# OnR-Partial b-Metric Spaces and Related Fixed Point Results with Applications

Authors: Muhammad Usman Ali; Yajing Guo; Fahim Uddin; Hassen Aydi; Khalil Javed; Zhenhua Ma
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we introduce the notion ofR-partial b-metric spaces and prove some related fixed point results in the context of this notion. We also discuss an example to validate our result. Finally, as applications, we evince the importance of our work by discussing some fixed point results on graphical-partial b-metric spaces and on partially-ordered-partial b-metric spaces.


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## Body

## 1. Introduction and Preliminaries

Due to the fact that fixed point theory plays a very crucial role for different mathematical models to obtain their solution existence and has a wide range of applications in different fields related to mathematics, this theory has intrigued many researchers.By the inception of the Banach fixed point theorem [1], researchers are continuously trying to get the generalizations of this classical result through different methodologies. For instance, Czerwik [2] introduced the notion of b-metric spaces, with a triangle inequality weaker than that of metric spaces, in a view to generalize the Banach contraction principle. Moving on the same sequel, Matthews [3] introduced the notion of a partial metric space, which was a part of the study for denotational semantics of dataflow networks and gave a generalized version of the Banach contraction principle. The concept of partial metric spaces was further extended to partial b-metric spaces by Shukla in [4]. A number of researchers took keen interest in the generalized version of the metric spaces some work is available in [5–27].Recently, Gordgi et al. [28] introduced the notion of orthogonal sets and gave a new extension for the classical Banach contraction principle. More details can be found in [29, 30].After looking into the structure of orthogonal metric spaces, introduced by [29, 30], and the binary relation used with a metric, [31, 32], we introduce the notion of R-partial b-metric spaces. We are also improving and generalizing the concept of orthogonal contractions in the sense of R-partial b-metric spaces and establish some fixed point theorems for the proposed contractions.Throughout this paper, we denote byℕ,ℝ,ℤ, and ℝ+ the set of natural numbers, real numbers, integer numbers, and nonnegative real numbers, respectively.Definition 1 (see [2].
LetHbe a nonempty set ands≥1.Suppose a mappingd:H×H⟶ℝ+satisfies the following conditions for allh,l,z∈H:
bM1dh,l=0 if and only if h=l;
bM2dh,l=dl,h;
bM3dh,l≤sdh,z+dz,l.
Thend is called a b-metric on H, and H,d is called a b-metric space with coefficient s.Definition 2 (see [3].
LetHbe a nonempty set. Letp:H×H⟶ℝ+satisfy the following for allh,l,z∈H:
pM1h=l if and only if ph,h=ph,l=pl,l;
pM2ph,h≤ph,l;
pM3ph,l=pl,h;
pM4ph,l≤ph,z+pz,l−pz,z.
ThenH,p is called a partial metric space.Definition 3 [4].
A partialb-metric onH≠∅is a functionσ:H×H⟶ℝ+such that for allh,l,z∈H, and for somes≥1, we have
σ1h=l if and only if σh,h=σh,l=σl,l;
σ2σh,h≤σh,l;
σ3σh,l=σl,h;
σ4σh,l≤sσh,z+σz,l−σz,z.
A partialb-metric space is denoted with H,σ,s. The number s is called the coefficient of H,σ,s.Remark 4 (see [4].
It is clear that every partial metric space is a partialb-metric space with coefficients=1and everyb-metric space is a partialb-metric space with the same coefficient and a zero self-distance. However, the converse of this fact need not hold.Example 1 [4].
LetH=ℝ+,p>1be a constant andσ:H×H⟶ℝ+be defined by(1)σh,l=h−lp+maxh,lpforallh,l∈H.
Then,H,σ,s is a partial b-metric space with coefficient s=2p>1, but it is neither a b-metric nor a partial metric space.Definition 5 [33].
LetHbe a nonempty set. A subsetRofH2is called a binary relation onH. Then, for anyh,l∈H, we say that"hisR-related tol", that is,hRl, or"hrelates tolunderR"if and only ifh,l∈R. h,l∉Rmeans that"his notR-related tol"or"his not related tolunderR".Definition 6 [33].
A binary relationRdefined on a nonempty setHis calledareflexive ifh,h∈R∀h∈H;
b irreflexive if h,h∉R for some h∈H;
c symmetric if h,l∈R implies l,h∈R∀h,l∈H;
d antisymmetric if h,l∈R and l,h∈R imply h=l∀h,l∈H;
e transitive if h,l∈R and l,z∈R imply h,z∈R∀h,l,z∈H;
f preorder if R is reflexive and transitive;
g partial order if R is reflexive, antisymmetric, and transitive.Definition 7 [32].
LetHbe a nonempty set and letRbe a binary relation onH.
(a) A sequencehn is called an R-sequence if
(2)∀n∈ℕ,hnRhn+1.
(b) A mapT:H⟶H is R-preserving if
(3)∀h,l∈H,hRlimpliesThRTl.Definition 8 [32].
LetH,dbe a metric space andRbe a binary relation onH. Then,H,d,Ris called anR-metric space.Definition 9 [31].
A mappingT:H⟶HisR-continuous ath0∈Hif for eachR-sequencehnn∈ℕinHwithhn⟶h0, we getThn⟶Th0. Thus,TisR-continuous onHifTisR-continuous at eachh0∈H.Definition 10 [31].
A mapT:H⟶His anR-contraction, if(4)dTh,Tl≤kdh,l,for all h,l∈H with hRl, where 0<k<1.Khalehoghli et al. [31] extended the result of Banach in the following way.Theorem 11 [31].
IfTis anR-preserving andR-continuousR-contraction on anR-completeR-metric space withh0∈Hsuch thath0Rlfor eachl∈H. Then,Thas a unique fixed point.


