# Strong Convergence Algorithms for Hierarchical Fixed Points Problems and Variational Inequalities

Authors: Gendai Gu; Shenghua Wang; Yeol Je Cho
Journal: Journal of Applied Mathematics
 (2011)

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## Abstract

We introduce a new iterative scheme that converges strongly to a
common fixed point of a countable family of nonexpansive mappings in a Hilbert space such
that the common fixed point is a solution of a hierarchical fixed point problem. Our results
extend the ones of Moudafi, Xu, Cianciaruso et al., and Yao et al.


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## Body

## 1. Introduction

LetH be a real Hilbert space and C a nonempty closed convex subset of H. A mapping T:C→C is called nonexpansive if one has (1.1)‖Tx-Ty‖≤‖x-y‖,∀x,y∈C.
If there exists a point x∈C such that x=Tx, then x is said to be a fixed point of T. We denote the set of all fixed points of T by F(T). It is well known that F(T) is closed and convex if T is nonexpansive.LetS:C→H be a mapping. The following problem is called a hierarchical fixed point problem: find x*∈F(T) such that (1.2)〈x*-Sx*,x*-x〉≤0,∀x∈F(T).It is known that the hierarchical fixed point problem (1.2) links with some monotone variational inequalities and convex programming problems (see [1]).In order to solve the hierarchical fixed point problem (1.2), Moudafi [2] introduced the following Krasnoselski-Mann algorithm: (1.3)xn+1=(1-αn)xn+αn(σnSxn+(1-σn)Txn),∀n≥0,
where {αn} and {σn} are two sequences in (0,1), and he proved that {xn} converges weakly to a fixed point of T which is a solution of the problem (1.2).Letf:C→C be a mapping. The mapping f is called a contraction if there exists a constant λ∈[0,1) such that ∥fx-fy∥≤λ∥x-y∥ for all x,y∈C. For obtaining a strong convergence result, Mainge and Moudafi in [3] and Marino and Xu in [4] introduced the following algorithm: (1.4)xn+1=(1-αn)f(xn)+αn(σnSxn+(1-σn)Txn),∀n≥0,
where S:C→C is a nonexpansive mapping and {αn} and {σn} are two sequences in (0,1), and they proved that {xn} converges strongly to a fixed point of T which is a solution of the problem (1.2). Recently, for solving the hierarchical fixed point problem (1.2), Cianciaruso et al. [5] also studied the following iterative scheme: (1.5)yn=βnSxn+(1-βn)xn,xn+1=αnf(xn)+(1-αn)Tyn,∀n≥0,
where {αn} and {βn} are two sequences in [0,1]. The authors proved some strong convergence results. Very recently, Yao et al. [1] introduced the following strong convergence iterative algorithm to solve the problem (1.2): (1.6)yn=βnSxn+(1-βn)xn,xn+1=PC[αnf(xn)+(1-αn)Tyn],∀n≥0,
where f:C→H is a contraction and {αn} and {βn} are two sequences in (0,1). Under some certain restrictions on parameters, the authors proved that the sequence {xn} generated by (1.6) converges strongly to z∈F(T), which is the unique solution of the following variational inequality: (1.7)〈(I-f)z,x-z〉≥0,∀x∈F(T).By changing the restrictions on parameters, the authors obtained another result on the iterative scheme (1.6), that is, the sequence {xn} generated by (1.6) converges strongly to a point x*∈F(T), which is the unique solution of the following variational inequality: (1.8)〈1τ(I-f)x*+(I-S)x*,y-x*〉≥0,∀y∈F(T),
where τ∈(0,∞) is a constant.LetS:C→H be a nonexpansive mapping and {Ti}i=1∞:C→C a countable family of nonexpansive mappings. In this paper, motivated and inspired by the results of Yao et al. [1] and Marino and Xu [4], we introduce and study the following iterative scheme: (1.9)yn=PC[βnSxn+(1-βn)xn],xn+1=PC[αnf(xn)+∑i=1n(αi-1-αi)Tiyn],∀n≥1,
where α0=1, {αn} is a strictly decreasing sequence in (0,1) and {βn} is a sequence in (0,1). Under some certain conditions on parameters, we first prove that the sequence {xn} generated by (1.9) converges strongly to x*∈⋂i=1∞F(Ti), which is the unique solution of the following variational inequality: (1.10)〈(I-f)x*,x-x*〉≥0,∀x∈⋂i=1∞F(Ti).
