A New Kind of Nonlinear Quasicontractions in Metric Spaces

Abstract

Starting from two extensions of the Banach contraction principle due to Ćirić (1974) and Wardowski (2012), in the present paper we introduce the concepts of Ćirić type $\psi F$-contraction and $\psi F$-quasicontraction on a metric space and give some sufficient conditions under which the respective mappings are Picard operators. Some known fixed point results from the literature can be obtained as particular cases.

1. Introduction and Preliminaries

The Banach contraction principle, known also as Banach-Caccioppoli-Picard, is an important tool in the theory of metric spaces, having a crucial role in the study of many diverse disciplines. Starting from The Banach contraction principle, metrical fixed point theory has developed intensively in recent decades, both by generalizing the contractions and the metric spaces, and by extending the applications. Our paper is part of this effort by providing some fixed point results for a new kind of contraction map. This results generalize some earlier ones.

Throughout this paper, $(X,\mathrm{d})$ stands for a given metric space. We will also denote by T a self-mapping on X.

We will denote by $\mathbb{R}$, ${\mathbb{R}}_{+}$ and $\mathbb{N}$ the set of all real numbers, all positive real numbers and all positive integers, respectively. We will also write ${\overline{\mathbb{R}}}_{+}={\mathbb{R}}_{+}\cup \{\infty \}$.

If $\nu ,\lambda \in {\overline{\mathbb{R}}}_{+}$, by $“\nu >\lambda ”$ we mean $\nu >\lambda$ if $\lambda \in {\mathbb{R}}_{+}$ and $\nu =\infty$ when $\lambda =\infty$.

If $⌀\ne M\subset X$, the diameter of M is defined by $\mathrm{diam}M={sup}_{x,y\in M}\mathrm{d}(x,y)$.

T is said to be a Picard Operator, (P.O.), if it has a unique fixed point $\xi \in X$ and $T^{n}x\underset{n}{⟶}\xi$ for every $x\in X$.

We call the orbit of T in some $x\in X$ the set $\mathcal{O}\left(x\right)=\{T^{k}x:k=0,1,2,\dots \}$, where $T^0=id$. If $x,y\in X$ we will also denote $\mathcal{O}(x,y)=\mathcal{O}\left(x\right)\cup \mathcal{O}\left(y\right)$.

We say that the space X is T-orbitally complete if every Cauchy sequence contained in $\mathcal{O}\left(x\right)$ for some $x\in X$ is convergent. It is clear that any complete metric space is T-orbitally complete.

In [1] Ćirić introduced the concept of quasi-contraction for a self-mapping T on X as follows: there exists $0<q<1$ such that

$$\mathrm{d}(Tx,Ty)\le q·\mathrm{diam}\{x,y,Tx,Ty\}\forall x,y\in X$$

(1)

and proved that, if the space X is T-orbitally complete, then T is a P.O. This fixed point result generalizes that one obtained by Hardy and Rogers [2].

In [3] Bessenyei called weak φ-quasicontraction (respectively strong φ-quasicontraction) a self-mapping T on a metric space X of bounded orbits and, for every $x,y\in X$,

$$\mathrm{d}(Tx,Ty)\le \phi \left(\mathrm{diam}\mathcal{O}(x,y)\right)$$

(2)

(resp.

$$\mathrm{d}(Tx,Ty)\le \phi \left(\mathrm{diam}\{x,y,Tx,Ty\}\right)),$$

(3)

where $\phi :{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ is an increasing and upper semicontinuous map which satisfies the properties $\phi \left(0\right)=0$ and $\phi \left(t\right)<t$ for all $t>0$ (called comparison function).

Bessenyei proved the following fixed point result.

Theorem 1

([3]). Any weak φ-quasicontraction on a complete metric space is a P.O.

Clearly, if a strong $\phi$-quasicontraction has bounded orbits, then it is a weak $\phi$-quasicontraction. Some example of a strong $\phi$-quasicontraction of unbounded orbits can be found in [4].

The fixed point result obtained by Ćirić [1] is a consequence of the previous theorem.

Bessenyei ([4] Theorems 1–3) given some sufficient conditions for the function $\phi$ under which T is a P.O. in complete metric spaces.

Recently, Mitrović and Hussain [5] extended the survey of Bessenyei in the setting of b-metric spaces.

In [6] Wardowski introduced an interesting generalization of Banach contraction, namely F-contraction. He considers the class $\mathfrak{F}$ of functions $F:{\mathbb{R}}_{+}\to \mathbb{R}$ satisfying the following three properties:

(F1) F is increasing;

(F2) for every sequence $\left(t_n\right)$ of positive numbers, ${lim}_n{t}_n=0$ if and only if $F\left(t_n\right)\to -\infty$;

(F3) ${lim}_{t\to 0^{+}}{t}^{\lambda }F\left(t\right)=0$ for some $\lambda \in (0,1)$.

A mapping $T:X\to X$ is say to be F-contraction if there exist $\tau >0$ and $F\in \mathfrak{F}$ such that

$$Tx\ne Ty\Rightarrow \tau +F\left(\mathrm{d}(Tx,Ty)\right)\le F\left(\mathrm{d}(x,y)\right).$$

Wardowski shows that, if $(X,\mathrm{d})$ is complete, then any F-contraction is a Picard operator.

