POINTS FOR NONEXPANSIVE NONSELF-MAPPING
RUDONG CHEN AND ZHICHUAN ZHUReceived 17 May 2006; Accepted 22 June 2006
LetCbe a closed convex subset of a uniformly smooth Banach spaceE, andT:C→E a nonexpansive nonself-mapping satisfying the weakly inwardness condition such that F(T)= ∅, and f :C→Ca fixed contractive mapping. Fort∈(0, 1), the implicit itera-tive sequence{xt}is defined byxt=P(t f(xt) + (1−t)Txt), the explicit iterative sequence
{xn}is given byxn+1=P(αnf(xn) + (1−αn)Txn), whereαn∈(0, 1) andPis a sunny
non-expansive retraction ofEontoC. We prove that{xt}strongly converges to a fixed point ofT ast→0, and{xn}strongly converges to a fixed point ofT asαnsatisfying
appro-priate conditions. The results presented extend and improve the corresponding results of Hong-Kun Xu (2004) and Yisheng Song and Rudong Chen (2006).
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.
1. Introduction
LetCbe a nonempty closed convex subset of a Banach spaceE, and LetT:C→C be a nonexpansive mapping (i.e.,Tx−T y ≤ x−yfor allx,y∈C). We use Fix(T) to denote the set of fixed points ofT; that is , Fix(T)= {x∈C:x=Tx}. Recall that a self-mapping f :C→Cis a contraction onCif there exists a constantβ∈(0, 1) such that
f(x)−f(y)≤βx−y, x,y∈C. (1.1)
Xu (see [6]) defined the following two viscosity iterations for nonexpansive mappings:
xt=t fxt+ (1−t)Txt, x∈C, (1.2)
xn+1=αnfxn+1−αnTxn, (1.3)
whereαnis a sequence in (0,1). Xu proved the strong convergence of{xt}defined by (1.2)
ast→0 and{xn}defined by (1.3) in both Hilbert space and uniformly smooth Banach
space.
Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 16470, Pages1–12
Recently, Song and Chen [2] proved ifCis a closed subset of a real reflexive Banach spaceEwhich admits a weakly sequentially continuous duality mapping fromEtoE∗, and ifT:C→Eis a nonexpansive nonself-mapping satisfying the weakly inward condi-tion,F(T)=φ, f :C→Cis a fixed contractive mapping, andPis a sunny nonexpansive retraction ofEontoC, then the sequences{xt}and{xn}defined by
xt=Pt fxt+ (1−t)Txt, (1.4) xn+1=Pαnfxn+1−αnTxn (1.5)
strongly converge to a fixed point ofT.
In this paper, we establish the strong convergence of both{xt}defined by (1.4) and
{xn}defined by (1.5) for a nonexpansive nonself-mappingTin a uniformly smooth Ba-nach space. Our results extend and improve the results in [2,6].
2. Preliminaries
LetEbe a real Banach space and letJdenote the normalized duality mapping fromEinto 2E∗given by
J(x)=f ∈E∗: x,f = xf,x = f ∀x∈E, (2.1)
whereE∗denotes the dual space ofEand ·,·denotes the generalized duality pairing. In the sequence, we will denote the single-valued duality mapping by j, andxn→xwill denote strong convergence of the sequence{xn}tox. In Banach spaceE, the following
result is well known [1,3] for allx,y∈E, for all j(x+y)∈J(x+y), for all j(x)∈J(x), x2+ 2y,j(x)≤ x+y2≤ x2+ 2y,j(x+y). (2.2)
Recall that the norm ofEis said to be Gˆateaux differentiable (andEis said to be smooth) if
lim
t→0
x+ty − x
t (2.3)
exists for eachx, yin its unit sphereU= {x∈E:x =1}. It is said to be uniformly Gˆateaux differentiable if, for eachy∈U, this limit is attained uniformly forx∈U. Fi-nally, the norm is said to be uniformly Fr´echet differentiable (andEis said to be uniformly smooth) if the limit in (2.3) is attained uniformly for (x,y)∈U×U. A Banach spaceE is said to be smooth if and only ifJis single valued. It is also well known that ifEis uni-formly smooth,Jis uniformly norm-to-norm continuous. These concepts may be found in [3].
