# Reconsiderations on the Equivalence of Convergence between Mann and Ishikawa Iterations for Asymptotically Pseudocontractive Mappings

Authors: Haizhen Sun; Zhiqun Xue
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

Our aim in this paper is to illustrate that the proof of main theorem of Rhoades and Şoltuz (2003) concerning the equivalence between the convergences of Ishikawa and Mann iterations for uniformlyL-Lipschitzian asymptotically pseudocontractive maps is incorrect and to provide its correct version.


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## Body

## 1. Introduction and Preliminary

In 2003, Rhoades and Şoltuz [1] proved the equivalence between convergences of Ishikawa and Mann iterations for an asymptotically pseudocontractive map. This result provided significant improvements of recent some important results. Their result is as follows.Theorem R-S (see [1, Theorem 8]).
LetB be a closed convex subset of an arbitrary Banach space X and (xn)n and (un)n defined by (3) and (4) with (αn)n and (βn)n satisfying (5). Let T be an asymptotically pseudocontractive and Lipschitzian map with L≥1 selfmap of B. Let x* be the fixed point of T. If u0=x0∈B, the following two assertions are equivalent:(i)
Mann type iteration (3) converges tox*∈F(T),(ii)
Ishikawa iteration (4) converges tox*∈F(T).However, after careful reading of the paper of Rhoades and Şoltuz [1], we find that there exists a serious gap in the proof of Theorem 8 of [1], which happens to be main theorem of the paper. Note: in the proof of Theorem 8 of [1] the following mistakes occurred. “Using (6) with x∶=xn+1,y∶=un+1” in line 19 of page 684 cannot obtain(1)∥(1+αn2)(xn+1-un+1)nnn+αn((αnknI-Tn)xn+1-(αnknI-Tn)un+1)(1+αn2)(xn+1-un+1)∥≥(1+αn2)∥xn+1-un+1∥.The reason is that the following conditions are not equivalent:(a1)
T is asymptotically pseudocontractive map,(a2)
∥x-y∥≤∥x-y+r((αnknI-Tn)x-(αnknI-Tn)y)∥, where αn and kn are from (1).The aim of this paper is for us to provide its correct version. For this, we need the following definitions and lemmas.Throughout this paper, suppose thatE is an arbitrary real Banach space and D is a nonempty closed convex subset of E. Let J denote the normalized duality mapping from E to 2E* defined by
(2)J(x)={f∈E*:〈x,f〉=∥x∥2=∥f∥2},∀x∈E,
where E*, 〈·,·〉, and j denote the dual space of E, the generalized duality pairing, and the single-valued normalized duality mapping, respectively.Definition 1 (see [1]).
LetT:D→D be a mapping.
T is called uniformly L-Lipschitz if there is a constant L>0 such that, for all x,y∈D,
(3)∥Tnx-Tny∥≤L∥x-y∥,∀n≥1.
T is called asymptotically nonexpansive with a sequence {kn}⊂[1,+∞) and limn→∞kn=1 if for each x,y∈D such that
(4)∥Tnx-Tny∥≤kn∥x-y∥,∀n≥1.
T is called asymptotically pseudocontractive map with a sequence {kn}⊂[1,+∞) and limn→∞kn=1 if, for each x,y∈D, there exists j(x-y)∈J(x-y) such that
(5)〈Tnx-Tny,j(x-y)〉≤kn∥x-y∥2,∀n≥1.Obviously, an asymptotically nonexpansive mapping is both asymptotically pseudocontractive and uniformlyL-Lipschitz. Conversely, it is not true in general.Definition 2 (see [2]).
For arbitrary givenu1,x1∈D, the sequences {un}n=1∞,{xn}n=1∞⊂D defined by
(6)un+1=(1-an)un+anTnun,n≥1,(7)yn=(1-bn)xn+bnTnxn,n≥1,xn+1=(1-an)xn+anTnyn,n≥1
are called modified Mann and Ishikawa iterations, respectively, where {an}, {bn} are two real sequences of [0,1] and satisfy some conditions.Lemma 3 (see [2]).
LetE be a real Banach space and J:E→2E* be a normalized duality mapping. Then
(8)∥x+y∥2≤∥x∥2+2〈y,j(x+y)〉,
for all x,y∈E and j(x+y)∈J(x+y).Lemma 4 (see [3]).
LetΦ:[0,+∞)→[0,+∞) be a strictly increasing and continuous function with Φ(0)=0, and let {δn}n=0∞, {λn}n=0∞, and {γn}n=0∞ be three nonnegative real sequences satisfying the following inequality:
(9)δn+12≤δn2-λnΦ(δn+1)+γn,n≥0,
where λn∈[0,1] with ∑n=0∞‍λn=∞, γn=o(λn). Then δn→0 as n→∞.


