The equivalence between Mann–Ishikawa iterations and multistep iteration

Abstract

We show that the convergence of Mann, Ishikawa iterations are equivalent to the convergence of a multistep iteration, for various classes of operators.

    Authors

    B.E. Rhoades

    S.M. Soltuz
    (Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy)

    Keywords

    Mann iteration; Ishikawa iteration; Strongly pseudocontractive; Strongly accretive map.

    References

    See the expanding block below.

    Paper coordinates

    B.E. Rhoades and Ş.M. Şoltuz, The equivalence between Mann-Ishikawa iterations and multistep iteration, Nonlinear Analysis: Theory, Methods & Applications, 58 (2004) no. 1-2, 219-228.
    doi: 10.1016/j.na.2003.11.013

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    The equivalence between Mann-Ishikawa iterations and multistep iteration

    B.E. Rhoades a, Stefan M. Soltuz b, * {}^{\mathrm{b},\text{ * }}
    a Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA
    b Str. Avram Iancu 13, Ap. 1, 400083 Cluj-Napoca, Romania
    Abstract

    We show that the convergence of Mann, Ishikawa iterations are equivalent to the convergence of a multistep iteration, for various classes of operators. © 2004 Elsevier Ltd. All rights reserved.

    Received 9 September 2003; accepted 21 November 2003

    MSC: 47H10

    Keywords: Mann iteration; Ishikawa iteration; Strongly pseudocontractive; Strongly accretive map

    1. Introduction

    Let XX be a Banach space, BB a nonempty, convex subset of XX, and TT a selfmap of BB. The two most popular iteration procedures for obtaining fixed points of TT, if they exist, are Mann iteration [5], defined by

    u1B,un+1=(1αn)un+αnTun,n1u_{1}\in B,\quad u_{n+1}=\left(1-\alpha_{n}\right)u_{n}+\alpha_{n}Tu_{n},\quad n\geqslant 1 (1.1)

    and Ishikawa iteration [4], defined by

    z1B,zn+1=(1αn)xn+αnTyn\displaystyle z_{1}\in B,\quad z_{n+1}=\left(1-\alpha_{n}\right)x_{n}+\alpha_{n}Ty_{n}
    yn=(1βn)xn+βnTzn,n1\displaystyle y_{n}=\left(1-\beta_{n}\right)x_{n}+\beta_{n}Tz_{n},\quad n\geqslant 1 (1.2)

    for certain choices of {αn},{βn}[0,1]\left\{\alpha_{n}\right\},\left\{\beta_{n}\right\}\subset[0,1].

    For XX a Hilbert space, BB a convex compact subset of X,TX,T a Lipschitzian pseudocontractive selfmap of BB, Ishikawa [4] was able to show that (1.2) converges strongly to the unique fixed point of TT in BB, provided that ( i\mathrm{i}^{\prime} ) 0αnβn10\leqslant\alpha_{n}\leqslant\beta_{n}\leqslant 1 for all n1n\geqslant 1, (ii) limβn=0\lim\beta_{n}=0, and (iii) n=1αnβn=\sum_{n=1}^{\infty}\alpha_{n}\beta_{n}=\infty. Previous attempts to establish the same result for Mann iteration had proved unsuccessful. Finally, in year 2000, in [1] an example was provided of a Lipschitzian pseudocontraction for which the Mann iteration fails to converge to the fixed point.

    Although condition (i’) was required in order to obtain the result of Ishikawa, it was noted that one could relax condition (i’) by replacing it with (i) 0αn,βn10\leqslant\alpha_{n},\beta_{n}\leqslant 1 and still obtain strong convergence for many different maps. Moreover, by proving a convergence theorem for this modified Ishikawa method, and then setting βn=0\beta_{n}=0 one obtained as a corollary the corresponding theorem for Mann iteration. The literature abounds with such papers.

    A reasonable conjecture is that the Ishikawa iteration methods satisfying (i) and the corresponding Mann iterations are equivalent for all maps for which either method provides convergence to a fixed point.

    In an attempt to verify this conjecture the authors, in a series of papers [9-14] have shown the equivalence for several classes of maps.

    In year 2000, M.A. Noor introduced in [7] the three-step procedure

    v1B,tn=(1γn)vn+γnTvn,\displaystyle v_{1}\in B,\quad t_{n}=\left(1-\gamma_{n}\right)v_{n}+\gamma_{n}Tv_{n},
    wn=(1βn)vn+βnTtn,\displaystyle w_{n}=\left(1-\beta_{n}\right)v_{n}+\beta_{n}Tt_{n},
    vn+1=(1αn)vn+αnTwn,n1.\displaystyle v_{n+1}=\left(1-\alpha_{n}\right)v_{n}+\alpha_{n}Tw_{n},\quad n\geqslant 1. (1.3)

    The presence of (1.3) raises an interesting question.
    Is there a map for which (1.3) converges to a fixed point, but for which (1.2), with (i’) fails to converge?

