- Research Article
- Open access
- Published:
Strong Convergence of an Iterative Method for Equilibrium Problems and Variational Inequality Problems
Fixed Point Theory and Applications volume 2009, Article number: 362191 (2009)
Abstract
We introduce an iterative method for finding a common element of the set of solutions of equilibrium problems, the set of solutions of variational inequality problems, and the set of fixed points of finite many nonexpansive mappings. We prove strong convergence of the iterative sequence generated by the proposed iterative algorithm to the unique solution of a variational inequality, which is the optimality condition for the minimization problem.
1. Introduction
Let be a real Hilbert space with inner product
and norm
, respectively. Suppose that
is nonempty, closed convex subset of
and
is a bifunction from
to
, where
is the set of real number. The equilibrium problem is to find a
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ1_HTML.gif)
The set of such solutions is denoted by EP(). Numerous problems in physics, optimization, and economics reduce to find a solution of equilibrium problem. Some methods have been proposed to solve the equilibrium problems in Hilbert space, see, for instance, Blum and Oettli [1], Combettes and Hirstoaga [2], and Moudafi [3].
A mapping is called monotone if
.
is called relaxed
-cocoercive, if there exist constants
and
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ2_HTML.gif)
when ,
is called
-strong monotone; when
,
is called relaxed
-cocoercive. Let
be a monotone operator, the variational inequality problem is to find a point
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ3_HTML.gif)
The set of solutions of variational inequality problem is denoted by . The variational inequality problem has been extensively studied in literatures, see, for example, [4, 5] and references therein.
Let be a strong positive bounded linear operator on
with coefficient
, that is, there exists a constant
such that
. A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive mapping on a real Hilbert space
:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ4_HTML.gif)
where T is a nonexpansive mapping on and
is a point on
.
A mapping from
into itself is called nonexpansive, if
. The set of fixed points of
is denoted by
. Let
be a finite family of nonexpansive mappings and
, define the mappings
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ5_HTML.gif)
where for all
. Such a mapping
is called
-mapping generated by
and
. We know that
is nonexpansive and
, see [6].
Let be a nonexpansive mapping and
is a contractive with coefficient
. Marino and Xu [7] considered the following general iterative scheme:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ6_HTML.gif)
They proved that converges strongly to
, where
is the metric projection from
onto
.
By combining equilibrium problems and (1.6), Plutbieng and Pumpaeng [8] proposed the following algorithm:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ7_HTML.gif)
They proved that if the sequences and
satisfy some appropriate conditions, then sequence
convergence to the unique solution
of the variational inequality
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ8_HTML.gif)
Motivated by [8], Colao et al. [9] introduced an iterative method for equilibrium problem and finite family of nonexpansive mappings
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ9_HTML.gif)
and proved that converges strongly to a point
and
also solves the variational inequality (1.8). For equilibrium problems, also see [10, 11].
On the other hand, let be a
-cocoercive mapping, for finding common element of the solution of variational inequality problems and the set of fixed point of nonexpansive mappings, Takahashi and Toyoda [12] introduced iterative scheme
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ10_HTML.gif)
They proved that converges weakly to
. Inspired by (1.10) and [13], Y. Yao and J.-C. Yao [14] given the following iterative process:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ11_HTML.gif)
and proved that converges strongly to
. By combining viscosity approximation method and (1.10), Chen et al. [15] introduced the process
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ12_HTML.gif)
and studied the strong convergence of sequence generated by (1.12). Motivated by (1.6), (1.11), and (1.12), Qin et al. [16] introduced the following general iterative process
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ13_HTML.gif)
and established a strong convergence theorem of to an element of
.
The purpose of this paper is to introduce the iterative process: and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ14_HTML.gif)
where is defined by (1.5),
is
-cocoercive, and
is a bounded linear operator. We should show that the sequences
converge strongly to an element of
. Our result extends the corresponding results of Qin et al. [16] and Colao et al. [9], and many others.
2. Preliminaries
Let be a real Hilbert space and
a nonempty, closed convex subset of
. We denote strong convergence of
to
by
and weak convergence by
. Let
is a mapping such that for every point
, there exists a unique
satisfying
, for all
.
is called the metric projection of
onto
. It is known that
is a nonexpansive mapping from
onto
. It is also known that
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ15_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ16_HTML.gif)
Let be a monotone mapping of
into
, then
if and only if
. The following result is useful in the rest of this paper.
