# Sequential ExtendedS-Metric Spaces and Relevant Fixed Point Results with Application to Nonlinear Integral Equations

Authors: Abdollah Karami; Shaban Sedghi; Vahid Parvaneh
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

In this paper, the notion of sequentialςp-metric spaces has been introduced as a generalization of usual S-metric spaces, Sb-metric spaces, SJS metric spaces, and specially of Sp-metric spaces. In view of this notion, we prove some fixed point theorems for some classes of ςp-rational Geraghty JS-contractions over such spaces. A supporting example and an application have been given in order to examine the validity of the obtained results.


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## Body

## 1. Introduction and Preliminaries

There is a huge number of extensions of Banach contraction principle. Some of them focus on using different forms of contractive conditions and some of them focus on the various generalized metric spaces. There are many interesting generalization of metric type spaces such asb-metric space [1], extended b-metric space [2], p-metric space [3], JS-metric spaces [4], C∗-algebra valued metric space [5], S-metric space, Sb-metric space [6, 7], Sp-metric space [8], and G-metric spaces [9]. Also, for considering and analyzing more generalization of the concept of metric spaces, one can consider the following works dealing with controlled metric spaces and generalized b-metric spaces [10, 11]. In the context of various metric type spaces which are the combination of the above mentioned spaces, several authors have proved different types of fixed point theorems [12]. Now, we present some definitions of some generalized metric spaces which are pertinent to our research.LetΩ be the class of all strictly increasing continuous functions ω:0,∞→0,∞ with ω−1t≤t≤ωt and ω−10=0=ω0.Definition 1 (see [3]).
LetO be a nonempty set. A function p:O×O→0,∞ is a p-metric if for some ω∈Ω and for all κ,κ′,κ″∈O,
(1)
pκ,κ′=0 if and only if κ=κ′(2)
pκ,κ′=pκ′,κ(3)
pκ,κ″≤ωpκ,κ′+pκ′,κ″ for all κ,κ′,κ″∈O
In this case, the pairO,p is called a p-metric space, or an extended b-metric space.
Ab-metric [1] is a p-metric with ωt=st, for some fixed s≥1. A p-metric space reduces to an ordinary metric space provided that ωt=t.
LetΛ be a nonempty set and JS:Λ×Λ→0,∞ be a mapping. For any κ∈Λ, let us define the set
(1)CJS,Λ,κ=κn⊂Λ:limn→∞JSκn,κ=0.Definition 2 (JS-metric space) (see [4]).
LetJS:Λ×Λ→0,∞ be a mapping so that
(1)
JSκ,κ′=0 implies κ=κ′(2)
for everyκ,κ′∈Λ, we have JSκ,κ′=JSκ′,κ(3)
ifκ,κ′∈Λ×Λ and κn∈CJS,Λ,κ then JSκ,κ′≤plimsupn→∞JSκn,κ′, for some p>0
The pairΛ,JS is called a generalized metric space, usually known as JS-metric space.
In [13], Roy et al. introduced a new type of extended b-metric spaces. First, let Cσp,Λ,κ, be defined as above, where JS is substituted by σp.Definition 3 (see [13]).
LetΛ be a nonempty set. A mapping σp:Λ×Λ→0,∞ is said to be a sequential p-metric if for all a,b∈Λ:
(a)
σpa,b=0 implies a=b(b)
σpa,b=σpb,a(c)
σpa,b≤ωlimsupn→∞σpan,b, where an∈Cσp,Λ,a and ω∈Ω so that ω−1s=s=ωs for s∈0,∞
The tripletΛ,σp,ω is called a sequential p-metric space. We represent a sequential p-metric space simply as Λ,σp.Example 4 (see [13]).
LetΛ=N and the metric σp:Λ2→0,∞ be defined by
(2)σp1,1=0;σpn,n=e−1,forn≥2;σp1,n=σpn,1=e1/n+1−1,forn≥2;σpn,m=σpm,n=emn−1,foralln,m≥2withn≠m.
σp is a sequential p-metric on Λ for ω1t=t for all t≥0 and ω2t=et−1 for all t≥0.Proposition 5 (see [13]).
IfΛ,JS is a JS-metric space, then JS is also a sequential p-metric on Λ.Proposition 6 (see [13]).
LetΛ,JS be a JS-metric space with coefficient p≥1. Let σpa,b≔ΓJSa,b, where Γ∈Ω. Then, σp is a sequential p-metric with ωt=Γpt for all t≥0.Sedghi et al. [14] have recommended the notion of a S-metric space as follows.Definition 7 (see [14]).
LetO be a nonempty set. An S-metric on O is a function S:O×O×O→0,∞ such that,
(3)Sκ,κ′,κ″≥0,and(1)
Sκ,κ′,κ″=0 if and only if κ=κ′=κ″(2)
Sκ,κ′,κ″≤Sκ,κ,a+Sκ′,κ′,a+Sκ″,κ″,a for all κ,κ′,κ″,a∈O
for eachκ,κ′,κ″,a∈O.
The pairO,S is nominated an S-metric space.
