Approximating fixed points for continuous functions on an arbitrary interval

Prasit Cholamjiak; Nattawut Pholasa

doi:10.1186/1029-242X-2013-214

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Approximating fixed points for continuous functions on an arbitrary interval

Prasit Cholamjiak¹Email author and
Nattawut Pholasa¹

Journal of Inequalities and Applications20132013:214

https://doi.org/10.1186/1029-242X-2013-214

Received: 22 November 2012

Accepted: 15 April 2013

Published: 29 April 2013

Abstract

In this research article, we introduce a new iterative method for solving a fixed point problem of continuous functions on an arbitrary interval. We then prove the convergence theorem of the proposed algorithm. We finally give numerical examples to compare the result with Mann, Ishikawa and Noor iterations. Our main results extend the corresponding results in the literature.

MSC:47H09, 47H10.

Keywords

continuous function convergence theorem fixed point iteration

1 Introduction

Let C be a closed interval on the real line and let $f : C \to C$ be a continuous function. A point $p \in C$ is called a fixed point of f if $f (p) = p$ .

One classical way to approximate a fixed point of a nonlinear mapping was introduced in 1953 by Mann [1] as follows: a sequence

{x_{n}}

defined by

x_{1} \in C

and

x_{n + 1} = (1 - α_{n}) x_{n} + α_{n} f (x_{n})

(1.1)

for all $n \geq 1$ , where ${α_{n}}$ is a sequence in $[0, 1]$ . Such an iteration process is known as Mann iteration. In 1991, Borwein and Borwein [2] proved the convergence theorem for a continuous function on the closed and bounded interval in the real line by using iteration (1.1).

Another classical iteration process was introduced by Ishikawa [3] as follows: a sequence

{x_{n}}

defined by

x_{1} \in C

and

\begin{aligned} y_{n} = (1 - β_{n}) x_{n} + β_{n} f (x_{n}), \\ x_{n + 1} = (1 - α_{n}) x_{n} + α_{n} f (y_{n}) \end{aligned}

(1.2)

for all $n \geq 1$ , where ${α_{n}}$ and ${β_{n}}$ are sequences in $[0, 1]$ . Such an iterative method is known as Ishikawa iteration. In 2006, Qing and Qihou [4] proved the convergence theorem of the sequence generated by iteration (1.2) for a continuous function on the closed interval in the real line (see also [5]).

In 2000, Noor [6] defined the following iterative scheme by

x_{1} \in C

and

\begin{aligned} z_{n} = (1 - μ_{n}) x_{n} + μ_{n} f (x_{n}), \\ y_{n} = (1 - β_{n}) x_{n} + β_{n} f (z_{n}), \\ x_{n + 1} = (1 - α_{n}) x_{n} + α_{n} f (y_{n}) \end{aligned}

(1.3)

for all $n \geq 1$ , where ${α_{n}}$ , ${β_{n}}$ and ${μ_{n}}$ are sequences in $[0, 1]$ . Such an iterative method is known as Noor iteration. Phuengrattana and Suantai [7] considered the convergence of Noor iteration for continuous functions on an arbitrary interval in the real line.

In this paper, motivated by the previous ones, we introduce a new modified iteration process for solving a fixed point problem for continuous functions on an arbitrary interval in the real line. Numerical examples are also presented to compare the result with Mann, Ishikawa and Noor iterations.

2 Main results

We begin this section by proving the following crucial lemmas.

Lemma 2.1 Let C be a closed interval on the real line (can be unbounded) and let

f : C \to C

be a continuous function. Let

{α_{n}}

,

{β_{n}}

,

{μ_{n}}

,

{γ_{n}}

and

{τ_{n}}

be sequences in

[0, 1]

with

0 \leq τ_{n} + β_{n} \leq 1

and

0 \leq γ_{n} + α_{n} \leq 1

. Let

{x_{n}}

be a sequence generated iteratively by

x_{1} \in C

and

\begin{matrix} z_{n} = (1 - μ_{n}) x_{n} + μ_{n} f (x_{n}), \\ y_{n} = (1 - τ_{n} - β_{n}) x_{n} + τ_{n} z_{n} + β_{n} f (z_{n}), \\ x_{n + 1} = (1 - γ_{n} - α_{n}) z_{n} + γ_{n} y_{n} + α_{n} f (y_{n}), n \geq 1, \end{matrix}

(2.1)

where $\sum_{n = 1}^{\infty} α_{n} = \infty$ , ${lim}_{n \to \infty} α_{n} = 0$ , $\sum_{n = 1}^{\infty} β_{n} < \infty$ and $\sum_{n = 1}^{\infty} μ_{n} < \infty$ .

