# Multiplication Operators on Weighted Nakano (sss)

Authors: Awad A. Bakery; Afaf R. Abou Elmatty; OM Kalthum S. K. Mohamed
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this article, we investigate the sufficient conditions on weighted Nakano sequence space to be premodular Banach (sss). We examine some topological and geometrical structures of the multiplication operators defined on weighted Nakano prequasi-normed (sss).


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## Body

## 1. Introduction

Summability is very important in mathematical models and has numerous implementations, such as normal series theory, approximation theory, ideal transformations, and fixed point theory. For more details, see [1–4]. By ℂℕ, c, ℓ∞, ℓr, and co, we signify the spaces of each, convergent, bounded, r-absolutely summable, and convergent to zero sequences of complex numbers. ℕ indicates the set of nonnegative integers. Let p = pn and βnn∈ℕ be sequences of positive reals, and the weighted Nakano sequence space [5] is defined by(1)ℓβpn=x=xk∈ℂℕ:σλx<∞,for someλ>0,where σx=∑k=0∞βkxkpk. And ℓβpn,⋅ is a Banach space, where x=infη>0:σx/η≤1.Whenpn∈ℓ∞, we have(2)ℓβpn=x=xk∈ℂℕ:σλx<∞,for anyλ>0.(1)
Ifβn=1, for all n ∈ℕ, then ℓβpn will reduce to ℓpn (see [6, 7])(2)
Ifβn=1/pn, for all n ∈ℕ, then ℓβpn will reduce to ℓpn which is studied by many authors (see [8–10])ByBW,Z, we will indicate the set of every operators which are linear and bounded between Banach spaces W and Z, and if W = Z, we write BW. The multiplication operators have a wide field of mathematics in functional analysis, for instance, in eigenvalue distributions theorem, geometric structure of Banach spaces, and theory of fixed point. For more details, see [11–16]. The s-numbers [17] have many examples such as the rth approximation number, denoted by αr(V), is defined by αr(V) = inf {V−B: B ∈BW,Z and rank(B) ≤ r}, and the rth Kolmogorov number, denoted by dr(V), is defined by dr(V) = infdimw ≤ r sup ω≤1 inf ν∈WVω−ν. The following notations will be used in the sequel:(3)Xs=XsW,Z,whereXsW,Z=V∈BW,ZSjVj=0∞∈X,Xapp=XappW,Z,whereXappW,Z=V∈BW,ZαjVj=0∞∈X,XKol=XKolW,Z,whereXKolW,Z=V∈BW,ZdjVJ=0∞∈X.A few of operator ideals in the class of Hilbert spaces or Banach spaces are defined by distinct scalar sequence spaces, such as the ideal of compact operatorsBc formed by (dr (V)) and c0. Pietsch [4] studied the quasi-ideals ℓrapp for r∈0,∞, the ideals of Hilbert Schmidt operators between Hilbert spaces constructed by ℓ2, and the ideals of nuclear operators generated by ℓ1. He explained that F¯=ℓrapp for r ∈ [1, ∞), where F¯ is the closed class of all finite rank operators, and the class ℓrapp became simple Banach and small [1]. The strictly inclusion ℓjappW,Z≠⊂BW,Z, whenever j > r > 0, W and Z are infinite-dimensional Banach spaces investigated through Makarov and Faried [2]. Faried and Bakery [18] gave a generalization of the class of quasi-operator ideal which is the prequasi-operator ideal, and they examined several geometric and topological structure of ℓMS and (ces(r))S. On sequence spaces, Mursaleen and Noman [19, 20] investigated the compact operators on some difference sequence spaces. The multiplication operators on (ces(r), ||.||) with the Luxemburg norm ||.|| elaborated by Komal et al. [21] and Bakery and Abou Elmatty [5] gave the sufficient (not necessary) conditions on ℓβpn such that Sℓβpn constructed a simple Banach prequi operator ideal. The prequasi-operator ideal Sℓβpnapp became small and strictly contained for different weights and powers. The aim of this article to explain some results of ℓβpnτ equipped with the prequasi-norm τ. Firstly, we give the sufficient conditions on weighted Nakano sequence space to form premodular Banach (sss). Secondly, we give the necessity and sufficient conditions on weighted Nakano sequence space equipped with the prequasi-norm such that the multiplication operator defined on it is bounded, approximable, invertible, Fredholm, and closed range operator.