## 2. Main Results

Let us begin this section with the definition ofR-partial b-metric spaces.Definition 12.
LetH≠∅andRbe a reflexive binary relation onH, denoted asH,R. A mapσR:H×H⟶ℝ+is called anR-partialb-metric on the setH, if the following conditions are satisfied for allh,l,z∈Hwith either (hRlorlRh), either (hRzorzRh) and either (zRlorlRz):
σR1h=l if and only if σRh,h=σRh,l=σRl,l;
σR2σRh,h≤σRh,l;
σR3σRh,l=σRl,h;
σR4σRh,l≤sσRh,z+σRz,l−σRz,z, where s≥1.
Then,H,R,σR,s is called R-partial b-metric space with the coefficient s≥1.Remark 13.
In the above definition, a setHis endowed with a reflexive binary relationRandσR:H×H⟶ℝ+satisfiesσR1-σR4only for those elements which are comparable under the reflexive binary relationR. Hence, theR-partialb-metric may not be a partialb-metric, but the converse is true.
The following simplest example shows that theR-partial b-metric with s≥1 need not to be a partial b-metric with s≥1.Example 2.
LetH=−1,−2,1,2and let the binary relation be defined byhRlif and only ifh=lorh,l>0. It is easy to prove thatσRh,l=max∣h∣,∣l∣is anR-partialb-metric onHwiths≥1, butσRis not a partialb-metric onHwiths≥1. Indeed, forh=−2andl=2, we haveσRh,h=σRh,l=σRl,l=2.
In the coming definitions, letH,R,σR,s be an R-partial b-metric space with the coefficient s≥1.Definition 14.
Lethnbe anR-sequence inH,R,σR,s, that is,hnRhn+1orhn+1Rhnfor eachn∈ℕ. Then(i)
hn is a convergent sequence to some h∈H if limn→∞σRhn,h=σRh,h and hnRh for each n≥k(ii)
hn is Cauchy if limn,m→∞σRhn,hm exists and is finiteDefinition 15.
H,R,σR,sis said to beR-complete if for every CauchyR-sequence inH, there ish∈Hwithlimn,m→∞σRhn,hm=limn→∞σRhn,h=σRh,handhnRhfor eachn≥k.Definition 16.
We say thatT:H⟶His anR-property map, if for any iterativeR-sequencehn:hn=Tnh,h∈HinH,R,σR,swithlimn→∞σRhn,h=σRh,h, hnRhfor somen≥kandlimn→∞σRhn,Th≤σRh,h, we have thathRThorThRh.Definition 17.
We say thatT:H⟶HisR-0-continuous ath∈Hif for eachR-sequencehninH,R,σR,swithlimn→∞σRhn,h=0, we havelimn→∞σRThn,Th=0. Also,TisR-0-continuous onHifTisR-0-continuous for eachh∈H.The following results help us to ensure the existence of fixed points for self maps. Throughout, we assume thatR is a preorder relation.Theorem 18.
