# θ‐ϕ‐Contraction on α,η‐Complete Rectangular b‐Metric Spaces

Authors: Abdelkarim Kari; Mohamed Rossafi; El Miloudi Marhrani; Mohamed Aamri
Journal: International Journal of Mathematics and Mathematical Sciences
 (2020)

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## Abstract

In this paper, we present some fixed point results for generalizedθ‐ϕ‐contraction in the framework of α,η−compete rectangular b‐metric spaces. Further, we establish some fixed point theorems for this type of mappings defined on such spaces. Our results generalize and improve many of the well-known results. Moreover, to support our main results, we give an illustrative example.


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## Body

## 1. Introduction

The well-known Banach contraction theory is one of the methods used, which states that ifX,d is a complete metric space and T:X⟶X is self-mapping with contraction, then T has a unique fixed point [1].In 2000, Branciari [2] introduced the notion of generalized metric spaces, for example, the triangle inequality is replaced by the inequality dx,y≤dx,u+du,v+dv,y for all pairwise distinct points x, y, u, and v∈X. Since then, several results have been proposed by many mathematicians on such spaces (see [3–8]).The concept of metric space, as an ambient space in fixed point theory, has been generalized in several directions, such as,b−metric spaces [9] and generalized metric spaces.Combining conditions are used for definitions ofb-metric and generalized metric spaces. Roshan et al. [10] announced the notion of rectangular b-metric space.Hussain et al. [11] introduced the concept of α−η−complete rectangular b−metric space and proved certain results of fixed point theory on such spaces.In this paper, we provide some fixed point results for generalizedθ−ϕ−contraction in the framework of α,η−compete rectangular b−metric spaces, and also we give two examples to support our results.


## 2. Preliminaries

Definition 1.
(see[10]). Let X be a nonempty set, s≥1 be a given real number, and let d: X×X⟶0,+∞ be a mapping such that for all x,y∈X and all distinct points u,v∈X, each distinct from x and y:(1)
dx,y=0, if only if x=y(2)
dx,y=dy,x(3)
dx,y≤sdx,u+du,v+dv,yb−rectangular inequality
Then,X,d is called a b−rectangular metric space.Lemma 1.
(see [10]). Let X,d be a rectangular b-metric space.(a)
Suppose that sequencesxn and yn in X are such that xn⟶x and yn⟶y as n⟶∞, with x≠y, xn≠x, and yn≠y for all n∈ℕ. Then, we have(1)1sdx,y≤limn⟶∞infdxn,yn≤limn⟶∞supdxn,yn≤sdx,y.(b)
Ify∈X and xn is a Cauchy sequence in X with xn≠xm for any m,n∈ℕ, m≠n, converging to x≠y, then(2)1sdx,y≤limn⟶∞infdxn,y≤limn⟶∞supdxn,y≤sdx,y,for all x∈X.Zheng et al. [12] introduced a new type of contractions called θ−ϕ-contractions in metric spaces and proved a new fixed point theorems for such mapping.Definition 2.
(see [13]). We denote by Θ the set of functions θ:]0,∞⟶1,∞[, satisfying the following conditions:θ1θ is increasingθ2 for each sequence xn∈]0,∞[,limn⟶∞θxn=1⇔limn⟶∞xn=0θ3θ is continuous on ]0,∞[Definition 3.
(see [12]). We denote by Φ the set of functions ϕ: 1,∞⟶1,∞ satisfying the following conditions:Φ1ϕ: 1,∞⟶1,∞ is nondecreasingΦ2 for each t>1, limn⟶∞ϕnt=1Φ3ϕ is continuous on 1,∞Lemma 2.
(see [12]). If ϕ∈Φ, then ϕ1=1 and φt<t for each t>1.
In 2014, Hussain et al. [14] introduced a weaker notion than the concept of completeness and called it α-completeness for metric spaces.Definition 4.
(see [14]). Let T:X⟶X and α,η :X×X⟶0,+∞. We say that T is a triangular α,η−admissible mapping ifT1αx,y≥1⟹αTx,Ty≥1, x,y∈XT2ηx,y≤1⟹ηTx,Ty≤1, x,y∈XT3αx,y≥1αy,z≥1⟹αx,z≥1 for all x,y,z∈XT4ηx,y≤1ηy,z≤1⟹ηx,z≤1 for all x,y,z∈XDefinition 5.
(see [14]). Let X,d be a b-rectangular metric space and let α,η:X×X⟶0,+∞ be two mappings. The space is said to be as follows:(a)
T is α−continuous mapping on X,d, if for given point x∈X and sequence xn in X, xn⟶x and αxn,xn+1≥1 for all n∈ℕ imply that Txn⟶Tx.(b)
T is η subcontinuous mapping on X,d, if for given point x∈X and sequence xn in X, xn⟶x and ηxn,xn+1≤1 for all n∈ℕ imply that Txn⟶Tx.(c)
T is α,η−continuous mapping on X,d, if for given point x∈X and sequence xn in X, xn⟶x and αxn,xn+1≥1 or ηxn,xn+1≤1 for all n∈ℕ imply that Txn⟶Tx.
The following definitions were given by Hussain et al. [11].Definition 6.
(see [11]). Let dX,d be a rectangular b-metric space and let α,η:X×X⟶0,+∞ be two mappings. The space X is said to be(a)
α−complete, if every Cauchy sequence xn in X with αxn,xn+1≥1 for all n∈ℕ converges in X(b)
η−subcomplete, if every Cauchy sequence xn in X with ηxn,xn+1≤1 for all n∈ℕ converges in X(c)
α,η−complete, if every Cauchy sequence xn in X with αxn,xn+1≥1 or ηxn,xn+1≤1 for all n∈ℕ converges in XDefinition 7.
(see [11]). Let X,d be a rectangular b-metric space and let α,η:X×X⟶0,+∞ be two mappings. The space X is said to be(a)
X,d is α-regular, if xn⟶x, where αxn,xn+1≥1 for all n∈ℕ implies αxn,x≥1 for all n∈ℕ(b)
X,d is η−subregular, if xn⟶x, where ηxn,xn+1≤1 for all n∈ℕ implies ηxn,x≤1 for all n∈ℕ(c)
X,d is α,η-regular, if xn⟶x, where αxn,xn+1≥1 or ηxn,xn+1≤1 for all n∈ℕ implies that αxn,x≥1 or ηxn,x≤1 for all n∈ℕ.


