# New Fixed Point Theorems forθ‐ϕ-Contraction on Rectangular b-Metric Spaces

Authors: Abdelkarim Kari; Mohamed Rossafi; El Miloudi Marhrani; Mohamed Aamri
Journal: Abstract and Applied Analysis
 (2020)

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## Abstract

The Banach contraction principle is the most celebrated fixed point theorem and has been generalized in various directions. In this paper, inspired by the concept ofθ‐ϕ-contraction in metric spaces, introduced by Zheng et al., we present the notion of θ‐ϕ-contraction in b-rectangular metric spaces and study the existence and uniqueness of a fixed point for the mappings in this space. Our results improve many existing results.


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## Body

## 1. Introduction

The Banach contraction principle is a fundamental result in fixed point theory [1]. Due to its importance and simplicity, several authors have obtained many interesting extensions and generalizations of the Banach contraction principle (see [2–4]).Many generalizations of the concept of metric spaces have been defined, and some fixed point theorems were proven in these spaces. In particular,b-metric spaces were introduced by Bakhtin [5] and Czerwik [6] as a generalization of metric spaces. Many mathematicians worked on this interesting space. For more, the reader can refer to [7–10].In 2000, generalized metric spaces were introduced by Branciari [11], in such a way that triangle inequality is replaced by the quadrilateral inequality dx,y≤dx,z+dz,u+du,y for all pairwise distinct points x,y,z, and u. Any metric space is a generalized metric space, but in general, generalized metric space might not be a metric space. Various fixed point results were established on such spaces (see [12–17] and references therein).Recently, George et al. [10] announced the notion of b-rectangular metric space; many authors initiated and studied many existing fixed point theorems in such spaces (see [18–23]).Very recently, Zheng et al. [24] introduced a new concept of θ‐ϕ-contractions and established some fixed point results for such mappings in complete metric spaces and generalized the results of Brower and Kannan. For more works related to theta-contractions, see [25–27].In this paper, we introduce a new notion of generalizedθ‐ϕ-contractions and establish some fixed point results for such mappings in complete b-rectangular metric spaces. The results presented in the paper extend the corresponding results of Kannan [3] and Reich [4] on b-rectangular metric spaces. Also, we derive some useful corollaries of these results.


## 2. Preliminaries

Definition 1 (see [7]).
LetX be a nonempty set and 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; and (3) dx,y≤sdx,u+du,v+dv,yb−rectangularinequality.ThenX,d is called a b-rectangular metric space.Example 2 (see [19]).
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:(1)dx,y=dy,xforallx,y∈X,dx,y=0⇔y=x,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−y∣2otherwise.ThenX,d is a b-rectangular metric space with coefficient s=3.Lemma 3 (see [20]).
LetX,d be a b-rectangular 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/sdx,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 1/sdx,y≤limn→∞infdxn,y≤limn→∞supdxn,y≤sdx,y, for all x∈XLemma 4.
LetX,d be a b-rectangular metric space and let xn be a sequence in X such that(2)limn→∞dxn,xn+1=limn→∞dxn,xn+2=0.
Ifxn is not a Cauchy sequence, then there exist ε>0 and two sequences mk and nk of positive integers such that(3)ε≤limk→∞infdxmk,xnk≤limk→∞supdxmk,xnk≤sε,ε≤limk→∞infdxnk,xmk+1≤limk→∞supdxnk,xmk+1≤sε,ε≤limk→∞infdxmk,xnk+1≤limk→∞supdxmk,xnk+1≤sε,εs≤limk→∞infdxmk+1,xnk+1≤limk→∞supdxmk+1,xnk+1≤s2ε.Proof.
Ifxn is not a Cauchy sequence, then there exist ε>0 and two sequences mk and nk of positive integers such that(4)mk>nk>k,ε≤dxmk,xnkanddxmk−1,xnk<ε,for all positive integers k. By the b-rectangular inequality, we have(5)ε≤dxmk,xnk≤sdxmk,xmk+1+dxmk+1,xmk−1+dxmk−1,xnk.
Taking the upper and lower limits ask→∞ in (5) and using (2) (4), we obtain(6)ε≤limk→∞infdxmk,xnk≤limk→∞supdxmk,xnk≤sε.
