# Kannan-Type Contractions on New Extendedb-Metric Spaces

Authors: Hassen Aydi; Muhammad Aslam; Dur-e-Shehwar Sagheer; Samina Batul; Rashid Ali; Eskandar Ameer
Journal: Journal of Function Spaces
 (2021)

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## Abstract

This article is focused on the generalization of some fixed point theorems with Kannan-type contractions in the setting of new extendedb-metric spaces. An idea of asymptotic regularity has been incorporated to achieve the new results. As an application, the existence of a solution of the Fredholm-type integral equation is presented.


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## Body

## 1. Introduction and Preliminaries

The existence of fixed points for some operators has a noteworthy contribution in many branches of applied and pure mathematics. The theory of a fixed point provides very valuable and effective tools in mathematics. It has a wide range of implications in nonlinear analysis and has been established in two directions. One is to change the space under consideration (see the works of Bakhtin [1], Jleli and Samet [2], Karapinar [3], Kamran et al. [4], etc.), and the other is to change the contraction conditions (see the works of Ćirić [5], Popescu [6], Rakotch [7], etc.).In 1922, the Polish mathematician Banach [8] established a remarkable result relevant to a metric fixed point theory, that is known as the Banach contraction principle (BCP). The work of Banach is well regarded and an adaptable consequence in the theory of fixed points. BCP laid a foundation of research in this field, which is further investigated by many researchers from 1922 till now. One of the prominent generalizations of the BCP was presented by Kannan [9]. Additional works on the existence of (common) fixed points can be seen in [10–14].In 1989, Bakhtin [1] introduced the notion of b-metric spaces. Later on, the concept of b-metric spaces was further used by Czerwik [15] to establish different fixed point results in b-metric spaces. The study of b-metric spaces endowed an imperative place in the fixed point theory with multiple aspects. Many mathematicians (Abdeljawad et al. [16, 17], Ali et al. [18], Akkouchi [19], Chifu and Karapinar [20], Kadelburg and Radenović [21], Parvaneh et al. [11], Gupta et al. [12], Mlaiki et al. [22], etc.) led the foundation to improve the fixed point theory in b-metric spaces. Another innovative task has been achieved by Kamran et al. [4] in 2017 by introducing the notion of an extended b-metric space, which generalizes the notion of a b-metric space. Some fixed point results are proved in this new setting; see for instance the work presented in [23–25].The work of Kannan [9] refined the concept of the Banach contraction mapping by introducing a new contraction, now known as Kannan contraction. The Kannan fixed point result has been further extended and generalized in the setup of b-metric spaces [15] and for generalized metric spaces [26].In 2019, the notion of a new extendedb-metric space has been initiated by Aydi et al. [27], where the control function depends on three variables. This fact is new since all precedent control functions depend on two variables. The objective of this work is at investigating Kannan-type contractions in the context of new extended b- metric spaces by extending the main results of Gornicki [28]. For this purpose, some basic concepts are needed in the sequel.Definition 1 (see [15]).
LetX be a nonempty set and s≥1 be a real number. A function db:X×X⟶ℝ is called a b-metric, if it satisfies the following for all j,κ,ℓ∈X:
(b1)
dbj,κ≥0(b2)
dbj,κ=0, if and only if j=κ(b3)
dbj,κ=dbκ,j(b4)
dbj,κ≤sdbj,ℓ+dbℓ,κThe pairX,db is called a b-metric space. If s=1, then, a b-metric space becomes a metric space. In 2017, Kamran et al. [4] generalized the b-metric space setting to an extended b-metric space (in the same direction, see also [29, 30]).Definition 2 (see [4]).
LetX be a nonempty set and θ:X×X⟶1,∞ be a function. The map dθ:X×X⟶ℝ is called an extended b-metric if for all j,κ,ℓ∈X, it satisfies the following axioms:
(i)
dθj,κ≥0(ii)
dθj,κ=0, if and only if j=κ(iii)
dθj,κ=dθκ,j(iv)
dθj,ℓ≤θj,ℓdθj,κ+dθκ,ℓIn 2019, Aydi et al. [27] introduced the notion of new extended b-metric spaces. Here, the control function depends on 3 variables.Definition 3 (see [27]).
LetX be a nonempty set and θ:X×X×X⟶1,∞ be a function. The map dθ:X×X⟶ℝ is called a new extended b-metric if for all j,κ,ℓ∈X, it satisfies the following axioms:
(i)
dθj,κ≥0(ii)
dθj,κ=0, if and only if j=κ(iii)
dθj,κ=dθκ,j(iv)
dθj,ℓ≤θj,κ,ℓdθj,κ+dθκ,ℓThe pairX,dθ is named to be a new extended b-metric space. If θj,κ,ℓ=s (for s≥1), we get Definition 1.Example 4 (see [27]).
