A Toeplitz-Like Operator with Rational Symbol Having Poles on the Unit Circle III: The Adjoint

Abstract

This paper contains a further analysis of the Toeplitz-like operators $T_\omega$ on $H^p$ with rational symbol $\omega$ having poles on the unit circle that were previously studied in Groenewald (Oper Theory Adv Appl 271:239–268, 2018; Oper Theory Adv Appl 272:133–154, 2019). Here the adjoint operator $T_\omega ^*$ is described. In the case where $p=2$ and $\omega$ has poles only on the unit circle ${\mathbb {T}}$, a description is given for when $T_\omega ^*$ is symmetric and when $T_\omega ^*$ admits a selfadjoint extension. If in addition $\omega$ is proper, it is shown that $T_\omega ^*$ coincides with the unbounded Toeplitz operator defined by Sarason (Integr Equ Oper Theory 61:281–298, 2008) and studied further by Rosenfeld (Classes of densely defined multiplication and Toeplitz operators with applications to extensions of RKHS’s, 2013; J Math Anal Appl 440:911–921, 2016).

1 Introduction

In this paper we proceed with our study of unbounded Toeplitz-like operators on $H^p$ with rational symbols that have poles on the unit circle ${\mathbb {T}}$ which was initiated in [4]. Our previous work on such Toeplitz-like operators focused on their Fredholm properties (in [4]) and the various parts of their spectra (in [5]). Here we determine properties of the adjoint operator and conditions under which the operator is symmetric and when it has a selfadjoint extension.

Before we can define our Toeplitz-like operators, some notation has to be introduced. We write ${\mathrm {Rat}}$ for the space of rational complex functions, ${\mathrm {Rat}}({\mathbb {T}})$ for the subspace of ${\mathrm {Rat}}$ consisting of rational complex functions with poles only on the unit circle ${\mathbb {T}}$, and ${\mathrm {Rat}}_0({\mathbb {T}})$ for the subspace of strictly proper functions in ${\mathrm {Rat}}({\mathbb {T}})$. Now let $\omega \in {\mathrm {Rat}}$, possibly with poles on ${\mathbb {T}}$. As in [4], we define the Toeplitz-like operator $T_\omega \, (H^p\rightarrow H^p)$, for $1<p<\infty$, via

$$\begin{aligned} {\mathrm {Dom}}(T_\omega )\!=\!\left\{ g\in H^p\! \mid \! \omega g\! =\! f \!+\! \rho \text{ with } \! f\in L^p\!, \rho \in Rat _0(\mathbb {T})\right\} ,\ T_\omega g = \mathbb {P}f. \nonumber \\ \end{aligned}$$

(1.1)

Here ${\mathbb {P}}$ is the Riesz projection of $L^p$ onto $H^p$. The operator $T_\omega$ is densely defined and closed. In case $\omega \in {\mathrm {Rat}}({\mathbb {T}})$, explicit formulas for the domain, kernel, range, and a complement of the range were obtained in [5], as an extension of a result in [4] for the case where $T_\omega$ is Fredholm. We recall these results in Sect. 2, as they will be frequently used throughout the paper.

If $\omega$ has no poles on ${\mathbb {T}}$, in fact for any $\omega \in L^\infty$, the adjoint of the Toeplitz operator $T_\omega$ on $H^p$ can be identified with the Toeplitz operator $T_{\omega ^*}$ on $H^{p'}$, with $1<p'<\infty$ so that $1/p + 1/p' =1$ and with $\omega ^*$ defined as $\omega ^*(z)={\overline{\omega (z)}}$ on ${\mathbb {T}}$. The identification of $(H^p)'$ and $H^{p'}$ goes via the usual pairing

$$\begin{aligned} \langle f,g \rangle _{p,p'}=\frac{1}{2\pi }\int _{{\mathbb {T}}} {\overline{g(z)}} f(z)\, dz\quad (f\in H^p, g\in H^{p'}). \end{aligned}$$

In the sequel we use the same notation for the similarly defined pairing between $L^p$ and $L^{p'}$ to identify $(L^p)'$ and $L^{p'}$, and in both cases the indices will often be omitted.

For the Toeplitz-like operators studied in this paper the situation is more complicated than for Toeplitz operators with $L^\infty$ symbols. However, we do obtain that $T_{\omega }^*$ can be identified with the restriction of the Toeplitz-like operator $T_{\omega ^*}$ on $H^{p'}$ to a dense subspace of its domain. Like for the operator $T_\omega$, in case $\omega$ is in ${\mathrm {Rat}}({\mathbb {T}})$ we obtain a more explicit description of $T_{\omega }^*$, which we present after introducing some further notation.

Throughout the paper ${\mathcal P}$ denotes the space of complex polynomials and ${\mathcal P}_k$, for any non-negative integer k, denotes the subspace of ${\mathcal P}$ of polynomials of degree at most k. The degree of a polynomial $r\in {\mathcal P}$ is denoted as ${\mathrm {deg} }(r)$. Given $r\in {\mathcal P}$ with ${\mathrm {deg} }(r)=k$, say $r(z)=r_0+z r_1+\cdots + z^k r_k$, we define the polynomial $r^\sharp$ by

$$\begin{aligned} r^\sharp (z)=z^k {\overline{r(1/{\overline{z}})}} =\overline{r_0}z^k + {\overline{r_1}} z^{k-1}+\cdots + {\overline{r_k}}. \end{aligned}$$

The following theorem is our first main result.

Theorem 1.1

Let $\omega = s/q\in {\mathrm {Rat}}$ with $s,q\in {\mathcal P}$ co-prime and $1<p<\infty$. Factor $s=s_-s_0s_+$ and $q=q_-q_0q_+$ with $s_-,q_-$ having roots only inside ${\mathbb {T}}$, $s_0,q_0$ having roots only on ${\mathbb {T}}$, and $s_+,q_+$ having roots only outside ${\mathbb {T}}$. Set $m=\deg (q)$, $n=\deg (s)$, $m_{\pm }={\mathrm {deg} }(q_{\pm })$, $n_{\pm }={\mathrm {deg} }(s_{\pm })$ $m_{0}={\mathrm {deg} }(q_{0})$, $n_{0}={\mathrm {deg} }(s_{0})$ and let $1<p'<\infty$ with $1/p + 1/p'=1$. Then

$$\begin{aligned} {\mathrm {Dom}}(T_\omega ^*) = (q_0)^\sharp H^{p'}\subset {\mathrm {Dom}}(T_{\omega ^*}) \quad \text{ and }\quad T_\omega ^* = T_{\omega ^*}|_{(q_0)^\sharp H^{p'}}. \end{aligned}$$

(1.2)

Furthermore, we have

$$\begin{aligned} \begin{aligned} {\mathrm {Ran}}(T_\omega ^*)&= T_{z^{m-n}(s_+)^\sharp /(q_+)^\sharp } Q_{n_0+n_--m_0-m_-} (s_0)^\sharp H^{p'},\\ {\mathrm {Ker}}(T_\omega ^*)&= \left\{ \frac{(q_-)^\sharp (q_0)^\sharp r}{(s_-)^\sharp } \mid \deg (r) < n_- - m_- - m_0 \right\} . \end{aligned} \end{aligned}$$

(1.3)

Here $Q_{k}=I_{H^{p'}}-P_{{\mathcal P}_{k-1}}$, with $P_{{\mathcal P}_{k-1}}$ the standard projection in $H^{p'}$ onto ${\mathcal P}_{k-1}\subset H^{p'}$ to be interpreted as 0 if $k\le 0$, i.e., $Q_{k}=I_{H^{p'}}$ if $k\le 0$. Thus, for $n_0+n_- \le m_0+m_-$ we have ${\mathrm {Ran}}(T_\omega ^*) = T_{z^{m-n}/(q_+)^\sharp } (s_+s_0)^\sharp H^{p'}$. Moreover,

$$\begin{aligned} \dim {\mathrm {Ker}}(T_\omega ^*) = \max \left\{ 0, \#\{\text {zeroes of } \omega \text { inside } {\mathbb {D}}\} - \#\{\text {poles of } \omega \text { in } \overline{{\mathbb {D}}}\} \right\} , \end{aligned}$$

where the multiplicities of the zeroes and poles are taken into account. Hence, $\dim {\mathrm {Ker}}(T_\omega ^*)$ is the maximum of 0 and $n_- -m_- - m_0$. In particular, $T_{\omega }^*$ is injective if and only if the number of poles of $\omega$ inside $\overline{{\mathbb {D}}}$ is greater than or equal to the number of zeroes of $\omega$ inside ${\mathbb {D}}$, multiplicities taken into account.

Before giving a proof of Theorem 1.1 in Sect. 4, we prove the specialization of this result for the case $\omega \in {\mathrm {Rat}}({\mathbb {T}})$ in Sect. 3. For this purpose we first provide a description of $T_{\omega ^*}$ in Sect. 2.

The injectivity result, but not the description of ${\mathrm {Ker}}(T_\omega ^*)$, can also be derived from general theory and results on $T_\omega$. Indeed, according to Theorem II.3.7 in [3], $T_\omega ^*$ is injective if and only if $T_\omega$ has dense range, so that the claim follows from Proposition 2.4 in [5]. More can be obtained in this way, since $H^p$, $1<p<\infty$, is reflexive. By Theorem II.2.14 of [3] it follows that $T_\omega ^{**}=T_\omega$, with the usual identifications of the dual spaces. Hence, applying the above to $T_\omega ^*$ we find that $T_\omega ^*$ has dense range if and only if $T_\omega$ is injective; see also Theorem II.4.10 in [3]. By Banach’s Closed Range Theorem, cf., [14], $T_\omega ^*$ has closed range if and only if $T_\omega$ has closed range. Again applying results from [5] now gives the following result.

Corollary 1.2

Let $\omega \in {\mathrm {Rat}}$ and $1<p<\infty$. Then $T_\omega ^*$ has closed range if and only if $\omega$ has no zeroes on ${\mathbb {T}}$, or equivalently, $\omega ^*$ has no zeroes on ${\mathbb {T}}$. Moreover, $T_\omega ^*$ has dense range if and only if

$$\begin{aligned} \# \left\{ \begin{array}{l} \text {poles of } \omega \text { inside }\overline{{\mathbb {D}}} \\ \text {multi. taken into account} \end{array}\right\} \le \# \left\{ \begin{array}{l} \text {zeroes of } \omega \text { inside }\overline{{\mathbb {D}}} \\ \text {multi. taken into account} \end{array}\right\} . \end{aligned}$$

Beyond Sect. 4, and in the remainder of this introduction, we only consider the case $p=2$ and $\omega \in {\mathrm {Rat}}({\mathbb {T}})$. By comparing the results on $T_\omega$ and $T_\omega ^*$ it is obvious $T_\omega$ cannot be selfadjoint, except when $\omega$ has no poles on ${\mathbb {T}}$. In Sect. 5 we describe in terms of $\omega$ when $T_\omega ^*$ is symmetric, in which case $T_\omega ^*\subset T_\omega$, and whenever $T_\omega ^*$ is symmetric we describe when $T_{\omega ^*}$ admits a selfadjoint extension. The following theorem collects some of the main results of Sect. 5; it follows directly from Theorem 5.1, Corollaries 5.2 and 5.7, Propositions 5.4 and 5.9.

Theorem 1.3

Let $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$ with $s,q\in {\mathcal P}$ co-prime. Consider $T_\omega$ on $H^2$. Then

$$\begin{aligned} T_\omega ^* \text{ is } \text{ symmetric }\quad \Longleftrightarrow \quad \omega ({\mathbb {T}})\subset {\mathbb {R}}. \end{aligned}$$

In particular, if $T_\omega ^*$ is symmetric, then $\deg (s)\le \deg (q)\le 2\deg (s)$. Furthermore, if $T_\omega ^*$ is symmetric, then $T_\omega ^*$ admits a selfadjoint extension if and only if the number of roots of $s-iq$ and $s+iq$ in ${\mathbb {D}}$, counting multiplicities, coincide. This happens in particular if $\omega ({\mathbb {T}})\ne {\mathbb {R}}$, but cannot happen in case $\deg (q)$ is odd.

Several other conditions for $T_\omega ^*$ to be symmetric and/or have a selfadjoint extension are derived in Sect. 5.

In [11] Sarason introduced and studied an unbounded Toeplitz-like operator with symbol in the Smirnov class. In Sect. 6 we show that if $\omega \in {\mathrm {Rat}}({\mathbb {T}})$ is proper, then the adjoint operator $T_\omega ^*$ is precisely a Toeplitz-like operator of the type studied by Sarason. Hence in this case our Toeplitz-like operator $T_\omega =T_\omega ^{**}$ coincides with the adjoint of the Toeplitz-like operator considered in [11]. Based on ideas in [11], we also show that $H(\overline{{\mathbb {D}}})$, the space of functions analytic on a neighborhood of $\overline{{\mathbb {D}}}$, is contained in ${\mathrm {Dom}}(T_\omega )$ and in fact is a core of $T_\omega$.

In the last section of [11], Sarason introduces a class of closed, densely defined Toeplitz-like operators on $H^2$ determined by algebraic properties, which was further investigated by Rosenfeld in [9, 10]. In particular, this class of Toeplitz-like operators contains the unbounded Toeplitz-like operator studied by Sarason and is closed under taking adjoints, and hence contains our Toeplitz-like operators with proper symbols in ${\mathrm {Rat}}({\mathbb {T}})$. In fact, we will show in Sect. 6 that $T_\omega$ is contained in the class of Toeplitz-like operators for any $\omega$ in ${\mathrm {Rat}}$.

