Abstract
Let 
be a nonempty closed convex subset of a Banach space 
with the dual 
, let 
be a continuous mapping, and let 
be a relatively nonexpansive mapping. In this paper, by employing the notion of generalized projection operator we study the variational inequality (for short, VI(
): find 
such that 
for all 
, where 
is a given element. By combining the approximate proximal point scheme both with the modified Ishikawa iteration and with the modified Halpern iteration for relatively nonexpansive mappings, respectively, we propose two modified versions of the approximate proximal point scheme L. C. Ceng and J. C. Yao (2008) for finding approximate solutions of the VI(
). Moreover, it is proven that these iterative algorithms converge strongly to the same solution of the VI(
), which is also a fixed point of 
.
be a real Banach space with the dual 
. As usually, 
denotes the duality pairing between 
and 
. In particular, if 
is a real Hilbert space, then 
denotes its inner product. Let 
be a nonempty closed convex subset of 
and 
be a mapping. Given 
, let us consider the following variational inequality problem (for short, 
): find an element 
such that
(1.1) has a (unique) solution 
. For any 
, define the following successive sequence in a uniformly convex and uniformly smooth Banach space 
:
is the normalized duality mapping on 
and 
is the generalized projection operator which assigns to an arbitrary point 
the minimum point of the functional 
with respect to 
. In [
, that is, 
as 
, if the following conditions hold:
is uniformly monotone, that is,
is a continuous strictly increasing function for all 
with 
;
has 
arbitrary growth, that is,
is a continuous nondecreasing function for all 
with 
. Note that solution methods for the problem (1.1) has also been studied in [
be a nonempty closed convex subset of a real Banach space 
with the dual 
. Assume that 
is a continuous mapping on 
and 
is a relatively nonexpansive mapping such that 
. The purpose of this paper is to introduce and study two new iterative algorithms (1.5) and (1.6) in a uniformly convex and uniformly smooth Banach space 
.
are sequences in 
, 
is a bounded sequence in 
, and 
is assumed to exist for each 
, 
is a sequence in 
, 
is a bounded sequence in 
, and 
is assumed to exist for each 
, 
.
generated by algorithm (1.5) and the sequence 
generated by algorithm (1.6) converge strongly to the same point 
, which is a solution of the 
. Our results represent the improvement, generalization, and development of the previously known results in the literature including Li [
for strong convergence.
be a Banach space with the dual 
. We denote by 
the normalized duality mapping from 
to 
defined by
denotes the generalized duality pairing. It is well known that if 
is smooth, then 
is single-valued and if 
is uniformly smooth, then 
is uniformly norm-to-norm continuous on bounded subsets of 
. We will still denote the single-valued duality mapping by 
.
is a nonempty closed convex subset of a Hilbert space 
and 
is the metric projection of 
onto 
, then 
is nonexpansive. This fact actually characterizes Hilbert spaces and hence, it is not available in more general Banach spaces. In this connection, Alber [
in a Banach space 
which is an analogue of the metric projection in Hilbert spaces.
is a smooth Banach space. Consider the functional defined as in [
, (2.2) reduces to 
.
is a mapping that assigns to an arbitrary point 
the minimum point of the functional 
; that is, 
, where 
is the solution to the minimization problem
follow from the properties of the functional 
and strict monotonicity of the mapping 
(see, e.g., [
.
be a closed convex subset of 
, and let 
be a mapping from 
into itself. A point 
in 
is called an asymptotically fixed point of 
[
contains a sequence 
which converges weakly to 
such that 
. The set of asymptotical fixed points of 
will be denoted by 
. A mapping 
from 
into itself is called relatively nonexpansive [
and 
for all 
and 
.
is called strictly convex if 
for all 
with 
and 
. It is said to be uniformly convex if 
for any two sequences 
such that 
and 
. Let 
be a unit sphere of 
. Then the Banach space 
is called smooth if
. It is also said to be uniformly smooth if the limit is attained uniformly for 
. Recall also that if 
is uniformly smooth, then 
is uniformly norm-to-norm continuous on bounded subsets of 
. A Banach space is said to have the Kadec-Klee property if for any sequence 
, whenever 
and 
, we have 
. It is known that if 
is uniformly convex, then 
has the Kadec-Klee property; see [
is a reflexive, strictly convex, and smooth Banach space, then for any 
, 
if and only if 
. It is sufficient to show that if 
, then 
. From (2.4), we have 
. This implies that 
. From the definition of 
, we have 
. Therefore, we have 
; see [
be a uniformly convex and smooth Banach space and let 
and 
be two sequences of 
. If 
and either 
or 
is bounded, then 
.
be a nonempty closed convex subset of a smooth Banach space 
and 
. Then, 
if and only if
be a reflexive, strictly convex, and smooth Banach space, let 
be a nonempty closed convex subset of 
and let 
. Then
be a strictly convex and smooth Banach space, let 
be a closed convex subset of 
, and let 
be a relatively nonexpansive mapping from 
into itself. Then 
is closed and convex.
be a smooth Banach space. Then the following inequality holds
be a uniformly convex and uniformly smooth Banach space, let 
be a nonempty closed convex subset of 
, let 
be a continuous mapping and, let 
be a relatively nonexpansive mapping such that 
. Assume that 
are sequences in 
and 
is a sequence in 
such that 
and 
. Define a sequence 
in 
by the following algorithm:
is assumed to exist for each 
, 
If 
is uniformly continuous and 
, then 
converges strongly to 
, which is a solution of the 
(1.1).
