Invariant means and lacunary sequence spaces of order (α, β)

Mohammad Ayman-Mursaleen; Md. Nasiruzzaman; Sunil K. Sharma; Qing-Bo Cai

doi:10.1515/dema-2024-0003

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Invariant means and lacunary sequence spaces of order (α, β)

Mohammad Ayman-Mursaleen , Md. Nasiruzzaman , Sunil K. Sharma and Qing-Bo Cai

Published/Copyright: June 7, 2024

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From the journal Demonstratio Mathematica Volume 57 Issue 1

Abstract

In this article, we use the notion of lacunary statistical convergence of order ( α , β ) to introduce new sequence spaces by lacunary sequence, invariant means defined by Musielak-Orlicz function ℳ = ( ℵ k ) . We also examine some topological properties and prove inclusion relations between newly constructed sequence spaces.

Keywords: difference sequence space; fractional difference operator; Orlicz function; Musielak-Orlicz function; lacunary sequence; invariant mean

MSC 2010: 40A05; 40A35; 40C05; 40F05; 46A45

1 Introduction and preliminaries

Consider a mapping σ from the set of positive integers into itself. A continuous linear functional ψ on l ∞ , is known to be an invariant mean or σ -mean if and only if

ψ ( ζ ) ≥ 0 when the sequence ζ = ( ζ k ) has ζ k ≥ 0 for all k ,
ψ ( e ) = 1 , where e = ( 1 , 1 , 1 , … ) , and
ψ ( ζ σ ( k ) ) = ψ ( ζ ) for all ζ ∈ l ∞ .

If ζ = ( ζ n ) , we can write T ζ = T ζ n = ( ζ σ ( n ) ) . It can be seen in [1] that

V σ = { ζ ∈ l ∞ : lim k t k n ( ζ ) = l , uniformly in n , l = σ − lim ζ } ,

where

t k n ( ζ ) = ζ n + ζ σ 1 n + ⋯ + ζ σ k n k + 1 .

In this case, σ is the translation mapping n → n + 1 , σ -mean is often called a Banach limit and V σ , the set of bounded sequences all of whose invariant means are equal, is the set of almost convergent sequences (see [2]).

The first edition of [3] introduced the concept of statistical convergence. Later, this concept was separately proposed by Fast [4] and Steinhaus [5], and some of its fundamental characteristics were investigated by Schoenberg [6], Salat [7], and Fridy [8]. This idea was expanded upon by Çolak [9] using the concept of α -density (for α = 1 , α -density lowered to natural density) and was given the name statistical convergence of order α . In [10,11], the lacunary statistical convergence of order α was studied. For more development of the topic, we refer to [12–15]. Şengül [16] recently offered an intriguing generalization of this idea by considering ( α , β ) on behalf of α and characterized statistical convergence of order ( α , β ) as follows.

The sequence ζ = ( ζ k ) is known to be statistically convergent of order ( α , β ) , briefly S α β -convergence to L if for each ε > 0 , we have

lim n → ∞ 1 n α ∣ { k ≤ n : ∣ ζ k − L ∣ ≥ ε } ∣ β = 0 ,

where ∣ { k ≤ n : k ∈ E } ∣ β represents the β th power of number of elements of E not exceeding n , and we write it as S α β − lim ζ k = L .

By a lacunary sequence θ = ( θ r ) where θ 0 = 0 , we mean an increasing sequence of positive integers with θ r − θ r − 1 → ∞ as r → ∞ . The intervals determined by θ will be represented by J r = ( θ r − 1 , θ r ] and t r = θ r − θ r − 1 . The space of lacunary strongly convergent sequences was developed by Freedman et al. [17] as

N θ = ζ ∈ w : lim r → ∞ 1 ϕ r ∑ k ∈ J r ∣ ζ k − l ∣ = 0 , for some l .

