# A Generalized Strong Convergence Algorithm in the Presence of Errors for Variational Inequality Problems in Hilbert Spaces

Authors: Mostafa Ghadampour; Donal O’Regan; Ebrahim Soori; Ravi P. Agarwal
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we study the strong convergence of an algorithm to solve the variational inequality problem which extends a recent paper (Thong et al., Numerical Algorithms. 78, 1045-1060 (2018)). We reduce and refine some of their algorithm conditions and we prove the convergence of the algorithm in the presence of some computational errors. Then, using the MATLAB software, the result will be illustrated with some numerical examples. Also, we compare our algorithm with some other well-known algorithms.


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## Body

## 1. Introduction

LetH be a real Hilbert space with the inner product .,. and the norm ., and C be a nonempty, closed, and convex subset of H. The variational inequality (VI) is to find a point x∈C such that
(1)Ax,y−x≥0,∀y∈C,where A is a mapping of C into H. The solution set of (1) is denoted by VIA,C. Variational inequalies arise in the study of network equilibriums, optimization problems, saddle point problem, Nash equilibrium problems in noncooperative games etc.; see, for example, [1–12] and the references therein.A new algorithm was proposed by Korpelevich [13] for solving the problem (VI) in the Euclidean space which is known as the extragradient method. Let x1 be an arbitrary element in H and consider
(2)yn=PCxn−λAxn,xn+1=PCxn−λAyn.where λ is a number in 0,1, PC is the Euclidean least distance projection of H onto C, and A:C⟶H is a monotone operator. The next algorithm (3) was introduced by Tseng [14] and applying the modified forward-backward (F-B) method is a good alternative to the extragradient method (TEGM):
(3)yn=PCxn−λAxn,xn+1=PXyn−λAyn−Axn,where X=C and X=H if A is Lipschitz continuous. The following algorithm (4) was proposed by Shehu and Iyiola [15] which a viscosity type subgradient extragradient method (VSEGM):
(4)yn=PCxn−λnAxn,Tn=z∈H:xn−λnAxn−yn,z−yn≤0,zn=PTnxn−λnAyn,xn+1=αnfxn+1−αnzn,where the operator A is monotone and Lipschitz continuous, f is a strict contraction mapping, l∈0,1,μ∈0,1, and λn=lmn where mn is the smallest nonnegative integer m such that
(5)λAxn−Ayn≤μrlmnxn,where rlmnx=x−PCx−lmnAx for all x∈C. Recently, the sequence produced by the following algorithm was introduced by Thong and Hieu [3] based on Tseng’s method (THEGM):
(6)yn=PCxn−λnAxn,zn=yn−λnAyn−Axn,xn+1=αnfxn+1−αnzn,where the operator A is monotone and Lipschitz continuous, γ>0,l∈0,1,μ∈0,1, and λn is chosen to be the largest λ∈γ,γl,γl2,⋯ satisfying
(7)λAxn−Ayn≤μxn−yn.In this paper, substituting a sequenceβn⊂0,1 of coefficients instead of the sequence 1−αn in the algorithm (6), we extend algorithm (6). Moreover, condition (7) will be removed just by a slight change in the coefficients λn. Also, a sequence of computational errors in our algorithm is considered. The strong convergence of the proposed algorithm to a point of the variational inequality VIC,A will be proved under the presence of computational errors. Finally, some examples will be presented which will examine the convergence of the proposed algorithm in different situations.


