# Some New Results of Interpolative Hardy–Rogers and Ćirić–Reich–Rus Type Contraction

Authors: Youssef Errai; El Miloudi Marhrani; Mohamed Aamri
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we present new concepts on completeness of Hardy–Rogers type contraction mappings in metric space to prove the existence of fixed points. Furthermore, we introduce the concept ofg-interpolative Hardy–Rogers type contractions in b-metric spaces to prove the existence of the coincidence point. Lastly, we add a third concept, interpolative weakly contractive mapping type, Ćirić–Reich–Rus, to show the existence of fixed points. These results are a generalization of previous results, which we have reinforced with examples.


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## Body

## 1. Introduction and Preliminaries

The theory of fixed points has known a lot of evolution. It has been given great merit and concern, thanks to its many uses in several fields of mathematics, such as differential equations, graph theory, and nonlinear analysis [1–8]. Besides, the emergence of the fixed theorem with Banach [4] in 1922 on complete normed space was followed by several improvements and generalizations of this theorem on two levels: the first level is related to the applications used, and the second to the spaces used in them. It first knew improvements with Kannan [9] in 1968, and later with other researchers such as Rus, Ćirić, Reich, Hardy, and Rogers. Afterwards, it took another turning with Karapinar [10] in 2018 in a new version, which has made several researchers pursue this field (see [11–19]). Thus, the concept has been applied in various spaces: metric space, b-metric space, rectangular b-metric spaces, and the Branciari distance. More recently, Errai et al. [14] have inserted g-interpolation over Ćirić–Reich–Rus type contraction. They have also introduced the concept of interpolative weakly contractive mapping, which makes us use these two concepts in this paper: the first concept on Hardy–Rogers type contraction and the second on Ćirić–Reich–Rus type contraction as a generalization of the previous findings, reinforced by various examples. This leads us to come up with some remarks. Before starting, we will take some basic concepts that we will use in this article.Definition 1.
(see [20, 21]). Let s≥1 be a given real number and Ϝ be a nonempty set. A function d:Ϝ×Ϝ⟶ℝ+ is a b-metric if the following conditions are met for all υ,y,z∈Ϝb1dυ,y=0 if and only if υ=yb2dυ,y=dy,υb3dυ,z≤sdυ,y+dy,z
The pairϜ,d is called a b-metric space.
It is worth mentioning thatb-metric spaces are a broader category than metric spaces.The definitions ofb-convergent and b-Cauchy sequences, as well as b-complete b-metric spaces, are defined in the same way as usual metric spaces (see, e.g., [22]).For the interesting examples and properties ofb-metric, see the following papers [23–25] as examples.Definition 2.
(see [26, 27]). Let υn be a sequence in a b-metric space Ϝ,d. S,h:Ϝ⟶Ϝ and υ∈Ϝ. υ is said to be coincidence point of pair S,h if Sυ=hυ.Definition 3.
(see [12, 13]). Let Ψ be the set of all nondecreasing functions ψ:0,∞⟶0,∞, with ∑k=0∞ψkt<∞ for all t>0. After that,(a)
ψt<t for each t>0(b)
ψ0=0Remark 1.
(see [22]). The following assertions apply in a b-metric space Ϝ,d:(1)
Eachb-convergent sequence is a b-Cauchy sequence.(2)
Ab-convergent sequence has a unique limit.(3)
In general, ab-metric is not continuous.To prove our results, the fact in the previous remark necessitates the following lemma regardingb-convergent sequences:Lemma 1.
(see [26]). Let Ϝ,d be a b-metric space with s≥1 , and assume that υn and yn are b-convergent to υ,y, respectively, so we have(1)1s2dυ,y≤liminfn⟶∞dυn,yn≤limsupn⟶∞dυn,yn≤s2dυ,y.
In particular, ifυ=y, then we have limn⟶∞dυn,yn=0. In addition, for each z∈Ϝ, we have(2)1sdυ,z≤liminfn⟶∞dυn,z≤limsupn⟶∞dυn,z≤sdυ,z.Lemma 2.
