# Solutions of Integral Equations via Fixed-Point Results on Orthogonal Gauge Structure

Authors: Mazhar Mehmood; Hassen Aydi; Muhammad Usman Ali; Fahimuddin; Abdullah Shoaib; Manuel De La Sen
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

The main outcome of this paper is to introduce the notion of orthogonal gauge spaces and to present some related fixed-point results. As an application of our results, we obtain existence theorems for integral equations.


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## Body

## 1. Introduction

Fixed-point theory is a very important tool for proving the existence and uniqueness of the solutions to various mathematical models, such as integral and partial differential equations, optimization, variational inequalities, and approximation theory. Fixed-point theory has also gained considerable importance in the recent past after the famous Banach contraction theorem [1]. Since then, there have been many results related to mappings satisfying various types of contractive inequalities [2–5]. Recently, Gordji et al. [6] introduced an exciting notion of the orthogonal sets after which, orthogonal metric spaces were introduced. The concept of a sequence, continuity, and completeness has been redefined in this space. Further, they gave an extension of the Banach fixed-point theorem on this newly described shape and also applied their theorem to show the existence of a solution for a differential equation, which cannot be applied by the Banach fixed-point theorem.On the other hand, many definitions and theorems in the literature do not require that all of the properties of a metric hold true. In the last decades, many concepts of generalized metrics (as controlled and double controlled metrics) have been introduced (see [7, 8]).Gauge spaces are characterized by the fact that the distance between two points may be zero even if the two points are distinct. For instance, Frigon [9] and Chis and Precup [10] gave a generalization of the Banach contraction principle on gauge spaces. In the same direction, many interesting results have been raised obtained by different authors in [11–17]. In 2013, Ali et al. [18] ensured the existence of fixed points for an integral operator via a fixed-point theorem on complete gauge spaces. In 2012, Wardowski [19] gave a new type of contractions, named as F-contractions, and established new related fixed-point results. This contraction type nicely generalizes the most famous Banach contraction condition. Later on, many researchers worldwide generalized this result (see [20–24]).To give the ongoing research a new direction, we have combined the above statements in two directions of research. For this, we apply the concept of orthogonality in gauge spaces and investigate the existence of solutions of integral equations through the fixed-point theorem on orthogonal complete gauge spaces.


## 2. Preliminaries and Basic Definitions

First, we include some basic definitions and theorems which are useful to understand the results presented in this paper.Wardowski [19] introduced the family ℱ of all functions F:0,∞⟶ℝ so that(F1: for all η1,η2∈0,∞ with η1<η2, we have Fη1<Fη2(F2): for each positive sequence Kn, limn⟶∞Kn=0 iff limn⟶∞FKn=−∞(F3): there is k∈0,1 so that limK⟶0+FK=0The following are elements inℱ:(i)
Fαυ=lnυ for υ>0(ii)
Fβυ=υ+lnυ for υ>0(iii)
Fγυ=−1/υ for υ>0Further, Wardowski [19] introduced F-contractions and the related fixed-point theorem in the following way.Theorem 1 (see [19]).
LetX,d be a complete metric space and let T:X⟶X be an F-contraction mapping, that is, there are F∈ℱ and τ>0 so that for all ς,υ∈X,dTς,Tυ>0 implies(1)τ+FdTς,Tυ≤Fdς,υ.
ThenT admits a unique fixed point in X.Minak et al. [25] generalized this result as follows.Theorem 2 (see [25]).
LetX,d be a complete metric space and T:X⟶X be a self-mapping. Suppose that there are F∈ℱ and τ>0 so that(2)τ+FdTς,Tυ≤Fmaxdς,υ,dς,Tς,dυ,Tυ,dς,Tυ+dυ,Tς2,for all ς,υ∈X, with dTς,Tυ>0. If F or T is continuous, then T admits a unique fixed point.Now, we explain the notion of a pseudometric.Definition 1 (see [26]).
LetX be a nonempty set. A function d:X×X⟶0,∞ is said to be a pseudometric on X if for all ϱ,ρ,σ∈X:(i)
dϱ,ϱ=0(ii)
dϱ,ρ=dρ,ϱ(iii)
dϱ,σ≤dϱ,ρ+dρ,σExample 1.
