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A New Hybrid Iterative Algorithm for FixedPoint Problems, Variational Inequality Problems, and Mixed Equilibrium Problems
Fixed Point Theory and Applications volumeÂ 2008, ArticleÂ number:Â 417089 (2008)
Abstract
We introduce a new hybrid iterative algorithm for finding a common element of the set of fixed points of an infinite family of nonexpansive mappings, the set of solutions of the variational inequality of a monotone mapping, and the set of solutions of a mixed equilibrium problem. This study, proves a strong convergence theorem by the proposed hybrid iterative algorithm which solves fixedpoint problems, variational inequality problems, and mixed equilibrium problems.
1. Introduction
Let be a nonempty closed convex subset of a real Hilbert space . Recall that a mapping is called contractive if there exists a constant such that for all . A mapping is said to be nonexpansive if for all . Denote the set of fixed points of by .
Let be a realvalued function and be an equilibrium bifunction, that is, for each . The mixed equilibrium problem (for short, MEP) is to find such that
In particular, if , this problem reduces to the equilibrium problem (for short, EP), which is to find such that
Denote the set of solutions of MEP by . The mixed equilibrium problems include fixedpoint problems, optimization problems, variational inequality problems, Nash equilibrium problems, and the equilibrium problems as special cases (see, e.g., [1â€“5]). Some methods have been proposed to solve the MEP and EP (see, e.g., [5â€“14]). In 1997, Combettes and Hirstoaga [13] introduced an iterative method of finding the best approximation to the initial data and proved a strong convergence theorem. Subsequently, S. Takahashi and W. Takahashi [8] introduced another iterative scheme for finding a common element of the set of solutions of EP and the set of fixedpoint points of a nonexpansive mapping. Yao et al. [12] considered an iterative scheme for finding a common element of the set of solutions of EP and the set of common fixed points of an infinite nonexpansive mappings. Very recently, Zeng and Yao [14] considered a new iterative scheme for finding a common element of the set of solutions of MEP and the set of common fixed points of finitely many nonexpansive mappings. Their results extend and improve many results in the literature.
Let of into be a nonlinear mapping. It is well known that the variational inequality problem is to find such that
The set of solutions of the variational inequality problem is denoted by . A mapping is called inversestrongly monotone if there exists a positive real number such that
Recently, some authors have proposed new iterative algorithms to approximate a common element of the set of fixed points of a nonxpansive mapping and the set of solutions of the variational inequality. For the details, see [15, 16] and the references therein.
Motivated by the recent works, in this paper we introduce a new hybrid iterative algorithm for finding a common element of the set of fixed points of an infinite family of nonexpansive mappings, the set of solutions of the variational inequality of a monotone mapping, and the set of solutions of a mixed equilibrium problem. We prove a strong convergence theorem by the proposed hybrid iterative algorithm which solves fixedpoint problems, variational inequality problems, and mixed equilibrium problems.
2. Preliminaries
Let be a real Hilbert space with inner product and norm . Let be a nonempty closed convex subset of . Then for any , there exists a unique nearest point in , denoted by such that
Such a is called the metric projection of onto . It is well known that is a nonexpansive mapping and satisfies
Moreover, is characterized by the following properties:
It is clear that
In this paper, for solving the mixed equilibrium problems for an equilibrium bifunction , we assume that satisfies the following conditions:
(H1) is monotone, that is, for all ;
(H2) for each fixed , is concave and upper semicontinuous;
(H3) for each , is convex.
A mapping is called Lipschitz continuous if there exists a constant such that
A differentiable function on a convex set is called:

(i)
convex if
(2.6)
where is the FrÃ©chet derivative of at ;

(ii)
strongly convex if there exists a constant such that
(2.7)
Let be a nonempty closed convex subset of a real Hilbert space , be a realvalued function, and be an equilibrium bifunction. Let be a positive number. For a given point , the auxiliary problem for MEP consists of finding such that
Let be the mapping such that for each , is the solution set of the auxiliary problem MEP, that is,
We first need the following important and interesting result.
Lemma 2.1 (see [14]).
Let be a nonempty closed convex subset of a real Hilbert space and let be a lower semicontinuous and convex functional. Let be an equilibrium bifunction satisfying conditions (H1)â€“(H3). Assume that

