# Synchronal Algorithm and Cyclic Algorithm for Hierarchical Fixed Point Problems and Variational Inequalities

Authors: Peichao Duan
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We propose synchronal algorithm and cyclic algorithm based on the general iterative method for solving a hierarchical fixed point problem. Under suitable parameters, the iterative sequence converges strongly to a common fixed point ofN nonexpansive mappings and also the unique solution of a variational inequality. The results
presented in this paper improve and extend the corresponding results reported recently by some authors. Furthermore, a numerical example is given to demonstrate the effectiveness of our iterative schemes.


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## Body

## 1. Introduction

LetH be a real Hilbert space with an inner product 〈,〉 and its induced norm ∥·∥. Let C be a nonempty, closed, and convex subset of H.LetT:C→H be a nonlinear mapping; we denote the set of fixed points of T by Fix(T) (i.e., Fix(T)={x∈C:Tx=x}). A mapping T:C→H is called k-Lipschitzian continuous if there exists a constant k>0 such that
(1)∥Tx-Ty∥≤k∥x-y∥,∀x,y∈C.
In particular, T is said to be a nonexpansive mapping if k=1. A mapping B is called η-strongly monotone on C, if there exists a constant η>0 such that
(2)〈Bx-By,x-y〉≥η∥x-y∥2,∀x,y∈C.A variational inequality (short for VI) is formulated as finding a pointx*∈C such that
(3)〈Bx*,x-x*〉≥0,∀x∈C.IfB is a monotone operator, then VI (3) is known as a monotone variational inequality. If the set C is replaced by the set of Fix(T) of fixed points of a mapping T, then the VI (3) is called a hierarchical variational inequality problem.Many practical problems in applied sciences such as signal processing [1], beamforming [2], and power control [3] are formulated as the monotone variational inequality with a fixed point constrained. In recent years, several authors paid attention toward this kind of problem. Some methods have been proposed to solve the hierarchical fixed point problems and variational inequalities; see for instance [4–10] and the references therein.In 2010, Tian [11] proposed a general iterative method and revealed the inner contact of the Yamada’s algorithm [12] and viscosity iterative algorithm; then he obtained the following result in a real Hilbert space.Theorem 1.
Letxn be generated by algorithm xn+1=αnγf(xn)+(I-αnμA)Txn with the sequence {αn} of parameters satisfying conditions (C1)–(C3):(C1)
αn→0,(C2)
∑n=0∞‍αn=∞,(C3)
either∑n=1∞‍|αn+1-αn|<∞ or limn→∞(αn+1/αn)=1.Then xn converges strongly to a fixed point x~ of T which solves the variational inequality
(4)〈(μA-γf)x~,x~-z〉≤0,∀z∈Fix(T).Recently, Yao et al. [10] investigated an iterative method for solving a hierarchical fixed point problem by
(5)yn=βnSxn+(1-βn)xn,xn+1=PC(αnf(xn)+(1-αn)Tyn),
where S, T are nonexpansive mapping with Fix(T)≠∅ and f is a contraction; the sequence converges strongly to the unique solution of the variational inequality
(6)x*∈Fix(T),〈(I-f)x*,p-x*〉≤0,∀p∈Fix(T).Very recently, on this basis, Wang and Xu [8] introduced a new modified iterative method for solving a hierarchical fixed point problem. To be more precise, they proposed the following algorithm:
(7)yn=βnSxn+(1-βn)xn,xn+1=PC(αnρUxn+(I-αnμF)Tyn),
where S,T are nonexpansive mappings with Fix(T)≠∅, U is a Lipschitzian mapping, and F is a Lipschitzian and strongly monotone operator. They proved the sequence generated by the above algorithm converges strongly to the unique solution of the variational inequality
(8)〈(ρU-μF)x*,p-x*〉≤0,∀p∈Fix(T).On the other hand, Tian and Di [13] established a synchronal algorithm and a cyclic algorithm for fixed point problems and variational inequalities. In 2012, Ceng et al. [4] proposed an iterative method to solve a special form of VI (3), where the constraint set is the set of common fixed points of N nonexpansive mappings T1,T2,…,TN.Motivated and inspired by the above works, in this paper, we combine the hybrid steepest descent algorithm and hierarchical variational inequalities to propose a synchronal algorithm and a cyclic algorithm involving finite family of nonexpansive mappings. Under certain assumptions, we will prove that the sequences converge strongly. Further an example will be given to demonstrate the effectiveness of our iterative schemes.


