Commutativity of Toeplitz Operators with Trigonometric Polynomial Symbols on Beurling Subspaces of Hardy–Sobolev Spaces

Abstract

We investigate the commutativity of Toeplitz operators with trigonometric polynomial symbols on the Beurling subspaces $z^k{H}_s^2$ of the Hardy–Sobolev spaces. An explicit formula for the commutator of two Toeplitz operators is first derived in terms of weighted shift operators. As a consequence, every such commutator is shown to have finite rank. It is then proved that if the degrees of the symbols are at most k, then the corresponding Toeplitz operators commute on the Beurling subspace $z^k{H}_s^2$. Moreover, this result is shown to be sharp by proving that, in general, the Beurling subspace $z^k{H}_s^2$ cannot be replaced by $z^{k-1}{H}_s^2$. Finally, these results generalize the corresponding commutativity theorem for Toeplitz operators on the classical Hardy space.

1. Introduction

Toeplitz operators constitute one of the central objects in operator theory and complex analysis. They have been extensively studied on a variety of spaces of analytic functions, including the Hardy, Bergman, and Dirichlet spaces, owing to their rich operator-theoretic structure and numerous applications [1,2,3,4]. Their close connection with function theory, Banach algebras, and operator algebras has made them a fundamental tool in the study of operators on spaces of analytic functions.

One of the fundamental problems in the theory of Toeplitz operators is to determine when two Toeplitz operators commute. This question plays a fundamental role in understanding the algebraic structure of Toeplitz operator algebras and has therefore been extensively investigated for a wide variety of symbol classes and analytic function spaces. The pioneering work of Brown and Halmos [5] characterized commuting Toeplitz operators with analytic symbols on the Hardy space. Axler and Čučković [6] completely characterized the commutativity of Toeplitz operators with harmonic symbols. Closely related to the commutativity problem is the study of finite-rank commutators, which reveals important rigidity phenomena for Toeplitz operators. In this direction, Guo, Sun and Zheng [7] completely characterized finite-rank commutators of Toeplitz operators with bounded harmonic symbols on the Bergman space. Ding and Zheng [8] obtained complete characterizations of finite-rank commutators for both Toeplitz and Hankel operators on the Hardy space. More recently, the study of commutativity and related algebraic properties of Toeplitz operators has continued to attract considerable attention. Examples include studies on the commutativity of Toeplitz and Hankel operators on the Hardy space over the n-torus [9], on the commutativity of Toeplitz operators on Bergman spaces of the unit disk [10], and on the skew commutativity of Toeplitz and Hankel operators [11].

Motivated by these developments, it is natural to investigate the commutativity problem on Beurling subspaces of the Hardy space. Recently, the author investigated the commutativity of Toeplitz operators with trigonometric polynomial symbols on Beurling subspaces of the classical Hardy space [12]. It was proved that if the degrees of the symbols do not exceed k, then the corresponding Toeplitz operators commute on the Beurling subspace $z^k{H}^2$. Moreover, the result was shown to be sharp in the sense that, in general, the subspace $z^k{H}^2$ cannot be replaced by $z^{k-1}{H}^2$.

The Hardy–Sobolev spaces $H_s^2$ form a natural weighted generalization of the classical Hardy space and constitute an important class of weighted Hardy spaces [3,13]. In recent years, several aspects of Toeplitz operators on Hardy–Sobolev spaces have been studied. Cao and He [14], He and Cao [15], and He, Huang, and Lee [16] investigated Toeplitz and dual Toeplitz operators on Hardy–Sobolev spaces, establishing results on compactness, Fredholmness, Toeplitz products, and related operator-theoretic properties. The parameter s includes the classical Hardy space as the special case $s=0$. Motivated by the above result on Beurling subspaces of the Hardy space, the purpose of the present paper is to investigate the corresponding commutativity problem for Toeplitz operators on Beurling subspaces of $H_s^2$. It is shown that the corresponding result extends to Hardy–Sobolev spaces and that its sharpness is preserved.