Definition

Consider a lacunary sequence θ = ( θ r ) . The sequence ζ = ( ζ k ) is S α β ( θ ) -statistically convergent (or lacunary statistically convergent of order ( α , β ) ) (see [18]) if there is a real number L such that

lim r → ∞ 1 ϕ r α ∣ { k ∈ J r : ∣ ξ k − L ∣ ≥ ε } ∣ β = 0 ,

where J r = ( θ r − 1 , θ r ] and ϕ r α represents the α th power ( ϕ r ) α of ϕ r , i.e., ϕ α = ( ϕ r α ) = ( ϕ 1 α , ϕ 2 α , … , ϕ r α , … ) and ∣ { k ≤ n : k ⊂ E } ∣ β represents the β th power of number of elements of E not exceeding n . In this case, the convergence is indicated by S α β ( θ ) − lim ξ k = L . S α β ( θ ) which will denote the set of all S α β ( θ ) -statistically convergent sequences. If α = β = 1 and θ = ( 2 r ) , then we will write S instead of S α β ( θ ) .

Kızmaz [19], who investigated the difference sequence spaces l ∞ ( Δ ) , c ( Δ ) and c 0 ( Δ ) , developed the idea of difference sequence spaces. Et and Çolak [20] introduced the spaces l ∞ ( Δ n ) , c ( Δ n ) , and c 0 ( Δ n ) to further expand the idea. The fractional difference operator defined by Baliarsingh [21] was as follows:

Assume that ξ = ( ξ k ) ∈ w and γ be a real number, then the fractional difference operator Δ ( γ ) is defined by

Δ ( γ ) ξ k = ∑ i = 0 k ( − γ ) i i ! ξ k − i ,

where ( − γ ) i denotes the Pochhammer symbol represented as

( − γ ) i = 1 , if γ = 0 or i = 0 , γ ( γ + 1 ) ( γ + 2 ) ⋯ ( γ + i − 1 ) , otherwise .

Numerous authors have used the ideas of n -normed spaces [22], difference sequences [23], the Orlicz function [24], and the Musielak-Orlicz function [25] to study sequence spaces along with their applications in approximation theory (see [26–32]). Mohiuddine et al. [33] and Sharma et al. [34] studied some sequence spaces of order ( α , β ) . In this article, we construct some new difference sequence spaces of lacunary statistical convergence defined by Musielak-Orlicz function as follows. We study topological properties of these new sequence spaces and obtain some inclusion results.

An Orlicz function ℵ is a function, ℵ : [ 0 , ∞ ) → [ 0 , ∞ ) , which is continuous, nondecreasing, and convex with ℵ ( 0 ) = 0 , ℵ ( x ) > 0 for x > 0 , and ℵ ( x ) → ∞ as x → ∞ . A sequence ℳ = ( ℵ k ) of Orlicz function is called a Musielak-Orlicz function [35].

Consider ℳ = ( ℵ k ) as a Musielak-Orlicz function, 0 < α ≤ β ≤ 1 and u = ( u k ) a bounded sequence of positive real numbers. In this article, we define the following sequence spaces:

w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 = ζ ∈ S ( n − X ) : lim r → ∞ 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β = 0 uniformly in n , for some ρ > 0 ,

w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ = ζ ∈ S ( n − X ) : lim r → ∞ 1 ϕ r α ∑ k ∈ I r ℵ k t k n ( Δ ( γ ) ζ k − l ) ρ , z 1 , … , z n − 1 u k β = 0 uniformly in n , for some l and ρ > 0 ,

and

w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ = ζ ∈ S ( n − X ) : sup r , n 1 ϕ r α ∑ k ∈ I r ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β < ∞ for some ρ > 0 .

If we take ℳ ( ζ ) = ζ , then the previous spaces become w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 , w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ , and w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ .

By taking u = ( u k ) = 1 , then the previous spaces become w α β [ ℳ , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 , w α β [ ℳ , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ and w α β [ ℳ , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ .

Throughout the article, the following inequality will be used. If 0 ≤ p k ≤ sup p k = K , D = max ( 1 , 2 K − 1 ) , then

(1.1) ∣ a k + b k ∣ p k ≤ D { ∣ a k ∣ p k + ∣ b k ∣ p k } ,

for all k and a k , b k ∈ C . Also ∣ a ∣ p k ≤ max ( 1 , ∣ a ∣ H ) for all a ∈ C .

2 Main results

Theorem 2.1

Let ℳ = ( ℵ k ) be a Musielak-Orlicz function and u = ( u k ) be a bounded sequence of positive real numbers. Then, the spaces w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 , w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ , and w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ are the linear spaces over C (where C is the field of complex number).

Proof

The proof is trivial, so we omit the details.□

Theorem 2.2

Let ℳ = ( ℵ k ) be a Musielak-Orlicz function and u = ( u k ) be a bounded sequence of positive real numbers. Then, w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 is a paranormed space with paranorm defined by

g ( ζ ) = inf ρ u r K : 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β 1 K ≤ 1 , r , n = 1 , 2 , … ,

where K = max ( 1 , sup k u k < ∞ ) .

Proof

Clearly, g ( ζ ) ≥ 0 for ζ = ( ζ k ) ∈ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 . Now, ℵ k ( 0 ) = 0 , and we have g ( 0 ) = 0 . Conversely, we suppose that g ( ζ ) = 0 , then

inf ρ u r K : 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β 1 K ≤ 1 , r , n = 1 , 2 , … = 0 .

This implies that for a given ε > 0 , there exists some ρ ε ( 0 < ρ ε < ε ) such that

1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ ε , z 1 , … , z n − 1 u k β 1 K ≤ 1 .

Thus,

1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ε , z 1 , … , z n − 1 u k β 1 K ≤ 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ ε , z 1 , … , z n − 1 u k β 1 K ≤ 1 .

For each k ∈ N , we suppose that ζ k ≠ 0 . This means that t k n ( Δ ( γ ) ζ k ) ≠ 0 . Let ε → 0 ⇒ t k n ( Δ ( γ ) ζ k ) ε , z 1 , … , z n − 1 → ∞ . It follows that

1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ε , z 1 , … , z n − 1 u k β 1 K → ∞ ,

which is a contradiction. Therefore, t k n ( Δ ( γ ) ζ k ) = 0 for each k , and thus, Δ ( γ ) ζ k = 0 for each k ∈ N . Let ρ 1 > 0 and ρ 2 > 0 be such that

1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ 1 , z 1 , … , z n − 1 u k β 1 K ≤ 1

and

1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ϑ k ) ρ 2 , z 1 , … , z n − 1 u k β 1 K ≤ 1 ,

for each r . Suppose that ρ = ρ 1 + ρ 2 . Therefore, by Minkowski’s inequality, we have

1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ( ζ k + ϑ k ) ) ρ , z 1 , … , z n − 1 u k β 1 K ≤ 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) + t k n ( Δ ( γ ) ϑ k ) ρ 1 + ρ 2 , z 1 , … , z n − 1 u k β 1 K ≤ 1 ϕ r α ∑ k ∈ J r ρ 1 ρ 1 + ρ 2 ℵ k t k n ( Δ ( γ ) ζ k ) ρ 1 , z 1 , … , z n − 1 + ρ 2 ρ 1 + ρ 2 ℵ k t k n ( Δ ( γ ) ϑ k ) ρ 2 , z 1 , … , z n − 1 u k β 1 K ≤ ρ 1 ρ 1 + ρ 2 1 ϕ r α ∑ k ∈ I r ℵ k t k n ( Δ ( γ ) ζ k ) ρ 1 , z 1 , … , z n − 1 u k β 1 K + ρ 2 ρ 1 + ρ 2 1 ϕ r α ∑ k ∈ J r ℵ k ‖ t k n ( Δ ( γ ) ϑ k ) ρ 2 , z 1 , … , z n − 1 ‖ u k β 1 K ≤ 1 .

