# Fixed-Point Theorems forθ−ϕ-Contraction in Generalized Asymmetric Metric Spaces

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

---

## Abstract

In the last few decades, a lot of generalizations of the Banach contraction principle had been introduced. In this paper, we present the notion ofθ-contraction and θ−ϕ-contraction in generalized asymmetric metric spaces to study the existence and uniqueness of the fixed point for them. We will also provide some illustrative examples. Our results improve many existing results.


---

## Body

## 1. Introduction

The problem of the existence of the solution of many mathematical models is equivalent to the existence of a fixed-point problem for a certain map. The study of fixed points, therefore, has a central role in many disciplines of applied sciences. The most essential and key part of the theory of fixed points is the existence of the solution of operator equations satisfying certain conditions, for example, Fredholm integral equations, Volterra integral equations, and two-point boundary-value problems in differential equations, as well as some eigenvalue problems [1–3]. A beautiful blend of analysis, topology, and geometry has laid down the foundation of the theory of fixed points.The Banach contraction principle [4] has become a powerful tool in modern analysis, and it is an important tool for solving existence problems in mathematics and physics. Many authors have established the theory of fixed points particularly in two directions: one by stating the conditions on the mapping T and second, taking the set X as a more general structure [5–8].Many generalizations of the concept of metric spaces are defined, and some fixed-point theorems are proved in these spaces. In particular, asymmetric metric spaces were introduced by Wilson [9] as metric spaces, but without the requirement that the asymmetric metric d has to satisfy dx,y=dy,x.Asymmetric metric spaces have numerous recent applications both in pure and applied mathematics, for example, in rate-independent models for plasticity [10], shape-memory alloys [11], models for material failure [12], and the questions of the existence and uniqueness of Hamilton–Jacobi equations [13].Many mathematicians worked on this interesting space. For more details, refer [14, 15].A. Branciari in [16] initiated the notions of a generalized metric space as a generalization of a metric space, where the triangular inequality of metric spaces was replaced by dx,y≤dx,u+du,v+dv,yquadrilateral inequality. Various fixed-point results were established on such spaces, see [4, 17–21] and references therein.Combining conditions used for definitions of asymmetric metric and generalized metric spaces, Piri et al. [22] announced the notions of the generalized asymmetric metric space.In this paper, we introduce the notion ofθ-contraction and θ−ϕ-contraction and establish some new fixed-point theorems for mappings in the setting of complete generalized asymmetric metric spaces. Our result generalizes, improves, and extends the corresponding results due to Kannan and Reich. Moreover, illustrative examples are presented to support the obtained results.


## 2. Preliminaries

In the following, we recollect some definitions which will be useful in our main results.Definition 1.
(see [16]). Let X 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(ii)
dx,y=dy,x for all distinct points x,y∈X(iii)
dx,y≤dx,u+du,v+dv,y (quadrilateral inequality)
Then,X,d is called a generalized metric space.Definition 2.
(see [22]). Let X 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(ii)
dx,y≤dx,u+du,v+dv,yquadrilateral inequality
Then,X,d is called a generalized asymmetric metric space.Definition 3.
(see [22]). Let X,d be a generalized asymmetric metric space and xnn∈ℕ be a sequence in X, and x∈X. Then,(i)
We say thatxnn∈ℕ forward (backward) converges to x if and only if(1)limn⟶+∞dx,xn=limn⟶+∞dxn,x=0.(ii)
We say thatxnn∈ℕ forward (backward) Cauchy if(2)limn,m⟶+∞dxn,xm=limn,m⟶+∞dxm,xn=0.Example 1.
LetX=A∪B, where A=0,2 and B=1/n,n∈ℕ∗ and d:X×X⟶0,+∞ defined by(3)d0,2=d2,0=1,d1n,0=1n,d0,1n=1,d1n,2=1,d2,1n=1n,d1n,1m=d1m,1n=1,for all n,m∈ℕ∗,n≠m. Then, X,d is a generalized asymmetric metric space. However, we have the following:(1)
X,d is not a metric space as d1/n,0≠d0,1/n, for all n>1(2)
X,d is not a asymmetric metric space as d2,0=1>1/2=d2,1/4+d1/4,0(3)
X,d is not a rectangular metric space as d1/n,2≠d2,1/n, for all n>1Remark 1.
LetX,d be as in Example 1 and 1/nn∈ℕ∗ be a sequence in X. However, we have the following:(i)
limn⟶+∞d1/n,0=0, limn⟶+∞d1/n,2=1 and limn⟶+∞d0,1/n=1, limn⟶+∞d2,1/n=0. Then, the sequence 1/n forward converges to 2 and backward converges to 0, so the limit is not unique.(ii)
limn⟶+∞d1/m,1/n=limn⟶+∞d1/m,1/n=1. So, forward (backward) convergence does not imply forward (backward) Cauchy.Lemma 1.
(see [22]). Let X,d be a generalized asymmetric metric space and xnn be a forward (or backward) Cauchy sequence with pairwise disjoint elements in X. If xnn forward converges to x∈X and backward converges to y∈X, then x = y.Definition 4.
