On kernels of Toeplitz operators

Abstract

We apply the theory of de Branges–Rovnyak spaces to describe kernels of some Toeplitz operators on the classical Hardy space $H^2$. In particular, we discuss the kernels of the operators $T_{{\bar{f}}/ f}$ and $T_{{\bar{I}}{\bar{f}}/ f}$, where f is an outer function in $H^2$ and I is inner such that $I(0)=0$. We also obtain a result on the structure of de Branges–Rovnyak spaces generated by nonextreme functions.

1 Introduction

Let $H^2$ denote the standard Hardy space on the unit disk $\mathbb {D}$. For $\varphi \in L^{\infty }(\partial \mathbb {D})$ the Toeplitz operator on $H^2$ is given by $T_{\varphi }f=P_{+}(\varphi f)$, where $P_{+}$ is the orthogonal projection of $L^2(\partial \mathbb {D})$ onto $H^2$. We will denote by $\mathcal {M}(\varphi )$ the range of $T_{\varphi }$ equipped with the range norm, that is, the norm that makes the operator $T_{\varphi }$ a coisometry of $H^2$ onto $\mathcal {M}(\varphi )$. For a nonconstant function b in the unit ball of $H^\infty$ the de Branges–Rovnyak space $\mathcal {H}(b)$ is the image of $H^2$ under the operator $(1-T_bT_{\bar{b}})^{1/2}$ with the corresponding range norm. The norm and the inner product in $\mathcal {H}(b)$ will be denoted by $\Vert \cdot \Vert _b$ and $\langle \cdot ,\cdot \rangle _b$. The space $\mathcal {H}(b)$ is a Hilbert space with the reproducing kernel

$$\begin{aligned} k_w^b(z)=\frac{1-\overline{b(w)}b(z)}{1-{\overline{w}}z}\qquad (z,w\in {\mathbb {D}}). \end{aligned}$$

In the case when b is an inner function the space $\mathcal {H}(b)$ is the well-known model space $K_b=H^2\ominus bH^2$.

If the function b fails to be an extreme point of the unit ball in $H^{\infty }$, that is, when $\log (1-|b|)\in L^1(\partial \mathbb D)$, we will say simply that b is nonextreme. In this case one can define an outer function a whose modulus on $\partial \mathbb {D}$ equals $\left( 1-|b|^2\right) ^{1/2}$. Then we say that the functions b and a form a pair (b, a). By the Herglotz representation theorem there exists a positive measure $\mu$ on $\partial \mathbb {D}$ such that

$$\begin{aligned} \frac{1+b(z)}{1-b(z)}=\int _{\partial \mathbb {D}}\frac{1+ e^{-i\theta }z}{1- e^{-i\theta }z}\,d\mu (e^{i\theta })+i{{\,\mathrm{Im}\,}}\frac{1+b(0)}{1-b(0)},\quad z\in \mathbb {D}. \end{aligned}$$

(1)

Moreover the function $\left| \frac{a}{1-b}\right| ^2$ is the Radon-Nikodym derivative of the absolutely continuous component of $\mu$ with respect to the normalized Lebesgue measure. If the measure $\mu$ is absolutely continuous the pair (b, a) is called special.

Recall that a function $f\in H^1$ is called rigid if and only if no other functions in $H^1$, except for positive scalar multiples of f have the same argument as f a.e. on $\partial \mathbb {D}$.

If (b, a) is a pair, then $\mathcal {M}(a)$ is contained contractively in $\mathcal {H}(b)$. If a pair (b, a) is special and $f=\frac{a}{1-b}$, then $\mathcal {M}(a)$ is dense in $\mathcal {H}(b)$ if and only if $f^2$ is rigid ( [20]). Spaces $\mathcal {H}(b)$ for nonextreme b have been studied in [2, 3, 5, 15, 16, 22], and [23].

The kernels of Toeplitz operators have been studied since the late 80’s. We mention that two recent survey articles [1, 8] and the book [4] contain a number of results on this topic.

The Hayashi theorem [12] (see also [21]) states that the kernel of a Toeplitz operator $T_{\varphi }$ is a subspace of $H^2$ of the form $\ker T_{\varphi }= fK_I$, where $K_I= H^2\ominus IH^2$ is the model space corresponding to the inner function I such that $I(0)=0$ and f is an outer function of unit $H^2$ norm that acts as an isometric multiplier from $K_I$ onto $f K_{I}$. Moreover, f can be expressed as $f=\frac{a}{1-Ib_0}$, where $(b_0,a)$ is a special pair and $\bigl (\frac{a}{1-b_0}\bigr )^2$ is a rigid function in $H^1$. Then we also have $\ker T_{\frac{\bar{I}\bar{f}}{f}}=fK_{I}$. In the recent paper [6] the authors considered the Toeplitz operator $T_{\frac{\bar{g}}{g}}$ where $g\in H^{\infty }$ is outer. Among other results, they described all outer functions g such that $\ker T_{\frac{\bar{g}}{g}}=K_{I}$. In Sect. 2 we describe all such functions g for which $\ker T_{\frac{\bar{g}}{g}}=f K_{I}$.

