# A Note on Lacunary Sequence Spaces of Fractional Difference Operator of Orderα,β

Authors: Qing-Bo Cai; Sunil K. Sharma; Mohammad Ayman Mursaleen
Journal: Journal of Function Spaces
 (2022)

---

## Abstract

In the present paper, we defined lacunary sequence spaces of fractional difference operator of orderα,β over n-normed spaces via Musielak-Orlicz function M=Ik. Our aim in this paper is to study some topological properties and inclusion relation between the spaces I−NαβA,M,u,θ,Δγ,·,⋯,·0, I−NαβA,M,u,θ,Δγ,·,⋯,·, and I−NαβA,M,u,θ,Δγ,·,⋯,·∞.


---

## Body

## 1. Introduction and Preliminaries

The concept of statistical convergence was introduced by Fast [1] and Schoenberg [2] independently. Many authors studied the concept of statistical convergence from the past few years we may refer to ([3–19]) and references therein.The sequenceξ=ξk is statistically convergent of order α to ℓ (see Çolak) if there is a complex number ℓ such that
(1)limn⟶∞1nαk≤n:ξk−ℓ≥ε=0.Let0<α≤β≤1. We define the α,β-density of the subset E of ℕ by
(2)δαβE=limn1nαk≤n:k∈Eβ,provided the limit exists, where k≤n:k∈Eβ denotes the βth power of number of elements of E not exceeding n ([20–22]).By a lacunary sequenceθ=θr, we mean a sequence of positive integers such that θ0=0, 0<θr<θr+1, and ϕr=θr−θr−1⟶∞ as r⟶∞. The intervals determined by θ will be denoted by Jr=θr−1,θr and tr=θr/θr−1. Freedman et al. [23] defined the space of lacunary strongly convergent sequences by
(3)Nθ=ξ∈w:limr⟶∞1ϕr∑k∈Jrξk−l=0,forsomel.Definition 1.
Letθ=θr be a lacunary sequence. The sequence ξ=ξk is Sαβθ-statistically convergent (or lacunary statistically convergent of order α,β) (see [20]) if there is a real number L such that
(4)limr⟶∞1ϕrαk∈Jr:ξk−L≥εβ=0,where Jr=θr−1,θr and ϕrα denotes the αth power ϕrα of ϕr, that is, ϕα=ϕrα=ϕ1α,ϕ2α,⋯,ϕrα,⋯. In this case, we write Sαβθ−limξk=L. The set of all Sαβθ-statistically convergent sequences is denoted by Sαβθ. If α=β=1 and θ=2r, then, we will write S instead of Sαβθ.
A familyI⊂2X of subsets of a nonempty set X is said to be an ideal in X if
(1)
ϕ∈I(2)
A,B∈I imply A∪B∈I(3)
A∈I, B⊂A imply B∈Iwhile an admissible ideal I of X further satisfies ξ∈I for each ξ∈X (see [24]).A sequenceξnn∈ℕ in X is said to be I-convergent to ξ∈X (see [24]), if Aε=n∈ℕ:ξn−ξ≥ε∈I, for each ε>0.A sequenceξnn∈ℕ in X is said to be I-bounded to ξ∈X if there exists an K>0 such that n∈ℕ:ξn>K∈I. Many authors studied the topological properties and applications of ideal, we refer to ([25–37]) and references therein.The concept of difference sequence spaces was introduced in [38] and further generalized in [39].In [40], Baliarsingh defined the fractional difference operator as follows:Letξ=ξk∈w and γ be a real number, then, the fractional difference operator Δγ is defined by
(5)Δγξk=∑i=0k−γii!ξk−i,where −γi denotes the Pochhammer symbol defined as
(6)−γi=1,ifγ=0ori=0,γγ+1γ+2⋯γ+i−1,otherwise,γγ+1γ+2⋯γ+i−1,otherwise.The concept of difference sequences, Orlicz function, Musielak-Orlicz function, andn-normed spaces was used by many authors and proves some topological properties (see [41–50]) and references therein. For details about n-normed spaces, we refer to ([51–55]), difference sequence spaces ([38, 39]), Orlicz function ([56–58]). Ideal convergence and fractional difference operator Δα has been studied in [59, 60]. We continue in this connection and construct new sequence spaces as follows.LetM=Ik be a Musielak-Orlicz function, u=uk be a bounded sequence of positive real numbers, and 0<α≤β≤1. We define the following sequence spaces in the present paper
(7)I−NαβA,M,u,θ,Δγ,·,⋯,·0=ξ∈w:I−limr1ϕrα∑k∈JrIkAkΔγξρ,x1,⋯,xn−1ukβ=0,forsomeρ>0,(8)I−NαβA,M,u,θ,Δγ,·,⋯,·=ξ∈w:I−limr1ϕrα∑k∈JrIkAkΔγξ−Lρ,x1,⋯,xn−1ukβ=0,forsomeLandρ>0,(9)I−NαβA,M,u,θ,Δγ,·,⋯,·∞=ξ∈w:1ϕrα∑k∈JrIkAkΔγξρ,x1,⋯,xn−1ukβisbounded,forsomeρ>0.If we takeMξ=ξ, the above spaces reduces to I−NαβA,u,θ,Δγ,·,⋯,·0, I−NαβA,u,θ,Δγ,·,⋯,·, and I−NαβA,u,θ,Δγ,·,⋯,·∞.If we takeu=uk=1, the above spaces reduces to I−NαβA,M,θ,Δγ,·,⋯,·0, I−NαβA,M,θ,Δγ,·,⋯,·, and I−NαβA,M,θ,Δγ,·,⋯,·∞.The following inequality will be used in the proceeding results. If0≤uk≤supuk=H, D=max1,2H−1, then
(10)rk+skuk≤Drkuk+skuk,for all k and rk,sk∈ℂ. Also ruk≤max1,rH for all r∈ℂ.