LetX=ℕ. Define dθ:X×X⟶ℝ by
(1)dθj,κ=0,⟺j=κ,1j,ifjisevenandκisodd,1κ,ifjisoddandκiseven,1,otherwise,where
(2)θj,κ,ℓ=1,ifj=ℓandκisevenorodd,jℓj+ℓ,ifj≠ℓ,jandℓareevenandκodd,κ2,ifj≠ℓ,jandℓareoddandκeven,32,ifj≠ℓ,j,ℓandκareallevenorallodd,j+κ1+jj1+κ,ifj≠ℓ,jiseven,ℓisoddandκiseven,ℓ+κℓ+1ℓκ+1,ifj≠ℓ,jisodd,ℓisevenandκiseven,2+ℓ1+ℓ,ifj≠ℓ,jisodd,ℓisevenandκisodd,j+1j,ifj≠ℓ,jiseven,ℓisoddandκisodd.
Here,X,dθ is a new extended b-metric space.On the other hand, by takingj=2p+1,ℓ=4p+1, and κ=2p, we have
(3)dθj,ℓdθj,κ+dθκ,ℓ=dθ2p+1,4p+1dθ2p+1,2p+dθ2p,4p+1=p.It is not possible to finds≥1 so that b4 holds. Thus, dθ is not a b-metric on X.Example 5.
ConsiderX=1,2,3. Take dθ:X×X⟶0,∞ as
(4)dθj,κ=j−κ2,where θ:X×X×X⟶1,∞ as θj,κ,ℓ=j+κ+2ℓ. Then, X,dθ is a new extended b-metric space.Proof.
The first three conditions are trivially verified. To check the triangular inequality, we proceed as follows:(5)dθ1,2≤θ1,3,2dθ1,3+dθ3,2≤1+3+44+1,so 1<40. Similarly, we can check the other two pairs.
Therefore, for allj,κ,ℓ in X, dθj,ℓ≤θj,κ,ℓdθj,κ+dθκ,ℓ.Definition 6.
LetX,dθ be a new extended b-metric space.
(i)
A sequencean⊂X is called convergent to a∈X if for ε>0, there is Nε∈ℕ such that limn⟶∞dθan,a<ε for all n≥Nε(ii)
A sequencean⊂X is called Cauchy if for ε>0, there is Nε∈ℕ such that limn⟶∞dθan,am<ε for all n,m≥Nε(iii)
The space is called complete if every Cauchy sequence inX is convergent in XDefinition 7.
LetX,dθ be a new extended b-metric space. Denote by Ba,r=b∈X:dθb,a<r and Ba,r=b∈X:dθb,a≤r the open and closed balls in X, respectively.
(i)
A subsetU of X is called open if for any u∈U, there exists an ε>0 such that Bu,ε⊂U(ii)
A subsetV of X is called closed if for any vn⊂V such that limn⟶∞vn=v, then v∈X
In this paper, we are going to prove some Kannan-type fixed point theorems in the setting of new extendedb-metric spaces. Some examples are also provided to make effective the obtained results.


## 2. Main Results

We define a Kannan-type fixed point contraction on new extendedb-metric spaces.Definition 8.
LetX,dθ be a new extended b-metric space. A mapping T:X⟶X is a Kannan-type contraction if there are K∈0,1/2 and 0≤L<1 such that
(6)dθTa,Tb≤Kdθa,Ta+dθb,Tb+Ldθb,Ta,foralla,b∈X.Our first main result is as follows:Theorem 9.
Consider a complete new extendedb-metric space X,dθ such that dθ is a continuous functional. Let T:X⟶X be a mapping such that there are ξ∈0,1/2 and L∈0,1 so that
(7)dθTa,Tb≤ξdθa,Ta+dθb,Tb+Ldb,Ta,∀a,b∈X.
Assume that(8)supm≥1limn⟶∞θan,an+1,am<1−ξξ,for each a0∈X, such that an=Tna0, n=1,2,⋯.
Then,T has only one fixed point v∈X. Further, for any a∈X, the iterative sequence Tna converges to v, and
(9)dθTan,v≤ξ1−Lξ1−ξndθTa0,a0,n=0,1,2,⋯.Proof.
For an arbitrary pointa0∈X, construct the iterative sequence
(10)an+1=Tan=Tn+1a0.
