# Some Fixed Point Results in Partial Rectangular Metric-Like Spaces

Authors: Eriola Sila; Sidite Duraj; Elida Hoxha
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we introduce a new concept of partial rectangular metric-like space and prove some results on the existence and uniqueness of a fixed point of a functionT:X⟶X, defined on a partial rectangular metric-like space X, which fulfills a nonlinear contractive condition using a comparison function and the diameter of the orbits. The obtained results generalize some previously acknowledged results in partial metric spaces, partial rectangular metric spaces, and rectangular metric-like spaces. The examples presented prove the usefulness of the introduced generalizations.


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## Body

## 1. Introduction

Fixed point theory has a wide application in functional analysis. Various authors [1–5] have contributed to this field expanding the metric spaces by changing their axioms or by improving the contractive conditions.Partial metric spaces were introduced by Mathew [6] in 1994. He studied their topologies and presented a result relying on the fixed points in these spaces related to Banach contraction. These results had been generalized subsequently by many other studies [7–11].In 2000, Branciari [12] introduced a new metric space defined as ‘rectangular metric space,’ replacing the triangle inequality in metric spaces with a new inequality called ‘the rectangular inequality.’ In his work, Branciari [12] studied fixed points for several types of functions satisfying certain contraction conditions. Many articles [13–16] were published in rectangular metric spaces inspired by this work, where the importance of these spaces and the applications in them are highlighted.Later in 2014, Shukla [17] devised the ‘partial rectangular metric spaces.’ He associated his work with applications of fixed point theory to the contractions of Banach, Kannan, Reich, Hardy-Rogers, and Chatterjea, respectively [18].In 2012, Amini-Harandi [19] defined the metric-like spaces and found some interesting results of fixed point theory in these spaces using the concepts given by Branciari [12, 19], on rectangular metric spaces and metric-like spaces respectively, Mlaiki et al. [20] in their work proposed a generalization of them, called “rectangular metric-like space” associated with respective theorems on the fixed points’ existence on them.Several distinguished results of fixed point theory on contractions including the diameter of the orbits had been extended on generalized metric spaces [21–23].Continuing in this way and trying to find an essential connection on these ideas, we propose the concept of partial rectangular metric-like spaces. As a first attempt, some results on the existence and uniqueness of fixed points in these new spaces using the diameter of orbits and comparison functions are displayed. Furthermore, some presented examples show that our results really generalize the previously known concepts and results in partial rectangular space and rectangular metric-like space.


