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An Algorithm for Finding a Common Solution for a System of Mixed Equilibrium Problem, Quasivariational Inclusion Problem, and Fixed Point Problem of Nonexpansive Semigroup
Journal of Inequalities and Applications volume 2010, Article number: 895907 (2010)
Abstract
We introduce a hybrid iterative scheme for finding a common element of the set of solutions for a system of mixed equilibrium problems, the set of common fixed point for nonexpansive semigroup, and the set of solutions of the quasi-variational inclusion problem with multivalued maximal monotone mappings and inverse-strongly monotone mappings in Hilbert space. Under suitable conditions, some strong convergence theorems are proved. Our results extend some recent results announced by some authors.
1. Introduction
Throughout this paper we assume that is a real Hilbert space, and
is a nonempty closed convex subset of
.
In the sequel, we denote the set of fixed points of by
.
A bounded linear operator is said to be strongly positive, if there exists a constant
such that

Let be a single-valued nonlinear mapping and
a multivalued mapping. The "so-called" quasi-variational inclusion problem (see, Chang [1, 2]) is to find an
such that

A number of problems arising in structural analysis, mechanics, and economics can be studied in the framework of this kind of variational inclusions (see, e.g., [3]).
The set of solutions of variational inclusion (1.2) is denoted by .
Special Case
If , where
is a nonempty closed convex subset of
, and
is the indicator function of
, that is,

then the variational inclusion problem (1.2) is equivalent to find such that

This problem is called Hartman-Stampacchia variational inequality problem (see, e.g., [4]). The set of solutions of (1.4) is denoted by .
Recall that a mapping is called
-inverse strongly monotone (see [5]), if there exists an
such that

A multivalued mapping is called monotone, if for all
,
, and
, then it implies that
. A multivalued mapping
is called maximal monotone, if it is monotone and if for any

(the graph of mapping ) implies that
.
Proposition 1.1 (see [5]).
Let be an
-inverse strongly monotone mapping, then
(a) is a
-Lipschitz continuous and monotone mapping;
(b)if is any constant in
, then the mapping
is nonexpansive, where
is the identity mapping on
.
Let be an equilibrium bifunction (i.e.,
), and let
be a real-valued function.
Recently, Ceng and Yao [6] introduced the following mixed equilibrium problem, that is, to find
such that

The set of solutions of (1.7) is denoted by , that is,

In particular, if , this problem reduces to the equilibrium problem, that is, to find
such that

Denote the set of solution of EP by .
On the other hand, Li et al. [7] introduced two steps of iterative procedures for the approximation of common fixed point of a nonexpansive semigroup on a nonempty closed convex subset
in a Hilbert space.
Very recently, Saeidi [8] introduced a more general iterative algorithm for finding a common element of the set of solutions for a system of equilibrium problems and of the set of common fixed points for a finite family of nonexpansive mappings and a nonexpansive semigroup.
Recall that a family of mappings is called a nonexpansive semigroup, if it satisfies the following conditions:
(a) for all
and
;
(b).
(c)the mapping is continuous, for each
.
Motivated and inspired by Ceng and Yao [6], Li et al. [7], Saeidi [8], and [9–13], the purpose of this paper is to introduce a hybrid iterative scheme for finding a common element of the set of solutions for a system of mixed equilibrium problems, the set of common fixed point for a nonexpansive semigroup, and the set of solutions of the quasi-variational inclusion problem with multivalued maximal monotone mappings and inverse-strongly monotone mappings in Hilbert space. Under suitable conditions, some strong convergence theorems are proved. Our results extend the recent results in Zhang et al. [5], S. Takahashi and W. Takahashi [14], Chang et al. [15], Ceng and Yao [6], Li et al. [7] and, Saeidi [8].
2. Preliminaries
In the sequel, we use and
to denote the weak convergence and strong convergence of the sequence
in
, respectively.
Definition 2.1.
Let be a multivalued maximal monotone mapping, then the single-valued mapping
defined by

is called the resolvent operator associated with, where
is any positive number, and
is the identity mapping.
Proposition 2.2 (see [5]).
The resolvent operator
associated with
is single-valued and nonexpansive for all
, that is,

The resolvent operator
is 1-inverse-strongly monotone, that is,

Definition 2.3.
A single-valued mapping is said to be hemicontinuous, if for any
, the mapping
converges weakly to
(as
).
It is well known that every continuous mapping must be hemicontinuous.
Lemma 2.4 (see [16]).
Let be a real Banach space,
the dual space of
a maximal monotone mapping, and
a hemicontinuous bounded monotone mapping with
, then the mapping
is a maximal monotone mapping.
For solving the equilibrium problem for bifunction let us assume that
satisfies the following conditions:
for all
;
is monotone, that is,
for all
;
for each ,
is concave and upper semicontinuous.
for each ,
is convex.
A map is called Lipschitz continuous, if there exists a constant
such that

