Open Access

# A Hybrid Iterative Scheme for Equilibrium Problems, Variational Inequality Problems, and Fixed Point Problems in Banach Spaces

Fixed Point Theory and Applications20092009:719360

DOI: 10.1155/2009/719360

Accepted: 10 April 2009

Published: 5 May 2009

## Abstract

The purpose of this paper is to introduce a new hybrid projection algorithm for finding a common element of the set of solutions of the equilibrium problem and the set of the variational inequality for an inverse-strongly monotone operator and the set of fixed points of relatively quasi-nonexpansive mappings in a Banach space. Then we show a strong convergence theorem. Using this result, we obtain some applications in a Banach space.

## 1. Introduction

Let be a real Banach space and let be the dual of . Let be a closed convex subset of . Let be an operator. The classical variational inequality problem for is to find such that
(1.1)

The set of solutions of (1.1) is denoted by . Such a problem is connected with the convex minimization problem, the complementarity, the problem of finding a point satisfying , and so on. First, we recall that

(1)an operator is called monotone if
(1.2)
(2)an operator is called -inverse-strongly monotone if there exists a constant with
(1.3)

Assume that

(C1) is -inverse-strongly monotone,

(C2) ,

(C3) for all and .

Iiduka and Takahashi [1] introduced the following algorithm for finding a solution of the variational inequality for an operator that satisfies conditions (C1)–(C3) in a -uniformly convex and uniformly smooth Banach space . For an initial point , define a sequence by
(1.4)

where is the duality mapping on , and is the generalized projection from onto . Assume that for some with where is the -uniformly convexity constant of . They proved that if is weakly sequentially continuous, then the sequence converges weakly to some element in where .

The problem of finding a common element of the set of the variational inequalities for monotone mappings in the framework of Hilbert spaces and Banach spaces has been intensively studied by many authors; see, for instance, [24] and the references cited therein.

Let be a bifunction. The equilibrium problem for is to find such that
(1.5)

The set of solutions of (1.5) is denoted by .

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

(A1) for all ;

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

(A3)for all ,
(1.6)

(A4)for all is convex and lower semicontinuous.

Recently, Takahashi and Zembayashi [5], introduced the following iterative scheme which is called the shrinking projection method:

(1.7)

where is the duality mapping on and is the generalized projection from onto . They proved that the sequence converges strongly to under appropriate conditions.

Very recently, Qin et al. [6] extend the iteration process (1.7) from a single relatively nonexpansive mapping to two relatively quasi-nonexpansive mappings:
(1.8)

Under suitable conditions over , and , they obtain that the sequence generated by (1.8) converges strongly to .

The problem of finding a common element of the set of fixed points and the set of solutions of an equilibrium problem in the framework of Hilbert spaces and Banach spaces has been studied by many authors; see [5, 716].

Motivated by Iiduka and Takahashi [1], Takahashi and Zembayashi [5], and Qin et al. [6], we introduce a new general process for finding common elements of the set of the equilibrium problem and the set of the variational inequality problem for an inverse-strongly monotone operator and the set of the fixed points for relatively quasi-nonexpansive mappings.

## 2. Preliminaries

Let be a real Banach space and let be the unit sphere of . A Banach space is said to be strictly convex if for any ,
(2.1)
It is also said to be uniformly convex if for each , there exists such that for any ,
(2.2)
It is known that a uniformly convex Banach space is reflexive and strictly convex; and we define a function called the modulus of convexity of as follows:
(2.3)
Then is uniformly convex if and only if for all . Let be a fixed real number with . A Banach space is said to be -uniformly convex if there exists a constant such that for all ; see [1719] for more details. A Banach space is said to be smooth if the limit
(2.4)
exists for all . It is also said to be uniformly smooth if the limit (2.4) is attained uniformly for . One should note that no Banach space is -uniformly convex for ; see [19]. It is well known that a Hilbert space is -uniformly convex, uniformly smooth. For each , the generalized duality mapping is defined by
(2.5)

for all . In particular, is called the normalized duality mapping. If is a Hilbert space, then , where is the identity mapping. It is also known that if is uniformly smooth, then is uniformly norm-to-norm continuous on each bounded subset of . See [20, 21] for more details.

Lemma 2.1 (See [18, 22]).

Let be a given real number with and a -uniformly convex Banach space. Then, for all , and ,
(2.6)

where is the generalized duality mapping of and is the -uniformly convexity constant of .

Let be a smooth Banach space. The function is defined by
(2.7)

for all . In a Hilbert space , we have for all .

Recall that a mapping is called nonexpansive if for all and relatively nonexpansive if satisfies the following conditions:

(1) , where is the set of fixed points of ;

(2) for all and ;

(3) , where is the set of all asymptotic fixed points of ;

see [10, 23, 24] for more details.

is said to be relatively quasi-nonexpansive if satisfies the conditions and . It is easy to see that the class of relatively quasi-nonexpansive mappings is more general than the class of relatively nonexpansive mappings [9, 25, 26].

