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Strong Convergence Theorem by Monotone Hybrid Algorithm for Equilibrium Problems, Hemirelatively Nonexpansive Mappings, and Maximal Monotone Operators
Fixed Point Theory and Applications volumeÂ 2008, ArticleÂ number:Â 617248 (2008)
Abstract
We introduce a new hybrid iterative algorithm for finding a common element of the set of fixed points of hemirelatively nonexpansive mappings and the set of solutions of an equilibrium problem and for finding a common element of the set of zero points of maximal monotone operators and the set of solutions of an equilibrium problem in a Banach space. Using this theorem, we obtain three new results for finding a solution of an equilibrium problem, a fixed point of a hemirelatively nonexpnasive mapping, and a zero point of maximal monotone operators in a Banach space.
1. Introduction
Let be a Banach space, let be a closed convex subset of , and let be a bifunction from to , where is the set of real numbers. The equilibrium problem is to find
The set of such solutions is denoted by .
In 2006, MartinezYanes and Xu [1] obtained strong convergence theorems for finding a fixed point of a nonexpansive mapping by a new hybrid method in a Hilbert space. In particular, Takahashi and Zembayashi [2] established a strong convergence theorem for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of a nonexpansive mapping in a uniformly convex and uniformly smooth Banach space. Very recently, Su et al. [3] proved the following theorem by a monotone hybrid method.
Theorem 1.1 (see Su et al. [3]).
Let be a uniformly convex and uniformly smooth real Banach space, let be a nonempty closed convex subset of , and let be a closed hemirelatively nonexpansive mapping such that . Assume that is a sequence in such that . Define a sequence in by the following:
where is the duality mapping on . Then, converges strongly to , where is the generalized projection from onto .
In this paper, motivated by Su et al. [3], we prove a strong convergence theorem for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of a hemirelatively nonexpansive mapping and for finding a common element of the set of zero points of maximal monotone operators and the set of solutions of an equilibrium problem in a Banach space by using the monotone hybrid method. Using this theorem, we obtain three new strong convergence results for finding a solution of an equilibrium problem, a fixed point of a hemirelatively nonexpnasive mapping, and a zero point of maximal monotone operators in a Banach space.
2. Preliminaries
Let be a real Banach space with dual . We denote by the normalized duality mapping from to defined by
where denotes the generalized duality pairing. It is well known that if is uniformly convex, then is uniformly continuous on bounded subsets of . In this case, is single valued and also one to one.
Let be a smooth, strictly convex, and reflexive Banach space and let be a nonempty closed convex subset of . Throughout this paper, we denote by the function defined by
Following Alber [4], the generalized projection from onto is defined by
The generalized projection from onto is well defined and single valued, and it satisfies
If is a Hilbert space, then and is the metric projection of onto .
If is a reflexive strict convex and smooth Banach space, then for if and only if . It is sufficient to show that if , then . From (2.4), we have . This implies . From the definition of , we have , that is, .
Let be a closed convex subset of and let be a mapping from into itself. We denote by the set of fixed points of . is called hemirelatively nonexpansive if for all and .
A point in is said to be an asymptotic fixed point of [5] if contains a sequence which converges weakly to such that the strong . The set of asymptotic fixed points of will be denoted by . A hemirelatively nonexpansive mapping from into itself is called relatively nonexpansive [1, 5, 6] if .
We need the following lemmas for the proof of our main results.
Lemma 2.1 (see Alber [4]).
Let be a nonempty closed convex subset of a smooth, strictly convex, and reflexive Banach space . Then,
Lemma 2.2 (see Alber [4]).
Let be a nonempty closed convex subset of a smooth, strictly convex, and reflexive Banach space, let , and let . Then,
Lemma 2.3 (see Kamimura and Takahashi [7]).
Let be a smooth and uniformly convex Banach space and let and be sequences in such that either or is bounded. If . Then .
Lemma 2.4 (see Xu [8]).
Let be a uniformly convex Banach space and let . Then, there exists a strictly increasing, continuous, and convex function such that and
where .
Lemma 2.5 (see Kamimura and Takahashi [7]).
Let be a smooth and uniformly convex Banach space and let . Then, there exists a strictly increasing, continuous, and convex function such that and
For solving the equilibrium problem, let us assume that a bifunction satisfies the following conditions:

(A1) for all

(A2) is monotone, that is, for all

(A3) for all ;

(A4) for all is convex.
Lemma 2.6 (see Blum and Oettli [9]).
Let be a closed convex subset of a smooth, strictly convex, and reflexive Banach space , let be a bifunction from to satisfying (A1)â€“(A4), let , and let . Then, there exists such that
Lemma 2.7 (see Takahashi and Zembayashi [10]).
Let be a closed convex subset of a uniformly smooth, strictly convex, and reflexive Banach space , let be a bifunction from to satisfying (A1)â€“(A4), and let , for . Define a mapping as follows:
Then, the following holds:

(1)
is single valued;

(2)
is a firmly nonexpansivetype mapping [11], that is, for all ,
(211) 
(3)
;

