# Onp-Hybrid Wardowski Contractions

Authors: Erdal Karapınar; Hassen Aydi; Andreea Fulga
Journal: Journal of Mathematics
 (2020)

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## Abstract

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


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## Body

## 1. Introduction

LetG represent the collection of functions G:0,∞⟶ℝ so that(i)
G1G is strictly increasing(ii)
G2 for each sequence ηn in 0,∞, limn⟶∞ηn=0 iff limn⟶∞Gηn=−∞(iii)
G3 there is k∈0,1 so that limn⟶∞ηkGη=0Definition 1 (see [1]).
A mappingT:ℳ,d⟶ℳ,d is called a Wardowski contraction if there exist τ>0 and G∈G such that for all ν,ω∈ℳ,(1)dTν,Tω>0⟹τ+GdTν,Tω≤Gdν,ω.Example 1 (see [1]).
The functionsG:0,∞⟶ℝ defined by(1)
Gx=lnx(2)
Gx=lnx+x(3)
Gx=−1/x(4)
Gx=lnx2+xbelong to G.
Wardowski [1] introduced a new proper generalization of Banach contraction. For other related papers in the literature, see [2–10]. The main result of Wardowski is as follows.Theorem 1 (see [1]).
Letℳ,d be a complete metric space, and let T:ℳ⟶ℳ be an G-contraction. Then, ϒ has a unique fixed point, say z, in ℳ and for any point σ∈ℳ, the sequence ϒjσ converges to z.Theorem 2 (see [11]).
Letℳ,d be a complete metric space and T:ℳ⟶ℳ be a given mapping such that(2)dTν,Tω≤σ1dν,ω+σ2dν,Tν+σ3dω,Tω+σ4dν,Tω+dω,Tν2,for all ν,ω∈ℳ, where σi, i=1,2,3,4, are nonnegative real numbers such that ∑i=14σi<1. Then, T admits a unique fixed point in ℳ.In the paper [12], the concept of interpolative Hardy–Rogers-type contractions was introduced.Definition 2.
(see [12]). On a metric space ℳ,d, a self-mapping T:ℳ⟶ℳ is an interpolative Hardy–Rogers-type contraction if there exist λ∈0,1 and σ1,σ2,σ3∈0,1 with σ1+σ2+σ3<1, such that(3)dTν,Tω≤λdν,ωσ1dν,Tνσ2dω,Tωσ3⋅dν,Tω+dω,Tν21−σ1−σ2−σ3,for all ν,ω∈ℳ/ϜTℳ, where ℳT=ζ∈ℳ:Tζ=ζ.Theorem 3 (see [12]).
Letℳ,d be a complete metric space and T be an interpolative Hardy–Rogers-type contraction. Then, T has a fixed point in ℳ.
The interpolation concept was used in other new papers related to fixed-point theory. For example, see [13–17]. 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

