# An Inertial Iterative Algorithm to Find Common Solution of a Split Generalized Equilibrium and a Variational Inequality Problem in Hilbert Spaces

Authors: Mohammad Farid; Rehan Ali; Watcharaporn Cholamjiak
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we introduce and study an iterative algorithm via inertial and viscosity techniques to find a common solution of a split generalized equilibrium and a variational inequality problem in Hilbert spaces. Further, we prove that the sequence generated by the proposed theorem converges strongly to the common solution of our problem. Furthermore, we list some consequences of our established algorithm. Finally, we construct a numerical example to demonstrate the applicability of the theorem. We emphasize that the result accounted in the manuscript unifies and extends various results in this field of study.


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## Body

## 1. Introduction

LetH1 and H2 be real Hilbert spaces with inner product ⋅,⋅ and norm ⋅. Let C and Q be nonempty closed convex subsets of H1 and H2, respectively. The variational inequality problem (in short, VIP) is to find x∗∈C such that(1)Bx∗,y−x∗≥0,∀y∈C,where B:C⟶H1 is a nonlinear mapping. The solution set of VIP (1) is denoted by Ω. It is introduced by Hartman and Stampacchia [1].In 1994, Blum and Oettli [2] introduced and studied the following equilibrium problem (in short, EP): find x∗∈C such that(2)G1x∗,y≥0,∀y∈C,where G1:C×C⟶ℝ is a bifunction. The solution set of EP (3) is denoted by Sol (EP(3)).In the last two decades, EP (2) has been generalized and extensively studied in many directions due to its importance; see, for example [3–7], for the literature on the existence and iterative approximation of solution of the various generalizations of EP (2).Censor et al. [8] introduced the split feasibility problem (in short, SPFP) in finite-dimensional Hilbert spaces for modelling of inverse problems that arise from phase retrievals and in medical image restoration as(3)findx∗∈Csuch thatBx∗∈Q,where B:H1⟶H2 is a bounded linear operator.In this paper, we consider the following split generalized equilibrium problem (in short,SPGEP):LetG1,b1:C×C⟶ℝ and G2,b2:Q×Q⟶ℝ be nonlinear mappings, and B:H1⟶H2 be a bounded linear operator, then SPGEP is to find x∗∈C such that(4)G1x∗,x+b1x,x∗−b1x∗,x∗≥0,∀x∈C,and such that(5)y∗=Bx∗∈QsolvesG2y∗,y+b2y,y∗−ϕy∗,y∗≥0,∀y∈Q.If we takeb1,b2≡0, then SPGEP becomes split equilibrium problem (in short, SPEP) as(6)G1x∗,x≥0,∀x∈C,and such that(7)y∗=Bx∗∈QsolvesG2y∗,y≥0,∀y∈Q.When looked separately, (4) is the generalized equilibrium problem (GEP) and we denote its solution set by Sol (GEP(4)). The SPGEP(4) and (5) constitute a pair of generalized equilibrium problems which have to be solved so that the image y∗=Bx∗ under a given bounded linear operator B of the solution x∗ of the GEP(4) in H1 is the solution of another GEP (4) in another space H2. We denote the solution set of GEP (5) by Sol (GEP (5)). The solution set of SPGEP (4) and (5) is denoted by Γ=p∈SolGEP4:Bp∈SolGEP5.SPGEP (4) and (5) generalize multiple-sets split feasibility problem. It also includes as special case, the split variational inequality problem, which is the generalization of split zero problems and split feasibility problems, see for details [9–12].In 2008, Mainge [13] introduced the following inertial Krasnosel’skiǐ–Mann algorithm by combining Krasnosel’skiǐ–Mann algorithm and the inertial extrapolation:(8)tn=xn+θnxn−xn−1,xn+1=1−ηntn+ηnTtn,,for each n≥1. He proved that the sequence xn generated by algorithm (8) converges weakly to a fixed point of T under some conditions. Recently, Bot et al. [14] studied the convergence analysis of the inertial Krasnosel’skiǐ–Mann algorithm for approximating a fixed point of nonexpansive mapping T by getting rid of some conditions used in the main result of Mainge [13]. Recently, Dong et al. [15, 16] introduced the inertial hybrid algorithm and established a strong convergence theorem for approximating a fixed point of nonexpansive mapping T in the setting of Hilbert space. For further study of some generalization of iterative algorithm (8), see for instance [17, 18]. Very recently, Monairah et al. [19] introduced and studied a hybrid iterative algorithm to approximate a common solution of generalized equilibrium problem, variational inequality problem, and fixed point problem in the framework of a 2 uniformly convex and uniformly smooth real Banach space. The inertial method has been studied by many researchers. The results and other related ones analyzed the convergence properties of inertial type algorithms and demonstrated their performance numerically on some imaging and data analysis problems, see for details [20–23].Motivated by the work given in [6, 13, 24], we propose an iterative algorithm via inertial and viscosity techniques to find a common solution of a split generalized equilibrium and a variational inequality problem in Hilbert spaces. We obtained the strong convergence for the proposed algorithm. Further, we give some consequences of the main result. Finally, we discuss a numerical example to demonstrate the applicability of the iterative algorithm. The method and result presented in this paper generalize and unify the previously known related methods and results. Our result can extend several iterative methods given in the literature.


