Equivalence of semistability of Picard, Mann, Krasnoselskij and Ishikawa iterations

Abstract

In this paper, we show that convergence of Picard, Mann, Krasnoselskij and Ishikawa iterations is equivalent in cone normed spaces. Also, we prove that semistability of these iterations is equivalent.

1 Introduction

Let $(E,{\parallel \cdot \parallel }_E)$ be a real Banach space. A subset $P\subseteq E$ is called a cone in E if it satisfies the following conditions:

• (i) P is closed, nonempty and $P\ne \{0\}$,

• (ii) $a,b\in \mathbb{R}$, $a,b\ge 0$ and $x,y\in P$ imply that $ax+by\in P$,

• (iii) $x\in P$ and $-x\in P$ imply that $x=0$.

The space E can be partially ordered by the cone P, by defining $x\le y$ if and only if $y-x\in P$. Also, we write $x\ll y$ if $y-x\in intP$, where intP denotes the interior of P. A cone P is called normal if there exists a constant $k>0$ such that $0\le x\le y$ implies ${\parallel x\parallel }_E\le k{\parallel y\parallel }_E$. The least positive number satisfying above is called the normal constant of P.

From now on, we suppose that E is a real Banach space, P is a cone in E and ≤ is a partial ordering with respect to P.

Lemma 1.1 ([1])

Let P be a normal cone and let $\{a_n\}$ and $\{b_n\}$ be sequences in E satisfying the following inequality:

$$a_{n+1}\le h{a}_n+b_n,$$

(1)

where $h\in (0,1)$ and $b_n\to 0$ as $n\to \mathrm{\infty }$. Then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

Definition 1.2 ([2])

Let X be a vector space over the field F. Assume that the function $p:X\to E$ having the properties:

• (i) $0\le p(x)$ for all x in X,

• (ii) $p(x+y)\le p(x)+p(y)$ for all x, y in X,

• (iii) $p(\alpha x)=|\alpha |p(x)$ for all $\alpha \in F$ and $x\in X$.

Then p is called a cone seminorm on X. A cone norm is a cone seminorm p such that

• (iv) $x=0$ if $p(x)=0$.

We will denote a cone norm by ${\parallel \cdot \parallel }_c$ and $(X,{\parallel \cdot \parallel }_c)$ is called a cone normed space. Also, $d_c(x,y)={\parallel x-y\parallel }_c$ defines a cone metric on X.

Definition 1.3 ([3])

Let $(X,{\parallel \cdot \parallel }_c)$ be a cone normed space. Then $A\subseteq X$ is called bounded above if there exists $c\in E$, $0\ll c$ such that ${\parallel x-y\parallel }_c\le c$ for all $x,y\in A$.

Definition 1.4 Let $(X,{\parallel \cdot \parallel }_c)$ be a cone normed space. Let $\{x_n\}$ be a sequence in X and $x\in X$. If for any $c\in E$ with $0\ll c$, there exists an integer $N\ge 1$ such that for all $n\ge N$, ${\parallel x_n-x\parallel }_c\ll c$, then we will say $\{x_n\}$ converges to x and we write ${lim}_{n\to \mathrm{\infty }}{x}_n=x$.

Definition 1.5 Let $(X,{\parallel \cdot \parallel }_c)$ be a cone normed space. Let $\{x_n\}$ be a sequence in X and $x\in X$. If for any $c\in E$ with $0\ll c$, there exists an integer $N\ge 1$ such that for all $n,m\ge N$, ${\parallel x_n-x_m\parallel }_c\ll c$, then $\{x_n\}$ is said to be a Cauchy sequence. If every Cauchy sequence is convergent in X, then X is called a cone Banach space.

Lemma 1.6 ([4])

Let $(X,d_c)$ be a cone metric space, P be a normal cone. Let $\{x_n\}$ be a sequence in X and $x\in X$. Then $\{x_n\}$ converges to x if and only if ${lim}_{n\to \mathrm{\infty }}{d}_c(x_n,x)=0$.

Lemma 1.7 Let $(X,{\parallel \cdot \parallel }_c)$ be a cone normed space over the real Banach space E with the cone P which is normal with the normal constant k. The mapping $N:X\to [0,\mathrm{\infty })$ defined by $N(x)={\parallel ({\parallel x\parallel }_c)\parallel }_E$ satisfies the following properties:

• (i) ${\parallel x\parallel }_c\le {\parallel y\parallel }_c$ implies $N(x)\le kN(y)$,

• (ii) $N(x+y)\le k[N(x)+N(y)]$ for all $x,y\in X$,

• (iii) $N(\alpha x)=|\alpha |N(x)$ for all $\alpha \in F$ and $x\in X$,

• (iv) $N(x-y)\le k[N(x-z_1)+\cdots +N(x-z_n)]$ for all $x,y,z_1,\dots ,z_n\in X$,

• (v) $x=0$ if and only if $N(x)=0$.

  Moreover, let A be a bounded above subset of X, then

• (vi) $\{N(x):x\in A\}$ is a bounded set.

Proof The proof is obvious. □

Definition 1.8 Let $(X,{\parallel \cdot \parallel }_c)$ be a cone normed space over the real Banach space E with the normal cone P. The mapping N, defined in Lemma 1.7, is called a norm type with respect to ${\parallel \cdot \parallel }_c$.

Lemma 1.9 Let $(X,{\parallel \cdot \parallel }_c)$ be a cone normed space over the real Banach space E with the normal cone P. Also, let $\{x_n\}$ be a sequence in X and $x\in X$. Then $\{x_n\}$ converges to x if and only if ${lim}_{n\to \mathrm{\infty }}N(x_n-x)=0$.

