1. Introduction
Let D denote the open unit disk and T denote the unit circle in the complex plane ℂ. For n ≥ 1, let Dn be the polydisk in ℂn, and let Tn be the n-torus which is the distinguished boundary of Dn. Also let dσ be the normalized Haar measure (or Lebesgue measure) on Tn. In the sequel, z will always denote a vector z=(z1,…,zn) in ℂn, and for m=(m1,…mn) ∈ ℤn, zm:=z1m1…znmn and |m|:=m1+…+mn. Also for λ ∈ ℤ we denote the n-tuple (λ,…,λ) as λ and zλ:=z1λ…znλ. In particular, z=z1…zn.
L2(Tn) is the space of Lebesgue square integrable functions on Tn with respect to dσ. Thus,
L2(Tn)={f:Tn ↦ ℂ | f(z)=∑m∈ℤnfmzm, ∑m∈ℤn|fm|2 < ∞} and for f,g ∈L2(Tn), 〈f,g〉:=∫Tnf(z)g(z)¯dσ(z). L∞(Tn) is the subspace of L2(Tn), consisting of all ϕ∈L2(Tn) having the following two properties:
1. ϕf∈L2(Tn) ∀ f∈ L2(Tn).
2. ∃ c>0 such that ‖ϕf‖ ≤ c‖f‖ ∀ f∈ L2(Tn).
For ϕ ∈ L∞(Tn), ‖ϕ‖∞ is defined as ‖ϕ‖∞=in f{c > 0 | ‖ϕf‖ ≤ c‖ f ‖ ∀ f∈ L2(Tn)}.
The Hardy space H2(Dn) consists of all holomorphic functions f on Dn such that sup0<r<1∫Tn| f(rz)|2dσ(z) < ∞. For every function f ∈ H2(Dn), the radial limit limr→1−f(rz) exists for almost every z ∈ Tn [20]. If we denote this radial limit by f(z), then the Hardy space H2(Dn) can be isometrically identified with the closure of the polynomials in L2(Tn). Thus we can now define P to be the orthogonal projection of L2(Tn) onto H2(Dn).
M. C. Ho [11] defined slant Toeplitz operators on L2(T) as those operators whose matrix representation with respect to an orthonormal basis can be obtained by eliminating every other row of a doubly infinite Toeplitz matrix. These types of operators appear frequently in wavelet analysis as their spectral properties have a connection with the smoothness of wavelets. The study of slant Toeplitz operators paved the way to the introduction of more new classes of operators over various function spaces, like kth order slant Toeplitz operators, essentially slant Toeplitz operators, weighted slant Toeplitz operators and so on. For relevant results on slant Toeplitz operators and their generalizations, we refer the reader to [1,3,5,10–13,17,18,23].
In this paper we define slant Toeplitz operators on L2(Tn), for n≥1. Motivated by the matricial definition of slant Toeplitz operators on L2(T), as given by M. C. Ho [11], we define a slant Toeplitz matrix of level-n, and subsequently show that an operator on L2(Tn) is a slant Toeplitz operator if and only if it can be represented as a slant Toeplitz matrix of level-n. For this we first define a Laurent matrix of level-n and establish its relation to a Laurent operator on L2(Tn).
2. Laurent matrix of level n
Definition 2.1. For ϕ ∈ L∞(Tn), the Laurent operator Mϕ is defined as the multiplication by ϕ. That is, Mϕ f=ϕ. f ∀ f ∈ L2(Tn).
Remark 2.2 (i). For ϕ ∈ L∞(Tn), we can easily show that Mϕ∗=Mϕ¯.
(ii) Let i be an integer such that 1 ≤ i ≤ n . If ϕ(z)=zi then Mϕ f=zi f ∀ f ∈ L2(Tn). In this case we denote Mϕ as Mzi.
Definition 2.3 (Laurent Matrix of Level n). Consider scalars {ak}k∈ℤn. A matrix of the type
is said to be Laurent matrix of level 1 and is denoted as
Lk2,…,kn(1).
A block matrix of the type
is said to be Laurent matrix of level 2 and is denoted by
Lk3,…,kn(2).
A block matrix of the type
is said to be Laurent matrix of level 3 and is denoted as Lk4,…,kn(3). Continuing, we get a block matrix of typecalled Laurent matrix of level n denoted as L(n).For example when n=4 we consider the sequence {ak}k∈ℤ4 given by
Then
where j={2, if k2 even;3, if k2 odd, k3 even;4, if k2,k3 odd, k4 even;5, if k2,k3,k4 odd.If we denote Lk2,k3,k4(1) as A,B,C,D for j taking values 2,3,4,5 respectively, then
if k3 is even;if k3 is odd but k4 is even; andif k3 and k4 are both odd. Let us denote these three matrices by E,F,G respectively. Then,if k4 is even; andif k4 is odd. If we denote these two matrices by I and J respectively, thenNotation to be followed:
1. ℤ,ℤ+ and ℕ denote the set of integers, set of non negative integers, and set of positive integers respectively.
2. For k ∈ ℤn, let ek(z):=zk. Then Q={ek}k∈ℤn is an orthonormal basis for L2(Tn), and {ek}k∈ℤ+n is an orthonormal basis for H2(Dn), as shown in [20].
3. For k=(k1,…,kn) ∈ ℤn, we say that k is even if each ki is even. Otherwise k is said to be odd.
4. S:={(t1,…,tn) ∈ ℤn : each ti is either 0 or 1}. If o(S) denotes the order of the set S, then o(S)=2n.
5. For j ∈ [1,n] ∩ ℤ, let ϵj:=(x1,…,xn) where xi=δij.
6. For j∈(1,n] ∩ ℤ, and arbitrarily fixed kj,kj+1,…,kn∈ℤ, let Q[kj,…,kn]={e(k1,…,kn):ki ∈ ℤ for 1≤i < j}. Then Q[kj,…,kn] is an orthonormal basis for L2(Tj−1).
7. For 1 < j≤n, and ki ∈ ℤ with j ≤ i ≤ n, we often use the notation Hj−1,(kj,…,kn) for the space L2(Tj−1) to convey that the basis being considered is Q[kj,…,kn].
