# Fixed Point Theorems for Generalizedθ−ϕ−Expansive Mapping in Rectangular Metric Spaces

Authors: Mohamed Rossafi; Abdelkarim Kari; El Miloudi Marhrani; Mohamed Aamri
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

In this paper, we present the notion ofθ−ϕ−expansive mapping in complete rectangular metric spaces and study various fixed point theorems for such mappings. The presented theorems extend, generalize, and improve many existing results in the literature.


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## Body

## 1. Introduction

The problem of fixed points of mapping with an adequate contractive condition has been the focal point of a rigorous research activity. It has an extensive applications in different areas such as nonlinear and adaptive control systems, parametrized estimation problems, fractal image decoding, and convergence of recurrent networks.In 1922, the Banach contraction principle [1] came into existence, which is a very famous theorem in nonlinear analysis and has many useful applications.Several mathematicians generalized the Banach contraction principle in two major directions, one by stating the conditions on the mappingT and second allowing the set X to have a more general structure [2–7].In 1984, Wang et al. [8] presented some interesting work on expansion mappings in metric spaces. Recently, Kumar et al. [9] introduced a new notion of generalized α,ψ−expansive mappings and established some fixed point theorems for such mappings in complete generalized metric spaces.In this paper, inspired by the idea ofθ−ϕ−contraction introduced by Zheng et al. [10] in metric spaces, we presented θ−ϕ−expansive mapping and establish various fixed point theorems for such mappings in complete rectangular metric spaces. Our theorems extend, generalize, and improve many existing results.


## 2. Preliminaries

In this section, we recall some definitions and results that will be used to prove our main results.By an expansion mapping [8] on a metric space X,d, we understand a mapping T:X⟶X satisfying for all x,y∈X:
(1)dTx,Ty≥kdx,y,where k is a real in1,+∞.In 2000 Branciari [2] introduced the concept of rectangular metric spaces as follows:Definition 1 (see [11]).
LetX be a nonempty set and d:X×X⟶ℝ+ be a mapping such that for all x,y∈X and for all distinct points u,v∈X, each of them different from x and y, one has
(i)
dx,y=0 if and only if x=y(2)dx,y=dy,x(ii)
dx,y≤dx,u+du,v+dv,y (the rectangular inequality)
Then,X,d is called a rectangular metric space.Definition 2 (see [12]).
LetT:X⟶X and α,η:X×X⟶0,+∞. We say that T is a triangular α,η-admissible mapping if
(T1) αx,y≥1⟹αTx,Ty≥1,x,y∈X