Since then, many authors extended and improved the result of Wardowski by simplifying the conditions for F or by considering generalized metric spaces. Piri and Kumam [7] and Secelean [8] proved that any F-contraction where F is continuous and satisfies (F1) and (F2) is a P.O.

Wardowski and Dung ([9], Definition 2.1) and Minak et al. ([10], Definition 2.1) defined Ćirić type generalized F-contraction (or F-weak contraction following Wardowski and Dung) for $T:X\to X$ as follows: there exist $F\in \mathfrak{F}$ and $\tau >0$ such that

$$\forall x,y\in X[\mathrm{d}(Tx,Ty)>0\Rightarrow \tau +F\left(\mathrm{d}(Tx,Ty)\right)\le F\left(maxM(x,y)\right)],$$

where ${\displaystyle M(x,y)=\left\{\mathrm{d}(x,y),\mathrm{d}(x,Tx),\mathrm{d}(y,Ty),\frac{1}{2}\left(\mathrm{d}(x,Ty)+\mathrm{d}(y,Tx)\right)\right\}}$.

In [9,10] it is proved that, if T is a Ćirić type generalized F-contraction on the complete metric space X and F is continuous, then T is a P.O.

Very recently, Proinov [11] considers two contractive type conditions:

$$\psi \left(\mathrm{d}(Tx,Ty)\right)\le \phi \left(\mathrm{d}(x,y)\right)\mathrm{for}\mathrm{all}x,y\in X\mathrm{with}\mathrm{d}(Tx,Ty)>0$$

and, respectively,

$$\psi \left(\mathrm{d}(Tx,Ty)\right)\le \phi \left(maxM(x,y)\right)\mathrm{for}\mathrm{all}x,y\in X\mathrm{with}\mathrm{d}(Tx,Ty)>0,$$

where $\psi ,\phi :{\mathbb{R}}_{+}\to \mathbb{R}$ are two functions such that $\phi <\psi$. Proinov does an extensive study to find sufficient conditions for the mappings $\psi$ and $\phi$ that assure that T is a P.O. The theorems proved in the above mentioned paper generalize and improve several existing fixed point results in the literature.

In what follows, we will denote by $\mathcal{F}$ the class of all nondecreasing mappings $F:(0,\nu )\to \mathbb{R}$ satisfying the following condition

(F4) $F(supA)\le supF\left(A\right)$, for every $A\subset (0,\nu )$ with $supA<\nu$, where $\nu >\mathrm{diam}X$.

Note that any nondecreasing and continuous map clearly satisfies (F4) while we can easily find nondecreasing and discontinuous functions satisfying (F4).

We also use the symbol $\mathsf{\Psi }$ to denote the family of all increasing functions $\psi :(-\infty ,\mu )\to (-\infty ,\mu )$ such that ${\psi }^n\left(t\right)\to -\infty$ for every $t\in (-\infty ,\mu )$, where $\mu =sup\left\{f\right(t):0<t<\nu \}$ and ${\psi }^n$ denotes the n-th composition of $\psi$.

We will need the following known results.

Lemma 1

([12] Lemma 3.2). Let $F:{\mathbb{R}}_{+}\to \mathbb{R}$ be a nondecreasing map and $\left(t_k\right)$ a sequence of positive real numbers. If $F\left(t_k\right)\to -\infty$, then $t_k\to 0$.

Lemma 2

([13] Lemma 3.1). If $\psi \in \mathsf{\Psi }$, then $\psi \left(t\right)<t$, for all $t\in (-\infty ,\mu )$.

Various examples of such functions $\psi$ can be obtained using the following lemma (see [13], Example 3.1).

Lemma 3

([13] Lemma 3.1). If $\psi :(-\infty ,\mu )\to (-\infty ,\mu )$ is an increasing and right continuous function such that $\psi \left(t\right)<t$ for all $t\in (-\infty ,\mu )$, then $\psi \in \mathsf{\Psi }$.

A mapping $T:X\to X$ is called $\psi F$-contraction (see [13]) if $F:(0,\nu )\to \mathbb{R}$ is nondecreasing, $\psi \in \mathsf{\Psi }$ and the following property holds

$$Tx\ne Ty\Rightarrow F\left(\mathrm{d}(Tx,Ty)\right)\le \psi F\left(\mathrm{d}(x,y)\right).$$

(4)

2. Main Results

Definition 1.

Let us consider $F\in \mathcal{F}$ and $\psi \in \mathsf{\Psi }$. A mapping $T:X\to X$ is said to be a Ćirić type $\psi F$-contraction if, for every $x,y\in X$, one has

$$Tx\ne Ty\Rightarrow F\left(\mathrm{d}(Tx,Ty)\right)\le \psi F\left(maxM(x,y)\right),$$

(5)

where ${\displaystyle M(x,y)=\left\{\mathrm{d}(x,y),\mathrm{d}(x,Tx),\mathrm{d}(y,Ty),\frac{1}{2}\left(\mathrm{d}(x,Ty)+\mathrm{d}(y,Tx)\right)\right\}}$.

