Dual Toeplitz operators on the orthogonal complement of the harmonic Bergman space

1.   Introduction

Throughout this paper, we use $\mathbb{D}$ to denote the open unit disk on the complex plane $\mathbb C$. Let

be the Lebesgue area measure on $\mathbb{D}$, normalized so that the measure of $\mathbb D$ is $1$. It is well-known that $L^2(\mathbb{D}, dA)$ is the Hilbert space of square integrable functions with the inner product

The Bergman space $L_a^2$ is the closed subspace of $L^2(\mathbb D, dA)$ consisting of analytic functions on $\mathbb D$, which is a reproducing kernel Hilbert space and its reproducing kernel (at $\lambda\in \mathbb D$) is given by

Letting $P$ be the orthogonal projection from $L^2(\mathbb{D}, dA)$ onto $L_a^2,$ we have

for every $f\in L^2(\mathbb{D}, dA)$ and $z\in \mathbb D$. For more details about the function theory and operator theory on the Bergman space, one can consult Zhu's book ^[1].

The harmonic Bergman space $L_h^2$ is the closed subspace of $L^2(\mathbb D, dA)$ consisting of all complex-valued harmonic functions on $\mathbb D$. Observe that $L_h^2$ can be decomposed as

where $\overline{L_a^2}$ denotes the complex conjugate of $L_a^2$. It is easy to check that the function

is the reproducing kernel (at $\lambda\in \mathbb D$) for the harmonic Bergman space $L_h^2$. Denoting the orthogonal projection from $L^2(\mathbb{D}, dA)$ onto $L_h^2$ by $Q$, then

for each $f\in L^2(\mathbb{D}, dA)$ and $z\in \mathbb D$. Using (1.5), we obtain that

for all $f\in L^2(\mathbb{D}, dA)$ and $z\in \mathbb D$. According to (1.7), routine calculations yield that

Additionally, we refer to the paper ^[2] for more knowledge about the harmonic Bergman space.

The Toeplitz operator and the Hankel operator with symbol $\varphi\in L^{\infty}(\mathbb{D}, dA)$ (the collection of all essentially bounded functions on the unit disk) on the harmonic Bergman space $L_h^2$ are defined by

and

respectively. Under the decomposition

the multiplication operator $M_{\varphi}$ with symbol $\varphi$ can be represented as

where the operator $S_\varphi$ is defined by

In fact, $S_\varphi$ is bounded and linear on $(L_h^2)^{\perp}$ when $\varphi\in L^\infty(\mathbb D, dA)$. The operator $S_\varphi$ is called the dual Toeplitz operator with symbol $\varphi.$ The following elementary properties of dual Toeplitz operators on $(L_a^2)^{\bot}$ can be founded in ^[3]:

(1) $\|S_\varphi\|\leqslant \|\varphi\|_\infty$;

(2) $S_{\varphi}^{*} = S_{\overline{\varphi}};$

(3) $S_{\alpha\varphi+\beta\psi} = \alpha S_{\varphi}+\beta S_{\psi}$ for all $\varphi, \psi\in L^{\infty}(\mathbb{D}, dA)$ and all complex constants $\alpha, \beta$;

one can easily verify that the above conclusions also hold for dual Toeplitz operators on $(L_h^2)^{\bot}$ using the definition of $S_{\varphi}$.

The concept of "dual Toeplitz operator" was first introduced and investigated by Stroethoff and Zheng on the orthogonal complement of the Bergman space; see ^[3,4]. Since then, researchers have extended the spectral theory and algebraic properties of dual Toeplitz operators on $(L_a^2)^{\bot}$ established in ^[3] to the setting of dual Toeplitz operators on the orthogonal complements of various function spaces. For instance, the Bergman space over the unit ball (^[5,6,7]), the Bergman space over the polydisk (^[8]), the Dirichlet space (^[9,10,11]), the harmonic Bergman space (^[12,13]), and the harmonic Dirichlet space (^[14]).

Recently, the investigation concerning dual Toeplitz operators on the orthogonal complement of the harmonic Bergman space $(L_h^2)^{\bot}$ has attracted the attention of many scholars. In 2015, Yang and Lu ^[13] obtained a complete characterization for the commuting dual Toeplitz operators on $(L_h^2)^{\bot}$ with bounded harmonic symbols. However, the corresponding commutativity problem for dual Toeplitz operators with nonharmonic symbols is still open. In 2021, Peng and Zhao ^[12] characterized the boundedness, compactness, spectral structure, and algebraic properties of dual Toeplitz operators on $(L_h^2)^{\perp}$. In addition, Wang and Zhao ^[15] established a necessary and sufficient condition for dual Toeplitz operators with nonharmonic symbols of the form

to be hypo-normal on $(L_h^2)^{\bot}$, where $n_1, n_2, m_1, m_2$ are nonnegative integers and $a, b$ are complex numbers.

