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# Fixed and periodic points of generalized contractions in metric spaces

Fixed Point Theory and Applications20132013:243

https://doi.org/10.1186/1687-1812-2013-243

• Accepted: 8 October 2013
• Published:

## Abstract

Wardowski (Fixed Point Theory Appl. 2012:94, 2012, doi:10.1186/1687-1812-2012-94) introduced a new type of contraction called F-contraction and proved a fixed point result in complete metric spaces, which in turn generalizes the Banach contraction principle. The aim of this paper is to introduce F-contractions with respect to a self-mapping on a metric space and to obtain common fixed point results. Examples are provided to support results and concepts presented herein. As an application of our results, periodic point results for the F-contractions in metric spaces are proved.

MSC:47H10, 47H07, 54H25.

## Keywords

• F-contraction
• property P
• property Q
• common fixed point

## 1 Introduction and preliminaries

The Banach contraction principle [1] is a popular tool in solving existence problems in many branches of mathematics (see, e.g., [24]). Extensions of this principle were obtained either by generalizing the domain of the mapping or by extending the contractive condition on the mappings [59]. Initially, existence of fixed points in ordered metric spaces was investigated and applied by Ran and Reurings [10]. Since then, a number of results have been proved in the framework of ordered metric spaces (see [1118]). Contractive conditions involving a pair of mappings are further additions to the metric fixed point theory and its applications (for details, see [1923]).

Recently, Wardowski [24] introduced a new contraction called F-contraction and proved a fixed point result as a generalization of the Banach contraction principle [1]. In this paper, we introduce an F-contraction with respect to a self-mapping on a metric space and obtain common fixed point results in an ordered metric space. In the last section, we give some results on periodic point properties of a mapping and a pair of mappings in a metric space. We begin with some basic known definitions and results which will be used in the sequel. Throughout this article, , ${\mathbb{R}}_{+}$, denote the set of natural numbers, the set of positive real numbers and the set of real numbers, respectively.

Definition 1 Let f and g be self-mappings on a set X. If $fx=gx=w$ for some x in X, then x is called a coincidence point of f and g and w is called a coincidence point of f and g. Furthermore, if $fgx=gfx$ whenever x is a coincidence point of f and g, then f and g are called weakly compatible mappings [22].

Let $C\left(f,g\right)=\left\{x\in X:fx=gx\right\}$ ($F\left(f,g\right)=\left\{x\in X:x=fx=gx\right\}$) denote the set of all coincidence points (the set of all common fixed points) of self-mappings f and g.

Definition 2 ([25])

Let $\left(X,d\right)$ be a metric space and $f,g:X\to X$. The mapping f is called a g-contraction if there exists $\alpha \in \left(0,1\right)$ such that
$d\left(fx,fy\right)\le \alpha d\left(gx,gy\right)$

holds for all $x,y\in X$.

In 1976, Jungck [25] obtained the following useful generalization of the Banach contraction principle.

Theorem 1 Let g be a continuous self-mapping on a complete metric space $\left(X,d\right)$. Then g has a fixed point in X if and only if there exists a g-contraction mapping $f:X\to X$ such that f commutes with g and $g\left(X\right)\subseteq f\left(X\right)$.

Let Ϝ be the collection of all mappings $F:{\mathbb{R}}_{+}\to \mathbb{R}$ that satisfy the following conditions:
1. (C1)

F is strictly increasing, that is, for all $\alpha ,\beta \in {\mathbb{R}}_{+}$ such that $\alpha <\beta$ implies that $F\left(\alpha \right).

2. (C2)

For every sequence ${\left\{{\alpha }_{n}\right\}}_{n\in \mathbb{N}}$ of positive real numbers, ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and ${lim}_{n\to \mathrm{\infty }}F\left({\alpha }_{n}\right)=-\mathrm{\infty }$ are equivalent.

3. (C3)
There exists $k\in \left(0,1\right)$ such that
$\underset{\alpha \to {0}^{+}}{lim}{\alpha }^{k}F\left(\alpha \right)=0.$

Definition 3 ([24])

Let $\left(X,d\right)$ be a metric space and $F\in Ϝ$. A mapping $f:X\to X$ is said to be an F-contraction on X if there exists $\tau >0$ such that
$d\left(fx,fy\right)>0\phantom{\rule{1em}{0ex}}\text{implies that}\phantom{\rule{1em}{0ex}}\tau +F\left(d\left(fx,fy\right)\right)\le F\left(d\left(x,y\right)\right)$
(1)

for all $x,y\in X$.

Note that every F-contraction is continuous (see [24]). We extend the above definition to two mappings.

Definition 4 Let $\left(X,d\right)$ be a metric space, $F\in Ϝ$ and $f,g:X\to X$. The mapping f is said to be an F-contraction with respect to g on X if there exists $\tau >0$ such that
$\tau +F\left(d\left(fx,fy\right)\right)\le F\left(d\left(gx,gy\right)\right)$
(2)

for all $x,y\in X$ satisfying $min\left\{d\left(fx,fy\right),d\left(gx,gy\right)\right\}>0$.