By changing the restrictions on parameters, we also prove that the sequence {xn} converges strongly to x*∈∩i=1∞F(Ti), which is the unique solution of the following variational inequality: (1.11)〈1τ(I-f)x*+(I-S)x*,y-x*〉≥0,∀y∈⋂i=1∞F(Ti),
where τ∈(0,∞) is a constant. It is easy to see that, if Ti=T for each i≥1 and S is a self-mapping of C into itself, then our algorithm (1.9) is reduced to (1.6) of Yao et al. [1] Also, our results extend the corresponding ones of Moudafi [6], Xu [7], and Cianciaruso et al. [5].


## 2. Preliminaries

LetH be a Hilbert space and C a nonempty closed convex subset of H. Let T be a nonexpansive mapping of C into itself such that F(T)≠∅. For all x̂∈F(T) and all x∈C, we have (2.1)‖x-x̂‖2≥‖Tx-Tx̂‖2=‖Tx-x̂‖2=‖Tx-x+(x-x̂)‖2=‖Tx-x‖2+‖x-x̂‖2+2〈Tx-x,x-x̂〉
and hence (2.2)‖Tx-x‖2≤2〈x-Tx,x-x̂〉,∀x̂∈F(T),x∈C.Letx∈H be an arbitrary point. There exists a unique nearest point in C, denoted by PCx, such that (2.3)‖PCx-x‖≤‖y-x‖,∀y∈C.
Moreover, we have the following: (2.4)z=PCx⟺〈x-z,z-y〉≥0,∀y∈C.LetI denote the identity operator of H, and let {xn} be a sequence in a Hilbert space H and x∈H. Throughout this paper, xn→x denotes that {xn} strongly converges to x and xn⇀x denotes that {xn} weakly converges to x.The following lemmas will be used in the next section.Lemma 2.1 (see [8]).
LetT:C→C be a nonexpansive mapping with  F(T)≠∅. If  {xn} is a sequence in  C weakly converging to a point  x∈C and  {(I-T)xn} converges strongly to a point  y∈C, then  (I-T)x=y. In particular, if  y=0, then  x∈F(T).Lemma 2.2 (see [9]).
Letf:C→H be a contraction with coefficient  λ∈[0,1) and  T:C→C a nonexpansive mapping. Then one has the following. (1)
The mapping(I-f) is strongly monotone with coefficient (1-λ), that is,
(2.5)〈x-y,(I-f)x-(I-f)y〉≥(1-λ)‖x-y‖2,∀x,y∈C.(2)
The mappingI-T is monotone, that is,
(2.6)〈x-y,(I-T)x-(I-T)y〉≥0,∀x,y∈C.Lemma 2.3 (see [10]).
Let{sn},  {cn} be the sequences of nonnegative real numbers, and let  {an}⊂(0,1). Suppose that  {bn} is a real number sequence such that
(2.7)sn+1≤(1-an)sn+bn+cn,∀n≥0.
Assume that ∑n=0∞cn<∞. Then the following results hold. (1)
Ifbn≤βan, where β≥0, then {sn} is a bounded sequence.(2)
If one has(2.8)∑n=0∞an=∞,limsup⁡n→∞bnan≤0, then limn→∞sn=0.


## 3. Main Results

Now, we give the main results in this paper.Theorem 3.1.
LetC be a nonempty closed convex subset of a real Hilbert space  H. Let  f:C→H be a  λ-contraction with  λ∈[0,1). Let  S:C→H be a nonexpansive mapping and  {Ti}i=1∞ a countable family of nonexpansive mappings of  C into itself such that  F=⋂i=1∞F(Ti)≠∅. Set  α0=1, and let  {αn}⊂(0,1) be a strictly decreasing sequence and  {βn}⊂(0,1) a sequence satisfying the following conditions:  (a)
limn→∞αn=0 and ∑n=1∞αn=∞,(b)
limn→∞(βn/αn)=0,(c)
∑n=1∞(αn-1-αn)<∞ and ∑n=1∞|βn-1-βn|<∞. Then the sequence {xn} generated by (1.9) converges strongly to a point x*∈F, which is the unique solution of the variational inequality
(3.1)〈(I-f)x*,x-x*〉≥0,∀x∈F.Proof.