Although many scholars have studied the properties of dual Toeplitz operators, there are few results on dual Toeplitz operators with nonharmonic symbols. In this paper, we try to study when the product of two dual Toeplitz operators with radial symbols equals zero and when two dual Toeplitz operators with nonharmonic symbols commute on the orthogonal complement of the harmonic Bergman space. As the function theory of $(L_h^2)^{\bot}$ is much more complicated than that of $(L_a^2)^{\bot}$, it is quite difficult to solve the zero-product problem and the commutativity problem mentioned above in general cases. In order to seek the breakthrough point of those two problems, in the present paper we consider some special radial symbols and quasi-homogeneous symbols, and give certain partial answers.

The organization of this paper is as follows. In Section 2, we will show that there exists an index $k$ such that $\varphi_k = 0$ a.e. on $\mathbb D$ if

in the case of

where $N$ is an arbitrary positive integer and each $a_{k, m}$ is a constant. Moreover, in Section 3 we give a characterization for

on $(L_h^2)^{\bot}$ if

and

where $a, b\in\mathbb{C}$ such that $a\neq b$ and $p_1$, $p_2$, $q_1$, $q_2$, $s$, $t$ are all nonnegative integers.

2.   The zero-product problem for dual Toeplitz operators

In this section, we investigate the zero-product problem concerning dual Toeplitz operators with some special symbols on the orthogonal complement of the harmonic Bergman space $(L_h^2)^{\bot}$ via the symbol map, which was established in Lemma 2.1 [12,Theorem 4.1]. For the sake of completeness, we state the result on the symbol map of dual Toeplitz operators on $(L_h^2)^{\bot}$ as follows:

Lemma 2.1. There is a contractive $C^*$-homomorphism $\rho$ from the dual Toeplitz algebra $\mathfrak{T}(L^\infty(\mathbb D, dA))$ to $L^\infty(\mathbb D, dA)$ such that $\rho(S_\varphi) = \varphi$ for each $\varphi\in L^\infty(\mathbb D, dA)$.

Let us begin with dual Toepitz operators with bounded harmonic symbols.

Proposition 2.2. Suppose that $\varphi_1, \varphi_2, \cdots, \varphi_{N}$ are $N$ bounded harmonic functions on the unit disk $\mathbb D$. If

then there exists $k\in \{1, 2, \cdots, N\}$ such that $\varphi_k = 0$.

Proof. By Lemma 2.1, we have that

if

Then, applying the uniqueness theorem of harmonic functions, we deduce that there exist some $k\in \{1, 2, \cdots, N\}$ such that $\varphi_k = 0$. □

In the following proposition, we solve the zero-product problem for two dual Toeplitz operators with symbols of the form $\sum\limits_{n = 0}^{\infty}c_n|z|^n$, where each $c_n$ is a complex constant.

Proposition 2.3. Suppose that

and

are two bounded functions on the unit disk $\mathbb D$. If $S_{\varphi}S_{\psi} = 0$, then $\varphi = 0$ or $\psi = 0$.

Proof. If $S_{\varphi}S_{\psi} = 0,$ then we have by Lemma 2.1 that $\varphi\psi = 0$. Note that

where

Thus, we have $c_k = 0$ for all $k\geqslant 0.$ In particular, $c_{0} = a_0b_0 = 0.$ Then, we consider the following three cases:

Case 1. $a_0 = 0, b_0\neq 0.$ In this case, we have that

Continuing this process, we obtain that

for $k = 1, 2, \cdots$. This yields that $a_n = 0$ for all $n\geqslant 0.$

Case 2. $a_0\neq 0, b_0 = 0.$ In this case, we get that

Using the same argument as the one used in Case 1, we conclude that

for $k = 1, 2, \cdots$. It follows that $b_m = 0$ for all $m\geqslant 0.$

Case 3. $a_0 = 0$ and $b_0 = 0.$ In this case, we have

and

Then, repeating the arguments used in the proof of Cases 1 and 2, we can show that $\varphi = 0$ or $\psi = 0$.

This completes the proof of Proposition 2.3. □

The next theorem shows that the previous proposition can be generalized to the case of the product of arbitrary finitely many dual Toeplitz operators.

Theorem 2.4. Suppose that

which are $N$ bounded functions on the unit disk $\mathbb D$. If

then there exist some $k\in \{1, 2, \cdots, N\}$ such that $\varphi_{k} = 0$.

Proof. Since

we again conclude by Lemma 2.1 that

on the disk $\mathbb D$. Observe that

and $\varphi_2\varphi_3\cdots \varphi_N$ can be written as follows:

Applying the conclusion of Proposition 2.3, we obtain that

If $\varphi_1 = 0$, then we are done. Otherwise, we have that

By using the same method as the one used in the previous paragraph, we deduce that

Then, repeating this process yields the desired result. □

3.   The commutativity problem for dual Toeplitz operators

In this section, we mainly study the commuting dual Toeplitz operators with some special quasi-homogeneous symbols on the orthogonal complement of the harmonic Bergman space $(L_h^2)^{\bot}$. To do so, we need the following lemma, which can be proven by direct calculations.

