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Existence of a ground state solution for a singular elliptic problem in unbounded domain and dimension 2
Nonlinear Analysis 98 (2014) 104–109
Contents lists available at ScienceDirect
Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
Existence of a ground state solution for a singular elliptic problem in unbounded domain and dimension 2 Rafael Abreu ∗ Universidade Estadual de Campinas, IMECC, Departamento de Matemática, Rua Sérgio Buarque de Holanda, 651, Campinas, SP, CEP 13083-859, Brazil
article
abstract
info
Article history: Received 23 September 2013 Accepted 15 December 2013 Communicated by Enzo Mitidieri
In this paper we prove the existence of ground state solutions for a class of singular elliptic problems in unbounded domains of R2 with Dirichlet boundary conditions and nonlinearity with subcritical growth. © 2013 Elsevier Ltd. All rights reserved.
Keywords: Elliptic problem Subcritical growth Lack of compactness Unbounded domains Trudinger–Moser inequality
1. Introduction In this paper, we are concerned with the existence of a ground state solution for the following class of elliptic problems:
−∆u + u = K (x) u∈
H01
(U ),
f (u)
|x|a
in U ,
(P)
where U = R2 or U = R2 \ Ω , Ω ⊂ R2 denoting a bounded set with smooth boundary, and 0≤a<
1 2
.
With respect to function K : U → R ∪ {−∞, +∞}, we suppose that
(K1 ) K is positive almost everywhere in U, that is, {x ∈ U : K (x) ≤ 0} has the Lebesgue measure zero; (K2 ) K ∈ L (U ). ∞
Furthermore, we assume that continuous function f : R → R satisfies the following hypotheses:
(f1 ) There exists β ≥ 1 such that f (s) = o(sβ ) as s → 0; ∗
Tel.: +55 019981844459. E-mail addresses: [email protected], [email protected].
0362-546X/$ – see front matter © 2013 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.na.2013.12.010
(1)
R. Abreu / Nonlinear Analysis 98 (2014) 104–109
105
(f2 ) For all α > 0, we have lim
|s|→+∞
f (s) eα|s|
2
= 0;
(f3 ) There exists θ > 2 such that 0 < θ F (s) ≤ sf (s), where F (s) =
s 0
∀s ∈ R \ {0},
f (τ )dτ .
We recall that a solution u of problem (P) is said to be ground state if I (u) = c , where I : H01 (U ) → R is the functional associated to (P) and c is the Mountain Pass level associated to I, both explicitly defined in Section 2. This definition of ground state solution is motivated by a characterization which can be found in [1]. Using this characterization, it is possible to prove that if u is a solution of (P) and I (u) = c, then I (u) = inf{I (ω) : ω ∈ H01 (U ) \ {0} is a solution of (P)}. There are in the literature some articles related to problem (P). In [2], Berestycki, Gallouet and Kavian studied the problem
−∆u + u = f (u) in R2 , u ∈ H 1 (R2 ),
where continuous function f has subcritical growth and satisfies f (s) = o(s) as s → 0. In their paper, they used a constrained minimization problem to obtain the existence of ground state solution. In [3], Cao studied the problem
−∆u + u = f (x, u) in R2 , u ∈ H 1 (R2 ).
He considered function f with critical growth and satisfying f (x, s) = o(s) as s → 0 and f (x, t ) → f¯ (t ) as |x| → +∞. In [4], Alves, Montenegro and Souto proved the existence of a ground state solution for the equation
−∆u = K (x)f (u) in RN , u ∈ D1,2 (RN ),
with N ≥ 3. They supposed that f is continuous with subcritical growth and K ∈ L∞ (RN ) ∩ Lr (RN ), for some r ≥ 1, is positive almost everywhere in RN . Motivated by the work [4] and its ideas, we prove the following result: Theorem 1.1. Let f be a function that satisfies (f1 ) − (f3 ) and σ be a number that satisfies
σ >
2−a 1 − 2a
.
