# OnJ-Cone Metric Spaces over a Banach Algebra and Some Fixed-Point Theorems

Authors: Jerolina Fernandez; Neeraj Malviya; Vahid Parvaneh; Hassen Aydi; Babak Mohammadi
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

In the present paper, we defineJ-cone metric spaces over a Banach algebra which is a generalization of Gpb-metric space (Gpb-MS) and cone metric space (CMS) over a Banach algebra. We give new fixed-point theorems assuring generalized contractive and expansive maps without continuity. Examples and an application are given at the end to support the usability of our results.


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## Body

## 1. Introduction

The notion of a generalized partialb-metric (for short, a Gpb-metric) space was introduced by Hussain et al. [1] in 2014 by generalizing the notions of a G-metric space (G-MS) and a partial b-metric space. They studied the related topological properties and provided fixed point theorems for some contractive maps.Huang and Zhang [2] generalized the notion of metric spaces to a CMS. Later, the interesting concept of a cone metric space over a Banach algebra (for short, CMS over a BA) was proposed by Liu and Xu [3], by replacing a CMS with a CMS over a BA. Motivated by these ideas, many authors further considered a CMS over a BA (see [4, 5]). Also, in [6], the authors introduced the concept of cone b2-metric space over Banach algebras which generalizes the notions of b2-metric space and cone metric spaces over Banach algebra. There are some very recent references (such as [7, 8]), where it is shown that the fixed point theory continues to provide useful tools for studying problems of Mathematical Physics.The present paper is organized as follows. In Section2, we recall some definitions. In Section 3, we generalize the concepts of a Gpb-metric space and a CMS over a Banach algebra by introducing a J-CMS over a Banach algebra (for short, J-CMS over a BA) with some examples. Section 4 is devoted to define generalized contractive and expansive maps. Finally, in Section 5 and Section 6, we prove some fixed point theorems for such certain contractive and expansive maps in the framework of a J-CMS over a BA. Our work generalizes and extends some interesting results of [9]. Two examples and an application are given to verify the strength of our main results.


## 2. Preliminaries

We start with some known concepts. Denote byY a real Banach algebra (for short, BA). From now on, assume that there is a unit element e∈Y. An element ω∈Y is called invertible if there is π∈Y so that ωπ=πω=e. We denote by ω−1 the inverse of ω. For more related details, one may check [10].Proposition 1 [10].
Assume that the spectral radiusSRω of an element ω∈Y is less than 1, that is,
(1)SRω=limn⟶∞ωn1/n=infωn1/n<1,
thene−ω is invertible, where e∈Y is unit. In addition,
(2)e−ω−1=∑i=0∞ωi.Remark 2 ([11]).
IfSRω<1, then ∥ωn∥⟶0 as n⟶∞.A setC⊂Y is said to be a cone if
(1)
C≠∅, closed and C≠θ(2)
αC+βC⊂C for all α,β≥0(3)
C2=CC⊂C(4)
−C∩C=θ,where θ is the null of Y. The partial ordering on Y is given as ω≼π iff π−ω∈C. We write ω≺π to indicate ω≼π, but ω≠π, while ω≪π stands for π−ω∈intC (here, intC is the interior of C).Definition 3 ([2, 3]).
LetM be a nonempty set. If d:M×M⟶Y verifies
(1)
θ≼dω,π for all ω,π∈M and dω,π=θ iff ω=π(2)
dω,π=dπ,ω for all ω,π∈M(3)
dω,π≼dω,ϖ+dϖ,π for all ω,π,ϖ∈M
thend is named as a cone metric on M, and M,d is said to be a CMS over the BA Y.Definition 4 ([1]).
LetM be a nonempty set and s≥1. Assume that the function Gpb:M×M×M⟶ℝ+ is so that:
Gpb1ω=π=ϖ  if  Gpbω,π,ϖ=Gpbϖ,ϖ,ϖ=Gpbπ,π,π=Gpbω,ω,ω
Gpb2Gpbω,ω,ω≤Gpbω,ω,π≤Gpbω,π,ϖ for all ω,π,ϖ∈M with ϖ≠π
Gpb3Gpbω,π,ϖ=Gpbpω,π,ϖ, where p is any permutation of ω,π and ϖ (symmetry in all three variables)
Gpb4Gpbω,π,ϖ≤sGpbω,a,a+Gpba,π,ϖ−Gpba,a,a+1−s/3Gpbω,ω,ω+Gpbπ,π,π+Gpbϖ,ϖ,ϖ for all ω,π,ϖ,a∈M (rectangle inequality).Then,Gpb is named as a Gpb-metric and M,Gpb is called a Gpb-metric space.Using (Gpb 4) and s≥1, we have
(3)Gpbω,π,ϖ≤sGpbω,a,a+Gpba,π,ϖ−Gpba,a,a.TheGpb-metric space Gpb is called symmetric if Gpbω,ω,π=Gpbω,π,π holds for all ω,π∈M. Otherwise, Gpb is an asymmetric Gpb−metric.Example 5 [1].
LetM=0,∞ and let Gpb:M×M×M⟶ℝ+ be given by Gpbω,π,ϖ=maxω,π,ϖp where p>1.Obviously,M,Gpb is a symmetric Gpb-metric space, which is not a G-metric space. In fact, if ω=π=ϖ>0, then Gpbω,π,ϖ=ωp>0. It is easy to see that Gpb1−Gpb4 are satisfied.


