# Iterative Computation for Solving the Variational Inequality and the Generalized Equilibrium Problem

Authors: Xiujuan Pan; Shin Min Kang; Young Chel Kwun
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

An iterative algorithm for solving the variational inequality and the generalized equilibrium problem has been introduced. Convergence result is given.


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## Body

## 1. Introduction

LetH be a real Hilbert space with inner product 〈·,·〉 and norm ∥·∥, respectively. Let C be a nonempty closed convex subset of H. Recall that a mapping A:C→H is said to be(i)
nonexpansive⇔∥Au-Av∥≤∥u-v∥, for all u,v∈C (we use Fix⁡(A) to denote the set of the fixed points of A);(ii)
firmly nonexpansive⇔∥Au-Av∥2≤〈u-v,Au-Av〉, for all u,v∈C;(iii)
L-Lipschitz ⇔ there exists a constant L>0 such that ∥Au-Av∥≤L∥u-v∥, for all u,v∈C;(iv)
monotone⇔〈u-v,Au-Av〉≥0, for all u,v∈C;(v)
strongly monotone⇔ there exists a constant ν>0 such that 〈u-v,Au-Av〉≥ν∥u-v∥2, for all u,v∈C;(vi)
inverse strongly monotone⇔ there exists ζ>0 such that 〈u-v,Au-Av〉≥ζ∥Au-Av∥2, for all u,v∈C;(vii)
ζ-inverse strongly ϕ-monotone ⇔〈ϕ(u)-ϕ(v),Au-Av〉≥ζ∥Au-Av∥2, for all u,v∈C and for some ζ>0, where ϕ:C→C is a nonlinear mapping.IfA is a multivalued mapping of H into 2H, then A is said to be a monotone operator on H⇔〈x-y,u-v〉≥0, for all x,y∈dom⁡(A), u∈Ax, and v∈Ay. A monotone operator A on H is said to be maximal if and only if its graph is not strictly contained in the graph of any other monotone operator on H.LetA,B, and ϕ be three nonlinear mappings on C. Let θ:C×C→R be a bifunction. Recall that the equilibrium problem is to find x†∈C such that
(1)θ(x†,y)+〈Ax†,y-x†〉≥0,∀y∈C.
The solution set of (1) is denoted by EP(θ,A). Now we know that the equilibrium theory provides us a natural, novel, and unified framework to study a wide class of problems arising in economics, finance, transportation, network and structural analysis, elasticity, and optimization. For related works, please refer to [1–3] and the references therein.Recall also that the variational inequality problem is to findu∈C,ϕ(u)∈C such that
(2)〈Bu,ϕ(v)-ϕ(u)〉≥0,∀ϕ(v)∈C.
The solution set of (2) is denoted by VI(B,ϕ,C). It is well known that variational inequality theory has emerged as an important tool in studying a wide class of obstacle, unilateral, free, moving, and equilibrium problems arising in several branches of pure and applied sciences in a unified and general framework. Several numerical methods have been developed for solving variational inequalities and related optimization problems. For related works, please see [4–14]. Noor [15] introduced an iterative scheme and studied the approximate solutions of variational inclusion in Hilbert spaces. Glowinski and Le Tallec [16] used the iterative schemes to find the approximate solutions of the elastoviscoplasticity problem, liquid crystal theory, and eigenvalue computation. In 1998, Haubruge et al. [17] studied the convergence analysis of the iterative schemes of Glowinski and Le Tallec and applied these schemes to obtain new splitting-type algorithms for solving variational inequalities, separable convex programming, and minimization of a sum of convex functions.Our main purpose in the present paper is to solve the following equilibrium problem and variational inequality problem: finding a pointx† such that
(3)x†∈VI(B,ϕ,C),ϕ(x†)∈EP(θ,A).
Our main motivations are inspired by the following two reasons.Firstly, it is still an interesting topic for solving the variational inequality problem and the equilibrium problem based on their applications in science and engineering. Secondly, the split problem of finding a pointx~ such that
(4)x†∈C,ϕ~(x†)∈D
has received much attention. For related works, please refer to [18–20]. However, we observe that the involved operator ϕ~ in (4) is a bounded liner operator. In this paper, we devote to study the problem (3), where the transformation ϕ is a nonlinear mapping. For this purpose, we introduce a new iterative algorithm. Consequently, strong convergence analysis is demonstrated.


