Let $X$$X$XX$X$ be a normed space and $T$$T$TT$T$ a selfmap of $X$$X$XX$X$. Let $x_0$$x_0$x_(0)x_{0}$x_0$ be a point of $X$$X$XX$X$, and assume that $x_{n+1}=f(T,x_n)$$x_{n+1}=f\left(T,x_n\right)$x_(n+1)=f(T,x_(n))x_{n+1}=f\left(T, x_{n}\right)$x_{n+1}=f(T,x_n)$ is an iteration procedure, involving $T$$T$TT$T$, which yields a sequence $\left\{x_n\right\}$$\left\{x_n\right\}${x_(n)}\left\{x_{n}\right\}$\left\{x_n\right\}$ of points from $X$$X$XX$X$. Suppose $\left\{x_n\right\}$$\left\{x_n\right\}${x_(n)}\left\{x_{n}\right\}$\left\{x_n\right\}$ converges to a fixed point $x^{\ast }$$x^{\ast }$x^(**)x^{*}$x^{\ast }$ of $T$$T$TT$T$. Let $\left\{{\xi }_n\right\}$$\left\{{\xi }_n\right\}${xi_(n)}\left\{\xi_{n}\right\}$\left\{{\xi }_n\right\}$ be an arbitrary sequence in $X$$X$XX$X$, and set ${ϵ}_n=\Vert {\xi }_{n+1}-f(T,{\xi }_n)\Vert$${ϵ}_n=‖{\xi }_{n+1}-f\left(T,{\xi }_n\right)‖$epsilon_(n)=||xi_(n+1)-f(T,xi_(n))||\epsilon_{n}=\left\|\xi_{n+1}-f\left(T, \xi_{n}\right)\right\|${ϵ}_n=\Vert {\xi }_{n+1}-f(T,{\xi }_n)\Vert$ for all $n\in \mathbb{N}$$n\in \mathbb{N}$n inNn \in \mathbb{N}$n\in \mathbb{N}$.

Definition 1.1 [1]. If $(\underset{n\to \mathrm{\infty }}{lim}{ϵ}_n=0)\Rightarrow (\underset{n\to \mathrm{\infty }}{lim}{\xi }_n=p)$$\left(\underset{n\to \mathrm{\infty }}{lim} {ϵ}_n=0\right)\Rightarrow \left(\underset{n\to \mathrm{\infty }}{lim} {\xi }_n=p\right)$(lim_(n rarr oo)epsilon_(n)=0)=>(lim_(n rarr oo)xi_(n)=p)\left(\lim _{n \rightarrow \infty} \epsilon_{n}=0\right) \Rightarrow\left(\lim _{n \rightarrow \infty} \xi_{n}=p\right)$(\underset{n\to \mathrm{\infty }}{lim}{ϵ}_n=0)\Rightarrow (\underset{n\to \mathrm{\infty }}{lim}{\xi }_n=p)$, then the iteration procedure $x_{n+1}=f(T,x_n)$$x_{n+1}=f\left(T,x_n\right)$x_(n+1)=f(T,x_(n))x_{n+1}=f\left(T, x_{n}\right)$x_{n+1}=f(T,x_n)$ is said to be $T$$T$TT$T$-stable with respect to $T$$T$TT$T$.