By different choices of mappings F in (1) and (2), one obtains a variety of contractions [24].

Example 1 Let ${F}_{1}:{\mathbb{R}}_{+}\to \mathbb{R}$ be given by ${F}_{1}\left(\alpha \right)=ln\left(\alpha \right)$. It is clear that $F\in Ϝ$. Suppose that $f:X\to X$ is an F-contraction with respect to a self-mapping g on X. From (2) we have
$\tau +ln\left(d\left(fx,fy\right)\right)\le ln\left(d\left(gx,gy\right)\right),$
which implies that
$d\left(fx,fy\right)\le {e}^{-\tau }d\left(gx,gy\right).$

Therefore an ${F}_{1}$-contraction map f with respect to g reduces to a g-contraction mapping.

Now we give an example of an F-contraction with respect to a self-mapping g on X which is not a g-contraction on X.

Example 2 Consider the following sequence of partial sums ${\left\{{S}_{n}\right\}}_{n\in \mathbb{N}}$ [[24], Example 2.5]:
$\begin{array}{c}{S}_{1}=1,\hfill \\ {S}_{2}=1+2,\hfill \\ {S}_{3}=1+2+3,\hfill \\ \cdots \hfill \\ {S}_{n}=1+2+\cdots +n=\frac{n\left(n+1\right)}{2},\phantom{\rule{1em}{0ex}}n\in \mathbb{N}.\hfill \end{array}$
Let $X=\left\{{S}_{n}:n\in \mathbb{N}\right\}$ and d be the usual metric on X. Let $f:X\to X$ and $g:X\to X$ be defined as
Let ${F}_{1}:{\mathbb{R}}_{+}\to \mathbb{R}$ be given by ${F}_{1}\left(\alpha \right)=ln\left(\alpha \right)$. As
$\underset{n\to \mathrm{\infty }}{lim}\frac{d\left(f{S}_{n},f{S}_{1}\right)}{d\left(g{S}_{n},g{S}_{1}\right)}=\underset{n\to \mathrm{\infty }}{lim}\frac{{S}_{n-1}-{S}_{1}}{{S}_{n+1}-{S}_{1}}=1,$
so f is not a g-contraction. If we take ${F}_{2}\left(\alpha \right)=ln\left(\alpha \right)+\alpha$, then ${F}_{2}\in Ϝ$ and f is an ${F}_{2}$-contraction with respect to a mapping g (taking $\tau =2$). Indeed, the following holds:
$\frac{d\left(f{S}_{n},f{S}_{1}\right)}{d\left(g{S}_{n},g{S}_{1}\right)}{e}^{d\left(f{S}_{n},f{S}_{1}\right)-d\left(g{S}_{n},g{S}_{1}\right)}=\frac{{S}_{n-1}-{S}_{1}}{{S}_{n+1}-{S}_{1}}{e}^{{S}_{n-1}-{S}_{1}-{S}_{n+1}+{S}_{1}}=\frac{{n}^{2}-n-2}{{n}^{2}+3n}{e}^{-4n-2}\le {e}^{-2}$
for all $n>1$. For all $m,n\in \mathbb{N}$ with $m>n>1$, we have
$\begin{array}{r}\frac{d\left(f{S}_{m},f{S}_{n}\right)}{d\left(g{S}_{m},g{S}_{n}\right)}{e}^{d\left(f{S}_{m},f{S}_{n}\right)-d\left(g{S}_{m},g{S}_{n}\right)}\\ \phantom{\rule{1em}{0ex}}=\frac{{S}_{m-1}-{S}_{n-1}}{{S}_{m+1}-{S}_{n+1}}{e}^{{S}_{m-1}-{S}_{n-1}-{S}_{m+1}+{S}_{n+1}}\\ \phantom{\rule{1em}{0ex}}=\frac{{m}^{2}+m-{n}^{2}-n}{{m}^{2}+3m-{n}^{2}-3n}{e}^{-2\left(m-n\right)}\le {e}^{-2}.\end{array}$

Definition 5 ([26], Dominance condition)

Let $\left(X,⪯\right)$ be a partially ordered set. A self-mapping f on X is said to be (i) a dominated map if $fx⪯x$ for each x in X, (ii) a dominating map if $x⪯fx$ for each x in X.

Example 3 Let $X=\left[0,1\right]$ be endowed with the usual ordering and $f,g:X\to X$ defined by $gx={x}^{n}$ for some $n\in \mathbb{N}$ and $fx=kx$ for some real number $k\ge 1$. Note that
$gx={x}^{n}\le x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x\le kx=fx$

for all x in X. Thus g is dominated and f is a dominating map.