Lemma 3.1. For any positive integers $s, m, n$ with $m > n$, we have

(1) $Q(|z|^sz^n \overline{z}^m) = \frac{2(m-n+1)}{s+2m+2}\overline{z}^{m-n};$

(2) $Q(|z|^s\overline{z}^{n}) = \frac{2(n+1)}{s+2n+2}\overline{z}^{n};$

(3) $Q(|z|^sz^{n}) = \frac{2(n+1)}{s+2n+2}z^{n};$

(4) $Q(z^n\overline{z}^m) = \frac{m-n+1}{m+1}\overline{z}^{m-n};$

(5) $Q(z^m\overline{z}^n) = \frac{m-n+1}{m+1}z^{m-n}.$

To study the commuting problem for the dual Toeplitz operators on the orthogonal complement of the harmonic Bergman space, we first consider the simplest radial symbol $|z|^n$.

Proposition 3.2. Suppose that

where $s, t$ are positive integers. Then,

if, and only if, $\varphi = \psi$.

Proof. The sufficiency is obvious, and we need only to prove the necessity. By (1.8) in Section 1, we see that

Since

we have

Elementary calculations give us that

and

Similarly, we have that

Combining (3.1), (3.3), and (3.4) gives

If $s\neq t$, then the coefficients of $|z|^t\overline{z}$ and $|z|^s\overline{z}$ are zero. This implies that

and

It follows that $s = t = 0$, which is a contradiction, completing the proof. □

The next theorem shows that Proposition 3.2 can be extended to a general case.

Theorem 3.3. Let

where $p, q, s, t$ are all nonnegative integers. Then, $S_{\varphi}S_{\psi} = S_{\psi}S_{\varphi}$ if, and only if, one of the following conditions holds:

$(1) \; \varphi = c\psi$ for some constant c;

$(2) \; \varphi$ and $\psi$ are both analytic;

$(3) \; \varphi$ and $\psi$ are both co-analytic;

$(4)$ Either $\varphi$ or $\psi$ is constant.

In order to simplify the proof of Theorem 3.3, we require the following lemma:

Lemma 3.4. Let

where $s, t, p, q\geqslant 0$, and $a, b\in\mathbb{C}.$ If there exist some $M$ such that $f(x)\equiv 0$ when $x > M,$ then $at-bq = 0.$

Proof. Since

and

we have

Let

Then, $g$ is a polynomial and (3.11) implies that $g$ has infinitely many zeros. Thus, $g\equiv 0$ and, hence, the coefficient of $x^3$ is zero, i.e., $at-bq = 0.$ □

Now we are ready to prove Theorem 3.3.

Proof of Theorem 3.3. The sufficiency is obvious, so we only need to show the necessity. Since

we have

for all integers $m$ satisfying

Notice that

where the third equality follows from Lemma 3.1. Moreover, we have

On the other hand,

It follows from (3.13) that

Next, we need to consider the following five cases:

Case 1. $p\neq s, \; p\neq0$, and $s\neq0$. Then, the coefficients of $|z|^{2p}\overline{z}^{m-s+t+q-p-1}$ and $|z|^{2s}\overline{z}^{m-s+t+q-p-1}$ are both zero, which implies that

By Lemma 3.4, we obtain that $t = q = 0.$ Hence, $(2)$ holds.

Case 2. $p = s, \; p\neq0$, and $s\neq0$. In this case, we have

This gives that the coefficient of $|z|^{2s}\overline{z}^{m-p+t+q-s-1}$ is zero, i.e.,

Using Lemma 3.4, we conclude that $t = q$. Hence, $(1)$ holds.

Case 3. $p\neq0$ and $s = 0$. Then, the coefficient of $|z|^{2p}\overline{z}^{m-s+t+q-p-1}$ is zero. Hence, we have

Applying Lemma 3.4, we get $t = 0,$ i.e., $\psi$ is constant.

Case 4. $s\neq0$ and $p = 0$. Then, the coefficients of $|z|^{2s}\overline{z}^{m-s+t+q-p-1}$ are zero, which is equivalent to

It follows from Lemma 3.4 that $q = 0$. This implies that $\varphi$ is constant.

Case 5. $s = p = 0$. This case is trivial.

This completes the proof of Theorem 3.3. □

In the rest of this section, we will study the commutativity problem for dual Toeplitz operators with symbols of the form $az^{p_1}\overline{z}^{q_1}+bz^{p_2}\overline{z}^{q_2}$ and $z^{s}\overline{z}^{t}$, where $p_1$, $p_2$, $q_1$, $q_2$, $s$, and $t$ are all nonnegative integers. To this end, we still require a number of lemmas as follows:

Lemma 3.5. Let

where $s, t, p_1, p_2, q_1, q_2$ are nonnegative integers. For $m\in\mathbb N$ large enough, we have

Proof. This can be proven easily by elementary computations. □

Lemma 3.6. Let $\varphi$ and $\psi$ be the functions as in Lemma 3.5. For $m\in\mathbb N$ large enough, we have

Proof. This is a direct conclusion of Lemma 3.5. □

In view of Lemma 3.5, we conclude that

for all $x\in (0, 1)$ if $S_{\varphi}S_{\psi} = S_{\psi}S_{\varphi}$, where the coefficients are given by:

Notice that (3.25) is equivalent to

Similarly, we have by Lemma 3.6 that

for all $x\in (0, 1)$ whenever

where

Clearly, (3.27) is equivalent to

Using the same method as the one used in the proof of Lemma 3.4, we list without proof five lemmas as follows:

Lemma 3.7. Let

where $s, t, p_2, q_2$ are all nonnegative numbers. If there exist some $M$ such that $f(x)\equiv 0$ when $x > M,$ then $t = q_2.$

Lemma 3.8. Let

where $s, t, p_1, p_2, q_1, q_2$ are all nonnegative numbers. If there exist some $M$ such that $f(x)\equiv 0$ when $x > M,$ then

Lemma 3.9. Let

where $s, t, p_1, p_2, q_1, q_2$ are all nonnegative numbers. If there exist some $M$ such that $f(x)\equiv 0$ when $x > M,$ then

Lemma 3.10. Let

where $s, t, p_2, q_2$ are all nonnegative numbers. If there exist some $M$ such that $f(x)\equiv 0$ when $x > M,$ then

Lemma 3.11. Let

where $s, t, p_1, p_2, q_1, q_2$ are all nonnegative numbers. If there exist some $M$ such that $f(x)\equiv 0$ when $x > M,$ then

We are now in the position to discuss the commutativity problem for dual Toeplitz operators with symbols

and

Before giving a complete answer to such a problem, we need to analyze various situations for the integers $s, t, p_j$, and $q_j$ with $j = 1, 2$. Based on Theorem 3.3, we assume that $a$ and $b$ are both nonzero in the following:

Proposition 3.12. Let

and

where $a, b\in\mathbb{C}$, $s, p_1, p_2$ are positive and $t$, $q_1$, $q_2$ are nonnegative. In the case of $s = p_1 = p_2$, we have that

if, and only if, $\varphi = c\psi$ for some constant $c.$

Proof. Clearly, we only need to prove the necessity. Based on Lemma 3.5, we will consider the following four cases:

Case 1. $p_1-q_1-p_2+q_2 = 0.$ Since $p_1 = p_2,$ we have $q_1 = q_2,$ which implies that

It follows from Theorem 3.3 that $q_1 = t$.

Case 2. $p_1-q_1-p_2+q_2 = -2s.$ By (3.26), we have that $\alpha_6 = 0$ and $\alpha_1+\alpha_2 = 0.$ Now, combining Lemmas 3.4 and 3.7 gives $t = q_1 = q_2.$

Case 3. $p_1-q_1-p_2+q_2 = 2s.$ Using (3.26) again, we deduce that $\alpha_3 = 0$ and $\alpha_4+\alpha_5 = 0$. Thus, we obtain by Lemmas 3.4 and 3.7 that $t = q_1 = q_2$.

Case 4. $p_1-q_1-p_2+q_2$ is not equal to $0$, not equal to $-2s$, and not equal to $2s.$ From (3.26), we conclude that $\alpha_3 = 0$ and $\alpha_6 = 0$, which implies $t = q_1 = q_2$.

This finishes the proof of Proposition 3.12. □

Proposition 3.13. Let

and

where $a, b\in\mathbb{C}$ such that $a\neq b$, $s, p_1, p_2$ are positive and $t, q_1, q_2$ are nonnegative. If $s = p_1\neq p_2$, then

if, and only if, $t = q_1 = q_2 = 0$.

Proof. In order to show the necessity, we need to consider the following five cases:

Case 1. $p_1-q_1-p_2+q_2 = 0.$ In this case, (3.26) can be rewritten as

Thus, $\alpha_5 = 0$ and $\alpha_3+\alpha_6 = 0.$ Applying Lemmas 3.4 and 3.8, we obtain $t = 0$ and

If $q_1 = 0$ or $q_2 = 0$, then $t = q_1 = q_2 = 0$ follows immediately.

Next, we consider the case that $q_1\neq 0$ and $q_2\neq 0$. Using

if, and only if,

we have by (3.28) that

Now we are going to analyze the following two sub-cases:

Sub-case 1.1. If $q_1 = q_2$, then $p_1-q_1-p_2+q_2 = 0$ implies $p_1 = p_2$. This is a contradiction.

Sub-case 1.2. If $q_1\neq q_2$, then $\beta_2 = \beta_5 = 0.$ However, Lemma 3.4 tells us that $s = 0,$ which is also a contradiction.

Therefore, we obtain that $t = q_1 = q_2 = 0$ in Case 1.

