The Split Feasibility Problem with Some Projection Methods in Banach Spaces

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.


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 H be a Hilbert space; let E be a strictly convex, reflexive, and smooth Banach space; let A be a bounded linear operator from H into E; let C and D be convex and closed subsets of H and E, respectively. Then, find a point z ∈ C ∩ A −1 ðDÞ. 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]: x n+1 = P C ðx n + rA T ðP D − IÞAx n Þ, where n ∈ ℕ and a linear operator A 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 A from E into F, where E and F are uniformly convex and smooth Banach spaces. In that paper, it is shown that z ∈ C ∩ A −1 ðDÞ is equivalent to where P C and P D are metric projections on subsets C of E and D of F, respectively; I E and I F are the identity mappings on E and F, respectively; J E and J F are duality mappings on E and F, respectively; r ∈ ð0, ∞Þ. Furthermore, the following convergence theorem is proved by the hybrid method with metric projections: Let E and F be uniformly convex and smooth Banach spaces; let C and D be nonempty, closed, and convex subsets of E and F, respectively; let J E and J F be duality mappings on E and F, respectively; let A be a bounded linear operator from E into F with A ≠ 0; let A * be the adjoint operator of A; let r ∈ ð0, ∞Þ. Suppose that C ∩ A −1 ðDÞ ≠ ∅. Let x 1 ∈ E and let fx n g be a sequence generated by for any n ∈ ℕ. Then, fx n g is strongly convergent to a point z 0 ∈ C ∩ A −1 ðDÞ for any r ∈ ð0, kAk −2 Þ, where z 0 = P C∩A −1 ðDÞ x 1 . In this paper, for uniformly convex and smooth Banach spaces E and F, we study the split feasibility problem of a bounded linear operator A from E to F. 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.
(T1) Let E be a Banach space, let E * be the topological dual space of E, and let J E be the duality mapping on E defined by for any for any x, y ∈ E with x ≠ y and kxk = kyk = 1 and for any y * ∈ J E y.
(T5) Let E be a Banach space and let E * be the topological dual space of E. Then, E is reflexive if and only if E * is reflexive.
(T6) Let E be a Banach space and let E * be the topological dual space of E. If E * is strictly convex, then E is smooth. Conversely, if E is reflexive and smooth, then E * is strictly convex.
(T7) Let E be a Banach space and let E * be the topological dual space of E. If E * is smooth, then E is strictly convex. Conversely, if E is reflexive and strictly convex, then, E * is smooth.
(T9) Let E be a Banach space, let E * be the topological dual space of E, and let J E be the duality mapping on E. If E has a Fréchet differentiable norm, then J E is norm-tonorm continuous.
(T10) Let E be a Banach space and let E * be the topological dual space of E. Then, E is uniformly smooth, that is, E has a uniformly Fréchet differentiable norm, if and only if E * is uniformly convex. Definition 1. Let E be a smooth Banach space, let J E be the duality mapping on E, and let V E be the mapping from E × E into ½0, ∞Þ defined by for any x, y ∈ E.
Since by (T3) J E is single-valued, V E is well-defined. It is obvious that x = y implies V E ðx, yÞ = 0. Conversely, by (T4), (T11) If E is also strictly convex, then V E ðx, yÞ = 0 implies x = y.
Let E be a strictly convex and smooth Banach space. By (T1) and (T3), J E is a bijective mapping from E onto J E ðEÞ. In particular, if E is also reflexive, then by (T2), J E is a bijective mapping from E onto E * . If E is strictly convex, reflexive, and smooth, then by (T5), (T6) and (T7) E * is also strictly convex, reflexive, and smooth. Furthermore, since E is reflexive, E * * = E holds and the duality mapping on E * is J −1 E . We use the following lemmas in this paper. The following is shown in [14].

Lemma 2.
Let E be a Banach space and let J E be the duality mapping on E. Then, hx − y, x * − y * i ≥ 0 for any x, y ∈ E, for any x * ∈ JðxÞ, and for any y * ∈ J E y. Furthermore, if E is strictly convex and smooth, then hx − y, x * − y * i = 0 if and only if x = y.
Definition 3. Let E be a strictly convex, reflexive, and smooth Banach space and let C be a nonempty, closed, and convex subset of E. We know that for any x ∈ E there exists a unique element z ∈ C such that kx − zk = min y∈C kx − yk. Such a z is denoted by P C x, and P C is called the metric projection of E onto C.

