Fixed-Point Theorems on Fuzzy Bipolar $\mathfrak{b}$-Metric Spaces

Abstract

In this manuscript, we establish some fixed-point theorems without continuity by using the triangular property on a fuzzy bipolar $\mathfrak{b}$-metric space as a generalized version and expansion of the well-known results. We also provide some examples and applications of the integral equation to the solution for our main results.

1. Introduction and Preliminaries

In 1960, the concept of continuous triangular norm was initiated by Schweizer and Sklar [1]. After that, Zadeh [2] initiated the theory of fuzzy sets in 1965. This concept of fuzziness was described by Kramosil and Michalek [3] in 1975 using the fuzzy metric space with the help of the continuous t-norm. Moreover, they were interested the notion of fuzziness to improve the well-known fixed-point results by Grabeic [4] in the sense of such spaces. After that, Gregori and Sapena [5] extended the fuzzy Banach contraction map to the fuzzy metric space. In 1994, the fuzzy metric space definition was modified by George and Veeramani [6]. Moreover, Shamas, Rehman, Aydi, Mahmood, and Ameer [7] described more results on the fixed-point results. Mutlu and Gurdal [8] introduced the concept of generalized metric space as a bipolar metric space. Bartwal, Dimri, and Prasad [9] proposed a fuzzy bipolar metric space and proved fixed-point results. Additional articles which relate to the concept can be seen in [10,11,12,13,14]. In this paper, we demonstrate some fixed-point results on a fuzzy bipolar $\mathfrak{b}$-metric space.

Now, we recall some basics

Definition 1

([9]). Let $\tilde{X}$ and $\tilde{Y}$ be two nonempty sets. A quadruple $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\ast )$ is called a fuzzy bipolar metric space (FB-space), where ∗ and ${\mathsf{Ϝ}}_{\varrho }$ are a continuous ⊺-norm and fuzzy set on $\tilde{X}\times \tilde{Y}\times (0,\infty )$, respectively, such that for all $⊺,\eta ,\delta >0$:

(FB1) 

${\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},⊺)>0$ for all $(♭,\widehat{x})\in \tilde{X}\times \tilde{Y}$;

(FB2) 

${\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},⊺)=1$ iff $♭=\widehat{x}$ for $♭\in \tilde{X}$ and $\widehat{x}\in \tilde{Y}$;

(FB3) 

${\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},⊺)={\mathsf{Ϝ}}_{\varrho }(\widehat{x},♭,⊺)$ for all $♭,\widehat{x}\in \tilde{X}\cap \tilde{Y}$;

(FB4) 

${\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_2,⊺+\eta +\delta )\ge {\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_1,⊺)\ast {\mathsf{Ϝ}}_{\varrho }({♭}_2,{\widehat{x}}_1,\eta )\ast {\mathsf{Ϝ}}_{\varrho }({♭}_2,{\widehat{x}}_2,\delta )$ for all ${♭}_1,{♭}_2\in \tilde{X}$ and ${\widehat{x}}_1,{\widehat{x}}_2\in \tilde{Y}$;

(FB5) 

${\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},.):[0,\infty )⟶[0,1]$ is left continuous;

(FB6) 

${\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},.)$ is nondecreasing $\forall ♭\in \tilde{X}$ and $\widehat{x}\in \tilde{Y}$.

Definition 2.

Let $\tilde{X}$ and $\tilde{Y}$ be two nonempty sets and $\mathfrak{s}\ge 1$ be a given real number. A five tuple $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ is called a fuzzy bipolar $\mathfrak{b}$-metric space (FBBMS), where ∗, ${\mathsf{Ϝ}}_{\varrho }$ is a continuous ⊺-norm and fuzzy set on $\tilde{X}\times \tilde{Y}\times (0,\infty )$, respectively, such that for all $⊺,\eta ,\delta >0$:

(FBB1) 

${\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},⊺)>0$ for all $(♭,\widehat{x})\in \tilde{X}\times \tilde{Y}$;

(FBB2) 

${\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},⊺)=1$ iff $♭=\widehat{x}$ for $♭\in \tilde{X}$ and $\widehat{x}\in \tilde{Y}$;

(FBB3) 

${\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},⊺)={\mathsf{Ϝ}}_{\varrho }(\widehat{x},♭,⊺)$ for all $♭,\widehat{x}\in \tilde{X}\cap \tilde{Y}$;

(FBB4) 

${\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_2,\mathfrak{s}(⊺+\eta +\delta ))\ge {\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_1,⊺)\ast {\mathsf{Ϝ}}_{\varrho }({♭}_2,{\widehat{x}}_1,\eta )\ast {\mathsf{Ϝ}}_{\varrho }({♭}_2,{\widehat{x}}_2,\delta )$ for all ${♭}_1,{♭}_2\in \tilde{X}$ and ${\widehat{x}}_1,{\widehat{x}}_2\in \tilde{Y}$;

(FBB5) 

${\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},.):[0,\infty )⟶[0,1]$ is left continuous;

(FBB6) 

${\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},.)$ is nondecreasing for all $♭\in \tilde{X}$ and $\widehat{x}\in \tilde{Y}$.

Definition 3.

Let $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ be a fuzzy bipolar $\mathfrak{b}$-metric space.

(A1) 

A point $♭\in \tilde{X}\cup \tilde{Y}$ is called a left, right and central point if $♭\in \tilde{X}$, $♭\in \tilde{Y}$, and both hold. Similarly, a sequence $\left\{{♭}_{\alpha }\right\}$, $\left\{{\widehat{x}}_{\alpha }\right\}$ on a set $\tilde{X}$, $\tilde{Y}$ are said to be a left and right sequence, respectively.

(A2) 

A sequence $\left\{{♭}_{\alpha }\right\}$ is convergent to a point ♭ if and only if $\left\{{♭}_{\alpha }\right\}$ is a left sequence, ♭ is a right point, and $\underset{\alpha \to \infty }{lim}{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },♭,⊺)=1$ for $⊺>0$, or $\left\{{♭}_{\alpha }\right\}$ is a right sequence, ♭ is a left point, and $\underset{\alpha \to \infty }{lim}{\mathsf{Ϝ}}_{\varrho }(♭,{♭}_{\alpha },⊺)=1$ for $⊺>0$.

(A3) 

A bisequence $(\left\{{♭}_{\alpha }\right\},\left\{{\widehat{x}}_{\alpha }\right\})$ is a sequence on the set $\tilde{X}\times \tilde{Y}$. If the sequences $\left\{{♭}_{\alpha }\right\}$ and $\left\{{\widehat{x}}_{\alpha }\right\}$ are convergent, then the bisequence $(\left\{{♭}_{\alpha }\right\},\left\{{\widehat{x}}_{\alpha }\right\})$ is said to be convergent. $(\left\{{♭}_{\alpha }\right\},\left\{{\widehat{x}}_{\alpha }\right\})$ is a Cauchy bisequence if $\underset{\alpha ,\mathrm{m}\to \infty }{lim}{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}},⊺)=1$ for $⊺>0$.

(A4) 

A fuzzy bipolar $\mathfrak{b}$-metric space is called complete if every Cauchy bisequence is convergent.

Lemma 1.

Let $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ be a fuzzy bipolar $\mathfrak{b}$-metric space. If $\mathfrak{u}\in \tilde{X}\cap \tilde{Y}$ is a limit of a sequence, then it is a unique limit of the sequence.

Definition 4.

Let $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ be a fuzzy bipolar $\mathfrak{b}$-metric space. The fuzzy bipolar $\mathfrak{b}$-metric space ${\mathsf{Ϝ}}_{\varrho }$ is $\mathfrak{b}$-triangular(BT) if

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_2,⊺)}-1& \le \mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_1,⊺)}-1\right)+\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_2,{\widehat{x}}_1,⊺)}-1\right)\hfill \\ \hfill & +\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_2,{\widehat{x}}_2,⊺)}-1\right)\hfill \end{array}$$

Lemma 2.

