# On Controlled Rectangular Metric Spaces and an Application

Authors: Nayab Alamgir; Quanita Kiran; Hassen Aydi; Yaé Ulrich Gaba
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we introduce the notion of controlled rectangular metric spaces as a generalization of rectangular metric spaces and rectangularb-metric spaces. Further, we establish some related fixed point results. Our main results extend many existing ones in the literature. The obtained results are also illustrated with the help of an example. In the last section, we apply our results to a common real-life problem in a general form by getting a solution for the Fredholm integral equation in the setting of controlled rectangular metric spaces.


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## Body

## 1. Introduction

The fixed point theory is a growing and exciting field of mathematics with a variety of variant applications in mathematical sciences, proposing newer applications in discrete dynamics and super fractals. The fixed point theory is a fundamental tool to various theoretical and applied fields, such as variational and linear inequalities, the approximation theory, nonlinear analysis, integral and differential equations and inclusions, the dynamic systems theory, mathematics of fractals, mathematical economics (game theory, equilibrium problems, and optimization problems), and mathematical modeling; see [1–3]. In particular, fixed point techniques have been applied in such diverse fields; see [4, 5]. There are particular real-life problems, whose statements are fairly easy to understand, which can be argued using some versions of fixed point theorems; see [6, 7].The notion of extendedb-metric spaces was introduced by Kamran et al. [9] as a generalization of metric spaces and b-metric spaces [10, 11]. This metric type space has been generalized in several directions (for instance, controlled metric spaces [12], double controlled metric spaces [13], and others [14–19]). In a different perception, Branciari [20] proposed rectangular metric spaces. In the same order, Asim et al. [21] included a control function to initiate the concept of extended rectangular b-metric spaces as a generalization of rectangular b-metric spaces [22]. In [23], Mlaiki et al. introduced controlled rectangular b-metric spaces, which generalize rectangular metric spaces and rectangular b-metric spaces.In this paper, our goal is to introduce the notion of controlled rectangular metric spaces, which is different from controlled rectangularb-metric spaces, and generalize rectangular metric spaces as well as rectangular b-metric spaces. Further, we prove some fixed point results on such spaces as a generalization of many preexisting results in the literature. Also, we give examples for the justification of our results. In the last, as an application, we give an existence theorem for the Fredholm integral equation in the setting of controlled rectangular metric spaces.


## 2. Preliminaries

In this section, we collect some basic concepts related to our main results.Definition 1 [22].
A mappingdζ:M×M⟶0,∞ on a nonempty set M is called a rectangular b-metric space, if there exists a constant s≥1 such that for all m1,m2∈M and all distinct μ1,μ2∈M different from m1andm2, the following axioms are satisfied:
(i)
dζm1,m2=0iffm1=m2(ii)
dζm1,m2=dζm2,m1(iii)
dζm1,m2≤sdζm1,μ1+dζμ1,μ2+dζμ2,m2
In this case, the pairM,dζ is called a rectangular b-metric space.Definition 2 [9].
LetM be a nonempty set and ζ:M×M→1,∞ be a mapping. Then, a mapping dζ:M×M⟶0,∞ is called an extended b-metric, if for all m1,m2,m3∈M, it satisfies the following axioms:
(i)
dζm1,m2=0iffm1=m2(ii)
dζm1,m2=dζm2,m1(iii)
dζm1,m3≤ζm1,m3dζm1,m2+dζm2,m3
The pairM,dζ is called an extended b-metric space.Definition 3 [21].
A mappingdζ:M×M⟶0,∞ on a nonempty set M is called an extended rectangular b-metric space, if for all m1,m2∈M and all distinct μ1,μ2∈M different from m1 and m2, the following axioms are satisfied:
(i)
dζm1,m2=0iffm1=m2(ii)
dζm1,m2=dζm2,m1(iii)
dζm1,m2≤ζm1,m2dζm1,μ1+dζμ1,μ2+dζμ2,m2where ζ:M×M⟶1,∞ is a mapping. In this case, the pair M,dζ is called an extended rectangular b-metric space.Note that the topology of rectangular metric spaces need not be Hausdorff. For more examples, see the papers of Sarma et al. [24] and Samet [25]. The topological structure of rectangular metric spaces is not compatible with the topology of classic metric spaces; see Example 7 in the paper of Suzuki [26]. In the same direction, extended rectangular b-metric spaces cannot be Hausdorff.Definition 4 [23].
