t_{1}>0$ such that $t_{\varepsilon}>t_{1}$ for $\epsilon>0$ small. Otherwise,
there exists $\epsilon_{n}\to 0$ as $n\to\infty$ such that
$t_{\epsilon_{n}}\to 0$ as $n\to\infty$. Thus,

$0<c_{\varepsilon}\leq\mathop{\max}\limits_{t\geq
0}{J_{\varepsilon}}(t{u_{\epsilon}})\to 0,$

which gives a contradiction, we can see that

$\displaystyle{J_{\varepsilon}}(t_{\epsilon}u_{\epsilon})\leq$
$\displaystyle\frac{at_{\epsilon}^{p}}{p}[u_{\epsilon}]_{s,p}^{p}+\frac{bt_{\epsilon}^{\theta
p}}{{\theta p}}[u_{\epsilon}]_{s,p}^{\theta
p}-\frac{t_{\epsilon}^{p_{s}^{*}}}{p_{s}^{*}}\|u_{\epsilon}\|_{{p_{s}^{*}}}^{p_{s}^{*}}+\frac{V_{2}t_{\epsilon}^{p}}{p}\|u_{\epsilon}\|_{p}^{p}-\frac{\lambda
t_{\epsilon}^{q}}{q}\|u_{\epsilon}\|_{q}^{q}$ $\displaystyle=:$ $\displaystyle
I_{\epsilon}(t_{\epsilon})+B_{\epsilon}-C_{\epsilon},$

where $I_{\epsilon}(t):=\frac{a}{p}{A_{\epsilon}}t^{p}+\frac{b}{{\theta
p}}A_{\epsilon}^{\theta}t^{\theta
p}-\frac{D_{\epsilon}}{p_{s}^{*}}t^{p_{s}^{*}}$. Therefore,

$c_{\epsilon}\leq\mathop{\sup}\limits_{t\geq
0}{I_{\epsilon}}(t)+B_{\epsilon}-C_{\epsilon}.$

Gathering the estimates in Lemma 2.7, we get

$\displaystyle c_{\varepsilon}\leq\mathop{\sup}\limits_{t\geq
0}{I_{\epsilon}}(t)+\left\\{\begin{array}[]{ll}O(\epsilon^{\frac{N-sp}{p-1}})-\frac{\lambda
t_{1}^{q}}{q}\|u_{\epsilon}\|_{q}^{q}&\text{if}\ {sp^{2}>N}\\\
O(\epsilon^{sp}(1+|\log\epsilon|))-\frac{\lambda
t_{1}^{q}}{q}\|u_{\epsilon}\|_{q}^{q}&\text{if}\ {N=sp^{2}}\\\
O(\epsilon^{sp})-\frac{\lambda
t_{1}^{q}}{q}\|u_{\epsilon}\|_{q}^{q}&\text{if}\
{sp^{2}<N}.\end{array}\right.$ (3.12)

Since ${\frac{N-sp}{p-1}}>0$ and $sp>0$, we obtain

$\mathop{\sup}\limits_{t\geq 0}{I_{\epsilon}}(t)\geq\frac{c_{\varepsilon}}{2}\
\ \rm{uniformly}\ \ \rm{for}\ \ \epsilon>0\ \ \rm{small}.$

Arguing as above, there exists $t_{3},t_{4}>0$ (independent of $\epsilon>0$)
such that

$\mathop{\sup}\limits_{t\geq
0}{I_{\epsilon}}(t)=\mathop{\sup}\limits_{t\in[t_{3},t_{4}]}{I_{\epsilon}}(t).$

By Lemmas 2.6 and 3.2, we deduce

$\displaystyle c_{\varepsilon}\leq$ $\displaystyle\sup\limits_{t\geq
0}H(S^{-\frac{sp-N}{sp^{2}}}t)+\left\\{\begin{array}[]{ll}O(\epsilon^{\frac{N-sp}{p-1}})-\frac{\lambda
t_{1}^{q}}{q}\|u_{\epsilon}\|_{q}^{q},&\text{if}\ {sp^{2}>N},\\\
O(\epsilon^{sp}(1+|\log\epsilon|))-\frac{\lambda
t_{1}^{q}}{q}\|u_{\epsilon}\|_{q}^{q},&\text{if}\ {N=sp^{2}},\\\
O(\epsilon^{sp})-\frac{\lambda
t_{1}^{q}}{q}\|u_{\epsilon}\|_{q}^{q},&\text{if}\
{sp^{2}<N},\end{array}\right.$ $\displaystyle\leq$ $\displaystyle
H(\hat{T})+\left\\{\begin{array}[]{ll}O(\epsilon^{\frac{N-sp}{p-1}})-\frac{\lambda
t_{1}^{q}}{q}\|u_{\epsilon}\|_{q}^{q},&\text{if}\ {sp^{2}>N},\\\
O(\epsilon^{sp}(1+|\log\epsilon|))-\frac{\lambda
t_{1}^{q}}{q}\|u_{\epsilon}\|_{q}^{q},&\text{if}\ {N=sp^{2}},\\\
O(\epsilon^{sp})-\frac{\lambda
t_{1}^{q}}{q}\|u_{\epsilon}\|_{q}^{q},&\text{if}\
{sp^{2}<N}.\end{array}\right.$

Case 1. $N>sp^{2}$. Then $q>\theta p>p>\frac{N(p-1)}{N-sp}$ and using Lemma
2.7, we have

$\displaystyle c_{\varepsilon}\leq
H(\hat{T})+O(\epsilon^{sp})-O(\lambda\epsilon^{N-\frac{N-sp}{p}q}).$

Using the fact $q>p$, we have $N-\frac{N-sp}{p}q<sp$. Thus, for $\epsilon>0$
small enough, we can infer that

$\displaystyle c_{\varepsilon}<H(\hat{T}).$

Case 2. $N=sp^{2}$. Thus $q>\theta p>p=\frac{N(p-1)}{N-sp}$, we have

$\displaystyle c_{\varepsilon}\leq
H(\hat{T})+O(\epsilon^{sp}(1+|\log\epsilon|))-O(\lambda\epsilon^{sp^{2}-s(p-1)q}|\log\epsilon|).$

Observing that $q>p$ yields

$\displaystyle\mathop{\lim}\limits_{\epsilon\to
0}\frac{\epsilon^{sp^{2}-s(p-1)q}|\log\epsilon|}{\epsilon^{sp}(1+|\log\epsilon|)}=\infty,$

we get the conclusion for $\epsilon$ small enough.

Case 3. $N<sp^{2}$. Then we can deduce that
$p_{s}^{*}-\frac{p}{p-1}>\frac{N(p-1)}{N-sp}$. Suppose that
$p_{s}^{*}>q>p_{s}^{*}-\frac{p}{p-1}$ and using Lemma 2.7, we have

$\displaystyle c_{\varepsilon}\leq
H(\hat{T})+O(\epsilon^{\frac{N-sp}{p-1}})-O(\lambda\epsilon^{N-\frac{N-sp}{p}q}).$

Using the fact $q>p_{s}^{*}-\frac{p}{p-1}$, we have
$N-\frac{N-sp}{p}q<{\frac{N-sp}{p-1}}$. Thus, for $\epsilon>0$ small enough,
one has

$\displaystyle c_{\varepsilon}<H(\hat{T}).$

In view of $q>\theta p$, we have to compare $\theta p$ and
$p_{s}^{*}-\frac{p}{p-1}$. Thus we define

$h(\theta)=\theta p-p_{s}^{*}+\frac{p}{p-1},\ \ \forall
1<\theta<\frac{N}{N-sp}.$

We can easily conclude that $h(\theta)\geq 0$ if
$\theta\in\left[\theta_{0},\frac{N}{N-sp}\right)$ and $h(\theta)<0$ if
$\theta\in(1,\theta_{0})$, where $\theta_{0}:=\frac{N}{N-sp}-\frac{1}{p-1}$.
Consequently, let $\theta\geq\theta_{0}$. Then $h(\theta)\geq 0$, which
implies $q>\theta p\geq p_{s}^{*}-\frac{p}{p-1}$. Arguing as above, we have
the conclusion for $\epsilon>0$ small enough.

Next, we consider the case $\theta<\theta_{0}$. Then $h(\theta)<0$ implies
$\theta p<p_{s}^{*}-\frac{p}{p-1}$. Now, we deal with the case $\theta p<q\leq
p_{s}^{*}-\frac{p}{p-1}$. For this purpose, we firstly compare $\theta p$ and
$\frac{N(p-1)}{N-sp}$. Then we define

$m(\theta)=\theta p-\frac{N(p-1)}{N-sp},\ \ \forall 1<\theta<\theta_{0}.$

Since $N<sp^{2}$, we have $\theta_{1}<\theta_{0}$, where
$\theta_{1}:=\frac{N(p-1)}{p(N-sp)}$. So we can easily deduce that
$m(\theta)<0$ if $\theta\in\left(1,\theta_{1}\right)$ and $m(\theta)\geq 0$ if
$\theta\in[\theta_{1},\theta_{0})$.

$(i)$ If $\theta\in[\theta_{1},\theta_{0})$, we have $m(\theta)\geq 0$, which
implies $p_{s}^{*}-\frac{p}{p-1}\geq q>\theta p\geq\frac{N(p-1)}{N-sp}$. We
get

$\displaystyle c_{\varepsilon}\leq
H(\hat{T})+O(\epsilon^{\frac{N-sp}{p-1}})-O(\lambda\epsilon^{N-\frac{N-sp}{p}q}),$

and choosing $\lambda=\epsilon^{-\mu}$, with
$\mu>N-\frac{N-sp}{p}q-\frac{N-sp}{p-1}$, we have the conclusion.

$(ii)$ If $\theta\in\left(1,\theta_{1}\right)$, we have $m(\theta)<0$, which
implies $\theta p<\frac{N(p-1)}{N-sp}$. We distinguish the following case

$\displaystyle\theta p<q<\frac{N(p-1)}{N-sp},\ \ q=\frac{N(p-1)}{N-sp}\ \
\mbox{and}\ \ \frac{N(p-1)}{N-sp}<q\leq p_{s}^{*}-\frac{p}{p-1}.$

When $\theta p<q<\frac{N(p-1)}{N-sp}$, then, for $\epsilon>0$ small, it holds

$\displaystyle c_{\varepsilon}\leq
H(\hat{T})+O(\epsilon^{\frac{N-sp}{p-1}})-O(\lambda\epsilon^{\frac{N-sp}{p(p-1)}q}),$

and noting that

$\displaystyle{\frac{N-sp}{p-1}}<{\frac{N-sp}{p(p-1)}q},$

we can take $\lambda=\epsilon^{-\mu}$, with
$\mu>{\frac{N-sp}{p(p-1)}q}-{\frac{N-sp}{p-1}}$, to obtain the thesis.
When $q=\frac{N(p-1)}{N-sp}$, then

$\displaystyle c_{\varepsilon}\leq
H(\hat{T})+O(\epsilon^{\frac{N-sp}{p-1}})-O(\lambda\epsilon^{\frac{N}{p}}|\log\epsilon|),$

and taking $\lambda=\epsilon^{-\mu}$, with
$\mu>\frac{N}{p}-{\frac{N-sp}{p-1}}$, we can deduce the assertion.
Finally, when $\frac{N(p-1)}{N-sp}<q\leq p_{s}^{*}-\frac{p}{p-1}$, arguing as
$(i)$, we get the conclusion. ∎

######  Lemma 3.4.

Every $(PS)_{c}$ sequence of $J_{\varepsilon}$ is bounded in
$\mathcal{W}_{\varepsilon}$.

###### Proof.

Let $\\{u_{n}\\}$ be a $(PS)$ sequence at level $c$, that is

$\displaystyle J_{\varepsilon}(u_{n})\rightarrow c\ \ \text{and}\ \
{{J^{\prime}}_{\varepsilon}}({u_{n}})\rightarrow 0,$

as $n\to\infty$. Using $(G_{3})$, we can deduce that

$\displaystyle c+o_{n}(1)=$ $\displaystyle
J_{\varepsilon}(u_{n})-\frac{1}{\mu}\left\langle{{{J^{\prime}}_{\varepsilon}}({u_{n}}),{u_{n}}}\right\rangle$
$\displaystyle=$
$\displaystyle\left({\frac{{\mu-p}}{{p\mu}}}\right)\|u_{n}\|_{\varepsilon}^{p}+\frac{1}{\mu}\int_{\mathbb{R}^{N}\setminus{\Lambda_{\varepsilon}}}{[g(\varepsilon
x,{u_{n}}){u_{n}}-\mu G}(\varepsilon x,{u_{n}})]dx$
$\displaystyle+\frac{1}{\mu}\int_{{\Lambda_{\varepsilon}}}{[g(\varepsilon
x,{u_{n}}){u_{n}}-\mu G}(\varepsilon x,{u_{n}})]dx+b(\frac{\mu-\theta
p}{\theta p\mu})[{u_{n}}]_{s,p}^{2p}$ $\displaystyle\geq$
$\displaystyle\left({\frac{{\mu-p}}{{p\mu}}}\right)\|u_{n}\|_{\varepsilon}^{p}+\frac{1}{\mu}\int_{\mathbb{R}^{N}\setminus{\Lambda_{\varepsilon}}}{[g(\varepsilon
x,{u_{n}}){u_{n}}-\mu G}(\varepsilon x,{u_{n}})]dx$ $\displaystyle\geq$
$\displaystyle\left({\frac{{\mu-p}}{{p\mu}}}\right)\|u_{n}\|_{\varepsilon}^{p}-\frac{\mu-p}{\mu}\int_{\mathbb{R}^{N}\setminus{\Lambda_{\varepsilon}}}{G}(\varepsilon
x,{u_{n}})dx$ $\displaystyle\geq$
$\displaystyle\left({\frac{{\mu-p}}{{p\mu}}}\right)\|u_{n}\|_{\varepsilon}^{p}-\left(\frac{\mu-p}{p\mu}\right)\int_{\mathbb{R}^{N}\setminus{\Lambda_{\varepsilon}}}\frac{V(\varepsilon
x)}{K}|u_{n}|^{p}dx$ $\displaystyle\geq$
$\displaystyle\left({\frac{{\mu-p}}{{p\mu}}}\right)\left(1-\frac{1}{K}\right)\|u_{n}\|_{\varepsilon}^{p}.$

Since $\mu>\theta p$ and $K>2$, we can conclude that
$\\{u_{n}\\}_{n\in\mathbb{N}}$ is bounded in $\mathcal{W}_{\varepsilon}$. ∎

######  Lemma 3.5.

There exists a sequence $\\{z_{n}\\}_{n\in\mathbb{N}}\subset\mathbb{R}^{N}$
and $R,\beta>0$ such that

$\int_{B_{R}(z_{n})}|u_{n}|^{p}\ dx\geq\beta>0.$

Moreover, $\\{z_{n}\\}_{n\in\mathbb{N}}$ is bounded in $\mathbb{R}^{N}$.