## 2. Preliminaries

Recall that given a nonempty, closed and convex subsetC of a real Hilbert space H, for any x∈H, there exists a unique nearest point in C, denoted by PCx, such that
(9)∥x-PCx∥≤∥x-y∥
for all y∈C. Such a PC is called the metric (or the nearest point) projection of H onto C. As we all know, y=PCx if and only if there holds the relation
(10)〈x-y,y-z〉≥0,∀z∈C.In the sequel, we will make use of the following lemmas in a real Hilbert spaceH.Lemma 2.
LetH be a real Hilbert space; the following inequalities hold:(i)
∥x+y∥2≤∥x∥2+2〈y,x+y〉, ∀x,y∈H,(ii)
∥tx+(1-t)y∥2≤t∥x∥2+(1-t)∥y∥2, ∀t∈[0,1], ∀x,y∈H.Lemma 3 (see [13]).
LetB:H→H be a k-Lipschitzian and η-strongly monotone operator on a Hilbert space H with k>0, η>0, 0<μ<2η/k2, and 0<t<1. Then S=(I-tμB):H→H is a contraction with contractive coefficient 1-tτ and τ=(1/2)μ(2η-μk2).Lemma 4 (see [5]).
LetV:C→H be an l-Lipschitz mapping with coefficient l≥0 and B:C→H a k-Lipschitzian continuous operator and η-strongly monotone operator with k>0, η>0. Then for 0<γ<μη/l,
(11)〈x-y,(μB-γV)x-(μB-γV)y〉≥(μη-γl)∥x-y∥2,x,y∈H.
That is, μB-γV is strongly monotone with coefficient μη-γl.Lemma 5 (see [9]).
Assume that{an} is a sequence of nonnegative real numbers such that
(12)an+1≤(1-γn)an+δn,
where {γn} is a sequence in (0,1) and {δn} is a sequence such that (i)
∑n=1∞γn=∞,(ii)
limsupn→∞(δn/γn)≤0 or ∑n=1∞|δn|<∞.Then, limn→∞an=0.Lemma 6 (see [14]).
LetH be a real Hilbert, and let Ti:H→H(i=1,2,…) be all nonexpansive mappings with ∩i=1∞Fix(Ti)≠∅. Let T=∑i=1∞‍ωiTi(i=1,2,…), where {ωi}⊂(0,1) such that ∑i=1∞‍ωi=1. Then T is a nonexpansive mapping with Fix(T)=∩i=1∞Fix(Ti).Lemma 7 (see [13]).
LetH be a Hilbert space, and let C be a nonempty closed convex subset of H and T:C→C a nonexpansive mapping with Fix(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.We adopt the following notations:(1)
xn⇀x stands for the weak convergence of {xn} to x,(2)
xn→x stands for the strong convergence of {xn} to x.


