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[mycorrespondingauthor]* Corresponding author: Rahul Mansotra

Fixed-Point Theorems in b-Metric Spaces via a Novel Simulation Function

Anuradha Gupta dishna2@yahoo.in    Rahul Mansotra mansotrarahul2@gmail.com Department of Mathematics Delhi College of Arts and Commerce Netaji nagar, New Delhi, 110023, India Department of Mathematics Faculty of Mathematical sciences University of Delhi New Delhi, 110007, India
Abstract

This paper introduces a new type of simulation function within the framework of bb-metric spaces, leading to the derivation of fixed-point results in this general setting. We explore the theoretical implications of these results and demonstrate their utility through a concrete example.

1 Introduction and Preliminaries

Fixed point theory has long been a cornerstone in both theoretical and applied mathematics, offering deep insights into the behavior of nonlinear systems and algorithms. One of its most powerful tools is the Banach Contraction Mapping Theorem bm3 , which provides a rigorous foundation for proving the existence and uniqueness of fixed points under specific conditions. The theorem also guarantees that certain methods will converge to the fixed point, making it an essential result in the study of mathematical structures and solutions. The exploration of fixed points extends across various mathematical disciplines, including optimization, differential equations, numerical analysis, control theory, and game theory.

In recent decades, there has been a growing interest in extending both classical and contemporary results from metric fixed point theory to a wider range of generalized metric spaces. This transition has not only broadened the scope of fixed point theory but also introduced new challenges and opportunities for research in more abstract settings. These advancements have led to a deeper understanding of how fixed point results can be adapted and applied beyond traditional metric spaces. A recent survey by Van An et al. bm23 offers a comprehensive exploration of these developments, highlighting key results and their potential implications in various mathematical contexts.

In most instances, this approach turned out to be remarkably straightforward, as the fixed point theorems developed in more general metric spaces could be easily derived from their counterparts in classical metric spaces through a systematic metrization process. This process is exemplified in several influential studies, including bm7 , bm8 , and bm10 .

However, there are certain generalized metric spaces, such as quasi metric spaces (often referred to as b-metric spaces within fixed point theory), where the transposition process typically leads to meaningful generalizations of fixed point theorems from traditional metric spaces. Bakhtin bm4 and Czerwik bm6 played a pivotal role in this development by extending the classical concept of metric space, introducing the more general notion of b-metric spaces, and thereby expanding the framework of fixed point theory and its related fields.

In 2014, Jleli and Samet bm14 introduced the concept of ϑ\vartheta-contractions, providing an important generalization of the Banach contraction principle in the framework of Branciari distance spaces bm4 . Later, Ahmad et al. bm13 refined the conditions on the auxiliary function ϑ\vartheta (say), leading to a comparable result in standard metric spaces. Alternatively, Khojasteh et al. bm17 established the concept of simulation functions with a view to consider a new class of contractions, called 𝒵\mathcal{Z}-contractions. Such family generalized, extended and improved several results that had been obtained in previous years. The simplicity and usefulness of these contractions have inspirited many researchers to diversify it further (see [ bm12 , bm15 , bm16 , bm19 , bm21 , and bm22 ]).

Building on the concept of simulation functions introduced by Khojasteh et al. bm17 , Cho et al. bm5 made a remarkable advancement in 2018 by introducing the ℒ\mathcal{L}-simulation function as a novel structure within the field. This groundbreaking contribution not only set a new standard but also sparked a wave of further research and development, highlighting the transformative influence of innovative methodologies on the evolution of simulation practices.
Throughout this article, 𝒳\mathcal{X} denotes a nonempty set, ℝ+\mathbb{R}^{+} represents the set of positive real numbers, ℕ\mathbb{N} stands for the set of positive integers, and ℕ0\mathbb{N}_{0} refers to the set of nonnegative integers.
This section begins with the following definition:

Definition 1.1

bm6 A map 𝔟:𝒳×𝒳→[0,∞)\mathfrak{b}:\mathcal{X}\times\mathcal{X}\rightarrow[0,\infty) is said to be bb-metric on 𝒳\mathcal{X} if there exist s≥1s\geq 1 such that for all x,y,z∈𝒳,𝔟x,y,z\in\mathcal{X},~{}\mathfrak{b} satisfies the following:

  1. (i)

    𝔟​(x,y)=0\mathfrak{b}(x,y)=0 if and only if x=yx=y;

  2. (ii)

    𝔟​(x,y)=𝔟​(y,x)\mathfrak{b}(x,y)=\mathfrak{b}(y,x);

  3. (iii)

    𝔟​(x,z)≤s​[𝔟​(x,y)+𝔟​(y,z)]\mathfrak{b}(x,z)\leq s[\mathfrak{b}(x,y)+\mathfrak{b}(y,z)].

Then, (𝒳,𝔟,s)(\mathcal{X},\mathfrak{b},s) is called a bb-metric space with coefficient ss.

Definition 1.2

bm6 Let (𝒳,𝔟,s)(\mathcal{X},\mathfrak{b},s) be a bb-metric space. Then:

  1. (i)

    A sequence (ak)(a_{k}) in 𝒳\mathcal{X} is said to be convergent if there is a∈𝒳a\in\mathcal{X} such that limk→∞𝔟​(ak,a)=0;\lim\limits_{k\rightarrow\infty}\mathfrak{b}(a_{k},a)=0;

  2. (ii)

    A sequence (ak)(a_{k}) in 𝒳\mathcal{X} is said to be cauchy in 𝒳\mathcal{X} if limk,m→∞𝔟​(ak,am)\lim\limits_{k,m\rightarrow\infty}\mathfrak{b}(a_{k},a_{m}) exists and is finite;

  3. (iii)

    𝒳\mathcal{X} is said to be complete if for every Cauchy sequence (ak)(a_{k}) in 𝒳\mathcal{X} there is a∈𝒳a\in\mathcal{X} such that limk,m→∞𝔟​(ak,am)=limk→∞𝔟​(ak,a)=0;\lim\limits_{k,m\rightarrow\infty}\mathfrak{b}(a_{k},a_{m})=\lim\limits_{k\rightarrow\infty}\mathfrak{b}(a_{k},a)=0;

  4. (iv)

    A function 𝒮:𝒳→𝒳\mathcal{S}:\mathcal{X}\to\mathcal{X} is said to be bb-continuous if for (an)⊆𝒳(a_{n})\subseteq\mathcal{X}, an→aa_{n}\to a in (𝒳,𝔟)(\mathcal{X},\mathfrak{b}) we have 𝒮​an→𝒮​a\mathcal{S}a_{n}\to\mathcal{S}a in (𝒳,𝔟)(\mathcal{X},\mathfrak{b}).

Following bm14 , Θ\Theta denotes the set of all mappings ϑ:(0,∞)→(1,∞)\vartheta:(0,\infty)\to(1,\infty) satisfies the following properties:
(a)(a) ϑ\vartheta is increasing;
(b)(b) for each sequence {an}⊆(0,∞),limn→∞ϑ​(an)=1⇔limn→∞an=0;\{a_{n}\}\subseteq(0,\infty),\lim\limits_{n\to\infty}\vartheta(a_{n})=1\iff\lim\limits_{n\to\infty}a_{n}=0;
(c)(c) there exist t∈(0,1)t\in(0,1) and d∈(0,∞]d\in(0,\infty] such that limx→0+ϑ​(x)−1xt=d.\lim\limits_{x\to 0^{+}}\dfrac{\vartheta(x)-1}{x^{t}}=d.
Further, Ahmad et al. bm13 replaced the condition (c)(c) with the following:
(d)(d) ϑ\vartheta is continuous.
The symbol Θ∗\Theta^{*} denotes the collection of all mappings satisfying conditions (a),(b)​ and ​(d)(a),(b)\text{ and }(d).

Accordingly, authors in bm13 established the following Fixed Point Theorem:

Theorem 1.3

Every ϑ\vartheta-contraction on a complete metric space has a unique fixed point.

Recently, Cho in bm5 introduced the ℒ\mathcal{L}-simulation function as follows:

Definition 1.4

A map ℒ\mathcal{L} from [1,∞)×[1,∞)[1,\infty)\times[1,\infty) to ℝ\mathbb{R} defines a ℒ\mathcal{L}-simulation function if for all a,b∈[1,∞)a,b\in[1,\infty), ℒ\mathcal{L} satisfies the following properties:

  1. (i)

    ℒ​(1,1)=1;\mathcal{L}(1,1)=1;

  2. (ii)

    ℒ​(a,b)<ba\mathcal{L}(a,b)<\dfrac{b}{a} for all a,b>1a,b>1;

  3. (iii)

    if (an)(a_{n}) and (bn)(b_{n}) are sequences in (1,∞)(1,\infty) such that 1<limn→∞an=limn→∞bn1<\lim\limits_{n\to\infty}a_{n}=\lim\limits_{n\to\infty}b_{n}, then lim supn→∞ℒ​(an,bn)<1\limsup\limits_{n\to\infty}\mathcal{L}(a_{n},b_{n})<1.