Denote byX=C0,∞,ℝ the set of continuous real-valued functions f:X⟶ℝ with t0∈0,∞. This point t0 then induces a pseudometric on X defined as df,g=ft0−gt0forf,g∈X.
In 2017, Gordji et al. [6] initiated the notion of an orthogonal set (or o-set).Definition 2 (see [6]).
LetX≠∅ and ⊥ be a binary relation defined on X×X. The pair X,⊥ is called an orthogonal set (or an o-set), if(3)∀y∈Xy⊥x0or∀y∈Xx0⊥y.
The elementx0 is said to be orthogonal. An orthogonal set may have more than one orthogonal element.Example 2 (see [6]).
LetX=ℤ. We write that ϱ⊥ρ if there is k∈ℤ so that ρ=kϱ. Note that 0⊥ϱ for each ϱ∈ℤ. Hence, X,⊥ is an o-set.Definition 3 (see [6]).
LetX,⊥ be an o-set. Any two elements ϱ,ρ∈X are said to be orthogonally related iff ϱ⊥ρ.Definition 4 (see [6]).
LetX,⊥ be an o-set. A mapping Γ:X⟶X is said to be ⊥-preserving if for all orthogonally related elements ϱ,ρ, we have Γϱ⊥Γρ.Definition 5 (see [6]).
LetX,⊥ be an orthogonal set. ϱnn∈ℕ is said to be an orthogonal sequence (briefly, an o-sequence) if(4)∀n,ϱn⊥ϱn+1or∀n,ϱn+1⊥ϱn.


## 3. Orthogonal Gauge Space

The simplest way of defining an orthogonal gauge space is that the gauge space defined on an o-set is called an orthogonal gauge space. The precise discussion is given below.LetX,⊥ be an o-set and d:X×X⟶0,∞ be a pseudometric on X, then X,d,⊥ is said to be an orthogonal pseudometric space (or an o-pseudometric space).Example 3.
LetX=C0,∞,ℝ be the set of continuous real-valued functions f:X⟶ℝ with t0∈0,∞. This element induces an orthogonal pseudometric on X defined by dϱt,ρt=ϱt0−ρt0forϱt,ρt∈X and the orthogonality on X is given as ϱ⊥ρ iff ϱtρt≥ϱt or ϱtρt≥ρt.Definition 6.
LetX,d,⊥ be an orthogonal pseudometric space. Then, the orthogonal d-ball (or d-ball) of radius μ>0 and centered at σ∈X is the set ℬσ,d,μ=ξ∈X:dσ,ξ<μ.
Note that when we say thatd is an o-pseudometric on X, it means that X,⊥ is an o-set and d is a pseudometric on X.Definition 7.
A familyℳ=dλ:λ∈U of o-pseudometrics on X is said to be separating if for each pair ϱ,ρ such that ϱ≠ρ, there is dλ∈ℳ with dλϱ,ρ≠0.Definition 8.
Letℳ=dλ:λ∈U be the family of o-pseudometrics on X. Then, the topology Tℳ having as a subbasis the set of balls(5)βℳ=ℬϱ,dλ,ε:ϱ∈X,dλ∈ℳ,ε>0is said to be the topology induced by the set ℳ of o-pseudometrics. The pair X,Tℳ,⊥ is called an orthogonal gauge space.Definition 9.
LetX,Tℳ,⊥ be an orthogonal gauge space with respect to the family ℳ=dδ:δ∈U of o-pseudometrics on X.(i)
An orthogonal sequenceυn converges to υ∈X if(6)∀ε>0,∃n0∈ℕso thatdδυn,υ<ε,∀n≥n0,∀δ∈U.(ii)
An orthogonal sequenceυn is Cauchy if(7)∀ε>0,∃n0∈ℕso thatdδυm,υn<ε,∀m,n≥n0,∀δ∈U.(iii)
X,Tℱ,⊥ is an o-complete gauge space iff each Cauchy o-sequence in X,Tℱ,⊥ converges to an element of X.(iv)
A subset ofX is called closed if it contains the limit of each convergent o-sequence.


## 4. Fixed-Point Results on Orthogonal Gauge Structure

In this section, we study the existence of fixed points for a mapping satisfying certain conditions on an orthogonal gauge structure. Throughout this article,U is a directed set and X is a nonempty o-set with an orthogonal element (say ωa) and also endowed with a separating o-complete gauge structure dλ:λ∈U of o-pseudometrics.Theorem 3.
