# On New Banach Sequence Spaces Involving Leonardo Numbers and the Associated Mapping Ideal

Authors: Taja Yaying; Bipan Hazarika; O. M. Kalthum S. K. Mohamed; Awad A. Bakery
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In the present study, we have constructed new Banach sequence spacesℓpL,c0L,cL, and ℓ∞L, where L=lv,k is a regular matrix defined by lv,k=lk/lv+2−v+2,0≤k≤v,0,k>v, for all v,k=0,1,2,⋯, where l=lk is a sequence of Leonardo numbers. We study their topological and inclusion relations and construct Schauder bases of the sequence spaces ℓpL,c0L, and cL. Besides, α-, β- and γ-duals of the aforementioned spaces are computed. We state and prove results of the characterization of the matrix classes between the sequence spaces ℓpL,c0L,cL, and ℓ∞L to any one of the spaces ℓ1,c0,c, and ℓ∞. Finally, under a definite functional ρ and a weighted sequence of positive reals r, we introduce new sequence spaces c0L,rρ and ℓpL,rρ. We present some geometric and topological properties of these spaces, as well as the eigenvalue distribution of ideal mappings generated by these spaces and s-numbers.


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## Body

## 1. Introduction and Preliminaries

Letω denote the set of all real- or complex-valued sequences. A linear subspace of ω is called a sequence space. Some of the well-known examples of sequence spaces are the space of absolutely p-summable sequences, the space of null sequences, the space of convergent sequences, and the space of bounded sequences, denoted by ℓp,c0,c, and ℓ∞, respectively. Here and afterwards, 1≤p<∞, unless stated otherwise. Let bs and cs denote the spaces of all bounded and convergent series, respectively. A Banach sequence space with continuous coordinates is called a BK-space. The spaces Z and ℓp are BK-spaces equipped with the supremum norm zℓ∞=supk∈ℕ0zk and the ℓp norm zℓp=∑k=0∞zkp1/p, respectively, where ℕ0 is the set of nonnegative integers and Z is any one of the spaces c0,c, or ℓ∞.LetA=av,k be an infinite matrix over the complex field ℂ. The A-transform of a sequence z=zk is a sequence Az=Azv=∑k=0∞av,kzk, provided that the series ∑k=0∞av,kzk exists, for each v∈ℕ0. In addition, if Z and U are two sequence spaces and Az∈U, for every sequence z∈Z, then the matrix A is said to define a matrix mapping from Z to U. The notation Z,U represents the family of all matrices that map from Z to U. Furthermore, the matrix A=av,k is called a triangle if av,v≠0 and av,k=0, for v<k. For any Z⊂ω, define ZA=z∈ω:Az∈ω. Then, ZA is a sequence space and is called the matrix domain of A in the space Z. It is well known that if Z is a BK-space and A is a triangle, then, the matrix domain ZA is also a BK-space under the norm zZA=AzZ. We refer to [1–13] for papers related to theory of sequence spaces and summability.


### 1.1. Some Special Integer Sequences and the Associated Sequence Spaces

We shall briefly highlight the literature concerning special integer sequences and the construction of the associated sequence spaces.Letfkk=0∞ be the sequence of Fibonacci numbers defined by the recurrence relation fv=fv−1+fv−2,v≥2, with f0=1 and f1=1. Several authors constructed different types of sequence spaces involving Fibonacci numbers. For instance, Kara [3] studied the Fibonacci sequence spaces ℓpF^≔ℓpF^ and ℓ∞F^≔ℓ∞F^ and examined certain topological and geometrical structures of these Banach sequence spaces, where F^=fv,k is a double band matrix of Fibonacci numbers defined by
(1)fv,k=−fv+1fv,k=v−1,fvfv+1,k=v,0,otherwise.v,k∈ℕ0. Besides, Basarir et al. [7] studied the sequence spaces ℓpF^≔ℓpF^,c0F^≔c0F^ and cF^≔cF^, where 0<p<1. The studies on Fibonacci sequence spaces are further strengthened by Kara and Basarir [4] by introducing the matrix domain ZF≔ZF, where Z represents any one of the sequence spaces ℓp,c0,c, or ℓ∞ and F=f~v,k is a regular matrix of Fibonacci numbers defined by
(2)f~v,k=fk2fvfv+1,0≤k≤v,0,k>v,for all v,k∈ℕ0. Furthermore, another regular matrix F¯=f¯v,k of Fibonacci numbers is defined by Debnath and Saha [1] as follows:
(3)f¯v,k=fkfv+2−1,0≤k≤v,0,k>v,for all v,k∈ℕ0. By using this matrix, Debnath and Saha [1] and Ercan and Bektas [2] defined and studied the matrix domains ℓpF¯≔ℓpF¯,c0F¯≔c0F¯,cF¯≔cF¯, and ℓ∞F¯≔ℓ∞F¯. More studies concerning construction of Banach sequence spaces involving Fibonacci numbers can be tracked in the literature that are generalization or extension of any one of the above discussed Fibonacci sequence spaces. We refer to [5, 6, 8, 9], for such studies.The number sequencetvv=0∞≔1,1,2,4,7,13,24,⋯ defined by the recurrence relation tv=tv−1+tv−2+tv−3,v≥3, with t0=t1=1 and t2=2, is called tribonacci sequence. Recently, Yaying and Hazarika [10] introduced tribonacci sequence spaces ℓpT≔ℓpT and ℓ∞T≔ℓ∞T, where T=t¯v,k is an infinite matrix of tribonacci numbers defined by
(4)t¯v,k=2tktv+2+tv−1,0≤k≤v,0,k>v,for all v,k∈ℕ0. Quite recently, Yaying and Kara [11] studied the matrix domains c0T≔c0T and cT≔cT. Moreover, Yaying et al. [12] studied Banach sequence spaces defined by the sequence of Padovan numbers pvv=0∞=1,1,1,2,2,3,4,5,7,⋯. Besides, A. M. Karakas and M. Karakas [13] also constructed BK-sequence spaces defined by using Lucas numbers l′vv=0∞=2,1,3,4,7,11,18,29,47,⋯.


### 1.2. Leonardo Numbers

The number sequence1,1,3,5,9,15,25,41,67,⋯ is termed as Leonardo sequence. Let lv,v=0,1,2,⋯, denote the vth Leonardo number. Then, the Leonardo numbers are defined by the following recurrence relation:
(5)lv=lv−1+lv−2+1,v≥2,withl0=l1=1.It is believed that Leonardo sequence is invented by Leonardo de Pisa, also known as Leonardo Fibonacci. But not much studies related to Leonardo numbers can be traced in the literature due to scarcity of research related to this integer sequence. Leonardo sequence has a very close relationship with the well-known Fibonacci sequencefvv=0∞ and the Lucus sequence l′vv=0∞:
(6)lv=2fv+1−1,lv=25l′v+l′v+2−1,v≥0.Quite recently, Catarino and Borges [14] studied basic properties of Leonardo numbers and established several interesting identities, some of which are listed below:
(7)∑k=0vlv=lv+2−v+2,v∈ℕ0,lv2−lv−rlv+r=lv−r+lv+r−2lv−−1v−rlv−1+12,v>r,r≥1Catalan′sidentity,lv2−lv−1lv+1=lv−1+lv−2+4−1v,v≥2Cassini′sidentity,lv=2xv+1−yv+1x−y−1,v≥0,wherex=1+52,y=1−52.Besides, Alp and Koçer [15] also established interesting relationships between Fibonacci, Lucus, and Leonardo numbers. Vieira et al. [16] worked in the matrix form of the Leonardo numbers and established several interesting relations. Moreover, Shannon [17] also worked on the extension and generalization of the Leonardo numbers.Inspired by the above studies, we define an infinite matrixL=lv,k involving Leonardo numbers and construct sequence spaces ℓpL,c0L,cL, and ℓ∞L. We study their topological and inclusion properties and obtain Schauder bases of the sequence spaces ℓpL,c0L, and cL. In Section 3, α-, β-, and γ-duals of these new spaces are determined. In Section 4, matrix classes from the space Z∈ℓpL,c0L,cL,ℓ∞L to any one of the spaces ℓ1,c0,c, and ℓ∞ are characterized. In Sections 5 and 6, we introduce new sequence spaces c0L,rρ and ℓpL,rρ under a definite functional ρ and weighted sequence of positive reals r and discuss certain geometric and topological properties of ℓpL,rρ and the eigenvalue distribution of mappings ideally generated by these spaces and s-numbers are presented.


## 1.1. Some Special Integer Sequences and the Associated Sequence Spaces

We shall briefly highlight the literature concerning special integer sequences and the construction of the associated sequence spaces.Letfkk=0∞ be the sequence of Fibonacci numbers defined by the recurrence relation fv=fv−1+fv−2,v≥2, with f0=1 and f1=1. Several authors constructed different types of sequence spaces involving Fibonacci numbers. For instance, Kara [3] studied the Fibonacci sequence spaces ℓpF^≔ℓpF^ and ℓ∞F^≔ℓ∞F^ and examined certain topological and geometrical structures of these Banach sequence spaces, where F^=fv,k is a double band matrix of Fibonacci numbers defined by
(1)fv,k=−fv+1fv,k=v−1,fvfv+1,k=v,0,otherwise.v,k∈ℕ0. Besides, Basarir et al. [7] studied the sequence spaces ℓpF^≔ℓpF^,c0F^≔c0F^ and cF^≔cF^, where 0<p<1. The studies on Fibonacci sequence spaces are further strengthened by Kara and Basarir [4] by introducing the matrix domain ZF≔ZF, where Z represents any one of the sequence spaces ℓp,c0,c, or ℓ∞ and F=f~v,k is a regular matrix of Fibonacci numbers defined by
(2)f~v,k=fk2fvfv+1,0≤k≤v,0,k>v,for all v,k∈ℕ0. Furthermore, another regular matrix F¯=f¯v,k of Fibonacci numbers is defined by Debnath and Saha [1] as follows:
(3)f¯v,k=fkfv+2−1,0≤k≤v,0,k>v,for all v,k∈ℕ0. By using this matrix, Debnath and Saha [1] and Ercan and Bektas [2] defined and studied the matrix domains ℓpF¯≔ℓpF¯,c0F¯≔c0F¯,cF¯≔cF¯, and ℓ∞F¯≔ℓ∞F¯. More studies concerning construction of Banach sequence spaces involving Fibonacci numbers can be tracked in the literature that are generalization or extension of any one of the above discussed Fibonacci sequence spaces. We refer to [5, 6, 8, 9], for such studies.The number sequencetvv=0∞≔1,1,2,4,7,13,24,⋯ defined by the recurrence relation tv=tv−1+tv−2+tv−3,v≥3, with t0=t1=1 and t2=2, is called tribonacci sequence. Recently, Yaying and Hazarika [10] introduced tribonacci sequence spaces ℓpT≔ℓpT and ℓ∞T≔ℓ∞T, where T=t¯v,k is an infinite matrix of tribonacci numbers defined by
(4)t¯v,k=2tktv+2+tv−1,0≤k≤v,0,k>v,for all v,k∈ℕ0. Quite recently, Yaying and Kara [11] studied the matrix domains c0T≔c0T and cT≔cT. Moreover, Yaying et al. [12] studied Banach sequence spaces defined by the sequence of Padovan numbers pvv=0∞=1,1,1,2,2,3,4,5,7,⋯. Besides, A. M. Karakas and M. Karakas [13] also constructed BK-sequence spaces defined by using Lucas numbers l′vv=0∞=2,1,3,4,7,11,18,29,47,⋯.


## 1.2. Leonardo Numbers

The number sequence1,1,3,5,9,15,25,41,67,⋯ is termed as Leonardo sequence. Let lv,v=0,1,2,⋯, denote the vth Leonardo number. Then, the Leonardo numbers are defined by the following recurrence relation:
(5)lv=lv−1+lv−2+1,v≥2,withl0=l1=1.It is believed that Leonardo sequence is invented by Leonardo de Pisa, also known as Leonardo Fibonacci. But not much studies related to Leonardo numbers can be traced in the literature due to scarcity of research related to this integer sequence. Leonardo sequence has a very close relationship with the well-known Fibonacci sequencefvv=0∞ and the Lucus sequence l′vv=0∞:
(6)lv=2fv+1−1,lv=25l′v+l′v+2−1,v≥0.Quite recently, Catarino and Borges [14] studied basic properties of Leonardo numbers and established several interesting identities, some of which are listed below:
(7)∑k=0vlv=lv+2−v+2,v∈ℕ0,lv2−lv−rlv+r=lv−r+lv+r−2lv−−1v−rlv−1+12,v>r,r≥1Catalan′sidentity,lv2−lv−1lv+1=lv−1+lv−2+4−1v,v≥2Cassini′sidentity,lv=2xv+1−yv+1x−y−1,v≥0,wherex=1+52,y=1−52.Besides, Alp and Koçer [15] also established interesting relationships between Fibonacci, Lucus, and Leonardo numbers. Vieira et al. [16] worked in the matrix form of the Leonardo numbers and established several interesting relations. Moreover, Shannon [17] also worked on the extension and generalization of the Leonardo numbers.Inspired by the above studies, we define an infinite matrixL=lv,k involving Leonardo numbers and construct sequence spaces ℓpL,c0L,cL, and ℓ∞L. We study their topological and inclusion properties and obtain Schauder bases of the sequence spaces ℓpL,c0L, and cL. In Section 3, α-, β-, and γ-duals of these new spaces are determined. In Section 4, matrix classes from the space Z∈ℓpL,c0L,cL,ℓ∞L to any one of the spaces ℓ1,c0,c, and ℓ∞ are characterized. In Sections 5 and 6, we introduce new sequence spaces c0L,rρ and ℓpL,rρ under a definite functional ρ and weighted sequence of positive reals r and discuss certain geometric and topological properties of ℓpL,rρ and the eigenvalue distribution of mappings ideally generated by these spaces and s-numbers are presented.


