Wardowski type fixed point theorems in complete metric spaces

Abstract

In this paper, we state and prove Wardowski type fixed point theorems in metric space by using a modified generalized F-contraction maps. These theorems extend other well-known fundamental metrical fixed point theorems in the literature (Dung and Hang in Vietnam J. Math. 43:743-753, 2015 and Piri and Kumam in Fixed Point Theory Appl. 2014:210, 2014, etc.). Examples are provided to support the usability of our results.

1 Introduction and preliminaries

One of the most well-known results in generalizations of the Banach contraction principle is the Wardowski fixed point theorem [3]. Before providing the Wardowski fixed point theorem, we recall that a self-map T on a metric space $(X,d)$ is said to be an F-contraction if there exist $F\in\mathcal{F}$ and $\tau\in (0, \infty)$ such that

$$ \forall x,y\in X, \quad \bigl[d(Tx, Ty)>0 \Rightarrow \tau+F \bigl(d(Tx,Ty)\bigr)\leq F\bigl(d(x,y)\bigr)\bigr], $$

(1)

where $\mathcal{F}$ is the family of all functions $F :(0,\infty) \to\mathbb{R}$ such that

• (F1) F is strictly increasing, i.e. for all $x,y\in\mathbb {R}_{+}$ such that $x < y$, $F (x) < F (y)$;

• (F2) for each sequence $\{\alpha_{n}\}_{n=1}^{\infty}$ of positive numbers, $\lim_{n\to\infty}\alpha_{n}=0$ if and only if $\lim_{n\to\infty}F(\alpha_{n})=-\infty$;

• (F3) there exists $k\in(0,1)$ such that $\lim_{\alpha\to 0^{+}}\alpha^{k}F(\alpha)=0$.

Obviously every F-contraction is necessarily continuous. The Wardowski fixed point theorem is given by the following theorem.

Theorem 1.1

[3]

Let $(X,d)$ be a complete metric space and let $T:X\rightarrow X$ be an F-contraction. Then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n\in\mathbb{N}}$ converges to $x^{*}$.

Later, Wardowski and Van Dung [4] have introduced the notion of an F-weak contraction and prove a fixed point theorem for F-weak contractions, which generalizes some results known from the literature. They introduced the concept of an F-weak contraction as follows.

Definition 1.2

Let $(X,d)$ be a metric space. A mapping $T:X\rightarrow X$ is said to be an F-weak contraction on $(X,d)$ if there exist $F\in\mathcal{F}$ and $\tau>0$ such that, for all $x,y\in X$,

$$d(Tx, Ty)> 0\quad \Rightarrow\quad \tau+F\bigl(d(Tx,Ty)\bigr)\leq F\bigl(M(x,y) \bigr), $$

where

$$ M(x,y)=\max\biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2}\biggr\} . $$

(2)

By using the notion of F-weak contraction, Wardowski and Van Dung [4] have proved a fixed point theorem which generalizes the result of Wardowski as follows.

Theorem 1.3

[4]

Let $(X,d)$ be a complete metric space and let $T:X\rightarrow X$ be an F-weak contraction. If T or F is continuous, then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n\in\mathbb{N}}$ converges to $x^{*}$.

Recently, by adding values $d(T^{2}x, x)$, $d(T^{2}x, T x)$, $d(T^{2}x, y)$, $d(T^{2}x,T y)$ to (2), Dung and Hang [1] introduced the notion of a modified generalized F-contraction and proved a fixed point theorem for such maps. They generalized an F-weak contraction to a generalized F-contraction as follows.

Definition 1.4

Let $(X,d)$ be a metric space. A mapping $T:X\rightarrow X$ is said to be a generalized F-contraction on $(X,d)$ if there exist $F\in\mathcal{F}$ and $\tau>0$ such that

$$ \forall x,y\in X, \quad \bigl[d(Tx, Ty)>0 \Rightarrow \tau+F\bigl(d(Tx,Ty)\bigr) \leq F\bigl(N(x,y)\bigr)\bigr], $$

where

$$\begin{aligned} N(x,y) =&\max\biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2}, \\ &\frac{d(T^{2}x,x)+d(T^{2}x,Ty)}{2},d\bigl(T^{2}x,Tx\bigr), d\bigl(T^{2}x,y \bigr),d\bigl(T^{2}x,Ty\bigr) \biggr\} . \end{aligned}$$

By using the notion of a generalized F-contraction, Dung and Hang have proved the following fixed point theorem, which generalizes the result of Wardowski and Van Dung [4].