IfCandDare nonempty subsets of a Banach spaceEsuch thatCis nonempty closed convex andD⊂C, then a mappingP:C→Dis called a retraction fromCtoDifP2=P.
It is easily known that a mappingP:C→Dis retraction, thenPx=x, for allx∈D. A mappingP:C→Dis called sunny if
wheneverPx+t(x−Px)∈Candt >0. A subsetDofCis said to be a sunny nonexpansive retract ofCif there exists a sunny nonexpansive retraction ofContoD. For more detail, see [1,3–5].
The following lemma is well known [3].
Lemma 2.1. LetCbe a nonempty convex subset of a smooth Banach spaceE,D∈C,J:E→ E∗the (normalized) duality mapping ofE, andP:C→Da retraction. Then the following
are equivalent:
(i) x−Px,j(y−Px) ≤0 for allx∈Candy∈D; (ii)Pis both sunny and nonexpansive.
LetCbe a nonempty convex subset of a Banach spaceE, then forx∈C, we define the inward set [4,5]:
IC(x)=y∈E:y=x+λ(z−x),z∈Candλ≥0. (2.5)
A mappingT:C→Eis said to be satisfying the inward condition ifTx∈IC(x) for all
x∈C.Tis also said to be satisfying the weakly inward condition if for eachx∈C,Tx∈ IC(x) (IC(x) is the closure ofIC(x)). ClearlyC⊂IC(x) and it is not hard to show thatIC(x)
is a convex set asCis. Using above these results and definitions, we can easily show the following lemma.
Lemma 2.2 ([2], Lemma 1.2). LetCbe a nonempty closed subset of a smooth Banach space E, letT:C→Ebe nonexpansive nonself-mapping satisfying the weakly inward condition, and letPbe a sunny nonexpansive retraction ofEontoC. ThenF(T)=F(PT).
Lemma 2.3 ([2], Lemma 2.1). Let Ebe a Banach space and letCbe a nonempty closed
convex subset ofE. Suppose thatT:C→Eis a nonexpansive mapping such that for each fixed contractive mapping f :C→C, andPis a sunny nonexpansive retraction ofEontoC. For eacht∈(0, 1),{xt}is defined by (1.4). Supposeu∈Cis a fixed point ofT, then
(i) xt−f(xt),j(xt−u) ≤0;
(ii){xt}is bounded.
Definition 2.4. μis called a Banach limit if μis a continuous linear functional on l∞ satisfying
(i)μ(e) =1=μ(1),e=(1, 1, 1,...);
(ii)μn(an)=μn(an+1), for allan∈(a0,a1,...)∈l∞;
(iii) lim infn→∞an≤μ(an)≤lim supn→∞an, for allan∈(a0,a1,...)∈l∞.
According to time and circumstances, we useμn(an) instead ofμ(a0,a1,...).
Further, we know the following result.
Lemma 2.5 ([3], Lemma 4.5.4). LetCbe a nonempty closed convex subset of a Banach space Ewith a uniformly Gˆateaux differentiable norm and let{xn}be a bounded sequence inE.
Letμbe a Banach limit andu∈C. Then
μnxn−u2=miny
∈Cμnxn−y 2
if and only if
μnx−u,Jxn−u≤0 (2.7)
for allx∈C.
3. Main results
Theorem 3.1. LetEbe a uniformly smooth Banach, suppose thatCis a nonempty closed
convex subset ofEandT:C→Eis a nonexpansive nonself-mapping satisfying the weakly inward condition andF(T)= ∅. Let f :C→Cbe a fixed contractive mapping, and let{xt}
be defined by (1.4), wherePis a sunny nonexpansive retraction ofEontoC. Then ast→0 {xt}converges strongly to some fixed pointqofTthatqis the unique solution inF(T) to the
following variational inequality:
(I−f)q,j(q−u)≤0 ∀u∈F(T). (3.1)
Proof. For allu∈F(T) byLemma 2.3(ii),{xt}is bounded, therefore the sets{Txt:t∈
(0, 1)} and{f(xt) :t∈(0, 1)} are also bounded. From xt=P(t f(xt) + (1−t)Txt), we
have
xt−PTxt=P
t fxt+ (1−t)Txt−PTxt
≤t fxt+ (1−t)Txt−Txt
=tTxt−fxt−→0 ast−→0.
