We introduce a new iterative scheme for finding the common element of the set of common fixed points of nonexpansive mappings, the set of solutions of an equilibrium problem, and the set of solutions of the variational inequality. We show that the sequence converges strongly to a common element of the above three sets under some parameters controlling conditions. Moreover, we apply our result to the problem of finding a common fixed point of a countable family of nonexpansive mappings, and the problem of finding a zero of a monotone operator. This main theorem extends a recent result of Yao et al. (2007) and many others.
1. Introduction
Let be a real Hilbert space with inner product and norm , and let be a closed convex subset of . Let be a bifunction of into , where is the set of real numbers. The equilibrium problem for is to find such that
The set of solutions of (1.1) is denoted by Given a mapping , let for all . Then if and only if for all that is, is a solution of the variational inequality. Numerous problems in physics, optimization, and economics reduce to find a solution of (1.1). In 1997, Flåm and Antipin [1] introduced an iterative scheme of finding the best approximation to initial data when is nonempty and proved a strong convergence theorem.
Let be a mapping. The classical variational inequality, denoted by , is to find such that
for all The variational inequality has been extensively studied in the literature. See, for example, [2, 3] and the references therein. A mapping of into is called inversestrongly monotone [4, 5] if there exists a positive real number such that
for all . It is obvious that any inversestrongly monotone mapping is monotone and Lipschitz continuous. A mapping of into itself is called nonexpansive if
for all . We denote by the set of fixed points of . For finding an element of , under the assumption that a set is nonempty, closed, and convex, a mapping is nonexpansive and a mapping is inversestrongly monotone, Takahashi and Toyoda [6] introduced the following iterative scheme:
for every where is a sequence in (0, 1), and is a sequence in . They proved that if , then the sequence generated by (1.5) converges weakly to some . Recently, motivated by the idea of Korpelevič's extragradient method [7], Nadezhkina and Takahashi [8] introduced an iterative scheme for finding an element of and the weak convergence theorem is presented. Moreover, Zeng and Yao [9] proposed some new iterative schemes for finding elements in and obtained the weak convergence theorem for such schemes. Very recently, Yao et al. [10] introduced the following iterative scheme for finding an element of under some mild conditions. Let be a closed convex subset of a real Hilbert space a monotone, Lipschitz continuous mapping, and a nonexpansive mapping of into itself such that Suppose that and are given by
where and satisfy some parameters controlling conditions. They proved that the sequence defined by (1.6) converges strongly to a common element of .
On the other hand, S. Takahashi and W. Takahashi [11] introduced an iterative scheme by the viscosity approximation method for finding a common element of the set of solution (1.1) and the set of fixed points of a nonexpansive mapping in a real Hilbert space. Let be a nonexpansive mapping. Starting with arbitrary initial define sequences and recursively by
They proved that under certain appropriate conditions imposed on and , the sequences and converge strongly to , where
Moreover, Aoyama et al. [12] introduced an iterative scheme for finding a common fixed point of a countable family of nonexpansive mappings in Banach spaces and obtained the strong convergence theorem for such scheme.
In this paper, motivated by Yao et al. [10], S. Takahashi and W. Takahashi [11] and Aoyama et al. [12], we introduce a new extragradient method (4.2) which is mixed the iterative schemes considered in [10–12] for finding a common element of the set of common fixed points of nonexpansive mappings, the set of solutions of an equilibrium problem, and the solution set of the classical variational inequality problem for a monotone Lipschitz continuous mapping in a real Hilbert space. Then, the strong convergence theorem is proved under some parameters controlling conditions. Further, we apply our result to the problem of finding a common fixed point of a countable family of nonexpansive mappings, and the problem of finding a zero of a monotone operator. The results obtained in this paper improve and extend the recent ones announced by Yao et al. results [10] and many others.
2. Preliminaries
Let be a real Hilbert space with norm and inner product and let be a closed convex subset of . For every point , there exists a unique nearest point in , denoted by , such that
is called the metric projection of onto It is well known that is a nonexpansive mapping of onto and satisfies
for every Moreover, is characterized by the following properties: and
for all . For more details, see [13]. It is easy to see that the following is true:
A setvalued mapping is called monotone if for all and imply . A monotone mapping is maximal if the graph of of is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping is maximal if and only if for for every implies . Let be a monotone map of into , Lipschitz continuous mapping and let be the normal cone to at , that is, for all . Define
Then is the maximal monotone and if and only if ; see [14].
The following lemmas will be useful for proving the convergence result of this paper.
Lemma 2.1 (see [15]).
Let be an inner product space. Then for all and with one has
Lemma 2.2 (see [16]).
Let and be bounded sequences in a Banach space and let be a sequence in with Suppose for all integers and Then,
Lemma 2.3 (see [17]).
Assume is a sequence of nonnegative real numbers such that
where is a sequence in and is a sequence in such that
(i) and
(ii) or
Then
Lemma 2.4 (see [12, Lemma 3.2]).
Let be a nonempty closed subset of a Banach space and let be a sequence of mappings of into itself. Suppose that . Then, for each , converges strongly to some point of . Moreover, let be a mapping of into itself defined by
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, for all
(A3)for each
(A4)for each is convex and lower semicontinuous.
The following lemma appears implicitly in [18].
Lemma 2.5 (see [18]).
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
The following lemma was also given in [1].
Lemma 2.6 (see [1]).
Assume that satisfies (A1)–(A4). For and , define a mapping as follows:
for all . Then, the following hold:
(i) is singlevalued;
(ii) is firmly nonexpansive, that is, for any
(iii)
(iv) is closed and convex.
3. Main Results
In this section, we prove a strong convergence theorem.
Theorem 3.1.
