Research Article
Open Access
Hybrid Approximate Proximal Point Algorithms for Variational Inequalities in Banach Spaces
- L. C. Ceng1, 2,
- S. M. Guu3 and
- J. C. Yao4Email author
Journal of Inequalities and Applications20092009:275208
https://doi.org/10.1155/2009/275208
© L. C. Ceng et al. 2009
Received: 10 March 2009
Accepted: 7 June 2009
Published: 14 July 2009
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
.
1. Introduction
Let
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
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)
Suppose that the
(1.1) has a (unique) solution
. For any
, define the following successive sequence in a uniformly convex and uniformly smooth Banach space
:
(1.1) has a (unique) solution
. For any
, define the following successive sequence in a uniformly convex and uniformly smooth Banach space
:
(1.2)
where
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 [1, Theorem?8.2], Alber proved that the above sequence converges strongly to the solution
, that is,
as
, if the following conditions hold:
(i)
is uniformly monotone, that is,
is uniformly monotone, that is,
(1.3)
where
is a continuous strictly increasing function for all
with
;
(ii)
has
arbitrary growth, that is,
has
arbitrary growth, that is,
(1.4)
where
is a continuous nondecreasing function for all
with
. Note that solution methods for the problem (1.1) has also been studied in [2–10].
Let
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
.
Algorithm 1.1.
(1.5)
where
are sequences in
,
is a bounded sequence in
, and
is assumed to exist for each
,
Algorithm 1.2.
(1.6)
where
is a sequence in
,
is a bounded sequence in
, and
is assumed to exist for each
,
.
In this paper, strong convergence results on these two iterative algorithms are established; that is, under appropriate conditions, both the sequence
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 [8], Zeng and Yao [9], Ceng and Yao [10], and Qin and Su [11].
Notation 1.
for strong convergence.2. Preliminaries
Let
be a Banach space with the dual
. We denote by
the normalized duality mapping from
to
defined by
be a Banach space with the dual
. We denote by
the normalized duality mapping from
to
defined by
(2.1)
where
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
.
Recall that if
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 [1] recently introduced a generalized projection operator
in a Banach space
which is an analogue of the metric projection in Hilbert spaces.
Next, we assume that
is a smooth Banach space. Consider the functional defined as in [1, 12] by
is a smooth Banach space. Consider the functional defined as in [1, 12] by
(2.2)
It is clear that in a Hilbert space
, (2.2) reduces to
.
The generalized projection
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
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
(2.3)
The existence and uniqueness of the operator
follow from the properties of the functional
and strict monotonicity of the mapping
(see, e.g., [13]). In a Hilbert space,
.
From [1], in uniformly convex and uniformly smooth Banach spaces, we have
(2.4)
Let
be a closed convex subset of
, and let
be a mapping from
into itself. A point
in
is called an asymptotically fixed point of
[14] if
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 [15–17] if
and
for all
and
.
A Banach space
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
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
(2.5)
exists for each
. 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 [18, 19] for more details.
Remark 2.1 ([11]).
If
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 [18, 19] for more details.
We need the following lemmas and proposition for the proof of our main results.
Lemma 2.2 (Kamimura and Takahashi [20]).
Let
be a uniformly convex and smooth Banach space and let
and
be two sequences of
. If
and either
or
is bounded, then
.
Lemma 2.3 (Alber [1]).
Let
be a nonempty closed convex subset of a smooth Banach space
and
. Then,
if and only if
be a nonempty closed convex subset of a smooth Banach space
and
. Then,
if and only if
(2.6)
Lemma 2.4 (Alber [1]).
Let
be a reflexive, strictly convex, and smooth Banach space, let
be a nonempty closed convex subset of
and let
. Then
be a reflexive, strictly convex, and smooth Banach space, let
be a nonempty closed convex subset of
and let
. Then
(2.7)
Lemma 2.5 (Matsushita and Takahashi [21]).