## 2. Preliminaries

In this section, we collect some concepts and results which are required for the presentation of the work. Let symbols⟶ and ⇀ denote strong and weak convergence, respectively.For every pointx∈H1, there exists a unique nearest point to x in C denoted by PCx such that(9)x−PCx≤x−y,∀y∈C.The mappingPC is called the metric projection of H1 onto C. It is well known that PC is nonexpansive and satisfies(10)x−y,PCx−PCy≥PCx−PCy2,∀x,y∈H1.Moreover,PCx is characterized by the fact that PCx∈C and(11)x−PCx,y−PCx≤0,∀y∈C.This implies that(12)x−y2≥x−PCx2+y−PCx2,∀x∈H1,∀y∈C.In a real Hilbert spaceH1, it is well known that(13)λx+1−λy2=λx2+1−λy2−λ1−λx−y2,∀x,y∈H1andλ∈0,1,(14)x+y2≤x2+2y,x+y,∀x,y∈H1.Definition 1 (see [25]).
A multivalued mappingM:H1⟶2H1 is called monotone if for all x,y∈H1, u∈Mx and v∈My such that(15)x−y,u−v≥0.Definition 2 (see [25]).
A multivalued monotone mappingM:H1⟶2H1 is maximal if the graphM, the graph of M, is not properly contained in the graph of any other monotone mapping.Remark 1.
It is known that a multivalued monotone mappingM is maximal if and only if for x,u∈H1×H1, x−y,u−v≥0, for every y,v∈GraphM implies that u∈Mx.Lemma 1 (see [26]).
Letxn and un be bounded sequences in a Banach space E and let βn be a sequence in 0,1 with 0<liminfn⟶∞βn≤limsupn⟶∞βn<1. Suppose xn+1=1−βnyn+βnxn for all integers n≥0 and limsupn⟶∞yn+1−yn-xn+1−xn≤0. Then,(16)limn⟶∞yn−xn=0.Lemma 2 (see [27]).
Letbn be a sequence of nonnegative real numbers such that there exists a subsequence bni of bn such that bni<bni+1, ∀i∈ℕ. Then, there exists a nondecreasing sequence mj of ℕ such that limj⟶∞mj=∞ and the following properties are satisfied by all (sufficiently large) numbers j∈ℕ(17)bmj≤bmj+1,bj≤bmj.In fact,mj is the largest number n in the set 1,2,3,…,j such that bn<bn+1.Lemma 3 (see [28]).
Assume thatD is a strongly positive self-adjoint bounded linear operator on a Hilbert space H1 with coefficient γ¯>0 and 0<ρ≤D−1. Then, I−ρD≤1−ργ¯.Lemma 4 (see [29]).
Assume thatan is a sequence of nonnegative real numbers such that(18)an+1≤1−γnan+δn,n≥0,where γn is a sequence in 0,1 and δn is a sequence in ℝ such that(i)
∑n=1∞γn=∞;(ii)
limsupn⟶∞δn/γn≤0or∑n=1∞δn<+∞.
Then,limn⟶∞an=0.Assumption 1.
LetG1:C×C⟶ℝandb1:C×C⟶ℝ be bimappings satisfying the following conditions:(1)
G1x,x=0,∀x∈C;(2)
G1 is monotone, i.e.,(19)G1x,y+G1y,x≤0,∀x,y∈C.(3)
For eachy∈C, x⟶G1x,y is weakly upper semicontinuous;(4)
For eachx∈C, y⟶G1x,y is convex and lower semicontinuous;(5)
b1.,. is weakly continuous and b1.,y is convex;(6)
b1 is skew-symmetric, i.e.,(20)b1x,x−b1x,y+b1y,y−b1y,x≥0,∀x,y∈C.Now, we defineTrG1,b1:H1⟶C as follows:(21)TrG1,b1z=x∈C:G1x,y+b1y,x−b1x,x+1ry−x,x−z≥0,∀y∈C,where r is a positive real number.Lemma 5 (see [30]).