If (b, a) is a special pair, $f=\frac{a}{1-b}$ and $b=Ib_0$, where I as above, then $f K_{I}\subset \ker T_{\frac{\bar{I}\bar{f}}{f}}$. In the next two sections we study the space $\ker T_{\frac{\bar{I}\bar{f}}{f}}\ominus f K_{I}$ and show that it is isometrically isomorphic to the orthogonal complement of $\mathcal {M}(a)$ in the de Branges–Rovnyak space $\mathcal {H}(b_0)$. We also give an example of a function f for which the space $\ker T_{\frac{\bar{I}\bar{f}}{f}}\ominus f K_{I}$ is one dimensional. In the last section we discuss the orthogonal complement of $\mathcal {M}(a)$ in ${\mathcal {H}}(b)$ and get a generalization of results obtained in [15] and [5] for the case when pairs are rational.

2 The kernel of $T_{\frac{{\bar{g}}}{g}}$

It is known that if g is an outer function in $H^2$, then the kernel of $T_{\frac{\bar{g}}{g}}$ is trivial if and only if $g^2$ is rigid (see e.g. [18]).

The finite dimensional kernels of Toeplitz operators were described by Nakazi [17]. Nakazi’s theorem says that $\dim \ker T_{\varphi }=n$ if and only if there exists an outer function $f\in H^2$ such that $f^2$ is rigid and $\ker T_{\varphi }=\{fp:\, p\in \mathcal {P}_{n-1}\},$ where $\mathcal {P}_{n-1}$ denotes the set of all polynomials of degree at most $n-1$.

Consider the following example.

Example

For $\alpha >-\frac{1}{2}$ set $g(z)=(1-z)^\alpha$, $z\in \mathbb {D}$. Then the kernel of $T_{\frac{\bar{g}}{g}}$ is trivial for $\alpha \in (-\frac{1}{2},\frac{1}{2}]$ and dimension of the kernel of $T_{\frac{{\overline{g}}}{g}}$ is n for $\alpha \in (n-\frac{1}{2},n+\frac{1}{2}], n=1,2,\ldots$, and

$$\begin{aligned} \ker T_{\frac{\overline{(1-z)^\alpha }}{(1-z)^\alpha }}=(1-z)^{\alpha -n}K_{z^n}. \end{aligned}$$

In the general case the kernels of Toeplitz operators are characterized by Hayashi’s theorem. To state this theorem we need some notation. We note that an outer function f having unit norm in $H^2$ $(\Vert f\Vert _2=1)$ can be written as

$$\begin{aligned} f=\frac{a}{1-b}, \end{aligned}$$

where a is an outer function, b is a function from the unit ball of $H^{\infty }$ such that $|a|^2+|b|^2=1$ a.e. on $\partial \mathbb {D}$. Following Sarason [p. 156] Sarason3 we call (b, a) the pair associated with f. Note also that b is a nonextreme point of the closed unit ball of $H^{\infty }$ and is given by

$$\begin{aligned} \frac{1+b(z)}{1-b(z)}=\frac{1}{2\pi }\int _{0}^{2\pi }\frac{1+ e^{-i\theta }z}{1- e^{-i\theta }z}|f(e^{i\theta })|^2d\theta ,\quad z\in \mathbb {D}. \end{aligned}$$

(2)

Let S denote the unilateral shift operator on $H^2$, i.e., $S=T_z$. A closed subspace M of $H^2$ is said to be nearly $S^*$-invariant if for every $f\in M$ vanishing at 0, we also have $S^*f\in M$. It is known that the kernels of Toeplitz operators are nearly $S^*$-invariant.

Nearly $S^*$-invariant spaces are characterized by Hitt’s theorem [14].

Hitt’s Theorem

The closed subspace M of $H^2$ is nearly $S^*$-invariant if and only if there exists a function f of unit norm and a model space $K_I=H^2\ominus IH^2$ such that $M= T_fK_I$, where I is an inner function vanishing at the origin, and $T_f$ acts isometrically on $K_I$.

