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Hindawi Publishing CorporationFixed Point Theory and Applications Volume 2007, Article ID 31056, 15 pages doi:10.1155/2007/31056 Research Article Weak and Strong Convergence of Multistep

Hindawi Publishing Corporation

Fixed Point Theory and Applications

Volume 2007, Article ID 31056, 15 pages

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

LetK be a nonempty subset of a real normed space E A self-mapping T : K → K is said to

be nonexpansive if Tx − T y  ≤  x − y for allx, y in K T is said to be asymptotically

nonexpansive if there exists a sequence{ r n }in [0,∞) with limn →∞ r n =0 such that T n x −

T n y  ≤(1 +r n) x − y for allx, y in K and n ∈ N

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

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 toL p-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

LetK be a nonempty closed subset of a normed space E, and let { T1,T2, , T N }:K →

K be N asymptotically nonexpansive mappings For a given x1 ∈ K and a fixed N ∈ N(N

denote the set of all positive integers), compute the sequence{ x n }by

x n+1 = x(N)

n = α(N)

n T n

N x(N −1)

n +β(N)

n x n+γ(N)

n u(N)

n ,

x(N −1)

n = α(N −1)

n T N n −1 x(N −2)

n +β(N −1)

n x n+γ(N −1)

n u(N −1)

n ,

x(3)

n = α(3)

n T n

3x(2)

n +β(3)

n x n+γ(3)

n u(3)

n ,

x(2)

n = α(2)

n T n

2x(1)

n +β(2)

n x n+γ(2)

n u(2)

n ,

x(1)

n = α(1)

n T1n x n+β(1)

n x n+γ(1)

n u(1)

n ,

(1.1)

where,{ u(1)n },{ u(2)n }, , { u(n N) }are bounded sequences inK and { α(n i) },{ β(n i) },{ γ(n i) }are appropriate real sequences in [0, 1] such thatα(n i)+β n(i)+γ n(i) =1 for eachi ∈ {1, 2, , N }

We now give some preliminaries and results which will be used in the rest of this paper

B S Thakur and J S Jung 3

A Banach spaceE is said to satisfy Opial’s condition if for each sequence x ninE, the

condition, that the sequencex n → x weakly, implies

lim sup

n →∞

x n − x< lim sup

n →∞

for ally ∈ E with y = x.

A Banach spaceE is said to have Kadec-Klee property if for every sequence { x n }inE,

x n → x weakly and  x n  →  x strongly together imply that x n − x  →0

We will make use of the following lemmas

Lemma 1.1 [2] Let E be a uniformly convex Banach space, let K be a nonempty closed convex subset of E, and let T : K → K be an asymptotically nonexpansive mapping Then,

I − T is demiclosed at zero, that is, for each sequence { x n } in K, if { x n } converges weakly to

q ∈ K and {(I − T)x n } converges strongly to 0, then (I − T)q = 0.

Lemma 1.2 [16] Let { a n } , { b n } , and { c n } be sequences of nonnegative real numbers satis-fying the inequality

a n+1 ≤1 +δ n



If∞

n =1 δ n < ∞ and∞

n =1 b n < ∞ , then lim n →∞ a n exists If, in addition, { a n } has a subse-quence which converges strongly to zero, then lim n →∞ a n = 0.

Lemma 1.3 [19] Let E be a uniformly convex Banach space and let b, c be two constants with 0 < b < c < 1 Suppose that { t n } is a real sequence in [b, c] and { x n } , { y n } are two sequences in E such that

lim sup

n →∞

x n  ≤ a,

lim sup

n →∞

y n  ≤ a,

lim

n →∞t n x n+

1− t n



y n  = a.

(1.4)

Then, lim n →∞  x n − y n  = 0, where a ≥ 0 is some constant.

Lemma 1.4 [12] Let E be a real reflexive Banach space such that its dual E ∗ has the Kadec-Klee property Let { x n } be a bounded sequence in E and p, q ∈ ω w(x n ), where ω w(x n ) denotes

the weak w-limit set of { x n } Suppose that lim n →∞  tx n+ (1− t)p − q  exists for all t ∈ [0, 1] Then p = 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 Let E be a real normed space and let K be a nonempty closed convex subset of

E Let { T1,T2, , T N }:K → K be N asymptotically nonexpansive mappings with sequences

{ r n(i) } such that∞

n =1r n(i) < ∞ , 1 ≤ i ≤ N Let { x n } be the sequence defined by ( 1.1 ) with

∞

n =1 γ(n i) < ∞ , 1 ≤ i ≤ N If F =N

i =1 F(T i)= ∅ , then lim n →∞  x n − p  exists for all p ∈ F.