Theorem 1.5

[1]

Let $(X,d)$ be a complete metric space and let $T:X\rightarrow X$ be a generalized F-contraction. If T or F is continuous, then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n\in\mathbb{N}}$ converges to $x^{*}$.

Very recently, Piri and Kumam [2] described a large class of functions by replacing the condition (F3) in the definition of F-contraction introduced by Wardowski with the following one:

(F3′):

F is continuous on $(0,\infty)$.

They denote by $\mathfrak{F}$ the family of all functions $F:\mathbb{R}_{+}\rightarrow \mathbb{R}$ which satisfy conditions (F1), (F2), and (F3′). Under this new set-up, Piri and Kumam proved some Wardowski and Suzuki type fixed point results in metric spaces as follows.

Theorem 1.6

[2]

Let T be a self-mapping of a complete metric space X into itself. Suppose there exist $F\in\mathfrak{F}$ and $\tau>0$ such that

$$\forall x,y\in X, \quad \bigl[d(Tx, Ty)>0 \Rightarrow \tau+F\bigl(d(Tx,Ty)\bigr) \leq F\bigl(d(x,y)\bigr)\bigr]. $$

Then T has a unique fixed point $x^{*}\in X$ and for every $x_{0}\in X$ the sequence $\{T^{n}x_{0}\}_{n=1}^{\infty}$ converges to $x^{*}$.

Theorem 1.7

[2]

Let T be a self-mapping of a complete metric space X into itself. Suppose there exist $F\in\mathfrak{F}$ and $\tau>0$ such that

$$\forall x,y\in X, \quad \biggl[\frac{1}{2}d(x, Tx)< d(x,y) \Rightarrow \tau +F\bigl(d(Tx,Ty)\bigr)\leq F\bigl(d(x,y)\bigr)\biggr]. $$

Then T has a unique fixed point $x^{*}\in X$ and for every $x_{0}\in X$ the sequence $\{T^{n}x_{0}\}_{n=1}^{\infty}$ converges to $x^{*}$.

The aim of this paper is to introduce the modified generalized F-contractions, by combining the ideas of Dung and Hang [1], Piri and Kumam [2], Wardowski [3] and Wardowski and Van Dung [4] and give some fixed point result for these type mappings on complete metric space.

2 Main results

Let $\mathfrak{F}_{G}$ denote the family of all functions $F:\mathbb{R}_{+}\rightarrow\mathbb{R}$ which satisfy conditions (F1) and (F3′) and $\mathcal{F}_{G}$ denote the family of all functions $F:\mathbb{R}_{+}\rightarrow\mathbb{R}$ which satisfy conditions (F1) and (F3).

Definition 2.1

Let $(X,d)$ be a metric space and $T:X\rightarrow X$ be a mapping. T is said to be modified generalized F-contraction of type (A) if there exist $F\in\mathfrak{F}_{G}$ and $\tau>0$ such that

$$ \forall x, y \in X,\quad \bigl[d(Tx, Ty)> 0 \Rightarrow \tau+F \bigl(d(Tx,Ty)\bigr)\leq F\bigl(M_{T}(x,y)\bigr)\bigr], $$

(3)

where

$$\begin{aligned} M_{T}(x,y) =& \max\biggl\{ d(x,y),\frac{d(x,Ty)+d(y,Tx)}{2}, \frac {d(T^{2}x,x)+d(T^{2}x,Ty)}{2},d\bigl(T^{2}x,Tx\bigr), \\ &d\bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr)+d(x,Tx), d(Tx,y)+d(y,Ty) \biggr\} . \end{aligned}$$

Remark 2.2

Note that $\mathfrak{F}\subseteq\mathfrak{F}_{W}$. Since, for $\beta\in(0,\infty)$, the function $F(\alpha)=\frac{-1}{\alpha+\beta}$ satisfies the conditions (F1) and (F3′) but it does not satisfy (F2), we have $\mathfrak{F}\subsetneq\mathfrak{F}_{W}$.

Definition 2.3

Let $(X,d)$ be a metric space and $T:X\rightarrow X$ be a mapping. T is said to be modified generalized F-contraction of type (B) if there exist $F\in\mathcal{F}_{G}$ and $\tau>0$ such that

$$ \forall x, y \in X,\quad \bigl[d(Tx, Ty)> 0\Rightarrow\tau+F\bigl(d(Tx,Ty)\bigr) \leq F\bigl(M_{T}(x,y)\bigr)\bigr]. $$

Remark 2.4

Note that $\mathcal{F}\subseteq\mathcal{F}_{W}$. Since, for $\beta\in (0,\infty)$, the function $F(\alpha)=\ln(\alpha+\beta)$ satisfies the conditions (F1) and (F3) but it does not satisfy (F2), we have $\mathcal{F}\subsetneq\mathcal{F}_{W}$.