## 3. Main Results

Definition 8.
LetdX,d be a α,η-rectangular b-metric space with parameter s>1 and let T be a self-mapping on X. Suppose that α,η:X×X⟶0,+∞ are two functions. We say that T is an α,η−θ−ϕ− contraction, if for all x,y∈X with αx,y≥1orηx,y≤1 and dTx,Ty>0, we have(3)θs2dTx,Ty≤ϕθβ1dx,y+β2dTx,x+β3dTy,y+β4dy,Tx,where θ∈Θ,ϕ∈Φ, βi≥0 for i∈1,2,3,4, ∑i=0i=4βi≤1, and β3<1/s.Definition 9.
LetX,d be a α,η-complete rectangular b−metric space and T:X⟶X be a mapping.(1)
T is said to be a θ−ϕ−Kannan-type contraction if there exist θ∈Θ and ϕ∈Φ with αx,y≥1 or ηx,y≤1 for any x,y∈X, Tx≠Ty, we have(4)θs2dTx,Ty≤ϕθdTx,x+dTy,y2s.(2)
T is said to be a θ−ϕ−Reich-type contraction if there exist θ∈Θ and ϕ∈Φ with αx,y≥1 or ηx,y≤1 for any x,y∈X, Tx≠Ty, we have(5)θs2dTx,Ty≤ϕθdx,y+dTx,x+dTy,y3s.(3)
T is said to be a Kannan-type mapping, that is, if there exists α∈]0,1/2s[ with αx,y≥1 or ηx,y≤1 for any x,y∈X, Tx≠Ty, we have(6)s2dTx,Ty≤αdTx,x+dy,Ty.(4)
T is said to be a Reich-type mapping, that is, if there exists λ∈]0,1/3s[ with αx,y≥1 or ηx,y≤1 for any x,y∈X, Tx≠Ty, we have(7)s2dTx,Ty≤λdx,y+dTx,x+dTy,y.Theorem 1.
LetX,d be a α,η−complete rectangular b-metric and let α,η:X×X⟶0,+∞ be two functions. Let T:X×X⟶X be a self-mapping satisfying the following conditions:(i)
T is a triangular α,η−admissible mapping(ii)
T is an α,η−θ−ϕ−contraction(iii)
There existsx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1(iv)
T is a α,η−continuous.
ThenT has a fixed point. Moreover, T has a unique fixed point when αx,y≥1 or ηx,y≤1 for all x,y∈X.Proof.
Letx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1.
Define a sequencexn by xn=Tnx0=Txn−1. Since T is a triangular α,η−admissible mapping, then αx0,x1=αx0,Tx0≥1⟹αTx0,Tx1≥1=αx1,x2 or ηx0,x1=ηx0,Tx0≤1⟹αTx0,Tx1≤1=αx1,x2.
Continuing this process, we haveαxn−1,xn≥1 or ηxn−1,xn≤1, for all n∈ℕ. By T3 and T4, one has(8)αxm,xn≥1orηxm,xn≤1,∀m,n∈ℕ,m≠n.
Suppose that there existsn0∈ℕ such that xn0=Txn0. Then, xn0 is a fixed point of T and the proof is finished. Hence, we assume that xn≠Txn, i.e., dxn−1,xn>0 for all n∈ℕ. We have(9)xn≠xm,∀m,n∈ℕ,m≠n.
Indeed, suppose thatxn=xm for some m=n+k>n, so we have(10)xn+1=Txn=Txm=xm+1.
Denotedm=dxm,xm+1. Then, (3) and Lemma 2 imply that(11)θdn=θdm≤θs2dm=θs2dTxm−1,Txm≤ϕθβ1dm−1+β2dm−1+β3dm<θβ1dm−1+β2dm−1+β3dm.
Asθ is increasing, so(12)dn=dm<β1dm−1+β2dm−1+β3dm.
Hence,(13)dm<β1+β21−β3dm−1.
Since(14)β1+β2+β3≤1.
Thus(15)dm<dm−1.
Continuing this process, we can prove thatdn=dm<dn, which is a contradiction. Thus, in the following, we can assume that (8) and (9) hold.
We shall prove that(16)limn⟶+∞dxn,xn+1=0andlimn⟶+∞dnn,xn+2=0.
SinceTis α,η−θ−ϕ−contraction, we get(17)θdn=θdTxn−1,Txn≤θs2dTxn−1,Txn≤ϕθβ1dn−1+β2dn−1+β3dn<θβ1dn−1+β2dn−1+β3dn.
