Let $X$ be a real Banach space, $D$ a nonempty, convex subset of $X$, and $T$ a selfmap of $D$, let $x_{0}=u_{0}\in D$. The Mann iteration, (see [5]), is defined by

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

where $\lambda\in(0,1)$.
Definition 1. [7] The operator $T:X\rightarrow X$ satisfies condition $Z$ (or is a quasicontraction) if and only if there exist real numbers $a,b,c$ satisfying $0<a<1,0<b,c<1/2$ such that for each pair $x,y$ in $X$, at least one condition is true

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  $\left(z_{1}\right)\|Tx-Ty\|\leq a\|x-y\|$,

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  $\left(z_{2}\right)\|Tx-Ty\|\leq b(\|x-Tx\|+\|y-Ty\|)$,

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  $\left(z_{3}\right)\|Tx-Ty\|\leq c(\|x-Ty\|+\|y-Tx\|)$.

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It is known, see Rhoades [8], that $\left(z_{1}\right),\left(z_{2}\right)$ and $\left(z_{3}\right)$ are independent conditions. Note that a map satisfying condition $Z$ is independent, see Rhoades [7], of the class of strongly pseudocontractive maps.

In [ $9,?$ ] the following conjecture was given: "if the Mann iteration converges, then so does the Ishikawa iteration". In a series of papers [9], [10], [11], [12], [13], Professor B. E. Rhoades and the author, we have given a positive answer to this Conjecture, showing the equivalence between Mann and Ishikawa iterations for strongly and uniformly pseudocontractive maps.

In [2], the following open question was given: "are Krasnoselskij iteration and Mann iteration equivalent (in the sense of [9]) for enough large classes of mappings?" We shall give a positive answer to this question: if Krasnoselskij iteration converges, then Mann (and the corresponding Ishikawa iteration) also converges and conversely, dealing with maps satisfying condition $Z$. Note that Professor B. E. Rhoades and the author have already given a positive answer in [15] for the class of pseudocontractive maps.

Lemma 1 [[18]]. Let $\left\{a_{n}\right\}$ be a nonnegative sequence which satisfies the following inequality

where $\lambda_{n}\in(0,1),\forall n\geq n_{0},\sum_{n=1}^{\infty}\lambda_{n}=\infty$, and $\sigma_{n}=o\left(\lambda_{n}\right)$. Then $\lim_{n\rightarrow\infty}a_{n}=0$.

Let $F(T)$ denote the fixed point set with respect to $D$ for the map $T$. Suppose that $x^{*}\in F(T)$.

Theorem 1. Let $X$ be a normed space, $D$ a nonempty, convex, closed subset of $X$ and $T:D\rightarrow D$ an operator satisfying condition $Z$. If $u_{0}=x_{0}\in D$, then the following are true: if the Mann iteration (1) converges to $x^{*}$, then the Krasnoselskij iteration (2) converges to $x^{*}$. Conversely, if the Krasnoselskij iteration (2) converges to $x^{*}$, then the Mann iteration (1) converges to $x^{*}$, provided that $\alpha_{n}\geq A>0,\forall n\in\mathbb{N}$.

Proof. Consider $x,y\in D$. Since $T$ satisfies condition $Z$, at least one of the conditions from $\left(z_{1}\right),\left(z_{2}\right)$ and $\left(z_{3}\right)$ is satisfied. If $\left(z_{2}\right)$ holds, then

thus

From $0\leq b<1$ one obtains,

If ( $z_{3}$ ) holds, then one gets,

hence,

Denote

to obtain

Finally, we get

Formula (4) was obtained as in [1].
We will prove the implication $(i)\Rightarrow(ii)$. Use (1) (2) and (4) with

to obtain

Denote

Since $\lim_{n\rightarrow\infty}\left\|u_{n}-x^{*}\right\|=0,T$ satisfies condition $Z$, and $x^{*}\in F(T)$, from (4) one has

Hence $\lim_{n\rightarrow\infty}\left\|u_{n}-Tu_{n}\right\|=0$; that is $\sigma_{n}=o\left(\lambda_{n}\right)$. Lemma 1 leads to $\lim_{n\rightarrow\infty}\left\|u_{n}-x_{n}\right\|=$ 0 . Use

to deduce

We will prove $(ii)\Rightarrow(i)$. That is, if Krasnoselskij iteration converges, then Mann iteration does converge. Use (4) with

to obtain

Denote

Since $\lim_{n\rightarrow\infty}\left\|x_{n}-x^{*}\right\|=0,T$ satisfies condition $Z$, and $x^{*}\in F(T)$, from (4) one has,

