- Research Article
- Open access
- Published:
A Hybrid Iterative Scheme for Equilibrium Problems, Variational Inequality Problems, and Fixed Point Problems in Banach Spaces
Fixed Point Theory and Applications volume 2009, Article number: 719360 (2009)
Abstract
The purpose of this paper is to introduce a new hybrid projection algorithm for finding a common element of the set of solutions of the equilibrium problem and the set of the variational inequality for an inverse-strongly monotone operator and the set of fixed points of relatively quasi-nonexpansive mappings in a Banach space. Then we show a strong convergence theorem. Using this result, we obtain some applications in a Banach space.
1. Introduction
Let be a real Banach space and let
be the dual of
. Let
be a closed convex subset of
. Let
be an operator. The classical variational inequality problem for
is to find
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ1_HTML.gif)
The set of solutions of (1.1) is denoted by . Such a problem is connected with the convex minimization problem, the complementarity, the problem of finding a point
satisfying
, and so on. First, we recall that
(1)an operator is called monotone if
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ2_HTML.gif)
(2)an operator is called
-inverse-strongly monotone if there exists a constant
with
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ3_HTML.gif)
Assume that
(C1) is
-inverse-strongly monotone,
(C2),
(C3) for all
and
.
Iiduka and Takahashi [1] introduced the following algorithm for finding a solution of the variational inequality for an operator that satisfies conditions (C1)–(C3) in a
-uniformly convex and uniformly smooth Banach space
. For an initial point
, define a sequence
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ4_HTML.gif)
where is the duality mapping on
, and
is the generalized projection from
onto
. Assume that
for some
with
where
is the
-uniformly convexity constant of
. They proved that if
is weakly sequentially continuous, then the sequence
converges weakly to some element
in
where
.
The problem of finding a common element of the set of the variational inequalities for monotone mappings in the framework of Hilbert spaces and Banach spaces has been intensively studied by many authors; see, for instance, [2–4] and the references cited therein.
Let be a bifunction. The equilibrium problem for
is to find
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ5_HTML.gif)
The set of solutions of (1.5) is denoted by .
For solving the equilibrium problem, let us assume that a bifunction satisfies the following conditions:
(A1) for all
;
(A2) is monotone, that is,
for all
;
(A3)for all ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ6_HTML.gif)
(A4)for all is convex and lower semicontinuous.
Recently, Takahashi and Zembayashi [5], introduced the following iterative scheme which is called the shrinking projection method:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ7_HTML.gif)
where is the duality mapping on
and
is the generalized projection from
onto
. They proved that the sequence
converges strongly to
under appropriate conditions.
Very recently, Qin et al. [6] extend the iteration process (1.7) from a single relatively nonexpansive mapping to two relatively quasi-nonexpansive mappings:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ8_HTML.gif)
Under suitable conditions over , and
, they obtain that the sequence
generated by (1.8) converges strongly to
.
The problem of finding a common element of the set of fixed points and the set of solutions of an equilibrium problem in the framework of Hilbert spaces and Banach spaces has been studied by many authors; see [5, 7–16].
Motivated by Iiduka and Takahashi [1], Takahashi and Zembayashi [5], and Qin et al. [6], we introduce a new general process for finding common elements of the set of the equilibrium problem and the set of the variational inequality problem for an inverse-strongly monotone operator and the set of the fixed points for relatively quasi-nonexpansive mappings.
2. Preliminaries
Let be a real Banach space and let
be the unit sphere of
. A Banach space
is said to be strictly convex if for any
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ9_HTML.gif)
It is also said to be uniformly convex if for each , there exists
such that for any
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ10_HTML.gif)
It is known that a uniformly convex Banach space is reflexive and strictly convex; and we define a function called the modulus of convexity of
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ11_HTML.gif)
Then is uniformly convex if and only if
for all
. Let
be a fixed real number with
. A Banach space
is said to be
-uniformly convex if there exists a constant
such that
for all
; see [17–19] for more details. A Banach space
is said to be smooth if the limit
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ12_HTML.gif)
exists for all . It is also said to be uniformly smooth if the limit (2.4) is attained uniformly for
. One should note that no Banach space is
-uniformly convex for
; see [19]. It is well known that a Hilbert space is
-uniformly convex, uniformly smooth. For each
, the generalized duality mapping
is defined by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ13_HTML.gif)
for all . In particular,
is called the normalized duality mapping. If
is a Hilbert space, then
, where
is the identity mapping. It is also known that if
is uniformly smooth, then
is uniformly norm-to-norm continuous on each bounded subset of
. See [20, 21] for more details.
