# A New Extension of the Rectangularb-Metric Spaces

Authors: Nabil Mlaiki; Mohamed Hajji; Thabet Abdeljawad
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

In this paper, we introduce a generalization of rectangularb-metric spaces, by changing the rectangular inequality as follows: Dζa,b≤ζa,b,u,vDζa,u+Dζu,v+Dζv,b, for all distinct a,b,u,v∈X. We prove some fixed point theorems, and we use our results to present a nice application in the last section of this paper.


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## Body

## 1. Introduction

It will not be an exaggeration if we say that Banach [1] in 1922 introduced in some way a new area in mathematics, which is called fixed point theory, and that is due to the fact that he proved the existence and uniqueness of a fixed point for self-contractive mappings in metric spaces. Since 1922, mathematicians around the world start to generalize his result either by changing the type of contractions or by generalizing the type of metric spaces (see [2–19]). The question here is what is the point of all these generalizations? Well, in fact, the answer to that is quite simple and that is the larger the class of functions or metrics, the more fields that results can be applied to, such as computer sciences and engineering.In this paper, and inspired by the work done in [20–27], we introduce the notion of controlled rectangular b-metric spaces as a generalization of the rectangular metric spaces and rectangular b-metric spaces. In the second section, we present some preliminaries; in the third section, we prove our main result; in the fourth section, we present an application of our result to polynomial equations; and in the closing section, we give a conclusion with some open questions.


## 2. Preliminaries

The concept of rectangular metric spaces was introduced by Branciari in [28] as follows.Definition 1 [28] (rectangular (or Branciari) metric spaces).
LetX be a nonempty set. A mapping Δ:X2⟶0,∞ is called a rectangular metric on X if for any x,y∈X and all distinct points u,v∈X\x,y; it satisfies the following conditions:
(R1) x=y if and only if Δx,y=0
(R2) Δx,y=Δy,x
(R3) Δx,y≤Δx,u+Δu,v+Δv,y
In this case, the pairX,Δ is called a rectangular metric space.In [29], George et al. introduced the concept of b-rectangular metric spaces as follows.Definition 2 [29] (rectangular b-metric spaces).
LetX be a nonempty set. A mapping B:X2⟶0,∞ is called a rectangular b-metric on X if there exists a constant a≥1 such that for any x,y∈X and all distinct points u,v∈X\x,y; it satisfies the following conditions:
(Rb1) x=y if and only if Bx,y=0
(Rb2) Bx,y=By,x
(Rb3) Bx,y≤aBx,u+Bu,v+Bv,y
In this case, the pairX,B is called a rectangular metric space.As a generalization of rectangularb-metric spaces, Abdeljawad et al. in [30] introduced the concept of extended Branciari b-distance spaces as follows.Definition 3 [30].
For a nonempty setS and a mapping ω:S×S⟶1,∞, we say that a function Bdist:S×S⟶0,∞ is called an extended Branciari b-distance if it satisfies
(i)
Bdistx,y=0 if and only if x=y(ii)
Bdistx,y=Bdisty,x(iii)
Bdistx,y≤ωx,yBdistx,u+Bdistu,v+Bdistv,yfor all x,y∈S and all distinct u,v∈S\x,y. The couple of the symbols S,Bdist denotes an extended Branciari b-distance space (shortly, Bdist-metric space).Now, we present the definition of controlled rectangularb-metric spaces.Definition 4.
LetX be a nonempty set, a function ζ:X4⟶1,∞, and Dζ:X2⟶0,∞. We say that X,Dζ is a controlled rectangular b-metric space if all distinct a,b,u,v∈X; we have
(1)
Dζa,b=0 if and only if a=b(2)
Dζa,b=Dζb,a(3)
Dζa,b≤ζa,b,u,vDζa,u+Dζu,v+Dζv,bNext, we present the topology of controlled rectangularb-metric spaces.Definition 5.
LetX,Dζ be a controlled rectangular b-metric space.
(1)
A sequencean is called Dζ-convergent in a controlled rectangular b-metric space X,Dζ, if there exists a∈X such that limn→∞Dζan,a=Dζa,a(2)
A sequencean is called Dζ-Cauchy if and only if limn,m→∞ρan,am exists and is finite(3)
A controlled rectangularb-metric space X,Dζ is called Dζ-complete if for every Dζ-Cauchy sequence an in X, there exists ν∈X, such that limn→∞Dζan,ν=limn,m→∞ρran,am=Dζν,ν(4)
Leta∈X define an open ball in a controlled rectangular b-metric space X,Dζ by Bζa,η=b∈X∣Dζa,b<ηNotice that rectangular metric spaces and rectangularb-metric spaces are controlled rectangular b-metric spaces, but the converse is not always true. In the following example, we present a controlled rectangular b-metric space which is not a rectangular metric space.Example 1.
LetX=Y∪Z, where Y=1/m∣misanaturalnumber and Z be the set of positive integers. We define Dζ:X2⟶0,∞ by
(1)Dζa,b=0,ifandonlyifa=b,2β,ifa,b∈Y,β2,otherwise,where β is a constant bigger than 0. Now, define ζ:X4⟶1,∞ by ζa,b,u,v=maxa,b,u,v+2β. It is not difficult to check that X,Dζ is a controlled rectangular b-metric space. However, X,Dζ is not a rectangular metric space; for instance, notice that Dζ1/2,1/3=2β>Dζ1/2,2+Dζ2,3+Dζ3,1/3=3β/2.


