# Kannan Contraction Operator on the Domain ofr.-Cesàro Matrix in ℓt. and Related Prequasi Ideal with an Application of Nonlinear Difference Equations

Authors: Awad A. Bakery; M. H. El Dewaik
Journal: Journal of Function Spaces
 (2021)

---

## Abstract

In this article, the sequence spaceΞr,tυ has been built by the domain of rl-Cesàro matrix in Nakano sequence space ℓtl, where t=tl and r=rl are sequences of positive reals with 1≤tl<∞, and υf=∑l=0∞∑z=0lrzfz/∑z=0lrztl, with f=fz∈Ξr,t. Some topological and geometric behavior of Ξr,tυ, the multiplication maps acting on Ξr,tυ, and the eigenvalues distribution of operator ideal constructed by Ξr,tυ and s-numbers have been examined. The existence of a fixed point of Kannan prequasi norm contraction mapping on this sequence space and on its prequasi operator ideal are investigated. Moreover, we indicate our results by some explanative examples and actions to the existence of solutions of nonlinear difference equations.


---

## Body

## 1. Introduction

As a remark of constant Lebesgue spaces, variable exponent Lebesgue spaces go again many years, and in successive centuries, variable Lebesgue and Sobolev spaces have been regularly studied. Next, many variable exponent real function spaces and complex function spaces have presented, for instance, Morrey spaces, Herz-Morrey spaces, Herz spaces, Hardy spaces, Besov spaces, Trieble-Lizorkin spaces, Fock spaces, Bessel potential spaces, and Bergman spaces with variable exponents. For three centuries, variable exponent function spaces have been extensively applied in approximation theory, image processing, and differential equations, and many variable exponent real function spaces and complex function spaces have shown. Thus far, the theory of variable exponent function spaces has pensively built upon on the boundedness of the Hardy-Littlewood maximal operator. This confines its technique in differential equations, optimization, and approximation. The spaces of all, bounded,r-absolutely summable and convergent to zero sequences of complex numbers will be denoted by CN, ℓ∞, ℓr, and c0. N=0,1,2,⋯. We denote the space of all, finite rank, approximable, and compact bounded linear maps from a Banach space P into a Banach space Q by BP,Q, FP,Q, AP,Q, and KP,Q, and if P=Q, we indicate BP, FP, AP, and KP, respectively. The ideal of all, finite rank, approximable, and compact maps are indicated by B, F, A, and K. We label el=0,0,⋯,1,0,0,⋯, as 1 lies at the lth coordinate, with l∈N.Definition 1 [1].
A functions:BP,Q⟶0,∞N is called an s-number, if the sequence sbXb=0∞, for any X∈BP,Q, satisfies the following setup:
(a)
X=s0X≥s1X≥s2X≥⋯≥0, with X∈BP,Q(b)
sl+b−1X1+X2≤slX1+sbX2, with X1,X2∈BP,Q and l, b∈N(c)
sbZYX≤ZsbYX, for all X∈BP0,P, Y∈BP,Q, and Z∈BQ,Q0, where P0 and Q0 are any two Banach spaces(d)
IfG∈BP,Q and γ∈C, then saγG=γsaG(e)
SupposerankX≤b, then sbX=0, for all X∈BP,Q(f)
sl≥bIb=0 or sl<bIb=1, where Ib indicates the unit operator on the b-dimensional Hilbert space ℓ2b
We give some examples ofs-numbers as follows:
(1)
Thel-th Kolmogorov number, dlX, where(1)dlX=infdimJ≤lsupf≤1infg∈JXf−g.(2)
Thel-th approximation number, αlX, where(2)αlX=infX−Y:Y∈BP,QandrankY≤l.Notations 2 [2].
(3)BVs≔BVsP,Q;PandQareBanachSpaces,whereBVsP,Q≔X∈BP,Q:saXa=0∞∈V.BVα≔BVαP,Q;PandQareBanachSpaces,whereBVαP,Q≔X∈BP,Q:αaXa=0∞∈V.BVd≔BVdP,QPandQareBanachSpaces,whereBVdP,Q≔X∈BP,Q:daXa=0∞∈V.Some of ideals in the class of Banach spaces or Hilbert spaces are generated by scalar sequence spaces. For example, the ideal of compact maps is constructed by the spacec0 and daX, for X∈BP,Q. Pietsch [3] investigated the quasi-ideals Bℓbα, for 0<b<∞. He examined that the ideals of nuclear maps and of Hilbert Schmidt maps between Hilbert spaces are formed by ℓ1 and ℓ2, respectively. He discussed that Fℓb are dense in Bℓb, and the algebra Bℓb, where 1≤b<∞, generated simple Banach space. Pietsch [4] established that Bℓbα, with 0<b<∞, is small. Makarov and Faried [5] proved that for every infinite dimensional Banach spaces P, Q, and r>b>0, hence BℓbαP,QÞBℓrαP,QÜBP,Q. Yaying et al. [6] constructed the sequence space, χrt, whose its r-Cesàro matrix in ℓt, with r∈0,1 and 1≤t≤∞. They studied the quasi Banach ideal of type χrt, with r∈0,1 and 1<t<∞. They introduced its Schauder basis, α−, β−, and γ− duals and determined certain matrix classes related to this sequence space. On sequence spaces, Basarir and Kara examined the compact maps on some Euler Bm-difference sequence spaces [7], some difference sequence spaces of weighted means [8], the Riesz Bm-difference sequence space [9], the B-difference sequence space derived by weighted mean [10], and the mth order difference sequence space of generalized weighted mean [11]. Mursaleen and Noman [12, 13] established the compact operators on some difference sequence spaces. Alotaibi et al. [14, 15] examined the compact operators on some Fibonacci difference sequence spaces and on a new sequence space related to ℓp spaces. The multiplication maps on Cesàro sequence spaces equipped with the Luxemburg norm investigated by Komal et al. [16]. İlkhan et al. [17], investigated the multiplication maps on Cesàro second order function spaces. Recently, many authors in the literature have investigated some nonabsolute type sequence spaces and brought current exquisite papers, for examples, Mursaleen and Noman [18] introduced the sequence space ℓpλ and ℓ∞λ of nonabsolute type and proved that the spaces ℓpλ and ℓpλ are linearly isomorphic for 0<p≤∞, ℓpλ is a p-normed space and a BK-space in the cases for 0<p<1 and 1≤p≤∞ and constructed the basis of the space ℓpλ for 1≤p<∞. In [19], they examined the α−, β−, and γ− duals of ℓpλ and ℓ∞λ of nonabsolute type, with 1≤p<∞. They detailed some related matrix classes and developed the properties of some other classes by means of a given basic lemma. On Cesàro summable sequences, Mursaleen and Basar [20] introduced some spaces of double sequences whose Cesàro transforms are bounded, convergent in the Pringsheim’s sense, null in the Pringsheim’s sense, both convergent in the Pringsheim’s sense and bounded, regularly convergent, and absolutely q-summable, respectively, and investigated some topological properties of those sequence spaces. The Banach fixed point theorem [21] gave many mathematicians the way to examine many generalizations for the contraction maps defined on the space or on the space itself. Kannan [22] investigated an example of a class of operators with the identical fixed point actions as contractions though that fails to be continuous. Ghoncheh [23] was the only one who examined Kannan operators in modular vector spaces. He showed that the existence of a fixed point of Kannan mapping in complete modular spaces that have Fatou property. Bakery and Mohamed [24] explored the concept of the prequasi norm on Nakano sequence space such that its variable exponent in 0,1. They explained the sufficient conditions on its equipped with the definite prequasi norm to generate prequasi Banach and closed space and examined the Fatou property of different prequasi norms on it. More, they showed the existence of a fixed point of Kannan prequasi norm contraction maps on it and on the prequasi Banach operator ideal constructed by s-numbers which belong to this sequence space. The next inequality will used in the sequel [25]: Assume ta≥1 and xa,za∈C, for all a∈N, and ℏ=supata, then
(4)xa+zata≤2ℏ−1xata+zata.The aim of this paper is arranged as follows: In Section3, we introduce the definition and some inclusion relations of the sequence space Ξr,tυ under the function υ. In Section 4, we investigate the enough setup on Ξr,t with definite function υ to form premodular private sequence space (pss), which gives that Ξr,tυ is a prequasi normed pss. In Section 5, we examine a multiplication map on Ξr,tυ and introduce the necessity and sufficient conditions on this sequence space such that the multiplication map is bounded, approximable, invertible, Fredholm, and closed range. In Section 6, firstly, we introduce the sufficient settings (not necessary) onΞr,tυ, such thatFis dense inBΞr,tυs. This investigates a negative answer of Rhoades [26] open problem about the linearity of s-type Ξr,tυ spaces. Secondly, we introduce the conditions on Ξr,tυ so that the components of prequasi ideal BΞr,ts are complete and closed. Thirdly, we investigate the sufficient settings on Ξr,tυ such that BΞr,tυα is strictly included for different weights and powers. We investigate the conditions for which the prequasi ideal BΞr,tυα is minimum. Fourthly, we explore the setting for which the Banach prequasi ideal BΞr,tυs is simple. Fifthly, we examine the sufficient setting on Ξr,tυ such that the class B which sequence of eigenvalues in Ξr,tυ equals BΞr,tυs. In Section 7, the existence of a fixed point of Kannan prequasi norm contraction mapping on this sequence space and on its prequasi operator ideal generated by Ξr,tυ, and s-numbers are presented. Additionally, in Section 8, we illustrate our results by some examples and applications to the existence of solutions of nonlinear difference equations. Finally, we introduce our conclusion in Section 9.