A fuzzy bipolar $\mathfrak{b}$-metric space ${\mathsf{Ϝ}}_{\varrho }$ is $\mathfrak{b}$-triangular.

Proof. 

Define ${\mathsf{Ϝ}}_{\varrho }:\tilde{X}\times \tilde{Y}\times (0,\infty )\to [0,1]$ given by

$$\begin{array}{c}\hfill {\mathsf{Ϝ}}_{\varrho }=\frac{⊺}{⊺+|♭-\widehat{x}{|}^2}\end{array}$$

for all $⊺>0$, $♭\in \tilde{X}$ and $\widehat{x}\in \tilde{Y}$.

Now,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_2,⊺)}-1& =\frac{|{♭}_1-{\widehat{x}}_2{|}^2}{⊺}=\frac{|{♭}_1-{\widehat{x}}_1+{\widehat{x}}_1-{♭}_2+{♭}_2-{\widehat{x}}_2{|}^2}{⊺}\hfill \\ \hfill & \le \frac{3|{♭}_1-{\widehat{x}}_1{|}^2}{⊺}+\frac{3|{♭}_2-{\widehat{x}}_1{|}^2}{⊺}+\frac{3|{♭}_2-{\widehat{x}}_2{|}^2}{⊺}\hfill \\ \hfill & =3\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_1,⊺)}-1\right)+3\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_2,{\widehat{x}}_1,⊺)}-1\right)\hfill \\ \hfill & +3\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_2,{\widehat{x}}_2,⊺)}-1\right),\hfill \end{array}$$

which implies that

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_2,⊺)}-1& \le 3\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_1,⊺)}-1\right)+3\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_2,{\widehat{x}}_1,⊺)}-1\right)\hfill \\ \hfill & +3\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_2,{\widehat{x}}_2,⊺)}-1\right),\mathrm{for}⊺>0.\hfill \end{array}$$

Hence, the fuzzy bipolar $\mathfrak{b}$-metric ${\mathsf{Ϝ}}_{\varrho }$ is $\mathfrak{b}$-triangular. □

In this manuscript, we prove fixed-point results on FBBMS with some applications.

2. Main Result

Now, we prove our first result:

Theorem 1.

Let $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ be a complete FBBMS with constant $\mathfrak{s}\ge 1$ and the mapping $\check{R}:\tilde{X}\cup \tilde{Y}\to \tilde{X}\cup \tilde{Y}$ such that

(B1) 

$\check{R}\left(\tilde{X}\right)\subseteq \tilde{X}$ and $\check{R}\left(\tilde{Y}\right)\subseteq \tilde{Y}$;

(B2) 

$\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}(♭),\check{R}\left(\widehat{x}\right),⊺)}-1\le \mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},⊺)}-1\right)$, for all $⊺>0$, $\mathfrak{j}\in (0,\frac{1}{s})$ with $\mathfrak{s}\mathfrak{j}<1$;

(B3) 

${\mathsf{Ϝ}}_{\varrho }$ is BT.

Then, $\check{R}$ has a $\mathrm{UFP}$ (unique fixed point).

Proof. 

Fix ${♭}_0\in \tilde{X}$ and ${\widehat{x}}_0\in \tilde{Y}$ and assume that $\check{R}\left({♭}_{\alpha }\right)={♭}_{\alpha +1}$ and $\check{R}\left({\widehat{x}}_{\alpha }\right)={\widehat{x}}_{\alpha +1}$ for all $\alpha \in \mathbb{N}\cup \left\{0\right\}$. Then,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1& =\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left({♭}_{\alpha -1}\right),\check{R}\left({\widehat{x}}_{\alpha -1}\right),⊺)}-1\hfill \\ \hfill & \le \mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha -1},{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & =\mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}{♭}_{\alpha -1},\check{R}{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le {\mathfrak{j}}^{\alpha }\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right).\hfill \end{array}$$

Letting $\alpha \to \infty$, we derive

$$\begin{array}{c}\hfill \underset{\alpha \to \infty }{lim}{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)=1,\mathrm{for}⊺>0.\end{array}$$

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\alpha },⊺)}-1& =\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left({♭}_{\alpha }\right),\check{R}\left({\widehat{x}}_{\alpha -1}\right),⊺)}-1\hfill \\ \hfill & \le \mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & =\mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}{♭}_{\alpha -1},\check{R}{\widehat{x}}_{\alpha -2},⊺)}-1\right)\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le {\mathfrak{j}}^{\alpha }\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_0,⊺)}-1\right).\hfill \end{array}$$

Letting $\alpha \to \infty$, we derive

$$\begin{array}{c}\hfill \underset{\alpha \to \infty }{lim}{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\alpha },⊺)=1,\mathrm{for}⊺>0.\end{array}$$

Let $\alpha <\mathrm{m}$, for $\alpha ,\mathrm{m}\in \mathbb{N}$. Then, since ${\mathsf{Ϝ}}_{\varrho }$ is BT, we obtain

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}},⊺)}-1\le & \mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1\right)+\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\alpha },⊺)}-1\right)\hfill \\ \hfill & +\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\mathrm{m}},⊺)}-1\right)\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le \mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1\right)+\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\alpha },⊺)}-1\right)\hfill \\ \hfill & +\cdots +{\mathfrak{s}}^{\mathrm{m}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\mathrm{m}-1},{\widehat{x}}_{\mathrm{m}-1},⊺)}-1\right)\hfill \\ \hfill & +{\mathfrak{s}}^{\mathrm{m}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\mathrm{m}},{\widehat{x}}_{\mathrm{m}-1},⊺)}-1\right)+{\mathfrak{s}}^{\mathrm{m}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\mathrm{m}},{\widehat{x}}_{\mathrm{m}},⊺)}-1\right)\hfill \\ \hfill & \le \mathfrak{s}{\mathfrak{j}}^{\alpha }\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)+\mathfrak{s}{\mathfrak{j}}^{\alpha }\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +\cdots +{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{\mathrm{m}-1}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)+{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{\mathrm{m}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{\mathrm{m}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & \le \mathfrak{s}{\mathfrak{j}}^{\alpha }(1+\mathfrak{s}\mathfrak{j}+{\mathfrak{s}}^2{\mathfrak{j}}^2+\cdots +{\mathfrak{s}}^{\mathrm{m}-1}{\mathfrak{j}}^{\mathrm{m}-\alpha })\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +\mathfrak{s}{\mathfrak{j}}^{\alpha }(1+\mathfrak{s}\mathfrak{j}+{\mathfrak{s}}^2{\mathfrak{j}}^2+\cdots +{\mathfrak{s}}^{\mathrm{m}-1}{\mathfrak{j}}^{\mathrm{m}-\alpha })\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & \le \frac{\mathfrak{s}{\mathfrak{j}}^{\alpha }}{1-\mathfrak{s}\mathfrak{j}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)+\frac{\mathfrak{s}{\mathfrak{j}}^{\alpha }}{1-\mathfrak{s}\mathfrak{j}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_1,{\widehat{x}}_0,⊺)}-1\right).\hfill \end{array}$$

Since $\mathfrak{s}\mathfrak{j}<1$ and $\alpha ,\mathrm{m}\to \infty$, we obtain

$$\begin{array}{c}\hfill \underset{\alpha ,\mathrm{m}\to \infty }{lim}{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}},⊺)=1,\mathrm{for}⊺>0.\end{array}$$

Thus, $(\left\{{♭}_{\alpha }\right\},\left\{{\widehat{x}}_{\alpha }\right\})$ is a Cauchy bisequence. Since $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ is a complete, it follows that the bisequence $(\left\{{♭}_{\alpha }\right\},\left\{{\widehat{x}}_{\alpha }\right\})$ is convergent. We know that the bisequence $(\left\{{♭}_{\alpha }\right\},\left\{{\widehat{x}}_{\alpha }\right\})$ is a biconvergent sequence. Then, $\left\{{♭}_{\alpha }\right\}\to \mathfrak{u}$ and $\left\{{\widehat{x}}_{\alpha }\right\}\to \mathfrak{u}$ for all $\mathfrak{u}\in \tilde{X}\cap \tilde{Y}$. By Lemma 1, both sequences $\left\{{♭}_{\alpha }\right\}$ and $\left\{{\widehat{x}}_{\alpha }\right\}$ have a unique limit. Since ${\mathsf{Ϝ}}_{\varrho }$ is BT, we prove that

