# New Tseng-Degree Gradient Method in Variational Inequality Problem

Authors: Zhuang Shan; Lijun Zhu; Long He; Danfeng Wu; Haicheng Wei
Journal: Mathematical Problems in Engineering
 (2020)

---

## Abstract

This paper focuses on the problem of variational inequalities with monotone operators in real Hilbert space. The Tseng algorithm constructed by Thong replaced a high-precision step. Thus, a new Tseng-like gradient method is constructed, and the convergence of the algorithm is proved, and the convergence performance is higher.


---

## Body

## 1. Introduction

LetH be the real Hilbert space, and ⋅,⋅ and ⋅ are respectively defined as the inner product and norm of space H. Let C be a nonempty closed convex subset of space H. Let T: H⟶H be an operator on the set C. The variational inequality problem (VIP) means to find a point x∗ on C such that(1)Tx∗,x−x∗≥0,∀x∈C.We useVIT,C to denote the solution set of problem (1). At present, the regular method and the projection method are the two main methods to solve the problem of variational inequality. Variational inequalities can also solve other problems such as equilibrium, fixed points, and optimization. In addition, it is also widely used in the industry, especially supply chain management and transportation. Supply chain management is a new type of management concept, which emphasizes the rapid market demand response, combat readiness management, high flexibility, low risk, cost-effectiveness, and other goals, attracting many theoretical and practical industry people to study and practice it. Some well-known international companies, such as Hewlett-Packard and Dell, have made great achievements in the practice of supply chain management. Therefore, the supply chain is an effective way for enterprises to adapt to global competition in the 21st century (see eg., [1–11].At present, the theoretical support of the variational inequality problem is gradually maturing, and people are gradually changing from the theoretical research of the problem to the construction of the variational inequality algorithm. In recent years, the variational inequality algorithm has developed rapidly, but the use conditions are harsh. The use of the Tseng algorithm does not require too harsh conditions, such as the strongly monotone or inverse strongly monotone of the problem and only one projection equation is used to approximate the solution of the problem. The specific method is as follows:(2)yn=PCxn−ρTxn,xn+1=yn−ρTyn−Txn.Thereρ∈0,1/L and PC is a projection operator in set C of Hilbert space H. In recent years, the Tseng algorithm has been widely recognized. Subsequently, Duong Viet and others improved the Tseng method. For the specific method, please see Lemma 7 [12].Although the Tseng algorithm constructed by Duong Viet et al. has many advantages, the shortcoming is that the step size is too simple. The work of this paper is to improve the convergence of the algorithm by optimizing the step size, making it more precise and practical on the basis of convergence. The step size adopted in this article is currently commonly used, and it has high accuracy itself, which can effectively improve the convergence speed of the algorithm. In addition, this paper popularizes the application field of the algorithm to make it more applicable.The structure of this paper is as follows: In Section2, we give some necessary definitions and lemmas. In Section 3, we give an algorithm for construction. We analyze and prove that the algorithm weakly converges to a certain point in VI (C, T) and give some inferences.


