# On Fixed Point Results in Controlled Metric Spaces

Authors: Jamshaid Ahmad; Abdullah Eqal Al-Mazrooei; Hassen Aydi; Manuel De la Sen
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this article, we introduce Reich type contractions and (α,F)-contractions in the class of controlled metric spaces and establish some new related fixed point theorems. Our results are generalizations of some known results of literature. Some examples and certain consequences are given to illustrate significance of presented results.


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## Body

## 1. Introduction and Preliminaries

In 1993, Czerwik [1] reintroduced a new class of generalized metric spaces, called as b-metric spaces, as generalizations of metric spaces.Definition 1 ([1]).
LetY be a nonempty set and s≥1. A function db:Y×Y⟶0,∞ is said to be a b-metric if for all ς,ω,τ∈Y,
(b1) dbς,ω=0 iff ς=ω
(b2) dbς,ω=dbω,ς for all ς,ω∈Y
(b3) dbς,τ≤sdbς,ω+dbω,τ
The pairY,db is then called a b-metric space. Subsequently, many fixed point results on such spaces were given (see [2–7]).
Kamran et al. [8] initiated the concept of extended b-metric spaces.Definition 2.
LetY be a nonempty set and p:Y×Y⟶1,∞ be a function. A function de:Y×Y⟶0,∞ is called an extended b-metric if for all ς,ω,τ∈Y,
(i)
deς,ω=0 iff ς=ω(ii)
deς,ω=deω,ς(iii)
deς,ω≤pς,ωdeς,τ+deτ,ω
The pairY,de is called an extended b-metric space.
Very recently, a new kind of a generalizedb-metric space was introduced by Mlaiki et al. [9].Definition 3.
LetY be a nonempty set and p:Y×Y⟶1,∞be a function. A function C:Y×Y⟶0,∞ is called a controlled metric if for all ς,ω,τ∈Y,
(i)
Cς,ω=0 iff ς=ω(ii)
Cς,ω=Cω,ς(iii)
Cς,ω≤pς,τCς,τ+pω,τCω,τ
The pairY,C is called a controlled metric space (see also [10]).
The Cauchy and convergent sequences in controlled metric type spaces are defined in this way.Definition 4 ([9]).
LetY,C be a controlled metric space and ςnn≥0 be a sequence in Y. Then,
(i)
The sequenceςn converges to some ς in Y; if for every ε>0, there exists N=Nε∈ℕ such that Cςn,ς<ε for all n≥N. In this case, we write limn→∞ςn=ς(ii)
The sequenceςn is Cauchy; if for every ε>0, there exists N=Nε∈ℕ such that Cςm,ςn<ε for all m,n≥N(iii)
The controlled metric spaceY,C is called complete if every Cauchy sequence is convergentDefinition 5 ([9]).
LetY,C be a controlled metric space. Let ς∈Y and ε>0.(i)
The open ballBς,ε is(1)Bς,ε=ω∈Y:Cς,ω⊲ε(ii)
The mappingΩ:Y⟶Y is said to be continuous at ς∈Y; if for all ε>0, there exists δ>0 such that ΩBς,δ⊆BΩς,ε
Very recently, Wardowski [11] introduced a new type of contractions, called F-contractions and established some new related fixed point theorems in the context of complete metric spaces.Definition 6.
LetF:ℝ+⟶ℝ be a function satisfying
(F1) F is strictly increasing, that is, for all t1,t2∈ℝ+ such that t1<t2 implies that Ft1<Ft2.
(F2) For every sequence tn of positive real numbers, limn→∞tn=0 and limn→∞Ftn=−∞ are equivalent.
(F3) There is h∈0,1 so that limt→0+thFt=0.
Let⊑ be the set of above functions F satisfying (F1)-(F3) (to be consistent with Wardowski [11]). A self-mapping Ω on the metric space Y,η is said to be an F-contraction if there are a function F satisfying (F1)-(F3) and a constant τ>0 so that
(2)ηΩς,Ωω>0⇒τ+FηΩς,Ωω≤Fης,ωfor all ς,ω∈Y.Theorem 7 [11].
LetY,η be a complete metric space and Ω:Y⟶Y be an F-contraction, then Ω admits a unique fixed point.
The authors in [11] manifested that a Banach contraction is a specific case of F-contractions, while there are many F-contractions which need not be a Banach contraction. For more details, we refer the readers to ([12–22]).In this paper, we first define Reich [23, 24] and (α,F)-contractions in the setting of controlled metric spaces and prove some new fixed point results. We also provide some examples to illustrate significance of the established results.