LetH,R,σR,sbe anR-completeR-partialb-metric space with the coefficients≥1and leth0∈Hbe such thath0Rlfor eachl∈H. LetT:H⟶Hbe anR-preserving and anR-property map satisfying the following(5)σRTh,Tl≤kσRh,lforallh,l∈HwithhRl,where k∈0,1/s. Then, T has a fixed point h∗∈H and σRh∗,h∗=0.Proof.
Ash0∈H is such that h0Rl for each l∈H, then by using the R-preserving nature of T, we construct an R-sequence hn such that hn=Thn−1=Tnh0 and hn−1Rhn for each n∈ℕ. We consider hn≠hn+1 for each n∈ℕ∪0. Thus, by (5), we get
(6)σRhn,hn+1=σRThn−1,Thn≤kσRhn−1,hn,for all n∈ℕ. This inequality yields
(7)σRhn,hn+1≤knσRh0,h1,for all n∈ℕ. To discuss the Cauchy criteria, we will consider an arbitrary integer n≥1,m≥1 with m>n and use σR4 along (7) in the following way.
(8)σRhn,hm≤sσRhn,hn+1+σRhn+1,hm−σRhn+1,hn+1≤sσRhn,hn+1+s2σRhn+1,hn+2+σRhn+2,hm−σRhn+2,hn+2≤sσRhn,hn+1+s2σRhn+1,hn+2+s3σRhn+2,hn+3+⋯+sm−nσRhm−1,hm≤sknσRh0,h1+s2kn+1σRh0,h1+s3kn+2σRh0,h1+⋯+sm−nkm−1σRh0,h1≤skn1+sk+sk2+⋯σRh0,h1=skn1−skσRh0,h1.
Ask∈0,1/s and s≥1, it follows from the above inequality that
(9)limn,m→∞σRhn,hm=0.
Therefore,hn is a Cauchy R-sequence. Since H is R-complete, there exists h∗∈H such that limn,m→∞σRhn,hm=limn→∞σRhn,h∗=σRh∗,h∗ and hnRh∗ for each n≥k (for some value of k). Thus, from above, we obtain 0=limn,m→∞σRhn,hm=limn→∞σRhn,h∗=σRh∗,h∗ and hnRh∗ for each n≥k. As hnRh∗ for each n≥k, from (5), we get
(10)σRThn,Th∗≤kσRhn,h∗.
This inequality and the above findings imply(11)limn→∞σRhn+1,Th∗≤σRh∗,h∗=0.
AsT is an R-property map, so we get h∗RTh∗ or Th∗Rh∗. Without any loss of generality, we take h∗RTh∗. Thus, by using σR4 with (5), we get the following for each n≥k(12)σRh∗,Th∗≤sσRh∗,hn+1+sσRhn+1,Th∗−σRhn+1,hn+1≤sσRh∗,hn+1+sσRThn,Th∗≤sσRh∗,hn+1+skσRhn,h∗.
Whenn tends to infinity, the above inequality yields σRh∗,Th∗=0. Hence, we get σRh∗,Th∗=0, σRh∗,h∗=0 and σRTh∗,Th∗=0. Therefore, h∗=Th∗, that is, h∗ is a fixed point of T.Remark 19.
Note that the fixed point ofTis unique if in the above theorem we add (I): for each fixed pointsh∗andl∗ofT, we haveh∗Rl∗orl∗Rh∗.
Sinceh∗ and l∗ are fixed points of T such that h∗Rl∗. Then, we have Tnh∗=h∗,Tnl∗=l∗ for all n∈ℕ. By the nature of h0, we obtain
(13)h0Rh∗andh0Rl∗.SinceT is R-preserving, we have
(14)Tnh0RTnh∗andTnh0RTnl∗,for all n∈ℕ. Therefore, by the triangle inequality and (5), we get
(15)σRh∗,l∗=σRTnh∗,Tnl∗=sσRTnh∗,Tnh0+σRTnh0,Tnl∗−σRTnh0,Tnh0≤sknσRh∗,h0+sknσRh0,l∗.Taking limit asn⟶∞ in the above inequality, we obtain
(16)σRh∗,l∗=0,and so