First,PFf is a contraction from C into itself with a constant λ and C is complete, and there exists a unique x*∈C such that x*=PFf(x*). From (2.4), it follows that x* is the unique solution of the problem (3.1).
Now, we prove that{xn} converges strongly to x*. To this end, we first prove that {xn} is bounded. Take p∈F. Then it follows from (1.9) that
(3.2)‖yn-p‖=‖PC[βnSxn+(1-βn)xn]-PCp‖≤‖βnSxn+(1-βn)xn-p‖≤βn‖Sxn-Sp‖+βn‖Sp-p‖+(1-βn)‖xn-p‖≤βn‖xn-p‖+βn‖Sp-p‖+(1-βn)‖xn-p‖=‖xn-p‖+βn‖Sp-p‖
and hence (note that {αn} is strictly decreasing)
(3.3)‖xn+1-p‖=‖PC[αnf(xn)+∑i=1n(αi-1-αi)Tiyn]-PCp‖≤‖αnf(xn)+∑i=1n(αi-1-αi)Tiyn-p‖=‖αn(f(xn)-p)+∑i=1n(αi-1-αi)(Tiyn-p)‖≤αn‖f(xn)-f(p)‖+αn‖f(p)-p‖+∑i=1n(αi-1-αi)‖Tiyn-p‖≤αnλ‖xn-p‖+αn‖f(p)-p‖+∑i=1n(αi-1-αi)‖yn-p‖≤αnλ‖xn-p‖+αn‖f(p)-p‖+∑i=1n(αi-1-αi)(‖xn-p‖+βn‖Sp-p‖)=αnλ‖xn-p‖+αn‖f(p)-p‖+(1-αn)(‖xn-p‖+βn‖Sp-p‖).
By condition (b), we can assume that βn≤αn for all n≥1. Hence, from above inequality, we get
(3.4)‖xn+1-p‖≤αnλ‖xn-p‖+αn‖f(p)-p‖+(1-αn)(‖xn-p‖+αn‖Sp-p‖)≤αnλ‖xn-p‖+αn‖f(p)-p‖+(1-αn)‖xn-p‖+αn‖Sp-p‖=[1-αn(1-λ)]‖xn-p‖+αn(‖f(p)-p‖+‖Sp-p‖).
For each n≥1, let an=αn(1-λ), bn=αn(∥f(p)-p∥+∥Sp-p∥), and cn=0. Then {an}, {bn}, and {cn} satisfy the condition of Lemma 2.3(1). Hence, it follows from Lemma 2.3(1) that {xn} is bounded and so are {f(xn)}, {yn}, {Tixn}, and {Tiyn} for all i≥1. Set un=αnf(xn)+∑i=1n(αi-1-αi)Tiyn for each n≥1. From (1.9), we have
(3.5)‖xn+1-xn‖=‖PCun-PCun-1‖≤‖un-un-1‖=‖∑i=1n(αi-1-αi)αn(f(xn)-f(xn-1))+(αn-αn-1)f(xn-1)+∑i=1n(αi-1-αi)(Tiyn-Tiyn-1)+(αn-1-αn)Tnyn-1‖≤αn‖f(xn)-f(xn-1)‖+∑i=1n(αi-1-αi)‖yn-yn-1‖+(αn-1-αn)(‖f(xn-1)‖+‖Tnyn-1‖)≤αnλ‖xn-xn-1‖+∑i=1n(αi-1-αi)‖yn-yn-1‖+(αn-1-αn)M=αnλ‖xn-xn-1‖+(1-αn)‖yn-yn-1‖+(αn-1-αn)M,
where M is a constant such that
(3.6)sup⁡n≥1{‖f(xn-1)‖+‖Tnyn-1‖+‖Sxn-1‖+‖xn-1‖}≤M.
From (1.9), we have
(3.7)‖yn-yn-1‖=‖PC(βnSxn+(1-βn)xn)-PC(βn-1Sxn-1+(1-βn-1)xn-1)‖≤‖(βnSxn+(1-βn)xn)-(βn-1Sxn-1+(1-βn-1)xn-1)‖=‖βn(Sxn-Sxn-1)+(βn-βn-1)Sxn-1+(1-βn)(xn-xn-1)+(βn-1-βn)xn-1‖≤‖xn-xn-1‖+|βn-βn-1|M.
