# A Cyclic Method for Solutions of a Class of Split Variational Inequality Problem in Banach Space

Authors: Bashir Ali; Aisha A. Adam; Yusuf Ibrahim
Journal: Abstract and Applied Analysis
 (2020)

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## Abstract

In this paper, a cyclic algorithm for approximating a class of split variational inequality problem is introduced and studied in some Banach spaces. A strong convergence theorem is proved. Some applications of the theorem are presented. The results presented here improve, unify, and generalize certain recent results in the literature.


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## Body

## 1. Introduction

LetC be a nonempty closed and convex subset of a real Banach space E, with dual E∗. Then, a mapping T:C⟶E is said to be(1)
Nonexpansive ifTx−Ty≤x−yfor anyx,y∈C.(2)
Demiclosed at zero if whenever a sequencevn in C converges weakly to u and vn−Tvn converges strongly to 0, then u∈FT.(3)
L-Lipschitz continuous onE if there exists L>0 such that(1)Tx−Ty≤Lx−y,for allx,y∈C.A mappingT:C⟶E∗ is said to be(1)
Monotone if(2)x−y,Tx−Ty≥0,for allx,y∈C.(2)
α-inverse strongly monotone if(3)x−y,Tx−Ty≥αTx−Ty2,for anyx,y∈C.(3)
Strongly monotone if(4)x−y,Tx−Ty≥αx−y2,for anyx,y∈C.Problem of the type findingu∈C such that(5)v−u,Tu≥0,for allv∈C,is called a variational inequality problem, and the set of solution of such problem is denoted by VIC,T.Variational inequality problems have played a crucial role in the study of several problems arising in physics, finance, economics, network analysis, optimization, medical image and structural analysis, and so on (see, for example, [1–5]). Variational inequality problems were formulated in the late 1960’s by Lions and Stampacchia [6]. Since then, various iterative algorithms for approximating solutions of such problems have been proposed by numerous researchers (see, for example, [7–12, 27]) and the references therein.In 1976, Korpelevch [14] introduced the following extragradient method for solving the variational inequality problem when the operator T is monotone and L-Lipschitz continuous in a finite dimensional Euclidean space ℝn,(6)xn+1=PCxn−λfPCxn−λTxn,for each n∈ℕ, λ∈0,1/L.The split feasibility problem in the finite dimensional Hilbert space was first introduced by Censor and Elfving [15] for modeling inverse problems which arise from phase retrievals and in medical image reconstruction.LetH1 and H2 be two real Hilbert spaces. Let C and Q be two nonempty closed convex subset of H1 and H2, respectively. The split feasibility problem is to find(7)u∈Csuch that Au∈Q.Assuming that the split feasibility problem is consistent (i.e., (7) has a solution), it is easy to see that x∈C solves (7) if and only if it solves the fixed point equation:(8)x=PCI+γA∗PQ−IAx,x∈C,where PC and PQ are the orthogonal projections onto C and Q, respectively, γ>0, and A∗ is the adjoint of A. To solve (8), Byrne [16] proposed the CQ algorithm which generates a sequence xn by(9)xn+1=PCI+γA∗PQ−IAxn,for each n∈ℕ, where γ∈0,2/λ, λ being the spectral radius of the operator A∗A.In 2010, Censor et al. [17] considered a new variational problem called split variational inequality problem (SVIP). It entails finding a solution of one variational inequality problem whose image under a bounded linear transformation is a solution of another variational inequality problem. The SVIP is formulated as(10)findu∈VIC,fsuch thatAu∈VIQ,g,where C and Q are the nonempty closed convex subsets of real Hilbert spaces H1 and H2, respectively, and A:H1⟶H2 is a bounded linear operator. They constructed the following iterative algorithm to solve such problem and proved a strong convergence theorem in a Hilbert space:(11)xn+1=PCI−λfxn+γA∗PQI−λg−IAxn,n∈ℕ,where γ∈0,1/L, L is the spectral radius of the operator A∗A, and A∗ is the adjoint of A.One can easily observe that split variational inequality has the split feasibility problem as a special case.Recently, Tian and Jiang [12], based on the work of Censor et al. [17], considered a class of SVIP which is to find(12)x∈VIC,fsuch thatAx∈FT,where C is a nonempty closed convex subset of a real Hilbert space H1, f:C⟶H1 is a monotone and k-Lipschitz continuous map, A:H1⟶H2 is a bounded linear map, and T:H2⟶H2 is a nonexpansive map. They proposed the following algorithm by combining the Korpelevich extragradient method and Byrne CQ algorithm:(13)x1=x∈C;yn=PCxn−γnA∗I−TAxn;tn=PCyn−λnfyn;xn+1=PCyn−λnftn.They obtained the following result.Theorem 1.
(see [12]). Let H1 and H2 be real Hilbert spaces. Let C be a nonempty, closed and convex subset of H1, A:H1⟶H2 be a bounded linear operator such that A≠0, f:C⟶H1 be a monotone and k-Lipschitz continuous map, and T:H2⟶H2 be a nonexpansive map. Setting Γ=z∈VIC,f:Az∈FT, assume that Γ≠∅. Let the sequence xn be generated by (13), where γn⊂a,b, a,b∈0,1/A2, and λn∈0,1/k. Then, the sequence xn converges weakly to a point z∈Γ.Remark 1.
Inspired by the results of Tian and Jiang [12], the authors raised the following motivational questions:Q1. Can the result of Tian and Jiang hold in a more general setting of Banach space than Hilbert?Q2. Can the result also be proved for a common fixed point of finite family of nonexpansive mapping?Q3. Can strong convergence theorem be proved?In this paper, the above questions are answered in affirmative. We study a cyclic algorithm in the setting of uniformly smooth which is also 2-uniformly convex real Banach space and 2-uniformly smooth real Banach space and prove its strong convergence to a solution of a variational inequality problem for a monotone K-Lipschitz continuous map whose image under a bounded linear operator is a common fixed point of a finite family of nonexpansive maps. Our theorems improve and extend the results of Tian and Jiang [12].


