# The Existence of Fixed Points for a Different Type of Contractions on Partialb-Metric Spaces

Authors: Muhammad Nazam; Hüseyin Işik; Khalil Javed; Muhammad Naeem; Muhammad Arshad
Journal: Journal of Mathematics
 (2021)

---

## Abstract

The aim of this study is to present fixed point results in the setting of partialb-metric spaces. A different type of contractions is used to prove fixed point results in the given space, which are real generalization of many well-known results. The readers are also provided with some very interesting examples to illustrate the feasibility of the proposed work.


---

## Body

## 1. Introduction and Preliminaries

The notion of metric space was initiated in 1906 by French Mathematician Frechet [1]. In metric fixed point theory, the Banach contraction principle [2] is one of the most fundamental tools to investigate the existence and uniqueness of solutions for contraction maps in a complete metric space. Since the inception of this principle, many authors studied fixed point theory vividly and enriched this field with different ideas. This classical result was generalized in different spaces, and different structures were attained using this topic and one may recall the existing notions, partial b-metric spaces [3], ℜ-partial b-metric spaces [4], fuzzy cone b-metric spaces [5], Gb-metric spaces [6], orthogonal partial b-metric spaces [7], orthogonal m-metric spaces [8], and several others. More details can be found in [9–13].One important and extensively used generalization of metric space is the notion ofb-metric spaces, which was introduced by Czerwik [14] and is defined as follows.Definition 1 (see [14]).
Letℑ be a nonempty set and s≥1 be a given real number. A function b:ℑ×ℑ⟶0,∞ is referred as a b-metric if the following conditions hold for all σ,υ,ς∈ℑ:(bi) bσ,ς=0⇔σ=ς(bii) bσ,ς=bς,σ(biii) bσ,ς≤sbσ,υ+bυ,ς
The tripletℑ,b,s is called a b-metric space.Example 1 (see [15]).
Letℑ=1,2,3,…∪∞ and define b:ℑ×ℑ⟶0,∞ by(1)bσ,ς=1σ−1ς,ifσ,ςare even orσς=∞,0,ifσ=ς,5,ifρ,υare odd orσ≠ς,2,otherwise.
Then,ℑ,b,s is a b-metric space with s=3.
Matthews [16], while working on networking, observed that a self-distance may not be zero, i.e., dσ,σ≠0. A loop is a good example of this case. He not only generalized the classical Banach fixed point theorem but also established some convergence criterion in this setting to ensure existence of a fixed point. His further investigations led him to the introduction of partial metric space, which is defined as follows.Definition 2 (see [16]).