Since ρ ’s are non-negative, so we have

g ( ζ + ϑ ) = inf ρ u r H : 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) + t k n ( Δ ( γ ) ϑ k ) ρ , z 1 , … , z n − 1 u k β 1 K ≤ 1 , r , n = 1 , 2 , … ≤ inf ρ 1 u r H : 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ 1 , z 1 , … , z n − 1 u k β 1 K ≤ 1 , r , n = 1 , 2 , … + inf ρ 2 u r H : 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ϑ k ) ρ 2 , z 1 , … , z n − 1 u k β 1 K ≤ 1 , r , n = 1 , 2 , … .

Thus,

g ( ζ + ϑ ) ≤ g ( ζ ) + g ( ϑ ) .

Finally, we have to show that the scalar multiplication is continuous. Suppose that ν be any complex number. Therefore, by definition,

g ( ν ζ ) = inf ρ u r H : 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) λ ζ k ) ρ , z 1 , … , z n − 1 u k β 1 K ≤ 1 , r , n = 1 , 2 , … .

Then,

g ( ν ζ k ) = inf ( ∣ ν ∣ t ) u r H : 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) t , z 1 , … , z n − 1 u k β 1 K ≤ 1 , r , n = 1 , 2 , … ,

where t = ρ ∣ ν ∣ . Since ∣ ν ∣ u r ≤ max ( 1 , ∣ ν ∣ sup u r ) , we have

g ( ν ζ ) ≤ max ( 1 , ∣ ν ∣ sup p r ) inf t u r H : 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) t , z 1 , … , z n − 1 u k β 1 K ≤ 1 , r , n = 1 , 2 , … .

From the aforementioned inequality, it follows that scalar multiplication is continuous.□

Theorem 2.3

Let ℳ = ( ℵ k ) be a Musielak-Orlicz function. Then,

w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ .

Proof

Now, w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ is obvious. We only need to show that w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ . For this, let w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ . Then, there exists some positive number ρ 1 such that

1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k − l ) ρ 1 , z 1 , … , z n − 1 u k β → 0 , as r → ∞ uniformly in n .

Define ρ = 2 ρ 1 . Since ℳ = ( ℵ k ) is non-decreasing, convex, and so using Inequality (1.1), we have

1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β ≤ D ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k − l ) ρ , z 1 , … , z n − 1 u k β + D ϕ r α ∑ k ∈ J r ℵ k l ρ 1 , z 1 , … , z n − 1 u k β ≤ D ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k − l ) ρ , z 1 , … , z n − 1 u k β + D max 1 , ℵ k l ρ 1 , z 1 , … , z n − 1 K .

Thus, ζ ∈ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ . □

Theorem 2.4

If sup k [ ℵ k ( q ) ] u k < ∞ for all q > 0 , then

w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ .

Proof

Let ζ ∈ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ . Using Inequality (1.1), we have

1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β ≤ D ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k − l ) ρ , z 1 , … , z n − 1 u k β + D ϕ r α ∑ k ∈ J r ℵ k l ρ , z 1 , … , z n − 1 u k β .

We can take sup k [ ℵ k ( q ) ] u k = Q as sup k [ ℵ k ( q ) ] u k < ∞ is given. Hence, we have ζ ∈ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ .□

Theorem 2.5

Consider that ℳ = ( ℵ k ) satisfies Δ 2 -condition for all k , then

w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ .

Proof

Let ζ ∈ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ . Then, we have

Q r = 1 ϕ r α ∑ k ∈ J r ‖ t k n ( Δ ( γ ) ζ k − l ) , z 1 , … , z n − 1 ‖ u k → ∞ , as r → ∞ .

Suppose ε > 0 and choose δ as 0 < δ < 1 such that ℵ k ( q ) < ε for 0 ≤ q ≤ δ for all k , we have

1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k − l ) ρ , z 1 , … , z n − 1 u k β = 1 ϕ r α ∑ k ∈ J r , ‖ t k n ( Δ ( γ ) ζ − l ) , z ‖ ≤ δ 1 ℵ k t k n ( Δ ( γ ) ζ k − l ) ρ , z 1 , … , z n − 1 u k β + 1 ϕ r α ∑ k ∈ J r , ‖ t k n ( Δ ( γ ) ζ − l ) , z ‖ > δ 2 ℵ k t k n ( Δ ( γ ) ζ k − l ) ρ , z 1 , … , z n − 1 u k β .