(2)
Let us suppose function K : U → R ∪ {−∞, +∞} satisfies (K1 ) − (K2 ) and
(K3 ) K ∈ Lr (U ), for some r >
4σ . σ (1−2a)+a−2
Then, (P) has a solution. Furthermore, if function f also satisfies
(f4 ) s →
f (s) s
is increasing in |s|,
then this solution is ground state. With the purpose of demonstrating Theorem 1.1, we use the Mountain Pass Theorem without the Palais–Smale condition and a version of the Trudinger–Moser inequality, which can be found in [5]. By the way, in this work, every time we mention the Trudinger–Moser inequality, we refer to this version proved by Adimurthi and Yang in [5]. An important point which also needs to be noted is that, in this class of problems, it is very common to try to obtain some appropriate convergence. We would like to emphasize that we obtain such convergence by using Vitali’s Theorem. According to our knowledge, the use of this theorem is not very common in this class of problems. We emphasize that the assumption a < 1/2 in (1) is necessary in our arguments. But the weight |x|−a is locally bounded if a < 2 and hence, the case a ∈ [1/2, 2) remains open. This paper is organized as follows: In Section 2, we introduce the variational formulation of the problem and some preliminary results. In Section 3, we demonstrate Theorem 1.1.
106
R. Abreu / Nonlinear Analysis 98 (2014) 104–109
2. Preliminary results We begin this section by giving some notations, definitions and basic lemmas. In the Sobolev space H01 (U ) we consider the norm given by
∥ u∥ =
21 |∇ u|2 + |u|2 dx ,
u ∈ H01 (U ),
U
and denote by I : H01 (U ) → R the energy functional defined by I ( u) =
1 2
∥u∥2 −
K (x)
F (u)
|x|a
U
s
where F (s) =
0
dx,
u ∈ H01 (U ),
(3)
f (τ )dτ . From conditions (f1 ) and (f2 ), it follows that there exist positive constants α1 and C1 such that
K (x) F (u) dx ≤ C1 ∥K ∥L∞ (U ) |x|a U
eα1 u − 1 2
|x|a
U
dx,
∀u ∈ H01 (U ).
Thus, I is well defined thanks to the Trudinger–Moser inequality. It is not difficult to check that I ∈ C 1 (H01 (U ), R). A straightforward calculation shows that I (u)v = ′
(∇ u∇v + uv) dx −
U
K (x) U
f (u)v
|x|a
dx,
∀u, v ∈ H01 (U ).
Hence, a critical point of (3) is a weak solution of (P). The first lemma is a consequence of the Trudinger–Moser inequality and is fundamental to show some convergences involving the Palais–Smale sequences. Lemma 2.1. Let θ be the number given by the condition (f3 ) and (un ) be a sequence satisfying lim sup ∥un ∥2 ≤ n→∞
2θ
θ −2
c,
for some c > 0. Then, there exist constants α > 0, t > 1 and C > 0, independent of n, such that
eα un − 1 2
t dx ≤ C ,
|x|a
U
for n large enough. Proof. Let
4π a θ − 2 1 . α= 1− σ 2 2θ c
(4)
From (1) and (2), we have that α > 0. Moreover, let 1
m :=
2
1 + σ.
2
2θ
θ −2
2−a
c.
Then,
2θ
θ −2
c
and hence, there exists n0 ∈ N such that
∥ un ∥ 2 < m ,
∀ n ≥ n0 .
From (1) and (2), we also have that 4 3
<
4σ 2 − a + 2σ + 2σ a
.
Let t ∈ R satisfying 4 3
4σ 2 − a + 2σ + 2σ a
.
(5)
R. Abreu / Nonlinear Analysis 98 (2014) 104–109
107
For a such t, it holds 4 3
at
4σ
2 − a + 2σ
2
.
Let β ∈ R such that t <β <
1−
at
4σ
2
2 − a + 2σ
= 1−
at
2
4π . αm
There exists a constant C = C (β) > 0 such that
e
α u2n
−1
e
2 βα m ∥uun ∥ n
dx ≤ C
|x|at
U
t
−1 dx
|x|at
U
for each n ≥ n0 . By the choice of β , we have
βα m < 1 −
at
2
4π .
Then, we conclude the proof by using the Trudinger–Moser inequality.