LetX be a nonempty o-set endowed with a separating o-complete gauge structure dλ:λ∈U of o-pseudometrics. Let T:X⟶X be a mapping with F∈ℱ and τ>0 so that(8)αx,y≥1⇒τ+FdλTx,Ty≤Faλdλx,y+bλdλx,Tx+cλdλy,Ty+eλdλx,Ty+Lλdλy,Tx,for all x,y∈X with x⊥y and for each λ∈U, whenever dλTx,Ty≠0 for λ∈U, where
aλ,bλ,cλ,eλ,Lλ≥0 and aλ+bλ+cλ+2eλ=1 for all λ∈U. Further, assume that(i)
T is ⊥-preserving(ii)
There is an elementω0∈X with ω0⊥Tω0 and αω0,Tω0≥1(iii)
For eachx,y∈X with x⊥y and αx,y≥1, we have αTx,Ty≥1(iv)
For any o-sequenceρn in X such that αρn,ρn+1≥1 for each n≥1 and ρn⟶ρasn⟶∞, we have αρn,ρ≥1 and ρn⊥ρ for each n≥1
Then,T possesses a fixed point.Proof.
Due to (ii), there isω0∈X with ω0⊥Tω0 and αω0,Tω0≥1, and by considering (iii), we get αTω0,T2ω0≥1. Moreover, we have Tω0⊥T2ω0, since T is ⊥-preserving. Repetition of the same arguments implies αTnω0,Tn+1ω0≥1 and Tnω0⊥Tn+1ω0 for each n∈ℕ. Consider ωn=Tnω0 for each n≥1. Then we say that ωn is an o-sequence with αωn,ωn+1≥1 for each n≥0. Also, note that if there is some m0∈ℕ so that ωm0=ωm0+1, then ωm0 is a fixed point of T. Thus, we assume there does not exist any such a natural number. As ω0∈X with ω0⊥ω1 and αω0,ω1≥1, then from (8), we have(9)τ+Fdλω1,ω2=τ+FdλTω0,Tω1≤Faλdλω0,ω1+bλdλω0,Tω0+cλdλω1,Tω1+eλdλω0,Tω1+Lλdλω1,Tω0=Faλdλω0,ω1+bλdλω0,ω1+cλdλω1,ω2+eλdλω0,ω2+Lλ.0≤Faλdλω0,ω1+bλdλω0,ω1+cλdλω1,ω2+eλdλω0,ω1+dλω1,ω2=Faλ+bλ+eλdλω0,ω1+cλ+eλdλω1,ω2,forallλ∈U.
SinceF is strictly increasing, from above we get(10)dλω1,ω2<aλ+bλ+eλdλω0,ω1+cλ+eλdλω1,ω2,that is,(11)1−cλ−eλdλω1,ω2<aλ+bλ+eλdλω0,ω1,for allλ∈U.
Sinceaλ+bλ+cλ+2eλ=1, we have(12)dλω1,ω2<dλω0,ω1,for allλ∈U.
Now, from (9), we have(13)τ+Fdλω1,ω2≤Fdλω0,ω1,for allλ∈U.
Also, we know thatω1⊥ω2 and αω1,ω2≥1. From (8), we get(14)τ+Fdλω2,ω3=τ+FdλTω1,Tω2≤Faλdλω1,ω2+bλdλω1,Tω1+cλdλω2,Tω2+eλdλω1,Tω2+Lλdλω2,Tω1=Faλdλω1,ω2+bλdλω1,ω2+cλdλω2,ω3+eλdλω1,ω3+Lλ.0≤Faλdλω1,ω2+bλdλω1,ω2+cλdλω2,ω3+eλdλω1,ω2+dλω2,ω3=Faλ+bλ+eλdλω1,ω2+cλ+eλdλω2,ω3,for allλ∈U.
SinceF is strictly increasing, again from above we get(15)dλω2,ω3<aλ+bλ+eλdλω1,ω2+cλ+eλdλω2,ω3,that is,(16)1−cλ−eλdλω2,ω3<aλ+bλ+eλdλω1,ω2,for allλ∈U.
Asaλ+bλ+cλ+2eλ=1, thus we have(17)dλω2,ω3<dλω1,ω2.
Now, from (14), we have(18)τ+Fdλω2,ω3≤Fdλω1,ω2.
