The Split Feasibility Problem with Some Projection Methods in Banach Spaces

Kawasaki, Toshiharu; Manaka, Hiroko

doi:https://doi.org/10.1155/2020/2913087

Abstract and Applied Analysis

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Research Article | Open Access

Volume 2020 | Article ID 2913087 | https://doi.org/10.1155/2020/2913087

The Split Feasibility Problem with Some Projection Methods in Banach Spaces

Toshiharu Kawasaki^1,2and Hiroko Manaka³

Academic Editor: Micah Osilike

Received24 Mar 2020

Revised27 Jun 2020

Accepted23 Jul 2020

Published28 Dec 2020

Abstract

In this paper, we study the split feasibility problem in Banach spaces. At first, we prove that a solution of this problem is a solution of the equivalent equation defined by using a metric projection, a generalized projection, and sunny generalized nonexpansive retraction, respectively. Then, using the hybrid method with these projections, we prove strong convergence theorems in mathematical programing in order to find a solution of the split feasibility problem in Banach spaces.

Dedicated to the late Professor Wataru Takahashi with our respect

1. Introduction

Bregman proposed a generalization for the cyclic metric projection method of computing points in the intersection of linear closed subspaces of a Hilbert space in [1], invented by von Neumann [2]. Alber and Butnariu achieve distinction of the study of this Bregman projection and the result of the properties. They used this cyclic Bregman projection method for finding the solution of the consistent convex feasibility problem of computing a common point of the closed convex subspaces in a reflexive Banach space [3]. Some fruitful results have been introduced with respect to the sequential algorithm with successive Bregman projection for computing a solution of the convex feasibility problem [4, 5] and so on. Ibaraki and Takahashi studied the properties of a generalized projection which is a special case of Bregman projection and a sunny generalized nonexpansive retraction in Banach spaces [6].

Alsulami, Latif, and Takahashi treated with the following convex feasibility problem [7]: Let be a Hilbert space; let be a strictly convex, reflexive, and smooth Banach space; let be a bounded linear operator from into ; let and be convex and closed subsets of and , respectively. Then, find a point . In particular, such a problem is called the split feasibility problem. Using the methods with metric projections in mathematical programing, they showed strong convergence theorems for finding a solution of the split feasibility problem. In the case of finite dimensional spaces, Byrne treated with the iterative algorithm [8]: , where and a linear operator is represented as a matrix which can be selected to impose consistency with measured data. With respect to examples in this case, there are results by Landweber [9] and Gordon, Bender, and Herman [10]. In [11], Takahashi treated with this problem of a linear bounded operator from into , where and are uniformly convex and smooth Banach spaces. In that paper, it is shown that is equivalent to where and are metric projections on subsets of and of , respectively; and are the identity mappings on and , respectively; and are duality mappings on and , respectively; . Furthermore, the following convergence theorem is proved by the hybrid method with metric projections: Let and be uniformly convex and smooth Banach spaces; let and be nonempty, closed, and convex subsets of and , respectively; let and be duality mappings on and , respectively; let be a bounded linear operator from into with ; let be the adjoint operator of ; let . Suppose that . Let and let be a sequence generated by for any . Then, is strongly convergent to a point for any , where .

In this paper, for uniformly convex and smooth Banach spaces and , we study the split feasibility problem of a bounded linear operator from to . First, we give the diversity of equivalent equations regarding equation (1) with respect to metric projections, generalized projections, and sunny generalized nonexpansive retractions, respectively. Then, using the hybrid methods with these projections, we prove the strong convergence theorems in mathematical programing in order to find a solution of the split feasibility problem in Banach spaces.

2. Preliminaries

We know that the following hold; for instance, see [12–14].

(T1) Let be a Banach space, let be the topological dual space of , and let be the duality mapping on defined by for any . Then, is strictly convex if and only if is injective; that is, implies .

(T2) Let be a Banach space, let be the topological dual space of , and let be the duality mapping on . Then, is reflexive if and only if is surjective; that is, .