A mappingdζ:M×M⟶0,∞ on a nonempty set M is called a controlled rectangular b-metric space, if for all distinct m1,m2,μ1,μ2∈M, the following axioms are satisfied:
(i)
dζm1,m2=0iffm1=m2(ii)
dζm1,m2=dζm2,m1(iii)
dζm1,m2≤ζm1,m2,μ1,μ2dζm1,μ1+dζμ1,μ2+dζμ2,m2where ζ:M4⟶1,∞ is a mapping. In this case, the pair M,dζ is called a controlled rectangular b-metric space.As a generalization of metric spaces, Mlaiki et al. in [12] introduced the concept of controlled metric spaces as follows.Definition 5 [12].
LetM be a nonempty set and ζ:M×M⟶1,∞. Then, a mapping dζ:M×M⟶0,∞ is called a controlled metric, if for all m1,m2,m3∈M, it satisfies the following axioms:
(i)
dζm1,m2=0iffm1=m2(ii)
dζm1,m2=dζm2,m1(iii)
dζm1,m3≤ζm1,m2dζm1,m2+ζm2,m3dζm2,m3
The pairM,dζ is called a controlled metric space.Note that Definition5 generalizes b-metric spaces and is different from Definition 2.Example 1 [12].
LetM=1,2,⋯. Define dζ:M×M⟶0,∞ as
(1)dζx,y=0,ifm1=m2,1m1,ifm1isevenandm2isodd,1m2,ifm1isoddandm2iseven,1,otherwise.
Hence,M,dζ is a controlled metric space, where ζ:M×M⟶1,∞ is defined as
(2)ζm1,m2=m1,ifm1isevenandm2isodd,m2,ifm1isoddandm2iseven,1,otherwise.


## 3. Main Results

In this section, we introduce the notion of controlled rectangular metric spaces. Also, we establish some fixed point results.Definition 6.
A mappingdζ:M×M⟶0,∞ on a nonempty set M is called a controlled rectangular metric space, if for all m1,m2∈M and all distinct μ1,μ2∈M different from m1 and m2, the following axioms are satisfied:
(i)
dζm1,m2=0iffm1=m2(ii)
dζm1,m2=dζm2,m1(iii)
dζm1,m2≤ζm1,μ1dζm1,μ1+ζμ1,μ2dζμ1,μ2+ζμ2,m2dζμ2,m2where ζ:M×M⟶1,∞ is a mapping, In this case, the pair M,dζ is called a controlled rectangular metric space.Remark 7.
(i)
Every rectangular metric space and rectangularb-metric is a controlled rectangular metric space(ii)
Clearly, Definition6 is different from Definition 4(iii)
Every controlled metric space is a controlled rectangular metric space, but its converse is not true in general. See the following exampleExample 2.
LetM=0,∞. Define a mapping dζ:M×M⟶0,∞ by
(3)dζm1,m2=0,ifm1=m2,1m1,ifm1≥1andm2∈0,1,1m2,ifm2≥1andm1∈0,1,1,otherwise.
Then,dζ is a controlled rectangular metric space, where ζ:M×M⟶1,∞ is a mapping defined as
(4)ζm1,m2=m1,ifm1,m2≥1,1,otherwise.
Clearly,M,dζ is not a controlled metric space if we take m1,m3≥1 and m2∈0,1. Then, dζm1,m3=1, dζm1,m2=1/m1, dζm2,m3=1/m3, ζm1,m2=1, and ζm2,m3=1. Here, the triangle inequality is not satisfied:
(5)dζm1,m3=1>ζm1,m2dζm1,m2+ζm2,m3dζm2,m3=1m1+1m3.Example 3.
LetM=1,2,3,4. Define dζ:M×M⟶o,∞ as
(6)dζ1,1=dζ2,2=dζ3,3=dζ4,4=0,dζ1,2=dζ2,1=dζ2,3=dζ3,2=dζ3,4=dζ4,3=dζ1,3=dζ3,1=80,dζ1,4=dζ4,1=1000,dζ2,4=dζ4,2=450.