## 2. Leonardo Sequence Spaces

Define an infinite matrixL=lv,k by
(8)lv,k=lklv+2−v+2,0≤k≤v,0,k>v,for all v,k∈ℕ0. Equivalently,
(9)L=1000⋯121200⋯1515350⋯110110310510⋯⋮⋮⋮⋮⋱.The inverse of the matrixL=lv,k is given by the matrix L−1=lv,k−1 defined by
(10)lv,k−1=−1v−klk+2−k+2lv,v≤k≤v+1,0,k>v,for all v,k∈ℕ0.Now, we define the following sequence spaces:(11)ℓpL≔z=zk∈ω:∑v=0∞∑k=0vlklv+2−v+2zkp<∞,c0L≔z=zk∈ω:limv⟶∞∑k=0vlklv+2−v+2zk=0,cL≔z=zk∈ω:limv⟶∞∑k=0vlklv+2−v+2zkexists,ℓ∞L≔z=zk∈ω:supv∈ℕ0∑k=0vlklv+2−v+2zk<∞,where the sequence w=wk defined by
(12)wv=Lzv=∑k=0vlklv+2−v+2zk,for each v∈ℕ0, which is known as the L-transform of the sequence z=zk. In what follows, the sequences z and w are related by (12). It is trivial that the above defined sequence spaces can be expressed in the form ZL=ZL, where Z represents any one of the spaces ℓp,c0,c, and ℓ∞. That is, ZL is the domain of the matrix L in the sequence space Z.We observe by the definition of the matrixL=lv,k that ∑k=0vlv,k=1. That is, supv∈ℕ0∑k=0vlv,k<∞ and limv⟶∞∑k=0∞lv,k=1. Additionally, limv⟶∞lv,k=0, for each k∈ℕ0. Thus, we conclude that the matrix L is regular.Theorem 1.
The following inclusion relations hold:(i)
Z⊂ZL, where Z is any one of the spaces ℓp,c0,c or ℓ∞(ii)
ℓpL⊂c0L⊂cL⊂ℓ∞L(iii)
ℓpL⊂ℓqL for 1≤p<qProof.
(i)
The inclusion part is trivial. Assume thatZ≔c and consider the sequence g=1,0,1,0,⋯. We observe that g∉c. However(13)Lgv=∑k=0vlklv+2−v+2gk=l0+l2+⋯+lvlv+2−v+2v∈ℕ0,which converges. Thus, g∈cL\c. In the similar manner strictness can be established for the other inclusions.
(ii)
It is known that the matrixL is regular and the inclusion ℓp⊂c0⊂c⊂ℓ∞ holds. These imply that the inclusion part holds. Now, consider the sequence h=1,1,1,1,⋯. Then, Lhv=∑k=0vlk/lv+2−v+2hk=1, for all v∈ℕ0. Thus, Lh∈c\c0. That is, h∈cL\c0L. This verifies the strictness of the inclusion c0L⊂cL. In the similar fashion, strictness of other inclusions can be established.(iii)
Assume that1≤p<q. Since L is regular and the inclusion ℓp⊂ℓq holds, therefore the desired inclusion holds. To prove the strictness part, we consider a sequence g=gk∈ℓq\ℓp. Define a sequence h=hk by hk=gklk+2−k+2−gk−1lk+1−k+1/lk,k∈ℕ0. Then, we get(14)Lhv=1lv+2−v+2∑k=0vlkhk=1lv+2−v+2∑k=0vgklk+2−k+2−gk−1lk+1−k+1=1lv+2−v+2gvlv+2−v+2=gv,for each v∈ℕ0, where the terms with negative subscripts are considered to be zero. Thus, we deduce that Lh=g∈ℓq\ℓp which implies h∈ℓqL\ℓpL. Thus there exists at least one sequence that is contained in ℓqL but not in ℓpL. Hence, the desired inclusion is strict. This completes the proof.Theorem 2.
We have the following results:(i)
The sequence spacesℓ∞L,c0L, and cL are BK-spaces equipped with the bounded norm zℓ∞=supk∈ℕ0Lzk(ii)
The sequence spaceℓpL is a BK-space equipped with the norm zℓp=∑v=0∞Lzvp1/pProof.
The proof is a routine exercise and so omitted.Theorem 3.
ZL≅Z, where Z is any one of the spaces ℓp,c0,c, or ℓ∞.Proof.
We present the proof for the spaceℓp. Define the mapping φ:ℓpL⟶ℓp by w=φz=Lz, for all z∈ℓpL. We observe that the mapping φ is linear and injective.
In view of the relation (12), we write
(15)zk=∑j=k−1k−1k−jlj+2−j+2lkwj‍,for each k∈ℕ0 and w=wk∈ℓp. Then,
(16)zℓpLp=∑v=0∞∑j=0kljlk+2−k+2zjp=∑v=0∞∑j=0kljlk+2−k+2∑u=j−1j−1j−ulu+2−u+2ljwup=∑v=0∞wp=wℓpp.
Thus,zℓpL=wℓp<∞. Thus, z∈ℓpL, and this implies that φ is surjective and norm preserving. Thus, ℓpL≅ℓp. In the similar manner, we can prove the existence of isomorphism between other given spaces. This completes the proof.Let us consider the following sequences:(17)g=1,1,−23,0,0,⋯,h=1,−3,23,0,0,⋯.Observe thatLg=1,1,0,0,0,⋯ and Lh=1,−1,0,0,0,⋯. Since L is linear, so Lg+h=2,0,0,0,⋯ and Lg−h=0,2,0,0,⋯. With some elementary calculation, we deduce that
(18)g+hℓpL2+g−hℓpL2=8≠221+1/p=2gℓpL2+hℓpL2.Thus, we realize that the norm·ℓpL violates the parallelogram identity for p≠2. This immediately allows us to write the following result.Theorem 4.
The sequence spaceℓpL is not a Hilbert space for p≠2.Proof.
The proof is immediate from the above discussion.We are well awarded that a matrix domainZA, where A is a triangle, has a basis, if and only if, Z has a basis (cf. [18]). Thus, in the light of Theorem 3, we have the following result:Theorem 5.
Define the sequencebk=bvkby(19)bvk=−1v−klk+2−k+2lv,v−1≤k≤v+1,0,k>vfor each fixed k∈ℕ0. Then,
(i)
The sequencebkk∈ℕ0 is the Schauder basis of the sequence spaces ℓpL and c0L, and every z in ℓpL or c0L is expressed uniquely in the form z=∑k=0∞bkwk, where w=wk is the L-transform of the sequence z=zk(ii)
The sequencee,b0,b1,b2,⋯ is the Schauder basis of the sequence space cL, and every z in cL is expressed uniquely in the form z=τe+∑k=0∞wk−τbk, where τ=limv⟶∞wv and e is the unit sequence(iii)
The sequence spaceℓ∞L has no Schauder basisCorollary 6.
The sequence spacesℓpL,c0L, and cL are separable spaces.


## 3.α-, β-, and γ-Duals

In this section, we obtain theα-, β-, and γ-duals of the sequence spaces ℓpL,c0L,cL, and ℓ∞L. Before proceeding, we recall the definitions of α-, β-, and γ-duals. Define the multiplier sequence space MZ,U by
(20)MZ,U≔d=dk∈ω:dz=dkzk∈U,∀z=zk∈U.In particular, ifU is ℓ1,cs or bs, then, the sets
(21)Zα=MZ,ℓ1,Zβ=MZ,cs,Zγ=MZ,bsare, respectively, termed as α-, β-, and γ-dual of the sequence space Z.We present Lemma7 which is essential to compute the dual spaces. In what follows, we denote the collection of all finite subsets of ℕ0 by N and q=p/p−1.Lemma 7 (see [19]).
The following statements hold:(i)
A=av,k∈ℓ1,ℓ1, if and only if, supk∈ℕ0∑v=0∞av,k<∞(ii)
A=av,k∈ℓp,ℓ1, if and only if
(22)supK∈N∑v=0∞∑k∈Kav,kq<∞(iii)
A=av,k∈c0,ℓ1=c,ℓ1=ℓ∞,ℓ1, if and only if, (22) holds with q=1(iv)
A=av,k∈ℓ1,ℓ∞, if and only if, supv,k∈ℕ0av,k<∞(v)
A=av,k∈ℓp,ℓ∞, if and only if,
(23)supv∈ℕ0∑k=0∞av,kq<∞(vi)
A=av,k∈c0,ℓ∞=c,ℓ∞=ℓ∞,ℓ∞, if and only if (23) holds with q=1Theorem 8.
Consider the following sets:(24)D1q≔d=dk∈ω:∑k=0∞lk+2−k+2lkdkq<∞,D1≔d=dk∈ω:supk∈ℕ0lk+2−k+2lkdk<∞.
Then,ℓ1Lα≔D1,ℓpLα≔D1q,andc0Lα=cLα=ℓ∞Lα≔D11, where 1<p<∞.Proof.
We observe that(25)dvzv=∑k=v−1v−1v−klk+2−k+2lvdvwk=Bwv,for each v∈ℕ0 and d=dk∈ω, where the matrix B=bv,k is defined by
(26)bv,k=−1v−klk+2−k+2lvdv,v−1≤k≤v,0,k>v,for all v,k∈ℕ0. We notice that the sequence dz=dvzv∈ℓ1 whenever z=zv∈c0L, if and only if, the sequence Bw∈ℓ1 whenever the sequence w=wv∈c0. We realize that the sequence d=dv∈c0Lα, if and only if, B∈c0,ℓ1. Thus, by employing Part (iii) of Lemma 7, we get that
(27)supV∈N∑k=0∞∑v∈Vbv,k<∞.
Moreover, for anyV∈N, (27) holds, if and only if,
(28)∑k=0∞lk+2−k+2lkdk<∞.
Consequently,c0Lα≔D11.
In the similar manner,α-dual of the other sequence spaces can be obtained by employing Part (i), Part (ii), and Part (iii) of Lemma 7.Theorem 9.
Consider the following sets:(29)D2q≔d=dk∈ω:∑k=0∞Δdklklk+2−k+2q<∞,D2≔d=dk∈ω:supk∈ℕ0Δdklklk+2−k+2<∞,D3≔d=dk∈ω:limk⟶∞lk+2−k+2lkdk=0,D4≔d=dk∈ω:limk⟶∞lk+2−k+2lkdkexists,where Δdk/lk=dk/lk−dk+1/lk+1. Then, ℓ1Lβ≔D1∩D2,ℓpLβ≔D1∩D2q,c0Lβ≔D1∩D21,cLβ≔D21∩D4, and ℓ∞Lβ≔D21∩D3, where 1<p<∞.Proof.
Letd=dk∈ω. Then, we have
(30)∑k=0vdkzk=∑k=0v∑j=k−1k−1k−jlj+2−j+2lkwjdk=∑k=0v−1dklk−dk+1lk+1lk+2−k+2wk+lv+2−v+2lvdvwv=∑k=0v−1Δdklklk+2−k+2wk+lv+2−v+2lvdvwv,for each v∈ℕ0. By employing Theorem 2 and Corollary 1 of Malkowsky and Savas [20], we get
(31)ℓ1Lβ≔d=dk∈ω:Δdklklk+2−k+2∈ℓ∞andlv+2−v+2lv∈ℓ∞,ℓpLβ≔d=dk∈ω:Δdklklk+2−k+2∈ℓq,lv+2−v+2lv∈ℓ∞,c0Lβ≔d=dk∈ω:Δdklklk+2−k+2∈ℓ1,lv+2−v+2lv∈ℓ∞,cLβ≔d=dk∈ω:Δdklklk+2−k+2∈ℓ1,lv+2−v+2lv∈c,ℓ∞Lβ≔d=dk∈ω:Δdklklk+2−k+2∈ℓ1,lv+2−v+2lv∈c0.
This completes the proof.Theorem 10.
We have the following results:(i)
ℓ1Lγ≔D1∩D2.(ii)
ℓpLγ≔D1∩D2q, where 1<p<∞(iii)
c0Lγ=cLγ=ℓ∞Lγ≔D1∩D21Proof.
It can be obtained by using relation (30) and Parts (iv), (v), and (vi) of Lemma 7, respectively.


