1 Introduction and preliminaries

In the nature of mathematics, there is the purpose of generalizing, expanding, and obtaining the most general forms of existing concepts and results. The concept of metric, which is the most fundamental and solid basis of the analysis study, has been constantly expanded and generalized with this motivation. Examples of the new metrics that have been put forward for this purpose can be counted as quasi-metric, b-metric, partial-metric, symmetric, D-metric, modular metric, fuzzy metric, soft-metric, G-metric, and so on. On the other hand, it was understood that not all of these newly defined metrics provide a new and original structure. For instance, G-metric can be reduced to semi-metric or cone metric to a standard metric. More examples can be given, but here we stop to focus on the main motivation. Two of the new and original generalizations of metric notions are b-metrics [1–16] and dislocated metrics [17–21]. Very recently, these two notions have emerged under the name of dislocated b-metric [22, 23].

Metric fixed point theory is a field of study that needs an abstract metric framework (see, for instance, [24–27]). Very recently Proinov [28] proved a fixed point theorem that not only unifies but also generalizes a number of well-known results in the framework of a standard metric space. In particular, he proved that Wardowski [29] and Jleli and Samet [30] results are not only equivalent to each other, but also they are a special case of one of the main results of [28].

In this paper, we improve the Proinov type contractions by involving certain rational expression to the corresponding contraction thought by Proinov [28]. After then, we prove fixed point theorems for these modified Proinov contractions in the framework of dislocated b-metrics. We bring forward illustrative examples to show the validity of the main results.

Let S be a nonempty set and \(\mathbb{N}=\{1,2,3,\ldots \}\). Some examples of rational contractivity conditions are shown in the following results (see also [31]).

Theorem 1

([32])

Let be a complete metric space and be a mapping such that there exist with such that

(1)

for all \(\mathsf {v},\mathsf {w}\in \mathsf {S}\). Then has a unique fixed point \(\mathsf {x}\in \mathsf {S}\), and the sequence converges to the fixed point x for all \(\mathsf {v}\in \mathsf {S}\).

Theorem 2

([33])

Let be a complete metric space and be a continuous mapping. If there exist with such that

(2)

for all distinct \(\mathsf {v},\mathsf {w}\in \mathsf {S}\), then possesses a unique fixed point in S.

Theorem 3

([28])

Let be a metric space and be a mapping such that

for all \(\mathsf {v}, \mathsf {w}\in \mathsf {S}\) with , where the functions \(\Psi, \Phi:(0,\infty )\rightarrow \mathbb{R}\) are such that the following conditions are satisfied:

  1. 1.

    Ψ is nondecreasing;

  2. 2.

    \(\Phi (\theta )<\Psi (\theta )\) for any \(\theta >0\);

  3. 3.

    \(\limsup_{\theta \rightarrow \theta _{0}+}\Phi ( \theta )<\Psi (\theta _{0}+)\) for any \(\theta _{0}>0\).

Then admits a unique fixed point.

Definition 4

([34])

A function \(\mathsf{d}_{l}:\mathsf {S}\times \mathsf{S}\rightarrow [ 0,\infty )\) is a dislocated-metric on S if it satisfies the conditions:

.:

\(\mathsf{d}_{l}(\mathsf {v},\mathsf {w})=0 \Rightarrow \mathsf {v}= \mathsf {w}\);

.:

symmetry: \(\mathsf{d}_{l} (\mathsf {w},\mathsf {v})=\mathsf{d}_{l}(\mathsf {v},\mathsf {w})\);

.:

the triangle inequality

$$\begin{aligned} \mathsf{d}_{l}(\mathsf{u},\mathsf{w} )\leq \mathsf{d}_{l} ( \mathsf{u}, \mathsf{v})+\mathsf{d}_{l}(\mathsf{v},\mathsf{w} ) \end{aligned}$$

for all \(\mathsf{u},\mathsf{v},\mathsf{w} \in \mathsf {S}\). In this case, the pair is a dislocated-metric space (shortly -MS).

