Asymmetric Truncated Toeplitz Operators of Rank One

Abstract

A truncated Toeplitz operator is a compression of the multiplicationoperator to a backward shift invariant subspace of the Hardy space $H^2$. Anasymmetric truncated Toeplitz operator is a compression of the multiplication operator that acts between two different backward shift invariant subspaces of $H^2$. All rank-one truncated Toeplitz operators have been described by Sarason. Here, we characterize all rank-one asymmetric truncated Toeplitz operators. This completes the description given by Łanucha for asymmetric truncated Toeplitz operators on finite-dimensional backward shift invariant subspaces.

1 Introduction

Denote by $H^2$ the Hardy space of the open unit disk $\mathbb {D}=\{z:|z|<1\}$ and let P be the orthogonal projection from $L^2(\partial \mathbb {D})$ onto $H^2$.

A classical Toeplitz operator $T_{\varphi }$ with symbol $\varphi \in L^{\infty }(\partial \mathbb {D})$ is defined on $H^2$ by

$$\begin{aligned} T_{\varphi }f=P(\varphi f). \end{aligned}$$

It is clear that a Toeplitz operator with symbol from $L^{\infty }(\partial \mathbb {D})$ is bounded. Note that if $\varphi \in L^{2}(\partial \mathbb {D})$, then the above definition produces an operator $T_{\varphi }$, densely defined on $H^{\infty }=H^2\cap L^{\infty }(\partial \mathbb {D})$. It is known that such $T_{\varphi }$ extends boundedly to $H^2$ if and only if $\varphi \in L^{\infty }(\partial \mathbb {D})$.

The operators $S=T_z$ and $S^{*}=T_{\overline{z}}$ are called the unilateral shift and the backward shift, respectively. Obviously, $Sf(z)=zf(z)$ and a simple computation reveals that

$$\begin{aligned} S^{*}f(z)=\frac{f(z)-f(0)}{z}. \end{aligned}$$

A truncated Toeplitz operator is a compression of a classical Toeplitz operator to a model space. Recall, that model spaces are the closed subspaces of $H^2$ of the form

$$\begin{aligned} K_{\alpha }=H^2\ominus \alpha H^2, \end{aligned}$$

where $\alpha$ is an inner function, i.e., $\alpha \in H^{\infty }$ and $|\alpha |=1$ a.e. on $\partial \mathbb {D}$. As aconsequence of the theorem of Beurling, model spaces are the typical non-trivial$S^{*}$-invariant subspaces of $H^2$.

Extensive study of the class of truncated Toeplitz operators began in 2007 with Sarason’s paper [10]. Generalizations of these operators, the so-called asymmetric truncated Toeplitz operators, were recently introduced in [2, 4, 5].

Let $\alpha$, $\beta$ be two inner functions. An asymmetric truncated Toeplitz operator $A_{\varphi }^{\alpha ,\beta }$ with symbol $\varphi \in L^2(\partial \mathbb {D})$ is given by

$$\begin{aligned} A_{\varphi }^{\alpha ,\beta }f=P_{\beta }(\varphi f),\quad f\in K_{\alpha }, \end{aligned}$$

where $P_{\beta }$ is the orthogonal projection from $L^2(\partial \mathbb {D})$ onto $K_{\beta }$. In particular, a truncated Toeplitz operator is the operator $A_{\varphi }^{\alpha }=A_{\varphi }^{\alpha ,\alpha }$.

Since $P_{\beta }(\varphi f)\in K_{\beta }$ for $f\in K_{\alpha }^{\infty }=K_{\alpha }\cap H^{\infty }$ and $K_{\alpha }^{\infty }$ is dense in $K_{\alpha }$, the operator $A_{\varphi }^{\alpha ,\beta }$ is densely defined. It is known that the boundedness of $\varphi$ is not necessary for $A_{\varphi }^{\alpha ,\beta }$ to extend boundedly to $K_{\alpha }$.

