# Existence and Uniqueness of Common Fixed Point on Complex Partialb-Metric Space

Authors: Arul Joseph Gnanaprakasam; Gunaseelan Mani; M. Aphane; Yaé U. Gaba
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this study, we establish the existence and uniqueness of common fixed point on complex partialb-metric space. An example and application to support our result is presented.
The fourth author Y. U. Gaba would like to dedicate this publication to his wife, Clémence A. Epse G., in celebration of her 33rd birthday


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## Body

## 1. Introduction

In 1989, Backhtin [1] and Czerwik [2] introduced the concept of b-metric spaces and provided a framework to extend the results in the classical setting of metric spaces which are known already. Azam et al. [3]introduced complex-valued metric spaces in 2011 and proved some common fixed-point theorems under the contraction condition. Then, in 2013, Rao et al. [4]introduced the definition of complex valued b-metric space and provided a method to extend the results. Later, in 2017, the concept of complex partial metric space was introduced by Dhivya and Marudai [5], and they proved common fixed-point theorems. Recently, Gunaseelan [6]introduced the concept of complex partial b-metric space in 2019. Many authors have discussed significant results and application on complex metric spaces [7–23]. In this study, we establish common fixed-point theorems on complex partial b-metric space using continuity property.


## 2. Preliminaries

LetC be the set of complex numbers and ζ1,ζ2,ζ3∈C. Define a partial order ≲ on C as follows:ζ1≲ζ2 if and only if Rζ1≤Rζ2 and Iζ1≤Iζ2.Then,ζ1≲ζ2 if one of the following properties is fulfilled:(i)
Rζ1=Rζ2, Iζ1<Iζ2(ii)
Rζ1<Rζ2, Iζ1=Iζ2(iii)
Rζ1<Rζ2, Iζ1<Iζ2(iv)
Rζ1=Rζ2, Iζ1=Iζ2In particular, we writeζ1⋨ζ2 if ζ1≠ζ2 and one of i,ii, and iii is fulfilled, and we write ζ1≺ζ2 if only iii is fulfilled.Definition 1.
(see [4]). Let H be a nonvoid set and let s≥1 be a given real number. A function ℓ:H×H⟶C is called a complex valued b-metric on H if, for all ϕ,μ,λ∈H, the following conditions are fulfilled:(i)
0≲ℓϕ,μ and ℓϕ,μ=0 if and only if ϕ=μ(ii)
ℓϕ,μ=ℓμ,ϕ(iii)
ℓϕ,μ≲sℓϕ,λ+ℓλ,μThe pairH,ℓ is called a complex valued b-metric space.Definition 2.
(see [5]). A complex partial metric on a nonvoid set H is a function ηcb:H×H⟶C+ such that, for all ϕ,μ,λ∈H,(i)
0≲ηcbϕ,ϕ≲ηcbϕ,μsmall self−distances(ii)
ηcbϕ,μ=ηcbμ,ϕsymmetry(iii)
ηcbϕ,ϕ=ηcbϕ,μ=ηcbμ,μif and only ifϕ=μequality(iv)
ηcbϕ,μ≲ηcbϕ,λ+ηcbλ,μ−ηcbλ,λtriangularity
A complex partial metric space is a pairH,ηcb such that H is a nonvoid set and ηcb is the complex partial metric on H.Definition 3.
(see [6]). A complex partial b-metric on a nonvoid set H is a function ℓcb:H×H⟶C+ such that, for all ϕ,μ,λ∈H,(i)
0≲ℓcbϕ,ϕ≲ℓcbϕ,μsmall self−distances(ii)
ℓcbϕ,μ=ℓcbμ,ϕsymmetry(iii)
ℓcbϕ,ϕ=ℓcbϕ,μ=ℓcbμ,μ⇔ϕ=μequality(iv)
∃s≥1such thatℓcbϕ,μ≲sℓcbϕ,λ+ℓcbλ,μ−ℓcbλ,λtriangularity
A complex partial b-metric space (b-CPMS) is a pairH,ℓcb such that H is a nonvoid set and ℓcb is the complex partial b-metric on H. The number s is called the coefficient of H,ℓcb.Definition 4.
(see [6]). Let H,ℓcb be a complex partial b-metric space with coefficient s. Let ϕα be any sequence in H and ϕ∈H. Then,(i)
The sequenceϕα is said to be convergent with respect to ℓcb and converges to ϕ if limα⟶∞ℓcbϕα,ϕ=ℓcbϕ,ϕ(ii)
The sequenceϕα is said to be Cauchy sequence in H,ℓcb iflimα,m⟶∞ℓcbϕα,ϕm exists and is finite(iii)
H,ℓcb is said to be a complete complex partial b-metric space if, for every Cauchy sequence ϕα in H, there exists ϕ∈H such that limα,m⟶∞ℓcbϕα,ϕm=limα⟶∞ℓcbϕα,ϕ=ℓcbϕ,ϕIn 2019, Gunaseelan [6] proved some fixed-point theorems on complex partial b-metric space as follows.Theorem 1.