## 2. Preliminaries

LetH be a real Hilbert space and C be a nonempty closed convex subset of H. As we all know, the following equation can be directly proved in Hilbert space.When∀x∈H, y∈H, and α∈R, we can get(3)x+y2=x2+2x,y+y2;αx+1−αy2=αx2+1−αy2−α1−αx−y2.Definition 1.
LetT be a operator on a Hilbert space, and then the operator T is a nonexpansive operator, if the operator T has the property(4)Tx−Ty≤cx−y,for  allx,y∈C,and for some constant c∈0,1.
LetPC be a projection operator in set C of Hilbert space H, then ∀x∈H. There is only one point PCx such that PCx−y≤x−y, ∀y∈C. It is easy to prove that PC is a nonexpansive operator.Lemma 1.
(see [13]). Let H be a real Hilbert space and C be a nonempty closed convex subset of H. For a given x∈H, z∈C. Then, ∀y∈C has the following relationship:(5)z=PCx⟺x−z,z−y≥0.Lemma 2.
(see [13]). Let H be a real Hilbert space and C be a nonempty closed convex subset of H, x∈H, then ∀y∈C. The following equation holds(i)
PCx−PCy2≤PCx−PCy,x−y,(ii)
PC,x−y2≤x−y2+PCx−x2.If the reader wants to know more about the nature of projections, please refer to part 3 in [13].Definition 2.
LetT: H⟶H, if ∀x,y∈H satisfies(6)Tx−Ty≤Sx−y,S>0.
Then, the operatorT is called S-Lipschitz continuous;(7)Tx−Ty,x−y≥0,
Then, the operatorT is called monotonic.Lemma 3.
(see [14]). Let φn, δn, and αn be sequences in 0,+∞ such that(8)φn+1≤φn+αnφn−φn−1,∀n≥1,∑n=1+∞δn<+∞,and there is a real number α, for any n∈N, such that 0≤αn≤α<1. Then,(i)
∑n=1+∞φn−φn−1+<+∞,where φn−φn−1+=maxφn−φn−1,0.(i)
There is a pointφ∗∈0,+∞ such that limn⟶+∞φn=φ∗.Lemma 4.
(Opial1967). Let ∅≠C⊂H and xn be the sequence in H such that the following two conditions hold:(i)
∀x∈C,limn⟶+∞xn−xexist;(ii)
Every point of weakly gathering in the sequencexn is in C.Then,xn weakly converges to a point x in C.Lemma 5.
(minty, please refer to reference [15], Lemma 7.1.7). Let T: C⟶H be a monotone continuous operator. Then, x∗ is the solution to problem (1) if and only if x∗ is a solution to the following problem:(9)findx∈Csuch thatTy,y−x≥0,∀y∈C.Remark 1.
The solution setVIC,T of variational inequality (1) is closed and convex.Lemma 6.
(see [15]). Let T:H⟶H be monotone and Lipschitz continuous on H with the constant S and VIC,T≠∅, ρ be a positive obtained real number such that ρ∈0,1/S. If ηn is a nonincreasing sequence, make 0≤ηn≤η and(10)η<1+8K−1−2K21−2K,where K:=1−ρS/1+ρS, then the sequence xn obtained by next algorithm weakly converges to VIC,T.(i)
Step 0: givenρ>0; Let x0,x1∈H be arbitrary(ii)
Step 1: setωn=xn+ηnxn−xn−1 and compute(11)yn=PCωn−ρTωn.
Whenyn=ωn, then stop iteration, and yn is the solution to the (VIP) problem; otherwise,(iii)
Step 2: compute(12)xn+1=yn−ρTyn−Tωn.Setn=n+1 and go to Step 1.Lemma 7.
(see [15]). Suppose that ρ∈0,1/S. Let xn be a sequence in H defined by(13)x0∈H,yn=PCxn−ρTxn,xn+1=yn−PCρTyn−ρTxn.Then, the sequencexn converges weakly to an element of VIC,T.


