Product and Commutativity of 𝑘th-Order Slant Toeplitz Operators

Abstract

The commutativity of 𝑘th-order slant Toeplitz operators with harmonic polynomial symbols, analytic symbols, and coanalytic symbols is discussed. We show that, on the Lebesgue space and Bergman space, necessary and sufficient conditions for the commutativity of 𝑘th-order slant Toeplitz operators are that their symbol functions are linearly dependent. Also, we study the product of two 𝑘th-order slant Toeplitz operators and give some necessary and sufficient conditions.

1. Preliminaries

Throughout this paper, 𝑘 is a fixed positive integer, and 𝑘≥2. Let 𝜑(𝑧)=∑^∞_𝑖=−∞𝑎_𝑖𝑧^𝑖 be a bounded measurable function on the unit circle 𝕋, where 𝑎_𝑖=⟨𝜑,𝑧^𝑖⟩ is the 𝑖th Fourier coefficient of 𝜑 and {𝑧^𝑖∶𝑖∈ℤ} is the usual basis of 𝐿²(𝕋), with ℤ being the set of integers. The 𝑘th-order slant Toeplitz operator 𝑈_𝜑 with symbol 𝜑 in 𝐿^∞(𝕋) is defined on 𝐿²(𝕋) as follows: 𝑈_𝜑𝑧^𝑙=^∞_𝑖=−∞𝑎_{𝑘𝑖−𝑙}𝑧^𝑖. (1)

In the past several decades, slant Toeplitz operators have played outstanding roles in wavelet analysis, curve and surface modelling, and dynamical systems (e.g., see [1–10]). For instance, Villemoes [7] has associated the Besov regularity of solution of the refinement equation with the spectral radius of an associated slant Toeplitz operators and has used the spectral radius of the slant Toeplitz operators to characterize the 𝐿_𝑝(1≤𝑝≤∞) regularity of refinable functions; Goodman et al. [6] have shown the connection between the spectral radii and conditions for the solutions of certain differential equations that in the Lipschitz classes. However, these mathematicians concentrated mainly on the applications, but these considerations serve as a source of motivation to introduce and study the properties of slant Toeplitz operators.

In 1995, Ho [11–14] began a systematic study of the slant Toeplitz operators on the Hardy space. In [15–17], the authors discussed some properties of 𝑘th-order slant Toeplitz operator. In [18, 19], the authors defined the slant Toeplitz operator and 𝑘th-order slant Toeplitz operator on the Bergman spaces, respectively, and studied some properties of these operators.

In this paper, properties of 𝑘th-order slant Toeplitz operators with harmonic polynomial symbols, analytic symbols, and coanalytic symbols are discussed. We show that, on the Lebesgue space and Bergman space, the necessary and sufficient conditions for the commutativity of 𝑘th-order slant Toeplitz operators are that their symbol functions are linearly dependent. Meanwhile, we study the product of two 𝑘th-order slant Toeplitz operators and give some necessary and sufficient conditions.

2. Commutativity of 𝑘th-Order Slant Toeplitz Operators on 𝐿²(𝕋)

In [17], we investigated the properties of 𝑘th-order slant Toeplitz operators on 𝐿²(𝕋) and have obtained that for𝜑,𝜓∈𝐿^∞(𝕋), 𝑈_𝜑 and 𝑈_𝜓 commute (essentially commute) if and only if 𝜑(𝑧^𝑘)𝜓−𝜑𝜓(𝑧^𝑘)=0.

Immediately we come up with the following problem.

Could the commutativity of two 𝑘th-order slant Toeplitz operators be fully characterized by their symbols?

The partial answer to the pervious problem has been obtained in [17]: for 𝜓∈𝐿^∞(𝕋) and 𝜑(𝑧)=∑^𝑘−1_𝑝=0𝑎_𝑝𝑧^𝑝 or 𝜑(𝑧)=∑⁰_{𝑝=−𝑘+1}𝑎_𝑝𝑧^𝑝, 𝑈_𝜑 and 𝑈_𝜓 commute (essentially commute) if and only if there exist scalars 𝛼 and 𝛽, not both zero, such that 𝛼𝜑+𝛽𝜓=0.

In this section the commutativity of 𝑘th-order slant Toeplitz operators with analytic symbols and harmonic symbols will be studied. First we discuss the commutativity of two 𝑘th-order slant Toeplitz operators with analytic symbols.

Proposition 1. Let 𝜑,𝜓∈𝐻^∞(𝕋), then the following statements are equivalent: (1.1)𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘); (1.2)there exist scalars 𝛼   and 𝛽, not both zero, such that 𝛼𝜑+𝛽𝜓=0.

Proof. Begin with the easy direction. First, suppose that (1.2) holds. Without loss of generality, let 𝛼≠0, so that 𝜑=−(𝛽/𝛼)𝜓. Thus, 𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘).
To prove the other direction of the proposition, suppose that (1.1) holds; that is, 𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘). Let 𝜑(𝑧)=∑^∞_𝑝=0𝑎_𝑝𝑧^𝑝 and 𝜓(𝑧)=∑^∞_𝑝=0𝑏_𝑝𝑧^𝑝, then ^∞_𝑝=0𝑎_𝑝𝑧^𝑘𝑝⋅^∞_𝑝=0𝑏_𝑝𝑧^𝑝=^∞_𝑝=0𝑎_𝑝𝑧^𝑝⋅^∞_𝑝=0𝑏_𝑝𝑧^𝑘𝑝, (2) that is, ^∞_{𝑝=0𝑘𝑖+𝑗=𝑝}𝑎_𝑖𝑏_𝑗𝑧^𝑝=^∞_{𝑝=0𝑖+𝑘𝑗=𝑝}𝑎_𝑖𝑏_𝑗𝑧^𝑝, (3) where 𝑖 and 𝑗 are both nonnegative integers. Hence, for all nonnegative integers 𝑖, 𝑗, and 𝑝, _{𝑘𝑖+𝑗=𝑝}𝑎_𝑖𝑏_𝑗=_{𝑖+𝑘𝑗=𝑝}𝑎_𝑖𝑏_𝑗. (4) Now we would give the proof in four cases.
Case I. Suppose that 𝑎₀𝑏₀≠0. Let 𝜆=𝑏₀/𝑎₀ and we continue the proof by the induction.
When 𝑝=1, from (4) we get that 𝑎₀𝑏₁=𝑎₁𝑏₀, which means that 𝑏₁=𝜆𝑎₁.
When 𝑝=2, from (4) we get that 𝑎₀𝑏₂+𝑎₁𝑏₀=𝑎₂𝑏₀+𝑎₀𝑏₁,if𝑎𝑘=2,₀𝑏₂=𝑎₂𝑏₀,if𝑘>2, (5) which means that 𝑏₂=𝜆𝑎₂, since 𝑏₁=𝜆𝑎₁.
Now suppose that 𝑏_𝑖=𝜆𝑎_𝑖 for all integers 𝑖 with 0≤𝑖≤𝑚. Then, observe the connection between 𝑎_𝑚+1 and 𝑏_𝑚+1. Let 𝑚+1=𝑘𝑙+𝑟, where 𝑙 and 𝑟 are nonnegative integers with 0≤𝑟≤𝑘−1.
When 𝑝=𝑚+1, from (4) we get 𝑎₀𝑏_𝑚+1+𝑎₁𝑏_{𝑚+1−𝑘}+𝑎₂𝑏_{𝑚+1−2𝑘}+⋯+𝑎_𝑙𝑏_{𝑚+1−𝑙𝑘}=𝑏₀𝑎_𝑚+1+𝑏₁𝑎_{𝑚+1−𝑘}+𝑏₂𝑎_{𝑚+1−2𝑘}+⋯+𝑏_𝑙𝑎_{𝑚+1−𝑙𝑘}. (6) From the assumption we have 𝑎₀𝑏_𝑚+1=𝑏₀𝑎_𝑚+1; that is; 𝑏_𝑚+1=𝜆𝑎_𝑚+1.
Hence, from the above discussion we get that 𝑏_𝑖=𝜆𝑎_𝑖 for all integers 𝑖 with 𝑖≥0 by the induction, which means that 𝜓(𝑧)=𝜆𝜑(𝑧). So, the required result holds.
Case II. Suppose that 𝑎₀=0 and 𝑏₀≠0. We want to show that 𝑎_𝑖=0 for all integers 𝑖 with 𝑖≥0; that is, 𝜑≡0. Suppose that there exists some 𝑎_𝑖 which is not zero. Without loss of generality, let 𝑎_𝑖=0 for all integers 𝑖 with 0≤𝑖≤𝑙−1 and 𝑎_𝑙≠0, where 𝑙≥1 is an integer, then 𝜑(𝑧)=𝑧^𝑙𝜑₁(𝑧) and 𝜑₁(𝑧)=∑^∞_𝑖=0𝑎_𝑖+𝑙𝑧^𝑖=̇∑^∞_𝑖=0𝑎^′_𝑖𝑧^𝑖.
Because 𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘), we can get that 𝑧^{(𝑘−1)𝑙}𝜑₁(𝑧^𝑘)𝜓(𝑧)=𝜑₁(𝑧)𝜓(𝑧^𝑘), which means that ^∞_{𝑝=0𝑘𝑖+𝑗=𝑝}𝑎^′_𝑖𝑏_𝑗𝑧^{𝑝+(𝑘−1)𝑙}=^∞_{𝑝=0𝑖+𝑘𝑗=𝑝}𝑎^′_𝑖𝑏_𝑗𝑧^𝑝. (7) Thus, we have 𝑎^′₀𝑏₀=0, and, so, since 𝑎^′₀=𝑎_𝑙≠0, we have 𝑏₀=0. This leads to a contradiction.
Hence, 𝜑≡0 and the required result holds.
Case III. Suppose that 𝑎₀≠0 and 𝑏₀=0. Similar to Case II, we can get that 𝜓≡0. So, the required result holds.
Case IV. Suppose that 𝑎₀=0 and 𝑏₀=0. If 𝜑≡0 or 𝜓≡0, then the required result holds. Otherwise, without loss of generality, let 𝑎_𝑖=0 for all integers 𝑖 with 0≤𝑖≤𝑙−1 and 𝑎_𝑙≠0, and let 𝑏_𝑗=0 for all integers 𝑗 with 0≤𝑗≤𝑚−1 and 𝑏_𝑚≠0, where 𝑙 and 𝑚 are positive integers. Then, 𝜑(𝑧)=𝑧^𝑙𝜑₁(𝑧) and 𝜓(𝑧)=𝑧^𝑚𝜓₁(𝑧).
If 𝑙=𝑚, then 𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘) is equal to 𝜑₁(𝑧^𝑘)𝜓₁(𝑧)=𝜑₁(𝑧)𝜓₁(𝑧^𝑘). The proof is similar to Case I.
If 𝑙>𝑚, then 𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘) is equal to 𝑧^{𝑘(𝑙−𝑚)}𝜑₁(𝑧^𝑘)𝜓₁(𝑧)=𝜑₁(𝑧)𝜓₁(𝑧^𝑘). The proof is similar to Case II.
If 𝑙<𝑚, then 𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘) is equal to 𝜑₁(𝑧^𝑘)𝜓₁(𝑧)=𝑧^{𝑘(𝑚−𝑙)}𝜑₁(𝑧)𝜓₁(𝑧^𝑘). The proof is similar to Case III.