If $x_{n} \to a$ , then a is a fixed point of f.

Proof Let $x_{n} \to a$ and suppose $a \neq f (a)$ . Then ${x_{n}}$ is bounded. So, ${f (x_{n})}$ is bounded by the continuity of f. So are ${y_{n}}$ , ${z_{n}}$ , ${f (y_{n})}$ and ${f (z_{n})}$ . Moreover, $z_{n} \to a$ since $x_{n} \to a$ and $μ_{n} \to 0$ . We also have $y_{n} \to a$ since $z_{n} \to a$ and $β_{n} \to 0$ .

From (2.1) we obtain

\begin{array}{rcl} x_{n + 1} & = & (1 - γ_{n} - α_{n}) z_{n} + γ_{n} y_{n} + α_{n} f (y_{n}) \\ = & z_{n} + γ_{n} (y_{n} - z_{n}) + α_{n} (f (y_{n}) - z_{n}) \\ = & (1 - μ_{n}) x_{n} + μ_{n} f (x_{n}) + γ_{n} (y_{n} - z_{n}) + α_{n} (f (y_{n}) - z_{n}) \\ = & x_{n} + μ_{n} (f (x_{n}) - x_{n}) + γ_{n} (y_{n} - z_{n}) + α_{n} (f (y_{n}) - z_{n}) \\ = & x_{n} + μ_{n} (f (x_{n}) - x_{n}) + γ_{n} ((1 - τ_{n} - β_{n}) (x_{n} - z_{n}) + β_{n} (f (z_{n}) - z_{n})) \\ + α_{n} (f (y_{n}) - z_{n}) \\ = & x_{n} + μ_{n} (f (x_{n}) - x_{n}) + γ_{n} ((1 - τ_{n} - β_{n}) μ_{n} (x_{n} - f (x_{n})) + β_{n} (f (z_{n}) - z_{n})) \\ + α_{n} (f (y_{n}) - z_{n}) \\ = & x_{n} + μ_{n} (1 - γ_{n} (1 - τ_{n} - β_{n})) (f (x_{n}) - x_{n}) + γ_{n} β_{n} (f (z_{n}) - z_{n}) \\ + α_{n} (f (y_{n}) - z_{n}) . \end{array}

(2.2)

Let

p_{k} = f (x_{k}) - x_{k}

,

q_{k} = f (z_{k}) - z_{k}

and

r_{k} = f (y_{k}) - z_{k}

. Then we observe

\begin{matrix} lim_{k \to \infty} p_{k} = lim_{n \to \infty} (f (x_{k}) - x_{k}) = f (a) - a \neq 0, \\ lim_{k \to \infty} q_{k} = lim_{n \to \infty} (f (z_{k}) - z_{k}) = f (a) - a \neq 0, \\ lim_{k \to \infty} r_{k} = lim_{n \to \infty} (f (y_{k}) - z_{k}) = f (a) - a \neq 0 . \end{matrix}

So, from (2.2) we obtain

\begin{array}{rcl} x_{n} & = & x_{1} + \sum_{k = 1}^{n} μ_{k} (1 - γ_{k} (1 - τ_{k} - β_{k})) (f (x_{k}) - x_{k}) + \sum_{k = 1}^{n} γ_{k} β_{k} (f (z_{k}) - z_{k}) \\ + \sum_{k = 1}^{n} α_{k} (f (y_{k}) - z_{k}) \\ = & x_{1} + \sum_{k = 1}^{n} μ_{k} (1 - γ_{k} (1 - τ_{k} - β_{k})) p_{k} + \sum_{k = 1}^{n} γ_{k} β_{k} q_{k} + \sum_{k = 1}^{n} α_{k} r_{k} . \end{array}