## 2. Preliminaries

The duality map of a Banach spaceE has the following properties:(1)
IfE is a reflexive, strictly convex, and smooth real Banach space, then J is single-valued and bijective. In this case, the inverse J−1:E∗⟶E is given by J−1=J∗ with J∗ being the duality mapping of E∗.(2)
In a Hilbert spaceH, the duality map J and its inverse J−1 are the identity maps on H.(3)
IfE is uniformly smooth and uniformly convex, then the dual space E∗ is also uniformly smooth and uniformly convex and the normalized duality map J and its inverse, J−1, are both uniformly continuous on bounded sets.LetE be a smooth real Banach space and ϕ:E×E⟶ℝ be defined by(14)ϕx,y=x2−2x,Jy+y2,∀x,y∈E.It is easy to see from the definition ofϕ that, in a real Hilbert space H, equation (14) reduces to ϕx,y=x−y2,∀x,y∈H.Furthermore, givenx,y,z∈E and τ∈0,1, we have the following properties (see, for example, [18]):P1:x−y2≤ϕx,y≤x+y2,P2:ϕx,y=ϕx,z+ϕz,y+2z−x,Jy−Jz,P3:ϕτx+1−τy,z≤τϕx,z+1−τϕy,z.Definition 1.
LetE be a smooth, strictly convex, and reflexive real Banach space and let C be a nonempty, closed, and convex subset of E. The map ΠC:E⟶C defined by x˜=ΠCx∈C such that ϕx˜,x=infy∈Cϕy,x is called the generalized projection of E onto C. Clearly, in a real Hilbert space H, the generalized projection ΠC coincides with the metric projection PC from H onto C.Definition 2.
LetE1 and E2 be two reflexive, strictly convex, and smooth Banach spaces. The collection of mappings A:E1⟶E2 is linear, and continuous is a normed linear space with norm defined by A=supx≤1Ax. The dual operator A∗:E2∗⟶E1∗ defined by A∗y∗,x=y∗,Ax∀x∈E1, y∗∈E2∗ is called the adjoint operator of A. The adjoint operator A∗ has the property A∗=A.Lemma 1 (see [19]).
LetC be a nonempty closed and convex subset of a smooth, strictly convex, and reflexive real Banach space E. Then,(1)
Ifx∈E, then x˜=ΠCx if and only if x˜−y,Jx−Jx˜≥0, for all y∈C,(2)
ϕy,x˜+ϕx˜,x≤ϕy,x,for allx∈E,y∈C.Lemma 2.
(see [20]). Let E be q-uniformly smooth Banach space. Then, there exists a constant dq>0 such that(15)x+yq≤xq+qy,jx+dqyq.Lemma 3.
(see [21]). Let E be a 2-uniformly convex and smooth real Banach space. Then, there exists a positive constant α such that(16)αx−y2≤ϕx,y,∀x,y∈E.Lemma 4.
(see [22]). Let C be a nonempty closed and convex subset of a reflexive space E and f, a monotone, and hemicontinuous map of C into E∗. Let B⊂E×E∗ be an operator defined by(17)Bu=fu+NCu,ifu∈C,∅,ifu∉C,where NCu is defined as(18)NCu=w∗∈E∗:u−z,w∗≥0,∀z∈C.
Then,B is maximal monotone and B−10=VIC,f.Lemma 5.
(see [23]). Let E be a uniformly convex and smooth real Banach space, and let xn and yn be two sequences of E. If either xn or yn is bounded and limn⟶∞ϕxn,yn=0, then limn⟶∞xn−yn=0.


