In this section, we assume that
is called a weak solution of the problem (1.1) if
and we know that the critical points of I are just the weak solutions of the problem (1.1).
is a separable and reflexive Banach space, then there exist (see [9
Now, we set
Lemma 3.1 Given , there is such that for all , .
Proof We prove the lemma by contradiction. Suppose that there exist and for every such that . Taking , we have for every and . Hence, is a bounded sequence, and we may suppose, without loss of generality, that in . Furthermore, for every since for all . This shows that . On the other hand, by the compactness of embedding , we conclude that . This proves the lemma. □
Now suppose that
. From (
), we know that
to be chosen posteriorly by Lemma 3.1, we have for all
Now, taking and noting that , if , we can choose such that , , and for every , , the proof is complete. □
Lemma 3.3 Suppose f satisfies (). Then given , there exist a subspace W of and a constant such that and .
Proof Let and be such that , and . First, we take with . Considering , we have . Let and be such that , and . Next, we take with . After a finite number of steps, we get such that , , and for all . Let , by construction, , and for every .
. Now, it suffices to verify that
From the condition (
, there is
such that for every
, a.e. x
where and . Observing that W is finite dimensional and we have , , the inequality is obtained by taking ; the proof is complete. □
Lemma 3.4 Suppose f satisfies (), then I satisfies the (PS) condition.
We suppose that
Noting that , , is bounded. By , Lemma 3.1, we know that I satisfies the (PS) condition. □
Proof of Theorem 2.1 First, we recall that , where and are defined in (3.1). Invoking Lemma 3.2, we find , and I satisfies with . Now, by Lemma 3.3, there is a subspace W of with and such that I satisfies (). Since and I is even, we may apply Lemma 2.4 to conclude that I possesses at least k pairs of nontrivial critical points. The proof is complete. □