By the obtained inequalities, we get(19)Fdλω2,ω3≤Fdλω1,ω2−τ≤Fdλω0,ω1−2τ,for allλ∈U.
Working on the same steps, we conclude that(20)Fdλωn,ωn+1≤Fdλω0,ω1−nτ,for eachn∈ℕand for allλ∈U.
Lettingn⟶∞ in (20), we get limn⟶∞Fdλωn,ωn+1=−∞ for all λ∈U. Thus, by property F2, we have limn⟶∞dλωn,ωn+1=0∀λ∈U. Let dλn=dλωn,ωn+1 for all λ∈U and for each n∈ℕ. Using F3, there is η∈0,1 so that(21)limn⟶∞dλnηFdλn=0,for allλ∈U.
From (20), we have(22)dλnηFdλn−dλnηFdλ0≤−dλnηnτ≤0,for eachn∈ℕandλ∈U.
Lettingn⟶∞ in (22), we get(23)limn⟶∞ndλnη=0,for allλ∈U.
This implies that there isn1∈ℕ with ndλnη≤1 for each n≥n1 and for all λ∈U. Thus, we have(24)dλn≤1n1/n,for eachn≥n1andλ∈U.
To ensure thatωn is a Cauchy o-sequence, take m,n∈ℕ with m>n>n1. Using (24) and triangular inequality, we have(25)dλωn,ωm≤dλωn,ωn+1+dλωn+1,ωn+2+⋯+dλωm−1,ωm=∑i=nm−1dλi≤∑i=n∞dλi≤∑i=n∞1i/1η,for allλ∈U.
The series∑i=n∞1/i1/n is convergent, so limn,m⟶∞dλωn,ωm=0 for all λ∈U. It yields that ωn is a Cauchy o-sequence. Since X is o-complete, there is ω∗∈X so that ωn⟶É∗ as n⟶∞. By (iv), we have αωn,ω∗≥1 and ωn⊥ω∗ for each n∈ℕ. We now claim that dλω∗,Tω∗=0 for all λ∈U. On contrary, suppose that there is λ0∈U with dλ0ω∗,Tω∗>0, then there exists n0∈ℕ such that dλ0ωn+1,Tω∗>0 for each n≥n0. Now, note that we have ωn,ω∗∈X with αωn,ω∗≥1, ωn⊥ω∗, and dλ0Tωn,Tω∗>0 for all n≥n0. Then, from (8), we get(26)τ+Fdλ0Tωn,Tω∗≤Faλ0dλ0ωn,ω∗+bλ0dλ0ωn,Tωn+cλ0dλ0ω∗,Tω∗+eλ0dλ0ωn,Tω∗+Lλ0dλ0ω∗,Tωn,for alln≥n0.
This implies that(27)dλ0Tωn,Tω∗<aλ0dλ0ωn,ω∗+bλ0dλ0ωn,Tωn+cλ0dλ0ω∗,Tω∗+eλ0dλ0ωn,Tω∗+Lλ0dλ0ω∗,Tωn,for alln≥n0.
Thus, by considering the triangular property and (27), we have for each n≥n0(28)dλ0ω∗,Tω∗≤dλ0ω∗,ωn+1+dλ0ωn+1,Tω∗=dλ0ω∗,ωn+1+dλ0Tωn,Tω∗<dλ0ω∗,ωn+1+aλ0dλ0ωn,ω∗+bλ0dλ0ωn,ωn+1+cλ0dλ0ω∗,Tω∗+eλ0dλ0ωn,Tω∗+Lλ0dλ0ω∗,ωn+1.
At the limitn⟶∞, one obtains(29)dλ0ω∗,Tω∗≤cλ0+eλ0dλ0ω∗,Tω∗<dλ0ω∗,Tω∗.
It is a contradiction, sodλω∗,Tω∗=0 for all λ∈U. As X is separating, we obtain ω∗=Tω∗.Example 4.
LetX=D0,100,ℝ be the collection of all twice differentiable real-valued functions. Take the metric(30)dnx,y=maxt∈0,nxt−ytfor each n∈1,2,3,…,100. For x,y∈X, consider x⊥y⇔xy=0. Let T:X⟶X be defined by(31)Txt=xt3+d2xtdt2,xt≥0∀t,dxtdt,otherwise.
