Characterizations of the product of asymmetric dual truncated Toeplitz operators

1.   Introduction

Let $\mathbb{D} = \{z: |z| < 1\}$ be the unit disk in the complex plane $\mathbb{C}$, its boundary the unit circle $\mathbb{T} = \{z: |z| = 1\}$. $L^2$ is the space of square-integrable functions on $\mathbb{T}$ with respect to the normalized Lebesgue measure d$\sigma$. It is known that $L^2$ is a Hilbert space with the inner product

Let $\ e_{n}(z) = z^n, z\in \mathbb{T}, n\in \mathbb{Z}$; then $\{e_{n}:n\in \mathbb{Z}\}$ forms a standard orthogonal basis for $L^2$. For each $h\in L^2$, it is well known that

and

The classical Hardy space $H^2$ is a closed subspace of $L^2$ consisting of $h$ with $h = \sum_{n\geq 0} \langle h, e_n \rangle e_n$. So $L^2 = H^2\oplus \overline{zH^2}$, where $\overline{zH^2} = \{ \overline{zf}: f\in H^2\}$. Since the evaluation at each point in $\mathbb{D}$ is continuous, then $H^2$ becomes a reproducing Hilbert space with the reproducing kernel given by

Let $P$ be the orthogonal projection from $L^2$ onto $H^2$; then

Denote $L^\infty$ and $H^\infty$ as the algebras of bounded functions in $L^2$ and $H^2$, respectively. Define the Toeplitz operator $T_{\varphi}$ on the Hardy space $H^2$ with symbol $\varphi\in L^\infty$ by

It is obvious that $T_\varphi$ is a bounded linear operator on $H^2$.

If $\theta \in H^\infty$ has $|\theta| = 1$ almost everywhere on the unit circle $\mathbb{T}$, then $\theta$ is called an inner function, and the corresponding model space $K_\theta$ is the orthogonal complement of $\theta H^2$ in $H^2$, i.e., $K_\theta = H^2\ominus\theta H^2$. It is known that for inner functions $u$ and $v$, $K_{uv} = K_u\oplus uK_v$. The shift operator $S$ is defined by $Sf(z) = zf(z)$; its adjoint operator is called the backward unilateral shift operator, which is $S^\ast f(z) = \frac{f(z)-f(0)}{z}$. The model space is an invariant subspace of the backward unilateral shift operator $S^\ast$, and also a reproducing kernel Hilbert space whose reproducing kernel is

Since $k_{w}^{\theta}$ is bounded, the set $K_{\theta}^{\infty} = K_{\theta}\cap H^{\infty}$ is dense in $K_{\theta}$.

For $\varphi\in L^\infty$, the truncated Toeplitz operator (TTO) $A_\varphi^\theta$ systematically studied by Sarason in ^[7] is defined on the model space $K_\theta$ by

where $P_\theta$ is the orthogonal projection from $L^2$ onto $K_\theta$. As is known to all, TTO is a natural generalization of Toeplitz matrices that appear in many contexts, such as in the study of finite-interval convolution equations, signal processing, control theory, probability, and diffraction problems ^[4,5,7]. Actually, $A_\varphi^\theta$ is the compression of $T_\varphi$ on the model space $K_\theta$, i.e., $A_\varphi^\theta = P_\theta T_{\varphi}|_{K_\theta}$. Sedlock ^[8] has ever defined the Sedlock class to study the product problem of truncated Toeplitz operators. For more information about model spaces and their operators, one is referred to ^[4].

Notice that

It is easy to see that $\{e_{-n}: n\geq 1\}$ and $\{\theta e_{n}: n\geq 0\}$ are standard orthonormal bases for $\overline{zH^{2}}$ and $\theta H^{2}$, respectively, and so

forms a standard orthonormal basis for $K_{\theta}^{\bot}$.

Let $Q_{\theta} = P_\theta^\perp = I - P_{\theta}$ be the orthogonal projection from $L^2$ onto $K_{\theta}^{\perp}$. So $Q_ \theta f = Qf + \theta P(\overline \theta f)$, where $Q = I - P$. In 2018, Ding and Sang ^[3] introduced the dual truncated Toeplitz operator (DTTO), which is defined on $K_{\theta}^{\perp}$ by

or written as

It is clear that $(D_{\varphi}^{\theta})^{*} = D_{\overline{\varphi}}^{\theta}$. Furthermore, for any complex constant $\lambda$, we have $D_{\lambda}^{\theta} = \lambda I$.