## 2. Preliminaries and Definitions

Definition 1 (see [4]).
An operatorV∈BW is called approximable if there are Dr ∈ FW, for every r ∈ ℕ and limr⟶∞ ||V—Dr|| = 0.
ByΥW,Z and BcW,Z, we will indicate the space of all approximable and compact operators from W to Z, respectively.Theorem 1 (see [4]).
If W is Banach space with dim (W) = ∞, thenFW≠⊂ΥW≠⊂BcW≠⊂BW.Definition 2 (see [22]).
An operator V∈BW is called Fredholm if dim (R(V))c < ∞, dim (kerV) < ∞, and R(V) is closed, where (R(V))c denotes the complement of range V.
The sequenceej = (0, 0, 1, 0, 0, ...) with 1 in the jth coordinate, for every j ∈ℕ, will be used in the sequel.Definition 3 (see [3]).
The space of linear sequence spaces Y is called (sss)(1)
Ifer ∈ Y with r ∈ ℕ,(2)
Letu = (ur) ∈ ℂℕ, v = (vr)∈ Y, and |ur|≤|vr|, for every r ∈ℕ, then u ∈ Y. This means Y be solid,”(3)
If (ur) urr=0∞∈Y, then ur/2r=0∞ ∈Y, wherever r/2 means the integral part of r/2Definition 4 (see [23]).
A subspace of the (sss)Yτ is called a premodular (sss) if there is a function: Y ⟶ [0, ∞) verifying the conditions:(i)
τ(y) ≥ 0, for each y ∈ Y and τ(y) = 0 ⇔ y = θ, where θ is the zero element of Y(ii)
There existsa ≥ 1 such that τ(ηy) ≤ a|η| τ(y), for all y ∈ Y, and η ∈ℂ(iii)
For someb ≥ 1, τ(y + z) ≤ b(τ(y) + τ(z)), for every y, z ∈ Y(iv)
|yr| ≤ |zr| withr ∈ ℕ, implies τ((yr)) ≤ τ((zr))(v)
For someb0 ≥ 1, τ ((yr)) ≤ τ(yr/2) ≤ b0τ ((yr))(vi)
Ify = yrr=0∞∈ Y, and d > 0, then there is r0 ∈ ℕ with τ(yrr=0∞) < d(vii)
There ist > 0 with τ (ν, 0, 0, 0, ...) ≥ t|ν| τ (1, 0, 0, 0, ...), for any ? ∈ ℂDefinition 5 (see [23]).
The (sss)Yτ is called prequasi-normed (sss) if τ satisfies the Parts (i)–(iii) of Definition 4, and when the space Y is complete under τ, then Yτ is called a prequasi-Banach (sss).Theorem 2 (see [23]).
A prequasi-norm (sss)Yτ, whenever it is premodular (sss).The inequality [24]: ai+biqi≤2H−1aiqi+biqi, where qi ≥ 0, for all i ∈ ℕ and H=max1,supiqi, will be used in the sequel.


## 3. Main Results

### 3.1. Prequasi-Norm onℓβpn

We investigate the sufficient conditions on the sequence spaceℓβpn equipped with prequasi-norm τ to form prequasi-Banach and closed (sss).Theorem 3.
ℓβpnτ, where τx=∑k=0∞βkxkpk1/H, for all x ∈ ℓβpn, is a premodular (sss), if the following conditions are satisfied:(a1)
The sequencePn ∈ ℝ+ℕ∩ ℓ∞ is increasing(a2)
Either (βn) is a monotonic decreasing or monotonic increasing such that there is C ≥ 1 for which β2n + 1 ≤ CβnProof.
Firstly, we have to proveℓβpn is a (sss):(1-i)
Letx, y ∈ ℓβpn. Since pn is bounded, we obtain(4)τX+Y=∑n=0∞βnxn+ynpn1/H≤∑n=0∞βnxnpn1/H+∑n=0∞βnynpn1/H=τX+τY<∞,then x + y ∈ ℓβpn.(1-ii)
Letλ ∈ ℂ and x ∈ ℓβpn. Since pn is bounded, we have(5)τλX=∑n=0∞βnλxnpn1/H≤nsupλpn/H∑n=0∞βnxnpn1/H≤aλτX<∞.(i)
Then,λ x ∈ ℓβpn Therefore, by using Parts (1-i) and (1-ii) we have that the space ℓβpn is linear. Also, en ∈ ℓβpn, for all n ∈ ℕ, since(6)τen=∑i=0∞βienipi1/H=βn1/H.(2)
Letxn≤yn, for all n ∈ ℕ and y ∈ ℓβpn. Since βn > 0, for all n ∈ ℕ, then τ(x) = ∑n=0∞βnxnpn1/H ≤ ∑n=0∞βnynpn1/H = τ(y) < ∞, we get x ∈ ℓβpn.(3)
Letxn ∈ ℓβpn and (βn) be an increasing sequence. There exists C ≥ 1 such that β2n + 1 ≤ Cβn and pn be an increasing; then, we have(7)τxn/2=∑n=0∞βnxn/2pn1/H=∑n=0∞β2nxnp2n+∑n=0∞β2n+1xnp2n+11/H≤∑n=0∞β2nxnpn+∑n=0∞β2n+1xnpn1/H≤2C1/H∑n=0∞βnxnpn1/H=2C1/Hτxn,then xn/2 ∈ ℓβpn. Secondly, we show that the functional τ on ℓβpn is a premodular:(i)
Evidently,τ (x) ≥ 0 and τ (x) = 0 ⇔ x = θ(ii)
There is a steadya = max {1, supn λpn/H−1} ≥ 1 such that τ (λx) ≤ a|λ| τ (x), for all x ∈ ℓβpnand λ ∈ ℂ(iii)
There existsK ≥ 1 such that τ (x + y) ≤ K (τ(x) + τ(y)), for all x, y ∈ ℓβpn(iv)
Clearly, from the proof part (2) of Theorem 3, the condition is clear since the weighted Nakano sequence space is solid(v)
It is obtained from (3) that b0=2C1/H ≥ 1(vi)
It is clear thatF¯ =  ℓβpn(vii)
There exists a steady 0 <ξ ≤ λp0/H−1, for λ ≠ 0 or ξ > 0, for λ = 0 such that τ (λ, 0, 0, 0, ...) ≥ ξ |λ| τ (1, 0, 0, 0, ...)Theorem 4.
If conditions (a1) and (a2) are satisfied, thenℓβpnτ, where τx=∑k=0∞βkxkpk1/H for all x ∈ ℓβpn, is a prequasi-Banach (sss).Proof.
Let the conditions be verified. By Theorem3, the space ℓβpnτ is a premodular (sss). Therefore, from Theorem 2, the space ℓβpnτ is a prequasi-normed (sss). To prove that ℓβpnτ is a prequasi-Banach (sss), suppose xn = xknk=0∞ be a Cauchy sequence in ℓβpnτ, then for every ε ∈ (0, 1), there exists a number n0∈ℕ such that, for all n, m ≥ n0, one has(8)τXn−Xm=∑j=0∞βjxjn−xjmpj1/H<ε.
Hence, forn, m ≥ n0 and j ∈ ℕ, we get xjn−xjm < ε.
So,xjm is a Cauchy sequence in ℂ for fixed j ∈ ℕ, and this gives limm⟶∞xjm=xj0 for j ∈ ℕ. Hence, τ (xn−x0) < ε, for all n ≥ n0. Finally, to prove that x0 ∈ ℓβpn, we have τ (x0) = τ (x0 − xn + xn) ≤ τ(xn–x0) + τ(xn) < ∞.
So, x0 ∈ ℓβpn. This means that ℓβpnτ is a prequasi-Banach (sss).Theorem 5.
If conditions (a1) and (a2) are satisfied, thenℓβpnτ, where τx=∑k=0∞βkxkpk1/H, for all x ∈ ℓβpn, is a prequasi-closed (sss).Proof.
Let the conditions be verified. By Theorem3, the space ℓβpnτ is a premodular (sss). Therefore, from Theorem 2, the space ℓβpnτ is a prequasi-normed (sss). To prove that ℓβpnτ is a prequasi-closed (sss), suppose Xn=xknk=0∞∈ ℓβpnτ and limn ⟶ ∞ τXn−X0=0, then for every ε ∈ (0, 1), there exists a number n0 ∈ ℕ such that for all n ≥ n0, one has(9)ε>τXn−X0=∑j=0∞βjxjn−xj0pj1/H≥βjxjn−xj0pj1/H.
Hence, forn ≥ n0 and j ∈ ℕ, we get xjn−xj0 < ε.
So,xjn is a convergent sequence in ℂ for fixed j ∈ ℕ, and this gives limm ⟶ ∞ xjm=xj0 for fixed j ∈ ℕ. Finally, to prove that x0 ∈ ℓβpn, we have τ (x0) = τ(x0 − xn + xn) ≤ τ(xn − x0) + τ(xn) < ∞.
So, x0 ∈ ℓβpn. This means that ℓβpnτ is a prequasi-closed (sss).


