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 inverse-strongly monotone if there exists such that

Such a mapping is also called -inverse-strongly monotone. It is known that each nonexpansive mapping is -inverse-strongly monotone and each -strictly pseudocontraction is -inverse-strongly 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 -inverse-strongly 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.

Let be a -inverse-strongly 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 well-known 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 set-valued 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 lower-semicontinuous 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 single-valued;

(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 single-valued. 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.

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 -inverse-strongly 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 -inverse-strongly 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 -inverse-strongly 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 -inverse-strongly 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).