# On the Shrinking Projection Method for the Split Feasibility Problem in Banach Spaces

Authors: Huanhuan Cui; Haixia Zhang
Journal: Journal of Mathematics
 (2021)

---

## Abstract

In this paper, we consider the split feasibility problem in Banach spaces. By applying the shrinking projection method, we propose an iterative method for solving this problem. It is shown that the algorithm under two different choices of the stepsizes is strongly convergent to a solution of the problem.


---

## Body

## 1. Introduction

In this paper, we consider the split feasibility problem [1]. It is very useful in dealing with problems arising from various applied disciplines (see, e.g., [2–6]). More precisely, the split feasibility problem requires to find a point x^∈ℝn satisfying the following property:(1)x^∈C,Ax^∈Q,where C and Q are nonempty closed convex subsets of ℝn and ℝm, respectively, and A is a linear operator.The split feasibility problem was first treated in Euclidean spaces and recently was extended to more general framework including Hilbert spaces and Banach spaces. In Hilbert spaces, Byrne [7] introduced the CQ algorithm:(2)xn+1=PCxn−rnA∗I−PQAxn,where rn>0 is a properly chosen stepsize, A∗ is the conjugate of A, I is the identity operator, and PC,PQ denote the metric projections onto the respective sets. By using Polyak’s gradient method, Wang [8] recently proposed another iterative algorithm:(3)xn+1=xn−rnI−PCxn+A∗I−PQAxn,where rn>0 is a properly chosen stepsize (see also [4, 7–13] for some related works). In the framework of Banach spaces, Schöpfer et al. [14] extended the CQ method as(4)xn+1=ΠCJX−1JXxn−rnA∗JYI−PQAxn,where rn is a positive parameter, JX,JY are, respectively, the duality mappings on X and Y, ΠC denotes the Bregman projection, and PC denotes the metric projection. The weak convergence of (4) is guaranteed if X is p-uniformly convex and uniformly smooth and JX is sequentially weak-to-weak continuous. Recently, Takahashi [15] suggested a novel way for the split feasibility problem:(5)zn=xn−rnJX∗A∗JYI−PQAxn,Cn=z∈C:zn−z,JXxn−zn≥0,Qn=z∈C:xn−z,JXx0−xn≥0,xn+1=PCn∩Qnx0,∀n∈ℕ.By applying the shrinking projection method, he [16] also proposed another method:(6)zn=xn−rnJX∗A∗JYI−PQAxn,Qn+1=z∈Qn:zn−z,JXxn−zn≥0,xn+1=PQn+1x0,∀n∈ℕ.Instead of weak convergence, Takahashi proved the strong convergence of both methods, under the assumption thatX is uniformly convex and smooth, which is clearly weaker than that used in [14]. Following the above works, Wang [10] recently proposed a new method, which generates a sequence as(7)zn=xn−rnJX∗JXI−PCxn+A∗JYI−PQAxn,Cn=z∈X:zn−z,JXxn−zn≥0,Qn=z∈X:xn−z,JXx0−xn≥0,xn+1=PCn∩Qnx0,∀n∈ℕ.Our aim of this paper is to continue the above works by constructing new iterative methods in Banach spaces. By applying ideas (6) and (7), we introduce a new iterative algorithm and propose two different choices of the stepsize. We show that if the spaces involved are smooth and uniformly convex, then the algorithm converges strongly under both choices of the stepsize. It is also worth noting that one choice of the stepsize does not need any a priori knowledge of the operator norm.