By ℒ\mathscr{L} we denote the family of all ℒ\mathcal{L}-simulation functions.

Hasanuzzaman et al. bm12 introduced the ℒ\mathcal{L}-contraction in metric space as follows:

Definition 1.5

Let (𝒳,d)(\mathcal{X},d) be a metric space. Then 𝒯:𝒳→𝒳\mathcal{T}:\mathcal{X}\to\mathcal{X} is called ℒ\mathcal{L}-contraction with respect to ℒ\mathcal{L} if there exist ℒ∈ℒ\mathcal{L}\in\mathscr{L} and ϑ∈Θ∗\vartheta\in\Theta^{*} such that

ℒ​(ϑ​(d​(𝒯​x,𝒯​y)),ϑ​(d​(x,y)))≥1\mathcal{L}(\vartheta(d(\mathcal{T}x,\mathcal{T}y)),\vartheta(d(x,y)))\geq 1

for all a,b∈𝒳a,b\in\mathcal{X} with d​(𝒯​x,𝒯​y)>0d(\mathcal{T}x,\mathcal{T}y)>0.

Inspired by Cho’s work bm5 on ℒ\mathcal{L}-contractions in metric spaces and the contributions of Gupta and Rohilla bm9 on simulation functions in bb-metric spaces, this article introduces the concept of 𝔸ℝ\mathbb{A}_{\mathbb{R}}-simulation functions. To underscore the importance and applicability of this concept, fixed point theorems are developed and substantiated with a comprehensive example that demonstrates its practical relevance.

2 Fixed Point Theorems Using 𝔍𝔸ℝ\mathfrak{J}_{{}_{\mathbb{A}_{\mathbb{R}}}}-Contractions

This section begins by introducing the essential definitions and notations that underpin the theorems and proofs presented.
Let 𝔉𝔠\mathfrak{F_{c}} denote the class of all the operators ℱc​ from ​[1,∞)×[1,∞)​ to ​ℝ\mathcal{F}_{c}\text{ from }[1,\infty)\times[1,\infty)\text{ to }\mathbb{R} such that for all a,b∈[1,∞)a,b\in[1,\infty), satisfying the following properties:

  1. (i)

    ℱc\mathcal{F}_{c} is continuous;

  2. (ii)

    ℱc​(x,y)≤x\mathcal{F}_{c}(x,y)\leq x;

  3. (iii)

    ℱc​(x,y)=x\mathcal{F}_{c}(x,y)=x implies that either x=1x=1 or y=1y=1;

  4. (iv)

    there exist c≥1c\geq 1 such that ℱc​(x,y)>c\mathcal{F}_{c}(x,y)>c implies that x>yx>y and ℱc​(x,x)≤c.\mathcal{F}_{c}(x,x)\leq c.

Example 2.1

ℱc​(x,y)=xy\mathcal{F}_{c}(x,y)=\frac{x}{y}. Here c=1c=1.

We define 𝔸ℝ\mathbb{A}_{\mathbb{R}}-simulation function in the following:

Definition 2.2

A map 𝔍\mathfrak{J} from [1,∞)×[1,∞)[1,\infty)\times[1,\infty) to ℝ\mathbb{R} defines a 𝔸ℝ\mathbb{A}_{\mathbb{R}}-simulation function if there exist s≥1s\geq 1, ℱc∈𝔉𝔠\mathcal{F}_{c}\in\mathfrak{F_{c}} and ϑ∈Θ∗\vartheta\in\Theta^{*} such that for all x,y∈(1,∞)x,y\in(1,\infty), 𝔍\mathfrak{J} satisfying the following properties:

  1. (i)

    𝔍​(x,y)<ℱc​(y,x)\mathfrak{J}(x,y)<\mathcal{F}_{c}(y,x);

  2. (ii)

    if (an)(a_{n}) and (bn)(b_{n}) are sequences in (0,∞)(0,\infty) such that

    0<lim infn→∞an≤s​(lim supn→∞bn)≤s2​(lim infn→∞an)<∞0<\liminf\limits_{n\to\infty}a_{n}\leq s(\limsup\limits_{n\to\infty}b_{n})\leq s^{2}(\liminf\limits_{n\to\infty}a_{n})<\infty

    and

    0<lim infn→∞bn≤s​(lim supn→∞an)≤s2​(lim infn→∞bn)<∞,0<\liminf\limits_{n\to\infty}b_{n}\leq s(\limsup\limits_{n\to\infty}a_{n})\leq s^{2}(\liminf\limits_{n\to\infty}b_{n})<\infty,

    then lim supn→∞𝔍​(ϑ​(an),ϑ​(bn))<c\limsup\limits_{n\to\infty}\mathfrak{J}(\vartheta(a_{n}),\vartheta(b_{n}))<c.

Let 𝒥\bf{\mathscr{J}} denote the set of all 𝔸ℝ\mathbb{A}_{\mathbb{R}}-simulation functions.

Example 2.3

Define 𝔍:[1,∞)×[1,∞)→ℝ\mathfrak{J}:[1,\infty)\times[1,\infty)\rightarrow\mathbb{R}, ℱc:[1,∞)×[1,∞)→ℝ\mathcal{F}_{c}:[1,\infty)\times[1,\infty)\rightarrow\mathbb{R} and ϑ:(0,∞)→(1,∞)\vartheta:(0,\infty)\to(1,\infty) by

𝔍​(x,y)=y4​x​ , ​ℱc​(y,x)=yx​ and ​ϑ​(x)=x+1.\mathfrak{J}(x,y)=\frac{y}{4x}\text{ , }\mathcal{F}_{c}(y,x)=\frac{y}{x}\text{ and }\vartheta(x)=x+1.

Note that 𝔍​(x,y)<ℱc​(y,x)\mathfrak{J}(x,y)<\mathcal{F}_{c}(y,x), for all x,y∈(1,∞)x,y\in(1,\infty) and c=1c=1. Further, If (an)(a_{n}) and (bn)(b_{n}) are sequences in (0,∞)(0,\infty) such that

0<lim infn→∞an≤4​(lim supn→∞bn)≤16​(lim infn→∞an)<∞0<\liminf\limits_{n\to\infty}a_{n}\leq 4(\limsup\limits_{n\to\infty}b_{n})\leq 16(\liminf\limits_{n\to\infty}a_{n})<\infty

and

0<lim infn→∞bn≤4​(lim supn→∞an)≤16​(lim infn→∞bn)<∞,0<\liminf\limits_{n\to\infty}b_{n}\leq 4(\limsup\limits_{n\to\infty}a_{n})\leq 16(\liminf\limits_{n\to\infty}b_{n})<\infty,

then lim supn→∞𝔍​(ϑ​(an),ϑ​(bn))=lim supn→∞(bn+14​(an+1))=lim supn→∞(bn+1)lim infn→∞4​(an+1)\limsup\limits_{n\to\infty}\mathfrak{J}(\vartheta(a_{n}),\vartheta(b_{n}))=\limsup\limits_{n\to\infty}(\dfrac{b_{n}+1}{4(a_{n}+1)})=\dfrac{\limsup\limits_{n\to\infty}(b_{n}+1)}{\liminf\limits_{n\to\infty}4(a_{n}+1)}.
Since lim supn→∞bn≤lim infn→∞4​an,lim supn→∞𝔍​(ϑ​(an),ϑ​(bn))<1\limsup\limits_{n\to\infty}b_{n}\leq\liminf\limits_{n\to\infty}4a_{n},~{}\limsup\limits_{n\to\infty}\mathfrak{J}(\vartheta(a_{n}),\vartheta(b_{n}))<1. Thus 𝔍\mathfrak{J} is a 𝔸ℝ\mathbb{A}_{\mathbb{R}}-simulation function.

Definition 2.4

An operator 𝒮\mathcal{S} from 𝒳\mathcal{X} to 𝒳\mathcal{X} defines a 𝔍𝔸ℝ\mathfrak{J}_{{}_{\mathbb{A}_{\mathbb{R}}}}-contraction if there exist 𝔍∈𝒥,ϑ∈Θ∗\mathfrak{J}\in\mathscr{J},\vartheta\in\Theta^{*} such that for all x,y∈𝒳x,y\in\mathcal{X} with 𝔟​(𝒮​x,𝒮​y)>0\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)>0 implies

𝔍​(ϑ​(𝔟​(𝒮​x,𝒮​y)),ϑ​(𝔟​(x,y)))≥c.\mathfrak{J}(\vartheta(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)),\vartheta(\mathfrak{b}(x,y)))\geq c. (2.1)
Theorem 2.5

Let (𝒳,𝔟,s)(\mathcal{X},\mathfrak{b},s) be a complete bb-metric space with coefficient s≥1s\geq 1 and 𝒮:𝒳→𝒳\mathcal{S}:\mathcal{X}\rightarrow\mathcal{X} be a given mapping. Suppose that 𝒮\mathcal{S} is a 𝔍𝔸ℝ\mathfrak{J}_{{}_{\mathbb{A}_{\mathbb{R}}}}-contraction. Then 𝒮\mathcal{S} has a unique fixed point.