Let
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.
Lemma 2.6 (Chang [7]).
Let
be a smooth Banach space. Then the following inequality holds
be a smooth Banach space. Then the following inequality holds
(2.8)
3. Main Results
Now we are in a position to prove the main theorems of this paper.
Theorem 3.1.
Let
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:
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:
(3.1)
where
is assumed to exist for each
,
If
is uniformly continuous and
, then
converges strongly to
, which is a solution of the
(1.1).
Proof.
First of all, let us show that
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
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
(3.2)
is equivalent to the one
(3.3)
Observe that there hold the following:
(3.4)
and
. Thus, we have
. Thus, we have
(3.5)
This implies that
. So,
is convex. Next let us show that
for all
. Indeed, we have for all
. So,
is convex. Next let us show that
for all
. Indeed, we have for all
(3.6)
So
for all
. Next let us show that
for all
. Next let us show that
(3.7)
We prove this by induction. For
, we have
. Assume that
. Since
is the projection of
onto
, by Lemma 2.3, we have
, we have
. Assume that
. Since
is the projection of
onto
, by Lemma 2.3, we have
(3.8)
As
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.
On the other hand, it follows from the definition of
that
. Since
, we have
(3.9)
Thus
is nondecreasing. Also from
and Lemma 2.4, it follows that
is nondecreasing. Also from
and Lemma 2.4, it follows that
(3.10)
for each
for each
. Consequently,
is bounded. Moreover, according to the inequality
for each
. Consequently,
is bounded. Moreover, according to the inequality
(3.11)
we conclude that
is bounded and so is
. Indeed, since
is relatively nonexpansive, we derive for each
is bounded and so is
. Indeed, since
is relatively nonexpansive, we derive for each
(3.12)
and hence
is bounded. Again from
we know that
is also bounded.
On account of the boundedness and nondecreasing property of
we deduce that
exists. From Lemma 2.4, we derive
(3.13)
for all
. This implies that
. So it follows from Lemma 2.2 that
. Since
, from the definition of
, we also have
. This implies that
. So it follows from Lemma 2.2 that
. Since
, from the definition of
, we also have
(3.14)
Observe that
(3.15)
On the other hand, since from (3.1) we have for each
(3.16)
utilizing Lemma 2.3 we obtain
. Thus, in terms of Lemmas 2.4 and 2.6 we conclude that
. Thus, in terms of Lemmas 2.4 and 2.6 we conclude that
(3.17)
Since
and
, we obtain
. Thus by Lemma 2.2 we have
. From
it follows that
is bounded. At the same time, observe that
and
, we obtain
. Thus by Lemma 2.2 we have
. From
it follows that
is bounded. At the same time, observe that
(3.18)
and hence
(3.19)
From
and the boundedness of
and
, we derive
. Note that
is uniformly continuous. Hence
by virtue of
. Since
and the boundedness of
and
, we derive
. Note that
is uniformly continuous. Hence
by virtue of
. Since
(3.20)
it is known that
is bounded. Consequently, from (3.15),
and
it follows that
is bounded. Consequently, from (3.15),
and
it follows that
(3.21)
Further, it follows from (3.14),
and
that
and
that
(3.22)
Utilizing Lemma 2.2, we obtain
(3.23)
Since
is uniformly norm-to-norm continuous on bounded subsets of
, we have
is uniformly norm-to-norm continuous on bounded subsets of
, we have
(3.24)
Furthermore, we have
(3.25)
It follows from
and
that
and
that
(3.26)
Noticing that
(3.27)
we have
(3.28)
From (3.24) and
, we obtain
, we obtain
(3.29)
Since
is also uniformly norm-to-norm continuous on bounded subsets of
, we obtain
is also uniformly norm-to-norm continuous on bounded subsets of
, we obtain
(3.30)
Observe that
(3.31)
Since
is uniformly continuous, it follows from (3.26), (3.30) and
that
.
