Research  Open  Published:
Rate of convergence of Mann, Ishikawa and Noor iterations for continuous functions on an arbitrary interval
Journal of Inequalities and Applicationsvolume 2013, Article number: 269 (2013)
Abstract
In this paper, we relax the control condition of convergence of SPiteration presented by Phuengrattana and Suantai (J. Comput. Appl. Math. 235:30063014, 2011). We compare the rate of convergence of Mann, Ishikawa and Noor iterations from another point of view and come to a different conclusion. Finally, we provide a numerical example for Ishikawa and Noor iterations, which supports our theoretical results.
MSC:47H05, 47H07, 47H10.
1 Introduction
Let E be a closed interval on the real line and let $f:E\to E$ be a continuous function. A point $p\in E$ is a fixed point of f if $f(p)=p$. We denote by $F(f)$ the set of fixed points of f. It is known that if E is also bounded, then $F(f)$ is nonempty.
Mann iteration (see [1]) is defined by ${u}_{1}\in E$ and
for all $n\ge 1$, where ${\{{\alpha}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is a sequence in $[0,1]$. Ishikawa iteration (see [2]) is defined by ${s}_{1}\in E$ and
for all $n\ge 1$, where ${\{{\alpha}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\beta}_{n}\}}_{n=1}^{\mathrm{\infty}}$ are sequences in $[0,1]$. Noor iteration (see [3]) is defined by ${w}_{1}\in E$ and
for all $n\ge 1$, where ${\{{\alpha}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\beta}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\gamma}_{n}\}}_{n=1}^{\mathrm{\infty}}$ are sequences in $[0,1]$. Clearly, Mann and Ishikawa iterations are special cases of Noor iteration, and Mann iteration is a special case of Ishikawa iteration.
In 1974, Rhoades [4] proved the convergence of Mann iteration for a class of continuous and nondecreasing functions on a closed unit interval, and then he [5] extended convergence results to Ishikawa iterations. Further, Borwein and Borwein [6] proved the convergence of Mann iteration of continuous functions on a bounded closed interval. Recently, Qing and Qihou [7] extended results in [6] to an arbitrary interval and to Ishikawa iteration and presented a necessary and sufficient condition for the convergence of Ishikawa iteration of continuous functions on an arbitrary interval.
Very recently, Phuengrattana and Suantai [8] introduced SPiteration as follows:
for all $n\ge 1$, where ${\{{\alpha}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\beta}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\gamma}_{n}\}}_{n=1}^{\mathrm{\infty}}$ are sequences in $[0,1]$, and it will be denoted by $SP({x}_{1},{\alpha}_{n},{\beta}_{n},{\gamma}_{n},f)$. They presented a necessary and sufficient condition for the convergence of SPiteration (1.4) of continuous functions on an arbitrary interval. They also compared the convergence speed of Mann, Ishikawa, Noor iterations and SPiteration and concluded that SPiteration is better than the others.
Inspired by the above work, in this paper, we compare the rate of convergence of Mann, Ishikawa and Noor iterations under the same computation cost and come to a different conclusion with Phuengrattana and Suantai [8]. We also present a numerical example for Ishikawa and Noor iterations, which verifies our theoretical results.
2 Convergence theorem
In this section, we present a new necessary and sufficient condition for the convergence of SPiteration (1.4), which relaxes the control condition presented by Phuengrattana and Suantai [8].
Phuengrattana and Suantai [8] proposed the following necessary and sufficient condition for the convergence of SPiteration (1.4) of continuous functions on an arbitrary interval.
Proposition 2.1 Let E be a closed interval on the real line and let $f:E\to E$ be a continuous function. For ${x}_{1}\in E$, let SPiteration ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ be defined by (1.4), where ${\{{\alpha}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\beta}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\gamma}_{n}\}}_{n=1}^{\mathrm{\infty}}$ are sequences in $[0,1]$ satisfying

