Fixed point theorems for generalized F-Suzuki-contraction mappings in complete b-metric spaces

Abstract

The aim of this work is to establish some new fixed point theorems for generalized F-Suzuki-contraction mappings in complete b-metric spaces.

1 Introduction and mathematical preliminaries

Recently, Wardowski [1] established a new contraction, the so-called F-contraction, and obtained a fixed point result as a generalization of the Banach contraction principle. After that Altun et al. [2] introduced the new concept of multivalued F-contraction mappings and gave some fixed point results. Wardowski and Dung [3] further generalized the concept of an F-contraction to an F-weak contraction and also obtained certain fixed point results. Dung and Hang [4] studied the notion of a generalized F-contraction and extended a fixed point theorem for such mappings. Recently Piri and Kumam [5] described a large class of functions by replacing condition (F3′) instead of the condition (F3) in the definition of F-contraction.

Following this direction of research, in this paper, we extend the fixed point results of Wardowski [1], Wardowski and Dung [3], Dung and Hang [4], and Piri and Kumam [5] by introducing a generalized F-Suzuki-contraction in b-metric spaces. We begin with some basic well-known definitions and results which will be used in the rest of this paper.

Throughout this paper, $\mathbb{N}_{0}$, $\mathbb{N}$, $\mathbb{R}_{+}$, $\mathbb{R}$ denote the set of nonnegative integer numbers, the set of natural numbers, the set of positive real numbers, and the set of real numbers, respectively.

Definition 1.1

Let $\mathcal{F}$ be the family of all functions $F :\mathbb {R}_{+} \to\mathbb{R}$ 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 $\lim_{\alpha\to 0^{+}}\alpha^{k}F(\alpha)=0$.

Definition 1.2

[1]

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

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

(1)

A new generalization of Banach contraction principle has been given by Wardowski [1] as follows.

Theorem 1.3

[1]

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 $\{T^{n}x\}_{n=1}^{\infty}$ converges to $x^{*}$.

In 2014, Wardowski and Dung [3] introduced the notion of an F-weak contraction and proved a related fixed point theorem as follows.

Definition 1.4

[3]

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 $x,y\in X$,

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

(2)

in which

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

Theorem 1.5

[3]

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 $\{T^{n}x\}_{n=1}^{\infty}$ converges to $x^{*}$.

Recall that a contraction conditions for a self-mapping T on a metric space $(X, d)$, usually contained at most five values $d(x, y)$, $d(x,Tx)$, $d(y,Ty)$, $d(x,Ty)$, $d(y,Tx)$ (for example see [6, 7]). Recently, by adding four new values $d(T^{2}x, x)$, $d(T^{2}x,Tx)$, $d(T^{2}x, y)$, $d(T^{2}x,Ty)$ to a contraction condition, Kumam et al. [8] stated a new generalization of the Ćirić fixed point theorem in [9]. Motivated and inspired by the idea of Kumam et al. [8], Dung and Hang [4] generalized the notion of a generalized F-contraction and proved some fixed point theorems for such maps. They gave examples to show that their result is a real generalization of Theorem 1.5 and some others in the literature.

Definition 1.6

[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, for all $x,y\in X$,

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

in which

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

(3)

Theorem 1.7

[4]

Let $(X,d)$ be a complete metric space and let $T:X\rightarrow X$ be a generalized F-contraction mapping. If T or F is continuous, then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n=1}^{\infty}$ converges to $x^{*}$.

In 2014, Piri and Kumam [5] described a large class of functions by replacing the condition (F3) in the definition of an F-contraction introduced by Wardowski [1] 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 1.8

[5]

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

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

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

Theorem 1.9

[5]

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

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

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 $x^{*}$.

