# The Equivalence of Convergence Results of Modified Mann and Ishikawa Iterations with Errors without Bounded Range Assumption

Authors: Zhiqun Xue; Yaning Wang; Haiyun Zhou
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

LetE be an arbitrary uniformly smooth real Banach space, let D be a nonempty closed convex subset of E, and let T:D→D be a uniformly generalized Lipschitz generalized asymptotically Φ-strongly pseudocontractive mapping with q∈F(T)≠∅. Let {an},{bn},{cn},{dn} be four real sequences in [0,1] and satisfy the conditions: (i) an+cn≤1, bn+dn≤1; (ii) an,bn,dn→0 as n→∞ and cn=o(an); (iii) Σn=0∞an=∞. For some x0,z0∈D, let {un},{vn},{wn} be any bounded sequences in D, and let {xn},{zn} be the modified Ishikawa and Mann iterative sequences with errors, respectively. Then the convergence of {xn} is equivalent to that of {zn}.


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## Body

## 1. Introduction and Preliminary

LetE be a real Banach space and let E* be its dual space. The normalized duality mapping J:E→2E* is defined by
(1.1)J(x)={f∈E*:〈x,f〉=∥x∥2=∥f∥2},∀x∈E,
where 〈·,·〉 denotes the generalized duality pairing. It is well known that(i)
ifE is a smooth Banach space, then the mapping J is single-valued;(ii)
J(αx)=αJ(x) for all x∈E and α∈ℜ;(iii)
ifE is a uniformly smooth Banach space, then the mapping J is uniformly continuous on any bounded subset of E. Throughout this paper, we denote that j is the single-valued normalized duality mapping, D is a nonempty closed convex subset of E, T:D→D is a mapping, and T0 is the unit mapping I.In 1972, Goebel and Kirk [1] introduced the class of asymptotically nonexpansive mappings as follows.Definition 1.1.
A mappingT is said to be asymptotically nonexpansive if for each x,y∈D(1.2)∥Tnx-Tny∥≤kn∥x-y∥,∀n≥0,
where {kn}⊂[1,+∞) with limn→∞kn=1.
Schu [2], in 1991, gave the definition of asymptotically pseudocontractive mappings and proved the correlation results.Definition 1.2.
The mappingT is called asymptotically pseudocontractive with the sequence {kn}⊂[1,+∞) if and only if limn→∞kn=1, and for all n∈N and all x,y∈D, there exists j(x-y)∈J(x-y) such that
(1.3)〈Tnx-Tny,j(x-y)〉≤kn∥x-y∥2.
It is easy to find that every asymptotically nonexpansive mapping is asymptotically pseudocontractive. However, the converse is not true in general. See example of [3].
Recently, Colao [4] combined the proof ideas of the papers of Chang [5] and C. E. Chidume and C. O. Chidume [6] and then showed the equivalent theorem results of the convergence between Mann and Ishikawa iterations with errors for generalized strongly asymptotically ϕ-pseudocontractive mapping with bounded range. In fact, he proved the following theorem.Theorem 1.3.
LetX be a uniformly smooth Banach space, and let T:X→X be generalized strongly asymptotically ϕ-pseudocontractive mapping with fixed point x* and bounded range. Let {xn} and {zn} be the sequences defined by (1.4) and (1.5), respectively,
(1.4)yn=(1-βn-δn)xn+βnTnxn+δnvn,n≥0,xn+1=(1-αn-γn)xn+αnTnyn+γnun,n≥0,(1.5)zn+1=(1-αn-γn)zn+αnTnzn+γnwn,n≥0,
where {αn},{γn},{βn},{δn}⊂[0,1] satisfy(H1)
limn→∞αn=limn→∞βn=limn→∞δn=0 and γn=o(αn),(H2)
Σn=1∞αn=∞,and the sequences {un},{vn},{wn} are bounded in X, then for any initial point z0,x0∈X, the following two assertions are equivalent.(1)
The modified Ishikawa iteration sequence with errors (1.4) converges to x*;(2)
The modified Mann iteration sequence with errors (1.5) converges to x*.The aim of this paper is to prove the equivalence of convergent results of above Ishikawa and Mann iterations with errors for generalized asymptoticallyΦ-strongly pseudocontractive mappings without bounded range assumptions in uniformly smooth real Banach spaces.  For this, we need the following concepts and lemmas.Definition 1.4 (see [4]).