## 3. Main Results

Theorem 6.
LetX,Dζbe a controlled rectangularb-metric space andTa self-mapping onX.If there exists0<δ<1, such thatDζTx,Ty≤δρx,yandsupm>1limn→∞ζxn,xn+1,xn+2,xm≤1/δ, thenThas a unique fixed point inX.Proof.
Letx0∈X and define the sequence xn as follows: x1=Tx0,x2=T2x0,⋯,xn=Tnx0,⋯. Now, by the hypothesis of the theorem, we have
(2)Dζxn,xn+1≤δDζxn−1,xn≤δ2Dζxn−2,xn−1≤⋯≤δnDζx0,x1.
Note that if we take the limit of the above inequality asn⟶∞, we deduce that Dζxn,xn+1⟶0 as n⟶∞. Now, consider Di=Dζxn+i,xn+i+1. Thus, for all n≥1, we have two cases.Case 1.
Letxn=xm for some integers n≠m. So, if for m>n, we have Tm−nxn=xn. Choose y=xn and p=m−n. Then, Tpy=y; that is, y is a periodic point of T. Thus, Dζy,Ty=DζTpy,Tp+1y≤kpDζy,Ty. Since δ∈0,1, we get Dζy,Ty=0, so y=Ty; that is, y is a fixed point of T.Case 2.
Suppose thatTnx≠Tmx for all integers n≠m. Let n<m be two natural numbers; to show that xn is a Dζ-Cauchy sequence, we need to consider two subcases:Subcase 1.
Assume thatm=n+2p+1. By property (R_3) of the controlled rectangular b-metric spaces, we have
(3)Dζxn,xn+2p+1≤ζxn,xn+1,xn+2,xn+2p+1Dζxn,xn+1+Dζxn+1,xn+2+Dζxn+2,xn+2p+1≤ζxn,xn+1,xn+2,xn+2p+1Dζxn,xn+1+Dζxn,xn+1,xn+2,xn+2p+1Dζxn+1,xn+2+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1Dζxn+2,xn+3+Dζxn+3,xn+4+Dζxn+4,xn+2p+1≤ζxn,xn+1,xn+2,xn+2p+1Dζxn,xn+1+ζxn,xn+1,xn+2,xn+2p+1Dζxn+1,xn+2+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1Dζxn+2,xn+3+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1Dζxn+3,xn+4+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1Dζxn+4,xn+2p+1≤⋯.
Thus,(4)Dζxn,xn+2p+1≤ζxn,xn+1,xn+2,xn+2p+1Dζxn,xn+1+ζxn,xn+1,xn+2,xn+2p+1Dζxn+1,xn+2+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1Dζxn+2,xn+3+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1Dζxn+3,xn+4+⋯+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1⋯ζxn+2p−2,xn+2p−1,xn+2p,xn+2p+1Dζxn+2p,xn+2p+1≤ζxn,xn+1,xn+2,xn+2p+1D0+ζxn,xn+1,xn+2,xn+2p+1D1+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1D2+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1D3+⋯+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1×⋯×⋯ζxn+2p−2,xn+2p−1,xn+2p,xn+2p+1D2p=ζxn,xn+1,xn+2,xn+2p+1D0+D1+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1D2+D3+⋯+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1×⋯×⋯ζxn+2p−2,xn+2p−1,xn+2p,xn+2p+1D2p−1+D2p.
