# An Implicit Relation Approach in Metric Spaces underw-Distance and Application to Fractional Differential Equation

Authors: Reena Jain; Hemant Kumar Nashine; Santosh Kumar
Journal: Journal of Function Spaces
 (2021)

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## Abstract

The purpose of this work is to introduce a new class of implicit relation and implicit type contractive condition in metric spaces underw-distance functional. Further, we derive fixed point results under a new class of contractive condition followed by three suitable examples. Next, we discuss results about weak well-posed property, weak limit shadowing property, and generalized w-Ulam-Hyers stability of the mappings of a given type. Finally, we obtain sufficient conditions for the existence of solutions for fractional differential equations as an application of the main result.


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## Body

## 1. Introduction and Preliminaries

In1996, Kada et al. [1] introduced the concept of a w-distance on a metric space and proved a generalized Caristi fixed point theorem, Ekeland’s ε-variational principle, and the nonconvex minimization theorem according to Mizoguchi and Takahashi [2].Definition 1 (see [1]).
LetΞ,d be a metric space. A function ω:Ξ×Ξ⟶0,∞ is called a w-distance on Ξ if it satisfies the following properties:
(W1) ωϑ,μ≤ωϑ,ν+ων,μ for any ϑ,ν,μ∈Ξ
(W2) ω is lower semicontinuous in its second variable, i.e., if ϑ∈Ξ and νn⟶ν∈Ξ, then ωϑ,ν≤liminfn⟶∞ωϑ,νn
(W3) For each ε>0, there exists a δ>0 such that ωμ,ϑ≤δ and ωμ,ν≤δ imply dϑ,ν≤εThe following examples show that aw-distance is not necessarily a metric.Example 1.
(1)
LetΞ,d be a metric space and c>0. Define w1ϑ,ν=cdϑ,ν and w2ϑ,ν=c for all ϑ,ν∈Ξ. Then, w1 and w2 satisfy (W1)–(W3). Obviously, w2 is not a metric since w2ϑ,ϑ=c≠0 for any ϑ∈Ξ(2)
Considerℝ as a metric space with the usual metric. Define(1)w3ϑ,ν=νforeveryϑ,ν∈ℝ×ℝThen,w3 is a w-distance on ℝ which is not a metric (since it is not symmetric). Note that ϑn=1,1,1,⋯ is a convergent sequence in ℝ but w3ϑn,ϑm=1 for all m,n∈ℕ.To prove the main theorem, we need the following lemma, proved by Kada et al. [1].Lemma 2.
LetΞ,d be a metric space and let ω be a w-distance on Ξ. Suppose that ϑn and νn are sequences in Ξ, αn and βn are sequences in 0,∞ converging to 0, and let ϑ,ν,μ∈Ξ. Then, the following assertions hold.
(i)
Ifωϑn,ν≤αn and ωϑn,μ≤βn for all n∈ℕ, then ν=μ. In particular, if ωϑ,ν=ωϑ,μ=0, then ν=μ(ii)
Ifωϑn,νn≤αn and ωϑn,ν≤βn for all n∈ℕ, then νn converges to ν(iii)
Ifωϑn,ϑm≤αn for all n,m∈ℕ with m>n, then ϑn is a Cauchy sequence(iv)
Ifων,ϑn≤αn for all n∈ℕ, then ϑn is a Cauchy sequenceLemma 3 (see [1, 3]).
Letω be a w-distance on a metric space Ξ,d and ϑn be a sequence in Ξ such that for each ε>0 there exists Nε∈ℕ such that m>n>Nε implies ωϑn,ϑm<ε, i.e., limm,n⟶∞ωϑn,ϑm=0. Then, ϑn is a Cauchy sequence.Recall that the setOθ0;I=Inθ0:n=0,1,2,⋯ is called the orbit of the self-map I at the point θ0∈Ξ.Definition 4.
LetΞ,d be metric spaces and I:Ξ⟶Ξ be a mapping. Then,
(1)
FixI=u∈Ξ∣u=Iu (the set of fixed points of I)(2)
a mappingI is called a Picard operator if there exists u∈Ξ such that FixI=u and Inx converges to u, for all x∈Ξ(3)
[4] a metric space Ξ is said to be I-orbitally complete if every Cauchy sequence contained in Ox;I (for some x in Ξ) converges in Ξ(4)
a mappingI is said to be orbitally U-continuous if, for some U⊂Ξ×Ξ, the following condition holds: for any x∈Ξ and a strictly increasing sequence ni of positive integers(2)limi⟶∞Inix=z∈Ξ,and Inix,z∈U for any i∈ℕ imply that
(3)limi⟶∞Ini+1x=Iz(5)
I is called orbitally continuous if, for any x∈Ξ and a strictly increasing sequence ni of positive integers, Inix⟶z∈Ξ as i⟶∞ implies that Ini+1x⟶Iz as i⟶∞In [5], Samet et al. defined the notion of α-admissible mapping which was further sharpened by Karapinar et al. [6] and extended in [7].Definition 5.
For a setΞ≠∅, let α:Ξ×Ξ⟶ℝ+ and I:Ξ⟶Ξ be two mappings. Then, I is said to be
(i)
[5] α-admissible if(4)ν,ϑ∈Ξwithαν,ϑ≥1⇒αIν,Iϑ≥1(ii)
[6] triangular α-admissible if I is α-admissible and(5)αν,ϑ≥1andαϑ,μ≥1implyαν,μ≥1Lemma 6 (see [6]).
LetI:Ξ⟶Ξ be a triangular α-admissible mapping. Assume that there exists ν0∈Ξ such that αν0,Iν0≥1. Define a sequence νn by νn+1=Iνn for n∈ℕ∗. Then, ανn,νm≥1 for all m,n∈ℕ with n<m.Similarly, we can state and prove the following lemma.Lemma 7.
LetI:Ξ⟶Ξ be a triangular α-admissible mapping. Assume that there exists ν0∈Ξ such that αIν0,ν0≥1. Define a sequence νn by νn+1=Iνn for n∈ℕ∗. Then, ανn,νm≥1 for all m,n∈ℕ with n>m.To the best of our knowledge, there is no fixed point result in the literature which has been derived by implicit type contractive relation in a metric space underw-distance. Also, w-distance is not necessarily a metric (examples are given above). Motivated by this fact, there is a need for introducing such type of contractive conditions. With this in mind, in Section 2, we introduce the notion of a new implicit relation and ω-implicit contractive mapping in the respective structure. Then, we establish unique fixed point results under aforesaid implicit contractive condition for α-admissible and orbitally continuous mappings on orbitally complete spaces. We demonstrate the results by three illustrative examples. We note that the symmetry condition and full completeness of the underlying space are not required. In Section 4, some new results on weak well-posed property, weak limit shadowing property, and generalized w-Ulam-Hyers stability of mappings of the mentioned type are discussed. In Section 4, a sufficient condition for the existence of solutions for fractional differential equations as an application of the main result is given.


