Ground state solution for a class of indefinite variational problems with critical growth
Claudianor O. Alves, Geilson F. Germano

TL;DR
This paper proves the existence of ground state solutions for a class of indefinite variational problems with critical growth, involving asymptotically periodic potentials and nonlinearities.
Contribution
It establishes ground state solutions for indefinite problems with critical growth under asymptotic periodicity conditions, extending previous results to more general settings.
Findings
Existence of ground state solutions proven.
Applicable to problems with asymptotically periodic potentials.
Handles nonlinearities with critical growth.
Abstract
In this paper we study the existence of ground state solution for an indefinite variational problem of the type where , and are continuous functions verifying some technical conditions and possesses a critical growth. Here, we will consider the case where the problem is asymptotically periodic, that is, is -periodic, goes to 0 at infinity and is asymptotically periodic.
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Ground state solution for a class of indefinite variational problems with critical growth
Claudianor O. Alves , Geilson F. Germano C. O. Alves was partially supported by CNPq/Brazil 301807/2013-2 and INCT-MAT, [email protected]. F. Germano was partially supported by CAPES, [email protected]
Universidade Federal de Campina Grande
Unidade Acadêmica de Matemática
CEP: 58429-900, Campina Grande - Pb, Brazil
Abstract
In this paper we study the existence of ground state solution for an indefinite variational problem of the type
[TABLE]
where , and are continuous functions verifying some technical conditions and possesses a critical growth. Here, we will consider the case where the problem is asymptotically periodic, that is, is -periodic, goes to 0 at infinity and is asymptotically periodic.
Mathematics Subject Classifications (2010): 35B33, 35A15, 35J15 .
Keywords: critical growth, variational methods, elliptic equations, indefinite strongly functional.
1 Introduction
In this paper we study the existence of ground state solution for an indefinite variational problem of the type
[TABLE]
where , are continuous functions verifying some technical conditions and has a critical growth. Here, we will consider the case where the problem is asymptotically periodic, that is, is -periodic, goes to 0 at infinity and is asymptotically periodic.
In [13], Kryszewski and Szulkin have studied the existence of ground state solution for an indefinite variational problem of the type
[TABLE]
where is a -periodic continuous function such that
[TABLE]
Related to the function , they assumed that is continuous, -periodic in with
[TABLE]
and
[TABLE]
for some , and where if and if . The above hypotheses guarantee that the energy functional associated with given by
[TABLE]
is well defined and belongs to . By , there is an equivalent inner product in such that
[TABLE]
where and corresponds to the spectral decomposition of with respect to the positive and negative part of the spectrum with , where and . In order to show the existence of solution for , Kryszewski and Szulkin introduced a new and interesting generalized link theorem. In [16], Li and Szulkin have improved this generalized link theorem to prove the existence of solution for a class of indefinite problem with being asymptotically linear at infinity.
The link theorems above mentioned have been used in a lot of papers, we would like to cite Chabrowski and Szulkin [5], do Ó and Ruf [8], Furtado and Marchi [9], Tang [30, 31] and their references.
Pankov and Pflüger [21] also have considered the existence of solution for problem with the same conditions considered in [13], however the approach is based on an approximation technique of periodic function together with the linking theorem due to Rabinowitz [22]. After, Pankov [20] has studied the existence of solution for problems of the type
[TABLE]
by supposing , and employing the same approach explored in [21]. In [20] and [21], the existence of ground state solution has been established by supposing that is and there is such that
[TABLE]
However, in [20], Pankov has found a ground state solution by minimizing the energy functional on the set
[TABLE]
The reader is invited to see that if is definite strongly, that is, when , the set is exactly the Nehari manifold associated with . Hereafter, we say that is called a ground state solution if
[TABLE]
In [25], Szulkin and Weth have established the existence of ground state solution for problem by completing the study made in [20], in the sense that, they also minimize the energy function on , however they have used more weaker conditions on , for example is continuous, -periodic in and satisfies
[TABLE]
for some and .
[TABLE]
[TABLE]
and
[TABLE]
The same approach has been used by Zhang, Xu and Zhang [36, 37] to study a class of indefinite and asymptotically periodic problem.
