Product of H-Toeplitz operator and Toeplitz operator on the Bergman space

1.   Introduction

Let $\mathbb{D}$ denote the unit disc in the complex plane. Let $L^{2}$ be the Hilbert space of all Lebesgue square integral functions with respect to the normalized area measure $dA$ on $\mathbb{D}$. The Bergman space $L_{a}^{2}$ is consisting of all holomorphic functions contained in $L^{2}$. It is well known that $L_{a}^{2}$ is a closed subspace of $L^{2}$ and has an orthogonormal basis $\{e_{n}\}_{n = 0}^{+\infty}$, where $e_{n}(w) = \sqrt{n+1}w^n$. The Bergman space is a reproducing Hilbert space with the reproducing kernel $K_z$, which is given explicitly by

Let $P$ be the orthogonal projection from $L^{2}$ onto $L_{a}^{2}$, then $P$ is given by

Denote $L^{\infty}$ as the set of all bounded measurable functions on $\mathbb{D}$. For $f\in L^\infty$, the Toeplitz operator $T_f$ with symbol $f$ is defined by

It is easy to see that $T_f$ is a bounded operator on the Bergman space.

Let $L_{h}^{2}$ be the harmonic Bergman space which is the collection of all harmonic functions in $L^{2}$. Define a unitary operator $K: L_{a}^{2}\rightarrow L_{h}^{2}$ by $K(e_{2n}) = e_n$ and $K(e_{2n+1}) = \overline{e_{n+1}}$, $n = 0, 1, 2, \cdots$. The H-Toeplitz operator $B_{f}$ with symbol $f\in L^\infty$ is defined by

Obviously $B_f$ is a bounded operator on the Bergman space.

Let $\mathcal{R}$ be the space of square integrable functions on $[0, 1]$ with respect to the measure $rdr$. It is clear that the functions in $\mathcal{R}$ are radial functions on $\mathbb{D}$. Since trigonometric polynomials are dense in $L^{2}$ and $e^{ik_1\theta}\mathcal{R}$ is orthogonal to $e^{ik_2\theta}\mathcal{R}$ for $k_1\neq k_2$, one can see that

So, for each $f\in L^{2}$, it can be written as (see ^[4])

where each $\varphi_{k}\in \mathcal{R}$ is bounded radial function when $f\in L^\infty$. Each function in $e^{ik\theta}\mathcal{R}$ is called a quasihomogeneous function with degree $k$.

In 1964, Brown and Halmos ^[1] showed that for Toeplitz operators on the Hardy space, $T_{f}T_{g} = T_{h}$ holds if and only if either $\bar f$ or $g$ is analytic and $h = fg$. In 1989, Zheng ^[2] showed that if $f, g$ are bounded harmonic functions such that $T_{f}T_{g} = T_{h}$ on the Bergman space, then either $\bar f$ or $g$ is analytic. The product problem on Toeplitz operators with general symbols turns out to be much more complicated. In ^[5] Louhichi and Zakariasy showed that if the product of two Toeplitz operators with the quasihomogeneous symbols on the Bergman space with the degree $p$ and $s$ respectively to be another Toeplitz operator, then the symbol functions must be quasihomogeneous with the degree $p+s$. In ^[6] Louhichi, Strouse and Zakariasy showd the relationship between the radial part of the quasihomogeneous symbols.

In 2007, Arora and Paliwal ^[7] started to study the H-Toeplitz on the Hardy space. Gupta and Singh expand this definition for Slant H-Toeplitz operators on the Hardy space ^[8] and for H-Toeplitz operator on the Bergman space ^[9]. In 2022, Liang et al. characterized the commuting of H-Toeplitz operators with quasihomogeneous symbols on the Bergman space, see ^[10].

Motivated by the mentioned works, in Section 3 of this paper we will characterize when the product of two H-Toeplitz operators to be another H-Toeplitz with one general and another quasihomogeneous symbols, see Theorems 3.1 and 3.3. Also, in Section 4 we will consider the product of Toeplitz operator and H-Toeplitz operator to be another H-Toeplitz with certain harmonic symbols, that is, when $T_fB_g = B_h$ or $B_fT_g = B_h$ holds for certain harmonic symbols $f, g, h$, see Theorems 4.1 and 4.2 respectively. With non-harmonic symbols, we consider a simple case which tells the answer of when $B_fT_g = B_h$ is not trivial, see Theorem 4.4.

