# Algorithms for Solving the Variational Inequality Problem over the Triple Hierarchical Problem

Authors: Thanyarat Jitpeera; Poom Kumam
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

This paper discusses the monotone variational inequality over the solution set of the variational
inequality problem and the fixed point set of a nonexpansive mapping. The strong convergence
theorem for the proposed algorithm to the solution is guaranteed under some suitable assumptions.


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## Body

## 1. Introduction

LetC be a closed convex subset of a real Hilbert space H with the inner product 〈·,·〉 and the norm ∥·∥. We denote weak convergence and strong convergence by notations ⇀ and →, respectively.A mappingA:H→H is said to be monotone if 〈Ax-Ay,x-y〉≥0,∀x,y∈H.A is said to be α-strongly monotone if there exists α>0 such that 〈Ax-Ay,x-y〉≥α∥x-y∥2,∀x,y∈H.   A is said to be β-inverse-strongly monotone if there exists β>0 such that 〈Ax-Ay,x-y〉≥β∥Ax-Ay∥2,∀x,y∈H.   A is said to be L-Lipschitz continuous if there exists L>0 such that ∥Ax-Ay∥≤L∥x-y∥,∀x,y∈H. A linear bounded operator A is said to be strongly positive on H if there exists γ¯>0 with the property 〈Ax,x〉≥γ¯∥x∥2,∀x∈H.Letf:C→C be a ρ-contraction if there exists ρ∈[0,1) such that
(1.1)‖f(x)-f(y)‖≤ρ‖x-y‖,∀x,y∈C.LetT:C→C be nonexpansive such that
(1.2)‖Tx-Ty‖≤‖x-y‖,∀x,y∈C.
A point x∈C is a fixed point of T provided Tx=x. Denote by F(T) the set of fixed points of T; that is, F(T)={x∈C:Tx=x}. If C is bounded closed convex and T is a nonexpansive mapping of C into itself, then F(T) is nonempty (see [1]). Let A be a nonlinear mapping. The Hartmann-Stampacchia variational inequality [2] is to finding x∈C such that
(1.3)〈Ax,y-x〉≥0,∀y∈C.
The set of solutions of (1.3) is denoted by VI(C,A). The variational inequality has been extensively studied in the literature [3, 4].We discuss the following variational inequality problem over the fixed point set of a nonexpansive mapping (see [5–12]), which is called the hierarchical problem. Let a monotone, continuous mapping A:H→H and a nonexpansive mapping T:H→H.
(1.4)Findx∈VI⁡(F(T),A)={x∈F(T):〈Ax,y-x〉≥0,∀y∈F(T)},F(T)≠∅.
This solution set is denoted by Ξ.We introduce the following variational inequality problem over solution set of variational inequality problem and the fixed point set of a nonexpansive mapping (see [13–16]), which is called the triple hierarchical problem (or the triple hierarchical constrained optimization problem (see also [13])). Let an inverse-strongly monotone A:H→H, a strongly monotone and Lipschitz continuous B:H→H, and a nonexpansive mapping T:H→H.