(T2) ηx,y≤1⟹ηTx,Ty≤1,x,y∈X
(T3) αx,y≥1αy,z≥1⟹αx,z≥1 for all x,y,z∈X
(T4) ηx,y≤1ηy,z≤1⟹ηx,z≤1 for all x,y,z∈XDefinition 3 (see [12]).
LetX,d be a rectangular metric space and let α,η:X×X⟶0,+∞ be two mappings.
(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 isη 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 isα,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
Recently, Hussain et al. [13] gave the following definitions.Definition 4 (see [13]).
LetdX,d be a rectangular 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−sup−complete, 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 5 (see [13]).
LetX,d be a rectangular metric space and let α,η:X×X⟶0,+∞ be two mappings. The space X,d 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
The following definitions were given by Jleli et al. in [14].Definition 6 (see [14]).
LetΘ be the family of all functions θ:0,+∞⟶1,+∞ such that
(θ1) θ is increasing, i.e., for all x,y∈ℝ+ such that x<y,θx<θy
θ2 For each sequence xn∈0,+∞(3)limn→0xn=0,ifandonlyiflimn→∞θxn=1
θ3θ is continuousDefinition 7 (see [14]).
LetX,d be a rectangular metric space and T:X⟶X be a mapping.
T is said to be a θ−contraction if there exist θ∈Θ and k∈0,1 such that for any x,y∈X,(4)dTx,Ty>0⟹θdTx,Ty≤θdx,yk.
Recently, Zheng et al. [10] considered the following family of functions and introduced the notion of θ−ϕ−contraction of metric spaces:Definition 8 (see [10]).
LetΦ be the family of all functions ϕ:1,+∞⟶1,+∞, such that
(θ1) ϕ is increasing
(θ2) For each t∈1,+∞, limn→∞ϕnt=1;
(θ3) ϕ is continuous.Lemma 9 (see [10]).
Ifϕ∈Φ. Then, ϕ1=1 and ϕt<t for all t>1.Definition 10 (see [10]).
LetX,d be a metric space and T:X⟶X be a mapping.
T is said to be a θ−ϕ−contraction if there exist θ∈Θ and ϕ∈Φ such that for any x,y∈X,
(5)dTx,Ty>0⟹θdTx,Ty≤ϕθNx,y,where
(6)Nx,y=maxdx,y,dx,Tx,dy,Ty.
Now, we recall the definition of generalizedα,ψ−expansive mapping.
LetΨ be the family of functions ψ:0,∞⟶0,∞ which satisfy the following:
(i)
ψ is upper semicontinuous(ii)
ψntn∈ℕ converges to 0 as n⟶∞, for all t>0(iii)
ψt<t, for any t>0Definition 11 (see [10]).
LetX,d be a generalized metric space and T:X⟶X be a given mapping. T is said to be a generalized α,ψ−expansive mapping of type I if there exists two functions α:X×X⟶0,∞and ψ∈Ψ such that for all x,y∈X,
(7)ψdTx,Ty≥αx,yMx,y,where
(8)Mx,y=maxdx,y,dx,Tx,dy,Ty.Definition 12 (see [10]).
LetX,d be a generalized metric space and T:X⟶X be a given mapping. T is said to be a generalized α,ψ−expansive mapping of type II if there exists two functions α:X×X⟶0,∞ and ψ∈Ψ such that for all x,y∈X,
(9)ψdTx,Ty≥αx,yNx,y,where
(10)Nx,y=mindx,y,dx,Tx+dy,Ty2.