Letvn be a sequence defined on a b-metric space Ϝ,d,s and meets the conditions:(i)
vn is b-convergent sequence in Ϝ,d,s(ii)
dvn+1,vn≤ψdvn,vn−1≤ψ2dvn−1,vn−2≤…≤ψndv1,v0, where ψ∈Ψ
Then,vn is a b-Cauchy sequence in Ϝ,d,s.Proof.
Letp∈ℕ\0; using the triangle inequality of the b-metric space and condition (ii), we have(3)dvn,vn+p≤sdvn,vn+1+dvn+1,vn+p≤sdvn,vn+1+sdvn+1,vn+2+sdvn+2,vn+p⋮≤sψndv0,v1+s2ψn+1dv0,v1+⋯+spψn+p−1dv0,v1≤sp∑k=nn+p−1ψkdv0,v1≤sp∑k=n∞ψkdv0,v1.
Since∑k=0∞ψkt<∞ for each t>0, then limn⟶∞∑k=n∞ψkdv0,v1=0, which implies for any finite integer p≥1:(4)limn⟶∞dvn,vn+p=0.
Then,vn is a b-Cauchy sequence in Ϝ,d,s. □


## 2. Results

The set of functionsξ:0,∞⟶0,∞ which satisfies ξt<t for all t>0 is denoted by Ξ.Definition 4.
Consider the metric spaceϜ,d. If there exist α,η,ω,θ∈0,1 with α+η+ω+θ>1, the self mapping T:Ϝ⟶Ϝ is named a ξ-interpolative Hardy–Rogers type contraction, such that(5)dTυ,Ty≤ξdυ,yαdυ,Tυηdy,Tyωdυ,Ty+dy,Tυ2θ,for all υ,y∈Ϝ\FixT, where FixT=a∈Ϝ|Ta=a and ξ∈Ξ.The following is our key finding:Theorem 1.
In a complete metric spaceϜ,d, a ξ-interpolative Hardy–Rogers type contraction T:Ϝ⟶Ϝ, we assume there exists υ∈Ϝ such that dυ,Tυ<1. Then, T has a fixed point in Ϝ.Proof.
Letυn be the sequence defined by υ0=υ and υn+1=Tυn for all integer n. If there exists n0 such that υn0=υn0+1, then υn0 is a fixed point of T. The proof is complete. Suppose that υn+1≠Tυn for all n≥0.
By substituting the valuesυ=υn−1 and y=υn in (5), we have(6)dυn,υn+1≤ξdυn−1,υnαdυn−1,υnηdυn,υn+1ωdυn−1,υn+1+dυn,υn2θ≤ξdυn−1,υnα+ηdυn,υn+1ωdυn−1,υn+dυn,υn+12θ.
Using the factξt<t for each t>0, we obtain(7)dυn,υn+1≤ξdυn−1,υnα+ηdυn,υn+1ωdυn−1,υn+dυn,υn+12θ<dυn−1,υnα+ηdυn,υn+1ωdυn−1,υn+dυn,υn+12θ.
Ifdυn−1,υn<dυn,υn+1 for some n≥1, then(8)dυn−1,υn+dυn,υn+12≤dυn,υn+1.
From (7), we obtain(9)dυn,υn+1<dυn−1,υnα+ηdυn,υn+1ω+θ.
Thus,(10)dυn,υn+11−ω−θ<dυn−1,υnα+η,which implies(11)dυn−1,υn1−ω−θ<dυn−1,υnα+η,which contradicts with 1−ω−θ<α+η and dυn−1,υn<1. Then, dυn−1,υn≥dυn,υn+1 for all n≥1, and the sequence dυn−1,υn+dυn,υn+1/2≤dυn−1,υn. From (7), we deduce(12)dυn,υn+11−ω<dυn−1,υnα+η+θfor alln≥1.
Sincedυ0,υ1<1, so there exists a real l∈0,1 such that dυ0,υ1≤l and l=dυ0,υ1+1/2.
By (12), we obtain(13)dυ1,υ2≤dυ0,υ1α+η+θ/1−ω≤lα+η+θ/1−ω.
Assume that there exists a realρn such that(14)dυn,υn+1≤lρnfor alln≥0.