We give some examples which are closed relatively quasi-nonexpansive; see [6].

Example 2.2.

Let be a uniformly smooth and strictly convex Banach space and be a maximal monotone mapping such that its zero set . Then, is a closed relatively quasi-nonexpansive mapping from onto and .

Example 2.3.

Let be the generalized projection from a smooth, strictly convex, and reflexive Banach space onto a nonempty closed convex subset of . Then, is a closed relatively quasi-nonexpansive mapping with .

Lemma 2.4 (Kamimura and Takahashi [27]).

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

Let be a nonempty closed convex subset of . If is reflexive, strictly convex and smooth, then there exists such that for and . The generalized projection defined by . The existence and uniqueness of the operator follows from the properties of the functional and strict monotonicity of the duality mapping ; for instance, see [20, 2730]. In a Hilbert space, is coincident with the metric projection.

Lemma 2.5 (Alber [28]).

Let be a nonempty closed convex subset of a smooth Banach space and . Then if and only if for all .

Lemma 2.6 (Alber [28]).

Let be a nonempty closed convex subset of a reflexive, strictly convex and smooth Banach space and let . Then
(2.8)

Lemma 2.7 (Qin et al. [6]).

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

Lemma 2.8 (Cho et al. [31]).

Let be a uniformly convex Banach space and let be a closed ball of . Then there exists a continuous strictly increasing convex function with such that
(2.9)

for all , and with .

Lemma 2.9 (Blum and Oettli [7]).

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
(2.10)

Lemma 2.10 (Qin et al. [6]).

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 all and , define a mapping as follows:
(2.11)

Then, the following hold:

(1) is single-valued;

(2) is a firmly nonexpansive-type mapping [32], that is, for all ,
(2.12)

(3) ;

(4) is closed and convex.

Lemma 2.11 (Takahashi and Zembayashi [14]).

Let be a closed convex subset of a smooth, strictly, and reflexive Banach space , let be a bifucntion from to satisfying (A1)–(A4), let . Then, for all and ,
(2.13)
We make use of the following mapping studied in Alber [28]:
(2.14)

for all and , that is, .

Lemma 2.12 (Alber [28]).

Let be a reflexive, strictly convex, smooth Banach space and let be as in (2.14). Then
(2.15)

for all and .

An operator of into is said to be hemicontinuous if for all , the mapping of into defined by is continuous with respect to the topology of . We define by the normal cone for at a point , that is,
(2.16)

Theorem 2.13 (Rockafellar [33]).

Let be a nonempty, closed convex subset of a Banach space and a monotone, hemicontinuous operator of into . Let be an operator defined as follows:
(2.17)

Then is maximal monotone and .

## 3. Strong Convergence Theorems

Theorem 3.1.

Let be a -uniformly convex, uniformly smooth Banach space, let be a nonempty closed convex subset of . Let be a bifunction from to satisfying (A1)–(A4), let be an operator of into satisfying (C1)–(C3), and let be two closed relatively quasi-nonexpansive mappings from into itself such that . For an initial point with and , define a sequence as follows:
(3.1)

where is the duality mapping on . Assume that , and are sequences in satisfying the restrictions:

(B1) ;

(B2) , ;

(B3) for some ;

(B4) for some with , where is the -uniformly convexity constant of .

Then, and converge strongly to .

Proof.

We divide the proof into eight steps.

Step 1.

Show that and are well defined.

It is obvious that is a closed convex subset of . By Lemma 2.7, we know that is closed and convex. From Lemma 2.10 , we also have is closed and convex. Hence is a nonempty, closed, and convex subset of ; consequently, is well defined.

Clearly, is closed and convex. Suppose that is closed and convex for . For all , we know is equivalent to
(3.2)

So, is closed and convex. By induction, is closed and convex for all . This shows that is well-defined.

Step 2.

Show that for all .

Put . First, we observe that for all and . Suppose for . Then, for all , we know from Lemma 2.6 and Lemma 2.12 that
(3.3)
Since and from (C1), we have
(3.4)
From Lemma 2.1 and (C3), we obtain
(3.5)
Replacing (3.4) and (3.5) into (3.3), we get
(3.6)
By the convexity of , for each , we obtain
(3.7)

This shows that ; consequently, . Hence for all .

Step 3.

Show that exists.

From and , we have
(3.8)
From Lemma 2.6, we have
(3.9)

Combining (3.8) and (3.9), we obtain that exists.

Step 4.

Show that is a Cauchy sequence in .

Since for , by Lemma 2.6, we also have
(3.10)
Taking , we obtain that . From Lemma 2.4, we have . Hence is a Cauchy sequence. By the completeness of and the closedness of , one can assume that as . Further, we obtain
(3.11)
Since , we have
(3.12)
as . Applying Lemma 2.4 to (3.11) and (3.12), we get
(3.13)
This implies that as . Since is uniformly norm-to-norm continuous on bounded subsets of , we also obtain
(3.14)

Step 5.