2.   Preliminaries

In this section, we will present some lemmas which will be used frequently.

The Mellin transform $\widehat{\varphi}$ of a function $\varphi\in L^{1}([0, 1], rdr)$ which plays an important role is defined by

It is clear that $\widehat{\varphi}$ is analytic on $\{w: {\rm Re}\, w > 2\}$. The following two lemmas have been proved in ^[10] which will be used often in the paper.

Lemma 2.1. Let $\varphi\in L^{1}([0, 1], rdr)$. If there exist a sequence of positive integers $\{n_k\}$ satisfying that $\sum\limits_{k = 1}^\infty \frac{1}{n_k} = \infty$ and $\widehat{\varphi}(n_k) = 0$ for all $k$, then $\varphi = 0$.

Lemma 2.2. Let $\phi$ be a bounded radial function and $p$ an integer. Then for any nonnegative integer $n$,

By Lemma 2.2, we obtain the following two lemmas immediately.

Lemma 2.3. Let $p$ be a nonnegative integer. Then for each nonnegative integer $n$,

Lemma 2.4. Suppose $f = \sum_{k\in\mathbb{Z}}e^{ik\theta}\varphi_{k}(r)$, $h = \sum_{s = -M}^{\infty}e^{is\theta}\psi_{s}(r)\in L^{\infty}$, where $M$ is a nonnegative integer. Then for nonnegative integer $n$,

and for $n\ge M$,

In ^[9], it is showed that the map $f\rightarrow B_{f}$ is one to one, then the following lemma holds.

Lemma 2.5. Suppose $f\in L^{\infty}$, then $B_{f} = 0$ if and only if $f = 0$.

3.   Product of H-Toeplitz operators

In this section, we focus on the product of two H-Toeplitz operators. Our aim here is to provide a sufficient and necessary condition for the product of two H-Toeplitz operators to be another H-Toeplitz operator with more general symbols.

Theorem 3.1. Let $p$ be an integer and $M$ a nonnegative integer. Suppose $\phi$ is a bounded radial function on $\mathbb{D}$ and $f, h\in L^{\infty}$ with $h = \sum_{s = -M}^{\infty}e^{is\theta}\psi_{s}(r)$. Then the following statements are equivalent:

(1) $B_{f}B_{e^{ip\theta}\phi} = B_h$,

(2) $B_{e^{ip\theta}\phi}B_{f} = B_h$,

(3) $f = h = 0$ or $\phi = h = 0$.

Proof. If (3) holds, then (1) and (2) hold clearly. Conversely, suppose one of (1) and (2) holds. If $\phi = 0$, then by Lemma 2.5 we can obtain $\phi = h = 0$ immediately. So in the following we assume $\phi\neq0$ and show that $f = h = 0$. For this end, we first write

where each $\varphi_{k}$ is bounded radial function. Choose $n$ satisfying $2n\geq M$. By Lemma 2.4,

"(1) $\Rightarrow$ (3)". Suppose $B_{f}B_{e^{ip\theta}\phi} = B_h$ and $\phi\ne 0$. We show the result in the following two cases.

Case 1. $p$ is even. Let $n > \max\{0, \frac{p}{2}+M+1\}$, by Lemmas 2.2 and 2.4, direct computations give that

Since (3.1) equals to (3.3), we obtain that

for any $k = -n-\frac{p}{2}, \cdots, n-\frac{p}{2}-M-1$. In other words, the above holds when $n > N_k = \max\{0, \frac p2+M+1, \frac p2+M+1+k, - \frac p2-k\}$ for each integer $k$. Set

By Lemma 2.1 and $\phi\ne 0$, we have $\sum_{n\in E_k}\frac{1}{n} = \infty$. For each fixed $k$, choose $n\in E_k$, then $\widehat{\varphi}_{k}(k+2n+p+2) = 0$ with $\sum_{n\in E_k}\frac{1}{k+2n+p+2} = \infty$. By Lemma 2.1 we get $\varphi_{k} = 0$ for each integer $k$. So we obtain $f = 0$ and hence $h = 0$.