## 3. Fixed Point Theorem on Rectangular Metric Spaces

We introduce a new notion of generalizedθ−ϕ−expansive mapping in the context of rectangular metric spaces as follows.Definition 13.
LetX,d be a rectangular metric space and T:X⟶X be a given mapping. T is said to be generalized θ−ϕ−expansive mapping if there exists two functions θ∈Θ and ϕ∈Φ such that
(11)Mx,y>0⟹ϕθdTx,Ty≥θMx,y,forallx,y∈X,where
(12)Mx,y=mindx,y,dx,Tx,dy,Ty,dx,Ty.Theorem 14.
LetX,d be a α,η-complete generalized metric space, and T:X⟶X be a bijective generalized θ−ϕ−expansive mapping satisfying the following conditions:
(i)
T−1 is a triangular α,η−admissible mapping(ii)
There existsx0∈X such that αx0,T−1x0≥1 or ηx0,T−1x0≤1(iii)
Tis a α,η−continuous
Then,T has a fixed point. Moreover, T has a unique fixed point when αz,u≥1 or ηz,u≤1 for all z,u∈FixT.Proof.
Letx0∈X such that αx0,T−1x0≥1 or ηx0,T−1x0≤1. We define the sequence xn in X by xn=Txn+1, for all n∈ℕ.
SinceT−1 is a triangular α,η−admissible mapping, then
(13)αx0,x1=αx0,T−1x0≥1⟹αT−1x0,T−1x1=αx1,x2≥1or
(14)ηx0,x1=ηx0,T−1x0≤1⟹ηT−1x0,T−1x1=ηx1,x2≤1.
Continuing this process, we have(15)αxn−1,xn≥1or
(16)ηxn−1,xn≤1,for all n∈ℕ. By (T3) and (T4), one has.
(17)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.
Step 1: We shall prove(18)limn→+∞dxn,xn+1=0.
Applying inequality (11) with x=xn and y=xn+1, we obtain
(19)θdxn−1,xn=θdTxn,Txn+1>ϕθdTxn,Txn+1≥θMxn,xn+1,where
(20)Mxn,xn+1=mindxn,xn+1,dxn,Txn,dxn+1,Txn+1,dxn,Txn+1=mindxn,xn+1,dxn,xn−1,dxn+1,xn,dxn,xn=mindxn,xn+1,dxn,xn−1.
If for somen,Mxn,xn+1=dxn,xn−1, then the inequality (19), we get
(21)θdxn−1,xn>ϕθdTxn,Txn+1≥θdxn−1,xn,
It is a contradiction. Hence,Mxn,xn+1=dxn,xn+1. Therefore,
(22)ϕθdTxn,Txn+1≥θdxn,xn+1.
Thus,(23)θdxn,xn+1≤ϕθdxn−1,xnforalln∈ℕ.
Continuing this process, we get(24)θdxn,xn+1≤ϕθdxn−1,xn≤⋯≤ϕnθdx0,x1.
Now, by (23) and (2.9), we deduce that
(25)dxn,xn+1<dxn−1,xn.
By the property ofθ and ϕ, it is evident that
(26)limn→+∞dxn,xn+1=0.
Step 2. Now, we shall prove(27)xn≠xm,forallm,n∈ℕ,m≠n.
On the contrary, assume thatxn=xm for some n=m+k>m. Indeed, suppose that xn=xm, so we have
(28)xn=Txn+1=Txm+1=xm.
Denotedn=dxn,xn+1. By inequality (25), we have
(29)dn<dn−1.
Continuing this process, we get(30)dm=dn<dn−1<⋯<dm,
which is a contradiction. Thus, (27) hold.
Step 3. We shall prove(31)limn→+∞dxn,xn+2=0.
Applying inequality (11) with x=xn, y=xn+2, we obtain
(32)θdxn−1,xn+1=θdTxn,Txn+2>ϕθdTxn,Txn+2≥θMxn,xn+2,where