## 3. AJ-CMS over a Banach Algebra

Here, we introduce the notion of aJ-CMS over the BA Y, as a generalization of a generalized partial b-CMS and a CMS over the BA Y.Definition 6.
LetM be a nonempty set and s≥1. Suppose that J:M×M×M⟶Y so that:
(J1) ω=π=ϖifJω,π,ϖ=Jϖ,ϖ,ϖ=Jπ,π,π=Jω,ω,ω
(J2) Jω,ω,ω≼Jω,ω,π≼Jω,π,ϖ for all ω,π,ϖ∈M with ϖ≠π
(J3) Jω,π,ϖ=Jpω,π,ϖ, where p is any permutation of ω,π and ϖ (symmetry in all three variables)
(J4) Jω,π,ϖ≼sJω,a,a+Ja,π,ϖ−Ja,a,a+1−s/3Jω,ω,ω+Jπ,π,π+Jϖ,ϖ,ϖ for all ω,π,ϖ,a∈M (rectangle inequality).Then,J is called a J-cone metric and M,J is called a J-CMS over the BA Y.Sinces≥1, from J4, we have
(4)Jω,π,ϖ≼sJω,a,a+Ja,π,ϖ−Ja,a,a.TheJ-CMS J is said to be symmetric if Jω,ω,π=Jω,π,π holds for all ω,π∈M. Otherwise, J is asymmetric.We now present some examples.Example 7.
LetY=CR10,∞ which is endowed with the norm ω=ω∞+∥ω′∥∞. Under the pointwise multiplication, Y is a real Banach algebra with unit e=1. Let C=ω,τ∈Y:ωt,τt≥0∀t∈0,∞. Moreover, P is not normal (see [12]).LetGpb:0,∞3⟶ℝ+ be given by Gpbω,π,ϖ=maxω,π,ϖp where p≥1. Let J:M×M×M⟶Y2 be given by
(5)Jω,π,ϖt=Gpbω,π,ϖ,αGpbω,π,ϖet.Obviously,M,Gpb is not a G-metric space. But M,J is a J-CMS over the BA Υ. Indeed, if ω=π=ϖ>0, then Jω,π,ϖt=ωp,αωpet≻θ. Hence, J1−J3 are satisfied. Now, we show that J4 holds. For all ω,π,ϖ,a∈M, we have
(6)maxω,π,ϖpet≼smaxω,ap+maxa,π,ϖp−apet+1−s3ωp+πp+ϖpet,(by Example 12 in [1]). Therefore,
(7)Jω,π,ϖ≼sJω,a,a+Ja,π,ϖ−Ja,a,a+1−s/3Jω,ω,ω+Jπ,π,π+Jϖ,ϖ,ϖ, for all ω,π,ϖ,a∈M (rectangle inequality).The following examples show that aJ-CMS over a BA need not be a Gb-CMS ([13]).Example 8.
LetM=0,1 and Y be the set of all real-valued continuously differentiable functions on M with the norm ω=ω∞+ω′∞. Define the multiplication in the usual way. Let C=ω∈Y:ωt≥0,t∈M. Clearly, C is a nonnormal cone and Y is a BA with a unit e=1. Consider J:M×M×M⟶Y as
(8)Jω,π,ϖt=maxω,π2+maxπ,ϖ2+maxϖ,ω2et,for all ω,π,ϖ∈M. Thus, M,J is a J-CMS over the BA Y, but it is not a Gb-CMS since Jω,ω,ωt=3ω2et≠θ.Example 9.
LetY=Cℝ10,1 and consider a norm on Y as ω=ω∞+ω′∞ for all ω∈Y. Let the multiplication on Y be the pointwise multiplication. Then, Y is a real unit BA with unit e=1. Set C=ω∈Y:ω≥0 which is a cone in Y. Let M=0,∞ and a>0 be any constant. Consider J:M×M×M⟶Y as
(9)Jω,π,ϖt=maxω,π,ϖ2+aet,for all ω,π,ϖ∈M. This M,J is a J-CMS over the BA Y, but it is not a Gb-CMS since Jω,ω,ωt=ω2+aet≠θ.Example 10.
LetY=Ca,b be the set of all continuous functions on a,b with the norm ω=ω∞+ω′∞. Taking the usual multiplication, then Y is a BA with a unit 1. Set C=ω∈Y:ωt≥0,t∈a,b and M=ℝ+. Consider J:M×M×M⟶Y by
(10)Jω,π,ϖt=maxω,π,ϖ2+ω−π2+π−ϖ2+ϖ−ω2et,for all ω,π,ϖ∈M. Then, M,J is a J-CMS over the BA Y. But it is not a Gb-CMS, because Jω,ω,ωt=ω2et≠θ.In the following,M,J is assumed to be a J-CMS over the BA Y.Lemma 11.
(a) IfJω,π,ϖ=θ, then ω=π=ϖ.
(b) Ifω≠π, then Jω,π,π≻θ.Definition 12.
For anω∈M and c>θ, the J-ball with center ω and radius c>θ is
(11)BJω,c=π∈M:Jω,ω,π≪Jω,ω,ω+c.Theθ-Cauchyness and convergence are given as follows.Definition 13.
A sequenceωn in M,J converges to ω∈M, whenever for each c≫θ, there is an integer N so that Jωn,ω,ω≪c for all n≥N.Definition 14.
A sequenceωn in M is called a θ-Cauchy sequence in M,J if for every c≫θ there is n0∈N such that Jωn,ωm,ωm≪c for all n,m≥n0.Definition 15.
M is said to be θ-complete if every θ-Cauchy sequence ωn in M,J is convergent to ω∈M so that Jω,ω,ω=θ.Definition 16.
LetM′,J′ and M′,J′ be two J-CMS over the BA Y. Then, a function f:M⟶M′ is called continuous at ω∈M iff for each convergent sequence ωn to ω, we have fωn is convergent to fω.


