# Fixed Point Problems in Cone Rectangular Metric Spaces with Applications

Authors: Muhammad Nazam; Anam Arif; Hasan Mahmood; Sang Og Kim
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we introduce an ordered implicit relation and investigate some new fixed point theorems in a cone rectangular metric space subject to this relation. Some examples are presented as illustrations. We obtain a homotopy result as an application. Our results generalize and extend several fixed point results in literature.


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## Body

## 1. Introduction

Many authors generalized the classical concept of metric space, by changing the metric conditions partially. Branciari [1] introduced rectangular metric space (RMS), where the triangular inequality condition of metric space was replaced by rectangular inequality. He also proved an analog of the Banach contraction principle in rectangular metric spaces. Azam and Arshad [2] mentioned some necessary conditions to get a unique fixed point for Kannan-type mappings in this context. Later, Karapinar et al. [3] investigated some fixed points for ψ,ϕ contractions on rectangular metric spaces. On the other hand, Di Bari and Vetro [4] used ψ,ϕ-weakly contractive condition to give an extension of the results in [3]. Subsequently, a number of authors were engrossed in rectangular metric spaces and proved the existence and uniqueness of fixed point theorems for certain types of mappings [2–6].The significance of the Banach contraction principle lies in the fact that it is a very essential tool to check the existence of solutions for differential equations, integral equations, matrix equations, and functional equations made by mathematical models of real-world problems. There has been a tendency for consistent theorists to improve both the underlying space and the contractive condition (explicit type) used by Banach [7] under the effect of one of the structures like order metric structure [8, 9], graphic metric structure [10, 11], multivalued mapping structure [12–14], α-admissible mapping structure [15], comparison functions, and auxiliary functions. The process of developing new fixed point theorems in the complete metric spaces is in progress under various new restrictions. In this regard, we can find very nice results by Debnath et al. that appeared in [10, 12, 14].Later on, Popa [16] introduced self-mappings satisfying implicit relation and obtained fixed points, under the effect of these functions. Popa [16–18] obtained some fixed point theorems in metric spaces. However, scrutiny into the fixed points of self-mappings satisfying implicit relations in order metric structure was made by Beg and Butt [19, 20], and some common fixed point theorems were established by Berinde and Vetro [21, 22] and Sedghi et al. [23]. Huang and Zhang [24] introduced cone metric by replacing real numbers with ordering Banach spaces and established a convergence criterion for sequences in cone metric space to generalize Banach fixed point theorem. Huang and Zhang [24] considered the concept of normal cone for their drawn outcomes; however, Rezapour and Hamlbarani [25] left the normality condition in some results by Huang. Many authors have investigated fixed point theorems and common fixed point theorems of self-mappings for normal and nonnormal cones in cone metric spaces (see [26–29]).Azam et al. [5] introduced the notion of a cone rectangular metric space and proved the Banach contraction principle in this context. In 2012, Rashwan [6] extended this idea as a continuation, which improved the results in [5]. The appealing nature of these spaces has enticed scrutiny into fixed point theorems for various contractions on cone rectangular metric spaces (see [5, 6]).In this paper, we continue the study initiated by Azam et al. [5] subject to an ordered implicit relation. Since every cone metric space is a cone rectangular metric space but not conversely, therefore we prefer to establish results in cone rectangular metric spaces. These results are supported by some examples and an application in homotopy theory.


## 2. Preliminaries

Definition 1.
A binary relationRover a setY≠ϕdefines a partial order ifRhas the following axioms:(1)
reflexive(2)
antisymmetric(3)
transitive
A set having partial orderR is known as a partially ordered set denoted by Y,R.In the present article,E stands for a real Banach space. Now, we present some definitions and relevant results, which will be required in the sequel.Definition 2 (see [24]).
A subsetPofEis called a cone if and only if the following conditions are satisfied:(1)
P is closed, nonempty, and P≠0(2)
ασ+βς∈P, for all σ,ς∈P and α,β∈ℝ such that α,β≥0(3)
P∩−P=0
GivenP⊂E, define the partial order ⪯ with respect to P as follows:
σ≤ς if and only if ς−σ∈P for all σ,ς∈E.(5)
σ≺ς represents that σ⪯ς but σ≠ς, while σ≪ς stands for ς−σ∈P∘ (interior of P).Definition 3 (see [24]).
The coneP⊆Eis called normal if, for allσ,ς∈P, there existsK>0such that(1)0⪯σ⪯ς⇒∥σ∥≤K∥ς∥.Throughout this paper, we assumeY=Y,R and ⪯ a partial order with respect to the cone P defined in E. If Y⊆E, then R and ⪯ are identical; otherwise, they are different.Definition 4 (see [24]).
