# Some New Aspects of Metric Fixed Point Theory

Authors: Khalil Javed; Fahim Uddin; Hüseyin Işık; Tareq M. Al-shami; Faizan Adeel; Muhammad Arshad
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

In this manuscript, we prove fixed point results inRm-metric spaces endowed with an amorphous binary relation. Moreover, we give an example to highlight the utility of our main results. Finally, we apply our result to examine the existence and uniqueness of the solution for a Fredholm integral equation.


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## Body

## 1. Introduction and Preliminaries

In metric fixed point theory, Banach contraction mapping principle [1] is one of the most fundamental tools to investigate the existence and uniqueness of solutions for contraction maps in a complete metric space. Since the appearance of this classical result, researchers have taken keen interest in generalizing and extending this result in different ways, either by improving contraction conditions or by relaxing the axioms of metric space. One may recall the existing notions of, namely, partial metric space [2], partial b-metric space [3], partially ordered S-metric space [4, 5], partial symmetric space [6], partial JS-metric space [7], metric-like space [8], b-metric space [9], rectangular metric space [10], and several others. In 2014, Asadi et al. [11] extended the concept of partial metric space and introduced the notion of m-metric spaces to investigate fixed point. This concept was extended in many different ways, such as mb-metric space [12], mv-metric space [13, 14], and rectangular m-metric space [15]. Also in this sequel, Patle et al. [16] extended the notion of m-metric by proving fixed point results for Nadler and Kannan type set valued mappings in m-metric spaces.Recently, Gordji et al. [17] introduced the notion of orthogonal sets and gave a new extension for the classical Banach contraction principle; more details can be found in [18–20]. Utilizing the structure of orthogonal metric spaces, which appeared in [18, 19], and the binary relation used with a metric, in [21], Ali et al. [22] introduced the notion of R-partial b-metric space and proved fixed point results in this space. Recently, Javed et al. [23] studied fixed point results in fuzzy b-metric space using ordered-theoretic relation, which can also be seen in the work of Alam et al. [24].Inspired by the metric structure used by Ali et al. [22] and using the concept of m-metric space, we introduce the notion of Rm-metric space. We improve and generalize some well-known results and establish fixed point theorems in the sense of Rm-metric space. We also impart some illustrative examples and a possible application for Fredholm integral equations to demonstrate the validity of our results.Definition 1 ([2]).
LetZ be a nonempty set. A function ρ:Z×Z⟶0,+∞ is called a partial metric pm, if the following conditions are satisfied:
(1)
ω=ν iff ρω,ω=ρω,ν=ρν,ν;(2)
ρω,ω≤ρω,ν, for all ω,ν∈Z;(3)
ρω,ν=ρν,ω, for all ω,ν∈Z,(4)
ρω,ν≤ρω,μ+ρμ,ν−ρμ,μ, for all ω,ν,μ∈Z.The pairZ,ρ is called a partial metric space (pms).Notation 2 [11].
(1)
mω,ν=minmω,ω,mν,ν(2)
Mω,ν=maxmω,ω,mν,νIn [11], a generalization of the pms was introduced as follows.Definition 3 ([11]).
LetZ be a nonempty set. A function m:Z×Z⟶0,+∞ is called an m-metric, if the following conditions are satisfied:
(1)
ω=ν iff mω,ω=mω,ν=mν,ν;(2)
mω,ν≤mω,ν, for all ω,ν∈Z;(3)
mω,ν=mν,ω, for all ω,ν∈Z;(4)
mω,ν−mω,ν≤mω,μ−mω,μ+mμ,ν−mμ,ν, for all ω,ν,μ∈Z.The pairZ,m is called an m-ms.Example 4.
LetZ=0,+∞. Then mω,ν=ω−ν on Z is an m-metric.Definition 5 ([25]).
