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Weak and Strong Convergence of Multistep Iteration for Finite Family of Asymptotically Nonexpansive Mappings

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Fixed Point Theory and Applications Volume 2007, Article ID 31056,15pages doi:10.1155/2007/31056

Research Article

Weak and Strong Convergence of Multistep Iteration for Finite Family of Asymptotically Nonexpansive Mappings

Balwant Singh Thakur and Jong Soo Jung Received 14 March 2007; Accepted 26 May 2007 Recommended by Nan-Jing Huang

Strong and weak convergence theorems for multistep iterative scheme with errors for finite family of asymptotically nonexpansive mappings are established in Banach spaces.

Our results extend and improve the corresponding results of Chidume and Ali (2007), Cho et al. (2004), Khan and Fukhar-ud-din (2005), Plubtieng et al.(2006), Xu and Noor (2002), and many others.

Copyright © 2007 B. S. Thakur and J. S. Jung. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, dis- tribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction and preliminaries

LetKbe a nonempty subset of a real normed spaceE. A self-mappingT:KKis said to be nonexpansive ifTxT yxyfor allx,yinK.T is said to be asymptotically nonexpansive if there exists a sequence{rn}in [0,) with limn→∞rn=0 such thatTnx Tny(1 +rn)xyfor allx,yinKandnN.

The class of asymptotically nonexpansive mappings which is an important generaliza- tion of that of nonexpansive mappings was introduced by Goebel and Kirk [6]. Iteration processes for nonexpansive and asymptotically nonexpansive mappings in Banach spaces including Mann [7] and Ishikawa [8] iteration processes have been studied extensively by many authors to solve the nonlinear operators as well as variational inequalities; see [1–22,25].

Noor [13] introduced a three-step iterative scheme and studied the approximate so- lution of variational inclusion in Hilbert spaces by using the techniques of updating the solution and auxiliary principle. Glowinski and Le Tallec [9] used three-step iterative schemes to find the approximate solutions of the elastoviscoplasticity problem, liquid crystal theory, and eigenvalue computation. It has been shown in [9] that the three-step

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iterative scheme gives better numerical results than the two-step and one-step approx- imate iterations. Thus, we conclude that the three-step scheme plays an important and significant role in solving various problems which arise in pure and applied sciences. Re- cently, Xu and Noor [5] introduced and studied a three-step scheme to approximate fixed points of asymptotically nonexpansive mappings in Banach space. Cho et al. [2] extended the work of Xu and Noor [5] to the three-step iterative scheme with errors in a Banach space and gave weak and strong convergence theorems for asymptotically nonexpansive mappings in a Banach space. Moreover, Suantai [20] gave weak and strong convergence theorems for a new three-step iterative scheme of asymptotically nonexpansive mappings.

More recently, Plubtieng et al. [4] introduced a three-step iterative scheme with errors for three asymptotically nonexpansive mappings and established strong convergence of this scheme to common fixed point of three asymptotically nonexpansive mappings. Very recently, Chidume and Ali [1] considered multistep scheme for finite family of asymptot- ically nonexpansive mappings and gave weak convergence theorems for this scheme in a uniformly convex Banach space whose the dual space satisfies the Kadec-Klee property.

They also proved a strong convergence theorem under some appropriate conditions on finite family of asymptotically nonexpansive mappings.

Inspired by the above facts, in this paper, a new multistep iteration scheme with errors for finite family of asymptotically nonexpansive mappings is introduced and strong and weak convergence theorems of this scheme to common fixed point of asymptotically non- expansive mappings are proved. In particular, our weak convergence theorem is proved in a uniformly convex Banach space whose the dual has a Kadec-Klee property. It is worth mentioning that there are uniformly convex Banach spaces, which have neither a Fr´echet differentiable norm nor Opial property; however, their dual does have the Kadec-Klee property. This means that our weak convergence result can apply not only toLp-spaces with 1< p <but also to other spaces which do not satisfy Opial’s condition or have a Fr´echet differentiable norm. Our theorems improve and generalize some previous results in [1–5,15,17–19]. Our iterative scheme is defined as below.