T is said to be a strong $\psi F$-quasicontraction if, for every $x,y\in X$,

$$Tx\ne Ty\Rightarrow F\left(\mathrm{d}(Tx,Ty)\right)\le \psi F\left(\mathrm{diam}\{x,y,Tx,Ty\}\right).$$

(6)

Similarly, T is called weak $\psi F$-quasicontraction if it has bounded orbits and, for every $x,y\in X$, we have

$$Tx\ne Ty\Rightarrow F\left(\mathrm{d}(Tx,Ty)\right)\le \psi F\left(\mathrm{diam}\mathcal{O}(x,y)\right).$$

(7)

Remark 1.

(1) It is obvious that, if T is a strong $\psi F$-quasicontraction, then $\mathrm{d}(Tx,Ty)\ne \mathrm{diam}\{x,y,Tx,Ty\}$ because otherwise, in (6) we would have

$$F\left(\mathrm{d}(Tx,Ty)\right)\le \psi F\left(\mathrm{diam}\{x,y,Tx,Ty\}\right)<F\left(\mathrm{d}(Tx,Ty)\right)$$

which is a contradiction. Therefore, in (6), one can assume that

$$\mathrm{diam}\{x,y,Tx,Ty\}=max\left\{\mathrm{d}\right(x,y),\mathrm{d}(x,Tx),\mathrm{d}(y,Ty),\mathrm{d}(x,Ty),\mathrm{d}(y,Tx\left)\right\}.$$

(2) If T is a strong $\psi F$-quasicontraction where $F:{\mathbb{R}}_{+}\to \mathbb{R}$, $F\left(t\right)=-1/t$ and $\psi :(-\infty ,0)\to (-\infty ,0)$, $\psi \left(t\right)=qt$, $q>1$, then T is a Ćirić quasicontraction (1) with $k=1/q$.

Example 1.

Let us consider $X=[0,\infty )$ endowed with Euclidean metric $\mathrm{d}(x,y)=|x-y|$ and the mappings $T:X\to X$, $Tx=x+\alpha$, $\alpha \ge 1$, $F:(0,\infty )\to \mathbb{R}$, $F\left(t\right)=lnt$, $\psi :(-\infty ,0)\to (-\infty ,0)$, $\psi \left(t\right)=t-e^{-t}$. Then $F\in \mathcal{F}$, $\psi \in \mathsf{\Psi }$ and T is a strong $\psi F$-quasicontraction.

Proof. 

Fix $x\ne y\in X$. It is easy to see that

$$\mathrm{diam}\{x,y,Tx,Ty\}=max\left\{|x-y|,\alpha ,|x-y|+\alpha \right\}=|x-y|+\alpha .$$

One has

$$F\left(\mathrm{d}(Tx,Ty)\right)\le \psi F\left(\mathrm{diam}\{x,y,Tx,Ty\}\right)$$

$$\iff ln|x-y|\le ln\left(|x-y|+\alpha \right)-\frac{1}{|x-y|+\alpha }$$

$$\iff ln\left(1+\frac{\alpha }{|x-y|}\right)-\frac{1}{|x-y|+\alpha }\ge 0.$$

The last inequality follows by considering the continuous function $f:{\mathbb{R}}_{+}\to \mathbb{R}$,

$$f\left(t\right)=ln(1+\frac{\alpha }{t})-\frac{1}{t+\alpha }$$

and seeing that ${lim}_{t\to 0^{+}}f\left(t\right)=\infty$, ${lim}_{t\to \infty }f\left(t\right)=0$ and f is decreasing. Therefore T is a strong $\psi F$-quasicontraction. □

Inspired by [3] we will adapt the next two results to our settings.

Lemma 4.

If $F\in \mathcal{F}$, $\psi \in \mathsf{\Psi }$ and $T:X\to X$ is a weak $\psi F$-quasicontraction, then, for every $n\in \mathbb{N}$, $T^n$ is a weak ${\psi }^{n}F$-quasicontraction.

Proof. 

Let $x,y\in X$ be such that $Tx\ne Ty$. Then $\mathrm{diam}\mathcal{O}(x,y)>0$ so the set $J_{xy}:=\left\{(k,l)\in \mathbb{N}\times \mathbb{N}:max\{\mathrm{d}(T^{k}x,T^{l}y),\mathrm{d}(T^{k}x,T^{l}x),\mathrm{d}(T^{k}y,T^{l}y)\}>0\right\}$ is nonempty.

Choose $(k,l)\in J_{x,y}$.