    The answer to that question is unknown, but we shall show in this paper that (1.3), (1.2) and (1.1) are equivalent for all classes of functions for which (1.3) has been used in [7,8]. In fact, we prove a more general result, by using a multi-step procedure of arbitrary fixed order p2p\geqslant 2, defined by

    xn+1=(1αn)xn+αnTyn1\displaystyle x_{n+1}=\left(1-\alpha_{n}\right)x_{n}+\alpha_{n}Ty_{n}^{1}
    yni=(1βni)xn+βniTyni+1,i=1,,p2,\displaystyle y_{n}^{i}=\left(1-\beta_{n}^{i}\right)x_{n}+\beta_{n}^{i}Ty_{n}^{i+1},\quad i=1,\ldots,p-2,
    ynp1=(1βnp1)xn+βnp1Txn.\displaystyle y_{n}^{p-1}=\left(1-\beta_{n}^{p-1}\right)x_{n}+\beta_{n}^{p-1}Tx_{n}. (1.4)

    The sequence {αn}\left\{\alpha_{n}\right\} is such that for all nn\in\mathbb{N}

    {αn}(0,1),limnαn=0,n=1αn=\left\{\alpha_{n}\right\}\subset(0,1),\quad\lim_{n\rightarrow\infty}\alpha_{n}=0,\quad\sum_{n=1}^{\infty}\alpha_{n}=\infty (1.5)

    and for all nn\in\mathbb{N}

    {βni}[0,1),1ip1,limnβn1=0\left\{\beta_{n}^{i}\right\}\subset[0,1),\quad 1\leqslant i\leqslant p-1,\quad\lim_{n\rightarrow\infty}\beta_{n}^{1}=0 (1.6)

    Taking p=3p=3 in (1.4) we obtain iteration (1.3). Taking p=2p=2 in (1.4) we obtain (1.2).

    The map J:X2XJ:X\rightarrow 2^{X^{*}} given by Jx:={fX:x,f=x2,f=x},xXJx:=\left\{f\in X^{*}:\langle x,f\rangle=\|x\|^{2},\|f\|=\|x\|\right\},\forall x\in X, is called the normalized duality mapping. The Hahn-Banach theorem assures that Jx,xXJx\neq\emptyset,\forall x\in X.

    Definition 1.1. A map T:BBT:B\rightarrow B is called strongly pseudocontractive if there exist k(0,1)k\in(0,1) and j(xy)J(xy)j(x-y)\in J(x-y) such that

    TxTy,j(xy)(1k)xy2,x,yB.\langle Tx-Ty,j(x-y)\rangle\leqslant(1-k)\|x-y\|^{2},\quad\forall x,y\in B. (1.7)

    A map S:XXS:X\rightarrow X is called strongly accretive if there exist k(0,1)k\in(0,1) and j(xy)J(xy)j(x-y)\in J(x-y) such that

    SxSy,j(xy)kxy2,x,yD(S).\langle Sx-Sy,j(x-y)\rangle\geqslant k\|x-y\|^{2},\quad\forall x,y\in D(S). (1.8)

    In (1.7) when k=0k=0, then TT is called pseudocontractive. In (1.8) when k=1,Sk=1,S is called accretive.

    Lemma 1.2 (Weng [15]). Let {an}\left\{a_{n}\right\} be a nonnegative sequence which satisfies the following inequality

    an+1(1λn)an+σna_{n+1}\leqslant\left(1-\lambda_{n}\right)a_{n}+\sigma_{n} (1.9)

    where λn(0,1),n,n=1λn=\lambda_{n}\in(0,1),\forall n\in\mathbb{N},\sum_{n=1}^{\infty}\lambda_{n}=\infty, and σn=o(λn)\sigma_{n}=o\left(\lambda_{n}\right). Then limnan=0\lim_{n\rightarrow\infty}a_{n}=0.
    The following Lemma is from [6].

    Lemma 1.3 (Morales and Jung [6]). If XX is a real Banach space, then the following relation is true:

    x+y2x2+2y,j(x+y),x,yX,j(x+y)J(x+y)\|x+y\|^{2}\leqslant\|x\|^{2}+2\langle y,j(x+y)\rangle,\quad\forall x,y\in X,\forall j(x+y)\in J(x+y) (1.10)

    2. Main results

    Theorem 2.1 Let XX be a real Banach space with a uniformly convex dual and BB a nonempty, closed, convex, bounded subset of XX. Let T:BBT:B\rightarrow B be a continuous and strongly pseudocontractive operator. If {αn}(0,1)\left\{\alpha_{n}\right\}\subset(0,1) satisfies (1.5) and {βni}[0,1)\left\{\beta_{n}^{i}\right\}\subset[0,1), i=1,,p1i=1,\ldots,p-1, satisfy (1.6) and u1=x1Bu_{1}=x_{1}\in B, then the following are equivalent:
    (i) the Mann iteration (1.1) converges to the fixed point of TT,
    (ii) the iteration (1.4) converges to the fixed point of TT.

