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Journal of Inequalities and Applications
Journal of Inequalities and Applications Cover Image
  • Research Article
  • Open Access

Hybrid Approximate Proximal Point Algorithms for Variational Inequalities in Banach Spaces

Journal of Inequalities and Applications20092009:275208

https://doi.org/10.1155/2009/275208

©  L. C. Ceng et al. 2009

  • Received: 10 March 2009
  • Accepted: 7 June 2009
  • Published:

Abstract

Let be a nonempty closed convex subset of a Banach space with the dual , let be a continuous mapping, and let be a relatively nonexpansive mapping. In this paper, by employing the notion of generalized projection operator we study the variational inequality (for short, VI( ): find such that for all , where is a given element. By combining the approximate proximal point scheme both with the modified Ishikawa iteration and with the modified Halpern iteration for relatively nonexpansive mappings, respectively, we propose two modified versions of the approximate proximal point scheme L. C. Ceng and J. C. Yao (2008) for finding approximate solutions of the VI( ). Moreover, it is proven that these iterative algorithms converge strongly to the same solution of the VI( ), which is also a fixed point of .

Keywords

  • Banach Space
  • Variational Inequality
  • Nonexpansive Mapping
  • Real Hilbert Space
  • Duality Pairing

1. Introduction

Let be a real Banach space with the dual . As usually, denotes the duality pairing between and . In particular, if is a real Hilbert space, then denotes its inner product. Let be a nonempty closed convex subset of and be a mapping. Given , let us consider the following variational inequality problem (for short, ): find an element such that
(1.1)
Suppose that the (1.1) has a (unique) solution . For any , define the following successive sequence in a uniformly convex and uniformly smooth Banach space :
(1.2)

where is the normalized duality mapping on and is the generalized projection operator which assigns to an arbitrary point the minimum point of the functional with respect to . In [1, Theorem?8.2], Alber proved that the above sequence converges strongly to the solution , that is, as , if the following conditions hold:

(i) is uniformly monotone, that is,
(1.3)

where is a continuous strictly increasing function for all with ;

(ii) has arbitrary growth, that is,
(1.4)

where is a continuous nondecreasing function for all with . Note that solution methods for the problem (1.1) has also been studied in [2–10].

Let be a nonempty closed convex subset of a real Banach space with the dual . Assume that is a continuous mapping on and is a relatively nonexpansive mapping such that . The purpose of this paper is to introduce and study two new iterative algorithms (1.5) and (1.6) in a uniformly convex and uniformly smooth Banach space .

Algorithm 1.1.

(1.5)

where are sequences in , is a bounded sequence in , and is assumed to exist for each ,

Algorithm 1.2.

(1.6)

where is a sequence in , is a bounded sequence in , and is assumed to exist for each , .

In this paper, strong convergence results on these two iterative algorithms are established; that is, under appropriate conditions, both the sequence generated by algorithm (1.5) and the sequence generated by algorithm (1.6) converge strongly to the same point , which is a solution of the . Our results represent the improvement, generalization, and development of the previously known results in the literature including Li [8], Zeng and Yao [9], Ceng and Yao [10], and Qin and Su [11].

Notation 1.

stands for weak convergence and for strong convergence.

2. Preliminaries

Let be a Banach space with the dual . We denote by the normalized duality mapping from to defined by
(2.1)

where denotes the generalized duality pairing. It is well known that if is smooth, then is single-valued and if is uniformly smooth, then is uniformly norm-to-norm continuous on bounded subsets of . We will still denote the single-valued duality mapping by .

Recall that if is a nonempty closed convex subset of a Hilbert space and is the metric projection of onto , then is nonexpansive. This fact actually characterizes Hilbert spaces and hence, it is not available in more general Banach spaces. In this connection, Alber [1] recently introduced a generalized projection operator in a Banach space which is an analogue of the metric projection in Hilbert spaces.

Next, we assume that is a smooth Banach space. Consider the functional defined as in [1, 12] by
(2.2)

It is clear that in a Hilbert space , (2.2) reduces to .

The generalized projection is a mapping that assigns to an arbitrary point the minimum point of the functional ; that is, , where is the solution to the minimization problem
(2.3)

The existence and uniqueness of the operator follow from the properties of the functional and strict monotonicity of the mapping (see, e.g., [13]). In a Hilbert space, .

From [1], in uniformly convex and uniformly smooth Banach spaces, we have
(2.4)

Let be a closed convex subset of , and let be a mapping from into itself. A point in is called an asymptotically fixed point of [14] if contains a sequence which converges weakly to such that . The set of asymptotical fixed points of will be denoted by . A mapping from into itself is called relatively nonexpansive [15–17] if and for all and .

A Banach space is called strictly convex if for all with and . It is said to be uniformly convex if for any two sequences such that and . Let be a unit sphere of . Then the Banach space is called smooth if
(2.5)

exists for each . It is also said to be uniformly smooth if the limit is attained uniformly for . Recall also that if is uniformly smooth, then is uniformly norm-to-norm continuous on bounded subsets of . A Banach space is said to have the Kadec-Klee property if for any sequence , whenever and , we have . It is known that if is uniformly convex, then has the Kadec-Klee property; see [18, 19] for more details.