## 2. Main Results

In this section, we study topological properties and prove some inclusion relations. In what follows, we will takeM=Ik a Musielak-Orlicz function and u=uk a bounded sequence of positive real numbers.Theorem 2.
The spacesI−NαβA,M,u,θ,Δγ,·,⋯,·, I−NαβA,M,u,θ,Δγ,·,⋯,·, and I−NαβA,M,u,θ,Δγ,·,⋯,·∞ are linear spaces.Proof.
Letξ1,ξ2∈I−NαβA,M,u,θ,Δγ,·,⋯,·0 and let μ,ν be scalars. Then, there exist two positive numbers ρ1 and ρ2 for ε>0(11)D1=r∈ℕ:1ϕrα∑k∈JrIkAkΔγξ1ρ1,x1,⋯,xn−1ukβ≥ε2D∈I,(12)D2=r∈ℕ:1ϕrα∑k∈JrIkAkΔγξ2ρ2,x1,⋯,xn−1ukβ≥ε2D∈I.
Letρ3=max2μρ1,2νρ2 and by inequality (1), we have
(13)1ϕrα∑k∈JrIkAkΔγμξ1+νξ2ρ3,x1,⋯,xn−1ukβ≤1ϕrα∑k∈JrIkμAkΔγξ1ρ3,x1,⋯,xn−1ukβ+1ϕrα∑k∈JrIkνAkΔγξ2ρ3,x1,⋯,xn−1ukβ≤1ϕrα∑k∈JrIkAkΔγξ1ρ1,x1,⋯,xn−1ukβ+1ϕrα∑k∈JrIkAkΔγξ2ρ2,x1,⋯,xn−1ukβ.
Now by (11) and (12), we get
(14)r∈ℕ:1ϕrα∑k∈JrIkAkΔγμξ1+νξ2ρ3,x1,⋯,xn−1ukβ>ε⊂D1∪D2.
Therefore,μξ1+νξ2∈I−NαβA,M,u,θ,Δγ,·,⋯,·0. Hence, I−NαβA,M,u,θ,Δγ,·,⋯,·0 is a linear space. On a similar way, we can prove that I−NαβA,M,u,θ,Δγ,·,⋯,· and I−NαβA,M,u,θ,Δγ,·,⋯,·∞ are linear spaces.Theorem 3.
The inclusionsI−NαβA,M,u,θ,Δγ,·,⋯,·0⊂I−NαβA,M,u,θ,Δγ,·,⋯,·⊂I−NαβA,M,u,θ,Δγ,·,⋯,·∞ hold.Proof.
The inclusionI−NαβA,M,u,θ,Δγ,·,⋯,·0⊂I−NαβA,M,u,θ,Δγ,·,⋯,· is obvious. We prove I−NαβA,M,u,θ,Δγ,·,⋯,·⊂I−NαβA,M,u,θ,Δγ,·,⋯,·∞. For this, let ξ∈I−NαβA,M,u,θ,Δγ,·,⋯,·. Then, there exists ρ1>0 such that for every ε>0(15)B1=r∈ℕ:1ϕrα∑k∈JrIkAkΔγξ−Lρ1,x1,⋯,xn−1ukβ≥ε∈I.
We putρ=2ρ1 and M=Ik is a Musielak-Orlicz function, we have
(16)IkAkΔγξρ,x1,⋯,xn−1≤IkAkΔγξ−Lρ1,x1,⋯,xn−1+IkLρ1,x1,⋯,xn−1.
Suppose thatr∉B1. Hence by above inequality and (1), we have
(17)1ϕrα∑k∈JrIkAkΔγξρ,x1,⋯,xn−1ukβ≤D1ϕrα∑k∈JrIkAkΔγξ−Lρ1,x1,⋯,xn−1ukβ+1ϕrα∑k∈JrIkLρ1,x1,⋯,xn−1ukβ<Dε+1ϕrα∑k∈JrIkLρ,x1,⋯,xn−1ukβ.