## 2. Preliminaries

LetX be a nonempty set.Definition 1.
(see [6]). The function p:X×X⟶R+ is a partial metric on X if for every x,y,z∈X, it satisfies the following conditions:(i)
p1x=y if and only if px,y=px,x=py,y(ii)
p2px,x≤px,y(iii)
p3px,y=py,x(iv)
p4px,y≤px,z+pz,y−pz,z
The pairX,p is called the partial metric space.Definition 2.
(see [12]). A function d:X×X⟶R+ is a rectangular metric on X if for every x,y∈X and for all distinct points u,v∈X−x,y, it satisfies the following conditions:(i)
R1x=y if and only if dx,y=0(ii)
R2dx,y=dy,x(iii)
R3dx,y≤dx,u+du,v+pv,y
The pairX,d is called the rectangular metric space.Definition 3.
(see [17]). A function ρ:X×X⟶R satisfying the conditions(i)
ρ1ρx,y≥0 for all x,y∈X(ii)
ρ2, x=y if and only if ρx,y=ρx,x=ρy,y for all x,y∈X(iii)
ρ3ρx,x≤ρx,y for all x,y∈X(iv)
ρ4ρx,y=ρy,x for all x,y∈X(v)
ρ5ρx,y≤ρx,w+ρw,z+ρz,y−ρw,w−ρz,z,for all x,y∈X and for all distinct points w,z∈X−x,yis called a partial rectangular metric on X. The pair X,pis called a partial rectangular metric space.Definition 4.
(see [19]). A function σ:X×X⟶R+ is called a metric-like on X if for each x,y,z∈X, it satisfies the following conditions:(i)
σ1 If σx,y=0 then x=y(ii)
σ2σx,y=σy,x(iii)
σ3σx,y≤σx,z+σz,y
The pairX,σ is called metric-like space.Definition 5.
(see [20]). The function ρr:X×X⟶R+ is called a partial rectangular metric-like in X if for every x,y∈X and for all distinct points u,v∈X−x,y, it satisfies the following conditions:(1)
ρrx,y=0 then x=y(2)
ρrx,y=ρry,x(3)
ρrx,y≤ρrx,u+ρru,v+ρrv,y
The pairX,ρr is called a rectangular metric-like space.Definition 6.
The functionρprl:X×X⟶R+ is called a partial rectangular metric-like if for all x,y∈X, the following conditions are satisfied:(1)
ρprlx,y=0 then x=y(2)
ρprlx,x≤ρprlx,y(3)
ρprlx,y=ρprly,x(4)
ρprlx,y≤ρprlx,u+ρprlu,v+ρprlv,y−ρprlu,u−ρprlv,v, for all u,v∈X−x,y
The pairX,ρprl is called a partial rectangular metric-like space.Remark 1.
We see that in a partial rectangular metric-like spaceX,ρprl, if x,y∈X and ρprlx,y=0, then x=y. The converse does not hold.
A rectangular metric-like space is not necessarily a partial rectangular metric-like space.Example 1.
LetX=R+and ρrl:X×X⟶R+ be such that(1)ρrlx,y=2x,x=y,maxx,y,x≠y.
Then,ρrl is a rectangular metric-like in X, but it is not a partial rectangular metric-like because it does not satisfy the second condition of Definition 6.
Ify<x, then ρrlx,x=2x>x=maxx,y=px,y.
Moreover,ρrl is not a rectangular metric since for every x≠0, ρrlx,x=2x≠0, and it is clearly not a partial rectangular metric.
Example2 shows that a partial rectangular metric-like space may not be a rectangular metric space.Example 2.
LetX=R+, then ρprl:X×X⟶R+ such that(2)ρprlx,y=2x,x=y,maxx,y+2,x≠y.is a partial rectangular metric-like in X, but it is not a rectangular metric.Definition 7.
LetX,ρprl be a partial rectangular metric-like space, xnn∈N a sequence in X and x∈X.(1)
Iflimn⟶+∞ρprlxn,x=ρprlx,x, then xnn∈N is called ρprl-convergent to the point x(2)
Iflimn,m⟶+∞ρprlxn,xm is finite, then xnn∈N is called ρprl-Cauchy in X,ρprl(3)
If eachρprl-Cauchy sequence xnn∈N in X,ρprl converges to a point x∈X (that is, if limn⟶+∞ρprlxn,x=limn,m⟶+∞ρprlxn,xm=ρprlx,x), then the space X,ρprl is called ρprl-completeDefinition 8.
(see [24]). A function φ:R+⟶R+ is called a comparison function if the following conditions hold:(i)
φ is an nondecreasing function(ii)
For eacht>0, 0<φt<t(iii)
For eacht>0, limn⟶+∞φnt=0
The set of comparison functions is denoted byΦ.