## 2. Definitions and Preliminaries

Lemma 3 [3].
AssumeU∈BP,Q and U∉AP,Q, then there exist operators X∈BP and Y∈BQ such that YUXeb=eb, for all b∈N.Definition 4 [3].
A Banach spaceV is called simple if the algebra BV contains one and only one nontrivial closed ideal.Theorem 5 [3].
IfV is a Banach space with dimV=∞, hence
(5)FVÞAVÞKVÞBV.Definition 6 [27].
An operatorU∈BV is called Fredholm if dimRangeUc<∞, dimkerU<∞, and RangeU is closed, where RangeUc indicates the complement of RangeU.Definition 7 [28].
A classW⊆B is called an operator ideal if each element WP,Q=W∩BP,Q verifies the following conditions:
(i)
IΩ∈W, if Ω indicates Banach space of one dimension(ii)
WP,Q is a linear space on C(iii)
AssumeX∈BP0,P, Y∈WP,Q, and Z∈BQ,Q0, then ZYX∈WP0,Q0, where P0 and Q0 are normed spacesDefinition 8 [2].
A mapΨ:W⟶0,∞ is called a prequasi norm on the operator ideal W, if it satisfies the following conditions:
(1)
For everyX∈WP,Q, ΨX≥0 and ΨX=0⇔X=0(2)
One hasE0≥1 such that ΨκX≤E0κΨX, with X∈WP,Q and κ∈C(3)
One hasG0≥1 such that ΨZ1+Z2≤G0ΨZ1+ΨZ2, for all Z1,Z2∈WP,Q(4)
One hasD0≥1 such that if X∈BP0,P, Y∈WP,Q, and Z∈BQ,Q0 hence ΨZYX≤D0ZΨYXTheorem 9 [2].
Every quasi norm on the idealW is a prequasi norm on the same ideal.Definition 10 [29].
The linear space of sequencesV is called a private sequence space pss, if it verifies the next setup:
(1)
eb∈V, with b∈N(2)
V is solid, i.e., for f=fb∈CN, g=gb∈V, and fb≤gb, with b∈N, then f∈V(3)
fb/2b=0∞∈V, where b/2 indicates the integral part of b/2, if fbb=0∞∈VTheorem 11 [29].
Suppose the linear sequence spaceV be a pss, then BVs be an operator ideal.Definition 12 [29].
A subspace of thepss is called a premodular pss, if there exists a map υ:V⟶0,∞ verifies the next setup:
(i)
For allf∈V, f=θ⇔υ∣f∣=0, and υf≥0, with θ is the zero vector of V(ii)
Assumef∈V and ρ∈C, one has E0≥1 with υρf≤∣ρ∣E0υf(iii)
υf+g≤G0υf+υg includes for some G0≥1, with f,g∈V(iv)
Ifb∈N, ∣fb∣≤∣gb∣, one has υ∣fb∣≤υ∣gb∣(v)
The inequality,υ∣fb∣≤υ∣fb/2∣≤D0υ∣fb∣ holds, for D0≥1(vi)
F¯=Vυ, where F indicates the space of all sequences with finite none zero coordinates(vii)
There areϖ>0 such that υρ,0,0,0,⋯≥ϖ∣ρ∣υ1,0,0,0,⋯, with ρ∈CDefinition 13 [29].
ThepssVυ is called a prequasi normed pss, if υ verifies the conditions (i)–(iii) of Definition 12. When V is complete equipped with υ, then Vυ is called a prequasi Banach pss.Theorem 14 [29].
Every premodularpssVυ is a prequasi normed pss.Theorem 15 [29].
The functionΨ is a prequasi norm on BVυs, where ΨY=υsbYb=0∞, for all Y∈BVυsP,Q, whenever Vυ is a premodular pss.Definition 16 [24].
A prequasi normυ on V satisfies the Fatou property, if for each sequence ta⊆Vυ with lima⟶∞υta−t=0 and all z∈Vυ then υz−t≤supjinfa≥jυz−ta.Definition 17 [24].
A prequasi normΨ on the ideal BVs, where ΨW=υsaWa=0∞, satisfies the Fatou property if for each sequence Waa∈N⊆BVsZ,M with lima⟶∞ΨWa−W=0 and all V∈BVsZ,M, then
(6)ΨV−W≤supainfi≥aΨV−Wi.Definition 18.
υWz−z+υWt−t, for all z,t∈Vυ.
An elementz∈Vυ is called a fixed point of W, if Wz=z.Definition 19 [24].
An operatorW:BVsZ,M⟶BVsZ,M is named a Kannan Ψ-contraction, if there exists λ∈0,1/2, so as to ΨWV−WT≤λΨWV−V+ΨWT−T, for all V,T∈BVsZ,M.Definition 20 [24].
IfVυ is a prequasi normed (sss), W:Vυ⟶Vυ and b∈Vυ. The operator W is called υ-sequentially continuous at b, if and only if, when lima⟶∞υta−b=0, then lima⟶∞υWta−Wb=0.Definition 21 [24].
For the prequasi normΨ on the ideal BVs, where ΨW=υsaWa=0∞, G:BVsZ,M⟶BVsZ,M, and B∈BVsZ,M. The operator G is called Ψ-sequentially continuous at B, if and only if, when limp⟶∞ΨWp−B=0, then limp⟶∞ΨGWp−GB=0.Definition 22 [29].
Ifω=ωk∈CN and Vυ is a prequasi normed pss. The operator Hω:Vυ⟶Vυ is called a multiplication operator on Vυ, when Hωf=ωbfb∈Vυ, with f∈Vυ. The multiplication operator is called created by ω, if Hω∈BVυ.Theorem 23 [30].
Assumes-type Vυ≔f=srX∈RN:X∈BP,Qandυf<∞. If BVυs is an operator ideal, then the next setups are confirmed:
(1)
F⊂s-type Vυ(2)
SupposesrX1r=0∞∈s-type Vυ and srX2r=0∞∈s-type Vυ, then srX1+X2r=0∞∈s-type Vυ(3)
Assumeλ∈C and srXr=0∞∈s-type Vυ, then ∣λ∣srXr=0∞∈s-type Vυ(4)
The sequence spaceVυ is solid, i.e., if srYr=0∞∈s-type Vυ and srX≤srY, for all r∈N and X,Y∈BP,Q, then srXr=0∞∈s-type Vυ


## 3. Main Results

### 3.1. The Sequence SpaceΞr,tυ

The definition and some inclusion relations of the sequence spaceΞr,tυ under the function υ in this section are presented.Definition 24.
Supposerl,tl∈R+N, where R+N be the space of all sequences of positive reals. The sequence space Ξr,tυ with the function υ is evident by:
(7)Ξr,tυ=f=fk∈CN:υρf<∞,forsomeρ>0,whereυf=∑l=0∞∑z=0lrzfz∑z=0lrztl.Theorem 25.
Assumetl∈R+N∩ℓ∞, one has
(8)Ξr,tυ=f=fk∈CN:υρf<∞,foranyρ>0.Proof.
Assumetl∈R+N∩ℓ∞, one has
(9)ΞΔ,rυ=f=fk∈CN:υρf<∞,forsomeρ>0=f=fk∈CN:∑l=0∞∑z=0lρrzfz∑z=0lrztl<∞,forsomeρ>0=f=fk∈CN:inflρtl∑l=0∞∑z=0lrzfz∑z=0lrztl<∞,forsomeρ>0=f=fk∈CN:∑l=0∞∑z=0lrzfz∑z=0lrztl<∞=f=fk∈CN:υρf<∞,foranyρ>0.Remark 26.
(1)
Assumetz=t, rz=rz, for all z∈N, 0<r≤1, and t≥1, the sequence space Ξr,t=χrt examined by Yaying et al. [6](2)
Iftz=t, rz=1, for all z∈N and t≥1, hence Ξr,t=cest, defined and studied by Ng and Lee [31]Theorem 27.
Pick uprl,tl∈R+N and 1≤tl<∞, then Ξr,tυ be nonabsolute type.Proof.
Letf=1,−1,0,0,0,, then f=1,1,0,0,0,. One has
(10)υf=1+r0−r1r0+r1t1+r0−r1r0+r1+r2t2+≠2+r0+r1r0+r1+r2t2+=υf.
Hence, the sequence spaceΞr,tυ is nonabsolute type.
Note that, we call the sequence spaceΞr,tυ as rl-generalized Cesàro sequence space of nonabsolute type since it is generated by the domain of rl-Cesàro matrix in ℓtl, where the rl-Cesàro matrix, Λr=λlzr, is defined as:
(11)λlzr=rz∑z=0lrz,0≤z≤l,0,z>l.
Therl-Cesàro matrix may be shown clearly as:
(12)Λr=1000⋯r0r0+r1r1r0+r100⋯r0r0+r1+r2r1r0+r1+r2r2r0+r1+r20⋯r0r0+r1+r2+r3r1r0+r1+r2+r3r2r0+r1+r2+r3r3r0+r1+r2+r3⋯⋮⋮⋮⋮⋱.Definition 28.
Pick uprl,tl∈R+N. The rl-generalized Cesàro sequence space of absolute type cesr,tφ is defined as:
(13)cesr,tφ=f=fk∈CN:φρf<∞,forsomeρ>0,whereφf=∑l=0∞∑z=0lrzfz∑z=0lrztl.Theorem 29.
Assumerl,tl∈R+N∩ℓ∞ with inflrl>0, then cesr,tφÜΞr,tυ.Proof.
Supposef∈cesr,tφ, as
(14)∑l=0∞∑z=0lrzfz∑z=0lrztl≤∑l=0∞∑z=0lrzfz∑z=0lrztl<∞.
Therefore,f∈Ξr,tυ. For tl∈1,∞N∩ℓ∞, we take g=−1z/rzz∈N; we have g∈Ξr,tυ and g∉cesr,tφ. For tl∈0,1N, we take h=1/r0,−1/r1,0,0,0,⋯; we have h∈Ξr,tυ and h∉cesr,tφ=0,0,⋯.


## 3.1. The Sequence SpaceΞr,tυ

The definition and some inclusion relations of the sequence spaceΞr,tυ under the function υ in this section are presented.Definition 24.
Supposerl,tl∈R+N, where R+N be the space of all sequences of positive reals. The sequence space Ξr,tυ with the function υ is evident by:
(7)Ξr,tυ=f=fk∈CN:υρf<∞,forsomeρ>0,whereυf=∑l=0∞∑z=0lrzfz∑z=0lrztl.Theorem 25.
Assumetl∈R+N∩ℓ∞, one has
(8)Ξr,tυ=f=fk∈CN:υρf<∞,foranyρ>0.Proof.
Assumetl∈R+N∩ℓ∞, one has
(9)ΞΔ,rυ=f=fk∈CN:υρf<∞,forsomeρ>0=f=fk∈CN:∑l=0∞∑z=0lρrzfz∑z=0lrztl<∞,forsomeρ>0=f=fk∈CN:inflρtl∑l=0∞∑z=0lrzfz∑z=0lrztl<∞,forsomeρ>0=f=fk∈CN:∑l=0∞∑z=0lrzfz∑z=0lrztl<∞=f=fk∈CN:υρf<∞,foranyρ>0.Remark 26.
(1)
Assumetz=t, rz=rz, for all z∈N, 0<r≤1, and t≥1, the sequence space Ξr,t=χrt examined by Yaying et al. [6](2)
Iftz=t, rz=1, for all z∈N and t≥1, hence Ξr,t=cest, defined and studied by Ng and Lee [31]Theorem 27.
Pick uprl,tl∈R+N and 1≤tl<∞, then Ξr,tυ be nonabsolute type.Proof.
Letf=1,−1,0,0,0,, then f=1,1,0,0,0,. One has
(10)υf=1+r0−r1r0+r1t1+r0−r1r0+r1+r2t2+≠2+r0+r1r0+r1+r2t2+=υf.
Hence, the sequence spaceΞr,tυ is nonabsolute type.
Note that, we call the sequence spaceΞr,tυ as rl-generalized Cesàro sequence space of nonabsolute type since it is generated by the domain of rl-Cesàro matrix in ℓtl, where the rl-Cesàro matrix, Λr=λlzr, is defined as:
(11)λlzr=rz∑z=0lrz,0≤z≤l,0,z>l.
Therl-Cesàro matrix may be shown clearly as:
(12)Λr=1000⋯r0r0+r1r1r0+r100⋯r0r0+r1+r2r1r0+r1+r2r2r0+r1+r20⋯r0r0+r1+r2+r3r1r0+r1+r2+r3r2r0+r1+r2+r3r3r0+r1+r2+r3⋯⋮⋮⋮⋮⋱.Definition 28.
Pick uprl,tl∈R+N. The rl-generalized Cesàro sequence space of absolute type cesr,tφ is defined as:
(13)cesr,tφ=f=fk∈CN:φρf<∞,forsomeρ>0,whereφf=∑l=0∞∑z=0lrzfz∑z=0lrztl.Theorem 29.
Assumerl,tl∈R+N∩ℓ∞ with inflrl>0, then cesr,tφÜΞr,tυ.Proof.
Supposef∈cesr,tφ, as
(14)∑l=0∞∑z=0lrzfz∑z=0lrztl≤∑l=0∞∑z=0lrzfz∑z=0lrztl<∞.
Therefore,f∈Ξr,tυ. For tl∈1,∞N∩ℓ∞, we take g=−1z/rzz∈N; we have g∈Ξr,tυ and g∉cesr,tφ. For tl∈0,1N, we take h=1/r0,−1/r1,0,0,0,⋯; we have h∈Ξr,tυ and h∉cesr,tφ=0,0,⋯.