## 4. Characterization of Matrix Classes

LetA=av,k be an infinite matrix over the field of complex numbers. Denote
(32)A~v≔lv+2−v+2lvAv=lv+2−v+2lvav,kk=0∞,foreachv∈ℕ0,(33)mv,k≔lk+2−k+2Δav,klk,forallv,k∈ℕ0.Now, we state the following result:Lemma 11.
LetZ denote either of the spaces ℓp or c0,1≤p≤∞. Then, A∈ZL,U, if and only if, A~∈MZ,c0, for each v∈ℕ0, and M∈Z,U, where A~ and M=mv,k are defined in (32) and (33), respectively.Proof.
It follows straightly from ([20], Theorem 3]).Lemma 12.
A∈cL,U, if and only if,
(34)A~v∈c,thatislimk⟶∞lv+2−v+2lvav,k=αvv∈N0,(35)M∈c0,U=ℓ∞,U,(36)αvv=0∞∈U.Proof.
Assume thatA∈cL,U. Then, Av∈cLβ which immediately shows the necessity of condition (35). Also, it is known that cL,U⊂c0L,U. Hence, by employing Lemma 11, we get that M∈c0,U. Thus, in the light of (30), we get that
(37)∑k=0∞av,kzk=∑k=0∞mv,kwk+zαv.
It is clear from the assumptions thatAz∈U and Mw∈U. These together yield αvv=0∞∈U.
Conversely, we assume that conditions (34), (35), and (36) hold. We realize that conditions (34) and (35) together imply that Av∈cLβ. Again condition (34) implies (37). By condition (35), Mw∈U, for all w∈c. This together with condition (36) implies that Az∈U, for all z∈cL. This proves that A∈cL,U.Now, using Lemmas11 and 12 together with the properties Mℓp,c0=Mc0,c0=ℓ∞1≤p<∞,Mc,c=c, and Mℓ∞,c0=c0, we deduce the following results:Corollary 13.
The following statements hold:(i)
A∈ℓ1L,ℓ∞, if and only if
(38)supk∈ℕ0lv+2−v+2lvav,k<∞,(39)supv,k∈ℕ0mv,k<∞(ii)
Let1<p<∞. Then A∈ℓpL,ℓ∞, if and only if, (38) holds, and
(40)supv∈ℕ0∑k=0∞mv,kq<∞(iii)
A∈c0L,ℓ∞, if and only if, (38) holds, and (40) holds with q=1(iv)
A∈cL,ℓ∞, if and only if, (34) and (40) hold with q=1, and
(41)supv∈ℕ0αv<∞,also holds(v)
A∈ℓ∞L,ℓ∞, if and only if,
(42)limk⟶∞lv+2−v+2lvav,k=0,forallv∈ℕ0,and (40) holds with q=1Corollary 14.
The following statements hold:(i)
A∈ℓ1L,c0, if and only if, (38) and (39) hold, and
(43)limv⟶∞mv,k=0,forallk∈ℕ0,also holds(ii)
Let1<p<∞. Then, A∈ℓpL,c0, if and only if, (38), (40), and (43) hold(iii)
A∈c0L,c0, if and only if, (38) and (40) hold with q=1, and (43) also holds(iv)
A∈cL,c0, if and only if, (34), (40) with q=1 and (43) hold, and
(44)limv⟶∞αv=0,also holds(v)
A∈ℓ∞L,c0, if and only if, (42) holds, and
(45)limv⟶∞∑k=0∞mv,k=0,also holdsCorollary 15.
The following statements hold:(i)
A∈ℓ1L,c, if and only if, (38), (39), and
(46)limv⟶∞mv,kexists,forallk∈ℕ0,also holds(ii)
Let1<p<∞. Then A∈ℓpL,c, if and only if, (38), (40) and (46) holds(iii)
A∈c0L,c, if and only if, (38) holds, (40) with q=1 and (46) hold(iv)
A∈cL,c, if and only if, (34), (40) with q=1 and (46) hold, and
(47)limv⟶∞αvexistsalso holds(v)
A∈ℓ∞L,c, if and only if, (42) and (46) hold, and
(48)∑k=0∞mv,kconvergesuniformlyinvalso holdsCorollary 16.
The following statements hold:(i)
A∈ℓ1L,ℓ1, if and only if, (38) holds and
(49)supv∈ℕ0∑v=0∞mv,k<∞,also holds(ii)
Let1<p<∞. Then A∈ℓpL,ℓ1, if and only if, (38) holds, and
(50)supV∈N∑k=0∞∑v∈Vmv,kq<∞,also holds(iii)
A∈c0L,ℓ1, if and only if, (38) holds, and (50) holds with q=1(iv)
A∈cL,ℓ1, if and only if, (34) and (50) hold with q=1, and
(51)∑v=0∞αv<∞,also holds(v)
A∈ℓ∞L,ℓ1, if and only if, (42) and (50) hold with q=1