(i)
${lim}_{n\to \mathrm{\infty}}{\alpha}_{n}=0$,

(ii)
${\sum}_{n=1}^{\mathrm{\infty}}{\alpha}_{n}=\mathrm{\infty}$,

(iii)
${\sum}_{n=1}^{\mathrm{\infty}}{\beta}_{n}<\mathrm{\infty}$ and

(iv)
${\sum}_{n=1}^{\mathrm{\infty}}{\gamma}_{n}<\mathrm{\infty}$.
Then ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is bounded if and only if ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to a fixed point of f.
Next proposition reveals that it is of interest to relax the control conditions on${\{{\alpha}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\beta}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{{\gamma}_{n}\}}_{n=1}^{\mathrm{\infty}}$ in Proposition 2.1.
Proposition 2.2 [8]
Let E be a closed interval on the real line and let $f:E\to E$ be a continuous and nondecreasing function such that $F(f)$ is nonempty and bounded. Let ${\{{\alpha}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\beta}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\gamma}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\alpha}_{n}^{\ast}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\beta}_{n}^{\ast}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\gamma}_{n}^{\ast}\}}_{n=1}^{\mathrm{\infty}}$ be sequences in $[0,1)$ such that ${\alpha}_{n}<{\alpha}_{n}^{\ast}$, ${\beta}_{n}<{\beta}_{n}^{\ast}$ and ${\gamma}_{n}<{\gamma}_{n}^{\ast}$ for all $n\ge 1$. Let ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{{x}_{n}^{\ast}\}}_{n=1}^{\mathrm{\infty}}$ be defined by $SP({x}_{1},{\alpha}_{n},{\beta}_{n},{\gamma}_{n},f)$ and $SP({x}_{1}^{\ast},{\alpha}_{n}^{\ast},{\beta}_{n}^{\ast},{\gamma}_{n}^{\ast},f)$, respectively. If ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to $p\in F(f)$, then ${\{{x}_{n}^{\ast}\}}_{n=1}^{\mathrm{\infty}}$ converges to p. Moreover, ${\{{x}_{n}^{\ast}\}}_{n=1}^{\mathrm{\infty}}$ is better than ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$, provided that ${x}_{1}^{\ast}={x}_{1}\in E$.
Consider the following threestep Mann iteration:
where ${\{{\lambda}_{3ni}\}}_{n=1}^{\mathrm{\infty}}$, $i=0,1,2$, are sequences in $[0,1]$.
Remark 2.1 Let ${\lambda}_{3n2}={\gamma}_{n}$, ${\lambda}_{3n1}={\beta}_{n}$, ${\lambda}_{3n}={\alpha}_{n}$ and ${v}_{1}={x}_{1}$, then (2.1) transforms into (1.4) with ${x}_{n}={v}_{3n2}$, ${z}_{n}={v}_{3n1}$, ${y}_{n}={v}_{3n}$, ${x}_{n+1}={v}_{3n+1}$. So, onestep SPiteration is exactly threestep Mann iteration.
Theorem 2.1 Let E be a closed interval on the real line and let $f:E\to E$ be a continuous function. For ${x}_{1}\in E$, let SPiteration ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ be defined by (1.4), where ${\{{\alpha}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\beta}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\gamma}_{n}\}}_{n=1}^{\mathrm{\infty}}$ are sequences in $[0,1]$ satisfying the conditions:

(i)
${\sum}_{n=1}^{\mathrm{\infty}}{\alpha}_{n}=\mathrm{\infty}$,

(ii)
${lim}_{n\to \mathrm{\infty}}{\alpha}_{n}=0$, ${lim}_{n\to \mathrm{\infty}}{\beta}_{n}=0$, ${lim}_{n\to \mathrm{\infty}}{\gamma}_{n}=0$.
Then ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is bounded if and only if ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to a fixed point of f.
Proof Let ${\lambda}_{3n2}={\gamma}_{n}$, ${\lambda}_{3n1}={\beta}_{n}$, ${\lambda}_{3n}={\alpha}_{n}$ and ${v}_{1}={x}_{1}$, then ${x}_{n}={v}_{3n2}$, ${z}_{n}={v}_{3n1}$, ${y}_{n}={v}_{3n}$, ${x}_{n+1}={v}_{3n+1}$. We divide the proof into three steps.
Step 1. By conditions (i)(ii), it is obvious that ${\{{\lambda}_{n}\}}_{n=1}^{\mathrm{\infty}}$ satisfies ${lim}_{n\to \mathrm{\infty}}{\lambda}_{n}=0$ and ${\sum}_{n=1}^{\mathrm{\infty}}{\lambda}_{n}=\mathrm{\infty}$. From Proposition 2.1, it follows that ${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is bounded if and only if ${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to a fixed point of f.
Step 2. Since ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is a subsequence of ${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}$, so ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is bounded if ${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is bounded. On the other hand, assume that ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is bounded, then ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ belongs to a bounded closed interval. By the continuity of f, we have that ${\{f({x}_{n})\}}_{n=1}^{\mathrm{\infty}}$ belongs to another bounded closed interval, and thus ${\{f({x}_{n})\}}_{n=1}^{\mathrm{\infty}}$ is bounded. Since ${z}_{n}=(1{\gamma}_{n}){x}_{n}+{\gamma}_{n}f({x}_{n})$, so ${\{{z}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is bounded, and thus ${\{f({z}_{n})\}}_{n=1}^{\mathrm{\infty}}$ is bounded. Similarly, since ${y}_{n}=(1{\beta}_{n}){z}_{n}+{\gamma}_{n}f({z}_{n})$, we have ${\{{y}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{f({y}_{n})\}}_{n=1}^{\mathrm{\infty}}$ are bounded. Since ${v}_{3n1}={z}_{n}$, ${v}_{3n}={y}_{n}$, ${v}_{3n+1}={x}_{n+1}$ for all $n\ge 1$, we obtain that ${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is bounded. So, ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is bounded if and only if ${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is bounded.
Step 3. Since ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is a subsequence of ${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}$, so ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to a fixed point of f if ${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to a fixed point of f. On the other hand, assume that ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to a fixed point p of f, then ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is bounded. From Step 2, it follows that ${\{f({x}_{n})\}}_{n=1}^{\mathrm{\infty}}$, ${\{{z}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{f({z}_{n})\}}_{n=1}^{\mathrm{\infty}}$ are bounded. Using (1.4), we have ${z}_{n}{x}_{n}={\gamma}_{n}(f({x}_{n}){x}_{n})$ and ${y}_{n}{z}_{n}={\beta}_{n}(f({z}_{n}){z}_{n})$. By the condition (ii), we get ${z}_{n}{x}_{n}\to 0$ and ${y}_{n}{z}_{n}\to 0$, so ${\{{z}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{{y}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converge to p. Since ${v}_{3n1}={z}_{n}$, ${v}_{3n}={y}_{n}$, ${v}_{3n+1}={x}_{n+1}$ for all $n\ge 1$, ${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to p. Therefore, ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to a fixed point of f if and only if ${\{{v}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to a fixed point of f. □
From Step 1, Step 2 and Step 3, the result follows.
Remark 2.2 The comparison of Proposition 2.1 and Theorem 2.1 implies that the conditions on parameters ${\{{\beta}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{{\gamma}_{n}\}}_{n=1}^{\mathrm{\infty}}$ are relaxed.
3 Rate of convergence
In this section, we compare the rate of convergence of Ishikawa and Noor iterations under the same computation cost.
In order to compare the rate of convergence, we use the following definition introduced by Rhoades [5].
Definition 3.1 Let E be a closed interval on the real line and let $f:E\to E$ be a continuous function. Suppose that ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{{y}_{n}\}}_{n=1}^{\mathrm{\infty}}$ are two iterations which converge to the fixed point p of f. Then ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is said to be better than ${\{{y}_{n}\}}_{n=1}^{\mathrm{\infty}}$ if
Phuengrattana and Suantai [8] obtained the following theorem on the relation of the convergence of Mann, Ishikawa and Noor iterations and SPiteration.
Proposition 3.1 Let E be a closed interval on the real line and $f:E\to E$ be a continuous and nondecreasing function such that $F(f)$ is nonempty and bounded. For ${u}_{1}={s}_{1}={w}_{1}={x}_{1}\in E$, let ${\{{u}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{s}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{w}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ be the sequences defined by (1.1)(1.4), respectively. Let ${\{{\alpha}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\beta}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\gamma}_{n}\}}_{n=1}^{\mathrm{\infty}}$ be sequences in $[0,1)$. Then the following are satisfied:

(i)
Ishikawa iteration ${\{{s}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to $p\in F(f)$ if and only if Mann iteration ${\{{u}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to p. Moreover, Ishikawa iteration is better than Mann iteration;

(ii)
Noor iteration ${\{{w}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to $p\in F(f)$ if and only if Ishikawa iteration ${\{{s}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to p. Moreover, Noor iteration is better than Ishikawa iteration;