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left(\mathfrak{u}\right),{\mathfrak{u}}_{\alpha },⊺)}-1& \le \mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left(\mathfrak{u}\right),\check{R}\left({\widehat{x}}_{\alpha }\right),⊺)}-1\right)+\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left({♭}_{\alpha }\right),\check{R}\left({\widehat{x}}_{\alpha }\right),⊺)}-1\right)\hfill \\ \hfill & +\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left({♭}_{\alpha }\right),{\mathfrak{u}}_{\alpha },⊺)}-1\right)\hfill \\ \hfill & \le \mathfrak{s}\mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\mathfrak{u},{\widehat{x}}_{\alpha },⊺)}-1\right)+\mathfrak{s}\mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1\right)\hfill \\ \hfill & +\mathfrak{s}\mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\mathfrak{u}}_{\alpha -1},⊺)}-1\right).\hfill \end{array}$$

Letting $\alpha \to \infty$, we obtain

$$\begin{array}{c}\hfill {\mathsf{Ϝ}}_{\varrho }(\check{R}\left(\mathfrak{u}\right),\mathfrak{u},⊺)=1.\end{array}$$

Therefore, $\check{R}\left(\mathfrak{u}\right)=\mathfrak{u}$. Let $\mathfrak{v}\in \tilde{X}\cap \tilde{Y}$ be another fixed point of $\check{R}$. Then,

$$\begin{array}{c}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\mathfrak{u},\mathfrak{v},⊺)}-1=\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left(\mathfrak{u}\right),\check{R}\left(\mathfrak{v}\right),⊺)}-1\le \mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\mathfrak{u},\mathfrak{v},⊺)}-1\right).\end{array}$$

Since $\mathfrak{j}\in (0,\frac{1}{s})$,

$$\begin{array}{c}\hfill {\mathsf{Ϝ}}_{\varrho }(\mathfrak{u},\mathfrak{v},⊺)=1.\end{array}$$

Hence, $\mathfrak{u}=\mathfrak{v}$. □

Example 1.

Consider $\tilde{X}=[0,1]$ and $\tilde{Y}=\left\{0\right\}\cup \mathbb{N}-\left\{1\right\}$ equipped with a continuous ⊺-norm. Define ${\mathsf{Ϝ}}_{\varrho }=\frac{⊺}{⊺+|♭-\widehat{x}{|}^2}$, ∀$⊺>0$, $♭\in \tilde{X}$, and $\widehat{x}\in \tilde{Y}$. Clearly, $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ is a complete FBBMS. Define $\check{R}:\tilde{X}\cup \tilde{Y}\to \tilde{X}\cup \tilde{Y}$ given by

$$\check{R}\left(\mathfrak{u}\right)=\left\{\begin{array}{cc}\frac{\mathfrak{u}+4}{5},\hfill & \mathit{if}\mathfrak{u}\in [0,1],\hfill \\ 0,\hfill & \mathit{if}\mathfrak{u}\in \mathbb{N}-\left\{1\right\},\hfill \end{array}\right.$$

∀$\mathfrak{u}\in \tilde{X}\cup \tilde{Y}$. Then,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}♭,\check{R}\widehat{x},⊺)}-1& =\frac{|\check{R}♭-\check{R}\widehat{x}{|}^2}{⊺}\hfill \\ \hfill & =\frac{|♭-\widehat{x}{|}^2}{25⊺}\hfill \\ \hfill & \le \frac{|♭-\widehat{x}{|}^2}{2⊺}\hfill \\ \hfill & =\frac{1}{2}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},⊺)}-1\right).\hfill \end{array}$$

Hence, all the axioms of Theorem 1 are fulfilled with $\mathfrak{j}=\frac{1}{2}\in (0,1)$. Therefore, $\check{R}$ has a $\mathrm{UFP}$, i.e., $♭=1$.

Theorem 2.

Let $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ be a complete FBBMS with constant $\mathfrak{s}\ge 1$ and the mapping $\check{R}:\tilde{X}\cup \tilde{Y}\to \tilde{X}\cup \tilde{Y}$ such that

(C1) 

$\check{R}\left(\tilde{X}\right)\subseteq \tilde{Y}$ and $\check{R}\left(\tilde{Y}\right)\subseteq \tilde{X}$;

(C2) 

$\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}(♭),\check{R}\left(\widehat{x}\right),⊺)}-1\le \mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},⊺)}-1\right)$, for all $⊺>0$, $\mathfrak{j}\in (0,\frac{1}{s})$ with ${\mathfrak{s}}^2{\mathfrak{j}}^2<1$;

(C3) 

${\mathsf{Ϝ}}_{\varrho }$ is BT.

Then, $\check{R}$ has a $\mathrm{UFP}$.

Proof. 

Fix ${♭}_0\in \tilde{X}$ and consider $\check{R}\left({♭}_{\alpha }\right)={\widehat{x}}_{\alpha }$ and $\check{R}\left({\widehat{x}}_{\alpha }\right)={♭}_{\alpha +1}$, $\forall \alpha \in \mathbb{N}\cup \left\{0\right\}$. Then,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1& =\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left({\widehat{x}}_{\alpha -1}\right),\check{R}\left({♭}_{\alpha }\right),⊺)}-1\hfill \\ \hfill & \le \mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & =\mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}{\widehat{x}}_{\alpha -1},\check{R}{♭}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le {\mathfrak{j}}^{2\alpha }\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right).\hfill \end{array}$$

Letting $\alpha \to \infty$, we derive

$$\begin{array}{c}\hfill \underset{\alpha \to \infty }{lim}{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)=1,\mathrm{for}⊺>0.\end{array}$$

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\alpha },⊺)}-1& =\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left({\widehat{x}}_{\alpha }\right),\check{R}\left({♭}_{\alpha }\right),⊺)}-1\hfill \\ \hfill & \le \mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & =\mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}{\widehat{x}}_{\alpha -1},\check{R}{♭}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le {\mathfrak{j}}^{2\alpha +1}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right).\hfill \end{array}$$

Letting $\alpha \to \infty$, we derive

$$\begin{array}{c}\hfill \underset{\alpha \to \infty }{lim}{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\alpha },⊺)=1,\mathrm{for}⊺>0.\end{array}$$

Let $\alpha <\mathrm{m}$, for $\alpha ,\mathrm{m}\in \mathbb{N}$. Then, since ${\mathsf{Ϝ}}_{\varrho }$ is BT, we obtain

