# A Simultaneous Iteration Algorithm for Solving Extended Split Equality Fixed Point Problem

Authors: Meixia Li; Xiping Kao; Haitao Che
Journal: Mathematical Problems in Engineering
 (2017)

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## Abstract

We study a kind of split equality fixed point problem which is an extension of split equality problem. We propose a kind of simultaneous iterative algorithm with a way of selecting the step length which does not need any a priori information about the operator norms and prove that the sequences generated by the iterative method converge weakly to the solution of this problem. Some numerical results are shown to confirm the feasibility and efficiency of the proposed methods.


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## Body

## 1. Introduction

LetC and Q be nonempty closed and convex subsets of the real Hilbert spaces H1 and H2, respectively. The split feasibility problem is to find(1)x∈Csuch thatAx∈Q.It can be used in various disciplines such as image restoration and radiation therapy treatment planning [1, 2]. These applications are in finite-dimensional Hilbert spaces [3, 4]. It also can be found in an infinite-dimensional real Hilbert space [5, 6].Recently, Moudafi [7] introduced a new split equality feasibility problem. Let H1, H2, and H3 be real Hilbert spaces. Let A:H1→H3 and B:H2→H3 be two bounded linear operators. The split equality feasibility problem is to find(2)x∈C,y∈Qsuch thatAx=By,which allows asymmetric and partial relations between the variables x and y. The interest is to cover many situations, for instance, applications in decomposition methods for PDEs, in game theory and in intensity-modulated radiation therapy (for short, IMRT).Moudafi [8] introduced the simultaneous iterative method to solve the split equality feasibility problem. Furthermore, Moudafi studied the fixed point formulation to avoid using the projection. Assume Fix(T) and Fix(S) are the sets of fixed points of T and S, respectively, where T:H1→H1 and S:H2→H2 are nonlinear operators such that Fix(T)≠∅ and Fix(S)≠∅. So the split equality fixed point problem is to find(3)x∈FixT,y∈FixSsuch thatAx=By,where T and S are firmly quasi-nonexpansive mappings. In order to find the solution of the split equality problem, Che and Li [9] proposed the following iterative algorithm:(4)un=xn-γnA∗Axn-Byn,xn+1=αnxn+1-αnTun,vn=yn+γnB∗Axn-Byn,yn+1=αnyn+1-αnSvn,and the weak convergence of the scheme (4) can also be established. Furthermore, Chang et al. [10] modified the iterative scheme (4) and provided a unified framework for solving this problem without using the projection. The framework is as follows.(5)un=xn-γnA∗Axn-Byn,xn+1=αnxn+1-αn1-ξI+ξT1-ηI+ηTun,vn=yn+γnB∗Axn-Byn,yn+1=αnyn+1-αn1-ξI+ξS1-ηI+ηSvn.They got the following conclusion that the sequence (xn,yn) generated by the above modification converges weakly to a solution of problem (3). Furthermore, some authors [11–14] studied the problems (1)–(3) in Banach space. They proposed effective algorithms and proved their convergence under some conditions.Recently, He and Sun [15] studied the problem of split convex feasibility and established a strongly convergent alternating algorithm. They proposed the following iterative algorithm to find(6)x∈C,y∈Qsuch thatAx=By∈K,via the formula(7)ωn=PKnAxn+Byn2,Kn=z∈H3:hAxn+Byn2+ςn,z-Axn+Byn2≤0,un=PCxn-ξA∗Axn-ωn,vn=PQyn-ξB∗Bxn-ωn,Cn+1×Qn+1=x,y∈Cn×Qn:un-x2+vn-y2≤xn-x2+yn-y2,fxn+μn,x-xn≤0,gyn+ηn,y-yn≤0,xn+1=PCn+1x1,yn+1=PQn+1y1,∀n∈N,where ξ∈(0,min(1/A2,1/B2)),  μn∈∂f(xn),  ηn∈∂g(yn), and  ςn∈∂h((Axn+Byn)/2). In some conditions such as Ω={(x,y)∈C×Q:Ax=By∈K}≠∅, they proved that the sequences (xn,yn)→(s,t),  (un,vn)→(s,t), and ωn→z∗≔As=Bt∈K, where (s,t)∈Ω.In this article, we study the following problem of extended split equality fixed point, which is to find(8)x∈FixG,y∈FixMsuch thatAx=By∈K.When K=H3, problem (8) is problem (3). Therefore, problem (8) is the extension of the split equality fixed point problem. We propose the simultaneous iterative algorithm for solving this problem, which avoids using the projection and the step length sequences do not depend on the operator norms A and B. Furthermore, we prove the sequences generated by the algorithm weakly converge to a solution of the extended split equality fixed point problem. Numerical examples show the feasibility and efficiency of this algorithm.