## 2. Preliminaries

In this section, some basic concepts are presented.LetH be a real Hilbert space with the inner product .,. and norm . and suppose that C is a nonempty closed convex subset of H and A:C⟶H is an operator. The operator A is said to be
(i)
Monotone if(8)Ax−Ay,x−y≥0,∀x,y∈C,(ii)
L-Lipchitz continuous if there exist L>0 such that(9)Ax−Ay≤Lx−y,∀x,y∈C.For the main results of this paper, we need the following useful lemmas.Lemma 1.
LetH be a real Hilbert space. Then, we have the following well-known results:
(10)x+y2=x2+2x,y+y2,∀x,y∈H.x+y2≤x2+2y,x+y,∀x,y∈H.Lemma 2 (Xu, see [16]).
Letan be a sequence of nonnegative real numbers satisfying the following relation:
(11)an+1≤1−ηnan+ηnσn+γn,n≥1,where
(a)
ηn⊂0,1,∑n=1∞ηn=∞(b)
limsupσn≤0(c)
γn≥0n≥1,∑n=1∞γn<∞
Then,an⟶0asn⟶∞.Lemma 3 (see [17]).
LetC be a closed and convex subset in a real Hilbert space H. Then, z=PCx if and only if x−z,y−z≤0∀y∈C.Lemma 4 (see [18]).
Letan be a sequence of nonnegative real numbers such that there exists a subsequence anj of an such that anj<anj+1 for all j∈ℕ. Then, there exists a nondecreasing sequence mk of ℕ such that limk⟶∞mk=∞ and the following properties are satisfied by all (sufficiently large) number k∈ℕ:
(12)amk≤amk+1andak≤amk+1.
In fact,mk is the largest number n in the set 1,2,⋯,k such that an<an+1.Lemma 5 (see [3]).
Letxn be a sequence generated by algorithm (3). Then,
(13)xn+1−p2≤xn−p2−1−μ2xn−yn∀p∈VIC,A.


