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

Anju Panwar¹, Ravi Parkash Bhokal^2*

¹Department of Mathematics, M. D. U. Rohtak, Haryana, India.
²Government College, Dujana, Jhajjar (Haryana), India.

DOI: 10.4236/oalib.1105245                

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.

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 ${\displaystyle {\sum }_{n=0}^{\infty }{\alpha }_n}=\infty$ is called the Mann iteration process [1] .

In 2013, Khan [2] 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 [2] showed that the rate of convergence of Picard-Mann hybrid iterative process is more than the Picard iteration scheme, Mann iteration scheme [1] and Ishikawa iterative schemes [3] .

In this direction Gursoy and Karakaya [4] , 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

$\{\begin{array}{l}{z}_n=\left(1-{\beta }_n\right)x_n+{\beta }_{n}T{x}_n,\\ y_n=\left(1-{\alpha }_n\right)T{x}_n+{\alpha }_{n}T{z}_n,\\ x_{n+1}=T{y}_n\end{array}$ (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 [4] 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 [5] , introduced a new hybrid iterative process as

$\{\begin{array}{l}{x}_0\in X,\\ y_n=T\left(1-{\beta }_n\right)x_n+{\beta }_{n}T{x}_n,\\ x_{n+1}=T\left(1-{\alpha }_n\right)y_n+{\alpha }_{n}T{y}_n\end{array}$ (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. [5] , that their iteration process converges faster than the iteration process of Gursoy and Karakaya [4] .

In 2016, Thakur et al. [6] 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

$\{\begin{array}{l}{z}_n=\left(1-{\beta }_n\right)x_n+{\beta }_{n}T{x}_n,\\ y_n=T\left(1-{\alpha }_n\right)x_n+{\alpha }_n{z}_n,\\ x_{n+1}=T{y}_n\end{array}$ (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 [6] it was claimed that the Thakur New Iteration Scheme has higher convergence rate than the iteration process of Karakaya et al. [7] .

In the recent work of Hussain et al. [8] , a new iteration scheme has been developed and it is claimed that it has better convergence rate than the iterative process Thakur et al. [6] . 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

$\{\begin{array}{l}{z}_n=\left(1-{\beta }_n\right)x_n+{\beta }_{n}T{x}_n,\\ y_n=T\left(1-{\alpha }_n\right)T{x}_n+{\alpha }_{n}T{z}_n,\\ x_{n+1}=T{y}_n\end{array}$ (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 [3] : 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

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

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

Definition 1.2 [9] : 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 $x,y\in X$ ,

$\Vert Tx-Ty\Vert =a\Vert x-y\Vert +b\Vert x-Ty\Vert$

Definition 1.3 [10] : 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 ${\phi }_n=\Vert z_{n+1}-f\left(T,z_n\right)\Vert ,n=0,1,2,\cdots$ , we have ${\lim }_{n\to \infty }{\phi }_n=0$ if and only if ${\lim }_{n\to \infty }{z}_n=q$ .

Definition 1.4 [7] : 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) $\lim {\inf }_{n\to \infty }\Vert x_n-x\Vert <\lim {\inf }_{n\to \infty }\Vert x_n-y\Vert$ ,

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

Lemma 1.5 [11] : 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 $n\in N$ , ${\displaystyle {\sum }_{n=1}^{\infty }{b}_n}=\infty$ and $\frac{b_n}{a_n}\to 0$ as $n\to \infty$ , then ${\lim }_{n\to \infty }{a}_n=0$ .

Lemma 1.6 [12] : 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 ${\lim }_{n\to \infty }{ϵ}_n=0$ . 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 ${\lim }_{n\to \infty }{a}_n=0$ .

Lemma 1.7 [13] : 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 $\underset{n\to \infty }{\lim \sup }\Vert a_n\Vert \le c$ , $\underset{n\to \infty }{\lim \sup }\Vert b_n\Vert \le c$ and $\underset{n\to \infty }{\lim \sup }\Vert g_n{a}_n+\left(1-g_n\right)b_n\Vert =c$ holds for some $c\ge 0$ . Then $\underset{n\to \infty }{\lim \sup }\Vert a_n-b_n\Vert =0$ .