LetH,ℓcb be any complete complex partial b-metric space with coefficient s≥1 and Δ:H⟶H be a mapping satisfying(1)ℓcbΔϕ,Δμ≲ϑmaxℓcbϕ,μ,ℓcbϕ,Δϕ,ℓcbμ,Δμ,for all ϕ,μ∈H, where ϑ∈0,1/s. Then, Δ has a unique fixed point ϕ∗∈H and ℓcbϕ∗,ϕ∗=0.We prove the existence and uniqueness of common fixed point on complex partial b-metric space, inspired by his work.


## 3. Main Results

Theorem 2.
LetH,ℓcb be a complete b-CPMS with the coefficient s≥1 and Δ,Ω:H⟶H be two continuous mappings such that(2)ℓcbΔϕ,Ωμ≲ϑmaxℓcbϕ,μ,ℓcbϕ,Δϕ,ℓcbμ,Ωμ,12ℓcbϕ,Ωμ+ℓcbμ,Δϕ,for all ϕ,μ∈H, where 0≤ϑ<1/s. Then, Δ and Ω have a unique common fixed point ϕ∗∈H and ℓcbϕ∗,ϕ∗=0.Proof.
Letϕ0∈H. Define(3)ϕ2α+1=Δϕ2αandϕ2α+2=Ωϕ2α+1,α=0,1,2,….
Then, by (1) and (2), we obtain(4)ℓcbϕ2α+1,ϕ2α+2=ℓcbΔϕ2α,Ωϕ2α+1,≲ϑmaxℓcbϕ2α,ϕ2α+1,ℓcbϕ2α,Δϕ2α,ℓcbϕ2α+1,Ωϕ2α+1,12ℓcbϕ2α,Ωϕ2α+1+ℓcbϕ2α+1,Δϕ2α,≲ϑmaxℓcbϕ2α,ϕ2α+1,ℓcbϕ2α,ϕ2α+1,ℓcbϕ2α+1,ϕ2α+2,12ℓcbϕ2α,ϕ2α+2+ℓcbϕ2α+1,ϕ2α+1,≲ϑmaxℓcbϕ2α,ϕ2α+1,ℓcbϕ2α+1,ϕ2α+2,12sℓcbϕ2α,ϕ2α+1+ℓcbϕ2α+1,ϕ2α+2−ℓcbϕ2α+1,ϕ2α+1+ℓcbϕ2α+1,ϕ2α+1,=ϑmaxℓcbϕ2α,ϕ2α+1,ℓcbϕ2α+1,ϕ2α+2,s2ℓcbϕ2α,ϕ2α+1+ℓcbϕ2α+1,ϕ2α+2.Case 1.
Ifmaxℓcbϕ2α,ϕ2α+1,ℓcbϕ2α+1,ϕ2α+2,\\s/2ℓcbϕ2α,ϕ2α+1+ℓcbϕ2α+1,ϕ2α+2=ℓcbϕ2α+1,ϕ2α+2, then we have(5)ℓcbϕ2α+1,ϕ2α+2≲ϑℓcbϕ2α+1,ϕ2α+2.
This impliesϑ≥1, which is a reductio ad absurdum.Case 2.
Ifmaxℓcbϕ2α,ϕ2α+1,ℓcbϕ2α+1,ϕ2α+2,s/2ℓcbϕ2α,ϕ2α+1+ℓcbϕ2α+1,ϕ2α+2=ℓcbϕ2α,ϕ2α+1, then we have(6)ℓcbϕ2α+1,ϕ2α+2≲ϑℓcbϕ2α,ϕ2α+1.
From the next step, we have(7)ℓcbϕ2α+2,ϕ2α+3≲ϑmaxℓcbϕ2α+1,ϕ2α+2,ℓcbϕ2α+2,ϕ2α+3,s2ℓcbϕ2α+1,ϕ2α+2+ℓcbϕ2α+2,ϕ2α+3.
We consider three cases.Case 3.
(8)ℓcbϕ2α+2,ϕ2n+3≲ϑℓcbϕ2α+2,ϕ2n+3,which implies ϑ≥1, is a reductio ad absurdum.Case 4.
(9)ℓcbϕ2α+2,ϕ2n+3≲ϑℓcbϕ2α+1,ϕ2α+2.
From (6) and (9), ∀α=0,1,2,…, we obtain(10)ℓcbϕα+1,ϕα+2≲ϑℓcbϕα,ϕα+1≲⋯≲ϑα+1ℓcbϕ0,ϕ1.
Forq,α∈ℕ, with q>α, we have(11)ℓcbϕα,ϕq≲sℓcbϕα,ϕα+1+ℓcbϕα+1,ϕq−ℓcbϕα+1,ϕα+1,≲sℓcbϕα,ϕα+1+ℓcbϕα+1,ϕq,≲sℓcbϕα,ϕα+1+s2ℓcbϕα+1,ϕα+2+ℓcbϕα+2,ϕq−ℓcbϕα+2,ϕα+2,≲sℓcbϕα,ϕα+1+s2ℓcbϕα+1,ϕα+2+ℓcbϕα+2,ϕq≲sℓcbϕα,ϕα+1+s2ℓcbϕα+1,ϕα+2+s3ℓcbϕα+2,ϕα+3,+⋯+sq−α−1ℓcbϕq−2,ϕq−1+sq−αℓcbϕq−1,ϕq.
