#### Abstract

We investigate some new topological properties of the multiplication operator on defined by Bilgin (The Punjab University Journal of Mathematics, vol. 30, pp. 67–77, 1997) equipped with the prequasi norm and the prequasi operator ideal formed by this sequence space and -numbers.

#### 1. Introduction

Summability, multiplication, and ideal operator theorems are very important in mathematical models and have numerous implementations, such as normal series theory, ideal transformations, geometry of Banach spaces, approximation theory, and fixed point theory. For more details, see [1–11]. By , , , and , we denote the spaces of every, bounded, r-absolutely summable, and convergent to zero sequences of complex numbers. indicates the set of nonnegative integers. For a sequence with , Bilgin [12] defined and studied the sequence space where .

The space is a Banach space with the Luxemburg norm

If , we have

Here and after, we denote , if , then , where was defined and investigated by Lim [13]. If and , for all , then , Lim [14] defined and determined its dual spaces and characterized some matrix classes. For any pair of Banach spaces and , the following notations will be used throughout the article: (the space of all bounded linear operators from into ), and if, we write , [15] (the-number sequence of operator ), (the approximation number sequence of operator ), (the space of all finite rank operators on ), (the space of all compact operators on ), and (the space of all approximable operators on ). For any sequence space , we will use the notations [16]

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 and . Pietsch [17] 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 examined that for , where is the closed class of all finite rank operators, and the class is a simple Banach and small [18]. The strict inclusion , whenever , and are infinite dimensional Banach spaces investigated through Makarov and Faried [19]. Faried and Bakery [16] investigated a generalization of the class of operator ideal which is the prequasi operator ideal; they studied several geometric and topological structures of and . On sequence spaces, Mursaleen and Noman [20, 21] investigated the compact operators on some difference sequence spaces. The multiplication operators on with the Luxemburg norm investigated by Komal et al. [22]. The point of this article is to explain some results of equipped with a prequasi norm . Firstly, we give the conditions on to be a Banach space. Secondly, we investigate the multiplication operator defined on . Finally, some geometric and topological structures of have been studied, such as closed, small, simple Banach, and . A strict inclusion relation of has been studied for different and .

#### 2. Definitions and Preliminaries

We will use , where 1 shows at the place, for all .

Lemma 1 [17]. *If and , then there are and such that , for all .*

*Definition 2 [17]. *A Banach space is called simple if the algebra has one and only one nontrivial closed ideal.

Theorem 3 [17]. *If is a Banach space with , then
*

*Definition 4 [23]. *An operator is called Fredholm if , , and , where is the complement of range .

*Definition 5 [24]. *A class of linear sequence spaces is called a special space of sequences (sss) if
(1), for every (2)if , , and , for all , then , i.e., “ is solid”(3)if , then , where denotes the integral part of

*Definition 6 [24]. *A subclass is called a premodular (sss) if there is a function verifying the conditions
(i) for each and , where is the zero vector of (ii)there is such that , for each and (iii)there is , , for all (iv)if , for every , then (v)there is , (vi)for all , and there is such that (vii)there is such that , for all

*Definition 7 [24]. *Let be a (sss). If there is a function verifying the parts (i), (ii), and (iii) of Definition 6. The space with is called prequasi normed (sss) and denoted by , which gives a class more general than the quasi normed (sss). If the space is complete with , then , is called a prequasi Banach (sss).

Theorem 8 [24]. *Every quasinorm (sss) is prequasi norm (sss).*

Theorem 9 [24]. *Every premodular (sss) is prequasi normed (sss).*

A generalization of the usual classes of operator ideal is the class prequasi operator ideal.

*Definition 10 [24]. *A function is called a prequasi norm on the class if it satisfies
(1)if , then and (2)there is such that , for every and (3)there is such that , for every (4)there is such that if , , and , then , where and are normed spaces

Theorem 11 [25]. *If is a premodular (sss), then , where .*

Theorem 12 [16]. *If is a premodular (sss), then is a prequasi norm on .*

The following inequality [26] will be used in the sequel: , where , , and for each .

#### 3. Main Results

##### 3.1. Premodular Banach (sss)

The conditions on and are such that the space is a premodular Banach (sss), where for all .

Theorem 13. *Let be an increasing with and , then the space is a premodular Banach (sss).*

*Proof. *(1-i) Assume . From , we get
hence .

(1-ii) If , , and from , one has
Therefore, , from (1-i) and (1-ii), the space is linear. Since with , we obtain
where be such that . Hence , for all .

(2) Let , for all and . Hence,
hence, .

