Mann-Ishikawa iterations and Mann-Ishikawa iterations with errors are equivalent models

Ştefan M. Şoltuz

Abstract

Mann-Ishikawa iterations and Mann-Ishikawa iterations with errors are equivalent models for several classes of operators.

Key words: Mann-Ishikawa iterations, Mann-Ishikawa iterations with errors

AMS subject classifications: 47 H 10
Received February 14, 2003
Accepted March 19, 2003

1. Preliminaries

Introduced in [4], Mann iteration is a viable method to approximate the fixed point of an operator, when Banach principle is not functional. Let $X$ be a Banach space, let $T:X\rightarrow X$ be a map. Let $x_{1}\in X$ . Mann iteration is given by:

x_{n+1}=\left(1-\alpha_{n}\right)x_{n}+\alpha_{n}Tx_{n}.

(1)

The sequence $\left(\alpha_{n}\right)_{n}\subset(0,1)$ is convergent, such that $\lim_{n\rightarrow\infty}\alpha_{n}=0$ , and $\sum_{n=1}^{\infty}\alpha_{n}=\infty$ . Ishikawa introduced later in [2] the following iteration,

	$\displaystyle x_{n+1}$	$\displaystyle=\left(1-\alpha_{n}\right)x_{n}+\alpha_{n}Ty_{n},$		(2)
	$\displaystyle y_{n}$	$\displaystyle=\left(1-\beta_{n}\right)x_{n}+\beta_{n}Tx_{n},\quad n=1,2,\ldots$

Sequences $\left(\alpha_{n}\right)_{n},\left(\beta_{n}\right)_{n}\subset(0,1)$ are convergent such that

\lim_{n\rightarrow\infty}\alpha_{n}=0,\lim_{n\rightarrow\infty}\beta_{n}=0,\text{ and }\sum_{n=1}^{\infty}\alpha_{n}=\infty.

In [2] the conditions on the above sequences were $0<\alpha_{n}\leq\beta_{n}<1$ . A better condition, introduced in [7], is $0<\alpha_{n},\beta_{n}<1$ . Now, letting $\beta_{n}=0,\forall n\in N$ from Ishikawa iteration (2), we get Mann iteration (1). Let us consider the following iteration, see [3]:

u_{n+1}=\left(1-\alpha_{n}\right)u_{n}+\alpha_{n}Tu_{n}+e_{n}.

(3)

⁰⁰footnotetext: • "T. Popoviciu" Institute of Numerical Analysis, P.O. Box 68-1, 3400 Cluj-Napoca, Romania, e-mail: soltuzul@yahoo.com, soltuz@itwm.fhg.de, stefansoltuz@personal.ro

Errors $\left(e_{n}\right)_{n}\subset X$ satisfy $\sum_{n=1}^{\infty}\left\|e_{n}\right\|<\infty$ . This iteration is known as Mann iteration with errors. In [3] Ishikawa iteration with errors is defined as

	$\displaystyle u_{n+1}$	$\displaystyle=\left(1-\alpha_{n}\right)u_{n}+\alpha_{n}Tv_{n}+p_{n}$		(4)
	$\displaystyle v_{n}$	$\displaystyle=\left(1-\beta_{n}\right)u_{n}+\beta_{n}Tu_{n}+q_{n},\quad n=1,2,\ldots$

Errors $\left(p_{n}\right)_{n},\left(q_{n}\right)_{n}$ and $\left(e_{n}\right)_{n}\subset X$ satisfy

\sum_{n=1}^{\infty}\left\|p_{n}\right\|<\infty,\lim_{n\rightarrow\infty}\left\|q_{n}\right\|=0,\sum_{n=1}^{\infty}\left\|e_{n}\right\|<\infty

(5)

where $\left(\alpha_{n}\right)_{n}$ and $\left(\beta_{n}\right)_{n}$ are the same as those from (1)and (2). When $e_{n}=0$ , respectively $p_{n}=q_{n}=0,\forall n\in N$ then we deal with Mann and Ishikawa iteration.

In [8] it was proven that for several classes of Lipschitzian operators, Mann and Ishikawa iteration methods are equivalent. We will prove further that Mann and Ishikawa iterations are equivalent models with Mann and Ishikawa iterations with errors. Thus the study of convergence of the above iterations is reduced to the study of Mann iteration, which is more convenient to be used.

Let us denote the identity map by $I$ .
Definition 1. Let $X$ be a real Banach space. A map $T:X\rightarrow X$ is called strongly pseudocontractive if there exists $k\in(0,1)$ such that we have

\|x-y\|\leq\|x-y+r[(I-T-kI)x-(I-T-kI)y]\|,

(6)

for all $x,y\in X$ , and $r>0$ .
The following lemma can be found in [3].
Lemma 1 [[3]]. Let $\left(a_{n}\right)_{n}$ be a nonnegative sequence which satisfies the following inequality

a_{n+1}\leq\left(1-\lambda_{n}\right)a_{n}+\sigma_{n}+w_{n}

(7)

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

2. Main result

Let us denote $F(T)=\left\{x^{*}:Tx^{*}=x^{*}\right\}$ . We are able now to give the following result:

Theorem 1. Let $X$ be a Banach space and let $T:X\rightarrow X$ be a Lipschitzian with $L\geq 1$ , strongly pseudocontractive map. If $u_{1}=x_{1}\in X$ , let $\lim_{n\rightarrow\infty}\alpha_{n}=$ 0 , $\lim_{n\rightarrow\infty}\beta_{n}=0$ , and $\sum_{n=1}^{\infty}\alpha_{n}=\infty$ , suppose that for iteration (4) the errors satisfy (5); then the following two assertions are equivalent:
(i) Ishikawa iteration (2) converges to $x^{*}\in F(T)$ ,
(ii) Ishikawa iteration with errors (4) converges to the same $x^{*}\in F(T)$ .

Proof. Corollary 1 from [1] assures that $F(T)\neq\emptyset$ ; strong pseudocontractivity assures the uniqueness of the fixed point.