Givenα:X×X⟶0,∞ as(32)αx,y=1,x,yboth are constant or linear functions,0,otherwise.
Suppose thatx,y∈X with x⊥y and αx,y=1. Then x and y are both constant functions with at least one of them is the zero function. Say x=0 and y≠0. Then, we have the following two cases:Case 1: forx=0 and y>0, we have Tx=0 and Ty=y/3. Thus, dnTx,Ty=y/3 and dnx,y=y for each n∈1,2,3,…,100.Case 2: forx=0 and y<0, we have Tx=0 and Ty=0. Thus, dnTx,Ty=0 and dnx,y=−y for each n∈1,2,3,…,100.
Takeaλ=1 and bλ=cλ=eλ=Lλ=0 for all λ∈1,2,3,…,100. Then(33)αx,y≥1⇒dλTx,Ty≤aλ3dλx,y≤aλexp1dλx,y.
By takingτ=1 and Ft=lnt, one can conclude that (8) holds for all x,y∈X with x⊥y and for each λ∈U=1,2,3,…,100.
Also, forx⊥y, we have xtyt=0 for each t, then at least one of them is the zero function, and hence, TxtTyt=0 for each t, that is, Tx⊥Ty. Thus, T is ⊥-preserving. Further, for w0=−1, Tw0=0, and so w0⊥Tw0 and αw0,Tw0=1. Furthermore, for all x,y∈X with x⊥y and αx,y=1, we know that x and y are constant functions, then Tx and Ty are also constant functions, and so αTx,Ty=1.
Moreover, for each o-sequenceρn in X with αρn,ρn+1=1 for each n≥1 and ρn⟶ρasn⟶∞, we have ρ=0. Therefore, αρn,ρ≥1 and ρn⊥ρ for each n≥1.
Consequently, all the conditions in Theorem3 are verified. One can conclude that T possesses a fixed point.Remark 1.
For the functions defined in the above example, note that (8) does not hold for every x,y∈X. It suffices to take xt=−5t and yt=−15t, then Txt=−5 and Tyt=−15. Also note that d1x,y=10 and d1Tx,Ty=10.Theorem 4.
LetX be a nonempty o-set endowed with a separating o-complete gauge structure dλ:λ∈U of o-pseudometrics. Let T:X⟶X be a self-mapping with F∈ℱ and τ>0 so that(34)αx,y≥1⇒τ+FdλTx,Ty≤Faλmaxdλx,y,dλx,Tx,dλy,Ty,dλx,Ty+dλy,Tx2+bλdλy,Tx,for all x,y∈X with x⊥y and for each λ∈U, whenever dλTx,Ty≠0 for λ∈U, where aλ,bλ are positive real numbers with aλ+bλ=1 for all λ∈U. Further, assume that(i)
T is ⊥-preserving(ii)
There isω0∈X with ω0⊥Tω0 and αω0,Tω0≥1(iii)
For eachx,y∈X with x⊥y and αx,y≥1, we have αTx,Ty≥1(iv)
For any o-sequenceρn in X so that αρn,ρn+1≥1 for each n∈ℕ and ρn⟶ρasn⟶∞, we have αρn,ρ≥1 and ρn⊥ρ for each n∈ℕ
Then,T admits a fixed point.Proof.
Using (ii), there isω0∈X with ω0⊥Tω0 and αω0,Tω0≥1, and by considering (iii), we get αTω0,T2ω0≥1. Moreover, we have Tω0⊥T2ω0, since T is ⊥-preserving. Repetition of the same arguments implies that αTnω0,Tn+1ω0≥1 and Tnω0⊥Tn+1ω0 for each n≥1. Consider ωn=Tnω0. Then ωn is an o-sequence with αωn,ωn+1≥1 for each n≥0. Also, note that if there is some m0≥1 such that ωm0=ωm0+1, then ωm0 is a fixed point of T. Thus, we assume that such a natural number does not exist. As ω0∈X with ω0⊥ω1 and αω0,ω1≥1, then from (34), we get(35)τ+Fdλω1,ω2=τ+FdλTω0,Tω1≤Faλmaxdλω0,ω1,dλω0,Tω0,dλω1,Tω1,dλω1,Tω0+dλω0,Tω12+bλdλω1,Tω0<Fmaxdλω0,ω1,dλω1,ω2,for allλ∈U.
