WEAK AND STRONG CONVERGENCE OF FINITE FAMILY WITH ERRORS OF NONEXPANSIVE NONSELF-MAPPINGS
S. PLUBTIENG AND K. UNGCHITTRAKOOL
Received 27 September 2005; Revised 5 May 2006; Accepted 8 May 2006
We are concerned with the study of a multistep iterative scheme with errors involving a finite family of nonexpansive nonself-mappings. We approximate the common fixed points of a finite family of nonexpansive nonself-mappings by weak and strong conver- gence of the scheme in a uniformly convex Banach space. Our results extend and improve some recent results, Shahzad (2005) and many others.
Copyright © 2006 S. Plubtieng and K. Ungchittrakool. This is an open access article dis- tributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is prop- erly cited.
1. Introduction
LetKbe a subset of a real normed linear spaceEand letTbe a self-mapping onK.T is said to be nonexpansive providedTx−T yx−yfor allx,y∈K.
Fixed-point iteration process for nonexpansive mappings in Banach spaces includ- ing Mann and Ishikawa iteration processes has been studied extensively by many au- thors to solve the nonlinear operator equations in Hilbert spaces and Banach spaces;
see [3,7,10,11,15,16]. Tan and Xu [15] introduced and studied a modified Ishikawa process to approximate fixed points of nonexpansive mappings defined on nonempty closed convex bounded subsets of a uniformly convex Banach spaceE. Five years later, Xu [18] introduced iterative schemes known as Mann iterative scheme with errors and Ishikawa iterative scheme with errors. Takahashi and Tamura [14] introduced and stud- ied a generalization of Ishikawa iterative schemes for a pair of nonexpansive mappings in Banach spaces. Recently, Khan and Fukhar-ud-din [6] extended their scheme to the modified Ishikawa iterative schemes with errors for two mappings and gave weak and strong convergence theorems. On the other hand, iterative techniques for approximat- ing fixed points of nonexpansive nonself-mappings have been studied by various au- thors; see [4,8,13,19]. Shahzad [12] introduced and studied an iteration scheme for
Hindawi Publishing Corporation Fixed Point Theory and Applications Volume 2006, Article ID 81493, Pages1–12 DOI10.1155/FPTA/2006/81493
approximating a fixed point of nonexpansive nonself-mappings (when such a fixed point exists) and gave some strong and weak convergence theorems for such mappings.
Inspired and motivated by these facts, we introduce and study a multistep iterative scheme with errors for a finite family of nonexpansive nonself-mappings. Our schemes can be viewed as an extension for two-step iterative schemes of Shahzad [12]. The scheme is defined as follows.
LetK be a nonempty closed convex subset of a uniformly convex Banach space E, which is also a nonexpansive retract ofE. And letT1,T2,. . .,TN:K→Ebe nonexpansive mappings, the following iteration scheme is studied:
x1n=Pα1nT1xn+β1nxn+γ1nu1n, x2n=Pα2nT2xn1+β2nxn+γ2nu2n,
... ...
xn+1=xNn =PαNnTNxnN−1 +βNnxn+γnNuNn
(1.1)
withx1∈K,n1, whereP is a nonexpansive retraction with respect to K and{α1n}, {α2n},. . .,{αNn},{β1n},{β2n},. . .,{βNn},{γ1n},{γ2n},. . .,{γNn}are sequences in [0, 1] withαin+ βin+γin=1 for alli=1, 2,. . .,N, and{u1n},{u2n},. . .,{uNn}are bounded sequences inK.
ForN=2,T1=T2≡T,βn=α1n,αn=α2n, andγ1n=γ2n≡0, then (1.1) reduces to the scheme for a mapping defined by Shahzad [12]:
x1=x∈K,
xn+1=P1−αnxn+αnTP1−βnxn+βnTxn, (1.2) where{αn},{βn}are sequences in [0, 1].
For N=2,T1,T2:K →K,T1=T, T2=S, and yn=x1n, then (1.1) reduces to the scheme with errors for two mappings defined by
x1=x∈K,
yn=α1nTxn+βn1xn+γ1nu1n, xn+1=x2n=α2nSyn+β2nxn+γ2nu2n,
(1.3)
where{α1n},{α2n},{β1n},{β2n},{γn1},{γ2n}are sequences in [0, 1] withα1n+β1n+γn1=1=α2n+ β2n+γ2nand{u1n},{u2n}are bounded sequences inK.
It is our purpose in this paper to establish several weak and strong convergence theorems of the multistep iterative scheme with errors for a finite family of nonexpansive nonself-mappings. More precisely, we prove weak convergence of these implicit iteration
processes in a uniformly convex Banach space which has the Kadec-Klee property. The results presented in this paper extend and improve the corresponding ones announced by Shahzad [12], and many others.
2. Preliminaries
In this section, we recall the well-known concepts and results.