Supposing Ishikawa iteration with errors (4) converges and taking $p_{n}=q_{n}=0,\forall n\in N$ , we get the convergence of (2). We will prove that the convergence of Ishikawa iteration (2) implies the convergence of Ishikawa iteration with errors (4). The proof is similar to the proof of Theorem 4 from [8]. We have

$\displaystyle x_{n}=$	$\displaystyle x_{n+1}+\alpha_{n}x_{n}-\alpha_{n}Ty_{n}$	(8)
$\displaystyle=$	$\displaystyle\left(1+\alpha_{n}\right)x_{n+1}+\alpha_{n}(I-T-kI)x_{n+1}+$
	$\displaystyle-(2-k)\alpha_{n}x_{n+1}+\alpha_{n}x_{n}+\alpha_{n}\left(Tx_{n+1}-Ty_{n}\right)$
$\displaystyle=$	$\displaystyle\left(1+\alpha_{n}\right)x_{n+1}+\alpha_{n}(I-T-kI)x_{n+1}+$
	$\displaystyle-(2-k)\alpha_{n}\left[\left(1-\alpha_{n}\right)x_{n}+\alpha_{n}Ty_{n}\right]+\alpha_{n}x_{n}+\alpha_{n}\left(Tx_{n+1}-Ty_{n}\right)$
$\displaystyle=$	$\displaystyle\left(1+\alpha_{n}\right)x_{n+1}+\alpha_{n}(I-T-kI)x_{n+1}+$
	$\displaystyle-(1-k)\alpha_{n}x_{n}+(2-k)\alpha_{n}^{2}\left(x_{n}-Ty_{n}\right)+\alpha_{n}\left(Tx_{n+1}-Ty_{n}\right)$

Also

$\displaystyle u_{n}=$	$\displaystyle u_{n+1}+\alpha_{n}u_{n}-\alpha_{n}Tv_{n}-p_{n}$	(9)
$\displaystyle=$	$\displaystyle\left(1+\alpha_{n}\right)u_{n+1}+\alpha_{n}(I-T-kI)u_{n+1}+$
	$\displaystyle-(2-k)\alpha_{n}u_{n+1}+\alpha_{n}u_{n}+\alpha_{n}\left(Tu_{n+1}-Tv_{n}\right)-p_{n}$
$\displaystyle=$	$\displaystyle\left(1+\alpha_{n}\right)u_{n+1}+\alpha_{n}(I-T-kI)u_{n+1}+$
	$\displaystyle-(2-k)\alpha_{n}\left[\left(1-\alpha_{n}\right)u_{n}+\alpha_{n}Tv_{n}+p_{n}\right]+$
	$\displaystyle+\alpha_{n}u_{n}+\alpha_{n}\left(Tu_{n+1}-Tv_{n}\right)-p_{n}$
$\displaystyle=$	$\displaystyle\left(1+\alpha_{n}\right)u_{n+1}+\alpha_{n}(I-T-kI)u_{n+1}+$
	$\displaystyle-(1-k)\alpha_{n}u_{n}+(2-k)\alpha_{n}^{2}\left(u_{n}-Tv_{n}\right)+\alpha_{n}\left(Tu_{n+1}-Tv_{n}\right)$
	$\displaystyle-p_{n}-(2-k)\alpha_{n}p_{n}$
$\displaystyle=$	$\displaystyle\left(1+\alpha_{n}\right)u_{n+1}+\alpha_{n}(I-T-kI)u_{n+1}+$
	$\displaystyle-(1-k)\alpha_{n}u_{n}+(2-k)\alpha_{n}^{2}\left(u_{n}-Tv_{n}\right)+\alpha_{n}\left(Tu_{n+1}-Tv_{n}\right)$
	$\displaystyle-p_{n}\left(1+(2-k)\alpha_{n}\right)$

From (8) and (9) we get

$\displaystyle x_{n}-u_{n}=$	$\displaystyle\left(1+\alpha_{n}\right)\left(x_{n+1}-u_{n+1}\right)$	(10)
	$\displaystyle+\alpha_{n}\left((I-T-kI)x_{n+1}-(I-T-kI)u_{n+1}\right)$
	$\displaystyle-(1-k)\alpha_{n}\left(x_{n}-u_{n}\right)+(2-k)\alpha_{n}^{2}\left(x_{n}-u_{n}-Ty_{n}+Tv_{n}\right)$
	$\displaystyle+\alpha_{n}\left(Tx_{n+1}-Tu_{n+1}-Ty_{n}+Tv_{n}\right)+p_{n}\left(1+(2-k)\alpha_{n}\right)$

Taking $\left(1+\alpha_{n}\right)\left(x_{n+1}-u_{n+1}\right)+\alpha_{n}\left((I-T-kI)x_{n+1}-(I-T-kI)u_{n+1}\right)$ in norm we have

		$\displaystyle\left\\|\left(1+\alpha_{n}\right)\left(x_{n+1}-u_{n+1}\right)+\alpha_{n}\left((I-T-kI)x_{n+1}-(I-T-kI)u_{n+1}\right)\right\\|$
		$\displaystyle=\left(1+\alpha_{n}\right)\left\\|\left(x_{n+1}-u_{n+1}\right)+\frac{\alpha_{n}}{1+\alpha_{n}}\left((I-T-kI)x_{n+1}-(I-T-kI)u_{n+1}\right)\right\\|$

and using (6) with $x:=x_{n+1}$ and $y:=u_{n+1}$ , we obtain

	$\displaystyle\left\\|\left(1+\alpha_{n}\right)\left(x_{n+1}-u_{n+1}\right)+\alpha_{n}\left((I-T-kI)x_{n+1}-(I-T-kI)u_{n+1}\right)\right\\|$
	$\displaystyle\geq\left(1+\alpha_{n}\right)\left\\|x_{n+1}-u_{n+1}\right\\|.$		(11)

Taking the norm in (10) and then using (11), we get

		$\displaystyle\left\\|x_{n}-u_{n}\right\\|$
		$\displaystyle\geq\left\\|\left(1+\alpha_{n}\right)\left(x_{n+1}-u_{n+1}\right)+\alpha_{n}\left((I-T-kI)x_{n+1}-(I-T-kI)u_{n+1}\right)\right\\|$
		$\displaystyle\quad-(1-k)\alpha_{n}\left\\|x_{n}-u_{n}\right\\|-(2-k)\alpha_{n}^{2}\left\\|x_{n}-u_{n}-Ty_{n}+Tv_{n}\right\\|$
		$\displaystyle\quad-\alpha_{n}\left\\|Tx_{n+1}-Tu_{n+1}-Ty_{n}+Tv_{n}\right\\|-\left\\|p_{n}\right\\|\left(1+(2-k)\alpha_{n}\right)$
		$\displaystyle\geq\left(1+\alpha_{n}\right)\left\\|x_{n+1}-u_{n+1}\right\\|-(1-k)\alpha_{n}\left\\|x_{n}-u_{n}\right\\|-(2-k)\alpha_{n}^{2}$
		$\displaystyle\quad\left\\|x_{n}-u_{n}-Ty_{n}+Tv_{n}\right\\|-\alpha_{n}\left\\|Tx_{n+1}-Tu_{n+1}-Ty_{n}+Tv_{n}\right\\|$
		$\displaystyle\quad-\left\\|p_{n}\right\\|\left(1+(2-k)\alpha_{n}\right).$