Put

$$\begin{aligned} \mathscr {T}(\alpha ,\beta )=\{A_{\varphi }^{\alpha ,\beta }\ :\ \varphi \in L^2(\partial \mathbb {D})\ \mathrm {and}\ A_{\varphi }^{\alpha ,\beta }\ \mathrm {is\ bounded}\} \end{aligned}$$

and $\mathscr {T}(\alpha )=\mathscr {T}(\alpha ,\alpha )$.

Sarason showed that the truncated Toeplitz operators of rank one are generated (in a sense to be explained below) by functions of the form

$$\begin{aligned} k_{w}^{\alpha }(z)=\frac{1-\overline{\alpha (w)}\alpha (z)}{1-\overline{w}z} \end{aligned}$$

and

$$\begin{aligned} \widetilde{k}_{w}^{\alpha }(z)=\frac{\alpha (z)-\alpha (w)}{z-w}. \end{aligned}$$

The function $k_{w}^{\alpha }$, called the reproducing kernel, is the kernel function for the point evaluation functional $f\mapsto f(w)$, which for $w\in \mathbb {D}$ is bounded on $K_{\alpha }$. In other words, $f(w)=\langle f,k_{w}^{\alpha }\rangle$ for every $f\in K_{\alpha }$ and each $w\in \mathbb {D}$. It is easy to check that $k_{w}^{\alpha }=P_{\alpha }k_w$, where $k_w(z)=(1-\overline{w}z)^{-1}$.

The function $\widetilde{k}_{w}^{\alpha }$ is the image of the reproducing kernel $k_{w}^{\alpha }$ under the map

$$\begin{aligned} C_{\alpha }f(z)=\alpha (z)\overline{zf(z)},\quad |z|=1. \end{aligned}$$

One can verify that $C_{\alpha }$ is an antilinear and isometric involution on $L^2(\partial \mathbb {D})$ which preserves $K_{\alpha }$. The map $C_{\alpha }$ is called a conjugation and, since $\widetilde{k}_{w}^{\alpha }=C_{\alpha }k_{w}^{\alpha }$, the function $\widetilde{k}_{w}^{\alpha }$ is often called the conjugate kernel.

It should be noted that the functional $f\mapsto f(w)$ is bounded on $K_{\alpha }$ also for $w\in \partial \mathbb {D}$ such that the function $\alpha$ has an angular derivative in the sense of Carathéodory (an ADC) at w. Recall that $\alpha$ is said to have an ADC at $w\in \partial \mathbb {D}$ if there exist complex numbers $\alpha (w)$ and $\alpha '(w)$ such that $\alpha (z)\rightarrow \alpha (w)\in \partial \mathbb {D}$ and $\alpha '(z)\rightarrow \alpha '(w)$ whenever z from $\mathbb {D}$ tends to w non-tangentially (that is, with $|z-w|/(1-|z|)$ bounded). In that case, every $f\in K_{\alpha }$ has a non-tangential limit f(w) at w, the functions ${k}_{w}^{\alpha }$ and $\widetilde{k}_{w}^{\alpha }$ belong to $K_{\alpha }$, and $f(w)=\langle f,k_{w}^{\alpha }\rangle$ (for more details, see for example [6, Thm. 7.4.1.]).

It was proved in [10] that the only rank-one operators in $\mathscr {T}(\alpha )$ are the scalar multiples of $\widetilde{k}_{w}^{\alpha } \otimes {k}_{w}^{\alpha }$ and ${k}_{w}^{\alpha } \otimes \widetilde{k}_{w}^{\alpha }$ for all $w\in \mathbb {D}$ and all $w\in \partial \mathbb {D}$ such that $\alpha$ has an ADC at w.