Theorem 2.4. For i=1,2,…,n, Mzi∗=Mz¯i and Mzi∗ek=ek−ϵi∀k ∈ ℤn.
Proof. Let f∈ L2(Tn) and f(z)=∑t∈ℤnatzt. Then for i=1,…,n and k ∈ ℤn, we have,
Thus, Mzi∗ek=ek−ϵi=z¯iek which implies that Mzi∗f(z)=z¯i f(z), so that Mzi∗=Mz¯i. □
Remark 2.5. (i)Mziek=ek+ϵi , for 1 ≤ i≤ n and k ∈ ℤn.
(ii) MziMzi∗=I=Mzi∗Mzi ∀ 1 ≤ i ≤ n.
Theorem 2.6. A bounded linear operator A on L2(Tn) can be represented as a Laurent matrix of level-n if and only if 〈Aek+ϵj,et+ϵj〉=〈Aek,et〉∀k, t ∈ ℤn, 1 ≤ j ≤ n.
Proof. Let us assume first that 〈Aek+ϵj,et+ϵj〉=〈Aek,et〉 ∀ k, t ∈ ℤn, 1 ≤ j ≤ n. Also let {aη,ξ}η,ξ∈ℤn be scalars such that
Step I: Let k=(k1,…,kn) ∈ ℤn and t=(t1,…,tn) ∈ ℤn be arbitrarily chosen. Keeping (k2,…,kn) and (t2,…,tn) fixed, if we vary k1 and t1, then 〈Aek+ϵ1,et+ϵ1〉=〈Aek,et〉 implies that at+ϵ1,k+ϵ1=at,k ∀ k1,t1∈ℤ. This means that A:H1,[k2,…,kn] ↦ H1,[t2,…,tn] can be represented as a Laurent matrix of level-1, which we denote as M(t2,…,tn)(k2,…,kn)1.
Step II: Here we keep (k3,…,kn) and (t3,…,tn) fixed and vary k2 and t2. Then 〈Aek+ϵ2,et+ϵ2〉=〈Aek,et〉 implies that M(t2+1,t3,…,tn)(k2+1,k3,…,kn)1=M(t2,…,tn)(k2,…,kn)1, and so A : H2,[k3,…,kn] ↦ H2,[t3,…,tn] can be represented as a Laurent matrix of level 2, say M(t3,…,tn)(k3,…,kn)2.
Step III: Keeping (k4,…,kn) and (t4,…,tn) fixed, and using the fact that 〈Aek+ϵ3,et+ϵ3〉=〈Aek,et〉, we get M(t3+1,t4,…,tn)(k3+1,k4,…,kn)2=M(t3,…,tn)(k3,…,kn)2, and so A : H3,[k4,…,kn] ↦ H3,[t4,…,tn] can be represented as a Laurent matrix of level 3. Continuing in the same way, after n steps we finally arrive at the fact that A:L2(Tn) ↦ L2(Tn) can be represented as a Laurent matrix of level-n.
To prove the converse we assume that A can be represented as a Toeplitz matrix of level-n. Hence for k=(k1,…,kn) and t=(t1,…,tn) arbitrarily chosen, A:Hj−1,[kj,…,kn] ↦ Hj−1[tj,…,tn] can be represented as a Toeplitz matrix of level (j−1), denoted as M(tj…,tn)(kj…,kn)j−1, for each j=2,…,n. This in turn implies that 〈Aek+ϵj,et+ϵj〉=〈Aek,et〉 for j=1,…,n−1.
Finally to show that 〈Aek+ϵn,et+ϵn〉=〈Aek,et〉 we consider the (n−1)-torus L2(Tn−1). For each kn ∈ ℤ,L2(Tn−1) is an isomorphic copy of Hn−1,(kn), and hence the n-torus L2(Tn) can be decomposed as L2(Tn)=⊕kn∈ℤHn−1,[kn]. Therefore, A:L2(Tn)↦L2(Tn) can be expressed as a matrix where the (tn,kn)th entry is Mtn,knn−1, and we also have Mtn+1,kn+1n−1=Mtn,knn−1. This finally implies that 〈Aek+ϵn,et+ϵn〉=〈Aek,et〉. □
Theorem 2.7. Let A be a bounded linear operator on L2(Tn). Then the following conditions are equivalent:
1. A can be represented as a Laurent matrix of level-n
2. A commutes with Mzj for 1≤j≤n
Proof. (1)⇒(2)
Since A can be represented as a Laurent matrix of level-n, so by Theorem 2.6 we have, 〈Aek+ϵj,et〉=〈Aek,et−ϵj〉 ∀ k, t ∈ ℤn, 1 ≤ j ≤ n. Applying Remark 2.5(i) and Theorem 2.4 to this relation we further get 〈AMzjek,et〉=〈MzjAek,et〉 ∀ k, t ∈ ℤn,1 ≤ j ≤ n. Thus, AMzj=MzjA ∀ 1 ≤ j ≤ n.
Suppose MziA=AMzi ∀1 ≤ i ≤ n. Then for k=(k1,…,kn) ∈ ℤn, we have MzkA=Mz1k1…MznknA=AMz1k1…Mznkn=AMzk.
As MzkA=AMzk ∀k∈ℤn, so taking k=ϵi we get MziA=AMzi for each i=1,…,n. Now for k, t∈ℤn, we have 〈Aek+ϵj,et+ϵj〉=〈AMzjek,Mzlet〉=〈MzjAek,Mzjet〉, and using Remark 2.5(ii) we get 〈Aek+ϵj,et+ϵj〉=〈Aek,et〉 for all k, t ∈ ℤn. Hence applying Theorem 2.6 we conclude that A can be represented as a Laurent matrix of level-n. □
Theorem 2.8. A bounded linear operator on L2(Tn) is a Laurent operator if and only if it can be represented as a Laurent matrix of level n.
Proof. For ϕ∈L∞(Tn), MϕMzi=MziMϕ ∀ 1 ≤ i ≤ n, and so by Theorem 2.7, Mϕ can be represented as a Laurent matrix of level n.
Conversely, let ϕ=Ae0, so that ϕ∈L2(Tn). We first show that ϕ is bounded.