## 2. Preliminaries

In this section, we recall some useful lemmas.Recall that the metric projectionprojC:H→C satisfies ∥u-projCu∥=inf⁡{∥u-v∥:v∈C}. The metric projection projC is a typical firmly nonexpansive mapping. The characteristic inequality of the projection is 〈u-projCu,v-projCu〉≤0, for all u∈H, v∈C.Assume thatθ:C×C→R is a bifunction which satisfies the following conditions:(C1)
θ(u,u)=0, for all u∈C;(C2)
θ is monotone; that is, θ(u,v)+θ(v,u)≤0, for all u,v∈C;(C3)
for eachu,v,w∈C, lim⁡t↓0θ(tw+(1-t)u,v)≤θ(u,v);(C4)
for eachu∈C, v↦θ(u,v) is convex and lower semicontinuous.Lemma 1 (see [2]).
LetC be a nonempty closed convex subset of a real Hilbert space H. Let θ:C×C→R be a bifunction which satisfies conditions (C1)–(C4). Let τ>0 and u∈C. Then, there exists w∈C such that
(5)θ(w,v)+1τ〈v-w,w-u〉≥0,∀v∈C.
Further, if Sτ(u)={w∈C:θ(w,v)+(1/τ)〈v-w,w-u〉≥0, for all v∈C}, then the following hold:(a)
Sτ is single-valued and Sτ is firmly nonexpansive;(b)
EP(θ) is closed and convex and EP(θ)=Fix⁡(Sτ).Lemma 2 (see [21]).
Let{xn} and {yn} be bounded sequences in a Banach space X and let {ηn} be a sequence in [0,1] with 0<liminf⁡n→∞ηn≤limsup⁡n→∞ηn<1. Suppose that xn+1=(1-ηn)yn+ηnxn, for all n≥0, and limsup⁡n→∞(∥yn+1-yn∥-∥xn+1-xn∥)≤0. Then, lim⁡n→∞∥yn-xn∥=0.Lemma 3 (see [22]).
Assume that the sequence{an} satisfies an≥0 and an+1≤(1-νn)an+ςnνn, where {νn} is a sequence in (0,1) and {ςn} is a sequence such that ∑n=1∞‍νn=∞ and limsup⁡n→∞ςn≤0(or ∑n=1∞‍|ςnνn|<∞). Then, lim⁡n→∞an=0.


## 3. Main Results

LetC be a nonempty closed convex subset of a real Hilbert space H. Let A:C→H be an η-inverse strongly monotone mapping. Let ϕ:C→C be a weakly continuous and ν-strongly monotone mapping such that R(ϕ)=C. Let B:C→H be a ζ-inverse strongly ϕ-monotone mapping. Let ϱ:C→H be an L-Lipschitz continuous mapping. Let θ:C×C→R be a bifunction which satisfies (C1)–(C4) in the above section. Let {ζn}⊂[0,1], {ηn}⊂[0,1], {ϖn}⊂(0,∞), and {τn}⊂(0,∞) be four real number sequences and let ς>0 be a constant.We useΔ to denote the solution set of (3). In order to solve (3), we introduce the following three-step algorithm.Algorithm 4.
Letx0∈C be an initial guess. Define the sequence {xn} as follows:
(6)un=projC[ϕ(xn)-ϖnBxn],n≥0,θ(zn,y)+〈Aun,y-zn〉+1τn〈y-zn,zn-un〉≥0,∀y∈C,ϕ(xn+1)=ηnϕ(xn)+(1-ηn)projC[ζnςϱ(xn)+(1-ζn)zn],n≥0.Theorem 5.
Suppose thatΔ≠∅. Assume that the following conditions are satisfied:(r1)
ϖn∈(a1,a2)⊂(0,2ζ) and lim⁡n→∞(ϖn+1-ϖn)=0;(r2)
τn∈(a3,a4)⊂(0,2η) and lim⁡n→∞(τn+1-τn)=0;(r3)
ηn∈[a5,a6]⊂(0,1);(r4)
lim⁡n→∞ζn=0 and ∑n‍ζn=∞;(r5)
ν∈(Lς,2ζ).
