# Eigenvalues of s-Type Operators on Prequasi NormedCt,p

Authors: Awad A. Bakery; OM Kalthum S. K. Mohamed
Journal: Journal of Function Spaces
 (2020)

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## Abstract

We investigate some new topological properties of the multiplication operator onCt,p 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 s-numbers.


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## Body

## 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 ℂℕ, ℓ∞, ℓr, and c0, we denote the spaces of every, bounded, r-absolutely summable, and convergent to zero sequences of complex numbers. ℕ indicates the set of nonnegative integers. For a sequence pr with infrpr>0, Bilgin [12] defined and studied the sequence space
(1)Ct,p=x=xn∈ℂℕ:hβx<∞forsomeβ>0,where hx=∑n=0∞∑k=2n−12n+1−2k+1−t∣xk∣/2npnandt≥0.The spaceCt,p is a Banach space with the Luxemburg norm
(2)x=infβ>0:hxβ≤1.Ifpn∈ℓ∞, we have
(3)Ct,p=xi∈ℂℕ:∑n=0∞∑k=2n−12n+1−2k+1−t∣xk∣2npn<∞.Here and after, we denotep=pn, if t=0, then Ct,p=cesp, where infnpn>0 was defined and investigated by Lim [13]. If t=0 and pn=p, for all n∈ℕ, then Ct,p=cesp, Lim [14] defined and determined its dual spaces and characterized some matrix classes. For any pair of Banach spacesT and U, the following notations will be used throughout the article: BT,U (the space of all bounded linear operators from T into U), and ifT=U, we write BT, srAr=0∞ [15] (thes-number sequence of operator A∈BT,U), αrAr=0∞ (the approximation number sequence of operator A∈BT,U), FT (the space of all finite rank operators on T), KT (the space of all compact operators on T), and AT (the space of all approximable operators on T). For any sequence space E, we will use the notations [16]
(4)Es≔EsT,U,whereEsT,U≔A∈BT,U:siAi=0∞∈E,Eα≔EαT,U,whereEαT,U≔A∈BT,U:αiAi=0∞∈E,Eλ≔EλT,U,whereEλT,U≔A∈BT,U:λnAn=0∞∈EandA−λrAI=0,forallr∈ℕ.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 operatorsK formed by drV and c0. Pietsch [17] studied the quasi-ideals ℓrα for r∈0,∞, the ideals of Hilbert Schmidt operators between Hilbert spaces constructed by ℓ2, and the ideals of nuclear operators generated by ℓ1. He examined that F¯=ℓrα for r∈1,∞, where F¯ is the closed class of all finite rank operators, and the class ℓrα is a simple Banach and small [18]. The strict inclusion ℓrαT,U⊂≠ℓjαT,U⊂≠BT,U, whenever j>r>0, T and U 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 ℓMs and cesrs. On sequence spaces, Mursaleen and Noman [20, 21] investigated the compact operators on some difference sequence spaces. The multiplication operators on cesr,∙ with the Luxemburg norm ∙ investigated by Komal et al. [22]. The point of this article is to explain some results of Ct,ph equipped with a prequasi norm h. Firstly, we give the conditions on Ct,ph to be a Banach space. Secondly, we investigate the multiplication operator defined on Ct,ph. Finally, some geometric and topological structures of Ct,phs have been studied, such as closed, small, simple Banach, and Ct,phs=Ct,phλ. A strict inclusion relation of Ct,phs has been studied for different p and t.