## 3. Main Results

Theorem 2.
LetE1 be a uniformly smooth and 2-uniformly convex real Banach space and E2 a 2-uniformly smooth real Banach space with smoothness constant d2∈0,1. Let C be a nonempty, closed, and convex subset of E1. Let f:C⟶E1∗ be a monotone and k-Lipschitz continuous map, and A:E1⟶E2 be a bounded linear operator with its adjoint A∗ such that A≠0. Let Ti:E2⟶E2, i=1,2,…,m be nonexpansive mappings. Setting Γ=z∈VIC,f:Az∈∩i=1mFTi and assuming Γ≠∅. Let a sequence xn be generated by(19)x1=x∈C=C1,yn=J−1Jxn−γA∗J2I−TnAxn,wheren=nmodm,tn=ΠCJ−1Jyn−λfyn,zn=ΠCJ−1Jyn−λftn,Cn+1=v∈Cn:ϕv,zn≤ϕv,yn≤ϕv,xn,,xn+1=ΠCn+1x1,where γ∈a,b, a,b∈0,1/d2∗A2, d2∗ being the smoothness constant of E1∗ as in Lemma 2, and λ∈0,α/k, α being a positive constant as in Lemma 3. Then, the sequence xn converges to a point z∈Γ.Proof.
We divide the proof into five steps.Step 1.
We show thatCn is closed and convex for any n≥1.
SinceC=C1, C1 is closed and convex.
AssumeCn is closed and convex for some n≥1. Since for any v∈Cn,(20)ϕv,yn≤ϕv,xn⟺2v,Jxn−Jyn≤xn2−yn2,ϕv,zn≤ϕv,yn⟺2v,Jyn−Jzn≤yn2−zn2,we have that Cn+1 is closed and convex. Therefore, Cn is closed and convex for any n≥1.Step 2.
We proveΓ⊂Cn for any n≥1.
Forn=1, Γ⊂C=C1.
AssumeΓ⊂Cn for some n≥1. Let u∈Γ, then(21)ϕu,yn=ϕu,J−1Jxn−γA∗J2I−TnAxn=u2+J−1Jxn+γA∗J2Tn−IAxn2−2u,Jxn+γA∗J2Tn−IAxn=u2+Jxn+γA∗J2Tn−IAxn2−2u,Jxn−2γAu,J2Tn−IAxn.
From the fact thatE1∗ is 2-uniformly smooth, we have(22)ϕu,yn≤u2+Jxn2+2γxn,A∗J2Tn−IAxn+d2∗γ2A2Tn−IAxn2−2u,Jxn−2γAu,J2Tn−IAxn=ϕu,xn+d2∗γ2A2Tn−IAxn2+2γAxn−Au,J2Tn−IAxn=ϕu,xn+d2∗γ2A2Tn−IAxn2+2γAxn−TnAxn+TnAxn−Au,J2Tn−IAxn=ϕu,xn+d2∗γ2A2Tn−IAxn2−2γTn−IAxn2+2γTnAxn−Au,J2Tn−IAxn.
Using the fact thatE2 is 2-uniformly smooth and Tn being nonexpansive, we have(23)2TnAxn−Au,J2Tn−IAxn≤d2TnAxn−Au2+TnAxn−Axn2−TnAxn−Au−TnAxn−Axn2≤d2−1Axn−Au2+Tn−IAxn2.
From (22) and (23), we get(24)ϕu,yn≤ϕu,xn+γ2d2∗A2Tn−IAxn2−2γTn−IAxn2+γd2−1Axn−Au2+γTn−IAxn2≤ϕu,xn−γ1−d2∗γA2Tn−IAxn2−γ1−d2Axn−Au2≤ϕu,xn−γ1−d2∗γA2Tn−IAxn2≤ϕu,xn.
Also by Lemma1, we have(25)ϕu,zn≤ϕu,J−1Jyn−λftn−ϕzn,J−1Jyn−λftn=u2−2u,Jyn−λftn−zn2+2zn,Jyn−λftn=ϕu,yn−ϕzn,yn+2u−zn,λftn=ϕu,yn−ϕzn,yn+2λu−tn,ftn+2λtn−zn,ftn.
By the fact thatu∈VIC,f and using property P2, we have(26)ϕu,zn≤ϕu,yn−ϕzn,yn+2λtn−zn,ftn=ϕu,yn−ϕzn,tn−ϕtn,yn+2zn−tn,Jyn−λftn−Jtn.
Also from the fact thattn=ΠCJ−1Jyn−λfyn, zn∈C, the Lipschitz continuity of f, Lemma 1, and Lemma 3, we obtain that(27)zn−tn,Jyn−λftn−Jtn=zn−tn,Jyn−λfyn−Jtn+λzn−tn,fyn−ftn≤λzn−tn,fyn−ftn≤kλzn−tnyn−tn≤kλ2zn−tn2+yn−tn2≤kλ2αϕzn,tn+ϕtn,yn.
Thus,(28)ϕu,zn≤ϕu,yn−ϕtn,yn−ϕzn,tn+kλαϕzn,tn+ϕtn,yn=ϕu,yn−1−kλαϕtn,yn+ϕzn,tn(29)≤ϕu,yn.
Hence,Γ⊂Cn for any n≥1.Step 3.
We shall show thatxn is a Cauchy sequence.