After a review bibliography, we have observed that there are few papers involving indefinite problem whose the nonlinearity has a critical growth. For example, the critical case for was considered in [5], [29] and [37] when is given by
[TABLE]
with being a function with subcritical growth and be a continuous function satisfying some conditions. For the case , we know only the paper [8] which considered the periodic case with having an exponential critical growth, namely there is such that
[TABLE]
Motivated by ideas found in Szulkin and Weth [25, 26] together with the fact that there are few papers involving critical growth for and and indefinite problem, we intend in the present paper to study the existence of ground state solution for , with the nonlinearity having critical growth and the problem being asymptotically periodic. Since we will work with the dimensions and , we will state our conditions in two blocks, however the conditions on and are the same for any these dimensions.
The conditions on and .
On the functions and , we have assumed the following conditions:
is continuous and -periodic.
.
is continuous and .
With relation to function , we have assumed the following conditions:
The dimension :
For this case, we suppose that is the form
[TABLE]
with and
and , where are continuous function, are -periodic, and are positive.
There is such that
[TABLE]
If , we assume that .
The dimension :
There exist functions such that
[TABLE]
where and are continuous functions, is -periodic with respect to , is nonnegative and satisfies the following condition: Given and , there exists such that
[TABLE]
as uniformly with respect to .
For each fixed , the functions and are increasing on and decreasing on .
There exist such that
[TABLE]
for all , where
[TABLE]
There exists such that for all .
for some positive continuous function with and .
An example of a function verifying is
[TABLE]
with and .
The above conditions imply that has a critical growth if or .
Our main theorem is the following:
Theorem 1.1
Assume that , , and hold. Then, problem has a ground state solution for . If , there is such that if , then problem has a ground state solution.
The Theorem 1.1 completes the study made in some of the papers above mentioned, in the sense that we are considering others conditions on and . For example, for the case , it completes the study made in [25], because the critical case was not considered for or , and the case asymptotically periodic was not also analyzed. The Theorem 1.1 also completes [8], because in that paper was proved the existence of a solution only for periodic case, while that we are finding ground state solution for the periodic and asymptotically periodic case by using a different method. Finally, the above theorem completes the main result of [29] and [36], because the authors considered only the case , and also the paper [5], because the dimension was not considered as well as the asymptotically periodic case. Moreover, in [5] and [29] the authors considered only the
[TABLE]
In Theorem 1.1 this condition was not assumed if .
Before concluding this introduction, we would like point out that the reader can find others interesting results involving indefinite variational problem in Jeanjean [12], Schechter [27, 28], Lin and Tang [17], Willem and Zou [34], Yang [35] and their references.
Notation: In this paper, we use the following notations:
- •
The usual norms in and will be denoted by and respectively.
- •
denotes (possible different) any positive constant.
- •
denotes the open ball with center and radius in .
- •
We say that in when
[TABLE]
- •
If is a mensurable function, the integral will be denoted by .
The plan of the paper is as follows: In Section 2 we will show some technical lemmas and prove the Theorem 1.1 for , while in Section 3 we will focus our attention to the dimension .
2 The case
In this section, our intention is to prove the Theorem 1.1 for the case . Some technical lemmas this section also are true for dimension and they will be used in Section 3.
In this section, our focus is the indefinite problem
[TABLE]
whose the energy functional given by
[TABLE]
is well defined, and its critical points are precisely weak solutions of (2.1). Here, is the bilinear form
[TABLE]
Note that the bilinear form is not positive definite, therefore it does not induce a norm. As in [25], there is an inner product in such that
[TABLE]
where and corresponds to the spectral decomposition of with respect to the positive and negative part of the spectrum with , where and . It is well known that is positive definite on , is negative definite on and the norm is an equivalent norm to the usual norm in , that is, there are such that
[TABLE]
Hereafter, we denote by the functional defined by
[TABLE]
or equivalently,
[TABLE]
Note that the critical points of are weak solutions of the periodic problem
[TABLE]
In the sequel, and denote the following sets
[TABLE]
and
[TABLE]
Therefore
[TABLE]
Moreover, we denote by and the real numbers
[TABLE]
2.1 Technical lemmas
In this section we are going to show some lemmas which will be used in the proof of main Theorem 1.1.
Lemma 2.1
If and where , and , then
[TABLE]
Proof. In the sequel, we fix
[TABLE]
and
[TABLE]
Then by a simple computation,
[TABLE]
Now, the proof follows by adapting the ideas explored in [25, Proposition 2.3].