(T3) Let be a Banach space and let be the duality mapping on . Then, is smooth if and only if is single-valued.

(T4) Let be a Banach space and let be the duality mapping on . Then, is strictly convex if and only if for any with and and for any .

(T5) Let be a Banach space and let be the topological dual space of . Then, is reflexive if and only if is reflexive.

(T6) Let be a Banach space and let be the topological dual space of . If is strictly convex, then is smooth. Conversely, if is reflexive and smooth, then is strictly convex.

(T7) Let be a Banach space and let be the topological dual space of . If is smooth, then is strictly convex. Conversely, if is reflexive and strictly convex, then, is smooth.

(T8) If is uniformly convex, that is, for any there exists such that and implies , then is reflexive.

(T9) Let be a Banach space, let be the topological dual space of , and let be the duality mapping on . If has a Fréchet differentiable norm, then is norm-to-norm continuous.

(T10) Let be a Banach space and let be the topological dual space of . Then, is uniformly smooth, that is, has a uniformly Fréchet differentiable norm, if and only if is uniformly convex.

Definition 1. Let be a smooth Banach space, let be the duality mapping on , and let be the mapping from into defined by for any .

Since by (T3) is single-valued, is well-defined. It is obvious that implies . Conversely, by (T4),

(T11) If is also strictly convex, then implies .

Let be a strictly convex and smooth Banach space. By (T1) and (T3), is a bijective mapping from onto . In particular, if is also reflexive, then by (T2), is a bijective mapping from onto . If is strictly convex, reflexive, and smooth, then by (T5), (T6) and (T7) is also strictly convex, reflexive, and smooth. Furthermore, since is reflexive, holds and the duality mapping on is .

We use the following lemmas in this paper. The following is shown in [14].

Lemma 2. Let be a Banach space and let be the duality mapping on . Then, for any , for any , and for any . Furthermore, if is strictly convex and smooth, then if and only if .

Definition 3. Let be a strictly convex, reflexive, and smooth Banach space and let be a nonempty, closed, and convex subset of . We know that for any there exists a unique element such that . Such a is denoted by , and is called the metric projection of onto .

The following holds.

Lemma 4. Let be a strictly convex and smooth Banach space, let be a nonempty closed subset of , and let be duality mapping on . Then, for any , if and only if for any .

Definition 5. Let be a strictly convex, reflexive, and smooth Banach space and let be a nonempty, closed, and convex subset of . We know that for any , there exists a unique element such that . Such a is denoted by , and is called the generalized projection of onto .

The following is shown in [15].

Lemma 6. Let be a strictly convex and smooth Banach space; let be a nonempty, closed, and convex subset of ; let be the duality mapping on . Then, the following hold. (i)For any , if and only if for any (ii) for any and for any

Definition 7. Let be a nonempty subset of a smooth Banach space . A mapping from into is said to be generalized nonexpansive [6] if the set of all fixed points of is nonempty and for any and for any fixed point of . Let be a nonempty subset of a Banach space . A mapping from onto is said to be sunny if for any and for any . A mapping from onto is called a retraction or a projection if for any .

The following are shown in [16].

Lemma 8. Let be a strictly convex, reflexive, and smooth Banach space and let be a nonempty and closed subset of . Then, the following are equivalent: (i)There exists a sunny generalized nonexpansive retraction of onto (ii)There exists a generalized nonexpansive retraction of onto (iii) is closed and convex.

Lemma 9. Let be a strictly convex, reflexive, and smooth Banach space, let be a nonempty and closed subset of , and . Suppose that there exists a sunny generalized nonexpansive retraction of onto . Then, the following are equivalent: (i)(ii).

The following are shown in [6].

Lemma 10. Let be a strictly convex and smooth Banach space and let be a nonempty and closed subset of . Suppose that there exists a sunny generalized nonexpansive retraction of onto . Then, the sunny generalized nonexpansive retraction is uniquely determined.