(17)h∗=l∗.Remark 20.
Note that the condition “leth0∈Hbe such thath0Rlfor eachl∈H”of Theorem18may be replaced with “leth0∈Hbe such thath0RTh0.”Example 3.
LetH=ℝand defineσR:H×H⟶ℝ+by(18)σRh,l=h−l2ifh,l≥0,0otherwise.
The relation onH is defined by hRl if and only if h=l or h,l≥0. Clearly, H,R,σR,4 is an R-complete partial b-metric space. Define a map T:H⟶H by
(19)Th=h4ifh≥0,0otherwise.
Then, it is very simple to verify the following:(1)
Ifh=l, then Th=Tl. While if h,l≥0, then Th,Tl≥0. Thus, T is an R-preserving map(2)
Suppose that for any iterativeR-sequence hn in H with limn→∞σRhn,h=σRh,h, hnRh for some n≥k, and limn→∞σRhn,Th≤σRh,h, then we get hRTh(3)
Considerh0≥0 any real number, then Th0≥0. Thus, we have h0,Th0≥0, that is, h0RTh0(4)
For eachh,l∈H with hRl, we have
case (a)h=l:
(20)σRTh,Tl=0=116×0=116σRh,l.case (b) h,l≥0:
(21)σRTh,Tl=h4−l42=116h−l2=116σRh,l.Hence, by Theorem18, T must has a fixed point.Example 4.
LetH=ℝand defineσR:H×H⟶ℝ+by(22)σRh,l=h−l2+maxh,l2ifh,l≥0,0otherwise.
The relation onH is defined by hRl if and only if h=l or h,l≥0.
Clearly,H,R,σR,4 is an R-complete partial b-metric space. Define a map T:H⟶H by
(23)Th=h6ifh≥0,−1otherwise.
Then, one can verify the following:(1)
Ifh=l, then Th=Tl. While if h,l≥0, then Th,Tl≥0. Thus, T is an R-preserving map(2)
Suppose that for any iterativeR-sequence hn in H with limn→∞σRhn,h=σRh,h, hnRh for some n≥k, and limn→∞σRhn,Th≤σRh,h, then we get hRTh(3)
Ifh0≥0 be some real number, then Th0≥0. Thus, we get h0,Th0≥0, that is, h0RTh0(4)
For eachh,l∈H with hRl, we have
Case (a) Ifh=l≥0, then Th=Tl≥0. Thus,
(24)σRTh,Tl=0+maxh6,l62=136maxh,l2=136σRh,l.
Case (b) Ifh=l<0, then Th=Tl=−1. Thus,
(25)σRTh,Tl=0=σRh,l.
Case (c) Ifh,l≥0, then Th,Tl≥0. Thus,
(26)σRTh,Tl=h6−l62+maxh6,l62=136σRh,l.Hence, by Theorem18, T must has a fixed point.Remark 21.
Note that the functionσRdefined in the above example is neither a metric nor ab-metric nor a partialb-metric onℝ. Indeed,σR4,1=25, σR4,−1=0, σR−1,1=0, σR−1,−1=0, that is,σ4andbM3do not exist.Theorem 22.
LetH,R,σR,sbe anR-completeR-partialb-metric space with the coefficients≥1and leth0∈Hbe such thath0Rlfor eachl∈H. LetT:H⟶Hbe anR-preserving andR-0-continuous map satisfying the following(27)σRTh,Tl≤kmaxσRh,l,σRh,Th,σRl,Tl,for all h,l∈H with hRl, hRTh, and lRTl, where k∈0,1/s. Also, let for each R-sequence hn in H with hnRa and hnRb, we have either aRb or bRa. Then, T has a fixed point h∗∈H and σRh∗,h∗=0.Proof.