If for somen, an=an+1, so an is a fixed point of T. Otherwise, assume that an≠an+1 for all n≥0. Since
(11)dθan,an+1=dθTan−1,Tan,one writes
(12)dθTan−1,Tan≤ξdθan−1,Tan−1+dθan,Tan+Ldθan,Tan−1.
Then,(13)dθan,an+1≤ξdθan−1,an+dθan,an+1+Ldθan,an.
That is,(14)dθan,an+1≤ξ1−ξdθan−1,an.
Continuing in this way, we have(15)dθan,an+1≤ξ1−ξndθa0,a1,(16)dθTan−1,Tan≤ξ1−ξndθa0,Ta0.
Letm,n∈ℕ be such that m>n. Applying triangular inequality, we get
(17)dθan,am≤θan,an+1,amdθan,an+1+dθan+1,am=θan,an+1,amdθan,an+1+θan,an+1,amdθan+1,am≤θan,an+1,amdθan,an+1+θan,an+1,amθan+1,an+2,amdθan+1,an+2+dθan+2,am≤θan,an+1,amdθan,an+1+θan,an+1,amθan+1,an+2,amdθan+1,an+2+⋯+θan,an+1,amθan+1,an+2,amθan+2,an+3,am⋯θam−2,am−1,amdθam−1,am.
Since(18)dθan,an+1≤ξ1−ξndθa0,a1n=0,1,2,⋯,one writes
(19)dθan,am≤θan,an+1,amξ1−ξndθa0,a1+θan,an+1,amθan+1,an+2,amξ1−ξn+1dθa0,a1+⋯+θan,an+1,amθan+1,an+2,amθan+2,an+3,am⋯θam−2,am−1,amξ1−ξm−1dθa0,a1≤θa1,a2,amθa2,a3,am⋯θan−1,an,amθan,an+1,amξ1−ξn+θa1,a2,amθa2,a3,am⋯θan,an+1,amθan+1,an+2,amξ1−ξn+1+⋯+θa1,a2,amθa2,a3,am⋯θan,an+1,amθan+1,an+2,am⋯θam−3,am−2,amθam−2,am−1,amξ1−ξm−1dθa0,a1.
Since(20)supm≥1limn,m⟶∞θan,an+1,amξ1−ξ<1,in view of (8), the series ∑n=1∞ξ/1−ξn∏i=1nθai,ai+1,am is convergent for each m∈ℕ by ratio test.
LetS=∑n=1∞ξ/1−ξn∏i=1nθai,ai+1,am and
(21)Sn=∑j=1nξ1−ξj∏i=1jθai,ai+1,am.
So, form>n, the above inequality implies that
(22)dθan,am≤dθa0,a1Sm−1−Sn−1.
That is,(23)limn⟶∞dθan,am=0.
Hence, the sequencean is a Cauchy sequence. By the completeness of X, there is v∈X such that limn⟶∞an=v.
We claim thatv is a fixed point of T. We have
(24)dθTan,Tv≤ξdθan,Tan+dθv,Tv+Ldθv,Tan.
That is,(25)dθan+1,Tv≤ξdθan,an+1+dθv,Tv+Ldθv,an+1.
Asn⟶∞, we have in view of the assumption that dθ is continuous,
(26)dθv,Tv≤ξdθv,Tv,which holds unless dθv,Tv=0, and so, Tv=v.
The uniqueness is as follows:
Letτ be another fixed point of T. We have
(27)0≤dθv,τ=dθTv,Tτ≤ξdθv,Tv+dθτ,Tτ+Ldθτ,Tv.
That is,(28)dθv,τ≤Ldθv,τ.
It is only possible ifdθv,τ=0. Thus, v∈X is the unique fixed point of T. Further, we have
(29)dθTan−1,Tan≤ξdθTan−2,Tan−1+dθTan−1,Tan+Ldθan,an.
Then,(30)dθTan−1,Tan≤ξ1−ξdθTan−2,Tan−1.
Also,(31)dθTan,v≤ξdθTan−1,Tan+dθv,Tv+Ldθv,Tan≤ξdθTan−1,Tan+Ldθv,Tan.
Using (15),
(32)dθTan,v≤ξξ1−ξndθTa0,a0+Ldθv,Tan.
That is,(33)dθTan,v≤ξ1−Lξ1−ξndθTa0,a0,n=0,1,2,⋯.The following examples illustrate Theorem9. We deal with noncompact sets.Example 10.
LetX=ℓ∞ be the space of all bounded sequences of real numbers, that is,
(34)ℓ∞=η=ηn:ηn≤Cη,∀n∈ℕ,where Cη∈ℝ may depend on the sequence η but does not depend on n. Take that
(35)dθη,ζ=supn∈ℕηn−ζn2,η=ηn,ζ=ζn,
are inX.