Case 2. $p_1-q_1-p_2+q_2 = 2s.$ In this case, we have $\alpha_3 = \alpha_4 = \alpha_5 = 0$. It follows from Lemmas 3.4 and 3.7 that $t = q_1 = q_2 = 0.$

Case 3. $p_1-q_1-p_2+q_2 = -2s.$ In this case, $\alpha_6 = 0$ and $\alpha_3+\alpha_4 = 0$. Thus, we have $t = q_2$ and $at-aq_1+bq_2 = 0$ by Lemmas 3.7 and 3.9. Next, we are going to show that $t = 0$. Suppose not, we assume that $t > 0.$ Using (3.28), we have

Sub-case 3.1. If $t+s\neq q_1$, then $\beta_4+\beta_5 = 0.$ However, Lemma 3.4 gives that $s = p_2$, which contradicts the assumption that $s\neq p_2.$

Sub-case 3.2. If $t+s = q_1$, then $\beta_2+\beta_4+\beta_5 = 0$. It follows from Lemma 3.10 that

Substituting $q_1 = s+t$, $s = p_1$ and $q_2 = t$ into the equation $p_1-q_1-p_2+q_2 = -2s$ gives $p_2 = 2s$. Thus, (3.38) is reduced to $(b-a)s = 0$, but $b-a\neq 0$ yields $s = 0$, which contradicts that $s > 0$.

Consequently, we get $t = 0$ and $q_1 = q_2 = 0$ in Case 3.

Case 4. $p_1-q_1-p_2+q_2 = -2p_2.$ By (3.26), we have

It follows that $\alpha_1+\alpha_2 = 0, \alpha_4 = 0, \alpha_6 = 0.$ Thus, we obtain by Lemmas 3.4 and 3.7 that $t = q_1 = q_2 = 0$.

Case 5. $p_1-q_1-p_2+q_2$ is not equal to $0, 2s, -2s$, or $-2p_2.$ In this case, we have that $\alpha_3 = \alpha_4 = \alpha_6 = 0.$ By Lemmas 3.4 and 3.7, we also obtain that $t = q_1 = q_2 = 0$.

This completes the proof of Proposition 3.13. □

Proposition 3.14. Let

and

where $a, b\in\mathbb{C}$, $s, p_1, p_2$ are positive and $t$, $q_1$, $q_2$ are nonnegative. In the case of $p_1 = p_2\neq s$, we have that $S_{\varphi}S_{\psi} = S_{\psi}S_{\varphi}$ if, and only if, $t = q_1 = q_2 = 0$.

Proof. Based on Lemma 3.5, we need to consider the following six cases:

Case 1. $p_1-q_1-p_2+q_2 = 0.$ Since $p_1 = p_2$, we have $q_1 = q_2$ and $\varphi(z) = (a+b)z^{p_1}\overline{z}^{q_2}.$ It follows from Theorem 3.3 that $t = q_1 = q_2 = 0$.

Case 2. $p_1-q_1-p_2+q_2 = 2s.$ In this case, using (3.26) we obtain

This implies that $\alpha_1+\alpha_6 = 0, \alpha_3 = 0$, and $\alpha_5 = 0$. Applying Lemmas 3.4, 3.7, and 3.9, we obtain $t = 0, t = q_1$, and $aq_1+bt-bq_2 = 0$. Hence, $t = q_1 = q_2 = 0.$

Case 3. $p_1-q_1-p_2+q_2 = 2p_1.$ In this case, we have $\alpha_3 = \alpha_4 = 0.$ By Lemmas 3.4 and 3.7, we obtain that $t = q_1$ and $q_2 = 0$. If $t > 0$, then we have by (3.28) that

Thus, $\beta_5+\beta_6 = 0$. Using Lemma 3.9, we get $p_2 = 0$, which is a contradiction. Hence, $t = 0$ and $q_1 = q_2 = 0.$

Case 4. $p_1-q_1-p_2+q_2=-2s.$ By (3.26), we obtain that $\alpha_2=\alpha_6=0$ and $\alpha_3+\alpha_4=0$. Combining Lemmas 3.4, 3.7, and 3.9 gives $t=0, t=q_2$, and $at-aq_1+bq_2=0$. This yields that $t=q_1=q_2=0.$

Case 5. $p_1-q_1-p_2+q_2=-2p_1.$ Now we have $\alpha_1=\alpha_6=0.$ By Lemmas 3.4 and 3.7, we obtain that $q_1=0$ and $t=q_2.$ Next, we will show $t=q_2=0.$ If not, we obtain by (3.28) that $\beta_4+\beta_5=0$, but Lemma 3.4 gives that $s=p_2$, which is a contradiction. Therefore, $t=q_1=q_2=0$.

Case 6. $p_1-q_1-p_2+q_2$ is not equal to $0, 2s, -2s, 2p_1$, or $-2p_1.$ In this case, we have $\alpha_3 = \alpha_6 = 0.$ Applying Lemma 3.4, we obtain $q_1 = t = q_2,$ which means $\varphi(z) = (a+b)z^{p_1}\overline{z}^{t}$. It follows from Theorem 3.3 again that $t = q_1 = q_2 = 0$.

This finishes the proof of Proposition 3.14. □

Proposition 3.15. Let

and

where $a, b\in\mathbb{C}$, $s, p_1, p_2$ are positive and $t, q_1, q_2$ are nonnegative. If $p_1, p_2, s$ are different from each other, then

if, and only if, $t = q_1 = q_2 = 0$.