Câmara ^[2] discussed the asymmetric dual truncated Toeplitz operator (ADTTO), which is defined on $K_{\theta}^{\perp}$ by

Also, $(D_{\varphi}^{\theta, \alpha})^{*} = D_{\overline{\varphi}}^{\alpha, \theta}$. DTTO and ADTTO, acting on these spaces, have realizations, for example, in long–distance communication links with several regenerators along the path that cancel low–frequency noise using high–pass filters, or in the description of wave propagation in the presence of finite–length obstacles.

In 1964, Brown and Halmos ^[1] proved that a bounded operator $A$ on $H^2$ is a Toeplitz operator if and only if $A-S^{*}AS = 0$. This operator equation plays a significant role in the study of Toeplitz operators and related topics. In 2007, Sarason proved a result on truncated Toeplitz operators in ^[7] that is similar to Brown and Halmos: Let $S_{\theta} = A_z^\theta$, the compressed shift operator on $K_ \theta$, then a bounded operator $A$ on $K_{\theta}$ is a truncated Toeplitz operator if and only if $A-S_{\theta}^{*}AS_{\theta}$ is at most a rank-2 operator, more precisely,

for some $\psi, \chi\in K_{\theta}$, where

In 2021, Gu ^[6] studied the dual truncated Toeplitz operators, and obtained that a bounded operator $A$ on $K_{\theta}^{\perp}$ is a dual truncated Toeplitz operator if and only if $A-D_{\overline{z}}^\theta AD_{z}^{\theta}$ is at most a rank-2 operator and

for some $\varphi \in L^{\infty}$, where $D_z^ \theta$ is the compressed shift on $K_ \theta^\perp$. This result is similar to Sarason's.

For the product problem of when two Toeplitz operators are another Toeplitz operator, Brown and Halmos in ^[1] established a necessary and sufficient condition based on the above operator equation characterization of the Toeplitz operator. In 2011, N. Sedlock did the same thing for truncated Toeplitz operators in ^[8].

Inspired by the above work, in this paper, we will establish an operator equation to obtain an equivalent characterization of the ADTTO, which is similar to Sarason's; see Theorem 3.1. Based on this, we follow a method taken in Sedlock's paper ^[8] and study the product problem of when the product of two ADTTOs with symbols in model spaces is another ADTTO; see Theorem 4.1.

2.   preliminaries

In what follows, for $g\in L^2(\mathbb{T})$, we write $g_n$ ($n\in\mathbb{Z}$) as the $n$-th Fourier coefficient $\langle h, e_n \rangle$ of $g$, unless otherwise stated.

The following lemmas come from ^[6].

Lemma 2.1. For any function $h\in K_{\theta}^{\perp}$, we have

Lemma 2.2. On $K_ \theta^\perp$, we have

We also need the following result, which says that an ADTTO satisfies an operator equation.

Lemma 2.3. The operator $D_{\varphi}^{\theta, \alpha}$ satisfies the following equation.

where

Proof. For $h\in K_{\theta}^{\bot}\ominus\text{span}\{e_{-1}\}$, using Lemma 2.1, direct calculations show that

Set $r = (D_{\varphi}^{\theta, \alpha}- D^{\alpha}_{\overline{z}}D_{\varphi}^{\theta, \alpha}D_{z}^{\theta})[\overline{z}]$, then for $h+ \overline{zc}$, where $h\in K_{\theta}^{\bot}\ominus\text{span}\{e_{-1}\}$ and $c$ is a constant, we have

Now we calculate $r = (D_{\varphi}^{\theta, \alpha} - D^{\alpha}_{\overline{z}}D_{\varphi}^{\theta, \alpha}D_{z}^{\theta})[\overline{z}]$. Since

thus, we obtain

Substituting the above into (2.1) proves the result. □

When $\theta = \alpha$, the above result becomes the following one, which was obtained by Gu ^[6].

Corollary 2.1. The operator $D_{\varphi}^{\theta}$ satisfies the following equation:

where

It is remarked passing that Câmara ^[2] also obtained that

It is not true. The following is a counterexample.

