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Article

On An Open Question in Controlled Rectangular b-Metric Spaces

1
Department of Mathematics, College of Science and Humanities in Alkharj, Prince Sattam bin Abdulaziz University, Al-Kharj 11942, Saudi Arabia
2
Department of Mathematics and Computer Science, St. Thomas College, Bhilai, Chhattisgarh 491022, India
3
École Nationale Polytechnique d’Oran Maurice Audin, School in Oran, B.P 1523 El M’naouer, Oran 31000, Algeria
4
SYSTEC—Research Center for Systems and Technologies, Department of Electrical Engineering, Faculty of Engineering, Porto University, 4200-465 Porto, Portugal
5
Faculty of Electrical Engineering, University of Banja Luka, Patre 5, 78000 Banja Luka, Bosnia and Herzegovina
6
Department of Mathematics and General Sciences, Prince Sultan University Riyadh, Riyadh 11586, Saudi Arabia
7
Department of Pharmaceutical Sciences, Faculty of Pharmacy, Vasile Goldiş Western University Arad, L. Rebreanu Street, no. 86, 310048 Arad, Romania
*
Authors to whom correspondence should be addressed.
Mathematics 2020, 8(12), 2239; https://doi.org/10.3390/math8122239
Submission received: 12 November 2020 / Revised: 11 December 2020 / Accepted: 15 December 2020 / Published: 18 December 2020
(This article belongs to the Special Issue Fixed Point, Optimization, and Applications)

Abstract

:
In this paper, we give an affirmative answer to an open question posed recently by Mlaiki et al. As a consequence of our results, we get some known results in the literature. We also give an application of our results to the existence of a solution of nonlinear fractional differential equations.

1. Introduction

Fréchet [1] introduced the axiomatic form of distance as L-space. Hausdorff [2] re-defined it as a metric space. In 1922, Banach [3] proved the existence and uniqueness of a fixed point for self-contractive mappings in a complete metric space. Inspired by the wide applications of Banach’s result, numerous extensions and generalizations of it appeared in the literature. One direction of such generalizations was by generalizing the concept of a metric space itself. Bakhtin [4], Branciari [5], George et al. [6], Mlaiki et al. [7], Abdeljawad et al. [8,9], and Shatanawi et al. [10] introduced the notion of a b-metric, rectangular metric, rectangular b-metric, controlled metric, double controlled metric, extended Branciari b-distance, and extended b-metric, respectively (see also [11,12,13]). The concept of a controlled rectangular b-metric space was introduced in Mlaiki et al. [14]. The following two theorems are the main results in [14].
Theorem 1.
Let ( X , D ξ ) be a D ξ -complete, controlled, rectangular b-metric space, T   : X X be a self map on X. If there exists 0 < δ < 1 , such that
D ξ ( T z , T t ) δ D ξ ( z , t ) ,
for all z , t X , and
sup m > 1 lim n ξ ( x n , x n + 1 , x n + 2 , x m ) 1 δ ,
then T has a unique fixed point in X .
Remark 1.
Note that, from Theorem 1 we obtain theorem 2.1 in [6].
Theorem 2.
Let ( X , D ξ ) be a D ξ -complete controlled rectangular b-metric space, T   : X X be a self map on X. If
D ξ ( T z , T t ) δ [ D ξ ( z , T z ) + D ξ ( t , T t ) ]
for all z , t E , wher δ ( 0 , 1 ) . Also, if
sup m > 1 lim n ξ ( x n , x n + 1 , x n + 2 , x m ) 1 δ ,
and for all u , v X we have
lim n ξ ( u , v , x n , x n + 1 ) 1 ,
then T has a unique fixed point in X .
Mlaiki et al. in the paper [14] gave the following open question.
Question 1.
Let ( X , D ξ ) be a controlled rectangular b-metric space, and T   : X X be a self map onX. Assume that for all distinct u , v , T u , T v X there exists δ ( 0 , 1 ) such that
D ξ ( T u , T v ) δ ξ ( u , v , T u , T v ) D ξ ( u , v ) .
What are the other hypotheses we should add so that T has a unique fixed point in the whole space X ?
The aim of this work is to provide an answer to the above question by providing other hypotheses required to prove the existence of a unique fixed point of the mapping T given in Question 1. We also provide a short proof of Theorem 1, and we replace condition (1) with
sup m , n , k N ξ ( x m , x n , x m + k , x n + k ) < + .
Also, we show that condition (2) is redundant and that condition (3) can be relaxed.

