1 Introduction

In recent times, abstract metric spaces have attracted much attention for several reasons, especially due to their possible applications. For instance, in the context of fuzzy regression, it is difficult to consider a genuine fuzzy metric because the triangle inequality among fuzzy numbers depends on a partial order that it is not commonly accepted (see [3, 16,17,18]). In this framework, it is usual to handle either real metrics or fuzzy semimetrics or, in general, mappings that associate to each two fuzzy numbers a mathematical object that can be interpreted as their distance.

In [14], Khojasteh et al. introduced a new family of auxiliary functions, called simulation functions, that became a new viewpoint in fixed point theory. After that, Roldán López de Hierro et al. [19] slightly modified the notion of a simulation function in order to explain the different behavior of its two arguments. Later, this study continued on several papers (see [20,21,22,23]).

In the context of fixed point theory, many extensions of the notion of a metric space have been introduced in recent years. All of them tried to weaken the axioms that defined the concept of metric space (see, for instance, [4, 5, 20,21,22,23,24]) following iterative schemes (see [8]).

In 1993, Czerwik [9] introduced the concept of b-metric spaces as a generalization of metric spaces by modifying the triangle inequality through a constant greater than or equal to one. Then, he was able to prove the contraction mapping principle in b-metric spaces, that is, a generalization of the Banach contraction principle in metric spaces. Afterward, several mathematicians studied many results in b-metric spaces (see [6, 7, 10]).

In this paper, we introduce the notion of an \(R_{s}\)-contraction in b-metric spaces by using a modification of the concept of R-function. We establish some fixed point results in the context of b-metric spaces, and we also give some example for supporting the main results. Our theorems modify and generalize the results given by Roldán López de Hierro et al. [20] and several well-known results given by some authors in metric spaces. Stability results for fixed point problems by using our main results are established.

2 Preliminaries

Throughout this paper, we denote by \({\mathbb {N}}\), \({\mathbb {R}}_{+}=\left[ 0,\infty \right) \) and \({\mathbb {R}}\) the sets of positive integers, nonnegative real numbers and real numbers, respectively. Given a bounded from below sequence \(\{r_{n}\}\) of real numbers, its limit inferior is

$$\begin{aligned} \liminf _{n\rightarrow \infty }~r_{n}=\sup \,\left\{ \,\inf \left\{ \,r_{m}:m\ge n\,\right\} :n\ge 1\,\right\} . \end{aligned}$$

Similarly, if \(\{r_{n}\}\) is a bounded from above sequence of real numbers, then its limit superior is

$$\begin{aligned} \limsup _{n\rightarrow \infty }~r_{n}=\inf \,\left\{ \,\sup \left\{ \,r_{m}:m\ge n\,\right\} :n\ge 1\,\right\} . \end{aligned}$$

In general, \(\liminf \limits _{n\rightarrow \infty }~r_{n}\le \limsup \limits _{n\rightarrow \infty }~r_{n}\).

If \(d:X\times X\rightarrow {\mathbb {R}}_{+}\) is a mapping, the range ofd is the nonempty set:

$$\begin{aligned} \text {ran}(d)=\left\{ \,d(x,y):x,y\in X\,\right\} \subset \left[ 0,\infty \right) . \end{aligned}$$

Let us recall some definitions in the context of b-metric spaces.

Definition 2.1

(Czerwik [9]) Let X be a nonempty set and let \(s\ge 1\) be a fixed real number. Suppose that the mapping \(d:X\times X\rightarrow {\mathbb {R}}_{+}\) satisfies the following conditions for all \(x,y,z,\in X\):

  1. 1.

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

  2. 2.

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

  3. 3.

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

Then (X, d) is called a b-metric space with the coefficients.

In the context of regression methodology, it is usual to handle the square of the Euclidean distance between two real numbers, that is, \(\left( x-y\right) ^{2}\), in order to evaluate the niceness of fit of several parametric models. However, this function is not a metric because it does not satisfy the triangle inequality (for instance, \(\left( 1-0\right) ^{2}>\left( 1-0.5\right) ^{2}+\left( 0.5-0\right) ^{2}\)). Nevertheless, it can be seen as a 2-metric space.

Example 2.2

Let \(X\subset {\mathbb {R}}\) be a nonempty subset of real numbers and let \(d:X\times X\rightarrow {\mathbb {R}}_{+}\) be the mapping defined by

$$\begin{aligned} d(x,y)=\left( x-y\right) ^{2} \end{aligned}$$

for all \(x,y\in X\). Then (X, d) is a b-metric space with coefficient \(s=2\).

Example 2.3

Let p be a given real number in the interval (0, 1). The space \(L_{p}[0,1] \) of all real functions \(x:[0,1]\rightarrow {\mathbb {R}}\) such that \(\int _{0} ^{1}|x(t)|^{p}dt<1\), together with the mapping \(d:L_{p}[0,1]\times L_{p}[0,1]\rightarrow {\mathbb {R}}_{+}\),

$$\begin{aligned} d(x,y)=\left( \int _{0}^{1}|x(t)-y(t)|^{p}dt\right) ^{1/p}\quad \text {for each}\quad x,y\in L_{p}[0,1], \end{aligned}$$

is a b-metric space with coefficient \(s=2^{\frac{1}{p}}\) (see [2, 25]).

Next, we give the notions of b-convergent sequence, b-Cauchy sequence and b-completeness in the framework of b-metric spaces.

Definition 2.4

(Boriceanu et al. [6]) Let (X, d) be a b-metric space and let \(\{x_{n}\}\) be a sequence in X. If there exists \(x\in X\) such that \(d(x_{n},x)\rightarrow 0\) as \(n\rightarrow \infty \), then a sequence \(\{x_{n}\}\) is called b-convergent. In this case, we write \(\underset{n\rightarrow \infty }{\lim }x_{n}=x\).

The limit of a b-convergent sequence in a b-metric space is unique.

Definition 2.5

([6]) Let (X, d) be a b-metric space and let \(\{x_{n}\}\) be a sequence in X. If \(d(x_{n},x_{m})\rightarrow 0\) as \(n,m\rightarrow \infty \), then a sequence \(\{x_{n}\}\) is called a b-Cauchy sequence.

Proposition 2.6

([6]) In a b-metric space (X, d), the following assertions hold:

\((p_1)\):

each b-convergent sequence is a b-Cauchy sequence;

\((p_2)\):

in general, a b-metric is not continuous.

Definition 2.7

([6]) Let (X, d) be a b-metric space. If every b-Cauchy sequence in Xb-converges, then (X, d) is called b-complete.