It has been proved by D. Sarason [18] that $T_f$ acts isometrically on $K_I$ if and only if I divides b (the first function in the pair associated with f). Consequently, the function f in Hitt’s theorem can be written as

$$\begin{aligned} f=\frac{a}{1-Ib_0}. \end{aligned}$$

The function $\frac{1+b_0(z)}{1-b_0(z)}$ has a positive real part and is the Herglotz integral of a positive measure on $\partial \mathbb {D}$ up to an additive imaginary constant,

$$\begin{aligned} \frac{1+b_0(z)}{1-b_0(z)}=\int _{\partial \mathbb {D}}\frac{1+ e^{-i\theta }z}{1- e^{-i\theta }z}\,d\mu (e^{i\theta })+ic. \end{aligned}$$

(3)

Clearly $b_0$ is also a nonextreme point of the closed unit ball of $H^{\infty }$ and $|a|^2+|b_0|^2=1$ a.e. on $\partial \mathbb {D}$. We remark that in view of (2) the pair (b, a) associated with an outer function $f\in H^2$ is special, while the pair $(b_0,a)$ need not to be special. We also put

$$\begin{aligned} f_0=\frac{a}{1-b_0} \end{aligned}$$

and note that $f_0\in H^2$ (see, e.g. [4, Theorem 23.1]).

Under the above notations Hayashi’s theorem reads as follows:

Hayashi’s Theorem

The nearly $S^*$-invariant space $M=T_fK_I$ is the kernel of a Toeplitz operator if and only if the pair $(b_0,a)$ is special and $f_0^2$ is a rigid function.

Moreover, it follows from Sarason’s proof of Hayashi’s theorem that if $M=T_fK_I$ is the kernel of a Toeplitz operator then it is the kernel of $T_{\frac{\bar{I}{\bar{f}}}{f}}$.

Recently E. Fricain, A. Hartmann and W. T. Ross [6] considered the Toeplitz operators $T_{\frac{\bar{g}}{g}}$ where $g\in H^{\infty }$ is outer. If $\ker T_{\frac{\bar{g}}{g}}$ is non-trivial, then by Hayashi’s theorem there exist the outer function f and the inner function I, $I(0)=0$, such that

$$\begin{aligned} \ker T_{\frac{\bar{g}}{g}}=fK_{I}. \end{aligned}$$

In the above mentioned paper [6] the authors described all outer functions $g\in H^{\infty }$ for which

$$\begin{aligned} \ker T_{\frac{\bar{g}}{g}}=K_I, \end{aligned}$$

where I is an inner function not necessarily satisfying $I(0)=0$.

We prove the following

Theorem 1

Assume that $g\in H^2$ is outer and $M= T_fK_{I}$ is the nearly $S^*$-invariant space, where I is an inner function such that $I(0)=0$, $(b_0I,a)$ is the pair associated with the outer function f, $(b_0,a)$ is special, and $f_0^2$ is rigid. Then $\ker T_{\frac{{\bar{g}}}{g}}=M$ if and only if

$$\begin{aligned} g= i\frac{I_1+I_2}{I_1-I_2}(1+ I)f, \end{aligned}$$

where $I_1$ and $I_2$ are inner and $I_1-I_2$ is outer.

Recall that the Smirnov class ${\mathcal {N}}^{+}$ consists of those holomorphic functions in $\mathbb {D}$ that are quotients of functions in $H^{\infty }$ in which the denominators are outer functions.

In the proof of Theorem 1, similarly to [6], we use the following result due to H. Helson [13].

Helson’s Theorem

The functions $f\in {\mathcal {N}}^{+}$ that are real almost everywhere on $\partial \mathbb {D}$ can be written as

$$\begin{aligned} f=i\,\frac{I_1+I_2}{I_1-I_2}, \end{aligned}$$

where $I_1$ and $I_2$ are inner and $I_1-I_2$ is outer.

We also apply a description of kernels in terms of $S^*$-invariant subspaces $K_I^p(|f|^p)$ of weighted Hardy spaces (in the case when $p=2$) considered by A. Hartmann and K. Seip in their paper [10] (see also [7]). For an outer function f in $H^2$ the weighted Hardy space is defined as follows

$$\begin{aligned} H^2(|f|^2)=\{g\in {\mathcal {N}}^{+}:\, \Vert g\Vert ^2_{2,f}=\frac{1}{2\pi }\int _0^{2\pi }|g(e^{it})|^2|f(e^{it})|^2dt<\infty \} \end{aligned}$$

and, for an inner function I, $K^2_I(|f|^2)= K_{I}(|f|^2)$ is given by

$$\begin{aligned} K_{I}(|f|^2)=\{g=I{\overline{\psi }}\in H^2(|f|^2):\, \psi \in H_0^2(|f|^2)\}, \end{aligned}$$

where $H_0^2(|f|^2)=zH^2(|f|^2)$.

