Abstract
We introduce a new iterative method for finding a common element of the set of solutions of a generalized equilibrium problem with a relaxed monotone mapping and the set of common fixed points of a countable family of nonexpansive mappings in a Hilbert space and then prove that the sequence converges strongly to a common element of the two sets. Using this result, we prove several new strong convergence theorems in fixed point problems, variational inequalities, and equilibrium problems.
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1. Introduction
Throughout this paper, let denote the set of all real numbers, let
denote the set of all positive integer numbers, let
be a real Hilbert space, and let
be a nonempty closed convex subset of
. Let
be a mapping. We call
nonexpansive if
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ1_HTML.gif)
The set of fixed points of is denoted by
. We know that the set
is closed and convex. Let
be a bifunction. The equilibrium problem for
is to find
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ2_HTML.gif)
The set of all solutions of the equilibrium problem is denoted by , that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ3_HTML.gif)
Some iterative methods have been proposed to find an element of ; see [1, 2].
A mapping is called inverse-strongly monotone if there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ4_HTML.gif)
Such a mapping is also called
-inverse-strongly monotone. It is known that each nonexpansive mapping is
-inverse-strongly monotone and each
-strictly pseudocontraction is
-inverse-strongly monotone; see [3, 4]. If there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ5_HTML.gif)
then is called a solution of the variational inequality. The set of all solutions of the variational inequality is denoted by
. It is known that
is closed and convex. Recently Takahashi and Toyoda [5] introduced an iterative method for finding an element of
; see also [6]. On the other hand, Plubtieng and Punpaeng [7] introduced an iterative method for finding an element of
; see also [8].
Consider a general equilibrium problem:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ6_HTML.gif)
The set of all solutions of the equilibrium problem is denoted by , that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ7_HTML.gif)
In the case of ,
coincides with
. In the case
,
coincides with
. Recently, S. Takahashi and W. Takahashi [9] introduced an iterative method to find an element of
. More precisely, they introduced the following iterative scheme:
,
, and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ8_HTML.gif)
where ,
, and
are three control sequences. They proved that
converges strongly to
.
A mapping is said to be relaxed
-
monotone if there exist a mapping
and a function
positively homogeneous of degree
, that is,
for all
and
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ9_HTML.gif)
where is a constant; see [10]. In the case of
for all
,
is said to be relaxed
-monotone. In the case of
for all
and
, where
and
,
is said to be
-monotone; see [11–13]. In fact, in this case, if
, then
is a
-strongly monotone mapping. Moreover, every monotone mapping is relaxed
-
monotone with
for all
and
.
In this paper, we consider a new general equilibrium problem with a relaxed monotone mapping:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ10_HTML.gif)
The set of all solutions of the equilibrium problem is denoted by , that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ11_HTML.gif)
In the case of , (1.10) is deduced to
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ12_HTML.gif)
The set of all solutions of (1.12) is denoted by , that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ13_HTML.gif)
In the case of ,
coincides with
. In the case of
and
,
coincides with
.
In this paper, we introduce a new iterative scheme for finding a common element of the set of solutions of a general equilibrium problem with a relaxed monotone mapping and the set of common fixed points of a countable family of nonexpansive mappings and then obtain a strong convergence theorem. More precisely, we introduce the following iterative scheme:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ14_HTML.gif)
where is a relaxed
-
monotone mapping,
is a
-inverse-strongly monotone mapping, and
is a countable family of nonexpansive mappings such that
,
, and
,
, and
are three control sequences. We prove that
defined by (1.14) converges strongly to
. Using the main result in this paper, we also prove several new strong convergence theorems for finding the elements of
,
,
, and
, respectively, where
is a nonexpansive mapping.
2. Preliminaries
Let be a
-inverse-strongly monotone mapping and let
denote the identity mapping of
. For all
and
, one has [6]
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ15_HTML.gif)
Hence, if , then
is a nonexpansive mapping of
into
.
For each point , there exists a unique nearest point of
, denoted by
, such that
for all
. Such a
is called the metric projection from
onto
. The well-known Browder's characterization of
ensures that
is a firmly nonexpansive mapping from
onto
, that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ16_HTML.gif)
Further, we know that for any and
,
if and only if
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ17_HTML.gif)
Let be a nonexpansive mapping of
into itself such that
. Then we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ18_HTML.gif)
which is obtained directly from the following:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ19_HTML.gif)
This inequality is a very useful characterization of . Observe what is more that it immediately yields that
is a convex closed set.
Let be a bifunction of
into
satisfying the following conditions:
for all
;
is monotone, that is,
for all
;
for each ,
;
for each ,
is convex and lower semicontinuous.
