# Multiplication Operators on Orlicz Generalized Difference (sss)

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

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## Abstract

In this article, we inspect the sufficient conditions on the Orlicz generalized difference sequence space to be premodular Banach (sss). We look at some topological and geometrical structures of the multiplication operators described on Orlicz generalized difference prequasi normed (sss).


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## Body

## 1. Introduction

The multiplication operators have a large subject of mathematics in functional analysis, namely, in eigenvalue distribution theorem, geometric structure of Banach spaces, and theory of fixed point. For more technicalities (see [1–6]), by ℂℕ, c, ℓ∞, ℓr, and c0, we mean the spaces of each, convergent, bounded, r-absolutely summable and convergent to zero sequences of complex numbers. ℕ displays the set of non-negative integers. Tripathy et al. [7] popularized and measured the forward and backward generalized difference sequence spaces:
(1)GΔnm=wk∈ℂℕ:Δnmwk∈G,GΔnm=wk∈ℂℕ:Δnmwk∈G,where m,n∈ℕ, G=ℓ∞, c, or c0, with
(2)Δnmwk=∑ν=0m−1νCνmwk+νn,andΔnmwk=∑ν=0m−1νCνmwk−νn,successively. When n=1, the generalized difference sequence spaces concentrated to GΔm defined and examined by Et and Çolak [8]. If m=1, the generalized difference sequence spaces diminished to GΔn constructed and studied by Tripathy and Esi [9]. While if n=1 and m=1, the generalized difference sequence spaces reduced to GΔ defined and investigated by Kizmaz [10].An Orlicz function [11] is a function ψ:0,∞→0,∞, which is convex, continuous, and nondecreasing with ψ0=0, ψu>0, for u>0 and ψu→∞, as u→∞. In [12], an Orlicz function ψ is called to satisfy the δ2-condition for each values of x≥0, if there is k>0, such that ψ2x≤kψx. The δ2-condition is equivalent to ψlx≤klψx, for every values of x and l>1. Lindentrauss and Tzafriri [13] used the idea of an Olicz function to construct the Orlicz sequence space:
(3)ℓψ=u∈ℂℕ:ρβu<∞,forsomeβ>0,whereρu=∑k=0∞ψ∣uk∣,ℓψ,. is a Banach space with the Luxemburg norm:
(4)u=infβ>0:ρuβ≤1.Every Orlicz sequence space includes a subspace that is isomorphic toc0 or ℓq, for some 1≤q<∞.Recently, different classes of sequences have been examined the usage of Orlicz functions via Et et al. [14], Mursaleen et al. [15–17], and Alotaibi et al. [18].Letr=rj∈ℝ+ℕ, where ℝ+ℕ denotes the space of sequences with positive reals, and we define the Orlicz backward generalized difference sequence space as follows:
(5)ℓψΔn+1mτ=w=wj∈ℂℕ:∃σ>0withτσw<∞,where τw=∑j=0∞ψΔn+1mwj, wj=0, for j<0, Δn+1m∣wj∣=Δn+1m−1∣wj∣−Δn+1m−1∣wj−1∣, and Δ0wj=wj, for all j,n,m∈ℕ. It is a Banach space, with
(6)w=infσ>0:τwσ≤1.Whenψw=wr, then ℓψΔn+1m=ℓrΔn+1m investigated via many authors (see [19–21]). By BW,Z, we will denote the set of every operators which are linear and bounded between Banach spaces W and Z, and if W=Z, we write BW. On sequence spaces, Basarir and Kara examined the compact operators on some Euler Bm-difference sequence spaces [22], some difference sequence spaces of weighted means [23], the Riesz Bm-difference sequence space [24], the B-difference sequence space derived by weighted mean [25], and the mth order difference sequence space of generalized weighted mean [26]. Mursaleen and Noman [27, 28] investigated the compact operators on some difference sequence spaces. The multiplication operators on cesr,. with the Luxemburg norm . elaborated by Komal et al. [29]. İlkhan et al. [30] studied the multiplication operators on Cesáro second order function spaces. Bakery et al. [31] examined the multiplication operators on weighted Nakano (sss). The aim of this article is to explain some results of ℓψΔn+1mτ equipped with the prequasi norm τ. Firstly, we accord the sufficient conditions on the Orlicz generalized difference sequence space to become premodular Banach (sss). Secondly, we provide with the necessity and sufficient conditions on the Orlicz generalized difference sequence space provided with the prequasi norm so that the multiplication operator defined on it is bounded, approximable, invertible, Fredholm, and closed range operator.


