A theorem of Brown–Halmos type for dual truncated Toeplitz operators

Abstract

In this paper, we investigate commuting dual truncated Toeplitz operators on the orthogonal complement of the model space $K^{2}_{u}.$ Let $f,g \in L^{\infty },$ if two dual truncated Toeplitz operators $D_{f}$ and $D_{g}$ commute, we obtain similar conditions of Brown–Halmos Theorem for Hardy-Toeplitz operators, that is, both f and g are analytic, or both f and g are co-analytic, or a nontrivial linear combination of f and g is constant. However, the first two conditions are not sufficient, one can easily construct two non-commuting dual truncated Toeplitz operators with analytic or co-analytic symbols. We prove that two bounded dual truncated Toeplitz operators $D_{f}$ and $D_{g}$ commute if and only if f, g, ${\bar{f}}(u-\lambda )$ and ${\bar{g}}(u-\lambda )$ all belong to $H^{2}$ for some constant $\lambda ;$ or ${\bar{f}},{\bar{g}}$, $f(u-\lambda )$ and $g(u-\lambda )$ all belong to $H^{2}$ for some constant $\lambda ;$ or a nontrivial linear combination of f and g is constant.

1 Introduction

Inspired by Sarason’s seminal paper [11], many work in the study of truncated Toeplitz operators has been done over the past ten years [1, 2, 6, 8]. In particular, the study of algebraic properties of Toeplitz operators is an active area of research [4, 12]. This paper aims to study the commutativity of two dual truncated Toeplitz operators, where the dual truncated Toeplitz operator is a newly defined operator on a Hilbert space of harmonic functions that closely relates to truncated Toeplitz operators.

We start by recalling a few basic definitions and facts. Let $H^{2}$ be the classical Hardy space of open unit disk ${\mathbb {D}}=\{z\in {\mathbb {C}}: |z|<1\}$ and $L^{2}=L^{2}({\mathbb {T}})$ be the usual Lebesgue space on the unit circle ${\mathbb {T}}=\{z\in {\mathbb {C}}: |z|=1\}$. The space $L^{\infty }$ is the collection of all essentially bounded measurable functions on ${\mathbb {T}},$ the space $H^{\infty }$ consists of all the functions that are analytic and bounded on ${\mathbb {D}}.$ Let P be the orthogonal projection from $L^{2}$ onto $H^{2}$. One defines for f and g in $L^{\infty }$ the Toeplitz operator $T_{f}$ and dual Toeplitz operator $S_{g}$ on $H^{2}$ and $(H^{2})^{\bot },$ respectively, as the following:

$$\begin{aligned} T_{f}x&=P(fx), \quad x\in H^{2},\\ S_{g}y&=(I-P)(gy), \quad y\in (H^{2})^{\bot }. \end{aligned}$$

To each non-constant inner function u and $f\in L^{2},$ the truncated Toeplitz operator $A_{f}$ is densely defined on model space $K^{2}_{u}=H^{2}\ominus uH^{2}$ by the formula

$$\begin{aligned} A_{f}x=P_{u}(fx), \quad x\in K^{2}_{u}. \end{aligned}$$

Here $P_{u}=P-M_{u}PM_{{\bar{u}}}$ is the orthogonal projection from $L^{2}$ onto $K^{2}_{u}$. Then we define the dual truncated Toeplitz operator $D_{f}$ on the orthogonal complement of $K^{2}_{u}$ by:

$$\begin{aligned} D_{f}y=(I-P_{u})(fy), \quad y \in (K_{u}^{2})^{\bot }. \end{aligned}$$

Clearly, $D_{f}^{*}=D_{{\bar{f}}}$ and $\Vert D_{f}\Vert =\Vert f\Vert _{\infty }$ [7, Property 2.1.].

Brown and Halmos [3, Theorem 9.] give a necessary and sufficient condition for the commutativity of two Toeplitz operators. By anti-unitary equivalence [9] of $T_{f}$ and $S_{{\bar{f}}},$ means $S_{f}S_{g}=S_{g}S_{f}$ if and only if $T_{f}T_{g}=T_{g}T_{f},$ one gets immediately that $S_{f}$ and $S_{g}$ are commuting if and only if either both f and g are analytic, or both f and g are co-analytic, or a nontrivial linear combination of f and g is constant.

Such equivalent property does not hold for truncated Toeplitz operators and dual truncated Toeplitz operators. It is easy to observe that two truncated Toeplitz operators with analytic symbols commute. I. Chalendar and D. Timotin [4] have a general criterion for the commutation of truncated Toeplitz operators. However, one can easily construct two non-commuting dual truncated Toeplitz operators with analytic symbols.

Example 1

Let u be an inner function with $u(0)=0.$z and zu are analytic functions. It is easy to check that $D_{z}D_{uz}{\bar{z}}=zu$ and $D_{uz}D_{z}{\bar{z}}=0.$ Hence $D_{z}D_{uz}\ne D_{uz}D_{z}.$

It’s not hard to understand the above example if one sees that $D_{f}$ lives on $(K_{u}^{2})^{\bot }=uH^{2}\oplus \overline{zH^{2}},$ a Hilbert space of harmonic functions, while $A_{f}$ is on $K_{u}^{2},$ a Hilbert space of analytic functions. Also, $D_{f}$ behaves differently from the Toeplitz operators on the harmonic Bergman space [5].

For the commuting problem of dual truncated Toeplitz operators, an interesting aspect is that we only need to consider the dual truncated Toeplitz operators with analytic symbols (see Theorem 2). Here, we turn this problem into a mixed commuting problem of three Hankel operators ($H_{{\bar{f}}}H^{*}_{{\bar{u}}}H_{{\bar{g}}}=H_{{\bar{g}}}H^{*}_{{\bar{u}}}H_{{\bar{f}}}$) that still remains unsolved in classic Hankel operator theory.

