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An Iterative Algorithm for Mixed Equilibrium Problems and Variational Inclusions Approach to Variational Inequalities
Fixed Point Theory and Applications volume 2010, Article number: 564361 (2010)
Abstract
We present an iterative algorithm for finding a common element of the set of solutions of a mixed equilibrium problem and the set of a variational inclusion in a real Hilbert space. Furthermore, we prove that the proposed iterative algorithms strongly converge to
which solves some variational inequality.
1. Introduction
Let be a nonempty closed convex subset of a real Hilbert space
. Let
be a nonlinear mapping, let
be a function, and let
be a bifunction of
into
. Now we consider the following mixed equilibrium problem:

The set of solution of problem (1.1) is denoted by EP.
If , then the mixed equilibrium problem (1.1) becomes the following mixed equilibrium problem:

which was considered by Ceng and Yao [1]. If , then the mixed equilibrium problem (1.1) becomes the following equilibrium problem:

which was studied by S. Takahashi and W. Takahashi [2]. If and
, then the mixed equilibrium problem (1.1) becomes the following equilibrium problem:

If for all
, the mixed equilibrium problem (1.1) becomes the following variational inequality problem:

The mixed equilibrium problems include fixed point problems, optimization problems, variational inequality problems, Nash equilibrium problems, and the equilibrium problems as special cases; see, for example, [3–8]. Some methods have been proposed to solve the mixed equilibrium problem and the equilibrium problem. In 1997, Flaim and Antipen [4] introduced an iterative method of finding the best approximation to the initial data and proved a strong convergence theorem. Subsequently, S. Takahashi and W. Takahashi [9] introduced another iterative scheme for finding a common element of the set of solutions of the equilibrium problem (1.2) and the set of fixed point points of a nonexpansive mapping. Furthermore, Yao et al. [10] introduced some new iterative schemes for finding a common element of the set of solutions of the equilibrium problem (1.2) and the set of common fixed points of finitely (infinitely) nonexpansive mappings. Very recently, Ceng and Yao [1] considered a new iterative scheme for finding a common element of the set of solutions of the mixed equilibrium problem and the set of common fixed points of finitely many nonexpansive mappings. Peng and Yao [11] developed a CQ method. They obtained some strong convergence results for finding a common element of the set of solutions of the mixed equilibrium problem (1.1) and the set of the variational inequality and the set of fixed points of a nonexpansive mapping. Their results extend and improve the corresponding results in [1, 9, 12].
Recall that a mapping is said to be
-inverse strongly monotone if there exists a constant
such that
for all
. A mapping
is strongly positive on
if there exists a constant
such that
.
Let be a single-valued nonlinear mapping and let
be a set-valued mapping. Now we concern the following variational inclusion, which is to find a point
such that

where is the zero vector in
. The set of solutions of problem (1.6) is denoted by
. If
, then problem (1.6) becomes the generalized equation introduced by Robinson [13]. If
, then problem (1.6) becomes the inclusion problem introduced by Rockafellar [14]. It is known that (1.6) provides a convenient framework for the unified study of optimal solutions in many optimization related areas including mathematical programming, complementarity, variational inequalities, optimal control, mathematical economics, equilibria, and game theory. Also various types of variational inclusions problems have been extended and generalized. Recently, Zhang et al. [15] introduced a new iterative scheme for finding a common element of the set of solutions to the problem (1.6) and the set of fixed points of nonexpansive mappings in Hilbert spaces. Peng et al. [16] introduced another iterative scheme by the viscosity approximate method for finding a common element of the set of solutions of a variational inclusion with set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an equilibrium problem, and the set of fixed points of a nonexpansive mapping. For some related works, please see [1, 2, 9–11, 13–34] and the references therein.
Inspired and motivated by the works in the literature, in this paper, we present an iterative algorithm for finding a common element of the set of solutions of a mixed equilibrium problem and the set of a variational inclusion in a real Hilbert space. Furthermore, we prove that the proposed iterative algorithms strongly converge to
which solves some variational inequality.
2. Preliminaries
Let be a real Hilbert space with inner product
and norm
. Let
be a nonempty closed convex subset of
. Then, for any
, there exists a unique nearest point in
, denoted by
, such that

Such a is called the metric projection of
onto
. We know that
is nonexpansive. Further, for
and
,

A set-valued mapping is called monotone if, for all
,
and
imply
. A monotone mapping
is maximal if its graph
is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping
is maximal if and only if, for
,
for every
implies
.
Let the set-valued mapping be maximal monotone. We define the resolvent operator
associated with
and
as follows:

where is a positive number. It is worth mentioning that the resolvent operator
is single-valued, nonexpansive, and 1-inverse strongly monotone and that a solution of problem (1.6) is a fixed point of the operator
for all
, see, for instance, [25].
Throughout this paper, we assume that a bifunction and a convex function
satisfy the following conditions:
(H1) for all
;
(H2) is monotone, that is,
for all
;
(H3)for each ,
is weakly upper semicontinuous;
(H4)for each ,
is convex and lower semicontinuous;
(H5)for each and
, there exists a bounded subset
and
such that for any
,

