# Strong Convergence of Viscosity Iteration Methods for Nonexpansive Mappings

Authors: Jong Soo Jung
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

We propose a new viscosity iterative scheme for finding fixed points of nonexpansive mappings
in a reflexive Banach space having a uniformly Gâteaux differentiable norm and satisfying that every weakly compact convex subset of the space has the fixed point property for nonexpansive mappings. Certain different control conditions for viscosity iterative scheme are given and strong convergence of viscosity iterative
scheme to a solution of a ceratin variational inequality is established.


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## Body

## 1. Introduction

LetE be a real Banach space and let C be a nonempty closed convex subset of E. Recall that a mapping f:C→C is a contraction on C if there exists a constant k∈(0,1) such that ∥f(x)-f(y)∥≤k∥x-y∥,x,y∈C. We use ΣC to denote the collection of mappings f verifying the above inequality. That is, ΣC={f:C→C|fisacontractionwithconstantk}. Let T:C→C be a nonexpansive mapping (recall that a mapping T:C→C is nonexpansive if ∥Tx-Ty∥≤∥x-y∥,x,y∈C) and let F(T) denote the set of fixed points of T; that is, F(T)={x∈C:x=Tx}.We consider the iterative scheme: forT a nonexpansive mapping, f∈ΣC and αn∈(0,1),
(1.1)xn+1=αnf(xn)+(1-αn)Txn,n≥0.
As a special case of (1.1), the following iterative scheme:
(1.2)zn+1=αnu+(1-αn)Tzn,n≥0,
where u,z0∈C are arbitrary (but fixed), has been investigated by many authors; see, for example, Browder [1], Chang [2], Cho et al. [3], Halpern [4], Lions [5], Reich [6, 7], Shioji and Takahashi [8], Wittmann [9], and Xu [10]. The authors above showed that the sequence {zn} generated by (1.2) converges strongly to a point in the fixed-point set F(T) under appropriate conditions on {αn} in Hilbert spaces or certain Banach spaces.The viscosity approximation method of selecting a particular fixed point of a given nonexpansive mapping in a Hilbert space was proposed by Moudafi [11] in 2000. In 2004, Xu [12] extended Theorem  2.2 of Moudafi [11] for the iterative scheme (1.1) to a Banach space setting by using the following conditions on the sequence {αn}:
(1.3)limn→∞αn=0,∑n=0∞‍αn=∞,∑n=0∞‍|αn+1-αn|<∞orlimn→∞αnαn+1=1.
For the iterative scheme (1.1) with generalized contractive mappings instead of contractions, we refer to [13].In 2005, Kim and Xu [14] provided a simpler modification of Mann iterative scheme in a uniformly smooth Banach space as follows:
(1.4)x0=x∈C,yn=βnxn+(1-βn)Txn,xn+1=αnu+(1-αn)yn,n≥0,
where u∈C is an arbitrary (but fixed) element, and {αn} and {βn} are two sequences in (0,1). They proved that the sequence {xn} generated by (1.4) converges to a fixed point of T under the following control conditions:(i)
limn→∞αn=0,limn→∞βn=0;(ii)
∑n=0∞αn=∞,∑n=0∞βn=∞;(iii)
∑n=0∞|αn+1-αn|<∞,∑n=0∞|βn+1-βn|<∞.Recently, Yao et al. [15] considered the following modified Mann iterative scheme in a uniformly smooth Banach space as the viscosity approximation method:
(1.5)x0=x∈C,yn=βnxn+(1-βn)Txn,xn+1=αnf(xn)+(1-αn)yn,n≥0,
and proved strong convergence of the sequence {xn} generated by (1.5) under certain different control conditions on {αn} and {βn}. In particular, their results remove the condition ∑n=0∞‍|αn+1-αn|<∞ imposed on {αn}.Very recently, Qin et al. [16] proposed the composite Halpern type iterative scheme in a uniformly smooth Banach space as follows:
(1.6)x0=x,u∈C,zn=γnxn+(1-γn)Txn,yn=βnxn+(1-βn)Tzn,xn+1=αnu+(1-αn)yn,n≥0,