## 2. Results on Reich Type Contractions

Theorem 8.
LetY,C be a complete controlled metric space. Let Ω:Y⟶Y be so that there are α,β,γ∈0,1 with k=α+β/1−γ<1,
(3)CΩς,Ωω≤αCς,ω+βCς,Ως+γCω,Ωωfor all ς,ω∈Y. For ς0∈Y, take ςn=Ωnς0. Assume that
(4)supm≥1limi→∞pςi+1,ςi+2pςi+1,ςmpςi,ςi+1<1k.
Suppose thatlimn→∞pςn,ς and limn→∞pς,ςn exist, are finite, and γlimn→∞pςn,ς<1 for every ς∈Y, then Ω possesses a unique fixed point.Proof.
The considered sequenceςn verifies ςn+1=Ωςn for all n∈ℕ. Obviously, if there exists n0∈ℕfor which ςn0+1=ςn0, then Ωςn0=ςn0, and the proof is finished. Thus, we suppose that ςn+1=ςn for every n∈ℕ. Thus, by (1), we have
(5)Cςn,ςn+1=CΩςn−1,Ωςn≤αCςn−1,ςn+βCςn−1,Ωςn−1+γCςn,Ωςn=αCςn−1,ςn+βCςn−1,ςn+γCςn,ςn+1which implies that
(6)Cςn,ςn+1≤α+β1−γCςn−1,ςn=kCςn−1,ςn.
Thus, we have(7)Cςn,ςn+1≤kCςn−1,ςn≤k2Cςn−2,ςn−1≤⋯≤knCς0,ς1.
For alln,m∈ℕn<m, we have
(8)Cςn,ςm≤pςn,ςn+1Cςn,ςn+1+pςn+1,ςmCςn+1,ςm≤pςn,ςn+1Cςn,ςn+1+pςn+1,ςmpςn+1,ςn+2Cςn+1,ςn+2+pςn+1,ςmpςn+2,ςmCςn+2,ςm≤pςn,ςn+1Cςn,ςn+1+pςn+1,ςmpςn+1,ςn+2Cςn+1,ςn+2+pςn+1,ςmpςn+2,ςmpςn+2,ςn+3Cςn+2,ςn+3+pςn+1,ςmpςn+2,ςmpςn+3,ςmCςn+3,ςm≤⋯≤pςn,ςn+1Cςn,ςn+1+∑i=n+1m−2∏j=n+1ipςj,ςmpςi,ςi+1Cςi,ςi+1+∏i=n+1m−1pςi,ςmCςm−1,ςm.
This implies that(9)Cςn,ςm≤pςn,ςn+1Cςn,ςn+1+∑i=n+1m−2∏j=n+1ipςj,ςmpςi,ςi+1Cςi,ςi+1+∏i=n+1m−1pςi,ςmpςm−1,ςmCςm−1,ςm≤pςn,ςn+1knCς0,ς1+∑i=n+1m−2∏j=n+1ipςj,ςmpςi,ςi+1kiCς0,ς1+∏i=n+1m−1pςi,ςmpςm−1,ςmkm−1Cς0,ς1=pςn,ςn+1knCς0,ς1+∑i=n+1m−1∏j=n+1ipςj,ςmpςi,ςi+1kiCς0,ς1.
Let(10)Sl=∑i=0l∏j=0ipςj,ςmpςi,ςi+1kiCς0,ς1.
Consider(11)υi=∏j=0ipςj,ςmpςi,ςi+1kiCς0,ς1.
We have(12)υi+1υi=pςi+1,ςmpςi+1,ςi+2pςi,ςi+1k.
In view of condition (4) and the ratio test, we ensure that the series ∑iυi converges. Thus, limn→∞Sn exists. Hence, the real sequence Sn is Cauchy.
Now, using (9), we get
(13)Cςn,ςm≤Cς0,ς1knpςn,ςn+1+Sm−1−Sn.
Above, we usedpς,ω≥1. Letting n,m⟶∞ in (13), we obtain
(14)limn,m→∞Cςn,ςm=0.