(see [22]). Let X,d be a generalized asymmetric metric space. X is said to be forward (backward) complete if every forward (backward) Cauchy sequence xnn in X forward (backward) converges to x∈X.Definition 5.
(see [22]). Let X,d be a generalized asymmetric metric space. X is said to be complete if X is forward and backward complete.The following definition was given by Jleli and Samet in [23].Definition 6.
(see [23]). Let Θ be the family of all functions θ: 0,+∞⟶1,+∞ such thatθ1θ is increasing,θ2 for each sequence xn∈0,+∞,(4)limn⟶0xn=0if and only iflimn⟶∞θxn=1.θ3θ is continuous.
Recently, Zheng et al. [24] introduced the new type of contractive mappings as follows:Definition 7.
(see [24]). Let Φ be the family of all functions ϕ: 1,+∞⟶1,+∞ such thatϕ1ϕ is increasingϕ2 for each t∈1,+∞, limn⟶+∞ϕnt=1ϕ3ϕ is continuousLemma 2.
(see [24]). If ϕ∈Φ, then ϕ1 = 1, and ϕt<t for all t∈1,+∞.Definition 8.
(see [24]). Let X,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.Theorem 1.
(see [22]). Let X,d be a generalized asymmetric metric space and T:X⟶X be a mapping. Suppose that there exist θ∈Θ and k∈0,1 such that, for any x,y∈X,(7)maxdTx,Ty,dTy,Tx>0⇒θdTx,Ty+dTy,Tx2≤θdx,y+dy,x2k.Then,T has a unique fixed point.


## 3. Main Result

Motivated and inspired by Piri et al. [22] and Zheng et al. [24], we define the notions of θ-contraction and θ−ϕ-contraction on the generalized asymmetric metric space, and we give some results on such space.Definition 9.
LetX,d be a generalized asymmetric metric 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, for any x,y∈X, we have(8)maxdx,y,dy,x>0⇒θdTx,Ty+dTy,Tx2≤θMx,yr,where(9)Mx,y=maxdx,y+dy,x2,dx,Tx+dTx,x2,dy,Ty+dTy,y2.(2)
T is said to be a θ−ϕ-contraction if there exist θ∈Θ and ϕ∈Φ such that, for any x,y∈X, we have(10)maxdx,y,dy,x>0⟹θdTx,Ty+dTy,Tx2≤ϕθMx,y,where(11)Mx,y=maxdx,y+dy,x2,dx,Tx+dTx,x2,dy,Ty+dTy,y2.(3)
T is said to be a θ−ϕ-Kannan-type contraction if there exist θ∈Θ and ϕ∈Φ such that(12)maxdTx,Ty,dTy,Tx>0,for anyx,y∈X, we have(13)θdTx,Ty+dTy,Tx2≤ϕθdTx,x+dx,Tx+dTy,y+dy,Ty4.(4)
T is said to be a θ−ϕ-Reich-type contraction if there exist θ∈Θ and ϕ∈Φ such that(14)maxdTx,Ty,dTy,Tx>0,for anyx,y∈X, we have(15)θdTx,Ty+dTy,Tx2≤ϕθdx,y+dy,x+dTx,x+dx,Tx+dTy,y+dy,Ty6.Theorem 2.
LetX,d be a complete generalized asymmetric metric space, and let T:X⟶X be a θ-contraction, i.e., there exist θ∈Θ and r∈0,1 such that, for any(16)x,y∈X,maxdTx,Ty,dTy,Tx>0⟹θdTx,Ty+dTy,Tx2≤θMx,yr.
Then,T has a unique fixed point.Proof.
Letx0∈X be a fixed point, and define a sequence xn by xn+1=Txn=Tn+1x0, for all n∈ℕ. If there exists n0∈ℕ such that dxn0,xn0+1=0 or dxn0+1,xn0=0, then proof is finished.
We can suppose thatdxn,xn+1>0 and dxn+1,xn>0 for all n∈ℕ; then, we have(17)maxdxn,xn+1,dxn+1,xn>0.
Substitutingx=xn−1 and y=xn, from (16), for all n∈ℕ, we have(18)θdxn,xn+1+dxn+1,xn2≤θMxn−1,xnr,∀n∈ℕ,where(19)Mxn−1,xn=maxdxn−1,xn+dxn,xn−12,dxn−1,xn+dxn,xn−12,dxn,xn+1+dxn+1,xn2=maxdxn−1,xn+dxn,xn−12,dxn,xn+1+dxn+1,xn2.
Now, we setDxn,xm=dxn,xm+dxm,xn.
Therefore,(20)Mxn−1,xn=maxDxn−1,xn2,Dxn,xn+12.
IfMxn−1,xn=Dxn,xn+1/2, then from the assumption of the theorem, we have(21)θDxn,xn+12≤θDxn,xn+12r<θDxn,xn+12,which is a contradiction. Hence, Mxn−1,xn=Dxn−1,xn/2.
Thus,(22)θDxn,xn+12≤θDxn−1,xn2r<θDxn−1,xn2.