(1.5)Findx∈VI⁡(Ξ,B)={x∈Ξ:〈Bx,y-x〉≥0,∀y∈Ξ},
where Ξ:=VI(F(T),A)≠∅.In 2009, Iiduka [13] introduced an iterative algorithm for the following triple hierarchical constrained optimization problem, the sequence {xn} defined by the iterative method below, with the initial guess x1∈H is chosen arbitrarily,
(1.6)yn=T(xn-λnA1xn),xn+1=yn-μαnA2yn,∀n≥0,
where αn∈(0,1] and λn∈(0,2α] satisfies certain conditions. Let A1:H→H be an inverse-strongly monotone, A2:H→H be a strongly monotone and Lipschitz continuous, and T:H→H be a nonexpansive mapping, then the sequence converges to strong analysis on (1.6).In 2011, Ceng et al. [17] studied the new following algorithms. For x0∈C is chosen arbitrarily, they defined a sequence {xn} iterative by
(1.7)xn+1=PC[λnγ(αnf(xn)+(1-αn)Sxn)+(I-λnμF)Txn],∀n≥0,
where the mapping S,T are nonexpansive mappings with F(T)≠∅. Let F:C→H be a Lipschitzian and strongly monotone operator and f:C→H be a contraction mapping satisfied some conditions. They proved that the proposed algorithms strongly converge to the minimum norm fixed point of T.Very recently, Yao et al. [18] studied the following algorithms. For x0∈C is chosen arbitrarily, let the sequence {xn} be generated iteratively by
(1.8)xn+1=βnxn+(1-βn)TPC[I-αn(A-γf)]xn,∀n≥0,
where the sequences {αn} and {βn} are two sequences in [0,1]. Then {xn} converges strongly to the unique solution of the variational inequality as follows. Find a point x*∈F(T) such that
(1.9)〈(A-γf)x*,x-x*〉≥0,∀x∈F(T),
where A:C→H is a strongly positive linear bounded operator, f:C→H is a ρ-contraction, and T:C→C is a nonexpansive mapping satisfied some suitable conditions. The solution (1.9) is denoted by Υ:=VI(F(T),A-γf):={x*∈F(T):〈(A-γf)x*,x-x*〉≥0,∀x∈F(T)}.In this paper, we introduce a new iterative algorithm for solving the triple hierarchical problem, which contain algorithms (1.6) and (1.8) as follows:
(1.10)yn=TPC[I-δn(A-γf)]xn,xn+1=αnu+βnxn+[(1-βn)I-αnμF]yn,∀n≥0.
The strong convergence for the proposed algorithms to the solution is solved under some assumptions. Our results generalize and improve the results of Ceng et al. [17], Iiduka [13], Yao et al. [18], and some authors.


## 2. Preliminaries

LetH be a real Hilbert space and C be a nonempty closed convex subset of H. The metric (or nearest point) projection from H onto C is the mapping PC:H→C which assigns to each point x∈C the unique point in PCx∈C satisfying the property
(2.1)‖x-PCx‖=inf⁡y∈C‖x-y‖=:d(x,C).
The following properties of projection are useful and pertinent to our purposes.Lemma 2.1.
Givenx∈H and z∈C,(a)
u=PCz⇔〈u-z,v-u〉≥0,∀v∈C,(b)
u=PCz⇔∥z-u∥2≤∥z-v∥2-∥v-u∥2,∀v∈C,(c)
PC is a firmly nonexpansive mapping of H onto C and satisfies
(2.2)‖PCx-PCy‖2≤〈PCx-PCy,x-y〉,∀x,y∈H.Consequently, PC is nonexpansive and monotone.Lemma 2.2.
There holds the following inequality in an inner product spaceH(2.3)‖x+y‖2≤‖x‖2+2〈y,x+y〉,∀x,y∈H.Lemma 2.3 (see [19]).
LetC be a closed convex subset of a real Hilbert space H and let T:C→C be a nonexpansive mapping. Then I-T is demiclosed at zero, that is,
(2.4)xn⇀x,xn-Txn⟶0
implies x=Tx.Lemma 2.4 (see [20]).
Each Hilbert spaceH satisfies Opial’s condition, that is, for any sequence {xn}⊂H with xn⇀x, the inequality
(2.5)liminf⁡n→∞‖xn-x‖<liminf⁡n→∞‖xn-y‖
hold for each y∈H with y≠x.Lemma 2.5 (see [21]).
Let{xn} and {yn} be bounded sequences in a Banach space X and let {βn} be a sequence in [0,1] with 0<liminfn→∞βn≤limsupn→∞βn<1. Suppose xn+1=(1-βn)yn+βnxn for all integers n≥0 and limsupn→∞(∥yn+1-yn∥-∥xn+1-xn∥)≤0. Then, limn→∞∥yn-xn∥=0.Lemma 2.6 (see [10]).
LetB:H→H be β-strongly monotone and L-Lipschitz continuous and let μ∈(0,2β/L2). For λ∈[0,1], define Tλ:H→H by Tλ(x):=x-λμB(x) for all x∈H. Then, for all x,y∈H,
(2.6)‖Tλ(x)-Tλ(y)‖≤(1-λτ)‖x-y‖
hold, where τ:=1-1-μ(2β-μL2)∈(0,1].Lemma 2.7 (see [22]).