and showed strong convergence of the sequence {xn} generated by (1.6) under the following control conditions:(i)
∑n=0∞αn=∞;(ii)
limn→∞αn=0,limn→∞βn=0and0<a≤γnforsomea∈(0,1);(iii)
∑n=0∞|αn+1-αn|<∞,∑n=0∞|βn+1-βn|<∞,∑n=0∞|γn+1-γn|<∞.In this paper, under the framework of a reflexive Banach space having a uniformly Gâteaux differentiable norm and satisfying that every weakly compact convex subset ofE has the fixed point property for nonexpansive mappings, we consider a new composite iterative scheme for a nonexpansive mapping T as the viscosity approximation method: for f∈ΣC and the initial guess x0=x∈C,
ISzn=γnxn+(1-γn)Txn,yn=βnxn+(1-βn)Tzn,xn+1=αnf(xn)+(1-αn)yn,n≥0,
where {αn}, {βn} and {γn} are sequences in (0,1). First, we prove under certain control conditions on the sequences {αn}, {βn} and {γn} different from those of Qin et al. [16] that the sequence {xn} generated by (IS) converges strongly to a fixed point of T, which is a solution of a certain variational inequality. Next we study the composite iterative scheme (IS) with the weakly contractive mapping instead of the contractions. The main results develop and complement the corresponding results of [2, 3, 8, 9, 11, 12, 15, 16]. In particular, if βn=0 for all n≥0 in (IS), then (IS) reduces a new viscosity iterative scheme for finding a fixed point of T:
(1.7)zn=γnxn+(1-γn)Txn,xn+1=αnf(xn)+(1-αn)Tzn,n≥0.


## 2. Preliminaries and Lemmas

LetE be a real Banach space with norm ∥·∥, and let E* be its dual. The value of f∈E* at x∈E will be denoted by 〈x,f〉. When {xn} is a sequence in E, then xn→x (resp., xn⇀x) will denote strong (resp., weak) convergence of the sequence {xn} to x.The (normalized) duality mappingJ from E into the family of nonempty (by Hahn-Banach theorem) weak-star compact subsets of its dual E* is defined by
(2.1)J(x)={f∈E*:〈x,f〉=∥x∥2=∥f∥2},
for each x∈E [17].The norm ofE is said to be Gâteaux differentiable (and E is said to be smooth) if
(2.2)limt→0∥x+ty∥-∥x∥t
exists for each  x,y in its unit sphere U={x∈E:∥x∥=1}. The norm is said to be uniformly Gâteaux differentiable if for y∈U, the limit is attained uniformly for x∈U. The space E is said to have a uniformly Fréchet differentiable norm (and E is said to be uniformly smooth) if the limit in (2.2) is attained uniformly for (x,y)∈U×U. It is known that E is smooth if and only if each duality mapping J is single-valued. It is also well-known that if E has a uniformly Gâteaux differentiable norm, J is uniformly norm-to-weak* continuous on each bounded subsets of E [17].LetC be a nonempty closed convex subset of E. C is said to have the fixed point property for nonexpansive mappings if every nonexpansive mapping of a bounded closed convex subset D of C has a fixed point in D. Let D be a subset of C. Then a mapping Q:C→D is said to be a retraction from C onto D if Qx=x for all x∈D. A retraction Q:C→D is said to be sunny if Q(Qx+t(x-Qx))=Qx for all x∈C and t≥0 with Qx+t(x-Qx)∈C. A subset D of C is said to be a sunny nonexpansive retract of C if there exists a sunny nonexpansive retraction of C onto D. In a smooth Banach space E, it is well-known [18, page 48] that Q is a sunny nonexpansive retraction from C onto D if and only if the following condition holds
(2.3)〈x-Qx,J(z-Qx)〉≤0,x∈C,z∈D.
We need the following lemmas for the proof of our main results. Lemma 2.1 was also given in Jung and Morales [19], Lemma 2.2 is Lemma  2 of Suzuki [20] and Lemma 2.3 is essentially Lemma  2 of Liu [21] (also see [10]).Lemma 2.1.
