Journal of Applied Mathematics

Volume 2014 (2014), Article ID 858679, 12 pages

http://dx.doi.org/10.1155/2014/858679

Research Article

Explicit Scheme for Fixed Point Problem for Nonexpansive Semigroup and Split Equilibrium Problem in Hilbert Space

Pei Zhou and Gou-Jie Zhao

School of Management, Tianjin University, Tianjin 300072, China

Received 1 May 2014; Revised 18 July 2014; Accepted 1 August 2014; Published 19 August 2014

Academic Editor: Filomena Cianciaruso

Copyright © 2014 Pei Zhou and Gou-Jie Zhao. 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 establish an iterative method for finding a common element of the set of fixed points of nonexpansive semigroup and the set of split equilibrium problems. Under suitable conditions, some strong convergence theorems are proved. Our works improve previous results for nonexpansive semigroup.

1. Introduction

Let be a real Hilbert space whose inner product and norm are denoted by and , respectively. Let be a nonempty closed convex subset of , and let be a bifunction of into which is the set of real numbers. The equilibrium problem introduced by Blum and Oettli [1] for is to find such that

The set of solutions of (1) is denoted by . Numerous problems in physics, optimization, and economics reduce to finding a solution of (1) (see [2–8]). The split equilibrium problem was introduced by Moudafi in [9]; he considers the following pair of equilibrium problems in different spaces: let and be two real Hilbert spaces, let and be nonlinear bifunctions, and let be a bounded linear operator, and consider the nonempty closed convex subsets and ; then the split equilibrium problem (SEP) is to find such that

and such that

The solution set of SEP (2)-(3) is denoted by .

Recall that mapping of into itself is called nonexpansive if

Let a family be a nonexpansive semigroup on , if it satisfies the following conditions:

(S1)  ;(S2) for each ;(S3) for each and ;(S4), is continuous.

The set of all the common fixed points of family is denoted by ; that is,

where is the set of fixed points of . It is well known that is closed and convex.

In 2010, Tian [10] introduced the following general iterative scheme for finding an element of set of solutions to the fixed point of nonexpansive mapping in a Hilbert space. Define the sequence by

where is -Lipschitzian and -strongly monotone operator. Then he proved that if the sequence satisfies appropriate conditions, the sequence generated by (6) converges strongly to the unique solution of the variational inequality

In 2011, Ceng et al. [11] added the metric project to the method of Tian (6) and studied the following explicit iterative scheme to find fixed points:

They prove the strong convergence of to a fixed point of the same variational inequality (7).

In 2008, Plubtieng and Punpaeng [12] introduced the following implicit iterative algorithm to prove a strong convergence theorem for fixed point problem with nonexpansive semigroup:

where is a continuous net and is a positive real divergent net.

In 2014, Kazmi and Rizvi [13] studied the following implicit iterative algorithm. Under some assumptions, they obtain some strong convergence theorem for EP (1) and the fixed point problem:

where and are the continuous nets in .

Motivated and inspired by [10–13], we introduce an explicit iterative scheme for finding a common element of the set of solutions SEP and fixed point for a nonexpansive semigroup in real Hilbert spaces. Starting with an arbitrary , define sequences and by

Under suitable conditions, some strong convergence theorems for approximating to these common elements are proved.

2. Preliminaries

This section collects some results that will be used in the proofs.

Let be a real Hilbert space with the inner product and the norm , respectively.

It is well known that, for all and , the following holds:

Let be a nonempty closed convex subset of . Then, for any , there exists a unique nearest point of , denoted by , such that for all . Such is called the metric projection from into . We know that is nonexpansive. It is also known that and

It is easy to see that (13) is equivalent to

Let be a nonlinear mapping. Recall the following definitions.