LetKbe a nonempty closed subset of a normed spaceE, and let{T1,T2,. . .,TN}:K KbeNasymptotically nonexpansive mappings. For a givenx1Kand a fixedNN(N denote the set of all positive integers), compute the sequence{xn}by

xn+1=x(N)n =α(N)n TNnx(Nn 1)+β(N)n xn+γ(N)n u(N)n , xn(N1)=α(Nn 1)TNn1x(Nn 2)+β(Nn 1)xn+γn(N1)u(Nn 1),

...

xn(3)=α(3)n T3nx(2)n +β(3)n xn+γ(3)n u(3)n , xn(2)=α(2)n T2nx(1)n +β(2)n xn+γ(2)n u(2)n , x(1)n =α(1)n T1nxn+β(1)n xn+γ(1)n u(1)n ,

(1.1)

where,{u(1)n },{u(2)n },. . .,{u(N)n }are bounded sequences inK and{α(i)n },{β(i)n },{γ(i)n }are appropriate real sequences in [0, 1] such thatα(i)n +βn(i)+γn(i)=1 for eachi∈ {1, 2,. . .,N}. We now give some preliminaries and results which will be used in the rest of this paper.

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A Banach spaceEis said to satisfy Opial’s condition if for each sequencexninE, the condition, that the sequencexnxweakly, implies

lim sup

n→∞

xnx<lim sup

n→∞

xny (1.2)

for allyEwithy=x.

A Banach spaceEis said to have Kadec-Klee property if for every sequence{xn}inE, xnxweakly andxnxstrongly together imply thatxnx0.

We will make use of the following lemmas.

Lemma 1.1 [2]. Let Ebe a uniformly convex Banach space, letK be a nonempty closed convex subset ofE, and letT:KK be an asymptotically nonexpansive mapping. Then, ITis demiclosed at zero, that is, for each sequence{xn}inK, if{xn}converges weakly to qKand{(IT)xn}converges strongly to 0, then (IT)q=0.

Lemma 1.2 [16]. Let{an},{bn}, and{cn}be sequences of nonnegative real numbers satis- fying the inequality

an+1 1 +δn

an+bn, n1. (1.3)

Ifn=1δn<andn=1bn<, then limn→∞anexists. If, in addition,{an}has a subse- quence which converges strongly to zero, then limn→∞an=0.

Lemma 1.3 [19]. LetEbe a uniformly convex Banach space and letb,cbe two constants with 0< b < c <1. Suppose that{tn}is a real sequence in [b,c] and{xn},{yn} are two sequences inEsuch that

lim sup

n→∞

xna, lim sup

n→∞

yna,

nlim→∞tnxn+1tn

yn=a.

(1.4)

Then, limn→∞xnyn =0, wherea0 is some constant.

Lemma 1.4 [12]. LetEbe a real reflexive Banach space such that its dualEhas the Kadec- Klee property. Let{xn}be a bounded sequence inEandp,qωw(xn), whereωw(xn) denotes the weak w-limit set of {xn}. Suppose that limn→∞txn+ (1t)pqexists for all t [0, 1]. Thenp=q.

2. Main results

In this section, we prove strong and weak convergence theorems for multistep iteration with errors in Banach spaces. In order to prove our main results, we need the following lemmas.

Lemma 2.1. LetEbe a real normed space and letK be a nonempty closed convex subset of E. Let{T1,T2,. . .,TN}:KKbeNasymptotically nonexpansive mappings with sequences {rn(i)} such thatn=1rn(i)<, 1iN. Let{xn}be the sequence defined by (1.1) with

n=1γ(i)n <, 1iN. IfF=N

i=1F(Ti)= ∅, then limn→∞xnpexists for allpF.