## 2. Definitions and Preliminaries

We will useer=0,0,⋯,1,0,0,⋯, where 1 shows at the rth place, for all r∈ℕ.Lemma 1 [17].
IfE∈BT,U and E∉AT,U, then there are G∈BT and D∈BU such that DEGer=er, for all r∈ℕ.Definition 2 [17].
A Banach spaceT is called simple if the algebra BT has one and only one nontrivial closed ideal.Theorem 3 [17].
IfT is a Banach space with dimT=∞, then
(5)FT⊂≠AT⊂≠KT⊂≠BT.Definition 4 [23].
An operatorA∈BT is called Fredholm if dimRAc<∞, dimkerA<∞, and RA¯=RA, where RAc is the complement of range A.Definition 5 [24].
A class of linear sequence spacesT is called a special space of sequences (sss) if
(1)
er∈T, for every r∈ℕ(2)
ift=tr∈ℂℕ, z=zr∈T, and ∣tr∣≤∣zr∣, for all r∈ℕ, then t∈T, i.e., “T is solid”(3)
iftrr=0∞∈T, then tr/2r=0∞∈T, where r/2 denotes the integral part of r/2Definition 6 [24].
A subclassTh is called a premodular (sss) if there is a function h:T⟶0,∞ verifying the conditions
(i)
ht≥0 for each t∈T and ht=0⇔t=θ, where θ is the zero vector of T(ii)
there isd≥1 such that hηt≤d∣η∣ht, for each t∈T and η∈ℂ(iii)
there isM≥1, ht+z≤Mht+hz, for all t,z∈T(iv)
if∣tr∣≤∣zr∣, for every r∈ℕ, then htr≤hzr(v)
there isM0≥1, htr≤htr/2≤M0htr(vi)
for allt=trr=0∞∈T, and ε>0 there is r0∈ℕ such that htrr=r0∞<ε(vii)
there isα>0 such that hη,0,0,0,⋯≥α∣η∣h1,0,0,0,⋯, for all η∈ℂDefinition 7 [24].
LetT be a (sss). If there is a function h:T⟶0,∞ verifying the parts (i), (ii), and (iii) of Definition 6. The space T with h is called prequasi normed (sss) and denoted by Th, which gives a class more general than the quasi normed (sss). If the space T is complete with h, then Th, 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 functionV:Ψ⟶0,∞ is called a prequasi norm on the class Ψ if it satisfies
(1)
ifA∈ΨT,U, then VA≥0 and VA=0⇔A=0(2)
there isd≥1 such that VηA≤d∣η∣VA, for every A∈ΨT,U and η∈ℂ(3)
there isM≥1 such that VA1+A2≤MVA1+VA2, for every A1,A2∈ΨT,U(4)
there isν≥1 such that if X∈BT0,T, Y∈ΨT,U, and Z∈BU,U0, then VZYX≤νZVYX, where T0 and U0 are normed spacesTheorem 11 [25].
IfEh is a premodular (sss), then FT,U¯=EhsT,U, where VA=hsrAr=0∞.Theorem 12 [16].
IfEh is a premodular (sss), then VB=hsrBr=0∞ is a prequasi norm on Ehs.The following inequality [26] will be used in the sequel: ar+brpr≤Harpr+brpr, where H=max1,2suprpr−1, 0≤pr≤suprpr<∞, and ar,br∈ℂ for each r∈ℕ.