Hence $\lim_{n\rightarrow\infty}\left\|x_{n}-Tx_{n}\right\|=0$, that is $\sigma_{n}=o\left(\lambda_{n}\right)$. Lemma 1 leads to $\lim_{n\rightarrow\infty}\left\|x_{n}-u_{n}\right\|=$ 0 . Thus,

The Ishikawa iteration is defined (see [3]) by

where $\left\{\alpha_{n}\right\}\subset(0,1),\left\{\beta_{n}\right\}\subset[0,1)$.
The following result is from [17].
Theorem 2 [[17]]. Let $X$ be a normed space, $D$ a nonempty, convex, closed subset of $X$ and $T:D\rightarrow D$ an operator satisfying condition $Z$. If $u_{0}=x_{0}\in D$, then the following are equivalent:
(i) the Mann iteration (1) converges to $x^{*}$,
(ii) the Ishikawa iteration (5) converges to $x^{*}$.

Theorems 1 and 2 lead to the following corollary.
Corollary 1. Let $X$ be a normed space, $D$ a nonempty, convex, closed subset of $X$ and $T:D\rightarrow D$ an operator satisfying condition $Z$. If $u_{0}=x_{0}\in D,\alpha_{n}\geq A>0,\forall n\in\mathbb{N}$, then the following are equivalent:
(i) the Mann iteration (1) converges to $x^{*}$,
(ii) the Ishikawa iteration (5) converges to $x^{*}$.
(iii) the Krasnoselskij iteration (2) converges to $x^{*}$.

For $v_{1}\in D$, Noor introduced in [6] the following three-step procedure,

The multi-step procedure of arbitrary fixed order $p\geq 2$, see [14], is defined by

where $\left\{\alpha_{n}\right\}\subset(0,1),\left\{\beta_{n}^{i}\right\}\subset[0,1),1\leq i\leq p-1$.
We shall generalize the above Theorem 2, see also [17], by proving that (7) and (1) are equivalent.

Theorem 3. Let $X$ be a normed space, $D$ a nonempty, convex, closed subset of $X$ and $T:D\rightarrow D$ an operator satisfying condition $Z$. If $u_{0}=x_{0}\in D$, then the following are equivalent:
(i) the Mann iteration (1) converges to $x^{*}$,
(ii) the iteration (7) converges to $x^{*}$.

Proof. We shall use (4) :

We will prove the implication $(i)\Rightarrow(ii)$. Suppose that $\lim_{n\rightarrow\infty}u_{n}=x^{*}$. Using $\lim_{n\rightarrow\infty}\left\|x_{n}-u_{n}\right\|=0$, and $0\leq\left\|x^{*}-x_{n}\right\|\leq\left\|u_{n}-x^{*}\right\|+\left\|x_{n}-u_{n}\right\|$ we get

Using now (1) (7) and (4) with

we have

Using (4) with $x:=u_{n},y:=y_{n}^{1}$, we have

Relations (8) and (9) lead to

Denote by

Since $\lim_{n\rightarrow\infty}\left\|u_{n}-x^{*}\right\|=0,T$ satisfies condition $Z$, and $x^{*}\in F(T)$, from (4) we obtain

Hence $\lim_{n\rightarrow\infty}\left\|u_{n}-Tu_{n}\right\|=0$; that is $\sigma_{n}=o\left(\lambda_{n}\right)$. Lemma 1 leads to $\lim_{n\rightarrow\infty}\left\|u_{n}-x_{n}\right\|=$ 0 .

We will prove now that if multistep iteration converges then Mann iteration does. Using (4) with

we obtain

The following relation holds

Substituting (12) in (11), we obtain

Denote by

Since $\lim_{n\rightarrow\infty}\left\|x_{n}-x^{*}\right\|=0,T$ satisfies condition $Z$, and $x^{*}\in F(T)$, from (4) we obtain

Note that $\beta_{n}^{i}\in[0,1),\forall n\geq 1,1\leq i\leq p-1$, and use (4) to obtain

Hence $\lim_{n\rightarrow\infty}\left\|x_{n}-Tx_{n}\right\|=0$ and $\lim_{n\rightarrow\infty}\left\|y_{n}^{1}-Ty_{n}^{1}\right\|=0$ that is $\sigma_{n}=o\left(\lambda_{n}\right)$. Lemma 1 and (13) lead to $\lim_{n\rightarrow\infty}\left\|x_{n}-u_{n}\right\|=0$. Thus, we get $\left\|x^{*}-u_{n}\right\|\leq\left\|x_{n}-u_{n}\right\|+\left\|x_{n}-x^{*}\right\|\rightarrow 0$.