(ProQuest: ... denotes non-US-ASCII text omitted.)
Bo Zhang 1 and Yufeng Lu 2
Recommended by Nikolai M. Vasilevski
1, College of Science, Dalian Ocean University, Dalian 116023, China
2, School of Mathematical Sciences, Dalian University of Technology, Dalian 116024, China
Received 24 May 2012; Revised 6 August 2012; Accepted 30 August 2012
This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Let dA (z ) denote the Lebesgue volume measure on the unit ball ... n of ... n normalized so that the measure of ... n equals 1. The Bergman space L a 2 ( ... n ) is the Hilbert space consisting of holomorphic functions on ... n that are also in L 2 ( ... n ,dA ) . Hence, for each z ∈ ... n , there exists an unique function K z ∈ L a 2 ( ... n ) with the following property: [figure omitted; refer to PDF] for all f ∈ L a 2 ( ... n ) . As is well known that the reproducing kernel K z is given by [figure omitted; refer to PDF]
Let P be the orthogonal projection from L 2 ( ... n ,dA ) onto L a 2 ( ... n ) . Given a function [straight phi] ∈ L ∞ ( ... n ,dA ) , the Toeplitz operator T [straight phi] : L a 2 ( ... n ) [arrow right] L a 2 ( ... n ) is defined by the formula [figure omitted; refer to PDF] for f ∈ L a 2 ( ... n ) . Since the Bergman projection P has norm 1 , it is clear that Toeplitz operators defined in this way are bounded linear operators on L a 2 ( ... n ) and || T [straight phi] || ...4; || [straight phi] || ∞ .
We now consider a more general class of Toeplitz operators. For F ∈ L 1 ( ... n ,dA ) , in analogy to ( 1.3) we define an operator T F on L a 2 ( ... n ) by [figure omitted; refer to PDF]
Since the Bergman projection P can be extended to L 1 ( ... n ,dA ) , the operator T F is well defined on H ∞ , the space of bounded analytic functions on ... n . Hence, T F is always densely defined on ... n . Since P is not bounded on L 1 ( ... n ,dA ) , it is well known that T F can be unbounded in general. In [ 1], Zhou and Dong gave the following definitions, which are based on the definitions on the unit disk in [ 2].
Definition 1.1.
Let F ∈ L 1 ( ... n ,dA ) .
(a) We say that F is a T -function if ( 1.4) defines a bounded operator on L a 2 ( ... n ) .
(b) If F is a T -function, we write T F for the continuous extension of the operator (it is defined on the dense subset H ∞ of L 2 ( ... n ) ) defined by ( 1.4). We say that T F is a Toeplitz operator if and only if T F is defined in this way.
(c) If there exists an r ∈ (0,1 ) , such that F is (essentially) bounded on the annulus {z :r < |z | <1 } , then we say F is "nearly bounded."
On the Bergman space of the unit ball, Grudsky et al. [ 3] gave necessary and sufficient conditions for boundedness of Toeplitz operators with radial symbols. These conditions give a characterization of the radial functions in L 1 ( ... n ,dA ) which correspond to bounded operators and furthermore show that the T -functions form a proper subset of L 1 ( ... n ,dA ) which contains all bounded and "nearly bounded" functions.
We denote the semicommutator and commutator of two Toeplitz operators T f and T g by [figure omitted; refer to PDF]
In the setting of the classical Hardy space, Brown and Halmos [ 4] gave a complete characterization for the product of two Toeplitz operators to be a Toeplitz operator. On the Bergman space of the unit disk, Ahern and Cuckovic [ 5] and Ahern [ 6] obtained a similar characterization for Toeplitz operators with bounded harmonic symbols. For general symbols, the situation is much more complicated. Louhichi et al. [ 2] gave necessary and sufficient conditions for the product of two Toeplitz operators with quasihomogeneous symbols to be a Toeplitz operator and Louhichi and Zakariasy made a further discussion in [ 7]. But it remains open to determine when the product of two Toeplitz operators is a Toeplitz operator on the Bergman space.