Remark 2.5

• (1) Every F-contraction is a modified generalized F-contraction.

• (2) Let T be a modified generalized F-contraction. From (3) for all $x,y\in X$ with $Tx\neq Ty$, we have

  $$\begin{aligned} F\bigl(d(Tx,Ty)\bigr) < &\tau+F\bigl(d(Tx,Ty)\bigr) \\ \leq& F\biggl(\max \biggl\{ d(x,y),\frac{d(x,Ty)+d(y,Tx)}{2},\frac {d(T^{2}x,x)+d(T^{2}x,Ty)}{2}, \\ &d \bigl(T^{2}x,Tx\bigr),d\bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr)+d(x,Tx), \\ & d(Tx,y)+d(y,Ty)\biggr\} \biggr). \end{aligned}$$

  Then, by (F1), we get

  $$\begin{aligned} d(Tx,Ty) < &\max \biggl\{ d(x,y),\frac{d(x,Ty)+d(y,Tx)}{2},\frac {d(T^{2}x,x)+d(T^{2}x,Ty)}{2},d \bigl(T^{2}x,Tx\bigr), \\ &d\bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr)+d(x,Tx), d(Tx,y)+d(y,Ty)\biggr\} , \end{aligned}$$

  for all $x,y\in X$, $Tx\neq Ty$.

The following examples show that the inverse implication of Remark 2.5(1) does not hold.

Example 2.6

Let $X=[0,2]$ and define a metric d on X by $d(x,y)=\mid x- y\mid$ and let $T: X\to X$ be given by

$$ Tx=\left \{ \textstyle\begin{array}{l@{\quad}l} 1, & x\in[0,2), \\ \frac{1}{2}, & x=2. \end{array}\displaystyle \right . $$

Obviously, $(X,d)$ is complete metric space. Since T is not continuous, T is not an F-contraction. For $x\in[0,2)$ and $y=2$, we have

$$ d(Tx,T2)=d\biggl(1,\frac{1}{2}\biggr)=\frac{1}{2}>0 $$

and

$$\begin{aligned}& \max\biggl\{ d(x,y),\frac{d(x,Ty)+d(y,Tx)}{2},\frac {d(T^{2}x,x)+d(T^{2}x,Ty)}{2},d\bigl(T^{2}x,Tx \bigr), \\& \qquad d\bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr)+d(x,Tx), d(Tx,y)+d(y,Ty) \biggr\} \\& \quad \geq d(Tx,y)+d(y,Ty) \\& \quad =d(1,2)+d\biggl(2,\frac{1}{2}\biggr) \\& \quad =\frac{5}{2}. \end{aligned}$$

Therefore

$$\begin{aligned} d(Tx,T2) \leq&\frac{1}{5}\max\biggl\{ d(x,y),\frac{d(x,Ty)+d(y,Tx)}{2}, \frac {d(T^{2}x,x)+d(T^{2}x,Ty)}{2},d\bigl(T^{2}x,Tx\bigr), \\ &d\bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr)+d(x,Tx), d(Tx,y)+d(y,Ty) \biggr\} . \end{aligned}$$

So, by choosing $F(\alpha)=\ln(\alpha)$ and $\tau=\ln\frac{1}{5}$ we see that T is modified generalized F-contraction of type (A) and type (B).

Example 2.7

Let $X=\{-2,-1,0,1,2\}$ and define a metric d on X by

$$ d(x,y)=\left \{ \textstyle\begin{array}{l@{\quad}l} 0, & \mbox{if }x=y, \\ 2, & \mbox{if }(x,y)\in\{(2,-2),(-2,2)\}, \\ 1, & \mbox{otherwise}. \end{array}\displaystyle \right . $$

Then $(X,d)$ is a complete metric space. Let $T : X \to X$ be defined by

$$T(-2) = T(-1) = T0 = -2, \qquad T1 = -1, \qquad T2 = 0. $$

First observe that

$$d(Tx,Ty)>0\quad \Leftrightarrow \quad \bigl[\bigl(x\in\{-2,-1,0\}\wedge y=1\bigr) \vee\bigl(x\in\{ -2,-1,0\}\wedge y=2\bigr)\vee(x=1,y=2)\bigr]. $$

Now we consider the following cases:

