Abstract and Applied Analysis

Volume 2016, Article ID 7826475, 10 pages

http://dx.doi.org/10.1155/2016/7826475

## Maximality Theorems on the Sum of Two Maximal Monotone Operators and Application to Variational Inequality Problems

Department of Mathematics, Virginia Polytechnic Institute and State University, Blacksburg, VA 24061, USA

Received 19 May 2016; Accepted 17 July 2016

Academic Editor: Lucas Jodar

Copyright © 2016 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.

#### Abstract

Let be a real locally uniformly convex reflexive Banach space with locally uniformly convex dual space . Let and be maximal monotone operators. The maximality of the sum of two maximal monotone operators has been an open problem for many years. In this paper, new maximality theorems are proved for under weaker sufficient conditions. These theorems improved the well-known maximality results of Rockafellar who used condition and Browder and Hess who used the quasiboundedness of and condition . In particular, the maximality of is proved provided that , where is a proper, convex, and lower semicontinuous function. Consequently, an existence theorem is proved addressing solvability of evolution type variational inequality problem for pseudomonotone perturbation of maximal monotone operator.

#### 1. Preliminaries

In what follows, the norm of spaces and will be denoted by . For and , pairing denotes value Let and be real Banach spaces. For operator , we define domain of by and range of by . We also use symbol for the graph of : . A single-valued 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 A multivalued operator is “bounded,” if it maps each bounded subset of into a bounded subset of . It is “finitely continuous,” if it is upper semicontinuous from each finite dimensional subspace of to the weak topology of . Throughout the paper, we use notations and in to denote the weak and strong convergence of sequence , respectively. Analogous notations are used for convergence of a sequence in . Let be a continuous and strictly increasing function such that as . The mapping defined by is called the “generalized duality mapping” associated with . If for all , is denoted by and is called “the normalized duality mapping.” As a consequence of the Hahn-Banach theorem, it is well-known that for all . Since and are locally uniformly convex, is single valued, bounded, monotone, and bicontinuous.

*Definition 1. *An operator is said to be (i)“monotone” if for every , , , and , one has ;(ii)“maximal monotone” if is monotone and for every ; that is, is maximal monotone if and only if is monotone and for every implying and

The following important lemma is due to Brézis et al. [1].

Lemma 2. *Let be a maximal monotone set in . If for all such that in and in and either orthen and as .*

Browder and Hess [2] introduced the following definitions. The original definition of single valued pseudomonotone operator is due to Brézis [3].

*Definition 3. *An operator is said to be “pseudomonotone" if the following conditions are satisfied: (i)For every , is nonempty, closed, convex, and bounded subset of .(ii) is finitely continuous; that is, for every and every weak neighborhood of in , there exists neighborhood of in such that .(iii)For each sequence with such that and one has that for every , there exists such that In particular, letting in place of in the above inequality, the pseudomonotonicity of implies

For basic properties of monotone type operators, the reader is referred to Browder and Hess [2] and Zeidler [4].

The main contribution of this work is to prove maximality of , where and are maximal monotone operators satisfying only one of the following conditions:(I)There exist and such that, for any bounded subset of and each , there exists number such thatfor all and , and for any bounded subset of , there exists number such thatfor all and .(II) is quasibounded and for a bounded subset of there exists number such thatfor all and . It is not difficult to see that (7) is satisfied if has nonempty interior and (8) is satisfied by , where is a proper, convex, and lower semicontinuous function with and . Furthermore, both conditions (7) and (8) are satisfied provided that , which is weaker than the well-known maximality condition due to Rockafellar [5]. In addition, condition (9) is satisfied if with and is quasibounded.

