Journal of Applied MathematicsVolume 2012 (2012), Article ID 401960, 20 pagesdoi:10.1155/2012/401960

Research Article

Hybrid Algorithm for Common Fixed Points of Uniformly Closed Countable Families of Hemirelatively Nonexpansive Mappings and Applications

Sumei Ai¹ and Yongfu Su²

¹Department of Mathematics, Cangzhou Normal University, Cangzhou 061001, China²Department of Mathematics, Tianjin Polytechnic University, Tianjin 300387, China

Received 27 November 2011; Accepted 15 December 2011

Academic Editor: Yonghong Yao

Copyright © 2012 Sumei Ai and Yongfu Su. 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.

The authors have obtained the following results: (1) the definition of uniformly closed countable family of nonlinear mappings, (2) strong convergence theorem by the monotone hybrid algorithm for two countable families of hemirelatively nonexpansive mappings in a Banach space with new method of proof, (3) two examples of uniformly closed countable families of nonlinear mappings and applications, (4) an example which is hemirelatively nonexpansive mapping but not weak relatively nonexpansive mapping, and (5) an example which is weak relatively nonexpansive mapping but not relatively nonexpansive mapping. Therefore, the results of this paper improve and extend the results of Plubtieng and Ungchittrakool (2010) and many others.

1. Introduction and Preliminaries

Let be a Banach space with the dual . We denote by the normalized duality mapping from to defined by where denotes the generalized duality pairing. It is well known that the normalized duality has the following properties: (1) if is smooth, then is single valued; (2) if is strictly convex, then is one-to-one (i.e., for all ); (3) if is reflexive, then is surjective; (4) if has Frchet differentiable norm, then is uniformly norm-to-norm continuous; (5) if is uniformly smooth, then is uniformly norm-to-norm continuous on each bounded subset of ; (7) if is a Hilbert space, then is the identity operator.

As we all know 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, consequently, it is not available in more general Banach spaces. In this connection, Alber [1] has recently introduced a generalized projection operator in a Banach space which is an analogue of the metric projection in Hilbert spaces.

Let be a smooth Banach space. Consider the function defined by Observe that, in a Hilbert space , (1.2) reduces to .

The generalized projection is a map that assigns to an arbitrary point the minimum point of the functional , that is, , where is the solution to the minimization problem existence and uniqueness of the operator follow from the properties of the functional and strict monotonicity of the mapping (see, e.g., [1, 2]). In Hilbert space, . It is obvious from the definition of function that

If is a reflexive strictly convex and smooth Banach space, then for , if and only if . It is sufficient to show that if , then . From (1.4), we have . This implies that . From the definitions of , we have . That is, ; see [3, 4] for more details.

Let be a closed convex subset of , and let be a mapping from into itself with nonempty set of fixed points. We denote by the set of fixed points of . is called hemi-relatively nonexpansive if for all and .

A point in is said to be an asymptotic fixed point of if contains a sequence which converges weakly to such that the strong . The set of asymptotic fixed points of will be denoted by . A hemi-relatively nonexpansive mapping from into itself is called relatively nonexpansive if (see, [5]).

A point in is said to be a strong asymptotic fixed point of if contains a sequence which converges strongly to such that the strong . The set of strong asymptotic fixed points of will be denoted by . A hemi-relatively nonexpansive mapping from into itself is called weak relatively nonexpansive if (see, [6]).

The following conclusions are obvious: (1) relatively nonexpansive mapping must be weak relatively nonexpansive mapping; (2) weak relatively nonexpansive mapping must be hemi-relatively nonexpansive mapping.

In this paper, we will give two examples to show that the inverses of above two conclusions are not hold.

In an infinite-dimensional Hilbert space, Mann's iterative algorithm has only weak convergence, in general, even for nonexpansive mappings. Hence in order to have strong convergence, in recent years, the hybrid iteration methods for approximating fixed points of nonlinear mappings has been introduced and studied by various authors.

In 2003, Nakajo and Takahashi [7] proposed the following modification of Mann iteration method for a single nonexpansive mapping in a Hilbert space : where is a closed convex subset of , and denotes the metric projection from onto a closed convex subset of . They proved that if the sequence is bounded above from one, then the sequence generated by (1.5) converges strongly to , where denote the fixed points set of .

The ideas to generalize the process (1.5) from Hilbert space to Banach space have recently been made. By using available properties on uniformly convex and uniformly smooth Banach space, Matsushita and Takahashi [8] presented their ideas as the following method for a single relatively nonexpansive mapping in a Banach space : where is the duality mapping on , and is the generalized projection from onto a nonempty closed convex subset . They proved the following convergence theorem.