Substituting (3.7) into (3.5), we get that
(3.8)‖xn+1-xn‖≤‖un-un-1‖≤αnλ‖xn-xn-1‖+(1-αn)[‖xn-xn-1‖+|βn-βn-1|M]+(αn-1-αn)M≤(1-(1-λ)αn)‖xn-xn-1‖+M[(αn-1-αn)+|βn-βn-1|].
Let an=(1-λ)αn, bn=0, and cn=(αn-1-αn)+|βn-βn-1|. Then conditions (a) and (c) imply that {an}, {bn}, and {cn} satisfy the condition of Lemma 2.3(2). Thus, by Lemma 2.3(2), we can conclude that
(3.9)lim⁡n→∞‖xn+1-xn‖=0.
Since Tixn∈C for each i≥1 and ∑i=1n(αi-1-αi)+αn=1, we have
(3.10)∑i=1n(αi-1-αi)Tixn+αnz∈C,∀z∈C.
Now, fixing a z∈F, from (1.9) we have
(3.11)∑i=1n(αi-1-αi)(xn-Tixn)=PCun+(1-αn)xn-(∑i=1n(αi-1-αi)Tixn+αnz)+αnz-xn+1=PCun-PC[∑i=1n(αi-1-αi)Tixn+αnz]+(1-αn)(xn-xn+1)+αn(z-xn+1).
Hence it follows that
(3.12)∑i=1n(αi-1-αi)〈xn-Tixn,xn-p〉=〈PCun-PC[∑i=1n(αi-1-αi)Tixn+αnz],xn-p〉+〈(1-αn)(xn-xn+1),xn-p〉+αn〈z-xn+1,xn-p〉≤‖un-∑i=1n(αi-1-αi)Tixn-αnz‖‖xn-p‖+(1-αn)‖xn-xn+1‖‖xn-p‖+αn‖z-xn+1‖‖xn-p‖=‖αn(f(xn)-z)+∑i=1n(αi-1-αi)(Tiyn-Tixn)‖‖xn-p‖+(1-αn)‖xn-xn+1‖‖xn-p‖+αn‖z-xn+1‖‖xn-p‖≤αn‖f(xn)-z‖‖xn-p‖+∑i=1n(αi-1-αi)‖yn-xn‖‖xn-p‖+(1-αn)‖xn-xn+1‖‖xn-p‖+αn‖z-xn+1‖‖xn-p‖≤αn‖f(xn)-z‖‖xn-p‖+∑i=1n(αi-1-αi)βn‖Sxn-xn‖‖xn-p‖+(1-αn)‖xn-xn+1‖‖xn-p‖+αn‖z-xn+1‖‖xn-p‖=αn‖f(xn)-z‖‖xn-p‖+(1-αn)βn‖Sxn-xn‖‖xn-p‖+(1-αn)‖xn-xn+1‖‖xn-p‖+αn‖z-xn+1‖‖xn-p‖≤(2αn+βn)M′+(1-αn)‖xn-xn+1‖‖xn-p‖,
where
(3.13)M′=sup⁡n≥1{‖f(xn)-z‖‖xn-p‖,‖Sxn-xn‖‖xn-p‖,‖z-xn+1‖‖xn-p‖}.
Now, from (2.2) and (3.12), we get
(3.14)12∑i=1n(αi-1-αi)‖xn-Tixn‖2≤∑i=1n(αi-1-αi)〈xn-Tixn,xn-p〉≤(2αn+βn)M′+(1-αn)‖xn-xn+1‖‖xn-p‖.
From (a), (b), (3.9), and (3.14), we have
(3.15)lim⁡n→∞∑i=1n(αi-1-αi)‖xn-Tixn‖=0.
Since (αi-1-αi)∥xn-Tixn∥≤∑i=1n(αi-1-αi)∥xn-Tixn∥ for each i≥1 and {αn} is strictly decreasing, one has
(3.16)lim⁡n→∞‖xn-Tixn‖=0,∀i≥1.
Next, we prove that(3.17)limsup⁡n→∞〈f(x*)-x*,xn-x*〉≤0.
Since {xn} is bounded, we can take a subsequence {xnk} of {xn} such that xnk⇀x′ and
(3.18)limsup⁡n→∞〈f(x*)-x*,xn-x*〉=lim⁡k→∞〈f(x*)-x*,xnk-x*〉.