2
Abstract and Applied Analysis The following holds.

Lemma 4.
Let E be a strictly convex and smooth Banach space, let C be a nonempty closed subset of E, and let J E be duality mapping on E. Then, for any ðx, zÞ ∈ E × C, z = P C x if and only if hz − y, J E ðx − zÞi ≥ 0 for any y ∈ C.
Definition 5. Let E be a strictly convex, reflexive, and smooth Banach space and let C be a nonempty, closed, and convex subset of E. We know that for any x ∈ E, there exists a unique element z ∈ C such that V E ðz, xÞ = min y∈C V E ðy, xÞ. Such a z is denoted by Π C x, and Π C is called the generalized projection of E onto C.
The following is shown in [15].

Lemma 6.
Let E be a strictly convex and smooth Banach space; let C be a nonempty, closed, and convex subset of E; let J E be the duality mapping on E. Then, the following hold.
for any x ∈ E and for any y ∈ C: [6] if the set of all fixed points of T is nonempty and for any x ∈ C and for any fixed point y of T. Let C be a nonempty subset of a Banach space E. A mapping R from E onto C is said to be sunny if for any x ∈ E and for any t ∈ ½0, ∞Þ. A mapping R from E onto C is called a retraction or a projection if Rx = x for any x ∈ C.
The following are shown in [16].

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

Lemma 9.
Let E be a strictly convex, reflexive, and smooth Banach space, let C be a nonempty and closed subset of E, and ðx, zÞ ∈ E × C. Suppose that there exists a sunny generalized nonexpansive retraction R C of E onto C. Then, the following are equivalent: The following are shown in [6].

Lemma 10.
Let E be a strictly convex and smooth Banach space and let C be a nonempty and closed subset of E. Suppose that there exists a sunny generalized nonexpansive retraction of E onto C. Then, the sunny generalized nonexpansive retraction is uniquely determined.
Lemma 11. Let E be a strictly convex and smooth Banach space, let C be a nonempty and closed subset of E, and let J E be the duality mapping on E. Suppose that there exists a sunny generalized nonexpansive retraction R C of E onto C. Then, the following hold.
for any x ∈ E and for any y ∈ C: for any x ∈ E. Then, J p is called the generalized duality mapping on E. In particular, J 2 = J E .
The following are shown in [17].

Lemma 15.
Let E be a uniformly convex and smooth Banach space and let ρ ∈ ð0, ∞Þ. Then, there exists a continuous, strictly increasing, and convex function g ρ from ½0, ∞Þ into ½0, ∞Þ such that g ρ ð0Þ = 0 and for any x, y ∈ B ρ ðEÞ.

