Toeplitz Kernels and Finite-Rank Commutators of Truncated Toeplitz Operators

Abstract

In this paper, using some properties about Toeplitz kernels, we present some results about finite-rank properties of the commutator $[A_f,~A_g]$. Firstly, we show that $[A_{B_n},~A_v^*]$ must have a finite rank on the model space $K_u^2$, where $B_n$ is a finite Blaschke product and v is an inner function. Next, we present that when $\text {ker}~T_{\overline{u}B_n}$ is an invariant subspace of $T_\phi ^*$, then $[A_{B_n},~A_\phi ^*]$ has a finite rank on $K_u^2$ for $\phi \in H^\infty$. Finally, we prove that $[A_{B_n},~A_\phi ^*]$ must have a finite rank on $K_u^2$ when $u=B_nu_1$ for an inner function $u_1$.

1 Introduction

Let ${\mathbb {D}}$ denote the open unit disk in the complex plane ${\mathbb {C}}$ and ${\mathbb {T}}$ denote the unit circle. Denote by $L^2=L^2({\mathbb {T}},~\text {d}m)$ the Hilbert space of square integrable functions with respect to the Lebesgue measure $\text {d}m$ on ${\mathbb {T}}$, normalized so that the measure of the entire circle is 1. Let $L^\infty$ be the space of essentially bounded functions on the unit circle. The Hardy space $H^2$ denotes the Hilbert space of all holomorphic functions in ${\mathbb {D}}$ having square-summable Taylor coefficients at the origin, and it will be identified with the space of boundary functions, the subspace of $L^2$ consisting of functions whose Fourier coefficients with negative indices vanish. Let $H^\infty$ denote the space of all bounded holomorphic functions in ${\mathbb {D}}$ and $C({\mathbb {T}})$ denote the space of all continuous functions on ${\mathbb {T}}$.

Every function in $H^2$, other than the constant function 0, can be factorized into the product of an inner function and an outer function. An inner function is a function $u\in H^\infty$ such that $|u(e^{i\theta })|=1$ almost everywhere with respect to the Lebesgue measure. Every inner function can be factorized into the product of a Blaschke product and a singular inner function. A Blaschke product is an analytic function $B\in H^\infty$ of the form

$$\begin{aligned} B(z)=z^m\prod _{k=1}^\infty \frac{\overline{z_k}}{z_k}\frac{z_k-z}{1-\overline{z_k}z}, \end{aligned}$$

where $\{z_k\}$ are zeros of B counting multiplicity which satisfy that $\sum \limits _k(1-|z_k|)< \infty .$ A nonconstant inner function that has no zeros in ${\mathbb {D}}$ is called a singular inner function $S_\mu$, which has the following form

$$\begin{aligned} S_\mu (z)=c \exp \left( -\int _0^{2\pi }\frac{e^{i\theta }+z}{e^{i\theta }-z}d\mu (\theta )\right) , \end{aligned}$$

where $\mu$ is a finite positive regular Borel measure on $[0,2\pi ]$, singular with respect to the Lebesgue measure and c is a constant of modulus 1. The function $F\in H^2$ is an outer function if F is a cyclic vector of the unilateral shift S. That is,

$$\begin{aligned} \bigvee _0^\infty \{S^kF\}=H^2. \end{aligned}$$

For more properties about Hardy spaces, we can refer to [14].

By Beurling’s theorem [4], the invariant subspace of the unilateral shift operator $Sf=zf$ on $H^2$ has the form $uH^2$, where u is an inner function. It is easy to check that $K_u^2=H^2\ominus uH^2$ is the invariant subspace of the backward shift operator $S^*$ on $H^2$, which is called the model space. Let P denote the orthogonal projection from $L^2$ onto $H^2$ and $P_u$ denote the orthogonal projection from $L^2$ onto $K_u^2$. For $f\in L^\infty$, the Toeplitz operator $T_f$ induced by the symbol f is defined on $H^2$ by

$$\begin{aligned} T_f g=P(f g), ~g\in H^2. \end{aligned}$$

Obviously, $T_f^*=T_{\overline{f}}$. Toeplitz operators acting on $H^2$ have very simple and natural matrix representations via infinite Toeplitz matrices that have constant entries on diagonals parallel to the main one. The Hankel operator $H_f$ induced by the symbol f is defined on $H^2$ by

$$\begin{aligned} H_f g=(I-P)(f g),~g\in H^2. \end{aligned}$$

Then $H^*_f h=P(\overline{f}h)$ for $h\in (H^2)^\perp$. Compressions of Toeplitz operators on $K_u^2$ are called truncated Toeplitz operators, for $\psi \in L^\infty$, which are defined by

$$\begin{aligned} A_\psi f=P_u(\psi f), ~f\in K_u^2. \end{aligned}$$

The function $\psi$ is called the symbol of $A_\psi$. Clearly, $A_\psi ^*=A_{\overline{\psi }}$.

Truncated Toeplitz operators represent a far reaching generalization of classical Toeplitz matrices. Although a particular case had appeared before in the literature, the general theory has been initiated in the seminal paper [21]. Since then, truncated Toeplitz operators have constituted an active area of research. We mention only a few relevant papers [5, 11, 17, 20] and so on. On the operator theory level, Nagy and others showed that $A_z$ is a model for a certain class of contraction operators [24]. Every contraction operator T on the Hilbert space H having defect indices (1, 1) and such that $\lim \limits _{n\rightarrow \infty }T^{*n}=0~ (\text {SOT})$ is unitarily equivalent to $A_z$ for an inner function u, where $\text {SOT}$ denotes the strong operator topology. In terms of function theory, Douglas et al. [9] in 1970 showed that functions in the model space and inner functions have the same analytic continuation neighborhood, that is, assuming that u could be analytically extended on $E\subseteq {\mathbb {T}}$, then all functions in the model space $K_u^2$ could be analytically extended on E. Therefore, the rational functions approximate the functions in the model space, the boundary value of functions in the model space, the angular derivative of inner functions and the relationship between them are deeply studied, refer to [2] and [3]. Thus, the research on truncated Toeplitz operators is of representative significance.

D. Sarason once proposed in [22]: what $f,~g\in H^2$ can make $T_fT_{\overline{g}}$ to be a bounded operator. From this, many scholars begin to study the boundedness of the product of Toeplitz operators and Hankel operators. By $T_{fg}-T_fT_g=H_{\overline{f}}^*H_g$, properties of commutators of Toeplitz operators also gradually begin to enter the research line of sight. The map $\tau :~f\rightarrow T_f$ is a contractive $*$-linear mapping from $L^\infty$ to $L(H^2)$, where $L(H^2)$ is the algebra of all bounded linear operators on $H^2$. But this mapping is not multiplicative. When f is a continuous function, by Hartman’s theorem (see Corollary 4.3.3 in [18]) we get that $T_{fg}-T_fT_g=H_{\overline{f}}^*H_g$ is compact for any $g\in L^\infty$. Thus, $\tau (fg)=\tau (f)\tau (g)+K$ for $f\in C({\mathbb {T}})$, where K is a compact operator. This makes the multiplicability problem very interesting, and many scholars begin to study the compactness of commutators of Toeplitz operators. In addition, the compactness of commutators of Toeplitz operators also originated from the research on Fredholm theory of Toeplitz operators in 1970s by Douglas, Sarason and others.

