# Control Fuzzy Metric Spaces via Orthogonality with an Application

Authors: Fahim Uddin; Khalil Javed; Hassen Aydi; Umar Ishtiaq; Muhammad Arshad
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this article, we are generalizing the concept of control fuzzy metric spaces by introducing orthogonal control fuzzy metric spaces. We prove some fixed point results in this setting. We provide nontrivial examples to show the validity of our main results and the introduced concepts. An application to fuzzy integral equations is also included. Our results generalize and improve several developments from the existing literature.


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## Body

## 1. Introduction and Preliminaries

Many authors studied fixed point theory explicitly, introduced and popularized lot of spaces, and made the area of fixed point theory more fascinating. In this connectedness, Bakhtin [1] and Czerwik [2] provided a generalization of metric spaces, named as a b-metric space. Zadeh [3] originated fuzzy sets. The formulation of metric spaces and fuzzy sets, named as fuzzy metric spaces, helped many authors in various ways. Nădăban [4] originated fuzzy b-metric spaces. Many authors [5–7] worked in fuzzy b-metric spaces.In [8], the authors introduced the concept of an extended fuzzy b-metric space as a generalization of fuzzy b-metric spaces. The work [9] originates the concept of controlled metric type spaces (see also [10]). Recently, in [11], the notion of controlled-type metric spaces has been generalized by a formulation of controlled fuzzy metric spaces, which are also generalizations of extended fuzzy b-metric spaces.Eshaghi et al. [12] introduced the notion of an orthogonal set and proved the Banach fixed point result. Many of the authors [13–15] continued working on orthogonal spaces. In this article, we are generalizing the concept of control fuzzy metric spaces [11]. Namely, we initiate the notion of orthogonal control fuzzy metric spaces.Let us first recall some basic definitions related to this manuscript.Definition 1 (see [4]).
A 4-tupleΖ,Δ,∗,u is called a fuzzy b-metric space if Ζ is an arbitrary (nonempty) set, ∗ is a continuous t-norm, and Δ is a fuzzy set on Ζ×Ζ×0,∞ satisfying the following conditions, for all ν,ω,ϰ∈Ζ,r,s>0 and for a given real numberu≥1:(B1)
Δν,ω,r>0(B2)
Δν,ω,r=1 if and only if ν=ω(B3)
Δν,ω,r=Δω,ν,r(B4)
Δν,ϰ,ur+s≥Δν,ω,r∗Δω,ϰ,ss(B5)
Δν,ω,⋅:0,∞⟶0,1 is continuousDefinition 2 (see [8]).
A 4-tupleΖ,Δα,∗,α is called an extended fuzzy b-metric space if Ζ is a (nonempty) set, where α:Ζ×Ζ⟶1,∞,∗ is a continuous t-norm, and Δα is a fuzzy set on Ζ×Ζ×0,∞, satisfying the following conditions, for all ν,ω,ϰ∈Ζandr,s>0:Δ1Δαν,ω,0=0Δ2Δαν,ω,r=1⟺ν=ωΔ3Δαν,ω,r=Δαω,ν,rΔ4Δαν,ϰ,αν,ϰr+s≥Δαν,ω,r∗Δαω,ϰ,sΔ5Δαν,ω,⋅:0,∞⟶0,1 is continuousDefinition 3 (see [11]).
A 4-tupleΖ,Δγ,∗ is called a control fuzzy metric space if Ζ is a (nonempty) set, γ:Ζ×Ζ⟶1,∞, where ∗ is a continuous t-norm and Δγ is a fuzzy set on Ζ×Ζ×0,∞, satisfying the following conditions, for all ν,ω,ϰ∈Ζandr,s>0:Δ1Δγν,ω,0=0Δ2Δγν,ω,r=1⟺ν=ωΔ3Δγν,ω,r=Δγω,ν,rΔ4Δγν,ϰ,r+s≥Δγν,ω,r/γν,ω∗Δγω,ϰ,s/γω,ϰΔ5Δγν,ω,⋅:0,∞⟶0,1 is continuousDefinition 4 (see [11]).
LetΖ be a set and let ζ:Ζ⟶Ζ and Oν=ν0,ζν0,ζ2ν0,…, for some ν0∈Ζ, be the orbit ofζ. A functionT:Ζ⟶Ζ is said to be ζ-orbitally lower semicontinuous at u∈Ζ if νn∈Oν0 such that νn⟶u, then we get Tu≥limn⟶∞infTνn.


## 2. Main Results

In this section, we introduce orthogonal control fuzzy metric spaces and prove some fixed point results.Definition 5.
