Nonlinear algorithms approach to split common solution problems

In this paper, we introduce some new iterative algorithms for the split common solution problems for equilibrium problems and fixed point problems of nonlinear mappings. Some examples illustrating our results are also given.

MSC: 47J25, 47H09, 65K10.

Throughout this paper, we assume that H is a real Hilbert space with zero vector θ, whose inner product and norm are denoted by and , respectively. Let K be a nonempty subset of H and T be a mapping from K into itself. The set of fixed points of T is denoted by . The symbols ℕ and ℝ are used to denote the sets of positive integers and real numbers, respectively.

Let C and K be nonempty subsets of real Banach spaces and , respectively. Let be a bounded linear mapping, T a mapping from C into itself with and f a bi-function from into R. The classical equilibrium problem is to find such that

The symbol is used to denote the set of all solutions of the problem (1.1), that is,

The equilibrium problem contains optimization problems, variational inequalities problems, saddle point problems, the Nash equilibrium problems, fixed point problems, complementary problems, bilevel problems, and semi-infinite problems as special cases and have many applications in mathematical program with equilibrium constraint; for detail, one can refer to [1-4] and references therein.

In this paper, we study the following split common solution problem (SCSP) for equilibrium problems and fixed point problems of nonlinear mappings A, T and f:

(SCSP) Find such that and which satisfies , . The solution set of (SCSP) is denoted by

Many authors had proposed some methods to find the solution of the equilibrium problem (1.1). As a generalization of the equilibrium problem (1.1), finding a common solution for some equilibrium problems and fixed point problems of nonlinear operators, it has been considered in the same subset of the same space; see [5-15]. However, some equilibrium problems and fixed point problems of nonlinear mappings always belong to different subsets of spaces in general. So the split common solution is very important for the research on generalized equilibriums problems and fixed point problems.

Example 1.1 Let , and . Let for all and for all . Let be define by for all . Clearly, A is a bounded linear operator, and . So .

Example 1.2 Let with the norm for and with the standard norm . Let and . Let for and for all . Then and A is a bounded and linear operator from into with . Now define a bi-function f as for all . Then f is a bi-function from into ℝ with .

Clearly, , . So .

Remark 1.1 It is worth to mention that the split common solution problem in Example 1.1 lies in two different subsets of the same space and the split common solution problem in Example 1.2 lies in two different subsets of the different space. So, Examples 1.1 and 1.2 also show that the split common solution problem is meaningful.

In this paper, we introduce a weak convergence algorithm and a strong convergence algorithm for the split common solution problem when the nonlinear operator T is a quasi-nonexpansive mapping. Some strong and weak convergence theorems are established. We also give some examples to illustrate our results.

We write to indicate that the sequence weakly converges to x and will symbolize strong convergence as usual.

A Banach space is said to satisfy Opial’s condition, if for each sequence in X which converges weakly to a point , we have

It is well known that any Hilbert space satisfies Opial’s condition.

Let K be a nonempty subset of real Hilbert spaces H. Recall that a mapping is said to be nonexpansive if for all and quasi-nonexpansive if and for all , .

Example 2.1 Let with the inner product defined by for all and the standard norm . Let and for all . Obviously, . It is easy to see that

and

Hence, T is a continuous quasi-nonexpansive mapping but not nonexpansive.

Definition 2.1 (see [16])

Let K be a nonempty closed convex subset of a real Hilbert space H and T a mapping from K into K. The mapping T is said to be demiclosed if, for any sequence which weakly converges to y, and if the sequence strongly converges to z, then .

Remark 2.1 In Definition 2.1, the particular case of demiclosedness at zero is frequently used in some iterative convergence algorithms, which is the particular case when , the zero vector of H; for more detail, one can refer to [16].

The following concept of zero-demiclosedness was introduced in [17].

Definition 2.2 (see [17])

Let K be a nonempty, closed, and convex subset of a real Hilbert space and T a mapping from K into K. The mapping T is called zero-demiclosed if in K satisfying and implies .

The following result was essentially proved in [17], but we give the proof for the sake of completeness.

Proposition 2.1LetKbe a nonempty, closed, and convex subset of a real Hilbert space with zero vectorθandTa mapping fromKintoK. Then the following statements hold.