2. Preliminaries

Definition 1
([14]). Let X be a non-empty set, ξ : X 4 [ 1 , ) and D ξ : X 2 [ 0 , ) be given mappings. We say that ( X , D ξ ) is a controlled rectangular b-metric space (or in short CRb-MS) if, for all distinct c , d , e , f X , we have
1. 
D ξ ( c , d ) = 0 if and only if c = d ;
2. 
D ξ ( c , d ) = D ξ ( d , c ) ;
3. 
D ξ ( c , d ) ξ ( c , d , e , f ) [ D ξ ( c , e ) + D ξ ( e , f ) + D ξ ( f , d ) ] .
Further, through the paper we will assume that ( X , D ξ ) is a CRb-MS. The following definition presents the topology of this space.
Definition 2
([14]).
1. 
A sequence { a n } is called D ξ -convergent in ( X , D ξ ) , if there exists a X such that lim n D ξ ( a n , a ) = 0 .
2. 
A sequence { a n } in ( X , D ξ ) is called D ξ -Cauchy if lim n , m D ξ ( a n , a m ) = 0 .
3. 
( X , D ξ ) is called D ξ -complete if every D ξ -Cauchy sequence { a n } in X is convergent in X.
4. 
For any c X and η > 0 , we define an open ball in ( X , D ξ ) with center c and radius η by B ξ ( c , η ) = { d X D ξ ( c , d ) < η } .
Remark 2.
Note that the condition (4) gives us the possibility to obtain, as a consequence of our result, a series of known results in a b-metric space, rectangular metric space, b-rectangular metric space, extended b-metric space, and extended b-rectangular metric space. Additionally, rectangular metric spaces (see Branciari in [5]), rectangular b -metric spaces (see George et al. [6]), and extended Branciari b-distance spaces (see Abdeljawad et al. in [9]) are all CRb-MS. (The converse is not necessarily true (see Example 1, [14]).
Remark 3.
In this paper, we consider that the sequence { x n } is defined by x n = T x n 1 , n N , where x 0 X .