LetC be a nonempty closed convex subset of Hilbert space H1. Let G1,b1:C×C⟶ℝ be nonlinear mappings satisfying Assumption 1. Assume that for each z∈H1 and for each x∈C, there exists a bounded subset Dx⊆C and zx∈C such that for any y∈C,Dx,(22)G1y,zx+b1zx,y−b1y,y+1rzx−y,y−z<0.
Let the mappingTrG1,b1 be defined by (21). Then, the following conclusions hold:(i)
TrG1,ϕ1z is nonempty for each z∈H1;(ii)
TrG1,ϕ1 is single-valued;(iii)
TrG1,ϕ1 is a firmly nonexpansive mapping, i.e., for all z1,z2∈H1,(23)TrG1,ϕ1z1−TrG1,ϕ1z22≤TrG1,ϕ1z1−TrG1,ϕ1z2,z1−z2.(iv)
FixTrG1,ϕ1=SolGEP4;(v)
SolGEP4 is closed and convex.
Further, assume thatG2:Q×Q⟶ℝandb2:Q×Q⟶ℝ satisfy Assumption 1. For s>0 and for all u∈H2, define a mapping TsG2,b2:H2⟶Q as follows:(24)TsG2,b2u=v∈Q:G2v,w+b2w,v−b2v,v+1sw−v,v−u≥0,∀w∈Q.
Then, we easily observe thatTsG2,b2 is nonempty, single-valued, firmly nonexpansive, fixTsG2,b2=SolGEP5, and SolGEP5 is closed and convex.Lemma 6 (see [30]).
LetG1 and b1 satisfy Assumption 1 and let the mapping TrG1,b1 be defined by (21). Let x1,x2∈H1 and r1,r2>0, then(25)Tr2G1,b1x2−Tr1G1,b1x1≤x2−x1+r2−r1r2Tr2G1,b1x2−x2.


## 3. Main Result

In this section, we prove a strong convergence theorem based on the proposed iterative algorithm to approximate a common solution ofSPGEP (4), (5), and VIP (1) (Algorithm 1).Algorithm 1: Iterative algorithm.
Initialization: choose x0,x1∈H1 to be arbitrary.Iterative Steps: given the current iterate xn, compute:Step 1. Compute tn=xn−θnxn−1−xnStep 2. Compute un=TrnG1,b1tn+λnB∗TrnG2,b2−IBtnStep 3. Compute vn=PCun−δnDunand calculate the next iteratexn+1 as follows:xn+1=ηngxn+1−ηnvnSetn≔n+1and go toStep 1.Theorem 1.
LetC and Q be two nonempty closed convex subsets of Hilbert spaces H1 and H2, respectively. Let B:H1⟶H2 be a bounded linear operator. Assume that G1:C×C⟶ℝ, G2:Q×Q⟶ℝ, b1:C×C⟶ℝ, and b2:Q×Q⟶ℝ are nonlinear mappings satisfying Assumption 1 and G2 is upper semicontinuous in the first argument. Assume that Θ=Γ∩Ω≠∅. Let g:H1⟶H1 be a contraction mapping with constant α∈0,1 and D:C⟶H1 be a τ-inverse strongly monotone mapping. Let xn be generated by Algorithm 1 and satisfy the following conditions:(i)
limn⟶∞ηn=0,∑n=0∞ηn=∞;(ii)
limn⟶∞θn/ηnxn−xn−1=0;(iii)
θn⊂0,θ,for someθ>0andδn⊂0,2τ;(iv)
liminfn⟶∞rn>0andlimn⟶∞rn+1−rn=0;(v)
λn⊂ℝ,such thata≤λn≤b<1/L,whereL=B2.
Then, the sequencexn converges strongly to some q∈Θ, where q=PΘgq.Proof.