Sinceθ is increasing, we deduce that dn<β1dn−1+β2dn−1+β3dn, and thus(18)dn<β1+β21−β3dn−1.
Sinceβ1+β2/1−β3≤1, then(19)dn<dn−1.
Therefore,dxn,xn+1 is monotone strictly decreasing sequence of nonnegative real numbers. Consequently, there exists α≥0, such that(20)limn⟶+∞dxn,xn+1=α,which again by (3) and (19) and property of θ, we have(21)1<θdn≤ϕθβ1dn−1+β2dn−1+β3dn≤ϕθdn−1≤ϕ2θdn−2≤…≤ϕnθd0=ϕnθdx0,x1.
By taking the limit asn⟶∞ in (21) and using Φ2, we have(22)1≤limn⟶+∞θdxn,xn+1≤ϕnθdx0,x1.
Then,limn⟶+∞θdxn,xn+1=1, by Θ2, we obtain(23)limn⟶+∞dxn,xn+1=0.
On the other hand,(24)θs2dxn,xn+2≤ϕθβ1dxn−1,xn+1+β2dxn−1,xn+β3dxn+1,xn+2+β4dxn+1,xn≤ϕθsβ1dxn−1,xn+2+sβ1dxn+2,xn+sβ1dxn,xn+1+β2dxn−1,xn+β3dxn+1,xn+2+β4dxn+1,xn,≤ϕθs2β1dxn−1,xn+s2β1dxn,xn+1+s2β1dxn+1,xn+2+sβ1dxn+2,xn+sβ1dxn,xn+1+β2dxn−1,xn+β3dxn+1,xn+2+β4dxn+1,xn.
Byθ1 and Lemma 2, we obtain(25)s2dxn,xn+2<s2β1dxn−1,xn+s2β1dxn,xn+1+s2β1dxn+1,xn+2+sβ1dxn+2,xn+sβ1dxn,xn+1+β2dxn−1,xn+β3dxn+1,xn+2+β4dxn+1,xn.
Therefore,(26)s2−sβ1dxn,xn+2<s2β1dxn−1,xn+s2β1dxn,xn+1+sβ1dxn+1,xn+2+sβ1dxn,xn+1+β2dxn−1,xn+β3dxn+1,xn+2+β4dxn+1,xn.
Taking the limit asn⟶∞ in (28) and using (23), since s2−sβ1>0, we have(27)limn⟶+∞dxn,xn+2=0.
Hence, (16) is proved.
Next, we show thatxn is an Cauchy sequence in X, if otherwise there exists an ε>0 for which we can find sequences of positive integers xnkk and xmkk of xn such that, for all positive integers k, nk>mk>k,(28)dxmk,xnk≥ε,(29)dxmk,xnk−1<ε.
From (30) and using the rectangular inequality, we get(30)ε≤dxmk,xnk≤sdxmk,xmk+1+sdxmk+1,xnk+1+sdxnk+1,xnk.
Taking the upper limit ask⟶∞ in (32) and using (16), we get(31)εslimn⟶+∞supdxmk+1,xnk+1.
Moreover,(32)dxmk,xnk≤sdxmk,xnk−1+sdxnk−1,xnk+1+sdxnk+1,xnk.
Then, from (23) and (31), we get(33)limn⟶+∞supdxmk,xnk≤sε.
On the other hand, we have(34)dxnk,xmk+1≤sdxnk,xnk−1+sdxnk−1,xmk+sdxxmk,xmk+1.
Then, from (23) and 31 we get(35)limn⟶+∞supdxnk,xmk+1≤sε.
Applying (3) with x=xmk and y=xnk, we have(36)θs2dxmk+1,xnk+1=θs2dTxmk,Txnk≤ϕθβ1dxmk,xnk+β2dxmk,Txmk+β3dxnk,Txnk+β4dxnk,Txmk.
Now taking the upper limit ask⟶∞ in (38) and using θ1,θ3,ϕ3, (23), (33), (35), (37), and Lemma 2, we have(37)θs2.εs=θε.s≤θs2limk⟶+∞supdxmk+1,xnk+1≤ϕθβ1limk⟶+∞supdxmk,xnk+β2limk⟶+∞supdxmk,Txmk+β3limk⟶+∞supdxnk,Txnk+β4limk⟶+∞supdxnk,Txmk≤ϕθβ1sε+β4sε<θsεβ1+β4.
Therefore,ε.s<sεβ1+β4 implies s<β1+β4, which is a contradiction.
Consequently,xn is a Cauchy sequence in α−η−complete rectangular b-metric space X,d. Since αxn−1,xn≥1 or ηxn−1,xn≤1, for all n∈ℕ.
This implies that the sequencexn converges to some z∈X. Suppose that z≠Tz. Then, we have all the assumption of Lemma 1 and T is α,η−continuous, then Txn⟶Tz as n⟶∞. Therefore,(38)1sdz,Tz≤limn⟶+∞supdxn,Txn=0.
Hence, we havedz,Tz=0 and so Tz=z. Thus, z is a fixed point of T.


