# Analysis of Subgradient Extragradient Iterative Schemes for Variational Inequalities

Authors: Danfeng Wu; Li-Jun Zhu; Zhuang Shan; Tzu-Chien Yin
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we investigate the monotone variational inequality in Hilbert spaces. Based on Censor’s subgradient extragradient method, we propose two modified subgradient extragradient algorithms with self-adaptive and inertial techniques for finding the solution of the monotone variational inequality in real Hilbert spaces. Strong convergence analysis of the proposed algorithms have been obtained under some mild conditions.


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## Body

## 1. Introduction

LetH be a real Hilbert space and S∈H be a nonempty closed convex subset. Let f:H⟶H be an operator. In this work, we investigate the following variational inequality problem (VIPs):(1)find a pointu‡∈S,s.t.fu‡,x−u‡≥0,∀x∈S.Denote bySolS,f the solution set of (1). The VIPs is an important tool to study various problems in the domain of mechanics, optimization, transportation, fixed point, economics equilibrium, contract problems in elasticity, and other branches of mathematics, see [1–17]. Therefore, VIPs have received much attention by many scholars, see [18–30]. There are a variety of methods to solve the VIPs, such as regularization method and projection method [31–39]. In this work, we focus on projection method.As we all know that the gradient projection algorithm is the simplest and oldest method ([40, 41]), the method is defined as follows:(2)uk+1=PSuk−γfuk,where PS:H⟶S is the metric projection and γ is some positive number.In order to obtain a convergent result, this algorithm requires that the operatorf is strongly monotone. In order to avoid the strong monotonicity hypothesis, Korpelevich [42] proposed the extragradient algorithm which is stated as follows:(3)xk=PSuk−γfuk,uk+1=PSuk−γfxk,where γ∈0,1/L and operator f is monotone and Lipschitz continuous in H.Note that the algorithm (3) calculates two projections on S in each iteration. If the set S is more complicated, there will be a lot of calculations. In order to overcome this difficulty, Censor et al. [43] constructed a half space with subdifferentiation and proposed subgradient extragradient method which is defined by(4)xk=PSuk−γfuk,Tk=z∈H|uk−γfuk−xk,z−xk≤0,uk+1=PTkuk−γfxk.Recently, Dong et al. [44] improved the algorithm (4) with self-adaptive stepsize which generates a sequence uk by the following form:(5)xk=PSuk−γkfuk,γkfuk−fxk≤σuk−xk,∀σ∈0,1,Tk=z∈Huk−γkfuk−xk,z−xk≤0,uk+1=PTkuk−τζkγkfxk,where ζk=uk−xk,κuk,xk/κuk,xk2 and κuk,xk=uk−xk−γkfuk−fxk.Weak convergence of Algorithm (5) has been obtained. Motivated and inspired by the above work, in this paper, we continue to investigate iterative algorithms for solving the monotone variational inequality in Hilbert spaces. We construct two modified subgradient extragradient algorithms for finding the solution of the monotone variational inequality. Our algorithms combine self-adaptive technique and inertial method. Under some mild conditions, we prove that the proposed algorithms converge strongly to a solution of the monotone variational inequality.The organizational structure of this paper is as follows. In Section2, we present some definitions and preliminary results, which will be used in further analysis of the proposed algorithms. In Section 3, we proposed two modified subgradient extragradient algorithms and prove strong convergence theorems.


## 2. Preliminaries

LetC be a nonempty closed convex subset of a real Hilbert space H. Use “⇀” and “⟶” to denote weak and strong convergence, respectively. Let xk be a sequence in H. We use ωwxk to denote the set of all weak cluster points of xk, i.e.,(6)ωwxk=x†:∃xki⊂xksuch thatxki⇀x†asi⟶∞.For∀u,v∈H, and λ∈R, the following results hold(7)u+v2≤u2+2v,u+v,(8)λu+1−λv2=λu2+1−λv2−λ1−λu−v2.Definition 1.