In the asymmetric case, the authors in [8] showed that the operators $\widetilde{k}_{w}^{\beta } \otimes {k}_{w}^{\alpha }$ and ${k}_{w}^{\beta } \otimes \widetilde{k}_{w}^{\alpha }$ belong to $\mathscr {T}(\alpha ,\beta )$ for every $w\in \mathbb {D}$ and every $w\in \partial \mathbb {D}$ such that both $\alpha$ and $\beta$ have an ADC at w. It is natural to ask: are the scalar multiples of these the only rank-one operators in $\mathscr {T}(\alpha ,\beta )$? This issue was addressed in [9]. The author in [9] proved that if $K_{\alpha }$ is one dimensional and $K_{\beta }$ has a finite dimension larger than two (or $K_{\beta }$ is one dimensional and $\dim K_{\alpha }>2$), then the answer is negative: there exists a rank-one operator in $\mathscr {T}(\alpha ,\beta )$ that is neither a scalar multiple of $\widetilde{k}_{w}^{\beta } \otimes {k}_{w}^{\alpha }$ nor a scalar multiple of ${k}_{w}^{\beta } \otimes \widetilde{k}_{w}^{\alpha }$. Nevertheless, it was also showed in [9] that the answer is positive for all other finite-dimensional cases (that is, whenever $\alpha$ and $\beta$ are two finite Blaschke products):

Theorem 1.1

([9, Thm. 4.5]) Let $\alpha$ and $\beta$ be two finite Blaschke products of degree $m>0$ and $n>0$, respectively. The only rank-one operators in $\mathscr {T}(\alpha ,\beta )$ are the non-zero scalar multiples of the operators $\widetilde{k}_w^{\beta }\otimes k_w^{\alpha }$ and $k_w^{\beta }\otimes \widetilde{k}_w^{\alpha }$, $w\in \overline{\mathbb {D}}$, if and only if either $mn\le 2$, or $m>1$ and $n>1$.

In this paper, we resolve the issue in the general case. In particular, we prove that if both $K_{\alpha }$ and $K_{\beta }$ have dimension larger than one (and not necessarily finite), then the only rank-one operators in $\mathscr {T}(\alpha ,\beta )$ are the non-zero scalar multiples of $\widetilde{k}_{w}^{\beta } \otimes {k}_{w}^{\alpha }$ and ${k}_{w}^{\beta } \otimes \widetilde{k}_{w}^{\alpha }$.

2 Rank-One Asymmetric Truncated Toeplitz Operators

In [10], Sarason gave several characterizations of the operators from $\mathscr {T}(\alpha )$. Forexample, he proved that a bounded linear operator A on $K_{\alpha }$ is a truncated Toeplitz operator if and only if $A-S_\alpha A S^*_\alpha$ is a rank-two operator of a special kind, where $S_{\alpha }=A_{z}^{\alpha }$ is the so-called compressed shift. Similar characterizations of asymmetric truncated Toeplitz operators were given in [2, 3] and [9] for some particular cases. Here, we use the following two characterizations of asymmetric truncated Toeplitz operators.

Theorem 2.1

([7, Thm. 2.1]) Let A be a bounded linear operator from $K_\alpha$ into $K_\beta$. Then, $A\in \mathscr {T(\alpha ,\beta )}$ if and only if there exist $\psi \in K_\beta$ and $\chi \in K_\alpha$ such that

$$\begin{aligned} A-S_\beta A S^*_\alpha =\psi \otimes k^\alpha _0 +k^\beta _0\otimes \chi . \end{aligned}$$

Corollary 2.2

([7, Cor. 2.7(c)]) Let A be a bounded linear operator from $K_\alpha$ into $K_\beta$. Then, $A\in \mathscr {T}(\alpha ,\beta )$ if and only if A is shift invariant, that is,

$$\begin{aligned} \left\langle AS f, S g\right\rangle =\left\langle A f,g \right\rangle \end{aligned}$$

for all $f\in K_\alpha$, $g\in K_\beta$ such that $Sf\in K_\alpha$, $Sg\in K_\beta$.

Note here that if $f\in K_\alpha$, then $Sf\in K_\alpha$ if and only if f is orthogonal to $\widetilde{k}_{0}^{\alpha }$ (see [10, p. 512]).