###### Proof.

By Lemma 3.4, we know that $\\{u_{n}\\}$ is bounded in
$\mathcal{W}_{\varepsilon}$. Then we may assume that $u_{n}\rightharpoonup u$
in $\mathcal{W}_{\varepsilon}$. Below we first prove that $\\{u_{n}\\}$ is
non-vanishing. Otherwise, assume that $\\{u_{n}\\}$ is vanishing. Then Lemma
2.2 implies that $u_{n}\to 0$ in $L^{q}(\mathbb{R}^{N})$ for any
$q\in(p,p_{s}^{*})$. From $(F_{1})$ and $(F_{2})$, it follows that

$\int_{\mathbb{R}^{N}}{F({u_{n}})}dx=\int_{\mathbb{R}^{N}}{f({u_{n}}){u_{n}}}dx=o_{n}(1)\
\ \text{as}\ \ n\to\infty.$

This implies that

$\int_{\mathbb{R}^{N}}G(\varepsilon
x,u_{n})dx=\frac{1}{{p_{s}^{*}}}\int_{{\Lambda_{\varepsilon}}\cup\\{{u_{n}}\leq
a\\}}{{{(u_{n}^{+})}^{p_{s}^{*}}}}dx+\frac{{{V_{0}}}}{{pK}}\int_{(\mathbb{R}^{N}\setminus\Lambda_{\varepsilon})\cap\\{{u_{n}}>a\\}}{|{u_{n}}{|^{p}}}dx+{o_{n}}(1)$
(3.15)

and

$\int_{\mathbb{R}^{N}}g(\varepsilon
x,u_{n})u_{n}dx=\int_{{\Lambda_{\varepsilon}}\cup\\{{u_{n}}\leq
a\\}}{{{(u_{n}^{+})}^{p_{s}^{*}}}}dx+\frac{{{V_{0}}}}{{K}}\int_{(\mathbb{R}^{N}\setminus\Lambda_{\varepsilon})\cap\\{{u_{n}}>a\\}}{|{u_{n}}{|^{p}}}dx+{o_{n}}(1).$
(3.16)

Using
$\left\langle{{J^{\prime}_{\varepsilon}}({u_{n}}),{u_{n}}}\right\rangle=o_{n}(1)$
and (3.16) we have

$\|u_{n}\|_{\varepsilon}^{p}+b[u_{n}]_{s,p}^{\theta
p}-\frac{{{V_{0}}}}{{K}}\int_{(\mathbb{R}^{N}\setminus\Lambda_{\varepsilon})\cap\\{{u_{n}}>a\\}}{|{u_{n}}{|^{p}}}dx=\int_{{\Lambda_{\varepsilon}}\cup\\{{u_{n}}\leq
a\\}}{{{(u_{n}^{+})}^{p_{s}^{*}}}}dx+{o_{n}}(1).$ (3.17)

Suppose that

$\int_{{\Lambda_{\varepsilon}}\cup\\{{u_{n}}\leq
a\\}}{{{(u_{n}^{+})}^{p_{s}^{*}}}}dx\rightarrow l^{p_{s}^{*}}.$

Note that $l>0$, otherwise (3.17) yields $\|u_{n}\|_{\varepsilon}\rightarrow
0$ as $n\to\infty$ which implies that $J_{\varepsilon}(u_{n})\rightarrow 0$,
and this is a contradiction because $c_{\varepsilon}>0$. Then, by (3.17) and
Sobolev inequality we obtain

$a{S}\left(\int_{{\Lambda_{\varepsilon}}\cup\\{{u_{n}}\leq
a\\}}{{{(u_{n}^{+})}^{p_{s}^{*}}}}dx\right)^{\frac{p}{p_{s}^{*}}}+bS^{\theta}\left(\int_{{\Lambda_{\varepsilon}}\cup\\{{u_{n}}\leq
a\\}}{{{(u_{n}^{+})}^{p_{s}^{*}}}}dx\right)^{\frac{\theta
p}{p_{s}^{*}}}\leq\int_{{\Lambda_{\varepsilon}}\cup\\{{u_{n}}\leq
a\\}}{{{(u_{n}^{+})}^{p_{s}^{*}}}}dx+o_{n}(1).$ (3.18)

Since $l>0$, by (3.18), we have

$a{S}l^{p}+bS^{\theta}l^{\theta p}-l^{p_{s}^{*}}\leq 0,$

which implies $H^{\prime}(l)\leq 0$ defined in (3.7), so we can deduce that
$l\geq{\hat{T}}$, where ${\hat{T}}$ is the unique maximum of $H$ defined in
Lemma 3.2. And we can know that

$\hat{H}({\hat{T}})=a{S}+bS^{\theta}{{\hat{T}}^{(\theta-1)p}}-{{\hat{T}}^{p_{s}^{*}-p}}=0,$

which implies

$a{S}{\hat{T}}^{p}+bS^{\theta}{{\hat{T}}^{\theta
p}}-{{\hat{T}}^{p_{s}^{*}}}=0.$ (3.19)

Therefore, from (3.15), (3.16), (3.18), $l\geq{\hat{T}}$ and Sobolev
inequality we can infer

$\displaystyle c+o_{n}(1)=$ $\displaystyle
J_{\varepsilon}(u_{n})-\frac{{1}}{{\theta
p}}\left\langle{{{J^{\prime}}_{\varepsilon}}({u_{n}}),{u_{n}}}\right\rangle$
$\displaystyle=$ $\displaystyle{\frac{(\theta-1)a}{\theta
p}}[u_{n}]_{s,p}^{p}+\frac{{(\theta-1)}}{{\theta
p}}\int_{\mathbb{R}^{N}}{V(\varepsilon x)|{u_{n}}|^{p}}dx+\frac{{1}}{{\theta
p}}\int_{\mathbb{R}^{N}}g(\varepsilon
x,u_{n})u_{n}dx-\int_{\mathbb{R}^{N}}G(\varepsilon x,u_{n})dx$
$\displaystyle\geq$ $\displaystyle{\frac{(\theta-1)a}{\theta
p}}[u_{n}]_{s,p}^{p}+\frac{{(\theta-1)}}{{\theta
p}}\int_{\mathbb{R}^{N}}{V(\varepsilon
x)|{u_{n}}|^{p}}dx-\frac{{(\theta-1)V_{0}}}{{\theta
pK}}\int_{\mathbb{R}^{N}}|{u_{n}}|^{p}dx$
$\displaystyle+\left(\frac{{1}}{{\theta
p}}-\frac{1}{{p_{s}^{*}}}\right)\int_{{\Lambda_{\varepsilon}}\cup\\{{u_{n}}\leq
a\\}}{{{(u_{n}^{+})}^{p_{s}^{*}}}}dx$ $\displaystyle\geq$
$\displaystyle\frac{(\theta-1)a}{\theta
p}{S}\left(\int_{{\Lambda_{\varepsilon}}\cup\\{{u_{n}}\leq
a\\}}{{{(u_{n}^{+})}^{p_{s}^{*}}}}dx\right)^{\frac{p}{p_{s}^{*}}}$
$\displaystyle+\left(\frac{{1}}{{\theta
p}}-\frac{1}{{p_{s}^{*}}}\right)\left(a{S}\left(\int_{{\Lambda_{\varepsilon}}\cup\\{{u_{n}}\leq
a\\}}{{{(u_{n}^{+})}^{p_{s}^{*}}}}dx\right)^{\frac{p}{p_{s}^{*}}}+bS^{\theta}\left(\int_{{\Lambda_{\varepsilon}}\cup\\{{u_{n}}\leq
a\\}}{{{(u_{n}^{+})}^{p_{s}^{*}}}}dx\right)^{\frac{\theta
p}{p_{s}^{*}}}\right)$ $\displaystyle=$
$\displaystyle\frac{(\theta-1)a}{\theta p}{S}l^{p}+\left(\frac{{1}}{{\theta
p}}-\frac{1}{{p_{s}^{*}}}\right)\left(a{S}l^{p}+bS^{\theta}l^{\theta
p}\right)$ $\displaystyle\geq$ $\displaystyle\frac{(\theta-1)a}{\theta
p}{S}{\hat{T}}^{p}+\left(\frac{{1}}{{\theta
p}}-\frac{1}{{p_{s}^{*}}}\right)\left(a{S}{\hat{T}}^{p}+bS^{\theta}{\hat{T}}^{\theta
p}\right)$ $\displaystyle=$
$\displaystyle\frac{a}{p}{S}{\hat{T}}^{p}+\frac{b}{\theta
p}S^{\theta}{\hat{T}}^{\theta p}-\frac{1}{p_{s}^{*}}{\hat{T}}^{p_{s}^{*}},$

which contradicts with Lemma 3.3.

Now, we show that $\\{z_{n}\\}$ is bounded in $\mathbb{R}^{N}$. Let us
consider the function $\eta_{R}\in\mathcal{C}^{\infty}(\mathbb{R}^{N})$
defined as $\eta_{R}(x)=0$ if $x\in B_{R}$, $\eta_{R}(x)=1$ if $x\notin
B_{2R}$, with $0\leq\eta_{R}\leq 1$ and $|\nabla\eta_{R}|\leq\frac{C}{R}$,
where $C$ is a constant independent of $R$. Since $\\{\eta_{R}u_{n}\\}$ is
bounded in $\mathcal{W}_{\varepsilon}$,
$\left\langle{{J^{\prime}_{\varepsilon}}({u_{n}}),{u_{n}}\eta_{R}}\right\rangle=o_{n}(1)$,
we get

$\displaystyle(a+b[{u_{n}}]_{s,p}^{(\theta-1)p})\iint_{\mathbb{R}^{2N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}\eta_{R}(x)}dxdy+\int_{\mathbb{R}^{N}}{V(\varepsilon
x)|{u_{n}}|^{p}}\eta_{R}(x)dx$ $\displaystyle=$ $\displaystyle
o_{n}(1)+\int_{\mathbb{R}^{N}}g(\varepsilon x,u_{n})u_{n}\eta_{R}(x)dx$
$\displaystyle-(a+b[{u_{n}}]_{s,p}^{(\theta-1)p})\iint_{\mathbb{R}^{2N}}\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p-2}}({u_{n}}(x)-{u_{n}}(y))({\eta_{R}}(x)-{\eta_{R}}(y))}}{{|x-y{|^{N+sp}}}}u_{n}(y)dxdy.$

Take $R>0$ such that $\Lambda_{\varepsilon}\subset B_{R}$. Then, using
$(G_{3})$-$(ii)$, we get

$\displaystyle\iint_{\mathbb{R}^{2N}}{a\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}\eta_{R}(x)}dxdy+\int_{\mathbb{R}^{N}}{V(\varepsilon
x)|{u_{n}}|^{p}}\eta_{R}(x)dx$ $\displaystyle\leq$ $\displaystyle
o_{n}(1)+\int_{\mathbb{R}^{N}}{\frac{V(\varepsilon
x)}{K}|{u_{n}}|^{p}\eta_{R}(x)}dx$
$\displaystyle-(a+b[{u_{n}}]_{s,p}^{(\theta-1)p})\iint_{\mathbb{R}^{2N}}\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p-2}}({u_{n}}(x)-{u_{n}}(y))({\eta_{R}}(x)-{\eta_{R}}(y))}}{{|x-y{|^{N+sp}}}}u_{n}(y)dxdy,$
(3.20)

which implies that

$\displaystyle(1-\frac{1}{K})V_{0}\int_{\mathbb{R}^{N}}|{{u_{n}}|^{p}}\eta_{R}(x)dx$
$\displaystyle\leq$ $\displaystyle
o_{n}(1)-(a+b[{u_{n}}]_{s,p}^{(\theta-1)p})\iint_{\mathbb{R}^{2N}}\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p-2}}({u_{n}}(x)-{u_{n}}(y))({\eta_{R}}(x)-{\eta_{R}}(y))}}{{|x-y{|^{N+sp}}}}u_{n}(y)dxdy.$

Exploiting the Hölder inequality and the boundedness on $\\{u_{n}\\}$ in
$\mathcal{W}_{\varepsilon}$ we have that

$\displaystyle\left|\iint_{\mathbb{R}^{2N}}\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p-2}}({u_{n}}(x)-{u_{n}}(y))({\eta_{R}}(x)-{\eta_{R}}(y))}}{{|x-y{|^{N+sp}}}}u_{n}(y)dxdy\right|$
$\displaystyle\leq$ $\displaystyle
C\left(\iint_{\mathbb{R}^{2N}}\frac{{|{\eta_{R}}(x)-{\eta_{R}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}|u_{n}(x)|^{p}dxdy\right)^{\frac{1}{p}}.$
(3.21)

Now, we show that

$\mathop{\lim}\limits_{R\to\infty}\mathop{\lim\sup}\limits_{n\to\infty}\iint_{\mathbb{R}^{2N}}{\frac{{|{\eta_{R}}(x)-{\eta_{R}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}|{u_{n}}(x){|^{p}}dxdy}=0.$

In fact, recalling that $0\leq\eta_{R}\leq 1$ and
$|\nabla\eta_{R}|\leq\frac{C}{R}$, and using polar coordinates, we obtain

$\displaystyle\int_{\mathbb{R}^{N}}{\frac{{|{\eta_{R}}(x)-{\eta_{R}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}|{u_{n}}(x){|^{p}}dxdy}$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{N}}{\int_{|y-x|>R}{\frac{{|{\eta_{R}}(x)-{\eta_{R}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}|{u_{n}}(x){|^{p}}dxdy+}}\int_{\mathbb{R}^{N}}{\int_{|y-x|\leq
R}{\frac{{|{\eta_{R}}(x)-{\eta_{R}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}|{u_{n}}(x){|^{p}}dxdy}}$
$\displaystyle\leq$ $\displaystyle
C\int_{\mathbb{R}^{N}}{|{u_{n}}(x){|^{p}}\left(\int_{|y-x|>R}\frac{{dy}}{{|x-y{|^{N+sp}}}}\right)}dx+\frac{C}{R^{p}}\int_{\mathbb{R}^{N}}|{u_{n}}(x){|^{p}}\left(\int_{|y-x|\leq
R}\frac{{dy}}{{|x-y{|^{N+sp-p}}}}\right)dx$ $\displaystyle\leq$ $\displaystyle
C\int_{\mathbb{R}^{N}}{|{u_{n}}(x){|^{p}}\left(\int_{|z|>R}\frac{{dy}}{{|z{|^{N+sp}}}}\right)}dx+\frac{C}{R^{p}}\int_{\mathbb{R}^{N}}{|{u_{n}}(x){|^{p}}\left(\int_{|z|\leq
R}\frac{{dy}}{{|z{|^{N+sp-p}}}}\right)}dx$ $\displaystyle\leq$ $\displaystyle
C\int_{\mathbb{R}^{N}}{|{u_{n}}(x){|^{p}}\left(\int_{R}^{\infty}{\frac{{{\rho^{N-1}}}}{{{\rho^{N+sp}}}}}d\rho\right)}dx+\frac{C}{R^{p}}\int_{\mathbb{R}^{N}}{|{u_{n}}(x){|^{p}}\left(\int_{0}^{R}{\frac{{{\rho^{N-1}}}}{{{\rho^{N+sp-p}}}}}d\rho\right)}dx$
$\displaystyle\leq$
$\displaystyle\frac{C}{{{R^{sp}}}}\int_{\mathbb{R}^{N}}|{u_{n}}(x)|^{p}dx+\frac{C}{{{R^{p}}}}R^{p-sp}\int_{\mathbb{R}^{N}}|{u_{n}}(x)|^{p}dx$
$\displaystyle\leq$
$\displaystyle\frac{C}{{{R^{sp}}}}\int_{\mathbb{R}^{N}}|{u_{n}}(x){|^{p}}dx\leq\frac{C}{{{R^{sp}}}},$
(3.22)

where in the last passage we used the boundedness of $\\{u_{n}\\}$ in
$\mathcal{W}_{\varepsilon}$. From the above estimate we can deduce that

$\displaystyle(1-\frac{1}{K})V_{0}\int_{\mathbb{R}^{N}}{|{u_{n}}|^{p}}\eta_{R}(x)dx\leq\frac{C}{{{R^{s}}}}+o_{n}(1),$

which implies that $\\{z_{n}\\}$ is bounded in $\mathbb{R}^{3}$. ∎

######  Lemma 3.6.

Let $\\{u_{n}\\}$ be a $(PS)_{c}$ sequence for ${J_{\varepsilon}}$, then for
each $\zeta>0$, there is a number $R_{1}=R(\zeta)>0$ such that

$\mathop{\lim\sup}\limits_{n\to\infty}\int_{\mathbb{R}^{N}\setminus{B_{{R_{1}}}}(0)}\left(a\int_{\mathbb{R}^{N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}dy+V(\varepsilon
x)u_{n}^{p}\right)dx<\zeta.$

###### Proof.

Let $\eta_{R}\in\mathcal{C}^{\infty}(\mathbb{R}^{N})$ be a cut-off function
such that $\eta_{R}(x)=0$ in $B_{R}$ and $\eta_{R}(x)=1$ in
$\mathbb{R}^{N}\setminus B_{2R}$, with $0\leq\eta_{R}(x)\leq 1$,
$|\nabla\eta_{R}|\leq\frac{C}{R}$ and $C$ is a constant independent on $R$.
Take $R>0$ such that $\Lambda_{\varepsilon}\subset B_{R}(0)$. In view of Lemma
3.4, $\\{u_{n}\\}$ is bounded. From
$\left\langle{{J^{\prime}_{\varepsilon}}({u_{n}}),{u_{n}}\eta_{R}}\right\rangle=o_{n}(1)$
and (3), we have

