In this section, we will present our result of the strong convergence of (1.9), but first, we need to prove, with different approach, the following lemma.

Lemma 4.1.

Let , ,,,,*μ*, and be as those in Theorem 3.2. If , and there exists a bounded sequence such that

Then,

Proof.

By the uniqueness of and with no loss of generality, we can choose such that

as . Let be the fixed point of the contraction

Then,

Now, by using Lemma 2.3, we have

Therefore,

Because , and are bounded, from (4.3) and (4.7), we conclude that

Moreover, we have

By Theorem 3.2, , as . So, using the boundedness of , we get

On the other hand, noticing that the sequence is bounded and the duality mapping is single-valued and norm to weak* uniformly continuous on bounded subsets of , we conclude that

Therefore, from (4.8) and (4.9), we obtain

This completes the proof.

Next, we prove the strong convergence of explicit iteration scheme (1.9).

Theorem 4.2.

Let be a real Banach space with a uniformly Gateaux differentiable norm, and let be a nonexpansive semigroup from into itself. Let also defined by (1.9) satisfies the following conditions:

(i),

(ii) for all .

Assume that is -strongly monotone and -Lipschitzian, and that is a sequence of positive numbers that . Assume also that the sequence in satisfies the control condition

If , then converges strongly to some fixed point , which is the unique solution in for the following variational inequality:

Proof.

Existence and uniqueness of the solution of is attained from Theorem 3.2. Now, we claim that is bounded. Indeed, taking a fixed , we have

Taking , , , and using Lemma 2.2, we conclude that is bounded and so is .

Next, we prove that converges strongly to the unique solution of . By definition of the algorithm and taking , we have

Taking , , and and using Lemma 4.1 together with Lemma 2.2 lead to , that is, in norm. This completes the proof.

Corollary 4.3.

Let be a real reflexive strictly convex Banach space with a uniformly Gateaux differentiable norm. Let also be a nonexpansive semigroup from into itself such that . Assume that defined by (1.9) satisfies condition in Theorem 4.2, then condition holds.

Proof.

Clearly, is a convex and continuous function. Because is a reflexive Banach space, according to [12, Theorem ], is nonempty. Also, by convexity and continuity of , the set is a closed convex subset of . Since for every and , we have

So, , and therefore . Suppose that . Because every nonempty closed convex subset of a strictly convex and reflexive Banach space is a Chebyshev set, according to [14, Corollary ], there exists a unique such that

On the other hand, for all , and is nonexpansive, so we get

that is, , by uniqueness of . Thus, . This completes the proof.

Corollary 4.4.

Let be a real Banach space, and let be a nonexpansive uniformly asymptotically regular semigroup from into itself. If is defined by (1.9), where satisfies , then condition in Theorem 4.2 holds.

Proof.

From (1.9), , and the boundedness of , we conclude that

as . On the other hand, the semigroup is uniformly asymptotically regular, , and is a bounded subset in , so for all , we have

Hence,

So, from (4.20), (4.21), and (4.22), we get

and it completes the proof.

Remark 4.5.

According to Corollaries 4.3 and 4.4, our assumptions are weaker than those of Song and Xu [11]. Also, noticing that for a contraction , the mapping is strongly monotone and Lipschitzian. So, by replacing by in (1.8) and (1.9), the following schemes are, respectively, obtained:

Remark 4.6.

In the same way and with the same conditions mentioned in Theorem 4.2, it's easy to see that the sequence defined by

converges strongly to the variational inequality .