Proof For any p ∈ F, we note that

x(1)

n − p  ≤ α(1)n T n

1x n − p+β(1)

n x n − p+γ(1)

n u(1)

n − p

≤ α(1)

n



1 +r nx n − p+β(1)

n x n − p+γ(1)

n u(1)

n − p

≤1 +r nx n − p+t(1)

n ,

(2.1)

wheret n(1)= γ(1)n  u(1)n − p  Since{ u(1)n }is bounded and∞

n =1γ(1)n < ∞, we can see that

∞

n =1 t n(1)< ∞ It follows from (2.1) that

x(2)

n − p  ≤ α(2)

n T n

2x(1)

n − p+β(2)

n x n − p+γ(2)

n u(2)

n − p

≤ α(2)

n



1 +r nx(1)

n − p+β(2)

n x n − p+γ(2)

n u(2)

n − p

≤ α(2)

n



1 +r n

1 +r nx n − p+t(1)

n



+β(2)

n x n − p+γ(2)

n u(2)

n − p

≤ α(2)

n



1 +r n 2 x n − p+α(2)

n t(1)

n



1 +r n

+β(2)

n x n − p+γ(2)

n u(2)

n − p

≤ α(2)n 

1 +r n 2 x n − p+α(2)

n t(1)n 

1 +r n

+β(2)n 

1 +r n

 2 x n − p+γ(2)

n u(2)

n − p

≤α(2)

n +β(2)

n



1 +r n 2 x n − p+α(2)

n t(1)

n



1 +r n

+γ(2)

n u(2)

n − p

≤1 +r n 2 x n − p+α(2)

n t(1)

n



1 +r n

+γ(2)

n u(2)

n − p

≤1 +r n 2 x n − p+t(2)

n ,

(2.2) wheret n(2)= α(2)n t(1)n (1 +r n) +γ(2)n  u(2)n − p  Since{ u(2)n }is bounded and∞

n =1 t(1)n < ∞,

we can see that∞

n =1t n(2)< ∞ Similarly, we see that

x(3)

n − p  ≤ α(3)

n



1 +r n

1 +r n 2 x n − p+t(2)

n



+β(3)

n x n − p+γ(3)

n u(3)

n − p

≤α(3)n +β(3)n 

1 +r n

 3 x n − p+α(3)

n t(2)n 

1 +r n



+γ(3)n u(3)

n − p

≤1 +r n 3 x n − p+α(3)

n t(2)

n



1 +r n

+γ(3)

n u(3)

n − p

≤1 +r n 3 x n − p+t(3)

n ,

(2.3) wheret n(3)= α(3)n t(2)n (1 +r n) +γ(3)n  u(3)n − p  Since{ u(3)n }is bounded and∞

n =1 t(2)n < ∞,

we can see that∞

n =1 t n(3)< ∞ Continuing the above process, we get

x n+1 − p  =  x(N)

n − p  ≤ 1 +r nNx n − p+t(N)

where { t n(N) } is nonnegative real sequence such that ∞

n =1t n(N) < ∞ By Lemma 1.2,

Lemma 2.2 Let E be a real uniformly convex Banach space and let K be a nonempty closed convex subset of E Let { T1,T2, , T N }:K → K be N asymptotically nonexpansive mappings

B S Thakur and J S Jung 5

with sequences { r n(i) } such that∞

n =1r n(i) < ∞ , 1 ≤ i ≤ N and let F =N

i =1F(T i)= ∅ Let

{ x n } be the sequence defined by ( 1.1 ) and some α, β ∈ (0, 1) with the following restrictions:

(i) 0< α ≤ α(n i) ≤ β < 1, 1 ≤ i ≤ N, for all n ≥ n0 for some n0 ∈ N ;

(ii)∞

n =1γ i

n < ∞ , 1 ≤ i ≤ N.

Then, lim n →∞  x n − T i x n  = 0.

