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# Iterative Schemes for Generalized Equilibrium Problem and Two Maximal Monotone Operators

*Journal of Inequalities and Applications*
**volumeÂ 2009**, ArticleÂ number:Â 896252 (2009)

## Abstract

The purpose of this paper is to introduce and study two new hybrid proximal-point algorithms for finding a common element of the set of solutions to a generalized equilibrium problem and the sets of zeros of two maximal monotone operators in a uniformly smooth and uniformly convex Banach space. We established strong and weak convergence theorems for these two modified hybrid proximal-point algorithms, respectively.

## 1. Introduction

Let be a real Banach space and its dual space. The normalized duality mapping is defined as

where denotes the generalized duality pairing. Recall that if is a smooth Banach space then is singlevalued. Throughout this paper, we will still denote by the single-valued normalized duality mapping. Let be a nonempty closed convex subset of , a bifunction from to , and a nonlinear mapping. The generalized equilibrium problem is to find such that

The set of solutions of (1.2) is denoted by . Problem (1.2) and similar problems have been extensively studied; see, for example, [1â€“11]. Whenever , problem (1.2) reduces to the equilibrium problem of finding such that

The set of solutions of (1.3) is denoted by . Whenever , problem (1.2) reduces to the variational inequality problem of finding such that

Whenever a Hilbert space, problem (1.2) was very recently introduced and considered by Kamimura and Takahashi [12]. Problem (1.2) is very general in the sense that it includes, as spacial cases, optimization problems, variational inequalities, minimax problems, the Nash equilibrium problem in noncooperative games, and others; see, for example, [13, 14]. A mapping is called nonexpansive if for all . Denote by the set of fixed points of , that is, . Iterative schemes for finding common elements of EP and fixed points set of nonexpansive mappings have been studied recently; see, for example, [12, 15â€“17] and the references therein.

On the other hand, a classical method of solving in a Hilbert space is the proximal point algorithm which generates, for any starting point , a sequence in by the iterative scheme

where is a sequence in , for each is the resolvent operator for , and is the identity operator on . This algorithm was first introduced by Martinet [14] and generally studied by Rockafellar [18] in the framework of a Hilbert space . Later many authors studied (1.5) and its variants in a Hilbert space or in a Banach space ; see, for example, [13, 19â€“23] and the references therein.

Let be a uniformly smooth and uniformly convex Banach space and let be a nonempty closed convex subset of . Let be a bifunction from to satisfying the following conditions (A1)â€“(A4) which were imposed in [24]:

(A1) for all ;

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

(A3)for all ;

(A4)for all is convex and lower semicontinuous.

Let be a maximal monotone operator such that

(A5).

The purpose of this paper is to introduce and study two new iterative algorithms for finding a common element of the set of solutions for the generalized equilibrium problem (1.2) and the set for maximal monotone operators in a uniformly smooth and uniformly convex Banach space . First, motivated by Kamimura and Takahashi [12, Theorem ], Ceng et al. [16, Theorem ], and Zhang [17, Theorem ], we introduce a sequence that, under some appropriate conditions, is strongly convergent to in Section 3. Second, inspired by Kamimura and Takahashi [12, Theorem ], Ceng et al. [16, Theorem ], and Zhang [17, Theorem ], we define a sequence weakly convergent to an element , where in Section 4. Our results represent a generalization of known results in the literature, including Takahashi and Zembayashi [15], Kamimura and Takahashi [12], Li and Song [22], Ceng and Yao [25], and Ceng et al. [16]. In particular, compared with Theorems and in [16], our results (i.e., Theorems 3.2 and 4.2 in this paper) extend the problem of finding an element of to the one of finding an element of . Meantime, the algorithms in this paper are very different from those in [16] (because of considering the complexity involving the problem of finding an element of ).

## 2. Preliminaries

In the sequel, we denote the strong convergence, weak convergence and weak* convergence of a sequence to a point by , and , respectively.

A Banach space is said to be strictly convex, if for all with . is said to be uniformly convex if for each there exists such that for all with . Recall that each uniformly convex Banach space has the Kadec-Klee property, that is,

The proof of the main results of Sections 3 and 4 will be based on the following assumption.