Using theb-rectangular inequality again, we have(7)ε≤dxnk,xmk+1≤sdxnk,xmk−1+dxmk−1,xmk+dxmk,xmk+1.
Taking the upper and lower limits ask→∞ in (7) and using (2) and (4), we obtain(8)ε≤limk→∞infdxnk,xmk+1≤limk→∞supdxnk,xmk+1≤sε.
Using theb-rectangular inequality again, we have(9)ε≤dxmk,xnk+1≤sdxmk,xmk−1+dxmk−1,xnk+dxnk,xnk+1.
Taking the upper and lower limits ask→∞ in (9) and using (2) and (4), we obtain(10)ε≤limk→∞infdxmk,xnk+1≤limk→∞supdxmk,xnk+1≤sε.
Using theb-rectangular inequality again, we have(11)dxmk+1,xnk+1≤sdxmk+1,xmk+dxmk,xnk+dxnk,xnk+1,(12)ε≤dxmk,xnk≤sdxmk,xmk+1+dxmk+1,xnk+1+dxnk+1,xnk.
Taking the upper and lower limits ask→∞ in (11) and (12) and using (2) (6), we obtain(13)εs≤limk→∞infdxmk+1,xnk+1≤limk→∞supdxmk+1,xnk+1≤s2ε.The following definition was given by Ding et al. in [13].Definition 5 (see [13]).
LetΘ be the family of all functions θ:0,+∞→1,+∞ such that θ1θ is increasing; θ2 for each sequence xn⊂0,+∞; limn→0xn=0ifandonlyiflimn→∞θxn=1; and θ3θ is continuous.In [21] Radenovic et al. presented the concept of θ‐ϕ-contractions on metric spaces.Definition 6 (see [21]).
LetΦ be the family of all functions ϕ:1,+∞→1,+∞, such that ϕ1ϕ is nondecreasing; ϕ2 for each t∈1,+∞, limn→∞ϕnt=1; and ϕ3ϕ is continuous.Lemma 7 (see [21]).
Ifϕ∈Φ, then ϕ1=1, and ϕt<t for all t∈1,∞.Definition 8 (see [21]).
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,(14)dTx,Ty>0⇒θdTx,Ty≤ϕθNx,y,where(15)Nx,y=maxdx,y,dx,Tx,dy,Ty.In [27], Zheng et al. proved the following nice result.Theorem 9 (see [21]).
LetX,d be a complete metric space and let T:X→X be a θ‐ϕ-contraction. Then, T has a unique fixed point.


## 3. Main Result

In this paper, using the idea introduced by Zheng et al., we present the conceptθ‐ϕ-contraction in b-rectangular metric spaces, and we prove some fixed point results for such spaces.Definition 10.
LetX,d be a b-rectangular metric space with parameter s>1 space and T:X→X be a mapping.(1)
T is said to be a θ-contraction if there exist θ∈Θ and r∈0,1 such that(16)dTx,Ty>0⇒θs2dTx,Ty≤θMx,yr,where(17)Mx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx.(2)
T is said to be a θ‐ϕ-contraction if there exist θ∈Θ and ϕ∈Φ such that(18)dTx,Ty>0⇒θs2dTx,Ty≤ϕθMx,y,where(19)Mx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx.(3)
T is said to be a θ‐ϕ-Kannan-type contraction if there exist θ∈Θ and ϕ∈Φ such that dTx,Ty>0, we have(20)θs2dTx,Ty≤φθdx,Tx+dy,Ty2.(4)
T is said to be a θ‐ϕ-Reich-type contraction if there exist θ∈Θ and ϕ∈Φ such that dTx,Ty>0, we have(21)θs2dTx,Ty≤ϕθdx,y+dx,Tx+dy,Ty3.Theorem 11.
LetX,d be a complete b-rectangular metric space and let T:X→X be an θ-contraction, i.e., there exist θ∈Θ and r∈0,1 such that for any x,y∈X, we have(22)dTx,Ty>0⇒θs2dTx,Ty≤θMx,yr.Then,T has a unique fixed point.Proof.
