Strong and Weak Convergence Theorems for Common Solutions of Generalized Equilibrium Problems and Zeros of Maximal Monotone Operators

Abstract

The purpose of this paper is to introduce and study two modified hybrid proximal-point algorithms for finding a common element of the solution set EP of a generalized equilibrium problem and the set for two maximal monotone operators and defined on a Banach space . Strong and weak convergence theorems for these two modified hybrid proximal-point algorithms are established.

1. Introduction

Let be a real Banach space with its dual . The mapping defined by

(1.1)

is called the normalized duality mapping. From the Hahn-Banach theorem, it follows that for each .

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,

(1.2)

It is well known that if is strictly convex, then is single-valued. In the sequel, we shall still denote the single-valued normalized duality mapping by . Let be a nonempty closed convex subset of , a bifunction, and a nonlinear mapping. Very recently, Zhang [1] considered and studied the generalized equilibrium problem of finding such that

(1.3)

The set of solutions of (1.3) is denoted by . Problem (1.3) and related problems have been studied and investigated extensively in the literature; See, for example, [2–12] and references therein. Whenever , problem (1.3) reduces to the equilibrium problem of finding such that

(1.4)

The set of solutions of (1.4) is denoted by . Whenever , problem (1.3) reduces to the variational inequality problem of finding such that

(1.5)

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

Whenever a Hilbert space, problem (1.3) was very recently introduced and considered by S. Takahashi and W. Takahashi [13]. Problem (1.3) is very general in the sense that it includes, as spacial cases, optimization problems, variational inequalities, minimax problems, Nash equilibrium problem in noncooperative games, and others; See, for example, [1, 2, 4, 6–9, 14–17] which are references therein.

A mapping is called nonexpansive if for all . Denote by the set of fixed points of , that is, . Very recently, W. Takahashi and K. Zembayashi [18] proposed an iterative algorithm for finding a common element of the solution set of the equilibrium problem (1.4) and the set of fixed points of a relatively nonexpansive mapping in a Banach space . They also studied the strong and weak convergence of the sequences generated by their algorithm. In particular, they proposed the following iterative algorithm:

(1.6)

where for all , and for some . They proved that the sequence generated by the above algorithm converges strongly to , where is the generalized projection of onto . They have also studied the weak convergence of the sequence generated by the following algorithm:

(1.7)

to , where .

Let be a nonempty closed convex subset of a uniformly smooth and uniformly convex Banach space . Let be an -inverse-strongly monotone mapping and a bifunction satisfying the following conditions:

(A1) for all ;

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

(A3) for all , ;

(A4) for all , is convex and lower semicontinuous.

Let be two relatively nonexpansive mappings such that . Let be the sequence generated by

(1.8)

Zhang [1] proved the strong convergence of the sequence to under appropriate conditions.

On the other hand, a classic 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

(1.9)

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 [19] and further studied by Rockafellar [20] in the framework of a Hilbert space . Later several authors studied (1.9) and its variants in the setting of a Hilbert space or in a Banach space ; See, for example, [15, 21–25] and references therein. Very recently, Li and Song [24] introduced and studied the following iterative scheme:

(1.10)

where and is the duality mapping on .

Algorithm (1.10) covers, as special cases, the algorithms introduced by Kohsaka and Takahashi [23] and Kamimura et al. [22] in a smooth and uniformly convex Banach space .

Let be a uniformly smooth and uniformly convex Banach space, and let be a nonempty closed convex subset of . Let be a maximal monotone operator such that:

(A5).

In addition, for each , define a mapping as follows:

(1.11)

for all .

Very recently, utilizing the ideas of the above algorithms in [15, 16, 18, 21, 22, 24], we [17] introduced two iterative methods for finding an element of and established the following strong and weak convergence theorems.

Theorem 1.1 (see [17]).

Suppose that conditions (A1)–(A5) are satisfied and let be chosen arbitrarily. Consider the sequence

(1.12)

where

(1.13)

is defined by (1.11), satisfy , , and satisfies . Then, the sequence converges strongly to , where is the generalized projection of onto .