For summation first in the aforementioned equation, we have ∑ 1 ≤ ε K and the second summation for all k , we write

‖ t k n ( Δ ( γ ) ζ k − l ) , z 1 , … , z n − 1 ‖ ≤ 1 + t k n ( Δ ( γ ) ζ k − l ) δ , z 1 , … , z n − 1 .

Now, this implies that

[ ℵ k ( ‖ t k n ( Δ ( γ ) ζ k − l ) , z 1 , … , z n − 1 ‖ ) ] < ℵ k 1 + t k n ( Δ ( γ ) ζ k − l ) δ , z 1 , … , z n − 1 ≤ 1 2 ( ℵ k ( 2 ) ) + 1 2 ℵ k ( 2 ) t k n ( Δ ( γ ) ζ k − l ) δ , z 1 , … , z n − 1 .

Since ℵ k satisfies Δ 2 -condition, we can have

[ ℵ k ( ‖ t k n ( Δ ( γ ) ζ k − l ) , z 1 , … , z n − 1 ‖ ) ] ≤ 1 2 L t k n ( Δ ( γ ) ζ k − l ) δ , z 1 , … , z n − 1 ℵ k ( 2 ) + 1 2 L t k n ( Δ ( γ ) ζ k − l ) δ , z 1 , … , z n − 1 ℵ k ( 2 ) = L t k n ( Δ ( γ ) ζ k − l ) δ , z 1 , … , z n − 1 ℵ k ( 2 ) .

So we write

1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k − l ) ρ , z 1 , … , z n − 1 u k β ≤ ε K + [ max ( 1 , L ℵ k ( 2 ) ) δ ] K Q r .

Now, r → ∞ implies that ζ ∈ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ . Hence, this completes the proof.□

Theorem 2.6

The following statements are equivalent:

w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ,
w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ,
sup r 1 ϕ r α ∑ k ∈ J r [ ℵ k ( q ) ] u k < ∞ , for all q > 0 .

Proof

(i) ⇒ (ii) We need to show that w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ⊂ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ . Let ζ ∈ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 . Then, there exists r ≥ r 0 , for ε > 0 , such that

1 ϕ r α ∑ k ∈ J r t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β < ε .

Hence, there exists K > 0 such that

sup r , n 1 ϕ r α ∑ k ∈ J r t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β < K ,

for all n and r . So we obtain ζ ∈ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ .

(ii) ⇒ (iii) Suppose that (iii) does not imply. Then, for some q > 0 ,

sup r 1 ϕ r α ∑ k ∈ J r [ ℵ k ( q ) ] u k = ∞ ,

and now, we can take a subinterval J r ( m ) of J r such that

(2.1) 1 ϕ r ( m ) α ∑ k ∈ J r ( m ) ℵ k 1 m u k β > m , m = 1 , 2 , … .

Let us define ζ = ( ζ k ) as follows: ζ k = 1 m if k ∈ J r ( m ) and ζ k = 0 if k ∉ J r ( m ) . Then, ζ ∈ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 but by equation (2.1), ζ ∉ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 , which is a contradiction. Thus, our supposition is wrong. Hence, (iii) must hold.

(iii) ⇒ (i) Suppose that (i) does not holds; thus, for ζ ∈ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ , we have

(2.2) sup r , n 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k = ∞ .

Let q = t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 for each k and fixed n , so that equation (2.2) becomes

sup r 1 ϕ r α ∑ k ∈ J r [ ℵ k ( q ) ] u k β = ∞ ,

which is a contradiction. Thus, our supposition is wrong. Hence, (i) must hold.□

Theorem 2.7

The following statements are equivalent:

w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ⊂ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ,
w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ⊂ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ ,
inf r ∑ k ∈ J r [ ℵ k ( q ) ] u k > 0 , for all q > 0 .