Proof. We first show that $p_1-q_1-p_2+q_2\neq 0$. If not, then we have by (3.26) that

It follows that $\alpha_2 = \alpha_1+\alpha_4 = 0$. Combining this with Lemma 3.4, we obtain that $t = 0$ and $aq_1+bq_2 = 0.$ If $q_1 = q_2$, then $p_1-q_1-p_2+q_2 = 0$ gives $p_1 = p_2$. This contradicts the hypothesis that $p_1\neq p_2$. If $q_1\neq q_2$, then we have by (3.28) that $\beta_2 = 0.$ Then, it follows from Lemma 3.4 that $s = 0$, which is a contradiction. Based on Lemma 3.5, we need only to consider the following five cases:

Case 1. $p_1-q_1-p_2+q_2 = 2s.$ In this case, we have $\alpha_3 = \alpha_1+\alpha_6 = 0.$ It follows from Lemmas 3.7 and 3.9 that $t = q_1$ and $aq_1+bt-bq_2 = 0$. If $t > 0$, then we obtain by (3.28) that

If $q_2 = 0,$ then $\beta_5+\beta_6 = 0.$ It follows from Lemma 3.9 that $-bs+bs-bp_2 = 0$, i.e., $p_2 = 0$. This is a contradiction. If $q_2\neq 0,$ then $\beta_6 = 0.$ By Lemma 3.7, we deduce that $s = p_2$, which is also a contradiction. Thus, $t = 0$. Moreover, $t = q_1 = q_2 = 0.$

Case 2. $p_1-q_1-p_2+q_2 = 2p_1.$ In this case, we have $\alpha_3 = \alpha_4 = 0.$ Using Lemmas 3.4 and 3.7, we obtain that $t = q_1$ and $q_2 = 0$. Now we are going to show $t = 0$. Suppose that $t > 0$. It follows from (3.28) that $\beta_5+\beta_6 = 0$. Using Lemma 3.9, we conclude that $p_2 = 0$, which is a contradiction. So, we have $t = q_1 = q_2 = 0.$

Case 3. $p_1-q_1-p_2+q_2 = -2s.$ In this case, we have

which gives that $\alpha_6 = \alpha_3+\alpha_4 = 0.$

Using Lemmas 3.7 and 3.9, we obtain that $t = q_2$ and $at-aq_1+bq_2 = 0$. If $t = 0,$ then we deduce that $q_1 = q_2 = 0$ immediately. If $t > 0,$ then substituting $t = q_2$ into (3.28), we get

If $q_1 = 0,$ then $\beta_2+\beta_3 = 0.$ Lemma 3.9 implies that $-as+as-ap_1 = 0,$ i.e., $p_1 = 0,$ which is a contradiction. If $q_1\neq 0$, then $\beta_3 = 0.$ It follows from Lemma 3.7 that $s = p_1$, which contradicts that $s\neq p_1$. Therefore, we have $t = q_1 = q_2 = 0$.

Case 4. $p_1-q_1-p_2+q_2 = -2p_2.$ Using (3.26), we have that $\alpha_1 = \alpha_6 = 0$. Moreover, it follows from Lemmas 3.4 and 3.7 that $q_1 = 0$ and $t = q_2$. If $t = 0,$ then $t = q_1 = q_2 = 0.$ If $t > 0$, then using (3.28), we get that $\beta_4+\beta_5 = 0$. However, Lemma 3.4 implies $s = p_2,$ which is a contradiction.

Case 5. $p_1-q_1-p_2+q_2$ is not equal to $2s$, $2p_1$, $-2s$, or $-2p_2$. In this case, we have $\alpha_3 = \alpha_6 = 0.$ It follows from Lemma 3.7 that $t = q_1 = q_2.$ If $t\neq 0$, then we have

If $-p_1+q_1+p_2-q_2 = 2t,$ then $\beta_4+\beta_5 = 0.$ Applying Lemma 3.4, we have $s = p_2,$ which is a contradiction. If $-p_1+q_1+p_2-q_2\neq2t,$ then $\beta_6 = 0$. Using Lemma 3.7, we obtain that $s = p_2$, which contradicts that $s\neq p_2$. Consequently, $t = q_1 = q_2 = 0$.

This finishes the proof of Proposition 3.15. □

Proposition 3.16. Let

and

where $a, b\in\mathbb{C}$. If $s = p_1 = 0$ and $p_2\geqslant 1$, then

if, and only if, $\psi$ is constant.

Proof. To show the necessity, we need to discuss the following two cases:

Case 1. $p_1-q_1-p_2+q_2 = -2p_2.$ In this case, we have $\alpha_4+\alpha_6 = 0$. By Lemma 3.9, we have that $t = 0$ and $\psi$ is a constant function.

Case 2. $p_1-q_1-p_2+q_2\neq-2p_2$. In this case, we have $\alpha_5 = 0$. Moreover, it follows from Lemma 3.4 that $t = 0$. This also implies that $\psi$ is constant. □

Proposition 3.17. Let

and

where $a, b\in\mathbb{C}$. If $p_1 = p_2 = 0$ and $s\geqslant 1$, then

if, and only if, $\varphi$ is constant.