## 4. Generalized Contractive and Expansive Maps

In this section, we consider contractive and expansive maps in aJ-CMS over a BA. Some examples are also presented.Definiton 17.
LetM,J be a J-CMS over the BA Y and C be a cone in Y. A map χ:M⟶M is named as a generalized contractive mapping if there is k∈C (with SRk<1) so that for all ω,π,ϖ∈M,
(12)Jχω,χπ,χϖ≼kJω,π,ϖ.Example 18.
LetY be a BA and C be a cone (as in Example 8) and let M=ℝ+. Define a mapping J:M×M×M⟶Y by
(13)Jω,π,ϖt=maxω,π,ϖ2+ω−π2+π−ϖ2+ϖ−ω2et,for all ω,π,ϖ∈M. Then, M,J is a J-CMS over the BA Y. Take χ:M⟶M by χω=ln1+ω/2. In view of ln1+u≤u for each u≥0, we have for all ω,π,ϖ∈M,
(14)Jχω,χπ,χϖt=maxln1+ω2,ln1+π2,ln1+ϖ22+ln1+ω2−ln1+π22+ln1+π2−ln1+ϖ22+ln1+ϖ2−ln1+ω22et≼maxω2,π2,ϖ22+ω2−π22+π2−ϖ22+ϖ2−ω22et≼14maxω,π,ϖ2+ω−π2+π−ϖ2+ϖ−ω2et=14Jω,π,ϖt,
wherek=1/4. Obviously, χ is a generalized contractive map on M.Definition 19.
LetM,J be a J-CMS over the BA Y and C be a cone in Y. A map χ:M⟶M is said to be an expansive mapping if for all ω,π,ϖ∈M,
(15)Jχω,χπ,χϖ≽kJω,π,ϖ,where k,k−1∈C and are called generalized contractive constants with SRk−1<1.Example 20.
LetY be a BA and C be a cone (as given in Example 9) and let M=ℝ+. As in Example 18, M,J is a J-CMS over the BA Y. Define χω=3ω+ω2 for all ω.
(16)Jχω,χπ,χϖt=max3ω+ω2,3π+π2,3ϖ+ϖ22+∣3ω+ω2−3π+π22+3π+π2−3ϖ+ϖ22+3ϖ+ϖ2−3ω+ω22et=max3ω+ω2,3π+π2,3ϖ+ϖ22+3ω−π+ω−πω+π2+3π−ϖ+π−ϖπ+ϖ2+3ϖ−ω+ϖ−ωϖ+ω2et=max3ω+ω2,3π+π2,3ϖ+ϖ22+ω−π23+ω+π2+π−ϖ23+π+ϖ2+ϖ−ω23+ϖ+ω2et≽max3ω,3π,3ϖ2+9ω−π2+9π−ϖ2+9ϖ−ω2et≽9maxω,π,ϖ2+ω−π2+π−ϖ2+ϖ−ω2et=9Jω,π,ϖt,
wherek=9. Clearly, χ is an expansive map in M.Definition 21 ([14]).
LetC be a solid cone in a Banach space E. A sequence un⊂C is said to be a c-sequence if for every c≫θ there is an integer N so that un≪c for all n>N.Lemma 22 ([5]).
IfE is a real Banach space with a solid cone C, then un is a c-sequence whenever un⊂C be a sequence with un⟶0∞.Lemma 23 ([10]).
LetY be a BA with a unit e and k∈Y. Then, limkn1/n exists, and the spectral radius SRk verifies
(17)SRk=limkn1/n=infnkn1/n.Then,ζe−k is invertible in Y provided that SRk<∣ζ∣. Moreover,
(18)ζe−k−1=∑i=0∞kiζi+1where the constantζ is complex.Lemma 24 ([10]).
LetY be a BA with a unit e and ϑ,ϖ∈Y. Then,
(19)SRϑ+ϖ≤SRϑ+SRϖ,SRϑϖ≤SRϑSRϖ,provided that ϑ commutes with ϖ.Lemma 25 ([12]).
LetE be a real Banach space with a solid cone C. Then,
(1)
ϑ≪κ, if ϑ≤ϖ≪κ(2)
ϑ=θ, if ϑ≪κ for every κ≫θLemma 26 ([11]).
LetC be a solid cone in the BA Y and suppose that k∈C. Then, kun is a c-sequence provided that un be a c-sequence in C.Lemma 27 ([5]).
LetY be a BA with a unit e and let k∈Y. Then,
(20)SRζe−k−1≤1ζ−SRk,where ζ is a complex constant with SRk<ζ.Lemma 28 ([5]).
LetC be a solid cone in the BA Y with a unit e and let ϑ,k,l∈C so that l≼k and ϑ≼lϑ. Then, ϑ=θ provided that SRk<1.In this paper, we prove some fixed-point theorems for generalized contractive and expansive maps in the setting of aJ-CMS over a BA.


