# General Iterative Algorithms for Hierarchical Fixed Points Approach to Variational Inequalities

Authors: Nopparat Wairojjana; Poom Kumam
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

This paper deals with new methods for approximating a solution to the fixed point problem; findx̃∈F(T), where H is a Hilbert space, C is a closed convex subset of H, f is a ρ-contraction from C into H, 0<ρ<1, A is a strongly positive linear-bounded operator with coefficient γ̅>0, 0<γ<γ̅/ρ, T is a nonexpansive mapping on C, and PF(T) denotes the metric projection on the set of fixed point of T. Under a suitable different parameter, we obtain strong convergence theorems by using the projection method which solves the variational inequality 〈(A-γf)x̃+τ(I-S)x̃,x-x̃〉≥0 for x∈F(T), where τ∈[0,∞). Our results generalize and improve the corresponding results of Yao et al. (2010) and some authors. Furthermore, we give an example which supports our main theorem in the last part.


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## Body

## 1. Introduction

Throughout this paper, we assume thatH is a real Hilbert space where inner product and norm are denoted by 〈·,·〉 and ∥·∥, respectively, and let C be a nonempty closed convex subset of H. A mapping T:C→C is called nonexpansive if
(1.1)‖Tx-Ty‖≤‖x-y‖,∀x,y∈C.We useF(T) to denote the set of fixed points of T, that is, F(T)={x∈C:Tx=x}. It is assumed throughout the paper that T is a nonexpansive mapping such that F(T)≠∅.Recall that a mappingf:C→H is a contraction on C if there exists a constant ρ∈(0,1) such that
(1.2)‖f(x)-f(y)‖≤ρ‖x-y‖,∀x,y∈C.A mappingA:H→H is called astrongly positive linear bounded operator on H if there exists a constant γ¯>0 with property
(1.3)〈Ax,x〉≥γ¯‖x‖2,∀x∈H.A mappingM:H→H is called a strongly monotone operator with α if
(1.4)〈x-y,Mx-My〉≥α‖x-y‖2,∀x,y∈H,
and M is called a monotone operator if
(1.5)〈x-y,Mx-My〉≥0,∀x,y∈H.
We easily prove that the mapping (I-T) is monotone operator, if T is nonexpansive mapping.The metric (or nearest point) projection from H onto C is mapping PC[·]:H→C which assigns to each point x∈C the unique point PC[x]∈C satisfying the property
(1.6)‖x-PC[x]‖=inf⁡y∈C‖x-y‖=:d(x,C).The variational inequality for a monotone operator, M:H→H over C, is to find a point in
(1.7)VI⁡(C,M):={x~∈C:〈x-x~,Mx~〉≥0,∀x∈C}.A hierarchical fixed point problem is equivalent to the variational inequality for a monotone operator over the fixed point set. Moreover, to find a hierarchically fixed point of a nonexpansive mapping T with respect to another nonexpansive mapping S, namely, we find x~∈F(T) such that
(1.8)〈x-x~,(I-S)x~〉≥0,∀x∈F(T).Iterative methods for nonexpansive mappings have recently been applied to solve a convex minimization problem; see, for example, [1–5] and the references therein. A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive mapping on a real Hilbert space H:
(1.9)min⁡x∈F(T)12〈Ax,x〉-〈x,b〉,
where b is a given point in H. In [5], it is proved that the sequence {xn} defined by the iterative method below, with the initial guess x0∈H chosen arbitrarily,
(1.10)xn+1=(I-αnA)Txn+αnb,n≥0,
converges strongly to the unique solution of the minimization problem (1.9) provided the sequence {αn} of parameters satisfies certain appropriate conditions.On the other hand, Moudafi [6] introduced the viscosity approximation method for nonexpansive mappings (see [7] for further developments in both Hilbert and Banach spaces). Starting with an arbitrary initial x0∈H, define a sequence {xn} recursively by
(1.11)xn+1=σnf(xn)+(1-σn)Txn,n≥0,
where {σn} is a sequence in (0,1). It is proved in [6, 7] that under certain appropriate conditions imposed on {σn}, the sequence {xn} generated by (1.11) strongly converges to the unique solution x* in C of the variational inequality
(1.12)〈(I-f)x*,x-x*〉≥0,x∈C.In 2006, Marino and Xu [8] introduced a general iterative method for nonexpansive mapping. Starting with an arbitrary initial x0∈H, define a sequence {xn} recursively by
(1.13)xn+1=ϵnγf(xn)+(I-ϵnA)Txn,n≥0.