Lemma 11. Let be a strictly convex and smooth Banach space, let be a nonempty and closed subset of , and let be the duality mapping on . Suppose that there exists a sunny generalized nonexpansive retraction of onto . Then, the following hold. (i)For any , if and only if for any (ii) for any and for any

Definition 12. Let . Define a mapping from into by for any . Then, is called the generalized duality mapping on . In particular, .

The following are shown in [17].

Lemma 13. Let be a Banach space. Then, the following are equivalent: (i) is uniformly convex;(ii)For any and for any , there exists a continuous, strictly increasing, and convex function from into such that and for any and for any (iii)For any and for any , there exists a continuous, strictly increasing, and convex function from into such that and for any , for any , and for any

Lemma 14. Let be a smooth Banach space. Then, the following are equivalent: (i) is uniformly smooth;(ii)For any and for any , there exists a continuous, strictly increasing, and convex function from into such that and for any (iii)For any and for any , there exists a continuous, strictly increasing, and convex function from into such that and for any

The following is shown in [18].

Lemma 15. Let be a uniformly convex and smooth Banach space and let . Then, there exists a continuous, strictly increasing, and convex function from into such that and for any .

3. Equivalent Conditions to the Existence of Solutions

In this section, we consider equivalent conditions to the existence of solutions of the split feasibility problem.

Theorem 16. Let and be strictly convex, reflexive, and smooth Banach spaces; let and be the identity mappings on and , respectively; let and be duality mappings on and , respectively; let and be nonempty, closed, and convex subsets of and , respectively; let be a bounded linear operator from into ; let be the adjoint operator of ; let . Suppose that . Consider the following condition: (i).

The following are equivalent to (i): (ii)(iii)(iv)(v)

Proof. The equivalence of (i) and (ii) is shown in [11, Lemma 3.1]. We show the rest.
Suppose that (i) holds. Since , holds. Therefore, and hence, Since , we obtain

Conversely, suppose that (iii), (iv), or (v) holds. Since these equations have the form of or , holds. We show .

In the case of (iii): By Lemma 4, we obtain for any . Therefore,

On the other hand, by Lemma 6, we obtain for any . Since , there exists . Putting and , and hold. Therefore, and hence,

By Lemma 2, we obtain , and hence, ; that is, .

In the case of (iv): By Lemma 6, we obtain for any . Therefore,

On the other hand, by Lemma 4, we obtain for any . Since , there exists . Putting and , and hold. Therefore, and hence,

Therefore, we obtain , and hence, ; that is, .

In the case of (v): By Lemma 6, we obtain for any . Therefore,

On the other hand, by Lemma 6, we obtain for any . Since , there exists . Putting and , and hold. Therefore, and hence,

By Lemma 2, we obtain , and hence, ; that is, .

Remark 17. Since in [11, Lemma 3.1] only the metric projection was used, only Lemma 4 was used for proving the equivalence between (i) and (ii). In Theorem 16, both of the metric projection and the generalize projection are used. Therefore, we have to use both of Lemmas 4 and 6 for proving the equivalence between (i) and (iii), and (iv) and (iv).

Theorem 18. Let and be strictly convex, reflexive, and smooth Banach spaces; let be the dual space of , let and be the identity mappings on and respectively; let and be duality mappings on and respectively; let and be nonempty, closed, and convex subsets of and respectively; let be a bounded linear operator from into ; let be the adjoint operator of ; let . Suppose that . Consider the following condition: (i).

Ifis closed, then the following are equivalent to (i):(vi);(vii).

Ifis closed, then the following are equivalent to (i): (viii);(ix).

Ifandare closed, then the following is equivalent to (i): (x).

Proof. Suppose that (i) holds. Since , hold. Therefore, and hence,

Since , we obtain

Conversely, suppose that (vi), (vii), (viii), (ix), or (x) holds. Since these equations have the form of , , or , holds. We show .