## 3. Main Results

### 3.1. Premodular Banach (sss)

The conditions ont and p are such that the space Ct,ph is a premodular Banach (sss), where hz=∑r=0∞∑k=2r−12r+1−2k+1−t∣zk∣/2rpr, for all z∈Ct,p.Theorem 13.
Letpr∈ℓ∞ be an increasing with p0>0 and t≥0, then the space Ct,ph is a premodular Banach (sss).Proof.
(1-i) Assumez,w∈Ct,p. From pr∈ℓ∞, we get
(6)hz+w=∑r=0∞∑k=2r−12r+1−2k+1−t∣zk+wk∣2rpr≤H∑r=0∞∑k=2r−12r+1−2k+1−t∣zk∣2rpr+∑r=0∞∑k=2r−12r+1−2k+1−t∣wk∣2rpr=Hhz+hw<∞,hence z+w∈Ct,p.
(1-ii) Ifη∈ℂ, z∈Ct,p, and from pr∈ℓ∞, one has
(7)hηz=∑r=0∞∑k=2r−12r+1−2k+1−t∣ηzk∣2rpr≤suprηpr∑r=0∞∑k=2r−12r+1−2k+1−t∣zk∣2rpr=suprηprhz<∞.
Therefore,ηz∈Ct,p, from (1-i) and (1-ii), the space Ct,p is linear. Since pr∈ℓ∞ with p0>0, we obtain
(8)hem=∑r=0∞∑k=2r−12r+1−2k+1−temk2rpr=m+1−t2r0pr0<∞,where r0∈ℕ be such that 2r0−1≤m≤2r0+1−2. Hence em∈Ct,p, for all m∈ℕ.
(2) Let∣zr∣≤∣wr∣, for all r∈ℕ and w∈Ct,p. Hence,
(9)hz=∑r=0∞∑k=2r−12r+1−2k+1−t∣zk∣2rpr≤∑r=0∞∑k=2r−12r+1−2k+1−t∣wk∣2rpr=hw<∞,hence, z∈Ct,p.
(3) Letzr∈Ct,p, we have
(10)hzr/2=∑r=0∞∑k=2r−12r+1−2k/2+1−t∣zk/2∣2rpr=∑r=0∞∑k=22r−122r+1−2k/2+1−t∣zk/2∣22rp2r+∑r=0∞∑k=22r+1−122r+2−2k/2+1−t∣zk/2∣22r+1p2r+1≤∑r=0∞∑k=22r−222r+1−2k/2+1−t∣zk/2∣22rpr+∑r=0∞∑k=22r+1−222r+2−2k/2+1−t∣zk/2∣22r+1pr≤∑r=0∞∑k=2r−12r+1−232rk+1−t∣zk∣22rpr+∑r=0∞∑k=2r−12r+1−232rk+1−t∣zk∣22r+1pr≤23suppr∑r=0∞∑k=2r−12r+1−2k+1−t∣zk∣2rpr,then zr/2∈Ct,p.
(i)
Evidently,hz≥0 and hz=0⇔z=θ(ii)
There isd=max1,suprηpr−1≥1 with hηz≤d∣η∣hz, for each z∈Ct,p and η∈ℂ(iii)
From condition (1), there isM≥1 with hz+w≤Mhz+hw for each z,w∈Ct,p(iv)
From condition (2), we haveCt,ph which is solid(v)
From condition (3), we findM0≥23suppr≥1(vi)
Clearly,F¯=Ct,p(vii)
There is0<α≤ηp0−1 with hη,0,0,0,⋯≥α∣η∣h1,0,0,0,⋯, for each η≠0 and α>0, when η=0
Therefore, the spaceCt,ph is premodular (sss). To prove that Ct,ph is a premodular Banach (sss), suppose zr=ziri=0∞ is a Cauchy sequence in Ct,ph; hence, for every ε∈0,1, there is r0∈ℕ such that for every r,v≥r0, one has
(11)hzr−zv=∑i=0∞∑j=2i−12i+1−2j+1−t∣zjr−zjv∣2ipi<εsupipi.
Hence, forr,v≥r0 and i∈ℕ, we get ∣zir−ziv∣<ε. So ziv is a Cauchy sequence in ℂ for fixed i∈ℕ; this gives limv→∞ziv=zi0 for fixed i∈ℕ. Hence, hzr−z0<εsupipi, for all r≥r0. Finally, to prove that z0∈Ct,p, we have
(12)hz0=hz0−zr+zr≤Hhzr−z0+hzn<∞,so z0∈Ct,p. This means that Ct,ph is a premodular Banach (sss).Corollary 14.
If0<p<∞, then cesph is a premodular Banach (sss), where hz=∑i=0∞∑j=2i−12i+1−2∣zj∣/2ip for every z∈cesp.


