# The Split Feasibility Problem with Some Projection Methods in Banach Spaces

Authors: Toshiharu Kawasaki; Hiroko Manaka
Journal: Abstract and Applied Analysis
 (2020)

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## 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.


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## Body

## 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 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−1D. 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]: xn+1=PCxn+rATPD−IAxn, 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−1D is equivalent to
(1)z=PCIE−rJE−1A∗JFIF−PDAz,where PC and PD are metric projections on subsets C of E and D of F, respectively; IE and IF are the identity mappings on E and F, respectively; JE and JF are duality mappings on E and F, respectively; r∈0,∞. 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 JE and JF 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∈0,∞. Suppose that C∩A−1D≠∅. Let x1∈E and let xn be a sequence generated by
(2)zn=IE−rJE−1A∗JFIF−PDAxn,Cn=z∈C∣zn−z,JExn−zn≥0,Qn=z∈C∣xn−z,JEx1−xn≥0,xn+1=PCn∩Qnx1for any n∈ℕ. Then, xn is strongly convergent to a point z0∈C∩A−1D for any r∈0,A−2, where z0=PC∩A−1Dx1.In this paper, for uniformly convex and smooth Banach spacesE 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.


## 2. Preliminaries

We know that the following hold; for instance, see [12–14].(T1) LetE be a Banach space, let E∗ be the topological dual space of E, and let JE be the duality mapping on E defined by
(3)JEx=x∗∈E∗∣x2=x,x∗=x∗2for any x∈E. Then, E is strictly convex if and only if JE is injective; that is, x≠y implies JEx∩JEy=∅.(T2) LetE be a Banach space, let E∗ be the topological dual space of E, and let JE be the duality mapping on E. Then, E is reflexive if and only if JE is surjective; that is, ⋃x∈EJEx=E∗.(T3) LetE be a Banach space and let JE be the duality mapping on E. Then, E is smooth if and only if JE is single-valued.(T4) LetE be a Banach space and let JE be the duality mapping on E. Then, E is strictly convex if and only if
(4)1−x,y∗>0for any x,y∈E with x≠y and x=y=1 and for any y∗∈JEy.(T5) LetE 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) LetE 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) LetE 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.(T8) IfE is uniformly convex, that is, for any ε∈0,2 there exists δ∈0,1 such that x=y=1 and x−y≥ε implies x+y/2≤1−δ, then E is reflexive.(T9) LetE be a Banach space, let E∗ be the topological dual space of E, and let JE be the duality mapping on E. If E has a Fréchet differentiable norm, then JE is norm-to-norm continuous.(T10) LetE 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.
LetE be a smooth Banach space, let JE be the duality mapping on E, and let VE be the mapping from E×E into 0,∞ defined by
(5)VEx,y=x2−2x,JEy+y2for any x,y∈E.Since by (T3)JE is single-valued, VE is well-defined. It is obvious that x=y implies VEx,y=0. Conversely, by (T4),(T11) IfE is also strictly convex, then VEx,y=0 implies x=y.LetE be a strictly convex and smooth Banach space. By (T1) and (T3), JE is a bijective mapping from E onto JEE. In particular, if E is also reflexive, then by (T2), JE 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 JE−1.We use the following lemmas in this paper. The following is shown in [14].Lemma 2.
LetE be a Banach space and let JE be the duality mapping on E. Then, x−y,x∗−y∗≥0 for any x,y∈E, for any x∗∈Jx, and for any y∗∈JEy. Furthermore, if E is strictly convex and smooth, then x−y,x∗−y∗=0 if and only if x=y.Definition 3.
LetE 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 x−z=miny∈Cx−y. Such a z is denoted by PCx, and PC is called the metric projection of E onto C.The following holds.Lemma 4.
LetE be a strictly convex and smooth Banach space, let C be a nonempty closed subset of E, and let JE be duality mapping on E. Then, for any x,z∈E×C, z=PCx if and only if z−y,JEx−z≥0 for any y∈C.Definition 5.
LetE 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 VEz,x=miny∈CVEy,x. Such a z is denoted by ΠCx, and ΠC is called the generalized projection of E onto C.The following is shown in [15].Lemma 6.
LetE be a strictly convex and smooth Banach space; let C be a nonempty, closed, and convex subset of E; let JE be the duality mapping on E. Then, the following hold.
(i)
For anyx,z∈E×C, z=ΠCx if and only if z−y,JEx−JEz≥0 for any y∈C;(ii)
VEy,ΠCx+VEΠCx,x≤VEy,x for any x∈E and for any y∈C.Definition 7.