    Proof. Corollary 1 of [2] assures the existence of a fixed point. The uniqueness of the fixed point comes from (1.7).

    Since BB is convex and bounded and TT is a selfmap of B,unBB,u_{n}\in B for each nn, and hence {un}\left\{u_{n}\right\} is bounded. The condition T:BBT:B\rightarrow B and the assumption that BB is bounded and convex lead us to conclusion {Tun}\left\{\left\|Tu_{n}\right\|\right\} is bounded.

    Denote

    P=supn{xn}P=\sup_{n\in\mathbb{N}}\left\{\left\|x_{n}\right\|\right\} (2.1)

    We will prove that {xn}\left\{\left\|x_{n}\right\|\right\} is bounded. Supposing now that

    xnBx_{n}\in B (2.2)

    we will prove that

    xn+1,yni(i=1,,p1)Bx_{n+1},y_{n}^{i}(i=1,\ldots,p-1)\in B (2.3)

    The fact that BB is a convex set, T:BBT:B\rightarrow B and relation (1.4) lead to

    ynp1=(1βnp1)xn+βnp1TxnB,y_{n}^{p-1}=\left(1-\beta_{n}^{p-1}\right)x_{n}+\beta_{n}^{p-1}Tx_{n}\in B, (2.4)

    similarly, we obtain

    ynp2=(1βnp2)xn+βnp2Tynp1By_{n}^{p-2}=\left(1-\beta_{n}^{p-2}\right)x_{n}+\beta_{n}^{p-2}Ty_{n}^{p-1}\in B (2.5)

    Recursively, we have

    yni=(1βni)xn+βniTyni+1B,i=1,,p3y_{n}^{i}=\left(1-\beta_{n}^{i}\right)x_{n}+\beta_{n}^{i}Ty_{n}^{i+1}\in B,\quad i=1,\ldots,p-3 (2.6)

    Thus yn1By_{n}^{1}\in B. Using the assumption T:BBT:B\rightarrow B we obtain that Tyn1BTy_{n}^{1}\in B. Hence

    xn+1=(1αn)xn+αnTyn1Bx_{n+1}=\left(1-\alpha_{n}\right)x_{n}+\alpha_{n}Ty_{n}^{1}\in B (2.7)

    because already xnBx_{n}\in B. Thus P<+P<+\infty. Set

    M:=\displaystyle M:= max{P,supn{Txn},supn{Tyni:1ip1},\displaystyle\max\left\{P,\sup_{n\in\mathbb{N}}\left\{\left\|Tx_{n}\right\|\right\},\sup_{n\in\mathbb{N}}\left\{\left\|Ty_{n}^{i}\right\|:1\leqslant i\leqslant p-1\right\},\right.
    supn{un},supn{Tun}}\displaystyle\left.\sup_{n\in\mathbb{N}}\left\{\left\|u_{n}\right\|\right\},\sup_{n\in\mathbb{N}}\left\{\left\|Tu_{n}\right\|\right\}\right\} (2.8)

    to obtain

    M<+M<+\infty (2.9)

    Because XX^{*} is uniformly convex the duality map is a single-valued map [3]. Using (1.1), (1.4), (1.7) and (1.10) with

    x:=(1αn)(xnun)\displaystyle x:=\left(1-\alpha_{n}\right)\left(x_{n}-u_{n}\right)
    y:=αn(Tyn1Tun)\displaystyle y:=\alpha_{n}\left(Ty_{n}^{1}-Tu_{n}\right)
    x+y=xn+1un+1\displaystyle x+y=x_{n+1}-u_{n+1} (2.10)