In his proof of [10, Thm. 5.1], Sarason uses the fact that every operator from $\mathscr {T}(\alpha )$ is $C_{\alpha }$-symmetric. Recall that an operator A on $K_\alpha$ is said to be $C_{\alpha }$-symmetric if $C_{\alpha }AC_{\alpha }=A^{*}$. In particular, $C_{\alpha }S_{\alpha }C_{\alpha }=S_{\alpha }^{*}$. In the case of asymmetric truncated Toeplitz operators it is known that $A\in \mathscr {T}(\alpha ,\beta )$ if and only if $C_{\beta }AC_{\alpha }\in \mathscr {T}(\alpha ,\beta )$ (see [9, p. 9]). So, here we take the approach from [9], that is, we use the following lemma (compare with [9, Lem. 4.4]; see also [10, pp. 503–504]).

Lemma 2.3

Let $\alpha$ and $\beta$ be two inner functions such that $K_{\alpha }$ and $K_{\beta }$ have dimension $m>1$ and $n>1$, respectively. Let $f\in K_{\alpha }$, $g\in K_{\beta }$ be two non-zero functions such that $g\otimes f$ belongs to $\mathscr {T}(\alpha ,\beta )$ and let $w\in \overline{\mathbb {D}}$. Then,

• (a) g is a scalar multiple of $k_w^{\beta }$ if and only if f is a scalar multiple of $\widetilde{k}_w^{\alpha }$,

• (b) g is a scalar multiple of $\widetilde{k}_w^{\beta }$ if and only if f is a scalar multiple of $k_w^{\alpha }$.

Proof

Let $f\in K_{\alpha }$, $g\in K_{\beta }$ be two non-zero functions such that $g\otimes f$ belongs to $\mathscr {T}(\alpha ,\beta )$.

We first show that if g is a scalar multiple of $k_w^{\beta }$ for some $w\in \overline{\mathbb {D}}$, then f is a scalar multiple of $\widetilde{k}_w^{\alpha }$. Actually, without loss of generality we can assume that $g=k_w^{\beta }$ for some $w\in \overline{\mathbb {D}}$.

Note here, that for $w\in \partial \mathbb {D},$ we automatically assume that $\beta$ has an ADC at w (this is implied by the fact that $k_w^{\beta }\in K_{\beta }$) and we must prove that $\alpha$ also has an ADC at w.

We consider two cases.

Case 1. $\beta (0)=0$.

Since the operator $k_w^{\beta }\otimes f$ belongs to $\mathscr {T}(\alpha ,\beta )$, it must be shift invariant in the sense of Corollary 2.2. That is,

$$\begin{aligned} \langle (k_w^{\beta }\otimes f)Sh_{\alpha },Sh_{\beta }\rangle =\langle (k_w^{\beta }\otimes f)h_{\alpha },h_{\beta }\rangle \end{aligned}$$

for all $h_{\alpha }\in K_{\alpha }$ and $h_{\beta }\in K_{\beta }$ such that $Sh_{\alpha }\in K_{\alpha }$ and $Sh_{\beta }\in K_{\beta }$. Equivalently,

$$\begin{aligned} \begin{aligned} \langle (k_w^{\beta }\otimes f)Sh_{\alpha },Sh_{\beta }\rangle -\langle (k_w^{\beta }\otimes f)h_{\alpha },h_{\beta }\rangle =\,&\langle Sh_{\alpha },f\rangle \langle k_w^{\beta },Sh_{\beta }\rangle -\langle h_{\alpha },f\rangle \langle k_w^{\beta },h_{\beta }\rangle \\ =\,&\langle h_{\alpha },\langle h_{\beta },S_{\beta }^{*}k_w^{\beta }\rangle S_{\alpha }^{*}f-\langle h_{\beta },k_w^{\beta }\rangle f\rangle =0 \end{aligned} \end{aligned}$$

for all $h_{\alpha }\in K_{\alpha }$ and $h_{\beta }\in K_{\beta },$ such that $h_{\alpha }\perp \widetilde{k}_{0}^{\alpha }$ and $h_{\beta }\perp \widetilde{k}_{0}^{\beta }$. But it follows from

$$\begin{aligned} S_{\beta }^{*}k_w^{\beta }=\overline{w}k_w^{\beta }-\overline{\beta (w)}\widetilde{k}_{0}^{\beta } \end{aligned}$$