If we assume thatmaxdλω0,ω1,dλω1,ω2=dλω1,ω2, then we have a contradiction with respect to (35). Thus, maxdλω0,ω1,dλω1,ω2=dλω0,ω1 for all λ∈U. Using (35), we have(36)τ+Fdλω1,ω2<Fdλω0,ω1,for allλ∈U.
Again, we know thatω1⊥ω2 and αω1,ω2≥1; then, from (34), we have(37)τ+Fdλω2,ω3=τ+FdλTω1,Tω2≤Faλmaxdλω1,ω2,dλω1,Tω1,dλω2,Tω2,dλω2,Tω1+dλω1,Tω22+bλdλω2,Tω1<Fmaxdλω1,ω2,dλω2,ω3,for allλ∈U.
If we assume thatmaxdλω1,ω2,dλω2,ω3=dλω2,ω3, then we have a contradiction to (37). Thus, maxdλω1,ω2,dλω2,ω3=dλω1,ω2∀λ∈U. Thus, from (37), we have(38)τ+Fdλω2,ω3<Fdλω1,ω2,for allλ∈U.
From (36) and (38), we have(39)Fdλω2,ω3<Fdλω0,ω1−2τ,for allλ∈U.
Working with the same steps, we obtain(40)Fdλωn,ωn+1<Fdλω0,ω1−nτ,for eachn∈ℕfor allλ∈U.
Lettingn⟶∞ in (40), we get limn⟶∞Fdλωn,ωn+1=−∞ for all λ∈U. Thus, by property F2, we have limn⟶∞dλωn,ωn+1=0. Let dλn=dλωn,ωn+1 for all λ∈U and for each n∈ℕ. From F3, there is η∈0,1 so that(41)limn⟶∞dλnηFdλn=0,for allλ∈U.
From (40), we have(42)dλnηFdλn−dλnηFdλ0<−dλnηnτ≤0for eachn∈ℕand for allλ∈U.
Lettingn⟶∞ in (42), we get(43)limn⟶∞ndλnη=0,∀λ∈U.
This implies that there isn1∈ℕ so that ndλnη≤1 for each n≥n1 and for all λ∈U. Thus, we have(44)dλn≤1n1/η,for eachn≥n1and for allλ∈U.
We claim thatωn is a Cauchy o-sequence. Take the integers m,n with m>n>n1. Using (44) and the triangular inequality, one writes(45)dλωn,ωm≤dλωn,ωn+1+dλωn+1,ωn+2+⋯+dλωm−1,ωm=∑i=nm−1dλi≤∑i=n∞dλi≤∑i=n∞1i1/η,for allλ∈U.
The series∑i=n∞1/i1/η converges, so limn,m⟶∞dλωn,ωm=0 for all λ∈U. That is, ωn is a Cauchy o-sequence. Since X is o-complete, there is ω∗∈X so that ωn⟶ω∗ as n⟶∞. By (iv), we have αωn,ω∗≥1 and ωn⊥ω∗ for each n∈ℕ. We now claim that dλω∗,Tω∗=0 for all λ∈U. On contrary, suppose that there is λ0∈U with dλ0ω∗,Tω∗>0. Then, there exists n0∈ℕ such that dλ0ωn+1,Tω∗>0 for each n≥n0. Now, note that ωn,ω∗∈X with αωn,ω∗≥1, ωn⊥ω∗ and dλ0Tωn,Tω∗>0 for all n≥n0. Then, from (34), we get(46)τ+Fdλ0Tωn,Tω∗≤Faλ0maxdλ0ωn,ω∗,dλ0ωn,Tωn,dλ0ω∗,Tω∗,dλ0ω∗,Tωn+dλ0ωn,Tω∗2+bλ0dλ0ω∗,Tωn.
This implies that(47)dλ0Tωn,Tω∗<aλ0maxdλ0ωn,ω∗,dλ0ωn,Tωn,dλ0ω∗,Tω∗,dλ0ω∗,Tωn+dλ0ωn,Tω∗2+bλ0dλ0ω∗,Tωn.
