Fixed points of conditionally F-contractions in complete metric-like spaces

Abstract

In this paper, we introduce the notion of a conditionally F-contraction in the setting of complete metric-like spaces and we investigate the existence of fixed points of such mappings. Our results unify, extend, and improve several results in the literature.

1 Introduction and preliminaries

Recently, Wardowski [1] introduced the notion of a F-contraction mapping and investigated the existence of fixed points for such mappings. The results of Wardowski [1] extend and unify several fixed point results in the literature including the celebrated Banach contraction mapping principle.

In this paper, we present the notion of conditionally F-contractions of various types and we investigate the existence of a fixed point for such mappings in metric-like spaces. We also present some criteria for the uniqueness of a fixed point.

Throughout the paper, $\mathbb{N}$ and $\mathbb{N}_{0}$ denote the set of positive integers and the set of nonnegative integers. Similarly, let $\mathbb{R}$, $\mathbb{R}^{+}$ and $\mathbb{R}^{+}_{0}$ represent the set of reals, positive reals, and the set of nonnegative reals, respectively.

Definition 1.1

[1]

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

Definition 1.2

[1]

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

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

(1)

Remark 1.3

From (F1) and (1) it is easy to conclude that every F-contraction is necessarily continuous.

Recently, Piri and Kumam [2] extended the result of Wardowski [1] by replacing the condition (F3) in Definition 1.1 with the following one:

(F3′):

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

Let $\mathfrak{F}$ denote the family of all functions $F:\mathbb{R}_{+}\rightarrow \mathbb{R}$ which satisfy conditions (F1), (F2) and (F3′).

Under this new set-up, they proved a fixed point result that generalized the result of Wardowski [1].

Definition 1.4

[2]

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

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

Wardowski and Van Dung [3] introduced the notion of a F-weak contraction and proved a fixed point theorem for F-weak contractions.

Definition 1.5

[3]

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

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

(2)

Wardowski and Van Dung [3] gave an example to show that their result was a proper extension of results in the literature.

There are many papers in the literature that generalize the notion of metric spaces as well as the Banach contraction mapping principle (see [4–11] and the references therein). The notion of a metric-like space was introduced by Hitzler [12] and re-introduced by Amini-Harandi in [13].

Definition 1.6

(See [12])

Let X be a non-empty set. A mapping $d: X\times X\rightarrow \mathbb{R}^{+}_{0}$ is said to be a metric-like (dislocated) on X, if for all $x,y,z\in X$ the following conditions are satisfied:

(D1):

if $d(x,y)=0$ then $x=y$.

(D2):

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

(D3):

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

The pair $(X,d)$ is called a dislocated (metric-like) space.

Notice that if we replace the condition (D3) with

(D3^∗):

$d(x,y) \leq d(x,z) +d(z,y) -d(z,z)$

in Definition 1.6 then, $(X,d)$ turns to be a partial metric space (PMS). Fore more details see e.g. [14–18].

Remark 1.7

(See [13])

Every partial metric is metric-like (dislocate).

A sequence $\{x_{n}\}_{n=1}^{\infty}$, in a metric-like space $(X, d )$,

• (a) converges to $x\in X$ if $\lim_{n\rightarrow\infty} d(x_{n},x)=d(x,x)$,

• (b) is called Cauchy in $(X, d)$, if $\lim_{n,m\rightarrow\infty} d(x_{n},x_{m})$ exists and is finite.

A metric-like space $(X,d )$ is said to be complete if and only if every Cauchy sequence $\{x_{n}\}_{n=1}^{\infty}$ in X converges to $x \in X$ so that

$$\lim_{n,m\rightarrow\infty}d(x_{n},x_{m})=\lim _{n\rightarrow\infty} d(x_{n},x)= d(x,x). $$

We recall some basic definitions and crucial results on the topic. In this paper, we follow the notation of Amini-Harandi [13].

Definition 1.8

(See [13])

Let $(X,d)$ be a metric-like space and let U be a subset of X. We say U is a d-open subset of X, if for all $x\in X$ there exists $r>0$ such that $B_{d}(x,r)\subseteq U$. Also, $V\subseteq X$ is a d-closed subset of X if $(X\backslash V)$ is a d-open subset of X.

