The equivalence between the T-stabilities of Picard-Banach and Mann-Ishikawa iterations

Stefan Soltuz
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Abstract

We show that T-stability of Picard-Banach and Mann-Ishikawa iterations are equivalent.

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Stefan M. Soltuz
Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy

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S.M. Şoltuz, The equivalence between the T-stabilities of Picard-Banach and Mann-Ishikawa iterations. Applied Math. E-Notes, 8 (2008), 109-114.

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2008 Soltuz The Equivalence Between The T-Stabilities Of

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Applied Mathematics E-Notes

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1607-2510

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1687-0425

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[1] S. Ishikawa, Fixed points by a new iteration method, Proc. Amer. Math. Soc., 44(1974), 147–150.
[2] A. M. Harder and T. Hicks, Stability results for fixed point iteration procedures, Math. Japonica, 33 (1988), 693–706.
[3] W. R. Mann, Mean value in iteration, Proc. Amer. Math. Soc., 4 (1953), 506–510.
[4] M. O. Osilike, Stability of the Mann and Ishikawa iteration procedures for φ-strong pseudocontractions and nonlinear equations of the φ-strongly accretive type, J.
Math. Anal. Appl., 227 (1998), 319–334.
[5] B. E. Rhoades and S.M. Soltuz, On the equivalence of Mann and Ishikawa iteration methods, Int. J. Math. Math. Sci., 2003(2003), 451–459.
[6] B. E. Rhoades and S. M. Soltuz, The equivalence between the T -stabilities of Mann and Ishikawa iterations, J. Math. Anal. Appl., 318(2006), 472-475.

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The Equivalence Between The T T TTT-Stabilities Of Picard-Banach And Mann-Ishikawa Iterations*

Ştefan M. Şoltuz † † ^(†){ }^{\dagger}†

Received 5 February 2007

Abstract

We show that T T TTT-stability of Picard-Banach and Mann-Ishikawa iterations are equivalent.

1 Introduction

Let X X XXX be a normed space and T T TTT a selfmap of X X XXX. Let x 0 x 0 x_(0)x_{0}x0 be a point of X X XXX, and assume that x n + 1 = f ( T , x n ) x n + 1 = f T , x n x_(n+1)=f(T,x_(n))x_{n+1}=f\left(T, x_{n}\right)xn+1=f(T,xn) is an iteration procedure, involving T T TTT, which yields a sequence { x n } x n {x_(n)}\left\{x_{n}\right\}{xn} of point from X X XXX. Suppose { x n } x n {x_(n)}\left\{x_{n}\right\}{xn} converges to a fixed point x ∗ x ∗ x^(**)x^{*}x∗ of T T TTT. Let { ξ n } ξ n {xi_(n)}\left\{\xi_{n}\right\}{ξn} be an arbitrary sequence in X X XXX, and set ϵ n = ‖ ξ n + 1 − f ( T , ξ n ) ‖ ϵ n = ξ n + 1 − f T , ξ n epsilon_(n)=||xi_(n+1)-f(T,xi_(n))||\epsilon_{n}=\left\|\xi_{n+1}-f\left(T, \xi_{n}\right)\right\|ϵn=‖ξn+1−f(T,ξn)‖ for all n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N.
DEFINITION 1. [2] If ( ( lim n → ∞ ϵ n = 0 ) ⇒ ( lim n → ∞ ξ n = p ) ) lim n → ∞   ϵ n = 0 ⇒ lim n → ∞   ξ n = p ((lim_(n rarr oo)epsilon_(n)=0)=>(lim_(n rarr oo)xi_(n)=p))\left(\left(\lim _{n \rightarrow \infty} \epsilon_{n}=0\right) \Rightarrow\left(\lim _{n \rightarrow \infty} \xi_{n}=p\right)\right)((limn→∞ϵn=0)⇒(limn→∞ξn=p)), then the iteration procedure x n + 1 = f ( T , x n ) x n + 1 = f T , x n x_(n+1)=f(T,x_(n))x_{n+1}=f\left(T, x_{n}\right)xn+1=f(T,xn) is said to be T T TTT-stable with respect to T T TTT.
REMARK 1. [2] In practice, such a sequence { ξ n } ξ n {xi_(n)}\left\{\xi_{n}\right\}{ξn} could arise in the following way. Let x 0 x 0 x_(0)x_{0}x0 be a point in X X XXX. Set x n + 1 = f ( T , x n ) x n + 1 = f T , x n x_(n+1)=f(T,x_(n))x_{n+1}=f\left(T, x_{n}\right)xn+1=f(T,xn). Let ξ 0 = x 0 ξ 0 = x 0 xi_(0)=x_(0)\xi_{0}=x_{0}ξ0=x0. Now x 1 = f ( T , x 0 ) x 1 = f T , x 0 x_(1)=f(T,x_(0))x_{1}=f\left(T, x_{0}\right)x1=f(T,x0). Because of rounding or discretization in the function T T TTT, a new value ξ 1 ξ 1 xi_(1)\xi_{1}ξ1 approximately equal to x 1 x 1 x_(1)x_{1}x1 might be obtained instead of the true value of f ( T , x 0 ) f T , x 0 f(T,x_(0))f\left(T, x_{0}\right)f(T,x0). Then to approximate ξ 2 ξ 2 xi_(2)\xi_{2}ξ2, the value f ( T , ξ 1 ) f T , ξ 1 f(T,xi_(1))f\left(T, \xi_{1}\right)f(T,ξ1) is computed to yields ξ 2 ξ 2 xi_(2)\xi_{2}ξ2, an approximation of f ( T , ξ 1 ) f T , ξ 1 f(T,xi_(1))f\left(T, \xi_{1}\right)f(T,ξ1). This computation is continued to obtain { ξ n } ξ n {xi_(n)}\left\{\xi_{n}\right\}{ξn} an approximate sequence of { x n } x n {x_(n)}\left\{x_{n}\right\}{xn}.
Consider e 0 = s 0 = t 0 = g 0 = h 0 e 0 = s 0 = t 0 = g 0 = h 0 e_(0)=s_(0)=t_(0)=g_(0)=h_(0)e_{0}=s_{0}=t_{0}=g_{0}=h_{0}e0=s0=t0=g0=h0. The Picard-Banach iteration is given by
(1) b n + 1 = T b n (1) b n + 1 = T b n {:(1)b_(n+1)=Tb_(n):}\begin{equation*} b_{n+1}=T b_{n} \tag{1} \end{equation*}(1)bn+1=Tbn
The two most popular iteration procedures for obtaining fixed points of T T TTT, when the Banach principle fails, are Mann iteration [3], defined by
(2) e n + 1 = ( 1 − α n ) e n + α n T e n (2) e n + 1 = 1 − α n e n + α n T e n {:(2)e_(n+1)=(1-alpha_(n))e_(n)+alpha_(n)Te_(n):}\begin{equation*} e_{n+1}=\left(1-\alpha_{n}\right) e_{n}+\alpha_{n} T e_{n} \tag{2} \end{equation*}(2)en+1=(1−αn)en+αnTen
and Ishikawa iteration [1], defined by
(3) s n + 1 = ( 1 − α n ) s n + α n T t n t n = ( 1 − β n ) s n + β n T s n (3) s n + 1 = 1 − α n s n + α n T t n t n = 1 − β n s n + β n T s n {:[(3)s_(n+1)=(1-alpha_(n))s_(n)+alpha_(n)Tt_(n)],[t_(n)=(1-beta_(n))s_(n)+beta_(n)Ts_(n)]:}\begin{align*} s_{n+1} & =\left(1-\alpha_{n}\right) s_{n}+\alpha_{n} T t_{n} \tag{3}\\ t_{n} & =\left(1-\beta_{n}\right) s_{n}+\beta_{n} T s_{n} \end{align*}(3)sn+1=(1−αn)sn+αnTtntn=(1−βn)sn+βnTsn
We have { α n } ⊂ ( 0 , 1 ) , { β n } ⊂ [ 0 , 1 ) α n ⊂ ( 0 , 1 ) , β n ⊂ [ 0 , 1 ) {alpha_(n)}sub(0,1),{beta_(n)}sub[0,1)\left\{\alpha_{n}\right\} \subset(0,1),\left\{\beta_{n}\right\} \subset[0,1){αn}⊂(0,1),{βn}⊂[0,1) and { α n } α n {alpha_(n)}\left\{\alpha_{n}\right\}{αn} usually satisfies
(4) lim n → ∞ α n = 0 , ∑ n = 1 ∞ α n = ∞ (4) lim n → ∞   α n = 0 , ∑ n = 1 ∞   α n = ∞ {:(4)lim_(n rarr oo)alpha_(n)=0","sum_(n=1)^(oo)alpha_(n)=oo:}\begin{equation*} \lim _{n \rightarrow \infty} \alpha_{n}=0, \sum_{n=1}^{\infty} \alpha_{n}=\infty \tag{4} \end{equation*}(4)limn→∞αn=0,∑n=1∞αn=∞
Recently, the equivalence between the T T TTT-stabilities of Mann and Ishikawa iterations was shown in [6]. In this note we shall prove the equivalence between T T TTT-stabilities of (1), (2) and (3).