Definition 2.1 (see [10]).
Let be a Banach space with the dual space
and let
be a nonempty subset of
. Let
and
be two mappings. The mapping
is said to be
-hemicontinuous if, for any fixed
, the function
defined by
is continuous at
.
Lemma 2.2.
Let be a Hilbert space and let
be a nonempty closed convex subset of
. Let
be an
-hemicontinuous and relaxed
-
monotone mapping. Let
be a bifunction from
to
satisfying (A1) and (A4). Let
and
. Assume that
(i) for all
;
(ii)for any fixed , the mapping
is convex.
Then the following problems (2.6) and (2.7) are equivalent:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ20_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ21_HTML.gif)
Proof.
Let be a solution of the problem (2.6). Since
is relaxed
-
monotone, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ22_HTML.gif)
Thus is a solution of the problem (2.7).
Conversely, let be a solution of the problem (2.7). Letting
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ23_HTML.gif)
then . Since
is a solution of the problem (2.7), it follows that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ24_HTML.gif)
The conditions (i), (ii), (A1), and (A4) imply that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ25_HTML.gif)
It follows from (2.10)-(2.11) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ26_HTML.gif)
Since is
-hemicontinuous and
, letting
in (2.12), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ27_HTML.gif)
for all . Therefore,
is also a solution of the problem (2.6). This completes the proof.
Definition 2.3 (see [14]).
Let be a Banach space with the dual space
and let
be a nonempty subset of
. A mapping
is called a KKM mapping if, for any
,
, where
denotes the family of all the nonempty subsets of
.
Lemma 2.4 (see [14]).
Let be a nonempty subset of a Hausdorff topological vector space
and let
be a KKM mapping. If
is closed in
for all
in
and compact for some
, then
.
Next we use the concept of KKM mapping to prove two basic lemmas for our main result. The idea of the proof of the next lemma is contained in the paper of Fang and Huang [10].
Lemma 2.5.
Let be a real Hilbert space and
be a nonempty bounded closed convex subset of
. Let
be an
-hemicontinuous and relaxed
-
monotone mapping, and let
be a bifunction from
to
satisfying (A1) and (A4). Let
. Assume that
(i) for all
;
(ii)for any fixed , the mapping
is convex and lower semicontinuous;
(iii) is weakly lower semicontinuous; that is, for any net
,
converges to
in
which implies that
.
Then problem (2.6) is solvable.
Proof.
Let . Define two set-valued mappings
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ28_HTML.gif)
We claim that is a KKM mapping. If
is not a KKM mapping, then there exist
and
,
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ29_HTML.gif)
By the definition of , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ30_HTML.gif)
It follows from (A1), (A4), and (ii) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ31_HTML.gif)
which is a contradiction. This implies that is a KKM mapping.
Now, we prove that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ32_HTML.gif)
For any given , taking
, then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ33_HTML.gif)
Since is relaxed
-
monotone, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ34_HTML.gif)
It follows that and so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ35_HTML.gif)
This implies that is also a KKM mapping. Now, since
is a convex lower-semicontinuous function, we know that it is weakly lower semicontinuous. Thus from the definition of
and the weak lower semicontinuity of
, it follows that
is weakly closed for all
. Since
is bounded closed and convex, we know that
is weakly compact, and so
is weakly compact in
for each
. It follows from Lemmas 2.2 and 2.4 that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ36_HTML.gif)
Hence there exists such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ37_HTML.gif)
This completes the proof.
Lemma 2.6.
Let be a real Hilbert space and let
be a nonempty bounded closed convex subset of
. Let
be an
-hemicontinuous and relaxed
-
monotone mapping and let
be a bifunction from
to
satisfying (A1), (A2), and (A4). Let
and define a mapping
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ38_HTML.gif)
for all . Assume that
(i), for all
;
(ii)for any fixed , the mapping
is convex and lower semicontinuous and the mapping
is lower semicontinuous;
(iii) is weakly lower semicontinuous;
(iv)for any ,
.
Then, the following holds:
(1) is single-valued;
(2) is a firmly nonexpansive mapping, that is, for all
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ39_HTML.gif)
(3);
(4) is closed and convex.
Proof.