Assume{an} is a sequence of nonnegative real numbers such that
(2.7)an+1≤(1-γn)an+δn,∀n≥0,
where {γn}⊂(0,1) and {δn} is a sequence in ℛ such that(i)
∑n=1∞‍γn=∞,(ii)
limsupn→∞(δn/γn)≤0 or ∑n=1∞‍|δn|<∞.Then limn→∞an=0.Remark 2.8.
IfA:C→H is a strongly positive linear bounded operator and f:C→H is a ρ-contraction, then for 0<γ<γ~/ρ, the mapping A-γf is strongly monotone. In fact, we have
(2.8)〈(A-γf)x-(A-γf)y,x-y〉=〈A(x-y),x-y〉-γ〈f(x)-f(y),x-y〉≥γ~‖x-y‖2-γγρ‖x-y‖2≥0.


## 3. Main Results

In this section, we introduce a new iterative algorithm for solving monotone variational inequality problem (whereA:C→H is a strongly positive linear bounded operator, f:C→H is a ρ-contraction) over solution set of variational inequality problem over the fixed point set of a nonexpansive mapping.Theorem 3.1.
LetC be a nonempty closed and convex subset of a real Hilbert space H. Let A:C→H be a strongly positive linear bounded operator, f:C→H be a ρ-contraction, and γ be a positive real number such that (γ¯-1)/ρ<γ<γ¯/ρ. Let F:C→C be κ-Lipschitzian and η-strongly monotone operators with constant κ and η>0, respectively. Let T:C→C be a nonexpansive mapping with F(T)≠∅. Let 0<μ<2η/κ2 and 0<γ<τ, where τ=1-1-μ(2η-μκ2). Assume that VI(Υ,F)≠∅, where Υ:=VI(F(T),A-γf). Suppose {xn} is a sequence generated by the following algorithm x0∈C arbitrarily and
(3.1)yn=TPC[I-δn(A-γf)]xn,xn+1=αnu+βnxn+[(1-βn)I-αnμF]yn,∀n≥0,
where {αn},{βn},{δn}⊂(0,1) satisfy the following conditions:(C1)αn≤κδn and βn<δn;(C2)∑n=1∞‍|αn+1-αn|<∞;(C3)∑n=1∞‍|βn+1-βn|<∞;(C4)limn→∞δn=0, ∑n=0∞‍δn=∞ and ∑n=1∞‍|δn+1-δn|<∞.
Then the sequence{xn} converges strongly to x*∈Υ, which is the unique solution of another variational inequality
(3.2)〈(I-μF)x*,x-x*〉≥0,∀x∈Υ.Proof.
We will divide the proof into four steps.
Step  1.  We will show {xn} is bounded. For any x*∈F(T), we have
(3.3)‖yn-x*‖=‖TPC[I-δn(A-γf)]xn-TPCx*‖≤‖[I-δn(A-γf)]xn-x*‖≤δn‖γf(xn)-γf(x*)‖+δn‖γf(x*)-Ax*‖+‖I-δnA‖‖xn-x*‖≤δnγρ‖xn-x*‖+δn‖γf(x*)-Ax*‖+(1-δnγ¯)‖xn-x*‖=[1-(γ¯-γρ)δn]‖xn-x*‖+δn‖γf(x*)-Ax*‖.
From (3.1), we deduce that
(3.4)‖xn+1-x*‖=‖αnu+βnxn+[(1-βn)I-αnμF]yn-x*‖≤αn‖u-μFx*‖+βn‖xn-x*‖+[(1-βn)I-αnμF]‖yn-x*‖≤αn‖u-μFx*‖+βn‖xn-x*‖+(1-βn-αnτ)‖yn-x*‖.