Case 2. $p$ is odd. Let $n > \max\{0, M+\frac{p+1}{2}\}$, by Lemmas 2.2 and 2.4 again, we have

Because (3.2) equals to (3.4), it follows that

where $k = -n-\frac{p+1}{2}, \cdots, n-M-\frac{p+1}{2}$. With the similar arguments as done in Case 1, we can obtain $f = 0$ and then $h = 0$. Therefore, (3) holds.

"(2) $\Rightarrow$ (3)". Suppose $B_{e^{ip\theta}\phi }B_{f} = B_h$ and $\phi\ne 0$. Let the integer $n > |p|+M+1$, we deduce (3) by the following two cases.

Case 1. $p\leq0$. By Lemmas 2.2 and 2.4, we may obtain that

Since (3.1) equals to (3.5), then we have $\widehat{\varphi}_{2k}(4n+2k+2)\widehat{\phi}(2n+2k+p+2) = 0$ for $k = -n-p, \ldots, n-p-M-1$, where $n > |p|+M+1$. As done in Case 1 of "(1) $\Rightarrow$ (3)", one may obtain $\varphi_{2k} = 0$ for any integer $k$. Also, by Lemmas 2.2 and 2.4 again, we get

Because (3.2) equals to (3.6), we have $\widehat{\varphi}_{2k+1}(4n+2k+5)\widehat{\phi}(2n+2k+p+4) = 0$ for $k = -n-p-1, \ldots, n-p-M-1$, where $n > |p|+M+1$. As done before, we then obtain $\varphi_{2k+1} = 0$ for each integer $k$. Thus we get $f = 0$, and hence $h = 0$. So (3) holds.

Case 2. $p > 0$. By Lemma 2.2 and (2.4), we have

Comparing (3.1) with (3.7), it gives that $\widehat{\varphi}_{2k}(4n+2k+2)\widehat{\phi}(2n+2k+p+2) = 0$ for $k = -n, \ldots, n-p-M-1$, where $n > p+M+1$. By using the same arguments as done in Case 1, we have $\widehat{\varphi}_{2k} = 0$ for any integer $k$. Also, by Lemma 2.2 and (2.4),

By comparing (3.2) with (3.8), it follows that $\widehat{\varphi}_{2k+1}(4n+2k+5)\widehat{\phi}(2n+2k+p+4) = 0$ for $k = -n-1, \ldots, n-p-M-1$, where $n > p+M+1$. Similarly we have $\widehat{\varphi}_{2k+1} = 0$ for any integer $k$. Above all, $f = 0$. Hence $h = 0$, so (3) holds. □

The following zero product problem holds immediately.

Corollary 3.2. Suppose $f\in L^{\infty}$ and $\phi$ is a bounded radial function. Let $p$ be an integer. Then the following statements are equivalent:

(1) $B_{f}B_{e^{ip\theta}\phi} = 0$,

(2) $B_{e^{ip\theta}\phi}B_{f} = 0$,

(3) $f = 0$ or $\phi = 0$.

Now we are ready to characterize the product of two H-Toeplitz operators to be another H-Toeplitz operator with harmonic and radial symbols.

Theorem 3.3. Suppose $f$ and $h$ are bounded harmonic functions on $\mathbb{D}$, $\phi$ is a bounded radial function. Then $B_{f}B_{\phi} = B_{h}$ if and only if $f = h = 0$ or $\phi = h = 0$.

Proof. The sufficiency is obvious, now we prove the necessity. First we write $f = f_{+}+ \overline{f_{-}}$ and $h = h_{+}+ \overline{h_{-}}$, where $f_{+} = \sum_{j = 0}^{\infty} a_jw^j$, $f_{-} = \sum_{s = 1}^{\infty} b_sw^s$, $h_{+} = \sum_{t = 0}^{\infty} c_tw^t$ and $h_{-} = \sum_{m = 1}^{\infty} d_mw^m$.