Letℑ be a nonempty set. A mapping ℷ:ℑ×ℑ⟶0,∞ is referred as a partial metric if, for all σ,ς,υ∈ℑ,ℷiσ=ς⇔ℷσ,σ=ℷσ,ς=ℷς,ςℷiiℷσ,σ≤ℷσ,ςℷiiiℷσ,ς=ℷς,σℷivℷσ,ς≤ℷσ,υ+ℷυ,ς−ℷυ,υ
Then, pairℑ,ℷ is referred as a partial metric space.Example 2 (see [16]).
Letℑ=0,∞ and define ℷ:ℑ×ℑ⟶0,∞ by(2)ℷσ,ς=maxσ,ςfor allσ,ς∈ℑ.
Then, the pairℑ,ℷ is a partial metric space.Example 3 (see [16]).
Letℑ=σ,ς:σ,ς∈0,∞ and define ℷ:ℑ×ℑ⟶0,∞ by(3)ℷσ,ς,υ,ρ=maxς,ρ−minσ,υ.
The pairℑ,ℷ is a partial metric space.Definition 3 (see [16]).
Letℑ,ℷ be a partial metric space and σκ be a sequence in ℑ. Then,(i)
σκ converges in ℑ⇔ℷς,ς=limκ⟶∞ℷς,σκ.(ii)
σκ is referred as a Cauchy sequence if limκ,j⟶∞ℷσκ,σj exists and is finite.(iii)
A partial metric space is complete if each Cauchy sequence converges inℑ satisfying(4)ℷς,ς=limκ,j⟶∞ℷσκ,σj.
The concept of partial metric space was further extended to partialb-metric space by Shukla [3] in 2014 by combining the partial metric space and b-metric space.Definition 4 (see [3]).
Letℑ be a nonempty set and s≥1 be a given real number. A mapping ℷb:ℑ×ℑ⟶0,∞ is referred as a partial b-metric if, for all σ,ς,υ∈ℑ,ℷbiσ=υ⇔ℷbυ,υ=ℷbυ,σ=ℷbσ,σℷbiiℷbυ,υ≤ℷbσ,υℷbiiiℷbς,σ=ℷbσ,ςℷbivℷbσ,ς≤sℷbσ,υ+ℷbυ,ς−ℷbυ,υ
The tripletℑ,ℷb,s is referred as a partial b-metric space with s∈1,∞.Definition 5 (see [3]).
Letℑ,ℷb,s be a partial b-metric space and σκ be a sequence in ℑ. Then,(i)
σκ converges in ℑ⇔ℷbς,ς=limκ⟶∞ℷbς,σκ.(ii)
σn is referred as a Cauchy sequence if limκ,j⟶∞ℷbσκ,σj exists and is finite.(iii)
A partialb-metric space is complete if each Cauchy sequence converges in ℑ such that(5)ℷbς,ς=limκ,j⟶∞ℷbσκ,σj.Remark 1.
(i)
In the partialb-metric space, ℷbσ,ς=0⇒σ=ς. The converse implication does not hold in general.(ii)
A partialb-metric is a generalization of a partial metric.Example 4 (see [3]).
Letℑ=0,∞ (with ξ>1) and define ℷb:ℑ×ℑ⟶0,∞ by(6)ℷbϰ,ς=maxϰ,ςξ+ϰ−ςξfor allϰ,ς∈ℑ.
ℷb is referred as a partial b-metric with s=2ξ. For ς>0, we have ℷbς,ς≠0, so ℷb is not a b-metric. On the contrary, for ϰ,σ,ς∈ℑ with ϰ>σ>ς, we have(7)ϰ−σξ+σ−ςξ<ϰ−ςξ.
Observe that(8)ℷbϰ,ς=ϰξ+σ−ςξ,ℷbϰ,σ+ℷbσ,ς−ℷbσ,σ=ϰξ+ϰ−σξ+ϰ−ςξ.
One can then write(9)ℷbϰ,σ+ℷbσ,ς−ℷbσ,σ<ℷbϰ,ς.
Thus,ℷb is not a partial metric.Lemma 1 (see [3]).
(1)
In a partialb-metric space ℑ,ℷb,s, σκ is a ℷb-Cauchy sequence ⇔ it is a b-Cauchy sequence in the b-metric space ℑ,b.(2)
A partialb-metric space ℑ,ℷb is ℷb-complete ⇔ the b-metric space ℑ,b is b-complete. Moreover, limκ⟶∞bσ,σκ=0⇔(10)limκ⟶∞ℷbσ,σκ=limκ,τ⟶∞ℷbστ,σκ=limκ⟶∞ℷbσ,σ.