Then,M,dζ is a controlled rectangular metric space with ζ:M×M⟶1,∞ defined as ζm1,m2=maxm1,m2+2, for all m1,m2∈M. However, M,dζ is not a rectangular metric space; for instance, notice
(7)dζ1,4=1000>dζ1,2+dζ2,3+dζ3,4=240.The concepts of convergence, Cauchyness, and completeness can simply be generalized in terms of controlled rectangular metric spaces.Definition 8.
LetM,dζ be a controlled rectangular metric space. Then,
(i)
A sequencemn in M,dζ is said to be convergent to m∈M, if limn⟶∞dζmn,m=0(ii)
A sequencemn in M,dζ is called a Cauchy sequence, if limn,r⟶∞dζmn,mr=0(iii)
M,dζ is called a complete controlled rectangular metric space, if every Cauchy sequence in M is convergent to some point of MDefinition 9.
LetM,dζ be a controlled rectangular metric space. Let m∈M and τ>0. Then,
(i)
The open ballBm,τ is define as(8)Bm,τ=m1∈M,dζm,m1<τ.(ii)
The mappingf:M⟶M is called continuous at m∈M, if for υ>0, there is ν>0 such that fBm,ν⊆Bfm,υ. Thus, if f is continuous at m, then for any sequence mn converging to m, we have limn⟶∞fmn=fmNote that a rectangularb-metric space is not continuous in general, and it is the same for controlled rectangular metric spaces.Lemma 10.
LetM,dζ be a controlled rectangular metric space and mn be a Cauchy sequence in M such that mn≠mr, whenever n≠r. If limn,r⟶∞ζmn,mr<∞ for all mn,mr∈M, then mn has a unique limit.Proof.
Suppose that a sequencemn in M has two limit points μ,ν∈M, that is, limn⟶∞mn=μ and limn⟶∞mn=ν. mn is a Cauchy sequence for mn≠mr, whenever n≠r. Hence, from condition (iii) of Definition 6, we have
(9)dζμ,ν≤ζμ,mndζμ,mn+ζmn,mrdζmn,mr+ζmr,νdζmr,ν⟶0asn,r⟶∞.
This implies that(10)dζμ,ν=0.
Hence,mn has a unique limit point in M.Definition 11.
LetM,dζ be a controlled rectangular metric space. Then,
(i)
For a mappingf:M⟶M, we define(11)Om,n=m,fm,⋯,fnm,Om,∞=m,fm,⋯,fnm,⋯,where m∈M and n∈ℕ. The Om,∞ is called an orbit of f.
(ii)
A mappingf:M⟶M is called f-orbitally continuous, if limk⟶∞fknm=m implies limk⟶∞ffknm=fm, where m∈M(iii)
A mappingf:M⟶M is called f-orbitally complete, if every Cauchy sequence in Om,∞ is convergent in MOur main result is similar to the Banach contraction principle in the setting of controlled rectangular metric spaces. Throughout this section, for a mappingf:M⟶M and m∈M, we consider an orbit Om,∞.Theorem 12.
Letf:M⟶M be a mapping on a controlled rectangular metric space M,dζ. Suppose that the following axioms hold:
(i)
For allm1,m2∈M, we have(12)dζfm1,fm2≤λdζm1,m2,where λ∈0,1.
(ii)
supq≥1limi→∞ζmi,mqζmi+1,mi+2/ζmi−1,miλ<1, for any mn∈M(iii)
M,dζ is f-orbitally complete(iv)
f is orbitally continuous(v)
For eachm∈M, limn⟶∞ςmn,μ and limn⟶∞ςμ,mn exist and are finite
Then,f has a unique fixed point in M.Proof.
Consider an arbitrary pointm0∈M, and define an iterative sequence mn over m0 as follows:
(13)m1=fm0,m2=fm1=ffm0=f2m0,⋯,mn=fnm0,⋯.
From equation (12), we have
(14)dζm1,m2=dζfm0,f2m0≤λdζm0,fm0=dζm0,m1.