Moreover, by using (9), we obtain(12)ℓcbϕα,ϕq≲sϑαℓcbϕ0,ϕ1+s2ϑα+1ℓcbϕ0,ϕ1+s3ϑα+2ℓcbϕ0,ϕ1+⋯+sq−α−1ϑq−2ℓcbϕ0,ϕ1+sq−αϑq−1ℓcbϕ0,ϕ1=∑i=1q−αsiϑi+α−1ℓcbϕ0,ϕ1.
Therefore,(13)ℓcbϕα,ϕq≤∑i=1q−αsi+α−1ϑi+α−1ℓcbϕ0,ϕ1=∑t=αq−1stϑtℓcbϕ0,ϕ1,≤∑i=α∞sϑtℓcbϕ0,ϕ1,=sϑα1−sϑℓcbϕ0,ϕ1.
Then, we have(14)ℓcbϕα,ϕq≤sϑα1−sϑℓcbϕ0,ϕ1|⟶0asα⟶∞.
Hence,ϕα is a Cauchy sequence in H.Case 5.
(15)ℓcbϕ2α+2,ϕ2α+3≲ϑs2ℓcbϕ2α+1,ϕ2α+2+ℓcbϕ2α+2,ϕ2α+3.
This implies that(16)ℓcbϕ2α+2,ϕ2n+3≲ϑs2−ϑsℓcbϕ2α+1,ϕ2α+2.
Sincea:=ϑs/2−ϑs<1, we get ℓcbϕα+1,ϕα+2≲aℓcbϕα,ϕα+1. Therefore, ϕαα∈ℕ is a Cauchy sequence in H.Case 6.
Ifmaxℓcbϕ2α,ϕ2α+1,ℓcbϕ2α+1,ϕ2α+2,s/2ℓcbϕ2α,ϕ2α+1+ℓcbϕ2α+1,ϕ2α+2=s/2ℓcbϕ2α,ϕ2α+1+ℓcbϕ2α+1,ϕ2α+2, then we have(17)ℓcbϕ2α+1,ϕ2α+2≲ϑs/2ℓcbϕ2α,ϕ2α+1+ℓcbϕ2α+1,ϕ2α+2.
Hence,(18)ℓcbϕ2α+1,ϕ2α+2≲ϑs2−ϑsℓcbϕ2α,ϕ2α+1.
For the next step, we have(19)ℓcbϕ2α+2,ϕ2α+3≲ϑmaxℓcbϕ2α+1,ϕ2α+2,ℓcbϕ2α+2,ϕ2α+3,s2ℓcbϕ2α+1,ϕ2α+2+ℓcbϕ2α+2,ϕ2α+3.
Then, we consider three cases.Case 7.
(20)ℓcbϕ2α+2,ϕ2α+3≲ϑℓcbϕ2α+2,ϕ2α+3,which implies ϑ≥1 and is a reductio ad absurdum.Case 8.
(21)ℓcbϕ2α+2,ϕ2n+3≲ϑℓcbϕ2α+1,ϕ2α+2.
Then, by (18) and (21), we get ℓcbϕα+1,ϕα+2≲γℓcbϕα,ϕα+1, where γ=maxϑ,ϑs/2−ϑs<1. Hence, ϕαα∈N is a Cauchy sequence in H.Case 9.
(22)ℓcbϕ2α+2,ϕ2n+3≲s2ℓcbϕ2α+1,ϕ2α+2+ℓcbϕ2α+2,ϕ2α+3.
Hence, we obtain(23)ℓcbϕ2α+2,ϕ2α+3≲ϑs2−ϑsℓcbϕ2α+1,ϕ2α+2.
By using (18) and (21) yields(24)ℓcbϕα+1,ϕα+2≲≀ℓcbϕα,ϕα+1,where 0≤≀=ϑs/2−ϑs<1.
Then,∀α=0,1,2,…, we obtain(25)ℓcbϕα+1,ϕα+2≲≀ℓcbϕα,ϕα+1≲⋯≲≀α+1ℓcbϕ0,ϕ1.
Forq,α∈ℕ, with q>α,(26)ℓcbϕα,ϕq≲sℓcbϕα,ϕα+1+ℓcbϕα+1,ϕq−ℓcbϕα+1,ϕα+1,≲sℓcbϕα,ϕα+1+ℓcbϕα+1,ϕq,≲sℓcbϕα,ϕα+1+s2ℓcbϕα+1,ϕα+2+ℓcbϕα+2,ϕq−ℓcbϕα+2,ϕα+2,≲sℓcbϕα,ϕα+1+s2ℓcbϕα+1,ϕα+2+ℓcbϕα+2,ϕq,≲sℓcbϕα,ϕα+1+s2ℓcbϕα+1,ϕα+2+s3ℓcbϕα+2,ϕα+3,+⋯+sq−α−1ℓcbϕq−2,ϕq−1+sq−αℓcbϕq−1,ϕq.