Substituting (3.3) into (3.4), we obtain
(3.5)‖xn+1-x*‖≤αn‖u-μFx*‖+βn‖xn-x*‖+(1-βn-αnτ)×{[1-(γ¯-γρ)δn]‖xn-x*‖+δn‖γf(x*)-Ax*‖}=αn‖u-μFx*‖+βn‖xn-x*‖+(1-βn-αnτ)[1-(γ¯-γρ)δn]‖xn-x*‖+(1-βn-αnτ)δn‖γf(x*)-Ax*‖≤αn‖u-μFx*‖+[1-(1-βn-αnτ)δn(γ¯-γρ)]‖xn-x*‖+(1-βn-αnτ)δn‖γf(q)-Aq‖≤κδn‖u-μFx*‖+[1-(1-βn-αnτ)δn(γ¯-γρ)]‖xn-x*‖+(1-βn-αnτ)δn‖γf(x*)-Ax*‖.
By induction, it follows that
(3.6)‖xn-x*‖≤max⁡{‖x0-x*‖+1γ¯-γρ‖γf(x*)-Ax*‖+1(1-βn-αnτ)(γ¯-γρ)κ‖u-μFx*‖},n≥0.
Therefore, {xn} is bounded and so are {yn},{Axn},{f(xn)}, and {F(xn)}.
Step  2.  We will show that limn→∞∥xn+1-xn∥=0,limn→∞∥yn-xn∥=0, and limn→∞∥yn-Tyn∥=0. From (3.1), we have
(3.7)‖yn+1-yn‖=‖TPC[I-δn+1(A-γf)]xn+1-TPC[I-δn(A-γf)]xn‖≤‖PC[I-δn+1(A-γf)]xn+1-PC[I-δn(A-γf)]xn‖≤‖[I-δn+1(A-γf)]xn+1-[I-δn(A-γf)]xn‖=‖δn+1(γf(xn+1)-γf(xn))+(δn+1-δn)γf(xn)A-γf{δn+1(γf(xn+1)-γf(xn))}+(I-δn+1A)(xn+1-xn)+(δn-δn+1)Axn‖≤δn+1γ‖f(xn+1)-f(xn)‖+(1-δn+1γ¯)‖xn+1-xn‖-γf+|δn+1-δn|(‖γf(xn)‖+‖Axn‖)≤δn+1γρ‖xn+1-xn‖+(1-δn+1γ¯)‖xn+1-xn‖+|δn+1-δn|(‖γf(xn)‖+‖Axn‖)=[1-(γ¯-γρ)δn+1]‖xn+1-xn‖+|δn+1-δn|(‖γf(xn)‖+‖Axn‖).
It follows that
(3.8)‖xn+2-xn+1‖=‖αn+1u+βn+1xn+1+[(1-βn+1)I-αn+1μF]yn+1-αnu-βnxn-[(1-βn)I-αnμF]yn‖≤|αn+1-αn|‖u‖+βn+1‖xn+1-xn‖+|βn+1-βn|‖xn‖βn+‖[(1-βn+1)I-αn+1μF]yn+1-[(1-βn+1)I-αn+1μF]yn‖βn+‖[(1-βn+1)I-αn+1μF]yn-[(1-βn)I-αnμF]yn‖≤|αn+1-αn|‖u‖+βn+1‖xn+1-xn‖+|βn+1-βn|‖xn‖βn+(1-βn+1-αn+1τ)‖yn+1-yn‖+(1-βn+1-αn+1μF-1+βn+αnμF)‖yn‖≤|αn+1-αn|‖u‖+βn+1‖xn+1-xn‖+|βn+1-βn|‖xn‖βn+(1-βn+1-αn+1τ)‖yn+1-yn‖+|βn+1-βn|‖yn‖+|αn+1-αn|τ‖yn‖=|αn+1-αn|(‖u‖+τ‖yn‖)+βn+1‖xn+1-xn‖+|βn+1-βn|(‖xn‖+‖yn‖)βn+(1-βn+1-αn+1τ)‖yn+1-yn‖≤|αn+1-αn|(‖u‖+τ‖yn‖)+βn+1‖xn+1-xn‖+|βn+1-βn|(‖xn‖+‖yn‖)βn+(1-βn+1-αn+1τ){[1-(γ¯-γρ)δn+1]‖xn+1-xn‖βn+|δn+1-δn|(‖γf(xn)‖+‖Axn‖)}≤|αn+1-αn|(‖u‖+τ‖yn‖)+βn+1‖xn+1-xn‖+|βn+1-βn|(‖xn‖+‖yn‖)βn+(1-βn+1-αn+1τ)[1-(γ¯-γρ)δn+1]‖xn+1-xn‖βn+|δn+1-δn|(‖γf(xn)‖+‖Axn‖)≤[1-(1-βn+1-αn+1τ)(γ¯-γρ)δn+1]‖xn+1-xn‖βn+|αn+1-αn|(‖u‖+τ‖yn‖)+|βn+1-βn|(‖xn‖+‖yn‖)βn+|δn+1-δn|(‖γf(xn)‖+‖Axn‖)≤[1-(1-βn+1-αn+1τ)(γ¯-γρ)δn+1]‖xn+1-xn‖βn+(|αn+1-αn|+|βn+1-βn|+|δn+1-δn|)M3,
where M3 is a constant such that
(3.9)sup⁡n≥0{(‖u‖+τ‖yn‖),(‖xn‖+‖yn‖),(‖γf(xn)‖+‖Axn‖)}≤M3.