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}},⊺)}-1\le & \mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1\right)+\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\alpha },⊺)}-1\right)\hfill \\ \hfill & +\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\mathrm{m}},⊺)}-1\right)\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le \mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1\right)+\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\alpha },⊺)}-1\right)\hfill \\ \hfill & +\cdots +{\mathfrak{s}}^{\mathrm{m}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\mathrm{m}-1},{\widehat{x}}_{\mathrm{m}-1},⊺)}-1\right)\hfill \\ \hfill & +{\mathfrak{s}}^{\mathrm{m}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\mathrm{m}},{\widehat{x}}_{\mathrm{m}-1},⊺)}-1\right)+{\mathfrak{s}}^{\mathrm{m}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\mathrm{m}},{\widehat{x}}_{\mathrm{m}},⊺)}-1\right)\hfill \\ \hfill & \le \mathfrak{s}{\mathfrak{j}}^{2\alpha }\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)+\mathfrak{s}{\mathfrak{j}}^{2\alpha +1}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +\cdots +{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{2\mathrm{m}-2}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)+{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{2\mathrm{m}-1}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{2\mathrm{m}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & \le \mathfrak{s}{\mathfrak{j}}^{2\alpha }(1+{\mathfrak{s}}^2{\mathfrak{j}}^2+\cdots +{\mathfrak{s}}^{\mathrm{m}-1}{\mathfrak{j}}^{2\mathrm{m}-2\alpha })\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +\mathfrak{s}{\mathfrak{j}}^{2\alpha +1}(1+{\mathfrak{s}}^2{\mathfrak{j}}^2+\cdots +{\mathfrak{s}}^{\mathrm{m}-1}{\mathfrak{j}}^{2\mathrm{m}-2\alpha -2})\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & \le \frac{\mathfrak{s}{\mathfrak{j}}^{2\alpha }}{1-{\mathfrak{s}}^{\mathfrak{2}}{\mathfrak{j}}^2}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)+\frac{\mathfrak{s}{\mathfrak{j}}^{2\alpha +1}}{1-{\mathfrak{s}}^{\mathfrak{2}}{\mathfrak{j}}^2}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right).\hfill \end{array}$$

Since ${\mathfrak{s}}^2{\mathfrak{j}}^2<1$, letting $\alpha ,\mathrm{m}\to \infty$, we obtain

$$\begin{array}{c}\hfill \underset{\alpha ,\mathrm{m}\to \infty }{lim}{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}},⊺)=1,\mathrm{for}⊺>0.\end{array}$$

Thus, $(\left\{{♭}_{\alpha }\right\},\left\{{\widehat{x}}_{\alpha }\right\})$ is a Cauchy bisequence. Since $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ is a complete, it follows that the bisequence $(\left\{{♭}_{\alpha }\right\},\left\{{\widehat{x}}_{\alpha }\right\})$ is convergent. We know that the bisequence $(\left\{{♭}_{\alpha }\right\},\left\{{\widehat{x}}_{\alpha }\right\})$ is a biconvergent sequence. Then, $\left\{{♭}_{\alpha }\right\}\to \mathfrak{u}$ and $\left\{{\widehat{x}}_{\alpha }\right\}\to \mathfrak{u}$ for all $\mathfrak{u}\in \tilde{X}\cap \tilde{Y}$. By Lemma 1, both sequences $\left\{{♭}_{\alpha }\right\}$ and $\left\{{\widehat{x}}_{\alpha }\right\}$ have a unique limit.

Now, we show that $\mathfrak{u}\in \tilde{X}\cap \tilde{Y}$ is a fixed point of $\check{R}$. Since ${\mathsf{Ϝ}}_{\varrho }$ is BT, we prove that

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left(\mathfrak{u}\right),{\mathfrak{u}}_{\alpha },⊺)}-1& \le \mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left(\mathfrak{u}\right),\check{R}\left({\widehat{x}}_{\alpha }\right),⊺)}-1\right)+\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left({♭}_{\alpha }\right),\check{R}\left({\widehat{x}}_{\alpha }\right),⊺)}-1\right)\hfill \\ \hfill & +\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left({♭}_{\alpha }\right),{\mathfrak{u}}_{\alpha },⊺)}-1\right)\hfill \\ \hfill & \le \mathfrak{s}\mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\mathfrak{u},{\widehat{x}}_{\alpha },⊺)}-1\right)+\mathfrak{s}\mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1\right)\hfill \\ \hfill & +\mathfrak{s}\mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\mathfrak{u}}_{\alpha -1},⊺)}-1\right).\hfill \end{array}$$

Letting $\alpha \to \infty$, we obtain

$$\begin{array}{c}\hfill {\mathsf{Ϝ}}_{\varrho }(\check{R}\left(\mathfrak{u}\right),\mathfrak{u},⊺)=1.\end{array}$$

Therefore, $\check{R}\left(\mathfrak{u}\right)=\mathfrak{u}$. Let $\mathfrak{v}\in \tilde{X}\cap \tilde{Y}$ be another fixed point of $\check{R}$. Then,

$$\begin{array}{c}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\mathfrak{u},\mathfrak{v},⊺)}-1=\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\left(\mathfrak{u}\right),\check{R}\left(\mathfrak{v}\right),⊺)}-1\le \mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\mathfrak{u},\mathfrak{v},⊺)}-1\right).\end{array}$$

Since $\mathfrak{j}\in (0,\frac{1}{s})$,

$$\begin{array}{c}\hfill {\mathsf{Ϝ}}_{\varrho }(\mathfrak{u},\mathfrak{v},⊺)=1.\end{array}$$

Hence, $\mathfrak{u}=\mathfrak{v}$. □

Example 2.

Let $\tilde{X}=\{0,1,2,7\}$ and $\tilde{Y}=\{0,\frac{1}{4},\frac{1}{2},\frac{5}{6},3\}$ be equipped with a continuous ⊺-norm. Define ${\mathsf{Ϝ}}_{\varrho }=\frac{⊺}{⊺+|♭-\widehat{x}{|}^2}$, ∀$⊺>0$, $♭\in \tilde{X}$, and $\widehat{x}\in \tilde{Y}$. Clearly, $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ is a complete FBBMS. Define $\check{R}:\tilde{X}\cup \tilde{Y}\to \tilde{X}\cup \tilde{Y}$ given by

$$\check{R}\left(\mathfrak{u}\right)=\left\{\begin{array}{cc}\frac{\mathfrak{u}+5}{6},\hfill & \mathit{if}\mathfrak{u}\in \{0,1,2,7\},\hfill \\ 0,\hfill & \mathit{if}\mathfrak{u}\in \{\frac{1}{4},\frac{1}{2},3\}.\hfill \end{array}\right.$$

Then,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}♭,\check{R}\widehat{x},⊺)}-1& =\frac{|\check{R}♭-\check{R}\widehat{x}{|}^2}{⊺}\hfill \\ \hfill & =\frac{|♭-\widehat{x}{|}^2}{36⊺}\hfill \\ \hfill & \le \frac{|♭-\widehat{x}{|}^2}{2⊺}\hfill \\ \hfill & =\frac{1}{2}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},⊺)}-1\right).\hfill \end{array}$$

Hence, all the axioms of Theorem 2 are fulfilled with $\mathfrak{j}=\frac{1}{2}\in (0,1)$. Therefore, $\check{R}$ has a $\mathrm{UFP}$, i.e., $♭=1$.

Theorem 3.

Let $\check{R}:(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )\leftrightarrows (\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$, where $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ is a complete FBBMS with $\mathfrak{s}\ge 1$, and let $\mathfrak{h}\in (0,\frac{1}{2})$ such that

(H1) 

$\check{R}\left(\tilde{X}\right)\subseteq \tilde{Y}$ and $\check{R}\left(\tilde{Y}\right)\subseteq \tilde{X}$;

(H2) 

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\widehat{x},\check{R}♭,⊺)}-1& \le \mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},⊺)}-1\right)+\gamma \left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(♭,\check{R}♭,⊺)}-1\right)\hfill \\ \hfill & +ı\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\widehat{x},\widehat{x},⊺)}-1\right)\hfill \end{array}$$

hold for all $♭\in \tilde{X}$ and $\widehat{x}\in \tilde{Y}$, where $\mathfrak{h},\gamma ,ı\ge 0$ such that $\mathfrak{h}+\gamma +ı<1$ and $\mathfrak{s}\left(\frac{\mathfrak{h}+ı}{1-\gamma }\right)\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)<1$;

(H3) 

${\mathsf{Ϝ}}_{\varrho }$ is BT.

Then, $\check{R}$ has a $\mathrm{UFP}$.

Proof. 