They proved that if the sequence {ϵn} of parameters satisfies appropriate conditions, then the sequence {xn} generated by (1.13) strongly converges to the unique solution x~=PF(T)(I-A+γf)x~ of the variational inequality
(1.14)〈(A-γf)x~,x-x~〉≥0,∀x∈F(T),
which is the optimality condition for the minimization problem
(1.15)min⁡x∈F(T)12〈Ax,x〉-h(x),
where h is a potential function for γf (i.e., h′(x)=γf(x) for x∈H).In 2010, Yao et al. [9] introduced an iterative algorithm for solving some hierarchical fixed point problem, starting with an arbitrary initial guess x0∈C, define a sequence {xn} iteratively by
(1.16)yn=βnSxn+(1-βn)xn,xn+1=PC[αnf(xn)+(1-αn)Tyn],∀n≥1.
They proved that if the sequences {αn} and {βn} of parameters satisfies appropriate conditions, then the sequence {xn} generated by (1.16) strongly converges to the unique solution z in H of the variational inequality
(1.17)z∈F(T),〈(I-f)z,x-z〉≥0,∀x∈F(T).In this paper we will combine the general iterative method (1.13) with the iterative algorithm (1.16) and consider the following iterative algorithm:
(1.18)yn=βnSxn+(1-βn)xn,xn+1=PC[αnγf(xn)+(I-αnA)Tyn],∀n≥1.
We will prove in Section 3 that if the sequences {αn} and {βn} of parameters satisfy appropriate conditions and limn→∞(βn/αn)=τ∈(0,∞) then the sequence {xn} generated by (1.18) converges strongly to the unique solution x~ in H of the following variational inequality
(1.19)x~∈F(T),〈1τ(A-γf)x~+(I-S)x~,x-x~〉≥0,∀x∈F(T).
In particular, if we take τ=0, under suitable difference assumptions on parameter, then the sequence {xn} generated by (1.18) converges strongly to the unique solution x~ in H of the following variational inequality
(1.20)x~∈F(T),〈(A-γf)x~,x-x~〉≥0,∀x∈F(T).
Our results improve and extend the recent results of Yao et al. [9] and some authors. Furthermore, we give an example which supports our main theorem in the last part.


## 2. Preliminaries

This section collects some lemma which can be used in the proofs for the main results in the next section. Some of them are known, others are not hard to derive.Lemma 2.1 (Browder [10]).
LetH be a Hilbert space, C be a closed convex subset of H, and T:C→C be a nonexpansive mapping with F(T)≠∅. If {xn} is a sequence in C weakly converging to x and if {(I-T)xn} converges strongly to y, then (I-T)x=y; in particular, if y=0 then x∈F(T).Lemma 2.2.
Letx∈H and z∈C be any points. Then one has the following:
(1)
Thatz=PC[x] if and only if there holds the relation:
(2.1)〈x-z,y-z〉≤0,∀y∈C.
(2)
Thatz=PC[x] if and only if there holds the relation:
(2.2)‖x-z‖2≤‖x-y‖2-‖y-z‖2,∀y∈C.
(3)
There holds the relation:(2.3)〈PC[x]-PC[y],x-y〉≥‖PC[x]-PC[y]‖2,∀x,y∈H.
Consequently, PC is nonexpansive and monotone.Lemma 2.3 (Marino and Xu [8]).
LetH be a Hilbert space, C be a closed convex subset of H, f:C→H be a contraction with coefficient 0<ρ<1, and T:C→C be nonexpansive mapping. Let A be a strongly positive linear bounded operator on a Hilbert space H with coefficient γ->0. Then, for 0<γ<γ-/ρ, for x,y∈C,  (1)
the mapping(I-f) is strongly monotone with coefficient (1-ρ), that is,
(2.4)〈x-y,(I-f)x-(I-f)y〉≥(1-ρ)‖x-y‖2,(2)
the mapping(A-γf) is strongly monotone with coefficient γ--γρ that is
(2.5)〈x-y,(A-γf)x-(A-γf)y〉≥(γ--γρ)‖x-y‖2.Lemma 2.4 (Xu [4]).
Assume that{an} is a sequence of nonnegative numbers such that
(2.6)an+1≤(1-γn)an+δn,∀n≥0,
where {γn} is a sequence in (0,1) and {δn} is a sequence in ℝ such that  (1)
∑n=1∞‍γn=∞,(2)
limsupn→∞(δn/γn)≤0 or ∑n=1∞‍|δn|<∞.  Then limn→∞an=0.Lemma 2.5 (Marino and Xu [8]).
AssumeA is a strongly positive linear bounded operator on a Hilbert space H with coefficient γ->0 and 0<α≤∥A∥-1. Then ∥I-αA∥≤1-αγ-.Lemma 2.6 (Acedo and Xu [11]).
LetC be a closed convex subset of H. Let {xn} be a bounded sequence in H. Assume that  (1)
The weakω-limit set ωw(xn)⊂C,(2)
For eachz∈C, limn→∞∥xn-z∥ exists.  Then {xn} is weakly convergent to a point in C.Notation
We use→ for strong convergence and ⇀ for weak convergence.


