### 4.1. -Mappings

Let be infinite mappings of into itself, and let be a nonnegative real sequence with , . For any , define a mapping of into itself as follows:

Nonexpansivity of each ensures the nonexpansivity of . The mapping is called a -mapping generated by and .

Throughout this section, we will assume that . Concerning defined by (4.1), we have the following useful lemmas.

Lemma 4.1 (see [4]).

Let be a nonempty closed convex subset of a a strictly convex, reflexive Banach space , a family of infinitely nonexpansive mapping with , and a real sequence such that , , then:

(1) is nonexpansive and for each ;

(2)for each and for each positive integer , the limit exists;

(3)the mapping define by

is a nonexpansive mapping satisfying , and it is called the -mapping generated by and .

From Remark 3.1 of Peng and Yao [29], we obtain the following lemma.

Lemma 4.2.

Let be a strictly convex, reflexive Banach space, a family of infinitely nonexpansive mappings with , and a real sequence such that , . Then sequence satisfies the -condition.

Applying Lemma 4.2 and Theorem 3.1, we obtain the following result.

Theorem 4.3.

Let be a reflexive Banach space which admits a weakly continuous duality mapping with gauge such that is invariant on . Let be a countable family of nonexpansive mappings with and an -contraction and a strongly positive bounded linear operator with coefficient and . Let the sequence be generated by the following:

where is defined by (4.1) and is a sequence in satisfying the conditions (C1), (C2), and (C3). Then converges strongly to in .

Applying Lemma 4.2 and Theorem 3.3, we obtain the following result.

Theorem 4.4.

Let , , , , , and be as in Theorem 4.3. Let the sequence be generated by the following:

then converges strongly to in .

Applying Lemma 4.2 and Theorem 3.4, we obtain the following result.

Theorem 4.5.

Let , , , , , and be as in Theorem 4.3. Let the sequence be generated by the following:

then converges strongly to in .

### 4.2. Accretive Operators

We consider the problem of finding a zero of an accretive operator. An operator is said to be accretive if for each and , there exists such that . An accretive operator is said to satisfy the range condition if for all , where is the domain of , is the identity mapping on , is the range of , and is the closure of . If is an accretive operator which satisfies the range condition, then we can define, for each , a mapping by , which is called the resolvent of . We know that is nonexpansive and for all . We also know the following [30]: for each and , it holds that

From the Resolvent identity, we have the following lemma.

Lemma 4.6.

Let be a Banach space and a nonempty closed convex subset of . Let be an accretive operator such that and . Suppose that is a sequence of such that and , then

(i)the sequence satisfies AKTT-condition,

(ii) for all and , where as .

Proof.

By the proof of Theorem 4.3 in [1] and applying Lemma 4.6 and Theorem 3.1, we obtain the following result.

Theorem 4.7.

Let be a reflexive Banach space which admits a weakly continuous duality mapping with gauge such that is invariant on . Let be an accretive operator such that . Assume that is a nonempty closed convex subset of such that and is an -contraction. Let be a strongly positive bounded linear operator with coefficient and . Suppose that is a sequence of such that . Let the sequence be generated by the following:

where is a sequence in satisfying the following conditions (C1), (C2), and (C3), then converges strongly to in .

Applying Lemma 4.6 and Theorem 3.3, we obtain the following result.

Theorem 4.8.

Let , , , , , , and be as in Theorem 4.7. Let be generated by the following:

then converges strongly to in .

Applying Lemma 4.6 and Theorem 3.4, we obtain the following result.

Theorem 4.9.

Let , , , , , , and be as in Theorem 4.7. Let be generated by the following:

Then converges strongly to in .

### 4.3. The Equilibrium Problems

Let be a real Hilbert space, and let be a bifunction of , where is the set of real numbers. The equilibrium problem for is to find such that

The set of solutions of (4.10) 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 (4.10). Some methods have been proposed to solve the equilibrium problem; see, for instance, Blum and Oettli [31] and Combettes and Hirstoaga [32]. For the purpose of 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 lemmas were also given in [31, 32], respectively.

Lemma 4.10 (see [31, Corollary 1]).

Let be a nonempty closed convex subset of , and let be a bifunction of satisfying . Let and , then there exists such that .

Lemma 4.11 (see [32, Lemma 2.12]).

Assume that satisfies . For and , define a mapping as follows:

then, the following hold:

(1) is single valued,

(2) is firmly nonexpansive, that is, for any ,

(3),

(4) is closed and convex.

Theorem 4.12.

Let be a real Hilbert space. Let be a bifunction from satisfying (A1)–(A4) and . Let be an -contraction, a strongly positive bounded linear operator with coefficient and . Let the sequences be generated by and

for all , where is a sequence in and satisfying the following conditions:

(C1),

(C2),

(C3),

(C4) and .

then and converge strongly to .

Proof.

Following the proof technique of Theorem 3.1, we only need, show that , for all . From (4.12), it follows that

On the other hand, from the definition of we have

Putting and in (4.14), we have

So, from (A2), we have

and hence,

then we have

and hence,

where is a constant satisfying . Substituting (4.19) in (4.13) yields

for some with (the definition ). By the assumptions on and and using Lemma 2.2, we conclude that

From the definition of and , it follows that

Combining (4.21) and (4.22), we have

From the definition of , it follows that

Putting in (4.14) and in (4.24), we have

So, from (A2), we have

and hence, for each ,

then

Since

then for each , we have from (4.23)

This completes the proof.

Applying Theorem 4.12, we can obtain the following result.

Corollary 4.13.

Let be a real Hilbert space. Let be a bifunction from satisfying (A1)–(A4) and . Let be an -contraction, a strongly positive bounded linear operator with coefficient and . Let the sequences be generated by and

for all , where is a sequence in and satisfying the following conditions:

(C1),

(C2),

(C3),

(C4) and ,

then and converge strongly to .

Proof.

We observe that for all . Then we rewrite the iterative sequence (4.31) by the following:

Let be the sequence given by and

Form Theorem 4.12, in . We claim that . Applying Lemma 2.6, we estimate

It follows from , , and Lemma 2.2 that as . Consequently, as required.