2 The Operator $T_{\omega ^*}$ for $\omega \in {\mathrm {Rat}}({\mathbb {T}})$

In this section we recall some results from [4, 5] on the operator $T_\omega$ for $\omega \in {\mathrm {Rat}}({\mathbb {T}})$ that we will use in the sequel, and apply them to the operator $T_{\omega ^*}$. Hence, throughout this section let $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$, with $s,q\in {\mathcal P}$ co-prime. We set $m=\deg (q)$ and $n={\mathrm {deg} }(s)$. Furthermore, factor $s=s_-s_0s_+$ with $s_-$, $s_0$ and $s_+$ polynomials having roots only inside, on, or outside ${\mathbb {T}}$, respectively. We then recall from Theorem 2.2 in [5] that

$$\begin{aligned} \begin{aligned} {\mathrm {Ker}}(T_\omega )&= \left\{ r/s_+ \mid \deg (r) < m - \deg (s_-s_0) \right\} ;\\ {\mathrm {Dom}}(T_\omega )&=qH^p+{\mathcal P}_{m-1}; \quad {\mathrm {Ran}}(T_\omega )=s H^p+{{\widetilde{{\mathcal P}}}}, \end{aligned} \end{aligned}$$

(2.1)

where ${\widetilde{{\mathcal P}}}$ is the subspace of ${\mathcal P}$ given by

$$\begin{aligned} {\widetilde{{\mathcal P}}}= \{ r\in {\mathcal P}\mid r q = r_1 s + r_2 \text{ for } r_1,r_2\in \mathcal {P}_{m-1}\}\subset {\mathcal P}_{n-1}. \end{aligned}$$

(2.2)

Furthermore, $H^p=\overline{{\mathrm {Ran}}(T_\omega )} + {\widetilde{{\mathcal Q}}}$ forms a direct sum decomposition of $H^p$, where

$$\begin{aligned} {\widetilde{{\mathcal Q}}}={\mathcal P}_{k-1}\quad \text{ with }\quad k=\max \{\deg (s_-)-m , 0\}, \end{aligned}$$

(2.3)

using the convention ${\mathcal P}_{-1}:=\{0\}$. Furthermore, the action of $T_\omega$ is as follows:

$$\begin{aligned}&T_{\omega }g =sh +{\widetilde{r}} \quad ({g= qh+r\in qH^p+{\mathcal P}_{m-1}={\mathrm {Dom}}(T_\omega )}),\\&\text{ where } {\widetilde{r}}\in {\mathcal P}_{n-1} \text{ is } \text{ such } \text{ that } rs={\widetilde{r}}q+r_2 \text{ for } \text{ some } r_2\in {\mathcal P}_{m-1}. \end{aligned}$$

We also recall from Lemma 5.3 in [4] that

$$\begin{aligned} T_{z^\kappa \omega }=T_{z^\kappa } T_{\omega }\ \ \text{ for } \text{ any } \text{ integer } \kappa \le 0. \end{aligned}$$

(2.4)

Recall that $\omega ^*$ is defined as $\omega ^*(z)={\overline{\omega (z)}}$ on ${\mathbb {T}}$, i.e., $\omega ^*(z)=\overline{s(z)}/\overline{q(z)}$. For $z\in {\mathbb {T}}$

$$\begin{aligned} \overline{q(z)}= \overline{q_0+zq_1+ \cdots +z^mq_m} = \overline{q_0}+\overline{q_1}\frac{1}{z}+\cdots +\overline{q_m}\frac{1}{z^m}= \frac{1}{z^m}q^\sharp (z). \end{aligned}$$

Hence $q^\sharp (z)=z^m {\overline{q(z)}}$, and likewise $s^\sharp (z)= z^n {\overline{s(z)}}$. Thus we have

$$\begin{aligned} \omega ^*(z)=\frac{z^{m-n}s^\sharp (z)}{q^\sharp (z)} \text{ if } m\ge n \quad \text{ and }\quad \omega ^*(z)=\frac{s^\sharp (z)}{z^{n-m}q^\sharp (z)} \text{ if } m< n. \end{aligned}$$

(2.5)

In fact, the formula $\omega ^*(z)=z^{m-n}s^\sharp (z)/q^\sharp (z)$ holds in both cases, but is not always a representation as the ratio of two polynomials. Note in particular that $\omega ^*\in {\mathrm {Rat}}({\mathbb {T}})$ in case $\omega$ is proper, while this need not be the case if $\omega$ is not proper. Thus, if $\omega$ is proper, the above formulas apply directly, while for the non-proper case, using (2.4) we can reduce certain questions to questions concerning the Toeplitz operator $T_{s^\sharp / q^\sharp }$ with symbol $s^\sharp / q^\sharp$ which is in ${\mathrm {Rat}}({\mathbb {T}})$.

A polynomial $r\ne 0$ is called self-inversive in case $r=\gamma r^\sharp$ for a constant $\gamma \in {\mathbb {C}}$, which necessarily is unimodular. In fact, $\gamma$ is the ratio $r_0/\overline{r_n}$ with $r_0=r(0)$ and $r_n$ the leading coefficient of r. By a theorem of Cohn [1], a polynomial r has all its roots on ${\mathbb {T}}$ if and only if r is self-inversive and its derivative has all its roots in the closed unit disc $\overline{{\mathbb {D}}}$. Hence, any polynomial with roots only on ${\mathbb {T}}$ is self-inversive. In particular, $q= \gamma q^\sharp$ and $s_0=\rho (s_0)^\sharp$ for unimodular constants $\gamma$ and $\rho$.

More generally, in the transformation $r\rightarrow r^\sharp$, the nonzero roots of r (including multiplicity) transfer along the unit circle via the map $\alpha \mapsto 1/{\overline{\alpha }}=|\alpha |^{-2} \alpha$, while the degree decreases by the multiplicity of 0 as a root of r. Consequently, in the factorization $s^\sharp = (s_+)^\sharp (s_0)^\sharp (s_-)^\sharp$, the polynomials $(s_+)^\sharp$, $(s_0)^\sharp$ and $(s_-)^\sharp$ contain the roots of $s^\sharp$ inside, on and outside ${\mathbb {T}}$, respectively, taking multiplicities into account. We write $(s_+)^\sharp$ rather than $s_+^\sharp$, etc., to avoid confusion with what one may interpret as $(s^\sharp )_+$.

We now apply the above to $T_{\omega ^*}$ acting on $H^{p'}$, $1<p'<\infty$, to fit better with the remainder of the paper.

Proposition 2.1

Let $\omega = s/q\in {\mathrm {Rat}}({\mathbb {T}})$, with $s,q\in {\mathcal P}$ co-prime, $m=\deg (q)$ and $n=\deg (s)$. Factor $s=s_-s_0s_+$ with $s_-$, $s_0$ and $s_+$ polynomials having roots only inside, on, or outside ${\mathbb {T}}$, respectively. Then for $T_{\omega ^*}$ on $H^{p'}$, with $1<p'<\infty$, we have

$$\begin{aligned}{\mathrm {Ker}}(T_{\omega ^*}) = \left\{ r_0/(s_-)^\sharp \mid \deg (r_0) < \deg (s_-) \right\} ,\ \ {\mathrm {Dom}}(T_{\omega ^*})=q^\sharp H^{p'}+{\mathcal P}_{m-1}. \end{aligned}$$

Moreover, we have

$$\begin{aligned} \begin{aligned} {\mathrm {Ran}}(T_{\omega ^*})&= z^{m-n} s^\sharp H^{p'}+{\widetilde{{\mathcal P}}}_*\quad \text{ if } m\ge n,\\ {\mathrm {Ran}}(T_{\omega ^*})&=T_{z^{m-n}}( s^\sharp H^{p'}+{\widetilde{{\mathcal P}}}_*)\quad \text{ if } m< n, \end{aligned} \end{aligned}$$

(2.6)

where for $m\ge n$ the subspace ${\widetilde{{\mathcal P}}}_*$ is given by

$$\begin{aligned} {\widetilde{{\mathcal P}}}_* = \{ r\in {\mathcal P}\mid r q^\sharp = z^{m-n} r_1 s^\sharp + r_2 \text{ for } r_1,r_2\in \mathcal {P}_{m-1}\}\subset {\mathcal P}_{m-n+\deg (s^\sharp )-1}, \end{aligned}$$

while for $m< n$ we have

$$\begin{aligned} {\widetilde{{\mathcal P}}}_* = \{ r\in {\mathcal P}\mid r q^\sharp = r_1 s^\sharp + r_2 \text{ for } r_1,r_2\in \mathcal {P}_{m-1}\}\subset {\mathcal P}_{\deg (s^\sharp )-1}. \end{aligned}$$

Furthermore, ${\mathrm {Ran}}(T_{\omega ^*})$ is dense in $H^{p'}$.

Proof

We separate the cases $m\ge n$ and $m< n$.

For $m\ge n$, we have $\omega ^*={\widetilde{s}}/{\widetilde{q}}\in {\mathrm {Rat}}({\mathbb {T}})$ with ${\widetilde{s}}=z^{m-n}s^\sharp$ and ${\widetilde{q}}=q^\sharp$. Hence ${\widetilde{s}}$ factors as ${\widetilde{s}}=(z^{m-n}(s_+)^\sharp ) (s_0)^\sharp (s_-)^\sharp$, where the factors have all their roots inside, on, or outside ${\mathbb {T}}$, respectively. Also, $\deg (q^\sharp )=\deg (q)$ and $\deg ((s_+)^\sharp )=\deg (s_+)$. So the formulas for ${\mathrm {Dom}}(T_{\omega ^*})$ and ${\mathrm {Ran}}(T_{\omega ^*})$ follow directly from (2.1), while the formula for ${\mathrm {Ker}}(T_{\omega ^*})$ follows because the bound on the degree of $r_0$ can be computed as

$$\begin{aligned} m-\deg (z^{m-n}(s_+)^\sharp (s_0)^\sharp )&\! =\! n- \deg ((s_+)^\sharp (s_0)^\sharp ) =n-\deg (s_+s_0)=\deg (s_-). \end{aligned}$$

Finally, a complement of the closure of ${\mathrm {Ran}}(T_{\omega ^*})$ is given by ${\mathcal P}_{k-1}$ with k the maximum of 0 and $\deg (z^{m-n}(s_+)^\sharp )-m=\deg ((s_+)^\sharp )-n\le 0$. Hence ${\mathcal P}_{-1}=\{0\}$. Thus $T_{\omega ^*}$ has dense range, as claimed.

In case $m< n$, we have $T_{\omega ^*}=T_{z^{m-n}}T_{s^\sharp / q^\sharp }$ and $s^\sharp /q^\sharp$ is in ${\mathrm {Rat}}({\mathbb {T}})$. Applying the above results for $T_\omega$ to $T_{s^\sharp / q^\sharp }$ directly gives the formulas for ${\mathrm {Dom}}(T_{\omega ^*})$ and ${\mathrm {Ran}}(T_{\omega ^*})$.

To see that the formula for ${\mathrm {Ker}}(T_{\omega ^*})$ holds, we follow the argumentation of the proof of Lemma 4.1 in [4]. For $g\in {\mathrm {Dom}}(T_{\omega ^*})={\mathrm {Dom}}(T_{s^\sharp /q^\sharp })$ to be in ${\mathrm {Ker}}(T_{\omega ^*})$ is equivalent to $T_{s^\sharp / q^\sharp } g \in {\mathcal P}_{n-m-1}$. In other words, by Lemma 3.2 in [4], to $s^\sharp g= q^\sharp {\widetilde{r}} +r_1$ with $r_1\in {\mathcal P}_{m-1}$ and ${\widetilde{r}}\in {\mathcal P}_{n-m-1}$, since then $T_{s^\sharp / q^\sharp }g= {\widetilde{r}}$. The latter happens precisely when $g=r/(s_-)^\sharp$ with $r\in {\mathcal P}_{\deg (s_-)-1}$. Indeed, in that case $\deg ((s_+)^\sharp (s_0)^\sharp r)<n$ which in the equation $(s_+)^\sharp (s_0)^\sharp r=s^\sharp g= q^\sharp {\widetilde{r}} +r_1$ corresponds to $\deg ({\widetilde{r}})< m-1$, as required. Finally, we note that a complement of ${\overline{{\mathrm {Ran}}(T_{s^\sharp /q^\sharp })}}$ in $H^{p'}$ is given by ${\mathcal P}_{k-1}$ with $k=\max \{0, \deg {s_+}^\sharp -m\}\le n-m$. Let $f\in H^{p'}$ and write $z^{n-m}f= h+r\in {\overline{{\mathrm {Ran}}(T_{s^\sharp /q^\sharp })}}+ {\mathcal P}_{k-1}$. Then $f=T_{z^{m-n}} z^{n-m}f=T_{z^{m-n}} (h+r)=T_{z^{m-n}} h\in T_{z^{m-n}} {\overline{{\mathrm {Ran}}(T_{s^\sharp /q^\sharp })}}\subset {\overline{{\mathrm {Ran}}( T_{z^{m-n}}T_{s^\sharp /q^\sharp })}}={\overline{{\mathrm {Ran}}(T_{\omega ^*})}}$. Thus also in this case ${\mathrm {Ran}}(T_{\omega ^*})$ is dense in $H^{p'}$. $\square$

We conclude this section with a lemma which will be of use in the sequel.