(see [10, Lem. 2.2(a)] and its Corollary) that for every $h_{\beta }\perp \widetilde{k}_{0}^{\beta },$ we have

$$\begin{aligned} \langle h_{\beta },S_{\beta }^{*}k_w^{\beta }\rangle =w\langle h_{\beta },k_w^{\beta }\rangle . \end{aligned}$$

Thus, the shift invariance of $k_w^{\beta }\otimes f$ means that

$$\begin{aligned} \langle k_w^{\beta },h_{\beta }\rangle \langle h_{\alpha },wS_{\alpha }^{*}f- f\rangle =0 \end{aligned}$$

(2.1)

for all $h_{\alpha }\perp \widetilde{k}_{0}^{\alpha }$ and $h_{\beta }\perp \widetilde{k}_{0}^{\beta }$.

Since the dimension of $K_{\beta }$ is greater than one, the functions $k_w^{\beta }$ and $\widetilde{k}_0^{\beta }$ are linearly independent. Indeed, if $k_w^{\beta }$ and $\widetilde{k}_0^{\beta }$ were linearly dependent, then so would be $\widetilde{k}_w^{\beta }$ and ${k}_0^{\beta }\equiv 1$. But, here it is easy to verify that if $\widetilde{k}_w^{\beta }=c{k}_0^{\beta }=c$, then $\beta (z)=cz$ and $K_{\beta }$ is one dimensional—a contradiction.

Therefore, there exists a function $h_{\beta }\in K_{\beta }$ orthogonal to $\widetilde{k}_0^{\beta }$ and such that $\langle k_w^{\beta },h_{\beta }\rangle \ne 0$. Putting such $h_{\beta }$ into (2.1), we see that

$$\begin{aligned} \langle h_{\alpha },wS_{\alpha }^{*}f- f\rangle =0 \end{aligned}$$

for all $h_{\alpha }\in K_{\alpha }$, $h_{\alpha }\perp \widetilde{k}_{0}^{\alpha }$. Thus,

$$\begin{aligned} (I_{K_{\alpha }}-wS_{\alpha }^{*})f=c \widetilde{k}_{0}^{\alpha } \end{aligned}$$

or, using $C_{\alpha }$-symmetry of $S_{\alpha }$,

$$\begin{aligned} (I_{K_{\alpha }}-\overline{w}S_{\alpha })C_{\alpha }f=\overline{c} k_{0}^{\alpha } \end{aligned}$$

(2.2)

for some complex number c (here $I_{K_{\alpha }}$ denotes the identity operator on $K_{\alpha }$).

If $c=0$ and $|w|<1$, then, by (2.2) and invertibility of $I_{K_{\alpha }}-\overline{w}S_{\alpha }$, we have $C_{\alpha }f=0$. This implies that $f=0$ which contradicts the assumption that f is non-zero.

If $c=0$ and $|w|=1$, then by (2.2),

$$\begin{aligned} S_{\alpha }C_{\alpha }f=wC_{\alpha }f, \end{aligned}$$

which again is not possible since $S_{\alpha }$ has no eigenvalues on $\partial \mathbb {D}$ ([10, Lem. 2.5]).

Therefore, $c\ne 0$ and without loss of generality we can assume that $c=1$. Hence, instead of (2.2) we consider

$$\begin{aligned} (I_{K_{\alpha }}-\overline{w}S_{\alpha })C_{\alpha }f=k_{0}^{\alpha }. \end{aligned}$$

(2.3)

As above, if $|w|<1$, then $I_{K_{\alpha }}-\overline{w}S_{\alpha }$ is invertible. Since $(I_{K_{\alpha }}-\overline{w}S_{\alpha })k_w^{\alpha }=k_0^{\alpha }$ (for $w=0$ this is obvious, for $w\ne 0$ see [10, Lem. 2.2(b)]), we have

$$\begin{aligned} f=C_{\alpha }(I_{K_{\alpha }}-\overline{w}S_{\alpha })^{-1} k_{0}^{\alpha }=C_{\alpha }k_w^{\alpha }=\widetilde{k}_w^{\alpha }. \end{aligned}$$