## 3. Synchronal Algorithm

Throughout the rest of this paper, we always assume thatV:C→H is an l-Lipschitzian mapping with coefficient l≥0 and B:C→H is a k-Lipschitzian continuous operator and η-strongly monotone with k>0, η>0. Let N≥1 be an integer. Let, for each 1≤i≤N, Ti:C→C be a nonexpansive mapping and S:C→C also nonexpansive. Assume that 0<μ<2η/k2 and 0<γ<μ(η-μk2/2)/α=τ/l.Define a mappingUn=βnS+(1-βn)I. Since S is nonexpansive, it is easy to get that Un is also nonexpansive. Consider the following mapping Gn on C defined by
(13)Gnx=PC(αnγV(x)+(I-αnμB)TUnx),∀x∈C,n∈ℕ,
where αn∈(0,1), T=∑i=1N‍ωiTi with ωi>0, and ∑i=1N‍ωi=1. By Lemmas 2 and 3, we obtain
(14)∥Gnx-Gny∥≤αnγ∥Vx-Vy∥∥Gnx-Gny∥hh+(1-αnτ)∥TUnx-TUny∥∥Gnx-Gny∥≤αnγl∥x-y∥+(1-αnτ)∥x-y∥∥Gnx-Gny∥=(1-αn(τ-γl))∥x-y∥.Since0<1-αn(τ-γl)<1, it follows that Gn is a contraction. Therefore, by the Banach contraction principle, Gn has a unique fixed point xnV∈C such that
(15)xnV=PC(αnγV(xnV)+(I-αnμB)TUnxnV).For simplicity, we will writexn for xnV provided that no confusion occurs. Next we prove that the sequence {xn} converges strongly to a point x*∈Ω=∩i=1NFix(Ti) which solves the variational inequality
(16)〈(γV-μB)x*,p-x*〉≤0,∀p∈Ω.
By the property of the projection, we can get x*=PΩ(I-μB+γV)x* equivalently.Theorem 8.
LetC be a nonempty, closed, and convex subset of a real Hilbert space H and V:C→H an l-Lipschitzian mapping with l≥0. Let N≥1 be an integer. Let, for each 1≤i≤N, Ti:C→C be a nonexpansive mapping and let S:C→C be also nonexpansive. Assume the set Ω=⋂i=1N‍Fix(Ti)≠∅. Let B:C→H be a k-Lipschitzian continuous operator and η-strongly monotone with k>0, η>0, 0<μ<2η/k2 and 0<γ<μ(η-μk2/2)/l=τ/l. Given x1∈C, let {xn} be the sequence generated by the following algorithm:
(17)yn=βnSxn+(1-βn)xn,xn+1=PC(αnγVxn+(I-μαnB)Tyn).
If {αn} and {βn} satisfy the following properties: (i)
{αn}⊂(0,1), limn→∞αn=0 and ∑n=1∞‍αn=∞,(ii)
{βn}⊂[0,1), limn→∞(βn/αn)=0,(iii)
∑n=1∞‍|αn+1-αn|<∞ and ∑n=1∞‍|βn+1-βn|<∞. Then, {xn} converges strongly to x*∈Ω, which solves the variational inequality (16).Proof.
The proof is divided into several steps.
Step  1. Show first that {xn} is bounded.
Take anyp∈Ω; we have
(18)∥yn-p∥=∥βn(Sxn-p)+(1-βn)(xn-p)∥∥yn-p∥≤βn∥Sxn-Sp∥+βn∥Sp-p∥∥yn-p∥jj+(1-βn)∥xn-p∥≤∥xn-p∥+βn∥Sp-p∥.
Further we get(19)∥xn+1-p∥=∥PC(αnγVxn+(I-μαnB)Tyn)-PCp∥≤∥αn(γVxn-μBp)+(I-μαnB)Tyn-(I-μαnB)p∥≤αn(∥γVxn-γVp∥+∥γVp-μBp∥)+(1-αnτ)∥Tyn-p∥≤αnlγ∥xn-p∥+αn∥γVp-μBp∥+(1-αnτ)∥yn-p∥=(1-αn(τ-lγ))∥xn-p∥+αn(τ-lγ)∥γVp-μBp∥+∥Sp-p∥τ-lγ≤max{∥xn-p∥,∥γVp-μBp∥+∥Sp-p∥τ-lγ}.
By induction, we obtain∥xn-p∥≤max{∥x1-p∥,(∥γVp-μBp∥+∥Sp-p∥)/(τ-lγ)}, n≥1. Hence, {xn} is bounded, so is {yn}. It follows from the Lipschitz continuity of B and V that {Bxn}, {Bun}, and {Vxn} are also bounded. From the nonexpansivity of T and S, it follows that {Txn}, {Sxn}, and {BTyn} are also bounded.
