# Onkth-Order Slant Weighted Toeplitz Operator

Authors: S. C. Arora; Ritu Kathuria
Journal: The Scientific World Journal
 (2013)

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## Abstract

Letβ={βn}n∈ℤ be a sequence of positive numbers with β0=1, 0<βn/βn+1≤1 when n≥0 and 0<βn/βn-1≤1 when n≤0. A kth-order slant weighted Toeplitz operator on L2(β) is given by Uϕ=WkMϕ, where Mϕ is the multiplication on L2(β) and Wk is an operator on L2(β) given by Wkenk(z)=(βn/βnk)en(z), {en(z)=zk/βk}k∈ℤ being the orthonormal basis for L2(β). In this paper, we define a kth-order slant weighted Toeplitz matrix and characterise Uϕ in terms of this matrix. We further prove some properties of Uϕ using this characterisation.


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## Body

## 1. Introduction

Toeplitz operators and slant Toeplitz operators [1] have been found immensely useful, especially in the study of prediction theory [2], wavelet analysis [3], and solution of differential equations [4]. Originally, these operators were defined and studied on the usual H2 and L2 spaces. During the past few decades, different generalisations of these spaces, like the weighted function spaces HwP and LwP and the weighted sequence spaces H2(β) and L2(β) have developed. Shields [5] has made a systematic study of the multiplication operator on L2(β). Lauric [6] has discussed the Toeplitz operator on H2(β). Motivated by these studies, we introduced and studied the notion of a slant weighted Toeplitz operator [7] on L2(β). In this paper, we study a kth-order slant weighted Toeplitz operator on the space L2(β). We begin with the following preliminaries.Letβ={βn}n∈ℤ be a sequence of positive numbers with β0=1, 0<(βn/βn+1)≤1 when n≥0 and 0<(βn/βn-1)≤1 when n≤0.Throughout the paper, we assume that for a fixed integerk≥2, βkn/βn≤M<∞. Consider the following spaces [5, 6]:
(1)L2(β)={f(z)=∑n=-∞∞‍anzn|an∈ℂ,mmm∥f∥β2=∑n=-∞∞‍|an|2βn2<∞},H2(β)={f(z)=∑n=0∞‍anzn|an∈ℂ,mmm∥f∥β2=∑n=0∞‍|an|2βn2<∞}.Then,(L2(β), ∥·∥β) is a Hilbert space [6] with an orthonormal basis given by {ek(z)=zk/βk}k∈ℤ and with an inner product defined by
(2)〈∑n=-∞∞‍anzn,∑n=-∞‍bnzn〉=∑n=-∞∞‍anb-nβn2.
Also, H2(β) is a subspace of L2(β). Further, the space
(3)L∞(β)={ϕ(z)=∑n=-∞∞‍anzn|ϕL2(β)⊆L2(β),∃c∈Rmsuchthat∥ϕf∥β≤c∥f∥β∀f∈L2(β)∑n=-∞∞‍anzn|}
is a Banach space with respect to the norm defined by
(4)∥ϕ∥∞=inf{c|∥ϕf∥β≤c∥f∥β∀f∈L2(β)}.
The mapping P:L2(β)→H2(β) is the orthogonal projection of L2(β) onto H2(β). For a given ϕ∈L∞(β), the induced weighted multiplication operator [5] is denoted by Mϕ and is given by Mϕ:L2(β)→L2(β) such that
(5)Mϕek(z)=1βk∑n=-∞∞‍anβn+ken+k(z)=1βk∑n=-∞∞‍an-kβnen(z).If we putϕ1(z)=z, then Mϕ1=Mz is the operator defined as Mzek(z)=wkek+1(z), where wk=βk+1/βk for all k∈ℤ, and it is known as a weighted shift [5].The slant weighted Toeplitz operator [7] Aϕ is an operator on L2(β) defined as Aϕ:L2(β)→L2(β) such that Aϕek(z)=(1/βk)∑n=-∞∞a2n-kβnen(z).IfW:L2(β)→L2(β) is such that
(6)We2n(z)=βnβ2nen(z),We2n-1(z)=0∀n∈ℤ,
then Aϕ can be alternately defined by
(7)Aϕ(f)=WMϕ(f)=W(ϕf)∀f∈L2(β).