LetEbe a real Banach space and letJbe the duality mapping. Then, for any givenx,y∈E,    one has(2.4)∥x+y∥2≤∥x∥2+2〈y,j(x+y)〉,
for all j(x+y)∈J(x+y).Lemma 2.2.
Let{xn}and{wn}be bounded sequences in a Banach spaceEand let{δn}be a sequence in[0,1]which satisfies the following condition:(2.5)0<lim infn→∞δn≤limsupn→∞δn<1.
suppose that
(2.6)xn+1=δnxn+(1-δn)wn,n≥0,limsupn→∞(∥wn+1-wn∥-∥xn+1-xn∥)≤0.
Then limn→∞∥wn-xn∥=0.Lemma 2.3.
Let{sn}be a sequence of nonnegative real numbers satisfying(2.7)sn+1≤(1-λn)sn+λnγn+δn,n≥0,
where {λn}, {γn}, and {δn} satisfy the following conditions:   (i)
{λn}⊂[0,1] and ∑n=0∞‍λn=∞ or, equivalently, ∏n=0∞‍(1-λn)=0;(ii)
limsupn→∞γn≤0 or ∑n=1∞‍λnγn<∞;(iii)
δn≥0(n≥0),∑n=0∞‍δn<∞.Then limn→∞sn=0.Recall that a mappingA:C→C is said to be weakly contractive if
(2.8)∥Ax-Ay∥≤∥x-y∥-ψ(∥x-y∥),∀x,y∈C,
where ψ:[0,+∞)→[0,+∞) is a continuous and strictly increasing function such that ψ is positive on (0,∞) and ψ(0)=0. As a special case, if ψ(t)=(1-k)t for t∈[0,+∞), where k∈(0,1), then the weakly contractive mapping A is a contraction with constant k. Rhoades [22] obtained the following result for weakly contractive mapping.Lemma 2.4 ([22, Theorem  2]).
Let(X,d)be a complete metric space and letAbe a weakly contractive mapping onX. ThenAhas a unique fixed pointpinX. Moreover, forx∈X, {Anx}converges strongly top.The following Lemma was given in [23, 24].Lemma 2.5.
Let{sn}and{γn}be two sequences of nonnegative real numbers and let{λn}be a sequence of positive numbers satisfying the conditions:(i)
∑n=0∞‍λn=∞;(ii)
limn→∞(γn/λn)=0.Let the recursive inequality,
(2.9)sn+1≤sn-λnψ(sn)+γn,n≥0,
be given, where ψ(t) is a continuous and strict increasing function on [0,+∞) with ψ(0)=0. Then limn→∞sn=0.


## 3. Main Results

First, we study a strong convergence theorem for a viscosity iterative scheme for the nonexpansive mapping with the contraction.ForT:C→C a nonexpansive mapping, t∈(0,1) and f∈ΣC, tf+(1-t)T:C→C defines a strict contraction mapping. Thus, by the Banach contraction mapping principle, there exists a unique fixed point xtf satisfying
Rxtf=tf(xtf)+(1-t)Txtf.For simplicity we will writext for xtf provided no confusion occurs.In 2006, the following result was given by Jung [25] (see also Xu [12] for the result in uniformly smooth Banach spaces).Theorem 3 J (see [25]).
LetE  be  a  reflexive Banach space having a uniformly Gâteaux  differentiable norm. Suppose that every weakly compact convex subset of  E has the fixed point property for nonexpansive mappings. Let C be a nonempty closed convex subset of E and let T be a nonexpansive mapping from C into itself with F(T)≠∅. Then {xt} defined by (R)   converges strongly to a point in F(T). If we define Q:ΣC→F(T) by
(3.1)Q(f):=limt→0+xt,f∈ΣC,
then Q(f) is the unique solution of the variational inequality
VI〈(I-f)(Q(f)),J(Q(f)-p)〉≤0,f∈ΣC,p∈F(T).Remark 3.1.
In Theorem J, iff(x)=u∈C is a constant, then (VI) become
(3.2)〈Qu-u,J(Qu-p)〉≤0,u∈C,p∈F(T).