By usingIkL/ρ1,x1,⋯,xn−1uk≤max1,IkL/ρ1,x1,⋯,xn−1H, we have
(18)1ϕrα∑k∈JrIkLρ,x1,⋯,xn−1ukβ<∞.
PutK=Dε+1/ϕrα∑k∈JrIkL/ρ,x1,⋯,xn−1ukβ. It follows that
(19)r∈ℕ:1ϕrα∑k∈JrIkAkΔγξρ,x1,⋯,xn−1ukβ>K∈I.This shows that ξ∈I−NαβA,M,u,θ,Δγ,·,⋯,·∞, which completes the proof.Theorem 4.
The spaceI−NαβA,M,u,θ,Δγ,·,⋯,·∞ is a paranormed space with paranorm defined by
(20)gx=infρ>0:1ϕrα∑k∈JrIkAkΔγξρ,x1,⋯,xn−1ukβ≤1.Proof.
Sincegξ=g−ξ and Ik0=0, we have g0=0. Let ξ1,ξ2∈I−NαβA,M,u,θ,Δγ,·,⋯,·∞. Let
(21)Bξ1=ρ>0:1ϕrα∑k∈JrIkAkΔγξ1ρ,x1,⋯,xn−1ukβ≤1,(22)Bξ2=ρ>0:1ϕrα∑k∈JrIkAkΔγξ2ρ,x1,⋯,xn−1ukβ≤1.
Letρ1∈Bξ1 and ρ2∈Bξ2 and ρ=ρ1+ρ2, we have
(23)1ϕrα∑k∈JrIkAkΔγξ1+ξ2ρ,x1,⋯,xn−1β≤ρ1ρ1+ρ21ϕrα∑k∈JrIkAkΔγξ1ρ1,x1,⋯,xn−1β+1ϕrα∑k∈JrIkAkΔγξ2ρ2,x1,⋯,xn−1β.
Thus,1/ϕrα∑k∈JrIkAkΔγξ1+ξ2/ρ1+ρ2,x1,⋯,xn−1ukβ≤1 and
(24)gξ1+ξ2≤infρ1+ρ2>0:ρ1∈Bξ1,ρ2∈Bξ2≤infρ1>0:ρ1∈Bξ1+infρ2>0:ρ2∈Bξ2=gξ1+gξ2.
Letσs⟶σ where σ,σs∈ℂ and let gξs−ξ⟶0 as s⟶∞. We have to show that gσsξs−σξ⟶0 as s⟶∞. Let
(25)Bξs=ρs>0:1ϕrα∑k∈JrIkAkΔγξsρs,x1,⋯,xn−1ukβ≤1,(26)Bξs−ξ=ρs′>0:1ϕrα∑k∈JrIkAkΔγξs−ξρs′,x1,⋯,xn−1ukβ≤1.
Ifρs∈Bξs and ρs′∈Bξs−ξ; then, we have
(27)1ϕrα∑k∈JrIkAkΔγσsξs−σξρs∣σs−σ∣+ρs′∣σ∣,x1,⋯,xn−1β≤1ϕrα∑k∈JrIkAkΔγσsξs−σξsρsσs−σ+ρs′σ+AkΔγσξs−σξρsσs−σ+ρs′σ,x1,⋯,xn−1β≤σs−σρsρsσs−σ+ρs′σ1ϕrα∑k∈JrIkAkΔγξsρs,x1,⋯,xn−1β+σρs′ρsσs−σ+ρs′σ1ϕrα∑k∈JrIkAkΔγξs−ξρs′,x1,⋯,xn−1β.
From the above inequality, it follows that(28)1ϕrα∑k∈JrIkAkΔγσsξs−σξρs∣σs−σ∣+ρs′∣σ∣,x1,⋯,xn−1ukβ≤1,and consequently,
(29)gσsξs−σξ≤infρsσs−σ+ρs′σ>0:ρs∈Bξs,ρs′∈Bξs−ξ≤σs−σ>0infρ>0:ρs∈Bξs+σ>0infρs′un/H:ρs′∈Bξs−ξ⟶0ass⟶∞,which completes the proof.Theorem 5.
LetM=Ik and M′=Ik′ be Musielak-Orlicz functions that satisfy the Δ2-condition. Then