## 3. Main Results

LetX,ρprl be a partial rectangular metric-like space, and let T:X⟶X be a function of X.IfA is a subset of X, then δA=supρprlx,y,ρprlx,x:x,y∈A is called a diameter of A. By Definition 5, if A=x⊂X, then δA=ρprlx,x may not be zero.For anyx∈X, we call the set Ox=x,Tx,T2x,…,Tnx,… by orbit of x.For eachx,y∈X, the union of the orbits is(3)Ox∪​Oy=x,Tx,T2x,…,Tnx,…,y,Ty,T2y,…,Tny,…,and the diameter can be expressed as(4)δOx=supρprlx,x,ρprlx,Tnx,ρprlTmx,Tnx,ρprlTnx,Tnx:n,m∈N.Theorem 1.
LetX,ρprl be a partial rectangular metric-like space. If for every x∈X, the orbit Ox is bounded and for the function T:X⟶X, there is a φ∈Φ, such that for every x,y∈X,(5)ρprlTpx,Tqy≤φδOx∪​Oy,where at least one of the numbers p or q is equal to 1, and then T has a unique fixed point.Proof.
Without any restriction, we assume thatq=1. Therefore, condition (5) of the theorem becomes(6)ρprlTpx,Ty≤φδOx∪​Oy,for everyx,y∈X.
We take any arbitraryu∈X. Firstly, we show that the sequence Tnun∈N is a ρprl-Cauchy one.
For that we have to show thatlimn,m⟶+∞ρprlTnu,Tmu is finite. We denote by r=maxp,1 and applying condition (6)
x=Tn+r−p+iu,y=Tn+r−1+ju in which n∈N,i,j∈N0 we get the inequality as follows:(7)ρprlTn+r+iu,Tn+r+ju≤φδOTn+r−p+iu∪​OTn+r−1+ju.
Sincen+r+i−p≥n+i and n+r+j−1≥n+j, then(8)δOTn+r−p+iu∪​OTn+r−1+ju≤δOTn+iu∪​OTn+ju≤δOTnu.
Since the functionφ is nondecreasing, then we get(9)ρprlTn+r+iu,Tn+r+ju≤φδOTn+r−p+iu∪​OTn+r−1+ju≤φδOTnu.
Using the following inequality:(10)δOTn+ru≤supρprlTn+r+iu,Tn+r+ju,ρprlTn+r+iu,Tn+r+iu:i,j∈N0.Condition 2 of Definition 6, and the relation (9), we get(11)δOTn+ru≤φδOTnu.
Applying this inequality, we obtain the following ones for anyn∈N:(12)δOTn⋅ru≤φδOTn−1⋅ru≤φ2δOTn−2⋅ru≤⋯≤φnδOu.
Since the orbitOuis bounded, we use the properties of φ to see that(13)limn⟶+∞φnδOu=0.
Using (12), it follows that limn⟶+∞δOTn⋅ru=0.
The sequenceδOTn⋅run∈N is a subsequence of δOTnun∈N, which is decreasing and bounded below by 0. Hence, the sequence δOTnun∈N converges to zero.
So, we get(14)limi,j⟶+∞ρprlTn+iu,Tn+ju=0.
Consequently, the sequenceTnun∈N is ρprl-Cauchy in X,ρprl. Since X,ρprl is complete, then there is any z∈X, z≠u such that(15)limn⟶+∞ρprlTnu,z=limn⟶+∞ρprlTn+iu,Tn+ju=ρprlz,z=0.
Meanwhile,(16)δOTn+pz∪​OTn+pu=supρprlTn+p+iz,Tn+p+ju,ρprlTn+p+iz,Tn+p+jzρprlTn+p+iu,Tn+p+ju,i,j∈N0.
Fori,j∈N0, we have(17)ρprlTn+p+iz,Tn+p+ju≤φδOTn+iz∪OTn+p−1+ju≤φδOTnz∪​OTnu,(18)ρprlTn+p+iz,Tn+p+jz≤φδOTn+iz∪​OTn+p−1+jz≤φδOTnz≤φδOTnz∪​OTnu,(19)ρprlTn+p+iu,Tn+p+ju≤φδOTn+iu∪​OTn+p−1+ju≤φδOTnu≤φδOTnz∪​OTnu.
From relationships (16)–(19), we get the inequality(20)δOTn+pz∪​OTn+pu≤φδOTnz∪​OTnu.
So,(21)δOTn⋅pz∪OTn⋅pu≤φδOTn−1pz∪OTn−1pu≤φ2δOTn−2pz∪​OTn−2pu≤⋯≤φnδOz∪Ou.
SinceOz and Ou are bounded in X, by using the third property of the function φ, we obtain 0≤limn⟶+∞δOTn⋅pz∪​OTn⋅pu≤limn⟶+∞φnδOz∪​Ou=0.