Proof Note that $\{{\parallel x_n-x\parallel }_c\}$ is a sequence in E and by Lemma 1.6, the proof is obvious. □

Definition 1.10 Let X be a cone normed space and $T:X\to X$ be a map for which there exist real numbers a, b, c satisfying $0<a<1$, $0<b<1/2$ and $0<c<1/2$. Then T is called a Zamfirescu operator with respect to $(a,b,c)$ if and only if for each pair $x,y\in X$, T satisfies at least one of the following conditions:

(Z1) ${\parallel Tx-Ty\parallel }_c\le a{\parallel x-y\parallel }_c$,

(Z2) ${\parallel Tx-Ty\parallel }_c\le b({\parallel x-Tx\parallel }_c+{\parallel y-Ty\parallel }_c)$,

(Z3) ${\parallel Tx-Ty\parallel }_c\le c({\parallel x-Ty\parallel }_c+{\parallel y-Tx\parallel }_c)$.

Usually, for simplicity, T is called a Zamfirescu operator if T is Zamfirescu with respect to some triple $(a,b,c)$ of scalers a, b and c with above restrictions. Also, T is called f-Zamfirescu operator if at least one of the relations (Z1), (Z2) and (Z3) hold for all $x\in X$ and for all $y\in F(T)$.

Remark 1.11 Let T be a Zamfirescu operator and $x,y\in X$ be arbitrary. Since T is Zamfirescu, at least one of the conditions (Z1), (Z2) and (Z3) is satisfied. If (Z2) holds, then

$$\begin{array}{rcl}{\parallel Tx-Ty\parallel }_c& \le & b({\parallel x-Tx\parallel }_c+{\parallel y-Ty\parallel }_c)\\ \le & b(2{\parallel x-Tx\parallel }_c+{\parallel y-x\parallel }_c+{\parallel Tx-Ty\parallel }_c).\end{array}$$

Thus we get

$$(1-b){\parallel Tx-Ty\parallel }_c\le b{\parallel x-y\parallel }_c+2b{\parallel x-Tx\parallel }_c.$$

Since $0<b<1$, we have

$${\parallel Tx-Ty\parallel }_c\le \frac{b}{1-b}{\parallel x-y\parallel }_c+\frac{2b}{1-b}{\parallel x-Tx\parallel }_c.$$

Similarly, if (Z3) holds, then we obtain

$${\parallel Tx-Ty\parallel }_c\le \frac{c}{1-c}{\parallel x-y\parallel }_c+\frac{2c}{1-c}{\parallel x-Tx\parallel }_c.$$

Hence

$${\parallel Tx-Ty\parallel }_c\le \delta {\parallel x-y\parallel }_c+2\delta {\parallel x-Tx\parallel }_c,$$

(2)

where $\delta :=max\{a,\frac{b}{1-b},\frac{c}{1-c}\}$ and $0<\delta <1$.

Definition 1.12 Let X be a cone normed space. A self-map T of X is called a quasi-contraction if for some constant $\lambda \in (0,1)$ and for every $x,y\in X$, there exists

$$u\in C(T;x,y)\equiv \{{\parallel x-y\parallel }_c,{\parallel x-Tx\parallel }_c,{\parallel y-Ty\parallel }_c,{\parallel y-Tx\parallel }_c,{\parallel x-Ty\parallel }_c\}$$

such that ${\parallel Tx-Ty\parallel }_c\le \lambda u$. If this inequality holds for all $x\in X$ and $y\in F(T)$, we say that T is a f-quasi-contraction.

Definition 1.13 Let X be a cone normed space, T be a self-map of X and $p_0=u_0=x_0=v_0\in X$. The Picard iteration is given by

$$p_{n+1}=T{p}_n.$$

(3)

For a sequence of self-maps ${\{T_n\}}_{n\in \mathbb{N}}$, the iteration $p_{n+1}=T_n{p}_n$ is called the Picard’s S-iteration.

Another two well-known iteration procedures for obtaining fixed points of T are Mann iteration defined by

$$u_{n+1}=(1-{\alpha }_n)u_n+{\alpha }_{n}T{u}_n$$

(4)

and Ishikawa iteration defined by

$$\begin{array}{r}{x}_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_{n}T{z}_n,\\ z_n=(1-{\beta }_n)x_n+{\beta }_{n}T{x}_n,\end{array}$$

(5)

where $\{{\alpha }_n\}\subseteq (0,1)$ and $\{{\beta }_n\}\subseteq [0,1)$. Also, the Krasnoselskij iteration is defined by

$$v_{n+1}=(1-\lambda )v_n+\lambda T{v}_n,$$

(6)

where $\lambda \in (0,1)$.

If T is a self-map of X, then by $F(T)$ we mean the set of fixed points of T. Also, ${\mathbb{N}}_0$ denotes the set of nonnegative integers, i.e., ${\mathbb{N}}_0=\mathbb{N}\cup \{0\}$.

Lemma 1.14 ([5])

Let $(X,d_c)$ be a complete cone metric space and P be a normal cone. Suppose that the mapping $T:X\to X$ satisfies the contractive condition

$$d_c(Tx,Ty)\le k{d}_c(x,y)$$

for all $x,y\in X$, where $k\in [0,1)$ is a constant. Then T has a unique fixed point in X and for each $x\in X$, the iterative sequence $\{T^{n}x\}$ converges to the fixed point.

Lemma 1.15 ([5])

Let $(X,d_c)$ be a complete cone metric space and P be a normal cone. Suppose that the mapping $T:X\to X$ satisfies the contractive condition

$$d_c(Tx,Ty)\le k(d_c(Tx,x)+d_c(Ty,y))$$

for all $x,y\in X$, where $k\in [0,1/2)$ is a constant. Then T has a unique fixed point in X and for each $x\in X$, the iterative sequence $\{T^{n}x\}$ converges to the fixed point.