(a) Tis zero-demiclosed if and only ifis demiclosed atθ;

(b) IfTis a nonexpansive mappings and there is a bounded sequencesuch thatas, thenTis zero-demiclosed.

Proof Obviously, the conclusion (a) holds. To see (b), since is bounded, there is a subsequence and such that . One can claim . Indeed, if , it follows from the Opial’s condition that

which is a contradiction. So and hence T is zero-demiclosed. □

Example 2.2 Let H, C, and T be the same as in Example 2.1. Let be a sequence in C. If and , then . Indeed, since T is continuous, we have and . Hence, T is zero-demiclosed.

Example 2.3 Let with the inner product defined by for all and the standard norm . Let . Let T be a mapping from C into C defined by

Then T is a discontinuous quasi-nonexpansive mapping but not zero-demiclosed.

Proof Obviously, , and T is a quasi-nonexpansive operator. On the other hand, let for all , then it is not hard to prove that , and . So T is not zero-demiclosed. □

Let and be two Hilbert spaces. Let and be two bounded linear operators. B is called the adjoint operator (or adjoint) of A, if for all , , B satisfies . It is known that the adjoint operator of a bounded linear operator on a Hilbert space always exists and is bounded linear and unique. Moreover, it is not hard to show that if B is an adjoint operator of A, then .

Example 2.4 Let with the norm for and with the norm for . Let and denote the inner product of and , respectively, where , , , . Let for . Then A is a bounded linear operator from into with . For , let . Then B is a bounded linear operator from into with . Moreover, for any and , , so B is an adjoint operator of A.

Let K be a closed and convex subset of a real Hilbert space H. For each point , there exists a unique nearest point in K, denoted by , such that , . The mapping is called the metric projection from H onto K. It is well known that has the following characterizations:

(i) for every .

(ii) for , and , , .

(iii) for all and .

The following lemmas are crucial in our proofs.

Lemma 2.1 (see [1])

LetKbe a nonempty, closed, and convex subset ofHandFbe a bi-function ofintoRsatisfying the following conditions.

(A1) for all;

(A2) Fis monotone, that is, for all;

(A3) for each, ;

(A4) for each, is convex and lower semicontinuous.

Letand. Then there existssuch that, for all.

Lemma 2.2 (see [3])

LetKbe a nonempty, closed, and convex subset ofHand letFbe a bi-function ofintoRsatisfying (A1)-(A4). For, define a mappingas follows:

for all. Then the following hold:

(i) is single-valued andfor anyandis closed and convex;

(ii) is firmly nonexpansive, that is, for any, .

Lemma 2.3 (see, e.g., [9])

LetHbe a real Hilbert space. Then the following hold.

(a) andfor all;

(b) for alland.

The following result is simple, but it is very useful in this paper; see also [18].

Lemma 2.4Let the mappingbe defined as (2.1). Then forand,

In particular, for anyand, that isis nonexpansive for any.

Proof For and , by (i) of Lemma 2.2, and for some . By the definition of , we have

and

So, combining (2.2), (2.3), and (A2), we get

or

or

or

or

which implies

and hence

In particular, the last inequality show that for any , is nonexpansive. The proof is completed. □

In this section, we first introduce a weak convergence iterative algorithms for the split common solution problem.

Theorem 3.1Letandbe two real Hilbert spaces andandbe two nonempty closed convex sets. Letbe zero-demiclosed quasi-nonexpansive mappings andbe bi-functions with. Letbe a bounded linear operator with its adjointB.

Givenand. Letandbe sequences generated by

wherewith, is a constant, is a projection operator fromintoCandsatisfiesfor. Thenand.

Proof Let . Then . For each , by Lemmas 2.2 and 2.3, we have

So,

By (b) of Lemma 2.3 and (3.2), for each , we get

Note that for any ,

so it follows from (3.1), (3.3), and (3.4) that

Since , , by (3.5), we obtain

and

The inequality (3.6) implies that exists. Further, from (3.6) and (3.7), we get

and

From (3.1) and (3.10), we have

Since exists, is bounded and hence has a weakly convergence subsequence . Assume that for some . Then , and by (3.10) and (3.11).