Ash0∈H is such that h0Rl for each l∈H, then by using the R-preserving nature of T, we obtain an R-sequence hn such that hn=Thn−1=Tnh0 and hn−1Rhn for each n∈ℕ. We take hn≠hn+1 for each n∈ℕ∪0. Then by (27), for each n∈ℕ, we get
(28)σRhn,hn+1=σRThn−1,Thn≤kmaxσRhn−1,hn,σRhn−1,Thn−1,σRhn,Thn=kmaxσRhn−1,hn,σRhn−1,hn,σRhn,hn+1=kmaxσRhn−1,hn,σRhn,hn+1.
IfmaxσRhn−1,hn,σRhn,hn+1=σRhn,hn+1, then from the above inequality, we obtain that σRhn,hn+1≤kσRhn,hn+1<σRhn,hn+1, which is a contradiction. Therefore, we must have maxσRhn−1,hn,σRhn,hn+1=σRhn−1,hn. Again, from the above inequality, we have
(29)σRhn,hn+1≤kσRhn−1,hn∀n∈ℕ.
On repeating this process, we obtain(30)σRhn,hn+1≤knσRh0,h1∀n∈ℕ.
Form,n∈ℕ with m>n, by σR4, we obtain
(31)σRhn,hm≤sσRhn,hn+1+σRhn+1,hm−σRhn+1,hn+1≤sσRhn,hn+1+s2σRhn+1,hn+2+σRhn+2,hm−σRhn+2,hn+2≤sσRhn,hn+1+s2σRhn+1,hn+2+s3σRhn+2,hn+3+⋯+sm−nσRhm−1,hm.
Using (30) in the above inequality, we obtain
(32)σRhn,hm≤sknσRh0,h1+s2kn+1σRh0,h1+s3kn+2σRh0,h1+⋯+sm−nkm−1σRh0,h1≤skn1+sk+sk2+⋯σRh0,h1=skn1−skσRh0,h1.
Ask∈0,1/s and s≥1, it follows from the above inequality that
(33)limn,m→∞σRhn,hm=0.
Therefore,hn is a Cauchy R-sequence. Since H is R-complete, there exists h∗∈H such that limn,m→∞σRhn,hm=limn→∞σRhn,h∗=σRh∗,h∗ and hnRh∗ for each n≥k. Thus, from above, we obtain 0=limn,m→∞σRhn,hm=limn→∞σRhn,h∗=σRh∗,h∗ and hnRh∗ for each n≥k. Since T is R-0-continuous, one gets that limn→∞σRhn,h∗=0, which leads to limn→∞σRThn,Th∗=0. Obviously, we have ThnRTh∗ for each n≥k. Thus, hnRTh∗ for each n>k. Since hnRh∗ and hnRTh∗ for each n>k, we have either h∗RTh∗ or Th∗Rh∗. By using σR4, we get the following for each n>k:
(34)σRh∗,Th∗≤sσRh∗,hn+1+sσRhn+1,Th∗−σRhn+1,hn+1.
Whenn tends to infinity, the above inequality yields σRh∗,Th∗=0. Hence, we get σRh∗,Th∗=0, σRh∗,h∗=0, and σRTh∗,Th∗=0. Therefore, we say that h∗=Th∗, i.e., h∗ is a fixed point of T.Remark 23.
Note that the fixed point ofTis unique if in the above result, we add the condition: for each fixed pointsh∗andl∗ofT, we haveh∗Rl∗orl∗Rh∗.
Sinceh∗=Th∗, we have l∗=Tl∗ and h∗Rl∗. From (27), we get
(35)σRh∗,l∗=σRTh∗,Tl∗≤kmaxσRh∗,l∗,σRh∗,Th∗,σRl∗,Tl∗=kmaxσRh∗,l∗,σRh∗,h∗,σRl∗,l∗=kσRh∗,l∗<σRh∗,l∗.It is a contradiction in the caseσRh∗,l∗≠0. Therefore, we must have σRh∗,l∗=0, that is, h∗=l∗.