Definition 1.10

[10]

Let X be a nonempty set and $s\geq1$ be a given real number. A mapping $d : X \times X\to\mathbb{R}^{+}$ is said to be a b-metric if for all $x, y, z \in X$ the following conditions are satisfied:

(bm₁):

$d(x, y) =0$ if and only if $x = y$;

(bm₂):

$d(x, y) = d(y,x)$;

(bm₃):

$d(x, z)\leq s[d(x, y) + d(y, z)]$.

In this case, the pair $(X, d)$ is called a b-metric space (with constant s).

Definition 1.11

[11]

Let $(X, d)$ be a b-metric space. A sequence $\{x_{n}\}_{n=1}^{\infty}$ in X is called:

• (A) Convergent if and only if there exists $x \in X$ such that $\lim_{ n\rightarrow\infty}d(x_{n}, x)=0$. In this case, we write $\lim_{ n\rightarrow} x_{n} = x$.

• (B) Cauchy if and only if $\lim_{ n,m\rightarrow\infty}d(x_{n}, x_{m})=0$.

Remark 1.12

[11]

Notice that in a b-metric space $(X, d)$ the following assertions hold:

• (A) a convergent sequence has a unique limit;

• (B) each convergent sequence is Cauchy;

• (C) in general, a b-metric is not continuous;

• (D) in general, a b-metric does not induce a topology on X.

Definition 1.13

[11]

The b-metric space $(X, d)$ is complete if every Cauchy sequence in X converges in X.

Definition 1.14

[12]

Let $(X, d_{X} )$ and $(Y, d_{Y} )$ be b-metric spaces; a mapping $f: X\to Y$ is called:

• (A) continuous at a point $x\in X$, if for every sequence $\{x_{n}\}_{n=1}^{\infty}$ in X such that $\lim_{ n\rightarrow} x_{n} = x$, then $\lim_{ n\rightarrow}f(x_{n})=f(x)$;

• (B) continuous on X, if it is continuous at each point $x\in X$.

2 Main results

We use $\mathfrak{F}_{G}$ to denote the set of all functions $F:\mathbb{R}_{+}\rightarrow\mathbb{R}$ which satisfy conditions (F1) and (F3′) and Ψ to denote the set of all functions $\psi:[0,\infty)\to[0,\infty)$ such that ψ is continuous and $\psi(t)=0$ if and only $t=0$.

Definition 2.1

Let $(X,d)$ be a b-metric space. A self-mapping $T: X\to X$ is said to be a generalized F-Suzuki-contraction if there exists $F\in \mathfrak{F}_{G}$ such that, for all $x,y\in X$ with $x\neq y$,

$$ \frac{1}{2s}d(x,Tx)< d(x,y)\quad \Rightarrow\quad F\bigl(s^{5}d(Tx,Ty) \bigr)\leq F\bigl(M_{T}(x,y)\bigr)-\psi\bigl(M_{T}(x,y) \bigr), $$

in which $\psi\in\Psi$ and

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

(4)

Theorem 2.2

Let $(X,d)$ be a complete b-metric space and $T : X \to X$ be a generalized F-Suzuki-contraction. Then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n=1}^{\infty}$ converges to $x^{*}$.

Proof

Take $x_{0}=x\in X$. Let $x_{n}=Tx_{n-1}$ for all $n\in\mathbb{N}$. If there exists $n\in\mathbb{N}$ such that $d(x_{n},Tx_{n})=0$ then $x=x_{n}$ becomes a fixed point of T, which completes the proof. So, in the rest of the proof, we assume that

$$ 0< d(x_{n},Tx_{n}),\quad \forall n\in \mathbb{N}. $$

(5)

Hence, we have

$$ \frac{1}{2s}d(x_{n},Tx_{n})< d(x_{n},Tx_{n})=d(x_{n},x_{n+1}), \quad \forall n\in\mathbb{N}. $$

(6)

So by the assumption of the theorem, we have

$$F\bigl(d(Tx_{n},Tx_{n+1})\bigr)\leq F\bigl(M_{T}(x_{n},x_{n+1}) \bigr)-\psi\bigl(M_{T}(x_{n},x_{n+1})\bigr). $$