### 3.1. Uniqueness

Letz,u∈ Fix T where z≠u and αz,u≥1 or ηz,u≤1.Applying (3) with x=z and y=u, we have(39)θdz,u=θdTz,Tu≤θs2dTz,Tu≤ϕθβ1dz,u+β2dz,Tz+β3du,Tu+β4dTz,u≤ϕθβ1dz,u+β4dTz,u≤ϕθdz,u.Sinceθ is increasing, therefore(40)dz,u<dz,u,which is a contradiction. Hence, z=u and T have a unique fixed point.Recall that a self-mappingT is said to have the property P, if FixT=FixTn for every n∈ℕ.Theorem 2.
Letα,η: X×X⟶ℝ+ be two functions and let X,d be an α,η−complete rectangular b-metric space. Let T:X⟶X be a mapping satisfying the following conditions:(i)
T is a triangular α,η−admissible mapping(ii)
T is an α,η−θ−ϕ−contraction(iii)
αz,Tz≥1 or ηz,Tz≤1, for all z∈ Fix T
ThenT has the property P.Proof.
Letz∈ Fix Tn for some fixed n>1. As αz,Tz≥1 or ηz,Tz≤1 and T is a triangular α,η-admissible mapping, then(41)αTz,T2z≥1orηT2z,Tz≤1.
Continuing this process, we have(42)αTnz,Tn+1z≥1orηTnz,Tn+1z≤1,for all n∈ℕ. By T3 and T4, we get(43)αTmz,Tnz≥1orηTmz,Tnz≤1,∀m,n∈ℕ,n≠m.
Assume thatz∉ Fix T, i.e., dz,Tz>0.
Applying (3) with x=Tn−1z and y=z, we get(44)dz,Tz=dTnz,Tz=dTTn−1z,Tz≤s2dTTn−1z,Tz,which implies that(45)θdTTn−1z,Tz≤ϕθβ1dTn−1z,z+β2dTn−1z,Tnz+β3dz,Tz+β4dz,Tnz<θβ1dTn−1z,z+β2dTn−1z,Tnz+β3dz,Tz+β4dz,Tnz=θβ1dTn−1z,Tnz+β2dTn−1z,Tnz+β3dz,Tz.
Sinceθ is increasing, therefore,(46)dz,Tz<β1+β21−β3dTn−1z,Tnz≤dTn−1z,Tnz,which is a contradiction as dTn−1z,Tnz⟶0 and dz,Tz>0.
Assuming the following conditions, we prove that Theorem2 still holds for T not necessarily continuous.Theorem 3.
Letα,η: X×X⟶ℝ+ be two functions and let dX,d be an α,η−complete rectangular b-metric space.
LetT:X⟶X be a mapping satisfying the following assertions:(i)
T is triangular α,η−admissible(ii)
T is α,η−θ−ϕ−contraction(iii)
There existsx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1(iv)
X,d is an α,η-regular rectangular b-metric space
ThenT has a fixed point. Moreover, T has a unique fixed point whenever αz,u≥1 or ηz,u≤1 for all z,u∈FixT.Proof.
Letx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1. Similar to the proof of Theorem 3, we can conclude that(47)αxn,xn+1≥1orηxn,xn+1≤1,andxn⟶zasn⟶∞,where xn+1=Txn.
Fromiv,αxn+1,z≥1orηxn+1,z≤1 holds for n∈ℕ.
Suppose thatTz=xn0+1=Txn0 for some n0∈ℕ. From Theorem 3, we know that the members of the sequence xn are distinct. Hence, we have Tz≠Txn, i.e., dTz,Txn>0 for all n>n0. Thus, we can apply (3), to xn and z for all n>n0 to get(48)θdTxn,Tz≤θs2dTxn,Tz≤ϕθβ1dxn,z+β2dxn,Txn+β3dz,Tz+β4dz,Txn.
By Lemma2 and θ1, we obtain(49)dTxn,Tz<β1dxn,z+β2dxn,Txn+β3dz,Tz+β4dz,Txn.
By taking the limit asn⟶∞ in (51), we have(50)limn⟶∞supdTxn,Tz≤β3dz,Tz.
Assume thatz≠Tz. Then, from Lemma 1,(51)1sdz,Tz≤limn⟶∞supdTxn,Tz≤β3dz,Tz.
By assumptionβ3<1/s, we have dz,Tz=0 and so z=Tz. Thus, z is a fixed point of T.
The proof of the uniqueness is similarly to that of Theorem3.
above theorems, if we takeϕt=tk, for some fixed k∈]0,1[, where β1=1andβ2=β3=β4=0. We obtain the following extension of Jamshaid et al. result (Theorem 1) [13] of α,η−complete rectangular b-metric space.Corollary 1.
Letα,η:X×X⟶0,+∞ be two functions and dX,d be an α,η−complete rectangular b-metric space and let T:X⟶X be self-mapping. Suppose for all x,y∈X with αx,y≥1 or ηx,y≤1 and dTx,Ty>0, we have(52)θs2dTx,Ty≤θdx,yk,where θ∈Θ and k∈]0,1[. If the mapping T satisfies the following assertions: point, if(i)
T is a triangular α,η−admissible mapping(ii)
There existsx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1(iii)
T is α,η-continuous or(iv)
is anα,η−regular rectangular b−metric space
ThenT has a fixed point. Moreover, T has a unique fixed point whenever \\αz,u≥1 or ηz,u≤1 for all z,u∈FixT.Proof.
Letϕt=tk, we prove that T is an α,η−θ−ϕ−contraction, Hence, T satisfies in assumption of Theorem 3 or 2 and is the unique fixed point of T.
It follows from Theorem3; we obtain the following fixed point theorems for θ−ϕ-Kannan-type contraction and θ−ϕ-Reich-type contraction.Theorem 4.
LetX,d be a α,η−complete rectangular b-metric space and let α,η:X×X⟶0,+∞ be two functions. Let T:X×X⟶X be a self-mapping satisfying the following conditions:(i)
T is a triangular α,η−admissible mapping(ii)
T is a α,η−θ−ϕ-Kannan-type contraction(iii)
There existsx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1(iv)
Tis a α,η−continuous
ThenT has a fixed point. Moreover, T has a unique fixed point when αx,y≥1 or ηx,y≤1 for all x,y∈X.Proof.
IfT is a α,η−θ−ϕ-Kannan-type contraction, thus there exist θ∈Θ and ϕ∈Φ with αx,y≥1 or ηx,y≤1 for any x,y∈X, Tx≠Ty, we have(53)θs2dTx,Ty≤ϕθdTx,x+dTy,y2s.
Therefore,(54)θs2dTx,Ty≤ϕθβ1dx,y+β2dTx,x+β3dTy,y+β4dTx,y,where β1=β4=0, β2=β3=1/2s, which implies that T is a α,η−θ−ϕ contraction Therefore, from Theorem 2, T has a unique fixed point.Theorem 5.
LetX,d be a α,η−complete rectangular b-metric space and let α,η:X×X⟶0,+∞ be two functions. Let T:X×X⟶X be a self-mapping satisfying the following conditions:(i)