If $\mathrm{d}(T^{k}x,T^{l}y)>0$, then $\mathrm{d}(T^{k-1}x,T^{l-1}y)>0$ and, by (7) and monotonicity of F and $\psi$, one has

$$F\left(\mathrm{d}(T^{k}x,T^{l}y)\right)=F\left(\mathrm{d}(T{T}^{k-1}x,T{T}^{l-1}y)\right)\le \psi F\left(\mathrm{diam}\mathcal{O}(T^{k-1}x,T^{l-1}y)\right)$$

$$\le \psi F\left(\mathrm{diam}\mathcal{O}(x,y)\right).$$

By a similar argument, if $\mathrm{d}(T^{k}x,T^{l}x)>0$ or $\mathrm{d}(T^{k}y,T^{l}y)>0$, we obtain

$$F\left(\mathrm{d}(T^{k}x,T^{l}x)\right)\le \psi F\left(\mathrm{diam}\mathcal{O}(x,y)\right)$$

respectively

$$F\left(\mathrm{d}(T^{k}y,T^{l}y)\right)\le \psi F\left(\mathrm{diam}\mathcal{O}(x,y)\right).$$

Using now (F4), it follows

$$\begin{array}{ccc}F\left(\mathrm{diam}\mathcal{O}(Tx,Ty)\right)\hfill & =\hfill & F\left(\underset{k,l\in \mathbb{N}}{sup}\{\mathrm{d}(T^{k}x,T^{l}y),\mathrm{d}(T^{k}x,T^{l}x),\mathrm{d}(T^{k}y,T^{l}y)\}\right)\hfill \\ \hfill & =\hfill & F\left(\underset{(k,l)\in J_{xy}}{sup}\{\mathrm{d}(T^{k}x,T^{l}y),\mathrm{d}(T^{k}x,T^{l}x),\mathrm{d}(T^{k}y,T^{l}y)\}\right)\hfill \\ \hfill & \le \hfill & \underset{(k,l)\in J_{xy}}{sup}F\left(max\{\mathrm{d}(T^{k}x,T^{l}y),\mathrm{d}(T^{k}x,T^{l}x),\mathrm{d}(T^{k}y,T^{l}y)\}\right)\hfill \\ \hfill & \le \hfill & \psi F\left(\mathrm{diam}\mathcal{O}(x,y)\right).\hfill \end{array}$$

(8)

For the last step of the proof we use the induction. For $n=1$, the assertion is obvious. Assume that it holds for some $n\in \mathbb{N}$. Let $x,y\in X$ be such that $T^{n+1}x\ne T^{n+1}y$. Hence $T^{n}x\ne T^{n}y$ and, by (8), one has

$$F\left(\mathrm{d}(T^{n+1}x,T^{n+1}y)\right)\le {\psi }^{n}F\left(\mathrm{diam}\mathcal{O}(Tx,Ty)\right)\le {\psi }^{n+1}F\left(\mathrm{diam}\mathcal{O}(x,y)\right)$$

completing the proof. □

Theorem 2.

Let T be a weak $\psi F$-quasicontraction. If X is T-orbitally complete, then T is a P.O.

Proof. 

Let any $x\in X$. We intend to prove that the sequence $\left(T^{n}x\right)$ is Cauchy. Let $J_x=\{(k,l)\in \mathbb{N}\times \mathbb{N};\mathrm{d}(T^{k}x,T^{l}x)>0\}$. If the set $J_x$ is finite, then the sequence $\left(T^{n}x\right)$ is stationary so it is Cauchy. Suppose that $J_x$ is infinite. Without loss of generality we can assume that $J_x=\mathbb{N}\times \mathbb{N}$. According to Lemma 4 one has (taking into account $\mathrm{diam}\mathcal{O}\left(x\right)>0$)

$$F\left(\mathrm{d}(T^{n}x,T^{n+p}x)\right)\le {\psi }^{n}F\left(\mathrm{diam}\mathcal{O}(x,T^{p}x)\right)={\psi }^{n}F\left(\mathrm{diam}\mathcal{O}\left(x\right)\right)\underset{n}{⟶}-\infty .$$

It follows from Lemma 1 that $\mathrm{diam}(T^{n}x,T^{m}x)\underset{n,m}{⟶}0$ hence the sequence $\left(T^{n}x\right)$ is Cauchy. By T-orbitally completeness of the space X we deduce that there is $\xi \in X$ such that $T^{n}x\to \xi$.

Take any $y\in X$. We claim that

$$T^{n}y\to \xi .$$

(9)

Indeed, if there is $N\in \mathbb{N}$ such that $T^{N}x=T^{N}y$ the assertion follows. Assuming that $T^{n}x\ne T^{n}y$ for all n and using again Lemma 4, one obtains

$$F\left(\mathrm{d}(T^{n}x,T^{n}y)\right)\le {\psi }^{n}F\left(\mathrm{diam}\mathcal{O}(x,y)\right)\underset{n}{⟶}-\infty$$

so, by Lemma 1, $\mathrm{d}(T^{n}x,T^{n}y)\to 0$. Next, since

$$\mathrm{d}(T^{n}y,\xi )\le \mathrm{d}(T^{n}y,T^{n}x)+\mathrm{d}(T^{n}x,\xi )\to 0,$$

we get the claim.