2 The Equivalence between T T TTT-Stabilities

Let X X XXX be a normed space and T : X → X T : X → X T:X rarr XT: X \rightarrow XT:X→X a map. Let { u n } , { p n } , { x n } , { y n } ⊂ X u n , p n , x n , y n ⊂ X {u_(n)},{p_(n)},{x_(n)},{y_(n)}sub X\left\{u_{n}\right\},\left\{p_{n}\right\},\left\{x_{n}\right\},\left\{y_{n}\right\} \subset X{un},{pn},{xn},{yn}⊂X be such that u 0 = p 0 = x 0 = y 0 u 0 = p 0 = x 0 = y 0 u_(0)=p_(0)=x_(0)=y_(0)u_{0}=p_{0}=x_{0}=y_{0}u0=p0=x0=y0, and consider
ε n := ‖ u n + 1 − ( 1 − α n ) u n − α n T u n ‖ δ n := ‖ p n + 1 − T p n ‖ ε n := u n + 1 − 1 − α n u n − α n T u n δ n := p n + 1 − T p n {:[epsi_(n):=||u_(n+1)-(1-alpha_(n))u_(n)-alpha_(n)Tu_(n)||],[delta_(n):=||p_(n+1)-Tp_(n)||]:}\begin{aligned} \varepsilon_{n}:=\left\|u_{n+1}-\left(1-\alpha_{n}\right) u_{n}-\alpha_{n} T u_{n}\right\| \\ \delta_{n}:=\left\|p_{n+1}-T p_{n}\right\| \end{aligned}εn:=‖un+1−(1−αn)un−αnTun‖δn:=‖pn+1−Tpn‖
For { β n } ⊂ [ 0 , 1 ) β n ⊂ [ 0 , 1 ) {beta_(n)}sub[0,1)\left\{\beta_{n}\right\} \subset[0,1){βn}⊂[0,1), we consider y n = ( 1 − β n ) x n + β n T x n y n = 1 − β n x n + β n T x n y_(n)=(1-beta_(n))x_(n)+beta_(n)Tx_(n)y_{n}=\left(1-\beta_{n}\right) x_{n}+\beta_{n} T x_{n}yn=(1−βn)xn+βnTxn, and
ξ n := ‖ x n + 1 − ( 1 − α n ) x n − α n T y n ‖ ξ n := x n + 1 − 1 − α n x n − α n T y n xi_(n):=||x_(n+1)-(1-alpha_(n))x_(n)-alpha_(n)Ty_(n)||\xi_{n}:=\left\|x_{n+1}-\left(1-\alpha_{n}\right) x_{n}-\alpha_{n} T y_{n}\right\|ξn:=‖xn+1−(1−αn)xn−αnTyn‖
DEFINITION 2. Definition 1 gives:
(i) The Ishikawa iteration (3), is said to be T T TTT-stable if and only if for all { α n } ⊂ ( 0 , 1 ) , { β n } ⊂ [ 0 , 1 ) , ∀ { x n } ⊂ X α n ⊂ ( 0 , 1 ) , β n ⊂ [ 0 , 1 ) , ∀ x n ⊂ X {alpha_(n)}sub(0,1),{beta_(n)}sub[0,1),AA{x_(n)}sub X\left\{\alpha_{n}\right\} \subset (0,1),\left\{\beta_{n}\right\} \subset[0,1), \forall\left\{x_{n}\right\} \subset X{αn}⊂(0,1),{βn}⊂[0,1),∀{xn}⊂X given, we have
lim n → ∞ ξ n = lim n → ∞ ‖ x n + 1 − ( 1 − α n ) x n − α n T y n ‖ = 0 ⇒ lim n → ∞ x n = x ∗ lim n → ∞   ξ n = lim n → ∞   x n + 1 − 1 − α n x n − α n T y n = 0 ⇒ lim n → ∞   x n = x ∗ lim_(n rarr oo)xi_(n)=lim_(n rarr oo)||x_(n+1)-(1-alpha_(n))x_(n)-alpha_(n)Ty_(n)||=0=>lim_(n rarr oo)x_(n)=x^(**)\lim _{n \rightarrow \infty} \xi_{n}=\lim _{n \rightarrow \infty}\left\|x_{n+1}-\left(1-\alpha_{n}\right) x_{n}-\alpha_{n} T y_{n}\right\|=0 \Rightarrow \lim _{n \rightarrow \infty} x_{n}=x^{*}limn→∞ξn=limn→∞‖xn+1−(1−αn)xn−αnTyn‖=0⇒limn→∞xn=x∗
The Mann iteration is said to be T T TTT-stable if and only if for all { α n } ⊂ ( 0 , 1 ) , ∀ { u n } ⊂ X α n ⊂ ( 0 , 1 ) , ∀ u n ⊂ X {alpha_(n)}sub(0,1),AA{u_(n)}sub X\left\{\alpha_{n}\right\} \subset(0,1), \forall\left\{u_{n}\right\} \subset X{αn}⊂(0,1),∀{un}⊂X given, we have
lim n → ∞ ε n = lim n → ∞ ‖ u n + 1 − ( 1 − α n ) u n − α n T u n ‖ = 0 ⇒ lim n → ∞ u n = x ∗ lim n → ∞   ε n = lim n → ∞   u n + 1 − 1 − α n u n − α n T u n = 0 ⇒ lim n → ∞   u n = x ∗ lim_(n rarr oo)epsi_(n)=lim_(n rarr oo)||u_(n+1)-(1-alpha_(n))u_(n)-alpha_(n)Tu_(n)||=0=>lim_(n rarr oo)u_(n)=x^(**)\lim _{n \rightarrow \infty} \varepsilon_{n}=\lim _{n \rightarrow \infty}\left\|u_{n+1}-\left(1-\alpha_{n}\right) u_{n}-\alpha_{n} T u_{n}\right\|=0 \Rightarrow \lim _{n \rightarrow \infty} u_{n}=x^{*}limn→∞εn=limn→∞‖un+1−(1−αn)un−αnTun‖=0⇒limn→∞un=x∗
(ii) The Picard iteration is said to be T T TTT-stable if and only if for all { p n } ⊂ X p n ⊂ X {p_(n)}sub X\left\{p_{n}\right\} \subset X{pn}⊂X given, we have
lim n → ∞ δ n = lim n → ∞ ‖ p n + 1 − T p n ‖ = 0 ⇒ lim n → ∞ p n = x ∗ lim n → ∞   δ n = lim n → ∞   p n + 1 − T p n = 0 ⇒ lim n → ∞   p n = x ∗ lim_(n rarr oo)delta_(n)=lim_(n rarr oo)||p_(n+1)-Tp_(n)||=0=>lim_(n rarr oo)p_(n)=x^(**)\lim _{n \rightarrow \infty} \delta_{n}=\lim _{n \rightarrow \infty}\left\|p_{n+1}-T p_{n}\right\|=0 \Rightarrow \lim _{n \rightarrow \infty} p_{n}=x^{*}limn→∞δn=limn→∞‖pn+1−Tpn‖=0⇒limn→∞pn=x∗