In our case, it is easy to see that the functional I in (3) verifies the Mountain Pass geometry; more exactly, we have the following lemma: Lemma 2.2. The functional I verifies the following conditions: (i) there exist constants η, ρ > 0 such that I (u) ≥ η,
∀u ∈ H01 (U ), ∥u∥ = ρ;
(ii) there exists e ∈ H01 (U ) such that ∥e∥ > ρ and I (e) < 0. Using a version of the Mountain Pass theorem without the Palais–Smale condition (see [1, Theorem 1.15]), we obtain the existence of a sequence (un ) in H01 (U ) satisfying I (un ) → c
and I ′ (un ) → 0,
as n → ∞,
(6)
where c = inf max I (γ (t )) > 0 γ ∈Γ t ∈[0,1]
and
Γ = γ ∈ C [0, 1], H01 (U ) : γ (0) = 0 and I (γ (1)) < 0 . 3. Proof of Theorem 1.1 In this section, we prove the existence of a ground state solution for (P). Proof of Theorem 1.1. We have that the Palais–Smale sequence (un ) is bounded in H01 (U ) and its weak limit u is a critical point of the functional I. We will show that u is nonzero. For this, we will use Vitali’s theorem to show that
K (x)
lim
n→∞
U
f (un )un
|x|a
K (x)
dx = U
f (u)u
|x|a
dx.
(7)
From (f1 ) and (f2 ) we have that, for all ε > 0 and all α > 0, there exist δ = δ(ε, α) > 0 and K = K (ε, α) > 0 such that
|f (s)| < ε|s|β , ∀ |s| < δ 2 |f (s)| < ε|s|3 eαs , ∀ |s| > K .
Thus, there exists C > 0 such that
|f (s)s| ≤ ε|s|β+1 + ε|s|4 eαs + max |f (s)s| δ≤|s|≤K 2 β+1 3 αs ≤ ε|s| + ε C |s| e − 1 + max |f (s)s| , 2
δ≤|s|≤K
∀s ∈ R.
108
R. Abreu / Nonlinear Analysis 98 (2014) 104–109
If A ⊂ U is a measurable set, then
K (x) A
|f (un )un | dx ≤ ε∥K ∥L∞ (U ) |x|a
β+1
U
|un | dx + ε C ∥K ∥L∞ (U ) |x|a
|un |3 U
eα un − 1 2
dx + max |f (s)s| |A| .
|x|a
δ≤|s|≤K
Here, |A| denotes the Lebesgue measure of the set A. We can consider α > 0 given by (4) and Lemma 2.1 implies that there exists t > 1 such that, up to a subsequence, eα un − 1 2
|x|a
∈ Lt (U ),
∀n ∈ N,
and there exists C > 0 such that
eα un − 1 2
t dx ≤ C ,
|x|a
U
∀n ∈ N.
By applying Holder’s inequality with exponents t and its conjugate t ′ and using continuous embedding, we have that there exist positive constants C1 and C2 such that
K (x) A
|f (un )un | dx ≤ ε C1 + ε C2 + max |f (s)s| |A| . δ≤|s|≤K |x|a
(8)
On the other hand, for all ε > 0 and all α > 0, there exists C = C (ε, α) > 0 such that
2 |f (s)s| ≤ ε|s|β+1 + C |s|3 eαs − 1 ,
∀s ∈ R.
Then, for R > 0 such that Ω ⊂ BR (0), we have
K (x)
U \BR (0)
|f (un )un | dx ≤ ε∥K ∥L∞ (U ) |x|a
β+1
|un | dx + C |x|a
U \BR (0)
U \BR (0)
K (x)|un |3
eα un − 1 2
|x|a
dx.
By considering α > 0 given by (4) and applying the Holder inequality with exponents r given by (K3 ), t given by (5), and 4 such that 1
+
r
1
+
t
1 4
= 1,
and using continuous embedding, we have that there exist positive constants C3 and C4 such that
|f (un )un | K (x) dx ≤ ε C3 + C4 |x|a U \BR (0)
U \BR (0)
1r
K (x) dx r
.
From (K3 ), we have that
U \BR (0)
K (x)r dx → 0
as R → +∞.