Hence by (2.3), Q reduces to the sunny nonexpansive retraction from C to F(T). Namely F(T) is a sunny nonexpansive retraction of C.
Using Theorem J, we have the following result.Theorem 3.2.
LetE be a reflexive Banach space having a uniformly Gâteaux differentiable norm. Suppose that every weakly compact convex subset of E has the fixed point property for nonexpansive mappings. Let C be a nonempty closed convex subset of E and let T be a nonexpansive mapping from C into itself with F(T)≠∅.Let{αn}, {βn}, and {γn} be sequences in (0,1) which satisfy the conditions:  (C1)
limn→∞αn=0, ∑n=0∞‍αn=∞;(C2)
limn→∞βn=0;(C3)
0<lim infn→∞γn≤limsupn→∞γn<1.Let f∈ΣC and the initial guess x0=x∈C be chosen arbitrarily. Let {xn} be the sequence generated by
ISzn=γnxn+(1-γn)Txn,yn=βnxn+(1-βn)Tzn,xn+1=αnf(xn)+(1-αn)yn,n≥0.
If limn→∞∥Tzn-zn∥=0, then {xn} converges strongly to Q(f)∈F(T), where Q(f) is the unique solution of the variational inequality
(3.3)〈(I-f)(Q(f)),J(Q(f)-p)〉≤0,f∈ΣC,p∈F(T).Proof.
We note that by Theorem J, there exists the unique solutionQ(f) of the variational inequality
(3.4)〈(I-f)(Q(f)),J(Q(f)-p)〉≤0,f∈ΣC,p∈F(T).
Namely, Q(f)=limt→0+xt where xt is defined by (R). We will show that xn→Q(f).
We proceed with the following steps.
Step 1.
We show that∥xn-z∥≤max{∥x0-z∥,(1/(1-k))∥f(z)-z∥} for all n≥0 and all z∈F(T) and so {xn}, {yn}, {zn}, {f(xn)}, {Txn}, and {Tzn} are bounded.
Indeed, letz∈F(T). Then, noting that
(3.5)∥zn-z∥≤γn∥xn-z∥+(1-γn)∥Txn-z∥≤∥xn-z∥,
we have
(3.6)∥yn-z∥=∥βnxn+(1-βn)(Tzn-z)∥≤βn∥xn-z∥+(1-βn)∥xn-z∥≤∥xn-z∥,
which yields that
(3.7)∥xn+1-z∥=∥αn(f(xn)-z)+(1-αn)(yn-z)∥≤αn(∥f(xn)-f(z)∥+∥f(z)-z∥)+(1-αn)∥xn-z∥≤αnk∥xn-z∥+αn∥f(z)-z∥+(1-αn)∥xn-z∥=(1-(1-k)αn)∥xn-z∥+αn∥f(z)-z∥≤max{∥xn-z∥,11-k∥f(z)-z∥}.
Using an induction, we obtain
(3.8)∥xn-z∥≤max{∥x0-z∥,11-k∥f(z)-z∥},
for all n≥0. Hence {xn} is bounded, and so are {yn}, {zn}, {Txn}, {Tzn}, and {f(xn)}.Step 2.
We show thatlimn→∞∥xn+1-yn∥=0 and limn→∞∥yn-Tzn∥=0. Indeed, it follows from condition (C1) and (C2) that
(3.9)∥xn+1-yn∥=αn∥f(xn)-yn∥→0(asn→∞),∥yn-Tzn∥=βn∥xn-Tzn∥→0(asn→∞).
Also from limn→∞∥Tzn-zn∥=0, we get
(3.10)∥yn-zn∥≤∥yn-Tzn∥+∥Tzn-zn∥→0(asn→∞).Step 3.
We show thatlimn→∞∥xn+1-xn∥=0. To this end, set δn=(1-αn)γn for n≥0. Then it follows from (C1) and (C3) that
(3.11)0<lim infn→∞δn≤limsupn→∞δn<1.
Define
(3.12)xn+1=δnxn+(1-δn)wn.