From the preceding Proposition 1, it is evident that Corollary 2 holds.

Corollary 2. Let 𝜑,𝜓∈𝐻^∞(𝕋), then the following statements are equivalent: (1.1)𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘); (1.2) there exist scalars 𝛼 and 𝛽, not both zero, such that 𝛼𝜑+𝛽𝜓=0.

From Theorem 2.8 in [17], Proposition 1, and Corollary 2, we can obtain the following Theorem 3.

Theorem 3. Let 𝜑,𝜓∈𝐻^∞(𝕋) or 𝜑, 𝜓∈𝐻^∞(𝕋), the following statements are equivalent: (1.1)𝑈_𝜑 and 𝑈_𝜓 commute; (1.2)𝑈_𝜑 and 𝑈_𝜓 essentially commute; (1.3)𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘); (1.4) there exist scalars 𝛼 and 𝛽, not both zero, such that 𝛼𝜑+𝛽𝜓=0.

Now we start to study the commutativity of two 𝑘th-order slant Toeplitz operators with harmonic symbols.

Proposition 4. Let 𝜑(𝑧)=∑^𝑛_𝑝=−𝑛𝑎_𝑝𝑧^𝑝 and 𝜓(𝑧)=∑^𝑛_𝑝=−𝑛𝑏_𝑝𝑧^𝑝, where 𝑎²_−𝑛+𝑏²_−𝑛≠0 and 𝑛 is a positive integer, then the following statements are equivalent: (1.1)𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘); (1.2)there exist scalars 𝛼   and 𝛽, not both zero, such that 𝛼𝜑+𝛽𝜓=0.

Proof. We begin with the easy direction. First, suppose that (1.2) holds and let 𝛼≠0 without lost of generality, so that 𝜑=−(𝛽/𝛼)𝜓. Thus, 𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘).
To prove the other direction of the proposition, suppose that (1.1) holds. Since 𝜑(𝑧)=∑^𝑛_𝑝=−𝑛𝑎_𝑝𝑧^𝑝 and 𝜓(𝑧)=∑^𝑛_𝑝=−𝑛𝑏_𝑝𝑧^𝑝, then 𝜑(𝑧^𝑘)=∑^𝑛_𝑝=−𝑛𝑎_p𝑧^𝑘𝑝, 𝜓(𝑧^𝑘)=∑^𝑛_𝑝=−𝑛𝑏_𝑝𝑧^𝑘𝑝 and 𝜑𝑧^𝑘𝜓(𝑧)=^𝑛_𝑝=−𝑛𝑎_𝑝𝑧^𝑘𝑝⋅^𝑛_𝑝=−𝑛𝑏_𝑝𝑧^𝑝=^{(𝑘+1)𝑛𝑛}_{𝑝=−(𝑘+1)𝑛𝑘𝑖+𝑗=𝑝}𝑎_𝑖𝑏_𝑗𝑧^𝑝,𝜑(𝑧)𝜓𝑧^𝑘=^𝑛_𝑝=−𝑛𝑎_𝑝𝑧^𝑝⋅^𝑛_𝑝=−𝑛𝑏_𝑝𝑧^𝑘𝑝=^{(𝑘+1)𝑛𝑛}_{𝑝=−(𝑘+1)𝑛𝑖+𝑘𝑗=𝑝}𝑎_𝑖𝑏_𝑗𝑧^𝑝, (8) where both 𝑖 and 𝑗 are integers with −𝑛≤𝑖,𝑗≤𝑛. Because 𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘), we can get that _{𝑘𝑖+𝑗=𝑝}𝑎_𝑖𝑏_𝑗=_{𝑖+𝑘𝑗=𝑝}𝑎_𝑖𝑏_𝑗, (9) for any integers 𝑝 with −(𝑘+1)𝑛≤𝑝≤(𝑘+1)𝑛.
Now we start to investigate the connection between 𝑎_𝑗 and 𝑏_𝑗 (𝑗=−𝑛,…,𝑛) by induction. Since 𝑎²_−𝑛+𝑏²_−𝑛≠0, without loss of generality, let 𝑎_−𝑛≠0 and let 𝜆=𝑏_−𝑛/𝑎_−𝑛, which also means that 𝑏_−𝑛=𝜆𝑎_−𝑛.
When 𝑝=−(𝑘+1)𝑛+1, then by (9), we can get that 𝑎_−𝑛𝑏_−𝑛+1=𝑎_−𝑛+1𝑏_−𝑛, which means that 𝑏_−𝑛+1=𝜆𝑎_−𝑛+1, since 𝑎_−𝑛≠0.
Suppose that 𝑏_−𝑛+𝑗=𝜆𝑎_−𝑛+𝑗 for any integers 𝑗 with 0≤𝑗≤𝑙−1, where 1≤𝑙≤2𝑛. Now we consider the connection between 𝑏_−𝑛+𝑙 and 𝑎_−𝑛+𝑙. Let 𝑙=𝑚𝑘+𝑟, where 𝑚 and 𝑟 are both nonnegative integers with 0≤𝑟≤𝑘−1.
When 𝑝=−(𝑘+1)𝑛+𝑙, by (9) we get that 𝑎_−𝑛𝑏_−𝑛+𝑙+𝑎_−𝑛+1𝑏_{−𝑛+𝑙−𝑘}+⋯+𝑎_−𝑛+𝑚𝑏_{−𝑛+𝑙−𝑚𝑘}=𝑏_−𝑛𝑎_−𝑛+𝑙+𝑏_−𝑛+1𝑎_{−𝑛+𝑙−𝑘}+⋯+𝑏_−𝑛+𝑚𝑎_{−𝑛+𝑙−𝑚𝑘}. (10) From the assumption we get that 𝑎_−𝑛𝑏_−𝑛+𝑙=𝑎_−𝑛+𝑙𝑏_−𝑛, which means that 𝑏_−𝑛+𝑙=𝜆𝑎_−𝑛+𝑙.
Hence, by the induction we obtain that 𝑏_𝑗=𝜆𝑎_𝑗(𝑗=−𝑛,…,𝑛); that is, 𝜓(𝑧)=∑^𝑛_𝑗=−𝑛𝜆𝑎_𝑗𝑧^𝑗=𝜆∑^𝑛_𝑗=−𝑛𝑎_𝑗𝑧^𝑗=𝜆𝜑(𝑧). So, the required result holds.

Lemma 5. Let 𝜑(𝑧)=∑^𝑛_𝑝=−𝑛𝑎_𝑝𝑧^𝑝, 𝜓(𝑧)=∑^𝑚_𝑝=−𝑚𝑏_𝑝𝑧^𝑝, and 𝑏_𝑚𝑏_−𝑚≠0, where 𝑛 and 𝑚 are integers and 𝑛>𝑚≥1. If 𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘), then 𝑎_𝑗=0 for any integers 𝑗 with 𝑚+1≤|𝑗|≤𝑛.

Proof. Since 𝜑(𝑧)=∑^𝑛_𝑝=−𝑛𝑎_𝑝𝑧^𝑝, 𝜓(𝑧)=∑^𝑚_𝑝=−𝑚𝑏_𝑝𝑧^𝑝, and 𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘), then ^{𝑘𝑛+𝑚}_{𝑝=−(𝑘𝑛+𝑚)𝑘𝑖+𝑗=𝑝}𝑎_𝑖𝑏_𝑗𝑧^𝑝=^{𝑛+𝑘𝑚}_{𝑝=−(𝑛+𝑘𝑚)𝑖+𝑘𝑗=𝑝}𝑎_𝑖𝑏_𝑗𝑧^𝑝. (11) Since 𝑛>𝑚≥1, we can get that for any integers 𝑝 with 𝑛+𝑘𝑚+1≤|𝑝|≤𝑘𝑛+𝑚, _{𝑘𝑖+𝑗=𝑝}𝑎_𝑖𝑏_𝑗=0, (12) for any integers 𝑝 with 0≤|𝑝|≤𝑛+𝑘𝑚, _{𝑘𝑖+𝑗=𝑝}𝑎_𝑖𝑏_𝑗=_{𝑖+𝑘𝑗=𝑝}𝑎_𝑖𝑏_𝑗. (13)
We want to show that 𝑎_𝑗=0 for any integers 𝑗 with 𝑚+1≤𝑗≤𝑛. Here are two cases: 𝑛<𝑚+𝑘 and 𝑛≥𝑚+𝑘. Let 𝑚=𝑙𝑘+𝑟, where 𝑙 and 𝑟 are nonnegative integers with 0≤𝑟≤𝑘−1.
First Case. If 𝑛<𝑚+𝑘, then 𝑘(𝑚+1)+𝑚=𝑘𝑚+𝑚+𝑘>𝑘𝑚+𝑛. Now we, continue the discussion by induction.
When 𝑝=𝑘𝑛+𝑚, by (12), we can get that 𝑎_𝑛𝑏_𝑚=0. So, 𝑎_𝑛=0, since 𝑏_−𝑚𝑏_𝑚≠0.
When 𝑝=𝑘𝑛+𝑚−𝑘, by (12) we can get that 𝑎_𝑛−1𝑏_𝑚+(𝑚−𝑘)⁺𝑎_𝑛𝑏_𝑚−𝑘=0, where (𝑚−𝑘)⁺=max{sgn(2𝑚−𝑘+1),0} and sgn is a sign function. So, 𝑎_𝑛−1=0, since 𝑏_−𝑚𝑏_𝑚≠0 and 𝑎_𝑛=0.
Suppose that 𝑎_𝑛−𝑗=0 for any integers 𝑗 with 0≤𝑗≤𝑡<𝑛−𝑚−1. Now, consider the value of 𝑎_{𝑛−𝑡−1}.
When 𝑝=𝑘𝑛+𝑚−𝑘(𝑡+1), by (12) we get that 𝑎_{𝑛−𝑡−1}𝑏_𝑚+⋯+𝑎_{𝑛−𝑡−1+𝜆}𝑏_{𝑚−𝜆𝑘}=0, (14) where 𝜆=𝑡+1,if2𝑚≥(𝑡+1)𝑘,𝜆=2𝑚𝑘,if2𝑚<(𝑡+1)𝑘, (15) and [2𝑚/𝑘] is the biggest integer which is not bigger than 2𝑚/𝑘. Then, by the assumption and (14), we get that 𝑎_{𝑛−𝑡−1}=0, since 𝑏_𝑚𝑏_−𝑚≠0.
Hence, from the above discussion we obtain that 𝑎_𝑗=0 for all integers 𝑗 with 𝑚+1≤𝑗≤𝑛 by the induction.
Second Case. If 𝑛≥𝑚+𝑘, then 𝑘(𝑚+1)+𝑚=𝑘𝑚+𝑚+𝑘≤𝑘𝑚+𝑛. Now we continue the discussion by induction.
When 𝑝=𝑘𝑛+𝑚, by (12), we can get that 𝑎_𝑛𝑏_𝑚=0. So, 𝑎_𝑛=0, since 𝑏_−𝑚𝑏_𝑚≠0.
Suppose that 𝑎_𝑛−𝑗 for any integers 𝑗 with 0≤𝑗≤𝑡<𝑛−𝑚−1. Now, consider the value of 𝑎_{𝑛−𝑡−1}.
If 𝑝=𝑘𝑛+𝑚−𝑘(𝑡+1)>𝑘𝑚+𝑛, by (12) we get that 𝑎_{𝑛−𝑡−1}𝑏_𝑚+⋯+𝑎_{𝑛−𝑡−1+𝜆}𝑏_{𝑚−𝜆𝑘}=0, (16) where 𝜆=𝑡+1,if2𝑚≥(𝑡+1)𝑘,𝜆=2𝑚𝑘,if2𝑚<(𝑡+1)𝑘, (17) and [𝑥] is the biggest integer which is not bigger than 𝑥. Then, by the assumption and (16), we get that 𝑎_{𝑛−𝑡−1}=0, since 𝑏_𝑚𝑏_−𝑚≠0.
If 𝑝=𝑘𝑛+𝑚−𝑘(𝑡+1)≤𝑘𝑚+𝑛, by (13), we get that 𝑎_{𝑛−𝑡−1}𝑏_𝑚+⋯+𝑎_{𝑛−𝑡−1+𝜆1}𝑏_{𝑚−𝜆1𝑘}=𝑏_𝑚𝑎_{𝑘𝑛−𝑘𝑚−𝑘𝑡−𝑘+𝑚}+⋯+𝑏_𝑚−𝜆2𝑎_{𝑘𝑛−𝑘𝑚−𝑘𝑡−𝑘+𝑚+𝜆2𝑘}, (18) where 𝜆₁=𝑡+1,if()𝜆2𝑚≥𝑡+1𝑘,₁=2𝑚𝑘,if()𝜆2𝑚<𝑡+1𝑘,₂=2𝑚,if𝜆𝑘𝑛+𝑘𝑚−𝑘𝑡−𝑘+𝑚≤𝑛,₂=𝑛+𝑘𝑚+𝑘𝑡+𝑘−𝑘𝑛−𝑚𝑘,if𝑘𝑛+𝑘𝑚−𝑘𝑡−𝑘+𝑚>𝑛, (19) and [𝑥] is the biggest integer which is not bigger than 𝑥. Then, by the assumption and (18), we get that 𝑎_{𝑛−𝑡−1}=0, since 𝑏_𝑚𝑏_−𝑚≠0.
Hence, from the above discussion we obtain that 𝑎_𝑗=0 for all integers 𝑗 with 𝑚+1≤𝑗≤𝑛 by the induction.
Similarly, we could get that 𝑎_𝑗=0 for all integers 𝑗 with −𝑛≤𝑗≤−(𝑚+1).