In 2015, the notions of simulation functions and \({\mathcal {Z}}\)-contraction mappings were introduced by Khojasteh et al. [14]. They also proved some existence and uniqueness theorems of fixed points for \({\mathcal {Z}}\)-contraction mappings in metric spaces.

Definition 2.8

(Khojasteh et al. [14]) A mapping \(\zeta :[0,\infty )\times [0,\infty )\rightarrow {\mathbb {R}}\) is called a simulation function if it satisfies the following conditions:

\((\zeta 1)\):

\(\zeta (0,0) = 0\);

\((\zeta 2)\):

\(\zeta (t,s)<s-t\) for all \(t, s > 0\);

\((\zeta 3)\):

if \(\{t_{n}\}\), \(\{s_{n}\}\) are sequences in \((0,\infty )\) such that \(\underset{n\rightarrow \infty }{\lim }t_{n}=\underset{n\rightarrow \infty }{\lim }s_{n}>0\), then

$$\begin{aligned} \underset{n\rightarrow \infty }{\limsup }~\zeta (t_{n},s_{n})<0. \end{aligned}$$

We denote by \({\mathcal {Z}}\) the class of all simulation functions.

Example 2.9

[14] Let \(\zeta :[0,\infty )\times [0,\infty )\rightarrow {\mathbb {R}}\) be defined by

$$\begin{aligned} \zeta (t,s)=\uppsi (s)-\phi (t) \end{aligned}$$

for all \(t,s\in [0,\infty )\), where \(\uppsi ,\phi :[0,\infty )\rightarrow [0,\infty )\) are two continuous functions such that \(\uppsi (t)=\phi (t)=0\) if and only if \(t=0\), and \(\uppsi (t)<t\le \phi (t)\) for all \(t>0\). Then \(\zeta \) is a simulation function.

Example 2.10

[14] Let \(\zeta : [0,\infty ) \times [0,\infty ) \rightarrow {\mathbb {R}}\) be defined by

$$\begin{aligned} \zeta (t,s) = s - \frac{f(t,s)}{g(t,s)}t \end{aligned}$$

for all \(t, s \in [0,\infty )\), where \(f, g : [0,\infty ) \rightarrow [0,\infty )\) are two continuous functions with respect to each variable such that \(f(t,s)>g(t,s)\) for all \(t, s > 0\). Then \(\zeta \) is a simulation function.

Example 2.11

[14] Let \(\zeta : [0,\infty ) \times [0,\infty ) \rightarrow {\mathbb {R}}\) be defined by

$$\begin{aligned} \zeta (t,s) = s-\varphi (s) - t \end{aligned}$$

for all \(t, s \in [0,\infty )\), where \(\varphi : [0,\infty ) \rightarrow [0,\infty )\) is a continuous function such that \(\varphi (t) = 0\) if and only if \(t=0\). Then \(\zeta \) is a simulation function.

Definition 2.12

[14] Let (X, d) be a metric space and \(\zeta \in {\mathcal {Z}}\). A mapping \(T:X\rightarrow X\) is called a \({\mathcal {Z}}\)-contraction with respect to\(\zeta \) if the following condition is satisfied

$$\begin{aligned} \zeta (d(Tx,Ty),d(x,y))\ge 0 \end{aligned}$$

for all \(x,y\in X\).

Recently, Roldán López de Hierro et al. [20] have introduced a new family of auxiliary functions, and they used it to present a new kind of contractive mappings. They also showed that this family contains several classes of contractive mappings.

Definition 2.13

(Roldán López de Hierro et al. [20]) Let \(A\subset {\mathbb {R}}\) be a nonempty set. A function \(\xi :A\times A\rightarrow {\mathbb {R}}\) is called an R-function if it satisfies the following two conditions:

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

If \(\{a_{n}\} \subset (0,\infty ) \cap A\) is a sequence such that

$$\begin{aligned} \xi (a_{n+1},a_{n}) > 0 \end{aligned}$$

for all \(n \in {\mathbb {N}}\), then \(\underset{n\rightarrow \infty }{ \lim } a_{n} = 0\).

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

If \(\{a_{n}\}, \{b_{n}\} \subset (0,\infty ) \cap A\) are two sequences converging to the same limit \(L \ge 0\) and verifying that \(L < a_{n}\) and

$$\begin{aligned} \xi (a_{n},b_{n}) > 0 \end{aligned}$$

for all \(n \in {\mathbb {N}}\), then \(L = 0\).

Denote by \(R_{A}\) the family of all R-functions whose domain is \(A \times A \).

In some cases, given a function \(\xi : A \times A \rightarrow {\mathbb {R}}\), we also consider the following property:

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

If \(\{a_{n}\}, \{b_{n}\} \subset (0,\infty ) \cap A\) are two sequences such that \(\underset{n\rightarrow \infty }{ \lim } b_{n} = 0\) and

$$\begin{aligned} \xi (a_{n},b_{n}) > 0 \end{aligned}$$

for all \(n \in {\mathbb {N}}\), then \(\underset{n\rightarrow \infty }{ \lim } a_{n} = 0\).

Lemma 2.14

[20] Every simulation function is an R-function that also verifies \((\xi _{3})\).

Example 2.15

[20] Let \(\xi : [0,1] \times [0,1] \rightarrow {\mathbb {R}}\) be a function defined by

$$\begin{aligned} \xi (t,s) = \lambda s - t \end{aligned}$$

for all \(t, s \in [0,1]\), where \(\lambda \in (0,1)\). Then \(\xi \) is an R-function, but it is not a simulation function.

Example 2.16

([20]) Let \(\xi :[0,\infty )\times [0,\infty )\rightarrow {\mathbb {R}}\) be a function defined by

$$\begin{aligned} \xi (t,s)=\left\{ \begin{array} [c]{ll} \frac{1}{2}s-t, &{} \text {if }\quad t<s;\\ 0, &{} \text {if}\quad t\ge s. \end{array} \right. \end{aligned}$$

Then \(\xi \) is an R-function on \([0,\infty )\) which also satisfies condition \((\xi _{3})\).

3 Main Results

In this section, we introduce a modified version of R-contractions based on the results given by Roldán López de Hierro et al. in [20].

Definition 3.1

Let \(s\ge 1\) and let \(A\subset {\mathbb {R}}\) be a nonempty set. A function \(\varsigma :A\times A\rightarrow {\mathbb {R}}\) is said to be a modified R-function if it satisfies the following conditions:

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

\(\varsigma (a,b)\le b-a\)   for every \(a,b\in A\cap (0,\infty )\).

