﻿ Fixed Point Results for K-Iteration Using Non-Linear Type Mappings

Open Access Library Journal
Vol.06 No.03(2019), Article ID:91276,14 pages
10.4236/oalib.1105245

Fixed Point Results for K-Iteration Using Non-Linear Type Mappings

Anju Panwar1, Ravi Parkash Bhokal2*

1Department of Mathematics, M. D. U. Rohtak, Haryana, India

2Government College, Dujana, Jhajjar (Haryana), India    Received: February 13, 2019; Accepted: March 17, 2019; Published: March 20, 2019

ABSTRACT

In this paper we establish convergence and stability results using general contractive condition, quasi-nonexpansive mapping and mean non expansive mapping for K-iteration process. We shall also generalize the K-iteration process for a pair of distinct mappings and with the help of example we claim that the generalized iteration process has better convergence rate than the K-iteration process for single mapping and some of the existing iteration processes. Suitable examples are given in the support of main results.

Subject Areas:

Mathematical Analysis

Keywords:

K-Iteration Process, Opial’s Condition, Mean Non-Expansive Mapping, Quasi Non-Expansive Mapping 1. Introduction and Preliminary Definitions

Let $\left(X,d\right)$ be a metric space and $T:X\to X$ be a self map defined on X. Let $F\left(T\right)=\left\{z\in X:Tz=z\right\}$ denote the set of fixed point of T. For ${x}_{0}\in X$ , the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ defined by

${x}_{n+1}=T{x}_{n},n\ge 0,$ (1.1)

is called the Picard iteration.

For ${x}_{0}\in X$ , the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ defined by

*Corrosponding author.

${x}_{n+1}=\left(1-{\alpha }_{n}\right){x}_{n}+{\alpha }_{n}T{x}_{n},n\ge 0,$ (1.2)

where ${\left\{{\alpha }_{n}\right\}}_{n=0}^{\infty }$ is a sequence in $\left[0,1\right]$ such that ${\sum }_{n=0}^{\infty }{\alpha }_{n}=\infty$ is called the Mann iteration process  .

In 2013, Khan  produced a new type of iteration process by introducing the concept of the following Picard-Mann hybrid iterative process for a single mapping T. For the initial value ${x}_{0}\in X$ , the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ defined by

${x}_{n+1}=T{y}_{n}$ ,

${y}_{n}=\left(1-{\alpha }_{n}\right){x}_{n}+{\alpha }_{n}T{x}_{n},n\ge 0,$ (1.3)

where ${\left\{{\alpha }_{n}\right\}}_{n=0}^{\infty }$ is a sequence in $\left[0,1\right]$ .

Khan  showed that the rate of convergence of Picard-Mann hybrid iterative process is more than the Picard iteration scheme, Mann iteration scheme  and Ishikawa iterative schemes  .

In this direction Gursoy and Karakaya  , gave new iteration process as follows:

For the initial value ${x}_{0}\in X$ , the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ defined by

$\left\{\begin{array}{l}{z}_{n}=\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}T{x}_{n},\\ {y}_{n}=\left(1-{\alpha }_{n}\right)T{x}_{n}+{\alpha }_{n}T{z}_{n},\\ {x}_{n+1}=T{y}_{n}\end{array}$ (1.4)

where ${\left\{{\alpha }_{n}\right\}}_{n=0}^{\infty }$ , ${\left\{{\beta }_{n}\right\}}_{n=0}^{\infty }$ is a sequence in $\left[0,1\right]$ is known as Picard-S iterative process. By giving appropriate example, Gursoy and Karakaya  proved that their iterative process has better convergence rate than Picard, Mann, Ishikawa, Noor and Normal-S iterative processes.

Karakaya et al. in their paper  , introduced a new hybrid iterative process as

$\left\{\begin{array}{l}{x}_{0}\in X,\\ {y}_{n}=T\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}T{x}_{n},\\ {x}_{n+1}=T\left(1-{\alpha }_{n}\right){y}_{n}+{\alpha }_{n}T{y}_{n}\end{array}$ (1.5)

where ${\left\{{\alpha }_{n}\right\}}_{n=0}^{\infty }$ , ${\left\{{\beta }_{n}\right\}}_{n=0}^{\infty }$ is a sequence in $\left[0,1\right]$ .

With the help of suitable example it was claimed by Karakaya et al.  , that their iteration process converges faster than the iteration process of Gursoy and Karakaya  .

In 2016, Thakur et al.  introduced a new iteration scheme called Thakur New Iteration Scheme as for the initial value ${x}_{0}\in X$ , the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ defined by

$\left\{\begin{array}{l}{z}_{n}=\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}T{x}_{n},\\ {y}_{n}=T\left(1-{\alpha }_{n}\right){x}_{n}+{\alpha }_{n}{z}_{n},\\ {x}_{n+1}=T{y}_{n}\end{array}$ (1.6)

where ${\left\{{\alpha }_{n}\right\}}_{n=0}^{\infty }$ , ${\left\{{\beta }_{n}\right\}}_{n=0}^{\infty }$ is a sequence in $\left[0,1\right]$ .

In  it was claimed that the Thakur New Iteration Scheme has higher convergence rate than the iteration process of Karakaya et al.  .

In the recent work of Hussain et al.  , a new iteration scheme has been developed and it is claimed that it has better convergence rate than the iterative process Thakur et al.  . This iteration process is called K-iteration process and is given as:

For the initial value ${x}_{0}\in X$ , the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ defined by

$\left\{\begin{array}{l}{z}_{n}=\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}T{x}_{n},\\ {y}_{n}=T\left(1-{\alpha }_{n}\right)T{x}_{n}+{\alpha }_{n}T{z}_{n},\\ {x}_{n+1}=T{y}_{n}\end{array}$ (1.7)

where ${\left\{{\alpha }_{n}\right\}}_{n=0}^{\infty }$ , ${\left\{{\beta }_{n}\right\}}_{n=0}^{\infty }$ is a sequence in $\left[0,1\right]$ .