Thus, the sequenceςn is Cauchy in the complete controlled metric space Y,C. So, there is some ς∗∈Y so that
(15)limn→∞Cςn,ς∗=0;that is, ςn⟶ς∗ as n⟶∞. Now, we will prove that ς∗ is a fixed point of Ω. By (3) and condition (iii), we get
(16)Cς∗,Ως∗≤pς∗,ςn+1Cς∗,ςn+1+pςn+1,Ως∗Cςn+1,Ως∗=pς∗,ςn+1Cς∗,ςn+1+pςn+1,Ως∗CΩςn,Ως∗≤pς∗,ςn+1Cς∗,ςn+1+pςn+1,Ως∗αCςn,ς∗+βCςn,Ωςn+γCς∗,Ως∗=pς∗,ςn+1Cς∗,ςn+1+pςn+1,Ως∗αCςn,ς∗+βCςn,ςn+1+γCς∗,Ως∗.
Taking the limit asn⟶∞ and using (5, 6) and the fact that limn→∞pςn,ς and limn→∞pς,ςn exist, are finite, we obtain that
(17)Cς∗,Ως∗≤γlimn→∞pςn+1,Ως∗Cς∗,Ως∗.
Suppose thatς∗≠Ως∗, having in mind that γlimn→∞pςn+1,Ως∗<1, so
(18)0<Cς∗,Ως∗≤γlimn→∞pςn+1,Ως∗Cς∗,Ως∗<Cς∗,Ως∗.
It is a contradiction. This yields thatς∗=Ως∗. The uniqueness of the fixed point follows easily. It completes the proof.Example 9.
ConsiderY=0,1,2. Take the controlled metric C defined as
(19)C0,1=12,C0,2=1120,C1,2=320,where p:Y×Y⟶1,∞ is symmetric such that
(20)p0,0=p1,1=p2,2=p1,2=1,p0,2=2,p0,1=32.
GivenΩ:Y⟶Y as
(21)Ω0=2andΩ1=Ω2=1.
Considerα=1/11 and β=γ=3/11. Take ς0=0, then ς1=2 and ςn=1 for all n≥2. Clearly, (4) is satisfied. On the other hand, note that (3) holds for all ς,ω∈Y. All other hypotheses of Theorem 8 are verified, and so Ω has a unique fixed point, which is u=1.Example 10.
LetY=0,1. Consider the controlled metric type C defined as
(22)Cς,ω=ς−ω2,where pς,ω=ς+ω+1 for ς,ω∈Y. Take Ως=ς2/4. Consider α=1/4 and β=γ=1/3. Take ς0=0; so, (4) is satisfied. Also, (3) holds. All conditions in Theorem 8 are fulfilled, and so, there is a unique fixed point, which is u=0.Corollary 11 (see. [9]).
LetY,C be a complete controlled metric space. Let Ω:Y⟶Y be s that there are α∈0,1 and
(23)CΩς,Ωω≤αCς,ωfor all ς,ω∈Y. For ς0∈Y, take ςn=Ωnς0. Assume that
supm≥1limi→∞pςi+1,ςi+2pςi+1,ςm/pςi,ςi+1<1/α.(36)
Suppose thatlimn→∞pςn,ς and limn→∞pς,ςn exist, are finite, and γlimn→∞pςn,ς<1 for every ς∈Y, then Ω possesses a unique fixed point.Proof.
Takingβ=γ=0 in Theorem 8.