Despite the close structural similarities between Hardy–Sobolev spaces and the classical Hardy space, the presence of the weight sequence ${\left\{{(n+1)}^s\right\}}_{n\ge 0}$ changes the behavior of the shift operators. In particular, the multiplication operators $T_z$ and $T_{\overline{z}}$ become weighted shift operators, and their iterates acquire nontrivial weights. Consequently, arguments that rely on the unweighted shift cannot be applied directly. This naturally leads to the question of whether the commutativity phenomenon established on the Hardy space persists in the Hardy–Sobolev setting.

The main contribution of the present paper is the extension of the commutativity result on Beurling subspaces of the Hardy space to Beurling subspaces of Hardy–Sobolev spaces. An explicit formula for the commutator of two Toeplitz operators with trigonometric polynomial symbols is first established in terms of weighted shift operators. As a consequence, every such commutator is shown to have finite rank. The commutator formula is then used to prove that Toeplitz operators whose symbols have degrees at most k commute on the Beurling subspace $z^k{H}_s^2$. Finally, it is shown that this commutativity result is sharp.

The remainder of the paper is organized as follows. Section 2 introduces the Hardy–Sobolev spaces and establishes the preliminary results needed throughout the paper. An explicit formula for the commutator of two Toeplitz operators with trigonometric polynomial symbols is derived in Section 3, where the finite-rank property of such commutators is also established. The main commutativity theorem and its sharpness are proved in Section 4. Section 5 presents several examples illustrating the main results. Finally, Section 6 concludes the paper with a discussion of possible directions for further research.

2. Preliminaries

In this section, we introduce the Hardy–Sobolev spaces, establish the notation used throughout the paper, and describe the shift representation of the Toeplitz operators $T_z$ and $T_{\overline{z}}$, which forms the foundation for the subsequent analysis.

Let $s\ge 0$. The Hardy–Sobolev space $H_s^2$ consists of all analytic functions

$$f\left(z\right)=\sum _{n\ge 0}{a}_n{z}^n$$

on the unit disc $\mathbb{D}$ such that

$${\parallel f\parallel }_{H_s^2}^2=\sum _{n\ge 0}{(n+1)}^{2s}{\left|a_n\right|}^2<\infty .$$

The space $H_s^2$ is a Hilbert space with inner product

$${\langle f,g\rangle }_{H_s^2}=\sum _{n\ge 0}{(n+1)}^{2s}{a}_n\overline{b_n},$$

where

$$f\left(z\right)=\sum _{n\ge 0}{a}_n{z}^n,g\left(z\right)=\sum _{n\ge 0}{b}_n{z}^n.$$

For $s=0$, the space $H_s^2$ reduces to the classical Hardy space $H^2$. Define

$$e_n^{\left(s\right)}\left(z\right)={\displaystyle \frac{z^n}{{(n+1)}^s}},n\ge 0.$$

Then ${\left\{e_n^{\left(s\right)}\right\}}_{n\ge 0}$ forms an orthonormal basis for $H_s^2$. For each $\mathcal{l}\ge 0$, let $P_{\mathcal{l}}$ denote the rank-one orthogonal projection onto $\mathrm{span}\left\{e_{\mathcal{l}}^{\left(s\right)}\right\}$; that is,

$$P_{\mathcal{l}}f={\langle f,e_{\mathcal{l}}^{\left(s\right)}\rangle }_{H_s^2}{e}_{\mathcal{l}}^{\left(s\right)},f\in H_s^2.$$

