Fixed and periodic points of generalized contractions in metric spaces

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.

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., [2–4]). Extensions of this principle were obtained either by generalizing the domain of the mapping or by extending the contractive condition on the mappings [5–9]. 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 [11–18]). Contractive conditions involving a pair of mappings are further additions to the metric fixed point theory and its applications (for details, see [19–23]).

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

Definition 2 ([25])

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

$$d(fx,fy)\le \alpha d(gx,gy)$$

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 $(X,d)$. 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 $g(X)\subseteq f(X)$.

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

• (C1) F is strictly increasing, that is, for all $\alpha ,\beta \in {\mathbb{R}}_{+}$ such that $\alpha <\beta$ implies that $F(\alpha )<F(\beta )$.

• (C2) For every sequence ${\{{\alpha }_n\}}_{n\in \mathbb{N}}$ of positive real numbers, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${lim}_{n\to \mathrm{\infty }}F({\alpha }_n)=-\mathrm{\infty }$ are equivalent.

• (C3) There exists $k\in (0,1)$ such that

  $$\underset{\alpha \to 0^{+}}{lim}{\alpha }^{k}F(\alpha )=0.$$

Definition 3 ([24])

Let $(X,d)$ 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

$$d(fx,fy)>0\text{implies that}\tau +F(d(fx,fy))\le F(d(x,y))$$

(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 $(X,d)$ 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

$$\tau +F(d(fx,fy))\le F(d(gx,gy))$$

(2)

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

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 $F_1(\alpha )=ln(\alpha )$. 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

$$\tau +ln(d(fx,fy))\le ln(d(gx,gy)),$$

which implies that

$$d(fx,fy)\le e^{-\tau }d(gx,gy).$$

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 ${\{S_n\}}_{n\in \mathbb{N}}$ [[24], Example 2.5]:

$$\begin{array}{c}{S}_1=1,\hfill \\ S_2=1+2,\hfill \\ S_3=1+2+3,\hfill \\ \cdots \hfill \\ S_n=1+2+\cdots +n=\frac{n(n+1)}{2},n\in \mathbb{N}.\hfill \end{array}$$

Let $X=\{S_n:n\in \mathbb{N}\}$ and d be the usual metric on X. Let $f:X\to X$ and $g:X\to X$ be defined as

$$f{S}_n=\{\begin{array}{cc}{S}_{n-1},\hfill & \text{if }n>1,\hfill \\ S_1,\hfill & \text{if }n=1,\hfill \end{array}g{S}_n=\{\begin{array}{cc}{S}_{n+1},\hfill & \text{if }n>1,\hfill \\ S_1,\hfill & \text{if }n=1.\hfill \end{array}$$

Let $F_1:{\mathbb{R}}_{+}\to \mathbb{R}$ be given by $F_1(\alpha )=ln(\alpha )$. As

$$\underset{n\to \mathrm{\infty }}{lim}\frac{d(f{S}_n,f{S}_1)}{d(g{S}_n,g{S}_1)}=\underset{n\to \mathrm{\infty }}{lim}\frac{S_{n-1}-S_1}{S_{n+1}-S_1}=1,$$

so f is not a g-contraction. If we take $F_2(\alpha )=ln(\alpha )+\alpha$, 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:

$$\frac{d(f{S}_n,f{S}_1)}{d(g{S}_n,g{S}_1)}{e}^{d(f{S}_n,f{S}_1)-d(g{S}_n,g{S}_1)}=\frac{S_{n-1}-S_1}{S_{n+1}-S_1}{e}^{S_{n-1}-S_1-S_{n+1}+S_1}=\frac{n^2-n-2}{n^2+3n}{e}^{-4n-2}\le e^{-2}$$

for all $n>1$. For all $m,n\in \mathbb{N}$ with $m>n>1$, we have

$$\begin{array}{r}\frac{d(f{S}_m,f{S}_n)}{d(g{S}_m,g{S}_n)}{e}^{d(f{S}_m,f{S}_n)-d(g{S}_m,g{S}_n)}\\ =\frac{S_{m-1}-S_{n-1}}{S_{m+1}-S_{n+1}}{e}^{S_{m-1}-S_{n-1}-S_{m+1}+S_{n+1}}\\ =\frac{m^2+m-n^2-n}{m^2+3m-n^2-3n}{e}^{-2(m-n)}\le e^{-2}.\end{array}$$

Definition 5 ([26], Dominance condition)

Let $(X,⪯)$ 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 $X=[0,1]$ 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

$$gx=x^n\le x\text{and}x\le kx=fx$$

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

Definition 6 Let $(X,⪯)$ 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 $(X,d,⪯)$ is called an ordered metric space if $(X,d)$ is a metric space and $(X,⪯)$ is a partially ordered set.