Letf:H⟶H be an operator. Recall that the operator f is said to be(i)
Monotone if(9)fu−fv,u−v≥0,∀u,v∈H.(ii)
Strongly monotone if there existsγ>0 s.t.(10)fu−fv,u−v≥γu−v2,∀u,v∈H.(iii)
L-Lipschitz continuous if there exists L>0 s.t.(11)fu−fv≤Lu−v,∀u,v∈H.
IfL<1, f is said to be L-contractive.
LetC be a nonempty closed convex subset of a real Hilbert space H. For any x∈H, there exists a unique point PCx∈C such that(12)x−PCx≤y−x,∀y∈C.
It is well known thatPC satisfies [45](13)x−PCx,y−PCx≤0,(14)x−PCx2+y−PCx2≤x−y2,for all x∈H and y∈C.Lemma 1 (see [46]).
Letbk be a real number sequence. Suppose that there exists a subsequence bkm of bk such that bkm<bkm+1 for all m∈ℕ. Define the sequence γk as follows:(15)γk=maxn∈ℕ|k¯0≤n≤k,bn≤bn+1,for each k≥k¯0. Then, the following inequality holds:(16)0≤bk≤bγk+1,for each k≥k¯0. Further, for all k≥k¯0, the sequence γk is nondecreasing and(17)limk⟶∞γk=+∞.Lemma 2 (see [33]).
Suppose that the sequenceδk of real numbers is nonnegative and there exists k0∈ℕ such that(18)δk+1≤1−γkδk+γklk,for each k≥k0, where the sequences γk and lk satisfy the following conditions:(19)γk⊂0,1,limk⟶∞γk=0,∑k=1∞γk=∞,limsupk⟶∞lk≤0.
Then,limk⟶∞δk=0.


## 3. Main Result

In this section, we present our main results.LetS be a nonempty closed convex subset of a real Hilbert space H. Suppose that the following three conditions are satisfied:(C1): the setSolS,f is not empty;(C2): the operatef is monotone;(C3): the operatef is L-Lipschitz continuous.Letσ,ρ∈0,1,τ∈0,2, and γ0>0 be four constants. Let θk,εk⊂0,1,andλk⊂a,b⊂0,1 be three sequences, satisfying(20)∑n=1∞θk=+∞,limk⟶∞θk=0,εk=oθk.Next, we introduce an iterative algorithm for solving (1).Lemma 3.
Ifxk=vk or κvk,xk=0 in Algorithm 1, then xk∈SolS,f.Algorithm 1: Strong convergence algorithm with contractive technique.
Initialization. Chooseu0,u1∈H arbitrarily.Step 1. Chooseρk s.t. 0≤ρk≤ρ¯k, where ρ¯k=minρ,εk/uk−uk−1,ifuk≠uk−1ρ,otherwiseCalculatevk=uk+ρkuk−uk−1Step 2. Calculatexk=PSvk−γkfvk,where,γk+1=minγk,σuk−xk/fuk−fxk,iffuk≠fxkγk,otherwiseStep 3. Construct the half spaceTk as follows Tk=z∈H|vk−γkfvk−xk,z−xk≤0Calculatetk=PTkvk−τζkγkfxkwhereζk=vk−xk,κvk,xk/κvk,xk2andκvk,xk=vk−xk−γkfvk−fxkStep 4. Computeuk+1=1−θk−λkvk+λktkIfxk=vk, then stop and xk∈SolS,f. Otherwise, set k:=k+1 and return to step 1.Proof.
Sincef is L-Lipschitz continuous, we obtain(21)κvk,xk=vk−xk−γkfvk−fxk≥vk−xk−γkfvk−fxk≥vk−xk−γkLvk−xk=1−γkLvk−xk,κvk,xk=vk−xk−γkfvk−fxk≤vk−xk+γkfvk−fxk≤vk−xk+γkLvk−xk=1+γkLvk−xk.
It follows that(22)1−γkLvk−xk≤κvk,xk≤1+γkLvk−xk.