Lemma 2.2

Let $r_1,r_2\in {\mathcal P}$. Set $n_i=\deg (r_i)$, $i=1,2$, and $n=\deg (r_1+r_2)$. Then

$$\begin{aligned} (r_1+r_2)^\sharp =z^{n-n_1} r_1^\sharp + z^{n-n_2} r_2^\sharp . \end{aligned}$$

In case $n<\max \{n_1,n_2\}$, then $n_1=n_2$ and 0 is a root of $r_1^\sharp +r_2^\sharp$ with multiplicity $n-n_1$, so that the left hand side in the above identity still is a polynomial without a root at 0.

Proof

By definition, for $z\in {\mathbb {T}}$ we have

$$\begin{aligned} (r_1+r_2)^\sharp (z)&= z^{n} (\overline{r_1(1/\overline{z})} + \overline{r_2(1/\overline{z})})=\\&=z^{n-n_1} z^{n_1}\overline{r_1(1/\overline{z})} + z^{n-n_2} z^{n_2} \overline{r_2(1/\overline{z})} \\&=z^{n-n_1} r_1^\sharp (z) + z^{n-n_2} r_2^\sharp (z).\\ \end{aligned}$$

$\square$

3 The Adjoint of $T_\omega$ for $\omega \in {\mathrm {Rat}}({\mathbb {T}})$

In this section we prove the first main result, Theorem 1.1, for the special case that $\omega \in {\mathrm {Rat}}({\mathbb {T}})$. In this case, the result specializes to the following theorem, which we prove in this section.

Theorem 3.1

Let $\omega = s/q\in {\mathrm {Rat}}({\mathbb {T}})$ with $s,q\in {\mathcal P}$ co-prime and $1<p<\infty$. Set $m={\mathrm {deg} }(q)$, $n={\mathrm {deg} }(s)$ and let $1<p'<\infty$ with $1/p + 1/p'=1$. Then

$$\begin{aligned} {\mathrm {Dom}}(T_\omega ^*) = q^\sharp H^{p'}\subset {\mathrm {Dom}}(T_{\omega ^*}) \quad \text{ and }\quad T_\omega ^* = T_{\omega ^*}|_{q^\sharp H^{p'}}. \end{aligned}$$

(3.1)

In fact, for $g=q^\sharp v\in q^\sharp H^{p'}$ we have $T_\omega ^*g =T_{z^{m-n}} s^\sharp v$. Moreover, factorize $s = s_- s_0 s_+$ with $s_-$, $s_0$ and $s_+$ polynomials having roots only inside, on, or outside ${\mathbb {T}}$, respectively. Then

$$\begin{aligned} \begin{aligned} {\mathrm {Ran}}(T_\omega ^*)&= T_{z^{m-n}} s^\sharp H^{p'},\\ {\mathrm {Ker}}(T_\omega ^*)&= \left\{ \frac{q^\sharp r}{(s_-)^\sharp } \mid \deg (r) < {\mathrm {deg} }(s_-) - m \right\} . \end{aligned} \end{aligned}$$

(3.2)

In particular, we have

$$\begin{aligned} \dim {\mathrm {Ker}}(T_\omega ^*) = \max \left\{ 0, \#\left\{ \text {zeroes of } \omega ^* \text { outside } {\mathbb {T}}\right\} - \#\left\{ \text {poles of } \omega ^* \text { on } {\mathbb {T}}\right\} \right\} , \end{aligned}$$

where the multiplicities of the zeroes and poles are taken into account. Thus $T_{\omega }^*$ is injective if and only if $\omega$ has at least as many poles inside ${\mathbb {T}}$ as zeroes inside ${\mathbb {T}}$ unequal to 0, multiplicities taken into account.

We first present some auxiliary lemmas. Throughout, let $1<p,p'<\infty$ such that $1/p+1/p^\prime =1$. We will consider $T_{\omega }$ as an operator with domain in $H^p$ and $T_{\omega ^*}$ as an operator with domain in $H^{p'}$.

Lemma 3.2

Let $\omega = s/q\in {\mathrm {Rat}}({\mathbb {T}})$ with $s,q\in {\mathcal P}$ co-prime, $m={\mathrm {deg} }(q)$ and $n={\mathrm {deg} }(s)$. Then

$$\begin{aligned} q^\sharp H^{p'} \subset {\mathrm {Dom}}(T_\omega ^*)\cap {\mathrm {Dom}}(T_{\omega ^*}) \quad \text{ and }\quad T_\omega ^*|_{q^\sharp H^{p'} }=T_{\omega ^*}|_{q^\sharp H^{p'} }. \end{aligned}$$

Moreover, for $g=q^\sharp v\in q^\sharp H^{p'}$, with $v\in H^{p'}$, we have $T_{\omega }^* g= T_{z^{m-n}}s^\sharp v$, and thus $T_\omega ^* (q^\sharp H^{p'})=T_{z^{m-n}} s^\sharp H^{p'}$.

Proof

The inclusion $q^\sharp H^{p'} \subset {\mathrm {Dom}}(T_{\omega ^*})$ follows from Proposition 2.1. Let g be in $q^\sharp H^{p'}$, say $g(z)=q^\sharp (z)v(z)$ for $v\in H^{p'}$. We show that for $f\in {\mathrm {Dom}}(T_\omega )$ we have $\langle T_w f,g \rangle _{p,p'}=\langle f,T_{\omega ^*} g \rangle _{p,p'}$. Let $f\in {\mathrm {Dom}}(T_\omega )$ and $h=T_{\omega }f\in H^p$, i.e., $sf=qh+r$ for some $r\in {\mathcal P}_{m-1}$, by [4, Lemma 2.3]. Then

$$\begin{aligned} \langle T_\omega f, g \rangle _{p,p'}= & {} \langle h, q^\sharp v\rangle _{p,p'}=\langle h, z^m \overline{q} v\rangle _{p,p'} =\langle qh, z^m v\rangle _{p,p'} = \langle sf-r, z^m v\rangle _{p,p'} \\= & {} \langle sf, z^m v\rangle _{p,p'} \quad (\text{ because } {\mathrm {deg} }(r)<m, v\in H^{p'})\\= & {} \langle f, z^m \overline{s}v\rangle _{p,p'} \!=\!\langle f, z^{m-n} s^\sharp v\rangle _{p,p'}\\= & {} \langle f, T_{z^{m-n}} s^\sharp v\rangle _{p,p'}\ \ (\text {because } f\in H^p). \end{aligned}$$

It remains to show that $T_{\omega ^*}g=T_{z^{m-n}} s^\sharp v$. If $m\ge n$, then $\omega ^*=z^{m-n}s^\sharp /q^\sharp$ is in ${\mathrm {Rat}}({\mathbb {T}})$ and $\omega ^* g=z^{m-n}s^\sharp v\in H^{p'}$, so that, $T_{\omega ^*}g=z^{m-n} s^\sharp v= T_{z^{m-n}} s^\sharp v$, by Lemma 2.3 in [4]. If $m<n$, we have $T_{\omega ^*} g=T_{z^{m-n}}T_{s^\sharp /q^\sharp }g=T_{z^{m-n}} s^\sharp v$. $\square$

Lemma 3.3

Let $\omega = s/q\in {\mathrm {Rat}}({\mathbb {T}})$ with $s,q\in {\mathcal P}$ co-prime, $m={\mathrm {deg} }(q)$ and $n={\mathrm {deg} }(s)$. Let $g\in {\mathrm {Dom}}(T_\omega ^*)$ and $k=T_\omega ^* g\in H^{p'}$. Then for any $r\in {\mathcal P}_{n-1}$ and $r_1\in {\mathcal P}_{m-1}$ so that

$$\begin{aligned} sr_1=qr +r_2 \text{ for } \text{ some } r_2 \in {\mathcal P}_{m-1} \end{aligned}$$

(3.3)

we have

$$\begin{aligned} \langle r_1,k \rangle _{p,p'}=\langle r,g \rangle _{p,p'}. \end{aligned}$$

Moreover, we have

$$\begin{aligned} z^{m-n}s^\sharp g- q^\sharp k\in {\mathcal P}_{m-1} \text{ if } m\ge n \text{ and } s^\sharp g- z^{n-m}q^\sharp k\in {\mathcal P}_{n-1} \text{ if } m< n. \end{aligned}$$

(3.4)

In particular, ${\mathrm {Dom}}(T_\omega ^*) \subset {\mathrm {Dom}}(T_{\omega ^*})$ and $T_\omega ^*=T_{{\omega ^*}}|_{{\mathrm {Dom}}(T_\omega ^*)}$.

Proof

Let $g\in {\mathrm {Dom}}(T_\omega ^*)$ and $k= T_\omega ^* g$. Hence $\langle T_\omega f,g \rangle _{p,p'}=\langle f,k \rangle _{p,p'}$ for each $f\in {\mathrm {Dom}}(T_\omega )$. Since $\omega \in {\mathrm {Rat}}({\mathbb {T}})$, we have ${\mathrm {Dom}}(T_\omega )=q H^p+{\mathcal P}_{m-1}$. Let $f=qh +r_1\in {\mathrm {Dom}}(T_\omega )$, with $h\in H^p$ and $r_1\in {\mathcal P}_{m-1}$. Then $T_\omega f= sh +r$ where $r\in {\mathcal P}_{n-1}$ is uniquely determined by (3.3). Thus

$$\begin{aligned} \langle sh,g \rangle +\langle r,g \rangle&\!=\!\langle sh+r,g \rangle \!=\!\langle T_\omega f,g \rangle \!=\!\langle f,k \rangle \!=\!\langle qh+r_1,k \rangle \!=\!\langle qh,k \rangle +\langle r_1,k \rangle . \end{aligned}$$

We obtain that

$$\begin{aligned} \langle sh,g \rangle -\langle qh,k \rangle =\langle r_1,k \rangle -\langle r,g \rangle . \end{aligned}$$

However, in choosing $f\in {\mathrm {Dom}}(T_\omega )$ we can choose $h\in H^p$ and $r_1\in {\mathcal P}_{m-1}$ independently, and in particular set one or the other equal to zero, so that

$$\begin{aligned}&\langle sh,g \rangle =\langle qh,k \rangle \ \ \ (h\in H^p),\\&\langle r_1,k \rangle =\langle r,g \rangle \ \ \ (r\in {\mathcal P}_{n-1},r_1\in {\mathcal P}_{m-1} \text{ as } \text{ in } (3.3)). \end{aligned}$$

The second identity proves the first claim of the lemma. From the first identity we obtain that

$$\begin{aligned} 0&= \langle h,{\overline{s}}g-{\overline{q}}k \rangle _{p,p'}=\langle h,z^{-n}s^\sharp g-z^{-m}q^\sharp k \rangle _{p,p'} \quad (h\in H^p). \end{aligned}$$

Thus ${\mathbb {P}}(z^{-n}s^\sharp g-z^{-m}q^\sharp k)=0$. On the other hand, for $l=\max \{m,n\}$ we have

$$\begin{aligned} z^l(z^{-n}s^\sharp g-z^{-m}q^\sharp k)=z^{l-n}s^\sharp g-z^{l-m}q^\sharp k\in H^{p'}. \end{aligned}$$

This can only occur if $z^{l-n}s^\sharp g-z^{l-m}q^\sharp k\in {\mathcal P}_{l-1}$, which proves the second claim.

To complete the proof, we show that $g\in {\mathrm {Dom}}(T_{\omega ^*})$ and $T_{\omega ^*}g=k$. For $m\ge n$ we have $\omega ^*\in {\mathrm {Rat}}({\mathbb {T}})$ and the first inclusion of (3.4) can be rewritten as

$$\begin{aligned} \omega ^* g= \left( \frac{z^{m-n} s^\sharp }{q^\sharp }\right) g= k +{\widetilde{r}}/q^\sharp ,\quad \text{ for } \text{ some } {\widetilde{r}}\in {\mathcal P}_{m-1}. \end{aligned}$$

Since $\deg (q^\sharp )=\deg (q)=m$, it now follows that $g\in {\mathrm {Dom}}(T_{\omega ^*})$ and $T_{\omega ^*}g=k$. In case $m<n$ we have $T_{\omega ^*}=T_{z^{m-n}}T_{s^\sharp /q^\sharp }$ and $s^\sharp /q^\sharp \in {\mathrm {Rat}}({\mathbb {T}})$. Now the second inclusion of (3.4) gives

$$\begin{aligned} \left( \frac{s^\sharp }{q^\sharp }\right) g= z^{n-m} k +{\widetilde{r}}/q^\sharp ,\quad \text{ for } \text{ some } {\widetilde{r}}\in {\mathcal P}_{n-1}. \end{aligned}$$

Write ${\widetilde{r}}= {\widetilde{r}}_1 q^\sharp + {\widetilde{r}}_2$ with ${\widetilde{r}}_2\in {\mathcal P}_{m-1}$. Then ${\widetilde{r}}/q^\sharp = {\widetilde{r}}_1+ {\widetilde{r}}_2/q^\sharp$ and $\deg ({\widetilde{r}}_1)<m-n$. Since ${\widetilde{r}}_2/q^\sharp \in {\mathrm {Rat}}_0({\mathbb {T}})$ it follows that $g\in {\mathrm {Dom}}(T_{s^\sharp /q^\sharp })={\mathrm {Dom}}(T_{\omega ^*})$ and $T_{s^\sharp /q^\sharp }g = z^{n-m} k +{\widetilde{r}}_1$. But then $T_{\omega ^*}g= T_{z^{m-n}}T_{s^\sharp /q^\sharp }g= T_{z^{m-n}}(z^{n-m} k +{\widetilde{r}}_1)=k$. $\square$

A special case of the following result was proven as part of the proof of Theorem 2.2 in [5].

Lemma 3.4

Let $r,{\widetilde{r}}\in {\mathcal P}$ be co-prime. Then $r H^p\cap {\widetilde{r}} H^p=r{\widetilde{r}} H^p$.