Observe that
(3.13)wn+1-wn=xn+2-δn+1xn+11-δn+1-xn+1-δnxn1-δn=αn+1f(xn+1)+(1-αn+1)yn+1-δn+1xn+11-δn+1-αnf(xn)+(1-αn)yn-δnxn1-δn=(αn+1f(xn+1)1-δn+1-αnf(xn)1-δn)+(1-αn+1)[γn+1xn+1+(1-γn+1)Txn+1]-δn+1xn+1+(1-αn+1)(yn+1-zn+1)1-δn+1-(1-αn)[γnxn+(1-γn)Txn]-δnxn+(1-αn)(yn-zn)1-δn=(αn+1f(xn+1)1-δn+1-αnf(xn)1-δn)+(1-αn+1)(1-γn+1)Txn+1+(1-αn+1)(yn+1-zn+1)1-δn+1-(1-αn)(1-γn)Txn+(1-αn)(yn-zn)1-δn=(αn+1f(xn+1)1-δn+1-αnf(xn)1-δn)+(Txn+1-Txn)-αn+11-δn+1Txn+1+αn1-δnTxn+(1-αn+1)(yn+1-zn+1)1-δn+1-(1-αn)(yn-zn)1-δn.
It follows from (3.13) that
(3.14)∥wn+1-wn∥-∥xn+1-xn∥≤αn+11-δn+1(∥f(xn+1)∥+∥Txn+1∥)+αn1-δn(∥f(xn)∥+∥Txn∥)+1-αn+11-δn+1∥yn+1-zn+1∥+1-αn1-δn∥yn-zn∥.
Since {f(xn)} and {Txn} are bounded, by (C1), (3.10), (3.11), and (3.14) we obtain that
(3.15)limsupn→∞(∥wn+1-wn∥-∥xn+1-xn∥)≤0.
Hence by Lemma 2.2, we have
(3.16)limn→∞∥wn-xn∥=0.
It follows from (3.11) and (3.12) that
(3.17)limn→∞∥xn+1-xn∥=0.Step 4.
We show thatlimn→∞∥xn-Txn∥=0. In fact, from (IS) it follows that
(3.18)∥Txn-xn∥≤∥xn-xn+1∥+∥xn+1-yn∥+∥yn-Tzn∥+∥Tzn-Txn∥≤∥xn-xn+1∥+∥xn+1-yn∥+∥yn-Tzn∥+(1-γn)∥Txn-xn∥.
So we have
(3.19)γn∥Txn-xn∥≤∥xn-xn+1∥+∥xn+1-yn∥+∥yn-Tzn∥.
Thus, from condition (C3), Steps 2, and 3, we have
(3.20)limn→∞∥Txn-xn∥=0.Step 5.
We show thatlimsupn→∞〈Q(f)-f(Q(f)),J(Q(f)-xn)〉≤0. To prove this, let a subsequence {xnj} of {xn} be such that
(3.21)limsupn→∞〈Q(f)-f(Q(f)),J(Q(f)-xn)〉=limj→∞〈Q(f)-f(Q(f)),J(Q(f)-xnj)〉
and xnj⇀q for some q∈E. From Step 4, it follows that limj→∞∥xnj-Txnj∥=0.
Now letQ(f)=limt→0+xt, where xt=tf(xt)+(1-t)Txt. Then we can write
(3.22)xt-xnj=t(f(xt)-xnj)+(1-t)(Txt-xnj).
Putting
(3.23)aj(t)=(1-t)2∥Txnj-xnj∥(2∥xt-xnj∥+∥Txnj-xnj∥)→0(j→∞)
by Step 4 and using Lemma 2.1, we obtain
(3.24)∥xt-xnj∥2≤(1-t)2∥Txt-xnj∥2+2t〈f(xt)-xnj,J(xt-xnj)〉≤(1-t)2(∥Txt-Txnj∥+∥Txnj-xnj∥)2+2t〈f(xt)-xt,J(xt-xnj)〉+2t∥xt-xnj∥2≤(1-t)2∥xt-xnj∥2+aj(t)+2t〈f(xt)-xt,J(xt-xnj)〉+2t∥xt-xnj∥2.