A 4-tupleΖ,θγ,∗,⊥ is called an orthogonal control fuzzy metric space if Ζ is an (nonempty) orthogonal set, γ:Ζ×Ζ⟶1,∞, where ∗ is a continuous t-norm and θγ is a fuzzy set on Ζ×Ζ×0,∞, satisfying the following conditions:θγ1θγν,ω,r>0,∀ν,ω∈Ζ,r>0such thatν⊥ωandω⊥νθγ2θγν,ω,r=1⟺ν=ω,∀ν,ω∈Ζ,r>0such thatν⊥ωandω⊥νθγ3θγν,ω,r=θγω,ν,r,∀ν,ω∈Ζ,r>0such thatν⊥ωandω⊥νθγ4θγν,ϰ,r+s≥θγν,ω,r/γν,ω∗θγω,ϰ,s/γω,ϰ,orθγν,ϰ,γν,ωγω,ϰr+s≥θγν,ω,r∗θγω,ϰ,s,∀ν,ω,ϰ∈Ζ,r,s>0 such that ν⊥ω, ω⊥ϰ, and ν⊥ϰθγ5θγν,ω,⋅:0,∞⟶0,1 is continuous, ∀ν,ω∈Ζ such that ν⊥ω and ω⊥ν
Now, we show that the following are equivalent:(i)
θγν,ϰ,r+s≥θγν,ω,r/γν,ω∗θγω,ϰ,s/γω,ϰ(ii)
θγν,s,γν,ωγω,ϰr+s≥θγν,ω,r∗θγω,ϰ,sProof.
(ii) ⟹ (i)(1)θγν,ϰ,γν,ωγω,ϰr+s=θγν,ϰ,γν,ωγω,ϰr+γν,ωγω,ϰs≥θγν,ω,γν,ωγω,ϰrγν,ωγω,ϰ∗θγω,ϰ,γν,ωγω,ϰsγν,ωγω,ϰ=θγν,ω,r∗θγω,ϰ,s.
Similarly, we can easily prove (i)⇒ (ii).Example 1.
LetΖ=−1,1,2,3,4,…=A∪B, where A=−1,1 and B=ℕ\1. Define a binary relation ⊥ by ν⊥ω⟺ν,ω∈ν,ω. Given θγ:Ζ×Ζ×0,∞⟶0,1 as(2)θγν,ω,r=1,ifν=ω,r+1/νr+1/ω,ifν∈Bandω∈A,r+1/ωr+1/ν,ifν∈Aandω∈B,r+1/maxν,ωr+1/minν,ω,if otherwise,with a continuous t-norm ∗ defined by r1∗r2=r1⋅r2. Given γ:Ζ×Ζ⟶1,∞ as(3)γν,ω=1,ifν,ω∈A,maxν,ω,otherwise.
Then,Ζ,θγ,∗,⊥ is an orthogonal control fuzzy metric space, but it is not a control fuzzy metric space.Proof.
θγ1,θγ2,θγ3, and θγ5 are obvious. Here, we prove θγ4:
(θγ4) θγν,ϰ,r+s≥θγν,ω,r/γν,ω∗θγω,ϰ,s/γω,ϰ,∀ν,ω,ϰ∈Ζ,r,s>0, such that ν⊥ω, ω⊥ϰ, and ν⊥ϰ
We have the following cases to proveθγ4.Case 1.
Ifϰ=ν, then θγν,ϰ,r+s=1. Also, θγν,ω,r/γν,ω≤1 and θγω,ϰ,s/γω,ϰ≤1.
This implies(4)θγν,ω,rγν,ω∗θγω,ϰ,sγω,ϰ≤1.Case 2.
Ifϰ=ω, then θγω,ϰ,s/γω,ϰ=1, and clearly, θγν,ϰ,r+s≥θγν,ω,r/γν,ω. This implies(5)θγν,ϰ,r+s≥θγν,ω,rγν,ω∗θγω,ϰ,sγω,ϰ.Case 3.
Ifϰ≠ν, ϰ≠ω, and ν=ω, then θγν,ω,r/γν,ω=1 and clearly,(6)θγν,ϰ,r+s≥θγω,ϰ,sγω,ϰ.
This implies(7)θγν,ϰ,r+s≥θγν,ω,rγν,ω∗θγω,ϰ,sγω,ϰ.Case 4.
Ifϰ≠ν, ϰ≠ω, andν≠ω, then we have the following cases:(1)
ν,ϰ∈A and ω∈B(2)
ω∈A and ν,ϰ∈B(3)
ω,ϰ∈A and ν∈B(4)
ν,ω∈A and ϰ∈B(5)
ϰ∈A and ν,ω∈B(6)
ν∈A and ω,ϰ∈B(7)
ν,b,ω∈A(8)
ν,b,ω∈BProof of (1).
Ifν,ϰ∈A and ω∈B, then(8)θγν,ϰ,r+s=r+s+1/maxν,ϰr+s+1/minν,ϰ.
Observe thatmaxν,ϰ=minν,ϰ=1. This implies θγν,ϰ,r+s=1.
On the contrary,(9)θγν,ω,rγν,ω=r/γν,ω+1/ωr/γν,ω+1/ν.
Observe thatγν,ω=ω; then,(10)θγν,ω,rγν,ω=νr+ννr+ω<1,θγω,ϰ,sγω,ϰ=s/γω,ϰ+1/ωs/γω,ϰ+1/ϰ.