Using (24), we obtain(27)ℓcbϕα,ϕq≲s≀αℓcbϕ0,ϕ1+s2≀α+1ℓcbϕ0,ϕ1+s3≀α+2ℓcbϕ0,ϕ1+⋯+sq−α−1≀q−2ℓcbϕ0,ϕ1+sq−α≀q−1ℓcbϕ0,ϕ1=∑i=1q−αsi≀i+α−1ℓcbϕ0,ϕ1.
Therefore,(28)ℓcbϕα,ϕq≤∑i=1q−αsi+α−1≀i+α−1ℓcbϕ0,ϕ1=∑t=αq−1st≀tℓcbϕ0,ϕ1≤∑i=α∞s≀tℓcbϕ0,ϕ1,=s≀α1−s≀ℓcbϕ0,ϕ1.
Hence, we have(29)ℓcbϕα,ϕq≤s≀α1−s≀ℓcbϕ0,ϕ1⟶0asα⟶∞.
Hence,ϕα is a Cauchy sequence in H. In all cases, ϕαα∈ℕ is a Cauchy sequence. Since H is complete, there exists ϕ∗∈H such that ϕα⟶ϕ∗ as α⟶∞ and(30)ℓcbϕ∗,ϕ∗=limα⟶∞ℓcbϕ∗,ϕα=limα⟶∞ℓcbϕα,ϕα=0.
By the continuity ofΔ, ϕ2α+1=Δϕ2α⟶Δϕ∗ as α⟶∞:(31)i.e.ℓcbΔϕ∗,Δϕ∗=limα⟶∞ℓcbΔϕ∗,Δϕ2α=limα⟶∞ℓcbΔϕ2α,Δϕ2α.
However,(32)ℓcbΔϕ∗,Δϕ∗=limα⟶∞ℓcbΔϕ2α,Δϕ2α=limα⟶∞ℓcbϕ2α+1,ϕ2α+1=0.
Next, we prove thatϕ∗ is a fixed point of Δ:(33)ℓcbΔϕ∗,ϕ∗≲ℓcbΔϕ∗,Δϕ2α+ℓcbΔϕ2α,ϕ∗−ℓcbΔϕ2α,Δϕ2α.
Asα⟶∞, we obtain ℓcbΔϕ∗,ϕ∗≤0. Thus, ℓcpΔϕ∗,ϕ∗=0. Hence, ℓcbϕ∗,ϕ∗=ℓcbϕ∗,Δϕ∗=ℓcbΔϕ∗,Δϕ∗=0 and Δϕ∗=ϕ∗. In the same way, we have ϕ∗∈H such that ϕα⟶ϕ∗ as α⟶∞ and(34)ℓcbϕ∗,ϕ∗=limα⟶∞ℓcbϕ∗,ϕα=limα⟶∞ℓcbϕα,ϕα=0.
By the continuity ofΔϕ2α+2=Ωϕ2α+1⟶Ωϕ∗ as α⟶∞,(35)i.e.ℓcbΩϕ∗,Ωϕ∗=limα⟶∞ℓcbΩϕ∗,Ωϕ2α+1=limα⟶∞ℓcbΩϕ2α+1,Ωϕ2α+1.
However,(36)ℓcbΩϕ∗,Ωϕ∗=limα⟶∞ℓcbΩϕ2α+1,Ωϕ2α+1=limα⟶∞ℓcbϕ2α+2,ϕ2α+2=0.
Next, we prove thatϕ∗ is a fixed point of Ω:(37)ℓcbΩϕ∗,ϕ∗≲ℓcbΩϕ∗,Ωϕ2α+1+ℓcbΩϕ2α+1,ϕ∗−ℓcbΩϕ2α+1,Δϕ2α+1.
Asα⟶∞, we obtain ℓcbΩϕ∗,ϕ∗≤0. Thus, ℓcpΩϕ∗,ϕ∗=0. Hence, ℓcbϕ∗,ϕ∗=ℓcbϕ∗,Ωϕ∗=ℓcbΩϕ∗,Ωϕ∗=0 and Ωϕ∗=ϕ∗. Therefore, Δ and Ω have a common fixed point ϕ∗.
Letμ∗∈H be another common fixed point for the mappings Δ and Ω. Then,(38)ℓcbϕ∗,μ∗=ℓcbΔϕ∗,Ωμ∗≲ϑmaxℓcbϕ∗,μ∗,ℓcbϕ∗,Δϕ∗,ℓcbμ∗,Ωμ∗,12ℓcbϕ∗,Ωμ∗+ℓcbμ∗,Δϕ∗,≲ϑmaxℓcbϕ∗,μ∗,ℓcbϕ∗,ϕ∗,ℓcbμ∗,μ∗12ℓcbϕ∗,μ∗+ℓcbμ∗,ϕ∗≲ϑℓcbϕ∗,μ∗.