    we obtain

    xn+1un+12\displaystyle\left\|x_{n+1}-u_{n+1}\right\|^{2} =(1αn)(xnun)+αn(Tyn1Tun)2\displaystyle=\left\|\left(1-\alpha_{n}\right)\left(x_{n}-u_{n}\right)+\alpha_{n}\left(Ty_{n}^{1}-Tu_{n}\right)\right\|^{2}
    (1αn)2xnun2+2αnTyn1Tun,J(xn+1un+1)\displaystyle\leqslant\left(1-\alpha_{n}\right)^{2}\left\|x_{n}-u_{n}\right\|^{2}+2\alpha_{n}\left\langle Ty_{n}^{1}-Tu_{n},J\left(x_{n+1}-u_{n+1}\right)\right\rangle
    =(1αn)2xnun2+2αnTyn1Tun,J(xn+1un+1)\displaystyle=\left(1-\alpha_{n}\right)^{2}\left\|x_{n}-u_{n}\right\|^{2}+2\alpha_{n}\left\langle Ty_{n}^{1}-Tu_{n},J\left(x_{n+1}-u_{n+1}\right)\right.
    J(yn1un)+2αnTyn1Tun,J(yn1un)\displaystyle\left.-J\left(y_{n}^{1}-u_{n}\right)\right\rangle+2\alpha_{n}\left\langle Ty_{n}^{1}-Tu_{n},J\left(y_{n}^{1}-u_{n}\right)\right\rangle
    \displaystyle\leqslant (1αn)2xnun2+2αn(1k)yn1un2\displaystyle\left(1-\alpha_{n}\right)^{2}\left\|x_{n}-u_{n}\right\|^{2}+2\alpha_{n}(1-k)\left\|y_{n}^{1}-u_{n}\right\|^{2}
    +2αnTyn1Tun,J(xn+1un+1)J(yn1un).\displaystyle+2\alpha_{n}\left\langle Ty_{n}^{1}-Tu_{n},J\left(x_{n+1}-u_{n+1}\right)-J\left(y_{n}^{1}-u_{n}\right)\right\rangle. (2.11)

    Set

    σn:=2αnTyn1Tun,J(xn+1un+1)J(yn1un).\sigma_{n}:=2\alpha_{n}\left\langle Ty_{n}^{1}-Tu_{n},J\left(x_{n+1}-u_{n+1}\right)-J\left(y_{n}^{1}-u_{n}\right)\right\rangle. (2.12)

    Proposition 12.3 of [3] assures that, when XX^{*} is uniformly convex, then JJ is singlevalued map and is uniformly continuous on every bounded set of XX. Since {Tyn1Tun}\left\{Ty_{n}^{1}-Tu_{n}\right\} is bounded, to have limnσn=0\lim_{n\rightarrow\infty}\sigma_{n}=0 is sufficient to prove that

    J(xn+1un+1)J(yn1un)0,(n).\displaystyle J\left(x_{n+1}-u_{n+1}\right)-J\left(y_{n}^{1}-u_{n}\right)\rightarrow 0,(n\rightarrow\infty). (2.13)
    (xn+1un+1)(yn1un)\displaystyle\left\|\left(x_{n+1}-u_{n+1}\right)-\left(y_{n}^{1}-u_{n}\right)\right\|
    =(xn+1yn1)(un+1un)\displaystyle\quad=\left\|\left(x_{n+1}-y_{n}^{1}\right)-\left(u_{n+1}-u_{n}\right)\right\|
    =αnxn+αnTyn1+βn1xnβn1Tyn2+αnunαnTun\displaystyle\quad=\left\|-\alpha_{n}x_{n}+\alpha_{n}Ty_{n}^{1}+\beta_{n}^{1}x_{n}-\beta_{n}^{1}Ty_{n}^{2}+\alpha_{n}u_{n}-\alpha_{n}Tu_{n}\right\|
    αn(xn+Tyn1+un+Tun)+βn1(xn+Tyn2)\displaystyle\quad\leqslant\alpha_{n}\left(\left\|x_{n}\right\|+\left\|Ty_{n}^{1}\right\|+\left\|u_{n}\right\|+\left\|Tu_{n}\right\|\right)+\beta_{n}^{1}\left(\left\|x_{n}\right\|+\left\|Ty_{n}^{2}\right\|\right)
    (αn+βn1)4M0, as n.\displaystyle\quad\leqslant\left(\alpha_{n}+\beta_{n}^{1}\right)4M\rightarrow 0,\quad\text{ as }n\rightarrow\infty. (2.14)

    The uniform continuity of J()J(\cdot) guarantees that (2.13) is satisfied.
    Relations (1.1), (1.4) and (1.10) with

    x:=(1βn1)(xnun)\displaystyle x:=\left(1-\beta_{n}^{1}\right)\left(x_{n}-u_{n}\right)
    y:=βn1(Tyn2un)\displaystyle y:=\beta_{n}^{1}\left(Ty_{n}^{2}-u_{n}\right)
    x+y=yn1un\displaystyle x+y=y_{n}^{1}-u_{n} (2.15)