Note thatγω,ϰ=ω; then,(11)θγω,ϰ,sγω,ϰ=ϰs+ϰϰs+ω<1.
This implies(12)θγν,ϰ,r+s≥θγν,ω,rγν,ω∗θγω,ϰ,sγω,ϰ.
Similarly, (2)–(8) are easily satisfied.
Now, we show thatθγ is not a control fuzzy metric space. Let ν,ω,ϰ∈A. Also, let ν=ϰ=1, ω=−1, and r,s>1; then,(13)θγν,ϰ,r+s=1.
On the contrary,(14)θγν,ω,rγν,ω=r+1r−1,r≠1,θγω,ϰ,sγω,ϰ=s+1s−1,s≠1.
This implies(15)1≥r+1r−1+s+1s−1.
This failsθγ4.Example 2.
Letℤ=A∪B, where A=−1,−2,−3,… and B=0,1,2,3,…. Define a binary relation ⊥ by ν⊥ω⟺ν+ω≥0. Define θγ:Ζ×Ζ×0,∞⟶0,1 by(16)θγν,ω,r=rr+maxν,ω,for all r>0andν,ω∈Ζ with a continuous t-norm ∗ defined by r1∗r2=r1⋅r2. Given γ:Ζ×Ζ⟶1,∞ as(17)γν,ω=1,ifν,ω∈Aorν=0orω=0,maxν,ω,otherwise.
Then,Ζ,θγ,∗,⊥ is an orthogonal control fuzzy metric space, but it is not a control fuzzy metric space.Proof.
First, we show thatθγ is an orthogonal control fuzzy metric space. θγ1,θγ3, and θγ5 are obvious. Here, we prove θγ2 and θγ4:
(θγ2) θγν,ω,r=1⟺ν=ω,∀ν,ω∈Ζ,r>0such thatν⊥ωandω⊥ν:(18)θγν,ω,r=1,⟺rr+maxν,ω=1,⟺r=r+maxν,ω,⟺maxν,ω=0,⟺ν=ω.
(θγ3) θγν,ω,r=θγω,ν,r,∀ν,ω∈Ζ,r>0such thatν⊥ωandω⊥ν:(19)θγν,ω,r=rr+maxν,ω=rr+maxω,ν=θγω,ν,r.
θγ4θγν,ϰ,γν,ωγω,ϰr+s≥θγν,ω,r∗θγω,ϰ,s,∀ν,ω,ϰ∈Ζ,r,s>0, such that ν⊥ω, ω⊥ϰ, and ν⊥ϰ:(20)⇒maxν,ϰ≤γν,ωmaxν,ω+γω,ϰmaxω,ϰ⇒rsmaxν,ϰ≤γν,ωrs+s2maxν,ω+γω,ϰrs+r2maxω,ϰ⇒rsmaxν,ϰ≤γν,ωr+ssmaxν,ω+γω,ϰs+rrmaxω,ϰ⇒rsmaxν,ϰ≤γν,ωγω,ϰr+ssmaxν,ωγω,ϰ+rmaxω,ϰγν,ω⇒rsmaxν,ϰ≤γν,ωγω,ϰr+ssmaxν,ω+rmaxω,ϰ⇒rsmaxν,ϰ≤γν,ωγω,ϰr+ssmaxν,ω+rmaxω,ϰ+maxν,ωmaxω,ϰ⇒γν,ωγω,ϰr+srs+rsmaxν,ϰ≤γν,ωγω,ϰr+srs+γν,ωγω,ϰr+ssmaxν,ω+rmaxω,ϰ+maxν,ωmaxω,ϰ⇒γν,ωγω,ϰr+srs+rsmaxν,ϰ≤γν,ωγω,ϰr+srs+smaxν,ω+rmaxω,ϰ+maxν,ωmaxω,ϰ⇒rsγν,ωγω,ϰr+s+maxν,ϰ≤γν,ωγω,ϰr+sr+maxν,ωs+maxω,ϰ⇒γν,ωγω,ϰr+sγν,ωγω,ϰr+s+maxν,ϰ≥rsr+maxν,ωs+maxω,ϰ⇒γν,ωγω,ϰr+sγν,ωγω,ϰr+s+maxν,ϰ≥rr+maxν,ω⋅ss+maxω,ϰ⇒θγν,ϰ,γν,ωγω,ϰr+s≥θγν,ω,r∗θγω,ϰ,s.
Now, we show thatθγ is not a control fuzzy metric space. Indeed,(21)θγν,ϰ,γν,ωγω,ϰr+s=γν,ωγω,ϰr+sγν,ωγω,ϰr+s+maxν,ϰθγν,ω,r=rr+maxν,ωθγω,ϰ,s=ss+maxω,ϰ.
This implies(22)γν,ωγω,ϰr+sγν,ωγω,ϰr+s+maxν,ϰ≥rr+maxν,ω⋅ss+maxω,ϰ.