Definition 8 Let $(X,⪯)$ 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 $\mathrm{\Delta }\subseteq X\times X$ by

$$\mathrm{\Delta }=\{(x,y)\in X\times X:x⪯y\text{ or }y⪯x\}.$$

Definition 9 An ordered metric space $(X,d,⪯)$ is said to have the sequential limit comparison property if for every non-decreasing sequence (non-increasing sequence) ${\{x_n\}}_{n\in \mathbb{N}}$ 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 $(X,⪯)$ 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 $f(X)\subseteq g(X)$. Assume that f is dominating and g is dominated. Then

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

• (b) $C(f,g)$ is well ordered if and only if $C(f,g)$ is a singleton.

• (c) f and g have a unique common fixed point if f and g are weakly compatible and $C(f,g)$ 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 $f(x_0)=g(x_1)$. As f is dominating and g is dominated, so we have

$$x_0⪯f{x}_0=g{x}_1⪯x_1.$$

Hence $(x_0,x_1)\in \mathrm{\Delta }$. 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 $(x_n,x_{n+1})\in \mathrm{\Delta }$ for every $n\in \mathbb{N}\cup \{0\}$. For the sake of simplicity, take

$${\gamma }_n=d(g{x}_n,g{x}_{n+1})$$

(3)

for all $n\in \mathbb{N}\cup \{0\}$. If there exists $n_0\in \mathbb{N}\cup \{0\}$ 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, $x_{n_0+1}\in C(f,g)$. Now we assume that $x_{n+1}\ne x_n$ for all $n\in \mathbb{N}\cup \{0\}$. As f is an F-contraction with respect to g on Δ, so we obtain

$$\begin{array}{rcl}F({\gamma }_n)& =& F(d(g{x}_n,g{x}_{n+1}))=F(d(f{x}_{n-1},f{x}_n))\\ \le & F(d(g{x}_{n-1},g{x}_n))-\tau \\ =& F(d(f{x}_{n-2},f{x}_{n-1}))-\tau \\ \le & F(d(g{x}_{n-2},g{x}_{n-1}))-2\tau \le \cdots \\ \le & F(d(g{x}_1,g{x}_2))-(n-1)\tau =F({\gamma }_1)-(n-1)\tau .\end{array}$$

That is,

$$F({\gamma }_n)\le F({\gamma }_1)-(n-1)\tau .$$

On taking limit as $n\to \mathrm{\infty }$, we obtain ${lim}_{n\to \mathrm{\infty }}F({\gamma }_n)=-\mathrm{\infty }$. Hence ${lim}_{n\to \mathrm{\infty }}{\gamma }_n=0$ by (C2). Now, by (C3), there exists $k\in (0,1)$ such that ${lim}_{n\to \mathrm{\infty }}{\gamma }_n^{k}F({\gamma }_n)=0$. Note that

$${\gamma }_n^{k}F({\gamma }_n)-{\gamma }_n^{k}F({\gamma }_1)\le {\gamma }_n^k(F({\gamma }_1)-(n-1)\tau )-{\gamma }_n^{k}F({\gamma }_1)=-{\gamma }_n^k(n-1)\tau \le 0.$$

(4)

Taking limit as $n\to \mathrm{\infty }$ in (4), we have ${lim}_{n\to \mathrm{\infty }}(n-1){\gamma }_n^k=0$. Consequently, ${lim}_{n\to \mathrm{\infty }}n{\gamma }_n^k=0$. Thus there exists $n_1$ in ℕ 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

$$\begin{array}{rcl}d(g{x}_n,g{x}_m)& \le & d(g{x}_n,g{x}_{n+1})+d(g{x}_{n+1},g{x}_{n+2})+\cdots +d(g{x}_{m-1},g{x}_m)\\ <& \sum _{i=n}^{\mathrm{\infty }}{\gamma }_i\le \sum _{i=n}^{\mathrm{\infty }}\frac{1}{i^{\frac{1}{k}}}<\mathrm{\infty }.\end{array}$$