Theorem 1.2 (see [17]).

Suppose that conditions (A1)–(A5) are satisfied and let be chosen arbitrarily. Consider the sequence

(1.14)

where is defined by (1.11), satisfy the conditions and , and satisfies . If is weakly sequentially continuous, then converges weakly to an element , where .

The purpose of this paper is to introduce and study two new iterative methods for finding a common element of the solution set of generalized equilibrium problem (1.3) and the set for maximal monotone operators and in a uniformly smooth and uniformly convex Banach space . Firstly, motivated by Theorem 1.1 and a result of Zhang [1], we introduce a sequence that converges strongly to under some appropriate conditions.

Secondly, inspired by Theorem 1.2 and a result of Zhang [1], we define a sequence that converges weakly to an element , where (Section 4).

Our results represent a generalization of known results in the literature, including those in [16–18, 24]. Our Theorems 3.1 and 4.2 are the extension and improvements of Theorems 1.1 and 1.2 in the following way:

(i)the problem of finding an element of includes the one of finding an element of as a special case;

(ii)the algorithms in this paper are very different from those in [17] because of considering the complexity involving the problem of finding an element of .

2. Preliminaries

Throughout the paper, we denote the strong convergence, weak convergence, and weak convergence of a sequence to a point by , and , respectively.

Assumption 2.1.

Let be a uniformly smooth and uniformly convex Banach space and let be a nonempty closed convex subset of . Let be an -inverse-strongly monotone mapping and let be a bifunction satisfying the conditions (A1)–(A4). 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

(2.1)

It is clear that in a Hilbert space , (2.1) 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

(2.2)

The existence and uniqueness of the operator follow from the properties of the functional and strict monotonicity of the mapping ; See, for example, [27]. In a Hilbert space, . From [26], in a smooth, strictly convex and reflexive Banach space , we have

(2.3)

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

(2.4)

Let be a mapping from into itself. A point in is called an asymptotic fixed point of [28] if contains a sequence which converges weakly to such that . The set of asymptotic fixed points of is denoted by . A mapping from into itself is called relatively nonexpansive [18, 29, 30] if and , for all and .

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 (2.3), we have , which implies that . From the definition of , we have and therefore, . For further details, we refer to [31].

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

Lemma 2.2 (see [32]).

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

Lemma 2.3 (see [26, 32]).

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

(2.5)

Lemma 2.4 (see [26, 32]).

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

(2.6)

Lemma 2.5 (see [33]).

Let be a reflexive, strictly convex and smooth Banach space and let be a multivalued operator. Then

(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 2.6 (see [34]).

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

(2.7)

for all and , where .

Lemma 2.7 (see [32]).

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

(2.8)

The following result is due to Blum and Oettli [14].

Lemma 2.8 (see [14]).

Let be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space , a bifunction satisfying conditions (A1)–(A4), and and . Then, there exists such that

(2.9)

Motivated by a result in [35] in a Hilbert space setting, Takahashi and Zembayashi [18] established the following lemma.

Lemma 2.9 (see [18]).

Let be a nonempty closed convex subset of a uniformly smooth, strictly convex and reflexive Banach space , and a bifunction satisfying conditions (A1)–(A4). For and , define a mapping as follows:

(2.10)

for all . Then

(i) is single-valued;

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

(2.11)

(iii);

(iv) is closed and convex.

Using Lemma 2.9, we have the following result.

Lemma 2.10 (see [18]).

Let be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space , a bifunction satisfying conditions (A1)–(A4), and . Then, for and ,

(2.12)

Utilizing Lemmas 2.8, 2.9, and 2.10, Zhang [1] derived the following result.

Proposition 2.11 (see [1]).

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, a bifunction satisfying conditions (A1)–(A4), and . Then

• (I) for , there exists such that

  

  (2.13)

• (II) if is additionally uniformly smooth and is defined as

  

  (2.14)

then the mapping has the following properties:

(i) is single-valued,

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

(2.15)

(iii),

(iv) is a closed convex subset of ,

(v) for all .

