Iterative Methods for Finding Common Solution of Generalized Equilibrium Problems and Variational Inequality Problems and Fixed Point Problems of a Finite Family of Nonexpansive Mappings

Abstract

We introduce a new method for a system of generalized equilibrium problems, system of variational inequality problems, and fixed point problems by using -mapping generated by a finite family of nonexpansive mappings and real numbers. Then, we prove a strong convergence theorem of the proposed iteration under some control condition. By using our main result, we obtain strong convergence theorem for finding a common element of the set of solution of a system of generalized equilibrium problems, system of variational inequality problems, and the set of common fixed points of a finite family of strictly pseudocontractive mappings.

1. Introduction

Let be a real Hilbert space, and let be a nonempty closed convex subset of . Let be a nonlinear mapping, and let be a bifunction. A mapping of into itself is called nonexpansive if for all . We denote by the set of fixed points of (i.e., ). Goebel and Kirk [1] showed that is always closed convex, and also nonempty provided has a bounded trajectory.

A bounded linear operator on is called strongly positive with coefficient if there is a constant with the property

(1.1)

The equilibrium problem for is to find such that

(1.2)

The set of solutions of (1.2) is denoted by . Many problems in physics, optimization, and economics are seeking some elements of , see [2, 3]. Several iterative methods have been proposed to solve the equilibrium problem, see, for instance, [2–4]. In 2005, Combettes and Hirstoaga [3] introduced an iterative scheme of finding the best approximation to the initial data when is nonempty and proved a strong convergence theorem.

The variational inequality problem is to find a point such that

(1.3)

The set of solutions of the variational inequality is denoted by VI, and we consider the following generalized equilibrium problem.

(1.4)

The set of such is denoted by , that is,

(1.5)

In the case of , . Numerous problems in physics, optimization, variational inequalities, minimax problems, the Nash equilibrium problem in noncooperative games reduce to find element of (1.5)

A mapping of into is called inverse-strongly monotone, see [5], if there exists a positive real number such that

(1.6)

for all .

The problem of finding a common fixed point of a family of nonexpansive mappings has been studied by many authors. The well-known convex feasibility problem reduces to finding a point in the intersection of the fixed point sets of a family of nonexpansive mapping (see [6, 7]).

The ploblem of finding a common element of and the set of all common fixed points of a family of nonexpansive mappings is of wide interdisciplinary interest and importance. Many iterative methods are purposed for finding a common element of the solutions of the equilibrium problem and fixed point problem of nonexpansive mappings, see [8–10].

In 2008, S.Takahashi and W.Takahashi [11] introduced a general iterative method for finding a common element of and . They defined in the following way:

(1.7)

where is an -inverse strongly monotone mapping of into with positive real number , and ,,, and proved strong convergence of the scheme (1.7) to , where in the framework of a Hilbert space, under some suitable conditions on , , and bifunction .

Very recently, in 2010, Qin, et al. [12] introduced a iterative scheme method for finding a common element of , and common fixed point of infinite family of nonexpansive mappings. They defined in the following way:

(1.8)

where is a contraction mapping and is -mapping generated by infinite family of nonexpansive mappings and infinite real number. Under suitable conditions of these parameters they proved strong convergence of the scheme (1.8) to , where .

In this paper, motivated by [11, 12], we introduced a general iterative scheme defined by

(1.9)

where and is -mapping generated by and . Under suitable conditions, we proved strong convergence of to , and is solution of

(1.10)

2. Preliminaries

In this section, we collect and give some useful lemmas that will be used for our main result in the next section.

Let be closed convex subset of a real Hilbert space , and let be the metric projection of onto , that is, for , satisfies the property

(2..1)

The following characterizes the projection .

Lemma 2.1 (see [13]).

Given and . Then if and only if there holds the inequality

(2..2)

Lemma 2.2 (see [14]).

Let be a sequence of nonnegative real numbers satisfying

(2..3)

where , satisfy the conditions

(1),

(2).

Then .

Lemma 2.3 (see [15]).

Let be a closed convex subset of a strictly convex Banach space . Let be a sequence of nonexpansive mappings on C. Suppose that is nonempty. Let be a sequence of positive numbers with . Then a mapping on defined by

(2..4)

for is well defined, nonexpansive, and hold.

Lemma 2.4 (see [16]).

Let be a uniformly convex Banach space, a nonempty closed convex subset of , and a nonexpansive mapping. Then is demiclosed at zero.

Lemma 2.5 (see [17]).

Let and be bounded sequences in a Banach space , and let be a sequence in with . Suppose that

(2..5)

for all integer and

(2..6)

Then .