Since

$$\begin{aligned}& \max \bigl\{ d(x_{n},x_{n+1}),d\bigl(T^{2}x_{n},x_{n+1} \bigr) \bigr\} \\& \quad \leq M_{T}(x_{n},x_{n+1}) \\& \quad =\max \biggl\{ d(x_{n},x_{n+1}),d(x_{n+2},x_{n+1}), \frac{d(x_{n},x_{n+2})}{2s},\frac {d(x_{n+2},x_{n})}{2s}, \\& \qquad d(x_{n+2},x_{n+1}),d(x_{n+1},x_{n+2}),d(x_{n},x_{n+1}) \biggr\} \\& \quad \leq\max \biggl\{ d(x_{n},x_{n+1}),d(x_{n+2},x_{n+1}), \frac {s[d(x_{n},x_{n+1})+d(x_{n+1},x_{n+2})]}{2s}, \\& \qquad \frac{s[d(x_{n},x_{n+1})+d(x_{n+1},x_{n+2})]}{2s}, d(x_{n+2},x_{n+1}),d(x_{n+1},x_{n+2}),d(x_{n},x_{n+1}) \biggr\} \\& \quad \leq\max \bigl\{ d(x_{n},x_{n+1}),d(x_{n+2},x_{n+1}) \bigr\} , \end{aligned}$$

we get

$$\begin{aligned} F\bigl(d(x_{n+1},x_{n+2})\bigr) \leq& F \bigl( \max \bigl\{ d(x_{n},x_{n+1}),d(x_{n+1},x_{n+2}) \bigr\} \bigr) \\ &{} -\psi \bigl( \max \bigl\{ d(x_{n},x_{n+1}),d(x_{n+1},x_{n+2}) \bigr\} \bigr). \end{aligned}$$

(7)

If $d(x_{n+1},x_{n+2})>d(x_{n},x_{n+1})$, then

$$\max \bigl\{ d(x_{n},x_{n+1}),d(x_{n+1},x_{n+2}) \bigr\} =d(x_{n+1},x_{n+2}), $$

so (7) becomes

$$F\bigl(d(x_{n+1},x_{n+2})\bigr)\leq F\bigl(d(x_{n+1},x_{n+2}) \bigr)-\psi\bigl(d(x_{n+1},x_{n+2})\bigr), $$

which is a contradiction (from (5) and the property of φ, we have $\psi(d(x_{n+1},x_{n+2}))>0$). Thus, we conclude that

$$\begin{aligned} F\bigl(d(x_{n+1},x_{n+2})\bigr)&\leq F \bigl(d(x_{n},x_{n+1})\bigr)-\psi\bigl(d(x_{n},x_{n+1}) \bigr) \\ &< F\bigl(d(x_{n},x_{n+1})\bigr). \end{aligned}$$

(8)

It follows from (8) and (F1) that

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

(9)

Therefore $\{d(x_{n+1},x_{n})\}_{n\in\mathbb{N}}$ is a nonnegative decreasing sequence of real numbers. Thus, there exists $\gamma\geq0$ such that $\lim_{n\rightarrow\infty}d(x_{n+1},x_{n})=\gamma$. Letting $n\rightarrow\infty$ in (8), we have

$$F(\gamma)\leq F(\gamma)-\psi(\gamma). $$

This implies that $\psi(\gamma) = 0$ and thus $\gamma= 0$. Consequently, we have

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

(10)

Now, we claim that $\{x_{n}\}_{n=1}^{\infty}$ is a Cauchy sequence. Arguing by contradiction, we assume that there exist $\epsilon>0$, and the sequences $\{p(n)\}_{n=1}^{\infty}$ and $\{q(n)\}_{n=1}^{\infty}$ of natural numbers such that, for all $n\in\mathbb{N}$,

$$ p(n)>q(n)>n,\quad d(x_{p(n)},x_{q(n)})\geq \epsilon,\qquad d(x_{p(n)-1},x_{q(n)})< \epsilon. $$

