#### 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).

#### 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 , *c*, , , and *c*_{o}, we signify the spaces of each, convergent, bounded, r-absolutely summable, and convergent to zero sequences of complex numbers. indicates the set of nonnegative integers. Let *=* and be sequences of positive reals, and the weighted Nakano sequence space [5] is defined bywhere . And is a Banach space, where

When , we have(1)If , for all *n ∈**,* then will reduce to (see [6, 7])(2)If , for all *n ∈*, then will reduce to which is studied by many authors (see [8–10])

By , 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 . 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 *α*_{r}(*V*)*,* is defined by *α*_{r}(*V*) = inf {*: B ∈* and rank(B) ≤ *r*}, and the rth Kolmogorov number, denoted by *d*_{r}(*V*), is defined by *d*_{r}(*V*) = inf_{dim}w _{≤} _{r} sup _{≤1} inf _{∈W}. The following notations will be used in the sequel:

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 operators formed by (*d*_{r} (*V*)) and *c*_{0}*.* Pietsch [4] studied the quasi-ideals for , the ideals of Hilbert Schmidt operators between Hilbert spaces constructed by , and the ideals of nuclear operators generated by . He explained that for *r ∈* [1, ∞), where is the closed class of all finite rank operators, and the class became simple Banach and small [1]. The strictly inclusion *,* 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 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 such that constructed a simple Banach prequi operator ideal. The prequasi-operator ideal became small and strictly contained for different weights and powers. The aim of this article to explain some results of equipped with the prequasi-norm . 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 operator is called approximable if there are Dr ∈ , for every *r* ∈ and limr⟶∞ ||V—Dr|| = 0.

By and , 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) = ∞, then .*

*Definition 2 (see [22]). *An operator V∈ 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 sequence *e*_{j} = (0, 0, 1, 0, 0, ...) with 1 in the jth coordinate, for every *j* ∈, will be used in the sequel.

*Definition 3 (see [3]). *The space of linear sequence spaces Y is called (sss)(1)If *e*_{r} ∈ Y with *r* ∈ ,(2)Let *u* = (*u*_{r}) ∈ , = ()∈ Y, and |*u*_{r}|*≤*||, for every *r* ∈, then *u* ∈ Y. This means Y be solid,”(3)If (*u*_{r}) , then ∈Y, wherever means the integral part of

*Definition 4 (see [23]). *A subspace of the (sss) is called a premodular (sss) if there is a function: ⟶ [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 exists *a* ≥ 1 such that *τ*(*ηy*) *≤* *a|η| τ*(*y*), for all *y* *∈* Y, and *η ∈*(iii)For some *b* ≥ 1, *τ*(*y* + *z*) ≤ *b*(*τ*(y) + *τ*(z)), for every *y*, z ∈ Y(iv)|yr| ≤ |zr| with *r* ∈ , implies *τ*((yr)) ≤ *τ*((zr))(v)For some *b*0 ≥ 1, *τ* ((yr)) ≤ *τ*() ≤ *b*0*τ* ((yr))(vi)If *y* = ∈ Y, and *d* > 0, then there is r0 ∈ with *τ*() < *d*(vii)There is *t* > 0 with *τ* (*ν*, 0, 0, 0, ...) ≥ *t*|*ν*| *τ* (1, 0, 0, 0, ...), for any *?* ∈

*Definition 5 (see [23]). *The (sss) 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 is called a prequasi-Banach (sss).

Theorem 2 (see [23]). *A prequasi-norm (sss) , whenever it is premodular (sss).*

The inequality [24]: where ≥ 0, for all *i* ∈ and , will be used in the sequel.

#### 3. Main Results

##### 3.1. Prequasi-Norm on

We investigate the sufficient conditions on the sequence space equipped with prequasi-norm to form prequasi-Banach and closed (sss).