In the case of (vi): By Lemma 4, we obtain for any . Therefore,

On the other hand, by Lemma 11, we obtain for any . Since , there exists . Putting and , and hold. Therefore, and hence,

By Lemma 2, we obtain , and hence, ; that is, .

In the case of (vii): By Lemma 6, we obtain for any . Therefore,

On the other hand, by Lemma 11, we obtain for any . Since , there exists . Putting and , and hold. Therefore, and hence,

By Lemma 2, we obtain , and hence, ; that is, .

In the case of (viii): By Lemma 11, we obtain for any . Therefore,

On the other hand, by Lemma 4, we obtain for any . Since , there exists . Putting and , and hold. Therefore, and hence,

Therefore, we obtain , and hence, ; that is, .

In the case of (ix): By Lemma 11, we obtain for any . Therefore,

On the other hand, by Lemma 6, we obtain for any . Since , there exists . Putting and , and hold. Therefore, and hence,

By Lemma 2, we obtain , and hence, ; that is, .

In the case of (x): By Lemma 11, we obtain for any . Therefore,

On the other hand, by Lemma 11, we obtain for any . Since , there exists . Putting and , and hold. Therefore, and hence,

By Lemma 2, we obtain , and hence, ; that is, .

4. Strong Convergence Theorems to Solutions

In this section, we consider strong convergence theorems to solutions of the split feasibility problem.

Lemma 19. Let be a uniformly convex and smooth Banach space and let . Put for any , where as in Lemma 13. Then, holds for any .

Proof. If , then for arbitrary , we obtain

In particular, the above equation holds for . If , then, since for any , we obtain

Therefore, we obtain

Lemma 20. Let be a uniformly smooth Banach space and let . Put for any and for any , where as in Lemma 14. Then, holds for any and for any .

Proof. If , then for arbitrary , we obtain

In particular, the above equation holds for . If , then, since , we obtain

Therefore, we obtain

We consider the following condition for a uniformly convex and smooth Banach space:

(1) .

We consider the following condition for a uniformly smooth Banach space:

(2) There exist and such that .

Example 1. We consider each of examples satisfying the conditions (1) and (2), respectively.

Let be a Banach space. The modulus of convexity of is defined by for any . Let . is said to be -uniformly convex if there exists a constant such that for any . Furthermore, we know that is uniformly convex if and only if for any . Let be the generalized duality mapping on a Banach space . By [17, Corollary 1], is -uniformly convex if and only if there exists a constant such that for any , for any , and for any . Therefore, if is -uniformly convex and smooth, then we can put for any . We obtain

The modulus of smoothness of is defined by for any. Let . is said to be -uniformly smooth if there exists a constant such that for any . Furthermore, we know that is uniformly smooth if and only if . Let be the generalized duality mapping on a smooth Banach space .

By [17, Corollary 1], is -uniformly smooth if and only if there exists a constant such that for any . Therefore, if is -uniformly smooth, then we can put for any . Let and let . Then, we obtain

Lemma 21. Let be a uniformly convex and smooth Banach space, let be a uniformly smooth Banach space, and let . Suppose that satisfies the condition (1) and satisfies the condition (2). Then, there exists such that for any and for any , where as in Lemma 19 and as in Lemma 20.

Proof. We obtain for any . Put

Then, we obtain for any and for any .

Theorem 22. Let be a uniformly convex and smooth Banach space; let be a strictly convex, reflexive, and uniformly smooth Banach space; let and be duality mappings on and , respectively; let and be nonempty, closed, and convex subsets of and , respectively; let be a bounded and linear operator from into with ; let be the adjoint operator of ; let . Suppose that satisfies the condition (2), satisfies the condition (1) and . Let and let be a sequence generated by for any , where . Then, there exists such that is strongly convergent to a point for any , where .

Proof. For the sake of the proof, we confirm the following facts. Since is uniformly convex, by (T8) is reflexive. Since is uniformly convex and smooth, by (T10) is uniformly smooth. Since is reflexive and uniformly smooth, by (T10) is uniformly convex. Since is strictly convex and reflexive, by (T7) is smooth.

