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Implicit eigenvalue problems for maximal monotone operators
Fixed Point Theory and Applications volume 2012, Article number: 178 (2012)
Abstract
We study the implicit eigenvalue problem of the form
where T is a maximal monotone multi-valued operator and the operator C satisfies condition () or (). In a regularization method by the duality operator, we use the degree theories of Kartsatos and Skrypnik upon conditions of C as well as Browder’s degree. There are two cases to consider: One is that C is demicontinuous and bounded with condition (); and the other is that C is quasibounded and densely defined with condition (). Moreover, the eigenvalue problem is also discussed.
1 Introduction and preliminaries
A general eigenvalue theory for maximal monotone operators has been developed in various ways with applications to partial differential equations; see [5–7, 10–12]. A key tool was topological degrees for appropriate classes of operators in, e.g., [2–4, 8, 9, 14–16] and the method of approach was in many cases to use regularization by means of the duality operator, while in [6, 11] the eigenvalue problem was solved by a transformed equation in terms of the approximant without using the regularization method.
Let X be a real reflexive Banach space with dual space and be a maximal monotone multi-valued operator. When the resolvents of T or the single-valued operator C are compact, the solvability of the nonlinear inclusion
was studied in [5, 6, 10, 12] by applying the Leray-Schauder degree theory for compact operators. More generally, implicit eigenvalue problems were considered in [7, 10], where the single-valued and compact case was dealt with in [7]. In direction of [10], we study the implicit eigenvalue problem of the form
where satisfies condition () or (). As in [12], we adopt property () about the solvability of the related equation
where is the Brezis-Crandall-Pazy approximant introduced in [1] and is the (normalized) duality operator with a gauge function ψ.
In the present paper, we divide our investigation into two cases to apply suitable degree theory. One case deals with demicontinuous bounded operators satisfying condition () with the aid of the most elementary degree theory of Skrypnik [14], in comparison with [10]. The other case is concerned with quasibounded densely defined operators satisfying condition (), where the degree theory of Kartsatos and Skrypnik [8, 9] for densely defined operators is used. In more concrete situations, the eigenvalue problem is discussed. We point out that Browder’s degree in [3] for the reduced simple operator under the homotopy plays a crucial role in the proof of our results presented here.
Let X be a real Banach space with its dual , Ω a nonempty subset of X, and Y another real Banach space. An operator is said to be bounded if F maps bounded subsets of Ω into bounded subsets of Y. F is said to be demicontinuous if for every and for every sequence in Ω with , we have . Here the symbol → (⇀) stands for strong (weak) convergence.
A multi-valued operator is said to be monotone if
where denotes the effective domain of T. T is said to be maximal monotone if it is monotone and it follows from and
that and .
We say that an operator satisfies condition () if for every sequence in with and
we have .
We say that satisfies condition () if for every sequence in with , and
we have , and .
We say that a multi-valued operator satisfies condition () on a set if for every sequence in M with and every sequence in with where , we have .
Throughout this paper, X will always be an infinite dimensional real reflexive Banach space which has been renormed so that X and its dual are locally uniformly convex.
A function is called a gauge function if ψ is continuous, strictly increasing, and as . An operator is called a duality operator with a gauge function ψ if
If ψ is the identity map I, then is called a normalized duality operator. It is known in [13] that is continuous, bounded, surjective, strictly monotone, maximal monotone and satisfies condition ().
Given a maximal monotone operator , and , there exist unique elements and such that
Two operators and defined by
are called the Brezis-Crandall-Pazy approximants. It is known in [1] that is continuous and bounded and is demicontinuous, bounded, and maximal monotone. It is easy to see that and for .
Let be an operator, where M is a subset of X. Then is said to be continuous in t uniformly with respect to if for every and for every sequence in with , we have uniformly with respect to .
We say that C satisfies condition () if for every and for every sequence in M with and
we have .
We say that C satisfies condition () if for every and for every sequence in M with , and
we have , and .
We often need the following demiclosedness property of maximal monotone operators given in [17].
Lemma 1.1 Let be a maximal monotone multi-valued operator. Then for every sequence in , in X and in , where , imply that and .
2 Implicit eigenvalue problem about demicontinuous operators
In this section, we are concerned with the implicit eigenvalue problem for perturbed maximal monotone operators in reflexive Banach spaces by applying the degree theories of Skrypnik and Browder for nonlinear operators of monotone type.
