#### Abstract

Let and be real Hilbert spaces, let be two nonempty closed convex sets, and let be two bounded linear operators. The split equality problem (SEP) is to find such that . Let ; consider a contraction with coefficient , a strongly positive linear bounded operator with coefficient , and is a -inverse strongly monotone mapping. Let , and be defined by restricting to is and restricting to is , that is, has the matrix form . It is proved that the sequence generated by the iterative method converges strongly to which solves the SEP and the following variational inequality: and for all . Moreover, if we take , then is a -inverse strongly monotone mapping, and the sequence generated by the iterative method converges strongly to which solves the SEP and the following variational inequality: for all .

#### 1. Introduction and Preliminaries

Throughout this paper, let be a real Hilbert space and a nonempty closed convex subset of , and let denote the identity operator on Hilbert space , the set of the fixed points of an operator . An operator on a Hilbert space is nonexpansive if, for each and in , . is said to be strongly positive, if there exists such that for all . is said to be inverse strongly monotone if there exists such that for all . A bounded linear operator on a Hilbert space is positive operator if and for all in the spectrum of or equivalent to for all .

Let denote the projection from onto a nonempty closed convex subset of ; that is, It is well known that is characterized by the inequality and is nonexpansive.

Let and be nonempty closed convex subsets of real Hilbert spaces and , respectively, and let be a bounded linear operator. The split feasibility problem (SFP) is to find a point satisfying the property The SFP was first introduced by Censor and Elfving [1], which attracts many authorsâ€™ attention due to its application in signal processing [1]. Various algorithms have been invented to solve it (see [2â€“7]).

Recently, Moudafi [8] proposes a new split equality problem (SEP): let , , and be real Hilbert spaces, let , be two nonempty closed convex sets, and let , be two bounded linear operators. Find , satisfying When , SEP reduces to the well-known SFP. In the paper [8], Moudafi gives the alternating CQ-algorithm and relaxed alternating CQ-algorithm iterative algorithm to solve the split equality problem. We must point out that the above algorithms converge weakly to a solution of the SEP.

We use to denote the solution set of SEP, and letting in define by , then has the matrix form The original problem can now be reformulated as finding with or, more generally, minimizing the function over . Therefore solving SEP (4) is equivalent to solving the following minimization problem: In the paper [9], we use the well-known Tychonov regularization to get some algorithms converge strongly to the minimum-norm solution of the SEP. Moreover, in the paper [9], we obtain that, for , . That is to say, the set of the solution of SEP is equal to the fixed points set of the nonexpansive mapping .

Recall that the variational inequality problem is to find such that , for all , where is closed, convex, and nonempty, and is a mapping on . The set of solutions of the variational inequality is denoted by . In the paper [9], we also get that if and only if is a solution of the variational inequality , for all .

In this paper, we consider a contraction on with coefficient , a strongly positive linear bounded operator with coefficient , and is a -inverse strongly monotone mapping. Let , . We prove that the sequence generated by the iterative method converges strongly to which solves the SEP and the following variational inequality: and for all under some mild conditions for , , and .

We now collect some elementary facts which will be used in the proofs of our main results.

Lemma 1. *Assuming is a positive operator, then is -inverse strongly monotone.*

*Proof. *Since is a positive operator on , by the functional calculus of positive operator (for more details see [10, Chapter 4]), there exists positive operator such that and .

Hence, for any , . It follows that . That is to say, is -inverse strongly monotone.

Lemma 2 (see [11, 12]). *Let be a Banach space, a closed convex subset of , and a nonexpansive mapping with . If is a sequence in weakly converging to and if converges strongly to , then .*

Lemma 3 (see [13]). *Let be a sequence of nonnegative real numbers, a sequence of real numbers in with , a sequence of nonnegative real numbers with , and a sequence of real numbers with . Suppose that
**
Then . *

Lemma 4 (see [14]). *Assuming is a strongly positive linear bounded operator on a Hilbert space with coefficient , then for . *

Lemma 5 (see [13]). *Let be a nonempty and closed subset of a Banach space , and let be a family of mappings of into itself which satisfies the condition: for each bounded subset of ,
**
Then, for each , converges strongly to a point in . Moreover, is defined by
**
Then, for each bounded subset of ,
**
and , if .*

#### 2. Main Results

Theorem 6. *Let denote the solutions set of the SEP, , a contraction on with coefficient , a strongly positive linear bounded operator with coefficient , and a -inverse strongly monotone mapping. Assume that , , and is the sequence generated by the following algorithm: ,
**
for all , where , and . If , , and are chosen such that for some with and*(i)*,*(ii)*, *(iii)*,*(v)(iv)*then the sequence converges strongly to which solves the SEP and the following variational inequality: and for all . *

*Proof. *First, we show that the sequence is bounded. Consider the mapping . Since is a -inverse strongly monotone mapping, we have that, for all ,
Since , we deduce that . That is to say, is nonexpansive.

Letting , , choosing , then and . is nonexpansive, which implies that
Since is strongly positive linear bounded operator, is nonexpansive, and is contraction, we have
It follows from induction that
Hence, is bounded, so are , , , and .

Next, we show that and . In fact, in view that and are nonexpansive, we can get that
Hence,
Based on
we deduce that
By Lemma 3, we have
On the other hand, since and we have
Therefore, we have
Since , , we can get that
Furthermore, by the property of the projection , we have
Then, we have
which yields that
Since , , and , we can get that
Next, since we can get that
Hence,

Let the mapping be defined by
By Lemma 5 we can get that and
Note that it is easy to check that is a contraction; by Banach contraction mapping principle, has a unique fixed point . By the property of the projection , we can get that
Choose a subsequence of such that
Since is bounded, there exists a subsequence of which converges weakly . Without loss of generality, we may assume that converges weakly . Note that, by , we can get converges weakly . Furthermore, ; by Lemmas 2 and 5, we obtain that .

Next we will prove that .

Choose , and let . Since , we can get . On the other hand, by the property of the projection and , we have
It follows that
Since and converges weakly , we can get that .

Note that is a -inverse strongly monotone mapping, and if , then for all . Hence, we can choose such that . By
we can get that
Hence
It follows that . We obtain that and , for all .

Finally, we prove that . Since , we have