The sequenceδOTn⋅pz∪​OTn⋅pun∈N is a subsequence of the decreasing and bounded below by zero sequence δOTnz∪​OTnun∈N, and it converges to zero; then,(22)limn⟶+∞δOTnz∪OTnu=0.
Taking into account the definition ofδOTnz∪​OTnu and its limit given by (22), the following limits are also obtained:(23)limi,j⟶+∞ρprlTn+iz,Tn+ju=0,limn⟶+∞ρprlTnz,Tnz=0,limn,m⟶+∞ρprlTnz,Tmz=0,limn,m⟶+∞ρprlTnu,Tmu=0,limn⟶+∞ρprlTnu,Tnu=0.
In the case whenTnu=Tn+1u=TTnu, then the function T has evidently a fixed point.
Otherwise, ifTnu≠Tn+1u, then(24)ρprlTnz,z≤ρprlTnz,Tnu+ρprlTnu,Tn+1u+ρprlTn+1u,z−ρprlTnu,Tnu−ρprlTn+1u,Tn+1u.
Furthermore, taking the limit ofρprlTnz,z for n⟶+∞ using (23), (14), and (15), we have(25)limn⟶+∞ρprlTnz,z=0=ρprlz,z.
Now, we have to prove thatz is a fixed point of the function T.
Indeed, sincez≠u, then Tnz≠Tnu for any n∈N; also,(26)ρprlTz,z≤ρprlTz,Tnz+ρprlTnz,Tnu+ρprlTnu,z−ρprlTnz,Tnz−ρprlTnu,TnuρprlTn+pz,Tz≤φδOTnz∪Oz≤φδOz.
Using the obtained results (23) and (25), we see that(27)δOz=supρprlTnz,Tmz,ρprlz,Tz,ρprlz,Tnz,ρprlTnz,Tz,ρprlz,z:n,m∈N=maxρprlz,Tz,ρprlTnz,Tz,n∈N.
Thus,(28)ρprlTn+pz,Tz≤φmaxρprlz,Tz,ρprlTnz,Tz,n∈N≤maxφρprlz,Tz,φρprlTnz,Tz.
Meanwhile,(29)ρprlTn⋅pz,Tz≤maxφρprlz,Tz,φρprlTn−1⋅pz,Tz≤maxφρprlz,Tz,φ2ρprlTn−2⋅pz,Tz≤⋯≤maxφρprlz,Tz,φnρprlz,Tz.
Sincelimn⟶+∞φnρprlz,Tz=0, then(30)limn⟶+∞ρprlTn⋅pz,Tz≤φρprlz,Tz.
From (25), we see that limn⟶+∞ρprlTn⋅pz,z=0 and using the obtained equation (30), we get the inequality(31)limn⟶+∞ρprlTn⋅pz,Tz=ρprlz,Tz≤φρprlz,Tz,which contradicts the main property φt<t of the comparison function.
So, finally we get that(32)ρprlz,Tz=0,Tz=z,and z is a fixed point of T.
To prove the uniqueness ofz, let us suppose that there is z1∈X, z1≠z such that Tz1=z1.
We see that(33)ρprlz,z1=ρprlTpz,Tz1≤φδOzUOz1=φmaxρprlz,z1,ρprlz,Tnz,ρprlz1,Tnz1,ρprlTnz,TmzρprlTnz1,Tmz1,ρprlTnz,Tmz1,n,mϵN,consequently ρprlz,z1≤φρprlz,z1<ρprlz,z1, and finally(34)ρprlz,z1=0,z=z1.
Puttingp=q=1,we have the following theorem.Corollary 1.
LetX,ρprl be a complete partial rectangular metric-like space. If a function T:X⟶X satisfies the condition, there exists φ∈Φ such that ρprlTz,Ty≤φOx∪​Oy, where x,y∈Xand x≠y, and for each x∈X, Ox is bounded,
Then, it has a unique fixed point. In the case of partial rectangular metric spaces, Theorem1 yields.Theorem 2.
LetX,ρ be a complete partial rectangular metric space and T:X⟶X a function such that for each x∈X, the orbit Ox is bounded and(35)there exitsφ∈Φsuch that for everyx,y∈XρTpz,Tqz≤φOx∪Oy,where at least p or q is equal to 1, and then the function T has a unique fixed point.The proof is analogous to that of Theorem1.Remark 2.
Theorem2 generalizes Theorem 19 in [17] and Corollary 1 of Theorem 4 in [25] because the contractive condition (5) is more general than the quasi-contractive condition with any constant λ∈0,1, since φt=λ⋅t∈Φ and for all x,y∈X(36)maxρx,y,ρx,Tx,ρy,Ty,ρx,Ty,ρy,Tx≤δ0x∪Oy.
