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http://www.aimspress.com/journal/Math

AIMS Mathematics, 5(4): 3125–3137.

DOI:10.3934/math.2020201

Received: 24 October 2019

Accepted: 10 March 2020

Published: 24 March 2020

Research article

Fixed point theorems in R−metric spaces with applications

Siamak Khalehoghli1, Hamidreza Rahimi1,∗and Madjid Eshaghi Gordji2

1Department of Mathematics, Central Tehran Branch, Islamic Azad University, Tehran, Iran

2Department of Mathematics, Semnan University, Semnan, Iran

*Correspondence: Email: rahimi@iauctb.ac.ir.

Abstract: The purpose of this paper is to introduce the notion of R-metric spaces and give a real

generalization of Banach ﬁxed point theorem. Also, we give some conditions to construct the Brouwer

ﬁxed point. As an application, we ﬁnd the existence of solution for a fractional integral equation.

Keywords: R-metric spaces; ﬁxed point; strong R-compact metric spaces; fractional integral

equations

Mathematics Subject Classiﬁcation: 54H25, 47H10

1. Introduction

Fixed point theory is a powerful and important tool in the study of nonlinear phenomena. It is an

interdisciplinary subject which can be applied in several areas of mathematics and other ﬁelds, like

game theory, mathematical economics, optimization theory, approximation theory, variational

inequality in [25], biology, chemistry, engineering, physics and etc. In 1886, Poincare was the ﬁrst to

work in this ﬁeld.Then Brouwer in [7] in 1912, proved ﬁxed point theorem for the f(x)=x. He also

proved ﬁxed point theorem for a square, a sphere and their n-dimentional counterparts which was

further extended by Kakutani in [17]. In 1922, Banach in [6] proved that a contraction mapping which

its domain is complete posses a unique ﬁxed point. The ﬁxed point theory as well as Banach

contraction principle has been studied and generalized in diﬀerent spaces for example: In 1969,

Nadler in [22] extended the Banach’s principle to set valued mappings in complete metric spaces. In

1990 ﬁxed point theory in modular function spaces was initiated by Khamsi, Kozlowski and Reich

in [18]. Modular metric spaces were introduced in [8, 9]. Fixed point theory in modular metric spaces

was studied by Abdou and Khamsi in [3]. In 2007, Huang and Zhang in [15] introduced cone metric

spaces which are generalizations of metric spaces and they extended Banach’s contraction principle to

such spaces, whereafter many authors (for examples in [1, 2, 4, 10, 12, 16, 34] and references therein)

studied ﬁxed point theorems in cone metric spaces. Moreover, in the case when the underlying cone is

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normal led Khamsi in [19] to introduce a new type of spaces which he called metric-type spaces,

satisfying basic properties of the associated space. Some ﬁxed point results were obtained in

metric-type spaces in the papers in [19]. The readers who are interested in hyperbolic type metrics

deﬁned on planar and multidimensional domains refer to [33]. In 2012, the notion of ordered spaces

and normed ordered spaces were introduced by Al-Rawashdeh et al. in [31] and get a ﬁxed point

theorem. The authors for example in [21, 26, 27] considered ﬁxed point theorems in E-metric spaces.

In 2017, M. Eshaghi et al. in [14], introduced a new class of generalization metric spaces which are

called orthogonal metric spaces. Subsequently they and many authors (for examples, in [5, 13, 28, 29])

gave an extension of Banach ﬁxed point theorem. It is still going on. At the end it is advised to those

who want to research the theory of ﬁxed point read the valuable references mentioned in [11,20].

2. R-metric spaces

Banach had been proved the following theorem in complete metric space X.

Theorem 2.1. Let (X,d)be a complete metric space and f :X→X be a mapping such that, for some

λ∈(0,1),

d(f(x),f(y)) ≤λd(x,y)

for all x,y∈X. Then f has a unique ﬁxed point in X.