### 3.2. Bounded Multiplication Operator onℓβpn

Here after, we investigate some topological and geometric structure of the multiplication operator acting onℓβpn. In this section, we examine the sufficient conditions on the sequence space ℓβpn equipped with prequasi-norm τ such that the multiplication operator defined on ℓβpnτℓβpnτ is bounded and isometry.Definition 6.
Letκ ∈ ℂℕ ∩ ℓ∞ and Wτ be a prequasi-normed (sss). An operator Vκ: Wτ ⟶ Wτ is called multiplication operator if Vκω = κω  = Krωrr=0∞ ∈ W, for all ω∈W. If Vκ ∈ BW, we name it a multiplication operator generated by κ.Theorem 6.
Letκ ∈ ℂℕ and conditions (a1) and (a2) be satisfied; then, κ ∈ ℓ∞, if and only if VK∈Bℓβpnτ, where τx=∑k=0∞βkxkpk1/H, for all x ∈ ℓβpn.Proof.
Let the conditions be satisfied. Assumeκ ∈ ℓ∞. Therefore, there is ε > 0 with |κr| ≤ ε, for every r ∈ ℕ. For x ∈ ℓβpnτ, since pr is bounded from above with pr > 0, for all r ∈ ℕ, then(10)τVKX=τKX=τKrXrr=0∞=∑r=0∞βrκrxrpr1/H≤∑r=0∞βr|εxr|pr1/H≤suprεpr/H∑r=0∞βr|xr|pr1/H=Dτx.
This gives Vκ ∈ Bℓβpnτ. Conversely, assume that Vκ ∈ Bℓβpnτ. Let us assume κ ∉ ℓ∞; hence, for each j ∈ ℕ, there is ij ∈ ℕ such that κij > j. Since pn ∈ ℝ+ℕ ∩ ℓ∞ is increasing, we have(11)τVKeij=τKeij=τKreijτr=0∞=∑r=0∞βrκreijrpr1/H=βijκijpij/H>βijjpij/H=βijjpij/Hτeij.
This shows that Vκ ∈Bℓβpnτ. Therefore, κ ∈ ℓ∞.Theorem 7.
Pick upκ ∈ ℂℕ and ℓβpnτ be a prequasi-normed (sss), with τ(x) = ∑k=0∞βkxkpk1/H, for all x ∈ ℓβpn. Then, κr = 1, for all r ∈ ℕ, if and only if, Vκ is an isometry.Proof.
Suppose |κr| = 1, for all r ∈ ℕ. Hence,(12)τVKX=τKX=τKrXrr=0∞=∑r=0∞βrκrxrpr1/H=∑r=0∞βrxrpr1/H=τx,for all x ∈ ℓβpnτ. Therefore, Vκ is an isometry. Conversely, Assume that |κi| < 1 for some i = i0. Since pr ∈ ℝ+ℕ ∩ ℓ∞ is increasing, we obtain(13)τVei0=τKei0=τKrei0rr=0∞=∑r=0∞βrκrei0rpr1/H<∑r=0∞βrei0rpr1/H=τei0.
When |κi0| > 1, we can prove that τVKei0 > τei0. Therefore, in both cases, we have a contradiction. So, |κr| = 1, for every r ∈ ℕ.