Then $K_{I}(|f|^2)$ is $S^*$-invariant and $fK_{I}(|f|^2)=\ker T_{\frac{{\bar{I}} {\bar{f}}}{f}}$ (see [10]).

Proof of Theorem 1

Assume that $\ker T_{\frac{{\bar{g}}}{g}}=fK_{I}$. Then

$$\begin{aligned} fK_I=\ker T_{\frac{{\bar{I}} {\bar{f}}}{f}}=\ker T_{\frac{{\bar{g}}}{g}}. \end{aligned}$$

Since $f\in \ker T_{\frac{{\bar{I}} {\bar{f}}}{f}}$, the last equalities imply that

$$\begin{aligned} \frac{{\overline{g}}f}{g}= {\overline{I}}_0{\bar{h}}, \end{aligned}$$

where $I_0$ is an inner function such that $I_0(0)=0$, and $h\in H^2$ is outer. This means that $|f(z)|=|h(z)|$ a.e. on $|z|=1$ and consequently $h(z)=cf(z)$, where c is a unimodular constant. Replacing $cI_0$ by $I_0$, we get

$$\begin{aligned} \frac{\overline{g}}{g}={\overline{I}}_0\frac{{\overline{f}}}{f}. \end{aligned}$$

(4)

It then follows

$$\begin{aligned} fK_{I}=\ker T_{\frac{{\bar{g}}}{g}}= \ker T_{\frac{ {\overline{I}}_0{\overline{f}}}{f}}= fK_{I_0}(|f|^2), \end{aligned}$$

which implies $I=I_0$ up to a unimodular constant. Indeed, these equalities imply that an analytic function h can be written in the form $h = f I_0{\overline{\psi }}_0$, where $\psi _0\in H^2_0(|f|^2)$, if and only if $h= f I{\overline{\psi }}$, where $\psi \in H^2_0$. Since $|\psi _0|= |\psi |$ a.e. on $|z|=1$ and $\psi _0\in {\mathcal {N}}^{+}$, we see that also $\psi _0\in H_0^2$. Hence $K_{I}=K_{I_0}$.

Consequently, equality (4) can be written as

$$\begin{aligned} \frac{{\bar{g}}}{g}=\frac{\overline{f(1+ I)}}{f(1+ I)} \quad \text {a.e. on } \partial \mathbb {D}, \end{aligned}$$

which means that the function $\frac{g}{f(1+ I)}$ is real a.e. on $\partial \mathbb {D}$. Since this function is in the Smirnov class ${\mathcal {N}}^{+}$, our claim follows from Helson’s theorem. To prove the other implication it is enough to observe that if

$$\begin{aligned} g=i\,\frac{I_1+I_2}{I_1-I_2}(1+ I)f, \end{aligned}$$

then

$$\begin{aligned} \frac{{\bar{g}}}{g}=\frac{{\overline{I}}{{\bar{f}}}}{f}. \end{aligned}$$

$\square$

3 The complement of $fK_I$ in $\ker T_{\frac{\bar{I}\bar{f}}{f}}$

It was noticed in [4, Corollary 30.21] that if f is an outer function of the unit norm, (b, a) is the pair associated with f, and I is an inner function vanishing at the origin that divides b, then

$$\begin{aligned} fK_{I}\subset \ker T_{\frac{\bar{I}{\bar{f}}}{f}} \end{aligned}$$

and, according to Hayashi’s theorem, the equality holds if and only if the pair $(b_0,a)$ is special and $f_0^2$ is rigid.

Recall that $\mathcal {M}(a)$ is dense in $\mathcal {H}(b_0)$ if and only if the pair $(b_0,a)$ is special and $f_0^2$ is a rigid function.

Theorem 2

Assume that $(Ib_0, a)$, where I is inner, and $I(0)=0$, is the pair associated with an outer function f. If the pair $(b_0,a)$ is not special or the function $f_0^2$ is not rigid, then for a positive integer k,

$$\begin{aligned} \dim \bigl (\ker T_{\frac{{\bar{I}} {\bar{f}}}{f}}\ominus fK_{I}\bigr )=k \end{aligned}$$

if and only if the codimension of $\overline{\mathcal {M}(a)}$ in the de Branges–Rovnyak space $\mathcal {H}(b_0)$ is k.