From (3.16) and Lemma 2.1, we conclude that x′∈F(Ti) for each i≥1, that is, x′∈⋂i=1∞F(Ti). Then
(3.19)lim⁡k→∞〈f(x*)-x*,xnk-x*〉=〈f(x*)-x*,x′-x*〉≤0.
By (2.4), we have
(3.20)〈PCun-un,PCun-x*〉≤0.
Also,
(3.21)∑i=1n(αi-1-αi)〈Tiyn-x*,xn+1-x*〉≤∑i=1n(αi-1-αi)‖Tiyn-x*‖‖xn+1-x*‖≤∑i=1n(αi-1-αi)‖yn-x*‖‖xn+1-x*‖=(1-αn)‖yn-x*‖‖xn+1-x*‖=(1-αn)‖PC(βnSxn+(1-βn)xn)-PCx*‖‖xn+1-x*‖≤(1-αn)‖βnSxn+(1-βn)xn-x*‖‖xn+1-x*‖=(1-αn)‖βn(Sxn-Sx*)+βn(Sx*-x*)+(1-βn)(xn-x*)‖‖xn+1-x*‖≤(1-αn)[βn‖xn-x*‖+βn‖Sx*-x*‖+(1-βn)‖xn-x*‖]‖xn+1-x*‖=(1-αn)‖xn-x*‖‖xn+1-x*‖+(1-αn)βn‖Sx*-x*‖‖xn+1-x*‖.
Thus,
(3.22)‖xn+1-x*‖2=〈PCun-un,PCun-x*〉+〈un-x*,xn+1-x*〉≤〈un-x*,xn+1-x*〉=αn〈f(xn)-f(x*),xn+1-x*〉+∑i=1n(αi-1-αi)〈Tiyn-x*,xn+1-x*〉+αn〈f(x*)-x*,xn+1-x*〉≤αnλ‖xn-x*‖‖xn+1-x*‖+(1-αn)‖xn-x*‖‖xn+1-x*‖+(1-αn)βn‖Sx*-x*‖‖xn+1-x*‖+αn〈f(x*)-x*,xn+1-x*〉=[1-αn(1-λ)]‖xn-x*‖‖xn+1-x*‖+(1-αn)βn‖Sx*-x*‖‖xn+1-x*‖+αn〈f(x*)-x*,xn+1-x*〉≤1-αn(1-λ)2[‖xn-x*‖2+‖xn+1-x*‖2]+(1-αn)βn‖Sx*-x*‖‖xn+1-x*‖+αn〈f(x*)-x*,xn+1-x*〉.
It follows that
(3.23)‖xn+1-x*‖2≤[1-2(1-λ)αn1+(1-λ)αn]‖xn-x*‖2+2(1-αn)βn1+(1-λ)αn‖Sx*-x*‖‖xn+1-x*‖+2αn1+(1-λ)αn〈f(x*)-x*,xn+1-x*〉=[1-2(1-λ)αn1+(1-λ)αn]‖xn-x*‖2+2(1-λ)αn1+(1-λ)αn×{(1-αn)βn(1-λ)αn‖Sx*-x*‖‖xn+1-x*‖+11-λ〈f(x*)-x*,xn+1-x*〉}.
Let an=2(1-λ)αn/(1+(1-λ))αn, bn=(2(1-λ)αn/(1+(1-λ)αn)){(1-αn)βn/(1-λ)αn∥Sx*-x*∥∥xn+1-x*∥+(1/1-λ)〈f(x*)-x*,xn+1-x*〉}, and cn=0 for all n≥1. Since
(3.24)limsup⁡n→∞{(1-αn)βn(1-λ)αn‖Sx*-x*‖‖xn+1-x*‖+11-λ〈f(x*)-x*,xn+1-x*〉}≤0,∑n=1∞αn=∞, and 2(1-λ)αn/(1+(1-λ)αn)≥(1-λ)αn, we have
(3.25)∑n=1∞an=∞,limsup⁡n→∞bnan=0,∑n=1∞cn=0.
Therefore, it follows from Lemma 2.3(2) that
(3.26)lim⁡n→∞‖xn-x*‖=0.
This completes the proof.Remark 3.2.
In (1.9), if f=0, then it follows that xn→x*=PF0. In this case, from (3.1), it follows that
(3.27)〈x*,x*-x〉≤0,∀x*∈F,
that is,
(3.28)‖x*‖2≤〈x*,x〉≤‖x*‖‖x‖,∀x∈F.