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Proof. For anypF, we note that

x(1)n pα(1)n T1nxnp+β(1)n xnp+γn(1)u(1)n p

α(1)n 1 +rnxnp+β(1)n xnp+γn(1)u(1)n p

1 +rnxnp+t(1)n ,

(2.1)

wheretn(1)=γ(1)n u(1)n p. Since{u(1)n }is bounded andn=1γ(1)n <, we can see that

n=1tn(1)<. It follows from (2.1) that

x(2)n pα(2)n T2nxn(1)p+β(2)n xnp+γn(2)u(2)n p

α(2)n 1 +rnxn(1)p+β(2)n xnp+γ(2)n u(2)n p

α(2)n 1 +rn1 +rnxnp+tn(1)+β(2)n xnp+γn(2)u(2)n p

α(2)n 1 +rn2xnp+α(2)n t(1)n 1 +rn

+β(2)n xnp+γ(2)n u(2)n p

α(2)n 1 +rn2xnp+α(2)n t(1)n 1 +rn +β(2)n 1 +rn

2xnp+γn(2)u(2)n p

α(2)n +β(2)n 1 +rn2xnp+α(2)n t(1)n 1 +rn+γ(2)n u(2)n p

1 +rn2xnp+α(2)n t(1)n 1 +rn+γ(2)n u(2)n p

1 +rn2xnp+t(2)n ,

(2.2) wheretn(2)=α(2)n t(1)n (1 +rn) +γ(2)n u(2)n p. Since{u(2)n }is bounded andn=1t(1)n <, we can see thatn=1tn(2)<. Similarly, we see that

xn(3)pα(3)n 1 +rn

1 +rn2xnp+t(2)n +βn(3)xnp+γ(3)n u(3)n p

α(3)n +β(3)n 1 +rn

3xnp+α(3)n t(2)n 1 +rn

+γ(3)n u(3)n p

1 +rn3xnp+α(3)n tn(2)1 +rn+γ(3)n u(3)n p

1 +rn3xnp+tn(3),

(2.3) wheretn(3)=α(3)n t(2)n (1 +rn) +γ(3)n u(3)n p. Since{u(3)n }is bounded andn=1t(2)n <, we can see thatn=1tn(3)<. Continuing the above process, we get

xn+1p=x(N)n p

1 +rnNxnp+tn(N), (2.4) where {tn(N)} is nonnegative real sequence such that n=1tn(N) <. By Lemma 1.2,

limn→∞xnpexists. This completes the proof.

Lemma 2.2. LetEbe a real uniformly convex Banach space and letKbe a nonempty closed convex subset ofE. Let{T1,T2,. . .,TN}:KKbeNasymptotically nonexpansive mappings

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with sequences{rn(i)}such thatn=1rn(i)<, 1iN and letF=N

i=1F(Ti)= ∅. Let {xn}be the sequence defined by (1.1) and someα,β(0, 1) with the following restrictions:

(i) 0< αα(i)n β <1, 1iN, for allnn0for somen0N; (ii)n=1γni <, 1iN.

Then, limn→∞xnTixn =0.

Proof. For any pF(T), it follows from Lemma 2.1 that limn→∞xnp exists. Let limn→∞xnp =afor somea0. We note that

xnN1p 1 +rn

N1xnp+t(Nn 1), n1, (2.5) where{tn(N1)}is nonnegative real sequence such thatn=1t(Nn 1)<. It follows that

lim sup

n→∞

x(Nn 1)plim sup

n→∞

1 +rnN1xnp+tnN1=lim

n→∞xnp=a (2.6) and so

lim sup

n→∞

TNnxn(N1)plim sup

n→∞

1 +rnxn(N1)p=lim sup

n→∞

x(Nn 1)pa.