From Proposition 4 and Lemma 5, it is evident that Proposition 6 holds.

Proposition 6. Let 𝜑(𝑧)=∑^𝑛_𝑝=−𝑛𝑎_𝑝𝑧^𝑝, 𝜓(𝑧)=∑^𝑚_𝑝=−𝑚𝑏_𝑝𝑧^𝑝 and 𝑏_𝑚𝑏_−𝑚≠0, where 𝑛,𝑚 are integers and 𝑛>𝑚≥1, then the following statements are equivalent: (1.1)𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘); (1.2)there exist scalars 𝛼   and 𝛽, not both zero, such that 𝛼𝜑+𝛽𝜓=0.

Theorem 7 is obvious from Theorem 2.8 in [17] and Proposition 6.

Theorem 7. Let 𝜑(𝑧)=∑^𝑛_𝑝=−𝑛𝑎_𝑝𝑧^𝑝, 𝜓(𝑧)=∑^𝑚_𝑝=−𝑚𝑏_𝑝𝑧^𝑝, and 𝑏_𝑚𝑏_−𝑚≠0, where 𝑛 and 𝑚 are integers and 𝑛>𝑚≥1, and the following statements are equivalent:(1.1)𝑈_𝜑 and 𝑈_𝜓 commute; (1.2)𝑈_𝜑 and 𝑈_𝜓 essentially commute; (1.3)𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧)𝜓(𝑧^𝑘); (1.4) there exist scalars 𝛼 and 𝛽, not both zero, such that 𝛼𝜑+𝛽𝜓=0.

3. Product of Two 𝑘th-Order Slant Toeplitz Operators on 𝐿²(𝕋)

In [15, 17], the authors have investigated properties of the product of two 𝑘th-order slant Toeplitz operators on 𝐿²(𝕋) and have obtained the following result.

Theorem 8 (see [15, 17]). Let 𝜑,𝜓∈𝐿^∞(𝕋), then the following statements are equivalent: (1.1)𝑈_𝜑𝑈_𝜓 is a 𝑘th-order slant Toeplitz operator; (1.2)𝑈_𝜑𝑈_𝜓 is a zero operator; (1.3)𝑈_𝜑𝑈_𝜓 is compact; (1.4)𝜑(𝑧^𝑘)𝜓(𝑧)=0.

In this section, we will describe properties of the product of two 𝑘th-order slant Toeplitz operators with analytic symbols and harmonic symbols on 𝐿²(𝕋) by their symbols. First, we start to discuss properties of two 𝑘th-order slant Toeplitz product with analytic symbols.

Proposition 9. Let 𝜑,𝜓∈𝐻^∞(𝕋), then 𝜑(𝑧^𝑘)𝜓(𝑧)=0 if and only if 𝜑=0 or 𝜓=0.

Proof. As we know, the “if” direction of the proposition is trivial.
Now suppose that 𝜑(𝑧^𝑘)𝜓(𝑧)=0. Since 𝜑,𝜓∈𝐻^∞(𝕋), we have 𝜑(𝑧^𝑘)𝜓(𝑧)=0, where 𝜑 and 𝜓 are the Poisson extensions of 𝜑 and 𝜓, respectively, and they are analytic on the unit disk 𝔻. Hence, we get that 𝜑(𝑧^𝑘) is identically 0 or 𝜓 is identically 0; that is, 𝜑 is identically 0 or 𝜓 is identically 0.

Similarly, we could obtain Corollary 10.

Corollary 10. Let 𝜑,𝜓∈𝐻^∞(𝕋), then 𝜑(𝑧^𝑘)𝜓(𝑧)=0 if and only if 𝜑=0 or 𝜓=0.

It is obvious that Theorem 11 holds from the preceding analysis.

Theorem 11. Let 𝜑,𝜓∈𝐻^∞(𝕋) or 𝜑,𝜓∈𝐻^∞(𝕋), then the following statements are equivalent: (1.1)𝑈_𝜑𝑈_𝜓 is a 𝑘th-order slant Toeplitz operator; (1.2)𝑈_𝜑𝑈_𝜓 is a zero operator; (1.3)𝑈_𝜑𝑈_𝜓 is compact; (1.4)𝜑(𝑧^𝑘)𝜓(𝑧)=0; (1.5)𝜑=0 or 𝜓=0.

Now, we start to discuss the properties of two 𝑘th-order slant Toeplitz product with harmonic symbols.

Proposition 12. Let 𝜑(𝑧)=∑^𝑛_𝑝=0𝑎_𝑝𝑧^𝑝 and 𝜓(𝑧)=∑^𝑚_𝑝=0𝑏_−𝑝𝑧^−𝑝, where 𝑛 and 𝑚 are both positive integers, then 𝜑(𝑧^𝑘)𝜓(𝑧)=0 if and only if 𝜑=0 or 𝜓=0.

Proof. We begin with the easy direction. First, suppose that 𝜑=0 or 𝜓=0; then it is clear that 𝜑(𝑧^𝑘)𝜓(𝑧)=0.
Without loss of generality, let 𝑎²_𝑛+𝑏²_−𝑚≠0. Otherwise, 𝑎²_𝑛+𝑏²_−𝑚=0; then we can consider the value of 𝑎_𝑖 and 𝑏_𝑗, where 𝑖 and 𝑗 are both integers with 0≤𝑖≤𝑛−1 and 0≤𝑗≤𝑚−1, since 𝜑(𝑧) and 𝜓 are both polynomial functions. There are four cases: (1)𝑎_𝑖=0 for all integers 𝑖 with 0≤𝑖≤𝑛; (2)𝑏_−𝑗=0 for all integers 𝑗 with 0≤𝑗≤𝑚; (3)𝑎_𝑠≠0 and 𝑎_𝑖=0 for all integers 𝑖 with 𝑠+1≤𝑖≤𝑛; (4)𝑏_−𝑡≠0 and 𝑏_−𝑗=0 for all integers 𝑗 with 𝑡+1≤𝑗≤𝑚, where 𝑠 and 𝑡 are both nonnegative integers with 𝑠≤𝑛−1 and 𝑡≤𝑚−1. If the first two cases hold, then the required result holds; if the latter two cases holds, then we have 𝜑(𝑧)=∑^𝑠_𝑝=0𝑎_𝑝𝑧^𝑝 and 𝑎²_𝑠+𝑏²_−𝑚≠0 or 𝜓(𝑧)=∑^𝑡_𝑝=0𝑏_−𝑝𝑧^−𝑝 and 𝑎²_𝑛+𝑏²_−𝑡≠0.
Now suppose that 𝜑(𝑧^𝑘)𝜓(𝑧)=0 and 𝑎²_𝑛+𝑏²_−𝑚≠0. Since 𝜑(𝑧)=∑^𝑛_𝑝=0𝑎_𝑝𝑧^𝑝, 𝜓(𝑧)=∑^𝑚_𝑝=0𝑏_−𝑝𝑧^−𝑝, then 𝜑𝑧^𝑘𝜓(𝑧)=^𝑛_𝑝=0𝑎_𝑝𝑧^𝑘𝑝⋅^𝑚_𝑝=0𝑏_−𝑝𝑧^−𝑝=^𝑘𝑛_{𝑝=−𝑚𝑘𝑖−𝑗=𝑝}𝑎_𝑖𝑏_−𝑗𝑧^𝑝, (20) where 𝑖 and 𝑗 are both integers with 0≤𝑖≤𝑛 and 0≤𝑗≤𝑚. Because 𝜑(𝑧^𝑘)𝜓(𝑧)=0, we get, for any integers 𝑝 with −𝑚≤𝑝≤𝑘𝑛, _{𝑘𝑖−𝑗=𝑝}𝑎_𝑖𝑏_−𝑗=0, (21) where 𝑖 and 𝑗 are both integers with 0≤𝑖≤𝑛 and 0≤𝑗≤𝑚. Since 𝑎²_𝑛+𝑏²_−𝑚≠0, yet we only obtain either 𝑎_𝑛≠0 or 𝑏_−𝑚≠0.
First Case. If 𝑎_𝑛≠0. Now we want to show that 𝜓≡0 by the induction.
When 𝑝=𝑘𝑛, by (21), we get that 𝑎_𝑛𝑏₀=0, which means that 𝑏₀=0, since 𝑎_𝑛≠0.
When 𝑝=𝑘𝑛−1, by (21), we get that 𝑎_𝑛𝑏₋₁=0, which means that 𝑏₋₁=0, since 𝑎_𝑛≠0.
Now suppose that 𝑏_−𝑙=0 for any integers 𝑙 with 0≤𝑙≤𝑡, where 𝑡 is an integer with 0≤𝑡≤𝑚−1. Considering the value of 𝑏_−𝑡−1, when 𝑝=𝑘𝑛−𝑡−1, by (21), we get that 𝑎_𝑛𝑏_−𝑡−1+𝑎_𝑛−1𝑏_{−𝑡−1+𝑘}+𝑎_𝑛−2𝑏_{−𝑡−1+2𝑘}+⋯+𝑎_𝑛−𝜆𝑏_{−𝑡−1+𝑘𝜆}=0, (22) where 𝜆=min{𝑛,[(𝑡+1)/𝑘]} and [𝑥] is the biggest integer which is not bigger than 𝑥. By the assumption and the above equation, we get that 𝑎_𝑛𝑏_−𝑡−1=0; that is, 𝑏_−𝑡−1=0, since 𝑎_𝑛≠0.
From the preceding analysis, by the induction, we can obtain that 𝑏_−𝑙=0 for any integers 𝑙 with 0≤𝑙≤𝑚. Hence, 𝜓≡0.
Second Case. If 𝑏_−𝑚≠0. Arguing as in the First case, we obtain that 𝜑≡0.