In the proof of this theorem we use some ideas from Sarason’s proof of Hayashi’s theorem. If a positive measure $\mu$ on the unit circle $\partial \mathbb {D}$ is as in (1) and $H^2(\mu )$ is the closure of the polynomials in $L^2(\mu )$, then an operator $V_b$ given by

$$\begin{aligned} (V_bq)(z)=(1-b(z))\int _{\partial \mathbb {D}}\frac{q(e^{i\theta })}{1-e^{-i\theta }z}\,d\mu (e^{i\theta }) \end{aligned}$$

(5)

is an isometry of $H^2(\mu )$ onto $\mathcal {H}(b)$. Furthermore, if (b, a) is a pair and $f=\frac{a}{1-b}$, then the operator $T_{1-b}T_{\bar{f}}$ is an isometry of $H^2$ into $\mathcal {H}(b)$. Its range is all of $\mathcal {H}(b)$ if and only if the pair (b, a) is special ( [21, III-6,7] and [4, Theorem 24.26]). We note that in the last case $d\mu (e^{i\theta })=\frac{1}{2\pi }|f(e^{i\theta })|^2d\theta$.

Proof of Theorem 2

Since the pair (b, a) is special, the operator $T_{1-b}T_{{\bar{f}}}$ is an isometry of $H^2$ onto $\mathcal {H}(b)$. Moreover, since I divides b, $T_{f}$ acts as an isometry on $K_{I}$ and $T_{1-b}T_{\bar{f}}(fK_I)=K_I$ ( [20]). Hence

$$\begin{aligned} \mathcal {H}(b)&=T_{1-b}T_{{\bar{f}}}(H^2)= T_{1-b}T_{\bar{f}}\bigl (\overline{T_{\frac{If}{{\bar{f}}}}(H^2)}\oplus \big (T_{\frac{If}{{\bar{f}}}}(H^2)\bigr )^\bot \bigr )\\&= \overline{I\mathcal {M}(a)}^b\oplus T_{1-b}T_{{\bar{f}}}(\ker T_{\frac{\bar{I}{\bar{f}}}{f}})\\ {}&= \overline{I\mathcal {M}(a)}^b\oplus T_{1-b}T_{{\bar{f}}} (fK_I) \oplus T_{1-b}T_{{\bar{f}}}(\ker T_{\frac{{\bar{I}}{\bar{f}}}{f}}\ominus fK_I)\\&= \overline{I\mathcal {M}(a)}^b\oplus K_I\oplus T_{1-b}T_{\bar{f}}(\ker T_{\frac{{\bar{I}}{\bar{f}}}{f}} \ominus fK_I), \end{aligned}$$

where $\overline{T_{\frac{If}{{\bar{f}}}}(H^2)}$ denotes the closure of $T_{\frac{If}{{\bar{f}}}}(H^2)$ in $H^2$ and $\overline{I\mathcal {M}(a)}^b$ denotes the closure of $I\mathcal {M}(a)$ in $\mathcal {H}(b)$. On the other hand,

$$\begin{aligned} \mathcal {H}(b) = \mathcal {H}(b_0I) =K_I\oplus I\mathcal {H}(b_0)= K_I\oplus I(\mathcal {H}(b_0)\ominus \overline{\mathcal {M}(a)}^{b_0})\oplus I\overline{\mathcal {M}(a)}^{b_0}. \end{aligned}$$

Since $T_{I}:\, \mathcal {H}(b_0)\rightarrow \mathcal {H}(Ib_0)$ is an isometry ( [20, Proposition 4]), $I\overline{(\mathcal {M}(a))}^{b_0}= \overline{I \mathcal {M}(a)}^{b}$. It then follows,

$$\begin{aligned} T_{1-b}T_{{\bar{f}}}(\ker T_{\frac{{\bar{I}}{\bar{f}}}{f}}\ominus fK_I)= I(\mathcal {H}(b_0)\ominus \overline{\mathcal {M}(a)}^{b_0}). \end{aligned}$$

(6)

$\square$

We remark that the orthogonal complement of $\mathcal {M}(a)$ in $\mathcal {H}(b)$ is discussed in Sect. 5.

4 The example

Let, as in the previous sections, f be an outer function in $H^2$ and let (b, a) be the pair associated with f. Let $b=Ib_0$, where I is an inner function such that $I(0)=0$ and $f_0=\frac{a}{1-b_0}$. Then $fK_I\subset \ker T_{\frac{\bar{I}\bar{f}}{f}}$ and equality holds if and only if the pair $(b_0,a)$ is special and $f_0^2$ is rigid. Moreover, if the pair $(b_0,a)$ is special and $f_0^2$ is rigid, then (b, a) is special and $f^2$ is rigid but the converse implication fails ( [18, p. 158]).