It is obvious that is closed and convex for any . We show that for any . Let . If or , then we obtain , and hence, that is, . If and , then

By Lemma 21, there exists such that for any and for any . Put

Since is uniformly smooth, by Lemma 20, we obtain

Since is uniformly convex and smooth, by Lemma 19, we obtain

By Lemma 6, we obtain

Therefore, we obtain that is, . We show that for any . Since it is obvious that . Suppose that there exists such that . Then, . Since , by Lemma 6, we obtain for any , and hence, the above inequality holds for any ; that is, we obtain . We obtain for any . Therefore, is well-defined.

It is obvious that is nonempty, closed, and convex. Therefore, is well-defined.

By the definition of , we obtain . By Lemma 6, for any . Let with . Since , we obtain and hence,

Since , we obtain for any . Since , we obtain . Since , we obtain and hence, is bounded and nondecreasing; that is, there exists the limit of . From (91), we obtain

By Lemma 15, is a Cauchy sequence. Since is complete, there exists such that . Since , by Lemma 6, we obtain for any . Since , the above inequality holds for any . Taking , since by (T9) is norm-to-norm continuous, we obtain for any . By Lemma 6, we obtain .

Theorem 23. Let be a uniformly convex and smooth Banach space; let be a strictly convex, reflexive, and uniformly smooth Banach space; let be the identity mapping on ; let and be duality mappings on and , respectively; let and be nonempty, closed, and convex subsets of and such that is closed, respectively; let be a bounded and linear operator from into with ; let be the adjoint operator of ; let . Suppose that satisfies the condition (2), satisfies the condition (1) and . Let and let be a sequence generated by for any , where . Then, there exists such that is strongly convergent to a point for any , where .

Proof. For the sake of the proof, we confirm the following facts. Since is uniformly convex, by (T8), is reflexive. Since is uniformly convex and smooth, by (T10), is uniformly smooth. Since is reflexive and uniformly smooth, by (T10), is uniformly convex. Since is strictly convex and reflexive, by (T7), is smooth. Since and have Fréchet differentiable norms, by (T9), and are norm-to-norm continuous. Therefore, and are closed.

It is obvious that is closed and convex for any . We show that for any . Let . If or , then we obtain , and hence, that is, . If and , then

By Lemma 21, there exists such that for any and for any . Put

Since is uniformly smooth, by Lemma 20, we obtain

Since is uniformly convex and smooth, by Lemma 19, we obtain

By Lemma 11, we obtain

Therefore, we obtain that is, . We show that for any . Since it is obvious that . Suppose that there exists such that . Then, . Since , by Lemma 11, we obtain for any , and hence, the above inequality holds for any ; that is, we obtain . We obtain for any . Therefore, is well-defined.

It is obvious that is nonempty, closed, and convex and is closed. Therefore, is well-defined.

By the definition of , we obtain . By Lemma 11, for any . Let with . Since , we obtain and hence,

Since , by Lemma 9, we obtain for any . Since , we obtain . Since , we obtain and hence, is bounded and nondecreasing; that is, there exists the limit of . From (109), we obtain

By Lemma 15, is a Cauchy sequence. Since by (T9) is norm-to-norm continuous, is also a Cauchy sequence. Since is complete, there exists such that . Since , by Lemma 11, we obtain for any . Since , the above inequality holds for any . Taking , since by (T9) is norm-to-norm continuous, we obtain for any . By Lemma 11, we obtain .

Remark 24. In this paper, we consider only two strong convergence theorems with respect to (v) in Theorem 16 and (x) in Theorem 18. The strong convergence theorem with respect to (ii) is shown in [11, Theorem 3.2]. Of course, we can consider other strong convergence theorems with respect to (iii), (iv) in Theorem 16, and (vi), (vii), (viii), and (ix) in Theorem 18. They will be described in the next paper.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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Copyright © 2020 Toshiharu Kawasaki and Hiroko Manaka. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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