## 3. Main Results

Theorem 3.1.
LetC be a nonempty closed convex subset of a real Hilbert space H. Let f:C→H be a ρ-contraction with ρ∈(0,1). Let S,T:C→C be two nonexpansive mappings with F(T)≠∅. Let A be a strongly positive linear bounded operator on H with coefficient γ->0. {αn} and {βn} are two sequences in (0,1) and 0<γ<γ-/ρ. Starting with an arbitrary initial guess x0∈C and {xn} is a sequence generated by
(3.1)yn=βnSxn+(1-βn)xn,xn+1=PC[αnγf(xn)+(I-αnA)Tyn],∀n≥1.
Suppose that the following conditions are satisfied:(C1)
limn→∞αn=0 and ∑n=1∞‍αn=∞,(C2)
limn→∞(βn/αn)=τ=0,(C3)
limn→∞(|αn-αn-1|/αn)=0 and limn→∞(|βn-βn-1|/βn)=0, or(C4)
∑n=1∞|αn-αn-1|<∞ and ∑n=1∞|βn-βn-1|<∞. Then the sequence {xn} converges strongly to a point x~∈H, which is the unique solution of the variational inequality:
(3.2)x~∈F(T),〈(A-γf)x~,x-x~〉≥0,∀x∈F(T).
Equivalently, one has PF(T)(I-A+γf)x~=x~.Proof.
We first show the uniqueness of a solution of the variational inequality (3.2), which is indeed a consequence of the strong monotonicity of A-γf. Suppose x-∈F(T) and x~∈F(T) both are solutions to (3.2), then 〈(A-γf)x-,x--x~〉≤0 and 〈(A-γf)x~,x~-x-〉≤0. It follows that
(3.3)〈(A-γf)x-,x--x~〉+〈(A-γf)x~,x~-x-〉=〈(A-γf)x-,x--x~〉-〈(A-γf)x~,x--x~〉=〈(A-γf)x--(A-γf)x~,x--x~〉≤0.
The strong monotonicity of A-γf (Lemma 2.3) implies that x-=x~ and the uniqueness is proved.
Next, we prove that the sequence{xn} is bounded. Since αn→0 and limn→∞(βn/αn)=0 by condition (C1) and (C2), respectively, we can assume, without loss of generality, that αn<∥A∥-1 and βn<αn for all n≥1. Take u∈F(T) and from (3.1), we have
(3.4)‖xn+1-u‖=‖PC[αnγf(xn)+(I-αnA)Tyn]-PC[u]‖≤‖αnγf(xn)+(I-αnA)Tyn-u‖≤αnγ‖f(xn)-f(u)‖+αn‖γf(u)-Au‖+‖(I-αnA)(Tyn-u)‖.
Since ∥I-αnA∥≤1-αnγ- and by Lemma 2.5, we note that
(3.5)‖xn+1-u‖≤αnγ‖f(xn)-f(u)‖+αn‖γf(u)-Au‖+(1-αnγ-)‖Tyn-u‖≤αnγρ‖xn-u‖+αn‖γf(u)-Au‖+(1-αnγ-)‖Tyn-Tu‖≤αnγρ‖xn-u‖+αn‖γf(u)-Au‖+(1-αnγ-)‖yn-u‖≤αnγρ‖xn-u‖+αn‖γf(u)-Au‖+(1-αnγ-)[βn‖Sxn-Su‖+βn‖Su-u‖+(1-βn)‖xn-u‖]≤αnγρ‖xn-u‖+αn‖γf(u)-Au‖+(1-αnγ-)[βn‖xn-u‖+βn‖Su-u‖+(1-βn)‖xn-u‖]=(1-αn(γ--γρ))‖xn-u‖+αn‖γf(u)-Au‖+(1-αnγ-)βn‖Su-u‖≤(1-αn(γ--γρ))‖xn-u‖+αn‖γf(u)-Au‖+βn‖Su-u‖≤(1-αn(γ--γρ))‖xn-u‖+αn‖γf(u)-Au‖+αn‖Su-u‖=(1-αn(γ--γρ))‖xn-u‖+αn[‖γf(u)-Au‖+‖Su-u‖]=(1-αn(γ--γρ))‖xn-u‖+αn(γ--γρ)‖γf(u)-Au‖+‖Su-u‖(γ--γρ).
By induction, we can obtain
(3.6)‖xn+1-u‖≤max⁡{‖x0-u‖,‖γf(u)-Au‖+‖Su-u‖(γ--γρ)},
which implies that the sequence {xn} is bounded and so are the sequences {f(xn)}, {Sxn}, and {ATyn}.