Lemma 1.9

(See [19])

Let $(X,d)$ be a metric-like space. Then:

• (A) if $d(x,y)=0$ then $d(x,x)=d(y,y)=0$;

• (B) if $\{x_{n}\}$ be a sequence such that $\lim_{n\to\infty}d(x_{n},x_{n+1})=0$, then we have,

  $$\lim_{n\to \infty}d(x_{n},x_{n})=\lim _{n\to\infty}d(x_{n+1},x_{n+1})=0; $$

• (C) if $x\neq y$ then $d(x,y)>0$;

• (D) $d(x,x)\leq\frac{2}{n}\sum_{i=1}^{i=n}d(x,x_{i})$ holds for all $x_{i},x\in X$ where $1\leq i\leq n$;

• (E) if $\{x_{n}\}$ is a sequence in a d-closed subset V of X with $x_{n}\to x$ as $n\to\infty$, then $x\in V$;

• (F) if $\{x_{n}\}$ is a sequence in X such that $x_{n}\to x$ as $n\to\infty$ and $d(x,x)=0$, then $\lim_{n\to\infty}d(x_{n},y)=d(x,y)$ for all $y\in X$.

Definition 1.10

Let $(X, d )$ and $(Y, \rho)$ be metric-like spaces and $\{x_{n}\}_{n=1}^{\infty}$ be a sequence in X such that $x_{n}\rightarrow x$. A mapping $f\colon X\to Y$ is said to be continuous at a point $x\in X$ if $f(x_{n}) \rightarrow f(x)$

2 Main results

We begin this section with the following definition.

Definition 2.1

Let $(X,d)$ be a metric-like space. A mapping $T:X\rightarrow X$ is said to be a conditionally F-contraction of type (A) if there exist $F\in\mathfrak{F}$ and $\tau>0$ such that, for all $x,y\in X$ with $d(Tx,Ty)>0$,

$$\begin{aligned} \frac{1}{2}d(x,Tx)< d(x,y)\quad\Rightarrow\quad\tau+F \bigl(d(Tx,Ty) \bigr)\leq F \bigl(M_{T}(x,y) \bigr), \end{aligned}$$

(3)

where

$$M_{T}(x,y)\max \biggl\{ d(x,y),d(x,Tx),d(y,Ty), \frac {d(x,Ty)+d(y,Tx)}{4} \biggr\} . $$

Theorem 2.2

Let $(X,d)$ is a complete metric-like space. If T is a conditionally F-contraction of type (A), then T has a fixed point $x^{*}\in X$.

Proof

Take $x \in X$ and construct a sequence $\{x_{n}\}$ as follows:

$$\begin{aligned} x_{n}=Tx_{n-1}=T^{n}x \quad\mbox{for all } n\in\mathbb{N} \mbox{ where } x_{0}=x. \end{aligned}$$

(4)

If there exists $n_{*} \in\mathbb{N}$ such that $d(x_{n_{*}},Tx_{{n_{*}}})=0$ then $x_{*}=x_{n_{*}}$ becomes a fixed point which completes the proof. Consequently, in the rest of the proof, we assume that, for every $n\in \mathbb{N}$,

$$\begin{aligned} 0< d(x_{n},Tx_{n}). \end{aligned}$$

(5)

Hence, from (5), we have

$$\begin{aligned} \frac{1}{2} d(x_{n},Tx_{n})< d(x_{n},Tx_{n})\quad \mbox{for all } n\in\mathbb{N}. \end{aligned}$$

(6)

Since T is conditionally F-contraction (note $d(Tx_{n},T(Tx_{n}))=d(x_{n+1}, Tx_{n+1})>0$), from the inequality (6), we have

$$\begin{aligned} \tau+F \bigl(d \bigl(Tx_{n},T^{2}x_{n} \bigr) \bigr)\leq{}& F \biggl(\max \biggl\{ d(x_{n},Tx_{n}),d(x_{n},Tx_{n}),d \bigl(Tx_{n},T^{2}x_{n} \bigr), \\ &{} \frac{d(x_{n},T^{2}x_{n})+d(Tx_{n},Tx_{n})}{4} \biggr\} \biggr) \\ \leq{}& F \biggl(\max \biggl\{ d(x_{n},Tx_{n}),d \bigl(Tx_{n},T^{2}x_{n} \bigr), \\ &{} \frac{d(x_{n},Tx_{n})+d(Tx_{n},T^{2}x_{n})+d(Tx_{n},x_{n})+d(x_{n},Tx_{n})}{4} \biggr\} \biggr) \\ ={}& F \biggl(\max \biggl\{ d(x_{n},Tx_{n}),d \bigl(Tx_{n},T^{2}x_{n} \bigr), \\ &{}\frac {3d(x_{n},Tx_{n})+d(Tx_{n},T^{2}x_{n})}{4} \biggr\} \biggr) \\ ={}& F \bigl(\max \bigl\{ d(x_{n},Tx_{n}),d \bigl(Tx_{n},T^{2}x_{n} \bigr) \bigr\} \bigr). \end{aligned}$$

(7)