The problem of determining when the semicommutator ( T f , T g ] or commutator [ T f , T g ] on the Bergman space has finite-rank seems to be far from solution. The analogous problem on the Hardy space has been completely solved (see [ 8, 9]). Guo et al. [ 10] completely characterized the finite-rank semicommutator or commutator of two Toeplitz operators with bounded harmonic symbols on the Bergman space of the unit disk and Luecking [ 11] showed that finite-rank Toeplitz operators on the Bergman space of the unit disk must be zero. Recently, Cuckovic and Louhichi [ 12] studied finite-rank semicommutators and commutators of Toeplitz operators with quasihomogeneous symbols and obtained different results from the case of harmonic Toeplitz operators.
Motivated by recent work of Cuckovic and Louhichi, Zhang et al., and Zhou and Dong (see [ 1, 12, 13]), we discuss the finite-rank commutator (semicommutator) of Toeplitz operators with more general symbols on the unit ball in this paper. Let p and s be two multi-indexes. A function f ∈ L 1 ( ... n ,dA ) is called a quasihomogeneous function of quasihomogeneous degree (p ,s ) if f is of the form ξ p ξ ¯ s [straight phi] (r ) for all ξ in the unit sphere ... n and some function [straight phi] (r ) defined on the interval [0,1 ) .
Let f and g be two nonconstant quasihomogeneous functions (with certain restrictions on their quasihomogeneous degree). In this paper, we investigate the following problems:
(1) Under what conditions does T f T g = T F hold for some quasihomogeneous function F ?
(2) Under what conditions does the semicommutator ( T f , T g ] have finite rank?
(3) Under what conditions does the commutator [ T f , T g ] have finite rank?
2. The Mellin Transform and Mellin Convolution
Main tool in this paper will be the Mellin transform. Recall that the Mellin transform [straight phi] ^ of a function [straight phi] ∈ L 1 ( [0,1 ] ,rdr ) is defined by the equation: [figure omitted; refer to PDF] It is easy to check that [figure omitted; refer to PDF] where p ...5;0 and z 0 ∈ ... .
For convenience, we denote [straight phi] ^ (z ) by [straight phi] ⋀ (z ) when the form of [straight phi] is complicated. It is clear that [straight phi] ^ is well defined on {z :Re (z ) >2 } and analytic on {z :Re (z ) >2 } . It is well known that the Mellin transform [straight phi] ^ is uniquely determined by its values on { n k } k ...5;0 , where n k ∈ ... and ∑ k ...5;0 ( 1 / n k ) = ∞ . The following classical theorem is proved in [ 14, page 102].
Theorem 2.1.
Assume that f is a bounded analytic function on {z :Re (z ) >0 } which vanishes at the pairwise distinct points z 1 , z 2 , ... , where
(i) inf { | z n | } >0 and
(ii) ∑ n ...5;1 ...Re ( 1 / z n ) = ∞ .
Then f vanishes identically on {z :Re (z ) >0 } .
Remark 2.2.
We will often use this theorem to show that if [straight phi] ∈ L 1 ( [0,1 ] ,rdr ) and if there exists a sequence { n k } k ...5;0 ⊂ ... such that [figure omitted; refer to PDF] then [straight phi] ^ (z ) =0 for all z ∈ {z :Re (z ) >2 } and so [straight phi] =0 .
If f and g are defined on the interval [0,1 ) , then their Mellin convolution is defined by [figure omitted; refer to PDF] The Mellin convolution theorem states that [figure omitted; refer to PDF] and that, if f and g are in L 1 ( [0,1 ] ,rdr ) , then so is f * M g .
3. Products of Toeplitz Operators with Quasihomogeneous Symbols
For any multi-index α = ( α 1 , ... , α n ) , where each α i is a nonnegative integer, we will write [figure omitted; refer to PDF] for z = ( z 1 , ... , z n ) ∈ ... n .