Setwn:=αnγf(xn)+(I-αnA)Tyn,n≥1. We get
(3.7)‖xn+1-xn‖=‖PC[wn+1]-PC[wn]‖≤‖wn+1-wn‖.
It follows that
(3.8)‖xn+1-xn‖≤‖(αnγf(xn)+(I-αnA)Tyn)-(αn-1γf(xn-1)+(I-αn-1A)Tyn-1)‖≤αnγ‖f(xn)-f(xn-1)‖+|αn-αn-1|‖γf(xn-1)-ATyn-1‖+(1-αnγ-)‖Tyn-Tyn-1‖≤αnγρ‖xn-xn-1‖+|αn-αn-1|‖γf(xn-1)-ATyn-1‖+(1-αnγ-)‖yn-yn-1‖.
By (3.7) and (3.8), we get
(3.9)‖xn+1-xn‖≤αnγρ‖wn-wn-1‖+|αn-αn-1|‖γf(xn-1)-ATyn-1‖+(1-αnγ-)‖yn-yn-1‖.
From (3.1), we obtain
(3.10)‖yn-yn-1‖=‖(βnSxn+(1-βn)xn)-(βn-1Sxn-1+(1-βn-1)xn-1)‖=‖βn(Sxn-Sxn-1)+(βn-βn-1)(Sxn-1-xn-1)+(1-βn)(xn-xn-1)‖≤‖xn-xn-1‖+|βn-βn-1|‖Sxn-1-xn-1‖≤‖xn-xn-1‖+|βn-βn-1|M,
where M is a constant such that
(3.11)sup⁡n∈N{‖γf(xn-1)-ATyn-1‖+‖Sxn-1-xn-1‖}≤M.
Substituting (3.10) into (3.8) to obtain
(3.12)‖xn+1-xn‖≤αnγρ‖xn-xn-1‖+|αn-αn-1|‖γf(xn-1)-ATyn-1‖+(1-αnγ-)[‖xn-xn-1‖+|βn-βn-1|M]≤αnγρ‖xn-xn-1‖+|αn-αn-1|M+(1-αnγ-)[‖xn-xn-1‖+|βn-βn-1|M]=(1-αn(γ--γρ))‖xn-xn-1‖+M[|αn-αn-1|+|βn-βn-1|]≤(1-αn(γ--γρ))‖wn-wn-1‖+M[|αn-αn-1|+|βn-βn-1|].
At the same time, we can write (3.12) as
(3.13)‖xn+1-xn‖≤(1-αn(γ--γρ))‖wn-wn-1‖+Mαn[|αn-αn-1|αn+|βn-βn-1|αn]≤(1-αn(γ--γρ))‖wn-wn-1‖+Mαn[|αn-αn-1|βn+|βn-βn-1|βn].
From (3.12), (C4), and Lemma 2.5 or from (3.13), (C3), and Lemma 2.5, we can deduce that ∥xn+1-xn∥→0, respectively.
From (3.1), we have
(3.14)‖xn-Txn‖≤‖xn-xn+1‖+‖xn+1-Txn‖=‖xn-xn+1‖+‖PC[wn]-PC[Txn]‖≤‖xn-xn+1‖+‖wn-Txn‖=‖xn-xn+1‖+‖αnγf(xn)+(I-αnA)Tyn-Txn‖≤‖xn-xn+1‖+αn‖γf(xn)-ATxn‖+(1-αnγ-)‖Tyn-Txn‖≤‖xn-xn+1‖+αn‖γf(xn)-ATxn‖+(1-αnγ-)‖yn-xn‖=‖xn-xn+1‖+αn‖γf(xn)-ATxn‖+(1-αnγ-)βn‖Sxn-xn‖.
Notice that αn→0, βn→0, and ∥xn+1-xn∥→0, so we obtain
(3.15)‖xn-Txn‖⟶0.
Next, we prove
(3.16)limsup⁡n→∞〈γf(z)-Az,xn-z〉≤0,
where z=PF(T)(I-A+γf)z. Since the sequence {xn} is bounded we can take a subsequence {xnk} of {xn} such that
(3.17)limsup⁡n→∞〈γf(z)-Az,xn-z〉=lim⁡k→∞〈γf(z)-Az,xnk-z〉,
and xnk⇀x~. From (3.15) and by Lemma 2.1, it follows that x~∈F(T). Hence, by Lemma 2.2(1) that
(3.18)limsup⁡n→∞〈γf(z)-Az,xn-z〉=lim⁡k→∞〈γf(z)-Az,xnk-z〉=〈γf(z)-Az,x~-z〉=〈(I-A+γf)z-z,x~-z〉≤0.