Case 1. Let $x\in\{-2,-1,0\}\wedge y=1$, then

$$\begin{aligned}& d(Tx,Ty)=d(-2,-1)=1,\qquad d(x,y)=d(x,1)=1,\qquad d(x,Tx)=d(x,-2)=0\vee1, \\& d(y,Ty)=d(1,-1)=1,\qquad \frac{d(x,Ty)+d(Tx,y)}{2}=\frac {d(x,-1)+d(-2,1)}{2}=\frac{1}{2} \vee1, \\& \frac{d(T^{2}x,x)+d(T^{2}x,Ty)}{2}=\frac{d(-2,x)+d(-2,-1)}{2}=\frac {1}{2}\vee1, \\& d\bigl(T^{2}x,Tx\bigr)=d(-2,-2)=0,\qquad d\bigl(T^{2}x,y \bigr)=d(-2,1)=1, \\& d\bigl(T^{2}x,Ty\bigr)=d(-2,-1)=1, \\& d\bigl(T^{2}x,Ty\bigr)+d(x,Tx)=d(-2,-1)+d(x,-2)=1\vee2, \\& d(Tx,y)+d(y,Ty)=d(-2,1)+d(1,-1)=2. \end{aligned}$$

Case 2. Let $x\in\{-2,-1,0\}\wedge y=2$, then

$$\begin{aligned}& d(Tx,Ty)=d(-2,0)=1,\qquad d(x,y)=d(x,2)=1\vee2,\qquad d(x,Tx)=d(x,-2)=0\vee1, \\& d(y,Ty)=d(2,0)=1,\qquad \frac{d(x,Ty)+d(Tx,y)}{2}=\frac {d(x,0)+d(-2,2)}{2}=1\vee \frac{3}{2}, \\& \frac{d(T^{2}x,x)+d(T^{2}x,Ty)}{2}=\frac{d(-2,x)+d(-2,0)}{2}=\frac {1}{2}\vee1, \\& d\bigl(T^{2}x,Tx\bigr)=d(-2,-2)=0,\qquad d\bigl(T^{2}x,y \bigr)=d(-2,2)=2, \\& d\bigl(T^{2}x,Ty\bigr)=d(-2,0)=1, \\& d\bigl(T^{2}x,Ty\bigr)+d(x,Tx)=d(-2,0)+d(x,-2)=1\vee2, \\& d(Tx,y)+d(y,Ty)=d(-2,2)+d(2,0)=3. \end{aligned}$$

Case 3. Let $x=1\wedge y=2$, then

$$\begin{aligned}& d(Tx,Ty)=d(-1,0)=1,\qquad d(x,y)=d(1,2)=1,\qquad d(x,Tx)=d(1,-1)=1, \\& d(y,Ty)=d(2,0)=1,\qquad \frac{d(x,Ty)+d(Tx,y)}{2}=\frac {d(1,0)+d(-1,2)}{2}=1, \\& \frac{d(T^{2}x,x)+d(T^{2}x,Ty)}{2}=\frac{d(-2,1)+d(-2,0)}{2}=1, \\& d\bigl(T^{2}x,Tx\bigr)=d(-2,-1)=1,\qquad d\bigl(T^{2}x,y \bigr)=d(-2,2)=2, \\& d\bigl(T^{2}x,Ty\bigr)=d(-2,0)=1,\qquad d\bigl(T^{2}x,Ty\bigr)+d(x,Tx)=d(-2,0)+d(1,-1)=2, \\& d(Tx,y)+d(y,Ty)=d(-1,2)+d(2,0)=2. \end{aligned}$$

In Case 1, we have

$$\begin{aligned} d(Tx,Ty)&=\max \biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2} \biggr\} \\ &=\max \biggl\{ \frac{d(T^{2}x,x)+d(T^{2}x,Ty)}{2},d\bigl(T^{2}x,Tx\bigr), d \bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr) \biggr\} =1. \end{aligned}$$

This proves that for all $F\in\mathcal{F}\cup\mathfrak{F}$, T is not an F-weak contraction and generalized F-contraction. Since every F-contraction is an F-weak contraction and a generalized F-contraction, T is not an F-contraction. However, we see that

$$\begin{aligned} d(Tx,T2) \leq&\frac{1}{2}\max\biggl\{ d(x,y),\frac{d(x,Ty)+d(y,Tx)}{2}, \frac {d(T^{2}x,x)+d(T^{2}x,Ty)}{2},d\bigl(T^{2}x,Tx\bigr), \\ &d\bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr)+d(x,Tx), d(Tx,y)+d(y,Ty)\biggr\} . \end{aligned}$$

Hence, by choosing $F(\alpha)=\ln(\alpha)$ and $\tau=\ln\frac{1}{2}$ we see that T is modified generalized F-contraction of type (A) and type (B).