Lemma 1.16 ([5])

Let $(X,d_c)$ be a complete cone metric space and P be a normal cone. Suppose that the mapping $T:X\to X$ satisfies the contractive condition

$$d_c(Tx,Ty)\le k(d_c(Tx,y)+d_c(Ty,x))$$

for all $x,y\in X$, where $k\in [0,1/2)$ is a constant. Then T has a unique fixed point in X and for each $x\in X$, the iterative sequence $\{T^{n}x\}$ converges to the fixed point.

Lemma 1.17 ([2])

Let T be a quasi-contraction with $0<\lambda <1/2$. Then T is a Zamfirescu operator.

Definition 1.18 Let $(X,{\parallel \cdot \parallel }_c)$ be a cone normed space and ${\{T_n\}}_n$ be a sequence of self-maps of X with ${\bigcap }_{n}F(T_n)\ne \mathrm{\varnothing }$. Let $x_0$ be a point of X and assume that $x_{n+1}=f(T_n,x_n)$ is an iteration procedure involving $\{T_n\}$, which yields a sequence $\{x_n\}$ of points from X. The iteration $x_{n+1}=f(T_n,x_n)$ is said to be $\{T_n\}$-semistable (or semistable with respect to $\{T_n\}$) if whenever $\{x_n\}$ converges to a fixed point q in ${\bigcap }_{n}F(T_n)$ and $\{y_n\}$ is a sequence in X with ${lim}_{n\to \mathrm{\infty }}{\parallel y_{n+1}-f(T_n,y_n)\parallel }_c=0$ and ${\parallel y_n-f(T_n,y_n)\parallel }_c=o(t_n)$ for some sequence $\{t_n\}\subseteq {\mathbb{R}}^{+}$, then $y_n\to q$.

The iteration $x_{n+1}=f(T_n,x_n)$ is said to be $\{T_n\}$ -stable (or stable with respect to $\{T_n\}$) if $\{x_n\}$ converges to a fixed point q in ${\bigcap }_{n}F(T_n)$ and whenever $\{y_n\}$ is a sequence in X with ${lim}_{n\to \mathrm{\infty }}{\parallel y_{n+1}-f(T_n,y_n)\parallel }_c=0$, we have $y_n\to q$.

Note that if $T_n=T$ for all n, then Definition 1.18 gives the definitions of T-semistability and T-stability respectively.

Lemma 1.19 ([2])

Let $(X,d_c)$ be a cone metric space, P be a normal cone and ${\{T_n\}}_{n\in {\mathbb{N}}_0}$ be a sequence of self-maps of X with ${\bigcap }_{n}F(T_n)\ne \mathrm{\varnothing }$. Suppose that there exist nonnegative bounded sequences $\{a_n\}$, $\{b_n\}$ with ${sup}_n{b}_n<1$ such that

$$d_c(T_{n}x,q)\le a_n{d}_c(x,T_{n}x)+b_n{d}_c(x,q)$$

for each $n\in {\mathbb{N}}_0$, $x\in X$ and $q\in {\bigcap }_{n}F(T_n)$. Then the Picard’s S-iteration is semistable with respect to $\{T_n\}$.

Lemma 1.20 ([2])

Let $(X,d_c)$ be a cone metric space, P be a normal cone and ${\{T_n\}}_{n\in {\mathbb{N}}_0}$ be a sequence of self-maps of X with ${\bigcap }_{n}F(T_n)\ne \mathrm{\varnothing }$. If for all $n\in {\mathbb{N}}_0$, $T_n$ is a f-Zamfirescu operator with respect to $({\alpha }_n,{\beta }_n,{\gamma }_n)$ with ${sup}_n{\gamma }_n<1/2$. Then the Picard’s S-iteration is semistable with respect to $\{T_n\}$.

Lemma 1.21 ([2])

Under the conditions of Lemma  1.22 if $T_n$ is a Zamfirescu operator for all n, then the Picard’s S-iteration is semistable with respect to ${\{T_n\}}_n$.

Lemma 1.22 ([2])

Let $(X,d_c)$ be a cone metric space, P be a normal cone and ${\{T_n\}}_{n\in {\mathbb{N}}_0}$ be a sequence of self-maps of X with ${\bigcap }_{n}F(T_n)\ne \mathrm{\varnothing }$. If for all $n\in {\mathbb{N}}_0$, $T_n$ is a f-quasi-contraction with ${\lambda }_n$ such that ${sup}_n{\lambda }_n<1$, then the Picard’s S-iteration is semistable with respect to  ${\{T_n\}}_n$.

For some other sources on these topics, we refer to [6–23].

2 Main results

Theorem 2.1 Let X be a cone normed space and P be a normal cone. Suppose that T is a Zamfirescu self-map of X and $q\in F(T)$. Then the following are equivalent:

• (i) the Picard iteration converges to q,

• (ii) the Mann iteration converges to q.