In order to prove that $\xi$ is a fixed point of T, will be enough to show that $\mathrm{diam}\mathcal{O}\left(\xi \right)=0$. Arguing by contradiction, assume that $\mathrm{diam}\mathcal{O}\left(\xi \right)>0$. We have

$$\mathrm{diam}\mathcal{O}\left(\xi \right)=\mathrm{diam}\left(\left\{\xi \right\}\cup \mathcal{O}\left(T\xi \right)\right)=\underset{k,l}{sup}\left\{\mathrm{d}(\xi ,T^k\xi ),\mathrm{d}(T^k\xi ,T^l\xi )\right\}.$$

Set $J_{\xi }=\{(k,l)\in \mathbb{N}\times \mathbb{N};T^k\xi \ne T^l\xi \}$.

If $J_{\xi }=\varnothing$, then $\mathrm{diam}\mathcal{O}\left(\xi \right)=\underset{n}{sup}\mathrm{d}(\xi ,T^n\xi )$.

Assume that $J_{\xi }\ne \varnothing$. Then, according to Lemma 2,

$$\begin{array}{ccc}F\left(\underset{k,l\in \mathbb{N}}{sup}\mathrm{d}(T^k\xi ,T^l\xi )\right)\hfill & =\hfill & F\left(\underset{(k,l)\in J_{\xi }}{sup}\mathrm{d}(T^k\xi ,T^l\xi )\right)\le \underset{(k,l)\in J_{\xi }}{sup}F\left(\mathrm{d}(T^k\xi ,T^l\xi )\right)\hfill \\ \hfill & \le \hfill & \underset{(k,l)\in J_{\xi }}{sup}\psi F\left(\mathrm{d}(T^{k-1}\xi ,T^{l-1}\xi )\right)\le \psi F\left(\mathrm{diam}\mathcal{O}\left(\xi \right)\right)\hfill \\ \hfill & <\hfill & F\left(\mathrm{diam}\mathcal{O}\left(\xi \right)\right).\hfill \end{array}$$

By the monotonicity of F we conclude that

$$\underset{k,l\in \mathbb{N}}{sup}\mathrm{d}(T^k\xi ,T^l\xi )<\mathrm{diam}\mathcal{O}\left(\xi \right)$$

(10)

so $\mathrm{diam}\mathcal{O}\left(\xi \right)\le \underset{n}{sup}\mathrm{d}(\xi ,T^n\xi )$. Since the converse inequality is obvious we conclude that $\mathrm{diam}\mathcal{O}\left(\xi \right)=\underset{n}{sup}\mathrm{d}(\xi ,T^n\xi )$.

By (9) one can find $K\in \mathbb{N}$ such that $\mathrm{diam}\mathcal{O}\left(\xi \right)=\mathrm{d}(\xi ,T^K\xi )$.

Write $\lambda =\underset{k,l\in \mathbb{N}}{sup}\mathrm{d}(T^k\xi ,T^l\xi )$ and choose $0<\epsilon <\mathrm{diam}\mathcal{O}\left(\xi \right)-\lambda$. By (9) there exists $n_{\epsilon }\in \mathbb{N}$ such that $\mathrm{d}(\xi ,T^n\xi )<\epsilon$ for all $n\ge n_{\epsilon }$. Consequently, for every $n\ge n_{\epsilon }$, one has

$$\mathrm{diam}\mathcal{O}\left(\xi \right)=\mathrm{d}(\xi ,T^K\xi )\le \mathrm{d}(\xi ,T^n\xi )+\mathrm{d}(T^n\xi ,T^K\xi )<\epsilon +\lambda <\mathrm{diam}\mathcal{O}\left(\xi \right).$$

This contradiction ensures that $\xi$ is a fixed point of T.

To complete the proof, suppose that ${\xi }_1\ne {\xi }_2$ are two fixed points of T. Then, for all n, $T^n{\xi }_1={\xi }_1$ and $T^n{\xi }_2={\xi }_2$. Hence

$$F\left(\mathrm{d}(T^n{\xi }_1,T^n{\xi }_2)\right)\le {\psi }^{n}F\left(\mathrm{diam}\mathcal{O}({\xi }_1,{\xi }_2)\right)\underset{n}{⟶}-\infty$$

so $\mathrm{d}({\xi }_1,{\xi }_2)=\mathrm{d}(T^n{\xi }_1,T^n{\xi }_2)⟶0$, that is ${\xi }_1={\xi }_2$.

The successive approximation of $\xi$ follows from (9). □

From Lemma 4 and Theorem 2 it follows obviously.

Corollary 1.

If T is a weak $\psi F$-quasicontraction on a T-orbitally complete metric space, then $T^n$ is a P.O. for every $n\in \mathbb{N}$.

In the following we will show that the class of $\psi F$-quasicontractions includes that of $\phi$-quasicontractions. We need first two preliminary results.

Lemma 5

([12] Lemma 3.2). Let $F:{\mathbb{R}}_{+}\to \mathbb{R}$ be an increasing mapping such that $infF=-\infty$. If $\left(t_n\right)$ is a sequence of positive real numbers with $\underset{n}{lim}{t}_n=0$, then $\underset{n}{lim}F\left(t_n\right)=-\infty$.

Lemma 6

([3]). If φ is a comparison function, then ${\phi }^n\left(t\right)\underset{n}{⟶}0$ for all $t\in {\mathbb{R}}_{+}$.

