Open Access

Wardowski type fixed point theorems in complete metric spaces

Fixed Point Theory and Applications20162016:45

https://doi.org/10.1186/s13663-016-0529-0

Received: 5 September 2015

Accepted: 9 March 2016

Published: 4 April 2016

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.

Keywords

fixed pointmetric space F-contraction

MSC

74H1054H25

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
  1. (F1)

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

     
  2. (F2)

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

     
  3. (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. (1)

    Every F-contraction is a modified generalized F-contraction.

     
  2. (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:
  1. (I)

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

     
  2. (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. (1)

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

     
  2. (2)

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

     
  3. (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. (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. (5)

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

     

Declarations

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, University of Bonab
(2)
Theoretical and Computational Science Center (TaCS-Center) & Department of Mathematics, Science Laboratory Building, Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT)
(3)
Department of Medical Research, China Medical University Hospital, China Medical University

References

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Copyright

© Piri and Kumam 2016