By the conditions (C2)–(C4) allow us to apply Lemma 2.7, we get
(3.10)lim⁡n→∞‖xn+1-xn‖=0.
On the other hand, we note that
(3.11)‖yn-Txn‖=‖TPC[I-δn(A-γf)]xn-Txn‖=‖TPC[I-δn(A-γf)]xn-TPCxn‖≤‖[I-δn(A-γf)]xn-xn‖≤δn‖(A-γf)xn‖,
by (C4), it follows that
(3.12)lim⁡n→∞‖yn-Txn‖=0.
From (3.7), we observe that
(3.13)‖yn+1-yn‖≤[1-(γ¯-γρ)δn+1]‖xn+1-xn‖+|δn+1-δn|(‖γf(xn)‖+‖Axn‖).
It follows that
(3.14)‖yn+1-yn‖-‖xn+1-xn‖≤(γ¯-γρ)δn+1‖xn+1-xn‖+|δn+1-δn|(‖γf(xn)‖+‖Axn‖).
From the conditions (C1)–(C4) and the boundedness of {xn},{f(xn)}, and {Axn}, which implies that
(3.15)limsup⁡n→∞(‖yn+1-yn‖-‖xn+1-xn‖)≤0.
Hence, by Lemma 2.5, we have
(3.16)lim⁡n→∞‖yn-xn‖=0.
From (3.12) and (3.16), we obtain
(3.17)lim⁡n→∞‖xn-Txn‖=0.
Step  3.  We will show that limsupn→∞〈un-x*,γf(x*)-Ax*〉≤0 is proven. Choose a subsequence {xni} of {xn} such that
(3.18)limsup⁡n→∞〈xn-x*,γf(x*)-Ax*〉=lim⁡i→∞〈xni-x*,γf(x*)-Ax*〉.
The boundedness of {xni} implies the existences of a subsequence {xnij} of {xni} and a point x^∈H such that {xnij} converges weakly to x^. We may assume without loss of generality that limi→∞〈xni,w〉=〈x^,w〉,w∈H. Assume x^≠T(x^). Since limn→∞∥xn-Txn∥=0 with F(T)≠∅ guarantee that
(3.19)liminf⁡i→∞‖xni-x^‖<liminf⁡i→∞‖xni-T(x^)‖=liminf⁡i→∞‖xni-T(xni)+T(xni)-T(x^)‖=liminf⁡i→∞‖T(xni)-T(x^)‖≤liminf⁡i→∞‖xni-x^‖,
which has a contradiction. Therefore, x^∈F(T). Since x*∈VI(Υ,F), then x*∈Υ:=VI(F(T),A-γf), it follows that
(3.20)limsup⁡n→∞〈xn-x*,γf(x*)-Ax*〉=lim⁡i→∞〈xni-x*,γf(x*)-Ax*〉=〈x^-x*,γf(x*)-Ax*〉≤0.
Setting un=[I-δn(A-γf)]xn and by (C4), we notice that
(3.21)‖un-xn‖≤δn‖(A-γf)‖⟶0,asn⟶∞.
Hence, we get
(3.22)limsup⁡n→∞〈un-x*,γf(x*)-Ax*〉≤0.