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

If \(\{a_{n}\}\subset (0,\infty )\cap A\) is a sequence such that

$$\begin{aligned} \varsigma (a_{n+1},a_{n})>0 \end{aligned}$$

for all \(n\in {\mathbb {N}}\), then \(\underset{n\rightarrow \infty }{\lim }a_{n}=0\).

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

If \(L \ge 0\) and \(\{a_{n}\},\{b_{n}\}\subset (0,\infty )\cap A\) are two sequences such that

$$\begin{aligned}&L<a_{n}<b_{n}\quad \text {for all }n\in {\mathbb {N}}, \end{aligned}$$
(3.1)
$$\begin{aligned}&\varsigma (a_{n},b_{n})>0\quad \text {for all }n\in {\mathbb {N}}\qquad \text {and} \end{aligned}$$
(3.2)
$$\begin{aligned}&\limsup _{n\rightarrow \infty }~b_{n}\le s\,L, \end{aligned}$$
(3.3)

then \(L=0\).

Remark 3.2

Notice that if \(L\ge 0\) and \(\{a_{n}\},\{b_{n}\}\subset (0,\infty )\cap A\) are two sequences satisfying the hypotheses of condition \(\left( \varsigma _{3}\right) \), then both of them are bounded, so they have limit inferior and superior as \(n\rightarrow \infty \), and the following inequalities hold:

$$\begin{aligned}&L\le \liminf _{n\rightarrow \infty }~a_{n}\le s\left( \limsup _{n\rightarrow \infty }~b_{n}\right) \le s^{2}\left( \liminf _{n\rightarrow \infty } ~a_{n}\right) \le s^{3}L, \end{aligned}$$
(3.4)
$$\begin{aligned}&L\le \liminf _{n\rightarrow \infty }~b_{n}\le s\left( \limsup _{n\rightarrow \infty }~a_{n}\right) \le s^{2}\left( \liminf _{n\rightarrow \infty } ~b_{n}\right) \le s^{3}L. \end{aligned}$$
(3.5)

For instance, the first line of inequalities follows from:

$$\begin{aligned} L\le \liminf _{n\rightarrow \infty }~a_{n}\le \limsup _{n\rightarrow \infty } ~b_{n}\le s\left( \limsup _{n\rightarrow \infty }~b_{n}\right) \le s\cdot sL\le s^{2}\left( \liminf _{n\rightarrow \infty }~a_{n}\right) \le s^{3}L, \end{aligned}$$

and the other one is similar.

Conversely, from inequalities (3.4)–(3.5), it can be deduced that

$$\begin{aligned}&L\le \liminf _{n\rightarrow \infty }~a_{n}\le sL,\qquad \frac{L}{s}\le \limsup _{n\rightarrow \infty }~a_{n}\le s^{2}L,\\&L\le \liminf _{n\rightarrow \infty }~b_{n}\le sL,\qquad \frac{L}{s}\le \limsup _{n\rightarrow \infty }~b_{n}\le s^{2}L,\\&\limsup _{n\rightarrow \infty }~b_{n}\le s\,\liminf _{n\rightarrow \infty }~a_{n},\qquad \limsup _{n\rightarrow \infty }~a_{n}\le s\,\liminf _{n\rightarrow \infty }~b_{n}. \end{aligned}$$

Next, we will introduce the idea of a weak continuity type for real-valued functions.

Definition 3.3

Let \(s\ge 1\). A function \(\uppsi : [0,\infty ) \rightarrow {\mathbb {R}}\) is said to be s-upper semi-continuous at the point\(t_{0} \ge 0\) if

$$\begin{aligned} \limsup _{n\rightarrow \infty }~\uppsi (r_{n})\le \uppsi (t_{0}) \end{aligned}$$

whenever \(\{r_{n}\}\) is a sequence in \([0,\infty )\) such that

$$\begin{aligned} t_{0}\le \liminf _{n\rightarrow \infty }~r_{n}\le \limsup _{n\rightarrow \infty }~r_{n}\le s\,t_{0}. \end{aligned}$$

Also, \(\uppsi \) is said to be s-upper semi-continuous on\([0,\infty )\) if it is s-upper semi-continuous at every point in \([0,\infty )\).

Remark 3.4

If \(\uppsi \) is s-upper semi-continuous at a point \(t_{0}\ge 0\) for some \(s\ge 1\), then it is also 1-upper semi-continuous at the point \(t_{0}\).

Given \(s\ge 1\), let denote by \(\Psi _{s}\) the set of all functions \(\uppsi :[0,\infty )\rightarrow [0,\infty ) \) satisfying the following conditions:

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

\(\uppsi \) is s-upper semi-continuous on \((0,\infty )\);

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

\(\uppsi (t)<t\) for all \(t > 0\).

Clearly, \(\Psi _{s}\subset \Psi _{1}\).

Now, we give a family of examples of modified R-functions by using the functions in the class \(\Psi _{s}\).

Example 3.5

Let \(s\ge 1\) be a given real number, let \(A = [0,\infty )\), let \(\uppsi \in \Psi _{s}\) be a function and let \(\varsigma _{s,A,\uppsi }:A\times A\rightarrow {\mathbb {R}}\) be the mapping defined by

$$\begin{aligned} \varsigma _{s,A,\uppsi }(a,b)=\uppsi (b)-s^{2}a \end{aligned}$$

for all \(a,b\in A\). We claim that \(\varsigma _{s,A,\uppsi }\) is a modified R-function. Indeed, for all \(a,b\in (0,\infty )\), we have

$$\begin{aligned} \varsigma _{s,A,\uppsi }(a,b)=\uppsi (b)-s^{2}a<b-a, \end{aligned}$$

which yields \((\varsigma _{1})\).