The mappingT is called generalized asymptotically Φ-strongly pseudocontractive if
(1.6)〈Tnx-Tny,j(x-y)〉≤kn∥x-y∥2-Φ(∥x-y∥),n≥0,
where j(x-y)∈J(x-y), {kn}⊂[1,+∞) is converging to one and Φ:[0,+∞)→[0,+∞) is strictly increasing continuous function with Φ(0)=0.Definition 1.5 (see [4]).
For arbitrary givenx0∈D, modified Ishikawa iterative process with errors {xn}n=0∞ defined by
(1.7)yn=(1-bn-dn)xn+bnTnxn+dnwn,n≥0,xn+1=(1-an-cn)xn+anTnyn+cnvn,n≥0,
where {vn},{wn} are any bounded sequences in D; {an},{bn},{cn},{dn} are four real sequences in [0,1] and satisfy an+cn≤1,bn+dn≤1, for all n≥0. If bn=dn=0, we define modified Mann iterative process with errors {zn} by
(1.8)zn+1=(1-an-cn)zn+anTnzn+cnun,n≥0,
where {un} is any bounded sequence in D.Lemma 1.6 (see [7]).
LetE be a uniformly smooth real Banach space and let J:E→2E* be a normalized duality mapping. Then
(1.9)∥x+y∥2≤∥x∥2+2〈y,J(x+y)〉,
for all x,y∈E.Lemma 1.7 (see [8]).
Let{ρn}n=0∞ be a nonnegative sequence which satisfies the following inequality:
(1.10)ρn+1≤(1-λn)ρn+σn,n≥0,
where λn∈[0,1] with ∑n=0∞‍λn=∞, σn=o(λn). Then ρn→0 as n→∞.


## 2. Main Results

First of all, we give a new concept.Definition 2.1.
A mappingT:D→D is called uniformly generalized Lipschitz if there exists a constant L>0 such that
(2.1)∥Tnx-Tny∥≤L(1+∥x-y∥),∀x,y∈D,∀n≥0.
It is mentioned to notice that if T has bounded range, then it is uniformly generalized Lipschitz. In fact, since R(Tn)⊆R(T), then supx∈D{∥Tnx∥}≤supx∈D{∥Tx∥}=M1, thus ∥Tnx-Tny∥≤2M1≤L(1+∥x-y∥), where L=2M1. On the contrary, it is not true in general (See [6]).In the following, we prove the main theorems of this paper.Theorem 2.2.
LetE be an arbitrary uniformly smooth real Banach space, let D be a nonempty closed convex subset of E, and let T:D→D be a uniformly generalized Lipschitz generalized asymptotically Φ-strongly pseudocontractive mapping with q∈F(T)≠∅. Let {an},{bn},{cn},{dn} be four real sequences in [0,1] and satisfy the following conditions:(i)
an+cn≤1, bn+dn≤1;(ii)
an,bn,dn→0 as n→∞ and cn=o(an);(iii)
Σn=0∞an=∞.For some x0,z0∈D, let {un},{vn},{wn} be any bounded sequences in D, and let {xn} and {zn} be Ishikawa and Mann iterative sequences with errors defined by (1.7) and (1.8), respectively. Then the following conclusions are equivalent:(1)
{xn} converges strongly to the unique fixed point q of T;(2)
{zn} converges strongly to the unique fixed point q of T.Proof.
(1)⇒(2) is obvious, that is, let bn=dn=0, (1.7) turns into (1.8). We only need to show that (2)⇒(1). Since T:D→D is a uniformly generalized Lipschitz generalized asymptotically Φ-strongly pseudocontractive mapping, then there exists a strictly increasing continuous function Φ:[0,+∞)→[0,+∞) with Φ(0)=0 such that
(2.2)〈Tnx-Tny,J(x-y)〉≤kn∥x-y∥2-Φ(∥x-y∥),
that is,
(2.3)〈(knI-Tn)x-(knI-Tn)y,J(x-y)〉≥Φ(∥x-y∥),(2.4)∥Tnx-Tny∥≤L(1+∥x-y∥),
for any x,y∈D. For convenience, denote k=supn{kn}.