Letx0∈X be an arbitrary point in X and define a sequence xn by(23)xn+1=Txn=Tn+1x0,for all n∈ℕ. If there exists n0∈ℕ such that dxn0,xn0+1=0, then the proof is finished.
We can suppose thatdxn,xn+1>0 for all n∈ℕ.
Substitutingx=xn−1 and y=xn, from (22), for all n∈ℕ, we have(24)θdxn,xn+1≤θs2dxn,xn+1≤θMxn−1,xnr,∀n∈ℕ,where(25)Mxn−1,xn=maxdxn−1,xn,dxn−1,xn,dxn,xn+1,dxn+1,xn+1=maxdxn−1,xn,dxn,xn+1.
IfMxn−1,xn=dxn,xn+1, by (24), we have(26)θdxn,xn+1≤θdxn,xn+1r<θdxn,xn+1,which is a contradiction. Hence, Mxn−1,xn=dxn−1,xn. Thus,(27)θdxn,xn+1≤θdxn−1,xnr.
Repeating this step, we conclude that(28)θdxn,xn+1≤θdxn−1,xnr≤θdxn−2,xn−1r2≤⋯≤θdx0,x1rn.
From (27) and using θ1, we get(29)dxn,xn+1<dxn−1,xn.
Therefore,dxn,xn+1n∈ℕ is a monotone strictly decreasing sequence of nonnegative real numbers. Consequently, there exists α≥0 such that(30)limn→∞dxn+1,xn=α.
Now, we claim thatα=0. Arguing by contradiction, we assume that α>0. Since dxn,xn+1n∈ℕ is a nonnegative decreasing sequence, then we have(31)dxn,xn+1≥α∀n∈ℕ.
By property ofθ, we get(32)1<θα≤θdx0,x1rn.
By lettingn→∞ in inequality (32), we obtain(33)1<θα≤1.
It is a contradiction. Therefore,(34)limn→∞dxn,xn+1=0.
Next, we shall prove that(35)limn→∞dxn,xn+2=0.
We assume thatxn≠xm for every n,m∈ℕ, n≠m. Indeed, suppose that xn=xm for some n=m+k with k>0 and using (29), we have(36)dxm,xm+1=dxn,xn+1<dxn−1,xn.
Continuing this process, we can that(37)dxm,xn+1=dxn,xn+1<dxm,xm+1.
It is a contradiction. Therefore,dxn,xm>0 for every n,m∈ℕ, n≠m.
Applying (22) with x=xn−1 and y=xn+1, we have(38)θdxn,xn+2=θdTxn−1,Txn+1≤θs2dTxn−1,Txn+1≤θMxn−1,xn+1r,where(39)Mxn−1,xn+1=maxdxn−1,xn+1,dxn−1,xn,dxn+1,xn+2,dxn+1,xn=maxdxn−1,xn+1,dxn−1,xn.
So, we get(40)θdxn,xn+2≤θmaxdxn−1,xn,dxn−1,xn+1r.
Takean=dxn,xn+2 and bn=dxn,xn+1. Thus, by (40), one can write(41)θan≤θmaxan−1,bn−1r.
Byθ1, we get(42)an<maxan−1,bn−1.
By (36), we have(43)bn≤bn−1≤maxan−1,bn−1.
It implies that(44)maxan,bn≤maxan−1,bn−1,∀n∈ℕ.
Therefore, the sequencemaxan−1,bn−1n∈ℕ is a nonnegative decreasing sequence of real numbers. Thus, there exists λ≥0 such that(45)limn→∞maxan,bn=λ.
By (34) assume that λ>0, we have(46)λ=limn→∞supan=limn→∞supmaxan,bn=limn→∞maxan,bn.
Taking thelimsupn→∞ in (40) and using the property of θ, we obtain(47)θlimn→∞supan≤θlimn→∞maxan−1,bn−1r<θlimn→∞maxan−1,bn−1.
Therefore,(48)θλ<θλ.
Byθ1, we get(49)λ<λ.
It is a contradiction. Therefore,(50)limn→∞dxn,xn+2=0.
