Abstract
The purpose of this paper is to introduce a split equilibrium problem (SEP) and find a solution of the equilibrium problem such that its image under a given bounded linear operator is a solution of another equilibrium problem. By using the iterative method, we construct some iterative algorithms to solve such problem in real Hilbert spaces and obtain some strong and weak convergence theorems. Finally, we point out that there exist many SEPs which need the use of new methods to solve them. Some examples are given to illustrate our results.
MSC:47J25, 47H09, 65K10.
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1 Introduction
Throughout this paper, the symbols and are used to denote the sets of positive integers and real numbers, respectively.
In this paper, we propose a new equilibrium problem, which is called a split equilibrium problem (SEP). Let and be two real Banach spaces. Let C be a closed convex subset of , K a closed convex subset of , and a bounded linear operator. f is a bi-function from into and g is a bi-function from into . The SEP is
and
If we consider only the problem (1.1), then (1.1) is a classical equilibrium problem. From (1.1) and (1.2), we can see that the SEP contains two equilibrium problems, and the image of a solution of one equilibrium problem under a given bounded linear operator is a solution of another equilibrium problem. Since many problems coming from physics, optimization, and economics reduce to find a solution of the equilibrium problem (1.1) (see, for instance, [1, 2]), the equilibrium problem (1.1) is very important in the field of applied mathematics. Some authors have proposed some methods to find the solution of the equilibrium problem (1.1). As a generalization of the equilibrium problem (1.1), when finding a common solution for some equilibrium problems, it has been considered in the same subset of the same space; see [3–5]. However, in general, some equilibrium problems always belong to different subsets of spaces, so the SEP is important and quite general. The SEP should enable us to split the solution between two different subsets of spaces so that the image of a solution point of one problem, under a given bounded linear operator, is a solution point of another problem. A special case of the SEP is the split variational inequality problem (SVIP); see [6].
For convenience, in this paper let , and denote the solution set of (1.1), (1.2) and the SEP, respectively.
Example 1.1 Let , and . Let for all , then A is a bounded linear operator. Let , and be defined by , , respectively. Clearly, and . So .
Example 1.2 Let with the standard norm and with the norm for some . and . Define a bi-function , where , , then f is a bi-function from into with . For each , let , then A is a bounded linear operator from into . In fact, it is also easy to verify that and for some and . Now define another bi-function g as follows: for all . Then g is a bi-function from into with .
Clearly, when , we have . So .
Remark 1.1 The SEP in Example 1.1 lies in two different subsets of the same space. While the SEP in Example 1.2 lies in two different subsets of the different space.
In this paper, we construct some iterative algorithms to solve the SEP. Some strong and weak convergence theorems are established. The results obtained in this paper can be reckoned as the new development of the equilibrium problem (1.1). Finally, we point out that there exist many SEPs which need the use of new methods to solve them. Some examples are given to illustrate our results.
2 Preliminaries
We assume that H is a real Hilbert space with zero vector θ whose inner product and norm are denoted by and , respectively; and we use symbols → and ⇀ to denote strong and weak convergence, respectively.
Let and be two Hilbert spaces. The operator A from into and the operator B from into are two bounded linear operators. B is called the adjoint operator of A, if for all , , B satisfies . Especially, if , then B reduces to the well-known adjoint operator of A.
Remark 2.1 It is easy to verify that the operator B, an adjoint operator of A, has the following characters:
-
(i)
; (ii) B is a unique adjoint operator of A.
Example 2.1 Let with the standard norm and with the norm for some . denotes the inner product of for some and denotes the inner product of for some , . Let , 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.
Example 2.2 Let with the norm for some and with the norm for some . Let and denote the inner product of and , respectively, where , , , . Let for , then A is a bounded linear operator from into with because . 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 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 characters:
-
(i)
for every .
-
(ii)
For , and , , .
-
(iii)
For and ,
(2.1)
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 each Hilbert space satisfies Opial’s condition.
The following results are crucial to our main results.
Lemma 2.1 (see [1])
Let K be a nonempty closed convex subset of H and F be a bi-function ofinto R satisfying the following conditions:
(A1) for all;
(A2) F is monotone, that is, for all;
(A3) for each,
(A4) for each, is convex and lower semi-continuous.