Step1. There exists x0∈D and x0≠Tx0 such that r0=(k+L)∥x0-q∥2+L∥x0-q∥∈R(Φ)(range of Φ).
Indeed, ifΦ(r)→+∞ as r→+∞, then r0∈R(Φ); if sup{Φ(r):r∈[0,+∞)}=r1<+∞ with r1<r0, then, for q∈D, there exists a sequence {νn} in D such that νn→q as n→∞ with νn≠q. Furthermore, there exists a natural number n0 such that (k+L)∥νn-q∥2+L∥νn-q∥<r1/2 for n≥n0, then we redefine x0,r0 such that x0=νn0,r0=(k+L)∥x0-q∥2+L∥x0-q∥∈R(Φ). Hence, it is to ensure that Φ-1(r0) is well defined.
Step2. For any n≥0, {xn} is a bounded sequence.
SetR=Φ-1(r0). From (2.3), we have
(2.5)〈kn(x0-q)-(Tnx0-q),J(x0-q)〉≥Φ(∥x0-q∥),
that is, (k+L)∥x0-q∥2+L∥x0-q∥≥Φ(∥x0-q∥). Thus, we obtain that ∥x0-q∥≤R. Denote
(2.6)B1={x∈D:∥x-q∥≤R},B2={x∈D:∥x-q∥≤2R},M=supn{∥vn-q∥}+supn{∥wn-q∥}.
Next, we want to prove that xn∈B1 for any n≥0 by induction. If n=0, then x0∈B1. Now we assume that it holds for some n, that is, xn∈B1. We prove that xn+1∈B1. Suppose that it is not the case, then ∥xn+1-q∥>R. Since J is uniformly continuous on bounded subset of E, then, for ϵ0=Φ(R/4)/24L(1+2R), there exists δ>0 such that ∥Jx-Jy∥<ϵ0 when ∥x-y∥<δ,forallx,y∈B2. Now denote
(2.7)τ0=min{R2[L(1+2R)+2R+M],R4[L(1+R)+2R+M],δ2[L(1+2R)+2R+M],Φ(R/4)24R2,Φ(R/4)24L(1+2R),Φ(R/4)48MR}.
Since an,bn,cn,dn→0 as n→∞, and cn=o(an), without loss of generality, we assume that 0≤an,bn,cn,dn≤τ0,cn<anτ0 for any n≥0. Then we obtain the following estimates:
(2.8)∥Tnxn-q∥≤L(1+∥xn-q∥)≤L(1+R),∥yn-q∥≤(1-bn-dn)∥xn-q∥+bn∥Tnxn-q∥+dn∥wn-q∥≤R+bnL(1+∥xn-q∥)+dnM≤R+bnL(1+R)+dnM≤R+τ0[L(1+R)+M]≤2R,∥Tnyn-q∥≤L(1+∥yn-q∥)≤L(1+2R),∥xn-Tnxn∥≤∥xn-q∥+∥Tnxn-q∥≤L+(1+L)∥xn-q∥≤L+(1+L)R,∥(xn-q)-(yn-q)∥≤bn∥xn-Tnxn∥+dn[∥wn-q∥+∥xn-q∥]≤bn[L+(1+L)R]+dn(M+R)≤τ0[L(1+R)+2R+M]≤τ0[L(1+2R)+2R+M]≤δ2<δ,∥xn-q∥≥∥xn+1-q∥-an∥Tnyn-xn∥-cn∥vn-xn∥≥∥xn+1-q∥-an[∥Tnyn-q∥+∥xn-q∥]-cn[∥xn-q∥+∥vn-q∥]>R-an[L(1+2R)+R]-cn(R+M)≥R-τ0[L(1+2R)+M+2R]≥R-R2=R2,∥yn-q∥≥∥xn-q∥-bn∥Tnxn-xn∥-dn∥xn-wn∥≥∥xn-q∥-bn[L+(1+L)R]-dn[∥xn-q∥+∥wn-q∥]≥∥xn-q∥-bn[L+(1+L)R]-dn(R+M)≥∥xn-q∥-τ0[L(1+R)+2R+M]>R2-R4=R4,∥xn+1-q∥≤(1-an-cn)∥xn-q∥+an∥Tnyn-q∥+cn∥vn-q∥≤R+τ0[L(1+2R)+M]≤2R,∥(xn+1-q)-(xn-q)∥≤an∥Tnyn-xn∥+cn∥un-xn∥≤an[∥Tnyn-q∥+∥xn-q∥]+cn[∥vn-q∥+∥xn-q∥]≤an[L(1+2R)+R]+cn(M+R)≤τ0[L(1+2R)+2R+M]≤δ2<δ.