(iii)
SPiteration ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to $p\in F(f)$ if and only if Noor iteration ${\{{w}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to p. Moreover, SPiteration is better than Noor iteration.
Remark 3.1 In above Proposition 3.1, Phuengrattana and Suantai [8] compared the rate of convergence of Mann, Ishikawa, Noor iterations and SPiteration and drew the conclusion that SPiteration is better than other iterations, Noor iteration is better than Ishikawa iteration and Ishikawa iteration is better than Mann iteration. However, we know from Remark 2.1 that onestep SPiteration is threestep Mann iteration. Clearly, the computation cost of onestep Ishikawa iteration and onestep Noor iteration equals to that of twostep Mann iteration and threestep Mann iteration, respectively. So, it seems to be more reasonable to compare the rate of convergence of Mann, Ishikawa and Noor iterations under the same computation cost. In this sense, from Proposition 3.1(iii), Mann iteration is better than Ishikawa and Noor iterations.
Next we compare the rate of convergence of Ishikawa and Noor iterations under the same computation cost. For purposes of comparison, we firstly define two iterations. Threestep Ishikawa iteration (denoted by IshikawaIII iteration) is defined by ${g}_{1}\in E$ and
for all $n\ge 1$, where ${\{{\rho}_{n,i}\}}_{n=1}^{\mathrm{\infty}}$, $i=0,1,2,3,4,5$ are sequences in $[0,1]$. Twostep Noor iteration (denoted by NoorII iteration) is defined by ${h}_{1}\in E$ and
for all $n\ge 1$, where ${\{{\rho}_{n,i}\}}_{n=1}^{\mathrm{\infty}}$, $i=0,1,2,3,4,5$ are sequences in $[0,1]$. Since IshikawaIII and NoorII iterations are both sixstep, their computation cost is same at every iteration.
Remark 3.2 It should be noted that IshikawaIII and NoorII iterations are not new iterations and we introduce them just for comparing the rate of convergence of Ishikawa and Noor iterations under the same computation cost.
Before proceeding with the main result, we present three lemmas (see Lemmas 3.2, 3.3 and 3.4 below). We first need to recall a lemma, which is used in the proof of Lemma 3.2.
Lemma 3.1 [8]
Let E be a closed interval on the real line and let $f:E\to E$ be a continuous and nondecreasing function. Let ${\{{\alpha}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\beta}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{\gamma}_{n}\}}_{n=1}^{\mathrm{\infty}}$ be sequences in $[0,1)$. Let ${\{{s}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{{w}_{n}\}}_{n=1}^{\mathrm{\infty}}$ be defined by (1.2) and (1.3), respectively. Then the following hold:

(i)
if $f({s}_{1})<{s}_{1}$, then $f({s}_{n})<{s}_{n}$ for all $n\ge 1$ and ${\{{s}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is nonincreasing;

(ii)
if $f({s}_{1})>{s}_{1}$, then $f({s}_{n})>{s}_{n}$ for all $n\ge 1$ and ${\{{s}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is nondecreasing;

(iii)
if $f({w}_{1})<{w}_{1}$, then $f({w}_{n})<{w}_{n}$ for all $n\ge 1$ and ${\{{w}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is nonincreasing;

(iv)
if $f({w}_{1})>{w}_{1}$, then $f({w}_{n})>{w}_{n}$ for all $n\ge 1$ and ${\{{w}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is nondecreasing.
Lemma 3.2 Let E be a closed interval on the real line and let $f:E\to E$ be a continuous and nondecreasing function. Let ${\{{\rho}_{n,i}\}}_{n=1}^{\mathrm{\infty}}$, $i=0,1,2,3,4,5$, be sequences in $[0,1)$. Let ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{g}_{n,2}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{{g}_{n,4}\}}_{n=1}^{\mathrm{\infty}}$ (resp. ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{{h}_{n,3}\}}_{n=1}^{\mathrm{\infty}}$) be defined by (3.1) (resp. (3.2)). Then the following hold:

(i)
if $f({g}_{1})<{g}_{1}$, then ${g}_{n}\ge {g}_{n,2}\ge {g}_{n,4}\ge {g}_{n+1}$;

(ii)
if $f({g}_{1})>{g}_{1}$, then ${g}_{n}\le {g}_{n,2}\le {g}_{n,4}\le {g}_{n+1}$;

