JMATHJournal of Mathematics2314-47852314-4629Hindawi10.1155/2020/16325261632526Research ArticleOn p-Hybrid Wardowski Contractionshttps://orcid.org/0000-0002-6798-3254KarapınarErdal123https://orcid.org/0000-0003-4606-7211AydiHassen45https://orcid.org/0000-0002-6689-0355FulgaAndreea6KocinacLjubisa1ETSI Division of Applied MathematicsThu Dau Mot UniversityThủ Dầu MộtBinh Duong ProvinceVietnamtdmu.edu.vn2Department of Medical ResearchChina Medical UniversityTaichungTaiwancmu.edu.cn3Department of MathematicsÇankaya UniversityAnkara 06790Turkeycankaya.edu.tr4Nonlinear Analysis Research GroupTon Duc Thang UniversityHo Chi Minh CityVietnamtdtu.edu.vn5Faculty of Mathematics and StatisticsTon Duc Thang UniversityHo Chi Minh CityVietnamtdtu.edu.vn6Department of Mathematics and Computer SciencesTransilvania University of BrasovBrasovRomaniaunitbv.ro202024820202020110520202507202024820202020Copyright © 2020 Erdal Karapınar et al.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The goal of this work is to introduce the concept of p-hybrid Wardowski contractions. We also prove related fixed-point results. Moreover, some illustrated examples are given.

1. Introduction

Let G represent the collection of functions G:0, so that

G1G is strictly increasing

G2 for each sequence ηn in 0,, limnηn=0 iff limnGηn=

G3 there is k0,1 so that limnηkGη=0

Definition 1 (see [<xref ref-type="bibr" rid="B1">1</xref>]).

A mapping T:,d,d is called a Wardowski contraction if there exist τ>0 and GG such that for all ν,ω,(1)dTν,Tω>0τ+GdTν,TωGdν,ω.

Example 1 (see [<xref ref-type="bibr" rid="B1">1</xref>]).

The functions G:0, defined by

Gx=ln x

Gx=lnx+x

Gx=1/x

Gx=lnx2+x

belong to G.

Wardowski  introduced a new proper generalization of Banach contraction. For other related papers in the literature, see . The main result of Wardowski is as follows.

Theorem 1 (see [<xref ref-type="bibr" rid="B1">1</xref>]).

Let ,d be a complete metric space, and let T: be an G-contraction. Then, ϒ has a unique fixed point, say z, in and for any point σ, the sequence ϒjσ converges to z.

Theorem 2 (see [<xref ref-type="bibr" rid="B11">11</xref>]).

Let ,d be a complete metric space and T: be a given mapping such that(2)dTν,Tωσ1dν,ω+σ2dν,Tν+σ3dω,Tω+σ4dν,Tω+dω,Tν2,for all ν,ω, where σi, i=1,2,3,4, are nonnegative real numbers such that i=14σi<1. Then, T admits a unique fixed point in .

In the paper , the concept of interpolative Hardy–Rogers-type contractions was introduced.

Definition 2.

(see ). On a metric space ,d, a self-mapping T: is an interpolative Hardy–Rogers-type contraction if there exist λ0,1 and σ1,σ2,σ30,1 with σ1+σ2+σ3<1, such that(3)dTν,Tωλdν,ωσ1dν,Tνσ2dω,Tωσ3dν,Tω+dω,Tν21σ1σ2σ3,for all ν,ω/ϜT, where T=ζ:Tζ=ζ.

Theorem 3 (see [<xref ref-type="bibr" rid="B12">12</xref>]).

Let ,d be a complete metric space and T be an interpolative Hardy–Rogers-type contraction. Then, T has a fixed point in .

The interpolation concept was used in other new papers related to fixed-point theory. For example, see . In this paper, we consider new contractive type self-mappings, named as p-hybrid Wardowski contractions. Our fixed-point results will be supported by concrete examples.