In what follows, for a bounded subset Ω of X, let and ∂ Ω denote the closure and the boundary of Ω in X, respectively. Following Browder [2, 3], a homotopy is said to be of class () if the following condition holds:
For every sequence in with and every sequence in with such that
we have and .
Kartsatos and Skrypnik [10] obtained the following result by using Browder’s degree in [4] for multi-valued operators. As in [12], we adopt property () in terms of the Brezis-Crandall-Pazy approximant so that we can apply the most essential degree theory of Skrypnik [14] for single-valued operators to prove in a direct method.
Theorem 2.1 Let Ω be a bounded open set in X with . Let be a maximal monotone multi-valued operator with and . Let Λ, , and be three positive numbers. Suppose that is demicontinuous, bounded and satisfies condition () such that for all and is continuous in t uniformly with respect to .
-
(a)
For a given , assume the following property:
() For every , there exists a such that the equation
has no solution in Ω. Then there exists a such that
-
(b)
If , T satisfies condition () on , and property () is fulfilled for all , then there exists a such that
Proof (a) We first claim that for any , there exists a such that
Assume on the contrary that for some and for every , the following holds:
Since and is injective by the strict monotonicity of J, we have for all . Thus, (2.2) holds for all .
Consider a homotopy defined by
Then H is of class (). To prove this, let be a sequence in with and be a sequence in with such that
Since and J are monotone, it follows from
that
and
By (2.3) and (2.5), we have
There are two cases to consider. If , then and so
Using (2.3), (2.4), and (2.7), we obtain
Since J satisfies condition (), we have
which implies by the demicontinuity of and C and the continuity of J
This means that . Now, let . We have
which implies
and hence by (2.6)
Since C satisfies condition (), we get from which , , and . Consequently, . We have just shown that the homotopy H is of class ().
We are now ready to apply the degree theory of Skrypnik [14, 15]. If denotes the Skrypnik degree, the homotopy invariance property of implies that
for all . The last equality follows from Theorem 3 in [3], based on the fact that is demicontinuous, bounded, injective, and satisfies condition () and for all . For every , the existence property of implies that there is a point x in Ω such that
which contradicts property (). Therefore, the first claim (2.1) is true.
In view of (2.1), let be a sequence in with and be the corresponding sequence in such that
Without loss of generality, we may suppose that
where , , and . To arrive at the conclusion of (a) in the next step, we may suppose that . In fact, when , we know that for all because is injective on and . For our aim, we will now show that
Assume on the contrary that there exists a subsequence of , denoted again by , such that
Note by (2.8) and the monotonicity of J that
and so
Hence, it follows from that
Let and be arbitrary. Since T is monotone, , and , we have
which implies
Since and are bounded and , we have
Combining (2.10) and (2.11), we get
Since T is maximal monotone, we have and . Letting in (2.12) yields a contradiction. Thus, (2.9) holds.
Since is continuous in t uniformly with respect to and , we have by (2.9)
by observing that
Since C satisfies condition (), it follows from (2.13) that
which implies by (2.8)
and as above
Since T is maximal monotone and , Lemma 1.1 implies that and
-
(b)
According to the statement (a), for a sequence in with , we can choose sequences in , in , and in with such that
(2.14)
We may suppose that
where , , , and . Note that . Indeed, if , then (2.14) implies which gives by condition () on , and therefore and , which contradicts the hypothesis that .
To show that , we first claim that
Assume the contrary. So, we may suppose that
Hence, it follows from (2.14) that and
which imply
For every and every , we obtain from the monotonicity of T that
which implies along with (2.16)
Since T is maximal monotone, we have and . Letting in (2.17), we have a contradiction. Thus, (2.15) is true.
As above, we can deduce from (2.15) that
Since C satisfies condition () and is demicontinuous, we obtain from (2.14) that
We conclude that and
This completes the proof. □
As a consequence of Theorem 2.1, we have the following result. When C is a compact operator, it was proved by Li and Huang [12] with the aid of the Leray-Schauder degree for compact operators.
Corollary 2.2 Let T, Ω, Λ, , and be as in Theorem 2.1. Suppose that is a demicontinuous bounded operator which satisfies condition ().