## 2. Implicit Relation forw-Distance on Metric Spaces

In this section, we introduce a modified version of implicit relation and examples discussed in [8, 9].LetΨ be the set of all continuous functions ψ:ℝ+6⟶ℝ satisfying the following conditions:(ψ1) ψ is nonincreasing in the fifth and sixth variables(ψ2) There exists ℏ∈0,1 such that for all ζ,ξ,μ≥0, k≥1.(ψ2a) ψkζ,ξ,ξ,ζ,ζ+ξ,μ≤0 implies that kζ≤ℏξ(ψ2b) ψkζ,ξ,ξ,ζ,μ,ζ+ξ≤0 implies that kζ≤ℏξ(ψ3) ψkζ,ζ,0,0,ζ,ζ>0 and ψkζ,0,0,ζ,ζ,0>0 for ζ>0andk≥1Example 2.
Letψℓ1,ℓ2,ℓ3,ℓ4,ℓ5,ℓ6=ℓ12−aℓ22−bℓ32+ℓ42/ℓ5+ℓ6+1, 0<a,b<1 and a+2b<1.
(ψ2a) For ζ,ξ,μ≥0, we have
(6)ψkζ,ξ,ξ,ζ,ζ+ξ,μ=k2ζ2−aξ2−bζ2+ξ21+ζ+ξ+μ≤0,which implies that k2ζ2≤aξ2+bζ2+ξ2. Then, k2ζ2≤a+b/1−bξ2. Hence, kζ≤ℏξ, ℏ=ka+b/k2−b<1 for k≥1
(ψ2b) Similarly as (ψ2a), if ψkζ,ξ,ξ,ζ,μ,ζ+ξ≤0, then kζ≤ℏξ for k≥1
(ψ3) For all ζ>0 and for k≥1, ψkζ,ζ,0,0,ζ,ζ=k−aζ2>0Example 3.
Letψℓ1,ℓ2,ℓ3,ℓ4,ℓ5,ℓ6=ℓ12−aℓ22−bℓ32+ℓ42/ℓ52+ℓ62+1,0<a,b<1 and a+2b<1.Similar to Example2.Example 4.
Letψℓ1,ℓ2,ℓ3,ℓ4,ℓ5,ℓ6=ℓ1−aℓ2−bℓ3−cℓ4ℓ5/ℓ5+ℓ6+1,0<a,b,c<1 and a+b+c<1.
(ψ2a) For ζ,ξ,μ≥0, we have
(7)ψkζ,ξ,ξ,ζ,ζ+ξ,μ=kζ−aξ−bξ−cζζ+ξ1+ζ+ξ+μ≤0,which implies that kζ−a+bξ≤cζ, that is, kζ≤ℏξ, ℏ=ka+b/k−c<1 for k≥1
(ψ2b) Similarly as (ψ2a), if ψkζ,ξ,ζ,ξ,0,μ≤0, then kζ≤ℏξ
(ψ3) ψℏζ,ζ,0,0,ζ,ζ=k−aζ>0 for all ζ>0 and for k≥1Example 5.
Letψℓ1,ℓ2,ℓ3,ℓ4,ℓ5,ℓ6=ℓ13−aℓ32ℓ42/ℓ2+ℓ5+ℓ6+1,0≤a<1.
(ψ2a) For ζ,ξ,μ≥0 and ψℏζ,ξ,ξ,ζ,μ,0=k3ζ3−aζ2ξ2/ξ+μ+1≤0.
Then,k3ζ≤aξ2/ξ+μ+1<aξ, that is, kζ≤a/k2ξ. Hence, kζ≤ℏξ, ℏ=a/k2<1 for k≥1.
(ψ2b) Similarly, if ψkζ,ξ,ζ,ξ,0,μ≤0, then kζ≤hξ
(ψ3) ψkζ,ζ,0,0,ζ,ζ=k3ζ3>0 for all ζ>0 for k≥1Example 6.
Letψℓ1,ℓ2,ℓ3,ℓ4,ℓ5,ℓ6=ℓ1−aℓ2−bℓ3ℓ6/ℓ5+ℓ6+1−cℓ4,0<a,b,c<1 and a+b+c<1.
(ψ2a) For ζ,ξ,μ≥0, we have
(8)ψkζ,ξ,ξ,ζ,ζ+ξ,μ=kζ−aξ−bζξ1+ζ+ξ+μ−cζ≤0,which implies that kζ−aξ−bζ≤cζ, that is, kζ≤ℏξ, ℏ=ka/k−b−c<1 for k≥1
(ψ2b) Similarly as (ψ2a), if ψkζ,ξ,ζ,ξ,0,μ≤0, then kζ≤ℏξ
(ψ3) ψkζ,ζ,0,0,ζ,ζ=k−aζ>0 for all ζ>0 for k≥1Now, we defineω-implicit contractive mapping in the metric space under w-distance using the above introduced implicit relation.Definition 8.
LetΞ,d be a metric space with w-distance ω, I:Ξ⟶Ξ be a given mapping, and α:Ξ×Ξ⟶ℝ+ be a functional. We say that I is an ω-implicit contractive mapping, if there exists a function ψ∈Ψ such that
(9)ψαϑ,νωIϑ,Iν,ωϑ,ν,ωϑ,Iϑ,ων,Iν,ωϑ,Iν,ωIϑ,ν≤0,for all ϑ,ν∈Ξ2.If (9) is satisfied for ϑ,ν∈Oθ0;I¯ (for some θ0∈Ξ), we say that I is an orbitally ω-implicit contractive mapping (at θ0).Now, we are equipped to state and prove our first main result as follows.Theorem 9.
LetΞ,d be a metric space with w-distance ω and I:Ξ⟶Ξ. Suppose that the following conditions hold:
(i)
There existsθ0∈Ξ such that αθ0,Iθ0≥1 and αIθ0,θ0≥1(ii)
I is a triangular α-admissible mapping(iii)
I is an orbitally ω-implicit contractive mapping(iv)
Ξ,ω is I-orbitally complete at θ0(v)
I is orbitally continuous