Then, the sequence{xn} generated by (6) converges strongly to x*∈Δ which solves the following variational inequality:
(7)〈ςϱ(x*)-ϕ(x*),ϕ(x)-ϕ(x*)〉≤0,∀x∈Δ.Proof.
First of all, we prove that the solution of the variational inequality (7) is unique. In fact, if x~∈Δ also solves (7), then we get
(8)〈ςϱ(x*)-ϕ(x*),ϕ(x~)-ϕ(x*)〉≤0,〈ςϱ(x~)-ϕ(x~),ϕ(x*)-ϕ(x~)〉≤0.
It follows that
(9)〈ςϱ(x~)-ϕ(x~)-ςϱ(x*)+ϕ(x*),ϕ(x*)-ϕ(x~)〉≤0.
So,
(10)∥ϕ(x*)-ϕ(x~)∥2≤ς〈ϱ(x*)-ϱ(x~),ϕ(x*)-ϕ(x~)〉≤ς∥ϱ(x*)-ϱ(x~)∥∥ϕ(x*)-ϕ(x~)∥,
which implies that
(11)∥ϕ(x*)-ϕ(x~)∥≤ς∥ϱ(x*)-ϱ(x~)∥.
Since ϕ is ν-strongly monotone, we have
(12)ν∥x*-x~∥2≤〈ϕ(x*)-ϕ(x~),x*-x~〉≤∥ϕ(x*)-ϕ(x~)∥∥x*-x~∥.
Thus,
(13)ν∥x*-x~∥≤∥ϕ(x*)-ϕ(x~)∥≤ς∥ϱ(x*)-ϱ(x~)∥≤ςL∥x*-x~∥.
Since ςL<ν, we deduce the contradiction. Therefore, x*=x~. So, the solution of variational inequality (7) is unique.
Letx#∈Δ. Hence, x#∈VI(B,ϕ,C) and ϕ(x#)∈EP(θ,A). Note that
(14)x#∈VI(B,ϕ,C)⟺ϕ(x#)=projC(ϕ(x#)-νBx#),∀ν>0.
Since ϖn>0, we have ϕ(x#)=projC[ϕ(x#)-ϖnBx#], for all n≥0. For u,v∈C, we have
(15)∥(ϕ(u)-ϖnBu)-(ϕ(v)-ϖnBv)∥2=∥ϕ(u)-ϕ(v)∥2-2ϖn〈Bu-Bv,ϕ(u)-ϕ(v)〉+ϖn2∥Bu-Bv∥2≤∥ϕ(u)-ϕ(v)∥2-2ϖnζ∥Bu-Bv∥2+ϖn2∥Bu-Bv∥2≤∥ϕ(u)-ϕ(v)∥2+ϖn(ϖn-2ζ)∥Bu-Bv∥2.
Hence,
(16)∥(ϕ(u)-ϖnBu)-(ϕ(v)-ϖnBv)∥≤∥ϕ(u)-ϕ(v)∥,
for all u,v∈C. Thus,
(17)∥un-ϕ(x#)∥=∥projC[ϕ(xn)-ϖnBxn]-projC[ϕ(x#)-ϖnBx#]∥≤∥(ϕ(xn)-ϖnBxn)-(ϕ(x#)-ϖnBx#)∥≤∥ϕ(xn)-ϕ(x#)∥.
