We will prove Theorem 1.1 by using Theorem 2.1. Firstly, we give the following three useful lemmas.
Lemma 3.1 Under the assumptions of Theorem 1.1, there exists large enough such that for all .
Proof Let , then there exists such that
Otherwise, for any positive integer n, there exists such that
for all k. Set , then and
for all k. Since , it follows from the compactness of the unit sphere of that there exists a subsequence, say , such that converges to some in . Hence, we have . By the equivalence of the norms on the finite-dimensional space , we have in , i.e.,
Thus there exist such that
In fact, if not, we have
for all positive integer n. This implies that
as , which gives a contradiction. Therefore, (3.4) holds.
and . By (3.2) and (3.4), we have
for all positive integer n. Let n be large enough such that and . Then we have
This implies that
for all large n, which is a contradiction with (3.3). Therefore, (3.1) holds.
For any , let . By condition (), for , there exists such that
Hence one has
for all with . It follows from ()-() and (3.1) that
for all with . Since , for large enough, we have . □
Lemma 3.2 Under the assumptions of Theorem 1.1, there exist , such that for all .
Proof Set . Then as . Indeed, it is clear that , so that , as . For every , there exists such that and . By the definition of , in E. Then it implies that in . Thus we have proved that . By (), we have
By (), for any , there exists such that
Therefore, there exists such that
Hence, for any , choose , by () and (3.5), we have
Let . Then, for any with , we have
uniformly for λ as . □
Lemma 3.3 Under the assumptions of Theorem 1.1, there exist as , such that , , where .
Proof It is easy to verify that (C1) and (C2) of Theorem 2.1 hold. By Lemmas 3.1, 3.2 and Theorem 2.1, we can obtain the result. □
Proof of Theorem 1.1 For the sake of notational simplicity, in what follows we always set for all . By Lemma 3.3, it suffices to prove that is bounded and possesses a strong convergent subsequence in E. If not, passing to a subsequence if necessary, we assume that as . In view of (), there exists such that
and combining (), we have
This implies that
Note that from (), . Let , then
By (), there exists such that
By (2.5), () and the Hölder inequality, one has
where is a constant independent of n. By (3.8) we obtain
combining this inequality with (3.6) and (3.7) yields that
as . Combining this with (3.9), we have
since . This is a contradiction. Therefore, is bounded in E. Without loss of generality, we may assume in E. Then in . Note that
Taking , we have , which means that in E and . Hence, has a critical point with . Consequently, we obtain infinitely many solutions since . □
Lemma 3.4 Under the assumptions of Theorem 1.2, there exists small enough such that and as uniformly for .
Proof For any , by using defined in Lemma 3.2, together with () and (), we have
for all with . Obviously, as . So and as uniformly for . □
Lemma 3.5 Under the assumptions of Theorem 1.2, there exists small enough such that for all .
Proof For any , by ()-() and the equivalence of the norms on the finite-dimensional space , we have
Since , for small enough, we can get for all . □
Proof of Theorem 1.2 It is easy to verify that (C1) and (D1) hold under the assumptions of Theorem 1.2. By Lemmas 3.4 and 3.5, the condition (D2) is also satisfied. Therefore, by Theorem 2.2 there exist , such that , as . In the following we show that is bounded. Indeed, note that
for some . Since , (3.10) yields that is bounded in E. By a standard argument, this yields a critical point of such that . Since as , we can obtain infinitely many critical points. □