2. Main Results

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

$\Vert Tx-q\Vert \le \delta \Vert x-q\Vert$ (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,

$\Vert x_{n+1}-q\Vert =\Vert T{y}_n-q\Vert \le \delta \Vert T{y}_n-q\Vert$ (2.2)

And

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

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

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

Using (2.4) in (2.3) we get,

$\begin{array}{c}\Vert y_n-q\Vert \le {\delta }^2\left[\left(1-{\alpha }_n\right)\Vert x_n-q\Vert +{\alpha }_n\left(1-{\beta }_n\right)\Vert x_n-q\Vert +{\alpha }_n{\beta }_n\delta \Vert x_n-q\Vert \right]\\ \le {\delta }^2\left(1-{\alpha }_n+{\alpha }_n\left(1-{\beta }_n\right)+{\alpha }_n{\beta }_n\delta \right)\Vert x_n-q\Vert \\ \le {\delta }^2\left(1-{\alpha }_n{\beta }_n\left(1-\delta \right)\right)\Vert x_n-q\Vert \end{array}$ (2.5)

Using (2.5) in (2.2) we get,

$\Vert x_{n+1}-q\Vert \le {\delta }^3\left(1-{\alpha }_n{\beta }_n\left(1-\delta \right)\right)\Vert x_n-q\Vert$

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

$\underset{n\to \infty }{\lim }\Vert x_{n+1}-q\Vert =0.$

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

Corollary 2.2: (Akewe and Okeke [14] ) Let X be a Banach space and $T:X\to X$ be a mapping satisfying the condition

$\Vert Tx-q\Vert \le \delta \Vert x-q\Vert$

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 [15] , Bosede and Rhoades [16] and Akewe and Okeke [14] .

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

$\Vert Tx-q\Vert \le \delta \Vert x-q\Vert$

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 ${\phi }_n=\Vert u_{n+1}-T{v}_n\Vert ,n=0,1,2,\cdots$ , 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 ${\lim }_{n\to \infty }{\phi }_n=0$ .

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

Now,

$\begin{array}{c}\Vert u_{n+1}-q\Vert =\Vert u_{n+1}-T{v}_n\Vert +\Vert T{v}_n-q\Vert \\ \le {\phi }_n+\delta \Vert v_n-q\Vert \end{array}$ (2.6)

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

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

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

Using (2.8) in (2.7) we get,

$\begin{array}{c}\Vert v_n-q\Vert \le {\delta }^2\left[\left(1-{\alpha }_n\right)\Vert u_n-q\Vert +{\alpha }_n\left(1-{\beta }_n\left(1-\delta \right)\right)\Vert u_n-q\Vert \right]\\ \le {\delta }^2\left(1-{\alpha }_n{\beta }_n\left(1-\delta \right)\right)\Vert u_n-q\Vert \end{array}$ (2.9)

Using (2.9) in (2.6) we get,

$\Vert u_{n+1}-q\Vert \le {\phi }_n+{\delta }^3\left(1-{\alpha }_n{\beta }_n\left(1-\delta \right)\right)\Vert u_n-q\Vert$ (2.10)

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

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

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

Now

$\begin{array}{c}{\phi }_n=\Vert u_{n+1}-T{v}_n\Vert \\ \le \Vert u_{n+1}-q\Vert +\Vert Tq-T{v}_n\Vert \\ \le \Vert u_{n+1}-q\Vert +\delta \Vert v_n-q\Vert \end{array}$ (2.11)

Substituting (2.9) in (2.11),

${\phi }_n\le \Vert u_{n+1}-q\Vert +{\delta }^3\left(1-{\alpha }_n{\beta }_n\left(1-\delta \right)\right)\Vert u_n-q\Vert$ (2.12)

Since ${\lim }_{n\to \infty }{u}_n=q$ , we have from (2.12) ${\lim }_{n\to \infty }{\phi }_n=0$ . 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

$\Vert Tx-q\Vert \le \delta \Vert x-q\Vert$ ,

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 ${\lim }_{n\to \infty }{x}_n=0$ .

Now

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

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

Taking limit $n\to \infty$ in (2.13), we have ${\lim }_{n\to \infty }{\phi }_n=0$ . 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 ${\displaystyle {\sum }_{n=0}^{\infty }{\alpha }_n{\beta }_n{\mu }_n}=\infty$ . 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,

$\begin{array}{c}\Vert z_n-q\Vert =\Vert \left(1-{\beta }_n\right)x_n+{\beta }_n{T}^n{x}_n-q\Vert \\ \le \left(1-{\beta }_n\right)\Vert x_n-q\Vert +{\beta }_n\Vert T^n{x}_n-q\Vert \\ \le \left(1-{\beta }_n\right)\Vert x_n-q\Vert +{\beta }_n\left(1+{\mu }_n\right)\Vert x_n-q\Vert \\ \le \left(1+{\beta }_n{\mu }_n\right)\Vert x_n-q\Vert \end{array}$ (2.15)