The fact that is nonempty is exactly the thesis of the previous lemma. We claim that
is single-valued. Indeed, for
and
, let
. Then,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ40_HTML.gif)
Adding the two inequalities, from (i) we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ41_HTML.gif)
From (A2), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ42_HTML.gif)
that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ43_HTML.gif)
Since is relaxed
-
monotone and
, one has
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ44_HTML.gif)
In (2.29) exchanging the position of and
, we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ45_HTML.gif)
that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ46_HTML.gif)
Now, adding the inequalities (2.30) and (2.32), by using (iv) we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ47_HTML.gif)
Hence,
Next we show that is firmly nonexpansive. Indeed, for
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ48_HTML.gif)
Adding the two inequalities and by (i) and (A2), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ49_HTML.gif)
that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ50_HTML.gif)
In (2.36) exchanging the position of and
, we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ51_HTML.gif)
Adding the inequalities (2.36) and (2.37), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ52_HTML.gif)
It follows from (iv) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ53_HTML.gif)
that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ54_HTML.gif)
This shows that is firmly nonexpansive.
Next, we claim that . Indeed, we have the following:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ55_HTML.gif)
Finally, we prove that is closed and convex. Indeed, Since every firm nonexpansive mapping is nonexpansive, we see that
is nonexpansive from (2). On the other hand, since the set of fixed points of every nonexpansive mapping is closed and convex, we have that
is closed and convex from (2) and (3). This completes the proof.
3. Main Results
In this section, we prove a strong convergence theorem which is our main result.
Theorem 3.1.
Let be a nonempty bounded closed convex subset of a real Hilbert space
and let
be a bifunction satisfying (A1), (A2), (A3), and (A4). Let
be an
-hemicontinuous and relaxed
-
monotone mapping, let
be a
-inverse-strongly monotone mapping, and let
be a countable family of nonexpansive mappings such that
. Assume that the conditions (i)–(iv) of Lemma 2.6 are satisfied. Let
and assume that
is a strictly decreasing sequence. Assume that
with some
and
with some
. Then, for any
, the sequence
generated by (1.14) converges strongly to
. In particular, if
contains the origin 0, taking
, then the sequence
generated by (1.14) converges strongly to the minimum norm element in
.
Proof.
We split the proof into following steps.
Step 1.
is closed and convex, the sequence
generated by (1.14) is well defined, and
for all
.
First, we prove that is closed and convex. It suffices to prove that
is closed and convex. Indeed, it is easy to prove the conclusion by the following fact:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ56_HTML.gif)
This implies that . Noting that
is a nonexpansive mapping for
and the set of fixed points of a nonexpansive mapping is closed and convex, we have that
is closed and convex.
Next we prove that the sequence generated by (1.14) is well defined and
for all
. It is easy to see that
is closed and convex for all
from the construction of
. Hence,
is closed and convex for all
. For any
, since
and
is nonexpansive, we have (note that
is strictly decreasing)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ57_HTML.gif)
So, for all
. Hence
, that is,
for all
. Since
is closed, convex, and nonempty, the sequence
is well defined.
Step 2.
and there exists
such that
as
.
From the definition of , we see that
for all
and hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ58_HTML.gif)
Noting that , we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ59_HTML.gif)
for all . This shows that
is increasing. Since
is bounded,
is bounded. So, we have that
exists.
Noting that and
for all
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ60_HTML.gif)
It follows from (3.5) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ61_HTML.gif)
By taking in (3.6), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ62_HTML.gif)
Since the limits of exists we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ63_HTML.gif)
that is, as
. Moreover, from (3.6) we also have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ64_HTML.gif)
This shows that is a Cauchy sequence. Hence, there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ65_HTML.gif)
Step 3.
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_IEq377_HTML.gif)
Since and
as
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ66_HTML.gif)
and hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ67_HTML.gif)
Note that can be rewritten as
for all
. Take
. Since
,
is
-inverse-strongly monotone, and
, we know that, for all
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ68_HTML.gif)
Using (1.14) and (3.13), we have (note that is strictly decreasing)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ69_HTML.gif)
and hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ70_HTML.gif)
Since and
are both bounded,
, and
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ71_HTML.gif)
Using Lemma 2.6, we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ72_HTML.gif)
So, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ73_HTML.gif)
From (3.18), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ74_HTML.gif)
and hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ75_HTML.gif)
By using and (3.16), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ76_HTML.gif)
Step 4.
, for all
It follows from the definition of scheme (1.14) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ77_HTML.gif)
that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ78_HTML.gif)
Hence, for any , one has
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ79_HTML.gif)
Since each is nonexpansive, by (2.4) we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ80_HTML.gif)
Hence, combining this inequality with (3.24), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ81_HTML.gif)
that is (noting that is strictly decreasing),
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ82_HTML.gif)
Since and
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ83_HTML.gif)
Step 5.
.
First we prove . Indeed, since
and
, we have
for each
. Hence,
.