$\displaystyle\iint_{\mathbb{R}^{2N}}{a\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}\eta_{R}(x)}dxdy+\left(1-\frac{1}{K}\right)\int_{\mathbb{R}^{N}}{V(\varepsilon
x)|{u_{n}}|^{p}}\eta_{R}(x)dx$ $\displaystyle\leq$ $\displaystyle
o_{n}(1)-(a+b[{u_{n}}]_{s,p}^{(\theta-1)p})\iint_{\mathbb{R}^{2N}}\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p-2}}({u_{n}}(x)-{u_{n}}(y))({\eta_{R}}(x)-{\eta_{R}}(y))}}{{|x-y{|^{N+sp}}}}u_{n}(y)dxdy.$
(3.23)

Gathering (3), (3) and (3) we infer that

$\mathop{\lim\sup}\limits_{n\to\infty}\int_{\mathbb{R}^{N}\setminus{B_{{R_{1}}}}(0)}\left(a\int_{\mathbb{R}^{N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}dy+V(\varepsilon
x)u_{n}^{p}\right)dx<\zeta.$

∎

######  Lemma 3.7.

Let $\\{u_{n}\\}$ be a $(PS)_{c}$ sequence for ${J_{\varepsilon}}$ such that
$u_{n}\rightharpoonup u$ in ${\mathcal{W}}_{\varepsilon}$, then

$\displaystyle\mathop{\lim}\limits_{n\to\infty}\int_{{B_{{R}}}(0)}\left(a\int_{\mathbb{R}^{N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}dy+V(\varepsilon
x)u_{n}^{p}\right)dx$ $\displaystyle=$
$\displaystyle\int_{{B_{{R}}}(0)}\left(a\int_{\mathbb{R}^{N}}{\frac{{|{u}(x)-{u}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}dy+V(\varepsilon
x)u^{p}\right)dx.$

###### Proof.

We define the cut-off function
$\eta_{\rho}\in\mathcal{C}^{\infty}(\mathbb{R}^{N})$ such that
$\eta_{\rho}(x)=0$ in $B_{\rho}$ and $\eta_{\rho}(x)=1$ in
$\mathbb{R}^{N}\setminus B_{2\rho}$, with $0\leq\eta_{\rho}(x)\leq 1$. Set

$\displaystyle{P_{n}}(x):$
$\displaystyle={M_{n}}\int_{\mathbb{R}^{N}}{\frac{\left[|u_{n}(x)-{u_{n}}(y)|^{p-2}(u_{n}(x)-{u_{n}}(y))-|u(x)-u(y)|^{p-2}(u(x)-u(y))\right]}{{|x-y{|^{N+sp}}}}}$
$\displaystyle\times{[({u_{n}}(x)-u(x))-({u_{n}}(y)-u(y))]}dy+V(\varepsilon
x)\left(|u_{n}|^{p-2}u_{n}-|u|^{p-2}u\right)(u_{n}-u),$

where $M_{n}:=a+b[u_{n}]_{s,p}^{(\theta-1)p}$.

Fix $R>0$ and choose $R<\rho$. We have

$\displaystyle{M_{n}}\iint_{\mathbb{R}^{2N}}{\frac{\left[|u_{n}(x)-{u_{n}}(y)|^{p-2}(u_{n}(x)-{u_{n}}(y))-|u(x)-u(y)|^{p-2}(u(x)-u(y))\right]}{{|x-y{|^{N+sp}}}}}$
$\displaystyle\times{[({u_{n}}(x)-u(x))-({u_{n}}(y)-u(y))]}\eta_{\rho}dxdy+\int_{\mathbb{R}^{N}}V(\varepsilon
x)\left(|u_{n}|^{p-2}u_{n}-|u|^{p-2}u\right)(u_{n}-u)\eta_{\rho}dx$
$\displaystyle=$
$\displaystyle{M_{n}}\iint_{\mathbb{R}^{2N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}\eta_{\rho}(x)dxdy+\int_{\mathbb{R}^{N}}V(\varepsilon
x)|{u_{n}}{|^{p}}\eta_{\rho}dx$
$\displaystyle+{M_{n}}\iint_{\mathbb{R}^{2N}}{\frac{{|{u}(x)-{u}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}\eta_{\rho}(x)dxdy+\int_{\mathbb{R}^{N}}V(\varepsilon
x)|{u}{|^{p}}\eta_{\rho}dx$
$\displaystyle-\left[\left({M_{n}}\iint_{\mathbb{R}^{2N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p-2}}}(u_{n}(x)-u_{n}(y))(u(x)-u(y))}{{|x-y{|^{N+sp}}}}}\eta_{\rho}(x)dxdy+\int_{\mathbb{R}^{N}}V(\varepsilon
x)|{u_{n}}{|^{p-2}}{u_{n}}u\eta_{\rho}dx\right)\right.$
$\displaystyle\left.-\left({M_{n}}\iint_{\mathbb{R}^{2N}}{\frac{{|{u}(x)-{u}(y){|^{p-2}}}(u(x)-u(y))(u_{n}(x)-u_{n}(y))}{{|x-y{|^{N+sp}}}}}\eta_{\rho}(x)dxdy+\int_{\mathbb{R}^{N}}V(\varepsilon
x)|{u}{|^{p-2}}uu_{n}\eta_{\rho}dx\right)\right].$

Setting

$\displaystyle
I_{n,\rho}^{1}:={M_{n}}\iint_{\mathbb{R}^{2N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}\eta_{\rho}(x)dxdy+\int_{\mathbb{R}^{N}}V(\varepsilon
x)|{u_{n}}{|^{p}}\eta_{\rho}dx-\int_{\mathbb{R}^{N}}g(\varepsilon
x,u_{n})u_{n}\eta_{\rho}dx.$ $\displaystyle
I_{n,\rho}^{2}:={M_{n}}\iint_{\mathbb{R}^{2N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p-2}}}(u_{n}(x)-u_{n}(y))(u(x)-u(y))}{{|x-y{|^{N+sp}}}}}\eta_{\rho}(x)dxdy$
$\displaystyle\qquad\qquad\qquad+\int_{\mathbb{R}^{N}}V(\varepsilon
x)|{u_{n}}{|^{p-2}}{u_{n}}u\eta_{\rho}dx-\int_{\mathbb{R}^{N}}g(\varepsilon
x,u_{n})u\eta_{\rho}dx.$ $\displaystyle
I_{n,\rho}^{3}:=\left({M_{n}}\iint_{\mathbb{R}^{2N}}{\frac{|u(x)-u(y)|^{p}}{{|x-y{|^{N+sp}}}}}\eta_{\rho}(x)dxdy+\int_{\mathbb{R}^{N}}V(\varepsilon
x)|u|^{p}\eta_{\rho}dx\right)$
$\displaystyle-\left({M_{n}}\iint_{\mathbb{R}^{2N}}{\frac{{|{u}(x)-{u}(y){|^{p-2}}}(u(x)-u(y))(u_{n}(x)-u_{n}(y))}{{|x-y{|^{N+sp}}}}}\eta_{\rho}(x)dxdy+\int_{\mathbb{R}^{N}}V(\varepsilon
x)|{u}{|^{p-2}}uu_{n}\eta_{\rho}dx\right)$ $\displaystyle
I_{n,\rho}^{4}:=\int_{\mathbb{R}^{N}}g(\varepsilon
x,u_{n})(u_{n}-u)\eta_{\rho}dx,$

obviously, we have

$\displaystyle
0\leq\int_{B_{R}(0)}{P_{n}}(x)dx\leq|I_{n,\rho}^{1}|+|I_{n,\rho}^{2}|+|I_{n,\rho}^{3}|+|I_{n,\rho}^{4}|.$

Next, we have to prove that

$\displaystyle\mathop{\lim}\limits_{\rho\to\infty}[\mathop{\lim\sup}\limits_{n\to\infty}|I_{n,\rho}^{1}|]=0,\qquad\qquad\mathop{\lim}\limits_{\rho\to\infty}[\mathop{\lim\sup}\limits_{n\to\infty}|I_{n,\rho}^{2}|]=0$

and

$\displaystyle\mathop{\lim}\limits_{\rho\to\infty}[\mathop{\lim\sup}\limits_{n\to\infty}|I_{n,\rho}^{3}|]=0,\qquad\qquad\mathop{\lim}\limits_{\rho\to\infty}[\mathop{\lim\sup}\limits_{n\to\infty}|I_{n,\rho}^{4}|]=0.$

Observe that

$\displaystyle
I_{n,\rho}^{1}=J^{\prime}_{\varepsilon}(u_{n})(u_{n}\eta_{\rho})-(a+b[u_{n}]_{s,p}^{(\theta-1)p})\iint_{\mathbb{R}^{2N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p-2}}}(u_{n}(x)-u_{n}(y))(\eta_{\rho}(x)-\eta_{\rho}(y))}{{|x-y{|^{N+sp}}}}}u_{n}(x)dxdy$

Since $\\{u_{n}\eta_{\rho}\\}$ bounded in $W^{s,p}(\mathbb{R}^{N})$, we have
$\left\langle{{J^{\prime}_{\varepsilon}}({u_{n}}),{u_{n}}\eta_{R}}\right\rangle=o_{n}(1)$.
Moreover, by Hölder inequality and the proof of (3), one has

$\displaystyle(a+b[u_{n}]_{s,p}^{(\theta-1)p})\iint_{\mathbb{R}^{2N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p-2}}}(u_{n}(x)-u_{n}(y))(\eta_{\rho}(x)-\eta_{\rho}(y))}{{|x-y{|^{N+sp}}}}}u_{n}(x)dxdy$
$\displaystyle\leq$ $\displaystyle
C_{1}\left(\iint_{\mathbb{R}^{2N}}\frac{{|{\eta_{\rho}}(x)-{\eta_{\rho}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}|u_{n}(x)|^{p}dxdy\right)^{\frac{1}{p}}\left(\iint_{\mathbb{R}^{2N}}\frac{{|u_{n}(x)-u_{n}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}dxdy\right)^{\frac{p-1}{p}}$
$\displaystyle\leq$ $\displaystyle
C_{2}\left(\iint_{\mathbb{R}^{2N}}\frac{{|(1-{\eta_{\rho}}(x))-(1-{\eta_{\rho}}(y)){|^{p}}}}{{|x-y{|^{N+sp}}}}|u_{n}(x)|^{p}dxdy\right)^{\frac{1}{p}}\rightarrow
0$

as $\rho\to\infty$, $n\to\infty$. Then,

$\displaystyle\mathop{\lim}\limits_{\rho\to\infty}[\mathop{\lim\sup}\limits_{n\to\infty}|I_{n,\rho}^{1}|]=0.$

We also have that

$\displaystyle
I_{n,\rho}^{2}=J^{\prime}_{\varepsilon}(u_{n})(u\eta_{\rho})-(a+b[u_{n}]_{s,p}^{(\theta-1)p})\iint_{\mathbb{R}^{2N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p-2}}}(u_{n}(x)-u_{n}(y))(\eta_{\rho}(x)-\eta_{\rho}(y))}{{|x-y{|^{N+sp}}}}}u(x)dxdy.$

Using
$\left\langle{{J^{\prime}_{\varepsilon}}({u_{n}}),{u}\eta_{R}}\right\rangle=o_{n}(1)$
and arguing as in the previous case, we can prove the second term of the right
side of the last equality tends to zero, as $n,\rho\to\infty$. Consequently,

$\displaystyle\mathop{\lim}\limits_{\rho\to\infty}[\mathop{\lim\sup}\limits_{n\to\infty}|I_{n,\rho}^{2}|]=0.$

On the other hand, from the weak convergence,

$\displaystyle\mathop{\lim}\limits_{\rho\to\infty}|I_{n,\rho}^{3}|=$
$\displaystyle\mathop{\lim}\limits_{\rho\to\infty}\left|\left({M_{n}}\iint_{\mathbb{R}^{2N}}{\frac{|u(x)-u(y)|^{p-2}(u(x)-u(y))[(u_{n}(x)-u_{n}(y))-(u(x)-u(y))]}{{|x-y{|^{N+sp}}}}}\eta_{\rho}(x)dxdy\right.\right.$
$\displaystyle\left.\left.+\int_{\mathbb{R}^{N}}V(\varepsilon
x)|u|^{p-2}u(u_{n}-u)\eta_{\rho}dx\right)\right|=0,$

where we used the fact that $M_{n}$ is bounded in $\mathbb{R}$. Below we shall
prove that

$\displaystyle\mathop{\lim}\limits_{\rho\to\infty}[\mathop{\lim\sup}\limits_{n\to\infty}|I_{n,\rho}^{4}|]=0.$

We first claim that

$\displaystyle{u_{n}}\to u\ \ \text{in}\ \
L^{p_{s}^{*}}(\Lambda_{\varepsilon}).$

Indeed, we may suppose that $|D^{s}u_{n}|^{p}\rightharpoonup\mu$ and
$|u_{n}|^{p_{s}^{*}}\rightharpoonup\nu$, where
$|D^{s}u_{n}|^{p}=\int_{\mathbb{R}^{N}}\frac{|u(x)-u(y)|^{p}}{|x-y|^{N+sp}}dy$,
$\mu$ and $\nu$ are bounded nonnegative measures in ${\mathbb{R}^{N}}$. By a
variant of the concentration-compactness lemma [30], which is proved by [7,
Lemma 2.4], we have an at most countable index set $\Gamma$ and sequence
$\\{x_{i}\\}\subset{\mathbb{R}^{N}}$, $\\{\mu_{i}\\}$,
$\\{\nu_{i}\\}\subset(0,\infty)$ such that

$\displaystyle\mu\geq|{D^{s}}u{|^{p}}+\sum\limits_{i\in\Gamma}{{\mu_{i}}}{\delta_{{x_{i}}}},\
\
\nu=|u{|^{p_{s}^{*}}}+\sum\limits_{i\in\Gamma}{{\nu_{i}}}{\delta_{{x_{i}}}},\
\ {\mu_{i}}\geq{S}{\nu_{i}}^{\frac{p}{{p_{s}^{*}}}}$ (3.24)

for all $i\in\Gamma$, where ${\delta_{{x_{i}}}}$ is the Dirac mass at
$x_{i}\in{\mathbb{R}^{N}}$. It suffices to show that
$\\{x_{i}\\}_{i\in\Gamma}\cap\Lambda_{\varepsilon}=\emptyset$. Suppose, by
contradiction, that ${x_{i}}\in\Lambda_{\varepsilon}$ for some $i\in\Gamma$.
Define $\varphi_{\rho}(x)=\varphi\left(\frac{x-x_{i}}{\rho}\right)$, where
$\varphi=1$ if $x\in B_{1}(0)$, $\varphi=0$ if $x\notin B_{2}(0)$, and
$|\nabla\varphi|_{\infty}\leq 2$. We suppose that $\rho$ is chosen in such way
that the support of $\varphi_{\rho}$ is contained in $\Lambda_{\varepsilon}$.
Since $\\{\varphi_{\rho}u_{n}\\}$ is bounded ,
$\left\langle{{J^{\prime}_{\varepsilon}}({u_{n}}),{u_{n}}\varphi_{\rho}}\right\rangle=o_{n}(1)$,
we have

$\displaystyle(a+b[u_{n}]_{s,p}^{(\theta-1)p})\int_{\mathbb{R}^{N}}(-\Delta)_{p}^{s}u_{n}(\varphi_{\rho}u_{n})dx+\int_{\mathbb{R}^{N}}V(\varepsilon
x)u_{n}^{p}\varphi_{\rho}dx$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{N}}g(\varepsilon
x,u_{n})u_{n}\varphi_{\rho}dx+o_{n}(1)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{N}}f(u_{n})u_{n}\varphi_{\rho}dx+\int_{\mathbb{R}^{N}}u_{n}^{p_{s}^{*}}\varphi_{\rho}dx+o_{n}(1).$

Since $\varphi_{\rho}$ has compact support, letting $n\to\infty$ and $\rho\to
0$ we obtain that $a{\mu_{i}}+b\mu_{i}^{\theta}\leq{\nu_{i}}$, which combining
with inequality that ${\mu_{i}}\geq{S}{\nu_{i}}^{\frac{p}{{p_{s}^{*}}}}$ we
infer that ${\nu_{i}}\geq
a{S}{\nu_{i}}^{\frac{p}{{p_{s}^{*}}}}+bS^{\theta}{\nu_{i}}^{\frac{{\theta
p}}{{p_{s}^{*}}}}$. Setting
${\nu^{\prime}_{i}}:=\nu_{i}^{\frac{1}{{p_{s}^{*}}}}$, we have