Therefore,(5)Dζxn,xn+2p+1≤ζxn,xn+1,xn+2,xn+2p+1δn+δn+1Dζx0,x1+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1δn+2+δn+3Dζx0,x1+⋯+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1×⋯×⋯ζxn+2p−2,xn+2p−1,xn+2p,xn+2p+1δn+2p−2+δn+2p−1Dζx0,x1≤ζxn,xn+1,xn+2,xn+2p+1δn+δn+1+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1δn+2+δn+3+⋯+ζxn,xn+1,xn+2,xn+2p+1ζxn+2,xn+3,xn+4,xn+2p+1×⋯×⋯ζxn+2p−2,xn+2p−1,xn+2p,xn+2p+1δn+2p−2+δn+2p−1Dζx0,x1=∑l=0p−1∏i=0lζxn+2i,xn+2i+1,xn+2i+2,xn+2p+1δn+2l+δn+2l+1Dζx0,x1=∑l=0p−1∏i=0lζxn+2i,xn+2i+1,xn+2i+2,xn+2p+11+δδn+2lDζx0,x1.
Now, using the fact thatδ<1, the above inequalities imply the following:
(6)Dζxn,xn+2p+1<∑l=0p−1∏i=0lζxn+2i,xn+2i+1,xn+2i+2,xn+2p+12δn+2lDζx0,x1.
Sincesupm>1limn→∞ζxn,xn+1,xn+2,xm≤1/δ, we deduce
(7)limn,p→∞Dζxn,xn+2p+1<∑l=0∞∏i=0lζxn+2i,xn+2i+1,xn+2i+2,xn+2p+12δn+2lDζx0,x1≤∑l=0∞1δl+12δn+2lDζx0,x1≤∑l=0∞2δn+l−1Dζx0,x1.
Note that the series∑l=0∞2δn+l−1Dζx0,x1 converges by the ratio test, which implies that Dζxn,xn+2p+1 converges as n,p⟶∞.Subcase 2.
m=n+2p. First of all, note that
(8)Dζxn,xn+2≤δDζxn−1,xn+1≤δ2Dζxn−2,xn≤⋯≤δnDζx0,x2,which leads us to conclude that Dζxn,xn+2⟶0 as n⟶∞. Similar to Subcase 1, we have
(9)Dζxn,xn+2p≤ζxn,xn+1,xn+2,xn+2pDζxn,xn+1+Dζxn+1,xn+2+Dζxn+2,xn+2p≤ζxn,xn+1,xn+2,xn+2pDζxn,xn+1+ζxn,xn+1,xn+2,xn+2pDζxn+1,xn+2+ζxn,xn+1,xn+2,xn+2pζxn+2,xn+3,xn+4,xn+2pDζxn+2,xn+3+Dζxn+3,xn+4+Dζxn+4,xn+2p≤ζxn,xn+1,xn+2,xn+2pDζxn,xn+1+ζxn,xn+1,xn+2,xn+2pDζxn+1,xn+2+ζxn,xn+1,xn+2,xn+2pζxn+2,xn+3,xn+4,xn+2pDζxn+2,xn+3+ζxn,xn+1,xn+2,xn+2ζxn+2,xn+3,xn+4,xn+2pDζxn+3,xn+4+ζxn,xn+1,xn+2,xn+2pζxn+2,xn+3,xn+4,xn+2pDζxn+4,xn+2p≤ζxn,xn+1,xn+2,xn+2pD0+ζxn,xn+1,xn+2,xn+2pD1+ζxn,xn+1,xn+2,xn+2pζxn+2,xn+3,xn+4,xn+2pD2+ζxn,xn+1,xn+2,xn+2pζxn+2,xn+3,xn+4,xn+2pD3+⋯+ζxn,xn+1,xn+2,xn+2pζxn+2,xn+3,xn+4,xn+2p×⋯×⋯ζxn+2p−3,xn+2p−2,xn+2p−1,xn+2pD2p+∏i=02p−2ζxn+2i,xn+2i+1,xn+2i+1,xn+2pDζxn+2p−2,xn+2p.
Hence,(10)Dζxn,xn+2p=∑l=0p−1∏i=0lζxn+2i,xn+2i+1,xn+2i+2,xn+2p+1δn+2l+δn+2l+1Dζx0,x1+∏i=02p−2ζxn+2i,xn+2i+1,xn+2i+1,xn+2pDζxn+2p−2,xn+2p=∑l=0p−1∏i=0lζxn+2i,xn+2i+1,xn+2i+2,xn+2p+11+δδn+2lDζx0,x1+∏i=02p−2ζxn+2i,xn+2i+1,xn+2i+1,xn+2pDζxn+2p−2,xn+2p≤∑l=0p−1∏i=0lζxn+2i,xn+2i+1,xn+2i+2,xn+2p+11+δδn+2lDζx0,x1+∏i=02p−2ζxn+2i,xn+2i+1,xn+2i+1,xn+2pδn+2p−2Dζx0,x2.
Sincesupm>1limn→∞ζxn,xn+1,xn+2,xm≤1/δ, we deduce
(11)limn,p→∞Dζxn,xn+2p≤limn,p→∞∑l=0p−11δl+11+δδn+2lDζx0,x1+δ2p−1δn+2p−2Dζx0,x2=limn,p→∞∑l=0p−11+δδn+l−1Dζx0,x1+δn−1Dζx0,x2≤∑m=0∞1+δδmDζx0,x1+δmDζx0,x2.