Proof Let $\{{\alpha }_n\}\subseteq (0,1)$ be given. We prove the implication $\text{(i)}\Rightarrow \text{(ii)}$. Suppose that ${lim}_{n\to \mathrm{\infty }}{p}_n=q$. Now, by using (3) and (4), we have

$$\begin{array}{rcl}{\parallel u_{n+1}-p_{n+1}\parallel }_c& \le & (1-{\alpha }_n){\parallel u_n-T{p}_n\parallel }_c+{\alpha }_n{\parallel T{u}_n-T{p}_n\parallel }_c\\ \le & (1-{\alpha }_n){\parallel u_n-p_n\parallel }_c+(1-{\alpha }_n){\parallel p_n-T{p}_n\parallel }_c+{\alpha }_n{\parallel T{u}_n-T{p}_n\parallel }_c\\ \le & (1-{\alpha }_n){\parallel u_n-p_n\parallel }_c+(1-{\alpha }_n)({\parallel p_n-q\parallel }_c+{\parallel T{p}_n-Tq\parallel }_c)\\ +{\alpha }_n{\parallel T{u}_n-T{p}_n\parallel }_c.\end{array}$$

(7)

Using (2) with $x:=p_n$, $y:=u_n$, we get

$$\begin{array}{rcl}{\parallel T{u}_n-T{p}_n\parallel }_c& \le & \delta {\parallel u_n-p_n\parallel }_c+2\delta {\parallel p_n-T{p}_n\parallel }_c\\ \le & \delta {\parallel u_n-p_n\parallel }_c+2\delta ({\parallel p_n-q\parallel }_c+{\parallel T{p}_n-Tq\parallel }_c).\end{array}$$

(8)

Using (2) with $x:=q$, $y:=p_n$, we obtain

$${\parallel T{p}_n-Tq\parallel }_c\le \delta {\parallel p_n-q\parallel }_c.$$

(9)

Relations (7), (8) and (9) lead to

$${\parallel u_{n+1}-p_{n+1}\parallel }_c\le (1-(1-\delta ){\alpha }_n){\parallel u_n-p_n\parallel }_c+(1-{\alpha }_n+2\delta {\alpha }_n)(1+\delta ){\parallel p_n-q\parallel }_c.$$

Set

$$\begin{array}{c}{a}_n:={\parallel u_n-p_n\parallel }_c,\hfill \\ b_n:=(1-{\alpha }_n+2\delta {\alpha }_n)(1+\delta ){\parallel p_n-q\parallel }_c,\hfill \\ h:=1-\underset{n}{sup}{\alpha }_n.\hfill \end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}{\parallel p_n-q\parallel }_c=0$, by using Lemma 1.1, we get

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel u_n-p_n\parallel }_c=0.$$

Thus

$$0\le {\parallel u_n-q\parallel }_c\le {\parallel u_n-p_n\parallel }_c+{\parallel p_n-q\parallel }_c\to 0,$$

as $n\to \mathrm{\infty }$. This completes the proof.

Now we prove $\text{(ii)}\Rightarrow \text{(i)}$. Suppose that ${lim}_{n\to \mathrm{\infty }}{\parallel u_n-q\parallel }_c=0$. Applying (3) and (4), we have

$$\begin{array}{rcl}{\parallel u_{n+1}-p_{n+1}\parallel }_c& \le & (1-{\alpha }_n){\parallel u_n-T{p}_n\parallel }_c+{\alpha }_n{\parallel T{u}_n-T{p}_n\parallel }_c\\ \le & (1-{\alpha }_n){\parallel u_n-T{u}_n\parallel }_c+{\parallel T{u}_n-T{p}_n\parallel }_c.\end{array}$$

(10)

Using (2) with $x:=u_n$, $y:=p_n$, we obtain

$${\parallel T{u}_n-T{p}_n\parallel }_c\le \delta {\parallel u_n-p_n\parallel }_c+2\delta {\parallel u_n-T{u}_n\parallel }_c.$$

(11)

Therefore, from (10) and (11), we get

$$\begin{array}{rcl}{\parallel u_{n+1}-p_{n+1}\parallel }_c& \le & \delta {\parallel u_n-p_n\parallel }_c+(1-{\alpha }_n+2\delta ){\parallel u_n-T{u}_n\parallel }_c\\ \le & \delta {\parallel u_n-p_n\parallel }_c+(1-{\alpha }_n+2\delta )({\parallel u_n-q\parallel }_c+{\parallel T{u}_n-Tq\parallel }_c)\\ \le & \delta {\parallel u_n-p_n\parallel }_c+(1-{\alpha }_n+2\delta )(1+\delta ){\parallel u_n-q\parallel }_c.\end{array}$$

(12)

Put

$$\begin{array}{c}{a}_n:={\parallel u_n-p_n\parallel }_c,\hfill \\ b_n:=(1-{\alpha }_n+2\delta )(1+\delta ){\parallel u_n-q\parallel }_c,\hfill \\ h:=\delta .\hfill \end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}{b}_n=0$, by Lemma 1.1 and relation (12), we get ${lim}_{n\to \mathrm{\infty }}{\parallel u_n-p_n\parallel }_c=0$. Thus

$$\parallel p_n-q\parallel \le {\parallel p_n-u_n\parallel }_c+{\parallel u_n-q\parallel }_c\to 0,$$

as $n\to \mathrm{\infty }$ and so the proof is complete. □

Theorem 2.2 Let X be a cone normed space and P be a normal cone. Suppose that T is a Zamfirescu self-map of X and $q\in F(T)$. Then the following are equivalent:

• (i) the Picard iteration converges to q,

• (ii) the Krasnoselskij iteration converges to q.

Proof For ${\alpha }_n=\lambda$, the Mann iteration reduces to the Krasnoselskij iteration. Now apply the proof of Theorem 2.1. □

Theorem 2.3 Let X be a cone normed space and P be a normal cone. Suppose that T is a Zamfirescu operator of X and $q\in F(T)$. Then the following are equivalent:

• (i) the Mann iteration converges to q,

• (ii) the Ishikawa iteration converges to q.