The results about compact or finite-rank (semi-)commutators of $T_f$ or $H_f$ are quite complete, corresponding results are summarized in [26]. But there are very few results for compact or finite-rank (semi-)commutators of truncated Toeplitz operators. In [11], Garcia concluded that $A_{fg}-A_fA_g$ is compact for $f,~g\in C({\mathbb {T}})$. In [7], authors described the kernels and ranks of commutators of truncated Toeplitz operators with symbols induced by finite Blaschke products. By $L^2=K_u^2\bigoplus (K_u^2)^\bot ,$ for $f\in L^\infty$, the multiplication operator $M_f$ is expressed as an operator matrix

$$\begin{aligned} M_f= \begin{pmatrix} A_f &{} B_{\overline{f}}^* \\ B_f &{} D_f \\ \end{pmatrix}, \end{aligned}$$

where $D_f$ denotes the dual truncated Toeplitz operator on $L^2\ominus K_u^2$ defined by

$$\begin{aligned} D_fh=(I-P_u)(fh),~h\in L^2\ominus K_u^2. \end{aligned}$$

It is easy to see that $D_f^*=D_{\overline{f}}.$ The truncated Hankel operator $B_f$ is defined by

$$\begin{aligned} B_f\varphi =(I-P_u)(f\varphi ),~\varphi \in K_u^2. \end{aligned}$$

Moreover, $B_f^*h=P_u(\overline{f}h)~\text {for}~h\in L^2\ominus K_u^2.$ By $M_f M_g=M_{fg},$ we get that

$$\begin{aligned} A_{fg }=A_f A_g+B_{\overline{f}}^*B_g. \end{aligned}$$

(1)

In [19], authors introduced sufficient and necessary conditions for $B_f$ to be of finite-rank or compact. By (1), we know some sufficient conditions for $A_f A_g-A_{fg}$ to be of finite-rank or compact. From this, we can study the compact commutator $[A_f,~A_g]$ by truncated Hankel operators $B_f$. In [26], authors of this article and others gave some results that commutators of truncated Toeplitz operators are compact or of finite-rank operators on model spaces. In our paper, using the structure of Toeplitz kernels, we present some results about finite-rank properties of commutators $[A_f,~A_g]$.

The paper is organized as follows. In Sect. 2, we recall some necessary definitions and properties about model spaces and truncated Toeplitz operators. In Sect. 3, we obtain that $[A_{B_n},~A_v^*]$ must have a finite rank on $K_u^2$, where $B_n$ is a finite Blaschke product and v is an inner function. In Sect. 4, using some properties of Toeplitz kernels, when $\text {ker}~T_{\overline{u}B_n}$ is an invariant subspace of $T_\phi ^*$, we show that $[A_{B_n},~A_\phi ^*]$ has a finite rank on $K_u^2$ for $\phi \in H^\infty$. In particular, we present that $[A_{B_n},~A_\phi ^*]$ has a finite rank on $K_u^2$ when $u=B_nu_1$ for an inner function $u_1$.

2 Preliminaries

In this section we introduce some basic properties of truncated Toeplitz operators. The reproducing kernel of $H^2$ at $\lambda \in {\mathbb {D}}$ is the function $k_\lambda (z)=\frac{1}{1-\overline{\lambda } z}$, and it is easy to check that the reproducing kernel of $K_u^2$ at $\lambda \in {\mathbb {D}}$ is the function

$$\begin{aligned} k_\lambda ^u(z)=(P_u k_\lambda )(z)=\frac{1-\overline{u(\lambda )}u(z)}{1-\overline{\lambda } z}. \end{aligned}$$

It is well known that $K_u^2$ carries a natural conjugation C, antiunitary, involution operator, defined by $Cf=\overline{zf}u$ for $f\in K_u^2$. We have that

$$\begin{aligned} \widetilde{k_\lambda ^u}(z)=(Ck_\lambda ^u)(z)=\frac{u(z)-u(\lambda )}{z-\lambda }, \end{aligned}$$

which is the conjugation reproducing kernel of $K_u^2$ at $\lambda \in {\mathbb {D}}$. That is,

$$\begin{aligned} {\widetilde{f}}(\lambda )=(Cf)(\lambda )=\langle \widetilde{k_\lambda ^u}, f\rangle , \end{aligned}$$

for $f\in K_u^2$. A bounded linear operator A on $K_u^2$ is called C-symmetric if

$$\begin{aligned} CAC=A^*. \end{aligned}$$

S. R. Garcia and M. Putinar introduced some properties of C-symmetric operators in [12], and they showed that all truncated Toeplitz operators are C-symmetric. More complex symmetric operators can be found in [13].

The systematic study of truncated Toeplitz operators was initiated by Sarason [21], and an intensive study revealed many interesting and different properties about these operators. For example, unbounded symbols may give bounded truncated Toeplitz operators, see Sarason’s example in [21]. For $f\in L^2$, Sarason in [21] proved that $A_f=0$ if and only if $f\in uH^2+\overline{uH^2}$. Therefore, the symbols of truncated Toeplitz operators are not unique.

It is well known that the model space $K_u^2$ is the kernel space of $T_{\overline{u}}$. There are many important connections between model spaces and Toeplitz kernels. In [6], authors presented some classical results about the relationship between Toeplitz kernels and model spaces. Using the relationship between Toeplitz kernels and model spaces, authors showed maximal vectors for model spaces by maximal vectors for Toeplitz kernels. Moreover, they also discussed the multiplier between Toeplitz kernels by the multiplier between model spaces.

For Toeplitz kernels, one classical result is the Coburn theorem (see Proposition 7.24 in [8]). It is said that either $\text {ker}~T_g=\{0\}$ or $\text {ker}~T_g^*=\{0\}$ for $g\in L^\infty .$ In 1986, Hayashi [15] showed that the kernel of $T_g$ can be written as $\phi K_\eta ^2$, where $\phi$ is an outer function and $\eta$ is an inner function with $\eta (0)=0$, and the function $\phi$ multiplies $K_\eta ^2$ isometrically onto $\text {ker}~T_g$. It is easy to check that the Toeplitz kernel is nearly $S^*$-invariant. In 1988, Hitt [16] showed that any nearly $S^*$-invariant of subspace M is of form $hK_\theta ^2$, where $h\in M$ meeting some conditions and $\theta$ is an inner function. Sarason [23] gave a new proof of Hitt’s theorem and presented better description of h and $\theta$. More about the research process of Toeplitz kernels can be found in [6].

The following lemma is well known and we provide a proof for the sake of completeness.