Proof.

Define a bifunction by

(2.16)

It is easy to verify that satisfies the conditions (A1)–(A4). Therefore, the conclusions (I) and (II) follow immediately from Lemmas 2.8, 2.9, and 2.10.

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 2.12 (see [23]).

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

(2.18)

Lemma 2.13 (see [36]).

Let and be two sequences of nonnegative real numbers such that 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 .

Theorem 3.1.

Suppose that Assumption 2.1 is satisfied. Let be chosen arbitrarily. Consider the sequence

(3.1)

where

(3.2)

is defined by (2.14), satisfy

(3.3)

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

Proof.

For the sake of simplicity, we define

(3.4)

so that

(3.5)

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

(3.6)

is equivalent to

(3.7)

Indeed, from (2.1) we deduce that there hold the following:

(3.8)

which combined with (3.6) yield that (3.6) is equivalent to (3.7). Thus we have

(3.9)

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

(3.10)

Then from Lemma 2.12, we obtain

(3.11)

Moreover, we have

(3.12)

and hence by Proposition 2.11,

(3.13)

So for all . Now, let us show that

(3.14)

We prove this by induction. For , we have . Assume that . Since is the projection of onto , by Lemma 2.3 we have

(3.15)

As by the induction assumption, the last inequality holds, in particular, for all . This, together with the definition of implies that . Hence (3.14) 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 2.4 that

(3.16)

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

(3.17)

so is bounded. From Lemma 2.4, we have

(3.18)

for each . This implies that

(3.19)

Step 4.

We claim that , , and .

Indeed, from , we have

(3.20)

Therefore, from and , it follows that .

Since and is uniformly convex and smooth, we have from Lemma 2.2 that

(3.21)

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

Let us set . Then, according to Lemma 2.5 and Proposition 2.11, 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 ,

(3.22)

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 2.6 there exists a continuous, strictly increasing, and convex function , with , such that

(3.23)

for and . So, we have that

(3.24)

and hence

(3.25)

for all . Consequently, we have

(3.26)

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

(3.27)

Therefore, from the properties of , we get

(3.28a)

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

(3.28b)

Observe first that

(3.29)

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

(3.30)

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

(3.31)

and hence

(3.32)

Since and , it follows from the boundedness of that . Thus, in terms of Lemma 2.2, we have that and so . Furthermore, it follows from (3.25) that

(3.33)

and hence

(3.34)

Since is uniformly norm-to-norm continuous on bounded subsets of , it follows from that . Thus from , , and the boundedness of both and , we deduce that . Utilizing the properties of , we have that . Since is uniformly norm-to-norm continuous on bounded subsets of , it follows that .

Step 5.

We claim that , where

(3.35)

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 (3.28a), (3.28b) and , we obtain that

(3.36)

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

(3.37)

Letting , we have that and . Then the maximality of the operators implies that and .

Next, let us show that . Since

(3.38)

from and Proposition 2.11 it follows that

(3.39)

Also, since

(3.40)

so we get

(3.41)

So, from (3.39), , and , we have .

Since is uniformly convex and smooth, we conclude from Lemma 2.2 that

(3.42)

From , , and (3.42), we have and .

Since is uniformly norm-to-norm continuous on bounded subsets of , from (3.42) we derive

(3.43)

From , it follows that

(3.44)

By the definition of , we have

(3.45)

where

(3.46)

Replacing by , we have from (A2) that

(3.47)

Since is convex and lower semicontinuous, it is also weakly lower semicontinuous. Letting in the last inequality, from (3.44) and (A4), we have

(3.48)

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

(3.49)

Dividing by , we have

(3.50)

Letting , from (A3) it follows that

(3.51)

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

Step 6.

We claim that converges strongly to .

Indeed, from and , it follows that

(3.52)

Since the norm is weakly lower semicontinuous, then

(3.53)

From the definition of , we have . Hence , and

(3.54)

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

Remark 3.2.

