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A New Strong Convergence Theorem for Equilibrium Problems and Fixed Point Problems in Banach Spaces
Fixed Point Theory and Applications volume 2011, Article number: 572156 (2011)
Abstract
We introduce a new iterative sequence for finding a common element of the set of fixed points of a relatively nonexpansive mapping and the set of solutions of an equilibrium problem in a Banach space. Then, we study the strong convergence of the sequences. With an appropriate setting, we obtain the corresponding results due to Takahashi-Takahashi and Takahashi-Zembayashi. Some of our results are established with weaker assumptions.
1. Introduction
Throughout this paper, we denote by and
the sets of positive integers and real numbers, respectively. Let
be a Banach space,
the dual space of
and
a closed convex subsets of
. Let
be a bifunction. The equilibrium problem is to find
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ1_HTML.gif)
The set of solutions of (1.1) is denoted by . The equilibrium problems include fixed point problems, optimization problems, variational inequality problems, and Nash equilibrium problems as special cases.
Let be a smooth Banach space and
the normalized duality mapping from
to
. Alber [1] considered the following functional
defined by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ2_HTML.gif)
Using this functional, Matsushita and Takahashi [2, 3] studied and investigated the following mappings in Banach spaces. A mapping is relatively nonexpansive if the following properties are satisfied:
(R1),
(R2) for all
and
,
(R3),
where and
denote the set of fixed points of
and the set of asymptotic fixed points of
, respectively. It is known that
satisfies condition (R3) if and only if
is demiclosed at zero, where
is the identity mapping; that is, whenever a sequence
in
converges weakly to
and
converges strongly to 0, it follows that
. In a Hilbert space
, the duality mapping
is an identity mapping and
for all
. Hence, if
is nonexpansive (i.e.,
for all
), then it is relatively nonexpansive.
Recently, many authors studied the problems of finding a common element of the set of fixed points for a mapping and the set of solutions of equilibrium problem in the setting of Hilbert space and uniformly smooth and uniformly convex Banach space, respectively (see, e.g., [4–21] and the references therein). In a Hilbert space , S. Takahashi and W. Takahashi [17] introduced the iteration as follows: sequence
generated by
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ3_HTML.gif)
for every , where
is nonexpansive,
and
are appropriate sequences in
, and
is an appropriate positive real sequence. They proved that
converges strongly to some element in
. In 2009, Takahashi and Zembayashi [19] proposed the iteration in a uniformly smooth and uniformly convex Banach space as follows: a sequence
generated by
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ4_HTML.gif)
for every ,
is relatively nonexpansive,
is an appropriate sequence in
, and
is an appropriate positive real sequence. They proved that if
is weakly sequentially continuous, then
converges weakly to some element in
.
Motivated by S. Takahashi and W. Takahashi [17] and Takahashi and Zembayashi [19], we prove a strong convergence theorem for finding a common element of the fixed points set of a relatively nonexpansive mapping and the set of solutions of an equilibrium problem in a uniformly smooth and uniformly convex Banach space.
2. Preliminaries
We collect together some definitions and preliminaries which are needed in this paper. We say that a Banach space is strictly convex if the following implication holds for
:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ5_HTML.gif)
It is also said to be uniformly convex if for any , there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ6_HTML.gif)
It is known that if is a uniformly convex Banach space, then
is reflexive and strictly convex. We say that
is uniformly smooth if the dual space
of
is uniformly convex. A Banach space
is smooth if the limit
exists for all norm one elements
and
in
. It is not hard to show that if
is reflexive, then
is smooth if and only if
is strictly convex.
Let be a smooth Banach space. The function
(see [1]) is defined by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ7_HTML.gif)
where the duality mapping is given by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ8_HTML.gif)
It is obvious from the definition of the function that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ9_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ10_HTML.gif)
for all and
. The following lemma is an analogue of Xu's inequality [22, Theorem 2] with respect to
.
Lemma 2.1.
Let be a uniformly smooth Banach space and
. Then, there exists a continuous, strictly increasing, and convex function
such that
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ11_HTML.gif)
for all ,
, and
.
It is also easy to see that if and
are bounded sequences of a smooth Banach space
, then
implies that
.
Lemma 2.2 (see [23, Proposition 2]).
Let be a uniformly convex and smooth Banach space, and let
and
be two sequences of
such that
or
is bounded. If
, then
.
Remark 2.3.
For any bounded sequences and
in a uniformly convex and uniformly smooth Banach space
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ12_HTML.gif)
Let be a nonempty closed convex subset of a reflexive, strictly convex, and smooth Banach space
. It is known that [1, 23] for any
, there exists a unique point
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ13_HTML.gif)
Following Alber [1], we denote such an element by
. The mapping
is called the generalized projection from
onto
. It is easy to see that in a Hilbert space, the mapping
coincides with the metric projection
. Concerning the generalized projection, the following are well known.