Lemma 2.2
Let be a compact subset, then there exists such that and .
Proof. Setting the functional
[TABLE]
with , we have
[TABLE]
Now, we apply [25, Lemma 2.2] to the functional to get the desired result.
Lemma 2.3
For all , the functional is weakly upper semicontinuous.
Proof. First of all, note that is weakly closed, because it is convex strongly closed. Now, we claim that the functional
[TABLE]
is weakly lower semicontinuous. Indeed, if on , then after passing to a subsequence a.e. in . Then by Fatou’s Lemma,
[TABLE]
[TABLE]
leading to
[TABLE]
Furthermore, the functional
[TABLE]
is weakly upper semicontinuous. In fact, since
[TABLE]
if with , then in and in . Thus,
[TABLE]
As , the result is proved.
Lemma 2.4
For each , is a singleton set and the element of this set is the unique global maximum of
Proof. The proof follows very closely the proof of [25, Lemma 2.6].
Lemma 2.5
There exists such that .
Proof. In what follows, let us fix and . For ,
[TABLE]
Thereby, the lemma follows by taking satisfying
[TABLE]
Lemma 2.6
The real number given in (2.8) is positive. In addition, if then .
Proof. By Lemma 2.5, there is such that
[TABLE]
For all , we know that , then by Lemma 2.4,
[TABLE]
from where it follows that
[TABLE]
In addition, for all ,
[TABLE]
implying that .
Next we will show a boundedness from above for which will be crucial in our approach. However, before doing this we need to prove two technical lemmas. The first one is true for and it has the following statement
Lemma 2.7
Consider and let and . Then there exists such that
[TABLE]
for all and with .
Proof. If the lemma does not hold, there are and satisfying
[TABLE]
Setting , we obtain
[TABLE]
Thus, passing to a subsequence if necessary,
[TABLE]
On the other hand,
[TABLE]
showing that is a bounded sequence in . As , there is such that for some subsequence (not renamed) in . Then by (2.10),
[TABLE]
which is absurd, since .
Lemma 2.8
Let be fixed. Then there are satisfying
[TABLE]
Proof. From Lemma 2.2,
[TABLE]
for some . Hence, there are and with and
[TABLE]
Next, we will prove that there exists such that
[TABLE]
Arguing by contradiction, suppose that for all
[TABLE]
Such supposition permit us to conclude that . On the other hand, recalling that
[TABLE]
we are leading to
[TABLE]
which is a contradiction. This completes the proof.
Now, we are ready to show the estimate from above involving the number given in (2.8)
Proposition 2.9
Assume the conditions of Theorem 1.1. If , then
[TABLE]
If , there is such that the estimate (2.14) holds for .
Proof. Since , it is enough to prove that
[TABLE]
If and , the estimate is made in [5, Proposition 4.2]. Next we will do the proof for and . To this end, we follow the same notation used in [5]. Let
[TABLE]
where and is such that
[TABLE]
From [33], we know that the estimates below hold
[TABLE]
and
[TABLE]
Adapting the same idea explored in [5, Proposition 4.2], for each we obtain
[TABLE]
where does not depend on . Now, arguing as in [1], we get
[TABLE]
implying that
[TABLE]
Moreover, in [1], we also find that
[TABLE]
from where it follows that there exists small enough verifying
[TABLE]
and so,
[TABLE]
for some small enough.
Now, we will consider the case . For each , the Lemma 2.8 guarantees the existence of satisfying
[TABLE]
Therefore, applying Lemma 2.7,
[TABLE]
where
[TABLE]
As
[TABLE]
there is such that
[TABLE]
showing the desired result.
Lemma 2.10
Let be a sequence verifying
[TABLE]
for some . Then, is bounded in .
Proof. In the sequel, let be the characteristic function on interval ,
[TABLE]
where . Fixing
[TABLE]
it follows that
[TABLE]
Note that
[TABLE]
for some sufficiently large. So
[TABLE]
Analogously,
[TABLE]
Since , the inequalities (2.17) and (2.18) give
[TABLE]
The last two inequalities lead to
[TABLE]
from where it follows
[TABLE]
for some . On the other hand,
[TABLE]
Thus,
[TABLE]
The same argument works to prove that
[TABLE]
Recalling that , the estimates (2.21) and (2.22) combined give
[TABLE]
On the other hand, we know that
[TABLE]
that is,
[TABLE]
where was fixed in . Now, (2.23) combines with (2.24) to give
[TABLE]
This concludes the verification of Lemma 2.10.
