We introduce a new iterative method for finding a common element of the set of solutions of a generalized equilibrium problem with a relaxed monotone mapping and the set of common fixed points of a countable family of nonexpansive mappings in a Hilbert space and then prove that the sequence converges strongly to a common element of the two sets. Using this result, we prove several new strong convergence theorems in fixed point problems, variational inequalities, and equilibrium problems.
1. Introduction
Throughout this paper, let denote the set of all real numbers, let denote the set of all positive integer numbers, let be a real Hilbert space, and let be a nonempty closed convex subset of . Let be a mapping. We call nonexpansive if
The set of fixed points of is denoted by . We know that the set is closed and convex. Let be a bifunction. The equilibrium problem for is to find such that
The set of all solutions of the equilibrium problem is denoted by , that is,
Some iterative methods have been proposed to find an element of ; see [1, 2].
A mapping is called inversestrongly monotone if there exists such that
Such a mapping is also called inversestrongly monotone. It is known that each nonexpansive mapping is inversestrongly monotone and each strictly pseudocontraction is inversestrongly monotone; see [3, 4]. If there exists such that
then is called a solution of the variational inequality. The set of all solutions of the variational inequality is denoted by . It is known that is closed and convex. Recently Takahashi and Toyoda [5] introduced an iterative method for finding an element of ; see also [6]. On the other hand, Plubtieng and Punpaeng [7] introduced an iterative method for finding an element of ; see also [8].
Consider a general equilibrium problem:
The set of all solutions of the equilibrium problem is denoted by , that is,
In the case of , coincides with . In the case , coincides with . Recently, S. Takahashi and W. Takahashi [9] introduced an iterative method to find an element of . More precisely, they introduced the following iterative scheme: , , and
where , , and are three control sequences. They proved that converges strongly to .
A mapping is said to be relaxed  monotone if there exist a mapping and a function positively homogeneous of degree , that is, for all and such that
where is a constant; see [10]. In the case of for all , is said to be relaxed monotone. In the case of for all and , where and , is said to be monotone; see [11–13]. In fact, in this case, if , then is a strongly monotone mapping. Moreover, every monotone mapping is relaxed  monotone with for all and .
In this paper, we consider a new general equilibrium problem with a relaxed monotone mapping:
The set of all solutions of the equilibrium problem is denoted by , that is,
In the case of , (1.10) is deduced to
The set of all solutions of (1.12) is denoted by , that is,
In the case of , coincides with . In the case of and , coincides with .
In this paper, we introduce a new iterative scheme for finding a common element of the set of solutions of a general equilibrium problem with a relaxed monotone mapping and the set of common fixed points of a countable family of nonexpansive mappings and then obtain a strong convergence theorem. More precisely, we introduce the following iterative scheme:
where is a relaxed  monotone mapping, is a inversestrongly monotone mapping, and is a countable family of nonexpansive mappings such that , , and , , and are three control sequences. We prove that defined by (1.14) converges strongly to . Using the main result in this paper, we also prove several new strong convergence theorems for finding the elements of , , , and , respectively, where is a nonexpansive mapping.
2. Preliminaries
Let be a inversestrongly monotone mapping and let denote the identity mapping of . For all and , one has [6]
Hence, if , then is a nonexpansive mapping of into .
For each point , there exists a unique nearest point of , denoted by , such that for all . Such a is called the metric projection from onto . The wellknown Browder's characterization of ensures that is a firmly nonexpansive mapping from onto , that is,
Further, we know that for any and , if and only if
Let be a nonexpansive mapping of into itself such that . Then we have
which is obtained directly from the following:
This inequality is a very useful characterization of . Observe what is more that it immediately yields that is a convex closed set.
Let be a bifunction of into satisfying the following conditions:
for all ;
is monotone, that is, for all ;
for each , ;
for each , is convex and lower semicontinuous.
Definition 2.1 (see [10]).
Let be a Banach space with the dual space and let be a nonempty subset of . Let and be two mappings. The mapping is said to be hemicontinuous if, for any fixed , the function defined by is continuous at .
Lemma 2.2.
Let be a Hilbert space and let be a nonempty closed convex subset of . Let be an hemicontinuous and relaxed  monotone mapping. Let be a bifunction from to satisfying (A1) and (A4). Let and . Assume that
(i) for all ;
(ii)for any fixed , the mapping is convex.
Then the following problems (2.6) and (2.7) are equivalent:
Proof.
Let be a solution of the problem (2.6). Since is relaxed  monotone, we have
Thus is a solution of the problem (2.7).
Conversely, let be a solution of the problem (2.7). Letting
then . Since is a solution of the problem (2.7), it follows that
The conditions (i), (ii), (A1), and (A4) imply that
It follows from (2.10)(2.11) that
Since is hemicontinuous and , letting in (2.12), we get
for all . Therefore, is also a solution of the problem (2.6). This completes the proof.