Hence, ∥J(xn-q)-J(yn-q)∥<ϵ0; ∥J(xn+1-q)-J(xn-q)∥<ϵ0.
Using Lemma1.6 and formulas above, we obtain
(2.9)∥xn+1-q∥2≤(1-an-cn)2∥xn-q∥2+2an〈Tnyn-q,J(xn+1-q)〉+2cn〈un-q,J(xn+1-q)〉≤(1-an)2∥xn-q∥2+2an〈Tnyn-q,J(xn+1-q)-J(xn-q)〉+2an〈Tnyn-q,J(xn-q)-J(yn-q)〉+2an〈Tnyn-q,J(yn-q)〉+2cn〈un-q,J(xn+1-q)〉≤(1-an)2∥xn-q∥2+2an∥Tnyn-q∥·∥J(xn+1-q)-J(xn-q)∥+2an∥Tnyn-q∥·∥J(xn-q)-J(yn-q)∥+2an[∥yn-q∥2-Φ(∥yn-q∥)]+2cn∥un-q∥·∥xn+1-q∥≤(1-an)2R2+4anL(1+2R)ϵ0+2an[∥yn-q∥2-Φ(∥yn-q∥)]+4cnMR,(2.10)∥yn-q∥2≤(1-bn-dn)2∥xn-q∥2+2bn〈Tnxn-q,J(yn-q)〉+2dn〈wn-q,J(yn-q)〉≤∥xn-q∥2+2bn〈Tnxn-q,J(yn-q)-J(xn-q)〉+2bn〈Tnxn-q,J(xn-q)〉+2dn∥wn-q∥·∥yn-q∥≤∥xn-q∥2+2bn∥Tnx-q∥·∥J(yn-q)-J(xn-q)∥+2bn[∥xn-q∥2-Φ(∥xn-q∥)]+2dn∥wn-q∥·∥yn-q∥≤R2+2bnL(1+R)ϵ0+2bnR2+4dnMR.
Substitute (2.10) into (2.9)
(2.11)∥xn+1-q∥2≤(1-an)2R2+4anL(1+2R)ϵ0+2an[R2+2bnL(1+R)ϵ0+2bnR2+4dnMR]-2anΦ(∥yn-q∥)+4cnMR≤R2+an2R2+4anL(1+2R)ϵ0+2an[2bnL(1+R)ϵ0+2bnR2+4dnMR]-2anΦ(R4)+4cnMR=R2+2an[an2R2+2L(1+2R)ϵ0+2bnL(1+R)ϵ0+2bnR2+4dnMR+2cnMRan]-2anΦ(R4)≤R2+2an[Φ(R/4)2-Φ(R4)]≤R2-Φ(R4)an≤R2,
this is a contradiction. Thus xn+1∈B1, that is, {xn} is a bounded sequence. So {yn},{Tnyn},{Tnxn} are all bounded sequences. Since ∥zn-q∥→0 as n→∞, without loss of generality, we let ∥zn-q∥≤1. Therefore, ∥xn-zn∥ is also bounded.
Step3. We want to prove ∥xn-zn∥→0 as n→∞.
SetM0=max{supn∥Tnyn-Tnzn∥,supn∥vn-un∥,supn∥xn-zn∥,supn∥Tnxn-xn∥,supn∥wn-xn∥,supn∥yn-zn∥,supn∥vn-xn∥}.