Then, there exists a pointϑ∗∈FixI. In addition, ωϑ∗,ϑ∗=0 provided αϑ∗,ϑ∗≥1 holds.Proof.
Letθ0∈Ξ be the point described in (i). Define a sequence θn by θn+1=Iθn for n≥1. If θn0=θn0+1 for some n0≥1, then obviously FixI≠∅. Hence, we suppose that θn≠θn+1 for all n≥1. First, we show that
(10)limn⟶∞ωInθ0,In+1θ0=0.
Using (ii) and Lemma6, we have αθn,θn+1≥1 for all n≥1. Then, for all n≥1, using (9) for ϑ=θn−1, ν=θn,
(11)ψαθn−1,θnωInθ0,In+1θ0,ωIn−1θ0,Inθ0,ωIn−1θ0,Inθ0,ωInθ0,In+1θ0,ωIn−1θ0,In+1θ0,ωInθ0,Inθ0≤0.
Denotingρn=ωInθ0,In+1θ0 for all n∈ℕ∗ and applying ψ1 in the fifth variable, we have
(12)ψαθn−1,θnρn,ρn−1,ρn−1,ρn,ρn−1+ρn,ωInθ0,Inθ0≤0.
It follows fromψ2a that there is ℏ∈0,1 such that
(13)ρn≤αθn,θn+1ρn≤ℏρn−1,and so,
(14)ρn≤ℏρn−1;that is, the sequence ρn is a nonincresing sequence of real numbers. Therefore, there exists ζ such that
(15)limn⟶∞ρn=limn⟶∞ωInθ0,In+1θ0=ζ,∀n∈ℕ.
Applying the limit in (12), by the continuity of ψ, we get
(16)ζ≤limn⟶∞αθn,θn+1ρn≤hζ,
a contradiction, and therefore,ζ=0.
ForαIθ0,θ0≥1, using condition (ii) and Lemma 7, we get αθn+1,θn≥1 for all n≥1. Using similar arguments as above, we can prove
(17)limn⟶∞ωIn+1θ0,Inθ0=0.
Next, we show thatInθ0 is a Cauchy sequence in Oθ0;I. For this, we show that
(18)limm,n⟶∞ωInθ0,Imθ0=0.
On the contrary, suppose that condition (18) does not hold. Then, we can find a δ>0 and increasing sequences mkk=1∞,nkk=1∞ of positive integers with mk>nk such that
(19)ωInkθ0,Imkθ0≥δ,forallk∈1,2,3,⋯.
By (10), there exists a k0∈ℕ, such that nk>k0 implies that
(20)ωInkθ0,Ink+1θ0<δ.
In view of the two last inequalities, we observe thatmk≠nk+1. We may assume that mk is the minimal index such that (19) holds, so that
(21)ωInkθ0,Irθ0<δ,forr∈nk+1,nk+2,⋯,mk−1.
Now, making use of (19), we get
(22)0<δ≤ωInkθ0,Imkθ0≤ωInkθ0,Imk−1θ0+ωImk−1θ0,Imkθ0<δ+ωImk−1θ0,Imkθ0.
Thus,(23)limk⟶∞ωInkθ0,Imkθ0=δ.
Using the triangle inequality, we have(24)ωInkθ0,Imkθ0≤ωInkθ0,Imk+1θ0+ωImk+1θ0,Imkθ0≤ωInkθ0,Ink+1θ0+ωInk+1θ0,Imk+1θ0+ωImk+1θ0,Imkθ0.
Taking the limit on both sides and making use of (10), (17), and (23), we obtain
(25)limk⟶∞ωInk+1θ0,Imk+1θ0≥δ.
Again, using the triangle inequality, we have(26)ωInk+1θ0,Imk+1θ0≤ωInk+1θ0,Imkθ0+ωImkθ0,Imk+1θ0≤ωInk+1θ0,Inkθ0+ωInkθ0,Imkθ0+ωImkθ0,Imk+1θ0.
Taking the limit on both sides and making use of (10), (17), and (23), we obtain
(27)limk⟶∞ωInk+1θ0,Imk+1θ0≤δ.
Combining (25) and (27), we have
(28)limk⟶∞ωInk+1θ0,Imk+1θ0=δ.
From Lemma6, we have αθnk,θmk≥1. Therefore, on applying condition (9), we get
(29)ψαθnk,θmkωInk+1θ0,Imk+1θ0,ωInkθ0,Imkθ0,ωInkθ0,Ink+1θ0,ωImkθ0,Imk+1θ0,ωInkθ0,Imk+1θ0,ωInk+1θ0,Imkθ0≤0.
Now applyingψ1 in the fifth and sixth variables, we have
(30)ψαθnk,θmkωInk+1θ0,Imk+1θ0,ωInkθ0,Imkθ0,ωInkθ0,Ink+1θ0,ωImkθ0,Imk+1θ0,ωInkθ0,Ink+1θ0+ωInk+1θ0,Imk+1θ0,ωInk+1θ0,Imk+1θ0+ωImk+1θ0,Imkθ0≤0.
Applying the limit and using continuity ofψ, we get
