Existence Theorems on Solvability of Constrained Inclusion Problems and Applications

Asfaw, Teffera M.

doi:https://doi.org/10.1155/2018/6953649

Abstract and Applied Analysis

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Volume 2018 | Article ID 6953649 | https://doi.org/10.1155/2018/6953649

Existence Theorems on Solvability of Constrained Inclusion Problems and Applications

Teffera M. Asfaw¹

Academic Editor: Gaston Mandata N’guérékata

Received10 Mar 2018

Accepted07 Jun 2018

Published12 Jul 2018

Abstract

Let be a real locally uniformly convex reflexive Banach space with locally uniformly convex dual space . Let be a maximal monotone operator and be bounded and continuous with . The paper provides new existence theorems concerning solvability of inclusion problems involving operators of the type provided that is compact or is of compact resolvents under weak boundary condition. The Nagumo degree mapping and homotopy invariance results are employed. The paper presents existence results under the weakest coercivity condition on . The operator is neither required to be defined everywhere nor required to be pseudomonotone type. The results are applied to prove existence of solution for nonlinear variational inequality problems.

1. Introduction: Preliminaries

In what follows, the norm of the spaces and will be denoted by For and , the pairing denotes the value Let and be real Banach spaces. For an operator , we define the domain of by , and the range of by . The symbol denotes the graph of given by An operator is “demicontinuous” if it is continuous from the strong topology of to the weak topology of . It is “compact” if it is strongly continuous and maps bounded subsets of to relatively compact subsets of An operator is “bounded” if it maps each bounded subset of into a bounded subset of . The mapping defined by is called the “normalized duality mapping”. It is known due to Hahn-Banach theorem that . In addition, the local uniform convexity of and implies that is single valued, bounded, monotone, bicontinuous, and of type .

Definition 1. An operator is said to be (i)“monotone” if for all and in ;(ii)“maximal monotone” if is monotone and for every implies and ;(iii)“coercive” if either is bounded or there exists a function such that as and for all and ;(iv)“expansive” if there exists such that for all , , , and .

It is well-known that a monotone operator is maximal monotone if and only if for every (cf. Theorem 2.2 [1]) and is single valued, monotone, and demicontinuous. For each , the operator , defined by , is the “Yosida approximant” of . It is bounded, continuous, and maximal monotone such that as , for every , where The operator defined by , is called the “Yosida resolvent” of . It is continuous, for every and for all , where is the convex hull of For each , for all , where denotes . A maximal monotone operator is called of compact resolvents if is compact for all . For further references on monotonicity theory, the reader is referred to Pascali and Sburlan [2], Barbu [1], Zeidler [3], Kenmochi [4], and the references therein. The following Lemma is due to Brèzis, Crandal, and Pazy [5].

Lemma 2. Let be a maximal monotone set in . If such that in , in , and then and as

The main objective of this paper is to establish sufficient conditions which guarantee existence of solution for inclusions of the type (where is maximal monotone of compact resolvents and is bounded and continuous with ) provided that there exists such that for all . Under this boundary condition, inclusion result of the type is included if is compact with and is arbitrary maximal monotone. The case where is expansive maximal monotone and is compact is also included. The operators of type are available in the form of lower order term in many nonlinear differential equations in appropriate function spaces.

The paper is organized as follows. In Section 2, the main existence results (Theorems 3 and 6) are proved. The arguments of the proofs of Theorem 3 are based on Nagumo homotopy invariance result. Theorem 6 follows as a result of Theorem 3. Nagumo [6] developed a degree theory in a setting of linear convex topological space for operators of the type , where is a compact operator, is the identity mapping on , and is a nonempty and open subset of . The important contributions of Nagumo are (i) is a nonempty and open subset of (not necessarily bounded) and (ii) the degree is invariant under the homotopy , provided that is compact and for all . For further references on Nagumo degree and related results, the reader is referred to the paper due to Nagumo [6, Theorem 5, 6, 7]. Throughout the paper stands for Nagumo degree. In Section 3, we demonstrated the applicability of the abstract results to prove existence of solution(s) for variational inequality problems.

Existence results concerning pseudomonotone perturbations of maximal monotone operators under coercivity condition can be found in the papers due to Kenmochi [4, 7, 8], Asfaw and Kartsatos [9], Le [10], Asfaw [11], and the references therein. For related results concerning existence of solution for inclusion problems of the type in , where is an everywhere defined bounded pseudomonotone operator under Leray-Schauder type boundary condition on , we cite the results due to Asfaw and Kartsatos [9, Theorem 11, Theorem 13, pp. 127-133]. Analogous result for possibly unbounded single multivalued pseudomonotone operator is due to Figueiredo [12]. However, the cases where is not necessarily pseudomonotone type is not studied earlier. It is the purpose of the present paper to address analogous result for , where is possibly not everywhere defined compact or is bounded and continuous, and is of compact resolvents.