Proof 2.6.

Let a0∈𝒳a_{0}\in\mathcal{X} and define 𝒮n​a0=an\mathcal{S}^{n}a_{0}=a_{n}, for all n∈ℕ0n\in\mathbb{N}_{0}. If 𝔟​(an,an+1)=0\mathfrak{b}(a_{n},a_{n+1})=0 then an=an+1=𝒮​ana_{n}=a_{n+1}=\mathcal{S}a_{n} becomes fixed point of 𝒮\mathcal{S}. From this point onward, we can consider that 𝔟​(an,an+1)≠0,\mathfrak{b}(a_{n},a_{n+1})\neq 0, for all n≥0n\geq 0. Put x=an​ and ​y=an+1x=a_{n}\text{ and }y=a_{n+1} in inequality (2.1)(2.1) then

c\displaystyle c ≤𝔍​(ϑ​(𝔟​(𝒮​an,𝒮​an+1)),ϑ​(𝔟​(an,an+1)))\displaystyle\leq\mathfrak{J}(\vartheta(\mathfrak{b}(\mathcal{S}a_{n},\mathcal{S}a_{n+1})),\vartheta(\mathfrak{b}(a_{n},a_{n+1})))
=𝔍​(ϑ​(𝔟​(an+1,an+2)),ϑ​(𝔟​(an,an+1)))\displaystyle=\mathfrak{J}(\vartheta(\mathfrak{b}(a_{n+1},a_{n+2})),\vartheta(\mathfrak{b}(a_{n},a_{n+1}))) (2.2)
<ℱc​(ϑ​(𝔟​(an,an+1)),ϑ​(𝔟​(an+1,an+2))).\displaystyle<\mathcal{F}_{c}(\vartheta(\mathfrak{b}(a_{n},a_{n+1})),\vartheta(\mathfrak{b}(a_{n+1},a_{n+2}))).

Thus, by the property of ℱc\mathcal{F}_{c}, we get ϑ​(𝔟​(an+1,an+2))<ϑ​(𝔟​(an,an+1))\vartheta(\mathfrak{b}(a_{n+1},a_{n+2}))<\vartheta(\mathfrak{b}(a_{n},a_{n+1})). Let us suppose that 𝔟​(an,an+1)<𝔟​(an+1,an+2)\mathfrak{b}(a_{n},a_{n+1})<\mathfrak{b}(a_{n+1},a_{n+2}). As ϑ\vartheta is increasing, ϑ​(𝔟​(an,an+1))≤ϑ​(𝔟​(an+1,an+2))\vartheta(\mathfrak{b}(a_{n},a_{n+1}))\leq\vartheta(\mathfrak{b}(a_{n+1},a_{n+2})), which is a contradiction. Thus, 𝔟​(an+1,an+2)≤𝔟​(an,an+1)\mathfrak{b}(a_{n+1},a_{n+2})\leq\mathfrak{b}(a_{n},a_{n+1}), for all n≥0.n\geq 0. So, (𝔟​(an,an+1))(\mathfrak{b}(a_{n},a_{n+1})) is a decreasing sequence of positive real numbers; hence limn→∞𝔟​(an,an+1)=a≥0\lim\limits_{n\to\infty}\mathfrak{b}(a_{n},a_{n+1})=a\geq 0. We will show that a=0.a=0. Suppose a>0a>0 then 0<a≤s​a≤s2​a<∞0<a\leq sa\leq s^{2}a<\infty. Let rn=𝔟​(an,an+1)r_{n}=\mathfrak{b}(a_{n},a_{n+1}), then

0<lim infn→∞rn+1≤s​(lim supn→∞rn)≤s2​(lim infn→∞rn)<∞0<\liminf\limits_{n\to\infty}r_{n+1}\leq s(\limsup\limits_{n\to\infty}r_{n})\leq s^{2}(\liminf\limits_{n\to\infty}r_{n})<\infty

and

0<lim infn→∞rn≤s​(lim supn→∞rn+1)≤s2​(lim infn→∞rn)<∞,0<\liminf\limits_{n\to\infty}r_{n}\leq s(\limsup\limits_{n\to\infty}r_{n+1})\leq s^{2}(\liminf\limits_{n\to\infty}r_{n})<\infty,

hence by the property of 𝔍\mathfrak{J}, we get lim supn→∞𝔍​(ϑ​(rn+1),ϑ​(rn))<c.\limsup\limits_{n\to\infty}\mathfrak{J}(\vartheta(r_{n+1}),\vartheta(r_{n}))<c.
Also, by inequality (2.2)(2.2), c≤lim supn→∞𝔍​(ϑ​(rn+1),ϑ​(rn))c\leq\limsup\limits_{n\to\infty}\mathfrak{J}(\vartheta(r_{n+1}),\vartheta(r_{n})), which leads to a contradiction. Hence limn→∞rn=0.\lim\limits_{n\to\infty}r_{n}=0. Now, we aim to prove that (an)(a_{n}) is a Cauchy sequence. Let us suppose that (an)(a_{n}) is not a Cauchy sequence in (𝒳,𝔟)(\mathcal{X},\mathfrak{b}). Then, there exist ε>0\varepsilon>0 and subsequences (ani)(a_{n_{i}}) and (ami)(a_{m_{i}}) of sequence (an)(a_{n}) such that nin_{i} is the smallest integer for which ni>mi>in_{i}>m_{i}>i with

𝔟​(ami,ani)≥ε​ and ​𝔟​(ami,ani−1)<ε.\mathfrak{b}(a_{m_{i}},a_{n_{i}})\geq\varepsilon\mbox{ and }\mathfrak{b}(a_{m_{i}},a_{n_{i}-1})<\varepsilon. (2.3)

Now, ε≤𝔟​(ami,ani)≤s​(𝔟​(ami,ani−1)+𝔟​(ani−1,ani))\varepsilon\leq\mathfrak{b}(a_{m_{i}},a_{n_{i}})\leq s(\mathfrak{b}(a_{m_{i}},a_{n_{i}-1})+\mathfrak{b}(a_{n_{i}-1},a_{n_{i}})), which implies that

ε≤lim infi→∞𝔟​(ami,ani)≤s​ε​ and ​ε≤lim supi→∞𝔟​(ami,ani)≤s​ε.\varepsilon\leq\liminf\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}},a_{n_{i}})\leq s\varepsilon\text{ and }\varepsilon\leq\limsup\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}},a_{n_{i}})\leq s\varepsilon. (2.4)

Note that ami≠ania_{m_{i}}\neq a_{n_{i}} as 𝔟​(ami,ani)≥ε\mathfrak{b}(a_{m_{i}},a_{n_{i}})\geq\varepsilon. Substitute x=ami−1​ and ​y=ani−1x=a_{m_{i}-1}\text{ and }y=a_{n_{i}-1} in inequality (2.1)(2.1), we get

c\displaystyle c ≤𝔍​(ϑ​(𝔟​(𝒮​an,𝒮​an+1)),ϑ​(𝔟​(an,an+1)))\displaystyle\leq\mathfrak{J}(\vartheta(\mathfrak{b}(\mathcal{S}a_{n},\mathcal{S}a_{n+1})),\vartheta(\mathfrak{b}(a_{n},a_{n+1})))
=𝔍​(ϑ​(𝔟​(ami,ani)),ϑ​(𝔟​(ami−1,ani−1)))\displaystyle=\mathfrak{J}(\vartheta(\mathfrak{b}(a_{m_{i}},a_{n_{i}})),\vartheta(\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1}))) (2.5)
<ℱc​(ϑ​(𝔟​(ami−1,ani−1)),ϑ​(𝔟​(ami,ani)))\displaystyle<\mathcal{F}_{c}(\vartheta(\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1})),\vartheta(\mathfrak{b}(a_{m_{i}},a_{n_{i}})))

which implies ϑ(𝔟(ami,ani))<ϑ(𝔟(ami−1,ani−1)))\vartheta(\mathfrak{b}(a_{m_{i}},a_{n_{i}}))<\vartheta(\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1}))). Let it be the case that 𝔟​(ami−1,ani−1)<𝔟​(ami,ani).\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1})<\mathfrak{b}(a_{m_{i}},a_{n_{i}}). As ϑ\vartheta is increasing, ϑ​(𝔟​(ami−1,ani−1))≤ϑ​(𝔟​(ami,ani))\vartheta(\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1}))\leq\vartheta(\mathfrak{b}(a_{m_{i}},a_{n_{i}})) which leads to a contradiction. Also, ε≤𝔟​(ami,ani)<𝔟​(ami−1,ani−1)≤\varepsilon\leq\mathfrak{b}(a_{m_{i}},a_{n_{i}})<\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1})\leq
s​(𝔟​(ami−1,ami)+𝔟​(ami,ani−1))s(\mathfrak{b}(a_{m_{i}-1},a_{m_{i}})+\mathfrak{b}(a_{m_{i}},a_{n_{i}-1})). Thus, using inequality (2.3)(2.3) and taking limit superior and limit inferior as ii goes to infinity, we get