Definition 6 Let $\left(X,⪯\right)$ be a partially ordered set. Two mappings $f,g:X\to X$ are said to be weakly increasing if $fx⪯gfx$ and $gx⪯fgx$ for all x in X (see [27]).

Definition 7 Let X be a nonempty set. Then $\left(X,d,⪯\right)$ is called an ordered metric space if $\left(X,d\right)$ is a metric space and $\left(X,⪯\right)$ is a partially ordered set.

Definition 8 Let $\left(X,⪯\right)$ be a partial ordered set, then x, y in X are called comparable elements if either $x⪯y$ or $y⪯x$ holds true. Moreover, we define $\mathrm{\Delta }\subseteq X×X$ by

Definition 9 An ordered metric space $\left(X,d,⪯\right)$ is said to have the sequential limit comparison property if for every non-decreasing sequence (non-increasing sequence) ${\left\{{x}_{n}\right\}}_{n\in \mathbb{N}}$ in X such that ${x}_{n}\to x$ implies that ${x}_{n}⪯x$ ($x⪯{x}_{n}$).

## 2 Common fixed point results in ordered metric spaces

We present the following theorem as a generalization of results in [25] and [[24], Theorem 2.1].

Theorem 2 Let $\left(X,⪯\right)$ be a partially ordered set such that there exists a metric d on X, and let $f:X\to X$ be an F-contraction with respect to $g:X\to X$ on Δ with $f\left(X\right)\subseteq g\left(X\right)$. Assume that f is dominating and g is dominated. Then
1. (a)

f and g have a coincidence point in X provided that $g\left(X\right)$ is complete and has the sequential limit comparison property.

2. (b)

$C\left(f,g\right)$ is well ordered if and only if $C\left(f,g\right)$ is a singleton.

3. (c)

f and g have a unique common fixed point if f and g are weakly compatible and $C\left(f,g\right)$ is well ordered.

Proof (a) Let ${x}_{0}$ be an arbitrary point of X. Since the range of g contains the range of f, there exists a point ${x}_{1}$ in X such that $f\left({x}_{0}\right)=g\left({x}_{1}\right)$. As f is dominating and g is dominated, so we have
${x}_{0}⪯f{x}_{0}=g{x}_{1}⪯{x}_{1}.$
Hence $\left({x}_{0},{x}_{1}\right)\in \mathrm{\Delta }$. Continuing this process, having chosen ${x}_{n}$ in X, we obtain ${x}_{n+1}$ in X such that
${x}_{n}⪯f{x}_{n}=g{x}_{n+1}⪯{x}_{n+1}.$
So, we obtain $\left({x}_{n},{x}_{n+1}\right)\in \mathrm{\Delta }$ for every $n\in \mathbb{N}\cup \left\{0\right\}$. For the sake of simplicity, take
${\gamma }_{n}=d\left(g{x}_{n},g{x}_{n+1}\right)$
(3)
for all $n\in \mathbb{N}\cup \left\{0\right\}$. If there exists ${n}_{0}\in \mathbb{N}\cup \left\{0\right\}$ for which ${x}_{{n}_{0}+1}={x}_{{n}_{0}}$, then $f{x}_{{n}_{0}}=g{x}_{{n}_{0}+1}$ implies that $f{x}_{{n}_{0}+1}=g{x}_{{n}_{0}+1}$, that is, ${x}_{{n}_{0}+1}\in C\left(f,g\right)$. Now we assume that ${x}_{n+1}\ne {x}_{n}$ for all $n\in \mathbb{N}\cup \left\{0\right\}$. As f is an F-contraction with respect to g on Δ, so we obtain
$\begin{array}{rcl}F\left({\gamma }_{n}\right)& =& F\left(d\left(g{x}_{n},g{x}_{n+1}\right)\right)=F\left(d\left(f{x}_{n-1},f{x}_{n}\right)\right)\\ \le & F\left(d\left(g{x}_{n-1},g{x}_{n}\right)\right)-\tau \\ =& F\left(d\left(f{x}_{n-2},f{x}_{n-1}\right)\right)-\tau \\ \le & F\left(d\left(g{x}_{n-2},g{x}_{n-1}\right)\right)-2\tau \le \cdots \\ \le & F\left(d\left(g{x}_{1},g{x}_{2}\right)\right)-\left(n-1\right)\tau =F\left({\gamma }_{1}\right)-\left(n-1\right)\tau .\end{array}$
That is,
$F\left({\gamma }_{n}\right)\le F\left({\gamma }_{1}\right)-\left(n-1\right)\tau .$
On taking limit as $n\to \mathrm{\infty }$, we obtain ${lim}_{n\to \mathrm{\infty }}F\left({\gamma }_{n}\right)=-\mathrm{\infty }$. Hence ${lim}_{n\to \mathrm{\infty }}{\gamma }_{n}=0$ by (C2). Now, by (C3), there exists $k\in \left(0,1\right)$ such that ${lim}_{n\to \mathrm{\infty }}{\gamma }_{n}^{k}F\left({\gamma }_{n}\right)=0$. Note that
${\gamma }_{n}^{k}F\left({\gamma }_{n}\right)-{\gamma }_{n}^{k}F\left({\gamma }_{1}\right)\le {\gamma }_{n}^{k}\left(F\left({\gamma }_{1}\right)-\left(n-1\right)\tau \right)-{\gamma }_{n}^{k}F\left({\gamma }_{1}\right)=-{\gamma }_{n}^{k}\left(n-1\right)\tau \le 0.$
(4)
Taking limit as $n\to \mathrm{\infty }$ in (4), we have ${lim}_{n\to \mathrm{\infty }}\left(n-1\right){\gamma }_{n}^{k}=0$. Consequently, ${lim}_{n\to \mathrm{\infty }}n{\gamma }_{n}^{k}=0$. Thus there exists ${n}_{1}$ in such that $n{\gamma }_{n}^{k}\le 1$ for all $n\ge {n}_{1}$, that is, ${\gamma }_{n}\le 1/{n}^{1/k}$ for all $n\ge {n}_{1}$. Now, for integers $m>n\ge 1$, we obtain
$\begin{array}{rcl}d\left(g{x}_{n},g{x}_{m}\right)& \le & d\left(g{x}_{n},g{x}_{n+1}\right)+d\left(g{x}_{n+1},g{x}_{n+2}\right)+\cdots +d\left(g{x}_{m-1},g{x}_{m}\right)\\ <& \sum _{i=n}^{\mathrm{\infty }}{\gamma }_{i}\le \sum _{i=n}^{\mathrm{\infty }}\frac{1}{{i}^{\frac{1}{k}}}<\mathrm{\infty }.\end{array}$
This shows that ${\left\{g{x}_{n}\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence in $g\left(X\right)$. As $g\left(X\right)$ is complete, so there exists q in $g\left(X\right)$ such that ${lim}_{n\to \mathrm{\infty }}g{x}_{n}=q$. Let $p\in X$ be such that $g\left(p\right)=q$. The sequential limit comparison property implies that $g{x}_{n+1}⪯q$. As ${x}_{n}⪯f{x}_{n}=g{x}_{n+1}⪯q=g\left(p\right)⪯p$ so $\left({x}_{n},p\right)\in \mathrm{\Delta }$. Hence from (2) we have
$F\left(d\left(g{x}_{n},fp\right)\right)=F\left(d\left(f{x}_{n-1},fp\right)\right)\le F\left(d\left(g{x}_{n-1},gp\right)\right)-\tau .$