Next, we shall prove thatxnn∈ℕ is a Cauchy sequence, i.e., limn→∞dxn,xm=0, for all n,m∈ℕ. Suppose to the contrary. By Lemma 4 Then, there is an ε>0 such that for an integer k there exists two sequences nk and mk such that [i)] ε≤limk→∞infdxmk,xnk≤limk→∞supdxmk,xnk≤sε, [ii)] ε≤limk→∞infdxnk,xmk+1≤limk→∞supdxnk,xmk+1≤sε, [iii)] ε≤limk→∞infdxmk,xnk+1≤limk→∞supdxmk,xnk+1≤sε, and [vi)] ε/s≤limk→∞infdxmk+1,xnk+1≤limk→∞supdxmk+1,xnk+1≤s2ε.
Now, using (i), (ii), and (34), we conclude that(51)limk→∞Mxmk,xnk=limk→∞maxdxmk,xnk,dxmk,xmk+1,dxnk,xnk+1,dxnk,xmk+1≤sε.
Now, applying (22) with x=xmk and y=xnk, we obtain(52)θs2dxmk+1,xnk+1≤θMxmk,xnkr.
Lettingk→∞ the above inequality and using θ3, (51) and (iv), we obtain(53)θεss2=θεs≤θs2limk→∞dxmk+1,xnk+1≤θlimk→∞Mxmk,xnkr.
Therefore,(54)θsε≤θsεr<θsε.
Sinceθ is increasing, we get(55)sε<sε,which is a contradiction. Then,(56)limn,m→∞dxm,xn=0.
Hence,xn is a Cauchy sequence in X. By completeness of X,d, there exists z∈X such that(57)limn→∞dxn,z=0.
Now, we show thatdTz,z=0; arguing by contradiction, we assume that(58)dTz,z>0.
Sincexn→z as n→∞ for all n∈N, then from Lemma 3, we conclude that(59)1sdz,Tz≤limn→∞supdTxn,Tz≤sdz,Tz.
Now, applying (22) with x=xn and y=z, we have(60)θs2dTxn,Tz≤θMxn,zr,∀n∈ℕ,where(61)Mxn,z=maxdxn,z,dxn,Txn,dz,Tz,dz,Txn.
Therefore,(62)θs2dTxn,Tz≤θmaxdxn,z,dxn,Txn,dz,Tz,dz,Txnr.
By lettingn→∞ in inequality (62), using (59) and θ3, we obtain(63)θs21sdz,Tz=θsdz,Tz≤θs2limn→∞dTxn,Tz≤θdz,Tzr<θdz,Tz.
Byθ1, we get(64)sdz,Tz<dz,Tz,
which implies that(65)dz,Tzs−1<0⇒s<1,
which is a contradiction. Hence,Tz=z.
Uniqueness: now, suppose that z,u∈X are two fixed points of T such that u≠z. Therefore, we have(66)dz,u=dTz,Tu>0.
Applying (22) with x=z and y=u, we have(67)θdz,u=θdTu,Tz≤θs2dTu,Tz≤θMz,ur,where(68)Mz,u=maxdz,u,dz,Tz,du,Tu,du,Tz=dz,u.
Therefore, we have(69)θdz,u≤θdz,ur<θdz,u,which implies that(70)dz,u<dz,u,which is a contradiction. Therefore, u=z.Corollary 12.
LetX,d be a complete b-rectangular metric space and T:X→X be the given mapping. Suppose that there exist θ∈Θ and k∈0,1 such that for any x,y∈X, we have(71)dTx,Ty>0⇒θs2dTx,Ty≤θdx,yk.Then,T has a unique fixed point.Example 13.
LetX=A∪B, where A=0,1/2,1/3,1/4 and B=1,2.