We have ‖ϕ‖=‖Ae0‖≤‖A‖ ‖e0‖. Again, ‖ϕ2‖=‖Ae0.ϕ‖ ≤ ‖A‖2‖e0‖. Repeating this argument inductively, we get ‖ϕm‖ ≤ ‖A‖m‖e0‖ for every positive integer m. This norm inequality can be expressed in the form ∫Tn|ϕ(z)‖A‖|2mdσ(z) ≤ ‖e0‖2=1, which implies that |ϕ(z)‖A‖| ≤ 1 almost everywhere. Thus |ϕ(z)| ≤ ‖A‖ almost everywhere which shows that ϕ is bounded. Hence ϕ∈L∞(Tn) and ‖ϕ‖∞ ≤ ‖A‖.
Next we will show that A=Mϕ. As A can be represented as a Laurent matrix of level n, so by Theorem 2.7, AMzk=MzkA ∀ k ∈ ℤn. Therefore,
Thus Af=ϕf for every polynomial f in z1,…,zn.
Let f∈L2(Tn). Then there exist polynomials fm in z1,…,zn such that fm→f. This implies that Afm→Af and ϕfm→ϕf, and since Afm=ϕfm∀m, so Af=ϕf. Equivalently, A=Mϕ. □
The following corollary follows immediately from Theorems 2.6–2.8.
Corollary 2.9. Let A be a bounded linear operator on L2(Tn). Then the following conditions are equivalent:
1. A is a Laurent operator
2. 〈Aek+ϵj,et+ϵj〉=〈Aek,et〉 ∀ k, t ∈ ℤn,1 ≤ j ≤ n
3. Mzi A=AMzi ∀ 1 ≤ i ≤ n.
The proof being obvious is omitted.
3. Toeplitz matrix of level n
Definition 3.1. For ϕ∈L∞(Tn), the Toeplitz operator Tϕ is defined as the compression of Mϕ to H2(Dn). That is, Tϕf=P(ϕ.f) ∀ f∈H2(Dn), where P is the orthogonal projection of L2(Tn) onto H2(Dn). Thus,
Toeplitz operators on H2(Dn) have been earlier discussed by several authors. For the convenience of the reader we mention here a few significant references [4,6–8,14,16,19,21]. In the present work we give a matricial definition of Toeplitz operators on H2(Dn). It may be mentioned that block Toeplitz operators with matrix symbol F∈L∞(ℂn×n) was first discussed in [9]. Again in [19] we come across block matrix representation of Toeplitz and Hankel operators on L2(Tn). In [15] the authors have used lexicographic ordering on ℤ+2 for matrix representation of Toeplitz operators on L2(Tn). These references motivated us to propose our definition of a Toeplitz matrix of level-n. However, our approach is significantly different from what is done in earlier works. We first propose the definition of a Toeplitz matrix of level n and then show that an operator on H2(Dn) is a Toeplitz operator if and only if it can be represented as a Toeplitz matrix of level n.
Definition 3.2 (Toeplitz Matrix of Level n). Consider a sequence of scalars {ak}k∈ℤn. A matrix of the type
is said to be a Toeplitz matrix of level 1, denoted as
Tk2,…,kn(1).
A block matrix of the type
is said to be Toeplitz matrix of level 2, denoted as Tk3,…,kn(2).Continuing, we get a block matrix of type
called Toeplitz matrix of level n denoted as T(n).Theorem 3.3. A bounded linear operator A on H2(Dn) can be represented as a Toeplitz matrix of level n if and only if 〈Aek+ϵj,et+ϵj〉=〈Aek,et〉 ∀ k,t ∈ Z+n and 1 ≤ j ≤ n.
Proof is similar to that of Theorem 2.6 and is therefore omitted.
Theorem 3.4. A necessary and sufficient condition that an operator on H2(Dn) be a Toeplitz operator is that it can be represented as a Toeplitz matrix of level n.
Proof. Let A=Tϕ=PMϕ for ϕ ∈ L∞(Tn). Then for k, t ∈ ℤ+n and 1 ≤ j ≤ n, we have
As Mϕ is a Laurent operator, so Corollary 2.9 implies that
Combining Eqs. (3.1) and (3.2) we get
Thus by Theorem 3.3, A can be represented as a Toeplitz matrix of level n.
Conversely, suppose A is a bounded linear operator on H2(Dn) which can be represented as a Toeplitz matrix of level n. Then by Theorem 3.3, we have
For m ∈ ℤ+, let Am:L2(Tn) ↦ L2(Tn) be defined as Am=Mzm∗APMzm where zm=z1m…znm. Then for k=(k1,…,kn) and t=(t1,…,tn) in ℤ+n, and m ∈ ℤ+, repeated application of Eq. (3.3) gives,
For k, t∈ ℤn, we choose N∈ℤ+, sufficiently large so that ki+N,ti+N ∈ ℤ+ ∀ 1 ≤ i ≤ n, and let k˜=(k1+N,…,kn+N), t˜=(t1+N,…,tn+N). Then by Eq. (3.4), we have ∀m ∈ ℤ+,
Also et˜=MzN et, ek˜=MzNek and Pek˜=ek˜. Using this in Eq. (3.5) we get
Thus ∀ k, t ∈ ℤn, ∃ N∈ ℤ+ such that
Therefore, if f and g are finite linear combinations of ek for k∈ℤn, then the sequence {〈Anf,g〉}n∈ℤ+ is convergent. Also as ‖An‖ ≤ ‖A‖ ∀ n ∈ ℤ+, so the sequence {An}n∈ℤ+ of operators on L2(Tn) is weakly convergent to a bounded operator A∞ on L2(Tn).
Claim: A∞ is a Laurent operator.
As in Eq. (3.6), for k,t∈ℤn and N∈ℤ+, we have
Taking limit as m→∞ , in Eqs. (3.6) and (3.7) we get
Thus 〈A∞ek,et〉=〈A∞ek+ϵi,et+ϵi〉∀ 1 ≤ i ≤ n, k, t ∈ ℤn, and hence Corollary 2.9 implies that A∞ is a Laurent operator, and the claim is established.
From Eq. (3.4) we have 〈Amek,et〉=〈Aek,et〉 ∀ k, t ∈ ℤ+n, and m ∈ ℤ+. Thus for f,g∈H2(Dn), we have 〈Amf,g〉=〈Af,g〉.