We can now state our main result.

Theorem 3

Let u be a nonconstant inner function and $f,g\in L^{\infty }$. Then

$$\begin{aligned} D_{f}D_{g}=D_{g}D_{f} \end{aligned}$$

if and only if one of the following cases holds:

• 1. f, g, ${\bar{f}}(u-\lambda )$ and ${\bar{g}}(u-\lambda )$ all belong to $H^{2}$   for some constant $\lambda ,$

• 2. ${\bar{f}},{\bar{g}}$, $f(u-\lambda )$ and $g(u-\lambda )$ all belong to $H^{2}$   for some constant $\lambda ,$

• 3. a nontrivial linear combination of f and g is constant.

2 Necessary condition

For f and g in $L^{2}$, let

$$\begin{aligned} f_{+}=Pf,~~~ f_{-}=P_{-}f=(I-P)f. \end{aligned}$$

Let A and B be bounded operators on a Hilbert space, then the commutator of A and B is define as

$$\begin{aligned}{}[A,B]=AB-BA. \end{aligned}$$

Define an operator V on $L^{2}$ by

$$\begin{aligned} Vf(w)={\overline{w}}\overline{f(w)} \end{aligned}$$

for $f\in L^{2}.$ It is easy to check that V is anti-unitary. The operator V satisfies the following properties:

$$\begin{aligned} V^{2}=I,~~VPV=I-P, ~~ V T_{f}=S_{{\bar{f}}}V . \end{aligned}$$

(1)

Let $K_{\lambda }=\frac{1}{1-w{\bar{\lambda }}}$ denote the reproducing kernel of $H^{2}$ at $\lambda$ and $k_{\lambda }=\frac{\sqrt{1-|\lambda |^{2}}}{1-w{\bar{\lambda }}}$ denote the normalized reproducing kernel of $H^{2}$ at $\lambda$. Let us state the well-known result of Brown and Halmos.

Theorem 1

[3, Theorem 9.] (Brown-Halmos Theorem.) For $\psi$ and $\varphi$ in $L^{\infty }.$ Then

$$\begin{aligned} T_{\psi }T_{\varphi }=T_{\varphi }T_{\psi } \end{aligned}$$

if and only if at least one of the following holds:

• 1. both $\psi$ and $\varphi$ are analytic,

• 2. both $\psi$ and $\varphi$ are co-analytic,

• 3. a nontrivial linear combination of $\psi$ and $\varphi$ is constant.

If $T_{f}$ and $T_{g}$ satisfy

$$\begin{aligned} T_{{\bar{z}}}[T_{f},T_{g}]T_{z}=[T_{f},T_{g}], \end{aligned}$$

it follows from [3, Theorem 6] and [13, Corollary 4.5] that $T_{f}$ and $T_{g}$ commute. The following lemma generalizes the above result for arbitrary inner function instead of function z.

Lemma 1

Let u be a nonconstant inner function. On the Hardy space $H^{2}$, for $f,g\in L^{\infty }$, if

$$\begin{aligned}{}[T_{f},T_{g}]=T_{{\bar{u}}}[T_{f},T_{g}]T_{u}, \end{aligned}$$

then either

• 1. both f and g are analytic, or

• 2. both f and g are co-analytic, or

• 3. a nontrivial linear combination of f and g is constant.

Proof

Assume $[T_{f},T_{g}]=T_{{\bar{u}}}[T_{f},T_{g}]T_{u},$ we have

$$\begin{aligned} T_{f}T_{g}-T_{g}T_{f}=T_{{\bar{u}}^{n}f}T_{u^{n}g}-T_{{\bar{u}}^{n}g}T_{u^{n}f} \end{aligned}$$

for each positive integer n. In [10, Lemma 2.1], Guo and Wang obtained $T_{{\bar{u}}^{n}}\rightarrow 0$ (SOT), for $F\in L^{2},$$P({\bar{u}}^{n}F)=P({\bar{u}}^{n}F_{+}),$ thus

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert P({\bar{u}}^{n}F)\Vert _{2}=0, \end{aligned}$$

and

$$\begin{aligned} \Vert (I-P)(u^{n}F)\Vert _{2}=\Vert VPVu^{n}F\Vert _{2} =\Vert VP{\bar{u}}^{n}(VF)\Vert _{2}\rightarrow 0,~~ \text {as}~~ n\rightarrow \infty . \end{aligned}$$

(2)

Write

$$\begin{aligned} \big \langle T_{{\bar{u}}^{n}f}T_{u^{n}g}k_{z},k_{z}\big \rangle&=\big \langle P[{\bar{u}}^{n}fP(u^{n}gk_{z})],k_{z}\big \rangle \\&=\big \langle {\bar{u}}^{n}fP(u^{n}gk_{z}),k_{z}\big \rangle \\&=\big \langle fgk_{z},k_{z}\big \rangle -\big \langle {\bar{u}}^{n}f(I-P)(u^{n}gk_{z}),k_{z}\big \rangle \\&=\big \langle fgk_{z},k_{z}\big \rangle -\big \langle (I-P)(u^{n}gk_{z}),{u}^{n}{\bar{f}}k_{z}\big \rangle . \end{aligned}$$

Note that the Cauchy-Schwarz inequality yields

$$\begin{aligned} |\langle (I-P)(u^{n}gk_{z}),{u}^{n}{\bar{f}}k_{z}\rangle |\le ||(I-P)(u^{n}gk_{z})||_{2}||{\bar{f}}k_{z}||_{2}, \end{aligned}$$

Using (2), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\langle T_{{\bar{u}}^{n}f}T_{u^{n}g}k_{z},k_{z}\rangle =\langle fgk_{z},k_{z}\rangle . \end{aligned}$$

Similarly,

$$\begin{aligned} \lim _{n\rightarrow \infty }\langle T_{{\bar{u}}^{n}g}T_{u^{n}f}k_{z},k_{z}\rangle =\langle fgk_{z},k_{z}\rangle . \end{aligned}$$