## 2. Preliminaries and Definitions

Definition 1 [32].
An operatorV∈BW is known as approximable if there are Dr∈FW, for every r∈ℕ and limr→∞V−Dr=0.ByϒW,Z, we will denote the space of all approximable operators from W to Z.Theorem 2 [32].
LetW be a Banach space with dimW=∞; then,
(7)FW⊂≠ϒW⊂≠BcW⊂≠BW.Definition 3 [33].
An operatorV∈BW is named Fredholm if dimRVc<∞, dimkerV<∞, and RV are closed, where RVc indicates the complement of range V.The sequenceej=0,0,...,1,0,0,⋯ with 1 in the jth coordinate, for all j∈ℕ, will be used in the sequel.Definition 4 [34].
The space of linear sequence spacesY is called (sss) if
(1)
er∈Y with r∈ℕ(2)
Letu=ur∈ℂℕ, v=vr∈Y, and ∣ur∣≤∣vr∣, for every r∈ℕ, then u∈Y. This means Y be solid(3)
Ifurr=0∞∈Y, then ur/2r=0∞∈Y, wherever r/2 indicates the integral part of r/2Definition 5 [35].
A subspace of the (sss)Yτ is named a premodular (sss) if there is a function τ:Y→0,∞ confirming the conditions:
(i)
τy≥0 for each y∈Y and τy=0⇔y=θ, where θ is the zero element of Y(ii)
There existsa≥1 such that τηy≤aητy, for all y∈Y, and η∈ℂ(iii)
For someb≥1, τy+z≤bτy+τz, for every y,z∈Y(iv)
yr≤zr with r∈ℕ implies τyr≤τzr(v)
For someb0≥1, τyr≤τyr/2≤b0τyi(vi)
Ify=yrr=o∞∈Y and d>0, then there is r0∈ℕ with τyrr=r0∞<d(vii)
There ist>0 with τν,0,0,0,⋯≥t∣ν∣τ1,0,0,0,⋯, for any ν∈ℂThe (sss)Yτ is known as prequasi normed (sss) if τ administers the parts (i)-(iii) of Definition 5 and when the space Y is complete under τ, then Yτ is named a prequasi Banach (sss).Theorem 6 [35].
A prequasi norm (sss)Yτ if it is premodular (sss).The inequality [36], ai+biqi≤Haiqi+biqi, where qi≥0 for all i∈ℕ, H=max1,2h−1 and h=supiqi, will be used in the sequel.