Let \(\{a_{n}\}\subset (0,\infty )\) be a sequence such that

$$\begin{aligned} \varsigma _{s,A,\uppsi }(a_{n+1},a_{n})>0 \end{aligned}$$

for all \(n\in {\mathbb {N}}\). Since \(\uppsi \in \Psi _{s}\), we get

$$\begin{aligned} 0&<\varsigma _{s,A,\uppsi }(a_{n+1},a_{n})\\&=\uppsi (a_{n})-s^{2}a_{n+1}\\&<a_{n}-s^{2}a_{n+1} \end{aligned}$$

and so

$$\begin{aligned} a_{n+1}\le s^{2}a_{n+1}\le a_{n}. \end{aligned}$$

Therefore, \(\{a_{n}\}\) is a decreasing sequence which is bounded from below by 0. Then, there exists \(r\ge 0\) such that

$$\begin{aligned} \lim _{n\rightarrow \infty }~a_{n}=r. \end{aligned}$$

Since \(0\le r\le a_{n}\), \(a_{n}\in A\) and \(A = [0,\infty )\), then \(r\in A\). In order to prove that \(r=0\), let suppose, by contradiction, that \(r>0\). Then, we have

$$\begin{aligned} 0&\le \limsup _{n\rightarrow \infty }~\varsigma _{s,A,\uppsi }(a_{n+1},a_{n})\\&=\limsup _{n\rightarrow \infty }\left( \uppsi (a_{n})-s^{2}a_{n+1}\right) \\&\le \limsup _{n\rightarrow \infty }~\uppsi (a_{n})-s^{2}\,\liminf _{n\rightarrow \infty }~a_{n+1}\\&\le \uppsi (r)-s^{2}r\\&<r-s^{2}r\\&=(1-s^{2})r\\&\le 0, \end{aligned}$$

which is a contradiction. Hence, \(r=0\) and so \(\underset{n\rightarrow \infty }{\lim }a_{n}=0\).

Finally, let \(L\ge 0\) and let \(\{a_{n}\},\{b_{n}\}\) be two sequences in \((0,\infty )\) such that

$$\begin{aligned}&L<a_{n}\le b_{n}\quad \text {for all }n\in {\mathbb {N}},\\&\varsigma _{s,A,\uppsi }(a_{n},b_{n})>0\quad \text {for all }n\in {\mathbb {N}} \qquad \text {and}\\&\limsup _{n\rightarrow \infty }~b_{n}\le s\,L. \end{aligned}$$

By contradiction, assume that \(L>0\). Since \(0\le L < a_{n}\), \(a_{n}\in A\) and \(A=[0,\infty )\), we get \(L\in A\). Furthermore,

$$\begin{aligned} L\le \liminf _{n\rightarrow \infty }~b_{n}\le \limsup _{n\rightarrow \infty } ~b_{n}\le s\,L. \end{aligned}$$

Since \(\uppsi \in \Psi _{s}\), we have \(\uppsi \left( L\right) <L\) and

$$\begin{aligned} \limsup _{n\rightarrow \infty }~\uppsi (b_{n})\le \uppsi (L). \end{aligned}$$

As a consequence,

$$\begin{aligned} 0&\le \limsup _{n\rightarrow \infty }~\varsigma _{s,A,\uppsi }(a_{n},b_{n})\\&=\limsup _{n\rightarrow \infty }~\left( \uppsi (b_{n})-s^{2}a_{n}\right) \\&\le \limsup _{n\rightarrow \infty }~\uppsi (b_{n})-s^{2}\,\liminf _{n\rightarrow \infty }~a_{n}\\&\le \uppsi (L)-s^{2}L\\&<L-s^{2}L\\&=(1-s^{2})L\\&\le 0, \end{aligned}$$

which is a contradiction. Therefore, \(L=0\) which proves \((\varsigma _{3})\). Hence, \(\varsigma _{s,A,\uppsi }\) is a modified R-function.

Lemma 3.6

Let \(\emptyset \ne A\subset {\mathbb {R}}\) and let \(\varsigma :A\times A\rightarrow {\mathbb {R}}\) be a function satisfying \((\varsigma _{1})\). If there exist two sequences \(\{a_{n}\},\{b_{n}\}\subset (0,\infty )\cap A\) such that \(\underset{n\rightarrow \infty }{\lim }b_{n}=0\) and

$$\begin{aligned} \varsigma (a_{n},b_{n})>0 \end{aligned}$$

for all \(n\in {\mathbb {N}}\), then

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }a_{n}=0. \end{aligned}$$

Proof

Suppose that \(\{a_{n}\},\{b_{n}\}\subset (0,\infty )\cap A\) are two sequences such that \(\underset{n\rightarrow \infty }{\lim }b_{n}=0\) and

$$\begin{aligned} \varsigma (a_{n},b_{n})>0 \end{aligned}$$

for all \(n\in {\mathbb {N}}\). Therefore,

$$\begin{aligned} 0<\varsigma (a_{n},b_{n})\le b_{n}-a_{n} \end{aligned}$$

and so \(0<a_{n}<b_{n}\) for all \(n\in {\mathbb {N}}\). Since \(\underset{n\rightarrow \infty }{\lim }b_{n}=0\), we get

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }a_{n}=0. \end{aligned}$$

\(\square \)

Next, we introduce an extended viewpoint of the notion of R-contraction mappings from a b-metric space into itself.

Definition 3.7

Let (X, d) be a b-metric space with coefficient \(s\ge 1\). A mapping \(T:X\rightarrow X\) is called an \(R_{s}\)-contraction mapping if there exist a subset \(A\subset {\mathbb {R}}\) and a modified R-function \(\varsigma :A\times A\rightarrow {\mathbb {R}}\) such that \(\text {ran} (d)\subset A\) and

$$\begin{aligned} \varsigma (d(Tx,Ty),d(x,y))>0 \end{aligned}$$
(3.6)

for all \(x,y\in X\) such that \(x\ne y\). In such a case, we will say that T is an \(R_{s}\)-contraction mapping with respect to\(\varsigma \).

Notice that condition \(\text {ran}(d)\subset A\) guarantees that A is nonempty.

Theorem 3.8

Every \(R_{s}\)-contraction from a complete b-metric space into itself has a unique fixed point.

Proof

Let (X, d) be a complete b-metric space with coefficient \(s\ge 1\), \(A \subset {\mathbb {R}}\) and let \(T:X\rightarrow X\) be an \(R_{s}\)-contraction with respect to \(\varsigma :A\times A\rightarrow {\mathbb {R}}\), where \(\text {ran}(d)\subset A\). Let \(x_{1}\in X\) be an initial point and let consider the Picard sequence \(\{x_{n}\}\subset X\) defined by \(x_{n+1}=Tx_{n}\) for all \(n\in {\mathbb {N}}\). If there is \(n_{0}\in {\mathbb {N}}\) such that \(x_{n_{0}+1}=x_{n_{0}}\), then \(x_{n_{0}}\) is a fixed point of T. Also, \(x_{n}=x_{n_{0}}\) for all \(n\ge n_{0}\), so \(\{x_{n}\}\)b-converges to the fixed point \(x_{n_{0}}\). In this case, the proof is finished.