## 4. Premodular Private Sequence Space

In this section, we investigate the sufficient conditions onΞr,t with definite function υ to form premodular pss, which gives that Ξr,t is a prequasi normed pss.Here and after, the space of all monotonic decreasing and monotonic increasing sequences of positive reals will be indicated byI↘ and I↗, respectively.Theorem 30.
Ξr,t is a pss, if the following conditions are satisfied:
(f1)tl∈I↗∩ℓ∞ with t0>1
(f2)rzz=0∞∈I↘ with infzrz>0 or rzz=0∞∈I↗∩ℓ∞, and there exists C≥1 such that r2z+1≤CrzProof.
(1-i)
Letf,g∈Ξr,t. We have(15)∑l=0∞∑z=0lrzfz+rzgz∑z=0lrztl≤2ℏ−1∑l=0∞∑z=0lrzfz∑z=0lrztl+∑l=0∞∑z=0lrzgz∑z=0lrztl<∞,therefore, f+g∈Ξr,t.
(1-ii)
Assumeρ∈C, f∈Ξr,t and since tl∈IZ∩ℓ∞, one has(16)∑l=0∞∑z=0lrzρfz∑z=0lrztl≤suplρtl∑l=0∞∑z=0lrzfz∑z=0lrztl<∞.
Therefore,ρf∈Ξr,t. According to conditions (1-i) and (1-ii), one has Ξr,t is a linear space.
Sincetl∈IZ∩ℓ∞ and t0>1, we have
(17)∑l=0∞∑z=0lrzebz∑z=0lrztl=∑l=b∞rb∑z=0lrztl≤suplrbtl∑l=b∞1∑z=0lrztl<∞.
Hence,eb∈Ξr,t, for all b∈N.
(1)
Assumefb≤gb, with b∈N and g∈Ξr,t. We get(18)∑l=0∞∑z=0lrzfz∑z=0lrztl≤∑l=0∞∑z=0lrzgz∑z=0lrztl<∞,
thenf∈Ξr,t.
(2)
Supposefz∈Ξr,t, with tl,rz∈IZ∩ℓ∞ and there exists C≥1 with r2z+1≤Crz, one has(19)∑l=0∞∑z=0lrzfz/2∑z=0lrztl=∑l=0∞∑z=02lrzfz/2∑z=02lrzt2l+∑l=0∞∑z=02l+1rzfz/2∑z=02l+1rzt2l+1≤∑l=0∞r2lfl+∑z=0lr2z+r2z+1fz∑z=0lrztl+∑l=0∞∑z=0lr2z+r2z+1fz∑z=0lrztl≤2ℏ−1∑l=0∞∑z=0lr2zfz∑z=0lrztl+∑l=0∞2C∑z=0lrzfz∑z=0lrztl+∑l=0∞2C∑z=0lrz∣fz∣∑z=0lrztl≤22ℏ−1+2ℏ−1+2ℏCℏ∑l=0∞∑z=0lrzfz∑z=0lrztl<∞,then fz/2∈Ξr,t.From Theorem11, one has the following theorem.Theorem 31.
If the conditions (f1) and (f2) are confirmed, thenBΞr,ts is an operator ideal.Theorem 32.
Ξr,tυ is a premodular pss, if the conditions (f1) and (f2) are verified.Proof.
(i)
Clearly,υf≥0 and υf=0⇔f=θ(ii)
One hasE0=max1,suplρtl−1≥1 with υρf≤E0ρυf, for each f∈Ξr,t and ρ∈C(iii)
The inequalityυf+g≤2ℏ−1υf+υg holds, with f,g∈Ξr,t(iv)
Obviously, from the proof part (31) of Theorem 30(v)
Clearly, the proof part (32) of Theorem 30 that D0≥22ℏ−1+2ℏ−1+2ℏCℏ≥1(vi)
Definitely,F¯=Ξr,t(vii)
There is0<ϖ≤suplρtl−1 with υρ,0,0,0,⋯≥ϖρυ1,0,0,0,⋯, for each ρ≠0 and ϖ>0, if ρ=0Theorem 33.
If the conditions (f1) and (f2) are verified, thenΞr,tυ is a prequasi Banach pss.Proof.
From Theorem32, the space Ξr,tυ is a premodular pss. From Theorem 14, the space Ξr,tυ is a prequasi normed pss. To prove that Ξr,tυ is a prequasi Banach pss, suppose fa=fzaz=0∞ be a Cauchy sequence in Ξr,tυ, then for every ε∈0,1, there exists a0∈N such that for every a,b≥a0, we have
(20)υfa−fb=∑l=0∞∑z=0lrzfza−fzb∑z=0lrztl<εℏ.Therefore, fora,b≥a0 and z∈N, one has fza−fzb<ε. Therefore, fzb is a Cauchy sequence in C; for constant z∈N, this implies limb⟶∞fzb=fz0, for constant z∈N. Then, υfa−f0<εℏ, for every a≥a0. Finally, to prove that f0∈Ξr,tυ, we obtain υf0≤2ℏ−1υfa−f0+υfa<∞, hence f0∈Ξr,tυ. This gives that Ξr,tυ is a prequasi Banach pss.According to Theorem23, we explain the following properties of the s-type Ξr,tυ.Theorem 34.
Assumes-type Ξr,tυ≔f=snX∈CN:X∈BP,Qandυf<∞. The following settings are satisfied:
(1)
We haves-type Ξr,tυ⊃F(2)
IfsrX1r=0∞∈s-type Ξr,tυ and srX2r=0∞∈s-type Ξr,tυ, then srX1+X2r=0∞∈s-type Ξr,tυ(3)
For allλ∈C and srXr=0∞∈s-type Ξr,tυ, then λsrXr=0∞∈s-type Ξr,tυ(4)
Thes-type Ξr,tυ is solid