## 2. Preliminaries

In what follows, we shall assume that the split feasibility problem is consistent, that is, its solution set denoted byS is nonempty. The notation “⟶” stands for strong convergence, “⇀” represents weak convergence, and ωwxn is the set of weak cluster points of a sequence xn. Let SX=x∈X:x=1 and BX=x∈X:x≤1, respectively, be the unit sphere and unit ball of X. T−10=x∈X:Tx=0 denotes the null-point set of an operator T defined on X. For x∈X, we let JXI−PCx=JXx−PCx and A∗JYI−PQAx=A∗JYAx−PQAx.Definition 1.
LetX be a real Banach space.(1)
The modulus of convexityδXε:0,2⟶0,1 is defined as(8)δXε=inf1−x+y2:x,y∈BX,x−y≥ε.(2)
The modulus of smoothnessρXτ:0,∞⟶0,∞ is defined by(9)ρXτ=supx+τy+x−τy2−1:x,y∈SX.(3)
The duality mappingJX:X⟶2X∗ is defined by(10)JXx=x∗∈X∗:x,x∗=x2=x∗2.Definition 2.
LetX be a real Banach space.(1)
X is called strictly convex if δX2=1.(2)
X is called smooth if limt⟶0x+ty−x/t exists for each x,y∈SX.(3)
X is called uniformly convex if δXε>0 for any ε∈0,2.(4)
X is called uniformly smooth if limτ⟶0ρXτ/τ=0.(5)
X is called p-uniformly convex if there exist p≥2 and a constant c>0 such that δXε≥cεp,∀ε∈0,2.Lemma 1 (see [17–19]).
IfX is uniformly convex, then X∗ is uniformly smooth; X is strictly convex and reflexive.Lemma 2 (see [17–19]).
Letxn be a sequence in X such that xn⇀x and xn⟶x as n⟶∞. If X is uniformly convex, then limnxn=x.Lemma 3 (see [17–19]).
LetM>0 and xn,yn be two sequences in X such that xn⟶M,yn⟶M and xn+yn⟶2M as n⟶∞. If X is uniformly convex, then limnxn−yn=0.Lemma 4 (see [17–19]).
LetJX be the duality mapping on X.(1)
JX is surjective if and only if X is reflexive.(2)
JX is injective if and only if X is strictly convex.(3)
JX is single-valued if and only if X is smooth.(4)
IfX is smooth, then JX is monotone, that is,(11)x−y,JXx−JXy≥0,∀x,y∈X.Moreover, ifX is further strictly convex, then JX is strictly monotone, that is,(12)x−y,JXx−JXy=0⇒x=y.(5)
IfX is reflexive, smooth, and strictly convex, then JX is one-to-one single-valued and JX−1=JX∗, where JX∗ is the duality mapping of X∗.The Bregman distance with respect to⋅ is given by(13)Δx,y=12x2−JXx,y+12y2.This notion goes back to Bregman [20] and now is successfully used in various optimization problems in Banach spaces (see, e.g., [21, 22]). In general, the Bregman distance is not a metric due to the absence of symmetry, but it has some distance-like properties.Definition 3.
LetC be a nonempty closed convex subset of X.(1)
The metric projectionPC:X⟶C is defined as(14)PCx:=argminy∈Cx−y,x∈X.(2)
The Bregman projectionΠC:X⟶C is defined as(15)ΠCx=argminy∈C△x,y,x∈X.
In Hilbert spaces, the metric and Bregman projections are the same, but in general they are completely different. More importantly, the metric projection cannot share the descent property as the Bregman projection in Banach spaces. We now collect some properties of metric projections.Lemma 5 (see [14]).
Letxn be a sequence in X and C⊆X be a nonempty closed convex subset. Then, for x∈X, the following holds.(1)
z−PCx,JXx−PCx≤0,∀z∈C.(2)
x−PCx2≤x−z,JXx−PCx,∀z∈C.(3)
Ifxn⇀x and xn−PCxn⟶0, then x∈C.