Definition 2.3 (see [14]).
Let be a Banach space with the dual space and let be a nonempty subset of . A mapping is called a KKM mapping if, for any , , where denotes the family of all the nonempty subsets of .
Lemma 2.4 (see [14]).
Let be a nonempty subset of a Hausdorff topological vector space and let be a KKM mapping. If is closed in for all in and compact for some , then .
Next we use the concept of KKM mapping to prove two basic lemmas for our main result. The idea of the proof of the next lemma is contained in the paper of Fang and Huang [10].
Lemma 2.5.
Let be a real Hilbert space and be a nonempty bounded closed convex subset of . Let be an hemicontinuous and relaxed  monotone mapping, and let be a bifunction from to satisfying (A1) and (A4). Let . Assume that
(i) for all ;
(ii)for any fixed , the mapping is convex and lower semicontinuous;
(iii) is weakly lower semicontinuous; that is, for any net , converges to in which implies that .
Then problem (2.6) is solvable.
Proof.
Let . Define two setvalued mappings as follows:
We claim that is a KKM mapping. If is not a KKM mapping, then there exist and , , such that
By the definition of , we have
It follows from (A1), (A4), and (ii) that
which is a contradiction. This implies that is a KKM mapping.
Now, we prove that
For any given , taking , then
Since is relaxed  monotone, we have
It follows that and so
This implies that is also a KKM mapping. Now, since is a convex lowersemicontinuous function, we know that it is weakly lower semicontinuous. Thus from the definition of and the weak lower semicontinuity of , it follows that is weakly closed for all . Since is bounded closed and convex, we know that is weakly compact, and so is weakly compact in for each . It follows from Lemmas 2.2 and 2.4 that
Hence there exists such that
This completes the proof.
Lemma 2.6.
Let be a real Hilbert space and let be a nonempty bounded closed convex subset of . Let be an hemicontinuous and relaxed  monotone mapping and let be a bifunction from to satisfying (A1), (A2), and (A4). Let and define a mapping as follows:
for all . Assume that
(i), for all ;
(ii)for any fixed , the mapping is convex and lower semicontinuous and the mapping is lower semicontinuous;
(iii) is weakly lower semicontinuous;
(iv)for any , .
Then, the following holds:
(1) is singlevalued;
(2) is a firmly nonexpansive mapping, that is, for all ,
(3);
(4) is closed and convex.
Proof.
The fact that is nonempty is exactly the thesis of the previous lemma. We claim that is singlevalued. Indeed, for and , let . Then,
Adding the two inequalities, from (i) we have
From (A2), we have
that is,
Since is relaxed  monotone and , one has
In (2.29) exchanging the position of and , we get
that is,
Now, adding the inequalities (2.30) and (2.32), by using (iv) we have
Hence,
Next we show that is firmly nonexpansive. Indeed, for , we have
Adding the two inequalities and by (i) and (A2), we get
that is,
In (2.36) exchanging the position of and , we get
Adding the inequalities (2.36) and (2.37), we have
It follows from (iv) that
that is,
This shows that is firmly nonexpansive.
Next, we claim that . Indeed, we have the following:
Finally, we prove that is closed and convex. Indeed, Since every firm nonexpansive mapping is nonexpansive, we see that is nonexpansive from (2). On the other hand, since the set of fixed points of every nonexpansive mapping is closed and convex, we have that is closed and convex from (2) and (3). This completes the proof.
3. Main Results
In this section, we prove a strong convergence theorem which is our main result.
Theorem 3.1.
Let be a nonempty bounded closed convex subset of a real Hilbert space and let be a bifunction satisfying (A1), (A2), (A3), and (A4). Let be an hemicontinuous and relaxed  monotone mapping, let be a inversestrongly monotone mapping, and let be a countable family of nonexpansive mappings such that . Assume that the conditions (i)–(iv) of Lemma 2.6 are satisfied. Let and assume that is a strictly decreasing sequence. Assume that with some and with some . Then, for any , the sequence generated by (1.14) converges strongly to . In particular, if contains the origin 0, taking , then the sequence generated by (1.14) converges strongly to the minimum norm element in .
Proof.
We split the proof into following steps.
Step 1.
is closed and convex, the sequence generated by (1.14) is well defined, and for all .
First, we prove that is closed and convex. It suffices to prove that is closed and convex. Indeed, it is easy to prove the conclusion by the following fact:
This implies that . Noting that is a nonexpansive mapping for and the set of fixed points of a nonexpansive mapping is closed and convex, we have that is closed and convex.