SinceΓ⊂Cn+1⊂Cn and xn+1=ΠCn+1x1⊂Cn, then by Lemma 1, we have that ϕxn+1,x1≤ϕu,x1 and also ϕxn,x1≤ϕxn+1,x1. Hence, ϕxn,x1 is nondecreasing. So, limn⟶∞ϕxn,x1 exists. By property P1, xn is bounded. Also, it follows from (24), (28), and the fact that A is a bounded linear operator that yn, zn, and Axn are bounded.
From Lemma1, we have that
ϕ
x
m
,
x
n
=
ϕ
x
m
,
Π
C
n
x
1
≤
ϕ
x
m
,
x
1
−
ϕ
x
n
,
x
1
⟶
0 as n,m⟶∞. Hence, xn is a Cauchy sequence.Step 4.
We show that(30)limn⟶∞xn−yn=limn⟶∞Tn−IAxn=limn⟶∞zn−tn=limn⟶∞zn−yn=0.
Sincexn+1∈Cn+1⊂Cn,(31)ϕxn,yn=ϕxn,xn+1+ϕxn+1,yn+2xn+1−xn,Jyn−Jxn+1≤ϕxn,xn+1+ϕxn+1,xn+2xn+1−xn,Jyn−Jxn+1=2xn−xn+1,Jxn−Jxn+1−2xn−xn+1,Jyn−Jxn+1.
Taking limit asn⟶∞, we have limn⟶∞ϕxn,yn=0.
Similarly,(32)ϕxn,zn≤2xn−xn+1,Jxn−Jxn+1−2xn−xn+1,Jzn−Jxn+1.
Taking limit asn⟶∞, we have limn⟶∞ϕxn,zn=0.
Sincexn is bounded, it follows from Lemma 5 that limn⟶∞xn−yn=0 and limn⟶∞xn−zn=0.
Now,(33)ϕu,xn−ϕu,yn=xn2−yn2−2u,Jxn−Jyn=xn−ynxn+yn+2u,Jxn−Jyn≤xn−ynxn+yn+2u,Jxn−Jyn.
Taking limit asn⟶∞, we have limn⟶∞ϕu,xn−ϕu,yn=0.
In a similar way, we also havelimn⟶∞ϕu,xn−ϕu,zn=0.
From (24), we obtain(34)0<γ−d2∗γ2A2Tn−IAxn2≤ϕu,xn−ϕu,yn.
Thus,(35)limn⟶∞Tn−IAxn2=0.
From (24) and (28),(36)ϕtn,yn≤11−kλ/αϕu,xn−ϕu,zn,(37)ϕzn,tn≤11−kλ/αϕu,xn−ϕu,zn,(35) and (36), respectively, implies that(38)limn⟶∞ϕtn,yn=0,limn⟶∞ϕzn,tn=0.
Sincezn and yn are bounded, by Lemma 5, we get(39)limn⟶∞tn−yn=0,(40)limn⟶∞zn−tn=0.
Thus, (39) and (40) imply(41)limn⟶∞zn−yn=0.Step 5.
We show thatxn converges to an element of Γ
Sincexn is a Cauchy sequence, we may assume that xn⟶x∗.
From the fact thatlimn⟶∞xn−yn=limn⟶∞xn−zn=limn⟶∞tn−yn=0, we obtain that yn⟶x∗, zn⟶x∗, and tn⟶x∗. Since A is a bounded linear operator, we have that Axn⟶Ax∗.
From (35), we have(42)limn⟶∞Tn−IAxn=0.
Thus, fori∈1,2,3,…,m, limn⟶∞Ti−IAxn=0.
SinceTi is nonexpansive for each i∈1,2,3,…,m, we have that I−Ti is demiclosed at 0 for i∈1,2,3,…,m. And therefore, Ax∗∈FTifor eachi∈1,2,3,…,m.
Thus,Ax∗∈∩i=1mFTi. Next, we show that x∗∈VIC,f
Define(43)Bv=fv+NCv,ifv∈C,∅,ifv∉C.
By Lemma4, B is a maximal monotone and 0∈Bv if and only if v∈VIC,f.
Letv,w∈GB. v,w∈GB⟹w∈Bv=fv+NCv⟹w−fv∈NCv.
Thus,v−p,w−fv≥0,∀p∈C. Since zn=ΠCJ−1Jyn−λftn and v∈C, we have by Lemma 4 that zn−v,Jyn−λftn−Jzn≥0.
Thus,(44)v−zn,Jzn−Jynλ+ftn≥0,n≥0.
Using the fact thatzn∈C and w−fv∈NCv, we have(45)v−zn,w≥v−zn,fv≥v−zn,fv−v−zn,Jzn−Jynλ+ftn=v−zn,fv−fzn+v−zn,fzn−ftn−v−zn,Jzn−Jynλ≥v−zn,fzn−ftn−v−zn,Jzn−Jynλ.
Using the fact thatJ is uniformly continuous on bounded sets and f is Lipschitz continuous, as n⟶∞, we have(46)v−x∗,w≥0.
SinceB is a maximal monotone, 0∈Bx∗, and hence x∗∈VIC,f, therefore x∗∈Γ.Corollary 1.