(11)

Observe that

$$\begin{aligned} \epsilon&\leq d(x_{p(n)},x_{q(n)})\leq s\bigl[d(x_{p(n)},x_{p(n)-1})+d(x_{p(n)-1},x_{q(n)}) \bigr] \\ &\leq s d(x_{p(n)},x_{p(n)-1})+s\epsilon. \end{aligned}$$

So from (10), we get

$$ \epsilon\leq\limsup_{n\rightarrow\infty}d(x_{p(n)},x_{q(n)}) \leq s\epsilon. $$

(12)

From the triangle inequality, we have

$$ \epsilon\leq d(x_{p(n)},x_{q(n)})\leq s \bigl[d(x_{p(n)},x_{q(n)+1})+d(x_{q(n)+1},x_{q(n)}) \bigr] $$

(13)

and

$$ d(x_{p(n)},x_{q(n)+1}) \leq s\bigl[d(x_{p(n)},x_{q(n)})+d(x_{q(n)},x_{q(n)+1}) \bigr]. $$

(14)

It follows from (10), (12), (13), and (14) that

$$ \frac{\epsilon}{s}\leq\limsup_{n\rightarrow\infty }d(x_{p(n)},x_{q(n)+1}) \leq s^{2}\epsilon. $$

(15)

Again, using above process, we get

$$ \frac{\epsilon}{s}\leq\limsup_{n\rightarrow\infty }d(x_{p(n)+1},x_{q(n)}) \leq s^{2}\epsilon. $$

(16)

From (15) and the inequality

$$d(x_{p(n)},x_{q(n)+1})\leq s\bigl[ d(x_{p(n)},x_{p(n)+1})+d(x_{p(n)+1},x_{q(n)+1}) \bigr], $$

we have

$$ \frac{\epsilon}{s^{2}}\leq\limsup_{n\rightarrow\infty }d(x_{p(n)+1},x_{q(n)+1}). $$

(17)

From (12) and the inequality

$$\begin{aligned} d(x_{p(n)+1},x_{q(n)+1})&\leq s\bigl[d(x_{p(n)+1},x_{q(n)})+d(x_{q(n)},x_{q(n)+1}) \bigr] \\ &\leq s^{2}\bigl[d(x_{p(n)+1},x_{p(n)})+d(x_{p(n)},x_{q(n)}) \bigr]+sd(x_{q(n)},x_{q(n)+1}), \end{aligned}$$

we have

$$ \limsup_{n\rightarrow\infty}d(x_{p(n)+1},x_{q(n)+1}) \leq s^{3}\epsilon. $$

(18)

It follows from (17) and (18) that

$$ \frac{\epsilon}{s^{2}}\leq\limsup_{n\rightarrow\infty }d(x_{p(n)+1},x_{q(n)+1}) \leq s^{3}\epsilon. $$

(19)

From (10) and (11), we can choose a positive integer $n_{1}\in\mathbb{N}$ such that

$$ \frac{1}{2s} d(x_{p(n)},Tx_{p(n)})< \frac{1}{2s} \epsilon< d(x_{p(n)},x_{q(n)}),\quad \forall n\geq n_{1}. $$

Therefore by assumption of theorem for every $n\geq n_{1}$, we have

$$ F\bigl(d(x_{p(n)+1},x_{q(n)+1})\bigr)\leq F \bigl(M_{T}(x_{p(n)},x_{q(n)})\bigr)-\psi \bigl(M_{T}(x_{p(n)},x_{q(n)})\bigr). $$

(20)