Theorem MT. Let be a uniformly convex and uniformly smooth Banach space, let be a nonempty closed convex subset of , let be a relatively nonexpansive mapping from into itself, and let be a sequence of real numbers such that and . Suppose that is given by (1.6), where is the duality mapping on . If is nonempty, then converges strongly to , where is the generalized projection from onto .

Recently, Plubtieng and Ungchittrakool [9] here proposed the following hybrid iteration method for a countable family of relatively nonexpansive mappings in a Banach space and proved the convergence theorem.

Theorem PU. Let be a uniformly smooth and uniformly convex Banach space and let and be two nonempty closed convex subsets of such that . Let be a sequence of relatively nonexpansive mappings from into such that is nonempty, and let be a sequence defined as follows: where satisfies either (a) for all and or(b). Suppose that for any bounded subset of there exists an increasing, continuous and convex function from into such that , and Let be a mapping from into defined by for all and suppose that Then , , and converge strongly to .

In this paper, the authors have obtained the following results: (1) the definition of uniformly closed countable family of nonlinear mappings, (2) strong convergence theorem by the monotone hybrid algorithm for a countable family of hemi-relatively nonexpansive mappings in a Banach space with new method of proof, (3) two examples of uniformly closed countable families of nonlinear mappings and applications, (4) an example which is hemi-relatively nonexpansive mapping but not weak relatively nonexpansive mapping, and (5) an example which is weak relatively nonexpansive mapping but not relatively nonexpansive mapping. Therefore, the results of this paper improve and extend the results of Plubtieng and Ungchittrakool [9] and many others.

We need the following definitions and lemmas.

Lemma 1.1 (Kamimura and Takahashi [10]). Let be a uniformly convex and smooth Banach space, and let and be two sequences of . If and either or is bounded, then .

Lemma 1.2 (Alber [1]). Let be a nonempty closed convex subset of a smooth Banach space and . Then, if and only if

Lemma 1.3 (Alber [1]). Let be a reflexive, strictly convex, and smooth Banach space, let be a nonempty closed convex subset of , and let . Then

The following lemma is not hard to prove.

Lemma 1.4. Let be a strictly convex and smooth Banach space, let be a closed convex subset of , and let be a hemi-relatively nonexpansive mapping from into itself. Then is closed and convex.

In this paper, we present the definition of uniformly closed for a sequence of mappings as follows.

Definition 1.5. Let be a Banach space, be a closed convex subset of , let be a sequence of mappings of into such that is nonempty. We say that is uniformly closed if whenever converges strongly to and as .

Lemma 1.6 (see [11, 12]). Let be a -uniformly convex Banach space with . Then, for all , , and , where is the generalized duality mapping from into , and is the -uniformly convexity constant of .

2. Main Results

Definition 2.1. Let be a Banach space, let be a nonempty closed convex subset of , and let , be two sequences of mappings. If for any convergence sequence , the following holds: Then and are said to satisfy the only asymptotically condition.

Theorem 2.2. Let be a uniformly convex and uniformly smooth Banach space, let be a nonempty closed convex subset of , let , be two uniformly closed sequences of hemi-relatively nonexpansive mappings satisfying the asymptotically condition such that , , and . Assume that , , and are four sequences in such that , and , for some constant . Define a sequence in by the following algorithm: Then converges strongly to .

Proof . We first show that is closed and convex for all . From the definitions of , it is obvious that is closed for all . Next, we prove that is convex for all . Since is equivalent to it is easy to get that is convex for all .Next, we show that for all . Indeed, for each , we have So, , which implies that for all .Since and , then we get the following: Therefore, is nondecreasing. On the other hand, by Lemma 1.3, we have for all and for all . Therefore, is also bounded. This together with (3.1) implies that the limit of exists. Put the following: From Lemma 1.3, we have, for any positive integer , that for all . This together with (3.6) implies that is, uniformly for all , holds. By using Lemma 1.1, we get that is, uniformly for all , holds. Then is a Cauchy sequence; therefore, there exists a point such that .Since , from the definition of , we have the following: This together with (2.10) and implies that Therefore, by using Lemma 1.1, we obtain the following: Since is uniformly norm-to-norm continuous on bounded sets, then we have the following: Noticing that which leads to From (2.15) and, , we obtain that Since is also uniformly norm-to-norm continuous on bounded sets, then we obtain that This together with (2.11) implies that Sine and satisfy the asymptotically condition, we also have Since and , are uniformly closed, we have Finally, we prove that , from Lemma 1.3, we have On the other hand, and , for all . Also from Lemma 1.3, we have By the definition of , we know that Combining (2.24) and (2.25), we know that . Therefore, it follows from the uniqueness of that . This completes the proof.