On the contrary, suppose that \(x_{n}\ne x_{n+1}\) for all \(n\in {\mathbb {N}}\), that is, \(d\left( x_{n},x_{n+1}\right) >0\) for all \(n\in {\mathbb {N}}\). As T is an \(R_{s}\)-contraction, for all \(n\in {\mathbb {N}}\),

$$\begin{aligned} 0<\varsigma \left( d\left( Tx_{n},Tx_{n+1}\right) ,d\left( x_{n} ,x_{n+1}\right) \right) =\varsigma \left( d\left( x_{n+1},x_{n+2}\right) ,d\left( x_{n},x_{n+1}\right) \right) , \end{aligned}$$

so condition \(\left( \varsigma _{2}\right) \) guarantees that

$$\begin{aligned} \lim _{n\rightarrow \infty }~d\left( x_{n},x_{n+1}\right) =0. \end{aligned}$$

Let us prove that \(x_{n}\ne x_{m}\) for all \(n,m\in {\mathbb {N}}\). Indeed, assume that there are \(n_{0},m_{0}\in {\mathbb {N}}\) such that \(n_{0}<m_{0}\) and \(x_{n_{0}}=x_{m_{0}}\). Let \(p_{0}=m_{0}-n_{0}\). Clearly, \(p_{0}\in {\mathbb {N}}\) and \(p_{0}\ge 2\). In this case, \(x_{n_{0}+p_{0}}=x_{m_{0}}=x_{n_{0}}\). Furthermore,

$$\begin{aligned} x_{n_{0}+2p_{0}}=x_{n_{0}+p_{0}+p_{0}}=x_{m_{0}+p_{0}}=T^{p_{0}}x_{m_{0} }=T^{p_{0}}x_{n_{0}}=x_{n_{0}+p_{0}}=x_{m_{0}}=x_{n_{0}}. \end{aligned}$$

By induction, it can be proved that \(x_{n_{0}+k\cdot p_{0}}=x_{n_{0}}\) for all \(k\in {\mathbb {N}}\). Therefore, the sequence \(\{d\left( x_{n_{0}+k\cdot p_{0} },x_{n_{0}+k\cdot p_{0}+1}\right) \}_{k\in {\mathbb {N}}}\) is constant because

$$\begin{aligned} d\left( x_{n_{0}+k\cdot p_{0}},x_{n_{0}+k\cdot p_{0}+1}\right) =d\left( x_{n_{0}},x_{n_{0}+1}\right) >0 \end{aligned}$$

for all \(k\in {\mathbb {N}}\). However, this is a contradiction because \(\underset{n\rightarrow \infty }{\lim }~d\left( x_{n},x_{n+1}\right) =0\). As a consequence, we have proved that

$$\begin{aligned} x_{n}\ne x_{m}\quad \text {for all }n,m\in {\mathbb {N}} . \end{aligned}$$
(3.7)

(In other words, T has not periodic points on \(\{x_{n}\}\), that is, there are not \(n,p\in {\mathbb {N}}\) such that \(T^{p}x_{n}=x_{n}\).)

Next we prove, by contradiction, that \(\{x_{n}\}\) is a b-Cauchy sequence on \(\left( X,d\right) \). If it were false, there would be \(\varepsilon >0\) and two subsequences \(\{x_{m(k)}\}_{k\in {\mathbb {N}}}\) and \(\{x_{n(k)}\}_{k\in {\mathbb {N}}}\) of \(\{x_{n}\}\) such that \(m\left( k\right) +1<n(k)<m\left( k+1\right) \) and

$$\begin{aligned} \varepsilon <d(x_{m(k)},x_{n(k)})\qquad \text {for all }k\in {\mathbb {N}}. \end{aligned}$$
(3.8)

Further, corresponding to each m(k), we can choose n(k) in such a way that it is the smallest integer with \(n(k)>m(k)\ge k\) satisfying (3.8). Hence, it implies that

$$\begin{aligned} d(x_{m(k)},x_{n(k)-1})\le \varepsilon \qquad \text {for all }k\in {\mathbb {N}} . \end{aligned}$$
(3.9)

Let us define \(L=\varepsilon >0\) and let us consider the sequences \(\{a_k\}\) and \(\{b_K\}\) defined by

$$\begin{aligned} a_{k}=d(x_{m(k)},x_{n(k)})\quad \text { and }\quad b_{k}=d(x_{m(k)-1} ,x_{n(k)-1})\qquad \text {for all }k\in {\mathbb {N}}. \end{aligned}$$

Clearly,

$$\begin{aligned} \{a_{k}\},\{b_{k}\}\subset \text {ran}(d)\cap \left( 0,\infty \right) \subset A\cap \left( 0,\infty \right) . \end{aligned}$$
(3.10)

As T is an \(R_{s}\)-contraction with respect to \(\varsigma \) and \(x_{m(k)-1}\ne x_{n(k)-1}\), we have

$$\begin{aligned} \varsigma \left( a_{k},b_{k}\right)&=\varsigma \left( d(x_{m(k)} ,x_{n(k)}),d(x_{m(k)-1},x_{n(k)-1})\right) \nonumber \\&=\varsigma \left( d(Tx_{m(k)-1},Tx_{n(k)-1}),d(x_{m(k)-1},x_{n(k)-1} )\right) >0 \end{aligned}$$
(3.11)

for all \(k\in {\mathbb {N}}\). For all \(k\in {\mathbb {N}}\), we have

$$\begin{aligned} L=\varepsilon<a_{k}=d(x_{m(k)},x_{n(k)})=d(Tx_{m(k)-1},Tx_{n(k)-1})<d(x_{m(k)-1} ,x_{n(k)-1})=b_{k}. \end{aligned}$$
(3.12)

Furthermore, the triangular inequality guarantees that

$$\begin{aligned} b_{k}&=d(x_{m(k)-1},x_{n(k)-1})\\&\le s[d(x_{m(k)-1},x_{m(k)})+d(x_{m(k)},x_{n(k)-1})]\\&<sd(x_{m(k)-1},x_{m(k)})+s\varepsilon . \end{aligned}$$

Taking limit superior as \(k\rightarrow \infty \) in the above inequality, we deduce that

$$\begin{aligned} \limsup _{k\rightarrow \infty }\,b_{k}\le s\varepsilon =sL. \end{aligned}$$
(3.13)

Taking into account (3.10), (3.11), (3.12) and (3.13), \(\{a_{k}\}\) and \(\{b_{k}\}\) are sequences satisfying the hypotheses of property \(\left( \varsigma _{3}\right) \). Hence, as \(\varsigma \) is a modified R-function, axiom \(\left( \varsigma _{3}\right) \) implies that \(\varepsilon =L=0\), but this is a contradiction because \(\varepsilon >0\). As a consequence, the sequence \(\{x_{n}\}\) is a b-Cauchy sequence in \(\left( X,d\right) \). Since (X, d) is a complete b-metric space, there exists \(u\in X\) such that \(\underset{n\rightarrow \infty }{\lim }x_{n}=u\), that is,