Ran et al. in [30] gave an extension of Banach ﬁxed point Theorem by weakened the contractivity

condition on elements that are comparable in the partial order:

Theorem 2.2. Let T be a partially ordered set such that every pair x,y∈T has a lower bound and an

upper bound. Furthermore, let d be a metric on T such that (T,d)is a complete metric space. If F is a

continuous, monotone (i.e., either order-preserving or order-reversing) map from T into T such that

∃0<c<1: d(F(x),F(y)) ≤cd(x,y),∀x≥y,

∃x0∈T : x0≤F(x0)or x0≥F(x0),

then F has a unique ﬁxed point x. Moreover, for every x ∈T , lim

n→∞ fn(x)=x.

Afterward Nieto et al. in [24] presented a new extension of Banach contractive mapping to partially

ordered sets, where some valuable applications and examples are given:

Theorem 2.3. Let (X,≤)be a partially ordered set and suppose that there exists a metric d in X such

that (X,d)is a complete metric space. Let f :X→X be a continuous and nondecreasing mapping

such that there exists k ∈[0,1) with

d(f(x),f(y)) ≤kd(x,y),∀x≥y.

If there exists x0∈X with x0≤f(x0), then f has a ﬁxed point.

Recently Eshaghi et al. in [14] Proved Banach contraction principle in orthogonal metric spaces

(X,d,⊥), where ⊥is a relation on X.

In this paper among other things, we try to present a real generalization of the mentioned Banach’s

contraction principle by introducing R-metric spaces, where Ris an arbitrary relation on X.We note

that in especial case Rcan be considered as R:=≤[24, 30], R:=⊥[14], or etc. Furthermore, in

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Brouwer and Kakutani theorems the existence of ﬁxed point is stated [17], and the Nash equilibrium

is proved only based on the existence of ﬁxed point [23]. Obtaining the exact value of the equilibrium

point is usually diﬃcult and only approximate value is obtained. If one can ﬁnd a suitable replacement

of Brouwer theorem which is determine the value of ﬁxed point then many problems on game theory

and economics can be solved. In this paper, we try to provide a structural method for ﬁnding a value

of ﬁxed point.

Deﬁnition 2.1. Suppose (X,d) is a metric space and Ris a relation on X. Then the triple (X,d,R) or in

brief Xis called R-metric space.

Example 2.1. Let (R,| |) be given and let R:=≤,R:=≥or R:=⊥, ..., then with each R, (R,| |,R) is

an R-metric space.

Example 2.2. Let X=[0,∞) equipped with Euclidean metric. Deﬁne xRy if xy ≤(x∨y) where

x∨y=xor y. Then (X,d,R) is an R-metric space.

We note that for a given speciﬁed metric space (X,d) and any relation Ron Xwe can consider an

R-metric space (X,d,R) .

Deﬁnition 2.2. A sequence {xn}in an R-metric space Xis called an R-sequence if xnRxn+kfor each

n,k∈N.

Deﬁnition 2.3. An R-sequence {xn}is said to converge to x∈Xif for every ε > 0 there is an integer N

such that d(xn,x)< ε if n≥N.

In this case, we write xn

R

−→ x.

Example 2.3. Suppose X=[1,2] with Euclidean metric be given. Let R:=≥and xn=1+1

nfor each

n∈N. Clearly {xn}is an R-sequence and xn

R

−→ 1. Note that xn=2−1

nis not an R-sequence.

Example 2.4. Let X=R. Consider P(R) with metric dist(A,B)=in f {|a−b|:a∈A and b ∈B}, let

R:=⊆.Deﬁne An=[1 +1

n,4−1

n], clearly {An}is an R-sequence and An

R

−→ (1,4).

Remark 2.1. Every subsequence of an R-sequence is an R-sequence too.

In the followings it is supposed that Xis an R-metric space.

Deﬁnition 2.4. Let E⊆X.x∈Xis called an R-limit point of Eif there exists an R-sequence {xn}in E

such that xn,xfor all n∈Nand xn

R

−→ x.