Example 2.1. When $\theta = \alpha, \ \theta_0\theta_1\neq0$ and $\varphi = \bar{z}$, then using Lemma 2.1 we obtain

and

so it is clear that

Using a proof similar to Lemma 2.3, we can show that the operator $D_{\varphi}^{\theta, \alpha}$ satisfies the following equations:

and

where $R_{\varphi}^{\alpha, \theta} = \ P_{\theta}^{\bot}[\overline{\varphi z}(\alpha-\alpha_0)].$ The first equation was also obtained by Câmara using a different method in ^[2].

3.   Characterization when an operator being an ADTTO

We denote $\mathcal{B}(K_{\theta}^{\bot}, K_{\alpha}^{\bot})$ as the set of all bounded linear operators from $K_{\theta}^{\bot}$ to $K_{\alpha}^{\bot}$. For $A\in \mathcal{B}(K_{\theta}^{\bot}, K_{\alpha}^{\bot})$, suppose $A = D_{\bar{z}}^{\alpha}AD_{z}^{\theta}$. Note that

then we have

and for each $n\geq0$, $m\geq1$,

The above observations will be used frequently in the proof of the following result.

Proposition 3.1. Let $A\in \mathcal{B}(K_{\theta}^{\bot}, K_{\alpha}^{\bot})$ and $A-D_{ \overline{z}}^{\alpha}AD_{z}^{\theta} = 0$.

$(a)$ If $\alpha_0 = 0$, then there exist $\psi\in H^{\infty}$ and $\omega\in L^{\infty}$ such that $A = F_{\omega}+G_{\psi}$, where $F_{\omega}$ and $G_{\psi}$ are defined by

for $h\in K_{\theta}^{\bot}$ and $\psi = P\omega$.

$(b)$ If $\alpha_0 \neq 0$, then there exist $\psi, \omega \in H^{\infty}$ such that $A = F_{\omega} + G_{\psi}$, where $F_{\omega}, G_{\psi}$ are defined as follows: for $h \in K_{\theta}^{\bot}$,

In particular, $A\theta = \overline{z \omega }+\alpha \psi$.

Proof. $(a)$ $\alpha_0 = 0$. For this case, $D_{z}^{\alpha}e_{-1} = 0$. There are two cases to be discussed:

(a1) $\theta_0 = 0$. In this case, $D_{z}^{\theta}e_{-1} = 0$ and $(D_{z}^{\theta})^{*} \theta = 0$. Then by (3.1) we have

so $A|_{\overline{zH^2}} = 0.$ Furthermore, for $n\geq0, \ m\geq1$, by (3.2) we obtain

So we can regard $A$ as a bounded operator from $\theta H^2$ to $\alpha H^2$. For $n\geq 0$, there exists a function sequence $\{h_n\}\subset H^2$, such that

Define a linear operator $B$ on $H^2$ by $Be_n = h_n$. Since $\theta$ and $\alpha$ are inner functions, $B$ is bounded, and $||B|| = \ ||A||$. For $A$,

So $h_n = P[\bar{z}h_{n+1}]$, and therefore

It implies that $B = T_{z}^{*}BT_{z}$. By the Brown-Halmos theorem ^[1], it tells that $B$ is a Toeplitz operator, so there exists $\omega \in L^{\infty}$ such that $B = T_{\omega}$. Hence $Ah = \alpha P[\omega \overline{\theta} Ph]$.

(a2) $\theta_0\neq0$. For this case, when $n\geq0$ and $m\geq1$, by (3.3) we may regard $A$ as a bounded operator from $K_{\theta}^{\bot}$ to $\alpha H^2$, so there exists $\psi \in H^2$ such that $A\theta = \alpha \psi$. Note that $D_z^\theta e_{-1} = \overline{\theta_0} \theta$, then for $n\geq 1$, by (3.1) again we have

Write $\psi = \sum\limits_{k \geq 0}a_k e_k$. Notice that

and

for $k\geq1$, combining with (3.4), we obtain

Therefore, for $h\in \overline{zH^2}$, we have $Ah = \overline{\theta_0}\alpha P[\psi h]$.