Repeating this step, we conclude that(23)θDxn,xn+12≤θDxn−1,xn2r<θDxn−1,xn2<⋯<θDx0,x12rn.
Fromθ1, we get(24)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(25)limn⟶∞Dxn,xn+1=α.
Now, we claim thatα=0. Arguing by the contraction, we assume that α>0. Since Dxn,xn+1n∈ℕ is a nonnegative decreasing sequence, we have(26)Dxn,xn+1≥α,∀n∈ℕ.
From the property ofθ, we get(27)1<θλ2≤θDx0,x12rn.
By lettingn⟶∞ in inequality (27), we obtain(28)1<θλ2≤1.
It is a contradiction. Therefore,(29)limn⟶∞Dxn,xn+1=0.
Next, we shall prove that(30)limn⟶∞dxn,xn+2=0,limn⟶∞dxn+2,xn=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.
So, from the assumption of the theorem, we get(31)θDxm,xm+12=θDxn,xn+12≤θDxn−1,xn2r<θDxn−1,xn2.
Sinceθ is increasing, we have(32)Dxn,xn+1<Dxn−1,xn.
Continuing this process, we can say that(33)Dxm,xm+1<Dxm,xm+1.
It is a contradiction. Therefore,(34)maxdxm,xn,dxn,xm>0,for every n,m∈ℕ, n≠m.
Substitutingx=xn and y=xn+2,(35)maxdxn,xn+2,dxn+2,xn>0.
Applying (16) with x=xn−1 and y=xn+1, we have(36)θdxn,xn+2+dxn+2,xn2≤θMxn−1,xn+1r,where(37)Mxn−1,xn+1=maxDxn−1,xn2,Dxn−1,xn+12,Dxn+1,xn+22=maxDxn−1,xn2,Dxn−1,xn+12.
So, we get(38)θDxn,xn+22≤θmaxDxn−1,xn2,Dxn−1,xn+12r.
Takean=Dxn,xn+2 and bn=Dxn,xn+1. Thus, by (38), one can write(39)θan2≤θmaxan−12,bn−12r.
Byθ1, we get(40)an<maxan−1,bn−1.
By (24), we have(41)bn≤bn−1≤maxan−1,bn−1,which implies that(42)maxan,bn≤maxan−1,bn−1,∀n∈ℕ.
Therefore, the sequencemaxan−1,bn−1n∈ℕ is monotone nonincreasing. Thus, there exists λ≥0 such that(43)limn⟶∞maxan,bn=λ.
By (29), we assume that λ>0; then, we get(44)limn⟶∞supan=limn⟶∞supmaxan,bn=limn⟶∞maxan,bn.
Takinglimsupn⟶∞ in (38) and using the properties of θ3, we obtain(45)θlimn⟶∞supan2<θlimn⟶∞maxan−12,bn−12.
Therefore,(46)θλ2<θλ2.
Byθ1, we get(47)λ<λ.
It is a contradiction. Therefore,(48)limn⟶∞Dxn+2,xn=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. Then, there is ε>0 such that, for an integer k, there exist two sequences nk and mkmk>nk>k such that(49)Dxmk,xnk≥ε,Dxmk−1,xnk<ε.
Now, using (29), (48), (49), and the quadrilateral inequality, we find(50)ε≤Dxmk,xnk≤Dxmk,xmk+1+Dxmk+1,xmk−1+Dxmk−1,xnk≤Dxmk,xmk+1+Dxmk+1,xmk−1+ε.
Then,(51)limk⟶∞Dmk,nk=ε.
Now, by the quadrilateral inequality, we have(52)Dxmk+1,xnk+1≤Dxmk+1,xmk+Dxmk,xnk+Dxnk,xnk+1,Dxmk,xnk≤Dxmk,xmk+1+Dxmk+1,xnk+1+Dxnk+1,xnk.
Lettingk⟶∞ in the above inequalities and using (29), (48), and (51), we obtain(53)limk⟶∞Dxmk+1,xnk+1=ε.
Therefore, by (29) and (51), we get that(54)limk⟶∞Mxmk,xnk=ε2.
By (53), there exists n0∈ℕ such that(55)Dxmk+1,xnk+1=dxmk+1,xnk+1+dxnk+1,xmk+1≥ε2,∀n≥n0.
Therefore,(56)maxdxmk+1,xnk+1,dxnk+1,xmk+1≥ε4,∀n≥n0.
So,(57)maxdTxmk,Txnk,dxnk,Txmk≥ε4,∀n≥n0.
Applying withx=xmk and y=xnk, we have(58)θDxmk+1,xnk+12≤θMxmk,xnkr.
Lettingk⟶∞ in the above inequality and using θ3, (53), and (54), we obtain(59)θlimk⟶∞Dxmk+1,xnk+12≤θlimk⟶∞Mxmk,xnkr.
Therefore,(60)θε2<θε2.
Sinceθ is increasing, we get(61)⇒ε<ε,which is a contradiction. Then,(62)limn,m⟶∞Dxm,xn=0.
Equivalently,(63)limn,m⟶∞dxm,xn=limn,m⟶∞dxn,xm=0.