Let

$$P_s:L^2\left(\mathbb{T}\right)⟶H_s^2$$

denote the analytic Fourier projection defined by

$$P_s\left({\displaystyle \sum _{n=-\infty }^{\infty }}{a}_n{z}^n\right)={\displaystyle \sum _{n=0}^{\infty }}{a}_n{z}^n,$$

for every Fourier series converging in $L^2\left(\mathbb{T}\right)$. For $\phi \in L^{\infty }\left(\mathbb{T}\right)$, the Toeplitz operator $T_{\phi }$ on $H_s^2$ is defined by

$$T_{\phi }f=P_s\left(\phi f\right),f\in H_s^2.$$

In particular, let

$$U:=T_z,V:=T_{\overline{z}}.$$

The operators U and V act on the orthonormal basis ${\left\{e_n^{\left(s\right)}\right\}}_{n\ge 0}$ by

$$U{e}_n^{\left(s\right)}={\left({\displaystyle \frac{n+2}{n+1}}\right)}^s{e}_{n+1}^{\left(s\right)},n\ge 0,$$

and

$$V{e}_n^{\left(s\right)}=\left\{\begin{array}{cc}0,\hfill & n=0,\hfill \\ {\left({\displaystyle \frac{n}{n+1}}\right)}^s{e}_{n-1}^{\left(s\right)},\hfill & n\ge 1.\hfill \end{array}\right.$$

Thus U is a forward-weighted shift and V is a backward-weighted shift on $H_s^2$.

For $p,q\ge 1$, the iterates of U and V satisfy

$$U^p{e}_n^{\left(s\right)}={\left({\displaystyle \frac{n+p+1}{n+1}}\right)}^s{e}_{n+p}^{\left(s\right)},n\ge 0,$$

(1)

and

$$V^q{e}_n^{\left(s\right)}=\left\{\begin{array}{cc}0,\hfill & n<q\hfill \\ {\left({\displaystyle \frac{n-q+1}{n+1}}\right)}^s{e}_{n-q}^{\left(s\right)},\hfill & n\ge q.\hfill \end{array}\right.$$

(2)

Proposition 1. 

Let $U=T_z$ and $V=T_{\overline{z}}$. Then for all integers $p,q\ge 1$,

$$T_{z^p}=U^p,T_{{\overline{z}}^q}=V^q.$$

Proof. 

Let $f\in H_s^2$. Since $z^{p}f$ is analytic, we have

$$T_{z^p}f=P_s\left(z^{p}f\right)=z^{p}f.$$

Since $U=T_z$, it follows by induction that

$$U^{p}f=z^{p}f.$$

Thus $T_{z^p}=U^p$.

For $q\ge 1$, we have

$$T_{{\overline{z}}^q}f=P_s\left({\overline{z}}^{q}f\right).$$

Since both operators are linear, it is enough to verify the identity on the orthonormal basis ${\left\{e_n^{\left(s\right)}\right\}}_{n\ge 0}$. If $n<q$, then

$$P_s\left({\overline{z}}^q{e}_n^{\left(s\right)}\right)=0,$$

and by the formula for the iterates of V in (2), we have $V^q{e}_n^{\left(s\right)}=0$. Thus

$$T_{{\overline{z}}^q}{e}_n^{\left(s\right)}=V^q{e}_n^{\left(s\right)}.$$

If $n\ge q$, then

$${\overline{z}}^q{e}_n^{\left(s\right)}={\overline{z}}^q{\displaystyle \frac{z^n}{{(n+1)}^s}}={\displaystyle \frac{z^{n-q}}{{(n+1)}^s}}.$$

Since

$$e_{n-q}^{\left(s\right)}={\displaystyle \frac{z^{n-q}}{{(n-q+1)}^s}},$$

it follows that

$${\overline{z}}^q{e}_n^{\left(s\right)}={\displaystyle \frac{z^{n-q}}{{(n+1)}^s}}={\left({\displaystyle \frac{n-q+1}{n+1}}\right)}^s{e}_{n-q}^{\left(s\right)}.$$

Thus

$$T_{{\overline{z}}^q}{e}_n^{\left(s\right)}=P_s\left({\overline{z}}^q{e}_n^{\left(s\right)}\right)={\left({\displaystyle \frac{n-q+1}{n+1}}\right)}^s{e}_{n-q}^{\left(s\right)}.$$