Some examples of suchS-metric spaces are
(1)
LetO be a normed space. Then, Sκ,κ′,κ″=κ′+κ″−2κ+κ′−κ″ is an S-metric on O(2)
LetO be a normed space. Then, Sκ,κ′,κ″=κ−κ″+κ′−κ″ is an S-metric on O(3)
LetO be a nonempty set and d be an ordinary metric on O. Then, Sκ,κ′,κ″=dκ,κ′+dκ′,κ″ is an S-metric on OLemma 8 (see [14]).
In anS-metric space, we have Sκ,κ,κ′=Sκ′,κ′,κ.Definition 9 (see [8]).
LetO be a nonempty set and ω∈Ω. Suppose that a mapping ςp:O×O×O→0,∞ satisfies:
(1)
ςpκ,κ′,κ″=0 iff κ=κ′=κ″(2)
ςpκ,κ′,κ″≤ωςpκ,κ,a+ςpκ′,κ′,a+ςpκ″,κ″,a for all κ,κ′,κ″,a∈O
Thenςp is called a modified S-metric, and the pair O,ςp is called a modified S-metric space.
AnS-metric space is an ςp-metric space where ωt=t and every Sb-metric space with parameter s≥1 is an ςp-metric space where ωt=st.Definition 10.
Aςp-metric is called symmetric if ςpκ,κ,κ′=ςpκ′,κ′,κ, for all κ,κ′∈O.Example 11 (see [8]).
LetO,S be an S-metric space. Then
(1)
ςpκ,κ′,κ″=eSκ,κ′,κ″−1 is an ςp-metric with ωt=et−1(2)
ςpκ,κ′,κ″=Sκ,κ′,κ″eSκ,κ′,κ″ is an ςp-metric with ωt=tetIn general, anςp-metric with nontrivial function ω need not to be jointly continuous in all its variables (see [15]). The following simple lemma is a modification of Lemma 1.11 of [8].Lemma 12.
LetO,ςp be an ςp-metric space.
(1)
Suppose thatκn and κn′ are ςp-convergent to κ and κ′, respectively. Then, we have(4)ω−2ςpκ,κ,κ′≤liminfn→∞ςpκn,κn,κn′≤limsupn→∞ςpκn,κn,κn′≤ω2ςpκ,κ,κ′.
In particular, ifκ=κ′, then, we have limn→∞ςpκn,κn,κn′=0.
(2)
Suppose thatςp-metric O,ςp is symmetric and κn is ςp-convergent to κ and κ″∈O is arbitrary. Then, we have(5)ω−1ςpκ,κ,κ″≤liminfn→∞ςpκn,κn,κ″≤limsupn→∞ςpκn,κn,κ″≤ωςpκ,κ,κ″.Proof.
(1) Using the rectangle inequality in theςp-metric space, it is easy to see that
(6)ςpκ,κ,κ′≤ω2ςpκ,κ,κn+ςpκ′,κ′,κn≤ω2ςpκ,κ,κn+ω2ςpκ′,κ′,κn′+ςpκn,κn,κn′,ςpκn,κn,κn′≤ω2ςpκn,κn,κ+ςpκn′,κn′,κ≤ω2ςpκn,κn,κ+ω2ςpκn′,κn′,κ′+ςpκ,κ,κ′.
The desired result is obtained via lettingn→∞ in the above inequalities and taking lower limit and the upper limit in the first and second inequality, respectively.
(2) Using the rectangle inequality, we see that(7)ω−1ςpκ,κ,κ″≤2ςpκ,κ,κn+ςpκn,κn,κ″,ςpκn,κn,κ″≤ω2ςpκn,κn,κ+ςpκ″,κ″,κ.
The desired result is obtained via takingliminf as n→∞ in the first inequality and the limsup as n→∞ in the second inequality, respectively.☐


## 2. Introduction to Sequentialςp-Metric Spaces

In this section, we introduce a new type of extendedςp-metric spaces. First, let us to define
(8)Cςp,O,a=an⊂O:limn→∞ςpa,a,an=0,whereςp:O×O×O→0,∞ is a given mapping and O is a nonempty set.Definition 13.
LetO be a nonempty set. A mapping ςp:O×O×O→0,∞ is said to be a sequential ςp-metric if for all a,b,c∈O:
(1)
ςpa,b,c=0 implies a=b=c(2)
ςpa,b,c≤ωlimsupn→∞ςpa,a,cn+limsupn→∞ςpb,b,cn, where cn∈Cςp,O,c and ω∈Ω so that ω−1s=s=ωs for s∈0,∞
The tripletO,ςp,ω is called a sequential ςp-metric space (SSPMS). It will be shown simply as O,ςp.The above definition is an extension of Definition13 of [13], that is, the definition of SJS-metric spaces.Example 14.
Let the tripletΛ,σp,ω be a sequential p-metric space. If we define ςpκ,κ′,κ″=σpκ,κ′+σpκ′,κ″, then, ςp will be a sequential ςp-metric with the same control function ω provided the ω is subadditive.According to the above example and Example15, we construct the following example of ςp-metrics.Example 15.