Then,X is a complete new extended b-metric space with θ:X×X×X⟶1,∞ being defined by
(36)θη,ζ,φ=supn∈ℕηn+ζn+φnηn+ζn+φn+1+3.
ConsiderT:X⟶X given as Tη=ηn−1/5,∀n=1,2,3,⋯. For each η,ζ∈X, we have
(37)dθTη,Tζ=supn∈ℕηn−15−ζn−152=supn∈ℕηn5−ζn52≤supn∈ℕ2ηn52+supn∈ℕ2ζn52=18supn∈ℕ16ηn52+18supn∈ℕ16ζn52=18supn∈ℕ4ηn52+18supn∈ℕ4ζn52≤18supn∈ℕ4ηn+152+18supn∈ℕ4ζn+152=18dθη,Tη+dθζ,Tζ.
Thus, (7) holds with ξ=1/8 and L∈0,1. Also, 1−ξ/ξ=7 and θη,ζ,φ<7 for all η,ζ,φ∈X. Hence, (8) holds, and by Theorem 9, T has a fixed point.Example 11.
LetX=Ca,b be the set of all real-valued continuous functions defined on a,b. Define
(38)dθx,y=supt∈a,bxt−yt2,∀x,y∈Ca,b,and θ:X×X×X⟶1,∞ as
(39)θx,y,z=supt∈a,bxt+yt+ztxt+yt+zt+1+2.Then, X is a complete new extended b-metric space, consider a mapping T:X⟶X given as
(40)Tηt=ηt−17,∀η∈Ca,b.
For allη,ζ∈X, we have
(41)dθTη,Tζ=supt∈a,bηt−17−ζt−172=supt∈a,bηt7−ζt72≤supt∈a,b2ηt72+supt∈a,b2ζt72=118supt∈a,b36ηt72+118supt∈a,b36ζt72=118supt∈a,b6ηt72+118supn∈ℕ6ζt72≤118supt∈a,b6ηt+152+118supt∈a,b6ζt+152<16dθη,Tη+dθζ,Tζ.
Thus, (7) holds with ξ=1/6 and L∈0,1. Also, 1−ξ/ξ=5 and θη,ζ,φ<5 for all η,ζ,φ∈X. That is, (8) holds. Since all the conditions of Theorem 9 are satisfied, T has a fixed point.Example 12.
ChooseX=1/4,1/16,⋯,1/4n,⋯∪0,1. Define θ:X×X×X⟶1,∞ and dθ:X×X⟶0,∞ by
(42)θx,y,z=x+y+z+2,dθx,y=x−y2.
LetT:X⟶X be given as
(43)Tu=14n+1,ifu=14n,n=0,1,2,3,⋯,u,ifu=0.
Then, for allx,y∈X with neither x=0 nor y=0, we have
(44)dθTx,Ty=14n+1−14m+12≤214n+12+214m+12=2934n+12+2934m+12=29dθx,Tx+dθy,Ty.
Ifx=0 and y≠0, then
(45)dθTx,Ty=0−14m+12=14m+12=2934m+12=29dθx,Tx+dθy,Ty.
Thus, (7) is satisfied for ξ=2/9 and for each L∈0,1. Also, 1−ξ/ξ=7/2. If a0=0, then, for the iterative sequence an=Tna0=0 for each n∈ℕ, we have limn,m⟶∞θan,an+1,am=limn,m⟶∞an+an+1+am+2<7/2. If a0≠0 (say a0=1/4k for some k∈0,1,2,⋯); then, for the iterative sequence an=Tna0=1/4k+n for each n∈ℕ, we have limn,m⟶∞θan,an+1,am=limn,m⟶∞an+an+1+am+2<7/2. Hence, Theorem 9 ensures the existence of a fixed point of T.Remark 13.
In the following, we ensure the completeness of the spaces given in precedent examples.(a)
Completeness ofℓ∞
LetX=ℓ∞ and let xm=ξ1m,ξ2m,ξ3m,⋯ be a Cauchy sequence in X. Define a metric on X as dθx,y=supjξj−ηj2, where x=ξj and y=ηj. Since xm is a Cauchy sequence, for ε>0, there is N∈ℕ such that for all m,n≥N,
(46)dθxn,xm=supjξjm−ξjn2<ε∀m,n≥N⟹ξjm−ξjn2<ε,∀jandm,n≥N.