3. An Answer to an Open Question

Theorem 3.
Let T   : X X be a mapping on D ξ -complete CRb-MS ( X , D ξ ) . Assume that
D ξ ( T u , T w ) δ ξ ( u , w , T u , T w ) D ξ ( u , w )
for all distinct u , w , T u , T w X , where δ [ 0 , 1 ) . If
sup m , n , k N i = 1 k ξ ( x m + k i , x n + k i , x m + k + 1 i , x n + k + 1 i ) < +
and
sup m , n , k N ξ ( x m , x n , x m + k , x n + k ) < + ,
then T has a unique fixed point in X.
Proof. 
Step 1. Let { x n } be a sequence defined as in Remark 3. If x n = x n + 1 then x n is a fixed point of T and we are done. So suppose that x n x n + 1 for all n 0 . Then x n x n + k for all n 0 , k 1 . Then, we assume x n x m for all distinct n , m N . Note that x n + k x m + k for all distinct n , m N x n + k , x m + k X \ { x n , x m } .
Step 2. From condition (6), we conclude that there exists M > 0 , such that
i = 0 k ξ ( x m + k i , x n + k i , x m + k + 1 i , x n + k + 1 i ) M
for all m , n , k N . Since δ [ 0 , 1 ) , we conclude that there exists n 1 N , such that
δ n 1 M < 1 .
Let k , m , m N , such that
min { k , m , n } n 1 .
Now, from Condition 3 of Definition 1, we have
D ξ ( x m , x n ) ξ ( x m , x n , x m + k , x n + k ) [ D ξ ( x m , x m + k ) + D ξ ( x m + k , x n + k ) + D ξ ( x n + k , x n ) ] .
Next, from Condition (5), we have
D ξ ( x m , x m + k ) δ ξ ( x m 1 , x m + k 1 , x m , x m + k ) D ξ ( x m 1 , x m + k 1 ) δ 2 ξ ( x m 1 , x m + k 1 , x m , x m + k ) ξ ( x m 2 , x m + k 2 , x m 1 , x m + k 1 ) D ξ ( x m 2 , x m + k 2 ) δ m i = 1 m ξ ( x m i , x m + k i , x m + 1 i , x m + k + 1 i ) D ξ ( x 0 , x k ) .
Similar as in (12), we get
D ξ ( x n , x n + k ) δ n i = 1 n ξ ( x n i , x n + k i , x n + 1 i , x n + k + 1 i ) D ξ ( x 0 , x k )
and
D ξ ( x m + k , x n + k ) δ k i = 1 k ξ ( x m + k i , x n + k i , x m + k + 1 i , x n + k + 1 i ) D ξ ( x m , x n ) .
Let
P m , n = i = 1 k ξ ( x m + k i , x n + k i , x m + k + 1 i , x n + k + 1 i ) ,
P m = i = 1 m ξ ( x m i , x m + k i , x m + 1 i , x m + k + 1 i ) ,
P n = i = 1 n ξ ( x n i , x n + k i , x n + 1 i , x n + k + 1 i ) .
From (11)–(14), we obtain
D ξ ( x m , x n ) ξ ( x m , x n , x m + k , x n + k ) [ δ m P m D ξ ( x 0 , x k ) + δ k P m , n D ξ ( x m , x n ) + δ n P n D ξ ( x k , x 0 ) ] .
Due to the conditions (7) and (9), we have
D ξ ( x m , x n ) ξ ( x m , x n , x m + k , x n + k ) ( δ m P m + δ n P n ) D ξ ( x 0 , x k ) 1 δ k ξ ( x m , x n , x m + k , x n + k ) P m , n .
Using condition (6), we obtain that { x n } is a D ξ -Cauchy. Thus, there exists x * X such that
lim n x n = x * .
We have
D ξ ( x * , T x * ) ξ ( x * , T x * , x n , x n + 1 ) [ D ξ ( x * , x n ) + D ξ ( x n , x n + 1 ) + D ξ ( x n + 1 , T x * ) ] ξ ( x * , T x * , x n , x n + 1 ) [ D ξ ( x * , x n ) + D ξ ( x n , x n + 1 ) + D ξ ( T x n , T x * ) ] ξ ( x * , T x * , x n , x n + 1 ) [ D ξ ( x * , x n ) + D ξ ( x n , x n + 1 ) + δ ξ ( x * , T x * , x n , x n + 1 ) D ξ ( x n , x * ) ] .
Since lim n D ξ ( x * , x n ) = 0 and lim n D ξ ( x n , x n + 1 ) = 0 , we conclude D ξ ( x * , T x * ) = 0 , i.e T x * = x * .
Step 3. Uniqueness. Let y * X such that T y * = y * . Then we have
D ξ ( x * , y * ) ξ ( x * , y * , x n , x n + 1 ) [ D ξ ( x * , x n ) + D ξ ( x n , x n + 1 ) + D ξ ( x n + 1 , y * ) ] .
Since lim n D ξ ( x * , x n ) = 0 , lim n D ξ ( x n , x n + 1 ) = 0 , lim n D ξ ( x n + 1 , y * ) = 0 we obtain D ξ ( x * , y * ) = 0 . Thus, x * = y * . □