LetΛ=ℕ and the metric ςp:Λ3→0,∞ be defined by
(9)ςp1,1,1=0;ςpn,n,n=2e−2,forn≥2;ςp1,n,n=ςp1,1,n=ςpn,n,1=ςpn,1,1=e1/n+1+e−2,n≥2;ςpn,1,n=ςp1,n,1=2e1/n+1−2,n≥2;ςpn,n,m=ςpm,n,n=ςpn,m,m=ςpm,m,n=emn+e−2,m,n≥2,m≠n,ςpm,n,m=ςpn,m,n=2emn−2,m,n≥2,m≠n,ςpm,n,o=emn−2+eno−2,m,n,o≥2,n≠m.
ςp is a sequential p-metric on Λ for ω1t=t.Proposition 16.
LetO,ςp be a SJS-metric space with coefficient k≥1. Let ςp′a,b,c=Γςpa,b,c, where Γ∈Ω. Then, ςp′ is a sequential ςp-metric for ωt=Γkt for all a,b,c∈O and t≥0.Proof.
Here, we show thatO,ςp′ satisfies all the conditions of Definition 25.
(a)
ςp′a,b,c=0 gives Γςpa,b,c=0. Then ςpa,b,c=Γ−10=0 which implies a=b=c(b)
For alla,b,c∈O, we have ςp′a,b,c=Γςpa,b,c≤Γklimsupn→∞ςpa,a,cn+klimsupn→∞ςpb,b,cn=ωlimsupn→∞ςpa,a,cn+limsupn→∞ςpb,b,cn where cn∈Cςp,O,c=Cςp′,O,c.☐Proposition 17.
LetO,ςp be a SSPMS with control function ω. Let ςp′a,b,c=Γςpa,b,c, where Γ∈Ω. Then, ςp′ is a sequential ςp-metric with ω′t=Γωt for all t≥0.Proof.
Here, we show thatO,ςp′ satisfies all the conditions of Definition 25.
(a)
ςp′a,b,c=0 gives Γςpa,b,c=0. Then ςpa,b,c=Γ−10=0 which implies a=b=c(b)
For alla,b,c∈O, we have ςp′a,b,c=Γςpa,b,c≤Γωlimsupn→∞ςpa,a,cn+ωlimsupn→∞ςpb,b,cn where cn∈Cςp,O,c=Cςp′,O,c.☐Definition 18.
LetO,ςp be a SSPMS. Also, let an be a sequence in O and a∈O.
(i)
an is said to be convergent and converges to a if an∈Cςp,O,a,(ii)
an is said to be Cauchy if limn,m→∞ςpan,an,am=0(iii)
O is said to be complete if any Cauchy sequence in O is convergentDefinition 19 (see [15]).
LetO,ςp and O′,ςp∗ be two sequential ςp-metric spaces. A mapping H:O→O′ is called continuous at a point a∈O if for any ε>0 there exists δε>0 such that for any l∈O, ςp∗Hl,Hl,Ha<ε whenever ςpl,l,a<δε. H is said to be continuous on O if H is continuous at each Point of O.Proposition 20.
In a SSPMSO,ςp if a sequence an is convergent, then, it converges to unique element in O.Proposition 21.
LetO,ςp be a SSPMS and an⊂O converges to some a∈O. Then, ςpa,a,a=0.Proof.
Sincean converges to a∈O, so limn→∞ςpan,an,a=0. Therefore, we have ςpa,a,a≤ω2limsupn→∞ςpa,a,an=ω0=0 which implies ςpa,a,a=0.Proposition 21.
Letan be a Cauchy sequence in a SSPMS O,ςp,ω such that ω−1 is continuous. If an has a convergent subsequence ank which converges to an a∈O, then, an also converges to a∈O.Proof.
From condition (3) of Definition25, we have
(10)ςpa,a,an≤ωlimsupk→∞ςpank,an,an,which implies that
(11)ω−1ςpa,a,an≤limsupk→∞ςpank,an,an,for all n∈N. Due to the Cauchyness of an, it follows that limn,k→∞ςpank,an,an=0 and thus ω−1ςpa,a,an→0 as n→∞ which implies that ςpa,a,an→0 as n→∞, since ω−1 is continuous. Hence, an converges to a∈O.Proposition 22.
In a SSPMSO,ςp if a self mapping H is continuous at a∈O, then Han∈Cςp,O,Ha for any sequence an∈Cςp,O,a.Proof.
Letε>0 be given. Since H is continuous at a, then for any ε>0 there exists δε>0 such that ςc,c,a<δε implies ςHc,Hc,Ha<ε. As an converges to a, so for δε>0, there exists N∈N such that ςa,a,an<δε for all n≥N. Therefore, for any n≥N, ςHa,Ha,Han<ε and thus Han→Ha as n→∞.☐Definition 23.
LetO,ςp be a SSPMS with supporting function ω. Define
(12)Bpa,ε≔b∈O:ςpa,a,b<ςpa,a,a+ε,Bpa,ε≔b∈O:ςpa,a,b⩽ςpa,a,a+ε,
for alla∈O and ε>0.Remark 24.
One can easily check that the collection(13)τςp≔∅∪U≠∅⊂O:foranya∈Uthereexistsε>0:Bpa,ε⊂U,
forms a topology onO.
Another approach to define a sequentialςp-metric is as follows:
Define(14)Cςp,O,a=an⊂O:limn→∞ςpan,an,a=0,where ςp:O×O×O→0,∞ is a given mapping, and O is a nonempty set.Definition 25.