That is,(47)ξjm−ξjn<ε1,∀jandm,n≥N,where ε1=ε is arbitrary. Hence, for every fixed j,ξj1,ξj2,⋯ is a Cauchy sequence of complex numbers and it converges, so ξjm⟶ξj as m⟶∞. Construct a sequence x=ξ1,ξ2,⋯ by using these infinitely many limits to show that x∈ℓ∞ and xm⟶x. From (48) with n⟶∞, we have
(48)ξjm−ξj<ε1,m≥N.
Sincexm=ξjm∈ℓ∞, there is a real number cm so that ξjm≤cm,∀j. Hence,
(49)ξj2=ξj−ξjm+ξjm2≤ξj−ξjm+ξjm2≤2ξj−ξjm2+ξjm2.
That is,(50)ξj≤2ε+2cm2.
So, (51) holds for each j. It implies that ξj is a bounded sequence of complex numbers. This leads to ξj∈ℓ∞. Also, from (48), we have
(51)dθxm,x=supjξjm−ξj2≤ε,m>N.
This implies thatxm⟶x∈ℓ∞ (ℓ∞ is endowed with the new extended metric dθ).
(b)
Completeness ofCa,b
LetX be the function space Ca,b, where a,b is any closed interval in ℝ. Define
(52)dθx,y=supt∈a,bxt−yt2,∀x,y∈Ca,b.
Letxm be a Cauchy sequence in Ca,b. Then, for each ε>0, there exists an N∈ℕ such that
(53)dθxm,xn=supt∈a,bxmt−xnt2<ε,∀m,n≥N.
Hence, for each fixedt0∈a,b, we have
(54)xmt0−xnt02<ε,∀m,n≥N.
That is,(55)xmt0−xnt0<ε=ε1,say∀m,n≥N.
This shows thatx1t0,x2t0,⋯ is a Cauchy sequence of real numbers; hence, it converges. That is, xmt0⟶xt0 as m⟶∞. In this way, we can associate to each t∈a,b a unique real number xt. This defines a function x (pointwise) in a,b. Further, we need to show that xm⟶x∈Ca,b. From (53), we have as n⟶∞(56)supt∈a,bxmt−xt2<ε,m≥N.
Hence, for eacht∈a,b and m>N, we have
(57)xmt−xt2<ε,which implies that xmt−xt<ε1. Thus, xmt converges uniformly to xt on the interval a,b. Since t⟶xmt is continuous and the convergence is uniform; hence, xt is continuous. This leads to the completeness of Ca,b with the new extended metric dθ.Theorem 14.
LetX,dθ be a complete new extended b-metric space where dθ is a continuous functional and M is a nonempty closed subset of X. Let T:M⟶M satisfy
(58)dθTa,Tb≤ξdθa,Ta+dθb,Tb+Ldθb,Ta,∀a,b∈MandL,ξ∈0,1.
Assume that there existα,β∈ℝ with 0<α<1 and β>0 such that for an arbitrary a∈M, there is μ∗∈M verifying
(59)dθμ∗,Tμ∗≤αdθa,Ta,dθμ∗,a≤βdθa,Ta.
Also, for an arbitrarya0∈M, assume that the sequence an=Tna0 verifies
(60)supm≥1limn⟶∞θan,an+1,am<1α.
Then,T has only one fixed point.Proof.
Leta0 be an arbitrary element of M. Consider the sequence an=Tna0 in M. We have
(61)dθTan+1,an+1≤αdθTan,an,dθan+1,an≤βdθTan,an,n=0,1,2,⋯.
Since(62)dθan+1,an=dθTan,an≤βdθTan,an,n=0,1,2,⋯,βdθTan,an≤βαdθTan−1,an−1≤βα2dθTan−2,an−2≤⋯≤βαndθTa0,a0,we have
(63)dθan+1,an≤βαndθTa0,a0.
Letm,n∈ℕ such that m>n. Applying triangular inequality, we get
(64)dθan,am≤θan,an+1,amdθan,an+1+dθan+1,am=θan,an+1,amdθan,an+1+θan,an+1,amdθan+1,am≤θan,an+1,amdθan,an+1+θan,an+1,amθan+1,an+2,amdθan+1,an+2+dθan+2,am≤θan,an+1,amdθan,an+1+θan,an+1,amθan+1,an+2,amdθan+1,an+2+⋯+θan,an+1,amθan+1,an+2,amθan+2,an+3,am⋯θam−2,am−1,amdθam−1,am.
Using (63), we get
(65)dθan,am≤θan,an+1,amβαndθTa0,a0+θan,an+1,amθan+1,an+2,amβαn+1dθTa0,a0+⋯+θan,an+1,amθan+1,an+2,amθan+2,an+3,am⋯θam−2,am−1,amβαm−1dθTa0,a0≤θa1,a2,amθa2,a3,am⋯θan−1,an,amθan,an+1amαn+θa1,a2,amθa2,a3,am⋯θan,an+1,amθan+1,an+2,amαn+1+⋯+θa1,a2,amθa2,a3,am⋯θan,an+1,amθan+1,an+2am⋯θam−2,am−1,amαm−1βdθTa0,a0.