Recursively, we have(15)dζmn,mn+1=dζfnm0,fn+1m0≤λdζfn−1m0,fnm0≤λ2dζfn−2m0,fn−1m0⋯≤λndζm0,m1.
That is,(16)dζmn,mn+1≤λndζm0,m1.
By taking limitn⟶∞, we get
(17)limn⟶∞dζmn,mn+1=0,(18)limn⟶∞dζmn+1,mn+2=0.Next, we show thatmn is a Cauchy sequence. For this, we will take the following two cases.Case 1.
Letρ be odd, that is, ρ=2r+1, where r≥1. Then, from condition (iii) of Definition 6 and equation (16) for n+ρ>n, we have
(19)dζmn,mn+2r+1≤ζmn,mn+1dζmn,mn+1+ζmn+1,mn+2dζmn+1,mn+2+ζmn+2,mn+2r+1×dζmn+2,mn+2r+1≤ζmn,mn+1λn+ζmn+1,mn+2λn+1dζm0,m1+ζmn+2,mn+2r+1dζmn+2,mn+2r+1≤ζmn,mn+1λn+ζmn+1,mn+2λn+1dζm0,m1+ζmn+2,mn+2r+1ζmn+2,mn+3×dζmn+2,mn+3+ζmn+2,mn+2r+1ζmn+3,mn+4dζmn+3,mn+4+ζmn+2,mn+2r+1×ζmn+4,mn+2r+1dζmn+4,mn+2r+1≤dζm0,m1ζmn,mn+1λn+ζmn+2,mn+3λn+1+dζm0,m1×ζmn+2,mn+2r+1ζmn+2,mn+3λn+2+ζmn+2,mn+2r+1ζmn+3,mn+4λn+3+⋯+dζm0,m1ζmn+2,mn+2r+1⋯ζmn+2r−2,mn+2n−1λn+2r−2+ζmn+2,mn+2r+1⋯ζmn+2r−1,mn+2rλn+2r−1+ζmn+2,mn+2r+1⋯ζmn+2r,mn+2r+1λn+2rdζm0,m1≤dζm0,m1ζmn,mn+1λn+ζmn+2,mn+3λn+1+∑i=1r−1∏j=1iζmn+2j,mn+2r+1×ζmn+2i,mn+2i+1λn+2idζm0,m1+∑i=1r−1∏j=1iζmn+2j+1,mn+2r+1ζmn+2i+1,mn+2i+2×λn+2i+1dζm0,m1+∏i=1r−1ζmn+2j,mn+2r+1λn+2rdζm0,m1.
As(20)∑i=1r−1∏j=1iζmn+2j,mn+2r+1ζmn+2i,mn+2i+1λn+2i≤∑i=1r−1∏j=1iζm2j,mn+2r+1ζm2i,mn+2i+1λ2i,(21)∑i=1r−1∏j=1iζmn+2j+1,mn+2r+1ζmn+2i+1,mn+2i+2λn+2i+1≤∑i=1r−1∏j=1iζm2j+1,mn+2r+1ζm2i+1,mn+2i+2λ2i+1,therefore, we obtain
(22)dζmn,mn+2r+1≤dζm0,m1ζmn,mn+1λn+ζmn+2,mn+3λn+1+∑i=1r−1∏j=1iζm2j,mn+2r+1×ζm2i,m2i+1λ2idζm0,m1+∑i=1r−1∏j=1iζm2j+1,mn+2r+1ζm2i+1,m2i+2×λ2i+1dζm0,m1+∏i=1r−1ζmn+2j,mn+2r+1λn+2rdζm0,m1.
Sincesupq≥1limi⟶∞ζmi,mqζmi+1,mi+2/ζmi−1,miλ<1, the series
(23)∑i=1∞∏j=1iζm2j,mn+2r+1ζm2i,m2i+1λ2i,∑i=1∞∏j=1iζm2j+1,mn+2r+1ζm2i+1,m2i+2λ2i+1,converge by the ratio test. Let
(24)S=∑i=1∞∏j=1iζm2j,mn+2r+1ζm2i,m2i+1λ2i,Sn=∑i=1n∏j=1iζm2j,mn+2r+1ζm2i,m2i+1λ2i,(25)S′=∑i=1∞∏j=1iζm2j+1,mn+2r+1ζm2i+1,m2i+2λ2i+1,Sn′=∑i=1n∏j=1iζm2j+1,mn+2r+1ζm2i+1,m2i+2λ2i+1.