Thus, for R > 0 large enough,
U \BR (0)
K (x)
|f (un )un | dx ≤ ε C3 + ε C4 . |x|a
(9)
Let us notice that (8) and (9) mean that the sequence
K (x)
|f (un )un | |x|a
is equi-integrable. Provided that K ( x)
|f (un (x))un (x)| |f (u(x))u(x)| → K (x) a.e. in U , |x|a |x|a
Vitali’s theorem implies (7). Since (un ) is a Palais–Smale sequence, (7) implies that lim ∥un ∥2 = lim
n→∞
n→∞
K (x) U
f ( un ) un
|x|
a
dx = U
K (x)
f (u)u
|x|a
dx.
R. Abreu / Nonlinear Analysis 98 (2014) 104–109
109
Recalling that u is a critical point of I, we conclude that
2
K (x)
lim ∥un ∥ =
n→∞
U
f (u)u
|x|a
dx = ∥u∥2 .
If u is zero, then lim ∥un ∥2 = 0.
n→∞
Since I ∈ C 1 H01 (U ), R , we have
I (un ) → 0, and it is a contradiction because I (un ) → c and c > 0. This way, we conclude that u is nonzero. Now, we will show that u is ground state. Since (un ) is a Palais–Smale sequence, we have that 2c = lim inf 2I (un ) = lim inf 2I (un ) − I ′ (un )un
n→∞
n→∞
K ( x)
= lim inf n→∞
U
(f (un )un − 2F (un )) dx. |x|a
By Fatou’s Lemma,
K (x)
2c ≥ U
(f (u)u − 2F (u)) dx. |x|a
Provided that u is a critical point of I, we have 2I (u) = 2I (u) − I (u)u = ′
K ( x) U
(f (u)u − 2F (u)) dx. |x|a
Hence, we can conclude that I (u) ≤ c. On the other hand, the condition (f4 ) implies that c = inf I (u) : u ∈ H01 (U ) \ {0} and I ′ (u)u = 0 ,
from where it follows that I (u) ≥ c. Thus, I (u) = c.
References [1] M. Willem, Minimax Theorems, Birkhauser, 1996. [2] H. Berestycki, T. Gallouet, O. Kavian, Equations de champs scalaires euclidiens non lineáires dans le plan, C. R. Acad. Sci.; Paris Ser. I Math. 297 (5) (1983) 307–310. [3] D.M. Cao, Nontrivial solution of semilinear elliptic equation with critical exponent in R2 , Commun. Partial Differential Equations 17 (1992) 407–435. [4] C.O. Alves, M. Montenegro, M.A.S. Souto, Existence of solution for two classes of elliptic problems in RN with zero mass, J. Differential Equations 252 (10) (2012) 5735–5750. [5] Adimurthi, Y. Yang, An interpolation of Hardy inequality and Trudinger–Moser inequality in RN and its applications, Int. Math. Res. Not. IMRN. (13) (2010) 2394–2426.
Contents lists available at ScienceDirect
Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
Existence of a ground state solution for a singular elliptic problem in unbounded domain and dimension 2 Rafael Abreu ∗ Universidade Estadual de Campinas, IMECC, Departamento de Matemática, Rua Sérgio Buarque de Holanda, 651, Campinas, SP, CEP 13083-859, Brazil
article
abstract
info
Article history: Received 23 September 2013 Accepted 15 December 2013 Communicated by Enzo Mitidieri
In this paper we prove the existence of ground state solutions for a class of singular elliptic problems in unbounded domains of R2 with Dirichlet boundary conditions and nonlinearity with subcritical growth. © 2013 Elsevier Ltd. All rights reserved.
Keywords: Elliptic problem Subcritical growth Lack of compactness Unbounded domains Trudinger–Moser inequality
1. Introduction In this paper, we are concerned with the existence of a ground state solution for the following class of elliptic problems:
−∆u + u = K (x) u∈
H01
(U ),
f (u)
|x|a
in U ,
(P)
where U = R2 or U = R2 \ Ω , Ω ⊂ R2 denoting a bounded set with smooth boundary, and 0≤a<
1 2
.
With respect to function K : U → R ∪ {−∞, +∞}, we suppose that
(K1 ) K is positive almost everywhere in U, that is, {x ∈ U : K (x) ≤ 0} has the Lebesgue measure zero; (K2 ) K ∈ L (U ). ∞
Furthermore, we assume that continuous function f : R → R satisfies the following hypotheses:
(f1 ) There exists β ≥ 1 such that f (s) = o(sβ ) as s → 0; ∗
Tel.: +55 019981844459. E-mail addresses: [email protected], [email protected].