## 3.1. Premodular Banach (sss)

The conditions ont and p are such that the space Ct,ph is a premodular Banach (sss), where hz=∑r=0∞∑k=2r−12r+1−2k+1−t∣zk∣/2rpr, for all z∈Ct,p.Theorem 13.
Letpr∈ℓ∞ be an increasing with p0>0 and t≥0, then the space Ct,ph is a premodular Banach (sss).Proof.
(1-i) Assumez,w∈Ct,p. From pr∈ℓ∞, we get
(6)hz+w=∑r=0∞∑k=2r−12r+1−2k+1−t∣zk+wk∣2rpr≤H∑r=0∞∑k=2r−12r+1−2k+1−t∣zk∣2rpr+∑r=0∞∑k=2r−12r+1−2k+1−t∣wk∣2rpr=Hhz+hw<∞,hence z+w∈Ct,p.
(1-ii) Ifη∈ℂ, z∈Ct,p, and from pr∈ℓ∞, one has
(7)hηz=∑r=0∞∑k=2r−12r+1−2k+1−t∣ηzk∣2rpr≤suprηpr∑r=0∞∑k=2r−12r+1−2k+1−t∣zk∣2rpr=suprηprhz<∞.
Therefore,ηz∈Ct,p, from (1-i) and (1-ii), the space Ct,p is linear. Since pr∈ℓ∞ with p0>0, we obtain
(8)hem=∑r=0∞∑k=2r−12r+1−2k+1−temk2rpr=m+1−t2r0pr0<∞,where r0∈ℕ be such that 2r0−1≤m≤2r0+1−2. Hence em∈Ct,p, for all m∈ℕ.
(2) Let∣zr∣≤∣wr∣, for all r∈ℕ and w∈Ct,p. Hence,
(9)hz=∑r=0∞∑k=2r−12r+1−2k+1−t∣zk∣2rpr≤∑r=0∞∑k=2r−12r+1−2k+1−t∣wk∣2rpr=hw<∞,hence, z∈Ct,p.
(3) Letzr∈Ct,p, we have
(10)hzr/2=∑r=0∞∑k=2r−12r+1−2k/2+1−t∣zk/2∣2rpr=∑r=0∞∑k=22r−122r+1−2k/2+1−t∣zk/2∣22rp2r+∑r=0∞∑k=22r+1−122r+2−2k/2+1−t∣zk/2∣22r+1p2r+1≤∑r=0∞∑k=22r−222r+1−2k/2+1−t∣zk/2∣22rpr+∑r=0∞∑k=22r+1−222r+2−2k/2+1−t∣zk/2∣22r+1pr≤∑r=0∞∑k=2r−12r+1−232rk+1−t∣zk∣22rpr+∑r=0∞∑k=2r−12r+1−232rk+1−t∣zk∣22r+1pr≤23suppr∑r=0∞∑k=2r−12r+1−2k+1−t∣zk∣2rpr,then zr/2∈Ct,p.
(i)
Evidently,hz≥0 and hz=0⇔z=θ(ii)
There isd=max1,suprηpr−1≥1 with hηz≤d∣η∣hz, for each z∈Ct,p and η∈ℂ(iii)
From condition (1), there isM≥1 with hz+w≤Mhz+hw for each z,w∈Ct,p(iv)
From condition (2), we haveCt,ph which is solid(v)
From condition (3), we findM0≥23suppr≥1(vi)
Clearly,F¯=Ct,p(vii)
There is0<α≤ηp0−1 with hη,0,0,0,⋯≥α∣η∣h1,0,0,0,⋯, for each η≠0 and α>0, when η=0
Therefore, the spaceCt,ph is premodular (sss). To prove that Ct,ph is a premodular Banach (sss), suppose zr=ziri=0∞ is a Cauchy sequence in Ct,ph; hence, for every ε∈0,1, there is r0∈ℕ such that for every r,v≥r0, one has
(11)hzr−zv=∑i=0∞∑j=2i−12i+1−2j+1−t∣zjr−zjv∣2ipi<εsupipi.
Hence, forr,v≥r0 and i∈ℕ, we get ∣zir−ziv∣<ε. So ziv is a Cauchy sequence in ℂ for fixed i∈ℕ; this gives limv→∞ziv=zi0 for fixed i∈ℕ. Hence, hzr−z0<εsupipi, for all r≥r0. Finally, to prove that z0∈Ct,p, we have
(12)hz0=hz0−zr+zr≤Hhzr−z0+hzn<∞,so z0∈Ct,p. This means that Ct,ph is a premodular Banach (sss).Corollary 14.
If0<p<∞, then cesph is a premodular Banach (sss), where hz=∑i=0∞∑j=2i−12i+1−2∣zj∣/2ip for every z∈cesp.