Now, by Lemma 2.2(1), we observe that
(3.19)〈PC[wn]-wn,PC[wn]-z〉≤0,
and so
(3.20)‖xn+1-z‖2=〈PC[wn]-z,PC[wn]-z〉=〈PC[wn]-wn,PC[wn]-z〉+〈wn-z,PC[wn]-z〉≤〈wn-z,PC[wn]-z〉=〈αnγf(xn)+(I-αnA)Tyn-z,xn+1-z〉≤αnγ‖f(xn)-f(z)‖‖xn+1-z‖+αn〈γf(z)-Az,xn+1-z〉+(1-αnγ-)‖Tyn-z‖‖xn+1-z‖≤αnγρ‖xn-z‖‖xn+1-z‖+αn〈γf(z)-Az,xn+1-z〉+(1-αnγ-)‖yn-z‖‖xn+1-z‖=αnγρ‖xn-z‖‖xn+1-z‖+αn〈γf(z)-Az,xn+1-z〉+(1-αnγ-)‖βnSxn+(1-βn)xn-z‖‖xn+1-z‖≤αnγρ‖xn-z‖‖xn+1-z‖+αn〈γf(z)-Az,xn+1-z〉+(1-αnγ-)[βn‖Sxn-Sz‖+βn‖Sz-z‖+(1-βn)‖xn-z‖]∥xn+1-z∥≤αnγρ‖xn-z‖‖xn+1-z‖+αn〈γf(z)-Az,xn+1-z〉+(1-αnγ-)[βn‖xn-z‖+βn‖Sz-z‖+(1-βn)‖xn-z‖]‖xn+1-z‖=(1-αn(γ--γρ))‖xn-z‖‖xn+1-z‖+αn〈γf(z)-Az,xn+1-z〉+(1-αnγ-)βn‖Sz-z‖‖xn+1-z‖≤[1-αn(γ--γρ)]2[‖xn-z‖2+‖xn+1-z‖2]+αn〈γf(z)-Az,xn+1-z〉+(1-αnγ-)βn‖Sz-z‖‖xn+1-z‖.
Hence, it follows that
(3.21)‖xn+1-z‖2≤1-αn(γ--γρ)1+αn(γ--γρ)‖xn-z‖2+2αn1+αn(γ--γρ)〈γf(z)-Az,xn+1-z〉+2(1-αnγ-)βn1+αn(γ--γρ)‖Sz-z‖‖xn+1-z‖=[2αn(γ--γρ)1+αn(γ--γρ)][βn(1-αnγ-)αn(γ--γρ)1αn(γ--γρ)〈γf(z)-Az,xn+1-z〉+βn(1-αnγ-)αn(γ--γρ)‖Sz-z‖‖xn+1-z‖]×[1-2αn(γ--γρ)1+αn(γ--γρ)]‖xn-z‖2.
We observe that
(3.22)limsup⁡n→∞[1αn(γ--γρ)〈γf(z)-Az,xn+1-z〉+βn(1-αnγ-)αn(γ--γρ)‖Sz-z‖‖xn+1-z‖]≤0.
Thus, by Lemma 2.4, xn→z as n→∞. This is completes.From Theorem3.1, we can deduce the following interesting corollary.Corollary 3.2 (Yao et al. [9]).
LetC be a nonempty closed convex subset of a real Hilbert space H. Let f:C→H be a ρ-contraction (possibly nonself) with ρ∈(0,1). Let S,T:C→C be two nonexpansive mappings with F(T)≠∅.  {αn} and {βn} are two sequences in (0,1). Starting with an arbitrary initial guess x0∈C and {xn} is a sequence generated by
(3.23)yn=βnSxn+(1-βn)xn,xn+1=PC[αnf(xn)+(1-αn)Tyn],∀n≥1.
Suppose that the following conditions are satisfied:  (C1)
limn→∞αn=0 and ∑n=1∞‍αn=∞,(C2)
limn→∞(βn/αn)=0,(C3)
limn→∞(|αn-αn-1|/αn)=0 and limn→∞(|βn-βn-1|/βn)=0, or(C4)
∑n=1∞‍|αn-αn-1|<∞ and ∑n=1∞‍|βn-βn-1|<∞. Then the sequence {xn} converges strongly to a point x~∈H, which is the unique solution of the variational inequality:
(3.24)x~∈F(T),〈(I-f)x~,x-x~〉≥0,∀x∈F(T).
Equivalently, one has PF(T)(f)x~=x~. In particular, if one takes f=0, then the sequence {xn} converges in norm to the Minimum norm fixed point x~ of T, namely, the point x~ is the unique solution to the quadratic minimization problem:
(3.25)z=argmin⁡x∈F(T)‖x‖2.Proof.
As a matter of fact, if we takeA=I and γ=1 in Theorem 3.1. This completes the proof.Under different conditions on data we obtain the following result.Theorem 3.3.