Since

$$\begin{aligned}& d(x_{p(n)},x_{q(n)}) \\& \quad \leq M_{T}(x_{p(n)},x_{q(n)}) \\& \quad = \max \biggl\{ d(x_{p(n)},x_{q(n)}),d(x_{p(n)+2},x_{q(n)}), \frac {d(x_{p(n)+1},x_{q(n)})+d(x_{p(n)},x_{q(n)+1})}{2s}, \\& \qquad \frac {d(x_{p(n)+2},x_{p(n)})+d(x_{p(n)+2},x_{q(n)+1})}{2s},d(x_{p(n)+2},x_{q(n)+1})+ d(x_{p(n)+2},x_{p(n)+1}), \\& \qquad d(x_{p(n)+2},x_{q(n)+1})+d(x_{p(n)+1},x_{p(n)}),d(x_{p(n)+1},x_{q(n)})+ d(x_{q(n)},x_{q(n)+1}) \biggr\} \\& \quad \leq\max \biggl\{ d(x_{p(n)},x_{q(n)}),s \bigl[d(x_{p(n)+2},x_{p(n)+1})+d(x_{p(n)+1},x_{q(n)}) \bigr], \\& \qquad \frac{d(x_{p(n)+1},x_{q(n)})+d(x_{p(n)},x_{q(n)+1})}{2s}, \\& \qquad \frac {s[d(x_{p(n)+2},x_{p(n)+1})+d(x_{p(n)+1},x_{p(n)})]+s[d(x_{p(n)+2},x_{p(n)+1})+d(x_{p(n)+1},x_{q(n)+1})]}{2s}, \\& \qquad s\bigl[d(x_{p(n)+2},x_{p(n)+1})+d(x_{p(n)+1},x_{q(n)+1}) \bigr]+d(x_{p(n)+2},x_{p(n)+1}), \\& \qquad s\bigl[d(x_{p(n)+2},x_{p(n)+1})+d(x_{p(n)+1},x_{q(n)+1}) \bigr]+d(x_{p(n)+1},x_{p(n)}), \\& \qquad d(x_{p(n)+1},x_{q(n)})+ d(x_{q(n)},x_{q(n)+1}) \biggr\} , \end{aligned}$$

taking the limit supremum as $n\rightarrow\infty$ on each side of the above inequality and using (12), (15), (16), and (19) we have

$$ \epsilon\leq\limsup_{n\rightarrow\infty} M_{T}(x_{p(n)},x_{q(n)}) \leq s^{3}\epsilon. $$

(21)

Also, we can show that

$$ \epsilon\leq\liminf_{n\rightarrow\infty} M_{T}(x_{p(n)},x_{q(n)}) \leq s^{3}\epsilon. $$

(22)

Taking the limit supremum as $n\rightarrow\infty$ in (20) and using (21) and (22), we get

$$\begin{aligned} F\bigl(s^{3}\epsilon\bigr)&=F\biggl(s^{5}\frac{\epsilon}{s^{2}} \biggr)\leq F\Bigl(\limsup_{n\rightarrow\infty}d(x_{p(n)+1},x_{q(n)+1}) \Bigr) \\ &\leq F\Bigl(\limsup_{n\rightarrow\infty}M_{T}(x_{p(n)},x_{q(n)}) \Bigr)-\psi\Bigl(\limsup_{n\rightarrow\infty}M_{T}(x_{p(n)},x_{q(n)}) \Bigr) \\ &\leq F\bigl(s^{3}\epsilon\bigr)-\psi(\epsilon), \end{aligned}$$

which is a contradiction with $\epsilon>0$, and it follows that $\{x_{n}\}$ is a Cauchy sequence in X. By completeness of $(X,d)$, $\{x_{n}\}_{n=1}^{\infty}$ converges to some point $x^{*}$ in X. Therefore,

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

(23)

We claim that, for every $n\in\mathbb{N}$,

$$ \frac{1}{2s}d(x_{n},Tx_{n})< d \bigl(x_{n},x^{*}\bigr) \quad \mbox{or}\quad \frac{1}{2s}d \bigl(Tx_{n},T^{2}x_{n}\bigr)< d\bigl(Tx_{n},x^{*} \bigr). $$