Definition 1. is said to be

(i)monotone if

(ii)strongly monotone if there exists a constant such that

for such a case, is said to be -strongly monotone,(iii)-inverse strongly monotone (-ism) if there exists a constant such that

(iv)-Lipschitz continuous if there exists a constant such that

Remark 2. Let , where is a -Lipschitz and -strongly monotone operator on with and is a Lipschitz mapping on with coefficient , . It is a simple matter to see that the operator is -strongly monotone over ; that is,

Lemma 3 (see [14]). Let be a nonexpansive mapping of a closed convex subset of a Hilbert space . If has a fixed point, then is demiclosed; that is, whenever the sequence of is weakly convergent to and is strongly convergent to , then .

Lemma 4 (see [15]). Let be a nonempty bounded closed convex subset of a Hilbert space and let be a nonexpansive semigroup on . Then, for each ,

Definition 5 (see [9]). A mapping is said to be averaged if it can be written as the average of the identity mapping and a nonexpansive mapping; that is,

where , is nonexpansive, and is the identity operator on .

Proposition 6 (see [9]).   

(i)If , where is averaged, is nonexpansive, and , then is averaged.(ii)The composite of finitely many averaged mappings is averaged.(iii)If is -ism, then, for , is -ism.(iv) is averaged if, and only if, its complement is -ism for some .

Assumption 7 (see [1]). For solving the equilibrium problem for a bifunction , let us assume that satisfies the following conditions:

(A1) for all ;(A2) is monotone, that is, for all ;(A3)for each ,

(A4)for each ,   is convex and lower semicontinuous.

Lemma 8 (see [2]). Let be a nonempty closed convex subset of , and let be a bifunction of   into satisfying (A1)–(A4). Let and . Then there exists such that

Define a mapping as follows:

for all . Then the following hold:

(1) is single-valued;(2) is firmly nonexpansive; that is, for any ,

(3);(4) is closed and convex.

Lemma 9 (see [16]). Let be a nonempty closed convex subset of a Hilbert space , and let be a bifunction. Let and . Then

Lemma 10 (see [17]). Assume that and . Let be a which is -strongly monotone and -Lipschitzian on with , . Let . Then the operator is contraction; that is, , where .

Lemma 11 (see [10]). Let be a Hilbert space, is a contractive mapping with constant . is -Lipschitzian and -strongly monotone operator with , . Then, for ,

That is, is strongly monotone with coefficient .

Lemma 12. Let be a real Hilbert space. Then the following well-known results hold:   ,

Lemma 13 (see [18]). Assume that is a sequence of nonnegative real numbers such that

where is a sequence in and is a sequence in such that

(i),(ii) or .

Then .

3. The General Explicit Iterative Method

Let be a contractive mapping with constant and let be -strongly monotone and -Lipschitzian with , . Let and . Let be a nonexpansive semigroup on such that . Assume and are the continuous nets of positive real numbers such that and .

In this section, we introduce the following explicit iterative scheme that the nets and are generated by

where , is the spectral radius of the operator , and is the adjoint of .

We prove the strong convergence of and to a fixed point of which solves the following variational inequality:

In the sequel, we denote by the sequence defined by

Next, we will prove first a lemma and then some corollaries to be used in the proofs for the main result of this section.

Theorem 14. Let and be two real Hilbert spaces and let and be nonempty closed subsets. Let be a bounded linear operator. Assume that and are the bifunctions satisfying Assumption 7 and is upper semicontinuous in the first argument. Let the sequences and be generated by (30), and suppose that the sequence satisfies the following conditions:

(a) and ;(b);(c)either or .

Then the sequences and converge strongly to , where , which is the unique solution of the variational inequality (31).

Proof. We divide the proof into several steps.

(i) is well defined.

For and , define a mapping by

According to Lemma 8, we can easily know that and both are firmly nonexpansive mappings and are averaged operators. From Proposition 6, we can obtain that the operator is averaged and hence nonexpansive. Following Lemma 9 and , we get

Since , we can claim that the mapping is a contraction mapping. Therefore, by Banach contraction principle, has the unique fixed point .

(ii) The sequences , , and are bounded.

Letting , we obtain that , , and .

From (30), we obtain

Since , we have

From (32), we have

Further, using (31) again, we obtain

It follows from (40) and induction that

Hence, the sequence is bounded and therefore , , and are also bounded.

