Algebraic Properties of Toeplitz Operators on the Polydisk

Abstract

We discuss some algebraic properties of Toeplitz operators on the Bergman space of the polydisk 𝔻^𝑛. Firstly, we introduce Toeplitz operators with quasihomogeneous symbols and property (P). Secondly, we study commutativity of certain quasihomogeneous Toeplitz operators and commutators of diagonal Toeplitz operators. Thirdly, we discuss finite rank semicommutators and commutators of Toeplitz operators with quasihomogeneous symbols. Finally, we solve the finite rank product problem for Toeplitz operators on the polydisk.

1. Introduction

Let 𝔻 be the open unit disk in the complex plane ℂ and its boundary the unit circle 𝕋. For a fixed positive integer 𝑛, the unit polydisk 𝔻^𝑛 and the torus 𝕋^𝑛 are the subsets of ℂ^𝑛 which are Cartesian products of 𝑛 copies 𝔻 and 𝕋, respectively. Let 𝑑𝑉(𝑧)=𝑑𝑉_𝑛(𝑧) denote the Lebesgue volume measure on the polydisk 𝔻^𝑛, normalized so that the measure of 𝔻^𝑛 equals 1. Let 𝐿^𝑝=𝐿^𝑝(𝔻^𝑛) denote the usual Lebesgue space. The Bergman space 𝐴²=𝐴²(𝔻^𝑛) is the Hilbert space consisting of holomorphic functions on 𝔻^𝑛 that are also in 𝐿²(𝔻^𝑛,𝑑𝑉(𝑧)). Since every point evaluation is a bounded linear functional on 𝐴², there corresponds to every 𝑧=(𝑧₁,…,𝑧_𝑛)∈𝔻^𝑛 a unique function 𝐾_𝑧∈𝐴² which has the following reproducing property: 𝑓(𝑧)=𝑓,𝐾_𝑧,𝑓∈𝐴², (1.1) where the notation ⟨⋅,⋅⟩ denotes the inner product in 𝐿². The function 𝐾_𝑧 is the well-known Bergman kernel and its explicit formula is given by 𝐾_𝑧(𝑤)=^𝑛_𝑗=111−𝑤_𝑗𝑧_𝑗²,𝑤∈𝔻^𝑛. (1.2) Here and elsewhere 𝑧_𝑗 denotes the 𝑗th component of 𝑧. The Bergman projection 𝑃 is defined for the Hilbert space orthogonal projection from 𝐿² onto 𝐴². Given a function 𝜑∈𝐿^∞(𝔻^𝑛,𝑑𝑉), the Toeplitz operator 𝑇_𝜑∶𝐴²→𝐴² is defined by the formula𝑇_𝜑𝑓(𝑧)=𝑃𝜑𝑓(𝑧)=_𝔻𝑛𝑓(𝑤)𝜑(𝑤)𝐾_𝑧(𝑤)𝑑𝑉(𝑤) (1.3) for all 𝑓∈𝐴². Since the Bergman projection 𝑃 has norm 1, it is clear that Toeplitz operators defined in this way are bounded linear operators on 𝐴² and ‖𝑇_𝜑‖≤‖𝜑‖_∞.

We now consider a more general class of Toeplitz operators. For 𝐹∈𝐿¹(𝔻^𝑛,𝑑𝑉), in analogy to (1.3) we define an operator 𝑇_𝐹 by𝑇_𝐹𝑓(𝑧)=_𝔻𝑛𝐹(𝑤)𝑓(𝑤)𝐾_𝑧(𝑤)𝑑𝑉(𝑤). (1.4)

Since the Bergman projection 𝑃 can be extended to 𝐿¹(𝔻^𝑛,𝑑𝑉), the operator 𝑇_𝐹 is well defined on 𝐻^∞, where 𝐻^∞ is the space of bounded holomorphic functions on 𝔻^𝑛. Hence, 𝑇_𝐹 is always densely defined on 𝐴²(𝔻^𝑛). Since 𝑃 is not bounded on 𝐿¹(𝔻^𝑛,𝑑𝑉), it is well known that 𝑇_𝐹 can be unbounded in general. This motivates the following definition, which is based on the definitions on unit ball in [1].

Definition 1.1. Let 𝐹∈𝐿¹(𝔻^𝑛,𝑑𝑉).(a)𝐹 is called a 𝑇-function if (1.4) defines a bounded operator on 𝐴².(b)If 𝐹 is a 𝑇-function, one writes 𝑇_𝐹 for the continuous extension of the operator (it is defined on the dense subset 𝐻^∞ of 𝐿²(𝔻^𝑛)) defined by (1.4). 𝑇_𝐹 is called a Toeplitz operator on 𝐴².(c)If there exist 𝑟_𝑗∈(0,1), 1≤𝑗≤𝑛, such that 𝐹 is (essentially) bounded on {𝑧=(𝑧₁,𝑧₂,…,𝑧_𝑛)∶𝑟_𝑗<|𝑧_𝑗|<1,1≤𝑗≤𝑛}, then one says 𝐹 is “nearly bounded.”

Notice that the 𝑇-functions form a proper subset of 𝐿¹(𝔻^𝑛,𝑑𝑉) which contains all bounded and “nearly bounded” functions. In this paper, the functions which we considered are all 𝑇-functions without special introduction. We denote the semicommutator and commutator of two Toeplitz operators 𝑇_𝑓 and 𝑇_𝑔 by 𝑇_𝑓,𝑇_𝑔=𝑇_𝑓𝑔−𝑇_𝑓𝑇_𝑔,𝑇_𝑓,𝑇_𝑔=𝑇_𝑓𝑇_𝑔−𝑇_𝑔𝑇_𝑓. (1.5)

The commuting problem and the finite-rank product problem for Toeplitz operators on the Hardy and Bergman spaces over various domains are some of the most interesting problems in operator theory.

For commuting problem, in 1963, Brown and Halmos [2] showed that two bounded Toeplitz operators 𝑇_𝜑 and 𝑇_𝜓 on the classical Hardy space commute if and only if (i) both 𝜑 and 𝜓 are analytic, (ii) both 𝜑 and 𝜓 are analytic, or (iii) one is a linear function of the other. On the Bergman space of the unit disk, some similar results were obtained for Toeplitz operators with bounded harmonic symbols or analytic symbols (see [2–4]). The problem of characterizing commuting Toeplitz operators with arbitrary bounded symbols seems quite challenging and is not fully understood until now. In recent years, by Mellin transform, some results with quasihomogeneous symbols (it is of the form 𝑒^𝑖𝑘𝜃𝜙, where 𝜙 is a radial function) or monomial symbols were obtained (see [5–7]). On the Hardy and Bergman spaces of several complex variables, the situation is much more complicated. On the unit ball, Toeplitz operators with pluriharmonic or quasihomogeneous symbols were studied in [1, 8–11]. On the polydisk, some results about Toeplitz operators with pluriharmonic symbols were obtained in [10, 12–14].

For finite-rank product problem, Luecking recently proved that a Toeplitz operator with measure symbol on the Bergman space of unit disk has finite rank if and only if its symbols are a linear combination of point masses (see [15]). In [16], Choe extended Luecking’s theorem to higher-dimensional cases. Using those results, Le studied finite-rank products of Toeplitz operators on the Bergman space of the unit disk and unit ball in [17, 18].

Motivated by recent work in [1, 5, 7, 17, 18], we define quasihomogeneous functions on the polydisk and study Toeplitz operators with quasihomogeneous symbols on the Bergman space of the polydisk. The present paper is assembled as follows. In Section 2, we introduce Mellin transform, Toeplitz operators with quasihomogeneous symbols and property (P). In Section 3, we study commutativity of certain quasihomogeneous Toeplitz operators and commutators of diagonal Toeplitz operators. In Sections 4 and 5, we prove that finite rank semicommutators and commutators of Toeplitz operators with quasihomogeneous symbols must be zero operator and we also solve the finite-rank product problem for Toeplitz operators on the Bergman space of the polydisk.

2. Mellin Transform, Toeplitz Operators with Quasihomogeneous Symbols and Property (P)

For any multi-index 𝛼=(𝛼₁,…,𝛼_𝑛)∈ℕ^𝑛 (here ℕ denotes the set of all nonnegative integers), we write 𝑎_𝛼=𝛼₁⋯𝛼_𝑛 and 𝑧^𝛼=𝑧^𝛼1₁⋯𝑧^𝛼𝑛_𝑛 for 𝑧=(𝑧₁,…,𝑧_𝑛)∈𝔻^𝑛. The standard orthonormal basis for 𝐴² is {𝑒_𝛼∶𝛼∈ℕ^𝑛}, where 𝑒_𝛼(𝑧)=𝛼₁+1⋯𝛼_𝑛+1𝑧^𝛼,𝛼∈ℕ^𝑛,𝑧∈𝔻^𝑛. (2.1)

For two 𝑛-tuples of integers 𝛼=(𝛼₁,…,𝛼_𝑛) and 𝛽=(𝛽₁,…,𝛽_𝑛), we define 𝛼≻𝛽 if 𝛼_𝑗>𝛽_𝑗 for all 0≤𝑗≤𝑛. Similarly, we write 𝛼≽𝛽 if 𝛼_𝑗≥𝛽_𝑗 for all 1≤𝑗≤𝑛 and 𝛼≽^╱𝛽 if otherwise. We also define 𝛼⟂𝛽 if 𝛼₁𝛽₁+⋯+𝛼_𝑛𝛽_𝑛=0 and 𝛼−𝛽=(𝛼₁−𝛽₁,…,𝛼_𝑛−𝛽_𝑛).

For any 𝑘=(𝑘₁,…,𝑘_𝑛)∈ℤ^𝑛, particularly we write 𝑘₁=(𝑘₁,…,𝑘₁) and put 𝑘^∗=(|𝑘₁|,…,|𝑘_𝑛|), 𝑘⁺=(1/2)(𝑘^∗+𝑘) and 𝑘⁻=(1/2)(𝑘^∗−𝑘). Then, 𝑘⁺,𝑘⁻≽0, 𝑘=𝑘⁺−𝑘⁻, and 𝑘⁺⟂𝑘⁻.