Theorem 2.8

Let $(X,d)$ be a complete metric space and $T:X\rightarrow X$ be a modified generalized F-contraction of type (A). Then T has a unique fixed point $x^{*}\in X$ and for every $x_{0}\in X$ the sequence $\{T^{n}x_{0}\}_{n\in\mathbb{N}}$ converges to $x^{*}$.

Proof

Let $x_{0}\in X$. Put $x_{n+1}=T^{n}x_{0}$ for all $n\in\mathbb{N}$. If, there exists $n\in\mathbb{N}$ such that $x_{n+1}=x_{n}$, then $Tx_{n}=x_{n}$. That is, $x_{n}$ is a fixed point of T. Now, we suppose that $x_{n+1}\neq x_{n}$ for all $n\in\mathbb{N}$. Then $d(x_{n+1}, x_{n})>0$ for all $n\in\mathbb{N}$. It follows from (3) that, for all $n\in\mathbb{N}$,

$$\begin{aligned}& \tau+F\bigl(d(Tx_{n-1},Tx_{n})\bigr) \\& \quad \leq F\biggl( \max\biggl\{ d(x_{n-1},x_{n}), \frac{d(x_{n-1},Tx_{n})+d(x_{n},Tx_{n-1})}{2}, \\& \qquad \frac {d(T^{2}x_{n-1},x_{n-1})+d(T^{2}x_{n-1},Tx_{n})}{2},d\bigl(T^{2}x_{n-1},Tx_{n-1} \bigr), \\& \qquad d\bigl(T^{2}x_{n-1},x_{n}\bigr),d \bigl(T^{2}x_{n-1},Tx_{n}\bigr)+d(x_{n-1},Tx_{n-1}), d(Tx_{n-1},x_{n})+d(x_{n},Tx_{n}) \biggr\} \biggr) \\& \quad =F\biggl( \max\biggl\{ d(x_{n-1},x_{n}), \frac{d(x_{n-1},x_{n+1})+d(x_{n},x_{n})}{2}, \\& \qquad \frac{d(x_{n+1},x_{n-1})+d(x_{n+1},x_{n+1})}{2},d(x_{n+1},x_{n+1}), \\& \qquad d(x_{n+1},x_{n}),d(x_{n+1},x_{n+1})+d(x_{n-1},x_{n}), d(x_{n},x_{n})+d(x_{n},x_{n+1}) \biggr\} \biggr) \\& \quad =F \bigl( \max \bigl\{ d(x_{n-1},x_{n}),d(x_{n},x_{n+1}) \bigr\} \bigr). \end{aligned}$$

(4)

If there exists $n\in\mathbb{N}$ such that $\max\{d(x_{n-1},x_{n}),d(x_{n},x_{n+1})\}=d(x_{n},x_{n+1})$ then (4) becomes

$$\tau+F\bigl(d(x_{n},x_{n+1})\bigr)\leq F\bigl(d(x_{n},x_{n+1}) \bigr). $$

Since $\tau>0$, we get a contradiction. Therefore

$$\max\bigl\{ d(x_{n-1},x_{n}),d(x_{n},x_{n+1}) \bigr\} =d(x_{n-1},x_{n}),\quad \forall n\in\mathbb{N}. $$

Thus, from (4), we have

$$\begin{aligned} F\bigl(d(x_{n},x_{n+1})\bigr) =&F \bigl(d(Tx_{n-1},Tx_{n})\bigr)\leq F\bigl(d(x_{n-1},x_{n}) \bigr)-\tau \\ < &F\bigl(d(x_{n-1},x_{n})\bigr). \end{aligned}$$

(5)

It follows from (5) and (F1) that

$$d(x_{n},x_{n+1})< d(x_{n-1},x_{n}), \quad \forall n\in\mathbb{N}. $$

Therefore $\{d(x_{n+1},x_{n})\}_{n\in\mathbb{N}}$ is a nonnegative decreasing sequence of real numbers, and hence

$$\lim_{n\rightarrow\infty}d(x_{n+1},x_{n})=\gamma\geq0. $$

Now, we claim that $\gamma=0$. Arguing by contradiction, we assume that $\gamma>0$. Since $\{d(x_{n+1},x_{n})\}_{n\in\mathbb{N}}$ is a nonnegative decreasing sequence, for every $n \in\mathbb{N}$, we have

$$ d(x_{n+1},x_{n})\geq\gamma. $$

(6)

From (6) and (F1), we get

$$\begin{aligned} \begin{aligned}[b] F(\gamma)\leq{}& F\bigl(d(x_{n+1},x_{n})\bigr) \leq F\bigl(d(x_{n-1},x_{n})\bigr)-\tau \\ \leq{}& F\bigl(d(x_{n-2},x_{n-1})\bigr)-2\tau \\ &\vdots \\ \leq{}& F\bigl(d(x_{0},x_{1})\bigr)-n \tau, \end{aligned} \end{aligned}$$