LetE1 be a uniformly smooth and 2-uniformly convex real Banach space and E2 a 2-uniformly smooth real Banach space with smoothness constant d2∈0,1. Let C be a nonempty, closed, and convex subset of E1. Let f:C⟶E1∗ be a monotone and k-Lipschitz continuous map and A:E1⟶E2 be a bounded linear operator with its adjoint A∗ such that A≠0. Let T:E2⟶E2 be a nonexpansive map. Let Γ=z∈VIC,f:Az∈FT≠∅. Let a sequence xn be generated by(47)x1=x∈C=C1,yn=J−1Jxn−γA∗J2I−TAxn,tn=ΠCJ−1Jyn−λfyn,zn=ΠCJ−1Jyn−λftn,Cn+1=v∈Cn:ϕv,yn≤ϕv,xn,ϕv,zn≤ϕv,yn,xn+1=ΠCn+1x1,where γ∈a,b, a,b∈0,1/d2∗A2, d2∗ being the smoothness constant of E1∗ as in Lemma 2, and λ∈0,α/k, α being a positive constant as in Lemma 3. Then, the sequence xn converges to a point z∈Γ.Proof.
The result followed from Theorem2 by setting Tn=T,∀n∈ℕ.Corollary 2.
LetH1 and H2 be real Hilbert spaces. Let C be a nonempty, closed, and convex subset of H1. A:H1⟶H2 be a bounded linear operator such that A≠0, f:C⟶H1 be a monotone and k-Lipschitz continuous map, and Ti:H2⟶H2, i=1,2,…,m, be a nonexpansive map. Let Γ=z∈VIC,f:Az∈FT≠∅. Let a sequence xn be generated by(48)x1=x∈C=C1,yn=xn−γA∗I−TnAxn,wheren=nmodm,tn=PCyn−λfyn,zn=PCyn−λftn,Cn+1=v∈Cn:yn−v≤xn−v,zn−v≤yn−v,xn+1=PCn+1x1,where γ⊂a,b, a,b∈0,1/A2, and λ∈0,1/k. Then, the sequence xn converges strongly to a point z∈Γ.Proof.
SettingE1=H1 and E2=H2 in Theorem 2, the result is as follows.Remark 2.
Corollary2 complements Theorem 1 (the result of Tian and Jiang [12]) in the sense that strong convergence of the sequence generated is obtained to a solution of the split variational inequality problem involving common fixed points of finite family of nonexpansive mappings, while weak convergence of the scheme is obtained in Theorem 1 involving only a single nonexpansive map. Though, additional condition of projecting the iterates on the half spaces Cn at each step is imposed.Lemma 6.
LetH be a real Hilbert space and Q be a nonempty closed convex subset of H. Let g:Q⟶H be an α- inverse strongly monotone mapping, that is, x−y,gx−gy≥αgx−gy2for anyx,y∈Q. Let μ∈0,2α. Then, the mapping PQI−μg is nonexpansive.Proof.
Letx,y∈H:(49)PQx−μgx−PQy−μgy2≤x−μgx−y−μgy2=x−y2−2μx−y,gx−gy+μ2gx−gy2≤x−y2−μ2−μαx−y,gx−gy≤x−y2.
Hence,PQI−μg is nonexpansive.Corollary 3.
LetH1 and H2 be real Hilbert spaces. Let C and Q be two nonempty, closed, and convex subset of H1 and H2, respectively. Let A:H1⟶H2 be a bounded linear operator such that A≠0, f:C⟶H1 be a monotone and k-Lipschitz continuous map, and g:H2⟶H2 be an inverse strongly monotone mapping. Setting Γ=z∈VIC,f:Az∈VIQ,g≠∅. Let a sequence xn be defined by(50)yn=xn−γA∗I−PQI−μgAxn,tn=PCyn−λfyn,zn=PCyn−λftn,Cn+1=v∈Cn:yn−v≤xn−v,zn−v≤yn−v,xn+1=PCn+1x1,where γ⊂a,b, a,b∈0,1/A2, λ∈01/k, and μ∈0,2α . Then, the sequence xn converges to a point z∈Γ.Proof.
SincePQI−μg is nonexpansive for μ∈0,2α and z∈VIQ,g if and only if z=PQI−μgz for μ>0, putting T=PQI−μg in Corollary 1, we get the desired result.