Again using Lemma1.6, we have
(2.12)∥xn+1-zn+1∥2≤(1-an-cn)2∥xn-zn∥2+2an〈Tnyn-Tnzn,J(xn+1-zn+1)〉+2cn〈vn-un,J(xn+1-zn+1)〉≤(1-an)2∥xn-zn∥2+2an〈Tnyn-Tnzn,J(xn+1-zn+1)-J(xn-zn)〉+2an〈Tnyn-Tnzn,J(xn-zn)-J(yn-zn)〉+2an〈Tnyn-Tnzn,J(yn-zn)〉+2cn∥vn-un∥·∥xn+1-zn+1∥≤(1-an)2∥xn-zn∥2+2anM0An+2anM0Bn+2an[∥yn-zn∥2-Φ(∥yn-zn∥)]+2cnM02,(2.13)∥yn-zn∥2≤∥xn-zn∥2+2bn〈Tnxn-xn,J(yn-zn)〉+2dn〈wn-xn,J(yn-zn)〉≤∥xn-zn∥2+2bnM02+2dnM02,
where An=∥J(xn+1-zn+1)-J(xn-zn)∥, Bn=∥J(xn-zn)-J(yn-zn)∥, and An,Bn→0 as n→∞.
Taking place (2.13) into (2.12), we have
(2.14)∥xn+1-zn+1∥2≤(1-an)2∥xn-zn∥2+2anM0An+2anM0Bn+2an[∥xn-zn∥2+2bnM02+2dnM02-Φ(∥yn-zn∥)]+2cnM02≤∥xn-zn∥2+an2M02+2anM0An+2anM0Bn+4anbnM02+4andnM02-2anΦ(∥yn-zn∥)+2cnM02=∥xn-zn∥2+2an[Cn-2anΦ(∥yn-zn∥)],
where Cn=anM02/2+M0An+M0Bn+2bnM02+2dnM02+cnM02/an→0 as n→∞.
Setinfn≥0Φ(∥yn-zn∥)/(1+∥xn+1-zn+1∥2)=λ, then λ=0. If it is not the case, we assume that λ>0. Let 0<γ<min{1,λ}, then Φ(∥yn-zn∥)/(1+∥xn+1-zn+1∥2)≥γ, that is, Φ(∥yn-zn∥)≥γ+γ∥xn+1-zn+1∥2≥γ∥xn+1-zn+1∥2. Thus, from (2.14) that
(2.15)∥xn+1-zn+1∥2≤∥xn-zn∥2+2an(Cn-γ∥xn+1-zn+1∥2),
which implies that
(2.16)∥xn+1-zn+1∥2≤11+2anγ∥xn-zn∥2+2anCn1+2anγ=(1-2anγ1+2anγ)∥xn-zn∥2+2anCn1+2anγ.
Let ρn=∥xn-zn∥2,λn=2anγ/(1+2anγ),σn=2anCn/(1+2anγ). Then we get that
(2.17)ρn+1≤(1-λn)ρn+σn.
Applying Lemma 1.7, we get that ρn→0 as n→∞. This is a contradiction and so λ=0. Therefore, there exists an infinite subsequence such that Φ(∥yni-zni∥)/(1+∥xni+1-zni+1∥2)→0 as i→∞. Since 0≤Φ(∥yni-zni∥)/(1+M02)≤Φ(∥yni-zni∥)/(1+∥xni+1-zni+1∥2), then Φ(∥yni-zni∥)→0 as i→∞. In view of the strictly increasing and continuity of Φ, we have ∥yni-zni∥→0 as i→∞. From (1.7), we have
(2.18)∥xni-zni∥≤∥yni-zni∥+bni∥xni-Txni∥+cni∥xni-wni∥→0,
as i→∞. Next we want to prove ∥xn-zn∥→0 as n→∞. Let forallε∈(0,1), there exists ni0 such that ∥xni-zni∥<ϵ,an,ani<min{ϵ/4L(1+M0),ϵ/8M0}, cn,cni<ϵ/16M0, bn,dn,bni,dni<ϵ/8M0,Cn,Cni<Φ(ϵ/4)/2, for any ni,n≥ni0. First, we want to prove ∥xni+1-zni+1∥<ϵ. Suppose it is not this case, then ∥xni+1-zni+1∥≥ϵ. Using (1.7), we may get the following estimates:
(2.19)∥xni-zni∥≥∥xni+1-zni+1∥-ani∥Tnyni-Tnzni∥-ani∥xni-zni∥-cni∥vni-uni∥-cni∥xni-zni∥≥ϵ-aniL(1+M0)-(ani+2cni)M0>ϵ2,(2.20)∥yni-zni∥≥∥xni-zni∥-bni∥Tnxni-xni∥-dni∥vni-xni∥≥ϵ2-(bni+dni)M0>ϵ4.