Let be a given real number with
and
a
-uniformly convex Banach space. Then, for all
,
and
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ14_HTML.gif)
where is the generalized duality mapping of
and
is the
-uniformly convexity constant of
.
Let be a smooth Banach space. The function
is defined by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ15_HTML.gif)
for all . In a Hilbert space
, we have
for all
.
Recall that a mapping is called nonexpansive if
for all
and relatively nonexpansive if
satisfies the following conditions:
(1), where
is the set of fixed points of
;
(2) for all
and
;
(3), where
is the set of all asymptotic fixed points of
;
see [10, 23, 24] for more details.
is said to be relatively quasi-nonexpansive if
satisfies the conditions
and
. It is easy to see that the class of relatively quasi-nonexpansive mappings is more general than the class of relatively nonexpansive mappings [9, 25, 26].
We give some examples which are closed relatively quasi-nonexpansive; see [6].
Example 2.2.
Let be a uniformly smooth and strictly convex Banach space and
be a maximal monotone mapping such that its zero set
. Then,
is a closed relatively quasi-nonexpansive mapping from
onto
and
.
Example 2.3.
Let be the generalized projection from a smooth, strictly convex, and reflexive Banach space
onto a nonempty closed convex subset
of
. Then,
is a closed relatively quasi-nonexpansive mapping with
.
Lemma 2.4 (Kamimura and Takahashi [27]).
Let be a uniformly convex and smooth Banach space and let
be two sequences of
. If
and either
or
is bounded, then
as
.
Let be a nonempty closed convex subset of
. If
is reflexive, strictly convex and smooth, then there exists
such that
for
and
. The generalized projection
defined by
. The existence and uniqueness of the operator
follows from the properties of the functional
and strict monotonicity of the duality mapping
; for instance, see [20, 27–30]. In a Hilbert space,
is coincident with the metric projection.
Lemma 2.5 (Alber [28]).
Let be a nonempty closed convex subset of a smooth Banach space
and
. Then
if and only if
for all
.
Lemma 2.6 (Alber [28]).
Let be a nonempty closed convex subset of a reflexive, strictly convex and smooth Banach space
and let
. Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ16_HTML.gif)
Lemma 2.7 (Qin et al. [6]).
Let be a uniformly convex, smooth Banach space, let
be a closed convex subset of
, let
be a closed and relatively quasi-nonexpansive mapping from
into itself. Then
is a closed convex subset of
.
Lemma 2.8 (Cho et al. [31]).
Let be a uniformly convex Banach space and let
be a closed ball of
. Then there exists a continuous strictly increasing convex function
with
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ17_HTML.gif)
for all , and
with
.
Lemma 2.9 (Blum and Oettli [7]).
Let be a closed convex subset of a smooth, strictly convex, and reflexive Banach space
, let
be a bifunction from
to
satisfying (A1)–(A4), and let
and
. Then, there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ18_HTML.gif)
Lemma 2.10 (Qin et al. [6]).
Let be a closed convex subset of a uniformly smooth, strictly convex, and reflexive Banach space
, and let
be a bifunction from
to
satisfying (A1)–(A4). For all
and
, define a mapping
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ19_HTML.gif)
Then, the following hold:
(1) is single-valued;
(2) is a firmly nonexpansive-type mapping [32], that is, for all
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ20_HTML.gif)
(3);
(4) is closed and convex.
Lemma 2.11 (Takahashi and Zembayashi [14]).
Let be a closed convex subset of a smooth, strictly, and reflexive Banach space
, let
be a bifucntion from
to
satisfying (A1)–(A4), let
. Then, for all
and
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ21_HTML.gif)
We make use of the following mapping studied in Alber [28]:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ22_HTML.gif)
for all and
, that is,
.
Lemma 2.12 (Alber [28]).
Let be a reflexive, strictly convex, smooth Banach space and let
be as in (2.14). Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ23_HTML.gif)
for all and
.