In the present work we shall generalize some convergence and stability results for K-iteration process. We shall also prove convergence and stability results for more general form of K-iteration process and K-iteration process for a pair of two distinct mappings.

Definition 1.1  : Let X be a real Banach space. The mapping $T:X\to X$ is said to be asymptotically quasi-nonexpansive if $F\left(T\right)\ne \varnothing$ and there exists a sequence $\left\{{\mu }_{n}\right\}\subset \left[0,\infty \right)$ with ${\mu }_{n}\to 0$ as $n\to \infty$ such that

$‖{T}^{n}x-q‖\le \left(1+{\mu }_{n}\right)‖x-q‖$ (1.8)

for all $x\in X,q\in F\left(T\right)$ and $n\ge 0$ .

Definition 1.2  : Let X be a real Banach space. The mapping $T:X\to X$ is said to be mean non-expansive if there exists two non negative real numbers $a,b$ such that $a+b\le 1$ and for all $x,y\in X$ ,

$‖Tx-Ty‖=a‖x-y‖+b‖x-Ty‖$

Definition 1.3  : Let ${\left\{{z}_{n}\right\}}_{n=0}^{\infty }$ be any sequence in X. Then the iterative process ${x}_{n+1}=f\left(T,{x}_{n}\right)$ which converges to a fixed point q, is said to be stable with respect to the mapping T if for ${\phi }_{n}=‖{z}_{n+1}-f\left(T,{z}_{n}\right)‖,n=0,1,2,\cdots$ , we have ${\mathrm{lim}}_{n\to \infty }{\phi }_{n}=0$ if and only if ${\mathrm{lim}}_{n\to \infty }{z}_{n}=q$ .

Definition 1.4  : A space X is said to satisfy Opial’s condition if for each sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ in X such that ${x}_{n}$ converges weakly to x we have for all $y\in X$ , $x\ne y$ following holds:

1) $\mathrm{lim}{\mathrm{inf}}_{n\to \infty }‖{x}_{n}-x‖<\mathrm{lim}{\mathrm{inf}}_{n\to \infty }‖{x}_{n}-y‖$ ,

2) $\mathrm{lim}{\mathrm{sup}}_{n\to \infty }‖{x}_{n}-x‖<\mathrm{lim}{\mathrm{sup}}_{n\to \infty }‖{x}_{n}-y‖$ .

Lemma 1.5  : Let ${\left\{{a}_{n}\right\}}_{n=0}^{\infty }$ and ${\left\{{b}_{n}\right\}}_{n=0}^{\infty }$ be non-negative real sequences satisfying the inequality:

${a}_{n+1}\le \left(1-{b}_{n}\right){a}_{n}+{b}_{n}$ ,

where ${b}_{n}\in \left(0,1\right)$ , for all $n\in N$ , ${\sum }_{n=1}^{\infty }{b}_{n}=\infty$ and $\frac{{b}_{n}}{{a}_{n}}\to 0$ as $n\to \infty$ , then ${\mathrm{lim}}_{n\to \infty }{a}_{n}=0$ .

Lemma 1.6  : Let $\delta$ be a real number such that $0\le \delta <1$ , and ${\left\{{ϵ}_{n}\right\}}_{n=0}^{\infty }$ be a sequence of positive numbers such that ${\mathrm{lim}}_{n\to \infty }{ϵ}_{n}=0$ . Then for any sequence of positive numbers ${\left\{{a}_{n}\right\}}_{n=0}^{\infty }$ satisfying ${a}_{n+1}\le \delta {a}_{n}+{ϵ}_{n},n=0,1,2,\cdots$ , we have ${\mathrm{lim}}_{n\to \infty }{a}_{n}=0$ .

Lemma 1.7  : Let X be a real Banach space and $\left\{{g}_{n}\right\}$ be any sequence in X such that $0<{g}_{n}<1$ for all $n\in N$ . Let ${\left\{{a}_{n}\right\}}_{n=0}^{\infty }$ and ${\left\{{b}_{n}\right\}}_{n=0}^{\infty }$ be non-negative real sequences satisfying $\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}‖{a}_{n}‖\le c$ , $\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}‖{b}_{n}‖\le c$ and $\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}‖{g}_{n}{a}_{n}+\left(1-{g}_{n}\right){b}_{n}‖=c$ holds for some $c\ge 0$ . Then $\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}‖{a}_{n}-{b}_{n}‖=0$ .

2. Main Results

Theorem 2.1: Let X be a Banach space and $T:X\to X$ be a mapping satisfying the condition

$‖Tx-q‖\le \delta ‖x-q‖$ (2.1)

where $q\in F,x\in X$ and $0\le \delta <1$ . Let ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ be the sequence defined by the K-iterative process given by (1.7). Then the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ converges strongly to $q\in F\left(T\right)$ .

Proof: From (1.7) and (2.1) we have,

$‖{x}_{n+1}-q‖=‖T{y}_{n}-q‖\le \delta ‖T{y}_{n}-q‖$ (2.2)

And

$\begin{array}{c}‖{y}_{n}-q‖=‖T\left(\left(1-{\alpha }_{n}\right)T{x}_{n}+{\alpha }_{n}T{z}_{n}\right)-q‖\\ \le \delta ‖\left(1-{\alpha }_{n}\right)T{x}_{n}+{\alpha }_{n}T{z}_{n}-q‖\\ \le \delta ‖\left(1-{\alpha }_{n}\right)\left(T{x}_{n}-q\right)+{\alpha }_{n}\left(T{z}_{n}-q\right)‖\\ \le \delta \left[\left(1-{\alpha }_{n}\right)‖T{x}_{n}-q‖+{\alpha }_{n}‖T{z}_{n}-q‖\right]\\ \le \delta \left[\left(1-{\alpha }_{n}\right)‖T{x}_{n}-q‖+{\alpha }_{n}‖T{z}_{n}-q‖\right]\\ \le {\delta }^{2}\left[\left(1-{\alpha }_{n}\right)‖{x}_{n}-q‖+{\alpha }_{n}‖{z}_{n}-q‖\right]\end{array}$ (2.3)

