# On the Existence of Solutions for the Critical Fractional Laplacian Equation inℝ

Authors: Zifei Shen; Fashun Gao
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We study existence of solutions for the fractional Laplacian equation-
Δ
s
u
+
V
x
u
=
u
2
*
s
-
2
u
+
f
x
,
u in ℝ
N, u
∈
H
s
(
R
N
), with critical exponent 2
*
s
=
2
N
/
(
N
-
2
s
), N
>
2
s, s
∈
0
,
1, where V
x
≥
0 has a potential well and f
:
ℝ
N
×
ℝ
→
ℝ is a lower order perturbation of the critical power u
2
*
s
-
2
u. By employing the variational method, we prove the existence of nontrivial solutions for the equation.


---

## Body

## 1. Introduction

In the last 20 years, the classical nonlinear Schrödinger equation has been extensively studied by many authors [1–10] and the references therein. We just mention some earlier work about it. Brézis and Nirenberg [1] proved that the critical problem with small linear perturbations can provide positive solutions. In [3], Rabinowitz proved the existence of standing wave solutions of nonlinear Schrödinger equations. Making a standing wave ansatz reduces the problem to that of studying a class of semilinear elliptic equations. Floer and Weinstein [10] proved that Schrödinger equation with potential V and cubic nonlinearity has standing wave solutions concentrated near each nondegenerate critical point of V.However, a great attention has been focused on the study of problems involving the fractional Laplacian recently. This type of operator seems to have a prevalent role in physical situations such as combustion and dislocations in mechanical systems or in crystals. In addition, these operators arise in modelling diffusion and transport in a highly heterogeneous medium. This type of problems has been studied by many authors [11–18] and the references therein.Servadei and Valdinoci [11–14] studied the problem
(1)
L
K
u
+
λ
u
+
|
u
|
2
*
(
s
)
-
2
u
+
f
(
x
,
u
)
=
0
in
Ω
,
u
=
0
in
ℝ
N
∖
Ω
,
where s
∈
(
0,1
), Ω is an open bounded set of ℝ
N, N
>
2
s, with Lipschitz boundary, λ
>
0 is a real parameter, and 2
*
(
s
)
=
2
N
/
(
N
-
2
s
) is a fractional critical Sobolev exponent. L
K is defined as follows:
(2)
L
K
u
(
x
)
=
1
2

∫
ℝ
N
‍
(
u
(
x
+
y
)

+

u
(
x
-
y
)

-
u
(
x
)
)
K
(
y
)
d
x
d
y
,
222222222222222222222222222222222222
x
∈
ℝ
N
.
Here K
:
ℝ
N
∖
{
0
}
→
(
0
,
+
∞
) is a function such that
(3)
m
K
∈
L
1
(
ℝ
N
)
,
where
m
(
x
)
=
min
{
|
x
|
2
,
1
}
;
there exists θ
>
0 such that K
(
x
)
≥
θ
|
x
|
-
(
N
+
2
s
) and K
(
x
)
=
K
(
-
x
) for any x
∈
ℝ
N
∖
{
0
}. They proved that problem (1) admits a nontrivial solution for any λ
>
0. They also studied the case f
(
x
,
u
)
≡
0 and K
(
x
)
=
|
x
|
-
(
N
+
2
s
), respectively.Felmer et al. [15] studied the following nonlinear Schrödinger equation with fractional Laplacian:
(4)
(
-
Δ
)
α
u
+
u
=
f
(
x
,
u
)
in
ℝ
N
,
u
>
0
in
ℝ
N
,
lim
|
x
|
→
∞
u
(
x
)
=
0
,
where 0
<
α
<
1, N
≥
2, and f
:
ℝ
N
×
ℝ
→
ℝ is superlinear and has subcritical growth with respect to u. The fractional Laplacian can be characterized as Ϝ
(
(
-
Δ
)
α
ϕ
)
(
ζ
)
=
|
ζ
|
2
α
Ϝ
(
ϕ
)
(
ζ
), where Ϝ is the Fourier transform. They gave the proof of existence of positive solutions and further analyzed regularity, decay, and symmetry properties of these solutions.In this paper, we consider the following problem:(5)
(
-
Δ
)
s
u
+
V
(
x
)
u
=
|
u
|
2
*
(
s
)
-
2
u
+
f
(
x
,
u
)
in
ℝ
N
,
u
∈
H
s
(
ℝ
N
)
,
with critical exponent 2
*
(
s
)
=
2
N
/
(
N
-
2
s
), N
>
2
s, s
∈
(
0,1
), where V
(
x
)
≥
0 has a potential well, where (
-
Δ
)
s is the fractional Laplace operator, which may be defined as
(6)
-
(
-
Δ
)
s
u
(
x
)
=
1
2
∫
ℝ
N
‍
u
(
x
+
y
)
+
u
(
x
-
y
)
-
2
u
(
x
)
|
y
|
N
+
2
s
d
y
,
x
∈
ℝ
N
.
H
s
(
ℝ
N
) is the usual fractional Sobolev space.f
:
ℝ
N
×
ℝ
→
ℝ is a lower order perturbation of the critical power |
u
|
2
*
(
s
)
-
2
u. Now we give our main assumptions. In order to find weak solutions of (5), we will assume the following general hypotheses:(
V
0
)
V
(
x
) is a continuous nonnegative function on ℝ
N and satisfying
(7)
lim
|
x
|
→
∞
V
(
x
)
=
+
∞
;
(
f
0
)
f
:
ℝ
N
×
ℝ
→
ℝ is a Carathéodory function;
(
f
1
)
lim
t
→
o
(
f
(
x
,
t
)
/
t
)
=
0 uniformly in x
∈
ℝ
N;
(
f
2
)
there exista
1, a
2
>
0 and q
∈
(
2
,
2
*
(
s
)
), 2
*
(
s
)
=
2
N
/
(
N
-
2
s
), such that
(8)
|
f
(
x
,
t
)
|
≤
a
1
+
a
2
|
t
|
q
-
1
a
.
e
.
x
∈
ℝ
N
,
t
∈
ℝ
;
(
f
3
)
sup{
|
f
(
x
,
t
)
|
:
a
.
e
.
x
∈
ℝ
N
,
|
t
|
≤
M
}
<
+
∞ for any M
>
0;
(
f
4
)
there existsμ
>
2 such that, for all t
>
0 and a
.
e
.
x
∈
ℝ
N,
(9)
0
<
μ
F
(
x
,
t
)
≤
t
f
(
x
,
t
)
,where F
(
x
,
t
)
=
∫
0
t
‍
f
(
x
,
τ
)
d
τ.The aim of this paper is to find solutions for (5) by variational methods. For this, we give the weak formulation of (5) by the following problem:
(10)
∫
ℝ
2
N
‍
(
u
(
x
)
-
u
(
y
)
)
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
u
(
x
)
φ
(
x
)
d
x
=
∫
ℝ
N
‍
|
u
(
x
)
|
2
*
(
s
)
-
2
u
(
x
)
φ
(
x
)
d
x
+
∫
ℝ
N
‍
f
(
x
,
u
(
x
)
)
φ
(
x
)
d
x
,
∀
φ
∈
H
s
(
ℝ
N
)
,
22222222222222222222222222
u
∈
H
s
(
ℝ
N
)
.
This problem represents the Euler-Lagrange equation of the function I
:
H
s
(
ℝ
N
)
→
ℝ defined as
(11)
I
(
u
)
=
1
2
∫
ℝ
2
N
‍
|
u
(
x
)
-
u
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
1
2
∫
ℝ
N
‍
V
(
x
)
|
u
(
x
)
|
2
d
x
-
1
2
*
(
s
)
∫
ℝ
N
‍
|
u
(
x
)
|
2
*
(
s
)
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
(
x
)
)
d
x
,
where F is defined as in (
f
4
). Critical points of I are weak solutions of (5). We will prove the existence of the critical points of the functional I.It is convenient to define(12)
S
s
∶
=
inf
u
∈
E
∖
{
0
}
S
s
(
u
)
,
(13)
S
r
,
s
∶
=
inf
u
∈
H
r
s
(
ℝ
N
)
∖
{
0
}
S
r
,
s
(
u
)
,
where E will be defined in Section 2 and for any u
∈
E
∖
{
0
}
(14)
S
s
(
u
)
∶
=
∫
ℝ
2
N