## 5. Main Results

Theorem 29.
LetM,J be a θ-complete symmetric J-CMS over the BA Y and χ:M⟶M be a mapping so that for all ω,π∈M(21)Jχω,χπ,χπ≼kJω,π,π,where SRk<1/s. Then, χ admits a unique fixed point.Proof.
Letω0 be an arbitrary point in M. Define a sequence ωn in M by ωn=χnω0. By (21), we have
(22)Jωn,ωn+1,ωn+1=Jχωn−1,χωn,χωn≼kJωn−1,ωn,ωn.Continuing in the same argument, we will get(23)Jωn,ωn+1,ωn+1≼knJω0,ω1,ω1.Moreover, for alln,m∈N;n<m, we have
(24)Jωn,ωm,ωm≼sJωn,ωn+1,ωn+1+Jωn+1,ωm,ωm−Jωn+1,ωn+1,ωn+1≼sJωn,ωn+1,ωn+1+sJωn+1,ωm,ωm≼sJωn,ωn+1,ωn+1+s2Jωn+1,ωn+2,ωn+2+Jωn+2,ωm,ωm−Jωn+2,ωn+2,ωn+2≼sJωn,ωn+1,ωn+1+s2Jωn+1,ωn+2,ωn+2+s2Jωn+2,ωm,ωm⋮≼sJωn,ωn+1,ωn+1+s2Jωn+1,ωn+2,ωn+2+⋯+sm−nJωm−1,ωm,ωm≼sknJω0,ω1,ω1+s2kn+1Jω0,ω1,ω1+⋯+sm−nkm−1Jω0,ω1,ω1=skne+sk+sk2+⋯+skm−n−1Jω0,ω1,ω1≼skn∑i=0∞skiJω0,ω1,ω1=e−sk−1sknJω0,ω1,ω1.In view of Remark2, sknJω0,ω1,ω1≼knJω0,ω1,ω1⟶0∞, by Lemma 22, we have sknJω0,ω1,ω1 is a c-sequence. Using Lemma 25 and Lemma 26, ωn is a θ-Cauchy sequence in M. By the θ-completeness of M, there is ρ∈M so that
(25)limJωn,ϱ,ϱ=limn,m⟶∞Jωn,ωm,ωm=Jϱ,ϱ,ϱ=θ.Furthermore, one has(26)Jϱ,χϱ,χϱ≼sJϱ,χωn,χωn+sJχωn,χϱ,χϱ−sJχωn,χωn,χωn≼sJϱ,χωn,χωn+sJχωn,χϱ,χϱ≼sJϱ,ωn+1,ωn+1+skJωn,ωn+1,ϱ.Hence,Jϱ,ωn+1,ωn+1 and Jωn,ωn+1,ϱ are c-sequences, then by using Lemma 25 and Lemma 26, we get that χϱ=ϱ. Thus, ϱ is a fixed point of χ. If χζ=ζ, then one writes
(27)Jϱ,ϱ,ζ=Jχϱ,χϱ,χζ=kJϱ,ϱ,ζ.That is,(28)e−kJϱ,ϱ,ζ≼θ.The multiplication by(29)e−k−1=∑i=0∞ki≥0,yields thatJϱ,ϱ,ζ≼θ. Thus, Jϱ,ϱ,ζ=θ, so ϱ=ζ, a contradiction. Hence, the fixed point is unique.Corollary 30.
LetM,J be a θ-complete symmetric J-CMS over the BA Y and χ:M⟶M be a mapping so that for all ω,π,ϖ∈M(30)Jχω,χπ,χϖ≼aJω,χω,ϖ+bJω,χω,π,where SRa+b<1/s. Then, χ has a unique fixed point.Proof.
