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.
Let , ,,,,μ, and be as those in Theorem 3.2. If , and there exists a bounded sequence such that
By the uniqueness of and with no loss of generality, we can choose such that
as . Let be the fixed point of the contraction
Now, by using Lemma 2.3, we have
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).
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:
(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:
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.
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.
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.
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.
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
So, from (4.20), (4.21), and (4.22), we get
and it completes the proof.
According to Corollaries 4.3 and 4.4, our assumptions are weaker than those of Song and Xu . 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:
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 .