For α = ( α 1 , ... , α n ) and β = ( β 1 , ... , β n ) , the notation α [succeeds, =] β means that [figure omitted; refer to PDF] and α [perpendicular] β means that [figure omitted; refer to PDF] We also define [figure omitted; refer to PDF] and obtain [figure omitted; refer to PDF]
It is known that [straight phi] is radial if and only if [straight phi] (Uz ) = [straight phi] (z ) for any unitary transform U of ... n . So we have [straight phi] (r ξ ) = [straight phi] (r η ) for ξ , η ∈ ... n and r ∈ [0,1 ) . That is, [straight phi] (z ) depends only on |z | . In this case, we denote [straight phi] by [straight phi] = [straight phi] (r ) for convenience. The definition of quasihomogeneous function on the unit disk has been given in many papers (see [ 2] or [ 7]), and a similar definition on the unit ball will be given in the following.
Definition 3.1.
Let p ,s [succeeds, =]0 . A function f ∈ L 1 ( ... n ,dA ) is called a quasihomogeneous function of quasihomogeneous degree (p ,s ) if f is of the form ξ p ξ ¯ s [straight phi] where [straight phi] is a radial function, that is, [figure omitted; refer to PDF] for any ξ in the unit sphere ... n and r ∈ [0,1 ) .
The following lemma is from [ 1] and we will use it often.
Lemma 3.2.
Let p , s be two multi-indexes and let [straight phi] be a bounded radial function on ... n . Then for any multi-index α , [figure omitted; refer to PDF]
Proposition 3.3.
Let p 1 , ... , p m , s 1 , ... , s m be multi-indexes and let [straight phi] 1 , ... , [straight phi] m be bounded radial functions on ... n . If the product ∏ i =0 m -1 T ξ p m -i ξ ¯ s m -i [straight phi] m -i is of finite rank, then there exists α 0 such that [figure omitted; refer to PDF]
Proof.
Denote by S the product of Toeplitz operators ∏ i =0 m -1 T ξ p m -i ξ ¯ s m -i [straight phi] m -i and let rank S =N . For multi-indexes α [succeeds, =] ∑ i =1 m ( p i + s i ) , we have [figure omitted; refer to PDF] where [figure omitted; refer to PDF] It follows that S ( z α ) = C ( α ,p ,s ) z α + ∑ i =1 m ... ( p i - s i ) , where α [succeeds, =] ∑ i =1 m ... ( p i + s i ) and C ( α ,p ,s ) is a constant dependent on α , p and s . Thus the set {S ( z α ) : α [succeeds, =] ∑ i =1 m ... ( p i + s i ) } = ⋁ α [succeeds, =] ∑ i =1 m ... ( p i + s i ) { z α + ∑ i =1 m ... ( p i - s i ) } contains at most N elements. Let n 0 = (N , ... ,N ) , then there exists α 0 [succeeds, =] n 0 + ∑ i =1 m ... ( p i + s i ) such that [figure omitted; refer to PDF]
Proposition 3.4.
p 1 , ... , p m , s 1 , ... , s m , [straight phi] 1 , ... , [straight phi] m are defined as in Proposition 3.3. The product ∏ i =0 m -1 T ξ p m -i ξ ¯ s m -i [straight phi] m -i is of finite rank if and only if [straight phi] i =0 for some i ∈ {1 , ... ,m } .
Proof.
Using Proposition 3.3and Theorem 2.1, we can easily get the result.
This result is analogous to Theorem 3.2 in [ 15], but we get it in a different way.
Similar to the proof of Proposition 3.3, we can get a result about finite-rank commutators (semicommutators).
Proposition 3.5.
Let p 1 , p 2 , s 1 , and s 2 be multi-indexes and let [straight phi] 1 , [straight phi] 2 be bounded radial functions on ... n . If the commutator [ T ξ p 1 ξ ¯ s 1 [straight phi] 1 - T ξ p 2 ξ ¯ s 2 [straight phi] 2 ] (or the semicommutator ( T ξ p 1 ξ ¯ s 1 [straight phi] 1 - T ξ p 2 ξ ¯ s 2 [straight phi] 2 ] ) is of finite rank, then there exists α 0 such that [ T ξ p 1 ξ ¯ s 1 [straight phi] 1 - T ξ p 2 ξ ¯ s 2 [straight phi] 2 ] ( z α ) =0 (or ( T ξ p 1 ξ ¯ s 1 [straight phi] 1 - T ξ p 2 ξ ¯ s 2 [straight phi] 2 ] ( z α ) =0 ) for α [succeeds, =] α 0 .