Assumption A.

Let be a uniformly smooth and uniformly convex Banach space and let be a nonempty closed convex subset of . Let be a bifunction from to satisfying the same conditions (A1)â€“(A4) as in Section 1. Let be two maximal monotone operators such that

(A5)â€².

Recall that if is a nonempty closed convex subset of a Hilbert space , then the metric projection of onto is nonexpansive. This fact actually characterizes Hilbert spaces and hence, it is not available in more general Banach spaces. In this connection, Alber [26] recently introduced a generalized projection operator in a Banach space which is an analogue of the metric projection in Hilbert spaces. Consider the functional defined as in [26] by

It is clear that in a Hilbert space , (2.2) reduces to .

The generalized projection is a mapping that assigns to an arbitrary point the minimum point of the functional ; that is, , where is the solution to the minimization problem

The existence and uniqueness of the operator follows from the properties of the functional and strict monotonicity of the mapping (see, e.g., [27]). In a Hilbert space, . From [26], in a smooth strictly convex and reflexive Banach space , we have

Moreover, by the property of subdifferential of convex functions, we easily get the following inequality:

Let be a mapping from into itself. A point in is called an asymptotically fixed point of if contains a sequence which converges weakly to such that [28]. The set of asymptotically fixed points of will be denoted by . A mapping from into itself is called relatively nonexpansive if and , for all and [15].

Observe that, if is a reflexive strictly convex and smooth Banach space, then for any if and only if . To this end, it is sufficient to show that if then . Actually, from (2.4), we have which implies that . From the definition of , we have and therefore, ; see [29] for more details.

We need the following lemmas for the proof of our main results.

Lemma 2.1 (Kamimura and Takahashi [12]).

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

Lemma 2.2 (Alber [26], Kamimura and Takahashi [12]).

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

Lemma 2.3 (Alber [26], Kamimura and Takahashi [12]).

Let be a nonempty closed convex subset of a smooth strictly convex and reflexive Banach space . Then

Lemma 2.4 (Rockafellar [18]).

Let be a reflexive strictly convex and smooth Banach space and let be a multivalued operator. Then there hold the following hold:

(i) is closed and convex if is maximal monotone such that ;

(ii) is maximal monotone if and only if is monotone with for all .

Lemma 2.5 (Xu [30]).

Let be a uniformly convex Banach space and let . Then there exists a strictly increasing, continuous, and convex function such that and

for all and , where .

Lemma 2.6 (Kamimura and Takahashi [12]).

Let be a smooth and uniformly convex Banach space and let . Then there exists a strictly increasing, continuous, and convex function such that and

The following result is due to Blum and Oettli [24].

Lemma 2.7 (Blum and Oettli [24]).

Let be a nonempty closed convex subset of a smooth strictly convex and reflexive Banach space , let be a bifunction from to satisfying (A1)â€“(A4), and let and . Then, there exists such that

Motivated by Combettes and Hirstoaga [31] in a Hilbert space, Takahashi and Zembayashi [15] established the following lemma.

Lemma 2.8 (Takahashi and Zembayashi [15]).

Let be a nonempty closed convex subset of a smooth strictly convex and reflexive Banach space , and let be a bifunction from to satisfying (A1)â€“(A4). For and , define a mapping as follows:

for all . Then, the following hold:

(i) is singlevalued;

(ii) is a firmly nonexpansive-type mapping, that is, for all ,

(iii);

(iv) is closed and convex.

Using Lemma 2.8, one has the following result.

Lemma 2.9 (Takahashi and Zembayashi [15]).

Let be a nonempty closed convex subset of a smooth strictly convex and reflexive Banach space , let be a bifunction from to satisfying (A1)â€“(A4), and let . Then, for and ,

Utilizing Lemmas 2.7, 2.8 and 2.9 as previously mentioned, Zhang [17] derived the following result.

Proposition 2.10 (Zhang [21, Lemma ]).