Hence

$$\begin{aligned} \langle [T_{f},T_{g}]k_{z},k_{z}\rangle =0. \end{aligned}$$

Since Berezin transform is one-to-one, $[T_{f},T_{g}]=0.$ By Brown-Halmos Theorem (see Theorem 1), it follows that either both f and g are all analytic, or f and g are all co-analytic, or a nontrivial linear combination of f and g is constant. $\square$

Recall that the Hankel operator $H_{f}$ with symbol $f\in L^{2}$ is densely defined by

$$\begin{aligned} H_{f}x=(I-P)(fx),~ \text {for} ~x\in H^{2}, \end{aligned}$$

(3)

and $H^{*}_{f}$ is defined by

$$\begin{aligned} H^{*}_{f}y=P({\bar{f}}y),~ \text {for} ~y\in [ H^{2}]^{\bot }. \end{aligned}$$

If $M_{f}$ is expressed as an operator matrix with respect to the decomposition $L^{2}=H^{2}\oplus \overline{zH^{2}},$ the result is of the form

$$\begin{aligned} M_f=\begin{pmatrix}T_{f}&{}{H^{*}_{{\overline{f}}}} \\ H_{f} &{} S_{f}\end{pmatrix}.\quad \end{aligned}$$

Since $M_{f}M_{g}=M_{fg},$ we have

$$\begin{aligned} T_{fg}&=T_{f}T_{g}+H^{*}_{{\bar{f}}}H_{g}; \end{aligned}$$

(4)

$$\begin{aligned} H_{fg}&=H_{f}T_{g}+S_{f}H_{g}. \end{aligned}$$

(5)

If $g\in H^{\infty }$, then $H_{g}=0$, (5) becomes

$$\begin{aligned} H_{fg}=H_{f}T_{g}. \end{aligned}$$

(6)

Similarly, if $f\in H^{\infty },$ (5) becomes

$$\begin{aligned} H_{fg}=S_{f}H_{g}. \end{aligned}$$

(7)

Define the unitary operator

$$\begin{aligned} U:L^{2}(=H^{2}\oplus \overline{zH^{2}})\rightarrow [K^{2}_{u}]^{\bot }(=uH^{2}\oplus \overline{zH^{2}}) \end{aligned}$$

by

$$\begin{aligned} U= \begin{pmatrix}M_u &{} 0\\ 0 &{} I\end{pmatrix}.\quad \end{aligned}$$

Clearly, $U^{*}$ maps $[K^{2}_{u}]^{\bot }$ to $L^{2}$ and equals

$$\begin{aligned} U^{*}= \begin{pmatrix}M_{{\bar{u}}} &{} 0\\ 0 &{} I\end{pmatrix}.\quad \end{aligned}$$

Next lemma gives a matrix representation of $D_{\phi }.$ The representation is useful in this paper and shows that the dual truncated Toeplitz operators on $[K^{2}_{u}]^{\bot }$ are closely related to the Toeplitz operators and Hankel operators on $H^{2}.$

Lemma 2

On $L^{2}=H^{2}\oplus \overline{zH^{2}},\phi \in L^{\infty },$

$$\begin{aligned} U^{*}D_{\phi }U= \begin{pmatrix}T_\phi &{} H^{*}_{u{\bar{\phi }}}\\ H_{u\phi } &{} S_{\phi } \end{pmatrix}.\quad \end{aligned}$$

(8)

Proof

If $f_{1}$ is in $H^{2},$ by the definition of $D_{\phi }$ we have

$$\begin{aligned} D_{\phi }uf_{1}= uP{\bar{u}}\phi u f_{1}+(I-P){\phi }uf_{1}. \end{aligned}$$

If $f_{2}$ is in $\overline{zH^{2}},$ similarly we have that

$$\begin{aligned} D_{\phi }f_{2}= uP{\bar{u}}\phi f_{2}+(I-P){\phi }f_{2}. \end{aligned}$$

Therefore for given $[f_{1},f_{2}]^{T}$ in $H^{2}\oplus \overline{zH^{2}}$ the above calculation gives

$$\begin{aligned} U^{*}D_{\phi }U\left( {\begin{array}{c}f_{1}\\ f_{2}\end{array}}\right)&=U^{*}D_{\phi }\left( {\begin{array}{c}uf_{1}\\ f_{2}\end{array}}\right) \\&=U^{*}\left( {\begin{array}{c}uP{\bar{u}}\phi u f_{1}+uP{\bar{u}}\phi f_{2}\\ (I-P){\phi }uf_{1}+(I-P){\phi }f_{2}\end{array}}\right) \\&=\left( {\begin{array}{c}P{\bar{u}}\phi u f_{1}+P{\bar{u}}\phi f_{2}\\ (I-P){\phi }uf_{1}+(I-P){\phi }f_{2}\end{array}}\right) \\&=\begin{pmatrix}T_\phi &{} H^{*}_{u{\bar{\phi }}}\\ H_{u\phi } &{} S_\phi \end{pmatrix}\left( {\begin{array}{c}f_{1}\\ f_{2}\end{array}}\right) . \end{aligned}$$

$\square$

By Lemmas 1 and 2, we now obtain the necessary conditions similar to Theorem 1. By Example  1, the conditions are not sufficient.

Theorem 2

Let $f,g\in L^{\infty }$. Assume $D_f D_g=D_{g}D_{f} ,$ then either

• 1. both f and g are analytic, or

• 2. both f and g are co-analytic, or

• 3. a nontrivial linear combination of f and g is constant.