LetZ be a nonempty set. A subset R of Z2 is called a binary relation on Z. Then, for any ω,ν∈Z, we say that “ω is R-related to ν,” that is, ωRν, or “ω relates to ν under R” iff ω,ν∈R. ω,ν∉R means that “ω is not R-related to ν” or “ν is not related to ω under R.”Definition 6 ([25]).
A binary relationR defined on a nonempty set Z is called
(a)
reflexive ifω,ω∈R∀ω∈Z;(b)
irreflexive ifν,ν∉R for some ν∈Z;(c)
symmetric ifω,ν∈R implies ν,ω∈R∀ω,ν∈Z;(d)
antisymmetric ifω,ν∈R and ν,ω∈R imply ω=ν∀ω,ν∈Z;(e)
transitive ifω,ν∈R and ν,μ∈R imply ω,μ∈R∀ω,ν,μ∈Z;(f)
preorder ifR is reflexive and transitive.Definition 7 ([21]).
LetZ,R be an R-set. A sequence ωi is called an R-sequence RSif
(1)∀i∈ℕ,ωiRωi+1or∀i∈ℕ,ωi+1Rωi.Definition 8 ([21]).
A mapg:Z⟶Z is R-continuousorRC in ω∈Z if for each RSωii∈ℕ in Z such that ωi⟶ω then gωi⟶gω. Also, g is said to be RC on Z, if g is RC for each ω∈Z.Definition 9 ([18]).
LetZ,d be a ms and R be a binary relation on Z. Then Z,d,R is called an R-ms.Definition 10 ([18]).
LetZ,R,d be an R-ms. Then Z is said to be R-complete, if every Cauchy RS is convergent.Definition 11 ([18]).
LetZ,R be an R-set. A map g:Z⟶Z is said to be R-preserving RP if gωRgν whenever ωRν. Also, g:Z⟶Z is said to be weakly RP, if gωRgν or gνRgω whenever ωRν.Definition 12 [21].
A mapg:Z⟶Z is an R-contraction, if
(2)dgω,gv≤kdω,v,for all ω,v∈Z with ωRv, where 0<k<1.


## 2. Main Results

We will start this section with the definition of anRm-ms, but first, we introduce the following notations, which would be helpful during the proof.Notation 13.
(1)
σRω,ν=minσRω,ω,σRν,ν;(2)
σRω,ν′=maxσRω,ω,σRν,ν.Definition 14.
LetZ≠∅ and R be a reflexive binary relation (br) on Z, denoted as Z,R. A map σR:Z×Z⟶ℝ+ is called an Rm-metric on Z, if the following conditions are satisfied, for all ω,ν,μ∈Z with either ωRν or νRω, either ωRμ or μRω and either μRν or νRμ:
σR1ω=ν iff σRω,ω=σRω,ν=σRν,ν;
σR2σRω,ν≤σRω,ν;
σR3σRω,ν=σRν,ω;
σR4σRω,ν−σRω,ν≤σRω,μ−σRω,μ+σRμ,ν−σRμ,ν.Then,Z,R,σR is called an Rm-ms.Remark 15.
In the above definition, a setZ is endowed with a reflexive br R and σR:Z×Z⟶ℝ+ satisfies σR1−σR4 only for those elements which are comparable under the reflexive br R. Hence, the Rm-ms may not be a m-metric, but the converse is true.The next example shows that theRm-ms does not need to be an m-ms.Example 16.
LetZ=ℝ and a br R be defined by: ωRν iff ω=ν or ω,ν>0. It is easy to prove that σRω,ν=max∣ω∣,∣ν∣ is an Rm-metric on Z, but σR is not an m-metric on Z; since for ω=−2 and ν=2, we have σRω,ω=σRω,ν=σRν,ν=2.Definition 17.