Therefore 〈PA∞f,g〉=〈A∞f,g〉=limm→∞〈Amf,g〉=〈Af,g〉 ∀ f,g ∈ H2(Dn). This implies that PA∞|H2(Dn)=A, where A∞ is a Laurent operator. Thus, A is a Toeplitz operator. □
Definition 3.5. For 1 ≤ j ≤ n, we define Uzj:H2(Dn) ↦ H2(Dn) as Uzj f(z)=zjf(z)∀ f∈ H2(Dn), z ∈ Tn.
Remark 3.6. Uzjek=ek+ϵj∀ k ∈ ℤ+n, 1 ≤ j ≤ n.
Theorem 3.7. For 1 ≤ j ≤ n and k=(k1,…,kn) ∈ ℤ+n, we have
Proof. Let f ∈ H2(Dn) where f(z)=∑m∈ℤ+n fmzm.
Then for 1 ≤ j ≤ n and k=(k1,…,kn) ∈ ℤ+n, we have
Remark 3.8. For 1≤j≤n and k=(k1,…,kn)∈ℤ+n, we have
Uzj∗Uzjek=ek and UzjUzj∗ek={ek if kj≠00 Otherwise
Theorem 3.9. Let A be a bounded linear operator on H2(Dn). Then A is a Toeplitz operator if and only if Uzj∗AUzj=A ∀ 1 ≤ j ≤ n.
Proof. If A is a Toeplitz operator on H2(Dn), then, by Theorem 3.4, A can be represented as a Toeplitz matrix of level-n. This implies, by Theorem 3.3, that 〈Aek+ϵj,et+ϵj〉=〈Aek,et〉 ∀ k,t ∈ ℤ+n, 1 ≤ j ≤ n. Applying Remark 3.6 to this relation we further deduce that 〈Uzj∗AUzjek,et〉=〈Aek,et〉 ∀ k,t ∈ ℤ+n,1 ≤ j ≤n, which in turn implies that Uzj∗AUzj=A ∀ 1≤j≤n.
Similarly, assuming Uzj∗AUzj=A ∀ 1 ≤ j ≤ n and applying Theorems 3.3, 3.4 and Remark 3.6 it can be shown that A is a Toeplitz operator on H2(Dn). □
4. Slant Toeplitz operator on L2(Tn)
Let us consider a sequence of scalars {ak}k∈ℤn . A matrix of the type
is said to be slant Toeplitz matrix of level-1, denoted as Ak2,…,kn(1). Observe that it is the matrix obtained by eliminating all odd rows of the Laurent matrix of level-1 namely Lk2,…,kn(1).Similarly if we eliminate all odd rows of the block Laurent matrix
we get the slant Toeplitz matrix of level-2, denotd by Ak3,…,kn(2) and given by
Continuing in this way, we eliminate the odd rows of the block Laurent matrix
to get the slant Toeplitz matrix of level-n, which isTheorem 4.1. If T is a bounded linear operator on L2(Tn), thenT can be represented as a slant Toeplitz matrix of level-n if and only if 〈Tek+2ϵj,et+ϵj〉=〈Tek,et〉 ∀ k,t ∈ ℤn, and 1 ≤ j ≤ n.
The proof follows as in Theorem 2.6, and is therefore omitted.
Theorem 4.2. A bounded linear operator T on L2(Tn) can be represented as a slant Toeplitz matrix of level-n if and only if MzjT=TMzj2 ∀ 1 ≤ j ≤ n.
Proof. Let j be an integer such that 1 ≤ j ≤ n. Then,
if and only if
〈MzjTek,et〉=〈T Mzj2ek,et〉,∀ k,t∈ℤn and
∀ 1 ≤ j ≤ nif and only if 〈Tek,et−ϵj〉=〈Tek+2ϵj,et〉,∀k,t∈ℤn and ∀ 1 ≤ j ≤ n
if and only if 〈Tek,et〉=〈Tek+2ϵj,et+ϵj〉,∀k,t∈ℤn and ∀ 1 ≤ j ≤ n
The result now follows from Theorem 4.1. □
Definition 4.3. W:L2(Tn)↦L2(Tn) is defined to be the linear operator such that for each k∈ℤn, Wek={ek2,if k is even;0,if k is odd.
So for f(z)=∑j∈ℤnajzj∈L2(Tn), we have Wf(z)=∑j∈ℤna2jzj.
Remark 4.4. and so
‖W‖=1(ii) W∗et=e2t∀ t∈ℤn, because 〈Wek,et〉=〈ek,e2t〉 ∀ k,t∈ℤn. So for f(z)=∑p∈ℤnapzp∈L2(Tn), we have W∗f(z)=∑p∈ℤnapz2p=f(z2).
Definition 4.5. For ϕ ∈ L∞(Tn), we define the slant Toeplitz operator Aϕ as W Mϕ. Hence, if ϕ is the constant function 1 then W=A1.
Proof. ‖Aϕ‖=‖W Mϕ‖≤‖W‖ ‖Mϕ‖=‖Mϕ‖=‖ϕ‖∞. □
Theorem 4.7. If Aϕ is a slant Toeplitz operator on L2(Tn), then
Theorem 4.8. A bounded linear operator A on L2(Tn) is a slant Toeplitz operator if and only if A can be represented as a slant Toeplitz matrix of level-n.
If A is a slant Toeplitz operator then by Theorems 4.1 and 4.7 it follows immediately that A can be represented as a slant Toeplitz matrix of level-n. However, for the converse implication we defer the proof to Section 6 as we first have to establish a few more results needed thereof.
5. Properties of W and W∗
Definition 5.1. Let Pe be the projection of L2(Tn) onto the closed span of {e2k}k∈ℤn in L2(Tn). Thus for j∈ℤn we have Peej={ej,if j is even;0,otherwise.
Or equivalently we have Pe(∑k∈ℤnakzk)=∑k∈ℤna2kz2k.
We make the following observations:
- (i)
W W∗=I, and so W is a co-isometry on L2(Tn).
- (ii)
W∗W=Pe, and so W is an isometry on Pe(L2(Tn)).