When , in Theorem 2.2, we obtain the following result.

Theorem 2.3. Let be a uniformly convex and uniformly smooth Banach space, let be a nonempty closed convex subset of , and let be a uniformly closed sequence of hemi-relatively nonexpansive mappings such that . Assume that is a sequences in [0,1] such that for some constant . Define a sequence in by the following algorithm: Then converges strongly to .

3. Applications for Equilibrium Problem

Let be a real Banach space, and let be the dual space of . Let be a closed convex subset of . Let be a bifunction from to . The equilibrium problem is to find such that The set of solutions of (1.2) is denoted by . Given a mapping let for all . Then, if and only if for all , that is, is a solution of the variational inequality. Numerous problems in physics, optimization, and economics reduce to find a solution of (1.2). Some methods have been proposed to solve the equilibrium problem in Hilbert spaces, see, for instance, [13–15].

For solving the equilibrium problem, let us assume that a bifunction satisfies the following conditions:(A1),(A2) is monotone, that is, ,(A3) for all , only ,(A4) for all , is convex and lower semicontinuous.

Lemma 3.1 (Blum and Oettli [13]). Let be a closed convex subset of a smooth, strictly convex, and reflexive Banach space , let be a bifunction from to satisfying (A1)–(A4), and let and . Then, there exists such that

Lemma 3.2 (Takahashi and Zembayashi [15]). Let be a closed convex subset of a uniformly smooth, strictly convex, and reflexive Banach space , and let be a bifunction from to satisfying (A1)–(A4). For , define a mapping as follows: for all . Then, the following hold: (1) is single-valued;(2) is a firmly nonexpansive-type mapping, that is, for all , (3);(4) is closed and convex;(5) is also a relatively nonexpansive mapping.

Lemma 3.3 (Takahashi and Zembayashi [15]). Let be a closed convex subset of a smooth, strictly convex, and reflexive Banach space , let be a bifunction from to satisfying (A1)–(A4), let , and let , , then the following holds:

Lemma 3.4. Let be a -uniformly convex with and uniformly smooth Banach space, and let be a nonempty closed convex subset of . Let be a bifunction from to satisfying (A1)–(A4). Let be a positive real sequence such that . Then the sequence of mappings is uniformly closed.

Proof. (1) Let be a convergent sequence in . Let for all , then Putting in (3.6) and in (3.7), we have So, from (A2), we have and hence Thus, we have which implies that By using Lemma 1.6, we obtain the following: Therefore, we get the following: On the other hand, for any , from , we have so that is bounded. Since , this together with (3.14) implies that is a Cauchy sequence. Hence is convergent.(2) By using Lemma 3.2, we know that (3) From (1) we know that, exists for all . So, we can define a mapping from into itself by It is obvious that is nonexpansive. It is easy to see that On the other hand, let and , we have By (A2) we know that Since and from (A4), we have , for all . Then, for and , Therefore, we have Letting and using (A3), we get the following: and hence . From above two respects, we know that .Next we show is uniformly closed. Assume and , from above results, we know that . On the other hand, from , we also get , so that . That is, the sequence of mappings is uniformly closed. This completes the proof.

Theorem 3.5. Let be a -uniformly convex with and uniformly smooth Banach space, and let be a nonempty closed convex subset of . Let and be two bifunctions from to satisfying (A1)–(A4). Assume that , , , and are four sequences in such that , , and for some constant . Let be a sequence generated by where and . Assume that the mappings and satisfy the asymptotically condition. Then converges strongly to .

Proof. By Lemma 3.4, , are uniformly closed; therefore, by using Theorem 2.2 and Lemma 3.2, we can obtain the conclusion of Theorem 3.5. This completes the proof.

When , in the Theorem 3.5, we obtain the following result.

Theorem 3.6. Let be a uniformly convex with and uniformly smooth Banach space, and let be a nonempty closed convex subset of . Let be a bifunction from to satisfying (A1)–(A4). Assume that is a sequences in such that for some constant . Define a sequence in by the following algorithm: where . Then converges strongly to .