(31)ψδlimk⟶∞αθnk,θmk,δ,0,0,δ,δ≤0,
a contradiction toψ3. Hence, Inθ0 must be a Cauchy sequence in Oθ0;I.
SinceΞ is I-orbitally complete, there exists a point ϑ∗∈Ξ such that limn⟶∞Inθ0=ϑ∗. We shall show that ϑ∗ is a fixed point of I.
Using the orbital continuity ofI (due to condition (v)), we have limn⟶∞IInθ0=Iϑ∗. Owing to the uniqueness of the limit, we obtain Iϑ∗=ϑ∗.
Finally, assume thatωϑ∗,ϑ∗>0. Then, by (9) for αϑ∗,ϑ∗≥1, we have
(32)ψαϑ∗,ϑ∗ωIϑ∗,Iϑ∗,ωϑ∗,ϑ∗,ωϑ∗,Iϑ∗,ωIϑ∗,ϑ∗,ωIϑ∗,ϑ∗,ωϑ∗,Iϑ∗≤0or
(33)ψαϑ∗,ϑ∗ωϑ∗,ϑ∗,ωϑ∗,ϑ∗,ωϑ∗,ϑ∗,ωϑ∗,ϑ∗,ωϑ∗,ϑ∗,ωϑ∗,ϑ∗≤0.
It follows fromψ2b (for ζ=ξ=μ=ϑ∗) that there is ℏ∈0,1 such that
(34)αϑ∗,ϑ∗ωϑ∗,ϑ∗≤ℏωϑ∗,ϑ∗,
a contradiction. Therefore,ωϑ∗,ϑ∗=0.☐Next, we have the following result.Theorem 10.
The conclusion of Theorem9 remains true if condition (v) is replaced by the following one:
(v’) For every ν∈Ξ with ν≠Iν, infωϑ,ν+ωϑ,Iϑ∣ϑ∈Ξ>0Proof.
Following the proof of Theorem9, we observe that the sequence Inθ0 is a Cauchy sequence, and so, there exists a point ϑ∗ in Ξ such that limn⟶∞Inθ0=ϑ∗. Since limm,n⟶∞ωInθ0,Imθ0=0, for each ε>0, there exists an Nε∈ℕ such that n>Nε implies ωINεθ0,Inθ0<ε. Since limn⟶∞Inθ0=ϑ∗ and ωϑ,· is lower semicontinuous,
(35)ωINεθ0,ϑ∗≤liminfn⟶∞ωINεθ0,Inθ0<ε.
Therefore,ωINεθ0,ϑ∗≤ε. Set ε=1/k, Nε=nk so that
(36)limk⟶∞ωInkθ0,ϑ∗=0.
Assume thatIϑ∗≠ϑ∗. Then, by the hypothesis (v’), we have
(37)0<infωϑ,ϑ∗+ωϑ,Iϑ∣ϑ∈Ξ≤infωInkθ0,ϑ∗+ωInkθ0,Ink+1θ0∣n∈ℕ⟶0,which contradicts our assumption. Therefore, Iϑ∗=ϑ∗.☐The last conclusion is derived as in the proof of Theorem9.In what follows, we give a sufficient condition for the uniqueness of the fixed point in Theorems9 and 10.Theorem 11.
In addition to the hypotheses of Theorem9 (or Theorem 10), if for all fixed points θ∗ and η∗ such that θ∗≠η∗∈Ξ, αθ∗,η∗≥1, then I has a unique fixed point.Proof.
Suppose thatθ∗ and η∗ are two fixed points of I such that θ∗≠η∗. Then, using (9) for αθ∗,η∗≥1,
(38)ψαθ∗,η∗ωIθ∗,Iη∗,ωθ∗,η∗,ωθ∗,Iθ∗,ωη∗,Iη∗,ωθ∗,Iη∗,ωIθ∗,η∗≤0,
i.e.,(39)ψαθ∗,η∗ωθ∗,η∗,ωθ∗,η∗,0,0,ωθ∗,η∗,ωθ∗,η∗≤0,
a contradiction toψ3, and thus, ωθ∗,η∗=0. Also, we have ωθ∗,θ∗=0. So, by using Lemma 2, we infer that θ∗=η∗, i.e., the fixed point of I is unique.☐By choosingψ∈Ψ from Examples 2–6, we have the following consequences.Corollary 12.
Let all the conditions of Theorems9 and 10 be satisfied, except that the assumption of orbitally ω-implicit contractive mapping for ψ∈Ψ is replaced by either of the form
(40)iαν,ϑωIν,Iϑ2≤aων,ϑ2+bων,Iν2+ωϑ,Iϑ21+ων,Iϑ+ωϑ,Iν,where 0<a,b<1, a+2b<1, αν,ϑa+b/αν,ϑ2−b<1, or
(41)iiαν,ϑωIν,Iϑ2≤aων,ϑ2+bων,Iν2+ωϑ,Iϑ21+ων,Iϑ2+ωϑ,Iν2,where 0<a,b<1, a+2b<1, αν,ϑa+b/αν,ϑ2−b<1, or
(42)iiiαν,ϑωIν,Iϑ≤aων,ϑ+bων,Iν+cωϑ,Iϑων,Iϑ1+ων,Iϑ+ωϑ,Iν,where 0<a,b,c<1, a+b+c<1, αν,ϑa+b/αν,ϑ−c<1, or
(43)ivαν,ϑωIν,Iϑ≤aων,ϑ+bων,Iνων,Iϑ1+ων,Iϑ+ωϑ,Iν+cωϑ,Iϑ,where 0<a,b,c<1, a+b+c<1, αν,ϑa+b/αν,ϑ−c<1.Then,FixI is a singleton.


