- Research Article
- Open Access
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Strong and Weak Convergence Theorems for Common Solutions of Generalized Equilibrium Problems and Zeros of Maximal Monotone Operators
Fixed Point Theory and Applications volume 2010, Article number: 590278 (2010)
Abstract
The purpose of this paper is to introduce and study two modified hybrid proximal-point algorithms for finding a common element of the solution set EP of a generalized equilibrium problem and the set for two maximal monotone operators
and
defined on a Banach space
. Strong and weak convergence theorems for these two modified hybrid proximal-point algorithms are established.
1. Introduction
Let be a real Banach space with its dual
. The mapping
defined by

is called the normalized duality mapping. From the Hahn-Banach theorem, it follows that for each
.
A Banach space is said to be strictly convex, if
for all
with
.
is said to be uniformly convex if for each
, there exists
such that
for all
with
. Recall that each uniformly convex Banach space has the Kadec-Klee property, that is,

It is well known that if is strictly convex, then
is single-valued. In the sequel, we shall still denote the single-valued normalized duality mapping by
. Let
be a nonempty closed convex subset of
,
a bifunction, and
a nonlinear mapping. Very recently, Zhang [1] considered and studied the generalized equilibrium problem of finding
such that

The set of solutions of (1.3) is denoted by . Problem (1.3) and related problems have been studied and investigated extensively in the literature; See, for example, [2–12] and references therein. Whenever
, problem (1.3) reduces to the equilibrium problem of finding
such that

The set of solutions of (1.4) is denoted by . Whenever
, problem (1.3) reduces to the variational inequality problem of finding
such that

The set of solutions of (1.5) is denoted by .
Whenever a Hilbert space, problem (1.3) was very recently introduced and considered by S. Takahashi and W. Takahashi [13]. Problem (1.3) is very general in the sense that it includes, as spacial cases, optimization problems, variational inequalities, minimax problems, Nash equilibrium problem in noncooperative games, and others; See, for example, [1, 2, 4, 6–9, 14–17] which are references therein.
A mapping is called nonexpansive if
for all
. Denote by
the set of fixed points of
, that is,
. Very recently, W. Takahashi and K. Zembayashi [18] proposed an iterative algorithm for finding a common element of the solution set of the equilibrium problem (1.4) and the set of fixed points of a relatively nonexpansive mapping
in a Banach space
. They also studied the strong and weak convergence of the sequences generated by their algorithm. In particular, they proposed the following iterative algorithm:

where for all
, and
for some
. They proved that the sequence
generated by the above algorithm converges strongly to
, where
is the generalized projection of
onto
. They have also studied the weak convergence of the sequence
generated by the following algorithm:

to , where
.
Let be a nonempty closed convex subset of a uniformly smooth and uniformly convex Banach space
. Let
be an
-inverse-strongly monotone mapping and
a bifunction satisfying the following conditions:
(A1) for all
;
(A2) is monotone, that is,
, for all
;
(A3) for all ,
;
(A4) for all ,
is convex and lower semicontinuous.
Let be two relatively nonexpansive mappings such that
. Let
be the sequence generated by

Zhang [1] proved the strong convergence of the sequence to
under appropriate conditions.
On the other hand, a classic method of solving in a Hilbert space
is the proximal point algorithm which generates, for any starting point
, a sequence
in
by the iterative scheme

where is a sequence in
,
for each
is the resolvent operator for
, and
is the identity operator on
. This algorithm was first introduced by Martinet [19] and further studied by Rockafellar [20] in the framework of a Hilbert space
. Later several authors studied (1.9) and its variants in the setting of a Hilbert space
or in a Banach space
; See, for example, [15, 21–25] and references therein. Very recently, Li and Song [24] introduced and studied the following iterative scheme:

where and
is the duality mapping on
.
Algorithm (1.10) covers, as special cases, the algorithms introduced by Kohsaka and Takahashi [23] and Kamimura et al. [22] in a smooth and uniformly convex Banach space .
Let be a uniformly smooth and uniformly convex Banach space, and let
be a nonempty closed convex subset of
. Let
be a maximal monotone operator such that:
(A5).
In addition, for each , define a mapping
as follows:

for all .
Very recently, utilizing the ideas of the above algorithms in [15, 16, 18, 21, 22, 24], we [17] introduced two iterative methods for finding an element of and established the following strong and weak convergence theorems.
Theorem 1.1 (see [17]).
Suppose that conditions (A1)–(A5) are satisfied and let be chosen arbitrarily. Consider the sequence