LetC be a nonempty closed convex subset of a real Hilbert space H. Let f:C→H be a ρ-contraction (possibly nonself) with ρ∈(0,1). Let S,T:C→C be two nonexpansive mappings with F(T)≠∅. Let A be a strongly positive linear bounded operator on a Hilbert space H with coefficient γ->0 and 0<γ<γ-/ρ. {αn} and {βn} are two sequences in (0,1). Starting with an arbitrary initial guess x0∈C and {xn} is a sequence generated by
(3.26)yn=βnSxn+(1-βn)xn,xn+1=PC[αnγf(xn)+(I-αnA)Tyn],∀n≥1.
Suppose that the following conditions are satisfied:  (C1)
limn→∞αn=0 and ∑n=1∞‍αn=∞,(C2)
limn→∞(βn/αn)=τ∈(0,∞),(C5)
limn→∞((|αn-αn-1|+|βn-βn-1|)/αnβn)=0,(C6)
there exists a constantK>0 such that (1/αn)|1/βn-1/βn-1|≤K. Then the sequence {xn} converges strongly to a point x~∈H, which is the unique solution of the variational inequality:
(3.27)x~∈F(T),〈1τ(A-γf)x~+(I-S)x~,x-x~〉≥0,∀x∈F(T).Proof.
First of all, we show that (3.27) has the unique solution. Indeed, let x- and x~ be two solutions. Then
(3.28)〈(A-γf)x~,x~-x-〉≤τ〈(I-S)x~,x--x~〉.
Analogously, we have
(3.29)〈(A-γf)x-,x--x~〉≤τ〈(I-S)x-,x~-x-〉.
Adding (3.28) and (3.29), by Lemma 2.3, we obtain
(3.30)(γ--γρ)‖x~-x-‖2≤〈(A-γf)x~-(A-γf)x-,x~-x-〉≤-τ〈(I-S)x~-(I-S)x-,x~-x-〉≤0,
and so x~=x-. From (C2), we can assume, without loss of generality, that βn≤(τ+1)αn for all n≥1. By a similar argument in Theorem 3.1, we have
(3.31)‖xn+1-u‖≤αnγρ‖xn-u‖+αn‖γf(u)-Au‖+(1-αnγ-)[‖xn-u‖+βn‖Su-u‖+(1-βn)‖xn-u‖]=(1-αn(γ--γρ))‖xn-u‖+αn‖γf(u)-Au‖+(1-αnγ-)βn‖Su-u‖≤(1-αn(γ--γρ))‖xn-u‖+αn‖γf(u)-Au‖+βn‖Su-u‖≤(1-αn(γ--γρ))‖xn-u‖+αn‖γf(u)-Au‖+(τ+1)αn‖Su-u‖=(1-αn(γ--γρ))‖xn-u‖+αn[‖γf(u)-Au‖+(τ+1)‖Su-u‖]=(1-αn(γ--γρ))‖xn-u‖+αn(γ--γρ)‖γf(u)-Au‖+(τ+1)‖Su-u‖(γ--γρ).
By induction, we obtain
(3.32)‖xn-u‖≤max⁡{‖x0-u‖,1γ--γρ[‖γf(u)-Au‖+(τ+1)‖Su-u‖]},
which implies that the sequence {xn} is bounded. Since (C5) implies (C4) then, from Theorem 3.1, we can deduce ∥xn+1-xn∥→0.
From (3.1), we note that
(3.33)xn+1=PC[wn]-wn+wn+yn-yn=PC[wn]-wn+αnγf(xn)+(Tyn-yn)+(yn-αnATyn).
Hence, it follows that
(3.34)xn-xn+1=(wn-PC[wn])+αn(Axn-γf(xn))+(yn-Tyn)+(xn-yn)+αn(ATyn-Axn)=(wn-PC[wn])+αn(A-γf)xn+(I-T)yn+βn(I-S)xn+αnA(Tyn-xn),
and so
(3.35)xn-xn+1(1-αn)βn=1(1-αn)βn(wn-PC[wn])+αn(1-αn)βn(A-γf)xn+1(1-αn)βn(I-T)yn+1(1-αn)(I-S)xn+αn(1-αn)βnA(Tyn-xn).
Set vn:=(xn-xn+1)/(1-αn)βn. Then, we have
(3.36)vn=1(1-αn)βn(wn-PC[wn])+αn(1-αn)βn(A-γf)xn+1(1-αn)βn(I-T)yn+1(1-αn)(I-S)xn+αn(1-αn)βnA(Tyn-xn).