(7)

for all $n \in\mathbb{N}$. Since $F(\gamma)\in\mathbb{R}$ and $\lim_{n\rightarrow\infty}[F(d(x_{0},x_{1}))-n \tau]=-\infty$, there exists $n_{1}\in\mathbb{N}$ such that

$$ F\bigl(d(x_{0},x_{1})\bigr)-n \tau< F(\gamma), \quad \forall n> n_{1}. $$

(8)

It follows from (7) and (8) that

$$F(\gamma)\leq F\bigl(d(x_{0},x_{1})\bigr)-n \tau< F(\gamma), \quad \forall n> n_{1}. $$

It is a contradiction. Therefore, we have

$$ \lim_{n\rightarrow\infty}d(x_{n},Tx_{n})= \lim_{n\rightarrow\infty} d(x_{n},x_{n+1})=0. $$

(9)

As in the proof of Theorem 2.1 in [2], we can prove that $\{x_{n}\}_{n=1}^{\infty}$ is a Cauchy sequence. So by completeness of $(X,d)$, $\{x_{n}\}_{n=1}^{\infty}$ converges to some point $x^{*}$ in X. Therefore,

$$ \lim_{n\rightarrow\infty}d\bigl(x_{n},x^{*}\bigr)=0. $$

(10)

Finally, we will show that $x^{*}=Tx^{*}$. We only have the following two cases:

• (I) $\forall n\in\mathbb{N}$, $\exists i_{n}\in\mathbb {N}$, $i_{n}> i_{n-1}$, $i_{0}=1$ and $x_{i_{n}+1} =Tx^{*}$,

• (II) $\exists n_{3}\in\mathbb{N}$, $\forall n\geq n_{3}$, $d(Tx_{n},Tx^{*})>0$.

In the first case, we have

$$x^{*}=\lim_{n\rightarrow\infty}x_{i_{n+1}} =\lim_{n\rightarrow\infty }Tx^{*}=Tx^{*}. $$

In the second case from the assumption of Theorem 2.8, for all $n\geq n_{3}$, we have

$$\begin{aligned}& \tau+F\bigl(d\bigl(x_{n+1},Tx^{*}\bigr)\bigr) \\& \quad =\tau+F\bigl(d\bigl(Tx_{n},Tx^{*}\bigr)\bigr) \\& \quad \leq F\biggl( \max\biggl\{ d\bigl(x_{n},x^{*}\bigr), \frac{d(x_{n},Tx^{*})+d(x^{*},Tx_{n})}{2}, \\& \qquad \frac{d(T^{2}x_{n},x_{n})+d(T^{2}x_{n},Tx^{*})}{2},d\bigl(T^{2}x_{n},Tx_{n} \bigr), \\& \qquad d\bigl(T^{2}x_{n},x^{*}\bigr),d\bigl(T^{2}x_{n},Tx^{*} \bigr)+d(x_{n},Tx_{n}), \\& \qquad d\bigl(Tx_{n},x^{*}\bigr)+d\bigl(x^{*},Tx^{*}\bigr)\biggr\} \biggr). \end{aligned}$$

(11)

From (F3′), (10), and taking the limit as $n\rightarrow\infty$ in (11), we obtain

$$ \tau+F\bigl(d\bigl(x^{*},Tx^{*}\bigr)\bigr)\leq F\bigl(d\bigl(x^{*},Tx^{*}\bigr)\bigr). $$

This is a contradiction. Hence, $x^{*}=Tx^{*}$. Now, let us to show that T has at most one fixed point. Indeed, if $x^{*},y^{*}\in X$ are two distinct fixed points of T, that is, $Tx^{*}=x^{*}\neq y^{*}=Ty^{*}$, then