Let ${♭}_0\in \tilde{X}$, and for each $\alpha \ge 0$, we define ${\widehat{x}}_{\alpha }=\check{R}{♭}_{\alpha }$ and ${♭}_{\alpha +1}=\check{R}{\widehat{x}}_{\alpha }$. Then,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1& =\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}{\widehat{x}}_{\alpha -1},\check{R}{♭}_{\alpha },⊺)}-1\hfill \\ \hfill & \le \mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1\right)+\gamma \left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },\check{R}{♭}_{\alpha },⊺)}-1\right)\hfill \\ \hfill & +ı\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}{\widehat{x}}_{\alpha -1},{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & =\mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1\right)+\gamma \left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1\right)\hfill \\ \hfill & +ı\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \end{array}$$

for all integers $\alpha \ge 1$. Then,

$$\begin{array}{c}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1\le \frac{\mathfrak{h}+ı}{1-\gamma }\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1\right),\end{array}$$

and

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1& =\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}{\widehat{x}}_{\alpha -1},\check{R}{♭}_{\alpha -1},⊺)}-1\hfill \\ \hfill & \le \mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha -1},{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & +\gamma \left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha -1},\check{R}{♭}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & +ı\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}{\widehat{x}}_{\alpha -1},{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & =\mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha -1},{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & +\gamma \left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha -1},{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & +ı\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1\right),\hfill \end{array}$$

so

$$\begin{array}{c}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1\le \frac{\mathfrak{h}+\gamma }{1-ı}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha -1},{\widehat{x}}_{\alpha -1},⊺)}-1\right).\end{array}$$

Consequently,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1& \le {\mathfrak{j}}^{\alpha }\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right),\hfill \\ \hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\alpha },⊺)}-1& \le {\mathfrak{j}}^{\alpha }\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right),\hfill \end{array}$$

for all $\alpha \ge 0$, where $\mathfrak{j}:=\left(\frac{\mathfrak{h}+ı}{1-\gamma }\right)\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)$. For each $\mathrm{m}>\alpha$,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}},⊺)}-1& \le \mathfrak{s}(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1+\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\alpha },⊺)}-1\hfill \\ \hfill & +\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\mathrm{m}},⊺)}-1)\hfill \\ & \le \left(\mathfrak{s}{\mathfrak{j}}^{\alpha }+\mathfrak{s}{\mathfrak{j}}^{\alpha }\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)\right)\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\mathrm{m}},⊺)}-1\right)\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le (\mathfrak{s}{\mathfrak{j}}^{\alpha }+\mathfrak{s}{\mathfrak{j}}^{\alpha }\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)+\cdots +{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{\mathrm{m}-1}\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)\hfill \\ \hfill & +{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{\mathrm{m}})\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & =(\mathfrak{s}{\mathfrak{j}}^{\alpha }+\cdots +{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{\mathrm{m}})\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +\left(\mathfrak{s}{\mathfrak{j}}^{\alpha }\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)+\cdots +{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{\mathrm{m}-1}\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)\right)\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & \le \frac{\mathfrak{s}{\mathfrak{j}}^{\alpha }}{1-\mathfrak{s}\mathfrak{j}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)+\frac{\mathfrak{s}{\mathfrak{j}}^{\alpha }}{1-\mathfrak{s}\mathfrak{j}}\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right).\hfill \end{array}$$

Since $\mathfrak{s}\mathfrak{j}<1$ and $\alpha ,\mathrm{m}\to \infty$, we obtain

$$\begin{array}{c}\hfill \underset{\alpha ,\mathrm{m}\to \infty }{lim}{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}},⊺)=1,\mathrm{for}⊺>0.\end{array}$$

Consequently, for each $\mathrm{m}<\alpha$

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}},⊺)}-1& \le \mathfrak{s}(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\mathrm{m}+1},{\widehat{x}}_{\mathrm{m}},⊺)}-1\hfill \\ \hfill & +\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\mathrm{m}+1},{\widehat{x}}_{\mathrm{m}+1})}-1+\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}+1})}-1)\hfill \\ & \le (\mathfrak{s}{\mathfrak{j}}^{\mathrm{m}}\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)+\mathfrak{s}{\mathfrak{j}}^{\mathrm{m}+1})\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}+1},⊺)}-1\right)\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le (\mathfrak{s}{\mathfrak{j}}^{\mathrm{m}}\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)+\mathfrak{s}{\mathfrak{j}}^{\mathrm{m}+1}\hfill \\ \hfill & +\cdots +{\mathfrak{s}}^{\alpha }{\mathfrak{j}}^{\alpha -1}\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)+{\mathfrak{s}}^{\alpha }{\mathfrak{j}}^{\alpha })\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & \le \left(\mathfrak{s}{\mathfrak{j}}^{\mathrm{m}}\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)+\cdots +{\mathfrak{s}}^{\alpha }{\mathfrak{j}}^{\alpha -1}\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)\right)\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +(\mathfrak{s}{\mathfrak{j}}^{\mathrm{m}+1}+\cdots +{\mathfrak{s}}^{\alpha }{\mathfrak{j}}^{\alpha })\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & \le \frac{\mathfrak{s}{\mathfrak{j}}^{\mathrm{m}}}{1-\mathfrak{s}\mathfrak{j}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\left(\frac{\mathfrak{h}+\gamma }{1-ı}\right)+\frac{\mathfrak{s}{\mathfrak{j}}^{\mathrm{m}+1}}{1-\mathfrak{s}\mathfrak{j}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right).\hfill \end{array}$$

Since $\mathfrak{s}\mathfrak{j}<1$ and $\alpha ,\mathrm{m}\to \infty$, we obtain

$$\begin{array}{c}\hfill \underset{\alpha ,\mathrm{m}\to \infty }{lim}{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}},⊺)=1,\mathrm{for}⊺>0.\end{array}$$

Therefore, $(\left\{{♭}_{\alpha }\right\},\left\{{\widehat{x}}_m\right\})$ is a Cauchy bisequence. Since $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ is complete, $\left\{{♭}_{\alpha }\right\}\to \varpi ,\left\{{\widehat{x}}_{\mathrm{m}}\right\}\to \varpi$, where $\varpi \in \tilde{X}\cup \tilde{Y}$. Moreover,

$$\begin{array}{c}\hfill \left\{\check{R}{♭}_{\alpha }\right\}=\left\{{\widehat{x}}_{\alpha }\right\}\to \varpi .\end{array}$$

On the other hand,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\varpi ,\check{R}{♭}_{\alpha },⊺)}-1& \le \mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },\varpi ,⊺)}-1\right)\hfill \\ \hfill & +\gamma \left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },\check{R}{♭}_{\alpha },⊺)}-1\right)+ı\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\varpi ,\varpi ,⊺)}-1\right),\hfill \end{array}$$

which in turn implies that $\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\varpi ,\varpi ,⊺)}-1\le \frac{1}{1-(\gamma +ı)}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\varpi ,\varpi ,⊺)}-1\right)$. Hence, $\check{R}\varpi =\varpi$. Let $\varrho$ be another fixed point of $\check{R}$. Then,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\varpi ,\varrho ,⊺)}-1=\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\varpi ,\check{R}\varrho ,⊺)}-1& \le \mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\varrho ,\varpi ,⊺)}-1\right)\hfill \\ \hfill & +\gamma \left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\varpi ,\check{R}\varpi ,⊺)}-1\right)\hfill \\ \hfill & +ı\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\varrho ,\varrho ,⊺)}-1\right)\hfill \\ \hfill & =\mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\varpi ,\varrho ,⊺)}-1\right).\hfill \end{array}$$

Consequently $\varpi =\varrho$. □

Next, we present a result which uses a Kannan type contraction [15].

Theorem 4.

Let $\check{R}:(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )\leftrightarrows (\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$, where $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ is a complete FBBMS with $\mathfrak{s}\ge 1$, and let $\mathfrak{h}\in (0,\frac{1}{2})$ such that the inequality

$$\begin{array}{c}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\widehat{x},\check{R}♭,⊺)}-1\le \mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(♭,\check{R}♭,⊺)}-1+\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\widehat{x},\widehat{x},⊺)}-1\right)\end{array}$$

holds for all $♭\in \tilde{X}$ and $\widehat{x}\in \tilde{Y}$ with $\mathfrak{s}{\mathfrak{j}}^2<1$. Then, the function $\check{R}:\tilde{X}\cup$$\tilde{Y}\to \tilde{X}\cup \tilde{Y}$ has a $\mathrm{UFP}$.