LetC be a nonempty subset of a smooth Banach space E. A mapping T from C into E is said to be generalized nonexpansive [6] if the set of all fixed points of T is nonempty and
(6)VETx,y≤VEx,yfor 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
(7)RRx+tx−Rx=Rxfor any x∈E and for any t∈0,∞. 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.
LetE 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 ofE onto C;(ii)
There exists a generalized nonexpansive retraction ofE onto C;(iii)
JEC is closed and convex.Lemma 9.
LetE be a strictly convex, reflexive, and smooth Banach space, let C be a nonempty and closed subset of E, and x,z∈E×C. Suppose that there exists a sunny generalized nonexpansive retraction RC of E onto C. Then, the following are equivalent:
(i)
z=RCx;(ii)
VEx,z=miny∈CVEx,y.The following are shown in [6].Lemma 10.
LetE 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.
LetE be a strictly convex and smooth Banach space, let C be a nonempty and closed subset of E, and let JE be the duality mapping on E. Suppose that there exists a sunny generalized nonexpansive retraction RC of E onto C. Then, the following hold.
(i)
For anyx,z∈E×C, z=RCx if and only if x−z,JEz−JEy≥0 for any y∈C.(ii)
VERCx,y+VEx,RCx≤VEx,y for any x∈E and for any y∈C.Definition 12.
Letp∈1,∞. Define a mapping Jp from E into E∗ by
(8)Jpx=x∗∈E∗∣x,x∗=xpandx∗=xp−1for any x∈E. Then, Jp is called the generalized duality mapping on E. In particular, J2=JE.The following are shown in [17].Lemma 13.
LetE be a Banach space. Then, the following are equivalent:
(i)
E is uniformly convex;(ii)
For anyp∈1,∞ and for any ρ∈0,∞, there exists a continuous, strictly increasing, and convex function gp,ρ from 0,∞ into 0,∞ such that gp,ρ0=0 and
(9)x+yp≥xp+py,x∗+gp,ρyfor any x,y∈BρE=defz∈E∣z≤ρ and for any x∗∈Jpx;(iii)
For anyp∈1,∞ and for any ρ∈0,∞, there exists a continuous, strictly increasing, and convex function gp,ρ from 0,∞ into 0,∞ such that gp,ρ0=0 and
(10)x−y,x∗−y∗≥gp,ρx−yfor any x,y∈BρE, for any x∗∈Jpx, and for any y∗∈Jpy.Lemma 14.
LetE be a smooth Banach space. Then, the following are equivalent:
(i)
E is uniformly smooth;(ii)
For anyq∈1,∞ and for any ρ∈0,∞, there exists a continuous, strictly increasing, and convex function gq,ρ∗ from 0,∞ into 0,∞ such that gq,ρ∗0=0 and
(11)x+yq≤xq+qy,Jqx+gq,ρ∗yfor any x,y∈BρE;(iii)
For anyq∈1,∞ and for any ρ∈0,∞, there exists a continuous, strictly increasing, and convex function gq,ρ∗ from 0,∞ into 0,∞ such that gq,ρ∗0=0 and
(12)x−y,Jqx−Jqy≤gq,ρ∗x−yfor any x,y∈BρE.The following is shown in [18].Lemma 15.
LetE be a uniformly convex and smooth Banach space and let ρ∈0,∞. Then, there exists a continuous, strictly increasing, and convex function gρ from 0,∞ into 0,∞ such that gρ0=0 and
(13)gρx−y≤VEx,yfor any x,y∈BρE.


## 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.
LetE and F be strictly convex, reflexive, and smooth Banach spaces; let IE and IF be the identity mappings on E and F, respectively; let JE and JF 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∈0,∞. Suppose that C∩A−1D≠∅. Consider the following condition:
(i)
z∈C∩A−1D.The following are equivalent to (i):(ii)
z=PCIE−rJE−1A∗JFIF−PDAz;(iii)
z=PCIE−rJE−1A∗JF−JFΠDAz;(iv)
z=ΠCJE−1JE−rA∗JFIF−PDAz;(v)
z=ΠCJE−1JE−rA∗JF−JFΠDAz.Proof.
The equivalence of (i) and (ii) is shown in [11, Lemma 3.1]. We show the rest.
Suppose that (i) holds. SinceAz∈D, PDAz=ΠDAz=Az holds. Therefore,
(14)JFIF−PDAz=JF−JFΠDAz=0,and hence,
(15)therightsideofiii=PCz,therightsidesofivandv=ΠCz.