Then, equation (22) takes the following form:
(26)dζmn,mn+2r+1≤dζm0,m1ζmn,mn+1λn+ζmn+2,mn+3λn+1+dζm0,m1Sr−1−Sn+1dζm0,m1+dζm0,m1Sr−1′−Sn+1′dζm0,m1+∏i=1r−1ζmn+2j,mn+2r+1λn+2rdζm0,m1.
By taking limitn⟶∞ in equation (26), we get
(27)limn⟶∞dζmn,mn+2r+1=0.Case 2.
Letρ be even, that is, ρ=2r, where r≥1. Then, from condition (iii) of Definition 6 and equation (16) for n+ρ>n, we have
(28)dζmn,mn+2r≤ζmn,mn+1dζmn,mn+1+ζmn+1,mn+2dζmn+1,mn+2+ζmn+2,mn+2r×dζmn+2,mn+2r≤ζmn,mn+1λn+ζmn+1,mn+2λn+1dζm0,m1+ζmn+2,mn+2rdζmn+2,mn+2r≤ζmn,mn+1λn+ζmn+1,mn+2λn+1dζm0,m1+ζmn+2,mn+2rζmn+2,mn+3×dζmn+2,mn+3+ζmn+2,mn+2rζmn+3,mn+4dζmn+3,mn+4+ζmn+2,mn+2r×ζmn+4,mn+2rdζmn+4,mn+2r≤dζm0,m1ζmn,mn+1λn+ζmn+2,mn+3λn+1+dζm0,m1×ζmn+2,mn+2rζmn+2,mn+3λn+2+ζmn+2,mn+2rζmn+3,mn+4λn+3+⋯+dζm0,m1ζmn+2,mn+2r⋯ζmn+2r−4,mn+2n−3λn+2r−4+ζmn+2,mn+2r⋯ζmn+2r−3,mn+2r−2λn+2r−3+ζmn+2,mn+2r⋯ζmn+2r−2,mn+2rλn+2r−2dζm0,m2≤dζm0,m1ζmn,mn+1λn+ζmn+2,mn+3λn+1+∑i=1r−2∏j=1iζmn+2j,mn+2r×ζmn+2i,mn+2i+1λn+2idζm0,m1+∑i=1r−2∏j=1iζmn+2j+1,mn+2rζmn+2i+1,mn+2i+2×λn+2i+1dζm0,m1+∏i=1r−1ζmn+2j,mn+2rλn+2r−2dζm0,m2.
As(29)∑i=1r−2∏j=1iζmn+2j,mn+2rζmn+2i,mn+2i+1λn+2i≤∑i=1r−2∏j=1iζm2j,mn+2rζm2i,m2i+1λ2i,(30)∑i=1r−2∏j=1iζmn+2j+1,mn+2r+1ζmn+2i+1,mn+2i+2λn+2i+1≤∑i=1r−2∏j=1iζm2j+1,mn+2r+1ζm2i+1,m2i+2λ2i+1,therefore, we obtain
(31)dζmn,mn+2r≤dζm0,m1ζmn,mn+1λn+ζmn+2,mn+3λn+1+∑i=1r−2∏j=1iζm2j,mn+2r×ζm2i,m2i+1λ2idζm0,m1+∑i=1r−2∏j=1iζm2j+1,mn+2rζm2i+1,m2i+2×λ2i+1dζm0,m1+∏i=1r−1ζmn+2j,mn+2rλn+2r−2dζm0,m2.
Sincesupq≥1limi⟶∞ζmi,mqζmi+1,mi+2/ζmi−1,miλ<1, the series
(32)∑i=1∞∏j=1iζm2j,mn+2rζm2i,m2i+1λ2i,∑i=1∞∏j=1iζm2j+1,mn+2rζm2i+1,m2i+2λ2i+1,converge by the ratio test. Let
(33)S=∑i=1∞∏j=1iζm2j,mn+2rζm2i,m2i+1λ2i,Sn=∑i=1n∏j=1iζm2j,mn+2rζm2i,m2i+1λ2i,(34)S′=∑i=1∞∏j=1iζm2j+1,mn+2rζm2i+1,m2i+2λ2i+1,Sn′=∑i=1n∏j=1iζm2j+1,mn+2rζm2i+1,m2i+2λ2i+1.