Since ${lim}_{n\to \mathrm{\infty }}d\left(g{x}_{n-1},gp\right)=0$, therefore by (C2) we have ${lim}_{n\to \mathrm{\infty }}F\left(d\left(g{x}_{n-1},gp\right)\right)=-\mathrm{\infty }$. Hence ${lim}_{n\to \mathrm{\infty }}F\left(d\left(g{x}_{n},fp\right)\right)=-\mathrm{\infty }$ implies that ${lim}_{n\to \mathrm{\infty }}d\left(g{x}_{n},fp\right)=0$. That is, ${lim}_{n\to \mathrm{\infty }}g{x}_{n}=fp$. Uniqueness of limit implies $fp=gp$, that is, $p\in C\left(f,g\right)$.

(b) Now suppose that $C\left(f,g\right)$ is well ordered. We prove that $C\left(f,g\right)$ is a singleton. Assume on the contrary that there exists another point w in X such that $fw=gw$ with $w\ne p$. Since $C\left(f,g\right)$ is well ordered, so $\left(w,p\right)\in \mathrm{\Delta }$. Now from (2) we have
$\tau \le F\left(d\left(gw,gp\right)\right)-F\left(d\left(fw,fp\right)\right)=0,$
a contradiction. Therefore $w=p$. Hence f and g have a unique coincidence point p in X. The converse follows immediately.
1. (c)

Now if f and g are weakly compatible mappings, then we have $fq=fgp=gfp=gq$, that is, q is the coincidence point of f and g. But q is the only point of coincidence of f and g, so $fq=gq=q$. Hence q is the unique common fixed point of f and g. □