and

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

$\begin{array}{l}\le \left(1+{\mu }_n\right)\left[\left(1-{\alpha }_n\right)\left(1+{\mu }_n\right)\Vert x_n-q\Vert +{\alpha }_n\left(1+{\mu }_n\right)\Vert z_n-q\Vert \right]\\ \le {\left(1+{\mu }_n\right)}^2\left[\left(1-{\alpha }_n\right)\Vert x_n-q\Vert +{\alpha }_n\Vert z_n-q\Vert \right]\\ \le {\left(1+{\mu }_n\right)}^2\left[\left(1-{\alpha }_n\right)\Vert x_n-q\Vert +{\alpha }_n\left(1+{\beta }_n{\mu }_n\right)\Vert x_n-q\Vert \right]\\ \le {\left(1+{\mu }_n\right)}^2\left(1-{\alpha }_n{\beta }_n{\mu }_n\right)\Vert x_n-q\Vert \end{array}$ (2.16)

Again using (2.14) we have,

$\begin{array}{c}\Vert x_{n+1}-q\Vert \le \Vert T^n{y}_n-q\Vert \\ \le \left(1+{\mu }_n\right)\Vert y_n-q\Vert \\ \le {\left(1+{\mu }_n\right)}^3\left(1-{\alpha }_n{\beta }_n{\mu }_n\right)\Vert x_n-q\Vert \end{array}$ (2.17)

By repeating the above process, we have the following inequalities

$\Vert x_{n+1}-q\Vert \le {\left(1+{\mu }_n\right)}^3\left(1-{\alpha }_n{\beta }_n{\mu }_n\right)\Vert x_n-q\Vert$

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

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

$\cdots$

$\Vert x_1-q\Vert \le {\left(1+{\mu }_0\right)}^3\left(1-{\alpha }_0{\beta }_0{\mu }_0\right)\Vert x_0-q\Vert$

So we can write,

$\Vert x_{n+1}-q\Vert \le {\left(1+{\mu }_0\right)}^{3\left(n+1\right)}\Vert x_0-q\Vert {\displaystyle {\prod }_{j=0}^n\left(1-{\alpha }_j{\beta }_j{\mu }_j\right)}$

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,

$\begin{array}{c}\Vert x_{n+1}-q\Vert \le {\left(1+{\mu }_0\right)}^{3\left(n+1\right)}\Vert x_0-q\Vert {\text{e}}^{-\left(1-{\alpha }_j{\beta }_j{\mu }_j\right)}\\ \le {\left(1+{\mu }_0\right)}^{3\left(n+1\right)}\Vert x_0-q\Vert {\text{e}}^{-{\displaystyle {\sum }_{j=0}^n{\alpha }_j{\beta }_j{\mu }_j}}\end{array}$ (2.18)

Taking limit $n\to \infty$ in (2.18), we have ${\lim }_{n\to \infty }\Vert x_n-q\Vert =0$ , 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 ${\displaystyle {\sum }_{n=0}^{\infty }{\alpha }_n{\beta }_n{\mu }_n}=\infty$ . 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 ${\phi }_n=\Vert u_{n+1}-f\left(T,x_n\right)\Vert .$

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

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

$\begin{array}{c}\Vert u_{n+1}-q\Vert =\Vert u_{n+1}-f\left(T,u_n\right)\Vert +\Vert f\left(T,u_n\right)-q\Vert \\ ={\phi }_n+\Vert T^n\left(T^n\left(1-{\beta }_n\right)T^n{u}_n+{\beta }_n{T}^n\left(\left(1-{\alpha }_n\right)u_n+{\alpha }_n{T}^n{u}_n\right)\right)-q\Vert \\ \le {\phi }_n+{\left(1+{\mu }_n\right)}^3\left(1-{\alpha }_n{\beta }_n{\mu }_n\right)\Vert x_n-q\Vert \end{array}$ (2.19)

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

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

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

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

Taking limit $n\to \infty$ both sides of (6) we have ${\lim }_{n\to \infty }{\phi }_n=0$ . 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 ${\lim }_{n\to \infty }\Vert x_n-q\Vert$ exists for some $q\in F$ .