We obtain

	$\displaystyle\left(1+\alpha_{n}\right)\left\\|x_{n+1}-u_{n+1}\right\\|$		(12)
	$\displaystyle\leq\left(1+(1-k)\alpha_{n}\right)\left\\|x_{n}-u_{n}\right\\|+(2-k)\alpha_{n}^{2}\left\\|x_{n}-u_{n}-Ty_{n}+Tv_{n}\right\\|$
	$\displaystyle+\alpha_{n}\left\\|Tx_{n+1}-Tu_{n+1}-Ty_{n}+Tv_{n}\right\\|+\left\\|p_{n}\right\\|\left(1+(2-k)\alpha_{n}\right)$
	$\displaystyle\leq\left(1+(1-k)\alpha_{n}\right)\left\\|x_{n}-u_{n}\right\\|+(2-k)\alpha_{n}^{2}\left\\|u_{n}-Tv_{n}\right\\|$
	$\displaystyle+(2-k)\alpha_{n}^{2}\left\\|x_{n}-Ty_{n}\right\\|+\alpha_{n}\left\\|Tx_{n+1}-Ty_{n}\right\\|$
	$\displaystyle+\alpha_{n}\left\\|Tu_{n+1}-Tv_{n}\right\\|+\left\\|p_{n}\right\\|\left(1+(2-k)\alpha_{n}\right).$

We aim to evaluate $\left\|u_{n}-Tv_{n}\right\|$ and $\left\|Tu_{n+1}-Tv_{n}\right\|$ :

	$\displaystyle\left\\|u_{n}-Tv_{n}\right\\|$	$\displaystyle\leq\left\\|u_{n}-x_{n}\right\\|+\left\\|x_{n}-Tx_{n}\right\\|+\left\\|Tx_{n}-Tv_{n}\right\\|$
		$\displaystyle\leq\left\\|x_{n}-u_{n}\right\\|+\left\\|x_{n}-Tx_{n}\right\\|+L\left\\|x_{n}-v_{n}\right\\|$

	$\displaystyle\left\\|x_{n}-v_{n}\right\\|$		(13)
	$\displaystyle=\left\\|\left(1-\beta_{n}\right)\left(x_{n}-u_{n}\right)+\beta_{n}\left(x_{n}-Tu_{n}\right)-q_{n}\right\\|$
	$\displaystyle\leq\left(1-\beta_{n}\right)\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left\\|x_{n}-Tu_{n}\right\\|+\left\\|q_{n}\right\\|$
	$\displaystyle\leq\left(1-\beta_{n}\right)\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left[\left\\|Tx_{n}-Tu_{n}\right\\|+\left\\|x_{n}-Tx_{n}\right\\|\right]+\left\\|q_{n}\right\\|$
	$\displaystyle\leq\left(1-\beta_{n}\right)\left\\|x_{n}-u_{n}\right\\|+\beta_{n}L\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left\\|x_{n}-Tx_{n}\right\\|+\left\\|q_{n}\right\\|$
	$\displaystyle=\left(1-\beta_{n}+\beta_{n}L\right)\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left\\|x_{n}-Tx_{n}\right\\|+\left\\|q_{n}\right\\|$
	$\displaystyle\leq L\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left\\|x_{n}-Tx_{n}\right\\|+\left\\|q_{n}\right\\|,$

because $1\leq L\Rightarrow 1-\beta_{n}+\beta_{n}L\leq L$ .
We have

		$\displaystyle\left\\|u_{n}-Tv_{n}\right\\|$
		$\displaystyle\leq\left\\|x_{n}-u_{n}\right\\|+\left\\|x_{n}-Tx_{n}\right\\|+L\left(L\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left\\|x_{n}-Tx_{n}\right\\|+\left\\|q_{n}\right\\|\right)$
		$\displaystyle\leq\left(1+L^{2}\right)\left\\|x_{n}-u_{n}\right\\|+\left(1+L\beta_{n}\right)\left\\|x_{n}-Tx_{n}\right\\|+L\left\\|q_{n}\right\\|.$

Now, $\left\|Tu_{n+1}-Tv_{n}\right\|$ satisfies

	$\displaystyle\left\\|Tu_{n+1}-Tv_{n}\right\\|$	$\displaystyle\leq L\left\\|u_{n+1}-v_{n}\right\\|=L\left\\|\left(1-\alpha_{n}\right)u_{n}+\alpha_{n}Tv_{n}-v_{n}+p_{n}\right\\|$
		$\displaystyle\leq L\left[\left(1-\alpha_{n}\right)\left\\|u_{n}-v_{n}\right\\|+\alpha_{n}\left\\|Tv_{n}-v_{n}\right\\|+\left\\|p_{n}\right\\|\right]$

Using (13) we evaluate:

	$\displaystyle\left\\|Tv_{n}-v_{n}\right\\|\leq$	$\displaystyle\left\\|Tv_{n}-Tx_{n}\right\\|+\left\\|Tx_{n}-x_{n}\right\\|+\left\\|x_{n}-v_{n}\right\\|$
	$\displaystyle\leq$	$\displaystyle(1+L)\left\\|x_{n}-v_{n}\right\\|+\left\\|Tx_{n}-x_{n}\right\\|$
	$\displaystyle\leq$	$\displaystyle(1+L)\left[L\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left\\|Tx_{n}-x_{n}\right\\|+\left\\|q_{n}\right\\|\right]+\left\\|Tx_{n}-x_{n}\right\\|$
	$\displaystyle=$	$\displaystyle(1+L)L\left\\|x_{n}-u_{n}\right\\|+\left[(1+L)\beta_{n}+1\right]\left\\|Tx_{n}-x_{n}\right\\|$
		$\displaystyle+(1+L)\left\\|q_{n}\right\\|$

and

	$\displaystyle\left\\|u_{n}-v_{n}\right\\|$	$\displaystyle=\left\\|u_{n}-\left(1-\beta_{n}\right)u_{n}-\beta_{n}Tu_{n}-q_{n}\right\\|=\beta_{n}\left\\|u_{n}-Tu_{n}\right\\|+\left\\|q_{n}\right\\|$
		$\displaystyle\leq\beta_{n}\left[\left\\|u_{n}-x_{n}\right\\|+\left\\|Tx_{n}-x_{n}\right\\|+\left\\|Tu_{n}-Tx_{n}\right\\|\right]+\left\\|q_{n}\right\\|$
		$\displaystyle\leq\beta_{n}\left((1+L)\left\\|x_{n}-u_{n}\right\\|+\left\\|Tx_{n}-x_{n}\right\\|\right)+\left\\|q_{n}\right\\|$

One obtains

	$\displaystyle\\|T$	$\displaystyle u_{n+1}-Tv_{n}\\|$
	$\displaystyle\leq$	$\displaystyle L\left[\left(1-\alpha_{n}\right)\left\\|u_{n}-v_{n}\right\\|+\alpha_{n}\left\\|Tv_{n}-v_{n}\right\\|+\left\\|p_{n}\right\\|\right]$
	$\displaystyle\leq$	$\displaystyle L\left\{\left(1-\alpha_{n}\right)\left(\beta_{n}\left((1+L)\left\\|x_{n}-u_{n}\right\\|+\left\\|Tx_{n}-x_{n}\right\\|\right)+\left\\|q_{n}\right\\|\right)\right.$
		$\displaystyle+\alpha_{n}\left((1+L)L\left\\|x_{n}-u_{n}\right\\|+\left[(1+L)\beta_{n}+1\right]\left\\|Tx_{n}-x_{n}\right\\|\right.$
		$\displaystyle\left.\left.+(1+L)\left\\|q_{n}\right\\|\right)+\left\\|p_{n}\right\\|\right\}$
	$\displaystyle=$	$\displaystyle\left(1-\alpha_{n}\right)\beta_{n}(1+L)L\left\\|x_{n}-u_{n}\right\\|+L\left(1-\alpha_{n}\right)\beta_{n}\left\\|Tx_{n}-x_{n}\right\\|$
		$\displaystyle+L\left(1-\alpha_{n}\right)\left\\|q_{n}\right\\|+\alpha_{n}(1+L)L^{2}\left\\|x_{n}-u_{n}\right\\|$
		$\displaystyle+\alpha_{n}L\left[(1+L)\beta_{n}+1\right]\left\\|Tx_{n}-x_{n}\right\\|+\alpha_{n}L(1+L)\left\\|q_{n}\right\\|+L\left\\|p_{n}\right\\|$
	$\displaystyle=$	$\displaystyle\left(L\left(1-\alpha_{n}\right)\beta_{n}(1+L)+\alpha_{n}(1+L)L^{2}\right)\left\\|x_{n}-u_{n}\right\\|$
		$\displaystyle+\left(\beta_{n}L\left(1-\alpha_{n}\right)+\alpha_{n}L\left[(1+L)\beta_{n}+1\right]\right)\left\\|Tx_{n}-x_{n}\right\\|$
		$\displaystyle+\left((1+L)\alpha_{n}+\left(1-\alpha_{n}\right)\right)L\left\\|q_{n}\right\\|+L\left\\|p_{n}\right\\|$

Also, we have

	$\displaystyle\left\\|u_{n}-Tv_{n}\right\\|\leq$	$\displaystyle\left\\|x_{n}-u_{n}\right\\|+\left\\|x_{n}-Tx_{n}\right\\|+L\left\\|x_{n}-v_{n}\right\\|$
	$\displaystyle\leq$	$\displaystyle\left\\|x_{n}-u_{n}\right\\|+\left\\|x_{n}-Tx_{n}\right\\|$
		$\displaystyle+L\left[L\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left\\|x_{n}-Tx_{n}\right\\|+\left\\|q_{n}\right\\|\right]$
	$\displaystyle=$	$\displaystyle\left(1+L^{2}\right)\left\\|x_{n}-u_{n}\right\\|+\left(1+\beta_{n}L\right)\left\\|x_{n}-Tx_{n}\right\\|+L\left\\|q_{n}\right\\|.$

Taking (12) with the above evaluations for $\left\|u_{n}-Tv_{n}\right\|,\left\|Tu_{n+1}-Tv_{n}\right\|$ , and
using the following inequalities $\left(1+\alpha_{n}\right)^{-1}\leq 1-\alpha_{n}+\alpha_{n}^{2},\left(1+\alpha_{n}\right)^{-1}\leq 1$ , we get

	$\displaystyle\left(1+\alpha_{n}\right)\\|$	$\displaystyle x_{n+1}-u_{n+1}\\|$
	$\displaystyle\leq$	$\displaystyle\left(1+(1-k)\alpha_{n}\right)\left\\|x_{n}-u_{n}\right\\|$
		$\displaystyle+(2-k)\alpha_{n}^{2}\left(\left(1+L^{2}\right)\left\\|x_{n}-u_{n}\right\\|+\left(1+\beta_{n}L\right)\left\\|x_{n}-Tx_{n}\right\\|+L\left\\|q_{n}\right\\|\right)$
		$\displaystyle+(2-k)\alpha_{n}^{2}\left\\|x_{n}-Ty_{n}\right\\|+\alpha_{n}\left\\|Tx_{n+1}-Ty_{n}\right\\|+$
		$\displaystyle+\alpha_{n}\left(L\left(1-\alpha_{n}\right)\beta_{n}(1+L)+\alpha_{n}(1+L)L^{2}\right)\left\\|x_{n}-u_{n}\right\\|$
		$\displaystyle+\alpha_{n}\left(\beta_{n}L\left(1-\alpha_{n}\right)+\alpha_{n}L\left[(1+L)\beta_{n}+1\right]\right)\left\\|x_{n}-Tx_{n}\right\\|$
		$\displaystyle+\alpha_{n}\left((1+L)\alpha_{n}+\left(1-\alpha_{n}\right)\right)L\left\\|q_{n}\right\\|$
		$\displaystyle+\alpha_{n}L\left\\|p_{n}\right\\|+\left\\|p_{n}\right\\|\left(1+(2-k)\alpha_{n}\right)$
	$\displaystyle\leq$	$\displaystyle\left\{\left(1+(1-k)\alpha_{n}\right)+(2-k)\alpha_{n}^{2}\left(1+L^{2}\right)\right.$
		$\displaystyle\left.+\alpha_{n}L(1+L)\left(\left(1-\alpha_{n}\right)\beta_{n}+\alpha_{n}L\right)\right\}\left\\|x_{n}-u_{n}\right\\|$
		$\displaystyle+\left\{(2-k)\alpha_{n}^{2}\left(1+\beta_{n}L\right)+\right.$
		$\displaystyle\left.+\alpha_{n}\left[\beta_{n}L\left(1-\alpha_{n}\right)+\alpha_{n}L\left[(1+L)\beta_{n}+1\right]\right]\right\}\left\\|x_{n}-Tx_{n}\right\\|$
		$\displaystyle+(2-k)\alpha_{n}^{2}\left\\|x_{n}-Ty_{n}\right\\|+L\alpha_{n}\left\\|x_{n+1}-y_{n}\right\\|+$
		$\displaystyle+\alpha_{n}L\left(\left(1-\alpha_{n}\right)+(1+L)\alpha_{n}+(2-k)\alpha_{n}\right)\left\\|q_{n}\right\\|$
		$\displaystyle+\left(1+(2-k)\alpha_{n}+\alpha_{n}L\right)\left\\|p_{n}\right\\|.$