Proof Let $\{{\alpha }_n\}\subseteq (0,1)$ and $\{{\beta }_n\}\subseteq [0,1)$ be given. We prove the implication $\text{(i)}\Rightarrow \text{(ii)}$. Suppose that ${lim}_{n\to \mathrm{\infty }}{u}_n=q$. Using

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel x_n-u_n\parallel }_c=0,$$

(13)

and

$$0\le {\parallel q-x_n\parallel }_c\le {\parallel u_n-q\parallel }_c+{\parallel x_n-u_n\parallel }_c,$$

we get ${lim}_{n\to \mathrm{\infty }}{x}_n=q$. The proof is complete if we prove relation (13).

Using (2), (4) and (5) with $x:=u_n$, $y:=z_n$, we have

$$\begin{array}{rcl}{\parallel u_{n+1}-x_{n+1}\parallel }_c& \le & {\parallel (1-{\alpha }_n)(u_n-x_n)+{\alpha }_n(T{u}_n-T{z}_n)\parallel }_c\\ \le & (1-{\alpha }_n){\parallel u_n-x_n\parallel }_c+{\alpha }_n{\parallel T{u}_n-T{z}_n\parallel }_c\\ \le & (1-{\alpha }_n){\parallel u_n-x_n\parallel }_c+{\alpha }_n\delta {\parallel u_n-z_n\parallel }_c+2{\alpha }_n\delta {\parallel u_n-T{u}_n\parallel }_c.\end{array}$$

(14)

Using (2) with $x:=u_n$, $y:=x_n$, we have

$$\begin{array}{rcl}{\parallel u_n-z_n\parallel }_c& \le & {\parallel (1-{\beta }_n)(u_n-x_n)+{\beta }_n(u_n-T{x}_n)\parallel }_c\\ \le & (1-{\beta }_n){\parallel u_n-x_n\parallel }_c+{\beta }_n{\parallel u_n-T{x}_n\parallel }_c\\ \le & (1-{\beta }_n){\parallel u_n-x_n\parallel }_c+{\beta }_n{\parallel u_n-T{u}_n\parallel }_c+{\beta }_n{\parallel T{u}_n-T{x}_n\parallel }_c\\ \le & (1-{\beta }_n){\parallel u_n-x_n\parallel }_c+{\beta }_n{\parallel u_n-T{u}_n\parallel }_c+{\beta }_n\delta {\parallel u_n-x_n\parallel }_c+2{\beta }_n\delta {\parallel u_n-T{u}_n\parallel }_c\\ =& (1-{\beta }_n(1-\delta )){\parallel u_n-x_n\parallel }_c+{\beta }_n(1+2\delta ){\parallel u_n-T{u}_n\parallel }_c.\end{array}$$

(15)

Relations (14) and (15) lead to

$$\begin{array}{rcl}{\parallel u_{n+1}-x_{n+1}\parallel }_c& \le & (1-{\alpha }_n){\parallel u_n-x_n\parallel }_c+{\alpha }_n\delta (1-{\beta }_n(1-\delta )){\parallel u_n-x_n\parallel }_c\\ +{\alpha }_n{\beta }_n\delta (1+2\delta )\parallel u_n-T{u}_n\parallel +2{\alpha }_n\delta {\parallel u_n-T{u}_n\parallel }_c\\ =& (1-{\alpha }_n(1-\delta (1-{\beta }_n(1-\delta )))){\parallel u_n-x_n\parallel }_c\\ +{\alpha }_n\delta ({\beta }_n(1+2\delta )+2){\parallel u_n-T{u}_n\parallel }_c.\end{array}$$

Put

$$\begin{array}{c}{a}_n:={\parallel u_n-x_n\parallel }_c,\hfill \\ b_n:={\alpha }_n\delta ({\beta }_n(1+2\delta )+2){\parallel u_n-T{u}_n\parallel }_c,\hfill \\ h:=1-\underset{n}{sup}{\alpha }_n.\hfill \end{array}$$

Note that ${lim}_{n\to \mathrm{\infty }}{\parallel u_n-q\parallel }_c=0$, T is Zamfirescu and $q\in F(T)$. By (2) we obtain

$$0\le {\parallel u_n-T{u}_n\parallel }_c\le {\parallel u_n-q\parallel }_c+{\parallel q-T{u}_n\parallel }_c\le (\delta +1){\parallel u_n-q\parallel }_c.$$

Hence ${lim}_{n\to \mathrm{\infty }}{\parallel u_n-T{u}_n\parallel }_c=0$; that is, ${lim}_{n\to \mathrm{\infty }}{b}_n=0$. Lemma 1.1 leads to

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel u_n-x_n\parallel }_c=0.$$

Now we will prove that $\text{(ii)}\Rightarrow \text{(i)}$. Using (2) with $x:=z_n$, $y:=u_n$, we obtain

$$\begin{array}{rcl}{\parallel x_{n+1}-u_{n+1}\parallel }_c& \le & {\parallel (1-{\alpha }_n)(x_n-u_n)+{\alpha }_n(T{z}_n-T{u}_n)\parallel }_c\\ \le & (1-{\alpha }_n){\parallel x_n-u_n\parallel }_c+{\alpha }_n{\parallel T{z}_n-T{u}_n\parallel }_c\\ \le & (1-{\alpha }_n){\parallel x_n-u_n\parallel }_c+{\alpha }_n\delta {\parallel z_n-u_n\parallel }_c+2{\alpha }_n\delta {\parallel z_n-T{z}_n\parallel }_c.\end{array}$$

(16)