From (12), we deduce(15)dυn+1,υn+21−ω<dυn,υn+1α+η+θ≤lα+η+θρn,which gives(16)dυn+1,υn+2<lρn+1,where ρn+1=α+η+θ/1−ωρn for all n≥1 with ρ0=1.
Sinceα+η+θ/1−ω>1; we have limn⟶∞ρn=+∞.
Consequently,(17)∑n=0+∞dυn,υn+1≤∑n=0+∞lρn,which is convergent, so υn is Cauchy sequence in Ϝ,d, and then it converges to some ν∈Ϝ. Suppose that ν≠Tν, we find by (5):(18)dυn+1,Tν≤ξdυn,ναdυn,υn+1ηdν,Tνωdυn,Tν+dν,υn+12θ<dυn,ναdυn,υn+1ηdν,Tνωdυn,Tν+dν,υn+12θ.
Passing the limit asn⟶+∞, we get dν,Tν≤0. So dν,Tν=0 which is a contradiction. Then, Tν=ν.Example 1.
LetE=0,3 be endowed with metric μ:E×E⟶0,∞, defined by(19)μu,v=0,ifu=v,2,ifu,v∈0,1andu≠v,3,otherwise.
LetT:E⟶E be defined as(20)Tu=13,ifu∈0,1;u3,ifu∈1,3;and the function ξt=2/7t2 for all t∈0,∞.
Chooseα=0.7; η=0.6; ω=0.8; and θ=0.5.
We haveμTu,Tv≤2 for all u,v∈E.
The following issues are discussed:
First case: ifu,v∈0,1 or u=v for all u,v∈0,3, we have ξt=2/7t2≥0 for all t∈0,∞, and μTu,Tv=0 for all u,v∈0,1 or u=v for all u,v∈0,3.
Consequently, in this case, inequality (5) is satisfied.
Second case: ifu,v∈1,3 and u≠v, we have(21)ξμu,vαμu,Tuημv,Tvωμu,Tv+μv,Tu2θ=ξ3α+η+ω+θ=27⋅35.2≥2.
Third case: ifu∈0,1 and v∈1,3 with u≠1/3, we have(22)ξμu,vαμu,Tuημv,Tvωμu,Tv+μv,Tu2θ=ξ3α+ω⋅2η−θ⋅5θ=27⋅33⋅20.2⋅5≥2.
Fourth case: ifu∈1,3 and v∈0,1 with v≠1/3, we have(23)ξμu,vαμu,Tuημv,Tvωμu,Tv+dv,Tu2θ=ξ3α+η⋅2ω−θ⋅5θ=27⋅32.6⋅20.6⋅5≥2.
Hence, in all cases, we have(24)μTu,Tv≤ξμu,vαμu,Tuημv,Tvωμu,Tv+μv,Tu2θ.for all u,v∈0,3\1/3.
As a result, all the conditions of Theorem1 are fulfilled, and T has a fixed point, u=1/3.Example 2.
LetϜ=a,q,m,r be endowed with the metric given in the following chart (Table 1).
Consider the self mappingT on Ϝ as(25)T:aqmrqqar.
Takeξt=3t−1/3t+1, for all t∈0,∞; α=0.4;η=0.8; ω=0.9; and θ=0.5.
We have(26)dTu,Tv≤ξdu,vαdu,Tuηdv,Tvωdu,Tv+dv,Tu2θ,for all u,v∈Ϝ\q,r.
Then,T has two fixed points, which are q and r.Table 1
Table of metricd for Example 2.
dx,yaqmra00.231q0.202.80.8m32.802r10.820If we replaceξt=λt with λ∈0,1 in Theorem 1, we get the following corollary.Corollary 1.
LetϜ,d be a complete metric space and T is self mapping on Ϝ such that(27)dTυ,Ty≤λdυ,yαdυ,Tυηdy,Tyωdυ,Ty+dy,Tυ2θis satisfied for all υ,y∈Ϝ\FixT, where FixT=a∈Ϝ|Ta=a and α,η,ω,θ,λ∈0,1 such that α+η+ω+θ>1.
If there existsυ∈Ϝ such that dx,Tx<1, then T has a fixed point in Ϝ.Definition 5.