	$\displaystyle\left\\|x_{n+1}-u_{n+1}\right\\|\leq$	$\displaystyle\left\{\left(1+(1-k)\alpha_{n}\right)\left(1-\alpha_{n}+\alpha_{n}^{2}\right)+(2-k)\alpha_{n}^{2}\left(1+L^{2}\right)\right.$
		$\displaystyle\left.+\alpha_{n}L(1+L)\left(\left(1-\alpha_{n}\right)\beta_{n}+\alpha_{n}L\right)\right\}\left\\|x_{n}-u_{n}\right\\|$
		$\displaystyle+\left\{(2-k)\alpha_{n}^{2}\left(1+\beta_{n}L\right)\right.$
		$\displaystyle\left.+\alpha_{n}\left[\beta_{n}L\left(1-\alpha_{n}\right)+\alpha_{n}L\left[(1+L)\beta_{n}+1\right]\right]\right\}\left\\|x_{n}-Tx_{n}\right\\|$
		$\displaystyle+(2-k)\alpha_{n}^{2}\left\\|x_{n}-Ty_{n}\right\\|+L\alpha_{n}\left\\|x_{n+1}-y_{n}\right\\|$
		$\displaystyle+\alpha_{n}L\left(\left(1-\alpha_{n}\right)+(1+L)\alpha_{n}+(2-k)\alpha_{n}\right)\left\\|q_{n}\right\\|$
		$\displaystyle+\left(1+(2-k)\alpha_{n}+\alpha_{n}L\right)\left\\|p_{n}\right\\|.$

That is

a_{n+1}\leq\gamma_{n}a_{n}+\sigma_{n}+w_{n},

where

	$\displaystyle a_{n}=$	$\displaystyle\left\\|x_{n}-u_{n}\right\\|,$
	$\displaystyle\gamma_{n}=$	$\displaystyle{\left[1+(1-k)\alpha_{n}\right]\left(1-\alpha_{n}+\alpha_{n}^{2}\right)+(2-k)\left(1+L^{2}\right)\alpha_{n}^{2}}$
		$\displaystyle+\alpha_{n}L(1+L)\left[\beta_{n}\left(1-\alpha_{n}\right)+L\alpha_{n}\right],$
	$\displaystyle w_{n}=$	$\displaystyle\left\\|p_{n}\right\\|\left(1+(2-k)\alpha_{n}+\alpha_{n}L\right),$
	$\displaystyle\sigma_{n}=$	$\displaystyle\alpha_{n}\left\{\left\{(2-k)\alpha_{n}\left(1+\beta_{n}L\right)\right.\right.$
		$\displaystyle\left.+\left[\beta_{n}L\left(1-\alpha_{n}\right)+\alpha_{n}L\left[(1+L)\beta_{n}+1\right]\right]\right\}\left\\|x_{n}-Tx_{n}\right\\|$
		$\displaystyle+(2-k)\alpha_{n}\left\\|x_{n}-Ty_{n}\right\\|+L\left\\|x_{n+1}-y_{n}\right\\|$
		$\displaystyle\left.+L\left\\|q_{n}\right\\|\left(\left(1-\alpha_{n}\right)+(3+L-k)\alpha_{n}\right)\right\}.$

Remark that $\gamma_{n}$ is the same as in formula (27) from [8]. The same motivation as in [8] leads us to

\gamma_{n}\leq 1-k^{2}\alpha_{n},\text{ from a sufficient large }n.

We get relation (7) with $\lambda_{n}:=k^{2}\alpha_{n}$

a_{n+1}\leq\left(1-\lambda_{n}\right)a_{n}+\sigma_{n}+w_{n}

Using (5) and using that Ishikawa iteration (2) converges i.e. $\lim_{n\rightarrow\infty}x_{n}=x^{*}$ , (more precisely using $\lim_{n\rightarrow\infty}\left\|x_{n+1}-y_{n}\right\|=0,\lim_{n\rightarrow\infty}\left\|x_{n}-Tx_{n}\right\|=0$ ), it is easy to see that $\sigma_{n}=o\left(\lambda_{n}\right)$ , and $\sum_{n=1}^{\infty}w_{n}<\infty$ . All the assumptions from Lemma 1 are satisfied, hence we have $\lim_{n\rightarrow\infty}a_{n}=0$ . That is

\lim_{n\rightarrow\infty}\left\|x_{n}-u_{n}\right\|=0

(14)

We suppose that $\lim_{n\rightarrow\infty}x_{n}=x^{*}$ . Relation (14) and the following inequality

\left\|u_{n}-x^{*}\right\|\leq\left\|x_{n}-u_{n}\right\|+\left\|x_{n}-x^{*}\right\|\rightarrow 0,(n\rightarrow\infty)

lead us to $\lim_{n\rightarrow\infty}u_{n}=x^{*}$ .
If we consider $\beta_{n}=0$ , in (2) and (4), then we have the following result
Theorem 2. Let $X$ be a Banach space and $T:X\rightarrow X$ be a Lipschitzian with $L\geq 1$ , strongly pseudocontractive map. If $u_{1}=x_{1}\in X$ , let $\lim_{n\rightarrow\infty}\alpha_{n}=0$ and $\sum_{n=1}^{\infty}\alpha_{n}=\infty$ , suppose that the errors satisfy (5), then the following two assertions are equivalent:
(i) Mann iteration (1) converges to $x^{*}\in F(T)$ ,
(ii) Mann iteration with errors (3) converges to the same $x^{*}\in F(T)$ .