Hence,xn is a forward and backward Cauchy sequence in X. By completeness of X,d, there exists z,u∈X such that(64)limx⟶∞dxn,z=limx⟶∞du,xn=0.
So, from Lemma1, we get z=u.
Now, we show thatdTz,z=0 = dz,Tz. Arguing by contradiction, we assume that(65)dTz,z>0,dz,Tz>0.
Therefore,(66)maxdTz,z,dz,Tz>0.
Now, by the quadrilateral inequality, we get(67)dTxn,Tz≤dTxn,xn+dxn,z+dz,Tz,(68)dz,Tz≤dz,xn+dxn,Txn+dTxn,Tz.
By lettingn⟶∞ in inequalities (67) and (68), we obtain(69)dz,Tz≤limn⟶∞dTxn,Tz≤dz,Tz.
Therefore,(70)limn⟶∞dTxn,Tz=dz,Tz.
On the contrary,(71)dTz,Txn≤dTz,z+dz,xn+dxn,Txn,(72)dTxn,Tz≤dTxn,xn+dxn,z+dz,Tz.
By lettingn⟶∞ in inequalities (71) and (72), we obtain(73)dTz,z≤limn⟶∞dTz,Txn≤dTz,z.
Therefore,(74)limn⟶∞dTz,Txn=dTz,z.
By (70) and from the definition of the limit, there exists n1∈ℕ such that(75)dTxn,Tz>dz,Tz>0,∀n≥n1.
Similarly, by (74), there exists n2∈ℕ such that(76)dTz,Txn>dTz,z>0,∀n≥n2.
LetN=maxn1,n2; we conclude(77)maxdTz,Txn,dTxn,Tz>0,∀n≥N.
Applying (16) with x=z and y=xn, we have(78)θDTz,Txn2≤θMz,xnr,∀n≥N,where(79)Mz,xn=maxDz,xn2,Dz,Tz2,Dxn,Txn2.
Therefore,(80)θDTz,Txn2≤θmaxDz,xn2,Dz,Tz2,Dxn,Txn2r.
By lettingn⟶∞ in inequality (80) and using θ3, we obtain(81)θlimn⟶∞DTz,Txn2=θDz,Tz2≤θlimn⟶∞maxDz,xn2,Dz,Tz2,Dxn,Txn2r<θlimn⟶∞maxDz,xn2,Dz,Tz2,Dxn,Txn2.
Byθ1, we get(82)Dz,Tz2<limn⟶∞maxDz,xn2,Dz,Tz2,Dxn,Txn2=Dz,Tz2,which is a contradiction. Hence, Tz=z.
Uniqueness: now, suppose thatz,u∈X are two fixed points of T such that u≠z. Therefore, we have(83)dTz,Tu=dz,u>0,dTu,Tz=du,z>0.
Therefore,(84)maxdTu,Tz,dTz,Tu>0.
Applying (16) with x=z and y=u, we have(85)θDTz,Tu2=θDz,u2≤θMz,ur,where(86)Mz,u=maxDz,u2,Dz,Tz2,Du,Tu2=Dz,u2.
Therefore, we have(87)θDz,u2≤θDz,u2r<θDz,u2.
Therefore,(88)Dz,u<Dz,u,is a contradiction. Therefore, u=z.Corollary 1.
(Theorem2; see [22]). Let X,d be a complete generalized asymmetric metric space and T:X⟶X be a given mapping. Suppose that there exist θ∈Θ and k∈0,1 such that, for any x,y∈X, we have maxdTx,Ty,dTy,Tx>0⟹θdTx,Ty+dTy,Tx/2≤θdx,y+dy,x/2k.Then,T has a unique fixed point.Example 2.
ConsiderX=0,1,2,3. Let d:X×X⟶0,+∞ be a mapping defined by the following:(i)
dx,y=0 if and only x=y(ii)
d0,1=d1,0=d2,1=d2,0=d3,0=d2,3=d3,1=1(iii)
d1,2=d0,2=2(iv)
d0,3=3, and d3,2=4(v)
d1,3=2
Clearly,X,d is not an asymmetric metric space. Indeed, d3,2=4>d3,0+d0,2=3.
However, it is a complete generalized asymmetric metric space.
LetT:X⟶X be given by(89)T0=T1=0,T2=1,T3=2.
Supposeθt=t+1 and k=12/13. Therefore, θ∈Θ and k∈0,1. First, observe that maxdTx,Ty,dTy,Tx>0⟺x=0,y=2, x=1,y=2, x=0,y=3, x=1,y=3, or x=2,y=3.
Forx=0,y=2, we have(90)dT0,T2+dT2,T02=d0,1+d1,02=1,M0,2=maxd0,2+d2,02,d0,T0+dT0,02,d2,T2+dT2,22=maxd0,2+d2,02,d0,0+d0,02,d2,1+d1,22=32.
Therefore,(91)θdT0,T2+dT2,T02=2,θmax32,0,32k=2.08.
So,(92)θdT0,T2+dT2,T02≤θM0,2k.
