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CONVERGENCE OF TWO-STEP ITERATIVE SCHEME WITH ERRORS
FOR TWO ASYMPTOTICALLY NONEXPANSIVE MAPPINGS
HAFIZ FUKHAR-UD-DIN and SAFEER HUSSAIN KHAN
Received 28 August 2003
A two-step iterative scheme with errors has been studied to approximate the common fixed points of two asymptotically nonexpansive mappings through weak and strong convergence in Banach spaces.
2000 Mathematics Subject Classification: 47H09, 49M05.
1. Introduction. In 1995, Liu [4] introduced iterative schemes with errors as follows. (a) For a nonempty subsetCof a normed spaceEandT:C→C, the sequence{xn}
inC, iteratively defined by
x1=x∈C,
xn+1=
1−an
xn+anT yn+un,
yn=
1−bn
xn+bnT xn+vn, n≥1,
(1.1)
where {an}, {bn} are sequences in[0,1] and {un}, {vn} are sequences inE
satisfying∞n=1un<∞,
∞
n=1vn<∞, is known as Ishikawa iterative scheme
with errors.
(b) WithE,C, andT as in (a), the sequence{xn}, iteratively defined by
x1=x∈C,
xn+1=
1−an
xn+anT xn+un, n≥1, (1.2)
where {an} is a sequence in [0,1] and {un} a sequence in E satisfying
∞
n=1un<∞, is known as Mann iterative scheme with errors.
In 1999, Huang [2] studied the above schemes for asymptotically nonexpansive map-pings. Recall that a mappingT:C→Cis asymptotically nonexpansive if there is a se-quence{kn} ⊂[1,∞)with limn→∞kn=1 andTnx−Tny ≤knx−yfor allx,y∈C and for alln∈N, whereNdenotes the set of positive integers.
Moreover, in 2001, Khan and Takahashi [3] approximated the fixed points of two asymptotically nonexpansive mappingsS,T:C→Cthrough the sequence{xn}given by
x1=x∈C,
xn+1=
1−an
xn+anSnyn,
yn=1−bnxn+bnTnxn,
(1.3)
Inspired and motivated by the study of the above schemes, we suggest a new iterative scheme{xn}inCconstructed through a pair of asymtotically nonexpansive mappings
S,T:C→Cgiven by
x1=x∈C,
xn+1=
1−an
xn+anSnyn+un,
yn=
1−bn
xn+bnTnxn+vn, n≥1,
(1.4)
where{an},{bn}are sequences in[0,1]with appropriate conditions and{un},{vn} are sequences inEwith∞n=1un<∞,∞n=1vn<∞.
It is to be noted here that each of the above schemes follows as a special case of our scheme.
2. Preliminaries. LetE be a Banach space with C as its nonempty convex subset. Throughout this paper,Ndenotes the set of positive integers andF(T )the set of fixed points of the mappingT. Now we list the following definitions and results used to prove the results in the next section.
Definition2.1. A mappingT:C→Cis uniformlyk-Lipschitzian if for somek >0, Tnx−Tny ≤kx−yfor allx,y∈Cand for alln∈N.
Definition2.2. A mappingT:C→Cis completely continuous if and only if{T xn} has a convergent subsequence for every bounded sequence{xn}inC.
Definition2.3. E is said to satisfy Opial’s condition [5] if for any sequence {xn}
inE,xn ximplies that lim supn→∞xn−x<lim supn→∞xn−yfor ally∈Ewith
y≠x.
Definition2.4. A mappingT:C→Eis called demiclosed with respect toy∈Eif for each sequence{xn}inC and eachx∈E,xn xandT xn→yimply thatx∈C andT x=y.
Lemma2.5[6]. Suppose thatEis a uniformly convex Banach space and0< p≤tn≤
q <1for alln∈N. Also, suppose that{xn}and{yn}are two sequences ofEsuch that lim supn→∞xn ≤r,lim supn→∞yn ≤r, andlimn→∞tnxn+(1−tn)yn =rhold for somer≥0. Thenlimn→∞xn−yn =0.
Lemma2.6[7]. Let{rn},{sn},{tn}be three nonnegative sequences satisfying
rn+1≤
1+sn
rn+tn ∀n≥1. (2.1)
If∞n=1sn<∞and
∞
n=1tn<∞, thenlimn→∞rnexists.
...