T is a triangular α,η−admissible mapping(ii)
T is a α,η−θ−ϕ−Reich-type contraction(iii)
There existsx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1(iv)
T is a α,η−continuous
ThenT has a fixed point. Moreover, T has a unique fixed point when αx,y≥1 or ηx,y≤1 for all x,y∈X.Proof.
IfT is a α,η−θ−ϕ−Reich-type contraction, thus there exist θ∈Θ and ϕ∈Φ with αx,y≥1 or ηx,y≤1 for any x,y∈X, Tx≠Ty, we have(55)θs2dTx,Ty≤ϕθdx,y+dTx,x+dTy,y3s.
Therefore,(56)θs2dTx,Ty≤ϕθβ1dx,y+β2dTx,x+β3dTy,y+β4dTx,y,where β1=β2=β3=1/3sandβ4=0, which implies that T is a α,η−θ−ϕ contraction. Therefore, from Theorem 3, T has a unique fixed point.Corollary 2.
LetX,d be a α,η−complete rectangular b-metric space and let α,η:X×X⟶0,+∞ be two functions. Let T:X×X⟶X be a self-mapping satisfying the following conditions:(i)
T is a triangular α,η−admissible mapping(ii)
T is a α,η−Kannan-type mapping(iii)
There existsx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1(iv)
Tis a α,η−continuous
ThenT has a fixed point. Moreover, T has a unique fixed point when αx,y≥1 or ηx,y≤1 for all x,y∈X. Then T has a unique fixed point x∈X.Proof.
Letθt=et for all t∈]0,+∞[, and ϕt=t2sα for all t∈1,+∞.
It is obvious thatθ∈Θ and ϕ∈Φ. We prove that T is a θ−ϕ-Kannan-type contraction:(57)θs2dTx,Ty=es2dTx,Ty≤eαdTx,x+dy,Ty=e2sαdTx,x+dy,Ty2s=edTx,x+dy,Ty2s2sα=ϕθdTx,x+dy,Ty2s.
Therefore, from Theorem3, T has a unique fixed point x∈X.Corollary 3.
LetX,d be a α,η−complete rectangular b-metric space and let α,η:X×X⟶0,+∞ be two functions. Let T:X×X⟶X be a self-mapping satisfying the following conditions:(i)
T is a triangular α,η−admissible mapping(ii)
T is a α,η−Reich-type mapping(iii)
There existsx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1(iv)
Tis a α,η−continuous
ThenT has a fixed point. Moreover, T has a unique fixed point when αx,y≥1 or ηx,y≤1 for all x,y∈X.Proof.
Letθt=et for all t∈]0,+∞[, and ϕt=t3sλ for all tt∈1,+∞.
We prove thatT is a θ−ϕ-Reich-type contraction:(58)θs2dTx,Ty=es2dTx,Ty≤e3λsdx,y+dTx,x+dy,Ty/3s=edx,y+dTx,x+dy,Ty3s3λs=ϕθdx,y+dTx,x+dy,Ty3.
Therefore, from Theorem3, T has a unique fixed point x∈X.Example 1.
Consider the setX=1,2,3,4. It is easy to check that the mapping d:X×X⟶0,+∞ given by(i)
dx,y=dy,x, dx,x=0 for all x,y∈X(ii)
d1,2=1/24, d1,3=3, d1,4=4(iii)
d2,3=5, d2,4=6, and d3,4=18
ClearlyX,d is a rectangular b−metric space with parameter s=2.
Define mappingT:X⟶X and α,η:X×X⟶0,+∞ by(59)T1=1,T2=1,T3=1,T4=2.(60)αx,y=x+ymaxx,y,(61)ηx,y=x−ymaxx,y.
Then,T is an α,η−continuous triangular α,η−admissible mapping.
Letθt=t+1, ϕt=2t+1/3, and β1=4/10, β2=1/10, β3=3/10, and β4=2/10. It is obvious that θ∈Θ and ϕ∈Φ. Evidently, αx,y≥1orx,y≤1 and dTx,Ty>0 are when, or x,y=3,4. Consider the following four possibilities:
Forx=1 and y=4, then(62)θs2dT1,T4=16+1=1.4,(63)ϕθβ1d1,4+β2d1,T1+β3d4,T4+β4d4,T1=ϕθ215=2.36.
Then(64)θs2dT1,T4≤ϕθβ1d1,4+β2d1,T1+β3d4,T4+β4d4,T1.
Forx=2 and y=4, then(65)θs2dT2,T4=16+1=1.4,(66)ϕθβ1d2,4+β2d2,T2+β3d4,T4+β4d4,T2=ϕθ3.65=2.27.
Then(67)θs2dT2,T4≤ϕθβ1d2,4+β2d2,T2+β3d4,T4+β4d4,T2.
Forx=3 and y=4, then(68)θs2dT3,T4=16+1=1.4,(69)ϕθβ1d3,4+β2d3,T3+β3d4,T4+β4d4,T3=ϕθ10.1=3.13.
Then(70)θs2dT3,T4≤ϕθβ1d3,4+β2d3,T3+β3d4,T4+β4d4,T3.
Hence,T satisfying the assumption of Theorems 3 and 1 is the unique fixed point of T.Example 2.
LetX=A∪B, where A=1/n:n∈2,3,4,5,6,7 and B=1,2. Define d:X×X⟶0,+∞ as follows:(71)dx,y=dy,xforallx,y∈X,dx,y=0⇔y=x,(72)d12,13=d14,15=d16,17=0.05,d12,14=d13,17=d15,16=0.08,d12,16=d13,14=d15,17=0.4,d12,15=d13,16=d14,17=0.24,d12,17=d13,15=d14,16=0.15,dx,y=x−y2otherwise.
Then,X,d is a rectangular b-metric space with coefficient s = 3. However, we have the following:(1)
X,d is not a metric space, as d1/5,1/7=0.4>0.29=d1/5,1/4+d1/4,1/7(2)
X,d is not a b-metric space for s = 3, as d1/3,1/4=0.4>0.39=3d1/3,1/2+d1/2,1/4(3)
X,d is not a rectangular metric space, as d1/5,1/7=0.4>0.28=d1/5,1/4+d1/4,1/2+d1/2,1/7
Define mappingT:X⟶X and α,η:X×X⟶0,+∞ by(73)Tx=x6,ifx∈1,2,1,ifx∈A,(74)αx,y=sinhx+y,ifx,y∈1,2,1ex+y,otherwise,(75)ηx,y=x+y4,ifx,y∈1,2,1+e−x+y,otherwise.
Then,T is an α,η−continuous triangular α,η−admissible mapping.
Letθt=t+1, ϕt=t+1/2 and taking β1=1 and β2=β3=β3=0. It is obvious that θ∈Θ and ϕ∈Φ. Evidently, αx,y≥1orx,y≤1 and dTx,Ty>0 are when x,y∈1,2 with x≠y.
Consider two cases:Case 1:x>y:(76)θs2dTx,Ty=3x6−y6+1,(77)ϕθdx,y=x−y2+1.On the other hand(78)θs2dTx,Ty−ϕθdx,y=6x6−y6−x−y2=12x6−y66−x56+x4y6+x3y26+x2y36+xy46+y56.Sincex,y∈1,2, then(79)6−x56+x4y6+x3y26+x2y36+xy46+y56≤0,which implies that(80)θs2dTx,Ty≤ϕθdx,y=ϕθβ1dx,y+β2dx,Tx+β3dy,Ty+β4dy,Tx.Case 2:y>x:(81)θs2dTx,Ty=3y6−x6+1,(82)ϕθdx,y=y−x2+1.
Similarly for Case 2, we conclude that(83)θs2dTx,Ty≤ϕθβ1dx,y+β2dx,Tx+β3dy,Ty+β4dy,Tx.
Hence, condition (3) is satisfied. Therefore, T has a unique fixed point z=1.