When $h\in \theta H^2$, similar to (a1), it can be shown that there exists $\omega \in L^\infty$ such that $Ah = \alpha P[\omega \overline{\theta} Ph]$. Hence, we obtain that

The above gives $A \theta = \alpha P\omega$, by $A \theta = \alpha\psi$ obtained before, we get $\psi = P\omega$.

(b) We first suppose $\alpha_0\neq 0$ and $\theta_0\neq 0$. In this case, by Lemma 2.1, we see that $D_{\overline{z}}^{\alpha}e_{-n} = e_{-(n+1)}$ for $n\geq 1$; $D_{ \overline z}^{\alpha} \alpha = \alpha_0 e_{-1}$; $D_{\overline{z}}^{\alpha} (\alpha e_{m+1}) = \alpha e_{m}$ for $m\geq 0$. So $D_{\overline{z}}^{\alpha}$ is invertible, and

Let

where $\omega, \ \psi\in H^2$.

Assume that $A\theta = \overline{z\omega}$, $\omega \in H^\infty$. For $n\geq 1$, like (3.4), it has

Therefore, for $g \in H^2$, we have

Because

we have

So

Therefore,

for $f\in H^2$.

For $h\in K_{\theta}^{\bot}$, $h = \overline{zg}+\theta f = Qh+Ph$, we get

denote the right side of the above equation as $F_{\omega}h$.

Now, assuming $A\theta = \alpha \psi$, where $\psi \in H^{\infty}$, and for $n \geq 1$, by (3.4) we have

Therefore,

for $g\in H^2$. On the other hand, for $n\geq 0$, we obtain that

Hence, $A[\theta f] = \alpha \psi \bar{\theta} \theta f$ for $f\in H^2$.

For $h\in K_{\theta}^{\bot}$, $h = \overline{zg}+\theta f = Qh+Ph$, it induces that

denote the right side of the above equation as $G_{\psi}h$.

Therefore, it follows that $A = F_{\omega}+G_{\psi}$ and $A\theta = \overline{z\omega}+\alpha\psi$. The proof is similar for the case of $\alpha_0\neq 0$ and $\theta_0 = 0$. □

By Proposition 3.1, we have the following result.

Corollary 3.1. Let $A\in \mathcal{B}(K_{\theta}^{\bot}, K_{\alpha}^{\bot})$.

$(a)$ If $\alpha_0 = 0$, then $A = 0$ if and only if $A-D_{ \overline{z}}^{\alpha}AD_{z}^{\theta} = 0$, $A\theta = 0$, and $A^{*}\alpha = 0$;

$(b)$ If $\alpha_0\neq 0$, then $A = 0$ if and only if $A-D_{\overline{z}}^{\alpha}AD_{z}^{\theta} = 0$ and $A\theta = 0$.

Proof. It only needs to show the sufficiency.

(a) $\alpha_0 = 0$. The proof is divided into the following two cases.

(a1) Suppose $\theta_0 = 0$. By Proposition 3.1, we have

where $h\in K_{\theta}^{\bot}$. Therefore, $A = PD^{\theta, \alpha}_{\omega \alpha \bar{\theta}}P$ and $A^{*} = P D^{\alpha, \theta}_{\overline{\omega \alpha} \theta}P$. Since $A\theta = \alpha P[\omega] = 0$ and $A^{*}\alpha = \theta P[\overline{\omega}] = 0$, we have $\omega = 0.$ Thus, $A = 0$.

(a2) Suppose $\theta_0\neq 0$. In this case, by Proposition 3.1 we have

and $A\theta = \alpha \psi$. So by $A \theta = 0$, we have $\psi = 0$. Then, $Ah = \alpha P[\omega \overline{\theta} Ph]$ for $h\in K_ \theta^\perp$. Similar to (a1), we can obtain $\omega = 0$. Hence, $A = 0$.

(b) $\alpha_0 \neq 0$. By Proposition 3.1, $A\theta = \overline{z\omega} + \alpha \psi = 0$, thus we have $\omega = 0$ and $\psi = 0$, which induces $A = 0$. □

Let

where

It follows from Lemma 2.3 that $A = D_{\varphi}^{\theta, \alpha}$ satisfies the above equation, then by Proposition 3.1, we can easily obtain the following theorem.