The last inequality implies
(3.25)〈xt-f(xt),J(xt-xnj)〉≤t2∥xt-xnj∥2+12taj(t).
It follows that
(3.26)limsupj→∞〈xt-f(xt),J(xt-xnj)〉≤t2M,
where M>0 is a constant such that M≥∥xt-xn∥2 for all n≥0 and t∈(0,1). Taking the limsup as t→0 in (3.26) and noticing the fact that the two limits are interchangeable due to the fact that J is uniformly continuous on bounded subsets of E from the strong topology of E to the weak* topology of E*, we have
(3.27)limsupj→∞〈Q(f)-f(Q(f)),J(Q(f)-xnj)〉≤0.
Indeed, letting t→0, from (3.26) we have
(3.28)limsupt→0limsupj→∞〈xt-f(xt),J(xt-xnj)〉≤0.
So, for any ε>0, there exists a positive number δ1 such that for any t∈(0,δ1),
(3.29)limsupj→∞〈xt-f(xt),J(xt-xnj)〉≤ε2.
Moreover, since xt→Q(f) as t→0, the set {xt-xnj} is bounded and the duality mapping J is norm-to-weak* uniformly continuous on bounded subset of E, there exists δ2>0 such that, for any t∈(0,δ2),
(3.30)|〈Q(f)-f(Q(f)),J(Q(f)-xnj)〉-〈xt-f(xt),J(xt-xnj)〉|=|〈Q(f)-f(Q(f)),J(Q(f)-xnj)-J(xt-xnj)〉+〈Q(f)-f(Q(f))-(xt-f(xt)),J(xt-xnj)〉|≤|〈Q(f)-f(Q(f)),J(xt-xnj)-J(Q(f)-xnj)〉|+∥Q(f)-f(Q(f))-(xt-f(xt))∥∥xt-xnj∥<ε2.
Choose δ=min{δ1,δ2}, we have for all t∈(0,δ) and j∈ℕ,
(3.31)〈Q(f)-f(Q(f)),J(Q(f)-xnj)〉<〈xt-f(xt),J(xt-xnj)〉+ε2,
which implies that
(3.32)limsupj→∞〈Q(f)-f(Q(f)),J(Q(f)-xnj)〉≤limsupj→∞〈xt-f(xt),J(xt-xnj)〉+ε2.
Since limsupj→∞〈xt-f(xt),J(xt-xnj)〉≤ε/2, we have
(3.33)limsupj→∞〈Q(f)-f(Q(f)),J(Q(f)-xnj)〉≤ε.
Since ε is arbitrary, we obtain that
(3.34)limsupj→∞〈Q(f)-f(Q(f)),J(Q(f)-xnj)〉≤0.Step 6.
We show thatlimn→∞∥xn-Q(f)∥=0. By using (IS), we have
(3.35)∥xn+1-Q(f)∥=∥αn(f(xn)-Q(f))+(1-αn)(yn-Q(f))∥.
Applying Lemma 2.1 and (3.6), we obtain
(3.36)∥xn+1-Q(f)∥2≤(1-αn)2∥yn-Q(f)∥2+2αn〈f(xn)-Q(f),J(xn+1-Q(f))〉≤(1-αn)2∥xn-Q(f)∥2+2αn〈f(xn)-f(Q(f)),J(xn+1-Q(f))〉+2αn〈f(Q(f))-Q(f),J(xn+1-Q(f))〉≤(1-αn)2∥xn-Q(f)∥2+2kαn∥xn-Q(f)∥∥xn+1-Q(f)∥+2αn〈f(Q(f))-Q(f),J(xn+1-Q(f))〉≤(1-2αn+αn2)∥xn-Q(f)∥2+kαn(∥xn-Q(f)∥2+∥xn+1-Q(f)∥2)+2αn〈f(Q(f))-Q(f),J(xn+1-Q(f))〉.