Now, letν=ω=ϰ=−1; then, γν,ω=γω,ϰ=1 and maxν,ϰ=maxν,ω=maxω,ϰ=−1. This implies that(23)r+sr+s−1≥rr−1⋅ss−1=rsr−1s−1,r,s≠1.
Takingr=s=2, we get a contradiction.Remark 1.
Every control fuzzy metric space is an orthogonal control fuzzy metric space, but the converse is not true.Remark 2.
Note that Example2 also holds for the t-norm :r1∗r2=minr1,r2.Definition 6.
LetΖ,θγ,∗,⊥ be an orthogonal control fuzzy metric space. Then, a sequence νn is said to be G-convergent to ν, where ν,νn∈Ζ if and only if limn⟶∞θγνn,ν,r=1 for any n>0 and for all r>0.Definition 7.
LetΖ,θγ,∗,⊥ be an orthogonal control fuzzy metric space. Then, a sequence νn is said to be a G-Cauchy sequence with νn∈Ζ if and only if limn⟶∞θγνn,νn+m,r=1 for all m>0 and r>0.Definition 8.
LetΖ,θγ,∗,⊥ be an orthogonal control fuzzy metric space; then, it is G-complete if and only if every G-Cauchy sequence is convergent.Definition 9.
ζ:Ζ⟶Ζ is ⊥-continuous at ν∈Ζ in an orthogonal control fuzzy metric space Ζ,θγ,∗,⊥ if for each orthogonal sequence νn in Ζ so that if limn⟶∞θγνn,ν,r exists and is finite for all r>0, then limn⟶∞θγζνn,ζν,r again exists and is finite for all r>0. Furthermore, ζ is ⊥-continuous if ζ is ⊥-continuous at each ν∈Ζ. Also, ζ is ⊥-preserving if ζν⊥ζω; hence, ν⊥ω.Remark 3.
It is not necessary that the limit of a convergent sequence will be unique in an orthogonal control fuzzy metric space.
For this, take a sequenceνn defined by νn=1−1/n for each integer n, and define an orthogonal control fuzzy metric space as in Example 2 with ν≥1. Also, in particular, take γν,ω=γω,ϰ=1; then,(24)limn⟶∞θγνn,ν,r=limn⟶∞rr+maxνn,ν=limn⟶∞rr+ν=θγν,ν,r,for all r>0. Observe that the sequence νn converges to all ν∈Ζ with ν≥1.Remark 4.
It is not necessary that the convergent sequence will be a Cauchy sequence in an orthogonal control fuzzy metric space.
For this, take a sequenceνn defined by νn=1+−1n for each integer n, and define an orthogonal control fuzzy metric space as in Example 2 with ν≥2. Also, in particular, take γν,ω=γω,ϰ=1; then,(25)limn⟶∞θγνn,ν,r=limn⟶∞rr+maxνn,ν=limn⟶∞rr+ν=θγν,ν,r,for all r>0. Observe that the sequence νn converges to all ν∈Ζ with ν≥2. However, limn⟶∞θγνn,νn+m,r does not exist.
Mihet [16] introduced a control function ψ. We generalize it as follows.Definition 10.
Letψ be the class of all mappings Ψ:0,1⟶0,1 such that Ψ is orthogonal continuous, nondecreasing, and Ψ£>£,for all£∈0,1. If Ψ∈ψ,thenΨ1=1andlimn⟶∞Ψn£=1,for all£∈0,1.Theorem 10.
LetΖ,θγ,∗,⊥ be an orthogonal G-complete control fuzzy metric space with γ:Ζ×Ζ⟶1,∞ such that(26)limr⟶∞θγν,ω,r=1,for all ν∈Ζ. Suppose that ζ:Ζ⟶Ζ is an ⊥-continuous, ⊥-contraction, and ⊥-preserving mapping so that(27)θγζν,ζω,kr≥θγν,ω,r,for all ν,ω∈Ζ,r>0, where k∈0,1. Also, assume that, for every ν∈Ζ,(28)limn⟶∞γνn,ω,limn⟶∞γω,νn,exist and are finite. Then, ζ has a unique fixed point in Ζ. Furthermore,(29)limn⟶∞θγζnu,u,r=θγu,u,r,for allu∈Ζandr>0.Proof.
SinceΖ,θγ,∗,⊥ is an orthogonal G-complete control fuzzy metric space, there exists ν0∈Ζ such that(30)ν0⊥ω,for allω∈Ζ.
This yields thatν0⊥ζν0. Consider(31)ν1=ζν0,ν2=ζ2ν0=ζν1,…,νn=ζnν0=ζνn−1.
Ifνn=νn−1, then νn is a fixed point of ζ. Suppose that νn≠νn−1 for all n∈ℕ. Since ζ is ⊥-preserving, νn is an orthogonal sequence. Since ζ is an ⊥-contraction, we have(32)θγνn,νn+1,r=θγζνn−1,ζνn,r≥θγνn−2,νn−1,rk≥⋯≥θγν0,ν1,rkn−1.