Lemma 1

If u and v are inner functions, then

$$\begin{aligned} \text {ker}~T_{\overline{u}v}=\{\varphi \in K_u^2:~v\varphi \in K_u^2\}. \end{aligned}$$

(2)

Moreover, $\text {ker}~T_{\overline{u}v}\ne \{0\}$ if and only if $v H^2\cap K_u^2\ne \{0\}.$

Proof

Denote $E=\{\varphi \in K_u^2:~v\varphi \in K_u^2\}.$ For $\phi \in E\subset K_u^2$, since the model space $K_u^2$ has the conjugation, there exists $\eta \in K_u^2$ such that $v\phi =u\overline{z\eta }.$ Then

$$\begin{aligned} T_{\overline{u}v}\phi =P(\overline{u}v\phi )=P(\overline{u}u\overline{z\eta })=P(\overline{z\eta })=0. \end{aligned}$$

It implies that

$$\begin{aligned} \phi \in \text {ker}~T_{\overline{u}v}~\text {and}~E\subseteq \text {ker}~T_{\overline{u}v}. \end{aligned}$$

For $\varphi \in \text {ker}~T_{\overline{u}v},$ we have that $T_{\overline{u}v}\varphi =P(\overline{u}v\varphi )=0,$ and there exists $x\in H^2$ such that

$$\begin{aligned} \overline{u}v\varphi =\overline{zx}. \end{aligned}$$

(3)

Then $v\varphi =u\overline{zx}\in H^2.$ Since

$$\begin{aligned} K_u^2=u\overline{zH^2}\cap H^2, \end{aligned}$$

(4)

we get that $v\varphi \in K_u^2.$ By (3), we have that $\varphi =u\overline{zxv}\in H^2.$ By (4), we conclude that $\varphi \in K_u^2~\text {and}~\varphi \in E.$ Thus $\text {ker}~T_{\overline{u}v} \subseteq E$ and we have proved that $\text {ker}~T_{\overline{u}v} =E.$

Suppose that $v H^2\cap K_u^2\ne \{0\}.$ There exists $0\ne h\in H^2$ such that $vh\in K_u^2$. Since $K_u^2$ has a conjugation, there is $0\ne g\in K_u^2$ such that $vh=u\overline{zg}.$ Then

$$\begin{aligned} T_{\overline{u}v}h=P(\overline{u}vh)=P(\overline{u}u\overline{zg})=0. \end{aligned}$$

It implies that $h\in \text {ker}~T_{\overline{u}v},$ and $\text {ker}~T_{\overline{u}v}\ne \{0\}$. By (2), it is easy to get that $v H^2\cap K_u^2\ne \{0\}$ when $\text {ker}~T_{\overline{u}v}\ne \{0\}$. The proof is completed. $\square$

3 The Finite-Rank Property of $[A_{B_n},~A_v^*]$ for an Inner Function v

By [11] we know that $A_fA_g-A_{fg}$ is compact for $f,~g\in C({\mathbb {T}})$. Thence $[A_{B_n},~A_{v}^*]$ must be compact for any finite Blaschke product $B_n$ and any inner function v. This makes us want to discuss when $[A_{B_n},~A_{v}^*]$ has a finite rank on $K_u^2$. In the following we show that $[A_{B_n},~A_{v}^*]$ must have a finite rank for any inner function v.

In the following we will frequently use the following relationship:

$$\begin{aligned} T_\psi ^*K_u^2\subseteq K_u^2~\text {and}~A_\psi ^*=T_\psi ^*|_{K_u^2}, \end{aligned}$$

for any $\psi \in H^\infty$ and an inner function u.

We use $\text {Hol}({\mathbb {D}})$ to denote the set of all holomorphic functions in ${\mathbb {D}}$. For a pair of inner functions v and $\eta$, we explore multipliers

$$\begin{aligned} {\mathcal {M}}(v,~\eta )=\{\phi \in \text {Hol}({\mathbb {D}}):~\phi K_v^2\subseteq K_\eta ^2\} \end{aligned}$$

between model spaces $K_v^2$ and $K_\eta ^2$.

Lemma 2

(Corollary 3.3 in [10]) If $v,~\eta$ are inner functions, then

$$\begin{aligned} {\mathcal {M}}(v,~\eta )\cap H^\infty =\text {ker}~T_{\overline{z\eta }v}\cap H^\infty \subseteq {\mathcal {M}}(v,~\eta )\subseteq \text {ker}~T_{\overline{z\eta }v}. \end{aligned}$$

A finite Blaschke product is a function of the form $B_n(z)=c\prod \limits _{i=1}^n\frac{z-z_i}{1-\overline{z_i}z}$ for $z_i\in {\mathbb {D}}$. The degree of a finite Blaschke product $B_n$ is its number of zeros.

Lemma 3

(Theorem 4.3 in [10]) If B is a finite Blaschke product and v is any inner function with the infinite degree, then ${\mathcal {M}}(B,~v)\cap H^\infty \ne \{0\}.$

Lemma 4

(Lemma 2.1 in [7]) Let $a_1,~a_2,~\cdot \cdot \cdot ,~a_n$ be points in ${\mathbb {D}}$ and put

$$\begin{aligned} B_n(z)=\prod \limits _{i=1}^n\frac{z-a_i}{1-\overline{a_i}z}, \end{aligned}$$

then

$$\begin{aligned} K_{B_n}^2=\frac{{\mathcal {P}}_{n-1}}{\prod \limits _{i=1}^n(1-\overline{a_i}z)}, \end{aligned}$$

where ${\mathcal {P}}_k$ denotes the set of all analytic polynomials with the degree less than or equal to k. In particular, $\text {dim}~K_{B_n}^2=n.$

Theorem 1

Let u be a nonconstant inner function and $K_u^2$ be the infinite dimensional model space. If $B_n$ is a finite Blaschke product with degree n, then $[A_{B_n},~A_{v}^*]$ has a finite rank on $K_u^2$ for any inner function v and $\text {rank}[A_{B_n},~A_{v}^*]\le 2n$. Moreover, if $\text {ker}~T_{\overline{u}B_n v}\ne \{0\}$, then

$$\begin{aligned} v \text {ker}~T_{\overline{u}B_n v}\subseteq \text {ker}~[A_{B_n },~A_{v}^*]~\text { and} ~\text {ran}~[A_{B_n },~A_{v}^*] \subseteq K_u^2\ominus B_n \text {ker}~T_{\overline{u}B_n v}. \end{aligned}$$

Proof

Since $zB_n$ is a finite Blaschke product, by Lemma 3, we get that

$$\begin{aligned} {\mathcal {M}}(zB_n,~u)\cap H^\infty \ne \{0\}. \end{aligned}$$

Then by Lemma 2,

$$\begin{aligned} \text {ker}~T_{\overline{zu}zB_n}=\text {ker}~T_{\overline{u}B_n}\ne \{0\}. \end{aligned}$$

By Lemma 1, we know that

$$\begin{aligned} \text {ker}~T_{\overline{u}B_n}=\{f\in K_u^2:~B_n f\in K_u^2\}. \end{aligned}$$