Next we will show that limsupn→∞〈xn+1-x*,x*-μFx*〉≤0 is proven. Choose a subsequence {xnk} of {xn} such that
(3.23)limsup⁡n→∞〈xn+1-x*,x*-μFx*〉=lim⁡k→∞〈xnk+1-x*,x*-μFx*〉.
The boundedness of {xnk} implies the existences of a subsequence {xnkl} of {xnk} and a point x¯∈H such that {xnkl} converges weakly to x¯. We may assume without loss of generality that limk→∞〈xnk,w〉=〈x¯,w〉,w∈H. Assume x¯≠T(x¯). By limn→∞∥xn-Txn∥=0 with F(T)≠∅ guarantee that
(3.24)liminf⁡k→∞‖xnk-x¯‖<liminf⁡k→∞‖xnk-T(x¯)‖=liminf⁡k→∞‖xnk-T(xnk)+T(xnk)-T(x¯)‖=liminf⁡k→∞‖T(xnk)-T(x¯)‖≤liminf⁡k→∞‖xnk-x¯‖,
which has a contradiction. Therefore, x¯∈F(T). From x*∈VI(Υ,F):=VI(VI(F(T),A-γf),F), we compute
(3.25)limsup⁡n→∞〈xn-x*,γx*-μFx*〉=lim⁡k→∞〈xnk-x*,x*-μFx*〉=〈x¯-x*,x*-μFx*〉≤0.
Using (3.10), we get
(3.26)limsup⁡n→∞〈xn+1-x*,x*-μFx*〉≤0.
Step  4.  Finally, we prove xn+1→x*. We observe that
(3.27)‖un-x*‖≤‖xn-x*‖+δn‖(A-γf)xn‖.
From (3.1), we compute
(3.28)‖xn+1-x*‖2=‖αnu+βnxn+[(1-βn)I-αnμF]yn-x*‖2=‖αn(u-x*)+αn(x*-μFx*)+βn(xn-x*)+[(1-βn)I-αnμF](yn-x*)‖2≤‖αn(u-x*)+βn(xn-x*)+αn(x*-μFx*)‖2+[1-βn-αnτ]‖yn-x*‖2≤αn2‖u-x*‖2+βn2‖xn-x*‖2+2αn〈x*-μFx*,xn+1-x*〉+[1-βn-αnτ]‖un-x*‖2≤αn‖u-x*‖2+βn‖xn-x*‖2+2αn〈x*-μFx*,xn+1-x*〉+[1-βn-αnτ]‖δn(γf(xn)-Ax*)+(I-δnA)(xn-x*)‖2≤κδn‖u-x*‖2+βn‖xn-x*‖2+2κδn〈x*-μFx*,xn+1-x*〉+[1-βn-αnτ][(1-δnγ~)2‖xn-x*‖2+2δn〈γf(xn)-Ax*,un-x*〉]≤κδn‖u-x*‖2+2κδn〈x*-μFx*,xn+1-x*〉+[1-2δnγ~(1-βn-αnτ)]‖xn-x*‖2+δn2γ~2(1-βn-αnτ)‖xn-x*‖2-αnτ‖xn-x*‖2+2δn〈γf(xn)-γfx*,un-x*〉+2δn〈γfx*-Ax*,un-x*〉≤κδn‖u-x*‖2+2κδn〈x*-μFx*,xn+1-x*〉+[1-2δnγ~]‖xn-x*‖2+δn2γ~2(1-βn-αnτ)‖xn-x*‖2+2δnγρ‖xn-x*‖‖un-x*‖+2δn〈γfx*-Ax*,un-x*〉≤[1-2δn(γ~-γρ)]‖xn-x*‖2+δn2γ~2(1-βn-αnτ)‖xn-x*‖2+2δn2γρ‖xn-x*‖‖(A-γf)xn‖+2δn〈γfx*-Ax*,un-x*〉+κδn‖u-x*‖2+2κδn〈x*-μFx*,xn+1-x*〉.
