In this paper $X$ denotes a real Banach space with $X^{*}$ strictly convex, $T:X\rightarrow X$ a map and let $x_{0},u_{0}\in X$. We consider the following iteration known as Mann iteration, ([9])

The sequence $\left\{\alpha_{n}\right\}\subset(0,1)$ satisfies $\lim_{n\rightarrow\infty}\alpha_{n}=0$, and $\sum_{n=1}^{\infty}\alpha_{n}=\infty$. We consider the following iteration known as Ishikawa iteration, ([8])

The sequences $\left\{\alpha_{n}\right\}\subset(0,1),\left\{\beta_{n}\right\}\subset[0,1)$ satisfy

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

We have

The following Remark is Proposition 12.3 from [7].

Remark 1.1.([7]) If $X$ is a real Banach space with $X$ * strictly convex then $J(\cdot)$ is a single map and uniformly continuous on all the bounded sets of $X$.

The following result is Lemma 1 from [10].
Lemma 1.2. If $X$ is a real normed space, then the following relation is true

The following definitions are from [3], [5] and [6].
Definition 1.3. Let $X$ be a normed space.
A map $T:X\rightarrow X$ is called weakly contractive map if for all $x,y\in X$, there exist $\psi:[0,+\infty)\rightarrow[0,+\infty)$ a continuous and strictly increasing map such that $\psi$ is positive on $(0,+\infty),\psi(0)=0$, and the following inequality is satisfied

A map $T:X\rightarrow X$ is called d-weakly contractive map if for all $x,y\in X$, there exist $j(x-y)\in J(x-y)$ and $\psi:[0,+\infty)\rightarrow[0,+\infty)$ a continuous and strictly increasing map such that $\psi$ is positive on $(0,+\infty),\psi(0)=0$, and the following inequality is satisfied

A map $T:X\rightarrow X$ is called $\psi$-uniformly pseudocontractive if there exist $j(x-y)\in J(x-y)$ and $\psi:[0,+\infty)\rightarrow[0,+\infty)$ a strictly increasing map such that $\psi$ is positive on $(0,+\infty),\psi(0)=0$ and the following inequality is satisfied

A map $C:X\rightarrow X$ is called $\psi$-uniformly accretive if there exist $j(x-y)\in J(x-y)$ and $\psi:[0,+\infty)\rightarrow[0,+\infty)$ a strictly increasing map such that $\psi$ is positive on $(0,+\infty)$, $\psi(0)=0$ and the following inequality is satisfied

We denote the identity map by $I$.
Remark 1.4. (i) If $T$ is a d-weakly contractive map, then $T$ is a $\psi$-uniformly pseudocontractive map.
(ii) The map $T$ is $\psi$-uniformly pseudocontractive if and only if $C:=(I-T)$ is $\psi$-uniformly accretive.

Proposition 1.5. If $T$ is a weakly contractive map, then $T$ is a $\psi$-uniformly pseudocontractive map.

Proof. Let $j(x-y)\in J(x-y)$. Using (1.5), (1.7) and (1.4) we get

Denote $\psi(a):=a\cdot\phi(a),\forall a\in[0,\infty)$ to obtain that $\psi$ is strictly increasing and positive.
The convergence of Mann iteration for a d-weakly contractive map in Hilbert spaces, was studied in [3]. It was shown in [5] that Mann iteration (1.1) for a d-weakly contractive map without a bounded range, converges in a Banach space more general then a Hilbert space. Also, it was shown in [6] that the same iteration for a $\psi$-uniformly pseudocontractive map without a bounded range, converges in a normed space.

If $T$ is a weakly contractive, then $T$ is a nonexpansive map. In this case the equivalence between Mann and Ishikawa iterations follows from Theorem 3 of the paper [11].

The above two motivations lead us to prove, in this note, the equivalence between Mann and Ishikawa iterations, (1.1) and (1.2), dealing with $\psi$-uniformly pseudocontractive maps without bounded range. As a corollary we obtain the convergence of Ishikawa iteration for the above operatorial classes. Also, we give a positive answer to the following conjecture, (see [11], page 452), "If Mann iteration converges, so does Ishikawa iteration".

For a $\psi$-uniformly pseudocontractive (respectively, $\psi$-uniformly accretive) map, the equivalence between Mann and Ishikawa iterations was shown also in Theorem 2.1 and Corollary 3.1 from [12]. There, in [12], the set $T(X)$ was assumed to be bounded. Removing the boundedness of the range, forces us to pay a price: both $\left\{\alpha_{n}\right\}$ and $\left\{\beta_{n}\right\}$ will depend on $T$ and $x^{*}$ ( see condition (2.1)).

Remark 1.6. Let $X$ be a normed space and $T:X\rightarrow X$ a uniformly continuous map. Then $I-T$ is a uniformly continuous map.

The following result is Proposition 2.1.2 from [4].
Proposition 1.7([4]) Let $X$ be a normed space and $T:X\rightarrow X$ be a uniformly continuous map. Then $T$ is bounded; i.e. it maps any bounded set into a bounded set.

Remark 1.6 and Proposition 1.7 lead to the following result.
Remark 1.8. Let $X$ be a normed space and $T:X\rightarrow X$ a uniformly continuous map. Then $I-T$ is bounded; i.e. it maps any bounded set into a bounded set.