(24)

Suppose, on the contrary, that there exists $m\in\mathbb{N}$ such that

$$ \frac{1}{2s}d(x_{m},Tx_{m})\geq d \bigl(x_{m},x^{*}\bigr) \quad \mbox{and}\quad \frac{1}{2s}d \bigl(Tx_{m},T^{2}x_{m}\bigr)\geq d \bigl(Tx_{m},x^{*}\bigr). $$

(25)

Therefore,

$$ 2sd\bigl(x_{m},x^{*}\bigr)\leq d(x_{m},Tx_{m}) \leq s\bigl[d\bigl(x_{m},x^{*}\bigr)+d\bigl(x^{*},Tx_{m} \bigr)\bigr], $$

which implies that

$$ d\bigl(x_{m},x^{*}\bigr)\leq d\bigl(x^{*},Tx_{m} \bigr). $$

(26)

From (9) and (26), we have

$$\begin{aligned} d\bigl(Tx_{m},T^{2}x_{m}\bigr)&< d(x_{m},Tx_{m})\leq sd\bigl(x_{m},x^{*}\bigr)+sd \bigl(x^{*},Tx_{m}\bigr) \\ &\leq2sd\bigl(x^{*},Tx_{m}\bigr). \end{aligned}$$

(27)

It follows from (25) and (27) that $d(Tx_{m},T^{2}x_{m})< d(Tx_{m},T^{2}x_{m})$. This is a contradiction. Hence, (24) holds. If part (I) of (24) is true, then we have

$$\begin{aligned} F\bigl(d\bigl(x_{n+1},Tx^{*}\bigr)\bigr)&=F\bigl(d \bigl(Tx_{n},Tx^{*}\bigr)\bigr) \\ &\leq F\bigl(M_{T}\bigl(x_{n},x^{*}\bigr)\bigr)-\psi \bigl(M_{T}\bigl(x_{n},x^{*}\bigr)\bigr). \end{aligned}$$

(28)

Since

$$\begin{aligned} d\bigl(x^{*},Tx^{*}\bigr) \leq& M_{T}\bigl(x_{n},x^{*}\bigr) \\ =& \max \biggl\{ d\bigl(x_{n},x^{*}\bigr),d\bigl(x_{n+2},x^{*} \bigr),\frac{d(x_{n+1},x^{*})+d(x_{n},Tx^{*})}{2s}, \\ &\frac{d(x_{n+2},x_{n})+d(x_{n+2},Tx^{*})}{2s},d\bigl(x_{n+2},Tx^{*}\bigr)+ d(x_{n+2},x_{n+1}), \\ &d\bigl(x_{n+2},Tx^{*}\bigr)+d(x_{n+1},x_{n}),d \bigl(x_{n+1},x^{*}\bigr)+d\bigl(x^{*},Tx^{*}\bigr) \biggr\} \\ \leq&\max \biggl\{ d\bigl(x_{n},x^{*}\bigr),d\bigl(x_{n+2},x^{*} \bigr),\frac{d(x_{n+1},x^{*})+d(x_{n},Tx^{*})}{2s}, \\ &\frac{s[d(x_{n+2},x_{n+1})+d(x_{n+1},x_{n})]+d(x_{n+2},Tx^{*})}{2s}, \\ &d\bigl(x_{n+2},Tx^{*}\bigr)+d(x_{n+2},x_{n+1}), \\ &d\bigl(x_{n+2},Tx^{*}\bigr)+d(x_{n+1},x_{n}),d \bigl(x_{n+1},x^{*}\bigr)+d\bigl(x^{*},Tx^{*}\bigr) \biggr\} , \end{aligned}$$