Recall that a function 𝜑 on 𝔻^𝑛 is radial if and only if 𝜑(𝑧) depends only on (|𝑧₁|,|𝑧₂|,…,|𝑧_𝑛|), that is, 𝜑(𝑒^𝑖𝜃1𝑧₁,𝑒^𝑖𝜃2𝑧₂,…,𝑒^𝑖𝜃𝑛𝑧_𝑛)=𝜑(𝑧₁,𝑧₂,…,𝑧_𝑛) for any 𝜃₁,𝜃₂,…,𝜃_𝑛∈𝑅. For any function 𝑓∈𝐿¹(𝔻^𝑛,𝑑𝑉), we define the radicalization of 𝑓 by rad𝑓𝑧₁,𝑧₂,…,𝑧_𝑛=1(2𝜋)^𝑛₀^2𝜋⋯₀^2𝜋𝑓𝑒^𝑖𝑡1𝑧₁,𝑒^𝑖𝑡2𝑧₂,…,𝑒^𝑖𝑡𝑛𝑧_𝑛𝑑𝑡₁⋯𝑑𝑡_𝑛. (2.2) Then, 𝑓 is radial if and only if rad(𝑓)=𝑓. For 𝛼∈ℕ^𝑛, we have 𝑇_rad(𝑓)𝑧^𝛼,𝑧^𝛼=1(2𝜋)^𝑛₀^2𝜋⋯₀^2𝜋𝑓𝑒^𝑖𝑡1𝑧₁,𝑒^𝑖𝑡2𝑧₂,…,𝑒^𝑖𝑡𝑛𝑧_𝑛𝑑𝑡₁⋯𝑑𝑡_𝑛𝑧^𝛼,𝑧^𝛼=1(2𝜋)^𝑛₀^2𝜋⋯₀^2𝜋_𝔻𝑛𝑓𝑒^𝑖𝑡1𝑧₁,𝑒^𝑖𝑡2𝑧₂,…,𝑒^𝑖𝑡𝑛𝑧_𝑛𝑧^𝛼𝑧^𝛼𝑑𝑉(𝑧)𝑑𝑡₁⋯𝑑𝑡_𝑛=1(2𝜋)^𝑛₀^2𝜋⋯₀^2𝜋𝑑𝑡₁⋯𝑑𝑡_𝑛𝔻𝑛𝑓𝑤₁,𝑤₂,…,w_𝑛𝑤^𝛼𝑤^𝛼(𝑤)=𝑇𝑑𝑉_𝑓𝑧^𝛼,𝑧^𝛼. (2.3)

The main tool in this paper will be the Mellin transform. which is defined by the equation 𝜑𝑧₁,…,𝑧_𝑛=^∞₀⋯^∞₀𝜑𝑠₁,…,𝑠_𝑛𝑠^𝑧1₁⁻¹⋯𝑠^𝑧𝑛_𝑛⁻¹𝑑𝑠₁⋯𝑑𝑠_𝑛. (2.4) We apply the Mellin transform to functions in 𝐿¹([0,1]^𝑛,𝑟₁⋯𝑟_𝑛𝑑𝑟₁⋯𝑑𝑟_𝑛); then, 𝜑𝑧₁,…,𝑧_𝑛=¹₀⋯¹₀𝜑𝑠₁,…,𝑠_𝑛𝑠^𝑧1₁⁻¹⋯𝑠^𝑧𝑛_𝑛⁻¹𝑑𝑠₁⋯𝑑𝑠_𝑛. (2.5) For convenience, we denote 𝜑(𝑧₁,…,𝑧_𝑛) by 𝜑^∧(𝑧₁,…,𝑧_𝑛) when the form of 𝜑 is complicated. It is clear that 𝜑 is well defined on 𝐼_𝑛={𝑧=(𝑧₁,𝑧₂,…,𝑧_𝑛)∶Re𝑧_𝑗>2,𝑗=1,2,…,𝑛}. Using the Hartogs theorem, for any function 𝜑∈𝐿¹([0,1]^𝑛,𝑟₁⋯𝑟_𝑛𝑑𝑟₁⋯𝑑𝑟_𝑛), the Mellin transform of 𝜑 is a bounded holomorphic function on 𝐼_𝑛.

By calculation, we can get𝜑𝑧+𝑝=𝜑𝑧₁+𝑝₁,𝑧₂+𝑝₂,…,𝑧_𝑛+𝑝_𝑛=𝑟^𝑝𝜑𝑧₁,𝑧₂,…,𝑧_𝑛=𝑟^𝑝𝜑(𝑧), (2.6) where 𝑝=(𝑝₁,𝑝₂,…,𝑝_𝑛)≽0, 𝑧=(𝑧₁,𝑧₂,…,𝑧_𝑛)∈𝐼_𝑛, and 𝑟^𝑝=𝑟^𝑝1₁𝑟^𝑝2₂⋯𝑟^𝑝𝑛_𝑛.

The quasihomogeneous functions have been defined in many spaces (see [5, 7]). In the following, we give a similar definition on the polydisk 𝔻^𝑛.

Definition 2.1. Let 𝑘∈ℤ^𝑛. A function 𝑓∈𝐿¹(𝔻^𝑛,𝑑𝑉) is called a quasihomogeneous function of degree 𝑘 if 𝑓 is of the form 𝜉^𝑘𝜑 where 𝜑 is a radial function, that is, 𝑓(𝑟𝜉)=𝜉^𝑘𝜑(𝑟) (2.7) for any 𝜉 in the torus 𝕋^𝑛 and 𝑟∈[0,1)^𝑛.

As in [19], for any 𝑛-tuple 𝑘∈ℤ^𝑛, let 𝐻_𝑘={𝑓∈𝐿²∶𝑓is a quasihomogeneous function of degree 𝑘}. It is clear that 𝐻_𝑘 is a closed subspace of 𝐿². By Lemma 3.2 in [19], 𝐿²=⨁_𝑠∈ℤ^𝑛𝐻_𝑠. In particular, for all 𝑧=(𝑟₁𝜉₁,…,𝑟_𝑛𝜉_𝑛)∈𝔻, if 𝑓∈𝐻_𝑘, that is, 𝑓(𝑟₁𝜉₁,…,𝑟_𝑛𝜉_𝑛)=𝜉^𝑘𝑓_𝑘(𝑟₁,…,𝑟_𝑛), then we conclude that 𝐿²(𝔻^𝑛,𝑑𝑉)=⨁_𝑘∈ℤ^𝑛𝜉^𝑘𝑓_𝑘(𝑟₁,…,𝑟_𝑛), 𝑓_𝑘∈ℜ, where ℜ={𝜑∶𝔻^𝑛→ℂradial|∫_[0,1]𝑛|𝜑(𝑟₁,…,𝑟_𝑛)|²∏^𝑛_𝑖=1𝑟_𝑖𝑑𝑟_𝑖<+∞}.

Lemma 2.2. Let 𝑘,𝑙∈ℤ^𝑛, and let 𝜑, 𝜓 be radial functions on 𝔻^𝑛, such that 𝜉^𝑘𝜑, 𝜉^𝑙𝜓, and 𝜉^𝑘+𝑙𝜑𝜓 are all 𝑇-functions. Then, the following equation holds for every 𝛼∈ℕ^𝑛: 𝑇_𝜉^𝑘_𝜑𝑧^𝛼=0if𝛼⪰^╱𝑘⁻,2^𝑛𝑎_1𝛼+𝑘+𝜑2𝛼+𝑘+2𝑧^𝛼+𝑘,if𝛼⪰𝑘⁻, (2.8) where 𝑎_1𝛼+𝑘+=(𝛼₁+𝑘₁+1)⋯(𝛼_𝑛+𝑘_𝑛+1) and 𝜑(2𝛼+𝑘+2)=𝜑(2𝛼₁+𝑘₁+2,2𝛼₂+𝑘₂+2,…,2𝛼_𝑛+𝑘_𝑛+2).

Using Lemma 2.2, we can get the following two results:𝑇_𝜉^𝑘_𝜑,𝑇_𝜉^𝑙_𝜓𝑧^𝛼=0if𝛼∈𝐸₁,2^𝑛𝑎_1𝛼+𝑚+𝜑𝜓2𝛼+𝑚+2𝑧^𝛼+𝑚if𝛼∈𝐸^𝑐₁∩𝐸^𝑐₂,2^𝑛𝑎_1𝛼+𝑚+𝜑𝜓2𝛼+𝑚+2−𝜆𝑧^𝛼+𝑚if𝛼∈𝐸₂, (2.9) where 𝑚=𝑘+𝑙, 𝜆=4^𝑛𝑎_1𝛼+𝑙+𝑎_1𝛼+𝑚+𝜓(2𝛼+𝑙+2)𝜑(2(𝛼+𝑙)+𝑘+2), 𝐸₁={𝛼∶𝛼≽^╱𝑚⁻}, 𝐸₂={𝛼∶𝛼≽𝑙⁻}∩{𝛼∶𝛼+𝑙≽𝑘⁻}, 𝐸^𝑐₁=ℕ^𝑛⧵𝐸₁, and 𝐸^𝑐₂=ℕ^𝑛⧵𝐸₂. It is easy to check that 𝐸₁∩𝐸₂=𝜙𝑇_𝜉^𝑘_𝜑,𝑇_𝜉^𝑙_𝜓𝑧^𝛼=0if𝛼∈𝐹^𝑐₁∩𝐹^𝑐₂,𝜆₁𝑧^{𝛼+𝑘+𝑙}if𝛼∈𝐹₁∩𝐹^𝑐₂,−𝜆₂𝑧^{𝛼+𝑘+𝑙}if𝛼∈𝐹^𝑐₁∩𝐹₂,𝜆₁−𝜆₂𝑧^{𝛼+𝑘+𝑙}if𝛼∈𝐹₁∩𝐹₂, (2.10) where 𝜆₁=4^𝑛𝑎_1𝛼+𝑙+𝑎_{1𝛼+𝑘+𝑙+}𝜓(2𝛼+𝑙+2)𝜑(2(𝛼+𝑙)+𝑘+2), 𝜆₂=4^𝑛𝑎_1𝛼+𝑘+𝑎_{1𝛼+𝑘+𝑙+}𝜑(2𝛼+𝑘+2)𝜓(2(𝛼+𝑘)+𝑙+2), 𝐹₁={𝛼∶𝛼≽𝑙⁻}∩{𝛼∶𝛼+𝑙≽𝑘⁻},𝐹₂={𝛼∶𝛼≽𝑘⁻}∩{𝛼∶𝛼+𝑘≽𝑙⁻},𝐹^𝑐₁=ℕ^𝑛⧵𝐹₁ and 𝐹^𝑐₂=ℕ^𝑛⧵𝐹₂.