(iii)
if $f({h}_{1})<{h}_{1}$, then ${h}_{n}\ge {h}_{n,3}\ge {h}_{n+1}$;

(iv)
if $f({h}_{1})>{h}_{1}$, then ${h}_{n}\le {h}_{n,3}\le {h}_{n+1}$.
Proof Since IshikawaIII (resp. NoorII) iteration is threestep Ishikawa (resp. twostep Noor) iteration, ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{g}_{n,2}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{g}_{n,4}\}}_{n=1}^{\mathrm{\infty}}$ (resp. ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$, ${\{{h}_{n,3}\}}_{n=1}^{\mathrm{\infty}}$) are subsequences of ${\{{s}_{n}\}}_{n=1}^{\mathrm{\infty}}$ (resp. ${\{{w}_{n}\}}_{n=1}^{\mathrm{\infty}}$). From Lemma 3.1, Lemma 3.2 follows. □
Lemma 3.3 Let E be a closed interval on the real line and let $f:E\to E$ be a continuous and nondecreasing function. Let ${\{{\rho}_{n,i}\}}_{n=1}^{\mathrm{\infty}}$, $i=0,1,2,3,4,5$, be sequences in $[0,1)$. Let ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$ be defined by (3.1) and (3.2), respectively. Then the following hold:

(i)
if $f({g}_{1})<{g}_{1}$, then $f({g}_{n})<{g}_{n}$ for all $n\ge 1$ and ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is nonincreasing;

(ii)
if $f({g}_{1})>{g}_{1}$, then $f({g}_{n})>{g}_{n}$ for all $n\ge 1$ and ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is nondecreasing;

(iii)
if $f({h}_{1})<{h}_{1}$, then $f({h}_{n})<{h}_{n}$ for all $n\ge 1$ and ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is nonincreasing;

(iv)
if $f({h}_{1})>{h}_{1}$, then $f({h}_{n})>{h}_{n}$ for all $n\ge 1$ and ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is nondecreasing.
Proof The sequence ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ (resp. ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$) can be considered as a subsequence of ${\{{s}_{n}\}}_{n=1}^{\mathrm{\infty}}$ (resp. ${\{{w}_{n}\}}_{n=1}^{\mathrm{\infty}}$), since IshikawaIII (resp. NoorII) iteration is threestep Ishikawa (resp. twostep Noor) iteration. So, Lemma 3.3 follows from Lemma 3.1.
For comparing the rate of convergence of Ishikawa and Noor iterations, here we make the following assumption:

(H)
${\rho}_{n,2}\le {\rho}_{n,1}\frac{{h}_{n}f({h}_{n,1})}{{h}_{n}f({h}_{n,2})}$ for all $n\ge 1$.
□
Lemma 3.4 Let E be a closed interval on the real line and let $f:E\to E$ be a continuous and nondecreasing function. Let ${\{{\rho}_{n,i}\}}_{n=1}^{\mathrm{\infty}}$, $i=0,1,2,\dots ,5$ be sequences in $[0,1)$ satisfying (H). For ${g}_{1}={h}_{1}\in E$, let ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$ be the sequences defined by (3.1) and (3.2), respectively. Then the following are satisfied:

(i)
if $f({g}_{1})<{g}_{1}$, then ${g}_{n}\le {h}_{n}$ for all $n\ge 1$;