Proof For any p ∈ F(T), it follows from Lemma 2.1 that limn →∞  x n − p  exists Let limn →∞  x n − p  = a for some a ≥0 We note that

x N −1

n − p  ≤ 1 +r n

N −1 x n − p+t(N −1)

where{ t n(N −1) }is nonnegative real sequence such that∞

n =1 t(n N −1) < ∞ It follows that lim sup

n →∞

x(N −1)

n →∞



1 +r nN −1x n − p+t N −1

n



=lim

n →∞x n − p  = a

(2.6) and so

lim sup

n →∞

T n

N x(N −1)

n →∞



1 +r nx(N −1)

n →∞

x(N −1)

n − p  ≤ a.

(2.7) Next, consider

T n

N x(N −1)

n − p + γ(N)

n



u(N)

n − x n  ≤  T N n x(N −1)

n − p+γ(N)

n u(N)

Thus,

lim sup

n →∞

T n

N x(N −1)

n − p + γ(N)

n



u(N)

Also,

x n − p + γ(N)

n



u(n N) − x n  ≤  x n − p+γ(N)

n u(N)

gives that

lim sup

n →∞

x n − p + γ(N)

n



u(n N) − x n  ≤ a, (2.11) and we observe that

x(N)

n − p = α(N)

n T N n x(N −1)

n − α(N)

n p + α(N)

n γ(N)

n u(N)

n − α(N)

n γ(N)

n x n+

1− α(N) n



x n

−1− α(N)

n



p − γ(N)

n x n+γ(N)

n u(N)

n − α(N)

n γ(N)

n u(N)

n +α(N)

n γ(N)

n x n

= α(n N)

T N n x n(N −1)− p + γ(n N)

u(n N) − x n



+

1− α(n N)

x n − p

−1− α(n N)

γ(n N) x n+

1− α(n N)

γ n(N) u(n N)

= α(n N)

T N n x n(N −1)− p + γ(n N)

u(n N) − x n



+

1− α(n N)

x n − p + γ n(N)

u(n N) − x n



.

(2.12)

a =lim

n →∞x(N)

n − p  =lim

n →∞α(N) n



T N n x(N −1)

n − p + γ(N)

n



u(N)

n − x n

+

1− α(N) n



x n − p + γ(N)

n



u(N)

By (2.9), (2.14), andLemma 1.3, we have

lim

n →∞T n

N x(N −1)

Now, we will show that limn →∞  T N n −1 x(n N −2) − x n  =0 For eachn ≥1,

x n − p  ≤  T n

N x(N −1)

n − x n+T n

N x(N −1)

n − p

≤T n

N x(N −1)

n − x n+

1 +r nx(N −1)

Using (2.14), we have

a =lim

n →∞x n − p  ≤lim inf

n →∞ x(N −1)

It follows that

a ≤lim inf

n →∞ x(N −1)

n →∞

x(N −1)

This implies that

lim

n →∞x(N −1)

On the other hand, we have

x(N −2)

n − p  ≤ 1 +r n

N −2x n − p+t(N −2)

n =1 t(n N −2) < ∞.Therefore,

lim sup

n →∞

x(N −2)

n − p  ≤lim sup

n →∞



1 +r n

N −2x n − p+t(N −2)

and hence,

lim sup

n →∞

T n

N −1x(N −2)

n →∞



1 +r nx(N −2)

Next, consider

T n

N −1 x(N −2)

n − p + γ(N −1)

n



u(N −1)

n − x n  ≤  T N n −1 x(N −2)

n − p+γ(N −1)

n u(N −1)

n − x n.

(2.22) Thus,

lim sup

n →∞

T n

N −1x(N −2)

n − p + γ(N −1)

n



u(N −1)

n − x n  ≤ a. (2.23)

B S Thakur and J S Jung 7 Also,

x n − p + γ(N −1)

n



u(N −1)

n − x n  ≤  x n − p+γ(N −1)

n u(N −1)

gives that

lim sup

n →∞

x n − p + γ(N −1)

n



u(N −1)

and we observe that

x n(N −1)− p = α(n N −1)T N n −1 x(n N −2)+

1− α(n N −1)

x n − γ(n N −1)x n

+γ(n N −1)u(n N −1)−1− α(n N −1)

p − α(n N −1)p

= α(N −1)

n



T n

N −1x(N −2)

n − p + γ(N −1)

n



u(N −1)

n − x n

+

1− α(N −1)

n



x n − p + γ(N −1)

n



u(N −1)

n − x n

,

(2.26)

and hence

a =lim

n →∞x(N −1)

n − p  =lim

n →∞α(N −1)

n



T N n −1 x n(N −2)− p + γ n(N −1)

u(n N −1)− x n



+

1− α(N −1)

n



x n − p + γ(N −1)

n



u(N −1)

n − x n. (2.27)