Since Φ is strictly increasing, then (2.20) leads to Φ(∥yni-zni∥)≥Φ(ϵ/4). From (2.14), we have
(2.21)∥xni+1-zni+1∥2≤∥xni-zni∥2+2ani[Cni-Φ(∥yni-zni∥)]<ϵ2+2ani[12Φ(ϵ4)-Φ(ϵ4)]≤ϵ2-Φ(ϵ4)ani≤ϵ2,
is a contradiction. Hence, ∥xni+1-zni+1∥<ϵ. Suppose that ∥xni+m-zni+m∥<ϵ holds. Repeating the above course, we can easily prove that ∥xni+m+1-zni+m+1∥<ϵ holds. Therefore, for any m and ni≥n0, we obtain that ∥xni+m-zni+m∥<ϵ, which means ∥xn-zn∥→0 as n→∞. This completes the proof.In order to make the existence of Theorem2.2 more meaningful, we give the following theorem.Theorem 2.3.
LetE be an arbitrary uniformly smooth real Banach space, let D be a nonempty closed convex subset of E, and let T:D→D be a uniformly generalized Lipschitz generalized asymptotically Φ-strongly pseudocontractive mapping with q∈F(T)≠∅. Let {an},{cn} be two real sequences in [0,1] and satisfy the conditions (i) an+cn≤1; (ii) an→0 as n→∞ and cn=o(an); (iii) Σn=0∞an=∞. For some z0∈D, let {un} be any bounded sequence in D and let {zn} be modified Mann iterative sequence with errors defined by (1.8). Then {zn} converges strongly to the unique fixed point q of T.Proof.
SinceT:D→D is a uniformly generalized Lipschitz generalized asymptotically Φ-strongly pseudocontractive mapping, then there exists a strictly increasing continuous function Φ:[0,+∞)→[0,+∞) with Φ(0)=0 such that
(2.22)〈(knI-Tn)x-(knI-Tn)y,J(x-y)〉≥Φ(∥x-y∥),(2.23)∥Tnx-Tny∥≤L(1+∥x-y∥),
for any x,y∈D.
Step1. There exists z0∈D and z0≠Tz0 such that r0=(k+L)∥z0-q∥2+L∥z0-q∥∈R(Φ), where k=supn{kn}. In fact, if Φ(r)→+∞ as r→+∞, then r0∈R(Φ); if sup{Φ(r):r∈[0,+∞)}=r1<+∞ with r1<r0, then, for q∈D, there exists a sequence {νn} in D such that νn→q as n→∞ with νn≠q. Furthermore, there exists a natural number n0 such that (k+L)∥νn-q∥2+L∥νn-q∥<(r1/2) for n≥n0, then we redefine z0,r0 such that z0=νn0,r0=(k+L)∥z0-q∥2+L∥z0-q∥∈R(Φ).
Step2. For any n≥0, {zn} is bounded.