Example 4 Let $X=\left[0,5\right]$ be endowed with usual metric and usual order. Define mappings $f,g:X\to X$ by
Clearly, g is dominated and f is dominating. Define $F:{\mathbb{R}}_{+}\to \mathbb{R}$ as $F\left(x\right)=ln\left(x\right)$. If $x\in \left[0,3\right)$ and $y\in \left[3,5\right)$, then
$\begin{array}{rcl}F\left(d\left(fx,fy\right)\right)& =& F\left(d\left(3,5\right)\right)=F\left(2\right)=ln\left(2\right)\approx 0.693\\ <& F\left(d\left(gx,gy\right)\right)=F\left(d\left(0,3\right)\right)\\ =& F\left(3\right)=ln\left(3\right)\approx 1.098.\end{array}$
Hence, for $\tau \in \left(0,0.40\right]$, inequality (2) is satisfied. Similarly, for $x\in \left[0,3\right)$ and $y=5$, we have
$\begin{array}{rcl}F\left(d\left(fx,fy\right)\right)& =& F\left(d\left(3,5\right)\right)=F\left(2\right)=ln\left(2\right)\approx 0.693\\ <& F\left(d\left(gx,gy\right)\right)=F\left(d\left(0,5\right)\right)\\ =& F\left(5\right)=ln\left(5\right)\approx 1.6094.\end{array}$
Hence, for $\tau \in \left(0,0.9164\right]$, inequality (2) is satisfied. We can take a $\tau \in \left(0,0.40\right]$ so that
$\tau +F\left(d\left(fx,fy\right)\right)\le F\left(d\left(gx,gy\right)\right)$

is satisfied for all $x,y\in \left[0,5\right]$, whenever $min\left\{d\left(fx,fy\right),d\left(gx,gy\right)\right\}>0$. Hence f is an F-contraction with respect to g on $\left[0,5\right]$. Hence all the conditions of Theorem 2 are satisfied. Moreover, $x=5$ is the coincidence point of f and g. Also note that f and g are weakly compatible and $x=5$ is the common fixed point of g and f as well.

Now we give a common fixed point result without imposing any type of commutativity condition for self-mappings f and g on X. Moreover, we relax the dominance conditions on f and g as well.

Theorem 3 Let $\left(X,⪯\right)$ be a partially ordered set such that there exists a complete metric d on X. If self-mappings f and g on X are weakly increasing and for some $\tau >0$ satisfy
$\tau +F\left(d\left(fx,gy\right)\right)\le F\left(d\left(x,y\right)\right)$
(5)

for all $\left(x,y\right)\in \mathrm{\Delta }$ such that $min\left\{d\left(fx,gy\right),d\left(x,y\right)\right\}>0$, then $F\left(f,g\right)\ne \mathrm{\varnothing }$, provided that X has the sequential limit comparison property. Further, f and g have a unique common fixed point if and only if $F\left(f,g\right)$ is well ordered.

Proof Let ${x}_{0}$ be an arbitrary point of X. Define a sequence ${\left\{{x}_{n}\right\}}_{n\in \mathbb{N}}$ in X as follows: ${x}_{2n+1}=f{x}_{2n}$ and ${x}_{2n+2}=g{x}_{2n+1}$. Since f and g are weakly increasing, we have ${x}_{2n+1}=f{x}_{2n}⪯gf{x}_{2n}=g{x}_{2n+1}={x}_{2n+2}$ and ${x}_{2n+2}=g{x}_{2n+1}⪯fg{x}_{2n+1}=f{x}_{2n+2}={x}_{2n+3}$. Hence $\left({x}_{2n+1},{x}_{2n+2}\right)\in \mathrm{\Delta }$ and $\left({x}_{2n+2},{x}_{2n+3}\right)\in \mathrm{\Delta }$ for every $n\in \mathbb{N}\cup \left\{0\right\}$. Now define
${\gamma }_{2n}=d\left({x}_{2n+1},{x}_{2n+2}\right)$
(6)
for all $n\in \mathbb{N}\cup \left\{0\right\}$. Using (5) the following holds for every $n\in \mathbb{N}\cup \left\{0\right\}$:
$\begin{array}{rcl}F\left({\gamma }_{2n}\right)& =& F\left(d\left({x}_{2n+1},{x}_{2n+2}\right)\right)=F\left(d\left(f{x}_{2n},g{x}_{2n+1}\right)\right)\\ \le & F\left(d\left({x}_{2n},{x}_{2n+1}\right)\right)-\tau =F\left({\gamma }_{2n-1}\right)-\tau .\end{array}$
Similarly,
$\begin{array}{rcl}F\left({\gamma }_{2n+1}\right)& =& F\left(d\left({x}_{2n+3},{x}_{2n+2}\right)\right)=F\left(d\left(f{x}_{2n+2},g{x}_{2n+1}\right)\right)\\ \le & F\left(d\left({x}_{2n+1},{x}_{2n+2}\right)\right)-\tau =F\left({\gamma }_{2n}\right)-\tau .\end{array}$
Therefore, for all $n\in \mathbb{N}\cup \left\{0\right\}$, we have
$\begin{array}{rcl}F\left({\gamma }_{n}\right)& \le & F\left({\gamma }_{n-1}\right)-\tau \le F\left({\gamma }_{n-2}\right)-2\tau \le \cdots \\ \le & F\left(d\left({x}_{1},{x}_{2}\right)\right)-n\tau =F\left({\gamma }_{0}\right)-n\tau .\end{array}$
Thus
$F\left({\gamma }_{n}\right)\le F\left({\gamma }_{0}\right)-n\tau .$
(7)
Taking limit as $n\to \mathrm{\infty }$ in (7), we get
$\underset{n\to \mathrm{\infty }}{lim}F\left({\gamma }_{n}\right)=-\mathrm{\infty }.$
By (C2) and (C3) we get ${lim}_{n\to \mathrm{\infty }}{\gamma }_{n}=0$ and $k\in \left(0,1\right)$ such that ${lim}_{n\to \mathrm{\infty }}{\gamma }_{n}^{k}F\left({\gamma }_{n}\right)=0$. Note that
${\gamma }_{n}^{k}F\left({\gamma }_{n}\right)-{\gamma }_{n}^{k}F\left({\gamma }_{0}\right)\le {\gamma }_{n}^{k}\left(F\left({\gamma }_{0}\right)-n\tau \right)-{\gamma }_{n}^{k}F\left({\gamma }_{0}\right)=-{\gamma }_{n}^{k}n\tau \le 0.$
(8)
By taking limit as $n\to \mathrm{\infty }$ in (8), we get ${lim}_{n\to \mathrm{\infty }}n{\gamma }_{n}^{k}=0$. This implies that there exists ${n}_{1}$ such that $n{\gamma }_{n}^{k}\le 1$ for all $n\ge {n}_{1}$. Consequently, we obtain ${\gamma }_{n}\le 1/{n}^{1/k}$ for all $n\ge {n}_{1}$. Now, for integers $m>n\ge 1$, we have
$d\left({x}_{n},{x}_{m}\right)\le d\left({x}_{n},{x}_{n+1}\right)+d\left({x}_{n+1},{x}_{n+2}\right)+\cdots +d\left({x}_{m-1},{x}_{m}\right)<\sum _{i=n}^{\mathrm{\infty }}{\gamma }_{i}\le \sum _{i=n}^{\mathrm{\infty }}\frac{1}{{i}^{\frac{1}{k}}}<\mathrm{\infty }.$
This shows that ${\left\{{x}_{n}\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence in X, so there exists p in X such that ${lim}_{n\to \mathrm{\infty }}{x}_{n}=p$. As X has the sequential limit comparison property, so $\left({x}_{n},p\right),\left({x}_{2n},p\right),\left({x}_{2n+1},p\right)\in \mathrm{\Delta }$. Therefore
$\underset{n\to \mathrm{\infty }}{lim}F\left(d\left({x}_{2n+1},gp\right)\right)=\underset{n\to \mathrm{\infty }}{lim}F\left(d\left(f{x}_{2n},gp\right)\right)\le F\left(d\left({x}_{2n},p\right)\right)-\tau .$