## 3. Illustrations

Example 7.
Consider the setΞ=0∪1/8n:n∈ℕ with the usual metric d. Define a w-distance ω:Ξ×Ξ⟶0,+∞ by ωϑ,ν=ν for all ϑ,ν∈Ξ.
Consider the self-mappingI on Ξ given by Iϑ=ϑ/8. Take θ0=1/8. It is simple to show that
(44)Oθ0;I⊂18k∣k∈ℕ∪0,Oθ0;I¯=Oθ0;I∪0,and that Ξ,ω is I-orbitally complete at θ0.
Define functionalα:Ξ×Ξ⟶0,+∞ as follows:
(45)αϑ,ν=2,ifϑ,ν∈Oθ0;I,0,otherwise.
Atθ0=1/8 in Oθ0;I, αθ0,Iθ0≥1 and αIθ0,θ0≥1. Also, I is a triangular α-admissible mapping in Oθ0;I.
Considering Example4, we can define ψ∈Ψ as
(46)ψℓ1,ℓ2,ℓ3,ℓ4,ℓ5,ℓ6=ℓ1−aℓ2−bℓ3−cℓ4ℓ51+ℓ5+ℓ6.Here,k=αν,ϑ≥1. One can easily check that ℏ=ka+b/k−b<1 for k≥1, 0<a,b,c<1 so that a+b+c<1 and that ψ belongs to the set Ψ. We will show that I is an orbitally ω-implicit contractive mapping.Takeϑ,ν∈Oθ0;I¯, and so, 0≤ϑ,ν≤1/8. Consider two cases.Case 1.
Ifϑ=0 and ν=1/8n, n∈ℕ, then (9) reduces to
(47)18n+1≤a·18n+c·182n+1+8n+1and is fulfilled for a=1/2, b=1/5=c. If ν=0 and ϑ=1/8n, n∈ℕ or ϑ=0, then (9) holds trivially.Case 2.
Letϑ,ν∈1/8n∣n∈ℕ. Then, (9) reduces to
(48)2·ν8≤a·ν+b·ϑ8+c·ν288+ν+ϑ
and is fulfilled fora=1/2, b=1/5=c.Thus,I is orbitally ω-implicit contractive mapping. Therefore, all the conditions of Theorem 9 are satisfied, and ϑ∗=0 is the unique fixed point of I in Oθ0;I¯.Example 8.
Consider the setΞ=0,+∞ with the usual metric d. Define a w-distance ω:Ξ×Ξ⟶0,+∞byωϑ,ν=ν for all ϑ,ν∈Ξ.
Consider the self-mappingI on Ξ given by
(49)Iϑ=1n+1,ifϑ=1n,0,ifϑ=0,1,otherwise.Takeθ0=1. It is simple to show that
(50)Oθ0;I⊂1n+1∣n∈ℕ∪0,Oθ0;I¯=Oθ0;I∪0,and that Ξ,ω is I-orbitally complete at θ0.Define a functionα:Ξ×Ξ⟶0,∞ as follows:
(51)αϑ,ν=1,ifϑ,ν∈Oθ0;I,0,otherwise.Atθ0=1 in Oθ0;I, αθ0,Iθ0≥1 and αIθ0,θ0≥1. Also, I is a triangular α-admissible mapping in Oθ0;I.Considering Example3, we can define ψ∈Ψ as
(52)ψℓ1,ℓ2,ℓ3,ℓ4,ℓ5,ℓ6=ℓ12−aℓ22−bℓ32+ℓ42ℓ52+ℓ62+1.Here,k=αν,ϑ≥1. One can easily check that ℏ=ka+b/k2−b<1 for k≥1, 0<a,b<1 so that a+2b<1 and that ψ belongs to the set Ψ. We will show that I is an orbitally ω-implicit contractive mapping.Takeϑ,ν∈Oθ0;I¯, and so, 0≤ϑ,ν≤1. Consider two cases.Case 1.
Ifϑ=0 and ν=1/n, n∈ℕ, then (9) reduces to
(53)α0,1mωI0,I1n2≤a·ω0,1n2+b·ω0,I02+ω1/n,I1/n21+ω0,I1/n2+ω1/n,I02,that is,
(54)1n+12≤a·1n2+b·1/n+121+1/n+12=an2+bn+12+1
and is fulfilled, for0<a,b<1 so that a+2b<1. If ν=0 and ϑ=1/n, n∈ℕ, or ϑ=0, then (9) holds trivially.Case 2.
Letϑ,ν∈1/n+1∣n∈ℕ∪0. Then, inequality (9) has the form
(55)α1n,1mωI1n,I1m2≤a·ω1n,1m2+b·ω1/n,I1/n2+ω1/m,I1/m21+ω1/n,I1/m2+ω1/m,I1/n2,that is,
(56)ω1n+1,1m+12≤a·ω1n,1m2+b·ω1/n,1/n+12+ω1/m,1/m+121+ω1/n,1/m+12+ω1/m,1/n+12,that is,
(57)1m+12≤a1m2+b·1/n+12+1/m+121+1/m+12+1/n+12=am2+b·n+12+m+12n+12+m+12+1
and is fulfilled, for0<a,b<1 so that a+2b<1.
Thus,I is an orbitally ω-implicit contractive mapping.
Ifν>0, we have ν≠Iν so that
(58)infωϑ,ν+ωϑ,Iϑ:ϑ∈Ξ=infν+Iϑ:ϑ∈Ξ>0.Thus, all the conditions of Theorem10 are satisfied and ϑ∗=0 is the unique fixed point of I in Oθ0;I¯.Example 9.
Consider the setΞ=0,1 with the usual metric d. Define a w-distance ω:Ξ×Ξ⟶0,+∞ by ωϑ,ν=2ϑ−ν for all ϑ,ν∈Ξ.
Consider the self-mappingI on Ξ given by
(59)Iϑ=12n+1,ifϑ=12n,12,otherwise.
Takeθ0=1/2. It is simple to show that
(60)Oθ0;I⊂12n∣n∈ℕ,Oθ0;I¯=Oθ0;I∪0,and that Ξ,ω is I-orbitally complete at θ0.
Define functionalα:Ξ×Ξ⟶0,+∞ as follows:
(61)αϑ,ν=1,ifϑ,ν∈Oθ0;I,0,otherwise.
Atθ0=1/2 in Oθ0;I, αθ0,Iθ0≥1 and αIθ0,θ0≥1. Also, I is a triangular α-admissible mapping in Oθ0;I.
Considering Example6, we can define ψ∈Ψ as
(62)ψℓ1,ℓ2,ℓ3,ℓ4,ℓ5,ℓ6=ℓ1−aℓ2−bℓ3ℓ6ℓ5+ℓ6+1−cℓ4,0<a,b,c<1 and a+b+c<1.Takeϑ,ν∈Oθ0;I¯, and so, 0≤ϑ,ν≤1/2. Consider two cases.Case 1.
Ifϑ=0 and ν=1/2n (or ν=0 and ϑ=1/2n), n∈ℕ, then (9) reduces to
(63)1−12n≤a·12n−1+b·1/2n1+1/2n+1/2n−1+c·1−12nand is fulfilled, for 0<a,b,c<1, a+b+c<1. If ν=0 and ϑ=0, then (9) holds trivially.Case 2.
Letϑ,ν∈1/2n∣n∈ℕ∪0. Consider ϑ=1/2n and ν=1/2mm>n. Then, the inequality (9) has the form
(64)212n+1−12m+1≤2a12n−12m+b21/2n−1/2n+1·21/2n−1/2m+11+21/2n−1/2m+1+21/2m−1/2n+1+2c12m−12m+1,which is equivalent to
(65)2n−2m2m+n∣≤a2n−2m2m+n−1+4b2m+1−2n2m+n+2m+1−2n+2n+1−2m+c2m,
which is obviously true for0<a,b,c<1 such that a+b+c<1.
It can be noted that for0≠I0, we have
(66)infω12n,0+ω12n,12n+1:n∈ℕ=inf22n+22n+1:n∈ℕ=0.Thus, all the conditions of Theorem10 are satisfied except (v’). Clearly, I has no fixed points in Oθ0;I¯.