## 5. Multiplication Operators onΞr,tυ

In this section, we examine the multiplication map on the prequasi normedpss, Ξr,tυ, and introduce the necessity and sufficient conditions on Ξr,tυ such that the multiplication operator is bounded, invertible, approximable, Fredholm, and closed range.Theorem 35.
Ifω∈CN, the conditions (f1) and (f2) are confirmed, then ω∈ℓ∞, if and only if, Hω∈BΞr,tv.Proof.
Assumeω∈ℓ∞. Therefore, there exists ν>0 such that ωb≤ν, for all b∈N. Suppose f∈Ξr,tυ, we have
(21)υHωf=υωf=∑l=0∞∑z=0lrzωzfz∑z=0lrztl≤∑l=0∞∑z=0lνrzfz∑z=0lrztl≤suplνtl∑l=0∞∑z=0lrzfz∑z=0lrztl=suplνtlυf.
Hence,Hω∈BΞr,tυ.
On the other hand, supposeHω∈BΞr,tυ and ω∉ℓ∞. Then, for every b∈N, there exist xb∈N such that ωxb>b. One has
(22)υHωexb=υωexb=∑l=0∞∑z=0lrzωzexbz∑z=0lrztl=∑l=xb∞rxbωxb∑z=0lrztl>∑l=xb∞brxb∑z=0lrztl>bt0υexb.
Therefore,Hω∉BΞr,tυ. Hence, ω∈ℓ∞.Theorem 36.
Ifω∈CN and Ξr,tυ is a prequasi normed pss. Then, ωb=g, for all b∈N and g∈C with g=1, if and only if, Hω is an isometry.Proof.
Assume the sufficient condition is satisfied. We have(23)υHωf=υωf=∑l=0∞∑k=0lrkωkfk∑z=0lrztl=∑l=0∞∑k=0lgrkfk∑z=0lrztl=υf,for all f∈Ξr,tυ. Hence, Hω is an isometry.
Assume the necessity condition is verified and∣ωb∣<1, for some b=b0. One has
(24)υHωeb0=υωeb0=∑l=0∞∑k=0lrkωkeb0k∑z=0lrztl=∑l=b0∞rb0ωb0∑z=0lrztl<∑l=b0∞rb0∑z=0lrztl=υeb0.
Also, ifωb0>1, clearly υHωeb0>υeb0, which is a contradiction for the two cases. Hence, ωb=1, for each b∈N.
The space of all sets with finite number of elements will be indicated byF.Theorem 37.
Ifω∈CN, the conditions (f1) and (f2) are verified. Then, Hω∈AΞr,tυ, if and only if, ωbb=0∞∈c0.Proof.
AssumeHω∈AΞr,tυ, hence Hω∈KΞr,tυ. Let limb⟶∞ωb≠0. Hence, one has ρ>0 so that the set Kρ=b∈N:ωb≥ρàF. Suppose αbb∈N⊂Kρ. Therefore, eαb:αb∈Kρ∈ℓ∞ is an infinite set in Ξr,tυ. As
(25)υHωeαa−Hωeαb=υωeαa−ωeαb=∑l=0∞∑k=0lrkωkeαak−eαbk∑z=0lrztl≥∑l=0∞∑k=0lrkρeαak−eαbk∑z=0lrztl≥inflρtlυeαa−eαb,for all αa,αb∈Kρ. Hence, eαb:αb∈Kρ∈ℓ∞, which cannot have a convergent subsequence under Hω. Then, Hω∉KΞr,tυ. Which gives Hω∉AΞr,tυ, this implies a contradiction. Therefore, limb⟶∞ωb=0. On the contrary, suppose limb⟶∞ωb=0. Hence, for each ρ>0, we have Kρ=b∈N:∣ωb∣≥ρ⊂F. Therefore, for all ρ>0, one has dimΞr,tυKρ=dimCKρ<∞. Then, Hω∈FΞr,tυKρ. Let ωa∈CN, for every a∈N, where
(26)ωab=ωb,b∈K1/a+1,0,otherwise.
Clearly,Hωa∈FΞr,tυB1/a+1 as dimΞr,tυB1/a+1<∞, for every a∈N. From tl∈IZ∩ℓ∞ with t0>1, one has
(27)υHω−Hωaf=υωb−ωabfbb=0∞=∑l=0∞∑b=0lrbωb−ωabfb∑z=0lrztl=∑l=0,l∈K1/a+1∞∑b=0lrbωb−ωabfb∑z=0lrztl+∑l=0,l∉K1/a+1∞∑b=0lrbωb−ωabfb∑z=0lrztl=∑l=0,l∉K1/a+1∞∑b=0lrbωbfb∑z=0lrztl≤1a+1t0∑l=0,l∉K1/a+1∞∑b=0lrbfb∑z=0lrztl<1a+1t0∑l=0∞∑b=0lrbfb∑z=0lrztl=1a+1t0υf.Therefore,Hω−Hωa≤1/a+1t0. This implies Hω is a limit of finite rank maps. Hence, Hω∈AΞr,tυ.Theorem 38.
Ifω∈CN, the conditions (f1) and (f2) are confirmed. Then, Hω∈KΞr,tυ, if and only if, ωbb=0∞∈c0.Proof.
Clearly, asAΞr,tυÞKΞr,tυ.Corollary 39.
If the conditions (f1) and (f2) are verified, thenKΞr,tυÞBΞr,tυ.Proof.
Sinceω=1,1, is created the multiplication map I on Ξr,tυ. This implies I∉KΞr,tυ and I∈BΞr,tυ.Theorem 40.
Pick upΞr,tυ be a prequasi Banach pss and Hω∈BΞr,tυ. Then, there exist α>0 and η>0 so that α<ωb<η, with b∈kerωc, if and only if, RangeHω is closed.Proof.
Let the sufficient setup be verified. Therefore, there existsρ>0 such that ∣ωb∣≥ρ, for each b∈kerωc. To prove that RangeHω is closed. Suppose g is a limit point of RangeHω. One has Hωfb∈Ξr,tυ, for all b∈N such that limb⟶∞Hωfb=g. Clearly, the sequence Hωfb is a Cauchy sequence. Since tl∈IZ∩ℓ∞ with t0>1, we have
(28)υHωfa−Hωfb=∑l=0∞∑k=0lrkωkfak−ωkfbk∑z=0lrztl=∑l=0,l∈kerωc∞∑k=0lrkωkfak−ωkfbk∑z=0lrztl+∑l=0,l∉kerωc∞∑k=0lrkωkfak−ωkfbk∑z=0lrztl≥∑l=0,l∈kerωc∞∑k=0lrkωkfak−ωkfbk∑z=0lrztl=∑l=0∞∣∑k=0lrkωkuak−ωkubk∣∑z=0lrztl>∑l=0∞∑k=0lrkρuak−ubk∑z=0lrztl≥inflρtlυua−ub,where
(29)uak=fak,k∈kerωc,0,k∉kerωc.
Therefore,ua is a Cauchy sequence in Ξr,tυ. Since Ξr,tυ is complete. Then, there exists f∈Ξr,tυ such that limb⟶∞ub=f. As Hω∈BΞr,tυ, one gets limb⟶∞Hωub=Hωf. But limb⟶∞Hωub=limb⟶∞Hωfb=g. Hence, Hωf=g. So g∈RangeHω. Then, RangeHω is closed. Next, let the necessity condition be verified. Therefore, there exists ρ>0 such that υHωf≥ρυf, with f∈Ξr,tυkerωc. When K=b∈kerωc:ωb<ρ≠∅, then for a0∈K, we obtain
(30)υHωea0=υωbea0bb=0∞=∑l=0∞∣∑b=0lrbωbea0b∣∑z=0lrztl<∑l=0∞∣∑b=0lrbea0bρ∣∑z=0lrztl≤suplρtlυea0,this implies a contradiction. Then, K=ϕ, one has ∣ωb∣≥ρ, with b∈kerωc. This shows the theorem.Theorem 41.
Ifω∈CN and Ξr,tυ is a prequasi Banach pss. Then, there exist α>0 and η>0 such that α<ωb<η, for all b∈N, if and only if, Hω∈BΞr,tυ is invertible.Proof.
Let the sufficient condition be confirmed. Ifκ∈CN with κb=1/ωb. From Theorem 35, the operators Hω and Hκ are bounded linear. One has Hω.Hκ=Hκ.Hω=I. Then, Hκ=Hω−1. After, assume Hω be invertible. Hence, RangeHω=Ξr,tυN. Therefore, RangeHω is closed. Hence, from Theorem 40, there exists α>0 such that ωb≥α, for all b∈kerωc. One gets kerω=∅, if ωb0=0, with b0∈N; this implies eb0∈kerHω which is an inconsistency, since kerHω is trivial. Hence, ωb≥α, for all b∈N. As Hω∈ℓ∞. By using Theorem 35, there exists η>0 such that ωb≤η, for all b∈N. Then, one obtains α≤ωb≤η, with b∈N.Theorem 42.
IfΞr,tυ is a prequasi Banach pss and Hω∈BΞr,tυ. Then, Hω is Fredholm operator, if and only if, (i) kerω⫋N is a finite and (ii) ωb≥ρ, with b∈kerωc.Proof.
Assume the sufficient setup be confirmed. SupposekerωÜN be an infinite, then eb∈kerHω, for all b∈kerω. As eb’s are linearly independent, we have that dimkerHω=∞; this gives an inconsistency. Therefore, kerωÜN must be finite. The setup (ii) follows from Theorem 40. After, assume the setups (i) and (ii) be satisfied. By using Theorem 40, the setup (ii) gives that RangeHω is closed. The condition (i) implies that dimkerHω<∞ and dimRangeHωc<∞. Then, Hω is Fredholm.


## 6. Prequasi Ideal Properties

In this section, firstly, we give the sufficient conditions (not necessary) onΞr,tυso thatFis dense inBΞr,tυs. This explains a negative answer of Rhoades [26] open problem about the linearity of s-type Ξr,tυ spaces. Secondly, for which setup on Ξr,tυ are BΞr,tυs complete and closed? Thirdly, we investigate the sufficient conditions on Ξr,tυ so that BΞr,tυα is strictly contained for different weights and powers. We introduce the conditions such that BΞr,tυα is minimum. Fourthly, we give the setup such that the Banach prequasi ideal BΞr,tυs is simple. Fifthly, we explore the sufficient setup on Ξr,tυ so that the space of all bounded linear operators which sequence of eigenvalues in Ξr,tυ equals BΞr,tυs.


### 6.1. Finite Rank Prequasi Ideal

Theorem 43.
BΞr,tυsP,Q=FP,Q¯, if the conditions f1 and f2 are confirmed. But the converse is not necessarily true.Proof.
To prove thatFP,Q¯⊆BΞr,tυsP,Q. Since el∈Ξr,tυ, for all l∈N and Ξr,tυ is a linear space. Assume Z∈FP,Q, we have slZl=0∞∈F. To prove that BΞr,tυsP,Q⊆FP,Q¯. We have ∑l=0∞1/∑z=0lrztl<∞. Suppose Z∈BΞr,tυsP,Q, one has slZl=0∞∈Ξr,tυ. Since υslZl=0∞<∞, assume ρ∈0,1, then there exists l0∈N−0 with υslZl=l0∞<ρ/2ℏ+3ηd, for some d≥1, where η=max1,∑l=l0∞1/∑z=0lrztl. As slZ is decreasing, one has
(31)∑l=l0+12l0∑j=0lrjs2l0Z∑z=0lrztl≤∑l=l0+12l0∑j=0lrjsjZ∑z=0lrztl≤∑l=l0∞∑j=0lrjsjZ∑z=0lrztl<ρ2ℏ+3ηd.
Then, there existsY∈F2l0P,Q such that rank Y≤2l0 and
(32)∑l=2l0+13l0∑j=0lrjZ−Y∑z=0lrztl≤∑l=l0+12l0∑j=0lrjZ−Y∑z=0lrztl<ρ2ℏ+3ηd,as tl∈IZ∩ℓ∞, one gets
(33)supl=l0∞∑j=0l0rjZ−Ytl<ρ22ℏ+2η.
Hence, we have(34)∑l=0l0∑j=0lrjZ−Y∑z=0lrztl<ρ2ℏ+3ηd.
From inequalities (4)–(34), we have
(35)dZ,Y=υslZ−Yl=0∞=∑l=03l0−1∑j=0lrjsjZ−Y∑z=0lrztl+∑l=3l0∞∑j=0lrjsjZ−Y∑z=0lrztl≤∑l=03l0∑j=0lrjZ−Y∑z=0lrztl+∑l=l0∞∑j=0l+2l0rjsjZ−Y∑j=0l+2l0rjtl+2l0≤∑l=03l0∑j=0lrjZ−Y∑z=0lrztl+∑l=l0∞∑j=0l+2l0rjsjZ−Y∑z=0lrztl≤3∑l=0l0∑j=0irjZ−Y∑z=0lrztl+∑l=l0∞∑j=02l0−1rjsjZ−Y+∑j=2l0l+2l0rjsjZ−Y∑z=0lrztl≤3∑l=0l0∑j=0lrjZ−Y∑z=0lrztl+2ℏ−1∑l=l0∞∑j=02l0−1rjsjZ−Y∑z=0lrztl+∑l=l0∞∑j=2l0l+2l0rjsjZ−Y∑z=0lrztl≤3∑l=0l0∑j=0lrjZ−Y∑z=0lrztl+2ℏ−1∑l=l0∞∑j=02l0−1rjZ−Y∑z=0lrztl+∑l=l0∞∑j=0lrj+2l0sj+2l0Z−Y∑z=0lrztl≤3∑l=0l0∑j=0lrjZ−Y∑z=0lrztl+2ℏ−1supl=l0∞∑j=02l0−1rjZ−Ytl∑l=l0∞∑z=0lrz−tl+2ℏ−1∑l=l0∞∑j=0lrjsjZ∑z=0lrztl<ρ.
Conversely, we have a counter example asI4∈BΞr,tυsP,Q, where r=0,0,0,0,1,1, and t=1,1,1,, but t0>1 is not satisfied. This gives the proof.