### 3.3. Approximable Multiplication Operator onℓβpn

In this section, we introduce the sufficient conditions on the sequence spaceℓβpn equipped with prequasi-norm τ such that the multiplication operator defined on ℓβpnτ is an approximable and compact. By card (A), we indicate the cardinality of the set A.Theorem 8.
Ifκ ∈ ℂℕ and ℓβpnτ is a prequasi-normed (sss), where τ(x) = ∑k=0∞βkxkpk1/H, for all x ∈ ℓβpn, then, Vκ ∈ Υℓβpnτ if and only if Krr=0∞ ∈ c0.Proof.
Assume that Vκ ∈ Υℓβpnτ. Therefore, Vκ ∈ BBcℓβpnτ, to prove that the sequence Krr=0∞ belongs to c0. Suppose Krr=0∞ ∉ c0. Hence, there is δ > 0 such that the set Aδ  = {r ∈ℕ: |κr| ≥ δ} has card Aδ = ∞. Assume ai ∈Aδ, for all i ∈ℕ. Hence, {eai: ai ∈ Aδ} is an infinite bounded set in ℓβpnτ. Let(14)τVKeai−VKeaj=τKeai−Keai=τκreair−eairr=0∞=∑r=0∞βrκreair−eajrpr1/H≥∑r=0∞βrδeair−eajrpr1/H≥infrδpr/Hτeai−eaj,for all ai,aj ∈ Aδ. This shows {eai: ai ∈ Bδ} ∈ ℓ∞ which cannot have a convergent subsequence under Vκ. This proves that Vk ∉ Bcℓβpnτ. Then, Vk ∉Υℓβpnτ, this gives a contradiction. So, limi ⟶ ∞ κi = 0. Conversely, let limi ⟶∞ κi = 0. Then, for each δ > 0, the set Aδ = {i ∈ℕ: |κi| ≥ δ} has card (Aδ) < ∞. Hence, for every δ > 0, the space ℓβpnτAδ = {x = (xi) ∈ ℂAδ} is finite dimensional. Then, Vκ|ℓβpnτAδ is a finite rank operator. For every i ∈ ℕ, dene κi∈ ℂℕ by (κ i)j = κj,j∈A1/i,0,otherwise.
It is clear thatVκi has rank Vκi <∞ as dim ℓβpnτA1/i < ∞, for each i ∈ ℕ. Therefore, since pn ∈ ℝ+ℕ∩ ℓ∞ is increasing, we obtain(15)τVk−VkiXτkj−kijXjj=0∞=∑j=0∞βjκj−κijxjpj1/H∑j=0,j∈A1/i∞βjκj−κijxjpj+∑j=0,j∈A1/i∞βjκi−κijxjpj1/H=∑j=0,j∉A1/i∞βjκjxjpj1/H≤supj1jpj/H∑j=0,j∉A1/i∞βjxipjpj/H>supj1jpj/Hτx=1jp0/Hτx.
This implies thatBc−VKi≤1/jp0/H and that Vκ is a limit of finite rank operators. Therefore, Vκ is an approximable operator.Theorem 9.
Letκ ∈ℂℕ and ℓβpnτ be a prequasi-normed (sss), where τ(x) = ∑r=0∞βrxrpr1/H, for all x ∈ ℓβpnτ. Then, Vκ ∈ Bcℓβpnτ, if and only if, Kii=0∞ ∈ c0.Proof.
It is simple so overlooked.Corollary 1.
Ifκ ∈ ℂℕ, conditions (a1) and (a2) are satisfied, then Bcℓβpnτ≠⊂Bℓβpnτ, where τ(x) = ∑r=0∞βrxrpr1/H, for all x ∈ℓβpn.Proof.
Since I is a multiplication operator onlβPnτ generated by κ = (1, 1, ...), therefore, I ∉ Bcℓβpnτ and I ∈ Bℓβpnτ.