(

|
u

(
x
)

-

u

(
y
)
|
2
/
|
x

-

y
|
N
+
2
s

)
‍
d
x
d
y

+

∫
ℝ
N
‍

V

(
x
)
|
u

(
x
)
|
2
d
x
(
∫
ℝ
N
‍
|
u
(
x
)
|
2
*
(
s
)
d
x
)
2
/
2
*
(
s
)
and H
r
s
(
ℝ
N
) will be defined in Section 2 and for any u
∈
H
r
s
(
ℝ
N
)
∖
{
0
}
(15)
S
r
,
s
(
u
)
∶
=
∫
ℝ
2
N
(
|
u
(
x
)
-
u
(
y
)
|
2
/
|
x
-
y
|
N
+
2
s
)
‍
d
x
d
y

+

∫
ℝ
N
‍
|
u
(
x
)
|
2
d
x
(
∫
ℝ
N
‍
|
u
(
x
)
|
2
*
(
s
)
d
x
)
2
/
2
*
(
s
)
.Now, we give our results as follows.Theorem 1.
LetN
>
2
s, s
∈
(
0,1
), V
(
x
), and f satisfy (
V
0
) and (f
0)–(f
4), respectively. Then, (5) possesses at least one nontrivial solution.We prove Theorem1 applying the mountain pass theorem to the functional I. Although the Palais-Smale sequences might lose compactness in the whole space ℝ
N, we cannot apply the mountain pass theorem directly. In [15], they used a comparison argument to over this difficulty. But we will use allow it to us to over this problem related to the lack of compactness and to show that the Palais-Smale condition holds true in a suitable range related to S
s.Theorem 2.
LetN
>
2
s, s
∈
(
0,1
), V
(
x
)
≡
1, and f satisfy (f
0)–(f
4). Then, (5) possesses at least one nontrivial radial symmetric solution.For the caseV
(
x
)
≡
1, the proof of the existence of solutions is similar to the proof of Theorem 1. To prove symmetry of solutions, we consider the subspace H
r
s
(
ℝ
N
) of H
s
(
ℝ
N
). H
r
s
(
ℝ
N
) consists of radial symmetric functions of H
s
(
ℝ
N
).This paper is organized as follows. In Section2, we give some preliminary results. In Section 3, we prove the geometry and Palais-Smale condition of the functional I and finish the proof of Theorem 1. In Section 4, we finish the proof of Theorem 2.


## 2. Preliminary Results

We consider the fractional Sobolev space(16)
H
s
(
ℝ
N
)
=
{
u
∈
L
2
(
ℝ
N
)
∣
∫
ℝ
2
N
‍
|
u
(
x
)
-
u
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
|
u
(
x
)
|
2
d
x
<
+
∞
∫
ℝ
2
N
‍
|
u
(
x
)
-
u
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
}
,
equipped with norm
(17)
∥
u
∥
H
=
(
∫
ℝ
2
N
‍
|
u
(
x
)
-
u
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
|
u
(
x
)
|
2
d
x
)
1
/
2
.
We denote a subspace of the fractional Sobolev space H
s
(
ℝ
N
) by E. This subspace E is defined as the completion of C
∞
(
ℝ
N
) with respect to the norm
(18)
∥
u
∥
E
=

(

∫
ℝ
2
N
‍

|
u
(
x
)