Now, fromθγ4, we have(33)θγνn,νn+m,r≥θγνn,νn+1,r2γνn,νn+1∗θγνn+1,νn+m,r2γνn+1,νn+m≥θγνn,νn+1,r2γνn,νn+1∗θγνn+1,νn+2,r22γνn+1,νn+mγνn+1,νn+2∗θγνn+2,νn+m,r22γνn+1,νn+mγνn+2,νn+m≥θγνn,νn+1,r2γνn,νn+1∗θγνn+1,νn+2,r22γνn+1,νn+mγνn+1,νn+2∗θγνn+2,νn+3,r23γνn+1,νn+mγνn+2,νn+mγνn+2,νn+3∗θγνn+3,νn+m,r23γνn+1,νn+mγνn+2,νn+mγνn+3,νn+m≥…≥θγνn,νn+1,r2γνn,νn+1∗∗i=n+1n+m−2θγνi,νi+1,r2m−2∏j=n+1iγνj,νn+mγνi,νi+1∗θγνn+m−1,νn+m,r2m−1∏i=n+1n+m−1γνi,νn+m≥θγν0,ν1,r2kn−1γνn,νn+1∗∗i=n+1n+m−2θγν0,ν1,r2m−1ki−1∏j=n+1iγνj,νn+mγνi,νi+1∗θγν0,ν1,r2m−1kn+m−1∏i=n+1n+m−1γνi,νn+m.
Now, taking limit asn⟶∞ in (33), in (32) together with (26), we have(34)limn⟶∞θγνn,νn+m,r≥1∗1∗⋯∗1=1,for all r>0 and m∈ℕ. Thus, νn is an orthogonal G-Cauchy sequence in Ζ. The completeness of Ζ,θγ,∗,⊥ implies the existence of u∈Ζ such that(35)limn⟶∞θγνn,u,r=1,for all r>0. Now, since ζ is an⊥-continuous mapping, one writes limθγνn+1,ζu,r=limθγζνn,ζu,r=1. For r>0 and from θγ4, we have(36)θγu,ζu,r≥θγu,νn+1,r2γu,νn+1∗θγνn+1,ζu,r2γνn+1,ζu=θγu,νn+1,r2γu,νn+1∗θγζνn,ζu,r2γνn+1,ζu.
Takingn⟶∞ in (36) and using (35), we get θγu,ζu,r=1 for all r>0, that is, ζu=u.
Now, for uniqueness, letw∈Ζ be another fixed point for ζ and let there exist r>0 such thatθγu,w,r≠1. We can obtain(37)ν0⊥u,ν0⊥w.
Sinceζ is an ⊥-preserving, this implies that(38)ζnν0⊥ζnu,ζnν0⊥ζnw,for alln∈ℕ.
From (27), we can derive(39)θγζnν0,ζnu,r≥θγζnν0,ζnu,kr≥θγν0,u,rknθγζnν0,ζnw,r≥θγζnν0,ζnw,kr≥θγν0,w,rkn.
We can write(40)θγu,w,r=θγζnu,ζnw,r≥θγζnν0,ζnu,r2γν0,u∗θγζnν0,ζnw,r2γν0,w≥θγν0,u,rkn2γν0,u∗θγν0,w,rkn2γν0,w,for all n∈ℕ. By taking limit as n⟶∞, we get θγu,w,r=1, for all r>0; hence, u=w.Corollary 1.
LetΖ,θγ,∗,⊥ be an orthogonal G-complete control fuzzy metric space. Let ζ:Ζ⟶Ζ be ⊥-contraction and ⊥-preserving. Also, assume that if νn is an O-sequence with νn⟶ν∈Ζ, then ν⊥νn for all n∈ℕ. Therefore, ζ has a unique fixed point ν∗∈Ζ. Furthermore, limn⟶∞θγζnν,ν∗,r=θγν∗,ν∗,r, for all ν∈Ζ and r>0.Proof.
We can prove alike as in the proof of Theorem 1 thatνn is a G-Cauchy sequence and converges to ν∗∈Ζ. Hence, ν∗⊥νn for all n∈ℕ. We get from (26) that(41)θγζν∗,νn+1,r=θγζν∗,ζνn,r≥θγζν∗,ζνn,kr≥θγν∗,νn,r,limn⟶∞θγζν∗,νn+1,r=1.
Then, we can write(42)θγν∗,ζν∗,r≥θγν∗,νn+1,r2γν∗,νn+1∗θγνn+1,ζν∗,r2γνn+1,ζν∗.
Taking limit asn⟶∞, we get θγν∗,ζν∗,r=1∗1=1, and hence, ζν∗=ν∗. The rest of proof is similar as in Theorem 1.Theorem 2.