(ii)
if $f({g}_{1})>{g}_{1}$, then ${g}_{n}\ge {h}_{n}$ for all $n\ge 1$.
Proof (i) We use mathematical induction. Firstly, it holds ${g}_{1}={h}_{1}$. Assume that ${g}_{k}\le {h}_{k}$. Thus $f({g}_{k})\le f({h}_{k})$. We obtain ${g}_{k,1}{h}_{k,1}=(1{\rho}_{k,0})({g}_{k}{h}_{k})+{\rho}_{k,0}(f({g}_{k})f({h}_{k}))\le 0$, so ${g}_{k,1}\le {h}_{k,1}$, which implies $f({g}_{k,1})\le f({h}_{k,1})$. Similarly, we get
which implies $f({g}_{k,2})\le f({h}_{k,2})$. From Lemma 3.2(i), it follows ${g}_{k,2}\le {g}_{k}$ and thus ${g}_{k,2}\le {h}_{k}$. So, we have ${g}_{k,3}{h}_{k,3}=(1{\rho}_{k,2})({g}_{k,2}{h}_{k})+{\rho}_{k,2}(f({g}_{k,2})f({h}_{k,2}))\le 0$, i.e.,
which implies $f({g}_{k,3})\le f({h}_{k,3})$. Using the condition (H) and (3.2), we obtain ${h}_{k,3}{h}_{k,2}={\rho}_{k,1}({h}_{k}f({h}_{k,1})){\rho}_{k,2}({h}_{k}f({h}_{k,2}))\ge 0$ and thus ${h}_{k,3}\ge {h}_{k,2}$. Combining with (3.3), we have
So, ${g}_{k,4}{h}_{k,4}=(1{\rho}_{k,3})({g}_{k,2}{h}_{k,3})+{\rho}_{k,3}(f({g}_{k,3})f({h}_{k,3}))\le 0$, i.e.,
which implies $f({g}_{k,4})\le f({h}_{k,4})$. From Lemma 3.2(i), it follows ${g}_{k,4}\le {g}_{k,2}$. Combining with (3.5), we get ${g}_{k,4}\le {h}_{k,3}$. We have ${g}_{k,5}{h}_{k,5}=(1{\rho}_{k,4})({g}_{k,4}{h}_{k,3})+{\rho}_{k,4}(f({g}_{k,4})f({h}_{k,4}))\le 0$, i.e., ${g}_{k,5}\le {h}_{k,5}$, which implies $f({g}_{k,5})\le f({h}_{k,5})$. Similarly, we obtain ${g}_{k+1}{h}_{k+1}={g}_{k,6}{h}_{k,6}=(1{\rho}_{k,5})({g}_{k,4}{h}_{k,3})+{\rho}_{k,5}(f({g}_{k,5})f({h}_{k,5}))\le 0$, i.e.,
By induction, we get ${g}_{n}\le {h}_{n}$ for all $n\ge 1$.