### 6.2. Banach and Closed Prequasi Ideal

Theorem 44.
Suppose the conditionsf1 and f2 be verified, then BΞr,tυs,Ψ be a prequasi Banach ideal, where ψX=υslXl=0∞.Proof.
SinceΞr,tυ is a premodular pss, then from Theorem 15, Ψ is a prequasi norm on BΞr,tυs. Let Xbb∈N be a Cauchy sequence in BΞr,tυsP,Q. Since BP,Q⊇BΞr,tυsP,Q, we have
(36)ΨXa−Xb=∑l=0∞∑z=0lrzszXa−Xb∑z=0lrztl≥Xa−Xbt0,then Xbb∈N is a Cauchy sequence in BP,Q. As BP,Q is a Banach space, hence there exists X∈BP,Q with limb⟶∞Xb−X=0. As slXbl=0∞∈Ξr,tυ, for all b∈N. Then, from Definition 12 conditions (ii), (iii), and (v), we obtain
(37)ΨX=∑l=0∞∑z=0lrzszX∑z=0lrztl≤2ℏ−1∑l=0∞∑z=0lrzsz/2X−Xb∑z=0lrztl+2ℏ−1∑l=0∞∑z=0lrzsz/2Xb∑z=0lrztl≤2ℏ−1∑l=0∞X−Xbtl+2ℏ−1D0∑l=0∞∑z=0lrzszXb∑z=0lrztl<∞.
Then,slXl=0∞∈Ξr,tυ, hence X∈BΞr,tυsP,Q.Theorem 45.
IfP and Q are normed spaces, the conditions f1 and f2 are confirmed; then, BΞr,tυs,Ψ is a prequasi closed ideal, where ΨX=υslXl=0∞.Proof.
SinceΞr,tυ is a premodular pss, from Theorem 15, Ψ is a prequasi norm on BΞr,tυs. Let Xb∈BΞr,tυsP,Q, for all b∈N and limb⟶∞ΨXb−X=0. Since BP,Q⊇BΞr,tυsP,Q, one gets
(38)ΨX−Xb=∑l=0∞∑z=0lrzszX−Xb∑z=0lrztl≥X−Xbt0,then Xbb∈N is a convergent sequence in BP,Q. As slXbl=0∞∈Ξr,tυ, for all b∈N. From Definition 12 conditions (ii), (iii), and (v), we have
(39)ΨX=∑l=0∞∑z=0lrzszX∑z=0lrztl≤2ℏ−1∑l=0∞∑z=0lrzsz/2X−Xb∑z=0lrztl+2ℏ−1∑l=0∞∑z=0lrzsz/2Xb∑z=0lrztl≤2ℏ−1∑l=0∞X−Xbtl+2ℏ−1D0∑l=0∞∑z=0lrzszXb∑z=0lrztl<∞.
We getslXl=0∞∈Ξr,tυ, so X∈BΞr,tυsP,Q.


### 6.3. Minimum Prequasi Ideal

Theorem 46.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are verified with 1<tl1<tl2 and 0<rl2/∑z=0lrz2≤rl1/∑z=0lrz1, for every l∈N, then
(40)BΞrl1,tl1υsP,QÞBΞrl2,tl2vsP,QÜBP,Q.Proof.
AssumeZ∈BΞrl1,tl1υsP,Q, then slZ∈Ξrl1,tl1υ. We have
(41)∑l=0∞∑z=0lrz2szZ∑z=0lrz2tl2<∑l=0∞∑z=0lrz1szZ∑z=0lrz1tl1<∞,then Z∈BΞrl2,tl2υsP,Q. Next, if we take slZl=0∞ with ∑z=0lrz1szZ=∑z=0lrz1/l+1tl1, one obtains Z∈BP,Q so that
(42)∑l=0∞∑z=0lrz1szZ∑z=0lrz1tl1=∑l=0∞1l+1=∞,∑l=0∞∑z=0lrz2szZ∑z=0lrz2tl2≤∑l=0∞∑z=0lrz1szZ∑z=0lrz1tl2=∑l=0∞1l+1tl2/tl1<∞.
Hence,Z∉BΞrl1,tl1υsP,Q and Z∈BΞrl2,tl2υsP,Q. Obviously, BΞrl2,tl2υsP,Q⊂BP,Q. After, if we choose slZl=0∞ so that ∑z=0lrz2szZ=∑z=0lrz2/l+1tl2. One gets Z∈BP,Q so that Z∉BΞrl2,tl2υsP,Q. This completes the proof.Theorem 47.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are verified, then BΞr,tυα is minimum.Proof.
Let the sufficient conditions be verified. Hence,BΞr,tα,Ψ, where ΨZ=∑l=0∞1/∑z=0lrz∑z=0lrzαzZtl, is a prequasi Banach ideal. Assume BΞr,tαP,Q=BP,Q, then there exists η>0 with ΨZ≤ηZ, for all Z∈BP,Q. From Dvoretzky’s theorem [32], for all b∈N, we have quotient spaces P/Yb and subspaces Mb of Q which can be operated onto ℓ2b by isomorphisms Vb and Xb with VbVb−1≤2 and XbXb−1≤2. If Ib is the identity map on ℓ2b, Tb is the quotient map from P to P/Yb, and Jb is the natural embedding map from Mb to Q. Assume mz be the Bernstein numbers [33] hence
(43)1=mzIb=mzXbXb−1IbVbVb−1≤XbmzXb−1IbVbVb−1=XbmzJbXb−1IbVbVb−1≤XbdzJbXb−1IbVbVb−1=XbdzJbXb−1IbVbTbVb−1≤XbαzJbXb−1IbVbTbVb−1,for 0≤l≤b. One gets
(44)∑z=0lrz≤∑z=0lXbrzαzJbXb−1IbVbTbVb−1⇒1≤XbVb−1tl∑z=0lrzαzJbXb−1IbVbTb∑z=0lrztl.
Therefore, for someρ≥1, we have
(45)b+1≤ρXbVb−1∑l=0b∑z=0lrzαzJbXb−1IbVbTb∑z=0lrztl⇒b+1≤ρXbVb−1ΨJbXb−1IbVbTb⇒b+1≤ρηXbVb−1JbXb−1IbVbTb⇒b+1≤ρηXbVb−1JbXb−1IbVbTb⇒b+1≤ρηXbVb−1JbXb−1IbVbTb=ρηXbVb−1Xb−1IbVb≤4ρη.
Hence, there is an inconsistency, whenb⟶∞. Hence, P and Q both cannot be infinite dimensional when BΞr,tαP,Q=BP,Q. This finishes the proof.Theorem 48.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are verified, then BΞr,td is minimum.


### 6.4. Simple Banach Prequasi Ideal

Theorem 49.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are verified with 1<tl1<tl2 and 0<rl2/∑z=0lrz2≤rl1/∑z=0lrz1, for all l∈N, then
(46)BBΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q=ABΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q.Proof.
SupposeX∈BBΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q and X∉ABΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q. From Lemma 3, one has Y∈BBΞrl2,tl2υsP,Q and Z∈BBΞrl1,tl1υsP,Q with ZXYIb=Ib. Then, for all b∈N, one has
(47)IbBΞrl1,tl1υsP,Q=∑l=0∞∑z=0lrz1szIb∑z=0lrz1tl1≤ZXYIbBΞrl2,tl2υsP,Q≤∑l=0∞∑z=0irz2szIj∑z=0lrz2tl2.
This contradicts Theorem46. Hence, X∈ABΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q, which completes the proof.Corollary 50.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are verified with 1<tl1<tl2 and 0<rl2/∑z=0lrz2≤rl1/∑z=0lrz1, for all l∈N, then
(48)BBΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q=KBΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q.Proof.
Obviously, sinceA⊂K.Theorem 51.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are confirmed, then BΞr,tυs is simple.Proof.
Suppose the closed idealKBΞr,tυsP,Q contains an operator X∉ABΞr,tυsP,Q. By using Lemma 3, there are Y,Z∈BBΞr,tυsP,Q with ZXYIb=Ib. This implies that IBΞr,tυsP,Q∈KBΞr,tυsP,Q. Hence, BBΞr,tυsP,Q=KBΞr,tυsP,Q. Therefore, BΞr,tυs is simple Banach space.


### 6.5. Eigenvalues ofs-Type Operators

Notation 52.
(49)BVsρ≔BVsρP,Q;PandQareBanachSpaces,whereBVsρP,Q≔X∈BP,Q:ρlXn=0∞∈VandX−ρlXIisnotinvertible,foralll∈N.Theorem 53.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are verified, then
(50)BΞr,tυsρP,Q=BΞr,tυsP,Q.Proof.
AssumeX∈BΞr,tυsρP,Q, then ρlXl=0∞∈Ξr,tυ and X−ρlXI=0, for every l∈N. One gets X=ρlXI, for each l∈N, then slX=slρlXI=∣ρlX∣, for all l∈N. Hence, slXl=0∞∈Ξr,tυ, so X∈BΞr,tυsP,Q.
Secondly, assumeX∈BΞr,tυsP,Q. Hence, slXl=0∞∈Ξr,tυ. Therefore, one can see
(51)∑l=0∞∑z=0lrzszX∑z=0lrztl≥∑l=0∞slXtl.
Hence,liml⟶∞slX=0. Suppose X−slXI−1 exists, for all l∈N. Then, X−slXI−1 exists and bounded, for all l∈N. Hence, liml⟶∞X−slXI−1=X−1 exists and bounded. Since BΞr,tυs,Ψ is a prequasi operator ideal, one obtains
(52)I=XX−1∈BΞr,tυsP,Q⇒slIl=0∞∈Ξr,t⇒liml⟶∞slI=0.
Therefore, there is a contradiction, asliml⟶∞slI=1. Then, X−slXI=0, for all l∈N. This implies X∈BΞr,tυsρP,Q. This confirms the proof.