LetΖ,θγ,∗,⊥ be an orthogonal G-complete control fuzzy metric space with γ:Ζ×Ζ⟶1,∞ so that(43)limr⟶∞θγν,ω,r=1,for all ν∈Ζ. If ζ:Ζ⟶Ζ is an ⊥-contraction and ⊥-preserving and satisfies(44)θγζν,ζ2ν,kr≥θγν,ζν,r,for all ν∈Oν,r>0,wherek∈0,1, thenζnν0⟶u. Furthermore, u is a fixed point of ζ if and only if ζν=θγν,ζν,r is ζ-orbitally lower semicontinuous at u.Proof.
SinceΖ,θγ,∗,⊥ is an orthogonal G-complete control fuzzy metric space, there exists ν0∈Ζ such that(45)ν0⊥ω,for allω∈Ζ.
This says thatν0⊥ζν0. Consider(46)ν1=ζν0,ν2=ζ2ν0=ζν1,…,νn=ζnν0=ζνn−1.
Ifνn=νn−1, then νn is a fixed point of ζ. Suppose that νn≠νn−1 for all n∈ℕ. Since ζ is ⊥-preserving, νn is an orthogonal sequence. Since ζ is an ⊥-contraction, we have(47)θγζnν0,ζn+1ν0,kr=θγνn,νn+1,kr≥θγνn−1,νn,rk≥⋯≥θγν0,ν1,rkn−1.
Now, fromθγ4, we have(48)θγνn,νn+m,r≥θγνn,νn+1,r2γνn,νn+1∗θγνn+1,νn+m,r2γνn+1,νn+m≥θγνn,νn+1,r2γνn,νn+1∗θγνn+1,νn+2,r22γνn+1,νn+mγνn+1,νn+2∗θγνn+2,νn+m,r22γνn+1,νn+mγνn+2,νn+m≥θγνn,νn+1,r2γνn,νn+1∗θγνn+1,νn+2,r22γνn+1,νn+mγνn+1,νn+2∗θγνn+2,νn+3,r23γνn+1,νn+mγνn+2,νn+mγνn+2,νn+3∗θγνn+3,νn+m,r23γνn+1,νn+mγνn+2,νn+mγνn+3,νn+m≥⋯≥θγνn,νn+1,r2γνn,νn+1∗∗i=n+1n+m−2θγνi,νi+1,r2m−2∏j=n+1iγνj,νn+mγνi,νi+1∗θγνn+m−1,νn+m,r2m−1∏i=n+1n+m−1γνi,νn+m≥θγν0,ν1,r2kn−1γνn,νn+1∗∗i=n+1n+m−2θγν0,ν1,r2m−1ki−1∏j=n+1iγνj,νn+mγνi,νi+1∗θγν0,ν1,r2m−1kn+m−1∏i=n+1n+m−1γνi,νn+m.
Now, taking limit asn⟶∞ in (48), we have(49)limn⟶∞θγνn,νn+m,r≥1∗1∗⋯∗1=1,for all r>0 and m∈ℕ. Thus, νn is an orthogonal G-Cauchy sequence in Ζ. From the completeness of Ζ,θγ,∗,⊥, there is νn⟶ζnν0=u. Assume that ζ is ζ-orbitally lower semicontinuous at u∈Ζ; then, we have(50)θγu,ζu,kr=limn⟶∞supθγζnν0,ζn+1ν0,kr≥limn⟶∞supθγν0,ν1,rkn−1=1.
Conversely, letu=ζuandνn∈Zwithνn⟶u; then, we obtain(51)ζu=θγu,ζu,kr=1≥limn⟶∞supζνn=θγζnν0,ζn+1ν0,kr.Theorem 3.
LetΖ,θγ,∗,⊥ be an orthogonal G-complete control fuzzy metric space and ζ:Ζ⟶Ζ be an ⊥-continuous, ⊥-contraction, and ⊥-preserving mapping so that(52)θγν,ω,r>0⇒θγζν,ζω,r≥Ψθγν,ω,r,for all ν,ω∈Ζ and r>0. Then, ζ has a unique fixed point in Ζ.Proof.
SinceΖ,θγ,∗,⊥ is an orthogonal G-complete control fuzzy metric space, there exists ν0∈Ζ such that(53)ν0⊥ω,for allω∈Ζ.
Thus,ν0⊥ζν0. Assume(54)ν1=ζν0,ν2=ζ2ν0=ζν1,…,νn=ζnν0=ζνn−1.
Ifνn=νn−1, then νn is a fixed point of ζ. Suppose that νn≠νn−1 for all n∈ℕ. Since ζ is ⊥-preserving, νn is an orthogonal sequence. Since ζ is an ⊥-contraction, we have(55)θγνn,νn+1,r=θγζνn−1,ζνn,r≥Ψθγνn−2,νn−1,r≥⋯≥Ψnθγν0,ν1,r.
