In this section, stand for different positive constants for .
By a direct computation, (f 1) and (f 2) imply that is even and satisfies
(3.1)
(3.2)
uniformly for .
(f 3±) implies that
uniformly for .
Lemma 3.1 Suppose that f satisfies (f 1)-(f 3±). Then the function φ satisfies the condition for any .
Proof First we define an operator
for any . By a direct computation, B is a bounded self-adjoint linear operator on E. Thus is also a self-adjoint linear operator on E. Then E has an orthogonal decomposition
Also there exists such that
(3.4)
(3.5)
So, set
and
for any , is defined in Section 2.
Let be any sequence such that
We first prove that is bounded. For any , (3.2) implies that there exists a constant such that
(3.6)
for all and . Since , we have
where , , . By (3.3±), there exists a constant such that
Therefore we have
Thus we have that is bounded. Using similar arguments, we can prove that is bounded. Consider . Arguing indirectly, we suppose is unbounded, then we have . According to the definition of , this implies that there are constants such that
(3.7)
By (3.7), we have
(3.8)
Then
(3.9)
By (3.8) and (3.3±), (3.9) is a contradiction. Hence is bounded. Therefore there exists a constant such that
So, we get is bounded, and going if necessary to a subsequence, we can assume that in E and in . Write and , then in E, in E and in .
In view of (3.6) and in , it is easy to verify
and
But then as , and
This yields in E. Similarly, in E and hence in E, that is, Φ satisfies the condition. Thus φ satisfies the condition. The proof of Lemma 3.1 is complete. □
Proof of Theorem 1.1 For , let , , , and
(3.10)
where and are from Section 2.
In order to obtain this theorem, we apply Theorem A to the functional . The proof of this theorem is divided into the following three steps.
Step 1. Φ satisfies ().
We first check that is -closed for any . Let be any sequence -converging to some . Write and , then in E and hence is bounded in the norm topology. Note that for and for , then for any and small , we have by (3.10) and (3.6)
which implies that Ψ is bounded from below on E. Consequently, combining and shows that is bounded in E by (2.3) and hence
(3.11)
Moreover, since , we have
It follows from (3.11) and the above inequality that is also bounded in E since all norms are equivalent in a finite dimensional space. Then () is bounded, and hence we can assume that converges weakly to in E. Thus we have . Note that is an equivalent norm on . By the lower semi-continuity of the norm, we get
that is, and hence is -closed.
Next, we prove that is continuous. To achieve this, it is sufficient to demonstrate that has the same property. Suppose in E, then converges uniformly to z on . Hence, for every given , we see that converges to in measure on . Moreover, by (3.6), one has
for all k and , where and denotes the natural norm of . Thus, the Vitali theorem is applicable and
for any . So, Φ satisfies ().
Step 2. Φ satisfies ().
By (3.1), there exist , such that
(3.12)
By Proposition 1.1 in [3], there is a positive constant ξ such that . Set small , then for each with , one has , and hence by (3.12)
Therefore,
and hence () holds.
Step 3. Φ satisfies ().
Let
Obviously, and . In order to obtain the desired conclusion, it is sufficient to prove that as on .
Let . By the definition of m, there exists a constant such that
for . Clearly, for any ,
(3.14)
Let . We claim that
(3.15)
Indeed, for , by (3.6), one has
which implies that (3.15) is true by the arbitrariness of ε. Then, for , by (2.3), (3.13) and (3.14), one has
Since and are finitely dimensional, (3.15) and the above estimate imply that as , . Hence () holds.
The proof of Theorem 1.1 is complete. □
Proof of Theorem 1.2 For , let , , , and
and
Then the conclusion is obtained by the same argument as in the proof of Theorem 1.1. The proof of Theorem 1.2. is complete. □
Example 3.1 Consider the following equation:
(3.16)
where . Let
Then
Let
Then
By a straightforward computation, we have
and
uniformly for .
Take , , , . By Theorem 1.2, we have , , for , , , ,
Then we have . By Theorem 1.2, (3.16) possesses at least two pairs 2π-periodic solutions.