## 4. Weak Well-Posedness, Weak Limit Shadowing, and Generalizedw-Ulam-Hyers Stability

The notion of well-posedness of a fixed point problem (fpp) has evoked much interest of several mathematicians, for example, Popa [10, 11]. In the paper [12], the authors defined a weak well-posed (wwp) property in metric space. In the following, we extend this notion to a w-distance in metric space.Definition 13.
LetΞ,d be a metric space and ω be a w-distance in Ξ. Let I:Ξ⟶Ξ be a mapping having a unique fixed point ϑ∗ such that αϑ∗,ϑ∗≥1. The fpp of I is said to be wwp with respect to ω if for any sequence θn in Ξ with limn⟶∞ωθn,Iθn=0 and limn,m⟶∞ωIθn,Iθm=0, one has limn⟶∞ωθn,ϑ∗=0.
To guarantee the wwp of a mappingI, we add the following additional condition for functions ψ∈Ψ and call the respective set Ψ′:
(ψ4) for all ζ,ξ>0, k≥1, ψkζ,ξ,0,0,ξ,ζ≤0 implies that there exists ℏ∈0,1 such that kζ≤ℏξExamples2–4 and 6 satisfy the condition (ψ4).Theorem 14.
LetΞ,d be a metric space and ω be a w-distance on Ξ. Suppose that all the hypotheses of Theorem 9 hold for ψ∈Ψ′. Then, the ffp for I is wwp.Proof.
Letθn be a sequence in Ξ such that limn⟶∞ωθn,Iθn=0 and limm,n⟶∞ωIθm,Iθn=0, for n>m. We obtain that
(67)ωθn,ϑ∗≤ωθn,Iθm+ωIθm,Iθn+ωIθn,ϑ∗.
Taking the limit asn⟶∞, we get
(68)limn⟶∞ωθn,ϑ∗≤limn⟶∞ωθn,Iθm+ωIθn,ϑ∗.
WLOG, we can assume that there exists a distinct subsequenceIθnk of Iθn. Otherwise, there exists θ0∈Ξ and n1∈N such that Iθn=θ0 for n≥n1. Since limn⟶∞ωθn,Iθn=0, we get limn⟶∞ωθn,θ0=0. If θ0≠ϑ∗, then θ0≠Iθ0 due to uniqueness of the fixed point of I. For n≥n1, we obtain θ0=Iθn≠Iθ0. From (9), we have
(69)ψαθn,θ0ωIθn,Iθ0,ωθn,θ0,ωθn,Iθn,ωθ0,Iθ0,ωθn,Iθ0,ωIθn,θ0≤0,
i.e.,(70)ψαθn,θ0ωθ0,Iθ0,ωθn,θ0,ωθn,θ0,ωθ0,Iθ0,ωθn,Iθ0,ωθ0,θ0≤0,
i.e.,(71)ψαθn,θ0ωθ0,Iθ0,ωθn,θ0,ωθn,θ0,ωθ0,Iθ0,ωθn,θ0+ωθ0,Iθ0,0≤0.
It follows from (ψ2a) that there is ℏ∈0,1 such that
(72)ωθ0,Iθ0≤αθn,θ0ωθ0,Iθ0≤ℏωθn,θ0,
which on applyingn⟶∞ gives ωθ0,Iθ0=0. Also, we have ωθ0,θ0=0. So, by Lemma 2, we get θ0=Iθ0, a contradiction. Hence, there exist m,q,n>n0m>q>n such that Iθm≠Iθq≠Iθn≠θn. Then,
(73)ωθn,Iθm≤ωθn,Iθn+ωIθn,Iθq+ωIθq,Iθm,
which⟶0 as n⟶∞. On replacing the value in (68), we get
(74)limn⟶∞ωθn,ϑ∗≤limn⟶∞ωIθn,ϑ∗.
From (9), we have
(75)ψαθn,ϑ∗ωIθn,Iϑ∗,ωθn,ϑ∗,ωθn,Iθn,ωϑ∗,Iϑ∗,ωθn,Iϑ∗,ωIθn,ϑ∗≤0,
i.e.,(76)ψαθn,ϑ∗ωIθn,ϑ∗,ωθn,ϑ∗,ωθn,Iθn,ωϑ∗,ϑ∗,ωθn,ϑ∗,ωIθn,ϑ∗≤0,
which on applying limit asn⟶∞ gives
(77)ψlimn⟶∞αθn,ϑ∗limn⟶∞ωIθn,ϑ∗,limn⟶∞ωθn,ϑ∗,0,0,limn⟶∞ωθn,ϑ∗,limn⟶∞ωIθn,ϑ∗≤0.
It follows from (ψ4) that there is ℏ∈0,1 such that