(30)I−NαβA,M,u,θ,Δγ,·,⋯,·0⊆I−NαβA,M′∘M,u,θ,Δγ,·,⋯,·0,(31)I−NαβA,M,u,θ,Δγ,·,⋯,·⊆I−NαβA,M′∘M,u,θ,Δγ,·,⋯,·,(32)I−NαβA,M,u,θ,Δγ,·,⋯,·∞⊆I−NαβA,M′∘M,u,θ,Δγ,·,⋯,·∞.Proof.
(i) Letξ∈I−NαβA,M,u,θ,Δγ,·,⋯,·0. Then, there exists K1>0 such that
(33)B1=r∈ℕ:1ϕrα∑k∈JrIkAkΔγξρ,x1,⋯,xn−1ukβ≥K1∈I,for ρ>0. Since M′ is a Musielak-Orlicz function which satisfies Δ2-condition, we have
(34)1ϕrα∑IkAkΔγξρ,x1,⋯,xn−1>δk∈JrIk′IkAkΔγξρ,x1,⋯,xn−1ukβ≤max1,K1δIk′2H1ϕrα∑IkAkΔγξρ,x1,⋯,xn−1>δk∈JrIkAkΔγξρ,x1,⋯,xn−1ukβ,for K≥1. By continuity of M′, we have
(35)1ϕrα∑IkAkΔγξρ,x1,⋯,xn−1≤δk∈JrIk′IkAkΔγξρ,x1,⋯,xn−1ukβ≤1ϕrα∑IkAkΔγξρ,x1,⋯,xn−1≤δk∈Jrεukβ≤1ϕrα∑IkAkΔγξρ,x1,⋯,xn−1≤δk∈Jrmaxεh,εHβ.
Supposer∉B1. Then, by using (34) and (35), we have
(36)1ϕrα∑k∈JrIk′IkAkΔγξρ,x1,⋯,xn−1ukβ=1ϕrα∑IkAkΔγξρ,x1,⋯,xn−1>δk∈JrIkAkΔγξρ,x1,⋯,xn−1ukβ+1ϕrα∑IkAkΔγξρ,x1,⋯,xn−1≤δk∈JrIkAkΔγξρ,x1,⋯,xn−1ukβ≤max1,K1δIk′2HK1+maxεh,εH=K2.Hence, r∉B2=r∈ℕ:1/ϕrα∑k∈JrIk′IkAkΔγξ/ρ,x1,⋯,xn−1ukβ>K2 and so B2⊂B1 which implies B2∈I. This shows that I−NαβA,M′∘M,u,θ,Δγ,·,⋯,·0. Hence, I−NαβA,M,u,θ,Δγ,·,⋯,·0⊆I−NαβA,M′∘M,u,θ,Δγ,·,⋯,·0. Similarly, we can prove (ii) and (iii) part.Corollary 6.
LetM=Ik satisfy Δ2-condition. Then,
(37)I−NαβA,u,θ,Δγ,·,⋯,·0⊆I−NαβA,M,u,θ,Δγ,·,⋯,·0,(38)I−NαβA,u,θ,Δγ,·,⋯,·⊆I−NαβA,M,u,θ,Δγ,·,⋯,·,(39)I−NαβA,u,θ,Δγ,·,⋯,·∞⊆I−NαβA,M,u,θ,Δγ,·,⋯,·∞.Proof.
If we putIkx=x and Ik′x=Ikx∀x∈0,∞ in Theorem 5, the result follows.Theorem 7.
LetM=Ik and M′=Ik′ be Musielak-Orlicz functions that satisfy the Δ2-condition. Then,
(40)I−NαβA,M,u,θ,Δγ,·,⋯,·0∩I−NαβA,M′,u,θ,Δγ,·,⋯,·0⊆I−NαβA,M′+M,u,θ,Δγ,·,⋯,·0,(41)I−NαβA,M,u,θ,Δγ,·,⋯,·∩I−NαβA,M′,u,θ,Δγ,·,⋯,·⊆I−NαβA,M′+M,u,θ,Δγ,·,⋯,·,(42)I−NαβA,M,u,θ,Δγ,·,⋯,·∞∩I−NαβA,M,u,θ,Δγ,·,⋯,·∞⊆I−NαβA,M′+M,u,θ,Δγ,·,⋯,·∞.Proof.
(i) Letξ∈I−NαβA,M,u,θ,Δγ,·,⋯,·0∩I−NαβA,M′,u,θ,Δγ,·,⋯,·0. Then, there exists K1>0 and K2>0 such that