## 3. Main Results

LetT:H⟶H be monotone and Lipschitz continuous on H with the constant S and VIC,T≠∅. First, we introduce Algorithm 1.Algorithm 1: New Tseng-degree gradient algorithm.
(i)
Step 0: givenρ0>0. Let x0,x1∈H be arbitrary and μ∈1/2,1(ii)
Step 1: setωn=xn+ηnxn−xn−1 and compute yn=PCωn−ρnTωn(iii)
Whenyn=ωn, then stop iteration, and yn is the solution to the (VIP) problem; otherwise,(iv)
Step 2: computexn+1=yn−ρnTyn−Tωn(v)
andρn+1=minμyn−ωn/Tyn−Tωn,ρn,Tyn−Tωn≠0,ρn,other(vi)
Setn=n+1 and go to Step 1Lemma 8.
Letρn be a sequence obtained by Algorithm 1, then the sequence xn decreases monotonically and its lower bound is minμ/S,ρ0, where S >0 is the Lipschitz condition constant.Proof.
According to the definition, the sequenceρn is monotonically nonincreasing, because T is a Lipschitz condition function. At that time, Tyn−Tωn≠0, and we have(14)μyn−ωnTyn−Tωn≥μyn−ωnSyn−ωn=μS.
Obviously, the sequenceρn has a lower bound minμ/S,ρ0.Theorem 1.
Let the sequencexn be obtained by Algorithm 1, then ∀p∈VIC,T have(15)xn+1−p2≤ωn−p2−1−ρn2μ2ρn+12yn−ωn2.
Proof(16)xn+1−p2=yn−ρnTyn−Tωn−p2=yn−p2+ρn2Tyn−Tωn2−2ρnyn−p,Tyn−Tωn=ωn−p2+ωn−yn2+2yn−p,Tyn−Tωn+ρn2Tyn−Tωn2−2ρnyn−p,Tyn−Tωn=ωn−p2+ωn−yn2−2yn−ωn,yn−ωn+2yn−ωn,yn−p+ρn2Tyn−Tωn2−2ρnyn−p,Tyn−Tωn=ωn−p2−‖ωn−yn‖2+2yn−ωn,yn−p+ρn2Tyn−Tωn2−2ρnyn−p,Tyn−Tωn.
Sinceyn=PCωn−ρnTωn, so(17)yn−ωn+ρnTωn,yn−p≤0,or(18)yn−ωn,yn−p≤−ρnTωn,yn−p,can be obtained from (16) and (18):(19)xn+1−p2≤ωn−p2−ωn−yn2−2ρnTωn,yn−p+ρn2Tyn−Tωn2−2ρnyn−p,Tyn−Tωn=ωn−p2−ωn−yn2+ρn2Tyn−Tωn2−2ρnyn−p,Tyn≤ωn−p2−ωn−yn2+ρn2μ2ρn+12yn−ωn2−2ρnyn−p,Tyn−Tp−2ρnyn−p,Tp≤ωn−p2−1−ρn2μ2ρn+12yn−ωn2.Theorem 2.
Letρn be a sequence generated by Algorithm 1. If ηn is a nonincreasing sequence, make 0≤ηn≤η and(20)η<1+8K−1−2K21−2K,where K:=1−μ2, then the sequence xn obtained by Algorithm 1 weakly converges to VIC,T.Proof.
The first step.
LetKn:=1−ρn2μ2/ρn+12/1+ρn2μ2/ρn+12≤1−ρn2μ2/ρn+12≤1−μ2=K.
Since1+8Kn−1−2Kn/21−2Kn monotonically increases in 0,1/2; ρn monotonically decreases, so ρn2/ρn+12≥1, and then there η<1+8Kn−1−2Kn/21−2Kn<1+8K−1−2K/21−2Kn.
From the second step, through the construction ofxn+1, we can get(21)xn+1−yn=yn−ρnTyn−Tωn−yn≤ρnTyn−Tωn≤ρnμρn+1yn−ωn.
So,(22)xn+1−ωn≤xn+1−yn+yn−ωn≤1+ρnμρn+1yn−ωn,thus having(23)yn−ωn≥11+ρnμ/ρn+1xn+1−ωn.
Letp∈VIC,T, through Theorem 1, and we can get(24)xn+1−p2≤ωn−p2−1−ρn2μ2ρn+12yn−ωn2.
From (23) and (24), we can get(25)xn+1−p2≤ωn−p2−1−ρn2μ2/ρn+121+ρn2μ2/ρn+12xn+1−ωn2=ωn−p2−1−ρnμ/ρn+11+ρnμ/ρn+1xn+1−ωn≤ωn−p2−Kxn+1−ωn2.
Through the construction ofωn, we can get(26)ωn−p2=xn+ηnxn−xn−1−p=1+ηnxn−p−ηnxn−1−p≤1+ηnxn−p2−ηnxn−1−p2+ηn1+ηnxn−xn−12.
From (25) and (26), we can get(27)xn+1−p2≤1+ηnxn−p2−ηnxn−1−p2+ηn1+ηnxn−xn−12≤1+ηnxn−p2−ηnxn−1−p2+2ηxn−xn−12.
On the other hand,(28)xn+1−ωn2=xn+1−xn−ηnxn−xn−12=xn+1−xn2+ηn2xn−xn−12−2ηnxn+1−xn,xn−xn−1≥xn+1−xn2+ηn2xn−xn−12−2ηnxn+1−xnxn−xn−1≥1−ηnxn+1−xn2+ηn2−ηnxn−xn−12.
From (25), (26), and (28), we get(29)xn+1−p2≤1+ηnxn−p2−ηnxn−1−p2+ηn1+ηnxn−xn−12−K1+ηnxn+1−xn2−Kηn2−ηnxn−xn−12=1+ηnxn−p2−ηnxn−1−p2−K1−ηnxn+1−xn2+ηn1+ηn−Kηn2−ηnxn−xn−12=1+ηnxn−p2−ηnxn−1−p2−γnxn+1−xn2+μnxn−xn−12,where γn:=K1−ηn and μn:=ηn1+ηn−Kηn2−ηn≥0.
LetΓn:=xn−p2−ηnxn−1−p2+μnxn−xn−12. From (29), we can obtain(30)Γn+1−Γn=xn+1−p2−1+ηn+1xn−p2+ηnxn−1−p2+μn+1xn+1−xn2−μnxn−xn−12≤xn+1−p2−1+ηnxn−p2+ηnxn−1−p2+μn+1xn+1−xn2−μnxn−xn−12≤−γn−μn+1xn+1−xn2.