Deﬁnition 2.5. The set of all R-limit points of Eis denoted by E0R.

Deﬁnition 2.6. E⊆Xis called R-closed, if E0R⊆E.

Deﬁnition 2.7. E⊆Xis called R-open, if ECis R-closed.

Theorem 2.4. E⊆X is R-open if and only if for any x ∈E and any R-sequence {xn}which xn

R

−→ x,

there exists N ∈Nsuch that xn∈E for all n ≥N.

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Proof. Let x∈E. Suppose there exists an R-sequence {xn}which xn

R

−→ xbut for every N∈N, there

exists a natural number n≥Nsuch that xn<E. Hence, we obtain an R-subsequence {xnN}of {xn},

which xnN

R

−→ x, and xnN<E. Therefore x∈EC. This contradicts x∈E.

Conversely, if x∈(EC)0R, then x<E. Hence (EC)0R⊂EC, so ECis R-closed. Thus Eis R-open.

Deﬁnition 2.8. Let E⊆X. The R-closure of Eis the set ER=E∪E0R.

Theorem 2.5. If E ⊂X, then

(a) ERis R-closed.

(b) E =ERif and only if E is R-closed.

Proof. (a) If x∈(ER)C, then xis neither a point of Enor an R-limit point of E. Let {xn}be an R-

sequence converging to x, then there exists N∈Nsuch that xn∈(ER)Cfor n≥N. Thus (ER)Cis

R-open so that ERis R-closed.

(b) If E=ER, (a) implies that Eis R-close. If Eis R-closed, then E0R⊂E[by deﬁnitions 2.6 and 2.8],

hence ER=E.

Example 2.5. Suppose X=Requipped with standard topology. Let R:=≤, let E=(0,1], then

ER=(0,1]. Hence Eis R-closed but it is not closed.

Theorem 2.6. The set {G⊆X|G is R-open}is a topology on X which is called an R-topology and is

denoted by τR

Proof. Trivially ϕand Xare R-open. Let {Uj}j∈Jbe a family of R-open sets. Put U=Sj∈JUj, let xbe

an R-limit point of UC=Tj∈JUC

j, hence there exists an R-sequence {xn}in UC\ {x}such that xn

R

−→ x.

Therefore for each j∈J,xis an R-limit point of UC

j. Since UC

jis R-closed x∈UC

j, so that x∈UC,

hence Uis R-open.

Let U1,...,Unbe R-open sets. Put W=Tn

j=1Ujand let xbe an R-limit point of WC=Sn

j=1UC

j,

hence there exists an R-sequence {xn}in WC\ {x}such that xn

R

−→ x. It is easy to show that there exists

1≤j≤nand a subsequence {xnk}of {xn}in UC

j\ {x}which xnk

R

−→ x. Since UC

jis R-closed x∈UC

j,

thus x∈WC. It follows that Wis R-open.

Corollary 2.1. Let X =Requipped with standard topology. Let R :=≤, then:

(a) The open intervals (a,b)(a =−∞ or b = +∞) form a basis for the standard topology on R,

(b) The intervals (a,b)(a =−∞ or b = +∞) and (a,b]form a basis for topology τR.

Theorem 2.7. Let τdbe a metric topology on X, then τd⊆τR.

Proof. Let G∈τd. Let xbe an R-limit point of GC. There exists an R-sequence {xn}in GC\ {x}such

that xn

R

−→ x. Since each R-sequence is a sequence hence xis a limit point of GC.GCis closed, thus

x∈GC. Therefore G∈τR.

Lemma 2.1. Let R :=X×X, then τR=τd.

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Proof. By theorem 2.7, τd⊆τR. Let Gis an R-open set and xis a limit point of GC. There exists a

sequence {xn}in GC\ {x}such that xn→x. Clearly {xn}is an R-sequence, hence xn

R

−→ x. It follows

that x∈GC. Thus τR⊆τd.