## 2. Preliminaries

In this paper, we recall some concepts, definitions, and conclusions, which are prepared for proving our main results. We writexn⇀x to indicate that the sequence {xn} converges weakly to x. xn→x implies that {xn} converges strongly to x. We denote by H1 a real Hilbert space with inner product 〈·,·〉 and induced norm ·.A mappingT:C→C is called(i)
quasi-nonexpansive, ifFix(T)≠∅,(9)Tx-x∗≤x-x∗,∀x∈C,x∗∈FixT,(ii)
quasi-pseudo-contractive, ifFix(T)≠∅,(10)Tx-x∗≤x-x∗+Tx-x,∀x∈C,x∗∈FixT.A mappingPC is said to be metric projection of H1 onto C if, for every point x∈H1, there exists a unique nearest point in C denoted by PCx such that(11)x-PCx≤x-y,∀y∈C.It is well known thatPC is a nonexpansive mapping and is characterized by the following properties:(12)PCx-PCy2≤x-y,PCx-PCy,∀x,y∈H1,x-PCx,y-PCx≤0,∀x∈H1,y∈C,x-y2≥x-PCx2+y-PCx2,∀x∈H1,y∈C,x-y-PCx-PCy2≥x-y2-PCx-PCy2,∀x,y∈H1.In the proof of our results, we need the following lemmas.Lemma 1 (see [16]).
LetH be a real Hilbert space; then the following conclusions hold.(13)x-y2=x2+y2-2x,y,∀x,y∈H,αx+1-αy2=αx2+1-αy2-α1-αx-y,∀x,y∈H,α∈0,1.Lemma 2 (see [16]).
LetH be a real Hilbert space and T:H→H be a L-Lipschitzian mapping with L≥1. Denote(14)G≔1-ξI+ξT1-ηI+ηT.If 0<ξ<η<1/(1+1+L2), then the following conclusions hold.(i)
Fix(T)=Fix(T((1-η)I+ηT))=Fix(G).(ii)
IfT is demiclosed at 0, then G is also demiclosed at 0.(iii)
In addition, ifT:H→H is quasi-pseudo-contractive, then the mapping G is quasi-nonexpansive; that is,(15)Gx-u∗≤x-u∗,∀x∈H,u∗∈FixT=FixG.