We divide the proof into several steps.Step 1. We show thatxn is bounded. Let q∈Θ=Γ∩Ω, then q=TrnG1,b1q and Bq=TrnG2,b2Bq. Applying the similar steps used in Theorem 1 [6], we obtain(26)un−q2≤tn−q2−λn1−λnLTrnG2,b2−IBtn2.Thus,(27)un−q≤tn−q.We estimate(28)tn−q=xn−θnxn−1−xn−q≤xn−q+θnxn−xn−1=xn−q+θnηnηnxn−xn−1.From condition (ii),∃N1>0 such that(29)θnηnxn−xn−1≤N1,∀n≥1.By (27)–(29), we have(30)un−q≤tn−q≤xn−q+ηnN1.Since the mappingI−δnB is nonexpansive, therefore(31)vn−q=PCI−δnDun−q≤I−λnDun−I−δnDp≤un−q.We estimate(32)xn+1−q=ηngxn+1−ηnvn−q=ηngxn−q+1−ηnvn−q≤ηngxn−q+1−ηnvn−q≤ηngxn−gq+ηngq−q+1−ηnvn−q≤ηnαxn−q+ηngq−q+1−ηnvn−q.Using (30) and (31) in the above inequality, we have(33)xn+1−q≤1−1−αηnxn−q+ηnN1+ηngq−q=1−1−αηnxn−q+ηnN1+1−αηnN1+gq−q1−α≤maxxn−q,N1+gq−q1−α≤⋯≤maxxn−q,N1+gq−q1−α.Thus,xn is bounded. Also, un, vn, and tn are bounded.Step 2. We show that1−ηnλn1−λnLTrnG2,b2−IBtn≤xn−q2−xn+1−q2+ηnN4, for someN4>0.We estimate(34)xn+1−q2≤ηngxn−q2+1−ηnvn−q2≤ηngxn−gq+gq−q2+1−ηnvn−q2≤ηnαxn−q+gq−q2+1−ηnvn−q2≤ηnxn−q+gq−q2+1−ηnvn−q2=ηnxn−q2+ηn2xn−qgq−q+gq−q2+1−ηnvn−q2≤ηnxn−q2+1−ηnun−q2+ηnN2,for someN2>0.Using (26) in the above inequality, we get(35)xn+1−q2≤ηnxn−q2+1−ηntn−q2−1−ηnλn1−λnLTrnG2,b2−IBtn2+ηnN2.From (30), we obtain(36)tn−q2≤xn−q+ηnN12=xn−q2+ηn2N1xn−q+ηnN12≤xn−q2+ηnN3,for someN3>0.By (35) and (36), we have(37)xn+1−q2≤ηnxn−q2+1−ηnxn−q2+ηnN3−1−ηnλn1−λnLTrnG2,b2−IBtn2+ηnN2=xn−q2+ηnN3−1−ηnλn1−λnLTrnG2,b2−IBtn2+ηnN2,which yields that(38)1−ηnλn1−λnLTrnG2,b2−IBtn2≤xn+1−q2−xn−q2+ηnN4,whereN4=N2+N3.Step 3. We show that(39)1−ηnun−tn2≤xn−q2−xn+1−q2+ηnN4+21−ηnλnun−qB∗TrnG2,b2−IBtn.Using the concept of firmly nonexpansive ofTrnG2,b2, we have(40)un−q2=TrnG1,ϕ1tn+λnB∗TrnG2,ϕ2−IBtn−q2=TrnG1,ϕ1tn+λnB∗TrnG2,ϕ2−IBtn−TrnG1,ϕ1q2≤un−q,tn+λnB∗TrnG2,ϕ2−IBtn−q=12un−q2+12tn+λnB∗TrnG2,ϕ2−IBtn−q2−12un−q−tn−λnB∗TrnG2,ϕ2−IBtn+q2=12un−q2+12tn−q+λnB∗TrnG2,ϕ2−IBtn2−12un−tn−λnB∗TrnG2,ϕ2−IBtn2=12un−q2+12tn−q2+12λn2B∗TrnG2,ϕ2−IBtn2+tn−q,λnB∗TrnG2,ϕ2−IBtn−12un−tn2−12λn2B∗TrnG2,ϕ2−IBtn2+un−tn,λnB∗TrnG2,ϕ2−IBtn=12un−q2+12tn−q2−12un−tn2+un−q,λnB∗TrnG2,ϕ2−IBtn,which implies that(41)un−q2≤tn−q2−un−tn2+2un−q,μnB∗TrnG2,ϕ2−IBtn≤tn−q2−un−tn2+2μnun−qB∗TrnG2,ϕ2−IBtn.Using (41) in (35), we get(42)xn+1−q2≤ηnxn−q2+1−ηntn−q2−1−ηnun−tn2+2μn1−ηnun−qB∗TrnG2,ϕ2−IBtn+ηnN2.Using (36), in the above inequality,(43)xn+1−q2≤ηnxn−q2+1−ηnxn−q2+ηnN3−1−ηnun−tn2+2μn1−ηnun−qB∗TrnG2,ϕ2−IBtn≤xn−q2−1−ηnun−tn2+2μn1−ηnun−qB∗TrnG2,ϕ2−IBtn+ηnN4,which implies(44)1−ηnun−tn2≤xn−q2−xn+1−q2+2μn1−ηnun−qB∗TrnG2,ϕ2−IBtn+ηnN4.Step 