Let be a smooth strictly convex and reflexive Banach space and let be a nonempty closed convex subset of . Let be an -inverse-strongly monotone mapping, let be a bifunction from to satisfying (A1)â€“(A4), and let . Then the following hold:

for , there exists such that

if is additionally uniformly smooth and is defined as

then the mapping has the following properties:

(i) is singlevalued,

(ii) is a firmly nonexpansive-type mapping, that is,

(iii),

(iv) is a closed convex subset of ,

(v)â€‰for all â€‰

Let be two maximal monotone operators in a smooth Banach space . We denote the resolvent operators of and by and for each , respectively. Then and are two single-valued mappings. Also, and for each , where and are the sets of fixed points of and , respectively. For each , the Yosida approximations of and are defined by and , respectively.It is known that

Lemma 2.11 (Kohsaka and Takahashi [13]).

Let be a reflexive strictly convex and smooth Banach space and let be a maximal monotone operator with . Then

Lemma 2.12 (Tan and Xu [32]).

Let and be two sequences of nonnegative real numbers satisfying the inequality: for all . If , then exists.

## 3. Strong Convergence Theorem

In this section, we prove a strong convergence theorem for finding a common element of the set of solutions for a generalized equilibrium problem and the set for two maximal monotone operators and .

Lemma 3.1.

Let be a reflexive strictly convex and smooth Banach space and let be a maximal monotone operator. Then for each , the following holds:

where and is the duality mapping on . In particular, whenever a real Hilbert space, is a nonexpansive mapping on .

Proof.

Since for each

we have that

Thus, from the monotonicity of it follows that

and hence

Theorem 3.2.

Suppose that Assumption A is fulfilled and let be chosen arbitrarily. Consider the sequence

where

is defined by (2.15), satisfy

and satisfies . Then, the sequence converges strongly to provided for any sequence with , where is the generalized projection of onto .

Remark 3.3.

In Theorem 3.2, if a real Hilbert space, then is a sequence of nonexpansive mappings on . This implies that as ,

In this case, we can remove the requirement that for any sequence with .

Proof of Theorem 3.2.

For the sake of simplicity, we define

so that

We divide the proof into several steps.

Step 1.

We claim that is closed and convex for each .

Indeed, it is obvious that is closed and is closed and convex for each . Let us show that is convex. For and , put . It is sufficient to show that . We first write for each . Next, we prove that

is equivalent to

Indeed, from (2.4) we deduce that the following equations hold:

which combined with (3.12) yield that (3.12) is equivalent to (3.13). Thus we have

This implies that . Therefore, is closed and convex.

Step 2.

We claim that for each and that is well defined.

Indeed, take arbitrarily. Note that is equivalent to

Then from Lemma 2.11 we obtain

Moreover, we have

where . So for all . Now, let us show that

We prove this by induction. For , we have . Assume that . Since is the projection of onto , by Lemma 2.2 we have

As by the induction assumption, the last inequality holds, in particular, for all . This, together with the definition of implies that . Hence (3.20) holds for all . So, for all . This implies that the sequence is well defined.

Step 3.

We claim that is bounded and that as .

Indeed,it follows from the definition of that . Since and , so for all ; that is, is nondecreasing. It follows from and Lemma 2.3 that

for each for each . Therefore, is bounded which implies that the limit of exists. Since

so is bounded. From Lemma 2.3, we have

for each . This implies that

Step 4.

We claim that , , and .

Indeed, from , we have

for all . Therefore, from and , it follows that . Since and is uniformly convex and smooth, we have from Lemma 2.1 that

and therefore, . Since is uniformly norm-to-norm continuous on bounded subsets of and , then .

