Weak and Strong Convergence Theorems for Equilibrium Problems and Countable Strict Pseudocontractions Mappings in Hilbert Space

Let be a nonempty closed convex subset of a Hilbert space and let be a self-mapping of . Then is said to be a strict pseudocontraction mappings if for all , there exists a constant such that

(1.1)

(if (1.1) holds, we also say that is a -strict pseudocontraction). We use to denote the set of fixed points of , to denote weak(strong) convergence, and to denote the set of .

Let be a bifunction where is the set of real numbers. Then, we consider the following equilibrium problem:

(1.2)

The set of such is denoted by . Numerous problems in physics, optimization, and economics can be reduced to find a solution of (1.2). Some methods have been proposed to solve the equilibrium problem (see [1–3]). Recently, S. Takahashi and W. Takahashi [4] introduced an iterative scheme by the viscosity approximation method for finding a common element of the set of solutions of the equilibrium problem and the set of fixed points of a nonexpansive mapping in Hilbert spaces. They also studied the strong convergence of the sequences generated by their algorithm for a solution of the EP which is also a fixed point of a nonexpansive mapping defined on a closed convex subset of a Hilbert space.

In this paper, thanks to the condition introduced by Aoyama et al. [5], We introduce two iterative sequence for finding a common element of the set of solutions of an equilibrium problems and the set of fixed points of a countable family of strict pseudocontractions mappings in Hilbert Space. Then we study the weak and strong convergence of the sequences. The additional condition is inspired by Marino and Xu [6] and Kim and Xu [7].

For solving the equilibrium problem, let us assume that the bifunction satisfies the following conditions (see [3]):

(A1)

(A2)

(A3) is upper-hemicontinuous, that is, for each ,

(A4) is convex and lower semicontinuous for each .

Let be a real Hilbert space. Then there hold the following well-known results:

(2.1)

If is a sequence in weakly convergent to , then

(2.2)

Recall that the nearest point projection from onto assigns to each its nearest point denoted by in ; that is, is the unique point in with the property

(2.3)

Given and , then if and only if there holds the following relation:

(2.4)

Lemma (see [6]).

Let be a nonempty closed convex subset of a real Hilbert space . Let : be a -strict pseudocontraction such that .

()(Demi-closed principle) is demi-closed on , that is, if and , then .

() satisfies the Lipschitz condition

(2.5)

()The fixed point set of is closed and convex so that the projection is well defined.

Lemma (see [5]).

Let be a nonempty closed convex subset of a Banach space and let be a sequence of mapping of into itself. Suppose Then, for each , converges strongly to some point of . Moreover, let be a mapping of into itself defined by

(2.6)

Then

Lemma (see [8]).

Let be a closed convex subset of . Let be a sequence in and . Let . If is such that and satisfies the condition

(2.7)

then .

Lemma (see [9]).

Let be a nonempty closed convex subset of . Let be a bifunction from satisfying (A1), (A2), (A3), and(A4). Then, for any and , there exists such that

(2.8)

Further, if , then the following holds:

() is single-valued;

() is firmly nonexpansive, that is,

(2.9)

()

() is closed and convex.

Theorem 3.1.

Let be a nonempty closed convex subset of a Hilbert space and let be a sequence of -strict pseudocontractions mappings on into itself with . Assume that Let be a bifunction satisfying (A1), (A2), (A3), (A4), and . Let and be sequence generated by and

(3.1)

Assume that for all , where is a small enough constant, and is a sequence in with Let for any bounded subset of and let be a mapping of into itself defined by and suppose that Then the sequences and converge weakly to an element of .

Proof.

Pick . Then from the definition of in Lemma 2.4, we have , and therefore . It follows from (3.1) that

(3.2)

Since for all , we get that is, the sequence is decreasing. Hence exists. In particular, is bounded. Since is firmly nonexpensive, is also bounded. Also (3.2) implies that

(3.3)

Taking the limit as yields that

(3.4)

Since is bounded, it follows that

(3.5)

We apply Lemma 2.2 to get

(3.6)

Next, we claim that . Indeed, let be an arbitrary element of . Then as above

(3.7)

and hence

(3.8)

Therefore, from (3.2), we have

(3.9)

and hence

(3.10)

So, from the existence of , we have

(3.11)

Next, we claim that . since is bounded and is reflexive, is nonempty. Let be an arbitrary element. Then a subsequence of converges weakly to . Hence, from (3.11) we know that As , we obtain that . Let us show . Since , we have

(3.12)

By (A2), we have

(3.13)

and hence

(3.14)

From (A4), we have

(3.15)

Then, for and , from (A1), and (A4), we also have

(3.16)

Taking and using (A3), we get

(3.17)

and hence . Since is a strict pseudocontraction mapping, by Lemma 2.1( ) we know that the mapping is demiclosed at zero. Note that and . Thus, . Consequently, we deduce that . Since is an arbitrary element, we conclude that .

To see that and are actually weakly convergent, we take Since exist for every , by (2.2), we have

(3.18)

Hence and proof is completed.

Theorem 4.1.

Let be a closed convex subset of a real Hilbert space . Let be a sequence of -strict pseudocontractions mappings on into itself with . Assume that Let be a bifunction satisfying (A1), (A2), (A3), (A4) and . For and , let and be sequence generated by and

(4.1)

Assume that for all , where is a small enough constant, and is a sequence in with and . Let for any bounded subset of and let be a mapping of into itself defined by Suppose that Then, converges strongly to .

Proof.

First, we show that is closed and convex. It is obvious that is closed and convex. Suppose that is closed and convex for some . For , we know that is equivalent to

(4.2)

So is closed and convex. Then, is closed and convex.

Next, we show by induction that for all . is obvious. Suppose that for some . Let . Putting for all , we know from (4.1) that

(4.3)

and hence . This implies that for all .

This implied that is well defined.

From , we have

(4.4)

Using , we have

(4.5)

Then, is bounded. So are and . In particular,

(4.6)

From and we have

(4.7)

Since is bounded, exists. From and we also have

(4.8)

In fact, from (4.8), we have

(4.9)

Since exists, we have that . On the other hand implies that

(4.10)

Further, we have

(4.11)

From (4.3), we have

(4.12)

On the other hand, we have

(4.13)

Then, we have

(4.14)

Therefore, we have

(4.15)

We apply Lemma 2.2 to get

(4.16)

Lastly, we show that the sequence converges to . Since is bounded and is reflexive, is nonempty. Let be an arbitrary element. Then a subsequence of converges weakly to . From Lemma 2.1 and (4.16), we obtain that . Next, we show . Let be an arbitrary element of . From and , we have

(4.17)

and hence

(4.18)

Therefore, we have

(4.19)

As in the proof of Theorem 3.1, we have

(4.20)

By (A2), we have

(4.21)

and hence

(4.22)

From (A4), we have

(4.23)

Then, for and , from (A1) and (A4), we also have

(4.24)

Taking and using (A3), we get

(4.25)

and hence . Lemma 2.3 and (4.6) ensure the strong convergence of to . This completes the proof.