Proof

Assume $D_{f}D_{g}=D_{g}D_{f}$. by the matrix representation (8) , we have

$$\begin{aligned} U^{*}D_{f}D_{g}U= \begin{pmatrix}T_{f}T_{g}+H^{*}_{u{\bar{f}}}H_{gu}&{} T_{f}H^{*}_{u{\bar{g}}}+H^{*}_{u{\bar{f}}}S_{g}\\ H_{fu}T_{g}+S_{f}H_{gu} &{} H_{fu}H^{*}_{{\bar{g}}u}+S_{f}S_{g}\end{pmatrix}\quad \end{aligned}$$

and

$$\begin{aligned} U^{*}D_{g}D_{f}U= \begin{pmatrix}T_{g}T_{f}+H^{*}_{u{\bar{g}}}H_{fu}&{} T_{g}H^{*}_{u{\bar{f}}}+H^{*}_{u{\bar{g}}}S_{f}\\ H_{gu}T_{f}+S_{g}H_{fu} &{} H_{gu}H^{*}_{{\bar{f}}u}+S_{g}S_{f}\end{pmatrix}.\quad \end{aligned}$$

Hence,

$$\begin{aligned} T_{f}T_{g}+H^{*}_{u{\bar{f}}}H_{gu}=T_{g}T_{f}+H^{*}_{u{\bar{g}}}H_{fu}. \end{aligned}$$

By (4), we have

$$\begin{aligned} T_{f}T_{g}-T_{g}T_{f}=T_{{\bar{u}}f}T_{ug}-T_{{\bar{u}}g}T_{uf}. \end{aligned}$$

By Lemma 1, then either both f and g are analytic, or f and g are co-analytic, or a nontrivial linear combination of f and g is constant. $\square$

3 Necessary and sufficient condition

Since $D_{f}^{*}=D_{{\bar{f}}}$, Theorem 2 shows that the study of commuting dual truncated Toeplitz operators can be reduced to the following question.

Problem 1

For which bounded analytic functions f and g, $D_{f}D_{g}=D_{g}D_{f}$?

In the case of analytic symbols, we translate Problem 1 into solving an equation about a Toeplitz operator and a Hankel operator.

Lemma 3

Let u be a nonconstant inner function and $f,g\in H^{\infty },$ the following statements are equivalent.

• 1. $D_{f}D_{g}=D_{g}D_{f}$ holds on $[K_{u}^{2}]^{\perp };$

• 2. $D_{{\bar{f}}}D_{{\bar{g}}}=D_{{\bar{g}}}D_{{\bar{f}}}$ holds on $[K_{u}^{2}]^{\perp };$

• 3. $H_{u{\bar{f}}}T_{{\bar{g}}}+S_{{\bar{f}}}H_{u{\bar{g}}}=H_{u{\bar{g}}}T_{{\bar{f}}}+S_{{\bar{g}}}H_{u{\bar{f}}}~~(H^{2} \rightarrow [H^{2}]^{\bot });$

• 4. $H_{u{\bar{f}}}T_{{\bar{g}}}-H_{{\bar{f}}}T_{u{\bar{g}}}=H_{u{\bar{g}}}T_{{\bar{f}}}-H_{{\bar{g}}}T_{u{\bar{f}}}~~(H^{2} \rightarrow [H^{2}]^{\bot });$

• 5. $H_{{\bar{f}}}H^{*}_{{\bar{u}}}H_{{\bar{g}}}=H_{{\bar{g}}}H^{*}_{{\bar{u}}}H_{{\bar{f}}}~~(H^{2} \rightarrow [H^{2}]^{\bot });$

Proof

$(1)\Leftrightarrow (2)$: Since $D^{*}_f=D_{{\bar{f}}}$, it is clear that (1) is equivalent to (2).

$(2)\Leftrightarrow (3)$: Assume $f,g\in H^{\infty }$, hence

$$\begin{aligned} U^{*}D_{{\bar{f}}}D_{{\bar{g}}}U= \begin{pmatrix}T_{{\bar{f}}}T_{{\bar{g}}}&{} 0\\ H_{{\bar{f}}u}T_{{\bar{g}}}+S_{{\bar{f}}}H_{{\bar{g}}u} &{} S_{{\bar{f}}}S_{{\bar{g}}}\end{pmatrix}\quad \end{aligned}$$

and

$$\begin{aligned} U^{*}D_{{\bar{g}}}D_{{\bar{f}}}U= \begin{pmatrix}T_{{\bar{g}}}T_{{\bar{f}}}&{} 0\\ H_{{\bar{g}}u}T_{{\bar{f}}}+S_{{\bar{g}}}H_{{\bar{f}}u} &{} S_{{\bar{g}}}S_{{\bar{f}}}\end{pmatrix}\quad \end{aligned}$$

By Brown-Halmos Theorem (see Theorem 1), we have

$$\begin{aligned} T_{{\bar{f}}}T_{{\bar{g}}}&=T_{{\bar{g}}}T_{{\bar{f}}},\\ S_{{\bar{f}}}S_{{\bar{g}}}&=S_{{\bar{g}}}S_{{\bar{f}}}. \end{aligned}$$

Hence

$$\begin{aligned} D_{{\bar{f}}}D_{{\bar{g}}}=D_{{\bar{g}}}D_{{\bar{f}}} \end{aligned}$$

if and only if

$$\begin{aligned} H_{u{\bar{f}}}T_{{\bar{g}}}+S_{{\bar{f}}}H_{u{\bar{g}}}=H_{u{\bar{g}}}T_{{\bar{f}}}+S_{{\bar{g}}}H_{u{\bar{f}}}. \end{aligned}$$

$(3)\Leftrightarrow (4)$: By (5), we have

$$\begin{aligned} S_{{\bar{f}}}H_{u{\bar{g}}}=H_{u{\overline{fg}}}-H_{{\bar{f}}}T_{u{\bar{g}}} \end{aligned}$$

and

$$\begin{aligned} S_{{\bar{g}}}H_{u{\bar{f}}}=H_{u{\overline{fg}}}-H_{{\bar{g}}}T_{u{\bar{f}}}, \end{aligned}$$

the result follows.