If there exists $n\in \mathbb {N}$ such that $\max\{d(x_{n},Tx_{n}),d(Tx_{n},T^{2}x_{n})\}=d(Tx_{n},T^{2}x_{n})$, then (7) becomes

$$\tau+F \bigl(d \bigl(Tx_{n},T^{2}x_{n} \bigr) \bigr) \leq F \bigl(d \bigl(Tx_{n},T^{2}x_{n} \bigr) \bigr), $$

which is a contradiction. Thus, we conclude that

$$\max \bigl\{ d(x_{n},Tx_{n}),d \bigl(Tx_{n},T^{2}x_{n} \bigr) \bigr\} =d(x_{n},Tx_{n}), $$

for all $n\in \mathbb {N}$. Hence, the inequality (7) turns into

$$\begin{aligned} F \bigl(d \bigl(Tx_{n},T^{2}x_{n} \bigr) \bigr)\leq F \bigl(d(x_{n},Tx_{n}) \bigr)-\tau \quad \mbox{for all } n\in \mathbb {N}, \end{aligned}$$

(8)

which is equivalent to

$$F \bigl(d(x_{n+1},Tx_{n+1}) \bigr)\leq F \bigl(d(x_{n},Tx_{n}) \bigr)-\tau\quad \mbox{for all } n\in \mathbb {N}. $$

Iteratively, we find that

$$ \begin{aligned}[b] F \bigl(d(x_{n},Tx_{n}) \bigr)&\leq F \bigl(d(x_{n-1},Tx_{n-1}) \bigr)-\tau \\ &\leq F \bigl(d(x_{n-2},Tx_{n-2}) \bigr)-2\tau \\ &\leq F \bigl(d(x_{n-3},Tx_{n-3}) \bigr)-3\tau \\ &\vdots \\ &\leq F \bigl(d(x_{0},Tx_{0}) \bigr)-n\tau. \end{aligned} $$

(9)

From (9), we obtain $\lim_{m\rightarrow\infty} F(d(x_{n},Tx_{n}))=-\infty$, which together with (F2) gives

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

(10)

Now, we claim that

$$\begin{aligned} \lim_{n,m\rightarrow\infty} d(x_{n},x_{m})=0. \end{aligned}$$

(11)

Arguing by contradiction, we assume that there exist $\epsilon>0$ and sequences $\{p(n)\}_{n=1}^{\infty}$ and $\{q(n)\}_{n=1}^{\infty}$ of natural numbers such that

$$\begin{aligned} p(n)>q(n)>n,\qquad d(x_{p(n)},x_{q(n)})\geq\epsilon,\qquad d(x_{p(n)-1},x_{q(n)})< \epsilon \quad\mbox{for all } n \in \mathbb{N}. \end{aligned}$$

(12)

From the triangle inequality, we get

$$\begin{aligned} \epsilon&\leq d(x_{p(n)},x_{q(n)})\leq d(x_{p(n)},x_{p(n)-1})+d(x_{p(n)-1},x_{q(n)}) \\ &\leq d(x_{p(n)},x_{p(n)-1})+\epsilon \\ &=d(x_{p(n)-1},Tx_{p(n)-1})+\epsilon\quad \mbox{for all } n \in \mathbb{N}. \end{aligned}$$

(13)

Thus from (10), (13), and the sandwich theorem, we get

$$\begin{aligned} \lim_{n\rightarrow\infty}d(x_{p(n)},x_{q(n)})= \epsilon. \end{aligned}$$

(14)

Again by the triangle inequality, $\mbox{ for all } n \in \mathbb{N}$, we have the following two inequalities:

$$\begin{aligned} d(x_{p(n)},x_{q(n)})\leq d(x_{p(n)},x_{p(n)+1})+ d(x_{p(n)+1},x_{q(n)+1})+d(x_{q(n)+1},x_{q(n)}) \end{aligned}$$

(15)

and

$$\begin{aligned} d(x_{p(n)+1},x_{q(n)+1})\leq d(x_{p(n)+1},x_{p(n)})+ d(x_{p(n)},x_{q(n)})+d(x_{q(n)},x_{q(n)+1}). \end{aligned}$$

(16)

Letting $n \to\infty$ in the inequalities (15) and (16), using (10) and (14), we obtain

$$\begin{aligned} \lim_{n\rightarrow\infty}d(x_{p(n)+1},x_{q(n)+1})= \epsilon. \end{aligned}$$

(17)

From (10) and (12), there exists $N_{2} \in\mathbb{N}$ such that

$$\begin{aligned} \frac{1}{2}d(x_{p(n)},Tx_{p(n)})< \frac{\epsilon}{2}< d(x_{p(n)},x_{q(n)}),\quad \forall n>N_{2}. \end{aligned}$$