A differentiable function on a convex set
is called
(i)-convex [6] if

where is the Fréchet derivative of
at
;
(ii)-strongly convex [6] if there exists a constant
such that

Let be an equilibrium bifunction satisfying the conditions (H1)–(H4). Let
be any given positive number. For a given point
, consider the following auxiliary problem for
(for short,
) to find
such that

where is a mapping, and
is the Fréchet derivative of a functional
at
. Let
be the mapping such that for each
,
is the set of solutions of
, that is,

Then the following conclusion holds.
Proposition 2.5 (see [6]).
Let be a nonempty closed convex subset of
a lower semicontinuous and convex functional. Let
be an equilibrium bifunction satisfying conditions (H1)–(H4). Assume that
is Lipschitz continuous with constant
such that

is affine in the first variable,
for each fixed ,
is continuous from the weak topology to the weak topology;
is
-strongly convex with constant
, and its derivative
is continuous from the weak topology to the strong topology;
for each , there exists a bounded subset
and
such that for any
, one has

Then the following hold:
is single-valued;
is nonexpansive if
is Lipschitz continuous with constant
such that
;
;
is closed and convex.
Lemma 2.6 (see [17]).
Let be a nonempty bounded closed convex subset of
, and let
be a nonexpansive semigroup on
, then for any

Lemma 2.7 (see [7]).
Let be a nonempty bounded closed convex subset of
, and let
be a nonexpansive semigroup on
. If
is a sequence in
such that
and
, then
.
3. The Main Results
In order to prove the main result, we first give the following lemma.
Lemma 3.1 (see [5]).
is a solution of variational inclusion (1.2) if and only if
, that is,

If
, then
is a closed convex subset in
.
In the sequel, we assume that satisfy the following conditions:
is a real Hilbert space,
is a nonempty closed convex subset;
is a strongly positive linear bounded operator with a coefficient
is a contraction mapping with a contraction constant
,
,
is an
-inverse-strongly monotone mapping, and
is a multivalued maximal monotone mapping;
is a nonexpansive semigroup;
is a finite family of bifunctions satisfying conditions (H1)–(H4), and
is a finite family of lower semicontinuous and convex functional;
is a finite family of Lipschitz continuous mappings with constant
such that

is affine in the first variable,
for each fixed ,
is sequentially continuous from the weak topology to the weak topology;
is a finite family of
-strongly convex with constant
, and its derivative
is not only continuous from the weak topology to the strong topology but also Lipschitz continuous with constant
.
In the sequel we always denote by the set of fixed points of the nonexpansive semi-group
,
the set of solutions to the variational inequality (1.2), and MEP(
) the set of solutions to the following auxiliary problem for a system of mixed equilibrium problems:

where

and ,
is the mapping defined by (2.8).
In the sequel we denote by for
and
.
Theorem 3.2.
Let be the same as above. Let
be a finite family of positive numbers,
, and
. If
and the following conditions are satisfied:
for each , there exists a bounded subset
and
such that for any

,
, and
, then
for each , there is a unique
such that

the sequence converges strongly to some point
, provided that
is firmly nonexpansive;
is the unique solution of the following variational inequality

Proof.
We observe that from condition (ii), we can assume, without loss of generality, that .
Since is a linear bounded self-adjoint operator on
, then

Since

this implies that is positive. Hence we have

For each given , let us define the mapping

Firstly we show that the mapping is a contraction. Indeed, for any
, we have

This implies that is a contraction mapping. Let
be the unique fixed point of
. Thus,

is well defined.
Letting ,
, and
, then

We divide the proof of Theorem 3.2 into 8 steps.
Step 1.
First prove that the sequences , and
are bounded.
-
(a)
Pick
, since
and
, we have
(3.14)
() Since
and
, we have
, and so

Letting ,
, we have

Similarly, we have

Form (3.5), (3.9), (3.14), (3.15), (3.16), and (3.17) we have

So, . This implies that
is a bounded sequence in
. Therefore
, and
are all bounded.
Step 2.
Next we prove that

Since , then

Hence

From condition (ii), we have

Let , then
is a nonempty bounded closed convex subset of
and
-invariant. Since
and
is bounded, there exists
such that
; it follows from Lemma 2.6 that

From (3.22) and (3.23), we have

Step 3.
Next we prove that

In fact, for any given and
, since
is firmly nonexpansive, we have

It follows that

From (3.5), we have

Since

and this together with (3.27) and (3.28), it yields

Simplifying it we have

Since and
, by condition (ii), it yields
.
Step 4.
Now we prove that for any given