## 3. Main Results

In this section, we assume that(i)
G:H1→H1 and M:H2→H2 are two L-Lipschitzian and quasi-pseudo-contractive mappings with L≥1,  Fix(G)≠∅, and Fix(M)≠∅.(ii)
H1, H2, and H3 be real Hilbert spaces. A:H1→H3 and B:H2→H3 are two bounded linear operators; A∗ and B∗ are their adjoint operators, respectively. Let K⊂H3 be nonempty closed convex set. We consider the extended split equality fixed point problem (8).Theorem 3.
LetH1, H2, and H3 be real Hilbert spaces and K be a closed convex level set(16)K=x∈H3:hx≤0,where h:H3→R is convex function which is subdifferentiable on K and its subdifferentials are bounded on bounded sets. A:H1→H3 and B:H2→H3 are two bounded linear operators with their adjoint operators A∗ and B∗, respectively, σ∈(0,1) is a parameter controlling step length, 0<γ<η<1/(1+1+L2), and {αn}⊂(0,1). Let {xn},  {yn},  {ωn}, {un}, and {vn} be sequences generated by(17)ωn=PKnAxn+Byn2,Kn=z∈H3:hAxn+Byn2+ςn,z-Axn+Byn2≤0,un=xn-ξnA∗Axn-ωn,xn+1=αnxn+1-αnTun,vn=yn-ξnB∗Byn-ωn,yn+1=αnyn+1-αnSvn,where (18)ςn∈∂hAxn+Byn2,ξn∈0,1-σAxn-ωn2+Byn-ωn2A∗Axn-ωn2+B∗Byn-ωn2,T≔1-γI+γG1-ηI+ηG,S≔1-γI+γM1-ηI+ηM, where G and M are demiclosed at 0. Assume Ω={(x,y)∈Fix(G)×Fix(M):Ax=By∈K}≠∅. Then (xn,yn)⇀(t,s),  (un,vn)⇀(t,s), and ωn⇀z∗≔At=Bs, where (t,s)∈Ω.Proof.
It is obvious thatK⊂Kn for any n∈N. Let (x∗,y∗)∈Ω; namely, x∗∈Fix(G),  y∗∈Fix(M), and ω∗=Ax∗=By∗∈K. By Lemma 1, we have(19)ωn-ω∗=PKnAxn+Byn2-PKnω∗2≤12Axn-ω∗+12Byn-ω∗2=12Axn-ω∗2+12Byn-ω∗2-14Axn-Byn2.(20)un-x∗2=xn-x∗-ξnA∗Axn-ωn2=xn-x∗2+ξn2A∗Axn-ωn2-2xn-x∗,ξnA∗Axn-ωn.By (20), we have(21)un-x∗2=xn-x∗2+ξn2A∗Axn-ωn2-ξnAxn-ω∗2-ξnAxn-ωn2+ξnωn-ω∗2.Similarly,(22)vn-y∗2=yn-y∗2+ξn2B∗Byn-ωn2-ξnByn-ω∗2-ξnByn-ωn2+ξnωn-ω∗2.By (21), (22), (17), and (19), we have(23)xn+1-x∗2+yn+1-y∗2=αnxn+1-αnTun-x∗2+αnyn+1-αnSvn-y∗2=αnxn+1-αnTun-1-αnx∗-αnx∗2+αnyn+1-αnSvn-1-αny∗-αny∗2=αnxn-x∗2+1-αnTun-x∗2-αn1-αnxn-Tun2+αnyn-y∗2+1-αnSvn-y∗2-αn1-αnyn-Svn2≤αnxn-x∗2+1-αnun-x∗2-αn1-αn1-αnxn-Tun2+αnyn-y∗2+1-αnvn-y∗2-αn1-αnyn-Svn2=αnxn-x∗2+1-αnxn-x∗2+1-αnξn2A∗Axn-ωn2-ξn1-αnAxn-ω∗2-ξn1-αnAxn-ωn2+2ξn1-αnωn-ω∗2-αn1-αnxn-Tun2+αnyn-y∗2+1-αnyn-y∗2+1-αnξn2B∗Byn-ωn2-ξn1-αnByn-ω∗2-ξn1-αnByn-ωn2-αn1-αnyn-Svn2≤αnxn-x∗2+1-αnxn-x∗2+1-αnξn2A∗Axn-ωn2-ξn1-αnAxn-ω∗2-ξn1-αnAxn-ωn2+ξn1-αnAxn-ω∗2+ξn1-αnByn-ω∗2-12ξn1-αnAxn-Byn2-αn1-αnxn-Tun2+αnyn-y∗2+1-αnyn-y∗2+1-αnξn2B∗Byn-ωn2-ξn1-αnByn-ω∗2-ξn1-αnByn-ωn2-αn1-αnyn-Svn2=xn-x∗2+yn-y∗2+1-αnξn2A∗Axn-ωn2+B∗Byn-ωn2-1-αnξnAxn-ωn2+Byn-ωn2-αn1-αnxn-Tun2+yn-Svn2-121-αnξnAxn-Byn2.Notice