## 3. Main Results

In this section, we prove a strong convergence theorem for finding a common element of the set of solutions of an equilibrium problem and a fixed point problem.Theorem 6.
LetC be a nonempty closed convex subset of a real Hilbert space H, A be a monotone, and L-Lipschitz continuous mapping on C and λ∈0,1 such that λL<1. Suppose that f:H⟶H is a contraction mapping with a constant ρ∈0,1. Let en⊆H be a sequence of computational errors, x0∈H be arbitrary and xn, yn, and zn be the sequences generated by
(14)yn=PCxn−λAxn,zn=yn−λAyn−Axn,xn+1=αnfxn+βnzn+en,where αn and βn are real sequences in 0,1 such that αn+βn≤1 for each n≥1. Also, assume the following conditions:
(15)Σn=1∞αnβn=∞,Σn=1∞1−αn−βn<∞,limn⟶∞1−αn−βnαn=limn⟶∞αn=0,Σn=1∞en<∞.
Then,(i)
VIC,A≠∅ if and only if xn is bounded and liminfn⟶∞xn−yn=0
SupposeVIC,A≠∅. Then,
(ii)
Iflimn⟶∞∥en∥/αn=0, then xn converges strongly to q=PVIC,A∘fq, where PVIC,A∘f:H⟶VIC,A is the mapping defined by PVIC,A∘fx=PVIC,Afx for each x∈HProof.
(i) Assume thatxn is a bounded sequence and liminfn⟶∞xn−yn=0. Then, there exists a subsequence ni⊂ℕ such that xni−yni⟶0 when i⟶∞. Note xni is a bounded sequence. Hence, there exists a subsequence xnik of xni such that xnik converges weakly to some x∈C. Now, noting that since ynik=PCxnik−λAxnik, for all z∈C, we have
(16)0≥xnik−λAxnik−ynik,z−ynik=xnik−ynik,z−ynik−λAxnik,z−ynik=xnik−ynik,z−ynik−λAxnik,z−xnik−λAxnik,xnik−ynik=xnik−ynik,z−ynik−λAxnik−Az,z−xnik−λAz,z−xnik−λAxnik,xnik−ynik≥xnik−ynik,z−ynik−λAz,z−xnik−λAxnik,xnik−ynik.
Now, we have(17)−λAz,z−xnik≤xnik−ynik,ynik−z+λAxnik,xnik−ynik.
Therefore,(18)−λAz,z−xnik≤ynik−xnikynik−z+λAxnikxnik−ynik.
Fromlimk⟶∞xnik−ynik=0, we have −λAz,z−x≤0 for all z∈C. Now, let y∈C and 0<t<1, and from the convexity C we have yt=ty+1−tx∈C. Therefore,
(19)0≤Ayt,yt−x=Ayt,ty−tx=tAyt,y−x.
Since0<t<1 then Ayt,y−x≥0 for all y∈C. Because the mapping A and multiplication are continuous, if t⟶0, then we have Ax,y−x≥0 for all y∈C, i.e; x∈VIC,A.
For the converse fixp∈VIC,A, using Lemma 5, we have
(20)zn−p2≤xn−p2−1−λL2xn−yn2.
Therefore,(21)zn−p≤xn−p.
Using the above inequality, we have(22)1xn+1−p=αnfxn+βnzn+en−p=αnfxn−p+βnzn−p−1−αn−βnp+en≤αnfxn−p+βnzn−p+1−αn−βnp+en≤αnfxn−fp+αnfp−p+βnxn−p+1−αn−βnp+en≤αnρxn−p+αnfp−p+βnxn−p+1−αn−βnp+en=αnρ+βnxn−p+αn1−ρ∥fp−p∥1−ρ+1−αn−βnp+en≤maxxn−p,∥fp−p∥1−ρ,p+en,so the sequence xn is bounded.
(ii) Letp∈VIC,A. From part (i), we have that xn is bounded. Then fxn, yn and zn are bounded. Now, using Lemma 5, we have