Sincesupm≥1limn,m⟶∞θan,an+1,amα<1, the series ∑n=1∞αn∏i=1nθai,ai+1,am is convergent for each m∈ℕ by ratio test.
Let(66)S=∑n=1∞αn∏i=1nθai,ai+1,am,Sn=∑j=1nαj∏i=1jθai,ai+1,am.
So, form>n, the above inequality implies that
(67)dθan,am≤dθa0,a1Sm−1−Sn−1.
Lettingn⟶∞, the sequence an is a Cauchy sequence. By the completeness of M, there is v∈M such that limn⟶∞an=v.
We will prove thatv is a fixed point of T. By using (58), we get
(68)dθTan,Tv≤ξdθan,Tan+dθv,Tv+Ldθv,Tan⟹dθan+1,Tv≤ξdθan,an+1+dθv,Tv+Ldθv,an+1.
Asn⟶∞, we have
(69)dθv,Tv≤ξdθv,Tv.
Hence,Tv=v.
The uniqueness is as follows:
Assume on contrary that there isτ≠v∈M so that Tτ=τ; then,
(70)0<dθv,τ=dθTv,Tτ≤ξdθv,Tv+dθτ,Tτ+Ldθτ,Tv=Ldθτ,v.
That is,dθv,τ<Ldθv,τ, which is a contradiction. Thus, v∈X is the unique fixed point of T.Remark 15.
To prove Theorem9 in new extended b-metric spaces, by using the following conditions:
(71)dθμ∗,Tμ∗≤αdθa,Ta,dθμ∗,a≤βdθa,Ta,we proceed as follows:
For anya∈X, take μ∗=Ta. Then,
(72)dθμ∗,Tμ∗=dθTa,Tμ∗≤ξdθa,Ta+dθμ∗,Tμ∗+Ldθμ∗,Ta1−ξdθμ∗,Tμ∗≤ξdθa,Tadθμ∗,Tμ∗≤ξ1−ξdθa,Ta,
where by assumptionξ/1−ξ<1 and dθμ∗,a=dθTa,a. Now, for arbitrary a0∈X, we can inductively define a sequence an+1=Tan. By Theorem 14, this sequence is convergent. So, limn⟶∞an=v. Thus, Tv=v.Also, for eacha∈X,
(73)dθTan−1,Tan≤ξdθTan−2,Tan−1+dθTan−1,Tan+Ldθan,Tan−1⟹dθTan−1,Tan≤ξ1−ξdθTan−2,Tan−1,dθTan,v≤ξdθTan−1,Tan+dθv,Tv+Ldθv,Tan≤ξdθTan−1,Tan+Ldθv,Tan⟹dθTan,v≤ξ1−Lξ1−ξndθTa,a,n=0,1,2,⋯.Theorem 16.
LetX,dθ be a complete new extended b-metric space such that dθ is a continuous functional. Let T:X⟶X satisfy
(74)dθTa,Tb≤αdθa,Ta+βdθb,Tb+γdθa,b+Ldθa,Tb,∀a,b∈X,
whereα,β,γ,L are nonnegative real numbers such that α+β+γ+L<1 and β+γ>0. Assume that for an arbitrary a0∈M, we have
(75)supm≥1limn⟶∞θan,an+1,am<1p,
wherep=β+γ/1−α and an=Tna0.
Then,T has only one fixed point.Proof.
For an arbitrarya0∈X, take the sequence Tna0. Substituting a=Tn−1a0=Tan−1=an and b=Tn−2a0=Tan−2=an−1 in (74), we obtain
(76)dθTan,Tan−1≤αdθTan−1,Tan+βdθTan−2,Tan−1+γdθTan−1,Tan−2+LdθTan−1,Tan−1.
That is,(77)1−αdθTan,Tan−1≤β+γdθTan−1,Tan−2.
Thus,(78)dθTan,Tan−1≤β+γ1−αdθTan−1,Tan−2.
Moreover,(79)dθTan,Tan−1≤pdθTan−1,Tan−2≤p2dθTan−2,Tan−3⋮≤pn−1dθTa1,Ta0,∀n>1.
Thus, we reach(80)dθTan,Tan−1≤pndθa0,a1,∀n∈ℕ.
By assumption on the parametersα, β, and γ, one has p=β+γ/1−α<1.