Since ${lim}_{n\to \mathrm{\infty }}d\left({x}_{2n},p\right)=0$, by (C2) we have ${lim}_{n\to \mathrm{\infty }}F\left(d\left({x}_{2n},p\right)\right)=-\mathrm{\infty }$. This implies ${lim}_{n\to \mathrm{\infty }}F\left(d\left({x}_{2n+1},gp\right)\right)=-\mathrm{\infty }$, which further implies that ${lim}_{n\to \mathrm{\infty }}d\left({x}_{2n+1},gp\right)=0$. Hence $d\left(p,gp\right)=0$ and $p=gp$. Similarly, we obtain $p=fp$. This shows that p is a common fixed point of g and f. Now suppose that $F\left(f,g\right)$ is well ordered. We prove that $F\left(f,g\right)$ is a singleton. Assume on the contrary that there exists another point q in X such that $q=fq=gq$ with $q\ne p$. Obviously, $\left(q,p\right)\in \mathrm{\Delta }$. So, from (5) we have $\tau \le F\left(d\left(q,p\right)\right)-F\left(d\left(fq,gp\right)\right)=0$, a contradiction. Therefore $q=p$. Hence g and f have a unique common fixed point p in X. The converse follows immediately. □

## 3 Periodic point results in metric spaces

If x is a fixed point of the self-mapping f, then x is a fixed point of ${f}^{n}$ for every $n\in \mathbb{N}$, but the converse is not true. In the sequel, we denote by $F\left(f\right)$ the set of all fixed points of f.

Example 5 Let $f:\left[0,1\right]\to \left[0,1\right]$ be given by
$f\left(x\right)=1-x.$
Then f has a unique fixed point $x=1/2$. Note that ${f}^{n}x=x$ holds for every even natural number n and x in $\left[0,1\right]$. On the other hand, define a mapping $g:\left[0,\pi \right]\to \left[0,\pi \right]$ as
$g\left(x\right)=cosx.$

Then g has the same fixed point as ${g}^{n}$ for every n.

Definition 10 The self-mapping f is said to have the property P if $F\left({f}^{n}\right)=F\left(f\right)$ for every $n\in \mathbb{N}$. A pair $\left(f,g\right)$ of self-mappings is said to have the property Q if $F\left(f\right)\cap F\left(g\right)=F\left({f}^{n}\right)\cap F\left({g}^{n}\right)$.

