#### Abstract

Let be a real reflexive Banach space and let be its dual. Let be open and bounded such that . Let be maximal monotone with and . Using the topological degree theory developed by Kartsatos and Quarcoo we study the eigenvalue problem where the operator is a single-valued of class . The existence of continuous branches of eigenvectors of infinite length then could be easily extended to the case where the operator is multivalued and is investigated.

#### 1. Preliminaries

In what follows we assume that is a real or complex Banach space and has been renormed such that it and its dual are locally uniformly convex. The normalized duality mapping is defined by The mapping is said to be “monotone” if for every , , and we have A monotone operator is “maximal monotone” if is maximal in , when is partially ordered by inclusion. In our setting, a monotone operator is maximal if and only if for every . If is maximal monotone operator, the operator is called the Yosida approximant of and the following, which can be found in [1, page 102] is true.

Lemma 1. *Let be a maximal monotone operator with and . Then*(i)* is a bounded maximal monotone mapping with for each ;*(ii)* in as for all , where denotes the element of minimum norm;*(iii)* as for all .**Also , where and satisfies: if , then in as , where denote the convex hull of the set .*

The proof of the next lemma can be found in Kartsatos and Skrypnik [2],

Lemma 2. *Let be maximal monotone such that and . Then the mapping is continuous on the set .*

An operator , with another Banach space, is bounded if it maps bounded subsets of onto bounded sets. It is compact if it is continuous and maps bounded subsets of onto relatively compact subsets of . It is demicontinuous if for every sequence such that we have .

We say that the operator satisfies condition on a set if for every sequence such that and any , with (some), we have . If , then we say that satisfies .

*Definition 3. *Let be a separable reflexive Banach space and let be a dense subspace of . A mapping is said to be of class if, for any sequence of finite dimensional subspaces of with , , with and
for all , we have , and .

If , then we call a mapping of class .

*Definition 4. *Let be a separable reflexive Banach space and let be a dense subspace of . A multivalued mapping is said to be of class if it satisfies the following conditions: (i) is bounded closed and convex for each ,(ii) is weakly upper semicontinuous in each finite dimensional space, that is, for each finite dimensional space of , , is upper semicontinuous in the weak topology,(iii)if for any sequence of finite dimensional subspaces of with , , and such that
for all and some , then we have , and .

If , then we call a mapping of class .

We will need the following two conditions:(*P*1) there exist a subspace of such that , and that the operator satisfies condition ;(*P*2) there exist a function which is nondecreasing such that

The following lemma can be found in Zeidler [3, page 915].

Lemma 5. *Let be maximal monotone. Thus the following are true:*(i)* and imply , and ;*(ii)* and imply , and .*

We will need the following lemma from Adhikari and Kartsatos [4].

Lemma 6. *Assume that the operators and are maximal monotone, with and . Assume further that is maximal monotone. Assume that there is a positive sequence such that , a sequence and a sequence such that and . Then the following are true:*(i)*the inequality
is impossible;*(ii)*if
then and . *

The proof of the following lemma is given in [5] proof of Theorem 7 but we will repeat it here for completeness and future reference.

Lemma 7. *Let be a quasibounded maximal monotone operator such that . Let and be such that
**
where are positive constants. Then there exists a number such that for all .*

*Proof. *Let
We have that
where . Thus,
From (11) we obtain
Now since and , we have that , which implies . We claim that is bounded, if not we may assume that and for all . Thus and
which implies that is bounded. Now since is strongly quasibounded, the boundedness of and imply the boundedness of , that is, a contradiction. It follows that is bounded and the proof is complete.

We denote by the duality mapping with gauge function . The function is continuous, strictly increasing such that and as . This mapping is continuous, bounded, surjective, strictly and maximal monotone, and satisfies condition . Also, for these facts we refer to Petryshyn [6, pages 32-33 and 132].

#### 2. The Eigenvalue Result

In this section we are using the topological degree developed by Kartsatos and Quarcoo in [7]. The following result will also improve Theorem 4 of Kartsatos and Skrypnik in [8] since we are no longer assuming that the perturbation is quasibounded.

Theorem 8. *Let be open and bounded with . Assume that the operator is maximal monotone with and . Assume that the operator , with , satisfies condition and (P2). Let , , and be positive numbers. Assume that* * there exists such that
has no solution in .**Then*(i)*there exists such that
*(ii)*if , satisfies on , and property is satisfied for every , then there exists such that .*

*Proof. *(i) Assume that is true and that (16) is false. We consider the homotopy inclusion
It is clear that this inclusion has no solution for , because and our assumption about (16). It is also true for because and the operator is strictly monotone, hence one-to-one. We are going to show that is an admissible homotopy for the degree in [4]. To this end, we set , and we recall the operators , , and we set . We have and , . We also set . We have and for . Let be an open and bounded subset of . We know that the equation
has no solution for any . Now we consider the equation
and we show that there exists such that
Assume that this is not true, then there exist with , , with , with , and
Clearly we cannot have for any , since and the operator is strictly monotone, hence one-to-one. Thus , for all . From (21) we have that
where is the bound of . Thus we have the boundedness of . Using Lemma 7, we have the boundedness of . We may thus assume that . From (21) we also have that the sequence is bounded and we may assume that .