## 2. Main Results

Now we prove the following theorem which is the main result of this paper.Theorem 5.
LetE be a real Banach space, D be a nonempty closed convex subset of E, and T:D→D be a uniformly L-Lipschitz asymptotically pseudocontractive mapping with a sequence {kn}⊂[1,+∞) such that limn→∞kn=1. Let {an},{bn} be two real numbers sequences in [0,1] and satisfy the conditions (i)  an,bn→0 as n→∞; (ii)  ∑n=1∞‍an=∞. For some u1,x1∈D, let {un} and {xn} be modified Mann and Ishikawa iterative sequences defined by (6) and (7), respectively. If F(T)={x∈D:Fx=x}≠∅, q∈F(T), and there exists a strictly increasing continuous function Φ:[0,+∞)→[0,+∞) with Φ(0)=0 such that
(*)〈Tnxn+1-Tnun+1,j(xn+1-un+1)〉≤kn∥xn+1-un+1∥2-Φ(∥xn+1-un+1∥),nnnnnnnnnnnnnnnnnnnnnnnnnnnn∀n≥1,
wherej(xn+1-un+1)∈J(xn+1-un+1), then the following two assertions are equivalent:(1-1)
the modified Mann iteration (6) converges strongly to the fixed point q of  T;(1-2)
the modified Ishikawa iteration (7) converges strongly to the fixed point q of  T.Proof.
We only need to prove (1-1)⇒ (1-2), that is, ∥un-q∥→0 as n→∞⇒∥xn-q∥→0 as n→∞. Without loss of generality, ∥un-q∥≤1. Since T:D→D is a uniformly L-Lipschitz, then ∥Tnx-Tny∥≤L∥x-y∥.
Step1. For any n≥0, {xn} is bounded.
Setun=q, for all n≥1, sup{kn:n≥0}=k, then (*):
(10)〈Tnxn+1-q,j(xn+1-q)〉≤k∥xn+1-q∥2-Φ(∥xn+1-q∥),∀n≥0.
And there existsx1∈D and x1≠Tx1 such that r0=(k+L)·∥x1-q∥2∈R(Φ). Indeed, 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 {ξn}⊂D such that ξn→q as n→∞ with ξn≠q. Hence there exists a natural number n0 such that (k+L)∥ξn-q∥2<r1/2 for n≥n0, and then we redefine x1=ξn0 and (k+L)∥x1-q∥2∈R(Φ).
SetR=Φ-1(r0), and then, from (*), we obtain that
(11)∥x1-q∥≤R.
DenoteB1={x∈D:∥x-q∥≤R}, B2={x∈D:∥x-q∥≤2R}. Next, we want to prove that xn∈B1. If n=1, then x1∈B1. Now assume that it holds for some n; that is, xn∈B1. We prove that xn+1∈B1. Suppose that it is not the case, and then ∥xn+1-q∥>R. Now denote
(12)τ0=min{11+2L,Φ(R)16R2(1+L)(1+2L)}.
Becausean,bn,kn-1→0 as n→∞, without loss of generality, let 0≤an,bn,kn-1≤τ0 for any n≥1. So we have
(13)∥yn-q∥≤(1-bn)∥xn-q∥+bn∥Tnxn-Tnq∥≤(1-bn+bnL)∥xn-q∥≤(1+bn+bnL)R≤2R,(14)∥xn+1-q∥≤(1-an)∥xn-q∥+an∥Tnyn-Tnq∥≤(1-an)∥xn-q∥+anL∥yn-q∥≤(1+an)∥xn-q∥+anL(1+bn+bnL)∥xn-q∥≤(1+an)∥xn-q∥+2anL∥xn-q∥≤2R,(15)∥xn+1-yn∥≤an∥xn-Tnyn∥+bn∥xn-Tnxn∥≤an(∥xn-q∥+∥Tnyn-Tnq∥)+bn(∥xn-q∥+∥Tnxn-Tnq∥)≤an(∥xn-q∥+L∥yn-q∥)+bn(1+L)∥xn-q∥≤an(1+2L)R+bn(1+L)R≤(an+bn)(1+2L)R≤Φ(R)8R(1+L),
so
(16)∥Tnxn+1-Tnyn∥≤LΦ(R)8R(1+L)<Φ(R)8R.
Using Lemma3 and the above formula, we obtain
(17)∥xn+1-q∥2=∥(1-an)(xn-q)+an(Tnyn-q)∥2≤(1-an)2∥xn-q∥2+2an〈Tnyn-q,j(xn+1-q)〉=(1-an)2∥xn-q∥2+2an〈Tnxn+1-q-Tnxn+1+Tnyn,j(xn+1-q)〉≤(1-an)2∥xn-q∥2+2an[kn∥xn+1-q∥2-Φ(∥xn+1-q∥)]+2an∥Tnxn+1-Tnyn∥·∥xn+1-q∥.
