On the Study of Fixed Points for a New Class of α-Admissible Mappings

Abstract

In this paper, we discuss the existence of fixed points for new classes of mappings. Some examples are presented to illustrate our results.

1. Introduction

The Banach contraction principle is one of the most famous and important results in metric fixed point theory. It is a useful tool in establishing existence results in nonlinear analysis. This principle has been extended and generalized by several authors in many directions (see e.g., [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15], and the references therein).

In [16], the author introduced the class of F-contractions, and established a fixed point result for this class of mappings, which generalizes the Banach contraction principle. The main result in [16] can be stated as follows.

Theorem 1.

Let $(X,d)$ be a complete metric space, and let $T:X\to X$ be a mapping satisfying

$$\tau +F\left(d\right(Tx,Ty\left)\right)\le F\left(d\right(x,y\left)\right),$$

(1)

for all $(x,y)\in X\times X$ with $d(Tx,Ty)>0$, where $\tau >0$ is a constant and $F:(0,+\infty )\to \mathbb{R}$ is a function satisfying

(a)

F is nondecreasing.

(b)

For every sequence $\left\{t_n\right\}\subset (0,+\infty )$, we have

$$\underset{n\to +\infty }{lim}F\left(t_n\right)=-\infty ⟺\underset{n\to +\infty }{lim}{t}_n=0.$$

(c)

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

Then T has a unique fixed point. Moreover, for any $x\in X$, the Picard sequence $\left\{T^{n}x\right\}$ converges to this fixed point.

Observe that, if $T:X\to X$ is a q-contraction for some $0<q<1$, i.e.,

$$d(Tx,Ty)\le qd(x,y),(x,y)\in X\times X,$$

then T satisfies (1) with $F\left(t\right)=lnt$, $t>0$, and $\tau =-lnq$. Therefore, the Banach contraction principle follows from Theorem 1.

For different extensions and generalizations of Theorem 1, we refer the reader to [17,18,19,20,21,22,23,24,25,26,27], and the references therein.

In [5], Ćirić introduced a class of mappings with a non-unique fixed point and he established the following fixed point result.

Theorem 2.

Let $(X,d)$ be a complete metric space, and let $T:X\to X$ be a continuous mapping satisfying

$$min\left\{d\right(Tx,Ty),d(x,Tx),d(y,Ty\left)\right\}-min\left\{d\right(x,Ty),d(y,Tx\left)\right\}\le qd(x,y),$$

(2)

for all $(x,y)\in X\times X$, where $0<q<1$ is a constant. Then, for any $x\in X$, the Picard sequence $\left\{T^{n}x\right\}$ converges to a fixed point of T.

An example was presented in [5] to show that the set of fixed points of mappings satisfying the condition of Theorem 2 contains in general more than one element.

In this paper, we first introduce the class of generalized Ćirić-contractions by combining the ideas in [5,16]. Next, a fixed point result is established for this class of mappings. Our result generalizes Theorem 2 and extends Theorem 1. Next, we introduce a more general class of mappings using the concept of $\alpha$-admissibility introduced in [28] (see also [29]). Our fixed point result for this class of mappings has several consequences. It is not only a generalization of Theorems 1 and 2, but generalizes most fixed point theorems dealing with F-contractions, linear contractions, and many others. Several examples are presented to illustrate this fact.

Throughout this paper, we denote by $\mathbb{N}$ the set of natural numbers, that is, $\mathbb{N}=\{0,1,2,\cdots \}$. We denote by ${\mathbb{N}}^{*}$ the set $\mathbb{N}\setminus \left\{0\right\}$. Let $T:X\to X$ be a certain self-mapping on X. For $n\in \mathbb{N}$, we denote by $T^n$ the nth-iterate of T (we suppose that $T^0$ is the identity mapping on X).

2. The Class of Generalized Ćirić-Contractions

Let $\Psi$ be the set of functions $\psi :[0,+\infty )\to (-\infty ,0)$ such that $\psi$ is upper semi-continuous from the right. We denote by $\Phi$ the set of functions $\phi :(0,+\infty )\to \mathbb{R}$ such that

(${\Phi }_1$)

$\phi$ is non-decreasing, i.e., $0<t<s\Rightarrow \phi \left(t\right)\le \phi \left(s\right)$.

(${\Phi }_2$)

For every sequence $\left\{t_n\right\}\subset (0,+\infty )$,

$$\underset{n\to +\infty }{lim}\phi \left(t_n\right)=-\infty$$

if and only if

$$\underset{n\to +\infty }{lim}{t}_n=0.$$

(${\Phi }_3$)

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

Let $(X,d)$ be a metric space. For a given mapping $T:X\to X$, let

$$M_T(x,y)=min\{d(Tx,Ty),d(x,Tx),d(y,Ty)\}-min\{d(x,Ty),d(y,Tx)\},(x,y)\in X\times X.$$

Definition 1.

A mapping $T:X\to X$ is said to be a generalized Ćirić-contraction, if there exists $(\phi ,\psi )\in \Phi \times \Psi$ such that

$$\phi \left(M_T(x,y)\right)\le \phi \left(d(x,y)\right)+\psi \left(d(x,y)\right),$$

(3)

for all $(x,y)\in X\times X$ with $M_T(x,y)>0$.

We have the following fixed point result.

Theorem 3.