Consequently,vk=xk⇔κvk,xk=0. Furthermore, if vk=xk or κvk,xk=0, we have(23)xk=PSxk−γkfxk.
Combining (13) and (23), we get(24)xk−γkfxk−xk,xk−z≥0,∀z∈S,which implies that(25)fxk,z−xk≥0,∀z∈S.
This completes the proof.Lemma 4.
The sequenceγkk=0∞ generated by Algorithm 1 is monotonically decreasing, and γk≤minγ0,σ/L for each k≥0.Proof.
Obviously, by the definition ofγk+1, we have γk is a monotonically decreasing sequence. Then, γk≥γ0,∀n>0. Since f is Lipschitz continuous, we have(26)fuk−fxk≤Luk−xk.
In the case offuk≠fxk, we have(27)σuk−xkfuk−fxk≥σL.
Obviously, the lower bound ofγk is minγ0,σ/L.
This completes the proof.Lemma 5.
Letζk be the sequence generated by Algorithm 1. Then, we have(28)ζk≥1−σ1+σ2.Proof.
Combining Lemma4 and Cauchy–Schwartz inequality, we have(29)vk−xk,κvk,xk=vk−xk,vk−xk−γkfvk−fxk=vk−xk2−γkvk−xk,fvk−fxk≥vk−xk2−γkvk−xkfvk−fxk≥vk−xk2−γkLvk−xk2=1−γkLvk−xk2≥1−σvk−xk2.
Sincef is monotone and Lipschitz continuous, then we obtain(30)κvk,xk2=vk−xk−γkfvk−fxk2=vk−xk2+γk2fvk−fxk2−2γkvk−xk,fvk−fxk≤vk−xk2+γk2L2vk−xk2=1+γk2L2vk−xk2≤1+σ2vk−xk2.
From (29) and (30), we have(31)ζk=vk−xk,κvk,xkκvk,xk2≥1−σ1+σ2.
This completes the proof.Lemma 6.
Letu‡∈SolS,f. Then,(32)tk−u‡2≤vk−u‡2−vk−tk−τζkκvk,xk2−τ2−τζk2κvk,xk2.Proof.
From (13) and Algorithm 1, we have(33)tk−u‡2≤PTkvk−τζkγkfxk−PTku‡2≤tk−u‡,vk−τζkγkfxk−u‡=12tk−u‡2+12vk−τζkγkfxk−u‡2−12tk−vk+τζkγkfxk2=12tk−u‡2+12vk−u‡2+12τ2ζk2γk2fxk2−vk−u‡,τζkγkfxk−12tk−vk2−12τ2ζk2γk2fxk2−tk−vk,τζkγkfxk=12tk−u‡2+12vk−u‡2−12tk−vk2−tk−u‡,τζkγkfxk.
It follows that(34)2tk−u‡2≤tk−u‡2+vk−u‡2−tk−vk2−2τζkγktk−u‡,fxk,or equivalently(35)tk−u‡2≤vk−u‡2−tk−vk2−2τζkγktk−u‡,fxk.
We deduce fromxk∈Candu‡∈SolS,f that fu‡,xk−u‡≥0. It follows from the monotonicity of operator f that fxk−fu‡,xk−u‡≥0. Then, fxk,xk−u‡≥0. It equates that fxk,tk−u‡≥fxk,tk−xk. Thus,(36)−2τζkγkfxk,tk−u‡≤−2τζkγkfxk,tk−xk.
On the other hand, combiningtk∈Tk and Algorithm 1, we obtain(37)κvk,xk,tk−xk≤γkfxk,tk−xk.
This implies that(38)−2τζkγkfxk,tk−u‡≤−2τζkκvk,xk,tk−xk.
Hence, we obtain(39)−2τζkγkfxk,tk−u‡≤−2τζkκvk,xk,tk−xk≤−2τζkκvk,xk,vk−xk+2τζkκvk,xk,vk−tk.