Then $B_n g\in K_u^2$ for $g\in \text {ker}~T_{\overline{u}B_n}.$ By $H^2=v H^2\oplus K_{v}^2,$ there exist $g_1\in H^2$ and $g_2\in K_{v}^2$ such that $g=v g_1+g_2.$ Then

$$\begin{aligned} \begin{aligned} {[}A_{B_n},~A_{v}^*{]}g&=(A_{B_n}A_{v}^*-A_{v}^*A_{B_n})g\\&=A_{B_n}T_{v}^*g-A_{v}^*A_{B_n}g\\&=P_u(B_n P(\overline{v}g))-P_u(\overline{v}P_u(B_n g))\\&=P_u(B_n P(\overline{v}g))-P_u(\overline{v}B_n g)\\&=P_u(B_n P(\overline{v}v g_1))-P_u(\overline{v}B_n v g_1)+P_u(B_n P(\overline{v}g_2))-P_u(\overline{v}B_n g_2)\\&=P_u(B_n g_1)-P_u(B_n g_1)+P_u(B_n P(\overline{v}g_2))-P_u(\overline{v}B_n g_2)\\&=P_u(B_n P(\overline{v}g_2))-P_u(\overline{v}B_n g_2). \end{aligned} \end{aligned}$$

Since $g_2\in K_{v}^2$, we get that $\overline{v}g_2\in \overline{zH^2}$ and $P(\overline{v}g_2)=0.$ Then

$$\begin{aligned}{}[A_{B_n},~A_{v}^*]g=-P_u(\overline{v}B_n g_2)=-P_uP(\overline{v}B_n g_2). \end{aligned}$$

We claim that $P(\overline{v}B_n K_{v}^2)\subseteq K_{B_n}^2$. In fact, for any $f\in K_{v}^2$ and $h\in H^2$, we have that

$$\begin{aligned} \langle P(\overline{v}B_n f),~B_n h \rangle =\langle f,~v h \rangle =0. \end{aligned}$$

Thence

$$\begin{aligned} \begin{aligned} {[}A_{B_n},~A_{v}^*{]}\mathrm{ker}~T_{\overline{u}B_n}\subseteq \mathrm{ran}~ P_uP_{B_n}. \end{aligned} \end{aligned}$$

By Lemma 4, we obtain that $\text {dim}~K_{B_n}^2=n$. Then $\text {dim}~(\text {ran}~P_uP_{B_n})\le n,$ and

$$\begin{aligned} \begin{aligned} \text {dim}~\left( [A_{B_n},~A_{v}^*]\text {ker}~T_{\overline{u}B_n}\right) \le n<\infty . \end{aligned} \end{aligned}$$

(5)

Since

$$\begin{aligned} \begin{aligned} K_u^2=\text {ker}~T_{\overline{u}B_n} \oplus (K_u^2\ominus \text {ker}~T_{\overline{u}B_n}), \end{aligned} \end{aligned}$$

(6)

in the following we consider the dimension of $[A_{B_n},~A_{v}^*](K_u^2\ominus \text {ker}~T_{\overline{u}B_n}).$ By

$$\begin{aligned} H^2=B_n H^2\oplus K_{B_n}^2, \end{aligned}$$

there exist $h_1\in H^2$ and $h_2\in K_{B_n}^2$ such that $h=B_n h_1+h_2$ for any $h\in H^2.$ Then

$$\begin{aligned} T_{\overline{B_n}u}h=P(\overline{B_n}uB_n h_1)+P(\overline{B_n}uh_2)=uh_1+P(\overline{B_n}uh_2). \end{aligned}$$

By Lemma 4, we know that $\text {dim}~K_{B_n}^2=n$. It follows that

$$\begin{aligned} \begin{aligned} \text {dim}~P(\overline{B_n}uK_{B_n}^2)\le n<\infty . \end{aligned} \end{aligned}$$

(7)

It is easy to check that $P(\overline{B_n}u\phi )\in K_u^2$ for any $\phi \in K_{B_n}^2$. Thus

$$\begin{aligned} \begin{aligned} T_{\overline{B_n}u}H^2\subseteq uH^2\oplus P(\overline{B_n}uK_{B_n}^2). \end{aligned} \end{aligned}$$

Consequently,

$$\begin{aligned} cl~(\text {ran}~T_{\overline{B_n}u})\subseteq uH^2\oplus P(\overline{B_n}uK_{B_n}^2), \end{aligned}$$

where the abbreviation “ cl " denotes the closure of a set. Since

$$\begin{aligned} (\text {ker}~T_{\overline{u}B_n})^\perp =cl~(\text {ran}~T_{\overline{B_n}u}), \end{aligned}$$

we get that

$$\begin{aligned} K_u^2\ominus \text {ker}~T_{\overline{u}B_n}=cl~(\text {ran}~T_{\overline{B_n}u})\cap K_u^2 \subseteq P(\overline{B_n}uK_{B_n}^2)\cap K_u^2. \end{aligned}$$

By $P(\overline{B_n}u\phi )\in K_u^2$ for any $\phi \in K_{B_n}^2$, we get that $P(\overline{B_n}uK_{B_n}^2)\subset K_u^2.$ It follows that

$$\begin{aligned} P(\overline{B_n}uK_{B_n}^2)\cap K_u^2 =P(\overline{B_n}uK_{B_n}^2). \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} K_u^2\ominus \text {ker}~T_{\overline{u}B_n}\subseteq P(\overline{B_n}uK_{B_n}^2). \end{aligned} \end{aligned}$$

(8)

By (7) and (8), we have that

$$\begin{aligned} \text {dim}~(K_u^2\ominus \text {ker}~T_{\overline{u}B_n})\le n<\infty . \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} \text {dim}~([A_{B_n},~A_{v}^*](K_u^2\ominus \text {ker}~T_{\overline{u}B_n}))\le n<\infty . \end{aligned} \end{aligned}$$

(9)

By (5), (6) and (9), we get that

$$\begin{aligned} \begin{aligned} \text {dim}~\left( [A_{B_n},~A_{v}^*]K_u^2\right)&=\text {dim}~\left( [A_{B_n},~A_{v}^*](\text {ker}~T_{\overline{u}B_n} \oplus (K_u^2\ominus \text {ker}~T_{\overline{u}B_n}))\right) \\&=\text {dim}~\left( [A_{B_n},~A_{v}^*]\text {ker}~T_{\overline{u}B_n}+[A_{B_n},~A_{v}^*](K_u^2\ominus \text {ker}~T_{\overline{u}B_n})\right) \\&\le \text {dim}~[A_{B_n},~A_{v}^*]\text {ker}~T_{\overline{u}B_n}+\text {dim}~[A_{B_n},~A_{v}^*](K_u^2\ominus \text {ker}~T_{\overline{u}B_n})\\&\le 2n< \infty . \end{aligned} \end{aligned}$$

Thus $[A_{B_n},~A_{v}^*]$ has a finite rank on $K_u^2$ and $\text {rank}[A_{B_n},~A_{v}^*]\le 2n$.