Note from (17) for n large enough (i.e. $n\geq N_{3}\geq N_{2}$ say) we have $d(Tx_{p(n)},Tx_{q(n)})=d(x_{p(n)+1},x_{q(n)+1})>0$. Since T is a conditionally F-contractive mapping of type (A), we have (with $n \geq N_{3}$)

$$\begin{aligned} &\tau+F \bigl(d(Tx_{p(n)},Tx_{q(n)}) \bigr) \\ & \quad\leq F \biggl(\max \biggl\{ d(x_{p(n)},x_{q(n)}),d(x_{p(n)},Tx_{p(n)}),d(x_{q(n)},Tx_{q(n)}), \\ &\qquad{} \frac{d(x_{p(n)},Tx_{q(n)})+d(x_{q(n)},Tx_{p(n)})}{4} \biggr\} \biggr) \\ & \quad\leq F \biggl(\max \biggl\{ d(x_{p(n)},x_{q(n)}),d(x_{p(n)},Tx_{p(n)}),d(x_{q(n)},Tx_{q(n)}), \\ & \qquad{}\frac {d(x_{p(n)},Tx_{p(n)})+d(x_{q(n)},Tx_{q(n)})+2d(x_{q(n)},x_{p(n)})}{4} \biggr\} \biggr) \\ & \quad\leq F \bigl( \max\{ d(x_{p(n)},x_{q(n)}),d(x_{p(n)},Tx_{p(n)}),d(x_{q(n)},Tx_{q(n)}), \\ & \qquad{}\max \bigl\{ d(x_{p(n)},x_{q(n)}),d(x_{p(n)},Tx_{p(n)}),d(x_{q(n)},Tx_{q(n)}) \bigr\} \bigr) \\ & \quad=F(\max \bigl\{ d(x_{p(n)},x_{q(n)}),d(x_{p(n)},Tx_{p(n)}),d(x_{q(n)},Tx_{q(n)}) \bigr\} . \end{aligned}$$

(18)

Letting $n \to\infty$ in the inequality above and using (10), (14), and (F3′) we obtain

$$\begin{aligned} \tau+ F(\varepsilon) \leq F(\varepsilon), \end{aligned}$$

(19)

a contradiction since $\tau>0$. Hence

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

Therefore, we conclude that $\lbrace x_{n}\rbrace_{n=1}^{\infty}$ is a Cauchy sequence in X. Now $(X, d)$ is a complete metric-like space, so there exists $x^{*} \in X$ such that

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

(20)

Now note

$$\begin{aligned} d \bigl(x^{*},Tx^{*} \bigr)&\leq d \bigl(x^{*},x_{n+1} \bigr)+d \bigl(x_{n+1},Tx^{*} \bigr) \\ &\leq d \bigl(x^{*},x_{n+1} \bigr)+d \bigl(x_{n+1},x^{*} \bigr)+d \bigl(x^{*},Tx^{*} \bigr) \\ &=2d \bigl(x^{*},x_{n+1} \bigr)+d \bigl(x^{*},Tx^{*} \bigr). \end{aligned}$$

Thus from (20) and the sandwich theorem, we get

$$\begin{aligned} \lim_{n\rightarrow\infty} \bigl[d \bigl(x^{*},x_{n+1} \bigr)+d \bigl(x_{n+1},Tx^{*} \bigr) \bigr]=d \bigl(x^{*},Tx^{*} \bigr). \end{aligned}$$

Now this and (20) yield

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

(21)

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

$$\begin{aligned} \frac{1}{2}d(x_{n},Tx_{n})< d \bigl(x_{n},x^{*} \bigr) \quad\mbox{or}\quad \frac {1}{2}d \bigl(Tx_{n},T^{2}x_{n} \bigr)< d \bigl(Tx_{n},x^{*} \bigr),\quad \forall n\in\mathbb{N}. \end{aligned}$$

(22)

Arguing by contradiction, we assume that there exists $m\in \mathbb{N}$ such that

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

(23)

Now from (8) and (F₁), we have

$$\begin{aligned} d \bigl(Tx_{m},T^{2}x_{m} \bigr)< d(x_{m},Tx_{m}). \end{aligned}$$

(24)

It follows from (23) and (24) that

$$\begin{aligned} d(x_{m},Tx_{m})&\leq d \bigl(x_{m},x^{*} \bigr)+d \bigl(x^{*},Tx_{m} \bigr) \\ &\leq\frac{1}{2}d(x_{m},Tx_{m})+\frac{1}{2}d \bigl(Tx_{m},T^{2}x_{m} \bigr) \\ &< \frac{1}{2}d(x_{m},Tx_{m})+\frac{1}{2}d(x_{m},Tx_{m}) \\ &=d(x_{m},Tx_{m}), \end{aligned}$$

which is a contradiction. Hence (22) holds.