Proof. Based on Lemma 3.5, we need to analyze the following three cases:

Case 1. $p_1-q_1-p_2+q_2 = 0.$ Since $p_1 = p_2 = 0$, we have $q_1 = q_2.$ By (3.26), we obtain that

This yields that $\alpha_1+\alpha_4 = 0.$ Moreover, Lemma 3.4 implies that $aq_1+bq_2 = 0$, i.e., $(a+b)q_1 = 0.$ If $a+b = 0,$ then

If $q_1 = 0$, then

Thus, $\varphi$ is a constant function.

Case 2. $p_1-q_1-p_2+q_2 = 2s.$ In this case, we have

It follows that $\alpha_4 = \alpha_2+\alpha_3 = 0.$ By Lemma 3.4, we get that $q_2 = 0$. Furthermore, we obtain by Lemma 3.9 that $-at+at-aq_1 = 0$, which implies $q_1 = 0.$ Thus, $\varphi$ is constant.

Case 3. $p_1-q_1-p_2+q_2$ is not equal to $2s$ and not equal to $0.$ In this case, we have $\alpha_1 = \alpha_5+\alpha_6 = 0.$ Now, Lemma 3.4 gives $q_1 = 0$ and Lemma 3.9 implies that $-bt+bt-bq_2 = 0$. This yields that $q_2 = 0$, so $\varphi$ is a constant function. □

Proposition 3.18. Let

and

where $a, b\in\mathbb{C}$. If $s = 0$, $p_1\neq 0$, and $p_2\neq0$, then

if, and only if, $\psi$ is a constant function.

Proof. To complete the proof, we need only to show the necessity and discuss the following three cases:

Case 1. $p_1-q_1-p_2+q_2 = 0.$ In this case, we have

If $p_1 = p_2$, then $p_1-q_1-p_2+q_2 = 0$ implies that $q_1 = q_2.$ Thus,

By Theorem 3.3, we have $t = 0$. If $p_1\neq p_2,$ then $\alpha_2 = \alpha_5 = 0.$ Lemma 3.4 implies that $t = 0.$ Thus, $\psi$ is constant.

Case 2. $p_1-q_1-p_2+q_2 = 2p_1.$ In this case, we have $\alpha_5 = 0.$ It follows from Lemma 3.4 that $t = 0,$ which means that $\psi$ is constant.

Case 3. $p_1-q_1-p_2+q_2$ is not equal to $0$ and not equal to $2p_1.$ Under this assumption, we have that $\alpha_4+\alpha_6 = 0$. Using Lemma 3.9, we obtain that $bq_2+bt-bq_2 = 0$, i.e., $t = 0.$ So, $\psi$ is constant. □

Proposition 3.19. Let

and

where $a, b\in\mathbb{C}$. In the cases of $p_1 = 0$, $s\neq 0$, and $p_2\neq0$, we have that

if, and only if, $q_1 = 0$ and one of the following conditions holds:

(1) $s = p_2$ and $t = q_2;$

(2) $s\neq p_2$ and $t = q_2 = 0.$

Proof. Based on Lemma 3.5, we need to consider the following four cases:

Case 1. $p_1-q_1-p_2+q_2 = 0.$ In this case, we obtain by (3.26) that

This yields that $\alpha_2+\alpha_3+\alpha_6 = 0$. Using Lemma 3.11, we have that

Let us show $q_1 = 0$ first. If $q_1 > 0$, then we observe that $t$ and $q_2$ cannot be equal to $0$ at the same time. Now we are going to discuss the following four sub-cases:

Sub-case 1.1. $t = 0$ and $q_2\neq0.$ Then, it follows from (3.28) that

If $q_1 = q_2$, then $p_1-q_1-p_2+q_2 = 0$ implies that $p_1 = p_2$. This is a contradiction. If $q_1\neq q_2$, then $\beta_2 = 0.$ Applying Lemma 3.4 gives $s = 0,$ which is also impossible.

Sub-case 1.2. $t\neq0$ and $q_2 = 0$. Substituting $p_1 = 0$ and $q_2 = 0$ into $p_1-q_1-p_2+q_2 = 0$ yields $q_1+p_2 = 0$. This is impossible, since $q_1\geqslant 0$ and $p_2\geqslant 1$.

Sub-case 1.3. $t = q_2$ and $q_2\neq0$. Using (3.51), we deduce that $q_1 = 0.$ This is a contradiction.

Sub-case 1.4. $t\neq q_2, t\neq0$ and $q_2\neq0.$ It follows from (3.28) that

If $t = q_1$, then $\beta_5 = 0$. Applying Lemma 3.4 gives that $s = 0$. This contradicts the assumption that $s\neq 0$. If $t\neq q_1,$ then $\beta_1+\beta_4 = 0.$ It follows from Lemma 3.4 that $ap_1+bp_2 = 0$. Combining this with $p_1 = 0$ yields $p_2 = 0$. This contradicts the assumption that $p_2\neq 0$.