LetYbe a nonempty set, anddc:Y×Y↦Esatisfies the following:
(d1)dcσ,ς⪰0, ∀σ,ς∈Y and dcσ,ς=0 if and only if σ=ς
(d2)dcσ,ς=dcς,σ
(d3)dcσ,ξ⪯dcσ,ς+dcς,ξ, ∀σ,ς,ξ∈Y
Then,dc is called a cone metric on Y, and Y,dc is then known as a cone metric space.Example 1 (see [5]).
LetY=ℝ, E=ℝ2, andP=σ,ς∈E:σ,ς≥0⊂ℝ2. Definedc:Y×Y→Eby(2)dcσ,ς=∣σ−ς∣,α∣σ−ς∣,where α≥0 is a constant. Then, dc defines a cone metric on Y.Proposition 5 (see [5]).
Consider a cone metric spaceY,dc, with coneP. Then, foru,v,w∈E ,we have(1)
Ifu⪯αuandα∈0,1, thenu=0(2)
If0⪯u≪vfor each0≪v ,thenu=0(3)
Ifu⪯vandv≪w, thenu≪wSurely, cone metric space “being space” generalizes metric space, because in cone metric space, the range of a metric function is an ordered vector space instead of real numbers. Although the set of real numbers is an ordered vector space, we can find many significant ordered vector spaces in the literature (see [25, 27, 28, 30]). In Theorem 1.4 and Lemma 2.1 that appeared in [31, 32], respectively, the authors developed a metric depending on a given cone metric and proved that a complete cone metric space is always a complete metric space, and then, this relationship between metric and cone metric led them to say that every contraction mapping in a cone metric space is essentially contraction mapping in a metric space.This paper addresses the fixed point results in the cone rectangular metric spaces. We know that every metric is a rectangular metric but rectangular metric needs not to be a metric (see [1–3, 33]). Also, we know that every cone metric is a cone rectangular metric but conversely does not hold in general (see [5, 6, 29]). In view of observations given in [5, 29], we infer that results in this paper are independent of what authors investigated in [31, 32]. The implicit relation and hence the contractive condition employed are even new and original in the rectangular metric space. The theorems in this paper are new in rectangular metric space, but we choose the cone rectangular metric space for the sake of the generality of our results.Definition 6 (see [24]).
Let A mappingdcr:Y×Y↦Eis said to be a cone rectangular metric if for allσ,ς,ξ,υ∈Ythe following conditions are satisfied:
(dR1)0⪯dcrσ,ς and dcrσ,ς=0 if and only if σ=ς
(dR2)dcrσ,ς=dcrς,σ
(dR3)dcrσ,υ⪯dcrσ,ς+dcrς,ξ+dcrξ,υ for all distinct ς,ξ∈Y\σ,υ
The cone rectangular metric space is denoted byY,dcr.Example 2 (see [5]).
LetE=ℝ2, P=σ,ς∈E:σ,ς≥0⊂ℝ2, Y=ℝ. Takeβ>0, and define(3)dcrσ,ς=0,0ifσ=ς,3β,3ifσ,ς∈1,2,σ≠ς,β,1ifσandςcannotbebothatatimein1,2,σ≠ς.
One can easily check thatY,dcr is a cone rectangular metric space, but not a cone metric space, since dcr1,2=3β,3≻dcr1,3+dcr3,2=2β,2.Definition 7 (see [1]).
LetEbe a real Banach space, Y,dcrbe a cone rectangular metric space andc∈Ewith0≪c.
(1)
A sequenceσn in Y,dcr is called a Cauchy sequence, if there exists a natural number N∈ℕ such that dcrσn,σm≪c for all n,m>N(2)
The sequenceσn is said to be convergent if there exists an N∈ℕ such that dcrσn,σ≪c for all n≥N and σ∈Y(3)
TheY,dcr is called complete if every Cauchy sequence converges in Y


## 3. Ordered Implicit Relations

Many authors have used implicit relations to establish fixed point results and have applied these results to solve nonlinear functional equations (see [19–22, 34, 35]).In this section, we define a new ordered implicit relation and explain it with an example. In the next section, we use it along with some other assumptions to develop some new fixed point theorems in the cone rectangular metric space.LetE,. be a real Banach space and BE,E be the space of all bounded linear operators T:E→E with the usual norm .1 defined in BE,E that is T1=supv∈ETv/vv≠0.In this section, generalizing the idea of [16], we define the following notion:LetL:E6→E be a continuous operator which satisfies the conditions given below:(L1) v5⪯υ5 and v6⪯υ6⇒Lv1,v2,v3,v4,v5,v6⪯Lv1,v2,v3,v4,υ5,υ6(L2) If Lv1,v2,v2,v1,v1+v2+v3,v1⪯0, then there exists an order preserving operator S∈BE,E with S1<1 such that v1⪯Sv2 and v3⪯Sv1 for all v1,v2,v3∈E or if Lv1,v2,v1,v2,v1,v1+v2+v3⪯0, then v2⪯Sv3 and v1⪯Sv2 for all v1,v2,v3∈E(L3) Lv,0,0,v,v,0≻0 whenever v>0LetG=L:E6→E∣LsatisfiesconditionsL1,L2,L3.Example 3.