Suppose, now, part (I) of (22) is satisfied and $d(x^{*},Tx^{*})>0$. Note from (21) there exists $N_{1} \in \mathbb {N}$ such that $d(Tx_{n},Tx^{*})=d(x_{n+1},Tx^{*})>0$ for $n \geq N_{1}$. Then from our assumption (with $n \geq N_{1}$) we have

$$\begin{aligned} \tau+F \bigl(d \bigl(x_{n+1},Tx^{*} \bigr) \bigr)={}& \tau+F \bigl(d \bigl(Tx_{n},Tx^{*} \bigr) \bigr) \\ \leq{}& F\biggl(\max \biggl\{ d \bigl(x_{n},x^{*} \bigr),d(x_{n},Tx_{n}),d \bigl(x^{*},Tx^{*} \bigr), \\ &{} \frac{d(x_{n},Tx^{*})+d(x^{*},Tx_{n})}{4} \biggr\} \biggr) \\ \leq{}& F\biggl(\max \biggl\{ d \bigl(x_{n},x^{*} \bigr),d(x_{n},Tx_{n}),d \bigl(x^{*},Tx^{*} \bigr), \\ &{} \frac{2d(x_{n},x^{*})+d(x^{*},Tx^{*})+d(x_{n},Tx_{n})}{4} \biggr\} \biggr). \end{aligned}$$

(25)

From (10) and (21), there exists $N_{3}\in \mathbb {N}$ (with $N_{3} \geq N_{1}$) such that for all $n\geq N_{3}$

$$\begin{aligned} &\max \biggl\{ d \bigl(x_{n},x^{*} \bigr),d(x_{n},Tx_{n}),d \bigl(x^{*},Tx^{*} \bigr), \frac {2d(x_{n},x^{*})+d(x^{*},Tx^{*})+d(x_{n},Tx_{n})}{4} \biggr\} \\ &\quad=d \bigl(x^{*},Tx^{*} \bigr). \end{aligned}$$

Now from (25), we get

$$\begin{aligned} \tau+F \bigl(d \bigl(x_{n+1},Tx^{*} \bigr) \bigr)\leq F \bigl(d \bigl(x^{*},Tx^{*} \bigr) \bigr), \quad\forall n\geq N_{3}. \end{aligned}$$

(26)

From (F3′) and (21), by taking the limit as $n\rightarrow\infty$ in (26), we obtain

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

which is a contradiction.

Now suppose part (II) of (22) is true, and $d(x^{*},Tx^{*})>0$. Note from (21) there exists $N_{2} \in \mathbb {N}$ such that $d(T(Tx_{n}),Tx^{*})=d(x_{n+2},Tx^{*})>0$ for $n \geq N_{2}$. Then from our assumption (with $n \geq N_{2}$) we have

$$\begin{aligned} \tau+F \bigl(d \bigl(x_{n+2},Tx^{*} \bigr) \bigr)={}& \tau+F \bigl(d \bigl(T^{2}x_{n},Tx^{*} \bigr) \bigr) \\ \leq{}& F\biggl(\max \biggl\{ d \bigl(Tx_{n},x^{*} \bigr),d \bigl(Tx_{n},T^{2}x_{n} \bigr),d \bigl(x^{*},Tx^{*} \bigr), \\ &{} \frac{d(Tx_{n},Tx^{*})+d(x^{*},T^{2}x_{n})}{4} \biggr\} \biggr) \\ \leq{}& F\biggl(\max \biggl\{ d \bigl(Tx_{n},x^{*} \bigr),d \bigl(Tx_{n},T^{2}x_{n} \bigr),d \bigl(x^{*},Tx^{*} \bigr), \\ &{} \frac{2d(Tx_{n},x^{*})+d(x^{*},Tx^{*})+d(Tx_{n},T^{2}x_{n})}{4} \biggr\} \biggr) \\ ={}& F\biggl(\max \biggl\{ d \bigl(x_{n+1},x^{*} \bigr),d(x_{n+1},Tx_{n+1}),d \bigl(x^{*},Tx^{*} \bigr), \\ &{} \frac{2d(x_{n+1},x^{*})+d(x^{*},Tx^{*})+d(x_{n+1},Tx_{n+1})}{4} \biggr\} \biggr). \end{aligned}$$

(27)