-
(a)
For a given , assume the following property:
() For every , there exists a such that the equation
has no solution in Ω. Then there exists a such that
-
(b)
If , T satisfies condition () on , and property () is fulfilled for all , then there exists a such that
Proof Define an operator by
By hypotheses on C, the operator is obviously demicontinuous, bounded, and satisfies condition (). Moreover, is continuous in t uniformly with respect to because is bounded. Apply Theorem 2.1 with . □
3 Implicit eigenvalue problem about densely defined operators
In this section, we study the implicit eigenvalue problem for densely defined perturbations of maximal monotone operators, based on the degree theories of Kartsatos and Skrypnik.
An operator is said to be uniformly quasibounded if for every there exists a constant such that for all and all with and , we have .
In a regularization method by the duality operator, we establish a new result on the existence of eigenvalues, by applying topological degree for densely defined operators in [8, 9].
Theorem 3.1 Let Ω be a bounded open set in X with and L a dense subspace of X. Let be a maximal monotone multi-valued operator with and . Let Λ, , and be positive numbers. Assume that is a single-valued operator with such that for all and is continuous in t uniformly with respect to . Assume further that
(c1) C is uniformly quasibounded,
(c2) C satisfies condition (), and
(c3) for every and for every and , the function , , is continuous on F, where denotes the set of all finite-dimensional subspaces of L.
Then the following statements hold:
-
(a)
For a given , assume the following property:
() For every , there exists a such that the equation
has no solution in . Then there is a such that
-
(b)
If , T satisfies condition () on , and property () is fulfilled for all , then there exists a such that
Proof (a) First, we prove that for every , there exists a such that
Assume on the contrary that for some and for every , the following holds:
Then (3.2) holds for all . For , the assertion is obvious.
For , we set and . To show that is an admissible homotopy in the sense of Definition 2.4 in [9], we have to check the following conditions on two families and . In fact, conditions on are automatically satisfied, with independent of t, due to the monotonicity of and .
() is uniformly strongly quasibounded with respect to , i.e., for every there exists a constant such that for all with and all ,
This follows trivially from (c1) and the fact that and are monotone and is bounded.
() For every pair of sequences in and in L such that , , and
where , , and , we have , and .
For this, there are two cases to consider. If , then the second inequality in (3.3) implies
and hence , , and . Now let . From the first inequality in (3.3) it follows that
and so
Combining this with
we have
In view of , we can find a subsequence of , denoted again by , such that and for some . Note that and . Hence (c2) implies that , , and and so . Thus, condition () is satisfied in both cases.
() For every and , the function , , is continuous.
Actually, is continuous on because in the notation of (c3) is continuous on F and is continuous on X. Consequently, we have shown that , , is an admissible homotopy.
Following Kartsatos and Skrypnik [8, 9], we can use the homotopy invariance property of the degree d with respect to the bounded open set Ω as follows:
for all . The latter follows from Theorem 3 in [3], noticing that is demicontinuous, bounded, strictly monotone and satisfies condition (). For every , the existence property of the degree implies that
which contradicts property (). Therefore, the first assertion (3.1) is true.
According to assertion (3.1), let be a sequence in with and be a sequence in such that
Without loss of generality, we may suppose that
where , , and . We may consider the case that as the conclusion of (a) does not hold for . In order to apply the condition (), we first show that
Assume on the contrary that there exists a subsequence of , denoted again by , such that
Note by (3.4) and the monotonicity of that
and so
Hence, it follows from that
For every and every , it is shown as in the proof of part (a) of Theorem 2.1 that
which implies in view of (3.6)
Since T is maximal monotone, we have and . Letting in (3.7), we have a contradiction. Thus, (3.5) holds.
As before, we can deduce from (3.5) that
Since C satisfies condition (), we have
Combining (3.4) and (3.8), we obtain
and
Since T is maximal monotone and , Lemma 1.1 implies that and
-
(b)
In view of (a), for a sequence in with , we take sequences in , in and in with such that
(3.9)
We may suppose that
where , , and . Note that . As in the proof of part (b) of Theorem 2.1, a similar argument shows that
and so in a usual way
Hence, it follows from (c2) that
which implies by (3.9)
Consequently, we obtain from the maximal monotonicity of T that and
This completes the proof. □
Remark 3.2 In an analogous way to Theorem 3.1, we can observe the eigenvalue problem for quasibounded perturbations of maximal monotone operators; see [10].
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Acknowledgements
This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (NRF-2011-0021-829).
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Kim, IS. Implicit eigenvalue problems for maximal monotone operators. Fixed Point Theory Appl 2012, 178 (2012). https://doi.org/10.1186/1687-1812-2012-178
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DOI: https://doi.org/10.1186/1687-1812-2012-178