Forx=1,y=2, we have(93)θdT1,T2+dT2,T12=θd0,1+d1,02=2,M1,2=maxd1,2+d2,12,d1,T1+dT1,12,d2,T2+dT2,22=32,which implies that(94)θmax32,1,32k=32+112/13=2.08.
Therefore,(95)θdT1,T2+dT2,T12≤θM1,2k.
Forx=0,y=3, we have(96)θdT0,T3+dT3,T02=θd0,2+d2,02=32+1=2.22,M0,3=maxd0,3+d3,02,d0,T0+dT0,02,d3,T3+dT3,32=max2,0,52=52.
On the contrary,(97)θmax2,0,52k=52+1k=2.39.
Then,(98)θdT0,T3+dT3,T02≤θM0,3k.
Forx=1,y=3, we have(99)θdT1,T3+dT3,T12=θd0,2+d2,02=32+1=2.22,M1,3=maxd1,3+d3,12,d1,T1+dT1,12,d3,T3+dT3,32=max32,1,52,θmax32,1,52k=2.39.
Therefore,(100)θdT1,T3+dT3,T12≤θM1,3k.
Forx=2, y=3, we have(101)θdT2,T3+dT3,T22=2.22,M2,3=maxd2,3+d3,22,d2,T2+dT2,22,d3,T3+dT3,32=52,θ52k=2.39.
Then,(102)θdT2,T3+dT3,T22≤θM2,3k.
Hence,T satisfies the assumption of the theorem, and z=0 is the unique fixed point of T.Example 3.
LetX=A∪B, where A=1/n:n∈3,4,5,6 and B=1/2,3/2.
Defined:X×X⟶0,+∞ as follows:(103)dx,y=dy,xforallx,y∈B,dx,y=0⟺y=xforallx,y∈X.d13,14=d15,16=0.3,d13,15=d14,16=0.2,d15,13=d16,14=0.35,d13,16=d13,16=0.6,dx,y=x−y,otherwise.
Then,X,d is a generalized asymmetric metric space. However, we have the following:(1)
X,d is not a metric space as d1/3,1/6=0.6>0.5=d1/3,1/4+d1/4,1/6(2)
X,d is not a generalized metric space as d1/6,1/4=0.35≠d1/4,1/6=0.2
Define mappingT:X⟶X by(104)Tx=x,ifx∈12,32,1,ifx∈A.
Then,Tx∈1/2,3/2. Let θt=etforallt∈0,+∞, and r=4/5. It is obvious that θ∈Θ and r∈0,1.
Consider the following possibilities:Case 1:x,y∈1/2,3/2 with x≠y, and assume that x>y.(105)DTx,Ty=dTx,Ty+dTy,Tx=x−y|+|y−x=2x−y,Dx,y=dx,y+dy,x=x−y+y−x=2x−y.Therefore,(106)θDTx,Ty2=ex−y,θDx,y24/5=ex−y4/5.On the contrary,(107)θDTx,Ty2−θDx,y24/5=ex−y−ex−y4/5≤0,which implies that(108)θDTx,Ty2≤θDx,y24/5≤θmaxDx,y2,Dx,Tx2,Dy,Ty24/5.Case 2:x∈1/2,3/2,y∈A or y∈1/2,3/2,x∈A. Therefore, Tx=x, Ty=1; then, dTx,Ty=x−1.
In this case, consider two possibilities:(i)
x≥1: then, x≥1. Therefore,(109)DTx,Ty=2x−1.So, we have(110)θDTx,Ty2=ex−1,Mx,y=maxDx,y2,Dx,Tx2,Dy,Ty2≥Dy,Ty2=1−y≥1−13=23,θ234/5=e2/34/5=e8/15.On the contrary,(111)θDTx,Ty2−θ234/5=ex−1−e8/15.Sincex∈1,3/2,(112)ex−1−e8/15≤0,which implies that(113)θDTx,Ty2≤θDy,Ty24/5≤θmaxDx,y2,Dx,Tx2,Dy,Ty24/5.(ii)
x<1: then, x<1. Therefore,(114)DTx,Ty=21−x.
So, we have(115)θDTx,Ty2=e1−x,Mx,y=maxDx,y2,Dx,Tx2,Dy,Ty2≥Dy,Ty2=1−y≥1−13=23,θ234/5=e2/34/5=e8/15.
On the contrary,(116)θDTx,Ty2−θ234/5=e1−x−e8/15.
Sincex∈1/2,1,(117)e1−x/2−e8/15≤0,which implies that(118)θDTx,Ty2≤θDy,Ty24/5≤θmaxDx,y2,Dx,Tx2,Dy,Ty24/5.
Hence,T satisfies the assumption of the theorem, and z=1 is the unique fixed point of T.Theorem 3.
LetX,d be a complete generalized asymmetric 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,(119)maxdTx,Ty,dTy,Tx>0⟹θdTx,Ty+dTy,Tx2≤ϕθMx,y,where(120)Mx,y=maxdx,y+dy,x2,dx,Tx+dTx,x2,dy,Ty+dTy,y2.Then,T has a unique fixed point.Proof.