Now, we calculate−2τζkκvk,xk,vk−xk and 2τζkκvk,xk,vk−tk separately. From the definition of ζk, we get(40)−2τζkκvk,xk,vk−xk=−2τζk2κvk,xk2.
Meanwhile,(41)2τζkκvk,xk,vk−tk=vk−tk2+τ2ζk2κvk,xk2−vk−tk−τζkκvk,xk2.
This implies that(42)−2τζkγkfxk,tk−u‡≤−2τζk2κvk,xk2+vk−tk2+τ2ζk2κvk,xk2−vk−tk−τζkκvk,xk2=vk−tk2−vk−tk−τζkκvk,xk2−τ2−τζk2κvk,xk2.
So, we get(43)tk−u‡2≤vk−u‡2−vk−tk−τζkκvk,xk2−τ2−τζk2κvk,xk2.
This completes the proof.Theorem 1.
The sequenceuk generated by Algorithm 1 converges strongly to u‡∈SolS,f.Proof.
We divide the proof into four claims.Claim 1. We prove the boundedness of the sequencesuk and tk. Indeed, from Algorithm 1 and Lemma 6, we get(44)uk+1−u‡=1−θk−λkvk+λktk−u‡=1−θk−λkvk−u‡+λktk−u‡−θku‡≤1−θk−λkvk−u‡+λktk−u‡+θku‡≤1−θk−λkvk−u‡+λktk−u‡+θku‡≤1−θkvk−u‡−λkvk−u‡+λkvk−u‡+θku‡=1−θkvk−u‡+θku‡.Combining Algorithm1 and (44), we obtain(45)uk+1−u‡=1−θkvk−u‡+θku‡=1−θkuk+ρkuk−uk−1−u‡+θku‡≤1−θkuk−u‡+ρk1−θkuk−uk−1+θku‡=1−θkuk−u‡+θkςk+u‡,where(46)ςk=1−θkρkθkuk−uk−1.Taking into accountεk=oθk and the definition of ρk, we get(47)limk⟶∞ςk=0.Then, the sequenceςk is bounded. Let M=supk≥1ςk+u‡. We obtain from (45) that(48)uk+1−u‡≤1−θkuk−u‡+θkM≤maxuk−u‡,M.For∀k≥k0, we have(49)uk+1−u‡≤maxuk−u‡,M.It follows that the sequenceuk is bounded. Therefore, the sequence tk is bounded.Claim 2. We prove that the following holds:(50)uk+1−u‡2≤1−θk2vk−u‡2−2λkθkvk−tk,uk+1−u‡+2θku‡,uk+1−u‡.Setzk=1−λkvk+λktk. Then, vk−zk=λkvk−tk. Therefore,(51)uk+1=1−θk−λkvk+λktk=zk−θkvk=1−θkzk−θkvk−zk=1−θkzk−θkλkvk−tk.From Lemma6, we have(52)tk−u‡≤vk−u‡,which implies that(53)zk−u‡2≤1−λkvk−u‡2+λktk−u‡2≤vk−u‡2.By (7), (51), and (53), we get(54)uk+1−u‡2=1−θkzk−θkλkvk−tk−u‡2=1−θkzk−u‡−θkλkvk−tk−θku‡2≤1−θk2zk−u‡2−2θkλkvk−tk,uk+1−u‡−2θku‡,uk+1−u‡≤1−θk2vk−u‡2−2θkλkvk−tk,uk+1−u‡−2θku‡,uk+1−u‡=1−θk2vk−u‡2−2θkλkvk−tk,uk+1−u‡+2θk−u‡,uk+1−u‡.Claim 3. By (8) and Algorithm 1, we obtain(55)vk−u‡2=uk+ρkuk−uk−1−u‡2=1+ρkuk−u‡−ρkuk−1−u‡2=1+ρkuk−u‡2−ρkuk−1−u‡2+ρk1+ρkuk−uk−12≤1+ρkuk−u‡2−ρkuk−1−u‡2+2ρkuk−uk−12=uk−u‡2+ρkuk−u‡2−uk−1−u‡2+2ρkuk−uk−12.Using Lemma (8) and (52), we get(56)uk+1−u‡2=1−θk−λkvk−u‡+λktk−u‡+θk−u‡2≤1−θk−λkvk−u‡2+λktk−u‡2+θku‡2≤1−θk−λkvk−u‡2+θku‡2+λkvk−u‡2−vk−tk−τζkκvk,xk2−τ2−τζk2κvk,xk2≤1−θkvk−u‡2+θku‡2−λkvk−tk−τζkκvk,xk2−λkτ2−τζk2κvk,xk2.From (55) and (56), we get(57)uk+1−u‡2≤1−θkuk+ρkuk−uk−1−u‡2+θku‡2−λkvk−tk−τζkκvk,xk2−λkτ2−τζk2κvk,xk2≤1−θkuk−u‡2+ρk1−θkuk−u‡2−uk−1−u‡2+2ρk1−θkuk−uk−12+θku‡2−λkvk−tk−τζkκvk,xk2−λkτ2−τζk2κvk,xk2≤uk−u‡2+ρk1−θkuk−u‡2−uk−1−u‡2+2ρk1−θkuk−uk−12+θku‡2−λkvk−tk−τζkκvk,xk2−λkτ2−τζk2κvk,xk2.Claim 4. Next, we will consider two different cases to prove the strong convergence of the sequenceuk−u‡2.Case 1. There exists anN∈ℕ s.t. uk+1−u‡2≤uk−u‡2, ∀k≥N. Obviously, the limit of the sequence uk−u‡2 exists which implies that limk⟶∞uk+1−uk=0. In (57), taking the limit as k⟶∞, we deduce(58)limk⟶∞κvk,xk=0,(59)limk⟶∞vk−tk−τζkκvk,xk2=0.On the other hand, we have(60)vk−tk≤vk−tk−τζkκvk,xk+τζkκvk,xk.So, we havelimk⟶∞vk−tk=0.Combining Lemma3 and (58), we obtain(61)limk⟶∞vk−xk=0.Now, we show thatwwuk⊂SolS,f. Choose p‡∈wwuk. It implies that there exists a subsequence unk of uk which converges weakly to p‡. Therefore, vnk⇀p‡. Due to limk⟶∞vk−xk=0, we obtain xnk⇀p‡∈S. By Algorithm 1, we have(62)xk−vk+γkfvk,u−xk≥0,∀u∈C.Sincef is monotone, we have(63)0≤xk−vk,u−xk+γkfvk,u−xk=xk−vk,u−xk+γkfvk,u−vk+γkfvk,vk−xk≤xk−vk,u−xk+γkfu,u−vk+γkfvk,vk−xk.Taking the limit in (63) as k⟶∞, we get(64)fu,u−p‡≥0,∀u∈C,which implies thatwwuk∈SolS,f.Setbk=uk−u‡2 for all k≥0. By (65) for q=u‡, we obtain(65)bk+1≤1−θkvk−q2+θk−2λkvk−tkuk+1−q+2−q,uk+1−q.We deduce from Algorithm1 that(66)vk−q2≤uk−q+ρkuk−uk−12=uk−q2+ρk2uk−uk−12+2ρkuk−quk−uk−1≤uk−q2+ρkuk−uk−12+2ρkuk−quk−uk−1≤bk+3Kρkuk−uk−1,where(67)K=supk≥1uk−uk−1,uk−q.By virtue of (65) and (66), we have(68)bk+1≤1−θkbk+δk,where(69)δk=θk3K1−θkρkθkuk−uk−1−2λkvk−tkuk+1−q+2−q,uk+1−q.So, we get(70)limsupk≥1−q,uk+1−q=supu‡∈wwuk−q,u‡−q≤0.From (70), we deduce that q∈PSolS,f0. Combining the property of projection, limk⟶∞vk−tk2=0 and limk⟶∞ρk/θkuk−uk−1=0, we have limsupk≥1δk≤0. By Lemma 2, we obtain bk=uk−u‡2⟶0k⟶∞. Therefore, the sequence uk converges strongly to u‡.Case 2. There exists a subsequencebki⊂bkk≥k˜0 s.t. bk˜i≤bk˜i+1 for ∀i≥0. From Lemma 2, we can deduce(71)bγk≤bγk+1,bk≤bγk+1,for eachk≥k˜0, where γk=maxn∈ℕ|k˜0≤n≤k,bn≤bn+1. Further, the sequence γkk≥k˜0 is nondecreasing (i.e., limk⟶∞γk=∞). Let bk=uk−u‡2. By (71) and Claim 3 for q=u‡, we obtain(72)λγkvγk−tγk−γζγkdvγk,xγk2+τ2−τζγk2dvγk,xγk2≤ργk1−θγkbγk−bγn−1+2ργk1−θγkuγk−uγn−12+θγkq2.