## 3. Results on (α,F)-Contractions

In 2012, Samet et al. [25] initiated the notion of α-admissible mappings and proved some related fixed point results in the context of complete metric spaces.Definition 12 ([25]).
LetY be a nonempty set, and α:Y×Y⟶0,∞ be a given function. A self-mapping Ω on Y is called α-admissible if
(24)ς,ω∈Y,ας,ω≥1⇒αΩς,Ωω≥1.Definition 13.
LetY,C be a controlled metric space. A mapping Ω:Y⟶Y is said to be an (α,F)-contraction if there are some α:Y×Y⟶ℝ+,F∈⊑, and τ>0 so that
(25)τ+ας,ωFCΩς,Ωω≤FCς,ωfor all ς,ω∈Y with CΩς,Ωω>0.Theorem 14.
LetY,C be a complete controlled metric space. Let Ω:Y⟶Y be an (α,F)-contraction so that
(i)
Ω is α-admissible(ii)
There isς0∈Y so that ας0,Ως0≥1(iii)
Ω is continuous(iv)
Forς0∈Y, define the Picard sequence ςn=Ωnς0 such that
(26)supm≥1limi→∞pςi+1,ςi+2pςi+1,ςmpςi,ςi+1<1.
Assume thatlimn→∞pςn,ς and limn→∞pς,ςn exist and are finite, for every ς∈Y, then Ω possesses a unique fixed point.Proof.
Letς0∈Y be such that ας0,Ως0≥1. We define a sequence ςn in Y by ςn+1=Ωςn for all n∈ℕ. Clearly, if there is n0 so that ςn0+1=ςn0, then the proof is finished. So, assume that ςn+1=ςn for each n∈ℕ, by using (i) and (ii), it is obvious that
(27)αςn,ςn+1≥1for all n∈ℕ. By (25), we have
(28)τ+FCςn,ςn+1=τ+FCΩςn−1,Ωςn≤τ+αςn,ςn+1FCΩςn−1,Ωςn.
SinceΩ is an (α,F)-contraction, we can write
(29)τ+FCςn,ςn+1≤τ+αςn,ςn+1FCΩςn−1,Ωςn≤FCςn−1,ςn.
Thus, from (29), we get
(30)FCςn,ςn+1≤FCςn−1,ςn−τ≤FCςn−2,ςn−1−2τ≤FCςn−3,ςn−2−3τ≤⋯≤FCς0,ς1.−nτ
Thus, by (29), we have
(31)FCςn,ςn+1≤≤FCς0,ς1−nτ.
Lettingn⟶∞ in (31), we get
(32)limn→∞FCςn,ςn+1=−∞.
By (F2), we get
(33)limn→∞Cςn,ςn+1=0.
Now, by (F3), there is h∈0,1 so that
(34)limn→∞Cςn,ςn+1hFCςn,ςn+1=0.
From (25), we have
(35)Cςn,ςn+1hFCςn,ςn+1−Cςn,ςn+1hFCς0,ςn+1≤−nτCςn,ςn+1h≤0.
On taking limit asn⟶∞, we obtain
(36)limn→∞nCςn,ςn+1h=0.
Hence,limn→∞n1/hCςn,ςn+1=0, and there exists n1∈ℕ such that n1/hCςn,ςn+1≤1 for all n≥n1. So, we have
(37)Cςn,ςn+1≤1n1/hfor all n≥n1. Consider the triangle inequality for q≥1 to have
(38)Cςn,ςn+q≤pςn,ςn+1Cςn,ςn+1+pςn+1,ςn+qCςn+1,ςn+q≤pςn,ςn+1Cςn,ςn+1+pςn+1,ςn+qpςn+1,ςn+2Cςn+1,ςn+2+pςn+1,ςn+qpςn+2,ςn+qCςn+2,ςn+q≤pςn,ςn+1Cςn,ςn+1+pςn+1,ςn+qpςn+1,ςn+2Cςn+1,ςn+2+pςn+1,ςn+qpςn+2,ςn+qpςn+2,ςn+3Cςn+2,ςn+3+pςn+1,ςn+qpςn+2,ςn+qpςn+3,ςn+qCςn+3,ςn+q≤⋯≤pςn,ςn+1Cςn,ςn+1+∑i=n+1n+q−2∏j=n+1ipςj,ςn+qpςi,ςi+1Cςi,ςi+1+∏i=n+1n+q−1pςi,ςn+qCςn+q−1,ςn+q.
It implies that(39)Cςn,ςn+q≤pςn,ςn+1Cςn,ςn+1+∑i=n+1n+q−2∏j=n+1ipςj,ςn+qpςi,ςi+1Cςi,ςi+1+∑i=n+1n+q−1pςi,ςn+qpςn+q−1,ςn+qCςn+q−1,ςn+q=pςn,ςn+1Cςn,ςn+1+∑i=n+1n+q−1∑j=n+1ipςj,ςn+qpςi,ςi+1Cςi,ςi+1≤pςn,ςn+1Cςn,ςn+1+∑i=n+1n+q−1∏j=0ipςj,ςn+qpςi,ςi+1Cςi,ςi+1≤pςn,ςn+1Cςn,ςn+1+∑i=n+1n+q−1∏j=0ipςj,ςn+qpςi,ςi+11i1/k.
