#### Abstract

We give sufficient conditions on a special space of sequences defined by Mohamed and Bakery (2013) such that the finite rank operators are dense in the complete space of operators whose approximation numbers belong to this sequence space. Hence, under a few conditions, every compact operator would be approximated by finite rank operators. We apply it on the sequence space defined by Tripathy and Mahanta (2003). Our results match those known for -absolutely summable sequences of reals.

#### 1. Introduction and Basic Definitions

By and , we will denote the spaces of all real sequences and all bounded linear operators between two Banach spaces into , respectively. In [1], Pietsch, by using the approximation numbers and* p*-absolutely summable sequences of real numbers, formed the operator ideals. In [2], Mohamed and Bakery have considered the space , when , which matches especially . A subclass of is an operator ideal if its components verify the following conditions:(i)The space of all finite rank operators is a subset of .(ii)The space is linear.(iii)For two Banach spaces and , if , , and , then . See [3, 4].

An Orlicz function is a function which is convex, positive, nondecreasing, and continuous, where and . An Orlicz function is said to satisfy -condition for all values of if there exists a constant , such that . Lindenstrauss and Tzafriri [5] used the idea of an Orlicz function to define Orlicz sequence spaces as follows: is a Banach space, where The space is an Orlicz sequence space with for .

*Remark 1. *For any Orlicz function , we have , for all with .

Let be the class of all subsets of that do not contain more than number of elements and let be a nondecreasing sequence of positive reals such that , for all . Tripathy and Mahanta [6] defined and studied the following sequence space: with the norm

Lemma 2. *(i) .**(ii) if and only if .*

As of late, different classes of sequences have been presented using Orlicz functions by Braha [7], Raj and Sharma Sunil [8], Raj et al. [9], and many others ([10–13]).

*Definition 3 (see [14, 15]). *A special space of sequences (sss) is a linear space with the following: (1) for all , where with 1 appearing at th place for all .(2) is .(3), if . A premodular (sss) is a (sss) and there is a function with the following: (i), for each and , where is the zero element of .(ii) satisfies -condition.(iii)For each , holds for some .(iv)The space is -solid; that is, , whenever , for all .(v)For some numbers , the inequality holds.(vi); that is, the set of all finite sequences is -dense in .(vii)For each , there is a constant such that .

Condition (ii) says that is continuous at . The function defines a metrizable topology in and the linear space enriched with this topology is denoted by .

*Definition 4 (see [16]). *Consider the following:where

Theorem 5 (see [2]). *If is a (sss), then is an operator ideal.*

We explain some results related to the operator spaces.

#### 2. Main Results

In this part, we give sufficient conditions on such that the finite rank operators are dense in the complete space of operators .

Lemma 6. *If is a premodular (sss) and is a decreasing sequence of positive reals, then *

*Proof. *By using Definition 3, conditions (iv) and (v), and since the elements of are decreasing, we get

Theorem 7. *Let be a premodular (sss); then , where .*

*Proof. *To prove that , since for each , from the linearity of and , then finitely many elements of are different from zero. Hence, . For the other inclusion , let and, from the definition of approximation numbers, there is , , with and also Since as , then as ; we have to prove that as , by taking , where is a natural number. From Definition 3, condition (iii), we have Since for , then By using Lemma 6, inequality (10), and Definition 3, condition (v), we get Now, using Lemma 6 and Definition 3, condition (v), we have Finally, we have to show that as . Since and is continuous at , we have as . Then, for each , there exists such that for all we have By taking for each and using inequality (13) and Definition 3, conditions (ii) and (iii), then we have where , and since , then for each there exists such that , for that we have This completes the proof.

We give here the sufficient conditions on the sequence spaces such that the class of all bounded linear operators between any arbitrary Banach spaces with ( in these sequence spaces form an ideal operator; the ideal of the finite rank operators in the class of Banach spaces is dense in .

Theorem 8. *Let be an Orlicz function satisfying -condition. Then,*(a)* is an operator ideal,*(b)*.*

*Proof. *We first prove that the space is a (sss).

(1) Let and ; then there exist and such that Let . Since is nondecreasing convex function with -condition, we have So, we get Thus, . Hence, is a linear space over the field of real numbers. Also, since and , we have for all .

(2) Let and with for each ; since is nondecreasing, then we get then .

(3) Let ; then we have then .

Finally, we have proved that the space with is a premodular (sss).

(i) Clearly, for all and .

(ii) Let and ; then for we have where . Thus,

Also, for , we have

(iii) Let ; then there exist and such that Let , and since is nondecreasing and convex, then we have Since ’s are nonnegative, we have (iv) Let for each , and since is nondecreasing, then we get thus, So, .

(v) Since we have So, .

(vi) For each , then we can find such that . This means the set of all finite sequences is -dense in .

(vii) For any , there exists a constant such that By using Theorems 5 and 7, the proof follows.

As special cases of the above theorem, we obtain the following corollaries.

Corollary 9. *If , one gets that *(a)* is an operator ideal,*(b)*.*

Corollary 10. *If and with , one gets that *(a)* is an operator ideal,*(b)*. See [1].*

Theorem 11. *If is a premodular (sss), then is complete.*

*Proof. *Let be a Cauchy sequence in ; then, by using Definition 3, condition (vii), and since , we get then is also a Cauchy sequence in . Since the space is a Banach space, then there exists such that , as , and since , for each , then from Definition 3, conditions (iii) and (v), and since is continuous at , we have we get , and then . This finishes the proof.

By applying Theorem 11 on , we can easily conclude the next corollaries.

Corollary 12. *Pick up an Orlicz function which satisfies -condition. Then, is continuous from the right at and is complete.*

Corollary 13. *Pick up an Orlicz function which satisfies -condition with . Then, is continuous from the right at and is complete.*

Corollary 14. * is complete if and .*

#### Competing Interests

The author declares that he has no competing interests.