LetO be a nonempty set. A mapping ςp:O×O×O→0,∞ is said to be a sequential ςp-metric if for all a,b,c∈O:
(1)
ςpa,b,c=0 implies a=b=c(2)
ςpa,b,c≤ωlimsupn→∞ςpcn,cn,a+limsupn→∞ςpcn,cn,b, where cn∈Cςp,O,c and ω∈Ω and ω−1s=s=ωs for s∈0,∞
The tripletO,ςp,ω is called a sequential ςp-metric space. It will be shown simply as O,ςp.In this paper, we investigate the existence of unique fixed point for some rational contractions of Jleli-Samet and Geraghty type in ordered sequentialςp-spaces. Our motivation is an interesting generalization of Banach contraction principle which is presented by Jleli et al. in [16].


## 3. Main Results

Note thatO,⪯ possesses the s.l.c.p. whenever κn is a nondecreasing sequence in O such that κn→U∈O, one has κn⪯U for all n∈ℕ. From now on, by SSPMS, we mean a sequential ςp-metric space and by SPRGJSC a ςp-rational Geraghty JS-contraction.LetG denotes the class of all functions G:0,∞→0,ω−11 satisfying the following condition:
(15)limsupn→∞Gtn=ω−11impliesthattn→0,asn→∞.The following are two examples of Geraghty functionsG.
(1)
Gt=e−tpetq for all t>0 and Gt=0 for t=0, where p,q>0(2)
Gt=1/1+tp for all t>0 and Gt=0 for t=0, where p>0In the above examples, we assumed thatωt=t. In the case that, for instance, ωt=sinht, then ω−11≃0.88, and one can multiple this amount to the above functions ω.Definition 26.
LetO,ςp be an ordered SSPMS. A mapping Q:O→O is called a SPRGJSC of type I if
(16)θςpQκ,Qκ′,Qκ″≤θMκ,κ′,κ″GMκ,κ′,κ″,
for someG∈G and for all comparable elements κ,κ′,κ″∈O, where
(17)Mκ,κ′,κ″=maxςpκ,κ′,κ″,ςpκ,Qκ,Qκςpκ′,Qκ′,Qκ″1+ςpκ,κ′,κ″,ςpκ,Qκ,Qκςpκ′,Qκ′,Qκ″1+ςpQκ,Qκ′,Qκ″.Theorem 27.
LetO,⪯,ςp,ω be an ordered ςp-complete SSPMS. Let Q:O→O be a ⪯-increasing mapping such that κ0⪯Qκ0 for some element κ0∈O. Suppose that Q be a SPRGJSC of type I. If
(I)
Q is continuous, or(II)
O,⪯ possesses the s.l.c.p
thenQ has a fixed point. Moreover, the set of fixed points of Q is well ordered if and only if FQ is a singleton.Proof.
Putκn=Qnκ0. Since κ0⪯Qκ0 and Q is ⪯-increasing, we obtain by induction that
(18)κ0⪯Qκ0⪯Q2κ0⪯⋯⪯Qnκ0⪯Qn+1κ0⪯.We will do the proof in the following steps.Step I.
We will show thatlimn→∞ςpκn,κn+1,κn+1=0. Without any loss of generality, we may assume that κn≠κn+1, for all n∈ℕ. Since κn⪯κn+1 for each n∈ℕ, then by (16), we have
(19)θςpκn,κn+1,κn+1=θςpQκn−1,Qκn,Qκn≤θMκn−1,κn,κnGMκn−1,κn,κn,because
(20)Mκn−1,κn,κn=maxςpκn−1,κn,κn,ςpκn−1,Qκn−1,Qκn−1ςpκn,Qκn,Qκn1+ςpκn−1,κn,κn,ςpκn−1,Qκn−1,Qκn−1ςpκn,Qκn,Qκn1+ςpQκn−1,Qκn,Qκn=maxςpκn−1,κn,κn,ςpκn−1,κn,κnςpκn,κn+1,κn+11+ςpκn−1,κn,κn,ςpκn−1,κn,κnςpκn,κn+1,κn+11+ςpκn,κn+1,κn+1≤maxςpκn−1,κn,κn,ςpκn,κn+1,κn+1,and if maxςpκn−1,κn,κn,ςpκn,κn+1,κn+1=ςpκn,κn+1,κn+1, then from (19), we have
(21)θςpκn,κn+1,κn+1≤θςpκn,κn+1,κn+1GMκn−1,κn,κn<θςpκn,κn+1,κn+1ω−11<θςpκn,κn+1,κn+1,which is a contradiction.
Hence,maxςpκn−1,κn,κn,ςpκn,κn+1,κn+1=ςpκn−1,κn,κn. So, from (19),
(22)θςpκn,κn+1,κn+1≤θςpκn−1,κn,κnGMκn−1,κn,κn<θςpκn−1,κn,κn.
That is,ςpκn,κn+1,κn+1 is a decreasing sequence, so, there exists r≥0 such that limn→∞ςpκn,κn+1,κn+1=r. We will prove that r=0. Suppose on contrary that r>0. Then, letting n→∞, from (22), we have
(23)θr≤θrlimn→∞GMκn−1,κn,κn,which implies that ω−11≤1≤limn→∞GMκn−1,κn,κn. Now, as G∈G we conclude that Mκn−1,κn,κn→0 which yields that r=0, a contradiction. Hence,
(24)limn→∞ςpκn,κn+1,κn+1=0.Step II.