Now, fromθγ4, we have(56)θγνn,νn+m,r≥θγνn,νn+1,r2γνn,νn+1∗θγνn+1,νn+m,r2γνn+1,νn+m≥θγνn,νn+1,r2γνn,νn+1∗θγνn+1,νn+2,r22γνn+1,νn+mγνn+1,νn+2∗θγνn+2,νn+m,r22γνn+1,νn+mγνn+2,νn+m≥θγνn,νn+1,r2γνn,νn+1∗θγνn+1,νn+2,r22γνn+1,νn+mγνn+1,νn+2∗θγνn+2,νn+3,r23γνn+1,νn+mγνn+2,νn+mγνn+2,νn+3∗θγνn+3,νn+m,r23γνn+1,νn+mγνn+2,νn+mγνn+3,νn+m≥⋯≥θγνn,νn+1,r2γνn,νn+1∗∗i=n+1n+m−2θγνi,νi+1,r2m−2∏j=n+1iγνj,νn+mγνi,νi+1∗θγνn+m−1,νn+m,r2m−1∏i=n+1n+m−1γνi,νn+m≥Ψnθγν0,ν1,r2γνn,νn+1∗∗i=n+1n+m−2Ψiθγν0,ν1,r2m−1∏j=n+1iγνj,νn+mγνi,νi+1∗Ψn+m−1θγν0,ν1,r2m−1∏i=n+1n+m−1γνi,νn+m.
Now, taking limit asn⟶∞ in (55 and 56), we have(57)limn⟶∞θγνn,νn+m,r≥1∗1∗⋯∗1=1,for all r>0 and m∈ℕ. Thus, νn is an orthogonal G-Cauchy sequence in Ζ. From the completeness of Ζ,θγ,∗,⊥, there exists u∈Ζ such that(58)limn⟶∞θγνn,u,r=1,for all r>0. Now, since ζ is an ⊥-continuous mapping, one gets limθγνn+1,ζu,r=limθγζνn,ζu,r=1 as n⟶∞. For r>0 and from θγ4, we have(59)θγu,ζu,r≥θγu,νn+1,r2γu,νn+1∗θγνn+1,ζu,r2γνn+1,ζu=θγu,νn+1,r2γu,νn+1∗θγζνn,ζu,r2γνn+1,ζu≥θγu,νn+1,r2γu,νn+1∗Ψθγνn,u,r2γνn+1,ζu.
Takingn⟶∞ in (59) and using (58), we get θγu,ζu,r=1 for all r>0, that is, ζu=u.
Now, for uniqueness, letw∈Ζ be another fixed point for ζ and let there exist r>0 such that u≠w. We can obtain(60)ν0⊥u,ν0⊥w.
Sinceζ is an ⊥-preserving, this implies(61)ζnν0⊥ζnu,ζnν0⊥ζnw,for alln∈ℕ.
We can derive(62)θγζnν0,ζnu,r≥θγζnν0,ζnu,kr≥Ψθγν0,u,r,θγζnν0,ζnw,r≥θγζnν0,ζnw,kr≥Ψθγν0,w,r.
We can write(63)θγu,w,r=θγζnu,ζnw,r≥θγζnν0,ζnu,kr∗θγζnν0,ζnw,kr≥Ψθγν0,u,r∗Ψθγν0,w,r≥θγν0,u,r∗θγν0,w,r,for all n∈ℕ. This is a contradiction; hence, u=w.Example 3.
LetΖ=ℤ=A∪B, where A=−1,−2,−3,…∪0,1 and B=2,3,4,…. Define a binary relation ⊥ byν⊥ω⟺ν,ω∈ν,ω. Define θγ:Ζ×Ζ×0,∞⟶0,1 by(64)θγν,ω,r=rr+maxν,ω,for all r>0andν,ω∈Ζ with a continuous t-norm ∗ defined by: r1∗r2=r1⋅r2. Define γ:Ζ×Ζ⟶1,∞ by(65)γν,ω=1,ifν,ω∈Aorν=0orω=0,maxν,ω,otherwise.
Then,Ζ,θγ,∗,⊥ is an orthogonal G-complete control fuzzy metric space. Observe that(66)limr⟶∞θγν,ω,r=1.
Now, we defineζ:Ζ⟶Ζ by(67)ζν=ν2,ifν∈A,1,ifν∈B,for all ν∈Ζ.Proof.
Observe that ifν⊥ω, then clearly ζν⊥ζω. Now, there are some cases to prove that the contraction is orthogonal for k∈1/2,1.(1)
Ifν,ω∈A, then ζν=v/2 and ζω=ω/2. We have(68)θγζν,ζω,kr=θγv2,ω2,kr=krkr+maxv/2,ω/2≥rr+maxν,ω=θγν,ω,r.(2)
Ifν,ω∈B, then ζν=1 and ζω=1. In this case,(69)θγζν,ζω,kr=θγ1,1,kr=krkr+max1,1≥rr+maxν,ω=θγν,ω,r.(3)
Ifν∈A and ω∈B, then ζν=v/2 and ζω=1. Here,(70)θγζν,ζω,kr=θγv2,1,kr=krkr+maxv/2,1≥rr+maxν,ω=θγν,ω,r.(4)
Ifν∈B and ω∈A, then ζν=1 and ζω=ω/2. This implies that(71)θγζν,ζω,kr=θγ1,ω2,kr=krkr+max1,ω/2≥rr+maxν,ω=θγν,ω,r.