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 E and F be strictly convex, reflexive, and smooth Banach spaces; let I E and I F be the identity mappings on E and F, respectively; let J E and J F be duality mappings on E and F, respectively; let C and D be nonempty, closed, and convex subsets of E and F, respectively; let A be a bounded linear operator from E into F; let A * be the adjoint operator of A; let r ∈ ð0, ∞Þ. Suppose that C ∩ A −1 ðDÞ ≠ ∅. Consider the following condition: The following are equivalent to (i): Proof. The equivalence of (i) and (ii) is shown in [11,Lemma 3.1]. We show the rest. Suppose that (i) holds. Since Az ∈ D, P D Az = Π D Az = Az holds. Therefore, and hence, the right side of iii ð Þ = P C z, Since z ∈ C, we obtain the right sides of iii ð Þ, iv ð Þ, and v ð Þ = z: Conversely, suppose that (iii), (iv), or (v) holds. Since these equations have the form of z = P C x or z = Π C x, z ∈ C holds. We show z ∈ A −1 ðDÞ.
In the case of (iii): By Lemma 4, we obtain for any y ∈ C. Therefore, On the other hand, by Lemma 6, we obtain for any v ∈ D. Since C ∩ A −1 ðDÞ ≠ ∅, there exists z 0 ∈ C ∩ A −1 ðDÞ. Putting y = z 0 and v = Az 0 , y ∈ C and v ∈ D hold. Therefore, and hence, By Lemma 2, we obtain Π D Az = Az, and hence, Az ∈ D; that is, z ∈ A −1 ðDÞ.
In the case of (iv): By Lemma 6, we obtain for any y ∈ C. Therefore, On the other hand, by Lemma 4, we obtain for any v ∈ D. Since C ∩ A −1 ðDÞ ≠ ∅, there exists z 0 ∈ C ∩ A −1 ðDÞ. Putting y = z 0 and v = Az 0 , y ∈ C and v ∈ D hold. Therefore, and hence, Therefore, we obtain P D Az = Az, and hence, Az ∈ D; that is, z ∈ A −1 ðDÞ.
In the case of (v): By Lemma 6, we obtain for any y ∈ C. Therefore, On the other hand, by Lemma 6, we obtain for any v ∈ D. Since C ∩ A −1 ðDÞ ≠ ∅, there exists z 0 ∈ C ∩ A −1 ðDÞ. Putting y = z 0 and v = Az 0 , y ∈ C and v ∈ D hold. Therefore, and hence, By Lemma 2, we obtain Π D Az = Az, and hence, Az ∈ D; that is, z ∈ A −1 ðDÞ.
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 E and F be strictly convex, reflexive, and smooth Banach spaces; let F * be the dual space of F, let I E and I F * be the identity mappings on E and F * , respectively; let J E and J F be duality mappings on E and F, respectively; let C and D be nonempty, closed, and convex subsets of E and F, respectively; let A be a bounded linear operator from E into F; let A * be the adjoint operator of A; let r ∈ ð0,∞Þ. Suppose that C ∩ A −1 ðDÞ ≠ ∅. Consider the following condition: If J F ðDÞ is closed, then the following are equivalent to (i): If J E ðCÞ is closed, then the following are equivalent to (i): If J E ðCÞ and J F ðDÞ are closed, then the following is equivalent to (i): Proof. Suppose that (i) holds. Since Az ∈ D, hold. Therefore, and hence, the right side of vi ð Þ = P C z, On the other hand, by Lemma 11, we obtain for any v ∈ J F ðDÞ. Since C ∩ A −1 ðDÞ ≠ ∅, there exists z 0 ∈ C ∩ A −1 ðDÞ. Putting y = z 0 and v = J F Az 0 , y ∈ C and v ∈ J F ðDÞ hold. Therefore, and hence, By Lemma 2, we obtain R J F ðDÞ J F Az = J F Az, and hence, J F Az ∈ J F ðDÞ; that is, z ∈ A −1 ðDÞ.
In the case of (viii): By Lemma 11, we obtain for any y ∈ J E ðCÞ. Therefore, On the other hand, by Lemma 4, we obtain for any v ∈ D. Since C ∩ A −1 ðDÞ ≠ ∅, there exists z 0 ∈ C ∩ A −1 ðDÞ. Putting y = J E z 0 and v = Az 0 , y ∈ J E ðCÞ and v ∈ D hold. Therefore, and hence, Therefore, we obtain P D Az = Az, and hence, Az ∈ D; that is, z ∈ A −1 ðDÞ.
In the case of (ix): By Lemma 11, we obtain Abstract and Applied Analysis for any y ∈ J E ðCÞ. Therefore, On the other hand, by Lemma 6, we obtain for any v ∈ D. Since C ∩ A −1 ðDÞ ≠ ∅, there exists z 0 ∈ C ∩ A −1 ðDÞ. Putting y = J E z 0 and v = Az 0 , y ∈ J E ðCÞ and v ∈ D hold. Therefore, and hence, By Lemma 2, we obtain P D Az = Az, and hence, Az ∈ D; that is, z ∈ A −1 ðDÞ.
In the case of (x): By Lemma 11, we obtain for any y ∈ J E ðCÞ. Therefore, On the other hand, by Lemma 11, we obtain for any v ∈ J F ðDÞ.
and v ∈ J F ðDÞ hold. Therefore, and hence, By Lemma 2, we obtain R J F ðDÞ J F Az = J F Az, and hence, J F Az ∈ J F ðDÞ; that is, z ∈ A −1 ðDÞ.