$$d\bigl(Tx^{*},Ty^{*}\bigr)=d\bigl(x^{*},y^{*}\bigr)>0. $$

It follows from (3) that

$$\begin{aligned} F\bigl(d\bigl(x^{*},y^{*}\bigr)\bigr) < &\tau+F\bigl(d\bigl(x^{*},y^{*}\bigr)\bigr) \\ =&\tau+F\bigl(d\bigl(Tx^{*},Ty^{*}\bigr)\bigr) \\ \leq& F\biggl( \max\biggl\{ d\bigl(x^{*},y^{*}\bigr),\frac{d(x^{*},Ty^{*})+d(y^{*},Tx^{*})}{2}, \frac {d(T^{2}x^{*},x^{*})+d(T^{2}x^{*},Ty^{*})}{2}, \\ &d\bigl(T^{2}x^{*},Tx^{*}\bigr),d\bigl(T^{2}x^{*},y^{*}\bigr),d \bigl(T^{2}x^{*},Ty^{*}\bigr)+d\bigl(x^{*},Tx^{*}\bigr), \\ &d\bigl(Tx^{*},y^{*}\bigr)+d\bigl(y^{*},Ty^{*}\bigr)\biggr\} \biggr) \\ =&F\biggl( \max\biggl\{ d\bigl(x^{*},y^{*}\bigr),\frac{d(x^{*},y^{*})+d(y^{*},x^{*})}{2}, \frac {d(x^{*},x^{*})+d(x^{*},y^{*})}{2}, \\ &d\bigl(x^{*},x^{*}\bigr),d\bigl(x^{*},y^{*}\bigr),d\bigl(x^{*},y^{*}\bigr)+d\bigl(x^{*},x^{*} \bigr), \\ &d\bigl(x^{*},y^{*}\bigr)+d\bigl(y^{*},y^{*}\bigr)\biggr\} \biggr) \\ =&F\bigl(d\bigl(x^{*},y^{*}\bigr)\bigr), \end{aligned}$$

which is a contradiction. Therefore, the fixed point is unique. □

Theorem 2.9

Let $(X,d)$ be a complete metric space and $T:X\rightarrow X$ be a continuous modified generalized F-contraction of type (B). Then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n\in\mathbb{N}}$ converges to $x^{*}$.

Proof

By using a similar method to that used in the proof of Theorem 2.8, we have

$$\begin{aligned} F\bigl(d(x_{n},x_{n+1})\bigr)&=F\bigl(d(Tx_{n-1},Tx_{n}) \bigr)\leq F\bigl(d(x_{n-1},x_{n})\bigr)-\tau \\ &< F\bigl(d(x_{n-1},x_{n})\bigr) \end{aligned}$$

and

$$ \lim_{n\rightarrow\infty}d(x_{n},Tx_{n})=\lim _{n\rightarrow\infty} d(x_{n},x_{n+1})=0. $$

As in the proof of Theorem 2.1 in [3], we can prove that $\{x_{n}\}_{n=1}^{\infty}$ is a Cauchy sequence. So, by completeness of $(X,d)$, $\{x_{n}\}_{n=1}^{\infty}$ converges to some point $x^{*}\in X$. Since T is continuous, we have

$$d\bigl(x^{*},Tx^{*}\bigr)=\lim_{n\rightarrow\infty} d(x_{n},Tx_{n})= \lim_{n\rightarrow\infty} d(x_{n},x_{n+1})=0. $$

Again by using similar method as used in the proof of Theorem 2.8, we can prove that $x^{*}$ is the unique fixed point of T. □

3 Some applications

Theorem 3.1

[2]

Let T be a self-mapping of a complete metric space X into itself. Suppose there exist $F\in\mathfrak{F}$ and $\tau>0$ such that

$$\forall x,y\in X, \quad \bigl[d(Tx, Ty)>0\Rightarrow\tau+F\bigl(d(Tx,Ty)\bigr) \leq F\bigl(d(x,y)\bigr)\bigr]. $$

Then T has a unique fixed point $x^{*}\in X$ and for every $x_{0}\in X$ the sequence $\{T^{n}x_{0}\}_{n=1}^{\infty}$ converges to $x^{*}$.

Proof

Since

$$\begin{aligned}& \max\biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2}\biggr\} \\& \quad \leq\max\biggl\{ d(x,y),\frac{d(x,Ty)+d(y,Tx)}{2},\frac {d(T^{2}x,x)+d(T^{2}x,Ty)}{2},d \bigl(T^{2}x,Tx\bigr), \\& \qquad d\bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr)+d(x,Tx), d(Tx,y)+d(y,Ty)\biggr\} , \end{aligned}$$

from (F1) and Theorem 2.8 the proof is complete. □

Theorem 3.2

[3]

Let $(X,d)$ be a complete metric space and let $T:X\rightarrow X$ be an F-contraction. Then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n\in\mathbb{N}}$ converges to $x^{*}$.

Proof

Since

$$\begin{aligned}& \max\biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2}\biggr\} \\& \quad \leq\max\biggl\{ d(x,y),\frac{d(x,Ty)+d(y,Tx)}{2},\frac {d(T^{2}x,x)+d(T^{2}x,Ty)}{2},d \bigl(T^{2}x,Tx\bigr), \\& \qquad d\bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr)+d(x,Tx), d(Tx,y)+d(y,Ty)\biggr\} . \end{aligned}$$

So from (F1) and Theorem 2.9 the proof is complete. □

Theorem 3.3

[4]

Let $(X,d)$ be a complete metric space and let $T:X\rightarrow X$ be an F-weak contraction. If T or F is continuous, then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n\in\mathbb{N}}$ converges to $x^{*}$.