3. Approximating common fixed points. We start with the following lemma.
Lemma3.1. LetEbe a normed space andCa nonempty bounded closed convex subset ofE. Let, fork >0,S andT be uniformlyk-Lipschitzian mappings ofCinto itself. Let {xn}be a sequence as defined in (1.4), where{un},{vn}are sequences inEsuch that limn→∞un =0=limn→∞vnand
lim n→∞xn−S
nx
n=0=lim n→∞xn−T
nx
n. (3.1)
Then
lim
n→∞xn−Sxn=0=nlim→∞xn−T xn. (3.2)
Proof. Takecn= xn−Tnxnanddn= xn−Snxn. Consider
xn+1−xn=an
Sny
n−xn+un ≤anSnyn−Snxn
+Snx
n−xn+un ≤ank1−bn
xn+bnTnxn+vn−xn+andn+un =ankbn
Tnxn−xn
+vn+andn+un ≤anbncnk+ankvn+andn+un ≤cnk+dn+kvn+un.
(3.3)
That is,
xn+1−xn≤cnk+dn+kvn+un. (3.4)
Next, consider
xn+1−Sxn+1=xn+1−Sn+1x n+1
+Sn+1x
n+1−Sxn+1 ≤dn+1+kxn+1−xn
+xn−Snxn
+Snxn−Snxn+1 ≤dn+1+kdn+k(k+1)xn+1−xn
≤dn+1+kdn+k(k+1)cnk+dn+kvn+un
(3.5)
by (3.4). Taking limsup on both sides in the above inequality, we obtain
lim sup n→∞
xn+1−Sxn+1≤0. (3.6)
That is,
lim
n→∞xn−Sxn=0. (3.7)
Similarly, we can prove that
lim
n→∞xn−T xn=0. (3.8)
Lemma3.2. LetEbe a uniformly convex Banach space andCits nonempty bounded closed convex subset. LetSandT be self-mappings ofCsatisfying
Snx−Sny≤knx−y,
Tnx−Tny≤knx−y, (3.9)
for alln∈N, where{kn} ⊂[1,∞)such thatn∞=1(kn−1) <∞. Let{xn}be as in (1.4) with{an},{bn}in[δ,1−δ]for someδ∈(0,1)and{un},{vn}inEwith∞n=1un<∞,
∞
n=1vn<∞. IfF(S)∩F(T )≠φ, thennlim→∞xn−Sxn =0=lim
n→∞xn−T xn.
Proof. Letp∈F(S)∩F(T ). Then
xn+1−p
=an
Snyn−p
+1−an
xn−p
+un ≤anknyn−p+1−anxn−p+un
=ankn1−bnxn+bnTnxn+vn−p+1−anxn−p+un =anknbn
Tnx n−p
+1−bn
xn−p
+vn+
1−anxn−p+un
≤anbnk2nxn−p+anknvn+an
1−bn
knxn−p+
1−anxn−p+un
=1+anbnk2n+an1−bnkn−anxn−p+anknvn+un.
(3.10)
Since{kn}is a bounded sequence, therefore there existsh >0 such thatkn≤hfor all
n≥1 so that
xn+1−p≤
1+anbnh
kn−1
+an
kn−1xn−p+anhvn+un. (3.11)
Takesn=anbnh(kn−1)+an
kn−1),tn=anhvn + un, andrn= xn−p. As
∞
n=1sn<∞and
∞
n=1tn<∞, so limn→∞xn−pexists byLemma 2.6. Let limn→∞xn− p =c, wherec≥0 is a real number. Assume thatc >0, as the result for the casec=0 is obviously true. NowTnxn−p ≤knx
n−pfor alln∈Ngives lim supn→∞Tnxn−
p ≤c. Also,
yn−p=bn
Tnx
n−p+1−bnxn−p+vn ≤xn−p+
kn−1
bnxn−p+vn
(3.12)
gives
lim sup n→∞
yn−p≤c. (3.13)
Next, consider
Sny
n−p+a−n1un≤knyn−p+a−n1un≤knyn−p+ 1
...