It then follows that
(3.37)∥xn+1-Q(f)∥2≤(1-2(1-k)αn1-kαn)∥xn-Q(f)∥2+αn21-kαn∥xn-Q(f)∥2+2αn1-kαn〈Q(f)-f(Q(f)),J(Q(f)-xn+1)〉≤(1-2(1-k)αn1-kαn)∥xn-Q(f)∥2+αn21-kαnM2+2αn1-kαn〈Q(f)-f(Q(f)),J(Q(f)-xn+1)〉,
where M=supn≥0∥xn-Q(f)∥. Put
(3.38)λn=2(1-k)αn1-kαn,γn=αn2(1-k)M2+11-k〈Q(f)-f(Q(f)),J(Q(f)-xn+1)〉.
From the condition (C1) and Step 5, it follows that λn→0, ∑n=0∞‍λn=∞, and limsupn→∞γn≤0. Since (3.37) reduces to
(3.39)∥xn+1-Q(f)∥2≤(1-λn)∥xn-Q(f)∥2+λnγn,
from Lemma 2.3 with δn=0, we conclude that limn→∞∥xn-Q(f)∥=0. This completes the proof.Corollary 3.3.
LetEbe a uniformly smooth Banach space. LetC, T,{αn},  {βn}, {γn}, f, x0, {zn}, {yn}, and{xn}be the same as in Theorem3.2. Then the conclusion of Theorem 3.2 still holds.Proof.
SinceE is a uniformly smooth Banach space, E is reflexive, the norm is uniformly Gâteaux differentiable, and every nonempty weakly compact convex subset of E has the fixed point property for nonexpansive mappings. Thus the conclusion of Corollary 3.3 follows from Theorem 3.2 immediately.Corollary 3.4.
LetCbe a nonempty closed convex subset of a uniformly smooth Banach spaceE. LetT:C→Cbe a nonexpansive mapping withF(T)≠∅. Let{αn},{βn}, and{γn}be three sequences in(0,1)which satisfy the control conditions (C1)–(C3)in Theorem3.2. Then for the initial guessx0∈Candu∈C, the sequence{xn}generated by (1.6) converges strongly to a fixed point ofTunder the assumptionlimn→∞∥Tzn-zn∥=0.Remark 3.5.
(1) In general, the condition (C3) in Theorem3.2 and the condition ∑n=0∞‍|γn+1-γn|<∞ of Qin et al. [16, Theorem  2.1] are not comparable; neither of them implies the other. Theorem 3.2 (and Corollary 3.4) removes the conditions ∑n=0∞‍|αn+1-αn|<∞ and ∑n=0∞‍|βn+1-βn|<∞ imposed on the control parameters {αn} and {βn} of Qin et al. [16, Theorem  2.1].
(2) Theorem3.2 (and Corollary 3.3) complements the corresponding results in Moudafi [11], Xu [12], and Yao et al. [15]. In particular, if βn=0 in (IS), then (IS) in Theorem 3.2 reduces a new one for finding a fixed point of T:
(3.40)zn=γnxn+(1-γn)Txn,xn+1=αnf(xn)+(1-αn)Tzn,n≥0.
(3) Corollary3.4 with βn=0 in (IS) develops the corresponding results of Shioji and Takahashi [8], Wittmann [9] without the condition ∑n=0∞‍|αn+1-αn|<∞ as well as the result of Chang [2] in which the condition limn→∞∥Txn-xn∥=0 was assumed.
Next, we consider the viscosity iterative scheme with the weakly contractive mappings instead of the contractions.Theorem 3.6.
LetE be a reflexive Banach space having a uniformly Gâteaux differentiable norm. Suppose that every weakly compact convex subset of  E has the fixed point property for nonexpansive mappings. Let Cbe a nonempty closed convex subset of Eand let  Tbe a nonexpansive mapping from  Cinto itself with  F(T)≠∅. Let {αn},{βn},and {γn}be sequences in (0,1)  which satisfy the conditions (C1)–(C3) in Theorem 3.2. Let  A:C→Cbe a weakly contractive mapping and let x0∈Cbe chosen arbitrarily. Let{xn}be the sequence generated by
(3.41)zn=γnxn+(1-γn)Txn,yn=βnxn+(1-βn)Tzn,xn+1=αnAxn+(1-αn)yn,n≥0.