LetZ,R,σR be an Rm-ms. Let ωn be an RS in Z,R,σR, that is, ωnRωn+1 or ωn+1Rωn for each n∈ℕ. Then, ωn is said to be
(i)
R-convergent to some ω∈Z, if(3)limi⟶∞σRωi,ω−σRωi,ω=0(ii)
R-Cauchy if(4)limi,j⟶∞σRωi,ωj−σRωi,ωj,limi,j⟶∞σRωi,ωj′−σRωi,ωjexist and are finite.Definition 18.
Z,R,σR is said to be R-complete Rm-ms, if for every R-Cauchy sequence in Z is convergent.Definition 19.
LetZ,R,σR be an Rm-ms. We say that g:Z⟶Z is RC at h∈Z if for each RSωi in Z with σRωi,ω⟶0, we have σRgωi,gω⟶0. Also, g is RC on Z if g is RC for each ω∈Z.The following results help us to ensure the existence of fixed pointfp for self-maps. Throughout, we assume that R is a preorder relation.Theorem 20.
LetZ,R,σR be an R-complete Rm-ms and g:Z⟶Z be RP and RC satisfying the following condition:
(5)σRgω,gν≤zσRω,ν,for allω,ν∈ZwithωRν,where z∈0,1.Then, g has a unique fp ω∈Z and σRω,ω=0.Proof.
Letω1=gω0,ω2=gω1,ω3=gω2,ω4=gω3,⋯,ωi+1=gωi, for all i∈ℕ. Since g is RP, ωi is RS. Then, by (5), we get
(6)σRωi,ωi+1=σRgωi−1,gωi≤ziσRω0,ω1,for all i∈ℕ. For any i,j∈ℕ and i<j, it follows that
(7)σRωi,ωj=σRgωi−1,gωj−1≤zσRωi−1,ωj−1≤zσRzσRωi−2,ωj−2≤z2σRωi−2,ωj−2≤⋯≤ziσRω1,ωj−i.
Hence,(8)σRωi,ωj−σRωi,ωj≤ziσRω0,ω1+σRω1,ωj−i≤ziσRω0,ω1+σRω1,ω2+σRω2,ωj−i≤ziσRω0,ω1+σRω1,ω2+⋯+σRωj−i−1,ωj−i≤ziσRω0,ω1+zσRω0,ω1+⋯+zj−1σRω0,ω1≤zi1+z+z2+⋯+zj−1σRω0,ω1≤zi1−zσRω0,ω1.
Asz∈0,1, it follows from the above inequality that
(9)σRωi,ωj−σRωi,ωj⟶0asi,j⟶∞.
Similarly,(10)σRωi,ωj′−σRωi,ωj⟶0asi,j⟶∞.
Thus,ωi is a R-Cauchy in Z. Since Z is R-complete, there exists ω∈Z such that
(11)σRωi,ω−σRωi,ω⟶0asi⟶∞.
Now, we show thatω is a fp of g. Consider
(12)σRω,gω≤limsupi⟶∞σRω,ωi+limsupi⟶∞σRωi,gω=limsupi⟶∞σRωi,gω=limsupi⟶∞σRgωi−1,gω≤limsupi⟶∞zσRωi−1,ω.
By usingσR4, we get
(13)σRω,gω≤limsupi⟶∞zσRωi−1,gω−σRωi−1,gω+σRgω,ω−σRgω,ω−σRgω,gω≤zσRgω,ω,implies
(14)σRω,gω=00≤σRgω,gω≤zσRω,ω=0.
Hence,(15)σRgω,gω=σRω,gω=σRω,ω.
ByσR1, we get gω=ω as desired.
Now, we show that ifω is a fp of g, then σRω,ω=0. Then,
(16)σRω,ω=σRgiω,giω≤zσRω,ω<σRω,ω,and hence, σRω,ω=0.
Finally, we show that fixed point ofg is unique. Assume that there exists ν∈Z such that gν=ν. Hence, we have giω=ω,giν=ν for all i∈ℕ. By the choice of ω0 in the first part of the proof, we obtain
(17)ω0Rωandω0RνorωRω0andνRω0.