ε≤lim infi→∞𝔟​(ami−1,ani−1)≤s​ε​ and ​ε≤lim supi→∞𝔟​(ami−1,ani−1)≤s​ε.\varepsilon\leq\liminf\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1})\leq s\varepsilon\text{ and }\varepsilon\leq\limsup\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1})\leq s\varepsilon. (2.6)

Using inequalities (2.3)(2.3), (2.4)(2.4) and (2.6)(2.6), we have

0<lim infi→∞𝔟​(ami,ani)≤s​ε≤s​(lim supi→∞𝔟​(ami−1,ani−1))≤s2​ε≤s2​(lim infi→∞𝔟​(ami,ani))<∞\displaystyle 0<\liminf\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}},a_{n_{i}})\leq s\varepsilon\leq s(\limsup\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1}))\leq s^{2}\varepsilon\leq s^{2}(\liminf\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}},a_{n_{i}}))<\infty
and
0<lim infi→∞𝔟​(ami−1,ani−1)≤s​ε≤s​(lim supi→∞𝔟​(ami,ani))≤s2​ε≤s2​(lim infi→∞𝔟​(ami−1,ani−1))<∞.\displaystyle 0<\liminf\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1})\leq s\varepsilon\leq s(\limsup\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}},a_{n_{i}}))\leq s^{2}\varepsilon\leq s^{2}(\liminf\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1}))<\infty.

Therefore, in light of the property of 𝔍\mathfrak{J}, we have

lim supi→∞𝔍​(ϑ​(𝔟​(ami,ani)),ϑ​(𝔟​(ami−1,ani−1)))<c.\limsup_{i\to\infty}\mathfrak{J}(\vartheta(\mathfrak{b}(a_{m_{i}},a_{n_{i}})),\vartheta(\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1})))<c.

Also, by inequality (2.5)(2.5), we get c≤lim supi→∞𝔍​(ϑ​(𝔟​(ami,ani)),ϑ​(𝔟​(ami−1,ani−1))),c\leq\limsup\limits_{i\to\infty}\mathfrak{J}(\vartheta(\mathfrak{b}(a_{m_{i}},a_{n_{i}})),\vartheta(\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1}))), which leads to a contradiction. Hence, limn→∞𝔟​(an,am)=0\lim\limits_{n\to\infty}\mathfrak{b}(a_{n},a_{m})=0.
Since (𝒳,𝔟,s)(\mathcal{X},\mathfrak{b},s) is a complete bb-metric space,

limn,m→∞𝔟​(an,am)=limn→∞𝔟​(an,z)=0, for some ​z∈𝒳.\lim\limits_{n,m\to\infty}\mathfrak{b}(a_{n},a_{m})=\lim\limits_{n\to\infty}\mathfrak{b}(a_{n},z)=0,\text{ for some }z\in\mathcal{X}.

We will show that zz is the unique fixed point of 𝒮\mathcal{S}. Suppose 𝔟​(an−1,z)≠0\mathfrak{b}(a_{n-1},z)\neq 0 and 𝔟​(an,𝒮​z)≠0\mathfrak{b}(a_{n},\mathcal{S}z)\neq 0 for infinitely many nn. Substitute x=an−1x=a_{n-1} and y=zy=z in inequality (2.1)(2.1), we get

c\displaystyle c ≤𝔍​(ϑ​(𝔟​(𝒮​an−1,𝒮​z)),ϑ​(𝔟​(an−1,z)))\displaystyle\leq\mathfrak{J}(\vartheta(\mathfrak{b}(\mathcal{S}a_{n-1},\mathcal{S}z)),\vartheta(\mathfrak{b}(a_{n-1},z)))
=𝔍​(ϑ​(𝔟​(an,𝒮​z)),ϑ​(𝔟​(an−1,z)))\displaystyle=\mathfrak{J}(\vartheta(\mathfrak{b}(a_{n},\mathcal{S}z)),\vartheta(\mathfrak{b}(a_{n-1},z)))
<ℱc​(ϑ​(𝔟​(an−1,z)),ϑ​(𝔟​(an,𝒮​z))),\displaystyle<\mathcal{F}_{c}(\vartheta(\mathfrak{b}(a_{n-1},z)),\vartheta(\mathfrak{b}(a_{n},\mathcal{S}z))),

hence, by the property of ℱc\mathcal{F}_{c}, we get ϑ​(𝔟​(an,𝒮​z))<ϑ​(𝔟​(an−1,z))\vartheta(\mathfrak{b}(a_{n},\mathcal{S}z))<\vartheta(\mathfrak{b}(a_{n-1},z)). Assume 𝔟​(an−1,z)<𝔟​(an,𝒮​z)\mathfrak{b}(a_{n-1},z)<\mathfrak{b}(a_{n},\mathcal{S}z). As ϑ\vartheta is increasing, ϑ​(𝔟​(an,𝒮​z))<ϑ​(𝔟​(an−1,z))\vartheta(\mathfrak{b}(a_{n},\mathcal{S}z))<\vartheta(\mathfrak{b}(a_{n-1},z)), which is a contradiction. Thus, 𝔟​(an,𝒮​z)≤𝔟​(an−1,z)\mathfrak{b}(a_{n},\mathcal{S}z)\leq\mathfrak{b}(a_{n-1},z), which implies that limn→∞𝔟​(an,𝒮​z)=0.\lim\limits_{n\to\infty}\mathfrak{b}(a_{n},\mathcal{S}z)=0. Now, 𝔟​(z,𝒮​z)≤s​(𝔟​(z,an)+𝔟​(an,𝒮​z))\mathfrak{b}(z,\mathcal{S}z)\leq s(\mathfrak{b}(z,a_{n})+\mathfrak{b}(a_{n},\mathcal{S}z)) which on applying limit, gives z=𝒮​zz=\mathcal{S}z. Finally, we will prove the uniqueness of the fixed point. Suppose ww be the another fixed point such that z≠wz\neq w. Then 𝔟​(z,w)>0\mathfrak{b}(z,w)>0. On substituting x=zx=z and y=wy=w in inequality (2.1)(2.1), we have

c≤𝔍​(ϑ​(𝔟​(𝒮​z,𝒮​w)),ϑ​(𝔟​(z,w)))<ℱc​(ϑ​(𝔟​(z,w)),ϑ​(𝔟​(z,w)))≤c,c\leq\mathfrak{J}(\vartheta(\mathfrak{b}(\mathcal{S}z,\mathcal{S}w)),\vartheta(\mathfrak{b}(z,w)))<\mathcal{F}_{c}(\vartheta(\mathfrak{b}(z,w)),\vartheta(\mathfrak{b}(z,w)))\leq c,

which is a contradiction. Hence zz becomes the unique fixed point.

Example 2.7.

Let 𝒳={1,2,3,4}\mathcal{X}=\{1,2,3,4\}. Define 𝔟:𝒳×𝒳→ℝ+\mathfrak{b}:\mathcal{X}\times\mathcal{X}\rightarrow\mathbb{R^{+}} given by
𝔟​(x,x)=0\mathfrak{b}(x,x)=0 for all x∈𝒳x\in\mathcal{X},
𝔟​(1,2)=𝔟​(2,1)=3\mathfrak{b}(1,2)=\mathfrak{b}(2,1)=3,
𝔟​(2,3)=𝔟​(3,2)=𝔟​(1,3)=𝔟​(3,1)=1\mathfrak{b}(2,3)=\mathfrak{b}(3,2)=\mathfrak{b}(1,3)=\mathfrak{b}(3,1)=1,
𝔟​(1,4)=𝔟​(4,1)=𝔟​(2,4)=𝔟​(4,2)=𝔟​(3,4)=𝔟​(4,3)=4.\mathfrak{b}(1,4)=\mathfrak{b}(4,1)=\mathfrak{b}(2,4)=\mathfrak{b}(4,2)=\mathfrak{b}(3,4)=\mathfrak{b}(4,3)=4.
Clearly, (𝒳,𝔟,s)(\mathcal{X},\mathfrak{b},s) is a complete bb-metric space with coefficient s=3.s=\sqrt{3}.
Also, define 𝔍:[1,∞)×[1,∞)→ℝ\mathfrak{J}:[1,\infty)\times[1,\infty)\rightarrow\mathbb{R}, ℱc:[1,∞)×[1,∞)→ℝ\mathcal{F}_{c}:[1,\infty)\times[1,\infty)\rightarrow\mathbb{R} and ϑ:(0,∞)→(1,∞)\vartheta:(0,\infty)\to(1,\infty) by

𝔍​(x,y)=y3​x​ , ​ℱc​(y,x)=yx​ and ​ϑ​(x)=x+1.\mathfrak{J}(x,y)=\frac{y}{\sqrt{3}x}\text{ , }\mathcal{F}_{c}(y,x)=\frac{y}{x}\text{ and }\vartheta(x)=x+1.