Remark 1.2 [1]. In practice, such a sequence $\left\{{\xi }_n\right\}$$\left\{{\xi }_n\right\}${xi_(n)}\left\{\xi_{n}\right\}$\left\{{\xi }_n\right\}$ could arise in the following way. Let $x_0$$x_0$x_(0)x_{0}$x_0$ be a point in $X$$X$XX$X$. Set $x_{n+1}=f(T,x_n)$$x_{n+1}=f\left(T,x_n\right)$x_(n+1)=f(T,x_(n))x_{n+1}=f\left(T, x_{n}\right)$x_{n+1}=f(T,x_n)$. Let ${\xi }_0=x_0$${\xi }_0=x_0$xi_(0)=x_(0)\xi_{0}=x_{0}${\xi }_0=x_0$. Now $x_1=f(T,x_0)$$x_1=f\left(T,x_0\right)$x_(1)=f(T,x_(0))x_{1}=f\left(T, x_{0}\right)$x_1=f(T,x_0)$. Because of rounding or discretization in the function $T$$T$TT$T$, a new value ${\xi }_1$${\xi }_1$xi_(1)\xi_{1}${\xi }_1$ approximately equal to $x_1$$x_1$x_(1)x_{1}$x_1$ might be obtained instead of the true value of $f(T,x_0)$$f\left(T,x_0\right)$f(T,x_(0))f\left(T, x_{0}\right)$f(T,x_0)$. Then to approximate $x_2$$x_2$x_(2)x_{2}$x_2$, the value $f(T,{\xi }_1)$$f\left(T,{\xi }_1\right)$f(T,xi_(1))f\left(T, \xi_{1}\right)$f(T,{\xi }_1)$ is computed to yield ${\xi }_2$${\xi }_2$xi_(2)\xi_{2}${\xi }_2$, an approximation of $f(T,{\xi }_1)$$f\left(T,{\xi }_1\right)$f(T,xi_(1))f\left(T, \xi_{1}\right)$f(T,{\xi }_1)$. This computation is continued to obtain $\left\{{\xi }_n\right\}$$\left\{{\xi }_n\right\}${xi_(n)}\left\{\xi_{n}\right\}$\left\{{\xi }_n\right\}$ an approximate sequence of $\left\{x_n\right\}$$\left\{x_n\right\}${x_(n)}\left\{x_{n}\right\}$\left\{x_n\right\}$.

Let $X$$X$XX$X$ be a normed space, $D$$D$DD$D$ a nonempty, convex subset of $X$$X$XX$X$, and $T$$T$TT$T$ a selfmap of $D$$D$DD$D$, let $p_0=e_0\in D$$p_0=e_0\in D$p_(0)=e_(0)in Dp_{0}=e_{0} \in D$p_0=e_0\in D$. The Mann iteration (see [2]) is defined by

where $\left\{{\alpha }_n\right\}\subset (0,1)$$\left\{{\alpha }_n\right\}\subset (0,1)${alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1)$\left\{{\alpha }_n\right\}\subset (0,1)$. The Ishikawa iteration is defined (see [3]) by

where $\left\{{\alpha }_n\right\}\subset (0,1),\left\{{\beta }_n\right\}\subset [0,1)$$\left\{{\alpha }_n\right\}\subset (0,1),\left\{{\beta }_n\right\}\subset [0,1)${alpha_(n)}sub(0,1),{beta_(n)}sub[0,1)\left\{\alpha_{n}\right\} \subset(0,1),\left\{\beta_{n}\right\} \subset[0,1)$\left\{{\alpha }_n\right\}\subset (0,1),\left\{{\beta }_n\right\}\subset [0,1)$. The Krasnoselskij iteration (see [4]) is defined by

where $\lambda \in (0,1)$$\lambda \in (0,1)$lambda in(0,1)\lambda \in(0,1)$\lambda \in (0,1)$. Recently, the equivalence between the $T$$T$TT$T$-stabilities of Mann and Ishikawa iterations, respectively, for modified Mann-Ishikawa iterations was shown in [5]. In the present paper, we shall prove the equivalence between the $T$$T$TT$T$-stabilities of the Krasnoselskij and the Mann iterations. Next, $\left\{u_n\right\},\left\{v_n\right\}\subset X$$\left\{u_n\right\},\left\{v_n\right\}\subset X${u_(n)},{v_(n)}sub X\left\{u_{n}\right\},\left\{v_{n}\right\} \subset X$\left\{u_n\right\},\left\{v_n\right\}\subset X$ are arbitrary.

(i) The Mann iteration (1.1) is said to be $T$$T$TT$T$-stable if and only if for all $\left\{{\alpha }_n\right\}\subset (0,1)$$\left\{{\alpha }_n\right\}\subset (0,1)${alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1)$\left\{{\alpha }_n\right\}\subset (0,1)$ and for every sequence $\left\{u_n\right\}\subset X$$\left\{u_n\right\}\subset X${u_(n)}sub X\left\{u_{n}\right\} \subset X$\left\{u_n\right\}\subset X$,