Show that .

Let . From (3.6) and Lemma 2.8, we know that there exists a continuous strictly increasing convex function with such that
(3.15)
This implies that
(3.16)
It follows from (3.13), (3.14), and (B ) that
(3.17)
By the property of , we also obtain that
(3.18)
Since is uniformly norm-to-norm continuous on bounded sets, so is . Then
(3.19)
In the same manner, we can show that
(3.20)
Again, by (3.15), we have
(3.21)
which yields that
(3.22)
From Lemma 2.6, Lemma 2.12, and (3.5), we have
(3.23)
It follows from Lemma 2.4 and (3.22) that
(3.24)
Hence as and
(3.25)
Combining (3.20) and (3.24), we also obtain
(3.26)

From (3.19), (3.26) and by the closedness of and , we get .

Step 6.

Show that .

From (3.15), we see
(3.27)
From (3.16), we observe
(3.28)
Note that . From (3.27) and Lemma 2.11, we have
(3.29)
From (3.28), we get . By Lemma 2.4, we obtain
(3.30)
as . Since , we have
(3.31)
as . From we have
(3.32)
By (A2), we have
(3.33)

From (A4) and , we get for all . For and . Define , then , which implies that . From (A1), we obtain that . Thus, . From (A3), we have for all . Hence .

Step 7.

Show that .

Define be as in (2.17). By Theorem 2.13, is maximal monotone and . Let . Since , we get . From , we have
(3.34)
On the other hand, since . Then, by Lemma 2.5, we have and thus
(3.35)
It follows from (3.34) and (3.35) that
(3.36)

where . By taking the limit as and from (3.24) and (3.25), we obtain . By the maximality of , we have and hence .

Step 8.

Show that .

From , we have
(3.37)
Since , we also have
(3.38)
By taking limit in (3.38), we obtain that
(3.39)

By Lemma 2.5, we can conclude that . Furthermore, it is easy to see that as . This completes the proof.

As a direct consequence of Theorem 3.1, we obtain the following results.

Corollary 3.2.

Let be a -uniformly convex and uniformly smooth Banach space, and let be a nonempty closed convex subset of . Let be a bifunction from to satisfying (A1)–(A4) and let be a closed relatively quasi-nonexpansive mapping from into itself such that . Assume that satisfies and for some . Then the sequence generated by (1.7) converges strongly to .

Proof.

Putting and in Theorem 3.1, we obtain the result.

Remark 3.3.

If in Theorem 3.1, then Theorem 3.1 reduces to Theorem of Qin et al. [6].

Remark 3.4.

Corollary 3.2 improves Theorem of Takahashi and Zembayashi [5] from the class of relatively nonexpansive mappings to the class of relatively quasi-nonexpansive mappings, that is, we relax the strong restriction: . Further, the algorithm in Corollary 3.2 is also simpler to compute than the one given in [14].

## 4. Applications

Next, we consider the problem of finding a zero point of an inverse-strongly monotone operator of into . Assume that satisfies the conditions:

(D1) is -inverse-strongly monotone,

(D2) .

Theorem 4.1.

Let be a -uniformly convex, uniformly smooth Banach space. Let be a bifunction from to satisfying (A1)–(A4), let be an operator of into satisfying (D1) and (D2), and let be two closed relatively quasi-nonexpansive mappings from into itself such that . For an initial point with and , define a sequence as follows:
(4.1)

where is the duality mapping on . Assume that , and are sequences in satisfying the conditions (B1)–(B4) of Theorem 3.1.

Then, and converge strongly to .

Proof.

Putting in Theorem 3.1, we have . We also have and then the condition (C3) of Theorem 3.1 holds for all and . So, we obtain the result.

Let be a nonempty, closed convex cone in , an operator of into . We define its polar in to be the set
(4.2)
Then the element is called a solution of the complementarity problem if
(4.3)

The set of solutions of the complementarity problem is denoted by .

Assume that is an operator satisfying the conditions:

(E1) is -inverse-strongly monotone,

(E2) ,

(E3) for all and .

Theorem 4.2.

Let be a -uniformly convex, uniformly smooth Banach space, and a nonempty, closed convex cone in . Let be a bifunction from to satisfying (A1)–(A4), let be an operator of into satisfying (E1)–(E3), and let be two closed relatively quasi-nonexpansive mappings from into itself such that . For an initial point with and , define a sequence as follows:
(4.4)

where is the duality mapping on . Assume that and are sequences in satisfying the conditions (B1)–(B4) of Theorem 3.1.

Then, and converge strongly to .

Proof.

From [20, Lemma ], we have . Hence, we obtain the result.

## Declarations

### Acknowledgments

The author would like to thank Professor Suthep Suantai and the referee for the valuable suggestions on the manuscript. The author was supported by the Commission on Higher Education and the Thailand Research Fund.

## Authors’ Affiliations

(1)
School of Science and Technology, Naresuan University at Phayao

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