## 4. Application to Equilibrium Problem

LetC be a nonempty closed convex subset of a real Banach space E and let F:C×C⟶ℝ be a bifunction.The equilibrium problem with respect toF and C is to find z∈C such that(51)Fz,y≥0,∀y∈C.The set of solutions of the equilibrium problem mentioned above is denoted byEPF. For solving the equilibrium problem, we assume that F satisfies the following conditions:(A1)Fx,x=0 for all x∈C;(A2)F is monotone, i.e., Fx,y+Fy,x≤0∀x,y∈C;(A3) for eachx,y,z∈C, limt↓0Ftz+1−tx,y≤Fx,y;(A4) for eachx∈C, y↦Fx,y is convex and lower semicontinous.Lemma 7.
(see [24]). Let E be a reflexive, strictly convex, and uniformly smooth Banach space and C be a nonempty closed convex subset of E. Let F:C×C⟶ℝ be a bifunction satisfying conditions (A1)–(A4), then for any x∈E and r>0, there exists a unique point z∈C such that(52)Fz,y+1ry−z,jz−jx≥0,∀y∈C.Lemma 8.
(see [24]). Let E be a reflexive, strictly convex, and smooth Banach space and C be a nonempty closed convex subset of E. Let F:C×C⟶ℝ be a bifunction satisfying conditions (A1)–(A4), then for any x∈E and r>0, define a mapping Tr:E⟶C by(53)Trx=x∈C:Fz,y+1ry−z,jz−jx≥0∀y∈C.
Then, the following holds:(1)
Tr is single valued;(2)
Tr is a firmly nonexpansive type, i.e.,(54)Trx−Try,JTrx−JTry≤Trx−Try,x−y,∀x,y∈E;(3)
FTr=EPF;(4)
EPF is closed and convex.Theorem 3.
LetE1 be a uniformly smooth and 2-uniformly convex real Banach space and E2 a 2-uniformly smooth real Banach space with smoothness constant d2∈0,1. Let C and Q be two nonempty, closed, and convex subsets of E1 and E2, respectively. Let A:E1⟶E2 be a bounded linear operator with its adjoint A∗ such that A≠0, f:C⟶E1∗ be a monotone and k-Lipschitz continuous map, and F:Q×Q⟶ be a bifunction satisfying conditions (A1)–(A4). Let Γ=z∈VIC,f:Az∈EPF≠∅. Let a sequence xn be generated by(55)x1=x∈C=C1,yn=J−1Jxn−γA∗J2I−TrAxn,tn=ΠCJ−1Jyn−λfyn,zn=ΠCJ−1Jyn−λftn,Cn+1=v∈Cn:ϕv,yn≤ϕv,xn,ϕv,zn≤ϕv,yn,xn+1=ΠCn+1x1,where γ∈a,b, a,b∈0,1/d2∗A2, d2∗ is the smoothness constant of E1∗ as in Lemma 2, λ∈0,α/k, α being a positive constant as in Lemma 3, and Tr is the resolvent of F for r>0. Then, the sequence xn converges to a point z∈Γ.Proof.
PuttingT=Tr in Corollary 1, we get the desired result.