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }d(x_{n},u)=0. \end{aligned}$$
(3.14)

In order to prove that u is a fixed point of T, notice that (3.7) guarantees the existence of \(n_{0}\in {\mathbb {N}}\) such that \(x_{n}\ne u\) and \(x_{n}\ne Tu\) for all \(n\ge n_{0}\). Hence, as T is an \(R_{s}\)-contraction with respect to \(\varsigma \), we obtain

$$\begin{aligned} 0<\varsigma \left( d(Tx_{n},Tu),d(x_{n},u)\right) =\varsigma \left( d(x_{n+1},Tu),d(x_{n},u)\right) \le d(x_{n},u)-d(x_{n+1},Tu), \end{aligned}$$

so \(d(x_{n+1},Tu)\le d(x_{n},u)\) for all \(n\ge n_{0}\). Therefore, \(\{x_{n}\}\) converges, at the same time, to u and to Tu. As the limit of a convergent sequence in a b-metric space is unique, we conclude that \(Tu=u\), that is, u is a fixed point of T.

Finally, we claim that the fixed point of T is unique. To prove it, let \(u,v\in X\) be two arbitrary distinct fixed points of T. Define \(a_{n}=d(u,v)>0\) for all \(n\in {\mathbb {N}}\). Then

$$\begin{aligned} \varsigma (a_{n+1},a_{n})=\varsigma (d(u,v),d(u,v))=\varsigma (d(Tu,Tv),d(u,v))>0 \end{aligned}$$

for all \(n\in {\mathbb {N}}\). By using \((\varsigma _{2})\), we have

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }a_{n}=0, \end{aligned}$$

which is a contradiction because \(\{a_{n}\}\) is a constant sequence. Therefore, T has a unique fixed point. \(\square \)

Corollary 3.9

Let (X, d) be a complete b-metric space with coefficient \(s\ge 1\) and let \(T:X\rightarrow X\) be a mapping. Assume that there is \(\uppsi \in \Psi _{s}\) such that

$$\begin{aligned} d\left( Tx,Ty\right) <\frac{\uppsi \left( d\left( x,y\right) \right) }{s^{2}} \end{aligned}$$

for all \(x,y\in X\) such that \(x\ne y\). Then, T has a unique fixed point.

Proof

Let \(A = [0,\infty )\). Then, \(\text {ran}(d)\subseteq A\). Let \(\varsigma :A\times A\rightarrow {\mathbb {R}}\) be the function given by:

$$\begin{aligned} \varsigma \left( a,b\right) =\uppsi \left( b\right) -s^{2}a, \end{aligned}$$

for all \(a,b\in A\). Then \(\varsigma \) is a modified R-function (recall Example 3.5). Furthermore, for all \(x,y\in X\) such that \(x\ne y\), we have that

$$\begin{aligned} \varsigma \left( d\left( Tx,Ty\right) ,d\left( x,y\right) \right) =\uppsi \left( d\left( x,y\right) \right) -s^{2}d\left( Tx,Ty\right) >0. \end{aligned}$$

Then T is an \(R_{s}\)-contraction with respect to \(\varsigma \). Theorem 3.8 concludes that T has a unique fixed point. \(\square \)

By using Theorem 3.8, it can be obtained some of the main results given by Roldán López de Hierro et al. in  [20] as follows.

Corollary 3.10

([20]) Let (X, d) be a complete metric space and let \(T:X\rightarrow X\) be an R-contraction with respect to an R-function \(\xi :A\times A\rightarrow {\mathbb {R}}\) satisfying the following condition:

$$\begin{aligned} \xi (a,b)\le b-a \end{aligned}$$

for all \(a,b\in A\cap (0,\infty )\). Then, T has a unique fixed point.

Remark 3.11

It is well known that the class of R-contractions unifies the following different classes of contractions:

  • \({\mathcal {Z}}\)-contractions [14];

  • manageable contractions [11];

  • Meir–Keeler contractions [15];

  • Geraghty’s contractions [13];

  • Dutta and Choudhury’s contractions [12].

Therefore, Theorem 3.8 is a very general fixed point theorem under a new contractive condition that extends and unifies the classes of all the above contractions.

Example 3.12

Let \(\{x_{n}\},\{y_{n}\}\) be defined by \(x_{n}=10n\) and \(y_{n}=10n+1 +\frac{1}{n}\) for all \(n\in {\mathbb {N}}\) and

$$\begin{aligned} X=\left\{ 0,\frac{1}{4}\right\} \cup \{x_{n}:n\in {\mathbb {N}}\}\cup \{y_{n}:n\in {\mathbb {N}}\}. \end{aligned}$$

Define the mapping \(d:X\times X\rightarrow [0,\infty )\) by

$$\begin{aligned} d(x,y)=\left( x-y\right) ^{2} \end{aligned}$$

for all \(x,y\in X\). Clearly, (X, d) is a complete b-metric space with \(s=2 \). Let \(T:X\rightarrow X\) be defined by

$$\begin{aligned} Tx=\left\{ \begin{array} [c]{ll} 0, &{} \text {if}\quad x\in \left\{ \,0,\dfrac{1}{4}\,\right\} \cup \{\,10n:n\in {\mathbb {N}}\,\},\\ \dfrac{1}{4}, &{} \text {if}\quad x\in \left\{ \,10n+1+\dfrac{1}{n}:n\in {\mathbb {N}}\,\right\} . \end{array} \right. \end{aligned}$$

Next, we will show that T is an \(R_{s}\)-contraction mapping. Notice that if \(x,y\in X\) are such that \(x\le y\), then one, and only one, of the following cases holds:

  1. 1.

    \(d(x,y)=0\Leftrightarrow x=y\);

  2. 2.

    \(d(x,y)=\frac{1}{16}\Leftrightarrow x=0\) and \(y=\frac{1}{4}\);

  3. 3.

    \(d(x,y)=\left( 1+\frac{1}{n}\right) ^{2}\Leftrightarrow x=x_{n}\) and \(y=y_{n}\), for some \(n\in {\mathbb {N}}\);

  4. 4.

    \(d(x,y)\ge 8^2\) in any other case.