By the above inequality and by virtue ofun →0 andkn→1 asn→ ∞, we get
lim sup n→∞
Sny
n−p+a−n1un≤c. (3.15)
Moreover,c=limn→∞xn+1−pmeans that
lim n→∞an
Sny
n−p+a−n1un+1−anxn−p=c. (3.16)
ApplyingLemma 2.5,
lim n→∞S
ny
n−xn+a−n1un=0. (3.17)
Thus
Sny
n−xn≤Snyn−xn+a−n1un+ 1
δun (3.18)
yields that
lim n→∞S
ny
n−xn=0. (3.19)
Also, then
xn−p≤xn−Sny
n+Snyn−p≤xn−Snyn+knyn−p (3.20)
implies that
c≤lim inf
n→∞ yn−p. (3.21)
By (3.13) and (3.21), we obtain
lim
n→∞yn−p=c. (3.22)
That is,
lim n→∞bn
Tnx
n−p+b−n1vn+1−bnxn−p=c. (3.23)
Again byLemma 2.5, we get
lim n→∞T
nx
n−xn+b−n1vn=0, (3.24)
which finally gives that
lim n→∞T
nx
n−xn=0. (3.25)
Now
Snx
n−xn≤Snxn−Snyn+Snyn−xn
≤knbnTnxn−xn+vn+Snyn−xn
implies, together with (3.19) and (3.25), that
lim n→∞S
nx
n−xn=0=lim n→∞T
nx
n−xn. (3.27)
Lemma 3.1now reveals that
lim
n→∞Sxn−xn=0=nlim→∞T xn−xn, (3.28)
which is as desired.
Theorem3.3. LetEbe a uniformly convex Banach space satisfying Opial’s condition and let C, S, T, and{xn}be as taken in Lemma 3.2. If F(S)∩F(T )≠φ, then {xn} converges weakly to a common fixed point ofS andT.
Proof. Letp∈F(S)∩F(T ). Then, as proved inLemma 3.2, limn→∞xn−pexists. Now we prove that {xn}has a unique weak subsequential limit in F(S)∩F(T ). To prove this, letw1andw2be weak limits of the subsequences{xni}and{xnj}of{xn}, respectively. ByLemma 3.2, limn→∞xn−Sxn =0 andI−Sis demiclosed with respect to zero byLemma 2.7; therefore, we obtainSw1=w1. Similarly,T w1=w1. Again, in the same way, we can prove thatw2∈F(S)∩F(T ). Next, we prove the uniqueness. For this, suppose thatw1≠w2; then by Opial’s condition,
lim
n→∞xn−w1=nlimi→∞xni−w1<nlim i→∞
xn
i−w2 =lim
n→∞xn−w2=nlimj→∞
xn
j−w2
< lim nj→∞
xn
j−w1=nlim→∞xn−w1,
(3.29)
a contradiction. Hence the proof is over.
Remark3.4. If we takeun=vn=0 for alln∈N, the above theorem reduces to [3, Theorem 1] of Khan and Takahashi. Moreover, [6, Theorem 2.1] of Schu becomes a special case of the above theorem whenun=vn=0 as well asT =I, the identity mapping.
Finally, we approximate common fixed points by the following strong convergence theorem.
Theorem3.5. LetE be a uniformly convex Banach space andCits bounded closed convex subset. LetS,T, and{xn}be as taken inLemma 3.2. IfF(S)∩F(T )≠φand eitherSorT is completely continuous, then{xn}converges strongly to a common fixed point of S andT.
Proof. Assume thatT:C→Cis completely continuous. Since{xn}is a bounded sequence andT is completely continuous, therefore{T xn}must have a convergent subsequence{T xni}. Hence by (3.28),{xn}must have a subsequence{xni}such that xni →q (say) inC asni→ ∞. Now continuity ofS and T gives that Sxni →Sq and T xni→T qasni→ ∞. Then, again by (3.28),Sq−q =0= T q−q. This yields that
...
Lemma 3.2, limn→∞xn−pexists for allp∈F(S)∩F(T ); therefore,{xn}must itself converge toq∈F(S)∩F(T ). Hence the proof.
Remark3.6. If we putT=I,vn=0 in the above theorem, then [2, Theorem 1] of Huang is obtained. When we take S =T in the above theorem, then [2, Theorem 2] of Huang follows except whenbn=0. Since a self-mapping with compact domain is completely continuous, therefore [3, Theorem 2] of Khan and Takahashi can also be obtained by puttingun=vn=0. It is also worth mentioning that the results presented in this paper are for two mappings while the results in Huang [2] are for one mapping only. Meanwhile, calculations in this paper are made much simpler as compared to Huang [2].
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Hafiz Fukhar-ud-din: Department of Mathematics, Islamia University, Bahawalpur, Pakistan E-mail address:[email protected]
Safeer Hussain Khan: Ghulam Ishaq Khan Institute of Engineering Sciences and Technology, Topi 23460, Swabi, North-Western Frontier Province (N.W.F.P.), Pakistan