In [4, vol. 2, pp. 541–542] the authors constructed a function h in $\ker T_{\frac{\bar{I}{\bar{f}}}{f}}$ which is not in $fK_{I}$ under the assumption that $f^2$ is not rigid. Here we consider the function f such that $f^2$ is rigid, the pair $(b_0,a)$ is special but $f^2_0$ is not rigid, and describe the space $\ker T_{\frac{{\bar{I}}{\bar{f}}}{f}} \ominus fK_{I}$.

Our example is a slight modification of the one given in see also [4, vol. 2, p. 494]. The corresponding functions f and $f_0$ are defined by taking $a(z)=\frac{1}{2}( 1+z)$, $b_0(z)=\frac{1}{2}z(1-z)$, and $I(z)=zB(z)$, where B(z) is a Blaschke product with zero sequence $\{r_n\}_{n=1}^{\infty }$ lying in $(-1,0)$ and converging to $-1$. It has been proved in [19, pp. 491–492] (see also [4, vol. 2, pp. 494–496] that $f^2$ is rigid while $f^2_0$ is not. Notice that the pair $(b_0,a)$ is rational and the point $-1$ is the only zero of the function a. It then follows from [15, Theorem 4.1] (see also [5]) that $\mathcal {M}(a)$ is a closed subspace of $\mathcal {H}(b_0)$ and

$$\begin{aligned} \mathcal {H}(b_0)= \mathcal {M}(a)\oplus {\mathbb {C}}k_{-1}^{b_0}, \end{aligned}$$

where

$$\begin{aligned} k_{-1}^{b_0}(z)=\frac{1-\overline{b_0(-1)}b_0(z)}{1+z}=\frac{2-z}{2}. \end{aligned}$$

Thus we see that

$$\begin{aligned} \mathcal {H}(b_0)\ominus \mathcal {M}(a)= \mathbb {C}k_{-1}^{b_0}. \end{aligned}$$

Moreover (6) implies that

$$\begin{aligned} T_{1-b}T_{{\bar{f}}}(\ker T_{\frac{{\bar{I}}{\bar{f}}}{f}}\ominus fK_I)= \mathbb {C}I k_{-1}^{b_0}. \end{aligned}$$

Our aim is to prove that

$$\begin{aligned} \ker T_{\frac{{\bar{I}}{\bar{f}}}{f}}\ominus fK_I=\mathbb {C}g, \end{aligned}$$

(7)

where the function $g\in H^2$ is given by $g=fk_{-1}(I+1),$ with $k_{-1}(z)=(1+z)^{-1}$, $z\in \mathbb {D}$.

For $\lambda$ in $\mathbb {D}$ let $k_\lambda$ denote the kernel function in $H^2$ for the functional of evaluation at $\lambda$, $k_\lambda (z)=(1-{\bar{\lambda }}z)^{-1}$. In the proof of (7) we will apply the following

Lemma

[9, Lemma 2]

• (i) $P_{+}\left( |f|^2Ik_\lambda \right) =\dfrac{Ik_\lambda }{1-b}+\dfrac{\overline{b_0(\lambda )}k_\lambda }{1-\overline{b(\lambda )}}$.

• (ii) $P_{+}\left( |f|^2k_\lambda \right) =\dfrac{k_\lambda }{1-b}+\dfrac{\overline{b(\lambda )}k_\lambda }{1-\overline{b(\lambda )}}$.

Since $I(r_n)=0$, (i) and (ii) in the Lemma yield

$$\begin{aligned} T_{1-b}T_{{\bar{f}}}(f I k_{r_n})&= Ik_{r_n}(1-\overline{b_0(r_n)}b_0)+ \overline{b_0(r_n)}k_{r_n}, \\ T_{1-b}T_{{\bar{f}}}(f k_{r_n})&=k_{r_n}. \end{aligned}$$

Hence

$$\begin{aligned} T_{1-b}T_{{\bar{f}}}(f k_{r_n}(I-\overline{b_0(r_n)}))=Ik_{r_n}(1-\overline{b_0(r_n)}b_0)=Ik_{r_n}^{b_0}. \end{aligned}$$

(8)

It follows from [4, Theorem 21.1] that

$$\begin{aligned} \Vert k_{r_n}^{b_0}- k_{-1}^{b_0}\Vert _{b_0}\underset{n\rightarrow \infty }{\longrightarrow }0. \end{aligned}$$

