Some characterizations of asymmetric truncated Toeplitz operators

TL;DR

This paper proves that all asymmetric truncated Toeplitz operators can be characterized using compressed shifts and specific rank-two operators, extending previous special-case results to all cases.

Contribution

The paper generalizes existing characterizations of asymmetric truncated Toeplitz operators to encompass all cases, not just special ones.

Findings

All asymmetric truncated Toeplitz operators can be characterized by compressed shifts.

The characterization involves rank-two operators of a specific form.

Previous results were limited to special cases, now extended universally.

Abstract

It was recently proved that in some special cases asymmetric truncated Toeplitz operators can be characterized in terms of compressed shifts and rank-two operators of special form. In this paper we show that such characterizations hold in all cases.

Equations

$$Sf(z)=z\cdot f(z).$$

$$S^{\ast }f(z)=\frac{f(z)-f(0)}{z}.$$

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

$$T_{\phi }f=P(\phi f).$$

$$A_{\phi }^{\alpha }f=P_{\alpha }(\phi f),$$

$$A_{\phi }^{\alpha ,\beta }f=P_{\beta }(\phi f).$$

$$\mathcal{T}(\alpha ,\beta )=\{A_{\phi }^{\alpha ,\beta }:\phi \in L^2\mathrm{and}{A}_{\phi }^{\alpha ,\beta }\mathrm{is}\mathrm{bounded}\}.$$

$$f(w)=⟨f,k_w^{\alpha }⟩,$$

$$k_w^{\alpha }(z)=\frac{1-\overline{\alpha (w)}\alpha (z)}{1-\overline{w}z}.$$

$$C_{\alpha }f(z)=\tilde{f}(z)=\alpha (z)\overline{z}\overline{f(z)},|z|=1.$$

$${\tilde{k}}_w^{\alpha }(z)=\frac{\alpha (z)-\alpha (w)}{z-w}.$$

$$A-S_{\beta }A{S}_{\alpha }^{\ast }=\psi \otimes k_0^{\alpha }+k_0^{\beta }\otimes \chi .$$

$$A_{\phi }^{\alpha ,\beta }-S_{\beta }{A}_{\phi }^{\alpha ,\beta }{S}_{\alpha }^{\ast }=\psi \otimes k_0^{\alpha }+k_0^{\beta }\otimes \chi$$

$$\psi =S_{\beta }{P}_{\beta }\left(\overline{z}\phi \right)\in K_{\beta },$$

$$\chi =P_{\alpha }\left(\overline{\phi }\right)\in K_{\alpha }.$$

$$S_{\alpha }^{\ast }f(z)=S^{\ast }f(z)=\frac{f(z)-f(0)}{z}$$

$$S_{\beta }^{\ast }g(z)=S^{\ast }g(z)=\frac{g(z)-g(0)}{z}$$

$$⟨S_{\beta }{A}_{\phi }^{\alpha ,\beta }{S}_{\alpha }^{\ast }f,g⟩$$

$$=⟨\overline{z}\phi f,S_{\beta }^{\ast }g⟩-f(0)⟨\overline{z}\phi ,S_{\beta }^{\ast }g⟩$$

$$=⟨\overline{z}\phi f,\frac{g-g(0)}{z}⟩-⟨f,k_0^{\alpha }⟩⟨S_{\beta }{P}_{\beta }(\overline{z}\phi ),g⟩$$

$$=⟨\overline{z}\phi f,\overline{z}g⟩-\overline{g(0)}⟨\overline{z}\phi f,\overline{z}⟩-⟨(S_{\beta }{P}_{\beta }(\overline{z}\phi )\otimes k_0^{\alpha })f,g⟩$$

$$=⟨A_{\phi }^{\alpha ,\beta }f,g⟩-⟨k_0^{\beta },g⟩⟨f,P_{\alpha }(\overline{\phi })⟩-⟨(S_{\beta }{P}_{\beta }(\overline{z}\phi )\otimes k_0^{\alpha })f,g⟩$$

$$=⟨A_{\phi }^{\alpha ,\beta }f,g⟩-⟨(k_0^{\beta }\otimes P_{\alpha }(\overline{\phi }))f,g⟩-⟨(S_{\beta }{P}_{\beta }(\overline{z}\phi )\otimes k_0^{\alpha })f,g⟩.$$