    lead to

    yn1un2\displaystyle\left\|y_{n}^{1}-u_{n}\right\|^{2} =(1βn1)(xnun)+βn1(Tyn2un)2\displaystyle=\left\|\left(1-\beta_{n}^{1}\right)\left(x_{n}-u_{n}\right)+\beta_{n}^{1}\left(Ty_{n}^{2}-u_{n}\right)\right\|^{2}
    (1βn1)2xnun2+2βn1Tyn2un,J(yn1un)\displaystyle\leqslant\left(1-\beta_{n}^{1}\right)^{2}\left\|x_{n}-u_{n}\right\|^{2}+2\beta_{n}^{1}\left\langle Ty_{n}^{2}-u_{n},J\left(y_{n}^{1}-u_{n}\right)\right\rangle
    xnun2+2βn1Tyn2unyn1un\displaystyle\leqslant\left\|x_{n}-u_{n}\right\|^{2}+2\beta_{n}^{1}\left\|Ty_{n}^{2}-u_{n}\right\|\left\|y_{n}^{1}-u_{n}\right\|
    xnun2+2βn1(Tyn2+un)(yn1+un)\displaystyle\leqslant\left\|x_{n}-u_{n}\right\|^{2}+2\beta_{n}^{1}\left(\left\|Ty_{n}^{2}\right\|+\left\|u_{n}\right\|\right)\left(\left\|y_{n}^{1}\right\|+\left\|u_{n}\right\|\right)
    xnun2+βn18M2\displaystyle\leqslant\left\|x_{n}-u_{n}\right\|^{2}+\beta_{n}^{1}8M^{2} (2.16)

    We already know that Tyn2M\left\|Ty_{n}^{2}\right\|\leqslant M and yn1M,n\left\|y_{n}^{1}\right\|\leqslant M,\forall n\in\mathbb{N}. Observe that we do not need further evaluations for yn3,,ynp1,xny_{n}^{3},\ldots,y_{n}^{p-1},x_{n}. This is the crucial point in this proof: starting the computations in (1.4), from xn+1x_{n+1} we do not need to evaluate more than two steps. The other steps are included in (2.4), (2.5), (2.6), and (2.7), to prove Tyn2M,n\left\|Ty_{n}^{2}\right\|\leqslant M,\forall n\in\mathbb{N}.

    Substituting (2.16) and (2.12) in (2.11), we obtain

    xn+1un+12\displaystyle\left\|x_{n+1}-u_{n+1}\right\|^{2}\leqslant (1αn)2xnun2+2αn(1k)xnun2\displaystyle\left(1-\alpha_{n}\right)^{2}\left\|x_{n}-u_{n}\right\|^{2}+2\alpha_{n}(1-k)\left\|x_{n}-u_{n}\right\|^{2}
    +σn+αnβn116M2(1k)\displaystyle+\sigma_{n}+\alpha_{n}\beta_{n}^{1}16M^{2}(1-k)
    =\displaystyle= (12kαn+αn2)xnun2+o(αn)\displaystyle\left(1-2k\alpha_{n}+\alpha_{n}^{2}\right)\left\|x_{n}-u_{n}\right\|^{2}+o\left(\alpha_{n}\right) (2.17)

    From (1.5) for all nn sufficiently large we have

    αnk\alpha_{n}\leqslant k (2.18)

    Substituting (2.18) into (2.17), we obtain

    12kαn+αn212kαn+kαn=1kαn.1-2k\alpha_{n}+\alpha_{n}^{2}\leqslant 1-2k\alpha_{n}+k\alpha_{n}=1-k\alpha_{n}. (2.19)

    Finally (2.17) becomes

    xn+1un+12(1kαn)xnun2+o(αn)\left\|x_{n+1}-u_{n+1}\right\|^{2}\leqslant\left(1-k\alpha_{n}\right)\left\|x_{n}-u_{n}\right\|^{2}+o\left(\alpha_{n}\right) (2.20)

    with

    an:=xnun2\displaystyle a_{n}:=\left\|x_{n}-u_{n}\right\|^{2}
    λn:=kαn(0,1)\displaystyle\lambda_{n}:=k\alpha_{n}\in(0,1) (2.21)

    and using Lemma 1.2, we obtain limnan=limnxnun2=0\lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}\left\|x_{n}-u_{n}\right\|^{2}=0, i.e.

    limnxnun=0\lim_{n\rightarrow\infty}\left\|x_{n}-u_{n}\right\|=0 (2.22)

    Suppose that limnun=x\lim_{n\rightarrow\infty}u_{n}=x^{*}. The inequality

    0xxnunx+xnun0\leqslant\left\|x^{*}-x_{n}\right\|\leqslant\left\|u_{n}-x^{*}\right\|+\left\|x_{n}-u_{n}\right\| (2.23)

    and (2.22), imply that limnxn=x\lim_{n\rightarrow\infty}x_{n}=x^{*}. Analogously limnxn=x\lim_{n\rightarrow\infty}x_{n}=x^{*} implies that limnun=x\lim_{n\rightarrow\infty}u_{n}=x^{*}.