We deduce from the definition ofbk that(73)bγk−bγk−1=uγk−q2−uγk−1−q2=uγk−q−uγk−1−quγk−q+uγk−1−q≤uγk−uγk−1uγk−q+uγk−1−q.
Combining (72) and (73), we have(74)λγkvγk−tγk−γζγkdvγk,uγk2+τ2−τζγk2dvγk,uγk2≤ργk1−θγkuγk−uγk−1uγk−q−uγk−1−q+2ργk1−θγkuγk−uγk−12+θγkq2.
Similarly, we haveργk1−θγkuγk−uγk−1⟶0. It follows that(75)wwuγk⊂SolS,f,limk⟶∞tγk−uγk2=limk⟶∞tγk−vγk2=0,and(76)bγk+1≤1−θγkbγk+θγk3K1−θγkργkθγkuγk−uγk−1−2λγkvγk−tγkuγk+1−q+2−q,uγk+1−q.
Sincebγk≤bγk+1 and θγk>0, from (76), we have(77)bγk≤3K1−θγkργkθγkuγk−uγk−1−2λγkvγk−tγkuγk+1−q+2−q,uγk+1−q.
Sinceq∈PSolS,f0 and wwuγk⊂SolS,f, we have limsupk⟶∞−q,uγk+1−q=supk⟶∞−q,u‡−q≤0. By (75), (77), and ργk/θγkuγk−uγk−1⟶0, we get(78)limsupk⟶∞bγk≤2supq∈wwuγk−q,u‡−q≤0.
It follows from (76) that(79)limsupk⟶∞bγk+1≤0orlimk⟶∞bγk+1=0.
Hence,limk⟶∞bk=0. Therefore, the sequence uk converges strongly to u‡. This completes the proof.
Suppose thatg:H⟶H is a ρ-contractive operator. Next, we propose an iterative algorithm with viscosity item.Theorem 2.
The sequenceuk generated by Algorithm 1 converges strongly to u‡=PSolS,fgu‡.Algorithm 2: Strong convergence algorithm with viscosity term.
Initialization. Chooseu0,u1∈H arbitrarily.Step 1. Chooseρk s.t. 0≤ρk≤ρ¯k, where ρ¯k=minρ,εk/uk−uk−1ifuk≠uk−1ρotherwiseCalculatevk=uk+ρkuk−uk−1Step 2. Calculatexk=PSvk−γkfvk,whereγk+1=minγk,σuk−xk/fuk−fxkiffuk≠fxkγkotherwiseStep 3. Construct the half spaceTk as follows Tk=z∈H|vk−γkfvk−xk,z−xk≤0.Calculatetk=PTkvk−τζkγkfxk,whereζk=vk−xk,κvk,xk/κvk,xk2andκvk,xk=vk,xk−γkfvk−fxk.Step 4. Computeuk+1=θkgvk+1−θktk.Ifxk=vk, then stop and xk∈SolS,f. Otherwise, set k:=k+1 and return to step 1.Proof.
