In order to prove our main results, we need the following lemmas.
Lemma 2.1 (see [13]).
Let
be bounded sequences in a Banach space
and
be a sequence in
which satisfies the following condition:
. Suppose that
for all
and
. Then,
.
Lemma 2.2 (see Xu [14]).
Assume that
is a sequence of nonnegative real numbers such that
, where
is a sequence in (0, 1) and
is a sequence in
such that
(i)
,
(ii)
or
.
Then
.
Lemma 2.3 (see [15] demiclosedness principle).
Let
be a nonempty closed convex subset of a reflexive Banach space
which satisfies Opial's condition, and suppose
is nonexpansive. Then the mapping
is demiclosed at zero, that is,
,
implies
.
Lemma 2.4 (see [16, Lemmas 3.1, 3.3]).
Let
be real smooth and strictly convex Banach space, and
be a nonempty closed convex subset of
which is also a sunny nonexpansive retraction of
. Assume that
is a nonexpansive mapping and
is a sunny nonexpansive retraction of
onto
, then
.
Lemma 2.5 (see [17, Lemma 2.2]).
Let
be a nonempty convex subset of a real
-uniformly smooth Banach space
and
be a
-strict pseudocontraction. For
, we define
. Then, as
,
,
is nonexpansive such that
.
Lemma 2.6 (see [12, Remark 2.6]).
When
is non-self-mapping, the Lemma 2.5 also holds.
Lemma 2.7 (see [12, Lemma 2.8]).
Assume that
is a strongly positive linear bounded operator on a smooth Banach space
with coefficient
and
. Then,
Lemma 2.8 (see [18, Lemma 2.3]).
Let
be an MKC on a convex subset
of a Banach space
. Then for each
, there exists
such that
Lemma 2.9.
Let
be a closed convex subset of a reflexive Banach space
which admits a weakly sequentially continuous duality mapping
from
to
. Let
be a nonexpansive mapping with
and
be a MKC,
is strongly positive linear bounded operator with coefficient
. Assume that
. Then the sequence
define by
converges strongly as
to a fixed point
of
which solves the variational inequality:
Proof.
The definition of
is well definition. Indeed, from the definition of MKC, we can see MKC is also a nonexpansive mapping. Consider a mapping
on
defined by
It is easy to see that
is a contraction. Indeed, by Lemma 2.8, we have
Hence,
has a unique fixed point, denoted by
, which uniquely solves the fixed point equation
We next show the uniqueness of a solution of the variational inequality (2.3). Suppose both
and
are solutions to (2.3), not lost generality, we may assume there is a number
such that
. Then by Lemma 2.8, there is a number
such that
. From (2.3), we know
Adding up (2.7) gets
Noticing that
Therefore
and the uniqueness is proved. Below, we use
to denote the unique solution of (2.3).
We observe that
is bounded. Indeed, we may assume, with no loss of generality,
, for all
, fixed
, for each
.
Case 1 (
).
In this case, we can see easily that
is bounded.
Case 2 (
).
In this case, by Lemmas 2.7 and 2.8, there is a number
such that
therefore,
. This implies the
is bounded.
To prove that
as
.
Since
is bounded and
is reflexive, there exists a subsequence
of
such that
. By
. We have
, as
. Since
satisfies Opial's condition, it follows from Lemma 2.3 that
. We claim
By contradiction, there is a number
and a subsequence
of
such that
. From Lemma 2.8, there is a number
such that
, we write
to derive that
It follows that
Therefore,
Using that the duality map
is single valued and weakly sequentially continuous from
to
, by (2.15), we get that
. It is a contradiction. Hence, we have
.
We next prove that
solves the variational inequality (2.3). Since
we derive that
Notice
It follows that, for
,
Now replacing
in (2.19) with
and letting
, noticing
for
, we obtain
. That is,
is a solution of (2.3); Hence
by uniqueness. In a summary, we have shown that each cluster point of
(at
) equals
, therefore,
as
.
Lemma 2.10 . (see, e.g., Mitrinović [19, page 63]).
Let
. Then the following inequality holds:
for arbitrary positive real numbers
,
.
Lemma 2.11.
Let
be a
-uniformly smooth Banach space which admits a weakly sequentially continuous duality mapping
from
to
and
be a nonempty convex subset of
. Assume that
is a countable family of
-strict pseudocontraction for some
and
such that
. Assume that
is a positive sequence such that
. Then
is a
-strict pseudocontraction with
and
.
Proof.
Let
and
. Then,
is a
-strict pseudocontraction with
. Indeed, we can firstly see the case of
.
which shows that
is a
-strict pseudocontraction with
. By the same way, our proof method easily carries over to the general finite case.
Next, we prove the infinite case. From the definition of
-strict pseudocontraction, we know
Hence, we can get
Taking
, from (2.24), we have
Consquently, for all
, if
,
and
, then
strongly converges. Let
we have
Hence,
So, we get
is
-strict pseudocontraction.
Finally, we show
. Suppose that
, it is sufficient to show that
. Indeed, for
, we have
where
. Hence,
for each
, this means that
.