## 3. Main Results

### 3.1. Prequasi Norm onℓψΔn+1m

In this section, we explain the conditions on the Orlicz backward generalized difference sequence space to form premodular Banach (sss).Definition 7.
The backward generalized differenceΔn+1m is named an absolute nondecreasing, if ∣xi∣≤∣yi∣, for all i∈ℕ, then Δn+1mxi≤Δn+1myi.Theorem 8.
Letψ be an Orlicz function fulfilling the δ2-condition and Δn+1m be an absolute nondecreasing, then the space ℓψΔn+1mτ can be a premodular Banach (sss), where
(8)τw=∑j=0∞ψΔn+1mwj,forallw∈ℓψΔn+1m.Proof.
(1-i) Assumev,w∈ℓψΔn+1m. Since ψ can be nondecreasing, convex, agreeable δ2-condition, and Δn+1m can be an absolute nondecreasing, then there is b>0 such that
(9)τv+w=∑i=0∞ψΔn+1mvi+wi≤∑i=0∞ψΔn+1mvi+Δn+1mwi≤12∑i=0∞ψ2Δn+1mvi+∑i=0∞ψ2Δn+1mwi≤b2τv+τw≤Bτv+τw<∞,for some B=max1,b/2. Then, v+w∈ℓψΔn+1m.
(1)
(1-ii) Supposeλ∈ℂ and v∈ℓψΔn+1m. Since ψ is fulfilling the δ2-condition, we obtain(10)τλv=∑r=0∞ψΔn+1mλvr≤d∣λ∣∑r=0∞ψΔn+1mvr≤Dλτv<∞,where D=max1,d. Then, λv∈ℓψΔn+1m. So, from parts (1-i) and (1-ii), the space ℓψΔn+1m is linear. Since er∈ℓq⊆ℓψ⊆ℓψΔn+1m, for every r∈ℕ and q≥1, hence, er∈ℓψΔn+1m, for each r∈ℕ.
(2)
Let∣xi∣≤∣yi∣, for every i∈ℕ and y∈ℓψΔn+1m. Since ψ is nondecreasing and Δn+1m is an absolute nondecreasing, therefore, we get(11)τx=∑i=0∞ψΔn+1mxi≤∑i=0∞ψΔn+1myi=τy<∞,hence x∈ℓψΔn+1m(3)
Supposevr∈ℓψΔn+1m, one has(12)τvr2=∑r=0∞ψΔn+1mvr2≤2∑r=0∞ψΔn+1mvr=2τv,then vr/2∈ℓψΔn+1m(i)
Evidently,τw≥0 and τw=0⇔w=θ(ii)
There isD≥1 where τηw≤D∣η∣τw, for every w∈ℓψΔn+1m and η∈ℂ(iii)
For someB≥1, we obtain τv+w≤Bτv+τw, for all v,w∈ℓψΔn+1m(iv)
Plainly from (2).(v)
From (3), we have thatb0=2≥1(vi)
It is apparent thatF¯=ℓψΔn+1m(vii)
Sinceψ is verifying the δ2-condition, there is ζ with 0<ζ≤ψ∣η∣/∣η∣ such that τη,0,0,0,⋯≥ζ∣η∣τ1,0,0,0,⋯, for each η≠0 and ζ>0, if η=0Therefore, the spaceℓψΔn+1mτ is premodular (sss). To show that ℓψΔn+1mτ is a premodular Banach (sss), Suppose xi=xkik=0∞ is a Cauchy sequence in ℓψΔn+1mτ, then for all ε∈0,1, there is i0∈ℕ such that for all i,j≥i0, we get
(13)τxi−xj=∑k=0∞ψΔn+1mxki−xkj<ψε.Sinceψ is nondecreasing; hence, for i,j≥i0 and k∈ℕ, we obtain
(14)Δn+1mxki−Δn+1mxkj<ε.Hence,Δn+1m∣xkj∣ is a Cauchy sequence in ℂ for fixed k∈ℕ, so limj→∞Δn+1mxkj=Δn+1mxk0 for fixed k∈ℕ. Therefore, τxi−x0<ψε, for each i≥i0. Finally, to explain that x0∈ℓψΔn+1m, we have
(15)τx0=τx0−xn+xn≤Bτxn−x0+τxn<∞.So,x0∈ℓψΔn+1m. This implies that ℓψΔn+1mτ is a premodular Banach (sss).Taking into consideration (Theorem6), we be over the following theorem.Theorem 9.
Ifψ is an Orlicz function satisfying the δ2-condition and Δn+1m is an absolute nondecreasing, then the space ℓψΔn+1mτ is prequasi Banach (sss), where
(16)τx=∑j=0∞ψ∣Δn+1mxj,forallx∈ℓψΔn+1m.Corollary 10.
If0<p<∞ and Δn+1m is an absolute nondecreasing, then ℓpΔn+1mτ is a premodular Banach (sss), where τx=∑i=0∞∣Δn+1m∣xip, for all x∈ℓpΔn+1m.