Sinceg is RP, we have
(18)giω0Rgiωandgiω0Rgiν,
or(19)giωRgiω0andgiνRgiω0,
for alli∈ℕ. Thus,
(20)σRω,ν=σRgiω,giν≤zσRω,ν<σRω,ν,which implies that σRω,ν=0 and so σRω,ω=σRω,ν=σRν,ν. Thus, ω=ν.☐Example 21.
LetZ=ℝ, and σR:Z×Z⟶ℝ+ be given by
(21)σRω,ν=ω−ν2,ifω,ν≥0,0,otherwise.
Define the binary relationR on Z by ωRν if ω=ν or ω,ν≥0. Then Z,R,σR is R-complete Rm-ms. Define g:Z⟶Z by
(22)gω=ω3,0≤ω≤3,0,otherwise.Then, it is very simple to verify the following:Ifω=ν, then gω=gν. If ω,ν≥0, then gω,gν≥0. Thus, g is RP.Also, one can see thatgω−gν2≤1/9ω−ν2 for all ω,ν∈Z with ωRν. Hence, the inequality (5) is satisfied with z=1/9. But g is not a contraction. Otherwise, for two points 11/4 and 20/4 and for all 0<c<1/3, we have g11/4−g20/42≤c11/4−20/42 and one can conclude that it is a contradiction.Letωi be an arbitrary RS in Z such that ωi converges to ω∈Z. Since the inequality (5) is satisfied, for each i∈ℕ, we have
(23)gωi−gω2≤19ωi−ω2.Asi goes to infinity, g is RC. But it can be easily seen that g is not continuous. Therefore, all the conditions of Theorem 20 are satisfied. Hence, we can conclude that g has a unique fixed point in Z which is 0.Remark 22.
Note that the functionσR defined in the above example is neither a metric, nor an m-metric on ℝ. Since σR4,1=9, σR4,−1=0, σR−1,1=0, and σR−1,−1=0.Theorem 23.
LetZ,R,σR be an R-complete Rm-ms and g:Z⟶Z be RP and RC satisfying the following condition:
(24)σRgω,gν≤zσRω,gω+σRν,gν,for all ω,ν∈ZwithωRν, where z∈0,1. Then g has a unique fp ω∈Z and σRω,ω=0.Proof.
Letω1=gω0,ω2=gω1,ω3=gω2,ω4=gω3,⋯, ωi+1=gωi, for all i∈ℕ. Since g is RP, then ωi is RS. Then by (24), we get
(25)σRωi,ωi+1=σRgωi−1,gωi≤zσRωi−1,gωi−1+σRωi,gωi=zσRωi−1,ωi+σRωi,ωi+1,for all i∈ℕ.σRωi,ωi+1≤zσRωi−1,ωi+zσRωi,ωi+1 which implies σRωi,ωi+1≤cσRωi−1,ωi, where c=z/1−z<1 as z∈0,1. By repeating this process,
(26)σRωi,ωi+1≤ciσRω0,ω1.
Thenlimi⟶∞σRωi,ωi+1=0. By (24), for any two natural number j>i, it follows that
(27)σRωi,ωj≤σRgωi−1,gωj−1≤zσRωi−1,gωi−1+σRωj−1,gωj−1=zσRωi−1,ωi+σRωj−1,ωj,and so limi⟶∞σRωi,ωj=0. For every ξ>0, we can find a natural number i0 such that σRωi,ωi+1<ξ/2 and σRωj,ωj+1<ξ/2, for all i,j>i0. Therefore, it follows that
(28)σRωi,ωj≤zσRωi−1,ωi+σRωj−1,ωj<zξ2+ξ2<ξ2+ξ2=ξ,for all i,j>i0, and σRωi,ωj⟶0 as i,j⟶∞(29)σRωi,ωj−σRωi,ωj<ξ,for alli,j>i0.
