Abstract

The main purpose of this paper is to study the split fixed point and equilibrium problems which includes fixed point problems, equilibrium problems, and variational inequality problems as special cases. A damped algorithm is presented for solving this split common problem. Strong convergence analysis is shown.

1. Introduction

Very recently, the split problems (e.g., the split feasibility problem, the split common fixed points problem, and the split variational inequality problem) have been studied extensively, see, for instance, [119]. Now we recall the related history. Let and be two Hilbert spaces and and two nonempty closed convex subsets. Let be a bounded linear operator. The split feasibility problem is to solve the inclusion: which arise in the field of intensity-modulated radiation therapy and was presented in [1]. The iteration is popular with . Further, Xu [3] suggested a single step regularized method. Dang and Gao [4] developed a damped projection algorithm. If and are the fixed point sets of mappings and , respectively, then (1) becomes a special case of the split common fixed point problem:

Censor and Segal [5] invented a scheme below to solve (2):

Cui et al., [6] extended the damped projection algorithm to the split common fixed point problems. Let be a bifunction. The equilibrium problem is to find such that

We will indicate with the set of solutions of (4).

In the present paper, our main purpose is to study the following split fixed point and equilibrium problem. where and are the sets of fixed points of two nonlinear mappings and , respectively; and are the solution sets of two equilibrium problems with bifunctions and , respectively, and is a bounded linear mapping. Denote the solution set of (5) by

We develop a damped algorithm to solve this split fixed point and equilibrium problem. Strong convergence of the suggested damped algorithm is demonstrated.

2. Concepts and Lemmas

Let be a real Hilbert space with inner product and norm , respectively. Let be a nonempty closed convex subset of . A mapping is called nonexpansive if for all . We call the metric projection if for each It is well known that the metric projection is firmly nonexpansive, that is, for all . Hence is also nonexpansive.

Lemma 1 (see [20]). Let be a nonempty closed convex subset of a real Hilbert space . Let be a bifunction which satisfies the following conditions: (H1) for all ; (H2) is monotone, that is, for all ; (H3) for each , ; (H4) for each , is convex and lower semicontinuous.
Let and . Then, there exists such that
Further, if , then the following hold: (i) is single-valued and is firmly nonexpansive, that is, for any , ; (ii) is closed and convex and .

Lemma 2 (see [21]). Let be a Hilbert space and a closed convex subset. Let be a nonexpansive mapping. Then, the mapping is demiclosed. That is, if is a sequence in such that weakly and strongly, then .

Lemma 3 (see [22]). Assume that 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

Let and be two Hilbert spaces and and two nonempty closed convex subsets. Let be a bounded linear operator with its adjoint . Let and let be two bifunctions satisfying the conditions (H1)–(H4) in Lemma 1. Let and be two nonexpansive mappings.

Algorithm 4. Let . Define a sequence as follows: where , and are three constants satisfying , , , and is a real number sequence in .

In the sequel, we assume that

Theorem 5. If satisfies , and , then generated by algorithm (12) converges strongly to which is the minimum-norm element in .

Proof. Let . Then, and . Set , and for all . Then . From Lemma 1, we know that and are firmly nonexpansive. Thus, we have Note that From (12) and (15), we have Observe that Since is the adjoint of , we have Using parallelogram law, we obtain From (16), (21) and (22), we have By (20) and (23), we deduce It follows from (19), we get The boundedness of the sequence yields.
Set . Then, we have Since , we derive by virtue of (18) and (26) that According to (17) and (27), we have It follows that Since is bounded, we can deduce is also bounded. From (29), we have Hence, Using the firmly-nonexpansivenessity of , we have Thus, we get It follows that This together with (30) and (C1) implies that Note that Hence, which implies that So, we get Since we get From (31), (35), and (41), we get Now, we show that Choose a subsequence of such that Notice that is bounded, we can choose of such that . Without loss of generality, we assume that . From the above conclusions, we derive that By Lemma 2, (39), and (41), we deduce and .
Next, we show that . Since , we have By the monotonicity of , we have and so Since , , we obtain . Thus, . For with and , let . We obtain . Hence, So, . And, thus, . This implies that . Similarity, we can prove that . To this end, we deduce and . That is to say, . Therefore, Finally, we prove . From (12), we have Applying Lemma 3 and (50) to (51), we deduce . The proof is completed.

Algorithm 6. Let arbitrarily define a sequence by the following: for all , where and is a real number sequence in .

Corollary 7. Suppose . If satisfies , , and , then the sequence generated by algorithm (52) converges strongly to which is the mum-norm element in .

Algorithm 8. Let arbitrarily define a sequence by the following: for all , where , and are three constants satisfying , , and is a real number sequence in .

Corollary 9. Suppose . If satisfies , , and , then the sequence generated by algorithm (53) converges strongly to which is the mum-norm element in .

Algorithm 10. Let arbitrarily define a sequence by the following: for all , where and is a real number sequence in .

Corollary 11. Suppose . If satisfies , , and , then the sequence generated by algorithm (54) converges strongly to which is the mum-norm element in .

Acknowledgment

Li-Jun Zhu was supported in part by NNSF of China (61362033), NZ13087, NGY2012097 and 2013xyz023.