On the other hand, by Formula (2),

$$V^q{e}_n^{\left(s\right)}={\left({\displaystyle \frac{n-q+1}{n+1}}\right)}^s{e}_{n-q}^{\left(s\right)}.$$

Hence

$$T_{{\overline{z}}^q}{e}_n^{\left(s\right)}=V^q{e}_n^{\left(s\right)}$$

for every $n\ge 0$. Since the two operators agree on the orthonormal basis, it follows that

$$T_{{\overline{z}}^q}=V^q.$$

□

3. A Commutator Formula for Toeplitz Operators

In this section, we establish basic properties of Toeplitz operators on Hardy–Sobolev spaces and derive auxiliary results that will be used in the proof of the main theorem. In particular, we derive an explicit formula for the commutator of two Toeplitz operators with trigonometric polynomial symbols. As applications, we establish the finite-rank property of the commutator.

Lemma 1. 

Let φ and ψ be trigonometric polynomials where

$$\phi \left(z\right)={\displaystyle \sum _{i=-n}^n}{a}_i{z}^i,\psi \left(z\right)={\displaystyle \sum _{j=-m}^m}{b}_j{z}^j.$$

Then the commutator of Toeplitz operators on $H_s^2$ is given by

$$[T_{\phi },T_{\psi }]={\displaystyle \sum _{p=1}^n}{\displaystyle \sum _{q=1}^m}(a_p{b}_{-q}-a_{-q}{b}_p)[U^p,V^q]$$

(3)

where $U=T_z$ and $V=T_{\overline{z}}$. Moreover, for $p,q\ge 1$,

$$[U^p,V^q]=\left\{\begin{array}{cc}{\displaystyle -U^{p-q}{\displaystyle \sum _{\mathcal{l}=0}^{q-1}}{P}_{\mathcal{l}}}\hfill & \mathit{if}p\ge q\hfill \\ {\displaystyle -V^{q-p}{\displaystyle \sum _{\mathcal{l}=q-p}^{q-1}}{P}_{\mathcal{l}}}\hfill & \mathit{if}q>p.\hfill \end{array}\right.$$

where $P_{\mathcal{l}}$ denotes the rank-one orthogonal projection onto $\mathrm{span}\left\{e_{\mathcal{l}}^{\left(s\right)}\right\}$; that is,

$$P_{\mathcal{l}}f={\langle f,e_{\mathcal{l}}^{\left(s\right)}\rangle }_{H_s^2}{e}_{\mathcal{l}}^{\left(s\right)},f\in H_s^2.$$

Proof. 

By Proposition 1, we may write

$$T_{\phi }={\displaystyle \sum _{i=-n}^n}{a}_i{T}_{z^i}={\displaystyle \sum _{p=0}^n}{a}_p{U}^p+{\displaystyle \sum _{q=1}^n}{a}_{-q}{V}^q,$$

and similarly,

$$T_{\psi }={\displaystyle \sum _{j=-m}^m}{b}_j{T}_{z^j}={\displaystyle \sum _{r=0}^m}{b}_r{U}^r+{\displaystyle \sum _{s=1}^m}{b}_{-s}{V}^s.$$

Hence

$$\begin{array}{cc}\hfill [T_{\phi },T_{\psi }]& =\left({\displaystyle \sum _{p=0}^n}{a}_p{U}^p+{\displaystyle \sum _{q=1}^n}{a}_{-q}{V}^q\right)\left({\displaystyle \sum _{r=0}^m}{b}_r{U}^r+{\displaystyle \sum _{s=1}^m}{b}_{-s}{V}^s\right)\hfill \\ \hfill & -\left({\displaystyle \sum _{r=0}^m}{b}_r{U}^r+{\displaystyle \sum _{s=1}^m}{b}_{-s}{V}^s\right)\left({\displaystyle \sum _{p=0}^n}{a}_p{U}^p+{\displaystyle \sum _{q=1}^n}{a}_{-q}{V}^q\right).\hfill \end{array}$$