## 6.1. Finite Rank Prequasi Ideal

Theorem 43.
BΞr,tυsP,Q=FP,Q¯, if the conditions f1 and f2 are confirmed. But the converse is not necessarily true.Proof.
To prove thatFP,Q¯⊆BΞr,tυsP,Q. Since el∈Ξr,tυ, for all l∈N and Ξr,tυ is a linear space. Assume Z∈FP,Q, we have slZl=0∞∈F. To prove that BΞr,tυsP,Q⊆FP,Q¯. We have ∑l=0∞1/∑z=0lrztl<∞. Suppose Z∈BΞr,tυsP,Q, one has slZl=0∞∈Ξr,tυ. Since υslZl=0∞<∞, assume ρ∈0,1, then there exists l0∈N−0 with υslZl=l0∞<ρ/2ℏ+3ηd, for some d≥1, where η=max1,∑l=l0∞1/∑z=0lrztl. As slZ is decreasing, one has
(31)∑l=l0+12l0∑j=0lrjs2l0Z∑z=0lrztl≤∑l=l0+12l0∑j=0lrjsjZ∑z=0lrztl≤∑l=l0∞∑j=0lrjsjZ∑z=0lrztl<ρ2ℏ+3ηd.
Then, there existsY∈F2l0P,Q such that rank Y≤2l0 and
(32)∑l=2l0+13l0∑j=0lrjZ−Y∑z=0lrztl≤∑l=l0+12l0∑j=0lrjZ−Y∑z=0lrztl<ρ2ℏ+3ηd,as tl∈IZ∩ℓ∞, one gets
(33)supl=l0∞∑j=0l0rjZ−Ytl<ρ22ℏ+2η.
Hence, we have(34)∑l=0l0∑j=0lrjZ−Y∑z=0lrztl<ρ2ℏ+3ηd.
From inequalities (4)–(34), we have
(35)dZ,Y=υslZ−Yl=0∞=∑l=03l0−1∑j=0lrjsjZ−Y∑z=0lrztl+∑l=3l0∞∑j=0lrjsjZ−Y∑z=0lrztl≤∑l=03l0∑j=0lrjZ−Y∑z=0lrztl+∑l=l0∞∑j=0l+2l0rjsjZ−Y∑j=0l+2l0rjtl+2l0≤∑l=03l0∑j=0lrjZ−Y∑z=0lrztl+∑l=l0∞∑j=0l+2l0rjsjZ−Y∑z=0lrztl≤3∑l=0l0∑j=0irjZ−Y∑z=0lrztl+∑l=l0∞∑j=02l0−1rjsjZ−Y+∑j=2l0l+2l0rjsjZ−Y∑z=0lrztl≤3∑l=0l0∑j=0lrjZ−Y∑z=0lrztl+2ℏ−1∑l=l0∞∑j=02l0−1rjsjZ−Y∑z=0lrztl+∑l=l0∞∑j=2l0l+2l0rjsjZ−Y∑z=0lrztl≤3∑l=0l0∑j=0lrjZ−Y∑z=0lrztl+2ℏ−1∑l=l0∞∑j=02l0−1rjZ−Y∑z=0lrztl+∑l=l0∞∑j=0lrj+2l0sj+2l0Z−Y∑z=0lrztl≤3∑l=0l0∑j=0lrjZ−Y∑z=0lrztl+2ℏ−1supl=l0∞∑j=02l0−1rjZ−Ytl∑l=l0∞∑z=0lrz−tl+2ℏ−1∑l=l0∞∑j=0lrjsjZ∑z=0lrztl<ρ.
Conversely, we have a counter example asI4∈BΞr,tυsP,Q, where r=0,0,0,0,1,1, and t=1,1,1,, but t0>1 is not satisfied. This gives the proof.


## 6.2. Banach and Closed Prequasi Ideal

Theorem 44.
Suppose the conditionsf1 and f2 be verified, then BΞr,tυs,Ψ be a prequasi Banach ideal, where ψX=υslXl=0∞.Proof.
SinceΞr,tυ is a premodular pss, then from Theorem 15, Ψ is a prequasi norm on BΞr,tυs. Let Xbb∈N be a Cauchy sequence in BΞr,tυsP,Q. Since BP,Q⊇BΞr,tυsP,Q, we have
(36)ΨXa−Xb=∑l=0∞∑z=0lrzszXa−Xb∑z=0lrztl≥Xa−Xbt0,then Xbb∈N is a Cauchy sequence in BP,Q. As BP,Q is a Banach space, hence there exists X∈BP,Q with limb⟶∞Xb−X=0. As slXbl=0∞∈Ξr,tυ, for all b∈N. Then, from Definition 12 conditions (ii), (iii), and (v), we obtain
(37)ΨX=∑l=0∞∑z=0lrzszX∑z=0lrztl≤2ℏ−1∑l=0∞∑z=0lrzsz/2X−Xb∑z=0lrztl+2ℏ−1∑l=0∞∑z=0lrzsz/2Xb∑z=0lrztl≤2ℏ−1∑l=0∞X−Xbtl+2ℏ−1D0∑l=0∞∑z=0lrzszXb∑z=0lrztl<∞.
Then,slXl=0∞∈Ξr,tυ, hence X∈BΞr,tυsP,Q.Theorem 45.
IfP and Q are normed spaces, the conditions f1 and f2 are confirmed; then, BΞr,tυs,Ψ is a prequasi closed ideal, where ΨX=υslXl=0∞.Proof.
SinceΞr,tυ is a premodular pss, from Theorem 15, Ψ is a prequasi norm on BΞr,tυs. Let Xb∈BΞr,tυsP,Q, for all b∈N and limb⟶∞ΨXb−X=0. Since BP,Q⊇BΞr,tυsP,Q, one gets
(38)ΨX−Xb=∑l=0∞∑z=0lrzszX−Xb∑z=0lrztl≥X−Xbt0,then Xbb∈N is a convergent sequence in BP,Q. As slXbl=0∞∈Ξr,tυ, for all b∈N. From Definition 12 conditions (ii), (iii), and (v), we have
(39)ΨX=∑l=0∞∑z=0lrzszX∑z=0lrztl≤2ℏ−1∑l=0∞∑z=0lrzsz/2X−Xb∑z=0lrztl+2ℏ−1∑l=0∞∑z=0lrzsz/2Xb∑z=0lrztl≤2ℏ−1∑l=0∞X−Xbtl+2ℏ−1D0∑l=0∞∑z=0lrzszXb∑z=0lrztl<∞.
We getslXl=0∞∈Ξr,tυ, so X∈BΞr,tυsP,Q.


## 6.3. Minimum Prequasi Ideal

Theorem 46.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are verified with 1<tl1<tl2 and 0<rl2/∑z=0lrz2≤rl1/∑z=0lrz1, for every l∈N, then
(40)BΞrl1,tl1υsP,QÞBΞrl2,tl2vsP,QÜBP,Q.Proof.
AssumeZ∈BΞrl1,tl1υsP,Q, then slZ∈Ξrl1,tl1υ. We have
(41)∑l=0∞∑z=0lrz2szZ∑z=0lrz2tl2<∑l=0∞∑z=0lrz1szZ∑z=0lrz1tl1<∞,then Z∈BΞrl2,tl2υsP,Q. Next, if we take slZl=0∞ with ∑z=0lrz1szZ=∑z=0lrz1/l+1tl1, one obtains Z∈BP,Q so that
(42)∑l=0∞∑z=0lrz1szZ∑z=0lrz1tl1=∑l=0∞1l+1=∞,∑l=0∞∑z=0lrz2szZ∑z=0lrz2tl2≤∑l=0∞∑z=0lrz1szZ∑z=0lrz1tl2=∑l=0∞1l+1tl2/tl1<∞.
Hence,Z∉BΞrl1,tl1υsP,Q and Z∈BΞrl2,tl2υsP,Q. Obviously, BΞrl2,tl2υsP,Q⊂BP,Q. After, if we choose slZl=0∞ so that ∑z=0lrz2szZ=∑z=0lrz2/l+1tl2. One gets Z∈BP,Q so that Z∉BΞrl2,tl2υsP,Q. This completes the proof.Theorem 47.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are verified, then BΞr,tυα is minimum.Proof.
Let the sufficient conditions be verified. Hence,BΞr,tα,Ψ, where ΨZ=∑l=0∞1/∑z=0lrz∑z=0lrzαzZtl, is a prequasi Banach ideal. Assume BΞr,tαP,Q=BP,Q, then there exists η>0 with ΨZ≤ηZ, for all Z∈BP,Q. From Dvoretzky’s theorem [32], for all b∈N, we have quotient spaces P/Yb and subspaces Mb of Q which can be operated onto ℓ2b by isomorphisms Vb and Xb with VbVb−1≤2 and XbXb−1≤2. If Ib is the identity map on ℓ2b, Tb is the quotient map from P to P/Yb, and Jb is the natural embedding map from Mb to Q. Assume mz be the Bernstein numbers [33] hence
(43)1=mzIb=mzXbXb−1IbVbVb−1≤XbmzXb−1IbVbVb−1=XbmzJbXb−1IbVbVb−1≤XbdzJbXb−1IbVbVb−1=XbdzJbXb−1IbVbTbVb−1≤XbαzJbXb−1IbVbTbVb−1,for 0≤l≤b. One gets
(44)∑z=0lrz≤∑z=0lXbrzαzJbXb−1IbVbTbVb−1⇒1≤XbVb−1tl∑z=0lrzαzJbXb−1IbVbTb∑z=0lrztl.
Therefore, for someρ≥1, we have
(45)b+1≤ρXbVb−1∑l=0b∑z=0lrzαzJbXb−1IbVbTb∑z=0lrztl⇒b+1≤ρXbVb−1ΨJbXb−1IbVbTb⇒b+1≤ρηXbVb−1JbXb−1IbVbTb⇒b+1≤ρηXbVb−1JbXb−1IbVbTb⇒b+1≤ρηXbVb−1JbXb−1IbVbTb=ρηXbVb−1Xb−1IbVb≤4ρη.
Hence, there is an inconsistency, whenb⟶∞. Hence, P and Q both cannot be infinite dimensional when BΞr,tαP,Q=BP,Q. This finishes the proof.Theorem 48.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are verified, then BΞr,td is minimum.


## 6.4. Simple Banach Prequasi Ideal

Theorem 49.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are verified with 1<tl1<tl2 and 0<rl2/∑z=0lrz2≤rl1/∑z=0lrz1, for all l∈N, then
(46)BBΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q=ABΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q.Proof.
SupposeX∈BBΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q and X∉ABΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q. From Lemma 3, one has Y∈BBΞrl2,tl2υsP,Q and Z∈BBΞrl1,tl1υsP,Q with ZXYIb=Ib. Then, for all b∈N, one has
(47)IbBΞrl1,tl1υsP,Q=∑l=0∞∑z=0lrz1szIb∑z=0lrz1tl1≤ZXYIbBΞrl2,tl2υsP,Q≤∑l=0∞∑z=0irz2szIj∑z=0lrz2tl2.
This contradicts Theorem46. Hence, X∈ABΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q, which completes the proof.Corollary 50.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are verified with 1<tl1<tl2 and 0<rl2/∑z=0lrz2≤rl1/∑z=0lrz1, for all l∈N, then
(48)BBΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q=KBΞrl2,tl2υsP,Q,BΞrl1,tl1υsP,Q.Proof.
Obviously, sinceA⊂K.Theorem 51.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are confirmed, then BΞr,tυs is simple.Proof.
Suppose the closed idealKBΞr,tυsP,Q contains an operator X∉ABΞr,tυsP,Q. By using Lemma 3, there are Y,Z∈BBΞr,tυsP,Q with ZXYIb=Ib. This implies that IBΞr,tυsP,Q∈KBΞr,tυsP,Q. Hence, BBΞr,tυsP,Q=KBΞr,tυsP,Q. Therefore, BΞr,tυs is simple Banach space.