Proposition 1.

Any weak φ-quasicontraction is a weak $\psi F$-quasicontraction.

Proof. 

Let $T:X\to X$ be a weak $\phi$-quasicontraction and consider an increasing and continuous function $F:{\mathbb{R}}_{+}\to \mathbb{R}$ such that $infF=-\infty$ (such functions F can be $lnt$, $lnt-t^{-1}$, ${(1-e^t)}^{-1}$, $-t^{\alpha }$, $-t^{\alpha }+t^{\beta }$, $\alpha <0$, $\beta >0$; more examples can be found in [12,13]). Set $\mu =supF\left({\mathbb{R}}_{+}\right)\in \overline{\mathbb{R}}$. It is obvious that $F\in \mathcal{F}$. Since $F:{\mathbb{R}}_{+}\to (-\infty ,\mu )$ is invertible, we can define $\psi :(-\infty ,\mu )\to (-\infty ,\mu )$ by

$$\psi =F\circ \phi \circ F^{-1}.$$

(11)

We claim that $\psi \in \mathsf{\Psi }$. Indeed, clearly $\psi$ is increasing as composition of increasing functions. Next, if $t\in (-\infty ,\mu )$ and $n\in \mathbb{N}$, then

$${\psi }^n\left(t\right)=F\phi F^{-1}\circ F\phi F^{-1}\circ \dots \circ F\phi F^{-1}\left(t\right)=F{\phi }^n{F}^{-1}\left(t\right).$$

It follows from Lemma 6 that ${\phi }^n{F}^{-1}\left(t\right)\underset{n}{⟶}0$ hence, by Lemma 5, $F{\phi }^n{F}^{-1}\left(t\right)\underset{n}{⟶}-\infty$. Consequently $\psi \in \mathsf{\Psi }$.

If $x,y\in X$ are such that $Tx\ne Ty$, then, by (2), (11) and the monotonicity of F, one has

$$F\left(\mathrm{d}(Tx,Ty)\right)\le F\phi \left(\mathrm{diam}\mathcal{O}(x,y)\right)=\psi F\left(\mathrm{diam}\mathcal{O}(x,y)\right)$$

as required. □

Remark 2.

Theorem 1 is a consequence of Proposition 1 and Theorem 2. Consequently, every quasi-contraction T on a T-orbitally complete metric space is a P.O. ([1], Theorem 1). This fact follows from the above and the boundedness of orbits (this property can be easily proved (for details see [3])). These theorems generalize a lot of other fixed point results.

In order to prove the next results we need the following lemma due to Turinici [14].

Lemma 7

([14], Proposition 3). Let us consider a sequence $\left(x_n\right)$ in the metric space X and let Δ be a dense subset of ${\mathbb{R}}_{+}$. If $\mathrm{d}(x_n,x_{n+1})\to 0$ and $\left(x_n\right)$ is not Cauchy, then there exist $\epsilon \in {\mathbb{R}}_{+}\setminus \mathsf{\Delta }$, $n_0\in \mathbb{N}$ and the sequences of natural numbers $\left(m_k\right),\left(n_k\right)$ such that

1. 

$\forall k\in \mathbb{N},k\le m_k<n_k\Rightarrow \mathrm{d}(x_{m_k},x_{n_k})>\epsilon$,

2. 

$\mathrm{d}(x_{m_k},x_{n_k})\to {\epsilon }^{+}$, $k\to \infty ,$

3. 

$\mathrm{d}(x_{m_k+p},x_{n_k+q})\to \epsilon$, $k\to \infty$, $p,q\in \{0,1\}$.

Definition 2

([15]). We say that a self-mapping T on X is asymptotically regular if, for each $x\in X$, ${lim}_n\mathrm{d}(T^{n}x,T^{n+1}x)=0$.

Theorem 3.

Assume that the mapping $\psi \in \mathsf{\Psi }$ is right continuous. Then any asymptotically regular strong $\psi F$-quasicontraction is a weak $\psi F$-quasicontraction.

Proof. 

Let T denote an asymptotically regular strong $\psi F$-quasicontraction.

Since (7) follows obviously from (6), it suffices to prove the boundedness of the orbits.

Fix $x\in X$ and denote $x_n=T^{n}x$ for $n=0,1,\dots$ We will prove that the sequence $\left(x_n\right)$ is Cauchy. As T is asymptotically regular, we can apply Lemma 7.