Again using (1.7) and (2.1) we get,

$\begin{array}{c}‖{z}_{n}-q‖=‖\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}T{x}_{n}-q‖\\ \le \left(1-{\beta }_{n}\right)‖{x}_{n}-q‖+{\beta }_{n}‖T{x}_{n}-q‖\\ \le \left(1-{\beta }_{n}\right)‖{x}_{n}-q‖+{\beta }_{n}\delta ‖{x}_{n}-q‖\end{array}$ (2.4)

Using (2.4) in (2.3) we get,

$\begin{array}{c}‖{y}_{n}-q‖\le {\delta }^{2}\left[\left(1-{\alpha }_{n}\right)‖{x}_{n}-q‖+{\alpha }_{n}\left(1-{\beta }_{n}\right)‖{x}_{n}-q‖+{\alpha }_{n}{\beta }_{n}\delta ‖{x}_{n}-q‖\right]\\ \le {\delta }^{2}\left(1-{\alpha }_{n}+{\alpha }_{n}\left(1-{\beta }_{n}\right)+{\alpha }_{n}{\beta }_{n}\delta \right)‖{x}_{n}-q‖\\ \le {\delta }^{2}\left(1-{\alpha }_{n}{\beta }_{n}\left(1-\delta \right)\right)‖{x}_{n}-q‖\end{array}$ (2.5)

Using (2.5) in (2.2) we get,

$‖{x}_{n+1}-q‖\le {\delta }^{3}\left(1-{\alpha }_{n}{\beta }_{n}\left(1-\delta \right)\right)‖{x}_{n}-q‖$

Since $0\le \delta <1,{\alpha }_{n}\in \left[0,1\right)$ and ${\sum }_{n=0}^{\infty }{\alpha }_{n}=\infty$ . Hence by using lemma (1.6), we have

$\underset{n\to \infty }{\mathrm{lim}}‖{x}_{n+1}-q‖=0.$

Hence the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ converges strongly to q.

Corollary 2.2: (Akewe and Okeke  ) Let X be a Banach space and $T:X\to X$ be a mapping satisfying the condition

$‖Tx-q‖\le \delta ‖x-q‖$

where $q\in F,x\in X$ and $0\le \delta <1$ . Let ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ be the sequence defined by the Picard-Mann hybrid iterative process given by (1.3). Then the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ converges strongly to q.

Remark 2.3: Theorem 2.1 gives generalization to many results in the literature by considering a wider class of contractive type operators and more general iterative process, including the results of Chidume  , Bosede and Rhoades  and Akewe and Okeke  .

Theorem 2.4: Let X be a Banach space and $T:X\to X$ be a mapping satisfying the condition

$‖Tx-q‖\le \delta ‖x-q‖$

where $q\in F,x\in X$ and $0\le \delta <1$ . Let ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ be the sequence defined by the K-iterative process given by (1.7). Then the iteration process (1.7) is T-stable.

Proof: By theorem 2.1, the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ converges strongly to q. Let ${\left\{{u}_{n}\right\}}_{n=0}^{\infty }$ , ${\left\{{v}_{n}\right\}}_{n=0}^{\infty }$ and ${\left\{{w}_{n}\right\}}_{n=0}^{\infty }$ be real sequences in X.

Let ${\phi }_{n}=‖{u}_{n+1}-T{v}_{n}‖,n=0,1,2,\cdots$ , where

${w}_{n}=\left(1-{\beta }_{n}\right){u}_{n}+{\beta }_{n}T{u}_{n}$ ,

${v}_{n}=T\left(\left(1-{\alpha }_{n}\right)T{u}_{n}+{\alpha }_{n}T{w}_{n}\right)$ ,

${u}_{n+1}=T{v}_{n}$ ,

and let ${\mathrm{lim}}_{n\to \infty }{\phi }_{n}=0$ .

We shall prove that ${\mathrm{lim}}_{n\to \infty }{u}_{n}=q$ .

Now,

$\begin{array}{c}‖{u}_{n+1}-q‖=‖{u}_{n+1}-T{v}_{n}‖+‖T{v}_{n}-q‖\\ \le {\phi }_{n}+\delta ‖{v}_{n}-q‖\end{array}$ (2.6)

$\begin{array}{c}‖{v}_{n}-q‖=‖T\left(\left(1-{\alpha }_{n}\right)T{u}_{n}+{\alpha }_{n}T{w}_{n}\right)-q‖\\ \le \delta ‖\left(1-{\alpha }_{n}\right)T{u}_{n}+{\alpha }_{n}T{w}_{n}-q‖\\ \le \delta ‖\left(1-{\alpha }_{n}\right)\left(T{u}_{n}-q\right)+{\alpha }_{n}\left(T{w}_{n}-q\right)‖\\ \le \delta \left[\left(1-{\alpha }_{n}\right)‖T{u}_{n}-q‖+{\alpha }_{n}‖T{w}_{n}-q‖\right]\\ \le \delta \left[\left(1-{\alpha }_{n}\right)‖T{u}_{n}-q‖+{\alpha }_{n}‖T{w}_{n}-q‖\right]\\ \le {\delta }^{2}\left[\left(1-{\alpha }_{n}\right)‖{u}_{n}-q‖+{\alpha }_{n}‖{w}_{n}-q‖\right]\end{array}$ (2.7)

Again using (1.7) and (2.1) we get,

$\begin{array}{c}‖{w}_{n}-q‖=‖\left(1-{\beta }_{n}\right){u}_{n}+{\beta }_{n}T{u}_{n}-q‖\\ \le \left(1-{\beta }_{n}\right)‖{u}_{n}-q‖+{\beta }_{n}‖T{u}_{n}-q‖\\ \le \left(1-{\beta }_{n}\right)‖{u}_{n}-q‖+{\beta }_{n}\delta ‖{u}_{n}-q‖\\ \le \left(1-{\beta }_{n}\left(1-\delta \right)\right)‖{u}_{n}-q‖\end{array}$ (2.8)