Let $(X,d)$ be a complete metric space, and let $T:X\to X$ be a continuous mapping. If T is a generalized Ćirić-contraction for some $(\phi ,\psi )\in \Phi \times \Psi$, then for any $x\in X$, the Picard sequence $\left\{T^{n}x\right\}$ converges to a fixed point of T.

Proof. 

Let $x\in X$ be fixed, and let $\left\{x_n\right\}\subset X$ be the sequence defined by

$$x_n=T^{n}x,n\in \mathbb{N}.$$

If $x_{p+1}=x_p$ for some $p\in \mathbb{N}$, then $x_p$ will be a fixed point of T. Therefore, we may assume that

$$d(x_n,x_{n+1})>0,n\in \mathbb{N}.$$

(4)

On the other hand, for every $n\in \mathbb{N}$, we have

$$\begin{array}{ccc}\hfill M_T(x_n,x_{n+1})& =& M_T(T^{n}x,T^{n+1}x)\hfill \\ & =& min\{d(T^{n+1}x,T^{n+2}x),d(T^{n}x,T^{n+1}x),d(T^{n+1}x,T^{n+2}x)\}\hfill \\ & & -min\{d(T^{n}x,T^{n+2}x),d(T^{n+1}x,T^{n+1}x)\}\hfill \\ & =& min\{d(x_{n+1},x_{n+2}),d(x_n,x_{n+1})\}.\hfill \end{array}$$

Therefore, from (4), we have

$$M_T(x_n,x_{n+1})>0,n\in \mathbb{N}.$$

From (3), we obtain

$$\phi \left(M_T(x_n,x_{n+1})\right)\le \phi \left(d(x_n,x_{n+1})\right)+\psi \left(d(x_n,x_{n+1})\right),n\in \mathbb{N}.$$

If for some $n\in \mathbb{N}$, we have $M_T(x_n,x_{n+1})=d(x_n,x_{n+1})$, then we obtain

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

that is,

$$0\le \psi \left(d(x_n,x_{n+1})\right),$$

which is a contradiction with the fact that $\psi \left(t\right)<0$, for all $t>0$. As a consequence, we have

$$M_T(x_n,x_{n+1})=d(x_{n+1},x_{n+2}),n\in \mathbb{N}.$$

Hence, we find

$$\phi \left(d(x_{n+1},x_{n+2})\right)\le \phi \left(d(x_n,x_{n+1})\right)+\psi \left(d(x_n,x_{n+1})\right),n\in \mathbb{N}.$$

(5)

Taking $n=0$ in (5), we obtain

$$\phi \left(d(x_1,x_2)\right)\le \phi \left(d(x_0,x_1)\right)+\psi \left(d(x_0,x_1)\right).$$

Taking $n=1$ in (5) and using the above inequality, we obtain

$$\begin{array}{ccc}\hfill \phi \left(d(x_2,x_3)\right)& \le & \phi \left(d(x_1,x_2)\right)+\psi \left(d(x_1,x_2)\right)\hfill \\ & \le & \phi \left(d(x_0,x_1)\right)+\psi \left(d(x_0,x_1)\right)+\psi \left(d(x_1,x_2)\right).\hfill \end{array}$$

Continuing this process, by induction we have

$$\phi \left(d(x_n,x_{n+1})\right)\le \phi \left(d(x_0,x_1)\right)+\sum _{i=0}^{n-1}\psi \left(d(x_i,x_{i+1})\right),n\in {\mathbb{N}}^{*}.$$

(6)

Next, let us denote by $\left\{u_n\right\}$ the real sequence defined by

$$u_n=d(x_n,x_{n+1}),n\in \mathbb{N}.$$

Observe that from (5), and using (${\Phi }_1$) and the fact that $\psi \left(t\right)<0$ for all $t>0$, we deduce that $\left\{u_n\right\}$ is a decreasing sequence. Therefore, there exists some $r\ge 0$ such that

$$u_n\downarrow r\mathrm{as}n\to +\infty .$$

Since $\psi$ is upper semi-continuous from the right, there exists some $N\in \mathbb{N}$ such that

$$\psi \left(u_p\right)<\psi \left(r\right)-\frac{\psi \left(r\right)}{2}=\frac{\psi \left(r\right)}{2},p\ge N.$$

(7)

Further, using (6) and the fact that $\psi \left(t\right)<0$ for all $t>0$, we obtain

$$\phi \left(u_n\right)\le \phi \left(u_0\right)+\sum _{i=N}^{n-1}\psi \left(u_i\right),n\ge N+1.$$

Therefore, from (7) we deduce that

$$\phi \left(u_n\right)\le \phi \left(u_0\right)+\frac{(n-N)}{2}\psi \left(r\right),n\ge N+1.$$

(8)

Let $n\to +\infty$ in (8) and we obtain

$$\underset{n\to +\infty }{lim}\phi \left(u_n\right)=-\infty ,$$

which implies from (${\Phi }_2$) that

$$\underset{n\to +\infty }{lim}{u}_n=0=r.$$

(9)

Next, we prove that $\left\{x_n\right\}$ is a Cauchy sequence. From (${\Phi }_3$) and (9), there exists some $k\in (0,1)$ such that

$$\underset{n\to +\infty }{lim}{u}_n^k\phi \left(u_n\right)=0.$$

(10)

Using (8), we obtain

$$u_n^k\phi \left(u_n\right)-u_n^k\phi \left(u_0\right)\le \frac{(n-N)}{2}\psi \left(r\right)u_n^k\le 0,n\ge N+1.$$