## 2.kth-Order Slant Weighted Toeplitz Operator

Suppose that the operatorWk:L2(β)→L2(β) is such that
(8)Wken(z)={βn/kβnen/k(z)ifnisdivisiblebyk,0otherwise,
Then the matrix of Wk is
(9)[⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯β0β00000000⋯⋯0000β1βk0⋯⋯⋯︸k-times⋯0000⋯⋯0β2β2k⋯⋯00⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯].
The adjoint of Wk is given by
(10)Wk*en(z)=βnβknekn(z)∀n∈ℤ.Definition 1 (see [8]).
For an integerk≥2, the kth order slant weighted Toeplitz operator Uϕ:L2(β)→L2(β) is such that Uϕ(f)=WkMϕ(f) for all f∈L2(β). Thus, Uϕej(z)=(1/βj)∑n=-∞∞ank-jβnen(z).
The(i,j)th entry of the matrix of Uϕ is given by
(11)〈Uϕej,ei〉=〈1βj∑n=-∞∞‍ank-jβnen(z),ei(z)〉=aik-jβiβj.
Hence, the matrix of Uϕ with respect to this basis is as follows:
(12)[⋯⋯⋯⋯⋯⋯⋯a-k+1β-1β-1a-kβ1β0a-k-1β-1β1a-k-2β-1β2⋯⋯a1β0β-1a0β0β0a-1β0β1a-2β0β2⋯⋯ak+1β1β-1akβ1β0ak-1β1β1ak-2β1β2⋯⋯a2k+1β2β-1a2kβ2β0a2k-1β2β1a2k-2β2β2⋯⋯⋯⋯⋯⋯⋯].Theorem 2.
The mappingϕ→Uϕ is linear and one to one.Proof.
For linearity, consider(13)Uαϕ+βψ=WkMαϕ+βψ=Wk(Mαϕ+Mβψ)=WkMαϕ+WkMβψ=αWkMϕ+βWkMψ=αUϕ+βUψ.
For checking one-to-one, let Uϕ=Uψ.
Then,Uϕ-Uψ=0. By linearity, we get Uϕ-ψ=0.
Hence,(14)Uϕ-ψen(z)=0,n∈ℤ,⟹WkMϕ-ψen(z)=0,n∈ℤ,⟹Wk(ϕ-ψ)en(z)=0,n∈ℤ.
This implies that either ϕ-ψ=0 or the degree of (ϕ-ψ)en(z) is not divisible by k. But since this is true for all n∈ℤ, the second possibility is ruled out. Hence, ϕ-ψ=0 or ϕ=ψ.Theorem 3.
Consider the following: (i)MzWk=WkMzk; (ii) MzmWk=WkMzkm, m∈ℤ.Proof.
(i) It is sufficient to prove that(15)MzWken(z)=WkMzken(z)∀n∈ℤ.
Suppose thatn is not a multiple of k. Then,
(16)MzWken(z)=Mz0=0=WkMzken(z).
Now, whenn=pk (multiple of k),
(17)MzWken(z)=Mzβpβpkep(z)=βp+1βpkep+1(z).
On the other hand,(18)WkMzken(z)=WkMzkepk(z)=Wkβ(p+1)kβpke(p+1)k(z)=βp+1βpkep+1(z).
Hence from (17) and (18), we get that
(19)MzWken(z)=WkMzken(z)whenevernisamultipleofk.
We therefore conclude that for all n∈ℤ,
(20)MzWken(z)=WkMzken(z)⟹MzWk=WkMzk.
(ii) We prove the result by induction onm.
Form=1, the result is true by part (i).
Suppose that the result is true form=l. Then,
(21)MzlWk=WkMzlk.
Now(22)Mz(l+1)kWk=MzMzlWk=MzWkMzlk=WkMzkMzlk=WkMz(l+1)k.
Hence by induction, the result is true for allm∈Z+.
Form=-1 and 1≤j≤k-1,
(23)MzmWkenk+j(z)=Mz-10=0=WkMz-kenk+j(z).
For m=-1 and j=nk (a multiple of k),
(24)MzmWkej(z)=Mz-1Wkenk(z)=Mz-1βnβnken(z)=βnβnkMz-1en(z)=βnβnkβn-1βnen-1(z)=βn-1βnken-1(z).
On the other hand,
(25)WkMz-1enk(z)=Wk(z-1znkβnk)=Wkz(n-1)kβnkβ(n-1)kβ(n-1)k=β(n-1)kβnkWke(n-1)k(z)=β(n-1)kβnkβn-1β(n-1)ken-1(z)=βn-1βnken-1(z).
From (23), (24), and (25), we conclude that
(26)Mz-1Wk=WkMz-k.
Hence, the result is true for m=-1 also. Therefore, by using induction, we can prove it for all negative integers m. The case when m=0 is trivially true.
Hence the theorem is true.Corollary 4.
Letϕ(z)=∑n=-∞∞anzn∈L∞(β). Then,
(27)Mϕ(z)Wk=WkMϕ(zk).Theorem 5.
Consider the following:Mϕ(z)Uψ(z)=Uϕ(zk)ψ(z).Proof.
The proof is as follows:(28)Mϕ(z)Uψ(z)=Mϕ(z)WkMψ(z)=WkMϕ(zk)Mψ(z)=WkMϕ(zk)ψ(z)=Uϕ(zk)ψ(z).