Finally, let us show 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
(3.32)
It follows from the definition of
that
and hence
that
and hence
(3.33)
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
. 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
(3.34)
Since
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
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
(3.35)
that is,
(3.36)
Letting
we conclude from (3.34) that
we conclude from (3.34) that
(3.37)
and hence
(3.38)
This shows that
is a solution of the
(1.1). This completes the proof.
Corollary 3.2 ([11, Theorem?2.1]).
Let
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:
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:
(3.39)
where
is the single-valued duality mapping on
. If
is uniformly continuous, then
converges strongly to
.
Proof.
In Theorem 3.1, we know from (3.1) and Lemma 2.3 that
(3.40)
is equivalent to
. Now, put
for all
. Then we have
. Now, put
for all
. Then we have
(3.41)
for all
. Thus algorithm (3.1) reduces to algorithm (3.39). By Theorem 3.1 we obtain the desired result.
Theorem 3.3.
Let
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:
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:
(3.42)
where
is assumed to exist for each
,
If
is uniformly continuous and
, then
converges strongly to
, which is a solution of the
(1.1).
Proof.
We only derive the difference. First, let us show that
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 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
(3.43)
is equivalent to
(3.44)
we know that
is convex. Next, let us show that
for each
. Indeed, we have for each
is convex. Next, let us show that
for each
. Indeed, we have for each
(3.45)
So
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
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
(3.46)
As in the proof of Theorem 3.1, we can deduce from
and
that
and hence
by Lemma 2.2. Further, it follows from
and the boundedness of
and
that
and
that
and hence
by Lemma 2.2. Further, it follows from
and the boundedness of
and
that
(3.47)
Since
, from the definition of
, we also have
, from the definition of
, we also have
(3.48)
It follows from (3.47) and
that
that
(3.49)
Utilizing Lemma 2.2, we have
(3.50)
Since
is uniformly norm-to-norm continuous on bounded subsets of
we have
is uniformly norm-to-norm continuous on bounded subsets of
we have
(3.51)
Note that
(3.52)
Therefore, from
we have
we have
(3.53)
Since
is also uniformly norm-to-norm continuous on bounded subsets of
, we obtain
is also uniformly norm-to-norm continuous on bounded subsets of
, we obtain
(3.54)
It follows that
(3.55)
Since
is uniformly continuous, it follows from (3.50) and (3.54) that
.
Finally, let us show 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
(3.56)
It follows from the definition of
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
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
(3.57)
Since
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
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
(3.58)
It follows from
that
that
(3.59)
Letting
we conclude from (3.34) that
we conclude from (3.34) that
(3.60)
This shows that
is a solution of the
(1.1). This completes the proof.
Corollary 3.4 ([11, Theorem?2.2]).
Let
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:
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:
(3.61)
where
is the single-valued duality mapping on
. If
is nonempty, then
converges strongly to
.
Proof.
In Theorem 3.3, we know from (3.42) and Lemma 2.3 that
(3.62)
is equivalent to
. Now, put
for all
. Then we have
. Now, put
for all
. Then we have
(3.63)
for all
. 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.
Declarations
Aknowledgments
The first author was partially supported by the National Science Foundation of China (10771141), Ph.D. Program Foundation of Ministry of Education of China (20070270004), and Science and Technology Commission of Shanghai Municipality grant (075105118), Leading Academic Discipline Project of Shanghai Normal University (DZL707), Shanghai Leading Academic Discipline Project (S30405) and Innovation Program of Shanghai Municipal Education Commission (09ZZ133). The third author was partially supported by the Grant NSF 97-2115-M-110-001.
Authors’ Affiliations
(1)
Scientific Computing Key Laboratory of Shanghai Universities
(2)
Department of Mathematics, Shanghai Normal University
(3)
Department of Business Administration, College of Management, Yuan-Ze University
(4)
Department of Applied Mathematics, National Sun Yat-Sen University
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