## 3. Convergence Analysis

To construct our algorithm, we need the following lemma. It converts the split feasibility problem to an equivalent null-point problem, which indeed amounts to a fixed point problem.Lemma 6 (see [10]).
LetT:=JX∗JXI−PC+A∗JYI−PQA. Then, S=T−10.By applying idea (5) to Lemma 6, we thus can propose the following algorithm for solving the split feasibility problem in Banach spaces. Choose x0∈X and Q0=X. Given xn, update xn+1 by the iteration formula:(16)zn=xn−rnJX∗JXI−PCxn+A∗JYI−PQAxn,Qn+1=z∈Qn:zn−z,JXxn−zn≥0,xn+1=PQn+1x0,∀n∈ℕ.Lemma 7.
Assume that bothX and Y are reflexive, smooth, and strictly convex. If rn is chosen so that 0<a≤rn≤1/1+A2, then for each n∈ℕ, the set Qn is nonempty, closed, and convex. Hence, the proposed algorithm is well defined.Proof.
It suffices to show thatQn is nonempty since it is clearly closed and convex. To this end, we now show by induction that S⊆Qn for all n∈ℕ and that S⊆Q1 is obvious. Suppose that S⊆Qk for some k∈ℕ. Take any z∈S. Then, we have z∈Qk. Furthermore, we have(17)xk−z,JXxk−zk=rkxk−z,JXI−PCxk+A∗JYI−PQAxk=rkxk−z,JXI−PCxk+rkAxk−Az,JYI−PQAxk,which from Lemma 5 yields(18)xk−z,JXxk−zk≥rkI−PCxk2+I−PQAxk2.
By a simple calculation, we have(19)xk−zk2=rk2JXI−PCxk+A∗JYI−PQAxk2≤rk2I−PCxk+AI−PQAxk2=rk2I−PCxk2+A2I−PQAxk2+2rk2AI−PCxkI−PQAxk≤rk2I−PCxk2+A2I−PQAxk2+rk2A2I−PCxk2+I−PQAxk2.
Hence, we have(20)xk−zk2≤rk21+A2I−PCxk2+I−PQAxk2.
It then follows from (18) that(21)zk−z,JXxk−zk=zk−xk,JXxk−zk+xk−z,JXxk−zk=−zk−xk2+xk−z,JXxk−zk≥rkI−PCxk2+I−PQAxk2−xk−zk2.
Substituting (20) into the above inequality, we have(22)zk−z,JXxk−zk≥rk1−rk1+A2I−PCxk2+I−PQAxk2.
By our choice ofrk, we see that(23)zk−z,JXxk−zk≥0.
This implies thatz∈Qk+1. Since z is chosen in S arbitrarily, we conclude that S⊆Qk+1. Consequently, S⊆Qn for all n∈ℕ. Now it is clear that the set Qn is nonempty, closed, and convex. Thus, the proposed algorithm is well defined.
Now let us state the convergence of the proposed algorithm.Theorem 1.
Assume thatX is uniformly convex and smooth and Y is reflexive, smooth, and strictly convex. If rn is chosen so that 0<a≤rn≤1/1+A2, then the sequence xn generated by (16) converges strongly to z^∈S, where z^=PSx0.Proof.
We first show the following equality:(24)limn⟶∞xn−zn=0.
To this end, letz∈S. From the previous lemma, it is clear that z∈Qn, xn+1∈Qn+1⊆Qn. Thus, we have for each n∈ℕ,(25)x0−xn=x0−PQnx0≤minx0−z,x0−xn+1.
This indicates thatx0−xn is nondecreasing and bounded above; thus, the limit of x0−xn exists. Now set M:=limnx0−xn. We have(26)limsupn⟶∞xn−x0+xn+1−x0≤limn⟶∞xn−x0+xn+1−x0=2M.
On the other hand, we have(27)liminfn⟶∞xn−x0+xn+1−x0=liminfn⟶∞2xn+xn+12−x0≥limn⟶∞2xn−x0=2M,where the inequality follows from the fact xn+xn+1/2∈Qn. Altogether, we have limnxn−x0+xn+1−x0=2M. Since X is uniformly convex, this yields that(28)limn⟶∞xn−xn+1=limn⟶∞xn−x0−xn+1−x0=0.
Furthermore, sincexn+1∈Qn+1, we have(29)zn−xn+1,JXxn−zn≥0,which clearly implies that(30)xn−zn2=xn−zn,JXxn−zn≤xn−xn+1,JXxn−zn≤xn−xn+1xn−zn.
Hence, we havexn−zn≤xn−xn+1⟶0, which yields (24).
We next show that every weak cluster point ofxn is a solution of the split feasibility problem. To this end, let x be any weak cluster point of xn and take a subsequence xnk of xn converging weakly to x. In view of (18) and (24), we have(31)limn⟶∞I−PCxn=limn⟶∞I−PQAxn=0.