$\displaystyle
a{S}{({\nu^{\prime}_{i}})^{p}}+bS^{\theta}{({\nu^{\prime}_{i}})^{\theta
p}}-{({\nu^{\prime}_{i}})^{p_{s}^{*}}}\leq 0,$

which implies $H^{\prime}({\nu^{\prime}_{i}})\leq 0$ defined in (3.7), so we
can deduce that ${\nu^{\prime}_{i}}\geq{\hat{T}}$, where ${\hat{T}}$ is the
unique maximum of $H$ defined in Lemma 3.2. And we can know that

$\hat{H}({\hat{T}})=a{S}+bS^{\theta}{{\hat{T}}^{(\theta-1)p}}-{{\hat{T}}^{p_{s}^{*}-p}}=0,$

which implies

$a{S}{\hat{T}}^{p}+bS^{\theta}{{\hat{T}}^{\theta
p}}-{{\hat{T}}^{p_{s}^{*}}}=0.$ (3.25)

Therefore, we can infer

$\displaystyle c+o_{n}(1)=$ $\displaystyle
J_{\varepsilon}(u_{n})-\frac{{1}}{{\theta
p}}\left\langle{{{J^{\prime}}_{\varepsilon}}({u_{n}}),{u_{n}}}\right\rangle$
$\displaystyle\geq$ $\displaystyle{\frac{(\theta-1)a}{\theta
p}}[u_{n}]_{s,p}^{p}+\left(\frac{{1}}{{\theta
p}}-\frac{1}{{p_{s}^{*}}}\right)\int_{{\Lambda_{\varepsilon}}}{{{(u_{n})}^{p_{s}^{*}}}}dx+\frac{{(\theta-1)}}{{\theta
p}}\int_{\mathbb{R}^{N}}{V(\varepsilon x)|{u_{n}}|^{p}}dx$
$\displaystyle+\frac{{1}}{{\theta
p}}\int_{{\mathbb{R}^{N}}\setminus{\Lambda_{\varepsilon}}}\left(g(\varepsilon
x,u_{n})u_{n}-\theta pG(\varepsilon x,u_{n})\right)dx$ $\displaystyle\geq$
$\displaystyle{\frac{(\theta-1)a}{\theta
p}}[u_{n}]_{s,p}^{p}+\frac{{(\theta-1)}}{{\theta
p}}\int_{\mathbb{R}^{N}}{V(\varepsilon
x)|{u_{n}}|^{p}}dx-\frac{{(\theta-1)}}{{\theta}}\int_{{\mathbb{R}^{N}}\setminus{\Lambda_{\varepsilon}}}G(\varepsilon
x,u_{n})dx$ $\displaystyle+\left(\frac{{1}}{{\theta
p}}-\frac{1}{{p_{s}^{*}}}\right)\int_{{\Lambda_{\varepsilon}}}{{{(u_{n})}^{p_{s}^{*}}}}dx$
$\displaystyle\geq$ $\displaystyle\frac{(\theta-1)a}{\theta
p}\int_{\Lambda_{\varepsilon}}|D^{s}u_{n}|^{p}\varphi_{\rho}dx+\left(\frac{{1}}{{\theta
p}}-\frac{1}{{p_{s}^{*}}}\right)\int_{{\Lambda_{\varepsilon}}}{{{(u_{n})}^{p_{s}^{*}}}}\varphi_{\rho}dx$
$\displaystyle+\left(\frac{(\theta-1)}{\theta p}-\frac{(\theta-1)}{\theta
pK}\right)\int_{{\mathbb{R}^{N}}\setminus{\Lambda_{\varepsilon}}}{V(\varepsilon
x)|{u_{n}}|^{p}}dx$ $\displaystyle\geq$
$\displaystyle{\frac{(\theta-1)a}{\theta p}}\mu_{i}+\left(\frac{{1}}{{\theta
p}}-\frac{1}{{p_{s}^{*}}}\right)\nu_{i}$ $\displaystyle\geq$
$\displaystyle{\frac{(\theta-1)a}{\theta
p}}{S}{\nu_{i}}^{\frac{p}{{p_{s}^{*}}}}+\left(\frac{{1}}{{\theta
p}}-\frac{1}{{p_{s}^{*}}}\right)\left(a{S}{\nu_{i}}^{\frac{p}{{p_{s}^{*}}}}+bS^{\theta}{\nu_{i}}^{\frac{{\theta
p}}{{p_{s}^{*}}}}\right)$ $\displaystyle=$
$\displaystyle\frac{(\theta-1)a}{\theta
p}{S}({\nu^{\prime}_{i}})^{p}+\left(\frac{{1}}{{\theta
p}}-\frac{1}{{p_{s}^{*}}}\right)\left(a{S}({\nu^{\prime}_{i}})^{p}+bS^{2}({\nu^{\prime}_{i}})^{\theta
p}\right)$ $\displaystyle\geq$ $\displaystyle\frac{(\theta-1)a}{\theta
p}{S}{\hat{T}}^{p}+\left(\frac{{1}}{{\theta
p}}-\frac{1}{{p_{s}^{*}}}\right)\left(a{S}{\hat{T}}^{p}+bS^{\theta}{\hat{T}}^{\theta
p}\right)$ $\displaystyle=$
$\displaystyle\frac{a}{p}{S}{\hat{T}}^{p}+\frac{b}{\theta
p}S^{\theta}{\hat{T}}^{\theta p}-\frac{1}{p_{s}^{*}}{\hat{T}}^{p_{s}^{*}},$

contracting with the assumption $c<c_{0}$. Then $u_{n}\to u$ in
$L^{p_{s}^{*}}(\Lambda_{\varepsilon})$. Observe that

$\displaystyle
I_{n,\rho}^{4}\leq\int_{({{\mathbb{R}^{N}}\setminus{\Lambda_{\varepsilon}}})\cap
B_{2\rho}(0)}|g(\varepsilon
x,u_{n})(u_{n}-u)|+\int_{{\Lambda_{\varepsilon}}\cap
B_{2\rho}(0)}|g(\varepsilon x,u_{n})(u_{n}-u)|.$

The Sobolev embedding theorem implies that $u_{n}\to u$ in
$L_{loc}^{q}({\mathbb{R}^{N}})$, $1\leq q<p_{s}^{*}$. By $(G_{3})$-$(ii)$, we
get $\int_{{({\mathbb{R}^{N}}\setminus{\Lambda_{\varepsilon}}})\cap
B_{2\rho}(0)}|g(\varepsilon x,u_{n})(u_{n}-u)|\to 0$, as $n\to\infty$.
Moreover, from $u_{n}\to u$ in $L^{p_{s}^{*}}(\Lambda_{\varepsilon})$ we
obtain

$\displaystyle\int_{{\Lambda_{\varepsilon}}\cap B_{2\rho}(0)}|g(\varepsilon
x,u_{n})(u_{n}-u)|\to 0,\ \ \mbox{as}\ \ n\to\infty.$

Then,

$\displaystyle\mathop{\lim}\limits_{\rho\to\infty}[\mathop{\lim\sup}\limits_{n\to\infty}|I_{n,\rho}^{4}|]=0.$

We obtain

$\displaystyle
0\leq\mathop{\lim\sup}\limits_{n\to\infty}\int_{B_{R}(0)}{P_{n}}(x)dx\leq\mathop{\lim\sup}\limits_{n\to\infty}(|I_{n,\rho}^{1}|+|I_{n,\rho}^{2}|+|I_{n,\rho}^{3}|+|I_{n,\rho}^{4}|)=0.$

Then, $\mathop{\lim\sup}\limits_{n\to\infty}\int_{B_{R}(0)}{P_{n}}(x)dx=0$.
Consequently, we obtain

$\displaystyle{M_{n}}\int_{B_{R}(0)}\int_{\mathbb{R}^{N}}{\frac{\left[|u_{n}(x)-{u_{n}}(y)|^{p-2}(u_{n}(x)-{u_{n}}(y))-|u(x)-u(y)|^{p-2}(u(x)-u(y))\right]}{{|x-y{|^{N+sp}}}}}$
$\displaystyle\times{[({u_{n}}(x)-u(x))-({u_{n}}(y)-u(y))]}dxdy+\int_{B_{R}(0)}V(\varepsilon
x)\left(|u_{n}|^{p-2}u_{n}-|u|^{p-2}u\right)(u_{n}-u)dx\rightarrow 0,$ (3.26)

as $n\to\infty$. By Lemma 2.4, for $p\geq 2$, we know that

$\displaystyle\int_{B_{R}(0)}\int_{\mathbb{R}^{N}}\frac{\left|[u_{n}(x)-{u_{n}}(y)]-[u(x)-u(y)]\right|^{p}}{|x-y|^{N+sp}}dxdy$
$\displaystyle\leq$ $\displaystyle
c_{p}\int_{B_{R}(0)}\int_{\mathbb{R}^{N}}{\frac{\left[|u_{n}(x)-{u_{n}}(y)|^{p-2}(u_{n}(x)-{u_{n}}(y))-|u(x)-u(y)|^{p-2}(u(x)-u(y))\right]}{{|x-y{|^{N+sp}}}}}$
$\displaystyle\times{[({u_{n}}(x)-u(x))-({u_{n}}(y)-u(y))]}dxdy\to 0.$

In a similar way, by (3), we get

$\displaystyle\int_{B_{R}(0)}V(\varepsilon x)|u_{n}-u|^{p}dx\leq$
$\displaystyle c_{p}\int_{B_{R}(0)}V(\varepsilon
x)\left(|u_{n}|^{p-2}u_{n}-|u|^{p-2}u\right)(u_{n}-u)dx.$

Now, we consider the case $1<p<2$. Since $u_{n}\rightharpoonup u$ in
${\mathcal{W}}_{\varepsilon}$, there exists $\kappa>0$ such that
$\|u_{n}\|_{\varepsilon}\leq\kappa$ for all $n\in\mathbb{N}$. Hence, by (3),
Simon inequality and Hölder inequality, we can see that

$\displaystyle\int_{B_{R}(0)}\int_{\mathbb{R}^{N}}\frac{\left|[u_{n}(x)-{u_{n}}(y)]-[u(x)-u(y)]\right|^{p}}{|x-y|^{N+sp}}dxdy$
$\displaystyle\leq$ $\displaystyle
C_{p}\int_{B_{R}(0)}\int_{\mathbb{R}^{N}}\frac{\left[|u_{n}(x)-{u_{n}}(y)|^{p-2}(u_{n}(x)-{u_{n}}(y))-|u(x)-u(y)|^{p-2}(u(x)-u(y))\right]^{\frac{p}{2}}}{|x-y|^{N+sp}}$
$\displaystyle\times\left[(u_{n}(x)-{u_{n}}(y)-(u(x)-u(y)))\right]^{\frac{p}{2}}\left(|u_{n}(x)-{u_{n}}(y)|^{p}+|u(x)-u(y)|^{p}\right)^{\frac{2-p}{2}}dxdy$
$\displaystyle\leq$ $\displaystyle
C_{p}(\int_{B_{R}(0)}\int_{\mathbb{R}^{N}}{\frac{\left[|u_{n}(x)-{u_{n}}(y)|^{p-2}(u_{n}(x)-{u_{n}}(y))-|u(x)-u(y)|^{p-2}(u(x)-u(y))\right]}{{|x-y{|^{N+sp}}}}}$
$\displaystyle\times{[({u_{n}}(x)-u(x))-({u_{n}}(y)-u(y))]}dxdy)^{\frac{p}{2}}\left([u_{n}]_{s,p}^{p}+[u]_{s,p}^{p}\right)^{\frac{2-p}{2}}$
$\displaystyle\leq$ $\displaystyle
C_{p}(\int_{B_{R}(0)}\int_{\mathbb{R}^{N}}{\frac{\left[|u_{n}(x)-{u_{n}}(y)|^{p-2}(u_{n}(x)-{u_{n}}(y))-|u(x)-u(y)|^{p-2}(u(x)-u(y))\right]}{{|x-y{|^{N+sp}}}}}$
$\displaystyle\times{[({u_{n}}(x)-u(x))-({u_{n}}(y)-u(y))]}dxdy)^{\frac{p}{2}}\left([u_{n}]_{s,p}^{\frac{p(2-p)}{2}}+[u]_{s,p}^{\frac{p(2-p)}{2}}\right)$
$\displaystyle\leq$ $\displaystyle
C^{\prime}_{p}(\int_{B_{R}(0)}\int_{\mathbb{R}^{N}}{\frac{\left[|u_{n}(x)-{u_{n}}(y)|^{p-2}(u_{n}(x)-{u_{n}}(y))-|u(x)-u(y)|^{p-2}(u(x)-u(y))\right]}{{|x-y{|^{N+sp}}}}}$
$\displaystyle\times{[({u_{n}}(x)-u(x))-({u_{n}}(y)-u(y))]}dxdy)^{\frac{p}{2}}\to
0.$

Then, we can also obtain

$\displaystyle\int_{B_{R}(0)}V(\varepsilon x)|u_{n}-u|^{p}dx\leq$
$\displaystyle C^{\prime\prime}_{p}\left(\int_{B_{R}(0)}V(\varepsilon
x)\left(|u_{n}|^{p-2}u_{n}-|u|^{p-2}u\right)(u_{n}-u)dx\right)^{\frac{p}{2}}\to
0.$

Consequently,

$\displaystyle\mathop{\lim}\limits_{n\to\infty}\int_{{B_{{R}}}(0)}\left(a\int_{\mathbb{R}^{N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}dy+V(\varepsilon
x)u_{n}^{p}\right)dx$ $\displaystyle=$
$\displaystyle\int_{{B_{{R}}}(0)}\left(a\int_{\mathbb{R}^{N}}{\frac{{|{u}(x)-{u}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}dy+V(\varepsilon
x)u^{p}\right)dx.$

∎

######  Lemma 3.8.

The function ${J_{\varepsilon}}$ verifies the $(PS)_{c}$ condition in
${\mathcal{W}}_{\varepsilon}$ at any level $c<c_{0}$.

###### Proof.

Let $\\{u_{n}\\}$ be a $(PS)_{c}$ sequence for $J_{\varepsilon}$. By Lemma 3.4
we see that $\\{u_{n}\\}$ is bounded in ${\mathcal{W}}_{\varepsilon}$. Passing
to a subsequence, we have $u_{n}\rightharpoonup u$ for some
$u\in{\mathcal{W}}_{\varepsilon}$. From Lemma 3.6, it follows that for
$\zeta>0$ given, there exists $R_{1}=R(\zeta)>C/\zeta$ with $C>0$ independent
on $\zeta$ such that

$\mathop{\lim\sup}\limits_{n\to\infty}\int_{\mathbb{R}^{N}\setminus{B_{{R_{1}}}}(0)}\left(a\int_{\mathbb{R}^{N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}dy+V(\varepsilon
x)u_{n}^{p}\right)dx<\zeta.$

Therefore, from Lemma 3.7, one has

$\displaystyle\|u\|_{\varepsilon}^{p}\leq$
$\displaystyle\mathop{\lim\inf}\limits_{n\to\infty}\|u_{n}\|_{\varepsilon}^{p}\leq\mathop{\lim\sup}\limits_{n\to\infty}\|u_{n}\|_{\varepsilon}^{p}$
$\displaystyle=$
$\displaystyle\mathop{\lim\sup}\limits_{n\to\infty}\left\\{\int_{{B_{{R_{1}}}}(0)}\left(a\int_{\mathbb{R}^{N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}dy+V(\varepsilon
x)u_{n}^{p}\right)dx\right.$
$\displaystyle+\left.\int_{\mathbb{R}^{N}\setminus{B_{{R_{1}}}}(0)}\left(a\int_{\mathbb{R}^{N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}dy+V(\varepsilon
x)u_{n}^{p}\right)dx\right\\}$ $\displaystyle=$
$\displaystyle\int_{{B_{{R_{1}}}}(0)}\left(a\int_{\mathbb{R}^{N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}dy+V(\varepsilon
x)u_{n}^{p}\right)dx$
$\displaystyle+\mathop{\lim\sup}\limits_{n\to\infty}\int_{\mathbb{R}^{N}\setminus{B_{{R_{1}}}}(0)}\left(a\int_{\mathbb{R}^{N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}dy+V(\varepsilon
x)u_{n}^{p}\right)dx$ $\displaystyle<$
$\displaystyle\int_{{B_{{R_{1}}}}(0)}\left(a\int_{\mathbb{R}^{N}}{\frac{{|{u_{n}}(x)-{u_{n}}(y){|^{p}}}}{{|x-y{|^{N+sp}}}}}dy+V(\varepsilon
x)u_{n}^{p}\right)dx+\zeta,$

where $R_{1}=R(\zeta)>C/\zeta$. Passing to the limit as $\zeta\to 0$ we have
$R_{1}\to\infty$, which implies that

$\displaystyle\|u\|_{\varepsilon}^{p}\leq$
$\displaystyle\mathop{\lim\inf}\limits_{n\to\infty}\|u_{n}\|_{\varepsilon}^{p}\leq\mathop{\lim\sup}\limits_{n\to\infty}\|u_{n}\|_{\varepsilon}^{p}\leq\|u\|_{\varepsilon}^{p}.$

Consequently, $\|u_{n}\|_{\varepsilon}\to\|u\|_{\varepsilon}$ and so $u_{n}\to
u$ in ${\mathcal{W}}_{\varepsilon}$. ∎

######  Proposition 3.9.

The function ${J_{\varepsilon}}$ restricted to $\mathcal{N}_{\varepsilon}$
satisfies the $(PS)_{c}$ condition in ${\mathcal{W}}_{\varepsilon}$ at any
level $c<c_{0}$.