Proof

Since

$$\begin{aligned}& \max\biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2}\biggr\} \\& \quad \leq\max\biggl\{ d(x,y),\frac{d(x,Ty)+d(y,Tx)}{2},\frac {d(T^{2}x,x)+d(T^{2}x,Ty)}{2},d \bigl(T^{2}x,Tx\bigr), \\& \qquad d\bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr)+d(x,Tx), d(Tx,y)+d(y,Ty)\biggr\} , \end{aligned}$$

if F is continuous, from (F1) and Theorem 2.8 the proof is complete. If T is continuous, from (F1) and Theorem 2.9 the proof is complete. □

Theorem 3.4

[1]

Let $(X,d)$ be a complete metric space and let $T:X\rightarrow X$ be a generalized F-contraction. If T or F is continuous, then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n\in\mathbb{N}}$ converges to $x^{*}$.

Proof

Since

$$\begin{aligned}& \max\biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2}, \\& \qquad \frac{d(T^{2}x,x)+d(T^{2}x,Ty)}{2},d\bigl(T^{2}x,Tx\bigr), d \bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr) \biggr\} \\& \quad \leq\max\biggl\{ d(x,y),\frac{d(x,Ty)+d(y,Tx)}{2},\frac {d(T^{2}x,x)+d(T^{2}x,Ty)}{2},d \bigl(T^{2}x,Tx\bigr), \\& \qquad d\bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr)+d(x,Tx), d(Tx,y)+d(y,Ty)\biggr\} , \end{aligned}$$

if F is continuous, from (F1) and Theorem 2.8 the proof is complete. If T is continuous, from (F1) and Theorem 2.9 the proof is complete. □

Theorem 3.5

Let $(X,d)$ be a complete metric space and let $T:X\rightarrow X$ be a function with the following property:

$$ d(Tx,Ty)\leq\alpha d(x,y)+\beta d(x,Tx)+\gamma d(y,Ty), $$

(12)

where α, β, and γ are nonnegative and satisfy $\alpha+\beta+\gamma<1$. Then T has a unique fixed point.

Proof

From (12), we have

$$\begin{aligned} d(Tx,Ty) \leq&(\alpha+\beta+\gamma)\max\biggl\{ d(x,y),\frac{d(x,Ty)+d(y,Tx)}{2}, \frac {d(T^{2}x,x)+d(T^{2}x,Ty)}{2}, \\ &d\bigl(T^{2}x,Tx\bigr),d\bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr)+d(x,Tx), d(Tx,y)+d(y,Ty)\biggr\} . \end{aligned}$$

Then if $d(Tx,Ty)>0$, we have

$$\begin{aligned}& \ln\frac{1}{\alpha+\beta+\gamma}+\ln\bigl(d(Tx,Ty)\bigr) \\& \quad \leq\ln\biggl( \max\biggl\{ d(x,y),\frac{d(x,Ty)+d(y,Tx)}{2},\frac {d(T^{2}x,x)+d(T^{2}x,Ty)}{2},d \bigl(T^{2}x,Tx\bigr), \\& \qquad d\bigl(T^{2}x,y\bigr),d\bigl(T^{2}x,Ty\bigr)+d(x,Tx), d(Tx,y)+d(y,Ty)\biggr\} \biggr). \end{aligned}$$

Therefore by taking $F(\alpha)=\ln(\alpha)$ and $\tau=\ln\frac{1}{\alpha+\beta+\gamma}$ in Theorem 2.8 or in Theorem 2.9 the proof is complete. □

Remark 3.6

Our theorems are extensions of the above theorems in the following aspects:

• (1) Theorem 2.8 gives all consequences of Theorem 2.1 of [2] without assumption (F2) used in its proof.

• (2) Theorem 2.9 gives all consequences of Theorem 2.1 of [3] without assumption (F2) used in its proof.

• (3) If in Theorem 3 of [1] F is continuous, Theorem 2.8 gives all consequences of Theorem 3 of [1] without assumptions (F2) and (F3) used in its proof.

• (4) If in Theorem 3 of [1] T is continuous, Theorem 2.9 gives all consequences of Theorem 3 of [1] without assumption (F2) used in its proof.

• (5) Because every F-weak contraction is a generalized F-contraction, (3) and (4) are also true for Theorem 2.4 of [4].