Remark 2.1 ([11]).

If is a reflexive, strictly convex, and smooth Banach space, then for any , if and only if . It is sufficient to show that if , then . From (2.4), we have . This implies that . From the definition of , we have . Therefore, we have ; see [18, 19] for more details.

We need the following lemmas and proposition for the proof of our main results.

Lemma 2.2 (Kamimura and Takahashi [20]).

Let be a uniformly convex and smooth Banach space and let and be two sequences of . If and either or is bounded, then .

Lemma 2.3 (Alber [1]).

Let be a nonempty closed convex subset of a smooth Banach space and . Then, if and only if
(2.6)

Lemma 2.4 (Alber [1]).

Let be a reflexive, strictly convex, and smooth Banach space, let be a nonempty closed convex subset of and let . Then
(2.7)

Lemma 2.5 (Matsushita and Takahashi [21]).

Let be a strictly convex and smooth Banach space, let be a closed convex subset of , and let be a relatively nonexpansive mapping from into itself. Then is closed and convex.

Lemma 2.6 (Chang [7]).

Let be a smooth Banach space. Then the following inequality holds
(2.8)

3. Main Results

Now we are in a position to prove the main theorems of this paper.

Theorem 3.1.

Let be a uniformly convex and uniformly smooth Banach space, let be a nonempty closed convex subset of , let be a continuous mapping and, let be a relatively nonexpansive mapping such that . Assume that are sequences in and is a sequence in such that and . Define a sequence in by the following algorithm:
(3.1)

where is assumed to exist for each , If is uniformly continuous and , then converges strongly to , which is a solution of the (1.1).

Proof.

First of all, let us show that and are closed and convex for each . Indeed, from the definition of and , it is obvious that is closed and is closed and convex for each . We claim that is convex. For any and any , put . It is sufficient to show that . Note that the inequality
(3.2)
is equivalent to the one
(3.3)
Observe that there hold the following:
(3.4)
and . Thus, we have
(3.5)
This implies that . So, is convex. Next let us show that for all . Indeed, we have for all
(3.6)
So for all . Next let us show that
(3.7)
We prove this by induction. For , we have . Assume that . Since is the projection of onto , by Lemma 2.3, we have
(3.8)

As by the induction assumption, the last inequality holds, in particular, for all . This together with the definition of implies that . Hence (3.7) holds for all . This implies that is well defined.

On the other hand, it follows from the definition of that . Since , we have

(3.9)
Thus is nondecreasing. Also from and Lemma 2.4, it follows that
(3.10)
for each for each . Consequently, is bounded. Moreover, according to the inequality
(3.11)
we conclude that is bounded and so is . Indeed, since is relatively nonexpansive, we derive for each
(3.12)

and hence is bounded. Again from we know that is also bounded.

On account of the boundedness and nondecreasing property of we deduce that exists. From Lemma 2.4, we derive

(3.13)
for all . This implies that . So it follows from Lemma 2.2 that . Since , from the definition of , we also have
(3.14)
Observe that
(3.15)
On the other hand, since from (3.1) we have for each
(3.16)
utilizing Lemma 2.3 we obtain . Thus, in terms of Lemmas 2.4 and 2.6 we conclude that
(3.17)
Since and , we obtain . Thus by Lemma 2.2 we have . From it follows that is bounded. At the same time, observe that
(3.18)
and hence
(3.19)
From and the boundedness of and , we derive . Note that is uniformly continuous. Hence by virtue of . Since
(3.20)
it is known that is bounded. Consequently, from (3.15), and it follows that
(3.21)
Further, it follows from (3.14), and that
(3.22)
Utilizing Lemma 2.2, we obtain
(3.23)
Since is uniformly norm-to-norm continuous on bounded subsets of , we have
(3.24)
Furthermore, we have
(3.25)
It follows from and that
(3.26)
Noticing that
(3.27)
we have
(3.28)
From (3.24) and , we obtain
(3.29)
Since is also uniformly norm-to-norm continuous on bounded subsets of , we obtain
(3.30)
Observe that
(3.31)

Since is uniformly continuous, it follows from (3.26), (3.30) and that .

Finally, let us show that converges strongly to , which is a solution of the (1.1). Indeed, assume that is a subsequence of such that . Then . Next let us show that and convergence is strong. Put . From and , we have . Now from weakly lower semicontinuity of the norm, we derive

(3.32)
It follows from the definition of that and hence
(3.33)
So we have . Utilizing the Kadec-Klee property of , we conclude that converges strongly to . Since is an arbitrarily weakly convergent subsequence of , we know that converges strongly to . Now observe that from (3.1) we have for each
(3.34)
Since is uniformly norm-to-norm continuous on bounded subsets of , from we infer that . Noticing that and is a continuous mapping, we obtain that and . Therefore, from it follows that
(3.35)
that is,
(3.36)
Letting we conclude from (3.34) that
(3.37)
and hence
(3.38)

This shows that is a solution of the (1.1). This completes the proof.