If $\phi = 0$, then the necessity holds. In the following we assume $\phi\neq0$. By Lemma 2.3,

and by Lemma 2.2,

Since (3.9) equals to (3.10), we have $c_t = 0$, $t\geq1$. Hence $h = c_0+ \overline{h_{-}}$. For the nonnegative integer $n$, direct calculations show that

and

where $d_0 = c_0$. By comparing (3.11) with (3.12), we then get $\widehat{\phi}(4n+2)a_{j} = 0$ for $j\geq 2n$. As we assume $\phi\neq0$, there must be a positive integer $n_0$ such that $\widehat{\phi}(4n_0+2)\neq0$. Thus $a_j = 0$ for any integer $j\ge 2n_0$. Then (3.11) becomes

Let $n\geq 2n_{0}$. Observe that the biggest degree of $w$ is $2n-1$ in (3.12), and $n+2n_0-1$ in (3.13), so we may obtain that $d_m = 0$ for $m = n-2n_0, \cdots, n+1$. Note that $n$ is any nonnegative integer with $n\geq 2n_{0}$, hence $d_m = 0$ for any integer $m\ge 0$. It follows that $h = 0$. By Corollary 3.2, we then obtain $f = 0$. □

4.   Product of Toeplitz operator and H-Toeplitz operator

In this section, we focus on the product of Toeplitz operator and H-Toeplitz operator to be another H-Toeplitz operator. First, we discuss the case of $T_fB_g = B_h$ with bounded harmonic symbols $f, g, h$. For this case we can apply the known result used for the product of two Toeplitz operators case on the Bergman space (see ^[3]).

Theorem 4.1. Suppose $f$, $g$ and $h$ are bounded harmonic functions. Then $T_{f}B_{g} = B_{h}$ if and only if one of the following statements holds:

(1) $f$ is a constant and $fg = h$.

(2) $f$ and $g$ are co-analytic and $fg = h$.

Proof. We notice a fact: for any nonnegative integer $n$, it has that

So by Corollary 1 in ^[3], we see that the above holds if and only if

with $f$, $g$ are both analytic or $f$, $g$ are both co-analytic or one of $f$ and $g$ is constant.

We first show the sufficiency. If (1) holds, then it is clear that $T_fB_g = B_h$. If (2) holds, then $h$ is also co-analytic, and so by Lemma 2.1, we have $T_{f}B_{g}(w^{2n+1}) = 0 = B_{h}(w^{2n+1})$ for any integer $n\ge 0$; on the other hand, by (4.1), we see that $T_fB_g(w^{2n}) = B_h(w^{2n})$ holds for each nonnegative integer $n$. Thus $T_fB_g = B_h$.

Now we show the necessity. As discussed before, when $T_fB_g = B_h$, then (4.2) holds and $f$ and $g$ are analytic, or $f$ and $g$ are co-analytic, or one of $f$ and $g$ is constant.

Case 1. Suppose $f$, $g$ are analytic. Then $h$ is also analytic by (4.2). We write $f = \sum_{j = 0}^{\infty} a_jw^j$, $g = \sum_{s = 0}^{\infty} b_sw^s$ and $h = \sum_{t = 0}^{\infty} c_tw^t$, then by Lemma 2.3, $T_{f}B_{g}(w^{2n+1}) = B_{h}(w^{2n+1})$ gives that

On comparing the coefficient of $w^{0}$ of both sides of (4.3), we get $c_{n+1} = a_0b_{n+1}$ for any nonnegative integer $n$. Therefore,

If $f(0)\ne 0$, then puting the above into (4.2) to get $h(0) = f(0)g(0)$, so $f(0)g = h$. By (4.2) again we obtain that $f$ is constant. If $f(0) = 0$, then (4.4) gives that $h$ is a constant. By (4.2) we see that $f$ and $g$ both are constants. Hence (1) holds.

Case 2. $f$ and $g$ are co-analytic and $fg = h$, this is (2).