4. We show that(45)xn+1−q2≤1−1−αηnxn−q2+1−αηn×21−αgq−q,xn+1−q+θnηnxn−xn−1N1−α,for someN>0.We estimate(46)tn−q2=xn+θnxn−xn−1−q2=xn−q2+2θnxn−q,xn−xn−1+θn2xn−xn−12≤xn−q2+2θnxn−qxn−xn−1+θn2xn−xn−12=xn−q2+θnxn−xn−12xn−q+θnxn−xn−1≤xn−q2+θnxn−xn−1N,for someN>0.Using (14), we calculate(47)xn+1−q2=ηngxn+1−ηnun−q2=ηngxn−gq+1−ηnun−q+ηngq−q2≤ηngxn−gq+1−ηnun−q2+2ηngq−q,xn+1−q≤ηngxn−gq2+1−ηnun−q2+2ηngq−q,xn+1−q≤ηnαxn−q2+1−ηnun−q2+2ηngq−q,xn+1−q≤ηnαxn−q2+1−ηntn−q2+2ηngq−q,xn+1−q.From (46) and (47), we have(48)xn+1−q2≤1−1−αηnxn−q2+θnxn−xn−1N+2ηngq−q,xn+1−q=1−1−αηnxn−q2+1−αηn×21−αgq−q,xn+1−q+θnηnxn−xn−1N1−α.Step 5. We show thatlimn⟶∞xn−q=0.
To show it, we have the two cases as follows:Case 1. There exists m∈ℕ such that xn+1−q2≤xn−q2,∀n≥m. This shows that limn⟶∞xn−q exists, and by step 2, we have(49)limn⟶∞TrnG2,b2−IBtn=0.Thanks to step 3 and (49), we obtain(50)limn⟶∞un−tn=0.Sincexn+1−un=ηnun−gxn, therefore(51)limn⟶∞xn+1−vn=0.Now,(52)xn−tn=θnxn−xn−1=θnηnxn−xn−1ηn⟶0,asn⟶∞.Next, prove thatlimn⟶∞un−vn=0.By (47), we have(53)xn+1−q2≤ηnαxn−q2+1−ηnvn−q2+2ηngq−q,xn+1−q=ηnαxn−q2+1−ηnvn−q2+2ηnζ,xn+1−q≤ηnαxn−q2+1−ηnvn−q2+2ηnϱ2.We setζ=fq−q and let ϱ>0 be a suitable constant with ϱ≥supnζ,xn−q in the above inequality. Thus,(54)xn+1−q2≤1−ηnPCun−λnBun−PCq−λnBq2+ηnxn−q2+2ηnϱ2≤1−ηnun−q2+λnλn−2γBun−Bq2+ηnxn−q2+2ηnϱ2≤1−ηnxn−x˜2+λnλn−2γBun−Bq2+ηnxn−x˜2+2ηnϱ2≤1−ηnλnλn−2γBun−Bq2+xn−q2+2ηnϱ2.This implies(55)1−ηnλn2γ−λnBun−Bq2≤xn−q2−xn+1−q2+2ηnϱ2.Thus,(56)limn⟶∞Bun−Bq=0.We compute(57)vn−q2=PCun−λnBun−PCq−λnBq2≤vn−q,un−λnBun−q−λnBq≤12vn−q2+un−λnBun−q−λnBq2−vn−un+λnBun−Bq2≤12vn−q2+un−q2−vn−un+λnBun−Bq2≤un−q2−vn−un2−λn2Bun−Bq2+2λnvn−un,Bvn−Bq≤un−q2−vn−un2+2λnvn−unBun−Aq≤xn−q2−vn−un2+2λnvn−unBun−Aq.By (53), we get(58)xn+1−q2≤1−λnvn−q2+λnxn−q2+2ηnϱ2≤1−λnxn−q2−vn−un2+2λnvn−unBun−Bq+λnxn−q2+2ηnϱ2,which implies(59)1−λnvn−un2≤xn−q2−xn+1−q2+21−λnλnvn−unBun−Bq+2ηnϱ2.Using (56) and the given conditions, we get(60)limn⟶∞vn−un=0.From (50)–(52) and (60), we have(61)xn+1−xn≤xn+1−un+un−tn+tn−xn⟶0,asn⟶∞.We prove thatlimsupn⟶∞g−Iq,xn−q≤0.Sinceun is bounded, there exists a subsequence uni of un which converges weakly to some p∈C. Without loss of generality, we can assume that uni⇀p such that(62)limsupn⟶∞g−Iq,un−q=limi⟶∞g−Iq,uni−q.We prove thatp∈Γ∩Ω.Sinceun=TrnG1,b1dn where dn≔tn+λnB∗TrnG2,b2−IBtn, we have(63)G1un,u+b1u,un−b1un,un+1rnu−un,un−dn≥0,∀u∈C,which implies that(64)b1u,un−b1un,un+1rnu−un,un−dn≥G1u,un,∀u∈Cusing monotonocity ofG1.Hence,(65)b1u,unk−b1unk,unk+u−unk,unk−dnkrnk≥G1u,unk,∀u∈C.Letut=1−tw+tu, for