Remark 2.2. Suppose (X,d)be a metric space and R :=X×X,then R-metric space (X,d,R)is

equivalent to metric space (X,d).

Deﬁnition 2.9. Let Rbe a relation on Rk.E⊆Rkis called R-convex if λx+(1 −λ)y∈E, whenever

x∈E,y∈E,xRy, and 0 <λ<1.

Lemma 2.2. Let E is a convex set, then E is R-convex.

Proof. Let x∈E,y∈E,xRy, and 0 <λ<1. By the deﬁnition of convex set λx+(1 −λ)y∈E.

The converse is not true.

Example 2.6. Let E=(0,1] ∪(2,4]. Deﬁne xRy if x,y∈(0,1] or x,y∈(2,4]. Clearly Eis R-convex

but it is not convex.

Deﬁnition 2.10. K⊆Xis called R-compact if every R-sequence {xn}in Khas a convergent

subsequence in K.

Example 2.7. Suppose the Euclidean metric space X=Rbe given. Let K=[0,1) and let R:=≥. Then

Kis R-compact.

Lemma 2.3. Suppose K ⊆X is compact, then K is R-compact.

Proof. Let {xn}be an R-sequence in K. It is clear that {xn}is a sequence too. So by theorem 3.6 (a) [32],

there exists a convergent subsequence {xnk}of {xn}in K. The theorem follows.

The converse is not true.

Example 2.8. Suppose the Euclidean metric space X=Rbe given. Let k=[0,1), and let Rdeﬁned

on Xby

xRy ⇐⇒

x≤y≤1

3

or

x=0i f y >1

3

.

Since Kis not closed so it is not compact. Let {xn}be an R-sequence in K. Then:

i) For all n∈N,xn=0, hence {xn}converges to 0.

ii) {xn}is increasing and bounded above to 1

3, therefore it is convergent.

It follows that Kis R-compact.

Example 2.9. Suppose X=Requipped with the Euclidean metric. Let K=(0,1]. Let R:=≤, and let

{xn}be an R-sequence in K. It is clear that {xn}is increasing and bounded above, hence it is convergent.

Therefore Kis R-compact but it is not compact.

Theorem 2.8. R-compact subsets of R-metric spaces are R-closed.

Proof. Let Kbe an R-compact subset of X. Let x∈K0R. Then there exists an R-sequence {xn}in K

such that xn

R

−→ x. By deﬁnition 2.10, {xn}has a convergent subsequence {xnk}in K, hence xnk

R

−→ x, so

that x∈K. The theorem follows.

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Theorem 2.9. An R-closed subset of an R-compact set, is R-compact.

Proof. Suppose F⊆K⊆X.Fis R-closed (relative to X), and Kis R-compact. Let {xn}be an R-

sequence in F, hence {xn}is an R-sequence in Ktoo. Since Kis R-compact, there exists a subsequence

{xnk}of {xn}such that xnk

R

−→ x. Since Fis R-closed x∈F.The theorem follows.

Corollary 2.2. Let F be R-closed and K be R-compact. Then F ∩K is R-compact.

Proof. By theorems 2.6 and 2.8, F∩Kis R-closed; since F∩K⊂K, theorem 2.9 shows that F∩Kis

R-compact.

Deﬁnition 2.11. A set K⊆Xis called strong R-compact, if each R-sequence {xn}in K, that has a

subsequence {xnk}which converges to x∗∈K, i.e, xnk

R

−→ x∗, then xnk+1

R

−→ x∗.

Example 2.10. Suppose X=Rwith standard topology be given, let R:=≤and let K=(0,10].

Suppose {xn}is an R-sequence, then each subsequence {xnk}of {xn}is increasing and bounded above,

so there exists x∗∈K, such that xnk

R

−→ x∗and xnk+1

R

−→ x∗. Therefore Kis a strong R-compact set.