Proof: We have

$\begin{array}{l}\Vert z_n-q\Vert =\Vert \left(1-{\beta }_n\right)x_n+{\beta }_{n}S{x}_n-q\Vert \\ \le \left(1-{\beta }_n\right)\Vert x_n-q\Vert +{\beta }_n\Vert S{x}_n-q\Vert \\ \le \left(1-{\beta }_n\right)\Vert x_n-q\Vert +{\beta }_n\left(a_1\Vert x_n-q\Vert +b_1\Vert x_n-q\Vert \right)\\ \le \left(1-{\beta }_n\right)\Vert x_n-q\Vert +{\beta }_n\left(a_1+b_1\right)\Vert x_n-q\Vert \\ \le \Vert x_n-q\Vert \end{array}$ (2.21)

Again using (2.20) and (2.21)

$\begin{array}{c}\Vert y_n-q\Vert =\Vert T\left(\left(1-{\alpha }_n\right)S{x}_n+{\alpha }_{n}T{z}_n\right)-q\Vert \\ \le a_2\Vert \left(\left(1-{\alpha }_n\right)S{x}_n+{\alpha }_{n}T{z}_n\right)-q\Vert +b_2\Vert \left(\left(1-{\alpha }_n\right)S{x}_n+{\alpha }_{n}T{z}_n\right)-q\Vert \\ \le \left(a_2+b_2\right)\Vert \left(\left(1-{\alpha }_n\right)S{x}_n+{\alpha }_{n}T{z}_n\right)-q\Vert \\ \le \left(1-{\alpha }_n\right)\Vert S{x}_n-q\Vert +{\alpha }_n\Vert T{z}_n-q\Vert \end{array}$

$\begin{array}{l}\le \left(1-{\alpha }_n\right)\left(a_1\Vert x_n-q\Vert +b_1\Vert x_n-q\Vert \right)+{\alpha }_n\left(a_2\Vert z_n-q\Vert +b_2\Vert z_n-q\Vert \right)\\ \le \left(1-{\alpha }_n\right)\left(a_1+b_1\right)\Vert x_n-q\Vert +{\alpha }_n\left(a_2+b_2\right)\Vert z_n-q\Vert \\ \le \left(1-{\alpha }_n\right)\Vert x_n-q\Vert +{\alpha }_n\Vert z_n-q\Vert \\ \le \Vert x_n-q\Vert \end{array}$ (2.22)

Again using (2.20) and (2.22)

$\begin{array}{c}\Vert x_{n+1}-q\Vert \le \Vert T{y}_n-q\Vert \\ \le a_2\Vert y_n-q\Vert +b_2\Vert y_n-q\Vert \\ \le \left(a_2+b_2\right)\Vert y_n-q\Vert \\ \le \Vert y_n-q\Vert \\ \le \Vert x_n-q\Vert \end{array}$ (2.23)

This shows that $\left\{\Vert x_n-q\Vert \right\}$ is non-increasing and bounded sequence for $q\in F$ . Hence ${\lim }_{n\to \infty }\Vert x_n-q\Vert$ 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 ${\lim }_{n\to \infty }\Vert S{x}_n-q\Vert ={\lim }_{n\to \infty }\Vert T{x}_n-q\Vert =0$ for some $q\in F$ . Then ${\lim }_{n\to \infty }\Vert T{x}_n-x_n\Vert =0$ .

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

${\lim }_{n\to \infty }\Vert x_n-q\Vert$ . Let ${\lim }_{n\to \infty }\Vert x_n-q\Vert =c$ . (2.24)

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

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

$\underset{n\to \infty }{\lim \sup }\Vert z_n-q\Vert \le \underset{n\to \infty }{\lim \sup }\Vert x_n-q\Vert =c$ (2.25)

Now

$\begin{array}{c}\Vert S{x}_n-q\Vert \le a_1\Vert x_n-q\Vert +b_1\Vert x_n-q\Vert \\ \le \left(a_1+b_1\right)\Vert x_n-q\Vert \le \Vert x_n-q\Vert \end{array}$

Implies that $\underset{n\to \infty }{\lim \sup }\Vert S{x}_n-q\Vert \le \underset{n\to \infty }{\lim \sup }\Vert x_n-q\Vert =c$ (2.26)

Now

$\begin{array}{c}\Vert x_{n+1}-q\Vert \le \Vert T{y}_n-q\Vert \le a_2\Vert y_n-q\Vert +b_2\Vert y_n-q\Vert \\ \le \left(a_2+b_2\right)\Vert y_n-q\Vert \le \Vert y_n-q\Vert \\ \le \Vert T\left(\left(1-{\alpha }_n\right)S{x}_n+{\alpha }_{n}T{z}_n\right)-q\Vert \\ \le a_2\Vert \left(\left(1-{\alpha }_n\right)S{x}_n+{\alpha }_{n}T{z}_n\right)-q\Vert +b_2\Vert \left(\left(1-{\alpha }_n\right)S{x}_n+{\alpha }_{n}T{z}_n\right)-q\Vert \\ \le \left(a_2+b_2\right)\Vert \left(\left(1-{\alpha }_n\right)S{x}_n+{\alpha }_{n}T{z}_n\right)-q\Vert \end{array}$