###### Proof.

Let $\\{u_{n}\\}\subset\mathcal{N}_{\varepsilon}$ be such that
$J_{\varepsilon}(u_{n})\to c$ and
$\|J^{\prime}_{\varepsilon}(u_{n})_{|_{\mathcal{N}_{\varepsilon}}}\|_{*}=o_{n}(1)$.
Then there exists $\\{\lambda_{n}\\}\subset\mathbb{R}$ such that

$\displaystyle
J^{\prime}_{\varepsilon}(u_{n})=\lambda_{n}G^{\prime}_{\varepsilon}(u_{n})+o_{n}(1),$

where
$J^{\prime}_{\varepsilon}(u)=\left\langle{{J^{\prime}_{\varepsilon}}({u}),{u}}\right\rangle$.
Moreover, since $u_{n}\in\mathcal{N}_{\varepsilon}$, we know that

$\displaystyle
0=\left\langle{{J^{\prime}_{\varepsilon}}({u_{n}}),{u_{n}}}\right\rangle=\lambda_{n}\left\langle{{G^{\prime}_{\varepsilon}}({u_{n}}),{u_{n}}}\right\rangle+o_{n}(1)\|u_{n}\|_{{\mathcal{W}}_{\varepsilon}}.$
(3.27)

Next, we will show that $\lambda_{n}\to 0$ as $n\to\infty$. Indeed, by
$(G_{4})$ and
$\left\langle{{J^{\prime}_{\varepsilon}}({u_{n}}),{u_{n}}}\right\rangle=o_{n}(1)$,
we deduce that

$\displaystyle\left\langle{{G^{\prime}_{\varepsilon}}({u_{n}}),{u_{n}}}\right\rangle=$
$\displaystyle p\|u_{n}\|_{\varepsilon}^{p}+\theta pb[u_{n}]_{s,p}^{\theta
p}-\int_{\mathbb{R}^{N}}g^{\prime}(\varepsilon
x,u_{n})u_{n}^{2}dx-\int_{\mathbb{R}^{N}}g(\varepsilon x,u_{n})u_{n}dx$
$\displaystyle=$
$\displaystyle-(\theta-1)p\|u_{n}\|_{\varepsilon}^{p}+\int_{\mathbb{R}^{N}}[(\theta
p-1)g(\varepsilon x,u_{n})u_{n}-g^{\prime}(\varepsilon x,u_{n})u_{n}^{2}]dx$
$\displaystyle\leq$
$\displaystyle-(\theta-1)p\|u_{n}\|_{\varepsilon}^{p}+(\theta-1)p\int_{\mathbb{R}^{N}}(1-\chi)(\varepsilon
x)\tilde{f}({u_{n}})u_{n}dx$ $\displaystyle\leq$
$\displaystyle-((\theta-1)p-\frac{(\theta-1)p}{K})\|u_{n}\|_{\varepsilon}^{p}.$

Thus, we may assume that
$\left\langle{{G^{\prime}_{\varepsilon}}({u_{n}}),{u_{n}}}\right\rangle\to
l<0$. It follows from (3.27) that $\lambda_{n}\to 0$ as $n\to\infty$ and then
we see that ${J^{\prime}_{\varepsilon}}({u_{n}})\to 0$ as $n\to\infty$. Hence,
$\\{u_{n}\\}$ is a $(PS)_{c}$ sequence for $J_{\varepsilon}$. ∎

##  4 the autonomous problem

In this section, we are concerned with the autonomous equation:

$\displaystyle(a+b\left[u\right]_{s,p}^{(\theta-1)p})(-\Delta)_{p}^{s}u+V_{0}{u^{p-1}}=f(u)+{u^{p_{s}^{*}-1}}\
\ $ $\displaystyle\mbox{in}\ {\mathbb{R}^{N}},\ \ \ u>0,\
u\in{W^{s,p}}(\mathbb{R}^{N}),$ (4.1)

where $V_{0}$ is given in $(V_{2})$.

The functional of (4.1) is denoted by

${J_{0}}(u)=\frac{1}{p}\|u\|_{0}^{p}+\frac{b}{{\theta
p}}[u]_{{}_{s,p}}^{\theta
p}-\int_{\mathbb{R}^{N}}{F(u)}dx-\frac{1}{p_{s}^{*}}\int_{\mathbb{R}^{N}}{(u^{+})^{p_{s}^{*}}}dx.$

The Nehari manifold corresponding to (4.1) is defined by

${\mathcal{N}}_{0}:=\left\\{u\in{{\mathcal{W}}_{0}\setminus\\{0\\}}:\left\langle{{J^{\prime}_{0}}(u),u}\right\rangle=0\right\\},$

where ${\mathcal{W}}_{0}={W^{s,p}}(\mathbb{R}^{N})$ is endowed with the norm
$\|u\|_{0}^{p}=a[u]_{s,p}^{p}+\int_{\mathbb{R}^{N}}{V_{0}}|u{|^{p}}dx$.

As (3.8), we have the following characterization of infimum of $J_{0}$ on
$\mathcal{N}_{0}$

${c_{V_{0}}}=\mathop{\inf}\limits_{u\in{\mathcal{W}_{0}}\setminus\\{0\\}}\mathop{\sup}\limits_{t\geq
0}{J_{0}}(tu)=\mathop{\inf}\limits_{u\in\mathcal{N}_{0}}{J_{0}}(u)>0,$ (4.2)

and as Lemma 3.3, we can show that $c_{V_{0}}<c_{0}$.

######  Lemma 4.1.

Let $\\{u_{n}\\}\subset\mathcal{W}_{0}$ be a $(PS)_{c_{V_{0}}}$ sequence of
${J_{0}}$. Then there exists $u\in\mathcal{W}_{0}\setminus\\{0\\}$ such that
$u_{n}\to u$ in $\mathcal{W}_{0}$. Moreover, $u$ is a positive ground state of
the equation (4.1).

###### Proof.

It is easy to check that $J_{0}$ has a mountain pass geometry, so there exists
a sequence $\\{u_{n}\\}\subset\mathcal{W}_{0}$ such that $J_{0}(u_{n})\to
c_{V_{0}}$ and $J^{\prime}_{0}(u_{n})\to 0$. Thus, we can easily see that
$\\{u_{n}\\}$ is bounded in $\mathcal{W}_{0}$ and we may assume that
$u_{n}\rightharpoonup u$ in $\mathcal{W}_{0}$ and
$[u_{n}]_{s,p}^{(\theta-1)p}\rightarrow B^{(\theta-1)p}$. Similar argument to
the proof of Lemma 3.5, we can deduce that the sequence $\\{u_{n}\\}$ is
nonvanishing, $i.e.$, there exists a sequence
$\\{x_{n}\\}\subset\mathbb{R}^{N}$ and $R_{0}>0$, $\beta_{0}>0$ such that

$\int_{B_{R}(x_{n})}|u_{n}|^{p}\ dx\geq\beta_{0}.$

Then we suppose that $u\neq 0$. Since
$\left\langle{{J^{\prime}_{0}}({u}),\varphi}\right\rangle=o_{n}(1)$, we can
deduce that for all $\varphi\in\mathcal{C}_{0}^{\infty}(\mathbb{R}^{N})$,

$\displaystyle
a\iint_{\mathbb{R}^{2N}}\frac{{|u(x)-u(y){|^{p-2}}(u(x)-u(y))(\varphi(x)-\varphi(y))}}{{|x-y{|^{N+sp}}}}dxdy+\int_{{\mathbb{R}^{N}}}V(\varepsilon
x)u^{p-1}\varphi dx$
$\displaystyle+bB^{(\theta-1)p}\iint_{{\mathbb{R}^{2N}}}\frac{{|u(x)-u(y){|^{p-2}}(u(x)-u(y))(\varphi(x)-\varphi(y))}}{{|x-y{|^{N+sp}}}}dxdy$
$\displaystyle=$
$\displaystyle\int_{{\mathbb{R}^{N}}}f(u)u\varphi+\int_{{\mathbb{R}^{N}}}(u^{+})^{p_{s}^{*}-1}\varphi$
(4.3)

where
$B^{(\theta-1)p}:=\mathop{\lim}\limits_{n\to\infty}[u_{n}]_{s,p}^{(\theta-1)p}$.
Let us note that $B^{(\theta-1)p}\geq[u]_{s,p}^{(\theta-1)p}$ by the weakly
semi-continuous of norm. If by construction
$B^{(\theta-1)p}>[u]_{s,p}^{(\theta-1)p}$, we may use (4) to deduce that
$\left\langle{{J^{\prime}_{0}}({u}),u}\right\rangle<0$. Moreover, conditions
$(F_{1})$ and $(F_{2})$ imply that $\left\langle{{J^{\prime}_{0}}({\tau
u}),\tau u}\right\rangle=0$. From $t_{0}\in(\tau,1)$, $(F_{3})$ and $(F_{4})$
we get

$\displaystyle c_{V_{0}}\leq$ $\displaystyle J_{0}(t_{0}u)-\frac{1}{\theta
p}\left\langle{{J^{\prime}_{0}}({t_{0}u}),t_{0}u}\right\rangle<J_{0}(u)-\frac{1}{\theta
p}\left\langle{{J^{\prime}_{0}}({u}),u}\right\rangle$ $\displaystyle\leq$
$\displaystyle\mathop{\lim\inf}\limits_{n\to\infty}[J_{0}(u_{n})-\frac{1}{\theta
p}\left\langle{{J^{\prime}_{0}}({u_{n}}),u_{n}}\right\rangle]=c_{V_{0}},$
(4.4)

which gives a contradiction. Therefore,
$B^{(\theta-1)p}=[u]_{s,p}^{(\theta-1)p}$ and we deduce that
$J^{\prime}_{0}(u)=0$. Using the fact that
$\left\langle{{J^{\prime}_{0}}({u}),u^{-}}\right\rangle=0$, where
$u^{-}=\min\\{u,0\\}$, by $f(t)=0$ for $t\leq 0$, and
$|x-y{|^{p-2}}(x-y)({x^{-}}-{y^{-}})\geq|{x^{-}}-{y^{-}}{|^{p}}$, where
$x^{-}=\min\\{x,0\\}$, we can deduce that

$\displaystyle\|u^{-}\|_{0}^{p}\leq$ $\displaystyle
a\iint_{\mathbb{R}^{2N}}\frac{{|u(x)-u(y){|^{p-2}}(u(x)-u(y))(u^{-}(x)-u^{-}(y))}}{{|x-y{|^{N+sp}}}}dxdy+\int_{{\mathbb{R}^{N}}}V(\varepsilon
x)|u|^{p-2}uu^{-}dx$ $\displaystyle\leq$
$\displaystyle\int_{{\mathbb{R}^{N}}}f(u)u^{-}+\int_{{\mathbb{R}^{N}}}(u^{+})^{p_{s}^{*}-1}u^{-}=0,$

which implies that $u^{-}=0$, that is $u\geq 0$. Moreover, proceeding as in
the proof of Lemma 6.1 below, we can see that $u\in
L^{\infty}({{\mathbb{R}^{N}}})$. By Corollary $5.5$ in [26], we deduce that
$u\in\mathcal{C}^{0,\alpha}({{\mathbb{R}^{N}}})$, and applying maximum
principle [18] we can conclude that $u>0$ in $\mathbb{R}^{N}$. Finally,
arguing as in (4) with $t_{0}=1$, we can show that $u$ is a positive ground
state solution to (4.1). ∎

######  Lemma 4.2.

The following limits holds

$\mathop{\lim\sup}\limits_{\varepsilon\to
0}{c_{\varepsilon}}\leq{c_{{V_{0}}}}.$

Moreover, $\mathop{\lim}\limits_{\varepsilon\to
0}{c_{\varepsilon}}={c_{{V_{0}}}}$.

###### Proof.

For any $\varepsilon>0$ we set
$\omega_{\varepsilon}(x):=\psi_{\varepsilon}(x)\omega(x)$, where $\omega$ is a
positive ground state solution, and $\psi_{\varepsilon}(x):=\psi(\varepsilon
x)$ with $\psi\in\mathcal{C}_{0}^{\infty}(\mathbb{R}^{N})$, $\psi\in[0,1]$,
$\psi(x)=1$ if $|x|\leq\frac{1}{2}$ and $\psi(x)=0$ if $|x|\geq 1$. Here we
assume that $supp(\psi)\subset B_{1}\subset\Lambda$. Using Lemma 2.2 and the
dominated convergence theorem we can see that $\omega_{\varepsilon}\to\omega$
in $\mathcal{W}_{0}$ and $J_{0}(\omega_{\varepsilon})\to
J_{0}(\omega)=c_{V_{0}}$ as $\varepsilon\to 0$. For each $\varepsilon>0$ there
exists $t_{\varepsilon}>0$ such that

$J_{\varepsilon}(t_{\varepsilon}\omega_{\varepsilon})=\mathop{\max}\limits_{t\geq
0}J_{\varepsilon}(t\omega_{\varepsilon}).$

Then, $J^{\prime}_{\varepsilon}(t_{\varepsilon}\omega_{\varepsilon})=0$ and
this implies that

$\displaystyle\frac{1}{t_{\varepsilon}^{(\theta-1)p}}\left(a[\omega_{\varepsilon}]_{s,p}^{p}+\int_{\mathbb{R}^{N}}V(\varepsilon
x)\omega_{\varepsilon}^{p}dx\right)+b[\omega_{\varepsilon}]_{s,p}^{\theta p}$
$\displaystyle=$
$\displaystyle\int_{{\mathbb{R}^{N}}}\frac{f(t_{\varepsilon}\omega_{\varepsilon})}{(t_{\varepsilon}\omega_{\varepsilon})^{\theta
p-1}}{\omega_{\varepsilon}^{\theta p}}dx+t_{\varepsilon}^{p_{s}^{*}-\theta
p}\int_{\mathbb{R}^{N}}|\omega_{\varepsilon}|^{p_{s}^{*}}dx.$ (4.5)

By $(F_{1})-(F_{4})$, $\omega\in\mathcal{W}_{0}$ and (4) it follows that
$t_{\varepsilon}\to 1$ as $\varepsilon\to 0$. On the other hand,

$\displaystyle c_{\varepsilon}\leq\mathop{\max}\limits_{t\geq
0}J_{\varepsilon}(t\omega_{\varepsilon})=J_{\varepsilon}(t_{\varepsilon}\omega_{\varepsilon})=J_{0}(t_{\varepsilon}\omega_{\varepsilon})+\frac{t_{\varepsilon}^{p}}{p}\int_{{\mathbb{R}^{N}}}(V(\varepsilon
x)-V_{0})\omega_{\varepsilon}^{p}dx.$

Since $V(\varepsilon x)$ is bounded on the support of $\omega_{\varepsilon}$,
by the dominated convergence theorem and the above inequality, we obtain the
thesis.