Proof

Let ${\widetilde{r}} f= rg$ with $f,g\in H^p$. Then $f=r \cdot g/{\widetilde{r}}\in H^p$, so we should show ${\widetilde{f}}:=g/{\widetilde{r}}\in H^p$, i.e., ${\widetilde{f}}$ analytic on ${\mathbb {D}}$ and $\int _{\mathbb {T}}|{\widetilde{f}}(z)|^p\, dz<\infty$.

Since $g\in H^p$, the function ${\widetilde{f}}$ can only fail to be analytic at the roots of ${\widetilde{r}}$ inside ${\mathbb {D}}$. However, if this were the case, then $f=r {\widetilde{f}}$ would also fail to be analytic in ${\mathbb {D}}$, since r and ${\widetilde{r}}$ are co-prime. Thus ${\widetilde{f}}$ is analytic on ${\mathbb {D}}$.

Divide ${\mathbb {T}}$ as ${\mathbb {T}}_1\cup {\mathbb {T}}_2$ with ${\mathbb {T}}_1\cap {\mathbb {T}}_2=\emptyset$ in such a way that ${\mathbb {T}}_1$ and ${\mathbb {T}}_2$ are both nonempty finite unions of line segments of ${\mathbb {T}}$ so that the interior of ${\mathbb {T}}_1$ contains the roots of r and the interior of ${\mathbb {T}}_2$ the roots of ${\widetilde{r}}$. Then $|{\widetilde{r}}(z)|> N_1$ on ${\mathbb {T}}_1$ and $|r(z)|> N_2$ on ${\mathbb {T}}_2$ for some $N_1,N_2>0$. Note that $f=r {\widetilde{f}}$ and $g={\widetilde{r}} {\widetilde{f}}$. We then obtain

$$\begin{aligned} \int _{{\mathbb {T}}_2} |{\widetilde{f}}(z)|^p\, dz&= \int _{{\mathbb {T}}_2} |f(z)/r(z)|^p\, dz \le N_2^{-p} \int _{{\mathbb {T}}_2} |f(z)|^p\, dz \le (2\pi N_2^{p})^{-1} \Vert f\Vert _{H^p}^p. \end{aligned}$$

Using $g={\widetilde{r}} {\widetilde{f}}$, one obtains similarly that $\int _{{\mathbb {T}}_1} |{\widetilde{f}}(z)|^p\, dz\le (2\pi N_1^{p})^{-1} \Vert g\Vert _{H^p}^p$. Thus $\int _{\mathbb {T}}|{\widetilde{f}}(z)|^p\, dz<\infty$. $\square$

Proof of Theorem 3.1

By Lemma 3.2, in order to prove (3.1), the formula for the action of $T_\omega ^{*}$ on $q^\sharp H^{p'}$ and for the range of $T_{\omega }^{*}$ in (3.2), it remains to show that ${\mathrm {Dom}}(T_\omega ^{*})\subset q^\sharp H^{p'}$.

View ${\mathcal P}$ and ${\mathcal P}_k$, $k=1,2,\ldots$, as subspaces of $H^{p}$ or $H^{p'}$, write $P_k$ for the projection onto ${\mathcal P}_{k-1}$ and set $Q_k=I-P_k$. Also, the standard $k\times k$ compression of a Toeplitz operator $T_{\phi }$ on $H^p$ (or $H^{p'}$) is denoted by $T_{\phi ,k}$, i.e., $T_{\phi ,k}=P_k T_{\phi }|_{{\mathcal P}_{k-1}}$. Now, the relation (3.3) between $r\in {\mathcal P}_{n-1}$ and $r_1\in {\mathcal P}_{m-1}$ can be rewritten as

$$\begin{aligned} T_s r_1-T_q r\in {\mathcal P}_{m-1}, \end{aligned}$$

or, equivalently, as

$$\begin{aligned} Q_mT_s P_m r_1 = Q_mT_s r_1 = Q_m T_q r = Q_m T_q P_n r. \end{aligned}$$

(3.5)

We now consider the cases $m\ge n$ and $m<n$ separately.

First assume $m\ge n$. We can then decompose $Q_mT_s P_m$ and $Q_m T_q P_n$ as

$$\begin{aligned} Q_mT_s P_m&=\left[ \begin{array}{cc}0&{} T_{s^\sharp ,n}^* T_{z^{m-n}}^*\\ 0 &{} 0\end{array}\right] :{\mathcal P}_{m-1}=\left[ \begin{array}{c}{\mathcal P}_{m-n}\\ T_{z^{m-n}} {\mathcal P}_{n-1}\end{array}\right] \rightarrow \left[ \begin{array}{c}{\mathcal P}_{n-1}\\ T_{z}^{n}H^p\end{array}\right] ,\\ Q_mT_q P_n&=\left[ \begin{array}{c}T_{q^\sharp ,n}^*\\ 0\end{array}\right] :{\mathcal P}_{n-1}\rightarrow \left[ \begin{array}{c}{\mathcal P}_{n-1}\\ T_{z^{n}}H^p\end{array}\right] . \end{aligned}$$

Hence, in this case the identity in (3.5) can be written as

$$\begin{aligned} T_{s^\sharp ,n}^* (T_{z^{m-n}}^* r_1) = T_{q^\sharp ,n}^* r. \end{aligned}$$

Since all Toeplitz matrices are upper triangular, we in fact have

$$\begin{aligned} T_{s^\sharp ,m}^* T_{z^{m-n},m}^* r_1= T_{q^\sharp ,m}^* r. \end{aligned}$$

Note that $T_{q^\sharp ,n}^*$ is invertible, because q has only roots on ${\mathbb {T}}$ so that $q(0)\ne 0$. We obtain that for given $r_1\in {\mathcal P}_{m-1}$, the polynomial $r\in {\mathcal P}_{n-1}$ that satisfies (3.3) is uniquely determined by

$$\begin{aligned} r&= (T_{q^\sharp ,m}^*)^{-1} T_{s^\sharp ,m}^* T_{z^{m-n},m}^* r_1 = T_{s^\sharp ,m}^*T_{z,m}^{*m-n}(T_{q^\sharp ,m}^*)^{-1} r_1, \end{aligned}$$

where the commutation of Toeplitz matrices can occur since they all have analytic symbols. Now take $r_1\in {\mathcal P}_{m-1}$ arbitrary, and define r as above, so that (3.3) holds. Then, by Lemma 3.3, we have

$$\begin{aligned} \langle r_1,P_m k \rangle _{{\mathcal P}_{m-1}}&=\langle r_1,k \rangle _{p,p'}=\langle r,g \rangle _{p,p'} =\langle r,P_m g \rangle _{{\mathcal P}_{m-1}}\\&=\langle T_{s^\sharp ,m}^*T_{z,m}^{*m-n}(T_{q^\sharp ,m}^*)^{-1} r_1,P_m g \rangle _{{\mathcal P}_{m-1}}\\&=\langle r_1,(T_{q^\sharp ,m})^{-1} T_{z,m}^{m-n} T_{s^\sharp ,m} P_m g \rangle _{{\mathcal P}_{m-1}}. \end{aligned}$$

Since $r_1\in {\mathcal P}_{m-1}$ is arbitrary, we have $P_m k=(T_{q^\sharp ,m})^{-1} T_{z,m}^{m-n} T_{s^\sharp ,m} P_m g$, and thus

$$\begin{aligned} P_m T_{q^\sharp } k= T_{q^\sharp ,m} P_m k= T_{z,m}^{m-n} T_{s^\sharp ,m} P_m g =P_m T_{z}^{m-n} T_{s^\sharp } g. \end{aligned}$$

This shows that $P_m q^\sharp k= P_m z^{m-n} s^\sharp g$. Together with the first inclusion in (3.4) we obtain that

$$\begin{aligned} q^\sharp k= z^{m-n} s^\sharp g. \end{aligned}$$

Since $q^\sharp$ and $z^{m-n} s^\sharp$ are co-prime, we can apply Lemma 3.4 to conclude $g\in q^\sharp H^{p'}$.

Now assume $m<n$. By [4, Lemma 2.4], we can write $\omega = \omega _0 + \omega _1$ uniquely with $\omega _0\in {\mathrm {Rat}}_0({\mathbb {T}})$ and $\omega _1\in {\mathrm {Rat}}$ without poles on ${\mathbb {T}}$, i.e, $\omega _1\in L^\infty ({\mathbb {T}})$. In fact $\omega _1\in {\mathcal P}$, since all poles of $\omega$ are on ${\mathbb {T}}$, and $\omega _0={\widetilde{s}}/q$ with ${\widetilde{s}}\in {\mathcal P}_{m-1}$. It now follows that ${\mathrm {Dom}}(T_{\omega _0}^*)=q^\sharp H^{p'}$, and since $T_{\omega _1}$ is bounded, ${\mathrm {Dom}}(T_\omega ^*)={\mathrm {Dom}}(T_{\omega _0}^*)=q^\sharp H^{p'}$. Furthermore, $T_{\omega }^*= T_{\omega _0}^*+ T_{\omega _1}^*|_{q^\sharp H^{p'}}=T_{\omega _0^*}|_{q^\sharp H^{p'}}+ T_{\omega _1^*}|_{q^\sharp H^{p'}}=T_{\omega ^*}|_{q^\sharp H^{p'}}$.

In the next part of the proof we prove the formula for ${\mathrm {Ker}}(T_{\omega ^*})$, without distinguishing between the proper and non-proper case. Let $g=q^\sharp v\in {\mathrm {Dom}}(T_\omega ^*)$ with $v\in H^{p'}$. Then $g\in {\mathrm {Ker}}(T_{\omega }^*)$ if and only if $g\in {\mathrm {Ker}}(T_{\omega ^*})$, i.e., $g=q^\sharp v=r_1/(s_-)^\sharp$ for $r_1\in {\mathcal P}_{\deg (s_-)-1}$, see Proposition 2.1. Thus $v=r_1/((s_-)^\sharp q^\sharp )\in {\mathrm {Rat}}\cap H^{p'}$. Then $v\in H^{p'}$ implies $r_1=q^\sharp r$, and $\deg (r)=\deg (r_1)-m<\deg (s_-)-m$. Hence $g=q^\sharp r/(s_-)^\sharp$ with $\deg (r)<\deg (s_-)-m$. That all such functions are in ${\mathrm {Ker}}(T_{\omega }^*)={\mathrm {Ker}}(T_{\omega ^*})\cap q^\sharp H^{p'}$ follows directly from the formula for ${\mathrm {Ker}}(T_{\omega ^*})$ obtained in Proposition 2.1. The formula for the dimension of ${\mathrm {Ker}}(T_\omega ^*)$ follows directly and the condition for injectivity follows since $\deg (s_-)^\sharp$ is equal to the number of nonzero roots of $s_-$, counting multiplicity. $\square$

4 The Adjoint of $T_{\omega }$: General Case

In the section we prove Theorem 1.1 in full generality. Hence let $\omega =s/q\in {\mathrm {Rat}}$ with $s,q\in {\mathcal P}$ co-prime. As in Theorem 1.1, factor $s=s_-s_0s_+$ and $q=q_-q_0q_+$ with $s_-,q_-$ having roots only inside ${\mathbb {T}}$, $s_0,q_0$ having roots only on ${\mathbb {T}}$, and $s_+,q_+$ having roots only outside ${\mathbb {T}}$. Set $m=\deg (q)$, $n=\deg (s)$, $m_{\pm }={\mathrm {deg} }(q_{\pm })$, $n_{\pm }={\mathrm {deg} }(s_{\pm })$, and $m_{0}={\mathrm {deg} }(q_{0})$, $n_{0}={\mathrm {deg} }(s_{0})$. By Lemma 5.1 in [4], and its proof, we can factor $\omega$ as $\omega =\omega _- (z^\kappa \omega _0) \omega _+$ with $\kappa =n_- - m_-$, $\omega _-=s_-/(z^{\kappa }q_-)$ having only poles and zeroes inside ${\mathbb {T}}$, $\omega _0=s_0/q_0$ having only poles and zeroes on ${\mathbb {T}}$, and $\omega _+=s_+/q_+$ having only poles and zeroes outside ${\mathbb {T}}$, and we have $T_{\omega }=T_{\omega _-} T_{z^\kappa \omega _0} T_{\omega _+}$. Moreover, $T_{\omega _-}$ and $T_{\omega _+}$ are bounded and boundedly invertible.