2. Main Results

Let ,d be a metric space and T be a self-mapping on this space. For p0 and κi0,i=1,2,3,4, such that i=14κi=1, we define the following expression:(4)ATpν,ω=κ1dν,ωp+κ2dν,Tνp+κ3dω,Tωp+κ4dω,Tν+dν,Tω2p1/p,for p>0,ν,ωdν,ωκ1dν,Tνκ2dω,Tωκ3dν,Tω+dω,Tν2κ4,for p=0,ν,ω/T.

On the other hand, let B represent the set of functions G:0, such that

GaG is strictly increasing

Gb there exists τ>0 such that τ+limtt0infGt>limtt0supGt, for every t0>0

Definition 3.

A mapping T:,d,d is called a p-hybrid Wardowski contraction, if there is GB such that(5)dTν,Tω>0 implies τ+GdTν,TωGATpν,ω,for every p>0.

In particular, if inequality (5) holds for p=0, we say the mapping T is a 0-hybrid Wardowski contraction.

Theorem 4.

A p-hybrid Wardowski contraction self-mapping on a complete metric space admits exactly one fixed point in .

Proof.

Taking an arbitrary point ν0, we consider the sequence νn defined by the relation νn=Tνn1, n1. According to this construction, it is easy to see that if there is n0 so that νn0=νn0+1=Tνn0, νn0 turns into a fixed point of T. We shall presume that for all n0,(6)νn+1νndνn+1,νn=dTνn,Tνn1>0.

On account of (4), for ν=νn and ω=νn1, we have that(7)ATpνn,νn1=κ1dνn,νn1p+κ2dνn,Tνnp+κ3dνn1,Tνn1p+κ4dνn,Tνn1+dνn1,Tνn2p1/p=κ1dνn,νn1p+κ2dνn,νn+1p+κ3dνn1,νnp+κ4dνn,νn+dνn1,νn+12p1/p,κ1dνn,νn1p+κ2dνn,νn+1p+κ3dνn1,νnp+κ4dνn1,νn+dνn,νn+12p.

Denoting by χn=dνn1,νn, we have(8)ATpνn,νn1=κ1+κ3χnp+κ2χn+1p+κ4χn+χn+12p1/p,and from (5), it follows that(9)τ+GdTνn1,TνnGATpνn1,νnGκ1dνn,νn1p+κ2dνn,νn+1p+κ3dνn1,νnp+κ4dνn1,νn+dνn,νn+121/p,which gives us(10)Gχn+1=Gdνn,νn+1=GdTνn1,TνnGκ1+κ3χnp+κ2χn+1p+κ4χn+χn+12p1/pτ.

If maxχn,χn+1=χn+1, then the above inequality becomes(11)Gχn+1Gκ1+κ2+κ3+κ4χn+1p1/pτ<Gχn+1,which is a contradiction. Consequently, maxχn,χn+1=χn and then there exists χ0 such that(12)limnχn=χ.

Supposing that χ>0, we have limnATpνn1,νn=χ and by Gb, we obtain(13)τ+Gχ+0Gχ+0,which is a contradiction. Therefore,(14)limndνn1,νn=0.

In order to prove that νn is a Cauchy sequence in ,d, we suppose that there exist ϵ>0 and the sequences nk,mk of positive integers, with nk>mk>k such that(15)dνnk,νmkε,dνnk1,νmk<ε,for any k.

Thus, we have(16)εdνnk,νmkdνnk,νnk1+dνnk1,νmk<dνnk,νnk1+ε.

When k, using (14) and (15), it follows(17)limkdνnk,νmk=ε.

By using the triangle inequality, we have(18)0dνnk+1,νmk+1dνnk,νmk,dνnk+1,νnk+dνmk,νmk+1,(19)limkdνnk+1,νmk+1dνnk,νmklimkdνnk+1,νnk+dνmk,νmk+1=0.So,(20)limkdνnk+1,νmk+1=limkdνnk,νmk=ϵ>0.

Moreover, since(21)ϵ=dνnk,νmkdνnk,νmk+1+dνmk+1,νmk,ϵ=dνnk,νmkdνnk,νnk+1+dνmk,νnk+1,we have(22)limndνnk,νmk+1=limndνmk,νnk+1=ε.