all t∈0,1. Since u∈C and w∈C, we get ut∈C and from (65), we have(66)0≤G1ut,unk−b1ut,unk+b1unk,unk−ut−unk,unk−tnkrnk+λnB∗TrnkG2,b2−IBtnkrnk.SinceB∗ is bounded linear, it follows from (49), (50), (60), and liminfrn>0 that unk−tnk/rnk⟶0 and B∗TrnkG2,b2−IBtnk/rnk⟶0 and so(67)b1ut,p−b1p,p≤G1ut,p.Now, fort>0,(68)0=G1ut,ut=tG1ut,ut+1−tG1ut,p≥tG1ut,ut+1−tb1ut,p−b1p,p≥tG1ut,ut+1−ttb1u,p−b1p,p≥G1ut,ut+1−tb1u,p−b1p,p.Lettingt⟶0, we have(69)G1p,u+b1u,p−b1p,p≥0,∀u∈C.This implies thatp∈SolGEP4.Next, we show thatBp∈SolGEP5. Since un−tn⟶0,un⇀p as n⟶∞ and tn is bounded, there exists a subsequence tnk of tn such that tnk⇀p and since B is a bounded linear operator so that Btnk⇀Bp.Now, settingvnk=Btnk−TrnkG2,b2Btnk, it follows that from (49), limk⟶∞vnk=0 and Btnk−vnk=TrnkG2,b2Btnk.Therefore, from Lemma5, we have(70)G2Btnk−vnk,z+b1z,unk−b1unk,unk+1rnkz−Btnk−vnk,Btnk−vnk−Btnk≥0,∀z∈Q.SinceG2 is upper semicontinuous in the first argument, taking limit superior to the above inequality as k⟶∞ and using condition, we obtain(71)G2Bp,z+b1z,unk−b1unk,unk≥0,∀z∈Q,which means thatBp∈SolGEP5 and hence p∈Γ.Next, we provep∈Ω. Since limn⟶∞un−vn=0 and limn⟶∞un−tn=0, there exist subsequences uni and vni of un and vn, respectively such that uni⇀p and vni⇀p.Define the mappingM as(72)Mz=Dz+NCp,ifp∈C,∅,ifp∉C,whereNCp≔v∈H1:p−u,v≥0,∀u∈C is the normal cone to C at p∈H1. In this case, the mapping M is maximal monotone and hence 0∈Mp mapping if and only if p∈SolVIP1. Let p,v∈graphM. Then, we have v∈Mp=Dp+NCp and hence v−Dp∈NCp. So, we have p−u,v−Dp≥0, for all u∈C. On the other hand, from vn=PCun−δnDun and p∈C, we have(73)un−δnDun−vn,vn−p≥0.This implies that(74)z−vn,vn−unδn+Dun≥0.Sincez−u,v−Dz≥0, for all z∈C and vni∈C, using monotonicity of D, we have(75)z−vni,v≥z−vni,Dp≥z−vni,Dz−z−vni,vni−uniδn+Duni=z−vni,Dz−Duni+z−vni,Dvni−Duni−z−vni,vni−uniδn≥z−vni,Dvni−Duniz−vni,vni−uniδn.SinceD is continuous therefore on taking limit i⟶∞, we have z−p,v≥0. Since M is maximal monotone, we have p∈M−10 and hence p∈Ω. Thus, p∈Γ∩Ω.Sinceq=PΘgq, therefore from (62),(76)limsupn⟶∞g−Iq,un−q=limi⟶∞g−Iq,p−q≤0.Using Lemma4, (76), and the given conditions in step 4, we get xn⟶q, where q=PΘgq.Case 2. There exists a subsequencexni−q2 of xn−q2 such that xni−q2<xni+1−q2,∀i∈ℕ. Thus, by Lemma 2, ∃ is a nondecreasing sequence mj of ℕ such that limj⟶∞mj=∞ and(77)xmj−q2≤xmj+1−q2,xj−q2≤xmj−q2.
By step 2, we get(78)1−ηmjλmj1−λmjLTrmjG2,b2−IBtmj≤xmj−q2−xmj+1−q2+αmjN4≤αmjN4.