It is easy to see that $\sum_{k = 1}^{\infty} μ_{k} (1 - γ_{k} (1 - τ_{k} - β_{k})) p_{k} < \infty$ since ${lim}_{k \to \infty} p_{k} \neq 0$ and $\sum_{k = 1}^{\infty} μ_{k} < \infty$ . Similarly, we have $\sum_{k = 1}^{\infty} γ_{k} β_{k} q_{k} < \infty$ since ${lim}_{k \to \infty} q_{k} \neq 0$ and $\sum_{k = 1}^{\infty} β_{k} < \infty$ . This shows that ${x_{n}}$ is a divergent sequence since ${lim}_{k \to \infty} r_{k} \neq 0$ and $\sum_{k = 1}^{\infty} α_{k} = \infty$ . This contradicts the convergence of ${x_{n}}$ . Hence $f (a) = a$ and a is a fixed point of f. □

Lemma 2.2 Let C be a closed interval on the real line (can be unbounded) and let

f : C \to C

be a continuous function. Let

{α_{n}}

,

{β_{n}}

,

{μ_{n}}

,

{γ_{n}}

and

{τ_{n}}

be sequences in

[0, 1]

with

0 \leq τ_{n} + β_{n} \leq 1

and

0 \leq γ_{n} + α_{n} \leq 1

. Let

{x_{n}}

be a sequence generated iteratively by

x_{1} \in C

and

\begin{matrix} z_{n} = (1 - μ_{n}) x_{n} + μ_{n} f (x_{n}), \\ y_{n} = (1 - τ_{n} - β_{n}) x_{n} + τ_{n} z_{n} + β_{n} f (z_{n}), \\ x_{n + 1} = (1 - γ_{n} - α_{n}) z_{n} + γ_{n} y_{n} + α_{n} f (y_{n}), n \geq 1, \end{matrix}

where $\sum_{n = 1}^{\infty} α_{n} = \infty$ , ${lim}_{n \to \infty} α_{n} = 0$ , $\sum_{n = 1}^{\infty} β_{n} < \infty$ and $\sum_{n = 1}^{\infty} μ_{n} < \infty$ .

If ${x_{n}}$ is bounded, then ${x_{n}}$ is convergent.

Proof Suppose

{x_{n}}

is not convergent. Let

a = {lim inf}_{n} x_{n}

and

b = {lim sup}_{n} x_{n}

. Then

a < b

. We first show that if

a < m < b

, then

f (m) = m

. Suppose

f (m) \neq m

. Without loss of generality, we suppose

f (m) - m > 0

. Since f is continuous, there exists δ with

0 < δ < b - a

such that for

| x - m | \leq δ

,

f (x) - x > 0 .

Since

{x_{n}}

is bounded and f is continuous,

{f (x_{n})}

is bounded. Hence

{z_{n}}

,

{y_{n}}

,

{f (z_{n})}

and

{f (y_{n})}

are all bounded. Using

\begin{matrix} x_{n + 1} - x_{n} = (1 - γ_{n} - α_{n}) (z_{n} - x_{n}) + γ_{n} (y_{n} - x_{n}) + α_{n} (f (y_{n}) - x_{n}), \\ y_{n} - x_{n} = τ_{n} (z_{n} - x_{n}) + β_{n} (f (z_{n}) - x_{n}), \\ z_{n} - x_{n} = μ_{n} (f (x_{n}) - x_{n}), \end{matrix}

we can easily show that

| z_{n} - x_{n} | \to 0

,

| y_{n} - x_{n} | \to 0

and

| x_{n + 1} - x_{n} | \to 0

. Thus there exists a positive integer N such that for all

n > N

,

| x_{n + 1} - x_{n} | < \frac{δ}{2}, | y_{n} - x_{n} | < \frac{δ}{2}, | z_{n} - x_{n} | < \frac{δ}{2} .

(2.3)

Since

b = {lim sup}_{n} x_{n} > m

, there exists

k_{1} > N

such that

x_{n_{k_{1}}} > m

. Let

n_{k_{1}} = k

, then

x_{k} > m

. For

x_{k}

, there exist two cases as follows.