letting $n\rightarrow\infty$ and using (23), we get

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

It follows from (28), (F3′), and the continuity of φ that

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

This yields $x^{*}=Tx^{*}$. If part (II) of (24) is true, using a similar method to the above, we get $x^{*}=Tx^{*}$. Hence, $x^{*}$ is a fixed point of T. Now we show that T has at most one fixed point. Indeed, if $x^{*},y^{*}\in X$ are two fixed points of T, such that $x^{*}\neq y^{*}$, then we have $0=\frac{1}{2s}d(x^{*},Tx^{*})< d(x^{*},y^{*})$ and from the assumption of the theorem, we obtain

$$\begin{aligned} F\bigl(d\bigl(x^{*},y^{*}\bigr)\bigr)&=F\bigl(d\bigl(Tx^{*},Ty^{*}\bigr)\bigr) \\ & \leq F\bigl(M_{T}\bigl(x^{*},y^{*}\bigr)\bigr)-\psi \bigl(M_{T}\bigl(x^{*},y^{*}\bigr)\bigr) \\ & =F\bigl( d\bigl(y^{*},x^{*}\bigr) \bigr)-\psi\bigl( d\bigl(y^{*},x^{*} \bigr)\bigr). \end{aligned}$$

This gives $\psi( d(y^{*},x^{*} ))\leq0$. Hence $y^{*}=x^{*}$. This completes the proof. □

The following two theorems can be obtained easily by repeating the steps in the proof of Theorem 2.2.

Theorem 2.3

Let $(X,d)$ be a complete b-metric space and $T : X \to X$ be a self-mapping such that, for every $x, y \in X$,

$$ \frac{1}{2s}d(x,Tx)< d(x,y)\quad \Rightarrow\quad F\bigl(d(Tx,Ty)\bigr) \leq F\bigl(M_{T}(x,y)\bigr)-\psi\bigl(N(x,y)\bigr), $$

where $N(x,y)$ is defined by (3) and ψ is defined as in Theorem  2.2. Then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n=1}^{\infty}$ converges to $x^{*}$.

Theorem 2.4

Let $(X,d)$ be a complete b-metric space and $T : X \to X$ be a self-mapping such that, for every $x, y \in X$,

$$ \frac{1}{2s}d(x,Tx)< d(x,y)\quad \Rightarrow \quad F\bigl(d(Tx,Ty)\bigr) \leq F\bigl(M_{T}(x,y)\bigr)-\psi\bigl(d(x,y)\bigr), $$

where $M_{T}(x,y)$ is defined by (4) and ψ is defined as in Theorem  2.2. Then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n=1}^{\infty}$ converges to $x^{*}$.

Since a b-metric space is a metric space when $s = 1$, so we obtain the following theorems.

Theorem 2.5

Let $(X,d)$ be a complete metric space and $T : X \to X$ be a generalized F-Suzuki-contraction. Then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n=1}^{\infty}$ converges to $x^{*}$.

Theorem 2.6

Let $(X,d)$ be a complete metric space and $T : X \to X$ be a self-mapping such that, for every $x, y \in X$,

$$ \frac{1}{2s}d(x,Tx)< d(x,y)\quad \Rightarrow \quad F\bigl(d(Tx,Ty)\bigr) \leq F\bigl(M_{T}(x,y)\bigr)-\psi\bigl(d(x,y)\bigr), $$

where $M_{T}(x,y)$ is defined by (4) and ψ is defined as in Theorem  2.2. Then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n=1}^{\infty}$ converges to $x^{*}$.

Theorem 2.7

[4]

Let $(X,d)$ be a complete metric space and let $T:X\rightarrow X$ be a generalized F-contraction. If F is continuous, then T has a unique fixed point $x^{*}\in X$ and for every $x\in X$ the sequence $\{T^{n}x\}_{n=1}^{\infty}$ converges to $x^{*}$.