For further details on these properties, we refer to [20, 28].

Let $\left(X,d\right)$ be a metric space and $f:X\to X$ be a self-mapping. The set $O\left(x\right)=\left\{x,fx,\dots ,{f}^{n}x,\dots \right\}$ is called the orbit of x [29]. A mapping f is called orbitally continuous at p if ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=p$ implies that ${lim}_{n\to \mathrm{\infty }}{f}^{n+1}x=fp$. A mapping f is orbitally continuous on X if f is orbitally continuous for all $x\in X$.

In this section we prove some periodic point results for self-mappings on complete metric spaces.

Theorem 4 Let X be a nonempty set such that there exists a complete metric d on X. Suppose that $f:X\to X$ satisfies
$\tau +F\left(d\left(fx,{f}^{2}x\right)\right)\le F\left(d\left(x,fx\right)\right)$
(9)

for some $\tau >0$ and for all x in X such that $d\left(fx,{f}^{2}x\right)>0$. Then f has the property P provided that f is orbitally continuous on X.

Proof First we show that $F\left(f\right)\ne \mathrm{\varnothing }$. Let ${x}_{0}\in X$. Define a sequence ${\left\{{x}_{n}\right\}}_{n\in \mathbb{N}}$ in X, such that ${x}_{n+1}=f{x}_{n}$, for all $n\in \mathbb{N}\cup \left\{0\right\}$. Denote ${\gamma }_{n}=d\left({x}_{n},{x}_{n+1}\right)$ for all $n\in \mathbb{N}\cup \left\{0\right\}$. If there exists ${n}_{0}\in \mathbb{N}\cup \left\{0\right\}$ for which ${x}_{{n}_{0}+1}={x}_{{n}_{0}}$, then $f{x}_{{n}_{0}}={x}_{{n}_{0}}$ and the proof is finished. Suppose that ${x}_{n+1}\ne {x}_{n}$ for all $n\in \mathbb{N}\cup \left\{0\right\}$. Using (9), we obtain
$\begin{array}{rcl}F\left({\gamma }_{n}\right)& =& F\left(d\left({x}_{n},{x}_{n+1}\right)\right)=F\left(d\left(f{x}_{n-1},{f}^{2}{x}_{n-1}\right)\right)\\ \le & F\left(d\left({x}_{n-1},f{x}_{n-1}\right)\right)-\tau =F\left(d\left(f{x}_{n-2},{f}^{2}{x}_{n-2}\right)\right)-\tau \\ \le & F\left(d\left({x}_{n-2},f{x}_{n-2}\right)\right)-2\tau \le \cdots \\ \le & F\left(d\left({x}_{1},{x}_{2}\right)\right)-\left(n-1\right)\tau \\ =& F\left(d\left(f{x}_{0},{f}^{2}{x}_{1}\right)\right)-\left(n-1\right)\tau \le F\left(d\left({x}_{0},{x}_{1}\right)\right)-n\tau \\ =& F\left({\gamma }_{0}\right)-n\tau \end{array}$
for every $n\in \mathbb{N}\cup \left\{0\right\}$. By taking limit as $n\to \mathrm{\infty }$ in the above inequality, we obtain that ${lim}_{n\to \mathrm{\infty }}F\left({\gamma }_{n}\right)=-\mathrm{\infty }$, which together with (C2) gives ${lim}_{n\to \mathrm{\infty }}{\gamma }_{n}=0$. From (C3), there exists $k\in \left(0,1\right)$ such that ${lim}_{n\to \mathrm{\infty }}{\gamma }_{n}^{k}F\left({\gamma }_{n}\right)=0$. Note that
$\begin{array}{rcl}{\gamma }_{n}^{k}F\left({\gamma }_{n}\right)-{\gamma }_{n}^{k}F\left({\gamma }_{0}\right)& \le & {\gamma }_{n}^{k}\left(F\left({\gamma }_{0}\right)-n\tau \right)-{\gamma }_{n}^{k}F\left({\gamma }_{0}\right)\\ =& -{\gamma }_{n}^{k}n\tau \le 0.\end{array}$
On taking limit as $n\to \mathrm{\infty }$, we get ${lim}_{n\to \mathrm{\infty }}n{\gamma }_{n}^{k}=0$. Hence there exists ${n}_{1}$ such that $n{\gamma }_{n}^{k}\le 1$ for all $n\ge {n}_{1}$. Consequently ${\gamma }_{n}\le 1/{n}^{1/k}$ for all $n\ge {n}_{1}$. Now, for integers $m>n\ge 1$ such that
$\begin{array}{rcl}d\left({f}^{n}{x}_{0},{f}^{m}{x}_{0}\right)& =& d\left({x}_{n},{x}_{m}\right)\le d\left({x}_{n},{x}_{n+1}\right)+d\left({x}_{n+1},{x}_{n+2}\right)+\cdots +d\left({x}_{m-1},{x}_{m}\right)\\ <& \sum _{i=n}^{\mathrm{\infty }}{\gamma }_{i}\le \sum _{i=n}^{\mathrm{\infty }}\frac{1}{{i}^{\frac{1}{k}}}<\mathrm{\infty }.\end{array}$
This shows that ${\left\{{f}^{n}{x}_{0}\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence. Since $\left\{{f}^{n}{x}_{0}:n\in \mathbb{N}\right\}\subseteq O\left({x}_{0}\right)\subseteq X$ and X is complete, which implies that there exists x in X such that ${lim}_{n\to \mathrm{\infty }}{f}^{n}{x}_{0}=x$. Since f is orbitally continuous at x, so $x={lim}_{n\to \mathrm{\infty }}{f}^{n}{x}_{0}=f\left({lim}_{n\to \mathrm{\infty }}{f}^{n-1}{x}_{0}\right)=fx$. Hence f has a fixed point and $F\left({f}^{n}\right)=F\left(f\right)$ is true for $n=1$. Now assume $n>1$. Suppose on the contrary that $u\in F\left({f}^{n}\right)$ but $u\notin F\left(f\right)$, then $d\left(u,fu\right)=\alpha >0$. Now consider
$\begin{array}{rcl}F\left(\alpha \right)& =& F\left(d\left(u,fu\right)\right)=F\left(d\left(f\left({f}^{n-1}u\right),{f}^{2}\left({f}^{n-1}u\right)\right)\right)\\ \le & F\left(d\left({f}^{n-1}u,{f}^{n}u\right)\right)-\tau \\ \le & F\left(d\left({f}^{n-2}u,{f}^{n-1}u\right)\right)-2\tau \le \cdots \\ \le & F\left(d\left(u,fu\right)\right)-n\tau .\end{array}$