Step  2. Show that
(20)limn→∞∥xn+1-xn∥=0.
By (17), we have
(21)∥xn+1-xn∥=∥PC(αnγVxn+(I-μαnB)Tyn)-PC(αn-1γVxn-1+(I-μαn-1B)Tyn-1)∥≤∥αnγVxn+(I-μαnB)Tyn-(αn-1γVxn-1+(I-μαn-1B)Tyn-1)∥≤αnγ∥Vxn-Vxn-1∥+∥(I-μαnB)Tyn-(I-μαnB)Tyn-1∥+∥(I-μαnB)Tyn-1-(I-μαn-1B)Tyn-1∥+|αn-αn-1|γ∥Vxn-1∥≤αnγl∥xn-xn-1∥+|αn-αn-1|(γl∥Vxn-1∥+μ∥BTyn-1∥)+(1-αnτ)∥yn-yn-1∥.
Observe that(22)∥yn-yn-1∥=∥βnSxn+(1-βn)xn-βn-1Sxn-1-(1-βn-1)xn-1∥≤∥βnSxn-βnSxn-1∥+∥(1-βn)xn-(1-βn)xn-1∥+∥βnSxn-1-βn-1Sxn-1∥+∥(1-βn)xn-1-(1-βn-1)xn-1∥≤βn∥xn-xn-1∥+(1-βn)∥xn-xn-1∥+|βn-βn-1|∥Sxn-1∥+|βn-βn-1|∥xn-1∥=∥xn-xn-1∥+|βn-βn-1|(∥Sxn-1∥+∥xn-1∥).
Together with (21) and (22), we get
(23)∥xn+1-xn∥≤αnγl∥xn-xn-1∥+|αn-αn-1|(γl∥Vxn-1∥+μ∥BTyn-1∥)+(1-αnτ)∥xn-xn-1∥+|βn-βn-1|(∥Sxn-1∥+∥xn-1∥)=(1-αn(τ-γl))∥xn-xn-1∥+|αn-αn-1|(γl∥Vxn-1∥+μ∥BTyn-1∥)+|βn-βn-1|(∥Sxn-1∥+∥xn-1∥)≤(1-αn(τ-γl))∥xn-xn-1∥+(|αn-αn-1|+|βn-βn-1|)M1,
where M1=supn{∥γlVxn-1∥+μ∥BTyn-1∥+∥Sxn-1∥+∥xn-1∥}.
By Lemma5, we obtain
(24)∥xn+1-xn∥⟶0.
Step  3. Show that
(25)∥xn-Txn∥⟶0.
Observe that(26)∥xn+1-Tyn∥=∥PC(αnγVxn+(I-μαnBTyn))-PCTyn∥≤αn∥γVxn-μBTyn∥⟶0.
From condition (i) and (ii), we obtain
(27)∥xn+1-Txn∥≤∥xn+1-Tyn∥+∥Tyn-Txn∥≤∥xn+1-Tyn∥+∥yn-xn∥=∥xn+1-Tyn∥+βn∥Sxn-xn∥⟶0.
Step  4. Show that
(28)limsupn→∞〈(γV-μB)x*,xn-x*〉≤0,
where x*=PΩ(I-μB+γV)x* is a unique solution of the variational inequality (16).
Indeed, take a subsequence{xnj} of {xn} such that
(29)limsupn→∞〈(γV-μB)x*,xn-x*〉=limj→∞〈(γV-μB)x*,xnj-x*〉.
Since{xnj} is bounded, there exists a subsequence {xnjk} of {xnj} which converges weakly to x^. Without loss of generality, we can assume xnj⇀x^ and
(30)∥xnj-Txnj∥⟶0.
By Lemma7, we have x^=Tx^. From Lemma 6, we get
(31)x^∈Fix(T)=⋂i=1N‍Fix(Ti).
Sincex*=PΩ(I-μB+γV)x*, it follows that
(32)limsupn→∞〈(γV-μB)x*,xn-x*〉=limj→∞〈(γV-μB)x*,xnj-x*〉=〈(γV-μB)x*,x^-x*〉≤0.
Step  5. Show that
(33)xn⟶x*.
Denotezn=αnγVxn+(I-μαnB)Tyn, then xn+1=PCzn. From (17), we have
(34)∥xn+1-x*∥2=∥PCzn-x*∥2=〈PCzn-zn,PCzn-x*〉+〈zn-x*,xn+1-x*〉≤〈zn-x*,xn+1-x*〉=〈(I-μαnB)Tyn-(I-μαnB)Tx*+αn(γVxn-μBx*),xn+1-x*〉≤(1-αnτ)∥yn-x*∥∥xn+1-x*∥+αn〈γVxn-γVx*,xn+1-x*〉+αn〈γVx*-μBx*,xn+1-x*〉≤(1-αnτ)(∥xn-x*∥+βn∥Sx*-x*∥)∥xn+1-x*∥+αnγl∥xn-x*∥∥xn+1-x*∥+αn〈γVx*-μBx*,xn+1-x*〉≤1-αn(τ-γl)2(∥xn-x*∥2+∥xn+1-x*∥2)+βn∥xn+1-x*∥∥Sx*-x*∥+αn〈γVx*-μBx*,xn+1-x*〉.
This implies that(35)∥xn+1-x*∥2≤1-αn(τ-γl)1+αn(τ-lγ)∥xn-x*∥2+21+αn(τ-γl)×(βn∥xn+1-x*∥∥Sx*-x*∥+αn〈γVx*-μBx*,xn+1-x*〉)≤(1-αn(τ-γl))∥xn-x*∥2+2βnM2+2αn〈γVx*-μBx*,xn+1-x*〉,
where M2=supn∥xn+1-x*∥∥Sx*-x*∥, n≥1. Put γn=αn(τ-lγ), δn=2βnM2+2αn〈γVx*-μBx*,xn+1-x*〉. It is easy to see that limsupn→∞δn/γn≤0. Hence by Lemma 5, the sequence {xn} converges strongly to x*.Remark 9.
LetN=1 in Theorem 8; we can get Theorem 3.1 of [8].Remark 10.
LetN=1, γ=1, μ=1, B=I and V, be a contraction in Theorem 8; it is easy to get the theorem of  [10].