Sincez∈C, we obtain
(16)therightsidesofiii,iv,andv=z.Conversely, suppose that (iii), (iv), or (v) holds. Since these equations have the form ofz=PCx or z=ΠCx, z∈C holds. We show z∈A−1D.In the case of (iii): By Lemma 4, we obtain
(17)0≤z−y,JEIE−rJE−1A∗JF−JFΠDAz−z=−rz−y,A∗JF−JFΠDAz=−rAz−Ay,JF−JFΠDAzfor any y∈C. Therefore,
(18)Az−Ay,JF−JFΠDAz≤0.On the other hand, by Lemma6, we obtain
(19)ΠDAz−v,JFAz−JFΠDAz≥0for any v∈D. Since C∩A−1D≠∅, there exists z0∈C∩A−1D. Putting y=z0 and v=Az0, y∈C and v∈D hold. Therefore,
(20)Az−Az0,JF−JFΠDAz≤0,ΠDAz−Az0,JFAz−JFΠDAz≥0,and hence,
(21)Az−ΠDAz,JFAz−JFΠDAz≤0.By Lemma2, we obtain ΠDAz=Az, and hence, Az∈D; that is, z∈A−1D.In the case of (iv): By Lemma 6, we obtain
(22)0≤z−y,JE−rA∗JFIF−PDAz−JEz=−rz−y,A∗JFIF−PDAz=−rAz−Ay,JFIF−PDAz,for any y∈C. Therefore,
(23)Az−Ay,JFIF−PDAz≤0.On the other hand, by Lemma4, we obtain
(24)PDAz−v,JFAz−PDAz≥0for any v∈D. Since C∩A−1D≠∅, there exists z0∈C∩A−1D. Putting y=z0 and v=Az0, y∈C and v∈D hold. Therefore,
(25)Az−Az0,JFIF−PDAz≤0,PDAz−Az0,JFAz−PDAz≥0,and hence,
(26)Az−PDAz,JFAz−PDAz=Az−PDAz2≤0.Therefore, we obtainPDAz=Az, and hence, Az∈D; that is, z∈A−1D.In the case of (v): By Lemma 6, we obtain
(27)0≤z−y,JEJE−1JE−rA∗JF−JFΠDAz−JEz=−rz−y,A∗JF−JFΠDAz=−rAz−Ay,JF−JFΠDAz,for any y∈C. Therefore,
(28)Az−Ay,JF−JFΠDAz≤0.On the other hand, by Lemma6, we obtain
(29)ΠDAz−v,JFAz−JFΠDAz≥0for any v∈D. Since C∩A−1D≠∅, there exists z0∈C∩A−1D. Putting y=z0 and v=Az0, y∈C and v∈D hold. Therefore,
(30)Az−Az0,JF−JFΠDAz≤0,ΠDAz−Az0,JFAz−JFΠDAz≥0,and hence,
(31)Az−ΠDAz,JFAz−JFΠDAz≤0.By Lemma2, we obtain ΠDAz=Az, and hence, Az∈D; that is, z∈A−1D.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.
LetE and F be strictly convex, reflexive, and smooth Banach spaces; let F∗ be the dual space of F, let IE and IF∗ be the identity mappings on E and F∗, respectively; let JE and JF 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∈0,∞. Suppose that C∩A−1D≠∅. Consider the following condition:
(i)
z∈C∩A−1D.IfJFDis closed, then the following are equivalent to (i):(vi)
z=PCIE−rJE−1A∗IF∗−RJFDJFAz;(vii)
z=ΠCJE−1JE−rA∗IF∗−RJFDJFAz.IfJECis closed, then the following are equivalent to (i):
(viii)
z=JE−1RJECJE−rA∗JFIF−PDAz;(ix)
z=JE−1RJECJE−rA∗JF−JFΠDAz.IfJECandJFDare closed, then the following is equivalent to (i):
(x)
z=JE−1RJECJE−rA∗IF∗−RJFDJFAz.Proof.