Let $n\to +\infty$, and using (9) and (10), we deduce that

$$\underset{n\to +\infty }{lim}n{u}_n^k=0.$$

Then there exists some $q\in \mathbb{N}$ such that

$$u_n<\frac{1}{n^{1/k}},n\ge q.$$

(11)

Using (11) and the triangle inequality, for $n\ge q$ and $m\in {\mathbb{N}}^{*}$, we have

$$d(x_n,x_{n+m})\le \sum _{i=n}^{n+m-1}{u}_i\le \sum _{i=n}^{+\infty }\frac{1}{i^{1/k}}.$$

The convergence of the Riemann series ${\sum }_n\frac{1}{n^{1/k}}$ (since $0<k<1$) yields $\left\{x_n\right\}$ is a Cauchy sequence. Since $(X,d)$ is complete, there exists some $\omega \in X$ such that

$$\underset{n\to +\infty }{lim}d(T^{n}x,\omega )=\underset{n\to +\infty }{lim}d(x_n,\omega )=0.$$

The continuity of T yields

$$\underset{n\to +\infty }{lim}d(T^{n+1}x,T\omega )=0.$$

Finally, the uniqueness of the limit implies that $\omega =T\omega$, i.e., $\omega$ is a fixed point of T. □

Let us give some examples to illustrate the result given by Theorem 3.

Example 1.

Let $(X,d)$ be a complete metric space, and let $T:X\to X$ be a continuous mapping. Let $F:(0,+\infty )\to \mathbb{R}$ be a function that belongs to Φ. Suppose that there exists a constant $\tau >0$ such that

$$\tau +F\left(M_T(x,y)\right)\le F\left(d(x,y)\right),$$

(12)

for all $(x,y)\in X\times X$ with $M_T(x,y)>0$. Then for any $x\in X$, the Picard sequence $\left\{T^{n}x\right\}$ converges to a fixed point of T. In order to prove this result, we apply Theorem 3 with $(\phi ,\psi )=(F,-\tau )$.

Example 2.

Suppose that all the assumptions of Theorem 2 are satisfied. Then T satisfies (3) with $\phi \left(t\right)=lnt$, $t>0$, and $\psi \equiv lnq$. Therefore, the result of Theorem 2 follows from Theorem 3.

Example 3.

Let

$$X=\left\{x_n=\frac{n(n+1)}{2}:n\in {\mathbb{N}}^{*}\right\}.$$

We endow X with the metric

$$d(x,y)=|x-y|,(x,y)\in X\times X.$$

Then $(X,d)$ is a complete metric space. Consider the mapping $T:X\to X$ defined by

$$T{x}_1=x_{1}andT{x}_{n+1}=x_n,n\in {\mathbb{N}}^{*}.$$

One observes easily that

$$\{(x,y)\in X\times X:M_T(x,y)>0\}=\{(x_n,x_{n+1}):n\in {\mathbb{N}}^{*}\}.$$

Furthermore, for all $n\in {\mathbb{N}}^{*}$, one has

$$\frac{M_T(x_n,x_{n+1})}{d(x_n,x_{n+1})}=\frac{n}{n+1}\to 1asn\to \infty ,$$

which shows that (2) is not satisfied. Hence Theorem 2 cannot be applied in this case. On the other hand, taking $\tau =1$ and

$$F\left(t\right)=t+lnt,t>0,$$

one obtains

$$\begin{array}{ccc}\hfill \tau +F\left(M_T(x_n,x_{n+1})\right)& =& 1+F\left(n\right)\hfill \\ & =& 1+n+lnn\hfill \\ & \le & 1+n+ln(n+1)\hfill \\ & =& F\left(d(x_n,x_{n+1})\right),\hfill \end{array}$$

for all $n\in {\mathbb{N}}^{*}$. Hence (12) is satisfied for all $(x,y)\in X\times X$ with $M_T(x,y)>0$. Therefore, by Example 1, one deduces that T has a fixed point $x^{*}\in X$. In this case, one observes that $x^{*}=x_1=1$.

3. A Larger Class of Mappings

In this part, we discuss the existence of fixed points for a larger class of mappings than the one studied in the previous section. First, let us recall some concepts introduced recently by Samet in [29] (see also [28]).

Let $(X,d)$ be a metric space, and let $\alpha :X\times X\to \mathbb{R}$ be a given function.

Definition 2.

Let $\left\{x_n\right\}\subset X$ be a given sequence. We say that $\left\{x_n\right\}$ is α-regular if

$$\alpha (x_n,x_{n+1})\ge 1,n\in \mathbb{N}.$$

Definition 3.

We say that $T:X\to X$ is α-admissible if

$$(x,y)\in X\times X,\alpha (x,y)\ge 1\Rightarrow \alpha (Tx,Ty)\ge 1.$$

Definition 4.

We say that $T:X\to X$ is α-continuous if for every α-regular sequence $\left\{x_n\right\}\subset X$ and $u\in X$,

$$\underset{n\to +\infty }{lim}d(x_n,u)=0$$

implies that there exists a sub-sequence $\left\{x_{n_k}\right\}$ of $\left\{x_n\right\}$ such that

$$\underset{k\to +\infty }{lim}d(T{x}_{n_k},Tu)=0.$$

Definition 5.

Let $\left\{x_n\right\}\subset X$ be a given sequence. We say that $\left\{x_n\right\}$ is α-Cauchy if

(i) 

$\left\{x_n\right\}$ is α-regular.

(ii) 

$\left\{x_n\right\}$ is a Cauchy sequence.