Then, equation (31) takes the following form:
(35)dζmn,mn+2r≤dζm0,m1ζmn,mn+1λn+ζmn+2,mn+3λn+1+dζm0,m1Sr−1−Sn+1+dζm0,m1Sr−1′−Sn+1′+∏i=1r−1ζmn+2j,mn+2rλn+2r−2dζm0,m1.
By taking limitn⟶∞ in equation (35), we get
(36)limn⟶∞dζmn,mn+2r=0.Hence, in both cases,limn⟶∞dζmn,mn+ρ=0, which shows that mn is a Cauchy sequence. As M is f-orbitally complete, so there exists μ∈M such that limn⟶∞mn=μ. Next, we show that μ is a fixed point of f. As f is orbitally continuous, so we have
(37)dζμ,fμ≤ζμ,mndζμ,mn+ζmn,mn+1dζmn,mn+1+ζmn+1,fμdζmn+1,fμ.Since for eachm∈M, limn⟶∞ζmn,m and limn⟶∞ζm,mn exist and are finite, so by taking limit n⟶∞ and using equation (17), we get
(38)limn⟶∞dζμ,fμ=0.Therefore,fμ=μ. Hence, μ is a fixed point of f. In view of Lemma 10, m is the unique fixed point of f.Example 4.
LetX=1,2. Define dζ:M×M⟶0,∞ by dζm1,m2=m1−m22. Then, M,dζ is a complete controlled rectangular metric space with ζ:M×M⟶1,∞ defined as ζm1,m2=3m1+2m2+5. Define a mapping f:M⟶M by
(39)fm=m4.
Clearly, all the axioms of Theorem12 are satisfied, and hence, m=0 is a fixed point of f.Corollary 13.
Letf:M⟶M be a mapping on a complete controlled rectangular metric space M,dζ. Suppose that the following axioms hold:
(i)
For allm1,m2∈M, we have(40)dζfm1,fm2≤λdζm1,m2,λ∈0,1.(ii)
supq≥1limi⟶∞ζmi,mqζmi+1,mi+2/ζmi−1,miλ<1, for any mn∈M(iii)
f is continuous
Then,f has a unique fixed point.Remark 14.
(i)
By puttingζm1,m2=s, for all m1,m2∈M in Theorem 12, we get Theorem 2.1 of George et al. [22](ii)
By puttingζm1,m2=1, for all m1,m2∈M in Theorem 12, we get the following corollary in view of Das and Dey [27]Corollary 15.
Letf:M⟶M be a mapping on a rectangular metric space M,dζ. Suppose that the following axioms hold:
(i)
For allm1,m2∈M, dζfm1,fm2≤λdζm1,m2, where λ∈0,1(ii)
M,dζ is f-orbitally complete(iii)
f is orbitally continuous
Then,f has a unique fixed point.Theorem 16.
Letf:M⟶M be a mapping on complete controlled rectangular metric space M,dζ, which satisfies the following axioms:
(i)
For allm1,m2∈M, we have(41)dζfm1,fm2≤λdζm1,fm1+dζm2,fm2,where λ∈0,1/2.
(ii)
supq≥1limi⟶∞ζmi,mqζmi+1,mi+2/ζmi−1,miλ<1, for any mn∈M, where λ≠1/ζm1,m2 for each m1,m2∈M(iii)
For eachm∈M, limn⟶∞ζmn,mn+1≤1, limn⟶∞ζmn,μ, and limn⟶∞ζμ,mn exist and are finite
Then,f has a unique fixed point in M.Proof.
Let us take an arbitrary elementm0∈M and choose m1=fm0 and m2=fm1. Then, from equation (41), we obtain
(42)dζm1,m2=dζfm0,fm1≤λdζm0,fm0+dζm1,fm1=λdζm0,m1+dζm1,m2.