-

u
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y

+

∫
ℝ
N

‍
V
(
x
)
|
u
(
x
)
|
2
d
x

)
1
/
2
.Obviously,(
E
,
∥
·
∥
E
) is a Hilbert space, with scalar product
(19)
〈
u
,
φ
〉
E
=
∫
ℝ
2
N
‍
(
u
(
x
)
-
u
(
y
)
)
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
u
(
x
)
φ
(
x
)
d
x
and E is continuously embedded in H
s
(
ℝ
N
).On the other hand, we also consider the subspaceH
r
s
(
ℝ
N
) of H
s
(
ℝ
N
). H
r
s
(
ℝ
N
) consists of radial symmetric functions of H
s
(
ℝ
N
) and has the same norm with H
s
(
ℝ
N
), and its norm is denoted by ∥
u
∥
H
r
s. H
r
s
(
ℝ
N
) is continuously embedded in H
s
(
ℝ
N
).The following two lemmas about the fractional Sobolev spaceH
s
(
ℝ
N
) are proved in [15].Lemma 3.
Let2
≤
q
≤
2
*
(
s
)
=
2
N
/
(
N
-
2
s
); then one has
(20)
∥
u
∥
L
q
(
ℝ
N
)
≤
C
∥
u
∥
H
∀
u
∈
H
s
(
ℝ
N
)
.
Consequently, the embedding E
↪
L
2
*
(
s
)
(
ℝ
N
) and H
r
s
(
ℝ
N
)
↪
L
2
*
(
s
)
(
ℝ
N
) are continuous. If further 2
≤
q
<
2
*
(
s
) and Ω
⊂
ℝ
N is a bounded domain, then the bound sequence {
u
k
}
⊂
H
s
(
ℝ
N
) has a convergent subsequence in L
q
(
Ω
).Thanks to Lemma3, we can define the constants S
s and S
r
,
s and get that S
s
>
0 and S
r
,
s
>
0.Lemma 4.
LetN
≥
2. Assume that {
u
k
} is bounded in H
s
(
ℝ
N
) and it satisfies
(21)
lim
k
→
∞
sup
ξ
∈
ℝ
N
∫
B
K
(
ξ
)
‍
|
u
k
(
x
)
|
2
d
x
=
0
,
where R
>
0. Then u
k
→
0 in L
q
(
ℝ
N
) for 2
<
q
<
2
*
(
s
).Lemma 5.
(a) If(
V
0
) holds true, then the embedding E
↪
L
ν
(
ℝ
N
) is compact for any ν
∈
[
2
,
2
*
(
s
)
).
(b) The embeddingH
r
s
(
ℝ
N
)
↪
L
q
(
ℝ
N
) is compact for any q
∈
(
2
,
2
*
(
s
)
).Proof.
(a) We give the proof as in [19].
Case
ν
=
2. We will show that u
n
→
0 strongly in L
2
(
ℝ
N
), whenever u
n
→
0 weakly in E. Indeed, let C
>
0 be such that ∥
u
∥
E
≤
C. Given ε
>
0, pick R
>
0 such that V
(
x
)
≥
2
C
2
/
ε for all |
x
|
≥
R and denote by B
R the ball of radius R in ℝ
N. Then we have that u
n
→
0 weakly in H
s
(
B
R
). The compact embedding H
s
(
B
R
)
↪
L
2
(
B
R
) implies that for some natural number n
0,
(22)
∫
B
R
‍
|
u
n
(
x
)
|
2
d
x
≤
ε
2
,
∀
n
>
n
0
.
On the other hand, by our choice of R
>
0, we have
(23)
2
ε
∫
ℝ
N
∖
B
R
|
u
n
(
x
)
|
2
d
x
≤
1
C
2
∫
ℝ
N
∖
B
R
V
(
x
)
|
u
n
(
x
)
|
2
d
x
≤
1
C
2
∥
u
n
∥
E
2
≤
1
.
Combining (22) and (23), we obtain that ∥
u
n
∥
L
2
(
ℝ
N
)
2
<
ε for all n
>
n
0.
Case
2
<
ν
<
2
*
(
s
). Using Lemma 3, together with the interpolation inequality (where 1
/
ν
=
(
σ
/
2
)
+
(
(
1
-
σ
)
/
2
*
(
s
)
)),
(24)
∥
u
∥
L
ν
(
ℝ
N
)
≤
∥
u
∥
L
2
(
ℝ
N
)
σ
∥
u
∥
L
2
*
(
s
)
(
ℝ
N
)
1
-
σ
,
∀
u
∈
L
2
(
ℝ
N
)
∩
L
2
*
(
s
)
(
ℝ
N
)
and by the fact that the embedding E
↪
L
2
(
ℝ
N
) is compact, we can obtain that the embedding E
↪
L
ν
(
ℝ
N
) is also compact.
(b) We give the proof as in Corollary  4.7.2 of [20].
For allx
∈
ℝ
N, for all R
>
0, let m
(
x
,
R
) be the largest number of disjoint balls with radius R and the centers lie on the same sphere with radius |
x
| centered at 0. It is easily seen that m
(
x
,
R
)
→
∞ as |
x
|
→
∞. By definition, for all u
∈
L
2
(
ℝ
N
) and for all r
>
0,
(25)
∫
B
r
(
x
)
‍
|
u
|
2
d
x
≤
m
(
x
,
r
)
-
1
∥
u
∥
L
2
(
ℝ
N
)
2
.
If {
u
j
} is a bounded sequence in H
r
s
(
ℝ
N
), for all ε
>
0, ∃
R
>
0, we have
(26)
sup
x
∈
ℝ
N
{
∫
B
r
(
x
)
‍
|
u
j
(
x
)
|
2
d
x
∣
|
x
|
≥
R
}
<
ε
.
We may assume that u
j
→
0 weakly in H
r
s
(
ℝ
N
); then, by Lemma 3, after a subsequence ∫
B
R
+
r
(
0
)
‍
|
u
n
j
|
2
d
x
→
0, it follows that
(27)
sup
x
∈
ℝ
N
{
∫
B
r
(
x
)
‍
|
u
n
j
(
x
)
|
2
d
x
∣
|
x
|
≤
R
}
⟶
0
.
By (26), (27), and Lemma 4, we have
(28)
u
n
j
⟶
0
in
L
q
(
ℝ
N
)
for q
∈
(
2
,
2
*
(
s
)
).In [11, Lemma  6], Servadei and Valdinoci proved the following result.Lemma 6.
Assume thatf satisfies (f
0)–(f
4). Then, for any ε
>
0, there exists δ
=
δ
(
ε
) such that a.e. x
∈
ℝ
N, and for any t
∈
ℝ
N
(29)
|
f
(
x
,
t
)
|
≤
2
ε
|
t
|
+
q
δ
(
ε
)
|
t
|
q
-
1
,
and so, as a consequence,
(30)
|
F
(
x
,
t
)
|
≤
ε
|
t
|
2
+
δ
(
ε
)
|
t
|
q
,
where F is defined as in (f
4) and q
∈
(
2
,
2
*
(
s
)
).In [17, Lemma  4], Servadei and Valdinoci proved the following result.Lemma 7.
Assume thatf satisfies (f
0)–(f
4). Then, there exist two positive measurable functions m
=
m
(
x
) and M
=
M
(
x
) such that a.e. x
∈
ℝ
N, and for any t
∈
ℝ,
(31)
F
(
x
,
t
)
≥
m
(
x
)
|
t
|
μ
-
M
(
x
)
,
where F is defined as in (f
4), 2
<
μ
<
2
*
(
s
), and m, M
∈
L
∞
(
ℝ
N
).