By using the ratio test, it is not difficult to see that the series(12)∑m=0∞1+δδmDζx0,x1+δmDζx0,x2,converges. Hence, Dζxn,xn+2p converges as n and p go toward ∞ Thus, by Subcases 1 and 2, we deduce that the sequence xn is a Dζ-Cauchy sequence. Since X,Dζ is a Dζ-complete extended rectangular b-metric space, we deduce that xn converges to some ν∈X. We claim that ν is a fixed point of T. Note that there exists an integer N such that xN=ν. Due to Case 2, Tnx≠ν for all n>N. Similarly, Tnx≠Tν for all n>N. Hence, we are in Case 1, so ν is a fixed point of T.Also, there exists an integerN such that TNx=Tν. Again, necessarily, Tnx≠ν and Tnx≠Tν for all n>N. Thus, Tν=ν. Therefore, we may assume that for all n, we have xn∈ν,Tν.(13)Dζν,Tν≤ζν,Tν,xn,xn+1Dζν,xn+Dζxn,xn+1+Dζxn+1,Tν≤Dζν,Tν,xn,xn+1Dζν,xn+Dζxn,xn+1+DζTxn,Tν≤ζν,Tν,xn,xn+1Dζν,xn+Dζxn,xn+1+δDζxn,ν.Now, taking the limit asn⟶∞, we deduce that Dζν,Tν=0; that is, Tν=ν and ν is a fixed point of T as desired.Finally, to show uniqueness assume, there exist two fixed points ofT say ν and μ such that ν≠μ. By the contractive property of T, we have
(14)Dζν,μ=DζTν,Tμ≤δDζν,μ<Dζν,μ,which leads us to a contradiction. Thus, T has a unique fixed point as required.Theorem 7.
LetX,Dζbe a complete extended rectangularb-metric space andTa self-mapping onXsatisfying the following condition; for alla,b∈X, there exists0<δ<1/2such that(15)DζTa,Tb≤δDζa,Ta+Dζb,Tb.
Also, if(16)supm>1limn→∞ζxn,xn+1,xn+2,xm≤1/δ,and for allu,v∈X, we have(17)limn→∞ζu,v,xn,xn+1≤1;then, T has a unique fixed point in X.Proof.
Letx0∈X and define the sequence xn as follows:
(18)x1=Tx0,x2=Tx1=T2x0,⋯,xn=Txn−1=Tnx0,⋯.
First of all, note that for alln≥1, we have
(19)Dζxn,xn+1≤δDζxn−1,xn+Dζxn,xn+1⇒1−kDζxn,xn+1≤δρxn−1,xn⇒Dζxn,xn+1≤δ1−δDζxn−1,xn.
Since0<δ<1/2, one can easily deduce that 0<δ/1−δ<1. So, let μ=δ/1−δ. Hence,
(20)Dζxn,xn+1≤μDζxn−1,xn≤μ2Dζxn−2,xn−1≤⋯≤μnDζx0,x1.
Therefore,(21)Dζxn,xn+1⟶0asn⟶∞.
Also, for alln≥1, we have
(22)Dζxn,xn+2≤δDζxn−1,xn+Dζxn+1,xn+2.
Thus, by using the fact thatDζxn,xn+1⟶0 as n⟶∞, we deduce that
(23)Dζxn,xn+2⟶0asn⟶∞.
Now, similar to the proof of Cases1 and 2 of Theorem 6, we deduce that the sequence xn is a Dζ-Cauchy sequence. Since X,Dζ is a Dζ-complete extended rectangular b-metric space, we conclude that xn converges to some ν∈X. Using the argument in the proof of Theorem 6, we may assume that for all n≥1, we have xn∈ν,Tν. Thus,
(24)Dζν,Tν≤ζν,Tν,xn,xn+1Dζν,xn+Dζxn,xn+1+Dζxn+1,Tν≤ζν,Tν,xn,xn+1Dζν,xn+Dζxn,xn+1+DζTxn,Tν≤ζν,Tν,xn,xn+1Dζν,xn+Dζxn,xn+1+δDζxn,Txn+δDζν,Tν.
Taking the limit of the above inequalities, we get(25)Dζν,Tν≤0+0+0+δDζν,Tν<Dζν,Tν.
Thus,Dζν,Tν=0 which implies that Tν=ν, and hence, ν is a fixed point of T. Finally, to show uniqueness, assume there exist two fixed points of T say ν and μ such that ν≠μ. By the contractive property of T, we have
(26)Dζν,μ=DζTν,Tμ≤δDζν,μ<Dζν,μ,which leads us to a contradiction. Thus, T has a unique fixed point as required.