Letℳ,d be a metric space and T be a self-mapping on this space. For p≥0 and κi≥0,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,forp>0,ν,ω∈ℳdν,ωκ1dν,Tνκ2dω,Tωκ3dν,Tω+dω,Tν2κ4,forp=0,ν,ω∈ℳ/ℳT.On the other hand, letB represent the set of functions G:0,∞⟶ℝ such that(i)
GaG is strictly increasing(ii)
Gb there exists τ>0 such that τ+limt⟶t0infGt>limt⟶t0supGt, for every t0>0Definition 3.
A mappingT:ℳ,d⟶ℳ,d is called a p-hybrid Wardowski contraction, if there is G∈B such that(5)dTν,Tω>0impliesτ+GdTν,Tω≤GATpν,ω,for everyp>0.
In particular, if inequality (5) holds for p=0, we say the mapping T is a 0-hybrid Wardowski contraction.Theorem 4.
Ap-hybrid Wardowski contraction self-mapping on a complete metric space admits exactly one fixed point in ℳ.Proof.
Taking an arbitrary pointν0∈ℳ, we consider the sequence νn defined by the relation νn=Tνn−1, n≥1. 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 n∈ℕ0,(6)νn+1≠νn⟺dνn+1,νn=dTνn,Tνn−1>0.
On account of (4), for ν=νn and ω=νn−1, we have that(7)ATpνn,νn−1=κ1dνn,νn−1p+κ2dνn,Tνnp+κ3dνn−1,Tνn−1p+κ4dνn,Tνn−1+dνn−1,Tνn2p1/p=κ1dνn,νn−1p+κ2dνn,νn+1p+κ3dνn−1,νnp+κ4dνn,νn+dνn−1,νn+12p1/p,≤κ1dνn,νn−1p+κ2dνn,νn+1p+κ3dνn−1,νnp+κ4dνn−1,νn+dνn,νn+12p.
Denoting byχn=dνn−1,νn, we have(8)ATpνn,νn−1=κ1+κ3χnp+κ2χn+1p+κ4χn+χn+12p1/p,and from (5), it follows that(9)τ+GdTνn−1,Tνn≤GATpνn−1,νn≤Gκ1dνn,νn−1p+κ2dνn,νn+1p+κ3dνn−1,νnp+κ4dνn−1,νn+dνn,νn+121/p,which gives us(10)Gχn+1=Gdνn,νn+1=GdTνn−1,Tνn≤Gκ1+κ3χnp+κ2χn+1p+κ4χn+χn+12p1/p−τ.
Ifmaxχn,χn+1=χn+1, then the above inequality becomes(11)Gχn+1≤Gκ1+κ2+κ3+κ4χn+1p1/p−τ<Gχn+1,which is a contradiction. Consequently, maxχn,χn+1=χn and then there exists χ≥0 such that(12)limn⟶∞χn=χ.
Supposing thatχ>0, we have limn⟶∞ATpνn−1,νn=χ and by Gb, we obtain(13)τ+Gχ+0≤Gχ+0,which is a contradiction. Therefore,(14)limn⟶∞dνn−1,νn=0.
In order to prove thatνn is a Cauchy sequence in ℳ,d, we suppose that there exist ϵ>0 and the sequences n∗k,m∗k of positive integers, with n∗k>m∗k>k such that(15)dνn∗k,νm∗k≥ε,dνn∗k−1,νm∗k<ε,for any k∈ℕ.
Thus, we have(16)ε≤dνn∗k,νm∗k≤dνn∗k,νn∗k−1+dνn∗k−1,νm∗k<dνn∗k,νn∗k−1+ε.
Whenk⟶∞, using (14) and (15), it follows(17)limk⟶∞dνn∗k,νm∗k=ε.
By using the triangle inequality, we have(18)0≤dνn∗k+1,νm∗k+1−dνn∗k,νm∗k,≤dνn∗k+1,νn∗k+dνm∗k,νm∗k+1,(19)limk⟶∞dνn∗k+1,νm∗k+1−dνn∗k,νm∗k≤limk⟶∞dνn∗k+1,νn∗k+dνm∗k,νm∗k+1=0.So,(20)limk⟶∞dνn∗k+1,νm∗k+1=limk⟶∞dνn∗k,νm∗k=ϵ>0.
Moreover, since(21)ϵ=dνn∗k,νm∗k≤dνn∗k,νm∗k+1+dνm∗k+1,νm∗k,ϵ=dνn∗k,νm∗k≤dνn∗k,νn∗k+1+dνm∗k,νn∗k+1,we have(22)limn⟶∞dνn∗k,νm∗k+1=limn⟶∞dνm∗k,νn∗k+1=ε.
So, the inequality(23)dTνn∗k,Tνm∗k=dνn∗k+1,νm∗k+1>0occurs for all k≥N, and using (5), there exists τ>0 such that(24)τ+Gdνn∗k+1,νm∗k+1≤GATpνn∗k,νm∗k,where(25)ATpνn∗k,νm∗k=κ1dνn∗k,νm∗kp+κ2dνn∗k,νn∗k+1p+κ3dνm∗k,νm∗k+1p+κ4dνn∗k,νm∗k+1+dνm∗k,νn∗k+12p1/p.
Moreover, since the functionG is increasing, we have(26)τ+liminfk⟶∞Gκ3+κ41/pdνn∗k+1,νm∗k+1≤τ+liminfk⟶∞GdTνn∗k,Tνm∗k≤liminfk⟶∞GATpνn∗k,νm∗k≤limsupn⟶∞GATpνn∗k,νm∗k.
And lettingk⟶∞,(27)τ+Gε+≤Gε+.
That is a contradiction, soε=0 and then, ε=0. Consequently, the sequence νn is Cauchy and by completeness of ℳ, it converges to some point ζ∈ℳ.
There exists a subsequenceνni such that Tνni=Tζ for all i∈ℕ; then,(28)dζ,Tζ=limi⟶∞dνni+1,Tζ=limi⟶∞dTνni,Tζ=0.
On the contrary, if there is a natural numberN such that Tνn≠Tζ for all n≥N, applying (5), for ν=νn and ω=ζ, we have(29)τ+Gκ3+κ41/pdνn+1,Tζ≤τ+Gdνn+1,Tζ=τ+GdTνn,Tζ≤GAνn,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 thatζ≠Tζ. Inasmuch as(31)limn⟶∞dνn,Tζ=dζ,Tζ,dlimn⟶∞Aνn,ζ=limn⟶∞κ1dνn,ζp+κ2dνn,νn+1p+κ3dζ,Tζp+κ4dνn,Tζ+dζ,Tνn2p1/p=κ3+κ41/pdζ,Tζ.