## 3. The Proof of Theorem1

In this section we study the mountain pass geometry and Palais-Smale condition in a suitable energy range and finish the proof of Theorem1. We consider the functional
(32)
I
(
u
)
=
1
2

∫
ℝ
2
N
‍
|
u
(
x
)
-
u
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
1
2

∫
ℝ
N
‍
V
(
x
)
|
u
(
x
)
|
2
d
x
-
1
2
*
(
s
)
∫
ℝ
N
‍
|
u
(
x
)
|
2
*
(
s
)
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
(
x
)
)
d
x
,
where F is defined as in (f
4).ThenI
∈
C
1
(
E
,
ℝ
) and critical points of I are solutions of
(33)
(
-
Δ
)
s
u
+
V
(
x
)
u
=
|
u
|
2
*
(
s
)
-
2
u
+
f
(
x
,
u
)
,
u
∈
H
s
(
ℝ
N
)
.
The Fréchet derivative of I is
(34)
〈
I
′
(
u
)
,
φ
〉
=
∫
ℝ
2
N
‍
(
u
(
x
)
-
u
(
y
)
)
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
u
(
x
)
φ
(
x
)
d
x
-
∫
ℝ
N
‍
|
u
(
x
)
|
2
*
(
s
)
-
2
u
(
x
)
φ
(
x
)
d
x
-
∫
ℝ
N
‍
f
(
x
,
u
(
x
)
)
φ
(
x
)
d
x
∀
φ
∈
E
.Proposition 8.
LetN
>
2
s, s
∈
(
0,1
), V
(
x
), and f satisfy (
V
0
) and (f
0)–(f
4), respectively. Then, there exist ρ
>
0 and β
>
0 such that for any u
∈
E with ∥
u
∥
E
=
ρ it results that I
(
u
)
≥
β.Proof.
Letu be a function in E. By Lemma 3 and (30), we get that, for any ε
>
0,
(35)
I
(
u
)
≥
1
2

∫
ℝ
2
N
‍
|
u
(
x
)
-
u
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
1
2