## 4.  Cyclic Algorithm

In this section, we consider the cyclic algorithm ofN nonexpansive mappings T1,T2,…,TN. Similarly, we can get that the mapping Gn on C defined by
(36)Gnx=PC(αnγV(x)+(I-αnμB)T[n]Unx),∀x∈C,n∈ℕ,
is a contraction, where T[n]=Ti with i=n(mod)N taking values in {1,2,…,N}.Theorem 11.
LetC be a nonempty, closed, and convex subset of a real Hilbert space H, and let V:C→H be an l-Lipschitzian mapping with l≥0. Let N≥1 be an integer. Let, for each 1≤i≤N, Ti:C→C be a nonexpansive mapping and let S:C→C be also nonexpansive. Assume the set Ω=⋂i=1N‍Fix(Ti)≠∅. Let B:C→H be a k-Lipschitzian continuous operator and η-strongly monotone with k>0, η>0, 0<μ<2η/k2, and 0<γ<μ(η-μk2/2)/l=τ/l. Given x1∈C, let {xn} be the sequence generated by the following algorithm:
(37)yn=βnSxn+(1-βn)xn,xn+1=PC(αnγVxn+(I-μαnB)T[n]yn).
If {αn} and {βn} satisfy the following properties:(i)
{αn}⊂(0,1), limn→∞αn=0 and ∑n=1∞‍αn=∞,(ii)
{βn}⊂[0,1), limn→∞(βn/αn)=0,(iii)
∑n=1∞‍|αn+1-αn|<∞ and ∑n=1∞‍|βn+1-βn|<∞. Assume in addition that
(38)⋂i=1N‍Fix(Ti)=Fix(TNTN-1⋯T2T1)=Fix(T1TNTN-1⋯T2)=⋯=Fix(T2T1TN⋯T3).
Then, {xn} converges strongly to x*∈Ω, which solves the variational inequality (16).Proof.
The proof is divided into several steps.
Step  1. Show first that {xn} is bounded.
The proof of Step 1 is similar to that of Theorem8.
Step  2. Show that
(39)limn→∞∥xn+N-xn∥=0.
By (37), we have
(40)∥xn+N+1-xn+1∥≤∥αn+NγVxn+N+(I-μαn+NB)T[n]yn+N-αnγVxn-(I-μαnB)T[n]yn∥≤αn+Nγ∥Vxn+N-Vxn∥+∥(I-μαn+NB)T[n]yn+N-(I-μαn+NB)T[n]yn∥+∥(I-μαn+NB)T[n]yn-(I-μαnB)T[n]yn∥+|αn+N-αn|γ∥Vxn∥≤αn+Nγl∥xn+N-xn∥+(1-αn+Nτ)∥yn+N-yn∥+|αn+N-αn|(γl∥Vxn∥+μ∥BT[n]yn∥).
Observe that(41)∥yn+N-yn∥=∥βn+NSxn+N+(1-βn+N)xn+N-βnSxn-(1-βn)xn∥≤∥βn+NSxn+N-βn+NSxn∥+∥(1-βn+N)xn+N-(1-βn+N)xn∥+∥βn+NSxn-βnSxn∥+∥(1-βn+N)xn-(1-βn)xn∥≤βn+N∥xn+N-xn∥+(1-βn+N)∥xn+N-xn∥+|βn+N-βn|∥Sxn∥+|βn+N-βn|∥xn∥=∥xn+N-xn∥+|βn+N-βn|(∥Sxn∥+∥xn∥).
Together with (40) and (41), we have
(42)∥xn+N+1-xn+1∥≤αn+Nγl∥xn+N-xn∥+|αn+N-αn|(γl∥Vxn∥+μ∥BT[n]yn∥)+(1-αn+Nτ)∥xn+N-xn∥+|βn+N-βn|(∥Sxn∥+∥xn∥)=(1-αn+N(τ-γl))∥xn+N-xn∥+|αn+N-αn|(γl∥Vxn∥+μ∥BT[n]yn∥)+|βn+N-βn|(∥Sxn∥+∥xn∥)≤(1-αn+N(τ-γl))∥xn+N-xn∥+(|αn+N-αn|+|βn+N-βn|)M3,
where M3=supn{∥γlVxn∥+μ∥BT[n]yn∥+∥Sxn∥+∥xn∥}.
By Lemma5, we get
(43)∥xn+N-xn∥⟶0.
Step  3. Show that
(44)∥xn-T[n+N-1]T[n+N-2]⋯T[n]xn∥⟶0.
Observe that(45)∥xn+1-T[n]yn∥=∥PC(αnγVxn+(I-μαn)BT[n]yn)-PCT[n]yn∥≤αn∥γVxn-μBT[n]yn∥⟶0.
From conditions (i) and (ii) of Theorem11, we obtain