The following example shows that Theorem2 is a proper generalization of Theorem 19 in [17].Example 3.
Let us takeX=0,+∞. We determine the function ρ:X×X⟶R+, putting(37)ρx,y=3,x,y∈1,2,maxx,y,otherwise.
The functionρ is a partial rectangular metric in X satisfying the conditions of Definition 1.
LetT:X⟶X be the function:(38)Tx=0,x=0,12n+1,12n+1<x≤12n,1,x>1,n=0,1,2,3,….
We see thatT satisfies the conditions of Theorem 2 with φx=Tx.
Firstly, we see that for allx∈X, 0<φx=Tx<x.
Furthermore,φ is a monotone nondecreasing function because for any x<y, examining all the cases for x and y, we see that φx=Tx≤Ty=φy.
For the last condition,ifx>1, thenφx=Tx=1, φ2x=T2x=T1=1/2 , φ3x=T3x=T1/2=1/22,…,φnx=Tnx=T1/2n−2=1/2n−1,ifx=1, then φx=φ1=T1=1/2, φ2x=1/22, ,…,,φnx=1/2n,if1/2n0+1<x≤1/2n0, for n0=0,1,2,3,…, then(39)φx=Tx=12n0+1,φ2x=12n0+2,…,φnx=12n0+n.
Considering all cases, we getlimn⟶+∞φnx=0, so φ∈Φ.
In the following, we show thatT satisfies the contractive condition in Theorem 2 for p,q=1. Firstly, we determine the orbits and their diameters:(i)
ifx>1, then Ox=x,1,1/2,1/22,……,,1/2n−1,…… and(40)δOx=3,x=2,x,x>1,x≠2.(ii)
ifx=1, then Ox=1,1/2,1/22,……,,1/2n,…… and δOx=1;(iii)
if1/2n0+1<x≤1/2n0 then Ox=x,1/2n0+1,1/2n0+2,……,1/2n0+n,… and δOx=x;(iv)
ifx=0, then O0=0,0,……, and δO0=0.
Secondly, we show the fulfillment of the contractive property.Case 1.
Ifx,y>1, then we have Tx=1,Ty=1, and ρTx,Ty=1, so(41)δOx∪Oy=ρx,y>1,ρTx,Ty=1=φδOx∪Oy.Case 2.
If1/2n0+1<x≤1/2n0 and y>1, then(42)Tx=12n0+1,Ty=1,ρTx,Ty=max12n0+1,1=1.
Also,(43)δOx∪Oy=ρx,y>1,ρTx,Ty=1=Ty=φy=φδOx∪Oy.Case 3.
Taking0<y<x≤1 such that 1/2n0+1<x≤1/2n0 and 1/2n1+1<y≤1/2n1 where n1≥n0, we have(44)Tx=12n0+1,Ty=12n1+1,ρTx,Ty=max12n0+1,12n1+1=12n0+1.
SinceρTx,Ty=1/2n0+1 and δOx∪​Oy=maxx,y=x, then(45)ρTx,Ty=12n0+1=φx=φδOx∪Oy.
An analogous proof is valid for0<x<y≤1.Case 4.
Ifx=0 and y>0, then Tx=0, δOx∪​Oy=δO0∪​Oy=y and(46)ρTx,Ty=Ty,ρTx,Ty=Ty=φy=φ,Ox∪Oy.
Therefore, the functionT satisfies all the conditions of Theorem 2, so it has x=0 a fixed point.
Meanwhile, we see that the functionT does not meet the conditions of Theorem 19 in [17].
Indeed, forx=0 and 0<y<1, we have(47)Ty=max0,Ty=maxT0,Ty=ρT0,Ty≤≤λ⋅maxρ0,y,ρ0,0,ρy,Ty,ρ0,Ty,ρy,T0=λ⋅maxρ0,y,ρy,Ty,ρ0,Ty==λ⋅maxy,maxy,Ty,Ty=λ⋅maxy,Ty.
So,Ty≤λ⋅maxy,Ty=λybecause λ∈0,1 and maxy,Ty cannot be Ty.
Thus, we already proved that for eachy>0, Ty≤λ⋅y.
Since0<y<1, there is any n∈N such that 1/2n+1<y≤1/2n.
Taking the limit fory⟶1/2n+1+ in the inequality Ty≤λy, we get 1/2n+1≤λ⋅1/2n+1.
The last inequality is not true for anyλ∈0,1 so it cannot be found any λ∈0,1 such that the condition of quasi-contraction of Theorem 19 in [17] for the function T is satisfied.Remark 3.