Now, we show that the sequenceκn is a ςp-Cauchy sequence. Suppose that κn is not ςp-Cauchy. Then for some ε>0 we can find two subsequences κmi and κni of κn such that mi and ni are the smallest index such that
(25)ni>mi>iandςpκmi,κni,κni≥ε.
This means that(26)ςpκmi−1,κni−1,κni−1<ε.
From the definition ofMκ,κ′,κ″ and the above limits,
(27)limsupi→∞Mκmi−1,κni−1,κni−1=limsupi→∞maxςpκmi−1,κni−1,κni−1,ςpκmi−1,Qκmi−1,Qκmi−1ςpκni−1,Qκni−1,Qκni−11+ςpκmi−1,κni−1,κni−1,ςpκmi−1,Qκmi−1,Qκmi−1ςpκni−1,Qκni−1,Qκni−11+ςpQκmi−1,Qκni−1,Qκni−1=limsupi→∞maxςpκmi−1,κni−1,κni−1,ςpκmi−1,κmi,κmiςpκni−1,κni,κni1+ςpκmi−1,κni−1,κni−1,ςpκmi−1,κmi,κmiςpκni−1,κni,κni1+ςpκmi,κni,κni≤ε.
Now, from (16) and the above inequalities, we have
(28)θε≤θlimsupi→∞ςpκmi,κni,κni≤limsupi→∞θMκmi−1,κni−1,κni−1GMκmi−1,κni−1,κni−1≤θεω−11<θε,
a contradiction. Therefore,κn is a ςp-Cauchy sequence. ςp-Completeness of O yields that κnςp-converges to a point U∈O.Step III.
U is a fixed point of Q.
First, letQ is continuous, so, we have
(29)U=limn→∞κn+1=limn→∞Qκn=fU.
Now, let (II) holds. Using the assumption onO, we have κn⪯U. Now, we show that U=fU. From definition of a sequential ςp-metric,
(30)ςpU,U,fU≤ωlimsupn→∞ςpκn+1,κn+1,fU≤2ωθ−1limsupn→∞θMκn,κn,Ulimsupn→∞GMκn,κn,U,where
(31)limn→∞Mκn,κn,U=limn→∞maxςpκn,κn,U,ςpκn,Qκn,Qκnςpκn,Qκn,fU1+ςpκn,κn,U,ςpκn,Qκn,Qκnςpκn,Qκn,fU1+ςpQκn,Qκn,fU=0.
Therefore, we deduce thatςpU,U,fU≤ω0=0, so, U≠V.
Finally, suppose that the set of fixed points ofQ be well ordered. Assume that U and V are two fixed points of Q such that U≠V. Then by (16), we have
(32)θςpU,V,V=θςpfU,fV,fV≤θMU,V,VGMU,V,V=θςpU,V,VGςpU,V,V<θςpU,V,Vω−11.
Because(33)MU,V,V=maxςpU,V,V,ςpU,U,UςpV,V,V1+ςpU,V,V=ςpU,V,V.So, we getlnθςpU,V,V<ω−11lnθςpU,V,V, a contradiction. Hence U=V, and Q has a unique fixed point. Conversely, if Q has a unique fixed point, then the set of fixed points of Q is a singleton which is well ordered.Corollary 28.
LetO,ςp,⪯ be an ordered ςp-complete SSPMS. Let Q:O→O be an ⪯-increasing mapping such that κ0⪯Qκ0 for some element κ0∈O. Suppose that
(34)θςpQκ,Qκ′,Qκ″≤θα1ςpκ,κ′,κ″+α2ςpκ,Qκ,Qκςpκ′,Qκ′,Qκ″1+ςpκ,κ′,κ″+α3ςpκ,Qκ,Qκςpκ′,Qκ′,Qκ″1+ςpQκ,Qκ′,Qκ″λ,where αi∈0,1, λ∈0,1, and ∑i=13αi<1. If
(I)
Q is continuous, or(II)
O,⪯ possesses the s.l.c.p
thenQ has a fixed point. Moreover, the set of fixed points of Q is well ordered if and only if Q has one and only one fixed point.Definition 29.
LetO,ςp,⪯ is an ordered SSPMS. A mapping Q:O→O is called a SPRGJSC of type II if there exists G∈G such that,
(35)θςpQκ,Qκ′,Qκ″≤Mκ,κ′,κ″GMκ,κ′,κ″,
for all comparable elementsκ,κ′,κ″∈O, where
(36)Mκ,κ′,κ″=maxςpκ,κ′,κ″,ςpκ,Qκ,Qκςpκ′,Qκ′,Qκ′1+ςpκ,κ′,κ′+ςpκ,κ,Qκ′+ςpκ′,κ′,Qκ,ςpκ′,Qκ′,Qκ′ςpκ″,Qκ″,Qκ″1+ςpκ′,Qκ″,Qκ″+ςpκ′,κ′,Qκ″+ςpκ″,κ″,Qκ′,ςpκ′,Qκ″,Qκ″ςpκ″,Qκ′,Qκ′1+ςpκ″,κ″,Qκ′+ςpκ′,Qκ′,Qκ′+ςpκ″,Qκ″,Qκ″.Theorem 30.