It is obvious that for α n := 0 , ∀ n ∈ N , β n := 0 , ∀ n ∈ N α n := 0 , ∀ n ∈ N , β n := 0 , ∀ n ∈ N alpha_(n):=0,AA n inN,beta_(n):=0,AA n inN\alpha_{n}:=0, \forall n \in \mathbb{N}, \beta_{n}:=0, \forall n \in \mathbb{N}αn:=0,∀n∈N,βn:=0,∀n∈N, one obtains ξ n = ε n = δ n ξ n = ε n = δ n xi_(n)=epsi_(n)=delta_(n)\xi_{n}=\varepsilon_{n}=\delta_{n}ξn=εn=δn.
THEOREM 1. Let X X XXX be a normed space and T : X → X T : X → X T:X rarr XT: X \rightarrow XT:X→X a map. If
(5) lim n → ∞ ‖ p n − T p n ‖ = 0 and lim n → ∞ ‖ u n − T u n ‖ = 0 (5) lim n → ∞   p n − T p n = 0  and  lim n → ∞   u n − T u n = 0 {:(5)lim_(n rarr oo)||p_(n)-Tp_(n)||=0" and "lim_(n rarr oo)||u_(n)-Tu_(n)||=0:}\begin{equation*} \lim _{n \rightarrow \infty}\left\|p_{n}-T p_{n}\right\|=0 \text { and } \lim _{n \rightarrow \infty}\left\|u_{n}-T u_{n}\right\|=0 \tag{5} \end{equation*}(5)limn→∞‖pn−Tpn‖=0 and limn→∞‖un−Tun‖=0
then the following are equivalent:
(i) for all { α n } ⊂ ( 0 , 1 ) α n ⊂ ( 0 , 1 ) {alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1){αn}⊂(0,1), the Mann iteration is T T TTT-stable,
(ii) the Picard iteration is T T TTT-stable.
PROOF. ( i ) ⇒ ( i i ) ( i ) ⇒ ( i i ) (i)=>(ii)(i) \Rightarrow(i i)(i)⇒(ii). Take lim n → ∞ δ n = 0 lim n → ∞   δ n = 0 lim_(n rarr oo)delta_(n)=0\lim _{n \rightarrow \infty} \delta_{n}=0limn→∞δn=0. Observe that
ε n = ‖ u n + 1 − ( 1 − α n ) u n − α n T u n ‖ ≤ ‖ u n + 1 − T u n ‖ + ( 1 − α n ) ‖ u n + 1 − u n ‖ + ( 1 − α n ) ‖ u n + 1 − T u n ‖ ≤ ( 2 − α n ) ‖ u n + 1 − T u n ‖ + ( 1 − α n ) ‖ u n + 1 − u n ‖ ≤ ( 2 − α n ) ‖ u n + 1 − T u n ‖ + ( 1 − α n ) ( ‖ u n + 1 − T u n ‖ + ‖ u n − T u n ‖ ) = ( 3 − 2 α n ) ‖ u n + 1 − T u n ‖ + ( 1 − α n ) ‖ u n − T u n ‖ = ( 3 − 2 α n ) δ n + ( 1 − α n ) ‖ u n − T u n ‖ → 0 ε n = u n + 1 − 1 − α n u n − α n T u n ≤ u n + 1 − T u n + 1 − α n u n + 1 − u n + 1 − α n u n + 1 − T u n ≤ 2 − α n u n + 1 − T u n + 1 − α n u n + 1 − u n ≤ 2 − α n u n + 1 − T u n + 1 − α n u n + 1 − T u n + u n − T u n = 3 − 2 α n u n + 1 − T u n + 1 − α n u n − T u n = 3 − 2 α n δ n + 1 − α n u n − T u n → 0 {:[epsi_(n)=||u_(n+1)-(1-alpha_(n))u_(n)-alpha_(n)Tu_(n)||],[ <= ||u_(n+1)-Tu_(n)||+(1-alpha_(n))||u_(n+1)-u_(n)||+(1-alpha_(n))||u_(n+1)-Tu_(n)||],[ <= (2-alpha_(n))||u_(n+1)-Tu_(n)||+(1-alpha_(n))||u_(n+1)-u_(n)||],[ <= (2-alpha_(n))||u_(n+1)-Tu_(n)||+(1-alpha_(n))(||u_(n+1)-Tu_(n)||+||u_(n)-Tu_(n)||)],[=(3-2alpha_(n))||u_(n+1)-Tu_(n)||+(1-alpha_(n))||u_(n)-Tu_(n)||],[=(3-2alpha_(n))delta_(n)+(1-alpha_(n))||u_(n)-Tu_(n)||],[ rarr0]:}\begin{aligned} \varepsilon_{n} & =\left\|u_{n+1}-\left(1-\alpha_{n}\right) u_{n}-\alpha_{n} T u_{n}\right\| \\ & \leq\left\|u_{n+1}-T u_{n}\right\|+\left(1-\alpha_{n}\right)\left\|u_{n+1}-u_{n}\right\|+\left(1-\alpha_{n}\right)\left\|u_{n+1}-T u_{n}\right\| \\ & \leq\left(2-\alpha_{n}\right)\left\|u_{n+1}-T u_{n}\right\|+\left(1-\alpha_{n}\right)\left\|u_{n+1}-u_{n}\right\| \\ & \leq\left(2-\alpha_{n}\right)\left\|u_{n+1}-T u_{n}\right\|+\left(1-\alpha_{n}\right)\left(\left\|u_{n+1}-T u_{n}\right\|+\left\|u_{n}-T u_{n}\right\|\right) \\ & =\left(3-2 \alpha_{n}\right)\left\|u_{n+1}-T u_{n}\right\|+\left(1-\alpha_{n}\right)\left\|u_{n}-T u_{n}\right\| \\ & =\left(3-2 \alpha_{n}\right) \delta_{n}+\left(1-\alpha_{n}\right)\left\|u_{n}-T u_{n}\right\| \\ & \rightarrow 0 \end{aligned}εn=‖un+1−(1−αn)un−αnTun‖≤‖un+1−Tun‖+(1−αn)‖un+1−un‖+(1−αn)‖un+1−Tun‖≤(2−αn)‖un+1−Tun‖+(1−αn)‖un+1−un‖≤(2−αn)‖un+1−Tun‖+(1−αn)(‖un+1−Tun‖+‖un−Tun‖)=(3−2αn)‖un+1−Tun‖+(1−αn)‖un−Tun‖=(3−2αn)δn+(1−αn)‖un−Tun‖→0