Proof. 

Let ${♭}_0\in \tilde{X}$, and for each $\alpha \ge 0$, we define ${\widehat{x}}_{\alpha }=\check{R}{♭}_{\alpha }$ and ${♭}_{\alpha +1}=\check{R}{\widehat{x}}_{\alpha }$. Then,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1& =\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}{\widehat{x}}_{\alpha -1},\check{R}{♭}_{\alpha },⊺)}-1\hfill \\ \hfill & \le \mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },\check{R}{♭}_{\alpha },⊺)}-1+\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}{\widehat{x}}_{\alpha -1},{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & =\mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1+\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \end{array}$$

for all integers $\alpha \ge 1$. Then,

$$\begin{array}{c}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1\le \frac{\mathfrak{h}}{1-\mathfrak{h}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1\right),\end{array}$$

and

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1& =\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}{\widehat{x}}_{\alpha -1},\check{R}{♭}_{\alpha -1},⊺)}-1\hfill \\ \hfill & \le \mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha -1},\check{R}{♭}_{\alpha -1},⊺)}-1+\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}{\widehat{x}}_{\alpha -1},{\widehat{x}}_{\alpha -1},⊺)}-1\right)\hfill \\ \hfill & =\mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha -1},{\widehat{x}}_{\alpha -1},⊺)}-1+\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1\right),\hfill \end{array}$$

so

$$\begin{array}{c}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1\le \frac{\mathfrak{h}}{1-\mathfrak{h}}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha -1},{\widehat{x}}_{\alpha -1},⊺)}-1\right).\end{array}$$

If we take $\mathfrak{j}:=\frac{\mathfrak{h}}{1-\mathfrak{h}}$, then we have $\mathfrak{j}\in (0,1)$ since $\mathfrak{h}\in (0,\frac{1}{2})$. Now,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1& \le {\mathfrak{j}}^{2\alpha }\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right),\hfill \\ \hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha -1},⊺)}-1& \le {\mathfrak{j}}^{2\alpha -1}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right).\hfill \end{array}$$

For each $\mathrm{m}>\alpha$,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}},⊺)}-1& \le \mathfrak{s}(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1+\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\alpha },⊺)}-1\hfill \\ \hfill & +\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\mathrm{m}},⊺)}-1)\hfill \\ & \le (\mathfrak{s}{\mathfrak{j}}^{2\alpha }+\mathfrak{s}{\mathfrak{j}}^{2\alpha +1})\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha +1},{\widehat{x}}_{\mathrm{m}},⊺)}-1\right)\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le (\mathfrak{s}{\mathfrak{j}}^{2\alpha }+\mathfrak{s}{\mathfrak{j}}^{2\alpha +1}+\cdots +{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{2\mathrm{m}-1}+{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{2\mathrm{m}})\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & =(\mathfrak{s}{\mathfrak{j}}^{2\alpha }+\cdots +{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{2\mathrm{m}})\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +(\mathfrak{s}{\mathfrak{j}}^{2\alpha +1}+\cdots +{\mathfrak{s}}^{\mathrm{m}}{\mathfrak{j}}^{2\mathrm{m}-1})\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & \le \frac{\mathfrak{s}{\mathfrak{j}}^{2\alpha }}{1-\mathfrak{s}{\mathfrak{j}}^2}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)+\frac{\mathfrak{s}{\mathfrak{j}}^{2\alpha +1}}{1-\mathfrak{s}{\mathfrak{j}}^2}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right).\hfill \end{array}$$

Since $\mathfrak{s}{\mathfrak{j}}^2<1$ and $\alpha ,\mathrm{m}\to \infty$, we obtain

$$\begin{array}{c}\hfill \underset{\alpha ,\mathrm{m}\to \infty }{lim}{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}},⊺)=1,\mathrm{for}⊺>0.\end{array}$$

Consequently, for each $\mathrm{m}<\alpha$

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}},⊺)}-1& \le \mathfrak{s}(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\mathrm{m}+1},{\widehat{x}}_{\mathrm{m}},⊺)}-1\hfill \\ \hfill & +\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\mathrm{m}+1},{\widehat{x}}_{\mathrm{m}+1})}-1+\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}+1})}-1)\hfill \\ & \le (\mathfrak{s}{\mathfrak{j}}^{2\mathrm{m}+1}+\mathfrak{s}{\mathfrak{j}}^{2\mathrm{m}+2})\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +\mathfrak{s}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}+1},⊺)}-1\right)\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le (\mathfrak{s}{\mathfrak{j}}^{2\mathrm{m}+1}+\mathfrak{s}{\mathfrak{j}}^{2\mathrm{m}+2}+\cdots +{\mathfrak{s}}^{\alpha }{\mathfrak{j}}^{2\alpha })\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +{\mathfrak{s}}^{\alpha }\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1\right)\hfill \\ \hfill & \le (\mathfrak{s}{\mathfrak{j}}^{2\mathrm{m}+1}+\mathfrak{s}{\mathfrak{j}}^{2\mathrm{m}+2}+\cdots +{\mathfrak{s}}^{\alpha }{\mathfrak{j}}^{2\alpha +1})\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & =(\mathfrak{s}{\mathfrak{j}}^{2\mathrm{m}+1}+\cdots +{\mathfrak{s}}^{\alpha }{\mathfrak{j}}^{2\alpha +1})\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & +(\mathfrak{s}{\mathfrak{j}}^{2\mathrm{m}+2}+\cdots +{\mathfrak{s}}^{\alpha }{\mathfrak{j}}^{2\alpha })\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)\hfill \\ \hfill & \le \frac{\mathfrak{s}{\mathfrak{j}}^{2\mathrm{m}+1}}{1-\mathfrak{s}{\mathfrak{j}}^2}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right)+\frac{\mathfrak{s}{\mathfrak{j}}^{2\mathrm{m}+2}}{1-\mathfrak{s}{\mathfrak{j}}^2}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_0,{\widehat{x}}_0,⊺)}-1\right).\hfill \end{array}$$

Since $\mathfrak{s}{\mathfrak{j}}^2<1$, as $\alpha ,\mathrm{m}\to \infty$, we obtain

$$\begin{array}{c}\hfill \underset{\alpha ,\mathrm{m}\to \infty }{lim}{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\mathrm{m}},⊺)=1,\mathrm{for}⊺>0.\end{array}$$

Therefore, $(\left\{{♭}_{\alpha }\right\},\left\{{\widehat{x}}_m\right\})$ is a Cauchy bisequence. Since $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\ast )$ is complete, $\left\{{♭}_{\alpha }\right\}\to \varpi ,\left\{{\widehat{x}}_{\mathrm{m}}\right\}\to \varpi$, where $\varpi \in \tilde{X}\cup \tilde{Y}$. Hence,

$$\begin{array}{c}\hfill \left\{\check{R}{♭}_{\alpha }\right\}=\left\{{\widehat{x}}_{\alpha }\right\}\to \varpi .\end{array}$$

On the other hand,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\varpi ,\check{R}{♭}_{\alpha },⊺)}-1& \le \mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },\check{R}{♭}_{\alpha },⊺)}-1+\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\varpi ,\varpi ,⊺)}-1\right)\hfill \\ \hfill & =\mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }({♭}_{\alpha },{\widehat{x}}_{\alpha },⊺)}-1+\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\varpi ,\varpi ,⊺)}-1\right);\hfill \end{array}$$

which in turn implies that $\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\varpi ,\varpi ,⊺)}-1\le \mathfrak{h}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\varpi ,\varpi ,⊺)}-1\right)$. Hence, $\check{R}\varpi =\varpi$. Let $\varrho$ be another fixed point of $\check{R}$. Then,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\varpi ,\varrho ,⊺)}-1=\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\varpi ,\check{R}\varrho ,⊺)}-1& \le \mathfrak{h}(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\varpi ,\check{R}\varpi ,⊺)}-1\hfill \\ \hfill & +\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}\varrho ,\varrho ,⊺)}-1)\hfill \\ \hfill & =\mathfrak{h}(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\varpi ,\varpi ,⊺)}-1\hfill \\ \hfill & +\frac{1}{{\mathsf{Ϝ}}_{\varrho }(\varrho ,\varrho ,⊺)-1})=0.\hfill \end{array}$$

Consequently, $\varpi =\varrho$. □

3. Applications

3.1. Application I

Consider the integral equation as follows.