Definition 6.

We say that $(X,d)$ is α-complete if every α-Cauchy sequence is convergent.

Next, we introduce the following class of mappings.

Let ${\mathcal{T}}_{\alpha }$ be the class of mappings $T:X\to X$ satisfying the following conditions:

(${\mathcal{T}}_1$)

T is $\alpha$-continuous.

(${\mathcal{T}}_2$)

There exists $(\phi ,\psi )\in \Phi \times \Psi$ such that for all $(x,y)\in X\times X$ with $d(Tx,Ty)>0$,

$$\alpha (x,y)exp\left(\phi \left(d\right(Tx,Ty\left)\right)\right)\le exp\left(\phi \left(d\right(x,y\left)\right)+\psi \left(d\right(x,y\left)\right)\right).$$

We now give some examples of mappings $T:X\to X$ that belong to the set ${\mathcal{T}}_{\alpha }$, for some $\alpha :X\times X\to \mathbb{R}$. Let $(X,d)$ be a metric space.

Proposition 1 (The class of generalized Ćirić-contractions).

Let $T:X\to X$ be a continuous mapping. If T is a generalized Ćirić-contraction, then there exists a function $\alpha :X\times X\to \mathbb{R}$ such that $T\in {\mathcal{T}}_{\alpha }$.

Proof. 

Let us consider the function $\alpha :X\times X\to \mathbb{R}$ defined by

$$\begin{array}{c}\hfill \alpha (x,y)=\left\{\begin{array}{ccc}1\hfill & \mathrm{if}\hfill & y=Tx,\hfill \\ 0\hfill & \mathrm{if}\hfill & y\ne Tx.\hfill \end{array}\right.\end{array}$$

(13)

Let $(x,y)\in X\times X$ be such that $d(Tx,Ty)>0$. We discuss two possible cases.

Case 1: $y\ne Tx$. In this case,

$$\alpha (x,y)exp\left(\phi \left(d\right(Tx,Ty\left)\right)\right)=0\le exp\left(\phi \left(d\right(x,y\left)\right)+\psi \left(d\right(x,y\left)\right)\right).$$

Case 2: $y=Tx$. In this case, we have

$$\begin{array}{ccc}\hfill M_T(x,y)& =& M_T(x,Tx)\hfill \\ & =& min\{d(Tx,T^{2}x),d(x,Tx)\}.\hfill \end{array}$$

Since $d(Tx,T^{2}x)=d(Tx,Ty)>0$, we have $d(x,Tx)>0$. Therefore, $M_T(x,y)>0$. Using the fact that T is a generalized Ćirić-contraction, we deduce that

$$\phi \left(M_T(x,Tx)\right)\le \phi \left(d(x,Tx)\right)+\psi \left(d(x,Tx)\right),$$

that is,

$$\phi (min\{d(Tx,T^{2}x),d(x,Tx)\})\le \phi \left(d(x,Tx)\right)+\psi \left(d(x,Tx)\right),$$

which yields (since $\psi \left(t\right)<0$, for all $t>0$)

$$\phi \left(d(Tx,T^{2}x)\right)\le \phi \left(d(x,Tx)\right)+\psi \left(d(x,Tx)\right).$$

Hence, we obtain

$$\alpha (x,Tx)exp\left(\phi \left(d(Tx,T^{2}x)\right)\right)\le exp\left(\phi \left(d\right(x,Tx\left)\right)+\psi \left(d\right(x,Tx\left)\right)\right).$$

Therefore, T satisfies (${\mathcal{T}}_2$) with $\alpha$ given by (13). Obviously, since T is continuous, then T is $\alpha$-continuous. Then T satisfies (${\mathcal{T}}_1$). As a consequence, we have $T\in {\mathcal{T}}_{\alpha }$. □

Proposition 2 (The class of F-contractions).

Let $T:X\to X$ be an F-contraction, for some $F\in \Phi$, that is, there exists a constant $\tau >0$ such that

$$\tau +F\left(d\right(Tx,Ty\left)\right)\le F\left(d\right(x,y\left)\right),$$

for all $(x,y)\in X\times X$ with $d(Tx,Ty)>0$. Then there exists a function $\alpha :X\times X\to \mathbb{R}$ such that $T\in {\mathcal{T}}_{\alpha }$.

Proof. 

Let

$$\alpha (x,y)=1,(x,y)\in X\times X.$$

(14)

Let $\phi =F$ and $\psi \equiv -\tau$. Then $(\phi ,\psi )\in \Phi \times \Psi$. Let $(x,y)\in X\times X$ be such that $d(Tx,Ty)>0$. Then

$$\phi \left(d\right(Tx,Ty\left)\right)\le \phi \left(d\right(x,y\left)\right)+\psi \left(d\right(x,y\left)\right),$$

which yields

$$\alpha (x,y)exp\left(\phi \left(d\right(Tx,Ty\left)\right)\right)\le exp\left(\phi \left(d\right(x,y\left)\right)+\psi \left(d\right(x,y\left)\right)\right).$$

Then T satisfies ${\mathcal{T}}_2$ with $\alpha$ given by (14). On the other hand, it can be easily seen that any F-contraction is continuous, so it is $\alpha$-continuous. Then T satisfies also ${\mathcal{T}}_1$. As a consequence, we have $T\in {\mathcal{T}}_{\alpha }$. □

Proposition 3.