Thus,(79)limn⟶∞TrmjG2,b2−IBtmj=0.
By step 3, we obtain(80)1−ηmjumj−tmj2≤xmj−q2−xmj+1−q2+ηmjN4+21−ηmjλmjumj−qB∗TrmjG2,b2−IBtmj≤ηmjN4+21−ηmjλmjumj−qB∗TrmjG2,b2−IBtmj.
Hence,(81)limn⟶∞umj−tmj=0.
By the similar steps of case 1, we get(82)limn⟶∞xmj+1−xmj=0,limsupj⟶∞g−Iq,xmj+1−q≤0.
By step 4, we have(83)xmj+1−q2≤1−1−αηmjxmj−q2+1−αηmj×21−αgq−q,xmj+1−q+σmjηmjxmj−xmj−1N1−α.
By (77) and (83), we have(84)xmj+1−q2≤1−1−αηmj+1xmj−q2+1−αηmj×21−αgq−q,xmj+1−q+σmjηmjxmj−xmj−1N1−α.
This implies(85)xmj+1−q2≤21−αgq−q,xmj+1−q+σmjηmjxmj−xmj−1N1−α.
Thus,(86)limsupj⟶∞xmj+1−q2≤0.
By (77) and (86), xj⟶q. This completes the proof.Moreover, we have the following consequences. If we takeθn=0, then Theorem 1 reduced to the following result without inertial as follows:Corollary 1.
LetC and Q be two nonempty closed convex subsets of Hilbert spaces H1 and H2, respectively. Let B:H1⟶H2 be a bounded linear operator. Assume that G1:C×C⟶ℝ, G2:Q×Q⟶ℝ, b1:C×C⟶ℝ, and b2:Q×Q⟶ℝ are nonlinear mappings satisfying Assumption 1 and G2 is upper semicontinuous in the first argument. Assume that Θ=Γ∩Ω≠∅. Let g:H1⟶H1 be a contraction mapping with constant α∈0,1 and D:C⟶H1 be a τ-inverse strongly monotone mapping. Let xn be generated by(87)x1∈H1un=TrnG1,b1tn+λnB∗TrnG2,b2−IBtnvn=PCun−δnDunxn+1=ηngxn+1−ηnvn,where the control sequence satisfies the following conditions:(i)
limn⟶∞ηn=0,∑n=0∞ηn=∞;(ii)
liminfn⟶∞rn>0anlimn⟶∞rn+1−rn=0;(iii)
λn⊂ℝsuch thata≤λn≤b<1/L,whereL=B2.
Then, the sequencexn converges strongly to some q∈Θ, where q=PΘgq.
Further, if we takeb1,b2≡0, then Theorem 1 reduced to the following result as follows:Corollary 2.
LetC and Q be two nonempty closed convex subsets of Hilbert spaces H1 and H2, respectively. Let B:H1⟶H2 be a bounded linear operator. Assume that G1:C×C⟶ℝ and G2:Q×Q⟶ℝ are nonlinear mappings satisfying Assumption 1 (1)–(4) and G2 is upper semicontinuous in first argument. Assume that Θ=ϒ∩Ω≠∅, where ϒ denotes the solution set of SPEP (6) and (7). Let g:H1⟶H1 be a contraction mapping with constant α∈0,1 and D:C⟶H1 be a τ-inverse strongly monotone mapping. Let xn be generated by(88)x1∈H1un=TrnG1tn+λnB∗TrnG2−IBtnvn=PCun−δnDunxn+1=ηngxn+1−ηnvn,where the control sequence satisfies the following conditions:(i)
limn⟶∞ηn=0,∑n=0∞ηn=∞;(ii)
limn⟶∞θn/ηnxn−xn−1=0;(iii)
θn⊂0,θ,for someθ>0andδn⊂0,2τ;(iv)
liminfn⟶∞rn>0andlimn⟶∞rn+1−rn=0;(v)
λn⊂ℝsuch thata≤λn≤b<1/L,whereL=B2.
Then, the sequencexn converges strongly to some q∈Θ, where q=PΘgq.


## 4. Numerical Illustration

Finally, to supporting our main theorem, we now give an example in infinitely dimensional spacesL20,1 such that ⋅ is L2-norm defined by x=∫01xt2dt where xt∈L20,1.Example 1.