Theorem 3.1. Let $A\in \mathcal{B}(K_{\theta}^{\bot}, K_{\alpha}^{\bot})$ and $\varphi \in L^{\infty}$.

$(a)$ If $\alpha_0 = 0$, then $A = D_{\varphi}^{\theta, \alpha}$ if and only if $A$ satisfies (3.5), $A\theta = D_{\varphi}^{\theta, \alpha}\theta$ and $A^{*}\alpha = (D_{\varphi}^{\theta, \alpha})^{*}\alpha;$

$(b)$ If $\alpha_0 \neq 0$, then $A = D_{\varphi}^{\theta, \alpha}$ if and only if $A$ satisfies (3.5) and $A\theta = D_{\varphi}^{\theta, \alpha}\theta.$

It is remarked that in ^[2], the authors also obtained the characterization for ADTTO with different presentations.

4.   The product problem of two ADTTOs for certain symbols

For the inner function $\theta$, we define a class of conjugation linear operators $C_{\theta}: L^2 \rightarrow L^2$ by

which satisfies that $\langle C_{\theta}f, C_{\theta}g \rangle = \langle g, f \rangle$, and $(C_{\theta})^2 = I$. According to the definition of $C_{\theta}$, we have

Hence, it is clear $C_{\theta}K_{\theta} = K_{\theta}, \ C_{\theta}(\theta H^2) = \overline{zH^2}$ and $C_{\theta}(\overline{zH^2}) = \theta H^2.$

Before we present the result of the product problem of when the product of two ADTTOs is another ADTTO for certain symbols, we first give the following lemma.

Lemma 4.1. Suppose $\varphi, \psi \in H^\infty$, $h\in L^\infty$, and $K_{\alpha} \subseteq K_{\gamma} \subseteq K_{\theta}$. If

then $h = \varphi \psi$.

Proof. First note that $K_{\alpha} \subseteq K_{\gamma} \subseteq K_{\theta}$ means that $\gamma/\alpha$, $\theta/\alpha$, and $\theta/\gamma$ all are inner functions.

Notice that $D_{\varphi}^{\gamma, \alpha}D_{\psi}^{\theta, \gamma}[\theta] = \varphi \psi \theta$ and $D_{h}^{\theta, \alpha}[\theta] = Q[h\theta]+ \alpha P[\overline{\alpha} h \theta]$, so it is obvious that $h\theta \in H^2$ and $P[\overline{\alpha} h \theta] = \overline{\alpha} \varphi \psi \theta$.

Let $\overline{\alpha} h \theta = \overline{\alpha} \varphi \psi \theta + \overline{zf} +\overline{zg}$, where $f \in K_{\alpha}$, $g \in \alpha H^2$, then

Because $h\theta \in H^2$, the above means that $C_\alpha g\in H^2$, which gives $g = 0$ since $C_\alpha g\in \overline{zH^2}$. Thus

Also, it is noted that

So $Q[h\overline{z}] = \varphi_0 \psi_0 \overline{z}$, and we see that $h\in H^2$ and $h_0 = \varphi_0 \psi_0$. Now by (4.1),

Since $\overline{z} \overline{\theta / \alpha} \overline{f} \in \overline{zH^2}$ and $h \in H^2$, it has $f = 0$ and hence $h = \varphi \psi$, so we obtain the desired conclusion. □

It is worth noting that we can use the result of Ding [3, Theorem 4.7] and Lemma 4.1 to obtain the following characterization of the product problem for DTTOs.

Corollary 4.1. For $\varphi, \psi \in H^\infty$, $D_\varphi^\theta D_\psi^\theta = D_h^\theta$ if and only if there exists $\lambda \in \mathbb{C}$ such that $\overline{\varphi}(\theta-\lambda), \overline{\psi}(\theta-\lambda), \overline{\varphi \psi}(\theta-\lambda)\in H^2$ or one of $\varphi$ and $\psi$ is a constant, in which case $h = \varphi \psi$.

We are ready to solve the product problem of two ADTTOs with certain analytic symbols.

Theorem 4.1. Let $\varphi, \psi \in K_{\alpha} \subseteq K_{\gamma} \subseteq K_{\theta}$ and $\varphi, \psi\in H^\infty$, $q^{\theta, \alpha} = \alpha P[\overline{\alpha} \theta \overline{z}]$.