LetϜ,d,s be a b-metric space and T,g:Ϝ⟶Ϝ be self mappings on Ϝ. We say that T is a g-interpolative Hardy–Rogers type contraction if there exist ψ∈Ψ and α,η,ω∈0,1 such that(28)dTυ,Ty≤ψdgυ,gyαdgυ,Tυηdgy,Tyωdgυ,Ty+dgy,Tυ2s1−α−η−ωis satisfied for all υ,y∈Ϝ such that Tυ≠gυ,Ty≠gy, and gυ≠gy.Theorem 2.
In ab-complete b-metric space Ϝ,d,s, if T is a g-interpolative Hardy–Rogers type contraction such that(1)
TϜ⊆gϜ(2)
gϜ is closed
Then,T and g have a coincidence point in Ϝ.Proof.
Letυ∈Ϝ, since TϜ⊆gϜ, we can inductively define a sequence υn such that(29)υ0=υ,andgυn+1=Tυn,for all integern.
If there existsn∈0,1,2,… such that gυn=Tυn, then υn is a coincidence point of g and T. Assume that gυn≠Tυn, for all n. By (28), we obtain(30)dTυn+1,Tυn≤ψdgυn+1,gυnαdgυn+1,Tυn+1ηdgυn,Tυnωdgυn+1,Tυn+dgυn,Tυn+12s1−α−η−ω=ψdTυn,Tυn−1αdTυn,Tυn+1ηdTυn−1,TυnωdTυn,Tυn+dTυn−1,Tυn+12s1−α−η−ω≤ψdTυn,Tυn−1α+ωdTυn,Tυn+1ηdTυn−1,Tυn+dTυn,Tυn+121−α−η−ω.
Using the factψt<t for each t>0, we obtain(31)dTυn+1,Tυn≤dTυn,Tυn−1α+ωdTυn,Tυn+1ηdTυn−1,Tυn+dTυn,Tυn+121−α−η−ω.
Suppose thatdTυn−1,Tυn<dTυn,Tυn+1 for some n≥1. Then,(32)dTυn−1,Tυn+dTυn,Tυn+12≤dTυn,Tυn+1.
Thus, from inequality (31), we have(33)dTυn+1,Tυn≤dTυn,Tυn−1α+ωdTυn,Tυn+11−α−ω.
This implies(34)dTυn+1,Tυnα+ω≤dTυn,Tυn−1α+ω.
So, we get(35)dTυn+1,Tυn≤dTυn,Tυn−1,which is a contradiction. Thus,(36)dTυn+1,Tυn≤dTυn,Tυn−1,for alln≥1.
This means that the positive sequencedTυn+1,Tυn is monotone decreasing, and consequently, there exists c≥0 such that limn⟶+∞dTυn+1,Tυn=c.
From (36), we have(37)dTυn,Tυn−1α+ωdTυn,Tυn+1ηdTυn−1,Tυn+dTυn,Tυn+121−α−η−ω≤dTυn,Tυn−1α+ωdTυn,Tυn−1ηdTυn−1,Tυn1−α−η−ω=dTυn,Tυn−1.
So, (31) together with the nondecreasing character of ψ, we obtain(38)dTυn+1,Tυn≤ψdTυn,Tυn−1.
Similar to the previous method, we find(39)dTυn+1,Tυn≤ψdTυn,Tυn−1≤ψ2dTυn−1,Tυn−2≤…≤ψndTυ1,Tυ0.
Lettingn⟶∞ in (39) and using the fact limn⟶∞ψnt=0 for each t>0, we deduce that c=0, that is,(40)limn⟶∞dTυn+1,Tυn=0.
By Lemma2, we deduce that Tυn is a b-Cauchy sequence, and consequently, gυn is also a b-Cauchy sequence. Let z∈Ϝ such that,(41)limn⟶∞dTυn,z=limn⟶∞dgυn+1,z=0.
And, sincez∈gϜ, there exists u∈Ϝ such that z=gu. We claim that u is a coincidence point of g and T. Thus, if we assume that gu≠Tu, we obtain(42)dTυn,Tu≤ψdgυn,guαdgυn,Tυnηdgu,Tuωdgυn,Tu+dgu,Tυn2s1−α−η−ω<dgυn,guαdgυn,Tυnηdgu,Tuωdgυn,Tu+dgu,Tυn2s1−α−η−ω.
