# Viscosity Method for Hierarchical Fixed Point Problems with an Infinite Family of Nonexpansive Nonself-Mappings

Authors: Yaqin Wang
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

A viscosity method for hierarchical fixed point problems is presented to solve variational inequalities, where the involved mappings are nonexpansive nonself-mappings. Solutions are sought in the set of the common fixed points of an infinite family of nonexpansive nonself-mappings. The results generalize and improve the recent results announced by many other authors.


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## Body

## 1.Introduction and Preliminaries

LetX a real Banach space and J be the normalized duality mapping from X into 2X* given by
(1)J(x)={x*∈X*:〈x,x*〉=∥x∥∥x*∥,∥x∥=∥x*∥}
for all x∈X, where X* denotes the dual space of X and 〈·,·〉 the generalized duality pairing between X and X*. If X=H is a Hilbert space, then J becomes the identity mapping on H. A point x∈C is a fixed point of T:C⊂X→X provided Tx=x. Denote by F(T) the set of fixed points of T; that is, F(T)={x∈C:Tx=x}.LetX be a normed linear space with dimX≥2. The modulus of smoothness of X is the function ρX:[0,+∞)→[0,+∞) defined by
(2)ρX(τ):=sup{∥x+y∥+∥x-y∥2-1:∥x∥=1,∥y∥=τ}.
The space X is said to be smooth if ρX(τ)>0,forallτ>0. It is well known that if X is smooth then J is single valued. A Banach space X is said to be strictly convex if ∥x∥=∥y∥=1,  x≠y, implies ∥x+y∥/2<1.LetC be a nonempty closed convex subset of a real Banach space X. Recall the following concepts.Definition 1.
(i) A mappingf:C→C is a ρ-contraction if ρ∈[0,1) and if the following property is satisfied
(3)∥f(x)-f(y)∥≤ρ∥x-y∥,∀x,y∈C.(ii)
A mappingT:C→E is nonexpansive provided
(4)∥Tx-Ty∥≤∥x-y∥,∀x,y∈C.(iii)
A mappingS:C→X is
(a)
accretive if for anyx,y∈C there exists j(x-y)∈J(x-y) such that
(5)〈Sx-Sy,j(x-y)〉≥0;(b)
β-strongly accretive if for any x,y∈C there exists j(x-y)∈J(x-y) such that
(6)〈Sx-Sy,j(x-y)〉≥β∥x-y∥2,
for some real constant β>0.
Noting that ifS:C→X is nonexpansive, then I-S is accretive; if f:C→C is a ρ-contraction, then I-f is (1-ρ)-strongly accretive. particulary, if X=H is a Hilbert space, then (strongly) accretive mappings become (strongly) monotone mappings.Definition 2.
LetC and D be nonempty subsets of a Banach space X such that C is nonempty closed convex and D⊂C. (i)
A mappingQ:C→D is called sunny, if Q(Qx+t(x-Qx))=Qx for each x∈C and t≥0 with Q(Qx+t(x-Qx))∈C.(ii)
A mappingQ:C→D is called a retraction from C to D if Q is continuous and F(Q)=D.(iii)
A subsetD of C⊂E is said to be a sunny nonexpansive retract of C if there exists a sunny nonexpansive retraction Q of C onto D. For details, see [1–3].
Note that ifX=H is a Hilbert space, Q becomes the projection on C, denoted by PC.