4. On Fixed Point Theorems of Banach and Kannan

Now, we give another variant of the Theorem 1.
Theorem 4.
Let T   : X X be a mapping on D ξ -complete CRb-MS ( X , D ξ ) . Suppose that for all distinct u , w , T u , T w X , there exists δ [ 0 , 1 ) , such that
D ξ ( T u , T w ) δ D ξ ( u , w )
and
sup m , n , k N ξ ( x m , x n , x m + k , x n + k ) < + .
Then, T has a unique fixed point in X.
Proof. 
Step 1. Same as in Theorem 3.
Step 2. From condition (19), we obtain that there exists M > 0 such that
ξ ( x m , x n , x m + k , x n + k ) M ,
for all m , n , k N . Since δ [ 0 , 1 ) we conclude that there exists n 1 N such that
δ n 1 M < 1 .
Let k , m , m N such that
min { k , m , n } n 1 .
Next, we have
D ξ ( x m , x n ) ξ ( x m , x n , x m + k , x n + k ) [ D ξ ( x m , x m + k ) + D ξ ( x m + k , x n + k ) + D ξ ( x n + k , x n ) ] .
Next, from condition (18), we have
D ξ ( x m , x m + k ) δ D ξ ( x m 1 , x m + k 1 ) δ m D ξ ( x 0 , x k ) .
Additionally, as in (24), we obtain
D ξ ( x n , x n + k ) δ n D ξ ( x 0 , x k )
and
D ξ ( x m + k , x n + k ) δ k D ξ ( x m , x n ) .
So, we obtain
D ξ ( x m , x n ) ξ ( x m , x n , x m + k , x n + k ) ( δ m + δ n ) D ξ ( x 0 , x k ) 1 δ k ξ ( x m , x n , x m + k , x n + k ) .
Next, using condition (19), we obtain that { x n } is a D ξ -Cauchy. Therefore, there exists x * X such that
lim n x n = x * .
Now let us prove that x * is a fixed point. We have
D ξ ( x * , T x * ) ξ ( x * , T x * , x n , x n + 1 [ D ξ ( x * , x n ) + D ξ ( x n , x n + 1 ) + D ξ ( x n + 1 , T x * ) ] ξ ( x * , T x * , x n , x n + 1 ) [ D ξ ( x * , x n ) + D ξ ( x n , x n + 1 ) + D ξ ( T x n , T x * ) ] ξ ( x * , T x * , x n , x n + 1 ) [ D ξ ( x * , x n ) + D ξ ( x n , x n + 1 ) + δ D ξ ( x n , x * ) ] .
Since
lim n D ξ ( x * , x n ) = 0 , lim n D ξ ( x n , x n + 1 ) = 0 ,
we conclude
D ξ ( x * , T x * ) = 0 .
Step 3. Uniqueness. Let y * be another fixed point of T. Then we have
D ξ ( x * , y * ) ξ ( x * , y * , x n , x n + 1 ) [ D ξ ( x * , x n ) + D ξ ( x n , x n + 1 ) + D ξ ( x n + 1 , y * ) ] .
Since
lim n D ξ ( x * , x n ) = 0 , lim n D ξ ( x n , x n + 1 ) = 0 , lim n D ξ ( x n + 1 , y * ) = 0 ,
we obtain D ξ ( x * , y * ) = 0 , so x * = y * . □
The next result is a version of Kannan’s theorem in controlled rectangular b-metric space.
Theorem 5.
Let T   : X X be a mapping on D ξ -complete CRb-MS ( X , D ξ ) , satisfying the following condition: For all z , t X , there exists δ [ 0 , 1 2 ) such that
D ξ ( T z , T t ) δ [ D ξ ( z , T z ) + D ξ ( t , T t ) ] .
Additionally, if
lim n ξ ( u , T u , x n , x n + 1 ) 1 δ ,
for all u X , then T has a unique fixed point in X .
Proof. 
Let { x n } be a sequence defined as in Remark 3. From condition (27), we obtain
d ( x n + 1 , x n ) δ [ D ξ ( x n , x n + 1 ) + D ξ ( x n 1 , x n ) ] ,
d ( x n + 1 , x n ) δ 1 δ D ξ ( x n , x n 1 ) ] ,
so,
d ( x n + 1 , x n ) β n D ξ ( x 1 , x 0 ) ] ,
for all n N , where β = δ 1 δ . Additionally, from condition (27), we have
d ( x m , x n ) δ [ D ξ ( x m 1 , x m ) + D ξ ( x n 1 , x n ) ] .
Using condition (29), we get
d ( x m , x n ) δ [ β m 1 D ξ ( x 0 , x 1 ) + β n 1 D ξ ( x 0 , x 1 ) ]
for all m , n N . Therefore, { x n } is D ξ -Cauchy. Let a * = lim n + x n . Since, X is controlled rectangular b-metric space we obtain
D ξ ( a * , T a * ) ξ ( a * , T a * , x n , x n + 1 ) [ D ξ ( a * , x n ) + D ξ ( x n , x n + 1 ) + D ξ ( x n + 1 , T a * ) ] ξ ( a * , T a * , x n , x n + 1 ) [ D ξ ( a * , x n ) + D ξ ( x n , x n + 1 ) + δ ( D ξ ( x n , x n + 1 ) + D ξ ( a * , T a * ) ] .
Now, from condition (28), we obtain T a * = a * . Uniqueness follows from condition (27). □
Remark 4.
Note that instead of condition (28) in Theorem 2 we can use the condition that T is continuous.
Corollary 1.
[15] (see also [16,17,18]) Let ( X , D ξ ) be a complete b-metric space and T   : X X such that
D ξ ( T s , T r ) λ D ξ ( s , r )
for all s , r X , where λ [ 0 , 1 ) . Then, T has a unique fixed point in X.
Remark 5.
As a consequence of Theorem 4, we obtain results for rectangular b-metric space (theorem 2.1 of [6]) and extended b-metric space (corollary 3.3 in [19] and theorem 2 in [20]).