As a byproduct of the last lemma, we have the corollaries below
Corollary 2.11
If is a (PS) sequence for , then is bounded. In addition, if in , then is a solution of
Corollary 2.12
* is coercive on , that is, as and .*
The Lemma 2.4 permits to consider a function
[TABLE]
The above function will be crucial in our approach. Next, we establish its continuity.
Lemma 2.13
The function is continuous.
Proof. Suppose in . Since
[TABLE]
without loss of generality, we may assume that .
There are and such that
[TABLE]
Note that is a compact set. Thereby, by Lemma 2.2, there exists such that in for all . Hence,
[TABLE]
showing that is a bounded sequence, and so, by Corollary 2.12, is a bounded sequence. The boundedness of implies that and are also bounded. Then, for some subsequence (not renamed),
[TABLE]
Recalling that , we obtain
[TABLE]
Thus, the Fatou’s Lemma combined with the weakly lower semicontinuous of the norm gives
[TABLE]
implying that
[TABLE]
From (2.26) and (2.27), in . Now, the Lemma 2.1 together with (2.27) guarantees that . Consequently,
[TABLE]
finishing the proof.
Hereafter, we consider the functional defined by . We know that is continuous by previous lemma. In the sequel, we denote by the restriction of to .
The next three results establish some important properties involving the functionals and and their proofs follow as in [25].
Lemma 2.14
, and
[TABLE]
Corollary 2.15
The following assertions hold:
- (a)
, and
[TABLE]
- (b)
* is a (PS)c sequence for if and only if is a (PS)c sequence for .*
- (c)
If is attained by , then .
Proposition 2.16
There exists a (PS) sequence for .
Our next lemma will be used to prove the existence of ground state solution for the periodic case.
Lemma 2.17
Let be a (PS)c sequence for functional given in (2.6) with . Then, there are and in satisfying
[TABLE]
In addition, if , the sequence is also a sequence for , and for some subsequence, in with .
Proof. By Corollary 2.11, the sequence is bounded in . Arguing by contradiction, we suppose that
[TABLE]
for some . Applying [23, Lemma 2.1], it follows that in , and so, by interpolation on the Lebesgue spaces, in for all . As
[TABLE]
we deduce that in . By a similar argument in . Hence
[TABLE]
Thereby, by continuity of , , which is absurd. Thus, there are and satisfying
[TABLE]
Recalling that for each there is such that
[TABLE]
we have
[TABLE]
finishing the proof of (2.29).
Now, assume and set . By a simple computation, we see that is also a sequence for with
[TABLE]
By Corollary 2.12, is bounded, and so, for some subsequence ( sill denoted by ), in for some . Suppose by contradiction and assume that
[TABLE]
By Concentration-Compactness Principle II due to Lions [15], there exist a countable set , and such that
[TABLE]
Now, our goal is showing that for all . First of all, note that
[TABLE]
On the other hand, setting , where is such that in , in and , with , we have that and is bounded in . So
[TABLE]
or equivalently
[TABLE]
By using the definition of and together with the last limit, we derive
[TABLE]
Now, taking the limit , we find
[TABLE]
By (2.32), . Then,
[TABLE]
If , the last inequality gives
[TABLE]
Thereby, by (2.33) and (2.34), if there exists such that , we would have
[TABLE]
which is absurd. Hence for all , so , and by (2.31), in . Consequently in which contradicts (2.30), showing that .
2.2 Proof of Theorem 1.1: The case .
The proof will be divided into two cases, more precisely, the Periodic Case and the Asymptotically Periodic Case.
1- The Periodic Case:
Proof. From Proposition 2.16, there exists a sequence for , where was given in (2.8). By Lemma 2.17, passing to a subsequence if necessary, and is a solution of problem (2.7), and so, . On the other hand
[TABLE]
[TABLE]
[TABLE]
From this, is a ground state solution for the problem .
**2- Asymptotically Periodic Case **
Proof. From definition of and , we have the inequality
[TABLE]
Next, our analysis will be divide into two cases, more precisely, and .