(ii)
Following the line of (i), we can show that ${g}_{n}\le {h}_{n}$ for all $n\ge 1$. □
Theorem 3.1 Let E be a closed interval on the real line and let $f:E\to E$ be a continuous and nondecreasing function such that $F(f)$ is nonempty and bounded. For ${g}_{1}={h}_{1}\in E$, let ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$ be the sequences defined by (3.1) and (3.2), respectively. Let ${\{{\rho}_{n,i}\}}_{n=1}^{\mathrm{\infty}}$, $i=0,1,\dots ,5$, be sequences in $[0,1)$ satisfying (H). Then IshikawaIII iteration ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to $p\in F(f)$ if and only if NoorII iteration ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to p. Moreover, IshikawaIII iteration is better than NoorII iteration.
Proof Firstly, if IshikawaIII iteration ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to $p\in F(f)$, then set ${\rho}_{n,i}=0$, $i=0,1,2,3$, and we get the convergence of Ishikawa iteration. On the other hand, assume that Ishikawa iteration ${\{{s}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to $p\in F(f)$. Let ${s}_{1}={g}_{1}$, ${\beta}_{3n2}={\rho}_{n,0}$, ${\alpha}_{3n2}={\rho}_{n,1}$, ${\beta}_{3n1}={\rho}_{n,2}$, ${\alpha}_{3n1}={\rho}_{n,3}$, ${\beta}_{3n}={\rho}_{n,4}$ and ${\alpha}_{3n}={\rho}_{n,5}$ for all $n\ge 1$, then ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is a subsequence of ${\{{s}_{n}\}}_{n=1}^{\mathrm{\infty}}$ and thus converges to p. So, IshikawaIII iteration ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to $p\in F(f)$ if and only if Ishikawa iteration ${\{{s}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to p. Similarly, we get NoorIII iteration ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to $p\in F(f)$ if and only if Noor iteration ${\{{w}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to p. From Theorem 3.7(ii) in [8], we have that Ishikawa iteration ${\{{s}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to $p\in F(f)$ if and only if Noor iteration ${\{{w}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to p. Therefore, IshikawaIII iteration ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to $p\in F(f)$ if and only if NoorII iteration ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$ converges to p.
Next we prove that IshikawaIII iteration ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is better than NoorII iteration ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$. Put $L=inf\{p\in E:p=f(p)\}$ and $U=sup\{p\in E:p=f(p)\}$. We divide our proof into the following three cases:
Case 1: ${g}_{1}={h}_{1}>U$,
Case 2: ${g}_{1}={h}_{1}<L$,
Case 3: $L\le {g}_{1}={h}_{1}\le U$.
Case 1: ${g}_{1}={h}_{1}>U$. By Proposition 3.5 in [8], we get $f({g}_{1})<{g}_{1}$ and $f({h}_{1})<{h}_{1}$. Using Lemma 3.4(i), we obtain ${g}_{n}\le {h}_{n}$ for all $n\ge 1$. Following the line of the proof of Theorem 3.7 in [8], we have $U\le {g}_{n}$ for all $n\ge 1$. Then we get $0\le {g}_{n}p\le {h}_{n}p$, and thus
We can see that IshikawaIII iteration ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is better than NoorII iteration ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$.
Case 2: ${g}_{1}={h}_{1}<L$. By Proposition 3.6 in [8], we get $f({g}_{1})>{g}_{1}$ and $f({h}_{1})>{h}_{1}$. Using Lemma 3.4(ii), we obtain ${g}_{n}\ge {h}_{n}$ for all $n\ge 1$. Following the line of the proof of Theorem 3.7 in [8], we have ${g}_{n}\le L$ for all $n\ge 1$. Then we get ${h}_{n}p\le {g}_{n}p\le 0$, and thus
We can see that IshikawaIII iteration ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is better than NoorII iteration ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$.
Case 3: $L\le {g}_{1}={h}_{1}\le U$. Suppose that $f({g}_{1})\ne {g}_{1}$. If $f({g}_{1})<{g}_{1}$, we have by Lemma 3.3(i) that ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is nonincreasing with limit p. By Lemma 3.4(i), we get $p\le {g}_{n}\le {h}_{n}$ for all $n\ge 1$. It follows that ${g}_{n}p\le {h}_{n}p$ for all $n\ge 1$. Hence, we obtain that IshikawaIII iteration ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is better than NoorII iteration ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$. If $f({g}_{1})>{g}_{1}$, we have by Lemma 3.3(ii) that ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is nondecreasing with limit p. By Lemma 3.4(ii), we have $p\ge {g}_{n}\ge {h}_{n}$ for all $n\ge 1$. It follows that ${g}_{n}p\le {h}_{n}p$ for all $n\ge 1$. Hence, we have that IshikawaIII iteration ${\{{g}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is better than NoorII iteration ${\{{h}_{n}\}}_{n=1}^{\mathrm{\infty}}$. □
Remark 3.3 From Theorem 3.1 and Proposition 3.1(iii), we come to a conclusion that, under the same computational cost, Mann iteration is better than Ishikawa and Noor iterations, Ishikawa iteration is better than Noor iteration if the condition (H) is satisfied.
Next, we present a numerical example. Set ${\rho}_{n,i}=\frac{1}{{n}^{2}+1}$, $i=0,1,3,4$, ${\rho}_{n,2}={\rho}_{n,1}\frac{{h}_{n}f({h}_{n,1})}{{h}_{n}f({h}_{n,2})}$, ${\rho}_{n,5}=\frac{1}{{n}^{0.2}+1}$, for which the condition (H) is obviously satisfied.
Example 3.4 Let $f:[0,8]\to [0,8]$ be defined by $f(x)=\frac{{x}^{2}+\sqrt{x}+8}{10}$. Then f is a continuous and nondecreasing function. Take initial points ${g}_{1}={h}_{1}=4$. Table 1 illustrates the comparison of the convergence rate of IshikawaIII and NoorII iterations to the exact fixed point $p=1$, and we observe that IshikawaII iteration is better than NoorII iteration, which verifies theoretical results.
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Acknowledgements
The research was supported by National Natural Science Foundation of China (No. 11201476), supported in part by the Foundation of Tianjin Key Lab for Advanced Signal Processing and it was also supported by Fundamental Research Funds for the Central Universities (No. ZXH2012K001).
The authors would like to express their thanks to the referees, whose careful reading and constructive suggestions led to improvements in the presentation of the results.
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Authors’ contributions
QLD made theoretical derivation and completed the paper. SH provided useful suggestions for the Theorem 3.1. XL participated in the program to calculate the numerical example. All authors read and approved the final manuscript.
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Keywords
 continuous functions
 convergence theorem
 fixed point
 nondecreasing functions
 rate of convergence