### 3.4. Fredholm Multiplication Operator onℓβpn

In this section, we give the sufficient conditions on the sequence spaceℓβpn equipped with prequasi-norm τ such that the multiplication operator defined on lβPnτ has closed range, invertible, and Fredholm.Theorem 10.
Ifκ ∈ ℂℕ, ℓβpnτ is prequasi-Banach (sss), where τ(x) = ∑r=0∞βrxrpr1/H, for all x ∈ ℓβpnτ, then κ is bounded away from zero on kerκ, if and only if, R(Vκ) is closed.Proof.
Let the sufficient condition be satisfied. Therefore, there isε > 0 with κi ≥ ε, for all i ∈ kerκc, to show R(Vκ) is closed. Assume d be a limit point of R(Vκ). Therefore, there is Vκxi in ℓβpnτ, for all i ∈ ℕ such that lim i⟶∞ Vκxi = d. Obviously, (Vκxi) is a Cauchy sequence. Since pn ∈ ℝ+ℕ∩ ℓ∞ is increasing, one has(16)τVkxi−Vkxj=∑r=0∞βrκrxir−κrxjrpr1/H=∑r=0,r∈kerκc∞βrκrxir−κrxjrpr+∑r=0,r∉kerκc∞βrκrxir−κrxjrpr1/H≥∑r=0,r∈kerκc∞βrκrxir−κrxjrpr1/H=∑r=0∞βrκryir−yjrpr1H>rinfεpr/H∑r=0∞βryir−yjrpr1/H,where(17)yir=xir,r∈kerκc,0,r∉kerκc.
This shows that (yi) is a Cauchy sequence inℓβpnτ. Since ℓβpnτ is complete, there is x∈ℓβpnτ such that limi ⟶ ∞yi = x. Since Vκ is continuous, then limi ⟶ ∞ Vκyi = Vκx. However, limi ⟶ ∞ Vκxi = lim i ⟶ ∞ Vκyi = d. Hence, Vκx = d. Therefore, d ∈ R(Vκ). This shows that R(Vκ) is closed. Conversely, let R(Vκ) be closed. Therefore, Vκ be bounded away from zero on ℓβpnτkerκc.. Hence, there exists ε > 0 such that τ (Vκx) ≥ ετ (x), for all x ∈ℓβpnτkerκc.
AssumeB = {r ∈kerκc: |κr|> }. If B≠ϕ, then for i0 ∈ B, we obtain(18)τVKen0=τKren0rr=0∞=∑r=0∞βrκren0rpr1/H<∑r=0∞βrεen0rpr1/H≤supεpr/Hr∑r=0∞βren0rpr1/H=supεpr/Hrτen0,and this gives contradiction. So, B = ϕ such that |κr| ≥ ε, for all r ∈ kerκc. This completes the proof of the theorem.Theorem 11.
Ifκ ∈ ℂℕ and ℓβpnτ is a prequasi-Banach (sss), with τω =  ∑r=0∞βrωrpr1/H, for all ω ∈ ℓβpn, there are b > 0 and B > 0 such that b <|κr| < B, for all r ∈ℕ, if and only if, Vκ ∈ Bℓβpnτ is invertible.Proof.
Let the conditions be verified. Defineγ ∈ CN by γr = 1/κr. From Theorem 6, we have Vκ, Vγ ∈ Bℓβpnτ and Vκ.Vγ = Vγ. Vκ = I. Then, Vγ is the inverse of Vκ. Conversely, let Vκ be invertible. Hence, R (Vκ) = ℓβpnτℕ. This implies R (Vκ) is closed. From Theorem 10, there is b > 0 such that |κr| ≥ b, for all r ∈ (ker (κ))c. Now, ker (κ) = ϕ, else κro = 0, for several ro ∈ ℕ, we get ero ∈ ker (Vκ). This gives a contradiction, since ker (Vκ) is trivial. So, |κr| ≥ b, for all r ∈ ℕ. Since Vκ is bounded, hence from Theorem 6, there is B > 0 such that |κr| ≤ B, for all r ∈ℕ. Therefore, we have shown that b ≤ |κr| ≤ B, for all r ∈ℕ.Theorem 12.
Ifκ ∈ ℂℕ and ℓβpnτ is a prequasi-Banach (sss), where τω = ∑r=0∞βrωrpr1/H, for all ω ∈ ℓβpn, then Vκ ∈ Bℓβpnτ is Fredholm operator, if and only if, (i) card (ker(κ)) < ∞ and (ii) |κr| ≥ ε, for all r ∈ (ker(κ))c.Proof.
Let Vκ be Fredholm. If card (ker(κ)) = ∞, hence en ∈ ker (Vκ), for all n ∈ ker(κ). Since en’s are linearly independent, this gives card (ker(Vκ)) = ∞. This is a contradiction. Therefore, card (ker(κ)) < ∞. By Theorem 10, condition (ii) is satisfied. Next, if the necessary conditions are verified, to show that Vκ is Fredholm. From Theorem 10, condition (ii) gives that R(Vκ) is closed. Condition (i) indicates that dim (ker(Vκ)) < ∞ and dim ((R(Vκ))c) < ∞. Therefore, Vκ is Fredholm.