This implies that(43)dζm1,m2≤λ1−λdζm0,m1,where ω=λ/1−λ<1, as λ<1/2. By recursively applying equation (41), we obtain
(44)dζmn,mn+1≤ωndζm0,m1.
Thus, by taking the limit in equation (44), we have
(45)limn→∞dζmn,mn+1=0.
Again from equation (41), we have
(46)dζmn,mn+2=dζfmn−1,fm1n+1≤λdζmn−1,fmn−1+dζmn+1,fmn+1=λdζmn−1,mn+dζmn+1,mn+2.
By using equation (45), we obtain
(47)limn→∞dζmn,mn+2=0.
Now, we will show thatmn is a Cauchy sequence. By following the same procedure as in the proof of Theorem 12 and using equations (45) and (47), we conclude that mn is a Cauchy sequence. As M is complete, so there exists μ∈M such that
(48)limn→∞dζmn,μ=0.
Next, we show thatμ is a fixed point of f. From condition (iii) of Definition 6, for any n∈ℕ, we have
(49)dζμ,fμ≤ζμ,mndζμ,mn+ζmn,mn+1dζmn,mn+1+ζmn+1,fμdζmn+1,fμ≤ζμ,mndζμ,mn+ζmn,mn+1dζmn,mn+1+ζmn+1,fμdζfmn,fμ≤ζμ,mndζμ,mn+ζmn,mn+1dζmn,mn+1+ζmn+1,fμλdζmn,fmn+dζμ,fμ.
λ≠1/ζm1,m2 for each m1,m2∈M. This implies that
(50)dζμ,fμ≤ζμ,mn1−λζmn+1,fμdζμ,mn+ζmn,mn+1+λζmn+1,fμ1−λζmn+1,fμdζmn,mn+1.
For eachm∈M, limn⟶∞ζmn,mn+1≤1, limn⟶∞ζmn,m, and limn⟶∞ζm,mn exist and are finite. Therefore, by taking limit n⟶∞ in equation (50) and using equations (47) and (48), we obtain
(51)dζμ,fμ=0,which implies that fμ=μ. For the uniqueness, let ν be another fixed point of f and μ≠ν. From equation (41), we obtain
(52)dζμ,ν=dζfμ,fν≤λdζμ,fμ+dζν,fν=λdζμ,μ+dζν,ν,where dζμ,μ=0 and dζν,ν=0. Hence, from the above inequality, we obtain dζμ,ν=0, that is, μ=ν, and μ is a unique fixed point of f.


## 4. Application

In this section, we will apply Corollary13 to prove the existence and uniqueness of a solution for the following Fredholm integral equation:
(53)mt=∫abτt,r,mrdr+vt,fort,r∈a,b,where v:a,b⟶ℝ and τ:a,b×a,b×ℝ⟶ℝ both are continuous functions. Let M=ℂa,b be the space of all continuous real-valued functions defined on the closed interval a,b. Consider
(54)dζm1,m2=supt∈a,bm1t−m2t2.Clearly,M,dζ is a complete controlled rectangular metric space with ζ:M×M⟶1,∞ defined as ζm1,m2=3m1t+2m2t+5. Next, we will prove our result as follows.Theorem 17.
For allm1,m2∈M and t,r∈a,b, the following condition holds:
(55)τt,r,m1r−τt,r,m2r≤12b−am1r−m2r.
Then, the integral equation (53) has a unique solution.Proof.
Definef:M⟶M by
(56)fmt=∫abτt,r,mrdr+vt,fort,r∈a,b,where v:a,b⟶ℝ and τ:a,b×a,b×ℝ⟶ℝ both are continuous functions. Clearly, γ is a fixed point of f, if and only if γ is a solution of the integral equation (53). For all m1,m2∈M, we have
(57)fm1t−fm2t2=∫abτt,r,m1r−τt,r,m2rdr2≤14b−a2sups∈a,bm1s−m2s2∫abds2≤14dζm1,m2.
It implies that(58)dζfm1,fm2≤14dζm1,m2,where λ=1/4∈0,1. Thus, all the conditions of Corollary 13 are satisfied. Hence, f has a unique fixed point; that is, the Fredholm integral equation (53) has a solution.



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*Source: 5564324-2021-04-30.xml*