From (10) and (21), there exists $N_{4}\in \mathbb {N}$ (with $N_{4} \geq N_{2}$) such that for all $n\geq N_{4}$

$$\begin{aligned} &\max \biggl\{ d \bigl(x_{n+1},x^{*} \bigr),d(x_{n+1},Tx_{n+1}),d \bigl(x^{*},Tx^{*} \bigr), \frac{2d(x_{n+1},x^{*})+d(x^{*},Tx^{*})+d(x_{n+1},Tx_{n+1})}{4} \biggr\} \\ &\quad =d \bigl(x^{*},Tx^{*} \bigr). \end{aligned}$$

From (27), we get

$$\begin{aligned} \tau+F \bigl(d \bigl(x_{n+2},Tx^{*} \bigr) \bigr)\leq F \bigl(d \bigl(x^{*},Tx^{*} \bigr) \bigr), \quad\forall n\geq N_{4}. \end{aligned}$$

(28)

From (F3) and (21), taking the limit as $n\rightarrow\infty$ in (28), we obtain

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

which is a contradiction. Hence, we conclude that $x^{*}$ is a fixed point of T. □

Definition 2.3

Let $(X,d)$ be a metric-like space. A mapping $T:X\rightarrow X$ is said to be a conditionally F-contraction of type (B) if there exist $F\in\mathfrak{F}$ and $\tau>0$ such that, for all $x,y\in X$ with $d(Tx,Ty)>0$,

$$\begin{aligned} \frac{1}{2}d(x,Tx)< d(x,y)\quad\Rightarrow\quad\tau+F \bigl(d(Tx,Ty) \bigr)&\leq F \bigl(\max \bigl\{ d(x,y),d(x,Tx),d(y,Ty) \bigr\} \bigr). \end{aligned}$$

(29)

Definition 2.4

Let $(X,d)$ be a metric-like space. A mapping $T:X\rightarrow X$ is said to be a conditionally F-contraction of type (C) if there exist $F\in\mathfrak{F}$ and $\tau>0$ such that, for all $x,y\in X$ with $d(Tx,Ty)>0$,

$$\begin{aligned} \frac{1}{2}d(x,Tx)< d(x,y)\quad\Rightarrow \quad\tau+F \bigl(d(Tx,Ty) \bigr)&\leq F \bigl(d(x,y) \bigr). \end{aligned}$$

(30)

Theorem 2.5

Let $(X,d)$ be a complete metric-like space. If T is a conditionally F-contraction of type (B), then T has a fixed point $x^{*}\in X$.

Proof

Following the proof in Theorem 2.2, we easily conclude the result. □

Theorem 2.6

Let $(X,d)$ be a complete metric-like space. If T is a conditionally F-contraction of type (C), then T has a fixed point $x^{*}\in X$.

Proof

It can easily be derived by following the proof in Theorem 2.2. □

Next we consider an example to illustrate our main result. We consider a mapping T which is not continuous, so not an F-contraction but it is a conditionally F-contraction of type (C).

Example 2.7

Consider $X=\{0,1,2\}$. Let $d:X\times X\to[0,\infty)$ be a mapping defined by

$$\begin{aligned}& d(0,0)=d(1,1)=0,\qquad d(2,2)=5/2, \\& d(0,2)=d(2,0)=2,\qquad d(1,2)=d(2,1)=3, \\& d(0,1)=d(1,0)=3/2. \end{aligned}$$

It is clear that d is a metric-like. Note that $d(2,2)\neq0$, so d is not a metric. Clearly, $(X,d)$ is a complete metric-like space. Let $T:X\to X$ be given by

$$T0=0=T1 \quad\mbox{and}\quad T2=1. $$

Suppose that $F(\alpha)=\frac{-1}{\alpha}+\alpha\in\mathfrak{F}$ and $\tau\in(0,1/2)$. Since T is not continuous, T is not a F-contraction by Remark 1.3.

We will consider the inequality

$$ \frac{1}{2}d(x,Tx)< d(x,y), $$

(31)

where $x, y\in X$ with $d(Tx,Ty)>0$ and the inequality

$$ \tau+F \bigl(d(Tx,Ty) \bigr)\leq F \bigl(d(x,y) \bigr) $$

(32)

for those $x,y \in X$ with $d(Tx,Ty)>0$ which satisfy (31).