Next, we show that . Noting that
, one obtains
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ84_HTML.gif)
Put for all
and
. Then, we have
. So, from (A2), (i), and (3.29) we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ85_HTML.gif)
Since , we have
. Further, from monotonicity of
, we have
. So, from (A4), (ii), and
-hemicontinuity of
we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ86_HTML.gif)
From (A1), (A4), (ii), and (3.31) we also have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ87_HTML.gif)
and hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ88_HTML.gif)
Letting , from (A3) and (ii) we have, for each
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ89_HTML.gif)
This implies that . Hence, we get
.
Finally, we show that . Indeed, from
and
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ90_HTML.gif)
Taking the limit in (3.35) and noting that as
, we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ91_HTML.gif)
In view of (2.3), one sees that . This completes the proof.
Corollary 3.2.
Let be a nonempty bounded closed convex subset of a Hilbert space
and let
be a bifunction satisfying (A1), (A2), (A3), and (A4). Let
be an
-hemicontinuous and relaxed
-
monotone mapping and let
be a nonexpansive mapping such that
. Assume that the conditions (i)–(iv) of Lemma 2.6 are satisfied. Assume that
with
,
with some
and
with
. Let
and let
be generated by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ92_HTML.gif)
Then the sequence converges strongly to
. In particular, if
contains the origin 0, taking
, the sequence
converges strongly to the minimum norm element in
.
Proof.
In Theorem 3.1, put ,
. Then, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ93_HTML.gif)
On the other hand, for all , we have that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ94_HTML.gif)
So, taking with
and choosing a sequence
of real numbers with
, we obtain the desired result by Theorem 3.1.
Corollary 3.3.
Let be a nonempty bounded closed convex subset of a Hilbert space
and let
be a bifunction satisfying (A1), (A2), (A3), and (A4). Let
be a monotone mapping and let
be a nonexpansive mapping such that
. Assume that
with
,
with some
and
with
. Let
and let
be generated by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ95_HTML.gif)
Then the sequence converges strongly to
. In particular, if
contains the origin 0, taking
, the sequence
converges strongly to the minimum norm element in
.
Proof.
In Corollary 3.2, put and
for all
. Then
is a monotone mapping and we obtain the desired result by Theorem 3.1.
Corollary 3.4.
Let be a closed convex subset of a Hilbert space
and let
be a bifunction satisfying (A1), (A2), (A3), and (A4). Let
be a
-inverse-strongly monotone mapping and let
be a nonexpansive mapping such that
. Assume that
with
,
with some
and
with
. Let
and let
be generated by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ96_HTML.gif)
Then the sequence converges strongly to
. In particular, if
contains the origin 0, taking
, the sequence
converges strongly to the minimum norm element in
.
Proof.
In Theorem 3.1, put ,
,
, and
. We obtain the desired result by Theorem 3.1.
Corollary 3.5.
Let be a closed convex subset of a Hilbert space
and let
be a bifunction satisfying (A1), (A2), (A3), and (A4). Let
be a nonexpansive mapping such that
. Assume that
with
,
with some
, and
with
. Let
and let
be generated by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ97_HTML.gif)
Then the sequence converges strongly to
. In particular, if
contains the origin 0, taking
, the sequence
converges strongly to the minimum norm element in
.
Proof.
In Corollary 3.4, by putting we obtain the desired result.
Corollary 3.6.
Let be a closed convex subset of a Hilbert space
and let
be a
-inverse-strongly monotone mapping. Let
be a nonexpansive mapping such that
. Assume that
with
,
with some
, and
with
. Let
and let
be generated by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ98_HTML.gif)
Then the sequence converges strongly to
. In particular, if
contains the origin 0, taking
, the sequence
converges strongly to the minimum norm element in
.
Proof.
In Theorem 3.1, put ,
,
,
, and
. Then, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F230304/MediaObjects/13663_2010_Article_1229_Equ99_HTML.gif)
Then, we obtain the desired result by Theorem 3.1.
Remark 3.7.
The novelty of this paper lies in the following aspects.
(i)A new general equilibrium problem with a relaxed monotone mapping is considered.
(ii)The definition of is of independent interest.
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Acknowledgment
This work was supported by the Natural Science Foundation of Hebei Province (A2010001482).
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Wang, S., Marino, G. & Wang, F. Strong Convergence Theorems for a Generalized Equilibrium Problem with a Relaxed Monotone Mapping and a Countable Family of Nonexpansive Mappings in a Hilbert Space. Fixed Point Theory Appl 2010, 230304 (2010). https://doi.org/10.1155/2010/230304
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DOI: https://doi.org/10.1155/2010/230304