## 3.1. Prequasi-Norm onℓβpn

We investigate the sufficient conditions on the sequence spaceℓβpn equipped with prequasi-norm τ to form prequasi-Banach and closed (sss).Theorem 3.
ℓβpnτ, where τx=∑k=0∞βkxkpk1/H, for all x ∈ ℓβpn, is a premodular (sss), if the following conditions are satisfied:(a1)
The sequencePn ∈ ℝ+ℕ∩ ℓ∞ is increasing(a2)
Either (βn) is a monotonic decreasing or monotonic increasing such that there is C ≥ 1 for which β2n + 1 ≤ CβnProof.
Firstly, we have to proveℓβpn is a (sss):(1-i)
Letx, y ∈ ℓβpn. Since pn is bounded, we obtain(4)τX+Y=∑n=0∞βnxn+ynpn1/H≤∑n=0∞βnxnpn1/H+∑n=0∞βnynpn1/H=τX+τY<∞,then x + y ∈ ℓβpn.(1-ii)
Letλ ∈ ℂ and x ∈ ℓβpn. Since pn is bounded, we have(5)τλX=∑n=0∞βnλxnpn1/H≤nsupλpn/H∑n=0∞βnxnpn1/H≤aλτX<∞.(i)
Then,λ x ∈ ℓβpn Therefore, by using Parts (1-i) and (1-ii) we have that the space ℓβpn is linear. Also, en ∈ ℓβpn, for all n ∈ ℕ, since(6)τen=∑i=0∞βienipi1/H=βn1/H.(2)
Letxn≤yn, for all n ∈ ℕ and y ∈ ℓβpn. Since βn > 0, for all n ∈ ℕ, then τ(x) = ∑n=0∞βnxnpn1/H ≤ ∑n=0∞βnynpn1/H = τ(y) < ∞, we get x ∈ ℓβpn.(3)
Letxn ∈ ℓβpn and (βn) be an increasing sequence. There exists C ≥ 1 such that β2n + 1 ≤ Cβn and pn be an increasing; then, we have(7)τxn/2=∑n=0∞βnxn/2pn1/H=∑n=0∞β2nxnp2n+∑n=0∞β2n+1xnp2n+11/H≤∑n=0∞β2nxnpn+∑n=0∞β2n+1xnpn1/H≤2C1/H∑n=0∞βnxnpn1/H=2C1/Hτxn,then xn/2 ∈ ℓβpn. Secondly, we show that the functional τ on ℓβpn is a premodular:(i)
Evidently,τ (x) ≥ 0 and τ (x) = 0 ⇔ x = θ(ii)
There is a steadya = max {1, supn λpn/H−1} ≥ 1 such that τ (λx) ≤ a|λ| τ (x), for all x ∈ ℓβpnand λ ∈ ℂ(iii)
There existsK ≥ 1 such that τ (x + y) ≤ K (τ(x) + τ(y)), for all x, y ∈ ℓβpn(iv)
Clearly, from the proof part (2) of Theorem 3, the condition is clear since the weighted Nakano sequence space is solid(v)
It is obtained from (3) that b0=2C1/H ≥ 1(vi)
It is clear thatF¯ =  ℓβpn(vii)
There exists a steady 0 <ξ ≤ λp0/H−1, for λ ≠ 0 or ξ > 0, for λ = 0 such that τ (λ, 0, 0, 0, ...) ≥ ξ |λ| τ (1, 0, 0, 0, ...)Theorem 4.
If conditions (a1) and (a2) are satisfied, thenℓβpnτ, where τx=∑k=0∞βkxkpk1/H for all x ∈ ℓβpn, is a prequasi-Banach (sss).Proof.
Let the conditions be verified. By Theorem3, the space ℓβpnτ is a premodular (sss). Therefore, from Theorem 2, the space ℓβpnτ is a prequasi-normed (sss). To prove that ℓβpnτ is a prequasi-Banach (sss), suppose xn = xknk=0∞ be a Cauchy sequence in ℓβpnτ, then for every ε ∈ (0, 1), there exists a number n0∈ℕ such that, for all n, m ≥ n0, one has(8)τXn−Xm=∑j=0∞βjxjn−xjmpj1/H<ε.
Hence, forn, m ≥ n0 and j ∈ ℕ, we get xjn−xjm < ε.
So,xjm is a Cauchy sequence in ℂ for fixed j ∈ ℕ, and this gives limm⟶∞xjm=xj0 for j ∈ ℕ. Hence, τ (xn−x0) < ε, for all n ≥ n0. Finally, to prove that x0 ∈ ℓβpn, we have τ (x0) = τ (x0 − xn + xn) ≤ τ(xn–x0) + τ(xn) < ∞.
So, x0 ∈ ℓβpn. This means that ℓβpnτ is a prequasi-Banach (sss).Theorem 5.
If conditions (a1) and (a2) are satisfied, thenℓβpnτ, where τx=∑k=0∞βkxkpk1/H, for all x ∈ ℓβpn, is a prequasi-closed (sss).Proof.
Let the conditions be verified. By Theorem3, the space ℓβpnτ is a premodular (sss). Therefore, from Theorem 2, the space ℓβpnτ is a prequasi-normed (sss). To prove that ℓβpnτ is a prequasi-closed (sss), suppose Xn=xknk=0∞∈ ℓβpnτ and limn ⟶ ∞ τXn−X0=0, then for every ε ∈ (0, 1), there exists a number n0 ∈ ℕ such that for all n ≥ n0, one has(9)ε>τXn−X0=∑j=0∞βjxjn−xj0pj1/H≥βjxjn−xj0pj1/H.
Hence, forn ≥ n0 and j ∈ ℕ, we get xjn−xj0 < ε.
So,xjn is a convergent sequence in ℂ for fixed j ∈ ℕ, and this gives limm ⟶ ∞ xjm=xj0 for fixed j ∈ ℕ. Finally, to prove that x0 ∈ ℓβpn, we have τ (x0) = τ(x0 − xn + xn) ≤ τ(xn − x0) + τ(xn) < ∞.
So, x0 ∈ ℓβpn. This means that ℓβpnτ is a prequasi-closed (sss).