Suppose that $\left(x_n\right)$ is not a Cauchy sequence. Let denote by $\mathsf{\Delta }$ the set of discontinuities of F. Due to the monotonicity of F, the set $\mathsf{\Delta }$ is at most countable hence it is dense in ${\mathbb{R}}_{+}$. According to Lemma 7, there exist $\epsilon >0$, $\epsilon \notin \mathsf{\Delta }$ and the sequences $\left(m_k\right),\left(n_k\right)$ such that

$$\mathrm{d}(x_{m_k},x_{n_k})\to {\epsilon }^{+},\mathrm{d}(x_{m_k+1},x_{n_k+1})\to \epsilon ,k\to \infty .$$

(12)

Therefore one obtains for each $k\in \mathbb{N}$

$$F\left(\mathrm{d}\right(x_{m_k+1},x_{n_k+1}\left)\right)$$

$$\le \psi F\left(max\{\mathrm{d}(x_{m_k},x_{n_k}),\mathrm{d}(x_{m_k},x_{m_k+1}),\mathrm{d}(x_{n_k},x_{n_k+1}),\mathrm{d}(x_{m_k},x_{n_k+1}),\mathrm{d}(x_{n_k},x_{m_k+1})\}\right)$$

From hypothesis and (12) it follows that

$$\mathrm{d}(x_{m_k},x_{n_k})\to {\epsilon }^{+},\mathrm{d}(x_{m_k},x_{m_k+1})\to 0,\mathrm{d}(x_{n_k},x_{n_k+1})\to 0,k\to \infty .$$

Since

$$\mathrm{d}(x_{m_k},x_{n_k+1})\le \mathrm{d}(x_{m_k},x_{n_k})+\mathrm{d}(x_{n_k},x_{n_k+1}),$$

$$\mathrm{d}(x_{n_k},x_{m_k+1})\le \mathrm{d}(x_{n_k},x_{m_k})+\mathrm{d}(x_{m_k},x_{m_k+1}),$$

we have

$$\epsilon \le \mathrm{d}(x_{m_k},x_{n_k+1})\underset{k}{⟶}\epsilon ,\epsilon \le \mathrm{d}(x_{n_k},x_{m_k+1})\underset{k}{⟶}\epsilon .$$

Accordingly

$$\epsilon \le max\left\{\mathrm{d}(x_{m_k},x_{n_k}),\mathrm{d}(x_{m_k},x_{m_k+1}),\mathrm{d}(x_{n_k},x_{n_k+1}),\mathrm{d}(x_{m_k},x_{n_k+1}),\mathrm{d}(x_{n_k},x_{m_k+1})\right\}\underset{k}{⟶}\epsilon .$$

Letting $k\to \infty$ and using the facts that F is continuous at $\epsilon$ and $\psi$ is right continuous we get

$$F\left(\epsilon \right)\le \psi \left(F\right(\epsilon \left)\right)<F\left(\epsilon \right),$$

which is a contradiction. Consequently the sequence $\left(T^{n}x\right)$ is Cauchy so it is bounded. This assures that $\mathcal{O}\left(x\right)$ is bounded. □

Remark 3.

The condition that T to be asymptotically regular in the above theorem can’t be dropped. Indeed, the mapping T from Example 1 is a strong $\psi F$-quasicontraction on the complete metric space X but it has no fixed points, hence, according to Theorem 2, it can’t be a weak $\psi F$-quasicontraction.

Corollary 2.

Assume that the mapping $\psi \in \mathsf{\Psi }$ is right continuous. Then any Ćirić type $\psi F$-contraction is a weak $\psi F$-quasicontraction.

Proof. 

According to Theorem 3 it suffices to show that T is asymptotically regular.

Fix $x\in X$ and denote $x_n=T^{n}x$ for $n=0,1,\dots$ We must prove that $\mathrm{d}(x_n,x_{n+1})\to 0$. Indeed, if $x_n=x_{n+1}$ for some $n\in \mathbb{N}$, then the assertion is obvious. Assume that $x_n\ne x_{n+1}$ for all $n\ge 1$. Then, by (5),

$$F\left(\mathrm{d}(x_n,x_{n+1})\right)$$

$$\le \psi F(max\left\{\mathrm{d}(x_n,x_{n-1}),\mathrm{d}(x_n,x_{n+1}),\mathrm{d}(x_{n-1},x_n),\frac{\mathrm{d}(x_n,x_n)+\mathrm{d}(x_{n-1},x_{n+1})}{2}\right\})$$

$$\le \psi F(max\left\{\mathrm{d}(x_n,x_{n-1}),\mathrm{d}(x_n,x_{n+1}),\frac{\mathrm{d}(x_{n-1},x_n)+\mathrm{d}(x_n,x_{n+1})}{2}\right\})$$

$$\le \psi F(max\left\{\mathrm{d}(x_n,x_{n-1}),\mathrm{d}(x_n,x_{n+1})\right\}),$$

where, in the last inequality, we used the relation

$$\frac{\mathrm{d}(x_{n-1},x_n)+\mathrm{d}(x_n,x_{n+1})}{2}\le max\{\mathrm{d}(x_{n-1},x_n),\mathrm{d}(x_n,x_{n+1})\}.$$