From (6), we have zn=Sτn(I-τnA)un, for all n≥0. Noting that ϕ(x#)∈EP(θ,A), we deduce ϕ(x#)=Sτn(I-τnA)ϕ(x#), for all n≥0. It follows from (17) that
(18)∥ϕ(xn+1)-ϕ(x#)∥≤ηn∥ϕ(xn)-ϕ(x#)∥+(1-ηn)∥projC[ζnςϱ(xn)+(1-ζn)zn]-ϕ(x#)∥≤ηn∥ϕ(xn)-ϕ(x#)∥+(1-ηn)∥ζn(ςϱ(xn)-ϕ(x#))+(1-ζn)(zn-ϕ(x#))∥≤ηn∥ϕ(xn)-ϕ(x#)∥+(1-ηn)[ζn∥ςϱ(xn)-ςϱ(x#)∥+ζn∥ςϱ(x#)-ϕ(x#)∥+(1-ζn)×∥Sτn(I-τnA)un-Sτn(I-τnA)ϕ(x#)∥ζn∥ςϱ(xn)-ςϱ(x#)∥+ζn∥ςϱ(x#)-ϕ(x#)∥]≤ηn∥ϕ(xn)-ϕ(x#)∥+(1-ηn)[ζnςL∥xn-x#∥+ζn∥ςϱ(x#)-ϕ(x#)∥+(1-ζn)∥un-ϕ(x#)∥]≤ηn∥ϕ(xn)-ϕ(x#)∥+(1-ηn)[ζnςLν∥ϕ(xn)-ϕ(x#)∥+ζn∥ςϱ(x#)-ϕ(x#)∥+(1-ζn)∥ϕ(xn)-ϕ(x#)∥ζnςLν∥ϕ(xn)-ϕ(x#)∥+ζn∥ςϱ(x#)-ϕ(x#)∥]=[1-(1-ςLν)(1-ηn)ζn]∥ϕ(xn)-ϕ(x#)∥+(1-ςLν)(1-ηn)ζn∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν.
An induction implies that
(19)∥ϕ(xn)-ϕ(x#)∥≤max⁡{∥ϕ(x0)-ϕ(x#)∥,∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν}.
Hence, {ϕ(xn)} is bounded. Since ϕ is ν-strongly monotone, we can deduce ν∥xn-x#∥≤∥ϕ(xn)-ϕ(x#)∥. So,
(20)∥xn-x#∥≤1ν∥ϕ(xn)-ϕ(x#)∥≤1νmax⁡{∥ϕ(x0)-ϕ(x#)∥,∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν}.
This implies that {xn} is bounded.
From (6), we have
(21)θ(zn,y)+1τn〈y-zn,zn-(un-τnAun)〉≥0,∀y∈C.
So,
(22)θ(zn,zn+1)+1τn〈zn+1-zn,zn-(un-τnAun)〉≥0.
Similarly,
(23)θ(zn+1,zn)+1τn+1〈zn-zn+1,zn+1-(un+1-τn+1Aun+1)〉≥0.
Hence,
(24)θ(zn,zn+1)+θ(zn+1,zn)+〈Aun-Aun+1,zn+1-zn〉+〈zn+1-zn,zn-unτn-zn+1-un+1τn+1〉≥0.
Since θ is monotone, we have
(25)θ(zn,zn+1)+θ(zn+1,zn)≤0.
So,
(26)〈Aun-Aun+1,zn+1-zn〉+〈zn+1-zn,zn-unτn-zn+1-un+1τn+1〉≥0.
Thus,
(27)τn〈Aun-Aun+1,zn+1-zn〉+〈zn+1-zn,zn-zn+1+zn+1-un-τnτn+1(zn+1-un+1)〉≥0.
It follows that
(28)∥zn+1-zn∥2≤τn〈Aun-Aun+1,zn+1-zn〉+〈zn+1-zn,un+1-un+(1-τnτn+1)(zn+1-un+1)〉=〈(I-τnA)un+1-(I-τnA)un,zn+1-zn〉+〈zn+1-zn,(1-τnτn+1)(zn+1-un+1)〉≤∥(I-τnA)un+1-(I-τnA)un∥∥zn+1-zn∥+|1-τnτn+1|∥zn+1-zn∥∥zn+1-un+1∥≤∥zn+1-zn∥(∥un+1-un∥+|1-τnτn+1|∥zn+1-un+1∥)
and hence
(29)∥zn+1-zn∥≤∥un+1-un∥+|1-τnτn+1|∥zn+1-un+1∥≤∥un+1-un∥+1a3|τn+1-τn|∥zn+1-un+1∥.