Thus $F\left(\alpha \right)\le {lim}_{n\to \mathrm{\infty }}F\left(d\left(u,fu\right)\right)-n\tau =-\mathrm{\infty }$. Hence $F\left(\alpha \right)=-\mathrm{\infty }$. By (C2) $\alpha =0$, a contradiction. So $u\in F\left(f\right)$. □

Theorem 5 Let $\left(X,⪯\right)$ be a partially ordered set such that there exists a complete metric d on X and f, g self-mappings on X. Further assume that f, g are weakly increasing and satisfy
$\tau +F\left(d\left(fx,gy\right)\right)\le F\left(d\left(x,y\right)\right)$

for some $\tau >0$, for all x, y in X such that $min\left\{d\left(fx,gy\right),d\left(x,y\right)\right\}>0$. Then f and g have the property Q provided that X has the sequential limit comparison property.

Proof By Theorem 3, f and g have a common fixed point. Suppose on the contrary that
$u\in F\left({f}^{n}\right)\cap F\left({g}^{n}\right)$
but $u\notin F\left(f\right)\cap F\left(g\right)$, then there are three possibilities (a) $u\in F\left(f\right)\setminus F\left(g\right)$, (b) $u\in F\left(g\right)\setminus F\left(f\right)$, (c) $u\notin F\left(f\right)$ and $u\notin F\left(g\right)$. Without loss of generality, let $u\notin F\left(g\right)$, that is, $d\left(u,gu\right)=\alpha >0$, so we get
$\begin{array}{rcl}F\left(\alpha \right)& =& F\left(d\left(u,gu\right)\right)=F\left(d\left(f\left({f}^{n-1}u\right),g\left({g}^{n}u\right)\right)\right)\\ \le & F\left(d\left({f}^{n-1}u,{g}^{n}u\right)\right)-\tau \\ \le & F\left(d\left({f}^{n-2}u,{g}^{n-1}u\right)\right)-2\tau \le \cdots \\ \le & F\left(d\left(u,gu\right)\right)-n\tau .\end{array}$

As ${lim}_{n\to \mathrm{\infty }}F\left(d\left(u,gu\right)\right)-n\tau =-\mathrm{\infty }$, so we have $F\left(\alpha \right)=-\mathrm{\infty }$. By (C2) $\alpha =0$, a contradiction. Hence $u\in F\left(g\right)\cap F\left(f\right)$. □

## Declarations

### Acknowledgements

The third author thanks for the support of the Ministry of Economy and Competitiveness of Spain, Grant MTM2012-37894-C02-01.

## Authors’ Affiliations

(1)
Department of Mathematics and Applied Mathematics, University of Pretoria, Hatfield, Pretoria, South Africa
(2)
Department of Mathematics, Syed Babar Ali School of Science and Engineering, Lahore University of Management Sciences, Lahore, 54792, Pakistan
(3)
Instituto Universitario de Matemática Pura y Aplicada, Universitat Politècnica de València, Valencia, 46022, Spain

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