This implies thatϕ∗=μ∗.Theorem 3.
LetH,ℓcb be a complete b-CPMS with the coefficient s≥1 and Δ,Ω:H⟶H be two continuous mappings such that(39)ℓcbΔϕ,Ωμ≲ϑmaxℓcbϕ,μ,ℓcbϕ,Δϕ,ℓcbμ,Ωμ,12ℓcbϕ,Ωμ+ℓcbμ,Δϕ,for all ϕ,μ∈H, where 0≤ϑ<1/s. Then, Δ and Ω have a unique common fixed point ϕ∗∈H and ℓcbϕ∗,ϕ∗=0.Proof.
Following from Theorem2, we can easily prove ϕα is a Cauchy sequence. Since H is complete, there exists ϕ∗∈H such that ϕα⟶ϕ∗ as α⟶∞.
Suppose thatℓcbϕ∗,Δϕ∗=λ>0.
Then, we estimate(40)λ=ℓcbϕ∗,Δϕ∗,≲sℓcbϕ∗,ϕ2i+2+ℓcbϕ2i+2,Δϕ∗−ℓcbϕ2i+2,ϕ2i+2,≲sℓcbϕ∗,ϕ2i+2+ℓcbϕ2i+2,Δϕ∗,≲sℓcbϕ∗,ϕ2i+2+sℓcbΩϕ2i+1,Δϕ∗,≲sℓcbϕ∗,ϕ2i+2+ϑsmaxℓcbϕ2i+1,ϕ∗,ℓcbϕ2i+1,Ωϕ2i+1,ℓcbϕ∗,Δϕ∗12ℓcbϕ2i+1,Δϕ∗+ℓcbϕ∗,Ωϕ2i+1,≲sℓcbϕ∗,ϕ2i+2+ϑsmaxℓcbϕ2i+1,ϕ∗,ℓcbϕ2i+1,ϕ2i+2,ℓcbϕ∗,Δϕ∗12ℓcbϕ2i+1,Δϕ∗+ℓcbϕ∗,ϕ2i+2,≲sℓcbϕ∗,ϕ2i+2+sϑℓcbϕ∗,Δϕ∗,≲sℓcbϕ∗,ϕ2i+2+sϑλ.
This yields(41)λ≤sℓcbϕ∗,ϕ2i+2+sϑλ.
Hence,ϑ≥1, which is a reductio ad absurdum. Then, ϕ∗=Δϕ∗. Similarly, we derive that ϕ∗=Ωϕ∗. Therefore, Δ and Ω have a common fixed point ϕ∗. Following from Theorem 2, we can easily prove uniqueness part.Theorem 4.
LetH,ℓcb be a complete b-CPMS with the coefficient s≥1 and Δ,Ω:H⟶H be two continuous mappings such that(42)ℓcbΔϕ,Ωμ≲ϑmaxℓcbϕ,μ,ℓcbϕ,Δϕℓcbμ,Ωμ1+ℓcbϕ,μ,ℓcbϕ,ΔϕℓcbΔϕ,Ωμ1+ℓcbϕ,μ,for all ϕ,μ∈H, where 0≤ϑ<1/s. Then, Δ and Ω have a unique common fixed point ϕ∗∈H and ℓcbϕ∗,ϕ∗=0.Proof.
Letϕ0∈H. Define(43)ϕ2α+1=Δϕ2αandϕ2α+2=Ωϕ2α+1,α=0,1,2,….
Then, by (42) and (43), we obtain(44)ℓcbϕ2α+1,ϕ2α+2=ℓcbΔϕ2α,Ωϕ2α+1,≲ϑmaxℓcbϕ2α,ϕ2α+1,ℓcbϕ2α,ϕ2α+1ℓcbΩϕ2α+1,Δϕ2α1+ℓcbϕ2α,ϕ2α+1,ℓcbϕ2α,Δϕ2α,ℓcbΔϕ2α,Ωϕ2α+11+ℓcbϕ2α,ϕ2α+1,≲ϑmaxℓcbϕ2α,ϕ2α+1,ℓcbϕ2α,ϕ2α+1ℓcbϕ2α+1,ϕ2α+21+ℓcbϕ2α,ϕ2α+1,ℓcbϕ2α,ϕ2α+1ℓcbϕ2α+1,ϕ2α+21+ℓcbϕ2α,ϕ2α+1,≲ϑmaxℓcbϕ2α,ϕ2α+1,ℓcbϕ2α+1,ϕ2α+2.
Ifmaxℓcbϕ2α,ϕ2α+1,ℓcbϕ2α+1,ϕ2α+2=ℓcbϕ2α+1,ϕ2α+2, then(45)ℓcbϕ2α+1,ϕ2α+2≲ϑℓcbϕ2α+1,ϕ2α+2.
This shows thatϑ≥1, which is a reductio ad absurdum. Therefore,(46)ℓcbϕ2α+1,ϕ2α+2≲ϑℓcbϕ2α,ϕ2α+1.