Using (2.8) in (2.7) we get,

$\begin{array}{c}‖{v}_{n}-q‖\le {\delta }^{2}\left[\left(1-{\alpha }_{n}\right)‖{u}_{n}-q‖+{\alpha }_{n}\left(1-{\beta }_{n}\left(1-\delta \right)\right)‖{u}_{n}-q‖\right]\\ \le {\delta }^{2}\left(1-{\alpha }_{n}{\beta }_{n}\left(1-\delta \right)\right)‖{u}_{n}-q‖\end{array}$ (2.9)

Using (2.9) in (2.6) we get,

$‖{u}_{n+1}-q‖\le {\phi }_{n}+{\delta }^{3}\left(1-{\alpha }_{n}{\beta }_{n}\left(1-\delta \right)\right)‖{u}_{n}-q‖$ (2.10)

Since $0\le \delta <1$ and since $0\le {\alpha }_{n},{\beta }_{n}\le 1$ we have by lemma (1.6)

$\underset{n\to \infty }{\mathrm{lim}}{u}_{n}=q.$

Conversely let ${\mathrm{lim}}_{n\to \infty }{u}_{n}=q$ . We shall show that ${\mathrm{lim}}_{n\to \infty }{\phi }_{n}=0$ .

Now

$\begin{array}{c}{\phi }_{n}=‖{u}_{n+1}-T{v}_{n}‖\\ \le ‖{u}_{n+1}-q‖+‖Tq-T{v}_{n}‖\\ \le ‖{u}_{n+1}-q‖+\delta ‖{v}_{n}-q‖\end{array}$ (2.11)

Substituting (2.9) in (2.11),

${\phi }_{n}\le ‖{u}_{n+1}-q‖+{\delta }^{3}\left(1-{\alpha }_{n}{\beta }_{n}\left(1-\delta \right)\right)‖{u}_{n}-q‖$ (2.12)

Since ${\mathrm{lim}}_{n\to \infty }{u}_{n}=q$ , we have from (2.12) ${\mathrm{lim}}_{n\to \infty }{\phi }_{n}=0$ . Hence the K-iteration scheme is T-stable.

From theorem 2.4, we have the following corollary.

Corollary 2.5: Let X be a Banach space and $T:X\to X$ be a mapping satisfying the condition

$‖Tx-q‖\le \delta ‖x-q‖$ ,

where $q\in F,x\in X$ and $0\le \delta <1$ . Let ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ be the sequence defined by the Picard-Mann hybrid iterative process given by (1.3). Then the iteration process (1.3) is T-stable.

Example 2.6: Let $X=\left[0,1\right]$ and consider the mapping $Tx=\frac{x}{2}$ . The clearly the mapping T satisfies the inequality (2.1). Now $F\left(T\right)=0$ . Now we claim that the K-iteration scheme (1.7) is T-stable. Let us take ${\alpha }_{n}={\beta }_{n}=\frac{1}{2}$ and consider the sequences ${x}_{n}={y}_{n}={z}_{n}=\frac{1}{n}$ . Then clearly ${\mathrm{lim}}_{n\to \infty }{x}_{n}=0$ .

Now

$\begin{array}{l}{\phi }_{n}=‖{x}_{n+1}-T{y}_{n}‖=‖{x}_{n+1}-\frac{{y}_{n}}{2}‖\\ =‖{x}_{n+1}-\frac{T\left(\left(1-{\alpha }_{n}\right)T{x}_{n}+{\alpha }_{n}T{z}_{n}\right)}{2}‖\\ =‖{x}_{n+1}-\frac{\left(1-{\alpha }_{n}\right)T{x}_{n}+{\alpha }_{n}T{z}_{n}}{4}‖\\ =‖{x}_{n+1}-\left(\frac{\left(1-{\alpha }_{n}\right){x}_{n}}{8}+\frac{{\alpha }_{n}{z}_{n}}{8}\right)‖\\ =‖{x}_{n+1}-\left(\frac{\left(1-{\alpha }_{n}\right){x}_{n}}{8}+\frac{{\alpha }_{n}\left(1-{\beta }_{n}\right){x}_{n}}{8}+\frac{{\alpha }_{n}{\beta }_{n}T{x}_{n}}{8}\right)‖\\ =‖{x}_{n+1}-\left(\frac{\left(1-{\alpha }_{n}\right){x}_{n}}{8}+\frac{{\alpha }_{n}\left(1-{\beta }_{n}\right){x}_{n}}{8}+\frac{{\alpha }_{n}{\beta }_{n}{x}_{n}}{16}\right)‖\\ =‖\frac{1}{n+1}-\left(\frac{1}{16n}+\frac{1}{32n}+\frac{1}{64n}\right)‖=‖\frac{1}{n+1}-\frac{1}{8n}‖\end{array}$

$=‖\frac{7n-1}{8n\left(n+1\right)}‖=‖\frac{7-\frac{1}{n}}{8\left(n+1\right)}‖$ (2.13)

Taking limit $n\to \infty$ in (2.13), we have ${\mathrm{lim}}_{n\to \infty }{\phi }_{n}=0$ . Hence the K-iteration process is T-stable.

Now we shall prove the convergence and stability results for asymptotically quasi-nonexpansive mapping by considering the more general form of K-iteration process as:

${z}_{n}=\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}{T}^{n}{x}_{n}$ ,

${y}_{n}={T}^{n}\left(\left(1-{\alpha }_{n}\right){T}^{n}{x}_{n}+{\alpha }_{n}{T}^{n}{z}_{n}\right)$ ,

${x}_{n+1}={T}^{n}{y}_{n}$ , where $n=0,1,2,\cdots$ , (2.14)

Theorem 2.7: Let H be a non-empty closed convex subset of a Banach space X and $T:H\to H$ be asymptotically quasi-nonexpansive mapping with real sequence ${\mu }_{n}\subseteq \left[0,\infty \right)$ . Let ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ be the sequence defined by the K-iterative process given by (2.14) and satisfies the assumption that ${\sum }_{n=0}^{\infty }{\alpha }_{n}{\beta }_{n}{\mu }_{n}=\infty$ . Then the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ converges strongly to some fixed point q of the mapping T.