5. Application in Fractional Differential Type Equation

In this section, using Theorem 3, we obtain existence and uniqueness solutions of nonlinear fractional differential equation system Caputo type (see, for example, Ref. [21])
C D θ ( ψ ( ι ) + χ ( ι , ψ ( ι ) ) ) = 0 , ι [ 0 , 1 ) , θ < 1 ψ ( 0 ) = ψ ( 1 ) = 0 ,
where ψ C ( [ 0 , 1 ] , R ) and C ( [ 0 , 1 ] , R ) is the set of all continuous functions from [ 0 , 1 ] to R , χ : [ 0 , 1 ] × R R is a continuous function, and C D θ is the Caputo derivative of order θ
C D θ ( ω ( ι ) ) : = 1 Γ ( n θ ) 0 ι ( ι κ ) n θ 1 ω ( n ) ( κ ) d κ , ( n 1 < θ < n ) ,
where ω : [ 0 , ) R is a continuous function, see [22,23,24]).
Let G be a Green function for the system (31), for α < 1 , as follows:
G ( ι , κ ) = ( ι ( 1 κ ) ) α 1 ( ι κ ) α 1 , i f 0 κ t 1 ; ( ι ( 1 κ ) ) α 1 Γ ( α ) , i f 0 ι κ 1 .
Let us give the first existence result of this section.
Theorem 6.
Let the nonlinear fractional differential Equation (31) and μ : R × R R be a given function, such that the following assumptions are true:
(i) 
There exists ψ 0 ( C [ 0 , 1 ] , R ) such that μ ( ψ 0 ( ι ) , 0 1 T ψ 0 ( ι ) d ι ) 0 for all ι [ 0 , 1 ] , where T : C ( [ 0 , 1 ] , R ) C ( [ 0 , 1 ] , R ) is defined as follows:
T ψ ( ι ) = 0 1 G ( ι , κ ) χ ( κ , x ( κ ) ) d κ
(ii) 
| χ ( ι , α ) χ ( ι , β ) | 1 τ ξ ( α , β , T α , T β ) | α β | for all ι [ 0 , 1 ] , τ > 1 and α , β R with μ ( α , β ) 0 where ξ : X 4 [ 1 , ) ;
(iii) 
for each ι [ 0 , 1 ] , if { ψ n } is a sequence in C ( [ 0 , 1 ] , R ) such that ψ n ψ in C ( [ 0 , 1 ] , R ) and μ ( x n ( ι ) , x n + 1 ( ι ) ) 0 for all n N , then μ ( ψ n ( t ) , ψ ( t ) ) 0 for all n N .
Then, the Equation (31) has a unique solution.
Proof. 
Let X = C ( [ 0 , 1 ] , R ) endowed with D ξ : X × X such that
D ξ ( ψ , φ ) = ψ φ = sup ι [ 0 , 1 ] ψ ( ι ) φ ( ι ) 2
and ξ : X 4 [ 1 , ) defined by ξ ( ψ , φ , T ψ , T φ ) = max { ψ , φ , T ψ , T φ } + 2 for any ψ , φ X . Clearly, ( X , D ξ ) is a D ξ -complete CRb-MS.
It is obvious the problem (31) can be resumed to finding an element ψ * X which is a fixed point for the operator T.
Let ψ , φ X , such that μ ( ψ ( ι ) , φ ( ι ) ) 0 for all ι [ 0 , 1 ] . By ( i ) and ( i i ) , we have true the following inequalities:
T ψ ( ι ) T φ ( ι ) 2 = 0 1 G ( ι , κ ) [ χ ( κ , ψ ( κ ) ) χ ( κ , φ ( κ ) ) ] d κ 2 0 1 G ( ι , κ ) d κ 2 0 1 χ ( κ , ψ ( κ ) ) χ ( κ , φ ( κ ) ) 2 d κ 1 τ 0 1 ξ 2 ( ψ ( κ ) , φ ( κ ) , T ψ ( κ ) , T φ ( κ ) ) ψ ( κ ) φ ( κ ) 2 d κ 0 1 G ( ι , κ ) d κ 2 .
Let ξ ( ψ ( κ ) , φ ( κ ) , T ψ ( κ ) , T φ ( κ ) ) = max κ [ 0 , 1 ] { ψ ( κ ) , φ ( κ ) , T ψ ( κ ) , T φ ( κ ) } + 2 = γ + 2 = λ be such that λ < τ .
Then, we have
T x ( ι ) T y ( ι ) 2 λ τ ψ ( ι ) φ ( ι ) 2 0 1 G ( ι , κ ) d κ 2 .
Then, taking supremum in both sides, for every ψ , φ X with μ ( ψ ( ι ) , φ ( ι ) ) > 0 and ι [ 0 , 1 ] , we obtain
D ξ ( ψ , φ ) = T ψ T φ λ τ ψ φ = λ τ D ξ ( ψ , φ ) .
It is easy to check that
i = 1 k ξ ( ψ m + k i ( κ ) , ψ n + k i ( κ ) , ψ m + k + 1 i ( κ ) , ψ n + k + 1 i ( κ ) ) = i = 1 k γ i < + ,