Let⪯be the partial order with respect to conePas defined in Section2and letE,.be a real Banach space. For allvi∈Ei=1to6, α∈0,1/3and1+α/2≤β≤1+α, defineL:E6→E, by(4)Lv1,v2,v3,v4,v5,v6=αv2+v3+v4−βv5+v6+v1.Then, the operatorL∈G:L1 Let v5⪯γ5 and v6⪯γ6, then γ5−v5∈P and γ6−v6∈P. Now, we show that Lv1,v2,v3,v4,v5,v6−Lv1,v2,v3,v4,γ5,γ6∈P. Given that γ5−v5∈P and γ6−v6∈P and by Definition 2 (2), we have
(5)Lv1,v2,v3,v4,v5,v6−Lv1,v2,v3,v4,γ5,γ6=αv2+v3+v4−βv5+v6+v1−αv2+v3+v4−βγ5+γ6+v1=βγ5−v5+γ6−v6∈P.Thus,Lv1,v2,v3,v4,γ5,γ6⪯Lv1,v2,v3,v4,v5,v6.L2 Let v1,v2,v3∈E be such that 0⪯v1,0⪯v2,0⪯v3. If Lv1,v2,v2,v1,v1+v2+v3,v1⪯0 then, we have −v1−αv2+v2+v1+βv1+v2+v3+v1∈P.So2β−α−1v1+β−2αv2+βv3∈P.By Definition2 (2), we have for 2β≥α+1 either
(6)2β−α−1v1+β−2αv2∈Por
(7)2β−α−1v1+βv3∈Por
(8)β−2αv2+βv3∈P.For (6), if v1=0 and v2≠0, then β−2αv2∈P. Thus, there exists S:E→E defined by Sv2=ηv2 (η=β−2α is a scalar) such that v1⪯Sv2. Now, if v2=0 and v1≠0, then, 2β−1−αv1∈P. So, there exists S:E→E defined by Sv1=ηv1 (η=2β−1−α is a scalar) such that v2⪯Sv1, for some α∈0,1/3. For if both v1≠0 and v2≠0, then we get an absurdity.For (7), if v3=0 and v1≠0, then 2β−α−1v1∈P. Thus, there exists S:E→E defined by Sv1=ηv1 (η=2β−α−1 is a scalar) such that v3⪯Sv1. Now, if v3≠0 and v1=0, then, βv3∈P. So, there exists S:E→E defined by Sv3=ηv3 (η=β is a scalar) such that v1⪯Sv3. For if both v1≠0 and v3≠0, then we get an absurdity. Similar arguments hold for (8).L3 Let v∈E be such that v>0 and consider, 0⪯Lv,0,0,v,v,0 then 1+α−βv∈P, which holds whenever v>0.Similarly, the operatorsL:E6→E defined by
(1)
Lv1,v2,v3,v4,v5,v6=v1−αv5+v6;α∈1,∞(2)
Lv1,v2,v3,v4,v5,v6=v3−αv5−βv6;α+β<1,α,β>0(3)
Lv1,v2,v3,v4,v5,v6=v1−αv3+v4−1−αβmaxv5,v6;α∈0,1andβ∈1/2,1(4)
Lv1,v2,v3,v4,v5,v6=v1−αv2forsomeα∈0,1(5)
Lv1,v2,v3,v4,v5,v6=v1−kv3+v4forsomek∈0,1/2 are members of GThe following remark is essential in the sequel.Remark 8.
IfS∈BE,E ,the Neumann seriesI+S+S2+⋯+Sn+⋯converges ifS1<1and diverges otherwise. Also, ifS1<1, then there existsλ>0such thatS1<λ<1and∥Sn∥1≤λn<1.


## 4. New Results

Recently, Popa [16] has employed implicit type contractive condition on self-mapping to obtain some fixed point theorems. Ran and Reurings [9] have presented an analog of Banach fixed point theorem for monotone self-mappings in an ordered metric space. Huang and Zhang [24] introduced the idea of cone metric spaces and obtained analogs of Banach fixed point theorem, Kannan fixed point theorem, and Chatterjea fixed point theorem in cone metric spaces. In this section, we prove some fixed point results for ordered implicit relations in a cone rectangular metric space which improves the results in [9, 16, 24]. We derive these results under two different partial orders: one defined in underlying set and the other in real Banach space.Theorem 9.