Lemma (see [11]).
Let be a nonempty closed convex subset of a real Hilbert space
. Let
be a bifunction and let
be a proper lower semicontinuous and convex function. For
and
, define a mapping
as follows:

for all . Assume that the conditions (H1)–(H5) hold. Then one has the following results:
(1)for each ,
and
is single-valued;
(2) is firmly nonexpansive, that is, for any
,

(3);
(4) is closed and convex.
Lemma (see [24]).
Let be a maximal monotone mapping and let
be a Lipschitz-continuous mapping. Then the mapping
is maximal monotone.
Lemma (see [34]).
Assume taht is a sequence of nonnegative real numbers such that
where
is a sequence in
and
is a sequence such that
(1);
(2) or
.
Then .
3. Main Results
In this section, we will prove our main result. First, we give some assumptions on the operators and the parameters. Subsequently, we introduce our iterative algorithm for finding a common element of the set of solutions of a mixed equilibrium problem and the set of a variational inclusion. Finally, we will show that the proposed algorithm has strong convergence.
Let be a nonempty closed convex subset of a real Hilbert space
. Let
be a lower semicontinuous and convex function and let
be a bifunction satisfying conditions (H1)–(H5). Let
be a strongly positive bounded linear operator with coefficient
and let
be a maximal monotone mapping. Let the mappings
be
-inverse strongly monotone and
-inverse strongly monotone, respectively. Let
and
be two constants such that
and
.
Now we introduce the following iteration algorithm.
Algorithm 3.1.
For given arbitrarily, compute the sequences
and
as follows:

where is a real sequence in
.
Now we study the strong convergence of the algorithm (3.1).
Theorem 3.2.
Suppose that . Assume the following conditions are satisfied:
(i);
(ii);
(iii).
Then the sequence generated by (3.1) converges strongly to
which solves the following variational inequality:

Proof.
Take . It is clear that

We divide our proofs into the following five steps:
(1)the sequences and
are bounded;
(2);
(3);
(4);
(5).
Proof.
Since is
-inverse strongly monotone and
is
-inverse strongly monotone, we have


It is clear that if and
, then
and
are all nonexpansive. Set
. It follows that

By Lemma 2.1, we have for all
. Then, we have

Hence, we have

Since is linear bounded self-adjoint operator on
, then

Observe that

that is to say is positive. It follows that

From (3.1), we deduce that

Therefore, is bounded. Hence,
,
, and
are all bounded.
Proof.
From (3.1), we have

Note that

Substituting (3.14) into (3.13), we get

Notice that This together with the last inequality and Lemma 2.3 implies that

Proof.
From (3.5) and (3.7), we get

By (3.1), we obtain

where is some constant satisfying
. From (3.17) and (3.18), we have

Thus,

which imply that

Proof.
Since is firmly nonexpansive, we have

Hence, we have

Since is 1-inverse strongly monotone, we have

which implies that

Thus, by (3.23) and (3.25), we obtain

Substitute (3.26) into (3.18) to get

Then we derive

So, we have

Proof.
We note that is a contraction. As a matter of fact,

for all . Hence
has a unique fixed point, say
. That is,
. This implies that
for all
. Next, we prove that

First, we note that there exists a subsequence of
such that

Since is bounded, there exists a subsequence
of
which converges weakly to
. Without loss of generality, we can assume that
.
We next show that . By
, we know that

It follows from (H2) that

Hence,

For and
, let
. From (3.35) we have

Since , we have
. Further, from the inverse strongly monotonicity of
, we have
. So, from (H4) and the weakly lower semicontinuity of
,
and
weakly, we have

From (H1), (H4), and (3.37), we also have

and hence

Letting , we have, for each
,

This implies that .
Next, we show that . In fact, since
is
-inverse strongly monotone,
is Lipschitz continuous monotone mapping. It follows from Lemma 2.2 that
is maximal monotone. Let
, that is,
. Again since
, we have
, that is,
. By virtue of the maximal monotonicity of
, we have

and so

It follows from ,
and
that

It follows from the maximal monotonicity of that
, that is,
. Therefore,
. It follows that

Proof.
First, we note that ; then for all
, we have
.
From (3.1), we have

that is,

where and
. It is easy to see that
and
. Hence, by Lemma 2.3, we conclude that the sequence
converges strongly to
. This completes the proof.
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Acknowledgment
The author was supported in part by NSC 98-2622-E-230-006-CC3 and NSC 98-2923-E-110-003-MY3.
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Liou, YC. An Iterative Algorithm for Mixed Equilibrium Problems and Variational Inclusions Approach to Variational Inequalities. Fixed Point Theory Appl 2010, 564361 (2010). https://doi.org/10.1155/2010/564361
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DOI: https://doi.org/10.1155/2010/564361