LetP:C→C a nonexpansive self-mapping on C and {Tn} be a countable family of nonexpansive nonself-mappings of C into X such that ℱ=⋂n=1∞F(Tn)≠∅. Then we consider the following problem: find hierarchically a common fixed point of the infinite family {Tn} with respect to a nonexpansive mapping P; namely, find x*∈ℱ, such that
(7)〈x*-Px*,J(x-x*)〉≥0,∀x∈ℱ.Particularly,if{Tn} is a finite family of nonexpansive nonself-mappings, problem (7) has been studied by Ceng and Petruşel [4]. If X=H and {Tn} is an infinite family of nonexpansive self-mappings, Problem (7) reduces to the following problem: find hierarchically a common fixed point of {Tn} with respect to a nonexpansive mapping P, namely, find x*∈ℱ, such that
(8)〈x*-Px*,x-x*〉≥0,∀x∈ℱ,
which was studied by Zhang et al. [5]. If X=H is a Hilbert space and Tn=T,foralln≥1, where T is a nonexpansive mapping on C, then problem (7) reduces to the following problem: finding hierarchically a fixed point of T with respect to another nonexpansive mapping P; namely, find x*∈F(T) such that
(9)〈x*-Px*,x-x*〉≥0,∀x∈F(T).Problem (7) includes many problems as special cases, so it is very important in the area of optimization and related fields, such as signal processing and image reconstruction (see [6–9]).In 2007, Moudafi [10] introduced the following Krasnoselski-Mann's algorithm in Hilbert spaces:
(10)xn+1=(1-αn)xn+αn(σnPxn+(1-σn)Txn),∀n≥0,
where {αn} and {σn} are two real sequences in (0,1) and T and P are two nonexpansive mappings of C into itself. Furthermore, he established a weak convergence result for Algorithm (10) for solving problem (9).Subsequently, Yao and Liou [11] derived a weak convergence result of algorithm (10) under the restrictions on parameters weaker than those in [10, Theorem 2.1].Recently, Marino and Xu [12] introduced the following explicit hierarchical fixed point algorithm in Hilbert spaces:
(11)xn+1=λnf(xn)+(1-λn)(αnVxn+(1-αn)Txn),∀n≥0,
where f is a contraction on C and V,T are two nonexpansive mappings of C into itself and proved that the sequence {xn} generated by (11) converges strongly to a solution of problem (9).Very recently, Zhang et al. [5] introduced the following iterative algorithm in order to find hierarchically a fixed point of Problem (8):
(12)x0∈C,xn+1=αnf(xn)+(1-αn)yn,yn=βnP(xn)+(1-βn)Txn,
where f:C→C is a contraction, P:C→C is a nonexpansive mapping, {Tn}:C→C is a countable family of nonexpansive mappings, and T:C→C is a mapping defined by
(13)T=∑n=1∞‍λnTn,λn≥0(n=1,2,…)with∑n=1∞‍λn=1.
Under suitable conditions on parameters {αn} and {βn}, they established some strong and weak convergence theorems. Note that, in [5], {Tn} is an infinite family of self-mappings and P is also a self-mapping. And they obtained the results in the setting of Hilbert spaces.Motivated and inspired by the above researches, in a reflexive Banach space which admits a weakly sequentially continuous duality mappingJ, we propose and analyze an iteration process for a countable family of nonexpansive nonself-mappings {Tn}:C→X and S:C→X is a nonexpansive nonself-mapping as follows:
(14)x0∈C,xn+1=Q(αnf(xn)+(1-αn)yn),yn=βnSxn+(1-βn)Txn,n≥0,
where Q is a sunny nonexpansive retraction of X onto C and establishes a convergence theorem. particularly, if X=H is a Hilbert space, we obtain some convergence results.To prove the main results, we need the following lemmas.Lemma 3 (see [1]).
LetC be a nonempty and convex subset of a smooth Banach space X, D⊂C, J:X→X* the normalized duality mapping of X, and Q:C→D a retraction. Then the following conditions are equivalent:(i)
〈x-Qx,J(y-Qx)〉≤0, for all x∈C and y∈D;(ii)
Q is both sunny and nonexpansive.Lemma 4 (see [13, Lemma 3.1, 3.3]).
LetX be a real smooth and strictly convex Banach space and C a nonempty closed and convex subset of X which is also a sunny nonexpansive retract of X. Assuming that T:C→X is a nonexpansive mapping and Q is a sunny nonexpansive retraction of X onto C, then F(T)=F(QT).Lemma 5 (see [1]).
LetX be a real Banach space and J:X→2X* the normalized duality mapping. Then for any x,y∈X, the following hold: (i)
∥x+y∥2≤∥x∥2+2〈y,j(x+y)〉,  for allj(x+y)∈J(x+y);(ii)
∥x∥2+2〈y,j(x)〉≤∥x+y∥2,   for all j(x)∈J(x).Lemma 6 (see [14]).
Let{an} and {bn} be two sequences of nonnegative real numbers satisfying
(15)∑n=0∞bn<∞,an+1≤an+bn,n=0,1,2,….
Then limn→∞an exists.Lemma 7 (see [15]).
