JAM Journal of Applied Mathematics 1687-0042 1110-757X Hindawi Publishing Corporation 274931 10.1155/2013/274931 274931 Research Article Reconsiderations on the Equivalence of Convergence between Mann and Ishikawa Iterations for Asymptotically Pseudocontractive Mappings Sun Haizhen Xue Zhiqun Sahu D. R. Department of Mathematics and Physics Shijiazhuang Tiedao University Shijiazhuang 050043 China stdu.edu.cn 2013 25 6 2013 2013 28 12 2012 11 06 2013 2013 Copyright © 2013 Haizhen Sun and Zhiqun Xue. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 uniformly L-Lipschitzian asymptotically pseudocontractive maps is incorrect and to provide its correct version.

1. Introduction and Preliminary

In 2003, Rhoades and Şoltuz  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 [<xref ref-type="bibr" rid="B1">1</xref>, Theorem 8]).

Let B 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 L1 selfmap of B. Let x* be the fixed point of T. If u0=x0B, the following two assertions are equivalent:

Mann type iteration (3) converges to x*F(T),

Ishikawa iteration (4) converges to x*F(T).

However, after careful reading of the paper of Rhoades and Şoltuz , we find that there exists a serious gap in the proof of Theorem 8 of , which happens to be main theorem of the paper. Note: in the proof of Theorem 8 of  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:

T is asymptotically pseudocontractive map,

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 that E 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)={fE*:x,f=  x2=  f2},xE, 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 [<xref ref-type="bibr" rid="B1">1</xref>]).

Let T:DD be a mapping.

T is called uniformly L-Lipschitz if there is a constant L>0 such that, for all x,yD, (3)Tnx-TnyLx-y,n1.

T is called asymptotically nonexpansive with a sequence {kn}[1,+) and limnkn=1 if for each x,yD such that (4)Tnx-Tnyknx-y,n1.

T is called asymptotically pseudocontractive map with a sequence {kn}[1,+) and limnkn=1 if, for each x,yD, there exists j(x-y)J(x-y) such that (5)Tnx-Tny,j(x-y)knx-y2,n1.

Obviously, an asymptotically nonexpansive mapping is both asymptotically pseudocontractive and uniformly L-Lipschitz. Conversely, it is not true in general.

Definition 2 (see [<xref ref-type="bibr" rid="B2">2</xref>]).

For arbitrary given u1,x1D, the sequences {un}n=1,{xn}n=1D defined by (6)un+1=(1-an)un+anTnun,n1,(7)yn=(1-bn)xn+bnTnxn,n1,xn+1=(1-an)xn+anTnyn,n1 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 [<xref ref-type="bibr" rid="B2">2</xref>]).

Let E be a real Banach space and J:E2E* be a normalized duality mapping. Then (8)x+y2x2+2y,j(x+y), for all x,yE and j(x+y)J(x+y).

Lemma 4 (see [<xref ref-type="bibr" rid="B4">3</xref>]).

Let Φ:[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+12δn2-λnΦ(δn+1)+γn,n0, where λn[0,1] with n=0λn=, γn=o(λn). Then δn0 as n.

2. Main Results

Now we prove the following theorem which is the main result of this paper.

Theorem 5.

Let E be a real Banach space, D be a nonempty closed convex subset of E, and T:DD be a uniformly L-Lipschitz asymptotically pseudocontractive mapping with a sequence {kn}[1,+) such that limnkn=1. Let {an},{bn} be two real numbers sequences in [0,1] and satisfy the conditions (i)  an,bn0 as n; (ii)  n=1an=. For some u1,x1D, let {un} and {xn} be modified Mann and Ishikawa iterative sequences defined by (6) and (7), respectively. If F(T)={xD:Fx=x}, qF(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)knxn+1-un+12-Φ(xn+1-un+1),nnnnnnnnnnnnnnnnnnnnnnnnnnnnn1,

where j(xn+1-un+1)J(xn+1-un+1), then the following two assertions are equivalent:

the modified Mann iteration (6) converges strongly to the fixed point q of  T;

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-q0 as nxn-q0 as n. Without loss of generality, un-q1. Since T:DD is a uniformly L-Lipschitz, then Tnx-TnyLx-y.

Step 1. For any n0, {xn} is bounded.

Set un=q, for all n1, sup{kn:n0}=k, then (*): (10)Tnxn+1-q,j(xn+1-q)kxn+1-q2-Φ(xn+1-q),n0.

And there exists x1D and x1Tx1 such that r0=(k+L)·x1-q2R(Φ). Indeed, if Φ(r)+ as r+, then, r0R(Φ); if sup{Φ(r):r[0,+)}=r1<+ with r1<r0, then, for qD, there exists a sequence {ξn}D such that ξnq as n with ξnq. Hence there exists a natural number n0 such that (k+L)ξn-q2<r1/2 for nn0, and then we redefine x1=ξn0 and (k+L)x1-q2R(Φ).

Set R=Φ-1(r0), and then, from (*), we obtain that (11)x1-qR.

Denote B1={xD:x-qR}, B2={xD:x-q2R}. Next, we want to prove that xnB1. If n=1, then x1B1. Now assume that it holds for some n; that is, xnB1. We prove that xn+1B1. 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)}.