In view of theproof of Theorem 29, we get the required result.Theorem 31.
LetM,J be a θ-complete symmetric J-CMS over the BA Y and χ:M⟶M be a mapping so that for all ω,π∈M,
(31)Jχω,χπ,χ2π≼aJω,χω,χ2ω+bJπ,χπ,χ2π+cJω,χω,χπ+dJπ,χπ,χ3ω,
wheresSRa+c+SRb+d<1 and s2SRc+d+SRbs<1. Then, χ admits a unique fixed point.Proof.
Takeω0∈M. We construct ωn in M so that
(32)ωn+1=χωn,foralln=0,1,2,⋯.Notice that ifωn′=ωn′+1 for some integer n′, then there is a fixed point. Suppose that ωn≠ωn+1 for all n∈N. That is,
(33)Jωn,ωn+1,ωn+2>θ.From (31), with ω=ωn−1 and π=ωn, we have
(34)Jχωn−1,χωn,χ2ωn≼aJωn−1,χωn−1,χ2ωn−1+bJωn,χωn,χ2ωn+cJωn−1,χωn−1,χωn+dJωn,χωn,χ3ωn−1, which implies that
(35)Jωn,ωn+1,ωn+2≼aJωn−1,ωn,ωn+1+bJωn,ωn+1,ωn+2+cJωn−1,ωn,ωn+1+dJωn,ωn+1,ωn+2,=a+cJωn−1,ωn,ωn+1+b+dJωn,ωn+1,ωn+2.Thus,(36)e−b−dJωn,ωn+1,ωn+2≼a+cJωn−1,ωn,ωn+1.SincesSRa+c+SRb+d<1 leads to SRb+d<1, by Lemma 23, e−b−d is invertible. So
(37)Jωn,ωn+1,ωn+2≼e−b−d−1a+cJωn−1,ωn,ωn+1.Putk=e−b−d−1a+c. Hence,
(38)Jωn,ωn+1,ωn+2≼kJωn−1,ωn,ωn+1≼⋯≼knJω0,ω1,ω2.By Lemma24 and Lemma 27, we have
(39)SRk=SRe−b−d−1a+c≤SRe−b−d−1SRa+c≤SRa+c1−SRb+d<1s.Thus,e−sk is invertible, so skn⟶0∞. Due to J2,
(40)Jωn,ωn,ωn+1≼Jωn,ωn+1,ωn+2, with ωn≠ωn+1, and by the symmetry property
(41)Jωn+1,ωn+1,ωn=Jωn+1,ωn,ωn=Jωn,ωn,ωn+1.Then, by (38) and (41), we have Jωn+1,ωn+1,ωn≼knJω0,ω1,ω2. Moreover, for all n,m∈N;n<m, we have
(42)Jωn,ωm,ωm≼sJωn,ωn+1,ωn+1+Jωn+1,ωm,ωm−Jωn+1,ωn+1,ωn+1≼sJωn,ωn+1,ωn+1+sJωn+1,ωm,ωm≼sJωn,ωn+1,ωn+1+s2Jωn+1,ωn+2,ωn+2+Jωn+2,ωm,ωm−Jωn+2,ωn+2,ωn+2≼sJωn,ωn+1,ωn+1+s2Jωn+1,ωn+2,ωn+2+s2Jωn+2,ωm,ωm⋮≼sJωn,ωn+1,ωn+1+s2Jωn+1,ωn+2,ωn+2+⋯+sm−nJωm−1,ωm,ωm≼sknJω0,ω1,ω2+s2kn+1Jω0,ω1,ω2+⋯+sm−nkm−1Jω0,ω1,ω2=skne+sk+sk2+⋯+skm−n−1Jω0,ω1,ω2≼skn∑i=0∞skiJω0,ω1,ω2=skne−sk−1Jω0,ω1,ω2.In