A New Iterative Method for Solving Equilibrium Problems and Fixed Point Problems for Infinite Family of Nonexpansive Mappings

Abstract

We introduce a new iterative scheme for finding a common element of the solutions sets of a finite family of equilibrium problems and fixed points sets of an infinite family of nonexpansive mappings in a Hilbert space. As an application, we solve a multiobjective optimization problem using the result of this paper.

1. Introduction

Let be a Hilbert space and be a nonempty, closed, and convex subset of . Let be a bifunction of into , where is the set of real numbers. The equilibrium problem for the bifunction is to find such that

(1.1)

The set of solutions of the above inequality is denoted by . Many problems arising from physics, optimization, and economics can reduce to finding a solution of an equilibrium problem.

In 2007, S. Takahashi and W. Takahashi [1] first introduced an iterative scheme by the viscosity approximation method for finding a common element of the solutions set of equilibrium problem and the set of fixed points of a nonexpansive mapping in a Hilbert space and proved a strong convergence theorem which is based on Combettes and Hirstoaga's result [2] and Wittmann's result [3]. More precisely, they obtained the following theorem.

Theorem 1.1 (see [1]).

Let be a nonempty closed and convex subset of . Let be a bifunction which satisfies the following conditions:

(A1) for all ;

(A2) is monotone, that is, for all ;

(A3)For all ,

(1.2)

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

Let be a nonexpansive mapping with , where denotes the set of fixed points of the mapping , and let be a contraction, if there exists a constant such that for all . Let and be the sequences generated by and

(1.3)

where and satisfy the following conditions:

(1.4)

Then the sequences and converge strongly to a point , where

(1.5)

is the metric projection of onto and denotes nearest point in from .

Recently, many results on equilibrium problems and fixed points problems in the context of the Hilbert space and Banach space are introduced (see, e.g., [4–8]).

Let be a nonlinear mapping. The variational inequality problem corresponding to the mapping is to find a point such that

(1.6)

The variational inequality problem is denoted by [9].

The mapping is called -Lipschitzian and -strongly monotone if there exist constants such that

(1.7)

(1.8)

respectively. It is well known that if is strongly monotone and Lipschitzian on , then has a unique solution. An important problem is how to find a solution of . Recently, there are many results to solve the (see, e.g., [10–14]).

Let be a nonempty closed and convex subset of a Hilbert space , be a countable family of nonexpansive mappings, and be bifunctions satisfying conditions (A1)–(A4) such that . Let . For each , define the mapping by

(1.9)

Lemma 2.5 (see below) shows that, for each , is firmly nonexpansive and hence nonexpansive and . Suppose that is a -Lipschitzian and -strong monotone operator and let . Assume that .

In this paper, motivated and inspired by the above research results, we introduce the following iterative process for finding an element in : for an arbitrary initial point ,

(1.10)

where , , is a strictly decreasing sequence in with , , with , and with . Then we prove that the iterative process defined by (1.10) strongly converge to an element , which is the unique solution of the variational inequality

(1.11)

As an application of our main result, we solve a multiobjective optimization problem.

2. Preliminaries

Let be a Hilbert space and a nonexpansive mapping of into itself such that . For all and , we have

(2.1)

and hence

(2.2)

It is well known that, for all and ,

(2.3)

which implies that

(2.4)

for all and with .

Let be a nonempty closed and convex subset of and, for any , there exists unique nearest point in , denoted by , such that

(2.5)

Moreover, we have the following:

(2.6)

Let denote the identity operator of and let be a sequence in a Hilbert space and . Throughout this paper, denotes that strongly converges to and denotes that weakly converges to .

We need the following lemmas for our main results.

Lemma 2.1 (see [15]).

Let be a nonempty closed and convex subset of a Hilbert space and a nonexpansive mapping from into itself. Then is demiclosed at zero, that is,

(2.7)

Lemma 2.2 (see [10, Lmma 3.1(b)]).

Let be a Hilbert space and be a nonexpansive mapping. Let be a mapping which is -Lipschitzian and -strong monotone on . Assume that and . Define a mapping by

(2.8)

Then for all , where .

If , Lemma 2.2 still holds.

Lemma 2.3 (see [16]).

Let , be the sequences of nonnegative real numbers and . Suppose that is a real number sequence such that

(2.9)

Assume that . Then the following results hold.

• (1) If for all , where , then is a bounded sequence.

• (2) If

  

  (2.10)

then .

Lemma 2.4 (see [17]).

Let C be a nonempty closed and convex subset of a Hilbert space H and be a bifunction which satisfies the conditions (A1)–(A4). Let and . Then there exists such that

(2.11)

Lemma 2.5 (see [2]).

Let be a Hilbert space and be a nonempty closed and convex subset of . Assume that satisfies the conditions (A1)–(A4). For all and , define a mapping as follows:

(2.12)

Then the following holds:

(1) is single-valued;

(2) is firmly nonexpansive, that is, for any ,

(2.13)

(3);

(4) is closed and convex.

The following lemma is an immediate consequence of an inner product.

Lemma 2.6.

Let H be a real Hilbert space. Then the following identity holds:

(2.14)

3. Main Results

First, we prove some lemmas as follows.

Lemma 3.1.

The sequence generated by (1.10) is bounded.

Proof.

Let for each . Lemma 2.5 shows that each is firmly-nonexpansive and hence nonexpansive. Hence, for each and , we have

(3.1)

(3.2)

By Lemma 2.2, we have

(3.3)

where . Therefore, by (3.2) and (3.3), we obtain (note that is strictly decreasing and )

(3.4)

By induction, we obtain . Hence is bounded and so are and for each . Since is -Lipschitzian, we have

(3.5)

which shows that is bounded. This completes the proof.