$$A_{\phi }^{\alpha ,\beta }f-S_{\beta }{A}_{\phi }^{\alpha ,\beta }{S}_{\alpha }^{\ast }f=(S_{\beta }{P}_{\beta }(\overline{z}\phi )\otimes k_0^{\alpha })f+(k_0^{\beta }\otimes P_{\alpha }(\overline{\phi }))f\text{for all }f\in K_{\alpha }^{\infty }.$$

$$⟨A_{\phi }^{\alpha ,\beta }f,g⟩=\sum _{n=1}^{\infty }⟨(S_{\beta }^n\psi \otimes S_{\alpha }^n{k}_0^{\alpha }+S_{\beta }^n{k}_0^{\beta }\otimes S_{\alpha }^n\chi )f,g⟩$$

$$P_{\alpha }(\overline{\phi })=\chi \text{and }{S}_{\beta }{P}_{\beta }(\overline{z}\phi )=S_{\beta }{P}_{\beta }(\overline{z}\psi )=S_{\beta }{S}_{\beta }^{\ast }\psi =\psi .$$

$$A_{\phi }^{\alpha ,\beta }-S_{\beta }{A}_{\phi }^{\alpha ,\beta }{S}_{\alpha }^{\ast }=\psi \otimes k_0^{\alpha }+k_0^{\beta }\otimes \chi ,$$

$$S_{\beta }^n{A}_{\phi }^{\alpha ,\beta }(S_{\alpha }^{\ast }{)}^n-S_{\beta }^{n+1}{A}_{\phi }^{\alpha ,\beta }(S_{\alpha }^{\ast }{)}^{n+1}=S_{\beta }^n\psi \otimes S_{\alpha }^n{k}_0^{\alpha }+S_{\beta }^n{k}_0^{\beta }\otimes S_{\alpha }^n\chi .$$

$$A_{\phi }^{\alpha ,\beta }=\sum _{n=1}^N(S_{\beta }^n\psi \otimes S_{\alpha }^n{k}_0^{\alpha }+S_{\beta }^n{k}_0^{\beta }\otimes S_{\alpha }^n\chi )+S_{\beta }^{N+1}{A}_{\phi }^{\alpha ,\beta }(S_{\alpha }^{\ast }{)}^{N+1}$$

$$\begin{array}{rl}⟨S_{\beta }^{N+1}& A_{\phi }^{\alpha ,\beta }(S_{\alpha }^{\ast }{)}^{N+1}f,g⟩\\ & =⟨A_{\overline{\chi }}^{\alpha ,\beta }(S_{\alpha }^{\ast }{)}^{N+1}f,(S_{\beta }^{\ast }{)}^{N+1}g⟩+⟨A_{\psi }^{\alpha ,\beta }(S_{\alpha }^{\ast }{)}^{N+1}f,(S_{\beta }^{\ast }{)}^{N+1}g⟩\\ & =⟨T_{\overline{\chi }}(S^{\ast }{)}^{N+1}f,(S^{\ast }{)}^{N+1}g⟩+⟨(S^{\ast }{)}^{N+1}f,T_{\overline{\psi }}(S^{\ast }{)}^{N+1}g⟩\\ & =⟨P({\overline{z}}^{N+1}\overline{\chi }f),(S^{\ast }{)}^{N+1}g⟩+⟨(S^{\ast }{)}^{N+1}f,P({\overline{z}}^{N+1}\overline{\psi }g)⟩\to 0\text{as }N\to 0\end{array}$$

$$\Vert P({\overline{z}}^{N+1}\overline{\chi }f){\Vert }_2\le \Vert f{\Vert }_{\infty }\cdot \Vert \chi {\Vert }_2,\Vert P({\overline{z}}^{N+1}\overline{\psi }g){\Vert }_2\le \Vert g{\Vert }_{\infty }\cdot \Vert \psi {\Vert }_2$$

$$⟨A_{\phi }^{\alpha ,\beta }f,g⟩=\sum _{n=1}^{\infty }⟨(S_{\beta }^n\psi \otimes S_{\alpha }^n{k}_0^{\alpha }+S_{\beta }^n{k}_0^{\beta }\otimes S_{\alpha }^n\chi )f,g⟩$$

$$⟨A_{\phi }^{\alpha ,\beta }f,g⟩=\sum _{n=1}^{\infty }⟨(S_{\beta }^n\psi \otimes S_{\alpha }^n{k}_0^{\alpha }+S_{\beta }^n{k}_0^{\beta }\otimes S_{\alpha }^n\chi )f,g⟩.$$