Proof: From the iterative process (2.14) we have,

$\begin{array}{c}‖{z}_{n}-q‖=‖\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}{T}^{n}{x}_{n}-q‖\\ \le \left(1-{\beta }_{n}\right)‖{x}_{n}-q‖+{\beta }_{n}‖{T}^{n}{x}_{n}-q‖\\ \le \left(1-{\beta }_{n}\right)‖{x}_{n}-q‖+{\beta }_{n}\left(1+{\mu }_{n}\right)‖{x}_{n}-q‖\\ \le \left(1+{\beta }_{n}{\mu }_{n}\right)‖{x}_{n}-q‖\end{array}$ (2.15)

and

$\begin{array}{c}‖{y}_{n}-q‖=‖{T}^{n}\left(\left(1-{\alpha }_{n}\right){T}^{n}{x}_{n}+{\alpha }_{n}{T}^{n}{z}_{n}\right)-q‖\\ \le \left(1+{\mu }_{n}\right)‖\left(1-{\alpha }_{n}\right){T}^{n}{x}_{n}+{\alpha }_{n}{T}^{n}{z}_{n}-q‖\\ \le \left(1+{\mu }_{n}\right)‖\left(1-{\alpha }_{n}\right)\left({T}^{n}{x}_{n}-q\right)+{\alpha }_{n}\left({T}^{n}{z}_{n}-q\right)‖\\ \le \left(1+{\mu }_{n}\right)\left[\left(1-{\alpha }_{n}\right)‖{T}^{n}{x}_{n}-q‖+{\alpha }_{n}‖{T}^{n}{z}_{n}-q‖\right]\end{array}$

$\begin{array}{l}\le \left(1+{\mu }_{n}\right)\left[\left(1-{\alpha }_{n}\right)\left(1+{\mu }_{n}\right)‖{x}_{n}-q‖+{\alpha }_{n}\left(1+{\mu }_{n}\right)‖{z}_{n}-q‖\right]\\ \le {\left(1+{\mu }_{n}\right)}^{2}\left[\left(1-{\alpha }_{n}\right)‖{x}_{n}-q‖+{\alpha }_{n}‖{z}_{n}-q‖\right]\\ \le {\left(1+{\mu }_{n}\right)}^{2}\left[\left(1-{\alpha }_{n}\right)‖{x}_{n}-q‖+{\alpha }_{n}\left(1+{\beta }_{n}{\mu }_{n}\right)‖{x}_{n}-q‖\right]\\ \le {\left(1+{\mu }_{n}\right)}^{2}\left(1-{\alpha }_{n}{\beta }_{n}{\mu }_{n}\right)‖{x}_{n}-q‖\end{array}$ (2.16)

Again using (2.14) we have,

$\begin{array}{c}‖{x}_{n+1}-q‖\le ‖{T}^{n}{y}_{n}-q‖\\ \le \left(1+{\mu }_{n}\right)‖{y}_{n}-q‖\\ \le {\left(1+{\mu }_{n}\right)}^{3}\left(1-{\alpha }_{n}{\beta }_{n}{\mu }_{n}\right)‖{x}_{n}-q‖\end{array}$ (2.17)

By repeating the above process, we have the following inequalities

$‖{x}_{n+1}-q‖\le {\left(1+{\mu }_{n}\right)}^{3}\left(1-{\alpha }_{n}{\beta }_{n}{\mu }_{n}\right)‖{x}_{n}-q‖$

$‖{x}_{n}-q‖\le {\left(1+{\mu }_{n-1}\right)}^{3}\left(1-{\alpha }_{n-1}{\beta }_{n-1}{\mu }_{n-1}\right)‖{x}_{n-1}-q‖$

$‖{x}_{n-1}-q‖\le {\left(1+{\mu }_{n-2}\right)}^{3}\left(1-{\alpha }_{n-2}{\beta }_{n-2}{\mu }_{n-2}\right)‖{x}_{n-2}-q‖$

$\cdots$

$‖{x}_{1}-q‖\le {\left(1+{\mu }_{0}\right)}^{3}\left(1-{\alpha }_{0}{\beta }_{0}{\mu }_{0}\right)‖{x}_{0}-q‖$

So we can write,

$‖{x}_{n+1}-q‖\le {\left(1+{\mu }_{0}\right)}^{3\left(n+1\right)}‖{x}_{0}-q‖{\prod }_{j=0}^{n}\left(1-{\alpha }_{j}{\beta }_{j}{\mu }_{j}\right)$

Since $1-x\le {\text{e}}^{-x}$ for all $x\in \left[0,1\right]$ . Now $1-{\alpha }_{j}{\beta }_{j}{\mu }_{j}<1$ , so we can write,

$\begin{array}{c}‖{x}_{n+1}-q‖\le {\left(1+{\mu }_{0}\right)}^{3\left(n+1\right)}‖{x}_{0}-q‖{\text{e}}^{-\left(1-{\alpha }_{j}{\beta }_{j}{\mu }_{j}\right)}\\ \le {\left(1+{\mu }_{0}\right)}^{3\left(n+1\right)}‖{x}_{0}-q‖{\text{e}}^{-{\sum }_{j=0}^{n}{\alpha }_{j}{\beta }_{j}{\mu }_{j}}\end{array}$ (2.18)

Taking limit $n\to \infty$ in (2.18), we have ${\mathrm{lim}}_{n\to \infty }‖{x}_{n}-q‖=0$ , that is the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ converges strongly to fixed point q of the mapping T.