## 3.1. Uniqueness

Letz,u∈ Fix T where z≠u and αz,u≥1 or ηz,u≤1.Applying (3) with x=z and y=u, we have(39)θdz,u=θdTz,Tu≤θs2dTz,Tu≤ϕθβ1dz,u+β2dz,Tz+β3du,Tu+β4dTz,u≤ϕθβ1dz,u+β4dTz,u≤ϕθdz,u.Sinceθ is increasing, therefore(40)dz,u<dz,u,which is a contradiction. Hence, z=u and T have a unique fixed point.Recall that a self-mappingT is said to have the property P, if FixT=FixTn for every n∈ℕ.Theorem 2.
Letα,η: X×X⟶ℝ+ be two functions and let X,d be an α,η−complete rectangular b-metric space. Let T:X⟶X be a mapping satisfying the following conditions:(i)
T is a triangular α,η−admissible mapping(ii)
T is an α,η−θ−ϕ−contraction(iii)
αz,Tz≥1 or ηz,Tz≤1, for all z∈ Fix T
ThenT has the property P.Proof.
Letz∈ Fix Tn for some fixed n>1. As αz,Tz≥1 or ηz,Tz≤1 and T is a triangular α,η-admissible mapping, then(41)αTz,T2z≥1orηT2z,Tz≤1.
Continuing this process, we have(42)αTnz,Tn+1z≥1orηTnz,Tn+1z≤1,for all n∈ℕ. By T3 and T4, we get(43)αTmz,Tnz≥1orηTmz,Tnz≤1,∀m,n∈ℕ,n≠m.
Assume thatz∉ Fix T, i.e., dz,Tz>0.
Applying (3) with x=Tn−1z and y=z, we get(44)dz,Tz=dTnz,Tz=dTTn−1z,Tz≤s2dTTn−1z,Tz,which implies that(45)θdTTn−1z,Tz≤ϕθβ1dTn−1z,z+β2dTn−1z,Tnz+β3dz,Tz+β4dz,Tnz<θβ1dTn−1z,z+β2dTn−1z,Tnz+β3dz,Tz+β4dz,Tnz=θβ1dTn−1z,Tnz+β2dTn−1z,Tnz+β3dz,Tz.
Sinceθ is increasing, therefore,(46)dz,Tz<β1+β21−β3dTn−1z,Tnz≤dTn−1z,Tnz,which is a contradiction as dTn−1z,Tnz⟶0 and dz,Tz>0.
Assuming the following conditions, we prove that Theorem2 still holds for T not necessarily continuous.Theorem 3.
Letα,η: X×X⟶ℝ+ be two functions and let dX,d be an α,η−complete rectangular b-metric space.
LetT:X⟶X be a mapping satisfying the following assertions:(i)
T is triangular α,η−admissible(ii)
T is α,η−θ−ϕ−contraction(iii)
There existsx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1(iv)
X,d is an α,η-regular rectangular b-metric space
ThenT has a fixed point. Moreover, T has a unique fixed point whenever αz,u≥1 or ηz,u≤1 for all z,u∈FixT.Proof.
Letx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1. Similar to the proof of Theorem 3, we can conclude that(47)αxn,xn+1≥1orηxn,xn+1≤1,andxn⟶zasn⟶∞,where xn+1=Txn.
Fromiv,αxn+1,z≥1orηxn+1,z≤1 holds for n∈ℕ.
Suppose thatTz=xn0+1=Txn0 for some n0∈ℕ. From Theorem 3, we know that the members of the sequence xn are distinct. Hence, we have Tz≠Txn, i.e., dTz,Txn>0 for all n>n0. Thus, we can apply (3), to xn and z for all n>n0 to get(48)θdTxn,Tz≤θs2dTxn,Tz≤ϕθβ1dxn,z+β2dxn,Txn+β3dz,Tz+β4dz,Txn.
By Lemma2 and θ1, we obtain(49)dTxn,Tz<β1dxn,z+β2dxn,Txn+β3dz,Tz+β4dz,Txn.
By taking the limit asn⟶∞ in (51), we have(50)limn⟶∞supdTxn,Tz≤β3dz,Tz.
Assume thatz≠Tz. Then, from Lemma 1,(51)1sdz,Tz≤limn⟶∞supdTxn,Tz≤β3dz,Tz.
By assumptionβ3<1/s, we have dz,Tz=0 and so z=Tz. Thus, z is a fixed point of T.
The proof of the uniqueness is similarly to that of Theorem3.
above theorems, if we takeϕt=tk, for some fixed k∈]0,1[, where β1=1andβ2=β3=β4=0. We obtain the following extension of Jamshaid et al. result (Theorem 1) [13] of α,η−complete rectangular b-metric space.Corollary 1.
Letα,η:X×X⟶0,+∞ be two functions and dX,d be an α,η−complete rectangular b-metric space and let T:X⟶X be self-mapping. Suppose for all x,y∈X with αx,y≥1 or ηx,y≤1 and dTx,Ty>0, we have(52)θs2dTx,Ty≤θdx,yk,where θ∈Θ and k∈]0,1[. If the mapping T satisfies the following assertions: point, if(i)