Let{an} be a sequence of nonnegative real numbers satisfying
(16)an+1≤(1-λn)an+λnbn+cn,∀n≥0,
where {λn},{bn} and {cn} satisfy the following conditions: (i)
{λn}⊂[0,1],∑n=0∞λn=∞ or, equivalently, Πn=0∞(1-λn)=0;(ii)
limsupn→∞bn≤0;(iii)
cn≥0(n≥0),∑n=0∞cn<∞.Then limn→∞an=0.If Banach spaceX admits sequentially continuous duality mapping J from weak topology to weak * topology, then by [16, Lemma 1] we get that duality mapping J is single-valued. In this case, duality mapping J is also said to be weakly sequentially continuous, that is, for each {xn}⊂X with xn⇀x, then J(xn)⇁Jx [16, 17].Recall that a Banach spaceX is said to be satisfying Opial’s condition if for any sequence {xn} in E, xn⇀x(n→∞) implies that
(17)limsupn→∞∥xn-x∥<limsupn→∞∥xn-y∥,∀y∈E,withy≠x.
By [16, Lemma 1], we know that if X admits a weakly sequentially continuous duality mapping, then X satisfies Opial’s condition.In the sequel, we also need the following lemmas.Lemma 8 (see [17]).
LetC be a nonempty, closed and convex subset of a reflexive Banach space X which satisfies Opial’s condition and T:C→X a nonexpansive mapping. Then the mapping I-T is demiclosed at zero, that is,
(18)xn⇀xxn-Txn⟶0implyx=Tx.
LetC be a nonempty and convex subset of a Banach space X. Then for x∈C, one defines the inward set IC(x) as follows [2, 3]:
(19)IC(x)={y∈X:y=x+λ(z-x),z∈C,λ≥0}.
A mapping T:C→X is said to satisfy the inward condition if Tx∈IC(x) for all x∈C. T is also said to satisfy the weakly inward condition if for each x∈C, Tx∈IC(x)¯(IC(x)¯ is the closure of IC(x)). Clearly C⊂IC(x) and it is not hard to show that IC(x) is a convex set if C does.Lemma 9 (see [18, Theorem 2.4]).
LetX be a reflexive Banach space which admits a weakly sequentially continuous duality mapping J from X to X*. Suppose C is a nonempty closed convex subset of X which is also a sunny nonexpansive retract of X, and T:C→X is a nonexpansive mapping satisfying the weakly inward condition and F(T)≠∅. Let {un} be defined by
(20)u0∈C,un+1=Q(αnf(un)+(1-αn)Tun),
where Q is a sunny nonexpansive retract of X onto C and αn∈(0,1) satisfy the following conditions: (i)
αn→0,asn→∞;(ii)
∑n=0∞‍αn=∞;(iii)
either∑n=0∞‍|αn+1-αn|<∞ or limn→∞(αn/αn+1)=1.Then {xn} converges strongly to a fixed point p of T such that p is the unique solution in F(T) to the following variational inequality:
(21)〈(I-f)p,j(p-u)〉≤0,∀u∈F(T).Remark 10.
If a Banach spaceX admits a sequentially continuous duality mapping J from weak topology to weak star topology, from Lemma 1 of [16] it follows that X is smooth. So for Lemma 9, if X is a reflexive and strictly convex Banach space which admits a weakly sequentially continuous duality mapping J, by Lemma 4, the weakly inward condition of T can be removed.


## 2. Main Results

Theorem 11.
LetX be a reflexive and strictly convex Banach space which admits a weakly sequentially continuous duality mapping J:X→X* and C a nonempty, closed and convex subset of X which is also a sunny nonexpansive retract of X. Let S:C→X be a nonexpansive nonself-mapping, f:C→C a contractive mapping with a contractive constant ρ∈(0,1) and Ti:C→X(i={1,2,…}) an infinite family of nonexpansive nonself-mappings such that ⋂i=1∞F(Ti)≠∅. Let T:C→X be defined by (13) and Q a sunny nonexpansive retraction of X onto C. Let {xn} be the sequence generated by (14), and {αn} and {βn} the sequences in(0,1)satisfying the following conditions: (i)
αn→0(n→∞),∑n=0∞‍αn=∞;(ii)
limn→∞(βn/αn)=0;(iii)
∑n=0∞‍|αn+1-αn|<∞.Then {xn} converges strongly to some point x*∈F(T)=⋂i=1∞F(Ti), which is the unique solution to the following variational inequality:
(22)〈(I-f)x*,J(x-x*)〉≥0,∀x∈F(T).Proof.
From condition (ii), without loss of generality, we can assume thatβn≤αn,foralln≥0.