Proof

(i) ⇒ (ii) : The proof is straightforward so we omit the details.

(ii) ⇒ (iii) Let us assume that (iii) does not hold. Then,

inf r 1 ϕ r α ∑ k ∈ J r [ ℵ k ( q ) ] u k β = 0 , for some q > 0 ,

and now, we can take a subinterval J r ( m ) of J r such that

(2.3) 1 ϕ r α ∑ k ∈ J r ( m ) [ ℵ k ( m ) ] u k β < 1 m , m = 1 , 2 , … .

Now, we take ζ k = m if k ∈ J r ( m ) and ζ k = 0 if k ∉ J r ( m ) . Therefore, by equation (2.3), ζ ∈ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 but ζ ∉ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ , which is a contradiction. Thus, our supposition is wrong. Therefore, (iii) holds.

(iii) ⇒ (i) It is trivial, so we omit the details.□

Theorem 2.8

The inclusion w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ ⊂ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 holds if and only if

(2.4) lim r → ∞ 1 ϕ r α ∑ k ∈ I r [ M k ( t ) ] u k β = ∞ .

Proof

Let w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ ⊂ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 . Let us assume that equation (2.4) does not hold. Therefore, there is a subinterval J r ( m ) of J r and a number q 0 > 0 , where q 0 = t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … z n − 1 for all k and n , such that

(2.5) 1 ϕ r ( m ) α ∑ k ∈ J r ( m ) [ ℵ k ( q 0 ) ] u k β ≤ M < ∞ , m = 1 , 2 , … .

Let us define ζ k = q 0 if k ∈ J r ( m ) and ζ k = 0 if k ∉ J r ( m ) . Then, by equation (2.5), ζ ∈ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ . But ζ ∉ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 . Hence, equation (2.5) holds.

Conversely, let us suppose that equation (2.5) holds and ζ ∈ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ . Then, we have

(2.6) 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β ≤ M < ∞ .

Now, suppose that ζ ∉ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 . Then, for ε > 0 and a subinterval J r i of J r , there is k 0 such that ‖ t k n ( Δ ( γ ) ζ k ) , z 1 , … , z n − 1 ‖ u k > ε for k ≥ k 0 . Thus, we have

ℵ k ε ρ u k ≤ ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k ,

which is a contradiction to equation (2.5) using equation (2.6). Hence, w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ ⊂ w α β [ u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 holds if and only if lim r → ∞ 1 ϕ r α ∑ k ∈ I r [ M k ( t ) ] u k β = ∞ . □

Theorem 2.9

The following inclusions hold for Musielak-Orlicz functions ℳ = ( ℵ k ) and ℳ ′ = ( ℵ k ′ ) :

w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ ∩ w α β [ ℳ ′ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞
⊂ w α β [ ℳ + ℳ ′ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ;
w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∩ w α β [ ℳ ′ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ
⊂ w α β [ ℳ + ℳ ′ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ;
w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ∩ w α β [ ℳ ′ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0
⊂ w α β [ ℳ + ℳ ′ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 .

Proof

Suppose that ζ ∈ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ ∩ w α β [ ℳ ′ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ . Then,

sup r , n 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β < ∞ uniformly in n

and

sup r , n 1 ϕ r α ∑ k ∈ J r ℵ k ′ t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β < ∞ uniformly in n .

Now, using Inequality (1.1), we obtain

( ℵ k + ℵ k ′ ) t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k ≤ D ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k + D ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k ⇒ 1 ϕ r α ∑ k ∈ J r ( ℵ k + ℵ k ′ ) t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β ≤ D ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β + D ϕ r α ∑ k ∈ J r ℵ k ′ t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 u k β uniformly in n .

This proves w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ ∩ w α β [ ℳ ′ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞

⊂ w α β [ ℳ + ℳ ′ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 .

Similarly, we prove

w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∩ w α β [ ℳ ′ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ⊂ w α β [ ℳ + ℳ ′ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ

and

□ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ∩ w α β [ ℳ ′ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ⊂ w α β [ ℳ + ℳ ′ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 .