## 2. Preliminary Results on Partialb-Metric Spaces

Following [17], we state the following.Definition 6.
A functionF:0,∞⟶−∞,∞ belongs to Δs ifWF1F is strictly increasingWF3 There exists ξ∈0,1 such that limσ⟶∞σξFσ=0
In the above definition, we omitted Wardowski’s second condition, that is,WF2. For every sequence σκ⊂0,∞,(11)limκ⟶∞σκ=0⇔limκ⟶∞Fσκ=−∞.Definition 7.
Letℑ,ℷb,s be a partial b-metric space. A mapping ψ:ℑ⟶ℑ is called an Fℷb-contraction if there are F∈Δbs and τ>0 so that(12)σ,ς∈ℑ,ℷbψσ,ψς>0⇒τ+Fsℷbψσ,ψς≤Fℷbσ,ς.Remark 2.
In Example5, there is an Fℷb-contraction mapping ψ:ℑ⟶ℑ, but it does not satisfy(13)ℷbψσ,ψς≤pℷbσ,ς,for all p∈0,1.Example 5.
Letℑ=σκ=2κ/2κ,κ∈ℕ and define ℷb:ℑ×ℑ⟶0,∞ by ℷbσ,ς=σ∨ς2. Then, the triplet ℑ,ℷb,s=2 is a partial b-metric space (with s=2). Consider ψ:ℑ⟶ℑ as(14)ψσ=σk−1=2κ−1/2κ−1,ifσ=σκwithk=1,2,…,0,ifσ=σ0=0.
We claim thatψ is an Fℷb-contraction in the partial b-metric space ℑ (with τ=1 and Fω=ω, which is in Δbs: WF1 and WF3 hold). For this, by this choice of F, we shall show that, for every σ,ς∈ℑ so that ψσ≠ψς, we have(15)2ℷbψσ,ψς−ℷbσ,ς≤−τ.
For this, letσ=σκ+k and σ=σκ. Then,(16)2ℷbψσκ+k,ψσκ−ℷbσκ+k,σκ=2ℷb2κ+k−1/2κ+k−1,2κ−1/2κ−1−ℷb2κ+k/2κ+k,2κ/2κ=2κ+k/2κ+k−1∨2κ/2κ−12−2κ+k/2κ+k∨2κ/2κ2=2κ2κ/2κ+k−1∨κ−12−2κ/2κ+k∨κ2=2κ2κ/2κ+k−12−2κ/2κ+k2=2κ2kκ+k−12−2kκ+k2=−2κ+k≤−1.
On the contrary,ψ is not a Banach contraction. Indeed,(17)limκ⟶∞2ℷbψσκ,ψσ0ℷbσκ,σ0=limκ⟶∞2σκ−1∨σ02σκ∨σ02=limκ⟶∞22κ−1/2κ−1∨022κ/2κ∨02=limκ⟶∞2κ/2κ−1∨022κ/2κ∨02=limκ⟶∞2κ/2κ−122κ/2κ2=limn⟶∞κ−12κ2=limκ⟶∞κ2+1−2κκ2=limκ⟶∞1+1κ2−2κ=1.
It shows thatℷbψσ,ψς≤pℷbσ,ς (for all p∈0,1) does not hold.
In [11], the authors introduced the following assumption:CH4 for all κ∈ℕ, τ>0τ+Fsσκ≤Fσκ−1⇒τ+Fsκσκ≤Fsκ−1σκ−1
Go back to Example5. We proceed as follows. Let τ+Fsσκ≤Fσκ−1, that is, τ+sσκ≤σκ−1. One writes(18)Fsκσκ+τ=sκσκ+τ=sκ−1sσκ+τ≤sκ−1σκ−1−τ+τ=τ+sκ−1σκ−1−τsκ−1=τ1−sκ−1+sκ−1σκ−1≤sκ−1σκ−1=Fsκ−1σκ−1.
Thus,CF4 holds.Definition 8.
Letan be a sequence in 0,∞ and bn be a sequence in 0,∞. We call bn∈Oan if there exists C>0 satisfying bn≤Can, for all n∈ℕ.Lemma 2.
Letσd be a sequence in the partial b-metric space ℑ,ℷb,s. Assume that(19)ℷbσd,σd+1∈∪Od−δ:δ>1+log2s.
Then,σd is a Cauchy sequence.Proof.