where

is defined by (1.11),
satisfy
,
, and
satisfies
. Then, the sequence
converges strongly to
, where
is the generalized projection of
onto
.
Theorem 1.2 (see [17]).
Suppose that conditions (A1)–(A5) are satisfied and let be chosen arbitrarily. Consider the sequence

where is defined by (1.11),
satisfy the conditions
and
, and
satisfies
. If
is weakly sequentially continuous, then
converges weakly to an element
, where
.
The purpose of this paper is to introduce and study two new iterative methods for finding a common element of the solution set of generalized equilibrium problem (1.3) and the set
for maximal monotone operators
and
in a uniformly smooth and uniformly convex Banach space
. Firstly, motivated by Theorem 1.1 and a result of Zhang [1], we introduce a sequence
that converges strongly to
under some appropriate conditions.
Secondly, inspired by Theorem 1.2 and a result of Zhang [1], we define a sequence that converges weakly to an element , where
(Section 4).
Our results represent a generalization of known results in the literature, including those in [16–18, 24]. Our Theorems 3.1 and 4.2 are the extension and improvements of Theorems 1.1 and 1.2 in the following way:
(i)the problem of finding an element of includes the one of finding an element of
as a special case;
(ii)the algorithms in this paper are very different from those in [17] because of considering the complexity involving the problem of finding an element of .
2. Preliminaries
Throughout the paper, we denote the strong convergence, weak convergence, and weak convergence of a sequence
to a point
by
,
and
, respectively.
Assumption 2.1.
Let be a uniformly smooth and uniformly convex Banach space and let
be a nonempty closed convex subset of
. Let
be an
-inverse-strongly monotone mapping and let
be a bifunction satisfying the conditions (A1)–(A4). Let
be two maximal monotone operators such that:
(A5).
Recall that if is a nonempty closed convex subset of a Hilbert space
, then the metric projection
of
onto
is nonexpansive. This fact actually characterizes Hilbert spaces and hence, it is not available in more general Banach spaces. In this connection, Alber [26] recently introduced a generalized projection operator
in a Banach space
which is an analogue of the metric projection in Hilbert spaces.
Consider the functional defined as in [26] by

It is clear that in a Hilbert space , (2.1) reduces to
.
The generalized projection is a mapping that assigns to an arbitrary point
the minimum point of the functional
, that is,
, where
is the solution to the minimization problem

The existence and uniqueness of the operator follow from the properties of the functional
and strict monotonicity of the mapping
; See, for example, [27]. In a Hilbert space,
. From [26], in a smooth, strictly convex and reflexive Banach space
, we have

Moreover, by the property of subdifferential of convex functions, we easily get the following inequality:

Let be a mapping from
into itself. A point
in
is called an asymptotic fixed point of
[28] if
contains a sequence
which converges weakly to
such that
. The set of asymptotic fixed points of
is denoted by
. A mapping
from
into itself is called relatively nonexpansive [18, 29, 30] if
and
, for all
and
.
Observe that, if is a reflexive, strictly convex and smooth Banach space, then for any
if and only if
. To this end, it is sufficient to show that if
, then
. Actually, from (2.3), we have
, which implies that
. From the definition of
, we have
and therefore,
. For further details, we refer to [31].
We need the following lemmas for the proof of our main results.
Lemma 2.2 (see [32]).
Let be a smooth and uniformly convex Banach space and let
and
be two sequences of
. If
and either
or
is bounded, then
.
Let be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space
,
and
. Then

Let be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space
. Then

Lemma 2.5 (see [33]).
Let be a reflexive, strictly convex and smooth Banach space and let
be a multivalued operator. Then
(i) is closed and convex if
is maximal monotone such that
;
(ii) is maximal monotone if and only if
is monotone with
for all
.
Lemma 2.6 (see [34]).
Let be a uniformly convex Banach space and let
. Then there exists a strictly increasing, continuous and convex function
such that
and

for all and
, where
.
Lemma 2.7 (see [32]).
Let be a smooth and uniformly convex Banach space and let
. Then there exists a strictly increasing, continuous, and convex function
such that
and

The following result is due to Blum and Oettli [14].
Lemma 2.8 (see [14]).
Let be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space
,
a bifunction satisfying conditions (A1)–(A4), and
and
. Then, there exists
such that