Following the same steps as given in Theorem14, one can show that an is a Cauchy sequence. By the completeness of X, there is v∈X such that Tna0⟶v. To prove Tv=v, replace a=Tna0 and b=v in (74). We have
(81)dθTn+1a0,Tv≤αdθTna0,Tn+1a0+βdθv,Tv+γdθTna0,v+LdθTna0,Tv.
Then,(82)dθan+1,Tv≤αdθTna0,Tn+1a0+βdθv,Tv+γdθan,v+LdθTv,an+1.
That is,(83)limn⟶∞dθan+1,Tv≤limn⟶∞αdθTna0,Tn+1a0+βdθv,Tv+γdθan,v+LdθTv,an+1.
We havedθv,Tv≤β+Ldθv,Tv, which holds unless Tv=v.
The uniqueness is as follows:
Letτ and v be two fixed points, such that τ≠v. Then, using inequality (74), we get
(84)dθTτ,Tv≤αdθτ,Tτ+βdθv,Tv+γdθτ,v+LdθTv,τdθτ,v≤L+γdθτ,v1≤L+γ,which is a contradiction. Hence, T has only one fixed point.Now, we use the concept of an asymptotically regular mapping [31, 32] in new extended b-metric spaces.Definition 17.
LetX,dθ be a new extended b-metric space. A mapping T:X⟶X satisfying the condition
(85)limn⟶∞dθTn+1a,Tna=0,foralla∈X,is called asymptotically regular.Example 18.
LetX=0∪1,3. Define T:X⟶X by T0=1 and Tx=0,for0<x≤3. Consider dθx,y=x−y2 and θx,y,z=x+y+z+11/x+y+z+1. We claim that T satisfies condition (7). Indeed,Case 1.
Ifx=y=0, then, (7) gives 0≤2ξ+L, which is true for all ξ∈0,1/2 and L∈0,1.Case 2.
Ifx∈1,3,y=0, then, (7) gives 1≤ξ1+x2, which is true for all ξ∈0,1/2 and L∈0,1.Case 3.
Ifx,y∈1,3, then, (7) implies that 0≤ξx2+y2+Ly2, which is true for all ξ∈0,1/2 and L∈0,1. Notice that T is fixed point free. The iterative sequence xn=Tn0 is not convergent, so T is not asymptotically regular.Theorem 19.
LetX,dθ be a complete new extended b-metric space such that dθ is a continuous functional. Let T:X⟶X be an asymptotically regular self mapping such that there is ξ<1 so that
(86)dθTa,Tb≤ξdθa,Ta+dθb,Tb,∀a,b∈X.
Then,T has only one fixed point v∈X.Proof.
Leta∈X and take an=Tna be defined inductively. Let m,n∈ℕ such that m>n; then, according to asymptotic regularity,
(87)dθTn+1a,Tm+1a≤ξdθTna,Tn+1a+dθTma,Tm+1a⟶0,asn⟶∞.
Thus, the sequenceTna is a Cauchy sequence. By the completeness of X, there is v∈X such that
(88)limn⟶∞Tna=v.
To prove thatv is a fixed point of T, we proceed as follows:
(89)dθTan,Tv≤ξdθan,Tan+dθv,Tv.
That is,(90)dθTan,Tv≤ξdθan,an+1+dθv,Tv.
Taking limitn⟶∞ and using the asymptotic regularity of T, one obtains
(91)dθv,Tv≤ξdθv,Tv,which implies that Tv=v.
To prove uniqueness, letτ be another fixed point of T. We have
(92)dθv,τ=dθTv,Tτ≤ξdθv,Tv+dθτ,Tτ.
This is true unlessdθv,τ=0, and so, v=τ. Hence, v is the only fixed point of T. Further, for each a∈X, the iterative sequence Tna converges to v.Remark 20.
It is noteworthy that if the mapping is asymptotically regular, then, the condition onθan,an+1,am can be relaxed.Theorem 21.
LetX,dθ be a complete new extended b-metric space such that dθ is a continuous functional. Let T:X⟶X be an asymptotically regular mapping such that there is 0<S<1 so that
(93)dθTa,Tb≤Sdθa,Ta+dθb,Tb+dθa,b,∀a,b∈X.
Then,T has only one fixed point v∈X provided that
(94)limn⟶∞S+Sθan,an+1,amθan+1,am+1,am1−Sθan,an+1,amθan+1,am+1,amexists for an=Tna,m>n and a is arbitrary in X.Proof.