view of Remark2, sknJω0,ω1,ω2≼sknJω0,ω1,ω2⟶0∞; by Lemma 22, we have sknJω0,ω1,ω2 is a c-sequence. Next, by using Lemma 25 and Lemma 26, ωn is a θ-Cauchy sequence in M. By the θ-completeness of M, there is ϖ∈M so that
(43)limJωn,ϖ,ϖ=limn,m⟶∞Jωn,ωm,ωm=Jϖ,ϖ,ϖ=θ.To see thatϖ is a fixed point of χ, we have
(44)Jϖ,χϖ,χ2ϖ≼sJϖ,χωn,χωn+Jχωn,χϖ,χ2ϖ−Jχωn,χωn,χωn=sJϖ,χωn,χωn+sJχωn,χϖ,χ2ϖ≼sJϖ,χωn,χωn+saJωn,χωn,χ2ωn+bJϖ,χϖ,χ2ϖ+cJωn,χωn,χϖ+dJϖ,χϖ,χ3ωn.Then,(45)Jϖ,χϖ,χ2ϖ≼sJϖ,ωn+1,ωn+1+saJωn,ωn+1,ωn+2+bJϖ,χϖ,χ2ϖ+cJωn,ωn+1,χϖ+dJϖ,χϖ,ωn+3≼sbJϖ,χϖ,χ2ϖ+csJϖ,ϖ,χϖ+sdJϖ,χϖ,ϖ,which implies that
(46)e−bsJϖ,χϖ,χ2ϖ≼sc+dJϖ,ϖ,χϖ.Sinces2SRc+d+SRbs<1 leads to SRbs<1, it follows by Lemma 23 that e−bs is invertible. So, Jϖ,χϖ,χ2ϖ≼e−bs−1sc+dJϖ,ϖ,χϖ. Put h=e−bs−1sc+d. Then
(47)Jϖ,χϖ,χ2ϖ≼hJϖ,ϖ,χϖ.Note that by Lemma24 and Lemma 27,
(48)SRh=SRe−bs−1sc+sd≤SRe−bs−1SRsc+sd≤SRsc+sd1−SRbs<1s.In view of (47), there is a arise 2 cases.Case I.
Ifχϖ=χ2ϖ, then
(49)Jϖ,χϖ,χϖ≼hJϖ,ϖ,χϖ≼hJϖ,χϖ,χϖbysymmetrye−hJϖ,χϖ,χϖ≼θ.We multiply by(50)e−h−1=∑i=0∞hi≥0,to get Jϖ,χϖ,χϖ=θ, which implies that ϖ=χϖ. Hence, ϖ is the fixed point of χ.Case II.
Ifχϖ≠χ2ϖ, then
(51)Jϖ,χϖ,χ2ϖ≼hJϖ,ϖ,χϖ≼hJϖ,χϖ,χϖbysymmetry≼Jϖ,χϖ,χ2ϖbyJ2e−hJϖ,χϖ,χ2ϖ≤θ.Again, we multiply by(52)e−h−1=∑i=0∞hi≥0, to have Jϖ,χϖ,χ2ϖ=θ. So, ϖ=χϖ=χ2ϖ. Hence, ϖ is a fixed point of χ. Suppose that w≠ϖ is such that χw=w. One writes
(53)Jϖ,ϖ,ϖ=Jχϖ,χϖ,χϖ=Jχϖ,χw,χwbysymmetry=Jχϖ,χw,χ2w≼aJϖ,χϖ,χ2ϖ+bJw,χw,χ2w+cJϖ,χϖ,χw+dJw,χw,χ3ϖ≼aJϖ,ϖ,ϖ+bJw,w,w+cJϖ,ϖ,w+dJw,w,ϖ≼aJϖ,ϖ,ϖ+bJw,w,w+cJϖ,ϖ,w+dJw,w,ϖbyJ2.Hence,(54)Jϖ,ϖ,w≼b+c+dJϖ,ϖ,w.SinceSRb+c+d≤s2SRc+d+SRbs<1, by Lemma 28, we acquire that Jϖ,ϖ,w=θ, that is, ϖ=w.