where ${\epsilon }_n:=\Vert u_{n+1}-(1-{\alpha }_n)u_n-{\alpha }_{n}T{u}_n\Vert$${\epsilon }_n:=‖u_{n+1}-\left(1-{\alpha }_n\right)u_n-{\alpha }_{n}T{u}_n‖$epsi_(n):=||u_(n+1)-(1-alpha_(n))u_(n)-alpha_(n)Tu_(n)||\varepsilon_{n}:=\left\|u_{n+1}-\left(1-\alpha_{n}\right) u_{n}-\alpha_{n} T u_{n}\right\|${\epsilon }_n:=\Vert u_{n+1}-(1-{\alpha }_n)u_n-{\alpha }_{n}T{u}_n\Vert$.
(ii) The Krasnoselskij iteration (1.3) is said to be $T$$T$TT$T$-stable if and only if for all $\lambda \in (0,1)$$\lambda \in (0,1)$lambda in(0,1)\lambda \in (0,1)$\lambda \in (0,1)$, and for every sequence $\left\{v_n\right\}\subset X$$\left\{v_n\right\}\subset X${v_(n)}sub X\left\{v_{n}\right\} \subset X$\left\{v_n\right\}\subset X$,

where ${\delta }_n:=\Vert v_{n+1}-(1-\lambda )v_n-\lambda T{v}_n\Vert$${\delta }_n:=‖v_{n+1}-(1-\lambda )v_n-\lambda T{v}_n‖$delta_(n):=||v_(n+1)-(1-lambda)v_(n)-lambda Tv_(n)||\delta_{n}:=\left\|v_{n+1}-(1-\lambda) v_{n}-\lambda T v_{n}\right\|${\delta }_n:=\Vert v_{n+1}-(1-\lambda )v_n-\lambda T{v}_n\Vert$.

Theorem 2.1. Let $X$$X$XX$X$ be a normed space and $T:X\to X$$T:X\to X$T:X rarr XT: X \rightarrow X$T:X\to X$ a map with bounded range and $\left\{{\alpha }_n\right\}\subset (0,1)$$\left\{{\alpha }_n\right\}\subset (0,1)${alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1)$\left\{{\alpha }_n\right\}\subset (0,1)$ satisfy $\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=\lambda ,\lambda \in (0,1)$$\underset{n\to \mathrm{\infty }}{lim} {\alpha }_n=\lambda ,\lambda \in (0,1)$lim_(n rarr oo)alpha_(n)=lambda,lambda in(0,1)\lim _{n \rightarrow \infty} \alpha_{n}=\lambda, \lambda \in(0,1)$\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=\lambda ,\lambda \in (0,1)$. Then the following are equivalent:
(i) the Mann iteration is $T$$T$TT$T$-stable,
(ii) the Krasnoselskij iteration is $T$$T$TT$T$-stable.

Proof. We prove that (i) ⇒ (ii). If $\underset{n\to \mathrm{\infty }}{lim}{\delta }_n=0$$\underset{n\to \mathrm{\infty }}{lim} {\delta }_n=0$lim_(n rarr oo)delta_(n)=0\lim _{n \rightarrow \infty} \delta_{n}=0$\underset{n\to \mathrm{\infty }}{lim}{\delta }_n=0$, then $\left\{v_n\right\}$$\left\{v_n\right\}${v_(n)}\left\{v_{n}\right\}$\left\{v_n\right\}$ is bounded. Set

Observe that $\Vert v_1\Vert \le {\delta }_0+(1-\lambda )\Vert v_0\Vert +\lambda \Vert T{v}_0\Vert \le {\delta }_0+M_1$$‖v_1‖\le {\delta }_0+(1-\lambda )‖v_0‖+\lambda ‖T{v}_0‖\le {\delta }_0+M_1$||v_(1)|| <= delta_(0)+(1-lambda)||v_(0)||+lambda||Tv_(0)|| <= delta_(0)+M_(1)\left\|v_{1}\right\| \leq \delta_{0}+(1-\lambda)\left\|v_{0}\right\|+\lambda\left\|T v_{0}\right\| \leq \delta_{0}+M_{1}$\Vert v_1\Vert \le {\delta }_0+(1-\lambda )\Vert v_0\Vert +\lambda \Vert T{v}_0\Vert \le {\delta }_0+M_1$. Set $M:=M_1+1/\lambda$$M:=M_1+1/\lambda$M:=M_(1)+1//lambdaM:=M_{1}+1 / \lambda$M:=M_1+1/\lambda$. Suppose that $\Vert v_n\Vert \le M$$‖v_n‖\le M$||v_(n)|| <= M\left\|v_{n}\right\| \leq M$\Vert v_n\Vert \le M$ to prove that $\Vert v_{n+1}\Vert \le M$$‖v_{n+1}‖\le M$||v_(n+1)|| <= M\left\|v_{n+1}\right\| \leq M$\Vert v_{n+1}\Vert \le M$. Remark that