Lettingn⟶∞ in inequality (29), we find that(32)τ+liminft⟶dζ,TζGκ3+κ41/pt≤τ+liminft⟶dζ,TζGt<liminft⟶dζ,TζGκ3+κ41/pt<limsupt⟶dζ,TζGκ3+κ41/pt,which contradicts Gb. Therefore, Tζ=ζ.
We claim now thatT admits only one fixed point. If there exists another point ξ∈ℳ, ξ≠ζ, such that ξ=Tξ, then dξ,ζ=dTξ,Tζ>0 and we have(33)τ+Gdξ,ζ=τ+GdTξ,Tζ≤GATpξ,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.
Letℳ=0,1 be endowed with the standard metric dν,ω=ν−ω. Let the mapping T:ℳ⟶ℳ be defined by T=x/8forx∈0,11/4forx=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:Forx,y∈[0,1,(34)ln43+lnGdx,yln4x−y24<lnx−y3=lnx−y291/2=lnκ1dx,y21/2<lnAT2x,y.Forx∈0,1 and y=1,(35)ln43+lnGdx,1=ln4x−224<ln79⋅34=ln4981d1,1421/2<lnAT2x,1.
Thus, all assumptions of Theorem4 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 functionG∈B 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 inℳ provided that for each sequence ηn in 0,∞, limn⟶∞ηn=0 iff limn⟶∞Gηn=−∞.Proof.
Following the same reasoning from the proof of the previous theorem, we can assume that for alln∈ℕ0,(38)νn+1≠νn⟺dνn+1,νn>0.
On account of (4), for ν=νn and ω=νn−1, we have that(39)AT0νn,νn−1=dνn,νn−1κ1dνn,Tνnκ2dνn−1,Tνn−1κ3dνn,Tνn−1+dνn−1,Tνn2κ4=dνn,νn−1κ1dνn,νn+1κ2dνn−1,νnκ3dνn,νn+dνn−1,νn+12κ4,≤dνn,νn−1κ1dνn,νn+1κ2dνn−1,νnκ3dνn−1,νn+dνn,νn+12κ4.
Using the same notation,χn=dνn−1,νn, and taking into account Ga, by (5), we have(40)τ+Gχn+1≤Gχnκ1+κ3χn+1κ2χn+χn+12κ4−τ.
We can remark that the casemaxχn,χn+1=χn+1, is not possible since the above inequality becomes(41)Gχn+1≤Gχn+1κ1+κ2+κ3+κ4−τ<Gχn+1,a contradiction. Therefore, χn>χn+1 for all n∈ℕ, and then, there exists χ≥0 such that(42)limn⟶∞χn=limn⟶∞dνn−1,νn=χ.
We claim thatχ=0. Indeed, if we suppose that χ>0, taking the limit as n⟶∞ in (40), we have(43)τ+Gχ+0≤Gχ+0,which contradicts G2. We conclude that(44)χ=limn∞dνn−1,νn=0.
Letn∈ℕ and j≥1 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)limn⟶∞AT0νn,νn+j=0.
Therefore,limn⟶∞GAT0νn,νn+j=−∞ and since(47)τ+limn⟶∞Gdνn+1,νn+j+1≤limn⟶∞GAT0νn,νn+j,we obtain that limn⟶∞Gdνn,νn+j=−∞ and so limn⟶∞dνn,νn+j=0. Thus, νn is a Cauchy sequence on a complete metric space ℳ,d and there exists ζ such that limn⟶∞νn=ζ. Of course, it easy to see that, for ν=νn and ω=ζ, we have(48)limn⟶∞AT0νn,ζ=0.
If we suppose that there is a subsequenceνns such that Tνns=Tζ, then we have(49)0=limn⟶∞dTνns,Tζ=limn⟶∞dνns+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∈ℕ, and by (5), we obtain(50)τ+GdTνn,Tζ≤GAT0νn,ζ.
Lettingn⟶∞ and taking into account the previous considerations, we have limn⟶∞GdTνn,Tζ=−∞ and then dζ,Tζ=limn⟶∞dTνn,Tζ=0. Consequently, ζ is a fixed point of T.Example 3.
Letℳ=x,y,z,t be a set endowed with the metric d:ℳ×ℳ⟶0,∞ (Table 1).
And the mappingT:ℳ⟶ℳ is defined as T:xyztxxtt.
First, we remark that Theorem1 is not satisfied, since for ν=y and ω=t,(51)dTy,Tt=dx,t=2>1=dy,t.
Hence, for anyτ>0 and G∈B, we can write(52)τ+GdTy,Tt>Gdy,t.
Choosingτ=ln4/3, κ1=κ2=7/16, κ3=κ4=1/16, and Gt=lnt, 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.Table 1
Definition of metric d.
dν,ωxyztx0332y3031z3302t2120


## 3. Consequences

C1 Considering Gt=lnt in Theorem 5 and σi=e−τκi, we obtain Theorem 2.C2 Considering Gt=lnt in Theorem 5 and λ=e−τ, we obtain Theorem 3.C3 Considering Gt=lnt in Theorem 4, λ=e−τ, and p=1, we obtain Theorem 3.



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