Next, since $T_{I}:\, \mathcal {H}(b_0)\rightarrow \mathcal {H}(Ib_0)=\mathcal {H}(b)$ is an isometry and $T_{1-b}T_{{\bar{f}}}$ is an isometry of $H^2$ onto $\mathcal {H}(b)$, we see that $\{fk_{r_n}(I-\overline{b_0(r_n)}) \}_{n\in \mathbb {N}}$ is a bounded sequence in $H^2$. So it contains a subsequence that converges weakly, say, to a function $g\in H^2$. Without loss of generality, we may assume that the sequence $\{fk_{r_n}(I-\overline{b_0(r_n)})\}$ itself converges weakly to g. Then for any point $z\in \mathbb {D}$,

$$\begin{aligned} g(z)&= \langle g, k_z\rangle = \lim \limits _{n\rightarrow \infty }\langle fk_{r_n}(I-\overline{b_0(r_n)}), k_z\rangle \\&=\lim _{n\rightarrow \infty }\frac{f(z)(I(z)-\overline{b_0(r_n)})}{1-r_nz}= \frac{f(z)(I(z)+1)}{1+z}. \end{aligned}$$

Now observe that since

$$\begin{aligned} \Vert fk_{r_n}(I-\overline{b_0(r_n)})\Vert _{2}=\Vert k_{r_n}^{b_0}\Vert _{b_0}\quad \text {and}\quad \Vert fk_{-1}(I+1)\Vert _{2}=\Vert k_{-1}^{b_0}\Vert _{b_0}, \end{aligned}$$

$fk_{r_n}(I-\overline{b_0(r_n)})\rightarrow fk_{-1}(I+1)$ in $H^2$ strongly. Finally, passing to the limit in (8) gives

$$\begin{aligned} T_{1-b}T_{{\bar{f}}}(fk_{-1}(I+1))=Ik_{-1}(1+b_0)=Ik_{-1}^{b_0}, \end{aligned}$$

which proves (7).

Remark

One can check directly that the function $g=fk_{-1}(I+1)$ is in $\ker T_{\frac{{\bar{I}}{\bar{f}}}{f}}\ominus fK_I$.

Indeed, we have

$$\begin{aligned}&T_{\frac{{\bar{I}}\bar{f}}{f}}\left( fk_{-1}(I+1)\right) =P_{+}\left( \bar{f}\bar{I}\frac{I+1}{1+z}\right) \\&\quad =P_{+}\left( \bar{f}\frac{\bar{z}(\bar{I}+1)}{\bar{z}+1}\right) =P_{+}\left( \bar{z}\bar{f}\overline{k_{-1}}(\bar{I}+1)\right) =0. \end{aligned}$$

To see that the functions $(fk_{r_n}I-b_0(r_n)fk_{r_n})$ are orthogonal to $fK_{I}$ note that a function $h\in H^2$ is in $K_{I}$ if and only if $h= h - I P_{+}({\bar{I}}h)$. So, we have to check that for any $h\in H^2$,

$$\begin{aligned} \langle fk_{r_n}I-b_0(r_n)fk_{r_n}, f(h-I P_{+}({\bar{I}}h))\rangle =0. \end{aligned}$$

Since the functions $\{k_{\lambda }, \lambda \in {\mathbb {D}}\}$ are dense in $H^2$, it is enough to show that for any $\lambda \in {\mathbb {D}}$,

$$\begin{aligned}&\langle fk_{r_n}I-b_0(r_n)fk_{r_n}, f(k_{\lambda }-I P_{+}(\bar{I}k_{\lambda }))\rangle \\&\quad =\langle fk_{r_n}I-b_0(r_n)fk_{r_n}, f(k_{\lambda }-\overline{I(\lambda )}Ik_{\lambda })\rangle = 0. \end{aligned}$$

Finally, the last equality follows from

$$\begin{aligned} \langle fk_{r_n}I,fk_{\lambda }\rangle&= \frac{I(\lambda )k_{r_n}(\lambda )}{1- b(\lambda )}+ b_0(r_n)k_{r_n}(\lambda ),\\ \langle fk_{r_n}I, -\overline{I(\lambda )}fIk_{\lambda }\rangle&= -\frac{I(\lambda )k_{r_n}(\lambda )}{1-b(\lambda )},\\ \langle -b_0(r_n)fk_{r_n}, fk_{\lambda }\rangle&= -\frac{b_0(r_n)k_{r_n}(\lambda )}{1-b(\lambda )},\\ \end{aligned}$$

and

$$\begin{aligned} \langle -b_0(r_n)fk_{r_n},-\overline{I(\lambda )}fIk_{\lambda }\rangle&= b_0(r_n)I(\lambda )\frac{b_0(\lambda )k_{r_n}(\lambda )}{1-b(\lambda )}= \frac{b_0(r_n)b(\lambda )k_{r_n}(\lambda )}{1-b(\lambda )}. \end{aligned}$$