$(a)$ If $\alpha_0 = 0$, then $D_{\varphi}^{\gamma, \alpha}D_{\psi}^{\theta, \gamma} = D_{h}^{\theta, \alpha}$ for some $h\in L^\infty$ if and only if $h = \varphi \psi \in K_{z\alpha}$.

$(b)$ If $\alpha_0 \neq 0$, then $D_{\varphi}^{\gamma, \alpha}D_{\psi}^{\theta, \gamma} = D_{h}^{\theta, \alpha}$ for some $h\in L^\infty$ if and only if $h = \varphi \psi \in K_{z\alpha}$ and

Proof. First, suppose that $D_{\varphi}^{\gamma, \alpha}D_{\psi}^{\theta, \gamma} = D_{h}^{\theta, \alpha}$. We notice that when $\varphi, \psi \in K_{\alpha} \subseteq K_{\gamma} \subseteq K_{\theta}$,

and

So by (3.5) we obtain

By Lemma 2.2, we have

Making use of (4.3) and (4.4), it follows that

where $\phi$ denotes

which satisfies $\phi\perp e_{-1}$. Also,

By (4.6) and the above equation, (4.5) becomes that

where $\Phi$ denotes

By Lemma 2.4, we see that $h = \varphi\psi$. In this case, we have

and

So

Now, by comparing the equalities (4.7) and (4.8), we obtain

and

The above is (4.2).

If $\alpha_0\ne 0$, then (4.9) gives that

Simple computation shows that it is

or

which is equivalent to that $\alpha \overline{\varphi\psi}\in H^2$. Thus $\varphi\psi\in\alpha\overline{ H^2} = zK_\alpha\oplus\overline{ H^2},$ which implies that $\varphi\psi\in K_{z\alpha}$. Hence, by Theorem 3.1, we see (b) holds.

If $\alpha_0 = 0$, then also $\theta_0 = 0$ since $\theta/\alpha$ is an inner function. In this case, the equality (4.9) holds naturally and (4.10) yields that $P[\overline{\theta z}\varphi \psi] = 0$, to obtain $\theta\overline{\varphi\psi}\in H^2$. On the other hand, it is easily seen that $(D_{\varphi}^{\gamma, \alpha}D_{\psi}^{\theta, \gamma})^{*}\alpha = (D_{\varphi\psi}^{\theta, \alpha})^{*}\alpha$ is

or $Q(\overline{\varphi\psi}\alpha) = 0$, that is, $\alpha \overline{\varphi \psi}\in H^2$. Also by Theorem 3.1, we have (a).

It is easy to see the converse holds. We finish the proof. □

We notice that when $\theta = \gamma = \alpha$, it has $q^{\theta, \alpha} = q^{\theta, \gamma} = q^{\gamma, \alpha} = 0$, so we can derive quickly a result: Let $\varphi, \psi \in K_{\theta}$, then $D_{\varphi}^{\theta}D_{\psi}^{\theta} = D_{h}^{\theta}$ if and only if $h = \varphi \psi \in K_{z\theta}$.

Obviously, it is a special case of Corollary 4.1 for the dual truncated Toeplitz operators.

The following corollary is also obvious.

Corollary 4.2. Let $\varphi, \psi \in K_{\alpha} \subseteq K_{\gamma} \subseteq K_{\theta}$ and $\varphi, \psi\in H^\infty$, then $D_{\varphi}^{\gamma, \alpha}D_{\psi}^{\theta, \gamma} = 0$ if and only if $\varphi = 0$ or $\psi = 0$.

5.   Conclusions

In this paper, we studied ADTTO acting on the orthogonal complement of two different model spaces. More precisely, we characterized when a given operator is an ADTTO with an operator equation. As applications of this result, we solved the product problem of two ADTTOs with certain analytic symbols. In future work, we will investigate the product problem of two ADTTOs with general symbols.

Author contributions

Zhenhui Zhu: Writing-original draft; Qi Wu: Writing-review and editing; Yong Chen: Supervision, Writing-review and editing, Funding acquisition. All authors have read and approved the final version of the manuscript for publication.

Use of Generative-AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The third author was supported by NSFC (12271134).

Conflict of interest

The authors declare that there are no conflicts of interest.