(46)∥xn+1-T[n]xn∥≤∥xn+1-T[n]yn∥+∥T[n]yn-T[n]xn∥≤∥xn+1-T[n]yn∥+∥yn-xn∥=∥xn+1-T[n]yn∥+βn∥Sxn-xn∥⟶0.
Recursively,(47)∥xn+N-T[n+N-1]xn+N-1∥⟶0,∥xn+N-1-T[n+N-2]xn+N-2∥⟶0.
Since everyT[n] is nonexpansive, it is easy to get
(48)∥T[n+N-1]xn+N-1-T[n+N-1]T[n+N-2]xn+N-2∥⟶0.
Similarly, we obtain(49)∥T[n+N-1]T[n+N-2]xn+N-2-T[n+N-1]T[n+N-2]T[n+N-3]xn+N-3∥⟶0⋮∥T[n+N-1]T[n+N-2]⋯T[n+1]xn+1-T[n+N-1]T[n+N-2]⋯T[n]xn∥⟶0.
Thus we get(50)∥xn+N-T[n+N-1]⋯T[n]xn∥≤∥xn+N-T[n+N-1]xn+N-1∥+∥T[n+N-1]xn+N-1-T[n+N-1]T[n+N-2]xn+N-2∥+⋯+∥T[n+N-1]T[n+N-2]⋯T[n+1]xn+1-T[n+N-1]⋯T[n]xn∥⟶0.
Since(51)∥xn-T[n+N-1]T[n+N-2]⋯T[n]xn∥≤∥xn-xn+N∥+∥xn+N-T[n+N-1]⋯T[n]xn∥,
we obtain (44).
Step  4. Show that
(52)limsupn→∞〈(γV-μB)x*,xn-x*〉≤0,
where x*=PΩ(I-μB+γV)x* is a unique solution of the variational inequality (16).
Indeed, take a subsequence{xnj} of {xn} such that
(53)limsupn→∞〈(γV-μB)x*,xn-x*〉=limj→∞〈(γV-μB)x*,xnj-x*〉.
Since{xnj} is bounded, there exists a subsequence {xnjk} of {xnj} which converges weakly to x^. Without loss of generality, we can assume xnj⇀x^ and
(54)∥xnj-T[nj+N-1]T[nj+N-2]⋯T[nj]xnj∥⟶0.
Notice that, for eachnj, T[nj+N-1]T[nj+N-2]⋯T[nj] is some permutation of the mappings T1T2⋯TN. Since T1T2⋯TN are finite, all the finite permutations are N!; there must be some permutation that appears infinite times. Without loss of generality, we can assume this permutation is TNTN-1⋯T1. We obtain
(55)∥xnj-TNTN-1⋯T1xnj∥⟶0.
Obviously,TNTN-1⋯T1 is nonexpansive. By Lemma 7, we have x^=TNTN-1⋯T1x^. Further by the assumption in Theorem 11, we get
(56)x^∈Fix(TNTN-1⋯T1)=⋂i=1N‍Fix(Ti).
Sincex*=PΩ(I-μB+γV)x*, it follows that
(57)limsupn→∞〈(γV-μB)x*,xn-x*〉=limj→∞〈(γV-μB)x*,xnj-x*〉=〈(γV-μB)x*,x^-x*〉≤0.
Step  5. Show that
(58)xn⟶x*.
Denotezn=αnγVxn+(I-μαnB)T[n]yn; then xn+1=PCzn. From (37), we have
(59)∥xn+1-x*∥2=∥PCzn-x*∥2=〈PCzn-zn,PCzn-x*〉+〈zn-x*,xn+1-x*〉≤〈zn-x*,xn+1-x*〉=〈(I-μαnB)T[n]yn-(I-μαnB)T[n]x*+αn(γVxn-μBx*),xn+1-x*〉≤(1-αnτ)∥yn-x*∥∥xn+1-x*∥+αn〈γVxn-γVx*,xn+1-x*〉+αn〈γVx*-μBx*,xn+1-x*〉≤(1-αnτ)(∥xn-x*∥+βn∥Sx*-x*∥)×∥xn+1-x*∥+αnγl∥xn-x*∥∥xn+1-x*∥+αn〈γVx*-μBx*,xn+1-x*〉≤1-αn(τ-γl)2(∥xn-x*∥2+∥xn+1-x*∥2)+βn∥xn-x*∥∥Sx*-x*∥+αn〈γVx*-μBx*,xn+1-x*〉.
This implies that(60)∥xn+1-x*∥2≤1-αn(τ-γl)1+αn(τ-γl)∥xn-x*∥2+21+αn(τ-γl)×(βn∥xn-x*∥∥Sx*-x*∥+αn〈γVx*-μBx*,xn+1-x*〉)≤(1-αn(τ-γl))∥xn-x*∥2+2βnM1+2αn〈γVx*-μBx*,xn+1-x*〉,
where M1=supn∥xn-x*∥∥Sx*-x*∥, n≥1. Put γn=αn(τ-lγ), δn=2βnM1+2αn〈γVx*-μBx*,xn+1-x*〉. It is easy to see that limsupn→∞δn/γn≤0. Hence, by Lemma 6, the sequence {xn} converges strongly to x*.