## 3.1. Prequasi Norm onℓψΔn+1m

In this section, we explain the conditions on the Orlicz backward generalized difference sequence space to form premodular Banach (sss).Definition 7.
The backward generalized differenceΔn+1m is named an absolute nondecreasing, if ∣xi∣≤∣yi∣, for all i∈ℕ, then Δn+1mxi≤Δn+1myi.Theorem 8.
Letψ be an Orlicz function fulfilling the δ2-condition and Δn+1m be an absolute nondecreasing, then the space ℓψΔn+1mτ can be a premodular Banach (sss), where
(8)τw=∑j=0∞ψΔn+1mwj,forallw∈ℓψΔn+1m.Proof.
(1-i) Assumev,w∈ℓψΔn+1m. Since ψ can be nondecreasing, convex, agreeable δ2-condition, and Δn+1m can be an absolute nondecreasing, then there is b>0 such that
(9)τv+w=∑i=0∞ψΔn+1mvi+wi≤∑i=0∞ψΔn+1mvi+Δn+1mwi≤12∑i=0∞ψ2Δn+1mvi+∑i=0∞ψ2Δn+1mwi≤b2τv+τw≤Bτv+τw<∞,for some B=max1,b/2. Then, v+w∈ℓψΔn+1m.
(1)
(1-ii) Supposeλ∈ℂ and v∈ℓψΔn+1m. Since ψ is fulfilling the δ2-condition, we obtain(10)τλv=∑r=0∞ψΔn+1mλvr≤d∣λ∣∑r=0∞ψΔn+1mvr≤Dλτv<∞,where D=max1,d. Then, λv∈ℓψΔn+1m. So, from parts (1-i) and (1-ii), the space ℓψΔn+1m is linear. Since er∈ℓq⊆ℓψ⊆ℓψΔn+1m, for every r∈ℕ and q≥1, hence, er∈ℓψΔn+1m, for each r∈ℕ.
(2)
Let∣xi∣≤∣yi∣, for every i∈ℕ and y∈ℓψΔn+1m. Since ψ is nondecreasing and Δn+1m is an absolute nondecreasing, therefore, we get(11)τx=∑i=0∞ψΔn+1mxi≤∑i=0∞ψΔn+1myi=τy<∞,hence x∈ℓψΔn+1m(3)
Supposevr∈ℓψΔn+1m, one has(12)τvr2=∑r=0∞ψΔn+1mvr2≤2∑r=0∞ψΔn+1mvr=2τv,then vr/2∈ℓψΔn+1m(i)
Evidently,τw≥0 and τw=0⇔w=θ(ii)
There isD≥1 where τηw≤D∣η∣τw, for every w∈ℓψΔn+1m and η∈ℂ(iii)
For someB≥1, we obtain τv+w≤Bτv+τw, for all v,w∈ℓψΔn+1m(iv)
Plainly from (2).(v)
From (3), we have thatb0=2≥1(vi)
It is apparent thatF¯=ℓψΔn+1m(vii)
Sinceψ is verifying the δ2-condition, there is ζ with 0<ζ≤ψ∣η∣/∣η∣ such that τη,0,0,0,⋯≥ζ∣η∣τ1,0,0,0,⋯, for each η≠0 and ζ>0, if η=0Therefore, the spaceℓψΔn+1mτ is premodular (sss). To show that ℓψΔn+1mτ is a premodular Banach (sss), Suppose xi=xkik=0∞ is a Cauchy sequence in ℓψΔn+1mτ, then for all ε∈0,1, there is i0∈ℕ such that for all i,j≥i0, we get
(13)τxi−xj=∑k=0∞ψΔn+1mxki−xkj<ψε.Sinceψ is nondecreasing; hence, for i,j≥i0 and k∈ℕ, we obtain
(14)Δn+1mxki−Δn+1mxkj<ε.Hence,Δn+1m∣xkj∣ is a Cauchy sequence in ℂ for fixed k∈ℕ, so limj→∞Δn+1mxkj=Δn+1mxk0 for fixed k∈ℕ. Therefore, τxi−x0<ψε, for each i≥i0. Finally, to explain that x0∈ℓψΔn+1m, we have
(15)τx0=τx0−xn+xn≤Bτxn−x0+τxn<∞.So,x0∈ℓψΔn+1m. This implies that ℓψΔn+1mτ is a premodular Banach (sss).Taking into consideration (Theorem6), we be over the following theorem.Theorem 9.
Ifψ is an Orlicz function satisfying the δ2-condition and Δn+1m is an absolute nondecreasing, then the space ℓψΔn+1mτ is prequasi Banach (sss), where
(16)τx=∑j=0∞ψ∣Δn+1mxj,forallx∈ℓψΔn+1m.Corollary 10.
If0<p<∞ and Δn+1m is an absolute nondecreasing, then ℓpΔn+1mτ is a premodular Banach (sss), where τx=∑i=0∞∣Δn+1m∣xip, for all x∈ℓpΔn+1m.