Let us set . Then, according to Lemma 2.4 and Proposition 2.10, we know that is a nonempty closed convex subset of such that . Fix arbitrarily. As in the proof of Step 2 we can show that , and . Hence it follows from the boundedness of that , and are also bounded. Let . Since is a uniformly smooth Banach space, we know that is a uniformly convex Banach space. Therefore, by Lemma 2.5 there exists a continuous, strictly increasing, and convex function with such that

for and . So, we have that

and hence

for all . Consequently we have

Since and is uniformly norm-to-norm continuous on bounded subsets of , we obtain . From , and we have

Therefore, from the properties of we get

recalling that is uniformly norm-to-norm continuous on bounded sunsets of . Next let us show that

Observe first that

Since , and is bounded, so it follows that . Also, observe that

Since and the sequences are bounded, so it follows that . Meantime, observe that

and hence

Since and , it follows from the boundedness of that . Thus, in terms of Lemma 2.1, we have that and so . Furthermore, since , from the uniform norm-to-norm continuity of on bounded subsets of , we obtain

Observe that

Thus, from (3.35) it follows that . Since is uniformly norm-to-norm continuous on bounded subsets of , it follows that .

Step 5.

We claim that , where

Indeed, since is bounded and is reflexive, we know that . Take arbitrarily. Then there exists a subsequence of such that . Hence it follows from , and that , and converge weakly to the same point . On the other hand, from (3.35), (3.36), and we obtain that

If and , then it follows from (2.17) and the monotonicity of the operators that for all

Letting , we have that and . Then the maximality of the operators implies that and . Next, let us show that . Since we have by (3.32)

from and Proposition 2.10 it follows that

Also, since

so we get

Thus from (3.47), , and , we have . Since is uniformly convex and smooth, we conclude from Lemma 2.1 that

From , and (3.51), we have and . Since is uniformly norm-to-norm continuous on bounded subsets of , from (3.51) we derive

From , it follows that

By the definition of , we have

where

Replacing by , we have from (A2) that

Since is convex and lower semicontinuous, it is also weakly lower semicontinuous. Letting in the last inequality, from (3.53) and (A4) we have

For with and , let . Since and , and hence . So, from (A1) we have

Dividing by , we have

Letting , from (A3) it follows that

Thus . Therefore, we obtain that by the arbitrariness of .

Step 6.

We claim that converges strongly to .

Indeed, from and , It follows that

Since the norm is weakly lower semicontinuous, we have

From the definition of , we have . Hence and

which implies that . Since has the Kadec-Klee property, then . Therefore, converges strongly to .

Remark 3.4.

In Theorem 3.2, put , and . Then, for all and , we have that

Moreover, the following hold:

and hence

In this case, the previous Theorem 3.2 reduces to [20, Theorem ].

## 4. Weak Convergence Theorem

In this section, we present the following algorithm for finding a common element of the set of solutions for a generalized equilibrium problem and the set for two maximal monotone operators and .

Let be chosen arbitrarily and consider the sequence generated by

where , and , is defined by (2.15).

Before proving a weak convergence theorem, we need the following proposition.

Proposition 4.1.

Suppose that Assumption A is fulfilled and let be a sequence defined by (4.1), where satisfy the following conditions:

Then, converges strongly to , where is the generalized projection of onto .

Proof.

We set and

so that

Then, in terms of Lemma 2.4 and Proposition 2.10, is a nonempty closed convex subset of such that . We first prove that is bounded. Fix . Note that by the first and third of (4.3), , and

Here, each is relatively nonexpansive. Then from Proposition 2.10 we obtain

and hence by Proposition 2.10, we have

Consequently, the last two inequalities yield that

for all . So, from , and Lemma 2.12, we deduce that exists. This implies that is bounded. Thus, is bounded and so are , and .

Define for all . Let us show that is bounded. Indeed, observe that

for each . This, together with the boundedness of , implies that is bounded and so is . Furthermore, from and (4.10) we have

Since is the generalized projection, then, from Lemma 2.3 we obtain

Hence, from (4.12), it follows that .

Note that , and is bounded, so that . Therefore, is a convergent sequence. On the other hand, from (4.10) we derive, for all ,

In particular, we have

Consequently, from and Lemma 2.3, we have

and hence

Let . From Lemma 2.6, there exists a continuous, strictly increasing, and convex function with such that

So, we have

Since is a convergent sequence, is bounded and is convergent, from the property of we have that is a Cauchy sequence. Since is closed, converges strongly to . This completes the proof.

Now, we are in a position to prove the following theorem.