Because an,bn,kn-10 as n, without loss of generality, let 0an,bn,kn-1τ0 for any n1. So we have (13)yn-q(1-bn)xn-q+bnTnxn-Tnq(1-bn+bnL)xn-q(1+bn+bnL)R2R,(14)xn+1-q(1-an)xn-q+anTnyn-Tnq(1-an)xn-q+anLyn-q(1+an)xn-q+anL(1+bn+bnL)xn-q(1+an)xn-q+2anLxn-q2R,(15)xn+1-ynanxn-Tnyn+bnxn-Tnxnan(xn-q+Tnyn-Tnq)+bn(xn-q+Tnxn-Tnq)an(xn-q+Lyn-q)+bn(1+L)xn-qan(1+2L)R+bn(1+L)R(an+bn)(1+2L)RΦ(R)8R(1+L), so (16)Tnxn+1-TnynLΦ(R)8R(1+L)<Φ(R)8R.

Using Lemma 3 and the above formula, we obtain (17)xn+1-q2=(1-an)(xn-q)+an(Tnyn-q)2(1-an)2xn-q2+2anTnyn-q,j(xn+1-q)=(1-an)2xn-q2+2anTnxn+1-q-Tnxn+1+Tnyn,j(xn+1-q)(1-an)2xn-q2+2an[knxn+1-q2-Φ(xn+1-q)]+2anTnxn+1-Tnyn·xn+1-q. Since 2knan0 as n, without loss of generality, let 1-2knan>0. Then (17) implies that (18)xn+1-q2(1-an)21-2knanxn-q2-2an1-2knanΦ(xn+1-q)+2an1-2knanTnxn+1-Tnyn·xn+1-qxn-q2+2an1-2knan×[[(kn-1)+an2]xn-q2nnnnnnn-Φ(xn+1-q)nnnnnnn+Tnxn+1-Tnyn·xn+1-q[(kn-1)+an2]]xn-q2+2an1-2knan×[Φ(R)R28R2(1+L)(1+2L)-Φ(R)+Φ(R)2R8R]xn-q2+2an1-2knan×[Φ(R)4-Φ(R)+Φ(R)4]R2-an1-2knanΦ(R)R2,

and this is a contradiction. Hence xn+1B1; that is, {xn} is a bounded sequence.

Step 2. We show that xn-q0 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+12=(1-an)(xn-un)+an(Tnyn-Tnun)2(1-an)2xn-un2+2anTnyn-Tnun,j(xn+1-un+1)(1-an)2xn-un2+2anTnyn-Tnxn+1,j(xn+1-un+1)+2anTnxn+1-Tnun+1,j(xn+1-un+1)+2anTnun+1-Tnun,j(xn+1-un+1)(1-an)2xn-un2+2anLyn-xn+1·xn+1-un+1+2an[knxn+1-un+12nnnnnnnnn-Φ(xn+1-un+1)xn+1-un+12]+2anLun+1-un·xn+1-un+1. Observe that (20)yn-xn+1=an(xn-Tnyn)-bn(xn-Tnxn)(an+bn+bnL)xn-q+anLyn-q(an+bn+bnL)xn-q+anL(1+bnL)xn-q(an+bn+bnL+anL+anbnL2)xn-qhn(xn-un+un-q)hn(xn-un+1),(21)un+1-un=an(Tnun-un)=anTnun-Tnq+q-unan(1+L)un-qan(1+L),

where hn=an+bn+bnL+anL+anbnL20 as n.

Substituting (20) and (21) into (19), we obtain (22)xn+1-un+12(1-an)2xn-un2+2Lanhn(xn-un+1)xn+1-un+1+2an(knxn+1-un+12-Φ(xn+1-un+1))+2anLan(1+L)xn+1-un+1(1-an)2xn-un2+Lanhn(xn-un2+2xn+1-un+12+1)+2an(knxn+1-un+12-Φ(xn+1-un+1))+L(1+L)an2+L(1+L)an2xn+1-un+12. Since an0 as n, without loss of generality, we may assume that (23)12<1-2Lanhn-2ankn-L(1+L)an2<1, for any n1. Then, (22) implies that (24)xn+1-un+12(1-an)2+Lanhn1-2Lanhn-2ankn-L(1+L)an2×xn-un2+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-un2+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-un2+2L(1+L)an2+2Lanhn-2anΦ(xn+1-un+1)xn-un2+[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+12δn2-λnΦ(δn+1)+γn. By Lemma 4, we obtain limnδn=0. That is, xn-un0 as n. From the inequality 0xn-qxn-un+un-q, we get xn-q0 as n. This completes the proof.

Remark 6.

The error in the proof of Theorem 8 of  has been pointed out and corrected, but it is not easy what the author really wants to obtain the proof of Theorem 8 in  at present.

Remark 7.

The proof method of Theorem 5 is quite different from that of  and others.

Acknowledgments

This work was supported by Hebei Provincial Natural Science Foundation (Grant no. A2011210033). And authors thank the reviewers for good suggestions and valuable comments of the paper.

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