Theorem 2.8: Let H be a non-empty closed convex subset of a Banach space X and $T:H\to H$ be asymptotically quasi-nonexpansive mapping with real sequence ${\mu }_{n}\subseteq \left[0,\infty \right)$ . Let ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ be the sequence defined by the K-iterative process given by (2.14) and satisfies the assumption that ${\sum }_{n=0}^{\infty }{\alpha }_{n}{\beta }_{n}{\mu }_{n}=\infty$ . Then the iterative process (2.14) is T-stable.

Proof: Let ${\left\{{u}_{n}\right\}}_{n=0}^{\infty }\subset X$ be any arbitrary sequence. Let the sequence generated by the iterative process (2.14) is ${x}_{n+1}=f\left(T,{x}_{n}\right)$ converging to the fixed point q.

Let ${\phi }_{n}=‖{u}_{n+1}-f\left(T,{x}_{n}\right)‖.$

We shall prove that ${\mathrm{lim}}_{n\to \infty }{\phi }_{n}=0$ if and only if ${\mathrm{lim}}_{n\to \infty }{u}_{n}=q$ .

First suppose ${\mathrm{lim}}_{n\to \infty }{\phi }_{n}=0$ . Now we have

$\begin{array}{c}‖{u}_{n+1}-q‖=‖{u}_{n+1}-f\left(T,{u}_{n}\right)‖+‖f\left(T,{u}_{n}\right)-q‖\\ ={\phi }_{n}+‖{T}^{n}\left({T}^{n}\left(1-{\beta }_{n}\right){T}^{n}{u}_{n}+{\beta }_{n}{T}^{n}\left(\left(1-{\alpha }_{n}\right){u}_{n}+{\alpha }_{n}{T}^{n}{u}_{n}\right)\right)-q‖\\ \le {\phi }_{n}+{\left(1+{\mu }_{n}\right)}^{3}\left(1-{\alpha }_{n}{\beta }_{n}{\mu }_{n}\right)‖{x}_{n}-q‖\end{array}$ (2.19)

where ${\alpha }_{n},{\beta }_{n}\in \left[0,1\right]$ , ${\mathrm{lim}}_{n\to \infty }{\phi }_{n}=0$ and ${\mathrm{lim}}_{n\to \infty }{\mu }_{n}=0$ .

Now using (2.19) together with lemma (1.5), we have ${\mathrm{lim}}_{n\to \infty }‖{u}_{n}-q‖=0$ that is ${\mathrm{lim}}_{n\to \infty }{u}_{n}=q$ .

Conversely let ${\mathrm{lim}}_{n\to \infty }{u}_{n}=q$ . we have

$\begin{array}{c}{\phi }_{n}=‖{u}_{n+1}-f\left(T,{u}_{n}\right)‖\\ \le ‖{u}_{n+1}-q‖+‖f\left(T,{u}_{n}\right)-q‖\\ \le ‖{u}_{n+1}-q‖+{\left(1+{\mu }_{n}\right)}^{3}\left(1-{\alpha }_{n}{\beta }_{n}{\mu }_{n}\right)‖{u}_{n}-q‖\end{array}$

Taking limit $n\to \infty$ both sides of (6) we have ${\mathrm{lim}}_{n\to \infty }{\phi }_{n}=0$ . Hence (2.14) is T-stable.

Now we shall prove the convergence results for mean non-expansive mapping by modifying the K-iteration process for two mappings as:

${z}_{n}=\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}S{x}_{n}$ ,

${y}_{n}=T\left(\left(1-{\alpha }_{n}\right)S{x}_{n}+{\alpha }_{n}T{z}_{n}\right)$ ,

${x}_{n+1}=T{y}_{n}$ , where $n=0,1,2,\cdots$ , (2.20)

Lemma 2.9: Let H be a non-empty closed convex subset of a Banach space X and $S,T:H\to H$ be two mean non-expansive mapping such that $F=F\left(T\right)\cap F\left(S\right)\ne \varphi$ . Let ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ be the sequence defined by the K-iterative process given by (2.20). Then ${\mathrm{lim}}_{n\to \infty }‖{x}_{n}-q‖$ exists for some $q\in F$ .

Proof: We have

$\begin{array}{l}‖{z}_{n}-q‖=‖\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}S{x}_{n}-q‖\\ \le \left(1-{\beta }_{n}\right)‖{x}_{n}-q‖+{\beta }_{n}‖S{x}_{n}-q‖\\ \le \left(1-{\beta }_{n}\right)‖{x}_{n}-q‖+{\beta }_{n}\left({a}_{1}‖{x}_{n}-q‖+{b}_{1}‖{x}_{n}-q‖\right)\\ \le \left(1-{\beta }_{n}\right)‖{x}_{n}-q‖+{\beta }_{n}\left({a}_{1}+{b}_{1}\right)‖{x}_{n}-q‖\\ \le ‖{x}_{n}-q‖\end{array}$ (2.21)

Again using (2.20) and (2.21)

$\begin{array}{c}‖{y}_{n}-q‖=‖T\left(\left(1-{\alpha }_{n}\right)S{x}_{n}+{\alpha }_{n}T{z}_{n}\right)-q‖\\ \le {a}_{2}‖\left(\left(1-{\alpha }_{n}\right)S{x}_{n}+{\alpha }_{n}T{z}_{n}\right)-q‖+{b}_{2}‖\left(\left(1-{\alpha }_{n}\right)S{x}_{n}+{\alpha }_{n}T{z}_{n}\right)-q‖\\ \le \left({a}_{2}+{b}_{2}\right)‖\left(\left(1-{\alpha }_{n}\right)S{x}_{n}+{\alpha }_{n}T{z}_{n}\right)-q‖\\ \le \left(1-{\alpha }_{n}\right)‖S{x}_{n}-q‖+{\alpha }_{n}‖T{z}_{n}-q‖\end{array}$