If  limn→∞∥Tzn-zn∥=0, then {xn} converges strongly to Q(Ax*)=x*∈F(T), where Q is a sunny nonexpansive retraction from C onto F(T).Proof.
It follows from Remark3.1 that F(T) is the sunny nonexpansive retract of C. Denote by Q the sunny nonexpansive retraction of C onto F(T). Then Q∘A is a weakly contractive mapping of C into itself. Indeed,
(3.42)∥Q(Ax)-Q(Ay)∥≤∥Ax-Ay∥≤∥x-y∥-ψ(∥x-y∥),∀x,y∈C.
Lemma 2.4 assures that there exists a unique element x*∈C such that x*=Q(Ax*). Such a x*∈C is an element of F(T).
Now we define an iterative scheme as follows:(3.43)un=γnwn+(1-γn)Twn,vn=βnwn+(1-βn)Tun,wn+1=αnAx*+(1-αn)vn,n≥0.
Let {wn} be the sequence generated by (3.37). Then, by taking un as zn in limn→∞∥Tzn-zn∥=0, Theorem 3.2 with f=Ax* a constant assures that {wn} converges strongly to Q(Ax*)=x* as n→∞. For any n≥0, observe that
(3.44)∥yn-vn∥≤βn∥xn-wn∥+(1-βn)∥Tzn-Tun∥≤βn∥xn-wn∥+(1-βn)∥zn-un∥≤βn∥xn-wn∥+(1-βn)∥γnxn+(1-γn)Txn-(γnwn+(1-γn)Twn)∥≤βn∥xn-wn∥+(1-βn)[γn∥xn-wn∥+(1-γn)∥xn-wn∥]=∥xn-wn∥
Then we have
(3.45)∥xn+1-wn+1∥≤αn∥Axn-Ax*∥+(1-αn)∥yn-vn∥≤αn(∥Axn-Awn∥+∥Awn-Ax*∥)+(1-αn)∥xn-wn∥≤αn∥xn-wn∥-αnψ(∥xn-wn∥)+αn(∥wn-x*∥-ψ(∥wn-x*∥))+(1-αn)∥xn-wn∥≤∥xn-wn∥-αnψ(∥xn-wn∥)+αn∥wn-x*∥.
Thus, for sn=∥xn-wn∥, we obtain the following recursive inequality:
(3.46)sn+1≤sn-αnψ(sn)+αn∥wn-x*∥.
Since ∥wn-x*∥→0, it follows from Lemma 2.5 that limn→∞∥xn-wn∥=0. Hence
(3.47)limn→∞∥xn-x*∥≤limn→∞(∥xn-wn∥+∥wn-x*∥)=0.
This completes the proof.Corollary 3.7.
LetE
be a uniformly smooth Banach space. Let
C, T, A, x0, {αn}, {βn}, {γn}, {zn}, {yn}, and
{xn}
be the same as in Theorem 3.6. Then the conclusion of Theorem 3.6 still holds.Remark 3.8.
(1) Theorem3.6 (and Corollary 3.7) develops and complements the corresponding results in Moudafi [11], Qin et al. [16], Shioji and Takahashi [8], Wittmann [9], Xu [10, 12], and Yao et al. [15].
(2) Even in the case ofβn=0 in Theorem 3.6, Theorem 3.6 appears to be independent of Theorem  5.6 of Wong et al. [13] in which the control conditions (C1) and ∑n=0∞‍|αn+1-αn|<∞ were utilized. In fact, it appears to be unknown whether a reflexive and strictly convex space satisfies the fixed point property for nonexpansive mappings.
(3) The merits of our results in this paper are that fewer restrictions are imposed on the control parameters{αn}, {βn}, and {γn}. All of our results can be viewed as a supplement to the results obtained by Qin et al. [16], Kim and Xu [14], and Yao et al. [15].
(4) The conclusions of Theorems3.2 and 3.6 still hold if E is assumed to be strictly convex instead of having the fixed point property for nonexpansive mappings.



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*Source: 573156-2009-06-14.xml*