Suppose that $\text {ker}~T_{\overline{u}B_n v}\ne \{0\}$. By Lemma 1, we get that $vB_n \phi \in K_u^2$ for any $\phi \in \text {ker}~T_{\overline{u}B_n v}$. Then there exists $\psi \in K_u^2$ such that

$$\begin{aligned} vB_n \phi =u\overline{z\psi }. \end{aligned}$$

That is, $v \phi =u\overline{z\psi B_n}.$ By $K_u^2=H^2\cap u\overline{zH^2}$, we have that $v\phi \in K_u^2$. Then

$$\begin{aligned} \begin{aligned} {[}A_{B_n},~A_{v}^*{]}v\phi&=(A_{B_n}A_{v}^*-A_{v}^*A_{B_n})v\phi \\&=A_{B_n}T_{v}^*v\phi -A_{v}^*A_{B_n}v\phi \\&=P_u(B_nP(\overline{v}v\phi ))-P_u(\overline{v}P_u(B_n v\phi ))\\&=P_u(B_n \phi )-P_u(\overline{v}B_n v\phi )\\&=P_u(B_n \phi )-P_u(B_n \phi )\\&=0. \end{aligned} \end{aligned}$$

Thus $v \text {ker}~T_{\overline{u}B_n v}\subseteq \text {ker}~[A_{B_n},~A_{v}^*]$. By the same way, we get that

$$\begin{aligned} B_n \text {ker}~T_{\overline{u}B_n v}\subseteq \text {ker}~[A_v,~A_{B_n}^*]. \end{aligned}$$

(10)

Since

$$\begin{aligned} K_u^2\ominus \text {ker}~[A_v,~A_{B_n}^*]=cl~(\text {ran}~[A_v,~A_{B_n}^*]^*)=cl~(\text {ran}~[A_{B_n},~A_{v}^*]), \end{aligned}$$

and $[A_{B_n},~A_{v}^*]$ has a finite rank, we obtain that

$$\begin{aligned} K_u^2\ominus \text {ker}~[A_v,~A_{B_n}^*]=\text {ran}~[A_{B_n},~A_{v}^*]. \end{aligned}$$

Then by (10),

$$\begin{aligned} \text {ran}~[A_{B_n},~A_{v}^*]\subseteq K_u^2\ominus B_n \text {ker}~T_{\overline{u}B_n v}. \end{aligned}$$

$\square$

In the following we give an example illustrating Theorem 1 and use $\text {span}\{h\}$ to denote the space generated by the function h.

Example 1

Let $B_1(z)=\frac{z-a}{1-\overline{a}z}$ for $a\in {\mathbb {D}}$ and $v(z)=\exp \frac{z+1}{z-1}$ be a singular inner function. If $u=B_1v$ and $K_u^2$ is the corresponding model space, then

$$\begin{aligned} {[}A_{B_1},~A_{v}^*{]}K_u^2\subseteq \text {span}\left\{ k_a\right\} \oplus \text {span}\left\{ P_u\left( B_1k_a \right) \right\} , \end{aligned}$$

and $\text {rank}[A_{B_1},~A_{v}^*]\le 2$, where $k_a(z)=\frac{1}{1-\overline{a}z}.$

Proof

By Lemma 4, we know that

$$\begin{aligned} K_{B_1}^2=\text {span}\left\{ k_a\right\} ~\text {and}~ \text {dim}~K_{B_1}^2=1. \end{aligned}$$

(11)

Since $u=B_1v$, we have that $K_{B_1}^2\subsetneq K_u^2$. Using the proof of Theorem 1 and (11), we get that

$$\begin{aligned} {[}A_{B_1},~A_v^*{]}\text {ker}~T_{\overline{u}B_1}\subseteq \text {ran}~ P_uP_{B_1}= \text {span}\left\{ P_u\left( k_a\right) \right\} =\text {span}\left\{ k_a\right\} =K_{B_1}^2. \end{aligned}$$

(12)

By (8) and (11), we have that

$$\begin{aligned} K_u^2\ominus \text {ker}~T_{\overline{u}B_1} \subseteq P(\overline{B_1}uK_{B_1}^2)= \text {span}\left\{ k_a\cdot \exp \frac{z+1}{z-1}\right\} . \end{aligned}$$

It implies that

$$\begin{aligned} \begin{aligned} {[}A_{B_1},~A_{v}^*{]}(K_u^2\ominus \text {ker}~T_{\overline{u}B_1})&\subseteq [A_{B_1},~A_{v}^*]\text {span}\left\{ k_a\cdot \exp \frac{z+1}{z-1}\right\} . \end{aligned} \end{aligned}$$

In fact,

$$\begin{aligned} \begin{aligned} {[}A_{B_1},~A_{v}^*{]}(k_a\cdot \exp \frac{z+1}{z-1})&=[A_{B_1},~A_{v}^*](v k_a)\\&=(A_{B_1}A_{\overline{v}}-A_{\overline{v}}A_{B_1})(vk_a)\\&=A_{B_1}P(\overline{v}vk_a)-A_{\overline{v}}P_u(B_1vk_a)\\&=A_{B_1}k_a-A_{\overline{v}}P_u(uk_a)\\&=P_u(B_1k_a ). \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} {[}A_{B_1},~A_{v}^*{]}(K_u^2\ominus \text {ker}~T_{\overline{u}B_1}) \subseteq \text {span}\left\{ P_u\left( B_1k_a \right) \right\} . \end{aligned}$$

(13)

By (12) and (13), we conclude that

$$\begin{aligned} \begin{aligned} {[}A_{B_1},~A_{v}^*{]}K_u^2&= [A_{B_1},~A_{v}^*](\text {ker}~T_{\overline{u}B_1} \oplus (K_u^2\ominus \text {ker}~T_{\overline{u}B_1}))\\&\subseteq K_{B_1}^2 \oplus \text {span}\left\{ P_u\left( B_1\cdot k_a \right) \right\} . \end{aligned} \end{aligned}$$

Then by (11),

$$\begin{aligned} \text {dim}~([A_{B_1},~A_{v}^*]K_u^2)\le 2. \end{aligned}$$

$\square$

Remark 1

From the proof of Theorem 1, we know that $P_uP_{B_n}$ has a finite rank because $B_n$ is the finite Blaschke product. In fact, we have the following claim.

Claim: The projection $P_uP_\theta$ has a finite rank if and only if u or $\theta$ is a finite Blaschke product.

Using the following lemma, we give a further proof of the claim.

Lemma 5

(Section 6 in [1]) Hankel operators $H_f$ have a finite rank if and only if $f\in \overline{b}H^\infty$, where b is the finite Blaschke product.