For solving the equilibrium problem for a bifunction , let us assume that satisfies the following conditions:

(A1)for all ;

(A2) is monotone, that is,,

(A3) for all ,

(2..7)

(A4) for all is convex and lower semicontinuous.

The following lemma appears implicitly in [2].

Lemma 2.6 (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

(2..8)

for all .

Lemma 2.7 (see [3]).

Assume that satisfies (A1)–(A4). For and , define a mapping as follows:

(2..9)

for all . Then, the following hold:

(1) is single-valued;

(2) is firmly nonexpansive, that is,

(2..10)

(3);

(4) is closed and convex.

In 2009, Kangtunyakarn and Suantai [18] defined a new mapping and proved their lemma as follows.

Definition 2.8.

Let be a nonempty convex subset of real Banach space. Let be a finite family of nonexpansive mappings of into itself. For each , let , where and . We define the mapping as follows:

(2..11)

This mapping is called -mapping generated by and .

Lemma 2.9.

Let be a nonempty closed convex subset of strictly convex. Let be a finite family of nonexpanxive mappings of into itself with , and let , , where , , for all,for all . Let be the mapping generated by and . Then .

Lemma 2.10.

Let be a nonempty closed convex subset of Banach space. Let be a finite family of nonexpansive mappings of into itself and ,  , where , and such that asfor and Moreover, for every , let and be the -mappings generated by and and and , respectively. Then for every .

Lemma 2.11 (see [19]).

Let be a nonempty closed convex subset of a Hilbert space , and let be defined by

(2..12)

with . Then if and only if .

3. Main Result

Theorem 3.1.

Let be a nonempty closed convex subset of a Hilbert space . Let and be two bifunctions from into satisfying (A1)–(A4), respectively. Let a -inverse strongly monotone mapping and be a -inverse strongly monotone mapping. Let be finite family of nonexpansive mappings with , where are defined by , . Let be a contraction with the coefficient . Let be the S-mappings generated by and , where , and and . Let be sequences generated by

(3.1)

where such that ,, ,,. Assume that

(i) and,

(ii),

(iii),

(iv),, ,,,,

(v), and, for all.

Then the sequence ,, , converge strongly to , and is solution of

(3.2)

Proof.

First, we show that , and are nonexpansive. Let . Since is -strongly monotone and for all, we have

(3.3)

Thus is nonexpansive. By using the same proof, we obtain that and are nonexpansive.

We will divide our proof into 6 steps.

Step 1.

We will show that the sequence is bounded. Since

(3.4)

then we have

(3.5)

By Lemma 2.7, we have . By the same argument as above, we obtaine that

Let . Then and . Hence

(3.6)

Again by Lemma 2.7, we have . Since , we have . By nonexpansiveness of , , , , we have

(3.7)

By induction we can prove that is bounded and so are ,,,. Without of generality, assume that there exists a bounded set such that

(3.8)

Step 2.

We will show that .

Putting , we have

(3.9)

From definition of , we have

(3.10)

By definition of , for , we have

(3.11)

By (3.11), we obtain that for each ,

(3.12)

This together with the condition (iv), we obtain

(3.13)

By (3.10), (3.13) and conditions (i), (ii), (iii), (iv), it implies that

(3.14)

From Lemma 2.5, (3.9), (3.14) and condition (ii), we have

(3.15)

From (3.9), we can rewrite

(3.16)

By (3.15), we have

(3.17)

On the other hand, we have

(3.18)

This implies that

(3.19)

By (3.17) and condition (ii), we have

(3.20)

Step 3.

Let ; we show that

(3.21)

From definition of , we have

(3.22)

(3.23)

By (3.23), we have

(3.24)

By (3.24), we have

(3.25)

From conditions (i)–(iii) and (3.17), we have

(3.26)

By using the same method as (3.26), we have

(3.27)

By nonexpansiveness of and (3.23), we have

(3.28)

By (3.28), we have

(3.29)

By (3.29), we have

(3.30)

From (3.17) and conditions (i)–(iii), we have

(3.31)

By using the same method as (3.31), we have

(3.32)

Step 4.

We will show that

(3.33)

Putting and , we will show that

(3.34)

Let ; by (3.28), we have

(3.35)

By nonexpansiveness of , we have

(3.36)

This implies

(3.37)

By using the same method as (3.37), we have

(3.38)

Substituting (3.37) and (3.38) into (3.35), we have

(3.39)

By (3.39), we have

(3.40)

It follows that

(3.41)

By conditions (i)–(iii), (3.41), (3.31), (3.32), and (3.17), we have

(3.42)

By using the same method as (3.42), we have

(3.43)

By nonexpansiveness of , we have

(3.44)

Hence, we have

(3.45)

By using the same method as (3.45), we have

(3.46)

Substituting (3.45) and (3.46) into (3.35), we have

(3.47)

By (3.47), we have

(3.48)

It follows that

(3.49)

From (3.17), (3.26), (3.27), and conditions (i)–(iii), we have

(3.50)

By using the same method as (3.50), we have

(3.51)

By (3.42) and (3.50), we have

(3.52)

By (3.43) and (3.51), we have

(3.53)

Since and , we have

(3.54)

By (3.52) and (3.53), we obtain

(3.55)

Note that

(3.56)

From (3.20) and (3.55), we have

(3.57)

Step 5.