## 4. Bounded Multiplication Operator onℓψΔn+1m

Here and after, we explain some geometric and topological structures of the multiplication operator reserve onℓψΔn+1m.Definition 11.
Letκ∈ℂℕ∩ℓ∞ and Wτ be a prequasi normed (sss). An operator Vκ:Wτ→Wτ is named multiplication operator if Vκw=κw=κrwrr=0∞∈W, for every w∈W. If Vκ∈BW, we call it a multiplication operator generated by κ.Theorem 12.
Ifκ∈ℂℕ, ψ is an Orlicz function verifying the δ2-condition, and Δn+1m is an absolute nondecreasing, then κ∈ℓ∞, if and only if, Vκ∈BℓψΔn+1mτ, where τx=∑r=0∞ψΔn+1mxr, for each x∈ℓψΔn+1m.Proof.
Assume the conditions can be satisfied. Letκ∈ℓ∞. So, there is ε>0 with ∣κr∣≤ε, for each r∈ℕ, for x∈ℓψΔn+1mτ. Since Δn+1m is an absolute nondecreasing and ψ is nondecreasing verifying the δ2-condition, then
(17)τVκx=τκx=τκrxrr=0∞=∑r=0∞ψΔn+1mκrxr≤∑r=0∞ψΔn+1mεxr≤dε∑r=0∞ψΔn+1mxr≤Dτx,where D=max1,dε. This implies Vκ∈BℓψΔn+1mτ. Inversely, suppose that Vκ∈BℓψΔn+1mτ. Let us suppose κ∉ℓ∞, hence, for all j∈ℕ, there is ij∈ℕ so as to κij>j. Since Δn+1m is an absolute nondecreasing and ψ is nondecreasing, one has
(18)τVκeij=τκeij=τκreijrr=0∞=∑r=0∞ψΔn+1mκreijr=ψΔn+1mκij>ψΔn+1mj=ψΔn+1mjτeij.
This proves thatVκ∉BℓψΔn+1mτ. Therefore, κ∈ℓ∞.Theorem 13.
Letκ∈ℂℕ and ℓψΔn+1mτ be a prequasi normed (sss), with τx=∑r=0∞ψΔn+1mxr, for all x∈ℓψΔn+1m. Then, ∣κr∣=1, for every r∈ℕ, if and only if, Vκ is an isometry.Proof.
Presume∣κr∣=1, for each r∈ℕ, we have
(19)τVκx=τκx=τκrxrr=0∞=∑r=0∞ψΔn+1mκrxr=∑r=0∞ψΔn+1mxr=τx,for each x∈ℓψΔn+1mτ. Therefore, Vκ is an isometry. Inversely, suppose that ∣κi∣<1, for some i=i0, given that Δn+1m is an absolute nondecreasing and ψ is nondecreasing, we get
(20)τVκei0=τκei0=τκrei0rr=0∞=∑r=0∞ψΔn+1mκrei0r<∑r=0∞ψΔn+1mei0r=τei0.Whileκi0>1, we can show that τVκei0>τei0. As a result, in both cases, we obtain a contradiction. Therefore, κr=1, for all r∈ℕ.