Deﬁnition 2.12. An R-sequence {xn}in Xis said to be an R-Cauchy sequence, if for every ε > 0 there

exists an integer Nsuch that d(xn,xm)< ε if n≥Nand m≥N. It is clear that xnRxmor xmRxn.

Lemma 2.4. (a) Every convergent R-sequence in X is an R-Cauchy sequence.

(b) Suppose K be an R-compact set and {xn}be a R-Cauchy sequence in K. Then {xn}converges to

x∈K.

Deﬁnition 2.13. Xis said to be R-complete if every R-Cauchy sequence in Xconverges to a point in

X.

Corollary 2.3. Every R-compact space is R-complete, but the converse is not true.

Example 2.11. Suppose X=Rwith standard topology be given and let R:=≤. It is easy to show that

Xis R-complete but it is not R-compact. Since the R-sequence {n}has no convergent subsequence.

Deﬁnition 2.14. Let f:X→Xbe a mapping. fis said to be R-continuous at x∈Xif for every

R-sequence {xn}in Xwith xn

R

−→ x, we have f(xn)−→ f(x). Also, fis said to be R-continuous on Xif

fis R-continuous in each x∈X.

Lemma 2.5. Every continuous mapping f :X→X, is R-continuous.

Proof. Since each R-sequence is a sequence.

The converse is not true.

Example 2.12. Suppose X=[0,1] equipped with standard topology and let f:X→Xbe a Dirichlet

mapping, i.e,

f(x)=(1i f x ∈Q∩[0,1]

0i f x ∈QC∩[0,1] .

Let Rbe an equality relation on X,then fis discontinuous at each point of Xbut fis R-continuous on

X.

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Example 2.13. Suppose the Euclidean metric space X=Rbe given. Deﬁne xRy if x,y∈(n+2

3,n+4

5)

for some n∈Zor x=0. Deﬁne f:X→Xby f(x)=[x]. Let x∈Xand {xn}be an arbitrary

R-sequence in Xsuch that converges to x, then the following cases are satisﬁed:

Case 1: If xn=0 for all n, then x=0, and f(xn)=0=f(x).

Case 2: If xn,0 for some n, then there exists m∈Zsuch that x∈[m+2

3,m+4

5], and f(xn)=m=

f(x).

Therefore fis R-continuous on X, but it is not continuous on X.

Deﬁnition 2.15. A mapping f:X→Xis said to be an R-contraction with Lipschutz constant 0 <λ<

1 if for all x,y∈Xsuch that xRy, we have

d(f(x),f(y)) ≤λd(x,y).

It is easy to show that every contraction is an R-contraction but the converse is not true. See the next

example.

Example 2.14. Let X=[0,0.99), and Xequipped with Euclidean metric. Let xRy if xy ∈ {x,y}for all

x,y∈X. Let f:X→Xbe a mapping deﬁned by

f(x)=(x2i f x ∈Q∩X

0i f x ∈QC∩X.

Suppose x=0.9, y=√2

2and 0 <λ<1, then

|f(x)−f(y)|=0.81 λ|0.9−√2

2|.

Hence fis not a contraction. Now, let xRy, therefore x=0 or y=0. Suppose x=0, thus

|f(x)−f(y)|=(y2i f y ∈Q∩X

0i f y ∈QC∩X.

Hence by choosing λ=0.99 it follows that

|f(x)−f(y)| ≤ y2≤λ|0−y|=λy.

So that fis R-contraction.

Deﬁnition 2.16. Let f:X→Xbe a mapping fis called R-preserving if xRy, then f(x)R f (y) for all

x,y∈X.

Example 2.15. Suppose X=Rwith standard topology be given and let R:=≥. Let f:X→Xbe a

mapping deﬁned by f(x)=x3. Let x1≥x2, then f(x1)=x3

1≥x3

2=f(x2). Hence fis R-preserving.

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3. The main results

In this section, it is proved two main theorems. The ﬁrst one is the real extension of one of the most

important theorem in mathematics which is named Banach contraction principle theorem 2.1, and the

second one is the version of Brouwer ﬁxed point theorem.