as n → ∞ n → ∞ n rarr oon \rightarrow \inftyn→∞. We know from ( i i iii ) that if lim n → ∞ ε n = 0 lim n → ∞   ε n = 0 lim_(n rarr oo)epsi_(n)=0\lim _{n \rightarrow \infty} \varepsilon_{n}=0limn→∞εn=0, then lim n → ∞ u n = x ∗ lim n → ∞   u n = x ∗ lim_(n rarr oo)u_(n)=x^(**)\lim _{n \rightarrow \infty} u_{n}=x^{*}limn→∞un=x∗, thus we have shown that if lim n → ∞ δ n = lim n → ∞ ‖ u n + 1 − T u n ‖ = 0 lim n → ∞   δ n = lim n → ∞   u n + 1 − T u n = 0 lim_(n rarr oo)delta_(n)=lim_(n rarr oo)||u_(n+1)-Tu_(n)||=0\lim _{n \rightarrow \infty} \delta_{n}=\lim _{n \rightarrow \infty}\left\|u_{n+1}-T u_{n}\right\|=0limn→∞δn=limn→∞‖un+1−Tun‖=0, then lim n → ∞ u n = x ∗ lim n → ∞   u n = x ∗ lim_(n rarr oo)u_(n)=x^(**)\lim _{n \rightarrow \infty} u_{n}=x^{*}limn→∞un=x∗.
For ( i i ) ⇒ ( i ) ( i i ) ⇒ ( i ) (ii)=>(i)(i i) \Rightarrow(i)(ii)⇒(i), take lim n → ∞ ε n = 0 lim n → ∞   ε n = 0 lim_(n rarr oo)epsi_(n)=0\lim _{n \rightarrow \infty} \varepsilon_{n}=0limn→∞εn=0. Observe that
δ n = ‖ p n + 1 − T p n ‖ ≤ ‖ p n + 1 − ( 1 − α n ) p n − α n T p n ‖ + ( 1 − α n ) ‖ p n − T p n ‖ ≤ ε n + ( 1 − α n ) ‖ p n − T p n ‖ → 0 δ n = p n + 1 − T p n ≤ p n + 1 − 1 − α n p n − α n T p n + 1 − α n p n − T p n ≤ ε n + 1 − α n p n − T p n → 0 {:[delta_(n)=||p_(n+1)-Tp_(n)||],[ <= ||p_(n+1)-(1-alpha_(n))p_(n)-alpha_(n)Tp_(n)||+(1-alpha_(n))||p_(n)-Tp_(n)||],[ <= epsi_(n)+(1-alpha_(n))||p_(n)-Tp_(n)||],[ rarr0]:}\begin{aligned} \delta_{n} & =\left\|p_{n+1}-T p_{n}\right\| \\ & \leq\left\|p_{n+1}-\left(1-\alpha_{n}\right) p_{n}-\alpha_{n} T p_{n}\right\|+\left(1-\alpha_{n}\right)\left\|p_{n}-T p_{n}\right\| \\ & \leq \varepsilon_{n}+\left(1-\alpha_{n}\right)\left\|p_{n}-T p_{n}\right\| \\ & \rightarrow 0 \end{aligned}δn=‖pn+1−Tpn‖≤‖pn+1−(1−αn)pn−αnTpn‖+(1−αn)‖pn−Tpn‖≤εn+(1−αn)‖pn−Tpn‖→0
as n → ∞ n → ∞ n rarr oon \rightarrow \inftyn→∞. We know from ( i i i i iii iii ) that if lim n → ∞ δ n = 0 lim n → ∞   δ n = 0 lim_(n rarr oo)delta_(n)=0\lim _{n \rightarrow \infty} \delta_{n}=0limn→∞δn=0, then lim n → ∞ p n = x ∗ lim n → ∞   p n = x ∗ lim_(n rarr oo)p_(n)=x^(**)\lim _{n \rightarrow \infty} p_{n}=x^{*}limn→∞pn=x∗, thus we have shown that if lim n → ∞ ε n = lim n → ∞ ‖ p n + 1 − ( 1 − α n ) p n − α n T p n ‖ = 0 lim n → ∞   ε n = lim n → ∞   p n + 1 − 1 − α n p n − α n T p n = 0 lim_(n rarr oo)epsi_(n)=lim_(n rarr oo)||p_(n+1)-(1-alpha_(n))p_(n)-alpha_(n)Tp_(n)||=0\lim _{n \rightarrow \infty} \varepsilon_{n}=\lim _{n \rightarrow \infty}\left\|p_{n+1}-\left(1-\alpha_{n}\right) p_{n}-\alpha_{n} T p_{n}\right\|=0limn→∞εn=limn→∞‖pn+1−(1−αn)pn−αnTpn‖=0, then lim n → ∞ p n = x ∗ lim n → ∞   p n = x ∗ lim_(n rarr oo)p_(n)=x^(**)\lim _{n \rightarrow \infty} p_{n}=x^{*}limn→∞pn=x∗.
REMARK 2. Note that no boundedness condition is needed in the above result. Note that lim n → ∞ ‖ u n − T u n ‖ = 0 lim n → ∞   u n − T u n = 0 lim_(n rarr oo)||u_(n)-Tu_(n)||=0\lim _{n \rightarrow \infty}\left\|u_{n}-T u_{n}\right\|=0limn→∞‖un−Tun‖=0 is used in order to prove that lim n → ∞ ε n = 0 lim n → ∞   ε n = 0 lim_(n rarr oo)epsi_(n)=0\lim _{n \rightarrow \infty} \varepsilon_{n}=0limn→∞εn=0, hence can not be avoided. Analogously, lim n → ∞ ‖ p n − T p n ‖ = 0 lim n → ∞   p n − T p n = 0 lim_(n rarr oo)||p_(n)-Tp_(n)||=0\lim _{n \rightarrow \infty}\left\|p_{n}-T p_{n}\right\|=0limn→∞‖pn−Tpn‖=0 is used in order to prove that lim n → ∞ δ n = 0 lim n → ∞   δ n = 0 lim_(n rarr oo)delta_(n)=0\lim _{n \rightarrow \infty} \delta_{n}=0limn→∞δn=0, hence can not be avoided.
THEOREM 2. Let X X XXX be a normed space and T : X → X T : X → X T:X rarr XT: X \rightarrow XT:X→X a map with bounded range. If
lim n → ∞ ‖ p n − T p n ‖ = 0 and lim n → ∞ ‖ x n − T x n ‖ = 0 lim n → ∞   p n − T p n = 0  and  lim n → ∞   x n − T x n = 0 lim_(n rarr oo)||p_(n)-Tp_(n)||=0" and "lim_(n rarr oo)||x_(n)-Tx_(n)||=0\lim _{n \rightarrow \infty}\left\|p_{n}-T p_{n}\right\|=0 \text { and } \lim _{n \rightarrow \infty}\left\|x_{n}-T x_{n}\right\|=0limn→∞‖pn−Tpn‖=0 and limn→∞‖xn−Txn‖=0
then the following are equivalent:
(i) for all { α n } ⊂ ( 0 , 1 ) α n ⊂ ( 0 , 1 ) {alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1){αn}⊂(0,1) and { β n } ⊂ [ 0 , 1 ) β n ⊂ [ 0 , 1 ) {beta_(n)}sub[0,1)\left\{\beta_{n}\right\} \subset[0,1){βn}⊂[0,1), satisfying (4), the Ishikawa iteration is T T TTT-stable,
(ii) the Picard iteration is T T TTT-stable.
PROOF. Let
M := max { sup x ∈ X { ‖ T ( x ) ‖ } , ‖ x 0 ‖ } . M := max sup x ∈ X   { ‖ T ( x ) ‖ } , x 0 . M:=max{s u p_(x in X){||T(x)||},||x_(0)||}.M:=\max \left\{\sup _{x \in X}\{\|T(x)\|\},\left\|x_{0}\right\|\right\} .M:=max{supx∈X{‖T(x)‖},‖x0‖}.
Since T T TTT has bounded range, we have M < ∞ M < ∞ M < ooM<\inftyM<∞.
We shall prove that ( i ) ⇒ ( i i ) ( i ) ⇒ ( i i ) (i)=>(ii)(i) \Rightarrow(i i)(i)⇒(ii). Take lim n → ∞ δ n = 0 lim n → ∞   δ n = 0 lim_(n rarr oo)delta_(n)=0\lim _{n \rightarrow \infty} \delta_{n}=0limn→∞δn=0. Observe that
ξ n = ‖ x n + 1 − ( 1 − α n ) x n − α n T y n ‖ ≤ ‖ x n + 1 − T x n ‖ + ‖ ( 1 − α n ) x n − α n T y n + T x n ‖ = ‖ x n + 1 − T x n ‖ + ‖ ( 1 − α n ) x n − α n T y n + T x n − α n T x n + α n T x n ‖ ≤ ‖ x n + 1 − T x n ‖ + ( 1 − α n ) ‖ x n − T x n ‖ + α n ‖ T x n − T y n ‖ = δ n + ( 1 − α n ) ‖ x n − T x n ‖ + 2 α n M → 0 ξ n = x n + 1 − 1 − α n x n − α n T y n ≤ x n + 1 − T x n + 1 − α n x n − α n T y n + T x n = x n + 1 − T x n + 1 − α n x n − α n T y n + T x n − α n T x n + α n T x n ≤ x n + 1 − T x n + 1 − α n x n − T x n + α n T x n − T y n = δ n + 1 − α n x n − T x n + 2 α n M → 0 {:[xi_(n)=||x_(n+1)-(1-alpha_(n))x_(n)-alpha_(n)Ty_(n)||],[ <= ||x_(n+1)-Tx_(n)||+||(1-alpha_(n))x_(n)-alpha_(n)Ty_(n)+Tx_(n)||],[=||x_(n+1)-Tx_(n)||+||(1-alpha_(n))x_(n)-alpha_(n)Ty_(n)+Tx_(n)-alpha_(n)Tx_(n)+alpha_(n)Tx_(n)||],[ <= ||x_(n+1)-Tx_(n)||+(1-alpha_(n))||x_(n)-Tx_(n)||+alpha_(n)||Tx_(n)-Ty_(n)||],[=delta_(n)+(1-alpha_(n))||x_(n)-Tx_(n)||+2alpha_(n)M],[ rarr0]:}\begin{aligned} \xi_{n} & =\left\|x_{n+1}-\left(1-\alpha_{n}\right) x_{n}-\alpha_{n} T y_{n}\right\| \\ & \leq\left\|x_{n+1}-T x_{n}\right\|+\left\|\left(1-\alpha_{n}\right) x_{n}-\alpha_{n} T y_{n}+T x_{n}\right\| \\ & =\left\|x_{n+1}-T x_{n}\right\|+\left\|\left(1-\alpha_{n}\right) x_{n}-\alpha_{n} T y_{n}+T x_{n}-\alpha_{n} T x_{n}+\alpha_{n} T x_{n}\right\| \\ & \leq\left\|x_{n+1}-T x_{n}\right\|+\left(1-\alpha_{n}\right)\left\|x_{n}-T x_{n}\right\|+\alpha_{n}\left\|T x_{n}-T y_{n}\right\| \\ & =\delta_{n}+\left(1-\alpha_{n}\right)\left\|x_{n}-T x_{n}\right\|+2 \alpha_{n} M \\ & \rightarrow 0 \end{aligned}ξn=‖xn+1−(1−αn)xn−αnTyn‖≤‖xn+1−Txn‖+‖(1−αn)xn−αnTyn+Txn‖=‖xn+1−Txn‖+‖(1−αn)xn−αnTyn+Txn−αnTxn+αnTxn‖≤‖xn+1−Txn‖+(1−αn)‖xn−Txn‖+αn‖Txn−Tyn‖=δn+(1−αn)‖xn−Txn‖+2αnM→0
as n → ∞ n → ∞ n rarr oon \rightarrow \inftyn→∞. Condition (i) assures that lim n → ∞ ξ n = 0 ⇒ lim n → ∞ x n = x ∗ lim n → ∞   ξ n = 0 ⇒ lim n → ∞   x n = x ∗ lim_(n rarr oo)xi_(n)=0=>lim_(n rarr oo)x_(n)=x^(**)\lim _{n \rightarrow \infty} \xi_{n}=0 \Rightarrow \lim _{n \rightarrow \infty} x_{n}=x^{*}limn→∞ξn=0⇒limn→∞xn=x∗. Thus, for a { x n } x n {x_(n)}\left\{x_{n}\right\}{xn} satisfying
lim n → ∞ δ n = lim n → ∞ ‖ x n + 1 − T x n ‖ = 0 lim n → ∞   δ n = lim n → ∞   x n + 1 − T x n = 0 lim_(n rarr oo)delta_(n)=lim_(n rarr oo)||x_(n+1)-Tx_(n)||=0\lim _{n \rightarrow \infty} \delta_{n}=\lim _{n \rightarrow \infty}\left\|x_{n+1}-T x_{n}\right\|=0limn→∞δn=limn→∞‖xn+1−Txn‖=0
we have shown that lim n → ∞ x n = x ∗ lim n → ∞   x n = x ∗ lim_(n rarr oo)x_(n)=x^(**)\lim _{n \rightarrow \infty} x_{n}=x^{*}limn→∞xn=x∗.
Conversely, we prove ( i i ) ⇒ ( i ) ( i i ) ⇒ ( i ) (ii)=>(i)(i i) \Rightarrow(i)(ii)⇒(i). Take lim n → ∞ ξ n = 0 lim n → ∞   ξ n = 0 lim_(n rarr oo)xi_(n)=0\lim _{n \rightarrow \infty} \xi_{n}=0limn→∞ξn=0. Observe that