Theorem 2.10

If 0 < u k ≤ v k and v k u k be bounded. Then, for each k, we have

(i) w α β [ ℳ , v , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] ∞ σ ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ ;

(ii) w α β [ ℳ , v , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ;

(iii) w α β [ ℳ , v , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 .

Proof

(i) Let ζ ∈ w α β [ ℳ , v , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ . Then,

sup r , n 1 ϕ r α ∑ k ∈ J r ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 v k β < ∞ uniformly in n .

Write μ k , n = ℵ k t k n ( Δ ( γ ) ζ k ) ρ , z 1 , … , z n − 1 v k β and λ k = u k v k . Since u k ≤ v k , 0 < λ < λ k ≤ 1 . Define x k , n = μ k , n , x k , n = 0 if μ k , n ≥ 1 and y k , n = μ k , n , y k , n = 0 if μ k , n ≥ 1 . So μ k , n = x k , n + y k , n and μ k , n λ k = x k , n λ k + y k , n λ k . Now, it follows that y k , n λ k ≤ x k , n ≤ y k , n and y k , n λ k ≤ y k , n λ . Therefore,

1 ϕ r α ∑ k ∈ J r μ k , n λ k = 1 ϕ r α ∑ k ∈ J r ( x k , n λ k + y k , n λ k ) ≤ 1 ϕ r α ∑ k ∈ J r y k , n + 1 ϕ r α ∑ k ∈ J r y k , n λ .

Since λ < 1 such that 1 λ > 1 , for each n and therefore by Holder’s inequality, we obtain

1 ϕ r α ∑ k ∈ J r y k , n λ = ∑ k ∈ J r 1 ϕ r α z k , n λ 1 ϕ r α 1 − λ ≤ ∑ k ∈ J r 1 ϕ r α y k , n λ 1 λ λ ∑ k ∈ J r 1 ϕ r α 1 − λ 1 ( 1 − λ ) 1 − λ = 1 ϕ r α ∑ k ∈ J r y k , n λ .

Thus, we have

1 ϕ r α ∑ k ∈ J r μ k , n λ k ≤ 1 ϕ r α ∑ k ∈ J r μ k , n + 1 ϕ r α ∑ k ∈ J r y k , n λ .

Thus, ζ ∈ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ . Hence,

w α β [ ℳ , v , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] ∞ σ ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ .

Similarly, we prove

w α β [ ℳ , v , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ

and

□ w α β [ ℳ , v , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 ⊂ w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 .

3 Conclusion

Baliarsingh [21] defined the fractional difference operator and Cai et al. [36] and Sharma et al. [34] studied some sequence spaces of order ( α , β ) . We have constructed here some new lacunary sequence spaces of fractional difference operator defined by Musielak-Orlicz function as w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ 0 , w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ , and w α β [ ℳ , u , Δ ( γ ) , θ , ‖ ⋅ , … , ⋅ ‖ ] σ ∞ and studied some topological properties. We also established some inclusion relation between these sequence spaces. Readers may find these spaces interesting of further scope to study their geometric properties, matrix transformations, and corresponding compact matrix operators [37].

mohdaymanm@gmail.com, nasir3489@gmail.com, sunilksharma42@gmail.com, qbcai@126.com

Acknowledgments

The first author is supported by JADD program (by UPM-UON).

Funding information: No funding was provided.
Author contributions: All authors contributed equally to writing this manuscript. All authors read and approved the manuscript.
Conflict of interest: The authors declare that they have no conflicts of interest.
Ethical approval: Not applicable.
Data availability statement: No data were used to support this study.

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Received: 2023-02-09

Revised: 2023-12-07

Accepted: 2024-02-12

Published Online: 2024-06-07

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/dema-2024-0003

Keywords for this article

difference sequence space; fractional difference operator; Orlicz function; Musielak-Orlicz function; lacunary sequence; invariant mean

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