Since 2knan→0 as n→∞, without loss of generality, let 1-2knan>0. Then (17) implies that
(18)∥xn+1-q∥2≤(1-an)21-2knan∥xn-q∥2-2an1-2knanΦ(∥xn+1-q∥)+2an1-2knan∥Tnxn+1-Tnyn∥·∥xn+1-q∥≤∥xn-q∥2+2an1-2knan×[[(kn-1)+an2]∥xn-q∥2nnnnnnn-Φ(∥xn+1-q∥)nnnnnnn+∥Tnxn+1-Tnyn∥·∥xn+1-q∥[(kn-1)+an2]]≤∥xn-q∥2+2an1-2knan×[Φ(R)R28R2(1+L)(1+2L)-Φ(R)+Φ(R)2R8R]≤∥xn-q∥2+2an1-2knan×[Φ(R)4-Φ(R)+Φ(R)4]≤R2-an1-2knanΦ(R)≤R2,
and this is a contradiction. Hencexn+1∈B1; that is, {xn} is a bounded sequence.
Step2. We show that ∥xn-q∥→0 as n→∞.
By Step 1, we obtain that{∥xn-un∥} is a bounded sequence, and denote M=supn{∥xn-un∥}. Applying (6), (7), and Lemma 3, we have
(19)∥xn+1-un+1∥2=∥(1-an)(xn-un)+an(Tnyn-Tnun)∥2≤(1-an)2∥xn-un∥2+2an〈Tnyn-Tnun,j(xn+1-un+1)〉≤(1-an)2∥xn-un∥2+2an〈Tnyn-Tnxn+1,j(xn+1-un+1)〉+2an〈Tnxn+1-Tnun+1,j(xn+1-un+1)〉+2an〈Tnun+1-Tnun,j(xn+1-un+1)〉≤(1-an)2∥xn-un∥2+2anL∥yn-xn+1∥·∥xn+1-un+1∥+2an[kn∥xn+1-un+1∥2nnnnnnnnn-Φ(∥xn+1-un+1∥)∥xn+1-un+1∥2]+2anL∥un+1-un∥·∥xn+1-un+1∥.
Observe that
(20)∥yn-xn+1∥=∥an(xn-Tnyn)-bn(xn-Tnxn)∥≤(an+bn+bnL)∥xn-q∥+anL∥yn-q∥≤(an+bn+bnL)∥xn-q∥+anL(1+bnL)∥xn-q∥≤(an+bn+bnL+anL+anbnL2)∥xn-q∥≤hn(∥xn-un∥+∥un-q∥)≤hn(∥xn-un∥+1),(21)∥un+1-un∥=∥an(Tnun-un)∥=an∥Tnun-Tnq+q-un∥≤an(1+L)∥un-q∥≤an(1+L),
wherehn=an+bn+bnL+anL+anbnL2→0 as n→∞.
Substituting (20) and (21) into (19), we obtain
(22)∥xn+1-un+1∥2≤(1-an)2∥xn-un∥2+2Lanhn(∥xn-un∥+1)∥xn+1-un+1∥+2an(kn∥xn+1-un+1∥2-Φ(∥xn+1-un+1∥))+2anLan(1+L)∥xn+1-un+1∥≤(1-an)2∥xn-un∥2+Lanhn(∥xn-un∥2+2∥xn+1-un+1∥2+1)+2an(kn∥xn+1-un+1∥2-Φ(∥xn+1-un+1∥))+L(1+L)an2+L(1+L)an2∥xn+1-un+1∥2.
Since an→0 as n→∞, without loss of generality, we may assume that
(23)12<1-2Lanhn-2ankn-L(1+L)an2<1,
for any n≥1. Then, (22) implies that
(24)∥xn+1-un+1∥2≤(1-an)2+Lanhn1-2Lanhn-2ankn-L(1+L)an2×∥xn-un∥2+Lanhn+L(1+L)an21-2Lanhn-2ankn-L(1+L)an2-2an1-2Lanhn-2ankn-L(1+L)an2Φ×(∥xn+1-un+1∥)=(1+2an(kn-1)+an2+3Lanhn+L(1+L)an21-2Lanhn-2ankn-L(1+L)an2)×∥xn-un∥2+Lanhn+L(1+L)an21-2Lanhn-2ankn-L(1+L)an2-2an1-2Lanhn-2ankn-L(1+L)an2Φ×(∥xn+1-un+1∥)≤(1+4an(kn-1)+2an2+2L(1+L)an2+6Lanhn)×∥xn-un∥2+2L(1+L)an2+2Lanhn-2anΦ(∥xn+1-un+1∥)≤∥xn-un∥2+[4an(kn-1)+2an2+2L(1+L)an2+6Lanhn]M2+2L(1+L)an2+2Lanhn-2anΦ(∥xn+1-un+1∥).
Let δn=∥xn-un∥, λn=2an, γn=[4an(kn-1)+2(1+L+L2)an2+6Lanhn]M2+2L(1+L)an2+2Lanhn=o(λn). Then (24) leads to
(25)δn+12≤δn2-λnΦ(δn+1)+γn.
By Lemma 4, we obtain limn→∞δn=0. That is, ∥xn-un∥→0 as n→∞. From the inequality 0≤∥xn-q∥≤∥xn-un∥+∥un-q∥, we get ∥xn-q∥→0 as n→∞. This completes the proof.Remark 6.
The error in the proof of Theorem 8 of [1] has been pointed out and corrected, but it is not easy what the author really wants to obtain the proof of Theorem 8 in [1] at present.Remark 7.
The proof method of Theorem5 is quite different from that of [1] and others.



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