(33)Mxn,xn+2=mindxn,xn+2,dxn,Txn,dxn+2,Txn+2,dxn,Txn+2=mindxn,xn+2,dxn,xn−1,dxn+2,xn+1,dxn,xn+1=mindxn,xn+2,dxn+1,xn+2.
Takean=dxn,xn+2 and bn=dxn+1,xn+2. Thus, by (32), one can write
(34)θan−1=θdxn−1,xn+1=θdTxn,Txn+2>ϕθdTxn,Txn+2≥θMxn,xn+2=θminan,bn.
Therefore,(35)an−1≥minan,bn.
Again, by (25),
(36)bn−1≥bn≥minan,bn,which implies that
(37)minan,bn≤minan−1,bn−1,∀n∈ℕ.
Then, the sequenceminan,bnn∈ℕ is monotone nonincreasing. Thus, there exists λ≥0 such that
(38)limn→∞minan,bn=λ.
Assume thatλ>0. By (26), we have
(39)limn→∞supan=limn→∞supminan,bn=limn→∞minan,bn=λ.
Taking thelimsupn⟶∞ in (32), and using θ3, ϕ3, and Lemma 9, we obtain
(40)θλ=θlimn→∞supan−1≥ϕθlimn→∞supan≥θlimn→∞minan,bn,which implies that
(41)θλ≥ϕθλ≥θλ.
Therefore,(42)θλ<θλ.
By (θ1), we get
(43)λ<λ.
It is a contradiction, then(44)limn→∞dxn,xn+2=0.
Step 4. We shall prove thatxn is a Cauchy sequence in X,d, that is,
(45)limn,m→∞dxn,xm=0foralln,m∈ℕ,m>n.
If otherwise there exists anε>0 for which we can find sequence of positive integers xnkk and xmkk of xn such that, for all positive integers k, nk>mk>k,(46)dxmk,xnk≥ε,(47)dxmk,xnk−1<ε.
Now, using (26), (44), (46), and the rectangular inequality, we find
(48)ε≤dxmk,xnk≤dxmk,xmk+1+dxmk+1,xmk−1+dxmk−1,xnk<dxmk,xmk+1+dxmk+1,xmk−1+ε.
Then,(49)limk→∞dxmk,xnk=ε.
Now, by rectangular inequality, we have(50)dxmk+1,xnk+1≤dxmk+1,xmk+dxmk,xnk+dxnk,xnk+1,dxmk,xnk≤dxmk,xmk+1+dxmk+1,xnk+1+dxnk+1,xnk,ε≤dxmk,xnk≤dxmk,xnk−1+dxnk−1,xnk+1+dxnk+1,xnk.
Lettingk⟶∞ in the above inequalities, using (26), (46), and (49), we obtain
(51)limk→∞dxmk+1,xnk+1=ε,(52)limk→∞dxmk,xnk−1=ε.
On the other hand,(53)Mxmk,xnk=mindxmk,xnk,dxmk,Txmk,dxnk,Txnk,dxmk,Txnk=mindxmk,xnk,dxmk,xmk−1,dxnk,xnk−1,dxmk,xnk−1.
Lettingk⟶∞ in the above inequalities and using (26), (49), and (52), we get that
(54)limk→∞Mxmk,xnk=ε.
By (51), let B=ε/2>0, from the definition of the limit, there exists n0∈ℕ such that
(55)∣dxmk+1,xnk+1−ε∣≤B∀n≥n0.
This implies that(56)dxmk+1,xnk+1≥B>0∀n≥n0,and by (54), let A=ε/2>0, from the definition of the limit, there exists n1∈ℕ such that
(57)Mxmk,xnk≥A>0∀n≥n1.
Applying (11) with x=xmk and y=xnk, we obtain
(58)ϕθdxmk+1,xnk+1≥θMdxmk,xnk.
Lettingk⟶∞, the above inequality and using (θ3) and ϕ3, we obtain
(59)ϕθlimk→∞dxmk+1,xnk+1≥θlimk→∞Mxmk,xnk.
Therefore,(60)ϕθε≥θε.
It is a contradiction with Lemma9, according to which
(61)ϕθε<θε.
Then,(62)limn,m→∞dxm,xn=0.
It follows thatxn is a Cauchy sequence in X. Since X,d is α,η-complete and