The main result due to Rockafellar [5] assumes condition (III) . It easily follows that (III) implies (I); that is, condition (I) is weaker than condition (III). Indeed, if , then there exists such that . For any bounded subset of , , and , the monotonicity of implies thatfor all and , where is an upper bounded for and ; that is, (7) holds. Similarly, it is not difficult to see that (III) implies (8). Therefore, Theorem 5 improves the well-known maximality result due to Rockafellar [5]. On the other hand, Theorem 10 improves the maximality result due to Browder and Hess [2] which required to be quasibounded and . Theorem 13 provides a new maximality result for , where is a proper, densely defined, convex, and lower semicontinuous function satisfying mild condition. As a consequence of this maximality result, an existence theorem for solvability of variational inequality problem involving operators of the type with respect to a closed convex subset of and the function is included in Theorem 16, where is a bounded pseudomonotone operator. These results are new and improve analogous results due to Asfaw and Kartsatos [6, Theorem , Corollary , Theorem , pp. 182–187].

The following lemma is useful in the squeal.

Lemma 4. *Let be maximal monotone, be bounded and there exist and such that , where and are the Yosida approximant of . Then, is bounded, where is the Yosida resolvent of .*

*Proof. *Let and be bounded. Let and be the Yosida approximant and resolvent of , respectively. It is well-known that , , and for all . Let . By the monotonicity of , we see thatfor all ; that is,for all . Since for all , it follows that is bounded if is bounded. Assume that is unbounded; that is, there exists a subsequence, denoted again by , such that . Dividing (12) by for all large , we see that that is, . Since is bounded and for all , it follows that is bounded.

#### 2. Main Results

The following theorem is one of the main results of the paper.

Theorem 5. *Let and be maximal monotone operators such that . Let be a bounded subset of . Suppose the following conditions are satisfied: *(i)*There exist and such that for any bounded subset of and each , there exists number such thatfor all and *(ii)*For any bounded subset of , there exists number such thatfor all and .**Then, is maximal monotone.*

*Proof. *Let be the Yosida approximants of . Since is maximal monotone, operator is surjective; that is, for each and , there exist and such thatfor all . Next, we show that is bounded. To this end, by choosing , and applying the monotonicity of and for all , we see that for all , where . This proves the boundedness of . Next, we show that is bounded. Since is bounded and is bounded, it follows that is bounded. Let and satisfy conditions (i) and (ii). Then, we obtainfor all , where is an upper bound for . Let be the Yosida resolvent of . It is well-known that , , and . In addition, for any and , by using the monotonicity of , boundedness of , and (16), we see that where is an upper bound for . By applying Lemma 4, we conclude that is bounded. For each , applying condition (15) and estimate in (18) yields for all , where which is a bounded subset of . Since for all , replacing instead of in the above inequality, we arrive at for all . Since whenever , it follows that there exists such that for all . Therefore, by applying the uniform boundedness theorem, we conclude that is bounded; that is, is bounded. Assume without loss of generality that (i.e., ), , and . On the other hand, we see that By applying (16) along with monotonicity , we obtainNext, we show that To this end, suppose this is false; that is, . Then, there exists a subsequence, denoted again by , such that . By applying Lemma 2, we conclude that and . However, this is impossible. Consequently, (16) implies that is, (24) implies By the maximality of along with Lemma 2, we conclude that , , and . Similarly, from (16), we obtain . However, this is impossible because ; that is, . As a result, we arrive at Since bounded demicontinuous of type , we conclude that and . Finally, letting in (16), we conclude that such that . Since is arbitrary, the surjectivity of is proved. Therefore, is maximal monotone. The proof is completed.

It is worth mentioning that Theorem 5 improves the result due to Chen et al. [7, Theorem , p. 25] because can be arbitrary instead of , and the side condition can be assumed to hold for all , where is a bounded subset of instead of assuming to hold for all and can be a number instead of using functions and from into with bounded and is upper bound for . In addition, Asfaw [8] used the degree theory developed by himself to prove maximality of sum , where is arbitrary and is densely defined which satisfies condition; that is, there exists a continuous strictly increasing function and, for each , there exists a number such that for all and . In addition, Theorem 5 improved the maximality result due to Asfaw [8, Corollary , p. 998]. For further results concerning useful homotopy invariance results, existence theorems, and examples of operators of type , the reader is referred to the paper due to Asfaw [8].

As a result of Theorem 5, the following corollaries hold.