Thus, for eachn≥n0, by considering the triangular property and (47), we have(48)dλ0ω∗,Tω∗≤dλ0ω∗,ωn+1+dλ0ωn+1,Tω∗=dλ0ω∗,ωn+1+dλ0Tωn,Tω∗<dλ0ω∗,ωn+1+aλ0maxdλ0ωn,ω∗,dλ0ωn,Tωn,dλ0ω∗,Tω∗,dλ0ω∗,Tωn+dλ0ωn,Tω∗2+bλ0dλ0ω∗,Tωn≤dλ0ω∗,ωn+1+aλ0maxdλ0ωn,ω∗,dλ0ωn,Tωn,dλ0ω∗,Tω∗,dλ0ω∗,ωn+dλ0ωn,Tωn+dλ0ωn,ω∗+dλ0ω∗,Tω∗2+bλ0dλ0ω∗,Tωn≤dλ0ω∗,ωn+1+aλ0dλ0ωn,ω∗+dλ0ωn,Tωn+dλ0ω∗,Tω∗+bλ0dλ0ω∗,Tωn.
Lettingn⟶∞ in the above inequality, we have(49)dλ0ω∗,Tω∗≤aλ0dλ0ω∗,Tω∗<dλ0ω∗,Tω∗.
It is a contradiction, sodλω∗,Tω∗=0 for all λ∈U. As X is separating, hence we obtain ω∗=Tω∗.
In the following corollaries, we assume thatX is a nonempty o-set with an orthogonal element (say ωa) and endowed with a separating o-complete gauge structure dλ:λ∈U. Further, assume that a directed graph G=V,E is defined on X so that the set of its vertices V coincides with X (i.e., V=X) and the set of edges E is so that Δ⊂E, where Δ=σ,σ:σ∈X. Moreover, G is supposed that it has no parallel edges.
The following corollaries can be obtained from our results by definingα:X×X⟶0,∞ as(50)αx,y=1,ifx,y∈E,0,otherwise.Corollary 1.
LetT:X⟶X be a mapping with F∈ℱ and τ>0 such that(51)x,y∈E⇒τ+FdλTx,Ty≤Faλdλx,y+bλdλx,Tx+cλdλy,Ty+eλdλx,Ty+Lλdλy,Tx,for each x,y∈X with x⊥y and for each λ∈U, whenever dλTx,Ty≠0 for λ∈U, where aλ,bλ,cλ,eλ,Lλ≥0 and aλ+bλ+cλ+2eλ=1 for all λ∈U. Further, assume that(i)
T is ⊥-preserving(ii)
There is an elementω0∈X with ω0⊥Tω0 and ω0,Tω0∈E(iii)
For eachx,y∈X with x⊥y and x,y∈E, we have Tx,Ty∈E(iv)
For any o-sequenceωn in X with ωn,ωn+1∈E for each n∈ℕ and ωn⟶xasn⟶∞, we have ωn,x∈E and ωn⊥x for each n∈ℕ
Then,T admits a fixed point.Corollary 2.
LetT:X⟶X be a mapping with F∈ℱ and τ>0 such that(52)x,y∈E⇒τ+FdλTx,Ty≤Faλmaxdλx,y,dλx,Tx,dλy,Ty,dλx,Ty+dλy,Tx2+bλdλy,Tx,for each x,y∈X with x⊥y and for each λ∈U, whenever dλTx,Ty≠0 for λ∈U, where aλ,bλ are positive real numbers with aλ+bλ=1 for all λ∈U. Further, assume that(i)
T is ⊥-preserving(ii)
There exists an elementω0∈X with ω0⊥Tω0 and ω0,Tω0∈E(iii)
For eachx,y∈X with x⊥y and x,y∈E, we have Tx,Ty∈E(iv)
For any o-sequenceωn in X such that ωn,ωn+1∈E for each n∈ℕ and ωn⟶xasn⟶∞, we have ωn,x∈E and ωn⊥x for each n∈ℕ
Then,T admits a fixed point.


## 5. Application to Integral Equations

Consider the following Volterra-type integral equation:(53)xt=gt+∫0ftKt,s,xsds,t∈I=0,∞,where(i)
g:I⟶1,∞ is continuous(ii)
f:I⟶0,∞ is continuous(iii)
K:I×I×ℝ⟶ℝ+=0,∞ is a continuous functionLetX=C0,∞,ℝ+ be the space of all real-valued continuous functions from 0,∞ into 0,∞. We can define orthogonality relation on X by(54)x⊥y⇔xtyt≥ytorxtyt≥xt.Define the family of pseudometrics asdnx,y=maxt∈0,nxt−yte−τt, for each n∈ℕ, where τ is a positive real number. Clearly, ℱ=dn:n∈ℕ defines a gauge structure on X, which is separating and o-complete.Theorem 6.