Lemma 3.2.

If the following conditions hold:

(3.6)

then

Proof.

For each , since each is nonexpansive, we have

(3.7)

By (3.7), we have

(3.8)

By the definition of the iterative sequence (1.10), we have

(3.9)

and hence

(3.10)

It follows from (3.8) and (3.10) that

(3.11)

where . Since is strictly decreasing, we have . Further, from the assumptions, it follows that

(3.12)

Therefore, by Lemma 2.3, we have . This completes the proof.

Lemma 3.3.

If the following conditions hold:

(3.13)

then for each .

Proof.

For any and , it follows from Lemma 2.5(2) that

(3.14)

and hence . Further, we have

(3.15)

Therefore, from (2.4) and (3.3), we have

(3.16)

It follows that

(3.17)

for each . Note that for . From the assumptions, Lemma 3.2, and the previous inequality, we conclude that as for each . Further, we have

(3.18)

This completes the proof.

Lemma 3.4.

If the following conditions hold:

(3.19)

then for all

Proof.

By the definition of the iterative sequence (1.10), we have

(3.20)

that is,

(3.21)

Hence, for any , we get

(3.22)

Since each is nonexpansive, by (2.2), we have

(3.23)

Hence, combining this inequality with (3.22), we get

(3.24)

which implies that (note that is a strictly decreasing sequence)

(3.25)

From Lemma 3.3, , and the inequality

(3.26)

we obtain

(3.27)

Therefore, from Lemma 3.2, (3.25), and (3.27), it follows that

(3.28)

This completes the proof.

Next we prove the main results of this paper.

Theorem 3.5.

Assume that the following conditions hold:

(3.29)

Then the sequence generated by (1.10) converges strongly to an element in , which is the unique solution of the variational inequality .

Proof.

Since , we can select an element , which implies that

(3.30)

First, we prove that

(3.31)

Since is bounded, there exists a subsequence of such that

(3.32)

Without loss of generality, we may further assume that for some . From Lemmas 3.4 and 2.1, we get for all . Hence we have . It follows from Lemma 2.5 that each is firmly nonexpansive and hence nonexpansive. Lemma 3.3 shows that as . Therefore, from Lemma 2.1, it follows that for each , which shows that . Lemma 2.5 shows that for each . Hence . By using the above argument, we conclude that

(3.33)

Noting that is a solution of the , we obtain

(3.34)

It follows from Lemma 2.6 that

(3.35)

Let and for all . Then, from the assumptions and (3.31), we have

(3.36)

Therefore, by applying Lemma 2.3 to (3.35), we conclude that the sequence strongly converges to a point .

In order to prove the uniqueness of solution of the , we assume that is another solution of . Similarly, we can conclude that converges strongly to a point . Hence , that is, is the unique solution of . This completes the proof.

As direct consequences of Theorem 3.5, we obtain the following corollaries.

Corollary 3.6.

Let be a nonempty closed and convex subset of a Hilbert space . For each let be bifunctions which satisfy conditions (A1)–(A4) such that . Let , and let be a strictly decreasing sequence with , , with , , and with . For an arbitrary initial , define the iterative sequence by

(3.37)

If the following conditions hold:

(3.38)

then the sequence converges strongly to an element .

Proof.

Put and for each in Theorem 3.5. Then we know that is -Lipschitzian and -strongly monotone, and . Therefore, by Theorem 3.5, we conclude the desired result.

Corollary 3.7.

Let be a nonempty closed and convex subset of a Hilbert space . Let be a countable family of nonexpansive mappings of such that and an operator which is -Lipschitzian and -strong monotone on . Let . Assume that . Let with be a strictly decreasing sequence, and with . For an arbitrary initial , define the iterative sequence by

(3.39)

where . If the following conditions hold:

(3.40)

then the sequence strongly converges to an element , which is the unique solution of the variational inequality

(3.41)

Proof.

Put for each and . Set in Theorem 3.5. Then, by (2.6), we have . Therefore, by Theorem 3.5, we conclude the desired result.

Remark 3.8.

• (1) Recently, many authors have studied the iteration sequences for infinite family of nonexpansive mappings. But our iterative sequence (1.10) is very different from others because we do not use -mapping generated by the infinite family of nonexpansive mappings and we have no any restriction with the infinite family of nonlinear mappings.

• (2) We do not use Suzuki's lemma [18] for obtaining the result that . However, many authors have used Suzuki's lemma [18] for obtaining the result that in the process of studying the similar algorithms. For example, see [5, 19, 20] and so on.

4. Application

In this section, we study a kind of multiobjective optimization problem based on the result of this paper. That is, we give an iterative sequence which solves the following multiobjective optimization problem with nonempty set of solutions:

(4.1)

where and are both convex and lower semicontinuous functions defined on a nonempty closed and convex subset of of a Hilbert space . We denote by the set of solutions of (4.1) and assume that .

We denote the sets of solutions of the following two optimization problems by and , respectively,

(4.2)

Obviously, if we find a solution , then one must have .

Now, let and be two bifunctions from to defined by and , respectively. It is easy to see that and , where denotes the set of solutions of the equilibrium problem:

(4.3)

respectively. In addition, it is easy to see that and satisfy the conditions (A1)–(A4). Therefore, by setting in Corollary 3.6, we know that, for any initial guess ,

(4.4)

By Corollary 3.6, we know that the sequence converges strongly to a solution , which is a solution of the multiobjective optimization problem (4.1).