Case 1: Let $x=0$. Now $d(T0,T0)=d(T0,T1)= d(0,0)=0$ so we need only consider $y=2$ in (31) and (32). Now (31) is true since

$$\frac{1}{2} d(0,T0)=0< d(0,2)=2. $$

Also note

$$d(T0,T2)= d(0,1)=\frac{3}{2}< 2=d(0,2). $$

Now inequality (32) is satisfied since

$$\begin{aligned} \tau+F \bigl(d(T0,T2) \bigr)&=\tau-\frac{1}{d(T0,T2)}+d(T0,T2) \\ &\leq\tau-\frac{1}{d(0,2)}+\frac{3}{2} \leq\frac{1}{2} - \frac {1}{d(0,2)}+ \frac{3}{2} \\ &= -\frac{1}{ d(0,2)}+2 = -\frac{1}{ d(0,2)}+d(0,2)=F \bigl(d(0,2) \bigr). \end{aligned}$$

Case 2: Let $x=1$. Now $d(T0,T1)=d(T1,T1)= d(0,0)=0$ so we need only consider $y=2$ in (31) and (32). Now (31) is true since

$$\frac{1}{2} d(1,T1)=0< d(1,2)=3. $$

Also note

$$d(T1,T2)= d(0,1)=\frac{3}{2}< 3=d(1,2). $$

Now inequality (32) is satisfied since

$$\begin{aligned} \tau+F \bigl(d(T1,T2) \bigr)&=\tau-\frac{1}{d(T1,T2)}+d(T1,T2) \\ &\leq\tau-\frac{1}{d(1,2)}+\frac{3}{2} \leq\frac{1}{2} - \frac{1}{d(1,2)}+ \frac{3}{2} = -\frac{1}{ d(1,2)}+2 \\ & \leq-\frac{1}{ d(1,2)}+3= -\frac{1}{ d(1,2)}+d(1,2)=F \bigl(d(1,2) \bigr). \end{aligned}$$

Case 3: Let $x=2$. Now $d(T2,T2)=0$ so we need only consider the case $y\in\{0,1\}$. Note $d(T2,T1)=d(1,0)>0$, $d(T2,T0)=d(1,0)>0$, and also note

$$\frac{1}{2}d(2,T2)=\frac{1}{2}d(1, 2)=\frac{3}{2} $$

and

$$d(2,0)=2,\qquad d(2,1)=3, $$

so (31) holds.

Note

$$d(T2,T0)= d(1,0)= \frac{3}{2} < 2=d(2,0) $$

and

$$d(T2,T1)= d(1,0)= \frac{3}{2} < 3=d(2,1), $$

so

$$\tau+F \bigl(d(T2,T0) \bigr) \leq F \bigl(d(2,0) \bigr) $$

follows as in Case 1 and

$$\tau+F \bigl(d(T2,T1) \bigr) \leq F \bigl(d(2,1) \bigr) $$

follows as in Case 2.

Hence T is a conditionally F-contraction of type (C). It is clear that 0 is the fixed point of T.

3 Consequences

In [14, 20], Matthews introduced the notion of partial metric, a generalization of a metric, as a part of the study of denotational semantics of dataflow networks.

Definition 3.1

(See [14])

Let X be a non-empty set. A mapping $p \colon X \times X\to \mathbb{R}^{+}_{0}$ is said to be a partial metric on X if for all $x, y, z \in X$ the following conditions are satisfied:

(p₁):

$x=y$ if and only if $p(x,x)=p(x,y)=p(y,y)$;

(p₂):

$p(x, x) \leq p(x,y)$;

(p₃):

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

(p₄):

$p(x, z)\leq p(x, y) + p(y, z)-p(y,y)$.

In this case, the pair $(X, p)$ is called a partial metric space (PMS).

Notice that the function $d_{p}:X\times X\rightarrow{\mathbb{R}}^{+}$ defined by $d_{p}(x,y)=2p(x,y)-p(x,x)-p(y,y)$ satisfies the conditions of a metric on X. Each partial metric p on X generates a $T_{0}$ topology $\tau_{p}$ on X, whose base is a family of open p-balls $\{ B_{p}(x,\varepsilon): x\in X, \varepsilon>0 \}$ where $B_{p}(x,\varepsilon)= \{ y\in X:p(x,y)\leq p(x,x)+\varepsilon\}$ for all $x\in X$ and $\varepsilon>0$. Consequently, it is easy to consider several topological concepts. A sequence $\{x_{n} \}$ in the PMS $(X,p)$ converges to the limit x if $p(x,x)=\lim_{n\rightarrow\infty} p(x,x_{n})$ and is said to be a Cauchy sequence if $\lim_{n,m\rightarrow\infty} p(x_{n},x_{m})$ exists and is finite. A PMS $(X,p)$ is called complete if every Cauchy sequence $\{x_{n} \}$ in X converges with respect to $\tau_{p}$, to a point $x\in X$ such that $p(x,x)=\lim_{n,m\rightarrow\infty} p(x_{n},x_{m})$. For more details, see e.g. [14, 20–40] and the related references therein.

Lemma 3.2

(See e.g. [21, 22])

Let $(X,p)$ be a complete PMS. Then

• (A) If $p(x,y)=0$ then $x=y$.

• (B) If $x\neq y$, then $p(x,y)>0$.