LetY,dcrbe a complete cone rectangular metric space andP⊂Ebe a cone. Letf:Y→Y. If there existT∈BE,E, identity operatorI:E→EandL∈Gsuch that, for all comparable elementsσ,κ∈Y(9)I−T2I+Tdcrσ,fσ⪯dcrσ,κimpliesLdcrfσ,fκ,dcrσ,κ,dcrσ,fσ,dcrκ,fκ,dcrσ,f2κ,dcrκ,f2σ⪯0and
(1)
there existsσ0∈Y such that σ0Rfσ0(2)
for allσ,κ∈Y, σRκ implies fσRfκ(3)
for everyσn⊆Y, dcrσn,σn+1⪯Tdcrσn−1,σn(4)
for a sequenceσn with σn→x∗ whose all sequential terms are comparable, we have σnRx∗ for all n∈ℕ and dcrx∗,fx∗⪯dcrx∗,f2x∗
Then,x∗=fx∗.Proof.
Letσ0∈Y be as assumed in (1). We construct a sequence σn by fσn−1=σn starting with σ0∈Y. Then, σ0Rσ1. By assumption (2), we have σ1Rσ2, σ2Rσ3⋯σn−1Rσn. For σ=σ0 and κ=σ1, we have by (9).
(10)I−T2I+Tdcrσ0,fσ0=I−T2I+Tdcrσ0,σ1⪯dcrσ0,σ1impliesLdcrfσ0,fσ1,dcrσ0,σ1,dcrσ0,fσ0,dcrσ1,fσ1,dcrσ0,f2σ1,dcrσ1,f2σ0⪯0,that is,
(11)Ldcrσ1,σ2,dcrσ0,σ1,dcrσ0,σ1,dcrσ1,σ2,dcrσ0,σ3,dcrσ1,σ2⪯0.
By (dR3), we have(12)dcrσ0,σ3⪯dcrσ0,σ1+dcrσ1,σ2+dcrσ2,σ3,and so we rewrite (11) employing condition (L1) as follows:
(13)Ldcrσ1,σ2,dcrσ0,σ1,dcrσ0,σ1,dcrσ1,σ2,dcrσ0,σ1+dcrσ1,σ2+dcrσ2,σ3,dcrσ1,σ2⪯0,and thus, by L2, there exists an order preserving operator S∈BE,E with S1<1 such that
(14)dcrσ1,σ2⪯Sdcrσ0,σ1,dcrσ2,σ3⪯Sdcrσ1,σ2.
Now, putσ=σ1 and κ=σ2 in (9) to have
(15)I−T2I+Tdcrσ1,fσ1=I−T2I+Tdcrσ1,σ2⪯dcrσ1,σ2impliesLdcrfσ1,fσ2,dcrσ1,σ2,dcrσ1,fσ1,dcrσ2,fσ2,dcrσ1,fσ3,dcrσ2,fσ2⪯0,that is,
(16)Ldcrσ2,σ3,dcrσ1,σ2,dcrσ1,σ2,dcrσ2,σ3,dcrσ1,σ4,dcrσ2,σ3⪯0.
By (dR3), we have(17)dcrσ1,σ4⪯dcrσ1,σ2+dcrσ2,σ3+dcrσ3,σ4,and L1 implies
(18)Ldcrσ2,σ3,dcrσ1,σ2,dcrσ1,σ2,dcrσ2,σ3,dcrσ1,σ2+dcrσ2,σ3+dcrσ3,σ4,dcrσ2,σ3⪯0.
ByL2, there exists S∈BE,E with S1<1 such that
(19)dcrσ3,σ4⪯Sdcrσ2,σ3⪯S2dcrσ1,σ2⪯S3dcrσ0,σ1.
By continuing this pattern, we can construct a sequenceσn such that σnRσn+1 with σn+1=fσn, and
(20)I−T2I+Tdcrσn−1,fσn−1=I−T2I+Tdcrσn−1,σn⪯dcrσn−1,σnimpliesdcrσn,σn+1⪯Sdcrσn−1,σn⪯S2dcrσn−2,σn−1⪯⋯⪯Sndcrσ0,σ1.
Form,n∈ℕ with m>n, we have by Remark 8(21)dcrσn,σm≤dcrσn,σn+1+dcrσn+1,σn+2+.⋯⋯+dcrσm−1,σm≤Sndcrσ0,σ1+Sn+1dcrσ0,σ1+⋯⋯+Sm−1dcrσ0,σ1=Sn+Sn+1+⋯+Sm−1dcrσ0,σ1,≤Sn1+S+⋯+Sm−n−1+⋯dcrσ0,σ1=SnI−S−1dcrσ0,σ1.