0362-546X/$ – see front matter © 2013 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.na.2013.12.010
(1)
R. Abreu / Nonlinear Analysis 98 (2014) 104–109
105
(f2 ) For all α > 0, we have lim
|s|→+∞
f (s) eα|s|
2
= 0;
(f3 ) There exists θ > 2 such that 0 < θ F (s) ≤ sf (s), where F (s) =
s 0
∀s ∈ R \ {0},
f (τ )dτ .
We recall that a solution u of problem (P) is said to be ground state if I (u) = c , where I : H01 (U ) → R is the functional associated to (P) and c is the Mountain Pass level associated to I, both explicitly defined in Section 2. This definition of ground state solution is motivated by a characterization which can be found in [1]. Using this characterization, it is possible to prove that if u is a solution of (P) and I (u) = c, then I (u) = inf{I (ω) : ω ∈ H01 (U ) \ {0} is a solution of (P)}. There are in the literature some articles related to problem (P). In [2], Berestycki, Gallouet and Kavian studied the problem
−∆u + u = f (u) in R2 , u ∈ H 1 (R2 ),
where continuous function f has subcritical growth and satisfies f (s) = o(s) as s → 0. In their paper, they used a constrained minimization problem to obtain the existence of ground state solution. In [3], Cao studied the problem
−∆u + u = f (x, u) in R2 , u ∈ H 1 (R2 ).
He considered function f with critical growth and satisfying f (x, s) = o(s) as s → 0 and f (x, t ) → f¯ (t ) as |x| → +∞. In [4], Alves, Montenegro and Souto proved the existence of a ground state solution for the equation
−∆u = K (x)f (u) in RN , u ∈ D1,2 (RN ),
with N ≥ 3. They supposed that f is continuous with subcritical growth and K ∈ L∞ (RN ) ∩ Lr (RN ), for some r ≥ 1, is positive almost everywhere in RN . Motivated by the work [4] and its ideas, we prove the following result: Theorem 1.1. Let f be a function that satisfies (f1 ) − (f3 ) and σ be a number that satisfies
σ >
2−a 1 − 2a
.
(2)
Let us suppose function K : U → R ∪ {−∞, +∞} satisfies (K1 ) − (K2 ) and
(K3 ) K ∈ Lr (U ), for some r >
4σ . σ (1−2a)+a−2
Then, (P) has a solution. Furthermore, if function f also satisfies
(f4 ) s →
f (s) s
is increasing in |s|,
then this solution is ground state. With the purpose of demonstrating Theorem 1.1, we use the Mountain Pass Theorem without the Palais–Smale condition and a version of the Trudinger–Moser inequality, which can be found in [5]. By the way, in this work, every time we mention the Trudinger–Moser inequality, we refer to this version proved by Adimurthi and Yang in [5]. An important point which also needs to be noted is that, in this class of problems, it is very common to try to obtain some appropriate convergence. We would like to emphasize that we obtain such convergence by using Vitali’s Theorem. According to our knowledge, the use of this theorem is not very common in this class of problems. We emphasize that the assumption a < 1/2 in (1) is necessary in our arguments. But the weight |x|−a is locally bounded if a < 2 and hence, the case a ∈ [1/2, 2) remains open. This paper is organized as follows: In Section 2, we introduce the variational formulation of the problem and some preliminary results. In Section 3, we demonstrate Theorem 1.1.
106
R. Abreu / Nonlinear Analysis 98 (2014) 104–109
2. Preliminary results We begin this section by giving some notations, definitions and basic lemmas. In the Sobolev space H01 (U ) we consider the norm given by
∥ u∥ =
21 |∇ u|2 + |u|2 dx ,
u ∈ H01 (U ),
U
and denote by I : H01 (U ) → R the energy functional defined by I ( u) =
1 2
∥u∥2 −
K (x)
F (u)
|x|a
U
s
where F (s) =
0
dx,
u ∈ H01 (U ),
(3)
f (τ )dτ . From conditions (f1 ) and (f2 ), it follows that there exist positive constants α1 and C1 such that
K (x) F (u) dx ≤ C1 ∥K ∥L∞ (U ) |x|a U
eα1 u − 1 2
|x|a
U
dx,
∀u ∈ H01 (U ).