Proof

If u or $\theta$ is a finite Blaschke product, it is obvious that $P_uP_\theta$ has a finite rank. Suppose that $P_uP_\theta$ has a finite rank. It is easy to get that

$$\begin{aligned} P_uP_\theta |_{H^2}=H_{\overline{u}}^*H_{\overline{u}}H_{\overline{\theta }}^*H_{\overline{\theta }}. \end{aligned}$$

(14)

By (14), we have that

$$\begin{aligned} H_{\overline{\theta }}H_{\overline{u}}^*H_{\overline{u}}H_{\overline{\theta }}^*H_{\overline{\theta }} H_{\overline{u}}^*H_{\overline{u}}H_{\overline{\theta }}^*= (H_{\overline{\theta }}H_{\overline{u}}^*)(H_{\overline{\theta }} H_{\overline{u}}^*)^*(H_{\overline{\theta }}H_{\overline{u}}^*)(H_{\overline{\theta }}H_{\overline{u}}^*)^* \end{aligned}$$

has a finite rank. Then $H_{\overline{\theta }}H_{\overline{u}}^*$ has a finite rank. Lemma 1 in [1] says that the operator $H_{\overline{\theta }}H_{\overline{u}}^*$ is compact if and only if $H_{\overline{u}}^*H_{\overline{\theta }}$ is compact. By the same way as in Lemma 1 in [1], we can get that $H_{\overline{\theta }}H_{\overline{u}}^*$ has a finite rank if and only if $H_{\overline{u}}^*H_{\overline{\theta }}$ has a finite rank. Using the fact that $H_{\overline{\theta }}H_{\overline{u}}^*$ has a finite rank if and only if $H_{\overline{\theta }}$ or $H_{\overline{u}}$ has a finite rank (see Theorem 4 in [1]), by Lemma 5, we get that $\overline{\theta }=\overline{b}h$ or $\overline{u}=\overline{b_1}h_1$, where b and $b_1$ are finite Blaschke products and $h,~h_1\in H^\infty$. It implies that $\theta$ or u is a finite Blaschke product. $\square$

For u and v inner functions, the compactness of $P_uP_v$ reflects the asymptotically orthogonal relationship of $K_u^2$ and $K_v^2$. We say that model spaces $K_u^2$ and $K_v^2$ are asymptotically orthogonal if $P_uP_v$ is a compact operator. Moreover, the following statements are equivalent.

• (a) Model spaces $K_u^2$ and $K_v^2$ are asymptotically orthogonal;

• (b) $T_uT_v^*-T_{u\overline{v}}$ is compact;

• (c) $H^\infty [\overline{u}]\cap H^\infty [\overline{v}]=H^\infty +C$, where $H^\infty [\overline{u}]$ denotes the Douglas algebra generated by $\overline{u}$ and $H^\infty$;

• (d) For each support set S, either $u|_s$ or $v|_s$ is a constant;

• (e) $\lim \limits _{|z|\rightarrow 1}\max (|u(z)|,~|v(z)|)=1$;

• (f) $H_{\overline{v}}|_{K_u^2}$ is compact;

• (g) $T_{\overline{v}}|_{K_u^2}$ is a compact perturbation of an isometry.

In fact, by $T_uT_v^*-T_{u\overline{v}}=-H_{\overline{u}}^*H_{\overline{v}}$, using the same meaning as in the proof that $P_uP_v$ has a finite rank, we get that $(a)\Leftrightarrow (b)$. The equivalence of (b) and (c) comes from Theorem 1 in [1] and [25]. By Lemma 3 in [1], we get that $(c)\Leftrightarrow (d)\Leftrightarrow (e)$. The proof of $(b)\Leftrightarrow (f)\Leftrightarrow (g)$ can be found in Theorem 2 in [1].

4 The Finite-Rank Property of $[A_{B_n},~A_\phi ^*]$ for $\phi \in H^\infty$

It is well known that $[A_{B_n},~A_\phi ^*]$ is compact when $B_n$ is a finite Blaschke product and $\phi \in H^\infty$. In this section assuming that $\text {ker}~T_{\overline{u}B_n}$ is an invariant subspace of $T_\phi ^*$, we present that $[A_{B_n},~A_\phi ^*]$ has a finite rank on $K_u^2$.

Theorem 2

Let u be a nonconstant inner function and $K_{u}^2$ be the infinite dimensional model space. If $T_\phi ^*(\text {ker}~T_{\overline{u}B_n})\subseteq \text {ker}~T_{\overline{u}B_n}$, then $[A_{B_n},~A_\phi ^*]$ has a finite rank on $K_u^2$ and $\text {rank}[A_{B_n},~A_\phi ^*]\le 2n$, where $\phi \in H^\infty$ and $B_n$ is a finite Blaschke product with degree n.

Proof

By Lemmas 2 and 3, we get that $\text {ker}~T_{\overline{u}B_n}\ne \{0\}.$ For $f\in \text {ker}~T_{\overline{u}B_n}$, by Lemma 1, we have that $B_nf\in K_u^2$. Then

$$\begin{aligned} \begin{aligned} {[}A_{B_n},~A_{\phi }^*{]}f&=(A_{B_n}A_{\phi }^*-A_{\phi }^*A_{B_n})f\\&=A_{B_n}T_{\phi }^*f-T_{\phi }^*A_{B_n}f\\&=P_u(B_n T_{\phi }^*f)-P(\overline{\phi }P_u(B_n f))\\&=P_u(B_n T_{\phi }^*f)-P(\overline{\phi }B_n f)\\&=P_u(B_n T_{\phi }^*f)-T_{B_n\overline{\phi }}f. \end{aligned} \end{aligned}$$

Since $T_\phi ^*(\text {ker}~T_{\overline{u}B_n})\subseteq \text {ker}~T_{\overline{u}B_n}$, by Lemma 1, we get that $B_n T_{\phi }^*f\in K_u^2$. Then

$$\begin{aligned} \begin{aligned} {[}A_{B_n},~A_{\phi }^*{]}f=B_n T_{\phi }^*f-T_{B_n\overline{\phi }}f=(T_{B_n} T_{\phi }^*-T_{B_n\overline{\phi }})f=-H_{\overline{B_n}}^*H_{\overline{\phi }}f. \end{aligned} \end{aligned}$$

It is easy to check that $\text {ker}~H_{\overline{B_n}}=B_n H^2$. Then

$$\begin{aligned} cl(\text {ran}~H_{\overline{B_n}}^*)=(\text {ker}~H_{\overline{B_n}})^\perp =H^2\ominus B_n H^2=K_{B_n}^2. \end{aligned}$$

By Lemma 4, we obtain that $\text {dim}~(K_{B_n}^2)=n$. It implies that

$$\begin{aligned} \begin{aligned} \text {dim}~[A_{B_n},~A_{\phi }^*](\text {ker}~T_{\overline{u}B_n})\le n< \infty . \end{aligned} \end{aligned}$$

(15)

Since

$$\begin{aligned} \begin{aligned} K_u^2=\text {ker}~T_{\overline{u}B_n} \oplus (K_u^2\ominus \text {ker}~T_{\overline{u}B_n}), \end{aligned} \end{aligned}$$

(16)

in the following we consider the dimension of $[A_{B_n},~A_{\phi }^*](K_u^2\ominus \text {ker}~T_{\overline{u}B_n}).$ By

$$\begin{aligned} H^2=B_nH^2\oplus K_{B_n}^2, \end{aligned}$$

there exist $h_1\in H^2$ and $h_2\in K_{B_n}^2$ such that $h=B_nh_1+h_2$ for any $h\in H^2.$ Then

$$\begin{aligned} T_{\overline{B_n}u}h=P(\overline{B_n}uB_nh_1)+P(\overline{B_n}uh_2)=uh_1+P(\overline{B_n}uh_2). \end{aligned}$$

We claim that

$$\begin{aligned} \begin{aligned} P(\overline{B_n}u\varphi )\in K_u^2~\text { for~ any}~ \varphi \in K_{B_n}^2. \end{aligned} \end{aligned}$$