and 
are closed and convex for each 
. Indeed, from the definition of 
and 
, it is obvious that 
is closed and 
is closed and convex for each 
. We claim that 
is convex. For any 
and any 
, put 
. It is sufficient to show that 
. Note that the inequality
. Thus, we have
. So, 
is convex. Next let us show that 
for all 
. Indeed, we have for all 
for all 
. Next let us show that
, we have 
. Assume that 
. Since 
is the projection of 
onto 
, by Lemma 2.3, we have
by the induction assumption, the last inequality holds, in particular, for all 
. This together with the definition of 
implies that 
. Hence (3.7) holds for all 
. This implies that 
is well defined.
that 
. Since 
, we have
is nondecreasing. Also from 
and Lemma 2.4, it follows that
for each 
. Consequently, 
is bounded. Moreover, according to the inequality
is bounded and so is 
. Indeed, since 
is relatively nonexpansive, we derive for each 
is bounded. Again from 
we know that 
is also bounded.
we deduce that 
exists. From Lemma 2.4, we derive
. This implies that 
. So it follows from Lemma 2.2 that 
. Since 
, from the definition of 
, we also have
. Thus, in terms of Lemmas 2.4 and 2.6 we conclude that
and 
, we obtain 
. Thus by Lemma 2.2 we have 
. From 
it follows that 
is bounded. At the same time, observe that
and the boundedness of 
and 
, we derive 
. Note that 
is uniformly continuous. Hence 
by virtue of 
. Since
is bounded. Consequently, from (3.15), 
and 
it follows that
and 
that
is uniformly norm-to-norm continuous on bounded subsets of 
, we have
and 
that
, we obtain
is also uniformly norm-to-norm continuous on bounded subsets of 
, we obtain
is uniformly continuous, it follows from (3.26), (3.30) and 
that 
.
converges strongly to 
, which is a solution of the 
(1.1). Indeed, assume that 
is a subsequence of 
such that 
. Then 
. Next let us show that 
and convergence is strong. Put 
. From 
and 
, we have 
. Now from weakly lower semicontinuity of the norm, we derive
that 
and hence
. Utilizing the Kadec-Klee property of 
, we conclude that 
converges strongly to 
. Since 
is an arbitrarily weakly convergent subsequence of 
, we know that 
converges strongly to 
. Now observe that from (3.1) we have for each 
is uniformly norm-to-norm continuous on bounded subsets of 
, from 
we infer that 
. Noticing that 
and 
is a continuous mapping, we obtain that 
and 
. Therefore, from 
it follows that
we conclude from (3.34) that
is a solution of the 
(1.1). This completes the proof.
be a uniformly convex and uniformly smooth Banach space, let 
be a nonempty closed convex subset of 
, and let 
be a relatively nonexpansive mapping such that 
. Assume that 
and 
are sequences in 
such that 
and 
. Define a sequence 
in 
by the following algorithm:
is the single-valued duality mapping on 
. If 
is uniformly continuous, then 
converges strongly to 
.
. Now, put 
for all 
. Then we have
. Thus algorithm (3.1) reduces to algorithm (3.39). By Theorem 3.1 we obtain the desired result.
be a uniformly convex and uniformly smooth Banach space, let 
be a nonempty closed convex subset of 
, let 
be a continuous mapping, and let 
be a relatively nonexpansive mapping such that 
. Assume that 
satisfies 
and 
satisfies 
. Define a sequence 
in 
by the following algorithm:
is assumed to exist for each 
, 
If 
is uniformly continuous and 
, then 
converges strongly to 
, which is a solution of the 
(1.1).
is closed and convex for each 
. From the definition of 
, it is obvious that 
is closed for each 
. We prove that 
is convex. Similarly to the proof of Theorem 3.1, since
is convex. Next, let us show that 
for each 
. Indeed, we have for each 
for all 
and 
. Similarly to the proof of Theorem 3.1, we also obtain 
for all 
. Consequently, 
for all 
. Therefore, the sequence 
generated by (3.42) is well defined. As in the proof of Theorem 3.1, we can obtain 
. Since 
, from the definition of 
, we also have
and 
that 
and hence 
by Lemma 2.2. Further, it follows from 
and the boundedness of 
and 
that
, from the definition of 
, we also have
that
is uniformly norm-to-norm continuous on bounded subsets of 
we have
we have
is also uniformly norm-to-norm continuous on bounded subsets of 
, we obtain
is uniformly continuous, it follows from (3.50) and (3.54) that 
.
converges strongly to 
, which is a solution of the 
(1.1). Indeed, assume that 
is a subsequence of 
such that 
. Then 
. Next let us show that 
and convergence is strong. Put 
. From 
and 
, we have 
. Now from weakly lower semicontinuity of the norm, we derive
that 
and hence 
. So, we have 
. Utilizing the Kadec-Klee property of 
, we conclude that 
converges strongly to 
. Since 
is an arbitrarily weakly convergent subsequence of 
, we know that 
converges strongly to 
. Now observe that from (3.1), we have for each 
is uniformly norm-to-norm continuous on bounded subsets of 
, from 
we infer that 
. Noticing that 
and 
is a continuous mapping, we obtain that 
and 
. Observe that
that
we conclude from (3.34) that
is a solution of the 
(1.1). This completes the proof.
be a uniformly convex and uniformly smooth Banach space, let 
be a nonempty closed convex subset of 
, and let 
be a relatively nonexpansive mapping. Assume that 
is a sequence in 
such that 
. Define a sequence 
in 
by the following algorithm:
is the single-valued duality mapping on 
. If 
is nonempty, then 
converges strongly to 
.
. Now, put 
for all 
. Then we have
. Thus algorithm (3.42) reduces to algorithm (3.61). Thus under the lack of the uniform continuity of 
it follows from (3.55) that 
. By the careful analysis of the proof of Theorem 3.3, we can obtain the desired result.