Now,(30)σRωi,ωi=σRgωi−1,gωi−1≤zσRωi−1,gωi−1+σRωi−1,gωi−1=zσRωi−1,ωi+σRωi−1,ωi≤2zσRωi−1,ωi⟶0asi⟶∞.
SoσRωi,ωi⟶0 which implies that
(31)σRωi,ωj−σRωi,ωj⟶0asi,j⟶∞.
Similarly,(32)σRωi,ωj′−σRωi,ωj⟶0asi,j⟶∞.
Thus,ωi is a R-Cauchy in Z. Since Z is R-complete, there exists ω∈Z such that
(33)σRωi,ω−σRωi,ω⟶0asi⟶∞.
Now, we show thatω is a fp of g. Consider,
(34)σRω,gω≤limsupi⟶∞σRω,ωi+limsupi⟶∞σRωi,gω=limsupn⟶∞σRωi,gω≤limsupi⟶∞zσRωi−1,gωi−1+σRω,gω≤limsupi⟶∞zσRωi−1,ωi+limsupi⟶∞zσRω,gω≤zσRω,gω,implies
(35)σRω,gω=00≤σRgω,gω≤2zσRω,ω=0.
Hence,(36)σRgω,gω=σRω,gω=σRω,ω.
ByσR1, we get gω=ω.
Now, we show that ifω is a fp of g, then σRω,ω=0. Consider,
(37)σRω,ω=σRgiω,giω≤zσRω,gω+σRω,gω=2zσRω,gω=2zσRω,ω,that is, σRω,ω=0.
Finally, we show that fixed point ofg is a unique. Assume that there exists ν∈Z such that gν=ν. Since g is RP, we have
(38)giω0Rgiωandgiω0Rgiν,or
(39)giωRgiω0andgiνRgiω0,for all i∈ℕ. Thus,
(40)σRω,ν=σRgiω,giν≤zσRω,gω+σRν,gν=zσRω,ω+σRν,ν=0,which implies that σRω,ν=0 and so σRω,ω=σRω,ν=σRν,ν. Thus, ω=ν.☐Theorem 24.
LetZ,R,σR be an R-complete Rm-ms and g:Z⟶Z be RP and RC satisfying the following condition:
(41)σRgω,gν≤zmaxσRω,ν,σRω,gω,σRν,gν,for all ω,ν∈ZwithωRν, where z∈0,1. Then, g has a unique fp ω∈Z and σRω,ω=0.Proof.
Letω1=gω0,ω2=gω1,⋯,ωi+1=gωi, for all i∈ℕ. Since g is RP, then ωi is RS. By (41), we get
(42)σRωi+1,ωi=σRgωi,gωi−1≤zmaxσRωi,ωi−1,σRωi,gωi,σRωi−1,gωi−1=zmaxσRωi,ωi−1,σRωi,ωi+1,σRωi−1,ωi=zmaxσRωi,ωi−1,σRωi,ωi+1.
IfmaxσRωi,ωi−1,σRωi,ωi+1=σRωi+1,ωi. Then, from the above inequality, we obtain that σRωi+1,ωi≤zσRωi+1,ωi, a contradiction. Hence, maxσRωi,ωi−1,σRωi,ωi+1=σRωi,ωi−1. From the above inequality, we have
(43)σRωi+1,ωi≤zσRωi,ωi−1.
On repeating this process, we obtain(44)σRωi+1,ωi≤ziσRω1,ω0,for all i≥0.(45)σRωi,ωj−σRωi,ωj≤ziσRω0,ω1+σRω1,ωj−i≤ziσRω0,ω1+σRω1,ω2+σRω2,ωj−i≤ziσRω0,ω1+σRω1,ω2+⋯+σRωj−i−1,ωj−i≤ziσRω0,ω1+zσRω0,ω1+⋯+zj−1σRω0,ω1≤zi1+z+z2+⋯+zj−1σRω0,ω1≤zi1−zσRω0,ω1.