Defined:X×X→0,+∞ as follows:(72)dx,y=dy,xforallx,y∈X,dx,y=0⇔y=x,d0,12=d12,13=0,16,d0,13=d13,14=0,04,d0,14=d12,14=0,25,dx,y=∣x−y∣2otherwise.Then,X,d is a b-rectangular metric space with coefficient s=3. However, we have the following: (1) X,d is not a metric space, as d0,1/4=0.25>0.08=d0,1/3+d1/3,1/4. (2) X,d is not a rectangular metric space, as d1/2,1/4=0.25>0.24=d1/2,0+d0,1/3+d1/3,1/4.Define mappingT:X→X by(73)Tx=14ifx∈1,2,13ifx∈A.Evidently,Tx∈X. Let θt=et,r=8/9. It is obvious that θ∈Θ and r∈0,1.Consider the following possibilities:(1)
x∈1,2, y∈A. Then,(74)Tx=14,Ty=13,dTx,Ty=0.04.On the other hand,(75)θs2dTx,Ty=e0.6,Mx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx≥dx,Tx=x−142≥1−142=342.Hence,(76)θ3428/9=e2/3≤θdx,Tx8/9≤θMx,y8/9.On the other hand,(77)e0.6−e2/3≤0,which implies that(78)θs2dTx,Ty≤ϕθdx,Tx8/9≤θmaxdx,y,dx,Tx,dy,Ty,dy,Tx8/9.(2)
Ifx,y∈1,2 or x,y∈A. Then,(79)Tx=Ty=14orTx=Ty=13,thendTx,Ty=0,which implies that(80)θs2dTx,Ty≤θmaxdx,y,dx,Tx,dy,Ty,dy,Tx8/9.Hence, condition (22) is satisfied. Therefore, T has a unique fixed point z=1/3.Theorem 14.
LetX,d be a complete b-rectangular metric space and T:X→X be a mapping. Suppose that there exist θ∈Θ and ϕ∈Φ such that for any x,y∈X,(81)dTx,Ty>0⇒θs2dTx,Ty≤ϕθMx,y,where(82)Mx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx.Then,T has a unique fixed point.Proof.
Letx0∈X be an arbitrary point in X and define a sequence xn by(83)xn+1=Txn=Tn+1x0,for all n∈ℕ. If there exists n0∈ℕ such that dxn0,xn0+1=0, then the proof is finished.
We can suppose thatdxn,xn+1>0 for all n∈ℕ.
Substitutingx=xn−1 and y=xn, from (81), for all n∈ℕ, we have(84)θdxn,xn+1≤θs2dxn,xn+1≤ϕθMxn−1,xn,∀n∈ℕ.
As in the proof of Theorem11, we conclude that(85)Mxn−1,xn=maxdxn−1,xn,dxn,xn+1.
If for somen, Mxn−1,xn=dxn,xn+1, it follows from (84), θ1, and using Lemma 7 we get(86)θdxn,xn+1≤ϕθdxn,xn+1.
It implies that(87)dxn,xn+1<dxn,xn+1,which is a contradiction. Hence, Mxn−1,xn=dxn−1,xn.
Therefore,(88)θdxn,xn+1≤ϕθdxn−1,xn<θdxn−1,xn.
Sinceθ is increasing, so(89)dxn+1,xn<dxn,xn−1.
Therefore,dxn+1,xnn∈ℕ is a monotone strictly decreasing sequence of nonnegative real numbers. Consequently, there exists α≥0 such that(90)limn→∞dxn+1,xn=α.
Now, we claim thatα=0. Arguing by contradiction, we assume that α>0. Since dxn,xn+1n∈ℕ is a nonnegative decreasing sequence, we have(91)dxn,xn+1≥α∀n∈ℕ.
Thus, we have(92)1<θα≤θdxn,xn+1≤ϕθdxn−1,xn≤⋯≤ϕnθdx0,x1.
By lettingn→∞ in inequality (92), using Φ2, we obtain(93)1<θα≤1.
It is a contradiction. Therefore,(94)limn→∞dxn,xn+1=0.
Next, we shall prove that(95)limn→∞dxn,xn+2=0.
We assume thatxn≠xm for every n,m∈ℕ,n≠m. Indeed, suppose that xn=xm for some n=m+k with k>0, so we have xn+1=Txn=Txm=xm+1.
By (89), we get(96)dxm,xm+1=dxn,xn+1<dxn−1,xn.
Continuing this process, we can that(97)dxm,xm+1<dxm,xm+1.
It is a contradiction. Therefore,(98)dxn,xm>0foreveryn,m∈ℕ,n≠m.
Applying (81) with x=xn−1 and y=xn+1, we have(99)θdxn,xn+2≤θs2dxn,xn+2≤ϕθMxn−1,xn+1,where(100)Mxn−1,xn+1=maxdxn−1,xn,dxn−1,xn+1,dxn+1,xn+2,dxn+1,xn=maxdxn−1,xn,dxn−1,xn+1.