that(24)xn+1-x∗2+yn+1-y∗2≤xn-x∗2+yn-y∗2+ξn1-αnξnA∗Axn-ωn2+B∗Byn-ωn2-Axn-ωn2+Byn-ωn2.Thanks to(25)ξn∈0,1-σAxn-ωn2+Byn-ωn2A∗Axn-ωn2+B∗Byn-ωn2, we obtain(26)xn+1-x∗2+yn+1-y∗2≤xn-x∗2+yn-y∗2.
Let(27)Xnx∗,y∗=xn-x∗2+yn-y∗2.We obtain(28)Xn+1x∗,y∗≤Xnx∗,y∗.This implies that Xn(x∗,y∗) is a nonincreasing sequence; hence limn→∞Xn(x∗,y∗) exists. As a result, {xn} and {yn} are bounded sequences. Rewrite (23) as(29)1-αnξnAxn-ωn2+Byn-ωn2-ξnA∗Axn-ωn2+B∗Byn-ωn2+αn1-αnxn-Tun2+yn-Svn2+121-αnξnAxn-Byn2≤xn-x∗2+yn-y∗2-xn+1-x∗2+yn+1-y∗2=Xnx∗,y∗-Xn+1x∗,y∗.Letting n→∞ and taking the limit in (29), we have(30)Axn-Byn⟶0,Tun-xn⟶0,Svn-yn⟶0,(31)Axn-ωn⟶0,Byn-ωn⟶0.Then,(32)limn→∞⁡un-xn=0,limn→∞⁡vn-yn=0,limn→∞⁡xn+1-xn=limn→∞⁡Tun-xn=0,limn→∞⁡yn+1-yn=limn→∞⁡Svn-yn=0,which imply that {xn} and {yn} are asymptotically regular. Furthermore, we get(33)limn→∞⁡Tun-un=0,limn→∞⁡Svn-vn=0.Since {xn} and {yn} are bounded sequences, there exist weakly convergent subsequences, say {xni}⊂{xn} such that xni⇀t; also {yni}⊂{yn} such that yni⇀s. The Opial property guarantees that the weakly subsequential limit of {(xn,yn)} is unique. So we have xn⇀t,  yn⇀s. Therefore un⇀t,  vn⇀s. Since G and M are demiclosed at 0, and from Lemma 2, by (33), we have Tt=t,  Ss=s, which imply that t∈Fix(G),  s∈Fix(M). Hence, (34)t,s∈FixG×FixM.Furthermore, since Axn-Byn⇀At-Bs, by using the weakly lower semicontinuity of squared norm, we have (35)At-Bs≤liminfn→∞⁡Axn-Byn=limn→∞⁡Axn-Byn=0;that is, (36)At=Bs.By (31), we have ωn⇀z∗≔At=Bs. Now, we prove z∗∈K. We know that {ςn} is bounded. There exists M¯>0, such that ςn≤M¯, where M¯ is a constant. Note that ωn=PKn((Axn+Byn)/2)∈Kn, and we have(37)hAxn+Byn2+ςn,ωn-Axn+Byn2≤0.Hence,(38)hAxn+Byn2≤-ςn,ωn-Axn+Byn2≤M¯ωn-Axn+Byn2.By the lower semicontinuity of h, (38), and (31), we obtain(39)hz∗≤liminfn→∞⁡hAxn+Byn2≤M¯liminfn→∞⁡ωn-Axn+Byn2=0.Thus z∗∈K. The proof is completed.


