An Algorithm for Finding a Common Solution for a System of Mixed Equilibrium Problem, Quasivariational Inclusion Problem, and Fixed Point Problem of Nonexpansive Semigroup

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

(1.1)

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

(1.2)

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,

(1.3)

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

(1.4)

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

(1.5)

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

(1.6)

(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

(1.7)

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

(1.8)

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

(1.9)

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].

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

(2.1)

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,

(2.2)

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

(2.3)

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

(2.4)

A differentiable function on a convex set is called

(i) -convex [6] if

(2.5)

where is the Fréchet derivative of at ;

(ii) -strongly convex [6] if there exists a constant such that

(2.6)

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

(2.7)

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,

(2.8)

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

(2.9)

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

(2.10)

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 .

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,

(3.1)

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:

(3.2)

where

(3.3)

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

(3.4)

, , and , then

for each , there is a unique such that

(3.5)

the sequence converges strongly to some point , provided that is firmly nonexpansive;

is the unique solution of the following variational inequality

(3.6)

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

(3.7)

Since

(3.8)

this implies that is positive. Hence we have

(3.9)

For each given , let us define the mapping

(3.10)

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

(3.11)

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

(3.12)

is well defined.

Letting , , and , then

(3.13)

We divide the proof of Theorem 3.2 into 8 steps.

Step 1.

First prove that the sequences , and are bounded.

1. (a)
   Pick , since and , we have

   

   (3.14)
    

( ) Since and , we have , and so

(3.15)

Letting , , we have

(3.16)

Similarly, we have

(3.17)

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

(3.18)

So, . This implies that is a bounded sequence in . Therefore , and are all bounded.

Step 2.

Next we prove that

(3.19)

Since , then

(3.20)

Hence

(3.21)

From condition (ii), we have

(3.22)

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

(3.23)

From (3.22) and (3.23), we have

(3.24)

Step 3.

Next we prove that

(3.25)

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

(3.26)

It follows that

(3.27)

From (3.5), we have

(3.28)

Since

(3.29)

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

(3.30)

Simplifying it we have

(3.31)

Since and , by condition (ii), it yields .

Step 4.

Now we prove that for any given

(3.32)

In fact, it follows from (3.15) that

(3.33)

Substituting (3.33) into (3.28), we obtain

(3.34)

Simplifying it, we have

(3.35)

Since , and are bounded, these imply that .

Step 5.

Next we prove that

(3.36)

In fact, since

(3.37)

for the purpose, it is sufficient to prove

(3.38)

( ) First we prove that . In fact, since

(3.39)

we have

(3.40)

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

(3.41)

Simplifying it we have

(3.42)

Since ,   , , and are bounded, these imply that .

( ) Next we prove that

(3.43)

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).

1. (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 .

    
2. (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

(3.44)

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

(3.45)

That is,

(3.46)

for all and , hence .

1. (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

(3.47)

By virtue of the maximal monotonicity of , we have

(3.48)

So,

(3.49)

Since , , and , we have

(3.50)

Since is maximal monotone, this implies that , that is, , and so .

1. (d)

   Now we prove that is the unique solution of variational inequality (3.6).

    

We first prove that .

Since

(3.51)

It follows that

(3.52)

Therefore,

(3.53)

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

(3.54)

we have

(3.55)

Hence for any we have,

(3.56)

then

(3.57)

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

(3.58)

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

(3.59)

So, we have

(3.60)

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

(3.61)

( ) First, we prove that

(3.62)

Indeed, there exists a subsequence of such that

(3.63)

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

(3.64)

( ) Now we prove that

(3.65)

From and (a), we have

(3.66)

Step 8.

Finally we prove that

(3.67)

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

(3.68)

This implies that

(3.69)

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

(3.70)

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,

(3.71)

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

(3.72)

Again, since , then . Therefore we have

(3.73)

The conclusion of Corollary 3.3 can be obtained from Theorem 3.2 immediately.

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:

(4.1)

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

(4.2)

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:

(4.3)

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

(4.4)

Since is the unique solution of (4.3), we have .

This completes the proof of Theorem 4.1.