## 6. Fixed-Point Results of Expansive Maps

Here, the case of expansive mappings in the setting of aJ-CMS over a BA is studied.Theorem 32.
LetM,J be a θ-complete J-CMS over the BA Y and χ:M⟶M be an onto mapping so that for all ω,π∈M,
(55)Jχω,χ2ω,χπ≽αJω,χω,π,where SRα−1<1/s. Then, χ possesses a unique fixed point.Proof.
Letω0∈M. Since χ is onto, there is ω1∈M so that ω0=χω1. By continuing this process, we get ωn=χωn+1, for all n∈N∪0. If ωn0=ωn0+1 for some n0∈N∪0, then ωn0 is a fixed point of χ. Assume that ωn≠ωn+1 for all n∈N. From (55), with ω=ωn+1 and π=ωn, we have
(56)Jωn,ωn−1,ωn−1=Jχωn+1,χ2ωn+1,χωn≽αJωn+1,χωn+1,ωn=αJωn+1,ωn,ωnwhich implies that
(57)Jωn+1,ωn,ωn≼hJωn,ωn−1,ωn−1≼⋯≼hnJω0,ω1,ω1,where h=α−1. Note that
(58)SRh=SRα−1<1s,so e−sh−1=∑i=0∞shi and shn⟶0∞. Following the lines of the proof of Theorem 29, we derive that ωn is a θ-Cauchy sequence. Since M,J is θ-complete, there is ϖ∈M so that ωn⟶ϖ, that is,
(59)limJωn,ϖ,ϖ=limn,m⟶∞Jωn,ωm,ωm=Jϖ,ϖ,ϖ=θ.
Sinceχ is onto, there is w∈M so that ϖ=χw. From (55) with ω=ωn+1 and π=w, we have
(60)Jωn,ωn−1,ϖ=Jχωn+1,χ2ωn+1,χw≽±αJωn+1,χωn+1,w=αJωn+1,ωn,w.
So,Jωn+1,ωn,w≼hJωn,ωn−1,ϖ.
Now,Jωn,ωn−1,ϖ is a c-sequence; then, by using Lemma 25 and Lemma 26, it follows that Jϖ,ϖ,w=θ. That is, ϖ=w. Then, ϖ=χw=χϖ. To prove the uniqueness, assume that ϖ≠u so that χu=u and χϖ=ϖ. Now, by
(61)Jϖ,ϖ,u=Jχϖ,χ2ϖ,χu≽αJϖ,χϖ,u≽αJϖ,ϖ,u≻Jϖ,ϖ,u,which is a contradiction. Hence, ϖ=u.Theorem 33.
LetM,J be a θ-complete J-CMS over the BA Y and χ:M⟶M be so that for all ω,π∈M,
(62)Jχω,χπ,χ2π≽αJω,χω,χ2ω,where SRα−1<1/s. Then, χ admits a unique fixed point.Proof.
Letω0∈M. Since χ is onto, there is ω1∈M so that ω0=χω1. Let ωn=χωn+1, for all n∈N∪0. In case ωn0=ωn0+1 for some n0∈N∪0, then ωn0 is a fixed point of χ. Suppose that ωn≠ωn+1 for all n∈N. From (62) with ω=ωn+1 and π=ωn, we have
(63)Jχωn+1,χωn,χ2ωn≽αJωn+1,χωn+1,χ2ωn+1,which implies that
(64)Jωn,ωn−1,ωn−2≽αJωn+1,ωn,ωn−1,or
(65)Jωn+1,ωn,ωn−1≼α−1Jωn,ωn−1,ωn−2.
Puth=α−1. It is evident that
(66)Jωn+1,ωn,ωn−1≼hJωn,ωn−1,ωn−2,where SRh<1/s. By the mimic of the proof of Theorem 36, we can show that ωn is a θ-Cauchy sequence. Since M,J is a θ-complete J-CMS over a BA, there is ϖ∈M such that
(67)limJωn,ϖ,ϖ=limn,m⟶∞Jωn,ωm,ωm=Jϖ,ϖ,ϖ=θ.Sinceχ is onto, there is w∈M so that ϖ=χw. From (62), we have for ω=w and π=ωn+1,
(68)Jϖ,ωn,ωn−1=Jχw,χωn+1,χ2ωn+1≽αJw,χw,χ2w.So,Jw,χw,χ2w≼hJϖ,ωn,ωn−1,whereh=α−1. Since Jϖ,ωn,ωn−1 is a c-sequence, by using Lemma 25 and Lemma 26, it follows that Jϖ,χw,χ2w=θ. That is, w=χw=χ2w.