By (6) and (16), we have
(30)∥un+1-un∥=∥projC[ϕ(xn+1)-ϖn+1Bxn+1]-projC[ϕ(xn)-ϖnBxn]∥≤∥[ϕ(xn+1)-ϖn+1Bxn+1]-[ϕ(xn)-ϖnBxn]∥≤∥ϕ(xn+1)-ϖn+1Bxn+1-(ϕ(xn)-ϖn+1Bxn)∥+|ϖn+1-ϖn|∥B(xn)∥≤∥ϕ(xn+1)-ϕ(xn)∥+|ϖn+1-ϖn|∥B(xn)∥.
Therefore,
(31)∥zn+1-zn∥≤∥ϕ(xn+1)-ϕ(xn)∥+|ϖn+1-ϖn|∥B(xn)∥+1a3|τn+1-τn|∥zn+1-un+1∥.
It follows that
(32)∥zn+1-zn∥-∥ϕ(xn+1)-ϕ(xn)∥≤|ϖn+1-ϖn|∥B(xn)∥+1a3|τn+1-τn|∥zn+1-un+1∥.
Since lim⁡n→∞(ϖn+1-ϖn)=0, lim⁡n→∞(τn+1-τn)=0, and the sequences {ϱ(xn)},{ϕ(xn)}, {zn}, {un}, and {Bxn} are bounded, we have
(33)limsup⁡n→∞(∥zn+1-zn∥-∥ϕ(xn+1)-ϕ(xn)∥)≤0.
From Lemma 2, we obtain
(34)lim⁡n→∞∥zn-ϕ(xn)∥=0.
Note that
(35)∥ϕ(xn+1)-ϕ(xn)∥≤(1-ηn)ζn∥ςϱ(xn)-ϕ(xn)∥+(1-ηn)(1-ζn)∥zn-ϕ(xn)∥.
Hence,
(36)lim⁡n→∞∥ϕ(xn+1)-ϕ(xn)∥=0.
This together with the ν-strong monotonicity of ϕ implies that
(37)lim⁡n→∞∥xn+1-xn∥=0.
From (18), we have
(38)∥ϕ(xn+1)-ϕ(x#)∥≤ηn∥ϕ(xn)-ϕ(x#)∥+(1-ηn)[ζnςL∥xn-x#∥+ζn∥ςϱ(x#)-ϕ(x#)∥+(1-ζn)∥un-ϕ(x#)∥]≤ηn∥ϕ(xn)-ϕ(x#)∥+(1-ηn)[ζnςLν∥ϕ(xn)-ϕ(x#)∥+ζn∥ςϱ(x#)-ϕ(x#)∥+(1-ζn)∥un-ϕ(x#)∥ζnςLν∥ϕ(xn)-ϕ(x#)∥+ζn∥ςϱ(x#)-ϕ(x#)∥]=[1-(1-ηn)(1-ςLζnν)]∥ϕ(xn)-ϕ(x#)∥+(1-ζn)(1-ηn)∥un-ϕ(x#)∥+(1-ηn)ζn(1-ςLν)∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν.
By the convexity of the norm and (17), we have
(39)∥ϕ(xn+1)-ϕ(x#)∥2≤[1-(1-ηn)(1-ςLζnν)]∥ϕ(xn)-ϕ(x#)∥2+(1-ζn)(1-ηn)∥un-ϕ(x#)∥2+(1-ηn)ζn(1-ςLν)(∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν)2≤[1-(1-ηn)(1-ςLζnν)]∥ϕ(xn)-ϕ(x#)∥2+(1-ζn)(1-ηn)×∥(ϕ(xn)-ϖnBxn)-(ϕ(x#)-ϖnBx#)∥2+(1-ηn)ζn(1-ςLν)(∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν)2≤[1-(1-ηn)(1-ςLζnν)]∥ϕ(xn)-ϕ(x#)∥2+(1-ζn)(1-ηn)×(∥ϕ(xn)-ϕ(x#)∥2+ϖn(ϖn-2ζ)∥Bxn-Bx#∥2)+(1-ηn)ζn(1-ςLν)(∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν)2≤∥ϕ(xn)-ϕ(x#)∥2+(1-ζn)(1-ηn)ϖn(ϖn-2ζ)∥Bxn-Bx#∥2+(1-ηn)ζn(1-ςLν)(∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν)2.