Now, consider(40)∑i=n+1n+q−1∏j=0ipςj,ςn+qpςi,ςi+11n1/k=∑i=n+1n+q−11i1/k∏j=0ipςj,ςn+qpςi,ςi+1≤∑i=n+1∞1i1/k∏j=0ipςj,ςn+qpςi,ςi+1=∑i=n+1∞UiVi,where
(41)Ui=1i1/k,Vi=pςi,ςi+1∏j=0ipςj,ςn+q.
Since1/k>0, ∑i=n+1∞1/i1/k converges and also Vii is increasing and bounded above, thus, limi→∞Vi, which is nonzero, exists. Hence, ∑i=n+1∞UiVin converges. Let us consider the partial sum
(42)Sq=∑i=0q∏j=0ipςj,ςn+qpςi,ςi+11i1/k.
Now, from (39), we have
(43)Cςn,ςn+q≤pςn,ςn+1Cςn,ςn+1+Sn+q−1−Sn.
By the ratio test and using the condition (26), we guarantee the existence of limn→∞Sn. Hence, the real sequence Sn is Cauchy. Now, taking the limit n⟶+∞ in (43), we get
(44)limn→∞Cςn,ςn+q=0;that is, ςn is a Cauchy sequence in Y,C, which is complete, so ςn converges to some u∈Y. We claim that Ωu=u. Since ςn⟶u as n⟶∞ and Ω is continuous, we have Ωςn⟶Ωu as n⟶∞. Thus, we have
(45)Cu,Ωu=limn→∞Cςn+1,Ωu=limn→∞CΩςn,Ωu=0,and hence, u=Ωu. Thus, u is a fixed point of Ω. Its uniqueness is obvious.Corollary 15.
LetY,C be a complete controlled metric space, and let Ω:Y⟶Y be continuous so that
(46)τ+FCΩς,Ωω≤FCς,ωfor all ς,ω∈Y. For ς0∈Y, take ςn=Ωnς0. Suppose that
(47)supm≥1limi→∞pςi+1,ςi+2pςi+1,ςmpςi,ςi+1<1.
Assume thatlimn→∞pςn,ς and limn→∞pς,ςn exist and are finite, for every ς∈Y, then Ω possesses a unique fixed point.Proof.
Takingα:Y×Y⟶0,∞ by ας,ω=1, for all ς,ω∈Y in Theorem 14.Corollary 16.
LetY,C be a complete extended b-metric space and Ω:Y⟶Y be a continuous, α-admissible and (α,F)-contraction so that there is ς0∈Y in order that ας0,Ως0≥1. Suppose that
(48)supm≥1limi→∞pςi+1,ςi+2pςi+1,ςmpςi,ςi+1<1.
If in addition,limn→∞pςn,ς and limn→∞pς,ςn exist and are finite, for every ς∈Y, then Ω has a unique fixed point.Corollary 17.
LetY,C be a complete b-metric space and Ω:Y⟶Y be a continuous, α-admissible and (α,F)-contraction so that there is ς0∈Y with ας0,Ως0≥1. Then, Ω has a unique fixed point.Proof.
Takingp:Y×Y⟶1,∞ by pς,ω=pω,τ, for all ς,ω,τ∈Y in Theorem 14.Corollary 18.
LetY,C be a complete metric space and Ω:Y⟶Y be a continuous, α-admissible and (α,F)-contraction so that there is ς0∈Y in order that ας0,Ως0≥1. Then, Ω has a unique fixed point.Proof.
Takingp:Y×Y⟶1,∞ by pς,ω=1, for all ς,ω∈Y in Theorem 14.Corollary 19 (see. [11]).
LetY,C be a complete metric space and Ω:Y⟶Y be an F-contraction. Then, Ω has a unique fixed point.Proof.
Takingα:Y×Y⟶0,∞ by ας,ω=1, for all ς,ω∈Y in Corollary 18.



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