As X is numerable, the range of d is also numerable. Hence, it can be described as

$$\begin{aligned} A=\text {ran}(d)=\left\{ 0,\frac{1}{16}\right\} \cup \left\{ \left( 1+\frac{1}{n}\right) ^{2}:n\in {\mathbb {N}}\right\} \cup \left\{ t_{k}:k\in {\mathbb {N}}\right\} , \end{aligned}$$

where \(t_{k}\ge 8^2\) for all \(k\in {\mathbb {N}}\). Let \(\varsigma :A\times A\rightarrow {\mathbb {R}}\) be the function given by:

$$\begin{aligned} \varsigma (u,v)=\left\{ \begin{array} [c]{ll} \dfrac{v}{2}, &{} \text {if }u=0;\\ v-4u, &{} \text {if }u, v>0. \end{array} \right. \end{aligned}$$

Next, we will show that \(\varsigma \) is a modified R-function.

(1):

It is easy to see that \(\varsigma (a,b) \le b - a\) for all \(a, b \in A \cap (0,\infty )\), that is, \((\varsigma _{1})\) holds.

(2):

Let \(\{a_n\} \subset (0,\infty ) \cap A\) is a sequence such that

$$\begin{aligned} \varsigma (a_{n+1},a_n) > 0 \end{aligned}$$

for all \(n \in {\mathbb {N}}\). Then

$$\begin{aligned} 0 < \varsigma (a_{n+1},a_n) = a_n - 4a_{n+1}. \end{aligned}$$
(3.15)

It yields that

$$\begin{aligned} a_{n+1}< 4a_{n+1} < a_n. \end{aligned}$$

Therefore, \(\{a_n\}\) is decreasing and bounded from below, and then there exists \(r \ge 0\) such that

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim } a_n = r. \end{aligned}$$

Assume that \(r > 0\). By using (3.15), we have

$$\begin{aligned} 0 \le \underset{n\rightarrow \infty }{\lim } (a_n - 4a_{n+1}) = r - 4r = -3r, \end{aligned}$$

which is a contradiction. Therefore, \(r=0\), and so

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim } a_n = 0. \end{aligned}$$
(3):

We will show that \((\varsigma _{3})\) holds. Assume that there are \(L \ge 0\) and two sequences \(\{a_{n}\},\{b_{n}\}\subset (0,\infty )\cap A\) satisfying

$$\begin{aligned}&L<a_{n}<b_{n}\quad \text {for all }n\in {\mathbb {N}},\\&\varsigma (a_{n},b_{n})>0\quad \text {for all }n\in {\mathbb {N}}\qquad \text {and}\\&\limsup _{n\rightarrow \infty }~b_{n}\le 2\,L. \end{aligned}$$

Suppose that \(L>0\). Then,

$$\begin{aligned} 0 < \varsigma (a_{n+1},a_n) = a_n - 4a_{n+1}. \end{aligned}$$

Taking limit superior as \(n\rightarrow \infty \), we have

$$\begin{aligned} 0&\le \limsup _{n\rightarrow \infty }\left( b_n-4a_{n}\right) \\&\le \limsup _{n\rightarrow \infty }\,b_{n}-4\liminf _{n\rightarrow \infty }\,a_{n}\\&\le 2\,L-4L\\&=-2\,L, \end{aligned}$$

which is a contradiction. Therefore, \(L=0\) and so \((\varsigma _{3})\) holds.

Hence, \(\varsigma \) is a modified R-function.

Next, we will show that T is an \(R_{s}\)-contraction mapping with respect to \(\varsigma \). Let \(x,y\in X\) be such that \(x\ne y\). We will divide this claim into the following cases:

Case A::

If \(d(Tx,Ty)=0\), then

$$\begin{aligned} \varsigma (d(Tx,Ty),d(x,y))&=\varsigma (0,d(x,y))\\&=\frac{d(x,y)}{2}\\&>0. \end{aligned}$$
Case B::

If \(d(Tx,Ty)>0\), we obtain

$$\begin{aligned} \{Tx,Ty\}=\left\{ 0,\frac{1}{4}\right\} \qquad \text {and}\qquad d(Tx,Ty)=\frac{1}{16}. \end{aligned}$$

Suppose that \(Tx=0\) and \(Ty=\frac{1}{4}\). Then, \(x\in \left\{ 0,\frac{1}{4}\right\} \cup \{10n:n\in {\mathbb {N}}\}\) and \(y\in \left\{ 10m+1 +\frac{1}{m}:m\in {\mathbb {N}}\right\} \). If \(x=x_{n}\) and \(y=y_{n}\) for some \(n\in {\mathbb {N}}\), we obtain

$$\begin{aligned} \varsigma (d(Tx,Ty),d(x,y))&=\varsigma \left( \frac{1}{16},\left( 1+\frac{1}{n}\right) ^{2}\right) \\&=\left( 1+\frac{1}{n}\right) ^{2}-\frac{1}{4}\\&>0. \end{aligned}$$

In other case, we get \(d(x,y)\ge 8^2\) and so

$$\begin{aligned} \varsigma (d(Tx,Ty),d(x,y))&=\varsigma \left( \frac{1}{16},d(x,y)\right) \\&=d(x,y)-\frac{1}{4}\\&\ge 8^2-\frac{1}{4}\\&>0. \end{aligned}$$

From Case A and Case B, we get

$$\begin{aligned} \varsigma (d(Tx,Ty),d(x,y))=\varsigma (d(Tx,Ty),d(x,y))>0 \end{aligned}$$

for all \(x,y\in X\) with \(x\ne y\). Then, T is an \(R_{s}\)-contraction with respect to \(\varsigma \).

Therefore, all the conditions of Theorem 3.8 are satisfied. Hence, we can conclude that T has a unique fixed point (which is a point 0).

4 Some Applications

In this section, we present some applications of the presented results in the previous section. We begin with the following definition:

Definition 4.1

Let (X, d) be a b-metric space with \(s \ge 1\) and \(T : X \rightarrow X\) be a given mapping. The fixed point equation

$$\begin{aligned} x = Tx \end{aligned}$$
(4.1)

is called generalized Ulam–Hyers stability if there exists \(\varphi : [0,\infty ) \rightarrow [0,\infty )\) which is increasing, s-upper semi-continuous and \(\varphi (0) = 0\) such that for every \(\varepsilon > 0\) and for each \(w^{*} \in X\) which is an \(\varepsilon \)-solution to the fixed point equation (4.1), that is, \(w^{*}\) satisfies

$$\begin{aligned} d(w^{*},Tw^{*}) \le \varepsilon , \end{aligned}$$
(4.2)

and there exists a solution \(x^{*} \in X\) of (4.1) such that

$$\begin{aligned} d(w^{*},x^{*}) \le \varphi (s\varepsilon ). \end{aligned}$$

Theorem 4.2

Let (X, d) be a b-metric space with \(s \ge 1\) and \(A =[0,\infty )\). Define \(\varsigma _{A} : A \times A \rightarrow {\mathbb {R}}\) and \(\gamma : [0,\infty ) \rightarrow [0,\infty )\) by

$$\begin{aligned} \varsigma _{A}(a,b) = \uppsi (b) - s^{2}a \text{ and } \gamma (t) = t - \frac{\uppsi (t)}{s} \end{aligned}$$

for all \(a, b, t \in A\), where \(\uppsi \in \)\(\Psi _{s}\) with \(\uppsi (0) = 0\) and \(\gamma \) is strictly increasing and onto. Suppose that \(T : X \rightarrow X\) is an \(R_{s}\)-contraction with respect to \(\varsigma _{A}\). Then, the following assertions hold.