So,
(40)(1-ηn)(1-ζn)ϖn(2ζ-ϖn)∥Bxn-Bx#∥2≤∥ϕ(xn)-ϕ(x#)∥2-∥ϕ(xn+1)-ϕ(x#)∥2+(1-ηn)ζn(1-ςLν)(∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν)2≤(∥ϕ(xn)-ϕ(x#)∥+∥ϕ(xn+1)-ϕ(x#)∥)×∥ϕ(xn+1)-ϕ(xn)∥+(1-ηn)ζn(1-ςLν)(∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν)2.
Since ζn→0, ∥ϕ(xn+1)-ϕ(xn)∥→0, and liminf⁡n→∞(1-ηn)(1-ζn)ϖn(2ζ-ϖn)>0, we obtain
(41)lim⁡n→∞∥Bxn-Bx#∥=0.
Note that
(42)∥un-ϕ(x#)∥2=∥projC[ϕ(xn)-ϖnBxn]-projC[ϕ(x#)-ϖnBx#]∥2≤〈ϕ(xn)-ϖnBxn-(ϕ(x#)-ϖnBx#),un-ϕ(x#)〉=12{∥ϕ(xn)-ϖnBxn-(ϕ(x#)-ϖnBx#)∥2+∥un-ϕ(x#)∥2-∥ϕ(xn)-ϖnBxn-(ϕ(x#)-ϖnBx#)-un+ϕ(x#)∥2}≤12{∥ϕ(xn)-ϕ(x#)∥2+∥un-ϕ(x#)∥2-∥ϕ(xn)-un-ϖn(Bxn-Bx#)∥2}=12{〈ϕ(xn)-un,Bxn-Bx#〉∥ϕ(xn)-ϕ(x#)∥2+∥un-ϕ(x#)∥2-∥ϕ(xn)-un∥2-ϖn2∥Bxn-Bx#∥2+2ϖn〈ϕ(xn)-un,Bxn-Bx#〉∥ϕ(xn)-ϕ(x#)∥2+∥un-ϕ(x#)∥2}.
It follows that
(43)∥un-ϕ(x#)∥2≤∥ϕ(xn)-ϕ(x#)∥2-∥ϕ(xn)-un∥2-ϖn2∥Bxn-Bx#∥2+2ϖn∥ϕ(xn)-un∥∥Bxn-Bx#∥.
From (39) and (43), we have
(44)∥ϕ(xn+1)-ϕ(x#)∥2≤[1-(1-ηn)(1-ςLζnν)]∥ϕ(xn)-ϕ(x#)∥2+(1-ζn)(1-ηn)∥un-ϕ(x#)∥2+(1-ηn)ζn(1-ςLν)(∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν)2≤[1-(1-ηn)(1-ςLζnν)]∥ϕ(xn)-ϕ(x#)∥2+(1-ζn)(1-ηn)∥ϕ(xn)-ϕ(x#)∥2-(1-ζn)(1-ηn)∥ϕ(xn)-un∥2+2(1-ζn)(1-ηn)ϖn∥ϕ(xn)-un∥∥Bxn-Bx#∥+(1-ηn)ζn(1-ςLν)(∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν)2≤∥ϕ(xn)-ϕ(x#)∥2-(1-ζn)(1-ηn)∥ϕ(xn)-un∥2+2ϖn∥ϕ(xn)-un∥∥Bxn-Bx#∥+(1-ηn)ζn(1-ςLν)(∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν)2.
Then, we obtain
(45)(1-ζn)(1-ηn)∥ϕ(xn)-un∥2≤(∥ϕ(xn)-ϕ(x#)∥+∥ϕ(xn+1)-ϕ(x#)∥)×∥ϕ(xn+1)-ϕ(xn)∥+2ϖn∥ϕ(xn)-un∥∥Bxn-Bx#∥+(1-ηn)ζn(1-ςLν)(∥ςϱ(x#)-ϕ(x#)∥1-ςL/ν)2.
Since lim⁡n→∞ζn=0, lim⁡n→∞∥ϕ(xn+1)-ϕ(xn)∥=0, and lim⁡n→∞∥Bxn-Bx#∥=0, we have
(46)lim⁡n→∞∥ϕ(xn)-un∥=0.