Now we are in a position to characterize when the product of two Toeplitz operators with some quasihomogeneous symbols equals a Toeplitz operator with quasihomogeneous symbols.
When n =1 , if p [perpendicular]s , then p =0 or s =0 . But when n ...5;2 , there exist nonzero multi-indexes p and s such that p [perpendicular]s . In this case, we have the following theorem.
Theorem 3.6.
Suppose p and s are two nonzero multi-indexes with p [perpendicular]s . Let [straight phi] 1 and [straight phi] 2 be bounded radial functions on ... n . If there exists a bounded radial function [straight phi] such that T ξ p [straight phi] 1 T ξ ¯ s [straight phi] 2 = T ξ p ξ ¯ s [straight phi] , then [straight phi] must be a solution of the equation [figure omitted; refer to PDF]
Proof.
Obviously, the equality T ξ p [straight phi] 1 T ξ ¯ s [straight phi] 2 z α = T ξ p ξ ¯ s [straight phi] z α holds for each monomial z α with the multi-index α .
Since p [perpendicular]s , α +p [succeeds, =]s is equivalent to α [succeeds, =]s . By Lemma 3.2, it is easy to check that [figure omitted; refer to PDF] where [figure omitted; refer to PDF] Since p [perpendicular]s , we have α ! ( α +p -s ) ! = ( α +p ) ! ( α -s ) ! . So ( 3.13) implies that [figure omitted; refer to PDF] As |s | ...5;1 , we have [figure omitted; refer to PDF] A direct calculation gives that r 2 |p | +2i ^ (2n +2 | α | -2 |s | ) = 1 / ( 2n +2 | α | -2 |s | +2 |p | +2i ) for 0 ...4;i ...4; |s | -1 . Then we have [figure omitted; refer to PDF] or equivalently, [figure omitted; refer to PDF] Combining the above equality with Remark 2.2, we get the conclusion.
In the following, we give some explicit examples in which Theorem 3.6is applied.
Example 3.7.
Suppose p = (1,0 ) , s = (0,1 ) , [straight phi] 1 =r , [straight phi] 2 =r , [straight phi] is a bounded radial function such that T ξ p r T ξ ¯ s r = T ξ p ξ ¯ s [straight phi] . Using Theorem 3.6, we can get that [straight phi] =1 .
Example 3.8.
Suppose s =0 , [straight phi] is a bounded radial function such that T ξ p [straight phi] 1 T [straight phi] 2 = T ξ p [straight phi] , then [straight phi] must be a solution of the equation [figure omitted; refer to PDF] where χ [0,1 ] is the characteristic function of the set [0,1 ] . For example, suppose p = (1,1 ) , [straight phi] 1 =r , [straight phi] 2 = r 4 , [straight phi] is a bounded radial function such that T ξ p r T r 4 = T ξ p [straight phi] , then it follows that |p | =2 and [straight phi] =4 r 2 -3r .
Louhichi et al. [ 2] showed that there exist two nontrivial quasihomogeneous Toeplitz operators on the Bergman space of the unit disk such that the product of those Toeplitz operators is also a nontrivial Toeplitz operator, for example T e i θ r T e -i θ r = T 1 -log ( 1 / r 2 ) . On weighted Bergman space of the unit ball ... n (n >2 ) , Vasilevski [ 16, 17] showed that there exist parabolic quasihomogeneous (It is clear that the quasihomogeneous function is also a parabolic quasihomogeneous function) symbol Toeplitz operators such that the finite product of those Toeplitz operators is also a Toeplitz operator of this type. However, on the unit ball ... n (n ...5;2 ) , if p and s are two nonzero multi-indexes which are not orthogonal, we can get that there exist no nontrivial [straight phi] 1 and [straight phi] 2 such that T ξ p [straight phi] 1 T ξ ¯ s [straight phi] 2 = T ξ p ξ ¯ s [straight phi] .
Theorem 3.9.