Setr=Φ-1(r0), we have ∥x0-q∥≤R. Let B1′={z∈D:∥z-q∥≤r},B2′={z∈D:∥z-q∥≤2r},M'=supn{∥un-q∥}. Next, we prove that zn∈B1′ for any n≥0 by induction. First z0∈B1′ is obvious. Suppose that zn∈B1′ holds. We prove that zn+1∈B1′. If it is not the case, then ∥zn+1-q∥>r. By uniformly continuity of J on bounded subset, we choose ϵ0=Φ(r/2)/16L(1+2r), there exists δ>0 such that ∥Jx-Jy∥<ϵ0 when ∥x-y∥<δ,for allx,y∈B2′. Now denote
(2.24)τ0=min{r2[L(1+r)+2r+M′],δ2[L(1+r)+2r+M′],Φ(r/2)8r2,Φ(r/2)24L(1+2r),Φ(r/2)16M′r}.
Since an,cn,kn-1→0 as n→∞, and cn=o(an), without loss of generality, let 0≤an,cn,kn-1≤τ0,cn<anτ0 for any n≥0. Then we have the following estimates from (1.8):
(2.25)∥zn-Tnzn∥≤∥zn-q∥+∥Tnzn-q∥≤r+L(1+r),∥zn-q∥≥∥zn+1-q∥-an∥Tnzn-zn∥-cn∥un-zn∥>r-an[r+L(1+r)]-cn(r+M′)≥r-τ0[L(1+r)+2r+M′]≥r2,∥zn+1-q∥≤(1-an-cn)∥zn-q∥+an∥Tnzn-q∥+cn∥un-q∥≤r+τ0[L(1+r)+M′]≤2r,∥(zn+1-q)-(zn-q)∥≤an∥Tnzn-zn∥+cn∥un-zn∥≤an[r+L(1+r)]+cn(r+M′)≤τ0[L(1+r)+2r+M′]≤δ2<δ.
Therefore, ∥J(zn+1-q)-J(zn-q)∥<ϵ0.
Using Lemma1.6 and formulas above, we obtain
(2.26)∥zn+1-q∥2≤(1-an)2∥zn-q∥2+2an〈Tnzn-q,J(zn+1-q)-J(zn-q)〉+2an〈Tnzn-q,J(zn-q)〉+2cn〈un-q,J(zn+1-q)〉≤(1-an)2∥zn-q∥2+2an∥Tnzn-q∥·∥J(zn+1-q)-J(zn-q)∥+2an[kn∥zn-q∥2-Φ(∥zn-q∥)]+2cn∥un-q∥·∥zn+1-q∥≤(1-an)2r2+4anL(1+2r)ϵ0+2an[kn∥zn-q∥2-Φ(∥zn-q∥)]+4cnM'r≤(1-an)2r2+4anL(1+2r)ϵ0+2an[knr2-Φ(r2)]+4cnM'r=r2+2an[an2r2+2L(1+2r)ϵ0+(kn-1)r2+2cnM′ran]-2anΦ(r2)≤r2+2an[Φ(r/2)2-Φ(r2)]≤r2-anΦ(r2)≤r2,
this is a contradiction. Thus zn+1∈B1′, that is, {zn} is a bounded sequence, so {Tnzn} is also bounded. Denote M0=supn{∥zn-q∥}+supn{∥Tnzn-q∥}+supn{∥un-q∥}.
Step3. We prove ∥zn-q∥→0 as n→∞.
Again using Lemma1.6, we have
(2.27)∥zn+1-q∥2≤(1-an-cn)2∥zn-q∥2+2an〈Tnzn-q,J(zn+1-q)〉+2cn〈un-q,J(zn+1-q)〉≤(1-an)2∥zn-q∥2+2an〈Tnzn-q,J(zn+1-q)-J(zn-q)〉+2an〈Tnzn-q,J(zn-q)〉+2cn∥un-q∥·∥zn+1-q∥≤(1-an)2∥zn-q∥2+2anM0Dn+2an[kn∥zn-q∥2-Φ(∥zn-q∥)]+2cnM02≤∥zn-q∥2+2an[(kn-1)M02+anM022+M0Dn+cnM02an-Φ(∥zn-q∥)]≤∥zn-q∥2+2an[En-Φ(∥zn-q∥)],
where
(2.28)Dn=∥J(zn+1-q)-J(zn-q)∥,En=(kn-1)M02+anM022+M0Dn+cnM02an,
and Dn,En→0 as n→∞.