2. Main Results

In this section, we prove the main existence theorem.

Theorem 3. Let be maximal monotone and with . Let . Suppose there exists such that and for all . Then (i) if is compact;(ii) if is expansive and is compact;(iii) if and is completely continuous;(iv) if is of compact resolvent and, is bounded and continuous.

Proof. Suppose the hypotheses hold. We divide the proof into two steps.
Step 1. Let be the subdifferential of the indicator function on and . The operator is maximal monotone with bounded domain because . For each , let and be the Yosida approximant and resolvent of , respectively. It is well-known that is bounded, continuous, and maximal monotone, and is bounded and continuous such that , , and for all . Let where is given by For each and , we shall show that is a compact operator. Let , i.e., and as . The continuity of and implies as . Let for all , i.e., for all . Fix . By applying the monotonicity of and boundedness of and , we obtain that for all ; i.e., for all , where is an upper bound for and . Thus, we get the boundedness of . Assume without loss of generality that as . By the monotonicity of , we arrive at Thus we conclude that as because is of type . Since and are continuous, it follows that ; i.e., we have We notice here that the above argument holds for any subsequence of (i.e., every subsequence of admits a convergent subsequence). Consequently, we conclude that converges ; i.e., the continuity of is proved. Next we assume is bounded in . The sequence is bounded because the sequences and are bounded. Since for all and is compact, we can extract a subsequence, denoted again by , such that as . We notice here that for all implies for all . By applying the argument used in the proof of continuity of , it follows that is bounded and admits a convergence subsequence. Thus the compactness of is proved. The same argument shows that is compact if is completely continuous with .
Next we fix temporarily and show that for all and sufficiently small . Suppose not, i.e., there exists , and such thatfor all . If for some , then it follows that ; i.e., . But this is impossible because . Assume for all and as . The boundedness of follows because of the boundedness of and is bounded and for all ; i.e., is bounded. Assume without loss of generality that , , and as . By the monotonicity of and (11), we claim that Suppose not, i.e., . Then there exists a subsequence that converges to as . The fact that , , and for all implies The maximality of along with Lemma 2 implies , , and as ; i.e., we get . However, this is impossible; i.e., the claim holds. The case implies as . However this is impossible. Assume that . By using in (11), we obtain that The condition on implies and as . The maximal monotonicity of together with Lemma 2 implies that and . The continuity of and implies as . Consequently, letting in (11) gives ; i.e., we obtain that This implies for some and . However, it is well-known thatAs a result, for some ; i.e., we get . However, this is impossible because of the hypothesis of the theorem. Therefore, for fixed , there exists such that for all and ; i.e., is an acceptable Nagumo homotopy.
Step 2. Fix . By using the admissible Nagumo homotopy obtained in Step 1 and applying the homotopy invariance properties of , we see that is independent of ; i.e., for all sufficiently small . Thus for each , there exists such that for all . The boundedness of and follows because of the boundedness of , , and . As a result we can easily follow the arguments used in the last part of Step 1 to conclude the existence of and such that ; i.e., for each , there exist and such that for all ; i.e., we get for all and for some and . However, applying the boundary hypothesis on , we conclude that for all and for all . Consequently, we arrive atfor all . The boundedness of implies that ; i.e., . This completes the proof of (i). Next we prove (ii). Suppose is expansive and is compact. By the compactness of we assume without loss of generality that ; i.e., . The expansiveness of implies as . The maximality of along with Lemma 2 yields and . As a result we conclude that ( i.e., ) because of the continuity of ; i.e., (ii) holds.
(iii) Suppose is completely continuous and . Assume by passing into a subsequence that as . The maximality of implies that is closed and convex; i.e., it is weakly closed and . The complete continuity of implies and ; i.e., and ; i.e., (iii) is proved.
(iv) Suppose is of resolvent compact and is bounded and continuous such that the boundary condition on holds. It is known due to Kartsatos [13, Lemma 3, pp. 1684] that is compact if and only if is compact, and is compact for all if is compact for some . As a result, the compactness of and is used equivalently. Since is bounded and is compact, it follows that is a compact operator; i.e., we can follow the arguments used in the proof of the first part of Theorem 3 to conclude that given by is a compact operator. By following exactly analogous arguments used in the proof of (i) through (iii) of this theorem, for each there exist and such that for all ; i.e., we get for all ; i.e., for all , where for all . By the compactness of , there exists a subsequence of , denoted again by , such that as . Assume without loss of generality (by passing into a subsequence) that and as . The maximal monotonicity of along with Lemma 2 implies that and . On the other hand, the continuity of implies ; i.e., ; i.e., . The proof is complete.