Corollary 6. *Let and be maximal monotone operators such that . Assume further that there exist and such that for each and bounded subset of , there exists number such that for all and . Then, is maximal monotone.*

*Proof. *Suppose and . By the monotonicity of , we see thatfor all and ; that is, condition (ii) of Theorem 5 is satisfied. Since (i) of Theorem 5 is assumed, the maximality of follows by Theorem 5.

Corollary 7. *Let and be maximal monotone operators such that . Let be a bounded subset of : *(i)*If there exist and number such that for all and , then is maximal monotone.*(ii)*Let be a proper, convex, and lower semicontinuous function. If , then is maximal monotone.*

*Proof. *(i) Let . Then, there exists such that . For each and , the monotonicity of implies for all and , where is an upper bound for ; that is, condition (14) is satisfied. Since (15) holds by the hypothesis, the maximality of follows Theorem 5.

(ii) Choose . By the definition of , we see that for all and . Since is proper, convex, and lower semicontinuous, there exist and number such that ; that is, for all ; that is, we have for all Let be a bounded subset of . Then, it follows that for all and , where is an upper bound for ; that is, (ii) of Theorem 5 is satisfied. Thus, the maximality of follows by Theorem 5.

*The following well-known result on maximality of the sum of two maximal monotone operators is due to Rockafellar [5]. The proof follows from the conclusion of Theorem 5, which gives a different proof of the maximality criterion due to Rockafellar [5].*

*Corollary 8. Let and be maximal monotone operators. If , then is maximal monotone.*

*Proof. *It is easy to see that condition implies conditions (i) and (ii) of Theorem 5. The maximality of follows by the conclusion of Theorem 5.

*Corollary 9. Let and be maximal monotone operators with . Let be a bounded subset of . Assume, further, that for each , there exists number such that for all and . Then is maximal monotone.*

*Proof. *For each , it follows that is maximal monotone; that is, is surjective. Thus, for each and , there exist and such that for all . The surjectivity of follows based on the arguments used in the proof of Theorem 5 by using in place of . The details are omitted here.

*In a recent paper by Chen et al. [7, Theorem , p. 27], the solvability of the sum of two maximal monotone operators and is given under the assumptions in Corollary 9, where , , is an upper bounded for , and are functions from into with to be bounded. However, Corollary 9 proves that the sum in Theorem due to Chen et al. [7] is maximal monotone and the conclusion of solvability of operator equation involving follows from results for single maximal monotone operator theory.*

*The second criterion of maximality for is given below. Here, we require to be quasibounded instead of assuming side conditions. Theorem 10 improves the well-known result due to Browder and Hess [2].*

*Theorem 10. Let and be maximal monotone operators. If is quasibounded and for any bounded subset of , there exists number such that for all and , then is maximal monotone.*

*Proof. *For each , let and be the Yosida approximant and resolvent of , respectively. It is well-known that and are everywhere defined bounded and continuous such that is maximal monotone. It is easy to see that for each , operator is maximal monotone. As a result, for each and , there exist unique and such that ; that is, for each , there exist and such thatfor all . We will show that is bounded. Choose and . Next using (38) and monotonicity of and , we get for all , where . This shows the boundedness of ; that is, is bounded. Next, we show that is bounded. By the monotonicity of , we see that for all , where is a suitable upper bound. By Lemma 4, we conclude that is bounded. Next, by the condition on , boundedness of , , and , we see thatfor all , where is bounded subset of , is a number corresponding to in the hypothesis and is an appropriate upper bound. Since is quasibounded and is bounded, we conclude that is bounded. Consequently, we arrive at the boundedness of . Assume without loss of generality that , , and . Since is maximal monotone, the argument used in the proof of Theorem 5 along with Lemma 2 gives As a result, (38) implies Since , , and , it follows that By Lemma 2, we conclude that , , and , that is, . Consequently, (38) implies Since is demicontinuous of type , we conclude that (i.e., ) and . Consequently, by using the maximality of and , we conclude that , , and such that . Since is arbitrary, we conclude that is surjective; that is, is maximal monotone. The proof is completed.