From (3.12) in Theorem 3.1 and (C6), we obtain
(3.37)‖xn+1-xn‖βn≤(1-αn(γ--γρ))‖xn-xn-1‖βn+M[|αn-αn-1|βn+|βn-βn-1|βn]=(1-αn(γ--γρ))‖xn-xn-1‖βn+(1-αn(γ--γρ))‖xn-xn-1‖βn-1-(1-αn(γ--γρ))‖xn-xn-1‖βn-1+M[|αn-αn-1|βn+|βn-βn-1|βn]=(1-αn(γ--γρ))‖xn-xn-1‖βn-1+(1-αn(γ--γρ))‖xn-xn-1‖[1βn-1βn-1]+M[|αn-αn-1|βn+|βn-βn-1|βn]≤(1-αn(γ--γρ))‖xn-xn-1‖βn-1+‖xn-xn-1‖|1βn-1βn-1|+M[|αn-αn-1|βn+|βn-βn-1|βn]≤(1-αn(γ--γρ))‖xn-xn-1‖βn-1+αnK‖xn-xn-1‖+M[|αn-αn-1|βn+|βn-βn-1|βn]≤(1-αn(γ--γρ))‖wn-wn-1‖βn-1+αnK‖xn-xn-1‖+M[|αn-αn-1|βn+|βn-βn-1|βn].
This together with Lemma 2.4 and (C2) imply that
(3.38)lim⁡n→∞‖xn+1-xn‖βn=lim⁡n→∞‖wn+1-wn‖βn=lim⁡n→∞‖wn+1-wn‖αn=0.
From (3.36), for z∈F(T), we have
(3.39)〈vn,xn-z〉=1(1-αn)βn〈wn-PC[wn],PC[wn-1]-z〉+αn(1-αn)βn〈(A-γf)xn,xn-z〉+1(1-αn)βn〈(I-T)yn,xn-z〉+1(1-αn)〈(I-S)xn,xn-z〉+αn(1-αn)βn〈A(Tyn-xn),xn-z〉=1(1-αn)βn〈wn-PC[wn],PC[wn]-z〉+1(1-αn)βn〈wn-PC[wn],PC[wn-1]-PC[wn]〉+αn(1-αn)βn〈(A-γf)xn-(A-γf)z,xn-z〉+αn(1-αn)βn〈(A-γf)z,xn-z〉+1(1-αn)〈(I-S)xn-(I-S)z,xn-z〉+1(1-αn)〈(I-S)z,xn-z〉+1(1-αn)βn〈(I-T)yn,xn-z〉+αn(1-αn)βn〈A(Tyn-xn),xn-z〉.
By Lemmas 2.2 and 2.3, we obtain
(3.40)〈vn,xn-z〉≥1(1-αn)βn〈wn-PC[wn],PC[wn-1]-PC[wn]〉+(γ--γρ)αn(1-αn)βn‖xn-z‖2+αn(1-αn)βn〈(A-γf)z,xn-z〉+1(1-αn)〈(I-S)z,xn-z〉+1(1-αn)βn〈(I-T)yn,xn-z〉+αn(1-αn)βn〈A(Tyn-xn),xn-z〉.
Now, we observe that
(3.41)‖xn-z‖2≤(1-αn)βn(γ--γρ)αn〈v,xn-z〉-βn(γ--γρ)αn〈(I-S)z,xn-z〉-1(γ--γρ)〈(A-γf)z,xn-z〉-1(γ--γρ)αn〈(I-T)yn,xn-z〉-1(γ--γρ)〈A(Tyn-xn),xn-z〉-1(γ--γρ)αn〈wn-PC[wn],PC[wn-1]-PC[wn]〉≤(1-αn)βn(γ--γρ)αn〈v,xn-z〉-βn(γ--γρ)αn〈(I-S)z,xn-z〉-1(γ--γρ)〈(A-γf)z,xn-z〉-1(γ--γρ)αn〈(I-T)yn,xn-z〉-1(γ--γρ)〈A(Tyn-xn),xn-z〉+‖wn-wn-1‖(γ--γρ)‖wn-PC[wn]‖.
From (C1) and (C2), we have βn→0. Hence, from (3.1), we deduce ∥yn-xn∥→0 and ∥xn+1-Tyn∥→0. Therefore,
(3.42)‖yn-Tyn‖≤‖yn-xn‖+‖xn-xn+1‖+‖xn+1-Tyn‖→0.
Sincevn→0, (I-T)yn→0, A(Tyn-xn)→0, and ∥wn-wn-1∥/(γ--γρ)→0, every weak cluster point of {xn} is also a strong cluster point. Note that the sequence {xn} is bounded, thus there exists a subsequence {xnk} converging to a point x~∈H. For all z∈F(T), it follows from (3.39) that
(3.43)〈(A-γf)xnk,xnk-z〉=(1-αnk)βnkαnk〈vnk,xnk-z〉-1αnk〈(I-T)ynk,xnk-z〉-βnkαnk〈(I-S)xnk,xnk-z〉-〈A(Tynk-xnk),xnk-z〉-1αnk〈wnk-PC[wnk],PC[wnk-1]-z〉≤(1-αnk)βnkαnk〈vnk,xnk-z〉-1αnk〈(I-T)ynk,xnk-z〉-βnkαnk〈(I-S)xnk,xnk-z〉-〈A(Tynk-xnk),xnk-z〉-1αnk〈wnk-PC[wnk],PC[wnk-1]-PC[wnk]〉-〈A(Tynk-xnk),xnk-z〉-1αnk〈wnk-PC[wnk],PC[wnk-1]-z〉≤(1-αnk)βnkαnk〈vnk,xnk-z〉-1αnk〈(I-T)ynk,xnk-z〉-βnkαnk〈(I-S)xnk,xnk-z〉-〈A(Tynk-xnk),xnk-z〉+‖wnk-wnk-1‖αnk‖wnk-PC[wnk]‖.