Expanding both products, we get

$$\begin{array}{cc}\hfill [T_{\phi },T_{\psi }]& ={\displaystyle \sum _{p=0}^n}{\displaystyle \sum _{r=0}^m}{a}_p{b}_r{U}^{p+r}+{\displaystyle \sum _{p=0}^n}{\displaystyle \sum _{s=1}^m}{a}_p{b}_{-s}{U}^p{V}^s+{\displaystyle \sum _{q=1}^n}{\displaystyle \sum _{r=0}^m}{a}_{-q}{b}_r{V}^q{U}^r\hfill \\ \hfill & +{\displaystyle \sum _{q=1}^n}{\displaystyle \sum _{s=1}^m}{a}_{-q}{b}_{-s}{V}^{q+s}-{\displaystyle \sum _{r=0}^m}{\displaystyle \sum _{p=0}^n}{b}_r{a}_p{U}^{r+p}-{\displaystyle \sum _{r=0}^m}{\displaystyle \sum _{q=1}^n}{b}_r{a}_{-q}{U}^r{V}^q\hfill \\ \hfill & -{\displaystyle \sum _{s=1}^m}{\displaystyle \sum _{p=0}^n}{b}_{-s}{a}_p{V}^s{U}^p-{\displaystyle \sum _{s=1}^m}{\displaystyle \sum _{q=1}^n}{b}_{-s}{a}_{-q}{V}^{s+q}.\hfill \end{array}$$

Since powers of U commute with each other and powers of V commute with each other, the first and the fifth sums cancel, and the fourth and the eighth sums cancel.

Also, the terms with $p=0$ or $r=0$ cancel because $U^0=I$ commutes with both $U^r$ and $V^q$. Therefore,

$$[T_{\phi },T_{\psi }]={\displaystyle \sum _{p=1}^n}{\displaystyle \sum _{q=1}^m}(a_p{b}_{-q}-a_{-q}{b}_p)U^p{V}^q-{\displaystyle \sum _{p=1}^n}{\displaystyle \sum _{q=1}^m}(a_p{b}_{-q}-a_{-q}{b}_p)V^q{U}^p.$$

Thus

$$[T_{\phi },T_{\psi }]={\displaystyle \sum _{p=1}^n}{\displaystyle \sum _{q=1}^m}(a_p{b}_{-q}-a_{-q}{b}_p)[U^p,V^q].$$

It remains to compute $[U^p,V^q]$. Let ${\left\{e_n^{\left(s\right)}\right\}}_{n\ge 0}$ be the orthonormal basis of $H_s^2$. If $p\ge q$, then for $n\ge q$ using the explicit formulas for $U^p{e}_n^{\left(s\right)}$ and $V^q{e}_n^{\left(s\right)}$ one checks that

$$U^p{V}^q{e}_n^{\left(s\right)}=V^q{U}^p{e}_n^{\left(s\right)},$$

and hence

$$[U^p,V^q]e_n^{\left(s\right)}=0.$$

If $0\le n\le q-1$, then $V^q{e}_n^{\left(s\right)}=0$ which implies that $U^p{V}^q{e}_n^{\left(s\right)}=0$, while a direct computation gives

$$V^q{U}^p{e}_n^{\left(s\right)}=U^{p-q}{e}_n^{\left(s\right)}.$$

Thus

$$[U^p,V^q]e_n^{\left(s\right)}=-U^{p-q}{e}_n^{\left(s\right)}.$$

It follows that

$$[U^p,V^q]=-U^{p-q}{\displaystyle \sum _{\mathcal{l}=0}^{q-1}}{P}_{\mathcal{l}}.$$