∫
ℝ
N
‍

V
(
x
)
|
u
(
x
)
|
2
d
x
-
1
2
*
(
s
)
∫
ℝ
N
‍
|
u
(
x
)
|
2
*
(
s
)
d
x
-
ε
∫
ℝ
N
‍
|
u
(
x
)
|
2
d
x
-
δ
(
ε
)
∫
ℝ
N
‍
|
u
(
x
)
|
q
d
x
=
1
2
∥
u
∥
E
2
-
ε
∥
u
∥
L
2
(
ℝ
N
)
2
-
1
2
*
(
s
)
∥
u
∥
L
2
*
(
s
)
(
ℝ
N
)
2
*
(
s
)
-
δ
(
ε
)
∥
u
∥
L
q
(
ℝ
N
)
q
≥
1
2
∥
u
∥
E
2
-
ε
C
1
∥
u
∥
E
2
-
1
2
*
(
s
)
C
2
∥
u
∥
E
2
*
(
s
)
-
δ
(
ε
)
C
3
∥
u
∥
E
q
.
Choosingε
>
0 such that (
1
/
2
)
-
ε
C
1
>
0, by (35), it easily follows that
(36)
I
(
u
)
≥
C
4
∥
u
∥
E
2
(
1
-
C
5
∥
u
∥
E
2
*
(
s
)
-
2
-
C
6
∥
u
∥
E
q
-
2
)
,
for suitable positive constants C
4, C
5, and C
6.
Now, letu
∈
E be such that ∥
u
∥
E
=
ρ
>
0. Since 2
*
(
s
)
>
q
>
2, we can choose ρ sufficiently small, so that
(37)
inf
u
∈
E
,
∥
u
∥
E
=
ρ
I
(
u
)
≥
C
4
ρ
2
(
1
-
C
5
ρ
2
*
(
s
)
-
2
-
C
6
ρ
q
-
2
)
=
:
β
>
0
.
Hence, Proposition8 is proved.Proposition 9.
There existsu
0
∈
E
∖
{
0
} with u
0
≥
0 a.e. in ℝ
N, such that
(38)
sup
t
≥
0
I
(
t
u
0
)
<
s
N
S
s
N
/
2
s
.Proof.
By definition ofS
s, we have that there exists u
0
∈
E
∖
{
0
} such that
(39)
∥
u
0
∥
E
2
∥
u
0
∥
L
2
*
(
s
)
(
ℝ
N
)
2
<
S
s
+
ε
,
for any ε
>
0. We have u
0
≥
0 a.e. in ℝ
N, or else we can take |
u
0
|
∈
E. Indeed, by triangle inequality, a.e. x, y
∈
ℝ
N,
(40)
|
|
u
0
(
x
)
|
-
|
u
0
(
y
)
|
|
≤
|
u
0
(
x
)
-
u
0
(
y
)
|
,
and so
(41)
∥
|
u
0
|
∥
E
2
≤
∥
u
0
∥
E
2
.
Thus,
(42)
∥
|
u
0
|
∥
E
2
∥
u
0
∥
L
2
*
(
s
)
(
ℝ
N
)
2
<
S
s
+
ε
.
It follows that there exists t
0
>
0 such that
(43)
max
t
≥
0
I
(
t
u
0
)
=
max
t
≥
0
(
t
2
2
∥
u
0
∥
E
2
-
t
2
*
(
s
)
2
*
(
s
)
∥
u
0
∥
L
2
*
(
s
)
(
ℝ
N
)
2
*
(
s
)
-
∫
ℝ
N
‍
F
(
x
,
t
u
0
(
x
)
)
d
x
t
2
2
∥
u
0
∥
E
2
-
t
2
*
(
s
)
2
*
(
s
)
)
=
t
0
2
2
∥
u
0
∥
E
2
-
t
0
2
*
(
s
)
2
*
(
s
)
∥
u
0
∥
L
2
*
(
s
)
(
ℝ
N
)
2
*
(
s
)
-
∫
ℝ
N
‍
F
(
x
,
t
0
u
0
(
x
)
)
d
x
≤
max
t
≥
0
(
t
2
2
∥
u
0
∥
E
2
-
t
2
*
(
s
)
2
*
(
s
)
∥
u
0
∥
L
2
*
(
s
)
(
ℝ
N
)
2
*
(
s
)
)
-
∫
ℝ
N
‍
F
(
x
,
t
0
u
0
(
x
)
)
d
x
=
s
N
(
∥
u
0
∥
E
2
∥
u
0
∥
L
2
*
(
s
)
(
ℝ
N
)
2
)
N
/
2
s
-
∫
ℝ
N
‍
F
(
x
,
t
0
u
0
(
x
)
)
d
x
<
s
N
(
S
s
+
ε
)
N
/
2
s
-
∫
ℝ
N
‍
F
(
x
,
t
0
u
0
(
x
)
)
d
x
,
thanks to (39). Since ∫
ℝ
N
‍
F
(
x
,
t
0
u
0
(
x
)
)
d
x
>
0, so we can choose ε
>
0 such that
(44)
sup
t
≥
0
I
(
t
u
0
)
<
s
N
S
s
N
/
2
s
.
Hence, Proposition9 is proved.Proposition 10.
LetN
>
2
s, s
∈
(
0,1
), V
(
x
), and f satisfy (
V
0
) and (f
0)–(f
4), respectively. Then, there exists e
∈
E such that e
≥
0 a.e. in ℝ
N, ∥
e
∥
E
>
ρ, and I
(
e
)
<
β, where ρ and β are given in Proposition 8.In particular, we can constructe as follows:
(45)
e
=
t
0
u
0
with u
0 as in (38) and t
0
>
0 large enough.Proof.
We fixu
∈
E such that ∥
u
∥
E
≠
0. By (31), we get
(46)
I
(
t
u
)
≤
1
2
t
2
∫
ℝ
2
N
‍
|
u
(
x
)
-
u
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
1
2
t
2
∫
ℝ
N
‍
V
(
x
)
|
u
(
x
)
|
2
d
x
-
1
2
*
(
s
)
t
2
*
(
s
)
∫
ℝ
N
‍
|
u
(
x
)
|
2
*
(
s
)
d
x
-
t
μ
∫
ℝ
N
‍
m
(
x
)
|
u
(
x
)
|
μ
-
M
(
x
)
d
x
=
1
2
t
2
∥
u
∥
E
2
-
1
2
*
(
s
)
t
2
*
(
s
)
∥
u
∥
L
2
*
(
s
)
(
ℝ
N
)
2
*
(
s
)
-
t
μ
C
1
∥
u
∥
L
μ
(
ℝ
N
)
μ
+
C
2
,
where C
1, C
2
≥
0 are constant. Since 2
*
(
s
)
>
2 and μ
>
2, passing to the limit as t
→
+
∞, we get that I
(
t
u
)
→
-
∞, so that the assertion follows taking e
=
t
u, with t sufficiently large.
In particular, we can takeu
=
u
0
∈
E, then e
=
t
0
u
0 with t
0 large enough.We easily see thatI
(
0
)
=
0
<
β, with β given in Proposition 8. Now, set
(47)
c
=
inf
T
∈
Γ
sup
u
∈
T
(
[
0,1
]
)
I
(
u
)
,
where
(48)
Γ
=
{
T
∈
C
(
[
0,1
]
;
E
)
:
T
(
0
)
=
0
,
T
(
1
)
=
e
}
,
with e
=
t
0
u
0 in Proposition 9.In [11, the proof of Theorem  1], Servadei and Valdinoci proved the following result.Proposition 11.
The constantc is  given in (47) such that
(49)
β
≤
c
<
s
N
S
s
N
/
2
s
,
where β is given in Proposition 8 and S
s is defined in formula (12).By [6, Theorem  2.2], we have a sequence u
j in E such that
(50)
I
(
u
j
)
⟶
c
,
(51)
sup
{
|
〈
I
′
(
u
j
)
,
φ
〉
|
:
φ
∈
E
,
∥
φ
∥
E
=
1
}
⟶
0
as j
→
+
∞.Proposition 12.
There existsu
∞
∈
E such that, up to a subsequence, ∥
u
j
-
u
∞
∥
E
→
0 as j
→
+
∞.Proof.
We proceed by steps.
Step  1. The sequence u
j is bounded in E.
Proof. For any j
∈
ℕ by (50) and (51) it easily follows that there exists C
1
>
0 such that
(52)
|
I
(
u
j
)
|
≤
C
1
,
|
〈
I
′
(
u
j
)
,
u
j
∥
u
j
∥
E
〉
|
≤
C
1
.
As a consequence of (52), we have
(53)
I
(
u
j
)
-
1
μ
〈
I
′
(
u
j
)
,
u
j
〉
≤
C
1
(
1
+
∥
u
j
∥
E
)
.
By (
f
2
) and (
f
4
), we have μ
<
2
*
(
s
) and
(54)
I
(
u
j
)
-
1
μ
〈
I
′
(
u
j
)
,
u
j
〉
=
(
1
2
-
1
μ
)
(
∫
ℝ
2
N
‍
|
u
j
(
x
)
-
u
j
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
|
u
j
(
x
)
|
2
d
x
∫
ℝ
2
N
‍
|
u
j
(
x
)
-
u
j
(
y
)
|
2
|
x
-
y
|
N
+
2
s
)
+
(
1
μ
-
1
2
*
(
s
)
)
∫
ℝ
N
‍
|
u
j
(
x
)
|
2
*
(
s
)
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
j
(
x
)
)
d
x
+
1
μ
∫
ℝ
N
‍
f
(
x
,
u
j
(
x
)
)
u
j
(
x
)
d
x
≥
(
1
2
-
1
μ
)
∥
u
j
∥
E
2
+
(
1
μ
-
1
2
*
(
s
)
)
∥
u
j
∥
L
2
*
(
s
)
(
ℝ
N
)
2
*
(
s
)
.
We obtain that u
j is bounded in E by 2
<
μ
<
2
*
(
s
), (53), and (54).
Step  2. Problem (10) admits a solution u
∞
∈
E.
Proof. By Step  1  and since E is a reflexive space, up to a subsequence, still denoted by u
j, there exists u
∞
∈
E such that u
j
→
u
∞ weakly in E; that is,
(55)
∫
ℝ
2
N
‍
(
u
j
(
x
)
-
u
j
(
y
)
)
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
u
j
(
x
)
φ
(
x
)
d
x
⟶
∫
ℝ
2
N
‍
(
u
∞
(
x
)
-
u
∞
(
y
)
)
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
u
∞
(
x
)
φ
(
x
)
d
x
as j
→
+
∞. By 2
<
μ
<
2
*
(
s
), (53), and (54), we have that u
j is bounded in L
2
*
(
s
)
(
ℝ
N
). Since L
2
*
(
s
)
(
ℝ
N
) is a reflexive space, up to a subsequence,
(56)
u
j
⟶
u
∞
weakly
in
L
2
*
(
s
)
(
ℝ
N
)
as j
→
+
∞, while by Lemma 5(a), up to a subsequence,
(57)
u
j
⟶
u
∞
in
L
ν
(
ℝ
N
)
,
(58)
u
j
⟶
u
∞
a
.
e
.
in
ℝ
N
as j
→
+
∞, for any ν
∈
[
2
,
2
*
(
s
)
). By (56) and since |
u
j
|
2
*
(
s
)
-
2
u
j is bounded in L
2
*
(
s
)
/
(
2
*
(
s
)
-
1
)
(
ℝ
N
), we have
(59)
|