## 3.2. Bounded Multiplication Operator onℓβpn

Here after, we investigate some topological and geometric structure of the multiplication operator acting onℓβpn. In this section, we examine the sufficient conditions on the sequence space ℓβpn equipped with prequasi-norm τ such that the multiplication operator defined on ℓβpnτℓβpnτ is bounded and isometry.Definition 6.
Letκ ∈ ℂℕ ∩ ℓ∞ and Wτ be a prequasi-normed (sss). An operator Vκ: Wτ ⟶ Wτ is called multiplication operator if Vκω = κω  = Krωrr=0∞ ∈ W, for all ω∈W. If Vκ ∈ BW, we name it a multiplication operator generated by κ.Theorem 6.
Letκ ∈ ℂℕ and conditions (a1) and (a2) be satisfied; then, κ ∈ ℓ∞, if and only if VK∈Bℓβpnτ, where τx=∑k=0∞βkxkpk1/H, for all x ∈ ℓβpn.Proof.
Let the conditions be satisfied. Assumeκ ∈ ℓ∞. Therefore, there is ε > 0 with |κr| ≤ ε, for every r ∈ ℕ. For x ∈ ℓβpnτ, since pr is bounded from above with pr > 0, for all r ∈ ℕ, then(10)τVKX=τKX=τKrXrr=0∞=∑r=0∞βrκrxrpr1/H≤∑r=0∞βr|εxr|pr1/H≤suprεpr/H∑r=0∞βr|xr|pr1/H=Dτx.
This gives Vκ ∈ Bℓβpnτ. Conversely, assume that Vκ ∈ Bℓβpnτ. Let us assume κ ∉ ℓ∞; hence, for each j ∈ ℕ, there is ij ∈ ℕ such that κij > j. Since pn ∈ ℝ+ℕ ∩ ℓ∞ is increasing, we have(11)τVKeij=τKeij=τKreijτr=0∞=∑r=0∞βrκreijrpr1/H=βijκijpij/H>βijjpij/H=βijjpij/Hτeij.
This shows that Vκ ∈Bℓβpnτ. Therefore, κ ∈ ℓ∞.Theorem 7.
Pick upκ ∈ ℂℕ and ℓβpnτ be a prequasi-normed (sss), with τ(x) = ∑k=0∞βkxkpk1/H, for all x ∈ ℓβpn. Then, κr = 1, for all r ∈ ℕ, if and only if, Vκ is an isometry.Proof.
Suppose |κr| = 1, for all r ∈ ℕ. Hence,(12)τVKX=τKX=τKrXrr=0∞=∑r=0∞βrκrxrpr1/H=∑r=0∞βrxrpr1/H=τx,for all x ∈ ℓβpnτ. Therefore, Vκ is an isometry. Conversely, Assume that |κi| < 1 for some i = i0. Since pr ∈ ℝ+ℕ ∩ ℓ∞ is increasing, we obtain(13)τVei0=τKei0=τKrei0rr=0∞=∑r=0∞βrκrei0rpr1/H<∑r=0∞βrei0rpr1/H=τei0.
When |κi0| > 1, we can prove that τVKei0 > τei0. Therefore, in both cases, we have a contradiction. So, |κr| = 1, for every r ∈ ℕ.


## 3.3. Approximable Multiplication Operator onℓβpn

In this section, we introduce the sufficient conditions on the sequence spaceℓβpn equipped with prequasi-norm τ such that the multiplication operator defined on ℓβpnτ is an approximable and compact. By card (A), we indicate the cardinality of the set A.Theorem 8.
Ifκ ∈ ℂℕ and ℓβpnτ is a prequasi-normed (sss), where τ(x) = ∑k=0∞βkxkpk1/H, for all x ∈ ℓβpn, then, Vκ ∈ Υℓβpnτ if and only if Krr=0∞ ∈ c0.Proof.
Assume that Vκ ∈ Υℓβpnτ. Therefore, Vκ ∈ BBcℓβpnτ, to prove that the sequence Krr=0∞ belongs to c0. Suppose Krr=0∞ ∉ c0. Hence, there is δ > 0 such that the set Aδ  = {r ∈ℕ: |κr| ≥ δ} has card Aδ = ∞. Assume ai ∈Aδ, for all i ∈ℕ. Hence, {eai: ai ∈ Aδ} is an infinite bounded set in ℓβpnτ. Let(14)τVKeai−VKeaj=τKeai−Keai=τκreair−eairr=0∞=∑r=0∞βrκreair−eajrpr1/H≥∑r=0∞βrδeair−eajrpr1/H≥infrδpr/Hτeai−eaj,for all ai,aj ∈ Aδ. This shows {eai: ai ∈ Bδ} ∈ ℓ∞ which cannot have a convergent subsequence under Vκ. This proves that Vk ∉ Bcℓβpnτ. Then, Vk ∉Υℓβpnτ, this gives a contradiction. So, limi ⟶ ∞ κi = 0. Conversely, let limi ⟶∞ κi = 0. Then, for each δ > 0, the set Aδ = {i ∈ℕ: |κi| ≥ δ} has card (Aδ) < ∞. Hence, for every δ > 0, the space ℓβpnτAδ = {x = (xi) ∈ ℂAδ} is finite dimensional. Then, Vκ|ℓβpnτAδ is a finite rank operator. For every i ∈ ℕ, dene κi∈ ℂℕ by (κ i)j = κj,j∈A1/i,0,otherwise.
It is clear thatVκi has rank Vκi <∞ as dim ℓβpnτA1/i < ∞, for each i ∈ ℕ. Therefore, since pn ∈ ℝ+ℕ∩ ℓ∞ is increasing, we obtain(15)τVk−VkiXτkj−kijXjj=0∞=∑j=0∞βjκj−κijxjpj1/H∑j=0,j∈A1/i∞βjκj−κijxjpj+∑j=0,j∈A1/i∞βjκi−κijxjpj1/H=∑j=0,j∉A1/i∞βjκjxjpj1/H≤supj1jpj/H∑j=0,j∉A1/i∞βjxipjpj/H>supj1jpj/Hτx=1jp0/Hτx.
This implies thatBc−VKi≤1/jp0/H and that Vκ is a limit of finite rank operators. Therefore, Vκ is an approximable operator.Theorem 9.
Letκ ∈ℂℕ and ℓβpnτ be a prequasi-normed (sss), where τ(x) = ∑r=0∞βrxrpr1/H, for all x ∈ ℓβpnτ. Then, Vκ ∈ Bcℓβpnτ, if and only if, Kii=0∞ ∈ c0.Proof.
It is simple so overlooked.Corollary 1.
Ifκ ∈ ℂℕ, conditions (a1) and (a2) are satisfied, then Bcℓβpnτ≠⊂Bℓβpnτ, where τ(x) = ∑r=0∞βrxrpr1/H, for all x ∈ℓβpn.Proof.
Since I is a multiplication operator onlβPnτ generated by κ = (1, 1, ...), therefore, I ∉ Bcℓβpnτ and I ∈ Bℓβpnτ.