An operator of
into
is said to be hemicontinuous if for all
, the mapping
of
into
defined by
is continuous with respect to the
topology of
. We define by
the normal cone for
at a point
, that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ24_HTML.gif)
Theorem 2.13 (Rockafellar [33]).
Let be a nonempty, closed convex subset of a Banach space
and
a monotone, hemicontinuous operator of
into
. Let
be an operator defined as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ25_HTML.gif)
Then is maximal monotone and
.
3. Strong Convergence Theorems
Theorem 3.1.
Let be a
-uniformly convex, uniformly smooth Banach space, let
be a nonempty closed convex subset of
. Let
be a bifunction from
to
satisfying (A1)–(A4), let
be an operator of
into
satisfying (C1)–(C3), and let
be two closed relatively quasi-nonexpansive mappings from
into itself such that
. For an initial point
with
and
, define a sequence
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ26_HTML.gif)
where is the duality mapping on
. Assume that
, and
are sequences in
satisfying the restrictions:
(B1);
(B2),
;
(B3) for some
;
(B4) for some
with
, where
is the
-uniformly convexity constant of
.
Then, and
converge strongly to
.
Proof.
We divide the proof into eight steps.
Step 1.
Show that and
are well defined.
It is obvious that is a closed convex subset of
. By Lemma 2.7, we know that
is closed and convex. From Lemma 2.10
, we also have
is closed and convex. Hence
is a nonempty, closed, and convex subset of
; consequently,
is well defined.
Clearly, is closed and convex. Suppose that
is closed and convex for
. For all
, we know
is equivalent to
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ27_HTML.gif)
So, is closed and convex. By induction,
is closed and convex for all
. This shows that
is well-defined.
Step 2.
Show that for all
.
Put . First, we observe that
for all
and
. Suppose
for
. Then, for all
, we know from Lemma 2.6 and Lemma 2.12 that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ28_HTML.gif)
Since and from (C1), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ29_HTML.gif)
From Lemma 2.1 and (C3), we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ30_HTML.gif)
Replacing (3.4) and (3.5) into (3.3), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ31_HTML.gif)
By the convexity of , for each
, we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ32_HTML.gif)
This shows that ; consequently,
. Hence
for all
.
Step 3.
Show that exists.
From and
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ33_HTML.gif)
From Lemma 2.6, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ34_HTML.gif)
Combining (3.8) and (3.9), we obtain that exists.
Step 4.
Show that is a Cauchy sequence in
.
Since for
, by Lemma 2.6, we also have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ35_HTML.gif)
Taking , we obtain that
. From Lemma 2.4, we have
. Hence
is a Cauchy sequence. By the completeness of
and the closedness of
, one can assume that
as
. Further, we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ36_HTML.gif)
Since , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ37_HTML.gif)
as . Applying Lemma 2.4 to (3.11) and (3.12), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ38_HTML.gif)
This implies that as
. Since
is uniformly norm-to-norm continuous on bounded subsets of
, we also obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ39_HTML.gif)
Step 5.
Show that .
Let . From (3.6) and Lemma 2.8, we know that there exists a continuous strictly increasing convex function
with
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ40_HTML.gif)
This implies that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ41_HTML.gif)
It follows from (3.13), (3.14), and (B) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ42_HTML.gif)
By the property of , we also obtain that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ43_HTML.gif)
Since is uniformly norm-to-norm continuous on bounded sets, so is
. Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ44_HTML.gif)
In the same manner, we can show that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ45_HTML.gif)
Again, by (3.15), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ46_HTML.gif)
which yields that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ47_HTML.gif)
From Lemma 2.6, Lemma 2.12, and (3.5), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ48_HTML.gif)
It follows from Lemma 2.4 and (3.22) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ49_HTML.gif)
Hence as
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ50_HTML.gif)
Combining (3.20) and (3.24), we also obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ51_HTML.gif)
From (3.19), (3.26) and by the closedness of and
, we get
.
Step 6.
Show that .
From (3.15), we see
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ52_HTML.gif)
From (3.16), we observe
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ53_HTML.gif)
Note that . From (3.27) and Lemma 2.11, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ54_HTML.gif)
From (3.28), we get . By Lemma 2.4, we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ55_HTML.gif)
as . Since
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ56_HTML.gif)
as . From
we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ57_HTML.gif)
By (A2), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ58_HTML.gif)
From (A4) and , we get
for all
. For
and
. Define
, then
, which implies that
. From (A1), we obtain that
. Thus,
. From (A3), we have
for all
. Hence
.