Also, the following relation holds:

$$\begin{array}{rcl}{\parallel z_n-u_n\parallel }_c& \le & {\parallel (1-{\beta }_n)(x_n-u_n)+{\beta }_n(T{x}_n-u_n)\parallel }_c\\ \le & (1-{\beta }_n){\parallel x_n-u_n\parallel }_c+{\beta }_n{\parallel T{x}_n-u_n\parallel }_c\\ \le & (1-{\beta }_n){\parallel x_n-u_n\parallel }_c+{\beta }_n{\parallel T{x}_n-x_n\parallel }_c+{\beta }_n{\parallel x_n-u_n\parallel }_c\\ \le & {\parallel x_n-u_n\parallel }_c+{\beta }_n{\parallel T{x}_n-x_n\parallel }_c.\end{array}$$

(17)

Substituting (17) in (16), we obtain

$$\begin{array}{c}{\parallel x_{n+1}-u_{n+1}\parallel }_c\hfill \\ \le (1-{\alpha }_n){\parallel x_n-u_n\parallel }_c+{\alpha }_n\delta ({\parallel x_n-u_n\parallel }_c+{\beta }_n{\parallel T{x}_n-x_n\parallel }_c)+2{\alpha }_n\delta \parallel z_n-T{z}_n\parallel \hfill \\ \le (1-(1-\delta ){\alpha }_n){\parallel x_n-u_n\parallel }_c+{\alpha }_n{\beta }_n\delta {\parallel T{x}_n-x_n\parallel }_c+2{\alpha }_n\delta {\parallel z_n-T{z}_n\parallel }_c.\hfill \end{array}$$

(18)

Put

$$\begin{array}{c}{a}_n:={\parallel x_n-u_n\parallel }_c,\hfill \\ b_n:={\alpha }_n{\beta }_n\delta {\parallel T{x}_n-x_n\parallel }_c+2{\alpha }_n\delta {\parallel z_n-T{z}_n\parallel }_c,\hfill \\ h:=1-\underset{n}{sup}{\alpha }_n.\hfill \end{array}$$

From ${lim}_{n\to \mathrm{\infty }}{\parallel x_n-q\parallel }_c=0$, T is Zamfirescu, $q\in F(T)$ and by (2) we obtain

$$0\le {\parallel x_n-T{x}_n\parallel }_c\le {\parallel x_n-q\parallel }_c+{\parallel q-T{x}_n\parallel }_c\le (\delta +1){\parallel x_n-q\parallel }_c,$$

and

$$\begin{array}{rcl}0& \le & {\parallel z_n-T{z}_n\parallel }_c\\ \le & {\parallel z_n-q\parallel }_c+{\parallel q-T{z}_n\parallel }_c\\ \le & (\delta +1){\parallel z_n-q\parallel }_c\\ \le & (\delta +1)[(1-{\beta }_n){\parallel x_n-q\parallel }_c+{\beta }_n{\parallel q-T{x}_n\parallel }_c]\\ \le & (\delta +1)[(1-{\beta }_n){\parallel x_n-q\parallel }_c+\delta {\beta }_n{\parallel q-x_n\parallel }_c]\\ \le & (\delta +1)(1-{\beta }_n(1-\delta )){\parallel q-x_n\parallel }_c.\end{array}$$

Hence ${lim}_{n\to \mathrm{\infty }}{\parallel x_n-T{x}_n\parallel }_c=0$ and ${lim}_{n\to \mathrm{\infty }}{\parallel z_n-T{z}_n\parallel }_c=0$; that is, ${lim}_{n\to \mathrm{\infty }}{b}_n=0$.

Lemma 1.1 and (18) lead to ${lim}_{n\to \mathrm{\infty }}{\parallel x_n-u_n\parallel }_c=0$. Thus, we get

$$\parallel q-u_n\parallel \le {\parallel x_n-u_n\parallel }_c+{\parallel x_n-q\parallel }_c\to 0,$$

and the proof is complete. □

Corollary 2.4 Let X be a cone Banach space, P be a normal cone and T be a Zamfirescu self-map of X. Then T has a unique fixed point in X and the Picard, Mann, Krasnoselskij and Ishikawa iterative sequences converge to the fixed point of T.

Corollary 2.5 Let X be a cone Banach space, P be a normal cone and T be a quasi-contraction mapping of X with $0<\lambda <1/2$. Then T has a unique fixed point in X and the Picard, Mann, Krasnoselskij and Ishikawa iterative sequences converge to the fixed point of T.

Theorem 2.6 Let X be a cone Banach space and P be a normal cone. Suppose that T is a self-map of X and that every Picard and Mann iteration converges to a fixed point of T. Then the following are equivalent:

• (i) the Picard iteration is semistable with respect to T,

• (ii) the Mann iteration is semistable with respect to T.

Proof Suppose that q is a fixed point of T such that every Picard and Mann iteration converges to q. Let $\{y_n\}$ be an arbitrary sequence in X. For $\text{(i)}\Rightarrow \text{(ii)}$, let

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{y}_n\parallel }_c=0$$

and ${\parallel y_n-T{y}_n\parallel }_c=o(t_n)$ for some $\{t_n\}\subseteq {\mathbb{R}}^{+}$. We have

$${\parallel y_{n+1}-T{y}_n\parallel }_c\le {\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{y}_n\parallel }_c+(1-{\alpha }_n){\parallel y_n-T{y}_n\parallel }_c\to 0$$

as $n\to \mathrm{\infty }$. By assumption (i), we get ${lim}_{n\to \mathrm{\infty }}{y}_n=q$.