Full text

1. Introduction

As usual, let $H^{2}$ denote the classical Hardy space. The space $H^{2}$ can be seen as a space of functions analytic in the unit disk $\mathbb{D}=\{z:|z|<1\}$ or as a closed subspace of $L^{2}:=L^{2}(\partial\mathbb{D})$. In the first case $H^{2}$ consists of functions analytic in $\mathbb{D}$ with square summable MacLaurin coefficients and in the second it consists of functions from $L^{2}$ such that their Fourier coefficients with negative indices vanish.

The unilateral shift $S$ on $H^{2}$ is the operator of multiplication by the independent variable, that is,

[TABLE]

The adjoint of $S^{*}$ of $S$ is called the backward shift. A simple verification shows that

[TABLE]

The famous Beurling theorem provides a characterization of all $S^{*}$-invariant subspaces of $H^{2}$. Namely, a closed nontrivial subspace of $H^{2}$ is $S^{*}$-invariant if and only if it is of the form

[TABLE]

where $\alpha$ is an inner function, i.e., $\alpha$ belongs to the algebra $H^{\infty}$ of bounded analytic functions and $|\alpha|=1$ a.e. on $\partial\mathbb{D}$. The space $K_{\alpha}$ is called the model space associated with $\alpha$.

The operators $S$ and $S^{*}$ are two examples of classical Toeplitz operators. Let $P$ denote the orthogonal projection from $L^{2}$ onto $H^{2}$. A Toeplitz operator $T_{\varphi}$ with symbol $\varphi\in L^{\infty}$ is defined on $H^{2}$ by

[TABLE]

Clearly, if $\varphi\in H^{\infty}$, then $T_{\varphi}$ is just the multiplication by $\varphi$. Also, the operator $T_{\varphi}$ is densely defined on bounded functions whenever $\varphi\in L^{2}$, and extends to a bounded operator on $H^{2}$ if and only if $\varphi\in L^{\infty}$. We have $S=T_{z}$ and $S^{*}=T_{\overline{z}}$.

Truncated Toeplitz operators are compressions of Toeplitz operators to model spaces. More precisely, a truncated Toeplitz operator $A_{\varphi}^{\alpha}$ with a symbol $\varphi\in L^{2}$ is defined on the model space $K_{\alpha}$ by

[TABLE]

where $P_{\alpha}$ is the orthogonal projection from $L^{2}$ onto $K_{\alpha}$. In particular, $S_{\alpha}=A^{\alpha}_{z}$ is called the compressed shift. Since $K_{\alpha}$ is $S^{*}$-invariant, it easily follows that $S^{*}_{\alpha}=S^{*}_{|K_{\alpha}}$.

The study of the class of truncated Toeplitz operators was started in 2007 with D. Sarason’s paper [8] (see [4]). Recently, the authors in [1] and [2, 3] initiated the study of so-called asymmetric truncated Toeplitz operators (see also [5, 6]).

Let $\alpha$, $\beta$ be two inner functions. An asymmetric truncated Toeplitz operator $A_{\varphi}^{\alpha,\beta}$ with a symbol $\varphi\in L^{2}$ is the operator from $K_{\alpha}$ into $K_{\beta}$ given by

[TABLE]

Clearly, $A_{\varphi}^{\alpha}=A_{\varphi}^{\alpha,\alpha}$. Put

[TABLE]

A bounded linear operator $T$ on $H^{2}$ is a Toeplitz operator if and only if $T-S^{*}TS=0$. D. Sarason [8] proved that a bounded linear operator $A$ on $K_{\alpha}$ is a truncated Toeplitz operator if and only if $A-S_{\alpha}^{*}AS_{\alpha}$ is an operator of a special kind and rank at most two. Similar characterization for the operators from $\mathscr{T}(\alpha,\beta)$ were proved for the case when $\beta$ divides $\alpha$ (that is, $\alpha/\beta$ is an inner function) [1], and for the case when $\alpha$ and $\beta$ are finite Blaschke products (in other words, when $K_{\alpha}$ and $K_{\beta}$ are finitely dimensional) [7].