Leta∈X and take an=Tna defined inductively. Let m,n∈ℕ such that m>n; then, using (93), we have
(95)dθTn+1a,Tm+1a≤SdθTna,Tn+1a+dθTma,Tm+1a+dθTna,Tma=SdθTna,Tn+1a+dθTma,Tm+1a+SdθTna,Tma≤SdθTna,Tn+1a+dθTma,Tm+1a+Sθan,an+1,amdθTna,Tn+1a+dθTn+1a,Tma=SdθTna,Tn+1a+dθTma,Tm+1a+Sθan,an+1,amdθTna,Tn+1a+Sθan,an+1,amdθTn+1a,Tma≤SdθTna,Tn+1a+dθTma,Tm+1a+Sθan,an+1,amdθTna,Tn+1a+Sθan,an+1,amθan+1,am+1,amdθTn+1a,Tm+1a+dθTm+1a,TmadθTn+1a,Tm+1a≤S+Sθan,an+1,am1−Sθan,an+1,amθan+1,am+1,amdθTna,Tn+1a+S+Sθan,an+1,amθan+1,am+1,am1−Sθan,an+1,amθan+1,am+1,amdθTm+1a,Tma⟶0,asn⟶∞.
Thus, the sequenceTna is a Cauchy sequence. By the completeness of X, there is v∈X such that
(96)limn⟶∞Tna=v.
Now, by using triangular inequality and (93), we get
(97)dθTan,Tv≤Sdθan,Tan+dθv,Tv+dθan,v,so
(98)dθan+1,Tv≤Sdθan,an+1+dθv,Tv+dθan,v.
At the limit,(99)limn⟶∞dθan+1,Tv≤limn⟶∞Sdθan,an+1+dθv,Tv+dθan,v.
Thus,dθv,Tv≤Sdθv,Tv, which is possible only if Tv=v.
The uniqueness is as follows:
Suppose that there isτ≠v∈M so that Tτ=τ, then
(100)dθTτ,Tv≤Sdθτ,Tτ+dθv,Tv+dθτ,v1≤S,
which a contradiction. Hence,T has only one fixed point. Thus, for each a∈X, Tna converges to v.


## 3. Application

LetX=Ca,b be the set of all real-valued continuous functions on a,b, and let dθ:X×X⟶0,∞ be defined as
(101)dθx,y=supt∈a,bxt−yt2,withθx,y,z=supt∈a,bxt+yt+ztxt+yt+zt+1+2.One can easily verify thatX is a complete new extended b-metric space. Consider the Fredholm integral equation
(102)xt=∫abKt,τ,xτdτ+ft,forallt∈a,b,where f:a,b⟶ℝ and K:a,b×a,b×ℝ⟶ℝ are continuous.Theorem 22.
LetX=Ca,b and let the operator I:X⟶X be defined by
(103)Ixt=∫abKt,τ,xτdτ+ft,forallt∈a,b.
Assume that the following condition holds for eachx,y∈X(104)Kt,τ,xτ−Kt,τ,yτ2≤ξ2b−axτ−Ixτ+yτ−Iyτ2,
for allt,τ∈a,b, where ξ∈0,1/2. Then, the integral equation (102) has a solution, provided that for every iterative sequence xn=Inx0, for each x0∈X, we have
(105)supm≥1limn⟶∞θxn,xn+1,xm<1−ξξ.Proof.
It is required to prove that the operatorI satisfies the conditions of Theorem 9. For this, we will use the following inequality for β>1:
(106)a+b2β≤aβ+bβ2.
Forx,y∈X, consider
(107)supt∈a,bIxt−Iyt2=∫abKt,τ,xτdτ+ft−∫abKt,τ,yτdτ−ft2=∫abKt,τ,xτ−Kt,τ,yτdτ2≤∫abKt,τ,xτ−Kt,τ,yτ2dτ≤∫abξ2b−axτ−Ixτ+yτ−Iyτ2dτ≤ξ2b−a∫ab2xτ−Ixτ2+yτ−Iyτ2dτ≤ξsupt∈a,bxt−Ixt2+supt∈a,byt−Iyt2.
That is,(108)dθIx,Iy≤ξdθx,Tx+dθy,Ty,∀x,y∈X.
This implies that (7) holds for L=0. Hence, by Theorem 9, the operator I has a fixed point, provided that for every iterative sequence xn=Inx0, for each x0∈X, we have supm≥1limn⟶∞θxn,xn+1,xm<1−ξ/ξ, that is, the Fredholm integral equation (102) has a solution.


## 4. Conclusion

(i)
The idea of new extendedb-metric spaces was elaborated with examples(ii)
Some results involving Kannan-type contractions on new extendedb-metric spaces are provided(iii)
Results presented by Gornicki [28] are generalized and modified



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