Let $T:X\to X$ be an orbitally continuous mapping, that is, for every $x\in X$, if

$$\underset{n\to +\infty }{lim}d(T^{n}x,u)=0,u\in X,$$

then

$$\underset{n\to +\infty }{lim}d(T{T}^{n}x,Tu)=0.$$

Suppose that there exist $F\in \Phi$ and a constant $\tau >0$ such that

$$\tau +F\left(d(Tx,Ty)\right)\le F\left(N_T(x,y)\right),$$

(15)

for all $(x,y)\in X\times X$ with $d(Tx,Ty)>0$, where

$$N_T(x,y)=max\left\{d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2}\right\}.$$

Then there exists a function $\alpha :X\times X\to \mathbb{R}$ such that $T\in {\mathcal{T}}_{\alpha }$.

Proof. 

Let $\alpha :X\times X\to \mathbb{R}$ be the function defined by (13). Let $\phi =F$ and $\psi \equiv -\tau$. Then $(\phi ,\psi )\in \Phi \times \Psi$. Let $(x,y)\in X\times X$ be such that $d(Tx,Ty)>0$. We discuss two possible cases.

Case 1. $y\ne Tx$. In this case,

$$\alpha (x,y)exp\left(\phi \left(d\right(Tx,Ty\left)\right)\right)=0\le exp\left(\phi \left(d\right(x,y\left)\right)+\psi \left(d\right(x,y\left)\right)\right).$$

Case 2. $y=Tx$. In this case,

$$N_T(x,y)=max\left\{d(x,Tx),d(Tx,T^{2}x),\frac{d(x,T^{2}x)}{2}\right\}.$$

On the other hand, by the triangle inequality, we have

$$\frac{d(x,T^{2}x)}{2}\le \frac{d(x,Tx)+d(Tx,T^{2}x)}{2}\le max\{d(x,Tx),d(Tx,T^{2}x)\}.$$

Therefore,

$$N_T(x,y)=max\left\{d(x,Tx),d(Tx,T^{2}x)\right\}.$$

Suppose that $N_T(x,y)=d(Tx,T^{2}x)$. Then by (15), we have

$$\tau +\phi \left(d(Tx,T^{2}x)\right)\le \phi \left(d(Tx,T^{2}x)\right),$$

which yields $\tau \le 0$, which is a contradiction. Then we have $N_T(x,y)=d(x,Tx)$. Again, by (15), we deduce that

$$\phi \left(d(Tx,T^{2}x)\right)\le \phi \left(d(x,Tx)\right)+\psi \left(d(x,Tx)\right),$$

which yields

$$\alpha (x,Tx)exp\left(\phi \left(d(Tx,T^{2}x)\right)\right)\le exp\left(\phi \left(d\right(x,Tx\left)\right)+\psi \left(d\right(x,Tx\left)\right)\right).$$

Then T satisfies ${\mathcal{T}}_2$ with $\alpha$ given by (13). Next, we prove that T is $\alpha$-continuous. Let $\left\{x_n\right\}\subset X$ be an $\alpha$-regular sequence. By the definition of $\alpha$, this means that

$$x_{n+1}=T{x}_n,n\in \mathbb{N},$$

that is,

$$x_n=T^n{x}_0,n\in \mathbb{N}.$$

Suppose that there exists $u\in X$ such that

$$\underset{n\to +\infty }{lim}d(x_n,u)=\underset{n\to +\infty }{lim}d(T^n{x}_0,u)=0.$$

Since T is orbitally continuous, we obtain

$$\underset{n\to +\infty }{lim}d(T{x}_n,Tu)=0.$$

Then T is $\alpha$-continuous, and it satisfies (${\mathcal{T}}_1$). As a consequence, we have $T\in {\mathcal{T}}_{\alpha }$. □

Remark 1.

Let $T:X\to X$ be a given mapping. Suppose that there exists a constant $0<q<1$ such that

$$d(Tx,Ty)\le q{N}_T(x,y),(x,y)\in X\times X.$$

It can be easily seen that T is orbitally continuous mapping, and it satisfies (15) with $\tau =-lnq$ and $F\left(t\right)=lnt$, $t>0$. Therefore, $T\in {\mathcal{T}}_{\alpha }$, where α is given by (13) and $(\varphi ,\psi )=(F,-lnq)$.

Proposition 4.

Let $T:X\to X$ be an orbitally continuous mapping. Suppose that there exist $F\in \Phi$ and a constant $\tau >0$ such that

$$\tau +F\left(d(Tx,Ty)\right)\le F\left({\mu }_T(x,y)\right),$$

(16)

for all $(x,y)\in X\times X$ with $d(Tx,Ty)>0$, where

$${\mu }_T(x,y)=max\left\{d(x,y),d(y,Ty)\frac{1+d(x,Tx)}{1+d(x,y)}\right\}.$$

Then there exists a function $\alpha :X\times X\to \mathbb{R}$ such that $T\in {\mathcal{T}}_{\alpha }$.

Proof. 

Let $\alpha :X\times X\to \mathbb{R}$ be the function defined by (13). Let $\phi =F$ and $\psi \equiv -\tau$. Then $(\phi ,\psi )\in \Phi \times \Psi$. Let $(x,y)\in X\times X$ be such that $d(Tx,Ty)>0$. We discuss two possible cases.