Note that $T_{\omega _-}T_{z^\kappa \omega _0}$ is closed and densely defined and ${{\mathrm {Ran}}}(T_{\omega _+})=H^p$, and thus by Corollary 1 in [12]

$$\begin{aligned} T_\omega ^*=T_{\omega _+}^* \left( T_{\omega _-}T_{z^\kappa \omega _0}\right) ^*. \end{aligned}$$

Furthermore, $T_{\omega _-}$ is bounded and $T_{z^\kappa \omega _0}$ is closed and densely defined. By Theorem 4 in [13] one has

$$\begin{aligned} \left( T_{\omega _-}T_{z^\kappa \omega _0}\right) ^* = T_{z^\kappa \omega _0}^*T_{\omega _-}^*. \end{aligned}$$

Combining this and using that $T_{\omega _+}^*=T_{\omega _+^*}$ and $T_{\omega _-}^*=T_{\omega _-^*}$ we see that

$$\begin{aligned} T_\omega ^*=T_{\omega _+}^*T_{z^\kappa \omega _0}^*T_{\omega _-}^*=T_{\omega _+^*}T_{z^\kappa \omega _0}^*T_{\omega _-^*} \quad \text{ on } {\mathrm {Dom}}(T_{\omega }^*). \end{aligned}$$

Note that

$$\begin{aligned} \omega _-^*&=\frac{(s_-)^\sharp }{(q_-)^\sharp },\quad \omega _0^*=z^{m_0-n_0}\frac{(s_0)^\sharp }{(q_0)^\sharp },\\ (z^\kappa \omega _0)^*&=z^{m_0-n_0-\kappa }\frac{(s_0)^\sharp }{(q_0)^\sharp },\quad \omega _+^*=z^{m_+-n_+}\frac{(s_+)^\sharp }{(q_+)^\sharp }. \end{aligned}$$

By construction, $\omega _-$ and $1/\omega _-$ are both anti-analytic. Consequently, $\omega _-^*$ and $1/\omega _-^*$ are both analytic functions. This implies $T_{\omega _-^*}^{\pm } (q_0)^\sharp H^{p'}\subset (q_0)^\sharp H^{p'}$, and thus $T_{\omega _-^*} (q_0)^\sharp H^{p'}= (q_0)^\sharp H^{p'}$. Since $T_{\omega _+^*}$ is invertible, to see that ${\mathrm {Dom}}(T_\omega ^*)=(q_0)^\sharp H^{p'}$ it suffices to show ${\mathrm {Dom}}(T_{z^\kappa \omega _0}^*)=(q_0)^\sharp H^{p'}$. For the case where $\kappa \ge 0$, so that $z^\kappa \omega _0\in {\mathrm {Rat}}({\mathbb {T}})$, this follows directly from Theorem 3.1. For $\kappa <0$, note that $T_{z^\kappa \omega _0}=T_{z^\kappa }T_{\omega _0}$, so that $T_{z^\kappa \omega _0}^*=T_{\omega _0}^*T_{z^\kappa }^*=T_{\omega _0}^*T_{z^{-\kappa }}$, again using Theorem 4 of [13]. Then $g\in {\mathrm {Dom}}(T_{z^\kappa \omega _0}^*)$ holds if and only if $z^{-\kappa } g\in {\mathrm {Dom}}(T_{\omega _0}^*)=(q_0)^\sharp H^{p'}$. By Lemma 3.4 this is the same as $g\in (q_0)^\sharp H^{p'}$, since $z^{-\kappa }$ and $q_0^\sharp$ are co-prime. Thus in both cases we arrive at ${\mathrm {Dom}}(T_\omega ^*)=(q_0)^\sharp H^{p'}$. Moreover, we also find that $T_{z^\kappa \omega _0}^*=T_{(z^\kappa \omega _0)^*}|_{(q_0)^\sharp H^{p'}}$, so that

$$\begin{aligned} T_\omega ^*=T_{\omega _+^*}T_{z^\kappa \omega _0}^*T_{\omega _-^*}=T_{\omega _+^*}T_{(z^\kappa \omega _0)^*}T_{\omega _-^*}|_{(q_0)^\sharp H^{p'}} =T_{\omega ^*}|_{(q_0)^\sharp H^{p'}}. \end{aligned}$$

Hence (1.2) holds.

Next we derive the formula for ${\mathrm {Ker}}(T_{\omega }^*)$. For $\kappa \ge 0$ we have $g\in {\mathrm {Ker}}(T_{\omega }^*)$ if and only if $T_{\omega _-^*}g\in {\mathrm {Ker}}(T_{z^\kappa \omega _0}^*)= (q_0)^\sharp {\mathcal P}_{\kappa -m_0-1}$, where the last identity follows by applying Theorem 3.1 to $z^\kappa \omega _0$. Thus $g\in {\mathrm {Ker}}(T_{\omega }^*)$ if and only if $((s_-)^\sharp /(q_-)^\sharp )g=(q_0)^\sharp r$, i.e., $g=(q_-)^\sharp (q_0)^\sharp r/(s_-)^\sharp$, for some $r\in {\mathcal P}_{\kappa -m_0-1}$, as claimed. For $\kappa <0$ we have $g\in {\mathrm {Ker}}(T_{\omega }^*)$ if and only if $z^{-\kappa } \omega _-^* g\in {\mathrm {Ker}}(T_{\omega _0}^*)$. However, ${\mathrm {Ker}}(T_{\omega _0}^*)=\{0\}$, by Theorem 3.1, so that ${\mathrm {Ker}}(T_{\omega }^*)=\{0\}$, in line with the formula in (1.3). The formula for the dimension of ${\mathrm {Ker}}(T_\omega ^*)$ follows directly.

Now we turn to the formula for ${\mathrm {Ran}}(T_\omega ^*)$. Note that

$$\begin{aligned} {\mathrm {Ran}}(T_\omega ^*)= T_{\omega _+^*} {\mathrm {Ran}}(T_{z^\kappa \omega _0}^* T_{\omega _-^*})= T_{\omega _+^*} {\mathrm {Ran}}(T_{z^\kappa \omega _0}^*). \end{aligned}$$

(4.1)

We first show that ${\mathrm {Ran}}(T_{z^\kappa \omega _0}^*)= T_{z^{m_0-n_0-\kappa }}(s_0)^\sharp H^{p'}$. Again, for the case $\kappa \ge 0$ this follows directly from Theorem 3.1. Assume $\kappa <0$. Then $T_{z^\kappa \omega _0}^*=T_{\omega _0}^* T_{z^{-\kappa }}$. Hence,

$$\begin{aligned} {\mathrm {Ran}}(T_{z^\kappa \omega _0}^*)&=T_{\omega _0}^* (z^{-\kappa } H^{p'}\cap {\mathrm {Dom}}(T_{\omega _0})) =T_{\omega _0}^* (z^{-\kappa } H^{p'}\cap (q_0)^\sharp H^{p'})\\&=T_{\omega _0}^* z^{-\kappa } (q_0)^\sharp H^{p'}. \end{aligned}$$

The last identity follows by Lemma 3.4. Now the action of $T_{\omega _0}^*$, as described in Theorem 3.1, shows that ${\mathrm {Ran}}(T_{z^\kappa \omega _0}^*)=T_{z^{m_0-n_0}}z^{-\kappa }(s_0)^\sharp H^{p'} =T_{z^{m_0-n_0-\kappa }}(s_0)^\sharp H^{p'}$. Since $1/q_+$ is analytic, $1/(q_+)^\sharp$ is anti-analytic, and therefore, independent of the sign of $m_+-n_+$, we have

$$\begin{aligned} T_{\omega _+^*}=T_{1/(q_+)^\sharp } T_{z^{m_+-n_+}} T_{(s_+)^\sharp }. \end{aligned}$$

Thus

$$\begin{aligned} {\mathrm {Ran}}(T_{\omega }^*)=T_{1/(q_+)^\sharp } T_{z^{m_+-n_+}} T_{(s_+)^\sharp } T_{z^{m_0-n_0-\kappa }}(s_0)^\sharp H^{p'}. \end{aligned}$$

Note that $T_{(s_+)^\sharp }$ and $T_{z^{m_0-n_0-\kappa }}$ need not commute, in case $m_0-n_0-\kappa <0$. However, we do have $T_{(s_+)^\sharp } T_{z^{m_0-n_0-\kappa }}=T_{z^{m_0-n_0-\kappa }} T_{(s_+)^\sharp } Q_{\kappa + n_0-m_0}$. Moreover, since $(s_+)^\sharp$ is analytic, $T_{(s_+)^\sharp } Q_{\kappa + n_0-m_0}=Q_{\kappa + n_0-m_0}T_{(s_+)^\sharp } Q_{\kappa + n_0-m_0}$ and we have

$$\begin{aligned} T_{z^{m_+-n_+}} T_{z^{m_0-n_0-\kappa }} Q_{\kappa + n_0-m_0}&=T_{z^{m_+-n_+ + m_0-n_0-\kappa }} Q_{\kappa + n_0-m_0}\\&=T_{z^{m-n}} Q_{\kappa + n_0-m_0}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} {\mathrm {Ran}}(T_{\omega }^*)&=T_{1/(q_+)^\sharp } T_{z^{m-n}} T_{(s_+)^\sharp }Q_{\kappa + n_0-m_0} (s_0)^\sharp H^{p'}\\&=T_{z^{m-n}(s_+)^\sharp /(q_+)^\sharp }Q_{\kappa + n_0-m_0} (s_0)^\sharp H^{p'}, \end{aligned}$$

again using that $1/(q_+)^\sharp$ is anti-analytic and $(s_+)^\sharp$ is analytic. This gives the general formula for ${\mathrm {Ran}}(T_\omega ^*)$. In case $\kappa + n_0-m_0\le 0$, we have $Q_{\kappa + n_0-m_0}=I$ and $T_{(s_+)^\sharp }Q_{\kappa + n_0-m_0} (s_0)^\sharp = (s_+s_0)^\sharp$, as claimed.

5 Symmetric Operators and Selfadjoint Extensions

For $\omega \in {\mathrm {Rat}}$, the second adjoint $T_\omega ^{**}$ is well-defined and $T_\omega ^{**} = T_\omega$, since $T_\omega$ is a closed, densely defined operator on a reflexive Banach space [7, Theorem III.5.24]. Now consider $\omega \in {\mathrm {Rat}}({\mathbb {T}})$ and $p=2$. From Theorem 1.1 it is obvious that $T_{\omega }\ne T_{\omega }^*$, except in the degenerate case where q is constant, since ${\mathrm {Dom}}(T_\omega )= q H^2 +{\mathcal P}_{\deg (q)-1}$ contains all polynomials while ${\mathrm {Dom}}(T_\omega ^*)= q^\sharp H^2$ only contains the polynomials that contain $q^\sharp$ as a factor. Consequently, $T_{\omega }$ cannot be selfadjoint. In this section we consider the question when $T_\omega ^*$ is symmetric, and, if this is the case, when does $T_\omega ^*$ have a selfadjoint extension L. The first topic is addressed in the following theorem.

Theorem 5.1

Let $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$ with $s,q\in {\mathcal P}$ co-prime. Set $n=\deg (s)$ and $m=\deg (q)$. Then the following are equivalent.

• (1) $T_\omega ^*$ is symmetric;

• (2) $\omega ({\mathbb {T}})\subset {\mathbb {R}}$;

• (3) $\omega (z)={\widetilde{\omega }}(-i\frac{z+1}{z-1})$ with ${\widetilde{\omega }}$ a real rational function with poles only on ${\mathbb {R}}$;

• (4) the essential spectrum $\sigma _\text {ess}(T_\omega )$ of $T_\omega$ is contained in ${\mathbb {R}}$;

• (5) $\omega$ is proper, $s=z^{m-n} {\widetilde{s}}$ with ${\widetilde{s}}$ self-inversive and $q_0 \overline{s_n}=\overline{q_m} s_{m-n}$ holds, where $s(z)=\sum _{k=0}^{n} s_k z^k$ and $q(z)=\sum _{k=0}^{m} q_k z^k$.

Moreover, if $T_\omega ^*$ is symmetric, then $T_\omega ^*\subset T_\omega$.

Proof

We first prove the equivalence of (1) and (2), and that (1) implies $T_\omega ^*\subset T_\omega$. Assume (2). For $z\in {\mathbb {T}}$, not a root of q, we have $\omega ^*(z)=\overline{\omega (z)}=\omega (z)$. Hence $\omega ^*=\omega$. Since q has only roots on ${\mathbb {T}}$, we have $q=\gamma q^\sharp$ for a unimodular constant $\gamma$. Hence $q H^2=q^\sharp H^2$. This shows $T^*_{\omega }=T_{\omega ^*}|_{q^\sharp H^2}=T_{\omega }|_{q H^2}\subset T_\omega$. Since $(T_\omega ^{*})^*=T_\omega$, it follows that $T_\omega ^*$ is symmetric and $T_\omega ^*\subset T_\omega$. Conversely, assume (1). Then we still have $q H^2=q^\sharp H^2$ and $T_{\omega }^*\subset (T_\omega ^*)^*=T_\omega$. Hence $T_\omega ^*=T_\omega |_{q H^2}$. In particular, we have $\omega ^* q=T_{\omega ^*} q=T_\omega ^* q=T_\omega q= \omega q$. This implies $\omega =\omega ^*$. Hence $\omega (z)=\overline{\omega (z)}$ for $z\in {\mathbb {T}}$, not a root of q. Thus $\omega ({\mathbb {T}})\subset {\mathbb {R}}$.

That (2) and (3) are equivalent follows simply because in (3) $\omega$ is the composition of ${\widetilde{\omega }}$ and the inverse Cayley transform, which maps the circle ${\mathbb {T}}$ bijectively onto ${\mathbb {R}}$. The fact that ${\widetilde{\omega }}$ is real rational, i.e., ${\widetilde{\omega }}={\widetilde{s}}/{\widetilde{q}}$ with ${\widetilde{s}}$ and ${\widetilde{q}}$ real polynomials, is equivalent to ${\widetilde{\omega }}({\mathbb {R}}):=\{{\widetilde{\omega }}(t) :t\in {\mathbb {R}},\ {\widetilde{q}}(t)\ne 0\}\subset {\mathbb {R}}$. Also, the equivalence of (2) and (4) is a direct consequence of the fact that $\sigma _\text {ess}(T_\omega )=\omega ({\mathbb {T}})$, by [5, Theorem 1.1].