Theorem 3. *, where , for all x ∈ , is a premodular (sss), if the following conditions are satisfied:*(a1)

*The sequence ∈ ∩ is increasing*(a2)

*Either (*

*β*_{n}) is a monotonic decreasing or monotonic increasing such that there is*C*≥ 1 for which*β*2_{n}+ 1 ≤ C*β*_{n}*Proof. *Firstly, we have to prove is a (sss):(1-i)Let *x*, y ∈ . Since is bounded, we obtainthen *x* + *y* ∈ .(1-ii)Let *λ* ∈ and *x* ∈ . Since is bounded, we have(i)Then, *λ* x ∈ Therefore, by using Parts (1-i) and (1-ii) we have that the space is linear. Also, en ∈ , for all *n* ∈ , since(2)Let , for all *n* ∈ and *y* ∈ . Since *β*_{n} > 0, for all *n* ∈ , then *τ*(*x*) = ≤ = *τ*(y) < ∞, we get *x* ∈ .(3)Let ∈ and (*β*_{n}) be an increasing sequence. There exists *C* ≥ 1 such that *β*2_{n} + 1 ≤ C*β*_{n} and be an increasing; then, we havethen ∈ . Secondly, we show that the functional *τ* on is a premodular:(i)Evidently, *τ* (*x*) ≥ 0 and *τ* (*x*) = 0 ⇔ *x* = *θ*(ii)There is a steady *a* = max {1, supn } ≥ 1 such that *τ* (*λ*x) ≤ *a*|*λ*| *τ* (*x*), for all *x* ∈ and *λ* ∈ (iii)There exists *K* ≥ 1 such that *τ* (*x* + *y*) ≤ K (*τ*(*x*) + *τ*(y)), for all *x*, y ∈ (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 ≥ 1(vi)It is clear that = (vii)There exists a steady 0 < ≤ , for *λ* 0 or > 0, for *λ* = 0 such that *τ* (*λ*, 0, 0, 0, ...) |*λ*| *τ* (1, 0, 0, 0, ...)

Theorem 4. *If conditions (a1) and (a2) are satisfied, then , where for all x ∈ , is a prequasi-Banach (sss).*

*Proof. *Let the conditions be verified. By Theorem 3, the space is a premodular (sss). Therefore, from Theorem 2, the space is a prequasi-normed (sss). To prove that is a prequasi-Banach (sss), suppose x_{n} = be a Cauchy sequence in , then for every *ε* ∈ (0, 1), there exists a number ∈ such that, for all *n*, *m* ≥ , one hasHence, for *n*, *m* ≥ and *j* ∈ , we get < *ε*.

So, is a Cauchy sequence in for fixed *j* ∈ , and this gives for *j* ∈ . Hence, *τ* (x_{n}−x_{0}) < *ε*, for all *n* ≥ . Finally, to prove that x_{0} ∈ , we have *τ* (x_{0}) = *τ* (x_{0} − x_{n} + x_{n}) ≤ *τ*(x_{n}–x_{0}) + *τ*(x_{n}) < ∞.

So, x_{0} ∈ . This means that is a prequasi-Banach (sss).

Theorem 5. *If conditions (a1) and (a2) are satisfied, then , where , for all x ∈ , is a prequasi-closed (sss).*

*Proof. *Let the conditions be verified. By Theorem 3, the space is a premodular (sss). Therefore, from Theorem 2, the space is a prequasi-normed (sss). To prove that is a prequasi-closed (sss), suppose ∈ and limn ⟶ ∞ , then for every *ε* ∈ (0, 1), there exists a number ∈ such that for all *n* ≥ , one hasHence, for *n* ≥ and *j* ∈ , we get < *ε*.

So, is a convergent sequence in for fixed *j* ∈ , and this gives limm ⟶ ∞ for fixed *j* ∈ . Finally, to prove that x_{0} ∈ , we have *τ* (x_{0}) = *τ*(x_{0} − x_{n} + x_{n}) ≤ *τ*(x_{n} − x_{0}) + *τ*(x_{n}) < ∞.

So, x_{0} ∈ . This means that is a prequasi-closed (sss).

##### 3.2. Bounded Multiplication Operator on

Here after, we investigate some topological and geometric structure of the multiplication operator acting on . In this section, we examine the sufficient conditions on the sequence space equipped with prequasi-norm *τ* such that the multiplication operator defined on is bounded and isometry.