(i)

$x_{k} > m + \frac{δ}{2}$ , then $x_{k + 1} > x_{k} - \frac{δ}{2} \geq m$ using (2.3). So, we have $x_{k + 1} > m$ .
(ii)

$m < x_{k} < m + \frac{δ}{2}$ , then $m - \frac{δ}{2} < y_{k} < m + δ$ and $m - \frac{δ}{2} < z_{k} < m + δ$ by (2.3). So, we obtain $| x_{k} - m | < \frac{δ}{2} < δ$ , $| y_{k} - m | < δ$ , $| z_{k} - m | < δ$ . Hence
$f (x_{k}) - x_{k} > 0, f (y_{k}) - y_{k} > 0, f (z_{k}) - z_{k} > 0 .$

(2.4)

We observe that

\begin{array}{rcl} y_{k} - z_{k} & = & (1 - τ_{k} - β_{k}) (x_{k} - z_{k}) + β_{k} (f (z_{k}) - z_{k}) \\ = & μ_{k} (1 - τ_{k} - β_{k}) (x_{k} - f (x_{k})) + β_{k} (f (z_{k}) - z_{k}) . \end{array}

(2.5)

From (2.2), (2.4) and (2.5), we have

\begin{array}{rcl} x_{k + 1} & = & x_{k} + μ_{k} (1 - γ_{k} (1 - τ_{k} - β_{k})) (f (x_{k}) - x_{k}) + γ_{k} β_{k} (f (z_{k}) - z_{k}) \\ + α_{k} (f (y_{k}) - z_{k}) \\ = & x_{k} + μ_{k} (1 - γ_{k} (1 - τ_{k} - β_{k})) (f (x_{k}) - x_{k}) + γ_{k} β_{k} (f (z_{k}) - z_{k}) \\ + α_{k} (f (y_{k}) - y_{k}) + α_{k} (y_{k} - z_{k}) \\ = & x_{k} + μ_{k} (1 - γ_{k} (1 - τ_{k} - β_{k})) (f (x_{k}) - x_{k}) + γ_{k} β_{k} (f (z_{k}) - z_{k}) \\ + α_{k} (f (y_{k}) - y_{k}) + α_{k} (μ_{k} (1 - τ_{k} - β_{k}) (x_{k} - f (x_{k})) + β_{k} (f (z_{k}) - z_{k})) \\ = & x_{k} + μ_{k} (1 - (γ_{k} + α_{k}) (1 - τ_{k} - β_{k})) (f (x_{k}) - x_{k}) + α_{k} (f (y_{k}) - y_{k}) \\ + β_{k} (γ_{k} + α_{k}) (f (z_{k}) - z_{k}) \\ > & x_{k} . \end{array}

Thus $x_{k + 1} > x_{k} > m$ . From (i) and (ii), we have $x_{k + 1} > m$ . Similarly, we get that $x_{k + 2} > m$ , $x_{k + 3} > m$ , …. Thus we have $x_{n} > m$ for all $n > k = n_{k_{1}}$ . So, $a = {lim}_{k \to \infty} x_{n_{k}} \geq m$ , which is a contradiction with $a < m$ . Thus $f (m) = m$ .

We next consider the following two cases.

(i)

There exists $x_{M}$ such that $a < x_{M} < b$ . Then $f (x_{M}) = x_{M}$ . It follows that
$z_{M} = (1 - μ_{M}) x_{M} + μ_{M} f (x_{M}) = x_{M}$

and
$\begin{array}{rcl} y_{M} & = & (1 - τ_{M} - β_{M}) x_{M} + τ_{M} z_{M} + β_{M} f (z_{M}) \\ = & (1 - τ_{M} - β_{M}) x_{M} + τ_{M} x_{M} + β_{M} f (x_{M}) \\ = & x_{M} . \end{array}$

Hence, we obtain
$\begin{array}{rcl} x_{M + 1} & = & (1 - γ_{M} - α_{M}) z_{M} + γ_{M} y_{M} + α_{M} f (y_{M}) \\ = & (1 - γ_{M} - α_{M}) x_{M} + γ_{M} x_{M} + α_{M} f (x_{M}) \\ = & x_{M} . \end{array}$

Similarly, we obtain $x_{M} = x_{M + 1} = x_{M + 2} = \dots$ . So, we conclude that $x_{n} \to x_{M}$ . Since there exists $x_{n_{k}} \to a$ , $x_{M} = a$ . This shows that $x_{n} \to a$ , which is a contradiction.
(ii)