Definition 5

([35])

Let \(\mathsf{s}\in [1,\infty )\) be a real number. A function \(\mathsf{b}:\mathsf {S}\times \mathsf{S}\rightarrow [ 0,\infty )\) is a b-metric on S if it satisfies the conditions:

\(\mathsf {b}_{1}\).:

\(\mathsf{b}(\mathsf {v},\mathsf {w})=0 \Leftrightarrow \mathsf {v}=\mathsf {w}\),

\(\mathsf {b}_{2}\).:

symmetry: \(\mathsf{b}(\mathsf {w},\mathsf {v})=\mathsf{b}(\mathsf {v},\mathsf {w})\)

\(\mathsf {b}_{3}\).:

the generalized version of the triangle inequality involving the number s

$$\begin{aligned} \mathsf{b}(\mathsf{u},\mathsf{w} )\leq \mathsf{s } \bigl[ \mathsf{ b}( \mathsf{u},\mathsf{v})+\mathsf{b}(\mathsf{v},\mathsf{w} )\mathsf{ } \bigr] \quad\text{for all }\mathsf{u},\mathsf{v},\mathsf{w} \in \mathsf {S}. \end{aligned}$$

In this case, the tripled \((\mathsf {S}, \mathsf {b}, \mathsf {s})\) forms a b-metric space (shortly b-MS).

Obviously, for \(\mathsf{s}=1\), we find the notion of metric space.

Definition 6

([36])

Let \(\mathsf{s}\in [1,\infty )\) be a real number(given). A function is a dislocated b-metric on S if it satisfies the conditions:

.:

;

.:

;

.:

for all \(\mathsf{u},\mathsf{v},\mathsf{w} \in \mathsf {S}\).

In this case, is a dislocated b-metric space (shortly -MS).

We mention that, when \(\mathsf {s}=1\), a -MS becomes a -MS.

Definition 7

([36])

A sequence \(\{ \mathsf {v}_{n} \} \) on a -MS is said to be:

  • -convergent to a point \(\mathsf {v}\in \mathsf {S}\) ⇔ ;

  • -Cauchy if and only if exists and tends to be finite.

Proposition 8

([36])

In a -MS the limit of a convergent sequence is unique.

Proposition 9

([36])

In a -MS every convergent sequence is -Cauchy.

In case every -Cauchy sequence is -convergent, we say that the space is a complete -MS. The next lemma will be useful in the sequel.

Lemma 10

Let a -MS , a mapping , and \(\mathsf {v}_{0}\) be arbitrary, but fixed point in S. If there exists \(\mathcal{C}\in [ 0,1 ) \) such that

(3)

for every \(n\in \mathbb{N}\), then the sequence is a -Cauchy sequence.

Proof

Let \(\mathsf {v}_{0}\) be an arbitrary point in S and the sequence \(\{ \mathsf {v}_{n} \} \) with

for \(n\in \mathbb{N}\cup \{ 0 \} \). Thus, by (3), we have

(4)

We split the proof in two cases, namely \(\mathsf {s}=1\) and \(\mathsf {s}>1\).

  1. 1.

    For \(\mathsf {s}=1\), becomes a dislocated metric and by ., for \(n< p\), we have

    Therefore, , that is, the sequence is Cauchy.

  2. 2.

    For \(\mathsf {s}>1\), we distinguish two sub-cases:

    1. (a)

      If \(\mathcal{C}\in [0,\frac{1}{\mathsf {s}})\), by and taking into account (4), we get

      that is, is -Cauchy.

    2. (b)

      If \(\mathcal{C}\in [\frac{1}{\mathsf {s}},1)\), then \(\mathcal{C}^{n}\rightarrow 0\), and we can find \(l\in \mathbb{N}\) such that \(\mathcal{C}^{n}<\frac{1}{\mathsf {s}}\). Therefore, by (a), the sequence is -Cauchy. But we have

      $$\begin{aligned} \{ \mathsf {v}_{n} \} = \{ \mathsf {v}_{0}, \mathsf {v}_{1},\ldots, \mathsf {v}_{l-1} \} \cup \{ \mathsf {v}_{l}, \mathsf {v}_{l+1},\ldots, \mathsf {v}_{l+n},\ldots \}, \end{aligned}$$

      and then the sequence is -Cauchy.

 □

2 Main results

Henceforth, we use the following notations:

$$\begin{aligned} \Theta = \bigl\{ \Phi, \Psi:(0,\infty )\rightarrow \mathbb{R} | \Phi (\theta )< \Psi (\theta ) \text{{ for every }}\theta \in (0,+ \infty ) \bigr\} \end{aligned}$$

and, respectively,

Let the functions \(R_{1}, R_{2}:\mathsf {S}\times \mathsf {S}\rightarrow [0,\infty )\) be defined by

where are nonnegative real numbers.

Theorem 11

Let be a complete -ms, \(\Psi, \Phi \in \Theta \), a number \(\alpha \in [1,\infty )\), and two continuous mappings such that, for every distinct \(\mathsf {v}, \mathsf {w}\in \mathsf {S}\) with , the following inequality

(5)

holds. Assume that:

\((\beta _{1})\):

and ;

\((\beta _{2})\):

Ψ is nondecreasing.