Case 1. $y\ne Tx$. In this case, we have

$$\alpha (x,y)exp\left(\phi \left(d\right(Tx,Ty\left)\right)\right)=0\le exp\left(\phi \left(d\right(x,y\left)\right)+\psi \left(d\right(x,y\left)\right)\right).$$

Case 2. $y=Tx$. In this case,

$${\mu }_T(x,y)=max\left\{d(x,Tx),d(Tx,T^{2}x)\right\}.$$

If ${\mu }_T(x,y)=d(Tx,T^{2}x)$, then by (16), we have

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

that is

$$\tau \le 0,$$

which is a contradiction. Therefore, ${\mu }_T(x,y)=d(x,Tx)$. Again, by (16), we deduce that

$$\phi \left(d(Tx,T^{2}x)\right)\le \phi \left(d(x,Tx)\right)+\psi \left(d(x,Tx)\right),$$

which yields

$$\alpha (x,Tx)exp\left(\phi \left(d(Tx,T^{2}x)\right)\right)\le exp\left(\phi \left(d\right(x,Tx\left)\right)+\psi \left(d\right(x,Tx\left)\right)\right).$$

Then T satisfies ${\mathcal{T}}_2$ with $\alpha$ given by (13). Since T is orbitally continuous, from the proof of Proposition 3, T is $\alpha$-continuous, and it satisfies ${\mathcal{T}}_1$. As a consequence, we have $T\in {\mathcal{T}}_{\alpha }$. □

Remark 2.

Let $T:X\to X$ be a given mapping. Suppose that there exists a constant $0<q<1$ such that

$$d(Tx,Ty)\le q{\mu }_T(x,y),(x,y)\in X\times X.$$

It can be easily seen that T is orbitally continuous mapping, and it satisfies (16) with $\tau =-lnq$ and $F\left(t\right)=lnt$, $t>0$. Therefore, $T\in {\mathcal{T}}_{\alpha }$, where α is given by (13) and $(\varphi ,\psi )=(F,-lnq)$.

Proposition 5 (The class of almost F-contractions).

Let $T:X\to X$ be an almost F-contraction (see [22]), that is, there exist $F\in \Phi$, $\tau >0$ and $L\ge 0$ such that

$$\tau +F\left(d\right(Tx,Ty\left)\right)\le F\left(d\right(x,y)+Ld(y,Tx\left)\right),$$

(17)

for all $(x,y)\in X\times X$ with $d(Tx,Ty)>0$. Then there exists a function $\alpha :X\times X\to \mathbb{R}$ such that $T\in {\mathcal{T}}_{\alpha }$.

Proof. 

Let $\alpha :X\times X\to \mathbb{R}$ be the function defined by (13). Let $\phi =F$ and $\psi \equiv -\tau$. Then $(\phi ,\psi )\in \Phi \times \Psi$. Let $(x,y)\in X\times X$ be such that $d(Tx,Ty)>0$. We discuss two possible cases.

Case 1. $y\ne Tx$. In this case, we have

$$\alpha (x,y)exp\left(\phi \left(d\right(Tx,Ty\left)\right)\right)=0\le exp\left(\phi \left(d\right(x,y\left)\right)+\psi \left(d\right(x,y\left)\right)\right).$$

Case 2. $y=Tx$. In this case, from (17), we have

$$\phi \left(d(Tx,T^{2}x)\right)\le \phi \left(d(x,y)\right)+\psi \left(d(x,y)\right),$$

which yields

$$\alpha (x,Tx)exp\left(\phi \left(d(Tx,T^{2}x)\right)\right)\le exp\left(\phi \left(d\right(x,Tx\left)\right)+\psi \left(d\right(x,Tx\left)\right)\right).$$

Then T satisfies ${\mathcal{T}}_2$ with $\alpha$ given by (13). Next, we shall prove that T is $\alpha$-continuous. Let $\left\{x_n\right\}\subset X$ be an $\alpha$-regular sequence, i.e.,

$$x_{n+1}=T{x}_n,n\in \mathbb{N}.$$

Suppose that there exists $u\in X$ such that

$$\underset{n\to +\infty }{lim}d(x_n,u)=0.$$

Let us define the set

$$\mathbb{I}=\{n\in \mathbb{N}:d(x_n,Tu)=0\}.$$

If $\left|\mathbb{I}\right|<+\infty$, then there exists some $N\in \mathbb{N}$ such that

$$d(x_{n+1},Tu)>0,n\ge N.$$

From (17) and (${\Phi }_1$), we have

$$d(x_{n+1},Tu)\le d(x_n,u)+Ld(u,x_{n+1}),n\ge N.$$

Let $n\to +\infty$ and we obtain

$$\underset{n\to +\infty }{lim}d(x_{n+1},Tu)=0.$$

If $\left|\mathbb{I}\right|=+\infty$, then there exists a sub-sequence $\left\{x_{n_k}\right\}$ of $\left\{x_n\right\}$ such that

$$d(x_{n_k},Tu)=0,k\in \mathbb{N}.$$

Therefore, we have

$$\underset{k\to +\infty }{lim}d(T{x}_{n_k},Tu)=\underset{k\to +\infty }{lim}d(x_{n_k+1},Tu)=0.$$

Then T is $\alpha$-continuous, and it satisfies ${\mathcal{T}}_1$. As a consequence, we have $T\in {\mathcal{T}}_{\alpha }$. □

Remark 3.