In the following few results we refer to the subset S of ℤn, the definition for which was already given in Section 2.
Lemma 5.2. Let S={(t1,…,tn) ∈ ℤn| each ti is either 0 or 1}. Then for k ∈ ℤn,k odd, there exist unique k˜ ∈ ℤn and 0≠t ∈ S such that k=2k˜+t.
Proof. Let k=(k1,…,kn) ∈ ℤn. For each 1 ≤ i ≤ n there exists k˜i ∈ ℤ such that
Let k˜=(k˜1,…,k˜n) and t=(t1,…,tn) where
Then k=2k˜+t where k˜∈ℤn, t∈S. Also k is odd implies ti=1 for at least one i, so that t is non-zero.
Uniqueness Suppose ∃ k′=(k′1,…,k′n) ∈ ℤn and t′=(t′1,…,t′n) ∈ S such that k=2k′+t′.
Let if possible t′≠t. Without loss of generality we suppose t′1≠t1, so that either t′1=1 and t1=0, or t1′=0 and t1=1.
Now t1′=1 implies k1=2k1′+t1′=2k1′+1, which is odd,
and t1=0 implies k1=2k˜1+t1=2k˜1, which is even. Thus we get a contradiction. Similarly, t′1=0 and t1=1 give us a contradiction.
Thus t′=t implies 2k′=2k˜ or k′=k˜. □
Remark 5.3. Every non zero entry in S is odd. Also ϵj ∈ S for j=1,2,…,n.
Lemma 5.4. For t,k∈S with t≠0,k≠0, we have t+k is even if and only if t=k.
Proof. Suppose t=k. Then t+k=2t, which is even.
Conversely, suppose t≠k. Then without loss of generality we assume that t1≠k1, where t=(t1,…,tn), k=(k1,…,kn). Thus, either t1=0 and k1=1; or t1=1 and k1=0. In any case,
t+k=(t1+k1,…,tn+kn), where t1+k1=1, and so t+k is odd.
Thus t≠k implies that t+k odd, or equivalently, t+k even implies that t=k. □
Theorem 5.5. W Mz2kW∗=Mzk if k∈ℤn, and W MztW∗=0 if t∈ℤn is odd.
Proof. For k,p∈ℤn, W Mz2kW∗ep=W Mz2ke2p=W e2(p+k)=e(p+k)=Mzkep. Now if t∈ℤn is odd, then by Lemma 5.2, 2p+t is odd, and so we have W MztW∗ep=W Mzte2p=W e(2p+t)=0. □
Corollary 5.6. W MzjW∗=0 ∀ 1 ≤ j ≤ n.
Proof. This follows from Theorem 5.5. Taking t=ϵj we get zt=zj where t is odd. Thus W Mzj−W∗=0. □
Proof. Let t=ϵ1+ϵ2+⋯+ϵn. Then t is odd and zt=z1⋯zn=z. So W MzW∗=0. □
Theorem 5.8. Let f,g∈L2(Tn) such that fg∈L2(Tn). Then
Proof. (i) W∗(eket)=W∗e(k+t)=e2(k+t)=e2ke2t=(W∗ek)(W∗et) for k,t∈ℤn. Thus for f,g∈L2(Tn), W∗(f g)=(W∗f)(W∗g), provided f g∈L2(Tn).
(ii) As WW∗=I so the result follows immediately from (i). □
Theorem 5.9. For ϕ∈L∞(Tn), WAϕ∗=Mψ where ψ=Wϕ¯.
Proof. Let ϕ(z)=∑t∈ℤnatzt, so that ϕ¯(z)=∑t∈ℤna¯tz−t. Then ψ(z)=Wϕ¯(z)=∑t∈ℤna¯tWz−t=∑t∈ℤna¯2tz−t and Mψ=∑t∈ℤna¯2tMz−t.
Again, Aϕ=W Mϕ implies Aϕ∗=Mϕ∗W∗=Mϕ¯W∗=∑t∈ℤna¯tMz−tW∗.
Hence by Theorem 5.5,
Corollary 5.10. If ϕ ∈ L∞(Tn), then Wϕ ∈ L∞(Tn).
Proof. Let ϕ∈L∞(Tn). Then by Theorem 5.9, MWϕ=W Aϕ¯∗, which is bounded on L2(Tn). Therefore W ϕ∈L∞(Tn). □
Lemma 5.11. Let f∈L2(Tn) and f(z)=∑k∈ℤnakzk. If for t∈S, we define ft(z)=∑k∈ℤna2k+tzk, then f(z)=∑t∈Sztft(z2). Also f0(z2)=Pef(z) and f0(z)=W f(z).
Proof. Let f(z)=∑k∈ℤnakzk. If g(z):=∑k∈ℤn,k evenakzk and h(z):=∑k∈ℤn,k oddakzk, then f(z)=g(z)+h(z). For t∈S define ft(z)=∑k∈ℤna2k+tzk. As f∈L2(Tn), so ft∈L2(Tn) ∀ t∈S. We have
Again, by Lemma 5.2,
h(z)=∑0≠t∈S∑k∈ℤna2k+tz2k+t=∑0≠t∈Szt(∑k∈ℤna2k+t(z2)k)=∑0≠t∈Sztft(z2). Thus we have, f(z)=f0(z2)+∑0≠t∈Sztft(z2)=∑t∈Sztft(z2).
Moreover, by Definition 4.3, we have Wg(z)=∑k∈ℤna2kzk=f0(z), and W h(z)=0, so that W f(z)=f0(z).
Again, by Definition 5.1, we have Peg(z)=g(z) and Peh(z)=0 so that Pef(z)=Peg(z)=g(z)=f0(z2). □
Theorem 5.12. Let f,g∈L2(Tn) such that one of f and g is in L∞(Tn). Then W(fg)=(Wf)(Wg)+∑0≠t∈Szt(W z¯tf)(W z¯tg).