Next we prove that the sequence generated by (1.14) is well defined and for all . It is easy to see that is closed and convex for all from the construction of . Hence, is closed and convex for all . For any , since and is nonexpansive, we have (note that is strictly decreasing)
So, for all . Hence , that is, for all . Since is closed, convex, and nonempty, the sequence is well defined.
Step 2.
and there exists such that as .
From the definition of , we see that for all and hence
Noting that , we get
for all . This shows that is increasing. Since is bounded, is bounded. So, we have that exists.
Noting that and for all , we have
It follows from (3.5) that
By taking in (3.6), we get
Since the limits of exists we get
that is, as . Moreover, from (3.6) we also have
This shows that is a Cauchy sequence. Hence, there exists such that
Step 3.
Since and as , we have
and hence
Note that can be rewritten as for all . Take . Since , is inversestrongly monotone, and , we know that, for all ,
Using (1.14) and (3.13), we have (note that is strictly decreasing)
and hence
Since and are both bounded, , and , we have
Using Lemma 2.6, we get
So, we have
From (3.18), we have
and hence
By using and (3.16), we have
Step 4.
, for all
It follows from the definition of scheme (1.14) that
that is,
Hence, for any , one has
Since each is nonexpansive, by (2.4) we have
Hence, combining this inequality with (3.24), we get
that is (noting that is strictly decreasing),
Since and , we have
Step 5.
.
First we prove . Indeed, since and , we have for each . Hence, .
Next, we show that . Noting that , one obtains
Put for all and . Then, we have . So, from (A2), (i), and (3.29) we have
Since , we have . Further, from monotonicity of , we have . So, from (A4), (ii), and hemicontinuity of we have
From (A1), (A4), (ii), and (3.31) we also have
and hence
Letting , from (A3) and (ii) we have, for each ,
This implies that . Hence, we get .
Finally, we show that . Indeed, from and , we have
Taking the limit in (3.35) and noting that as , we get
In view of (2.3), one sees that . This completes the proof.
Corollary 3.2.
Let be a nonempty bounded closed convex subset of a Hilbert space and let be a bifunction satisfying (A1), (A2), (A3), and (A4). Let be an hemicontinuous and relaxed  monotone mapping and let be a nonexpansive mapping such that . Assume that the conditions (i)–(iv) of Lemma 2.6 are satisfied. Assume that with , with some and with . Let and let be generated by
Then the sequence converges strongly to . In particular, if contains the origin 0, taking , the sequence converges strongly to the minimum norm element in .
Proof.
In Theorem 3.1, put , . Then, we have
On the other hand, for all , we have that
So, taking with and choosing a sequence of real numbers with , we obtain the desired result by Theorem 3.1.
Corollary 3.3.
Let be a nonempty bounded closed convex subset of a Hilbert space and let be a bifunction satisfying (A1), (A2), (A3), and (A4). Let be a monotone mapping and let be a nonexpansive mapping such that . Assume that with , with some and with . Let and let be generated by
Then the sequence converges strongly to . In particular, if contains the origin 0, taking , the sequence converges strongly to the minimum norm element in .
Proof.
In Corollary 3.2, put and for all . Then is a monotone mapping and we obtain the desired result by Theorem 3.1.
Corollary 3.4.
Let be a closed convex subset of a Hilbert space and let be a bifunction satisfying (A1), (A2), (A3), and (A4). Let be a inversestrongly monotone mapping and let be a nonexpansive mapping such that . Assume that with , with some and with . Let and let be generated by
Then the sequence converges strongly to . In particular, if contains the origin 0, taking , the sequence converges strongly to the minimum norm element in .
Proof.
In Theorem 3.1, put , , , and . We obtain the desired result by Theorem 3.1.
Corollary 3.5.
Let be a closed convex subset of a Hilbert space and let be a bifunction satisfying (A1), (A2), (A3), and (A4). Let be a nonexpansive mapping such that . Assume that with , with some , and with . Let and let be generated by
Then the sequence converges strongly to . In particular, if contains the origin 0, taking , the sequence converges strongly to the minimum norm element in .
Proof.
In Corollary 3.4, by putting we obtain the desired result.
Corollary 3.6.
Let be a closed convex subset of a Hilbert space and let be a inversestrongly monotone mapping. Let be a nonexpansive mapping such that . Assume that with , with some , and with . Let and let be generated by
Then the sequence converges strongly to . In particular, if contains the origin 0, taking , the sequence converges strongly to the minimum norm element in .
Proof.
In Theorem 3.1, put , , , , and . Then, we have
Then, we obtain the desired result by Theorem 3.1.
Remark 3.7.
The novelty of this paper lies in the following aspects.
(i)A new general equilibrium problem with a relaxed monotone mapping is considered.
(ii)The definition of is of independent interest.
Acknowledgment
This work was supported by the Natural Science Foundation of Hebei Province (A2010001482).
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