We argue . Since T is a zero-demiclosed mapping, by (3.9) and , we obtain . Applying Lemma 2.2, for any . We claim . If , since as from (3.10) and applying Opial’s condition, we have

which lead to a contradiction. So , and hence we show .

Now, we prove converges weakly to . Otherwise, if there exists other subsequence of which is denoted by such that with . Then, by Opial’s condition,

This is a contradiction. Hence, converges weakly to an element .

Finally, we prove converges weakly to . Since , we have as . Thus, by (3.10), we obtain as . The proof is completed. □

Corollary 3.1Letandbe two real Hilbert spaces. Letbe a zero-demiclosed quasi-nonexpansive mapping withandbe a bi-function with. Letbe a bounded linear operator with its adjointB. Given. Letandbe sequences generated by

whereandwith. Supposeand the control coefficient sequencesatisfiesfor. Then the sequenceconverges weakly to an elementandweakly to.

Next, we introduce a strong convergence algorithm for the split common solution problem.

Theorem 3.2Letandbe two nonempty, closed, and convex sets, zero-demiclosed quasi-nonexpansive mappings anda bi-function with. Letbe a bounded linear operator with the adjointB. Given, and. Letandbe sequences generated by

wherewith, is a projection operator fromintoCandis a constant, satisfiesfor, thenand.

Proof First, we claim for . In fact, let . Following the same argument as in Theorem 3.1, we have

and

By (3.13), (3.14), and (3.15), we get

Notice , . It follows from (3.16) that

and hence for all . Hence, and for all .

Now, we prove is a closed convex set for each . It is not hard to verify that is closed for each , so it suffices to verify that is convex for each . Indeed, let . For any , since

we have . Similarly, we also have , which implies . Hence, we show that is a convex set for each .

Notice that and , then for with . It follows that exists. Hence is bounded, which yields that and are bounded. For any with , from and the character (iii) of the projection operator P, we have

Since exists, by (3.17), we have , which implies that is a Cauchy sequence.

Let . One claim . Firstly, by , from (3.13) we have

and

Setting , from (3.16) again, we have

So

and

Let . Since as , Lemma 2.4 and equation (3.21) imply that

So , which say that . On the other hand, since by (3.19) and , we have . Notice that T is zero-demiclosed quasi-nonexpansive mappings, by (3.20), , namely, . So . From (3.21), we also have converges strongly to . The proof is completed. □

Corollary 3.2Letandbe two real Hilbert spaces. Letbe a zero-demiclosed quasi-nonexpansive mappings withandbe a bi-function with. Letbe a bounded linear operator with the adjoint B. Given, , and. Letandbe sequences generated by

wherewith, andis a constant. Suppose thatand the control coefficient sequencesatisfiesfor, then the sequenceconverges strongly to an elementandconverges strongly to.

Example 3.1 Let with the inner product defined by for all and the standard norm . Let and for all . From Examples 2.1 and 2.2, we know that T is an zero-demiclosed quasi-nonexpansive mapping with .

Let and for all , then f satisfies the condition (A1)-(A4) and . Let for all , then A is a bounded linear operator with B (the adjoint of A) =A and .

Obviously, , so . Let , and be sequences generated by

where, and for all , is a projection operator from into C. Then the sequence converges strongly to and converges strongly to .

Proof

(i) Firstly, for given for , we prove that for any , there exists a unique sequence in K such that

Because (3.24) is equivalent with

while (3.25) is true if and only if for all . So the conclusion is true.

(ii) Secondly, it is not hard to compute for all . Hence,

(iii) By (i) and (ii), for , we can rewrite the algorithm (3.23) as follows:

and

As in Example 2.1, we easily obtain . Hence, from (3.26) and (3.27), we get

which shows . Since , , we obtain .

 □

The authors declare that they have no competing interests.

Both authors contributed equally and significantly in writing this paper. Both authors read and approved the final manuscript.

The authors would like to express their sincere thanks to the anonymous referee for their valuable comments and useful suggestions in improving the paper. The first author was supported by the Natural Science Foundation of Yunnan Province (2010ZC152). The second author was supported partially by Grant no. NSC 100-2115-M-017-001 of the National Science Council of the Republic of China.

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