Finally, we prove (2) $\Leftrightarrow$ (5). Since $q=\gamma q^\sharp$, we have

$$\begin{aligned} \omega ^*=z^{m-n} \frac{s^\sharp }{q^\sharp }= z^{m-n} \gamma \frac{s^\sharp }{q}. \end{aligned}$$

Thus, we have $\omega =\omega ^*$ if and only if $z^{m-n} \gamma s^\sharp = s$. Hence (2) is equivalent to $z^{m-n}\gamma s^\sharp = s$. Now assume (2). Since $\deg (s^\sharp )\le \deg (s)$, the identity $z^{m-n}\gamma s^\sharp = s$ can only occur if $m\ge n$, i.e., if $\omega$ is proper. The identity also shows that $s= z^{m-n} {\widetilde{s}}$ for ${\widetilde{s}}=\gamma s^\sharp$. On the other hand, $s^\sharp =(z^{m-n} {\widetilde{s}})^\sharp = {\widetilde{s}}^\sharp$. Thus ${\widetilde{s}}= \gamma s^\sharp = \gamma {\widetilde{s}}^\sharp$, which shows ${\widetilde{s}}$ is self-inversive, with constant $\gamma$. Note that $\gamma =q_0/\overline{q_m}$. Also, we have $s_0=\cdots =s_{m-n-1}=0$ and ${\widetilde{s}}(z)=\sum _{k=0}^{2n-m} s_{m-n+k} z^k$. Since ${\widetilde{s}}$ is self-inversive, ${\widetilde{s}}=\delta {\widetilde{s}}^\sharp$ with $\delta =s_{m-n}/\overline{s_n}$. But also $\delta =\gamma$, so $s_{m-n}/\overline{s_n}=q_0/\overline{q_m}$. Thus $q_0 \overline{s_n}=\overline{q_m} s_{m-n}$. Hence (5) holds. Conversely, assume (5). Reversing the above argument, it follows that $q_0 \overline{s_n}=\overline{q_m} s_{m-n}$ implies ${\widetilde{s}}=\delta {\widetilde{s}}^\sharp$ with $\delta =\gamma$. Thus $\gamma s^\sharp =\gamma {\widetilde{s}}^\sharp = {\widetilde{s}}$. This implies $s =z^{m-n} {\widetilde{s}}= z^{m-n} \gamma s^\sharp$, and hence (2). $\square$

Corollary 5.2

Let $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$ with $s,q\in {\mathcal P}$ co-prime.Assume $T_\omega ^*$ is symmetric. Then $\deg (s)\le \deg (q)\le 2\deg (s)$.

Proof

By Theorem 5.1 condition (5) holds with $m=\deg (q)$ and $n=\deg (s)$. Since ${\widetilde{s}}$ is self-inversive, we have ${\widetilde{s}}(0)\ne 0$. Consequently, 0 would be a non-removable singularity of $s=z^{m-n}{\widetilde{s}}$ in case $m<n$, which gives a contradiction. Hence $m\ge n$. Furthermore, comparing the degrees on both sides of $s=z^{m-n}{\widetilde{s}}$ yields, $n=m-n+\deg ({\widetilde{s}})\ge m-n$. Hence $m\le 2n$. $\square$

When $T_{\omega }^*$ is symmetric, it need not be the case that $T_{\omega }^*$ has a selfadjoint extension. In Proposition 5.4 below we characterize when $T_{\omega }^*$ does have a selfadjoint extension. However, we first give a concrete example that shows this does not always happen.

Example 5.3

In [6] Helson considered the functions $\omega _k(z)=\left( -i\frac{z+1}{z-1}\right) ^k$ for $k\in {\mathbb {N}}$. For all k we have $\omega _k({\mathbb {T}})\subset {\mathbb {R}}$, see Theorem 5.1 (3) above, hence $T_{\omega _k}^*$ is symmetric by Theorem 5.1. In fact, for k even $\omega _k({\mathbb {T}})={\mathbb {R}}_+$, while for k odd we have $\omega _k({\mathbb {T}})={\mathbb {R}}$. We show that $T_{\omega _k}^*$ does not have a selfadjoint extension for $k=1$. In Example 5.8 we return to this example for general k.

For $k=1$ we have $\omega (z)=\omega _1(z) = -i\frac{z+1}{z-1}$. Hence ${\mathrm {Dom}}(T_\omega ) = (z-1)H^2 + {\mathbb {C}}$ and ${\mathrm {Dom}}(T_\omega ^*) = (z-1)H^2$. Suppose $T_\omega ^*$ has a selfadjoint extension L. Then $L=L^*$ and thus $T_{\omega }^*\subset L=L^* \subset T_{\omega }^{**}=T_\omega$. Since $T_\omega$ is not selfadjoint, the inclusions are strict. Hence ${\mathrm {Dom}}(T_\omega ^*)\subset {\mathrm {Dom}}(L)\subset {\mathrm {Dom}}(T_\omega )$, with strict inclusions. However, the complement of ${\mathrm {Dom}}(T_\omega ^*)$ in ${\mathrm {Dom}}(T_\omega )$ is one-dimensional, hence not both inclusions can be strict. Thus $T_\omega$ does not admit a selfadjoint extension.

Proposition 5.4

Let $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$, with $s,q\in {\mathcal P}$ coprime, be such that $T_\omega ^*$ is symmetric. Then $T_\omega ^*$ admits a selfadjoint extension if and only if the number of roots of $s-iq$ and $s+iq$ in ${\mathbb {D}}$, counting multiplicities, coincide.

Proof

The operator $T_\omega ^*$ is an adjoint, and hence closed, and by assumption symmetric. Following definition X.2.12 from [2] we define the deficiency subspaces of $T_\omega ^*$ as the spaces

$$\begin{aligned} \mathcal {L}_+= & {} \mathrm{Ker\, } (T_\omega ^{**}-i)=(\mathrm{Ran\, }(T_\omega ^*+i))^\perp , \\ \mathcal {L}_-= & {} \mathrm{Ker\, } (T_\omega ^{**}+i)=(\mathrm{Ran\, }(T_\omega ^*-i))^\perp , \end{aligned}$$

and the deficiency indices as the integers $n_{\pm }=\mathrm{dim\, }\mathcal {L}_{\pm }$. Since $T_{\omega }^{**}=T_\omega$, we have

$$\begin{aligned} n_+=\mathrm{dim\, }\mathrm{Ker\, } (T_\omega -i) \quad \text{ and }\quad n_-=\mathrm{dim\, }\mathrm{Ker\, } (T_\omega +i). \end{aligned}$$

Also, we have $T_\omega \pm i= T_{\omega \pm i}$. By item (b) of Theorem X.2.20 in [2], $T_\omega$ has a selfadjoint extension if and only if $n_+=n_-$. Note that $\omega \pm i = (s\pm i q)/q$. We now apply Corollary 4.2 from [4] to $T_{\omega \pm i}$, to obtain that $n_\pm$ is equal to the maximum of 0 and the difference of m and the number of roots of $s\pm i q$ in $\overline{{\mathbb {D}}}$, counting multiplicities. However, since $T_\omega ^*$ is symmetric, $\omega$ is proper so the number of roots cannot exceed m. Note also that $\omega ({\mathbb {T}})\subset {\mathbb {R}}$, so $s\pm i q$ cannot have roots on ${\mathbb {T}}$. It thus follows that $T_\omega ^*$ has a selfadjoint extension if and only if the number of roots in ${\mathbb {D}}$ of $s-iq$ and $s+iq$, counting multiplicities, coincide, as claimed. $\square$

Since $T_\omega ^*$ is never selfadjoint for $\omega \in {\mathrm {Rat}}({\mathbb {T}})$ having at least one pole on ${\mathbb {T}}$, the formulas for $n_\pm$ in the above proof along with item (a) of Theorem X.2.20 in [2] directly give the following corollary.

Corollary 5.5

Let $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$, with $s,q\in {\mathcal P}$ coprime, be such that $T_\omega ^*$ is symmetric. Then $s+i q$ or $s-iq$ must have a root in ${\mathbb {D}}$.

Proposition 5.4 can be rephrased in terms of the index of the operators $T_{\omega \pm i}$.

Proposition 5.6

Let $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$, with $s,q\in {\mathcal P}$ coprime, be such that $T_\omega ^*$ is symmetric. Then $T_{\omega +i}$ and $T_{\omega - i}$ are both Fredholm and $T_\omega ^*$ admits a selfadjoint extension if and only if the Fredholm indices of $T_{\omega +i}$ and $T_{\omega - i}$ coincide.

Proof

This follows directly from Proposition 5.4 and Theorem 1.1 of [4] applied to $\omega + i$ and $\omega - i$, using that $\omega \pm i=(s\pm i q)/q$. $\square$

Corollary 5.7

Let $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$, with $s,q\in {\mathcal P}$ coprime, be such that $T_\omega ^*$ is symmetric. Assume $\omega ({\mathbb {T}})\ne {\mathbb {R}}$. Then $T_\omega ^*$ admits a selfadjoint extension.

Proof

The Fredholm index of $T_{\omega -\lambda }$ is constant with respect to $\lambda \in {\mathbb {C}}$ on the connected components of ${\mathbb {C}}$ separated by the essential spectrum of $T_\omega$, which is equal to $\omega ({\mathbb {T}})$; see [5, Theorem 1.1]. Hence if $\omega ({\mathbb {T}})\ne {\mathbb {R}}$, but $\omega ({\mathbb {T}})\subset {\mathbb {R}}$ since $T_\omega ^*$ is symmetric, then i and $-i$ are in the same connected component and thus $T_{\omega +i}$ and $T_{\omega -i}$ have the same index. The conclusion now follows from Proposition 5.6. $\square$

Example 5.8

We return to the functions $\omega _k(z)=\left( -i\frac{z+1}{z-1}\right) ^k$ considered in Example 5.3. Since $\omega _k({\mathbb {T}})={\mathbb {R}}_+$ for k even, we obtain directly from Corollary 5.7 that $T_{\omega _k}^*$ admits a selfadjoint extension in case k is even.

For odd values of k we have $\omega _k({\mathbb {T}})={\mathbb {R}}$, and thus no conclusion can be drawn from Corollary 5.7. To deal with the odd case we resort to Proposition 5.4. Take $s(z)=(-i)^k(z+1)^k$ and $q=(z-1)^k$ and write k as $k=2l+1$. The polynomials $s\pm iq$ are given by

$$\begin{aligned} s(z)\pm iq(z)&=i\left( (-1)^{l+1}(z+1)^{2l+1}\pm (z-1)^{2l+1}\right) \\&=i\left( (-1)^{l+1}\sum _{j=0}^{2l+1}\left( {\begin{array}{c}2l+1\\ j\end{array}}\right) z^j \pm \sum _{j=0}^{2l+1} \left( {\begin{array}{c}2l+1\\ j\end{array}}\right) z^j (-1)^{2l+1-j}\right) \\&=i\sum _{j=0}^{2l+1} \left( {\begin{array}{c}2l+1\\ j\end{array}}\right) z^j \left( (-1)^{l+1}\pm (-1)^{2l+1-j}\right) \\&=i \sum _{j=0}^{2l+1}\left( {\begin{array}{c}2l+1\\ j\end{array}}\right) z^j \left( (-1)^{l+1}\pm (-1)^{j-1}\right) . \end{aligned}$$

For odd values of l one obtains:

$$\begin{aligned} s(z)-iq(z)&= -2i \left( \left( {\begin{array}{c}2l+1\\ 0\end{array}}\right) +\cdots + \left( {\begin{array}{c}2l+1\\ 2l-2\end{array}}\right) z^{2l-2} + \left( {\begin{array}{c}2l+1\\ 2l\end{array}}\right) z^{2l} \right) ,\\ s(z)+iq(z)&= 2i \left( \left( {\begin{array}{c}2l+1\\ 1\end{array}}\right) z +\cdots + \left( {\begin{array}{c}2l+1\\ 2l-1\end{array}}\right) z^{2l-1} + \left( {\begin{array}{c}2l+1\\ 2l+1\end{array}}\right) z^{2l+1} \right) \\&= 2i z \left( \left( {\begin{array}{c}2l+1\\ 2l\end{array}}\right) +\cdots + \left( {\begin{array}{c}2l+1\\ 2\end{array}}\right) z^{2-2} + \left( {\begin{array}{c}2l+1\\ 0\end{array}}\right) z^{2l} \right) \end{aligned}$$

Observe that $s+iq$ is of the form $izp_+(z^2)$ where $p_+$ is a real polynomial of degree 2l and that $s-ig$ is of the form $i p_-(z^2)$ where $p_-$ is a real polynomial of degree 2l. Because $p_+$ and $p_-$ are real polynomials and the fact that $z^2$ is the variable rather than z itself, the nonzero roots of $zp_+(z^2)$ come either in pairs (z and $-z$) for real nonzero roots or in quadruples ($z, {\bar{z}}, -z, -{\bar{z}}$) for nonreal roots, while zero appears as a simple root. Similarly, the roots of $p_-(z^2)$ come in pairs (z and $-z$) or quadruples ($z, {\bar{z}}, -z, -{\bar{z}}$) and there is no root at zero. Hence $s+iq$ has an odd number of roots inside the unit disc, and $s-iq$ has an even number of roots inside the unit disc, so that the indices $n_+$ and $n_-$ can never coincide. One further observes that $p_-=p_+^\sharp$. In a similar way, for even values of l the polynomial $s+iq$ will have an even number of roots inside the unit disc and $s-iq$ will have an odd number of roots inside the unit disc. Hence, in all cases where k is odd, $T_\omega ^*$ does not have a selfadjoint extension.

We now present a proposition that rephrases the criteria of Proposition 5.4 in terms of the roots of $s+iq$ (or $s-iq$) only. The observation that $T_{\omega _k}^*$ in Example 5.8 has no selfadjoint extension follows as a special case. In general, $T_\omega ^*$ cannot have a selfadjoint extension whenever $\deg (q)$ is odd for any $\omega \in {\mathrm {Rat}}({\mathbb {T}})$.