Proposition 13. Let 𝜑(𝑧)=∑^𝑛_𝑝=0𝑎_𝑝𝑧^𝑝 and 𝜓(𝑧)=∑^𝑚_𝑝=−𝑚𝑏_𝑝𝑧^𝑝, where 𝑛 and 𝑚 are positive integers and 𝑎²_𝑛+𝑏²_𝑚+𝑏²_−𝑚≠0, then the following statements are equivalent: (1.1)𝜑(𝑧^𝑘)𝜓(𝑧)=0; (1.2)𝜑(𝑧)𝜓(𝑧^𝑘)=0; (1.3)𝜑=0   or 𝜓=0.

Proof. We begin with the easy direction. First, suppose that 𝜑=0 or 𝜓=0; then it is clear that (1.1) and (1.2) hold.
Now suppose that (1.1) holds. Since 𝜑(𝑧)=∑^𝑛_𝑝=0𝑎_𝑝𝑧^𝑝, 𝜓(𝑧)=∑^𝑚_𝑝=−𝑚𝑏_𝑝𝑧^𝑝, then 𝜑𝑧^𝑘𝜓(𝑧)=^𝑛_𝑝=0𝑎_𝑝𝑧^𝑘𝑝⋅^𝑚_𝑝=−𝑚𝑏_𝑝𝑧^𝑝=^{𝑘𝑛+𝑚}_{𝑝=−𝑚𝑘i+𝑗=𝑝}𝑎_𝑖𝑏_𝑗𝑧^𝑝, (23) where 𝑖 and 𝑗 are both integers with 0≤𝑖≤𝑛 and −𝑚≤𝑗≤𝑚. Because 𝜑(𝑧^𝑘)𝜓(𝑧)=0, we get that for any integers 𝑝 with −𝑚≤𝑝≤𝑘𝑛+𝑚, _{𝑘𝑖+𝑗=𝑝}𝑎_𝑖𝑏_𝑗=0, (24) where 𝑖 and 𝑗 are both integers with 0≤𝑖≤𝑛 and −𝑚≤𝑗≤𝑚. Since 𝑎²_𝑛+𝑏²_𝑚+𝑏²_−𝑚≠0, yet we can obtain that 𝑎_𝑛≠0, 𝑏_−𝑚≠0, or 𝑏_𝑚≠0.
First Case. If 𝑎_𝑛≠0. Now we start to continue the proof by induction.
When 𝑝=𝑘𝑛+𝑚, by (24), we get that 𝑎_𝑛𝑏_𝑚=0, which means that 𝑏_𝑚=0, since 𝑎_𝑛≠0.
When 𝑝=𝑘𝑛+𝑚−1, by (24), we get that 𝑎_𝑛𝑏_𝑚−1=0, which means that 𝑏_𝑚−1=0, since 𝑎_𝑛≠0.
Now suppose that 𝑏_𝑙=0 for any integers 𝑙 with 𝑡≤𝑙≤𝑚, where 𝑡 is an integer with 0<𝑡≤𝑚. Considering the value of 𝑏_𝑡−1, when 𝑝=𝑘𝑛+𝑡−1, by (24), we get that 𝑎_𝑛𝑏_𝑡−1+⋯+𝑎_𝑛−𝜆𝑏_{𝑡−1+𝑘𝜆}=0, (25) where 𝜆=min{𝑛,[(𝑚−𝑡+1)/2]} and [𝑥] is the biggest integer which is not bigger than 𝑥. By the assumption and the above equation, we get that 𝑎_𝑛𝑏_𝑡−1=0; that is, 𝑏_𝑡−1=0, since 𝑎_𝑛≠0.
From the preceding analysis, by the induction we can obtain that 𝑏_𝑙=0 for any integers 𝑙 with 0≤𝑙≤𝑚. Then, by Proposition 12 we get that 𝜓≡0, since 𝜑 is not 0 identically.
Second Case. If 𝑏_𝑚≠0. In the following, we will continue the proof by induction.
When 𝑝=𝑘𝑛+𝑚, by (24), we get that 𝑎_𝑛𝑏_𝑚=0, which means that 𝑎_𝑛=0, since 𝑏_𝑚≠0.
When 𝑝=𝑘𝑛+𝑚−𝑘, by (24), we get that 𝑎_𝑛−1𝑏_𝑚+(𝑚−𝑘)⁺𝑎_𝑛𝑏_𝑚−𝑘=0, where (𝑚−𝑘)⁺=max{0,sgn(2𝑚−𝑘+1)} and sgn is a sign function. So, 𝑎_𝑛−1=0, since 𝑎_𝑛=0 and 𝑏_𝑚≠0.
Now, suppose that 𝑎_𝑙=0 for any integers 𝑙 with 𝑡≤𝑙≤𝑛, where 𝑡 is an integer with 0<𝑡≤𝑛. Considering the value of 𝑎_𝑡−1, when 𝑝=𝑚+𝑘𝑡−𝑘, by (24), we get that 𝑏_𝑚𝑎_𝑡−1+⋯+𝑏_{𝑚−𝑘𝜆}𝑎_{𝑡−1+𝜆}=0, (26) where 𝜆=min{𝑛−𝑡+1,[2𝑚/𝑘]} and [𝑥] is the biggest integer which is not bigger than 𝑥. By the assumption and the above equation, we get that 𝑏_𝑚𝑎_𝑡−1=0; that is, 𝑎_𝑡−1=0, since 𝑏_𝑚≠0.
From the preceding analysis, by the induction we can obtain that 𝑎_𝑙=0 for any integers 𝑙 with 0≤𝑙≤𝑛, which means that 𝜑≡0.
Third Case. If 𝑏_−𝑚≠0. A computation analogous to the one done in the second case from which we can get that 𝜑≡0.
From the above analysis, we have that (1.3) holds.
Arguing as in the pervious discussion, we obtain that if (1.2) holds, then (1.3) is true.

Proposition 14. Let 𝜑(𝑧)=∑^𝑛_𝑝=−𝑛𝑎_𝑝𝑧^𝑝, 𝜓(𝑧)=∑^𝑚_𝑝=−𝑚𝑏_𝑝𝑧^𝑝, where 𝑛 and 𝑚 are both positive integers and 𝑎²_𝑛+𝑏²_𝑚+𝑎²_−𝑛+𝑏²_−𝑚≠0, then 𝜑(𝑧^𝑘)𝜓(𝑧)=0 if and only if 𝜑=0 or 𝜓=0.

Proof. We begin with the easy direction. First, suppose that 𝜑=0 or 𝜓=0; then it is clear that 𝜑(𝑧^𝑘)𝜓(𝑧)=0.
Now suppose that 𝜑(𝑧^𝑘)𝜓(𝑧)=0. Since 𝜑(𝑧)=∑^𝑛_𝑝=−𝑛𝑎_𝑝𝑧^𝑝, 𝜓(𝑧)=∑^𝑚_𝑝=−𝑚𝑏_𝑝𝑧^𝑝, then 𝜑𝑧^𝑘𝜓(𝑧)=^𝑛_𝑝=−𝑛𝑎_𝑝𝑧^𝑘𝑝⋅^𝑛_𝑝=−𝑛𝑏_𝑝𝑧^𝑝=^{𝑘𝑛+𝑚}_{𝑝=−(𝑘𝑛+𝑚)𝑘𝑖+𝑗=𝑝}𝑎_𝑖𝑏_𝑗𝑧^𝑝, (27) where 𝑖 and 𝑗 are both integers with −𝑛≤𝑖≤𝑛 and −𝑚≤𝑗≤𝑚. Because 𝜑(𝑧^𝑘)𝜓(𝑧)=0, we get that, for any integers 𝑝 with −(𝑘𝑛+𝑚)≤𝑝≤𝑘𝑛+𝑚, _{𝑘𝑖+𝑗=𝑝}𝑎_𝑖𝑏_𝑗=0, (28) where 𝑖 and 𝑗 are both integers with −𝑛≤𝑖≤𝑛 and −𝑚≤𝑗≤𝑚.
Since 𝑎²_𝑛+𝑏²_𝑚+𝑎²_−𝑛+𝑏²_−𝑚≠0, without loss of generality, suppose that 𝑎_𝑛≠0. We want to continue the proof by induction.
When 𝑝=𝑘𝑛+𝑚, by (28), we get that 𝑎_𝑛𝑏_𝑚=0, which means that 𝑏_𝑚=0, since 𝑎_𝑛≠0, when 𝑝=𝑘𝑛+𝑚−1, by (28), we get that 𝑎_𝑛𝑏_𝑚−1=0, which means that 𝑏_𝑚−1=0, since 𝑎_𝑛≠0.
Now suppose that 𝑏_𝑙=0 for any integers 𝑙 with 𝑡≤𝑙≤𝑚, where 𝑡 is an integer with 0<𝑡≤𝑚. Considering the value of 𝑏_𝑡−1. When 𝑝=𝑘𝑛+𝑡−1, by (28) we get that 𝑎_𝑛𝑏_𝑡−1+⋯+𝑎_𝑛−𝜆𝑏_{𝑡−1+𝑘𝜆}=0, (29) where 𝜆=min{2𝑛,[(𝑚−𝑡+1)/𝑘]} and [𝑥] is the biggest integer which is not bigger than 𝑥. By the assumption and the above equation, we get that 𝑎_𝑛𝑏_𝑡−1=0; that is, 𝑏_𝑡−1=0, since 𝑎_𝑛≠0.
From the preceding analysis, by the induction we can obtain that 𝑏_𝑙=0 for any integers 𝑙 with 0≤𝑙≤𝑚. Then, by Proposition 13 we get that 𝜓≡0, since 𝜑 is not 0 identically and 𝜑(𝑧^𝑘)𝜓(𝑧)=0 is equivalent to 𝜑(𝑧^𝑘)𝜓(𝑧)=𝜑(𝑧^𝑘)⋅𝜓(𝑧)=0.

Form Proposition 3.4 and Theorem 3.1 in [15, 17], we will obtain the following theorem which describes the product of two slant Toeplitz operators with harmonic symbols.

Theorem 15. Let 𝜑(𝑧)=∑^𝑛_𝑝=−𝑛𝑎_𝑝𝑧^𝑝, 𝜓(𝑧)=∑^𝑚_𝑝=−𝑚𝑏_𝑝𝑧^𝑝, where 𝑛 and 𝑚 are both positive integers and 𝑎²_𝑛+𝑏²_𝑚+𝑎²_−𝑛+𝑏²_−𝑚≠0, then the following statements are equivalent:(1.1)𝑈_𝜑𝑈_𝜓 is a 𝑘th-order slant Toeplitz operator; (1.2)𝑈_𝜑𝑈_𝜓 is a zero operator; (1.3)𝑈_𝜑𝑈_𝜓 is compact; (1.4)𝜑(𝑧^𝑘)𝜓(𝑧)=0; (1.5)𝜑=0 or 𝜓=0.