*Definition 6. *Let *κ* ∈ ∩ ℓ∞ and W_{τ} be a prequasi-normed (sss). An operator V_{κ}: W_{τ} ⟶ W_{τ} is called multiplication operator if V_{κ} = *κ* = ∈ W, for all ∈W. If V*κ* ∈ , we name it a multiplication operator generated by *κ*.

Theorem 6. *Let κ ∈ and conditions (a1) and (a2) be satisfied; then, κ ∈ ℓ∞, if and only if , where , for all x ∈ .*

*Proof. *Let the conditions be satisfied. Assume *κ* ∈ ℓ∞. Therefore, there is *ε* > 0 with || ≤ *ε*, for every *r* ∈ . For *x* ∈ , since is bounded from above with > 0, for all *r* ∈ , thenThis gives V_{κ} ∈ . Conversely, assume that V_{κ} ∈ . Let us assume *κ* ∉ ℓ∞; hence, for each *j* ∈ , there is ∈ such that > *j*. Since ∈ ∩ ℓ∞ is increasing, we haveThis shows that V_{κ} ∈. Therefore, *κ* ∈ ℓ∞.

Theorem 7. *Pick up κ ∈ and be a prequasi-normed (sss), with τ(x) = , for all x ∈ . Then, = 1, for all r ∈ , if and only if, Vκ is an isometry.*

*Proof. *Suppose |*κ*_{r}| = 1, for all *r* ∈ . Hence,for all *x* ∈ . Therefore, V_{κ} is an isometry. Conversely, Assume that |*κ*i| < 1 for some *i* = *i*0. Since ∈ ∩ ℓ∞ is increasing, we obtainWhen || > 1, we can prove that > . Therefore, in both cases, we have a contradiction. So, |*κ*_{r}| = 1, for every *r* ∈ .

##### 3.3. Approximable Multiplication Operator on

In this section, we introduce the sufficient conditions on the sequence space equipped with prequasi-norm *τ* such that the multiplication operator defined on is an approximable and compact. By card (A), we indicate the cardinality of the set A.

Theorem 8. *If κ ∈ and is a prequasi-normed (sss), where τ(x) = , for all x ∈ , then, V_{κ} ∈ if and only if ∈ c_{0}.*

*Proof. *Assume that V_{κ} ∈ . Therefore, V_{κ} ∈ , to prove that the sequence belongs to c_{0}. Suppose ∉ c_{0}. Hence, there is *δ* > 0 such that the set = {r ∈: |*κ*_{r}| ≥ *δ*} has card = ∞. Assume ai ∈, for all *i* ∈. Hence, {: ai ∈ A*δ*} is an infinite bounded set in . Letfor all ∈ A*δ*. This shows {: ∈ B*δ*} ∈ ℓ∞ which cannot have a convergent subsequence under V_{κ}. This proves that V_{k} ∉ . Then, V_{k} ∉, this gives a contradiction. So, limi ⟶ ∞ *κ*_{i} = 0. Conversely, let limi ⟶∞ *κ*_{i} = 0. Then, for each *δ* > 0, the set = {i ∈: |*κ*_{i}| ≥ *δ*} has card (A*δ*) < ∞. Hence, for every *δ* > 0, the space = {*x* = (xi) ∈ } is finite dimensional. Then, V*κ*| is a finite rank operator. For every *i* ∈ , dene *κ*i∈ by (*κ i*)*j* =

It is clear that has rank <∞ as dim < ∞, for each *i* ∈ . Therefore, since ∈ ∩ ℓ∞ is increasing, we obtainThis implies that and that V_{κ} is a limit of finite rank operators. Therefore, V_{κ} is an approximable operator.

Theorem 9. *Let κ ∈ and be a prequasi-normed (sss), where τ(x) = , for all x ∈ . Then, V_{κ} ∈ , if and only if, ∈ c_{0}.*

*Proof. *It is simple so overlooked.

Corollary 1. *If κ ∈ , conditions (a1) and (a2) are satisfied, then , where τ(x) = , for all x ∈.*

*Proof. *Since I is a multiplication operator on generated by *κ* = (1, 1, ...), therefore, I ∉ and I ∈ .

##### 3.4. Fredholm Multiplication Operator on

In this section, we give the sufficient conditions on the sequence space equipped with prequasi-norm such that the multiplication operator defined on has closed range, invertible, and Fredholm.