## 4. Some Properties of Multiplication Operators onCt,ph

We investigate the necessity and sufficient conditions ont and p such that the multiplication operator defined on the prequasi normed (sss), Ct,ph, becomes an element of BCt,ph, ACt,ph, KCt,ph, closed range, and Fredholm operator.Definition 15.
Pick upζ∈ℂℕ∩ℓ∞, and Th is a prequasi normed (sss). An operator Aζ:Th⟶Th is called a multiplication operator, if Aζz=ζz=ζkzkk=0∞, for all z∈T. It is called a multiplication operator generated by ζ, if Aζ∈BTh is a multiplication operator.Theorem 16.
Ifζ∈ℂℕ, pn∈ℓ∞ is increasing with p0>0 and t≥0, then
(13)ζ∈ℓ∞⇔Aζ∈BCt,ph.Proof.
Letζ∈ℓ∞. Therefore, there is D>0 such that ∣ζr∣≤D, for every r∈ℕ. Since pr∈ℓ∞ and for z∈Ct,ph, we have
(14)hAζz=hζz=∑i=0∞∑j=2i−12i+1−2j+1−tζjzj2ipi≤∑i=0∞∑j=2i−12i+1−2j+1−tD∣zj∣2ipi≤supiDpihz,this implies that Aζ∈BCt,ph. Conversely, let Aζ∈BCt,ph, to show that ζ∈ℓ∞. For, if ζ∉ℓ∞, then for all r∈ℕ, there are ir∈ℕ such that ζir>r. Hence,
(15)hAζeir=hζeir=∑i=0∞∑j=2i−12i+1−2j+1−t∣ζj∣eirj2ipi=ir+1−t∣ζir∣2i0pi0>ir+1−tr2i0pi0=rpi0heir,where i0∈ℕ be such that 2i0−1≤ir≤2i0+1−2. This shows that Aζ∉BCt,ph. So, ζ∈ℓ∞.Theorem 17.
Assumeζ∈ℂℕ and Ct,ph is a prequasi normed (sss). ∣ζr∣=1, for every r∈ℕ if and only if Aζ is an isometry.Proof.
If∣ζr∣=1, for every r∈ℕ. Therefore, we have
(16)hAζz=∑i=0∞∑j=2i−12i+1−2j+1−tζjzj2ipi=∑i=0∞∑j=2i−12i+1−2j+1−t∣zj∣2ipi=hz,for all z∈Ct,ph. Hence, Aζ is an isometry. Conversely, let Aζ be an isometry and ∣ζr∣<1, for all r∈ℕ. Then,
(17)hAζz=∑i=0∞∑j=2i−12i+1−2j+1−tζjzj2ipi<∑i=0∞∑j=2i−12i+1−2j+1−t∣zj∣2ipi=hz.
Also if∣ζr∣>1, then we get hζz>hz. We have a contradiction in the two cases. Therefore, ∣ζr∣=1, for every r∈ℕ.ByND, we indicate the cardinality of the set D.Theorem 18.
Forζ∈ℂℕ, pr∈ℓ∞ is increasing with p0>0, t≥0, and Aζ∈BCt,ph. Then,
(18)Aζ∈ACt,ph⇔ζrr=0∞∈c0.Proof.
LetAζ∈ACt,ph; hence Aζ∈KCt,ph. To show that ζrr=0∞∈c0. Assume ζrr=0∞∉c0. Therefore, the set Bδ=r∈ℕ:∣ζr∣≥δ has NBδ=∞, for δ>0. Suppose dr∈Bδ, for all r∈ℕ. Therefore, E=edr:dr∈Bδ⊆ℓ∞∩Ct,ph with NE=∞. We have
(19)hAζedr−Aζedm=hζedr−ζedm=∑i=0∞∑j=2i−12i+1−2j+1−t∣ζjedrj−edmj∣2ipi≥∑i=0∞∑j=2i−12i+1−2j+1−t∣δedrj−edmj∣2ipi≥infiδpi∑i=0∞∑j=2i−12i+1−2j+1−t∣edrj−edmj∣2ipi=infiδpihedr−edm,for all dr,dm∈Bδ. This proves AζE is not a convergent subsequence. This shows that Aζ∉KCt,ph; hence, Aζ∉ACt,ph. This gives a contradiction. So, limr→∞ζr=0. Conversely, suppose limr→∞ζr=0. Hence, for all δ>0, we have NBδ<∞. Therefore, for all δ>0 the space
(20)Ct,phBδ=z=zr∈ℂBδ:hz<∞,is a finite dimensional. Hence, Aζ∣Ct,phBδ∈FCt,phBδ. Let r∈ℕ and define ζr∈ℂℕ as
(21)ζrm=ζm,m∈B1/r,0,otherwise.
Obviously,Aζr∈FCt,phB1/r since dimCt,phB1/r<∞, for all r∈ℕ. Since pi∈ℓ∞ is increasing with p0>0, we have
(22)hAζ−Aζrz=∑i=0,i∈B1/r∞∑j=2i−12i+1−2j+1−t∣ζj−ζrjzj∣2ipi+∑i=0,i∉B1/r∞∑j=2i−12i+1−2j+1−t∣ζj−ζnjzj∣2ipi=∑i=0,i∈B1/r∞∑j=2i−1,j∉B1/r2i+1−2j+1−t∣ζjzj∣2ipi+∑i=0,i∉B1/r∞∑j=2i−1,j∉B1/r2i+1−2j+1−t∣ζjzj∣2ipi<1rp0∑i=0∞∑j=2i−12i+1−2j+1−t∣zj∣2ipi=1rp0hz.