T is a triangular α,η−admissible mapping(ii)
There existsx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1(iii)
T is α,η-continuous or(iv)
is anα,η−regular rectangular b−metric space
ThenT has a fixed point. Moreover, T has a unique fixed point whenever \\αz,u≥1 or ηz,u≤1 for all z,u∈FixT.Proof.
Letϕt=tk, we prove that T is an α,η−θ−ϕ−contraction, Hence, T satisfies in assumption of Theorem 3 or 2 and is the unique fixed point of T.
It follows from Theorem3; we obtain the following fixed point theorems for θ−ϕ-Kannan-type contraction and θ−ϕ-Reich-type contraction.Theorem 4.
LetX,d be a α,η−complete rectangular b-metric space and let α,η:X×X⟶0,+∞ be two functions. Let T:X×X⟶X be a self-mapping satisfying the following conditions:(i)
T is a triangular α,η−admissible mapping(ii)
T is a α,η−θ−ϕ-Kannan-type contraction(iii)
There existsx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1(iv)
Tis a α,η−continuous
ThenT has a fixed point. Moreover, T has a unique fixed point when αx,y≥1 or ηx,y≤1 for all x,y∈X.Proof.
IfT is a α,η−θ−ϕ-Kannan-type contraction, thus there exist θ∈Θ and ϕ∈Φ with αx,y≥1 or ηx,y≤1 for any x,y∈X, Tx≠Ty, we have(53)θs2dTx,Ty≤ϕθdTx,x+dTy,y2s.
Therefore,(54)θs2dTx,Ty≤ϕθβ1dx,y+β2dTx,x+β3dTy,y+β4dTx,y,where β1=β4=0, β2=β3=1/2s, which implies that T is a α,η−θ−ϕ contraction Therefore, from Theorem 2, T has a unique fixed point.Theorem 5.
LetX,d be a α,η−complete rectangular b-metric space and let α,η:X×X⟶0,+∞ be two functions. Let T:X×X⟶X be a self-mapping satisfying the following conditions:(i)
T is a triangular α,η−admissible mapping(ii)
T is a α,η−θ−ϕ−Reich-type contraction(iii)
There existsx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1(iv)
T is a α,η−continuous
ThenT has a fixed point. Moreover, T has a unique fixed point when αx,y≥1 or ηx,y≤1 for all x,y∈X.Proof.
IfT is a α,η−θ−ϕ−Reich-type contraction, thus there exist θ∈Θ and ϕ∈Φ with αx,y≥1 or ηx,y≤1 for any x,y∈X, Tx≠Ty, we have(55)θs2dTx,Ty≤ϕθdx,y+dTx,x+dTy,y3s.
Therefore,(56)θs2dTx,Ty≤ϕθβ1dx,y+β2dTx,x+β3dTy,y+β4dTx,y,where β1=β2=β3=1/3sandβ4=0, which implies that T is a α,η−θ−ϕ contraction. Therefore, from Theorem 3, T has a unique fixed point.Corollary 2.
LetX,d be a α,η−complete rectangular b-metric space and let α,η:X×X⟶0,+∞ be two functions. Let T:X×X⟶X be a self-mapping satisfying the following conditions:(i)
T is a triangular α,η−admissible mapping(ii)
T is a α,η−Kannan-type mapping(iii)
There existsx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1(iv)
Tis a α,η−continuous
ThenT has a fixed point. Moreover, T has a unique fixed point when αx,y≥1 or ηx,y≤1 for all x,y∈X. Then T has a unique fixed point x∈X.Proof.
Letθt=et for all t∈]0,+∞[, and ϕt=t2sα for all t∈1,+∞.
It is obvious thatθ∈Θ and ϕ∈Φ. We prove that T is a θ−ϕ-Kannan-type contraction:(57)θs2dTx,Ty=es2dTx,Ty≤eαdTx,x+dy,Ty=e2sαdTx,x+dy,Ty2s=edTx,x+dy,Ty2s2sα=ϕθdTx,x+dy,Ty2s.
Therefore, from Theorem3, T has a unique fixed point x∈X.Corollary 3.
LetX,d be a α,η−complete rectangular b-metric space and let α,η:X×X⟶0,+∞ be two functions. Let T:X×X⟶X be a self-mapping satisfying the following conditions:(i)