Consequently,(43)1sdz,Tu≤sdz,guαs2dz,zηdgu,Tuωdz,Tu+dgu,z21−α−η−ω=0,which is a contradiction. This implies that(44)Tu=z=gu.
Then,u is a coincidence point in Ϝ of T and g.Example 3.
Let the setϜ=a,r,i,m and a function d:Ϝ×Ϝ⟶0,∞ defined as follows (Table 2)
One can check that the functiond is a b-metric for s = 2 by doing a simple calculation. Self-mappings g,T on Ϝ are defined as(45)g:arimamii,T:arimimmi.
Chooseα=0.4;η=0.1;ω=0.7; and ψt=5t−1/5t+1 for all t∈0,∞.
For allu,v∈Ϝ\r,m, it is obvious that g,T fulfills (28). Furthermore, r and m are two coincidence points of g and T.Table 2
Table of metricd for Example 3.
dx,yarima0449/169/4r401/161/4i49/161/1601/16m9/41/41/160Example 4.
LetϜ=0,+∞ and d:Ϝ×Ϝ⟶0,∞ be defined by(46)dυ,y=υ+y2,ifυ≠y;0,ifυ=y.
Then,Ϝ,d is a complete b-metric space.
Define two self-mappingsT and g on Ϝ by gυ=υ2, for all υ∈Ϝ and(47)Tx=1,ifυ∈0,2;e−υ,ifυ∈2,+∞.
T is a g-interpolative Hardy–Rogers type contraction for α=0.3; η=0.6; and ω=0.4. Let ψt=3/5t2, for all t∈0,+∞.
Hence, the following issues are discussed.
First case: ifυ,y∈0,2 or υ=y for all υ∈0,+∞, this is obvious.
Second case: ifυ∈0,2\1 and y∈2,+∞.
Consider(48)Mυ,y=dgυ,gyαdgυ,Tυηdgy,Tyωdgυ,Ty+dgy,Tυ2s1−α−η−ω.
We have(49)dTυ,Ty=1+e−υ2≤1+e−22and(50)ψMυ,y=ψdgυ,gyαdgυ,Tυηdgy,Tyωdgυ,Ty+dgy,Tυ2s1−α−η−ω=ψυ2+y22αυ2+12ηy2+e−y2ωυ2+e−υ2+y2+1241−α−η−ω≥ψ42α⋅12η⋅42ω5241−α−η−ω=ψ41.7⋅5−0.6=3⋅43.4⋅5−2.2≥1+e−22.
Therefore,(51)dTυ,Ty≤ψdgυ,gyαdgυ,Tυηdgy,Tyωdgυ,Ty+dgy,Tυ2s1−α−η−ω.
Third case: ifυ∈2,+∞ and y∈0,2\1.
We have(52)dTυ,Ty=e−υ+12≤1+e−22,and(53)ψMυ,y=ψdgυ,gyαdgυ,Tυηdgy,Tyωdgυ,Ty+dgy,Tυ2s1−α−η−ω=ψυ2+y22αυ2+e−υ2ηy2+12ωυ2+12+y2+e−υ241−α−η−ω≥ψ42α⋅42η⋅12ω52+0241−α−η−ω=3⋅44.2⋅5−2.2≥1+e−22.
Therefore,(54)dTυ,Ty≤ψdgυ,gyαdgυ,Tυηdgy,Tyωdgυ,Ty+dgy,Tυ2s1−α−η−ω.
Fourth case: ifυ,y∈2,+∞ and υ≠y, we have(55)dTυ,Ty=e−υ+e−y2≤2e−4.
Using the property ofψ, we get(56)ψMυ,y=ψdgυ,gyαdgυ,Tυηdgy,Tyωdgυ,Ty+dgy,Tυ2s1−α−η−ω=ψυ2+y22αυ2+e−υ2ηy2+e−y2ωυ2+e−y2+y2+e−υ241−α−η−ω≥ψ82α⋅42η⋅42ω42+4241−α−η−ω=35⋅29.8≥2e−4.