## 6.5. Eigenvalues ofs-Type Operators

Notation 52.
(49)BVsρ≔BVsρP,Q;PandQareBanachSpaces,whereBVsρP,Q≔X∈BP,Q:ρlXn=0∞∈VandX−ρlXIisnotinvertible,foralll∈N.Theorem 53.
IfP and Q are Banach spaces with dimP=dimQ=∞ and the conditions f1 and f2 are verified, then
(50)BΞr,tυsρP,Q=BΞr,tυsP,Q.Proof.
AssumeX∈BΞr,tυsρP,Q, then ρlXl=0∞∈Ξr,tυ and X−ρlXI=0, for every l∈N. One gets X=ρlXI, for each l∈N, then slX=slρlXI=∣ρlX∣, for all l∈N. Hence, slXl=0∞∈Ξr,tυ, so X∈BΞr,tυsP,Q.
Secondly, assumeX∈BΞr,tυsP,Q. Hence, slXl=0∞∈Ξr,tυ. Therefore, one can see
(51)∑l=0∞∑z=0lrzszX∑z=0lrztl≥∑l=0∞slXtl.
Hence,liml⟶∞slX=0. Suppose X−slXI−1 exists, for all l∈N. Then, X−slXI−1 exists and bounded, for all l∈N. Hence, liml⟶∞X−slXI−1=X−1 exists and bounded. Since BΞr,tυs,Ψ is a prequasi operator ideal, one obtains
(52)I=XX−1∈BΞr,tυsP,Q⇒slIl=0∞∈Ξr,t⇒liml⟶∞slI=0.
Therefore, there is a contradiction, asliml⟶∞slI=1. Then, X−slXI=0, for all l∈N. This implies X∈BΞr,tυsρP,Q. This confirms the proof.


## 7. Kannan Contraction Operator

Theorem 54.
The functionυf=∑l=0∞∑z=0lrzfz/∑z=0lrztl1/ℏ confirms the Fatou property, for every f∈Ξr,t, if the conditions f1 and f2 are verified.Proof.
Letgb⊆Ξr,tυ with limb⟶∞υgb−g=0. Since the space Ξr,tυ is a prequasi closed space, hence g∈Ξr,tυ. Therefore, for each f∈Ξr,tυ, one has
(53)υf−g=∑l=0∞∑z=0lfz−gzrz∑z=0lrztl1/ℏ≤∑l=0∞∑z=0lfz−gzbrz∑z=0lrztl1/ℏ+∑l=0∞∑z=0lgzb−gzrz∑z=0lrztl1/ℏ≤supjinfb≥jυf−gb.Theorem 55.
The functionυf=∑l=0∞∑z=0lrzfz/∑z=0lrztl does not satisfy the Fatou property, for every f∈Ξr,t, if the conditions f1 and f2 are verified.Proof.
Letgb⊆Ξr,tυ with limb⟶∞υgb−g=0. Since the space Ξr,tυ is a prequasi closed space, hence g∈Ξr,tυ. Therefore, for each f∈Ξr,tυ, one gets
(54)υf−g=∑l=0∞∑z=0lfz−gzrz∑z=0lrztl≤2ℏ−1∑l=0∞∑z=0lfz−gzbrz∑z=0lrztl+∑l=0∞∑z=0lgzb−gzrz∑z=0lrztl≤2ℏ−1supjinfb≥jυf−gb.
Hence,υ does not satisfy the Fatou property.
Now, we give the sufficient conditions onΞr,tυ equipped with definite prequasi norm such that there exists a unique fixed point of Kannan contraction operator.Theorem 56.
If the conditionsf1 and f2 are verified, and W:Ξr,tυ⟶Ξr,tυ is Kannan υ-contraction operator, where υf=∑l=0∞∑z=0lrzfz/∑z=0lrztl1/ℏ, for all f∈Ξr,t, hence W has a unique fixed point.Proof.
Assumef∈Ξr,t, hence Wpf∈Ξr,t. As W is a Kannan υ-contraction operator, one has
(55)υWp+1f−Wpf≤λυWp+1f−Wpf+υWpf−Wp−1f⇒υWp+1f−Wpf≤λ1−λυWpf−Wp−1f≤λ1−λ2υWp−1f−Wp−2f≤≤λ1−λpυWf−f.
Hence, for allp,q∈N with q>p, one can see
(56)υWpf−Wqf≤λυWpf−Wp−1f+υWqf−Wq−1f≤λλ1−λp−1+λ1−λq−1υWf−f.
Therefore,Wpf is a Cauchy sequence in Ξr,tυ. As the space Ξr,tυ is prequasi Banach space. Hence, there is g∈Ξr,tυ such that limp⟶∞Wpf=g. To prove that Wg=g. Since υ has the Fatou property, one has
(57)υWg−g≤supiinfp≥iυWp+1f−Wpf≤supiinfp≥iλ1−λpυWf−f=0,so Wg=g. Therefore, g is a fixed point of W. To show that the fixed point is unique. Let we have two different fixed points b,g∈Ξr,tυ of W. Hence, we have
(58)υb−g≤υWb−Wg≤ξυWb−b+υWg−g=0.
Therefore,b=g.Corollary 57.
If the conditionsf1 and f2 are verified, and W:Ξr,tυ⟶Ξr,tυ is Kannan υ-contraction operator, where υf=∑l=0∞∑z=0lrzfz/∑z=0lrztl1/ℏ, for every f∈Ξr,t, then W has one and only one fixed point b with υWpf−b≤λλ/1−λp−1υWf−f.Proof.
From Theorem56, there exists a unique fixed point b of W. Hence, we have
(59)υWpf−b=υWpf−Wb≤λυWpf−Wp−1f+υWb−b=λλ1−λp−1υWf−f.Theorem 58.
If the conditionsf1 and f2 are verified, and W:Ξr,tυ⟶Ξr,tυ, where υf=∑l=0∞∑z=0lrzfz/∑z=0lrztl, for every f∈Ξr,t. The point g∈Ξr,tυ is the only fixed point of W, if the following conditions are confirmed:
(a)
W is Kannan υ-contraction operator(b)
W is υ-sequentially continuous at g∈Ξr,tυ(c)
One hasv∈Ξr,tυ so that the sequence of iterates Wpv has a subsequence Wpiv converging to gProof.
Let the sufficient conditions be verified. Assumeg be not a fixed point of W, hence Wg≠g. From the conditions (b) and (c), we have
(60)limpi⟶∞υWpif−g=0andlimpi⟶∞υWpi+1f−Wg=0.
As the operatorW is Kannan υ-contraction, one gets
(61)0<υWg−g=υWg−Wpi+1f+Wpif−g+Wpi+1f−Wpif≤22ℏ−2υWpi+1v−Wg+22ℏ−2υWpiv−g+2ℏ−1λλ1−λpi−1υWf−f.
Aspi⟶∞, we have a contradiction. Therefore, g is a fixed point of W. To prove that the fixed point g is unique. Let we have two different fixed points g,b∈Ξr,tυ of W. Then, one obtains
(62)υg−b≤υWg−Wb≤λυWg−g+υWb−b=0.
Hence,g=b.Example 59.
AssumeT:Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ⟶Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ, where υp=∑t=0∞∑x=0tx+2/x+1px/∑x=0tx+2/x+12t+3/t+2, with p∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞ and
(63)Tp=p4,υp∈0,1,p5,υp∈1,∞.
As for eachp,q∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with υp,υq∈0,1, one gets
(64)υTp−Tq=υp4−q4≤1274υ3p4+υ3q4=1274υTp−p+υTq−q.
For everyp,q∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with υp,υq∈1,∞, one can see
(65)υTp−Tq=υp5−q5≤1644υ4p5+υ4q5=1644υTp−p+υTq−q.
For allp,q∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with υp∈0,1 and υq∈1,∞, one has
(66)υTp−Tq=υp4−q5≤1274υ3p4+1644υ4q5≤1274υ3p4+υ4q5=1274υTp−p+υTq−q.Hence, the operatorT is Kannan υ-contraction. As υ verifies the Fatou property. From Theorem 56, the operator T has a unique fixed point θ∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ.Assumepa⊆Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with lima⟶∞υpa−p0=0, where p0∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with υp0=1. As the prequasi norm υ is continuous, one has
(67)lima⟶∞υTpa−Tp0=lima⟶∞υpa4−p05=υp020>0.Therefore,T is not υ-sequentially continuous at p0. Hence, the operator T is not continuous at p0.Assumeυp=∑t=0∞∑x=0tx+2/x+1px/∑x=0tx+2/x+12t+3/t+2, for all p∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞.As for eachp,q∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with υp,υq∈0,1, one obtains
(68)υTp−Tq=υp4−q4≤227υ3p4+υ3q4=227υTp−p+υTq−q.Ifp,q∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with υp,υq∈1,∞, one has
(69)υTp−Tq=υp5−q5≤14υ4p5+υ4q5=14υTp−p+υTq−q.For allp,q∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with υp∈0,1 and υq∈1,∞, one gets
(70)υTp−Tq=υp4−q5≤227υ3p4+14υ4q5≤227υ3p4+υ4q5=227υTp−p+υTq−q.Hence, the operatorT is Kannan υ-contraction, and Trp=p/4r,υp∈0,1,p/5r,υp∈1,∞.Obviously,T is υ-sequentially continuous at θ∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ, and Trp has a subsequence Trjp converging to θ. From Theorem 58, the vector θ∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ is the only fixed point of T.Example 60.
SupposeT:Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ⟶Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ, with υp=∑t=0∞∑x=0tx+2/x+1px/∑x=0tx+2/x+12t+3/t+2, for every p∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞ and
(71)Tp=14e1+p,p0∈−∞,13,13e1,p0=13,14e1,p0∈13,∞.
As for everyp,q∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with p0,q0∈−∞,1/3, one has
(72)υTp−Tq=υ14p0−q0,p1−q1,p2−q2,⋯≤227υ3p4+υ3q4≤227υTp−p+υTq−q.
For everyp,q∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with p0,q0∈1/3,∞, then for every ε>0 one gets
(73)υTp−Tq=0≤ευTp−p+υTq−q.For eachp,q∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with p0∈−∞,1/3 and q0∈1/3,∞, one can see
(74)υTp−Tq=υp4≤127υ3p4=127υTp−p≤127υTp−p+υTq−q.Hence, the operatorT is Kannan υ-contraction. Clearly, T is υ-sequentially continuous at 1/3e1∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ, and there exists p∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with p0∈−∞,1/3 so that the sequence of iterates Trp=∑a=1r1/4ae1+1/4rp has a subsequence Trjp=∑a=1rj1/4ae1+1/4rjp converging to 1/3e1. From Theorem 58, the operator T has one fixed point 1/3e1∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ. Recall that T is not continuous at 1/3e1∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ.Supposeυp=∑t=0∞∑x=0tx+2/x+1px/∑x=0tx+2/x+12t+3/t+2, for every p∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞. As for every p,q∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with p0,q0∈−∞,1/3, one has
(75)υTp−Tq=υ14p0−q0,p1−q1,p2−q2,⋯≤1274υ3p4+υ3q4≤1274υTp−p+υTq−q.For everyp,q∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with p0,q0∈1/3,∞, then for every ε>0, one can see
(76)υTp−Tq=0≤ευTp−p+υTq−q.For everyp,q∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ with p0∈−∞,1/3 and q0∈1/3,∞, one has
(77)υTp−Tq=υp4≤1274υ3p4=1274υTp−p≤1274υTp−p+υTq−q.Hence, the operatorT is Kannan υ-contraction. As υ verifies the Fatou property. From Theorem 56, the operator T has a unique fixed point 1/3e1∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞υ.We examine the existence of a fixed point of Kannan contraction operator in the prequasi Banach operator ideal constructed byΞr,tυ and s-numbers.Theorem 61.
The prequasi normΨW=∑l=0∞∑z=0lrzszW/∑z=0lrztl1/h does not satisfy the Fatou property, for all W∈BΞr,tυsP,Q, if the conditions f1 and f2 are confirmed.Proof.
Assume the setup be satisfied andWpp∈N⊆BΞr,tυsP,Q with limp⟶∞ΨWp−W=0. Since the space BΞr,tυs is a prequasi closed ideal. Therefore, W∈BΞr,tυsP,Q. Hence, for every V∈BΞr,tυsP,Q, we have
(78)ΨV−W=∑l=0∞∑z=0lrzszV−W∑z=0lrztl1/ℏ≤∑l=0∞∑z=0lrzsz/2V−Wi∑z=0lrztl1/ℏ+∑l=0∞∑z=0lrzsz/2W−Wi∑z=0lrztl1/ℏ≤21/ℏsuppinfi≥p∑l=0∞∑z=0lrzszV−Wi∑z=0lrztl1/ℏ.
Then,Ψ does not satisfy the Fatou property.Theorem 62.
Suppose the conditionsf1 and f2 be confirmed and G:BΞr,tυsP,Q⟶BΞr,tυsP,Q, where ΨW=∑l=0∞∑z=0lrzszW/∑z=0lrztl1/ℏ, for all W∈BΞr,tυsP,Q. The point A∈BΞr,tυsP,Q is the unique fixed point of G, if the following conditions are satisfied:
(a)
G is Kannan Ψ-contraction mapping(b)
G is Ψ-sequentially continuous at a point A∈BΞr,tυsP,Q(c)
We haveB∈BΞr,tυsP,Q such that the sequence of iterates GpB has a subsequence GpiB converging to AProof.
Assume the sufficient conditions be confirmed. LetA be not a fixed point of G, hence GA≠A. From the setups (b) and (c), we get
(79)limpi⟶∞ΨGpiB−A=0andlimpi⟶∞ΨGpi+1B−GA=0.
SinceG is Kannan Ψ-contraction operator, one has
(80)0<ΨGA−A=ΨGA−Gpi+1B+GpiB−A+Gpi+1B−GpiB≤21/ℏΨGpi+1B−GA+22/ℏΨGpiB−A+22/ℏλλ1−λpi−1ΨGB−B.
Aspi⟶∞, we have a contradiction. Therefore, A is a fixed point of G. To show that the fixed point A is unique. Let we have two different fixed points A,D∈BΞr,tυsP,Q of G. Hence, one can see
(81)ΨA−D≤ΨGA−GD≤λΨGA−A+ΨGD−D=0.
Hence,A=D.Example 63.
LetM:SΞt+1/t+2t=0∞,2t+3/t+2t=0∞υP,Q⟶SΞt+1/t+2t=0∞,2t+3/t+2t=0∞υP,Q, where ΨH=∑t=0∞∑x=0tx+1/x+2sx/∑x=0tx+1/x+22t+3/t+2, for all H∈SΞt+1/t+2t=0∞,2t+3/t+2t=0∞υP,Q and
(82)MH=H6,ΨH∈0,1,H7,ΨH∈1,∞.As for everyH1,H2∈SΞt+1/t+2t=0∞,2t+3/t+2t=0∞υ with ΨH1,ΨH2∈0,1, one has
(83)ΨMH1−MH2=ΨH16−H26≤21254Ψ5H16+Ψ5H26=21254ΨMH1−H1+ΨMH2−H2.For everyH1,H2∈SΞt+1/t+2t=0∞,2t+3/t+2t=0∞υ with ΨH1,ΨH2∈1,∞, one gets
(84)ΨMH1−MH2=ΨH17−H27≤22164Ψ6H17+Ψ6H27=22164ΨMH1−H1+ΨMH2−H2.For everyH1,H2∈SΞt+1/t+2t=0∞,2t+3/t+2t=0∞υ with ΨH1∈0,1 and ΨH2∈1,∞, one has
(85)ΨMH1−MH2=ΨH16−H27≤21254Ψ5H16+22164Ψ6H27≤21254ΨMH1−H1+ΨMH2−H2.Hence, the operatorM is Kannan Ψ-contraction and MrH=H/6r,ΨH∈0,1,H/7r,ΨH∈1,∞.Obviously,M is Ψ-sequentially continuous at the zero operator Θ∈SΞt+1/t+2t=0∞,2t+3/t+2t=0∞υ, and MrH has a subsequence MrjH converging to Θ. From Theorem 62, the zero operator Θ∈SΞt+1/t+2t=0∞,2t+3/t+2t=0∞υ is the only fixed point of M. Let Ha⊆SΞt+1/t+2t=0∞,2t+3/t+2t=0∞υ be so that lima⟶∞ΨHa−H0=0, where H0∈SΞt+1/t+2t=0∞,2t+3/t+2t=0∞υ with ΨH0=1. As the prequasi norm Ψ is continuous, one obtains
(86)lima⟶∞ΨMHa−MH0=lima⟶∞ΨH06−H07=ΨH042>0.Therefore,M is not Ψ-sequentially continuous at H0. Hence, the operator M is not continuous at H0.