4. Commutativity of 𝑘th-Order Slant Toeplitz Operators on Bergman Spaces

Let ℂ be the complex plane and let 𝔻 be the unit disk in ℂ. Let 𝑑𝐴 be the area measure on 𝔻 normalized so that ∫_𝔻1𝑑𝐴=1. Let 𝐴²(𝔻) be the space of analytic functions in 𝐿²(𝔻,𝑑𝐴) which consists of Lebesgue measurable functions 𝑓 on 𝔻 with _𝔻𝑓(𝑧)²𝑑𝐴(𝑧)<+∞. (30) It is well known that 𝐴²(𝔻) is a closed subspace of the Hilbert space 𝐿²(𝔻,𝑑𝐴) with the inner product ⟨⋅,⋅⟩ and {𝑧^𝑖∶𝑖∈ℤ⁺} are the orthogonal basis in 𝐴²(𝔻), where ℤ⁺ is the set of nonnegative integers. Let 𝐿^∞(𝔻) be the Banach space of Lebesgue measurable functions 𝑓 on 𝔻 with 𝑓_∞=esssup𝑓(𝑧)∶𝑧∈𝔻<+∞. (31)

Let 𝑃 denote the orthogonal projection from 𝐿²(𝔻,𝑑𝐴) onto 𝐴²(𝔻). For 𝑓∈𝐿^∞(𝔻), the Toeplitz operator 𝑇_𝑓 on 𝐴²(𝔻) is defined by 𝑇_𝑓𝑔=𝑃𝑓𝑔,𝑔∈𝐴²(𝔻), (32) and the 𝑘th-order slant Toeplitz operator 𝐵_𝑓 on 𝐴²(𝔻) is defined by 𝐵_𝑓=𝑊_𝑘𝑇_𝑓, (33) where 𝑊_𝑘 is a bounded linear operator on 𝐴²(𝔻) which is defined as 𝑊_𝑘𝑧^𝑖=𝑧^𝑖/𝑘,if𝑖isdivisibleby𝑘,0,otherwise. (34)

In this section we will investigate the commutativity of 𝑘th-order slant Toeplitz operators with coanalytic symbols and harmonic symbols on Bergman space 𝐴²(𝔻). First, we study the commutativity of 𝑘th-order slant Toeplitz operators with coanalytic symbols.

Lemma 16. Let 𝑞≥0 be an integer, let 𝑓,𝑔∈𝐻^∞(𝔻), both of which are not 0 identically. If 𝑓𝑊^∗_𝑘𝑔=𝑔𝑊^∗_𝑘𝑓, the following statements are equivalent: (1.1)𝑓^(𝑖)(0)=0 for any integers 𝑖 with 0≤𝑖≤𝑞 and 𝑓^(𝑞+1)(0)≠0; (1.2)𝑔^(𝑖)(0)=0 for any integers 𝑖 with 0≤𝑖≤𝑞 and 𝑔^(𝑞+1)(0)≠0.

Proof. First, suppose that (1.1) holds. Since 𝑓∈𝐻^∞(𝔻), we get that 𝑓(𝑧)=𝑧^𝑞+1𝑓₁(𝑧) and (𝑊^∗_𝑘𝑓)(𝑧)=𝑧^{𝑘𝑞+𝑘}𝑓₂(𝑧), where 𝑓₁(0)≠0, 𝑓₂(0)≠0, and 𝑓₁,𝑓₂∈𝐻^∞(𝔻). Because 𝑓𝑊^∗_𝑘𝑔=𝑔𝑊^∗_𝑘𝑓 and 𝑔∈𝐻^∞(𝔻), we get that 𝑓₁(𝑧)𝑊^∗_𝑘𝑔(𝑧)=𝑧^{(𝑘−1)(𝑞+1)}𝑓₂(𝑧)𝑔(𝑧), (35) so 𝑔(0)=0. Since 𝑔 is not 0 identically, without loss of generality, take 𝑔^(𝑗)(0)=0 for any integers 𝑗 with 0≤𝑗≤𝑝 and 𝑔^(𝑝+1)(0)≠0, then 𝑔(𝑧)=𝑧^𝑝+1𝑔₁(𝑧) and (𝑊^∗_𝑘𝑔)(𝑧)=𝑧^𝑘(𝑝+1)𝑔₂(𝑧), where 𝑔₁(0)≠0, 𝑔₂(0)≠0, and 𝑔₁,𝑔₂∈𝐻^∞(𝔻); so, by (35) we get that 𝑓₁(𝑧)𝑔₂(𝑧)𝑧^{(𝑘−1)(𝑝+1)}=𝑧^{(𝑘−1)(𝑞+1)}𝑓₂(𝑧)𝑔₁(𝑧), (36) so 𝑝=𝑞. Otherwise, if 𝑝>𝑞, then by the above equation we get that 𝑓₁(𝑧)𝑔₂(𝑧)𝑧^{(𝑘−1)(𝑝−𝑞)}=𝑓₂(𝑧)𝑔₁(𝑧), which means that 𝑓₂(0)𝑔₁(0)=0, that leads a contradiction; if 𝑝<𝑞, then by the above equation we get that 𝑓₁(𝑧)𝑔₂(𝑧)=𝑧^{(𝑘−1)(𝑞−𝑝)}𝑓₂(𝑧)𝑔₁(𝑧), which means that 𝑓₁(0)𝑔₂(0)=0, that leads a contradiction. Hence (1.2) holds.
Similarly, we can obtain the other direction of the Lemma.

Theorem 17. Let 𝜑,𝜓∈𝐻^∞(𝔻), then the following statements are equivalent:(1.1)𝐵_𝜑 and 𝐵_𝜓 commute; (1.2)there exist scalars 𝛼 and 𝛽, not both zero, such that 𝛼𝜑+𝛽𝜓=0.

Proof. First suppose that (1.2) holds; then it is obvious that 𝐵_𝜑 and 𝐵_𝜓 commute.
Now suppose that (1.1) holds. So, we get that 𝐵^∗_𝜑𝐵^∗_𝜓(1)=𝐵^∗_𝜓𝐵^∗_𝜑(1); that is, 𝜑𝑊^∗_𝑘𝜓=𝜓𝑊^∗_𝑘𝜑.
Now we continue the discussion in three cases.
First Case. If 𝜑≡0 or 𝜓≡0. It is obvious that the required result holds.
Second Case. If 𝜑(0)≠0 and 𝜓(0)≠0. Since 𝜑,𝜓∈𝐻^∞(𝔻), let 𝜑(𝑧)=∑^∞_𝑝=0𝑎_𝑝𝑧^𝑝 and 𝜓(𝑧)=∑^∞_𝑝=0𝑏_𝑝𝑧^𝑝, then 𝑎₀≠0, 𝑏₀≠0, and 𝑊^∗_𝑘𝜑(𝑧)=∑^∞_𝑝=0𝑎_𝑝((𝑘𝑝+1)/(𝑝+1))𝑧^𝑘𝑝 and 𝑊^∗_𝑘𝜓(𝑧)=∑^∞_𝑝=0𝑏_𝑝((𝑘𝑝+1)/(𝑝+1))𝑧^𝑘𝑝, so 𝜓𝑊^∗_𝑘𝜑(𝑧)=^∞_𝑝=0𝑏_𝑝𝑧^𝑝⋅^∞_𝑝=0𝑎_𝑝𝑘𝑝+1𝑧𝑝+1^𝑘𝑝=^∞_{𝑝=0𝑘𝑖+𝑗=𝑝}𝑘𝑖+1𝑎𝑖+1^𝑖𝑏_𝑗𝑧^𝑝,𝜑𝑊^∗_𝑘𝜓(𝑧)=^∞_𝑝=0𝑎_𝑝𝑧^𝑝⋅^∞_𝑝=0𝑏_𝑝𝑘𝑝+1𝑧𝑝+1^𝑘𝑝=^∞_{𝑝=0𝑖+𝑘𝑗=𝑝}𝑘𝑗+1𝑎𝑗+1^𝑖𝑏_𝑗𝑧^𝑝. (37) Because 𝜑𝑊^∗_𝑘𝜓=𝜓𝑊^∗_𝑘𝜑, we get that ^∞_{𝑝=0𝑘𝑖+𝑗=𝑝}𝑘𝑖+1𝑎𝑖+1^𝑖𝑏_𝑗𝑧^𝑝=^∞_{𝑝=0𝑖+𝑘𝑗=𝑝}𝑘𝑗+1𝑎𝑗+1^𝑖𝑏_𝑗𝑧^𝑝, (38) that is, for any integers 𝑝≥0, _{𝑘𝑖+𝑗=𝑝}𝑘𝑖+1𝑎𝑖+1^𝑖𝑏_𝑗=_{𝑖+𝑘𝑗=𝑝}𝑘𝑗+1𝑎𝑗+1^𝑖𝑏_𝑗, (39) where 𝑖 and 𝑗 are both nonnegative integers. In the following, we want to continue the proof by the induction.
When 𝑝=0, by (39), we get that 𝑎₀𝑏₀=𝑎₀𝑏₀, so 𝑏₀=(𝑏₀/𝑎₀)𝑎₀, since 𝑎₀≠0. Let 𝜆=𝑏₀/𝑎₀, then 𝑏₀=𝜆𝑎₀.
When 𝑝=1, by (39), we get that 𝑎₀𝑏₁=𝑎₁𝑏₀, so 𝑏₁=𝜆𝑎₁.
Suppose that 𝑏_𝑗=𝜆𝑎_𝑗 for any integers 𝑗 with 0≤𝑗≤𝑙, where 𝑙 is a nonnegative integer. Now, consider the connection between 𝑎_𝑙+1 and 𝑏_𝑙+1.
When 𝑝=𝑙+1, by (39), we get that _{𝑘𝑖+𝑗=𝑙+1}𝑘𝑖+1𝑎𝑖+1^𝑖𝑏_𝑗=_{𝑖+𝑘𝑗=𝑙+1}𝑘𝑗+1𝑎𝑗+1^𝑖𝑏_𝑗, (40) that is, 𝑎₀𝑏_𝑙+1+⋯+𝑘𝜆+1𝑎𝜆+1^𝜆𝑏_{𝑙+1−𝜆𝑘}=𝑏₀𝑎_𝑙+1+⋯+𝑘𝜆+1𝑏𝜆+1^𝜆𝑎_{𝑙+1−𝜆𝑘}, (41) where 𝜆=[(𝑙+1)/𝑘] and [𝑥] is the biggest integer which is not bigger than 𝑥. By this assumption, we can obtain that 𝑎₀𝑏_𝑙+1=𝑏₀𝑎_𝑙+1; that is, 𝑏_𝑙+1=𝜆𝑎_𝑙+1, since 𝑎₀≠0.
Hence, by the induction, we obtain that 𝑏_𝑗=𝜆𝑎_𝑗 for any nonnegative integers 𝑗 from the pervoius discussion; that is, 𝜓(𝑧)=∑^∞_𝑝=0𝜆𝑎_𝑝𝑧^𝑝=𝜆𝜑(𝑧). So, the required result holds.
Third Case. If 𝜑 and 𝜓 are both not 0 identically, and 𝜑(0)=0 or 𝜓(0)=0. Without loss of generality, take 𝜑^(𝑖)(0)=0 for any integers 0≤𝑖≤𝑖₁ and 𝜑^(𝑖1+1)(0)≠0. By Lemma 16, we get that 𝜓^(𝑖)(0)=0 for any integers 0≤𝑖≤𝑖₁ and 𝜓^(𝑖1+1)(0)≠0, and 𝜑(𝑧)=𝑧^𝑖1+1𝜑₁(𝑧),𝜓(𝑧)=𝑧^𝑖1+1𝜓₁(𝑧), (42) where 𝜑₁,𝜓₁∈𝐻^∞(𝔻), and 𝜑₁(0)≠0, 𝜓₁(0)≠0, so 𝜓𝑊^∗_𝑘(𝜑)=𝑧^{(𝑘+1)(𝑖1+1)}𝜓₁𝑊^∗_𝑘(𝜑₁) and 𝜑𝑊^∗_𝑘(𝜓)=𝑧^{(𝑘+1)(𝑖1+1)}𝜑₁𝑊^∗_𝑘(𝜓₁). Since 𝜓𝑊^∗_𝑘(𝜑)=𝜑𝑊^∗_𝑘(𝜓), yet we can get that 𝜓₁𝑊^∗_𝑘(𝜑₁)=𝜑₁𝑊^∗_𝑘(𝜓₁). Since 𝜑₁(0)≠0 and 𝜓₁(0)≠0, yet by the second case we get that 𝜓₁=𝜆₁𝜑₁, where 𝜆₁=𝜓₁(0)/𝜑₁(0). So, 𝜓(𝑧)=𝑧^𝑖1+1𝜓₁(𝑧)=𝜆₁𝑧^𝑖1+1𝜑₁(𝑧)=𝜆₁𝜑₁(𝑧). The required result holds.