LetEbe a real Banach space. A subsetKofEis said to be a retract ofEif there exists a continuous mapP:E→Ksuch thatPx=xfor allx∈K. A mapP:E→Eis said to be a retraction ifP2=P. It follows that if a mapPis a retraction, thenP y=yfor all yin the range ofP. A mappingT:K→Eis called demiclosed with respect toy∈Eif for each sequence{xn}inKand eachx∈E,xnxandTxn→yimply thatx∈KandTx=y. A Banach spaceEis said to satisfy Opial’s condition [9] if for any sequence{xn}inE,xnx implies that
lim sup
n→∞
xn−x<lim sup
n→∞
xn−y (2.1)
for ally∈Ewithx=y. A Banach spaceEis said to have the Kadec-Klee property if for every sequence{xn}inE,xnxandxn → xtogether implyxn−x →0. A family {Ti:i=1, 2,. . .,N}ofNnonself-mappings ofK(i.e.,Ti:K→E) withF= Ni=1F(Ti)=∅ is said to satisfy condition (B) onKif there is a nondecreasing functionf : [0,∞)→[0,∞) with f(0)=0 andf(r)>0 for allr∈(0,∞) such that for allx∈K,
1maxiNx−Tixfd(x,F). (2.2) The family{Ti:i=1, 2,. . .,N} is said to satisfy condition (AN) if (2.2) is replaced by 1/NNi=1x−Tixf(d(x,F)) for allx∈K. Note that condition (B) reduces to condi- tion (AN) whenx−T1x = x−T2x = ··· = x−TNx.
A mappingT:K→Eis called semicompact if any sequence{xn}inKsatisfyingxn− Txn →0 asn→ ∞has a convergent subsequence.
Lemma 2.1 (Tan and Xu [15]). Let{sn},{tn}be two nonnegative sequences satisfying
sn+1sn+tn, ∀n1. (2.3)
If∞n=1tn<∞, then limn→∞snexists. Moreover, if there exists a subsequence{snj}of{sn} such thatsnj→0 as j→ ∞, thensn→0 asn→ ∞.
Lemma 2.2 (Xu [17]). Let p >1 andR >0 be two fixed numbers andEa Banach space.
ThenEis uniformly convex if and only if there exists a continuous, strictly increasing, and convex function g: [0,∞)→[0,∞) withg(0)=0 such thatλx+ (1−λ)ypλxp+ (1−λ)yp−Wp(λ)g(x−y) for allx,y∈BR(0)= {x∈E:xR}, andλ∈[0, 1], whereWp(λ)=λ(1−λ)p+λp(1−λ).
Lemma 2.3 (Kaczor [5]). LetEbe a real reflexive Banach space such that its dualE∗has the Kadec-Klee property. Let{xn}be a bounded sequence inEandx∗,y∗∈ωw(xn); hereωw(xn)
denote the set of all weak subsequential limits of{xn}. Suppose limn→∞txn+(1−t)x∗−y∗ exists for allt∈[0, 1]. Thenx∗=y∗.
Lemma 2.4 (Browder [1]). LetEbe a uniformly convex Banach space,Ka nonempty closed convex subset ofE, andT:K→Ea nonexpansive mapping. ThenI−T is demiclosed at zero.
3. Main results
In this section, we prove weak and strong convergence theorems of the iterative scheme given in (1.1) for a finite family of nonexpansive mappings in a Banach space. In order to prove our main results, the following lemmas are needed.
Lemma 3.1. Let E be a uniformly convex Banach space andK a nonempty closed con- vex subset of E which is also a nonexpansive retract of E. Let T1,T2,. . .,TN:K→E be nonexpansive mappings. Let {xn} be the sequence defined by (1.1) with ∞n=1γin<∞ for eachi=1, 2,. . .,N. If Ni=1F(Ti)=∅, then limn→∞xn−x∗exists for allx∗∈ Ni=1F(Ti).
Proof. For eachn1, we note that
x1n−x∗α1nT1xn−x∗+β1nxn−x∗+γ1nu1n−x∗ α1nxn−x∗+β1nxn−x∗+γ1nu1n−x∗ xn−x∗+dn0,
(3.1)
wheredn0=γ1nu1n−x∗. Since∞n=1γ1n<∞,∞n=1d0n<∞. Next, we note that x2n−x∗α2nT2xn1−x∗+β2nxn−x∗+γ2nu2n−x∗
α2nxn1−x∗+β2nxn−x∗+γ2nu2n−x∗
=
α2n+β2nxn−x∗+α2ndn0+γn2u2n−x∗ xn−x∗+dn1,
(3.2)
wheredn1=α2ndn0+γ2nu2n−x∗. Since∞n=1dn0<∞and∞n=1γ2n<∞,∞n=1d1n<∞. Simi- larly, we have
x3n−x∗α3nxn2−x∗+β3nxn−x∗+γ3nu3n−x∗ α3nxn−x∗+dn1+β3nxn−x∗+γ3nu3n−x∗ xn−x∗+α3ndn1+γ3nu3n−x∗=xn−x∗+dn2,
(3.3)
wheredn2=α3ndn1+γ3nu3n−x∗, so∞n=1d2n<∞.