Conversely, we prove $\text{(ii)}\Rightarrow \text{(i)}$. Let ${lim}_{n\to \mathrm{\infty }}{\parallel y_{n+1}-T{y}_n\parallel }_c=0$ and ${\parallel y_n-T{y}_n\parallel }_c=o(t_n)$ for some $\{t_n\}\subseteq {\mathbb{R}}^{+}$. We have

$$\begin{array}{c}{\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{y}_n\parallel }_c\hfill \\ \le {\parallel y_{n+1}-T{y}_n\parallel }_c+(1-{\alpha }_n){\parallel y_{n+1}-y_n\parallel }_c+(1-{\alpha }_n){\parallel y_{n+1}-T{y}_n\parallel }_c\hfill \\ \le (2-{\alpha }_n){\parallel y_{n+1}-T{y}_n\parallel }_c+(1-{\alpha }_n){\parallel y_{n+1}-y_n\parallel }_c\hfill \\ \le (2-{\alpha }_n){\parallel y_{n+1}-T{y}_n\parallel }_c+(1-{\alpha }_n)({\parallel y_{n+1}-T{y}_n\parallel }_c+{\parallel y_n-T{y}_n\parallel }_c)\hfill \\ =(3-2{\alpha }_n){\parallel y_{n+1}-T{y}_n\parallel }_c+(1-{\alpha }_n){\parallel y_n-T{y}_n\parallel }_c\to 0\hfill \end{array}$$

as $n\to \mathrm{\infty }$. Thus ${lim}_{n\to \mathrm{\infty }}{y}_n=q$ and so the Picard iteration is semistable with respect to T. □

Theorem 2.7 Let X be a cone Banach space and P be a normal cone. Suppose that T is a self-map of X and that every Picard and Krasnoselskij iteration converges to a fixed point of T. Then the following are equivalent:

• (i) the Picard iteration is semistable with respect to T,

• (ii) the Krasnoselskij iteration is semistable with respect to T.

Proof In Theorem 2.7, put ${\alpha }_n=\lambda$. Then by the same method used in the proof of Theorem 2.7, we can complete the proof. □

Theorem 2.8 Let X be a cone Banach space and P be a normal cone. Suppose that $\{{\alpha }_n\}$ in Ishikawa iteration procedure satisfies ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, T is a self-map of X with bounded above range and also every Picard and Ishikawa iterative sequence converges to a fixed point of T. Then the following are equivalent:

• (i) the Picard iteration is semistable with respect to T,

• (ii) the Ishikawa iteration is semistable with respect to T.

Proof Suppose that q is a fixed point of T such that every Picard and Ishikawa iterative sequence converges to q. Let $\{y_n\}\subseteq X$ and $\{{\beta }_n\}\subseteq [0,1)$ be given and set

$$\begin{array}{c}{s}_n:=(1-{\beta }_n)y_n+{\beta }_{n}T{y}_n,\hfill \\ {\gamma }_n:={\parallel y_{n+1}-T{y}_n\parallel }_c,\hfill \\ {\delta }_n:={\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{s}_n\parallel }_c,\hfill \\ M:=sup\{N(Tx):x\in X\},\hfill \end{array}$$

where N is the norm type with respect to $({\parallel \cdot \parallel }_c)$. It is assumed that T has bounded above range and so, by Lemma 1.7, $M<\mathrm{\infty }$.

Now we prove that $\text{(i)}\Rightarrow \text{(ii)}$. Let ${lim}_{n\to \mathrm{\infty }}{\delta }_n=0$ and ${\parallel y_n-T{y}_n\parallel }_c=o(t_n)$ for some $\{t_n\}\subseteq {\mathbb{R}}^{+}$. Observe that

$$\begin{array}{rcl}{\gamma }_n& =& {\parallel y_{n+1}-T{y}_n\parallel }_c\\ \le & {\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{s}_n\parallel }_c+{\parallel (1-{\alpha }_n)y_n+{\alpha }_{n}T{s}_n-T{y}_n\parallel }_c\\ \le & {\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{s}_n\parallel }_c+{\alpha }_n({\parallel y_n\parallel }_c+{\parallel T{s}_n\parallel }_c)+{\parallel y_n-T{y}_n\parallel }_c\\ \le & {\delta }_n+{\alpha }_n({\parallel y_n-T{y}_n\parallel }_c+{\parallel T{y}_n\parallel }_c+{\parallel T{s}_n\parallel }_c)+{\parallel y_n-T{y}_n\parallel }_c\\ \le & {\delta }_n+{\alpha }_n({\parallel T{y}_n\parallel }_c+{\parallel T{s}_n\parallel }_c)+(1+{\alpha }_n){\parallel y_n-T{y}_n\parallel }_c.\end{array}$$

By Lemma 1.7 we have

$$\begin{array}{rcl}N({\gamma }_n)& \le & kN({\delta }_n+{\alpha }_n({\parallel T{y}_n\parallel }_c+{\parallel T{s}_n\parallel }_c)+(1+{\alpha }_n){\parallel y_n-T{y}_n\parallel }_c)\\ \le & kN({\delta }_n)+2kM{\alpha }_n+k(1+{\alpha }_n)N(y_n-T{y}_n)\to 0\end{array}$$

as $n\to \mathrm{\infty }$ (here k is the normal constant of P). So, by Lemma 1.9, ${lim}_{n\to \mathrm{\infty }}{\gamma }_n=0$ and the condition (i) assures that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$. Thus the Ishikawa iteration is semistable with respect to T.