So, the inequality(23)dTνnk,Tνmk=dνnk+1,νmk+1>0occurs for all kN, and using (5), there exists τ>0 such that(24)τ+Gdνnk+1,νmk+1GATpνnk,νmk,where(25)ATpνnk,νmk=κ1dνnk,νmkp+κ2dνnk,νnk+1p+κ3dνmk,νmk+1p+κ4dνnk,νmk+1+dνmk,νnk+12p1/p.

Moreover, since the function G is increasing, we have(26)τ+liminfkGκ3+κ41/pdνnk+1,νmk+1τ+liminfkGdTνnk,TνmkliminfkGATpνnk,νmklimsupnGATpνnk,νmk.

And letting k,(27)τ+Gε+Gε+.

That is a contradiction, so ε=0 and then, ε=0. Consequently, the sequence νn is Cauchy and by completeness of , it converges to some point ζ.

There exists a subsequence νni such that Tνni=Tζ for all i; then,(28)dζ,Tζ=limidνni+1,Tζ=limidTνni,Tζ=0.

On the contrary, if there is a natural number N such that TνnTζ for all nN, applying (5), for ν=νn and ω=ζ, we have(29)τ+Gκ3+κ41/pdνn+1,Tζτ+Gdνn+1,Tζ=τ+GdTνn,TζGAνn,ζ,where(30)ATpνn,ζ=κ1dνn,ζp+κ2dνn,Tνnp+κ3dζ,Tζp+κ4dνn,Tζ+dζ,Tνn2p1/p,=Gκ1dνn,ζp+κ2dνn,νn+1p+κ3dζ,Tζp++κ4dνn,Tζ+dζ,Tνn2p1/p.

We suppose that ζTζ. Inasmuch as(31)limndνn,Tζ=dζ,Tζ,dlimnAνn,ζ=limnκ1dνn,ζp+κ2dνn,νn+1p+κ3dζ,Tζp+κ4dνn,Tζ+dζ,Tνn2p1/p=κ3+κ41/pdζ,Tζ.

Letting n in inequality (29), we find that(32)τ+liminftdζ,TζGκ3+κ41/ptτ+liminftdζ,TζGt<liminftdζ,TζGκ3+κ41/pt<limsuptdζ,TζGκ3+κ41/pt,which contradicts Gb. Therefore, Tζ=ζ.

We claim now that T admits only one fixed point. If there exists another point ξ, ξζ, such that ξ=Tξ, then dξ,ζ=dTξ,Tζ>0 and we have(33)τ+Gdξ,ζ=τ+GdTξ,TζGATpξ,ζ=Gκ1dξ,ζp+κ2dξ,Tξp+κ3dζ,Tζp+κ4dξ,Tζ+dζ,Tξ2p1/p,=Gκ1dξ,ζp+κ2dξ,ξp+κ3dζ,ζp+κ4dξ,ζ+dζ,ξ2p1/p,=Gκ1+κ41/pdξ,ζ,Gdξ,ζ,which is a contradiction.

Example 2.

Let =0,1 be endowed with the standard metric dν,ω=νω. Let the mapping T: be defined by T=x/8for x0,11/4for x=1. Take p=2, τ=ln4/3, κ1=1/9, κ2=κ4=6/81, κ3=60/81, and Gt=lnt. Then, we have the following:

For x,y[0,1,

(34)ln43+lnGdx,yln4xy24<lnxy3=lnxy291/2=lnκ1dx,y21/2<lnAT2x,y.

For x0,1 and y=1,

(35)ln43+lnGdx,1=ln4x224<ln7934=ln4981d1,1421/2<lnAT2x,1.

Thus, all assumptions of Theorem 4 hold, and T has a unique fixed point. On the other hand, for x=7/8 and y=1, we have(36)dT78,T1=d764,14=964>18=d78,1.

Thus, it is not a Wardowski contraction, since for every function GB and τ>0(37)τ+GdT78,T1>Gd78,1.

Theorem 5.

A 0-hybrid Wardowski contraction self-mapping on a complete metric space admits a fixed point in provided that for each sequence ηn in 0,, limnηn=0 iff limnGηn=.

Proof.