Let be a closed convex subset of a real Hilbert space . Let be a bifunction from to satisfying (A1)–(A4), a monotone Lipschitz continuous mapping and let be a sequence of nonexpansive mappings of into itself such that Let the sequences , and be generated by
where , , and satisfy the following conditions:
(C1),
(C2)
(C3)
(C4)
(C5).
Suppose that for any bounded subset of . Let be a mapping of into itself defined by and suppose that . Then the sequences , and converge strongly to the same point , where .
Proof.
Let . Since is a contraction with , we obtain
Therefore, is a contraction of into itself, which implies that there exists a unique element such that . Then we divide the proof into several steps.
Step 1 ( is bounded).
Indeed, put for all . Let . From (2.5) we have . Also it follows from (2.4) that
Since is monotone and is a solution of the variational inequality problem , we have
This together with (3.3) implies that
From (2.3), we have
so that
Hence it follows from (3.5) and (3.7) that
Since as , there exists a positive integer such that , when . Hence it follows from (3.8) that
Observe that
and hence
Thus, we can calculate
It follows from induction that
Therefore, is bounded. Hence, so are , and .
Step 2 ().
Indeed, we observe that for any ,
which implies that
Thus
On the other hand, from and we note that
Putting in (3.17) and in (3.18), we have
So, from (A2), we have
and hence
Without loss of generality, let us assume that there exists a real number such that for all Then, we have
and hence
where . It follows from (3.16) and the last inequality that
Setting , we obtain for all . Thus, we have
It follows from (3.24) that
Combining (3.25) and (3.26), we have
This together with (C1)–(C5) and implies that
Hence, by Lemma 2.2, we obtain as . It then follows that
By (3.23) and (3.24), we also have
Step 3 ().
Indeed, pick any , to obtain
Therefore, . From Lemma 2.1 and (3.9), we obtain, when , that
and hence
It now follows from the last inequality, (C1), (C2), (C3) and (3.29), that
Noting that
Thus
We note that
Using (3.37), we have
so that
This implies that
It now follows from (3.36) and (3.40) that
Applying Lemma 2.4 and (3.41), we have
It follows from the last inequality and (3.36) that
Step 4 ().
Indeed, we choose a subsequence of such that
Without loss of generality, we may assume that converges weakly to . From we obtain Now, we will show that . Firstly, we will show . Indeed, we observe that , and
From (A2), we also have
and hence
From and we get . Since it follows by (A4) that for all For with and let Since and we have and hence So, from (A1) and (A4), we have
and hence . From (A3), we have for all , and hence By the Opial's condition, we can obtain that Next we will show that . Let
Then is maximal monotone (see [14]). Let . Since and we have . On the other hand, from , we have
that is,
Therefore, we obtain
Noting that as , is Lipschitz continuous and (3.52), we obtain
Since is maximal monotone, we have , and hence . Hence The property of the metric projection implies that
Step 5 ().
Indeed, we observe that
which implies that
Setting , we have . Applying Lemma 2.3 to (3.56), we conclude that converges strongly to . Consequently, and converge strongly to . This completes the proof.
As in [12, Theorem 4.1], we can generate a sequence of nonexpansive mappings satisfying condition for any bounded subset of by using convex combination of a general sequence of nonexpansive mappings with a common fixed point.
Corollary 3.2.
Let be a closed convex subset of a real Hilbert space . Let be a bifunction from to satisfying (A1)–(A4), a monotone, Lipschitz continuous mapping and let be a family of nonnegative numbers with indices with such that
(i) for all ;
(ii) for every ;
(iii).
Let be a sequence of nonexpansive mappings of into itself with Let and and be the sequences generated by
where , and satisfy the following conditions:
(C1),
(C2)
(C3)
(C4)
(C5).
Then the sequences , and converge strongly to the same point , where .
Setting and in Theorem 3.1, we have the following result.
Corollary 3.3 (see [10, Theorem 3.1]).
Let be a closed convex subset of a real Hilbert space . Let be a monotone, Lipschitz continuous mapping, and let be a nonexpansive mapping of into itself such that Suppose and are given by
where are sequences in satisfying the following conditions:
(i),
(ii)
(iii)
(iv).
Then converges strongly to
Proof.
Put for all and in Theorem 3.1. Thus, we have . Then the sequence generated in Corallary 3.3 converges strongly to .
4. Applications
In this section, we consider the problem of finding a zero of a monotone operator. A multivalued operator with domain and range is said to be monotone if for each and we have . A monotone operator is said to be maximal if its graph is not properly contained in the graph of any other monotone operator. Let denote the identity operator on and let be a maximal monotone operator. Then we can define, for each , a nonexpansive singlevalued mapping by . It is called the resolvent (or the proximal mapping) of . We also define the Yosida approximation by . We know that and for all . We also know that for all ; see, for instance, Rockafellar [19] or Takahashi [20].
Lemma 4.1 (the resolvent identity).
For , there holds the identity
By using Theorem 3.1 and Lemma 4.1, we may obtain the following improvement.
Theorem 4.2.
Let be a maximal monotone operator. Let be a bifunction from to satisfying (A1)–(A4), a monotone Lipschitz continuous mapping of into such that and a contraction of into itself with coefficient . Let the sequences and be generated by
where , and satisfy the following conditions:
(C1),
(C2)
(C3)
(C4)
(C5).
Then converges strongly to .
Proof.
We first verify that for any bounded subset of . Let be a bounded subset of . Since for each is bounded. It follows from Lemma 4.1 that
Thus
for each and where . Hence we get
By the assumption that , we obtain for some . Since , we obtain that for all . Since for all , we have . Therefore, by Theorem 3.1, converges strongly to .
Acknowledgments
The author would like to thank the referees for reading this paper carefully, providing valuable suggestions and comments, and pointing out a major error in the original version of this paper. This research was partially supported by the Commission on Higher Education.
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