The following result is from [8].
Theorem 3 [[8]]. Let $K$ be a closed convex (not necessarily bounded) subset of an arbitrary Banach space $X$ and let $T$ be a Lipschitzian pseudocontractive selfmap of K Let us consider Mann iteration and Ishikawa iteration with the same initial point and with the conditions $\lim_{n\rightarrow\infty}\alpha_{n}=0,\lim_{n\rightarrow\infty}\beta_{n}=0$ , and $\sum_{n=1}^{\infty}\alpha_{n}=\infty$ . Let $x^{*}\in F(T)$ . Then the following conditions are equivalent:
(i) Mann iteration (1) converges to $x^{*}\in F(T)$ ,
(ii) Ishikawa iteration (2) converges to $x^{*}\in F(T)$ .

Take $K:=X$ in the above result. Theorem 1, Theorem 2 and Theorem 3 lead us to the conclusion:

Corollary 1. In the same assumptions as in Theorem 1 we have the equivalence between the convergences of (1), (2), (3) and (4).

3. The equivalence for strongly accretive and accretive maps

The map $J:X\rightarrow 2^{X^{*}}$ given by

Jx:=\left\{x^{*}\in X^{*}:\left\langle x,x^{*}\right\rangle=\|x\|^{2},\left\|x^{*}\right\|=\|x\|\right\},\forall x\in X,

is called the normalized duality mapping. Let us denote the identity map by $I$ .
Definition 2. Let $X$ be a Banach space. A map $T:X\rightarrow X$ is called strongly pseudocontractive if there exists $k\in(0,1)$ and $j(x-y)\in J(x-y)$ such that

\operatorname{Re}\langle Tx-Ty,j(x-y)\rangle\leq k\|x-y\|^{2},

for all $x,y\in X$ . This is equivalent with (6).
A map $S:X\rightarrow X$ is called strongly accretive if there exists $\gamma\in(0,1)$ and $j(x-y)\in J(x-y)$ such that

\operatorname{Re}\langle Sx-Sy,j(x-y)\rangle\geq\gamma\|x-y\|^{2}.

for all $x,y\in X$ .
A map $S:X\rightarrow X$ is called accretive if there exists $j(x-y)\in J(x-y)$ such that

\operatorname{Re}\langle Sx-Sy,j(x-y)\rangle\geq 0.

for all $x,y\in X$ .
Let us denote the identity map by $I$ .
Remark 1. Map $T$ is a strongly pseudocontractive map with $k\in(0,1)$ if and only if $(I-T)$ is a strongly accretive map with $(1-k)$ .

Let us consider the following operator equation

Sx=f

where $S$ is a strongly accretive map and $f$ is given. Consider the map $Tx=f+(I-S)x,\forall x\in X$ . A fixed point for $T$ will be a solution for the equation $Sx=f;$ such a solution exists, see [6]. We consider iterations (2) and (4) with $f+(I-S)x$ instead of $Tx$ .

	$\displaystyle x_{n+1}$	$\displaystyle=\left(1-\alpha_{n}\right)x_{n}+\alpha_{n}\left(f+(I-S)y_{n}\right),$		(15)
	$\displaystyle y_{n}$	$\displaystyle=\left(1-\beta_{n}\right)x_{n}+\beta_{n}\left(f+(I-S)x_{n}\right),n=1,2,\ldots,$

and

	$\displaystyle u_{n+1}$	$\displaystyle=\left(1-\alpha_{n}\right)u_{n}+\alpha_{n}\left(f+(I-S)v_{n}\right)+p_{n}$		(16)
	$\displaystyle v_{n}$	$\displaystyle=\left(1-\beta_{n}\right)u_{n}+\beta_{n}\left(f+(I-S)u_{n}\right)+q_{n},\quad n=1,2,\ldots$

Sequences $\left(\alpha_{n}\right)_{n},\left(\beta_{n}\right)_{n}\subset(0,1)$ , are convergent such that $\lim_{n\rightarrow\infty}\alpha_{n}=0,\lim_{n\rightarrow\infty}\beta_{n}=$ 0 , and $\sum_{n=1}^{\infty}\alpha_{n}=\infty$ . Errors $\left(p_{n}\right)_{n},\left(q_{n}\right)_{n}$ satisfy (5).

Theorem 1 assures that the Ishikawa iteration and Ishikawa iteration with errors are equivalent models for a strongly pseudocontractive map. Using Remark 1, observe that if $S$ is Lipschitzian and strongly accretive, then the map $Tx=f+(I-S)x$ is Lipschitzian strongly pseudocontractive. We obtain

Theorem 4. Let $X$ be a Banach space, let $S:X\rightarrow X$ be a Lipschitzian with $L\geq 1$ , a strongly accretive map. If $u_{1}=x_{1}\in B$ , then the following two assertions are equivalent:
(i) Ishikawa iteration (15) converges to $x^{*}\in F(T)$ , which is the solution of $Sx=f$ ,
(ii) Ishikawa iteration with errors (16) converges to the same $x^{*}\in F(T)$ , which is the solution of $Sx=f$ .

When we take $\beta_{n}=0$ , we get a similar result for Mann iteration and Mann iteration with errors. Thus Mann iteration with errors (3) is equivalent with Mann iteration (1), when we take $Tx=f+(I-S)x,\forall x\in X$ .

According to observation from [8], Theorem 3 holds if we take the above operator $T$ , with $S$ strongly accretive. Theorem 4 and the equivalence between Mann iteration with errors (3) and Mann iteration (1) lead us to the following conclusion

Corollary 2. In the same assumptions as in Theorem 4 we have the equivalence between the convergences of (1), (2), (3) and (4) for a Lipschitzian strongly accretive map $S$ , and $Tx=f+(I-S)x,\forall x\in X$ .