If , then from
we obtain , which is a contradiction to . Hence it follows that . Since is bounded, we may assume that and we have .

From (21) it follows that
Thus
Now
hence
where the last inequality follows from Lemma 6. Now since is of class , it follows that , and . Also we have that
and using Lemma 6 we get that and . This is a contradiction since , we have . We have shown that is an admissible homotopy for our degree. We can now work as in Theorem 3 in [8] in order to show that .

Thus
where the last equality follows from Theorem 3, (i) of [9].

Consequently, the inclusion has a solution in for each . In particular, this says that has a solution in for every . This of course contradicts condition and finishes the proof of (i).

(ii) Let , be such that, for some ,
again, for any is not possible. Since , we have by property that
where is the bound of . This show that is bounded and further we obtain the boundedness of by Lemma 2. We may thus assume that , , , then .

If , then (30) implies that
Since satisfies , it follows that . Now, by the demiclosedness of (see Lemma 6) we obtain that , and this contradicts . Hence, . Repeating the proof of (i) starting from (21), we get again that
and since is of class , we have , . Using the demiclosedness of , we obtain , and . The proof is now complete.

#### 3. Continuous Branches of Eigenvectors

In this section we are interested in showing that the results obtained in previous sections could be used in order to obtain the existence of continuous branches of eigenvectors. We need the following definition.

*Definition 9. *Let , be given and consider the problem
An eigenvector is solution of (34) for some eigenvalue with . We say that the nonzero eigenvectors of the problem (34) form a continuous branch of infinite length if there exists such that, for every , the sphere contains at least one nonzero eigenvector of (34).

Theorem 10. *Assume that the operator is maximal monotone with and . Assume that the operator with is of class and satisfies . Let be a positive number. Assume that implies , satisfies and* * there exists and such that
**Then the nonzero eigenvectors of problem (34) form a continuous branch of infinite length with corresponding eigenvalues .*

*Proof. *Let be given. Let be so small that . Then
, which implies that the inclusion
has no solution for any . Since and is of type , Theorem 8 implies the existence of a solution , for some . The same argument can be repeated for any . This complete the proof.

*Remark 11. *This result is also true when is assumed to be multivalued. We have the following theorem without proof.

Theorem 12. *Assume that the operator is maximal monotone with and . Assume that the operator with is of class and satisfies . Let be a positive number. Assume that implies , satisfies , and** there exists and such that
**
Then the nonzero eigenvectors of problem (34) form a continuous branch of infinite length with corresponding eigenvalues .*

In the following result, we assume that the operator is defined and bounded on all of . In this case we demonstrate the fact that the assumption may be replaced by the assumption on .

Theorem 13. *Assume that the operator is maximal monotone and bounded with and . Assume that the operator , with satisfies condition and . Assume that implies , satisfies and there exist such that
**
Then the nonzero eigenvectors of the problem (34) form a continuous branch of infinite length.*

*Proof. *We show that the problem (34) has eigenvectors on the set for every . To this end, let fix , and show the existence of such that
If this is not true, then there exist a sequence of such that and one of the following is true:(i)the degree is not well-defined; (ii).

In the case (i) there exist eigenvectors such that
In the case (ii) there exist eigenvectors such that (41) holds. Thus in either case, there exist a sequence such that (41) is true. But this leads to a contradiction because , while the set lie in a bounded set. Thus, (40) is true for some . We consider the homotopy
either there exist and such that
or
that is a contradiction. The last equality in (44) follows from Theorem 3, (i) in [2]. It follows that (43) is true. Naturally, we must have in (43) because otherwise . This cannot happen because we already have and is one-to-one. From (43) we obtain sequences , such that
We may assume that . Again, cannot contain a subsequence such that as because the sequence lies in a bounded set and as . Also we have that the sequence is bounded. Thus we may assume that . Since the sequence is bounded, we may assume that . Again, otherwise and the property of would imply that . Since is demiclosed, Lemma 6 would imply . This is a contradiction to our assumption that implies . It follows that and we may also assume that for all .

Now proceeding as in the proof of Theorem 8, it is easy to see that
Now since is of class , we conclude that , and . Thus since also , and the demiclosedness of , we obtain and . Since is arbitrary, the nonzero eigenvectors of problem (34) form a continuous branch of infinite length.