This shows that ${\{g{x}_n\}}_{n\in \mathbb{N}}$ is a Cauchy sequence in $g(X)$. As $g(X)$ is complete, so there exists q in $g(X)$ such that ${lim}_{n\to \mathrm{\infty }}g{x}_n=q$. Let $p\in X$ be such that $g(p)=q$. The sequential limit comparison property implies that $g{x}_{n+1}⪯q$. As $x_n⪯f{x}_n=g{x}_{n+1}⪯q=g(p)⪯p$ so $(x_n,p)\in \mathrm{\Delta }$. Hence from (2) we have

$$F(d(g{x}_n,fp))=F(d(f{x}_{n-1},fp))\le F(d(g{x}_{n-1},gp))-\tau .$$

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

(b) Now suppose that $C(f,g)$ is well ordered. We prove that $C(f,g)$ is a singleton. Assume on the contrary that there exists another point w in X such that $fw=gw$ with $w\ne p$. Since $C(f,g)$ is well ordered, so $(w,p)\in \mathrm{\Delta }$. Now from (2) we have

$$\tau \le F(d(gw,gp))-F(d(fw,fp))=0,$$

a contradiction. Therefore $w=p$. Hence f and g have a unique coincidence point p in X. The converse follows immediately.

• (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 $X=[0,5]$ be endowed with usual metric and usual order. Define mappings $f,g:X\to X$ by

$$gx=\{\begin{array}{cc}0\hfill & \text{if }x\in [0,3),\hfill \\ 3\hfill & \text{if }x\in [3,5),\hfill \\ 5\hfill & \text{if }x=5,\hfill \end{array}fx=\{\begin{array}{cc}3\hfill & \text{if }x\in [0,3),\hfill \\ 5\hfill & \text{if }x\in [3,5].\hfill \end{array}$$

Clearly, g is dominated and f is dominating. Define $F:{\mathbb{R}}_{+}\to \mathbb{R}$ as $F(x)=ln(x)$. If $x\in [0,3)$ and $y\in [3,5)$, then

$$\begin{array}{rcl}F(d(fx,fy))& =& F(d(3,5))=F(2)=ln(2)\approx 0.693\\ <& F(d(gx,gy))=F(d(0,3))\\ =& F(3)=ln(3)\approx 1.098.\end{array}$$

Hence, for $\tau \in (0,0.40]$, inequality (2) is satisfied. Similarly, for $x\in [0,3)$ and $y=5$, we have

$$\begin{array}{rcl}F(d(fx,fy))& =& F(d(3,5))=F(2)=ln(2)\approx 0.693\\ <& F(d(gx,gy))=F(d(0,5))\\ =& F(5)=ln(5)\approx 1.6094.\end{array}$$

Hence, for $\tau \in (0,0.9164]$, inequality (2) is satisfied. We can take a $\tau \in (0,0.40]$ so that

$$\tau +F(d(fx,fy))\le F(d(gx,gy))$$

is satisfied for all $x,y\in [0,5]$, whenever $min\{d(fx,fy),d(gx,gy)\}>0$. Hence f is an F-contraction with respect to g on $[0,5]$. 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 $(X,⪯)$ 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

$$\tau +F(d(fx,gy))\le F(d(x,y))$$

(5)

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

Proof Let $x_0$ be an arbitrary point of X. Define a sequence ${\{x_n\}}_{n\in \mathbb{N}}$ 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 $(x_{2n+1},x_{2n+2})\in \mathrm{\Delta }$ and $(x_{2n+2},x_{2n+3})\in \mathrm{\Delta }$ for every $n\in \mathbb{N}\cup \{0\}$. Now define

$${\gamma }_{2n}=d(x_{2n+1},x_{2n+2})$$

(6)

for all $n\in \mathbb{N}\cup \{0\}$. Using (5) the following holds for every $n\in \mathbb{N}\cup \{0\}$:

$$\begin{array}{rcl}F({\gamma }_{2n})& =& F(d(x_{2n+1},x_{2n+2}))=F(d(f{x}_{2n},g{x}_{2n+1}))\\ \le & F(d(x_{2n},x_{2n+1}))-\tau =F({\gamma }_{2n-1})-\tau .\end{array}$$