Remark 2. If $S$ is an accretive map, then $T=f-S$ is a strongly pseudocontractive map.

Proof. For all $x,y\in X$ and $j(x-y)\in J(x-y)$ , we have

	$\displaystyle\operatorname{Re}\langle Sx-Sy,j(x-y)\rangle$	$\displaystyle\geq 0\Leftrightarrow$
	$\displaystyle\operatorname{Re}\langle(f-T)x-(f-T)y,j(x-y)\rangle$	$\displaystyle\geq 0\Leftrightarrow$
	$\displaystyle-\operatorname{Re}\langle Tx-Ty,j(x-y)\rangle$	$\displaystyle\geq 0\Leftrightarrow$
	$\displaystyle\operatorname{Re}\langle Tx-Ty,j(x-y)\rangle$	$\displaystyle\leq 0\leq k\\|x-y\\|^{2},\forall k\in(0,1).$

Let us consider the following operator equation

x+Sx=f

where $S$ is a strongly accretive map and $f$ is given. The existence of the solution for $x+Sx=f$ follows from [5]. It is clear that $x+Sx$ is Lipschitzian if $S$ is. Consider the map $Tx=f-Sx,\forall x\in X$ . A fixed point for $T$ will be a solution for the equation $Sx=f$ . Using Remark 2, if $S$ is an accretive map, then $T$ is strongly pseudocontractive. Now let us consider iterations (2) with $Tx=f-Sx$

	$\displaystyle x_{n+1}$	$\displaystyle=\left(1-\alpha_{n}\right)x_{n}+\alpha_{n}\left(f-Sy_{n}\right)$		(17)
	$\displaystyle y_{n}$	$\displaystyle=\left(1-\beta_{n}\right)x_{n}+\beta_{n}\left(f-Sx_{n}\right),\quad n=1,2,\ldots,$

and the Ishikawa iteration with errors (4):

	$\displaystyle u_{n+1}$	$\displaystyle=\left(1-\alpha_{n}\right)u_{n}+\alpha_{n}\left(f-Sv_{n}\right)+p_{n}$		(18)
	$\displaystyle v_{n}$	$\displaystyle=\left(1-\beta_{n}\right)u_{n}+\beta_{n}\left(f-Su_{n}\right)+q_{n},\quad n=1,2,\ldots.$

Sequences $\left(\alpha_{n}\right)_{n},\left(\beta_{n}\right)_{n}\subset(0,1)$ are convergent such that $\lim_{n\rightarrow\infty}\alpha_{n}=0$ , $\lim_{n\rightarrow\infty}\beta_{n}=0$ , and $\sum_{n=1}^{\infty}\alpha_{n}=\infty$ . The errors verify (5).

Theorem 1 assures that Ishikawa iteration and Ishikawa iteration with errors are equivalent models for a strongly pseudocontractive map. According to Remark 2, the map $Tx=f-Sx,\forall x\in X$ , is (Lipschitzian) strongly pseudocontractive map when $S$ is a (Lipschitzian) accretive. These arguments lead us to the following conclusion

Theorem 5. Let $X$ be a Banach space, let $S:X\rightarrow X$ be a Lipschitzian with $L\geq 1$ , accretive map. If $u_{1}=x_{1}\in X$ , then the following two assertions are equivalent:
(i) Ishikawa iteration (17) converges to $x^{*}\in F(T)$ , which is the solution of $x+Sx=f$
(ii) Ishikawa iteration with errors (18) converges to the same $x^{*}\in F(T)$ , which is the solution of $x+Sx=f$ .

According to second observation from [8], Theorem 1 holds if we take $Tx=f-Sx,\forall x\in X$ , with $S$ accretive. Also Theorem 5 and the remark concerning the equivalence between Mann iteration with errors and Mann iteration for $Tx=f-Sx,\forall x\in X$ , lead us to

Corollary 3. In the same assumptions as in Theorem 5 we have the equivalence between the convergences of (1), (2), (3) and (4) for a Lipschitzian accretive map $S$ , and $Tx=f-Sx,\forall x\in X$ .

4. The multivalued case

In the multivalued case the following definitions hold
Definition 3. Let $X$ be a real Banach space. A map $T:X\rightarrow 2^{X}$ is called strongly pseudocontractive if there exists $k\in(0,1)$ and $j(x-y)\in J(x-y)$ such that

\langle\xi-\theta,j(x-y)\rangle\leq k\|x-y\|^{2}

for all $x,y\in X,\xi\in Tx,\theta\in Ty$ .
Let $S:X\rightarrow 2^{X}$ , the map $S$ is called strongly accretive if there exists $\gamma\in(0,1)$ and $j(x-y)\in J(x-y)$ such that

\langle\xi-\theta,j(x-y)\rangle\geq\gamma\|x-y\|^{2}

for all $x,y\in X,\xi\in Tx,\theta\in Ty$ , etc.
We remark that all results from this paper hold in the multivalued case, provided that these multivalued maps admit single valued selections.

Acknowledgment. The author is indebted to referee and to Ioana C. Şoltuz for carefuly reading and making useful, pertinent suggestions.

References

[1] K. Deimling, Zeroes of accretive operators, Manuscripta Math. 13(1974), 365374.
[2] S. Ishikawa, Fixed points by a new iteration method, Proc. Amer. Math. Soc. 44(1974), 147-150.
[3] L.-S. Liu, Ishikawa and Mann iterative process with errors for nonlinear strongly accretive mappings in Banach spaces, J. Math. Anal. Appl. 194(1995), 114-125.
[4] W. R. Mann, Mean value in iteration, Proc. Amer. Math. Soc. 4(1953), 506510.
[5] R. H. Martin Jr, A global existence theorem for autonomous differential equations in Banach spaces, Proc. Amer. Math. Soc. 26(1970), 307-314.
[6] C. Morales, Surjectivity theorems for multi-valued mappings of accretive type, Comm. Math. Univ. Carolinae 26(1985), 397-413.
[7] B. E. Rhoades, Comments on two fixed point iteration methods, J. Math. Anal. Appl. 56(1976), 741-750.
[8] B. E. Rhoades, Ş. M. Şoltuz, On the equivalence of Mann and Ishikawa iteration methods, International Journal of Mathematics and Mathematical Science 33(2003), 451-459.