Therefore, the point x* is the unique solution to the quadratic minimization problem
(3.29)x*=argmin⁡x∈F‖x‖2.In Theorem3.1, if Ti=T for all i≥1, then we get the following result.Corollary 3.3.
LetC be a nonempty closed convex subset of a real Hilbert space  H. Let  f:C→H be a  λ-contraction with  λ∈[0,1). Let  S:C→H be a nonexpansive mapping and  T:C→C a nonexpansive mapping such that  F(T)≠∅. Let  x1∈C and define a sequence  {xn} by
(3.30)yn=PC[βnSxn+(1-βn)xn],xn+1=PC[αnf(xn)+(1-αn)Tyn],∀n≥1,
where {αn}⊂(0,1) and {βn}⊂(0,1) are two sequences satisfying the following conditions:  (a)
limn→∞αn=0 and ∑n=1∞αn=∞,(b)
limn→∞(βn/αn)=0,(c)
∑n=1∞|αn-1-αn|<∞ and ∑n=1∞|βn-1-βn|<∞. Then the sequence {xn} converges strongly to a point x*∈F(T), which is the unique solution of the variational inequality
(3.31)〈(I-f)x*,x-x*〉≥0,∀x∈F(T).In Corollary3.3, if S is a self-mapping of C into itself, then we get the following result.Corollary 3.4.
LetC be a nonempty closed convex subset of a real Hilbert space  H. Let  f:C→H be a  λ-contraction with  λ∈[0,1). Let  S:C→C be a nonexpansive mapping and  T:C→C a nonexpansive mapping such that  F(T)≠∅. Let  x1∈C and define a sequence  {xn} by
(3.32)yn=βnSxn+(1-βn)xn,xn+1=PC[αnf(xn)+(1-αn)Txn],∀n≥1,
where the sequences {αn}⊂(0,1) and {βn}⊂(0,1) are two sequences satisfying the following conditions:   (a)
limn→∞αn=0 and ∑n=1∞αn=∞,(b)
limn→∞(βn/αn)=0,(c)
∑n=1∞|αn-1-αn|<∞ and ∑n=1∞|βn-1-βn|<∞. Then the sequence {xn} converges strongly to a point x*∈F(T), which is the unique solution of the variational inequality
(3.33)〈(I-f)x*,x-x*〉≥0,∀x∈F(T).By changing the restrictions on parameters in Theorem3.1, we obtain the following.Theorem 3.5.
LetC be a nonempty closed convex subset of a real Hilbert space  H. Let  f:C→H be a  λ-contraction with  λ∈[0,1). Let  S:C→C be a nonexpansive mapping and  {Ti}i=1∞:C→C a countable family of nonexpansive mappings such that  F=⋂i=1∞F(Ti)≠∅. Set  α0=1. Let  x1∈C and define a sequence  {xn} by
(3.34)yn=βnSxn+(1-βn)xn,xn+1=PC[αnf(xn)+∑i=1n(αi-1-αi)Tiyn],∀n≥1,
where {αn}⊂(0,1) is a strictly decreasing sequence and {βn}⊂(0,1) is a sequence satisfying the following conditions: (a)
limn→∞αn=0, and ∑n=1∞αn=∞,(b)
limn→∞(βn/αn)=τ∈(0,∞),(c)
∑n=1∞(αn-1-αn)<∞ and ∑n=1∞|βn-1-βn|<∞,(d)
limn→∞((αn-1-αn)+|βn-βn-1|)/αnβn=0,(e)
there exists a constantK>0 such that (1/αn)|1/βn-1/βn-1|≤K. Then the sequence {xn} generated by (3.34) converges strongly to a point x*∈F, which is the unique solution of the variational inequality
(3.35)〈1τ(I-f)x*+(I-S)x*,x-x*〉≥0,∀x∈F.Proof.
First, the proof of Theorem  3.2 of [1] shows that (3.35) has the unique solution. By a similar argument as in that of Theorem 3.1, we can conclude that {xn} is bounded, ∥xn+1-xn∥→0 and ∥xn-Tixn∥→0 as n→∞. Note that conditions (a) and (b) imply that βn→0 as n→∞. Hence we have
(3.36)‖yn-xn‖=βn‖Sxn-xn‖⟶0(n⟶∞).