Assume firstly . Let be a ground state solution of (2.7) for the periodic case and such that
[TABLE]
Then,
[TABLE]
implying that with . By Corollary 2.15, part (c), we deduce that is a ground state solution of (2.1).
Now, assume and let be a sequence for given by Proposition 2.16. By Lemma 2.10, is a bounded sequence, then for some subsequence (still denoted by ) in . We claim that . Indeed, if it is easy to see that
[TABLE]
In addiction, by , we also have
[TABLE]
Arguing as in Lemma 2.17, we derive that in , and so,
[TABLE]
Hence
[TABLE]
that is, is a sequence for . By Proposition 2.9,
[TABLE]
Then, Proposition 2.17 guarantees the existence of such that in and . Consequently
[TABLE]
which is absurd, proving that . Now, we repeat the same argument explored in the periodic case to conclude that is a ground state solution of (2.1).
3 The case
In this section we are going to show the existence of ground state solution for the following indefinite problem
[TABLE]
by assuming and . Since we will work with exponential critical growth, in the next subsection we recall some facts involving this type of growth.
3.1 Results involving exponential critical growth
The exponential critical growth on is motivated by the following estimates proved by Trudinger [32] and Moser [19].
Lemma 3.1
(Trudinger-Moser inequality for bounded domains)* Let be a bounded domain. Given any , we have*
[TABLE]
Moreover, there exists a positive constant such that
[TABLE]
The next result is a version of the Trudinger-Moser inequality for whole , and its proof can be found in Cao [4] ( see also Ruf [24] ).
Lemma 3.2
(Trudinger-Moser inequality for unbounded domains)* For all , we have*
[TABLE]
Moreover, if and , then there exists a positive constant such that
[TABLE]
The Trudinger-Moser inequalities will be strongly utilized throughout this section in order to deduce important estimates. The reader can find more recent results involving this inequality in [6], [10], [11], [18] and references therein
In the sequel, we state some technical lemmas found in [3] and [7], which will be essential to carry out the proof of our results.
Lemma 3.3
Let and . Then, for every each , there exists a constant such that
[TABLE]
Lemma 3.4
Let be a sequence such that a.e. in and is bounded in . Then, in for all , and so,
[TABLE]
3.2 Technical Lemmas
In this subsection we have used the same notations of Section 2, however we will recall some of them for the convenience of the reader. In what follows, we denote by the energy functional given by
[TABLE]
where is the bilinear form
[TABLE]
It is well known that with
[TABLE]
Therefore critical points of are solutions of (3.35). Moreover, we can rewrite the functional of the form
[TABLE]
In what follows, we also consider the -functional
[TABLE]
or equivalently
[TABLE]
whose the critical points are weak solutions of periodic problem
[TABLE]
As in Section 2, we will consider the sets
[TABLE]
[TABLE]
Hence
[TABLE]
Moreover, we fix the real numbers
[TABLE]
Lemma 3.5
If and where and such that , then
[TABLE]
Proof. The proof follows as in Lemma 2.1.
Lemma 3.6
Let be a compact subset, then there exists such that and .
Proof. Fix and repeat the argument used in the proof of Lemma 2.2.
Lemma 3.7
For all , the functional is weakly upper semicontinuous.
Proof. See proof of Lemma 2.3.
Lemma 3.8
For all , is a singleton set and the element of this set is the unique global maximum of
Proof. See proof of Lemma 2.4.
In the proof of next lemma the fact that has an exponential critical growth brings some difficulty and we will do its proof.
Lemma 3.9
There exists such that .
Proof. Given and , there is such that
[TABLE]
Then, for all , the Lemmas 3.2 and 3.3 lead to
[TABLE]
By Lemma 3.2, if ,
[TABLE]
So,
[TABLE]
Hence, decreasing if necessary and fixing small enough, we get
[TABLE]
Lemma 3.10
The real number is positive. In addition, if then .
Proof. See proof of Lemma 2.6
The next lemma shows that sequences of are bounded, as we are working with the exponential critical growth the arguments explored in Section 2 does not work in this case and a new proof must be done.
Lemma 3.11
If is a sequence such that
[TABLE]
for some , then is bounded in and is bounded in .
Proof. First of all, note that
[TABLE]
Hence, is bounded. Recalling that for all and , it follows that is bounded in . On the other hand, we know that
[TABLE]
and so,
[TABLE]
where .