For all $n, x_{n} \leq a$ or $x_{n} \geq b$ . Since $b - a > 0$ and ${lim}_{n \to \infty} | x_{n + 1} - x_{n} | = 0$ , there exists $\bar{N}$ such that $| x_{n + 1} - x_{n} | < \frac{(b - a)}{2}$ for $n > \bar{N}$ . So, it is always that $x_{n} \leq a$ for $n > \bar{N}$ ; or it is always that $x_{n} \geq b$ for $n > \bar{N}$ . If $x_{n} \leq a$ for $n > \bar{N}$ , then $b = {lim}_{l \to \infty} x_{n_{l}} \leq a$ , which is a contradiction with $a < b$ . If $x_{n} \geq b$ for $n > \bar{N}$ , then $a = {lim}_{k \to \infty} x_{n_{k}} \geq b$ , which is a contradiction with $a < b$ . Thus we conclude that $x_{n} \to a$ . This completes the proof. □

We are now ready to prove the main results of this paper.

Theorem 2.3 Let C be a closed interval on the real line (can be unbounded) and let

f : C \to C

be a continuous function. Let

{α_{n}}

,

{β_{n}}

,

{μ_{n}}

,

{γ_{n}}

and

{τ_{n}}

be sequences in

[0, 1]

with

0 \leq τ_{n} + β_{n} \leq 1

and

0 \leq γ_{n} + α_{n} \leq 1

. Let

{x_{n}}

be a sequence generated iteratively by

x_{1} \in C

and

\begin{matrix} z_{n} = (1 - μ_{n}) x_{n} + μ_{n} f (x_{n}), \\ y_{n} = (1 - τ_{n} - β_{n}) x_{n} + τ_{n} z_{n} + β_{n} f (z_{n}), \\ x_{n + 1} = (1 - γ_{n} - α_{n}) z_{n} + γ_{n} y_{n} + α_{n} f (y_{n}), n \geq 1, \end{matrix}

where $\sum_{n = 1}^{\infty} α_{n} = \infty$ , ${lim}_{n \to \infty} α_{n} = 0$ , $\sum_{n = 1}^{\infty} β_{n} < \infty$ and $\sum_{n = 1}^{\infty} μ_{n} < \infty$ .

If ${x_{n}}$ is bounded, then ${x_{n}}$ converges to a fixed point of f.

Proof Let ${x_{n}}$ be a bounded sequence. Then, by Lemma (2.2), ${x_{n}}$ is a convergent sequence. Hence, by Lemma (2.1), it converges to a fixed point of f. □

As a direct consequence of Theorem 2.3, we obtain the following result.

Theorem 2.4 Let C be a closed interval on the real line (can be unbounded) and let

f : C \to C

be a continuous function. Let

{α_{n}}

,

{β_{n}}

,

{μ_{n}}

,

{γ_{n}}

and

{τ_{n}}

be sequences in

[0, 1]

with

0 \leq τ_{n} + β_{n} \leq 1

and

0 \leq γ_{n} + α_{n} \leq 1

. Let

{x_{n}}

be a sequence generated iteratively by

x_{1} \in C

and

\begin{matrix} z_{n} = (1 - μ_{n}) x_{n} + μ_{n} f (x_{n}), \\ y_{n} = (1 - τ_{n} - β_{n}) x_{n} + τ_{n} z_{n} + β_{n} f (z_{n}), \\ x_{n + 1} = (1 - γ_{n} - α_{n}) z_{n} + γ_{n} y_{n} + α_{n} f (y_{n}), n \geq 1, \end{matrix}

where $\sum_{n = 1}^{\infty} α_{n} = \infty$ , ${lim}_{n \to \infty} α_{n} = 0$ , $\sum_{n = 1}^{\infty} β_{n} < \infty$ and $\sum_{n = 1}^{\infty} μ_{n} < \infty$ .

Then ${x_{n}}$ converges to a fixed point of f if and only if ${x_{n}}$ is bounded.