Thus, I is well defined thanks to the Trudinger–Moser inequality. It is not difficult to check that I ∈ C 1 (H01 (U ), R). A straightforward calculation shows that I (u)v = ′
(∇ u∇v + uv) dx −
U
K (x) U
f (u)v
|x|a
dx,
∀u, v ∈ H01 (U ).
Hence, a critical point of (3) is a weak solution of (P). The first lemma is a consequence of the Trudinger–Moser inequality and is fundamental to show some convergences involving the Palais–Smale sequences. Lemma 2.1. Let θ be the number given by the condition (f3 ) and (un ) be a sequence satisfying lim sup ∥un ∥2 ≤ n→∞
2θ
θ −2
c,
for some c > 0. Then, there exist constants α > 0, t > 1 and C > 0, independent of n, such that
eα un − 1 2
t dx ≤ C ,
|x|a
U
for n large enough. Proof. Let
4π a θ − 2 1 . α= 1− σ 2 2θ c
(4)
From (1) and (2), we have that α > 0. Moreover, let 1
m :=
2
1 + σ.
2
2θ
θ −2
2−a
c.
Then,
2θ
θ −2
c
and hence, there exists n0 ∈ N such that
∥ un ∥ 2 < m ,
∀ n ≥ n0 .
From (1) and (2), we also have that 4 3
<
4σ 2 − a + 2σ + 2σ a
.
Let t ∈ R satisfying 4 3
4σ 2 − a + 2σ + 2σ a
.
(5)
R. Abreu / Nonlinear Analysis 98 (2014) 104–109
107
For a such t, it holds 4 3
at
4σ
2 − a + 2σ
2
.
Let β ∈ R such that t <β <
1−
at
4σ
2
2 − a + 2σ
= 1−
at
2
4π . αm
There exists a constant C = C (β) > 0 such that
e
α u2n
−1
e
2 βα m ∥uun ∥ n
dx ≤ C
|x|at
U
t
−1 dx
|x|at
U
for each n ≥ n0 . By the choice of β , we have
βα m < 1 −
at
2
4π .
Then, we conclude the proof by using the Trudinger–Moser inequality.
In our case, it is easy to see that the functional I in (3) verifies the Mountain Pass geometry; more exactly, we have the following lemma: Lemma 2.2. The functional I verifies the following conditions: (i) there exist constants η, ρ > 0 such that I (u) ≥ η,
∀u ∈ H01 (U ), ∥u∥ = ρ;
(ii) there exists e ∈ H01 (U ) such that ∥e∥ > ρ and I (e) < 0. Using a version of the Mountain Pass theorem without the Palais–Smale condition (see [1, Theorem 1.15]), we obtain the existence of a sequence (un ) in H01 (U ) satisfying I (un ) → c
and I ′ (un ) → 0,
as n → ∞,
(6)
where c = inf max I (γ (t )) > 0 γ ∈Γ t ∈[0,1]
and
Γ = γ ∈ C [0, 1], H01 (U ) : γ (0) = 0 and I (γ (1)) < 0 . 3. Proof of Theorem 1.1 In this section, we prove the existence of a ground state solution for (P). Proof of Theorem 1.1. We have that the Palais–Smale sequence (un ) is bounded in H01 (U ) and its weak limit u is a critical point of the functional I. We will show that u is nonzero. For this, we will use Vitali’s theorem to show that
K (x)
lim
n→∞
U
f (un )un
|x|a
K (x)
dx = U
f (u)u
|x|a
dx.
(7)
From (f1 ) and (f2 ) we have that, for all ε > 0 and all α > 0, there exist δ = δ(ε, α) > 0 and K = K (ε, α) > 0 such that
|f (s)| < ε|s|β , ∀ |s| < δ 2 |f (s)| < ε|s|3 eαs , ∀ |s| > K .
Thus, there exists C > 0 such that
|f (s)s| ≤ ε|s|β+1 + ε|s|4 eαs + max |f (s)s| δ≤|s|≤K 2 β+1 3 αs ≤ ε|s| + ε C |s| e − 1 + max |f (s)s| , 2
δ≤|s|≤K
∀s ∈ R.
