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Journal of Applied Mathematics

Volume 2014 (2014), Article ID 346517, 10 pages

http://dx.doi.org/10.1155/2014/346517
Research Article

Strong Convergence to a Solution of a Variational Inequality Problem in Banach Spaces

1Department of Information Science, Toho University, Miyama, Funabashi, Chiba 274-8510, Japan

2Sundai Preparatory School, Surugadai, Kanda, Chiyoda-ku, Tokyo 101-8313, Japan

Received 22 January 2014; Accepted 14 May 2014; Published 9 June 2014

Academic Editor: Luigi Muglia

Copyright © 2014 Yasunori Kimura and Kazuhide Nakajo. 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

We consider the variational inequality problem for a family of operators of a nonempty closed convex subset of a 2-uniformly convex Banach space with a uniformly Gâteaux differentiable norm, into its dual space. We assume some properties for the operators and get strong convergence to a common solution to the variational inequality problem by the hybrid method proposed by Haugazeau. Using these results, we obtain several results for the variational inequality problem and the proximal point algorithm.

1. Introduction

Let and be the set of all positive integers and the set of all real numbers, respectively. Throughout this paper, is a real Banach space with norm and is the dual of . For and , let be the value of at . Suppose that is a nonempty closed convex subset of and is a monotone operator of into ; that is, holds for all . Then, we consider the variational inequality problem [1], that is, the problem of finding an element such that

The set of all solutions to the variational inequality problem for is denoted by . For , we say that is -inverse strongly monotone [25] if

Haugazeau [6] introduced a sequence generated by the hybrid method by the following way. Let be a family of mappings of a real Hilbert space into itself with , where is the set of all fixed points of . Let be a sequence generated by for each , where is the metric projection of onto . He proved a strong convergence theorem when for every , where is the metric projection of onto a nonempty closed convex subset of for each with . Later, Solodov and Svaiter [7], Bauschke and Combettes [8], Nakajo and Takahashi [9], and many researchers studied the hybrid method in a real Hilbert space. In a real Banach space, Kamimura and Takahashi [10], Ohsawa and Takahashi [11], Kohsaka and Takahashi [12], Matsushita and Takahashi [13], Matsushita et al. [14], Nakajo et al. [15], and several researchers studied the hybrid method.

In a real Hilbert space , Iiduka et al. [16] considered a sequence generated by the following hybrid method: for each , where is an -inverse strongly monotone operator of into with , is the metric projection of onto a nonempty closed convex subset of , and . They proved that converges strongly to ; see also [17, 18]. In a -uniformly convex and uniformly smooth Banach space , Iiduka and Takahashi [19] proved the following.

Theorem 1 (Iiduka and Takahashi [19]). Let be an -inverse strongly monotone operator of into with and for some , where is a positive constant satisfying that for every . Let be a sequence generated by for each , where is the generalized projection of onto and for . Then, converges strongly to .

Motivated by [19], we propose a new family of operators and prove strong convergence theorems of the sequence generated by these mappings. Using these results, we get several additional results for the problem of variational inequalities and the proximal point algorithm.

2. Preliminaries

Throughout this paper, we write to indicate that a sequence converges weakly to and will symbolize strong convergence. We denote by the unit sphere of a Banach space ; that is, .

We define the modulus of convexity of as follows: is a function of into such that for every . is said to be uniformly convex if for each . Let . is said to be -uniformly convex if there exists a constant such that for every . It is obvious that a -uniformly convex Banach space is uniformly convex. is said to be strictly convex if for all with . We know that a uniformly convex Banach space is strictly convex and reflexive. The duality mapping of is defined by for every . It is also known that if is strictly convex and reflexive, then the duality mapping of is bijective and is the duality mapping of . is said to be smooth if the limit exists for every . The norm of is said to be uniformly Gâteaux differentiable if, for each , the limit (8) is attained uniformly for . is said to be uniformly smooth if the limit (8) is attained uniformly for . We know that the duality mapping of is single-valued if and only if is smooth. It is also known that if is uniformly smooth, then the duality mapping is uniformly continuous on bounded subsets of and if the norm of is uniformly Gâteaux differentiable, then is norm-to-weak uniformly continuous on bounded subsets of ; see [20, 21] for more details. The following is proved by Xu [22]; see also [23].

Theorem 2 (Xu [22]). Let be a smooth Banach space. Then, the following are equivalent. (i) is -uniformly convex.(ii)There exists a constant such that holds for each .

Remark 3. In the case where is a real Hilbert space, is the identity mapping and we can choose .

Let be a smooth Banach space. The function is defined by for every . It is obvious that for each and for all . It is also known that if is strictly convex and smooth, then, for , if and only if ; see also [13]. We have the following result from Theorem 2.

Lemma 4. Let be a -uniformly convex and smooth Banach space. Then, for each , holds, where is a constant in Theorem 2.

Proof. Let . By Theorem 2, we have which is the desired result.