By (19), there exist δ>1+log2s and D>0 satisfying(20)ℷbσd,σd+1≤Dd−δ∀d∈ℕ.
Note that2s<2δ. Choose μ∈ℕ satisfying 2s1+1/μ<2δ; then, we have(21)2μs1+μ<2μδ⇒2μ1−δs1+μ<1.
Define a mappingψ:ℕ⟶ℕ by(22)ψd=2+2μd−12μ−1.
Fora,c,d∈ℕ satisfying ψa≤c<d≤ψa+1, we have(23)ℷbσc,σd≤sψd−c∑j=cd−1ℷbσj,σj+1≤sψψa+1−ψa∑j=ψaψa+1−1ℷbσj,σj+1=sψ2μa∑j=ψaψa+1−1ℷbσj,σj+1=sμa∑j=ψaψa+1−1ℷbσj,σj+1≤Dsμa∑j=ψaψa+1−1j−δ≤Dsμa∫ψa−1ψa+1−1t−δdt≤Dsμa∫ψa−2+1/2μ−1ψa+1−2+1/2μ−1t−δdt=Dsμa1−δt1−δψa−2+1/2μ−1ψa+1−2+1/2μ−1=Dsμa1−2μ1−δδ−12μ−1δ−12μa1−δ=Ms21−δμa,where M=D1−2μ1−δ2μ−1δ−1/δ−1>0. Since(24)s21−δμ=sμ+12μ1−δs<1s≤1,we have(25)limμ⟶∞Ms21−δμa=0.
Fora,l,c,d∈ℕ such that a<l,ψa≤c<ψa+1 and ψl<d≤ψl+1, then, from (21) and (23),(26)ℷbσc,σd≤sℷbσc,σψa+1+sℷbσψa+1,σd−ℷbσψa+1,σψa+1≤sℷbσc,σψa+1+ssℷbσψa+1,σψa+2+sℷbσψa+2,σd−sℷbσψa+2,σψa+2−ℷbσψa+1,σψa+1=sℷbσc,σψa+1+s2ℷbσψa+1,σψa+2+s2ℷbσψa+2,σd−s2ℷbσψa+2,σψa+2−ℷbσψa+1,σψa+1≤sMs21−δμa+s2Ms21−δμa+1+s3Ms21−δμa+2+s3ℷbσψa+3,σd−s2ℷbσψa+3,σψa+3−s1Ms21−δμa+1−s0Ms21−δμa≤∑κ=02sκ+1Ms21−δμa+κ+s3ℷbσψa+3,σd−∑κ=01sκMs21−δμa+κ−s2ℷbσψa+3,σψa+3≤....≤∑κ=0l−a−1sκ+1Ms21−δμa+κ+sl−aℷbσψl,σd−∑vl−k−1sκMs21−δμa+κ−sl−kℷbσψa+κ+1,σψa+κ+1≤∑κ=0l−a−1sκ+1Ms21−δμa+κ+sl−aMs21−δμl−∑κ=0l−a−1sκMs21−δμa+κ−sl−aMs21−δμa+κ≤∑κ=0l−asκ+1Ms21−δμa+κ−∑κ=0l−asκMs21−δμa+κ≤∑κ=0∞sκ+1Ms21−δμa+κ−∑κ=0∞sκMs21−δμa+κ=Ms21−δμa+κs1−sμ+12μ1−δ−11−sμ+12μ1−δ.
By (25), σd is a Cauchy sequence.Lemma 3 (see [9]).
Letσn be a decreasing sequence in 0,∞. Assume that there is a mapping F: 0,∞⟶−∞,∞, τ∈0,∞, and k∈0,1 satisfying WF3 so that(27)nτ+Fsnσn≤Fσ0.
Then,σn∈On1/k.
Generally, a partialb-metric is discontinuous. Following Lemma 3, this statement is justified.Lemma 4.
Consider the convergent sequenceσϵ in the partial b-metric space ℑ,ℷb,s. Suppose that limϵ⟶∞σϵ=σ and σ,ς∈ℑ; then,(28)1sℷbσ,ς≤liminfε⟶∞ℷbσε,ς≤limsupε⟶∞ℷbσε,ς≤sℷbσ,ς.Proof.
Sinceℑ,ℷb,s is a partial b-metric space, by ℷbiv, we obtain(29)ℷbσ,ς≤sℷbσ,σε+ℷbσε,ς−ℷbσε,σε≤sℷbσ,σε+ℷbσε,ς⇒1sℷbσ,ς−ℷbσ,σε≤ℷbσε,ς.
Takingliminf, we obtain(30)1sℷbσ,ς≤liminfε⟶∞ℷbσε,ς.
Again, byℷbiv, we obtain(31)ℷbσε,ς≤sℷbσε,σ+ℷbσ,ς−ℷbσ,σ≤sℷbσε,σ+ℷbσ,ς⇒ℷbσε,ς≤sℷbσε,σ+sℷbσ,ς.
Takinglimsup,(32)limsupε⟶∞ℷbσε,ς≤sℷbσ,ς.
Also,(33)liminfε⟶∞ℷbσε,ς≤limsupε⟶∞ℷbσε,ς.
Combining (30), (32), and (33), we have(34)1sℷbσ,ς≤liminfε⟶∞ℷbσε,ς≤limsupε⟶∞ℷbσε,ς≤sℷbσ,ς.