We will show that

(3.58)

where . To show this inequality, take subsequence of such that

(3.59)

Since is bounded, there exists a subsequence of which converges weakly to . Without loss of generality, we can assume that . Since is closed convex, is weakly closed. So, we have . Let us show that . We first show that . From (3.42), we have . Since , for any , we have

(3.60)

From (A2), we have

(3.61)

This implies that

(3.62)

Put for all and . Then, we have . So, from (3.62), we have

(3.63)

Since , we have . Further, from monotonicity of , we have . So, from (A4), we have

(3.64)

From (A1), (A4), and (3.64), we also have

(3.65)

Thus

(3.66)

Letting , we have, for each ,

(3.67)

This implies that

(3.68)

From (3.43), we have . Since , for any , we have

(3.69)

From (A2), we have

(3.70)

This implies that

(3.71)

Put for all and . Then, we have . So, from (3.71) we have

(3.72)

Since , we have . Further, from monotonicity of , we have . So, from (A4), we have

(3.73)

From (A1), (A4), and (3.64), we also have

(3.74)

hence

(3.75)

Letting , we have, for each ,

(3.76)

This implies that

(3.77)

Define a mapping by

(3.78)

where . From Lemma 2.3, we have that is nonexpansive with

(3.79)

Next, we show that

(3.80)

By nonexpansiveness of and , we have

(3.81)

where . From (3.17), (3.42), (3.43), (3.55), and condition (iii), we have . Since , it follows from (3.80) that, . By Lemma 2.4, we obtain that

(3.82)

Assume that . Using Opial property, (3.57) and Lemma 2.10 we have

(3.83)

This is a contradiction, so we have

(3.84)

From (3.68), (3.77) (3.82), and (3.84), we have . Since is contraction with the coefficient , has a unique fixed point. Let be a fixed point of , that is . Since and , we have

(3.85)

Step 6.

Finally, we will show that as . By nonexpansiveness of , we can show that . Then

(3.86)

we have

(3.87)

By Step 5, (3.87), and Lemma 2.2, we have , where . It easy to see that sequences , , and converge strongly to .

4. Application

Using our main theorem (Theorem 3.1), we obtain the following strong convergence theorems involving finite family of -strict pseudocontractions.

To prove strong convergence theorem in this section, we need definition and lemma as follows.

Definition 4.1.

A mapping is said to be a -strongly pseudo contraction mapping, if there exist such that

(4.1)

Lemma 4.2 (see [20]).

Let be a nonempty closed convex subset of a real Hilbert space and a -strict pseudo contraction. Define by for each . Then, as    is nonexpansive such that .

Theorem 4.3.

Let be a nonempty closed convex subset of a Hilbert space . Let and be two bifunctions from into satisfying (A1)–(A4), respectively. Let is a -inverse strongly monotone mapping and be a -inverse strongly monotone mapping. Let be a finite family of -psuedo contractions with , where are defined by , for all. Define a mapping by , for all . Let be a contraction with the coefficient . Let be the S-mappings generated by and , where , and for all for all and for allfor all. Let be sequences generated by

(4.2)

where ,, such that ,,,,. Assume that

(i),

(ii),

(iii),

(iv),,,,,,

(v)and, for all.

Then the sequence , , , converges strongly to , and is solution of

(4.3)

Proof.

For every , by Lemma 4.2, we have is nonexpansive mappings. From Theorem 3.1, we can concluded the desired conclusion.

Theorem 4.4.

Let be a nonempty closed convex subset of a Hilbert space . Let and be two bifunctions from into satisfying (A1)–(A4), respectively. Let be a -inverse strongly monotone mapping. Let be a finite family of -strict pseudo contractions with , where defined by . Define a mapping by , . Let a contraction with the coefficient . Let be the S-mappings generated by and , where , and for allfor all and for allfor all . Let be sequences generated by

(4.4)

where ,, such that ,,. Assume that

(i)and,

(ii),

(iii),

(iv)and, for all.

Then the sequence ,, converges strongly to , and is solution of

(4.5)

Proof.

For every , by Lemma 4.2, we have that is nonexpansive mappings, putting , , , , and . From Theorem 3.1, we can conclude the desired conclusion.