$(4)\Leftrightarrow (5)$: Since (4) and (6),

$$\begin{aligned} H_{u{\bar{f}}}T_{{\bar{g}}}-H_{{\bar{f}}}T_{u{\bar{g}}}&=H_{{\bar{f}}}T_{u}T_{{\bar{g}}}-H_{{\bar{f}}}T_{u{\bar{g}}}\\&=H_{{\bar{f}}}(T_{u}T_{{\bar{g}}}-T_{u{\bar{g}}})\\&=-H_{{\bar{f}}}H^{*}_{{\bar{u}}}H_{{\bar{g}}}. \end{aligned}$$

Similarly,

$$\begin{aligned} H_{u{\bar{g}}}T_{{\bar{f}}}-H_{{\bar{g}}}T_{u{\bar{f}}} =-H_{{\bar{g}}}H^{*}_{{\bar{u}}}H_{{\bar{f}}}. \end{aligned}$$

$\square$

Example 2

Let $u=\theta \eta ,$ both $\theta$ and $\eta$ are inner functions. For $\lambda \in {\mathbb {D}},$

$$\begin{aligned} H_{{\bar{\theta }}}H^{*}_{{\bar{u}}}H_{{\bar{\eta }}}k_\lambda&=H_{{\bar{\theta }}}P\theta \eta (I-P){\bar{\eta }}k_\lambda \\&=H_{{\bar{\theta }}}P\theta k_\lambda -H_{{\bar{\theta }}}P\theta \eta P{\bar{\eta }} k_\lambda \\&=0-\overline{\eta (\lambda )}H_{{\bar{\theta }}}\theta \eta k_\lambda =0, \end{aligned}$$

Similarly,

$$\begin{aligned} H_{{\bar{\eta }}}H^{*}_{{\bar{u}}}H_{{\bar{\theta }}}k_\lambda =0. \end{aligned}$$

Since by $\{k_{\lambda }\}_{\lambda \in {\mathbb {D}}}$ is dense in $H^{2}$ and Lemma 3, we have $D_{\theta }D_{\eta }=D_{\eta }D_{\theta }$.

Lemma 4

Let $f,g\in H^{\infty }$ and f is not constant. Assume there exists a constant $\lambda$ such that $(u-\lambda ){\bar{f}}\in H^{2}$, then $D_{f}D_{g}=D_{g}D_{f}$ implies that $(u-\lambda ){\bar{g}}\in H^{2}$.

Proof

Assume $(u-\lambda ){\bar{f}}\in H^{2}$ and $D_{f}D_{g}=D_{g}D_{f}.$ By Lemma 3, we have $H_{{\bar{f}}}H^{*}_{{\bar{u}}}H_{{\bar{g}}}=H_{{\bar{g}}}H^{*}_{{\bar{u}}}H_{{\bar{f}}}.$ Also, $H_{{\bar{f}}}H^{*}_{\overline{u-\lambda }}H_{{\bar{g}}}=H_{{\bar{g}}}H^{*}_{\overline{u-\lambda }}H_{{\bar{f}}},$ Note that

$$\begin{aligned} H_{(u-\lambda ){\bar{f}}{\bar{g}}} =H_{{\bar{f}}}T_{(u-\lambda ){\bar{g}}}+S_{{\bar{f}}}H_{(u-\lambda ){\bar{g}}} =H_{{\bar{g}}}T_{(u-\lambda ){\bar{f}}}+S_{{\bar{g}}}H_{(u-\lambda ){\bar{f}}}. \end{aligned}$$

An easy computation gives

$$\begin{aligned} H_{(u-\lambda ){\bar{f}}{\bar{g}}}-H_{{\bar{f}}}H_{\overline{u-\lambda }}^{*}H_{{\bar{g}}}&=H_{{\bar{f}}}T_{(u-\lambda ){\bar{g}}}+S_{{\bar{f}}}H_{(u-\lambda ){\bar{g}}}-H_{{\bar{f}}}H_{{\bar{u}}-{\bar{\lambda }}}^{*}H_{{\bar{g}}}\\&=H_{{\bar{f}}}(T_{(u-\lambda ){\bar{g}}}-H_{\overline{u-\lambda }}^{*}H_{{\bar{g}}})+S_{{\bar{f}}}H_{(u-\lambda ){\bar{g}}}\\&=H_{{\bar{f}}}T_{(u-\lambda )}T_{{\bar{g}}}+S_{{\bar{f}}}H_{(u-\lambda ){\bar{g}}}\\&=H_{(u-\lambda ){\bar{f}}}T_{{\bar{g}}}+S_{{\bar{f}}}H_{(u-\lambda ){\bar{g}}}.\\ \end{aligned}$$

Similarly,

$$\begin{aligned} H_{(u-\lambda ){\bar{f}}{\bar{g}}}-H_{{\bar{g}}}H_{\overline{u-\lambda }}^{*}H_{{\bar{f}}}&=H_{(u-\lambda ){\bar{g}}}T_{{\bar{f}}}+S_{{\bar{g}}}H_{(u-\lambda ){\bar{f}}}. \end{aligned}$$

Thus we have

$$\begin{aligned} H_{(u-\lambda ){\bar{f}}}T_{{\bar{g}}}+S_{{\bar{f}}}H_{(u-\lambda ){\bar{g}}} =H_{(u-\lambda ){\bar{g}}}T_{{\bar{f}}}+S_{{\bar{g}}}H_{(u-\lambda ){\bar{f}}}. \end{aligned}$$

(9)

Since ${\bar{f}}(u-\lambda )\in H^{2}, H_{(u-\lambda ){\bar{f}}}=0,$ (9) becoms

$$\begin{aligned} S_{{\bar{f}}}H_{(u-\lambda ){\bar{g}}}=H_{(u-\lambda ){\bar{g}}}T_{{\bar{f}}}. \end{aligned}$$