Suppose that (i) holds. SinceAz∈D,
(32)PDAz=ΠDAz=AzandRJFDJFAz=JFAzhold. Therefore,
(33)JFIF−PDAz=JF−JFΠDAz=IF∗−RJFDJFAz=0,and hence,
(34)therightsideofvi=PCz,therightsideofvii=ΠCz,therightsidesofviii,ix,andx=JE−1RJECJEz.Sincez∈C, we obtain
(35)therightsidesofvi,vii,viii,ix,andx=z.Conversely, suppose that (vi), (vii), (viii), (ix), or (x) holds. Since these equations have the form ofz=PCx, z=ΠCx, or z=JE−1RJECx, z∈C holds. We show z∈A−1D.In the case of (vi): By Lemma 4, we obtain
(36)0≤z−y,JEIE−rJE−1A∗IF∗−RJFDJFAz−z=−rz−y,A∗IF∗−RJFDJFAz=−rAz−Ay,IF∗−RJFDJFAz,for any y∈C. Therefore,
(37)Az−Ay,IF∗−RJFDJFAz≤0.On the other hand, by Lemma11, we obtain
(38)JF−1RJFDJFAz−JF−1v,JFAz−RJFDJFAz≥0for any v∈JFD. Since C∩A−1D≠∅, there exists z0∈C∩A−1D. Putting y=z0 and v=JFAz0, y∈C and v∈JFD hold. Therefore,
(39)Az−Az0,IF∗−RJFDJFAz≤0,JF−1RJFDJFAz−Az0,JFAz−RJFDJFAz≥0,and hence,
(40)Az−JF−1RJFDJFAz,JFAz−RJFDJFAz≤0.By Lemma2, we obtain RJFDJFAz=JFAz, and hence, JFAz∈JFD; that is, z∈A−1D.In the case of (vii): By Lemma 6, we obtain
(41)0≤z−y,JEJE−1JE−rA∗IF∗−RJFDJFAz−JEz=−rz−y,A∗IF∗−RJFDJFAz=−rAz−Ay,IF∗−RJFDJFAz,for any y∈C. Therefore,
(42)Az−Ay,IF∗−RJFDJFAz≤0.On the other hand, by Lemma11, we obtain
(43)JF−1RJFDJFAz−JF−1v,JFAz−RJFDJFAz≥0for any v∈JFD. Since C∩A−1D≠∅, there exists z0∈C∩A−1D. Putting y=z0 and v=JFAz0, y∈C and v∈JFD hold. Therefore,
(44)Az−Az0,IF∗−RJFDJFAz≤0,JF−1RJFDJFAz−Az0,JFAz−RJFDJFAz≥0,and hence,
(45)Az−JF−1RJFDJFAz,JFAz−RJFDJFAz≤0.By Lemma2, we obtain RJFDJFAz=JFAz, and hence, JFAz∈JFD; that is, z∈A−1D.In the case of (viii): By Lemma 11, we obtain
(46)0≤JE−1JEz−JE−1y,JE−rA∗JFIE−PDAz−JEz=−rz−JE−1y,A∗JFIE−PDAz=−rAz−AJE−1y,JFIE−PDAz,for any y∈JEC. Therefore,
(47)Az−AJE−1y,JFIE−PDAz≤0.On the other hand, by Lemma4, we obtain
(48)PDAz−v,JFAz−PDAz≥0for any v∈D. Since C∩A−1D≠∅, there exists z0∈C∩A−1D. Putting y=JEz0 and v=Az0, y∈JEC and v∈D hold. Therefore,
(49)Az−Az0,JFIE−PDAz≤0,PDAz−Az0,JFAz−PDAz≥0,and hence,
(50)Az−PDAz,JFAz−PDAz=Az−PDAz2≤0.Therefore, we obtainPDAz=Az, and hence, Az∈D; that is, z∈A−1D.In the case of (ix): By Lemma 11, we obtain
(51)0≤JE−1JEz−JE−1y,JE−rA∗JF−JFΠDAz−JEz=−rz−JE−1y,A∗JF−JFΠDAz=−rAz−AJE−1y,JF−JFΠDAz,for any y∈JEC. Therefore,
(52)Az−AJE−1y,JF−JFΠDAz≤0.On the other hand, by Lemma6, we obtain
(53)ΠDAz−v,JFAz−JFΠDAz≥0for any v∈D. Since C∩A−1D≠∅, there exists z0∈C∩A−1D. Putting y=JEz0 and v=Az0, y∈JEC and v∈D hold. Therefore,
(54)Az−Az0,JF−JFΠDAz≤0,ΠDAz−Az0,JFAz−JFΠDAz≥0,and hence,
(55)Az−ΠDAz,JFAz−JFΠDAz≤0.By Lemma2, we obtain PDAz=Az, and hence, Az∈D; that is, z∈A−1D.In the case of (x): By Lemma 11, we obtain
(56)0≤JE−1JEz−JE−1y,JE−rA∗IF∗−RJFDJFAz−JEz=−rz−JE−1y,A∗IF∗−RJFDJFAz=−rAz−AJE−1y,IF∗−RJFDJFAz,for any y∈JEC. Therefore,
(57)Az−AJE−1y,IF∗−RJFDJFAz≤0.On the other hand, by Lemma11, we obtain
(58)JF−1RJFDJFAz−JF−1v,JFAz−RJFDJFAz≥0for any v∈JFD. Since C∩A−1D≠∅, there exists z0∈C∩A−1D. Putting y=JEz0 and v=JFAz0, y∈JEC and v∈JFD hold. Therefore,
(59)Az−Az0,IF∗−RJFDJFAz≤0,JF−1RJFDJFAz−Az0,JFAz−RJFDJFAz≥0,and hence,
(60)Az−JF−1RJFDJFAz,JFAz−RJFDJFAz≤0.By Lemma2, we obtain RJFDJFAz=JFAz, and hence, JFAz∈JFD; that is, z∈A−1D.