Now we are in a position to discuss the commutativity of slant Toeplitz operators with harmonic symbols.

Theorem 18. Let 𝜑(𝑧)=∑^𝑛_𝑝=0𝑎_𝑝𝑧^𝑝+∑^𝑛_𝑘=1𝑎_−𝑝𝑧^𝑝 and 𝜓(𝑧)=∑^𝑛_𝑝=0𝑏_𝑝𝑧^𝑝+∑^𝑛_𝑘=1𝑏_−𝑝𝑧^𝑝, where 𝑎²_−𝑛+𝑏²_−𝑛≠0 and 𝑛≥1 is an integer, then the following statements are equivalent: (1.1)𝐵_𝜑 and 𝐵_𝜓 commute; (1.2)there exist scalars 𝛼 and 𝛽, not both zero, such that 𝛼𝜑+𝛽𝜓=0.

Proof. First suppose that (1.2) holds. It is obvious that 𝐵_𝜑 and 𝐵_𝜓 commute.
Now suppose that (1.1) holds. Let 𝜑₁(𝑧)=∑^𝑛_𝑝=0𝑎_𝑝𝑧^𝑝, 𝜑₂(𝑧)=∑^𝑛_𝑝=1𝑎_−𝑝𝑧^𝑝, 𝜓₁(𝑧)=∑^𝑛_𝑝=0𝑏_𝑝𝑧^𝑝, and 𝜓₂(𝑧)=∑^𝑛_𝑝=1𝑏_−𝑝𝑧^𝑝, then 𝜑=𝜑₁+𝜑₂ and 𝜓=𝜓₁+𝜓₂. Since 𝐵_𝜑 and 𝐵_𝜓 commute, we have 𝑇_𝜑𝑊^∗_𝑘𝑇_𝜓𝑊^∗_𝑘1=𝑇_𝜓𝑊^∗_𝑘𝑇_𝜑𝑊^∗_𝑘1, that is, 𝜑₂𝑊^∗_𝑘𝜓₂+𝜓₁(0)𝜑₂+𝑃𝜑₁𝑊^∗_𝑘𝜓₂=𝜓₂𝑊^∗_𝑘𝜑₂+𝜑₁(0)𝜓₂+𝑃𝜓₁𝑊^∗_𝑘𝜑₂. (43) Then, by (43), we get that for any integers 𝑝 with 𝑘𝑛+1≤𝑝≤𝑘𝑛+𝑛, _{𝑖+𝑘𝑗=𝑝}𝑎_−𝑖𝑏_−𝑗𝑘𝑗+1𝑗+1=_{𝑘𝑖+𝑗=p}𝑎_−𝑖𝑏_−𝑗𝑘𝑖+1,𝑖+1 (44) where 𝑖 and 𝑗 are positive integers which are not bigger than 𝑛. Since 𝑎²_−𝑛+𝑏²_−𝑛≠0, without loss of generality, take 𝑏_−𝑛≠0. Now we continue the proof by the induction.
When 𝑝=𝑘𝑛+𝑛, by (44), we get that 𝑎_−𝑛𝑏_−𝑛((𝑘𝑛+1)/(𝑛+1))=𝑎_−𝑛𝑏_−𝑛((𝑘𝑛+1)/(𝑛+1)), so 𝑎_−𝑛=(𝑎_−𝑛/𝑏_−𝑛)𝑏_−𝑛, since 𝑏_−𝑛≠0. Let 𝜆=𝑎_−𝑛/𝑏_−𝑛, then 𝑎_−𝑛=𝜆𝑏_−𝑛.
When 𝑝=𝑘𝑛+𝑛−1, by (44), we get that 𝑎_−𝑛+1𝑏_−𝑛((𝑘𝑛+1)/(𝑛+1))=𝑎_−𝑛𝑏_−𝑛+1((𝑘𝑛+1)/(𝑛+1)), so 𝑎_−𝑛+1=𝜆𝑏_−𝑛+1, since 𝑏_−𝑛≠0.
Suppose that 𝑎_−𝑛+𝑖=𝜆𝑏_−𝑛+𝑖 for any integers 𝑖 with 0≤𝑖≤𝑙<𝑛−1. Now, consider the connection between 𝑎_{−𝑛+𝑙+1} and 𝑏_{−𝑛+𝑙+1}.
When 𝑝=𝑘𝑛+𝑛−𝑙−1, by (44), we get that 𝑎_{−𝑛+𝑙+1}𝑏_−𝑛2𝑛+1𝑛+1+⋯+𝑎_{−𝑛+𝑙+1−𝑘𝛾}𝑏_−𝑛+𝛾𝑘𝛾+1=𝑘+1𝑏_{−𝑛+𝑙+1}𝑎_−𝑛2𝑛+1𝑛+1+⋯+𝑏_{−𝑛+𝑙+1−𝑘𝛾}𝑎_−𝑛+𝛾𝑘𝛾+1,𝑘+1 (45) where 𝛾=[(𝑙+1)/𝑘] and [𝑥] is the biggest integer which is not bigger than 𝑥. From the assumption we obtain that 𝑎_{−𝑛+𝑙+1}𝑏_−𝑛((2𝑛+1)/(𝑛+1))=𝑏_{−𝑛+𝑙+1}𝑎_−𝑛((2𝑛+1)/(𝑛+1)), so 𝑎_{−𝑛+𝑙+1}=𝜆𝑏_{−𝑛+𝑙+1}, since 𝑏_−𝑛≠0.
From the pervous discussion, by the induction we obtain that 𝑎_−𝑛+𝑖=𝜆𝑏_−𝑛+𝑖 for any integers 𝑖 with 0≤𝑖≤𝑛−1. Hence, 𝜑₂(𝑧)=∑^𝑛_𝑝=1𝑎_−𝑝𝑧^𝑝=∑^𝑛_𝑝=1𝜆𝑏_−𝑝𝑧^𝑝=𝜆𝜓₂(𝑧).
Since 𝜑₂(𝑧)=𝜆𝜓₂(𝑧), by (43), we get that 𝜓₁(0)𝜑₂+𝑃𝜑₁𝑊^∗_𝑘𝜓₂=𝜑₁(0)𝜓₂+𝑃𝜓₁𝑊^∗_𝑘𝜑₂. (46) Then, 𝜓₁(0)𝜑₂+𝑃𝜑₁𝑊^∗_𝑘𝜓₂,𝑧^𝑘𝑛=𝑊^∗_𝑘𝜓₂,𝑧^𝑘𝑛𝜑₁=𝑏_−𝑛𝑎₀1=𝑛+1𝜑₁(0)𝜓₂+𝑃𝜓₁𝑊^∗_𝑘𝜑₂,𝑧^𝑘𝑛=𝑊^∗_𝑘𝜑₂,𝑧^𝑘𝑛𝜓₁=𝑎_−𝑛𝑏₀1,𝑛+1 (47) and we get that 𝑏_−𝑛𝑎₀=𝑎_−𝑛𝑏₀; that is, 𝑎₀=𝜆𝑏₀. So, by (46) we have that 𝜓₁(0)𝜑₂=𝜑₁(0)𝜓₂ and 𝑃(𝜑₁𝑊^∗_𝑘𝜓₂)=𝑃(𝜓₁𝑊^∗_𝑘𝜑₂), that is, 𝑃(𝜆𝜓₁−𝜑₁𝑊^∗_𝑘𝜓₂)=0. Hence, for any integers 𝑙 with 𝑘𝑛−𝑛≤𝑙≤𝑘𝑛−1, we have 0=𝑧^𝑙,0=𝑧^𝑙,𝑃𝜆𝜓₁−𝜑₁𝑊^∗_𝑘𝜓₂=𝑧^𝑙,𝜆𝜓₁−𝜑₁𝑊^∗_𝑘𝜓₂=𝜆𝜓₁−𝜑₁𝑧^𝑙,𝑊^∗_𝑘𝜓₂, (48) that is, ^𝑛_𝑝=0𝜆𝑏_𝑝−𝑎_𝑝𝑧^𝑝+𝑙,^𝑛_𝑝=1𝑏_−𝑝𝑘𝑝+1𝑧𝑝+1^𝑘𝑝=0. (49) Since 𝑎₀=𝜆𝑏₀, we have ^𝑛_𝑝=1𝜆𝑏_𝑝−𝑎_𝑝𝑧^𝑝+𝑙,^𝑛_𝑝=1𝑏_−𝑝𝑘𝑝+1𝑧𝑝+1^𝑘𝑝=0. (50) When 𝑙=𝑘𝑛−1, by (50), we get that (𝜆𝑏₁−𝑎₁)𝑏_−𝑛(1/(𝑛+1))=0, so 𝑎₁=𝜆𝑏₁, since 𝑏_−𝑛≠0. Then we have ^𝑛_𝑝=2𝜆𝑏_𝑝−𝑎_𝑝𝑧^𝑝+𝑙,^𝑛_𝑝=1𝑏_−𝑝𝑘𝑝+1𝑧𝑝+1^𝑘𝑝=0. (51)
Suppose that 𝑎_𝑗=𝜆𝑏_𝑗 for any integers 𝑗 with 0≤𝑗≤𝑠, where 0≤𝑠≤𝑛−1. By (50), we get that ^𝑛_𝑝=𝑠+1𝜆𝑏_p−𝑎_𝑝𝑧^𝑝+𝑙,^𝑛_𝑝=1𝑏_−𝑝𝑘𝑝+1𝑧𝑝+1^𝑘𝑝=0. (52) Now consider the connection between 𝑎_𝑠+1 and 𝑏_𝑠+1.
When 𝑙=𝑘𝑛−𝑠−1, by (52), we get that (𝜆𝑏_𝑠+1−𝑎_𝑠+1)𝑏_−𝑛(1/(𝑛+1))=0, so 𝑎_𝑠+1=𝜆𝑏_𝑠+1, since 𝑏_−𝑛≠0.
Form the above discussion, by the induction we can obtain that 𝑎_𝑗=𝜆𝑏_𝑗 for any integers 𝑗 with 0≤𝑗≤𝑛, so 𝜑₁(𝑧)=∑^𝑛_𝑝=0𝑎_𝑝𝑧^𝑝=∑^𝑛_𝑝=0𝜆𝑏_𝑝𝑧^𝑝=𝜆𝜓₁(𝑧).
Since 𝜑₂=𝜆𝜓₂, we have 𝜑=𝜑₁+𝜑₂=𝜆𝜓₁+𝜆𝜓₂=𝜆𝜓. Hence, the required result holds.