Hence

$$\begin{aligned} S_{{\bar{f}}}H_{(u-\lambda ){\bar{g}}}1=H_{(u-\lambda ){\bar{g}}}T_{{\bar{f}}}1. \end{aligned}$$

This implies that

$$\begin{aligned} ({\bar{f}}-\overline{f(0)})P_{-}{(u-\lambda ){\bar{g}}}=0. \end{aligned}$$

Since f is not constant, $P_{-}{(u-\lambda ){\bar{g}}}=0$. Hence $(u-\lambda ){\bar{g}}\in H^{2}.$$\square$

Lemma 5

Let u be a nonconstant inner function and $f,g\in H^{\infty }$. Then

$$\begin{aligned} D_{f}D_{g}=D_{g}D_{f} \end{aligned}$$

if and only if one of the following cases holds:

• 1. ${\bar{f}}(u-\lambda )$ and ${\bar{g}}(u-\lambda )$ both belong to $H^{2}$   for some constant $\lambda .$

• 2. a nontrivial linear combination of f and g is constant.

Proof

Assume that ${\bar{f}}(u-\lambda )$ and ${\bar{g}}(u-\lambda )$ are both analytic for some constant $\lambda$. For $z\in {\mathbb {D}}$,

$$\begin{aligned} H_{{\bar{f}}}H^{*}_{{\overline{u}}}H_{{\bar{g}}}k_{z}&=H_{{\bar{f}}}H^{*}_{\overline{u-\lambda }}H_{{\bar{g}}}k_{z}\\&=H_{{\bar{f}}}P(u-\lambda )[{\bar{g}}-\overline{g(z)}]k_{z}\\&=H_{{\bar{f}}}(u-\lambda )[{\bar{g}}-\overline{g(z)}]k_{z}\\&=(I-P){\overline{fg}}(u-\lambda )k_{z}. \end{aligned}$$

Similarly,

$$\begin{aligned} H_{{\bar{g}}}H^{*}_{{\bar{u}}}H_{{\bar{f}}}k_{z} =(I-P){\overline{fg}}(u-\lambda )k_{z}. \end{aligned}$$

Since by $\{k_{\lambda }\}_{\lambda \in {\mathbb {D}}}$ is dense in $H^{2},$$H_{{\bar{f}}}H^{*}_{{\bar{u}}}H_{{\bar{g}}}=H_{{\bar{g}}}H^{*}_{{\bar{u}}}H_{{\bar{f}}}.$ Hence $D_{f}D_{g}=D_{g}D_{f}$ by Lemma 3.

Conversely, by Lemma 3, $D_{f}D_{g}=D_{g}D_{f}$ implies that

$$\begin{aligned} H_{{\bar{f}}}T_{u{\bar{g}}}-H_{u{\bar{f}}}T_{{\bar{g}}}=H_{{\bar{g}}}T_{u{\bar{f}}}-H_{u{\bar{g}}}T_{{\bar{f}}}. \end{aligned}$$

Easy calculations give

$$\begin{aligned}&\big (H_{{\bar{f}}}(1\otimes 1)T_{u{\bar{g}}}-H_{u{\bar{f}}}(1\otimes 1)T_{{\bar{g}}}-H_{{\bar{g}}}(1\otimes 1)T_{u{\bar{f}}}+H_{u{\bar{g}}}(1\otimes 1)T_{{\bar{f}}}\big )T_{z}\\&\quad =H_{{\bar{f}}}(I-T_{z}T_{{\bar{z}}})T_{u{\bar{g}}}T_{z} -H_{u{\bar{f}}}(I-T_{z}T_{{\bar{z}}})T_{{\bar{g}}}T_{z} -H_{{\bar{g}}}(I-T_{z}T_{{\bar{z}}})T_{u{\bar{f}}}T_{z}\\&\qquad +H_{u{\bar{g}}}(I-T_{z}T_{{\bar{z}}})T_{{\bar{f}}}T_{z}\\&\quad =H_{{\bar{f}}}T_{u{\bar{g}}}T_{z}-H_{{\bar{f}}}T_{z}T_{{\bar{z}}}T_{u{\bar{g}}}T_{z} -H_{u{\bar{f}}}T_{{\bar{g}}}T_{z}+H_{u{\bar{f}}}T_{z}T_{{\bar{z}}}T_{{\bar{g}}}T_{z} -H_{{\bar{g}}}T_{u{\bar{f}}}T_{z}\\&\qquad +H_{{\bar{g}}}T_{z}T_{{\bar{z}}}T_{u{\bar{f}}}T_{z} +H_{u{\bar{g}}}T_{{\bar{f}}}T_{z}-H_{u{\bar{g}}}T_{z}T_{{\bar{z}}}T_{{\bar{f}}}T_{z}\\&\quad =H_{{\bar{f}}}T_{u{\bar{g}}}T_{z}-H_{{\bar{f}}}T_{z}T_{u{\bar{g}}} -H_{u{\bar{f}}}T_{{\bar{g}}}T_{z}+H_{u{\bar{f}}}T_{z}T_{{\bar{g}}} -H_{{\bar{g}}}T_{u{\bar{f}}}T_{z}+H_{{\bar{g}}}T_{z}T_{u{\bar{f}}}\\&\qquad +H_{u{\bar{g}}}T_{{\bar{f}}}T_{z}-H_{u{\bar{g}}}T_{z}T_{{\bar{f}}}\\&\quad =H_{{\bar{f}}}T_{u{\bar{g}}}T_{z}-S_{z}H_{{\bar{f}}}T_{u{\bar{g}}} -H_{u{\bar{f}}}T_{{\bar{g}}}T_{z}+S_{z}H_{u{\bar{f}}}T_{{\bar{g}}} -H_{{\bar{g}}}T_{u{\bar{f}}}T_{z}+S_{z}H_{{\bar{g}}}T_{u{\bar{f}}}\\&\qquad +H_{u{\bar{g}}}T_{{\bar{f}}}T_{z}-S_{z}H_{u{\bar{g}}}T_{{\bar{f}}}\\&\quad =(H_{{\bar{f}}}T_{u{\bar{g}}} -H_{u{\bar{f}}}T_{{\bar{g}}} -H_{{\bar{g}}}T_{u{\bar{f}}} +H_{u{\bar{g}}}T_{{\bar{f}}})T_{z}\\&\qquad -S_{z}(H_{{\bar{f}}}T_{u{\bar{g}}} -H_{u{\bar{f}}}T_{{\bar{g}}} -H_{{\bar{g}}}T_{u{\bar{f}}} +H_{u{\bar{g}}}T_{{\bar{f}}})\\&\quad =0. \end{aligned}$$