Next, we prove that limsup⁡n→∞〈ςϱ(x*)-ϕ(x*),un-ϕ(x*)〉≤0, where x* is the unique solution of (7). Let {uni} be a subsequence of {un} such that
(47)limsup⁡n→∞〈ςϱ(x*)-ϕ(x*),un-ϕ(x*)〉=lim⁡i→∞〈ςϱ(x*)-ϕ(x*),uni-ϕ(x*)〉=lim⁡i→∞〈ςϱ(x*)-ϕ(x*),ϕ(xni)-ϕ(x*)〉.
By the boundedness of {xni}, there exists a subsequence {xnij} of {xni} which converges weakly to some point z∈C. Without loss of generality, we may assume that xni⇀z. From the weak continuity of ϕ, we deduce ϕ(xni)⇀ϕ(z). Next, we prove z∈Δ. We firstly show z∈EP(θ,A). Noting that zn=Sτn(un-τnAun), for any y∈C, we have
(48)θ(zn,y)+1τn〈y-zn,zn-(un-τnAun)〉≥0.
Since θ is monotone, we have
(49)1τn〈y-zn,zn-(un-τnAun)〉≥θ(y,zn),∀y∈C.
Hence,
(50)〈y-zni,zni-uniτni+Auni〉≥θ(y,zni),∀y∈C.
Let vt=ty+(1-t)z, for all t∈(0,1] and y∈C. We have vt∈C. So, from (50) we have
(51)〈vt-zni,Avt〉≥〈vt-zni,Avt〉-〈vt-zni,zni-uniτni+Auni〉+θ(vt,zni)=〈vt-zni,Avt-Azni〉+〈vt-zni,Azni-Auni〉-〈vt-zni,zni-uniτni〉+θ(vt,zni).
Note that ∥Azni-Auni∥≤(1/η)∥zni-uni∥→0. Further, from monotonicity of A, we have 〈vt-zni,Avt-Azni〉≥0. Letting i→∞ in (51), we have 〈vt-z,Avt〉≥θ(vt,z). This together with (C1) and (C4) implies that
(52)0=θ(vt,vt)≤tθ(vt,y)+(1-t)θ(vt,z)≤tθ(vt,y)+(1-t)〈vt-z,Avt〉=tθ(vt,y)+(1-t)t〈y-z,Avt〉
and hence 0≤θ(vt,y)+(1-t)〈Avt,y-z〉. Letting t→0, we have 0≤θ(z,y)+〈y-z,Az〉. This implies that z∈EP(θ,A). Next, we prove z∈VI(B,ϕ,C). Set
(53)Rv={Bv+NC(v),v∈C,∅,v∉C.
It is well known that R is maximal ϕ-monotone. Let (v,w)∈G(R) (the graph of R). Since w-Bv∈NC(v) and xn∈C, we have 〈ϕ(v)-ϕ(xn),w-Bv〉≥0. Noting that un=projC[ϕ(xn)-ϖnBxn], we get
(54)〈ϕ(v)-un,un-[ϕ(xn)-ϖnBxn]〉≥0.
It follows that
(55)〈ϕ(v)-un,un-ϕ(xn)ϖn+Bxn〉≥0.
Then,
(56)〈ϕ(v)-ϕ(xni),w〉≥〈ϕ(v)-ϕ(xni),Bv〉≥〈ϕ(v)-ϕ(xni),Bv〉-〈ϕ(v)-uni,uni-ϕ(xni)ϖni〉-〈ϕ(v)-uni,Bxni〉=〈ϕ(v)-ϕ(xni),Bv-Bxni〉+〈ϕ(v)-ϕ(xni),Bxni〉-〈ϕ(v)-uni,uni-ϕ(xni)ϖni〉-〈ϕ(v)-uni,Bxni〉≥-〈ϕ(v)-uni,uni-ϕ(xni)ϖni〉-〈ϕ(xni)-uni,Bxni〉.