Letand. Then, there existssuch that
Lemma 2.2 (see [7])
Let K be a nonempty closed convex subset of H and let F be a bi-function ofinto R satisfying (A 1)-(A 4). Forand, define a mappingas follows:
for all. Then the following hold:
-
(i)
is single-valued;
-
(ii)
is firmly non-expansive, that is, for any,
-
(iii)
for;
-
(iv)
is closed and convex.
Lemma 2.3 (see [3])
Let H be a real Hilbert space. Then for anyandwith, , we have
In particular, we have
-
(1)
for alland;
-
(2)
the mapdefined byis convex.
Lemma 2.4 (see, e.g., [8])
Let H be a real Hilbert space. Then the following hold:
-
(a)
for all;
-
(b)
for all.
Lemma 2.5 Let K be a nonempty closed convex subset of H. For, let the mappingbe the same as in Lemma 2.2. Then forand,
Proof For and , by (i) of Lemma 2.2, we can let and . By the definition of , we have
and
Taking in (2.3) and in (2.4), we have
and
Since the bi-function F satisfies the condition (A2), from (2.5) and (2.6) we have
which implies that
Thus, we have
this implies that
so
namely,
The proof is completed. □
3 Main results
In this section, we will solve the SEP which satisfies the conditions (A1)-(A4).
Theorem 3.1 (Weak convergence theorem)
Let C be a nonempty closed convex subset ofand K a nonempty closed convex subset of, whereandare two real Hilbert spaces. denotes a finite index set. For any, is a bi-function with. Letbe a bounded linear operator with the adjoint B anda bi-function with. Letand () be sequences generated by
wherewith, is a projection operator frominto C andis a constant. Suppose that, then the sequencesand () converge weakly to an element, whileconverges weakly to.
Proof For each and each , let : be defined by (2.2), then for all by Lemma 2.2. Again let : be defined by (2.2), then for all . So (3.1) can be rewritten as follows:
Let be a point such that and , namely, . By Lemma 2.2 and Lemma 2.4, it follows that
hence
Applying Lemma 2.3, we get
(3.3) and (3.4) imply that
Again from Lemma 2.2, we have
By (b) of Lemma 2.4 and (3.6), for each , we have
We also have
From (3.2), (3.5)-(3.8), we have
Notice , . It follows from (3.9) that
and
(3.10) implies exists. Further, from (3.10)-(3.11),
Again from (3.5), we have
which yields that
and
Because exists, which implies is bounded, hence has a weakly convergence subsequence . Assume that for some . Then , and by (3.13) and (3.15).
Now we prove or, to be more precise, we prove and . By Lemma 2.2, for any , , , and . For , since by (3.14), we have for . Otherwise, if for all , then by Opial’s condition, we have
this is a contradiction. So this shows that . Similarly, we can prove .
Finally, we prove and converge weakly to , while converges weakly to . Firstly, if there exists other subsequence of which is denoted by such that with , then, by Opial’s condition,
This is a contradiction. Hence and converge weakly to , respectively.
On the other hand, by (3.15) we also have . Notice that by (3.12), so we have and . We obtain the desired result. □
Corollary 3.1 Let C be a nonempty closed convex subset ofand K a nonempty closed convex subset of, whereandare two real Hilbert spaces. is a bi-function with. Letbe a bounded linear operator with the adjoint B anda bi-function with. Letandbe sequences generated by
wherewith, is a projection operator frominto C andis a constant. Suppose that, then the sequencesandconverge weakly to an element, whileconverges weakly to.
Theorem 3.2 (Strong convergence theorem)
Let C be a nonempty closed convex subset ofand K a nonempty closed convex subset of, whereandare two real Hilbert spaces. denotes a finite index set. For any, is a bi-function with. Letbe a bounded linear operator with the adjoint B anda bi-function with. Letandand () be sequences generated by
wherewith, is a projection operator frominto C andis a constant. Suppose that, then the sequencesand () converge strongly to an element, whileconverges strongly to.
Proof By Lemma 2.2, for all , and for all . We claim for . In fact for . Indeed, let , it follows from (3.7) and (3.8) that
and
From (3.17)-(3.19) we have
Notice , . It follows from (3.20) that
this shows for all , so and for .