The first equality follows from $1\otimes 1=I-T_{z}T_{{\bar{z}}},$ and the fourth equality follows from (6) and (7).

Thus we have

$$\begin{aligned} (H_{{\bar{f}}}1)\otimes (T_{{\bar{z}}{\bar{u}}g}1)-(H_{u{\bar{f}}}1)\otimes (T_{{\bar{z}}g}1)=(H_{{\bar{g}}1})\otimes (T_{{\bar{z}}{\bar{u}}f}1)-(H_{u{\bar{g}}}1)\otimes (T_{{\bar{z}}f}1). \end{aligned}$$

(10)

Since

$$\begin{aligned} T_{{\bar{z}}{\bar{u}}g} 1=P {\bar{z}}{\bar{u}}g1=PVu{\bar{g}}=V P_{-}u{\bar{g}}=V H_{u{{\bar{g}}}}1, \end{aligned}$$

Similarly,

$$\begin{aligned} T_{{\bar{z}}g}1&=VH_{{\bar{g}}}1,\\ T_{{\bar{z}}{\bar{u}}f}1&=V H_{u{{\bar{f}}}}1,\\ T_{{\bar{z}}f}1&=VH_{{\bar{f}}}1. \end{aligned}$$

Now (10) becomes

$$\begin{aligned} (H_{{\bar{f}}}1)\otimes (V H_{u{{\bar{g}}}}1)-(H_{u{\bar{f}}}1)\otimes (VH_{{\bar{g}}}1)=(H_{{\bar{g}}}1)\otimes (V H_{u{{\bar{f}}}}1)-(H_{u{\bar{g}}}1)\otimes (VH_{{\bar{f}}}1). \end{aligned}$$

(11)

Obviously, if f or g is constant, then condition (2) hold. Assume that neither of f and g is constant.

Case 1.

Assume that $\left\{ H_{{\bar{f}}}1, H_{u {\bar{f}}} 1\right\}$ is linearly dependent. Hence, there exist a complex number $\lambda$ such that

$$\begin{aligned} H_{u{\bar{f}}}1&=\lambda H_{{\bar{f}}} 1. \end{aligned}$$

So, ${\bar{f}}(u-\lambda )\in H^{2}.$ By Lemma 4, we have ${\bar{g}}(u-\lambda )\in H^{2}.$ Condition (1) would hold.

Case 2.

Assume that $\left\{ H_{{\bar{f}}}1, H_{u {\bar{f}}} 1\right\}$ is linearly independent. Then $\left\{ H_{{\bar{g}}} 1, H_{u {\bar{g}}} 1\right\}$ is also linearly independent. If $\left\{ H_{{\bar{g}}} 1, H_{u {\bar{g}}} 1\right\}$ is linearly dependent, which contradicts to Lemma 4. Since V is anti-unitary, $\left\{ VH_{{\bar{g}}} 1, VH_{u {\bar{g}}} 1\right\}$ is linearly independent, by Gram-Schmidt procedure, there exist a nonzero function $x_{0}$ in $span\left\{ VH_{{\bar{g}}} 1, VH_{u {\bar{g}}} 1\right\}$ such that

$$\begin{aligned} \langle VH_{u{\bar{g}}}1,x_{0}\rangle&=1,\\ \langle VH_{{\bar{g}}}1,x_{0}\rangle&=0. \end{aligned}$$

Applying operator Eq. (11) to $x_{0}$ gives

$$\begin{aligned} H_{{\bar{f}}}1=\langle x_{0},VH_{u{\bar{f}}}1\rangle H_{{\bar{g}}}1 -\langle x_{0},VH_{{\bar{f}}}1\rangle H_{u{\bar{g}}}1. \end{aligned}$$

Hence,

$$\begin{aligned} H_{{\bar{f}}}1\in span\left\{ H_{{\bar{g}}} 1, H_{u {\bar{g}}} 1\right\} . \end{aligned}$$

Similarly,

$$\begin{aligned}&H_{u{\bar{f}}}1\in span\left\{ H_{{\bar{g}}} 1, H_{u {\bar{g}}} 1\right\} ,\\&H_{{\bar{g}}}1\in span\left\{ H_{{\bar{f}}} 1, H_{u {\bar{f}}} 1\right\} ,\\&H_{u{\bar{g}}}1\in span\left\{ H_{{\bar{f}}} 1, H_{u {\bar{f}}} 1\right\} . \end{aligned}$$

Therefore

$$\begin{aligned} span\left\{ H_{{\bar{g}}} 1, H_{u {\bar{g}}} 1\right\} =span\left\{ H_{{\bar{f}}} 1, H_{u {\bar{f}}} 1\right\} , \end{aligned}$$

there exist constants $a_{11}, a_{12}, a_{21}, a_{22}$ such that

$$\begin{aligned} \begin{aligned} H_{{\bar{f}}}1&=a_{11}H_{{{\bar{g}}}}1 +{a_{12}}H_{u{\bar{g}}}1,\\ H_{u{\bar{f}}}1&=a_{21}H_{{{\bar{g}}}}1 +{a_{22}}H_{u{\bar{g}}}1. \end{aligned} \end{aligned}$$