Now assume that $q>p$. If $n\ge q$, then we have

$$U^p{V}^q{e}_n^{\left(s\right)}=V^q{U}^p{e}_n^{\left(s\right)},$$

and hence

$$[U^p,V^q]e_n^{\left(s\right)}=0.$$

If $0\le n\le q-p-1$, then $U^p{V}^q{e}_n^{\left(s\right)}=0$ since $V^q{e}_n^{\left(s\right)}=0$. Also, since $q>n+p$, we have

$$V^q{U}^p{e}_n^{\left(s\right)}={\left({\displaystyle \frac{n+p+1}{n+1}}\right)}^s{V}^q{e}_{n+p}^{\left(s\right)}=0,$$

so again

$$[U^p,V^q]e_n^{\left(s\right)}=0.$$

Finally, if $q-p\le n\le q-1$, then $V^q{e}_n^{\left(s\right)}=0$, and hence $U^p{V}^q{e}_n^{\left(s\right)}=0$. On the other hand, a straightforward computation shows that

$$V^q{U}^p{e}_n^{\left(s\right)}=V^{q-p}{e}_n^{\left(s\right)}.$$

Therefore,

$$[U^p,V^q]e_n^{\left(s\right)}=-V^{q-p}{e}_n^{\left(s\right)}.$$

It follows that

$$[U^p,V^q]=-V^{q-p}{\displaystyle \sum _{\mathcal{l}=q-p}^{q-1}}{P}_{\mathcal{l}}.$$

This completes the proof. □

Proposition 2. 

Let

$$\phi \left(z\right)={\displaystyle \sum _{i=-n}^n}{a}_i{z}^i,\psi \left(z\right)={\displaystyle \sum _{j=-m}^m}{b}_j{z}^j$$

be trigonometric polynomials. Then

$$\mathrm{Ran}[T_{\phi },T_{\psi }]\subseteq \mathrm{span}\{e_0^{\left(s\right)},e_1^{\left(s\right)},\dots ,e_{k-1}^{\left(s\right)}\},$$

where $k=\max \{n,m\}$. In particular, $[T_{\phi },T_{\psi }]$ has finite rank.

Proof. 

By Lemma 1,

$$[T_{\phi },T_{\psi }]={\displaystyle \sum _{p=1}^n}{\displaystyle \sum _{q=1}^m}(a_p{b}_{-q}-a_{-q}{b}_p)[U^p,V^q].$$

Thus, it suffices to determine the range of each commutator $[U^p,V^q]$. If $p\ge q$, then by Lemma 1,

$${\displaystyle [U^p,V^q]=-U^{p-q}{\displaystyle \sum _{\mathcal{l}=0}^{q-1}}{P}_{\mathcal{l}}.}$$

Hence

$$\mathrm{Ran}[U^p,V^q]\subseteq \mathrm{span}\{e_{p-q}^{\left(s\right)},e_{p-q+1}^{\left(s\right)},\cdots ,e_{p-1}^{\left(s\right)}\}\subseteq \mathrm{span}\{e_0^{\left(s\right)},e_1^{\left(s\right)},\dots ,e_{p-1}^{\left(s\right)}\}.$$

Since $p\le n\le k$, we obtain

$$\mathrm{Ran}[U^p,V^q]\subseteq \mathrm{span}\{e_0^{\left(s\right)},e_1^{\left(s\right)},\dots ,e_{k-1}^{\left(s\right)}\}.$$

If $q>p$, then again by Lemma 1,

$${\displaystyle [U^p,V^q]=-V^{q-p}{\displaystyle \sum _{\mathcal{l}=q-p}^{q-1}}{P}_{\mathcal{l}}.}$$

Therefore,

$$\mathrm{Ran}[U^p,V^q]\subseteq \mathrm{span}\{e_0^{\left(s\right)},e_1^{\left(s\right)},\dots ,e_{p-1}^{\left(s\right)}\}\subseteq \mathrm{span}\{e_0^{\left(s\right)},e_1^{\left(s\right)},\dots ,e_{k-1}^{\left(s\right)}\},$$

since $p\le n\le k$. Thus, in both cases,

$$\mathrm{Ran}[U^p,V^q]\subseteq \mathrm{span}\{e_0^{\left(s\right)},e_1^{\left(s\right)},\dots ,e_{k-1}^{\left(s\right)}\}.$$