## 3.4. Fredholm Multiplication Operator onℓβpn

In this section, we give the sufficient conditions on the sequence spaceℓβpn equipped with prequasi-norm τ such that the multiplication operator defined on lβPnτ has closed range, invertible, and Fredholm.Theorem 10.
Ifκ ∈ ℂℕ, ℓβpnτ is prequasi-Banach (sss), where τ(x) = ∑r=0∞βrxrpr1/H, for all x ∈ ℓβpnτ, then κ is bounded away from zero on kerκ, if and only if, R(Vκ) is closed.Proof.
Let the sufficient condition be satisfied. Therefore, there isε > 0 with κi ≥ ε, for all i ∈ kerκc, to show R(Vκ) is closed. Assume d be a limit point of R(Vκ). Therefore, there is Vκxi in ℓβpnτ, for all i ∈ ℕ such that lim i⟶∞ Vκxi = d. Obviously, (Vκxi) is a Cauchy sequence. Since pn ∈ ℝ+ℕ∩ ℓ∞ is increasing, one has(16)τVkxi−Vkxj=∑r=0∞βrκrxir−κrxjrpr1/H=∑r=0,r∈kerκc∞βrκrxir−κrxjrpr+∑r=0,r∉kerκc∞βrκrxir−κrxjrpr1/H≥∑r=0,r∈kerκc∞βrκrxir−κrxjrpr1/H=∑r=0∞βrκryir−yjrpr1H>rinfεpr/H∑r=0∞βryir−yjrpr1/H,where(17)yir=xir,r∈kerκc,0,r∉kerκc.
This shows that (yi) is a Cauchy sequence inℓβpnτ. Since ℓβpnτ is complete, there is x∈ℓβpnτ such that limi ⟶ ∞yi = x. Since Vκ is continuous, then limi ⟶ ∞ Vκyi = Vκx. However, limi ⟶ ∞ Vκxi = lim i ⟶ ∞ Vκyi = d. Hence, Vκx = d. Therefore, d ∈ R(Vκ). This shows that R(Vκ) is closed. Conversely, let R(Vκ) be closed. Therefore, Vκ be bounded away from zero on ℓβpnτkerκc.. Hence, there exists ε > 0 such that τ (Vκx) ≥ ετ (x), for all x ∈ℓβpnτkerκc.
AssumeB = {r ∈kerκc: |κr|> }. If B≠ϕ, then for i0 ∈ B, we obtain(18)τVKen0=τKren0rr=0∞=∑r=0∞βrκren0rpr1/H<∑r=0∞βrεen0rpr1/H≤supεpr/Hr∑r=0∞βren0rpr1/H=supεpr/Hrτen0,and this gives contradiction. So, B = ϕ such that |κr| ≥ ε, for all r ∈ kerκc. This completes the proof of the theorem.Theorem 11.
Ifκ ∈ ℂℕ and ℓβpnτ is a prequasi-Banach (sss), with τω =  ∑r=0∞βrωrpr1/H, for all ω ∈ ℓβpn, there are b > 0 and B > 0 such that b <|κr| < B, for all r ∈ℕ, if and only if, Vκ ∈ Bℓβpnτ is invertible.Proof.
Let the conditions be verified. Defineγ ∈ CN by γr = 1/κr. From Theorem 6, we have Vκ, Vγ ∈ Bℓβpnτ and Vκ.Vγ = Vγ. Vκ = I. Then, Vγ is the inverse of Vκ. Conversely, let Vκ be invertible. Hence, R (Vκ) = ℓβpnτℕ. This implies R (Vκ) is closed. From Theorem 10, there is b > 0 such that |κr| ≥ b, for all r ∈ (ker (κ))c. Now, ker (κ) = ϕ, else κro = 0, for several ro ∈ ℕ, we get ero ∈ ker (Vκ). This gives a contradiction, since ker (Vκ) is trivial. So, |κr| ≥ b, for all r ∈ ℕ. Since Vκ is bounded, hence from Theorem 6, there is B > 0 such that |κr| ≤ B, for all r ∈ℕ. Therefore, we have shown that b ≤ |κr| ≤ B, for all r ∈ℕ.Theorem 12.
Ifκ ∈ ℂℕ and ℓβpnτ is a prequasi-Banach (sss), where τω = ∑r=0∞βrωrpr1/H, for all ω ∈ ℓβpn, then Vκ ∈ Bℓβpnτ is Fredholm operator, if and only if, (i) card (ker(κ)) < ∞ and (ii) |κr| ≥ ε, for all r ∈ (ker(κ))c.Proof.
Let Vκ be Fredholm. If card (ker(κ)) = ∞, hence en ∈ ker (Vκ), for all n ∈ ker(κ). Since en’s are linearly independent, this gives card (ker(Vκ)) = ∞. This is a contradiction. Therefore, card (ker(κ)) < ∞. By Theorem 10, condition (ii) is satisfied. Next, if the necessary conditions are verified, to show that Vκ is Fredholm. From Theorem 10, condition (ii) gives that R(Vκ) is closed. Condition (i) indicates that dim (ker(Vκ)) < ∞ and dim ((R(Vκ))c) < ∞. Therefore, Vκ is Fredholm.



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