LetO,ςp,⪯ be an ordered ςp-complete SSPMS. Let Q:O→O be a ⪯-increasing mapping such that κ0⪯Qκ0 for an element κ0∈O. Suppose that Q be a SPRGJSC of type II. If
(I)
Q is continuous, or(II)
O,⪯ possesses the s.l.c.p
thenQ possesses a fixed point. Moreover, the set of fixed points of Q is well ordered if and only if Q has one and only one fixed point.Proof.
Letκn=Qnκ0.Step I.
We will show thatlimn→∞ςpκn,κn+1,κn+1=0. Since κn⪯κn+1 for each n∈ℕ, then by (35), we have
(37)θςpκn,κn+1,κn+1≤θςpκn,κn+1,κn+1=θςpQκn−1,Qκn,Qκn+1≤Mκn−1,κn,κn+1≤Gςpκn−1,κn,κn+1ςpκn−1,κn,κn+1GMκn−1,κn,κn+1<ςpκn−1,κn,κn+1ω−11≤ςpκn−1,κn,κn+1,because
(38)Mκn−1,κn,κn=maxςpκn−1,κn,κn,ςpκn−1Qκn−1,Qκn−1ςpκn,Qκn,Qκn1+ςpκn−1,κn,κn+ςpκn−1,κn−1,Qκn+ςpκn,κn,Qκn−1,ςpκn,Qκn,Qκnςpκn,Qκn,Qκn1+ςpκn,Qκn,Qκn+ςpκn,κn,Qκn+ςpκn,κn,Qκn,ςpκn,Qκn,Qκnςpκn,Qκn,Qκn1+ςpκn,κn,Qκn+ςpκn,Qκn,Qκn+ςpκn,Qκn,Qκn=maxςpκn−1,κn,κn,ςpκn−1,κn,κnςpκn,κn+1,κn+11+ςpκn−1,κn,κn+ςpκn−1,κn−1,κn+1+ςpκn,κn,κn,ςpκn,κn+1,κn+1ςpκn,κn+1,κn+11+ςpκn,κn+1,κn+1+ςpκn,κn,κn+1+ςpκn,κn,κn+1,ςpκn,κn+1,κn+1ςpκn,κn+1,κn+11+ςpκn,κn,κn+1+ςpκn,κn+1,κn+1+ςpκn,κn+1,κn+1≤maxςpκn−1,κn,κn,ςpκn,κn+1,κn+1.
Therefore,ςpκn,κn+1,κn+1 is decreasing. Similar to what have been done in Theorem 27, we have
(39)limn→∞ςpκn−1,κn,κn=0.Step II.
Now, we prove that the sequenceκn is a ςp-Cauchy sequence. In other case, i.e., that κn is not a ςp-Cauchy sequence, for an ε>0, we can find two subsequences κmi and κni of κn such that mi and ni are the smallest index for which
(40)ni>mi>iandςpκmi,κni,κni≥ε.
This means that(41)ςpκmi−1,κni−1,κni−1<ε.
From the definition ofMκ,κ′,κ″ and the above limits,
(42)limsupi→∞Mκmi−1,κni−1,κni−1=limsupi→∞maxςpκmi−1,κni−1,κni−1,ςpκmi−1,Qκmi−1,Qκmi−1ςpκni−1,Qκni−1,Qκni−11+ςpκmi−1,κni−1,κni−1+ςpκmi−1,κmi−1,Qκni−1+ςpκni−1,κni−1,Qκmi−1,ςpκni−1,Qκni−1,Qκni−1ςpκni−1,κni−1,Qκni−11+ςpκni−1,Qκni−1,Qκni−1+ςpκni−1,κni−1,Qκni−1+ςpκni−1,κni−1,Qκni−1,ςpκni−1,Qκni−1,Qκni−1ςpκni−1,Qκni−1,Qκni−11+ςpκni−1,κni−1,Qκni−1+ςpκni−1,Qκni−1,Qκni−1+ςpκni−1,Qκni−1,Qκni−1≤ε
Now, from (35) and the above inequalities, we have
(43)θε≤θlimsupi→∞ςpκmi,κni,κni≤limsupi→∞θMκmi−1,κni−1,κniGMκmi−1,κni−1,κni−1≤θεω−11,
which implies thatω−11≤limsupi→∞GMκmi,κni−1,κni−1. Now, as G∈G, we conclude that κn is a ςp-Cauchy sequence. ςp-Completeness of O yields that κnςp-converges to a point U∈O.Step III.
U is a fixed point of Q.
WhenQ is continuous, the proof is straightforward.
Now, let (II) holds. We leave the proof as it is similar to the proof of step III of Theorem27.Corollary 31.