Let p , s be two nonzero multi-indexes which are not orthogonal. Given n ...5;2 , and let [straight phi] 1 and [straight phi] 2 be two bounded radial functions on ... n . If there exists a bounded radial function [straight phi] such that T ξ p [straight phi] 1 T ξ ¯ s [straight phi] 2 = T ξ p ξ ¯ s [straight phi] , then [straight phi] 1 =0 or [straight phi] 2 =0 .
Proof.
If T ξ p [straight phi] 1 T ξ ¯ s [straight phi] 2 = T ξ p ξ ¯ s [straight phi] , as in Theorem 3.6, we can get [figure omitted; refer to PDF] where [figure omitted; refer to PDF] We claim that there exists M 0 such that [figure omitted; refer to PDF]
Since p is not orthogonal to s , without loss of generality, we can suppose p 1 s 1 ...0;0 .
Case 1 ( n =2 ). We denote p = ( p 1 , p 2 ) , s = ( s 1 , s 2 ) . For M > s 1 + s 2 +1 , let α M = ( s 1 ,M - s 1 ) and β M = ( s 1 +1 ,M - s 1 -1 ) .
Let F (M ) = c ( β M ) / c ( α M ) -1 . Then [figure omitted; refer to PDF] where N = ( s 1 + s 2 ) ( p 2 - s 2 ) ( s 1 +1 ) ( p 1 +1 ) - ( s 1 + p 1 +1 ) s 1 ( s 1 - p 2 + s 2 ) .
Therefore, the equation F (M ) =0 has two solutions at most. It means that there exists M 0 ...5; |s | +1 such that F (M ) ...0;0 for any M ...5; M 0 . Thus for each M ...5; M 0 , we have | α M | = | β M | =M and c ( α M ) ...0;c ( β M ) .
Case 2 (n ...5;3 ) . Given a multi-index γ = ( γ 1 , γ 2 , γ 3 , ... , γ n ) [succeeds, =]s , let α M = ( s 1 ,M - s 1 , γ 3 , ... , γ n ) and β M = ( s 1 +1 ,M - s 1 -1 , γ 3 , ... , γ n ) for M > s 1 + s 2 +1 . As in Case 1, we can find an integer M 0 with M 0 ...5; |s | +1 such that | α M | = | β M | =M and c ( α M ) ...0;c ( β M ) for any M ...5; M 0 .
Using ( 3.20), we have [figure omitted; refer to PDF] for M > M 0 . As d (M ) ...0;0 and c ( α M ) ...0;c ( β M ) , it is easy to see that ( 3.22) holds.
Let E 1 = {M ∈ ... :M ...5; M 0 and [straight phi] ^ 1 (2n +2M -2 |s | + |p | ) =0 } and E 2 = {M ∈ ... :M ...5; M 0 and [straight phi] ^ 2 (2n +2M - |s | ) =0 } . Then {M ∈ ... :M ...5; M 0 } = E 1 ... ... E 2 . Since [figure omitted; refer to PDF] we know that at least one of the series ∑ M ∈ E 1 ... ( 1 / M ) and ∑ M ∈ E 2 ... ( 1 / M ) diverges. Hence it follows from Remark 2.2that [straight phi] 1 =0 or [straight phi] 2 =0 .
4. Finite-Rank Semicommutator
On the unit ball ... n (n ...5;2 ) , we will show that the semicommutator of two Toeplitz operators with some quasihomogeneous symbols is of finite rank if and only if it is zero.
Theorem 4.1.
Let p , s be two multi-indexes, and let [straight phi] 1 , [straight phi] 2 be two integrable radial functions on ... n such that [straight phi] 1 , ξ p ξ ¯ s [straight phi] 2 , and ξ p ξ ¯ s [straight phi] 1 [straight phi] 2 are T-functions. If the semicommutator ( T [straight phi] 1 , T ξ p ξ ¯ s [straight phi] 2 ] has finite rank, then it is equal to zero.
Proof.