$\begin{array}{l}\le \left(1-{\alpha }_n\right)\Vert S{x}_n-q\Vert +{\alpha }_n\Vert T{z}_n-q\Vert \\ \le \left(1-{\alpha }_n\right)\left(a_1\Vert x_n-q\Vert +b_1\Vert x_n-q\Vert \right)+{\alpha }_n\left(a_2\Vert z_n-q\Vert +b_2\Vert z_n-q\Vert \right)\\ \le \left(1-{\alpha }_n\right)\left(a_1+b_1\right)\Vert x_n-q\Vert +{\alpha }_n\left(a_2+b_2\right)\Vert z_n-q\Vert \\ \le \left(1-{\alpha }_n\right)\Vert x_n-q\Vert +{\alpha }_n\Vert z_n-q\Vert \\ \le \Vert x_n-q\Vert -{\alpha }_n\Vert x_n-q\Vert +{\alpha }_n\Vert z_n-q\Vert \end{array}$

$\Rightarrow \frac{\Vert x_{n+1}-q\Vert -\Vert x_n-q\Vert }{{\alpha }_n}=\Vert z_n-q\Vert -\Vert x_n-q\Vert$

and hence

$\Vert x_{n+1}-q\Vert -\Vert x_n-q\Vert \le \frac{\Vert x_{n+1}-q\Vert -\Vert x_n-q\Vert }{{\alpha }_n}=\Vert z_n-q\Vert -\Vert x_n-q\Vert$

which implies that $\Vert x_{n+1}-q\Vert \le \Vert z_n-q\Vert$ (2.27)

Taking limit inferior in (2.27) we obtain

$c\le \underset{n\to \infty }{\lim \inf }\Vert z_n-q\Vert$ (2.28)

From (2.20) and (2.28) we have

$\begin{array}{c}c=\underset{n\to \infty }{\lim }\Vert z_n-q\Vert \\ =\underset{n\to \infty }{\lim }\Vert \left(1-{\beta }_n\right)x_n+{\beta }_{n}S{x}_n-q\Vert \\ =\underset{n\to \infty }{\lim }\Vert {\beta }_n\left(S{x}_n-q\right)+\left(1-{\beta }_n\right)\left(x_n-q\right)\Vert \end{array}$ (2.29)

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

Now,

$\Vert T{x}_n-q\Vert \le a_2\Vert x_n-q\Vert +b_2\Vert x_n-q\Vert \le \Vert x_n-q\Vert$

$\Rightarrow \underset{n\to \infty }{\lim \sup }\Vert T{x}_n-q\Vert \le \underset{n\to \infty }{\lim \sup }\Vert x_n-q\Vert \le c$ (2.30)

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

$C=\underset{n\to \infty }{\lim }\Vert {\beta }_n\left(T{x}_n-q\right)+\left(1-{\beta }_n\right)\left(x_n-q\right)\Vert$ (2.31)

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

${\lim }_{n\to \infty }\Vert T{x}_n-x_n\Vert =0.$

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, ${\lim }_{n\to \infty }\Vert T{x}_n-x_n\Vert =0$ .

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) ${\lim }_{n\to \infty }\Vert x_n-q_1\Vert$ and ${\lim }_{n\to \infty }\Vert x_n-q_2\Vert$ both exist. Using Opial’s condition we have,

$\begin{array}{c}\underset{n\to \infty }{\lim }\Vert x_n-q_1\Vert \le \underset{n\to \infty }{\lim }\Vert x_{n_m}-q_1\Vert <\underset{n\to \infty }{\lim }\Vert x_{n_m}-q_2\Vert \\ ={\lim }_{n\to \infty }\Vert x_n-q_2\Vert ={\lim }_{n\to \infty }\Vert x_{n_i}-q_2\Vert \\ <{\lim }_{n\to \infty }\Vert x_{n_i}-q_1\Vert \le {\lim }_{n\to \infty }\Vert x_n-q_1\Vert \end{array}$

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

$\begin{array}{c}\Vert x_{n_i}-Tq\Vert =\Vert x_{n_i}-T{x}_{n_i}\Vert +\Vert T{x}_{n_i}-Tq\Vert \\ \le \Vert x_{n_i}-T{x}_{n_i}\Vert +a_2\Vert x_{n_i}-q\Vert +b_2\Vert x_{n_i}-q\Vert \\ \le \Vert x_{n_i}-T{x}_{n_i}\Vert +\Vert x_{n_i}-q\Vert \end{array}$ (2.32)

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

In [8] 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 [8] 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.

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