LetO,ςp,⪯ be an ordered ςp-complete SSPMS. Let Q:O→O be a ⪯-increasing mapping so that κ0⪯Qκ0 for an element κ0∈O. Suppose that
(44)θςpQκ,Qκ′,Qκ″≤θα1ςpκ,κ′,κ″+α2ςpκ,Qκ,Qκςpκ′,Qκ′,Qκ′1+ςpκ,κ′,κ′+ςpκ,κ,Qκ′+ςpκ′,κ′,Qκ+α3ςpκ′,Qκ′,Qκ′ςpκ″,Qκ″,Qκ″1+ςpκ′,Qκ″,Qκ″+ςpκ′,κ′,Qκ″+ςpκ″,κ″,Qκ′+α4ςpκ′,Qκ″,Qκ″ςpκ″,Qκ′,Qκ′1+ςpκ″,κ″,Qκ′+ςpκ′,Qκ′,Qκ′+ςpκ″,Qκ″,Qκ″λ
whereαi∈0,1, λ∈0,1, and ∑i=14αi<1. If
(I)
Q is continuous, or(II)
O,⪯ possesses the s.l.c.p
thenQ admits a fixed point. In addition, the set of fixed points of Q is well ordered if and only if FixQ is a singleton.The Banach contraction principle is immediately obtained from the mentioned corollaries28, 34, and 31.Definition 32.
LetO,ςp,⪯ is an ordered SSPMS. A mapping Q:O→O is called a SPGJSC if there exists G∈G, such that
(45)θςpQκ,Qκ′,Qκ″≤θMκ,κ′,κ″GMκ,κ′,κ″,
for all comparable elementsκ,κ′,κ″∈O, where
(46)Mκ,κ′,κ″=maxςpκ,κ′,κ″,ςpκ′,Qκ″,Qκ″,ςpκ″,Qκ′,Qκ′.Theorem 33.
LetO,ςp,⪯ be an ordered ςp-complete SSPMS. Let Q:O→O be an ⪯-increasing mapping such that κ0⪯Qκ0 for some element κ0∈O. Suppose that Q be a SPGJSC. If
(III)
Q is continuous, or(IV)
O,⪯ possesses the s.l.c.p
thenQ has a fixed point. Moreover, the set of fixed points of Q is well ordered if and only if Q has one and only one fixed point.Proof.
Assume thatκn=Qnκ0. Since κ0⪯Qκ0 and Q is an ⪯-increasing function, we obtain by induction that
(47)κ0⪯Qκ0⪯Q2κ0⪯⋯⪯Qnκ0⪯Qn+1κ0⪯.We shall do the proof as follows.Step I.
We will show thatlimn→∞ςpκn,κn+1,κn+1=0. Since κn⪯κn+1 for each n∈ℕ, then by (45), we have
(48)θςpκn,κn+1,κn+1=θςpQκn−1,Qκn,Qκn≤θMκn−1,κn,κnGMκn−1,κn,κn≤θςpκn−1,κn,κnGMκn−1,κn,κn<θςpκn−1,κn,κnω−11≤θςpκn−1,κn,κn,because
(49)Mκn−1,κn,κn=maxςpκn−1,κn,κn,ςpκn,Qκn,Qκn,ςpκn,Qκn,Qκn=maxςpκn−1,κn,κn,ςpκn,κn+1,κn+1,ςpκn,κn+1,κn+1≤maxςpκn−1,κn,κn,ςpκn,κn+1,κn+1,
and it is easy to see thatmaxςpκn−1,κn,κn,ςpκn,κn+1,κn+1=ςpκn−1,κn,κn, so from (48), we conclude that ςpκn,κn+1,κn+1 is decreasing. Then, there exists r≥0 such that limn→∞ςpκn,κn+1,κn+1=r.
We will prove thatr=0. This can be done as in the previous theorem. So,
(50)limn→∞ςpκn−1,κn,κn=0,holds true.Step II.
Now, we prove that the sequenceκn is a ςp-Cauchy sequence. Suppose that κn is not a ςp-Cauchy sequence. Then for some ε>0, we can find two subsequences κmi and κni of κn such that mi and ni are the smallest index for which
(51)ni>mi>iandςpκmi,κni,κni≥ε,
i.e.,(52)ςpκmi−1,κni−1,κni−1<ε.
From the definition ofMκ,κ′,κ″ and the above limits,
(53)limsupi→∞Mκmi−1,κni−1,κni−1=limsupi→∞maxςpκmi−1,κni−1,κni−1,ςpκni−1,Qκni−1,Qκni−1,ςpκni−1,Qκni−1,Qκni−1≤ε.
Now, from (45) and the above inequalities, we have
(54)θε≤θlimsupi→∞ςpκmi,κni,κni≤limsupi→∞θMκmi−1,κni−1,κni−1limsupi→∞GMκmi−1,κni−1,κni−1≤θεlimsupi→∞GMκmi−1,κni−1,κni−1,
which implies that1<limsupi→∞GMκmi,κni−1,κni−1≤ω−11. Now, as G∈G, we conclude that κn is a ςp-Cauchy sequence. ςp-Completeness of O yields that κnςp-converges to a point U∈O.Step III.
U is a fixed point of Q. This step is proved as the proof of step III of Theorem 27.Corollary 34.