Moreover, by (3.8), (4.2) and $J_{0}(u)\leq J_{\varepsilon}(u)$ for any
$u\in\mathcal{W}_{0}$, we have $c_{V_{0}}\leq c_{\varepsilon}$. Then
$c_{V_{0}}\leq\mathop{\lim\inf}\limits_{\varepsilon\to 0}{c_{\varepsilon}}$.
Therefore, $\mathop{\lim}\limits_{\varepsilon\to
0}{c_{\varepsilon}}={c_{{V_{0}}}}$. ∎

##  5 Multiple solutions for the Modified problem

We consider $\delta>0$ such that $M_{\delta}\subset\Lambda$. Let $w$ be a
positive ground state solution of problem (4.1), and choose
$\psi\in\mathcal{C}_{c}^{\infty}(\mathbb{R}^{+},[0,1])$ satisfying $\psi(s)=1$
if $0\leq s\leq\frac{\delta}{2}$, and $\psi(s)=0$ if $s\geq\delta$.

For each $y\in M=\left\\{x\in\Lambda:V(x)=V_{0}\right\\}$, we define

${\Upsilon_{\varepsilon,y}}(x)=\psi(|\varepsilon
x-y|)w\left(\frac{{\varepsilon x-y}}{\varepsilon}\right).$

Then there exists $t_{\varepsilon}>0$ such that $\mathop{\max}\limits_{t\geq
0}{J_{\varepsilon}}(t{\Upsilon_{\varepsilon,y}})={J_{\varepsilon}}({t_{\varepsilon}}{\Upsilon_{\varepsilon,y}})$.
Define $\Phi_{\varepsilon}:M\rightarrow\mathcal{N}_{\varepsilon}$ by
$\Phi_{\varepsilon}(y)=t_{\varepsilon}\Upsilon_{\varepsilon,y}$.

######  Lemma 5.1.

The following limits hold: $\mathop{\lim}\limits_{\varepsilon\to
0}{J_{\varepsilon}(\Phi_{\varepsilon}(y))}={c_{{V_{0}}}}$ uniformly in $y\in
M$.

###### Proof.

Suppose by condition that there exist $\delta_{0}$, $y_{n}\in M$ and
$\varepsilon_{n}\to 0$ such that

$\displaystyle\left|J_{\varepsilon_{n}}(\Phi_{\varepsilon_{n}}(y_{n}))-c_{V_{0}}\right|\geq\delta_{0}.$
(5.1)

Note that for each $n\in\mathbb{N}$ and for all $z\in
B_{\frac{\delta}{\varepsilon_{n}}}(0)$, we have $\varepsilon_{n}z\in
B_{\delta}(0)$. So

$\varepsilon_{n}z+y_{n}\in B_{\delta}(0)\subset M_{\delta}\subset\Lambda.$

Using the definition of $\Phi_{\varepsilon_{n}}$ and making the change of the
variable $z=x-\frac{y_{n}}{\varepsilon_{n}}$, we have

$\displaystyle A_{n}:=$
$\displaystyle\frac{1}{{t_{{\varepsilon_{n}}}^{(\theta-1)p}}}\|\Upsilon_{\varepsilon_{n},y_{n}}\|_{\varepsilon_{n}}^{p}+b[\Upsilon_{\varepsilon_{n},y_{n}}]_{s,p}^{\theta
p}$ $\displaystyle=$ $\displaystyle\frac{1}{{t_{{\varepsilon_{n}}}^{\theta
p-1}}}\int_{\mathbb{R}^{N}}f({t_{{\varepsilon_{n}}}}\psi(|{\varepsilon_{n}}z|)w(z))\psi(|{\varepsilon_{n}}z|)w(z)dz+t_{{\varepsilon_{n}}}^{p_{s}^{*}-\theta
p}\int_{\mathbb{R}^{N}}[\psi(|{\varepsilon_{n}}z|)w(z)]^{p_{s}^{*}}dz:=B_{n}.$
(5.2)

By using Lemma 2.3, we have

$\displaystyle\|\Upsilon_{\varepsilon_{n},y_{n}}\|_{\varepsilon_{n}}^{p}\rightarrow\|w\|_{0}^{p}.$
(5.3)

And by Lebesgue’s theorem, we can verify that

$\displaystyle\int_{\mathbb{R}^{N}}f(\Upsilon_{\varepsilon_{n},y_{n}})\Upsilon_{\varepsilon_{n},y_{n}}\rightarrow\int_{\mathbb{R}^{N}}f(w)w,\
\
\int_{\mathbb{R}^{N}}|\Upsilon_{\varepsilon_{n},y_{n}}|^{p_{s}^{*}}\rightarrow\int_{\mathbb{R}^{N}}|w|^{p_{s}^{*}}.$
(5.4)

If $t_{\varepsilon_{n}}\rightarrow\infty$, then we get from (5) that

$\displaystyle
B_{n}\geq\int_{B_{\delta/2}(0)}t_{{\varepsilon_{n}}}^{p_{s}^{*}-\theta
p}[\psi(|{\varepsilon_{n}}z|)w(z)]^{p_{s}^{*}}\geq
t_{{\varepsilon_{n}}}^{p_{s}^{*}-\theta
p}k^{p_{s}^{*}}|{B_{\delta/2}(0)}|\rightarrow\infty$ (5.5)

as $\varepsilon_{n}\rightarrow 0$, where
$k:=\mathop{\inf}\limits_{x\in{B_{\delta/2}}(0)}w(x)$. However, $A_{n}$ in (5)
is bounded. This yields a contradiction. Hence, $\\{t_{\varepsilon_{n}}\\}$ is
bounded and assume that $t_{\varepsilon_{n}}\rightarrow t_{0}\geq 0$. Since
$t_{\varepsilon_{n}}\Upsilon_{\varepsilon_{n},y_{n}}\in\mathcal{N}_{\varepsilon_{n}}$,
we can easily know
$t_{\varepsilon_{n}}\|\Upsilon_{\varepsilon_{n},y_{n}}\|_{{\varepsilon_{n}}}\geq\rho>0$.
With the use of (5.3), we have $t_{0}>0$. Since $t_{\varepsilon_{n}}\to
t_{0}$, passing limit in (5) we have

$\displaystyle\frac{1}{{t_{0}^{p}}}\|w\|_{0}^{p}+b[w]_{s,p}^{\theta
p}=\int_{\mathbb{R}^{N}}\frac{f(t_{0}w)}{{(t_{0}w)^{\theta p-1}}}w^{\theta
p}dz+t_{0}^{p_{s}^{*}-\theta p}\int_{\mathbb{R}^{N}}|w|^{p_{s}^{*}}dz.$

Observing that $w$ is a positive ground state solution of (4.1), then by
$(F_{4})$ we know that $t_{0}=1$. Letting $n\to\infty$ and recalling
$t_{\varepsilon_{n}}\rightarrow 1$, using (5.3) and (5.4) we get

$\displaystyle\mathop{\lim}\limits_{n\to\infty}{J_{{\varepsilon_{n}}}}({\Phi_{{\varepsilon_{n}}}})={J_{0}}(w)={c_{{V_{0}}}},$

which contradicts with (5.1). The ends the proof. ∎

For any $\delta>0$, let $\rho=\rho(\delta)>0$ be such that $M_{\delta}\subset
B_{\rho}(0)$. Let $\chi:\mathbb{R}^{N}\rightarrow\mathbb{R}^{N}$. Define
$\chi(x)=x$ if $|x|<\rho$ and $\chi(x)=\frac{\rho x}{|x|}$ if $|x|\geq\rho$.
Finally, we define
$\beta_{\varepsilon}:\mathcal{N}_{\varepsilon}\rightarrow\mathbb{R}^{N}$ by

${\beta_{\varepsilon}}(u):=\frac{\int_{\mathbb{R}^{N}}\chi(\varepsilon
x)|u(x)|^{p}dx}{\int_{\mathbb{R}^{N}}|u(x)|^{p}dx}.$

Since $M\subset B_{\rho}(0)$, by the definition of $\chi$ and Lebesgue’s
theorem, arguing as in Lemma $3.13$ in [6], we conclude that

$\mathop{\lim}\limits_{\varepsilon\to
0}{\beta_{\varepsilon}(\Phi_{\varepsilon}(y))}=y\ \ \ \mbox{uniformly}\
\mbox{for}\ \ y\in M.$ (5.6)

######  Lemma 5.2.

Let $\varepsilon_{n}\rightarrow 0$ and
$\\{u_{n}\\}\subset\mathcal{N}_{\varepsilon_{n}}$ be such that
$J_{\varepsilon_{n}}(u_{n})\rightarrow c_{V_{0}}$. Then there exists a
sequence $\\{y_{n}\\}\subset\mathbb{R}^{N}$ such that
$v_{n}(x)=u_{n}(x+y_{n})$ has a convergent subsequence in $\mathcal{W}_{0}$.
Moreover, up to a subsequence, ${{\tilde{y}}_{n}}:={\varepsilon_{n}}{y_{n}}\to
y_{0}\in M$.

###### Proof.

Arguing as in the proof of Lemma 3.4, $\\{u_{n}\\}$ is bounded in
$\mathcal{W}_{0}$ and non-vanishing. Then, there exist $R,\delta^{\prime}>0$
and $y_{n}\in\mathbb{R}^{N}$ such that
$\int_{B_{R}(y_{n})}u_{n}^{p}dx>\delta^{\prime}$. Define
$v_{n}(x)=u_{n}(x+y_{n})$, then passing to a subsequence, we assume
$v_{n}\rightharpoonup v\neq 0$ in $\mathcal{W}_{0}$. Let $t_{n}>0$ be such
that ${\tilde{v}_{n}}={t_{n}}{v_{n}}\in\mathcal{N}_{0}$. Set
${{\tilde{y}}_{n}}:={\varepsilon_{n}}{y_{n}}$. Then

$c_{V_{0}}\leq
J_{0}({\tilde{v}_{n}})=J_{0}({t_{n}}{v_{n}})=J_{0}({t_{n}}{u_{n}})\leq
J_{\varepsilon_{n}}({t_{n}}{u_{n}})\leq
J_{\varepsilon_{n}}({u_{n}})=c_{V_{0}}+o_{n}(1).$

Then $J_{0}({\tilde{v}_{n}})\rightarrow c_{V_{0}}$. Moreover,
$\\{\tilde{v}_{n}\\}$ is bounded in $\mathcal{W}_{0}$ and
$\tilde{v}_{n}\rightharpoonup\tilde{v}$. We assume that $t_{n}\to t^{*}>0$.
From the uniqueness of the weak limit we have that $\tilde{v}=t^{*}v\not\equiv
0$. This combined with Lemma 4.1 implies that
$\tilde{v}_{n}\rightarrow\tilde{v}$ in $\mathcal{W}_{0}$. Consequently,
$v_{n}\to v$ in $\mathcal{W}_{0}$ as $n\rightarrow\infty$.

Next, we claim that $\\{{\tilde{y}}_{n}\\}$ is bounded. Indeed, suppose by
contradiction that $|{\tilde{y}}_{n}|\to\infty$. Choose $R>0$ such that
$\Lambda\subset B_{R}(0)$, then for $n$ large enough, we have
$|{\tilde{y}}_{n}|>2R$, and for each $x\in B_{\frac{R}{\varepsilon_{n}}}(0)$,
we have

$|\varepsilon_{n}x+{\tilde{y}}_{n}|>|{\tilde{y}}_{n}|-|\varepsilon_{n}x|\geq
2R-R=R.$

Hence, using $v_{n}\rightarrow v$ in $\mathcal{W}_{0}$ and Lebesgue’s theorem,
we obtain

$\displaystyle a[v_{n}]_{s,p}^{p}+\int_{{\mathbb{R}^{N}}}V_{0}v_{n}^{p}dx\leq$
$\displaystyle
a[v_{n}]_{s,p}^{p}+\int_{{\mathbb{R}^{N}}}V(\varepsilon_{n}x+{\tilde{y}}_{n})v_{n}^{p}dx+b[v_{n}]_{s,p}^{\theta
p}$ $\displaystyle=$
$\displaystyle\int_{B_{\frac{R}{\varepsilon_{n}}}(0)}g(\varepsilon_{n}x+{\tilde{y}}_{n},v_{n})v_{n}dx+\int_{{\mathbb{R}^{N}}\setminus{B_{\frac{R}{\varepsilon_{n}}}(0)}}g(\varepsilon_{n}x+{\tilde{y}}_{n},v_{n})v_{n}dx$
$\displaystyle\leq$
$\displaystyle\frac{V_{0}}{K}\int_{\mathbb{R}^{N}}v_{n}^{p}dx+\int_{{\mathbb{R}^{N}}\setminus{B_{\frac{R}{\varepsilon_{n}}}(0)}}\left(f(v_{n})v_{n}+v_{n}^{p_{s}^{*}}\right)dx$
$\displaystyle\leq$
$\displaystyle\frac{V_{0}}{K}\int_{\mathbb{R}^{N}}v_{n}^{p}dx+o_{n}(1).$

Then $v_{n}\to 0$ in $\mathcal{W}_{0}$, which contradicts to $v\not\equiv 0$.
Therefore, $\\{{\tilde{y}}_{n}\\}$ is bounded and we may assume that
${\tilde{y}}_{n}\to y_{0}\in\mathbb{R}^{N}$. If $y_{0}\notin{\bar{\Lambda}}$,
as before we can infer that $v_{n}\to 0$ in $\mathcal{W}_{0}$. This is
impossible. Thus we have that $y_{0}\in{\bar{\Lambda}}$.

In order to prove that $V(y_{0})=V_{0}$, we can suppose by contradiction that
$V_{0}<V(y_{0})$. Then from $\tilde{v}_{n}\rightarrow\tilde{v}$ in
$\mathcal{W}_{0}$ and the invariance of $\mathbb{R}^{N}$ by translation, to
obtain

$\displaystyle c_{V_{0}}=$ $\displaystyle J_{0}(\tilde{v})$ $\displaystyle<$
$\displaystyle\mathop{\lim\inf}\limits_{n\to\infty}\left\\{\frac{a}{p}[\tilde{v}_{n}]_{s,p}^{p}+\frac{1}{p}\int_{{\mathbb{R}^{N}}}V(\varepsilon_{n}(x+{y}_{n}))|\tilde{v}_{n}|^{p}dx+\frac{b}{\theta
p}[\tilde{v}_{n}]_{s,p}^{\theta
p}-\int_{{\mathbb{R}^{N}}}F(\tilde{v}_{n})dx-\frac{1}{p_{s}^{*}}|\tilde{v}_{n}^{+}|^{p_{s}^{*}}dx\right\\}$
$\displaystyle\leq$
$\displaystyle\mathop{\lim\inf}\limits_{n\to\infty}\left\\{\frac{at_{n}^{p}}{p}[u_{n}]_{s,p}^{p}+\frac{t_{n}^{p}}{p}\int_{{\mathbb{R}^{N}}}V(\varepsilon_{n}z)|u_{n}|^{p}dx+\frac{bt_{n}^{\theta
p}}{\theta p}[u_{n}]_{s,p}^{\theta
p}-\int_{{\mathbb{R}^{N}}}G(\varepsilon_{n}z,t_{n}u_{n})dx\right\\}$
$\displaystyle=$
$\displaystyle\mathop{\lim\inf}\limits_{n\to\infty}J_{\varepsilon_{n}}(t_{n}u_{n})\leq\mathop{\lim\inf}\limits_{n\to\infty}J_{\varepsilon_{n}}(u_{n})=c_{V_{0}},$

which yields a contradiction. Therefore, $V(y_{0})=V_{0}$ and $z_{0}\in M$.
The condition $(V_{2})$ shows that $y_{0}\notin\partial M$, so, $y_{0}\in M$.
The proof is complete. ∎

Let $h(\varepsilon)$ be any positive function satisfying $h(\varepsilon)\to 0$
as $\varepsilon\to 0$. Define the set

$\Sigma_{\varepsilon}=\left\\{u\in\mathcal{N}_{\varepsilon}:J_{\varepsilon}(u)\leq
c_{V_{0}}+h(\varepsilon)\right\\}.$

For any $y\in M$, we deduce from Lemma 5.1 that
$h(\varepsilon)=\left|J_{\varepsilon}(\Phi_{\varepsilon}(y)))-c_{V_{0}}\right|\rightarrow
0$ as $\varepsilon\rightarrow 0$. Thus
$\Phi_{\varepsilon}(y)\in\Sigma_{\varepsilon}$ and
$\Sigma_{\varepsilon}\neq\emptyset$ for $\varepsilon>0$. Moreover, as in Lemma
$3.14$ in [6], we have the following lemma

######  Lemma 5.3.

For any $\delta>0$, there holds that $\mathop{\lim}\limits_{\varepsilon\to
0}\mathop{\sup}\limits_{u\in\Sigma_{\varepsilon}}dist({\beta_{\varepsilon}}(u)-{M_{\delta}})=0$.