Note that 𝔍​(x,y)<ℱc​(y,x)\mathfrak{J}(x,y)<\mathcal{F}_{c}(y,x), for all x,y∈(1,∞)x,y\in(1,\infty) and c=1c=1. Further, If (an)(a_{n}) and (bn)(b_{n}) are sequences in (0,∞)(0,\infty) such that

0<lim infn→∞an≤3​(lim supn→∞bn)≤3​(lim infn→∞an)<∞0<\liminf\limits_{n\to\infty}a_{n}\leq\sqrt{3}(\limsup\limits_{n\to\infty}b_{n})\leq 3(\liminf\limits_{n\to\infty}a_{n})<\infty

and

0<lim infn→∞bn≤3​(lim supn→∞an)≤3​(lim infn→∞bn)<∞,0<\liminf\limits_{n\to\infty}b_{n}\leq\sqrt{3}(\limsup\limits_{n\to\infty}a_{n})\leq 3(\liminf\limits_{n\to\infty}b_{n})<\infty,

then lim supn→∞𝔍​(ϑ​(an),ϑ​(bn))=lim supn→∞(bn+13​(an+1))=lim supn→∞(bn+1)lim infn→∞3​(an+1)\limsup\limits_{n\to\infty}\mathfrak{J}(\vartheta(a_{n}),\vartheta(b_{n}))=\limsup\limits_{n\to\infty}(\dfrac{b_{n}+1}{\sqrt{3}(a_{n}+1)})=\dfrac{\limsup\limits_{n\to\infty}(b_{n}+1)}{\liminf\limits_{n\to\infty}\sqrt{3}(a_{n}+1)}.
As lim supn→∞bn≤lim infn→∞3​an.\limsup\limits_{n\to\infty}b_{n}\leq\liminf\limits_{n\to\infty}\sqrt{3}a_{n}. Hence, lim supn→∞𝔍​(ϑ​(an),ϑ​(bn))<1\limsup\limits_{n\to\infty}\mathfrak{J}(\vartheta(a_{n}),\vartheta(b_{n}))<1. Thus 𝔍\mathfrak{J} is a 𝔸ℝ\mathbb{A}_{\mathbb{R}}-simulation function. Moreover, define 𝒮:𝒳→𝒳\mathcal{S}:\mathcal{X}\rightarrow\mathcal{X} by

𝒮​x={3, when ​x≠4,1, else .\mathcal{S}x=\begin{cases}3,\text{ when }x\neq 4,\\ 1,\text{ else }.\end{cases}

We will now verify that 𝒮\mathcal{S} is a 𝔍𝔸ℝ\mathfrak{J}_{{}_{\mathbb{A}_{\mathbb{R}}}}-contraction. Note that

𝔟​(𝒮​x,𝒮​y)={𝔟​(1,3)=1,if ​x=4,y≠4,𝔟​(1,1)=0,if ​x=4,y=4,𝔟​(3,3)=0,if ​x≠4,y≠4,𝔟​(3,1)=1,if ​x≠4,y=4,\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)=\begin{cases}\mathfrak{b}(1,3)=1,&\text{if }x=4,y\neq 4,\\ \mathfrak{b}(1,1)=0,&\text{if }x=4,y=4,\\ \mathfrak{b}(3,3)=0,&\text{if }x\neq 4,y\neq 4,\\ \mathfrak{b}(3,1)=1,&\text{if }x\neq 4,y=4,\end{cases}

hence 𝔟​(𝒮​x,𝒮​y)>0\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)>0 if and only if x=4,y≠4,x=4,y\neq 4, and x≠4,y=4x\neq 4,y=4. Now, if x=4,y≠4,x=4,y\neq 4, and x≠4,y=4,x\neq 4,y=4, then 𝔟​(x,y)=4\mathfrak{b}(x,y)=4 and 𝔟​(𝒮​x,𝒮​y)=1.\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)=1. Further, for all x,y∈𝒳x,y\in\mathcal{X} with 𝔟​(𝒮​x,𝒮​y)>0,\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)>0, we have

𝔍​(ϑ​(𝔟​(𝒮​x,𝒮​y)),ϑ​(𝔟​(x,y)))=ϑ​(𝔟​(x,y))3​ϑ​(𝔟​(𝒮​x,𝒮​y))=𝔟​(x,y)+13​(𝔟​(𝒮​x,𝒮​y)+1)=4+13​(1+1)=52​3>1\mathfrak{J}(\vartheta(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)),\vartheta(\mathfrak{b}(x,y)))=\dfrac{\vartheta(\mathfrak{b}(x,y))}{\sqrt{3}\vartheta(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y))}=\dfrac{\mathfrak{b}(x,y)+1}{\sqrt{3}(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)+1)}=\dfrac{4+1}{\sqrt{3}(1+1)}=\dfrac{5}{2\sqrt{3}}>1

Hence, 𝒮\mathcal{S} is a 𝔍𝔸ℝ\mathfrak{J}_{{}_{\mathbb{A}_{\mathbb{R}}}}-contraction. By Theorem 2.52.5, 𝒮\mathcal{S} has a unique fixed point 33.

Theorem 2.8.

Let (𝒳,𝔟,s)(\mathcal{X},\mathfrak{b},s) be a complete b-metric space with coefficient s≥1s\geq 1 and 𝒮:𝒳→𝒳\mathcal{S}:\mathcal{X}\rightarrow\mathcal{X} be a 𝔟\mathfrak{b}-continuous self-mapping. Suppose 𝔍∈𝒥\mathfrak{J}\in\bf{\mathscr{J}}, ϑ∈Θ∗\vartheta\in\Theta^{*} and satisfies

𝔍​(ϑ​(𝔟​(𝒮​x,𝒮​y)),ϑ​(max⁡{𝔟​(x,y),𝔟​(x,𝒮​x),𝔟​(y,𝒮​y),𝔟​(𝒮​x,y)+𝔟​(x,𝒮​y)2​s}))≥c,\mathfrak{J}\left(\vartheta(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)),\vartheta(\max\{\mathfrak{b}(x,y),\mathfrak{b}(x,\mathcal{S}x),\mathfrak{b}(y,\mathcal{S}y),\frac{\mathfrak{b}(\mathcal{S}x,y)+\mathfrak{b}(x,\mathcal{S}y)}{2s}\})\right)\geq c, (2.7)

for all 𝒮​x≠𝒮​y,x,y∈𝒳\mathcal{S}x\neq\mathcal{S}y,~{}x,y\in\mathcal{X}. Then 𝒮\mathcal{S} has a unique fixed point.

Proof 2.9.

Proceeding in the similar manner as the proof of Theorem 2.52.5, substitute x=an,y=an+1x=a_{n},y=a_{n+1} in inequality (2.7)(2.7), we get

𝔍​(ϑ​(𝔟​(𝒮​an,𝒮​an+1)),ϑ​(max⁡{𝔟​(an,an+1),𝔟​(an,𝒮​an),𝔟​(an+1,𝒮​an+1),𝔟​(𝒮​an,an+1)+𝔟​(an,𝒮​an+1)2​s}))=\displaystyle\mathfrak{J}(\vartheta(\mathfrak{b}(\mathcal{S}a_{n},\mathcal{S}a_{n+1})),\vartheta(\max\{\mathfrak{b}(a_{n},a_{n+1}),\mathfrak{b}(a_{n},\mathcal{S}a_{n}),\mathfrak{b}(a_{n+1},\mathcal{S}a_{n+1}),\frac{\mathfrak{b}(\mathcal{S}a_{n},a_{n+1})+\mathfrak{b}(a_{n},\mathcal{S}a_{n+1})}{2s}\}))=
𝔍​(ϑ​(𝔟​(an+1,an+2)),ϑ​(max⁡{𝔟​(an,an+1),𝔟​(an,an+1),𝔟​(an+1,an+2),𝔟​(an+1,an+1)+𝔟​(an,an+2)2​s}))≥c.\displaystyle\hskip 28.45274pt\mathfrak{J}(\vartheta(\mathfrak{b}(a_{n+1},a_{n+2})),\vartheta(\max\{\mathfrak{b}(a_{n},a_{n+1}),\mathfrak{b}(a_{n},a_{n+1}),\mathfrak{b}(a_{n+1},a_{n+2}),\frac{\mathfrak{b}(a_{n+1},a_{n+1})+\mathfrak{b}(a_{n},a_{n+2})}{2s}\}))\geq c.