Let 𝐺 be a region in complex plane ℂ and 𝑓 holomorphic on 𝐺. If {𝑧_𝑘}^∞_𝑘=1 has a limit point in 𝐺, such that 𝑓(𝑧_𝑘)=0, then 𝑓≡0. For functions of several complex variables, the above conclusion does not hold. For example, 𝑓(𝑧₁,𝑧₂)=𝑧₁𝑧₂ is analytic on bidisk {(𝑧₁,𝑧₂)∶|𝑧₁|<1,|𝑧₂|<1}, point sequence {(0,1/𝑘)}, 𝑘=2,3,…, has a limit point (0,0), and 𝑓(0,1/𝑘)=0, but 𝑓 is not a zero function on the bidisk. So we need the following definition, which is given in [9, 17].

For any 1≤𝑗≤𝑛, let 𝜎_𝑗∶ℕ×ℕ^𝑛−1→ℕ^𝑛 be the map defined by the formula 𝜎_𝑗(𝑠,(𝛼₁,…,𝛼_𝑛−1))=(𝛼₁,…,𝛼_𝑗−1,𝑠,𝛼_𝑗,…,𝛼_𝑛−1) for all 𝑠∈ℕ and (𝛼₁,…,𝛼_𝑛−1)∈ℕ^𝑛−1. If 𝑀 is a subset of ℕ^𝑛 and 1≤𝑗≤𝑛, we define 𝑀_𝑗=𝛼=𝛼₁,…,𝛼_𝑛−1∈ℕ^𝑛−1∶_{𝑠∈ℕ,𝜎𝑗𝑠,}𝛼_∈𝑀1𝑠+1=∞. (2.11)

As in [9, 17], we say that 𝑀 has property (P) if one of the following statements holds:(1)𝑀=∅, (2)𝑀≠∅,𝑛=1, and ∑_𝑠∈𝑀1/𝑠<∞, or (3)𝑀≠∅,𝑛≥2, and, for any 1≤𝑗≤𝑛, the set 𝑀_𝑗 has property (P) as a subset of ℕ^𝑛−1.

Let 𝑀 and 𝑁 be two sets that have property (P). It is not difficult to check that the following statements hold:(1)𝑀⋂𝑁 and 𝑀⋃𝑁 have property (P);(2)ℕ^𝑛⧵𝑀 do not have property (P).

Lemma 2.3. If 𝜑∈𝐿¹([0,1]^𝑛,𝑟₁⋯𝑟_𝑛𝑑𝑟₁⋯𝑑𝑟_𝑛) and 𝑍(𝜑)={𝛼∈ℕ^𝑛∶𝜑(𝛼)=0} does not have property (𝑃), then 𝜑 is identically zero.

Proof. By the Müntz theorem, we can prove that it is true when 𝑛=1 (see [7] for more details). Suppose that the conclusion of the lemma holds whenever 𝑛≤𝑁, where 𝑁 is a positive integer. Consider the case 𝑛=𝑁+1. Since 𝑍(𝜑) does not have property (P), there must be a 1≤𝑗≤𝑁+1, such that 𝑍(𝜑)_𝑗 does not have property (P). Without loss of generality, taking 𝑗=𝑁+1, then, 𝑍(𝜑)_𝑁+1≠∅. For each 𝑟̃∈𝑍(𝜑)_𝑁+1, ∑_𝑠∈ℕ,𝜑₍𝑟̃_,𝑠)=01/(𝑠+1)=∞. So 𝜑(𝑟,̃𝑧_𝑁+1)=0, for all 𝑧_𝑁+1∈𝐼₁. For every 𝜆∈𝐼₁, let 𝜑_𝜆(𝑧^′)=𝜑(𝑧^′,𝜆); then, 𝜑_𝜆 is an analytic function on 𝐼_𝑁 and 𝑍(𝜑_𝜆)=𝑍(𝜑)_𝑁+1, which does not have property (P). By the induction hypothesis, we have 𝜑(𝑧^′,𝜆)=0,𝑧^′∈𝐼_𝑁. Thus, 𝜑(𝑧)=0 on 𝐼_𝑁+1. Therefore, 𝜑 is identically zero.

Theorem 2.4. Let 𝑝=(𝑝₁,𝑝₂,…,𝑝_𝑛)∈ℤ^𝑛, and let 𝑓 be a 𝑇-function. Then, the following statements hold.(i)If 𝐸_𝑝={𝛼;⟨𝑇_𝑓𝑧^𝛼+𝑝,𝑧^𝛼⟩=0forall𝛼≽𝑝⁻} does not have property (P), then ⟨𝑇_𝑓𝑧^𝛼+𝑝,𝑧^𝛼⟩=0 for all 𝛼≽𝑝⁻.(ii)Let 𝐸, 𝐸^′⊆ℕ^𝑛 be the sets that have property (P). If ⟨𝑇_𝑓𝑧^𝛼,𝑧^𝛽⟩=0 for all 𝛼∈ℕ^𝑛⧵𝐸, 𝛽∈ℕ^𝑛⧵𝐸^′, then 𝑓(𝑧)≡0 for almost all 𝑧∈𝔻^𝑛.(iii)If ⟨𝑇_𝑓𝑧^𝛼,𝑧^𝛽⟩=0forall𝛼+𝑝≠𝛽, then 𝑓 is a quasihomogeneous function of degree 𝑝.