(3.2)
This implies that
limt
→0
xt−PTxt=0. (3.3)
Assumetn→0, setxn:=xtn, and defineg:C→Rbyg(x)=μnxn−x2,x∈C, whereμn
is a Banach limit on∞. Let
K= x∈C:g(x)=min
y∈Cμnxn−y 2
. (3.4)
It is easily seen that K is a nonempty closed convex bounded subset of E, since (note xn−Txn →0)
g(Tx)=μnxn−Tx2=μnTxn−Tx2≤μnxn−x2=g(x). (3.5)
It follows thatT(K)⊂K, that is,K is invariant underT. Since a uniformly smooth Ba-nach space has the fixed point property for nonexpansive mappings,Thas a fixed point, sayq, inK. FromLemma 2.5we get
For allq∈F(T), we havet f(xt) + (1−t)q=P[t f(xt) + (1−t)q], then
xt−
t fxt+ (1−t)q
=Pt fxt+ (1−t)Txt−Pt fxt+ (1−t)q
≤(1−t)Txt−q≤(1−t)xt−q.
(3.7)
Hence from (2.2) and the above inequality we get xt−
t fxt+ (1−t)q2
=(1−t)xt−q+txt−fxt2
≥(1−t)2x
t−q2+ 2t(1−t)xt−fxt,jxt−q.
(3.8)
Therefore
xt−fxt,jxt−q≤0. (3.9)
Then
0≥xt−fxt,jxt−q
=xt−q2+q−f(q),jxt−q+f(q)−fxt,jxt−q
≥(1−β)xt−q2+q−f(q),jxt−q.
(3.10)
We get
xt−q2
≤ 1 1−β
f(q)−q,jxt−q. (3.11)
Now applying Banach limit to the above inequality, we get
μnxt−q2≤μn
1 1−β
f(q)−q,jxt−q. (3.12)
Letx= f(q) in (3.6), and noting (3.12), we have
μnxt−q2≤0, (3.13)
that is,
μnxn−q2=0 (3.14)
and then exists a subsequence which is still denoted by{xn}such that
xn−→q, n−→ ∞. (3.15)
We have proved that for any sequence{xtn}in {xt:t∈(0, 1)}, there exists a
the entire net{xt}converges toq, suppose that there exists another sequence{xsk} ⊂ {xt}
such thatxsk→p, assk→0, then we also havep∈F(T) (using limt→0xt−PTxt =0).
Next we show p=q andq is the unique solution inF(T) to the following variational inequality:
(I−f)q,j(q−u) ∀u∈F(T). (3.16)
Since the sets{xt−u}and{xt−f(xt)}are bounded and the uniform smoothness ofE
implies that the duality mapJ is norm-to-norm uniformly continuous on bounded sets ofE, for anyu∈F(T), byxsk→p(sk→0), we have
(I−f)xsk−(I−f)p−→0, sk−→0, xs
k−f
xsk
,jxsk−u
−(I−f)p,j(p−u)
=xsk−f
xsk
−(I−f)p,jxsk−u
−(I−f)p,jxsk−u
−j(p−u)
≤(I−f)xsk−(I−f)pxsk−u
+(I−f)p,jxsk−u
−j(p−u)−→0 assk−→0.
(3.17)
Therefore, notingLemma 2.3(i), for anyu∈F(T), we get
(I−f)p,j(p−u)=lim
sk→0
xsk−f
xsk
,jxsk−u
≤0. (3.18)
Similarly, we also can show
(I−f)q,j(q−u)=xtn−f
xtn
,jxtn−u
≤0. (3.19)
Interchangeqanduto obtain
(I−f)p,j(p−q)≤0. (3.20)
Interchangepanduto obtain
(I−f)q,j(q−p)≤0. (3.21)
This implies that
(p−q)−f(p)−f(q),j(p−q)≤0, (3.22)
that is,
p−q2≤βp−q2. (3.23)
This is a contradiction, so we must haveq=p.
FromTheorem 3.1we can get the following corollary directly.