Therefore,(57)dTυ,Ty≤ψdgυ,gyαdgυ,Tυηdgy,Tyωdgυ,Ty+dgy,Tυ2s1−α−η−ω.
For allυ,y∈Ϝ\1, it is obvious that g,T fulfills (28). Furthermore, one is a coincidence point of g and T.The two previous examples lead us to the following remark.Remark 2.
In the Theorem2, T and g do not need a fixed point, just as T and g accept a coincidence point and are not necessarily unique.Definition 6.
In a metric spaceϜ,d, we say that a self mapping T:Ϝ⟶Ϝ is an interpolative weakly contractive mapping type Ćirić–Reich–Rus, if there exists a constant α,η∈0,1 such that(58)ζdTυ,Ty≤ζRυ,y−φRυ,y,for all υ,y∈Ϝ\FixT, where(59)FixT=a∈X|Ta=a,Rυ,y=dυ,yαdυ,Tυηdy,Ty1−α−η,φ:0,∞⟶0,∞ is a lower semicontinuous function with φt=0 if and only if t=0, ζ:0,∞⟶0,∞ is a continuous monotone nondecreasing function with ζt=0 if and only if t=0.Theorem 3.
In a complete metric spaceϜ,d, if T:Ϝ⟶Ϝ is a interpolative weakly contractive mapping type Ćirić–Reich–Rus, then T has a fixed point.Proof.
For anyυ0∈Ϝ, we consider a sequence υn by υ=υ0 and υn+1=Tυn, n=0,1,2,…
If there existsn0∈ℕ such that υn0+1=υn0, then υn0 is clearly a fixed point in Ϝ. Otherwise, υn+1≠υn for each n≥0.
Substitutingυ=υn and y=υn−1 in (58), we obtain that(60)ζdυn+1,υn≤ζdυn,υn−1αdυn,υn+1ηdυn−1,υn1−α−η−φdυn,υn−1αdυn,υn+1ηdυn−1,υn1−α−η≤ζdυn,υn−11−ηdυn,υn+1η.
Using property of functionζ, we get(61)dυn+1,υn≤dυn,υn−11−ηdυn,υn+1η.
We derive(62)dυn+1,υn1−η≤dυn,υn−11−η.
Therefore,(63)dυn+1,υn≤dυn,υn−1,for alln≥1.
It follows that the positive sequencedυn+1,υn is decreasing. Eventually, there exists c≥0 such that limn⟶∞dυn+1,υn=c.
Takingn⟶∞ in inequality (60), we obtain(64)ζc≤ζc−φc.
We deduce thatc=0. Hence,(65)limn⟶∞dυn+1,υn=0.
Therefore,υn is a Cauchy sequence. Suppose not, then there exists a real number ε>0, for any k∈ℕ,∃mk≥nk≥k such that(66)dυmk,υnk≥εanddυmk−1,υnk<ε.
Puttingυ=υnk−1 and y=υmk−1 in (58) and using (66), we obtain(67)ζε≤ζdυmk,υnk≤ζRυmk−1,υnk−1−φRυmk−1,υnk−1,where(68)Rυmk−1,υnk−1=dυmk−1,υnk−1αdυmk−1,υmkηdυnk−1,υnk1−α−ηand(69)dυmk−1,υnk−1≤dυmk−1,υnk+dυnk,υnk−1≤ε+dυnk,υnk−1.
Lettingk⟶∞ and using (65), we conclude(70)limk⟶∞Rυmk−1,υnk−1=0.
Then,(71)ζε≤ζ0−φ0=0,which is contradiction with ε>0. Thus, υn is a Cauchy sequence, and since Ϝ,d is complete, we obtain τ∈Ϝ such that limn⟶∞dυn,τ=0 and assume that Tτ≠τ, we have(72)ζdυn+1,Tτ≤ζRυn,τ−φRυn,τfor alln,where(73)Rυn,τ=dυn,ταdυn,υn+1ηdτ,Tτ1−α−η.
Using (65), we get(74)limn⟶∞Rυn,τ=0.