## 3. Onγ,F-Weak Contractions

We shall investigate the existence of a fixed point forγ,F-weak contraction mappings in partial b-metric spaces. We shall also provide some examples in support of main results.Definition 9.
Letψ:ℑ⟶ℑ and αs:ℑ×ℑ⟶0,∞ be two mappings and ℑ,ℷb,s≥1 be a partial b-metric space. Such ψ is αs-admissible if(35)αsσ,ς≥s2⇒αsψσ,ψς≥s2for allσ,ς∈ℑ.Example 6.
Letℑ=0,∞. Define ψ:ℑ⟶ℑ and αs:ℑ×ℑ⟶0,∞ by(36)ψσ=lnσ,for allσ∈ℑ,αsσ,ς=2,ifσ≥ς,0,ifσ<ς.
Then,ψ is αs-admissible.Definition 10.
Letℑ,ℷb,s be a partial b-metric space and ψ:ℑ⟶ℑ and αs:ℑ×ℑ⟶0,∞ be two mappings. Then, ψ is triangular αs-admissible if(1)
αsσ,ς≥s2 implies αsψσ,ψς≥s2(2)
αsσ,ς≥s2, αsς,υ≥s2 implies αsσ,υ≥s2, for all σ,ς,υ∈ℑExample 7.
Letℑ=0,∞, ψσ=σ3, and αsσ,ς=eσ−ς. Here, ψ is triangular αs-admissible. Indeed, if αsσ,ς=eσ−ς≥s2, then σ≥ς, which implies ψσ≥ψς. That is, αsψσ,ψς=eψσ−ψς≥s2. Also, if αsσ,ς≥s2 and αsς,υ≥s2, then σ−ς≥0,ς−υ≥0. That is, σ−υ≥0, so αsσ,υ=eσ−υ≥s2.Remark 3.
For the givenαs, there is γ:ℑ×ℑ⟶0,∞ defined by γσ,ς=αsσ,ς/s2 having the following conditions:(1)
ψ is αs-admissible iff ψ is γ-admissible, i.e.,(37)γσ,ς≥1⇒γψσ,ψς≥1,for allσ,ς∈ℑ.(2)
The partialb-metric space is αs-complete iff it is γ-complete.Definition 11.
Letℑ,ℷb,s be a partial b-metric space and ψ,ϕ:ℑ⟶ℑ and γ:ℑ×ℑ⟶0,∞ be three mappings. Then, the pair ψ,ϕ is weakly γ-admissible if(38)γσ,ς≥1⇒γψσ,ϕψσ≥1,γϕς,ψϕς≥1,for allσ,ς∈ℑ.Definition 12.
Letℑ,ℷb,s be a partial b-metric space and ψ,ϕ:ℑ⟶ℑ and γ:ℑ×ℑ⟶0,∞ be three mappings. The pair ψ,ϕ is triangular weakly γ-admissible if(1)
γσ,ς≥1⇒γψσ,ϕψσ≥1 and γϕς,ψϕς≥1, for all σ,ς∈ℑ(2)
γσ,ς≥1,γς,υ≥1 implies γσ,υ≥1, for all σ,ς,υ∈ℑExample 8.
Letℑ=0,∞ and define ψ,ϕ:ℑ⟶ℑ by(39)ψσ=σ1/2ifσ∈0,1,1,ifσ∈1,∞,ϕσ=σ,ifσ∈0,1,1,ifσ∈1,∞.
Defineγ:ℑ×ℑ⟶0,∞ by(40)γσ,ς=eσ−ς,ifσ,ς∈0,1,0,ifσ,ς∈1,∞.
Ifγσ,ς≥1 and γς,υ≥1, then(41)σ−ς≥0,ς−υ≥0⇒σ−υ≥0⇒γσ,υ≥1.
Next, ifγσ,ς≥1, then(42)γψσ,ϕψσ=γσ1/2,ϕσ1/2=γσ1/2,σ1/2=eσ1/2−σ1/2≥1,γϕς,ψϕς=γς,ψς=γς,ς1/2=eς−ς1/2≥1.Definition 13.
Consider the partialb-metric space ℑ,ℷb,s and two mappings γ:ℑ×ℑ⟶0,∞ and ψ:ℑ⟶ℑ. ψ is γ-continuous if for given υ∈ℑ and a sequence σκ so that(43)limκ⟶∞ℷbσκ,υ=0,γσκ,σκ+1≥1⇒limκ⟶∞ℷbψσκ,ψυ=0,for allκ∈ℕ.Example 9.
Letℑ=0,∞ and define a partial b-metric ℷb:ℑ×ℑ⟶0,∞ by(44)ℷbσ,ς=σ∨ς2,for allσ,ς∈ℑ.
Defineγ:ℑ×ℑ⟶0,∞ and ψ:ℑ⟶ℑ by(45)γσ,ς=σ2+ς2+1,ifσ,μ∈0,1,0,otherwise,ψω=sinπω,ifω∈0,1,cosπω+2,ifω∈1,∞.
ψ is not continuous because(46)limω⟶1−ψω≠limω⟶1+ψω.
Next, we prove thatψ is γ-continuous. For given υ=0∈0,∞ and σκ=1/κ, we obtain(47)limκ⟶∞ℷbσκ,υ=limκ⟶∞ℷb1κ,0=limκ⟶∞1κ∨0=0,γ1κ,1κ+1=1+1κ2+1κ+12≥1.
Now,(48)limκ⟶∞ℷbψσκ,ψυ=limκ⟶∞ℷbψ1κ,ψ0=limκ⟶∞sinπκ∨sin0=limκ⟶∞sinπκ=0.