Let S be the semicommutator ( T [straight phi] 1 , T ξ p ξ ¯ s [straight phi] 2 ] . If S is finite rank, using Proposition 3.5, there exists α 0 [succeeds, =]0 such that [figure omitted; refer to PDF] Therefore, Lemma 3.2implies that [figure omitted; refer to PDF] for all α [succeeds, =] α 0 . Since [figure omitted; refer to PDF] the above equation is equivalent to [figure omitted; refer to PDF] Note that ( r |p | [straight phi] 1 * M r |s | [straight phi] 2 ^ ) and ( r |p | * M r |s | [straight phi] 1 [straight phi] 2 ^ ) are both analytic on the right half-plane {z :Rez >2 } and the sequence {2n +2 | α | + |p | -2 |s | } α [succeeds, =] α 0 is arithmetic. Then Remark 2.2implies that [figure omitted; refer to PDF] Hence, T [straight phi] 1 T ξ p ξ ¯ s [straight phi] 2 = T ξ p ξ ¯ s [straight phi] 1 [straight phi] 2 . The proof is complete.
Next we will consider when the semicommutator of two quasihomogeneous Toeplitz operators is a finite-rank operator.
Remark 4.2.
If the semicommutator ( T ξ p [straight phi] 1 , T ξ s [straight phi] 2 ] is of finite rank, following the same process as in Theorem 4.1, we can prove that it must be zero.
On the unit disk, Cuckovic and Louhichi [ 12] gave an example to show that there exists a nonzero finite rank semicommutator ( T e ip θ f , T e -is θ g ] , where f and g are radial functions. However, the situation on the unit ball ... n (n ...5;2 ) is different. Let [straight phi] 1 , [straight phi] 2 be two integrable radial functions on ... n (n ...5;2 ) and p , s be two multi-indexes. Then we will prove that ( T ξ p [straight phi] 1 , T ξ ¯ s [straight phi] 2 ] is a finite-rank operator if and only if ( T ξ p [straight phi] 1 , T ξ ¯ s [straight phi] 2 ] =0 . Now, we begin with the case that p and s are not orthogonal.
Theorem 4.3.
Let p ,s be two multi-indexes which are not orthogonal, and let [straight phi] 1 , [straight phi] 2 be two integrable radial functions on ... n (n ...5;2 ) such that ξ p [straight phi] 1 , ξ ¯ s [straight phi] 2 and ξ p ξ ¯ s [straight phi] 1 [straight phi] 2 are T-functions. If the semicommutator ( T ξ p [straight phi] 1 , T ξ ¯ s [straight phi] 2 ] has finite rank, then [straight phi] 1 =0 or [straight phi] 2 =0 .
Proof.
Let S be the semicommutator ( T ξ p [straight phi] 1 , T ξ ¯ s [straight phi] 2 ] . If S is of finite rank, using Proposition 3.5, we can get that there exists α 0 such that [figure omitted; refer to PDF]
Lemma 3.2gives that [figure omitted; refer to PDF] for all α [succeeds, =] α 0 .
Since p is not orthogonal to s , from the proof of Theorem 3.9and using ( 4.7), we can get that there exists M 0 such that [figure omitted; refer to PDF] Analogous to the proof of Theorem 3.9, it is easy to get [straight phi] 1 =0 or [straight phi] 2 =0 .
Next, we will show that there exists no nontrivial finite-rank semicommutator ( T ξ p [straight phi] 1 , T ξ ¯ s [straight phi] 2 ] in the case that p [perpendicular]s .
Theorem 4.4.
Let p , s be two multi-indexes with p [perpendicular]s , and let [straight phi] 1 and [straight phi] 2 be two integrable radial functions on ... n such that ξ p [straight phi] 1 , ξ ¯ s [straight phi] 2 and ξ p ξ ¯ s [straight phi] 1 [straight phi] 2 are T-functions. The semicommutator ( T ξ p [straight phi] 1 , T ξ ¯ s [straight phi] 2 ] has finite rank if and only if ( T ξ p [straight phi] 1 , T ξ ¯ s [straight phi] 2 ] =0 .
Proof.