		$\displaystyle\left\\|x_{n}-u_{n}\right\\|$
		$\displaystyle\geq\left\\|\left(1+\alpha_{n}\right)\left(x_{n+1}-u_{n+1}\right)+\alpha_{n}\left((I-T-kI)x_{n+1}-(I-T-kI)u_{n+1}\right)\right\\|$
		$\displaystyle\quad-(1-k)\alpha_{n}\left\\|x_{n}-u_{n}\right\\|-(2-k)\alpha_{n}^{2}\left\\|x_{n}-u_{n}-Ty_{n}+Tv_{n}\right\\|$
		$\displaystyle\quad-\alpha_{n}\left\\|Tx_{n+1}-Tu_{n+1}-Ty_{n}+Tv_{n}\right\\|-\left\\|p_{n}\right\\|\left(1+(2-k)\alpha_{n}\right)$
		$\displaystyle\geq\left(1+\alpha_{n}\right)\left\\|x_{n+1}-u_{n+1}\right\\|-(1-k)\alpha_{n}\left\\|x_{n}-u_{n}\right\\|-(2-k)\alpha_{n}^{2}$
		$\displaystyle\quad\left\\|x_{n}-u_{n}-Ty_{n}+Tv_{n}\right\\|-\alpha_{n}\left\\|Tx_{n+1}-Tu_{n+1}-Ty_{n}+Tv_{n}\right\\|$
		$\displaystyle\quad-\left\\|p_{n}\right\\|\left(1+(2-k)\alpha_{n}\right).$

	$\displaystyle\left(1+\alpha_{n}\right)\left\\|x_{n+1}-u_{n+1}\right\\|$		(12)
	$\displaystyle\leq\left(1+(1-k)\alpha_{n}\right)\left\\|x_{n}-u_{n}\right\\|+(2-k)\alpha_{n}^{2}\left\\|x_{n}-u_{n}-Ty_{n}+Tv_{n}\right\\|$
	$\displaystyle+\alpha_{n}\left\\|Tx_{n+1}-Tu_{n+1}-Ty_{n}+Tv_{n}\right\\|+\left\\|p_{n}\right\\|\left(1+(2-k)\alpha_{n}\right)$
	$\displaystyle\leq\left(1+(1-k)\alpha_{n}\right)\left\\|x_{n}-u_{n}\right\\|+(2-k)\alpha_{n}^{2}\left\\|u_{n}-Tv_{n}\right\\|$
	$\displaystyle+(2-k)\alpha_{n}^{2}\left\\|x_{n}-Ty_{n}\right\\|+\alpha_{n}\left\\|Tx_{n+1}-Ty_{n}\right\\|$
	$\displaystyle+\alpha_{n}\left\\|Tu_{n+1}-Tv_{n}\right\\|+\left\\|p_{n}\right\\|\left(1+(2-k)\alpha_{n}\right).$

	$\displaystyle\left\\|x_{n}-v_{n}\right\\|$		(13)
	$\displaystyle=\left\\|\left(1-\beta_{n}\right)\left(x_{n}-u_{n}\right)+\beta_{n}\left(x_{n}-Tu_{n}\right)-q_{n}\right\\|$
	$\displaystyle\leq\left(1-\beta_{n}\right)\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left\\|x_{n}-Tu_{n}\right\\|+\left\\|q_{n}\right\\|$
	$\displaystyle\leq\left(1-\beta_{n}\right)\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left[\left\\|Tx_{n}-Tu_{n}\right\\|+\left\\|x_{n}-Tx_{n}\right\\|\right]+\left\\|q_{n}\right\\|$
	$\displaystyle\leq\left(1-\beta_{n}\right)\left\\|x_{n}-u_{n}\right\\|+\beta_{n}L\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left\\|x_{n}-Tx_{n}\right\\|+\left\\|q_{n}\right\\|$
	$\displaystyle=\left(1-\beta_{n}+\beta_{n}L\right)\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left\\|x_{n}-Tx_{n}\right\\|+\left\\|q_{n}\right\\|$
	$\displaystyle\leq L\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left\\|x_{n}-Tx_{n}\right\\|+\left\\|q_{n}\right\\|,$

	$\displaystyle\left\\|Tv_{n}-v_{n}\right\\|\leq$	$\displaystyle\left\\|Tv_{n}-Tx_{n}\right\\|+\left\\|Tx_{n}-x_{n}\right\\|+\left\\|x_{n}-v_{n}\right\\|$
	$\displaystyle\leq$	$\displaystyle(1+L)\left\\|x_{n}-v_{n}\right\\|+\left\\|Tx_{n}-x_{n}\right\\|$
	$\displaystyle\leq$	$\displaystyle(1+L)\left[L\left\\|x_{n}-u_{n}\right\\|+\beta_{n}\left\\|Tx_{n}-x_{n}\right\\|+\left\\|q_{n}\right\\|\right]+\left\\|Tx_{n}-x_{n}\right\\|$
	$\displaystyle=$	$\displaystyle(1+L)L\left\\|x_{n}-u_{n}\right\\|+\left[(1+L)\beta_{n}+1\right]\left\\|Tx_{n}-x_{n}\right\\|$
		$\displaystyle+(1+L)\left\\|q_{n}\right\\|$

	$\displaystyle\left\\|u_{n}-v_{n}\right\\|$	$\displaystyle=\left\\|u_{n}-\left(1-\beta_{n}\right)u_{n}-\beta_{n}Tu_{n}-q_{n}\right\\|=\beta_{n}\left\\|u_{n}-Tu_{n}\right\\|+\left\\|q_{n}\right\\|$
		$\displaystyle\leq\beta_{n}\left[\left\\|u_{n}-x_{n}\right\\|+\left\\|Tx_{n}-x_{n}\right\\|+\left\\|Tu_{n}-Tx_{n}\right\\|\right]+\left\\|q_{n}\right\\|$
		$\displaystyle\leq\beta_{n}\left((1+L)\left\\|x_{n}-u_{n}\right\\|+\left\\|Tx_{n}-x_{n}\right\\|\right)+\left\\|q_{n}\right\\|$

Mann-Ishikawa iterations and Mann-Ishikawa iterations with errors are equivalent models

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Mann-Ishikawa iterations and Mann-Ishikawa iterations with errors are equivalent models

Abstract

1. Preliminaries

2. Main result

3. The equivalence for strongly accretive and accretive maps

4. The multivalued case

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