Strong Convergence Theorems to Solutions
In this section, we consider strong convergence theorems to solutions of the split feasibility problem.
Let E be a Banach space. The modulus of convexity δ E of E is defined by for any x, y ∈ E, for any x * ∈ J p x, and for any y * ∈ J p y. Therefore, if E is 2-uniformly convex and smooth, then we can put g 2,ρ ðrÞ = cr 2 for any ρ ∈ ð0, ∞Þ. We obtain The modulus of smoothness ρ E of E is defined by for anyt ∈ ð0, ∞Þ. Let q ∈ ð1, ∞Þ. E is said to be q-uniformly smooth if there exists a constant c ∈ ð0, ∞Þ such that ρ E ðtÞ ≤ ct q for any t ∈ ð0, ∞Þ. Furthermore, we know that E is uniformly smooth if and only if lim t↓0 ðρ E ðtÞ/tÞ = 0. Let for any x, y ∈ E. Therefore, if E is 2-uniformly smooth, then we can put g * 2,ρ ðrÞ = cr 2 for any ρ ∈ ð0, ∞Þ. Let r 0 ∈ ð0, ∞Þ and let α ∈ ð1, 2. Then, we obtain Lemma 21. Let E be a uniformly convex and smooth Banach space, let F be a uniformly smooth Banach space, and let M ∈ ð0, ∞Þ. Suppose that E satisfies the condition ( * 1) and F satisfies the condition ( * 2). Then, there exists r M ∈ ð0, ∞Þ such that gðρÞr − Mg * ðr, ρÞ > 0 for any r ∈ ð0, r M Þ and for any ρ ∈ ð0, ∞Þ, where gðρÞ as in Lemma 19 and g * ðr, ρÞ as in Lemma 20.
Theorem 22. Let E be a uniformly convex and smooth Banach space; let F be a strictly convex, reflexive, and uniformly smooth Banach space; let J E and J F be duality mappings on E and F, respectively; let C and D be nonempty, closed, and convex subsets of E and F, respectively; let A be a bounded and linear operator from E into F with A ≠ 0; let A * be the adjoint operator of A; let r ∈ ð0, ∞Þ. Suppose that E * satisfies the condition ( * 2), F * satisfies the condition ( * 1) and C ∩ A −1 ðDÞ ≠ ∅. Let x 1 ∈ E and let fx n g be a sequence generated by for any n ∈ ℕ, where D 0 = C. Then, there exists r ∥A∥ 2 ∈ ð0, ∞Þ such that fx n g is strongly convergent to a point z 0 ∈ C ∩ A −1 ðDÞ for any r ∈ ð0, r ∥A∥ 2 Þ, where z 0 = Π C∩A −1 ðDÞ x 1 .
Proof. For the sake of the proof, we confirm the following facts. Since E is uniformly convex, by (T8) E is reflexive. Since E is uniformly convex and smooth, by (T10) E * is uniformly smooth. Since F is reflexive and uniformly smooth, by (T10) F * is uniformly convex. Since F is strictly convex and reflexive, by (T7) F * is smooth.
It is obvious that C n ∩ D n is closed and convex for any n ∈ ℕ. We show that C ∩ A −1 ðDÞ ⊂ C n for any n ∈ ℕ. Let z ∈ C ∩ A −1 ðDÞ. If Ax n = 0 or Ax n ∈ D, then we obtain z n = x n , and hence, that is, z ∈ C n . If Ax n ≠ 0 and Ax n ∉ D, then By Lemma 21, there exists r ∥A∥ 2 ∈ ð0, ∞Þ such that gðρÞr − ∥A∥ 2 g * ðr, ρÞ > 0 for any r ∈ ð0, r ∥A∥ 2 Þ and for any ρ ∈ ð0, ∞Þ. Put ρ n ∈ max ∥x n ∥ ∥A * J F − J F Π D ð Þ Ax n ∥ + r, ∥Ax n ∥ ∥J F Ax n − J F Π D Ax n ∥ , , ∞