It follows that, for all i≥1,
(3.37)‖yn-Tixn‖≤‖yn-xn‖+‖xn-Tixn‖⟶0(n⟶∞).
Now, it follows from (3.36) and (3.37) that, for all i≥1,
(3.38)‖yn-Tiyn‖≤‖yn-Tixn‖+‖Tixn-Tiyn‖≤‖yn-Tixn‖+‖xn-yn‖⟶0(n⟶∞).
From (3.8), we get
(3.39)‖xn+1-xn‖βn≤‖un-un-1‖βn≤(1-(1-λ)αn)‖xn-xn-1‖βn+M[|βn-βn-1|βn+αn-1-αnβn]=(1-(1-λ)αn)‖xn-xn-1‖βn-1+(1-(1-λ)αn)‖xn-xn-1‖(1βn-1βn-1)+M[|βn-βn-1|βn+αn-1-αnβn].
Note that
(3.40)(1-(1-λ)αn)(1βn-1βn-1)≤αn1αn|1βn-1βn-1|≤αnK.
Hence, from (3.39), we have
(3.41)‖xn+1-xn‖βn≤‖un-un-1‖βn≤(1-(1-λ)αn)‖xn-xn-1‖βn-1+αnK‖xn-xn-1‖+M[|βn-βn-1|βn+αn-1-αnβn]≤(1-(1-λ)αn)‖un-1-un-2‖βn-1+αnK‖xn-xn-1‖+M[|βn-βn-1|βn+αn-1-αnβn].
Let an=(1-λ)αn and bn=αnK∥xn-xn-1∥+M[|βn-βn-1|/βn+(αn-1-αn)/βn]. From conditions (a) and (d), we have
(3.42)∑n=1∞an=∞,lim⁡n→∞bnan=0.
By Lemma 2.3(2), we get
(3.43)lim⁡n→∞‖xn+1-xn‖βn=0,lim⁡n→∞‖un-un-1‖βn=lim⁡n→∞‖un-un-1‖αn=0.
From (3.34), we have
(3.44)xn+1=PCun-un+αnf(xn)+∑i=1n(αi-1-αi)(Tiyn-yn)+(1-αn)yn.
Hence it follows that
(3.45)xn-xn+1=(1-αn)xn+αnxn-[PCun-un+αnf(xn)+∑i=1n(αi-1-αi)(Tiyn-yn)+(1-αn)yn]=(1-αn)βn(xn-Sxn)+(un-PCun)+∑i=1n(αi-1-αi)(yn-Tiyn)+αn(xn-f(xn))
and hence
(3.46)xn-xn+1(1-αn)βn=xn-Sxn+1(1-αn)βn(un-PCun)+1(1-αn)βn∑i=1n(αi-1-αi)(yn-Tiyn)+αn(1-αn)βn(xn-f(xn)).
Let vn=(xn-xn+1)/(1-αn)βn. For any z∈F, we have
(3.47)〈vn,xn-z〉=1(1-αn)βn〈un-PCun,PCun-1-z〉+αn(1-αn)βn〈(I-f)xn,xn-z〉+〈xn-Sxn,xn-z〉+1(1-αn)βn∑i=1n(αi-1-αi)〈yn-Tiyn,xn-z〉.
By Lemma 2.2, we have
(3.48)〈xn-Sxn,xn-z〉=〈(I-S)xn-(I-S)z,xn-z〉+〈(I-S)z,xn-z〉≥〈(I-S)z,xn-z〉,(3.49)〈(I-f)xn,xn-z〉=〈(I-f)xn-(I-f)z,xn-z〉+〈(I-f)z,xn-z〉≥(1-λ)‖xn-z‖2+〈(I-f)z,xn-z〉,(3.50)〈yn-Tiyn,xn-z〉=〈(I-Ti)yn-(I-Ti)z,xn-yn〉+〈(I-Ti)yn-(I-Ti)z,yn-z〉≥〈(I-Ti)yn-(I-Ti)z,xn-yn〉=βn〈(I-Ti)yn,xn-Sxn〉,i≥1.
By (2.4), we have
(3.51)〈un-PCun,PCun-1-z〉=〈un-PCun,PCun-1-PCun〉+〈un-PCun,PCun-z〉≥〈un-PCun,PCun-1-PCun〉.