Step 7.
Show that .
Define be as in (2.17). By Theorem 2.13,
is maximal monotone and
. Let
. Since
, we get
. From
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ59_HTML.gif)
On the other hand, since . Then, by Lemma 2.5, we have
and thus
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ60_HTML.gif)
It follows from (3.34) and (3.35) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ61_HTML.gif)
where . By taking the limit as
and from (3.24) and (3.25), we obtain
. By the maximality of
, we have
and hence
.
Step 8.
Show that .
From , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ62_HTML.gif)
Since , we also have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ63_HTML.gif)
By taking limit in (3.38), we obtain that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ64_HTML.gif)
By Lemma 2.5, we can conclude that . Furthermore, it is easy to see that
as
. This completes the proof.
As a direct consequence of Theorem 3.1, we obtain the following results.
Corollary 3.2.
Let be a
-uniformly convex and uniformly smooth Banach space, and let
be a nonempty closed convex subset of
. Let
be a bifunction from
to
satisfying (A1)–(A4) and let
be a closed relatively quasi-nonexpansive mapping from
into itself such that
. Assume that
satisfies
and
for some
. Then the sequence
generated by (1.7) converges strongly to
.
Proof.
Putting and
in Theorem 3.1, we obtain the result.
Remark 3.3.
If in Theorem 3.1, then Theorem 3.1 reduces to Theorem
of Qin et al. [6].
Remark 3.4.
Corollary 3.2 improves Theorem of Takahashi and Zembayashi [5] from the class of relatively nonexpansive mappings to the class of relatively quasi-nonexpansive mappings, that is, we relax the strong restriction:
. Further, the algorithm in Corollary 3.2 is also simpler to compute than the one given in [14].
4. Applications
Next, we consider the problem of finding a zero point of an inverse-strongly monotone operator of into
. Assume that
satisfies the conditions:
(D1) is
-inverse-strongly monotone,
(D2).
Theorem 4.1.
Let be a
-uniformly convex, uniformly smooth Banach space. Let
be a bifunction from
to
satisfying (A1)–(A4), let
be an operator of
into
satisfying (D1) and (D2), and let
be two closed relatively quasi-nonexpansive mappings from
into itself such that
. For an initial point
with
and
, define a sequence
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ65_HTML.gif)
where is the duality mapping on
. Assume that
, and
are sequences in
satisfying the conditions (B1)–(B4) of Theorem 3.1.
Then, and
converge strongly to
.
Proof.
Putting in Theorem 3.1, we have
. We also have
and then the condition (C3) of Theorem 3.1 holds for all
and
. So, we obtain the result.
Let be a nonempty, closed convex cone in
,
an operator of
into
. We define its polar in
to be the set
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ66_HTML.gif)
Then the element is called a solution of the complementarity problem if
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ67_HTML.gif)
The set of solutions of the complementarity problem is denoted by .
Assume that is an operator satisfying the conditions:
(E1) is
-inverse-strongly monotone,
(E2),
(E3) for all
and
.
Theorem 4.2.
Let be a
-uniformly convex, uniformly smooth Banach space, and
a nonempty, closed convex cone in
. Let
be a bifunction from
to
satisfying (A1)–(A4), let
be an operator of
into
satisfying (E1)–(E3), and let
be two closed relatively quasi-nonexpansive mappings from
into itself such that
. For an initial point
with
and
, define a sequence
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F719360/MediaObjects/13663_2009_Article_1170_Equ68_HTML.gif)
where is the duality mapping on
. Assume that
and
are sequences in
satisfying the conditions (B1)–(B4) of Theorem 3.1.
Then, and
converge strongly to
.
Proof.
From [20, Lemma ], we have
. Hence, we obtain the result.
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The author would like to thank Professor Suthep Suantai and the referee for the valuable suggestions on the manuscript. The author was supported by the Commission on Higher Education and the Thailand Research Fund.
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Cholamjiak, P. A Hybrid Iterative Scheme for Equilibrium Problems, Variational Inequality Problems, and Fixed Point Problems in Banach Spaces. Fixed Point Theory Appl 2009, 719360 (2009). https://doi.org/10.1155/2009/719360
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DOI: https://doi.org/10.1155/2009/719360