Proof. By Lemma 5.11, we have, f(z)=f0(z2)+∑0≠t∈Sztft(z2) and g(z)=g0(z2)+∑0≠t∈Sztgt(z2), where f0(z2)=Pef(z), and g0(z2)=Peg(z). Hence,
Applying Remark 4.4(ii), Theorem 5.8(ii) and Lemma 5.11 in that order, we get,
As f0(z2)(∑0≠t∈Sztgt(z2)) and g0(z2)(∑0≠t∈Sztft(z2)) are expressions that involve only odd powers of z, so by Definition 4.3,
Again, (∑0≠t∈Sztgt(z2))(∑0≠t∈Sztft(z2))=∑0≠t∈S∑0≠t∈Szt+kgt(z2)fk(z2), where t+k is even iff t=k, by Lemma 5.4. Therefore by Definition 4.3,
As g(z)=∑k∈Szkgk(z2), so for any 0≠t∈S, we have
ztgt(z2)=g(z)−∑k∈S,k≠tzkgk(z2)⇒gt(z2)=z¯tg(z)−∑k∈S,k≠tzk−tgk(z2). Therefore,
Similarly,
So, Eqs. (5.1)–(5.6) together imply,
Remark 5.13. The above relation is correct provided the expression on the RHS is in L2(Tn) i.e. (Wf)(Wg) and (W z¯tf)(W z¯tg) should be L2(Tn) functions. This is guaranteed by Corollary 5.10.
6. Properties of Aϕ and Aϕ*
Lemma 6.1. For a bounded linear operator A on L2(Tn), we have MzjA=AMzj2 ∀ 1 ≤ j ≤n if and only if MzkA=AMz2k ∀ k∈ℤn.
Proof. Suppose MzkA=AMz2k ∀ k ∈ ℤn. In this expression if we put k=ϵj for any integer j with 1 ≤ j ≤ n, we get MzjA=AMzj2. Conversely, if Mzj=AMzj2∀ 1 ≤ j ≤ n, then for k=(k1,…,kn) ∈ ℤn we have MzkA=Mz1k1…Mz1knA=AMz12k1…Mzn2kn=Mz2k. □
Theorem 6.2. A bounded linear operator A on L2(Tn) is a slant Toeplitz operator if and only if MzkA=AMz2k ∀ k ∈ ℤn. In such a case we have A=Aϕ with ϕ(z)=∑t∈Sz¯t(Aet)(z2).
Proof. If A is a slant Toeplitz operator, then the result follows from Theorems 4.8, 4.2 and Lemma 6.1.
Conversely, let A be a bounded linear operator on L2(Tn) and suppose MzkA=AMz2k∀ k∈ℤn.
We prove the result in the following steps:
(i) For t∈S, define ϕt(z)=z¯t(Aet)(z2). We will show that ϕt∈L∞(Tn).
(ii) Define ϕ(z)=∑t∈Sϕt(z). We will show that A=Aϕ.
Proof of step (i): Let h∈L2(Tn) with h(z)=∑k∈ℤnδkzk. If ξ(z):=h(z2), then ξ∈L2(Tn), ξ(z)=∑k∈ℤnδkz2k and ‖ξ‖=‖h‖. For t∈S we have
Also, ‖MAet.h‖=‖Aet.h‖=‖AMzt.ξ‖≤‖A‖ ‖Mz‖t‖ξ‖=‖A‖ ‖h‖
Therefore MAet is bounded and so Aet∈L∞(Tn)∀ t∈S.
This implies that ϕt∈L∞(Tn) ∀ t∈S.
Proof of step (ii): As ϕt∈L∞(Tn) ∀ t∈S, so ϕ∈L∞(Tn). Let F∈L2(Tn). So by Lemma 5.11, F(z)=F0(z2)+∑0≠t∈SztFt(z2), where W F(z)=F0(z). Again applying Theorem 5.12 we get
By Definition 4.3, we have Wϕt(z)=0 for 0≠t∈S, and so
Also, by applying Lemma 5.4 we get
Thus, AϕF=AF ∀ F∈L2(Tn) which implies that Aϕ=A. □
We are now in a position to complete the proof of Theorem 4.8, stated in Section 4.
Proof of Theorem 4.8. Suppose A is a bounded linear operator on L2(Tn) such that A can be represented as a slant Toeplitz matrix of level-n. Then by Theorem 4.2 we have MzjA=AMzj2 ∀ 1≤j≤n. This together with Lemma 6.1 implies that MzkA=AMz2k ∀ k∈ℤn. Hence, applying Theorem 6.2, we conclude that A is a slant Toeplitz operator.
Theorem 6.3. AϕAϕ∗=Mψ where ψ=W(|ϕ|2).
Proof. AϕAϕ∗=W MϕMϕ∗W∗=W M|ϕ|2W∗=W(W M|ϕ|2)∗=W A|ϕ|2∗=Mψ, where ψ=W(|ϕ|2), by Theorem 5.9. □
Corollary 6.4. ‖Aϕ‖2=‖ψ‖∞, where ψ=W(|ϕ|2).
Proof. ‖Aϕ‖2=‖AϕAϕ∗‖=‖Mψ‖=‖ψ‖∞. □
Corollary 6.5. Aϕ∗ is an isometry iff W|ϕ|2=1.
Theorem 6.6. The map τ:L∞(Tn)↦B(L2(Tn)) defined as τ(ϕ)=Aϕ is linear and injective. Here B(L2(Tn) denotes the algebra of all bounded linear operators on L2(Tn).
Proof. For ϕ,ψ∈L∞(Tn) and λ∈ℂ,
τ(λϕ+ψ)=Aλϕ+ψ=WMλϕ+ψ=λ(WMϕ)+(WMψ)=λAϕ+Aψ=λτ(ϕ)+τ(ψ). Therefore τ is linear.
Next to show τ is injective: Suppose τ(ϕ)=0. This implies Aϕ=WMϕ=0. Let ϕ(z)=∑k∈ℤnakzk. Then by Lemma 5.11, we have ϕ(z)=∑ξ∈Szξϕξ(z2) where ϕξ(z)=∑k∈ℤna2k+ξzk. So for t∈S we have ϕ(z)zt=∑ξ∈Szξ+tϕξ(z2), and since by Lemma 5.4, ξ+t is even only for ξ=t, so W(zξ+tϕξ(z2))=0 if ξ≠t. Hence we have
Therefore, W Mϕ=0 implies that ϕt(z)=0 ∀ t∈S, from which we can conclude that ϕ(z)=0. □
Theorem 6.7. MϕAψ is a slant Toeplitz operator and MϕAψ=Aϕ(z2).ψ(z).