We want to prove is a closed convex set for . It is easy to verify that is closed for , so it suffices to verify is convex for . In fact, let , for each , we have
namely, . Similarly, we have , this shows and is a convex set for .
By (iv) of Lemma 2.2, Ω is a closed convex set, so there exists a unique element . Since , we have , which shows that is bounded. So are and . Notice that and , then
It follows that exists.
For some with , from and (2.1), we have
By (3.22)-(3.23) we have , so is a Cauchy sequence. Let .
Next we prove . Firstly, by , from (3.17) we have
Again from (3.20), we have
So
Notice , hence from (3.24) and (3.5), we obtain
Since , (3.27) and Lemma 2.5 imply that for ,
which yields that for all , further . Since A is a bounded linear operator, by . Then for , by (3.26) and Lemma 2.5 we have
hence, for . Thus we have proved , namely, converges strongly to . Notice (3.27), we also have () converges strongly to .
Since by (3.24), we have by . Again from (3.26) we have
hence . The proof is completed. □
Corollary 3.2 Let C be a nonempty closed convex subset ofand K a nonempty closed convex subset of, whereandare two real Hilbert spaces. is a bi-function with. Letbe a bounded linear operator with the adjoint B anda bi-function with. Letandandbe sequences generated by
wherewith, is a constant. Suppose that. Then the sequencesandconverge strongly to an element, whileconverges strongly to.
If and in Theorem 3.1 and Theorem 3.2, we have the following corollaries.
Corollary 3.3 Letandbe two real Hilbert spaces. denotes a finite index set. For any, is a bi-function with. Letbe a bounded linear operator with the adjoint B anda bi-function with. Letandbe sequences generated by
wherewith, is a constant. Suppose that. Then the sequencesand () converge weakly to an element, whileconverges weakly to.
Corollary 3.4 Letandbe two real Hilbert spaces. is a bi-function with. Letbe a bounded linear operator with the adjoint B anda bi-function with. Letandbe sequences generated by
wherewith, is a constant. Suppose that. Then the sequencesandconverge weakly to an element, whileconverges weakly to.
Corollary 3.5 Letandbe two real Hilbert spaces. denotes a finite index set. For any, is a bi-function with. Letbe a bounded linear operator with the adjoint B anda bi-function with. Letandand () be sequences generated by
wherewith, is a constant. Suppose that. Then the sequencesand () converge strongly to an element, whileconverges strongly to.
Corollary 3.6 Letandbe two real Hilbert spaces. is a bi-function with. Letbe a bounded linear operator with the adjoint B anda bi-function with. Letandandbe sequences generated by
wherewith, is a constant. Suppose that. Then the sequencesandconverge strongly to an element, whileconverges strongly to.
Remark 3.1 Since Example 1.1 and Example 1.2 satisfy the conditions of Corollary 3.1 and Corollary 3.2, the SEPs in Example 1.1 and Example 1.2 can be solved by the algorithm (3.16) and (3.28).
Remark 3.2 The results of this paper provide some solution algorithms for some SEPs; however, there are still some SEPs which cannot be solved by the results of this paper. The following examples belong to the case.
Example 3.1 Let and with the norm for some . and . Define a bi-function , where , , then f is a bi-function from into with . For each , let , then A is a bounded linear operator from into . Now define another bi-function g as follows: for all . Then g is a bi-function from into with .
Clearly, when with and , we have . So . However, because the bi-function f does not satisfy the conditions (A1)-(A4), the SEP in this example cannot be solved by Corollary 3.1 or Corollary 3.2.
Example 3.2 Let , , A and B be the same as Example 2.2. Let and . Define a bi-function , where , , then f is a bi-function from into with . Define another bi-function , where , , then .
Clearly, when , we have . So . However, since all f and g do not satisfy the conditions (A1)-(A4), we cannot use the results obtained in this paper to solve the SEP in this example.
4 Conclusion
There are still many SEPs which do not satisfy the conditions (A1)-(A4), so we need to develop some new methods to solve these problems in the future.
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Acknowledgement
The author was supported by the Natural Science Foundation of Yunnan Province (2010ZC152).
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He, Z. The split equilibrium problem and its convergence algorithms. J Inequal Appl 2012, 162 (2012). https://doi.org/10.1186/1029-242X-2012-162
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DOI: https://doi.org/10.1186/1029-242X-2012-162