## 5. Application to Maximal Monotone Operator

A set valued mappingB⊂E×E∗ with domain DB=x∈E:Bx≠∅ and range RB=∪Bx:x∈DB is said to be monotone if x−y,x∗−y∗≥0 whenever x,x∗,y,y∗∈B. A monotone operator B⊂E×E∗ is said to be maximal monotone if its graph GB=x,y:y∈Bx is not properly contained in the graph of any other monotone mapping. We know that if B is a maximal monotone, then the zero of B, B−10=x∈E:0∈Bx is closed and convex. If E is a smooth, strictly convex, and reflexive Banach space, then a monotone operator B:E⟶E∗ is maximal if and only if RJ+rB=E∗ for each r>0. Let E be a smooth, strictly convex, and reflexive Banach space, C be a nonempty closed convex subset of E, and B⊂E×E∗ be a monotone operator satisfying(56)DB⊂C⊂J−1J+rB,for all r>0. If B is a maximal monotone, then (56) holds for C=DB¯ and we can define the resolvent Sr:C⟶DB by(57)Srx=x∈E:Jx∈Jz+rBz,for all x∈C, i.e., Sr=J+rB−1J. We know the following (see [23, 25–27]):(1)
Sr is single valued,(2)
FSr=B−10, where FSr is the set of fixed points of Sr,(3)
Sr is a firmly nonexpansive type, i.e.,(58)Srx−Sry,JSrx−JSry≤Srx−Sry,x−y,∀x,y∈E.Theorem 4.
LetE1 be a uniformly smooth and 2-uniformly convex real Banach space and E2 a 2-uniformly smooth real Banach space with smoothness constant d2∈0,1. Let C and Q be two nonempty, closed, and convex subsets of E1 and E2, respectively; A:E1⟶E2 be a bounded linear operator with its adjoint A∗ such that A≠0; f:C⟶E1∗ be a monotone and k-Lipschitz continuous map; and B:E⟶E∗ be a maximal monotone operator. Let Γ=z∈VIC,f:Az∈B−10≠∅. Let a sequence xn be generated by(59)x1=x∈C=C1,yn=J−1Jxn−γA∗J2I−SrAxn,tn=ΠCJ−1Jyn−λfyn,zn=ΠCJ−1Jyn−λftn,Cn+1=v∈Cn:ϕv,yn≤ϕv,xn,ϕv,zn≤ϕv,yn,xn+1=ΠCn+1x1,where γ∈a,b, a,b∈0,1/d2∗A2, d2∗ being the smoothness constant of E1∗ as in Lemma 2, λ∈0,α/k, α being a positive constant as in Lemma 3, and Sr is the resolvent of B for r>0. Then, the sequence xn converges to a point z∈Γ.Proof.
PuttingT=Sr in Corollary 1, we get the desired result.