*In addition, Theorem 10 improves maximality result due to Asfaw and Kartsatos [6, Corollary , p. 187] using quasiboundedness of instead of strong quasiboundedness of with and weaker side condition on instead of the one used by the authors. As a consequence of Theorem 10, we get the following corollary.*

*Corollary 11. Let and be maximal monotone operators. Suppose one of the following conditions holds: (i) is quasibounded and .(ii) or is bounded.Then is maximal monotone.*

*Proof. *(i) By choosing , applying the monotonicity of givesfor all and . That is, condition on in Theorem 10 is satisfied. Therefore, the maximality of follows by applying Theorem 10.

(ii) If is bounded, one can apply Lemma 4 to conclude that is bounded. The maximality of follows by following the arguments used in the proof of Theorems 5 and 10.

*Theorem 10 or Corollary 11 improves the following well-known maximality result due to Browder and Hess [2, Theorem , p. 284].*

*Corollary 12. Let and be maximal monotone operators. If is quasibounded and , then is maximal monotone.*

*Proof. *The proof follows as a particular case of Corollary 11.

*The following theorem gives a maximality result for perturbed operator , where satisfies mild conditions.*

*Theorem 13. Let be maximal monotone. Let be proper, densely defined, convex, and lower semicontinuous function. Assume, further, that there exists a nondecreasing continuous function satisfying for all Then is maximal monotone. The same conclusion holds if is quasibounded and .*

*Proof. *Fix . Let . Let and be the Yosida approximant and resolvent of , respectively. For each , it follows that is maximal monotone; that is, for each , is surjective. As a result, for each , there exist and such that for all . We will show that , , and are bounded. The boundedness of and follows by using the arguments in the proof of Theorem 5 and applying Lemma 4, respectively. Next, we show that is bounded. Fixing , using the boundedness of and definition of , it follows that for all and . Since is convex and lower semicontinuous, there exist and number such that for all . As a result of this and the condition on , we get for all , where is an upper bound for sequence and is an upper bound for Since is dense in and is continuous, for each we have for all ; that is, using and in place of simultaneously, we arrive at for all ; that is, we see that for each , sequence is bounded. The boundedness of follows by applying the uniform boundedness principle. Consequently, we conclude that is bounded. The proof of the surjectivity of is established following the arguments used in the proof of Theorems 5 and 10. The details are omitted here.

*The following result provides solvability of variational inequality problem , where is a nonempty, closed, and convex subset of . We will recall the definition of solvability of variational inequality problem as given in the following definition.*

*Definition 14. *Let be a nonempty, closed, and convex subset of The variational inequality problem, denoted by is said to be “solvable” in if there exist , , and such that for all .

*For any nonempty, bounded, convex, and open subset of , Definition 14 implies that problem is not solvable in if there exists such that for all , , and Since is a dense subset of , it is not difficult to see that the solvability of inclusion in implies the solvability of problem VIP in . If , we denote problem VIP just by VIP*

*In what follows, we will use the following useful lemma due to Asfaw and Kartsatos [6, Lemma ]. It worth mentioning here that Lemma 15 is useful because the global variational inequality problem is solvable based on the solvability of local problem in provided that it has no solution in .*

*Lemma 15. Let be a nonempty, closed, and convex subset of and . Let be a nonempty, open, and convex subset of Then, problem is solvable in provided that problem is solvable in *

*Next we prove the following result.*

*Theorem 16. Let be a nonempty, closed, and convex subset of . Let be maximal monotone, be a proper, densely defined, convex, and lower semicontinuous function, and be bounded pseudomonotone. Assume, further, that there exists nondecreasing continuous function satisfying for all . Let . Suppose one of the following conditions holds: (i) is bounded.(ii) is unbounded and there exists and such that for all , , and .Then variational inequality problem is solvable in .*

*Proof. *Suppose (i) holds; that is, is bounded. Let . By Theorem 13, we have the maximality of . To prove that is solvable, it is sufficient to show that is solvable. Let be the Yosida approximant of . Since is bounded, it is well-known that is surjective bounded pseudomonotone. Thus, for each , there exist and such that for all . Since and are bounded, it follows that and are bounded. Assume without loss of generality that , , and . By following the arguments used in the proofs of Theorems 5–13 along with Lemma 2, it follows that , , and and Since is pseudomonotone, for each , there exists such that that is, we have