## 5. Mapping Ideal

In this section, we constructs-type mapping ideals on Leonardo sequence spaces c0L,rρ and ℓpL,rρ. By B, we denote the class of all bounded linear mappings between any two Banach spaces. In particular, BX,Y denote the class of all bounded linear mappings acting from Banach space X to Banach space Y. We note down certain notations and definitions before moving to our results:Definition 17 (see [21, 22]).
Letω+ represent the set of nonnegative real sequences. Then, s-number is a mapping s:BX,Y⟶ω+ that satisfies the following settings:
(i)
ϕ=s0ϕ≥s1ϕ≥s2ϕ≥⋯≥0, for each ϕ∈BX,Y(ii)
sa+b−1ϕ+ψ≤saϕ+sbψ, for each ϕ,ψ∈BX,Y and a,b∈ℕ0(iii)
saϕθψ≤ϕsaθψ, for all ϕ∈BX0,X,θ∈BX,Y, and ψ∈BY,Y0, where X0 and Y0 are any two Banach sequence spaces(iv)
Letϕ∈BX,Y and ν∈ℂ. Then, saνϕ=νsaϕ(v)
Ifrankϕ≤a, then saϕ=0 for all ϕ∈BX,Y(vi)
svIa=0 for v≥a or svIa=1 for v<a, where Ia denotes the identity mapping on the a-dimensional Hilbert space ℓ2a
In an assorted illustration ofs-numbers, we intimate the next settings:
(1)
Thea-th Kolmogorov number, denoted by daX, is defined as
(52)daX=infdimJ≤asupf≤1infg∈JXf−g(2)
Thea-th approximation number, denoted by αaX, is defined as
(53)αaX=infX−Y:Y∈BX,Y,rankY≤aDefinition 18 (see [23]).
LetW⊂B and denote WX,Y=W∩BX,Y. Then, W is known as a mapping ideal if it satisfies the following settings:
(i)
ID∈W, where D is a Banach sequence space of one dimension(ii)
WX,Y is a linear space over ℂ(iii)
Ifψ∈BX0,X,θ∈BX,Y and ϕ∈BY0,Y, then ϕθψ∈BX0,Y0, where X0 and Y0 are any two normed spacesDefinition 19 (see [24]).
A prequasi norm on the idealW is a mapping μ:W⟶ω+ satisfying the following settings:
(i)
μϕ≥0 and μϕ=0 if and only if ϕ=0, for all ϕ∈WX,Y(ii)
There existsm0≥1 such that μςϕ≤m0ςμϕ, for all ϕ∈WX,Y(iii)
There existsn0≥1 such that μϕ+ψ≤n0μϕ+μψ, for all ϕ,ψ∈WX,Y(iv)
There existsp0≥1 such that μϕθψ≤p0ϕμθψ whenever ψ∈BX0,X,θ∈BX,Y and ϕ∈BY0,YDefinition 20 (see [24]).
The subspaceZ⊂ω is said to be a private sequence space (or in short pss) if it satisfies the following settings:
(i)
ev∈Z, for each v∈ℕ0, where ev denotes the sequence with 1 in the vth position and 0 elsewhere(ii)
Ifg=gv∈ω,h=hv∈Z and gv≤hv, for v∈ℕ0, then g∈Z(iii)
gv/2∈Z whenever gv∈Z, where v/2 denotes the integral part of v/2Definition 21 (see [24]).
A subspace of thepss is said to be a premodular pss, if there is a function υ:Z⟶0,∞ satisfying the following conditions:
(i)
For everyj∈Z, j=0⇔υj=0, and υj≥0, with 0 is the zero vector of Z(ii)
Ifj∈Z and ρ∈ℂ, then there are E0≥1 with υρj≤ρE0υj(iii)
υh+j≤G0υh+υj holds for some G0≥1, with f,g∈Z(iv)
Assumex∈ℕ0, hx≤jx, we have υhx≤υjx(v)
The inequality,υjx≤υjx/2≤D0υjx verifies, for D0≥1(vi)
C¯=Zυ, where C¯ denotes the closure of the space of all sequences with infinite zero coordinates(vii)
We haveη>0 such that υρ,0,0,0,⋯≥ηρυ1,0,0,0,⋯, with ρ∈ℂDefinition 22 (see [24]).
ThepssZυ is said to be a prequasi normed pss, if υ confirms the setups (i)-(iii) of Definition 21. If Z is complete equipped with υ, then Zυ is called a prequasi Banach pss.Lemma 23 (see [24]).
Every premodularpss is a prequasi normed pss.
In what follows, we will use the following inequality:(54)g+hp≤2p−1gp+hp,where 1≤p<∞ and g,h∈ℂ. For detailed studies concerning s-numbers and mapping ideals, we refer to [23–28].Definition 24.
We define the following sequence spaces:(55)ℓpL,rρ1≔z=zk∈ω:ρ1z=∑v=0∞∑k=0vrklklv+2−v+2zkp<∞,c0L,rρ2≔z=zk∈ω:limv⟶∞∑k=0vrklklv+2−v+2zk=0,where r=rk∈ω+ and ρ2z=supv∈ℕ0∑k=0vrklk/lv+2−v+2zk.
ByI↗ and I↘, we will denote the space of all monotonic increasing and decreasing sequences of positive reals, respectively.Theorem 25.
c0L,r is a pss, whenever rklkk=0∞∈I↘ or rklkk=0∞∈I↗∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk.Proof.
(i)
Letg,h∈c0L,r, we obtain
(56)supv∈ℕ0∑k=0vrklklv+2−v+2gk+hk≤supv∈ℕ0∑k=0vrklklv+2−v+2gk+supv∈ℕ0∑k=0vrklklv+2−v+2hk<∞
Thus,g+h∈c0L,r.
Assume thatς∈ℂ and g∈c0L,r. Then, we have
(57)supv∈ℕ0∑k=0vrklklv+2−v+2ςgk=ςsupv∈ℕ0∑k=0vrklklv+2−v+2gk<∞.
Thusc0L,r is a linear space. Moreover
(58)supv∈ℕ0∑k=0vrklklv+2−v+2eak=ralasupv∈ℕ01lv+2−v+2<∞.
This impliesea∈c0L,r, for each a∈ℕ0.(ii)
Assume thatgk≤hk, for all k∈ℕ0 and h∈c0L,r. Then, we have
(59)supv∈ℕ0∑k=0vrklklv+2−v+2gk≤supv∈ℕ0∑k=0vrklklv+2−v+2hk<∞
This concludes thatg∈c0L,r.(iii)
Letgk∈c0L,r, rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk. Then, we have
(60)supv∈ℕ0∑k=0vrklklv+2−v+2gk2≤supv∈ℕ0∑k=02vrklkl2v+2−2v+2gk2+supv∈ℕ0∑k=02v+1rklkl2v+3−2v+3gk2≤supv∈ℕ01l2v+2−2v+2r2vl2vgv+∑k=0vr2kl2kgk+r2k+1l2k+1gk+supv∈ℕ01l2v+3−2v+3∑k=0vr2kl2kgk+r2k+1l2k+1gk≤supv∈ℕ0Clv+2−v+2∑k=0vrklkgk+supv∈ℕ02Clv+2−v+2∑k=0vrklkgk+supv∈ℕ02Clv+2−v+2∑k=0vrklkgk≤5Csupv∈ℕ01lv+2−v+2∑k=0vlkgk<∞
Thus,gk/2∈c0L,r.
This completes the proof.Theorem 26.
ℓpL,r is a pss, whenever 1<p<∞,rklkk=0∞∈I↗ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk.Proof.
(i)
Letg,h∈ℓpL,r. By using (54), we obtain
(61)∑v=0∞∑k=0vrklklv+2−v+2gk+hkp≤2p−1∑v=0∞∑k=0vrklklv+2−v+2gkp+∑v=0∞∑k=0vrklklv+2−v+2hkp<∞
Thus,g+h∈ℓpL,r.
Assume thatς∈ℂ and g∈ℓpL,r. Then, we have
(62)∑v=0∞∑k=0vrklklv+2−v+2ςgkp=ςp∑v=0∞∑k=0vrklklv+2−v+2gkp<∞.
Thus,ℓpL,r is a linear space. Moreover,
(63)∑v=0∞∑k=0vrklklv+2−v+2eakp=raplap∑v=r∞1lv+2−v+2p<∞.
This impliesea∈ℓpL,r for each a∈ℕ0.(ii)
Assume thatgk≤hk, for all k∈ℕ0 and h∈ℓpL,r. Then, we have
(64)∑v=0∞∑k=0vrklklv+2−v+2gkp≤∑v=0∞∑k=0vrklklv+2−v+2hkp<∞
This concludes thatg∈ℓpL,r.(iii)
Letgk∈ℓpL,r, rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk. Then, we have
(65)∑v=0∞∑k=0vrklklv+2−v+2gk/2p=∑v=0∞∑k=02vrklkl2v+2−2v+2gk/2p+∑v=0∞∑k=02v+1rklkl2v+3−2v+3gk/2p≤∑v=0∞1l2v+2−2v+2r2vl2vgv+∑k=0vr2kl2kgk+r2k+1l2k+1gkp+∑v=0∞1l2v+3−2v+3∑k=0vr2kl2kgk+r2k+1l2k+1gkp≤2p−1∑v=0∞Clv+2−v+2∑k=0vrklkgkp+∑v=0∞2Clv+2−v+2∑k=0vrklkgkp+∑v=0∞2Clv+2−v+2∑k=0vrklkgkp≤22p−1+2p+2p−1Cp∑v=0∞1lv+2−v+2∑k=0vlkgkp<∞
Thus,gk/2∈ℓpL,r.
This completes the proof.Define the setsBZsX,Y, BZαX,Y, and BZdX,Y by
(66)BZsX,Y≔ϕ∈BX,Y:saϕ∈Z,BZαX,Y≔ϕ∈BX,Y:αaϕ∈Z,BZdX,Y≔ϕ∈BX,Y:daϕ∈Z,where X and Y are any two Banach sequence spaces. We denote BZs≔BZsX,Y., BZα≔BZαX,Y, and BZd≔BZdX,Y, respectively.Lemma 27 (see [24]).
Let the linear sequence spaceZ be a pss. Then, BZs is a mapping ideal.Theorem 28.
Letrklkk=0∞∈I↘ or rklkk=0∞∈I↗∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk. Then, Bc0L,rs is a mapping ideal.Proof.
It follows straightly from Lemma27.Theorem 29.
Let1<p<∞,rklkk=0∞∈I↘ or rklkk=0∞∈I↗∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk. Then, BℓpL,rs is a mapping ideal.Proof.
It follows straightly from Lemma27.Theorem 30.
Letrklkk=0∞∈I↘ or rklkk=0∞∈I↗∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk. Then, c0L,rρ is a premodular pss.Proof.
(i)
Clearly, for allg∈c0L,rρ that ρg≥0 and ρg=0, if and only if, g=0(ii)
For anyε≥1. Then ραg≤εαρg, for all g∈c0L,r and α∈ℂ(iii)
Observe thatρg+h≤ρg+ρh, for all g,h∈c0L,r(iv)
We haveρgk≤ρhk, whenever gk≤hk (see Proof Part (ii), Theorem 25).(v)
It is immediate from Proof Part (iii) of Theorem25 that ρgk≤ρgk/2≤δρgk with δ=5C.
(67)C¯=c0L,r(vi)
We have, whenα≠0 then 0<γ≤1, for ρα,0,0,⋯≥γαρ1,0,0,⋯ and when α=0 then γ>0
This completes the proof.Theorem 31.
Let1<p<∞,rklkk=0∞∈I↘ or rklkk=0∞∈I↘∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk. Then, ℓpL,rρ is a premodular pss.Proof.
(i)
Clearly, for allg∈ℓpL,rρ that ρg≥0 and ρg=0, if and only if, g=0(ii)
Letε=max1,αp−1≥1. Then, ραg≤εαρg, for all g∈ℓpL,r and α∈ℂ(iii)
Observe thatρg+h≤2p−1ρg+ρh, for all g,h∈ℓpL,r(iv)
We haveρgk≤ρhk, whenever gk≤hk (see Proof Part (ii), Theorem 26).(v)
It is immediate from Proof Part (iii) of Theorem26 that ρgk≤ρgk/2≤δρgk with δ=22p−1+2p+2p−1Cp(68)C¯=ℓpL,r(vi)
We have, whenα≠0 then 0<γ<αp−1, for ρα,0,0,⋯≥γαρ1,0,0,⋯, and when α=0 then γ>0
This completes the proof.Theorem 32.
Assume thatrklkk=0∞∈I↘ or rklkk=0∞∈I↘∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk. Then, c0L,rρ is a prequasi Banach pss.Proof.
In view of Theorem30 and Lemma 23, it is enough to prove that every Cauchy sequence in c0L,rρ is convergent in c0L,rρ. We assume that gm=gkm is a Cauchy sequence in c0L,rρ. Then, for all ε∈0,1 there exists n0∈ℕ0 such that
(69)ρgm−gn=supv=0∞1lv+2−v+2∑k=0vlkgkm−gnk<ε,for all m,n≥n0. This implies that gm−gn<ε, for all m,n≥n0. Thus, gm is a Cauchy sequence in ℂ. Since ℂ is complete, limm⟶∞gkm=gk, for a fixed k∈ℕ0. This yields, by using (69), that ρgm−g<ε, for all m≥n0. Besides, we have ρg≤ρgm−g+ρgm<∞. This concludes that g∈c0L,rρ. Thus, c0L,rρ is a prequasi Banach pss.Theorem 33.
Assume that1<p<∞,rklkk=0∞∈I↘orrklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk. Then, ℓpL,rρ is a prequasi Banach pss.Proof.
In view of Theorem31 and Lemma 23, it is enough to prove that every Cauchy sequence in ℓpL,rρ is convergent in ℓpL,rρ. We assume that gm=gkm is a Cauchy sequence in ℓpL,rρ. Then, for all ε∈0,1 there exists n0∈ℕ0 such that
(70)ρgm−gn=∑v=0∞1lv+2−v+2∑k=0vlkgkm−gnk<εp,for all m,n≥n0. This implies that gm−gn<ε, for all m,n≥n0. Thus, gm is a Cauchy sequence in ℂ. Since ℂ is complete, limm⟶∞gkm=gk, for a fixed k∈ℕ0. This yields, by using (70), that ρgm−g<εp, for all m≥n0. Besides, we have ρg≤2p−1ρgm−g+ρgm<∞. This concludes that g∈ℓpL,rρ. Thus ℓpL,rρ is a prequasi Banach pss.Theorem 34 (see [27]).
Supposes−type Eρ≔h=sxH∈ω+:H∈BX,Yandρh<∞.IfBEρs is a mapping ideal, then the following conditions are verified:
(1)
C⊂s−typeEρ(2)
SupposesxH1x=0∞∈s−typeEρ and sxH2x=0∞∈s−typeEρ; then sxH1+H2x=0∞∈s−typeEρ(3)
Assumeλ∈ℂ and sxHx=0∞∈s−typeEρ; then λsxHx=0∞∈s−typeEρ(4)
The sequence spaceEρ is solid; i.e., if sxJx=0∞∈s−typeEρ and sxH≤sxJ, for all x∈ℕ0 and H,J∈BX,Y; then sxHx=0∞∈s−typeEρ
In view of Theorem34, we construct the next properties of the s−typec0L,rρ and the s−typeℓpL,rρ.Theorem 35.
Lets−typec0L,rρ≔f=snX∈ω+:X∈BX,Yandρf<∞. The next conditions are established:
(1)
One hass−typec0L,rρ⊃C(2)
SupposesrX1r=0∞∈s−typec0L,rρ and srX2r=0∞∈s−typec0L,rρ; then srX1+X2r=0∞∈s−typec0L,rρ(3)
Assumeλ∈ℂ and srXr=0∞∈s−typec0L,rρ; hence λsrXr=0∞∈s−typec0L,rρ(4)
Thes−typec0L,rρ is solidTheorem 36.
Lets−typeℓpL,rρ≔f=snX∈ω+:X∈BX,Yandρf<∞. The next conditions are established:
(1)
One hass−typeℓpL,rρ⊃C(2)
SupposesrX1r=0∞∈s−typeℓpL,rρ and srX2r=0∞∈s−typeℓpL,rρ; then srX1+X2r=0∞∈s−typeℓpL,rρ(3)
Assumeλ∈ℂ and srXr=0∞∈s−typeℓpL,rρ; hence λsrXr=0∞∈s−typeℓpL,rρ(4)
Thes−typeℓpL,rρ is solid


## 6. Characteristics of the Prequasi Ideal

The conventions listed below will be followed throughout the article; if the species is preowned, we will give it to you.Conventions 1.
Please see the following conventions:
F: the ideal of finite rank mappings between any arbitrary Banach spaces
A: the ideal of approximable mappings between any arbitrary Banach spaces
K: the ideal of compact mappings between any arbitrary Banach spaces
FX,Y: the space of finite rank mappings from a Banach space X into a Banach space Y
FX: the space of finite rank mappings from a Banach space X into itself
AX,Y: the space of approximable mappings from a Banach space X into a Banach space Y
AX: the space of approximable mappings from a Banach space X into itself
KX,Y: the space of compact mappings from a Banach space X into a Banach space Y
KX: the space of compact mappings from a Banach space X into itselfLemma 37 (see [28]).
IfM∈BX,Y and M∉AX,Y, then there are operators Q∈BX and L∈BY so that LMQex=ex, for all x∈ℕ0.Definition 38 (see [28]).
A Banach spaceE is called simple if the algebra BE includes one and only one nontrivial closed ideal.Theorem 39 (see [28]).
SupposeE is a Banach space with dimE=∞; then
(71)FE⊂AE⊂KE⊂BE.In this section, firstly, we introduce the enough setups (not necessary) onc0L,rρ and ℓpL,rρ such that F¯=Bc0L,rρs and F¯=BℓpL,rρs. This investigates a negative answer of Rhoades [29] open problem about the linearity of s−typec0L,rρ and ℓpL,rρ spaces. Secondly, for which conditions on c0L,rρ and ℓpL,rρ, are Bc0L,rρs and BℓpL,rρs closed and complete? Thirdly, we explain the enough setups on c0L,rρ and ℓpL,rρ such that Bc0L,rρs and BℓpL,rρs are strictly contained for different weights and powers. We offer the setups so that BℓpL,rρα is minimum. Fourthly, we introduce the conditions so that the Banach prequasi ideal Bc0L,rρs and BℓpL,rρs are simple Banach spaces. Fifthly, we investigate the enough conditions on c0L,rρ and ℓpL,rρ such that the space of all bounded linear operators which sequence of eigenvalues in c0L,rρ and ℓpL,rρ equal Bc0L,rρs and BℓpL,rρs, respectively.