Letx0∈X be a fixed point, and define a sequence xn by(121)xn+1=Txn=Tn+1x0,foralln∈ℕ.
If there existsn0∈ℕ such that dxn0,xn0+1=0 or dxn0+1,xn0=0, then proof is finished.
We can suppose thatdxn,xn+1>0 and dxn+1,xn>0 for all n∈ℕ; then, we have(122)maxdxn,xn+1,dxn+1,xn>0.
Substitutingx=xn−1 and y=xn, from (119), for all n∈ℕ, we have(123)θdxn,xn+1+dxn+1,xn2≤ϕθMxn−1,xn,∀n∈ℕ,where(124)Mxn−1,xn=dxn−1,xn+dxn,xn−12,dxn−1,xn+dxn,xn−12,dxn,xn+1+dxn+1,xn2=dxn−1,xn+dxn,xn−12,dxn,xn+1+dxn+1,xn2.
Now, we setDxn,xm=dxn,xm+dxm,xn.
Therefore,(125)Mxn−1,xn=Dxn−1,xn2,Dxn,xn+12,if for some n, Mxn−1,xn=Dxn,xn+1/2.
From (123), θ1, and using Lemma 2, we get(126)θDxn,xn+12≤ϕθDxn,xn+12,which implies that(127)Dxn,xn+1<Dxn,xn+1,which is contradiction. Hence, Mxn−1,xn=Dxn−1,xn/2.
Therefore,(128)θDxn,xn+12≤ϕθDxn−1,xn2<θDxn−1,xn2.
Asθ is increasing,(129)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(130)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(131)Dxn+1,xn≥α,∀n∈ℕ.
Thus, we have(132)1<θλ2≤ϕθDxn,xn+12≤⋯≤ϕnθDx0,x12.
By lettingn⟶∞ in inequality (132) and using θ3 and ϕ3, we obtain(133)1<θλ2≤1.
It is a contradiction. Therefore,(134)limn⟶∞Dxn+1,xn=0.
Next, we shall prove that(135)limn⟶∞dxn,xn+2=0,limn⟶∞dxn+2,xn=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.
So, from the assumption of the theorem, we get(136)θDxm,xm+12=θDxn,xn+12≤ϕθDxn−1,xn2<θDxn−1,xn2.
Sinceθ is increasing,(137)Dxn,xm+1<Dxn−1,xn.
Continuing this process, we can get that(138)Dxm,xn+1<Dxm,xm+1,which is a contradiction. Therefore,(139)maxdxm,xn,dxn,xm>0,foreveryn,m∈ℕ,n≠m.
Substitutingx=xn and y=xn+2,(140)maxdxn,xn+2,dxn+2,xn>0.
Applying (119) with x=xn−1 and y=xn+1, we have(141)θdxn,xn+2+dxn+2,xn2≤ϕθMxn−1,xn+1,where(142)Mxn−1,xn+1=maxDxn−1,xn2,Dxn−1,xn+12,Dxn+1,xn+22=maxDxn−1,xn2,Dxn−1,xn+12.
Therefore,(143)θDxn,xn+22≤ϕθmaxDxn−1,xn2,Dxn−1,xn+12.
So, from Lemma2, we have(144)θDxn,xn+22<θmaxDxn−1,xn2,Dxn−1,xn+12.
Takean=Dxn,xn+2 and bn=Dxn,xn+1. Thus, from (144) and θ1, we have(145)an<maxan−1,bn−1.
Again by (137),(146)bn≤bn−1≤maxan−1,bn−1.
Therefore,(147)maxan,bn≤maxan−1,bn−1,∀n∈ℕ.
Then, the sequencemaxan,bnn is monotone nonincreasing, so it converges to some β≥0 such that(148)limn⟶∞maxan,bn=β.
By (134) and assuming that β>0, we have(149)limn⟶∞supan=limn⟶∞supmaxan,bn=limn⟶∞maxan,bn.
Takinglimsupn⟶∞ in (144) and using θ3, we obtain(150)θlimn⟶∞supan<θlimn⟶∞maxan−1,bn−1.
Therefore,(151)θβ2<θβ2.
Fromθ1, we get(152)β<β,which is a contradiction. Therefore,(153)limn⟶∞Dxn+2,xn=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. Then, there is ε>0 such that, for an integer k, there exist two sequences nk and mk, mk>nk>k, such that(154)Dxmk,xnk≥ε,Dxmk−1,xnk<ε.
As in the proof of Theorem2, we conclude that(155)limk⟶∞Dxmk,xnk=ε,limk⟶∞Dxmk+1,xnk+1=ε.
Hence, there existsn0∈ℕ such that(156)maxdTxmk,Txnk,dTxnk,Txmk≥ε4,∀n≥n0.
SinceT is a θ−ϕ-contraction, we derive(157)θDxmk+1,xnk+12≤ϕθMxmk,xnk,where(158)Mxmk,xnk=maxDxmk,xnk2,Dxmk,xmk+12,Dxnk,xnk+12.
As in the proof of Theorem2, we have(159)limk⟶∞Mxmk,xnk=ε2.