## 4. Strong Convergence Theorems to Solutions

In this section, we consider strong convergence theorems to solutions of the split feasibility problem.Lemma 19.
LetE be a uniformly convex and smooth Banach space and let u,v∈E. Put
(61)gρ=g2,max∥u∥,∥v∥∥u−v∥1ifu=vandρ∈0,max∥u∥,∥v∥∥u−v∥,g2,ρ1ifu=vorρ∈max∥u∥,∥v∥∥u−v∥,∞for any ρ∈0,∞, where g2,ρ as in Lemma 13. Then,
(62)u−v,JEu−JEv≥gρ∥u−v∥2holds for any ρ∈0,∞.Proof.
Ifu=v, then for arbitrary gρ, we obtain
(63)u−v,JEu−JEv=0=gρ∥u−v∥2.In particular, the above equation holds forgρ=g2,ρ1. If u≠v, then, since 1/(∥u−v∥u,1/∥u−v∥v∈BρE for any ρ∈max∥u∥,∥v∥/∥u−v∥,∞, we obtain
(64)1∥u−v∥u−1∥u−v∥v,1∥u−v∥JEu−1∥u−v∥JEv≥g2,max∥u∥,∥v∥/∥u−v∥1∥u−v∥u−1∥u−v∥vifρ∈0,max∥u∥,∥v∥∥u−v∥,g2,ρ1∥u−v∥u−1∥u−v∥vifρ∈max∥u∥,∥v∥∥u−v∥,∞=gρ.Therefore, we obtain(65)u−v,JEu−JEv≥gρ∥u−v∥2.Lemma 20.
LetE be a uniformly smooth Banach space and let u,v∈E. Put
(66)g∗r,ρ=g2,∥u∥/∥v∥+r∗rifv≠0andρ∈0,∥u∥∥v∥+r,g2,ρ∗rifv=0orρ∈∥u∥∥v∥+r,∞for any r∈0,∞ and for any ρ∈0,∞, where g2,ρ∗ as in Lemma 14. Then,
(67)v,JEu−rv−JEu≥−g∗r,ρrv2holds for any r∈0,∞ and for any ρ∈0,∞.Proof.
Ifv=0, then for arbitrary g∗r,ρ, we obtain
(68)v,JEu−rv−JEu=0=−g∗r,ρrv2.In particular, the above equation holds forg∗r,ρ=g2,ρ∗r. If v≠0, then, since 1/vu−rv,1/vu∈B∥u∥/∥v∥+rE, we obtain
(69)−rvv,1vJEu−rv−1vJEu≤g2,∥u∥/∥v∥+r∗−rvvifρ∈0,uv+r,g2,ρ∗−rvvifρ∈uv+r,∞=g∗r,ρ.Therefore, we obtain(70)v,JEu−rv−JEu≥−g∗r,ρrv2.We consider the following condition for a uniformly convex and smooth Banach space:(∗1) infρ∈0,∞g2,ρ1>0.We consider the following condition for a uniformly smooth Banach space:(∗2) There exist r0∈0,∞ and α∈1,∞ such that supr∈0,r0,ρ∈0,∞g2,ρ∗r/rα<∞.Example 1.
We consider each of examples satisfying the conditions (∗1) and (∗2), respectively.LetE be a Banach space. The modulus of convexity δE of E is defined by
(71)δEε=inf1−12x+yx=y=1,x−y=εfor any ε∈0,2. Let p∈1,∞. E is said to be p-uniformly convex if there exists a constant c∈0,∞ such that δEε≥cεp for any ε∈0,2. Furthermore, we know that E is uniformly convex if and only if δEε>0 for any ε∈0,2. Let Jp be the generalized duality mapping on a Banach space E. By [17, Corollary 1], E is p-uniformly convex if and only if there exists a constant c∈0,∞ such that
(72)x−y,x∗−y∗≥cx−ypfor any x,y∈E, for any x∗∈Jpx, and for any y∗∈Jpy. Therefore, if E is 2-uniformly convex and smooth, then we can put g2,ρr=cr2 for any ρ∈0,∞. We obtain
(73)infρ∈0,∞g2,ρ1=c>0.The modulus of smoothnessρE of E is defined by
(74)ρEt=sup12x+y+x−y−1x=1,y=tfor anyt∈0,∞. Let q∈1,∞. E is said to be q-uniformly smooth if there exists a constant c∈0,∞ such that ρEt≤ctq for any t∈0,∞. Furthermore, we know that E is uniformly smooth if and only if limt↓0ρEt/t=0. Let Jq be the generalized duality mapping on a smooth Banach space E.By [17, Corollary 1;′], E is q-uniformly smooth if and only if there exists a constant c∈0,∞ such that
(75)x−y,Jqx−Jqy≤cx−yqfor any x,y∈E. Therefore, if E is 2-uniformly smooth, then we can put g2,ρ∗r=cr2 for any ρ∈0,∞. Let r0∈0,∞ and let α∈1,2. Then, we obtain
(76)supr∈0,r0,ρ∈0,∞g2,ρ∗rrα=cr02−α<∞.Lemma 21.