Since $[T_{\phi },T_{\psi }]$ is a finite sum of such operators, it follows that

$$\mathrm{Ran}[T_{\phi },T_{\psi }]\subseteq \mathrm{span}\{e_0^{\left(s\right)},e_1^{\left(s\right)},\dots ,e_{k-1}^{\left(s\right)}\}.$$

Hence $[T_{\phi },T_{\psi }]$ has finite rank. □

Corollary 1. 

Let φ and ψ be trigonometric polynomials. Then the commutator

$$[T_{\phi },T_{\psi }]$$

has finite rank on $H_s^2$. In particular, $[T_{\phi },T_{\psi }]$ is compact.

Proof. 

The conclusion follows immediately from Proposition 2. □

Corollary 2. 

Let φ and ψ be trigonometric polynomials of degrees n and m, respectively, and suppose that $k=\max \{n,m\}$. If $r\ge k$, then

$$[T_{\phi },T_{\psi }]f=0\mathit{for}\mathit{every}f\in z^r{H}_s^2.$$

Proof. 

Let $f\in z^r{H}_s^2$ where $r\ge k$. Since every function in $z^r{H}_s^2$ has its first r Fourier coefficients equal to zero, it follows that $P_{\mathcal{l}}f=0$ for every $0\le \mathcal{l}\le k-1$. If $p\ge q$, then by Lemma 1,

$$[U^p,V^q]=-U^{p-q}{\displaystyle \sum _{\mathcal{l}=0}^{q-1}}{P}_{\mathcal{l}},$$

and since $q-1\le k-1$, it follows that $[U^p,V^q]f=0$.

If $q>p$, then by Lemma 1,

$$[U^p,V^q]=-V^{q-p}{\displaystyle \sum _{\mathcal{l}=q-p}^{q-1}}{P}_{\mathcal{l}},$$

and again all indices satisfy $\mathcal{l}\le q-1\le k-1$, so $[U^p,V^q]f=0$. Therefore every term in the sum in (3) vanishes on f, and hence

$$[T_{\phi },T_{\psi }]f=0\mathit{for}\mathit{every}f\in z^r{H}_s^2.$$

□

4. Main Results

In this section, we apply results established in the previous section to prove the main commutativity theorem for Toeplitz operators with trigonometric polynomials symbols on Beurling subspaces of Hardy–Sobolev spaces. We also show that the obtained result is sharp.

Theorem 1. 

Let φ and ψ be trigonometric polynomials of degrees n and m, respectively, and suppose that $k=\max \{n,m\}$. Then

$$[T_{\phi },T_{\psi }]f=0\mathit{for}\mathit{every}f\in z^k{H}_s^2.$$

Proof. 

By Corollary 2, the commutator $[T_{\phi },T_{\psi }]$ vanishes on $z^r{H}_s^2$ for every $r\ge k$. Taking $r=k$ gives the desired conclusion. □

Remark 1. 

The conclusion of Theorem 1 is sharp in general. Indeed, let

$$\phi \left(z\right)=z^k,\psi \left(z\right)={\overline{z}}^k.$$

Then

$$[T_{\phi },T_{\psi }]=[U^k,V^k].$$

By Lemma 1,

$$[U^k,V^k]=-{\displaystyle \sum _{\mathcal{l}=0}^{k-1}}{P}_{\mathcal{l}}.$$

Hence

$$[U^k,V^k]e_{k-1}^{\left(s\right)}=-e_{k-1}^{\left(s\right)}\ne 0.$$

Since $e_{k-1}^{\left(s\right)}\in z^{k-1}{H}_s^2$, it follows that $[T_{\phi },T_{\psi }]$ does not vanish on $z^{k-1}{H}_s^2$. Therefore, in general, the Beurling subspace $z^k{H}_s^2$ appearing in Theorem 1 is optimal.