The following corollary holds.

Corollary 4. Let and be as given in Theorem 3. The conclusions in Theorem 3 hold if (a)the boundary condition for all is replaced by for all and some , where for ;(b)there exist and such that for all and ;(c) and is replaced by a nonempty, closed, bounded, and convex subset of such that and is replaced by , and for all , where is the subdifferential of the indicator function on .

Proof. (a) The proof for the analogous result under the boundary condition involving follows because the operator inherits all the properties of ; i.e., is bounded, continuous, monotone, and of type .
(b) The side condition in (b) is equivalent to for all because of the definition of for all , , and the fact that for all . This implies that the condition in (b) is equivalent to the boundary condition in Theorem 3.
(c) Suppose for all . The maximal monotonicity of follows because . It is not difficult to see that boundary condition on implies that there exists such that for all and . The proof of (c) follows based on the arguments of the proofs of (i) through (iii) of Theorem 3 by using instead of . The details are omitted here. To prove (iv) of Theorem 3 under (c), we shall show that is resolvent compact. Let be a sequence in such that as and for all ; i.e., for all . Let and such that for all . The boundedness of follows because is bounded. Assume without loss of generality that . Since is closed and convex (i.e., it is weakly closed), we have . In addition, the condition yields for all , where is an upper bound for . Since , it follows that is strongly quasibounded, which yields the boundedness of , i.e., is bounded. Assume without loss of generality that and as . By following the arguments used in the proof of Theorem 3 and using the fact that is bounded of type , we get as and ; i.e., . Since this convergence holds for each subsequence of , we conclude that is continuous. Next we assume is bounded. Then the compactness of implies the existence of a subsequence, denoted by , such that Thus the compactness of follows; i.e., is compact for all . The remaining proof follows as in the arguments of the proof of Theorem 3. The detail is omitted here.

Next we give the following surjectivity result.

Corollary 5. Let and be as given in Theorem 3. Assume that is coercive; i.e., there exists such that Then the following conclusions hold. (i) if is compact.(ii) if is expansive and is compact.(iii) if is completely continuous.(iv) if is of resolvent compact and is bounded and continuous.

Proof. Let for and some . It is known that is bounded, monotone, continuous, and of type . Fix . It is enough to show that the boundary condition in Theorem 3 holds. The coercivity of impliesfor all and . The coercivity of implies that the right side in (35) approaches to as ; i.e., there exists such that for all and . Thus we have for all and ; i.e., we get for all . Consequently, the conclusions (i) through (iv) follows based on the conclusion of Theorem 3.

It is worth noticing that Theorem 3 is new in the sense that the conclusion required only the boundary condition for all . For analogous results under such boundary condition, the reader is referred to Figueiredo [12] (for single multivalued pseudomonotone operator ) and Asfaw and Kartsatos [9, Theorem 11 and Theorem 13] (for pseudomonotone perturbations of maximal monotone operator).

Next we prove the following result.

Theorem 6. Let be maximal monotone and be such that . Assume that , , and such thatfor all with , andThen the following hold. (i) if is compact.(ii) is surjective if is completely continuous and (iii) is surjective if is expansive and is compact;(iv) is surjective if is of compact resolvent and is bounded and continuous.

Proof. Fix and . Let be defined by , . It is well-known that is bounded, continuous, maximal monotone, and of type . Then applying (39) givesfor all and , where As a result, we getas . The operator is maximal monotone. By using the operators and in Corollary 5, we conclude that . Then there exist and such that as . The compactness of implies that (for some subsequence By applying the maximality of together with Lemma 2 and following the arguments used in the proof of Theorem 3 we arrive at . The surjectivity of follows because is arbitrary. Next we give the proof of (i). Since for all , for each , there exist and such thatfor all ; i.e., (39) givesfor all . We shall show that is bounded. Suppose not, i.e., there exists a subsequence, denoted again by , such that as . Dividing (46) by for all large implies for all large ; i.e., (45) impliesfor all large ; i.e., we get However, this is a contradiction to (40). As a result the sequence is bounded. Letting in (45) gives . The proofs of (ii), (iii), and (iv) follow based on the arguments used in the proof of (i) and Theorem 3. The details are omitted here.