In this paper we provide such characterizations of the operators from $\mathscr{T}(\alpha,\beta)$ for all $\alpha,\beta$.

2. Characterizations of asymmetric truncated Toeplitz operators

Recall that a model space $K_{\alpha}$ is a reproducing kernel Hilbert space. That is to say that for every $f$ in the model space $K_{\alpha}$ and each $w\in\mathbb{D}$,

[TABLE]

where the reproducing kernel function $k_{w}^{\alpha}$ is of the form

[TABLE]

Observe that since $k_{w}^{\alpha}\in H^{\infty}$, the set $K_{\alpha}^{\infty}=K_{\alpha}\cap H^{\infty}$ is dense in $K_{\alpha}$.

A conjugate kernel is the function $\widetilde{k}_{w}^{\alpha}=C_{\alpha}{k}_{w}^{\alpha}$, where $C_{\alpha}:L^{2}\to L^{2}$ is given by

[TABLE]

It can be seen that $C_{\alpha}$, as defined on $L^{2}$, is an antilinear isometric involution (a map with such property is called a conjugation). It can also be verified that the conjugation $C_{\alpha}$ preserves $K_{\alpha}$. Therefore $\widetilde{k}_{w}^{\alpha}\in K_{\alpha}$ for all $w\in\mathbb{D}$ and a simple computation gives

[TABLE]

Theorem 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

[TABLE]

The proof of Theorem 2.1 follows the proof given by D. Sarason for truncated Toeplitz operators [8, Theorem 4.1] and requires some auxiliary lemmas.

Lemma 2.2**.**

For every $\varphi\in L^{2}$ the equality

[TABLE]

holds on $H^{2}$, where

[TABLE]

Proof.

Let $f\in K_{\alpha}^{\infty}$ and $g\in K_{\beta}^{\infty}$. The functions

[TABLE]

and

[TABLE]

clearly belong to $H^{2}\cap L^{\infty}=H^{\infty}$. Therefore,

[TABLE]

From this

[TABLE]

For an arbitrary $f\in K_{\alpha}$ there exists $\{f_{n}\}\in K_{\alpha}^{\infty}$ such that $f_{n}\to f$ in $H^{2}$ norm and both $\varphi f_{n}\to\varphi f$ and $\varphi S^{*}_{\alpha}f_{n}\to\varphi S^{*}_{\alpha}f$ in $L^{1}$. Hence, $A_{\varphi}^{\alpha,\beta}f_{n}\to A_{\varphi}^{\alpha,\beta}f$ and $A_{\varphi}^{\alpha,\beta}S^{*}_{\alpha}f_{n}\to A_{\varphi}^{\alpha,\beta}S^{*}_{\alpha}f$ uniformly on compact subsets of $\mathbb{D}$, which implies that (2.2) holds for all $f\in K_{\alpha}$. ∎

Lemma 2.3**.**

If $\varphi=\overline{\chi}+\psi$, where $\chi\in K_{\alpha}$ and $\psi\in K_{\beta}$, $\psi(0)=0$, then the equality

[TABLE]

holds for all $f\in K_{\alpha}^{\infty}$ and $g\in K_{\beta}^{\infty}$.

Proof.

Note that if $\varphi=\overline{\chi}+\psi$, where $\chi\in K_{\alpha}$ and $\psi\in K_{\beta}$, $\psi(0)=0$, then

[TABLE]

By Lemma 2.2,

[TABLE]

and for every $n\geq 0$,

[TABLE]

From this

[TABLE]

for every $N\geq 1$.

If $f\in K_{\alpha}^{\infty}$ and $g\in K_{\beta}^{\infty}$, then

[TABLE]

since

[TABLE]

and $(S^{*})^{N}\to 0$ in the strong operator topology. Therefore

[TABLE]

for all $f\in K_{\alpha}^{\infty}$ and $g\in K_{\beta}^{\infty}$.