We divide the proof into 4 claims.Claim 1. We prove the boundedness of the sequencesgvk,xk and tk. From Algorithm 1, we get(80)vk−u‡=uk−ρkuk−uk−1−u‡≤uk−u‡+ρkuk−uk−1=uk−u‡+θkρkθkuk−uk−1.From Algorithm1, we obtain ρk/θkuk−uk−1⟶0,k⟶∞. Then, ∃M1>0 s.t.(81)ρkθkuk−uk−1≤M1,∀k>0.By Algorithm1 and (81), we have(82)uk+1−u‡=θkgvk+1−θktk−u‡=θkgvk−u‡+1−θktk−u‡≤θkgvk−u‡+1−θktk−u‡≤θkgvk−gu‡+θkgu‡−u‡+1−θktk−u‡≤θkρvk−u‡+θkgu‡−u‡+1−θkvk−u‡≤1−1−ρθkvk−u‡+θkgu‡−u‡≤1−1−ρθkuk+ρkuk−uk−1−u‡+θkgu‡−u‡≤1−1−ρθkuk−u‡+1−1−ρθkρkuk−uk−1+θkgu‡−u‡≤1−1−ρθkuk−u‡+1−1−ρθkθkρkθkuk−uk−1+θkgu‡−u‡.From (81) and (82), we have(83)uk+1−u‡≤1−1−ρθkuk−u‡+1−1−ρθkθkρkθkuk−uk−1+θkgu‡−u‡≤1−1−ρθkuk−u‡+1−1−ρθkθkM1+θkgu‡−u‡≤1−1−ρθkuk−u‡+1−ρθk1−1−ρθkM1+gu‡−u‡1−ρ≤1−1−ρθkuk−u‡+1−ρθkM1+gu‡−u‡1−ρ≤maxuk−u‡,M1+gu‡−u‡1−ρ≤⋯≤maxuk−u‡,M1+gu‡−u‡1−ρ.It is obvious that the sequenceuk is bounded. Furthermore, the sequences gvk,xk and tk are bounded.Claim 2. From (7) and Algorithm 1, we have(84)uk+1−u‡2=θkgvk+1−θktk−u‡2=θkgvk−gu‡+1−θktk−u‡+θkgu‡−u‡2≤θkgvk−gu‡+1−θktk−u‡2+2θkgu‡−u‡,uk+1−u‡≤θkgvk−gu‡2+1−θktk−u‡2+2θkgu‡−u‡,uk+1−u‡≤θkρvk−u‡2+1−θkvk−u‡2+2θkgu‡−u‡,uk+1−u‡=1−1−ρθkvk−u‡2+2θkgu‡−u‡,uk+1−u‡.Claim 3. By (8) and (55), we obtain(85)uk+1−u‡2=θkgvk−u‡+1−θktk−u‡2≤θkgvk−u‡2+1−θktk−u‡2−θk1−θkgvk−tk2≤θkgvk−u‡2+1−θktk−u‡2≤θkgvk−u‡2+1−θkvk−u‡2−1−θkvk−tk−τζkκvk,xk2−1−θkτ2−τζk2κvk,xk2.From (85) and (55), we obtain(86)uk+1−u‡2≤θkgvk−u‡2+1−θkuk−u‡2+1−θkρkuk−u‡2−uk−1−u‡2+21−θkρkuk−uk−12−1−θkvk−tk−τζkκvk,xk2−1−θkτ2−τζk2κvk,xk2≤θkgvk−u‡2+uk−u‡2+1−θkρkuk−u‡2−uk−1−u‡2+21−θkρkuk−uk−12−1−θkvk−tk−τζkκvk,xk2−1−θkτ2−τζk2κvk,xk2.This implies that(87)1−θkvk−tk−τζkκvk,xk2+1−θkτ2−τζk2κvk,xk2−1−θkρkuk−u‡2−uk−1−u‡2−21−θkρkuk−uk−12≤uk−u‡2−uk+1−u‡2+θkgvk−u‡2.Claim 4. According to Claim 3, we can see that there are two possible cases.Case 1. There exists anN∈ℕ, s.t. uk+1−u‡2≤uk−u‡2 for ∀k>N. It follows that limk⟶∞uk−u‡ exists. From (86) and limk⟶∞θk=0, we