u
j
|
2
*
(
s
)
-
2

u
j

⟶

|

u
∞
|
2
*
(
s
)
-
2
u
∞
weakly
in
L
2
*
(
s
)
/
(
2
*
(
s
)
-
1
)

(
ℝ
N

)
as j
→
+
∞. By the proof of Lemma 6 [11, Lemma  6], we get
(60)
∫
ℝ
N
‍
f
(
x
,
u
(
x
)
)
d
x
≤
2
ε
∫
B
σ
‍
|
u
(
x
)
|
d
x
+
q
δ
(
ε
)
∫
ℝ
N
∖
B
σ
‍
|
u
(
x
)
|
q
-
1
d
x
.
Moreover, we have, by takingε
=
1,
(61)
∫
ℝ
N
‍
f
(
x
,
u
(
x
)
)
d
x
≤
C
1
+
C
2
∫
ℝ
N
∖
B
σ
‍
|
u
(
x
)
|
q
-
1
d
x
.
By (61), u
j being bounded in E, and Lemma 5(a), we obtain
(62)
f
(
·
,
u
j
(
·
)
)
is
bounded
in
L
q
/
(
q
-
1
)
(
ℝ
N
)
.
Since L
q
/
(
q
-
1
)
(
ℝ
N
) is a reflexive space, we get
(63)
f

(
·
,
u
j
(
·
)
)