Corollary 3.2 ([11, Theorem?2.1]).

Let be a uniformly convex and uniformly smooth Banach space, let be a nonempty closed convex subset of , and let be a relatively nonexpansive mapping such that . Assume that and are sequences in such that and . Define a sequence in by the following algorithm:
(3.39)

where is the single-valued duality mapping on . If is uniformly continuous, then converges strongly to .

Proof.

In Theorem 3.1, we know from (3.1) and Lemma 2.3 that
(3.40)
is equivalent to . Now, put for all . Then we have
(3.41)

for all . Thus algorithm (3.1) reduces to algorithm (3.39). By Theorem 3.1 we obtain the desired result.

Theorem 3.3.

Let be a uniformly convex and uniformly smooth Banach space, let be a nonempty closed convex subset of , let be a continuous mapping, and let be a relatively nonexpansive mapping such that . Assume that satisfies and satisfies . Define a sequence in by the following algorithm:
(3.42)

where is assumed to exist for each , If is uniformly continuous and , then converges strongly to , which is a solution of the (1.1).

Proof.

We only derive the difference. First, let us show that is closed and convex for each . From the definition of , it is obvious that is closed for each . We prove that is convex. Similarly to the proof of Theorem 3.1, since
(3.43)
is equivalent to
(3.44)
we know that is convex. Next, let us show that for each . Indeed, we have for each
(3.45)
So for all and . Similarly to the proof of Theorem 3.1, we also obtain for all . Consequently, for all . Therefore, the sequence generated by (3.42) is well defined. As in the proof of Theorem 3.1, we can obtain . Since , from the definition of , we also have
(3.46)
As in the proof of Theorem 3.1, we can deduce from and that and hence by Lemma 2.2. Further, it follows from and the boundedness of and that
(3.47)
Since , from the definition of , we also have
(3.48)
It follows from (3.47) and that
(3.49)
Utilizing Lemma 2.2, we have
(3.50)
Since is uniformly norm-to-norm continuous on bounded subsets of we have
(3.51)
Note that
(3.52)
Therefore, from we have
(3.53)
Since is also uniformly norm-to-norm continuous on bounded subsets of , we obtain
(3.54)
It follows that
(3.55)

Since is uniformly continuous, it follows from (3.50) and (3.54) that .

Finally, let us show that converges strongly to , which is a solution of the (1.1). Indeed, assume that is a subsequence of such that . Then . Next let us show that and convergence is strong. Put . From and , we have . Now from weakly lower semicontinuity of the norm, we derive

(3.56)
It follows from the definition of that and hence . So, we have . Utilizing the Kadec-Klee property of , we conclude that converges strongly to . Since is an arbitrarily weakly convergent subsequence of , we know that converges strongly to . Now observe that from (3.1), we have for each
(3.57)
Since is uniformly norm-to-norm continuous on bounded subsets of , from we infer that . Noticing that and is a continuous mapping, we obtain that and . Observe that
(3.58)
It follows from that
(3.59)
Letting we conclude from (3.34) that
(3.60)

This shows that is a solution of the (1.1). This completes the proof.

Corollary 3.4 ([11, Theorem?2.2]).

Let be a uniformly convex and uniformly smooth Banach space, let be a nonempty closed convex subset of , and let be a relatively nonexpansive mapping. Assume that is a sequence in such that . Define a sequence in by the following algorithm:
(3.61)

where is the single-valued duality mapping on . If is nonempty, then converges strongly to .

Proof.

In Theorem 3.3, we know from (3.42) and Lemma 2.3 that
(3.62)
is equivalent to . Now, put for all . Then we have
(3.63)

for all . Thus algorithm (3.42) reduces to algorithm (3.61). Thus under the lack of the uniform continuity of it follows from (3.55) that . By the careful analysis of the proof of Theorem 3.3, we can obtain the desired result.

Declarations

Aknowledgments

The first author was partially supported by the National Science Foundation of China (10771141), Ph.D. Program Foundation of Ministry of Education of China (20070270004), and Science and Technology Commission of Shanghai Municipality grant (075105118), Leading Academic Discipline Project of Shanghai Normal University (DZL707), Shanghai Leading Academic Discipline Project (S30405) and Innovation Program of Shanghai Municipal Education Commission (09ZZ133). The third author was partially supported by the Grant NSF 97-2115-M-110-001.

Authors’ Affiliations

(1)
Scientific Computing Key Laboratory of Shanghai Universities, Shanghai, 200234, China
(2)
Department of Mathematics, Shanghai Normal University, Shanghai, 200234, China
(3)
Department of Business Administration, College of Management, Yuan-Ze University, Chung-Li, Taoyuan Hsien, 330, Taiwan
(4)
Department of Applied Mathematics, National Sun Yat-Sen University, Kaohsiung, 804, Taiwan

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© L. C. Ceng et al. 2009

This article is published under license to BioMed Central Ltd. 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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