Lemma 2.1 (see [17]).
Assume is a sequence of nonegative real number such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ17_HTML.gif)
where is a sequence in
, and
is a sequence in
such that
(1),
(2) or
.
Then, .
Lemma 2.2 (see [18]).
Let be bounded sequences in Banach space
satisfying
and
. Let
be a sequence in
with
. Then,
.
Lemma 2.3.
For all , there holds the inequality
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ18_HTML.gif)
Lemma 2.4 (see [7]).
Assume that is a strong positive linear bounded operator on a Hilbert space
with coefficient
and
. Then
.
For solving the equilibrium problem for a bifunction , we assume that
satisfies the following conditions:
(A1) for all
;
(A2) is monotone:
for all
;
(A3)for all ;
(A4)for all is convex and lower semicontinuous.
The following result is in Blum and Oettli [1].
Lemma 2.5 (see [1]).
Let be a nonempty closed convex subset of a Hilbert space
, let
be a bifunction from
into
satisfying (A1)–(A4), let
, and let
. Then there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ19_HTML.gif)
We also know the following lemmas.
Lemma 2.6 (see [19]).
Let be a nonempty closed convex subset of Hilbert space
, let
be a bifunction from
to
satisfying (A1)–(A4), let
, and let
, define a mapping
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ20_HTML.gif)
for all . Then, the following holds:
(1) is single-valued;
(2) is firmly nonexpansive-type mapping, that is, for all
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ21_HTML.gif)
(3);
(4) is closed and convex.
A monotone operator is said to be maximal monotone if its graph
is not properly contained in the graph of any other monotone operators. Let
be a monotone mapping of
into
and let
be the normal cone for
at a point
, that is
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ22_HTML.gif)
Define
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ23_HTML.gif)
It is known that in this case is maximal monotone, and
if and only if
.
3. Strong Convergence Theorem
Theorem 3.1.
Let be a real Hilbert space and
be a nonempty closed convex subset of
.
a finite family of nonexpansive mappings from
into itself and
a bifunction satisfying (A1)–(A4). Let
be relaxed
-cocoercive and
-Lipschitzian. Let
be an
-contraction with
and
a strong positive linear bounded operator with coefficient
,
is a constant with
. Let sequences
,
be in
and
be in
,
is a constant in
. Assume
and
(i),
;
(ii),
;
(iii) for some
with
and
;
(iv).
Then the sequence generated by (1.14) converges strongly to
and
solves the variational inequality
, that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ24_HTML.gif)
Proof.
Without loss of generality, we can assume . Then from Lemma 2.4 we know
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ25_HTML.gif)
Since is relaxed
-cocoercive and
-Lipschitzian and (iii) holds, we know from [14] that for all
and
, the following holds:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ26_HTML.gif)
We divide the proof into several steps.
Step 1.
is bounded.
Take , notice that
and form Lemma 2.6(2) that
is nonexpansive, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ27_HTML.gif)
Since , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ28_HTML.gif)
Then we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ29_HTML.gif)
Thus From (3.5) we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ30_HTML.gif)
hence is bounded, so is
,
.
Step 2.
.
Let , where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ31_HTML.gif)
Then we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ32_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ33_HTML.gif)
Next we estimate ,
and
. At first
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ34_HTML.gif)
Put , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ35_HTML.gif)
By recursion we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ36_HTML.gif)
for some . Similarly, we also get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ37_HTML.gif)
Since
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ38_HTML.gif)
Put in the first inequality and
in the second one, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ39_HTML.gif)
Adding both inequality, by (A2) we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ40_HTML.gif)
therefore, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ41_HTML.gif)
which implies that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ42_HTML.gif)
Hence we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ43_HTML.gif)
so, by (3.20) and the property , we arrive at
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ44_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ45_HTML.gif)
Therefore, by (3.14) and (3.20) we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ46_HTML.gif)
Now submitting (3.11), (3.13), and (3.23) into (3.9), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ47_HTML.gif)
Thus conditions (ii), (iii), and (iv) imply that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ48_HTML.gif)
Then, Lemma 2.2 yields
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ49_HTML.gif)
Step 3.
for
.