Similarly, we obtain(47)ℓcbϕ2α+2,ϕ2α+3≲ϑℓcbϕ2α+1,ϕ2α+2.
From (46) and (47), ∀α=0,1,2,…, we obtain(48)ℓcbϕα+1,ϕα+2≲ϑℓcbϕα,ϕα+1≲⋯≲ϑα+1ℓcbϕ0,ϕ1.
Forq,α∈ℕ, with q>α, we have(49)ℓcbϕα,ϕq≲sℓcbϕα,ϕα+1+ℓcbϕα+1,ϕq−ℓcbϕα+1,ϕα+1,≲sℓcbϕα,ϕα+1+ℓcbϕα+1,ϕq,≲sℓcbϕα,ϕα+1+s2ℓcbϕα+1,ϕα+2+ℓcbϕα+2,ϕq−ℓcbϕα+2,ϕα+2,≲sℓcbϕα,ϕα+1+s2ℓcbϕα+1,ϕα+2+ℓcbϕα+2,ϕq,≲sℓcbϕα,ϕα+1+s2ℓcbϕα+1,ϕα+2+s3ℓcbϕα+2,ϕα+3,+⋯+sq−α−1ℓcbϕq−2,ϕq−1+sq−αℓcbϕq−1,ϕq.
By using (48), we obtain(50)ℓcbϕα,ϕq≲sϑαℓcbϕ0,ϕ1+s2ϑα+1ℓcbϕ0,ϕ1+s3ϑα+2ℓcbϕ0,ϕ1+⋯+sq−α−1ϑq−2ℓcbϕ0,ϕ1+sq−αϑq−1ℓcbϕ0,ϕ1,=∑i=1q−αsiϑi+α−1ℓcbϕ0,ϕ1.
Therefore,(51)ℓcbϕα,ϕq≤∑i=1q−αsi+α−1ϑi+α−1ℓcbϕ0,ϕ1=∑i=1q−αstϑtℓcbϕ0,ϕ1≤∑i=α∞sϑtℓcbϕ0,ϕ1,=sϑα1−sϑℓcbϕ0,ϕ1.
Hence, we have(52)ℓcbϕα,ϕq≤sϑα1−sϑℓcbϕ0,ϕ1⟶0asα⟶∞.
Hence,ϕα is a Cauchy sequence in H. Since H is complete, there exists ϕ∗∈H such that ϕα⟶ϕ∗ as α⟶∞ and(53)ℓcbϕ∗,ϕ∗=limα⟶∞ℓcbϕ∗,ϕα=limα⟶∞ℓcbϕα,ϕα=0.
SinceΩ is continuous, we obtain(54)ϕ∗=limα⟶∞ϕ2α+2=limα⟶∞Ωϕ2α+1=Ωlimα⟶∞ϕ2α+1=Ωϕ∗.
Similarly, we derive thatϕ∗=Δϕ∗. Then, Δ and Ω have a common fixed point. Let μ∗∈H be another common fixed point for the mappings Δ and Ω. Then,(55)ℓcbϕ∗,μ∗=ℓcbΔϕ∗,Ωμ∗,≲ϑmaxℓcbϕ∗,μ∗,ℓcbϕ∗,Δϕ∗ℓcbμ∗,Ωμ∗1+ℓcbϕ∗,μ∗,ℓcbϕ∗,Δϕ∗ℓcbΩμ∗,Δϕ∗1+ℓcbϕ∗,μ∗,≲ϑℓcbϕ∗,μ∗.
This implies thatϕ∗=μ∗.Theorem 5.
LetH,ℓcb be a complete b-CPMS with the coefficient s≥1 and Δ,Ω:H⟶H be two continuous mappings such that(56)ℓcbΔϕ,Ωμ≲ϑmaxℓcbϕ,μ,ℓcbϕ,Δϕℓcbμ,Ωμ1+ℓcbϕ,μ,ℓcbϕ,ΔϕℓcbΔϕ,Ωμ1+ℓcbϕ,μ,for all ϕ,μ∈H, where 0≤ϑ<1/s. Then, Δ and Ω have a unique common fixed point ϕ∗∈H and ℓcbϕ∗,ϕ∗=0.Proof.
Following from Theorem5, we can easily prove ϕα is a Cauchy sequence. Since H is complete, there exists ϕ∗∈H such that ϕα⟶ϕ∗ as α⟶∞ and(57)ℓcbϕ∗,ϕ∗=limα⟶∞ℓcbϕ∗,ϕα=limα⟶∞ℓcbϕα,ϕα=0.
Suppose thatℓcbϕ∗,Δϕ∗=λ>0.
