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Fixed point theorems in R-metric spaces with applications

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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.
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 Rmetric 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 dierent 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 :XX be a mapping such that, for some
λ(0,1),
d(f(x),f(y)) λd(x,y)
for all x,yX. 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,yT 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),xy,
x0T : x0F(x0)or x0F(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 :XX be a continuous and nondecreasing mapping
such that there exists k [0,1) with
d(f(x),f(y)) kd(x,y),xy.
If there exists x0X with x0f(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 dicult 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 (xy) where
xy=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,kN.
Definition 2.3. An R-sequence {xn}is said to converge to xXif for every ε > 0 there is an integer N
such that d(xn,x)< ε if nN.
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
nN. Clearly {xn}is an R-sequence and xn
R
1. Note that xn=21
nis not an R-sequence.
Example 2.4. Let X=R. Consider P(R) with metric dist(A,B)=in f {|ab|:aA and b B}, let
R:=.Define An=[1 +1
n,41
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 EX.xXis called an R-limit point of Eif there exists an R-sequence {xn}in E
such that xn,xfor all nNand xn
R
x.
Definition 2.5. The set of all R-limit points of Eis denoted by E0R.
Definition 2.6. EXis called R-closed, if E0RE.
Definition 2.7. EXis called R-open, if ECis R-closed.
Theorem 2.4. EX 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 xnE for all n N.
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Proof. Let xE. Suppose there exists an R-sequence {xn}which xn
R
xbut for every NN, there
exists a natural number nNsuch that xn<E. Hence, we obtain an R-subsequence {xnN}of {xn},
which xnN
R
x, and xnN<E. Therefore xEC. This contradicts xE.
Conversely, if x(EC)0R, then x<E. Hence (EC)0REC, so ECis R-closed. Thus Eis R-open.
Definition 2.8. Let EX. The R-closure of Eis the set ER=EE0R.
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 NNsuch that xn(ER)Cfor nN. 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 E0RE[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 {GX|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}jJbe a family of R-open sets. Put U=SjJUj, let xbe
an R-limit point of UC=TjJUC
j, hence there exists an R-sequence {xn}in UC\ {x}such that xn
R
x.
Therefore for each jJ,xis an R-limit point of UC
j. Since UC
jis R-closed xUC
j, so that xUC,
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
1jnand a subsequence {xnk}of {xn}in UC
j\ {x}which xnk
R
x. Since UC
jis R-closed xUC
j,
thus xWC. 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 τdτR.
Proof. Let Gτd. 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
xGC. Therefore GτR.
Lemma 2.1. Let R :=X×X, then τR=τd.
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Proof. By theorem 2.7, τdτR. Let Gis an R-open set and xis a limit point of GC. There exists a
sequence {xn}in GC\ {x}such that xnx. Clearly {xn}is an R-sequence, hence xn
R
x. It follows
that xGC. Thus τRτd.
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.ERkis called R-convex if λx+(1 λ)yE, whenever
xE,yE,xRy, and 0 <λ<1.
Lemma 2.2. Let E is a convex set, then E is R-convex.
Proof. Let xE,yE,xRy, and 0 <λ<1. By the definition of convex set λx+(1 λ)yE.
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. KXis 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
xy1
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 nN,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 xK0R. 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 xK. The theorem follows.
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Theorem 2.9. An R-closed subset of an R-compact set, is R-compact.
Proof. Suppose FKX.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 xF.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, FKis R-closed; since FKK, theorem 2.9 shows that FKis
R-compact.
Definition 2.11. A set KXis called strong R-compact, if each R-sequence {xn}in K, that has a
subsequence {xnk}which converges to xK, 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 xK, such that xnk
R
xand 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 nNand mN. 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
xK.
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:XXbe a mapping. fis said to be R-continuous at xXif 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 xX.
Lemma 2.5. Every continuous mapping f :XX, 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:XXbe 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 nZor x=0. Define f:XXby f(x)=[x]. Let xXand {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 mZsuch 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:XXis said to be an R-contraction with Lipschutz constant 0 <λ<
1 if for all x,yXsuch 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,yX. Let f:XXbe a mapping defined by
f(x)=(x2i f x QX
0i f x QCX.
Suppose x=0.9, y=2
2and 0 <λ<1, then
|f(x)f(y)|=0.81 λ|0.92
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 QX
0i f y QCX.
Hence by choosing λ=0.99 it follows that
|f(x)f(y)| ≤ y2λ|0y|=λy.
So that fis R-contraction.
Definition 2.16. Let f:XXbe a mapping fis called R-preserving if xRy, then f(x)R f (y) for all
x,yX.