Proof. (i) By direct computation, we have 𝑇_𝑓𝑧^𝛼+𝑝,𝑧^𝛼=_𝔻𝑛𝑓(𝑧)𝑧^𝛼+𝑝𝑧^𝛼(𝑧)=1𝑑𝑉𝜋^𝑛_[0,1]^𝑛_[0,2𝜋]^𝑛𝑓𝑟₁𝑒^𝑖𝜃1,…,𝑟_𝑛𝑒^{𝑖𝜃𝑛𝑛}_𝑗=1𝑟^{2𝛼𝑗+|𝑝𝑗}_𝑗^𝑛|+1_𝑗=1𝑒^{𝑖𝜃𝑗𝑝𝑗}𝑑𝜃₁⋯𝑑𝜃_𝑛𝑑𝑟₁⋯𝑑𝑟_𝑛. (2.12) Let 𝐹𝑟₁,…,𝑟_𝑛=1𝜋^𝑛_[0,2𝜋]^𝑛𝑓𝑟₁𝑒^𝑖𝜃1,…,𝑟_𝑛𝑒^{𝑖𝜃𝑛𝑛}_𝑗=1𝑒^{𝑖𝜃𝑗𝑝𝑗}𝑑𝜃₁⋯𝑑𝜃_𝑛. (2.13) Then, 𝐹∈𝐿¹([0,1]^𝑛,𝑟₁⋯𝑟_𝑛𝑑𝑟₁⋯𝑑𝑟_𝑛). In fact, ‖𝐹(𝑟₁,…,𝑟_𝑛)‖_𝐿¹_([0,1]^𝑛_{,𝑟1⋯𝑟𝑛𝑑𝑟1⋯𝑑𝑟𝑛)}≤‖𝑓‖_𝐿¹_(𝔻^𝑛_,𝑑𝑉). Therefore, equality (2.12) shows that 𝐹(2𝛼₁+|𝑝₁|+2,…,2𝛼_𝑛+|𝑝_𝑛|+2)=0 for any 𝛼=(𝛼₁,…,𝛼_𝑛)∈ℕ^𝑛. That is, 𝑍(𝐹) does not have property (P). Thus, Lemma 2.3 implies that 𝐹≡0 and ⟨𝑇_𝑓𝑧^𝛼+𝑝,𝑧^𝛼⟩=0 for all 𝛼≽𝑝⁻.
(ii) For each 𝑙=(𝑙₁,…,𝑙_𝑛)∈ℕ^𝑛, ⟨𝑇_𝑓𝑧^𝛼,𝑧^𝛼+𝑙⟩=0 for 𝛼∈(ℕ^𝑛⧵𝐸)∩(ℕ^𝑛⧵(𝐸^′−𝑙)). Since 𝐸 and 𝐸^′ have property (P), the subset (ℕ^𝑛⧵𝐸)∩(ℕ^𝑛⧵(𝐸^′−𝑙))=(ℕ^𝑛⧵(𝐸∪(𝐸^′−𝑙)) does not have property (P). By (i), we have ⟨𝑇_𝑓𝑧^𝛼,𝑧^𝛼+𝑙⟩=0 for 𝛼∈ℕ^𝑛. It is easy to prove that ⟨𝑇_𝑓𝑧^𝛼+𝑙,𝑧^𝛼⟩=0 for 𝛼∈ℕ^𝑛. So ⟨𝑇_𝑓𝑧^𝛼,𝑧^𝛽⟩=0 for all 𝛼,𝛽∈ℕ^𝑛, that is, 𝑇_𝑓=0 and 𝑓(𝑧)≡0 for almost all 𝑧∈𝔻^𝑛.
(iii) Since rad𝜉^𝑝𝑓𝑧₁,…,𝑧_𝑛=1(2𝜋)^𝑛_[0,2𝜋]^𝑛𝜉^𝑝𝑓𝑒^𝑖𝑡1𝑧₁,𝑒^𝑖𝑡2𝑧₂,…,𝑒^𝑖𝑡𝑛𝑧_𝑛𝑑𝑡₁⋯𝑑𝑡_𝑛=1(2𝜋)^𝑛_[0,2𝜋]^𝑛𝑒^{−𝑖(𝑡1+𝜃1)𝑝1}⋯𝑒^{−𝑖(𝑡𝑛+𝜃𝑛)𝑝𝑛}𝑓𝑟₁𝑒^{𝑖(𝑡1+𝜃1)},…,𝑟_𝑛𝑒^{𝑖(𝑡𝑛+𝜃𝑛)}𝑑𝑡₁⋯𝑑𝑡_𝑛=1(2𝜋)^𝑛_[0,2𝜋]^𝑛𝑒^{−𝑖(∑𝑛𝑗=1𝑡𝑗𝑝𝑗)}𝑓𝑟₁𝑒^𝑖𝑡1,…,𝑟_𝑛𝑒^𝑖𝑡𝑛𝑑𝑡₁⋯𝑑𝑡_𝑛, (2.14) we have _𝔻𝑛𝜉^𝑝rad𝜉^𝑝𝑓(𝑧)𝑧^𝛼𝑧^𝛽(𝑧)=1𝑑𝑉(2𝜋)^𝑛_[0,2𝜋]^𝑛_[0,1]^𝑛_[0,2𝜋]^𝑛𝑒^{−𝑖(∑𝑛𝑗=1𝜃𝑗(𝑝𝑗+𝛼𝑗−𝛽𝑗))}𝑓𝑟₁𝑒^𝑖𝑡1,…,𝑟_𝑛𝑒^𝑖𝑡𝑛𝑑𝜃₁⋯𝑑𝜃_𝑛×^𝑛_𝑗=1𝑟^{𝛼𝑗+𝛽𝑗}_𝑗⁺¹𝑒^{−𝑖(∑𝑛𝑗=1𝑡𝑗𝑝𝑗)}𝑑𝑟₁⋯𝑑𝑟_𝑛𝑑𝑡₁⋯𝑑𝑡_𝑛. (2.15) If 𝛼+𝑝≠𝛽, then ∫_𝔻𝑛[𝜉^𝑝rad(𝜉^𝑝𝑓)](𝑧)𝑧^𝛼𝑧^𝛽𝑑𝑉(𝑧)=0. Otherwise, if 𝛼+𝑝=𝛽, then _𝔻𝑛𝜉^𝑝rad𝜉^𝑝𝑓(𝑧)𝑧^𝛼𝑧^𝛽(𝑧)=1𝑑𝑉(2𝜋)^𝑛_[0,2𝜋]^𝑛_[0,1]^𝑛𝑓𝑟₁𝑒^𝑖𝑡1,…,𝑟_𝑛𝑒^{𝑖𝑡𝑛𝑛}_𝑗=1𝑟^{𝛼𝑗+𝛽𝑗}_𝑗⁺¹𝑒^{−𝑖(∑𝑛𝑗=1𝑡𝑗𝑝𝑗)}𝑑𝑟₁⋯𝑑𝑟_𝑛𝑑𝑡₁⋯𝑑𝑡_𝑛=_𝔻𝑛𝑓(𝑧)𝑧^𝛼𝑧^𝛽𝑑𝑉(𝑧). (2.16) Thus, we get ∫_𝔻𝑛[𝜉^𝑝rad(𝜉^𝑝𝑓)](𝑧)𝑧^𝛼𝑧^𝛽𝑑𝑉(𝑧)=∫_𝔻𝑛𝑓(𝑧)𝑧^𝛼𝑧^𝛽𝑑𝑉(𝑧) for any 𝛼,𝛽∈ℤ^𝑛. So 𝜉^𝑝rad(𝜉^𝑝𝑓)=𝑓, this means that there exists 𝜑(𝑟) such that 𝜉^𝑝𝑓(𝑧)=𝜑(𝑟), that is, 𝑓(𝑧)=𝜉^𝑝𝜑(𝑟) is a quasihomogeneous function of degree 𝑝.

Remark 2.5. Let 𝑓 be as in Theorem 2.4. Then, 𝑇_𝑓=∑_𝛼∈ℕ^𝑛𝑤(𝑓,𝛼,𝛼+𝑝)𝑒_𝛼+𝑝⊗𝑒_𝛼, where 𝑤𝑓,𝛼,𝛼+𝑝=𝑇_𝑓𝑒_𝛼,𝑒_𝛼+𝑝=√𝑎_1𝛼+𝑎_{1𝛼+𝑝+𝔻𝑛}𝜉^𝑝𝜑(𝑟)𝑧^𝛼𝑧^𝛼+𝑝(𝑧)=√𝑑𝑉𝑎_1𝛼+𝑎_1𝛼+𝑝+𝜑2𝛼+𝑝+2. (2.17) Recall that a densely defined operator on 𝐴²(𝔻^𝑛) is said to be diagonal if it is diagonal with respect to the standard orthonormal basis. In particular, for 𝑓∈𝐿^∞(𝔻^𝑛), 𝑇_𝑓 is diagonal if and only if rad(𝑓)=𝑓. In this case, 𝑇_𝑓=∑_𝛼∈ℕ^𝑛𝑤(𝑓,𝛼)𝑒_𝛼⊗𝑒_𝛼, where 𝑤(𝑓,𝛼)=⟨𝑇_𝑓𝑒_𝛼,𝑒_𝛼⟩=𝑎_1𝛼+𝑓(2𝛼+2).

3. Commutativity of Toeplitz Operators

In this section, we study the commutativity of the Toeplitz operators with some special quasihomogeneous symbols and give the characterizations, respectively.

Theorem 3.1. Let 𝑔=𝜉^𝑝𝜑(𝑟)∈𝐿²(𝔻^𝑛) be a quasihomogeneous function of degree 𝑝 and 𝑓=∑_𝑘∈ℤ^𝑛𝜉^𝑘𝑓_𝑘(𝑟₁,…,𝑟_𝑛)∈𝐿²(𝔻^𝑛). Then, 𝑇_𝑓𝑇_𝑔=𝑇_𝑔𝑇_𝑓 if and only if 𝑇_𝜉^𝑘_𝑓𝑘𝑇_𝑔=𝑇_𝑔𝑇_𝜉^𝑘_𝑓𝑘 for any 𝑘∈ℤ^𝑛.
Moreover, the following statements hold. (i)If 𝑄₁={𝛼∶𝛼+𝑝≽^╱0}∩{𝛼∶𝛼+𝑘≽0}∩{𝛼∶𝛼+𝑘+𝑝≽0}≠∅, then, for each 𝛼∈𝑄₁, 𝜑(2𝛼+2𝑘+𝑝+2)𝑓_𝑘(2𝛼+𝑘+2)=0. (ii)If 𝑄₂={𝛼∶𝛼+𝑘≽^╱0}∩{𝛼∶𝛼+𝑝≽0}∩{𝛼∶𝛼+𝑘+𝑝≽0}≠∅, then, for each 𝛼∈𝑄₂, 𝜑(2𝛼+𝑝+2)𝑓_𝑘(2𝛼+2𝑝+𝑘+2)=0.

Proof. Note that, for 𝛼∈ℕ^𝑛, 𝑇_𝑔𝑇_𝑓𝑒_𝛼=_{𝑘∈ℤ𝑛}𝑇_𝑔𝑇_𝜉^𝑘_𝑓𝑘𝑒_𝛼=_{𝑘∈ℤ𝑛𝛽∈ℕ𝑛}𝑇_𝑔𝑇_𝜉^𝑘_𝑓𝑘𝑒_𝛼,𝑒_𝛽𝑒_𝛽=_{𝑘∈ℤ𝑛}𝑇_𝑔𝑇_𝜉^𝑘_𝑓𝑘𝑒_𝛼,𝑒_{𝛼+𝑘+𝑝}𝑒_{𝛼+𝑘+𝑝}. (3.1) The second equality follows that ⟨𝑇_𝑔𝑇_𝜉^𝑘_𝑓𝑘𝑒_𝛼,𝑒_𝛽⟩=0, when 𝛽≠𝛼+𝑘+𝑝.
Similarly, 𝑇_𝑓𝑇_𝑔𝑒_𝛼=_{𝑘∈ℤ𝑛}𝑇_𝜉^𝑘_𝑓𝑘𝑇_𝑔𝑒_𝛼=_{𝑘∈ℤ𝑛𝛽∈ℕ𝑛}𝑇_𝜉^𝑘_𝑓𝑘𝑇_𝑔𝑒_𝛼,𝑒_𝛽𝑒_𝛽=_{𝑘∈ℤ𝑛}𝑇_𝜉^𝑘_𝑓𝑘𝑇_𝑔𝑒_𝛼,𝑒_{𝛼+𝑘+𝑝}𝑒_{𝛼+𝑘+𝑝}. (3.2) Since {𝑒_𝛼} are the standard orthogonal basis and ⟨𝑇_𝑔𝑇_𝜉^𝑘_𝑓𝑘𝑒_𝛼,𝑒_𝛽⟩=⟨𝑇_𝜉^𝑘_𝑓𝑘𝑇_𝑔𝑒_𝛼,𝑒_𝛽⟩=0 for 𝛽≠𝛼+𝑘+𝑝, it is easy to check that the following statements are equal:(I)𝑇_𝑓𝑇_𝑔=𝑇_𝑔𝑇_𝑓;(II)⟨𝑇_𝑔𝑇_𝜉^𝑘_𝑓𝑘𝑒_𝛼,𝑒_{𝛼+𝑘+𝑝}⟩=⟨𝑇_𝜉^𝑘_𝑓𝑘𝑇_𝑔𝑒_𝛼,𝑒_{𝛼+𝑘+𝑝}⟩, 𝛼∈ℕ^𝑛;(III)𝑇_𝑔𝑇_𝜉^𝑘_𝑓𝑘𝑒_𝛼=𝑇_𝜉^𝑘_𝑓𝑘𝑇_𝑔𝑒_𝛼, 𝛼∈ℕ^𝑛;(IV)𝑇_𝑔𝑇_𝜉^𝑘_𝑓𝑘=𝑇_𝜉^𝑘_𝑓𝑘𝑇_𝑔.
Furthermore, 𝑇_𝑔𝑇_𝜉^𝑘_𝑓𝑘𝑒_𝛼,𝑒_{𝛼+𝑘+𝑝}=𝑇_𝜉^𝑘_𝑓𝑘𝑒_𝛼,𝑒_𝛼+𝑘𝑇_𝑔𝑒_𝛼+𝑘,𝑒_{𝛼+𝑘+𝑝}=0,𝛼+𝑘⪰^╱0or𝛼+𝑘+𝑝^╱√⪰0,𝑎_1𝛼+𝑎_{1𝛼+𝑘+𝑝+}𝑎_1𝛼+𝑘+𝜑2𝛼+2𝑘+𝑝+2𝑓_𝑘2𝑇2𝛼+𝑘+,𝛼+𝑘⪰0,𝛼+𝑘+𝑝⪰0,_{𝜉𝑘𝑓𝑘}𝑇_𝑔𝑒_𝛼,𝑒_{𝛼+𝑘+𝑝}=𝑇_𝑔𝑒_𝛼,𝑒_𝛼+𝑝𝑇_𝜉^𝑘_𝑓𝑘𝑒_𝛼+𝑝,𝑒_{𝛼+𝑘+𝑝}=0,𝛼+𝑝⪰^╱0or𝛼+𝑘+𝑝^╱√⪰0,𝑎_1𝛼+𝑎_{1𝛼+𝑘+𝑝+}𝑎_1𝛼+𝑝+𝑓_𝑘2𝛼+2𝑝+𝑘+2𝜑2𝛼+𝑝+2,𝛼+𝑝⪰0,𝛼+𝑘+𝑝⪰0. (3.3) Thus, the statements (i) and (ii) hold.