Lemma 19. Let 𝜑(𝑧)=∑^𝑛_𝑝=0𝑎_𝑝𝑧^𝑝+∑^𝑛_𝑝=1𝑎_−𝑝𝑧^𝑝 and 𝜓(𝑧)=∑^𝑚_𝑝=0𝑏_𝑝𝑧^𝑝+∑^𝑚_𝑝=1𝑏_−𝑝𝑧^𝑝, where 𝑛 and 𝑚 are integers with 𝑛>𝑚≥1 and 𝑏_−𝑚≠0. If 𝐵_𝜑 and 𝐵_𝜓 commute, then 𝑎_−𝑗=0 for any integers 𝑗 with 𝑚+1≤𝑗≤𝑛.

Proof. Let 𝜑₁(𝑧)=∑^𝑛_𝑝=0𝑎_𝑝𝑧^𝑝, 𝜑₂(𝑧)=∑^𝑛_𝑝=1𝑎_−𝑝𝑧^𝑝, 𝜓₁(𝑧)=∑^𝑚_𝑝=0𝑏_𝑝𝑧^𝑝 and 𝜓₂(𝑧)=∑^𝑚_𝑝=1𝑏_−𝑝𝑧^𝑝, then 𝜑=𝜑₁+𝜑₂ and 𝜓=𝜓₁+𝜓₂. Since 𝐵_𝜑 and 𝐵_𝜓 commute, we have 𝑇_𝜑𝑊^∗_𝑘𝑇_𝜓𝑊^∗_𝑘1=𝑇_𝜓𝑊^∗_𝑘𝑇_𝜑𝑊^∗_𝑘1; that is, 𝜑₂𝑊^∗_𝑘𝜓₂+𝜓₁(0)𝜑₂+𝑃𝜑₁𝑊^∗_𝑘𝜓₂=𝜓₂𝑊^∗_𝑘𝜑₂+𝜑₁(0)𝜓₂+𝑃𝜓₁𝑊^∗_𝑘𝜑₂,_{𝑘𝑚+𝑛𝑝=𝑘+1𝑖+𝑘𝑗=𝑝}𝑎_−𝑖𝑏_−𝑗𝑘𝑗+1𝑧𝑗+1^𝑝+𝜓₁(0)^𝑛_𝑝=1𝑎_−𝑝𝑧^𝑝+^𝑘𝑚_{𝑝=0𝑘𝑗−𝑖=𝑝}𝑎_𝑖𝑏_−𝑗𝑘𝑗−𝑖+1𝑧𝑘𝑗+1^𝑝=^{𝑘𝑛+𝑚}_{𝑝=𝑘+1𝑘𝑖+𝑗=𝑝}𝑎_−𝑖𝑏_−𝑗𝑘𝑖+1𝑧𝑖+1^𝑝+𝜑₁(0)^𝑚_𝑝=1𝑏_−𝑝𝑧^𝑝+^𝑘𝑛_{𝑝=0𝑘𝑖−𝑗=𝑝}𝑏_𝑗𝑎_−𝑖𝑘𝑖−𝑗+1𝑧𝑘𝑖+1^𝑝. (53)
Since 𝑛>𝑚, let 𝑛=𝑚+𝑟, where 𝑟 is a positive integer. Then, by the pervoius equation, we get that _{𝑘𝑖+𝑗=𝑘𝑛+𝑚}𝑎_−𝑖𝑏_−𝑗𝑘𝑖+1𝑖+1=0,thatis,𝑎_−𝑛𝑏_−𝑚𝑘𝑛+1𝑛+1=0, (54) so 𝑎_−𝑛=0, since 𝑏_−𝑚≠0.
Suppose that 𝑎_−𝑛+𝑡=0 for any integers 𝑡 with 0≤𝑡≤𝑟−2. Now consider the value of 𝑎_{−𝑛+𝑡+1}. By the assumption and the above equation we get that ^{𝑘𝑚+𝑛−𝑡−1}_{𝑝=𝑘+1𝑖+𝑘𝑗=𝑝}𝑎_−𝑖𝑏_−𝑗𝑘𝑗+1𝑧𝑗+1^𝑝+𝜓₁(0)^{𝑛−𝑡−1}_𝑝=1𝑎_−𝑝𝑧^𝑝+^𝑘𝑚_{𝑝=0𝑘𝑗−𝑖=𝑝}𝑎_𝑖𝑏_−𝑗𝑘𝑗−𝑖+1𝑧𝑘𝑗+1^𝑝=^{𝑘𝑛+𝑚−𝑘𝑡−𝑘}_{𝑝=𝑘+1𝑘𝑖+𝑗=𝑝}𝑎_−𝑖𝑏_−𝑗𝑘𝑖+1𝑧𝑖+1^𝑝+𝜑₁(0)^𝑚_𝑝=1𝑏_−𝑝𝑧^𝑝+^{𝑘𝑛−𝑘𝑡−𝑘}_{𝑝=0𝑘𝑖−𝑗=𝑝}𝑏_𝑗𝑎_−𝑖𝑘𝑖−𝑗+1𝑧𝑘𝑖+1^𝑝. (55) Since 𝑡+1<𝑟, we can get that ∑_{𝑘𝑖+𝑗=𝑘𝑛+𝑚−𝑘𝑡−𝑘}𝑎_−𝑖𝑏_−𝑗((𝑘𝑖+1)/(𝑖+1))=0; that is, 𝑎_{−𝑛+𝑡+1}𝑏_−𝑚𝑘(𝑛−𝑡−1)+1𝑛−𝑡−1+⋯+𝑎_{−𝑛+𝑡+1−𝜆}𝑏_{−𝑚+𝑘𝜆}𝑘(𝑛−𝑡−1+𝜆)+1𝑛−𝑡−1+𝜆=0, (56) where 𝜆=min{𝑡+1,[2𝑚/𝑘]} and [2𝑚/𝑘] is the biggest integer which is not bigger than 2𝑚/𝑘. Since 𝑏_−𝑚≠0, by assumption, we get that 𝑎_{−𝑛+𝑡+1}=0.
From the preceding discussion, by the induction we can get that 𝑎_−𝑗=0 for any integers 𝑗 with 𝑚+1≤𝑗≤𝑛.

Theorem 20. Let 𝜑(𝑧)=∑^𝑛_𝑝=0𝑎_𝑝𝑧^𝑝+∑^𝑛_𝑝=1𝑎_−𝑝𝑧^𝑝 and 𝜓(𝑧)=∑^𝑚_𝑝=0𝑏_𝑝𝑧^𝑝+∑^𝑚_𝑝=1𝑏_−𝑝𝑧^𝑝, where 𝑎²_−𝑛+𝑏²_−𝑚≠0, 𝑛 and 𝑚 are integers with 𝑛>𝑚≥1, then the following statements are equivalent: (1.1)𝐵_𝜑 and 𝐵_𝜓 commute; (1.2) there exist scalars 𝛼 and 𝛽, not both zero, such that 𝛼𝜑+𝛽𝜓=0.