(63)αxn−1,xn≥1orηxn−1,xn≤1,for all n∈ℕ, the there exists z∈X such that
(64)limx→∞dxn,z=0.
Step 5. We show thatdTz,z=0 arguing by contradiction, we assume that
(65)dTz,z>0.
By rectangular inequality, we get(66)dTxn,Tz≤dTxn,xn+dxn,z+dz,Tz,(67)dz,Tz≤dz,xn+dxn,Txn+dTxn,Tz.
By lettingn⟶∞ in inequality (66) and (67), we obtain
(68)dz,Tz≤limn→∞dTxn,Tz≤dz,Tz.
Therefore,(69)limn→∞dTxn,Tz=dz,Tz.
SinceT is α,η-continuous, then Txn⟶Tz i.e limn→∞dTxn,Tz=0. Hence, dTz,z=0, so Tz=z.
Step 6. Uniqueness. Now, suppose thatz,u∈X are two fixed points of T such that u≠z and αz,u≥1 or ηz,u≤1. Therefore, we have
(70)dTz,Tu=dz,u>0.
Applying (11) with x=z and y=u, we have
(71)ϕθdTu,Tz≥θMz,u,where
(72)Mz,u=mindz,u,dz,Tz,du,Tu,dz,Tu=dz,u.
Therefore, we have(73)θdz,u>ϕθdu,z≥θdz,u.
It is a contradiction. Therefore,u=z.Theorem 15.
Letα,η: X×X⟶ℝ+ be two functions and let X,d be a α,η−complete rectangular metric space. Let T:X⟶X be a bijective, mapping satisfying the following conditions:
(i)
T−1 is a triangular α,η-admissible mapping(ii)
T is a generalized α,η−θ−ϕ-expansive mapping(iii)
αz,T−1z≥1 or ηz,T−1z≤1, for all z∈FixT
Then,T has the property P.Proof.
Letz∈FixTn for some fixed n>1. As αz,T−1z≥1 or ηz,T−1z≤1 and T−1 is a triangular α,η-admissible mapping, then
(74)αT−1z,T−2z≥1orηT−2z,T−1z≤1.
Continuing this process, we have(75)αT−nz,T−n−1z≥1orηT−nz,T−n−1z≤1,for all n∈ℕ. By T3 and T4, we get
(76)αT−mz,T−nz≥1orηT−mz,T−nz≤1,∀m,n∈ℕ,n≠m.
SinceT is a bijective mapping, then T−nz=z=Tnz for all n∈ℕ and z∈FixT. Therefore,
(77)αTmz,Tnz≥1orηTmz,Tnz≤1,∀m,n∈ℕ,n≠m.
Assume thatz∉FixT, i.e., dz,Tz>0. Then, we have
(78)dz,Tz=dTnz,Tz=dTTn−1z,Tz.
Applying (11) with x=z and y=Tn−1z, we obtain
(79)ϕθdz,Tz=ϕθdTTn−1z,Tz≥θMTn−1z,z,where
(80)MTn−1z,z=mindz,Tn−1z,dz,Tz,dTn−1z,TTn−1z,dTn−1z,Tz
Lettingn⟶∞ in (80), we obtain
(81)limn→+∞MTn−1z,z=dz,Tz.
Now, using (θ3) and ϕ3, we get
(82)ϕθdz,Tz≥θdz,Tz.
It is a contradiction. Then,z∈FixT.Example 16.
LetX=1,+∞ and d:X×X⟶0,+∞ defined by
(83)dx,y=x−y.
Then,X,d is a metric space and rectangular metric space. Define mapping T:X⟶X and α,η:X×X⟶0,+∞ by
(84)Tx=x4,αx,y=x+ymaxx,y,ηx,y=x−ymaxx,y.
Then,T is an α,η−continuous triangular α,η−admissible mapping and T is a bijective mapping.
Letθt=t+1,ϕt=t+1/2.
Evidently, (αx,y≥1orx,y≤1 and mindx,y,dx,Tx,dy,Ty,dy,Ty>0 are when x≠y≠1.Consider two cases:Case 1.
x>y>1.
(85)dTx,Ty=x4−y4,θdTx,Ty=x4−y4+1=x2+y2x+yx−y+1.Thus,(86)ϕθdTx,Ty=x2+y2x+yx−y4+1.We have(87)θdx,y=x−y.On the other hand,(88)θdx,y−ϕθdTx,Ty=x−y+1−x2+y2x+yx−y4−1=x−y−x2+y2x+yx−y4.Sincex,y∈1,+∞, then