## 4. Consequent Results

In this section, we give some corollaries, which are easily obtained from Theorem3.IfG=M, we have the following corollary.Corollary 4.
LetH1,  H2, and H3 be real Hilbert spaces and K be a closed convex level set (40)K=x∈H3:hx≤0,where h:H3→R is convex function which is subdifferentiable on K and its subdifferentials are bounded on bounded sets. A:H1→H3 and B:H2→H3 are two bounded linear operators with their adjoint operators A∗ and B∗, respectively, σ∈(0,1) is a parameter controlling step length, 0<γ<η<1/(1+1+L2), and {αn}⊂(0,1). Let {xn},  {yn},  {ωn},  {un}, and {vn} be sequences generated by(41)ωn=PKnAxn+Byn2,Kn=z∈H3:hAxn+Byn2+ςn,z-Axn+Byn2≤0,un=xn-ξnA∗Axn-ωn,xn+1=αnxn+1-αnTun,vn=yn-ξnB∗Byn-ωn,yn+1=αnyn+1-αnTvn,where (42)ςn∈∂hAxn+Byn2,ξn∈0,1-σAxn-ωn2+Byn-ωn2A∗Axn-ωn2+B∗Byn-ωn2,T≔1-γI+γG1-ηI+ηG,where G is demiclosed at 0. Assume Ω={(x,y)∈Fix(G)×Fix(G):Ax=By∈K}≠∅. Then (xn,yn)⇀(t,s),(un,vn)⇀(t,s) and ωn⇀z∗≔At=Bs, where (t,s)∈Ω.IfA=B, we have the following corollary.Corollary 5.
LetH1 and H3 be real Hilbert spaces and K be a closed convex level set (43)K=x∈H3:hx≤0,where h:H3→R is convex function which is subdifferentiable on K and its subdifferentials are bounded on bounded sets. A:H1→H3 is a bounded linear operator with its adjoint operator A∗, σ∈(0,1) is a parameter controlling step length, 0<γ<η<1/(1+1+L2), and {αn}⊂(0,1). Let {xn},  {yn},  {ωn},  {un}, and {vn} be sequences generated by(44)ωn=PKnAxn+Ayn2,Kn=z∈H3:hAxn+Ayn2+ςn,z-Axn+Ayn2≤0,un=xn-ξnA∗Axn-ωn,xn+1=αnxn+1-αnTun,vn=yn-ξnA∗Ayn-ωn,yn+1=αnyn+1-αnSvn,where (45)ςn∈∂hAxn+Byn2,ξn∈0,1-σAxn-ωn2+Byn-ωn2A∗Axn-ωn2+B∗Byn-ωn2,T≔1-γI+γG1-ηI+ηG,S≔1-γI+γM1-ηI+ηM,where G and M are demiclosed at 0. Assume Ω={(x,y)∈Fix(G)×Fix(M):Ax=Ay∈K}≠∅. Then (xn,yn)⇀(s,t),  (un,vn)⇀(s,t), and ωn⇀z∗≔As=At, where (s,t)∈Ω.IfG=M and A=B, we have the following corollary.Corollary 6.
LetH1 and H3 be real Hilbert spaces and K be a closed convex level set (46)K=x∈H3:hx≤0,where h:H3→R is convex function which is subdifferentiable on K and its subdifferentials are bounded on bounded sets. A:H1→H3 is a bounded linear operator with its adjoint operator A∗, and σ∈(0,1) is a parameter controlling step length, 0<γ<η<1/(1+1+L2), and {αn}⊂(0,1). Let {xn},  {yn},  {ωn},  {un}, and {vn} be sequences generated by(47)ωn=PKnAxn+Ayn2,Kn=z∈H3:hAxn+Ayn2+ςn,z-Axn+Ayn2≤0,un=xn-ξnA∗Axn-ωn,xn+1=αnxn+1-αnTun,vn=yn-ξnA∗Ayn-ωn,yn+1=αnyn+1-αnTvn,where (48)ςn∈∂hAxn+Byn2,ξn∈0,1-σAxn-ωn2+Byn-ωn2A∗Axn-ωn2+B∗Byn-ωn2,T≔1-γI+γG1-ηI+ηG,where G is demiclosed at 0. Assume Ω={(x,y)∈Fix(G)×Fix(G):Ax=Ay∈K}≠∅. Then (xn,yn)⇀(t,s),  (un,vn)⇀(t,s), and ωn⇀z∗≔At=As, where (t,s)∈Ω.IfG=M=A=B is an identity operator, we have the following corollary.Corollary 7.
LetH1 and H3 be real Hilbert spaces and K be a closed convex level set (49)K=x∈H3:hx≤0,where h:H3→R is convex function which is subdifferentiable on K and its subdifferentials are bounded on bounded sets. σ∈(0,1) is a parameter controlling step length, 0<γ<η<1/(1+1+L2), and {αn}⊂(0,1). Let {xn},  {yn},  {ωn},  {un}, and {vn} be sequences generated by(50)ωn=PKnxn+yn2,Kn=z∈H3:hxn+yn2+ςn,z-xn+yn2≤0,un=1-ξxn+ξωn,xn+1=αnxn+1-αnun,vn=1-ξyn+ξωn,yn+1=αnyn+1-αnvn,where ςn∈∂h((xn+yn)/2),  ξ∈(0,1-σ). Assume Ω={(x,y)∈H1×H1:x=y∈K}≠∅. Then (xn,yn)⇀(t,s),  (un,vn)⇀(t,s), and ωn⇀z∗≔t=s, where (t,s)∈Ω.