## 8. Application to the Existence of Solutions of Nonlinear Difference Equations

Summable equations as (87) are examined by Salimi et al. [34], Agarwal et al. [35], and Hussain et al. [36]. In this section, we search for a solution to (87) in Ξr,tυ, where the conditions f1 and f2 are verified and υf=∑l=0∞∑z=0lrzfz/∑z=0lrztl1/ℏ, for every f∈Ξr,t. Consider the summable equations
(87)fz=pz+∑m=0∞Az,mgm,fm,and suppose W:Ξr,tυ⟶Ξr,tυ be defined by
(88)Wfzz∈N=pz+∑m=0∞Az,mgm,fmz∈N.Theorem 64.
The summable equation (87) has a solution in Ξr,tυ, if A:N2⟶R, g:N×R⟶R, d:N⟶R, and p:N⟶R, suppose there is a number λ such that suplλtl/ℏ∈0,1/2 and for every l∈N, one has
(89)∑z=0l∑m∈NAz,mgm,fm−gm,dmrz≤λ∑z=0lpz−fz+∑m=0∞Az,mgm,fmrz+∑z=0lpz−dz+∑m=0∞Az,mgm,dmrz.Proof.
Assume the setup be verified. Let the mappingW:Ξr,tυ⟶Ξr,tυ defined by equation (88). One has
(90)υWf−Wd=∑l=0∞∑z=0lWfz−Wdzrz∑z=0lrztl1/ℏ=∑l=0∞∑z=0l∑m∈NAz,mgm,fm−gm,dmrz∑z=0lrztl1/ℏ≤suplλtl/ℏ∑l=0∞∑z=0lpz−fz+∑m=0∞Az,mgm,fmrz∑z=0lrztl1/ℏ+suplλtl/ℏ∑l=0∞∑z=0lpz−dz+∑m=0∞Az,mgm,dmrz∑z=0lrztl1/ℏ=suplλtl/ℏυWf−f+υWd−d.
From Theorem56, one gets a solution of equation (87) in Ξr,tυ.Example 65.
Pick up the sequence spaceΞt+2/t+1t=0∞,2t+3/t+2t=0∞ϕ, where υf=∑t=0∞∑x=0tx+2/x+1fx/∑x=0tx+2/x+12t+3/t+2, for all f∈Ξt+2/t+1t=0∞,2t+3/t+2t=0∞. Consider the nonlinear difference equations:
(91)fz=e−3z+6+∑m=0∞−1z+mfz−2pfz−1q+m2+1,with p,q,f−2,f−1>0 and assume W:Ξt+2/t+1t=0∞,2t+3/t+2t=0∞⟶Ξt+2/t+1t=0∞,2t+3/t+2t=0∞, defined by
(92)Wfzz=0∞=e−3z+6+∑m=0∞−1z+mfz−2pfz−1q+m2+1z=0∞.Obviously, there exists a numberλ so that suplλ2l+3/2l+4∈0,1/2 and for every l∈N, one obtains
(93)∑z=0l∑m=0∞−1zfz−2pfz−1q+m2+1−1m−−1mz+2z+1≤λ∑z=0le−3z+6−fz+∑m=0∞−1z+mfz−2pfz−1q+m2+1z+2z+1+λ∑z=0le−3z+6−dz+∑m=0∞−1z+mdz−2pdz−1q+m2+1z+2z+1.From Theorem64, the nonlinear difference equations (91) has a solution in Ξt+2/t+1t=0∞,2t+3/t+2t=0∞.


## 9. Conclusion

In this article, we present some topological and geometric properties ofΞr,tυ, of the multiplication maps acting on Ξr,tυ, of the class BΞr,tυs, and of the class BΞr,tυsρ. We explain the existence of a fixed point of Kannan contraction map acting on these spaces. Some several numerical experiments are introduced to illustrate our results. More, some successful applications to the existence of solutions of nonlinear difference equations are discussed. This article has a number of advantages for researchers such as studying the fixed points of any contraction maps on this prequasi normed sequence space which is a generalization of the quasi normed sequence spaces, a new general space of solutions for many difference equations, examining the eigenvalue problem in this new settings and note that the closed operator ideals are certain to play an important function in the principle of Banach lattices.



---

*Source: 9928947-2021-05-03.xml*