This proves that∥Aζ−Aζr∥≤1/rp0 and Aζ=limr→∞Aζr. Therefore, Aζ∈ACt,ph.According to Theorem3, it is easy to prove the following theorem.Theorem 19.
Forζ∈ℂℕ, pr∈ℓ∞ is increasing with p0>0, t≥0, and Aζ∈BCt,ph. Then,
(23)Aζ∈KCt,ph⇔ζrr=0∞∈c0.Corollary 20.
Ifpi∈ℓ∞ is increasing with p0>0 and t≥0, we have
(24)KCt,ph⊂≠BCt,ph.Proof.
The operatorI=Aζ on Ct,ph is a multiplication operator generated by ζ=1,1,⋯. Therefore, I∉KCt,ph and I∈BCt,ph.Theorem 21.
SupposeCt,ph is a prequasi Banach (sss) and Aζ∈BCt,ph. Then, there is δ>0 such that ∣ζr∣≥δ, for all r∈kerζc, if and only if, Aζ has a closed range.Proof.
Let the sufficient condition be verified, to show thatRAζ¯=RAζ. If f is a limit point of RAζ, then there is Aζxr∈Ct,ph, for each r∈ℕ with limr→∞Aζxr=f. Clearly, the sequence Aζxr is a Cauchy sequence. Since h is nondecreasing, one can see
(25)hAζxr−Aζxm=∑i=0∞∑j=2i−12i+1−2j+1−t∣ζjxrj−ζjxmj∣2ipi=∑i=0,i∈kerζc∞∑j=2i−12i+1−2j+1−t∣ζjxrj−ζjxmj∣2ipi+∑i=0,i∉kerζc∞∑j=2i−12i+1−2j+1−t∣ζjxrj−ζjxmj∣2ipi≥∑i=0,i∈kerζc∞∑j=2i−12i+1−2j+1−t∣ζjxrj−ζjxmj∣2ipi=∑i=0∞∑j=2i−12i+1−2j+1−t∣ζjyrj−ζjymj∣2ipi>∑i=0∞∑j=2i−12i+1−2j+1−t∣εyrj−εymj∣2ipi=hεyr−ym,where
(26)yrk=xrk,k∈kerζc,0,k∉kerζc.
This proves thatyr is a Cauchy sequence in Ct,ph. But Ct,ph is complete. Therefore, there exists x∈Ct,ph such that limr→∞yr=x. In view of the continuity of Aζ, hence limr→∞Aζyr=Aζx. But limn→∞Aζxn=limn→∞Aζyn=f. Therefore, Aζx=f. Hence f∈RAζ. Therefore, RAζ¯=RAζ. Let the necessary condition be satisfied. Therefore, there is ε>0 such that hAζx≥hεx, for all x∈Ct,phkerζc. Let D=k∈kerζc:∣ζk∣<ε. If D≠ϕ, then for n0∈D, since h is nondecreasing, we have
(27)hAζen0=∑i=0∞∑j=2i−12i+1−2j+1−tζjen0j2ipi<∑i=0∞∑j=2i−12i+1−2j+1−tε∣en0j∣2ipi=hεen0,which gives a contradiction. So, D=ϕ with ∣ζr∣≥ε, for every r∈kerζc.Theorem 22.
Pick upζ∈ℂℕ, and Ct,ph is a prequasi Banach (sss). There is d>0 and D>0 with d<ζr<D, for every r∈ℕ, if and only if, Aζ∈BCt,ph is invertible.Proof.
Let the condition be true. Defineη∈ℂℕ by ηn=1/ζn. Then, by Theorem 16, Aζ and Tη are bounded linear operators. Also Aζ·Aη=Aη·Aζ=I. Hence, Aζ is the inverse of Aη. Conversely, let Aζ be invertible. Hence, RAζ=Ct,phℕ. Therefore, RAζ¯=RAζ. Hence, by Theorem 21, there is d>0 with ∣ζr∣≥d, for every r∈kerζc. Now, kerζ=ϕ; otherwise, ζr0=0; for some r0∈ℕ, we have er0∈kerAζ; this implies a contradiction, since kerAζ=ϕ. So, ∣ζr∣≥d, for every r∈ℕ. Since Aζ∈BCt,ph, from Theorem 16, there is D>0 with ∣ζr∣≤D, for every r∈ℕ. Therefore, we get that d≤∣ζr∣≤D, for each r∈ℕ.Theorem 23.
For a prequasi Banach (sss)Ct,ph. The operator Aζ is Fredholm, if and only if, (i) Nkerζ<∞ and (ii) ∣ζr∣≥ε, for each r∈kerζc.Proof.
Let the sufficient condition be true andNkerζ=∞, so er∈kerAζ, for every r∈kerζ. We have dimkerAζ=∞, since en’s are linearly independent. This implies a contradiction. Hence, Nkerζ<∞. From Theorem 21, the condition (ii) follows. Next, if the necessary conditions are true, from Theorem 21, the condition (ii) gives that RAζ¯=RAζ. By condition (i), we have dimkerAζ<∞ and dimRAζc<∞. This implies that Aζ is Fredholm.