Case 3. If $g$ is constant, then for any nonnegative integer $n$, $0 = T_{f}B_{g}(w^{2n+1}) = B_{h}(w^{2n+1})$. It follows from the right side of (4.3) that $h = h(0)$. Thus by (4.2), we see that $f$ is constant. This is a special case of (1).

Case 4. If $f$ is constant, then it is easy to see that (1) holds. □

Now we discuss the case of $B_fT_g = B_h$ with bounded harmonic symbols $f, g, h$. Although we only prove the case when $g = w^p$, the obtained result tells us that it may hold only in the trivial case.

Theorem 4.2. Suppose $f$ and $h$ are bounded harmonic functions on $\mathbb{D}$. Let $p$ be a nonnegative integer. Then $B_{f}T_{w^p} = B_{h}$ if and only if one of the following statements holds:

(1) $p = 0$, $f = h$.

(2) $p\neq0$, $f = h = 0$.

Proof. The sufficiency is obvious, now we prove the necessity. Suppose $B_{f}T_{w^p} = B_{h}$. If $p = 0$, we obtain $f = h$ immediately. In the following we suppose $p\neq0$.

Write $f$ and $h$ as $f_{+}+ \overline{f_{-}}$ and $h_{+}+ \overline{h_{-}}$ respectively, where $f_{+} = \sum_{j = 0}^{\infty} a_jw^j$, $f_{-} = \sum_{s = 1}^{\infty} b_sw^s$, $h_{+} = \sum_{t = 0}^{\infty} c_tw^t$ and $h_{-} = \sum_{m = 1}^{\infty} d_mw^m$. We show the result by two cases.

Case 1. $p$ is even. For any nonnegative integer $n$, by Lemma 2.3, we have

and

Write $h_{+}^{1} = \sum_{t = \frac{p}{2}}^{\infty}c_{t}w^{t}$ and $h_{+}^{2} = \sum_{t = 0}^{\frac{p}{2}-1}c_{t}w^{t}$, then $h_{+} = h_{+}^{1}+h_{+}^{2}$. Because (4.5) equals to (4.6), so for each nonnegative integer $n$, we have

that is,

Hence $f_{+} = h_{+}^{1} = 0$. Also we have

Since (4.7) equals to (4.8), we get $c_t = 0$ for $t = 1, 2, \cdots, \frac{p}{2}-1$. Now $h = c_0+\overline{h_{-}}$. Putting $f_+ = 0$ and $h = c_0+\overline{h_-}$ into (4.5) and (4.6) respectively, we then get

which shows that $b_s = 0$ for $s = 1, 2, \cdots, \frac{p}{2}-1$. For any nonnegative integer $n$, the coefficients of $w^n$ in the above equation gives that

Hence $c_0 = b_{\frac{p}{2}} = 0$. Now $h_{+} = 0$ and (4.9) can be rewritten as

for any integer $n\ge 1$. Now for fixed integer $m: 1\le m\le n$, comparing the coefficients of $w^{n-m}$, we get

Similarly we may get $d_{m} = b_{\frac{p}{2}+m} = 0$ for $1\le m\le n$. Since $n$ is any nonnegative integer, so we obtain that $d_{m} = b_{\frac{p}{2}+m} = 0$ for any integer $m\ge 1$. Therefore $h_- = f_- = 0$ and then it follows that $f = h = 0$.

Case 2. $p$ is odd. By Lemma 2.3, for any nonnegative integer $n$, $B_{f}T_{w^p}(w^{2n+1}) = B_{h}(w^{2n+1})$ gives that

For fixed $s: 1\le s\le n+\frac{p+1}{2}$, comparing the coefficients of $w^{n+\frac{p+1}{2}-s}$ in the above induces

Since $\lim\limits_{n\rightarrow \infty}c_{2n+\frac{p+1}{2}-s+1} = 0$, then $b_{s} = 0$. Because $s$ is any term of $1, 2, \cdots, n+\frac{p+1}{2}$ and $n$ is any nonnegative integer, it implies that $b_s = 0$ for any $s\geq1$. Hence, $f_{-} = 0$. By (4.10), one can get that