Hence, it is an ⊥-contraction. Now, we show that it is not a contraction. Letν,ω∈A, then ζν=ν/2 and ζω=ω/2. Here,(72)θγζν,ζω,kr=θγν2,ω2,kr=krkr+maxν/2,ω/2.
Letν=ω=−2,k=9/10andr=10, so(73)θγζν,ζω,kr=99+max−1,−1≤1010+max−2,−2=θγν,ω,r,which implies θγζν,ζω,kr≤θγν,ω,r. This is wrong.
Iflimn⟶∞θγνn,ν,r is finite and exists, then also limn⟶∞θγζνn,ζν,r is finite and exists. This implies that it is ⊥-continuous. Also, observe that(74)limn⟶∞γνn,ω,limn⟶∞γω,νn,are finite and exist. All circumstances of Theorem 1 are fulfilled and 0 is the unique fixed point of ζ.


## 3. An Application to a Fuzzy Integral Equation

In this section, we utilize Theorem 1 to examine the existence and uniqueness of a solution of a fuzzy Fredholm-type integral equation of second kind.LetΖ=Ce,g,ℝ be the set of all continuous real-valued functions defined on e,g.Now, we consider the fuzzy Fredholm-type integral equation of the second kind:(75)νl=fj+β∫egFl,jνldj,forl,j∈e,g.where β>0,fj is a fuzzy function of j∈e,g and F∈Ζ. Define θγ by(76)θγνl,ωl,r=supl∈e,grr+maxνl,ωl,forν,ω∈Ζandr>0,with a continuous t-norm ∗ defined by r1∗r2=r1⋅r2. Define γ:Ζ×Ζ⟶1,∞ by(77)γν,ω=1,ifν,ω∈Aorν=0orω=0,maxν,ω,otherwise.Then,Ζ,θγ,∗⊥ is an orthogonal G-complete control fuzzy metric space.Theorem 4.
Assume thatmaxFl,jνl,Fl,jωl≤maxνl,ωl for ν,ω∈Ζ, k∈0,1, and ∀l,j∈e,g.
Also, consider∫egdj=g−e≤k<1. Let ζ:Ζ⟶Ζ be(i)
⊥-preserving(ii)
⊥-contraction(iii)
⊥-continuous
Then, the fuzzy Fredholm-type integral equation of second kind in equation (75) has a unique solution.Proof.
Defineζ:Ζ⟶Ζ by(78)ζνl=fj+β∫egFl,jeldj,for alll,j∈e,g.(i)
Take orthogonality asνl⊥ωl⟺νlωl∈νl,ωl. We see that νl and ζνl belong to Ζ. So, if νl⊥ωl, then clearly ζνl⊥ζωl.(ii)
Observe that the existence of a fixed point of the operatorζ is equivalent to the existence of a solution of the Fredholm-type integral Equation (75).(iii)
Note that(79)maxFl,jνl,Fl,jωl≤maxνl,ωl⇒fj+β∫egmaxFl,jνl,Fl,jωl≤fj+β∫egmaxνl,ωl.(iv)Now, for allν,ω∈Ζ, we have(80)θγζνl,ζωl,kr=supl∈e,gkrkr+maxζνl,ζωl=supl∈e,gkrkr+max∫egFl,jνldj,∫egFl,jωldj=supl∈e,gkrkr+∫egmaxFl,jνl,Fl,jωldj≥supl∈e,gkrkr+∫egmaxνl,ωldj=supl∈e,gkrkr+maxνl,ωl∫egdj≥supl∈e,gkrkr+kmaxνl,ωl≥rr+maxνl,ωl=θγνl,ωl,r.(v)
Hence,ζ is an ⊥-contraction.(vi)
Supposeνn is an orthogonal sequence in Ζ such that νn converges to ν∈Ζ. Because ζ is ⊥-preserving, ζνn is an orthogonal sequence for each n∈ℕ. From (ii), we have(81)θγνl,ωl,kr≥θγνl,ωl,r.
Aslimn⟶∞θγνl,ωl,r is finite and exists, for all r>0, it is clear that limn⟶∞θγνl,ωl,kr is finite and exists.
Hence,ζ is ⊥-continuous.
Therefore, all circumstances of Theorem 1 are fulfilled. Hence, the operatorζ has a unique fixed point. This says that the fuzzy Fredholm-type integral equation (75) has a unique solution.Corollary 2.
LetΖ,θγ,∗ be a G-complete control fuzzy metric space. Define ζ:Ζ⟶Ζ by(82)ζνl=fj+β∫egFl,jeldj,for alll,j∈e,g.
Suppose the following conditions hold:(I)
maxFl,jνl,Fl,jωl≤maxνl,ωl for ν,ω∈Ζ, k∈0,1 and ∀l,j∈e,g(II)
∫egdj=g−e≤k<1
Then, the integral equation (75) has a solution.Proof.
We can prove it easily from Theorem4.



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