LetH1=H2=L20,1 and C=Q=xt∈L20,1:∫01txtdt≤1. Define mappings as follows:(i)
bounded linear operatorB:H1⟶H2 by Bxt=3xt,∀xt∈L20,1;(ii)
contraction mappingg:H1⟶H1 by gxt=αxt where α∈0,1;(iii)
nonlinear mappingsG1:C×C⟶ℝ, G2:Q×Q⟶ℝ by(89)G1xt,yt=G2xt,yt=yt−xt,xt,∀xt,yt∈L20,1,(iv)
nonlinear mappingsb1:C×C⟶ℝ and b2:Q×Q⟶ℝ by(90)b1xt,yt=b2xt,yt=yt,yt,∀xt,yt∈L20,1.(v)
τ-inverse strongly monotone mapping D:C⟶H1 by(91)Dxt=B∗I−PQBxt,∀xt∈C.It is obvious thatG1,G2,b1,b2 satisfy Assumption 1 and G2 is upper semicontinuous in the first argument by the definition of the inner product .,.. On the other hand, we consider(92)0≤G1x,y+b1y,x−b1x,x+1ry−x,x−z=y−x,x+1ry−x,x−z=y−x,1+rx−z.This implies thatTrG1,b1zt=PCzt/1+r,∀zt∈L20,1 where(93)PCxt=1−t,xtt2t+xt,ift,xt>1,xt,ift,xt≤1.Similarly, we haveTrG2,b2zt=PCzt/1+r,∀zt∈L20,1. For the experiments in this section, we use the Cauchy error xn+1−xn2<10−5 for the stopping criterion. We split considering all of the performances of our algorithm in five cases.Case I: we start computation by comparing of the algorithm with different parametersθ¯n where(94)θn=θ¯n,ifn≤N,xn≠xn−1,ηnnxn−xn−1,ifn>N,xn≠xn−1,ηn,otherwise,whereN is the number of iteration that we want to stop. We choose rn=0.1, λn=δn=0.1, ηn=1/n+1, and α=0.1 and initializations x0=sint and x1=sint/2. Then, the results are presented as follows:Case II: we compare the performance of the algorithm with different parametersrn by setting θ¯n=0.5, λn=δn=0.1, ηn=1/n+1, and α=0.1 and initializations x0=sint and x1=sint/2. Then, the results are presented as follows:Case III: we compare the performance of the algorithm with different parametersηn by setting θ¯n=0.5, rn=1, λn=δn=0.1, and α=0.1 and initializations x0=sint and x1=sint/2. Then, the results are presented as follows:Case IV: we compare the performance of the algorithm with different parametersλn and δn by setting θ¯n=0.5, rn=1, ηn=1/n+1, and α=0.1 and initializations x0=sint and x1=sint/2. Then, the results are presented as follows:Case V: we compare the performance of the algorithm with different parametersα by setting θ¯n=0.5, rn=1, ηn=1/n+1, and λn=δn=0.001 and initializations x0=sint and x1=sint/2. Then, the results are presented as follows:From Tables1–5 and Figures 1–5, we noticed that in all the above 5 cases, selecting θ¯n=0.5, rn=1, ηn=1/n+1, λn=δn=0.001, and α=0.1 yields the best results.Table 1
Numerical results ofθ¯n.
θ¯n00.51/n1/n2min1/n2xn−xn−1,0.5No. of iters15414104CPU time (s)4.1067131.8626123.7644222.9221322.023393Table 2
Numerical results ofrn.
rn0.0010.010.1110No. of iters1514435CPU time (s)4.1691213.9272341.9149401.6785312.077360Table 3
Numerical results ofαn.
ηn1/n+11/10n+11/20n+11/100n+11/nNo. of ites.38886CPU time (s)1.6785312.7107362.7281622.7001513.312465Table 4
Numerical results ofλn and μn.
λn=δn10.10.010.0010.0001No. of iters33333CPU time (s)1.6674501.6727321.6980851.6639791.678559Table 5
Numerical results ofα.
α0.0010.10.50.90.999No. of iters73877CPU time (s)2.5128961.6639792.6916932.5058022.631734Figure 1
The Cauchy error plotting number of iterations for different parametersθ¯n.Figure 2
The Cauchy error plotting number of iterations for different parametersrn.Figure 3
The Cauchy error plotting number of iterations for different parametersηn.Figure 4
The Cauchy error plotting number of iterations for different parametersλn and δn.Figure 5
The Cauchy error plotting number of iterations for different parametersα.


## 5. Conclusion

In this paper, we developed an iterative algorithm via inertial and viscosity techniques to find a common solution of a split generalized equilibrium and a variational inequality problem in Hilbert spaces. Further, we study the convergence analysis of our main result and point out some consequences. Finally, we constructed a numerical example to demonstrate the applicability of theorem and compared the performance of algorithm by taking different parameters.



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