Theorem 3.2. Let 𝑓₁, 𝑓₂ be quasihomogeneous functions of degree 𝑝 and −𝑠, where 𝑝≽𝑠≻0. If [𝑇_𝑓1,𝑇_𝑓2]=0, then 𝑓₁=0 or 𝑓₂=0.

Proof. If 𝑓₁,𝑓₂ are quasihomogeneous functions of degree 𝑝 and −𝑠, then there exist radial functions 𝜑₁ and 𝜑₂, such that 𝑓₁=𝜉^𝑝𝜑₁ and 𝑓₂=𝜉^𝑠𝜑₂. If 𝑇_𝑓1𝑇_𝑓2=𝑇_𝑓2𝑇_𝑓1, (2.10) implies that, for all 𝛼∈ℕ^𝑛, 𝑎_{1𝛼−𝑠+}𝜑₂2𝛼−𝑠+2𝜑₁2𝛼−2𝑠+𝑝+2=𝑎_1𝛼+𝑝+𝜑₁2𝛼+𝑝+2𝜑₂2𝜑2𝛼+2𝑝−𝑠+,if𝛼⪰𝑠,₁2𝛼+𝑝+2𝜑₂2𝛼+2𝑝−𝑠+2=0,if𝛼⪰^╱𝑠. (3.4) We claim that there exist {𝜆_𝑘}^∞_𝑘=1 with ∑1/𝜆_𝑘=∞ such that 𝜑₁2𝜆_𝑘+𝑝₁+2,𝑧^′𝜑₂2𝜆_𝑘+2𝑝₁−𝑠₁+2,𝑧^′=0,forany𝑧^′∈𝐼_𝑛−1. (3.5) It follows that 𝑍(𝜑₁𝜑₂) do not have property (P). So we can get 𝑓₁=0 or 𝑓₂=0 by Lemma 2.3.
We only need to prove the claim. Since 𝑠≻0, there exists 1≤𝑗≤𝑛, such that 𝑠_𝑗≥1. Without losing generality, suppose that 𝑗=1. Let 𝜆₀=𝑠₁−1≥0; then, 𝛼⁰=(𝜆₀,𝛼′)≽^╱𝑠 and 𝜑₁(2𝜆₀+𝑝₁+2,𝛼′+𝑝′+2)𝜑₂(2𝜆₀+𝑝₁+2,𝛼′+2𝑝′−𝑠′+2)=0 where 𝛼′∈ℕ^𝑛−1, 𝑝′=(𝑝₂,…,𝑝_𝑛) and 𝑠′=(𝑠₂,…,𝑠_𝑛). Denote 𝐸₀={𝑧′∈𝐼_𝑛−1∶𝜑₁(2𝜆₀+𝑝₁+2,𝑧′)=0} and 𝐹₀={𝑧′∈𝐼_𝑛−1∶𝜑₁(2𝜆₀+𝑝₁+2,𝑧′)=0}. Note that at least one of the sets 𝐸₀ and 𝐹₀ does not have property (P). Since 𝜑₁(2𝜆₀+𝑝₁+2,𝑧′)=0 and 𝜑₂(2𝜆₀+𝑝₁+2,𝑧′)=0 are analytic on 𝐼_𝑛−1, Lemma 2.3 shows that 𝐸₀=𝐼_𝑛−1 or 𝐹₀=𝐼_𝑛−1.
Case 1. If 𝐸₀=𝐼_𝑛−1, then 𝜑₁2𝛼⁰+𝑠+𝑝+2𝜑₂2𝛼⁰2=𝑎+𝑠+2𝑝−𝑠+^{1𝛼−𝑠+}𝑎_1𝛼+𝑝+𝜑₁2𝜑⁰+𝑝+2𝜑₂2𝛼⁰+𝑠+2=0, (3.6) where 𝛼⁰=(𝜆₀+𝑠,𝛼^′) with 𝛼^′∈𝑁^𝑛−1. Let 𝜆₁=𝜆₀+𝑠. Denote by 𝐸₁={𝑧^′∈𝐼_𝑛−1∶𝜑₁(2𝜆₁+𝑝₁+2,𝑧^′)=0} and 𝐹₁={𝑧^′∈𝐼_𝑛−1∶𝜑₁(2𝜆₁+𝑝₁+2,𝑧^′)=0}. Then, at least one of the sets 𝐸₁ and 𝐹₁ does not have property (P). By Lemma 2.3 again, we have 𝐸₁=𝐼_𝑛−1 or 𝐹₁=𝐼_𝑛−1. Thus 𝜑₁(𝜆₁,𝑧^′)𝜑₂(𝜆₁,𝑧^′)=0,forany𝑧^′∈𝐼_𝑛−1.Case 2. If 𝐹₀=𝐼_𝑛−1, then𝜑₁2𝛼⁰+𝑝+𝑝+2𝜑₂2𝛼⁰2=𝑎+𝑝+2𝑝−𝑠+_{1𝛼−𝑠+}𝑎_1𝛼+𝑝+𝜑₁2𝛼⁰+3𝑝−2𝑠+2𝜑₂2𝛼⁰+2𝑝−𝑠+2=0. (3.7) By the same technique, we can get that (3.5) holds when 𝜆₁=𝜆₀+𝑝₁.
Similarly, we can find a sequence 𝜆_𝑘=𝜆₀+𝑢(𝑘)𝑝₁+𝑣(𝑘)𝑠₁, where the functions 𝑢(𝑘)=1 or 0, 𝑣(𝑘)=1 or 0, and 𝑢(𝑘)+𝑣(𝑘)=1 for 𝑘∈ℤ⁺. Then, (i) 𝜆₁≥min{𝑝₁,𝑠₁}, (ii) ∑_𝑘∈ℕ1/𝜆_𝑘=+∞, and (iii) for every 𝑘≥1, 𝜆_𝑘 satisfies (3.5). So we complete the proof.

For 𝑛=2, we have the following results.

Theorem 3.3. Let 𝑔(𝑟)=𝑟^𝑚1₁𝑟^𝑚2₂, where 𝑚₁≥0 and 𝑚₂≥0. Let 𝑓(𝑟₁,𝑟₂)∈𝐿^∞([0,1]²), 𝑝=(𝑝₁,𝑝₂)∈ℕ² and 𝑝₁⋅𝑝₂≠0. Then, 𝑇_𝜉^𝑝_𝑔𝑇_𝜉^p_𝑓=𝑇_𝜉^𝑝_𝑓𝑇_𝜉^𝑝_𝑔 if and only if there exists an analytic function on ℂ, such that the function 𝜓((𝑧₂𝑝₁−𝑧₁𝑝₂)/(𝑝²₁+𝑝²₂))/((𝑧₁+𝑚₁)(𝑧₂+𝑚₂)) is bounded on 𝐼₂ and 𝑓𝑧₁,𝑧₂=𝜓𝑧₂𝑝₁−𝑧₁𝑝₂/𝑝²₁+𝑝²₂𝑧₁+𝑚₁𝑧₂+𝑚₂,∀𝑧=𝑧₁,𝑧₂∈𝐼₂. (3.8)