Corollary 3.2. LetEbe a uniformly smooth space, supposeCis a nonempty closed convex
subset ofE,T:C→Eis a nonexpansive mapping satisfying the weakly inward condition, andF(T)= ∅. Let f :C→Cbe a fixed contractive mapping fromCtoC.{xt}is defined by xt=t fxt+ (1−t)PTxt, (3.24)
wherePis a sunny nonexpansive retraction ofEontoC, thenxt converges strongly to some fixed pointqofT ast→0 andqis the unique solution inF(T) to the following variational inequality:
(I−f)q,j(q−u) ∀u∈F(T). (3.25)
Lemma 3.3 ([6], Lemma 2.1). Let{αn}be a sequence of nonnegative real numbers satisfying
the property
αn+1≤1−γnαn+δn ∀n≥0, (3.26)
where{γn} ∈(0, 1) andδnis a sequence inRsuch that:
(i) limn→∞γn=0 and∞n=0γn= ∞;
(ii) either∞n=0δn<+∞or lim supn→∞(δn/γn)≤0,
then limn→∞αn=0.
Theorem 3.4. LetEbe a uniformly smooth Banach space, suppose thatCis a nonempty
closed convex subset ofE,T:C→Eis a nonexpansive nonself-mapping satisfying the weakly inward condition, andF(T)= ∅. Let f :C→Cbe a fixed contractive mapping, and{xn}
is defined by (1.5), wherePis a sunny nonexpansive retraction ofEontoC, andαn∈(0, 1)
satisfies the following conditions: (i)αn→0, asn→ ∞; (ii)∞n=0αn= ∞;
(iii) either∞n=0|αn+1−αn|<∞or limn→∞(αn+1/αn)=1.
Thenxnconverges strongly to a fixed pointqofTsuch thatqis the unique solution inF(T) to the following variational inequality:
(I−f)q,j(q−u)≤0 ∀u∈F(T). (3.27)
Proof. First we show{xn}is bounded. Takeu∈F(T), it follows that
xn+1−u=P
1−αnTxn+αnfxn−Pu
≤1−αnTxn+αnfxn−u
≤1−αnTxn−u+αnfxn−f(u)+f(u)−u
≤1−αnxn−u+αnβxn−u+f(u)−u
=1−(1−β)αnxn−u+αnf(u)−u
≤max xn−u, 1
1−βf(u)−u
.
By induction,
xn−u≤max x0−u, 1
1−βf(u)−u
, n≥0, (3.29)
and{xn}is bounded, so are{Txn}and{f(xn)}. We claim that
xn+1−xn−→0 asn−→ ∞. (3.30)
Indeed we have (for some appropriate constantM >0) xn+1−xn=Pα
nfxn+1−αnTxn−Pαn−1fxn−1+1−αn−1Txn−1
≤αnfxn+1−αnTxn−αn−1f
xn−1
−1−αn−1
Txn−1
≤1−αnTxn−Txn−1
+αn−αn−1
fxn−1
−Txn−1
+αnfxn−fxn−1
≤1−αnxn−xn−1[3pt] +Mαn−αn−1+βαnxn−xn−1
=1−(1−β)αnxn−xn−1[3pt] +Mαn−αn−1.
(3.31)
ByLemma 3.3we havexn+1−xn →0, asn→ ∞. We now show that
xn−PTxn−→0. (3.32)
In fact,
xn+1−PTxn=P
αnfxn+1−αnTxn−PTxn
≤αnfxn−Txn.
(3.33)
This follows from (3.30) that
xn−PTxn≤xn−xn+1+xn+1−PTxn
≤xn−xn+1+αnfxn−Txn−→0 asn−→ ∞. (3.34)
Letq=limt→0xt, where{xt}is defined inCorollary 3.2, we get thatqis the unique
solu-tion inF(T) to the following variational inequality:
(I−f)q,j(q−u)≤0 ∀u∈F(T). (3.35)
We next show that
lim sup
n→∞
FormCorollary 3.2, letxt=t f(xt) + (1−t)PTxt, indeed we can write
xt−xn=tfxt−xn+ (1−t)PTxt−xn. (3.37)
Noting (3.32), putting
an(t)=xn−PTxnxn−PTxn+ 2xn−xt−→0 asn−→ ∞, (3.38)
and using (2.2), we obtain
xt−xn2
≤(1−t)2PTx
t−xn2+ 2tfxt−xn,jxt−xn
≤(1−t)2PTx
t−PTxn+PTxn−xn2+ 2tfxt−xt,jxt−xn
+ 2txt−xn2≤(1−t)2xt−xn2+ (1−t)2xn−PTxn2
+ 2(1−t)2PTxn−xnxt−xn+ 2tfxt−xt,jxt−xn+ 2txt−xn2
≤1 +t2x
t−xn2+an(t) + 2tfxt−xt,jxt−xn.