Similarly,

$$\begin{array}{rcl}F({\gamma }_{2n+1})& =& F(d(x_{2n+3},x_{2n+2}))=F(d(f{x}_{2n+2},g{x}_{2n+1}))\\ \le & F(d(x_{2n+1},x_{2n+2}))-\tau =F({\gamma }_{2n})-\tau .\end{array}$$

Therefore, for all $n\in \mathbb{N}\cup \{0\}$, we have

$$\begin{array}{rcl}F({\gamma }_n)& \le & F({\gamma }_{n-1})-\tau \le F({\gamma }_{n-2})-2\tau \le \cdots \\ \le & F(d(x_1,x_2))-n\tau =F({\gamma }_0)-n\tau .\end{array}$$

Thus

$$F({\gamma }_n)\le F({\gamma }_0)-n\tau .$$

(7)

Taking limit as $n\to \mathrm{\infty }$ in (7), we get

$$\underset{n\to \mathrm{\infty }}{lim}F({\gamma }_n)=-\mathrm{\infty }.$$

By (C2) and (C3) we get ${lim}_{n\to \mathrm{\infty }}{\gamma }_n=0$ and $k\in (0,1)$ such that ${lim}_{n\to \mathrm{\infty }}{\gamma }_n^{k}F({\gamma }_n)=0$. Note that

$${\gamma }_n^{k}F({\gamma }_n)-{\gamma }_n^{k}F({\gamma }_0)\le {\gamma }_n^k(F({\gamma }_0)-n\tau )-{\gamma }_n^{k}F({\gamma }_0)=-{\gamma }_n^{k}n\tau \le 0.$$

(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

$$d(x_n,x_m)\le d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})+\cdots +d(x_{m-1},x_m)<\sum _{i=n}^{\mathrm{\infty }}{\gamma }_i\le \sum _{i=n}^{\mathrm{\infty }}\frac{1}{i^{\frac{1}{k}}}<\mathrm{\infty }.$$

This shows that ${\{x_n\}}_{n\in \mathbb{N}}$ 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 $(x_n,p),(x_{2n},p),(x_{2n+1},p)\in \mathrm{\Delta }$. Therefore

$$\underset{n\to \mathrm{\infty }}{lim}F(d(x_{2n+1},gp))=\underset{n\to \mathrm{\infty }}{lim}F(d(f{x}_{2n},gp))\le F(d(x_{2n},p))-\tau .$$

Since ${lim}_{n\to \mathrm{\infty }}d(x_{2n},p)=0$, by (C2) we have ${lim}_{n\to \mathrm{\infty }}F(d(x_{2n},p))=-\mathrm{\infty }$. This implies ${lim}_{n\to \mathrm{\infty }}F(d(x_{2n+1},gp))=-\mathrm{\infty }$, which further implies that ${lim}_{n\to \mathrm{\infty }}d(x_{2n+1},gp)=0$. Hence $d(p,gp)=0$ and $p=gp$. Similarly, we obtain $p=fp$. This shows that p is a common fixed point of g and f. Now suppose that $F(f,g)$ is well ordered. We prove that $F(f,g)$ 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, $(q,p)\in \mathrm{\Delta }$. So, from (5) we have $\tau \le F(d(q,p))-F(d(fq,gp))=0$, 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 $F(f)$ the set of all fixed points of f.

Example 5 Let $f:[0,1]\to [0,1]$ be given by

$$f(x)=1-x.$$

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 $[0,1]$. On the other hand, define a mapping $g:[0,\pi ]\to [0,\pi ]$ as

$$g(x)=cosx.$$

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

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

Let $(X,d)$ be a metric space and $f:X\to X$ be a self-mapping. The set $O(x)=\{x,fx,\dots ,f^{n}x,\dots \}$ 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

$$\tau +F\left(d(fx,f^{2}x)\right)\le F(d(x,fx))$$

(9)