Proof. As in the proof of Theorem 3.2, it is easy to check that 𝑇_𝜉^𝑝_𝑔𝑇_𝜉^𝑝_𝑓=𝑇_𝜉^𝑝_𝑓𝑇_𝜉^𝑝_𝑔 if and only if 𝑇_𝜉^𝑝_𝑔𝑇_𝜉^𝑝_𝑓𝑒_𝛼,𝑒_𝛼+2𝑝=𝑇_𝜉^𝑝_𝑓𝑇_𝜉^𝑝_𝑔𝑒_𝛼,𝑒_𝛼+2𝑝, (3.9) which is equal to 𝑔2𝛼+𝑝+2𝑓2𝛼+3𝑝+2=𝑓2𝛼+𝑝+2𝑔2𝛼+3𝑝+2,∀𝛼∈ℕ². (3.10)
Suppose that there is a function 𝜓 as in this theorem.
Note that 𝑔(𝑧)=_[0,1]2𝑟^{1𝑚+𝑧−}𝑑𝑟₁𝑑𝑟₂=1𝑧₁+𝑚₁𝑧₂+𝑚₂≠0,∀𝑧∈𝐼₂, (3.11) and (2𝛼₂+3𝑝₂+2)𝑝₁−(2𝛼₁+3𝑝₁+2)𝑝₂=(2𝛼₂+𝑝₂+2)𝑝₁−(2𝛼₁+𝑝₁+2)𝑝₂, for any 𝛼=(𝛼₁,𝛼₂)∈ℕ². Then, it is easy to check that equality (3.10) holds, that is, 𝑇_𝜉^𝑝_𝑔𝑇_𝜉^𝑝_𝑓=𝑇_𝜉^𝑝_𝑓𝑇_𝜉^𝑝_𝑔.
Conversely, if 𝑇_𝜉^𝑝_𝑔 and 𝑇_𝜉^𝑝_𝑓 commute, we will structure an analytic 𝜓 which satisfies the conditions in this theorem.
Since 𝑔(𝑧)≠0 for all 𝑧∈𝐼₂, the function 𝑓(𝑧)/𝑔(𝑧) is analytic on 𝐼₂. Note that |𝑟^𝑧−1|<1 for 0<𝑟_𝑗<1, 𝑗=1,2 and 𝑧∈𝐼₂. Thus, 𝑓(𝑧)≤_𝔻2𝑓𝑑𝑉(𝑧)=𝑓_𝐿1([0,1]²₎,𝑔(𝑧)≤𝑔_𝐿1([0,1]²₎. (3.12) Fix 𝛼₀∈ℕ², and let 𝑧₀=2𝛼₀+𝑝+2, then, 𝑓𝑧₀𝑔𝑧₀=𝑓𝑧₀+2𝑝𝑔𝑧₀𝑓𝑧+2𝑝=⋯=⁰+2𝑘𝑝𝑔𝑧₀+2𝑘𝑝,𝑘=0,1,2,…. (3.13) Combining this with Lemma 2.3, we can get that the above equality holds for any 𝑧₀∈𝐼₂. Let 𝑝^⟂=(−𝑝₂,𝑝₁); then, 𝑝⟂𝑝^⟂. For each 𝑧₀∈𝐼₂, there exist 𝜇₁,𝜇₂∈ℂ such that 𝑧₀=𝜇₁𝑝+𝜇₂𝑝^⟂. So 𝑓𝜇₁𝑝+𝜇₂𝑝^⟂𝑔𝜇₁𝑝+𝜇₂𝑝^⟂=𝑓𝜇₁+2𝑝+𝜇₂𝑝^⟂𝑔𝜇₁+2𝑝+𝜇₂𝑝^⟂𝑓𝜇=⋯=¹+2𝑘𝑝+𝜇₂𝑝^⟂𝑔𝜇₁+2𝑘𝑝+𝜇₂𝑝^⟂,𝑘=0,1,2,…. (3.14) Put 𝐹(𝜆)=𝑓𝜆𝑝+𝜇₂𝑝^⟂𝑔𝜆𝑝+𝜇₂𝑝^⟂−𝑓𝜇₁𝑝+𝜇₂𝑝^⟂𝑔𝜇₁𝑝+𝜇₂𝑝^⟂, (3.15) then, 𝐹(𝜆) is analytic on {z∈ℂ∶Re(𝑧𝑝₁−𝜇₂𝑝₂)>2andRe(𝑧𝑝₂+𝜇₂𝑝₁)>2} and |𝐹(𝜆)|≤𝑓_𝐿1𝑚₁+𝜆𝑝₁−𝜇₂𝑝₂⋅𝑚₂+𝜆𝑝₂+𝜇₂𝑝₁+𝐶₁≤𝑓_𝐿1𝐷₁|𝜆|²+𝐷₂|𝜆|+𝐷₃, (3.16) where 𝐶₁,𝐷₁,𝐷₂,𝐷₃ are all constants. Since 𝐹(𝜇₁+2𝑘)=0 and ∑^+∞_𝑘=01/2𝑘=+∞, the set {𝜇₁+2𝑘∶𝑘=0,1,2,…}⊆𝑍(𝐹). Thus 𝐹(𝜆)≡0. That is 𝑓𝜆𝑝+𝜇₂𝑝^⟂=𝑓𝜇₁𝑝+𝜇₂𝑝^⟂𝑔𝜇₁𝑝+𝜇₂𝑝^⟂𝑔𝜆𝑝+𝜇₂𝑝^⟂. (3.17) For each 𝜇∈ℂ, there exists 𝜆₀∈ℂ such that Re(𝜆₀𝑝₁−𝜇𝑝₂)>2andRe(𝜆₀𝑝₂+𝜇𝑝₁)>2; then, let 𝜓(𝜇)=𝑓(𝜆₀𝑝+𝜇𝑝^⟂)/𝑔(𝜆₀𝑝+𝜇𝑝^⟂). By equality (3.17), we conclude that the function 𝜓 is well defined. Since the function 𝑓/𝑔 is analytic on 𝐼₂, we can prove that 𝜓 is an analytic function on ℂ. Let 𝑧₁=𝜆𝑝₁−𝜇₂𝑝₂,𝑧₂=𝜆𝑝₂+𝜇₂𝑝₁; (3.18) Then, 𝑧𝜆=¹𝑝₁−𝑧₂𝑝₂𝑝²₁+𝑝²₂,𝜇₂=𝑧₂𝑝₁−𝑧₁𝑝₂𝑝²₁+𝑝²₂. (3.19) So (3.17) is equal to 𝑓𝑧₁,𝑧₂=𝜓𝑧₂𝑝₁−𝑧₁𝑝₂/𝑝²₁+𝑝²₂𝑧₁+𝑚₁𝑧₂+𝑚₂, (3.20) where |𝜓((𝑧₂𝑝₁−𝑧₁𝑝₂)/(𝑝²₁+𝑝²₂))/((𝑧₁+𝑚₁)(𝑧₂+𝑚₂))|≤‖𝑓‖_𝐿¹_([0,1]²₎ and (𝑧₁,𝑧₂)∈𝐼₂. This completes the proof.

Corollary 3.4. Let 𝑓,𝑔 be as in Theorem 3.3 and 𝑝∈ℕ⁺; then, the following statements hold:(i)𝑇_{𝑒1𝑖𝑝𝜃𝑔}𝑇_{𝑒1𝑖𝑝𝜃𝑓}=𝑇_{𝑒1𝑖𝑝𝜃𝑓}𝑇_{𝑒1𝑖𝑝𝜃𝑔} if and only if 𝑓=𝑟^𝑚1₁𝜑(𝑟₂), where 𝜑∈𝐿^∞([0,1]);(ii)𝑇_{𝑒2𝑖𝑝𝜃𝑔}𝑇_{𝑒2𝑖𝑝𝜃𝑓}=𝑇_{𝑒2𝑖𝑝𝜃𝑓}𝑇_{𝑒2𝑖𝑝𝜃𝑔} if and only if 𝑓=𝑟^𝑚2₂𝜑(𝑟₁), where 𝜑∈𝐿^∞([0,1]).

Proof. (i) By (3.17) we have _[0,1]2𝑓𝑟₁,𝑟₂𝑟^𝑧1₁⁻¹𝑟^𝑧2₂⁻¹𝑑𝑟₁𝑑𝑟₂𝑧=𝜓²𝑝_[0,1]²𝑟^𝑚1+𝑧1₁⁻¹𝑟^𝑚2+z2₂⁻¹𝑑𝑟₁𝑑𝑟₂, (3.21) that is, _[0,1][0,1]𝑓𝑟₁,𝑟₂𝑟^𝑧2₂⁻¹𝑑𝑟₂−𝑟^𝑚1₁𝜓𝑧₂𝑝_[0,1]𝑟^𝑚2+𝑧2₂⁻¹𝑑𝑟₂𝑟^𝑧1₁⁻¹𝑑𝑟₁=0. (3.22) Then, _[0,1]𝑓𝑟₁,𝑟₂𝑟^𝑧2₂⁻¹𝑑𝑟₂=𝑟^𝑚1₁𝜓𝑧₂𝑝1𝑚₂+𝑧₂. (3.23) It follows that there exists 𝜑=𝜑(𝑟₂)∈𝐿^∞([0,1]) such that 𝑓=𝑟^𝑚1₁𝜑(𝑟₂).
On the other hand, if 𝑓=𝑟^𝑚1₁𝜑(𝑟₂), then 𝑓(𝑧)𝑔=1𝑧+2𝑝,0𝑚₁+𝑧₁⋅𝜑𝑧₂⋅1𝑚₁+𝑧₁⋅1+2𝑝𝑚₂+𝑧₂,𝑓𝑔(𝑧)=1𝑧+2𝑝,0𝑚₁+𝑧₁⋅𝜑𝑧+2𝑝²⋅1𝑚₁+𝑧₁⋅1𝑚₂+𝑧₂. (3.24) Thus, we have 𝑇_𝜉^𝑝_𝑔𝑇_𝜉^𝑝_𝑓=𝑇_𝜉^𝑝_𝑓𝑇_𝜉^𝑝_𝑔.
(ii) Can also be proved in the same way.