(i)
is Lipschitz continuous with constant such that

(a)
for all,

(b)
is affine in the first variable,

(c)
for each fixed, is sequentially continuous from the weak topology to the weak topology;

(ii)
is strongly convex with constantand its derivativeis sequentially continuous from the weak topology to the strong topology;

(iii)
for each, there exist a bounded subsetandsuch that for any,
(2.10)
Then there hold the following:

(i)
is singlevalued;

(ii)
is nonexpansive if is Lipschitz continuous with constant such that and
(2.11)
where for ;

(iii)
;

(vi)
is closed and convex.
We also need the following lemmas for proving our main results.
Lemma 2.2 (see [17]).
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 [18]).
Assume is a sequence of nonnegative real numbers such that , where is a sequence in and is a sequence such that

(1)
;

(2)
or .
Then .
3. Iterative Algorithm and Strong Convergence Theorems
In this section, we first introduce a new iterative algorithm. Consequently, we will establish a strong convergence theorem for this iteration algorithm. To be more specific, let be infinite mappings of into itself and let be real numbers such that for every . For any , define a mapping of into itself as follows:
Such is called the mapping generated by and . For the iterative algorithm for a finite family of nonexpansive mappings, we refer the reader to [19].
We have the following crucial Lemmas 3.1 and 3.2 concerning which can be found in [20]. Now we only need the following similar version in Hilbert spaces.
Lemma 3.1.
Let be a nonempty closed convex subset of a real Hilbert space . Let be nonexpansive mappings of into itself such that is nonempty, and let be real numbers such that for any . Then for every and , the limit exists.
Lemma 3.2.
Let be a nonempty closed convex subset of a real Hilbert space . Let be nonexpansive mappings of into itself such that is nonempty, and let be real numbers such that for any . Then .
The following remark [12] is important to prove our main results.
Remark 3.3.
Using Lemma 3.1, one can define a mapping of into itself as for every . If is a bounded sequence in , then we have
Throughout this paper, we will assume that for every .
Now we introduce the following iteration algorithm.
Algorithm 3.4.
Let be a constant. Let be a lower semicontinuous and convex functional and let be an equilibrium bifunction. Let be a inversestrongly monotone mapping and be the mapping defined by (3.1). Let be a contraction of into itself with coefficient and given arbitrarily. Suppose that the sequences and are generated iteratively by
where , , and are three sequences in , and is a sequence in .
Now we study the strong convergence of the hybrid iterative algorithm (3.3).
Theorem 3.5.
Let be a nonempty closed convex subset of a real Hilbert space and let be a lower semicontinuous and convex functional. Let be an equilibrium bifunction satisfying conditions (H1)â€“(H3) and let be an infinite family of nonexpansive mappings of into itself. Let be a inversestrongly monotone mapping such that . Suppose , , and are three sequences in with . Assume that

(i)
is Lipschitz continuous with constant such that

(a)
for all,

(b)
is affine in the first variable,

(c)
for each fixed, is sequentially continuous from the weak topology to the weak topology;

(ii)
is strongly convex with constantand its derivativeis not only sequentially continuous from the weak topology to the strong topology but also Lipschitz continuous with constantsuch that;

(iii)
for each; there exist a bounded subsetandsuch that for any,
(3.4)