Suppose that $\underset{n\to \mathrm{\infty }}{lim}{\delta }_n=0$$\underset{n\to \mathrm{\infty }}{lim} {\delta }_n=0$lim_(n rarr oo)delta_(n)=0\lim _{n \rightarrow \infty} \delta_{n}=0$\underset{n\to \mathrm{\infty }}{lim}{\delta }_n=0$ to note that

Condition (i) assures that if $\underset{n\to \mathrm{\infty }}{lim}{\epsilon }_n=0$$\underset{n\to \mathrm{\infty }}{lim} {\epsilon }_n=0$lim_(n rarr oo)epsi_(n)=0\lim _{n \rightarrow \infty} \varepsilon_{n}=0$\underset{n\to \mathrm{\infty }}{lim}{\epsilon }_n=0$, then $\underset{n\to \mathrm{\infty }}{lim}{v}_n=x^{\ast }$$\underset{n\to \mathrm{\infty }}{lim} v_n=x^{\ast }$lim_(n rarr oo)v_(n)=x^(**)\lim _{n \rightarrow \infty} v_{n}=x^{*}$\underset{n\to \mathrm{\infty }}{lim}{v}_n=x^{\ast }$. Thus, for a $\left\{v_n\right\}$$\left\{v_n\right\}${v_(n)}\left\{v_{n}\right\}$\left\{v_n\right\}$ satisfying

we have shown that $\underset{n\to \mathrm{\infty }}{lim}{v}_n=x^{\ast }$$\underset{n\to \mathrm{\infty }}{lim} v_n=x^{\ast }$lim_(n rarr oo)v_(n)=x^(**)\lim _{n \rightarrow \infty} v_{n}=x^{*}$\underset{n\to \mathrm{\infty }}{lim}{v}_n=x^{\ast }$.
Conversely, we prove (ii) ⇒ (i). First, we prove that $\left\{u_n\right\}$$\left\{u_n\right\}${u_(n)}\left\{u_{n}\right\}$\left\{u_n\right\}$ is bounded. Since $\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=\lambda$$\underset{n\to \mathrm{\infty }}{lim} {\alpha }_n=\lambda$lim_(n rarr oo)alpha_(n)=lambda\lim _{n \rightarrow \infty} \alpha_{n}= \lambda$\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=\lambda$, for $\beta \in (0,1)$$\beta \in (0,1)$beta in(0,1)\beta \in(0,1)$\beta \in (0,1)$ given, there exists $n_0\in N$$n_0\in N$n_(0)in Nn_{0} \in N$n_0\in N$, such that $1-{\alpha }_n\le \beta$$1-{\alpha }_n\le \beta$1-alpha_(n) <= beta1-\alpha_{n} \leq \beta$1-{\alpha }_n\le \beta$, for all $n\ge n_0$$n\ge n_0$n >= n_(0)n \geq n_{0}$n\ge n_0$. Set $M_1:=max\{\underset{x\in X}{sup}\Vert Tx\Vert ,\Vert u_0\Vert \}$$M_1:=max\left\{\underset{x\in X}{sup} \Vert Tx\Vert ,‖u_0‖\right\}$M_(1):=max{s u p_(x in X)||Tx||,||u_(0)||}M_{1}:= \max \left\{\sup _{x \in X}\|T x\|,\left\|u_{0}\right\|\right\}$M_1:=max\{\underset{x\in X}{sup}\Vert Tx\Vert ,\Vert u_0\Vert \}$ and $M:=n_0+1+\beta /(1-\beta )+M_1$$M:=n_0+1+\beta /(1-\beta )+M_1$M:=n_(0)+1+beta//(1-beta)+M_(1)M:=n_{0}+1+\beta /(1-\beta)+M_{1}$M:=n_0+1+\beta /(1-\beta )+M_1$ to obtain