(12)

Replacing above formulas in (11) yields

$$\begin{aligned}&(H_{{\bar{g}}}1)\otimes (\overline{{a_{11}}}VH_{u{\bar{g}}}1 -\overline{a_{21}}VH_{{{\bar{g}}}}1-VH_{u{\bar{f}}}1)\\&\quad =(H_{u{\bar{g}}}1)\otimes (\overline{{a_{12}}}VH_{u{\bar{g}}}1 -\overline{a_{22}}VH_{{{\bar{g}}}}1+VH_{{\bar{f}}}1). \end{aligned}$$

Since $\left\{ H_{{\bar{g}}} 1, H_{u {\bar{g}}} 1\right\}$ is linearly independent,

$$\begin{aligned} VH_{{\bar{f}}}1&=-\overline{{a_{12}}}VH_{u{\bar{g}}}1 +\overline{a_{22}}VH_{{{\bar{g}}}}1,\\ VH_{u{\bar{f}}}1&=\overline{{a_{11}}}VH_{u{\bar{g}}}1 -\overline{a_{21}}VH_{{{\bar{g}}}}1, \end{aligned}$$

which simplifies to

$$\begin{aligned} \begin{aligned} H_{{\bar{f}}}1&=a_{22}H_{{{\bar{g}}}}1 -{a_{12}}H_{u{\bar{g}}}1,\\ H_{u{\bar{f}}}1&=-a_{21}H_{{{\bar{g}}}}1 +{a_{11}}H_{u{\bar{g}}}1. \end{aligned} \end{aligned}$$

(13)

Combining (12) and (13) gives

$$\begin{aligned} a_{11}&=a_{22},\\ a_{12}&=-a_{12},\\ a_{21}&=-a_{21}. \end{aligned}$$

Then, $a_{12}=a_{21}=0.$ Let $a_{11}=a_{22}=c,$ we have $H_{{\bar{f}}}1=c H_{{{\bar{g}}}}1,$ and hence

$$\begin{aligned} {\bar{f}}-c{\bar{g}} \in H^{\infty }. \end{aligned}$$

Since $f,g\in H^{\infty }$, ${\bar{f}}-c{\bar{g}}$ is a constant, condition (2) would hold. $\square$

Corollary 1

If $f,g\in K^{2}_{u}\cap H^{\infty },$ then $D_{f}D_{g}=D_{g}D_{f}.$

Proof

In fact, $K^{2}_{u}=H^{2}\cap u\overline{zH^{2}}\subseteq H^{2}\cap u\overline{H^{2}}.$ Since $f,g\in K_{u}^{2}$, there exist $f_{1},g_{1}\in H^{2}$ such that $f=u{\bar{f}}_{1}$, $g=u{\bar{g}}_{1}$. Thus $u{\bar{f}}=u({\bar{u}}f_{1})=f_{1}\in H^{2}$ and $u{\bar{g}}=u({\bar{u}}g_{1})=g_{1}\in H^{2}$. Hence $D_{f}D_{g}=D_{g}D_{f}$ by Lemma 5. $\square$

Combining Theorem 2 and Lemma 3, we get our main result.

Theorem 3

Let u be a nonconstant inner function and $f,g\in L^{\infty }$.Then

$$\begin{aligned} D_{f}D_{g}=D_{g}D_{f} \end{aligned}$$

if and only if one of the following cases holds:

• 1. f, g, ${\bar{f}}(u-\lambda )$ and ${\bar{g}}(u-\lambda )$ all belong to $H^{2}$ for some constant $\lambda ,$

• 2. ${\bar{f}},{\bar{g}}$, $f(u-\lambda )$ and $g(u-\lambda )$ all belong to $H^{2}$ for some constant $\lambda ,$

• 3. a nontrivial linear combination of f and g is constant.

On classical Hardy space $H^{2},$ assume that any nontrivial linear combination of f and g is not constant, then $T_{f}T_{g}=T_{g}T_{f}$ implies that $T_{f}T_{g}=T_{fg}$. In general, $T_{f}T_{g}=T_{fg}$ does not implies that $T_{f}T_{g}=T_{g}T_{f}$. But on Hilbert spaces of harmonic functions $(K_{u}^{2})^{\perp }=uH^{2}\oplus \overline{zH^{2}}$, $D_{f}D_{g}=D_{fg}$ implies that $D_{f}D_{g}=D_{g}D_{f}$ by the following theorem. However, the converse is not true.

Theorem 4

[7, Theorem 4.7] Let $f,g\in L^{\infty }$ and u be a nonconstant inner function. Then $D_{f}D_{g}=D_{fg}$ if and only if one of the following cases holds:

• 1. $f,g,{\bar{f}}(u-\lambda ),{\bar{g}}(u-\lambda )$ and ${\bar{f}}{\bar{g}}(u-\lambda )$ all belong to $H^{2}$ for some constant $\lambda .$

• 2. ${\bar{f}},{\bar{g}},f(u-\lambda ),g(u-\lambda )$ and $fg(u-\lambda )$ all belong to $H^{2}$ for some constant $\lambda .$

• 3. either f or g is constant.

Example 3

Let $u=\theta \eta ,$ where $\theta$ and $\eta$ are nonconstant inner functions. Let $f=u$,$g=\theta$. Since $u{\bar{f}}=1\in H^{2}$ and $u{\bar{g}}=\eta \in H^{2},D_{f}D_{g}=D_{g}D_{f}$ by Theorem  3. But $u{\bar{f}}{\bar{g}}={\bar{\eta }}$ is not analytic, by the above theorem, we have $D_{f}D_{g}\ne D_{fg}$.