Proof. For each k∈ℤn we have Mzk(MϕAψ)=MϕMzkAψ=(MϕAψ)Mz2k. So, by Theorem 6.2, MϕAψ is a slant Toeplitz operator. Again, as Aψ is slant Toeplitz, so MzkAψ=AψMz2k∀ k∈ℤn which implies that Mϕ(z)Aψ=AψMϕ(z2)∀ϕ∈L∞(Tn).
Therefore, MϕAψ=AψMϕ(z2)=WMψMϕ(z2)=WMψ(z)ϕ(z2)=Aψ(z)ϕ(z2). □
Theorem 6.8. AψMϕ=MϕAψ if and only if ψ(z)ϕ(z)=ψ(z)ϕ(z2) for any z∈Tn. In particular, if ψ is invertible, then AψMϕ=MϕAψ if and only if ϕ(z)= constant.
Proof. AψMϕ=W MψMϕ=WMψϕ=Aψϕ, and by Theorem 6.7, MϕAψ=Aϕ(z2).ψ(z). Thus AψMϕ=MϕAψ if and only if Aψ(z)ϕ(z)=Aϕ(z2)|ψ(z). In other words, AψMϕ=MϕAψ if and only if ψ(z)ϕ(z)=ψ(z)ϕ(z2) for any z∈Tn.
If ψ is invertible, then there exists ψ−1∈L∞(Tn) such that ψψ−1=I=ψ−1ψ, so that ϕ(z)ψ(z)=ϕ(z2)ψ(z) if and only if ϕ(z)=ϕ(z2).
Now if ϕ(z)=∑k∈ℤnakzk, then ϕ(z2)=∑k∈ℤnakz2k, and soϕ(z)=ϕ(z2) implies ∑k∈ℤn, k oddakzk+∑0≠k∈ℤn(a2k−ak)z2k=0
This gives ak=0 ∀ k odd and a2k=ak ∀ 0≠k∈ℤn
Therefore ak=0 ∀ 0≠k∈ℤn, which yields ϕ(z)=a0, a constant.
On the other hand, if ϕ(z) is constant then ϕ(z)=ϕ(z2), and so we have AψMϕ=MϕAψ. □
Theorem 6.9. WAϕ is a slant Toeplitz operator if and only if ϕ=0.
Proof. The result is obvious if ϕ=0. Conversely suppose WAϕ is a slant Toeplitz operator with ϕ(z)=∑τ∈ℤnδτzτ. Then by Theorem 4.7, we get 〈WAϕek+2ϵj,et+ϵj〉=〈WAϕek,et〉 ∀k,t∈ℤn, 1≤j≤n.
Now 〈WAϕek+2ϵj,et+ϵj〉=〈Aϕek,e2t+ϵj〉=〈W Mϕek,e2t+ϵj〉=〈Mϕek,e4t+2ϵj〉, and similarly, 〈WAϕek,et〉=〈Mϕek,e4t〉.
So, WAϕ is slant Toeplitz implies that 〈Mϕek,e4t+2ϵj〉=〈Mϕek,e4t〉, which yields 〈ϕ(z)zk,z4t+2ϵj〉=〈ϕ(z)zk,z4t〉. Consequently 〈Στδτzτ+k,z4t+2ϵj〉=〈Στδτzτ+k,z4t〉 showing that δ4t−k+2ϵj=δ4t−k ∀k,t∈ℤn, 1≤j≤n.
Thus δk+2ϵj=δk ∀ k∈ℤn, 1≤j≤n.
So, for each k∈ℤn, and j=1,2,…,n, we have δk=δk+2ϵj=δk+4ϵj=δk+6ϵj=⋯ But |k+2λϵj|→∞ as λ→∞, and as ϕ∈L∞(Tn) so δk+2 λϵj→0 as λ→∞.
Hence δk=0. As this is true for all k∈ℤn, so we must have ϕ=0. □
Theorem 6.10. For ϕ,ψ∈L∞(Tn) the following are equivalent:
(i) AϕAψ is a slant Toeplitz operator.
Proof. (i) ⇔ (ii) Using Theorem 6.7 we get AϕAψ=W MϕAψ=W Aϕ(z2).ψ(z). Therefore by Theorem 6.9, AϕAψ is a slant Toeplitz operator if and only if ϕ(z2).ψ(z)=0.
(ii) ⇔ (iii)
As AϕAψ=WAϕ(z2).ψ(z), so AϕAψ=0 implies that W Aϕ(z2).ψ(z) is a slant Toeplitz operator. Therefore, by Theorem 6.9, ϕ(z2).ψ(z)=0. Conversely, ϕ(z2).ψ(z)=0 implies Mϕ(z2).ψ(z)=0, and so AϕAψ=WAϕ(z2).ψ(z)=W2Mϕ(z2).ψ(z)=0. □
The following theorem gives a necessary and sufficient condition for two slant Toeplitz operators to commute.
Theorem 6.11. For ϕ,ψ∈L∞(Tn),AϕAψ=AψAϕ if and only if ϕ(z2)ψ(z)=ψ(z2)ϕ(z) for any z∈ℤn.
Proof. Suppose AϕAψ=AψAϕ. Now AϕAψ=WMϕAψ and by Theorem 6.7, MϕAψ=Aϕ(z2)ψ(z). Thus, AϕAψ=AψAϕ implies W Aϕ(z2)ψ(z)=W Aψ(z2)ϕ(z). In view of Theorem 6.6 we can infer that W Aϕ(z2)ψ(z)−ψ(z2)ϕ(z)=0, which in turn implies that W Aϕ(z2)ψ(z)−ψ(z2)ϕ(z) is a slant Toeplitz operator. Therefore by Theorem 6.9 we get ϕ(z2)ψ(z)−ψ(z2)ϕ(z)=0 which implies ϕ(z2)ψ(z)=ψ(z2)ϕ(z).
Conversely, suppose ϕ(z2)ψ(z)=ψ(z2)ϕ(z). Then by Theorem 6.9, we have W Aϕ(z2)ψ(z)−ψ(z2)ϕ(z) is slant Toeplitz operator and equals zero. This together with Theorems 6.6 and 6.7 implies that AϕAψ=AψAϕ. □
Theorem 6.12. Aϕ2=Aϕ if and only if ϕ=0. In other words, there are no non-zero idempotents among slant Toeplitz operators.