The converse of Theorem 1 is false in general, as shown in Example 3.

5. Examples

In this section, we present several examples illustrating the main theorem and demonstrating the sharpness of the obtained result.

Example 1. 

Let

$$\phi \left(z\right)=z^5-z^3+z^2-\sqrt{2}i,\psi \left(z\right)=z^4+z^3-z^2-z+1.$$

Then $\max \left\{\mathrm{deg}\right(\phi ),\mathrm{deg}(\psi \left)\right\}=5$, and since both φ and ψ are analytic polynomials, by Theorem 1, we have

$$[T_{\phi },T_{\psi }]f=0\mathit{for}\mathit{every}f\in z^5{H}_s^2.$$

Example 2. 

Let $p=4$ and $q=2$. Then, by Lemma 1,

$$[U^4,V^2]=-U^2(P_0+P_1).$$

Applying the above identity to the basis vectors, we obtain

$$[U^4,V^2]e_0^{\left(s\right)}=-e_2^{\left(s\right)},[U^4,V^2]e_1^{\left(s\right)}=-e_3^{\left(s\right)},$$

while

$$[U^4,V^2]e_n^{\left(s\right)}=0,n\ge 2.$$

Thus,

$$\mathrm{Ran}[U^4,V^2]=\mathrm{span}\{e_2^{\left(s\right)},e_3^{\left(s\right)}\},$$

which illustrates the conclusion of Proposition 2.

Example 3. 

Let

$$\phi \left(z\right)=z^3-z+3,\psi \left(z\right)=z^2+i.$$

Then $\max \left\{\mathrm{deg}\right(\phi ),\mathrm{deg}(\psi \left)\right\}=3$. Since both φ and ψ are analytic polynomials, we have

$$T_{\phi }{T}_{\psi }=T_{\phi \psi }=T_{\psi \phi }=T_{\psi }{T}_{\phi }.$$

Equivalently,

$$[T_{\phi },T_{\psi }]=0\mathit{on}{H}_s^2.$$

Hence,

$$[T_{\phi },T_{\psi }]f=0\mathit{for}\mathit{every}f\in z^2{H}_s^2.$$

However, $2<\max \left\{\mathrm{deg}\right(\phi ),\mathrm{deg}(\psi \left)\right\}=3$. Thus, commutativity on the Beurling subspace $z^2{H}_s^2$ does not imply that $\max \left\{\mathrm{deg}\right(\phi ),\mathrm{deg}(\psi \left)\right\}\le 2$.

6. Concluding Remarks

(i)

The present paper establishes an explicit formula for the commutator of two Toeplitz operators with trigonometric polynomial symbols on Hardy–Sobolev spaces. As a consequence, every such commutator is shown to have finite rank, leading to the main commutativity theorem on Beurling subspaces. The latter extends the corresponding commutativity result for the classical Hardy space $H^2$ established in [12]. In particular, the Hardy space case is recovered by taking $s=0$, since $H_0^2=H^2$. This shows that the commutativity phenomenon is robust under the passage from the classical Hardy space to its weighted Hardy–Sobolev analog, despite the presence of nontrivial weighted shift operators.

(ii)

The proofs presented in this paper are based primarily on the weighted shift representations of the Toeplitz operators $T_z$ and $T_{\overline{z}}$. This suggests that similar commutativity results may hold for more general weighted Hardy spaces determined by suitable weight sequences. Investigating this problem appears to be a natural direction for future research.

(iii)

A second natural direction for future research is to consider broader classes of symbols. The present paper is restricted to Toeplitz operators with trigonometric polynomial symbols. It would be of interest to investigate whether the commutativity results obtained here remain valid for bounded harmonic or more general $L^{\infty }$ symbols. This problem appears to require different techniques and remains an interesting topic for future research.