(17)

In fact, for any $\psi \in H^2$, we have that

$$\begin{aligned} \langle P(\overline{B_n}u\varphi ), u\psi \rangle =\langle \varphi , B_n \psi \rangle =0. \end{aligned}$$

Thus

$$\begin{aligned} T_{\overline{B_n}u}H^2\subseteq uH^2\oplus P(\overline{B_n}uK_{B_n}^2). \end{aligned}$$

This implies that

$$\begin{aligned} cl~(\text {ran}~T_{\overline{B_n}u})\subseteq uH^2\oplus P(\overline{B_n}uK_{B_n}^2). \end{aligned}$$

Since $(\text {ker}~T_{\overline{u}B_n})^\perp =cl~(\text {ran}~T_{\overline{B_n}u}),$ we get that

$$\begin{aligned} \begin{aligned} K_u^2\ominus \text {ker}~T_{\overline{u}B_n}=cl~(\text {ran}~T_{\overline{B_n}u})\cap K_u^2 \subseteq (uH^2\oplus P(\overline{B_n}uK_{B_n}^2))\cap K_u^2. \end{aligned} \end{aligned}$$

(18)

We claim that

$$\begin{aligned} \begin{aligned} (uH^2\oplus P(\overline{B_n}uK_{B_n}^2))\cap K_u^2\subseteq P(\overline{B_n}uK_{B_n}^2). \end{aligned} \end{aligned}$$

(19)

For $g\in (uH^2\oplus P(\overline{B_n}uK_{B_n}^2))\cap K_u^2$, there are $g_1\in H^2$ and $g_2\in P(\overline{B_n}uK_{B_n}^2)$ such that

$$\begin{aligned} g=ug_1+g_2\in K_u^2, \end{aligned}$$

we have that

$$\begin{aligned} 0=\langle ug_1, ug_1+g_2\rangle =\Vert g_1\Vert ^2+\langle ug_1,g_2\rangle . \end{aligned}$$

By (17), we obtain that $\langle ug_1,g_2\rangle =0$. Then $\Vert g_1\Vert ^2=0$ and $g_1=0$. This implies that $g=g_2$ and (19) holds. By (17), (18) and (19), we obtain that

$$\begin{aligned} K_u^2\ominus \text {ker}~T_{\overline{u}B_n}\subseteq P(\overline{B_n}uK_{B_n}^2). \end{aligned}$$

By Lemma 4, we get that $\text {dim}~K_{B_n}^2=n$ and $\text {dim}~(K_u^2\ominus \text {ker}~T_{\overline{u}B_n})\le n<\infty .$ Then

$$\begin{aligned} \begin{aligned} \text {dim}~([A_{B_n},~A_{B_\phi }^*](K_u^2\ominus \text {ker}~T_{\overline{u}B_n}))\le n<\infty . \end{aligned} \end{aligned}$$

(20)

By (15), (16) and (20), we get that

$$\begin{aligned} \begin{aligned} \text {dim}~([A_{B_n},~A_{\phi }^*]K_u^2)&=\text {dim}~\left( [A_{B_n},~A_{\phi }^*](\text {ker}~T_{\overline{u}B_n} \oplus (K_u^2\ominus \text {ker}~T_{\overline{u}B_n}))\right) \\&=\text {dim}~\left( [A_{B_n},~A_{\phi }^*]\text {ker}~T_{\overline{u}B_n}+[A_{B_n},~A_{\phi }^*](K_u^2\ominus \text {ker}~T_{\overline{u}B_n})\right) \\&\le \text {dim}~[A_{B_n},~A_{\phi }^*]\text {ker}~T_{\overline{u}B_n}+\text {dim}~[A_{B_n},~A_{\phi }^*](K_u^2\ominus \text {ker}~T_{\overline{u}B_n})\\&\le 2n< \infty . \end{aligned} \end{aligned}$$

Thus $[A_{B_n},~A_{\phi }^*]$ has a finite rank on $K_u^2$ and $\text {rank}[A_{B_n},~A_\phi ^*]\le 2n$.

$\square$

Remark 2

For the condition of Theorem 2, we give the following explanation. For any $f\in \text {ker}~T_{\overline{u}B_n}$, we have that $T_\phi ^*(\text {ker}~T_{\overline{u}B_n})\subseteq \text {ker}~T_{\overline{u}B_n}$ if and only if $B_nT_{\phi }^*f\in K_u^2$ if and only if $P_u(\phi {\widetilde{f}})\in B_n\text {ker}~T_{\overline{u}B_n}$, where ${\widetilde{f}}=C(f)$.

Proof

By $\text {ker}~T_{\overline{u}B_n}=\{f\in K_u^2:~B_nf\in K_u^2\},$ it is easy to get that $\text {ker}~T_{\overline{u}B_n}$ is invariant under $T_\phi ^*$ if and only if $B_nT_{\phi }^*f\in K_u^2$.

Suppose that $B_nT_{\phi }^*f\in K_u^2$ for $f\in \text {ker}~T_{\overline{u}B_n}$. There exists $g\in K_u^2$ such that

$$\begin{aligned} B_nT_{\phi }^*f=u\overline{zg}. \end{aligned}$$

By $K_u^2=H^2\cap u\overline{zH^2}$, we get that

$$\begin{aligned} \begin{aligned} T_{\phi }^*f=P(\overline{\phi }f)=u\overline{zgB_n}\in K_u^2. \end{aligned} \end{aligned}$$

(21)

There is $g_1\in H^2$ such that

$$\begin{aligned} \overline{\phi }f=u\overline{zgB_n}+\overline{zg_1}. \end{aligned}$$

Since $f\in K_u^2$, there exists $f_1\in K_u^2$ such that $f=u\overline{zf_1}$. Then $\overline{\phi }u\overline{zf_1}=u\overline{zgB_n}+\overline{zg_1}.$ This implies that

$$\begin{aligned} \phi f_1=gB_n+ug_1~\text { and} ~P_u(\phi f_1)=P_u(gB_n). \end{aligned}$$

By (21), we have that

$$\begin{aligned} B_ng\in K_u^2~\text { and} ~g\in \text {ker}~T_{\overline{u}B_n}. \end{aligned}$$

Then $P_u(\phi f_1)=gB_n.$ By $f=u\overline{zf_1},$ we get that $f_1=u\overline{zf}=C(f)={\widetilde{f}}.$ Thus, the function $P_u(\phi {\widetilde{f}})$ belongs to $B_n\text {ker}~T_{\overline{u}B_n}.$

Suppose that $P_u(\phi {\widetilde{f}})\in B_n\text {ker}~T_{\overline{u}B_n}$ for $f\in \text {ker}~T_{\overline{u}B_n}$. There exist functions $g_1\in \text {ker}~T_{\overline{u}B_n}$ and $g_2\in H^2$ such that

$$\begin{aligned} \phi {\widetilde{f}}=\phi C(f)=B_ng_1+ug_2. \end{aligned}$$

Then

$$\begin{aligned} \overline{\phi }u\overline{zC(f)}=u\overline{zB_ng_1}+\overline{zg_2}. \end{aligned}$$