Setinfn≥0Φ(∥zn-q∥)/(1+∥zn+1-q∥2)=λ, then λ=0. If it is not the case, we assume that λ>0. Let 0<γ<min{1,λ}, then Φ(∥zn-q∥)/(1+∥zn+1-q∥2)≥γ, that is, Φ(∥zn-q∥)≥γ+γ∥zn+1-q∥2≥γ∥zn+1-q∥2. Thus, from (2.14) that
(2.29)∥zn+1-q∥2≤∥zn-q∥2+2an(En-γ∥zn+1-q∥2),
which implies that
(2.30)∥zn+1-q∥2≤11+2anγ∥zn-q∥2+2anEn1+2anγ=(1-2anγ1+2anγ)∥zn-q∥2+2anEn1+2anγ.
Let ρn=∥zn-q∥2,λn=2anγ/(1+2anγ),σn=2anEn/(1+2anγ). Then we get that
(2.31)ρn+1≤(1-λn)ρn+σn.
Applying Lemma 1.7, we get that ρn→0 as n→∞. This is a contradiction and so λ=0. Therefore, there exists an infinite subsequence such that Φ(∥zni-q∥)/(1+∥zni+1-q∥2)→0 as i→∞. Since 0≤Φ(∥zni-q∥)/(1+M02)≤Φ(∥zni-q∥)/(1+∥zni+1-q∥2), then Φ(∥zni-q∥)→0 as i→∞. In view of the strictly increasing and continuity of Φ, we have ∥zni-q∥→0 as i→∞. Let ε∈(0,1)  be any given, there exists ni0 such that ∥zni-q∥<ϵ,ani,an<min{ϵ/4L(1+M0),ϵ/8M0}, cni,cn<ϵ/16M0, Eni,En<Φ(ϵ/2)/2, for any ni,n≥ni0. First, we want to prove ∥zni+1-q∥<ϵ. Suppose it is not this case, then ∥zni+1-q∥≥ϵ. Using (1.8), we may get the following estimates:
(2.32)∥zni-q∥≥∥zni+1-q∥-ani∥Tnzni-q∥-ani∥zni-q∥-cni∥uni-q∥≥ϵ-aniL(1+M0)-(ani+2cni)M0>ϵ2.
Since Φ is strictly increasing, then (2.32) leads to Φ(∥zni-q∥)≥Φ(ϵ/2). From (2.27), we have
(2.33)∥zni+1-q∥2≤∥zni-q∥2+2ani[Eni-Φ(∥zni-q∥)]<ϵ2+2ani[12Φ(ϵ2)-Φ(ϵ2)]≤ϵ2-Φ(ϵ2)ani≤ϵ2,
is a contradiction. Hence, ∥zni+1-q∥<ϵ. Suppose that ∥zni+m-q∥<ϵ holds. Repeating the above course, we can easily prove that ∥zni+m+1-q∥<ϵ holds. Therefore, for any m and ni≥n0, we obtain that ∥zni+m-q∥<ϵ, which means ∥zn-q∥→0 as n→∞. This completes the proof.Theorem 2.4.
LetE be an arbitrary uniformly smooth real Banach space, let D be a nonempty closed convex subset of E, and let T:D→D be a uniformly generalized Lipschitz generalized asymptotically Φ-strongly pseudocontractive mapping with q∈F(T)≠∅. Let {an},{bn},{cn},{dn} be four real sequences in [0,1] and satisfy the conditions (i) an+cn≤1, bn+dn≤1; (ii) an,bn,dn→0 as n→∞ and cn=o(an); (iii) Σn=0∞an=∞. For some x0∈D, let {vn},{wn} be two arbitrary bounded sequences in D, and let {xn} be Ishikawa iterative sequence with errors defined by (1.7). Then (1.7) converges strongly to the unique fixed point q of T.Proof.
By Theorems2.3 and 2.2, we obtain directly the result of Theorem 2.4.Remark 2.5.
Our Theorem2.2 extends and improves Theorem 3.1 of [4] from the bounded range of T to uniformly generalized Lipschitz mapping, and the proof course of Theorem 2.2 is quite different from that of [4].



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*Source: 909187-2012-09-25.xml*