Theorem 5.

Let us consider the integral equation

$$\begin{array}{c}\hfill ♭(\mathit{j})=\varrho \left(\mathsf{ȷ}\right)+{\int }_{{\mathcal{E}}_1\cup {\mathcal{E}}_2}\Omega (\mathsf{ȷ},\widehat{x},♭\left(\widehat{x}\right))d\widehat{x},\mathsf{ȷ}\in {\mathcal{E}}_1\cup {\mathcal{E}}_2,\end{array}$$

(1)

where ${\mathcal{E}}_1\cup {\mathcal{E}}_2$ is a Lebesgue measurable set. Suppose

(T1) 

$\Omega :({\mathcal{E}}_1^2\cup {\mathcal{E}}_2^2)\times [0,\infty )\to [0,\infty )$ and $b\in L^{\infty }\left({\mathcal{E}}_1\right)\cup L^{\infty }\left({\mathcal{E}}_2\right)$;

(T2) 

There is a continuous function $\theta :{\mathcal{E}}_1^2\cup {\mathcal{E}}_2^2\to [0,\infty )$ and $\mathfrak{j}\in (0,1)$ satisfying

$$\begin{array}{c}\hfill |\Omega (\mathsf{ȷ},\widehat{x},♭\left(\widehat{x}\right))-\Omega (\mathsf{ȷ},\widehat{x},\widehat{x}\left(\widehat{x}\right))|\le \sqrt{\mathfrak{j}\theta (\mathsf{ȷ},\widehat{x})}\left(\right|♭\left(\mathsf{ȷ}\right)-\widehat{x}\left(\mathsf{ȷ}\right)\left|\right),\end{array}$$

for $\mathsf{ȷ},\widehat{x}\in {\mathcal{E}}_1^2\cup {\mathcal{E}}_2^2$;

(T3) 

${\int }_{{\mathcal{E}}_1\cup {\mathcal{E}}_2}\theta (\mathsf{ȷ},\widehat{x})d\widehat{x}\le 1$.

Then, the integral equation has a unique solution in $L^{\infty }\left({\mathcal{E}}_1\right)\cup L^{\infty }\left({\mathcal{E}}_2\right)$.

Proof. 

Consider $\tilde{X}=L^{\infty }\left({\mathcal{E}}_1\right)$ and $\tilde{Y}=L^{\infty }\left({\mathcal{E}}_2\right)$ two normed linear spaces, where ${\mathcal{E}}_1,{\mathcal{E}}_2$ are Lebesgue measurable sets and $m({\mathcal{E}}_1\cup {\mathcal{E}}_2)<\infty$.

• Let ${\mathsf{Ϝ}}_{\varrho }:\tilde{X}\times \tilde{Y}\times (0,\infty )\to [0,1]$ given by

  $$\begin{array}{c}\hfill {\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x},⊺)=\frac{⊺}{⊺+|♭-\widehat{x}{|}^2}.\end{array}$$

  for all $♭\in \tilde{X},\widehat{x}\in \tilde{Y}$. Then, $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\star )$ is a complete FBBMS.
• Define $\check{R}:L^{\infty }\left({\mathcal{E}}_1\right)\cup L^{\infty }\left({\mathcal{E}}_2\right)\to L^{\infty }\left({\mathcal{E}}_1\right)\cup L^{\infty }\left({\mathcal{E}}_2\right)$ given by

  $$\begin{array}{c}\hfill \check{R}(♭\left(\mathsf{ȷ}\right))=\varrho \left(\mathsf{ȷ}\right)+{\int }_{{\mathcal{E}}_1\cup {\mathcal{E}}_2}\Omega (\mathsf{ȷ},\widehat{x},♭\left(\widehat{x}\right))d\widehat{x},\mathsf{ȷ}\in {\mathcal{E}}_1\cup {\mathcal{E}}_2.\end{array}$$

Now,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}♭\left(\mathsf{ȷ}\right),\check{R}\widehat{x}\left(\mathsf{ȷ}\right),⊺)}-1& =\frac{|\check{R}♭\left(\mathsf{ȷ}\right)-\check{R}\widehat{x}{\left(\mathsf{ȷ}\right)|}^2}{⊺}\hfill \\ \hfill & =\frac{|\varrho \left(\mathsf{ȷ}\right)+{\int }_{{\mathcal{E}}_1\cup {\mathcal{E}}_2}\Omega (\mathsf{ȷ},\widehat{x},♭\left(\widehat{x}\right))d\widehat{x}-\left(\varrho \left(\mathsf{ȷ}\right)+{\int }_{{\mathcal{E}}_1\cup {\mathcal{E}}_2}\Omega (\mathsf{ȷ},\widehat{x},♭\left(\widehat{x}\right))d\widehat{x}\right){|}^2}{⊺}\hfill \\ \hfill & =\frac{|{\int }_{{\mathcal{E}}_1\cup {\mathcal{E}}_2}(\Omega (\mathsf{ȷ},\widehat{x},♭\left(\widehat{x}\right))-\Omega (\mathsf{ȷ},\widehat{x},♭\left(\widehat{x}\right)))d\widehat{x}{|}^2}{⊺}\hfill \\ \hfill & \le \frac{{\int }_{{\mathcal{E}}_1\cup {\mathcal{E}}_2}\mathfrak{j}\theta (\mathsf{ȷ},\widehat{x})\left(\right|♭\left(\mathsf{ȷ}\right)-\widehat{x}\left(\mathsf{ȷ}\right){|}^2)d\widehat{x}}{⊺}\hfill \\ \hfill & \le \frac{\mathfrak{j}|♭\left(\mathsf{ȷ}\right)-\widehat{x}{\left(\mathsf{ȷ}\right)|}^2}{⊺}\hfill \\ \hfill & =\mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(♭\left(\mathsf{ȷ}\right),\widehat{x}\left(\mathsf{ȷ}\right),⊺)}-1\right).\hfill \end{array}$$

Hence, all the axioms of Theorem 1 are fulfilled and consequently, Equation (1) has a unique solution. □

3.2. Application II

We recall many important definitions from fractional calculus theory [16,17]. For a function $♭\in \mathcal{C}[0,1]$, the order $\delta >0$ of Riemann–Liouville fractional derivative is

$$\begin{array}{c}\hfill \frac{1}{\mathsf{\Gamma }(\vartheta -\delta )}\frac{d^{\vartheta }}{d{\mathsf{ȷ}}^{\vartheta }}{\int }_0^{\mathsf{ȷ}}\frac{♭\left(\mathfrak{e}\right)d\mathfrak{e}}{{(\mathsf{ȷ}-\mathfrak{e})}^{\delta -\vartheta +1}}={\mathcal{D}}^{\delta }♭\left(\mathsf{ȷ}\right).\end{array}$$

(2)

From (2), the right-hand side is pointwise defined on $[0,1]$, where $\left[\delta \right]$ and $\mathsf{\Gamma }$ are the integer part of the number $\delta$ and the Euler gamma function.