We only need to prove the necessity. Let S = ( T ξ p [straight phi] 1 , T ξ ¯ s [straight phi] 2 ] be of finite rank. Since p [perpendicular]s , it is easy to see that α +p [succeeds, =]s if and only if α [succeeds, =]s for multi-indexes α . By Lemma 3.2, the following statements hold:
(i) if α +p [succeeds, =] s , then S ( z α ) =0 ;
(ii) if α [succeeds, =]s , then [figure omitted; refer to PDF]
Combining (i) and (ii) with the assumption that S has finite rank, we get that there exists α 0 [succeeds, =]s such that [figure omitted; refer to PDF] To finish the proof, we will prove that S ( z α ) =0 for α [succeeds, =]s .
Note that α ! ( α +p -s ) ! = ( α +p ) ! ( α -s ) ! for p [perpendicular]s and α [succeeds, =]s . Then for each α [succeeds, =]s , (ii) implies that S ( z α ) =0 if and only if [figure omitted; refer to PDF] that is, [figure omitted; refer to PDF] Since χ [0,1 ] ^ (n ) = 1 / n , the preceding equality is equivalent to [figure omitted; refer to PDF] for all α [succeeds, =]s .
By ( 4.10), we obtain that the equality ( 4.13) holds for all α [succeeds, =] α 0 . It is easy to see that {2n +2 | α | } α [succeeds, =] α 0 is arithmetic. Therefore, by Remark 2.2, we have [figure omitted; refer to PDF] for all Rez ...5;2 .
In particular if z =2n +2 | α | , with α [succeeds, =]s , we have [figure omitted; refer to PDF] It follows that the equality ( 4.13) holds for all α [succeeds, =]s . So S ( z α ) =0 for all α [succeeds, =]s . Hence, the proof is complete.
Example 4.5.
Let p , s be two multi-indexes, [straight phi] is a bounded radial function and a p , b s ∈ ... . If ( T [straight phi] , T a p z p ] is of finite rank, using Theorem 4.1, we obtain that ( T [straight phi] , T a p z p ] =0 . If p is not orthogonal to s and ( T a p z p , T b s z ¯ s ] is of finite rank, so it follows from Theorem 4.3that a p =0 or b s =0 . But if p [perpendicular]s , there exist a p 0 ...0;0 and b s 0 ...0;0 such that ( T a p 0 z p , T b s 0 z ¯ s ] =0 . In particular, suppose p = (0,0 ) , s = (2,2 ) , a p 0 = b s 0 =1 , a direct calculation gives that p [perpendicular]s and T z p T z ¯ s = T z p z ¯ s , that is, ( T z p , T z ¯ s ] =0 .
5. Finite-Rank Commutators
In this section, let [straight phi] 1 , [straight phi] 2 be two integrable radial functions on ... n . We now pass to investigate the commutator of two quasihomogeneous Toeplitz operators and consider when [ T [straight phi] 1 , T ξ p ξ ¯ s [straight phi] 2 ] , [ T ξ p [straight phi] 1 , T ξ s [straight phi] 2 ] , or [ T ξ p [straight phi] 1 , T ξ ¯ s [straight phi] 2 ] have finite rank, respectively.
Theorem 5.1.
Let p ,s be two multi-indexes with p [perpendicular]s , and let [straight phi] 1 , [straight phi] 2 be two integrable radial functions on ... n such that [straight phi] 1 and ξ p ξ ¯ s [straight phi] 2 are T-functions. If [straight phi] 1 is not a constant, then [ T [straight phi] 1 , T ξ p ξ ¯ s [straight phi] 2 ] is of finite rank if and only if |p | = |s | or [straight phi] 2 =0 .
Proof.
Let S be the commutator [ T [straight phi] 1 , T ξ p ξ ¯ s [straight phi] 2 ] . By Lemma 3.2, S ( z α ) =0 if and only if [figure omitted; refer to PDF] for α [succeeds, =]s . If S is of finite rank, using Proposition 3.5, there exists α 0 [succeeds, =]p +s such that [figure omitted; refer to PDF]
Since p [perpendicular]s , by Theorem 2.1and following the same process as in Theorem 4.4, we get S =0 . Using Theorem 4.4 in [ 1], we have S =0 if and only if |p | = |s | or [straight phi] 2 =0 .
Conversely, if |p | = |s | or [straight phi] 2 =0 , then we can easily show that S ( z α ) =0 for each multi-index α , which implies that T [straight phi] 1 and T ξ p ξ ¯ s [straight phi] 2 commute.