Then, we estimate(58)λ=ℓcbϕ∗,Δϕ∗,≲sℓcbϕ∗,ϕ2i+2+ℓcbϕ2i+2,Δϕ∗−ℓcbϕ2i+2,ϕ2i+2,≲sℓcbϕ∗,ϕ2i+2+ℓcbΔϕ∗,ϕ2i+2,≲sℓcbϕ∗,ϕ2i+2+sℓcbΔϕ∗,Ωϕ2i+1,≲sℓcbϕ∗,ϕ2i+2+ϑsmaxℓcbϕ∗,ϕ2i+1,ℓcbϕ∗,Δϕ∗ℓcbϕ2i+1,Ωϕ2i+11+ℓcbϕ∗,ϕ2i+1,ℓcbϕ∗,Δϕ∗ℓcbΔϕ∗,Ωϕ2i+11+ℓcbϕ∗,ϕ2i+1,≲sℓcbϕ∗,ϕ2i+2+ϑsmaxℓcbϕ∗,ϕ2i+1,ℓcbϕ∗,Δϕ∗ℓcbϕ2i+1,ϕ2i+21+ℓcbϕ∗,ϕ2i+1ℓcbϕ∗,Δϕ∗ℓcbΔϕ∗,ϕ2i+21+ℓcbϕ∗,ϕ2i+1,≲sℓcbϕ∗,ϕ2i+2+sϑℓcbϕ∗,Δϕ∗2,≲sℓcbϕ∗,ϕ2i+2+sϑλ2.
This yields(59)λ≤sℓcbϕ∗,ϕ2i+2+sϑλ2.
Hence,ϑ≥1, which is a reductio ad absurdum. Then, ϕ∗=Δϕ∗. Similarly, we derive that ϕ∗=Ωϕ∗. Therefore, Δ and Ω have a common fixed point ϕ∗. Following from Theorem 5, we can easily prove the uniqueness part.Example 3.5.
LetH=1,2,3,4 be endowed with the partial order ϕ≲μ iff μ≤ϕ. We define ℓcb:H×H⟶C+ in Tables 1 and 2.
It is easy to verify thatH,ℓcb is a complete b-CPMS with the coefficient s≥1 for x∈0,π/2. Define Δ,Ω:H⟶H by Δϕ=1:(60)Ωϕ=1ifϕ∈1,2,3,2ifϕ=4.
Clearly,Δ and Ω are continuous functions. Now, for ϑ=1/9, we consider the following cases:(A)
Ifϕ=1 and μ∈H−4, then Δϕ=Ωμ=1. Hence, all the conditions of Theorem 2 are fulfilled.(B)
Ifϕ=1 and μ=4, then Δϕ=1 and Ωμ=2:(61)ℓcbΔϕ,Ωμ=e2ix≲9ϑei2x=ϑmax9ei2x,0,9ei2x,12ei2x+9ei2x=ϑmaxℓcbϕ,μ,ℓcbϕ,Δϕ,ℓcbμ,Ωμ12ℓcbϕ,Ωμ+ℓcbμ,Δϕ.(C)
Ifϕ=2 and μ=4, thenΔϕ=1 and Ωμ=2:(62)ℓcbΔϕ,Ωμ=e2ix≲9ϑei2x=ϑmax9ei2x,ei2x,9ei2x,120+9ei2x=ϑmaxℓcbϕ,μ,ℓcbϕ,Δϕ,ℓcbμ,Ωμ12ℓcbϕ,Ωμ+ℓcbμ,Δϕ.(D)
Ifϕ=3 and μ=4, then Δϕ=1 and Ωμ=2:(63)ℓcbΔϕ,Ωμ=e2ix≲9ϑei2x=ϑmax9ei2x,ei2x,9ei2x,12ei2x+9ei2x=ϑmaxℓcbϕ,μ,ℓcbϕ,Δϕ,ℓcbμ,Ωμ12ℓcbϕ,Ωμ+ℓcbμ,Δϕ.(E)
Ifϕ=4 and μ=4, then Δϕ=2 and Ωμ=2:(64)ℓcbΔϕ,Ωμ=e2ix≲9ϑei2x=ϑmax9ei2x,9ei2x,9ei2x,129ei2x+9ei2x=ϑmaxℓcbϕ,μ,ℓcbϕ,Δϕ,ℓcbμ,Ωμ12ℓcbϕ,Ωμ+ℓcbμ,Δϕ.
All the conditions of Theorem1, with ϑ=1/9<1, are fulfilled. Therefore, Δ and Ω have a unique common fixed point 1.Table 1
Example of unique common fixed point.
ϕ,μℓcbϕ,μ(1,1), (3,3)0(1,2), (2,1), (1,3), (3,1), (2,3), (3,2), (2,2)e2ix(1,4), (4,1), (2,4), (4,2), (3,4), (4,3), (4,4)9e2ixTable 2
Example of no common fixed point.
ϕ,μℓcbϕ,μ(1,1), (3,3)0(1,2), (2,1), (1,3), (3,1), (2,3), (3,2), (2,2)e2ix(1,4), (4,1), (2,4), (4,2), (3,4), (4,3), (4,4)e2ixExample 3.6.
LetH=P∪Q, where P=1,2 and Q=3,4 be endowed with the partial order ϕ≲μ iff μ≤ϕ. Define ℓcb:H×H⟶C+ by ℓcbϕ,μ=ϕ−μ2+2e2ix, for all ϕ,μ∈P or ϕ∈P and μ∈Q or ϕ∈Q,μ∈P.