Example 2.15. Suppose X=Rwith standard topology be given and let R:=. Let f:XXbe a
mapping defined by f(x)=x3. Let x1x2, then f(x1)=x3
1x3
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 :XX be R-continuous, R-contraction with Lipschutz constant λand R-preserving.
Suppose there exists x0X 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)=xfor all x X.
Proof. Let x1=f(x0), x2=f(x1)=f2(x0),..., xn=f(xn1)=fn(x0),...,for all nN. let n,mN,
and n<m, put k=mn. 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(xn1),f(xn))
λd(xn1,xn)≤ ·· · ≤ λnd(x1,x0),
for all nN. If m,nN, and nm, then
d(xn,xm)d(xn,xn+1)+· ·· +d(xm1,xm)
λnd(x0,x1)+· ·· +λm1d(x0,x1)
λn
1λd(x1,x0).
Hence d(xn,xm)0 as m,n→ ∞. Therefore {xn}is an R-Cauchy sequence. Since Xis R-complete,
there exists xXsuch 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 xis a fixed point of f.
To prove the uniqueness, let yXbe a fixed point of f, then x0R f (y)=y. Hence (xn=fn(x0))Ry
for all nN. 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 nN.
Therefore
d(x,fn(x)) =d(fn(x),fn(x)) d(fn1(x),fn1(x0)) +d(fn1(x0),fn1(f(x)))
λn1d(x,x0)+λn1d(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 :XX
be a mapping such that for some λ(0,1),
d(f(x),f(y)) λd(x,y)
for all x,yX. Then f has a unique fixed point in X.
Proof. Suppose that R:=X×X. Fix x0X. Clearly, x0Ry for all yf(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:XXbe a mapping defined by
f(x)=(x2
3:x1
3
0 : x>1
3
.
Let {xn}be an R-sequence in X. It is obvious that there exists NNsuch that xn=0 for all nN,
nN. 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
3λ|0y|=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 yf(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:EEfrom 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 :EE is R-continuous and R-preserving. Suppose x0E be given such that x0Ry for all
yf(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 nN. Let n,mN
where n<m, put k=mn.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:EEbe 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 xn1=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:EEbe 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) (tu)α1x(u)du +f(t),tI=[0,T]
f(t)1,tI
(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=uC(I,R) : u(t)>0,tI. We define the following relation Rin X:
xRy x(t)y(t)(x(t)y(t)),tI.
Define
kxkr=max
tIert|x(t)|,xX,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 tI, then x0Rx for all
xX.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 tI, the definition of Rimplies xn(t)xn+k(t)(xn(t)xn+k(t)) for each
n,kN. Since xn(t)>0 for all nN, there exists an R-subsequence {xni}of {xn}such that xni(t)1
for all iN. Clearly {xni}also converges to x, therefore x1, hence xX.
Define F:XXby
Fx(t)=a(t)
Γ(α)Zt
0
b(u) (tu)α1x(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 tI,Fx(t)=a(t)
Γ(α)Rt
0b(u) (tu)α1x(u)du +f(t)1 which implies that
Fx(t)Fy(t)≥ Fy(t). Thus FxRFy.
AIMS Mathematics Volume 5, Issue 4, 3125–3137.
3135
Step (2) Fis R- contraction.
Proof. Let xRy and tI,
ert|F x(t)− Fy(t)| ≤ ert a(t)
Γ(α)Zt
0
b(u) (tu)α1|x(u)y(u)|du
ert a(t)
Γ(α)Zt
0
b(u) (tu)α1erueru |x(u)y(u)|du
ert a(t)
Γ(α)kxykrZt
0
b(u) (tu)α1erudu
1
Γ(α)kak kbk kxykr
1
rα1Zt
0
rα1(tu)α1eruertdu .
Put w=r(tu), then 0 wrt, we get
ert|F x(t)− Fy(t)| ≤ 1
Γ(α)kak kbk kxykr
1
rαZrt
0
wα1ewdw
(kak kbk1
rα)kxykr.
Since tis an arbitrary element of I, by (4.2) it follows that 0 λ=(kakkbk1
rα)<1, hence kF x−F ykr
λkxykr.
Step (3) Fis R-continuous.
Proof: Let {xn}be an R-sequence converging to xX. Using the first part of the proof x(t)1 for all
tI, hence xn(t)x(t)xn(t) for all nNand all tI, therefore xnRx. By step (2), we have
ert|F xn(t)− F x(t)| ≤ λkxnxkr.
Thus
kF xn− F xkrλkxnxkr.
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 Rmetric 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 dierential 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.
AIMS Mathematics Volume 5, Issue 4, 3125–3137.
3136
Conflict of interest
The authors declare no conflict of interest.
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is an open access article distributed under the
terms of the Creative Commons Attribution License
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