Hence,ψ is γ-continuous.Definition 14.
Letℑ,ℷb,s be a partial b-metric space and γ be defined above. The partial b-metric space is γ-complete iff every Cauchy sequence σκ in ℑ satisfying γσκ,σκ+1≥1 is convergent to σ∈ℑ, for all κ∈ℕ.Remark 4.
Ifℑ,ℷb,s is a complete partial b-metric space, then ℑ,ℷb,s is a γ-complete partial b-metric space, but not conversely.Example 10.
Letℑ=0,∞ and ℷb:ℑ×ℑ⟶0,∞ be the partial b-metric defined by(49)ℷbσ,ς=σ∨ς2,for allσ,ς∈ℑ.
Defineγ:ℑ×ℑ⟶0,∞ by(50)γσ,ς=eℷbσ,ς,ifσ,ς∈2,5,0,otherwise.
We can show thatℑ,ℷb,s is not a complete partial b-metric space, but ℑ,ℷb,s is a γ-complete partial b-metric space. Indeed, σκ is a Cauchy sequence in ℑ satisfying γσκ,σκ+1≥1 for all κ∈ℕ; then, σκ∈2,5. Since 2,5 is a closed subset of −∞,∞, we can check that 2,5,ℷb,2 is a complete partial b-metric space, and then, there is υ∈2,5 such that σκ⟶υ as κ⟶∞.Definition 15.
Letℑ,ℷb,s be a partial b-metric space and γ:ℑ×ℑ⟶0,∞ be a mapping. ℑ,ℷb,s is called γ-regular if, for every sequence σκ⊂ℑ satisfying γσκ,σκ+1≥1 and σκ⟶υ as κ⟶∞, we have γσκ,υ≥1, for all κ∈ℕ.Definition 16.
Consider the partialb-metric space ℑ,ℷb,s. Then, the pair of self-mappings ψ,ϕ forms a γ,F-weak contraction if there exist F∈Δbs and τ>0 such that(51)τ+Fsγσ,ςℷbψσ,ϕς≤FM1σ,ς,for allσ,ς∈ℑ,with γσ,ς≥1, whenever(52)minγσ,ςℷbψσ,ϕς,M1σ,ς>0,where(53)M1σ,ς=maxℷbσ,ς,ℷbσ,ψσ,ℷbς,ϕς,ℷbσ,ϕς+ℷbς,ψσ2s.Theorem 1.
Let the pair of self-mappingsψ,ϕ be a γ,F-weak contraction defined on a γ-complete partial b-metric space ℑ,ℷb,s. Let k∈0,1/1+log2s. Suppose that(1)
The pairψ,ϕ is weakly γ-admissible.(2)
There existsσ0∈ℑ such that γσ0,ψσ0≥1.(3)
(a) Eitherℑ is γ-regular and F is continuous.(b) One ofψ,ϕ is γ-continuous.
Then, we have a sequenceσκ in ℑ such that σκ⟶υ∈ℑ. If γυ,υ≥1, then υ is a common fixed point of the pair ψ,ϕ. Moreover, if ω is another common fixed point of the pair ψ,ϕ satisfying γυ,ω≥1, then υ=ω.Proof.
Forσ1,σ2∈ℑ such that σ1≠σ2, then M1σ1,σ2>0. Let σ0∈ℑ be as in (2). Consider the iterative sequence σκ in ℑ such that σ1=ψσ0 and σ2=ϕσ1, and generally, σ2κ+1=ψσ2κ,σ2κ=ϕσ2κ−1, for all κ∈ℕ∪0.
Using the weaklyγ-admissibility, we have(54)γψσ0,ϕψσ0=γσ1,σ2≥1,γψσ1,ϕψσ1=γσ2,σ3≥1,γψσ2,ϕψσ2=γσ3,σ4≥1,γψσ3,ϕψσ3=γσ4,σ5≥1.
Continuing in this way,(55)γψσ2κ,ϕψσ2κ=γσ2κ+1,σ2κ+2≥1,γψσ2κ−1,ϕψσ2κ−1=γσ2κ,σ2κ+1≥1.
Hence,γσκ,σκ+1≥1, for all κ∈ℕ∪0. If ℷbψσ2κ,ϕσ2κ+1=0, then σ2κ is a common fixed point of ψ,ϕ. Let ℷbψσ2κ,ϕσ2κ+1>0. From the contractive condition (51),(56)Fsℷbσ2κ,σ2κ+1≤Fsγσ2κ,σ2κ+1ℷbψσ2κ,ϕσ2κ+1≤FM1σ2κ,σ2κ+1−τ∀κ∈ℕ∪0,where(57)M1σ2κ,σ2κ+1=maxℷbσ2κ,σ2κ+1,ℷbσ2κ+1,ϕσ2κ,ℷbσ2κ+1,ϕσ2κ+1,ℷbσ2κ,ϕσ2κ+1+ℷbσ2κ+1,ψσ2κ2s=maxℷbσ2κ,σ2κ+1,ℷbσ2κ,σ2κ+1,ℷbσ2κ+1,σ2κ+2,ℷbσ2κ,σ2κ+2+ℷbσ2κ+1,σ2κ+12s≤maxℷbσ2κ,σ2κ+1,ℷbσ2κ+1,σ2κ+2.