108
R. Abreu / Nonlinear Analysis 98 (2014) 104–109
If A ⊂ U is a measurable set, then
K (x) A
|f (un )un | dx ≤ ε∥K ∥L∞ (U ) |x|a
β+1
U
|un | dx + ε C ∥K ∥L∞ (U ) |x|a
|un |3 U
eα un − 1 2
dx + max |f (s)s| |A| .
|x|a
δ≤|s|≤K
Here, |A| denotes the Lebesgue measure of the set A. We can consider α > 0 given by (4) and Lemma 2.1 implies that there exists t > 1 such that, up to a subsequence, eα un − 1 2
|x|a
∈ Lt (U ),
∀n ∈ N,
and there exists C > 0 such that
eα un − 1 2
t dx ≤ C ,
|x|a
U
∀n ∈ N.
By applying Holder’s inequality with exponents t and its conjugate t ′ and using continuous embedding, we have that there exist positive constants C1 and C2 such that
K (x) A
|f (un )un | dx ≤ ε C1 + ε C2 + max |f (s)s| |A| . δ≤|s|≤K |x|a
(8)
On the other hand, for all ε > 0 and all α > 0, there exists C = C (ε, α) > 0 such that
2 |f (s)s| ≤ ε|s|β+1 + C |s|3 eαs − 1 ,
∀s ∈ R.
Then, for R > 0 such that Ω ⊂ BR (0), we have
K (x)
U \BR (0)
|f (un )un | dx ≤ ε∥K ∥L∞ (U ) |x|a
β+1
|un | dx + C |x|a
U \BR (0)
U \BR (0)
K (x)|un |3
eα un − 1 2
|x|a
dx.
By considering α > 0 given by (4) and applying the Holder inequality with exponents r given by (K3 ), t given by (5), and 4 such that 1
+
r
1
+
t
1 4
= 1,
and using continuous embedding, we have that there exist positive constants C3 and C4 such that
|f (un )un | K (x) dx ≤ ε C3 + C4 |x|a U \BR (0)
U \BR (0)
1r
K (x) dx r
.
From (K3 ), we have that
U \BR (0)
K (x)r dx → 0
as R → +∞.
Thus, for R > 0 large enough,
U \BR (0)
K (x)
|f (un )un | dx ≤ ε C3 + ε C4 . |x|a
(9)
Let us notice that (8) and (9) mean that the sequence
K (x)
|f (un )un | |x|a
is equi-integrable. Provided that K ( x)
|f (un (x))un (x)| |f (u(x))u(x)| → K (x) a.e. in U , |x|a |x|a
Vitali’s theorem implies (7). Since (un ) is a Palais–Smale sequence, (7) implies that lim ∥un ∥2 = lim
n→∞
n→∞
K (x) U
f ( un ) un
|x|
a
dx = U
K (x)
f (u)u
|x|a
dx.
R. Abreu / Nonlinear Analysis 98 (2014) 104–109
109
Recalling that u is a critical point of I, we conclude that
2
K (x)
lim ∥un ∥ =
n→∞
U
f (u)u
|x|a
dx = ∥u∥2 .
If u is zero, then lim ∥un ∥2 = 0.
n→∞
Since I ∈ C 1 H01 (U ), R , we have
I (un ) → 0, and it is a contradiction because I (un ) → c and c > 0. This way, we conclude that u is nonzero. Now, we will show that u is ground state. Since (un ) is a Palais–Smale sequence, we have that 2c = lim inf 2I (un ) = lim inf 2I (un ) − I ′ (un )un
n→∞
n→∞
K ( x)
= lim inf n→∞
U
(f (un )un − 2F (un )) dx. |x|a
By Fatou’s Lemma,
K (x)
2c ≥ U
(f (u)u − 2F (u)) dx. |x|a
Provided that u is a critical point of I, we have 2I (u) = 2I (u) − I (u)u = ′
K ( x) U
(f (u)u − 2F (u)) dx. |x|a
Hence, we can conclude that I (u) ≤ c. On the other hand, the condition (f4 ) implies that c = inf I (u) : u ∈ H01 (U ) \ {0} and I ′ (u)u = 0 ,
from where it follows that I (u) ≥ c. Thus, I (u) = c.
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