## 3. Matrix Characterisation ofUϕ

Definition 6.
Letwn=βn+1/βn for all n∈ℤ. Then, the kth order slant weighted Toeplitz matrix corresponding to the weight sequence 〈wn〉 is a bilaterally infinite matrix 〈λij〉 such that
(29)λi+1,j+k=wiwjwj+1⋯wj+k-1λi,j,i,j=0,±1,±2,….
We have proved earlier [8] that U is a kth order slant weighted Toeplitz operator if and only if MzU=UMzk.
Next, we prove a characterisation of thekth order slant weighted Toeplitz operator in terms of the matrix previously defined.Theorem 7.
A necessary and sufficient condition that an operatorU on L2(β) is a kth order slant weighted Toeplitz operator is that its matrix with respect to the orthonormal basis {ek(z)=zk/βk}k∈ℤ is a kth order slant weighted Toeplitz matrix.Proof.
Suppose thatU is a kth order slant weighted Toeplitz operator. Then, the corresponding matrix 〈λij〉 is given by
(30)λij=〈Uej,ei〉=aki-jβiβj.
Further,
(31)λi+1,j+k=〈Uej+k,ei+1〉=aki+k-j-kβi+1βj+k=aki-jβi+1βj+k=βi+1βj+kβjβiλi,j=wiwjwj+1⋯wj+k-1λi,j,
where wn=βn+1/βn for all n∈ℤ. Thus, the matrix of  U is a kth order slant weighted Toeplitz matrix. Conversely, we assume that U is an operator on L2(β) whose matrix is a kth order slant weighted Toeplitz matrix. This means that for all i,j∈ℤ, we have
(32)〈Uej,ei〉=wjwj+1⋯wj+k-1wi〈Uej+k,ei+1〉.
Now,
(33)〈MzUej,ei〉=〈Uej,Mz*ei〉=〈Uej,wi-1ei-1〉=wi-1〈Uej,ei-1〉=wi-1wjwj+1⋯wj+k-1wi-1×〈Uej+k,ei〉(using(32))=〈UMzkej,ei〉,∀i,j∈ℤ.
Hence MzU=UMzk. Therefore, we conclude that U is a kth order slant weighted Toeplitz operator.Next, consider the operatorS:L2(β)→L2(β) given by Sej=(1/wj)ej+1. Then, S*ej=(1/wj-1)ej-1.Now,S is bounded as 〈wn〉 is always positive and bounded.Lemma 8.
Consider the following:S*=Mz-1.Proof.
The proof is as follows:(34)S*Mzej=S*wjej+1=wjwjej=ej,MzS*ej=Mz1wj-1ej-1=wj-1wj-1ej=ej.Theorem 9.
A bounded operatorU on L2(β) is a kth order slant weighted Toeplitz operator if and only if  U=Mz-1UMzk, where Mz and Mzk, are the multiplication operators induced by z and zk, respectively.Proof.
LetU be a kth order slant weighted Toeplitz operator on L2(β). Then, from (32) we get that
(35)〈Uej,ei〉=wjwj+1⋯wj+k-1wi〈Uej+k,ei+1〉=〈UMzkej,Sei〉=〈S*UMzkej,ei〉=〈Mz-1UMzkej,ei〉,i,j=0,±1,±2,….
Thus, U=Mz-1UMzk.
For the converse part, we assume thatU is a bounded operator on L2(β) satisfying U=Mz-1UMzk for some fixed integer k≥2. Then, for all i,j∈ℤ,
(36)〈Uej,ei〉=〈Mz-1UMzkej,ei〉=〈S*UMzkej,ei〉=〈UMzkej,Sei〉wjwj+1⋯wj+k-1wi〈Uej+k,ei+1〉.
The previous equation shows that the matrix corresponding to U is a kth order slant weighted Toeplitz matrix. Hence, by Theroem 7,   U is a kth order slant weighted Toeplitz operator.Corollary 10.
For a fixed integerk≥2, the set of all kth order slant weighted Toeplitz operators is weakly closed and hence strongly closed.Proof.
We assume that for each positive integern, Un is a kth order slant weighted Toeplitz operator and Un→U weakly. Then, for f,g∈L2(β), we get that 〈Unf,g〉→〈Uf,g〉.
From the previous theorem, this implies that(37)〈Mz-1UnMzkf,g〉=〈S*UnMzkf,g〉=〈UnMzkf,Sg〉=〈Unzk·f,Sg〉⟶〈Uzk·f,Sg〉=〈Mz-1UMzkf,g〉.
But, for each n, Un=Mz-1UnMzk.
Hence,U=Mz-1UMzk. Therefore, U is a kth order slant weighted Toeplitz operator.


## 4. Compression toH2(β)

Definition 11.
The compression of akth order slant weighted Toeplitz operator on H2(β) is denoted by Vϕ and defined as Vϕej(z)=(1/βj)∑n=0∞ank-jβnen(z).j=0,1,2,….Alternatively,(38)Vϕ=PUϕ|H2(β)=PWkMϕ|H2(β)=WkPMϕ|H2(β)(sincePreducesWk[8])=WkTϕ,
where Tϕ:H2(β)→H2(β) is the Toeplitz operator on H2(β) induced by ϕ∈L∞(β). The matrix of Vϕ is unilaterally infinite and has the form
(39)[a0β0β0a-1β0β1a-2β0β2⋯akβ1β0ak-1β1β1ak-2β1β2⋯a2kβ2β0a2k-1β2β1a2k-2β2β2⋯⋯⋯⋯⋯]
for a given ϕ=∑n=-∞∞anzn.Further, we observe the following.(i)
The mappingϕ→Vϕ is linear and one to one.(ii)
TϕWk=PMϕWk=PWkMϕ(zk)=PUϕ(zk)=Vϕ(zk).



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