Then . Moreover, if , then the set has exactly one element.

Proof

For an arbitrary (but fixed) point \(\mathsf {v}_{0}\in \mathsf {S}\), let \(\{ \mathsf {v}_{n} \} \) be the sequence defined as follows:

(6)

for all \(n\in \mathbb{N}_{0}\). First of all, we claim that \(\mathsf {v}_{n}\neq \mathsf {v}_{n+1}\) for any \(n\in \mathbb{N}_{0}\). Indeed, if we can find \(l_{0}\in \mathbb{N}\) such that \(\mathsf {v}_{l_{0}}=\mathsf {v}_{l_{0}+1}=\mathsf {v}_{l_{0}+2}=\mathsf {x}\), then .

Under this assumption, and letting \(\mathsf {v}=\mathsf {v}_{2n}\) and \(\mathsf {w}=\mathsf {v}_{2n+1}\) in (5), because the functions \(\Psi, \Phi \) belong to Θ, we have

Taking (\(\beta _{1}\)) into account, we get

or

(7)

where , holds due to the first assumption in \((\beta _{1})\).

In the same way, replacing in (5) v with \(\mathsf {v}_{2n-1}\) and w with \(\mathsf {v}_{2n}\), and keeping in mind , we have

(8)

which leads us to

(9)

Consequently, (7) and (9) show us that

(10)

for any \(n\in \mathbb{N}\), where . By Lemma 10 it follows that \(\{ \mathsf {v}_{n} \} \) is a Cauchy sequence. Thus, exists and is finite. Moreover, since the -ms is complete, we get that there exists \(\mathsf {x}\in \mathsf {S}\) such that \(\lim_{n\rightarrow \infty }\mathsf {v}_{n}=\mathsf {x}\) and

(11)

Since the mappings and are supposed to be continuous, we have

that is, . If we suppose that there exist such that \(\mathsf {x}\neq \mathsf {y}\), by (5) and since \(\Psi, \Phi \in \Theta \), we have

where

However, applying and taking into account ,

Moreover, by \((\beta _{2})\) we get

which is a contradiction. Therefore, and from it follows that \(\mathsf {x}=\mathsf {y}\), that is, the set has exactly one element. □

Corollary 12

Let be a complete -ms, \(\Psi, \Phi \in \Theta \), a number \(\alpha \in [1,\infty )\), and a continuous mapping such that, for every distinct \(\mathsf {v}, \mathsf {w}\in \mathsf {S}\) with , the following inequality

(12)

holds, where for nonnegative real numbers,

for all \(\mathsf {v},\mathsf {w}\in \mathsf {S}, \mathsf {v}\neq \mathsf {w}\). Assume that:

\((\beta _{1})\):

and ;

\((\beta _{2})\):

Ψ is nondecreasing.

Then . Moreover, if , then the set has exactly one element.

Proof

Let in Theorem 11. □

Theorem 13

Let be a complete -ms, \(\Psi, \Phi \in \Theta \), a number \(\alpha \in [1,\infty )\), and two mappings such that, for every distinct \(\mathsf {v}, \mathsf {w}\in \mathsf {S}\) with , the following inequality

(13)

holds. Assume that:

\((\beta _{1})\):

, , , ;

\((\beta _{2})\):

Ψ is nondecreasing.

Then . Moreover, if , then the set has exactly one element.

Proof

Let \(\mathsf {v}_{0}\in \mathsf {S}\) be a chosen point and \(\{ \mathsf {v}_{n} \} \) be the sequence defined by (6) in the proof of Theorem 11. Thus, following the same arguments, we can assume that and from (13) we get

Since by \((\beta _{2})\) Ψ is nondecreasing, we deduce that

which is equivalent to

(14)

where by \((\beta _{1})\). Similarly, taking \(\mathsf {v}=\mathsf {v}_{2n}\) and \(\mathsf {w}=\mathsf {v}_{2n-1}\) in (5) and keeping in mind , we get

(15)

However, from relations (14), (15), together with Lemma 10, we find that \(\{ \mathsf {v}_{n} \} \) is a Cauchy sequence in a complete -ms. Therefore, there exists \(\mathsf {x}\in \mathsf {S}\) such that

(16)

Without loss of generality, we can suppose that \(\mathsf {x}\neq \mathsf {v}_{n}\) for any \(n\in \mathbb{N}\). Supposing that , by (5), we have

or, taking \((\beta _{2})\) into account,

(17)

However, since

we obtain

(18)

On the other hand,

and then

which contradicts our assumption . Thus, we get , that is, . Moreover, if we suppose that , since ,

or, keeping in mind \((\beta _{2})\)

which is a contradiction. Therefore, which implies by that . That is, .