δ n = ‖ p n + 1 − T p n ‖ ≤ ‖ p n + 1 − ( 1 − α n ) p n − α n T y n ‖ + ‖ ( 1 − α n ) p n + α n T y n − T p n ‖ ≤ ‖ p n + 1 − ( 1 − α n ) p n − α n T y n ‖ + α n ( ‖ p n ‖ + ‖ T y n ‖ ) + ‖ p n − T p n ‖ ≤ ε n + α n ( ‖ p n ‖ + M ) + ‖ p n − T p n ‖ → 0 δ n = p n + 1 − T p n ≤ p n + 1 − 1 − α n p n − α n T y n + 1 − α n p n + α n T y n − T p n ≤ p n + 1 − 1 − α n p n − α n T y n + α n p n + T y n + p n − T p n ≤ ε n + α n p n + M + p n − T p n → 0 {:[delta_(n)=||p_(n+1)-Tp_(n)||],[ <= ||p_(n+1)-(1-alpha_(n))p_(n)-alpha_(n)Ty_(n)||+||(1-alpha_(n))p_(n)+alpha_(n)Ty_(n)-Tp_(n)||],[ <= ||p_(n+1)-(1-alpha_(n))p_(n)-alpha_(n)Ty_(n)||+alpha_(n)(||p_(n)||+||Ty_(n)||)+||p_(n)-Tp_(n)||],[ <= epsi_(n)+alpha_(n)(||p_(n)||+M)+||p_(n)-Tp_(n)||],[ rarr0]:}\begin{aligned} & \delta_{n}=\left\|p_{n+1}-T p_{n}\right\| \\ & \leq\left\|p_{n+1}-\left(1-\alpha_{n}\right) p_{n}-\alpha_{n} T y_{n}\right\|+\left\|\left(1-\alpha_{n}\right) p_{n}+\alpha_{n} T y_{n}-T p_{n}\right\| \\ & \leq\left\|p_{n+1}-\left(1-\alpha_{n}\right) p_{n}-\alpha_{n} T y_{n}\right\|+\alpha_{n}\left(\left\|p_{n}\right\|+\left\|T y_{n}\right\|\right)+\left\|p_{n}-T p_{n}\right\| \\ & \leq \varepsilon_{n}+\alpha_{n}\left(\left\|p_{n}\right\|+M\right)+\left\|p_{n}-T p_{n}\right\| \\ & \rightarrow 0 \end{aligned}δn=‖pn+1−Tpn‖≤‖pn+1−(1−αn)pn−αnTyn‖+‖(1−αn)pn+αnTyn−Tpn‖≤‖pn+1−(1−αn)pn−αnTyn‖+αn(‖pn‖+‖Tyn‖)+‖pn−Tpn‖≤εn+αn(‖pn‖+M)+‖pn−Tpn‖→0
as n → ∞ n → ∞ n rarr oon \rightarrow \inftyn→∞. Note that lim n → ∞ ‖ p n − T p n ‖ = 0 lim n → ∞   p n − T p n = 0 lim_(n rarr oo)||p_(n)-Tp_(n)||=0\lim _{n \rightarrow \infty}\left\|p_{n}-T p_{n}\right\|=0limn→∞‖pn−Tpn‖=0 and using the boundedness of { T p n } T p n {Tp_(n)}\left\{T p_{n}\right\}{Tpn} we obtain the boundedness of { p n } p n {p_(n)}\left\{p_{n}\right\}{pn}. Condition (ii) assures that
lim n → ∞ δ n = 0 ⇒ lim n → ∞ x n = x ∗ lim n → ∞   δ n = 0 ⇒ lim n → ∞   x n = x ∗ lim_(n rarr oo)delta_(n)=0=>lim_(n rarr oo)x_(n)=x^(**)\lim _{n \rightarrow \infty} \delta_{n}=0 \Rightarrow \lim _{n \rightarrow \infty} x_{n}=x^{*}limn→∞δn=0⇒limn→∞xn=x∗
Thus, for a { p n } p n {p_(n)}\left\{p_{n}\right\}{pn} satisfying lim n → ∞ ξ n = lim n → ∞ ‖ p n + 1 − ( 1 − α n ) p n − α n T y n ‖ = 0 lim n → ∞   ξ n = lim n → ∞   p n + 1 − 1 − α n p n − α n T y n = 0 lim_(n rarr oo)xi_(n)=lim_(n rarr oo)||p_(n+1)-(1-alpha_(n))p_(n)-alpha_(n)Ty_(n)||=0\lim _{n \rightarrow \infty} \xi_{n}=\lim _{n \rightarrow \infty}\left\|p_{n+1}-\left(1-\alpha_{n}\right) p_{n}-\alpha_{n} T y_{n}\right\|=0limn→∞ξn=limn→∞‖pn+1−(1−αn)pn−αnTyn‖=0, we have shown that lim n → ∞ p n = x ∗ lim n → ∞   p n = x ∗ lim_(n rarr oo)p_(n)=x^(**)\lim _{n \rightarrow \infty} p_{n}=x^{*}limn→∞pn=x∗.
Theorems 1 and 2 lead to the following result.
COROLLARY 1. Let X X XXX be a normed space and T : X → X T : X → X T:X rarr XT: X \rightarrow XT:X→X a map with bounded range. If
lim n → ∞ ‖ p n − T p n ‖ = 0 , lim n → ∞ ‖ x n − T x n ‖ = 0 and lim n → ∞ ‖ u n − T u n ‖ = 0 , lim n → ∞   p n − T p n = 0 , lim n → ∞   x n − T x n = 0  and  lim n → ∞   u n − T u n = 0 , lim_(n rarr oo)||p_(n)-Tp_(n)||=0,lim_(n rarr oo)||x_(n)-Tx_(n)||=0" and "lim_(n rarr oo)||u_(n)-Tu_(n)||=0,\lim _{n \rightarrow \infty}\left\|p_{n}-T p_{n}\right\|=0, \lim _{n \rightarrow \infty}\left\|x_{n}-T x_{n}\right\|=0 \text { and } \lim _{n \rightarrow \infty}\left\|u_{n}-T u_{n}\right\|=0,limn→∞‖pn−Tpn‖=0,limn→∞‖xn−Txn‖=0 and limn→∞‖un−Tun‖=0,
then the following are equivalent:
(i) for all { α n } ⊂ ( 0 , 1 ) α n ⊂ ( 0 , 1 ) {alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1){αn}⊂(0,1) and { β n } ⊂ [ 0 , 1 ) β n ⊂ [ 0 , 1 ) {beta_(n)}sub[0,1)\left\{\beta_{n}\right\} \subset[0,1){βn}⊂[0,1), satisfying (4), the Ishikawa iteration is T T TTT-stable,
(ii) for all { α n } ⊂ ( 0 , 1 ) α n ⊂ ( 0 , 1 ) {alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1){αn}⊂(0,1), satisfying (4), the Mann iteration is T T TTT-stable,
(iii) the Picard iteration is T T TTT-stable.