Since {xn},{Axn},{f(xn)}, and {Fxn} are all bounded, we can choose a constant M4>0 such that
(3.29)sup⁡n≥01γ~-γρ{(1-βn-αnτ)γ~22‖xn-x*‖+γρ‖xn-x*‖‖(A-γf)xn‖}≤M4.
It follows that
(3.30)‖xn+1-x*‖2≤[1-2(γ~-γρ)δn]‖xn-x*‖2+2(γ~-γρ)δnξn,
where
(3.31)ξn:=δnM4+1γ~-γρ〈γfx*-Ax*,un-x*〉+κγ~-γρ‖u-x*‖2+κγ~-γρ〈x*-μFx*,xn+1-x*〉.
By the conditions (C1), (C4), (3.22), and (3.26), we get
(3.32)limsup⁡n→∞ξn≤0.
Now, applying Lemma 2.7 and (3.30), we conclude that xn→x*. This completes the proof.Next, the following example shows that all conditions of Theorem3.1 are satisfied.Example 3.2.
For instance, letαn=n/(n2+1),βn=1/2n and δn=1/n. Then, clearly the sequences {αn},{βn}, {δn} satisfy the following condition (C1):
(3.33)nn2+1<κ1n,12n<1n.
We will show that the condition (C2) is achieved. Indeed, we have
(3.34)∑n=1∞|αn+1-αn|=∑n=1∞|n+1(n+1)2+1-nn2+1|=∑n=1∞|(n+1)(n2+1)-n(n2+2n+2)(n2+2n+2)(n2+1)|=∑n=1∞|1-n-n2n4+2n3+3n2+2n+2|.
The sequence {αn} satisfies the condition (C2) by p-series. Next, we will show that the condition (C3) is achieved. We compute
(3.35)∑n=1∞|βn+1-βn|=∑n=1∞|12(n+1)-12n|≤|12⋅1-12⋅2|+|12⋅2-12⋅3|+|12⋅3-12⋅4|+⋅⋅⋅=12.
The sequence {βn} satisfies the condition (C3). Finally, we will show that the condition (C4) is achieved. We compute
(3.36)lim⁡n→∞δn=lim⁡n→∞1n=0,∑n=1∞δn=∑n=1∞1n=∞,∑n=1∞|δn+1-δn|=∑n=1∞|1n+1-1n|≤|11-12|+|12-13|+|14-13|+⋯=1.
The sequence {δn} satisfies the condition (C4).Corollary 3.3.
LetC be a nonempty closed and convex subset of a real Hilbert space H. Let A:C→H be a strongly positive linear bounded operator, f:C→H be a ρ-contraction, and γ be a positive real number such that (γ¯-1)/ρ<γ<γ¯/ρ. Let T:C→C be a nonexpansive mapping with F(T)≠∅. Assume that Υ:=VI(F(T),A-γf)≠∅. Suppose {xn} is a sequence generated by the following algorithm x0∈C arbitrarily and
(3.37)yn=TPC[I-δn(A-γf)]xn,xn+1=αnu+βnxn+(1-βn-αn)yn,
where {αn},{βn},{δn}⊂(0,1) satisfy the following conditions (C1)–(C4). Then the sequence {xn} converges strongly to x*∈F(T), which is the unique solution of variational inequality
(3.38)〈(A-γf)x*,x-x*〉≥0,∀x∈F(T).Proof.
Puttingμ=2 and F≡I/2 in Theorem 3.1, we can obtain desired conclusion immediately.Corollary 3.4.
LetC be a nonempty closed and convex subset of a real Hilbert space H. Let T:C→C be a nonexpansive mapping with F(T)≠∅. Suppose {xn} is a sequence generated by the following algorithm x0∈C arbitrarily and
(3.39)yn=TPC(1-δn)xn,xn+1=αnu+βnxn+(1-βn-αn)yn,∀n≥0,
where {αn},{βn},{δn}⊂(0,1) satisfy the following conditions (C1)–(C4). Then the sequence {xn} converges strongly to x*∈F(T).Proof.
Puttingf≡0 and A≡I in Corollary 3.3, we can obtain desired conclusion immediately.Remark 3.5.
Our results generalize and improve the recent results of Iiduka [13] and Yao et al. [18].



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*Source: 827156-2012-07-24.xml*