Conversely, we prove $\text{(ii)}\Rightarrow \text{(i)}$. Let ${lim}_{n\to \mathrm{\infty }}{\gamma }_n=0$ and ${\parallel y_n-T{y}_n\parallel }_c=o(t_n)$ for some $\{t_n\}\subseteq {\mathbb{R}}^{+}$. We have

$$\begin{array}{rcl}{\delta }_n& =& {\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{s}_n\parallel }_c\\ \le & {\parallel y_{n+1}-T{y}_n\parallel }_c+{\parallel T{y}_n-(1-{\alpha }_n)y_n-{\alpha }_{n}T{s}_n\parallel }_c\\ =& {\parallel y_{n+1}-T{y}_n\parallel }_c+{\parallel T{y}_n-(1-{\alpha }_n)y_n-{\alpha }_{n}T{s}_n+{\alpha }_{n}T{y}_n-{\alpha }_{n}T{y}_n\parallel }_c\\ \le & {\gamma }_n+(1-{\alpha }_n){\parallel y_n-T{y}_n\parallel }_c+{\alpha }_n{\parallel T{y}_n-T{s}_n\parallel }_c.\end{array}$$

By Lemmas 1.7 and 1.9, we get

$$\begin{array}{rcl}N({\delta }_n)& \le & kN({\gamma }_n+(1-{\alpha }_n){\parallel y_n-T{y}_n\parallel }_c+{\alpha }_n{\parallel T{y}_n-T{s}_n\parallel }_c)\\ \le & kN({\gamma }_n)+k(1-{\alpha }_n)N(y_n-T{y}_n)+k{\alpha }_{n}N(T{y}_n-T{s}_n)\\ \le & kN({\gamma }_n)+k(1-{\alpha }_n)N(y_n-T{y}_n)+2kM{\alpha }_n\to 0\end{array}$$

as $n\to \mathrm{\infty }$, where k is the normal constant of P. So ${lim}_{n\to \mathrm{\infty }}{\delta }_n=0$ and by assumption (ii), we have ${lim}_{n\to \mathrm{\infty }}{y}_n=q$. Thus the Picard iteration is semistable with respect to T. □

Theorem 2.9 Let X be a cone Banach space and P be a normal cone. Suppose that $\{{\alpha }_n\}$ in Mann and Ishikawa procedures satisfies ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, T is a self-map of X with bounded above range and also every Mann and Ishikawa iterative sequence converges to a fixed point of T. Then the following are equivalent:

• (i) the Mann iteration is T-stable,

• (ii) the Ishikawa iteration is T-stable.

Proof Let q be a fixed point of T and every Mann and Ishikawa iterative sequence converge to q. Suppose that k is the normal constant of P and put

$$M:=sup\{N(Tx):x\in X\},$$

where N is the norm type with respect to ${\parallel \cdot \parallel }_c$. Since T has bounded above range, then $M<\mathrm{\infty }$. Now let $\{y_n\}$ be an arbitrary sequence in X. We prove $\text{(i)}\Rightarrow \text{(ii)}$. For this suppose that

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{s}_n\parallel }_c=0,$$

where $s_n=(1-{\beta }_n)y_n+{\beta }_{n}T{y}_n$ and $\{{\beta }_n\}\subseteq [0,1)$. We show that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$. Note that

$${\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{y}_n\parallel }_c\le {\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{s}_n\parallel }_c+{\parallel {\alpha }_{n}T{s}_n-{\alpha }_{n}T{y}_n\parallel }_c.$$

By Lemma 1.7 and Lemma 1.9, we obtain

$$N(y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{y}_n)\le kN(y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{s}_n)+2kM{\alpha }_n\to 0,$$

as $n\to \mathrm{\infty }$ and so

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{y}_n\parallel }_c=0.$$

Condition (i) assures that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$. Thus the Ishikawa iteration is T-stable.

Conversely, we prove $\text{(ii)}\Rightarrow \text{(i)}$. Suppose that

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{y}_n\parallel }_c=0.$$

We show that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$. Put

$$s_n:=(1-{\beta }_n)y_n+{\beta }_{n}T{y}_n,$$

and observe that

$${\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{s}_n\parallel }_c\le {\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{y}_n\parallel }_c+{\parallel {\alpha }_{n}T{y}_n-{\alpha }_{n}T{s}_n\parallel }_c.$$

By Lemma 1.7 and Lemma 1.9, we obtain

$$N(y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{s}_n)\le kN(y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{y}_n)+2kM{\alpha }_n\to 0$$

as $n\to \mathrm{\infty }$ and hence ${lim}_{n\to \mathrm{\infty }}{\parallel y_{n+1}-(1-{\alpha }_n)y_n-{\alpha }_{n}T{s}_n\parallel }_c=0$. By assumption (ii), we get ${lim}_{n\to \mathrm{\infty }}{y}_n=q$ and the proof is complete. □

Corollary 2.10 Let $(X,{\parallel \cdot \parallel }_c)$ be a cone normed space, P be a normal cone and T be a self-map of X and $q\in F(T)$. Suppose that there exist nonnegative real numbers a and b with $b<1$ such that

$${\parallel Tx-q\parallel }_c\le a{\parallel x-Tx\parallel }_c+b{\parallel x-q\parallel }_c$$

for each $x\in X$. Assume that for $\{{\alpha }_n\}\subseteq (0,1)$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, and let every Picard, Mann, Krasnoselskij and Ishikawa iterative sequence converge to q. Then the Picard, Mann and Krasnoselskij iterations are T-semistable. Moreover, if T has bounded above range, then the Ishikawa iteration is T-semistable.

Corollary 2.11 Let $(X,{\parallel \cdot \parallel }_c)$ be a cone normed space, P be a normal cone and T be a f-Zamfirescu or quasi-contraction self-map of X and $q\in F(T)$. Assume that $\{{\alpha }_n\}$ in Mann and Ishikawa iteration procedures satisfies $\{{\alpha }_n\}\subseteq (0,1)$ and ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$. Also, let every Picard, Mann, Krasnoselskij and Ishikawa iterative sequence converge to q. Then the Picard, Mann and Krasnoselskij iterations are T-semistable. Moreover, if T has bounded above range, then the Ishikawa iteration is T-semistable.