SinceS1<1, so, Sn→0 as n→∞. Thus, limn→∞dcrσn,σm=0 which implies that σn is a Cauchy sequence in Y. Since Y,dcr is a complete cone rectangular metric space, so, there exists x∗∈Y such that σn→x∗ as n→∞. Equivalently, there exists a natural number N2 such that
(22)dcrσn,x∗≪cforalln≥N2.
We claim that(23)I−T2I+Tdcrσn,fσn⪯dcrσn,x∗.
We assume against our claim that(24)I−T2I+Tdcrσn,fσn≻dcrσn,x∗,I−T2I+Tdcrσn+2,fσn+2≻dcrσn+2,x∗forsomen∈ℕ.
By (dR3), (9), and assumption (3), we have
(25)dcrσn,fσn≤dcrσn,x∗+dcrx∗,σn+2+dcrσn+2,σn+1≺I−T2I+Tdcrσn,fσn+I−T2I+Tdcrσn+2,fσn+2+Tdcrσn,fσn≺I−T2I+Tdcrσn,fσn+I−T2I+TT2dcrσn,fσn+Tdcrσn,fσn,I−Tdcrσn,fσn≺I−T2I+TI+T2dcrσn,fσndcrσn,fσn≺I−T−1I−T2I+TI+T2dcrσn,fσn≺I−T4dcrσn,fσn.
Thus,T4dcrσn,fσn≺0, which is an absurdity. Hence, for each n≥1, we have
(26)I−T2I+Tdcrσn,fσn⪯dcrσn,x∗.
Assume thatdcrx∗,fx∗>0. As σn−1Rσn and by the assumption (4), we have σnRx∗ for all n∈ℕ and then by (9), we get
(27)Ldcrfσn,fx∗,dcrσn,x∗,dcrσn,fσn,dcrx∗,fx∗,dcrσn−1,f2x∗,dcrf2σn,fσn⪯0.
Lettingn→∞ and in view of assumption (4) and (26), we have
(28)Ldcrx∗,fx∗,dcrx∗,x∗,dcrx∗,x∗,dcrx∗,fx∗,dcrx∗,f2x∗,dcrx∗,x∗⪯0,Ldcrx∗,fx∗,0,0,dcrx∗,fx∗,dcrx∗,f2x∗,0⪯0.
ByL1, we have
(29)Ldcrx∗,fx∗,0,0,dcrx∗,fx∗,dcrx∗,fx∗,0⪯0.
This is a contradiction toL3. Thus, dcrx∗,fx∗=0. Hence, dcrx∗,fx∗=0. It follows from (dR1) that x∗=fx∗.Theorem 10.
LetY,dcrbe a complete cone rectangular metric space andfbe a self-mapping onY. If for all comparable elementsσ,κ∈Y, there existT∈BE,E, identity operatorI:E→EandL∈Gsuch that(30)I−T2I+Tdcrσ,fσ⪯dcrσ,κimpliesLdcrfσ,fκ,dcrσ,κ,dcrσ,fσ,dcrκ,fκ,dcrσ,f2κ,dcrκ,f2σ⪯0,and
(1)
there existsσ0∈Y such that fσ0Rσ0(2)
for anyσ,κ∈Y, σRκ implies fκRfσ(3)
for everyσn⊆Y, dcrσn,σn+1⪯Tdcrσn−1,σn(4)
for a sequenceσn with σn→x∗ whose all sequential terms are comparable, we have σnRx∗ for all n∈ℕ and dcrx∗,fx∗⪯dcrx∗,f2x∗Then,f has a fixed point in Y.Proof.
Letσ0 be in Y as assumed in (1). Define the sequence σn by σn=fσn−1 for all n. Since σ1=fσ0Rσ0 and by assumption (2) σ1=fσ0Rfσ1=σ2, repeated application of assumption (2) gives us either σnRσn−1 or σn−1Rσn for each n. By (30), we have for σ=σ1 and κ=σ0(31)I−T2I+Tdcrfσ0,σ0=I−T2I+Tdcrσ1,σ0⪯dcrσ1,σ0impliesLdcrfσ1,fσ0,dcrσ1,σ0,dcrσ1,fσ1,dcrσ0,fσ0,dcrσ1,f2σ0,dcrσ0,f2σ1≤0⇒Ldcrσ1,σ2,dcrσ0,σ1,dcrσ1,σ2,dcrσ0,σ1,dcrσ1,σ2,dcrσ0,σ3≤0.