## 5. Numerical Examples

In this section, we give an example to show some insight into the behavior of the algorithm presented in this paper. The whole codes are written in Matlab 7.0. All the numerical results are carried out on a personal Lenovo ThinkPad computer with Intel(R) Core(TM) i7-6500U CPU 2.50 GHz and RAM 8.00 GB.LetH1=R2,  H2=R3.  A∈R3×2 and B∈R3×3 are as follows.(51)A=0.81470.91340.90580.63240.12700.0975,B=0.27850.96490.95720.54690.15760.48540.95750.97060.8003.LetK={x∈Rn∣h(x)=x13+x22+x3-2≤0}, where x=x1,x2,x3T. For convenience, we take σ=0.2, αn=1/3+1/2n, and G=M=(1/2)I. We choose Axn-Byn≤10-6 as the stopping criterion.Figure1 presents the behaviors of Axn-Byn in different initial points such as(52)x=1,1T,y=1,1,1T,x=5.2378,2.6487T,y=6.8357,43.6327,17.3853T,x=0.5221,19.0936T,y=86.1193,192.3197,152.4829T,x=0.0588,5.4403T,y=14.1190,12.9026,11.0462T.It is easy to see that the presentation reveals that Ax=By. Table 1 shows the number of iterations and the CPU time for the above four initial points. We denote Iter. and Sec. as the number of iterations and the CPU time in seconds, respectively.Table 1
Number of iterations and the CPU time for different initial points.
Initial point
Iter.
Sec.
x
=
1,1
T,
y
=
1,1
,
1
T
27
0.01
x
=
5.2378,2.6487
T,
y
=
6.8357,43.6327,17.3853
T
42
0.016
x
=
0.5221,19.0936
T,
y
=
86.1193,192.3197,152.4829
T
50
0.017
x
=
0.0588,5.4403
T,
y
=
14.1190,12.9026,11.0462
T
38
0.012Figure 1
The behaviors ofAxn-Byn for different initial points.Furthermore, for testing the stationary property of iterative numbers, we carry out 500 experiments for different initial points which are presented randomly, such as(53)x=rand⁡2,1,y=rand⁡3,1,x=10∗rand⁡2,1,y=10∗rand⁡3,1,x=1000∗rand⁡2,1,y=600∗rand⁡3,1,x=100000∗rand⁡2,1,y=60000∗rand⁡3,1,separately. In these cases, we take σ=0.2,  αn=0.4 and choose Axn-wn+Byn-wn≤10-6 as the stopping criterion. Figure 2 illustrates the behaviors of iterative numbers for different initial points, which reveals the stationary property of iterative numbers of the algorithm.Figure 2
The behaviors of iterative numbers for different initial points.



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*Source: 9737062-2017-04-06.xml*