## 7. Examples

The following examples demonstrate the results obtained in aJ-CMS over Banach algebra Y.Example 34.
Take onY=Cℝ10,1 the norm ω=ω∞+ω′∞ for ω∈Y. Consider in Y a just pointwise multiplication. Then, Y is a real unit BA with unit e=1. Set the cone C=ω∈Y:ω≥0 in Y. Moreover, C is not normal (see [12]). Let M=0,∞. Take J:M×M×M⟶Y as
(69)Jω,π,ϖt=θ,ifω=π=ϖmaxω,π,ϖ2et,otherwise=θ,ifω=π=ϖmaxω2,π2,ϖ2et,otherwisebe a J-CMS over a BA on M. Define χ:M⟶M by χω=ω/5sinω/5. We have
(70)Jχω,χπ,χπt=Jω5sinω5,π5sinπ5,π5sinπ5t≼maxω52,π52,π52et=125maxω2,π2et.Therefore,Jχω,χπ,χπt≼1/25Jω,ω,πt. Then, the contractive condition (21) holds, and θ is the unique fixed point of χ. That is, the conditions of Theorem 36 hold.Example 35.
LetY be a BA and C be a cone (the same ones as those in Example 8), and let M=ℝ+. Consider J:M×M×M⟶Y as in Example 8. Then, M,J is a J-CMS over the BA Y. Define χ:M⟶M by χω=2ω. Choose α=4, then all the conditions of Theorem 6 hold and θ is the unique fixed point of χ. Indeed,
(71)Jχω,χ2ω,χπt=maxχω,χ2ω2+maxχ2ω,χπ2+maxχπ,χω2et=max2ω,4ω2+max4ω,2π2+max2π,2ω2et=16ω2+4max2ω,π2+4maxω,π2et=44ω2+max2ω,π2+maxω,π2et,Jω,χω,πt=maxω,χω2+maxχω,π2+maxπ,ω2et,=maxω,2ω2+max2ω,π2+maxπ,ω2et,=4ω2+max2ω,π2+maxω,π2et,and so, Jχω,χ2ω,χπt≽4Jω,χω,πt.


## 8. Application to the Existence of a Solution of Integral Equations

Denote byC0,T,ℝ the class of all continuous functions on 0,T (where T>0).LetY=C0,T be considered with the norm ω=ω∞+ω′∞. Take the usual multiplication, then Y is a BA with the unit 1. Set C=ω∈Y:ωt≥0,t∈a,b.LetJ be the J-cone metric given as
(72)Jρ,ϱ,σt=supt∈0,Tρt−ϱt2+ϱt−σt2+ρt−ϱt2et,for all ρ,ϱ,σ∈C0,T,ℝ. Note that C0,T,ℝ,J is a complete J-CMS over BA C0,T,ℝ.Theorem 36.
LetK, f, and a be continuous mappings so that:
(i)
supt∈0,T∣Kt,s∣=M<1/T, where K∈C0,T×0,T,ℝ(ii)
a∈C0,T,ℝ(iii)
f∈C0,T×ℝ,ℝ(iv)
fs,ρs−fs,ρs≤ζρs−ϱsfor all ρ,ϱ∈C0,T,ℝ,t∈0,T, where ζ∈0,1/s. Then, the integral equation
(73)ρt=∫0TKt,sfs,ρsds+at,t∈0,T,admits a unique solution in C0,T,ℝ.Proof.
LetF:C0,T,ℝ⟶C0,T,ℝ be defined by
(74)Fρt=∫0TKt,sfs,ρsds+at,for all t∈0,T. We have
(75)JFρ,Fϱ,Fϱt=supt∈0,TFρt−Fϱt2+Fϱt−Fϱt2+Fρt−Fρt2et=2supt∈0,TFρt−Fρt2et=2supt∈0,T∫0TKt,s2fs,ρs−fs,ρs2dset≼M2T2sups∈0,Tζρs−ρs2et=ζJρ,ϱ,ϱ.Then, all the conditions of Theorem29 hold, and hence, there is a unique fixed point of F. So, (73) has a unique solution.



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