Corollary 2.5 Let

f : [a, b] \to [a, b]

be a continuous function. Let

{α_{n}}

,

{β_{n}}

,

{μ_{n}}

,

{γ_{n}}

and

{τ_{n}}

be sequences in

[0, 1]

with

0 \leq τ_{n} + β_{n} \leq 1

and

0 \leq γ_{n} + α_{n} \leq 1

. Let

{x_{n}}

be a sequence generated iteratively by

x_{1} \in [a, b]

and

\begin{matrix} z_{n} = (1 - μ_{n}) x_{n} + μ_{n} f (x_{n}), \\ y_{n} = (1 - τ_{n} - β_{n}) x_{n} + τ_{n} z_{n} + β_{n} f (z_{n}), \\ x_{n + 1} = (1 - γ_{n} - α_{n}) z_{n} + γ_{n} y_{n} + α_{n} f (y_{n}), n \geq 1, \end{matrix}

where $\sum_{n = 1}^{\infty} α_{n} = \infty$ , ${lim}_{n \to \infty} α_{n} = 0$ , $\sum_{n = 1}^{\infty} β_{n} < \infty$ and $\sum_{n = 1}^{\infty} μ_{n} < \infty$ .

Then ${x_{n}}$ converges to a fixed point of f.

If we take $τ_{n} = γ_{n} = 0$ , then we obtain the following result.

Corollary 2.6 Let C be a closed interval on the real line (can be unbounded) and let

f : C \to C

be a continuous function. Let

{α_{n}}

,

{β_{n}}

and

{μ_{n}}

be sequences in

[0, 1]

. Let

{x_{n}}

be a sequence generated iteratively by

x_{1} \in C

and

\begin{matrix} z_{n} = (1 - μ_{n}) x_{n} + μ_{n} f (x_{n}), \\ y_{n} = (1 - β_{n}) x_{n} + β_{n} f (z_{n}), \\ x_{n + 1} = (1 - α_{n}) z_{n} + α_{n} f (y_{n}), n \geq 1, \end{matrix}

where $\sum_{n = 1}^{\infty} α_{n} = \infty$ , ${lim}_{n \to \infty} α_{n} = 0$ , $\sum_{n = 1}^{\infty} β_{n} < \infty$ and $\sum_{n = 1}^{\infty} μ_{n} < \infty$ .

Then ${x_{n}}$ converges to a fixed point of f if and only if ${x_{n}}$ is bounded.

If we take $τ_{n} + β_{n} = 1$ and $γ_{n} + α_{n} = 1$ , then we obtain the following result.

Corollary 2.7 Let C be a closed interval on the real line (can be unbounded) and let

f : C \to C

be a continuous function. Let

{α_{n}}

,

{β_{n}}

and

{μ_{n}}

be sequences in

[0, 1]

. Let

{x_{n}}

be a sequence generated iteratively by

x_{1} \in C

and

\begin{matrix} z_{n} = (1 - μ_{n}) x_{n} + μ_{n} f (x_{n}), \\ y_{n} = (1 - β_{n}) z_{n} + β_{n} f (z_{n}), \\ x_{n + 1} = (1 - α_{n}) y_{n} + α_{n} f (y_{n}), n \geq 1, \end{matrix}

where $\sum_{n = 1}^{\infty} α_{n} = \infty$ , ${lim}_{n \to \infty} α_{n} = 0$ , $\sum_{n = 1}^{\infty} β_{n} < \infty$ and $\sum_{n = 1}^{\infty} μ_{n} < \infty$ .

Then ${x_{n}}$ converges to a fixed point of f if and only if ${x_{n}}$ is bounded.

Remark 2.8 Corollary 2.7 extends the main result obtained in [8] from the modified Ishikawa iteration to the modified Noor iteration.

3 Numerical examples

In this section, we give numerical examples to demonstrate the convergence of the algorithm defined in this paper. For convenience, we call the iteration (2.1) the CP-iteration.

Example 3.1 Let $f : [1, \infty) ⟶ [1, \infty)$ be defined by $f (x) = \sqrt{0.9 ln x + 1}$ . Then f is a continuous function. Use the initial point $x_{1} = 3$ and the control conditions $α_{n} = \frac{1}{{(n + 1)}^{0.5}}$ , $β_{n} = \frac{1}{{(n + 1)}^{2.5}}$ , $μ_{n} = \frac{1}{{(n + 1)}^{1.5}}$ , $τ_{n} = \frac{1}{n + 1}$ and $γ_{n} = \frac{1}{7}$ .