######  Lemma 5.4.

[12] Let $I$ be a $C^{1}$-functional defined on a $C^{1}$-Finsler manifold
$\nu$. If $I$ is bounded from below and satisfies the $(PS)$ condition, then
$I$ has at least $cat_{\nu}(\nu)$ distinct critical points.

######  Lemma 5.5.

[9] Let $\Gamma$, $\Omega^{+}$, $\Omega^{-}$ be closed sets with
$\Omega^{-}\subset\Omega^{+}$. Let $\Phi:\Omega^{-}\rightarrow\Gamma$,
$\beta:\Gamma\rightarrow\Omega^{+}$ be two continuous maps such that
$\beta\circ\Phi$ is homotopically equivalent to the embedding
$Id:\Omega^{-}\rightarrow\Omega^{+}$. Then $cat_{\Gamma}(\Gamma)\geq
cat_{\Omega^{+}}(\Omega^{-})$.

######  Proposition 5.6.

For any $\delta>0$ such that $M_{\delta}\subset\Lambda$, there exists
${{\bar{\varepsilon}}_{\delta}}>0$ such that, for any
$\varepsilon\in(0,{{\bar{\varepsilon}}_{\delta}})$, problem (3.1) has at least
$cat_{M_{\delta}}(M)$ positive solutions.

###### Proof.

Given $\delta>0$ such that $M_{\delta}\subset\Lambda$, we can use Lemma 5.1,
Lemma 5.3 and (5.6) to obtain ${{\bar{\varepsilon}}_{\delta}}>0$ such that for
any $\varepsilon\in(0,{{\bar{\varepsilon}}_{\delta}})$, the diagram

$M\xrightarrow{{\Phi_{\varepsilon}}}{\Sigma_{\varepsilon}}\xrightarrow{\beta_{\varepsilon}}{M_{\delta}}$

is well defined. In view of (5.6), for $\varepsilon$ small enough, we can
denote by $\beta_{\varepsilon}(\Phi_{\varepsilon}(y))=y+\theta(y)$ for $y\in
M$, where $|\theta(y)|<\delta^{\prime}/2$ uniformly for $y\in M$. Define
$S(t,y)=y+(1-t)\theta(y)$. Thus $S:[0,1]\times M\rightarrow M_{\delta}$ is
continuous. Obviously, $S(0,y)=\beta_{\varepsilon}(\Phi_{\varepsilon}(y))$ and
$S(1,y)=y$ for all $y\in M$. That is,
$\beta_{\varepsilon}\circ{{\Phi_{\varepsilon}}}$ is homotopically equivalent
to $Id:M\rightarrow M_{\delta}$. By Lemma 5.5, we obtain that

$\mathop{cat}\limits_{\Sigma_{\varepsilon}}({\Sigma_{\varepsilon}})\geq\mathop{cat}\limits_{M_{\delta}}({M_{\delta}}).$

Since $c_{V_{0}}\leq c_{0}$, we can use the definition of
$\Sigma_{\varepsilon}$ and Proposition 3.9 to conclude that $J_{\varepsilon}$
satisfies $(PS)$ condition in $\Sigma_{\varepsilon}$ for all small
$\varepsilon>0$. Therefore, Lemma 5.4 proves at least
$\mathop{cat}\limits_{\Sigma_{\varepsilon}}({\Sigma_{\varepsilon}})$ critical
points of $J_{\varepsilon}$ restricted to ${\Sigma_{\varepsilon}}$. Using the
same arguments as in the proof of Proposition 3.9, we can conclude that a
critical point of the functional $J_{\varepsilon}$ on
$\mathcal{N}_{\varepsilon}$ is $\mathcal{W}_{\varepsilon}$ and therefore a
week solution for the problem the theorem is proved. ∎

##  6 Proof of the main result

In the last section, we provide the proof of our main result. Firstly, we
develop a iteration scheme [33] which will be the main key to deduce that the
solutions to (3.4) are indeed solutions to (3.1).

######  Lemma 6.1.

Let $\varepsilon_{n}\to 0$ and $\\{u_{n}\\}\in\mathcal{N}_{\varepsilon_{n}}$
be a solution to (3.4). Then $v_{n}(x)=u_{n}(\cdot+y_{n})$ satisfies $v_{n}\in
L^{\infty}(\mathbb{R}^{N})$ and there exists $C>0$ such that

$|v_{n}|_{\infty}\leq C\ \ \text{for}\ \text{all}\ n\in\mathbb{N},$

where $y_{n}$ was given by Lemma 5.2. Moreover

$\mathop{\lim}\limits_{|x|\to\infty}{v_{n}}(x)=0\ \ \text{uniformly}\
\text{in}\ n\in\mathbb{N}.$

###### Proof.

For any $L>0$ and $\beta>1$, let us define the function

$\gamma(v_{n}):=\gamma_{L,\beta}(v_{n})=v_{n}v_{L,n}^{p(\beta-1)}\in\mathcal{W}_{\varepsilon},$

where $v_{L,n}=\min\\{L,v_{n}\\}$. Since $\gamma$ is an increasing function,
we have

$(a-b)(\gamma(a)-\gamma(b))\geq 0\ \ \text{for}\ \text{any}\ \
a,b\in\mathbb{R}.$

Let us introduce the following functions

$\mathcal{Q}(t):=\frac{|t|^{p}}{p},\qquad\text{and}\qquad\Gamma(t):=\int_{0}^{t}(\gamma^{\prime}(\tau))^{\frac{1}{p}}d\tau.$

Then, applying Jensen’s inequality we get for all $a,b\in\mathbb{R}$ such that
$a>b$

$\displaystyle\mathcal{Q}^{\prime}(a-b)(\gamma(a)-\gamma(b))=$
$\displaystyle(a-b)^{p-1}(\gamma(a)-\gamma(b))=(a-b)^{p-1}\int_{a}^{b}\gamma^{\prime}(t)dt$
$\displaystyle=$
$\displaystyle(a-b)^{p-1}\int_{a}^{b}(\Gamma^{\prime}(t))^{p}dt\geq\left(\int_{a}^{b}(\Gamma^{\prime}(t))dt\right)^{p}.$

The same argument works when $a\leq b$. Therefore

$\displaystyle\mathcal{Q}^{\prime}(a-b)(\gamma(a)-\gamma(b))\geq\left|\Gamma(a)-\Gamma(b)\right|^{p}\
\ \text{for}\ \text{any}\ \ a,b\in\mathbb{R}.$ (6.1)

From (6.1), we can see that

$\displaystyle\left|\Gamma(v_{n})(x)-\Gamma(v_{n})(y)\right|^{p}\leq|v_{n}(x)-v_{n}(y)|^{p-2}(v_{n}(x)-v_{n}(y))\left((v_{n}v_{L,n}^{p(\beta-1)})(x)-(v_{n}v_{L,n}^{p(\beta-1)})(y)\right).$
(6.2)

Choosing $\gamma(v_{n})=v_{n}v_{L,n}^{p(\beta-1)}$ as a test function in (3.4)
and using (6.2) we get

$\displaystyle
a[\Gamma(v_{n})]_{s,p}^{p}+\int_{\mathbb{R}^{N}}V_{n}(x)|v_{n}|^{p}v_{L,n}^{p(\beta-1)}dx$
$\displaystyle\leq$
$\displaystyle(a+b[v_{n}]_{s,p}^{(\theta-1)p})\iint_{\mathbb{R}^{2N}}\frac{{|{{{v}}_{n}}(x)-{v_{n}}(y){|^{p-2}}({{{v}}_{n}}(x)-{v_{n}}(y))}}{{|x-y{|^{N+sp}}}}\left({({v_{n}}v_{L,n}^{p(\beta-1)})(x)-({v_{n}}v_{L,n}^{p(\beta-1)})(y)}\right)dxdy$
$\displaystyle+\int_{\mathbb{R}^{N}}V_{n}(x)|v_{n}|^{p}v_{L,n}^{p(\beta-1)}dx$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{N}}g_{n}(v_{n})v_{n}v_{L,n}^{p(\beta-1)}dx,$
(6.3)

where we use the notations $V_{n}(x):=V(\varepsilon_{n}(x+y_{n}))$ and
$g_{n}(v_{n}):=g(\varepsilon_{n}(x+y_{n}),v_{n})$. Observing that

$\displaystyle\Gamma(v_{n})\geq\frac{1}{\beta}v_{n}v_{L,n}^{\beta-1}$

and by Lemma 2.1, we have

$\displaystyle[\Gamma(v_{n})]_{s,p}^{p}\geq
C_{*}^{-1}|\Gamma(v_{n})|_{p_{s}^{*}}^{p}\geq\left(\frac{1}{\beta}\right)^{p}C_{*}^{-1}\|v_{n}v_{L,n}^{\beta-1}\|_{p_{s}^{*}}^{p}.$
(6.4)

On the other hand, by $(G_{1})$ and $(G_{2})$, for any $\xi>0$, there exists
$C_{\xi}>0$ such that

$\displaystyle|g_{n}(v_{n})|\leq\xi|v_{n}|^{p-1}+C_{\xi}|v_{n}|^{p_{s}^{*}-1}.$
(6.5)

Thus, taking $\xi\in(0,V_{1})$, using (6.4) and (6.5), we can see that (6)
yields

$\displaystyle\|w_{L,n}\|_{p_{s}^{*}}^{p}\leq
C\beta^{p}\int_{\mathbb{R}^{N}}|v_{n}|^{p_{s}^{*}}v_{L,n}^{p(\beta-1)}dx,$
(6.6)

where $w_{L,n}=v_{n}v_{L,n}^{\beta-1}$ . Now, we take
$\beta=\frac{p_{s}^{*}}{p}$ and fix $R>0$. Noting that $0\leq v_{L,n}\leq
v_{n}$, we can infer that

$\displaystyle\int_{\mathbb{R}^{N}}|v_{n}|^{p_{s}^{*}}v_{L,n}^{p(\beta-1)}dx=$
$\displaystyle\int_{\mathbb{R}^{N}}|v_{n}|^{p_{s}^{*}-p}|v_{n}|^{p}v_{L,n}^{p_{s}^{*}-p}dx$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{N}}|v_{n}|^{p_{s}^{*}-p}(v_{n}v_{L,n}^{\frac{p_{s}^{*}-p}{p}})^{p}dx$
$\displaystyle\leq$
$\displaystyle\int_{v_{n}<R}R^{p_{s}^{*}-p}v_{n}^{p_{s}^{*}}dx+\int_{v_{n}>R}|v_{n}|^{p_{s}^{*}-p}(v_{n}v_{L,n}^{\frac{p_{s}^{*}-p}{p}})^{p}dx$
$\displaystyle\leq$
$\displaystyle\int_{v_{n}<R}R^{p_{s}^{*}-p}v_{n}^{p_{s}^{*}}dx$
$\displaystyle+\left(\int_{v_{n}>R}|v_{n}|^{p_{s}^{*}}dx\right)^{{\frac{p_{s}^{*}-p}{p_{s}^{*}}}}\left(\int_{v_{n}>R}(v_{n}v_{L,n}^{\frac{p_{s}^{*}-p}{p}})^{p_{s}^{*}}\right)^{\frac{p}{p_{s}^{*}}}.$
(6.7)

Since $\\{v_{n}\\}$ is bounded in $L^{p_{s}^{*}}(\mathbb{R}^{N})$, we can see
that for any $R$ sufficiently large

$\displaystyle\left(\int_{v_{n}>R}|v_{n}|^{p_{s}^{*}}dx\right)^{{\frac{p_{s}^{*}-p}{p_{s}^{*}}}}\leq\frac{1}{2C\beta^{p}}$
(6.8)

putting together (6.6), (6) and (6.8), we get

$\displaystyle\left(\int_{v_{n}>R}(v_{n}v_{L,n}^{\frac{p_{s}^{*}-p}{p}})^{p_{s}^{*}}\right)^{\frac{p}{p_{s}^{*}}}\leq
C\beta^{p}R^{p_{s}^{*}-p}\int_{\mathbb{R}^{N}}v_{n}^{p_{s}^{*}}dx<\infty,$

and taking the limiting as $L\to\infty$, we obtain $v_{n}\in
L^{\frac{(p_{s}^{*})^{2}}{p}}({\mathbb{R}^{N}})$.

Now, noting that $0\leq v_{L,n}\leq v_{n}$ and letting $L\to\infty$ in (6.6)
we have $|v_{n}|_{\beta p_{s}^{*}}^{\beta p}\leq
C\beta^{p}\int_{\mathbb{R}^{N}}v_{n}^{p_{s}^{*}+p(\beta-1)}dx$, from which we
deduce that

$\displaystyle\left(\int_{\mathbb{R}^{N}}v_{n}^{\beta
p_{s}^{*}}dx\right)^{\frac{1}{p_{s}^{*}(\beta-1)}}\leq
C\beta^{\frac{1}{\beta-1}}\left(\int_{\mathbb{R}^{N}}v_{n}^{p_{s}^{*}+p(\beta-1)}dx\right)^{\frac{1}{p(\beta-1)}}.$

From $m\geq 1$, we define $\beta_{m+1}$ inductively so that
$p_{s}^{*}+p(\beta_{m+1}-1)=p_{s}^{*}\beta_{m}$ and
$\beta_{1}=\frac{p_{s}^{*}}{p}$. Then, we have

$\displaystyle\left(\int_{\mathbb{R}^{N}}v_{n}^{\beta_{m+1}p_{s}^{*}}dx\right)^{\frac{1}{p_{s}^{*}(\beta_{m+1}-1)}}\leq
C\beta_{m+1}^{\frac{1}{\beta_{m+1}-1}}\left(\int_{\mathbb{R}^{N}}v_{n}^{p_{s}^{*}\beta_{m}}dx\right)^{\frac{1}{p_{s}^{*}(\beta_{m}-1)}}.$

Set
$D_{m}:=\left(\int_{\mathbb{R}^{N}}v_{n}^{p_{s}^{*}\beta_{m}}dx\right)^{\frac{1}{p_{s}^{*}(\beta_{m}-1)}}$.
Using a standard iteration argument, we can find $C_{0}>0$ independent of $m$
such that

$\displaystyle{D_{m+1}}\leq\prod\limits_{k=1}^{m}{{{(C{\beta_{k+1}})}^{\frac{1}{{{\beta_{k+1}}-1}}}}}{D_{1}}\leq{C_{0}}{D_{1}}.$

Passing to the limit as $m\to\infty$ we get $|v_{n}|_{\infty}\leq K$ for all
$n\in\mathbb{N}$.

Now, we note that $v_{n}$ is solution to

$\displaystyle(-\Delta)_{p}^{s}v_{n}=\left[g(\varepsilon_{n}(x+y_{n}),v_{n})-V(\varepsilon_{n}(x+y_{n}))v_{n}^{p-1}\right](a+b[v_{n}]_{s,p}^{(\theta-1)p})^{-1}=:h_{n}\
\ \rm{in}\ \ \mathbb{R}^{N}.$

Moreover, $h_{n}\in L^{\infty}(\mathbb{R}^{N})$ and $|v_{n}|_{\infty}\leq C$
for all $n\in\mathbb{N}$. Indeed, this last inequality is a consequence of the
growth assumption on $g$. Then, $|v_{n}|_{\infty}\leq C$ and $a\leq
a+b[v_{n}]_{s,p}^{(\theta-1)p}\leq C$ for all $n\in\mathbb{N}$. Then using
Corollary $5.5$ in [26], we can deduce that
$v_{n}\in\mathcal{C}^{0,\alpha}(\mathbb{R}^{N})$ for some $\alpha>0$. Since
$v_{n}\to v$ in $\mathcal{W}_{0}$ (see Lemma 5.2), we can infer that

$\mathop{\lim}\limits_{|x|\to\infty}{v_{n}}(x)=0\ \ \text{uniformly}\
\text{in}\ n\in\mathbb{N}.$

∎

Now, we give the proof of Theorem 1.2.