The following result, stated below, is Lemma 3.1 from [1]. In [1], the map $\psi$ is assumed to be continuous in order to obtain an estimate for the convergence rate of the sequence $\left\{\lambda_{n}\right\}$. Another proof for the Lemma 3.1 can be found in ([2], pages 12-13). The same lemma, without the continuity assumption on $\psi$, appears in [6].

Lemma 1.9.([1]) Let $\left\{\lambda_{n}\right\}$ and $\left\{\gamma_{n}\right\}$ be sequences of nonnegative numbers and $\left\{\alpha_{n}\right\}$ a sequence of positive numbers satisfying the conditions $\sum_{n=1}^{\infty}\alpha_{n}=+\infty$ and $\left(\gamma_{n}/\alpha_{n}\right)\rightarrow 0$
as $n\rightarrow+\infty$. Suppose that

is satisfied, where $\psi:[0,+\infty)\rightarrow[0,+\infty)$ is a strictly increasing map such that $\psi$ is positive on $(0,+\infty)$, with $\psi(0)=0$. Then $\lim_{n\rightarrow\infty}\lambda_{n}=0$.

Let $F(T)$ denote the fixed point set of $T$.
Theorem 2.1. Let $X$ be a real Banach space with $X^{*}$ stricly convex. If $T:X\rightarrow X$ is a $\psi$-uniformly pseudocontractive and uniformly continuous map with $x^{*}\in F(T)$, $x_{0}=u_{0}\in X$ and there exists a constant $d_{0}:=d_{0}\left(T,x^{*}\right)\in(0,1)$, which depends on $T$ and $x^{*}$, such that $\left\{\alpha_{n}\right\},\left\{\beta_{n}\right\}$ satisfy

and (1.3), then the following are equivalent:
(i) the Mann iteration (1.1) converges to the $x^{*}\in F(T)$,
(ii) the Ishikawa iteration (1.2) converges to the same $x^{*}$.

Proof. The fixed point $x^{*}$ is unique. If not, then there exists at least another fixed point $y^{*}\in F(T)$, with $x^{*}\neq y^{*}$. Relation (1.9) leads to

The implication (ii) $\Rightarrow$ (i) is obvious, by setting, in (1.2), $\beta_{n}=0$, for all $n\in\mathbb{N}$. We will prove the implication (i) $\Rightarrow$ (ii). Suppose that $\lim_{n\rightarrow\infty}u_{n}=x^{*}$. If

then

and it follows that

Thus, to complete the proof it suffices to verify relation (2.3).

With $A:=(I-T)$ in (1.9), we have

Taking $x:=x_{n}$ and $y:=u_{n}$ in (2.6) we obtain

Choose $R>0$ such that $\left\{u_{n}:n\in\mathbb{N}\right\}\subset B_{R}\left(x^{*}\right)$ and $x_{0}\in B_{2R}\left(x^{*}\right)$. Remark 1.8 assures that $A\left(B_{2R}\left(x^{*}\right)\right)$ is bounded. Denote

Since the map $J(\cdot)$ is uniformly continuous on bounded subsets of $X$, with

there exists a $\delta_{1}>0$ such that $\|x-y\|\leq\delta_{1}$ implies $\|J(x)-J(y)\|\leq\varepsilon$.
The map $T(\cdot)$ is also uniformly continuous. Thus for the same $\varepsilon$, there exits a $\delta_{2}>0$ such that $\|x-y\|\leq\delta_{2}$ implies $\|Tx-Ty\|\leq\varepsilon$.

We shall prove by induction that $\left\{x_{n}\right\}$ is bounded. We know that $0=\left\|x_{0}-u_{0}\right\|\leq R$. Suppose that $\left\|x_{k}-u_{k}\right\|\leq R,\forall k\in\{1,\ldots,n\}$. We shall prove that

Assume that $\left\|x_{n}-u_{n}\right\|\leq R$ and that

From $\left\|x_{k}-u_{k}\right\|\leq R,\forall k\in\{1,\ldots,n\}$ we know

From (2.12), we have $x_{n}\in B_{2R}\left(x^{*}\right)$ and the following inequality satisfied

$\operatorname{Using}\operatorname{diam}\left(A\left(B_{2R}\left(x^{*}\right)\right)\right)\leq\sigma$ and $x_{n}\in B_{2R}\left(x^{*}\right)$, (i. e. $\left\|Ax_{n}\right\|\leq\sigma$ ), we get

Such a $S>0$ exists because

and $A$ is a bounded map.
For all $n\in\mathbb{N}$, we have

Set

Defining

it follows that, for all $n\in\mathbb{N}$, using (2.1) and (2.16), that

From (1.1) and (1.2),

From (2.20), using (2.11), (2.8), (2.14) and the first evaluation from (2.19),

Using the induction assumption,

Thus we get

By setting (1.6),

we obtain

We again apply (1.6) with

to obtain,

Substituting (2.27) into (2.25) and using (2.21) we have

Setting

and

and using (2.8) and (2.23),

Using (2.14) and (2.19) we obtain

From the uniform continuity of $J(\cdot)$,

Relation (2.19) leads to

Since $T$ is uniformly continuous,

Substituting (2.33), (2.35) (with $\varepsilon$ given by (2.9)), and (2.23) in (2.31) we obtain