Let rn=𝔟​(an,an+1)r_{n}=\mathfrak{b}(a_{n},a_{n+1}), Then 𝔍​(ϑ​(rn+1),ϑ​(max⁡{rn,rn,rn+1,0+𝔟​(an,an+2)2​s}))≥c.\mathfrak{J}(\vartheta(r_{n+1}),\vartheta(\max\{r_{n},r_{n},r_{n+1},\dfrac{0+\mathfrak{b}(a_{n},a_{n+2})}{2s}\}))\geq c. Using the property of 𝔍\mathfrak{J}, it follows that 𝔍​(ϑ​(rn+1),ϑ​(max⁡{rn,rn+1,𝔟​(an,an+2)2​s}))<ℱc​(ϑ​(max⁡{rn,rn+1,𝔟​(an,an+2)2​s}),ϑ​(rn+1)),\mathfrak{J}(\vartheta(r_{n+1}),\vartheta(\max\{r_{n},r_{n+1},\dfrac{\mathfrak{b}(a_{n},a_{n+2})}{2s}\}))<\mathcal{F}_{c}(\vartheta(\max\{r_{n},r_{n+1},\dfrac{\mathfrak{b}(a_{n},a_{n+2})}{2s}\}),\vartheta(r_{n+1})), using the property of ℱc\mathcal{F}_{c}, it follows that ϑ​(rn+1)<ϑ​(max⁡{rn,rn+1,𝔟​(an,an+2)2​s})\vartheta(r_{n+1})<\vartheta(\max\{r_{n},r_{n+1},\dfrac{\mathfrak{b}(a_{n},a_{n+2})}{2s}\}). Considering that max⁡{rn,rn+1,𝔟​(an,an+2)2​s}<rn+1\max\{r_{n},r_{n+1},\dfrac{\mathfrak{b}(a_{n},a_{n+2})}{2s}\}<r_{n+1}. As ϑ\vartheta is increasing, ϑ​(max⁡{rn,rn+1,𝔟​(an,an+2)2​s})≤ϑ​(rn+1)\vartheta(\max\{r_{n},r_{n+1},\dfrac{\mathfrak{b}(a_{n},a_{n+2})}{2s}\})\leq\vartheta(r_{n+1}), which leads to a contradiction. Hence, rn+1<max⁡{rn,rn+1,𝔟​(an,an+2)2​s}r_{n+1}<\max\{r_{n},r_{n+1},\dfrac{\mathfrak{b}(a_{n},a_{n+2})}{2s}\}. Further, rn+1<max⁡{rn,rn+1,𝔟​(an,an+2)2​s}≤max⁡{rn,rn+1,rn+rn+12}=rnr_{n+1}<\max\{r_{n},r_{n+1},\dfrac{\mathfrak{b}(a_{n},a_{n+2})}{2s}\}\leq\max\{r_{n},r_{n+1},\dfrac{r_{n}+r_{n+1}}{2}\}=r_{n}. Thus, (rn)(r_{n}) is a decreasing sequence of positive reals; hence limn→∞rn=a≥0\lim\limits_{n\to\infty}r_{n}=a\geq 0. Following the steps in Theorem 2.52.5, we get limn→∞rn=0.\lim\limits_{n\to\infty}r_{n}=0. Now, we aim to prove that (an)(a_{n}) is a Cauchy sequence. Assume on contrary, that there exists an ε>0\varepsilon>0 such that subsequences (ani)(a_{n_{i}}) and (ami)(a_{m_{i}}) of sequence (an)(a_{n}) such that nin_{i} is the smallest integer for which

ni>mi>i,𝔟​(ami,ani)≥ε​and​𝔟​(ami,ani−1)<ε.n_{i}>m_{i}>i,~{}\mathfrak{b}(a_{m_{i}},a_{n_{i}})\geq\varepsilon~{}\text{and}~{}\mathfrak{b}(a_{m_{i}},a_{n_{i}-1})<\varepsilon.

Now, ε≤𝔟​(ami,ani)≤s​(𝔟​(ami,ani−1)+𝔟​(ani−1,ani))\varepsilon\leq\mathfrak{b}(a_{m_{i}},a_{n_{i}})\leq s(\mathfrak{b}(a_{m_{i}},a_{n_{i}-1})+\mathfrak{b}(a_{n_{i}-1},a_{n_{i}})), which implies that

ε≤lim infi→∞𝔟​(ami,ani)≤s​ε​ and ​ε≤lim supi→∞𝔟​(ami,ani)≤s​ε.\varepsilon\leq\liminf\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}},a_{n_{i}})\leq s\varepsilon\text{ and }\varepsilon\leq\limsup\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}},a_{n_{i}})\leq s\varepsilon. (2.8)

Note ami≠ani​ as ​𝔟​(ami,ani)≥εa_{m_{i}}\neq a_{n_{i}}\text{ as }\mathfrak{b}(a_{m_{i}},a_{n_{i}})\geq\varepsilon. Substitute x=amix=a_{m_{i}} and y=aniy=a_{n_{i}} in inequality (2.7)(2.7), we get

c\displaystyle c ≤𝔍​(ϑ​(𝔟​(ami,ani)),ϑ​(max⁡{𝔟​(ami−1,ani−1),rmi,rni,𝔟​(ami,ani−1)+𝔟​(ami−1,ani)2​s}))\displaystyle\leq\mathfrak{J}(\vartheta(\mathfrak{b}(a_{m_{i}},a_{n_{i}})),\vartheta(\max\{\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1}),r_{m_{i}},r_{n_{i}},\frac{\mathfrak{b}(a_{m_{i}},a_{n_{i}-1})+\mathfrak{b}(a_{m_{i}-1},a_{n_{i}})}{2s}\})) (2.9)
<ℱc​(ϑ​(max⁡{𝔟​(ami−1,ani−1),rmi,rni,𝔟​(ami,ani−1)+𝔟​(ami−1,ani)2​s}),ϑ​(𝔟​(ami,ani))).\displaystyle<\mathcal{F}_{c}(\vartheta(\max\{\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1}),r_{m_{i}},r_{n_{i}},\frac{\mathfrak{b}(a_{m_{i}},a_{n_{i}-1})+\mathfrak{b}(a_{m_{i}-1},a_{n_{i}})}{2s}\}),\vartheta(\mathfrak{b}(a_{m_{i}},a_{n_{i}}))).

By the property (i​v)(iv) of ℱc\mathcal{F}_{c}, we have

ϑ​(𝔟​(ami,ani))<ϑ​(max⁡{𝔟​(ami−1,ani−1),rmi,rni,𝔟​(ami,ani−1)+𝔟​(ami−1,ani)2​s}).\vartheta(\mathfrak{b}(a_{m_{i}},a_{n_{i}}))<\vartheta(\max\{\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1}),r_{m_{i}},r_{n_{i}},\frac{\mathfrak{b}(a_{m_{i}},a_{n_{i}-1})+\mathfrak{b}(a_{m_{i}-1},a_{n_{i}})}{2s}\}).

Suppose max⁡{𝔟​(ami−1,ani−1),rmi,rni,𝔟​(ami,ani−1)+𝔟​(ami−1,ani)2​s}<𝔟​(ami,ani).\max\{\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1}),r_{m_{i}},r_{n_{i}},\dfrac{\mathfrak{b}(a_{m_{i}},a_{n_{i}-1})+\mathfrak{b}(a_{m_{i}-1},a_{n_{i}})}{2s}\}<\mathfrak{b}(a_{m_{i}},a_{n_{i}}). As ϑ\vartheta is increasing,
ϑ​(max⁡{𝔟​(ami−1,ani−1),rmi,rni,𝔟​(ami,ani−1)+𝔟​(ami−1,ani)2​s})≤ϑ​(𝔟​(ami,ani))\vartheta(\max\{\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1}),r_{m_{i}},r_{n_{i}},\dfrac{\mathfrak{b}(a_{m_{i}},a_{n_{i}-1})+\mathfrak{b}(a_{m_{i}-1},a_{n_{i}})}{2s}\})\leq\vartheta(\mathfrak{b}(a_{m_{i}},a_{n_{i}})), which is a contradiction. Hence, 𝔟​(ami,ani)≤max⁡{𝔟​(ami−1,ani−1),rmi,rni,𝔟​(ami,ani−1)+𝔟​(ami−1,ani)2​s}.\mathfrak{b}(a_{m_{i}},a_{n_{i}})\leq\max\{\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1}),r_{m_{i}},r_{n_{i}},\dfrac{\mathfrak{b}(a_{m_{i}},a_{n_{i}-1})+\mathfrak{b}(a_{m_{i}-1},a_{n_{i}})}{2s}\}. Now, consider the following three cases:
Case (i)(i): If 𝔟​(ami,ani)<rmi\mathfrak{b}(a_{m_{i}},a_{n_{i}})<r_{m_{i}} or rni~{}r_{n_{i}} holds for infinitely many ii then limi→∞𝔟​(ami,ani)=0\lim\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}},a_{n_{i}})=0, which is contrary to inequality (2.8)(2.8).
Case (i​i)(ii): If 𝔟​(ami,ani)<𝔟​(ami−1,ani−1)\mathfrak{b}(a_{m_{i}},a_{n_{i}})<\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1})~{} holds for infinitely many ii then it follows from Theorem 2.52.5, (an)(a_{n}) is a Cauchy sequence.
Case (i​i​i)(iii): Let pi=12​s​(𝔟​(ami,ani−1)+𝔟​(ami−1,ani))p_{i}=\dfrac{1}{2s}(\mathfrak{b}(a_{m_{i}},a_{n_{i}-1})+\mathfrak{b}(a_{m_{i}-1},a_{n_{i}})). If 𝔟​(ami,ani)<pi\mathfrak{b}(a_{m_{i}},a_{n_{i}})<p_{i} holds for infinitely many ii then pi≤12​(𝔟​(ami,ani)+2​rni−1+𝔟​(ami−1,ani−1))p_{i}\leq\dfrac{1}{2}(\mathfrak{b}(a_{m_{i}},a_{n_{i}})+2r_{n_{i}-1}+\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1})). Hence 𝔟​(ami,ani)<pi≤12​(𝔟​(ami,ani)+2​rni−1+𝔟​(ami−1,ani−1))\mathfrak{b}(a_{m_{i}},a_{n_{i}})<p_{i}\leq\dfrac{1}{2}(\mathfrak{b}(a_{m_{i}},a_{n_{i}})+2r_{n_{i}-1}+\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1})). Also, 𝔟​(ani−1,ami−1)≤s​(𝔟​(ami−1,ami)+𝔟​(ami,ani−1))\mathfrak{b}(a_{n_{i}-1},a_{m_{i}-1})\leq s(\mathfrak{b}(a_{m_{i}-1},a_{m_{i}})+\mathfrak{b}(a_{m_{i}},a_{n_{i}-1})) hence, lim supi→∞𝔟​(ami−1,ani−1)≤s​ε.\limsup\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}-1},a_{n_{i}-1})\leq s\varepsilon. By inequality (2.8)(2.8) and limn→∞rn=0\lim\limits_{n\to\infty}r_{n}=0, we get