### 6.1. Finite Rank Prequasi Ideal

Theorem 40.
Bc0L,rρsX,Y=FX,Y¯; suppose the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are satisfied. But the converse is not necessarily true.Proof.
To investigate thatFX,Y¯⊆Bc0L,rρsX,Y, as el∈c0L,rρ, for every l∈ℕ0,c0L,rρ is a linear space. Let Z∈FX,Y, one gets slZl=0∞∈C. To explain that Bc0L,rρsX,Y⊆FX,Y¯, assume Z∈Bc0L,rρsX,Y, we obtain slZl=0∞∈c0L,rρ. Since ρslZl=0∞<∞, let ρ∈0,1; hence, there is l0∈ℕ0−0 with ρslZl=l0∞<ρ/16d, for some d≥1. Since slZ∈I], we get
(72)supl=l0+12l0∑j=0lrjljs2l0Zll+2−l+2≤supl=l0+12l0∑j=0lrjljsjZll+2−l+2≤supl=l0∞∑j=0lrjljsjZll+2−l+2<ρ16d.
Hence, there isY∈F2l0X,Y so that rankY≤2l0 and
(73)supl=2l0+13l0∑j=0lrjljZ−Yll+2−l+2≤supl=l0+12l0∑j=0lrjljZ−Yll+2−l+2<ρ16d,we have
(74)∑j=0l0rjljZ−Y<ρ16.
Therefore, one has(75)supl=0l0∑j=0lrjljZ−Yll+2−l+2<ρ16d.
In view of inequalities (72)–(75), and rklkk=0∞∈I], one gets
(76)dZ,Y=ρslZ−Yl=0∞≤supl=03l0−1∑j=0lrjljsjZ−Yll+2−l+2+supl=3l0∞∑j=0lrjljsjZ−Yll+2−l+2≤supl=03l0∑j=0lrjljsjZ−Yll+2−l+2+supl=l0∞∑j=0l+2l0rjljsjZ−Yll+2l0+2−l+2l0+2≤supl=03l0∑j=0lrjljsjZ−Yll+2−l+2+supl=l0∞∑j=0l+2l0rjljsjZ−Yll+2−l+2≤3supl=0l0∑j=0lrjljZ−Yll+2−l+2+supl=l0∞∑j=02l0−1rjljsjZ−Y+∑j=2l0l+2l0rjljsjZ−Yll+2−l+2≤3supl=0l0∑j=0lrjljZ−Yll+2−l+2+supl=l0∞∑j=02l0−1rjljsjZ−Yll+2−l+2+supl=l0∞∑j=2l0l+2l0rjljsjZ−Yll+2−l+2≤3supl=0l0∑j=0lrjljZ−Yll+2−l+2+supl=l0∞∑j=02l0−1rjljsjZ−Yll+2−l+2+supl=l0∞∑j=0lrj+2l0lj+2l0sj+2l0Z−Yll+2−l+2≤3supl=0l0∑j=0lrjljZ−Yll+2−l+2+2∑j=0l0ljrjZ−Y+supl=l0∞∑j=0lrjljsjZll+2−l+2<ρ.
On the opposite side, one has a negative example asI3∈Bc0L,rρsX,Y, where r=0,0,0,1,0,1,0,. This shows the proof.Theorem 41.
BℓpL,rρsX,Y=FX,Y¯; suppose the setups 1<p<∞,rklkk=0∞∈I↘orrklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are confirmed. But the converse is not necessarily true.Proof.
To investigate thatFX,Y¯⊆BℓpL,rρsX,Y, as el∈ℓpL,rρ, for every l∈ℕ0, ℓpL,rρ is a linear space. Let Z∈FX,Y; one gets slZl=0∞∈C. To explain that BℓpL,rρsX,Y⊆FX,Y¯, assume Z∈BℓpL,rρsX,Y; we obtain slZl=0∞∈ℓpL,rρ. Since ρslZl=0∞<∞, let ρ∈0,1, hence, there is l0∈ℕ0−0 with ρslZl=l0∞<ρ/2p+3ηd, for some d≥1, where η=max1,∑l=l0∞1/ll+2−l+2p. Since slZ∈I], we get
(77)∑l=l0+12l0∑j=0lrjljs2l0Zll+2−l+2p≤∑l=l0+12l0∑j=0lrjljsjZll+2−l+2p≤∑l=l0∞∑j=0lrjljsjZll+2−l+2p<ρ2p+3ηd.
Hence, there isY∈F2l0X,Y so that rankY≤2l0 and
(78)∑l=2l0+13l0∑j=0lrjljZ−Yll+2−l+2p≤∑l=l0+12l0∑j=0lrjljZ−Yll+2−l+2p<ρ2p+3ηd.
Since1<p<∞, we have
(79)∑j=0l0rjljZ−Yp<ρ22p+2η.
Therefore, one has(80)∑l=0l0∑j=0lrjljZ−Yll+2−l+2p<ρ2p+3ηd.
In view of inequalities (54), (77)–(80), and rklkk=0∞∈I], one gets
(81)dZ,Y=ρslZ−Yl=0∞=∑l=03l0−1∑j=0lrjljsjZ−Yll+2−l+2p+∑l=3l0∞∑j=0lrjljsjZ−Yll+2−l+2p≤∑l=03l0∑j=0lrjljsjZ−Yll+2−l+2p+∑l=l0∞∑j=0l+2l0rjljsjZ−Yll+2l0+2−l+2l0+2p≤∑l=03l0∑j=0lrjljsjZ−Yll+2−l+2p+∑l=l0∞∑j=0l+2l0rjljsjZ−Yll+2−l+2p≤3∑l=0l0∑j=0lrjljZ−Yll+2−l+2p+∑l=l0∞∑j=02l0−1rjljsjZ−Y+∑j=2l0l+2l0rjljsjZ−Yll+2−l+2p≤3∑l=0l0∑j=0lrjljZ−Yll+2−l+2p+2p−1∑l=l0∞∑j=02l0−1rjljsjZ−Yll+2−l+2p+∑l=l0∞∑j=2l0l+2l0rjljsjZ−Yll+2−l+2p≤3∑l=0l0∑j=0lrjljZ−Yll+2−l+2p+2p−1∑l=l0∞∑j=02l0−1rjljsjZ−Yll+2−l+2p+∑l=l0∞∑j=0lrj+2l0lj+2l0sj+2l0Z−Yll+2−l+2p≤3∑l=0l0∑j=0lrjljZ−Yll+2−l+2p+2p∑j=0l0ljrjZ−Yp∑l=l0∞1ll+2−l+2p+2p−1∑l=l0∞∑j=0lrjljsjZll+2−l+2p<ρ.
On the opposite side, one has a negative example asI4∈Bℓ0.5L,rρsX,Y, where r=0,0,0,0,1,1,. This shows the proof.


### 6.2. Banach and Closed Prequasi Ideal

Theorem 42 (see [24]).
The functionΨ is a prequasi norm on BEρs, where ΨY=ρsbYb=0∞, for every Y∈BEρsX,Y, if Eρ is a premodular pss.Theorem 43.
If the setupsrklkk=0∞∈I↘orrklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk with r0>0 are satisfied, then Bc0L,rρs,Ψ is a prequasi Banach ideal, where ψX=ρslXl=0∞.Proof.
Asc0L,rρ is a premodular pss, hence from Theorem 42, Ψ is a prequasi norm on Bc0L,rρs. Suppose Xbb∈ℕ0 is a Cauchy sequence in Bc0L,rρsX,Y. As BX,Y⊇Bc0L,rρsX,Y, one obtains
(82)ΨXa−Xb=supl∈ℕ0∑j=0lrjljsjXa−Xbll+2−l+2≥r0Xa−Xb.
Hence,Xbb∈ℕ0 is a Cauchy sequence in BX,Y. Since BX,Y is a Banach space, then there is X∈BX,Y with limb⟶∞Xb−X=0. Since slXbl=0∞∈c0L,rρ, every b∈ℕ0. According to Definition 21 setups (ii), (iii), and (v), one gets
(83)ΨX=supl∈ℕ0∑j=0lrjljsjXll+2−l+2≤supl∈ℕ0∑j=0lrjljsj/2X−Xbll+2−l+2+supl∈ℕ0∑j=0lrjljsj/2Xbll+2−l+2≤supl∈ℕ0∑j=0lrjljX−Xbll+2−l+2+D0supl∈ℕ0∑j=0lrjljsjXbll+2−l+2<∞.
Therefore,slXl=0∞∈c0L,rρ; then X∈Bc0L,rρsX,Y.Theorem 44.
If the setups1<p<∞,rklkk=0∞∈I↘orrklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk with r0>0 are confirmed; then BℓpL,rρs,Ψ is a prequasi Banach ideal, where ψX=ρslXl=0∞.Proof.
AsℓpL,rρ is a premodular pss, hence from Theorem 42, Ψ is a prequasi norm on BℓpL,rρs. Suppose Xbb∈ℕ0 is a Cauchy sequence in BℓpL,rρsX,Y. As BX,Y⊇BℓpL,rρsX,Y, one obtains
(84)ΨXa−Xb=∑l=0∞∑j=0lrjljsjXa−Xbll+2−l+2p≥r0Xa−Xbp.
Hence,Xbb∈ℕ0 is a Cauchy sequence in BX,Y. Since BX,Y is a Banach space, then there is X∈BX,Y with limb⟶∞Xb−X=0.Since slXbl=0∞∈ℓpL,rρ, every b∈ℕ0. According to Definition 21 setups (ii), (iii), and (v), one gets
(85)ΨX=∑l=0∞∑j=0lrjljsjXll+2−l+2p≤2p−1∑l=0∞∑j=0lrjljsj/2X−Xbll+2−l+2p+2p−1∑l=0∞∑j=0lrjljsj/2Xbll+2−l+2p≤2p−1∑l=0∞∑j=0lrjljX−Xbll+2−l+2p+2p−1D0∑l=0∞∑j=0lrjljsjXbll+2−l+2p<∞.
Therefore,slXl=0∞∈ℓpL,rρ; then X∈BℓpL,rρsX,Y.Theorem 45.
AssumeX, Y are normed spaces; the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞; and there exists C≥1 such that r2k+1l2k+1≤Crklk with r0>0 are satisfied; then Bc0L,rρs,Ψ is a prequasi closed ideal, where ΨX=ρslXl=0∞.Proof.
Asc0L,rρ is a premodular pss, by using Theorem 42, Ψ is a prequasi norm on Bc0L,rρs. Assume Xb∈Bc0L,rρsX,Y, every b∈ℕ0 and limb⟶∞ΨXb−X=0. As BX,Y⊇Bc0L,rρsX,Y, we have
(86)ΨX−Xb=supl∈ℕ0∑j=0lrjljsjX−Xbll+2−l+2≥r0X−Xb.
Hence,Xbb∈ℕ0 is a convergent sequence in BX,Y. Since slXbl=0∞∈c0L,rρ, for every b∈ℕ0. In view of Definition 21 setups (ii), (iii), and (v), one has
(87)ΨX=supl∈ℕ0∑j=0lrjljsjXll+2−l+2≤supl∈ℕ0∑j=0lrjljsj/2X−Xbll+2−l+2+supl∈ℕ0∑j=0lrjljsj/2Xbll+2−l+2≤supl∈ℕ0∑j=0lrjljX−Xbll+2−l+2+D0supl∈ℕ0∑j=0lrjljsjXbll+2−l+2<∞.
We getslXl=0∞∈c0L,rρ, so X∈Bc0L,rρsX,Y.Theorem 46.
AssumeX, Y are normed spaces; the setups 1<p<∞,rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞; and there exists C≥1 such that r2k+1l2k+1≤Crklk with r0>0 are satisfied; hence, BℓpL,rρs,Ψ is a prequasi closed ideal, where ΨX=ρslXl=0∞.Proof.
AsℓpL,rρ is a premodular pss, by using Theorem 42, Ψ is a prequasi norm on BℓpL,rρs. Assume Xb∈BℓpL,rρsX,Y, for every b∈ℕ0 and limb⟶∞ΨXb−X=0. As BX,Y⊇BℓpL,rρsX,Y, we have
(88)ΨX−Xb=∑l=0∞∑j=0lrjljsjX−Xbll+2−l+2p≥r0X−Xbp.
Hence,Xbb∈ℕ0 is a convergent sequence in BX,Y. Since slXbl=0∞∈ℓpL,rρ, every b∈ℕ0. In view of Definition 21 setups (ii), (iii), and (v), one has
(89)ΨX=∑l=0∞∑j=0lrjljsjXll+2−l+2p≤2p−1∑l=0∞∑j=0lrjljsj/2X−Xbll+2−l+2p+2p−1∑l=0∞∑j=0lrjljsj/2Xbll+2−l+2p≤2p−1∑l=0∞∑j=0lrjljX−Xbll+2−l+2p+2p−1D0∑l=0∞∑j=0lrjljsjXbll+2−l+2p<∞.
We getslXl=0∞∈ℓpL,rρ, so X∈BℓpL,rρsX,Y.