Theorem 3.1. Let X be an R-complete metric space (not necessarily complete metric space) and 0<

λ < 1. Let f :X→X be R-continuous, R-contraction with Lipschutz constant λand R-preserving.

Suppose there exists x0∈X such that x0Ry for all y ∈f(X). Then f has a unique ﬁxed point x∗. Also,

f is a Picard operator, that is, lim

n→∞ fn(x)=x∗for all x ∈X.

Proof. Let x1=f(x0), x2=f(x1)=f2(x0),..., xn=f(xn−1)=fn(x0),...,for all n∈N. let n,m∈N,

and n<m, put k=m−n. We have x0R f k(x0) since fis R-preserving (xn=fn(x0))R(fn+k(x0)=xm).

Hence {xn}is an R-sequence. On the other hand, fis R-contraction, thus we have

d(xn,xn+1)=d(f(xn−1),f(xn)) ≤

λd(xn−1,xn)≤ ·· · ≤ λnd(x1,x0),

for all n∈N. If m,n∈N, and n≤m, then

d(xn,xm)≤d(xn,xn+1)+· ·· +d(xm−1,xm)

≤λnd(x0,x1)+· ·· +λm−1d(x0,x1)

≤λn

1−λd(x1,x0).

Hence d(xn,xm)−→ 0 as m,n→ ∞. Therefore {xn}is an R-Cauchy sequence. Since Xis R-complete,

there exists x∗∈Xsuch that xn

R

−→ x∗. On the other hand, fis R-continuous so f(xn)R

−→ f(x∗), therefore

f(x∗)=f(lim

n→∞ xn)=lim

n→∞ f(xn)=lim

n→∞ xn+1=x∗. Thus x∗is a ﬁxed point of f.

To prove the uniqueness, let y∗∈Xbe a ﬁxed point of f, then x0R f (y∗)=y∗. Hence (xn=fn(x0))Ry∗

for all n∈N. Therefore, by triangle inequality, we have

d(x∗,y∗)=d(fn(x∗),fn(y∗)) ≤d(fn(x∗),fn(x0)) +d(fn(x0),fn(y∗))

≤λnd(x∗,x0)+λnd(x0,y∗)−→ 0,

as n→ ∞. Thus it follows that x∗=y∗.

Finally, let xbe an arbitrary element of X. We have x0R f (x), hence fn(x0)R f n+1(x) for all n∈N.

Therefore

d(x∗,fn(x)) =d(fn(x∗),fn(x)) ≤d(fn−1(x∗),fn−1(x0)) +d(fn−1(x0),fn−1(f(x)))

≤λn−1d(x∗,x0)+λn−1d(x0,f(x)) −→ 0,

as n→ ∞. Hence lim

n→∞ fn(x)=x∗.

We now show that our theorem is an extension of Banach contraction principle.

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Corollary 3.1. (Banach contraction principle) Let (X,d)be a complete metric space and f :X→X

be a mapping such that for some λ∈(0,1),

d(f(x),f(y)) ≤λd(x,y)

for all x,y∈X. Then f has a unique ﬁxed point in X.

Proof. Suppose that R:=X×X. Fix x0∈X. Clearly, x0Ry for all y∈f(X). Since Xis complete, it is

R-complete. It is clear that fis R-preserving, R-contraction and R-continuous. Using Theorem 3.1, f

has a unique ﬁxed point in X.

The following example, shows that our theorem is a real extension of Banach contraction principle.

Moreover, the next example is not satisﬁed [24, Theorem 2.1] and satisfy Theorem 3.1 in this paper.

Example 3.1. Let X=[0,1) and let the metric on Xbe the Euclidean metric. Deﬁne xRy if xy ∈ {x,y},

that is, x=0 or y=0. Let f:X→Xbe a mapping deﬁned by

f(x)=(x2

3:x≤1

3

0 : x>1

3

.