Proof

Since $N(x,y)\leq M_{T}(x,y)$, so from (F1) and by taking $\psi=\tau$ in Theorem 2.5 the proof is complete. □

Theorem 2.8

[5]

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

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

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

Proof

Since $d(x,y)\leq M_{T}(x,y)$, from (F1) and by taking $\psi=\tau$ and $s=1$ in Theorem 2.2 the proof is complete. □

Example 2.9

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

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

Then $(X, d)$ is a b-metric space with coefficient $s=2$. But it is not a metric space since the triangle inequality is not satisfied. Let $T : X \to X$ be defined by

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

First observe that

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

Now we consider the following cases:

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

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

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

$$\begin{aligned}& d(Tx,Ty)=d(2,1)=1,\qquad d(x,y)=d(x,-1)=1, \qquad d(x,Tx)=d(x,0)=0 \vee1, \\& d(y,Ty)=d(-1,1)=4,\qquad \frac{d(x,Ty)+d(Tx,y)}{2}=\frac {d(x,1)+d(0,-1)}{2}=\frac{1}{2} \vee1, \\& \frac{d(T^{2}x,x)+d(T^{2}x,Ty)}{2}=\frac{d(0,x)+d(0,1)}{2}=\frac {1}{2}\vee1, \\& d(x,Ty)=d(x,1)= 1,\qquad d(Tx,y)=d(0,-1)=1, \\& d\bigl(T^{2}x,Tx\bigr)=d(0,0)=0,\qquad d \bigl(T^{2}x,y\bigr)=d(0,-1)=1, \\& d\bigl(T^{2}x,Ty\bigr)+d(x,Tx)=d(0,1)+d(x,0)=1\vee2 \\& d(Tx,y)+d(y,Ty)=d(0,-1)+d(-1,1)=5. \end{aligned}$$

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

$$\begin{aligned}& d(Tx,Ty)=d(-2,1)=1,\qquad d(x,y)=d(1,-1)=4,\qquad d(x,Tx)=d(1,-2)=1, \\& d(y,Ty)=d(-1,1)=4,\qquad \frac{d(x,Ty)+d(Tx,y)}{2}=\frac {d(1,1)+d(-2,-1)}{2}= \frac{1}{2}, \\& \frac{d(T^{2}x,x)+d(T^{2}x,Ty)}{2}=\frac {d(0,1)+d(0,1)}{2}=1, \\& d(x,Ty)=d(1,1)=0,\qquad d(Tx,y)=d(-2,-1)=1, \\& d\bigl(T^{2}x,Tx\bigr)=d(0,-2)=1,\qquad d\bigl(T^{2}x,y\bigr)=d(0,-1)=1, \\& d\bigl(T^{2}x,Ty\bigr)+d(x,Tx)=d(0,1)+d(1,-2)=2, \\& d(Tx,y)+d(y,Ty)=d(-2,-1)+d(-1,1)=5. \end{aligned}$$

In Case 1, we have

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

This proves that for all $F\in\mathcal{F}$, T is not an F-weak contraction, generalized F-contraction, and F-contraction. Hence Theorem 1.3, Theorem 1.5, and Theorem 2.7 are not applicable for this example. However, we see that, for all $x,y\in X$,

$$\frac{1}{2}d(x,Tx)< d(x,y),\qquad d(Tx,Ty)=1, \quad \mbox{and}\quad M_{T}(x,y)\geq2. $$

Since

$$\begin{aligned} \ln\bigl(d(Tx,Ty)\bigr)&\leq\ln\bigl(M_{T}(x,y)\bigr)+\ln\biggl( \frac{1}{2}\biggr) \\ &\leq\ln\bigl(M_{T}(x,y)\bigr)-\frac{68}{100}. \end{aligned}$$

So by taking $F(t)=\ln(t)$ and $\varphi(t)=\frac{1}{20}t$, we have

$$F\bigl( d(Tx,Ty)\bigr)\leq F\bigl( M_{T}(x,y)\bigr)-\varphi \bigl(M_{T}(x,y)\bigr). $$

Hence T satisfies the assumption of Theorem 2.2.