If $|w|=1$, then applying $C_{\alpha }$ to both sides of (2.3) and using $C_{\alpha }$-symmetry of $S_{\alpha }$ we get

$$\begin{aligned} f(z)-w\cdot \frac{f(z)-f(0)}{z}=\frac{\alpha (z)-\alpha (0)}{z},\quad z\in \mathbb {D}. \end{aligned}$$

Hence,

$$\begin{aligned} \frac{\alpha (z)-(\alpha (0)+wf(0))}{z-w}=f(z)\in H^2 \end{aligned}$$

and by [6, Thm. 7.4.1.] $\alpha$ has an ADC at w with the non-tangential limit $\alpha (w)$ equal to $\alpha (0)+wf(0)$, $\widetilde{k}_w^{\alpha }$ belongs to $K_{\alpha }$ and

$$\begin{aligned} f(z)=\frac{\alpha (z)-\alpha (w)}{z-w}=\widetilde{k}_w^{\alpha }. \end{aligned}$$

This completes the proof in the first case.

Case 2. $\beta (0)\ne 0$.

For any two complex numbers a, b and functions $f\in K_{\alpha }$, $g\in K_{\beta }$ define

$$\begin{aligned} J_{a}^{\alpha }f(z)=\frac{\sqrt{1-|a|^2}}{1-\overline{a}\alpha (z)}\cdot f(z)\quad \mathrm {and}\quad J_{b}^{\beta }g(z)=\frac{\sqrt{1-|b|^2}}{1-\overline{b}\beta (z)}\cdot g(z). \end{aligned}$$

It is known that $J_{a}^{\alpha }$ is a unitary map from $K_{\alpha }$ onto $K_{\alpha _a}$, where

$$\begin{aligned} \alpha _a(z)=\frac{a-\alpha (z)}{1-\overline{a}\alpha (z)}, \end{aligned}$$

and that $J_{b}^{\beta }$ is a unitary map from $K_{\beta }$ onto $K_{\beta _b}$, where

$$\begin{aligned} \beta _b(z)=\frac{b-\beta (z)}{1-\overline{b}\beta (z)} \end{aligned}$$

(see [10, pp. 521–523] for more details). It was also proved in [8] that $A\in \mathscr {T}(\alpha ,\beta )$ if and only if $J_{b}^{\beta }A(J_{a}^{\alpha })^{-1}\in \mathscr {T}(\alpha _a,\beta _b)$ ([8, Prop. 2.5]).

Take $a=0$ and $b=\beta (0)$. Clearly, $\alpha _0=\alpha$ and $J_{0}^{\alpha }=I_{K_{\alpha }}$. Therefore, since $k_w^{\beta }\otimes f$ belongs to $\mathscr {T}(\alpha ,\beta )$, the rank-one operator

$$\begin{aligned} J_{b}^{\beta }(k_w^{\beta }\otimes f)(J_{a}^{\alpha })^{-1}=(J_{b}^{\beta }k_w^{\beta })\otimes f \end{aligned}$$

belongs to $\mathscr {T}(\alpha ,\beta _b)$. A simple computation reveals that

$$\begin{aligned} J_{b}^{\beta }k_w^{\beta }=\frac{1-b\overline{\beta (w)}}{\sqrt{1-|b|^2}}\cdot k_w^{\beta _b} \end{aligned}$$

(this can also be seen using Lemma 2.4 from [8]). Hence, the rank-one operator $k_w^{\beta _b}\otimes f$ belongs to $\mathscr {T}(\alpha ,\beta _b)$, and, since $\beta _b(0)=0$, f is a scalar multiple of $\widetilde{k}_w^{\alpha }$ by Case 1.

This completes the proof in the second case and the proof of the first implication in (a).