First we prove that the sequence{xn} is bounded.
In fact, for anyu∈F(T), we have
(23)∥xn+1-u∥=∥Q(αnf(xn)+(1-αn)yn)-Qu∥≤αn∥f(xn)-u∥+(1-αn)∥βnSxn+(1-βn)Txn-u∥≤αn(ρ∥xn-u∥+∥f(u)-u∥)+(1-αn)(βn∥Sxn-u∥+(1-βn)∥Txn-u∥)≤(1-αn(1-ρ))∥xn-u∥+αn∥f(u)-u∥+(1-αn)βn∥Su-u∥≤(1-αn(1-ρ))∥xn-u∥+αn(∥f(u)-u∥+∥Su-u∥)≤max{∥xn-u∥,∥f(u)-u∥+∥Su-u∥1-ρ}.
By induction,
(24)∥xn+1-u∥≤max{∥x0-u∥,∥f(u)-u∥+∥Su-u∥1-ρ}.
Thus {xn} is bounded, so {Sxn} and {Txn} are also bounded.
Next we prove that∥xn-un∥→0,asn→∞, where the sequence {un} is defined by
(25)u0=x0∈C,un+1=Q(αnf(un)+(1-αn)Tun).
By Lemma9 and Remark 10, {un} converges strongly to some point x*∈F(T), which is the unique solution to the following variational inequality:
(26)〈(I-f)x*,j(x*-x)〉≤0,∀x∈F(T).
Furthermore, we obtain
(27)∥xn+1-un+1∥≤∥Q(αnf(xn)+(1-αn)yn)-Q(αnf(un)+(1-αn)Tun)∥≤∥αn(f(xn)-f(un))+(1-αn)(yn-Tun)∥≤αnρ∥xn-un∥+(1-αn)×(βn∥Sxn-Tun∥+(1-βn)∥Txn-Tun∥)≤(1-αn(1-ρ))∥xn-un∥+(1-αn)βnM≤(1-αn(1-ρ))∥xn-un∥+βnM,
where M=supn≥0∥Sxn-Tun∥. It follows from conditions (i)-(ii) and Lemma 7 we have limn→∞∥xn-un∥=0. Since as n→∞, un→x*∈F(T), we get xn→x*(n→∞), which is the unique solution to the variational inequality (22).Remark 12.
Theorem11 extends Theorem 2.1 in [5] from the following aspects: (i) from Hilbert spaces to reflexive and strictly convex Banach spaces which admits a weakly sequentially continuous duality mapping; (ii) for the infinite family of mappings {Ti} from self-mappings to nonself-mappings. In addition, the existence of the sunny nonexpansive retraction has been proved in [19, Theorem 3.10].Remark 13.
If we take(28)αn=1(1+n)α,βn=1(1+n)β,0<α<β<1,
then since |αn+1-αn|≈1/nα+1 and |βn+1-βn|≈1/nβ+1(as n→∞), it is not hard to find that the conditions (i)–(iii) are satisfied. For details, see [12, Remark 3.2].In the sequel, we consider the result in the setting of Hilbert spaces.Theorem 14.
LetH be a Hilbert space and C a nonempty, closed and convex subset of H. Let S:C→H be a nonexpansive nonself-mapping, f:C→C a contractive mapping with a contractive constant ρ∈(0,1), and Ti:C→H(i={1,2,…}) an infinite family of nonexpansive nonself-mappings such that F(T)=⋂i=1∞F(Ti)≠∅. Let {xn} be the sequence generated by (14) and {αn} and {βn} the sequences in(0,1)satisfying the following conditions: (i)
αn→0,∑n=0∞‍αn=∞;(ii)
limn→∞(βn/αn)=τ∈(0,+∞);(iii)
limn→∞((|βn-βn-1|+|αn-αn-1|)/αnβn)=0;(iv)
there exists a constantK>0 such that 1/αn|(1/βn)-(1/βn-1)|≤K for all n>0.Then {xn} converges strongly to some point x*∈F(T), which is the unique solution to the following variational inequality:
(29)〈1τ(I-f)x*+(I-S)x*,x-x*〉≥0,∀x∈F(T).Proof.
By condition (ii), without loss of generality, we can assume thatβn≤(τ+1)αn,foralln≥0. Similar to the proof of (24), for any u∈F(T), we have
(30)∥xn+1-u∥≤max{∥x0-u∥,(τ+1)(∥f(u)-u∥+∥Su-u∥)1-ρ}.