From Sub-cases 1.1–1.4, we conclude that $q_1 = 0$. Combining this with (3.51) implies that $t = q_2$. Furthermore, if $s = p_2$, then we obtain (1); otherwise, if $s\neq p_2$, then $\alpha_5 = 0$. It follows from Lemma 3.4 that $t = q_2 = 0$, which gives (2).

Case 2. $p_1-q_1-p_2+q_2 = -2s.$ In this case, we obtain by (3.26) that

Next, we will consider the following three sub-cases:

Sub-case 2.1. $p_2 = 2s.$ It follows that $\alpha_1+\alpha_5 = \alpha_6 = 0.$ From Lemma 3.4, we get that

and

If $t = 0,$ then $q_1$ must be zero, since $a\neq0.$ If $t\neq0,$ then $q_1\neq 0$. By (3.28), we obtain that $\beta_3 = 0.$ Using Lemma 3.7, we obtain that $s = p_1$, which is a contradiction. Hence, we have $t = 0.$ Combining this and (3.55) gives that $t = q_1 = q_2 = 0.$

Sub-case 2.2. $p_2 = s.$ In this case, we have that $\alpha_1 = \alpha_6 = 0.$ Using Lemmas 3.4 and 3.7, we obtain $q_1 = 0$ and $t = q_2.$

Subcase 2.3. $p_2\neq 2s$ and $p_2\neq s.$ In this case, we have $\alpha_1 = \alpha_5 = \alpha_6 = 0$. It follows from Lemmas 3.4 and 3.7 that $t = q_1 = q_2 = 0$.

Case 3. $p_1-q_1-p_2+q_2 = -2p_2.$ It follows that $\alpha_1 = 0.$ Using Lemma 3.4, we have $q_1 = 0.$ Then we can deduce that (1) or (2) holds by using Theorem 3.3.

Case 4. $p_1-q_1-p_2+q_2$ is not equal to $0$, $-2s$, or $-2p_2.$ In this case, we obtain that $\alpha_2+\alpha_3 = 0.$ Lemma 3.9 implies that $-at+at-aq_1 = 0 = -aq_1$, which yields $q_1 = 0.$ Finally, using Theorem 3.3 again, we get the desired results.

This completes the proof of Proposition 3.19. □

Now we are ready to state and prove the main result of this section.

Theorem 3.20. Let

and

where $a, b\in\mathbb{C}\backslash\{0\}$ such that $a\neq b$, $p_j, q_j$, $s$, and $t$ are nonnegative integers, $j = 1, 2$. Then,

if, and only if, one of the following conditions holds:

$\textrm{(i)}\; \psi$ is a constant function;

$\textrm{(ii)}$ Both $\varphi$ and $\psi$ are analytic;

$\textrm{(iii)}$ Both $\varphi$ and $\psi$ are co-analytic;

$\textrm{(iv)}$ There exist $\alpha, \beta\in\mathbb{C},$ not both zero, such that $\varphi = \alpha\psi+\beta$.

Proof. Obviously, it is sufficient to show the necessity. To do so, we divide the proof into nine steps as follows:

(1) $s = p_1 = p_2$ (Proposition 3.12);

(2) $s = p_1$ and $p_1 \neq p_2$ (Proposition 3.13);

(3) $p_1 = p_2$ and $p_2\neq s$ (Proposition 3.14);

(4) $s$, $p_1$, $p_2$ are pairwise different (Proposition 3.15);

(5) $s = p_1 = 0$ and $p_2\neq0$ (Proposition 3.16);

(6) $p_1 = p_2 = 0$ and $s\neq0$ (Proposition 3.17);

(7) $s = 0$, $p_1\neq 0$, and $p_2\neq0$ (Proposition 3.18);

(8) $p_1 = 0$, $s\neq 0$, and $p_2\neq0$ (Proposition 3.19);

(9) $s = p_1 = p_2 = 0$ (this situation is trivial).

This finishes the proof of Theorem 3.20. □

4.   Conclusions

In this research, we conduct a study of dual Toeplitz operators on the orthogonal complement of the harmonic Bergman space and obtain that:

$(1)$ Suppose that

which are $N$ bounded functions on the unit disk $\mathbb D$. If

then there exist some $k\in \{1, 2, \cdots, N\}$ such that $\varphi_{k} = 0$;

$(2)$ Let

and

where $a, b\in\mathbb{C}\backslash\{0\}$ such that $a\neq b$, $p_j, q_j$, $s$, and $t$ are nonnegative integers, $j = 1, 2$. Then,

if, and only if, one of the following conditions holds:

(i) $\psi$ is a constant function;

(ii) Both $\varphi$ and $\psi$ are analytic;

(iii) Both $\varphi$ and $\psi$ are co-analytic;

(iv) There exist $\alpha, \beta\in\mathbb{C},$ not both zero, such that $\varphi = \alpha\psi+\beta$.

Use of AI tools declaration

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

Acknowledgments

We are grateful to Professor Xianfeng Zhao (Chongqing University) for providing useful suggestions. This work was partially supported by NSFC (grant number: 12371125) and Chongqing Natural Science Foundation (cstc2020jcyj-msxmX0700).

Conflict of interest

The author declares that she has no conflict of interest.