Let {xn}be an R-sequence in X. It is obvious that there exists N∈Nsuch that xn=0 for all n≥N,

n∈N. Hence xn

R

−→ 0. Therefore Xis R-complete but not complete. fis R-continuous but not

continuous. Suppose xRy, thus x=0 or y=0. Let x=0, then

|f(x)−f(y)|=(y2

3i f y ≤1

3

0i f y >1

3

.

Hence by choosing λ=0.99 it follows that

|f(x)−f(y)| ≤ y2

3≤λ|0−y|=λy.

So fis R-contraction. Let xRy, then x=0 or y=0, hence f(x)=0 or f(y)=0, so f(x)R f (y), that is,

fis R-preserving. Let y∈f(X), then 0Ry. Therefore using Theorem 3.1, fhas a unique ﬁxed point in

X. However, by Banach contraction principle and [24, Theorem 2.1], we can not ﬁnd any ﬁxed point

of fin X.

As, we know in Brouwer theorem a continuous mapping f:E→Efrom a convex and compact,

set E(⊆Rn) into Ehas a ﬁxed point without mentioning how to ﬁnd the ﬁxed point.

In our theorem, we omit the convexity and substitute the compactness with strong R-compactness

to show how the ﬁxed point can be constructed.

Theorem 3.2. Suppose X =Rnequipped with standard topology and let E ⊆X be a strong R-compact

set. Let f :E→E is R-continuous and R-preserving. Suppose x0∈E be given such that x0Ry for all

y∈f(E). Then f has a ﬁxed point.

Proof. Let x1=f(x0), x2=f(x1)=f(f(x0)) =f2(x0),..., xn=fn(x0),...,for all n∈N. Let n,m∈N

where n<m, put k=m−n.x0R f k(x0), hence fn(x0)R f n+k(x0)=fm(x0), i.e, xnR xm, therefore, the

sequence {xn}is an R-sequence. Since Eis R-compact, there exists a convergent subsequence {xnk}

such that xnk

R

−→ x∗. Thus f(xnk)R

−→ f(x∗), therefore f(x∗)=lim

k→∞ f(xnk)=lim

k→∞ xnk+1=x∗.

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Example 3.2. Suppose X=Rwith standard topology be given. Let R:=≤and E=(0,1]. Let

f:E→Ebe a mapping deﬁned by f(x)=x+1

2. Let x0=1

2, then x1=f(x0)=3

4,x2=f(x1)=7

8,

x3=f(x2)=15

16 ,x4=f(x3)=31

32 , . . . . Hence we obtain an R-sequence {xn}such that xn→1=x∗.

Thus f(x∗)=f(1) =1+1

2=1=x∗.

Example 3.3. Suppose X=Rwith standard topology be given. Let R:=≤and let E=(−1,2] ∪(3,4].

Let f:E→Ebe a mapping deﬁned by f(x)=x+1

4. Put x0=0, then x1=f(x0)=f(0) =1

4,

x2=f(x1)=f(1

4)=5

42,x3=f(x2)=f(5

42)=21

43,x4=f(x3)=f(21

43)=85

44, . . . . It is easy to show that

xn

R

−→ 1

3=x∗. Hence the ﬁxed point is 1

3.

In the above example Eis neither compact nor convex but fhas a ﬁxed point.

4. Application to fractional integral equations

Our aim here is to apply Theorem 3.1 to prove the existence and uniqueness of solution for the

following fractional integral equation of the type

x(t)=a(t)

Γ(α)Rt

0b(u) (t−u)α−1x(u)du +f(t),t∈I=[0,T]

f(t)≥1,t∈I

(4.1)

where 0 <α<1 and a,b,f: [0,T]→Rare continuous functions.

Theorem 4.1. Under the above conditions, for all T >0the fractional integral equation (4.1) has a

unique solution.