IfM1σ2κ,σ2κ+1=ℷbσ2κ+1,σ2κ+2, then(58)Fsℷbσ2κ+1,σ2κ+2≤Fℷbσ2κ+1,σ2κ+2−τ,which contradicts WF1. Therefore,(59)Fsℷbσ2κ+1,σ2κ+2≤Fℷbσ2κ,σ2κ+1−τ,κ∈ℕ∪0.
Similarly, we have(60)Fsℷbσ2κ+2,σ2κ+3≤Fℷbσ2κ+1,σ2κ+2−τ,κ∈ℕ∪0.
From (59) and (60), we have(61)Fsℷbσκ,σκ+1≤Fℷbσκ−1,σκ−τ,κ∈ℕ.
Letbκ=ℷbσκ,σκ+1, for each κ∈ℕ∪0; then, from (61) and CH4,(62)τ+Fsκbκ≤Fsκ−1bκ−1,κ∈ℕ.
Continuing in this way, we obtain(63)Fsκbκ≤Fb0−κτ,κ∈ℕ.
So,bκ∈Oκ−1/k by Lemma 3. Since 1/k∈1+log2s,∞, by Lemma 2, σκ is a Cauchy sequence. We know that ℑ is a γ-complete partial b-metric space; then, there is υ∈ℑ satisfying σ2κ+1⟶υ and σ2κ+2⟶υ as κ⟶∞. Using γ-continuity of ψ, one writes(64)υ=limκ⟶∞σκ=limκ⟶∞σ2κ+1=limκ⟶∞σ2κ+2=limκ⟶∞ψσ2κ+1=ψlimκ⟶∞σ2κ+1=ψυ.
Ifℷbυ,ϕυ>0 as γυ,υ≥1, then by the contractive condition (51), we have(65)τ+Fsℷbυ,ϕυ≤τ+Fsγυ,υℷbψυ,ϕυ≤FM1υ,υ=Fℷbυ,ϕυ.
It is a contradiction. Hence,(66)ℷbυ,ϕυ=0⇒υ=ϕυ.
Thus, we haveψυ=ϕυ=υ. Hence, ψ,ϕ has a common fixed point υ.
Ifℑ is γ-regular and F is continuous, then there are two different cases. Firstly, if there is a subsequence σnκκ∈ℕ⊂σnn∈ℕ satisfying(67)σnκ=ψυ,ifκ∈2n,ϕυ,ifκ∈2n+1.
Then,(68)υ=limκ⟶∞σnκ=limκ⟶∞ψυ=ψυ,υ=limκ⟶∞σnκ=limκ⟶∞ϕυ=ϕυ.
It is completed. Second, if there is no such subsequence ofσκκ∈N, then there is μ0∈ℕ such that, for each κ≥μ0, we obtain(69)ℷbψσ2κ,ϕυ>0,ℷbϕσ2κ+1,ψυ>0.
Sinceℑ is γ-regular, one writes(70)γσ2κ+1,υ≥1,γσ2κ,υ≥1.
By the contractive condition (51), we have(71)τ+Fsγσ2κ,υℷbψσ2κ,ϕυ≤Fmaxℷbσ2κ,υ,ℷbσ2κ,ψσ2κ,ℷbυ,ϕυ,ℷbσ2κ,ϕυ+ℷbυ,ψσ2κ2s.
Now, we prove thatℷbυ,ϕυ=0. Suppose, on the contrary, that ℷbυ,ϕυ=q>0. Put λκ=ℷbσκ,υ=0∀κ∈ℕ. Since limκ⟶∞σκ=υ, there exists μ1∈ℕ satisfying for each κ≥μ1, both λκ<q/2 and bκ<q/2 hold. Consequently, by (71), we obtain(72)τ+Fsγσ2κ,υℷbψσ2κ,ϕυ≤Fmaxλ2κ,b2κ,q,ℷbσ2κ,ϕv+λ2κ+12s≤Fmaxλ2κ,b2κ,q,sλ2κ+sq+λ2κ+12s≤Fmaxq2,q2,q,sq/2+sq+q/22s=Fq.
Then, for everyκ≥maxμ0,μ1, we obtain(73)τ+Fsγσ2κ,υℷbψσ2κ,ϕυ≤Fℷbυ,ϕυ.
SinceF is increasing and continuous, by Lemma 4 and (73), we have(74)τ+Fℷbυ,ϕυ≤τ+Fsγσ2κ,υliminfκ⟶∞ℷbψσ2κ,ϕυ≤τ+liminfκ⟶∞Fsγσ2κ,υℷbψσ2κ,ϕυ≤Fℷbυ,ϕυ.
The above result shows thatτ≤0, which is a contradiction. Then, ℷbυ,ϕυ=0, and hence, υ=ϕυ. Similarly, we can show that υ=ψυ, and consequently, υ is a common fixed point of ψ and ϕ.
Now, we prove that self-mappingsψ,ϕ have a unique common fixed point. Let υ and ω be two common fixed points of ψ and ϕ. Then, ψυ=υ≠ω=ϕω. It follows that(75)ℷbψυ,ϕω=ℷbυ,ω>0.
Sinceγυ,ω≥1, by contractive condition, one writes(76)τ+Fsγυ,ωℷbψυ,ϕω≤Fmaxℷbυ,ω,ℷbψυ,υ,ℷbϕω,ω,ℷbϕω,υ+ℷbψυ,ω2s=Fℷbυ,ω≤Fsγυ,ωℷbυ,ω.
It shows thatτ≤0, which is a contradiction. Hence, ψ and ϕ have a unique common fixed point.Example 11.