Motivated by a result in [35] in a Hilbert space setting, Takahashi and Zembayashi [18] established the following lemma.
Lemma 2.9 (see [18]).
Let be a nonempty closed convex subset of a uniformly smooth, strictly convex and reflexive Banach space
, and
a bifunction satisfying conditions (A1)–(A4). For
and
, define a mapping
as follows:

for all . Then
(i) is single-valued;
(ii) is a firmly nonexpansive-type mapping, that is, for all
,

(iii);
(iv) is closed and convex.
Using Lemma 2.9, we have the following result.
Lemma 2.10 (see [18]).
Let be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space
,
a bifunction satisfying conditions (A1)–(A4), and
. Then, for
and
,

Utilizing Lemmas 2.8, 2.9, and 2.10, Zhang [1] derived the following result.
Proposition 2.11 (see [1]).
Let be a smooth, strictly convex and reflexive Banach space and let
be a nonempty closed convex subset of
. Let
be an
-inverse-strongly monotone mapping,
a bifunction satisfying conditions (A1)–(A4), and
. Then
-
(I)
for
, there exists
such that
(2.13)
-
(II)
if
is additionally uniformly smooth and
is defined as
(2.14)
then the mapping has the following properties:
(i) is single-valued,
(ii) is a firmly nonexpansive-type mapping, that is,

(iii),
(iv) is a closed convex subset of
,
(v) for all
.
Proof.
Define a bifunction by

It is easy to verify that satisfies the conditions (A1)–(A4). Therefore, the conclusions (I) and (II) follow immediately from Lemmas 2.8, 2.9, and 2.10.
Let be two maximal monotone operators in a smooth Banach space
. We denote the resolvent operators of
and
by
and
for each
, respectively. Then
and
are two single-valued mappings. Also,
and
for each
, where
and
are the sets of fixed points of
and
, respectively. For each
, the Yosida approximations of
and
are defined by
and
, respectively. It is known that

Lemma 2.12 (see [23]).
Let be a reflexive, strictly convex and smooth Banach space, and let
be a maximal monotone operator with
. Then,

Lemma 2.13 (see [36]).
Let and
be two sequences of nonnegative real numbers such that
for all
. If
, then
exists.
3. Strong Convergence Theorem
In this section, we prove a strong convergence theorem for finding a common element of the set of solutions for a generalized equilibrium problem and the set for two maximal monotone operators
and
.
Theorem 3.1.
Suppose that Assumption 2.1 is satisfied. Let be chosen arbitrarily. Consider the sequence

where

is defined by (2.14),
satisfy

and satisfies
. Then, the sequence
converges strongly to
, where
is the generalized projection of
onto
.
Proof.
For the sake of simplicity, we define

so that

We divide the proof into several steps.
Step 1.
We claim that is closed and convex for each
.
Indeed, it is obvious that is closed and
is closed and convex for each
. Let us show that
is convex. For
and
, put
. It is sufficient to show that
. We first write
for each
. Next, we prove that

is equivalent to

Indeed, from (2.1) we deduce that there hold the following:

which combined with (3.6) yield that (3.6) is equivalent to (3.7). Thus we have

This implies that . Therefore,
is closed and convex.
Step 2.
We claim that for each
and that
is well defined.
Indeed, take arbitrarily. Note that
is equivalent to

Then from Lemma 2.12, we obtain

Moreover, we have

and hence by Proposition 2.11,

So for all
. Now, let us show that

We prove this by induction. For , we have
. Assume that
. Since
is the projection of
onto
, by Lemma 2.3 we have

As by the induction assumption, the last inequality holds, in particular, for all
. This, together with the definition of
implies that
. Hence (3.14) holds for all
. So,
for all
. This implies that the sequence
is well defined.
Step 3.
We claim that is bounded and that
as
.
Indeed, it follows from the definition of that
. Since
and
, so
for all
, that is,
is nondecreasing. It follows from
and Lemma 2.4 that

for each for each
. Therefore,
is bounded, which implies that the limit of
exists. Since

so is bounded. From Lemma 2.4, we have

for each . This implies that

Step 4.
We claim that ,
, and
.
Indeed, from , we have

Therefore, from and
, it follows that
.
Since and
is uniformly convex and smooth, we have from Lemma 2.2 that

and, therefore, . Since
is uniformly norm-to-norm continuous on bounded subsets of
and
, then
.
Let us set . Then, according to Lemma 2.5 and Proposition 2.11, we know that
is a nonempty closed convex subset of
such that
. Fix
arbitrarily. As in the proof of Step 2, we can show that
,