Since ∥ϕ(xni)-uni∥→0 and ϕ(xni)⇀ϕ(z), we deduce that 〈ϕ(v)-ϕ(z),w〉≥0 by taking i→∞ in (56). Thus, z∈R-10 by the maximal ϕ-monotonicity of R. Hence, z∈VI(B,ϕ,C). Therefore, z∈Δ. From (47), we obtain
(57)limsup⁡n→∞〈ςϱ(x*)-ϕ(x*),un-ϕ(x*)〉=lim⁡i→∞〈ςϱ(x*)-ϕ(x*),ϕ(xni)-ϕ(x*)〉=〈ςϱ(x*)-ϕ(x*),ϕ(z)-ϕ(x*)〉≤0.
Set yn=projC[ζnςϱ(xn)+(1-ζn)zn], for all n≥0. Then, we have
(58)∥yn-ϕ(x*)∥2≤〈ζnςϱ(xn)+(1-ζn)zn-ϕ(x*),yn-ϕ(x*)〉≤ζnς〈ϱ(xn)-ϱ(x*),yn-ϕ(x*)〉+ζn〈ςϱ(x*)-ϕ(x*),yn-ϕ(x*)〉+(1-ζn)〈zn-ϕ(x*),yn-ϕ(x*)〉≤ζnLςν∥ϕ(xn)-ϕ(x*)∥∥yn-ϕ(x*)∥+ζn〈ςϱ(x*)-ϕ(x*),yn-ϕ(x*)〉+(1-ζn)∥zn-ϕ(x*)∥∥yn-ϕ(x*)∥≤ζn(ςLν)∥ϕ(xn)-ϕ(x*)∥∥yn-ϕ(x*)∥+ζn〈ςϱ(x*)-ϕ(x*),yn-ϕ(x*)〉+(1-ζn)∥ϕ(xn)-ϕ(x*)∥∥yn-ϕ(x*)∥=[1-(1-Lςν)ζn]∥ϕ(xn)-ϕ(x*)∥∥yn-ϕ(x*)∥+ζn〈ςϱ(x*)-ϕ(x*),yn-ϕ(x*)〉=1-(1-Lς/ν)ζn2∥ϕ(xn)-ϕ(x*)∥2+12∥yn-ϕ(x*)∥2+ζn〈ςϱ(x*)-ϕ(x*),yn-ϕ(x#)〉.
It follows that
(59)∥yn-ϕ(x*)∥2≤[1-(1-Lςν)ζn]∥ϕ(xn)-ϕ(x*)∥2+2ζn〈ςϱ(x*)-ϕ(x*),yn-ϕ(x*)〉.
Therefore,
(60)∥ϕ(xn+1)-ϕ(x*)∥2≤ηn∥ϕ(xn)-ϕ(x*)∥2+(1-ηn)∥yn-ϕ(x*)∥2≤ηn∥ϕ(xn)-ϕ(x*)∥2+(1-ηn)[1-(1-ςLν)ζn]∥ϕ(xn)-ϕ(x*)∥2+2(1-ηn)ζn〈ςϱ(x*)-ϕ(x*),yn-ϕ(x*)〉=[1-(1-ςLν)(1-ηn)ζn]∥ϕ(xn)-ϕ(x*)∥2+2(1-ηn)ζn〈ςϱ(x*)-ϕ(x*),yn-ϕ(x*)〉=[1-(1-ςLν)(1-ηn)ζn]∥ϕ(xn)-ϕ(x*)∥2+(1-ςLν)(1-ηn)ζn×(21-ςL/ν〈ςϱ(x*)-ϕ(x*),yn-ϕ(x*)〉)=(1-νn)∥ϕ(xn)-ϕ(x*)∥2+ςnνn,
where νn=(1-ςL/ν)(1-ηn)ζn and ςn=(2/(1-ςL/ν))〈ςϱ(x*)-ϕ(x*),yn-ϕ(x*)〉. It is easily seen that ∑n‍νn=∞. Since
(61)∥yn-un∥≤∥yn-zn∥+∥zn-un∥≤ζn∥ςϕ(xn)-zn∥+∥zn-un∥⟶0
and by limsup⁡n→∞〈ςϱ(x*)-ϕ(x*),un-ϕ(x*)〉≤0, we get limsup⁡n→∞ςn≤0. We can therefore apply Lemma 3 to conclude that ϕ(xn)→ϕ(x*) and xn→x*. This completes the proof.



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*Source: 478172-2014-05-27.xml*