Note that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ50_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ51_HTML.gif)
hence by , we know
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ52_HTML.gif)
By Lemma 2.3, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ53_HTML.gif)
for some . Submitting (3.28) into (3.30), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ54_HTML.gif)
which implies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ55_HTML.gif)
hence from conditions (i), (iii), and (3.26), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ56_HTML.gif)
Similarly, submitting (3.29) into (3.30), we also have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ57_HTML.gif)
Step 4.
.
By (2.2) we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ58_HTML.gif)
hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ59_HTML.gif)
Submitting (3.36) into (3.30), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ60_HTML.gif)
This implies that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ61_HTML.gif)
Hence by Step 3 we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ62_HTML.gif)
Put , where
. We have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ63_HTML.gif)
Hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ64_HTML.gif)
so, we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ65_HTML.gif)
Submitting into (3.30), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ66_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ67_HTML.gif)
which implies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ68_HTML.gif)
So, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ69_HTML.gif)
which together with (3.39) gives
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ70_HTML.gif)
Since and
is firmly nonexpansive, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ71_HTML.gif)
which implies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ72_HTML.gif)
which together with (3.30) gives
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ73_HTML.gif)
So
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ74_HTML.gif)
Now (3.47) and condition (i) imply that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ75_HTML.gif)
Since
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ76_HTML.gif)
Then we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ77_HTML.gif)
hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ78_HTML.gif)
Note that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ79_HTML.gif)
thus from (3.47)–(3.55), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ80_HTML.gif)
Step 5.
, where
is the unique solution of variational inequality
Take a subsequence of
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ81_HTML.gif)
Since is bounded, without loss of generality, we assume
itself converges weakly to a point
. We should prove
.
First, let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ82_HTML.gif)
with the same argument as used in [14], we can derive , since
is maximal monotone, we know
.
Next, from (A2), for all we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ83_HTML.gif)
in particular
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ84_HTML.gif)
Condition (A4) implies that is weakly semicontinuous, then from (3.52) and let
we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ85_HTML.gif)
Replacing by
with
, using (A1) and (A4), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ86_HTML.gif)
Divide by in both side yields
, let
, by (A3) we conclude
. Therefore,
.
Finally, from we know that
. Assume
, that is,
. Since Hilbert space satisfies Opial's condition, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ87_HTML.gif)
this is a contradiction, thus , therefore,
. So we know
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ88_HTML.gif)
Step 6.
The sequence converges strongly to
.
From the definition of and Lemmas 2.3, and 2.4, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ89_HTML.gif)
which implies that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ90_HTML.gif)
Since from condition (i) we have and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ91_HTML.gif)
so, by Lemma 2.1, we conclude . This completes the proof.
Putting and
for all
in Theorem 3.1, we obtain the following corollary.
Corollary 3.2.
Let be a real Hilbert space and
be a nonempty closed convex subset of
.
a finite family of nonexpansive mappings from
into itself. Let
be relaxed
-cocoercive and
-Lipschitzian. Let
be an
-contraction with
and
a strong positive linear bounded operator with coefficient
,
be a constant with
. Let sequences
,
in
and
be a constant in
. Assume
and
(i),
;
(ii),
for some
with
and
;
(iii).
Then the sequence generated by
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ92_HTML.gif)
converges strongly to and
solves the variational inequality
, that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ93_HTML.gif)
Putting and
,
,
in Theorem 3.1, we obtain the following corollary.
Corollary 3.3.
Let be a real Hilbert space and
be a nonempty closed convex subset of
.
a finite family of nonexpansive mappings from
into itself and
a bifunction satisfying (A1)–(A4). Let
be an
-contraction with
and
a strong positive linear bounded operator with coefficient
,
is a constant with
. Let sequences
,
in
and
in
. Assume
and
(i),
;
(ii),
.
Then the sequence generated by
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ94_HTML.gif)
converges strongly to and
solves the variational inequality
, that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362191/MediaObjects/13663_2008_Article_1135_Equ95_HTML.gif)
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Li, H.Y., Li, H.Z. Strong Convergence of an Iterative Method for Equilibrium Problems and Variational Inequality Problems. Fixed Point Theory Appl 2009, 362191 (2009). https://doi.org/10.1155/2009/362191
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DOI: https://doi.org/10.1155/2009/362191