TakeX=C0,∞,ℝ+. Define the operator T:X⟶X by(55)Txt=gt+∫0ftKt,s,xsds,t∈I=0,∞,where g:I⟶1,∞, f:I⟶0,∞, and K:I×I×ℝ⟶ℝ+=0,∞ are continuous functions. Also, assume that there are τ>0 and γ:X⟶0,∞ so that for all x,y∈X with xtyt≥ytorxtyt≥xt and t,s∈0,n, we have(56)Kt,s,x−Kt,s,y≤e−τγx+ydnx,yfor eachn∈ℕ.
Moreover,(57)∫0ft1γxs+ysds≤eτt,for each t∈I. Then, (53) admits at least one solution.Proof.
Forx∈X, take(58)Txt=gt+∫0ftKt,s,xsds≥1,for every t∈I=0,∞. Note that TxtTyt≥Tyt for every t∈I. Hence, we say that if x⊥y, then Tx⊥Ty. Also, note that for each ω0∈X, we have ω0tTω0t≥ω0t for every t∈I, that is, ω0⊥Tω0.
Now, for allx,y∈X with xtyt≥ytorxtyt≥xt and t∈0,n for each n≥1, we have(59)Txt−Tyt≤∫0ftKt,s,xs−Kt,s,ysds≤∫0fte−τγxs+ysdnx,yds=e−τdnx,y∫0ft1γxs+ysds≤eτte−τdnx,y.
Thus, we have(60)Txt−Tyte−τt≤e−τdnx,y.
This implies that(61)dnTx,Ty≤e−τdnx,y.
One writes(62)lndnTx,Ty≤lne−τdnx,y.
That is,(63)τ+lndnTx,Ty≤lndnx,yfor eachn∈ℕ.
Therefore, it can be concluded that Theorem4 is applied for the operator T with the choice of ας,υ=1 for all ς,υ∈X, aλ=1 and bλ=cλ=eλ=Lλ=0 for each λ∈ℕ and Fx=lnx. Hence, T possesses a fixed point, i.e., (53) admits at least one solution.
Take(64)xt=gt+∫abKt,s,xsds,t∈I=0,∞,where(i)
g:I⟶ℝ is continuous(ii)
K:I×I×ℝ⟶ℝ is continuous
The above equation is a Fredholm-type integral equation.Theorem 7.
LetX=C0,∞,ℝ and let the operator T:X⟶X be defined by(65)Txt=gt+∫abKt,s,xsds,t∈I=0,∞,where g:I⟶ and K:I×I×ℝ⟶ℝ are continuous functions (a<b). Assume that(i)
If(66)xtyt≥ytorxtyt≥xt,then we have(67)TxtTyt≥TytorTxtTyt≥Txt.(ii)
There areτ>0 and γ:X⟶0,∞ so that for each x,y∈X with xtyt≥ytorxtyt≥xt and t,s∈0,n, we have(68)Kt,s,x−Kt,s,y≤e−τγx+ydnx,yfor eachn∈ℕ.Moreover,(69)∫ab1γxs+ysds≤1.(iii)
There isω0∈X so that ω0tTω0t≥Tω0torω0tTω0t≥ω0t.(iv)
For any sequenceωn in X with ωnωn+1≥ωn or ωnωn+1≥ωn+1 for each n∈ℕ and ωn⟶x, we have ωnx≥ωn or ωnx≥x.
Then, (64) admits at least one solution.
LetX=C0,∞,ℝ be the set of all real-valued continuous functions. Again, define orthogonality relation on X by(70)x⊥y⇔xtyt≥ytorxtyt≥xtand family of pseudometrics given as dnx,y=maxt∈0,nxt−yte−τt, for each n∈ℕ, where τ is a positive real number. Note that ℱ=dn:n∈ℕ defines a gauge structure on X, which is separating and o-complete, so the conclusion of this theorem can be obtained from Theorem 4 by taking ας,υ=1 for all ς,υ∈X, aλ=1 and bλ=cλ=eλ=Lλ=0 for each λ∈ℕ and Fx=lnx.



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*Source: 8387262-2021-01-28.xml*