(89)x−y−x2+y2x+yx−y4≤0,which implies that(90)ϕθdTx,Ty≥θdx,y≥θmindx,y,dx,Tx,dy,Ty,dy,Ty.Case 2.
y>x>1.(91)dTx,Ty=y4−x4,θdTx,Ty=y4−x4+1=y2+x2y+xy−x+1.Thus,(92)ϕθdTx,Ty=x2+y2x+yy−x4+1.We have(93)θdx,y=y−x.On the other hand(94)θdx,y−ϕθdTx,Ty=x−y+1−x2+y2x+yy−x4−1=y−x−x2+y2x+yy−x4.Sincex,y∈1,+∞, then
(95)y−x−x2+y2x+yy−x4≤0,which implies that(96)ϕθdTx,Ty≥θdx,y≥θmindx,y,dx,Tx,dy,Ty,dy,Ty.Hence, condition (11) is satisfied. Therefore, T has a unique fixed point z=1.Theorem 17.
Letα,η:X×X⟶ℝ+ be two functions and let dX,d be a α,η−complete rectangular metric space. Let T:X⟶X be a bijective mapping satisfying the following assertions:
(i)
T−1 is triangular α,η−admissible(ii)
T is a generalized α,η−θ−ϕ−expansive mapping(iii)
There existsx0∈X such that αx0,T−1x0≥1orηx0,T−1x0≤1(iv)
X,dis aα,η-regular rectangular metric spaceThen,T 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.
Let x0∈X such that αx0,T−1x0≥1 or ηx0,T−1x0≤1. Similar to the proof of Theorem 14, we can conclude that
(97)αxn,xn+1≥1orηxn,xn+1≤1,xn⟶z,asn⟶∞,and from inequality (69), we have
(98)limn→∞dTxn,Tz=dz,Tz.
From (iv)αxn,z≥1orηxn,z≤1 hold for n∈ℕ.
Suppose thatTz=xn0−1=Txn0 for some n0∈ℕ∗. From Theorem 14, 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 (11), to xn and z for all n>n0 to get
(99)ϕθdTz,Txn≥θMz,xn,∀n≥n0,where
(100)Mz,xn=mindz,xn,dz,Tz,dxn,Txn,dz,Txn=mindz,xn,dz,Tz,dxn,xn−1,dz,xn−1.
Therefore,(101)ϕθdTz,Txn≥θmindz,xn,dz,Tz,dxn,xn−1,dz,xn−1.
By lettingn⟶∞ in inequality (101), we obtain
(102)limn→∞ϕθdTz,Txn≥limn→∞θmindz,xn,dz,Tz,dxn,xn−1,dz,xn−1.
Sinceθ and ϕ are continuous functions and limn→+∞Mz,xn=dz,Tz, we conclude that
(103)ϕθdz,Tz≥θdz,Tz.which implies that
(104)dz,Tz<dz,Tz.
It is a contradiction. Hence,Tz=z.
The proof of the uniqueness is similarly to that of Theorem14.Corollary 18.
Letα,η:X×X⟶0,+∞ be two functions, X,d be a α,η-complete rectangular metric space and T:X⟶X be a bijective mapping. Suppose that for all x,y∈X with αx,y≥1 or ηx,y≤1 and Mx,y>0 we have
(105)θdTx,Tyk≥θMx,y,where
(106)Mx,y=mindx,y,dx,Tz,dy,Ty,dx,Ty.and k∈0,1, θ∈Θ. Then, T has a fixed point, if
(i)
T−1 is a triangular α,η−admissible mapping(ii)
there existsx0∈X such that αx0,T−1x0≥1 or ηx0,T−1x0≤1(iii)
Tis a α,η−continuous or(iv)
X,d is an α,η-regular rectangular metric space
Moreover,T has a unique fixed point when αz,u≥1 or ηz,u≤1 for all z,u∈FixT.