By (dR3), we have(32)dcrσ0,σ3≤dcrσ0,σ1+dcrσ1,σ2+dcrσ2,σ3,and then using L1, we obtain
(33)Ldcrσ1,σ2,dcrσ0,σ1,dcrσ1,σ2,dcrσ0,σ1,dcrσ1,σ2,dcrσ0,σ1+dcrσ1,σ2+dcrσ2,σ3⪯0.
ByL2, there exists an order preserving operator S∈BE,E with S1<1 such that
(34)dcrσ1,σ2≤Sdcrσ0,σ1,dcrσ2,σ3≤Sdcrσ1,σ2.
Using2fσ0Rfσ1, take σ=σ1 and κ=σ2 in (30), we have
(35)I−T2I+Tdcrσ1,fσ1=I−T2I+Tdcrσ1,σ2⪯dcrσ1,σ2impliesLdcrfσ1,fσ2,dcrσ1,σ2,dcrσ1,fσ1,dcrσ2,fσ2,dcrσ1,f2σ2,dcrσ2,f2σ1⪯0⇒Ldcrσ2,σ3,dcrσ1,σ2,dcrσ1,σ2,dcrσ2,σ3,dcrσ1,σ4,dcrσ2,σ3⪯0.
By (dR3),L1 and L2, we get
(36)dcrσ3,σ4≤Sdcrσ2,σ3⪯S2dcrσ1,σ2⪯S3dcrσ0,σ1.
By continuing the pattern, we construct a sequenceσn such that
(37)dcrσn,σn+1⪯Sdcrσn−1,σn⪯S2dcrσn−2,σn−1⪯⋯⪯Sndcrσ0,σ1.
Hence, by the same reasoning as in the proof of Theorem9, we have x∗=fx∗.Theorem 11.
LetY,dcrbe a complete cone rectangular metric space andfbe a monotone self-mapping onY. If for all comparable elementsσ,κ∈Y, there existT∈BE,E, identity operatorI:E→E,andL∈Gsuch that(38)I−T2I+Tdcrσ,fσ≤dcrσ,κimpliesLdcrfσ,fκ,dcrσ,κ,dcrσ,fσ,dcrκ,fκ,dcrσ,f2κ,dcrκ,f2σ⪯0,and
(1)
there existsσ0∈Y such that either σ0Rfσ0 or fσ0Rσ0(2)
for everyσn⊆Y, dcrσn,σn+1⪯Tdcrσn−1,σn(3)
for a sequenceσn with σn→x∗ whose all sequential terms are comparable, we have σnRx∗ for all n∈ℕ and dcrx∗,fx∗°dcrx∗,f2x∗Then,f has a fixed point in Y.Proof.
This proof follows the same pattern as given in the previous two proofs, so, we omit it.Remark 12.
(1)
In Theorem9, Theorem 10, and Theorem 11, uniqueness of the fixed point of f can be attained by assuming that for every pair of elements σ,κ∈X, there exists either an upper bound or lower bound of σ,κ(2)
The cone is taken as nonnormal in the above theoremsTheorem 13.
LetY,dcrbe a complete cone rectangular metric space andfbe a monotone self-mapping onY. If for all comparable elementsσ,κ∈Y, there existT∈BE,E, identity operatorI:E→E,andL∈Gsuch that(39)I−T2I+Tdcrσ,fσ≤dcrσ,κimpliesdcrfσ,fκ⪯Sdcrσ,κ.
Moreover, if(1)
there existsσ∈Y such that σ0Rfσ0 or fσ0Rσ0(2)
for a sequenceσn with σn→x∗ whose all sequential terms are comparable, we have σnRx∗ for all n∈ℕ and dcrx∗,fx∗≤dcrx∗,f2x∗
Then, there existsx∗∈Y such that x∗=fx∗.Proof.
DefineSv=qv for all v∈E and q∈0,1, then S∈BE,E with ∥S∥1<1 also define implicit relation by
(40)Lσ1,σ2,σ3,σ4,σ5,σ6=σ1−ασ2forsomeα∈0,1.
The proof follows by the application of Theorem9.The following examples illustrate the main theorem.Example 4.
LetE=ℝ3,∥·∥be a real Banach space. With∥σ∥=max∣σ1∣,∣σ2∣,∣σ3∣, thenE,.is a real Banach space. Define the partial order⪯onEby(41)σ,ξ,ν⪰0⇔σ≥0,ξ≥0,ν≥0.
DefineP=σ,ξ,ν∈ℝ3:σ,ξ,ν≥0, then, it is a cone in E.
LetY=0,0,0,1/4,0,0,1/4,1/4,0,1/4,1/4,1/4⊂E, define order on Y by R≕≤ and define f:Y→Y such that
(42)fσ=0,0,0ifσ∈0,0,0,14,14,0,14,0,0,14,14,0ifσ=14,14,14.