It is easy to verify thatH,ℓcb is a complete b-CPMS with the coefficient s≥1 for x∈0,π/2. Define Δ,Ω:H⟶H by(65)Δϕ=1ifϕ∈1,32,2ifϕ∈32,2∪Q,(66)Ωϕ=2ifϕ∈1,32,4ifϕ∈32,2∪Q.
Clearly,Δ and Ω are not continuous functions. Now, we consider the following cases:(A)
Ifϕ,μ∈1,3/2, thenΔϕ=1 and Ωμ=2:(67)ℓcbΔϕ,Ωμ=e2ix≲133ei2x=ϑmaxℓcbϕ,μ,ℓcbϕ,Δϕ,ℓcbμ,Ωμ12ℓcbϕ,Ωμ+ℓcbμ,Δϕ.(B)
Ifϕ∈1,3/2 and μ∈3/2,2∪Q, then Δϕ=1 and Ωμ=4:(68)ℓcbΔϕ,Ωμ=e2ix≲133ei2x=ϑmaxℓcbϕ,μ,ℓcbϕ,Δϕ,ℓcbμ,Ωμ12ℓcbϕ,Ωμ+ℓcbμ,Δϕ.(C)
Ifμ∈1,3/2 and ϕ∈3/2∪Q, thenΔϕ=2 and Ωμ=2:(69)ℓcbΔϕ,Ωμ=e2ix≲313ei2x=ϑmaxℓcbϕ,μ,ℓcbϕ,Δϕ,ℓcbμ,Ωμ12ℓcbϕ,Ωμ+ℓcbμ,Δϕ.(D)
Ifμ,ϕ∈3/2,2∪Q, then Δϕ=2 and Ωμ=4:(70)ℓcbΔϕ,Ωμ=e2ix≲313ei2x=ϑmaxℓcbϕ,μ,ℓcbϕ,Δϕ,ℓcbμ,Ωμ12ℓcbϕ,Ωμ+ℓcbμ,Δϕ.
All the conditions of Theorems2 and 3, with ϑ<1, are fulfilled except continuous mapping. Therefore, Δ and Ω have no common fixed point.Remark 1.
In view of the fact in Theorems2 and 3, we cannot drop the continuous mapping.


## 4. Application

Consider the following systems of integral equations:(71)υs=∫cdT1s,ℵ,υℵdℵ,(72)ϱs=∫cdT2s,ℵ,ϱℵdℵ,where(i)
υs and ϱs are unknown variables for each s∈J=c,d,  d>c≥0(ii)
T1s,ℵ and T2s,ℵ are deterministic kernels defined for s,ℵ∈J=c,dLetH=CJ,Rα be the set of continuous functions defined on J. Define ℓcb:H×H⟶C+ by(73)ℓcbυ,ϱ=υs−ϱs2+2,∀υ,ϱ∈H. Then, ℓcb is a complete b-CPMS. Define partial order ≲ given by(74)υ,ϱ∈H,υ≲ϱif and onlyυs≥ϱs,∀s∈J.Theorem 6.
Assume that(A)
T1,T2:J×J×Rα⟶Rα are continuous functions satisfying(75)T1s,ℵ,υℵ−T2s,ℵ,ϱs≲Sυ,ϱb−aet−2b−a,∀t>0,where(76)Sυ,ϱ=maxℓcbυ,ϱ,ℓcbυ,Δυ,ℓcbϱ,Ωϱ12ℓcbυ,Ωϱ+ℓcbϱ,Δυ.Then, systems (71) and (72) have a unique common solution.Proof.
Forυ,ϱ∈CJ,Rα and s∈J, define the continuous mappings Δ,Ω:H⟶H by(77)Δυs=∫cdT1s,ℵ,υℵdℵ,(78)Ωϱs=∫cdT2s,ℵ,ϱℵdℵ.
Then,(79)ℓcbΔυs,Ωϱs=Δυs−Ωϱs2+2=∫cdT1s,ℵ,υℵ−T2s,ℵ,s2dℵ+2≲∫cdSυ,ϱb−aet−2b−adℵ+2=Sυ,ϱet=ϑSυ,ϱ=ϑmaxℓcbυ,ϱ,ℓcbυ,Δυ,ℓcbϱ,Ωϱ12ℓcbυ,Ωϱ+ℓcbϱ,Δυ.
Hence, all the conditions of Theorem2 are fulfilled for 0<ϑ=1/et<1/s with t>0. Therefore, integrals (71) and (72) have a unique common solution.


## 5. Conclusion

In this paper, we proved common fixed-point theorems on complex partial b-metric space. An illustrative example and application on complex partial b-metric space is given. Recently, Khalehoghli et al. [24, 25] introduced R-metric spaces and obtained a generalization of Banach fixed-point theorem. It is an interesting open problem to study the relation R instead of complex partial b-metric space and obtain common fixed-point results on R-complete complex partial b-metric spaces.



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