Now, from (3.47)–(3.51) it follows that
(3.52)〈vn,xn-z〉≥1(1-αn)βn〈un-PCun,PCun-1-PCun〉+αn(1-αn)βn〈(I-f)z,xn-z〉+〈(I-S)z,xn-z〉+1(1-αn)∑i=1n(αi-1-αi)〈(I-Ti)yn,xn-Sxn〉+(1-λ)αn(1-αn)βn‖xn-z‖2.
Observe that (3.52) implies that
(3.53)‖xn-z‖2≤(1-αn)βn(1-λ)αn[〈vn,xn-z〉-〈(I-S)z,xn-z〉]-11-λ〈(I-f)z,xn-z〉-βn(1-λ)αn∑i=1n(αi-1-αi)〈(I-Ti)yn,xn-Sxn〉+‖un-1-un‖(1-λ)αn‖un-PCun‖.
Since vn→0, yn-Tiyn→0 for all i≥1, and ∥un-1-un∥/αn→0 as n→∞, every weak cluster point of {xn} is also a strong cluster point. Since {xn} is bounded, there exists a subsequence {xnk} of {xn} converging to a point x*∈C. Note that xn-Tixn→0 as n→∞ for all i≥1. By the demiclosed principle for a nonexpansive mapping, we have x*∈F(Ti) for all i≥1 and so x*∈F=⋂i=1∞F(Ti). From (3.47), (3.48), (3.50), and (3.51), it follows that, for all z∈F,
(3.54)〈(I-f)xnk,xnk-z〉=(1-αnk)βnkαnk〈vnk,xnk-z〉-1αnk〈unk-PCunk,PCunk-1-z〉-(1-αnk)βnkαnk〈xnk-Sxnk,xnk-z〉-1αnk∑i=1nk(αi-1-αi)〈ynk-Tiynk,xnk-z〉≤(1-αnk)βnkαnk〈vnk,xnk-z〉-1αnk‖unk-PCunk‖‖unk-1-unk‖-(1-αnk)βnkαnk〈(I-S)z,xnk-z〉-βnkαnk∑i=1nk(αi-1-αi)〈(I-Ti)ynk,xnk-Sxnk〉.
Since vn→0, (I-Ti)yn→0 for all i≥1, and ∥un-un-1∥/αn=0, letting k→∞ in (3.54), we obtain
(3.55)〈(I-f)x*,x*-z〉≤-τ〈(I-S)z,x*-z〉,∀z∈F.
Since (3.35) has the unique solution, it follows that ωw(xn)={x̅}. Since every weak cluster point of {xn} is also a strong cluster point, we conclude that xn→x̅ as n→∞. This completes the proof.In Theorem3.5, if Ti=T for each i≥1, then we have the following result, which is Theorem  3.2 of Yao et al. [1].Corollary 3.6.
LetC be a nonempty closed convex subset of a real Hilbert space  H. Let  f:C→H be a  λ-contraction with  λ∈[0,1). Let  S:C→C be a nonexpansive mapping and  T:C→C a nonexpansive mapping such that  F(T)≠∅. Let  x1∈C and define a sequence  {xn} by
(3.56)yn=βnSxn+(1-βn)xn,xn+1=PC[αnf(xn)+(1-αn)Tyn],∀n≥1,
where {αn}⊂(0,1) and {βn}⊂(0,1) are the sequences satisfying the following conditions: (a)
limn→∞αn=0 and ∑n=1∞αn=∞,(b)
limn→∞(βn/αn)=τ∈(0,∞),(c)
∑n=1∞(αn-1-αn)<∞ and ∑n=1∞|βn-1-βn|<∞,(d)
limn→∞((αn-1-αn)+|βn-βn-1|)/αnβn=0,(e)
there exists a constantK>0 such that (1/αn)|(1/βn)-1/βn-1|≤K. Then the sequence {xn} generated by (3.56) converges strongly to a point x*∈F(T), which is the unique solution of the variational inequality
(3.57)〈1τ(I-f)x*+(I-S)x*,x-x*〉≥0,∀x∈F(T).Remark 3.7.
In (1.9), if S=I and f is a self-contraction of C, then we get
(3.58)xn+1=αnf(xn)+(1-αn)Tyn,∀n≥0,
which is well known as the viscosity method studied by Moudafi [6] and Xu [7]. If S and f are both self-mappings of C in (1.9), then we get the algorithm of Cianciaruso et al. [5].



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*Source: 164978-2011-09-08.xml*