Lettingn⟶∞ in (72), we get(75)ζdτ,Tτ≤ζ0−φ0=0,which is a contradiction. Thus, Tτ=τ. □Example 5.
Consider the spaceϜ=a,r,i,m equipped with the metric defined by the values of the following table (Table 3).
Consider the self mappingT on Ϝ as(76)T:arimimim.
Forζt=et−1 and φt=3/2t−1 for all t∈0,∞, taking α=0.2 and η=0.7, we have(77)ζdTu,Tv≤ζdu,vαdu,Tuηdv,Tv1−α−η−φdu,vαdu,Tuηdv,Tv1−α−η,for all u,v∈Ϝ\i,m.
Then,T possesses two fixed points: i and m.Table 3
Table of metricd for Example 5.
dx,yarima0278r2056i7501m8610Example 6.
Let the setΛ=0,5 and a function ϱ:Λ×Λ⟶0,∞ be defined as follows:(78)ϱυ,y=0,ifυ=y,5,ifυ,y∈0,1andυ≠y,3,otherwise.
Then,Λ,ϱ is a complete metric space.
LetT: Λ⟶Λ be defined as(79)Tx=0,ifυ∈0,1;3,ifυ∈1,5.
Chooseζt=t2 and φt=1/3t for all t∈0,+∞, taking α=0.4 and η=0.3, the following issues are discussed:
First case: ifυ=y or υ,y∈0,1, or υ,y∈1,5\3 with υ≠y, this is obvious.
Second case: Ifυ∈0,1 and y∈1,5\3, we have(80)ζϱTυ,Ty=ζϱ0,3=ζ3=9,and(81)ϱυ,yαϱυ,Tυηϱy,Ty1−α−η=3α⋅5η⋅31−α−η=3⋅53η.
Therefore,(82)ζϱυ,yαϱυ,Tυηϱy,Ty1−α−η−φϱυ,yαϱυ,Tυηϱy,Ty1−α−η=53η9⋅53η−1≥9=ζ3=ζϱTυ,Ty.Third case: ifυ∈1,5\3 and y∈0,1, we have(83)ζϱTυ,Ty=ζϱ3,0=9and(84)ϱυ,yαϱυ,Tυηϱy,Ty1−α−η=3α⋅3η⋅51−α−η=5⋅35α+η.Thus,(85)ζϱυ,yαϱυ,Tυηϱy,Ty1−α−η−φϱυ,yαϱυ,Tυηϱy,Ty1−α−η=35α+η25⋅35α+η−53≥9=ζ3=ζϱTυ,Ty.Hence,(86)ζdTu,Tv≤ζϱυ,yαϱυ,Tυηϱy,Ty1−α−η−φϱυ,yαϱυ,Tυηϱy,Ty1−α−η,for all υ,y∈Λ\0,3.Then,T possesses two fixed points: 0 and 3.The two previous examples lead us to the next remark.Remark 3.
IfT is an interpolative weakly contractive mapping type Ćirić–Reich–Rus, T accepts a fixed point that is not necessarily a single one.We have the following corollary ifζt=t in Theorem 3:Corollary 2.
LetϜ,d be complete metric space and T:Ϝ⟶Ϝ self mapping on Ϝ. If there exists a constant α,η∈0,1 such that(87)dTυ,Ty≤dυ,yαdυ,Tυηdy,Ty1−α−η−φdυ,yαdυ,Tυηdy,Ty1−α−η,for all υ,y∈Ϝ and υ≠Tυ,y≠Ty, where φ:0,∞⟶0,∞ is a lower semicontinuous function with φt=0 if and only if t=0, then T has a fixed point.If we useφt=1−λt for a constant λ,η∈0,1 in Corollary 2, then we get the following corollary.Corollary 3.
LetϜ,d be complete metric space and T:Ϝ⟶Ϝ self mapping on Ϝ. If there exists a constant α,η∈0,1 such that(88)dTυ,Ty≤λdυ,yαdυ,Tυηdy,Ty1−α−η,for all υ,y∈Ϝ and υ≠Tυ,y≠Ty, then T has a fixed point.



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*Source: 9992783-2021-05-20.xml*