Letℑ=0,∞ and define ℷb,γ:ℑ×ℑ⟶0,∞ by(77)ℷbσ,ς=σ∨ς2,γσ,ς=K≥1,ifσ≥ς,0,ifσ<ς.
So,ℑ is a γ-complete partial b-metric space with s=2.
Takeψ,ϕ:ℑ⟶ℑ as(78)ψσ=ln1+σ6,ϕσ=ln1+σ7.
Definitely,ψ and ϕ are γ-continuous. To prove that pair ψ,ϕ is weakly γ-admissible, let σ,ς∈ℑ be such that ς=ψσ, so ς=ln1+σ/6. We have(79)ψσ=ς=lnσ6+1≥ln1+ς=ln1+σ/67=ln1+ς7=ϕς=ϕψσ.
Thus,γψσ,ϕψσ≥1. Now, let υ,ς∈ℑ be such that υ=ϕς, so υ=ln1+ς/7. Since(80)ϕς=ln1+ς7≥ln1+ς=ln1+ς/76=ln1+υ6=ψυ=ψϕς,we get that γϕς,ψϕς≥1. Hence, the pair ψ,ϕ is weakly γ-admissible.
Next, for allσ,ς∈ℑ with σ≥ς, choose ξ such that ξ/2K>1+log2s; then,(81)2γσ,ςℷbψσ,ϕς=2Kψσ∨ϕς2=2Klnσ6+1∨ln1+ς72=2Kln1+ς62≤2Kln1+σ62≤2Kξℷbσ,ς≤2KξM1σ,ς.
The above result may be written as(82)lnξ2k+ln2γσ,ςℷbψσ,ϕς+1≤lnM1σ,ς+1.
DefineF:0,∞⟶−∞,∞ by(83)Fυ=lnυ+1,for allυ∈0,∞.
Then,F∈Δbs. Hence, for all σ,ς∈ℑ such that ℷbψσ,ϕς>0 and τ=lnξ/2K, we obtain(84)τ+F2γσ,ςℷbψσ,ϕς≤FM1σ,ς.
Consequently, the contractive condition (51) holds for all σ,ς∈ℑ. Hence, all the assumptions of Theorem 1 are verified. Thus, ψ and ϕ have a unique common fixed point which is υ=0.Definition 17.
Letℑ,ℷb,s be a partial b-metric space. Then, the self-mappings ψ,ϕ:ℑ⟶ℑ form a Hardy–Rogers γ,F-contraction if there exist F∈Δbs and τ>0 satisfying(85)τ+Fsγσ,ςℷbψσ,ϕς≤FRσ,ς,∀σ,ς∈ℑ,with γσ,ς≥1, whenever(86)minγσ,ςℷbψσ,ϕς,Rσ,ς>0,where(87)Rσ,ς=ξ1ℷbσ,ς+ξ2ℷbσ,ψσ+ξ3ℷbς,ϕς+ξ4ℷbσ,ϕς+ℷbψσ,ς,such that ξ1+ξ2+ξ3+2sξ4=1.Theorem 2.
Given the pairψ,ϕ:ℑ⟶ℑ of a Hardy–Rogers-type γ,F-contraction defined on a γ-complete partial b-metric space ℑ,ℷb,s. Consider k∈0,1/1+log2s. Suppose that(1)
The pairψ,ϕ is weakly γ-admissible(2)
There existsσ0∈ℑ such that γσ0,ψσ0≥1(3)
(a)ℑ is γ-regular and F is continuous(b) Either one ofψ,ϕ is γ-continuous
Then, we have a sequenceσn in ℑ such that σn⟶υ∈ℑ. If γυ,υ≥1, then υ is a common fixed point of the pair ψ,ϕ. Moreover, if ω is another fixed point of the pair ψ,ϕ satisfying γυ,w≥1, then υ=ω.Proof.
We have(88)Rς,σ=ξ1ℷbς,σ+ξ2ℷbψσ,σ+ξ3ℷbϕς,ς+ξ4ℷbϕς,σ+ℷbς,ψσ=ξ1ℷbς,σ+ξ2ℷbψσ,σ+ξ3ℷbϕς,ς+2sξ4ℷbσ,ϕς+ℷbς,ϕσ2s≤ξ1M1σ,ς+ξ2M1σ,ς+ξ3M1σ,ς+2sξ4M1σ,ς=ξ1+ξ2+ξ3+2sξ4M1σ,ς=M1σ,ς.
Inequality (85) implies (51), so the rest of proof follows from Theorem 1.


## 4. Conclusion

This paper contains a comparative study on partialb-metric spaces with examples. The obtained results generalize corresponding ones in the literature. We developed a methodology to obtain fixed point theorems in partial b-metric spaces. This paper may lead many researchers of this subject to investigate new fixed point theorems in partial b-metric spaces.



---

*Source: 5158552-2021-08-14.xml*