By (2.23), (2.25), andLemma 1.3, we have

lim

n →∞T n

N −1 x(N −2)

Similarly, by using the same argument as in the proof above, we have

lim

n →∞T n

N −1x(N −2)

Continuing similar process, we have

lim

n →∞T N − i x(N − i −1)

Now,

T n

1x n − p + γ(1)

n



u(1)

n − x n  ≤  T n

1x n − p+γ(1)

n u(1)

Thus,

lim sup

n →∞

T n

1x n − p + γ(1)

n



u(1)

Also,

x n − p + γ(1)

n



u(1)

n − x n  ≤  x n − p+γ(1)

n u(1)

gives that

lim sup

n →∞

x n − p + γ(1)

n



u(1)

and hence,

a =lim

n →∞x(1)

n − p  =lim

n →∞α(1)

n



T1n x n − p + γ(1)n 

u(1)n − x n



+

1− α(1)

n



x n − p + γ(1)

n



u(1)

By (2.32), (2.34), andLemma 1.3, we have

lim

n →∞T n

and this implies that

x n+1 − x n  =  α(N)

n T N n x(N −1)

n +

1− α(N)

n − γ(N) n



x n+γ(N)

n u(N)

n − x n

≤ α(n N)T n

N x(n N −1)− x n+γ(N)

n u(N)

n − x n  −→ ∞, asn −→ ∞ (2.37)

Thus, we have

T n

N x n − x n  ≤  T N n x n − T N n x(n N −1)+T n

N x(n N −1)− x n

≤1 +r nx n − x(N −1)

n +T n

N x(N −1)

n − x n

=1 +r nx n − α(N −1)

n T n

N −1x(N −2)

n +

1− α(N −1)

n − γ(N −1)

n



x n

+γ(N −1)

n u(N −1)

n +T n

N x(N −1)

n − x n

≤1 +r n

α(N −1)

n x n − T n

N −1x(N −2)

n +γ(N −1)

n u(N −1)

n − x n

+T n

N x(N −1)

n − x n  −→ ∞, asn −→ ∞,

(2.38)

and we have

T N x n − x n  ≤  x n+1 − x n+x n+1 − T n+1

N x n+1

+T n+1

N x n+1 − T N n+1 x n+T n+1

N x n − T N x n

≤x n+1 − x n+x n+1 − T n+1

N x n+1

+

1 +r n+1x n+1 − x n+

1 +r1T n

N x n − x n.

(2.39)

It follows from (2.37), (2.38), and (2.39) that

lim

B S Thakur and J S Jung 9 Next, we consider

T n

N −1 x n − x n  ≤  T N n −1 x n − T N n −1 x(N −2)

n +T n

N −1 x(N −2)

n − x n

≤1 +r nx n − x(N −2)

n +T n

N −1 x(N −2)

n − x n

≤1 +r n

α(N −2)

n x n − T n

N −2 x(N −3)

n +γ(N −2)

n u(N −2)

n − x n

+T n

N −1 x(N −2)

n − x n  −→ ∞, asn −→ ∞,

(2.41)

T N −1x n − x n  ≤  x n+1 − x n+x n+1 − T n+1

N −1 x n+1

+T n+1

N −1x n+1 − T n+1

N −1x n+T n+1

N −1x n − T N −1x n

≤x n+1 − x n+x n+1 − T n+1

N −1 x n+1

+

1 +r n+1x n+1 − x n+

1 +r1T n

N −1xn − x n.

(2.42)

It follows from (2.37), (2.41) and the above inequality that

lim

Continuing similar process, we have

lim

n →∞T N − i x n − x n  =0, 0≤ i ≤(N −2). (2.44) Now,

T1x n − x n  ≤  x n+1 − x n+x n+1 − T n+1

1 x n+1

+T n+1

1 x n+1 − T n+1

1 x n+T n+1

1 x n − T1x n

≤x n+1 − x n+x n+1 − T n+1

1 x n+1

+

1 +r n+1x n+1 − x n+

1 +r1T n

1x n − x n.