Proposition 5.9

Let $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$, with $s,q\in {\mathcal P}$ coprime, be such that $T_\omega ^*$ is symmetric. Set $l_{\pm }=m-\deg (s\pm iq)$ and define

$$\begin{aligned} k_{\pm ,1}=\# \left\{ \begin{array}{l} \text {zeroes of } \omega \pm i \text { inside }{\mathbb {T}} \\ \text {multi. taken into account}\!\!\!\! \end{array}\right\} ,\ k_{\pm ,2}=\# \left\{ \begin{array}{l} \text {zeroes of } \omega \pm i \text { outside }{\mathbb {T}} \\ \text {multi. taken into account} \end{array}\right\} . \end{aligned}$$

Then

$$\begin{aligned} T_\omega ^* \text{ has } \text{ a } \text{ selfadjoint } \text{ extension } \ \Leftrightarrow \ l_+ + k_{+,2}= k_{+,1} \ \Leftrightarrow \ l_- + k_{-,2}= k_{-,1}. \end{aligned}$$

In particular, if $T_\omega ^*$ has a selfadjoint extension, then $\deg (q)$ must be even.

The basis for the proof of Proposition 5.9 lies in the following lemma, which clarifies the relation between $s+iq$ and $s-iq$ under the assumption that $T_\omega ^*$ is symmetric.

Lemma 5.10

Let $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$, with $s,q\in {\mathcal P}$ coprime, be such that $T_\omega ^*$ is symmetric. Set $l_{\pm }=\deg (q)-\deg (s\pm iq)$ and let $\gamma$ be the unimodular constant such that $q=\gamma q^\sharp$. Then

$$\begin{aligned} s\pm iq=\gamma z^{l_{\mp }}(s\mp iq)^\sharp . \end{aligned}$$

(5.1)

Moreover, we have $l_\pm =0$ if and only if $\omega (0)=\pm i$. In particular, only one of $l_+$ and $l_-$ can be nonzero.

Proof

Since $T_\omega ^*$ is symmetric, by assumption, $\omega$ has the properties listed in Theorem 5.1. In particular, $\omega$ is proper, $m:=\deg (q)\ge \deg (s)=:n$, and $s=z^{m-n} {\widetilde{s}}$ with ${\widetilde{s}}$ self-inversive and the unimodular constants that establish the self-inversiveness of ${\widetilde{s}}$ and q coincide (equivalently, $q_0 \overline{s_n}=\overline{q_m}s_{m-n}$).

Note that $\deg (s\pm iq)\ne m$ occurs precisely when $\deg (s)=\deg (q)$ and the leading coefficients $s_m$ and $q_m$ of s and q, respectively, satisfy $s_m\pm i q_m=0$, i.e., $s_m/q_m= \mp i$. Since $m=n$, the identity $q_0 \overline{s_n}=\overline{q_m}s_{m-n}$ shows $\omega (0)=s_0/q_0=\overline{s_m}/\overline{q_m}$. Hence $\deg (s\pm iq)\ne m$ holds if and only if $\omega (0)=\overline{\mp i}=\pm i$, as claimed.

We first prove (5.1) for the case $\omega (0)=0$. So assume $\omega (0)=0$, or equivalently, $s(0)=0$. In this case $l_+=l_-=0$. Since $s=z^{m-n}{\widetilde{s}}$ and ${\widetilde{s}}(0)\ne 0$ (because ${\widetilde{s}}$ is self-inversive), we have $m>n$. Also note that $m-n$ is equal to the multiplicity of 0 as a root of s. We now employ Lemma 2.2, using that $\deg (s+iq)=m=\deg (iq)$, to obtain

$$\begin{aligned} \gamma (s\mp iq)^{\sharp }&= z^{\deg (s+iq)-\deg (s)} \gamma s^\sharp \mp (-i) \gamma q^\sharp =z^{m-n} \gamma {\widetilde{s}}^\sharp \pm i q\\&= z^{m-n} {\widetilde{s}} \pm i q =s\pm iq. \end{aligned}$$

Hence (5.1) holds.

Now assume $\omega (0)\ne 0$, i.e., $s(0)\ne 0$. In that case $s={\widetilde{s}}$. Hence s is self-inversive with the same constant $\gamma$ that establishes the self-inversiveness of q. This also yields $m=n$. Since s and q are self-inversive with the same constant $\gamma$, we have

$$\begin{aligned} \overline{s_{m-k}} q_{k} =\overline{q_{m-k}} \overline{s_{m-k}} \gamma = \overline{q_{m-k}} s_{k}\quad \text{ for } k=0,\ldots , m. \end{aligned}$$

Hence for all k we have

$$\begin{aligned} \overline{s_{m-k}}(s_k+i q_k)= & {} s_k (\overline{s_{m-k}} + i \overline{q_{m-k}}) \ \ \text{ and }\ \ \overline{q_{m-k}}(s_k+i q_k)\\= & {} q_k (\overline{s_{m-k}} + i \overline{q_{m-k}}). \end{aligned}$$

In case $s_{m-k}=0$ and $q_{m-k}=0$, also $s_k=0$ and $q_k=0$, since $s_k=\gamma \overline{s_{m-k}}$ and $q_k=\gamma \overline{q_{m-k}}$, and thus $s_k+i q_k=0=\gamma (\overline{s_{m-k}} + i \overline{q_{m-k}})$. If either $s_{m-k}\ne 0$ or $q_{m-k}\ne 0$, divide the first identity by $\overline{s_{m-k}}$ or the second identity by $\overline{q_{m-k}}$ to arrive at $s_k+i q_k=\gamma (\overline{s_{m-k}} + i \overline{q_{m-k}})$. Hence

$$\begin{aligned} s_k+i q_k=\gamma (\overline{s_{m-k} - i q_{m-k}}) \quad \text{ for } k=0,\ldots , m. \end{aligned}$$

(5.2)

Thus $s_k+i q_k=0$ if and only if $s_{m-k} - i q_{m-k}=0$. It follows that 0 is a root of $s\pm iq$ with multiplicity $l_\mp$. Comparing coefficients, it follows that the identities in (5.1) correspond to the identities in (5.2). Hence (5.1) holds. $\square$

Proof of Proposition 5.9

Since $T_\omega ^*$ is assumed to be symmetric, (5.1) holds. Together with the fact that the $\sharp$ operator reflects roots over ${\mathbb {T}}$, this implies that the number of roots of $s\pm iq$ inside ${\mathbb {T}}$ are equal to $l_\pm$ plus the number of roots of $s\mp iq$ outside ${\mathbb {T}}$, counting multiplicities. In other words, we have

$$\begin{aligned} k_{+,1}=l_- + k_{-,2} \quad \text{ and }\quad k_{-,1}=l_+ + k_{+,2}. \end{aligned}$$

(5.3)

By Proposition 5.6, $T_{\omega }^*$ has a selfadjoint extension if and only if $s+iq$ and $s-iq$ have an equal number of roots inside ${\mathbb {T}}$, again counting multiplicities, equivalently, $k_{+,1}=k_{-,1}$. Given (5.3), it follows that $k_{+,1}=k_{-,1}$ is equivalent to $k_{+,1}=l_+ + k_{+,2}$, and likewise to $k_{-,1}=l_- + k_{-,2}$. This proves the two criteria for $T_{\omega }^*$ to have a selfadjoint extension.

By Lemma 5.10, either $l_+=0$ or $l_-=0$. Say $l_+=0$. Since $s+iq$ cannot have roots on ${\mathbb {T}}$, we have $\deg (q)=\deg (s+iq)=k_{+,1}+k_{+,2}$. If $T_\omega ^*$ admits a selfadjoint extension, then we have $k_{+,1}=l_+ + k_{+,2}=k_{+,2}$. Hence $\deg (q)= 2 k_{+,1}$ is even. For $l_-=0$ the arguments goes similarly. $\square$

Combining the fact that $T_\omega ^*$ cannot have a selfadjoint extension in case $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$, s, q co-prime, and $\deg (q)$ odd with Corollary 5.7 immediately yields the following result.

Corollary 5.11

Let $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$, with $s,q\in {\mathcal P}$ co-prime, be such that $T_\omega ^*$ is symmetric and $\deg (q)$ is odd. Then $\omega ({\mathbb {T}})={\mathbb {R}}$.

The next example shows that also with $\deg (q)$ even it can occur that $T_\omega ^*$ does not admit a selfadjoint extension.

Example 5.12

Let $\omega =s/q$ with

$$\begin{aligned} s(z)=i(1+az+z^2),\ \text{ for } \text{ some } 0\ne a\in {\mathbb {R}}, \quad \text{ and }\quad q(z)=1-z^2. \end{aligned}$$

Then $m=n$ and

$$\begin{aligned} s^\sharp =-s,\quad q^\sharp =-q. \end{aligned}$$

So $T_\omega ^*$ is symmetric by Theorem 5.1 (5). Also, we have

$$\begin{aligned} (s+iq)(z)=i(2+az)\quad \text{ and }\quad (s-iq)(z)=iz(a+2z). \end{aligned}$$

Hence the number of roots of $s-iq$ inside ${\mathbb {D}}$ is 1 if $|a|\ge 2$ and 2 if $0\ne |a|< 2$, while the number of roots of $s+iq$ inside ${\mathbb {D}}$ is 1 if $|a|> 2$ and 0 if $0\ne |a|\le 2$. Thus $T_\omega ^*$ admits a selfadjoint extension if and only if $|a|>2$.

6 Comparison with the Unbounded Toeplitz Operator Defined by Sarason

The Smirnov class $N^+$ consists of quotients $\frac{b}{a}$ with a and b $H^\infty$-functions such that the denominator a is an outer function. The function $\varphi = \frac{b}{a}\in N^+$ is said to be in canonical form if $a(0) > 0$ and $\vert a\vert ^2 + \vert b\vert ^2 = 1$ on ${\mathbb {T}}$. By Proposition 3.1 of [11], every function $\varphi \in N^+$ can be uniquely written in canonical form.

In [11], Sarason investigated an unbounded Toeplitz operator $T_{\varphi }^\text {Sa}$ with symbol $\varphi$ in $N^+$, which is defined by

$$\begin{aligned} {\mathrm {Dom}}(T_\varphi ^\text {Sa}) = \{ f\in H^2 : \varphi f \in H^2 \},\quad T_\varphi ^\text {Sa} f = \varphi f\quad (f\in {\mathrm {Dom}}(T_\varphi ^{Sa})). \end{aligned}$$

More generally, $T_\varphi ^\text {Sa}$ can be defined in this way for any holomorphic function $\varphi$ on ${\mathbb {D}}$, but for $T_\varphi ^\text {Sa}$ to be densely defined, $\varphi$ must be in $N^+$ [11, Lemma 5.2].

Let $\varphi = \frac{b}{a}\in N^+$ be the canonical representation of $\varphi$. Then it is shown in Proposition 5.3 of [11] that ${\mathrm {Dom}}(T_\varphi ^\text {Sa}) = aH^2$. The adjoint of the operator $T_\varphi ^\text {Sa}$ is motivated by the action of the conjugate transpose of the matrix representation of $T_\varphi ^\text {Sa}$, which is lower triangular. The domain of the adjoint operator is shown to contain the space $H(\overline{{\mathbb {D}}})$ of functions that are analytic on some neighborhood of the closed unit disc $\overline{{\mathbb {D}}}$, and the adjoint is equal to the closure of the operator on $H(\overline{{\mathbb {D}}})$; see [11, Lemmas 6.1 and 6.4].

Let $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$ with $s,q\in {\mathcal P}$ co-prime. Set $n={\mathrm {deg} }(s)$ and $m = {\mathrm {deg} }(q)$. Assume $\omega$ is proper, i.e., $n\le m$. Then $\omega ^* (z) = z^{m-n}s^\sharp /q^\sharp \in {\mathrm {Rat}}({\mathbb {T}})$. Since $q^\sharp$ has zeroes only on ${\mathbb {T}}$ it is outer and thus $\omega ^* \in N^+$. While in general $T_\omega$ and $T_\omega ^\text {Sa}$ are different, the following proposition shows that $T_\omega$ coincides with $T_{\omega ^*}^\text {Sa}$, and hence $T_\omega =T_\omega ^{**}=T_{\omega ^*}^\text {Sa}$. Without the properness assumption, $\omega ^*$ is not in $N^+$, because $\omega ^*$ has a pole at 0, and hence $T_{\omega ^*}^\text {Sa}$ is not defined.

Proposition 6.1

Let ${\widetilde{\omega }}={\widetilde{s}}/{\widetilde{q}}\in {\mathrm {Rat}}({\mathbb {T}})$ with ${\widetilde{s}},{\widetilde{q}}\in {\mathcal P}$ co-prime. Then ${\mathrm {Dom}}(T_{\widetilde{\omega }}^\text {Sa})={\widetilde{q}}H^2$ and $T_{\widetilde{\omega }}^\text {Sa}=T_{{\widetilde{\omega }}}|_{{\widetilde{q}}H^2}$. In particular, if $\omega \in {\mathrm {Rat}}({\mathbb {T}})$ is proper, then $T_{\omega }^*=T_{\omega ^*}^\text {Sa}$.