## 6. Application to Constrained Convex Minimization Problem

The problem of findingy∈C such that(60)ϕy=minx∈Cϕx,where C is a nonempty closed convex subset of H and ϕ is a real-valued convex function is called constraint convex minimization problem. We denote the set of solution of the constraint convex minimization problem by argminx∈Cϕx.Lemma 9.
(see [12]). Let H be a real Hilbert space and C be a nonempty closed convex subset of H. Let ϕ be a convex function of H into ℝ. If ϕ is differentiable, then z is a solution of the constraint convex minimization problem if and only if z∈VIC,▽ϕ.Theorem 5.
LetH1 and H2 be real Hilbert spaces. Let C and Q be two nonempty, closed, and convex subset of H1 and H2, respectively. Let A:H1⟶H2 be a bounded linear operator such that A≠0, and f:C⟶H1 be a monotone and k-Lipschitz continuous. Let ϕ:H2⟶ℝ be a differentiable convex function and suppose ▽ϕ is α-inverse strongly monotone mapping. Setting Γ=z∈VIC,f:Az∈argminy∈Qϕy, assume that Γ≠∅. Let a sequence(61)yn=xn−γA∗I−PQI−μ▽ϕAxn,tn=PCyn−λfyn,zn=PCyn−λftn,Cn+1=v∈Cn:yn−v≤xn−v,zn−v≤yn−v,xn+1=PCn+1x1,where γ⊂a,b, a,b∈0,1/A2, λ∈0,1/k, and μ∈0,2α . Then, the sequence xn converges to a point z∈Γ.Proof.
PuttingT=▽ϕ in Corollary 3, by Lemma 9 we get the desired result.


## 7. Application to Split Minimization Problem

LetH1 and H2 be real Hilbert spaces. Let C and Q be two nonempty, closed, and convex subset of H1 and H2, respectively, and ϕ1:H1⟶ℝ, ϕ2:H2⟶ℝ be two convex functions. Let AH1⟶H2 be a bounded linear operator.The problem of findingz satisfying the conditions(62)z∈argminx∈Cϕ1xsuch thatAz∈argminy∈Qϕ2sy,is called the split minimization problem.Theorem 6.
LetH1 and H2 be real Hilbert spaces. Let C and Q be two nonempty, closed, and convex subset of H1 and H2, respectively. Let A:H1⟶H2 be a bounded linear operator such that A≠0, and ϕ1:H1⟶ℝ and ϕ2:H2⟶ℝ be two differentiable convex functions. Suppose that ▽ϕ1 is k-Lipschitz continuous and ▽ϕ2 is α-inverse strongly monotone mapping, let Γ=z∈argminx∈Cϕ1x:Az∈argminy∈Qϕ2y≠∅. Let a sequence(63)yn=xn−γA∗I−PQI−μ▽ϕ1Axn,tn=PCyn−λ▽ϕ2yn,zn=PCyn−λ▽ϕ2tn,Cn+1=v∈Cn:yn−v≤xn−v,zn−v≤yn−v,xn+1=PCn+1x1,where γ⊂a,b, a,b∈0,1/A2, λ∈0,1/k, and μ∈0,2α. Then, the sequence xn converges to a point z∈Γ.Proof.
Sinceϕ1 is convex, we have ▽ϕ1 is monotone. Putting f=▽ϕ1 and g=▽ϕ2 in Corollary 3 and by Lemma 9, we get the desired result.


## 8. Numerical Example

In this section, we present a numerical example to show the convergence of a sequence generated by our algorithm. LetE1=E2=ℝ, C=0,∞.Setm=2 and Let T0,T1:C⟶C be defined by(64)T0x=23x,∀x∈C,T1x=12x,∀x∈C.Then,T0 and T1 are nonexpansive.Letf:C⟶C be defined by(65)fx=13x,∀x∈C.Then,f is monotone and VIC,f=0.LetA:C⟶C be defined by(66)Ax=12x,∀x∈C.Then,A is a bounded linear operator and A2=1/4, A∗y=1/2y.Whenz∈VIC,f, Az=0∈FT0∩FT1.So,Γ=z∈VIC,f:Az∈FT0∩FT1≠∅.Clearly,Γ=0. Taking λ=3/n2 and γ=4/n2. It follows from Theorem 2 that a sequence xn is generated by the following algorithm:(67)yn=xn−γA∗I−TnAxn,tn=PCyn−λfyn,zn=PCyn−λftn,Cn+1=v∈Cn:v≤xn2−zn22xn−2zn,xn+1=PCn+1x1=xn2−zn22xn−2zn,converges strongly to 0∈Γ (see Figure 1).Figure 1
The table and graph of sequencexn generated by (67) versus number of iterations n:=0,1,…,15 with an initial choice of x0=1.0000.



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*Source: 7151953-2020-04-10.xml*