LetO,ςp,⪯ be an ordered ςp-complete SSPMS. Let Q:O→O be an ⪯-increasing mapping such that κ0⪯Qκ0 for an element κ0∈O. Suppose that
(55)θςpQκ,Qκ′,Qκ″≤θα1ςpκ,κ′,κ″+α2ςpκ′,Qκ″,Qκ″+α3ςpκ″,Qκ′,Qκ′λ,
whereαi∈0,1, λ∈0,1 and ∑i=13αi<1. If
(I)
Q is continuous, or(II)
O,⪯ possesses the s.l.c.p
thenQ has a fixed point. Moreover, the set of fixed points of Q is well ordered if and only if FixQ is a singleton.The Banach contraction principle is immediately obtained from the above corollary.Example 35.
LetO=0,∞ and ςp on O be defined as ςpκ,κ′,κ″=κ−κ′+κ′−κ″2, for all κ,κ′,κ″∈O. Let the ordering “⪯” on O be as follows:
(56)κ⪯κ′⇔κ′≤κ,∀κ,κ′∈O.
Define self-mapQ on O by
(57)Qκ=14lnκ2+1+κ=14sinh−1κ.DefineG:0,1→0,1/2) as Gt=1/16t and θ:0,∞→1,∞ as θt=cosht.Using the mean value theorem for the functionsinh−1, we have
(58)coshςpQκ,Qκ′,Qκ″=coshQκ−Qκ′2+Qκ′−Qκ″2=cosh14sinh−1κ−14sinh−1κ′2+14sinh−1κ′−14sinh−1κ″2≤cosh116κ−κ′2+κ′−κ″2≤coshςpκ,κ′,κ″GMκ,κ′,κ″≤coshMκ,κ′,κ″GMκ,κ′,κ″.Note that for allx≥0 one can see that cosh1/16x≤coshx1/16x. Thus, (16) is satisfied for all κ,κ′,κ″∈O. Therefore, all the conditions of Theorem 27 hold true. Moreover, 0 is a fixed point of Q.


## 4. Existence of a Solution for an Integral Equation

We consider the following integral equation:(59)κt=∫abΓt,s,κsds+γt,where b>a≥0. Our aim in this part is to present the existence of a solution for (59) which is an element of O=Ca,b as an application of Theorem 30.The above equation can be changed as follows.LetQ:O→O be defined by
(60)Qκt=∫abΓt,s,κsds+γt,for all κ∈O and for all t∈a,b. Clearly, existence of a solution for (59) is equivalent to the existence of a fixed point of Q.Let(61)dU,V=maxt∈a,bUt−Vtp=U−V∞p.LetO be equipped with the sequential ςp-metric given by
(62)ςpU,V,W=ϝdU,V+dV,W,for all U,V,W∈O where ϝ:0,∞→0,∞ is a strictly increasing continuous function with t≤ϝt for t≥0 and ϝ0=0 which is a ςp-complete SSPMS with control function ωt=ϝ2p−1t. Let the partial ordered ⪯ given by κ⪯κ′⇔κt≤κ′t, for all t∈a,b, be defined on O. O,⪯ has the sequential limit comparison property [17].Now, we state the following consequence.Theorem 36.
Suppose that(i)
Γ:a,b×a,b×R→R and γ:a,b→R are continuous(ii)
∣Γt,r,κr−Γt,r,κ′r∣≤∣κr−κ′r∣
for allt,r∈a,b and for all κ,κ′∈O with κ⪯κ′;
(iii)
for all positive valuesλ,μ,ν we have(63)θϝλμ+λν≤θϝμ+ν,(iv)
for some functionα:a,b→ℝ one has(64)αt≤∫abΓt,s,αsds+γt.Then, the integral equations (59) has a solution κ∈O.Proof.
Letκ,κ′∈O be such that κ⪰κ′. From condition (ii), for all t∈a,b, we have
(65)θϝQκt−Qκ′tp+Qκ′t−Qκ″tp≤θϝ∫ab∣Γt,s,κs−Γt,s,κ′s∣dsp+∫abΓt,s,κ′s−Γt,s,κ″sdsp≤θϝa−bpq∫abΓt,s,κs−Γt,s,κ′spds+a−bpq∫abΓt,s,κ′s−Γt,s,κ″spdsss≤θϝa−bpq∫abκs−κ′spds+a−bpq∫abκ′s−κ″s∞pds≤θϝa−b2pqκ−κ′∞p+a−b2pqκ′−κ″p≤θϝκ−κ′∞p+κ′−κ″p.
Therefore, taking the sup ona,b, we have
(66)θςpQκ,Qκ′,Qκ″≤θMκ,κ′,κ″,where
(67)Mκ,κ′,κ″=maxςpκ,κ′,κ″,ςpκ,κ,Qκςpκ′,κ′,Qκ′1+ϝςpκ,κ′,κ′+ςpκ,κ,Qκ′,ςpκ′,κ′,Qκ′ςpκ″,κ″,Qκ″1+ϝςpκ′,κ″,κ″+ςpκ′,κ′,Qκ″,ςpκ′,κ″,κ″ςpκ′,κ′,κ″1+ςpκ′,Qκ′,Qκ′+ςpκ″,Qκ″,Qκ″,☐So, from Theorem30, there exists κ∈O, a fixed point of Q which is a solution of (59).


## 5. Conclusions

In this paper, we worked on a space which fails the commutativity property, usual rectangular property, and continuous property. So, we are very restricted in such spaces.



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*Source: 9910861-2021-06-21.xml*