Proof. Let X=u∈C(I,R) : u(t)>0,∀t∈I. We deﬁne the following relation Rin X:

xRy ⇐⇒ x(t)y(t)≥(x(t)∨y(t)),∀t∈I.

Deﬁne

kxkr=max

t∈Ie−rt|x(t)|,x∈X,f or r >(kak·kbk)1

α(4.2)

It is easy to show that (X,d) is an R-metric space. Suppose x0(t)≡1 for all t∈I, then x0Rx for all

x∈X.Let {xn} ⊆ Xbe a Cauchy R-sequence. It is easy to show that {xn}is converging to an element

xuniformly in C(I,R). Fix t∈I, the deﬁnition of Rimplies xn(t)xn+k(t)≥(xn(t)∨xn+k(t)) for each

n,k∈N. Since xn(t)>0 for all n∈N, there exists an R-subsequence {xni}of {xn}such that xni(t)≥1

for all i∈N. Clearly {xni}also converges to x, therefore x≥1, hence x∈X.

Deﬁne F:X→Xby

Fx(t)=a(t)

Γ(α)Zt

0

b(u) (t−u)α−1x(u)du +f(t).

Clearly the ﬁxed points of Fare the solutions of (4.1) .

It is enough to prove the following three steps:

Step (1) Fis R-preserving.

Proof. Let xRy and t∈I,Fx(t)=a(t)

Γ(α)Rt

0b(u) (t−u)α−1x(u)du +f(t)≥1 which implies that

Fx(t)Fy(t)≥ Fy(t). Thus FxRFy.

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3135

Step (2) Fis R- contraction.

Proof. Let xRy and t∈I,

e−rt|F x(t)− Fy(t)| ≤ e−rt a(t)

Γ(α)Zt

0

b(u) (t−u)α−1|x(u)−y(u)|du

≤e−rt a(t)

Γ(α)Zt

0

b(u) (t−u)α−1erue−ru |x(u)−y(u)|du

≤e−rt a(t)

Γ(α)kx−ykrZt

0

b(u) (t−u)α−1erudu

≤1

Γ(α)kak kbk kx−ykr

1

rα−1Zt

0

rα−1(t−u)α−1erue−rtdu .

Put w=r(t−u), then 0 ≤w≤rt, we get

e−rt|F x(t)− Fy(t)| ≤ 1

Γ(α)kak kbk kx−ykr

1

rαZrt

0

wα−1e−wdw

≤(kak kbk1

rα)kx−ykr.

Since tis an arbitrary element of I, by (4.2) it follows that 0 ≤λ=(kakkbk1

rα)<1, hence kF x−F ykr≤

λkx−ykr.

Step (3) Fis R-continuous.

Proof: Let {xn}be an R-sequence converging to x∈X. Using the ﬁrst part of the proof x(t)≥1 for all

t∈I, hence xn(t)x(t)≥xn(t) for all n∈Nand all t∈I, therefore xnRx. By step (2), we have

e−rt|F xn(t)− F x(t)| ≤ λkxn−xkr.

Thus

kF xn− F xkr≤λkxn−xkr.

Therefore Fxn

R

−→ F x.

Now applying theorem 3.1 it follows that the fractional integral equation (4.1) has a unique solution.

5. Conclusion

In this paper, by introducing a new and applicable metric space which was called R−metric space,

we have shown a real generalization of one of the most important theorems in mathematics which is

named Banach ﬁxed point Theorem. The results have immediately applied to show the existence and

uniqueness of fractional integral equations and can be extended to ordinary diﬀerential equations and

etc. Moreover, the results we have obtained suggest a constructive method to ﬁnd the value of Brouwer

ﬁxed point which is very important in many ﬁelds like game theory, mathematical economics, physics,

chemistry. engineering and etc.

Acknowledgments

The authors thank the referees for the careful reading and for the very useful comments, suggestions

and remarks that contributed to the improvement of the initial version of the manuscript.

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3136

Conﬂict of interest

The authors declare no conﬂict of interest.

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