As a last step, we claim that x is the unique fixed point of the mappings and . Indeed, if we suppose that there exists another point such that \(\mathsf {x}\neq \upsilon \), by (13) we have

Since the function Ψ is supposed to be nondecreasing, it follows that

which is a contradiction. Therefore, the set has exactly one element. □

Corollary 14

Let be a complete -ms, \(\Psi, \Phi \in \Theta \), a number \(\alpha \in [1,\infty )\), and a mapping such that, for every \(\mathsf {v}, \mathsf {w}\in \mathsf {S}\) with , the following inequality

(19)

holds, where for nonnegative real numbers,

Assume that:

\((\beta _{1})\):

, , and ;

\((\beta _{2})\):

Ψ is nondecreasing.

Then . Moreover, if , then the set has exactly one element.

Proof

Let in Theorem 13. □

Example 15

Let the set and the function be defined by

m

n

p

q

m

0

2

5

7

n

2

6

8

5

p

5

8

0

1

q

7

5

1

0

Obviously, is a -metric, with \(\mathsf {s}=2\). Let be two mappings, where and . We have, in this case,

v

m

n

p

q

p

q

p

p

q

q

p

p

5

5

0

1

7

5

0

1

  

v

m

n

p

q

 

p

q

p

p

w

     

m

q

 

1

0

1

1

n

q

 

1

0

1

1

p

p

 

0

1

0

0

q

p

 

0

1

0

0

Letting the functions \(\Psi, \Phi \in \Theta \), \(\Psi (\theta )=\theta \), \(\Phi (\theta )=\frac{3}{4}\theta \) and the numbers \(\alpha =2\), , , , we can easily see that assumptions \((\beta _{1})\) and \((\beta _{2})\) in Theorem 13 are satisfied. We show that (13) is satisfied for any pair , where

(the other cases are excluded by the hypotheses of Theorem 13).

  • \((\mathsf {v},\mathsf {w})=(m,n)\)

  • \((\mathsf {v},\mathsf {w})=(n,p)\)

The other cases are discussed similarly.

Thus .

Theorem 16

Let be a complete -ms, the functions \(\Psi, \Phi \in \Theta \), a number \(\alpha \in [1,\infty )\), , and two mappings such that, for every distinct \(\mathsf {v}, \mathsf {w}\in \mathsf {S}\) with , the following inequality

(20)

holds, where

(21)

Assume that:

\((\beta _{1})\):

;

\((\beta _{2})\):

Ψ is nondecreasing.

Then the set has exactly one element.

Proof

Let us take in (20), \(\mathsf {v}=\mathsf {v}_{2n}\) and \(\mathsf {w}=\mathsf {v}_{2n+1}\), where the sequence \(\{ \mathsf {v}_{n} \} \) is defined as in Theorem 11. We have

with

Furthermore, taking \((\beta _{2})\) and the above relation into account, we get

which implies

(22)

Similarly, taking \(\mathsf {v}=\mathsf {v}_{2n}\), respectively \(\mathsf {w}=\mathsf {v}_{2n-1}\), we obtain

(23)

Now, choosing (by assumption \((\beta _{1})\)), we have for any \(n\in \mathbb{N}\). Therefore, Lemma 10 leads us to the conclusion that \(\{ \mathsf {v}_{n} \} \) is a Cauchy sequence. Thus, since the space is complete, there exists \(\mathsf {x}\in \mathsf {S}\) such that

(24)

Supposing that , we have

(25)

Moreover, without loss of generality, we can assume that for any \(n\in \mathbb{N}\), and then from (20) we get

or, by \((\beta _{2})\),

Returning in (25), we have

Letting \(n\rightarrow \infty \) and keeping in mind (24), we get

which is a contradiction. Thus, and from we have .