Theorem 4.2.

Suppose that Assumption A is fulfilled and let be a sequence defined by (4.1), where satisfy the following conditions:

and satisfies . If is weakly sequentially continuous, then converges weakly to , where .

Proof.

We consider the notations (4.3). As in the proof of Proposition 4.1, we have that , and are bounded sequences. Let

From Lemma 2.5 and as in the proof of Theorem 3.2, there exists a continuous, strictly increasing, and convex function with such that

for and . Observe that for ,

Hence,

Consequently, the last two inequalities yield that

Thus, we have

By the proof of Proposition 4.1, it is known that is convergent; since , and , then we have

Taking into account the properties of , as in the proof of Theorem 3.2, we have

since is uniformly norm-to-norm continuous on bounded subsets of . Note that . Hence, from the uniform norm-to-norm continuity of on bounded subsets of we obtain . Also, observe that

From it follows that . Since is uniformly norm-to-norm continuous on bounded subsets of , we have

Now let us show that

Indeed, from (4.10) we get

which, together with , yields that

From (4.9) it follows that

Note that

Since and , we obtain , which, together with , yields that

We have from (4.8) that

which, together with , yields that

Also from (4.7) it follows that

which, together with , yields that

Similarly from (4.6) it follows that

which, together with , yields that

On the other hand, let us show that

Indeed, let . From Lemma 2.6, there exists a continuous, strictly increasing, and convex function with such that

Since and , we deduce from Proposition 2.10 that for ,

This implies that

Since is uniformly norm-to-norm continuous on bounded subsets of , from the properties of we obtain

Note that

Since , it follows from (4.31) and (4.35) that and . Also, observe that

and hence

Thus, from , and , it follows that . In terms of Lemma 2.1, we derive .

Next, let us show that , where .

Indeed, since is bounded, there exists a subsequence of such that . Hence it follows from (4.31), (4.35), and that both and converge weakly to the same point . Furthermore, from and (4.31) we have that

If and , then it follows from (2.17) and the monotonicity of the operators that for all

Letting , we obtain that

Then the maximality of the operators implies that .

Now, by the definition of , we have

where . Replacing by , we have from (A2) that

Since is convex and lower semicontinuous, it is also weakly lower semicontinuous. Letting in the last inequality, from (4.35) and (A4) we have

For , with , and , let . Since and , then and hence . So, from (A1) we have

Dividing by , we get . Letting , from (A3) it follows that . So, . Therefore, . Let . From Lemma 2.2 and , we get

From Proposition 4.1, we also know that . Note that . Since is weakly sequentially continuous, then as . In addition, taking into account the monotonicity of , we conclude that . Hence

From the strict convexity of , it follows that . Therefore, , where . This completes the proof.

Remark 4.3.

In Theorem 4.2, put , , and . Then, for all and , we have that

Moreover, the following hold:

In this case, Algorithm (4.1) reduces to the following one:

Corollary 4.4.

Suppose that conditions (A1)â€“(A5) are fulfilled and let be a sequence defined by (4.65), where is defined in Lemma 2.8, satisfy the conditions and , and satisfies . If is weakly sequentially continuous, then converges weakly to , where .

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## Acknowledgments

The first author (http://zenglc@hotmail.com) was partially supported by the National Science Foundation of China (10771141), Ph. D. Program Foundation of Ministry of Education of China (20070270004), and Science and Technology Commission of Shanghai Municipality grant (075105118), Leading Academic Discipline Project of Shanghai Normal University (DZL707), Shanghai Leading Academic Discipline Project (S30405) and Innovation Program of Shanghai Municipal Education Commission (09ZZ133). The third author (http://yaojc@math.nsysu.edu.tw) was partially supported by the Grant NSF 97-2115-M-110-001.

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Zeng, L.C., Lin, Y.C. & Yao, J.C. Iterative Schemes for Generalized Equilibrium Problem and Two Maximal Monotone Operators.
*J Inequal Appl* **2009**, 896252 (2009). https://doi.org/10.1155/2009/896252

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DOI: https://doi.org/10.1155/2009/896252