(iii) Consider that .

According to (35) and Lemma 12, we obtain

From (40), we obtain

Since is bounded, , and , we obtain that

From (30), we have

Hence, we obtain

It follows from (40) and (44) that

where . From (46), we obtain

Since is bounded, , and and considering (42), we can claim that

(iv) Consider that .

From (30) and Lemma 13, we have

where

From (32), we obtain

Following (48) and (50), we obtain

Using (30) again, we obtain

where

Since , , and are bounded, we can claim that and are bounded. We can deduce that , and , and .

Following conditions (a)–(c), , , and Lemma 13, we obtain that

(v) Consider that .

From (30) and (32), we obtain

Since and and are bounded, we obtain

By (54) and (56), we get

Next, we have

On the other hand, by (30), we have

So without loss of generality, we may assume that is nonexpansive semigroup on , and, by Lemma 4, we obtain

From (58), (59), and (60), we have

(vi) Consider that .

Since is bounded, there exists a subsequence of which converges weakly to . From (47), we obtain which converges weakly to . From (58), follows. We show . According to (30) and (A2),

and hence

Since and , from (A4), it follows that for all . For with and , let ; then we get . So, from (A1) and (A4) we have

and hence . From (A3), we have for all . Therefore, .

Since and is a bounded linear operator, we obtain . Let . Following (42), we obtain and . Then from Lemma 8, we get

Since is upper semicontinuous in the first argument, taking to above inequality as and using , we obtain that

which means that and hence .

(vii) Consider that .

According to (30), letting

we can observe that

Observe that

Hence, for each , we can obtain

Taking limit in (70) and noticing that , , and , we obtain that

which implies .

(viii) Consider that .

From (30), we have , and, for any given ,

Since is the metric projection from onto , we obtain

It follows from (72) and (37) that

which hence implies that

where

It is easily seen that and (due to (a), (b), (c), and (72)). According to Lemma 13, we can conclude that . This completes the proof.

Corollary 15. Let and be two real Hilbert spaces and let and be nonempty closed subsets. Let be a bounded linear operator. Assume that and are the bifunctions satisfying Assumption 7 and is upper semicontinuous in the first argument. Let the sequences and be generated by

Suppose that the sequence satisfies the following conditions: (a), (b), and (c) (the same in Theorem 14). Then the sequences and converge strongly to , where , which is the unique solution of the variational inequality (31).

If is a strongly positive bounded linear self-adjoint operator on with constant such that , we can easily obtain the following corollary.

Corollary 16. Let and be two real Hilbert spaces and let and be nonempty closed subsets. Let be a bounded linear operator. Assume that and are the bifunctions satisfying Assumption 7 and is upper semicontinuous in the first argument. Let the sequences and be generated by

Suppose that the sequence satisfies the following conditions: (a), (b), and (c) (the same in Theorem 14). Then the sequences and converge strongly to , where , which is the unique solution of the variational inequality

4. Application

In this section, we introduce an example of numerical test to illustrate the algorithm given in Corollary 16.

Example 17. Let , the set of all real numbers, with the inner product defined by , , and induced usual norm . Assume that and are defined by , , and , . It is easy to claim that the bifunctions and satisfy Assumption 7 and the bifunction is upper semicontinuous. Next, we can find the formula of . From Lemma 8, we can claim that is single-valued; for any ,

Letting , we obtain that

and so

Similarly, we can also obtain that

For all , we can define mapping , ,   , and the nonexpansive mapping is satisfied as . It is claimed that the mapping is contraction with constant , is a bounded linear operator on with adjoint operator and , and is a strongly positive bounded linear self-adjoint operator with constant on . On the other hand, we can take which satisfies . Hence, it is easy to observe that , , and . Furthermore, we can obtain that . Consequently, . Therefore, all the assumptions in Corollary 16 are satisfied. We can obtain the following numerical algorithm:

where and . Then, by Corollary 16, the sequence converges to a solution of Example 17. For a number , using the MATLAB, we generated a sequence as in Figure 1.

Figure 1: Convergence of iterative sequence .

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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