Lemma 2.4 (see [23, Propositions 4 and 5]).
Let be a nonempty closed convex subset of a reflexive, strictly convex and smooth Banach space
,
, and
. Then,
(a) if and only if
for all
,
(b) for all
.
Remark 2.5.
The generalized projection mapping above is relatively nonexpansive and
.
Let be a reflexive, strictly convex and smooth Banach space. The duality mapping
from
onto
coincides with the inverse of the duality mapping
from
onto
, that is,
. We make use of the following mapping
studied in Alber [1]
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ14_HTML.gif)
for all and
. Obviously,
for all
and
. We know the following lemma (see [1] and [24, Lemma 3.2]).
Lemma 2.6.
Let be a reflexive, strictly convex and smooth Banach space, and let
be as in (2.10). Then,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ15_HTML.gif)
for all and
.
Lemma 2.7 (see [25, Lemma 2.1]).
Let be a sequence of nonnegative real numbers. Suppose that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ16_HTML.gif)
for all , where the sequences
in
and
in
satisfy conditions:
,
, and
. Then,
.
Lemma 2.8 (see [26, Lemma 3.1]).
Let be a sequence of real numbers such that there exists a subsequence
of
such that
for all
. Then, there exists a nondecreasing sequence
such that
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ17_HTML.gif)
for all . In fact,
.
For solving the equilibrium problem, we usually assume that a bifunction satisfies the following conditions:
(A1) for all
,
(A2) is monotone, that is,
, for all
,
(A3)for all ,
,
(A4)for all ,
is convex and lower semicontinuous.
The following lemma gives a characterization of a solution of an equilibrium problem.
Lemma 2.9 (see [19, Lemma 2.8 ]).
Let be a nonempty closed convex subset of a reflexive, strictly convex, and uniformly smooth Banach space
. Let
be a bifunction satisfying conditions (A1)–(A4). For
, define a mapping
so-called the resolvent of
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ18_HTML.gif)
for all . Then, the following hold:
(i) is single-valued,
(ii) is a firmly nonexpansive-type mapping [27], that is, for all
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ19_HTML.gif)
(iii),
(iv) is closed and convex,
Lemma 2.10 (see [4, Lemma 2.3]).
Let be a nonempty closed convex subset of a Banach space
,
a bifunction from
satisfying conditions (A1)–(A4) and
. Then,
if and only if
for all
.
Remark 2.11 (see [27]).
Let be a nonempty subset of a smooth Banach space
. If
is a firmly nonexpansive-type mapping, then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ20_HTML.gif)
for all and
. In particular,
satisfies condition (R2).
Lemma 2.12 (see [3, Proposition 2.4]).
Let be a nonempty closed convex subset of a strictly convex and smooth Banach space
and
a relatively nonexpansive mapping. Then,
is closed and convex.
3. Main Results
In this section, we prove a strong convergence theorem for finding a common element of the fixed points set of a relatively nonexpansive mapping and the set of solutions of an equilibrium problem in a uniformly convex and uniformly smooth Banach space.
Theorem 3.1.
Let be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space
and
a bifunction satisfying conditions (A1)–(A4) and
a relatively nonexpansive mapping such that
. Let
and
be sequences generated by
,
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ21_HTML.gif)
for all , where
satisfying
and
,
, and
. Then,
and
converge strongly to
.
Proof.
Note that can be rewritten as
. Since
is nonempty, closed, and convex, we put
. Since
,
, and
satisfy condition (R2), by (2.6), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ22_HTML.gif)
and so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ23_HTML.gif)
By induction, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ24_HTML.gif)
for all . This implies that
is bounded and so are
,
, and
. Put
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ25_HTML.gif)
Then, . Using Lemma 2.6 gives
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ26_HTML.gif)
Let be a function satisfying the properties of Lemma 2.1, where
. Then, by Remark 2.11 and (3.6), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ27_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ28_HTML.gif)
where for all
. Notice that
satisfying
and
.
The rest of the proof will be divided into two parts.
Case 1.
Suppose that there exists such that
is nonincreasing. In this situation,
is then convergent. Then,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ29_HTML.gif)
It follows from (3.7) and that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ30_HTML.gif)
Since ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ31_HTML.gif)
Consequently, by Remark 2.3,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ32_HTML.gif)
From (2.6) and , we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ33_HTML.gif)
This implies that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ34_HTML.gif)
Therefore,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ35_HTML.gif)
Since is bounded and
is reflexive, we choose a subsequence
of
such that
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ36_HTML.gif)
Then, . Since
and
, by Remark 2.3,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ37_HTML.gif)
Notice that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ38_HTML.gif)
Replacing by
, we have from (A2) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ39_HTML.gif)
Letting , we have from (3.17) and (A4) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ40_HTML.gif)
From Lemma 2.10, we have . Since
satisfies condition (R3) and (3.15),
. It follows that
. By Lemma 2.4(a), we immediately obtain that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ41_HTML.gif)
Since ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ42_HTML.gif)
It follows from Lemma 2.7 and (3.8) that . Then,
and so
.