$\begin{array}{l}\le \left(1-{\alpha }_{n}\right)\left({a}_{1}‖{x}_{n}-q‖+{b}_{1}‖{x}_{n}-q‖\right)+{\alpha }_{n}\left({a}_{2}‖{z}_{n}-q‖+{b}_{2}‖{z}_{n}-q‖\right)\\ \le \left(1-{\alpha }_{n}\right)\left({a}_{1}+{b}_{1}\right)‖{x}_{n}-q‖+{\alpha }_{n}\left({a}_{2}+{b}_{2}\right)‖{z}_{n}-q‖\\ \le \left(1-{\alpha }_{n}\right)‖{x}_{n}-q‖+{\alpha }_{n}‖{z}_{n}-q‖\\ \le ‖{x}_{n}-q‖\end{array}$ (2.22)

Again using (2.20) and (2.22)

$\begin{array}{c}‖{x}_{n+1}-q‖\le ‖T{y}_{n}-q‖\\ \le {a}_{2}‖{y}_{n}-q‖+{b}_{2}‖{y}_{n}-q‖\\ \le \left({a}_{2}+{b}_{2}\right)‖{y}_{n}-q‖\\ \le ‖{y}_{n}-q‖\\ \le ‖{x}_{n}-q‖\end{array}$ (2.23)

This shows that $\left\{‖{x}_{n}-q‖\right\}$ is non-increasing and bounded sequence for $q\in F$ . Hence ${\mathrm{lim}}_{n\to \infty }‖{x}_{n}-q‖$ exists.

Lemma 2.10: Let be a non-empty closed convex subset of a Banach space and $S,T:H\to H$ be two mean non-expansive mapping such that $F=F\left(T\right)\cap F\left(S\right)\ne \varphi$ . Let ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ be the sequence defined by the K-iterative process given by (2.20). Also consider that ${\mathrm{lim}}_{n\to \infty }‖S{x}_{n}-q‖={\mathrm{lim}}_{n\to \infty }‖T{x}_{n}-q‖=0$ for some $q\in F$ . Then ${\mathrm{lim}}_{n\to \infty }‖T{x}_{n}-{x}_{n}‖=0$ .

Proof: Let $q\in F$ . In lemma (2.9) we have proved the existence of

${\mathrm{lim}}_{n\to \infty }‖{x}_{n}-q‖$ . Let ${\mathrm{lim}}_{n\to \infty }‖{x}_{n}-q‖=c$ . (2.24)

W.L.O.G. let $c>0$ .

Now from (2.20) and (2.24) we have,

$\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}‖{z}_{n}-q‖\le \underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}‖{x}_{n}-q‖=c$ (2.25)

Now

$\begin{array}{c}‖S{x}_{n}-q‖\le {a}_{1}‖{x}_{n}-q‖+{b}_{1}‖{x}_{n}-q‖\\ \le \left({a}_{1}+{b}_{1}\right)‖{x}_{n}-q‖\le ‖{x}_{n}-q‖\end{array}$

Implies that $\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}‖S{x}_{n}-q‖\le \underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}‖{x}_{n}-q‖=c$ (2.26)

Now

$\begin{array}{c}‖{x}_{n+1}-q‖\le ‖T{y}_{n}-q‖\le {a}_{2}‖{y}_{n}-q‖+{b}_{2}‖{y}_{n}-q‖\\ \le \left({a}_{2}+{b}_{2}\right)‖{y}_{n}-q‖\le ‖{y}_{n}-q‖\\ \le ‖T\left(\left(1-{\alpha }_{n}\right)S{x}_{n}+{\alpha }_{n}T{z}_{n}\right)-q‖\\ \le {a}_{2}‖\left(\left(1-{\alpha }_{n}\right)S{x}_{n}+{\alpha }_{n}T{z}_{n}\right)-q‖+{b}_{2}‖\left(\left(1-{\alpha }_{n}\right)S{x}_{n}+{\alpha }_{n}T{z}_{n}\right)-q‖\\ \le \left({a}_{2}+{b}_{2}\right)‖\left(\left(1-{\alpha }_{n}\right)S{x}_{n}+{\alpha }_{n}T{z}_{n}\right)-q‖\end{array}$

$\begin{array}{l}\le \left(1-{\alpha }_{n}\right)‖S{x}_{n}-q‖+{\alpha }_{n}‖T{z}_{n}-q‖\\ \le \left(1-{\alpha }_{n}\right)\left({a}_{1}‖{x}_{n}-q‖+{b}_{1}‖{x}_{n}-q‖\right)+{\alpha }_{n}\left({a}_{2}‖{z}_{n}-q‖+{b}_{2}‖{z}_{n}-q‖\right)\\ \le \left(1-{\alpha }_{n}\right)\left({a}_{1}+{b}_{1}\right)‖{x}_{n}-q‖+{\alpha }_{n}\left({a}_{2}+{b}_{2}\right)‖{z}_{n}-q‖\\ \le \left(1-{\alpha }_{n}\right)‖{x}_{n}-q‖+{\alpha }_{n}‖{z}_{n}-q‖\\ \le ‖{x}_{n}-q‖-{\alpha }_{n}‖{x}_{n}-q‖+{\alpha }_{n}‖{z}_{n}-q‖\end{array}$

$⇒\frac{‖{x}_{n+1}-q‖-‖{x}_{n}-q‖}{{\alpha }_{n}}=‖{z}_{n}-q‖-‖{x}_{n}-q‖$

and hence

$‖{x}_{n+1}-q‖-‖{x}_{n}-q‖\le \frac{‖{x}_{n+1}-q‖-‖{x}_{n}-q‖}{{\alpha }_{n}}=‖{z}_{n}-q‖-‖{x}_{n}-q‖$

which implies that $‖{x}_{n+1}-q‖\le ‖{z}_{n}-q‖$ (2.27)

Taking limit inferior in (2.27) we obtain

$c\le \underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}‖{z}_{n}-q‖$ (2.28)

From (2.20) and (2.28) we have

$\begin{array}{c}c=\underset{n\to \infty }{\mathrm{lim}}‖{z}_{n}-q‖\\ =\underset{n\to \infty }{\mathrm{lim}}‖\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}S{x}_{n}-q‖\\ =\underset{n\to \infty }{\mathrm{lim}}‖{\beta }_{n}\left(S{x}_{n}-q\right)+\left(1-{\beta }_{n}\right)\left({x}_{n}-q\right)‖\end{array}$ (2.29)

Now from (2.24), (2.26), (2.29) and lemma (1.7), we have $\underset{n\to \infty }{\mathrm{lim}}‖S{x}_{n}-{x}_{n}‖=0$ .