## 5. Banach and Closed Prequasi Ideal

We explain the sufficient conditions ont and p such that Ct,phs,V is Banach and a closed prequasi ideal, where hx=∑i=0∞∑j=2i−12i+1−2j+1−t∣xj∣/2ipi, for all x∈Ct,p and VA=hsnAn=0∞.Theorem 24.
LetT, U be Banach spaces, pr∈ℓ∞ be increasing with p0>0 and t≥0, then Ct,phs,V is a prequasi Banach ideal.Proof.
Let the sufficient conditions be satisfied. From Theorem13, the space Ct,ph is a premodular (sss). Therefore, by Theorem 12, the function VA=hsnAn=0∞ is a prequasi norm on Ct,phs. Let Ai be a Cauchy sequence in Ct,phsT,U, since BT,U⊇Ct,phsT,U; we get
(28)VAi−Aj=∑n=0∞∑m=2n−12n+1−2m+1−tsmAi−Aj2npn≥Ai−Ajp0.
Hence,Ajj∈ℕ is a Cauchy sequence in the Banach space BT,U. Therefore, there is A∈BT,U such that limi→∞Ai−A=0, since srAir=0∞∈Ct,ph for every i∈ℕ. By using parts (ii), (iii), and (iv) of Definition (6), one can see
(29)VA=∑i=0∞∑j=2i−12i+1−2j+1−tsjA2ipi=∑i=0∞∑j=2i−12i+1−2j+1−tsjA−Am+Am2ipi≤M∑i=0∞∑j=2i−12i+1−2j+1−tsj/2A−Am2ipi+M∑i=0∞∑j=2i−12i+1−2j+1−tsj/2Am2ipi≤M∑i=0∞∑j=2i−12i+1−2j+1−t∥Am−A∥2ipi+MM0∑i=0∞∑j=2i−12i+1−2j+1−tsjAm2ipi<ε.
Therefore,srAr=0∞∈Ct,ph; hence, A∈Ct,phsT,U.Theorem 25.
LetT, U be normed spaces and pr∈ℓ∞ be increasing with p0>0 and t≥0, then Ct,phs,V is a prequasi closed ideal.Proof.
Let the sufficient conditions be satisfied. From Theorem13, the space Ct,ph is a premodular (sss). Therefore, by Theorem 12, the function VA=hsnAn=0∞ is a prequasi norm on Ct,phs. Assume Am∈Ct,phsT,U for every m∈ℕ and limm→∞VAm−A=0, since BT,U⊇Ct,phsT,U, we have
(30)VA−Aj=∑n=0∞∑m=2n−12n+1−2m+1−tsmA−Aj2npn≥A−Ajp0.
Hence,Ajj∈ℕ is a convergent sequence in BT,U. So, srAjr=0∞∈Ct,ph for each j∈ℕ. By using parts (ii), (iii), and (iv) of Definition (6), we have
(31)VA=∑i=0∞∑j=2i−12i+1−2j+1−tsjA2ipi=∑i=0∞∑j=2i−12i+1−2j+1−tsjA−Am+Am2ipi≤M∑i=0∞∑j=2i−12i+1−2j+1−tsj/2A−Am2ipi+M∑i=0∞∑j=2i−12i+1−2j+1−tsj/2Am2ipi≤M∑i=0∞∑j=2i−12i+1−2j+1−t∥Am−A∥2ipi+MM0∑i=0∞∑j=2i−12i+1−2j+1−tsjAm2ipi<ε.
Therefore,srAr=0∞∈Ct,ph; hence, A∈Ct,phsT,U.