have(88)limk⟶∞κvk,xk=0,limk⟶∞vk−tk−τζkκvk,xk=0.Note that(89)vk−tk2≤vk−tk−τζkκvk,xk+τζkκvk,xk.So,(90)limk⟶∞vk−tk=0.Similarly, we can obtain(91)wwuk⊂SolS,f.Setbk=uk−q2 for all k≥0. By (84) for q=u‡, we get(92)bk+1≤1−1−ρθkvk−u‡2+2θkgu‡−u‡,uk+1−u‡.It follows from (66) and (92) that(93)bk+1≤1−1−ρθkbk+δk,where(94)δk=3K1−1−ρθkρkuk−uk−1+2θkgq−q,uk+1−q.Then,(95)limsupk≥1gq−q,uk+1−q=limsupu‡∈wwukgq−q,u‡−q≤0.Hence, we deduceq∈PSolS,f0. Combining the property of projection, limk⟶∞vk−tk2=0 and limk⟶∞ρk/θkuk−uk−1=0, we have limsupk≥1δk≤0. By Lemma 2, we obtain bk=uk−u‡2⟶0k⟶∞. Therefore, the sequence uk converges strongly to u‡.Case 2. There exists a subsequencebki⊂bkk≥k˜0, s.t. bk˜i≤bk˜i+1 for each i≥0. From Lemma 1, we deduce that(96)bγk≤bγk+1,bk≤bγk+1,for allk≥k˜0, where γk=maxn∈ℕ|k˜0≤n≤k,bn≤bn+1. Therefore, the sequence γkk≥k˜0 is nondecreasing (i.e., limk⟶∞γk=∞). By (96) and Claim 3 for q=u‡, we obtain(97)1−θγkvγk−tγk−τζγkκvγk,xγk2+1−θγkτ2−τζγk2κvγk,xγk2≤uγk−q2−uγn+1−u‡2+θγkgvγk−q2+1−θγkργkbγk−bγk−1+21−θγkργkuγk−uγk−12.
From (73) and (97), we get(98)1−θγkvγk−tγk−τζγkκvγk,xγk2+1−θγkτ2−τζγk2κvγk,xγk2≤uγk−u‡2−uγk+1−u‡2+θγkgvγk−u‡2+21−θγkργkuγk−uγn−12+1−θγkργkuγk−uγn−1uγk−q+uγn−1−q.
Using Claim 1 and (98), we have limk⟶∞1−θγkργkuγk−uγk−1=0. Therefore,(99)wwuγk⊂SolS,f,limk⟶∞tγk−uγk2=limk⟶∞tγk−vγk2=0,bγk+1≤1−1−ρθγkbγk+3K1−1−ρθγkργkuγk−uγk−1+2θγkgq−q,uγk+1−q.
Sincebγk≤bγk+1 and bγk≥0, we receive(100)bγk≤3K1−1−ρθγkργkuγk−uγk−1+2θγkgq−q,uγk+1−q.
Note thatq∈PSolS,f0 and wwuγk⊂SolS,f. By the property of projection, we have(101)limsupk⟶∞gq−q,uγk+1−q=supu‡∈wwuγkgq−q,u‡−q≤0.
Since1−θγkργkuγk−uγn−1⟶0, we deduce(102)limsupk⟶∞aγk≤2supu‡∈wwxγkgq−q,u‡−q≤0.
So,limsupk⟶∞bγk+1≤0 or limk⟶∞bγk+1=0. Hence, limk⟶∞bk=0 which implies that the sequence uk converges strongly to u‡. This completes the proof.



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*Source: 2218666-2021-11-05.xml*