0<lim infi→∞𝔟​(ami,ani)≤s​(lim supi→∞pi)≤s2​ε≤s2​(lim infi→∞𝔟​(ami,ani))<∞0<\liminf\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}},a_{n_{i}})\leq s(\limsup\limits_{i\to\infty}p_{i})\leq s^{2}\varepsilon\leq s^{2}(\liminf\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}},a_{n_{i}}))<\infty

and

0<lim infi→∞pi≤s​ε≤s​(lim supi→∞𝔟​(ami,ani))≤s2​ε≤s2​(lim infi→∞pi)<∞.0<\liminf\limits_{i\to\infty}p_{i}\leq s\varepsilon\leq s(\limsup\limits_{i\to\infty}\mathfrak{b}(a_{m_{i}},a_{n_{i}}))\leq s^{2}\varepsilon\leq s^{2}(\liminf\limits_{i\to\infty}p_{i})<\infty.

Hence by the property of 𝔍\mathfrak{J}, we get lim supi→∞𝔍​(ϑ​(𝔟​(ami,ani)),ϑ​(pi))<c,\limsup\limits_{i\to\infty}\mathfrak{J}(\vartheta(\mathfrak{b}(a_{m_{i}},a_{n_{i}})),\vartheta(p_{i}))<c, which is a contradiction to the inequality (2.9)(2.9). Hence (an)(a_{n}) is a Cauchy sequence.
Since (𝒳,𝔟,s)(\mathcal{X},\mathfrak{b},s) is complete bb-metric space,

limn,m→∞𝔟​(an,am)=limn→∞𝔟​(an,z)=0,for some​z∈𝒳.\lim\limits_{n,m\to\infty}\mathfrak{b}(a_{n},a_{m})=\lim_{n\to\infty}\mathfrak{b}(a_{n},z)=0,~{}\text{for some}~{}z\in\mathcal{X}.

We will prove that zz becomes the unique fixed point of 𝒮\mathcal{S}. Also, 𝔟​(𝒮​z,z)≤s​(𝔟​(𝒮​z,𝒮​an−1)+𝔟​(𝒮​an−1,z)).\mathfrak{b}(\mathcal{S}z,z)\leq s(\mathfrak{b}(\mathcal{S}z,\mathcal{S}a_{n-1})+\mathfrak{b}(\mathcal{S}a_{n-1},z)). Using continuity of 𝒮\mathcal{S} and limn→∞an=z\lim\limits_{n\to\infty}a_{n}=z, we get 𝔟​(𝒮​z,z)=0.\mathfrak{b}(\mathcal{S}z,z)=0. Hence, zz becomes the fixed point of 𝒮.\mathcal{S}. Let ww be another fixed point of 𝒮\mathcal{S} such that z≠wz\neq w. On substitution of x=zx=z and y=wy=w in inequality (2.7)(2.7), we have

c\displaystyle c ≤𝔍​(ϑ​(𝔟​(𝒮​z,𝒮​z)),ϑ​(max⁡{𝔟​(z,z),𝔟​(z,𝒮​z),𝔟​(x,𝒮​z),𝔟​(𝒮​z,z)+𝔟​(z,𝒮​z)2​s}))\displaystyle\leq\mathfrak{J}\left(\vartheta(\mathfrak{b}(\mathcal{S}z,\mathcal{S}z)),\vartheta(\max\{\mathfrak{b}(z,z),\mathfrak{b}(z,\mathcal{S}z),\mathfrak{b}(x,\mathcal{S}z),\frac{\mathfrak{b}(\mathcal{S}z,z)+\mathfrak{b}(z,\mathcal{S}z)}{2s}\})\right)
<ℱc​(ϑ​(max⁡{𝔟​(z,z),𝔟​(z,z),𝔟​(z,z),𝔟​(z,z)+𝔟​(z,z)2​s}),ϑ​(𝔟​(z,z)))\displaystyle<\mathcal{F}_{c}\left(\vartheta(\max\{\mathfrak{b}(z,z),\mathfrak{b}(z,z),\mathfrak{b}(z,z),\frac{\mathfrak{b}(z,z)+\mathfrak{b}(z,z)}{2s}\}),\vartheta(\mathfrak{b}(z,z))\right)
<ℱc(ϑ(max{𝔟(z,z),0,0,𝔟​(z,z)s},ϑ(𝔟(z,z)))\displaystyle<\mathcal{F}_{c}\left(\vartheta(\max\{\mathfrak{b}(z,z),0,0,\frac{\mathfrak{b}(z,z)}{s}\},\vartheta(\mathfrak{b}(z,z))\right)
<ℱc​(ϑ​(𝔟​(z,z)),ϑ​(𝔟​(z,z)))\displaystyle<\mathcal{F}_{c}\left(\vartheta(\mathfrak{b}(z,z)),\vartheta(\mathfrak{b}(z,z))\right)
≤c,\displaystyle\leq c,

which is a contradiction. Hence zz becomes the unique fixed point.

Example 2.10.

Let 𝒳={1,2,3,4}\mathcal{X}=\{1,2,3,4\}. Define 𝔟:𝒳×𝒳→ℝ+\mathfrak{b}:\mathcal{X}\times\mathcal{X}\rightarrow\mathbb{R^{+}} given by
𝔟​(x,x)=0\mathfrak{b}(x,x)=0 for all x∈𝒳x\in\mathcal{X},
𝔟​(1,2)=𝔟​(2,1)=3\mathfrak{b}(1,2)=\mathfrak{b}(2,1)=3,
𝔟​(2,3)=𝔟​(3,2)=𝔟​(1,3)=𝔟​(3,1)=1\mathfrak{b}(2,3)=\mathfrak{b}(3,2)=\mathfrak{b}(1,3)=\mathfrak{b}(3,1)=1,
𝔟​(1,4)=𝔟​(4,1)=15,\mathfrak{b}(1,4)=\mathfrak{b}(4,1)=15,
𝔟​(2,4)=𝔟​(4,2)=𝔟​(3,4)=𝔟​(4,3)=4.\mathfrak{b}(2,4)=\mathfrak{b}(4,2)=\mathfrak{b}(3,4)=\mathfrak{b}(4,3)=4.
Clearly, (𝒳,𝔟,s)(\mathcal{X},\mathfrak{b},s) is a complete bb-metric space with coefficient s=3.s=3.
Also, define 𝔍:[1,∞)×[1,∞)→ℝ\mathfrak{J}:[1,\infty)\times[1,\infty)\rightarrow\mathbb{R}, ℱc:[1,∞)×[1,∞)→ℝ\mathcal{F}_{c}:[1,\infty)\times[1,\infty)\rightarrow\mathbb{R} and ϑ:(0,∞)→(1,∞)\vartheta:(0,\infty)\to(1,\infty) by

𝔍​(x,y)=y3​x​ , ​ℱc​(y,x)=yx​ and ​ϑ​(x)=x+1.\mathfrak{J}(x,y)=\frac{y}{3x}\text{ , }\mathcal{F}_{c}(y,x)=\frac{y}{x}\text{ and }\vartheta(x)=x+1.

