1. Introduction

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

(1.1)

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

(1.2)

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

(1.3)

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

(1.4)

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, 15–17] 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

(1.5)

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, 19–23] 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)–(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,

(2.1)

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)–(A4) as in Section 1. Let be two maximal monotone operators such that

(A5)′.

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

(2.2)

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

(2.3)

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

(2.4)

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

(2.5)

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

(2.6)

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

(2.7)

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

(2.8)

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

(2.9)

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)–(A4), and let and . Then, there exists such that

(2.10)

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)–(A4). For and , define a mapping as follows:

(2.11)

for all . Then, the following hold:

(i) is singlevalued;

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

(2.12)

(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)–(A4), and let . Then, for and ,

(2.13)

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)–(A4), and let . Then the following hold:

for , there exists such that

(2.14)

if is additionally uniformly smooth and is defined as

(2.15)

then the mapping has the following properties:

(i) is singlevalued,

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

(2.16)

(iii),

(iv) is a closed convex subset of ,

(v) for all  

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

(2.17)

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

(2.18)

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:

(3.1)

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

Proof.

Since for each

(3.2)

we have that

(3.3)

Thus, from the monotonicity of it follows that

(3.4)

and hence

(3.5)

Theorem 3.2.

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

(3.6)

where

(3.7)

is defined by (2.15), satisfy

(3.8)

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 ,

(3.9)

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

(3.10)

so that

(3.11)

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

(3.12)

is equivalent to

(3.13)

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

(3.14)

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

(3.15)

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

(3.16)

Then from Lemma 2.11 we obtain

(3.17)
(3.18)

Moreover, we have

(3.19)

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

(3.20)

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

(3.21)

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

(3.22)

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

(3.23)

so is bounded. From Lemma 2.3, we have

(3.24)

for each . This implies that

(3.25)

Step 4.

We claim that , , and .

Indeed, from , we have

(3.26)

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

(3.27)

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

(3.28)

for and . So, we have that

(3.29)
(3.30)

and hence

(3.31)
(3.32)

for all . Consequently we have

(3.33)

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

(3.34)

Therefore, from the properties of we get

(3.35)

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

(3.36)

Observe first that

(3.37)

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

(3.38)

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

(3.39)

and hence

(3.40)

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

(3.41)

Observe that

(3.42)

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

(3.43)

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

(3.44)
(3.45)

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

(3.46)

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)

(3.47)

from and Proposition 2.10 it follows that

(3.48)

Also, since

(3.49)

so we get

(3.50)

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

(3.51)

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

(3.52)

From , it follows that

(3.53)

By the definition of , we have

(3.54)

where

(3.55)

Replacing by , we have from (A2) that

(3.56)

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

(3.57)

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

(3.58)

Dividing by , we have

(3.59)

Letting , from (A3) it follows that

(3.60)

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

Step 6.

We claim that converges strongly to .

Indeed, from and , It follows that

(3.61)

Since the norm is weakly lower semicontinuous, we have

(3.62)

From the definition of , we have . Hence and

(3.63)

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

(3.64)

Moreover, the following hold:

(3.65)

and hence

(3.66)

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

(4.1)

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:

(4.2)

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

Proof.

We set and

(4.3)

so that

(4.4)

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

(4.5)

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

(4.6)
(4.7)

and hence by Proposition 2.10, we have

(4.8)
(4.9)

Consequently, the last two inequalities yield that

(4.10)

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

(4.11)

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

(4.12)

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

(4.13)

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 ,

(4.14)

In particular, we have

(4.15)

Consequently, from and Lemma 2.3, we have

(4.16)

and hence

(4.17)

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

(4.18)

So, we have

(4.19)

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:

(4.20)

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

(4.21)

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

(4.22)

for and . Observe that for ,

(4.23)

Hence,

(4.24)
(4.25)

Consequently, the last two inequalities yield that

(4.26)

Thus, we have

(4.27)

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

(4.28)

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

(4.29)

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

(4.30)

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

(4.31)

Now let us show that

(4.32)

Indeed, from (4.10) we get

(4.33)

which, together with , yields that

(4.34)

From (4.9) it follows that

(4.35)

Note that

(4.36)

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

(4.37)

We have from (4.8) that

(4.38)

which, together with , yields that

(4.39)

Also from (4.7) it follows that

(4.40)

which, together with , yields that

(4.41)

Similarly from (4.6) it follows that

(4.42)

which, together with , yields that

(4.43)

On the other hand, let us show that

(4.44)

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

(4.45)

Since and , we deduce from Proposition 2.10 that for ,

(4.46)

This implies that

(4.47)

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

(4.48)

Note that

(4.49)
(4.50)

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

(4.51)

and hence

(4.52)

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

(4.53)

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

(4.54)

Letting , we obtain that

(4.55)

Then the maximality of the operators implies that .

Now, by the definition of , we have

(4.56)

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

(4.57)

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

(4.58)

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

(4.59)

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

(4.60)

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

(4.61)

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

(4.62)
(4.63)

Moreover, the following hold:

(4.64)

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

(4.65)

Corollary 4.4.

Suppose that conditions (A1)–(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 .