(2.45)

It follows from (2.36), (2.37) and the above inequality that

lim

and hence,

lim

n →∞T N − i x n − x n  =0, 0≤ i ≤(N −1). (2.47)

We recall the following definitions:

(i) A mappingT : K → K with F(T) = ∅ is said to satisfy condition (A) [21] onK

if there exists a nondecreasing function f : [0, ∞)→[0,∞) with f (0) =0 and

f (r) > r for all r ∈(0,∞) such that for allx ∈ K  x − Tx  ≥ f (d(x, F)), where d(x, F(T)) =inf{ x − p :p ∈ F(T) }

(ii) A finite family{ T1, , T N }ofN self mappings of K with F =i =1 N F(T i)= ∅

is said to satisfy condition (B) on K [1] if there exist f and d as in (i) such that

max1≤ i ≤ N  x − T i x  ≥ f (d(x, F)) for all x ∈ K.

(iii) A finite family{ T1, , T N }ofN self mappings of K with F =i =1N F(T i)= ∅

is said to satisfy condition (C) on K [1] if there exist f and d as in (i) such that

(1/N)N

i =1 x − T i x  ≥ f (d(x, F)) for all x ∈ K.

Note that condition (B) reduces to condition (A) whenT i = T, for all i =1, 2, , N.

It is well known that every continuous and demicompact mapping must satisfy condi-tion (A) (see [21]) Since every completely continuous mappingT : K → K is continuous

and demicompact, it satisfies condition (A) Therefore, to study strong convergence of

{ x n }defined by (1.1), we use condition (B) instead of the complete continuity of map-pingsT1,T2, , T N

Theorem 2.3 Let E be a real uniformly convex Banach space and K let be a nonempty closed convex subset of E Let { T1, , T N }:K → K be N asymptotically nonexpansive mappings with sequences { r n(i) } such that∞

n =1 r n(i) < ∞ for all 1 ≤ i ≤ N and F =N

i =1 F(T i)= ∅ Suppose that { T1,T2, , T N } satisfies condition (B) Let { x n } be the sequence defined by ( 1.1 ) and some α, β ∈ (0, 1) with the following restrictions:

(i) 0< α ≤ α(n i) ≤ β < 1, 1 ≤ i ≤ N ∀ n ≥ n0 for some n0 ∈ N ;

(ii)∞

n =1 γ i

n < ∞ , 1 ≤ i ≤ N.

Then, { x n } converges strongly to a common fixed point of the mappings { T1, , T N } Proof ByLemma 2.1, we see that limn →∞  x n − p exists for allp ∈ F Let lim n →∞  x n −

p  = a for some a ≥0 Without loss of generality, if a =0, there is nothing to prove Assume thata > 0, as proved inLemma 2.1, we have

x n+1 − p  =  x(n N) − p  ≤ 1 +r n

Nx n − p+t(N)

where{ t n(N) }is nonnegative real sequence such that∞

n =1 t(n N) < ∞ This gives that

d

x n+1,F

≤1 +r nN

d

x n,F

+t(N)

ApplyingLemma 1.2to the above inequality, we obtain that limn →∞ d(x n,F) exists Also,

byLemma 2.2, limn →∞  x n − T i x n  =0, for alli =1, 2, , N Since { T1,T2, , T N } satis-fies condition (B), we conclude that limn →∞ d(x n,F) =0

Next, we show that{ x n }is a Cauchy sequence Since limn →∞ d(x n,F) =0, given any

ε > 0, there exists a natural number n0such thatd(x n,F) < ε/3 for all n ≥ n0 So, we can findp ∗ ∈ F such that  x n0− p ∗  < ε/2 For all n ≥ n0andm ≥1, we have

x n+m − x n  ≤  x n+m − p ∗+x n − p ∗  ≤  x n0− p ∗+x n

0− p ∗< ε

2+

ε

2= ε.

(2.50) This shows that{ x n } is a Cauchy sequence and so is convergent since E is complete.

Let limn →∞ x n = q ∗ Thenq ∗ ∈ K It remains to show that q ∗ ∈ F Let ε1 > 0 be given.

Then, there exists a natural numbern1 such that x n − q ∗  < ε1/4 for all n ≥ n1 Since limn →∞ d(x n,F) =0, there exists a natural numbern2 ≥ n1such that for alln ≥ n2we have

d(x n,F) < ε1/5 and in particular we have d(x n,F) ≤ ε1/5 Therefore, there exists w ∗ ∈ F