and
ξ ( ψ m ( κ ) , ψ n ( κ ) , ψ m + k ( κ ) , ψ n + k ( κ ) ) < + ,
for all k , m , n n 0 , with n 0 N .
For 0 < δ = 1 τ < 1 and λ < τ , we conclude that all the conditions of the Theorem 3 are accomplished. Then, the system (31) has a unique solution. □
Let us consider the following type of integral equation
x ( ι ) = q ( ι ) + 0 t P ( ι , κ ) g ( κ , x ( κ ) ) d κ , ι [ 0 , 1 ] , x ( ι ) X ,
where g ( ι , x ( ι ) ) : [ 0 , 1 ] × R R , q ( ι ) : [ 0 , 1 ] R are two bounded continuous functions and P : [ 0 , 1 ] × [ 0 , 1 ] [ 0 , ) is a function such that P ( ι , · ) L 1 ( [ 0 , 1 ] ) for all ι [ 0 , 1 ] .
Then, our aim is to establish the existence of a solution for the previous integral Equation (34) using Theorem 4. In this frame, let us give the following result.
Theorem 7.
Let X = C ( [ 0 , 1 ] , R ) be the set of all continuous real-valued functions defined on [ 0 , 1 ] . Let T   : X X be an operator defined by:
T x ( ι ) = q ( ι ) + 0 ι P ( ι , κ ) g ( κ , x ( κ ) ) d κ , ι [ 0 , 1 ] , x ( ι ) X .
Suppose the following conditions hold:
(i) 
The functions q ( ι ) : [ 0 , 1 ] R and g ( ι , x ( ι ) ) : [ 0 , 1 ] × R R are continuous;
(ii) 
P : [ 0 , 1 ] × [ 0 , 1 ] [ 0 , ) is a function such that P ( ι , · ) L 1 ( [ 0 , 1 ] ) for all ι [ 0 , 1 ] and, for every κ [ 0 , 1 ] , we have:
0 ι P ( ι , κ ) d κ < 1 .
(iii) 
g ( κ , x ( κ ) ) g ( κ , y ( κ ) ) 1 α x ( κ ) y ( κ ) , for all x , y X and α > e 2 τ ι .
Then, the Equation (34) has a unique solution.
Proof. 
Let X = C ( [ 0 , 1 ] , R ) endowed with the Bielecki norm
D ξ = x B = sup ι [ 0 , 1 ] x ( ι ) e τ ι w i t h τ > 1
and ξ : X 4 [ 1 , ) defined by ξ ( x , y , T x , T y ) = max { x , y , T x , T y } + 2 for any x , y X . It is easy to check that ( X , D ξ ) is a D ξ -complete CRb-MS.
Then, our problem (34) can be resumed to finding an element x * X which is a fixed point for the operator T.
We have the estimation
T x ( ι ) T y ( ι ) 2 0 ι [ P ( ι , κ ) g ( κ , x ( κ ) ) P ( ι , κ ) g ( κ , y ( κ ) ) ] d κ 2 0 ι P ( ι , κ ) [ g ( κ , x ( κ ) ) g ( κ , y ( κ ) ) ] 2 d κ 0 ι P ( ι , κ ) d κ 2 0 ι [ g ( κ , x ( κ ) ) g ( κ , y ( κ ) ) ] 2 d κ 1 α 0 ι x ( κ ) y ( κ ) 2 d κ 0 ι P ( ι , κ ) d κ 2 = e 2 τ ι α e 2 τ ι 0 ι x ( κ ) y ( κ ) 2 e 2 τ ι d κ 0 ι P ( ι , κ ) e τ ι d κ 2 .
Thus, we obtain
T x ( ι ) T y ( ι ) 2 e 2 τ ι α e 2 τ ι 0 ι x ( κ ) y ( κ ) 2 e 2 τ ι d κ 0 ι P ( ι , κ ) e τ ι d κ 2 .
Taking supremum in the inequality (36), we have
[ sup ι [ 0 , 1 ] T x ( ι ) T y ( ι ) e τ ι ] 2 1 α e 2 τ ι [ sup ι [ 0 , 1 ] x ( ι ) y ( ι ) e τ ι ] 2 0 ι sup ι [ 0 , 1 ] P ( ι , κ ) e τ ι d κ 2 .
Using the hypothesis ( i i ) , we have
D ξ ( x , y ) = T x T y B 1 α e τ ι x y B = 1 α e τ ι D ξ ( x , y ) .
It is easy to check that ξ ( x m ( κ ) , x n ( κ ) , x m + k ( κ ) , x n + k ( κ ) ) < + , for all k , m , n n 0 , with n 0 N .
Then, for 0 < δ = 1 α e τ ι < 1 , we have true all the conditions of Theorem 4. In this condition, we can conclude that Equation (34) has a unique solution. □