Suppose $\underset{n\to \mathrm{\infty }}{lim}{\mathcal{E}}_n=0$$\underset{n\to \mathrm{\infty }}{lim} {\mathcal{E}}_n=0$lim_(n rarr oo)E_(n)=0\lim _{n \rightarrow \infty} \mathcal{E}_{n}=0$\underset{n\to \mathrm{\infty }}{lim}{\mathcal{E}}_n=0$. Observe that

Condition (ii) assures that if $\underset{n\to \mathrm{\infty }}{lim}{\delta }_n=0$$\underset{n\to \mathrm{\infty }}{lim} {\delta }_n=0$lim_(n rarr oo)delta_(n)=0\lim _{n \rightarrow \infty} \delta_{n}=0$\underset{n\to \mathrm{\infty }}{lim}{\delta }_n=0$, then $\underset{n\to \mathrm{\infty }}{lim}{v}_n=x^{\ast }$$\underset{n\to \mathrm{\infty }}{lim} v_n=x^{\ast }$lim_(n rarr oo)v_(n)=x^(**)\lim _{n \rightarrow \infty} v_{n}=x^{*}$\underset{n\to \mathrm{\infty }}{lim}{v}_n=x^{\ast }$. Thus, for a $\left\{u_n\right\}$$\left\{u_n\right\}${u_(n)}\left\{u_{n}\right\}$\left\{u_n\right\}$ satisfying

we have shown that $\underset{n\to \mathrm{\infty }}{lim}{u}_n=x^{\ast }$$\underset{n\to \mathrm{\infty }}{lim} u_n=x^{\ast }$lim_(n rarr oo)u_(n)=x^(**)\lim _{n \rightarrow \infty} u_{n}=x^{*}$\underset{n\to \mathrm{\infty }}{lim}{u}_n=x^{\ast }$.
Remark 2.2. Let $X$$X$XX$X$ be a normed space and $T:X\to X$$T:X\to X$T:X rarr XT: X \rightarrow X$T:X\to X$ a map with bounded range and $\left\{{\alpha }_n\right\}\subset (0,1)$$\left\{{\alpha }_n\right\}\subset (0,1)${alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1)$\left\{{\alpha }_n\right\}\subset (0,1)$ satisfy $\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=\lambda ,\lambda \in (0,1)$$\underset{n\to \mathrm{\infty }}{lim} {\alpha }_n=\lambda ,\lambda \in (0,1)$lim_(n rarr oo)alpha_(n)=lambda,lambda in(0,1)\lim _{n \rightarrow \infty} \alpha_{n}=\lambda, \lambda \in(0,1)$\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=\lambda ,\lambda \in (0,1)$. If the Mann iteration is not $T$$T$TT$T$-stable, then the Krasnoselskij iteration is not $T$$T$TT$T$-stable, and conversely.

Example 2.3. Let $T:[0,1)\to [0,1)$$T:[0,1)\to [0,1)$T:[0,1)rarr[0,1)T:[0,1) \rightarrow[0,1)$T:[0,1)\to [0,1)$ be given by $Tx=x^2$$Tx=x^2$Tx=x^(2)T x=x^{2}$Tx=x^2$, and $\lambda =1/2$$\lambda =1/2$lambda=1//2\lambda=1 / 2$\lambda =1/2$. Then the Krasnoselskij iteration converges to the unique fixed point $x^{\ast }=0$$x^{\ast }=0$x^(**)=0x^{*}=0$x^{\ast }=0$, and it is not $T$$T$TT$T$-stable.

The Krasnoselskij iteration converges because, supposing $F:=\underset{n}{sup}{p}_n<1$$F:=\underset{n}{sup} p_n<1$F:=s u p_(n)p_(n) < 1F:=\sup _{n} p_{n}<1$F:=\underset{n}{sup}{p}_n<1$, the sequence $p_n\to 0$$p_n\to 0$p_(n)rarr0p_{n} \rightarrow 0$p_n\to 0$, as we can see from