By $C(f)=u\overline{zf}$, we have that

$$\begin{aligned} \overline{\phi }u\overline{zu\overline{zf}}=u\overline{zB_ng_1}+\overline{zg_2}. \end{aligned}$$

That is,

$$\begin{aligned} \overline{\phi }f=u\overline{zB_ng_1}+\overline{zg_2}. \end{aligned}$$

This implies that

$$\begin{aligned} P(\overline{\phi }f)=P(u\overline{zB_ng_1}). \end{aligned}$$

Since $g_1\in \text {ker}~T_{\overline{u}B_n}$, we get that $B_ng_1\in K_u^2$ and $u\overline{zB_ng_1}\in K_u^2$. Then

$$\begin{aligned} P(\overline{\phi }f)=u\overline{zB_ng_1}~ \text {and}~ B_nT_\phi ^*f=u\overline{zg_1}. \end{aligned}$$

By $K_u^2=H^2\cap u\overline{zH^2}$, we have that $B_nT_\phi ^*f=u\overline{zg_1}\in K_u^2.$ The proof is completed. $\square$

Corollary 1

Let u be a nonconstant inner function and $K_{u}^2$ be the infinite dimensional model space. If $u=B_nu_1$ for a finite Blaschke $B_n$ and an inner function $u_1$, then $[A_{B_n},~A_\phi ^*]$ has a finite rank on $K_u^2$ for $\phi \in H^\infty$.

Proof

Since $u=B_nu_1$, we get that

$$\begin{aligned} T_\phi ^*(\text {ker}~T_{\overline{u}B_n})=T_\phi ^*(\text {ker}~T_{\overline{u_1}})=T_\phi ^* K_{u_1}^2\subseteq K_{u_1}^2=\text {ker}~T_{\overline{u_1}}=\text {ker}~T_{\overline{u}B_n}. \end{aligned}$$

By Theorem 2, we obtain that $[A_{B_n},~A_\phi ^*]$ has a finite rank on $K_u^2$. $\square$

Example 2

Let $B_1(z)=\frac{z-a}{1-\overline{a}z}$ for $a\in {\mathbb {D}}$ and $v(z)=\exp \frac{z+1}{z-1}$ be a singular inner function. If $u=B_1v$ and $K_u^2$ is the corresponding model space, then the following statements hold.

• (a) If $\phi =1+B_1$ is an outer function, then

  $$\begin{aligned}{}[A_{B_1},~A_{\phi }^*]K_u^2\subseteq \text {span}\left\{ k_a\right\} \oplus \text {span}\left\{ P_u\left( B_1P(\overline{B_1}vk_a) \right) \right\} , \end{aligned}$$

  and $\text {rank}[A_{B_1},~A_{\phi }^*]\le 2$, where $k_a(z)=\frac{1}{1-\overline{a}z}.$

• (b) If $\phi =1+v$ is an outer function, then

  $$\begin{aligned} {[}A_{B_1},~A_{\phi }^*{]}K_u^2\subseteq \text {span}\left\{ k_a\right\} \oplus \text {span}\left\{ P_u\left( B_1k_a \right) \right\} , \end{aligned}$$

  and $\text {rank}[A_{B_1},~A_{\phi }^*]\le 2$, where $k_a(z)=\frac{1}{1-\overline{a}z}.$

Proof

By Lemma 4, we know that

$$\begin{aligned} K_{B_1}^2=\text {span}\left\{ k_a\right\} ~\text {and}~ \text {dim}~K_{B_1}^2=1, \end{aligned}$$

(22)

Since $u=B_1v$, we have that $K_{B_1}^2\subsetneq K_u^2$. Using the proof of Theorem 2, we get that

$$\begin{aligned} {[}A_{B_1},~A_{\phi }^*{]}\text {ker}~T_{\overline{u}B_1}\subseteq cl(\text {ran}~H_{\overline{B_1}}^*)=K_{B_1}^2, \end{aligned}$$

(23)

and

$$\begin{aligned} \begin{aligned} {[}A_{B_1},~A_{\phi }^*{]}(K_u^2\ominus \text {ker}~T_{\overline{u}B_1})\subseteq [A_{B_1},~A_{\phi }^*]P(\overline{B_1}uK_{B_1}^2) = [A_{B_1},~A_{\phi }^*]\text {span}\left\{ vk_a \right\} .\qquad \end{aligned} \end{aligned}$$

(24)

• (a) If $\phi =1+B_1$, then

  $$\begin{aligned} \begin{aligned} {[}A_{B_1},~A_{\phi }^*{]}(vk_a)&=(A_{B_1}A_{\overline{1+B_1}}-A_{\overline{1+B_1}}A_{B_1})(vk_a)\\&=A_{B_1}P(\overline{1+B_1}vk_a)-A_{\overline{1+B_1}}P_u(B_1vk_a)\\&=A_{B_1}(vk_a)+A_{B_1}P(\overline{B_1}vk_a)\\&=P_u\left( B_1P(\overline{B_1}vk_a)\right) . \end{aligned} \end{aligned}$$

  By (23) and (24), we conclude that

  $$\begin{aligned} \begin{aligned} {[}A_{B_1},~A_{\phi }^*{]}K_u^2&= [A_{B_1},~A_{\phi }^*](\text {ker}~T_{\overline{u}B_1} \oplus (K_u^2\ominus \text {ker}~T_{\overline{u}B_1}))\\&\subseteq K_{B_1}^2 \oplus \text {span}\left\{ P_u\left( B_1P(\overline{B_1}vk_a) \right) \right\} . \end{aligned} \end{aligned}$$

  Then by (22),

  $$\begin{aligned} \text {dim}~([A_{B_1},~A_{\phi }^*]K_u^2)\le 2. \end{aligned}$$

• (b) If $\phi =1+v$, then

  $$\begin{aligned} \begin{aligned} {[}A_{B_1},~A_{\phi }^*{]}(vk_a)&=(A_{B_1}A_{\overline{1+v}}-A_{\overline{1+v}}A_{B_1})(vk_a)\\&=A_{B_1}P(\overline{1+v}vk_a)-A_{\overline{1+v}}P_u(B_1vk_a)\\&=A_{B_1}(vk_a)+A_{B_1}P(\overline{v}vk_a)\\&=P_u\left( B_1k_a\right) . \end{aligned} \end{aligned}$$

  By (23) and (24), we conclude that

  $$\begin{aligned} \begin{aligned} {[}A_{B_1},~A_{\phi }^*{]}K_u^2&= [A_{B_1},~A_{\phi }^*](\text {ker}~T_{\overline{u}B_1} \oplus (K_u^2\ominus \text {ker}~T_{\overline{u}B_1}))\\&\subseteq K_{B_1}^2 \oplus \text {span}\left\{ P_u\left( B_1k_a\right) \right\} . \end{aligned} \end{aligned}$$

  Then by (22),

  $$\begin{aligned} \text {dim}~([A_{B_1},~A_{\phi }^*]K_u^2)\le 2. \end{aligned}$$

$\square$