Applying (119) with x=xmk and y=xnk, we have(160)θDxmk+1,xnk+12≤ϕθmaxDxmk,xnk2,Dxmk,xmk+12,Dxnk,xnk+12.
By lettingk⟶∞ in inequality (160) and using θ1,θ3,ϕ3, and Lemma 2, we obtain(161)θlimk⟶∞Dxmk+1,xnk+12≤ϕθlimk⟶∞maxDxmk,xnk2,Dxmk,xmk+12,Dxnk,xnk+12,which implies that(162)ε<ε,which is a contradiction. Therefore,(163)limn,m⟶∞Dxm,xn=0.
Thus,(164)limn,m⟶∞dxm,xn=limn,m⟶∞dxn,xm=0.
Hence,xn is a forward and backward Cauchy sequence in X. By completeness of X,d, there exists z,u∈X such that(165)limn⟶∞dxn,z=limn⟶∞du,xn=0.
So, from Lemma1, we get z=u.
Now, we show thatdTz,z=0 = dz,Tz. Arguing by contradiction, we assume that(166)dTz,z>0,dz,Tz>0.
Therefore,(167)maxdTz,z,dz,Tz>0.
As in the proof of Theorem2, we conclude that(168)limn⟶∞dTz,Txn=dTz,z,(169)limn⟶∞dTxn,Tz=dz,Tz.
By (168) and (169), there exists q∈ℕ such that(170)maxdTz,Txn,dTxn,Tz>0,∀n≥q.
SinceT is a θ−ϕ-contraction, we derive(171)θDTz,Txn2≤ϕθMz,xn,∀n≥q,where(172)Mz,xn=maxDz,xn2Dz,Tz2,Dxn,Txn2,which implies that(173)θDTz,Txn2≤ϕθmaxDz,xn2Dz,Tz2,Dxn,Txn2.
By lettingn⟶∞ in inequality (172) and using θ1,θ3,ϕ3, and Lemma 2, we obtain(174)θlimn⟶∞DTz,Txn2≤ϕθlimn⟶∞maxDz,xn2Dz,Tz2,Dxn,Txn2<θlimn⟶∞maxDz,xn2,Dz,Tz2,Dxn,Txn2.
Therefore,(175)Dz,Tz<Dz,Tz,which is a contradiction. Thus, z=Tz. Thus, T has a fixed point.
Uniqueness: letz,u∈ FixT, where z≠u. Then,(176)dTz,Tu=dz,u>0,dTu,Tz=du,z>0.
Therefore,(177)maxdTz,Tu,dTu,Tz>0.
From the assumption of the theorem, we get(178)θDTz,Tu2=θDz,u2≤ϕθMz,u,where(179)Mz,u=maxDz,u2,Dz,Tz2,Du,Tu2=Dz,u2.
Therefore, we have(180)θDz,u2≤ϕθDz,u2<θDz,u2,which implies that Dz,u<Dz,u. It is a contradiction. Therefore, u=z.
From Theorem3, we obtain the following fixed-point theorems for θ−ϕ-Reich-type contraction and θ−ϕ-Kannan-type contraction.Theorem 4.
LetX,d be a complete generalized asymmetric space and T:X⟶X be a θ−ϕ-Kannan-type contraction; then, T has a unique fixed point.Proof.
SinceT is a θ−ϕ-Kannan-type contraction, there exist θ∈Θ and ϕ∈Φ such that(181)θdTx,Ty+dTy,Tx2≤ϕθdTx,x+dx,Tx+dTy,y+dy,Ty4≤ϕθmaxdx,Tx+dTx,x2,dy,Ty+dTy,y2≤ϕθmaxdx,y+dy,x2,dTx,x+dx,Tx2,dy,Ty+dTy,y2.
Therefore,T is a θ−ϕ-contraction. As in the proof of Theorem 3, we conclude that T has a unique fixed point.Theorem 5.
LetX,d be a complete generalized asymmetric space and T:X⟶X be a θ−ϕ-Reich-type contraction.Then,T has a unique fixed point.Proof.
SinceT is a θ−ϕ-Reich-type contraction, there exist θ∈Θ and ϕ∈Φ such that(182)θdTx,Ty+dTy,Tx2≤ϕθdx,y+dy,x+dTx,x+dx,Tx+dTy,y+dy,Ty6≤ϕθmaxdx,y+dy,x2,dTx,x+dx,Tx2,dy,Ty+dTy,y2.
Therefore,T is a θ−ϕ-contraction. As in the proof of Theorem 3, we conclude that T has a unique fixed point.Corollary 2.
LetX,d be a complete 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,(183)maxdTx,Ty,dTy,Tx>0⟹dTx,Ty+dTy,Tx2≤αdTx,x+dx,Tx+dTy,y+dy,Ty2.Then,T has a unique fixed point.Proof.
Letθt=et for all t∈0,+∞ and ϕt=t2α for all t∈1,+∞. We prove that T is a θ−ϕ-Kannan-type contraction. Indeed,(184)θdTx,Ty+dTy,Tx2=edTx,Ty+dTy,Tx/2≤eαdTx,x+dx,Tx+dTy,y+dy,Ty/2=e2αdTx,x+dx,Tx+dTy,y+dy,Ty/4=edTx,x+dx,Tx+dTy,y+dy,Ty/42α=ϕθdTx,x+dx,Tx+dTy,y+dy,Ty4.