Thus {xn} is bounded. Furthermore, {f(xn)},{Txn},{yn},{Sxn} are all bounded. Put un=αnf(xn)+(1-αn)yn and M=supn≥0{∥f(xn)∥+∥yn∥,∥Txn∥+∥Sxn∥}. So {un} and {PC(un)} are also bounded.
Step1. We prove that ∥xn+1-xn∥→0(n→∞).
From (14), we obtain
(31)∥xn+1-xn∥=∥PC(un)-PC(un-1)∥≤∥un-un-1∥≤αn∥f(xn)-f(xn-1)∥+(1-αn)∥yn-yn-1∥+|αn-αn-1|(∥f(xn-1)∥+∥yn-1∥)≤αnρ∥xn-xn-1∥+(1-αn)×∥yn-yn-1∥+|αn-αn-1|M,(32)∥yn-yn-1∥≤βn∥Sxn-Sxn-1∥+(1-βn)∥Txn-Txn-1∥+|βn-βn-1|(∥Sxn-1∥+∥Txn-1∥)≤∥xn-xn-1∥+|βn-βn-1|M.
Substituting (32) into (31), we have
(33)∥xn+1-xn∥≤(1-αn(1-ρ))∥xn-xn-1∥+αn(|αn-αn-1|+|βn-βn-1|)Mαn.
By conditions (i), (iii), and Lemma 7, we have ∥xn+1-xn∥→0(n→∞).
Step2. We prove that ωw(xn)⊂F(T), where ωw(xn) is the ω-limit point set of {xn} in the weak topology:
(34)∥xn+1-QTxn∥≤αn∥f(xn)∥+βn∥Sxn∥+(αn+βn+αnβn)∥Txn∥.
Noting that αn→0 and βn→0, we have ∥xn+1-QTxn∥→0(n→∞). Then from Step 1 we have ∥xn-QTxn∥→0(n→∞). Furthermore, it follows from Lemmas 4 and 8 that ωw(xn)⊂F(QT)=F(T), where Q=PC.
Step3. We show that ∥xn+1-xn∥/βn→0(n→∞).
It follows from (31) and (33) that
(35)∥xn+1-xn∥βn≤∥un-un-1∥βn≤(1-αn(1-ρ))∥xn-xn-1∥βn-1+(1-αn(1-ρ))∥xn-xn-1∥|1βn-1βn-1|+(|αn-αn-1|+|βn-βn-1|)Mβn≤(1-αn(1-ρ))∥xn-xn-1∥βn-1+αn∥xn-xn-1∥K+(|αn-αn-1|+|βn-βn-1|)Mαnβnαn.
By conditions (i) and (iii), ∥xn-xn-1∥→0(n→∞), and Lemma 7, we have
(36)∥xn+1-xn∥βn⟶0(n⟶∞).
Thus from (35), we get
(37)∥un-un-1∥βn⟶0(n⟶∞).
Step4. We show that {xn} converges strongly to some point x′∈F(T), which is the unique solution of (29).
SettingWn=βnS+(1-βn)T, we have
(38)xn+1=PC(un)-un+αnf(xn)+(1-αn)Wnxn.
Then
(39)xn-xn+1=un-PC(un)+αn(I-f)xn+(1-αn)(I-Wn)xn.
Letting vn=(xn-xn+1)/(1-αn)βn, from condition (i) and (36), we have vn→0(n→∞). Noting that I-Wn is monotone and I-f is (1-ρ)-strongly monotone, for any x*∈F(T), from Lemma 3 we obtain
(40)〈vn,xn-x*〉=1(1-αn)βn〈un-PC(un),xn-x*〉+αn(1-αn)βn〈(I-f)xn,xn-x*〉+1βn〈(I-Wn)xn,xn-x*〉=1(1-αn)βn〈un-PC(un),xn-x*〉+αn(1-αn)βn〈(I-f)xn,xn-x*〉+1βn〈(I-Wn)xn-(I-Wn)x*,xn-x*〉+1βn〈(I-Wn)x*,xn-x*〉=1(1-αn)βn〈un-PC(un),PC(un)-x*〉+1(1-αn)βn×〈un-PC(un),-(PC(un)-x*)+(PC(un-1)-x*)〉+αn(1-αn)βn〈(I-f)xn-(I-f)x*,xn-x*〉+αn(1-αn)βn〈(I-f)x*,xn-x*〉+1βn〈(I-Wn)xn-(I-Wn)x*,xn-x*〉+1βn〈(I-Wn)x*,xn-x*〉≥1(1-αn)βn〈un-PC(un),PC(un-1)-PC(un)〉+αn(1-ρ)(1-αn)βn∥xn-x*∥2+αn(1-αn)βn〈(I-f)x*,xn-x*〉+〈(I-S)x*,xn-x*〉.