To prove the first implication in (b) assume that $g=\widetilde{k}_{w}^{\beta }$ for some $w\in \overline{\mathbb {D}}$. Then the rank-one operator

$$\begin{aligned} {k}_{w}^{\beta }\otimes C_{\alpha }f =C_{\beta }(g\otimes f)C_{\alpha } \end{aligned}$$

belongs to $\mathscr {T}(\alpha ,\beta )$ [9, p. 9], and it follows from the part of (a) proved above that $C_{\alpha }f$ is a scalar multiple of $\widetilde{k}_{w}^{\alpha }$ and f is a scalar multiple of ${k}_{w}^{\alpha }$.

To complete the proof of (a) assume that $f=\widetilde{k}_{w}^{\alpha }$. Then,

$$\begin{aligned} \widetilde{k}_{w}^{\alpha }\otimes g=(g\otimes \widetilde{k}_{w}^{\alpha })^{*} \end{aligned}$$

belongs to $\mathscr {T}(\beta ,\alpha )$ [2, Lem. 3.2], and it follows from the part of (b) proved above that g is a scalar multiple of ${k}_{w}^{\beta }$.

Similarly, if $f={k}_{w}^{\alpha }$, then

$$\begin{aligned} {k}_{w}^{\alpha }\otimes g=(g\otimes {k}_{w}^{\alpha })^{*} \end{aligned}$$

belongs to $\mathscr {T}(\beta ,\alpha )$ and it follows from (a) that g is a scalar multiple of $\widetilde{k}_{w}^{\beta }$. This completes the proof of the lemma. $\square$

Using Lemma 2.3, one can modify the proof of [10, Thm. 5.1(c)] to show the following.

Theorem 2.4

Let $\alpha$ and $\beta$ be two inner functions such that $K_{\alpha }$ and $K_{\beta }$ have dimension $m>1$ and $n>1$, respectively. Then the only rank-one operators in $\mathscr {T}(\alpha ,\beta )$ are the non-zero scalar multiples of the operators $\widetilde{k}_w^{\beta }\otimes k_w^{\alpha }$ and $k_w^{\beta }\otimes \widetilde{k}_w^{\alpha }$ where $w\in \mathbb {D}$ or $w\in \partial \mathbb {D}$ and $\alpha$ and $\beta$ have an ADC at w.

Note that neither the above nor [9] includes the case when one of the model spaces is one dimensional and the other is infinite dimensional.

Let us now assume that $\alpha$, $\beta$ are two inner functions such that $\dim K_{\alpha }=1$ and $\dim K_{\beta }=+\infty$.

Clearly, every bounded linear operator A from $K_{\alpha }$ into $K_{\beta }$ is of rank one. Moreover, since $K_{\alpha }$ is spanned by $k_{0}^{\alpha }$, $A-S_\beta A S^*_\alpha$ is a rank-one operator of the form $\psi \otimes k^\alpha _0$ for some $\psi \in K_\beta$. So here, by Theorem 2.1, all bounded linear operators from $K_{\alpha }$ into $K_{\beta }$ are asymmetric truncated Toeplitz operators of rank one.

As the finite-dimensional case suggests, one can expect that here there are more rank-one operators in $\mathscr {T}(\alpha ,\beta )$ than just scalar multiples of $\widetilde{k}_w^{\beta }\otimes k_w^{\alpha }$ and $k_w^{\beta }\otimes \widetilde{k}_w^{\alpha }$. We now prove that this is indeed the case. We do this in two steps: (1) we show that if every rank-one operator from $\mathscr {T}(\alpha ,\beta )$ is a scalar multiple of $\widetilde{k}_w^{\beta }\otimes k_w^{\alpha }$ or $k_w^{\beta }\otimes \widetilde{k}_w^{\alpha }$ for some $w\in \overline{\mathbb {D}}$, then every non-zero function from $K_{\beta }$ must be a scalar multiple of a reproducing kernel or a conjugate kernel; (2) we prove that since $\dim K_{\beta }=+\infty$, there must exist $f\in K_{\beta }$ that is neither a scalar multiple of a reproducing kernel nor a scalar multiple of a conjugate kernel.