4. Applications for Maximal Monotone Operators

In this section, we apply our above results to prove some strong convergence theorem concerning maximal monotone operators in a Banach space .

Let be a multivalued operator from to with domain and range . An operator is said to be monotone if for each and , . A monotone operator is said to be maximal if it's graph is not properly contained in the graph of any other monotone operator. We know that if is a maximal monotone operator, then is closed and convex. The following result is also wellknown.

Theorem 4.1. Let be a reflexive, strictly convex, and smooth Banach space, and let be a monotone operator from to . Then is maximal if and only if . for all .

Let be a reflexive, strictly convex, and smooth Banach space, and let be a maximal monotone operator from to . Using Theorem 4.1 and strict convexity of , we obtain that for every and , there exists a unique such that

Then we can define a single-valued mapping by and such a is called the resolvent of , we know that for all .

Theorem 4.2. Let be a uniformly convex and uniformly smooth Banach space, let be a maximal monotone operator from to , and let be a resolvent of for . Then for any sequence such that , is a uniformly closed sequence of hemi-relatively nonespansive mappings.

Proof. Firstly, we show that is uniformly closed. Let be a sequence such that and . Since is uniformly norm-to-norm continuous on bounded sets, we obtain that It follows from and the monotonicity of that for all and . Letting , we have for all and . Therefore, from the maximality of , we obtain that for all , that is, .Next we show that is a hemi-relatively nonexpansive mapping for all . For any and , from the monotonicity of , we have This implies that is a hemi-relatively nonexpansive mapping for all . This completes the proof.

Theorem 4.3. Let be a uniformly convex and uniformly smooth Banach space, let and be two maximal monotone operators from to with nonempty common zero point set , let be a resolvent of and a resolvent of for . Assume that , , , and are four sequences in such that , , , and for some constant . Let be a sequence generated by where . Assume that the mappings and satisfy the asymptotically condition. Then converges strongly to .

Proof. From Theorem 4.2, is uniformly closed countable family of hemi-relatively nonexpansive mappings, on the other hand, , by using Theorem 2.2, we can obtain the conclusion of Theorem 4.3.

When in Theorem 4.3, we obtain the following result.

Theorem 4.4. Let be a uniformly convex and uniformly smooth Banach space, let be a maximal monotone operator from to with nonempty zero point set , and let be a resolvent of for . Assume that is a sequence in such that for some constant . Define a sequence in by the following algorithm: where . Then converges strongly to .

5. Examples

Firstly, we give an example which is hemi-relatively nonexpansive mapping but not weak relatively nonexpansive mapping.

Example 5.1. Let and be a any element of . We define a mapping as follows: for . Next we show that is a hemi-relatively nonexpansive mapping but no weak relatively nonexpansive mapping. First, it is obvious that . In addition, it is easy to see that This implies that for all . It follows from above inequality that for all and . That is for all and ; hence is a hemi-relatively nonexpansive mapping. Finally, we show that is not weak relatively nonexpansive mapping. In fact that, letting from the definition of , we have which implies that and () as . That is but .Next, we give an example which is weak relatively nonexpansive mapping but not relatively nonexpansive mapping.

Example 5.2. Let , where It is well known that is a Hilbert space, so that . Let be a sequence defined by where for all . Define a mapping as follows:

Conclusion 1. converges weakly to .

Proof. For any , we have as . That is, converges weakly to .

Conclusion 2. is not a Cauchy sequence, so that, it does not converge strongly to any element of .

Proof. In fact, we have for any . Then is not a Cauchy sequence.

Conclusion 3. has a unique fixed point 0, that is, .

Proof . The conclusion is obvious.

Conclusion 4. is an asymptotic fixed point of .

Proof. Since converges weakly to and as , so that, is an asymptotic fixed point of .

Conclusion 5. has a unique strong asymptotic fixed point 0, so that .

Proof . In fact that, for any strong convergent sequence such that and as , from Conclusion 2, there exists sufficiently large nature number such that , for any . Then for , and it follows from that and hence .

Conclusion 6. is a weak relatively nonexpansive mapping.

Proof. Since is a Hilbert space, we have From Conclusion 2, we have , then is a weak relatively nonexpansive mapping.

Conclusion 7. is not a relatively nonexpansive mapping.

Proof. From Conclusions 3 and 4, we have , so that is not a relatively nonexpansive mapping.

Acknowledgment

This project is supported by the National Natural Science Foundation of China under Grant (11071279).

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