## 5. Numerical Result

In this section, we consider the following simple example to demonstrate the effectiveness, realization, and convergence of the algorithms in Theorems8 and 11.Example 12.
LetH=ℝ, C=[1/4,+∞). Define T1:x↦x, T2:x↦x+π/4-arctanx, S:x↦sinx. Take B=I with Lipschitz constant k=1 and strongly monotone constant η=1, Vx=2x, ∀x∈H, with Lipschitz coefficient l=2. Give the parameters αn=1/2n; βn=1/n2 for every n≥1; fix μ=1 and γ=1/4. Then by Theorems 8 and 11, respectively, the sequence {xn} is generated by
(61)yn=1n2sin(xn)+(1-1n2)xn,xn+1=PC(14nxn+(1-12n)Tyn),(62)yn=1n2sin(xn)+(1-1n2)xn,xn+1=PC(14nxn+(1-12n)T[n]yn).
As n→∞, we have {xn}→x*=1.Letωi=1/2, i=1,2; then we have Tx=(1/2)(x+x+π/4-arctanx), T[n]x=x if n is odd and T[n]x=x+π/4-arctanx if n is even. Put zn=(1/4n)xn+(1-1/2n)Tyn; then (61) is equivalent to
(63)xn+1={zn,ifzn∈C,14,ifzn∈-C.
Using the same method to treat (62), we can get similar equation as the above formula.Now we turn to numerical simulation using the algorithms (17) and (37), respectively. Take the initial guess x1=2; using software Matlab R2012, we obtain the numerical experiment results in Tables 1 and 2.Table 1
x
1
=
2.
n (iterative number)
x
n (iterative point)
Errors (n)
50
0.9901
9.9 × 10−3
500
0.9990
9.9852 × 10−4
2000
0.9999
9.9985 × 10−5Table 2
x
1
=
2.
n (iterative number)
x
n (iterative point)
Errors (n)
50
0.9902
9.8 × 10−3
500
0.9990
9.9814 × 10−4
2000
0.9999
9.9981 × 10−5From the computer programming point of view, the algorithms are easier to implement in this paper.



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*Source: 750473-2013-10-01.xml*