∎

Proof of Theorem 2.1.

If $A\in\mathscr{T}(\alpha,\beta)$, then $A=A_{\varphi}^{\alpha,\beta}$ for some $\varphi\in L^{2}$ and it satisfies (2.1) by Lemma 2.2.

Assume now that $A$ is a bounded linear operator from $K_{\alpha}$ into $K_{\beta}$ such that (2.1) holds for $\psi\in K_{\beta}$ and $\chi\in K_{\alpha}$. Without any loss of generality we can assume that $\psi(0)=0$. Indeed, if this was not the case we would replace $\psi$ and $\chi$ with $\psi-ck^{\beta}_{0}$ and $\chi+\overline{c}k^{\alpha}_{0}$, respectively, for $c=\psi(0)/(1-|\beta(0)|^{2})$.

Define $\varphi=\overline{\chi}+\psi$. By Lemma 2.3, for every $f\in K_{\alpha}^{\infty}$ and $g\in K_{\beta}^{\infty}$,

[TABLE]

But an argument similar to the one given in the proof of Lemma 2.2 shows that

[TABLE]

and so $A=A_{\varphi}^{\alpha,\beta}\in\mathscr{T}(\alpha,\beta)$. ∎

Corollary 2.4**.**

Let $A$ be a bounded linear operator form $K_{\alpha}$ into $K_{\beta}$. Then $A\in\mathscr{T}(\alpha,\beta)$ if and only if

[TABLE]

for some $\psi\in K_{\beta}$ and $\chi\in K_{\alpha}$.

Proof.

Recall first that the linear operator $A$ belongs to $\mathscr{T}(\alpha,\beta)$ if and only if $B=C_{\beta}AC_{\alpha}$ belongs to $\mathscr{T}(\alpha,\beta)$ (see [7, p. 9]).

If $A\in\mathscr{T}(\alpha,\beta)$, then $C_{\beta}AC_{\alpha}\in\mathscr{T}(\alpha,\beta)$. By Theorem 2.1, there exist functions $\chi_{0}\in K_{\alpha}$ and $\psi_{0}\in K_{\beta}$ such that

[TABLE]

Thus (using the symmetry of compressed shifts) we get

[TABLE]

and $A$ satisfies (2.3) with

[TABLE]

The other implication can be proved in a similar way. ∎

Conditions from Corollary 2.4 can also be formulated in terms of modified compressed shifts.

Corollary 2.5**.**

Let $A$ be a bounded linear operator form $K_{\alpha}$ into $K_{\beta}$. Then $A\in\mathscr{T}(\alpha,\beta)$ if and only if one (and all) of the following conditions holds

1. (a)

$A-S_{\beta,b}AS^{*}_{\alpha,a}=\psi\otimes k^{\alpha}_{0}+k^{\beta}_{0}\otimes\chi$; 2. (b)

$A-S^{*}_{\beta,b}AS_{\alpha,a}=\psi\otimes\widetilde{k}^{\alpha}_{0}+\widetilde{k}^{\beta}_{0}\otimes\chi$;

for some $a,b\in\mathbb{C}$, $\psi\in K_{\beta}$ and $\chi\in K_{\alpha}$ (possibly different for different conditions), where

[TABLE]

are the modified compressed shifts.

Proof.

The proof uses Corollary 2.4 and is analogous to the proof given in [8, Theorem 7.1]. The details are therefore left to the reader. ∎

Corollary 2.6**.**

$\mathscr{T}(\alpha,\beta)$* is closed in the weak operator topology.*

Proof.

See [8], [1] or [7]. ∎

Using Corollary 2.4 we can now characterize the operators from $\mathscr{T}(\alpha,\beta)$ in terms of shift invariance. The following corollary was first noted in [1] (for the case when $\beta$ divides $\alpha$).

Corollary 2.7**.**

Let $A$ be a bounded linear operator form $K_{\alpha}$ into $K_{\beta}$. Then $A\in\mathscr{T}(\alpha,\beta)$ if and only if it has the following property

[TABLE]

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

Proof.

See [7], [1]. ∎