LetE be a uniformly convex and smooth Banach space, let F be a uniformly smooth Banach space, and let M∈0,∞. Suppose that E satisfies the condition (∗1) and F satisfies the condition (∗2). Then, there exists rM∈0,∞ such that gρr−Mg∗r,ρ>0 for any r∈0,rM and for any ρ∈0,∞, where gρ as in Lemma 19 and g∗r,ρ as in Lemma 20.Proof.
We obtain(77)gρr−Mg∗r,ρ≥infρ∈0,∞g2,ρ1r−Msupr∈0,r0,ρ∈0,∞g2,ρ∗rrαrα=rinfρ∈0,∞g2,ρ1−Msupr∈0,r0,ρ∈0,∞g2,ρ∗rrαrα−1for any r∈0,r0. Put
(78)rM=mininfρ∈0,∞g2,ρ1Msupr∈0,r0,ρ∈0,∞g2,ρ∗r/rα1/α−1,r0.Then, we obtaingρr−Mg∗r,ρ>0 for any r∈0,rM and for any ρ∈0,∞.Theorem 22.
LetE be a uniformly convex and smooth Banach space; let F be a strictly convex, reflexive, and uniformly smooth Banach space; let JE and JF 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∈0,∞. Suppose that E∗ satisfies the condition (∗2), F∗ satisfies the condition (∗1) and C∩A−1D≠∅. Let x1∈E and let xn be a sequence generated by
(79)zn=JE−1JE−rA∗JF−JFΠDAxn,Cn=z∈C∣zn−z,JExn−JEzn≥0,Dn=z∈Dn−1∣xn−z,JEx1−JExn≥0,xn+1=ΠCn∩Dnx1for any n∈ℕ, where D0=C. Then, there exists r∥A∥2∈0,∞ such that xn is strongly convergent to a point z0∈C∩A−1D for any r∈0,r∥A∥2, where z0=ΠC∩A−1Dx1.Proof.
For the sake of the proof, we confirm the following facts. SinceE 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 thatCn∩Dn is closed and convex for any n∈ℕ. We show that C∩A−1D⊂Cn for any n∈ℕ. Let z∈C∩A−1D. If Axn=0 or Axn∈D, then we obtain zn=xn, and hence,
(80)zn−z,JExn−JEzn=0;that is, z∈Cn. If Axn≠0 and Axn∉D, then
(81)zn−z,JExn−JEzn=rJE−1JExn−rA∗JF−JFΠDAxn−xn,A∗JF−JFΠDAxn+xn−z,A∗JF−JFΠDAxn=rJE−1JExn−rA∗JF−JFΠDAxn−xn,A∗JF−JFΠDAxn+Axn−ΠDAxn,JF−JFΠDAxn+ΠDAxn−Az,JF−JFΠDAxn=rJE−1JExn−rA∗JF−JFΠDAxn−JE−1JExn,A∗JF−JFΠDAxn+JF−1JFAxn−JF−1JFΠDAxn,JFAxn−JFΠDAxn+ΠDAxn−Az,JFAxn−JFΠDAxn.By Lemma21, there exists r∥A∥2∈0,∞ such that gρr−∥A∥2g∗r,ρ>0 for any r∈0,r∥A∥2 and for any ρ∈0,∞. Put
(82)ρn∈max∥xn∥∥A∗JF−JFΠDAxn∥+r,∥Axn∥∥JFAxn−JFΠDAxn∥,∥ΠDAxn∥∥JFAxn−JFΠDAxn∥,∞.SinceE∗ is uniformly smooth, by Lemma 20, we obtain
(83)JE−1JExn−rA∗JF−JFΠDAxn−JE−1JExn,A∗JF−JFΠDAxn≥−g∗r,ρnr∥A∗JF−JFΠDAxn∥2≥−∥A∥2g∗r,ρnr∥JF−JFΠDAxn∥2.SinceF∗ is uniformly convex and smooth, by Lemma 19, we obtain
(84)JF−1JFAxn−JF−1JFΠDAxn,JFAxn−JFΠDAxn≥gρnJFAxn−JFΠDAxn2.By Lemma6, we obtain