Therefore,(101)θxn,xn+2≤ϕθmaxdxn−1,xn,dxn−1,xn+1,
which implies that(102)dxn,xn+2<maxdxn−1,xn,dxn−1,xn+1.
Takean=dxn,xn+2 and bn=dxn,xn+1. By (102), we have(103)an<maxan−1,bn−1.
Again by (89), we get(104)bn≤bn−1≤maxan−1,bn−1.
Therefore,(105)maxan,bn≤maxan−1,bn−1,∀n∈ℕ.
Then, the sequencemaxan,bnn is monotone nonincreasing, so it converges to some β≥0 such that(106)limn→∞maxan,bn=β.
By (94) assume thatβ>0, we have(107)β=limn→∞supan=limn→∞supmaxan,bn=limn→∞maxan,bn.
Taking thelimsupn→∞ in (101) and using θ3, ϕ3, and Lemma 7, we obtain(108)θlimn→∞supan≤ϕθlimsupn→∞maxan−1,bn−1,which implies that(109)θβ≤ϕθβ.
Therefore,(110)β<β,which is a contradiction. Therefore,(111)limn→∞dxn,xn+2=0.Next, we shall prove that xnn∈ℕ is a Cauchy sequence, i.e., limn→∞dxn,xm=0, for all n,m∈ℕ. Suppose to the contrary, then there is an ε>0 such that for an integer k, there exist two sequences nk and mkmk>nk>k, such that [i)] ε≤limk→∞infdxmk,xnk≤limk→∞supdxmk,xnk≤sε, [i)] ε≤limk→∞infdxnk,xmk+1≤limk→∞supdxnk,xmk+1≤sε, [iii)] ε≤limk→∞infdxmk,xnk+1≤limk→∞supdxmk,xnk+1≤sε, and [vi)] ε/s≤limk→∞infdxmk+1,xnk+1≤limk→∞supdxmk+1,xnk+1≤s2ε.
SinceT is a θ‐ϕ-contraction, applying (81) with x=xmk and y=xnk, we have(112)θdxmk+1,xnk+1≤θs2dxmk+1,xnk+1≤φθMxmk,xnk.
As in the proof of Theorem 11, we have(113)Mxmk,xnk=maxdxmk,xnk,dxmk,xmk+1,xnk,xnk+1,dxnk,xmk+1,(114)limk→∞Mxmk,xnk≤sε.
By lettingk→∞ in inequality (112) and using θ1, θ3, Φ3, vi), (114) and Lemma 7, we obtain(115)θs2εs=θsε≤θs2limk→∞dxmk+1,xnk+1≤φθlimk→∞Mxmk,xnk.
Therefore,(116)θsε<θsε.
It is a contradiction. Therefore,(117)limn,m→∞dxm,xn=0.
Hence,xn is a Cauchy sequence in X. By completeness of X,d, there exists z in X such that(118)limx→∞dxn,z=0.
Now, we show thatdTz,z=0 arguing by contradiction, we assume that(119)dTz,z>0.
As in the proof of Theorem 11, we conclude that(120)1sdz,Tz≤limn→∞supdTxn,Tz≤sdz,Tz.
SinceT is a θ‐ϕ-contraction, applying (81) with x=xn and y=z, we conclude that(121)θs2dTxn,Tz≤ϕθMxn,z,where(122)Mxn,z=maxdxn,z,dxn,Txn,dz,Tz,dz,Txn.
This implies that(123)θs2dTxn,Tz≤ϕθmaxdxn,z,dxn,Txn,dz,Tz,dz,Txn.
By lettingn→∞ in inequality (123) and using θ3,Φ3, (120) and Lemma 7, we obtain(124)θs21sdz,Tz=θsdz,Tz≤θs2limn→∞dTxn,Tz≤ϕθlimn→∞maxdxn,z,dxn,Txn,dz,Tz,dz,Txn=ϕθdz,Tz<θdz,Tz.
Asθ is increasing, then we deduced that(125)dz,Tz<sdz,Tz.