Proof

We first show ${\mathrm {Dom}}(T_{{\widetilde{\omega }}}^\text {Sa})={\widetilde{q}}H^2$. Let ${\widetilde{\omega }}=a/b$ be the canonical form of ${\widetilde{\omega }}$. As noted above, ${\mathrm {Dom}}(T_{{\widetilde{\omega }}}^\text {Sa})=a H^2$. By the Fejér-Riesz Theorem there is a polynomial r such that on ${\mathbb {T}}$ we have $\vert r\vert ^2 = \vert {\widetilde{s}}\vert ^2 + \vert {\widetilde{q}}\vert ^2$, r has no roots in ${\mathbb {D}}$ and $\text {arg}(r(0))=\text {arg}({\widetilde{q}}(0))$. The latter is possible since ${\widetilde{q}}(0)\ne 0$ and implies ${\widetilde{q}}(0)/r(0)>0$. Note that r also has no roots on ${\mathbb {T}}$, since ${\widetilde{s}}$ and ${\widetilde{q}}$ are co-prime. It follows that ${\widetilde{q}}/r$ and ${\widetilde{s}}/r$ are both $H^\infty$-functions, ${\widetilde{q}}/r$ is outer and ${\widetilde{q}}(0)/r(0)>0$. Hence $a={\widetilde{q}}/r$ and $b={\widetilde{s}}/r$, by the uniqueness of the canonical form. Also, since all the roots of r are outside ${\mathbb {T}}$, $r^{-1} H^2=H^2$, so that $a H^2= {\widetilde{q}} H^2$.

Now let $f\in {\mathrm {Dom}}(T_{\widetilde{\omega }}^\text {Sa})$, say $f= {\widetilde{q}}h$ with $h\in H^2$. Then $T_{\widetilde{\omega }}^\text {Sa} f={\widetilde{\omega }} f= {\widetilde{s}} h$. On the other hand, the fact that ${\widetilde{\omega }} f= {\widetilde{s}} h$ and ${\widetilde{s}} h\in H^2$ shows $T_{\widetilde{\omega }} f ={\mathbb {P}}{\widetilde{s}} h={\widetilde{s}} h$. Hence $T_{\widetilde{\omega }}^\text {Sa}=T_{{\widetilde{\omega }}}|_{{\widetilde{q}}H^2}$. $\square$

Next we employ some of the ideas from [11] to derive the following result. Recall that for a Hilbert space operator $T:{\mathrm {Dom}}(T)\rightarrow {\mathcal {H}}$ a linear submanifold ${\mathcal {D}}\subset {\mathrm {Dom}}(T)$ is called a core in case the graph $G(T|_{{\mathcal {D}}})$ of $T|_{{\mathcal {D}}}$ is dense in the graph G(T) of T; cf., page 166 in [7].

Theorem 6.2

Let $\omega \in {\mathrm {Rat}}({\mathbb {T}})$. Then $H(\overline{{\mathbb {D}}})$ is contained in ${\mathrm {Dom}}(T_\omega )$. If $\omega$ is proper, then $H(\overline{{\mathbb {D}}})$ is a core of $T_\omega$.

Proof of $H(\overline{{\mathbb {D}}})\subset {\mathrm {Dom}}(T_\omega )$. Write $\omega =\frac{s}{q}\in {\mathrm {Rat}}_0({\mathbb {T}})$ with $s,q\in {\mathcal P}$ coprime. Let $f\in H(\overline{{\mathbb {D}}})$. Then there exists a $R>1$ such that f is still analytic on an open neighborhood of the closed disc with radius R. Set ${\widetilde{f}}(z)=f(Rz)$, ${\widetilde{q}}(z)=q(Rz)$ and ${\widetilde{s}}(z)=s(Rz)$. Then ${\widetilde{f}}\in H^2$ and ${\widetilde{q}}$ is a polynomial with no roots on ${\mathbb {T}}$ and $\deg (q)=\deg ({\widetilde{q}})$. By Theorem 3.1 in [4], $H^2={\widetilde{q}} H^2+ {\mathcal P}_{\deg (q)-1}$. Thus ${\widetilde{s}}{\widetilde{f}}={\widetilde{q}}{\widetilde{h}}+{\widetilde{r}}$ for some ${\widetilde{h}}\in H^2$ and ${\widetilde{r}}\in {\mathcal P}$ with $\deg ({\widetilde{r}})<\deg (q)$. Now set $r(z)={\widetilde{r}}(z/R)$ and $h(z)={\widetilde{h}}(z/R)$. Then $r\in {\mathcal P}$ with $\deg (r)=\deg ({\widetilde{r}})<\deg (q)$ and $h\in H^2$, even $h\in H(\overline{{\mathbb {D}}})$. Also, we have $s f =qh +r$. Thus $f\in {\mathrm {Dom}}(T_\omega )$. $\square$

Before proving the second claim of Theorem 6.2 it is useful to consider the value of $T_\omega$ when applied to the evaluation functional or reproducing kernel element $k_\lambda (z) = (1 - \overline{\lambda }z)^{-1}$, where $\lambda \in {\mathbb {D}}$. Note that $k_\lambda \in H(\overline{{\mathbb {D}}})$, hence $k_\lambda \in H^2$, and $k_\lambda$ has the reproducing kernel property for $H^2$:

$$\begin{aligned} \text {span}\{k_\lambda :\lambda \in {\mathbb {D}}\} \text{ dense } \text{ in } H^2 \quad \text{ and }\quad \langle h,k_\lambda \rangle =h(\lambda ) \quad (h\in H^2,\ \lambda \in {\mathbb {D}}). \end{aligned}$$

See [8] for a recent account of the theory of reproducing kernel Hilbert spaces and further references.

Lemma 6.3

Let $\omega =s/q\in {\mathrm {Rat}}({\mathbb {T}})$, with $s,q\in {\mathcal P}$ co-prime, be proper. Then

$$\begin{aligned} T_\omega k_\lambda = \overline{\omega ^*(\lambda )} k_\lambda \quad (\lambda \in {\mathbb {D}}). \end{aligned}$$

Proof

Suppose $g = T_\omega k_\lambda$ then $s(z)(1-\overline{\lambda }z)^{-1} = q(z)g(z) + r(z)$, where $r\in {\mathcal P}_{m - 1}$. Here $m=\deg (q)$. Hence $(1-\overline{\lambda }z) g=(s+(1-\overline{\lambda }z)r)/q$ is in ${\mathrm {Rat}}({\mathbb {T}})$ as well as in $H^2$. This can only occur if $(1-\overline{\lambda }z) g$ is a polynomial, i.e., $g= k_\lambda {\widetilde{r}}$ for some ${\widetilde{r}}\in {\mathcal P}$. Thus $s+(1-\overline{\lambda }z)r=q {\widetilde{r}}$. Since $\omega$ is proper, the degree of the left hand side is at most m. But then ${\widetilde{r}}$ is constant, say with value ${\widetilde{c}}$. This shows $T_\omega k_\lambda ={\widetilde{c}} k_\lambda$.

To determine ${\widetilde{c}}$ we evaluate the identity $s+(1-\overline{\lambda }z)r=q {\widetilde{c}}$ at $1/\overline{\lambda }$. This gives $s(1/\overline{\lambda })= q(1/\overline{\lambda }){\widetilde{c}}$. Note that

$$\begin{aligned} s^\sharp (\lambda )=\lambda ^n \overline{s(1/\overline{\lambda })} \quad \text{ and }\quad q^\sharp (\lambda )=\lambda ^m \overline{q(1/\overline{\lambda })}, \end{aligned}$$

where $n=\deg (s)$. Hence

$$\begin{aligned} s(1/\overline{\lambda })= \overline{\lambda }^{-n} \overline{s^\sharp (\lambda )} \quad \text{ and }\quad q(1/\overline{\lambda })= \overline{\lambda }^{-m} \overline{q^\sharp (\lambda )}. \end{aligned}$$

This gives

$$\begin{aligned} {\widetilde{c}} =\frac{\overline{\lambda }^{-n} \overline{s^\sharp (\lambda )}}{\overline{\lambda }^{-m} \overline{q^\sharp (\lambda )}} =\overline{\left( \frac{\lambda ^{m-n}s^\sharp (\lambda )}{q^\sharp (\lambda )} \right) }=\overline{\omega ^*(\lambda )}. \end{aligned}$$

$\square$

Proof of Theorem 6.2

It remains to prove that $H(\overline{{\mathbb {D}}})$ is a core for $T_\omega$ in case $\omega$ is proper. So, assume $\omega$ is proper. We need to show that the graph of $T_\omega |_{H(\overline{{\mathbb {D}}})}$ is dense in the graph of $T_\omega$. In other words, let $f,g\in H^2$ with (f, g) perpendicular to $G(T_\omega |_{H(\overline{{\mathbb {D}}})})$, then we need to show (f, g) is perpendicular to $G(T_\omega )$. Since $k_\lambda \in H(\overline{{\mathbb {D}}})$, for $\lambda \in {\mathbb {D}}$, we have

$$\begin{aligned} 0&= \langle (f,g),(k_\lambda , T_\omega k_\lambda ) \rangle =\langle f,k_\lambda \rangle + \langle g,\overline{\omega ^*(\lambda )} k_\lambda \rangle =f(\lambda )+ \omega ^*(\lambda ) g(\lambda )\ (\lambda \in {\mathbb {D}}). \end{aligned}$$

Hence $\omega ^* g=-f$. In particular, $\omega ^* g\in H^2$. Thus $g\in {\mathrm {Dom}}(T_{\omega ^*}^\text {Sa})={\mathrm {Dom}}(T_\omega ^*)$ and $T_{\omega }^* g =-f$, by Proposition 6.1. For any $h\in {\mathrm {Dom}}(T_\omega )$ we have

$$\begin{aligned}&\langle (f,g),(h,T_\omega h) \rangle =\langle (-T_{\omega }^* g,g),(h,T_\omega h) \rangle =-\langle T_\omega ^* g,h \rangle +\langle g,T_\omega h \rangle =0.\\ \end{aligned}$$

$\square$

In Section 8 of [11], Sarason introduced the class of closed, densely defined operators T on $H^2$ which satisfy

• (1) $T_z\, {\mathrm {Dom}}(T) \subset {\mathrm {Dom}}(T)$;

• (2) $T_z^* T T_z =T$;

• (3) $f\in {\mathrm {Dom}}(T)$, $f(0)=0 \Rightarrow T_z^* f \in {\mathrm {Dom}}(T)$.

This class of operators was further studied by Rosenfeld in [9, 10] in which he referred to such operators as Sarason–Toeplitz operators. The operators $T^\text {Sa}_\varphi$, for $\varphi \in N^+$, are Sarason–Toeplitz operators, and the class of operators is closed under taking adjoints, by Proposition 2.1 in [10]. Hence, by Proposition 6.1, $T_\omega$ is a Sarason–Toeplitz operator whenever $\omega \in {\mathrm {Rat}}({\mathbb {T}})$ is proper. We show that in fact $T_\omega$ is a Sarason–Toeplitz operator for any $\omega \in {\mathrm {Rat}}$.

Proposition 6.4

Let $\omega \in {\mathrm {Rat}}$. Then $T_\omega$ on $H^2$ is a Sarason–Toeplitz operator.

Proof

First consider $\omega \in {\mathrm {Rat}}({\mathbb {T}})$. That $T_\omega$ satisfies (1) and (2) was proved in [4, Lemma 2.3]. We claim that $T_{z}^*\, {\mathrm {Dom}}(T_\omega )\subset {\mathrm {Dom}}(T_\omega )$. Write $\omega =s/q$ with $s,q\in {\mathcal P}$ co-prime. Then ${\mathrm {Dom}}(T_\omega )=q H^2 + {\mathcal P}_{\deg (q)-1}$. Let $f= q h +r\in {\mathrm {Dom}}(T_\omega )$ with $h\in H^2$ and $r\in {\mathcal P}$, $\deg (r)<\deg (q)$. Then $T_z^* f= q T_z^* h + h(0) T_z^* q + T_z^* r$, which is in $q H^2+{\mathcal P}_{\deg (q)-1}={\mathrm {Dom}}(T_\omega )$. Hence $T_\omega$ is a Sarason–Toeplitz operator in case $\omega \in {\mathrm {Rat}}({\mathbb {T}})$.

Now take $\omega \in {\mathrm {Rat}}$ arbitrarily. By Lemma 5.1 in [4], see also Sect. 4 above, $\omega = \omega _- z^\kappa \omega _0 \omega _+$ with $\kappa \in {\mathbb {Z}}$, and $\omega _-$, $\omega _0$ and $\omega _+$ in ${\mathrm {Rat}}$ with zeroes and poles only inside, on or outside ${\mathbb {T}}$, respectively. In particular, $\omega _0\in {\mathrm {Rat}}({\mathbb {T}})$, $\omega _-$ and $\omega _-^{-1}$ are both anti-analytic, and $\omega _+$ and $\omega _+^{-1}$ are both analytic. Also, $T_\omega =T_{\omega _-} T_{z^\kappa \omega _0} T_{\omega _+}$. Note that $z^\kappa \omega _0\in {\mathrm {Rat}}({\mathbb {T}})$ in case $\kappa \ge 0$ and $T_{z^\kappa \omega _0}=T_{z^\kappa }T_{\omega _0}$ in case $\kappa <0$ (by [4, Lemma 5.3]). In both cases it now easily follows that $T_{z^\kappa \omega _0}$ is a Sarason–Toeplitz operator. The claim for $T_\omega$ follows since $T_{\omega _+}^{\pm 1} T_z=T_z T_{\omega _+}^{\pm 1}$ and $T_{\omega _-}^{\pm 1} T_z^*=T_z^* T_{\omega _-}^{\pm 1}$. $\square$

In fact, by the same arguments one can show that $T_\omega$ on $H^p$, $1<p<\infty$, satisfies (1)–(3) in case $T_z^*$ is replaced by $T_{z^{-1}}$.