(a):

The fixed point equation (4.1) is generalized Ulam–Hyers stability.

(b):

The fixed point equation (4.1) is well posed, that is, if \(\{x_{n}\} \subset X\) are such that \(\underset{n\rightarrow \infty }{\lim } d(x_{n},Tx_{n}) = 0\) and \(s > 1\), then

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim } x_{n} = x^{*}, \end{aligned}$$

where \(x^{*}\) is a unique fixed point of T.

(c):

If \(f : X \rightarrow X\) is a mapping such that \(d(Tx,fx) \le \eta \) for all \(x \in X\), where \(\eta \ge 0\), then

$$\begin{aligned} d(x^{*},y^{*}) \le \gamma ^{-1}(s\eta ) \end{aligned}$$

where \(x^*\) is a fixed point of T and \(y^{*}\) is a fixed point of f.

Proof

From Example 3.5, we have \(\varsigma _{A}\) a modified R-function. Since T is an \(R_{s}\)-contraction with respect to \(\varsigma _{A}\), we get

$$\begin{aligned} 0 < \varsigma _{A}(d(Tx,Ty),d(x,y)) = \uppsi (d(x,y)) - s^{2}d(Tx,Ty) \end{aligned}$$

and so

$$\begin{aligned} d(Tx,Ty) < \frac{\uppsi (d(x,y))}{s^{2}} \end{aligned}$$
(4.3)

for all \(x, y \in X\) with \(x \ne y\).

  1. (a)

    By Theorem 3.8, we get T which has a unique fixed point, say, a point \(x^{*}\). Let \(\varepsilon > 0\) and \(w^{*} \in X\) be an \(\varepsilon \)-solution to the fixed point equation (4.1). Using (4.3), we have

    $$\begin{aligned} d(x^{*},w^{*})&= d(Tx^{*},w^{*})\\&\le s[d(Tx^{*},Tw^{*}) + d(Tw^{*},w^{*})]\\&\le s\left[ \frac{\uppsi (d(x^{*},w^{*}))}{s^{2}} + \varepsilon \right] . \end{aligned}$$

    This implies that

    $$\begin{aligned} d(x^{*},w^{*}) - \frac{\uppsi (d(x^{*},w^{*}))}{s} \le s \varepsilon . \end{aligned}$$

    Therefore,

    $$\begin{aligned} \gamma (d(x^{*},w^{*}))&= d(x^{*},w^{*}) - \frac{\uppsi (d(x^{*},w^{*}))}{s}\\&\le s \varepsilon \end{aligned}$$

    and so

    $$\begin{aligned} d(x^{*},w^{*}) \le \gamma ^{-1}(s \varepsilon ). \end{aligned}$$

    Hence, the fixed point equation (4.1) is generalized Ulam–Hyers stability.

  2. (b)

    Since \(\{x_{n}\} \subset X\) such that \(\underset{n\rightarrow \infty }{\lim } d(x_{n},Tx_{n}) = 0\), \(x^{*}\) is a unique fixed point of T, and using (4.3), we obtain

    $$\begin{aligned} d(x_{n},x^{*})&\le s[d(x_{n},Tx_{n}) + d(Tx_{n},x^{*})]\\&= s[d(x_{n},Tx_{n}) + d(Tx_{n},Tx^{*})]\\&\le s\left[ d(x_{n},Tx_{n}) + \frac{\uppsi (d(x_{n},x^{*}))}{s^{2}}\right] \\&\le sd(x_{n},Tx_{n}) + \frac{d(x_{n},x^{*})}{s} \end{aligned}$$

    for all \(n\in {\mathbb {N}}\). This implies that

    $$\begin{aligned} \left( 1-\frac{1}{s}\right) d(x_{n},x^{*}) \le s d(x_{n},Tx_{n}) \end{aligned}$$

    for all \(n\in {\mathbb {N}}\). Taking limit as \(n \rightarrow \infty \) in above inequality, we get

    $$\begin{aligned} \underset{n\rightarrow \infty }{\lim } d(x_{n},x^{*}) = 0 \end{aligned}$$

    and so

    $$\begin{aligned} \underset{n\rightarrow \infty }{\lim } x_{n} = x^{*}. \end{aligned}$$

    Hence, the fixed point equation (4.1) is well posed.

  3. (c)

    Let \(\eta \ge 0\). Suppose that \(f : X \rightarrow X\) is a mapping such that \(d(Tx,fx) \le \eta \) for all \(x \in X\). By using (4.3), we have

    $$\begin{aligned} d(x,x^{*})&\le s[d(x,Tx) + d(Tx,x^{*})]\\&= s[d(x,Tx) + d(Tx,Tx^{*})]\\&\le s[d(x,Tx) + \frac{\uppsi (d(x,x^{*}))}{s^{2}}] \end{aligned}$$

    for all \(x\in X\). This implies that

    $$\begin{aligned} d(x,x^{*}) - \frac{\uppsi (d(x,x^{*}))}{s} \le s d(x,Tx). \end{aligned}$$

    Therefore,

    $$\begin{aligned} \gamma (d(x,x^{*}))&= d(x,x^{*}) - \frac{\uppsi (d(x,x^{*}))}{s}\\&\le s d(x,Tx). \end{aligned}$$

    and so

    $$\begin{aligned} d(x,x^{*}) \le \gamma ^{-1}(s d(x,Tx)). \end{aligned}$$

    For \(x:=y^{*}\), we have

    $$\begin{aligned} d(y^{*},x^{*})&\le \gamma ^{-1}(s d(y^{*},Ty^{*}))\\&= \gamma ^{-1}(s d(fy^{*},Ty^{*}))\\&\le \gamma ^{-1}(s \eta ). \end{aligned}$$

This completes the proof. \(\square \)

Remark 4.3

By using the variety of choices in the class of modified R-function, we can prove the similar results with the above theorem. It becomes to the useful tool for proving many applications of the main results in this work.