### 6.3. Minimum Prequasi Ideal

Theorem 47.
SupposeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I] or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are confirmed with 0<rl2≤rl1, for all l∈ℕ0, hence
(90)Bc0L,rl1ρsX,Y⊂Bc0L,rl2ρsX,Y⊂BX,Y.Proof.
LetZ∈Bc0L,rl1ρsX,Y; then slZ∈c0L,rl1ρ. One obtains
(91)supl∈ℕ0∑j=0lrj2ljsjZll+2−l+2≤supl∈ℕ0∑j=0lrj1ljsjZll+2−l+2<∞.
Then,Z∈Bc0L,rl2ρsX,Y. Next, if we choose slZl=0∞ with ∑j=0lrj1ljsjZ=ll+2−l+2 and ∑j=0lrj1ljsjZ=ll+2−l+2/l+1ll+2−l+2/l+1, one gets Z∈BX,Y such that Z∉Bc0L,rl1ρsX,Y and Z∈Bc0L,rl2ρsX,Y.
Clearly,Bc0L,rl2ρsX,Y⊂BX,Y. Next, if we put slZl=0∞ such that ∑j=0lrj2ljsjZ=ll+2−l+2. We have Z∈BX,Y such that Z∉Bc0L,rl2ρsX,Y. This explains the proof.Theorem 48.
SupposeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘orrklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are confirmed with 1<h1<h2 and 0<rl2≤rl1 for all l∈ℕ0; hence,
(92)Bℓh1L,rl1ρsX,Y⊂Bℓh2L,rl2ρsX,Y⊂BX,Y.Proof.
LetZ∈Bℓh1L,rl1ρsX,Y; then slZ∈ℓh1L,rl1ρ. One obtains
(93)∑l=0∞∑j=0lrj2ljsjZll+2−l+2h2<∑l=0∞∑j=0lrj1ljsjZll+2−l+2h1<∞.
ThenZ∈Bℓh2L,rl2ρsX,Y. Next, if we choose slZl=0∞ with ∑j=0lrj1ljsjZ=ll+2−l+2/l+1h1ll+2−l+2/l+1h1, one gets Z∈BX,Y such that
(94)∑l=0∞∑j=0lrj1ljsjZll+2−l+2h1=∑l=0∞1l+1=∞,∑l=0∞∑j=0lrj2ljsjZll+2−l+2h2≤∑l=0∞∑j=0lrj1ljsjZll+2−l+2h2=∑l=0∞1l+1h2h1<∞.
Therefore,Z∉Bℓh1L,rl1ρsX,Y and Z∈Bℓh2L,rl2ρsX,Y.
Clearly,Bℓh2L,rl2ρsX,Y⊂BX,Y. Next, if we put slZl=0∞ such that ∑j=0lrj2ljsjZ=ll+2−l+2/l+1h2. We have Z∈BX,Y such that Z∉Bℓh2L,rl2ρsX,Y. This explains the proof.Theorem 49.
LetX and Y be Banach spaces with dimX=dimY=∞, and the setups 1<p<∞,rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are established with ∑j=0lrjlj/ll+2−l+2l∈ℕ0∉ℓp; then, BℓpL,rρα is minimum.Proof.
Suppose the enough setups are confirmed; thenBℓpL,rρα,Ψ, where ΨZ=∑l=0∞∑j=0lrjljαjZ/ll+2−l+2p, is a prequasi Banach ideal. Suppose BℓpL,rραX,Y=BX,Y; hence, there is η>0 with ΨZ≤ηZ, for every Z∈BX,Y. According to Dvoretzky’s Theorem [23], for every b∈ℕ0, one obtains quotient spaces X/Yb and subspaces Mb of Y which can be mapped onto ℓ2b by isomorphisms Vb and Xb with VbVb−1≤2 and XbXb−1≤2. Let Ib be the identity operator on ℓ2b and Tb be the quotient operator from X onto X/Yb, and Jb is the natural embedding operator from Mb into Y. Suppose mz is the Bernstein numbers [26]; then
(95)1=mzIb=mzXbXb−1IbVbVb−1≤XbmzXb−1IbVbVb−1=XbmzJbXb−1IbVbVb−1≤XbdzJbXb−1IbVbVb−1=XbdzJbXb−1IbVbTbVb−1≤XbαzJbXb−1IbVbTbVb−1,for 0≤l≤b. We have
(96)∑j=0lrjljll+2−l+2≤∑z=0lXbrzlzαzJbXb−1IbVbTbVb−1ll+2−l+2⟹∑j=0lrjljll+2−l+2p≤XbVb−1p∑z=0lrzlzαzJbXb−1IbVbTbll+2−l+2p.
Hence, for someρ≥1, one gets
(97)∑l=0b∑j=0lrjljll+2−l+2p≤ρXbVb−1∑l=0b∑z=0lrzlzαzJbXb−1IbVbTbll+2−l+2p⟹∑l=0b∑j=0lrjljll+2−l+2p≤ρXbVb−1ΨJbXb−1IbVbTb⟹∑l=0b∑j=0lrjljll+2−l+2p≤ρηXbVb−1JbXb−1IbVbTb⟹∑l=0b∑j=0lrjljll+2−l+2p≤ρηXbVb−1JbXb−1IbVbTb=ρηXbVb−1Xb−1IbVb≤4ρη.
Therefore, we have a contradiction, ifb⟶∞. Then, X and Y both cannot be infinite dimensional if BℓpL,rραX,Y=BX,Y. This shows the proof.By the same manner, we can easily conclude the next theorem.Theorem 50.
LetX and Y be Banach spaces with dimX=dimY=∞, and the setups 1<p<∞,rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are established with ∑j=0lrjlj/ll+2−l+2l∈ℕ0∉ℓp; then BℓpL,rρd is minimum.


### 6.4. Simple Banach Prequasi Ideal

Theorem 51.
SupposeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are confirmed with 0<rl2≤rl1, for all l∈ℕ0; then
(98)BBc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y=ABc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y.Proof.
LetX∈BBc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y and X∉ABc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y. In view of Lemma 37, there are Y∈BBc0L,rl2ρsX,Y and Z∈BBc0L,rl1ρsX,Y with ZXYIb=Ib. Therefore, for every b∈ℕ0, we get
(99)IbBc0L,rl1ρsX,Y=supl∈ℕ0∑j=0lrj1ljsjIbll+2−l+2≤ZXYIbBc0L,rl2ρsX,Y≤supl∈ℕ0∑j=0lrj2ljsjIbll+2−l+2.
This contradicts Theorem47. Then X∈ABc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y, which finishes the proof.Corollary 52.
AssumeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are established with 0<rl2≤rl1, for all l∈ℕ0; then
(100)BBc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y=KBc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y.Proof.
Clearly,A⊂K.Theorem 53.
SupposeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are confirmed with 1<h1<h2and0<rl2≤rl1, for all l∈ℕ0; then
(101)BBℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y=ABℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y.Proof.
LetX∈BBℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y and X∉ABℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y. In view of Lemma 37, there are Y∈BBℓh2L,rl2ρsX,Y and Z∈BBℓh1L,rl1ρsX,Y with ZXYIb=Ib. Therefore, for every b∈ℕ0, we get
(102)IbBℓh1L,rl1ρsX,Y=∑l=0∞∑j=0lrj1ljsjIbll+2−l+2h1≤ZXYIbBℓh2L,rl2ρsX,Y≤∑l=0∞∑j=0lrj2ljsjIbll+2−l+2h2.
This contradicts Theorem48. Then X∈ABℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y, which finishes the proof.Corollary 54.
AssumeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are established with 1<h1<h2 and 0<rl2≤rl1, for all l∈ℕ0; then
(103)BBℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y=KBℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y.Proof.
Clearly,A⊂K.Theorem 55.
SupposeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are confirmed with 1<h<∞ and 0<rl2≤rl1, for all l∈ℕ0, then
(104)BBc0L,rl2ρsX,Y,BℓhL,rl1ρsX,Y=ABc0L,rl2ρsX,Y,BℓhL,rl1ρsX,Y.Proof.
LetX∈BBc0L,rl2ρsX,Y,BℓhL,rl1ρsX,Y and X∉ABc0L,rl2ρsX,Y,BℓhL,rl1ρsX,Y. In view of Lemma 37, there are Y∈BBc0L,rl2ρsX,Y and Z∈BBℓhL,rl1ρsX,Y with ZXYIb=Ib. Therefore, for every b∈ℕ0, we get
(105)IbBℓhL,rl1ρsX,Y=∑l=0∞∑j=0lrj1ljsjIbll+2−l+2h≤ZXYIbBc0L,rl2ρsX,Y≤supl∈ℕ0∑j=0lrj2ljsjIbll+2−l+2.
This contradictsℓhL,rl1⊂c0L,rl2. Then X∈ABc0L,rl2ρsX,Y,BℓhL,rl1ρsX,Y, which finishes the proof.Theorem 56.
LetX and Y be Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are satisfied; hence, Bc0L,rρs is simple.Proof.
Assume the closed idealKBc0L,rρsX,Y includes an operator X∉ABc0L,rρsX,Y. In view of Lemma 37, we have Y,Z∈BBc0L,rρsX,Y with ZXYIb=Ib. This gives that IBc0L,rρsX,Y∈KBc0L,rρsX,Y. Then, BBc0L,rρsX,Y=KBc0L,rρsX,Y. Hence, Bc0L,rρs is simple Banach space.Theorem 57.
LetX and Y be Banach spaces with dimX=dimY=∞, and the setups 1<p<∞,rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are satisfied; hence, BℓpL,rρs is simple.Proof.
Assume the closed idealKBℓpL,rρsX,Y includes an operator X∉ABℓpL,rρsX,Y. In view of Lemma 37, we have Y,Z∈BBℓpL,rρsX,Y with ZXYIb=Ib. This gives that IBℓpL,rρsX,Y∈KBℓpL,rρsX,Y. Then, BBℓpL,rρsX,Y=KBℓpL,rρsX,Y. Hence, BℓpL,rρs is simple Banach space.


### 6.5. Eigenvalues of S-Type Operators

Conventions 2.
Please see the following conventions:(106)BEsρ≔BEsρX,Y;XandYareBanachSpaces,whereBEsρX,Y≔X∈BX,Y:ρlXn=0∞∈EandX−ρlXIisnotinvertible,foralll∈ℕ0Theorem 58.
LetX and Y be Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are verified with infl∈ℕ0∑j=0l‍rjlj/ll+2−l+2>0; then Bc0L,rρsρX,Y=Bc0L,rρsX,Y.Proof.
LetX∈Bc0L,rρsρX,Y; hence, ρlXl=0∞∈c0L,rρ and X−ρlXI=0, for all l∈ℕ0. We have X=ρlXI, for all l∈ℕ0; hence, slX=slρlXI=ρlX, for every l∈ℕ0. Therefore, slXl=0∞∈c0L,rρ; then X∈Bc0L,rρsX,Y.
Secondly, supposeX∈Bc0L,rρsX,Y. Then slXl=0∞∈c0L,rρ. Hence, we have
(107)liml⟶∞∑j=0lrjljsjXll+2−l+2≥infl∈ℕ0∑j=0lrjljll+2−l+2liml⟶∞slX.
Therefore,liml⟶∞slX=0. Assume X−slXI−1 exists, for every l∈ℕ0. Hence, X−slXI−1 exists and bounded, for every l∈ℕ0. Then, liml⟶∞X−slXI−1=X−1 exists and bounded. As Bc0L,rρs,Ψ is a prequasi operator ideal, we get
(108)I=XX−1∈Bc0L,rρsX,Y⟹slIl=0∞∈c0L,rρ⟹liml⟶∞slI=0.
So we have a contradiction, sinceliml⟶∞slI=1. Hence, X−slXI=0, for every l∈ℕ0. This gives X∈Bc0L,rρsρX,Y. This shows the proof.Theorem 59.
LetX and Y be Banach spaces with dimX=dimY=∞, and the setups 1<p<∞,rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are satisfied with infl∑j=0lrjlj/ll+2−l+2p>0; then BℓpL,rρsρX,Y=BℓpL,rρsX,Y.Proof.
AssumeX∈BℓpL,rρsρX,Y; hence, ρlXl=0∞∈ℓpL,rρ and X−ρlXI=0, for all l∈ℕ0. We have X=ρlXI, for all l∈ℕ0; hence, slX=slρlXI=ρlX, for every l∈ℕ0. Therefore, slXl=0∞∈ℓpL,rρ; then X∈BℓpL,rρsX,Y.
Secondly, supposeX∈BℓpL,rρsX,Y. Then slXl=0∞∈ℓpL,rρ. Hence, we have
(109)∑l=0∞∑j=0lrjljsjXll+2−l+2p≥infl∑j=0lrjljll+2−l+2p∑l=0∞slXp.
Therefore,liml⟶∞slX=0. Assume X−slXI−1 exists, for every l∈ℕ0. Hence, X−slXI−1 exists and bounded, for every l∈ℕ0. Then, liml⟶∞X−slXI−1=X−1 exists and bounded. As BℓpL,rρs,Ψ is a prequasi operator ideal, we get
(110)I=XX−1∈BℓpL,rρsX,Y⟹slIl=0∞∈ℓpL,rρ⟹liml⟶∞slI=0.
So we have a contradiction, sinceliml⟶∞slI=1. Hence, X−slXI=0, for every l∈ℕ0. This gives X∈BℓpL,rρsρX,Y. This shows the proof.