Define the functiondcr:Y×Y→E by
(43)dcrσ1,σ2=0,0,0ifσ1=σ2,14,14,14ifσ1,σ2∈0,0,0,14,14,0,14,14,14,14,0,0ifσ1≠σ2cannotbebothatatimein0,0,0,14,14,0,14,14,14.It is a cone rectangular metric space but not a cone metric space. The assumptions (1)–(4) in Theorem9 can be verified immediately. Define Tσ=σ/3, then, T∈BE,E with ∥T∥1≤1. Now, we verify condition (9). For if σ=0,0,0 and κ=1/4,0,0, then
(44)dcrσ,fσ=dcrfσ,fκ=dcrσ,f2κ=0,0,0,dcrκ,f2σ=14,0,0=dcrσ,κ,and thus,(45)I−T2I+Tdcrσ,fσ=0,0,0,αdcrσ,f2κ+dcrκ,f2σ=α0,0,0+14,0,0=α14,0,0:α≥1.For ifσ=1/4,0,0 and κ=1/4,1/4,0, then dcrσ,fσ=dcrσ,κ=dcrσ,f2κ=1/4,0,0;dcrκ,f2σ=1/4,1/4,1/4;dcrfσ,fκ=0,0,0;I−T2I+Tdcrσ,fσ=16/271/4,0,0 and
(46)αdcrσ,f2κ+dcrκ,f2σ=α14,0,0+14,14,14=α24,14,14.Finally, forσ=1/4,1/4,0 and κ=1/4,1/4,1/4, we have dcrσ,fσ=dcrfσ,fκ=dcrσ,f2κ=dcrσ,κ=dcrκ,f2σ=1/4,1/4,1/4;I−T2I+Tdcrσ,fσ=16/271/4,1/4,1/4 and
(47)αdcrσ,f2κ+dcrκ,f2σ=α14,14,14+14,14,14=α24,24,24.Define the functionL by
(48)Ldcrfσ,fκ,dcrσ,κ,dcrσ,fσ,dcrκ,fκ,dcrσ,f2κ,dcrκ,f2σ=dcrfσ,fκ−αdcrσ,f2κ+dcrκ,f2σ.As a consequence, we haveI−T2I+Tdcrσ,fσ⪯dcrσ,κ implies
(49)Ldcrfσ,fκ,dcrσ,κ,dcrσ,fσ,dcrκ,fκ,dcrσ,f2κ,dcrκ,f2σ⪯0.Thus, this example explains Theorem9 well. Note that 0,0,0 is a fixed point of f.Example 5.
LetE=CR11,2and∥g∥E=∥g∥∞+∥g′∥∞,where∥g∥∞=supt∈1,2∣gt∣andP=gt∈E:gt>0,t∈1,2. For eachK≥1, takegt=tandκt=t2K. By definition∥g∥=2and∥κ∥=2K+1. Define the partial order≤by
for anyg,κ∈E;g⪯κ⇔gt≤κt∀t∈1,2.
Then,g⪯κ implies K∥g∥≤∥κ∥. Hence, P is a nonnormal cone. Define T:E→E such that
(50)Tgt=12∫1tgsds.
Then,T∈BE,E. Let Y=1,2,3 and f:Y→Y such that f1=f2=1 and f3=2, f is monotone with respect to usual order (≤) and also assumption (1)–(2) in Theorem 13 are verified. Define dcr by
(51)dcrσ,ςt=0ifσ=ς,et3ifσ,ς∈1,2,etotherwise.
The vector-valued functiondcr is a cone rectangular metric space but not a cone metric space. We are left to verify the contractive condition only. For this, if σ=1 and ς=2, then
(52)dcrσ,fσ=0;dcrfσ,fς=0;dcrς,fς=et3,and for σ=2, ς=3, we have
(53)dcrσ,ς=et;dcrσ,fσ=et3;dcrfσ,fς=et3;dcrς,fς=et.
Thus,(54)I−TI−T2dcrσ,fσ=et31−t21−t24,Tdcrσ,ςt=tet2,Tdcrς,fςt=tet2.
Ast∈1,2 so t2≤4⇒t2/4≤1 and 1−t2/4≥0. Hence,
(55)I−TI−T2dcrσ,fσ⪯dcrσ,ςimpliesdcrfσ,fς⪯Tdcrσ,ς.
Note that1 is a fixed point of f.


## 5. A Homotopy Result

In what follows, we derive a homotopy result by applying Theorem13.Theorem 14.