Therefore,s<1. It is a contradiction. So, z=Tz. Thus, T has a fixed point.
Uniqueness: let z,u∈ fix T where z≠u. Then, from(126)dTz,Tu>0.
Applying (81) with x=z and y=u, we have(127)θdz,u=θdTu,Tz≤θs2dTu,Tz≤φθMz,u,where(128)Mz,u=maxdz,u,dz,Tz,du,Tu,du,Tz=dz,u.
Therefore, we have(129)θdz,u≤φθdz,u<θdz,u.
This implies thatdz,u<dz,u. It is a contradiction. Therefore, u=z.Following from Theorem14, we obtain the fixed point theorems for the θ‐ϕ-Kannan-type contraction and the θ‐ϕ-Reich-type contraction. The results presented in the paper improve and extend the corresponding results due to the Kannan-type contraction and Reich-type contraction on rectangular b-metric space.Theorem 15.
LetX,d be a complete b-rectangular metric space and T:X→X be a Kannan-type contraction, then T has a unique fix.Proof.
SinceT is a Kannan-type contraction, then there exist θ∈Θ and ϕ∈Φ such that(130)θs2dTx,Ty≤ϕθdTx,x+dTy,y2≤ϕθmaxdx,Tx,dy,Ty≤ϕθmaxdx,y,dx,Tx,dy,Ty,dy,Tx.Therefore,T is θ‐ϕ-contraction. As in the proof of Theorem 14 we conclude that T has a unique fixed point.Theorem 16.
LetX,d be a complete b-rectangular metric space and T:X→X be a Reich-type contraction. Then, T has a unique fixed point.Proof.
SinceT is a Reich-type contraction, then there exist θ∈Θ and ϕ∈Φ such that(131)θs2dTx,Ty≤ϕθdx,y+dTx,x+dTy,y3≤ϕθmaxdx,y,dx,Tx,dy,Ty,dy,Tx.Therefore,T is a θ‐ϕ-contraction. As in the proof of Theorem 14 we conclude that T has a unique fixed point.Corollary 17.
LetX,d be a complete b-rectangular metric space and T:X→X be a Kannan type mapping, i.e., there exists α∈0,1/2 such that for all x,y∈X,(132)dTx,Ty>0⇒s2dTx,Ty≤αdTx,x+dTy,y.Then,T has a unique fixed point.Proof.
Letθt=et for all t∈0,+∞, and ϕt=t2α for all t∈1,+∞. Clearly ϕ∈Φ and θ∈Θ. We prove that T is a θ‐ϕ-Kannan-type contraction. Indeed,(133)θs2dTx,Ty=es2dTx,Ty≤eαdTx,x+dTy,y=e2αdTx,x+dTy,y2=edTx,x+dTy,y22α=ϕθdTx,x+dTy,y2.As in the proof of Theorem15, T has a unique fixed point x∈X.Corollary 18.
LetX,d be a complete b-rectangular metric space and T:X→X be a Reich-type mapping, i.e., there exists λ∈0,1/3 such that for all x,y∈X,(134)dx,y>0⇒s2dTx,Ty≤λdx,y+dTx,x+dTy,y.Then,T has a unique fixed point.Proof.
Letθt=et for all t∈0,+∞ and ϕt=t3λ for all t∈1,+∞.
We prove thatT is a θ‐ϕ-Reich-type contraction. Indeed,(135)θs2dTx,Ty=es2dTx,Ty≤eλdx,y+dTx,x+dTy,y=e3λdx,y+dTx,x+dTy,y3=ϕθdx,y+dTx,x+dTy,y3.
As in the proof of Theorem16, T has a unique fixed point x∈X.Corollary 19.
(Theorem11 Let X,d be a complete b-rectangular metric space and T:X→X be a mapping. Suppose that there exist θ∈Θ and r∈0,1 such that for any x,y∈X,(136)dTx,Ty>0⇒θs2dTx,Ty≤θMx,yr,where(137)Mx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx.Then,T has a unique fixed point.Proof.
By takingϕt=tr, with r∈0,1, obvious ϕ∈Φ, then we conclude that T is a θ‐ϕ-contraction. As in the proof of Theorem 14, T has a unique fixed point.Very recently, Kari et al. in [8] proved the result (Theorem 1) on α,η-complete rectangular b-metric spaces. In this paper, we prove this result in complete rectangular b-metric spaces.Corollary 20.