Letting k→∞, we obtain
(3.44)〈(A-γf)x~,x~-z〉≤-τ〈(I-S)x~,x~-z〉,∀z∈F(T).
By Lemma 2.6 and (3.27) having the unique solution, it follows that ωw(xn)={x~}. Therefore, xn→x~ as n→∞. This completes the proof.From Theorem3.3, we can deduce the following interesting corollary.Corollary 3.4 (Yao et al. [9]).
LetC be a nonempty closed convex subset of a real Hilbert space H. Let f:C→H be a ρ-contraction (possibly nonself) with ρ∈(0,1). Let S,T:C→C be two nonexpansive mappings with F(T)≠∅. {αn} and {βn} are two sequences in (0,1) Starting with an arbitrary initial guess x0∈C and {xn} is a sequence generated by
(3.45)yn=βnSxn+(1-βn)xn,xn+1=PC[αnf(xn)+(1-αn)Tyn],∀n≥1.
Suppose that the following conditions are satisfied:  (C1)
limn→∞αn=0 and ∑n=1∞‍αn=∞,(C2)
limn→∞(βn/αn)=τ∈(0,∞),(C5)
limn→∞((|αn-αn-1|+|βn-βn-1|)/αnβn)=0,(C6)
there exists a constantK>0 such that (1/αn)|1/βn-1/βn-1|≤K. Then the sequence {xn} converges strongly to a point x~∈H, which is the unique solution of the variational inequality:
(3.46)x~∈F(T),〈1τ(I-f)x~+(I-S)x~,x-x~〉≥0,∀x∈F(T).Proof.
As a matter of fact, if we takeA=I and γ=1 in Theorem 3.3 then this completes the proof.Corollary 3.5 (Yao et al. [9]).
LetC be a nonempty closed convex subset of a real Hilbert space H. Let S,T:C→C be two nonexpansive mappings with F(T)≠∅. {αn} and {βn} are two sequences in (0,1). Starting with an arbitrary initial guess x0∈C and suppose {xn} is a sequence generated by
(3.47)yn=βnSxn+(1-βn)xn,xn+1=PC[(1-αn)Tyn],∀n≥1.
Suppose that the following conditions are satisfied:  (C1)
limn→∞αn=0 and ∑n=1∞‍αn=∞,(C2)
limn→∞(βn/αn)=1,(C5)
limn→∞((|αn-αn-1|+|βn-βn-1|)/αnβn)=0,(C6)
there exists a constantK>0 such that (1/αn)|1/βn-1/βn-1|≤K. Then the sequence {xn} converges strongly to a point x~∈H, which is the unique solution of the variational inequality:
(3.48)x~∈F(T),〈(I-S2)x~,x-x~〉≥0,∀x∈F(T).Proof.
As a matter of fact, if we takeA=I, f=0, and γ=1 in Theorem 3.3 then this is completes the proof.Remark 3.6.
Prototypes for the iterative parameters are, for example,αn=n-θ and βn=n-ω (with θ,ω>0). Since |αn-αn-1|≈n-θ and |βn-βn-1|≈n-ω, it is not difficult to prove that (C5) is satisfied for 0<θ,ω<1 and (C6) is satisfied if θ+ω≤1.Remark 3.7.
Our results improve and extend the results of Yao et al. [9] by taking A=I and γ=1 in Theorems 3.1 and 3.3.The following is an example to support Theorem3.3.Example 3.8.
LetH=ℝ,C=[-1/4,1/4],T=I,S=-I,A=I, f(x)=x2, PC=I,βn=1/n, αn=1/n for every n∈ℕ, we have τ=1 and choose γ-=1/2,  ρ=1/3 and γ=1. Then {xn} is the sequence
(3.49)xn+1=xn2n+(1-1n)(1-2n)xn,
and xn→x~=0 as n→∞, where x~=0 is the unique solution of the variational inequality
(3.50)x~∈F(T)=[-14,14],〈(3x~-x~2),x-x~〉≥0,∀x∈F(T)=[-14,14].



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*Source: 174318-2012-07-12.xml*