Theorem 10. *If κ ∈ , is prequasi-Banach (sss), where τ(x) = , for all x ∈ , then κ is bounded away from zero on , if and only if, R(Vκ) is closed.*

*Proof. *Let the sufficient condition be satisfied. Therefore, there is > 0 with ≥ , for all *i* ∈ *c*, to show R(V*κ*) is closed. Assume *d* be a limit point of R(V*κ*). Therefore, there is V_{κ}xi in , for all *i* ∈ such that lim *i*⟶∞ V*κ*xi = *d*. Obviously, (V_{κ}xi) is a Cauchy sequence. Since ∈ ∩ ℓ∞ is increasing, one haswhereThis shows that (yi) is a Cauchy sequence in . Since is complete, there is *x* such that limi ⟶ ∞yi = *x*. Since V_{κ} is continuous, then limi ⟶ ∞ V_{κ}yi = V_{κ}x. However, limi ⟶ ∞ V_{κ}xi = lim *i* ⟶ ∞ V_{κ}yi = *d*. Hence, V_{κ}x = *d*. Therefore, *d* ∈ R(V*κ*). This shows that R(V*κ*) is closed. Conversely, let R(V*κ*) be closed. Therefore, V_{κ} be bounded away from zero on . Hence, there exists > 0 such that (*V*_{κ}*x*) ≥ (*x*), for all *x* ∈.

Assume *B* = {r ∈: |*κ*_{r}|> }. If B*ϕ*, then for ∈ B, we obtainand this gives contradiction. So, *B* = *ϕ* such that |*κ*_{r}| ≥ , for all *r* ∈ . This completes the proof of the theorem.

Theorem 11. *If κ ∈ and is a prequasi-Banach (sss), with τ = , for all ∈ , there are b > 0 and B > 0 such that b <|κ_{r}| < B, for all r ∈, if and only if, V_{κ} ∈ is invertible.*

*Proof. *Let the conditions be verified. Define *γ* ∈ CN by *γ*r = . From Theorem 6, we have V_{κ}, V_{γ} ∈ and V_{κ}.V_{γ} = V_{γ}. V_{κ} = I. Then, V_{γ} is the inverse of V_{κ}. Conversely, let V_{κ} be invertible. Hence, *R* (V_{κ}) = . This implies *R* (V_{κ}) is closed. From Theorem 10, there is *b* > 0 such that |*κ*_{r}| ≥ *b*, for all *r* ∈ (ker (*κ*))c. Now, ker (*κ*) = *ϕ*, else = 0, for several r_{o} ∈ , we get ∈ ker (V_{κ}). This gives a contradiction, since ker (V_{κ}) is trivial. So, |*κ*_{r}| ≥ *b*, for all *r* ∈ . Since V*κ* is bounded, hence from Theorem 6, there is *B* > 0 such that |*κ*_{r}| ≤ B, for all *r* ∈. Therefore, we have shown that *b* ≤ |*κ*_{r}| ≤ B, for all *r* ∈.

Theorem 12. *If κ ∈ and is a prequasi-Banach (sss), where τ = , for all ∈ , then Vκ ∈ is Fredholm operator, if and only if, (i) card (ker(κ)) < ∞ and (ii) |κ_{r}| ≥ , for all r ∈ (ker(κ))c.*

*Proof. *Let V_{κ} be Fredholm. If card (ker(*κ*)) = ∞, hence en ∈ ker (V*κ*), for all *n* ∈ ker(*κ*). Since en’s are linearly independent, this gives card (ker(V*κ*)) = ∞. 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 V_{κ} is Fredholm. From Theorem 10, condition (ii) gives that R(V_{κ}) is closed. Condition (i) indicates that dim (ker(V_{κ})) < ∞ and dim ((R(V_{κ}))c) < ∞. Therefore, V_{κ} is Fredholm.

#### Data Availability

No data were used to support this study.

#### Conflicts of Interest

The authors declare that they have no competing interests.

#### Authors’ Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

#### Acknowledgments

This work was funded by the University of Jeddah, Saudi Arabia, under Grant no. UJ-02-054-DR. The authors, therefore, acknowledge with thanks the University technical and financial support.