⟶

f

(
·
,
u
∞
(
·
)
)
weakly
in
L
q
/
(
q
-
1
)
(
ℝ
N
)
as j
→
+
∞. It is easily seen that
(64)
∫
ℝ
N
‍
f
(
x
,
u
j
(
x
)
)
φ
(
x
)
d
x
⟶
∫
ℝ
N
‍
f
(
x
,
u
∞
(
x
)
)
φ
(
x
)
d
x
∀
φ
∈
L
q
(
ℝ
N
)
as j
→
+
∞ and so, in particular,
(65)
∫
ℝ
N
‍
f
(
x
,
u
j
(
x
)
)
φ
(
x
)
d
x
⟶
∫
ℝ
N
‍
f
(
x
,
u
∞
(
x
)
)
φ
(
x
)
d
x
∀
φ
∈
E
as j
→
+
∞. Since (51) holds true, for any φ
∈
E,
(66)
0
⟵
〈
I
′
(
u
j
)
,
φ
〉
=
∫
ℝ
2
N
‍
(
u
j
(
x
)
-
u
j
(
y
)
)
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
u
j
(
x
)
φ
(
x
)
d
x
-
∫
ℝ
N
‍
|
u
j
(
x
)
|
2
*
(
s
)
-
2
u
j
(
x
)
φ
(
x
)
d
x
-
∫
ℝ
N
‍
f
(
x
,
u
j
(
x
)
)
φ
(
x
)
d
x
.
Passing to the limit in this expression asj
→
+
∞ and taking into account (55), (57), (59), and (65), we get
(67)
∫
ℝ
2
N
‍
(
u
∞
(
x
)
-
u
∞
(
y
)
)
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
u
∞
(
x
)
φ
(
x
)
d
x
-
∫
ℝ
N
‍
|
u
∞
(
x
)
|
2
*
(
s
)
-
2
u
∞
(
x
)
φ
(
x
)
d
x
-
∫
ℝ
N
‍
f
(
x
,
u
∞
(
x
)
)
φ
(
x
)
d
x
=
0
for any φ
∈
E; that is, u
∞ is a solution of problem (10).
Step  3. The following equality holds true:
(68)
I
(
u
∞
)
=
s
N
∫
ℝ
N
‍
|
u
∞
(
x
)
|
2
*
(
s
)
d
x
+
1
2
∫
ℝ
N
‍
f
(
x
,
u
∞
(
x
)
)
u
∞
(
x
)
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
∞
(
x
)
)
d
x
≥
0
.
Proof. By Step  2, taking φ
=
u
∞
∈
E as a test function in (10), we have
(69)
∫
ℝ
2
N
‍
|
u
∞
(
x
)
-
u
∞
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
|
u
∞
(
x
)
|
2
d
x
=
∫
ℝ
N
‍
|
u
∞
(
x
)
|
2
*
(
s
)
d
x
+
∫
ℝ
N
‍
f
(
x
,
u
∞
(
x
)
)
u
∞
(
x
)
d
x
so that
(70)
I
(
u
∞
)
=
s
N
∫
ℝ
N
‍
|
u
∞
(
x
)
|
2
*
(
s
)
d
x
+
1
2
∫
ℝ
N
‍
f
(
x
,
u
∞
(
x
)
)
u
∞
(
x
)
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
∞
(
x
)
)
d
x
≥
0
.
The last inequality follows from assumption (
f
4
).
Now, we conclude the proof of Proposition12.
We writev
j
:
=
u
j
-
u
∞; then, v
j
→
0 weakly in E. Moreover, since (58) holds true, by the Brézis-Lieb Lemma, we get
(71)
∫
ℝ
2
N
‍
|
u
j
(
x
)
-
u
j
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
|
u
j
(
x
)
|
2
d
x
=
∫
ℝ
2
N
‍
|
v
j
(
x
)
-
v
j
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
|
v
j
(
x
)
|
2
d
x
+
∫
ℝ
2
N
‍
|
u
∞
(
x
)
-
u
∞
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
|
u
∞
(
x
)
|
2
d
x
+
∘
(
1
)
,
∫
ℝ
N
‍
|
u
j
(
x
)
|
2
*
(
s
)
d
x
=
∫
ℝ
N
‍
|
v
j
(
x
)
|
2
*
(
s
)
d
x
+
∫
ℝ
N
‍
|
u
∞
(
x
)
|
2
*
(
s
)
d
x
+
∘
(
1
)
.
Then,
(72)
c
⟵
I
(
u
j
)
=
1
2
∫
ℝ
2
N
‍
|
u
j
(
x
)
-
u
j
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
1
2
∫
ℝ
N
‍
V
(
x
)
|
u
j
(
x
)
|
2
d
x
-
1
2
*
(
s
)
∫
ℝ
N
‍
|
u
j
(
x
)
|
2
*
(
s
)
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
j
(
x
)
)
d
x
=
1
2
∫
ℝ
2
N
‍
|
v
j
(
x
)
-
v
j
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
1
2
∫
ℝ
N
‍
V
(
x
)
|
v
j
(
x
)
|
2
d
x
+
1
2
∫
ℝ
2
N
‍
|
u
∞
(
x
)
-
u
∞
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
1
2
∫
ℝ
N
‍
V
(
x
)
|
u
∞
(
x
)
|
2
d
x
-
1
2
*
(
s
)
∫
ℝ
N
‍
|
v
j
(
x
)
|
2
*
(
s
)
d
x
-
1
2
*
(
s
)
∫
ℝ
N
‍
|
u
∞
(
x
)
|
2
*
(
s
)
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
∞
(
x
)
)
d
x
+
∘
(
1
)
=
I
(
u
∞
)
+
1
2
∫
ℝ
2
N
‍
|
v
j
(
x
)
-
v
j
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
1
2
∫
ℝ
N
‍
V
(
x
)
|
v
j
(
x
)
|
2
d
x
-
1
2
*
(
s
)
∫
ℝ
N
‍
|
v
j
(
x
)
|
2
*
(
s
)
d
x
+
∘
(
1
)
,
(73)
〈
I
′
(
u
j
)
,
u
j
〉
=
∫
ℝ
2
N
‍
|
u
j
(
x
)
-
u
j
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
|
u
j
(
x
)
|
2
d
x
-
∫
ℝ
N
‍
|
u
j
(
x
)
|
2
*
(
s
)
d
x
-
∫
ℝ
N
‍
f
(
x
,
u
j
(
x
)
)
u
j
(
x
)
d
x
=
∫
ℝ
2
N
‍
|
v
j
(
x
)
-
v
j
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
|
v
j
(
x
)
|
2
d
x
+
∫
ℝ
2
N
‍
|
u
∞
(
x
)
-
u
∞
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
|
u
∞
(
x
)
|
2
d
x
-
∫
ℝ
N
‍
|
v
j
(
x
)
|
2
*
(
s
)
d
x
-
∫
ℝ
N
‍
|
u
∞
(
x
)
|
2
*
(
s
)
d
x
-
∫
ℝ
N
‍
f
(
x
,
u
∞
(
x
)
)
u
∞
(
x
)
d
x
+
∘
(
1
)
=
〈
I
′
(
u
∞
)
,
u
∞
〉