(iv)
, , , , and .
Let be a contraction of into itself and given arbitrarily. Then the sequence generated by (3.3) converges strongly to , where provided that is firmly nonexpansive.
Proof.
We first note that is a contraction with coefficient . Then for all . Therefore is a contraction of into itself which implies that there exists a unique element such that .
Next we divide the following proofs into several steps.
Step 1 ( , , and are bounded).
Let . From the definition of , we know that . It follows that
For all and , we note that
which implies that is nonexpansive.
Set for all . From (2.4), we have that . It follows from (3.6) that
Hence we obtain that
Therefore is bounded, so are and .
Step 2 ().
Setting for all . It follows that
which implies that
Now we estimate and .
From (3.1), since and are nonexpansive, we have
where is a constant such that .
At the same time, we observe that
Since and , from the nonexpansivity of , we get
Substituting (3.11)â€“(3.13) into (3.10), we have
Since , and , we have
Hence by Lemma 2.2, we have
Consequently,
Step 3 ().
Note that . Then we have
For , noting that is firmly nonexpansive, we have
and hence
So, we have
that is,
From (3.6), we obtain that
Then we have
which implies that
We note that
Then we derive
Hence
which implies that
Since , we have
Combining the above inequality, (3.18)â€“(3.29), and Remark 3.3, we have
Step 4 (, where ).
To show the above inequality, we can choose a subsequence of such that
Since is bounded, there exists a subsequence of which converges weakly to . Without loss of generality, we can assume that . From , we obtain .
First, we show . Assume that . Since and , by Opial's condition, we have
which is a contradiction. Hence we get . By the same argument as that in the proof of [21, Theorem 3.1], we can prove that ; and by the same argument as that in the proof of [14, Theorem 4.1], we also can prove that . Hence .
Since and , we have
Step 5 (, where ).
From (3.3), we have
that is,
It is easy to see that and
Applying Lemma 2.3 and (3.34) to (3.36), we conclude that as . This completes the proof.
Concerning , we give the following remark.
Remark 3.6.
For each , we denote and . Then for all , we have
Taking in (3.38) and in (3.39), and adding up these two inequalities, we obtain
Note that and . Hence from (3.40), we deduce
which implies that
Since is strongly monotone with constant , then from (3.42), we conclude that
Take , , and . Then from (3.43), we have
This indicates that is firmly nonexpansive.
Corollary 3.7.
Let be a nonempty closed convex subset of a real Hilbert space and let be a lower semicontinuous and convex functional. Let be an equilibrium bifunction satisfying conditions (H1)â€“(H3). Let be a inversestrongly monotone mapping such that . Suppose , , and are three sequences in with . Assume that

(i)
is Lipschitz continuous with constant such that

(a)
for all,

(b)
is affine in the first variable,

(c)
for each fixed, is sequentially continuous from the weak topology to the weak topology;

(ii)
isstrongly convex with constantand its derivativeis not only sequentially continuous from the weak topology to the strong topology but also Lipschitz continuous with constantsuch that;

(iii)
for each, there exist a bounded subsetandsuch that, for any,
(3.45)

(iv)
, , , , and .
Let be a contraction of into itself and given arbitrarily. Let the sequences , , and be generated iteratively by
Then the sequence generated by (3.46) converges strongly to , where provided that is firmly nonexpansive.
Proof.
Take for all , and for all in (3.1). Then for all . The conclusion follows immediately from Theorem 3.5. This completes the proof.
Corollary 3.8.
Let be a nonempty closed convex subset of a real Hilbert space . Let be an infinite family of nonexpansive mappings of into itself. Let be a inversestrongly monotone mapping such that . Suppose , , and are three sequences in with , . Assume that

(i)
and ;

(ii)
;

(iii)
and
Let be a contraction of into itself and given arbitrarily. Then the sequence , generated iteratively by
converges strongly to , where .
Proof.
Set and for all and put . Take and for all . Then we have . Hence the conclusion follows. This completes the proof.
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Acknowledgments
The authors are extremely grateful to the anonymous referees and Professor T. Suzuki for their useful comments and suggestions. The first author was partially supposed by National Natural Science Foundation of China Grant no. 10771050. The second author was partially supposed by the Grant no. NSC 962221E230003.
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Yao, Y., Liou, YC. & Yao, JC. A New Hybrid Iterative Algorithm for FixedPoint Problems, Variational Inequality Problems, and Mixed Equilibrium Problems. Fixed Point Theory Appl 2008, 417089 (2008). https://doi.org/10.1155/2008/417089
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DOI: https://doi.org/10.1155/2008/417089