Therefore, as in the proof of Theorem4, T has a unique fixed point x∈X.Corollary 3.
LetX,d be a complete 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,(185)maxdTx,Ty,dTy,Tx>0,and we have(186)dTx,Ty+dTy,Tx2≤λdx,y+dy,x+dTx,x+dx,Tx+dTy,y+dy,Ty2.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.(187)θdTx,Ty+dTy,Tx2=edTx,Ty+dTy,Tx/2≤eλdx,y+dy,x+dTx,x+dx,Tx+dTy,y+dy,Ty/2=e3λdx,y+dy,x+dTx,x+dx,Tx+dTy,y+dy,Ty/6=ϕθdx,y+dy,x+dTx,x+dx,Tx+dTy,y+dy,Ty6.
Therefore, as in the proof of Theorem5, T has a unique fixed point x∈X.Corollary 4 (Theorem2).
LetX,d be a complete generalized asymmetric space and T:X⟶X be a given mapping. Suppose that there exist θ∈Θ and r∈0,1 such that, for any x,y∈X,(188)maxdTx,Ty,dTy,Tx>0⟹θdTx,Ty+dTy,Tx2≤θMx,yr,where(189)Mx,y=maxdx,y+dy,x2,dx,Tx+dTx,x2,dy,Ty+dTy,y2.Then,T has a unique fixed point.Proof.
Ifϕt=tr, with r∈0,1, we prove that T is a θ−ϕ-contraction. Therefore, as in the proof of Theorem 3, T has a unique fixed point.Example 4.
ConsiderX=1,2,3,4. Let d:X×X⟶0,+∞ be a mapping defined by the following:(i)
dx,y=0⟺x=y.(ii)
d1,2=3, d2,1=1.(iii)
d2,3=d3,2=d1,3=d3,1=1.(iv)
d1,4=d4,1=d4,2=d2,4=d3,4=4.(v)
d4,3=3.
Clearly,X,d is not an asymmetric metric space, from d1,2=3>d1,3+d3,2=2.
However, it is a complete generalized asymmetric metric space. LetT:X⟶X be given by(190)T1=T2=T3=1,T4=3.
Supposeθt=t+1 and φt=2t+1/3. Therefore, θ∈Θ and ϕ∈Φ.
First, observe thatmaxdTx,Ty,dTy,Tx>0⟺x=4,y=1, x=4,y=2, or x=4,y=3; for x=4,y=1, we have(191)dT4,T1+dT1,T42=d3,1+d1,32=1.
On the contrary,(192)maxd4,T4+dT4,42,d1,T1+d1,T12,d4,1+d1,42=4.
Therefore,(193)θdT4,T1+dT1,T42=2,ϕθ4=2.33.
So,(194)θDT4,T12≤ϕθM4,1.
Forx=4, y=2, we have(195)θdT4,T2+dT2,T42=2,maxd4,T4+dT4,42,d2,T2+d2,T22,d4,2+d2,42=4.
Therefore,(196)φθ4=2.33.
So,(197)θDT4,T22≤ϕθM4,2.
Forx=4, y=3, we have(198)θdT4,T3+dT3,T42=θd3,1+d1,32=2,maxd4,T4+dT4,42,d3,T3+d3,T32,d4,3+d3,42=72.
Therefore,(199)ϕθ72=2.24.
So,(200)θDT4,T32≤ϕθM4,3.
Hence,T satisfies the assumption of the theorem, and z=1 is the unique fixed point of T.Example 5.
LetX=1,+∞ and d:X×X⟶0,+∞ defined by(201)dx,y=y−xify≥x,dx,y=12x−yify<x.
Clearly,X,d is not metric, asymmetric metric, but it is a complete generalized asymmetric metric space.
LetT:X⟶X be given by(202)Tx=x.
Letθt=t+1 and ϕt=t+1/2. It is obvious that θ∈Θ and ϕ∈Φ.
Consider the following possibilities:Case 1:y≥x. We have(203)dx,y=y−x,dy,x=12y−x.Then,(204)Dx,y2=34y−x,θDx,y2=32y−x+1.So,(205)ϕθDx,y2=34y−x+1,θDTx,Ty2=32y−x+1.On the contrary,(206)θDx,y2−ϕθDTx,Ty2=32y−x−34y−x=342y−x−y−x=34y−x2−y+x.Sincex,y∈1,+∞,(207)2−y+x≤0,which implies that(208)θDTx,Ty2≤ϕθDx,y2≤ϕθmaxDx,y2,Dx,Tx2,Dy,Ty2.Case 2:y<x. Similar to case 1, we conclude that(209)θDTx,Ty2≤ϕθmaxDx,y2,Dx,Tx2,Dy,Ty2.
Hence,T satisfies the assumption of the theorem, and z=1 is the unique fixed point of T.



---

*Source: 8867020-2020-11-12.xml*