Step 1. Let $f\in K_{\beta }{\setminus }\{0\}$. Then, $f\otimes k_0^{\alpha }\in \mathscr {T}(\alpha ,\beta )$ and there exist $w\in \overline{\mathbb {D}}$ and $c\ne 0$ such that

$$\begin{aligned} f\otimes k_0^{\alpha }=c(\widetilde{k}_w^{\beta }\otimes {k}_w^{\alpha })\quad \mathrm {or}\quad f\otimes k_0^{\alpha }=c({k}_w^{\beta }\otimes \widetilde{k}_w^{\alpha }). \end{aligned}$$

Since $\dim K_{\alpha }=1$, we can replace $k_0^{\alpha }$ in the above with $k_w^{\alpha }$ or $\widetilde{k}_w^{\alpha }$ (multiplying both sides of the assumed equality by a constant if necessary). Then it follows easily that f is a scalar multiple of either $\widetilde{k}_w^{\beta }$ or ${k}_w^{\beta }$.

Step 2. Here, we use [1, Prop. 2.8]. Fix $w_1,w_2\in \mathbb {D}$, $w_1\ne w_2$ and let$f=k_{w_1}^{\beta }+k_{w_2}^{\beta }\in K_{\beta }{\setminus }\{0\}$. Since $\dim K_{\beta }=+\infty$, by [1, Prop. 2.8], the functions $k_{w_1}^{\beta }$, $k_{w_2}^{\beta }$ and $\widetilde{k}_{w}^{\beta }$ are linearly independent for any fixed $w\in \mathbb {D}$ and for any fixed $w\in \partial \mathbb {D}$ such that $\beta$ has an ADC at w. Thus, f cannot be a scalar multiple of $\widetilde{k}_{w}^{\beta }$. A similar reasoning shows that f cannot be a scalar multiple of ${k}_{w}^{\beta }$ for any $w\in \overline{\mathbb {D}}$.

We proved that if $\dim K_{\alpha }=1$ and $\dim K_{\beta }=+\infty$, then there is a rank-one operator from $\mathscr {T}(\alpha ,\beta )$ that is neither a scalar multiple of $\widetilde{k}_w^{\beta }\otimes k_w^{\alpha }$ nor a scalar multiple of $k_w^{\beta }\otimes \widetilde{k}_w^{\alpha }$. The case $\dim K_{\alpha }=+\infty$ and $\dim K_{\beta }=1$ can be treated analogously.

Theorem 2.5

Let $\alpha$ and $\beta$ be two inner functions such that $\dim K_{\alpha }=m$ and $\dim K_{\beta }=n$ (with m or n possibly infinite). The only rank-one operators in $\mathscr {T}(\alpha ,\beta )$ are the non-zero scalar multiples of the operators $\widetilde{k}_w^{\beta }\otimes k_w^{\alpha }$ and $k_w^{\beta }\otimes \widetilde{k}_w^{\alpha }$, $w\in \overline{\mathbb {D}}$, if and only if either $mn\le 2$, or $m>1$ and $n>1$.

Proof

Assume that the only rank-one operators in $\mathscr {T}(\alpha ,\beta )$ are the non-zero scalar multiples of the operators $\widetilde{k}_w^{\beta }\otimes k_w^{\alpha }$ and $k_w^{\beta }\otimes \widetilde{k}_w^{\alpha }$. If both m and n are finite, then $\alpha$ and $\beta$ are two finite Blaschke products of degree m and n, respectively, and the claim follows from Theorem 1.1. If m or n is infinite, then $m>1$ and $n>1$ as already observed before.

To complete the proof, assume that either $mn\le 2$, or $m>1$ and $n>1$. If both m and n are finite, then the only rank-one operators in $\mathscr {T}(\alpha ,\beta )$ are the non-zero scalar multiples of the operators $\widetilde{k}_w^{\beta }\otimes k_w^{\alpha }$ and $k_w^{\beta }\otimes \widetilde{k}_w^{\alpha }$ by Theorem 1.1. If m or n is infinite, then clearly $m>1$ and $n>1$ and the claim follows from Theorem 2.4. $\square$