In this section, we establish the existence of a solution for a nonlinear implicit eigenvalue problem under normalized conditions by using the Browder degree theory for class ().
Recall that a mapping is of class () if the following condition holds:
For any sequence in with and any sequence in with for which
we have and ; see [2, 6].
As mentioned in the introduction, Kartsatos and Skrypnik [4] gave the following result provided that property () is fulfilled. In a more concrete situation, we adopt the normalization method considered in [3].
Theorem 2.1 Let Ω be a bounded open set in X with . Let be a maximal monotone operator such that the closure of Ω is included in the interior of with and T satisfies condition () on . Assume that is demicontinuous, bounded and satisfies condition () such that for all and is continuous in t uniformly with respect to . Further assume that
(c1) There exists a positive number such that the weak sequential closure of the set
does not contain the origin 0, where
(c2) , where .
Then the following statements hold:
-
(a)
For each , there exists a point in such that
-
(b)
If and T satisfies condition () on ∂ Ω, then the implicit eigenvalue problem
has a solution in .
Proof (a) For our aim, we use the Browder degree given in [6]. Let ε be any positive number. We first prove that there is a number such that
(2.1)
Assume the contrary. For a sequence in with , the following occurs:
For each , either there exists a point such that , in view of , or there exists a point such that . Thus, we get a sequence in such that
(2.2)
This implies
for sufficiently large n. Since the sequence is bounded in the reflexive Banach space , we may suppose that converges weakly to some . Using (c1), it is clear that . It follows from (2.2) that
(2.3)
However, by (c2) implies , which is a contradiction to condition (). Hence assertion (2.1) holds.
Next, we consider a mapping given by
Then H is of class (). To prove this, let be any sequence in with and be any sequence in with such that
(2.4)
Since the operators T and J are monotone, it follows from
(2.5)
that
(2.6)
By (2.4) and (2.6), we have
(2.7)
There are two cases to consider. If , then and so
(2.8)
Since (2.5) implies
it follows from (2.4) and (2.8) that
Since J satisfies condition (), we obtain
which implies
on observing that T is demicontinuous on . This means that . If , we have
and hence by (2.7),
Since C satisfies condition (), we get from which , , and . Consequently, . We have just shown that the mapping H is of class ().
We are now ready to apply the degree theory of Browder [6, 7]. Then we have
The last equality is based on Theorem 3 in [6] because the operator is strictly monotone and demicontinuous on and satisfies condition (). On the other hand, it is shown in (2.1) that
Hence, in view of Theorem 4 in [6], there exist and such that
It follows from the injectivity of that . Consequently, if we let and , then we have and .
-
(b)
Let be a sequence in such that . According to statement (a), there exists a sequence in such that
If we set and , it can be written in the form
(2.9)
Without loss of generality, we may suppose that
(2.10)
where , , and . Note that belongs to . In fact, if , then implies . Since T satisfies condition () on ∂ Ω, we obtain that and therefore , which contradicts the hypothesis that . If , then (c2) implies and so . As in (2.3), a similar argument proves that converges weakly to some nonzero vector, which contradicts condition (). Thus we have shown that .
For the next aim, we now show that
(2.11)
Assume that (2.11) is false. Then there exists a subsequence of , denoted again by , such that
Hence we obtain from (2.9) that
Noticing by (2.9) and (2.10) that , we get
(2.12)
For every , we have by the monotonicity of T
which implies along with (2.12)
(2.13)
By the maximal monotonicity of T, we have and . Letting in (2.13), we get a contradiction. Therefore, (2.11) is true.
Since , it follows from (2.11) and
that
Since C satisfies condition () and , we have , which implies . Hence we obtain from (2.9) that . By Lemma 1.1, we conclude that . This completes the proof. □
Remark 2.2 In the proof of Theorem 2.1, the demicontinuity of T on is needed to show that H is of class (). This is guaranteed under an additional condition . Actually, local boundedness of T on implies the demicontinuity of T; see, e.g., [5].