(3.39)
The last inequality implies
fx
t−xt,jxn−xt≤t
2xt−xn
2
+ 1
2tan(t). (3.40)
Froman(t)→0 asn→ ∞we get
lim sup
n→∞
fxt−xt,jxn−xt≤M·t
2, (3.41)
whereM >0 is a constant such thatM≥ xt−xn2for alln≥0 andt∈(0, 1). By letting t→0 in (3.41) we have
lim
t→0
lim sup
n→∞
fxt−xt,jxn−xt≤0. (3.42)
On the one hand, for allε >0,∃δ1such thatt∈(0,δ1),
lim sup
n→∞
fxt−xt,jxn−xt≤ε
On the other hand,{xt}strongly converges toq, ast→0, the set{xt−xn}is bounded, and
the duality mapJis norm-to-norm uniformly continuous on bounded sets of uniformly smooth spaceE; fromxt→q(t→0), we get
f(q)−q−
fxt−xt−→0, t−→0,
f(q)−q,j
xn−q−fxt−xt,jxn−xt
=f(q)−q,jxn−q−jxn−xt+f(q)−q−fxt−xt,jxn−xt
≤f(q)−qjxn−q−jxn−xt
+f(q)−q−fxt−xtxn−xt−→0, t−→0.
(3.43)
Hence for the aboveε >0,∃δ2, such that for allt∈(0,δ2), for alln, we have
f(q)−q,j
xn−q−fxt−xt,jxn−xt≤ ε
2. (3.44)
Therefore, we have
f(q)−q,jxn−q≤fxt−xt,jxn−xt+ε
2. (3.45)
Noting (**) and takingδ=min{δ1,δ2}, for allt∈(0,δ), we have
lim sup
n→∞
f(q)−q,jxn−q
≤lim sup
n→∞
fxt−xt,jxn−xt+ ε
2
≤ε 2+
ε 2=ε.
(3.46)
Sinceεis arbitrary, we get
lim sup
n→∞
f(q)−q,jxn−q≤0. (3.47)
Finally we showxn→q. Indeed
By (2.2) we have xn+1−q2
=xn+1−αnfxn+1−αnq+αnfxn−q2
≤xn+1−Pαnfxn+1−αnq2+ 2αnfxn−q,jxn+1−q
≤Pαnfxn+1−αnTxn−Pαnfxn+1−αnq2
+ 2αnfxn−q,jxn+1−q
≤1−αn2Txn−q2+ 2αnfxn−f(q),jxn+1−q
+ 2αnf(q)−q,jxn+1−q
≤1−αn2xn−q2+ 2αnf(q)−fxnxn+1−q
+ 2αnf(q)−q,jxn+1−q
≤1−αn2xn−q2+αnf(q)−fxn2+xn+1−q2
+ 2αnf(q)−q,jxn+1−q.
(3.49)
Therefore, we have
1−αnxn+1−q2
≤1−αn2xn−q2+αnβ2xn−q2+ 2αnf(q)−q,jxn+1−q.
(3.50)
That is, xn+1−q2
≤
1−1−β2 1−αnαn
xn−q+ α 2 n
1−αn
xn−q2
+ 2αn 1−αn
f(q)−q,jxn+1−q
≤1−γnxn−q2+λγnαn+ 2 1−β2γn
f(q)−q,jxn+1−q,
(3.51)
whereγn=((1−β2)/(1−αn))αnandλis a constant such thatλ >(1/(1−β2))xn−q2.
Hence, xn+1−q2
≤1−γnxn−q2+γn
λαn+ 2
1−β2
f(q)−q,jxn+1−q. (3.52)
It is easily seen thatγn→0,∞n=1γn= ∞, and (noting (3.36))
lim sup
n→∞
λαn+ 2
1−β2
f(q)−q,jxn+1−q≤0. (3.53)
ApplyingLemma 3.3onto (3.52), we havexn→q.
Acknowledgment
This work is supported by the National Science Foundation of China, Grants 10471033 and 10271011.
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Rudong Chen: Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, China
E-mail address:[email protected]
Zhichuan Zhu: Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, China