Example 3.2 Let $f : [1, \infty) ⟶ [1, \infty)$ be defined by $f (x) = 0.2 \sqrt{x - 1} + \sqrt{x}$ . Then f is a continuous function. Use the initial point $x_{1} = 5$ and the control conditions $α_{n} = \frac{1}{{(n + 1)}^{0.5}}$ , $β_{n} = \frac{1}{{(n + 1)}^{2}}$ , $μ_{n} = \frac{1}{{(n + 1)}^{1.1}}$ , $τ_{n} = \frac{1}{n + 1}$ and $γ_{n} = \frac{1}{5}$ .

Remark 3.3 From Table 1, Figure 1, Table 2 and Figure 2, we observe that the sequence generated by the CP-iteration converges to a fixed point faster than that of Mann, Ishikawa and Noor iterations.

Figure 1
Convergence behavior of Mann, Ishikawa, Noor and CP iterations for the function given in Example 3.1.

Figure 2
Convergence behavior of Mann, Ishikawa, Noor and CP iterations for the function given in Example 3.2.

Table 1

Comparison of the convergence rate of Mann, Ishikawa, Noor and CP iterations for the function given in Example 3.1

n	Mann	Ishikawa	Noor	CP-iteration
n	$x_{n}$	$x_{n}$	$x_{n}$	$x_{n}$	$\| f (x_{n}) - x_{n} \|$
1	3.000000	3.000000	3.000000	3.000000	1.589769
10	1.074110	1.071538	1.071437	1.043308	0.024408
20	1.015478	1.014946	1.014925	1.008428	0.004658
30	1.004821	1.004656	1.004650	1.002531	0.001394
40	1.001822	1.001760	1.001757	1.000934	0.000514
50	1.000777	1.000750	1.000749	1.000392	0.000216
60	1.000360	1.000348	1.000348	1.000180	0.000099
70	1.000178	1.000172	1.000172	1.000088	0.000048
80	1.000093	1.000089	1.000089	1.000045	0.000025
90	1.000050	1.000048	1.000048	1.000024	0.000013
100	1.000028	1.000027	1.000027	1.000014	0.000007

Table 2

Comparison of the convergence rate of Mann, Ishikawa, Noor and CP iterations for the function given in Example 3.2

n	Mann	Ishikawa	Noor	CP-iteration
n	$x_{n}$	$x_{n}$	$x_{n}$	$x_{n}$	$\| f (x_{n}) - x_{n} \|$
1	5.000000	5.000000	5.000000	5.000000	2.363932
10	1.529828	1.504824	1.501931	1.359649	0.073668
20	1.293319	1.284893	1.283943	1.224654	0.023218
30	1.223891	1.219794	1.219335	1.187049	0.011033
40	1.193599	1.191253	1.190991	1.171120	0.006204
50	1.177752	1.176279	1.176115	1.162994	0.003827
60	1.168542	1.167561	1.167452	1.158380	0.002505
70	1.162807	1.162126	1.162050	1.155568	0.001710
80	1.159055	1.158568	1.158513	1.153766	0.001205
90	1.156509	1.156152	1.156112	1.152565	0.000870
100	1.154730	1.154463	1.154433	1.151740	0.000641
110	1.153458	1.153255	1.153233	1.151160	0.000480
120	1.152530	1.152374	1.152357	1.150743	0.000365
130	1.151842	1.151721	1.151707	1.150438	0.000281
140	1.151325	1.151230	1.151219	1.150212	0.000218
150	1.150931	1.150856	1.150847	1.150042	0.000171
160	1.150628	1.150568	1.150561	1.149912	0.000136
170	1.150393	1.150345	1.150339	1.149813	0.000108
180	1.150209	1.150169	1.150165	1.149735	0.000087
190	1.150063	1.150031	1.150027	1.149675	0.000070
200	1.149947	1.149920	1.149918	1.149627	0.000057

Declarations

Acknowledgements

The authors wish to thank Professor Suthep Suantai for the valuable guidance and suggestion. The first author was supported by the Thailand Research Fund, the Commission on Higher Education, and University of Phayao under Grant No. MRG5580016.

Authors’ Affiliations

(1)

School of Science, University of Phayao

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