Claim 3.12
* is a bounded sequence.*
Indeed, by a direct computation, there exists such that
[TABLE]
Moreover, by [8, Lemma 2.11],
[TABLE]
Now, the Lemma 3.2 combined with the above inequalities for and leads to
[TABLE]
As is bounded in , the last inequality yields is bounded. Consequently, there exists satisfying
[TABLE]
Thereby, by (3.37),
[TABLE]
Analogously, there is such that
[TABLE]
The inequalities (3.40) and (3.41) combine to give
[TABLE]
for some . Hence,
[TABLE]
from where it follows that is bounded.
As a byproduct of the last lemma we have the corollary below
Corollary 3.13
* is coercive on , that is, as .*
Moreover, we also have the following result
Corollary 3.14
If is a (PS) sequence for , then is bounded. In addition, if in , then is a solution of
Proof. The corollary follows applying the Lemmas 3.2 and 3.4.
As in Section 2, the Lemma 3.8 permits to define a function
[TABLE]
Now, we invite the reader to observe that the same approach used in Section 2 works to guarantee that the proposition below holds
Proposition 3.15
There exists a (PS) sequence for .
Our next proposition is crucial when has an exponential critical growth.
Proposition 3.16
Fixed , there is such that for , where was given in (2.5).
Proof. Let with and set
[TABLE]
where
[TABLE]
with given in Lemma 2.7. Then, a straightforward computation leads to
[TABLE]
Thereby, by and Lemma 2.7,
[TABLE]
From the last inequality there is such that
[TABLE]
finishing the proof.
Proposition 3.17
Fix and . Then, there exist a sequence and such that
[TABLE]
Moreover, increasing if necessary, the sequence can be chosen in .
Proof. Suppose by contradiction that the lemma does not hold for some . Then, by a lemma due to Lions [14],
[TABLE]
Define . Since for all , from Lemma 3.10 we have , and so,
[TABLE]
On the other hand, we also know that
[TABLE]
As and , we derive that
[TABLE]
By [2, Proposition 2.3], we have . Therefore, passing to the limit in (3.42) as , we obtain
[TABLE]
which contradicts the Proposition 3.16. Thus, there are and such that
[TABLE]
Now, we repeat the same idea explored in Lemma 2.17 to conclude the proof.
3.3 Proof of Theorem 1.1: The case .
As in Section 2, the proof will be divided into two cases, the Periodic Case and the Asymptotically Periodic Case.
3.4 Periodic Case
Proof. First of all, we recall there is a sequence for which must be bounded. Thus, there is such that for some subsequence of , still denoted by itself, we have
[TABLE]
and
[TABLE]
Moreover, by Lemma 3.11 the sequence is bounded in . Therefore, by Lemma 3.4,
[TABLE]
If we combine the Lemma 3.2 with the density of in , we see that is a critical point of , that is,
[TABLE]
Moreover, by Fatou’s Lemma, we also have
[TABLE]
If , we must have
[TABLE]
showing that , and so, is a ground state solution.
If , we can apply Lemma 3.17 to get a sequence and real numbers verifying
[TABLE]
Setting , a direct computation gives that is also a for . Moreover, for some subsequence, there is such that
[TABLE]
showing that . Therefore, arguing as above, is a ground state solution for .
3.5 The Asymptotically Periodic Case
Proof. First of all, we recall that , and so, . As in Section 2, we will consider the cases and . The first one follows as in Section 2, and we will omit its proof.
In what follows, we are considering and be a sequence for which was given in Lemma 3.15. The sequence is bounded by Lemma 3.11. Thus, there is and a subsequence of , still denoted by itself, such that in . Suppose by contradiction . Repeating the arguments explored in the case , we have
[TABLE]
From , given and such that
[TABLE]
it must exist satisfying
[TABLE]
Therefore, by Lemma 3.2
[TABLE]
leading to
[TABLE]
A similar argument works to prove that
[TABLE]
The above limits yield
[TABLE]
Arguing as in the periodic case, without loss of generality, we can assume that
[TABLE]
Thus, . On the other hand, by Fatou’s Lemma,
[TABLE]
which is absurd, because we are supposing . Thereby, and since is bounded in , we can conclude that is a ground state solution of .
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