Asz∈0,1, it follows from the above inequality that
(46)σRωi,ωj−σRωi,ωj⟶0asi,j⟶∞.
Similarly,(47)σR′ωi,ωj−σRωi,ωj⟶0asi,j⟶∞.
Thus,ωi is a R-Cauchy in Z. Since Z is R-complete, there exists ω∈Z such that
(48)σRωi,ω−σRωi,ω⟶0asi⟶∞.
Now, we show thatω is a fp of g. Consider
(49)σRω,gω≤limsupi⟶∞σRω,ωi+limsupi⟶∞σRωi,gω=limsupi⟶∞σRωi,gω=limsupi⟶∞σRgωi−1,gω≤limsupi⟶∞zmaxσRωi−1,ω,σRωi−1,gωi−1,σRω,gω≤zmaxlimsupi⟶∞σRωi−1,ω,limsupi⟶∞ωi−1,ωi,limsupi⟶∞σRω,gω≤zσRω,gω.
So,(50)σRω,gω=00≤σRgω,gω≤zmaxσRω,ω,σRω,gω,σRω,gω=σRω,ω=0.
Thus,(51)σRgω,gω=σRω,gω=σRω,ω.
ByσR1, we get gω=ω as desired.
Now, we show that ifω is a fp of g, then σRω,ω=0. Then,
(52)σRω,ω=σRgiω,giω≤zmaxσRω,ω,σRω,gω,σRω,gω<σRω,ω,and hence, σRω,ω=0.
Finally, we show that fixed point ofg is a unique. Assume that there exists ν∈Z such that gν=ν. Hence, we have giω=ω,giν=ν for all i∈ℕ. By the choice of ω0 in the first part of the proof, we obtain
(53)ω0Rωandω0RνorωRω0andνRω0.
Sinceg is RP, we have
(54)giω0Rgiωandgiω0Rgiν,or
(55)giωRgiω0andgiνRgiω0,for all i∈ℕ. Thus,
(56)σRω,ν=σRgiω,giν≤zmaxσRω,ν,σRω,gω,σRν,gν=zmaxσRω,ν,σRω,ω,σRν,ν=zσRω,ν<σRω,ν,which is a contradiction. Hence, σRω,ν=0 and so σRω,ω=σRω,ν=σRν,ν. Thus, ω=ν.☐


## 3. Application

Within this section, we are attempting to apply Theorem20 to investigate the presence and uniqueness of solution for a Fredholm integral equation. The space of all continuous real valued functions defined on I=0,1 is considered to be Z=CI,ℝ with the Rm-metric given by
(57)σRωr,νr=supr∈Iωr+νr2,for all ω,ν∈Z. It can also be equipped with a relation given by ω,ν∈Z,ωRν⇔ω≤ν. Then, Z,R,σR is R-complete Rm-ms.Consider the following Fredholm integral equation:(58)ωr=∫01Jr,s,ωrds,r,s∈I,where J∈CI×I×R,R.Theorem 25.
If there existsz∈0,1), such that
(59)Jr,s,ωr+Jr,s,νr≤zωr+νr,r,s∈I,for all ω,ν∈Z. Then the integral equation (58) has a unique solution.Proof.
Defineg:Z⟶Zby
(60)gωr=∫01Jr,s,ωrds,r,s∈I.
Observe that the presence of fixed point of an operatorg is identical to that of solution of the integral equation (58). Now, for all ω,ν∈Z with ωRν,
(61)σRgω,gν=supr∈Igωr+gνr2=supr∈I∫01Jr,s,ωr+Jr,s,νr2ds≤supr∈I∫01Jr,s,ωr+Jr,s,νr2ds≤zsupr∈I∫01ωr+νr2ds≤zsupr∈Iωr+νr2∫01ds≤zσRω,ν.
Thus, all the conditions of Theorem20 are fulfilled. Therefore, the operator g has a unique fp, meaning that the integral equation (58) has a unique solution.☐



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