Note that 𝔍​(x,y)<ℱc​(y,x)\mathfrak{J}(x,y)<\mathcal{F}_{c}(y,x), for all x,y∈(1,∞)x,y\in(1,\infty) and c=1c=1. Further, If (an)(a_{n}) and (bn)(b_{n}) are sequences in (0,∞)(0,\infty) such that

0<lim infn→∞an≤3​(lim supn→∞bn)≤32​(lim infn→∞an)<∞0<\liminf\limits_{n\to\infty}a_{n}\leq 3(\limsup\limits_{n\to\infty}b_{n})\leq 3^{2}(\liminf\limits_{n\to\infty}a_{n})<\infty

and

0<lim infn→∞bn≤3​(lim supn→∞an)≤32​(lim infn→∞bn)<∞,0<\liminf\limits_{n\to\infty}b_{n}\leq 3(\limsup\limits_{n\to\infty}a_{n})\leq 3^{2}(\liminf\limits_{n\to\infty}b_{n})<\infty,

then lim supn→∞𝔍​(ϑ​(an),ϑ​(bn))=lim supn→∞(bn+13​(an+1))=lim supn→∞(bn+1)lim infn→∞3​(an+1)\limsup\limits_{n\to\infty}\mathfrak{J}(\vartheta(a_{n}),\vartheta(b_{n}))=\limsup\limits_{n\to\infty}(\dfrac{b_{n}+1}{3(a_{n}+1)})=\dfrac{\limsup\limits_{n\to\infty}(b_{n}+1)}{\liminf\limits_{n\to\infty}3(a_{n}+1)}.
As lim supn→∞bn≤lim infn→∞3​an.\limsup\limits_{n\to\infty}b_{n}\leq\liminf\limits_{n\to\infty}3a_{n}. Hence, lim supn→∞𝔍​(ϑ​(an),ϑ​(bn))<1\limsup\limits_{n\to\infty}\mathfrak{J}(\vartheta(a_{n}),\vartheta(b_{n}))<1. Thus 𝔍\mathfrak{J} is a 𝔸ℝ\mathbb{A}_{\mathbb{R}}-simulation function. Moreover, define 𝒮:𝒳→𝒳\mathcal{S}:\mathcal{X}\rightarrow\mathcal{X} by

𝒮​x={3, when ​x≠41, else .\mathcal{S}x=\begin{cases}3,\text{ when }x\neq 4\\ 1,\text{ else }\end{cases}.

We will now verify that 𝒮\mathcal{S} is a 𝔍𝔸ℝ\mathfrak{J}_{{}_{\mathbb{A}_{\mathbb{R}}}}-contraction. Note that

𝔟​(𝒮​x,𝒮​y)={𝔟​(1,3)=1,if ​x=4,y≠4,𝔟​(1,1)=0,if ​x=4,y=4,𝔟​(3,3)=0,if ​x≠4,y≠4,𝔟​(3,1)=1,if ​x≠4,y=4,\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)=\begin{cases}\mathfrak{b}(1,3)=1,&\text{if }x=4,y\neq 4,\\ \mathfrak{b}(1,1)=0,&\text{if }x=4,y=4,\\ \mathfrak{b}(3,3)=0,&\text{if }x\neq 4,y\neq 4,\\ \mathfrak{b}(3,1)=1,&\text{if }x\neq 4,y=4,\end{cases}

hence 𝔟​(𝒮​x,𝒮​y)>0\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)>0 if and only if x=4,y≠4,x=4,y\neq 4, and x≠4,y=4x\neq 4,y=4. Now, consider the following cases:
Case 1: if x=4x=4 and y≠4,y\neq 4, then

1<15+13​(1+1)=𝔟​(1,4)+13​(𝔟​(𝒮​x,𝒮​y)+1)=𝔟​(𝒮​x,x)+13​(𝔟​(𝒮​x,𝒮​y)+1)≤𝔍(ϑ(𝔟(𝒮x,𝒮y)),ϑ(Ms(x,y))=Ms​(x,y)+13​(𝔟​(𝒮​x,𝒮​y)+1)1<\dfrac{15+1}{3(1+1)}=\dfrac{\mathfrak{b}(1,4)+1}{3(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)+1)}=\dfrac{\mathfrak{b}(\mathcal{S}x,x)+1}{3(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)+1)}\leq\mathfrak{J}(\vartheta(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)),\vartheta(M_{s}(x,y))=\dfrac{M_{s}(x,y)+1}{3(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)+1)}

where Ms​(x,y)=max⁡{𝔟​(x,y),𝔟​(x,𝒮​x),𝔟​(y,𝒮​y),𝔟​(𝒮​x,y)+𝔟​(x,𝒮​y)2​s}M_{s}(x,y)=\max\{\mathfrak{b}(x,y),\mathfrak{b}(x,\mathcal{S}x),\mathfrak{b}(y,\mathcal{S}y),\dfrac{\mathfrak{b}(\mathcal{S}x,y)+\mathfrak{b}(x,\mathcal{S}y)}{2s}\}. Hence, 𝒮\mathcal{S} is a 𝔍𝔸ℝ\mathfrak{J}_{{}_{\mathbb{A}_{\mathbb{R}}}}-contraction.
Case 2:2: if x=1x=1, and y=4y=4, then

1<15+13​(1+1)=𝔟​(1,4)+13​(𝔟​(𝒮​x,𝒮​y)+1)=𝔟​(x,y)+13​(𝔟​(𝒮​x,𝒮​y)+1)≤𝔍(ϑ(𝔟(𝒮x,𝒮y)),ϑ(Ms(x,y))=Ms​(x,y)+13​(𝔟​(𝒮​x,𝒮​y)+1)1<\dfrac{15+1}{3(1+1)}=\dfrac{\mathfrak{b}(1,4)+1}{3(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)+1)}=\dfrac{\mathfrak{b}(x,y)+1}{3(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)+1)}\leq\mathfrak{J}(\vartheta(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)),\vartheta(M_{s}(x,y))=\dfrac{M_{s}(x,y)+1}{3(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)+1)}

where Ms​(x,y)=max⁡{𝔟​(x,y),𝔟​(x,𝒮​x),𝔟​(y,𝒮​y),𝔟​(𝒮​x,y)+𝔟​(x,𝒮​y)2​s}M_{s}(x,y)=\max\{\mathfrak{b}(x,y),\mathfrak{b}(x,\mathcal{S}x),\mathfrak{b}(y,\mathcal{S}y),\dfrac{\mathfrak{b}(\mathcal{S}x,y)+\mathfrak{b}(x,\mathcal{S}y)}{2s}\}. Hence, 𝒮\mathcal{S} is a 𝔍𝔸ℝ\mathfrak{J}_{{}_{\mathbb{A}_{\mathbb{R}}}}-contraction.
Case 3:3: if x∈{2,3}x\in\{2,3\}, and y=4y=4, then

1<15+13​(1+1)=𝔟​(1,4)+13​(𝔟​(𝒮​x,𝒮​y)+1)=𝔟​(𝒮​y,y)+13​(𝔟​(𝒮​x,𝒮​y)+1)≤𝔍(ϑ(𝔟(𝒮x,𝒮y)),ϑ(Ms(x,y))=Ms​(x,y)+13​(𝔟​(𝒮​x,𝒮​y)+1)1<\dfrac{15+1}{3(1+1)}=\dfrac{\mathfrak{b}(1,4)+1}{3(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)+1)}=\dfrac{\mathfrak{b}(\mathcal{S}y,y)+1}{3(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)+1)}\leq\mathfrak{J}(\vartheta(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)),\vartheta(M_{s}(x,y))=\dfrac{M_{s}(x,y)+1}{3(\mathfrak{b}(\mathcal{S}x,\mathcal{S}y)+1)}

where Ms​(x,y)=max⁡{𝔟​(x,y),𝔟​(x,𝒮​x),𝔟​(y,𝒮​y),𝔟​(𝒮​x,y)+𝔟​(x,𝒮​y)2​s}M_{s}(x,y)=\max\{\mathfrak{b}(x,y),\mathfrak{b}(x,\mathcal{S}x),\mathfrak{b}(y,\mathcal{S}y),\dfrac{\mathfrak{b}(\mathcal{S}x,y)+\mathfrak{b}(x,\mathcal{S}y)}{2s}\}. Hence, 𝒮\mathcal{S} is a 𝔍𝔸ℝ\mathfrak{J}_{{}_{\mathbb{A}_{\mathbb{R}}}}-contraction.
Therefore, in all cases, 𝒮\mathcal{S} is a 𝔍𝔸ℝ\mathfrak{J}_{{}_{\mathbb{A}_{\mathbb{R}}}}-contraction. So, by Theorem 2.72.7, 𝒮\mathcal{S} has a unique fixed point 33.

Remark 2.11.

Only increasing property of ϑ\vartheta and property (i​v)(iv) of ℱc\mathcal{F}_{c} is used throughout.

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