Let $T:X\to X$ be a mapping that belongs to the class of Berinde mappings (see [2]), that is, there exist $0<q<1$ and $\ell \ge 0$ such that

$$d(Tx,Ty)\le qd(x,y)+\ell d(y,Tx),(x,y)\in X\times X.$$

It can be easily seen that T is an almost F-contraction with $F\left(t\right)=lnt$, $t>0$, and $(\tau ,L)=(-lnq,\ell /q)$. Therefore, $T\in {\mathcal{T}}_{\alpha }$, where α is given by (13) and $(\varphi ,\psi )=(F,-lnq)$.

Now, we state and prove the main result of this section.

Theorem 4.

Let $(X,d)$ be a metric space, and let $T:X\to X$ be a given mapping. Suppose that

(i) 

There exists $\alpha :X\times X\to \mathbb{R}$ such that $(X,d)$ is α-complete.

(ii) 

There exists $(\phi ,\psi )\in \Phi \times \Psi$ such that $T\in {\mathcal{T}}_{\alpha }$.

(iii) 

T is α-admissible.

(iv) 

There exists some $x_0\in X$ such that $\alpha (x_0,T{x}_0)\ge 1$.

Then there exists a sub-sequence $\left\{T^{n_k}{x}_0\right\}$ of $\left\{T^n{x}_0\right\}$ that converges to a fixed point of T.

Proof. 

Let $\left\{x_n\right\}$ be the Picard sequence defined by

$$x_n=T^n{x}_0,n\in \mathbb{N}.$$

Without loss of generality, we may suppose that

$$d(x_n,x_{n+1})>0,n\in \mathbb{N}.$$

From (${\mathcal{T}}_2$), we have

$$\alpha (x_{n-1},x_n)exp\left(\phi \left(d(x_n,x_{n+1})\right)\right)\le exp\left(\phi \left(d(x_{n-1},x_n)\right)+\psi \left(d(x_{n-1},x_n)\right)\right),n\in {\mathbb{N}}^{*}.$$

On the other hand, from $\left(iii\right)$ and $\left(iv\right)$, we have

$$\alpha (x_{n-1},x_n)\ge 1,n\in {\mathbb{N}}^{*}.$$

(18)

Therefore, we obtain

$$exp\left(\phi \left(d(x_n,x_{n+1})\right)\right)\le exp\left(\phi \left(d(x_{n-1},x_n)\right)+\psi \left(d(x_{n-1},x_n)\right)\right),n\in {\mathbb{N}}^{*},$$

which yields

$$\phi \left(d(x_n,x_{n+1})\right)\le \phi \left(d(x_{n-1},x_n)\right)+\psi \left(d(x_{n-1},x_n)\right),n\in {\mathbb{N}}^{*}.$$

Next, following the same argument as in the proof of Theorem 3, we can prove that $\left\{x_n\right\}$ is a Cauchy sequence. Moreover, from (18), $\left\{x_n\right\}$ is $\alpha$-Cauchy. Since $(X,d)$ is $\alpha$-complete, there exists some $\omega \in X$ such that

$$\underset{n\to +\infty }{lim}d(x_n,\omega )=0.$$

From (${\mathcal{T}}_1$), there exists a sub-sequence $\left\{x_{n_k}\right\}$ of $\left\{x_n\right\}$ such that

$$\underset{k\to +\infty }{lim}d(x_{n_k+1},T\omega )=0.$$

The uniqueness of the limit yields $T\omega =\omega$, i.e., $\omega$ is a fixed point of T. □

Remark 4.

From the proof of Theorem 4, it can be easily seen that if we replace (${\mathcal{T}}_1$) by the continuity of T, then the Picard sequence $\left\{T^n{x}_0\right\}$ converges to a fixed point of T.

Next, we will show that most fixed point results from the literature involving F-contraction mappings follow easily from Theorem 4.

The following lemma will be used later.

Lemma 1.

Let $T:X\to X$ be a given mapping. Let $\alpha :X\times X\to \mathbb{R}$ be the function defined by (13). Then T is α-admissible.

Proof. 

Let $(x,y)\in X\times X$ be such that $\alpha (x,y)\ge 1$. By the definition of $\alpha$, this means that $y=Tx$. Then $Ty=T^{2}x$, which yields $\alpha (Tx,Ty)=1$. This proves that T is $\alpha$-admissible. □

Corollary 1.

Theorem 4 ⇒ Theorem 3.

Proof. 

Suppose that all the assumptions of Theorem 3 are satisfied. By Proposition 1, we know that $T\in {\mathcal{T}}_{\alpha }$, where $\alpha :X\times X\to \mathbb{R}$ is given by (13). Since $(X,d)$ is complete, then it is $\alpha$-complete. From Lemma 1, T is $\alpha$-admissible. From the definition of $\alpha$, we have $\alpha (x,Tx)=1$, for all $x\in X$. Therefore, all the assumptions of Theorem 4 are satisfied. In particular $\left(iv\right)$ is satisfied for every $x\in X$. Taking in consideration Remark 4, we obtain that for any $x\in X$, the Picard sequence $\left\{T^{n}x\right\}$ converges to a fixed point of T. □

Corollary 2.

Theorem 4 ⇒ Theorem 1.

Proof. 

It follows from Proposition 2, Lemma 1 and Remark 4. □

Corollary 3.

Let $(X,d)$ be a complete metric space, and let $T:X\to X$ be an orbitally continuous mapping. Suppose that there exist $F\in \Phi$ and a constant $\tau >0$ such that (15) is satisfied. Then, for any $x\in X$, there exists a sub-sequence $\left\{T^{n_k}x\right\}$ of $\left\{T^{n}x\right\}$ such that $\left\{T^{n}x\right\}$ converges to a fixed point of T.