Proof. Suppose Aϕ2=Aϕ. Then Aϕ2=Aϕ.Aϕ is slant Toeplitz, so that by Theorem 6.10 we have Aϕ2=0. Thus Aϕ=0=WAϕ. This in turn implies that WAϕ is slant Toeplitz, and by Theorem 6.9 we get ϕ=0.
Conversely, ϕ=0⇒Mϕ=0⇒Aϕ=W Mϕ=0⇒Aϕ2=Aϕ. □
Theorem 6.13. Aϕ∗ is a slant Toeplitz operator if and only if ϕ=0.
Proof. If ϕ=0 then Aϕ=0=Aϕ∗ and so Aϕ∗ is slant Toeplitz.
Conversely, suppose Aϕ∗ is slant Toeplitz. Then, by Theorem 4.7,
〈Aϕ∗ek+2ϵj,et+ϵj〉=〈Aϕ∗ek,et〉 ∀ k,t∈ℤn, 1≤j≤n. This implies that
〈Aϕet+ϵj,ek+2ϵj〉=〈Aϕet,ek〉 ∀ k,t ∈ ℤn,1≤j≤n, which in turn implies 〈W Mϕet+ϵj,ek+2ϵj〉=〈W Mϕet,ek〉∀ k,t∈ℤn, 1≤j≤n.
Taking ϕ(z)=∑τ∈ℤnaτzτ in the above expression, we get
〈∑τaτzτ+t+ϵj,z2k+4ϵj〉=〈∑τaτzτ+t,z2k〉 ∀ k,t∈ℤn, 1≤j≤n, which yields a2k−t+3ϵj=a2k−t∀ k,t∈ℤn,1≤j≤n
Thus, ak+3ϵj=ak ∀ k∈ℤn,1 ≤ j ≤ n, and as in Theorem 6.9, this implies that ϕ=0. □
Corollary 6.14. There is no non-zero self adjoint slant Toeplitz operator.
Theorem 6.15. Aϕ is hyponormal if and only if ϕ=0.
Proof. Aϕ is hyponormal if and only if ‖Aϕf‖2≥‖Aϕ∗f‖2 ∀ f∈L2(Tn). Equivalently, ‖W(ϕ(z)f(z))‖2≥‖ϕ(z)¯W∗f(z)‖2∀ f∈L2(Tn)
Let ϕ(z)=∑k∈ℤnakzk. If ϕ=0 then clearly Aϕ=0 is hyponormal. For the converse we consider the following two cases.
Case 1: Let f(z)=1=e0(z). Then by Remark 4.4(ii), we have W∗f(z)=1
Therefore, ϕ(z)=∑k∈ℤna2kz2k.
Case 2: Let f(z)=z so that W∗f(z)=z2, and
Thus, ϕ=0. □
Theorem 6.16. A slant Toeplitz operator cannot be an isometry.
Proof. Let, if possible Aϕ be an isometry. Then Aϕ∗Aϕ=I. Or equivalently, ‖Aϕf‖=‖f‖ ∀ f∈L2(Tn)
By Theorem 6.3, AϕAϕ∗=Mψ where ψ=W|ϕ|2
Also by Corollary 6.4, ‖ψ‖∞=‖Aϕ‖2=‖Aϕ∗Aϕ‖=1. Thus, ‖W|ϕ|2‖∞=1
Let ϕ(z)=∑k∈ℤnakzk. Then ‖ϕ‖=(∑k∈ℤn|ak|2)12=[∑t∈S(∑k∈ℤn|a2k+t|2)]12, by Lemma 5.2. Therefore,
For t∈S,Aϕz¯t=W Mϕz¯t=W∑k∈ℤnakzk−t=W∑τ∈ℤnaτ+tzτ. So by Remark 4.4(i), Aϕz¯t=∑τ∈ℤna2τ+tzτ. Combining this along with the fact that ‖Aϕz¯t‖2=‖z¯t‖2=1, we get ‖∑τ∈ℤna2τ+tzτ‖2=1, which implies that ∑τ∈ℤn|a2τ+t|2=1.
Using this in Eq. (6.2) we get ‖W|ϕ|2‖∞=2h2, since o(S)=2n.
So, ‖W|ϕ|2‖∞=1 means 2n2=1, which is only possible if n=0. n being a positive integer this is not possible, and so we conclude that Aϕ cannot be an isometry. □
Theorem 6.17. Aϕ is compact if and only if ϕ=0.
Proof. We know that the set of compact operators on a Hilbert space H is an ideal of the space of bounded linear operators on H. Thus,
Aϕ is compact ⇒AϕMzt is compact ⇒(AϕMzt)∗ is compact ∀ t∈S.
For t∈S,(AϕMzt)∗=(WMϕMzt)∗=(W Mztϕ)∗=Aztϕ∗.
So, W(AϕMzt)∗=W Aztϕ∗=Mψ, where ψ=W(ztϕ¯) by Theorem 5.9
But AϕMzt is compact implies that W(AϕMzt)∗=Mψ is compact, and consequently ψ=0.
Hence 〈W(ztϕ¯),zm〉=〈ψ,zm〉=0 ∀ m∈ℤn, and so 〈ϕ¯,z2m+t〉=0 ∀m∈ℤn, t∈S. Thus, if ϕ(z)=∑k∈ℤnakzk, then we must have a2m+t=0 ∀ m∈ℤn, t∈S, and now, by Lemma 5.2, we get ak=0 ∀ k∈ℤn, so that ϕ=0. □
7. Remarks
In Section 3 we have shown that an operator on H2(Dn) is a Toeplitz operator if and only if it can be represented as a Toeplitz matrix of level n. Using this relation, we can define the operator Bϕ on H2(Dn) as Bϕ=WTϕ, calling it the compression of the slant Toeplitz operator to the Hardy space of the polydisk. A study of the properties of Bϕ vis-a-vis that of Aϕ should yield interesting results. For H2(D), similar studies have been conducted in [2], Section 3 [18], and [22].
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