## 6. Smallness and Simplicity of the Prequasi Ideal

We examine here some inclusion relations and topological structures ofCt,phs.Theorem 26.
LetT or U be finite dimensional Banach spaces, tq>tp>0 and increasing sequences pn∈ℓ∞, qn∈ℓ∞ with 0<pr<qr, for all r∈ℕ, then
(32)Ctp,phsT,UÞCtq,qhsT,UÜBT,U.Proof.
Let the conditions be satisfied, ifA∈Ctp,phsT,U, then sjA∈Ctp,p. We have
(33)∑i=0∞∑j=2i−12i+1−2j+1−tqsjA2iqi<∑i=0∞∑j=2i−12i+1−2j+1−tpsjA2ipi<∞,hence, A∈Ctq,qhsT,U. Next, if we take sjAj=0∞ such that ∑j=2i−12i+1−2j+1−tpsjA=2ii+1−1/pi, one can find A∈BT,U with
(34)∑i=0∞∑j=2i−12i+1−2j+1−tpsjA2ipi=∑i=0∞1i+1=∞,∑i=0∞∑j=2i−12i+1−2j+1−tqsjA2iqi≤∑i=0∞∑j=2i−12i+1−2j+1−tpsjA2iqi=∑i=0∞1i+1qi/pi<∞.
Hence,A∉Ctp,phsT,U and A∈Ctq,qhsT,U. Clearly, Ctq,qhsT,U⊂BT,U. Secondly, by choosing sjAj=0∞ such that ∑j=2i−12i+1−2j+1−tqsjA=2ii+1−1/qi. Hence, there is A∈BT,U with A∉Ctq,qhsT,U. This completes the proof.We study in the next theorem thatCt,phs is not small.Theorem 27.
IfT and U are infinite dimensional Banach spaces, pr∈ℓ∞ is increasing with p0>0 and t>0, then Ct,phs is not small.Proof.
Let the conditions be satisfied. Therefore,Ct,phs is a prequasi operator ideal. For all A∈BT,U, there is C>0 such that ∥A∥≤C; we have
(35)hA=∑i=0∞∑j=2i−12i+1−2j+1−tsjIA2ipi≤supnCpn∑i=0∞∑j=2i−12i+1−2j+1−tsjI2ipi≤supnCpn∑i=0∞12itp0<∞.
Hence,A∈Ct,phsT,U which gives Ct,phsT,U=BT,U. This finishes the proof.We investigate here the simplicity ofCt,phs.Theorem 28.
For any finite dimensional Banach spacesT or U, tq>tp>0 and increasing sequences pr∈ℓ∞, qr∈ℓ∞ with 0<pr<qr, for every r∈ℕ, then
(36)BCtq,qhs,Ctp,phs=ACtq,qhs,Ctp,phs.Proof.
Let there isD∈BCtq,qhs,(C(tp,p))hs with D∉ACtq,qhs,(C(tp,p))hs. By Lemma 1, we have A∈BCtq,qhs and B∈BCtp,phs with BDAIk=Ik. Then, it follows, for all k∈ℕ that
(37)IkCtp,phs=∑i=0∞∑j=2i−12i+1−2j+1−tpsjIk2ipi≤BDAIkCtq,qhs≤∑i=0∞∑j=2i−12i+1−2j+1−tqsjIk2iqi.By Theorem26, we get a contradiction. Therefore, D∈ACtq,qhs,(C(tp,p))hs.Theorem 29.
For any finite dimensional Banach spacesT or U. If pr∈ℓ∞ is increasing with p0>1, then Ct,phs is simple.Proof.
Assume there isD∈KCt,phs with D∉ACt,phs. By Lemma 1, we have A,B∈BCt,phs with BDAIk=Ik. This means that ICt,phs∈KCt,ps. Consequently, BSCt,phs=KCt,phs. Therefore, ACt,phs is the only nontrivial closed ideal in BCt,phs.We study the approximation of the prequasi idealCt,phs.Theorem 30.
For any finite dimensional Banach spacesT or U. If pr∈ℓ∞ is increasing with p0>0, then FT,U¯=Ct,phsT,U.Proof.
SinceCt,ph is a premodular (sss), then from Theorem 11, we have FT,U¯=Ct,phsT,U. Conversely, since I4∈SC1,1h but p0=0. This gives a counter example.


## 7. Spectrum of s-Type Operators

We explain the spectrum of the prequasi idealCt,phs.Theorem 31.
For any finite dimensional Banach spacesT or U. If pr∈ℓ∞ is increasing with p0>0, then
(38)Ct,phλT,U=Ct,phsT,U.Proof.
LetA∈Ct,phλT,U, then λrAr=0∞∈Ct,p and A−λnAI=0, for every r∈ℕ. Therefore, A=λrAI, for every r∈ℕ, then srA=srλrAI=λrA, for every r∈ℕ. Hence srAr=0∞∈Ct,p, so A∈Ct,phsT,U. Therefore, the sequence srAr=0∞ is the eigenvalues of A. Secondly, let the sufficient conditions be verified and A∈Ct,phsT,U. Therefore, srAr=0∞∈Ct,p. Hence, we have
(39)∑i=0∞∑j=2i−12i+1−2j+1−tsjA2ipi≥∑i=0∞siApi.
Hence,siAi=0∞∈ℓpi, so limi→∞siA=0. Assume A−siAI−1 exists, for every i∈ℕ. Therefore, A−siAI−1 exists and is bounded, for every i∈ℕ. So, limi→∞A−siAI−1=A−1 exists and is bounded. From the prequasi operator ideal of Ct,ps,V, we obtain
(40)I=AA−1∈Ct,phsT,U⇒siIi=0∞∈Ct,p⇒limi→∞siI=0.
We have a contradiction, sincelimi→∞siI=1. Therefore, A−siAI=0, for every i∈ℕ. This gives A∈Ct,phλT,U.



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