LetE,.be a real Banach space andP⊂Ebe a cone. LetY,dcrbe a rectangular cone metric space andU⊂Yis open. Assume that there existsT∈BE,EandTP⊂P. Let the operatorh:U¯×0,1→Ysatisfy ((39)) and the condition (1) of Theorem13in the first variable and(1)
σ≠hσ,θ for every σ∈∂U (∂U denotes the boundary of U in Y)(2)
there existsM≥0 such that(56)dcrhσ,θ,hσ,μ≤Mθ−μfor every σ∈U¯ and μ,θ∈0,1(3)
For someσ∈U, if there exists κ with dcrσ,κ≤r, then σRκ, where r is radius of open ball in U
Ifh·,0 has a fixed point in the open set U, then h·,1 also has a fixed point in the open set U.Proof.
Let(57)B=θ∈0,1σ=hσ,θ;forsomeσ∈U.
Define the relation⪯ in E by u⪯vifandonlyifu≤v for all u,v∈E. Next, 0∈B, since h.,0 has a fixed point in the open set U. So B is nonempty. Since dcrσ,hσ,θ=dcrσ,κ, I−T2I+Tdcrσ,hσ,θ≤dcrσ,κ for all σRκ, then by Theorem 13, we have
(58)dcrhσ,θ,hκ,θ≤Sdcrσ,κ.
Firstly, we show thatB is closed in 0,1. For this, let θnn=1∞⊆B with θn→θ∈0,1 as n→∞. It is necessary to prove that θ∈B. Since θn∈B for n∈ℕ, there exists σn∈U with σn=hσn,θn. Since hσ,· is monotone, so, for n,m∈ℕ, we have σmRσn. Since
(59)I−T2I+Tdcrσn,hσm,θm=I−T2I+Tdcrσn,σm⪯dcrσn,σm,we have
(60)dcrhσn,θm,hσm,θm≤Sdcrσn,σm,dcrσn,σm+1=dcrhσn,θn,hσm+1,θm+1⪯dcrhσn,θn,hσn,θm+dcrhσn,θm,hσn,θm+1+dcrhσn,θm+1,hσm+1,θm+1,dcrσn,σm≤Mθn−θm+Mθm−θm+1+Sdcrσn,σm,dcrσn,σm≤M1−S∣θn−θm∣+∣θm−θm+1∣.
Sinceθnn=1∞ is a RMS Cauchy sequence in 0,1, we have
(61)limn,m→∞dcrσn,σm=0,that is, dcrσn,σm≪c, whenever n,m→∞. Hence, σn is a Cauchy sequence in Y. Since Y is a complete cone rectangular metric space, there exists σ∈U¯ with limn→∞dcrσn,σ≪c. Hence, σnRσ for all n∈ℕ. Now consider
(62)dcrσn,hσ,θ=dcrhσn,θn,hσ,θ⪯dcrhσn,θn,hσn,θ+dcrhσn,θ,hσn+1,θ+dcrhσn+1,θ,hσ,θ,dcrσn,hσ,θ≤Mθn−θ+Mθn+1−θ+Sdcrσn+1,σ.
So we have(63)limn→∞dcrσn,hσ,θ=0.
Thus,dcrσ,hσ,θ=0. Hence, θ∈B and so B is RMS closed in 0,1.
Next, we show thatB is open in 0,1. For this, let θ2∈B. Then, we have the existence of σ2∈U with hσ2,θ2=σ2. Since U is open, there exists r>0 such that Bσ2,r⊆U. Now, assume
(64)l=dcrσ2,∂U=infdcrσ2,ξ:ξ∈∂U.
Then,r=l>0. Fix ε>0 with ε<1−Tl/2M. Let θ∈θ1−ε,θ1+ε and θ1∈θ2−ε,θ2+ε. Then,
(65)σ∈Bσ2,r¯=σ∈X:dcrσ,σ2≤r,asσRσ2.
Consider(66)dcrhσ,θ,σ2=dcrhσ,θ,hσ2,θ2≤dcrhσ,θ,hσ,θ1+dcrhσ,θ1,hσ,θ2+dcrhσ,θ2,hσ2,θ2,dcrhσ,θ,σ2≤Mθ1−θ+Mθ2−θ1+Sdcrσ2,σ≤Mε+Mε+Sl=2Mε+Sl<l.
Thus, for every fixedθ∈θ2−ε,θ2+ε, h·,θ:Bσ,r¯→Bσ,r¯ has a fixed point in U¯ and can be deduced by applying Theorem 13. But this fixed point should be in U as in the previous case. Hence, θ∈B for any θ∈θ2−ε,θ2+ε and so B is open in 0,1. Thus, we showed that B is RMS open as well as RMS closed in 0,1 and by connectedness, B=0,1. Hence, h·,1 has a fixed point in U.



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*Source: 8021234-2020-09-10.xml*