## 5. Approximable Multiplication Operator onℓψΔn+1m

In this section, we investigate the sufficient conditions on the Orlicz backward generalized difference sequence space equipped with prequasi normτ so that the multiplication operator acting on ℓψΔn+1m is an approximable and compact.By cardA, we denote the cardinality of the set A.Theorem 14.
Ifκ∈ℂℕ and ℓψΔn+1mτ is a prequasi normed (sss), where τx=∑r=0∞ψΔn+1mxr, for all x∈ℓψΔn+1m, then Vκ∈ϒℓψΔn+1mτ if and only if κrr=0∞∈c0.Proof.
LetVκ∈ϒℓψΔn+1mτ. So, Vκ∈BcℓψΔn+1mτ, to show that κrr=0∞∈c0. Assume κrr=0∞∉c0, therefore there is δ>0 so that Aδ=r∈ℕ:∣κr∣≥δ has cardAδ=∞. Suppose ai∈Aδ, for each i∈ℕ, then eai:ai∈Aδ is an infinite bounded set in ℓψΔn+1mτ. Suppose
(21)τVκeai−Vκeaj=τκeai−κeaj=τκreair−eajrr=0∞=∑r=0∞ψΔn+1mκreair−eajr≥∑r=0∞ψΔn+1mδeair−eajr=τδeai−δeaj,for every ai,aj∈Aδ. This proves eai:ai∈Bδ∈ℓ∞ which cannot have a convergent subsequence under Vκ. This gives that Vκ∉BcℓψΔn+1mτ. Then, Vκ∉ϒℓψΔn+1mτ, and this gives a contradiction. So, limi→∞κi=0. Contrarily, assume limi→∞κi=0, then for all δ>0, the set Aδ=i∈ℕ:∣κi∣≥δ has card Aδ<∞. So, for all δ>0, the space
(22)ℓψΔn+1mτAδ=x=xi∈ℂAδis finite dimensional. Then, Vκ∣ℓψΔn+1mτAδ is a finite rank operator. For all i∈ℕ, illustrate κi∈ℂℕ by
(23)κij=κj,j∈A1i0,otherwise.
Evidently,Vκi has rank Vκi<∞ as dimℓψΔn+1mτA1/i<∞, for all i∈ℕ. Hence, since Δn+1m is an absolute nondecreasing and ψ is convex and nondecreasing, we obtain
(24)τVκ−Vκix=τκj−κijxjj=0∞=∑j=0∞ψΔn+1mκj−κijxj=∑j=0,j∈A1i∞ψΔn+1mκj−κijxj+∑j=0,j∉A1i∞ψΔn+1mκj−κijxj=∑j=0,j∉A1i∞ψΔn+1mκjxj≤1i∑j=0,j∉A1i∞ψΔn+1mxj<1i∑j=0∞ψΔn+1mxj=1iτx.
This gives thatVκ−Vκi≤1/i, and that Vκ is a limit of finite rank operators. So, Vκ is an approximable operator.Theorem 15.
Pick upκ∈ℂℕ and ℓψΔn+1mτ be a prequasi normed (sss), where τx=∑r=0∞ψΔn+1mxr, for every x∈ℓψΔn+1m. Therefore, Vκ∈BcℓψΔn+1mτ, if and only if,κii=0∞∈c0.Proof.
Clearly, since every approximable operator is compact.Corollary 16.
Ifκ∈ℂℕ, ψ is an Orlicz function satisfying the δ2-condition, and Δn+1m is an absolute nondecreasing, then BcℓψΔn+1mτ⊂≠BℓψΔn+1mτ, where τx=∑r=0∞ψΔn+1mxr, for all x∈ℓψΔn+1m.Proof.
In view ofI that is a multiplication operator on ℓψΔn+1mτ generated by κ=1,1,. So, I∉BcℓψΔn+1mτ and I∈BℓψΔn+1mτ.