+

∫
ℝ
2
N

‍
|
v
j
(
x
)
-
v
j
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
|
v
j
(
x
)
|
2
d
x
-
∫
ℝ
N
‍
|
v
j
(
x
)
|
2
*
(
s
)
d
x
+
∘
(
1
)
.
By〈
I
′
(
u
∞
)
,
u
∞
〉
=
0 and 〈
I
′
(
u
j
)
,
u
j
〉
→
0, we get
(74)
∫
ℝ
2
N
‍
|
v
j
(
x
)
-
v
j
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
V
(
x
)
|
v
j
(
x
)
|
2
d
x
⟶
b
,
∫
ℝ
N
‍
|
v
j
(
x
)
|
2
*
(
s
)
d
x
⟶
b
.
By the definition of S
s, we have
(75)
∥
v
j
∥
E
2
≥
S
s
∥
v
j
∥
L
2
*
(
s
)
(
ℝ
N
)
2
and so b
≥
S
s
b
2
/
2
*
(
s
). Either b
=
0 or b
≥
S
s
N
/
2
s. If b
=
0, the proof is complete. Assuming that b
≥
S
s
N
/
2
s, we obtain, from (49), (68), and (72),
(76)
s
N
S
s
N
/
2
s
≤
(
1
2
-
1
2
*
(
s
)
)
b
≤
c
<
s
N
S
s
N
/
2
s
which is a contradiction. Thus b
=
0 and
(77)
∥
u
j
-
u
∞
∥
E
⟶
0
as j
→
+
∞. This ends the proof of Proposition 12.We have finished the proof of Theorem1 by Propositions 8, 10, and 12 and the mountain pass theorem.


## 4. The Proof of Theorem2

In this section we consider the caseV
(
x
)
≡
1, study the mountain pass geometry and Palais-Smale condition in a suitable energy range, and finish the proof of Theorem 2. We consider the functional
(78)
I
1
(
u
)
=
1
2
∫
ℝ
2
N
‍
|
u
(
x
)
-
u
(
y
)
|
2
|
x
-
y
|
N
+
2
s
d
x
d
y
+
1
2
∫
ℝ
N
‍
|
u
(
x
)
|
2
d
x
-
1
2
*
(
s
)
∫
ℝ
N
‍
|
u
(
x
)
|
2
*
(
s
)
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
(
x
)
)
d
x
,
where F is defined as in (
f
4
). Then I
1
∈
C
1
(
E
,
ℝ
) and critical points of I
1 are solutions of
(79)
(
-
Δ
)
s
u
+
u
=
|
u
|
2
*
(
s
)
-
2
u
+
f
(
x
,
u
)
,
u
∈
H
s
(
ℝ
N
)
.
The Fréchet derivative of I
1 is
(80)
〈
I
1
′
(
u
)
,
φ
〉
=
∫
ℝ
2
N
‍
(
u
(
x
)
-
u
(
y
)
)
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
2
s
d
x
d
y
+
∫
ℝ
N
‍
u
(
x
)
φ
(
x
)
d
x
-
∫
ℝ
N
‍
|
u
(
x
)
|
2
*
(
s
)
-
2
u
(
x
)
φ
(
x
)
d
x
-
∫
ℝ
N
‍
f
(
x
,
u
(
x
)
)
φ
(
x
)
d
x
for any φ
∈
H
r
s
(
ℝ
N
).Similar to the proof of Theorem1, we have the following conclusions.Proposition 13.
LetN
>
2
s, s
∈
(
0,1
), and f satisfy (f
0)–(f
4). Then, there exist ρ
>
0 and β
>
0 such that for any u
∈
H
r
s
(
ℝ
N
) with ∥
u
∥
H
r
s
=
ρ it results that I
1
(
u
)
≥
β.Proposition 14.
There existsu
0
∈
H
r
s
(
ℝ
N
)
∖
{
0
} with u
0
≥
0 a.e. in ℝ
N, such that
(81)
sup
t
≥
0
I
1
(
t
u
0
)
<
s
N
S
r
,
s
N
/
2
s
.Proposition 15.
LetN
>
2
s, s
∈
(
0,1
), and f satisfy (f
0)–(f
4). Then, there exists e
∈
H
r
s
(
ℝ
N
) such that e
≥
0 a.e. in ℝ
N, ∥
e
∥
H
r
s
>
ρ, and I
1
(
e
)
<
β, where ρ and β are given in Proposition 13.In particular, we can constructe as follows:
(82)
e
=
t
0
u
0
with u
0 as in (81) and t
0
>
0 large enough.We easily see thatI
1
(
0
)
=
0
<
β, with β given in Proposition 13. Now, set
(83)
c
=
inf
T
∈
Γ
sup
u
∈
T
(
[
0,1
]
)
I
1
(
u
)
,
where
(84)
Γ
=
{
T
∈
C
(
[
0,1
]
;
H
r
s
(
ℝ
N
)
)
:
T
(
0
)
=
0
,
T
(
1
)
=
e
}
,
with e
=
t
0
u
0 in Proposition 13.Proposition 16.
The constantc is given in (83) such that
(85)
β
≤
c
<
s
N
S
r
,
s
N
/
2
s
,
where β is given in Proposition 13 and S
r
,
s is defined in formula (13).By [6, Theorem  2.2], we have a sequence u
j in H
r
s
(
ℝ
N
) such that
(86)
I
1
(
u
j
)
⟶
c
,
sup
{
|
〈
I
1
′
(
u
j
)
,
φ
〉
|
:
φ
∈
H
r
s
(
ℝ
N
)
,
∥
φ
∥
H
r
s
=
1
}
⟶
0
as j
→
+
∞.Proposition 17.
There existsu
∞
∈
H
r
s
(
ℝ
N
) such that, up to a subsequence, ∥
u
j
-
u
∞
∥
H
r
s
→
0 as j
→
+
∞.We have finished the proof of Theorem2 by Propositions 13, 15, and 17 and the mountain pass theorem.


## Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.



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*Source: 143741-2014-01-12.xml*