## 4. Application

In closing, we present the following application for our results.Theorem 8.
For any natural numberκ≥3, the equation(27)sκ+1=κ4−1sκ+1+κ4s,has a unique real solution.Proof.
First of all, note that if∣s∣>1, Equation (3.1) does not have a solution. So, let X=−1,1, and for all s,r∈X, let Dζs,r=∣s−r∣ and ζs,r,u,v=maxs,r,u,v+2. It is not difficult to see that X,Dζ is a Dζ-complete controlled rectangular b-metric space. Now, let
(28)Ts=sκ+1κ4−1sκ+κ4.
Notice that sinceκ≥2, we can deduce that κ4≥6. Thus,
(29)DζTs,Tr=sκ+1κ4−1sκ+κ4−rκ+1κ4−1rκ+κ4=sκ−rκκ4−1sκ+κ4κ4−1rκ+κ4≤s−rκ4≤s−r6=16Dζs,r.
Hence,(30)DζTs,Tr≤δDζs,r,δ=16.
On the other hand, notice that for alls0∈X, we have
(31)sn=Tns0≤2κ4.
Thus,(32)supn≥1limi→∞ζsi,si+1,si+2,sn=2κ4≤2<6=1δ.
Finally, note thatT satisfies all the hypothesis of Theorem 6. Therefore, T has a unique fixed point in X, which implies that Equation (3.1) has a unique real solution as desired.Example 2.
The following equation(33)s100+1=99999999s101+100000000s,has a unique real solution.Proof.
The proof is a direct consequence of Theorem6, by taking κ=100.


## 5. Conclusion

In closing, we would like to bring to the readers’ attention to the following open questions:Question 1.
LetX,Dζ be a controlled rectangular b-metric space and T a self-mapping on X. Also, assume that for all distinct s,r,Ts,Tr∈X, there exists δ∈0,1 such that
(34)DζTs,Tr≤δζs,r,Ts,TrDζs,r.
What are the other hypotheses we should add so thatT has a unique fixed point in the whole space X?Question 2.
LetX,Dζ be a controlled rectangular b-metric space and T a self-mapping on X. Also, assume that for all distinct s,r,Ts,Tr∈X, there exists δ∈0,1 such that
(35)DζTs,Tr≤ζs,r,Ts,TrDζs,Ts+Dζr,Tr.What are the other hypotheses we should add so that T has a unique fixed point in the whole space X?



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