## 6. Fredholm Multiplication Operator onℓψΔn+1m

In this section, we introduce the sufficient conditions on the sequence spaceℓψΔn+1m equipped with prequasi norm τ so that the multiplication operator acting on it has closed range, invertible, and Fredholm.Theorem 17.
Letκ∈ℂℕ, ℓψΔn+1mτ, be prequasi Banach (sss), where τx=∑r=0∞ψΔn+1mxr, for all x∈ℓψΔn+1m, and Vκ∈BℓψΔn+1mτ. Then, κ be bounded away from zero on kerκc, if and only if, RVκ is closed.Proof.
Suppose the sufficient condition be satisfied, so, there isε>0 with κi≥ε, for every i∈kerκc, to prove that RVκ is closed. Let d be a limit point of RVκ. Hence, there is Vκxi in ℓψΔn+1mτ, for each i∈ℕ so that limi→∞Vκxi=d. Clearly, Vκxi is a Cauchy sequence. As Δn+1m is an absolute nondecreasing and ψ is nondecreasing, we have
(25)τVκxi−Vκxj=∑r=0∞ψΔn+1mκrxir−κrxjr=∑r=0,r∈kerκc∞ψΔn+1mκrxir−κrxjr+∑r=0,r∉kerκc∞ψΔn+1mκrxir−κrxjr≥∑r=0,r∈kerκc∞ψΔn+1mκrxir−xjr=∑r=0∞ψΔn+1mκryir−yjr>∑r=0∞ψΔn+1mεyir−yjr=τεyn−ym,where
(26)yir=xir,r∈kerκc0,r∉kerκc.
This gives thatyi is a Cauchy sequence in ℓψΔn+1mτ. As ℓψΔn+1mτ is complete, there is x∈ℓψΔn+1mτ so that limi→∞yi=x. As Vκ is continuous, then limi→∞Vκyi=Vκx. Although limi→∞Vκxi=limi→∞Vκyi=d, therefore, Vκx=d. So, d∈RVκ. This implies that RVκ is closed. Inversely, assume RVκ be closed, hence, Vκ be bounded away from zero on ℓψΔn+1mτkerκc. So, there is ε>0 so that τVκx≥ετx, for every x∈ℓψΔn+1mτkerκc.
LetB=r∈kerκc:∣κr∣<ε as Δn+1m is an absolute nondecreasing and ψ is nondecreasing verifying the δ2-condition, if B≠φ; then for i0∈B, one has
(27)τVκei0=τκrei0rr=0∞=∑r=0∞ψΔn+1mκren0r<∑r=0∞ψΔn+1mεen0r≤dετen0,for some d≥1. This implies a contradiction. Therefore, B=φ so that κr≥ε, for each r∈kerκc. This completes the proof of the theorem.Theorem 18.
Letκ∈ℂℕ and ℓψΔn+1mτ be a prequasi Banach (sss), with τw=∑r=0∞ψΔn+1mwr, for every w∈ℓψΔn+1m. There are b>0 and B>0 so that b<κr<B, for every r∈ℕ, if and only if, Vκ∈BℓψΔn+1mτ be invertible.Proof.
Assume the conditions be established, defineγ∈ℂℕ by γr=1/κr, from Theorem 12, we obtain Vκ,Vγ∈BℓψΔn+1mτ and Vκ.Vγ=Vγ.Vκ=I. Therefore, Vγ is the inverse of Vκ. Conversely, let Vκ be invertible. Hence, RVκ=ℓψΔn+1mτℕ. This gives RVκ which is closed. From Theorem 17, there is b>0 so that ∣κr∣≥b, for each r∈kerκc. Now, kerκ=φ, else κr0=0, for several r0∈ℕ, we have er0∈kerVκ. This implies a contradiction, as kerVκ is trivial. Therefore, ∣κr∣≥a, for every r∈ℕ. Because Vκ is bounded, so from Theorem 12, there is B>0 so that ∣κr∣≤B, for each r∈ℕ. Hence, we have proved that b≤∣κr∣≤B, for every r∈ℕ.Theorem 19.
Pick upκ∈ℂℕ and ℓψΔn+1mτ be a prequasi Banach (sss), where τw=∑r=0∞ψΔn+1mwr, for every w∈ℓψΔn+1m. Then, Vκ∈BℓψΔn+1mτ be the Fredholm operator, if and only if, (i) card kerκ<∞ and (ii) ∣κr∣≥ε, for each r∈kerκc.Proof.
AssumeVκ be Fredholm, Let card kerκ=∞. Therefore, en∈kerVκ, for every n∈kerκ. As en’s is linearly independent, this implies card kerVκ=∞. This gives a contradiction. Hence, card kerκ<∞. From Theorem 17, condition (ii) is verified. Next, if the necessary conditions are satisfied, to prove that Vκ is Fredholm, from Theorem 17, condition (ii) implies that RVκ is closed. Condition (i) gives that dimkerVκ<∞ and dimRVκc<∞. So, Vκ is Fredholm.



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*Source: 6627966-2020-12-03.xml*