(85)ΠDAxn−Az,JFAxn−JFΠDAxn≥0.Therefore, we obtain(86)zn−z,JExn−JEzn≥gρnr−A2g∗r,ρnJF−JFΠDAxn2≥0;that is, z∈Cn. We show that C∩A−1D⊂Dn for any n∈ℕ. Since
(87)D1=z∈C∣x1−z,JEx1−JEx1≥0=C,it is obvious that C∩A−1D⊂D1. Suppose that there exists k∈ℕ such that C∩A−1D⊂Dk. Then, C∩A−1D⊂Ck∩Dk. Since xk+1=ΠCk∩Dkx1, by Lemma 6, we obtain
(88)xk+1−z,JEx1−JExk+1≥0for any z∈Ck∩Dk, and hence, the above inequality holds for any z∈C∩A−1D; that is, we obtain C∩A−1D⊂Dk+1. We obtain C∩A−1D⊂Dn for any n∈ℕ. Therefore, xn is well-defined.It is obvious thatC∩A−1D is nonempty, closed, and convex. Therefore, ΠC∩A−1D is well-defined.By the definition ofDn, we obtain xn=ΠDnx1. By Lemma 6,
(89)VEz,xn+VExn,x1≤VEz,x1for any z∈Dn. Let m∈ℕ with m≥n. Since xm∈Dm⊂Dn, we obtain
(90)VExm,xn+VExn,x1≤VExm,x1,and hence,
(91)VExm,xn≤VExm,x1−VExn,x1.Sincexn=ΠDnx1, we obtain
(92)VExn,x1≤VEz,x1for any z∈Dn. Since xn+1=ΠCn∩Dnx1, we obtain xn+1∈Cn∩Dn. Since xn+1∈Dn, we obtain
(93)VExn,x1≤VExn+1,x1,and hence, VExn,x1 is bounded and nondecreasing; that is, there exists the limit of VExn,x1. From (91), we obtain
(94)limm,n→∞VExm,xn=0.By Lemma15, xn is a Cauchy sequence. Since E is complete, there exists z0∈E such that limn→∞xn=z0. Since xn+1=ΠCn∩Dnx1, by Lemma 6, we obtain
(95)xn+1−z,JEx1−JExn+1≥0for any z∈Cn∩Dn. Since C∩A−1D⊂Cn∩Dn, the above inequality holds for any z∈C∩A−1D. Taking n⟶∞, since by (T9) JE is norm-to-norm continuous, we obtain
(96)z0−z,JEx1−JEz0≥0for any z∈C∩A−1D. By Lemma 6, we obtain z0=ΠC∩A−1Dx1.Theorem 23.
LetE be a uniformly convex and smooth Banach space; let F be a strictly convex, reflexive, and uniformly smooth Banach space; let IF∗ be the identity mapping on F∗; let JE and JF be duality mappings on E and F, respectively; let C and D be nonempty, closed, and convex subsets of E and F such that JFD is closed, 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∈0,∞. Suppose that E∗ satisfies the condition (∗2), F∗ satisfies the condition (∗1) and C∩A−1D≠∅. Let x1∈E and let xn be a sequence generated by
(97)zn=JE−1JE−rA∗IF∗−RJFDJFAxn,Cn=z∈C∣zn−z,JExn−JEzn≥0,Dn=z∈Dn−1∣xn−z,JEx1−JExn≥0,xn+1=JE−1RJECn∩DnJEx1for any n∈ℕ, where D0=C. Then, there exists r∥A∥2∈0,∞ such that xn is strongly convergent to a point z0∈C∩A−1D for any r∈0,r∥A∥2, where z0=JE−1RJEC∩A−1DJEx1.Proof.