Let be a nonempty closed convex subset of a strictly convex, reflexive, and smooth Banach space and let . Then, there exists a unique element such that

We denote by and call the generalized projection of onto ; see [10, 24, 25]. We have the following well-known results [10, 24, 25] for the generalized projection.

Lemma 5. Let be a nonempty convex subset of a smooth Banach space E, , and . Then, if and only if for all .

Let be a nonempty closed convex subset of a strictly convex and reflexive Banach space and let . Then, there exists a unique element such that . Putting , we call the metric projection of onto ; see [26]. We have the following result for the metric projection; see [20] for more details.

Lemma 6. Let be a nonempty closed convex subset of a strictly convex, reflexive, and smooth Banach space , , and . Then, if and only if for all .

An operator is said to be monotone if for every . Notice that we often identify a set-valued operator with its graph; if and only if .

A monotone operator is said to be maximal if the graph of is not properly contained in the graph of any other monotone operator. It is easy to see that a monotone operator is maximal if and only if, for , for every implies that . We know the following result.

Theorem 7 (Rockafellar [27]; see also [28]). Let be a strictly convex, reflexive, and smooth Banach space and let be a monotone operator of into . Then, is maximal if and only if for all , where is the range of .

From this fact, we also know that if is a strictly convex, reflexive, and smooth Banach space and is a maximal monotone operator of into , then, for any and , there exists a unique element such that , where is the domain of . We define by for every and , and such is called the resolvent of ; see [21, 29] for more details.

3. Main Results

Let be a nonempty closed convex subset of a strictly convex, reflexive, and smooth Banach space and a family of operators of into satisfying the following:(i) ;(ii) for all , , and ;(iii)there exists a sequence in such that and for every , , and ;(iv)for all , ;(v)for every bounded sequence , , and with , if , then there exists a subsequence of such that .

Let us observe some properties of the mappings and the subsets deduced from the assumptions above.

First, we know that, for any , the image of by is a singleton. Indeed, for , we have by the condition (iii). On the other hand, since , it follows that . Thus, we get for all .

Next, if we assume , then we have . Indeed, the inclusion is trivial. To show the opposite inclusion, let and . By the condition (iii), we have , which implies ; that is, for all . Hence, we get .

We also know that is closed and convex. Indeed, for and , let . By the condition (iii), and hold for all . Thus we get which implies that for each . Since and for every and , we have for all and ; that is, for each . Hence, is convex.

To see being closed, let be a sequence in such that . Since we have for every from the condition (iii), we get for all . Since for each and , we obtain for every and ; that is, for all . Therefore, is closed.

Now, we get the following result by the hybrid method using the generalized projections.

Theorem 8. Let be a nonempty closed convex subset of a -uniformly convex Banach space whose norm is uniformly Gâteaux differentiable, and let be a sequence of operators of into satisfying the conditions (i)–(v). Let be a sequence in such that and , where is the constant in Theorem 2. Let and a sequence in generated by

for each . Then, converges strongly to .

Proof. It is obvious that is closed and convex for every . Since if and only if , we have that is closed and convex for all . Next, we show that, for , implies that . Let . We have for every . Since and by Lemma 5, we get . Further, by Lemma 4, we obtain

Using and the condition (iii), we have and thus for any . Since a sequence satisfies for all , , and , we can choose a positive sequence such that

So, we obtain

Since , we have . Using and the condition (ii), we have . Thus, we get that is, . From this fact, we get that for every and is well defined. Indeed, is given and since , we have . Assume that is well defined and for some . There exists a unique element and we get for all by Lemma 5. Since , we have for every ; that is, . Since , we have . Hence, we obtain . By mathematical induction, we get for every and is well defined. Since and , we have for every , which implies that is bounded. Further, since , we have for all . Thus, there exists and

By Lemma 4, we get

Using , we have for all . From the condition (iii), we have which implies that for every and . Since , using the condition (iv) and the boundedness of , we get that is bounded. Since (22)–(24) hold and is bounded, we have , which implies that

by Lemma 4. From (23) and (27), we also have . Using the facts that for all , , the condition (iv) holds, , , and are bounded, and the duality mapping is norm-to-weak uniformly continuous on bounded subset of , we obtain

Since (17) holds for every , it follows from (29) and that

From the condition (v), there exists a subsequence of such that . Since the norm of is weakly lower semicontinuous, we get which implies and

Using , , and Lemma 5, we have which implies that for all . From (32), we get and by Lemma 4, we obtain , which is the desired result.

Next, we have the following result by the hybrid method using the metric projections.

Theorem 9. Assume that , , , , and are the same as in Theorem 8. Let and a sequence in generated by for each . Then, converges strongly to .

Proof. is closed convex for every . As in the proof of Theorem 8, we have that is closed and convex. We also get that, for , implies that . By this fact, we obtain for every and is well defined. Indeed, is given and since . Assume that is well defined and for some . There exists a unique element and we get for all by Lemma 6. Since , we have for every ; that is, . Since , we also have . Thus, we obtain . By mathematical induction, we get for every and is well defined.