6. Conclusions

In this paper, we gave an affirmative answer to an open question posed recently by Mlaiki et al. [Adv. Math. Phys., 2020]. As a consequence of our results, we get some known results in the literature. We also gave an application of our results to the existence of a solution of nonlinear fractional differential equations. We believe that our results will have an impact on other researchers in this field.

Author Contributions

Investigation, R.G., A.B., S.B., Z.D.M., N.M. and L.G.; Methodology, R.G., Z.D.M., L.G.; Software, Z.D.M.; Supervision, R.G., Z.D.M., and L.G. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Acknowledgments

The authors are thankful to the Deanship of Scientific Research at Prince Sattam bin Abdulaziz University, Al-Kharj, Kingdom of Saudi Arabia, for supporting this research.

Conflicts of Interest

The authors declare no conflict of interest.

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George, R.; Belhenniche, A.; Benahmed, S.; Mitrović, Z.D.; Mlaiki, N.; Guran, L. On An Open Question in Controlled Rectangular b-Metric Spaces. Mathematics 2020, 8, 2239. https://doi.org/10.3390/math8122239

AMA Style

George R, Belhenniche A, Benahmed S, Mitrović ZD, Mlaiki N, Guran L. On An Open Question in Controlled Rectangular b-Metric Spaces. Mathematics. 2020; 8(12):2239. https://doi.org/10.3390/math8122239

Chicago/Turabian Style

George, Reny, Abdelkader Belhenniche, Sfya Benahmed, Zoran D. Mitrović, Nabil Mlaiki, and Liliana Guran. 2020. "On An Open Question in Controlled Rectangular b-Metric Spaces" Mathematics 8, no. 12: 2239. https://doi.org/10.3390/math8122239

APA Style

George, R., Belhenniche, A., Benahmed, S., Mitrović, Z. D., Mlaiki, N., & Guran, L. (2020). On An Open Question in Controlled Rectangular b-Metric Spaces. Mathematics, 8(12), 2239. https://doi.org/10.3390/math8122239

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