3 Applications

The following example is from [2] and [4]. For sake of completeness we give here the whole proof.
EXAMPLE 1. Let T : [ 0 , 1 ] → [ 0 , 1 ] , T x = x T : [ 0 , 1 ] → [ 0 , 1 ] , T x = x T:[0,1]rarr[0,1],Tx=xT:[0,1] \rightarrow[0,1], T x=xT:[0,1]→[0,1],Tx=x.
  • [2] Picard iteration converges but is not T T TTT-stable. Then every point in ( 0,1 ] is a fixed point of T T TTT. Let b 0 b 0 b_(0)b_{0}b0 be a point in ( 0,1 ], then b n + 1 = T b n = T n b 0 = b 0 b n + 1 = T b n = T n b 0 = b 0 b_(n+1)=Tb_(n)=T^(n)b_(0)=b_(0)b_{n+1}=T b_{n}=T^{n} b_{0}=b_{0}bn+1=Tbn=Tnb0=b0. Thus lim n → ∞ b n = b 0 lim n → ∞   b n = b 0 lim_(n rarr oo)b_(n)=b_(0)\lim _{n \rightarrow \infty} b_{n}=b_{0}limn→∞bn=b0. Take p 0 = 0 p 0 = 0 p_(0)=0p_{0}=0p0=0 and p n = 1 n p n = 1 n p_(n)=(1)/(n)p_{n}=\frac{1}{n}pn=1n. Thus
δ n = | p n + 1 − T p n | = 1 n ( n + 1 ) → 0 δ n = p n + 1 − T p n = 1 n ( n + 1 ) → 0 delta_(n)=|p_(n+1)-Tp_(n)|=(1)/(n(n+1))rarr0\delta_{n}=\left|p_{n+1}-T p_{n}\right|=\frac{1}{n(n+1)} \rightarrow 0δn=|pn+1−Tpn|=1n(n+1)→0
but lim n → ∞ p n = 0 ≠ b 0 lim n → ∞   p n = 0 ≠ b 0 lim_(n rarr oo)p_(n)=0!=b_(0)\lim _{n \rightarrow \infty} p_{n}=0 \neq b_{0}limn→∞pn=0≠b0.
  • [4] Mann iteration converges but is not T T TTT-stable. Let e 0 e 0 e_(0)e_{0}e0 be a point in ( 0 , 1 ] ( 0 , 1 ] (0,1](0,1](0,1], then e n + 1 = ( 1 − α n ) e n + α n e n = e n = … = e 0 e n + 1 = 1 − α n e n + α n e n = e n = … = e 0 e_(n+1)=(1-alpha_(n))e_(n)+alpha_(n)e_(n)=e_(n)=dots=e_(0)e_{n+1}=\left(1-\alpha_{n}\right) e_{n}+\alpha_{n} e_{n}=e_{n}=\ldots=e_{0}en+1=(1−αn)en+αnen=en=…=e0. Take u 0 = e 0 , u n = 1 n + 1 u 0 = e 0 , u n = 1 n + 1 u_(0)=e_(0),u_(n)=(1)/(n+1)u_{0}=e_{0}, u_{n}=\frac{1}{n+1}u0=e0,un=1n+1 to obtain
ε n = | u n + 1 − ( 1 − α n ) u n − α n T u n | = | 1 n + 2 − ( 1 − α n ) 1 n + 1 − α n 1 n + 1 | = | 1 n + 2 − 1 n + 1 | = 1 ( n + 1 ) ( n + 2 ) → 0 ε n = u n + 1 − 1 − α n u n − α n T u n = 1 n + 2 − 1 − α n 1 n + 1 − α n 1 n + 1 = 1 n + 2 − 1 n + 1 = 1 ( n + 1 ) ( n + 2 ) → 0 {:[epsi_(n)=|u_(n+1)-(1-alpha_(n))u_(n)-alpha_(n)Tu_(n)|=|(1)/(n+2)-(1-alpha_(n))(1)/(n+1)-alpha_(n)(1)/(n+1)|],[=|(1)/(n+2)-(1)/(n+1)|=(1)/((n+1)(n+2))rarr0]:}\begin{aligned} \varepsilon_{n} & =\left|u_{n+1}-\left(1-\alpha_{n}\right) u_{n}-\alpha_{n} T u_{n}\right|=\left|\frac{1}{n+2}-\left(1-\alpha_{n}\right) \frac{1}{n+1}-\alpha_{n} \frac{1}{n+1}\right| \\ & =\left|\frac{1}{n+2}-\frac{1}{n+1}\right|=\frac{1}{(n+1)(n+2)} \rightarrow 0 \end{aligned}εn=|un+1−(1−αn)un−αnTun|=|1n+2−(1−αn)1n+1−αn1n+1|=|1n+2−1n+1|=1(n+1)(n+2)→0
but lim n → ∞ u n = 0 ≠ e 0 lim n → ∞   u n = 0 ≠ e 0 lim_(n rarr oo)u_(n)=0!=e_(0)\lim _{n \rightarrow \infty} u_{n}=0 \neq e_{0}limn→∞un=0≠e0.
EXAMPLE 2. Let T : [ 0 , ∞ ) → [ 0 , ∞ ) T : [ 0 , ∞ ) → [ 0 , ∞ ) T:[0,oo)rarr[0,oo)T:[0, \infty) \rightarrow[0, \infty)T:[0,∞)→[0,∞) be given by T x = x 3 T x = x 3 Tx=(x)/(3)T x=\frac{x}{3}Tx=x3. Then the Mann iteration converges to the fixed point of x ∗ = 0 x ∗ = 0 x^(**)=0x^{*}=0x∗=0 but is not T T TTT-stable, and applying Theorem 1, the Picard iteration is not T T TTT-stable while it converges.
(i) Mann iteration converges because the sequence e n → 0 e n → 0 e_(n)rarr0e_{n} \rightarrow 0en→0 as we can see:
e n + 1 = ( 1 − α n ) e n + α n e n 3 = ( 1 − 2 α n 3 ) e n = ∏ k = 1 n ( 1 − 2 α k 3 ) e 0 ≤ exp ⁡ ( − 2 3 ∑ k = 1 n α k ) → 0 e n + 1 = 1 − α n e n + α n e n 3 = 1 − 2 α n 3 e n = ∏ k = 1 n   1 − 2 α k 3 e 0 ≤ exp ⁡ − 2 3 ∑ k = 1 n   α k → 0 {:[e_(n+1)=(1-alpha_(n))e_(n)+alpha_(n)(e_(n))/(3)=(1-(2alpha_(n))/(3))e_(n)],[=prod_(k=1)^(n)(1-(2alpha_(k))/(3))e_(0) <= exp(-(2)/(3)sum_(k=1)^(n)alpha_(k))rarr0]:}\begin{aligned} e_{n+1} & =\left(1-\alpha_{n}\right) e_{n}+\alpha_{n} \frac{e_{n}}{3}=\left(1-\frac{2 \alpha_{n}}{3}\right) e_{n} \\ & =\prod_{k=1}^{n}\left(1-\frac{2 \alpha_{k}}{3}\right) e_{0} \leq \exp \left(-\frac{2}{3} \sum_{k=1}^{n} \alpha_{k}\right) \rightarrow 0 \end{aligned}en+1=(1−αn)en+αnen3=(1−2αn3)en=∏k=1n(1−2αk3)e0≤exp⁡(−23∑k=1nαk)→0
the last inequality is true because 1 − x ≤ exp ⁡ ( − x ) , ∀ x ≥ 0 1 − x ≤ exp ⁡ ( − x ) , ∀ x ≥ 0 1-x <= exp(-x),AA x >= 01-x \leq \exp (-x), \forall x \geq 01−x≤exp⁡(−x),∀x≥0, and ∑ α n = + ∞ ∑ α n = + ∞ sumalpha_(n)=+oo\sum \alpha_{n}=+\infty∑αn=+∞ supplied by (4).
(ii) Mann iteration is not T T TTT-stable. Take u n = n n + 1 u n = n n + 1 u_(n)=(n)/(n+1)u_{n}=\frac{n}{n+1}un=nn+1, note that u n → 1 ≠ x ∗ = 0 u n → 1 ≠ x ∗ = 0 u_(n)rarr1!=x^(**)=0u_{n} \rightarrow 1 \neq x^{*}=0un→1≠x∗=0, and ε n = ‖ u n + 1 − ( 1 − α n ) u n − α n T u n ‖ → 0 ε n = u n + 1 − 1 − α n u n − α n T u n → 0 epsi_(n)=||u_(n+1)-(1-alpha_(n))u_(n)-alpha_(n)Tu_(n)||rarr0\varepsilon_{n}=\left\|u_{n+1}-\left(1-\alpha_{n}\right) u_{n}-\alpha_{n} T u_{n}\right\| \rightarrow 0εn=‖un+1−(1−αn)un−αnTun‖→0 because
ε n = | n + 1 n + 2 − ( 1 − α n ) n n + 1 − α n n 3 ( n + 1 ) | = 3 + 2 α n n 2 + 4 α n n 3 ( n + 1 ) ( n + 2 ) ε n = n + 1 n + 2 − 1 − α n n n + 1 − α n n 3 ( n + 1 ) = 3 + 2 α n n 2 + 4 α n n 3 ( n + 1 ) ( n + 2 ) {:[epsi_(n)=|(n+1)/(n+2)-(1-alpha_(n))(n)/(n+1)-alpha_(n)(n)/(3(n+1))|],[=(3+2alpha_(n)n^(2)+4alpha_(n)n)/(3(n+1)(n+2))]:}\begin{aligned} \varepsilon_{n} & =\left|\frac{n+1}{n+2}-\left(1-\alpha_{n}\right) \frac{n}{n+1}-\alpha_{n} \frac{n}{3(n+1)}\right| \\ & =\frac{3+2 \alpha_{n} n^{2}+4 \alpha_{n} n}{3(n+1)(n+2)} \end{aligned}εn=|n+1n+2−(1−αn)nn+1−αnn3(n+1)|=3+2αnn2+4αnn3(n+1)(n+2)
(iii) Picard iteration converges to fixed point x ∗ = 0 x ∗ = 0 x^(**)=0x^{*}=0x∗=0, because b n + 1 = T b n = T n b 0 = b 0 3 n → 0 b n + 1 = T b n = T n b 0 = b 0 3 n → 0 b_(n+1)=Tb_(n)=T^(n)b_(0)=(b_(0))/(3^(n))rarr0b_{n+1}=T b_{n}=T^{n} b_{0}= \frac{b_{0}}{3^{n}} \rightarrow 0bn+1=Tbn=Tnb0=b03n→0.
REMARK. Take again T : [ 0 , ∞ ) → [ 0 , ∞ ) , T x = x 3 T : [ 0 , ∞ ) → [ 0 , ∞ ) , T x = x 3 T:[0,oo)rarr[0,oo),Tx=(x)/(3)T:[0, \infty) \rightarrow[0, \infty), T x=\frac{x}{3}T:[0,∞)→[0,∞),Tx=x3, and x n = n n + 1 x n = n n + 1 x_(n)=(n)/(n+1)x_{n}=\frac{n}{n+1}xn=nn+1 to note that lim n → ∞ ξ n = 0 lim n → ∞   ξ n = 0 lim_(n rarr oo)xi_(n)=0\lim _{n \rightarrow \infty} \xi_{n}=0limn→∞ξn=0 and lim n → ∞ x n = 1 ≠ x ∗ = 0 lim n → ∞   x n = 1 ≠ x ∗ = 0 lim_(n rarr oo)x_(n)=1!=x^(**)=0\lim _{n \rightarrow \infty} x_{n}=1 \neq x^{*}=0limn→∞xn=1≠x∗=0, and to conclude that Ishikawa iteration is not T T TTT-stable. Remark (analogously to Mann iteration, see also [5]) that it converges while T T TTT is a contraction.