For the sake of the proof, we confirm the following facts. SinceE 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. Since E∗ and F∗ have Fréchet differentiable norms, by (T9), JE−1 and JF−1 are norm-to-norm continuous. Therefore, JEC and JECn∩Dn are closed.It is obvious thatCn∩Dn is closed and convex for any n∈ℕ. We show that C∩A−1D⊂Cn for any n∈ℕ. Let z∈C∩A−1D. If Axn=0 or Axn∈D, then we obtain zn=xn, and hence,
(98)zn−z,JExn−JEzn=0;that is, z∈Cn. If Axn≠0 and Axn∉D, then
(99)zn−z,JExn−JEzn=rJE−1JExn−rA∗IF∗−RJFDJFAxn−xn,A∗IF∗−RJFDJFAxn+xn−z,A∗IF∗−RJFDJFAxn=rJE−1JExn−rA∗IF∗−RJFDJFAxn−xn,A∗IF∗−RJFDJFAxn+Axn−JF−1RJFDJFAxn,IF∗−RJFDJFAxn+JF−1RJFDJFAxn−Az,IF∗−RJFDJFAxn=rJE−1JExn−rA∗IF∗−RJFDJFAxn−JE−1JExn,A∗IF∗−RJFDJFAxn+JF−1JFAxn−JF−1RJFDJFAxn,JFAxn−RJFD,JFAxn+JF−1RJFDJFAxn−JF−1JFAz,JFAxn−RJFD,JFAxn.By Lemma21, there exists r∥A∥2∈0,∞ such that gρr−∥A∥2g∗r,ρ>0 for any r∈0,r∥A∥2 and for any ρ∈0,∞. Put
(100)ρn∈maxxnA∗IF∗−RJFDJFAxn+r,AxnJFAxn−RJFDJFAxn,∥RJFDJFAxn∥∥JFAxn−RJFDJFAxn∥,∞.SinceE∗ is uniformly smooth, by Lemma 20, we obtain
(101)JE−1JExn−rA∗IF∗−RJFDJFAxn−JE−1JExn,A∗IF∗−RJFDJFAxn≥−g∗r,ρnrA∗IF∗−RJFDJFAxn2≥−∥A∥2g∗r,ρnrIF∗−RJFDJFAxn2.SinceF∗ is uniformly convex and smooth, by Lemma 19, we obtain
(102)JF−1JFAxn−JF−1RJFDJFAxn,JFAxn−RJFDJFAxn≥gρnJFAxn−RJFDJFAxn2.By Lemma11, we obtain
(103)JF−1RJFDJFAxn−JF−1JFAz,JFAxn−RJFDJFAxn≥0.Therefore, we obtain(104)zn−z,JExn−JEzn≥gρnr−A2g∗r,ρnIF∗−RJFDJFAxn2≥0;that is, z∈Cn. We show that C∩A−1D⊂Dn for any n∈ℕ. Since
(105)D1=z∈C∣x1−z,JEx1−JEx1≥0=C,it is obvious that C∩A−1D⊂D1. Suppose that there exists k∈ℕ such that C∩A−1D⊂Dk. Then, C∩A−1D⊂Ck∩Dk. Since xk+1=JE−1RJECk∩DkJEx1, by Lemma 11, we obtain
(106)xk+1−JE−1z,JEx1−JExk+1≥0for any z∈JECk∩Dk, and hence, the above inequality holds for any z∈JEC∩A−1D; that is, we obtain C∩A−1D⊂Dk+1. We obtain C∩A−1D⊂Dn for any n∈ℕ. Therefore, xn is well-defined.It is obvious thatJEC∩A−1D is nonempty, closed, and convex and C∩A−1D is closed. Therefore, RJEC∩A−1D is well-defined.By the definition ofDn, we obtain xn=JE−1RJEDnJEx1. By Lemma 11,
(107)VE∗JExn,z+VE∗JEx1,JExn≤VE∗JEx1,zfor any z∈JEDn. Let m∈ℕ with m≥n. Since xm∈Dm⊂Dn, we obtain
(108)VE∗JExn,JExm+VE∗JEx1,JExn≤VE∗JEx1,JExm,and hence,
(109)VE∗JExn,JExm≤VE∗JEx1,JExm−VE∗JEx1,JExn.Sincexn=JE−1RJEDnJEx1, by Lemma 9, we obtain
(110)VE∗JEx1,JExn≤VE∗JEx1,zfor any z∈JEDn. Since xn+1=JE−1RJECn∩DnJEx1, we obtain xn+1∈Cn∩Dn. Since xn+1∈Dn, we obtain
(111)VE∗JEx1,JExn≤VE∗JEx1,JExn+1,and hence, VE∗JEx1,JExn is bounded and nondecreasing; that is, there exists the limit of VE∗JEx1,JExn. From (109), we obtain
(112)limm,n→∞VE∗JExn,JExm=0.By Lemma15, JExn is a Cauchy sequence. Since by (T9) JE−1 is norm-to-norm continuous, xn is also a Cauchy sequence. Since E is complete, there exists z0∈E such that limn→∞xn=z0. Since xn+1=JE−1RJECn∩DnJEx1, by Lemma 11, we obtain
(113)xn+1−JE−1z,JEx1−JExn+1≥0for any z∈JECn∩Dn. Since C∩A−1D⊂Cn∩Dn, the above inequality holds for any z∈JEC∩A−1D. Taking n⟶∞, since by (T9) JE is norm-to-norm continuous, we obtain
(114)z0−JE−1z,JEx1−JEz0≥0for any z∈JEC∩A−1D. By Lemma 11, we obtain z0=JE−1RJEC∩A−1DJEx1.Remark 24.
In this paper, we consider only two strong convergence theorems with respect to (v) in Theorem16 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.



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