(23)zn−p2≤xn−p2−1−λL2xn−yn2∀p∈VIC,A,
(note that our sequencezn in the algorithm (14) replaces xn+1 in Lemma 5).☐Sinceαn+βn+1−αn−βn=1, by the convexity of .2 and the relation x+y2=x2+2x,y+y2, we conclude that
(24)xn+1−p2=αnfxn+βnzn+en−p2=αnfxn+en−p+βnzn+en−p+1−αn−βnen−p2≤αnfxn+en−p2+βnzn+en−p2+1−αn−βnen−p2≤αnfxn−p2+1−αnzn−p2+1−αn−βnp2+αn+βn+1−αn−βnen2+2αnfxn−αnp+βnzn−βnp−1−αn−βnp,en≤αnfxn−p2+1−αnxn−p2−1−αn1−λL2xn−yn2+1−αn−βnp2+en2+2αnfxn+βnzn−p,enby3.3≤αnfxn−p2+xn−p2−1−αn1−λL2xn−yn2+1−αn−βnp2+en2+2αnfxn+βnzn−pen.Therefore,(25)1−αn1−λL2xn−yn2≤xn−p2−xn+1−p2+αnfxn−p2+1−αn−βnp2+en2+2αnfxn+βnzn−pen.Also, we have,(26)xn+1−xn=αnfxn+βnzn+en−xn=αnfxn−xn+βnzn−xn−1−αn−βnxn+en≤αnfxn−xn+βnzn−xn+1−αn−βnxn+en=αnfxn−xn+βnyn−λAyn−Axn−xn+1−αn−βnxn+en≤αnfxn−xn+βnyn−xn+βnλLyn−xn+1−αn−βnxn+en.Note thatPVIC,A∘f is a contraction mapping. Then, by the Banach contraction principle, there exists a unique element q∈H such that q=PVIC,A∘fq. Now, we claim that xn converges strongly to q=PVIC,A∘fq. It is enough to consider two cases:Case 1.
Suppose there exists somen0∈ℕ such that ∥xn+1−q∥2≤∥xn−q∥2 for all n≥n0. Then, limn⟶∞∥xn−q∥ exists. From (25) and our assumptions, limn⟶∞∥xn−yn∥=0. From the conditions (b), (c), and (d), limn⟶∞1−αn−βn=limn⟶∞αn=limn⟶∞en=0. Then, from (26) and the boundedness of the sequences ∥xn∥, ∥fxn−xn∥ and βn, we obtain that
(27)limn⟶∞xn+1−xn=0.
Now, we have(28)xn+p−xn≤xn+p−xn+p−1+xn+p−1−xn+p−2+⋯+xn+1−xn,for all p∈ℕ. Note that from (27), it is concluded that limn⟶∞xn+p−i−xn+p−i+1=0, for all 0≤i≤p−1 and p∈ℕ. Then, from (28), limn⟶∞∥xn+p−xn∥=0, for all p∈ℕ, i.e., xn is a Cauchy sequence in the Hilbert space H; therefore, xn is convergent. Now we show that xn converges strongly to q. Note that
(29)xn+1−q2=αnfxn+βnzn+en−q2=αnfxn+en−q+βnzn+en−q+1−αn−βnen−q2≤βn2zn+en−q2+2αnfxn+en−q+1−αn−βnen−q,xn+1−qbyLemma2.1=βn2zn+en−q2+2αnfxn+en−q,xn+1−q+21−αn−βnen−q,xn+1−q=βn2zn+en−q2+2αnfxn−q,xn+1−q+2αnen,xn+1−q+21−αn−βnen−q,xn+1−q≤βn2zn+en−q2+2αnρxn−qxn+1−q+2αnfq−q,xn+1−q+2αnen,xn+1−q+21−αn−βnen−q,xn+1−q≤βn2zn−q2+2βn2en,zn+en−q+2αnρxn−q2byLemma2.1+2αnfq−q,xn+1−q+2αnen,xn+1−q+21−αn−βnen−q,xn+1−q≤1−αn2xn−q2+2αnρxn−q2+2αnfq−q,xn+1−q+2βn2en,zn+en−q+2αnen,xn+1−q+21−αn−βnen−q,xn+1−q=1−2αn1−ρxn−q2+2αn1−ραnxn−q221−ρ+fq−q,xn+1−q1−ρ+2βn2en,zn+en−q+2αnen,xn+1−q+21−αn−βnen−q,xn+1−q≤1−2αn1−ρxn−q2+2αn1−ραnxn−q221−ρ+fq−q,xn+1−q1−ρ+2βn2enzn+en−q+2αnenxn+1−q+21−αn−βnen−qxn+1−q.Sincexn is strongly convergent, so xn⟶z for some z∈C, then we conclude that xn⇀z. Also, as in the proof of part (i), we conclude z∈VIC,A. Therefore, from Lemma 3(30)limsupn⟶∞fq−q,xn+1−q=fq−q,z−q≤0.Next, we consider the sequencesηn, an,σn and γn in (3.8) as follows:
(31)ηn≔2αn1−ρ,an≔xn−q2,σn≔αnxn−q221−ρ+fq−q,xn+1−q1−ρ,γn≔2βn2enzn+en−q+2αnenxn+1−q+21−αn−βnen−qxn+1−q.From the above, we have(32)an+1≤1−ηnan+ηnσn+γn.Fromc, there exists an integer n1∈ℕ such that ηn⊂0,1, for each n≥n1. Without loss of generality, we may assume that ηn⊂0,1, for each n≥1. From condition a, note ∞=Σn=1∞αnβn≤Σn=1∞αn, so Σn=1∞αn=∞, and hence, Σn=1∞ηn=21−ρΣn=1∞αn=∞. Therefore, condition a of Lemma 2 holds. From c, the boundedness of xn−q2 and (3.9), we have
(33)limsupn⟶∞σn=limsupn⟶∞αnxn−q221−ρ+fq−q,xn+1−q1−ρ≤limsupn⟶∞αnxn−q221−ρ+limsupn⟶∞fq−q,xn+1−q1−ρ≤0.Therefore, conditionb of Lemma 2 holds. Next, note γn≥0 and zn+en−q, xn+1−q, and en−q are bounded. Hence, there exist some positive constants M1, M2, and M3 such that, zn+en−q≤M1,xn+1−q≤M2 and en−q≤M3, for each n≥1. Then, from b and d, we have
(34)Σn=1∞γn=Σn=1∞2βn2enzn+en−q+2αnenxn+1−q+21−αn−βnen−qxn+1−q≤Σn=1∞2βn2enM1+2αnenM2+21−αn−βnM3M2≤2M1Σn=1∞en+2M2Σn=0∞en+2M3M2Σn=1∞1−αn−βn<∞.Therefore, conditionc of Lemma 2 holds. Then, from Lemma 2, it follows that an⟶0 as n⟶∞, i.e., limn⟶∞xn−q2=0.Case 2.
Suppose there exists a subsequencexnj−q2 of xn−q2 such that xnj−q2<xnj+1−q2 for all j∈ℕ. Now from Lemma 4, there exists a nondecreasing sequence mk of ℕ such that limk⟶∞mk=∞ and the following inequalities hold for all k∈ℕ:
(35)xmk−q2≤xmk+1−q2andxk−q2≤xmk+1−q2.
Now, from (25), we have
(36)1−αmk1−λL2xmk−ymk2≤xmk−p2−xmk+1−p2+αmkfxmk−p2+1−αmk−βmkp2+emk2+2αmkfxmk+βmkzmk−pemk.
Hence,limk⟶∞xmk−ymk=0, therefore from (26) (adjusted) limk⟶∞xmk+1−xmk=0.
From (29) (adjusted), we conclude that
(37)xmk+1−q2≤1−2αmk1−ρxmk−q2+2αmk1−ραmkxmk−q221−ρ+fq−q,xmk+1−q1−ρ+2βmk2emkzmk+emk−q+2αmkemkxmk+1−q+21−αmk−βmkemk−qxmk+1−q≤1−2αmk1−ρxmk+1−q2+2αmk1−ραmkxmk−q221−ρ+fq−q,xmk+1−q1−ρ+2βmk2emkzmk+emk−q+2αmkemkxmk+1−q+21−αmk−βmkemk−qxmk+1−q.
Therefore, we have(38)xk−q2≤xmk+1−q2≤αmkxmk−q221−ρ+fq−q,xmk+1−q1−ρ+emkαmkM1+1−αmk−βmkαmkM2,where
(39)M1=βmk21−ρzmk+emk−q+2αmkxmk+1−q,M2=11−ρemk−qxmk+1−q.
From our assumptionslimk⟶∞αmk=0, limk⟶∞1−αmk−βmk/αmk=0, limk⟶∞∥emk∥/αmk=0, and (30) (adjusted), we conclude limsupk⟶∞∥xk−q∥2=0. Hence, xk⟶q which completes the proof of part (ii).Open problem 1. Can we remove the conditionliminfn⟶∞∥xn−yn∥=0 in (i) in Theorem 6?