## 4. Fixed Point Theorem on Rectangular Metric Spaces Endowed with a Partial Order

Definition 19 (see [4]).
LetX,d,≺¯ be an ordered rectangular metric space and T:X⟶X be a mapping.
(1)
X,d is said to be O-complete, if every Cauchy xn in X with xn≺¯xn+1 for all n∈ℕ or xn±xn+1 for all n∈ℕ converges in X(2)
X,d is said to be O-regular, if for each sequence xn in X, xn⟶x and xn≻¯xn+1 for all n∈ℕ or xn±xn+1 for all n∈ℕ imply that xn≺¯x or xn±x, respectively(3)
T is said to be O-continuous, if for given x∈X and sequence xn with xn≻¯xn+1 or xn≺¯xn+1 for all n∈ℕ, xn⟶x⇒Txn⟶TxDefinition 20.
LetX,d,≺¯ be an ordered rectangular metric spaces and T:X⟶X be a mapping. We say that T aids to be an ordered θ−ϕ−expansive mapping, if for all x,y∈Xwithx≺¯yorx≻¯y such that
(107)Mx,y>0⟹ϕθdTx,Ty≥θMx,y,where
(108)Mx,y=mindx,y,dx,Tx,dy,Ty,dx,Ty.Theorem 21.
LetX,d,≺¯ be an O-complete partially ordered rectangular metric space. Let T:X⟶X be a bijective self mapping on X satisfying the following assertions:
(i)
T−1 is monotone(ii)
T is an ordered θ−ϕ−expansive mapping(iii)
there existsx0∈X such that x0≺¯T−1x0 or x0≻¯T−1x0(iv)
eitherT is O-continuous or(v)
X,d is O-regular
Then,T has a fixed point. Moreover, T has a unique fixed point whenever z≺¯u or z≻¯u for all z,u∈FixT.Proof.
Define the mappingα:X×X⟶0,+∞ by
(109)αx,y=1,ifx≺¯y,0,otherwise,and the mapping η:X×X⟶0,+∞ by
(110)ηx,y=1,ifx≻¯y,0,otherwise.
Using condition (iii), we have(111)x0≺¯T−1x0⟹αx0,T−1x0≥1orx0≻¯T−1x0⟹ηx0,T−1x0≤1.
Owing to the monotonicity ofT−1, we get
(112)αx,y≥1⟹x≺¯y⟹T−1x°T−1y⟹αT−1x,T−1y≥1orηx,y≤1⟹x≻¯y⟹T−1x±T−1y⟹ηT−1x,T−1y≤1
Therefore,T1 and T2 hold.
On the other hand, if(113)αx,y≥1αx,y≥1⟹x≺¯y,y≺¯zor
(114)ηx,y≤1ηx,y≤1⟹x≻¯y,y≻¯z.
SinceX,d be an O-complete partially ordered rectangular metric space, we conclude that
(115)x≺¯zorx≻¯z⟹αx,z≥1orηx,z≤1
Thus, (T3) and (T4) hold. This shows that T−1 is a triangular α,η−admissible mapping then
(116)αxn,xn+1≥1orηxn,xn+1≤1.
Now, ifT is O-continuous, then
xn≺¯xn+1 or xn≻¯xn+1⟹αxn,xn+1≥1 or ηxn,xn+1≤1 and xn⟶z as n⟶∞ with z∈X⟹Txn⟶Tx.
The existence and uniqueness of a fixed point follows from Theorem14.
Now, suppose that followX,d,≺¯ is O-regular. Let xn be a sequence such that
(117)nn°xornn≻¯x,which implies that
(118)αxn,x≥1orηxn,x≤1,for all n and xn⟶x as n⟶∞. This shows that X,d is α,η−regular. Thus, the existence and uniqueness of fixed point from Theorem 17.Corollary 22.
LetX,d,≺¯ be an O-complete partially ordered rectangular metric spaces.
LetT:X⟶X be a bijective self mapping on X be such that T−1 is a monotone mapping. Suppose that there exists k∈0,1 such that
(119)kdTx,Ty≥dx,y,for all x,y∈X with x≺¯y or x≻¯y. Suppose also the following conditions hold:
(i)
There existsx0∈X such that x0≺¯T−1x0 or x0≻¯T−1x0(ii)
EitherT is O-continuous or(iii)
X is O-regular
Then,T has a fixed point. Moreover, T has a unique fixed point whenever z≺¯u or z≻¯u for all z,u∈FixT.Proof.
By takingϕt=tk and θt=et, obviously that ϕ∈Φ and θ∈Θ and T is a θ−ϕ-expansive mapping. As in the proof of Theorem 21, we get the result.


## 5. Conclusion

In this article, we have proved some fixed point theorems for theθ−ϕ−expansive mapping in the settings of complete rectangular metric spaces. These results are extensions of the related results announced in [6, 7, 9].



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*Source: 6642723-2021-03-12.xml*