    For p=2p=2 we get the following result from [10].
    Theorem 2.2 (Rhoades and Soltuz [10]). Let XX be a real Banach space with a uniformly convex dual and BB a nonempty, closed, convex, bounded subset of XX. Let T:BBT:B\rightarrow B be a continuous and strongly pseudocontractive operator. Then for u1=x1Bu_{1}=x_{1}\in B the following are equivalent:
    (i) the Mann iteration (1.1) converges to the fixed point of TT,
    (ii) the Ishikawa iteration (1.2) converges to the fixed point of TT.

    Theorems 2.1 and 2.2 lead to the following result.
    Corollary 2.3. Let XX be a real Banach space with a uniformly convex dual and BB a nonempty, closed, convex, bounded subset of XX. Let T:BBT:B\rightarrow B be a continuous
    and strongly pseudocontractive operator. Then for u1=x1Bu_{1}=x_{1}\in B the following are equivalent:
    (i) the Mann iteration (1.1) converges to the fixed point of TT,
    (ii) the Ishikawa iteration (1.2) converges to the fixed point of TT,
    (iii) the iteration (1.4) converges to the fixed point of TT.

    For p=3p=3, from Theorems 2.1 and 2.2, we have the following result:
    Corollary 2.4. Let XX be a real Banach space with a uniformly convex dual and BB a nonempty, closed, convex, bounded subset of XX. Let T:BBT:B\rightarrow B be a continuous and strongly pseudocontractive operator. Then for u1=x1Bu_{1}=x_{1}\in B the following are equivalent:
    (i) the Mann iteration (1.1) converges to the fixed point of TT,
    (ii) the Ishikawa iteration (1.2) converges to the fixed point of TT,
    (iii) the Noor iteration (1.3) converges to the fixed point of TT.

    Remark 2.5. (i) If BB is not bounded then Theorem 2.1 holds only supposing that {xn}\left\{x_{n}\right\} is bounded.
    (ii) If the Mann iteration converges to a point, it is clear that this point is a fixed point of TT. Thus we can omit the discussion of the existence of a fixed point in the proof of Theorem 2.1.
    (iii) If T(B)T(B) is bounded then {xn}\left\{x_{n}\right\} is bounded.

    Comments (i) and (ii) already been discussed in [10].
    Proof. We prove part (iii). Let

    M:=max{supxBTx,x1}M:=\max\left\{\sup_{x\in B}\|Tx\|,\left\|x_{1}\right\|\right\} (2.24)

    Then x1M\left\|x_{1}\right\|\leqslant M and supposing xnM\left\|x_{n}\right\|\leqslant M, we have

    xn+1(1αn)xn+αnM(1αn)M+αnM=M\left\|x_{n+1}\right\|\leqslant\left(1-\alpha_{n}\right)\left\|x_{n}\right\|+\alpha_{n}M\leqslant\left(1-\alpha_{n}\right)M+\alpha_{n}M=M (2.25)

    3. Further equivalences

    Let II denote the identity map.
    Remark 3.1. Let T,S:XX,fXT,S:X\rightarrow X,f\in X given. Then
    (i) A fixed point for the map Tx=f+(IS)x,xXTx=f+(I-S)x,\forall x\in X is a solution for Sx=fSx=f.
    (ii) A fixed point for Tx=fSxTx=f-Sx is a solution for x+Sx=fx+Sx=f.

    Remark 3.2 (Rhoades and Soltuz [10]). (i) The operator TT is a (strongly) pseudocontractive map if and only if ( ITI-T ) is (strongly) accretive.
    (ii) If SS is an accretive map then T=fST=f-S is strongly pseudocontractive map.

    We consider iterations (1.1) and (1.4), with Tx=f+(IS)xTx=f+(I-S)x and p2,{αn},{βni}(0,1),i=1,,p1p\geqslant 2,\left\{\alpha_{n}\right\},\left\{\beta_{n}^{i}\right\}\subset(0,1),i=1,\ldots,p-1 satisfying (1.5) and (1.6)

    un+1=(1αn)un+αn(f+(IS)un),\displaystyle u_{n+1}=\left(1-\alpha_{n}\right)u_{n}+\alpha_{n}\left(f+(I-S)u_{n}\right), (3.1)
    xn+1=(1αn)xn+αn(f+(IS)yn1),\displaystyle x_{n+1}=\left(1-\alpha_{n}\right)x_{n}+\alpha_{n}\left(f+(I-S)y_{n}^{1}\right),
    yni=(1βni)xn+βni(f+(IS)yni+1),i=1,,p2,\displaystyle y_{n}^{i}=\left(1-\beta_{n}^{i}\right)x_{n}+\beta_{n}^{i}\left(f+(I-S)y_{n}^{i+1}\right),\quad i=1,\ldots,p-2,
    ynp1=(1βnp1)xn+βnp1(f+(IS)xn).\displaystyle y_{n}^{p-1}=\left(1-\beta_{n}^{p-1}\right)x_{n}+\beta_{n}^{p-1}\left(f+(I-S)x_{n}\right). (3.2)

    Theorems 2.1 and 2.2, Remark 2.5(i), Remark 3.1(i), Remark 3.2(i) and Corollary 2.4 lead to the following result.