By continuing the above method, there exists a nonnegative real sequence{dni−1}such that∞n=1dni−1<∞and
xni−x∗xn−x∗+dni−1, ∀n1,∀i=1, 2,. . .,N. (3.4) Thusxn+1−x∗ = xNn −x∗xn−x∗+dnN−1for alln∈N. Hence, byLemma 2.1,
limn→∞xn−x∗exists. This completes the proof.
Lemma 3.2. LetEbe a uniformly convex Banach space andK a nonempty closed convex subset ofEwhich is also a nonexpansive retract ofE. LetT1,T2,. . .,TN:K→Ebe nonex- pansive mappings. Let{xn}be the sequence defined by (1.1) with∞n=1γni <∞and{αin} ⊆ [ε, 1−ε] for alli=1, 2,. . .,N, for someε∈(0, 1). If Ni=1F(Ti)=∅, then limn→∞xn− Tixn =0 for alli=1, 2,. . .,N.
Proof. Letx∗∈ Ni=1F(Ti). Then, by Lemma 3.1, limn→∞xn−x∗exists. Let limn→∞
xn−x∗ =r. Ifr=0, then by the continuity of eachTi the conclusion follows. Sup- pose thatr >0. Firstly, we are now to show that limn→∞TNxn−xn =0. Since{xn}and {uin}are bounded for alli=1, 2,. . .,N, there existsR >0 such thatxn−x∗+γin(uin−xn), Tixni−1−x∗+γin(uin−xn)∈BR(0) for alln1 and for alli=1, 2,. . .,N. UsingLemma 2.2, we have
xNn −x∗2αNnTNxNn−1+βNnxn+γNnuNn−x∗2
=αNnTNxNn−1−x∗+γNnuNn−xn+1−αNnxn−x∗+γNnuNn −xn2 αNnTNxNn−1−x∗+γnNuNn−xn2+1−αNnxn−x∗+γNnuNn −xn2
−W2
αNngTNxNn−1−xn
αNnxNn−1−x∗+γnNuNn −xn2+1−αNnxn−x∗+γNnuNn −xn2
−W2
αNngTNxNn−1−xn
αNnxn−x∗+dNn−2+γNnuNn −xn2 +1−αNnxn−x∗+dnN−2+γnNuNn −xn2
−W2
αNngTNxNn−1−xn
=xn−x∗+λNn−22−W2
αNngTNxnN−1−xn,
(3.5) whereλNn−2:=dNn−2+γNnuNn −x∗. Observe thatε3W2(αNn) now (3.5) implies that ε3g(TNxNn−1−xn)xn−x∗2− xn+1−x∗2+ρnN−2, where ρNn−2:=2λNn−2xn− x∗2+ (λNn−2)2. Since∞n=1dNn−2<∞and∞n=1γNn−2<∞, we get∞n=1ρNn−2<∞. This implies that limn→∞g(TNxNn−1−xn)=0. Sinceg is strictly increasing and continuous
at 0, it follows that limn→∞TNxNn−1−xn =0. Note that
xn−x∗xn−TNxNn−1+TNxNn−1−x∗ xn−TNxNn−1+xNn−1−x∗,
(3.6)
for alln1. Thusr=limn→∞xn−x∗lim infn→∞xNn−1−x∗lim supn→∞xnN−1− x∗rand therefore limn→∞xNn−1−x∗ =r. Using the same argument in the proof above, we have
xNn−1−x∗2αNn−1xNn−2−x∗+γNn−1uNn−1−x∗2 +1−αNn−1xn−x∗+γNn−1uNn−1−x∗2
−W2
αNn−1gTN−1xnN−2−xn
αNn−1xn−x∗+dnN−3+γnN−1uNn−1−x∗2 +1−αNn−1xn−x∗+dnN−3+γNn−1uNn−1−x∗2
−W2
αNn−1gTN−1xnN−2−xn xn−x∗2+ρNn−3−W2
αNn−1gTN−1xnN−2−xn.
(3.7)
This implies thatε3g(TN−1xNn−2−xn)xn−x∗2− xNn−1−x∗2+ρNn−3and there- fore limn→∞TN−1xnN−2−xn =0. Thus, we have
xn−TNxnxn−TNxnN−1+TNxnN−1−TNxn xn−TNxnN−1+xNn−1−xn
=xn−TNxNn−1+PαNn−1TN−1xNn−2+βNn−1xn+γnN−1uNn−1−Pxn xn−TNxnN−1+αNn−1xn−TN−1xNn−2+γnN−1uNn−1−xn.
(3.8) Since limn→∞xn−TNxnN−1 =0, limn→∞xn−TN−1xNn−2 =0, and∞n=1γNn−1<∞, it follows that limn→∞xn−TNxn =0. Similarly, by using the same argument as in the proof above, we have limn→∞xn−TN−2xNn−3 =limn→∞xn−TN−3xNn−4 =,. . .,= limn→∞xn−T2x1n=0. This implies that limn→∞xn−TN−1xn=limn→∞xn−TN−2xn = ,. . .,=limn→∞xn−T3xn =0. It remains to show that
limn→∞xn−T1xn=0, limn→∞xn−T2xn=0. (3.9)