In [13], Kartsatos proved that if is compact and is maximal monotone with provided that there exists such that as and such thatfor all , , and andIt is worth mentioning here that (50) implies that for each there exists such that for all ; i.e., (50) implies (39) and (40) implies (51). On the other hand, Theorem 6 due to Kartsatos [13] holds if is used instead of . The proof follows based on the proof of (i) of Theorem 6 by using instead of . However, it is worth noticing that condition (39) is natural in applications than that of (50).

3. Applications

In this section, we demonstrate the applicability of the result to prove existence of solution for variational inequality problems.

Example 7. Let be a nonempty, bounded, and open subset of with and is of smooth boundary. Let and be a nonempty, closed, and convex subset of with nonempty interior. Let be proper, convex, and lower semicontinuous function, and . Let be defined byif and , and otherwise. It is known that is given by for each , where domain of given by and where is the outward normal on and is the canonical base in Assume, further, that () be Carathéodory function; i.e., is measurable for almost all and is continuous for almost all ;()there exist and such that for all Let be defined by for . In the following theorem, we prove the solvability of variational inequality problem ; i.e., we find such thatfor all .

Theorem 8. Let and be as in (52). Suppose and are satisfied. If for or for , then the problem is solvable.

Proof. Let be the subdifferential of in the sense of convex analysis. It is well-known that is an -accretive operator (equivalently, is maximal monotone) with compact resolvents (cf. Proposition 2.2.2 and Proposition 2.2.3 [14]). By applying conditions and , it follows that is bounded and continuous. Let be the subdifferential of the indicator function on . The -accretivity (maximal monotonicity) of follows because of the condition . In addition, it is not difficult to see that is of resolvent compact -accretive operator. Next we show that satisfies the boundary condition in Theorem 3. Let be the functional generated by . Choose . By applying the definition of and conditions and , we arrive at for all and . As a result, we obtain the estimatefor and . Since (if ) or (if ), the right side of (60) approaches to as and as ; i.e., the coercivity condition on is satisfied. Thus for each we conclude that the inclusion problem is solvable; i.e., the problem is solvable. This completes the proof.

Next we present the following example.

Example 9. Let be a nonempty, bounded, and open subset of with smooth boundary and , , , , , and . Let and be defined byfor , where Let denote the inner product in ; i.e., for and ; we have with identified with . The norm of is the normed induced from . It is well-known that is maximal monotone, surjective, , and for all . Since is compactly embedded in , it follows that is a compact operator; i.e., for each the resolvent is compact. Assume, further, that () is Carathéodory function; i.e., is measurable for almost all , and is continuous for almost all ;()there exist and such that for all ;() is a proper, convex, and lower semicontinuous function;()there exists a nonnegative constant such that , where We notice here that the set defined in () is nonempty if is nondecreasing for almost all such that for all . Actually, condition implies that the unit ball in is contained in with . In Theorem 10 below, we prove existence of solution(s) for variational inequality problem denoted by , i.e., finding such thatfor all , where is a nonempty, closed, and convex subset of with .

Theorem 10. Let be as given in (62) and . Assume, further, that and conditions through are satisfied. Then the problem is solvable.

Proof. Let , be as given in (62), and be defined by The operator is linear maximal monotone, surjective, one to one, and densely defined. The density of in implies . As a result, we conclude that is maximal monotone (cf. Rockafellar [15]). The arguments used in the proof of (c) of Corollary 4 imply that is resolvent compact. In addition, the continuity of follows based on the conditions and . The proof follows if we prove the coercivity of . Let us denote again the restriction of on by . The definition on as given in implies that for all . The conditions on imply that there exist and such that for all . Choose . The monotonicity of and definition of imply for all . We notice that because . In addition, the right side of the above inequality approaches as ; i.e., we get that is, is coercive. Consequently, we conclude that is surjective. Thus for each , the problem in (67) admits at least one weak solution. The proof is completed.

The reader can find plenty of resolvent compact maximal monotone operators in the paper due to Brèzis and Nirenberg [16] and in the books due to Vrabie [14], Barbu [1, 17], Showalter [18], and the references therein.

Data Availability

No data were used to support this study.

Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of the paper.

Acknowledgments

The author is grateful to Virginia Tech for funding the article processing fee.

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Copyright © 2018 Teffera M. Asfaw. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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