    Corollary 3.3. Let XX be a real Banach space with a uniformly convex dual and S:XXS:X\rightarrow X be a continuous and strongly accretive operator and let {xn}\left\{x_{n}\right\} given by (3.2) be bounded. If {αn}(0,1)\left\{\alpha_{n}\right\}\subset(0,1) satisfies (1.5) and {βni}[0,1),i=1,,p1\left\{\beta_{n}^{i}\right\}\subset[0,1),i=1,\ldots,p-1, satisfy (1.6) and u1=x1Bu_{1}=x_{1}\in B, then the following are equivalent:
    (i) the Mann iteration (3.1) converges to the solution of Sx=fSx=f,
    (ii) the Ishikawa iteration (1.2) with Tx=f+(IS)xTx=f+(I-S)x, converges to the solution of Sx=fSx=f,
    (iii) the iteration (3.2) converges to the solution of Sx=fSx=f,
    (iv) the Noor iteration (1.3) with Tx=f+(IS)xTx=f+(I-S)x, converges to the solution of Sx=fSx=f.

    We consider iterations (1.1) and (1.4), with Tx=fSxTx=f-Sx and p2,{αn},{βni}(0,1),i=1,,p1p\geqslant 2,\left\{\alpha_{n}\right\},\left\{\beta_{n}^{i}\right\}\subset(0,1),i=1,\ldots,p-1 satisfying (1.5) and (1.6)

    un+1=(1αn)un+αn(fSun),\displaystyle u_{n+1}=\left(1-\alpha_{n}\right)u_{n}+\alpha_{n}\left(f-Su_{n}\right), (3.3)
    xn+1=(1αn)xn+αn(fSyn1),\displaystyle x_{n+1}=\left(1-\alpha_{n}\right)x_{n}+\alpha_{n}\left(f-Sy_{n}^{1}\right),
    yni=(1βni)xn+βni(fSyni+1),i=1,,p2,\displaystyle y_{n}^{i}=\left(1-\beta_{n}^{i}\right)x_{n}+\beta_{n}^{i}\left(f-Sy_{n}^{i+1}\right),\quad i=1,\ldots,p-2,
    ynp1=(1βnp1)xn+βnp1(fSxn).\displaystyle y_{n}^{p-1}=\left(1-\beta_{n}^{p-1}\right)x_{n}+\beta_{n}^{p-1}\left(f-Sx_{n}\right). (3.4)

    Theorems 2.1 and 2.2, Remark 2.5(i), Remark 3.1(ii), Remark 3.2(ii), and Corollary 2.4 lead to the following result.

    Corollary 3.4. Let XX be a real Banach space with a uniformly convex dual and S:XXS:X\rightarrow X be a continuous and accretive operator and let {xn}\left\{x_{n}\right\} given by (3.4) be bounded. If {αn}(0,1)\left\{\alpha_{n}\right\}\subset(0,1) satisfies (1.5) and {βni}[0,1),i=1,,p1\left\{\beta_{n}^{i}\right\}\subset[0,1),i=1,\ldots,p-1, satisfy (1.6) and u1=x1Bu_{1}=x_{1}\in B, then the following are equivalent:
    (i) the Mann iteration (3.3) converges to the solution of x+Sx=fx+Sx=f,
    (ii) the Ishikawa iteration (1.2) with Tx=fSxTx=f-Sx, converges to the solution of x+Sx=fx+Sx=f,
    (iii) the iteration (3.4) converges to the solution of x+Sx=fx+Sx=f,
    (iv) the Noor iteration (1.3) with Tx=fSxTx=f-Sx, converges to the solution of x+Sx=fx+Sx=f.

    4. The equivalence between TT-stabilities

    All the arguments for the equivalence between TT-stabilities of Mann, Ishikawa, Multistep and Noor iterations are similar to those from [13]. Let us denote by F(T)={xB:x=T(x)}F(T)=\left\{x^{*}\in B:x^{*}=T\left(x^{*}\right)\right\}. Suppose that xF(T)x^{*}\in F(T). The following nonnegative sequences are well-defined for all nn\in\mathbb{N} :

    εn\displaystyle\varepsilon_{n} :=xn+1(1αn)xnαnTyn1\displaystyle:=\left\|x_{n+1}-\left(1-\alpha_{n}\right)x_{n}-\alpha_{n}Ty_{n}^{1}\right\| (4.1)
    δn\displaystyle\delta_{n} :=un+1(1αn)unαnTun\displaystyle:=\left\|u_{n+1}-\left(1-\alpha_{n}\right)u_{n}-\alpha_{n}Tu_{n}\right\| (4.2)

    Definition 4.1. If limnεn=0\lim_{n\rightarrow\infty}\varepsilon_{n}=0, (respectively limnδn=0\lim_{n\rightarrow\infty}\delta_{n}=0 ) implies that limnxn=x\lim_{n\rightarrow\infty}x_{n}=x^{*}, (respectively limnun=x\lim_{n\rightarrow\infty}u_{n}=x^{*} ), then (1.1) (respectively (1.4)) is said to be TT-stable.