Proof. First suppose that (1.2) holds. It is obvious that 𝐵_𝜑 and 𝐵_𝜓 commute.
Now suppose that (1.1) holds. Since 𝑎²_−𝑛+𝑏²_−𝑚≠0, we can get that 𝑎_−𝑛≠0 or 𝑏_−𝑚≠0.
If 𝑎_−𝑛≠0, by Theorem 18, we can get the required result.
If 𝑏_−𝑚≠0, by Lemma 19, we can get that 𝜑(𝑧)=∑^𝑛_𝑝=0𝑎_𝑝𝑧^𝑝+∑^𝑚_𝑘=1𝑎_−𝑝𝑧^𝑝. Let 𝜑₁(𝑧)=∑^𝑛_𝑝=0𝑎_𝑝𝑧^𝑝, 𝜑₂(𝑧)=∑^𝑚_𝑝=1𝑎_−𝑝𝑧^𝑝, 𝜓₁(𝑧)=∑^𝑚_𝑝=0𝑏_𝑝𝑧^𝑝, and 𝜓₂(𝑧)=∑^𝑚_𝑝=1𝑏_−𝑝𝑧^𝑝, then 𝜑=𝜑₁+𝜑₂ and 𝜓=𝜓₁+𝜓₂.
Since 𝐵_𝜑 and 𝐵_𝜓 commute, we have 𝑇_𝜑𝑊^∗_𝑘𝑇_𝜓𝑊^∗_𝑘1=𝑇_𝜓𝑊^∗_𝑘𝑇_𝜑𝑊^∗_𝑘1; that is, 𝜑₂𝑊^∗_𝑘𝜓₂+𝜓₁(0)𝜑₂+𝑃𝜑₁𝑊^∗_𝑘𝜓₂=𝜓₂𝑊^∗_𝑘𝜑₂+𝜑₁(0)𝜓₂+𝑃𝜓₁𝑊^∗_𝑘𝜑₂. (57) By (57) we get that for any integers 𝑝 with 𝑘𝑚+1≤𝑝≤𝑘𝑚+𝑚, _{𝑖+𝑘𝑗=𝑝}𝑎_−𝑖𝑏_−𝑗𝑘𝑗+1𝑗+1=_{𝑘𝑖+𝑗=𝑝}𝑎_−𝑖𝑏_−𝑗𝑘𝑖+1,𝑖+1 (58) where 𝑖 and 𝑗 are positive integers with −𝑚≤𝑖≤𝑛 and −𝑚≤𝑗≤𝑚. Now we continue the proof by the induction.
When 𝑝=𝑘𝑚+𝑚, by (58), we get that 𝑎_−𝑚𝑏_−𝑚((𝑘𝑚+1)/(𝑚+1))=𝑎_−𝑚𝑏_−𝑚((𝑘𝑚+1)/(𝑚+1)), so 𝑎_−𝑚=(𝑎_−𝑚/𝑏_−𝑚)𝑏_−𝑚, since 𝑏_−𝑚≠0. Let 𝜆=(𝑎_−𝑚/𝑏_−𝑚), then 𝑎_−𝑚=𝜆𝑏_−𝑚.
When 𝑝=𝑘𝑚+𝑚−1, by (58), we get that 𝑎_−𝑚+1𝑏_−𝑚((𝑘𝑚+1)/(𝑚+1))=𝑎_−𝑚𝑏_−𝑚+1((𝑘𝑚+1)/(𝑚+1)), so 𝑎_−𝑚+1=𝜆𝑏_−𝑚+1, since 𝑏_−𝑚≠0.
Suppose that 𝑎_−𝑚+𝑖=𝜆𝑏_−𝑚+𝑖 for any integers 𝑖 with 0≤𝑖≤𝑙<𝑚−1. Now consider the connection between 𝑎_{−𝑚+𝑙+1} and 𝑏_{−𝑚+𝑙+1}.
When 𝑝=𝑘𝑚+𝑚−𝑙−1, by (58), we get that 𝑎_{−𝑚+𝑙+1}𝑏_−𝑚𝑘𝑚+1+𝑚+1+⋯𝑎_{−𝑚+𝑙+1−𝑟𝑘}𝑏_−𝑚+𝑟𝑘(𝑚−𝑟)+1=𝑚−𝑟+1𝑏_{−𝑚+𝑙+1}𝑎_−𝑚𝑘𝑚+1+𝑚+1+⋯𝑏_{−𝑚+𝑙+1−𝑟𝑘}𝑎_−𝑚+𝑟𝑘(𝑚−𝑟)+1.𝑚−𝑟+1 (59) From the assumption we obtain that 𝑎_{−𝑚+𝑙+1}𝑏_−𝑚((𝑘𝑚+1)/(𝑚+1))=𝑏_{−𝑚+𝑙+1}𝑎_−𝑚((𝑘𝑚+1)/(𝑚+1)), so 𝑎_{−𝑚+𝑙+1}=𝜆𝑏_{−𝑚+𝑙+1}, since 𝑏_−𝑚≠0.
From the pervious discussion, by the induction we obtain that 𝑎_−𝑚+𝑖=𝜆𝑏_−𝑚+𝑖 for any integers 𝑖 with 0≤𝑖≤𝑚−1. Hence, 𝜑₂(𝑧)=∑^𝑚_𝑝=1𝑎_−𝑝𝑧^𝑝=∑^𝑚_𝑝=1𝜆𝑏_−𝑝𝑧^𝑝=𝜆𝜓₂(𝑧).
Since 𝜑₂(𝑧)=𝜆𝜓₂(𝑧), by (57), we get that 𝜓₁(0)𝜑₂+𝑃𝜑₁𝑊^∗𝜓₂=𝜑₁(0)𝜓₂+𝑃𝜓₁𝑊^∗𝜑₂. (60) Then, 𝜓₁(0)𝜑₂+𝑃𝜑₁𝑊^∗𝜓₂,𝑧^𝑘𝑚=𝑊^∗𝜓₂,𝑧^𝑘𝑚𝜑₁=𝑏_−𝑚𝑎₀1=𝑚+1𝜑₁(0)𝜓₂+𝑃𝜓₁𝑊^∗𝜑₂,𝑧^𝑘𝑚=𝑊^∗𝜑₂,𝑧^𝑘𝑚𝜓₁=𝑎_−𝑚𝑏₀1,𝑚+1 (61) and we get that 𝑏_−𝑚𝑎₀=𝑎_−𝑚𝑏₀; that is, 𝑎₀=𝜆𝑏₀. So, by (60), we have that 𝜓₁(0)𝜑₂=𝜑₁(0)𝜓₂ and 𝑃(𝜑₁𝑊^∗𝜓₂)=𝑃(𝜓₁𝑊^∗_𝑘𝜑₂); that is, 𝑃((𝜆𝜓₁−𝜑₁)𝑊^∗_𝑘𝜓₂)=0. Hence, for any integers 𝑙 with 𝑘𝑚−𝑚≤𝑙≤𝑘𝑚−1, we have 0=𝑧^𝑙,0=𝑧^𝑙,𝑃𝜆𝜓₁−𝜑₁𝑊^∗_𝑘𝜓₂=𝑧^𝑙,𝜆𝜓₁−𝜑₁𝑊^∗_𝑘𝜓₂=𝜆𝜓₁−𝜑₁𝑧^l,𝑊^∗_𝑘𝜓₂, (62) that is, ^𝑚_𝑝=0𝜆𝑏_𝑝−𝑎_𝑝𝑧^𝑝+𝑙−^𝑛_𝑝=𝑚+1𝑎_𝑝𝑧^𝑝+𝑙,^𝑚_𝑝=1𝑏_−𝑝𝑘𝑝+1𝑧𝑝+1^𝑘𝑝=0. (63) Since 𝑎₀=𝜆𝑏₀, we have ^𝑚_𝑝=1𝜆𝑏_𝑝−𝑎_𝑝𝑧^𝑝+𝑙−^𝑛_𝑝=𝑚+1𝑎_𝑝𝑧^𝑝+𝑙,^𝑚_𝑝=1𝑏_−𝑝𝑘𝑝+1𝑧𝑝+1^𝑘𝑝=0. (64)
When 𝑙=𝑘𝑚−1, by (64), we get that (𝜆𝑏₁−𝑎₁)𝑏_−𝑚((𝑘𝑚+1)/(𝑚+1))=0, so 𝑎₁=𝜆𝑏₁, since 𝑏_−𝑚≠0. Then, we have ^𝑚_𝑝=2𝜆𝑏_𝑝−𝑎_𝑝𝑧^𝑝+𝑙−^𝑛_𝑝=𝑚+1𝑎_𝑝𝑧^𝑝+𝑙,^𝑚_𝑝=1𝑏_−𝑝𝑘𝑝+1𝑧𝑝+1^𝑘𝑝=0. (65) Suppose that 𝑎_𝑗=𝜆𝑏_𝑗 for any integers 𝑗 with 0≤𝑗≤𝑠, where 0≤𝑠≤𝑚−1. By (64), we get that ^𝑚_𝑝=𝑠+1𝜆𝑏_𝑝−𝑎_𝑝𝑧^𝑝+𝑙−^𝑛_𝑝=𝑚+1𝑎_𝑝𝑧^𝑝+𝑙,^𝑚_𝑝=1𝑏_−𝑝𝑘𝑝+1𝑧𝑝+1^𝑘𝑝=0. (66) Now consider the connection between 𝑎_𝑠+1 and 𝑏_𝑠+1.
When 𝑙=𝑘𝑚−𝑠−1, by (66), we get that (𝜆𝑏_𝑠+1−𝑎_𝑠+1)𝑏_−𝑚((𝑘𝑚+1)/(𝑚+1))=0, so 𝑎_𝑠+1=𝜆𝑏_𝑠+1, since 𝑏_−𝑚≠0.
From the pervious discussion, by the induction, we can obtain that 𝑎_𝑗=𝜆𝑏_𝑗 for any integers 𝑗 with 0≤𝑗≤𝑚. Hence, 𝜑₁(𝑧)=^𝑛_𝑝=0𝑎_𝑝𝑧^𝑝=^𝑚_𝑝=0𝜆𝑏_𝑝𝑧^𝑝+^𝑛_𝑝=𝑚+1𝑎_𝑝𝑧^𝑝=𝜆𝜓₁(𝑧)+^𝑛_𝑝=𝑚+1𝑎_𝑝𝑧^𝑝. (67) Let 𝜑₃(𝑧)=∑^𝑛_𝑝=𝑚+1𝑎_𝑝𝑧^𝑝, then 𝜑=𝜑₁+𝜑₂=𝜆𝜓₁+𝜑₃+𝜆𝜓₂=𝜆𝜓+𝜑₃, since 𝜑₂=𝜆𝜓₂.
Now we want to show that 𝜑₃≡0. Because 𝐵_𝜑𝐵_𝜓=𝐵_𝜓𝐵_𝜑, yet we get that 𝐵_𝜑3𝐵_𝜓=𝐵_𝜓𝐵_𝜑3, which means that 𝑇_𝜑3𝑊^∗_𝑘𝑇_𝜓𝑊^∗_𝑘=𝑇_𝜓𝑊^∗_𝑘𝑇_𝜑3𝑊^∗_𝑘. (68) Since 𝑚≥1, we have 𝑇_𝜑3𝑊^∗_𝑘𝑇_𝜓𝑊^∗_𝑘1=0; that is, 𝑇_𝜑3𝑊^∗_𝑘[𝜓₂+𝜓₁(0)]=0. Then, we get that for any integers 𝑙 with 0≤𝑙≤𝑘𝑚−𝑚−1, ⟨𝑇_𝜑3𝑊^∗_𝑘[𝜓₂+𝜓₁(0)],𝑧^𝑙⟩=0; that is, ^𝑚_𝑗=1𝑏_−𝑗𝑘𝑗+1𝑧𝑗+1^𝑘𝑗+𝜓₁(0),^𝑛_𝑖=𝑚+1𝑎_𝑖𝑧^𝑖+𝑙=0. (69) By (69), we can get that for any integers 𝑙 with max{0,𝑘𝑚−𝑛}≤𝑙≤𝑘𝑚−𝑚−1, 𝑏_−𝑚𝑎_{𝑘𝑚−𝑙}(1/(𝑚+1))=0; that is, 𝑎_{𝑘𝑚−𝑙}=0, since 𝑏_−𝑚≠0.
If 𝑛≤𝑘𝑚, then 𝜑₃≡0, which means that 𝜑=𝜆𝜓, so the required result holds.
If 𝑛≥𝑘𝑚+1, then 𝜑₃(𝑧)=∑^𝑛_{𝑝=𝑘𝑚+1}𝑎_𝑝𝑧^𝑝. So, by (68), we can get that 𝑇_𝜑3𝑊^∗_𝑘𝑇_𝜓𝑊^∗_𝑘(𝑧^𝑚)=0, that is, 𝑇_𝜑3𝑊^∗_𝑘[𝜓₂𝑧^𝑘𝑚+𝑃(𝜓₁𝑧^𝑘𝑚)]=0. So we get that for any integers 𝑙 with max{0,𝑘²𝑚+𝑘𝑚−𝑛}≤𝑙≤𝑘²𝑚−1, 𝑊^∗_𝑘𝜓₂𝑧^𝑘𝑚+𝑃𝜓₁𝑧^𝑘𝑚,^𝑛_{𝑖=𝑘𝑚+1}𝑎_𝑖𝑧^𝑖+𝑙=0. (70) By (70), we can get that for any integers 𝑙 with max{0,𝑘²𝑚+𝑘𝑚−𝑛}≤𝑙≤𝑘²𝑚−1, 𝑏_−𝑚𝑎_𝑘²_{𝑚+𝑘𝑚−𝑙}(1/(𝑘²𝑚+𝑘𝑚+1))=0, that is, 𝑎_𝑘²_{𝑚+𝑘𝑚−𝑙}=0, since 𝑏_−𝑚≠0.
If 𝑛≤𝑘²𝑚+𝑘𝑚, then 𝜑₃≡0, which means that 𝜑=𝜆𝜓, so, the required result holds.
If 𝑛≥𝑘²𝑚+𝑘𝑚+1, then 𝜑₃(𝑧)=∑^𝑛_{𝑝=𝑘2𝑚+𝑘𝑚+1}𝑎_𝑝𝑧^𝑝. Then successively by the pervoius method, we can get that 𝑎_𝑖=0 for all integers 𝑖 with 𝑚+1≤𝑖≤𝑛, which means that 𝜑=𝜆𝜓. So, the required result holds.

Acknowledgments

The authors thank the referees for several suggestions that improved the paper. This research is supported by NSFC, Items nos. 11271059 and 11226120.