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
.