## 6.1. Finite Rank Prequasi Ideal

Theorem 40.
Bc0L,rρsX,Y=FX,Y¯; suppose the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are satisfied. But the converse is not necessarily true.Proof.
To investigate thatFX,Y¯⊆Bc0L,rρsX,Y, as el∈c0L,rρ, for every l∈ℕ0,c0L,rρ is a linear space. Let Z∈FX,Y, one gets slZl=0∞∈C. To explain that Bc0L,rρsX,Y⊆FX,Y¯, assume Z∈Bc0L,rρsX,Y, we obtain slZl=0∞∈c0L,rρ. Since ρslZl=0∞<∞, let ρ∈0,1; hence, there is l0∈ℕ0−0 with ρslZl=l0∞<ρ/16d, for some d≥1. Since slZ∈I], we get
(72)supl=l0+12l0∑j=0lrjljs2l0Zll+2−l+2≤supl=l0+12l0∑j=0lrjljsjZll+2−l+2≤supl=l0∞∑j=0lrjljsjZll+2−l+2<ρ16d.
Hence, there isY∈F2l0X,Y so that rankY≤2l0 and
(73)supl=2l0+13l0∑j=0lrjljZ−Yll+2−l+2≤supl=l0+12l0∑j=0lrjljZ−Yll+2−l+2<ρ16d,we have
(74)∑j=0l0rjljZ−Y<ρ16.
Therefore, one has(75)supl=0l0∑j=0lrjljZ−Yll+2−l+2<ρ16d.
In view of inequalities (72)–(75), and rklkk=0∞∈I], one gets
(76)dZ,Y=ρslZ−Yl=0∞≤supl=03l0−1∑j=0lrjljsjZ−Yll+2−l+2+supl=3l0∞∑j=0lrjljsjZ−Yll+2−l+2≤supl=03l0∑j=0lrjljsjZ−Yll+2−l+2+supl=l0∞∑j=0l+2l0rjljsjZ−Yll+2l0+2−l+2l0+2≤supl=03l0∑j=0lrjljsjZ−Yll+2−l+2+supl=l0∞∑j=0l+2l0rjljsjZ−Yll+2−l+2≤3supl=0l0∑j=0lrjljZ−Yll+2−l+2+supl=l0∞∑j=02l0−1rjljsjZ−Y+∑j=2l0l+2l0rjljsjZ−Yll+2−l+2≤3supl=0l0∑j=0lrjljZ−Yll+2−l+2+supl=l0∞∑j=02l0−1rjljsjZ−Yll+2−l+2+supl=l0∞∑j=2l0l+2l0rjljsjZ−Yll+2−l+2≤3supl=0l0∑j=0lrjljZ−Yll+2−l+2+supl=l0∞∑j=02l0−1rjljsjZ−Yll+2−l+2+supl=l0∞∑j=0lrj+2l0lj+2l0sj+2l0Z−Yll+2−l+2≤3supl=0l0∑j=0lrjljZ−Yll+2−l+2+2∑j=0l0ljrjZ−Y+supl=l0∞∑j=0lrjljsjZll+2−l+2<ρ.
On the opposite side, one has a negative example asI3∈Bc0L,rρsX,Y, where r=0,0,0,1,0,1,0,. This shows the proof.Theorem 41.
BℓpL,rρsX,Y=FX,Y¯; suppose the setups 1<p<∞,rklkk=0∞∈I↘orrklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are confirmed. But the converse is not necessarily true.Proof.
To investigate thatFX,Y¯⊆BℓpL,rρsX,Y, as el∈ℓpL,rρ, for every l∈ℕ0, ℓpL,rρ is a linear space. Let Z∈FX,Y; one gets slZl=0∞∈C. To explain that BℓpL,rρsX,Y⊆FX,Y¯, assume Z∈BℓpL,rρsX,Y; we obtain slZl=0∞∈ℓpL,rρ. Since ρslZl=0∞<∞, let ρ∈0,1, hence, there is l0∈ℕ0−0 with ρslZl=l0∞<ρ/2p+3ηd, for some d≥1, where η=max1,∑l=l0∞1/ll+2−l+2p. Since slZ∈I], we get
(77)∑l=l0+12l0∑j=0lrjljs2l0Zll+2−l+2p≤∑l=l0+12l0∑j=0lrjljsjZll+2−l+2p≤∑l=l0∞∑j=0lrjljsjZll+2−l+2p<ρ2p+3ηd.
Hence, there isY∈F2l0X,Y so that rankY≤2l0 and
(78)∑l=2l0+13l0∑j=0lrjljZ−Yll+2−l+2p≤∑l=l0+12l0∑j=0lrjljZ−Yll+2−l+2p<ρ2p+3ηd.
Since1<p<∞, we have
(79)∑j=0l0rjljZ−Yp<ρ22p+2η.
Therefore, one has(80)∑l=0l0∑j=0lrjljZ−Yll+2−l+2p<ρ2p+3ηd.
In view of inequalities (54), (77)–(80), and rklkk=0∞∈I], one gets
(81)dZ,Y=ρslZ−Yl=0∞=∑l=03l0−1∑j=0lrjljsjZ−Yll+2−l+2p+∑l=3l0∞∑j=0lrjljsjZ−Yll+2−l+2p≤∑l=03l0∑j=0lrjljsjZ−Yll+2−l+2p+∑l=l0∞∑j=0l+2l0rjljsjZ−Yll+2l0+2−l+2l0+2p≤∑l=03l0∑j=0lrjljsjZ−Yll+2−l+2p+∑l=l0∞∑j=0l+2l0rjljsjZ−Yll+2−l+2p≤3∑l=0l0∑j=0lrjljZ−Yll+2−l+2p+∑l=l0∞∑j=02l0−1rjljsjZ−Y+∑j=2l0l+2l0rjljsjZ−Yll+2−l+2p≤3∑l=0l0∑j=0lrjljZ−Yll+2−l+2p+2p−1∑l=l0∞∑j=02l0−1rjljsjZ−Yll+2−l+2p+∑l=l0∞∑j=2l0l+2l0rjljsjZ−Yll+2−l+2p≤3∑l=0l0∑j=0lrjljZ−Yll+2−l+2p+2p−1∑l=l0∞∑j=02l0−1rjljsjZ−Yll+2−l+2p+∑l=l0∞∑j=0lrj+2l0lj+2l0sj+2l0Z−Yll+2−l+2p≤3∑l=0l0∑j=0lrjljZ−Yll+2−l+2p+2p∑j=0l0ljrjZ−Yp∑l=l0∞1ll+2−l+2p+2p−1∑l=l0∞∑j=0lrjljsjZll+2−l+2p<ρ.
On the opposite side, one has a negative example asI4∈Bℓ0.5L,rρsX,Y, where r=0,0,0,0,1,1,. This shows the proof.


## 6.2. Banach and Closed Prequasi Ideal

Theorem 42 (see [24]).
The functionΨ is a prequasi norm on BEρs, where ΨY=ρsbYb=0∞, for every Y∈BEρsX,Y, if Eρ is a premodular pss.Theorem 43.
If the setupsrklkk=0∞∈I↘orrklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk with r0>0 are satisfied, then Bc0L,rρs,Ψ is a prequasi Banach ideal, where ψX=ρslXl=0∞.Proof.
Asc0L,rρ is a premodular pss, hence from Theorem 42, Ψ is a prequasi norm on Bc0L,rρs. Suppose Xbb∈ℕ0 is a Cauchy sequence in Bc0L,rρsX,Y. As BX,Y⊇Bc0L,rρsX,Y, one obtains
(82)ΨXa−Xb=supl∈ℕ0∑j=0lrjljsjXa−Xbll+2−l+2≥r0Xa−Xb.
Hence,Xbb∈ℕ0 is a Cauchy sequence in BX,Y. Since BX,Y is a Banach space, then there is X∈BX,Y with limb⟶∞Xb−X=0. Since slXbl=0∞∈c0L,rρ, every b∈ℕ0. According to Definition 21 setups (ii), (iii), and (v), one gets
(83)ΨX=supl∈ℕ0∑j=0lrjljsjXll+2−l+2≤supl∈ℕ0∑j=0lrjljsj/2X−Xbll+2−l+2+supl∈ℕ0∑j=0lrjljsj/2Xbll+2−l+2≤supl∈ℕ0∑j=0lrjljX−Xbll+2−l+2+D0supl∈ℕ0∑j=0lrjljsjXbll+2−l+2<∞.
Therefore,slXl=0∞∈c0L,rρ; then X∈Bc0L,rρsX,Y.Theorem 44.
If the setups1<p<∞,rklkk=0∞∈I↘orrklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk with r0>0 are confirmed; then BℓpL,rρs,Ψ is a prequasi Banach ideal, where ψX=ρslXl=0∞.Proof.
AsℓpL,rρ is a premodular pss, hence from Theorem 42, Ψ is a prequasi norm on BℓpL,rρs. Suppose Xbb∈ℕ0 is a Cauchy sequence in BℓpL,rρsX,Y. As BX,Y⊇BℓpL,rρsX,Y, one obtains
(84)ΨXa−Xb=∑l=0∞∑j=0lrjljsjXa−Xbll+2−l+2p≥r0Xa−Xbp.
Hence,Xbb∈ℕ0 is a Cauchy sequence in BX,Y. Since BX,Y is a Banach space, then there is X∈BX,Y with limb⟶∞Xb−X=0.Since slXbl=0∞∈ℓpL,rρ, every b∈ℕ0. According to Definition 21 setups (ii), (iii), and (v), one gets
(85)ΨX=∑l=0∞∑j=0lrjljsjXll+2−l+2p≤2p−1∑l=0∞∑j=0lrjljsj/2X−Xbll+2−l+2p+2p−1∑l=0∞∑j=0lrjljsj/2Xbll+2−l+2p≤2p−1∑l=0∞∑j=0lrjljX−Xbll+2−l+2p+2p−1D0∑l=0∞∑j=0lrjljsjXbll+2−l+2p<∞.
Therefore,slXl=0∞∈ℓpL,rρ; then X∈BℓpL,rρsX,Y.Theorem 45.
AssumeX, Y are normed spaces; the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞; and there exists C≥1 such that r2k+1l2k+1≤Crklk with r0>0 are satisfied; then Bc0L,rρs,Ψ is a prequasi closed ideal, where ΨX=ρslXl=0∞.Proof.
Asc0L,rρ is a premodular pss, by using Theorem 42, Ψ is a prequasi norm on Bc0L,rρs. Assume Xb∈Bc0L,rρsX,Y, every b∈ℕ0 and limb⟶∞ΨXb−X=0. As BX,Y⊇Bc0L,rρsX,Y, we have
(86)ΨX−Xb=supl∈ℕ0∑j=0lrjljsjX−Xbll+2−l+2≥r0X−Xb.
Hence,Xbb∈ℕ0 is a convergent sequence in BX,Y. Since slXbl=0∞∈c0L,rρ, for every b∈ℕ0. In view of Definition 21 setups (ii), (iii), and (v), one has
(87)ΨX=supl∈ℕ0∑j=0lrjljsjXll+2−l+2≤supl∈ℕ0∑j=0lrjljsj/2X−Xbll+2−l+2+supl∈ℕ0∑j=0lrjljsj/2Xbll+2−l+2≤supl∈ℕ0∑j=0lrjljX−Xbll+2−l+2+D0supl∈ℕ0∑j=0lrjljsjXbll+2−l+2<∞.
We getslXl=0∞∈c0L,rρ, so X∈Bc0L,rρsX,Y.Theorem 46.
AssumeX, Y are normed spaces; the setups 1<p<∞,rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞; and there exists C≥1 such that r2k+1l2k+1≤Crklk with r0>0 are satisfied; hence, BℓpL,rρs,Ψ is a prequasi closed ideal, where ΨX=ρslXl=0∞.Proof.
AsℓpL,rρ is a premodular pss, by using Theorem 42, Ψ is a prequasi norm on BℓpL,rρs. Assume Xb∈BℓpL,rρsX,Y, for every b∈ℕ0 and limb⟶∞ΨXb−X=0. As BX,Y⊇BℓpL,rρsX,Y, we have
(88)ΨX−Xb=∑l=0∞∑j=0lrjljsjX−Xbll+2−l+2p≥r0X−Xbp.
Hence,Xbb∈ℕ0 is a convergent sequence in BX,Y. Since slXbl=0∞∈ℓpL,rρ, every b∈ℕ0. In view of Definition 21 setups (ii), (iii), and (v), one has
(89)ΨX=∑l=0∞∑j=0lrjljsjXll+2−l+2p≤2p−1∑l=0∞∑j=0lrjljsj/2X−Xbll+2−l+2p+2p−1∑l=0∞∑j=0lrjljsj/2Xbll+2−l+2p≤2p−1∑l=0∞∑j=0lrjljX−Xbll+2−l+2p+2p−1D0∑l=0∞∑j=0lrjljsjXbll+2−l+2p<∞.
We getslXl=0∞∈ℓpL,rρ, so X∈BℓpL,rρsX,Y.