By Lemma5, x∈C. On the other hand, for any x∗∈X∗, it follows that(32)limk⟶∞Axnk,x∗=limk⟶∞xnk,A∗x∗=x,A∗x∗=Ax,x∗,implying Axnk⇀Ax. By Lemma 5, Ax∈Q. Altogether, x∈S. Since x is arbitrary, we obtain the desired conclusion.
Finally, we prove thatxn converges strongly to z^. Now take any x∈ωwxn. Then, x∈S, and there exists a subsequence xnk of xn converging weakly to x. It then follows that(33)x0−z^=x0−PSx0≤x0−x≤limk⟶∞x0−xnk=limk⟶∞x0−PQnkx0≤x0−z^,where the first and the last inequalities follow from the property of metric projections and the second one follows from the weak lower semicontinuity of the norm. Hence,(34)z^=x,limk⟶∞x0−xnk=x0−z^.
Sincex is chosen arbitrarily, this implies that ωwxn is exactly a single-point set, that is, xn converges weakly to z^. Note that x0−xnk⇀x0−z^. By Lemma 2, the uniform convexity implies limkxnk=z^. Since xn converges weakly, this yields limnxn=z^ as desired.
As we see from the previous theorem, the choice ofrn is related to A. Thus, to implement this algorithm, one has to compute the norm A, which is generally not easy in practice. In what follows, we introduce another choice of rn, which ultimately has no relation with A. By applying an idea in [4], we can propose another choice of the parameter rn as follows:(35)rn=I−PCxn2+I−PQAxn2JXI−PCxn+A∗JYI−PQAxn2.
Now let us state the convergence ofxn under this choice of rn.Theorem 2.
Assume thatX is uniformly convex and smooth and Y is reflexive, smooth, and strictly convex. Then, the proposed algorithm with (35) is well defined. Moreover, the sequence xn generated by (16) converges strongly to z^∈S, where z^=PSx0.Proof.
We first show that the algorithm under (35) is well defined. To this end, it suffices to show that for each n∈ℕ, Qn is nonempty. We now prove this by induction. Suppose that S⊆Qk for some k∈ℕ. Take any z∈S. Then, we have z∈Qk. Furthermore, by Lemma 5, we have(36)xk−z,JXxk−zk=rkxk−z,JXI−PCxk+A∗JYI−PQAxk=rkxk−z,JXI−PCxk+rkAxk−Az,JYI−PQAxk≥rkI−PCxk2+I−PQAxk2,and(37)xk−zk2=rk2JXI−PCxk+A∗JYI−PQAxk2=rkI−PCxk2+I−PQAxk2.
Consequently, we have(38)zk−z,JXxk−zk=zk−xk,JXxk−zk+xk−z,JXxk−zk≥rkI−PCxk2+I−PQAxk2−zk−xk2=0.
This implies thatz∈Qk+1. Since z is chosen in S arbitrarily, we conclude that S⊆Qk+1. Consequently, S⊆Qn for all n∈ℕ. Now it is clear that the set Qn is nonempty, closed, and convex. Thus, the proposed algorithm is well defined.
We now prove that the sequencexn generated by (16) converges strongly to z^∈S. From the proof of the previous theorem, it suffices to verify that (31) still holds. Similarly, we obtain limnxn−zn=0. From (37), we have(39)limn⟶∞rnI−PCxn2+I−PQAxn2=0.
On the other hand, we see that(40)rn=I−PCxn2+I−PQAxn2JXI−PCxn+A∗JYI−PQAxn2≥I−PCxn2+I−PQAxn2JXI−PCxn+AJYI−PQAxn2≥I−PCxn2+I−PQAxn21+A2I−PCxn2+I−PQAxn2=11+A2>0.
This together with (39) yields (31) as desired. Hence, the proof is complete.Remark 1.
It is worth noting that our algorithm is new even in Hilbert spaces. Indeed, in Hilbert spaces, our algorithm is reduced to(41)zn=xn−rnI−PCxn+A∗I−PQAxn,Qn+1=z∈Qn:zn−z,xn−zn≥0,xn+1=PQn+1x0,∀n∈ℕ.



---

*Source: 6689983-2021-01-13.xml*