Following the same reasoning from the proof of the previous theorem, we can assume that for all n0,(38)νn+1νndνn+1,νn>0.

On account of (4), for ν=νn and ω=νn1, we have that(39)AT0νn,νn1=dνn,νn1κ1dνn,Tνnκ2dνn1,Tνn1κ3dνn,Tνn1+dνn1,Tνn2κ4=dνn,νn1κ1dνn,νn+1κ2dνn1,νnκ3dνn,νn+dνn1,νn+12κ4,dνn,νn1κ1dνn,νn+1κ2dνn1,νnκ3dνn1,νn+dνn,νn+12κ4.

Using the same notation, χn=dνn1,νn, and taking into account Ga, by (5), we have(40)τ+Gχn+1Gχnκ1+κ3χn+1κ2χn+χn+12κ4τ.

We can remark that the case maxχn,χn+1=χn+1, is not possible since the above inequality becomes(41)Gχn+1Gχn+1κ1+κ2+κ3+κ4τ<Gχn+1,a contradiction. Therefore, χn>χn+1 for all n, and then, there exists χ0 such that(42)limnχn=limndνn1,νn=χ.

We claim that χ=0. Indeed, if we suppose that χ>0, taking the limit as n in (40), we have(43)τ+Gχ+0Gχ+0,which contradicts G2. We conclude that(44)χ=limndνn1,νn=0.

Let n and j1 now; we have(45)AT0νn,νn+j=dνn,νn+jκ1dνn,Tνnκ2dνn+j,Tνn+jκ3dνn,Tνn+j+dνn+j,Tνn2κ4=dνn,νn+jκ1dνn,νn+1κ2dνn+j,νn+j+1κ3dνn,νn+j+1+dνn+j,νn+12κ4=0.

And taking into account (44),(46)limnAT0νn,νn+j=0.

Therefore, limnGAT0νn,νn+j= and since(47)τ+limnGdνn+1,νn+j+1limnGAT0νn,νn+j,we obtain that limnGdνn,νn+j= and so limndνn,νn+j=0. Thus, νn is a Cauchy sequence on a complete metric space ,d and there exists ζ such that limnνn=ζ. Of course, it easy to see that, for ν=νn and ω=ζ, we have(48)limnAT0νn,ζ=0.

If we suppose that there is a subsequence νns such that Tνns=Tζ, then we have(49)0=limndTνns,Tζ=limndνns+1,Tζ=dζ,Tζ,which means that ζ is a fixed point of T. Therefore, we can assume that dTνn,Tζ>0 for every n, and by (5), we obtain(50)τ+GdTνn,TζGAT0νn,ζ.

Letting n and taking into account the previous considerations, we have limnGdTνn,Tζ= and then dζ,Tζ=limndTνn,Tζ=0. Consequently, ζ is a fixed point of T.

Example 3.

Let =x,y,z,t be a set endowed with the metric d:×0, (Table 1).

And the mapping T: is defined as T:xyztxxtt.

First, we remark that Theorem 1 is not satisfied, since for ν=y and ω=t,(51)dTy,Tt=dx,t=2>1=dy,t.

Hence, for any τ>0 and GB, we can write(52)τ+GdTy,Tt>Gdy,t.

Choosing τ=ln4/3, κ1=κ2=7/16, κ3=κ4=1/16, and Gt=ln t, for ν=y and ω=z, we have(53)ln43+lndTy,Tz=ln43dx,t=ln83=0,980829253<1,04792915=ln37/1637/1621/1621/16,<lndy,z5/16dy,x5/16dz,t5/16dy,t+dz,x21/16,=lndy,z5/16dy,Ty5/16dz,Tz5/16dy,Tz+dz,Ty21/16,=lnAT0y,z.

Definition of metric d.

dν,ωxyzt
x0332
y3031
z3302
t2120
3. Consequences

C1 Considering Gt=lnt in Theorem 5 and σi=eτκi, we obtain Theorem 2.

C2 Considering Gt=lnt in Theorem 5 and λ=eτ, we obtain Theorem 3.

C3 Considering Gt=lnt in Theorem 4, λ=eτ, and p=1, we obtain Theorem 3.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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