Thus we have
(41)∥xn-x*∥2≤(1-αn)βnαn(1-ρ)〈vn,xn-x*〉-(1-αn)βnαn(1-ρ)〈(I-S)x*,xn-x*〉-1αn(1-ρ)〈un-PC(un),PC(un-1)-PC(un)〉-1(1-ρ)〈(I-f)x*,xn-x*〉≤(1-αn)βnαn(1-ρ)∥vn∥∥xn-x*∥-(1-αn)βnαn(1-ρ)〈(I-S)x*,xn-x*〉+1(1-ρ)∥un-PC(un)∥∥un-1-unαn∥-1(1-ρ)〈(I-f)x*,xn-x*〉.
Since βn≤(τ+1)αn, by (37) we have
(42)∥un-un-1∥αn⟶0(n⟶∞).
Combining condition (ii), vn→0(n→∞), (41), and (42), every weak cluster point of {xn} is also a strong cluster point. From (40), we obtain
(43)〈(I-f)xn,xn-x*〉=(1-αn)βnαn〈vn,xn-x*〉-1αn〈un-PC(un),xn-x*〉-(1-αn)αn〈(I-Wn)xn,xn-x*〉=(1-αn)βnαn〈vn,xn-x*〉-1αn〈un-PC(un),PC(un)-x*〉-1αn〈un-PC(un),PC(un-1)-PC(un)〉-(1-αn)αn×〈(I-Wn)xn-(I-Wn)x*,xn-x*〉-(1-αn)αn〈(I-Wn)x*,xn-x*〉≤(1-αn)βnαn∥vn∥∥xn-x*∥+1αn∥un-PC(un)∥PC(un-1)-PC(un)∥-(1-αn)βnαn〈(I-S)x*,xn-x*〉≤(1-αn)βnαn∥vn∥∥xn-x*∥+∥un-1-un∥αn∥un-PC(un)∥-(1-αn)βnαn〈(I-S)x*,xn-x*〉.
Note that the sequence {xn} is bounded; thus there exists a subsequence {xnj} converging to a point x′∈H. From Step 2, we have x′∈F(T). Then it follows from the above inequality, (42), and vn→0(n→∞) that
(44)〈(I-f)x′,x′-x*〉≤-τ〈(I-S)x*,x′-x*〉,∀x*∈F(T).
Replacing x* with x′+μ(x*-x′), where μ∈(0,1) and x*∈F(T), we have
(45)〈(I-f)x′,x′-x*〉≤-τ〈(I-S)(x′+μ(x*-x′)),x′-x*〉,∀x*∈F(T).
Letting μ→0, we have
(46)〈(I-f)x′,x′-x*〉≤-τ〈(I-S)x′,x′-x*〉,∀x*∈F(T).
If there exists another subsequence {xnj′} of {xn} converging to a point x′′∈H. From Step 2, we also have x′′∈F(T). Then from (46) we obtain
(47)〈(I-f)x′,x′-x′′〉≤-τ〈(I-S)x′,x′-x′′〉
and, via interchanging x′ and x′′,
(48)〈(I-f)x′′,x′′-x′〉≤-τ〈(I-S)x′′,x′′-x′〉.
Adding up these two inequalities yields
(49)(1-ρ)∥x′-x′′∥2≤〈(I-f)x′-(I-f)x′′,x′-x′′〉≤0,
which implies x′=x′′. Then {xn} converges strongly to x′∈F(T), which is the solution to the following variational inequality:
(50)〈1τ(I-f)x′+(I-S)x′,x-x′〉≥0,∀x∈F(T).
Since I-f is (1-ρ)-strongly monotone and I-S is monotone, it is easy to see that the above variational inequality has a unique solution.Remark 15.
Theorem14 extends Theorem 3.2 in [12] from the following aspects: (i) from a nonexpansive mapping T to an infinite family of nonexpansive mappings {Ti}; (ii) from self-mappings to nonself-mappings.



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*Source: 574215-2013-04-22.xml*