Proof of Theorem 1.2. We begin by proving that there exists
$\varepsilon_{0}>0$ such that for any $\varepsilon\in(0,\varepsilon_{0})$ and
any mountain pass solution $u_{\varepsilon}\in\mathcal{W}_{\varepsilon}$ of
(3.4), it holds

$\displaystyle|u_{\varepsilon}|_{L^{\infty}(\mathbb{R}^{N})\setminus{\Lambda_{\varepsilon}}}<a.$
(6.9)

Assume by contradiction that for some subsequence $\\{\varepsilon_{n}\\}$ such
that $\varepsilon_{n}\to 0$, we find
$u_{n}:=u_{\varepsilon_{n}}\in\mathcal{W}_{\varepsilon_{n}}$ such that
$J_{\varepsilon_{n}}(u_{n})=c_{\varepsilon_{n}}$,
$J^{\prime}_{\varepsilon_{n}}(u_{n})=0$ and

$\displaystyle|u_{\varepsilon}|_{L^{\infty}({\mathbb{R}^{N}}\setminus{\Lambda_{\varepsilon_{n}}})}\geq
a.$ (6.10)

From Lemma 5.2, there exists $\\{y_{n}\\}\subset\mathbb{R}^{N}$ such that
$v_{n}=u_{n}(\cdot+y_{n})\to v$ in $\mathcal{W}_{0}$ and
$\varepsilon_{n}y_{n}\to y_{0}$ for $y_{0}\in\Lambda$ such that
$V(y_{0})=V_{0}$.

Now, if we choose $r>0$ such that $B_{r}(y_{0})\subset
B_{2r}(y_{0})\subset\Lambda$, we can see that
$B_{\frac{r}{\varepsilon_{n}}}\left(\frac{y_{0}}{\varepsilon_{n}}\right)\subset\Lambda_{\varepsilon_{n}}$.
Then, for any $y\in B_{\frac{r}{\varepsilon_{n}}}(y_{n})$, it holds
$|y-\frac{y_{0}}{\varepsilon_{n}}|\leq|y-y_{n}|+|y_{n}-\frac{y_{0}}{\varepsilon_{n}}|<\frac{1}{\varepsilon_{n}}(r+o_{n}(1))<\frac{2r}{\varepsilon_{n}}$
for $n$ sufficiently large. Hence, for there values of $n$ we have

$\displaystyle{\mathbb{R}^{N}}\setminus{\Lambda_{\varepsilon_{n}}}\subset{\mathbb{R}^{N}}\setminus{B_{\frac{r}{\varepsilon_{n}}}(y_{n})}$
(6.11)

for any $n$ big enough. Using Lemma 6.1, we can see that

$\displaystyle v_{n}(x)\to 0\ \ \mbox{as}\ \ |x|\to\infty\ \ \mbox{uniformly}\
\mbox{in}\ n\in\mathbb{N}.$ (6.12)

Therefore, there exists $R>0$ such that $v_{n}(x)<a$ for $|x|\geq R$ $\forall
n\in\mathbb{N}$, from which $u_{n}(x)=v_{n}(x-y_{n})<a$ for
$x\in\mathbb{R}^{N}\setminus B_{R}(y_{n})$ and $\forall n\in\mathbb{N}$. On
the other hand, by (6.11), there exists $\nu\in\mathbb{N}$ such that for any
$n\geq\nu$, it holds
${\mathbb{R}^{N}}\setminus{\Lambda_{\varepsilon_{n}}}\subset{\mathbb{R}^{N}}\setminus{B_{\frac{r}{\varepsilon_{n}}}(y_{n})}\subset\mathbb{R}^{N}\setminus
B_{R}(y_{n})$, which gives $u_{n}(x)<a$, $\forall
x\in{\mathbb{R}^{N}}\setminus{\Lambda_{\varepsilon_{n}}}$. This last fact
contradicts (6.10) and thus (6.9) is verified.

Now, let $u_{\varepsilon}$ be a nonnegative solution to (3.4), since
$u_{\varepsilon}$ satisfies (6.9) for any $\varepsilon\in(0,\varepsilon_{0})$,
it follows from the definition $g$ that $u_{\varepsilon}$ is a solution to
(3.4), and $v_{\varepsilon}=u_{\varepsilon}(x/\varepsilon)$ is a solution to
(1.1) for any $\varepsilon\in(0,\varepsilon_{0})$.

Finally, we study the behavior of maximum points of solutions to problem
(3.4). Taking $\varepsilon_{n}\to 0^{+}$ and consider a sequence
$\\{u_{\varepsilon_{n}}\\}\subset\mathcal{W}_{\varepsilon_{n}}$ of solution to
$(\ref{1.2})$. We first notice that, by $(G_{1})$, there exists
$\gamma\in(0,a)$ such that

$\displaystyle
g(\varepsilon_{n}x,t)t=f(t)t+t^{p_{s}^{*}}\leq\frac{V_{1}}{K}t^{p}\ \
\text{for}\ \text{any}\ \ x\in{\mathbb{R}^{N}},\ \ 0\leq t\leq\gamma.$ (6.13)

Arguing as before, we can find $R>0$

$\displaystyle|u_{\varepsilon_{n}}|_{L^{\infty}({\mathbb{R}^{N}}\setminus{B_{R}(y_{n})})}<\gamma.$
(6.14)

Moreover, up to subsequence, we may assume that

$\displaystyle|u_{\varepsilon_{n}}|_{L^{\infty}({B_{R}(y_{n})})}\geq\gamma.$
(6.15)

Indeed, if (6.15) does not hold, in view of (6.14) we can see that
$|u_{\varepsilon_{n}}|_{\infty}<\gamma$. Then, from
$\left\langle{{J^{\prime}_{\varepsilon}}({u_{\varepsilon_{n}}}),{u_{\varepsilon_{n}}}}\right\rangle=o_{n}(1)$
and (6.13)

$\|u_{\varepsilon_{n}}\|_{\varepsilon_{n}}^{p}\leq\|u_{\varepsilon_{n}}\|_{\varepsilon_{n}}^{p}+b[u_{\varepsilon_{n}}]_{s,p}^{\theta
p}=\int_{\mathbb{R}^{N}}g(\varepsilon_{n}x,u_{\varepsilon_{n}})u_{\varepsilon_{n}}dx\leq\frac{V_{1}}{K}\int_{\mathbb{R}^{N}}u_{\varepsilon_{n}}^{p},$

which yields $\|u_{\varepsilon_{n}}\|_{\varepsilon_{n}}=0$, and this is a
contradiction. Hence, (6.15) holds true. In the light of (6.14) and (6.15), we
can deduce that the maximum points $p_{n}\in{\mathbb{R}^{N}}$ of
$u_{\varepsilon_{n}}$ belongs to $B_{R}(y_{n})$. Thus, $p_{n}=y_{n}+q_{n}$ for
some $q_{n}\in B_{R}$. Consequently,
$\eta_{\varepsilon_{n}}={\varepsilon_{n}}y_{n}+\varepsilon_{n}q_{n}$ is the
maximum point of
$v_{\varepsilon_{n}}=u_{\varepsilon_{n}}(x/{\varepsilon_{n}})$, we conclude
that the maximum point $\eta_{\varepsilon_{n}}$ of $v_{\varepsilon_{n}}$ is
given by $\eta_{\varepsilon_{n}}:=\varepsilon_{n}y_{n}+\varepsilon_{n}q_{n}$.
Since $\\{q_{n}\\}\subset B_{R}$ is bounded and $\varepsilon_{n}y_{n}\to
y_{0}$ with $V(y_{0})=V_{0}$, from the continuity of $V$, we can deduce that
$\mathop{\lim}\limits_{n\to\infty}V(\eta_{\varepsilon_{n}})=V(y_{0})=V_{0}$.

Next, we give a decay estimate for $v_{\varepsilon_{n}}$. For this purpose,
using Lemma $7.1$ in [17], we can find a continuous positive function $w$ and
a constant $C>0$ such that for large $|x|>R_{0}$ it holds that

$\displaystyle 0<w(x)\leq\frac{C}{1+|x|^{N+sp}}$ (6.16)

and

$\displaystyle(-\Delta)_{p}^{s}w+\frac{V_{1}}{2(a+bA_{1}^{(\theta-1)p})}w^{p-1}\geq
0\ \ \rm{in}\ \ \mathbb{R}^{N}\setminus B_{R_{1}},$ (6.17)

for some suitable $R_{1}>0$, and $A_{1}>0$ is such that
$a+b[v_{n}]_{s,p}^{(\theta-1)p}\leq a+bA_{1}^{(\theta-1)p}$ for all
$n\in\mathbb{N}$. Using $(G_{1})$ and (6.12), we can find $R_{2}>0$
sufficiently large such that

$\displaystyle(-\Delta)_{p}^{s}{v_{n}}+\frac{V_{1}}{2(a+bA_{1}^{(\theta-1)p})}{v_{n}}^{p-1}\leq$
$\displaystyle(-\Delta)_{p}^{s}{v_{n}}+\frac{V_{1}}{2(a+b[v_{n}]_{s,p}^{(\theta-1)p})}{v_{n}}^{p-1}$
$\displaystyle=$
$\displaystyle\frac{1}{2(a+b[v_{n}]_{s,p}^{(\theta-1)p})}\left[g(\varepsilon_{n}x+\varepsilon_{n}y_{n},v_{n})-(V(\varepsilon_{n}x+\varepsilon_{n}y_{n})-\frac{V_{1}}{2}){v_{n}}^{p-1}\right]$
$\displaystyle\leq$
$\displaystyle\frac{1}{2(a+b[v_{n}]_{s,p}^{(\theta-1)p})}\left[g(\varepsilon_{n}x+\varepsilon_{n}y_{n},v_{n})--\frac{V_{1}}{2}{v_{n}}^{p-1}\right]$
$\displaystyle\leq$ $\displaystyle 0,\ \ \mbox{in}\ \
\mathbb{R}^{N}\setminus{B_{R_{2}}}.$ (6.18)

Thanks to the continuity of $v_{n}$ and $w$. there exists $C_{1}>0$ such that

$\psi_{n}:=v_{n}-C_{1}w\leq 0\ \ \mbox{for}\ \ |x|=R_{3},$

where $R_{3}:=\max\\{R_{1},R_{2}\\}$ .

Taking $\phi_{n}=\max\\{\psi_{n},0\\}\in W^{s,p}(B_{R}^{c})$ as a test
function in (6) and using (6.17) with $\tilde{w}=C_{1}w$, we can deduce that

$\displaystyle 0\geq$
$\displaystyle\iint_{\mathbb{R}^{2N}}\frac{{|{v_{n}}(x)-{v_{n}}(y){|^{p-2}}({v_{n}}(x)-{v_{n}}(y))({\phi_{n}}(x)-{\phi_{n}}(y))}}{{|x-y{|^{N+sp}}}}dxdy$
$\displaystyle+\frac{V_{1}}{2(a+bA_{1}^{(\theta-1)p})}\int_{\mathbb{R}^{N}}v_{n}^{p-1}\phi_{n}dx$
$\displaystyle\geq$
$\displaystyle\iint_{\mathbb{R}^{2N}}\frac{\mathcal{G}_{n}(x,y)}{{|x-y{|^{N+sp}}}}({\phi_{n}}(x)-{\phi_{n}}(y))dxdy+\frac{V_{1}}{2(a+bA_{1}^{(\theta-1)p})}\int_{\mathbb{R}^{N}}[v_{n}^{p-1}-{{\tilde{w}}^{p-1}}]\phi
dx,$ (6.19)

where
${\mathcal{G}_{n}(x,y)}:={|{v_{n}}(x)-{v_{n}}(y){|^{p-2}}({v_{n}}(x)-{\tilde{w}}(y))}-{|{\tilde{w}}(x)-{\tilde{w}}(y){|^{p-2}}({\tilde{w}}(x)-{\tilde{w}}(y))}$.
Therefore, if we prove that

$\displaystyle\iint_{\mathbb{R}^{2N}}\frac{\mathcal{G}_{n}(x,y)}{{|x-y{|^{N+sp}}}}({\phi_{n}}(x)-{\phi_{n}}(y))dxdy\geq
0,$ (6.20)

it follows from (6) that

$0\geq\frac{V_{1}}{2(a+bA_{1}^{(\theta-1)p})}\int_{v_{n}\geq{\tilde{w}}}[v_{n}^{p-1}-{\tilde{w}}^{p-1}](v_{n}-{\tilde{w}})dx\geq
0,$

which implies that

$\\{x\in\mathbb{R}^{N}:|x|\geq R_{3}\ \ and\ \
v_{n}\geq{\tilde{w}}\\}=\emptyset.$

To obtain our purpose, we first note that for all $s,d\in\mathbb{R}$ it holds

$|d|^{p-2}d-|c|^{p-2}c=(p-1)(d-c)\int_{0}^{1}|c+t(d-c)|^{p-2}dt.$

Taking $d=v_{n}(x)-v_{n}(y)$ and $c={\tilde{w}}(x)-{\tilde{w}}(y)$ we can see
that

$|d|^{p-2}d-|c|^{p-2}c=(p-1)(d-c)I(x,y),$

where $I(x,y)\geq 0$ stands for the integral. Now, recalling that

$(x-y)(x^{+}-y^{+})\geq|x^{+}-y^{+}|^{2}\ \ \mbox{for}\ \mbox{all}\ \
x,y\in\mathbb{R},$

one has

$\displaystyle(d-c)(\phi_{n}(x)-\phi_{n}(y))=$
$\displaystyle[(v_{n}-{\tilde{w}})(x)-(v_{n}-{\tilde{w}})(y)][(v_{n}-{\tilde{w}})^{+}(x)-(v_{n}-{\tilde{w}})^{+}(y)]$
$\displaystyle\geq$
$\displaystyle|(v_{n}-{\tilde{w}})^{+}(x)-(v_{n}-{\tilde{w}})^{+}(y)|^{2},$

which gives $(|d|^{p-2}d-|c|^{p-2}c)(\phi_{n}(x)-\phi_{n}(y))\geq 0$, that is
(6.20) holds.

Therefore, $\psi_{n}\leq 0$ in $B_{R_{3}}^{c}$, which implies that $v_{n}\leq
C_{1}w$ in $B_{R_{3}}^{c}$, that is $v_{n}(x)\leq C\frac{1}{1+|x|^{N+sp}}$ in
$B_{R_{3}}^{c}$. Therefore, we have

$\displaystyle v_{\varepsilon_{n}}(x)=$
$\displaystyle{u_{\varepsilon_{n}}}\left(\frac{x}{\varepsilon_{n}}\right)=v_{n}\left(\frac{x}{\varepsilon_{n}}-y_{n}\right)\leq
C_{1}w\left(\frac{x}{\varepsilon_{n}}-y_{n}\right)$ $\displaystyle\leq$
$\displaystyle\frac{C}{1+|\frac{x}{\varepsilon_{n}}-y_{n}|^{N+sp}}$
$\displaystyle=$
$\displaystyle\frac{C{\varepsilon_{n}}^{N+sp}}{{\varepsilon_{n}}^{N+sp}+|x-{\varepsilon_{n}}y_{n}|^{N+sp}}$
$\displaystyle\leq$
$\displaystyle\frac{C{\varepsilon_{n}}^{N+sp}}{{\varepsilon_{n}}^{N+sp}+|x-\eta_{\varepsilon_{n}}|^{N+sp}}.$

The proof is complete. ∎

Acknowledgements:

The research bas been supported by National Natural Science Foundation of
China 11971392, Natural Science Foundation of Chongqing, China cstc2021ycjh-
bgzxm0115, Fundamental Research Funds for the Central Universities
XDJK2020B047, and the Chongqing Graduate Student Research Innovation Project
CYS21098.

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