(43)B1=r∈ℕ:1ϕrα∑k∈JrIkAkΔγξρ,x1,⋯,xn−1ukβ≥K1∈I,(44)B2=r∈ℕ:1ϕrα∑k∈JrIk′AkΔγξρ,x1,⋯,xn−1ukβ≥K2∈I,for some ρ>0. Let r∉B1∪B2. Then, we have
(45)1ϕrα∑k∈JrIk+Ik′AkΔγξρ,x1,⋯,xn−1ukβ≤D1ϕrα∑k∈JrIkAkΔγξρ,x1,⋯,xn−1ukβ+1ϕrα∑k∈JrIk′AkΔαξρ,x1,⋯,xn−1ukβ<K1+K2.
r∉B=r∈ℕ:1/ϕrα∑k∈JrIk′+IkAkΔγξ/ρ,x1,⋯,xn−1ukβ>K. We have B1∪B2∈I and so B⊂B1∪B2 which implies B∈I. This shows that x∈I−NαβA,M′+M,u,θ,Δγ,·,⋯,·0. Hence, I−NαβA,M,u,θ,Δγ,·,⋯,·0∩I−NαβA,M′,u,θ,Δγ,·,⋯,·0⊆I−NαβA,M′+M,u,θ,Δγ,·,⋯,·0. Similarly, we can prove (ii) and (iii) part of the theorem.Theorem 8.
Let0<uk≤vk and vk/uk be bounded. Then, the following inclusions hold
(46)I−NαβA,M,v,θ,Δγ,·,⋯,·0⊆I−NαβA,M,u,θ,Δγ,·,⋯,·0,(47)I−NαβA,M,v,θ,Δγ,·,⋯,·⊆I−NαβA,M,u,θ,Δγ,·,⋯,·.Proof.
(i) Letξ∈I−NαβA,M,v,θ,Δγ,·,⋯,·0. Write sk=IkAkΔγξ/ρ,x1,⋯,xn−1vk and λk=uk/vk, so that 0<λ<λk≤1. By using Hölder inequality, we have
(48)1ϕrα∑k∈Jrskλkβ=1ϕrα∑sk≥1k∈Jrskλkβ+1ϕrα∑sk<1k∈Jrskλkβ≤1ϕrα∑sk≥1k∈Jrskβ+1ϕrα∑sk<1k∈Jrskλβ=1ϕrα∑sk≥1k∈Jrskβ+∑sk<1k∈Jr1ϕrαskλ1ϕrα1−λβ≤1ϕrα∑sk≥1k∈Jrskβ+∑sk<1k∈Jr1ϕrαskλ1/λλβ∑sk<1k∈Jr1ϕrα1−λ1/λ−11−λβ≤1ϕrα∑sk≥1k∈Jrskβ+1ϕrα∑sk<1k∈Jrskλβ.
Hence, for everyε>0, we have
(49)r∈ℕ:1ϕrα∑k∈JrIkAkΔγξρ,x1,⋯,xn−1ukβ≥ε⊂r∈ℕ:1ϕrα∑sk≥1k∈JrIkAkΔγξρ,x1,⋯,xn−1ukβ≥ε2∪r∈ℕ:1ϕrα∑sk<1k∈JrIkAkΔγξρ,x1,⋯,xn−1vkβ≥ε21/λ.
This implies thatr∈ℕ:1/ϕrα∑k∈IrIkAkΔγξ/ρ,x1,⋯,xn−1ukβ≥ε∈I and so ξ∈I−NαβA,M,u,θ,Δγ,·,⋯,·0. Hence, I−NαβA,M,v,θ,Δγ,·,⋯,·0⊆I−NαβA,M,u,θ,Δγ,·,⋯,·0. Similarly, we can prove I−NαβA,M,v,θ,Δγ,·,⋯,·⊆I−NαβA,M,u,θ,Δγ,·,⋯,·.Corollary 9.
If0<infuk≤1. Then, the following inclusions hold:
(50)I−NαβA,M,θ,Δγ,·,⋯,·0⊆I−NαβA,M,u,θ,Δγ,·,⋯,·0(51)I−NαβA,M,θ,Δγ,·,⋯,·⊆I−NαβA,M,u,θ,Δγ,·,⋯,·.Proof.
The proof follows from Theorem8.



---

*Source: 2779479-2022-03-15.xml*