So, $c_{n+1} = \cdots = c_{2n+\frac{p+1}{2}+1} = 0$. Because $n$ is any nonnegative integer, thus $c_t = 0$ for $t\geq1$. Now we obtain that the left side of (4.11) is also zero. Therefore $f_{+} = 0$. Above all, $f = 0$. By Lemma 2.5, we have $h = 0$. □

For the case of $B_fT_g = B_h$ with non-harmonic symbols, it becomes much complicated. So we only focus on the simple case with $f$ and $g$ both are radial functions and $h$ is a general one. Even for such simple case, the obtained relation of $f, g$ and $h$ is not explicit, but it still tells that $B_fT_g = B_h$ holds with nontrivial case which is different from the previous result.

We need the following lemma which is proved in ^[5].

Lemma 4.3. Let $p$ be an integer and $\psi$ a bounded radial function on $\mathbb{D}$. Then for any nonnegative integer $n$,

Theorem 4.4. Suppose $\phi$ and $\psi$ are bounded radial functions on $\mathbb{D}$, $h\in L^{\infty}$. Then $B_{\phi}T_{\psi} = B_{h}$ if and only if $h$ is a radial function and a solution of the equation

Proof. We first show the necessity. Write $h$ as $h = \sum_{k\in\mathbb{Z}}e^{ik\theta}\varphi_{k}(r),$ where each $\varphi_{k}$ is bounded radial function. For any nonnegative integer $n$, by Lemmas 2.2 and 4.3, it follows from $B_{\phi}T_{\psi}(w^{2n}) = B_{h}(w^{2n})$ that

Hence for $n\geq 0$, we have $\widehat{\varphi}_{k}(k+2n+2) = 0$, $k\neq 0$. Note that $\sum_{n = 0}^{\infty} \frac{1}{k+2n+2} = \infty$, so by Lemma 2.1, we get $\varphi_{k} = 0$ for all $k\neq 0$, which means that $h$ is a radial function. Furthermore, we see that $B_{\phi}T_{\psi}(w^{2n+1}) = 0 = B_{h}(w^{2n+1})$, so (4.13) becomes

It implies that $h$ is a solution of the equation

The sufficiency is obvious by the above arguments. □

We note that the Eq (4.12) has a nontrivial solution

where $a$ and $c$ are any constants.

5.   Conclusions

In this research, it obtains the following characterizations for the product of H-Toeplitz operators and Toeplitz operators with certain symbols on the Bergman space.

(1) Let $p$ be an integer and $M$ a nonnegative integer. Suppose $\phi$ is a bounded radial function on $\mathbb{D}$ and $f, h\in L^{\infty}$ with $h = \sum_{s = -M}^{\infty}e^{is\theta}\psi_{s}(r)$. Then $B_{f}B_{e^{ip\theta}\phi} = B_h$ if and only if $B_{e^{ip\theta}\phi}B_{f} = B_h$, and if and only if $f = h = 0$ or $\phi = h = 0$.

(2) Suppose $f$ and $h$ are bounded harmonic functions on $\mathbb{D}$, $\phi$ is a bounded radial function. Then $B_{f}B_{\phi} = B_{h}$ if and only if $f = h = 0$ or $\phi = h = 0$.

(3) Suppose $f$, $g$ and $h$ are bounded harmonic functions. Then $T_{f}B_{g} = B_{h}$ if and only if $f$ is a constant and $fg = h$, or, $f$ and $g$ are co-analytic and $fg = h$.

(4) Suppose $f$ and $h$ are bounded harmonic functions on $\mathbb{D}$. Let $p$ be a nonnegative integer. Then $B_{f}T_{w^p} = B_{h}$ if and only if $p = 0$ and $f = h$, or, $p\neq0$ and $f = h = 0$.

Use of AI tools declaration

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

Acknowledgments

The authors would like to thank the referees for many helpful comments and suggestions.

The first author was supported by NSFC (Nos.12101108) and the second author was supported by NSFC (Nos. 11771401, 12271134).

Conflict of interest

The authors declare that they have no competing interests.