(4)
is closed and convex.
Lemma 2.8 (see Takahashi and Zembayashi [10]).
Let be a closed convex subset of a smooth, strictly convex, and reflexive Banach space and let be a bifunction from to satisfying (A1)â€“(A4). Then, for and , and ,
Lemma 2.9 (see Su et al. [3]).
Let be a strictly convex and smooth real Banach space, let be a closed convex subset of , and let be a hemirelatively nonexpansive mapping from into itself. Then, is closed and convex.
Recall that an operator in a Banach space is called closed, if , then .
3. Strong Convergence Theorem
Theorem 3.1.
Let be a uniformly convex and uniformly smooth real Banach space, let be a nonempty closed convex subset of , let be a bifunction from to satisfying (A1)â€“(A4), and let be a closed hemirelatively nonexpansive mapping such that . Define a sequence in by the following:
for every , where is the duality mapping on are sequences in such that and for some . Then, converges strongly to , where is the generalized projection of onto .
Proof.
First, we can easily show that and are closed and convex for each .
Next, we show that for all . Let . Putting for all , from Lemma 2.8, we have relatively nonexpansive. Since are relatively nonexpansive and is hemirelatively nonexpansive, we have
Hence, we have
Next, we show that for all . We prove this by induction. For , we have
Suppose that , by Lemma 2.2, we have
As , by the induction assumptions, the last inequality holds, in particular, for all . This, together with the definition of , implies that . So, is well defined.
Since and for all , we have
Therefore, is nondecreasing. In addition, from the definition of and Lemma 2.2, . Therefore, for each , we have
Therefore, and are bounded. This, together with (3.6), implies that the limit of exists. From Lemma 2.1, we have, for any positive integer ,
Therefore,
From (3.9), we can prove that is a Cauchy sequence. Therefore, there exists a point such that converges strongly to .
Since , we have
Therefore, we have
From Lemma 2.3, we have
So, we have
Since is uniformly normtonorm continuous on bounded sets, we have
Let Since is a uniformly smooth Banach space, we know that is a uniformly convex Banach space. Therefore, from Lemma 2.4, there exists a continuous, strictly increasing, and convex function with , such that
for , and . So, we have that for ,
Therefore, we have
Since
we have
From , we have
Therefore, from the property of , we have
Since is uniformly normtonorm continuous on bounded sets, we have
Since is a closed operator and , then is a fixed point of .
On the other hand,
So, we have from (3.19) that
From Lemma 2.3, we have that
From and , we have .
From (3.25), we have
From , we have
By , we have
From (A2), we have that
From (3.27) and (A4), we have
For with and , let . We have . So, from (A1), we have
Dividing by , we have
Letting , from (A3), we have
Therefore, . Finally, we prove that . From Lemma 2.1, we have
Since and , for all , we get from Lemma 2.1 that
By the definition of , it follows that and , whence . Therefore, it follows from the uniqueness of that . This completes the proof.
Corollary 3.2.
Let be a uniformly convex and uniformly smooth real Banach space, let be a nonempty closed convex subset of , and let be a bifunction from to satisfying (A1)â€“(A4). Define a sequence in by the following:
for every , where is the duality mapping on and for some . Then, converges strongly to .
Proof.
Putting in Theorem 3.1, we obtain Corollary 3.2.
Corollary 3.3.
Let be a uniformly convex and uniformly smooth real Banach space, let be a nonempty closed convex subset of , and let be a closed hemirelatively nonexpansive mapping. Define a sequence in by the following:
for every , where is the duality mapping on are sequences in such that . Then, converges strongly to .
Proof.
Putting for all and for all in Theorem 3.1, we obtain Corollary 3.3.
Corollary 3.4.
Let be a uniformly convex and uniformly smooth real Banach space, let be a nonempty closed convex subset of , let be a bifunction from to satisfying (A1)â€“(A4), and let be a closed relatively nonexpansive mapping such that . Define a sequence in by the following:
for every , where is the duality mapping on are sequences in such that and for some . Then, converges strongly to .
Proof.
Since every relatively nonexpansive mapping is a hemirelatively one, Corollary 3.4 is implied by Theorem 3.1.
Remark 3.5 (see Rockafellar [12]).
Let be a reflexive, strictly convex, and smooth Banach space and let be a monotone operator from to . Then, is maximal if and only if for all .
Let be a reflexive, strictly convex, and smooth Banach space and let be a maximal monotone operator from to . Using Remark 3.5 and strict convexity of , we obtain that for every and , there exists a unique such that If , then we can define a singlevalued mapping by , and such a is called the resolvent of . We know that for all and is relatively nonexpansive mapping (see [2] for more details). Using Theorem 3.1, we can consider the problem of strong convergence concerning maximal monotone operators in a Banach space.
Theorem 3.6.
Let be a uniformly convex and uniformly smooth real Banach space, let be a nonempty closed convex subset of , let be a bifunction from to satisfying (A1)â€“(A4), and let be a resolvent of and a closed mapping such that , where . Define a sequence in by the following:
for every , where is the duality mapping on , is a sequences in such that and for some , Then, converges strongly to .
Proof.
Since is a closed relatively nonexpansive mapping and , from Corollary 3.4, we obtain Theorem 3.6.
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Acknowledgment
This work is supported by Tianjin Natural Science Foundation in China Grant no. 06YFJMJC12500.
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Cheng, Y., Tian, M. Strong Convergence Theorem by Monotone Hybrid Algorithm for Equilibrium Problems, Hemirelatively Nonexpansive Mappings, and Maximal Monotone Operators. Fixed Point Theory Appl 2008, 617248 (2008). https://doi.org/10.1155/2008/617248
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DOI: https://doi.org/10.1155/2008/617248