## 6.3. Minimum Prequasi Ideal

Theorem 47.
SupposeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I] or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are confirmed with 0<rl2≤rl1, for all l∈ℕ0, hence
(90)Bc0L,rl1ρsX,Y⊂Bc0L,rl2ρsX,Y⊂BX,Y.Proof.
LetZ∈Bc0L,rl1ρsX,Y; then slZ∈c0L,rl1ρ. One obtains
(91)supl∈ℕ0∑j=0lrj2ljsjZll+2−l+2≤supl∈ℕ0∑j=0lrj1ljsjZll+2−l+2<∞.
Then,Z∈Bc0L,rl2ρsX,Y. Next, if we choose slZl=0∞ with ∑j=0lrj1ljsjZ=ll+2−l+2 and ∑j=0lrj1ljsjZ=ll+2−l+2/l+1ll+2−l+2/l+1, one gets Z∈BX,Y such that Z∉Bc0L,rl1ρsX,Y and Z∈Bc0L,rl2ρsX,Y.
Clearly,Bc0L,rl2ρsX,Y⊂BX,Y. Next, if we put slZl=0∞ such that ∑j=0lrj2ljsjZ=ll+2−l+2. We have Z∈BX,Y such that Z∉Bc0L,rl2ρsX,Y. This explains the proof.Theorem 48.
SupposeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘orrklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are confirmed with 1<h1<h2 and 0<rl2≤rl1 for all l∈ℕ0; hence,
(92)Bℓh1L,rl1ρsX,Y⊂Bℓh2L,rl2ρsX,Y⊂BX,Y.Proof.
LetZ∈Bℓh1L,rl1ρsX,Y; then slZ∈ℓh1L,rl1ρ. One obtains
(93)∑l=0∞∑j=0lrj2ljsjZll+2−l+2h2<∑l=0∞∑j=0lrj1ljsjZll+2−l+2h1<∞.
ThenZ∈Bℓh2L,rl2ρsX,Y. Next, if we choose slZl=0∞ with ∑j=0lrj1ljsjZ=ll+2−l+2/l+1h1ll+2−l+2/l+1h1, one gets Z∈BX,Y such that
(94)∑l=0∞∑j=0lrj1ljsjZll+2−l+2h1=∑l=0∞1l+1=∞,∑l=0∞∑j=0lrj2ljsjZll+2−l+2h2≤∑l=0∞∑j=0lrj1ljsjZll+2−l+2h2=∑l=0∞1l+1h2h1<∞.
Therefore,Z∉Bℓh1L,rl1ρsX,Y and Z∈Bℓh2L,rl2ρsX,Y.
Clearly,Bℓh2L,rl2ρsX,Y⊂BX,Y. Next, if we put slZl=0∞ such that ∑j=0lrj2ljsjZ=ll+2−l+2/l+1h2. We have Z∈BX,Y such that Z∉Bℓh2L,rl2ρsX,Y. This explains the proof.Theorem 49.
LetX and Y be Banach spaces with dimX=dimY=∞, and the setups 1<p<∞,rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are established with ∑j=0lrjlj/ll+2−l+2l∈ℕ0∉ℓp; then, BℓpL,rρα is minimum.Proof.
Suppose the enough setups are confirmed; thenBℓpL,rρα,Ψ, where ΨZ=∑l=0∞∑j=0lrjljαjZ/ll+2−l+2p, is a prequasi Banach ideal. Suppose BℓpL,rραX,Y=BX,Y; hence, there is η>0 with ΨZ≤ηZ, for every Z∈BX,Y. According to Dvoretzky’s Theorem [23], for every b∈ℕ0, one obtains quotient spaces X/Yb and subspaces Mb of Y which can be mapped onto ℓ2b by isomorphisms Vb and Xb with VbVb−1≤2 and XbXb−1≤2. Let Ib be the identity operator on ℓ2b and Tb be the quotient operator from X onto X/Yb, and Jb is the natural embedding operator from Mb into Y. Suppose mz is the Bernstein numbers [26]; then
(95)1=mzIb=mzXbXb−1IbVbVb−1≤XbmzXb−1IbVbVb−1=XbmzJbXb−1IbVbVb−1≤XbdzJbXb−1IbVbVb−1=XbdzJbXb−1IbVbTbVb−1≤XbαzJbXb−1IbVbTbVb−1,for 0≤l≤b. We have
(96)∑j=0lrjljll+2−l+2≤∑z=0lXbrzlzαzJbXb−1IbVbTbVb−1ll+2−l+2⟹∑j=0lrjljll+2−l+2p≤XbVb−1p∑z=0lrzlzαzJbXb−1IbVbTbll+2−l+2p.
Hence, for someρ≥1, one gets
(97)∑l=0b∑j=0lrjljll+2−l+2p≤ρXbVb−1∑l=0b∑z=0lrzlzαzJbXb−1IbVbTbll+2−l+2p⟹∑l=0b∑j=0lrjljll+2−l+2p≤ρXbVb−1ΨJbXb−1IbVbTb⟹∑l=0b∑j=0lrjljll+2−l+2p≤ρηXbVb−1JbXb−1IbVbTb⟹∑l=0b∑j=0lrjljll+2−l+2p≤ρηXbVb−1JbXb−1IbVbTb=ρηXbVb−1Xb−1IbVb≤4ρη.
Therefore, we have a contradiction, ifb⟶∞. Then, X and Y both cannot be infinite dimensional if BℓpL,rραX,Y=BX,Y. This shows the proof.By the same manner, we can easily conclude the next theorem.Theorem 50.
LetX and Y be Banach spaces with dimX=dimY=∞, and the setups 1<p<∞,rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are established with ∑j=0lrjlj/ll+2−l+2l∈ℕ0∉ℓp; then BℓpL,rρd is minimum.


## 6.4. Simple Banach Prequasi Ideal

Theorem 51.
SupposeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are confirmed with 0<rl2≤rl1, for all l∈ℕ0; then
(98)BBc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y=ABc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y.Proof.
LetX∈BBc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y and X∉ABc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y. In view of Lemma 37, there are Y∈BBc0L,rl2ρsX,Y and Z∈BBc0L,rl1ρsX,Y with ZXYIb=Ib. Therefore, for every b∈ℕ0, we get
(99)IbBc0L,rl1ρsX,Y=supl∈ℕ0∑j=0lrj1ljsjIbll+2−l+2≤ZXYIbBc0L,rl2ρsX,Y≤supl∈ℕ0∑j=0lrj2ljsjIbll+2−l+2.
This contradicts Theorem47. Then X∈ABc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y, which finishes the proof.Corollary 52.
AssumeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are established with 0<rl2≤rl1, for all l∈ℕ0; then
(100)BBc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y=KBc0L,rl2ρsX,Y,Bc0L,rl1ρsX,Y.Proof.
Clearly,A⊂K.Theorem 53.
SupposeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are confirmed with 1<h1<h2and0<rl2≤rl1, for all l∈ℕ0; then
(101)BBℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y=ABℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y.Proof.
LetX∈BBℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y and X∉ABℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y. In view of Lemma 37, there are Y∈BBℓh2L,rl2ρsX,Y and Z∈BBℓh1L,rl1ρsX,Y with ZXYIb=Ib. Therefore, for every b∈ℕ0, we get
(102)IbBℓh1L,rl1ρsX,Y=∑l=0∞∑j=0lrj1ljsjIbll+2−l+2h1≤ZXYIbBℓh2L,rl2ρsX,Y≤∑l=0∞∑j=0lrj2ljsjIbll+2−l+2h2.
This contradicts Theorem48. Then X∈ABℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y, which finishes the proof.Corollary 54.
AssumeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are established with 1<h1<h2 and 0<rl2≤rl1, for all l∈ℕ0; then
(103)BBℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y=KBℓh2L,rl2ρsX,Y,Bℓh1L,rl1ρsX,Y.Proof.
Clearly,A⊂K.Theorem 55.
SupposeX and Y are Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are confirmed with 1<h<∞ and 0<rl2≤rl1, for all l∈ℕ0, then
(104)BBc0L,rl2ρsX,Y,BℓhL,rl1ρsX,Y=ABc0L,rl2ρsX,Y,BℓhL,rl1ρsX,Y.Proof.
LetX∈BBc0L,rl2ρsX,Y,BℓhL,rl1ρsX,Y and X∉ABc0L,rl2ρsX,Y,BℓhL,rl1ρsX,Y. In view of Lemma 37, there are Y∈BBc0L,rl2ρsX,Y and Z∈BBℓhL,rl1ρsX,Y with ZXYIb=Ib. Therefore, for every b∈ℕ0, we get
(105)IbBℓhL,rl1ρsX,Y=∑l=0∞∑j=0lrj1ljsjIbll+2−l+2h≤ZXYIbBc0L,rl2ρsX,Y≤supl∈ℕ0∑j=0lrj2ljsjIbll+2−l+2.
This contradictsℓhL,rl1⊂c0L,rl2. Then X∈ABc0L,rl2ρsX,Y,BℓhL,rl1ρsX,Y, which finishes the proof.Theorem 56.
LetX and Y be Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are satisfied; hence, Bc0L,rρs is simple.Proof.
Assume the closed idealKBc0L,rρsX,Y includes an operator X∉ABc0L,rρsX,Y. In view of Lemma 37, we have Y,Z∈BBc0L,rρsX,Y with ZXYIb=Ib. This gives that IBc0L,rρsX,Y∈KBc0L,rρsX,Y. Then, BBc0L,rρsX,Y=KBc0L,rρsX,Y. Hence, Bc0L,rρs is simple Banach space.Theorem 57.
LetX and Y be Banach spaces with dimX=dimY=∞, and the setups 1<p<∞,rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are satisfied; hence, BℓpL,rρs is simple.Proof.
Assume the closed idealKBℓpL,rρsX,Y includes an operator X∉ABℓpL,rρsX,Y. In view of Lemma 37, we have Y,Z∈BBℓpL,rρsX,Y with ZXYIb=Ib. This gives that IBℓpL,rρsX,Y∈KBℓpL,rρsX,Y. Then, BBℓpL,rρsX,Y=KBℓpL,rρsX,Y. Hence, BℓpL,rρs is simple Banach space.


## 6.5. Eigenvalues of S-Type Operators

Conventions 2.
Please see the following conventions:(106)BEsρ≔BEsρX,Y;XandYareBanachSpaces,whereBEsρX,Y≔X∈BX,Y:ρlXn=0∞∈EandX−ρlXIisnotinvertible,foralll∈ℕ0Theorem 58.
LetX and Y be Banach spaces with dimX=dimY=∞, and the setups rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are verified with infl∈ℕ0∑j=0l‍rjlj/ll+2−l+2>0; then Bc0L,rρsρX,Y=Bc0L,rρsX,Y.Proof.
LetX∈Bc0L,rρsρX,Y; hence, ρlXl=0∞∈c0L,rρ and X−ρlXI=0, for all l∈ℕ0. We have X=ρlXI, for all l∈ℕ0; hence, slX=slρlXI=ρlX, for every l∈ℕ0. Therefore, slXl=0∞∈c0L,rρ; then X∈Bc0L,rρsX,Y.
Secondly, supposeX∈Bc0L,rρsX,Y. Then slXl=0∞∈c0L,rρ. Hence, we have
(107)liml⟶∞∑j=0lrjljsjXll+2−l+2≥infl∈ℕ0∑j=0lrjljll+2−l+2liml⟶∞slX.
Therefore,liml⟶∞slX=0. Assume X−slXI−1 exists, for every l∈ℕ0. Hence, X−slXI−1 exists and bounded, for every l∈ℕ0. Then, liml⟶∞X−slXI−1=X−1 exists and bounded. As Bc0L,rρs,Ψ is a prequasi operator ideal, we get
(108)I=XX−1∈Bc0L,rρsX,Y⟹slIl=0∞∈c0L,rρ⟹liml⟶∞slI=0.
So we have a contradiction, sinceliml⟶∞slI=1. Hence, X−slXI=0, for every l∈ℕ0. This gives X∈Bc0L,rρsρX,Y. This shows the proof.Theorem 59.
LetX and Y be Banach spaces with dimX=dimY=∞, and the setups 1<p<∞,rklkk=0∞∈I↘ or rklkk=0∞∈IZ∩ℓ∞, and there exists C≥1 such that r2k+1l2k+1≤Crklk are satisfied with infl∑j=0lrjlj/ll+2−l+2p>0; then BℓpL,rρsρX,Y=BℓpL,rρsX,Y.Proof.
AssumeX∈BℓpL,rρsρX,Y; hence, ρlXl=0∞∈ℓpL,rρ and X−ρlXI=0, for all l∈ℕ0. We have X=ρlXI, for all l∈ℕ0; hence, slX=slρlXI=ρlX, for every l∈ℕ0. Therefore, slXl=0∞∈ℓpL,rρ; then X∈BℓpL,rρsX,Y.
Secondly, supposeX∈BℓpL,rρsX,Y. Then slXl=0∞∈ℓpL,rρ. Hence, we have
(109)∑l=0∞∑j=0lrjljsjXll+2−l+2p≥infl∑j=0lrjljll+2−l+2p∑l=0∞slXp.
Therefore,liml⟶∞slX=0. Assume X−slXI−1 exists, for every l∈ℕ0. Hence, X−slXI−1 exists and bounded, for every l∈ℕ0. Then, liml⟶∞X−slXI−1=X−1 exists and bounded. As BℓpL,rρs,Ψ is a prequasi operator ideal, we get
(110)I=XX−1∈BℓpL,rρsX,Y⟹slIl=0∞∈ℓpL,rρ⟹liml⟶∞slI=0.
So we have a contradiction, sinceliml⟶∞slI=1. Hence, X−slXI=0, for every l∈ℕ0. This gives X∈BℓpL,rρsρX,Y. This shows the proof.



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