Analogously, we have

or, by \((\beta _{2})\),

On the other hand, supposing that , we have

Combining the above inequalities and taking limit as \(n\rightarrow \infty \), we obtain , which is a contradiction. Therefore, , and then . Thus, x is a common fixed point for and , that is, and it remains to show that the set is in fact reduced to a single point. On the contrary, let with \(\upsilon \neq \mathsf {x}\). Replaced in (20), we have

and, due to \((\beta _{2})\),

which is a contradiction. Therefore, it follows that \(\mathsf {x}=\upsilon \) and the set has exactly one element. □

Example 17

Let \(\mathsf {S}= \{ 2,4,5,7 \} \) and two self-mappings be defined on S by

v

2

4

5

7

5

5

5

4

4

5

5

5

Let be the -metric on S (with \(\mathsf {s}=2\)) given by

Considering the functions \(\Psi, \Phi \in \Theta \) as in Example (15) and letting \(\alpha =2\), , , we have the following cases:

  • \((\mathsf {v}, \mathsf {w})=(4,2)\)

  • \((\mathsf {v}, \mathsf {w})=(5,2)\)

  • \((\mathsf {v}, \mathsf {w})=(7,2)\)

  • \((\mathsf {v}, \mathsf {w})=(7,4)\)

  • \((\mathsf {v}, \mathsf {w})=(7,5)\)

The other cases are excluded by the hypothesis of Theorem 16. Therefore, .

Corollary 18

Let be a complete -ms, \(\Psi, \Phi \in \Theta \), a number \(\alpha \in [1,\infty )\), , and a mapping such that, for every distinct \(\mathsf {v}, \mathsf {w}\in \mathsf {S}\) with , the following inequality

(26)

holds, where

(27)

Assume that:

\((\beta _{1})\):

;

\((\beta _{2})\):

Ψ is nondecreasing.

Then the set has exactly one element.

Proof

Let in Theorem 16. □

3 Consequences

Taking particular functions Ψ and Φ, we obtain as consequences some known results. For example, let \(\Phi (\theta )=\beta (\theta )\Psi (\theta )\) for all \(\theta >0\) and \(\beta:[0,\infty )\rightarrow [0,\frac{1}{\mathsf {s}})\).

Corollary 19

Let be a complete -ms, a number \(\alpha \in [1,\infty )\), and two continuous mappings such that, for every distinct \(\mathsf {v}, \mathsf {w}\in \mathsf {S}\) with , the following inequality

(28)

holds. Assume that:

\((\beta _{1})\):

and ;

\((\beta _{3})\):

\(\Psi:(0,\infty )\rightarrow (0,\infty )\) is nondecreasing;

\((\beta _{4})\):

\(\beta:(0,\infty )\rightarrow (0,\frac{1}{\mathsf {s}})\) satisfies \(\limsup_{\theta \rightarrow \theta _{0}}\beta ( \theta )<\frac{1}{\mathsf {s}}\) for any \(\theta _{0}>0\).

Then . Moreover, if , then the set has exactly one element.

Corollary 20

Let be a complete -ms, a number \(\alpha \in [1,\infty )\), and two mappings such that, for every distinct \(\mathsf {v}, \mathsf {w}\in \mathsf {S}\) with , the following inequality

(29)

holds. Assume that:

\((\beta _{1})\):

, , , and ;

\((\beta _{3})\):

\(\Psi:(0,\infty )\rightarrow (0,\infty )\) is nondecreasing;

\((\beta _{4})\):

\(\beta:(0,\infty )\rightarrow (0,\frac{1}{\mathsf {s}})\) satisfies \(\limsup_{\theta \rightarrow \theta _{0}}\beta ( \theta )<\frac{1}{\mathsf {s}}\) for any \(\theta _{0}>0\).

Then . Moreover, if , then the set has exactly one element.

Corollary 21

Let be a complete -ms, a number \(\alpha \in [1,\infty )\), and two mappings such that, for every distinct \(\mathsf {v}, \mathsf {w}\in \mathsf {S}\) with , the following inequality

(30)

holds. Assume that:

\((\beta _{1})\):

, ;

\((\beta _{3})\):

\(\Psi:(0,\infty )\rightarrow (0,\infty )\) is nondecreasing;

\((\beta _{4})\):

\(\beta:(0,\infty )\rightarrow (0,\frac{1}{\mathsf {s}})\) satisfies \(\limsup_{\theta \rightarrow \theta _{0}}\beta ( \theta )<\frac{1}{\mathsf {s}}\) for any \(\theta _{0}>0\).

Then the set has exactly one element.

Considering \(\Phi (\theta )=\kappa \Psi (\theta )\) or \(\Phi (\theta )=\kappa \cdot \theta \) for all \(\theta >0\) in Theorems 11, 13 or 16, other consequences can be listed. On the other hand, many other corollaries can be deduced considering or letting \(\mathsf {s}=1\).