Since and , we have for every and, hence, is bounded. Using and Theorem 2, we have for each , which implies that there exists and

Since , by Lemma 5, we have which implies that from Lemma 4. As in the proof of Theorem 8, is bounded. Thus, we get that is also bounded by the boundedness of and so is . Since , we have ; that is, for all . By the boundedness of , , , and with (38) and Lemma 4, we have

As in the proof of Theorem 8, using (17) and (29), we have

From the condition (v), there exists a subsequence of such that . Since the norm of is weakly lower semicontinuous and (36) holds, we get which implies that and

Using the facts that and and by Theorem 2, we obtain for all . By (45), we have which is the desired result.

Remark 10. Even though we replace the definition of in Theorems 8 and 9 with the theorems are still valid.

4. The Variational Inequality Problem for Monotone Operators

Let be a countable set and a mapping. Nakajo et al. [30] propose the condition (NST) as follows: satisfies the condition (NST) if there exists a subsequence of such that, for any , there is with for all sufficiently large . Using the condition (NST), we get the following result for the variational inequality problem by Theorem 8.

Theorem 11. Let be a nonempty closed convex subset of a -uniformly convex and uniformly smooth Banach space , a countable set, and a family of operators of into such that (i) ;(ii) is an inverse strongly monotone operator for each ; that is, there exists such that for every and , the inequality holds;(iii)for all , .

Suppose that the index mapping satisfies the condition (NST) and . Let be a sequence in such that and , where is the constant in Theorem 2. Let and a sequence in generated by for each . Then, converges strongly to .

Proof. We apply Theorem 8 with for all . Then, the conditions (i)–(iv) are satisfied, and we will verify the condition (v). Let be a bounded sequence in with with , and such that . By the condition (NST), there exists a weakly convergent subsequence of such that, for any , there is with for all sufficiently large . Let and fix . There exists such that for every sufficiently large . We consider a subsequence of for all and denote it by again. We have for all , which implies that . Let . By Lemma 5, we have for every sufficiently large and . Since for each sufficiently large , , , , , and the duality mapping is uniformly continuous on bounded subset of , we have for all . Since is inverse strongly monotone, we have for every , which implies that that is, . From (53), for all . Therefore, for every ; that is, . Hence, the condition (v) is satisfied. Consequently, we obtain by Theorem 8.

As in the proof of Theorem 11, we get the following result for the variational inequality problem by Theorem 9.

Theorem 12. Assume that , , , , , , , , and are the same as Theorem 11. Let and a sequence in generated by for each . Then, converges strongly to .

Remark 13. In Theorems 11 and 12, under the assumption that , we have for all . Indeed, is trivial. Let , , and . From the condition (ii), we have which implies that . On the other hand, from . So, we obtain ; that is, . Therefore, for all . Now suppose that instead of the condition (i) and . By the argument mentioned above and Remark 10, Theorems 11 and 12 hold under the conditions (i) and (ii) and we get the result of [19].

Remark 14. We know that, for a continuously Fréchet differentiable and convex functional on a Banach space , if is Lipschitz continuous with constant , then is -inverse strongly monotone operator; see [2, 19]. So, we can apply Theorems 11 and 12 and Remark 13 to such a functional; see [19].

5. The Proximal Point Algorithm

Let be a strictly convex, reflexive, and smooth Banach space, a maximal monotone operator with , , and for all , where is the resolvent of . Then, is well defined as a mapping of into for all . We also have

In fact, since is equivalent to and . Let and . Since and , we have which implies that . By Theorem 8 and Remark 10, we get the following result using the index mapping which satisfies the condition (NST).

Theorem 15. Let be a countable set, a -uniformly convex Banach space whose norm is uniformly Gâteaux differentiable, and a family of maximal monotone operators of into such that . Let be a sequence in with and a sequence in such that and , where is the constant in Theorem 2. Let and a sequence in generated by for each , where the index mapping satisfies the condition (NST) and is the resolvent of . Then, converges strongly to .

Proof. Suppose that for every and in Theorem 8. Then, we have that is a mapping of into with , the condition (iii) is satisfied with for all , and the conditions (ii) and (iv) hold by for all and all . Let be a bounded sequence in , , and with . Assume that

By the condition (NST), there exists a subsequence of such that, for any , there is with for all sufficiently large . Let and . As in the proof of Theorem 11, there exists such that for every sufficiently large and we get . Let . Since we obtain for each . As is a maximal monotone operator, for every . So, we get . Therefore, the condition (v) holds. So, we get conclusion by Theorem 8 and Remark 10.

As in the proof of Theorem 15, we get the following result from Theorem 9 and Remark 10.

Theorem 16. Assume that , , , , , , , , and are the same as Theorem 15. Let and let be a sequence in generated by for each . Then, converges strongly to .

Let be a proper, lower semicontinuous, and convex function. Then, it is known that the subdifferential of defined by for all is a maximal monotone operator [31, 32]. Moreover, when is strictly convex, reflexive, and smooth, we know that, for the resolvent of , for every and and ; see [21] for more details. Now, we have the following results from Theorems 15 and 16.

Theorem 17.