In fact, it follows from (3.15) that

Substituting (3.33) into (3.28), we obtain

Simplifying it, we have

Since , and
are bounded, these imply that
.
Step 5.
Next we prove that

In fact, since

for the purpose, it is sufficient to prove

() First we prove that
. In fact, since

we have

Substituting (3.40) into (3.28), it yields that

Simplifying it we have

Since ,
,
, and
are bounded, these imply that
.
() Next we prove that

In fact, since , so
. This together with (3.25) shows that
.
Step 6.
Next we prove that there exists a subsequence of
such that
, and
is the unique solution of the variational inequality (3.6).
-
(a)
We first prove that
. In fact, since
is bounded, there exists a subsequence
of
such that
. From Lemma 2.7 and Step 2, we obtain
.
-
(b)
Now we prove that
.
Since and noting Step 3, without loss of generality, we may assume that
. Hence for any
and for any
, we have

By the assumptions and by condition (H2) we know that the function and the mapping
both are convex and lower semicontinuous, hence they are weakly lower semicontinuous. These together with
and
, we have

That is,

for all and
, hence
.
-
(c)
Now we prove that
.
In fact, since is
-inverse-strongly monotone, it follows from Proposition 1.1 that
is a
-Lipschitz continuous monotone mapping and
(where
is the domain of
). It follows from Lemma 2.4 that
is maximal monotone. Let
, that is,
. Since
and noting Step 3, without loss of generality, we may assume that
; in particular, we have
. From
, we can prove that
. Again since
, we have

By virtue of the maximal monotonicity of , we have

So,

Since ,
, and
, we have

Since is maximal monotone, this implies that
, that is,
, and so
.
-
(d)
Now we prove that
is the unique solution of variational inequality (3.6).
We first prove that
.
Since

It follows that

Therefore,

Now, replacing in (3.53) with
and letting
and
, we have
.
Next we prove that
is the unique solution of the variational inequality (3.6).
Since

we have

Hence for any we have,

then

It is easily seen that is monotone. Thus from (3.57) we have that

Now, in (3.58) replacing by
and letting
and
, from (3.36), we have

So, we have

It follows from [18, Theorem ] that the solution of the variational inequality (3.6) is unique, that is,
is a unique solution of (3.6).
Step 7.
Next we prove that

() First, we prove that

Indeed, there exists a subsequence of
such that

We may also assume that . This together with (3.22) and (3.36) shows that
. Since
, we have
. Again by the same method as given in Step 6 we can prove that
. So, we have

() Now we prove that

From and (a), we have

Step 8.
Finally we prove that

Indeed, from (3.5), (3.15), and (3.17), we have

This implies that

Combining (3.61) and (3.69), we obtain that .
This completes the proof of Theorem 3.2.
Corollary 3.3.
Let be the same as in Theorem 3.2. Let
be a finite family of positive parameter,
and
. If
and conditions (i) and (ii) in Theorem 3.2 are satisfied, then
for each there is a unique
such that

the sequence converges strongly to some point
, provided that
is firmly nonexpansive;
is the unique solution of variational inequality (3.6).
Proof.
Taking in Theorem 3.2, where
is the indicator function of
, that is,

then the variational inclusion problem (1.2) is equivalent to variational inequality (1.4), that is, to find such that

Again, since , then
. Therefore we have

The conclusion of Corollary 3.3 can be obtained from Theorem 3.2 immediately.
4. Applications to Optimization Problem
Let be a real Hilbert space,
a nonempty closed convex subset of
a strongly positive linear bounded operator with a constant
, and
a nonexpansive mapping. In this section we will utilize the results presented in Section 3 to study the following optimization problem:

where is the set of fixed points of
in
and
is a potential function for
(i.e.,
,
), where
is a contractive mapping with a contractive constant
. We have the following theorem.
Theorem 4.1.
Let be the same as above. Let
be sequences in
satisfying condition (ii) in Theorem 3.2. If
is a nonempty compact subset of
, then for each
there is a unique
such that

and the sequence converges strongly to some point
which is the unique minimal point of optimization problem (4.1).
Proof.
Taking ,
,
,
,
in Corollary 3.3, hence we have
,
,
,
,
,
,
. Hence from Corollary 3.3 we know that the sequence
defined by (4.2) converges strongly to some point
which is the unique solution of the following variational inequality:

Since is nonexpansive, then
is convex. Again by the assumption that
is compact, therefore it is a compact and convex subset of
, and
is a continuous mapping. By virtue of the well-known Weierstrass theorem, there exists a point
which is a minimal point of optimization problem (4.1). As is known to all, (4.3) is the optimality necessary condition [19] for the optimization problem (4.1). Therefore we also have

Since is the unique solution of (4.3), we have
.
This completes the proof of Theorem 4.1.
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Liu, M., Chang, S. & Zuo, P. An Algorithm for Finding a Common Solution for a System of Mixed Equilibrium Problem, Quasivariational Inclusion Problem, and Fixed Point Problem of Nonexpansive Semigroup. J Inequal Appl 2010, 895907 (2010). https://doi.org/10.1155/2010/895907
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DOI: https://doi.org/10.1155/2010/895907
Keywords
- Variational Inequality
- Nonexpansive Mapping
- Multivalued Mapping
- Maximal Monotone
- Common Fixed Point