Hence it follows from the boundedness of that
, and
are also bounded. Let
. Since
is a uniformly smooth Banach space, we know that
is a uniformly convex Banach space. Therefore, by Lemma 2.6 there exists a continuous, strictly increasing, and convex function
, with
, such that

for and
. So, we have that

and hence

for all . Consequently, we have

Since and
is uniformly norm-to-norm continuous on bounded subsets of
, we obtain
. From
and
, we have

Therefore, from the properties of , we get

recalling that is uniformly norm-to-norm continuous on bounded subsets of
. Next let us show that

Observe first that

Since , and
is bounded, so it follows that
. Also, observe that

Since , and the sequences
are bounded, so it follows that
. Meantime, observe that

and hence

Since and
, it follows from the boundedness of
that
. Thus, in terms of Lemma 2.2, we have that
and so
. Furthermore, it follows from (3.25) that

and hence

Since is uniformly norm-to-norm continuous on bounded subsets of
, it follows from
that
. Thus from
,
, and the boundedness of both
and
, we deduce that
. Utilizing the properties of
, we have that
. Since
is uniformly norm-to-norm continuous on bounded subsets of
, it follows that
.
Step 5.
We claim that , where

Indeed, since is bounded and
is reflexive, we know that
. Take
arbitrarily. Then there exists a subsequence
of
such that
. Hence it follows from
,
, and
that
and
converge weakly to the same point
. On the other hand, from (3.28a), (3.28b) and
, we obtain that

If and
, then it follows from (2.17) and the monotonicity of the operators
that for all

Letting , we have that
and
. Then the maximality of the operators
implies that
and
.
Next, let us show that . Since

from and Proposition 2.11 it follows that

Also, since

so we get

So, from (3.39), , and
, we have
.
Since is uniformly convex and smooth, we conclude from Lemma 2.2 that

From ,
, and (3.42), we have
and
.
Since is uniformly norm-to-norm continuous on bounded subsets of
, from (3.42) we derive

From , it follows that

By the definition of , we have

where

Replacing by
, we have from (A2) that

Since is convex and lower semicontinuous, it is also weakly lower semicontinuous. Letting
in the last inequality, from (3.44) and (A4), we have

For , with
, and
, let
. Since
and
, then
and hence
. So, from (A1) we have

Dividing by , we have

Letting , from (A3) it follows that

So, . Therefore, we obtain that
by the arbitrariness of
.
Step 6.
We claim that converges strongly to
.
Indeed, from and
, it follows that

Since the norm is weakly lower semicontinuous, then

From the definition of , we have
. Hence
, and

which implies that . Since
has the Kadec-Klee property, then
. Therefore,
converges strongly to
.
Remark 3.2.
In Theorem 3.1, let ,
, and
. Then, for all
and
, we have that

Moreover, there hold the following

and hence

In this case, Theorem 3.1 reduces to [17, Theorem ].
4. Weak Convergence Theorem
In this section, we present the following algorithm for finding a common element of the solution set of a generalized equilibrium problem and the set for two maximal monotone operators
and
.
Let be chosen arbitrarily and consider the sequence
generated by

where , and
is defined by (2.14).
Before proving a weak convergence theorem, we need the following proposition.
Proposition 4.1.
Suppose that Assumption 2.1 is fulfilled and let be a sequence defined by (4.1), where
satisfy the following conditions:

Then, converges strongly to
, where
is the generalized projection of
onto
.
Proof.
We set and

so that

Then, in terms of Lemma 2.5 and Proposition 2.11, is a nonempty closed convex subset of
such that
. We first prove that
is bounded. Fix
. Note that by the first and third of (4.3),
and

Here, each is relatively nonexpansive. Then from Proposition 2.11, we obtain


and hence by Proposition 2.11

 

Consequently, the last two inequalities yield that

for all . So, from
,
, and Lemma 2.13, we deduce that
exists. This implies that
is bounded. Thus,
is bounded and so are
,
,
, and
.
Define for all
. Let us show that
is bounded. Indeed, observe that

for each . This, together with the boundedness of
, implies that
is bounded and so is
. Furthermore, from
and (4.6e), we have

Since is the generalized projection, then, from Lemma 2.4 we obtain

Hence, from (4.8), it follows that .
Note that ,
, and
is bounded, so that
. Therefore,
is a convergent sequence. On the other hand, from (4.6e) we derive, for all
,