(3) Let , we have
then .
(i)Evidently, and (ii)There is with , for each and (iii)From condition (1), there is with for each (iv)From condition (2), we have which is solid(v)From condition (3), we find (vi)Clearly, (vii)There is with , for each and , when Therefore, the space is premodular (sss). To prove that is a premodular Banach (sss), suppose is a Cauchy sequence in ; hence, for every , there is such that for every , one has
Hence, for and , we get . So is a Cauchy sequence in for fixed ; this gives for fixed . Hence, , for all . Finally, to prove that , we have
so . This means that is a premodular Banach (sss).

Corollary 14. *If , then is a premodular Banach (sss), where for every .*

#### 4. Some Properties of Multiplication Operators on

We investigate the necessity and sufficient conditions on and such that the multiplication operator defined on the prequasi normed (sss), , becomes an element of , , , closed range, and Fredholm operator.

*Definition 15. *Pick up , and is a prequasi normed (sss). An operator is called a multiplication operator, if , for all . It is called a multiplication operator generated by , if is a multiplication operator.

Theorem 16. *If , is increasing with and , then
*

*Proof. *Let . Therefore, there is such that , for every . Since and for , we have
this implies that Conversely, let , to show that . For, if , then for all , there are such that . Hence,
where be such that . This shows that . So, .

Theorem 17. *Assume and is a prequasi normed (sss). , for every if and only if is an isometry.*

*Proof. *If , for every . Therefore, we have
for all . Hence, is an isometry. Conversely, let be an isometry and , for all . Then,
Also if , then we get . We have a contradiction in the two cases. Therefore, , for every .

By , we indicate the cardinality of the set .

Theorem 18. *For , is increasing with , , and . Then,
*

*Proof. *Let ; hence . To show that . Assume . Therefore, the set has , for . Suppose , for all . Therefore, with . We have
for all . This proves is not a convergent subsequence. This shows that ; hence, . This gives a contradiction. So, . Conversely, suppose . Hence, for all , we have . Therefore, for all the space
is a finite dimensional. Hence, . Let and define as
Obviously, since , for all . Since is increasing with , we have
This proves that and . Therefore, .

According to Theorem 3, it is easy to prove the following theorem.

Theorem 19. *For , is increasing with , , and . Then,
*

Corollary 20. *If is increasing with and , we have
*

*Proof. *The operator on is a multiplication operator generated by . Therefore, and .

Theorem 21. *Suppose is a prequasi Banach (sss) and . Then, there is such that , for all , if and only if, has a closed range.*

*Proof. *Let the sufficient condition be verified, to show that . If is a limit point of , then there is , for each with . Clearly, the sequence is a Cauchy sequence. Since is nondecreasing, one can see
where
This proves that is a Cauchy sequence in . But is complete. Therefore, there exists such that . In view of the continuity of , hence . But . Therefore, . Hence . Therefore, . Let the necessary condition be satisfied. Therefore, there is such that , for all . Let . If , then for , since is nondecreasing, we have
which gives a contradiction. So, with , for every .

Theorem 22. *Pick up , and is a prequasi Banach (sss). There is and with , for every , if and only if, is invertible.*

*Proof. *Let the condition be true. Define by . Then, by Theorem 16, and are bounded linear operators. Also . Hence, is the inverse of . Conversely, let be invertible. Hence, . Therefore, . Hence, by Theorem 21, there is with , for every . Now, ; otherwise, ; for some , we have ; this implies a contradiction, since . So, , for every . Since , from Theorem 16, there is with , for every . Therefore, we get that , for each .

Theorem 23. *For a prequasi Banach (sss) . The operator is Fredholm, if and only if, (i) and (ii) , for each .*

*Proof. *Let the sufficient condition be true and , so , for every . We have , since ’s are linearly independent. This implies a contradiction. Hence, . From Theorem 21, the condition (ii) follows. Next, if the necessary conditions are true, from Theorem 21, the condition (ii) gives that . By condition (i), we have and . This implies that is Fredholm.

#### 5. Banach and Closed Prequasi Ideal

We explain the sufficient conditions on and such that is Banach and a closed prequasi ideal, where , for all and .

Theorem 24. *Let , be Banach spaces, be increasing with and , then is a prequasi Banach ideal.*

*Proof. *Let the sufficient conditions be satisfied. From Theorem 13, the space is a premodular (sss). Therefore, by Theorem 12, the function is a prequasi norm on . Let be a Cauchy sequence in , since ; we get
Hence, is a Cauchy sequence in the Banach space . Therefore, there is such that , since for every . By using parts (ii), (iii), and (iv) of Definition (6), one can see
Therefore, ; hence, .

Theorem 25. *Let , be normed spaces and be increasing with and , then is a prequasi closed ideal.*

*Proof. *Let the sufficient conditions be satisfied. From Theorem 13, the space is a premodular (sss). Therefore, by Theorem 12, the function is a prequasi norm on . Assume for every and , since , we have
Hence, is a convergent sequence in . So, for each . By using parts (ii), (iii), and (iv) of Definition (6), we have