• (C) A sequence $\{x_{n}\}$ is a Cauchy sequence in the PMS $(X,p)$ if and only if it is a Cauchy sequence in the metric space $(X,d_{p})$.

• (D) A PMS $(X,p)$ is complete if and only if the metric space $(X,d_{p})$ is complete. Moreover,

  $$ \lim_{n\rightarrow\infty}d_{p}(x,x_{n})=0 \quad\Leftrightarrow\quad p(x,x) =\lim_{n\rightarrow\infty} p(x,x_{n})=\lim _{n,m\rightarrow\infty} p(x_{n},x_{m}). $$

  (33)

• (E) Assume $x_{n}\rightarrow z$ as $n\rightarrow\infty$ in a PMS $(X,p)$ such that $p(z,z)=0$. Then $\lim_{n\rightarrow\infty} p(x_{n},y)=p(z,y)$ for every $y \in X$.

Now we derive the analog of Theorem 2.2 in the context of partial metric spaces. In fact, in the following theorem we conclude not only the existence of a fixed point of the given mapping but also the uniqueness.

Theorem 3.3

Let $(X,p)$ be a complete partial metric space and let $T:X\rightarrow X$ be a self-mapping. Suppose that there exist $F\in\mathfrak{F}$ and $\tau>0$ such that, for all $x,y\in X$ with $p(Tx,Ty)>0$,

$$\begin{aligned} \begin{aligned} &\frac{1}{2}p(x,Tx)< p(x,y)\\ &\quad\Rightarrow\quad\tau+F \bigl(p(Tx,Ty) \bigr)\leq F \biggl(\max \biggl\{ p(x,y),p(x,Tx),p(y,Ty), \frac{p(x,Ty)+p(y,Tx)}{2} \biggr\} \biggr). \end{aligned} \end{aligned}$$

(34)

Then T has a unique fixed point $x^{*}\in X$.

Proof

Since every partial metric space is a metric-like space, we obtain the proof by following the proof in Theorem 2.2. Note that the expression $\frac{p(x,Ty)+p(y,Tx)}{4}$ in the inequality (3) is replaced by $\frac{p(x,Ty)+p(y,Tx)}{2}$ in the inequality (34). This difference arises due to assumptions (p₂) and (p₄) of partial metric spaces. Hence, taking (p₂) and (p₄) into account, following the proof in Theorem 2.2 yields the existence of a fixed point ($x^{*}\in X$) of T.

We now show the uniqueness of the fixed point of T. Suppose there is another fixed point $y^{*}\in X$ of T, such that $x^{*}\neq y^{*}$. Thus from Lemma 3.2, we have $p(x^{*},y^{*})>0$. From (p₂), we have

$$\frac{1}{2}p \bigl(x^{*},Tx^{*} \bigr)=\frac{1}{2}p \bigl(x^{*},x^{*} \bigr)< p \bigl(x^{*},x^{*} \bigr)\leq p \bigl(x^{*},y^{*} \bigr). $$

Thus, from (p₂), we obtain (note $p(Tx^{*},Ty^{*})=p(x^{*},y^{*})>0$)

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

This is a contradiction, and hence $x^{*}=y^{*}$. □

The following two theorems can be obtained easily by repeating the steps in the proof of Theorem 3.3.

Theorem 3.4

Let $(X,p)$ be a complete partial metric space and let $T:X\rightarrow X$ be a self-mapping. Suppose that there exist $F\in\mathfrak{F}$ and $\tau>0$ such that, for all $x,y\in X$ with $p(Tx,Ty)>0$,

$$\begin{aligned} \frac{1}{2}p(x,Tx)< p(x,y)\quad\Rightarrow\quad\tau+F \bigl(p(Tx,Ty) \bigr)\leq F \bigl(\max \bigl\{ p(x,y),p(x,Tx),p(y,Ty) \bigr\} \bigr). \end{aligned}$$

(35)

Then T has a unique fixed point $x^{*}\in X$.

Theorem 3.5

Let $(X,p)$ be a complete partial metric space and let $T:X\rightarrow X$ be a self-mapping. Suppose that there exist $F\in\mathfrak{F}$ and $\tau>0$ such that, for all $x,y\in X$ with $p(Tx,Ty)>0$,

$$\begin{aligned} \frac{1}{2}p(x,Tx)< p(x,y)\quad\Rightarrow\quad\tau+F \bigl(p(Tx,Ty) \bigr)\leq F \bigl(p(x,y) \bigr). \end{aligned}$$

(36)

Then T has a unique fixed point $x^{*}\in X$.

Remark 3.6

On can also easily conclude that the analog of Theorem 3.3-Theorem 3.5 in the context of metric spaces since each metric space is a partial metric space.