Because0≤ηn≤ηn+1≤η, so(31)γn−μn+1=K1−ηn−ηn+11+ηn+1+Kηn+12−ηn+1≥K1+ηn+1−ηn+11+ηn+1+Kηn+12−ηn+1≥K1−ηn−η1+η+Kη2−η≥−1−Kη2−1+2Kη+K.
Combining (30) and (31), we get(32)Γn+1−Γn≤−δxn+1−xn2,where δ:=−1−Kη2−1+2Kη+K, δ>0 can be obtained from (20).
Easy to prove(33)Γn+1−Γn≤0.
Therefore, the sequenceΓn is nonincreasing.
On the other hand, due toμn≥0, there is(34)Γn=xn−p2−ηnxn−1−p2+μnxn−xn−12≥xn−p2−ηnxn−1−p2.
This means(35)xn−p2≤ηnxn−1−p2+Γn≤ηxn−1−p2+Γ1≤⋯≤ηnx0−p2+Γ11+⋯+ηn−1≤ηnx0−p2+Γ11−η.
Similarly,(36)Γn+1=xn+1−p2−ηn+1xn−p2+μn+1xn+1−xn2≥xn+1−p2−ηn+1xn−p2≥−ηn+1xn−p2.
From (35) and (36), we can see that(37)−Γn+1≤ηn+1xn−p2≤ηxn−p2≤ηn+1x0−p2+ηΓ11−η.
According to (32), we get the following inequality:(38)δ∑n=1kxn+1−xn2≤Γ1−Γk+1≤ηk+1x0−p2+Γ11−η≤x0−p2+Γ11−η.
In other words,∑n=1+∞xn+1−xn2<+∞, xn+1−xn2⟶0 can be obtained from the convergence series property, that is, xn+1−xn⟶0. So,(39)xn+1−ωn2=xn+1−xn2+ηn2xn−xn−12−2ηnxn+1−xn,xn−xn−1.
So,xn+1−ωn⟶0. By (27) and Lemma 3, we have(40)limn⟶+∞xn−p2=l,and from (26), we can get(41)ωn−p2=xn−p2+ηnxn−p2−xn−1−p2+ηn1+ηnxn−xn−12.
According to the sequence,ηn is bounded, so(42)limn⟶+∞ωn−p2=l.
From this, we get thatxn, ωn, and zn are all bounded sequences, and then there(43)0≤xn−ωn≤xn−xn+1+xn+1−ωn⟶0,n⟶+∞.
Now, we prove that the sequencexn weakly converges to p∈VIC,T. In fact, because the sequence xn is bounded, we assume that there is a subset of the sequence xn, which makes it weakly converge to z∈H. Without loss of generality, we still use xn below to represent the subset. That is, xn⇀z, where “⇀” represents weak convergence.
Because ofxn−ωn⟶0, there is a subset to make it weakly converge to z. We still use ωn below to represent the sequence, which is ωn⇀z. Similarly, the following uses yn to represent the subset of the original number sequence yn. Since T is a monotone operator and yn=PCωn−ρnTωn, ∀x∈C, there is(44)0≤yn−ωn+ρnTωn,x−yn=yn−ωn,x−yn+ρnTωn,x−yn=yn−ωn,x−yn+ρnTωn,x−ωn+ρnTωn,ωn−yn≤yn−ωn,x−yn+ρnTωn,x−ωn+ρnTx,ωn−yn.
Letn⟶+∞, then ∀x∈C has(45)Tx,x−z≥0.
By Lemma4, we can get z∈VIC,T, so we have proved the following:(a)
∀p∈VIC,T, then limn⟶∞xn−p2 limits exist;(b)
Each weak convergence point of the logarithmic sequencexn is in VIC,T.
From Lemma3, we know that the sequence xn weakly converges to p∈VIC,T.Corollary 1.
Givenρ0>0 , suppose that xn be a sequence in H defined by(46)x0,x1∈H,μ∈12,1,yn=PCxn−ρnTxn,xn+1=yn−PCρnTyn−ρnTxn,where(47)ρn+1=minμyn−ωnTyn−Tωn,ρn,Tyn−Tωn≠0,ρn,other.Then, the sequencexn converges weakly to an element of ∈VIC,T.Lemma 9.
Letπ>0. A point x∈C is a solution of VIC,T if and only if x∈FixPCId−πT (see [16]).According to Theorem2 and Lemma 9, we get the following two inferences.Corollary 2.
Letρn be a sequence obtained by Algorithm 1, and π>0 and FixPCId−πT≠∅. If ηn is a nonincreasing sequence, make 0≤ηn≤η and η<1+8K−1−2K/21−2K, where K:=1−μ2, then the sequence xn obtained by Algorithm 1 weakly converges to FixPCId−πT.Corollary 3.
Givenρ0>0, suppose that xn be a sequence in H defined by(48)x0,x1∈H,μ∈12,1,yn=PCxn−ρnTxn,xn+1=yn−PCρnTyn−ρnTxn,where(49)ρn+1=minμyn−ωnTyn−Tωn,ρn,Tyn−Tωn≠0,ρn,other.Then, the sequencexn converges weakly to an element of FixPCId−πT.The new Tseng-like gradient method is convergent. It is known that the algorithm has good accuracy due to the selection of step size. Finally, it also can be known that the algorithm has good applicability through inference.



---

*Source: 9832579-2020-08-05.xml*