Assuming that $max\left\{\mathrm{d}(x_n,x_{n-1}),\mathrm{d}(x_n,x_{n+1})\right\}=\mathrm{d}(x_n,x_{n+1})$ we deduce by the above that

$$F\left(\mathrm{d}(x_n,x_{n+1})\right)\le \psi F\left(\mathrm{d}(x_n,x_{n+1})\right)<F\left(\mathrm{d}(x_n,x_{n+1})\right)$$

a contradiction. Hence $max\left\{\mathrm{d}(x_n,x_{n-1}),\mathrm{d}(x_n,x_{n+1})\right\}=\mathrm{d}(x_n,x_{n-1})$. Consequently

$$F\left(\mathrm{d}(x_n,x_{n+1})\right)\le \psi F\left(\mathrm{d}(x_n,x_{n-1})\right)<F\left(\mathrm{d}(x_n,x_{n-1})\right)$$

and so

$$\mathrm{d}(x_n,x_{n+1})<\mathrm{d}(x_{n-1},x_n)$$

for all $n=1,2,\dots$ Inductively we get

$$F\left(\mathrm{d}(x_n,x_{n+1})\right)\le {\psi }^{n}F\left(\mathrm{d}(x,Tx)\right)\underset{n}{⟶}-\infty .$$

Thus $\mathrm{d}(x_n,x_{n+1})\to 0^{+}.$ □

From Corollary 2 and Theorem 2 it follows:

Corollary 3.

If T is a Ćirić type $\psi F$-contraction on the T-orbitally complete metric space and ψ is right continuous, then it is a P.O.

Taking $\psi \left(t\right)=t-\tau$, $\tau >0$, in the previous corollary, we obtain:

Corollary 4.

Any Ćirić type generalized F-contraction T on a metric space X, where $F\in \mathcal{F}$, is a Ćirić type $\psi F$-contraction. Moreover, if X is T-orbitally complete, then T is a P.O.

Remark 4.

The results from ([9], Theorem 2.4) and ([10], Theorem 2.2) can be obtained from Corollary 4, without assuming that F satisfies (F2) and (F3). In addition, it is enough to suppose that F satisfies (F4) instead of F continuous. The above-mentioned results are also generalizations of other ones known in literature (some of these can be found in [7,9,10,11,16]).

Proposition 2.

Let T be a $\psi F$-contraction on the metric space X and assume that at least one of the following assertions holds:

(a)

F is bounded from above;

(b)

$\mathcal{O}\left(x\right)$ is bounded for some $x\in X$;

(c)

ψ is right continuous.

Then T is a weak $\psi F$-quasicontraction. If, further, the space X is T-orbitally complete, then T is a P.O.

Proof. 

Since (7) follows obviously from (4), we will prove the boundedness of the orbits. The proof is straightforward using some standard argumentations.

Under the hypothesis $\left(a\right)$, choose $x\in X$. If $\mathcal{O}\left(x\right)$ is finite, the assertion is obvious. Assume that $\mathcal{O}\left(x\right)$ is infinite and choose $n,p\in \mathbb{N}$ such that $T^{n}x\ne T^{n+p}x$. Then

$$\begin{array}{ccc}F\left(\mathrm{d}(T^{n}x,T^{n+p}x)\right)\hfill & \le \hfill & \psi F\left(\mathrm{d}(T^{n-1}x,T^{n+p-1}x)\right)\le \dots \hfill \\ \hfill & \le \hfill & {\psi }^{n}F\left(\mathrm{d}(x,T^{p}x)\right)\le {\psi }^n(supF)\underset{n}{⟶}-\infty .\hfill \end{array}$$

Thus, by Lemma 1, $\mathrm{d}(T^{n}x,T^{n+p}x)\underset{n,p}{⟶}0$, hence $\mathcal{O}\left(x\right)$ is bounded.

In case $\left(b\right)$, let $x\in X$ be such that $\mathcal{O}\left(x\right)$ is bounded and take any $y\in X$. If $T^{n}y=T^{n}x$ for all $n\in \mathbb{N}$, then the conclusion follows. Assume that there exists $n\in \mathbb{N}$ such that $\mathrm{d}(T^{n}y,T^{n}x)>0$. One has

$$F\left(\mathrm{d}(T^{n}y,T^{n}x)\right)\le \dots \le {\psi }^{n}F\left(\mathrm{d}(x,y)\right)\underset{n}{⟶}-\infty$$

hence, as before, the sequence $\left(\mathrm{d}(T^{n}y,T^{n}x)\right)$ converges to 0 so one can find $M>0$ such that $\mathrm{d}(T^{n}y,T^{n}x)\le M$ for all $n\in \mathbb{N}$. The conclusion follows from the inequalities

$$\mathrm{d}(T^{n}y,T^{m}y)\le \mathrm{d}(T^{n}y,T^{n}x)+\mathrm{d}(T^{n}x,x)+\mathrm{d}(x,T^{m}x)+\mathrm{d}(T^{m}x,T^{m}y)\le 2M+2\mathrm{diam}\mathcal{O}\left(x\right),$$

for every $m,n\in \mathbb{N}\cup \left\{0\right\}$.

The case $\left(c\right)$ is a consequence of Corollary 2.

The last part of the statement follows from Theorem 2. □

Remark 5.

Taking $\psi \left(t\right)=t-\tau$, $\tau >0$, in Proposition 2 we obtain immediately ([8], Theorem 3.9), ([7], Theorem 2.1) (without imposing the condition (F2′) for F).

Open problem: To find (sufficient) conditions on the functions F and $\psi$ that guarantee that any strong $\psi F$-quasicontraction is a weak $\psi F$-quasicontraction (see Theorem 3).