Remark 5.2.
The same as in Theorem 5.1, we can easily prove if the commutator [ T ξ p [straight phi] 1 , T ξ s [straight phi] 2 ] is of finite rank, then it must be a zero operator.
Next, we give some examples.
Example 5.3.
Suppose that f (z ) = a p z s , where a p ...0;0 and s ...0;0 . Let g = ξ k [straight phi] ...0;0 be a T -function, where [straight phi] is a radial function. Then [ T f , T g ] is of finite rank if and only if T f and T g commute. By Theorem 4.9 in [ 1], we can also get [ T f , T g ] is of finite rank if and only if g is a monomial.
On the unit disk, if the commutator [ T e ip θ f , T e -is θ g ] has finite rank N , then N is at most equal to the quasihomogeneous degree s and a nonzero finite rank commutator has been given in [ 12]. On the unit ball ... n (n ...5;2 ) , we will show that the commutator [ T ξ p [straight phi] 1 , T ξ ¯ s [straight phi] 2 ] has finite rank if and only if T ξ p [straight phi] 1 commutes with T ξ ¯ s [straight phi] 2 if and only if [straight phi] 1 =0 or [straight phi] 2 =0 .
Theorem 5.4.
Let p , s be two nonzero multi-indexes, and let [straight phi] 1 , [straight phi] 2 be two integrable radial functions on ... n (n ...5;2 ) such that ξ p [straight phi] 1 and ξ ¯ s [straight phi] 2 are T-functions. If the commutator [ T ξ p [straight phi] 1 , T ξ ¯ s [straight phi] 2 ] has finite rank, then [straight phi] 1 =0 or [straight phi] 2 =0 .
Proof.
Let S denote the commutator [ T ξ p [straight phi] 1 , T ξ ¯ s [straight phi] 2 ] . Applying Lemma 3.2, we get [figure omitted; refer to PDF] for α [succeeds, =]s . If S is finite rank, using Proposition 3.5, there exists α 0 [succeeds, =]s such that [figure omitted; refer to PDF]
If p [perpendicular]s , then we have α ! ( α +p -s ) ! = ( α +p ) ! ( α -s ) ! . Combining ( 5.4) and ( 5.3), we have [figure omitted; refer to PDF] for α [succeeds, =] α 0 . Analogous to the proof of Theorem 4.8 in [ 1], it is not difficult to get that [straight phi] 1 =0 or [straight phi] 2 =0 .
On the other hand, if p is not orthogonal to s , then ( 5.3) and ( 5.4) imply that [figure omitted; refer to PDF] for α [succeeds, =] α 0 , where [figure omitted; refer to PDF] Following the same process as in Theorem 3.9, we get [straight phi] 1 =0 or [straight phi] 2 =0 , as desired.
Acknowledgment
The authors thank the referee for several suggestions that improved the paper. This research is supported by NSFC, Item Number: 10971020.
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Copyright © 2012 Bo Zhang and Yufeng Lu. Bo Zhang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
We consider when the product of two Toeplitz operators with some quasihomogeneous symbols on the Bergman space of the unit ball equals a Toeplitz operator with quasihomogeneous symbols. We also characterize finite-rank semicommutators or commutators of two Toeplitz operators with quasihomogeneous symbols.
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Neither ProQuest nor its licensors make any representations or warranties with respect to the translations. The translations are automatically generated "AS IS" and "AS AVAILABLE" and are not retained in our systems. PROQUEST AND ITS LICENSORS SPECIFICALLY DISCLAIM ANY AND ALL EXPRESS OR IMPLIED WARRANTIES, INCLUDING WITHOUT LIMITATION, ANY WARRANTIES FOR AVAILABILITY, ACCURACY, TIMELINESS, COMPLETENESS, NON-INFRINGMENT, MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. Your use of the translations is subject to all use restrictions contained in your Electronic Products License Agreement and by using the translation functionality you agree to forgo any and all claims against ProQuest or its licensors for your use of the translation functionality and any output derived there from. Hide full disclaimer