References

[1] S. Ishikawa, Fixed points by a new iteration method, Proc. Amer. Math. Soc., 44(1974), 147-150.
[2] A. M. Harder and T. Hicks, Stability results for fixed point iteration procedures, Math. Japonica, 33 (1988), 693-706.
[3] W. R. Mann, Mean value in iteration, Proc. Amer. Math. Soc., 4 (1953), 506-510.
[4] M. O. Osilike, Stability of the Mann and Ishikawa iteration procedures for ϕ ϕ phi\phiϕ-strong pseudocontractions and nonlinear equations of the ϕ ϕ phi\phiϕ-strongly accretive type, J. Math. Anal. Appl., 227 (1998), 319-334.
[5] B. E. Rhoades and Ş. M. Şoltuz, On the equivalence of Mann and Ishikawa iteration methods, Int. J. Math. Math. Sci., 2003(2003), 451-459.
[6] B. E. Rhoades and Ş. M. Şoltuz, The equivalence between the T T TTT-stabilities of Mann and Ishikawa iterations, J. Math. Anal. Appl., 318(2006), 472-475.
  1. *Mathematics Subject Classifications: 47H10
    † † ^(†){ }^{\dagger}† Institute of Numerical Analysis of Romanian Academy, P.O. Box 68-1, Cluj-Napoca, Romania, and Departamento de Matematicas, Universidad de los Andes, Carrera 1 No. 18A-10, Bogota, Colombia.
2008

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