In [6], Čučkovi𝑡𝑒𝑥𝑡^′𝑐 and Rao showed that if 𝑓,𝑔∈𝐿^∞(𝔻) and 𝑔 is a nonconstant radial function, then 𝑇_𝑓𝑇_𝑔=𝑇_𝑔𝑇_𝑓 implies that 𝑓 is a radial function. However, this is not true if 𝑓,𝑔∈𝐿^∞(𝔻^𝑛), where 𝑛≥2. For example, 𝑔(𝑧)=𝑔(𝑧₁,…,𝑧_𝑗−1,|𝑧_𝑗|,𝑧_𝑗+1,…,𝑧_𝑛) and 𝑓(𝑧)=𝑓(|𝑧₁|,…,|𝑧_𝑛|) only for |𝑧_𝑗|, and it is clear that 𝑇_𝑔𝑇_𝑓=𝑇_𝑓𝑇_𝑔, but 𝑔(𝑧) may not be a radial function. Let 𝐺={𝑔∈𝐿^∞(𝔻^𝑛)∶𝑔isradialandfor𝑓∈L^∞(𝔻^𝑛),𝑇_𝑓𝑇_𝑔=𝑇_𝑔𝑇_𝑓impliesthat𝑓isradial}(if 𝑛=1, this set is exactly the set of all non-constant bounded radial functions). In the following, we can give a complete description of 𝐺.

Theorem 3.5. 𝐺={𝑔(𝑧) is a bounded radial function: for each 𝑘=(𝑘₁,…,𝑘_𝑛)≠0,𝑎_(2𝑧)𝑔(2𝑧)≠𝑎_{(2𝑧+2𝑘)}𝑔(2𝑧+2𝑘),where𝑧∈𝐼_𝑛and𝑧+𝑘∈𝐼_𝑛}.

Proof. Suppose that 𝑓(𝑧)=∑_𝑘∈ℤ^𝑛𝜉^𝑘𝑓_𝑘(𝑟₁,…,𝑟_𝑛)∈𝐿^∞(𝔻^𝑛) and 𝑔 is a radial function. Lemma 2.2 shows that 𝑇_𝑓𝑇_𝑔𝑧^𝛼=_{𝑘∈ℤ𝑛}2^𝑛𝑎_1𝛼+𝑔2𝛼+2𝐶_𝑘𝑓_𝑘2𝛼+𝑘+2𝑧^𝛼+𝑘,𝑇_𝑔𝑇_𝑓𝑧^𝛼=_{𝑘∈ℤ𝑛}𝐶_𝑘𝑓_𝑘2𝛼+𝑘+22^𝑛𝑎_1𝛼+𝑘+𝑔2𝛼+2𝑘+2𝑧^𝛼+𝑘, (3.25) where 𝛼∈ℕ^𝑛 and 𝐶_𝑘=0if𝛼⪰^╱𝑘⁻,2^𝑛𝑎_1𝛼+𝑘+if𝛼⪰𝑘⁻. (3.26) It follows that 𝑇_𝑓𝑇_𝑔=𝑇_𝑔𝑇_𝑓 if and only if 𝑎_1𝛼+𝑔2𝛼+2𝐶_𝑘𝑓_𝑘2𝛼+𝑘+2=𝐶_𝑘𝑓_𝑘2𝛼+𝑘+2𝑎_1𝛼+𝑘+𝑔2𝛼+2𝑘+2, (3.27) for any 𝑘∈ℤ^𝑛. Let 𝐸₁={𝛼∶(𝑎_1𝛼+𝑔−𝑎_1𝛼+𝑘+𝑟^𝑘𝑔)^∧(2𝛼+2)=0} and 𝐸₂={𝛼∶𝑟^𝑘𝑓_𝑘(2𝛼+2)=0}. Then 𝐸₁∪𝐸₂={𝛼;𝛼≽𝑘⁻}. The commutativity of 𝑇_𝑓 and 𝑇_𝑔 is equivalent to that at least one of 𝐸₁ and 𝐸₂ does not have property (P); then, Lemma 2.3 shows that 𝑓_𝑘≡0 or 𝑎_(2𝑧)𝑔(2𝑧)≠𝑎_{(2𝑧+2𝑘)}𝑔(2𝑧+2𝑘), where 𝑧∈𝐼_𝑛and𝑧+𝑘∈𝐼_𝑛. The rest of the proof is obvious.

Remark 3.6. In Theorem 3.5, particularly if 𝑔 is a radial function such that 𝑔(𝑧)=∏^𝑛_𝑗=1𝑔_𝑗(𝑧_𝑗) or 𝑔(𝑧)=∑^𝑛_𝑗=1𝑔_𝑗(𝑧_𝑗), where each 𝑔_𝑗(𝑧_𝑗)(1≤𝑗≤𝑛) is a non-constant radial function, then foreach𝑘=(𝑘₁,⋯,𝑘_𝑛)≠0,𝑎_(2𝑧)𝑔(2𝑧)≠𝑎_{(2𝑧+2𝑘)}𝑔(2𝑧+2𝑘),where𝑧∈𝐼_𝑛and𝑧+𝑘∈𝐼_𝑛, so 𝑔∈𝐺. It follows that 𝐺 is nonempty.

4. Finite Rank Semicommutators and Commutators

Recall that Čučković and Louhichi (see [5]) have found some nonzero finite rank semicommutators of quasihomogeneous symbol Toeplitz operators on the Bergman space of unit disk. In this section, we will show that the finite rank semicommutators and commutators of Toeplitz operators with quasihomogeneous symbols must be zero on 𝐴²(𝔻^𝑛) with 𝑛≥2. Our idea is mainly from [17].

Theorem 4.1. Let 𝑘,𝑙∈ℤ^𝑛 with 𝑛≥2, 𝑘+𝑙=𝑚, and let 𝜑, 𝜓 be radial functions such that 𝑓₁=𝜉^𝑘𝜑, 𝑓₂=𝜉^𝑙𝜓, and 𝜉^𝑚𝜑𝜓 are all 𝑇-functions. If the semicommutator (𝑇_𝑓1,𝑇_𝑓2] has finite rank, then it must be zero.

Proof. Let 𝑆 denote the semicommutator (𝑇_𝑓1,𝑇_𝑓2]. For 𝛼∈ℕ^𝑛, if 𝑆 is finite rank, by equality (2.9), we have that there exists 𝛼⁰≽𝑘⁻+𝑙⁻ such that 𝑆𝑧^𝛼=2^𝑛𝑎_1𝛼+𝑚+2^𝑛𝑎_1𝛼+𝑙+𝜓2𝜑2()2−22𝛼+𝑙+𝛼+𝑙+𝑘+𝜑𝜓2𝛼+𝑚+=0for𝛼⪰𝛼⁰, (4.1) which is equivalent to 𝑟^𝑘++𝑙+𝜑^∧2𝛼+𝑙−𝑘⁻−𝑙⁻+2𝑟^{𝑘−+𝑙−}𝜓^∧2𝛼+𝑙−𝑘⁻−𝑙⁻+2=𝑟^{𝑙++𝑙−∧}2𝛼+𝑙−𝑘⁻−𝑙⁻+2𝑟^𝑘𝜑𝜓^∧2𝛼+𝑙−𝑘⁻−𝑙⁻+2, (4.2) for 𝛼≽𝛼⁰. Combining this with Lemma 2.3, we get 𝑟^𝑘++𝑙+𝜑^∧(𝑧)𝑟^{𝑘−+𝑙−}𝜓^∧(𝑧)=𝑟^{𝑙++𝑙−∧}(𝑧)𝑟^𝑘𝜑𝜓^∧(𝑧),for𝑧∈𝐼_𝑛. (4.3) Hence, 𝑆𝑧^𝛼=0∀𝛼∈𝐸₂. (4.4)
In the following, we only need to prove that 𝜑𝜓(𝛼)=0 for all 𝛼∈𝐸^𝑐₁∩𝐸^𝑐₂.
If 𝐸^𝑐₁∩𝐸^𝑐₂≠∅, there is a 𝑗  (0≤𝑗≤𝑛) such that 𝑚⁻_𝑗<𝑙⁻_𝑗. Without loss of generality, assume that 𝑗=1. Then, {(𝛼₁,𝑚₂+𝛼₂,…,𝑚_𝑛+𝛼_𝑛)∶𝑚⁻₁≤𝛼₁≤𝑙⁻₁,𝛼_𝑗≥0,𝑗=2,…,𝑛}⊆𝐸₂. For each 𝑚⁻₁≤𝛼₁, let 𝐹_𝑎1(𝑟₂,…,𝑟_𝑛)=∫¹₀(𝜑𝜓)(𝑟₁,𝑟₁,…,𝑟_𝑛)𝑟^𝑎1₁⁻¹𝑑𝑟₁. Since 𝑍(𝐹_𝑎1)⊇{(𝑚₂+𝛼₂,…,𝑚_𝑛+𝛼_𝑛)∶(𝛼₂,…,𝛼_𝑛)∈𝑁^𝑛−1} does not have property (P), we have 𝐹_𝑎1≡0. Therefore, 𝜑𝜓(𝑎₁,𝑎₂,…,𝑎_𝑛)=0 for 𝑚⁻₁≤𝑎₁≤𝑙⁻₁ and 𝑎_𝑗≥𝑚⁻_𝑗,𝑗=2,…,𝑛. So 𝑆(𝑧^𝛼)=0 for 𝛼∈𝐸^𝑐₁∩𝐸^𝑐₂.
This completes the proof.

We now pass to the commutator of two quasihomogeneous Toeplitz operators. Here the situation is the same as for the semicommutator.

Theorem 4.2. Let 𝑘,𝑙∈ℤ^𝑛 with 𝑛≥2, and let 𝜑, 𝜓 be radial functions such that 𝑓₁=𝜉^𝑘𝜑 and 𝑓₂=𝜉^𝑙𝜓 are both 𝑇-functions. The commutator [𝑇_𝑓1,𝑇_𝑓2] has finite rank if and only if it is a zero operator.