Now,

$‖T{x}_{n}-q‖\le {a}_{2}‖{x}_{n}-q‖+{b}_{2}‖{x}_{n}-q‖\le ‖{x}_{n}-q‖$

$⇒\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}‖T{x}_{n}-q‖\le \underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}‖{x}_{n}-q‖\le c$ (2.30)

Using the conditions of the lemma in (2.30), we can write

$C=\underset{n\to \infty }{\mathrm{lim}}‖{\beta }_{n}\left(T{x}_{n}-q\right)+\left(1-{\beta }_{n}\right)\left({x}_{n}-q\right)‖$ (2.31)

Using (2.24), (2.30), (2.31) along with the lemma (1.7), we have

${\mathrm{lim}}_{n\to \infty }‖T{x}_{n}-{x}_{n}‖=0.$

Theorem 2.11: Let H be a non-empty closed convex subset of a Banach space X satisfying Opial’s condition and S, T and ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ be same as defined in the lemma (2.10) .Then the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ converges weakly to some $q\in F$ .

Proof: From lemma (2.10) we have, ${\mathrm{lim}}_{n\to \infty }‖T{x}_{n}-{x}_{n}‖=0$ .

Since X is uniformly convex and hence it is reflexive so there exists a subsequence $\left\{{x}_{{n}_{m}}\right\}$ of $\left\{{x}_{n}\right\}$ such that $\left\{{x}_{{n}_{m}}\right\}$ converges weakly to some ${q}_{1}\in F$ . Since H is closed so ${q}_{1}\in H$ . Now we claim the weak convergence of $\left\{{x}_{n}\right\}$ to ${q}_{1}$ . Let it is not true, then there exists a subsequence of $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ which converges weakly to ${q}_{2}$ and let ${q}_{1}\ne {q}_{2}$ . Also ${q}_{2}\in F$ . Now from lemma (2.9) ${\mathrm{lim}}_{n\to \infty }‖{x}_{n}-{q}_{1}‖$ and ${\mathrm{lim}}_{n\to \infty }‖{x}_{n}-{q}_{2}‖$ both exist. Using Opial’s condition we have,

$\begin{array}{c}\underset{n\to \infty }{\mathrm{lim}}‖{x}_{n}-{q}_{1}‖\le \underset{n\to \infty }{\mathrm{lim}}‖{x}_{{n}_{m}}-{q}_{1}‖<\underset{n\to \infty }{\mathrm{lim}}‖{x}_{{n}_{m}}-{q}_{2}‖\\ ={\mathrm{lim}}_{n\to \infty }‖{x}_{n}-{q}_{2}‖={\mathrm{lim}}_{n\to \infty }‖{x}_{{n}_{i}}-{q}_{2}‖\\ <{\mathrm{lim}}_{n\to \infty }‖{x}_{{n}_{i}}-{q}_{1}‖\le {\mathrm{lim}}_{n\to \infty }‖{x}_{n}-{q}_{1}‖\end{array}$

This is a contradiction, so we must have ${q}_{1}={q}_{2}$ . Thus the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ converges weakly to some $q\in F$ .

Theorem 2.12: Let H be a non-empty closed compact subset of a Banach space X and S, T and ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ be same as defined in the lemma (2.10). Then the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ converges strongly to some $q\in F$ .

Proof: Since H is compact and hence it is sequentially compact. So there exists a subsequence $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ which converges to $q\in H$ .

Now

$\begin{array}{c}‖{x}_{{n}_{i}}-Tq‖=‖{x}_{{n}_{i}}-T{x}_{{n}_{i}}‖+‖T{x}_{{n}_{i}}-Tq‖\\ \le ‖{x}_{{n}_{i}}-T{x}_{{n}_{i}}‖+{a}_{2}‖{x}_{{n}_{i}}-q‖+{b}_{2}‖{x}_{{n}_{i}}-q‖\\ \le ‖{x}_{{n}_{i}}-T{x}_{{n}_{i}}‖+‖{x}_{{n}_{i}}-q‖\end{array}$ (2.32)

Taking limit $n\to \infty$ in (2.32) we have, $Tq=q$ that is $q\in F$ . We have earlier proved that ${\mathrm{lim}}_{n\to \infty }‖{x}_{n}-q‖$ exists for $q\in F$ . Hence the sequence ${\left\{{x}_{n}\right\}}_{n=0}^{\infty }$ converges strongly to some $q\in F$ .

In  it is proves that the K-iteration process converges faster than Picard-S, Thakur-New and Vatan two-step iterative process. Now we shall compare the rate of convergence the K-iteration process defined in  and our new modified K-iteration process for two mappings.

Table 1. Iterative values of K-iteration process and Modified K-iteration process.

Example 2.13: Let $S,T:\left[0,3\right]\to \left[0,3\right]$ be two mappings defined by $T\left(x\right)=\frac{x+2}{2}$ and $s\left(x\right)={\left(x+2\right)}^{\frac{1}{2}}$ . Let ${\alpha }_{n},{\beta }_{n}$ be the sequences defined by ${\alpha }_{n}={\beta }_{n}=\frac{1}{4}$ . Let the initial approximation be ${x}_{0}=2.25$ . Clearly S, T has

unique common fixed point 2. The convergence pattern of K-iteration process and modified K-iteration process is shown in Table 1.

Clearly we can conclude from Table 1, that the modified K-iteration process has better rate of convergence than the k-iteration process.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

Cite this paper

Panwar, A. and Bhokal, R.P. (2019) Fixed Point Results for K-Iteration Using Non-Linear Type Mappings. Open Access Library Journal, 6: e5245. https://doi.org/10.4236/oalib.1105245

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