(78)limn⟶∞ωIθn,ϑ∗<ℏlimn⟶∞ωθn,ϑ∗.☐Combining (74) and (78), we get limn⟶∞ωθn,ϑ∗=0 as ℏ<1.The limit shadowing property of fixed point problems has been discussed in the papers [13, 14]. We define weak limit shadowing property (wlsp) in metric spaces under w-distance.Definition 15.
LetΞ,d be a complete metric space and ω be a w-distance in Ξ. Let I:Ξ⟶Ξ be a mapping. The fpp of I is said to have wlsp in Ξ if assuming that θn in Ξ satisfies ωθn,Iθn⟶0 as n⟶∞ and ωIθn,Iθm⟶0, it follows that there exists x∈Ξ such that ωθn,Inϑ∗⟶0 as n⟶∞FixI=ϑ∗.Theorem 16.
In addition to the hypotheses of Theorem9 (or Theorem 10) if θn in Ξ is such that limn⟶∞ωθn,Iθn=0, limn,m⟶∞ωIθn,Iθm=0 and ϑ∗∈FixI, then I has the wlsp.Proof.
Sinceϑ∗ is a fixed point of I, we have ωϑ∗,Iϑ∗=0 and let θn in Ξ such that limn⟶∞ωθn,Iθn=0, limn,m⟶∞ωIθn,Iθm=0, then by virtue of Theorem 9, we have limn⟶∞ωθn,ϑ∗=0, and therefore, we can write limn⟶∞ωθn,Inϑ∗=0.☐Next, we define generalized-ω-Ulam-Hyers stability (G-ω-UHS) of fixed point problem (fpp) in metric spaces under w-distance.Definition 17.
LetΞ,d be a metric space and ω be a w-distance on Ξ. Let I:Ξ⟶Ξ be a mapping. The the fixed point equation (FPE)
(79)ϑ=Iϑ,ϑ∈Ξis said to be G-ω-UHS in the setting of metric spaces under w-distance, if there exists an increasing function ϕ:ℝ+⟶ℝ+, continuous at 0, with ϕ0=0, such that for each ε>0 and an ε-solution υ∈Ξ, that is,
(80)ωυ,Iυ≤ε,
there exists a solutionϑ∗∈Ξ of (79) such that
(81)ωυ,ϑ∗≤ϕε.Ifϕξ=bξ for all ξ∈ℝ+, where b>0, then FPE (79) is said to be ω-UHS in the framework of metric spaces under w-distance.Remark 18.
Ifp=d, then Definition 17 reduces to the notion of GUHS in metric spaces. Also, if ϕξ=bξ for all ξ∈ℝ+, where b>0, then it reduces to the notion of UHS in metric spaces. Finally, if ωϑ,υ=ϑ−υ, then it reduces to the classical UHS.Theorem 19.
LetΞ,d be a metric space and ω be a w-distance on Ξ. Suppose that all the hypotheses of Theorem 9 hold, using the contraction condition in the form (42). Then, the FPE (79) is G-ω-UHS.Proof.
Following Theorem9, we have Iρ∗=ρ∗; that is, ρ∗∈Ξ is a solution of the FPE (79) with ωρ∗,Iρ∗=0. Let ε>0 and σ∗∈Ξ be an ε-solution of (79), that is,
(82)ωσ∗,Iσ∗≤ε.
Sinceωρ∗,Iρ∗=ωρ∗,ρ∗=0≤ε, ρ∗ and σ∗ are ε-solutions. Now,
(83)ωρ∗,σ∗≤ωρ∗,Iρ∗+ωIρ∗,Iσ∗+ωIσ∗,σ∗≤ωIρ∗,Iσ∗+ε.
From the contractive condition (42) for I, we get
(84)ωIρ∗,Iσ∗≤aωρ∗,σ∗+bωρ∗,Iρ∗+cωσ∗,Iσ∗ωρ∗,Iσ∗1+ωρ∗,Iσ∗+ωσ∗,Iρ∗≤aωρ∗,σ∗+cεωρ∗,Iσ∗1+ωρ∗,Iσ∗+ωσ∗,Iρ∗≤aωρ∗,σ∗+cε.
Therefore, from (83), we obtain
(85)ωρ∗,σ∗≤aωρ∗,σ∗+cε+ε=aωρ∗,σ∗+1+cε,which implies that
(86)ωρ∗,σ∗1−a≤1+cε,
i.e.,(87)ωρ∗,σ∗≤1+c1−aε=ϕεas1+c1−a>0.☐Thus, the inequality (81) holds, and therefore, the FPE (79) is G-ω-UHS in the metric spaces under w-distance.