## 3. Applications to Graphical Partialb-Metric Spaces and Partially-Ordered-Partial b-Metric Spaces

In this section, we define a directed graphG on H, denoted by G=VH,EH, with the vertex set VH=H and the edge set EH such that EH⊂H×H and h,h:h∈H⊂EH. Also, EH has no parallel edge. Note that hPl denotes the path between h and l, that is, there exists a finite sequence kii=0j, for some finite j, such that k0=h, kj=l, and ki,ki+1∈EH for i∈0,1,⋯,j−1.Definition 24.
LetH≠∅be associated the above-definedG, denoted asH,G. A mapσG:H×H→ℝ+is called aG-partialb-metric on the setH, if the following conditions are satisfied for allh,l,z∈HwithhPlandz∈hPl:
σG1h=l if and only if σGh,h=σGh,l=σGl,l;
σG2σGh,h≤σGh,l;
σG3σGh,l=σGl,h;
σG4σGh,l≤sσGh,z+σGz,l−σGz,z, where s≥1.
Then,H,G,σG,s is called a G -partial b-metric space with the coefficient s≥1.Remark 25.
IfhPlandz∈hPl, then we gethPzandzPl. Also note ifhPzandzPl, then we havehPl.
Thus,P is a preorder relation on H. Therefore, H,G,σG,s is also an R-partial b-metric space.Definition 26.
Lethnbe aG-sequence inH,G,σG,s, that is,hnPhn+1orhn+1Phnfor eachn. Then, we say that(i)
hn is a convergent sequence to h∈H if limn→∞σGhn,h=σGh,h and hnPh for each n≥k(ii)
hn is Cauchy if limn,m→∞σGhn,hm exists and is finiteDefinition 27.
H,G,σG,sis said to beG-complete if for each CauchyG-sequence inHthere ish∈Hwithlimn,m→∞σGhn,hm=limn→∞σGhn,h=σGh,handhnPhfor eachn≥k.Note that for a mapT:H→H, the G-0-continuity and G-property are defined in the same way as explained in the last section.Theorem 28.
LetH,G,σG,sbe aG-completeG-partialb-metric space with the coefficients≥1and leth0∈Hbe such thath0Plfor eachl∈H. LetT:H⟶Hbe an edge preserving (ifh,l∈EH, thenTh,Tl∈EH) and aG-property map satisfying the following(36)σGTh,Tl≤kσGh,lforallh,l∈HwithhPl,where k∈0,1/s. Then, T has a fixed point h∗∈H and σGh∗,h∗=0.
By Remark25, we know that P is a preorder relation on H and H,G,σG,s is an R-partial b-metric space. Also, an edge preserving map is path preserving. Thus, all the conditions of Theorem 18 hold. Hence, T has a fixed point.
In the following, we obtain partially-ordered-partialb-metric spaces from R-partial b-metric spaces, by considering ° as a partial order on H.Definition 29.
LetH≠∅be associated with a partial order°, denoted asH°. Given a mapσ°:H×H→ℝ+. If the following conditions are satisfied for allh,l,z∈Hwithh°landh°z°l:
σ°1h=l if and only if σ°h,h=σ°h,l=σ°l,l;
σ°2σ°h,h≤σ°h,l;
σ°3σ°h,l=σ°l,h;
σ°4σ°h,l≤sσ°h,z+σ°z,l−σ°z,z, where s≥1,
thenH,G,σ°,s is called a partially-ordered-partial b-metric space with the coefficient s≥1.As we discussed in the above, we state the following result.Theorem 30.
LetH,G,σ°,sbe an°-complete partially-ordered-partialb-metric space with the coefficients≥1and leth0∈Hbe such thath0°lfor eachl∈H. LetT:H⟶Hbe order preserving (ifh°lthenTh°Tl), and an°-property map satisfying the following:(37)σ°Th,Tl≤kσ°h,lforallh,l∈Hwithh°l,where k∈0,1/s. Then, T has a fixed point h∗∈H and σ°h∗,h∗=0.Remark 31.
≼−completeness is defined in the same way asG-completeness.


## 4. Conclusion

By combining the concepts of orthogonality and the binary relation, we introduced the notion ofR-partial b-metric spaces. We presented some related fixed point results. Some illustrated examples and an application to graphical partial b-metric spaces and partially-ordered-partial b-metric spaces have been provided. As perspectives, it would be interesting to consider in this setting more generalized contraction mappings involving simulation functions or more control functions.



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