Proof. 

It follows from Proposition 3, Lemma 1, and Theorem 4. □

Remark 5.

By Remark 4, if we replace the assumption T is orbitally continuous with T is continuous, then for any $x\in X$, the Picard sequence $\left\{T^{n}x\right\}$ converges to a fixed point of T. Such a result was established by Wardowski and Van Dung in [27].

Corollary 4.

Let $(X,d)$ be a complete metric space, and let $T:X\to X$ be an orbitally continuous mapping. Suppose that there exist $F\in \Phi$ and a constant $\tau >0$ such that (16) is satisfied. Then, for any $x\in X$, there exists a sub-sequence $\left\{T^{n_k}x\right\}$ of $\left\{T^{n}x\right\}$ such that $\left\{T^{n}x\right\}$ converges to a fixed point of T.

Proof. 

It follows from Proposition 4, Lemma 1, and Theorem 4. □

The next result was established by Minak et al. [22].

Corollary 5.

Let $(X,d)$ be a complete metric space, and let $T:X\to X$ be an almost F-contraction, that is, there exit $F\in \Phi$, $\tau >0$ and $L\ge 0$ such that (17) is satisfied. Then, for any $x\in X$, there exists a sub-sequence $\left\{T^{n_k}x\right\}$ of $\left\{T^{n}x\right\}$ such that $\left\{T^{n}x\right\}$ converges to a fixed point of T.

Proof. 

It follows from Proposition 5, Lemma 1, and Theorem 4. □

Next, we will show that we can deduce easily from Theorem 4 several fixed point results in partially ordered metric spaces.

Corollary 6.

Let $(X,d)$ be a complete metric space, and let $T:X\to X$ be continuous mapping. Suppose that X is partially ordered by a certain binary relation ⪯. Suppose that

(i) 

T is non-decreasing with respect to ⪯, i.e.,

$$Tx⪯Ty,$$

for all $(x,y)\in X\times X$ with $x⪯y$.

(ii) 

There exists $x_0\in X$ such that $x_0⪯T{x}_0$.

(iii) 

There exist $F\in \Phi$ and $\tau >0$ such that

$$\tau +F\left(d\right(Tx,Ty\left)\right)\le F\left(d\right(x,y\left)\right),$$

for all $(x,y)\in X\times X$ with $x⪯y$ and $d(Tx,Ty)>0$.

Then $\left\{T^n{x}_0\right\}$ converges to a fixed point of T.

Proof. 

Let $\alpha :X\times X\to \mathbb{R}$ be the function defined by

$$\begin{array}{c}\hfill \alpha (x,y)=\left\{\begin{array}{ccc}1\hfill & \mathrm{if}\hfill & x⪯y,\hfill \\ 0\hfill & \mathrm{if}\hfill & x⋠y.\hfill \end{array}\right.\end{array}$$

(19)

From $\left(i\right)$ and the definition of $\alpha$, it can be easily seen that T is $\alpha$-admissible. Since T is continuous, it is $\alpha$-continuous. Since $(X,d)$ is complete, it is $\alpha$-complete. On the other hand, from $\left(iii\right)$, we have

$$exp\left(F\left(d\right(Tx,Ty\left)\right)\right)\le exp\left(F\left(d\right(x,y\left)\right)-\tau \right),$$

for all $(x,y)\in X\times X$ with $x⪯y$ and $d(Tx,Ty)>0$. Let $(\phi ,\psi )=(F,-\tau )$. Then $(\phi ,\psi )\in \Phi \times \Psi$. Further, by the definition of $\alpha$, for all $(x,y)\in X\times X$ with $\left(d\right(Tx,Ty)>0$, we have

$$\alpha (x,y)exp\left(\phi \left(d\right(Tx,Ty\left)\right)\right)\le exp\left(\phi \left(d\right(x,y\left)\right)+\psi \left(d\right(x,y\left)\right)\right).$$

Therefore, $T\in {\mathcal{T}}_{\alpha }$, where $\alpha$ is given by (19). Note that by (ii), we have $\alpha (x_0,T{x}_0)=1$. Applying Theorem 4 and taking in consideration Remark 4, we obtain the desired result. □

Corollary 7

(Ran–Reurings fixed point theorem [13]). Let $(X,d)$ be a complete metric space, and let $T:X\to X$ be continuous mapping. Suppose that X is partially ordered by a certain binary relation ⪯. Suppose that

(i) 

T is non-decreasing with respect to ⪯.

(ii) 

There exists $x_0\in X$ such that $x_0⪯T{x}_0$.

(iii) 

There exists $0<q<1$ such that for all $(x,y)\in X\times X$ with $x⪯y$,

$$d(Tx,Ty)\le qd(x,y).$$

Then $\left\{T^n{x}_0\right\}$ converges to a fixed point of T.

Proof. 

We have observe that T satisfies the condition $\left(iii\right)$ of Corollary 6 with $F\left(t\right)=lnt$, $t>0$, and $\tau =-lnq$. Therefore, the result follows immediately from Corollary 6. □

Remark 6.

Note that several other fixed point results can be deduced from Theorem 4. For example, we mention the Banach fixed point theorem, the Berinde fixed point theorem [2], the Dass–Gupta fixed point theorem [7], the Chatterjea fixed point theorem [4], the Kannan fixed point theorem [11], the Reich fixed point theorem [14], the Hardy–Rogers fixed point theorem [8], etc.