## 5. Application

Fractional differential/integral equations (FDE/DIE) have been extensively studied as an application of fixed point theory. In fact, to get the unique solution of FDE, one has to apply the Banach fixed point theorem or its variants. There are different types of FDEs in the literature, but the FDEs in the Caputo sense are the easiest to apply. The main advantage of Caputo derivative is that the derivative of the constant function is0 while most of the other fractional derivatives do not have such an important property. This property helps in initial value problems to apply fixed point theorems. In paper [15], the existence of solutions for some Atangana-Baleanu fractional differential equations in the Caputo sense have been discussed. Some other FDE-related work can be seen in [16, 17].The Caputo derivative of fractional orderβ is defined as
(88)cDβpρ=1Γn−β∫0ρρ−σn−β−1pnσdsn−1<β<n,n=β+1,where p:0,∞⟶ℝ is a continuous function, β denotes the integer part of the positive real number β, and Γ is the gamma function.In this section, we discuss the existence of solutions of following nonlinear fractional differential equation (FDE) [18] as an application of Theorem 9.Consider the nonlinear FDE(89)cDβϑρ=ℏρ,ϑρ0<ρ<1,1<β≤2,with the integral boundary conditions
(90)ϑ0=0,ϑ1=∫0ηϑσds0<η<1,where J=0,1, ϑ∈CJ,ℝ, and ℏ:J×ℝ⟶ℝ is a continuous function.LetΞ=CJ,ℝ be endowed with the usual distance and w-distance ω:Ξ×Ξ⟶0,+∞ by
(91)ωϑ,ν=maxρ∈Jϑρ+νρ.Theorem 20.
LetI:Ξ⟶Ξ be the operator defined by
(92)Iϑρ=1Γβ∫0ρρ−σβ−1ℏσ,ϑσdσ−2ρ2−η2Γβ∫011−σβ−1ℏσ,ϑσdσ+2ρ2−η2Γβ∫0η∫0σσ−ςβ−1ℏz,ϑzdςdσ,for ϑ∈Ξ, ρ∈J. Also, let ζ:ℝ×ℝ⟶ℝ be a given function. Assume that
(F1) ℏ:J×ℝ⟶ℝ is a continuous function, nondecreasing in the second variable
(F2) there exists ϑ0∈Ξ such that ζϑ0ρ,Iϑ0ρ≥0, ζIϑ0ρ,ϑ0ρ≥0 for all ρ∈J
(F3) for ϑ,ν,υ∈Ξ, ζϑρ,νρ≥0 and ζϑρ,υρ≥0 for all ρ∈J implies that ζϑρ,υρ≥0 for all ρ∈J
(F4) there exist 0<a,b,c<1 with a+b+c<1 such that for ϑ,ν∈Ξ and ρ∈J we have
(93)ζϑ,νℏρ,ϑρ+ℏρ,νρ≤λ×Θϑ,νρ,where
(94)Θϑ,νρ=+c·a·ϑρ+νρ+b·ϑρ+Iϑρϑρ+Iϑρϑρ+Iνρ1+maxρ∈Jϑρ+Iνρ+maxρ∈Jνρ+Iϑρ,and λ=Γβ+1/5. Then, the problem (89)–(90) has at least one solution ϑ∗∈Ξ.Proof.
Define a functionα:Ξ2⟶0,∞ by
(95)αϑ,ν=1,ifforζϑρ,νρ≥0,forallρ∈J,γ,otherwise,where γ∈0,1. It is easy to check that the assumption (F2)-(F3) implies the condition (i)-(ii) of Theorem 9, respectively.
Letϑ,ν∈Ξ, then for each ρ∈J, by the definition (92) of operator I, we have
(96)ζϑ,νIϑρ+Iνρ=ζϑ,ν1Γβ∫0ρρ−σβ−1ℏσ,ϑσdσ−2ρ2−η2Γβ∫011−σβ−1ℏσ,ϑσdσ+2ρ2−η2Γβ∫0η∫0σσ−ςβ−1ℏς,ϑzdςdσ+1Γβ∫0ρρ−σβ−1ℏσ,νσdσ+2ρ2−η2Γβ∫011−σβ−1ℏσ,νσdσ−2ρ2−η2Γβ∫0η∫0σσ−ςβ−1ℏς,νςdςdσ≤ζϑ,ν1Γβ∫0ρρ−σβ−1ℏσ,ϑσ+ℏσ,νσdσ+2ρ2−η2Γβ∫011−σβ−1ℏs,ϑσ+ℏσ,νσdσ+2ρ2−η2Γβ∫0η∫0σσ−ςβ−1ℏς,ϑς+ℏς,νςdςdσ,that is,
(97)ζϑ,νIϑρ+Iνρ≤1Γβ∫0ρρ−σβ−1λ×Θϑ,νρdσ+2ρ2−η2Γβ∫011−σβ−1λ×Θϑ,νρdσ+2ρ2−η2Γβ∫0η∫0σσ−ςβ−1λ×Θϑ,νρdςdσ≤λ×ΘIϑ,ν×maxρ∈J1Γβ∫0ρρ−σβ−1dσ+2ρ2−η2Γβ∫011−σβ−1dσ+2ρ2−η2Γβ∫0η∫0σσ−ςβ−1dςdσ.
After easy calculations, we get(98)ζϑ,νIϑρ+Iνρ≤ΘIϑ,ν.
This implies that(99)αϑ,νωIϑ,Iν=αϑ,νmaxt∈IIϑρ+Iνρ≤ΘIϑ,ν
for allϑ,ν∈Ξ, where
(100)ΘIϑ,ν=a·ωϑ,ν+b·ωϑ,Iϑ+c·ωϑ,Iϑωϑ,Iν1+ωϑ,Iν+ων,Iϑ.
If we consider(101)ψℓ1,ℓ2,ℓ3,ℓ4,ℓ5,ℓ6=ℓ1−aℓ2−bℓ3−cℓ4ℓ51+ℓ5+ℓ6,where 0<a,b,c<1 so that a+b+c<1, then ψ∈Ψ and the condition (iii) of Theorem 9 is satisfied. Therefore, all the requirements of Theorem 9 are fulfilled, and we conclude that there is a fixed point ϑ∗∈Ξ of the operator I. It is well known (see, e.g., [18], Theorem 3.1) that in this case ϑ∗ is also a solution of the integral equation (92) and the FDE (89) with condition (90).☐


## 6. Further Work

On the line of our work, the following two types of FDE can also be discussed:(102)Dβϑρ+ℏρ,ϑρ=00≤ρ≤1,1<β,with the two-point boundary conditions
(103)ϑ0=0,ϑ1=0,whereℏ:I=0,1×ℝ⟶ℝ is a continuous function.
(104)Dαϑρ+Dβϑt=ℏρ,ϑρ0≤t≤1,0<β<α<1,with the two-point boundary conditions
(105)ϑ0=0,ϑ1=0,where ℏ:I=0,1×ℝ⟶ℝ is a continuous function.


## 7. Conclusion

In this work, a new class of implicit relation and implicit type contractive condition under aforesaid implicit relation in the metric spaces underw-distance functional have been introduced. Next, some basic fixed point results under respective contractive conditions followed by three suitable examples have been discussed. Further, we have discussed weak well-posed property, weak limit shadowing property, and generalized w-Ulam-Hyers stability in the underlying spaces. Finally, sufficient conditions for the existence of solutions for the fractional differential equation as an application of the established result have been discussed.



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