LetdX,d be a complete b-rectangular metric space with parameter s>1, and let T be self-mapping on X. If for all x,y∈X, we have(138)dTx,Ty>0⇒θs2.dTx,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. Then, T has a unique fixed point.Proof.
We prove thatT is a θ‐ϕ-contraction. Indeed,(139)θs2.dTx,Ty≤ϕθβ1dx,y+β2dTx,x+β3dTy,y+β4dy,Tx≤ϕθβ1+β2+β3+β4maxdx,y,dTx,x,dTy,y,dy,Tx≤ϕθmaxdx,y,dTx,x,dTy,y,dy,Tx.
As in the proof of Theorem14, T has a unique fixed point.Example 21.
LetX=A∪B, where A=1/n:n∈3,4,5,6 and B=1/2,3/2. Define d:X×X→0,+∞ as follows:(140)dx,y=dy,xforallx,y∈X,dx,y=0⇔y=x,d13,14=d14,15=0,1,d13,15=d14,16=0,05,d13,16=d15,16=0,5,dx,y=∣x−y∣2otherwise.
Then,X,d is a b-rectangular metric space with coefficient s=3. However we have the following: [1)]X,dis not a metric space, as d1/5,1/6=0.5>0.15=d1/5,1/4+d1/4,1/6. [2)] X,dis not ab-metric space fors=3, as d1/5,1/6=0.5>0.45=3d1/5,1/4+d1/4,1/6. [3)] X,dis not a rectangular metric space, as d1/5,1/6=0.5>0.2=d1/5,1/3+d1/3,1/4+d1/4,1/6. Define mapping T:X→X by(141)Tx=x+45ifx∈12,32,1ifx∈A.Then,Tx∈1/2,3/2. Let θt=t+1,ϕt=t+1/2. It is obvious that θ∈Θ and ϕ∈Φ.Consider the following possibilities:(142)dTx,Ty=x−y52.Case 1.
x,y∈1/2,3/2, with x≠y and assume that x>y.Therefore,(143)θs2dTx,Ty=35x−y+1,ϕθdx,y=x−y2+1.On the other hand,(144)θs2dTx,Ty−ϕθdx,y=6x−y−5x−y10=110x−y6−5x+y.Sincex,y∈1/2,3/2, then(145)6−56≤6−5x+y≤6−102≤0,which implies that(146)θs2dTx,Ty≤ϕθdx,y≤ϕθmaxdx,y,dx,Tx,dy,Ty,dy,Tx.Case 2.
x∈1/2,3/2,y∈A, or y∈1/2,3/2,x∈A.Therefore,Tx=x+4/5, Ty=1, then dTx,Ty=∣x−1/5∣2.In this case, consider two possibilities:(1)
x≥1: then x≥1. Therefore,(147)dTx,Ty=x−152.So, we have(148)θs2dTx,Ty=35x−1+1,Mx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx≥dy,Ty=1−y2≥1−132=232,ϕθ232=13+1.On the other hand,(149)θs2dTx,Ty−ϕθd1,13=35x−1−13=1159x−14.Sincex∈1,3/2, then(150)1159x−14≤0.This implies that(151)θs2dTx,Ty≤ϕθdy,Ty≤ϕθd1,13≤ϕθdy,Ty≤ϕθmaxdx,y,dx,Tx,dy,Ty,dy,Tx.(2)
x<1: then x<1. Therefore,(152)dTx,Ty=1−x52=1−x52.So, we have(153)θs2dTx,Ty=351−x+1,Mx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx≥232,ϕθ232=13+1.On the other hand,(154)θs2dTx,Ty−ϕθd1,13=351−x−13=1154−9x.Sincex∈1/2,1, then(155)1154−9x≤0.This implies that(156)θs2dTx,Ty≤ϕθdy,Ty≤ϕθmaxdx,y,dx,Tx,dy,Ty,dy,Tx.Hence, condition (81) is satisfied. Therefore, T has a unique fixed point z=1.



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*Source: 8833214-2020-10-20.xml*