In Theorem 3.1, let , , and . Then, for all and , we have that

(3.55)

Moreover, there hold the following

(3.56)

and hence

(3.57)

In this case, Theorem 3.1 reduces to [17, Theorem ].

4. Weak Convergence Theorem

In this section, we present the following algorithm for finding a common element of the solution set of 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 (2.14).

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

Proposition 4.1.

Suppose that Assumption 2.1 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 2.5 and Proposition 2.11, 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 2.11, we obtain

(4.6a)

(4.6b)

and hence by Proposition 2.11

(4.6c)

 

(4.6d)

Consequently, the last two inequalities yield that

(4.6e)

for all . So, from , , and Lemma 2.13, 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

(4.7)

for each . This, together with the boundedness of , implies that is bounded and so is . Furthermore, from and (4.6e), we have

(4.8)

Since is the generalized projection, then, from Lemma 2.4 we obtain

(4.9)

Hence, from (4.8), it follows that .

Note that , , and is bounded, so that . Therefore, is a convergent sequence. On the other hand, from (4.6e) we derive, for all ,

(4.10)

In particular, we have

(4.11)

Consequently, from and Lemma 2.4, we have

(4.12)

and hence

(4.13)

Let . From Lemma 2.7, there exists a continuous, strictly increasing, and convex function with such that

(4.14)

So, we have

(4.15)

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 2.1 is fulfilled and let be a sequence defined by (4.1), where satisfy the following conditions:

(4.16)

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

(4.17)

From Lemma 2.6 and as in the proof of Theorem 3.1, there exists a continuous, strictly increasing, and convex function with such that

(4.18)

for and . Observe that for ,

(4.19)

Hence,

(4.20)

Consequently, the last two inequalities yield that

(4.21)

Thus, we have

(4.22)

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

(4.23)

Taking into account the properties of , as in the proof of Theorem 3.1, we have

(4.24)

since is uniformly norm-to-norm continuous on bounded subsets of .

Now let us show that

(4.25)

Indeed, from (4.6e) we get

(4.26)

which, together with , yields that

(4.27)

From (4.6d) it follows that

(4.28)

which, together with , yields that

(4.29)

From (4.6c) it follows that

(4.30)

which, together with , yields that

(4.31)

From (4.6c) it follows that

(4.32)

which together with

(4.33)

yields that

(4.34)

From (4.6a) it follows that

(4.35)

which, together with , yields that

(4.36)

On the other hand, let us show that

(4.37)

Indeed, let . From Lemma 2.7, there exists a continuous, strictly increasing, and convex function with such that

(4.38)

Since and , we deduce from Proposition 2.11 that for ,

(4.39)

This implies that

(4.40)

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

(4.41)

Note that

(4.42)

Since , it follows from (4.24) and (4.41) that and .

Also, observe that

(4.43)

and hence

(4.44)

Thus, from , and , it follows that . In terms of Lemma 2.2, we derive .

Next, let us show that , where .

Indeed, since is bounded, there exists a subsequence of such that . Hence it follows from (4.24), (4.41), and that and converge weakly to the same point . Furthermore, from and (4.24), we have that

(4.45)

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

(4.46)

Letting , we obtain that

(4.47)

Then the maximality of the operators implies that .

Now, by the definition of , we have

(4.48)

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

(4.49)

Since is convex and lower semicontinuous, it is also weakly lower semicontinuous. Letting in the last inequality, from (4.41) and (A4), we have

(4.50)

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

(4.51)

Dividing by , we get . Letting , from (A3) it follows that . So, . Therefore, . Let . From Lemma 2.3 and , we get

(4.52)

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

(4.53)

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

Remark 4.3.

Compared with the algorithm of Theorem 1.2, the above algorithm (4.1) can be applied to find an element of . But, the algorithm of Theorem 1.2 cannot be applied. Therefore, algorithm (4.1) develops and improves the algorithm of Theorem 1.2.