In particular, we have

Consequently, from and Lemma 2.4, we have

and hence

Let . From Lemma 2.7, there exists a continuous, strictly increasing, and convex function
with
such that

So, we have

Since is a convergent sequence,
is bounded and
is convergent; from the property of
, we have that
is a Cauchy sequence. Since
is closed,
converges strongly to
. This completes the proof.
Now, we are in a position to prove the following theorem.
Theorem 4.2.
Suppose that Assumption 2.1 is fulfilled and let be a sequence defined by (4.1), where
satisfy the following conditions:

and satisfies
. If
is weakly sequentially continuous, then
converges weakly to
, where
.
Proof.
We consider the notations (4.3). As in the proof of Proposition 4.1, we have that , and
are bounded sequences. Let

From Lemma 2.6 and as in the proof of Theorem 3.1, there exists a continuous, strictly increasing, and convex function with
such that

for and
. Observe that for
,

Hence,

Consequently, the last two inequalities yield that

Thus, we have

By the proof of Proposition 4.1, it is known that is convergent; since
,
,
, and
, then we have

Taking into account the properties of , as in the proof of Theorem 3.1, we have

since is uniformly norm-to-norm continuous on bounded subsets of
.
Now let us show that

Indeed, from (4.6e) we get

which, together with , yields that

From (4.6d) it follows that

which, together with , yields that

From (4.6c) it follows that

which, together with , yields that

From (4.6c) it follows that

which together with

yields that

From (4.6a) it follows that

which, together with , yields that

On the other hand, let us show that

Indeed, let . From Lemma 2.7, there exists a continuous, strictly increasing, and convex function
with
such that

Since and
, we deduce from Proposition 2.11 that for
,

This implies that

Since is uniformly norm-to-norm continuous on bounded subsets of
, from the properties of
, we obtain

Note that

Since , it follows from (4.24) and (4.41) that
and
.
Also, observe that

and hence

Thus, from , and
, it follows that
. In terms of Lemma 2.2, we derive
.
Next, let us show that , where
.
Indeed, since is bounded, there exists a subsequence
of
such that
. Hence it follows from (4.24), (4.41), and
that
and
converge weakly to the same point
. Furthermore, from
and (4.24), we have that

If and
, then it follows from (2.17) and the monotonicity of the operators
that for all

Letting , we obtain that

Then the maximality of the operators implies that
.
Now, by the definition of , we have

where . Replacing
by
, we have from (A2) that

Since is convex and lower semicontinuous, it is also weakly lower semicontinuous. Letting
in the last inequality, from (4.41) and (A4), we have

For , with
, and
, let
. Since
and
, then
and hence
. So, from (A1), we have

Dividing by , we get
. Letting
, from (A3) it follows that
. So,
. Therefore,
. Let
. From Lemma 2.3 and
, we get

From Proposition 4.1, we also know that . Note that
. Since
is weakly sequentially continuous, then
as
. In addition, taking into account the monotonicity of
, we conclude that
. Hence

From the strict convexity of , it follows that
. Therefore,
, where
. This completes the proof.
Remark 4.3.
Compared with the algorithm of Theorem 1.2, the above algorithm (4.1) can be applied to find an element of . But, the algorithm of Theorem 1.2 cannot be applied. Therefore, algorithm (4.1) develops and improves the algorithm of Theorem 1.2.
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Acknowledgments
In this research, the first author was partially supported by the Leading Academic Discipline Project of Shanghai Normal University (DZL707), Innovation Program of Shanghai Municipal Education Commission Grant (09ZZ133), National Science Foundation of China (10771141), Ph.D. Program Foundation of Ministry of Education of China (20070270004), Science and Technology Commission of Shanghai Municipality Grant (075105118), and Shanghai Leading Academic Discipline Project (S30405). The Fourth author was partially supported by a grant NSC 98-2115-M-110-001.
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Zeng, LC., Ansari, Q., Shyu, D. et al. Strong and Weak Convergence Theorems for Common Solutions of Generalized Equilibrium Problems and Zeros of Maximal Monotone Operators. Fixed Point Theory Appl 2010, 590278 (2010). https://doi.org/10.1155/2010/590278
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DOI: https://doi.org/10.1155/2010/590278
Keywords
- Hilbert Space
- Banach Space
- Convex Function
- Equilibrium Problem
- Lower Semicontinuous