Theorem1 holds also in the rectangular metric-like space, and consequently it is a generalization of Theorem 2, Theorem 3, and Theorem 4 in [20].Remark 4.
Theorem1 can also be formulated and proved in the metric-like space; furthermore, it is a generalization of Theorem 2.4 in [19]. To prove this, we show in the following that our contractive condition is more general than the contractive condition in Theorem 2.4 in [19].
The contractive condition in Theorem 2.4 in [19] is(48)σTx,Ty≤φrx,y,for eachx,y∈X,where(49)rx,y=maxσx,y,σx,Tx,σy,Ty,σx,Ty,σy,Tx,σx,x,σy,y,and φ satisfies the following:a1φ:0,+∞⟶0,+∞ is nondecreasingb1φt<t, for all t>0c1lims⟶t+φs<t, for all t>0d1limt⟶∞t−φt=∞
Since(50)δOx∪Oy=maxσx,Tix,σy,Tiy,σTix,Tjx,σTiy,Tjy,σx,x,σy,y,σTix,Tix,σTjy,Tjy,σTix,Tjy,i,jεN,we have Mx,y≤δOx∪​Oy.
It remains to show that if the functionφ satisfies the conditions a1,b1,c1, and d1, then it fulfills also the conditions a,b,andc of Definition 7.
It is clear that if the conditionsa1andb1 hold for the function φ, then the conditions a and b of Definition 7 also hold.
To show that the conditionc is fulfilled, we take φnt=φφn−1t<φn−1t for each n∈N, t>0, so the sequence φntnϵNis a decreasing one and bounded below by 0. Consequently, it is a convergent sequence and limn⟶+∞φnt=t∗≥0; hence, infφnt,n∈N≥0. If t∗>0, since φ it is monotone nondecreasing, we have that it has one-sided limits in t• and limt⟶t∗−φt≤φt∗≤limt⟶t∗+φt.
From the fulfillment of the conditionc1, an ε>0 can be found, such that limt⟶t∗+φt<ε<t∗. Therefore, for that ε, there is any δ>0 such that for t∗<t<δ, we have φt∗≤limt⟶t∗+φt≤φt<ε. For that δ, there is any p∈N such that for all naturals n>p, t∗≤φnt<δ, so(51)φt∗≤φφnt≤ε<t∗,soφn+1t<ε<t∗,which contradicts the supposition t∗>0, being t∗=infφnt,nεN.
Consequently, it remains thatt∗=0; thus, limn⟶∞φnt=0, and we get that the condition of contractivity in Theorem 1 is more general than the one in Theorem 2.5 [19].


## 4. Conclusions

This paper defines the partial rectangular metric-like spaces, which generalize the rectangular metric-like spaces [20] and rectangular metric space [12]. In addition, they are generalizations of partial metric spaces.The main result of this paper is Theorem1, which provides the existence of a unique fixed point for the functions T:X⟶X that satisfy condition (5) of contractivity with respect to a comparison function φ and the orbits’ diameters of T.It generalizes Theorems 2, 3, and 4 in [20] and Theorem 2.4 in [19], as it is shown in Remark 4, where we have proved that our condition of contractivity is really more general than the analogous one at Theorem 2.4 in [19]. Example 3.5 shows that Theorem 1 is a generalization of Theorem 19 in [17] and Corollary 1 of Theorem 4 in [25].



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