Proof. Let 𝑆 denote the commutator [𝑇_𝑓1,𝑇_𝑓2]. For 𝛼∈ℕ^𝑛, if 𝑆 has finite rank 𝑁, by equality (2.10), we have that there exists 𝛼⁰≽𝑘⁻+𝑙⁻ such that 𝑆𝑧^𝛼=4^𝑛𝑎_{1𝛼+𝑘+𝑙+}𝑎_1𝛼+𝑙+𝜓2𝛼+𝑙+2𝜑2(𝛼+𝑙)+𝑘+2−𝑎_1𝛼+𝑘+𝜑2𝛼+𝑘+2𝜓2(𝛼+𝑘)+𝑙+2𝑧^𝛼=0 (4.5) for 𝛼≽𝛼⁰. As in the proof of Theorem 4.1, the above equation implies that 𝜒_[0,1]𝑛^∧2𝜓2𝜑2()2=𝜒2𝛼+2𝑘+2𝛼+𝑙+𝛼+𝑙+𝑘+_[0,1]𝑛∧2𝛼+2𝑙+2𝜑2𝛼+𝑘+2𝜓2(𝛼+𝑘)+𝑙+2 (4.6) for 𝛼∈𝐹₁∩𝐹₂. Hence, 𝑆𝑧^𝛼=0∀𝛼∈𝐹₄. (4.7) For 𝛼∈𝐹₁∩𝐹^𝑐₂ or 𝛼∈𝐹^𝑐₁∩𝐹₂, following the same way as above, we can also prove that 𝑆(𝑧^𝛼)=0.
This completes the proof.

5. Finite Rank Products of Toeplitz Operators

In [17], the author showed that under certain conditions on the bounded operators 𝑆₁ and 𝑆₂ on 𝐴²(𝔹_𝑛), if 𝑓∈𝐿²(𝔹_𝑛), such that 𝑆₂𝑇_𝑓𝑆₁ is a finite-rank operator, then 𝑓 must be zero almost everywhere on 𝔹_𝑛. On the Bergman space of the polydisk, using the same method as in [17], we can prove Theorem 5.1. Using Theorem 5.1, we get two useful theorems for Toeplitz operators with quasihomogeneous symbols.

Theorem 5.1. Let 𝑆₁, 𝑆₂ be two bounded operators on 𝐴²(𝔻^𝑛). Suppose that there is a set 𝕊⊆ℕ^𝑛 which has property (P), such that ker(𝑆₂)⊆ℳ and 𝒩⊆ran(𝑆₁). Here, ℳ (resp., 𝒩) is the linear subspace of 𝐴²(𝔻^𝑛) spanned by {𝑧^𝑚,𝑚∈𝕊} (resp., {𝑧^𝑚,𝑚∈ℕ^𝑛⧵𝕊}). Suppose that 𝑓∈𝐿²(𝔻^𝑛) such that the operator 𝑆₂𝑇_𝑓𝑆₁ has finite rank; then 𝑓 is the zero function.

Theorem 5.2. Let 𝑀 and 𝑊 be two positive integers. Let 𝑓₁,…,𝑓_𝑀 and 𝑔₁,…,𝑔_𝑊 be quasihomogeneous functions, none of which is the zero function. If 𝑓∈𝐿² such that the operator 𝑇_𝑓𝑀⋯𝑇_𝑓1𝑇_𝑓𝑇_𝑔𝑊⋯𝑇_𝑔1 has finite rank, then 𝑓 is the zero function.

Proof. Let 𝑆₂=𝑇_𝑓𝑀⋯𝑇_𝑓1 and 𝑆₁=𝑇_𝑔𝑊⋯𝑇_𝑔1. Suppose that 𝑓_𝑗=𝜉^𝑝𝑗𝜑_𝑗(𝑟), 1≤𝑗≤𝑀, and 𝑔_𝑙=𝜉^𝑞𝑙𝜓_𝑙(𝑟), 1≤𝑙≤𝑊, where 𝑝_𝑗,𝑞_𝑙∈ℤ^𝑛. By Lemma 2.2, for 𝛼≽∑^𝑤_𝑙=1𝑞⁻_𝑙, we have 𝑆₁𝑧^𝛼=2^{𝑊2𝑛𝑊}_𝑗=1𝑎_𝛼+∑^𝑗_{𝑙=1𝑞𝑙+1}𝜓_𝑗2𝛼+^𝑗_𝑙=1𝑞_𝑙−𝑞_𝑗+2𝑧^{𝛼+∑𝑊𝑗=1𝑞𝑗}. (5.1) Define 𝒥={𝛼∈ℕ^𝑛∶𝛼≽^╱∑^𝑊_𝑙=1𝑞⁻_𝑙}⋃(⋃^𝑊_𝑗=1{𝛼∈ℕ^𝑛∶𝜓_𝑗(2(𝛼+∑^𝑗_𝑙=1𝑞_𝑙)−𝑞_𝑗+2)=0}). Since none of the functions 𝜓₁,…,𝜓_𝑊 is the zero function, the set 𝒥 has property (P).
For 𝛼∈ℕ^𝑛⧵𝒥, we see that 𝑆₁(𝑧^𝛼)≠0. Suppose that 𝜑∈𝐴²_𝛼 such that 𝑆₁(𝜑)=0; then, 0=𝑆₁𝜑=𝑆_{1𝛼∈ℕ𝑛}𝜑,𝑧^𝛼𝑧^𝛼=_{𝛼∈ℕ𝑛}𝜑,𝑧^𝛼𝑆₁𝑧^𝛼. (5.2) So (5.1) implies that for any 𝛼∈ℕ^𝑛⧵𝒥, ⟨𝜑,𝑧^𝛼⟩=0. Therefore ker(𝑆₁) is contained in the closure of the linear span of {𝑧^𝛼∶𝛼∈𝒥} in 𝐴²_𝛼. Now suppose that ℐ=𝛼∶𝛼⪰^╱𝑊_𝑙=1𝑞⁻_𝑙ℕ^𝑛𝒥+^𝑊_𝑗=1𝑞_𝑗. (5.3) Then the set ℐ has property (P) and, for any 𝛼∈ℕ^𝑛⧵ℐ, 𝛽=𝛼−∑^𝑊_𝑗=1𝑞_𝑗 belongs to ℕ^𝑛⧵𝒥. Equality (5.1) implies that 𝑧^𝛼=𝑧^{𝛽+∑𝑊𝑗=1𝑞𝑗} is a multiple of 𝑆₁𝑧^𝛽. So the linear span of {𝑧^𝛼∶𝛼∈ℕ^𝑛⧵ℐ} is contained in the range of 𝑆₁. So there exist subsets 𝒥 and ℐ of ℕ^𝑛 that have property (P) such that ker(𝑆₂) is contained in the closure in 𝐴²_𝛼 of Span({𝑧^𝛼∶𝛼∈𝒥}) and Span({𝑧^𝛼∶𝛼∈ℕ^𝑛⧵ℐ} is a subspace of 𝑆₁(𝐴²_𝛼). Let 𝕊=ℐ⋃𝒥; then, Theorem 5.1 implies that 𝑓 is the zero function.

Theorem 5.3. Suppose that the function 𝑓(𝑧)∈𝐿²(𝔻^𝑛) has the expansion 𝑓(𝑧)=_𝑘⪯𝑀𝜉^𝑘𝑓_𝑘𝑟₁,…,𝑟_𝑛, (5.4) and 𝑓_𝑀(𝑙)≠0 for all 𝑙≽𝑙₀, where 𝑙₀∈ℕ^𝑛, if there is a function 𝑔(𝑧)∈𝐿²(𝔻^𝑛), such that 𝑇_𝑔𝑇_𝑓 has finite rank; then 𝑔=0.

Proof. For 𝛼∈ℕ^𝑛, 𝑇_𝑓𝑧^𝛼=_{𝑀⪰𝑘⪰−𝛼}2^𝑛𝑎_1𝛼+𝑘+𝑓_𝑘2𝛼+𝑘+2𝑧^𝛼+𝑘=2^𝑛𝑎_1𝛼+𝑀+𝑓_𝑀2𝛼+𝑀+2𝑧^𝛼+𝑀+_{𝑀⪰𝑘⪰−𝛼,𝑘≠𝑀}2^𝑛𝑎_1𝛼+𝑘+𝑓_𝑘2𝛼+𝑘+2𝑧^𝛼+𝑘. (5.5) By hypothesis, there exists 𝛼₀∈ℕ^𝑛, such that, for any 𝛼≽𝛼₀, 𝑓_𝑀(2𝛼+𝑀+2)≠0; then, 𝑓_𝑀(2𝛼₀+𝑀+2)≠0. Thus, we have 𝑧^𝛼0+𝑀∈Span𝑇_𝑓𝑧^𝛼0,𝑧^𝛼∶0⪯𝛼⪯𝛼₀+𝑀,𝛼≠𝛼₀+𝑀. (5.6)Considering the same argument, we get, for all 𝑙≽0, 𝑧^{𝛼0+𝑀+𝑙}∈Span𝑇_𝑓𝑧^𝛽,𝑧^𝛼∶𝛼₀⪯𝛽⪯𝛼₀+𝑙,0⪯𝛼⪯𝛼₀+𝑀,𝛼≠𝛼₀+𝑀. (5.7)
Now suppose that 𝑇_𝑔𝑇_𝑓 has finite rank, and let {𝜑₁,…,𝜑_𝑁} be the set that spans 𝑇_𝑔𝑇_𝑓(𝒫), where 𝒫 is the space of all holomorphic polynomials in the variable 𝑧. Then, for any 𝑙∈ℕ^𝑛, we see that 𝑇_𝑔𝑧^{𝛼0+𝑀+𝑙} is a linear combination of {𝜑₁,…,𝜑_𝑁}⋃{𝑇_𝑔(𝑧^𝛼),where0⪯𝛼⪯𝛼₀+𝑀and𝛼≠𝛼₀+𝑀}, and it follows that 𝑇_𝑔 is a finite-rank operator. By Theorem 2.4, we conclude that 𝑔 is the zero function.

Acknowledgments

The authors thank the referee for several suggestions that improved the paper. This research is supported by NSFC, Item number 10971020.