Case 2.
Suppose that there exists a subsequence of
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ43_HTML.gif)
for all . Then, by Lemma 2.8, there exists a nondecreasing sequence
such that
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ44_HTML.gif)
for all . From (3.7) and
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ45_HTML.gif)
Using the same proof of Case 1, we also obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ46_HTML.gif)
From (3.8), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ47_HTML.gif)
Since , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ48_HTML.gif)
In particular, since , we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ49_HTML.gif)
It follows from (3.26) that . This together with (3.27) gives
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ50_HTML.gif)
But for all
, we conclude that
, and
.
From two cases, we can conclude that and
converge strongly to
and the proof is finished.
Applying Theorem 3.1 and [28, Theorem 3.2], we have the following result.
Theorem 3.2.
Let be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space
,
a bifunction satisfying conditions (A1)–(A4), and
a sequence of relatively nonexpansive mappings such that
. Let
and
be sequences generated by (3.1), where
is defined by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ51_HTML.gif)
Then, and
converge strongly to
.
Setting and
in Theorem 3.1, we have the following result.
Corollary 3.3.
Let be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space
and
a relatively nonexpansive mapping. Let
and
be sequences generated by
,
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ52_HTML.gif)
for all , where
satisfying
and
,
. Then,
and
converge strongly to
.
Letting in Corollary 3.3, we have the following result.
Corollary 3.4.
Let be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space
and
a relatively nonexpansive mapping. Let
be a sequence in
defined by
,
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ53_HTML.gif)
for all , where
satisfying
and
,
. Then
converges strongly to
.
Let be the identity mapping in Theorem 3.1, we also have the following result.
Corollary 3.5.
Let be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space
and
a bifunction satisfying conditions (A1)–(A4) such that
. Let
and
be sequences generated by
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ54_HTML.gif)
for all , where
satisfying
and
,
, and
. Then,
and
converge strongly to
.
4. Deduced Theorems in Hilbert Spaces
In Hilbert spaces, every nonexpansive mappings are relatively nonexpansive, and is the identity operator. We obtain the following result.
Theorem 4.1.
Let be a nonempty closed convex subset of a Hilbert space
,
a bifunction satisfying conditions (A1)–(A4), and
a nonexpansive mapping such that
. Let
be a sequence in
defined by
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ55_HTML.gif)
for all , where
is the resolvent of
,
satisfying
and
,
, and
. Then,
converges strongly to
.
Remark 4.2.
In Theorem 4.1, we have the same conclusion if the mapping is only quasinonexpansive (i.e.,
and
for all
and
) such that
is demiclosed at zero.
Letting in Theorem 4.1, we have the following result.
Corollary 4.3.
Let be a nonempty closed convex subset of a Hilbert space
and
a nonexpansive mapping such that
. Let
be a sequence in
defined by
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ56_HTML.gif)
for all , where
satisfying
,
, and
. Then,
converges strongly to
.
Let be the identity mapping in Theorem 4.1, we have the following result.
Corollary 4.4.
Let be a nonempty closed convex subset of a Hilbert space
and
a bifunction satisfying conditions (A1)–(A4). Let
be a sequence in
defined by
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ57_HTML.gif)
for all , where
is the resolvent of
,
satisfying
,
, and
. Then
converges strongly to
.
Proof.
We may assume without loss of generality that for all
. Setting
and
for all
, we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ58_HTML.gif)
, and
. Applying Theorem 4.1,
converges strongly to
.
Remark 4.5.
Corollary 4.4 improves and extends [29, Corollary 5.3]. More precisely, the conditions and
are removed.
Applying Corollary 4.4 and [30, Theorem 8], we have the following result.
Corollary 4.6.
Let be a nonempty closed convex subset of a Hilbert space
,
a bifunction satisfying conditions (A1)–(A4), and
a contraction of
into itself. Let
be a sequence in
defined by
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F572156/MediaObjects/13663_2010_Article_1410_Equ59_HTML.gif)
for all , where
is the resolvent of
,
satisfying
and
and
. Then,
converges strongly to
.
Remark 4.7.
Corollary 4.6 improves and extends [16, Corollary 3.4]. More precisely, the conditions and
are removed.
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Nilsrakoo, W. A New Strong Convergence Theorem for Equilibrium Problems and Fixed Point Problems in Banach Spaces. Fixed Point Theory Appl 2011, 572156 (2011). https://doi.org/10.1155/2011/572156
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DOI: https://doi.org/10.1155/2011/572156