T is a triangular α,η−admissible mapping(ii)
T is a α,η−Reich-type mapping(iii)
There existsx0∈X such that αx0,Tx0≥1 or ηx0,Tx0≤1(iv)
Tis a α,η−continuous
ThenT has a fixed point. Moreover, T has a unique fixed point when αx,y≥1 or ηx,y≤1 for all x,y∈X.Proof.
Letθt=et for all t∈]0,+∞[, and ϕt=t3sλ for all tt∈1,+∞.
We prove thatT is a θ−ϕ-Reich-type contraction:(58)θs2dTx,Ty=es2dTx,Ty≤e3λsdx,y+dTx,x+dy,Ty/3s=edx,y+dTx,x+dy,Ty3s3λs=ϕθdx,y+dTx,x+dy,Ty3.
Therefore, from Theorem3, T has a unique fixed point x∈X.Example 1.
Consider the setX=1,2,3,4. It is easy to check that the mapping d:X×X⟶0,+∞ given by(i)
dx,y=dy,x, dx,x=0 for all x,y∈X(ii)
d1,2=1/24, d1,3=3, d1,4=4(iii)
d2,3=5, d2,4=6, and d3,4=18
ClearlyX,d is a rectangular b−metric space with parameter s=2.
Define mappingT:X⟶X and α,η:X×X⟶0,+∞ by(59)T1=1,T2=1,T3=1,T4=2.(60)αx,y=x+ymaxx,y,(61)ηx,y=x−ymaxx,y.
Then,T is an α,η−continuous triangular α,η−admissible mapping.
Letθt=t+1, ϕt=2t+1/3, and β1=4/10, β2=1/10, β3=3/10, and β4=2/10. It is obvious that θ∈Θ and ϕ∈Φ. Evidently, αx,y≥1orx,y≤1 and dTx,Ty>0 are when, or x,y=3,4. Consider the following four possibilities:
Forx=1 and y=4, then(62)θs2dT1,T4=16+1=1.4,(63)ϕθβ1d1,4+β2d1,T1+β3d4,T4+β4d4,T1=ϕθ215=2.36.
Then(64)θs2dT1,T4≤ϕθβ1d1,4+β2d1,T1+β3d4,T4+β4d4,T1.
Forx=2 and y=4, then(65)θs2dT2,T4=16+1=1.4,(66)ϕθβ1d2,4+β2d2,T2+β3d4,T4+β4d4,T2=ϕθ3.65=2.27.
Then(67)θs2dT2,T4≤ϕθβ1d2,4+β2d2,T2+β3d4,T4+β4d4,T2.
Forx=3 and y=4, then(68)θs2dT3,T4=16+1=1.4,(69)ϕθβ1d3,4+β2d3,T3+β3d4,T4+β4d4,T3=ϕθ10.1=3.13.
Then(70)θs2dT3,T4≤ϕθβ1d3,4+β2d3,T3+β3d4,T4+β4d4,T3.
Hence,T satisfying the assumption of Theorems 3 and 1 is the unique fixed point of T.Example 2.
LetX=A∪B, where A=1/n:n∈2,3,4,5,6,7 and B=1,2. Define d:X×X⟶0,+∞ as follows:(71)dx,y=dy,xforallx,y∈X,dx,y=0⇔y=x,(72)d12,13=d14,15=d16,17=0.05,d12,14=d13,17=d15,16=0.08,d12,16=d13,14=d15,17=0.4,d12,15=d13,16=d14,17=0.24,d12,17=d13,15=d14,16=0.15,dx,y=x−y2otherwise.
Then,X,d is a rectangular b-metric space with coefficient s = 3. However, we have the following:(1)
X,d is not a metric space, as d1/5,1/7=0.4>0.29=d1/5,1/4+d1/4,1/7(2)
X,d is not a b-metric space for s = 3, as d1/3,1/4=0.4>0.39=3d1/3,1/2+d1/2,1/4(3)
X,d is not a rectangular metric space, as d1/5,1/7=0.4>0.28=d1/5,1/4+d1/4,1/2+d1/2,1/7
Define mappingT:X⟶X and α,η:X×X⟶0,+∞ by(73)Tx=x6,ifx∈1,2,1,ifx∈A,(74)αx,y=sinhx+y,ifx,y∈1,2,1ex+y,otherwise,(75)ηx,y=x+y4,ifx,y∈1,2,1+e−x+y,otherwise.
Then,T is an α,η−continuous triangular α,η−admissible mapping.
Letθt=t+1, ϕt=t+1/2 and taking β1=1 and β2=β3=β3=0. It is obvious that θ∈Θ and ϕ∈Φ. Evidently, αx,y≥1orx,y≤1 and dTx,Ty>0 are when x,y∈1,2 with x≠y.
Consider two cases:Case 1:x>y:(76)θs2dTx,Ty=3x6−y6+1,(77)ϕθdx,y=x−y2+1.On the other hand(78)θs2dTx,Ty−ϕθdx,y=6x6−y6−x−y2=12x6−y66−x56+x4y6+x3y26+x2y36+xy46+y56.Sincex,y∈1,2, then(79)6−x56+x4y6+x3y26+x2y36+xy46+y56≤0,which implies that(80)θs2dTx,Ty≤ϕθdx,y=ϕθβ1dx,y+β2dx,Tx+β3dy,Ty+β4dy,Tx.Case 2:y>x:(81)θs2dTx,Ty=3y6−x6+1,(82)ϕθdx,y=y−x2+1.
Similarly for Case 2, we conclude that(83)θs2dTx,Ty≤ϕθβ1dx,y+β2dx,Tx+β3dy,Ty+β4dy,Tx.
Hence, condition (3) is satisfied. Therefore, T has a unique fixed point z=1.



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