## 4. Numerical Example

In this section, the algorithm (14) is illustrated with some examples.Example 1.
Putαn=1/n+1,βn=1−1/n+1,en=0,H=L20,1,A≡I,Fx≡1,λ=1/2,C=B0,1=f∈L20,1:f≤1.
Then, from the algorithm (14), we have the following sequences:
(40)gn=PCfn−λAfn=PCfn2,kn=gn−λAgn−Afn=12fn+gn,fn+1=αnFfn+βnkn+en=1n+1+nn+1kn,where PC is as follows
(41)PBz,ρx=xx−z≤ρ,z+ρx−zx−zx−z>ρ,where ρ>0 (see [19]). We have,
(42)VIC,A=f∈C:Af,g−f≥0,∀g∈C=f∈C:f,g−f≥0,∀g∈C.
Obviously,0∈VIC,A; hence, VIC,A≠∅.Now, withf1=2 and using the MATLAB software, we see that fn converges to 0 (Figure 1).Figure 1
Example1. Convergence behavior fn in Example 1.In the following example, using the MATLAB software, we compare some similar algorithms and their convergence speed and behavior. In particular, the TEGM algorithm (3), VSEGM algorithm (4), THEGM algorithm (6), and the algorithm (14) are compared. We see that the algorithm (14) has a higher convergence speed than the other algorithms (Figure 2).Example 2.
Letαn=1/n,βn=1−1/n−1/n2,λ=1/2,en=0,H=ℝ,A≡I,fx=x/2,andC=0,∞, x1=2 (Figure 2).Figure 2
The number of iteration for convergence to zero is, respectively, 15, 27, 25, and 27 in Example2.Now, we examine the convergence of the sequencesxn,yn,andzn in Theorem 6 in the following example.Example 3.
Putαn=1/n,βn=1−1/n−1/n2,λ=1/2,en=1/n2,H=ℝ,A≡I,f≡1,andC=0,∞, x1=2.
Then, we have(43)yn=PCxn−λAxn=PC12xn,zn=yn−λAyn−Axn=12yn+xn,xn+1=αnfxn+βnzn+en=1n+1−1n−1n2zn+1n2.
Hence,(44)VIC,A=x∈C:Ax,y−x≥0,∀y∈C=x∈0,∞:x,y−x≥0,∀y∈C=0.
Therefore,q=PVIA,C∘fq=P0∘q/2=0, and now, by Theorem 6, the sequence xn converges strongly to 0.In the following example, we examine the case thatVIC,A=∅, and the sequence generated by the algorithm (14) is divergent (Table 1 and Figure 3).Example 4.
Putαn=1/n,βn=1−1/n,λ=1/2,en=0,H=ℝ,A≡−1,f≡1,andC=0,∞, x1=2.
Then, we have(45)yn=PCxn−λAxn=PCxn+12,zn=yn−λAyn−Axn=yn,xn+1=αnfxn+βnzn+en=1n+1−1nPCxn+12.
Hence,(46)VIC,A=x∈C:Ax,y−x≥0,∀y∈C=x∈0,∞:−1,y−x≥0,∀y∈C=∅.
Now, we will prove by induction that, for alln∈ℕ(47)yn=PCxn+12=xn+12.
Whenn=1, then we have x1=2≥−1/2.
Induction step: suppose forn=k the inequality xk≥−1/2 holds.Table 1
Numerical results of convergence forx1=2 in Example 3.
nxnynzn1211.520.50.250.37530.44290.22140.3321930.00050.00020.0003940.00050.00020.0003950.00040.00020.00031850.00010.00010.00011860.00010.00010.00011870.00010.00010.00012770.00010.00000.00002780.00010.00000.00002790.00000.00000.0000Figure 3
Convergence behaviorxn in Example 3.Then,(48)xk+1=1k+1−1kPCxk+12=1k+1−1kxk+12=1k+1−1kxk+1−1k12≥1k+1−1k−12+1−1k12≥−12.Hence,yn=xn+1/2. Consequently limn⟶∞xn−yn=1/2≠0.Next, we show thatxn is an unbounded sequence. Note x3=5/4>3/4.Induction step: suppose forn=k the inequality xk>k/4 holds.Then,(49)xk+1=1k+1−1kxk+12>1k+1−1kk4+12=1k+k4−14+12−12k=12k+k4+14>k+14,thus xn is an unbounded sequence (Figure 4).Figure 4
Convergence behaviorxn in Example 4.


## 5. Conclusions

In this paper, we proposed a Tseng-type viscosity algorithm based on the viscosity method, which is an extension of the Thong et al.’s algorithm [3]. We showed that the sequence generated by the proposed algorithm strongly converges to an element ofVIC,A.The following are the results in this paper:(i)
We extended the results of Thong et al.’s. method [3] and provided necessary and sufficient conditions for the VIC,A to be nonempty(ii)
In our generated sequencexn, in the algorithm (14), we put a sequence of computational errors en and proved the convergence of the sequence in the presence of computational errors(iii)
We provided some numerical examples to compare our algorithm with the algorithms TEGM, VSEGM, and THEGM



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