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

Proof First we show that $F(f)\ne \mathrm{\varnothing }$. Let $x_0\in X$. Define a sequence ${\{x_n\}}_{n\in \mathbb{N}}$ in X, such that $x_{n+1}=f{x}_n$, for all $n\in \mathbb{N}\cup \{0\}$. Denote ${\gamma }_n=d(x_n,x_{n+1})$ for all $n\in \mathbb{N}\cup \{0\}$. If there exists $n_0\in \mathbb{N}\cup \{0\}$ 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 $n\in \mathbb{N}\cup \{0\}$. Using (9), we obtain

$$\begin{array}{rcl}F({\gamma }_n)& =& F(d(x_n,x_{n+1}))=F\left(d(f{x}_{n-1},f^2{x}_{n-1})\right)\\ \le & F(d(x_{n-1},f{x}_{n-1}))-\tau =F\left(d(f{x}_{n-2},f^2{x}_{n-2})\right)-\tau \\ \le & F(d(x_{n-2},f{x}_{n-2}))-2\tau \le \cdots \\ \le & F(d(x_1,x_2))-(n-1)\tau \\ =& F\left(d(f{x}_0,f^2{x}_1)\right)-(n-1)\tau \le F(d(x_0,x_1))-n\tau \\ =& F({\gamma }_0)-n\tau \end{array}$$

for every $n\in \mathbb{N}\cup \{0\}$. By taking limit as $n\to \mathrm{\infty }$ in the above inequality, we obtain that ${lim}_{n\to \mathrm{\infty }}F({\gamma }_n)=-\mathrm{\infty }$, which together with (C2) gives ${lim}_{n\to \mathrm{\infty }}{\gamma }_n=0$. From (C3), there exists $k\in (0,1)$ such that ${lim}_{n\to \mathrm{\infty }}{\gamma }_n^{k}F({\gamma }_n)=0$. Note that

$$\begin{array}{rcl}{\gamma }_n^{k}F({\gamma }_n)-{\gamma }_n^{k}F({\gamma }_0)& \le & {\gamma }_n^k(F({\gamma }_0)-n\tau )-{\gamma }_n^{k}F({\gamma }_0)\\ =& -{\gamma }_n^{k}n\tau \le 0.\end{array}$$

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

$$\begin{array}{rcl}d(f^n{x}_0,f^m{x}_0)& =& d(x_n,x_m)\le d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})+\cdots +d(x_{m-1},x_m)\\ <& \sum _{i=n}^{\mathrm{\infty }}{\gamma }_i\le \sum _{i=n}^{\mathrm{\infty }}\frac{1}{i^{\frac{1}{k}}}<\mathrm{\infty }.\end{array}$$

This shows that ${\{f^n{x}_0\}}_{n\in \mathbb{N}}$ is a Cauchy sequence. Since $\{f^n{x}_0:n\in \mathbb{N}\}\subseteq O(x_0)\subseteq X$ 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 $x={lim}_{n\to \mathrm{\infty }}{f}^n{x}_0=f({lim}_{n\to \mathrm{\infty }}{f}^{n-1}{x}_0)=fx$. Hence f has a fixed point and $F(f^n)=F(f)$ is true for $n=1$. Now assume $n>1$. Suppose on the contrary that $u\in F(f^n)$ but $u\notin F(f)$, then $d(u,fu)=\alpha >0$. Now consider

$$\begin{array}{rcl}F(\alpha )& =& F(d(u,fu))=F\left(d(f\left(f^{n-1}u\right),f^2\left(f^{n-1}u\right))\right)\\ \le & F\left(d(f^{n-1}u,f^{n}u)\right)-\tau \\ \le & F\left(d(f^{n-2}u,f^{n-1}u)\right)-2\tau \le \cdots \\ \le & F(d(u,fu))-n\tau .\end{array}$$

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

Theorem 5 Let $(X,⪯)$ 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

$$\tau +F(d(fx,gy))\le F(d(x,y))$$

for some $\tau >0$, for all x, y in X such that $min\{d(fx,gy),d(x,y)\}>0$. 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 $u\notin F(f)\cap F(g)$, then there are three possibilities (a) $u\in F(f)\setminus F(g)$, (b) $u\in F(g)\setminus F(f)$, (c) $u\notin F(f)$ and $u\notin F(g)$. Without loss of generality, let $u\notin F(g)$, that is, $d(u,gu)=\alpha >0$, so we get

$$\begin{array}{rcl}F(\alpha )& =& F(d(u,gu))=F\left(d(f\left(f^{n-1}u\right),g\left(g^{n}u\right))\right)\\ \le & F\left(d(f^{n-1}u,g^{n}u)\right)-\tau \\ \le & F\left(d(f^{n-2}u,g^{n-1}u)\right)-2\tau \le \cdots \\ \le & F(d(u,gu))-n\tau .\end{array}$$

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