    Remark 4.2 (Rhoades and Soltuz [13]). Let XX be a normed space, BXB\subset X be a nonempty, convex, closed subset and T:BBT:B\rightarrow B be continuous map. If the Mann (respectively (1.4)) iteration converges, then limnδn=0\lim_{n\rightarrow\infty}\delta_{n}=0 (respectively limnεn=0\lim_{n\rightarrow\infty}\varepsilon_{n}=0 ).

    Theorem 4.3. Let XX be a real Banach space with a uniformly convex dual and BB a nonempty, closed, convex, bounded subset of XX. Let T:BBT:B\rightarrow B be a continuous and strongly pseudocontractive operator. If {αn}(0,1)\left\{\alpha_{n}\right\}\subset(0,1) satisfies (1.6) and {βni}[0,1)\left\{\beta_{n}^{i}\right\}\subset[0,1), i=1,,p1i=1,\ldots,p-1, satisfy (1.5) and u1=x1Bu_{1}=x_{1}\in B, then the following are equivalent:
    (i) the Mann iteration (1.1) is TT-stable,
    (ii) the iteration (1.4) is TT-stable.

    Proof. The equivalence (i) \Leftrightarrow (ii) means that limnεn=0limnδn=0\lim_{n\rightarrow\infty}\varepsilon_{n}=0\Leftrightarrow\lim_{n\rightarrow\infty}\delta_{n}=0. The implication limnεn=0limnδn=0\lim_{n\rightarrow\infty}\varepsilon_{n}=0\Rightarrow\lim_{n\rightarrow\infty}\delta_{n}=0 is obvious by setting βni=0,i{1,,p1},n\beta_{n}^{i}=0,i\in\{1,\ldots,p-1\},\forall n\in\mathbb{N}, in (1.4) and using (4.2). Conversely, suppose that (1.1) is TT-stable. Using Definition 4.1 we obtain

    limnδn=0limnun=x\lim_{n\rightarrow\infty}\delta_{n}=0\Rightarrow\lim_{n\rightarrow\infty}u_{n}=x^{*} (4.3)

    Theorem 2.1 assures that limnun=x\lim_{n\rightarrow\infty}u_{n}=x^{*} leads us to limnxn=x\lim_{n\rightarrow\infty}x_{n}=x^{*}. Using Remark 4.2 we have limnεn=0\lim_{n\rightarrow\infty}\varepsilon_{n}=0. Thus we get limnδn=0limnεn=0\lim_{n\rightarrow\infty}\delta_{n}=0\Rightarrow\lim_{n\rightarrow\infty}\varepsilon_{n}=0.

    Analogously, we can prove the equivalence between TT-stabilities for the strongly accretive and accretive cases with Tx=f+(IS)xTx=f+(I-S)x, respectively Tx=fSxTx=f-Sx.

    Corollary 4.4. Let XX be a real Banach space with a uniformly convex dual and S:XXS:X\rightarrow X be a continuous and strongly accretive operator and let {xn}\left\{x_{n}\right\} given by (3.2) be bounded. If {αn}(0,1)\left\{\alpha_{n}\right\}\subset(0,1) satisfies (1.5) and {βni}[0,1),i=1,,p1\left\{\beta_{n}^{i}\right\}\subset[0,1),i=1,\ldots,p-1, satisfy (1.6) and u1=x1Bu_{1}=x_{1}\in B, then the following are equivalent:
    (i) the Mann iteration (3.1) is TT-stable,
    (ii) the iteration (3.2) is TT-stable.

    Corollary 4.5. Let XX be a real Banach space with a uniformly convex dual and S:XXS:X\rightarrow X be a continuous and accretive operator and let {xn}\left\{x_{n}\right\} given by (3.4) be bounded. If {αn}(0,1)\left\{\alpha_{n}\right\}\subset(0,1) satisfies (1.5) and {βni}[0,1),i=1,,p1\left\{\beta_{n}^{i}\right\}\subset[0,1),i=1,\ldots,p-1, satisfy (1.6) and u1=x1Bu_{1}=x_{1}\in B, then the following are equivalent:
    (i) the Mann iteration (3.3) is TT-stable,
    (ii) the iteration (3.4) is TT-stable.

    The authors are indebted to referee for carefully reading the paper and for making useful suggestions.

    References

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    2004

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