(2.7) Next, consider

TNnx(Nn 1)p+γn(N)u(N)n xnTNnxn(N1)p+γn(N)u(N)n xn. (2.8) Thus,

lim sup

n→∞

TNnx(Nn 1)p+γ(N)n u(N)n xna. (2.9)

Also,

xnp+γ(N)n u(N)n xnxnp+γ(N)n u(N)n xn (2.10) gives that

lim sup

n→∞

xnp+γ(N)n u(N)n xna, (2.11)

and we observe that

x(N)n p=α(N)n TNnx(Nn 1)α(N)n p+α(N)n γ(N)n u(N)n α(N)n γ(N)n xn+1α(N)n xn

1α(N)n pγ(N)n xn+γn(N)u(N)n α(N)n γn(N)u(N)n +α(N)n γ(N)n xn

=α(N)n TNnxn(N1)p+γ(N)n u(N)n xn

+1α(N)n xnp

1α(N)n γ(N)n xn+1α(N)n γn(N)u(N)n

=α(N)n TNnxn(N1)p+γ(N)n u(N)n xn

+1α(N)n xnp+γn(N)u(N)n xn

. (2.12)

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Therefore, a=lim

n→∞x(N)n p=lim

n→∞α(N)n TNnx(Nn 1)p+γ(N)n u(N)n xn

+1α(N)n xnp+γ(N)n u(N)n xn. (2.13) By (2.9), (2.14), andLemma 1.3, we have

nlim→∞TNnx(Nn 1)xn=0. (2.14) Now, we will show that limn→∞TNn1x(Nn 2)xn =0. For eachn1,

xnpTNnx(Nn 1)xn+TNnx(Nn 1)p

TNnx(Nn 1)xn+1 +rnx(Nn 1)p. (2.15) Using (2.14), we have

a=lim

n→∞xnplim inf

n→∞ x(Nn 1)p. (2.16) It follows that

alim inf

n→∞ xn(N1)plim sup

n→∞

x(Nn 1)pa. (2.17)

This implies that

nlim→∞x(Nn 1)p=a. (2.18) On the other hand, we have

xn(N2)p 1 +rn

N2xnp+t(Nn 2), n1, (2.19) wheren=1t(Nn 2)<.Therefore,

lim sup

n→∞

x(Nn 2)plim sup

n→∞

1 +rn

N2xnp+tn(N2)=a, (2.20)

and hence,

lim sup

n→∞

TNn1x(Nn 2)plim sup

n→∞

1 +rnx(Nn 2)pa. (2.21)

Next, consider

TNn1xn(N2)p+γn(N1)u(Nn 1)xnTNn1x(Nn 2)p+γ(Nn 1)u(Nn 1)xn. (2.22) Thus,

lim sup

n→∞

TNn1xn(N2)p+γn(N1)u(Nn 1)xna. (2.23)

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Also,

xnp+γ(Nn 1)u(Nn 1)xnxnp+γ(Nn 1)u(Nn 1)xn (2.24) gives that

lim sup

n→∞

xnp+γ(Nn 1)u(Nn 1)xna, (2.25)

and we observe that

xn(N1)p=α(Nn 1)TNn1x(Nn 2)+1α(Nn 1)xnγ(Nn 1)xn

+γ(Nn 1)u(Nn 1)

1α(Nn 1)pα(Nn 1)p

=α(Nn 1)TNn1xn(N2)p+γ(Nn 1)u(Nn 1)xn +1α(Nn 1)xnp+γ(Nn 1)u(Nn 1)xn,

(2.26)

and hence a=lim

n→∞x(Nn 1)p=lim

n→∞α(Nn 1)TNn1xn(N2)p+γn(N1)u(Nn 1)xn

+1α(Nn 1)xnp+γ(Nn 1)u(Nn 1)xn. (2.27) By (2.23), (2.25), andLemma 1.3, we have

nlim→∞TNn1xn(N2)xn=0. (2.28) Similarly, by using the same argument as in the proof above, we have

nlim→∞TNn1xn(N2)xn=0. (2.29) Continuing similar process, we have

nlim→∞TNix(Nn i1)xn=0, 0i(N2). (2.30) Now,

T1nxnp+γ(1)n u(1)n xnT1nxnp+γ(1)n u(1)n xn. (2.31) Thus,

lim sup

n→∞

T1nxnp+γn(1)u(1)n xna. (2.32)

Also,

xnp+γ(1)n u(1)n xnxnp+γ(1)n u(1)n xn (2.33)

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