Consider the following FDE (fractional differential equation)

$$\begin{array}{cc}\hfill & {}^{\mathfrak{e}}{\mathcal{D}}^{♭}♭\left(\mathsf{ȷ}\right)+\ell (\mathsf{ȷ},♭\left(\mathsf{ȷ}\right))=0,1\le \mathsf{ȷ}\le 0,2\le ♭>1;\hfill \\ \hfill & ♭\left(0\right)=♭\left(1\right)=0,\hfill \end{array}$$

(3)

where $\ell :[0,1]\times \mathbb{R}\to \mathbb{R}$ is a continuous function and ${}^{\mathfrak{e}}{\mathcal{D}}^{♭}$ represents the Caputo fractional derivative of order ♭, and it is defined by

$$\begin{array}{c}\hfill {}^{\mathfrak{e}}{\mathcal{D}}^{♭}=\frac{1}{\mathsf{\Gamma }(\vartheta -♭)}{\int }_0^{\zeta }\frac{{♭}^{\vartheta }\left(\mathfrak{e}\right)d\mathfrak{e}}{{(\mathsf{ȷ}-\mathfrak{e})}^{♭-\vartheta +1}}\end{array}$$

Let $\tilde{X}=(\mathcal{C}[0,1],[0,\infty ))=\{\mathfrak{g}:[0,1]\to [0,\infty ):\mathfrak{g}\mathrm{be}\mathrm{a}\mathrm{continuous}\mathrm{function}\}$ and $\tilde{Y}=(\mathcal{C}[0,1],(-\infty ,0])=\{\mathfrak{g}:[0,1]\to (-\infty ,0]:\mathfrak{g}\mathrm{be}\mathrm{a}\mathrm{continuous}\mathrm{function}\}$. Consider ${\mathsf{Ϝ}}_{\varrho }:\tilde{X}\times \tilde{Y}\times (0,\infty )\to {\mathbb{R}}^{+}$ given by

$$\begin{array}{c}\hfill {\mathsf{Ϝ}}_{\varrho }(♭,{♭}^{{}^{\prime }},⊺)=\frac{⊺}{⊺+{sup}_{\mathsf{ȷ}\in [0,1]}{|♭\left(\mathsf{ȷ}\right)-\widehat{x}\left(\mathsf{ȷ}\right)|}^2}.\end{array}$$

for all $(♭,\widehat{x})\in \tilde{X}\times \tilde{Y}$. Then, $(\tilde{X},\tilde{Y},{\mathsf{Ϝ}}_{\varrho },\mathfrak{s},\star )$ is a complete FBBMS.

Theorem 6.

Let us consider the nonlinear FDE (3). Suppose that the following conditions are satisfied:

(i) 

We can find $\mathfrak{j}\in (0,1)$ and $(♭,\widehat{x})\in \tilde{X}\times \tilde{Y}$ such that

$$\begin{array}{c}\hfill |\ell (\mathsf{ȷ},♭)-\ell (\mathsf{ȷ},\widehat{x})|\le \sqrt{\mathfrak{j}}|♭\left(\mathsf{ȷ}\right)-\widehat{x}\left(\mathsf{ȷ}\right)|;\end{array}$$

(ii) 

$$\begin{array}{c}\hfill \underset{\mathsf{ȷ}\in [0,1]}{sup}{\int }_0^1\left|\mathcal{G}(\mathsf{ȷ},\mathfrak{e})\right|d\check{R}\le 1.\end{array}$$

Then, Equation (3) has a unique solution in $\tilde{X}\cup \tilde{Y}$.

Proof. 

The given Equation (3) is equivalent to the succeeding integral equation

$$\begin{array}{c}\hfill ♭\left(\mathsf{ȷ}\right)={\int }_0^1\mathcal{G}(\mathsf{ȷ},\mathfrak{e})\ell (\check{R},♭\left(\mathfrak{e}\right))d\mathfrak{e},\end{array}$$

where

$$\begin{array}{c}\hfill \mathcal{G}(\mathsf{ȷ},\mathfrak{e})=\left\{\begin{array}{cc}\frac{{\left[\mathsf{ȷ}(1-\mathfrak{e})\right]}^{♭-1}-{(\mathsf{ȷ}-\mathfrak{e})}^{♭-1}}{\mathsf{\Gamma }(♭)},\hfill & 0\le \mathfrak{e}\le \mathsf{ȷ}\le 1,\hfill \\ \frac{{\left[\mathsf{ȷ}(1-\mathfrak{e})\right]}^{♭-1}}{\mathsf{\Gamma }(♭)},\hfill & 0\le \mathsf{ȷ}\le \mathfrak{e}\le 1.\hfill \end{array}\right.\end{array}$$

Define $\check{R}:\tilde{X}\cup \tilde{Y}\to \tilde{X}\cup \tilde{Y}$ defined by

$$\begin{array}{c}\hfill \check{R}♭\left(\mathsf{ȷ}\right)={\int }_0^1\mathcal{G}(\mathsf{ȷ},\mathfrak{e})\ell (\check{R},♭\left(\mathfrak{e}\right))d\mathfrak{e}.\end{array}$$

Now,

$$\begin{array}{cc}\hfill |\check{R}♭\left(\mathsf{ȷ}\right)-\check{R}\widehat{x}{\left(\mathsf{ȷ}\right)|}^2& =|{\int }_0^1\mathcal{G}(\mathsf{ȷ},\mathfrak{e})\ell (\check{R},♭\left(\mathfrak{e}\right))d\mathfrak{e}-{\int }_o^1\mathcal{G}(\mathsf{ȷ},\mathfrak{e})\ell (\check{R},\widehat{x}\left(\mathfrak{e}\right))d\mathfrak{e}{|}^2\hfill \\ \hfill & \le {\int }_0^1{\left|\mathcal{G}(\mathsf{ȷ},\mathfrak{e})\right|}^{2}d\mathfrak{e}·{\int }_0^1|\ell (\check{R},♭\left(\mathfrak{e}\right))-\ell (\check{R},\widehat{x}\left(\mathfrak{e}\right)){|}^{2}d\mathfrak{e}\hfill \\ \hfill & \le \mathfrak{j}|♭\left(\mathsf{ȷ}\right)-\widehat{x}\left(\mathsf{ȷ}\right){|}^2.\hfill \end{array}$$

Taking the supremum on both sides, we obtain

$$\begin{array}{c}\hfill {\mathsf{Ϝ}}_{\varrho }(\check{R}♭,\check{R}\widehat{x})\le \mathfrak{j}{\mathsf{Ϝ}}_{\varrho }(♭,\widehat{x}).\end{array}$$

Consequently,

$$\begin{array}{cc}\hfill \frac{1}{{\mathsf{Ϝ}}_{\varrho }(\check{R}♭\left(\mathsf{ȷ}\right),\check{R}\widehat{x}\left(\mathsf{ȷ}\right),⊺)}-1& =\frac{{sup}_{\mathsf{ȷ}\in [0,1]}{|\check{R}♭\left(\mathsf{ȷ}\right)-\check{R}\widehat{x}\left(\mathsf{ȷ}\right)|}^2}{⊺}\hfill \\ \hfill & \le \frac{\mathfrak{j}{sup}_{\mathsf{ȷ}\in [0,1]}{|♭\left(\mathsf{ȷ}\right)-\widehat{x}\left(\mathsf{ȷ}\right)|}^2}{⊺}\hfill \\ \hfill & =\mathfrak{j}\left(\frac{1}{{\mathsf{Ϝ}}_{\varrho }(♭\left(\mathsf{ȷ}\right),\widehat{x}\left(\mathsf{ȷ}\right),⊺)}-1\right).\hfill \end{array}$$

Therefore, all the axioms of Theorem 1 are fulfilled and consequently, the fractional differential Equation (3) has a unique solution. □

4. Conclusions

In this manuscript, we proved fixed-point results without continuity and by using the BT property on FBBMS with some applications. In 2017, Mehmood, Ali, Ionescu, and Kamran [18] introduced the extended fuzzy $\mathfrak{b}$-metric space and proved some fixed-point theorems. We give an open problem to introduce the extended fuzzy bipolar $\mathfrak{b}$-metric space and prove some fixed-point theorems.