5 A remark on orthogonal complement of $\mathcal {M}(a)$ in $\mathcal {H}(b)$

In this section we continue to assume that b is nonextreme. Recall that if the pair (b, a) is special and $f^2=\left( \frac{a}{1-b}\right) ^2$is not rigid or the pair (b, a) is not special, then the space $\mathcal {M}(a)$ is not dense in $\mathcal {H}(b)$ ( [20]). In such a case let $\mathcal {H}_0(b)$ denote the orthogonal complement of $\mathcal {M}(a)$ in $\mathcal {H}(b)$. Let Y be the restriction of the shift operator S to $\mathcal {H}(b)$ and let $Y_0$ be the compression of Y to the subspace $\mathcal {H}_0(b)$. Then the spectrum of $Y_0$ is contained in the unit circle. Moreover, if $z_0\in \partial {\mathbb {D}}$ and k is a positive integer, then $\ker (Y^*-\bar{z_0})^k$, which actually equals $\ker (Y^*_0-\bar{z_0})^k$, lies in ${\mathcal {H}}_0(b)$. The necessary and sufficient conditions for $\mathcal {H}_0(b)$ to have finite dimension are given in Chapter X of [21] (see also [4, Theorem 29.11]). In particular, the dimension of $\mathcal {H}_0(b)$ is N if and only if the operator $Y_0$ has distinct eigenvalues $z_1$, $z_2$, ..., $z_s$ with their algebraic multiplicities $n_1$, ..., $n_s$, $N=n_1+n_2+\dots +n_s$. Then $\bar{z_1}$, $\bar{z_2}$, ..., $\bar{z_s}$ are the eigenvalues of $Y^*_0$ with the same multiplicities, i.e., $\dim \ker (Y_0-z_j)^{n_j}=\dim \ker (Y^*_0-\bar{z_j})^{n_j}$ and $\mathcal {H}_0(b)$ is the direct sum of the subspaces $\ker (Y^*_0-\bar{z_j})^{n_j}$, $j=1,2,\ldots , s$.

On the other hand, if $z_0$ is a point of $\partial {\mathbb {D}}$ and b has an angular derivative in the sense of Carathéodory at $z_0$, then the function given by

$$\begin{aligned} k_{z_0}^b(z)= \dfrac{1-\overline{b(z_0)}b(z)}{1-\bar{z_0}z}, \end{aligned}$$

(9)

where $b(z_0)$ is the nontangential limit of b at $z_0$, is in $\mathcal {H}(b)$ (see [21, VI-4,5], [4, Theorem 21.1]). In this section we actually show that $k_{z_0}^b$ is in $\mathcal {H}_0(b)$.

Here we consider the case when the eigenspaces corresponding to eigenvalues $z_1$, $z_2$, ..., $z_s$ are one dimensional and show that then the space ${\mathcal {H}}_0(b)$ is spanned by the functions $k_{z_1}^b$, $k_{z_2}^b$, ..., $k_{z_s}^b$.

For $|\lambda |=1$ let $\mu _{\lambda }$ denote the measure for which equality in (1) holds when b is replaced by ${\bar{\lambda }}b$. If we put $F_{\lambda }(z)= \frac{a}{1-{\overline{\lambda }}b}$, then the Radon-Nikodym derivative of the absolutely continuous component of $\mu _{\lambda }$ is $|F_{\lambda }|^2$. Note also that $\mathcal {H}(b)=\mathcal {H}({{\bar{\lambda }}} b)$.

In the proof of our main result in this section we use the following theorem proved in [21, X-13].

Sarason’s Theorem

Let $z_0$ be a point of $\partial \mathbb {D}$ and $\lambda$ a point of $\partial \mathbb {D}$ such that the measure $\mu _\lambda$ is absolutely continuous. The following conditions are equivalent.

• (i) $\bar{z_0}$ is an eigenvalue of $Y^*$.

• (ii) The function $\dfrac{F_\lambda (z)}{1-\bar{z_0}z}$ is in $H^2$.

• (iii) The function b has an angular derivative in the sense of Carathéodory at $z_0$.

In view of remark in Sect. 3 under the assumption of Sarason’s Theorem the operator $T_{1-{\bar{\lambda }}b}T_{{\overline{F}}_\lambda }$ is an isometry of $H^2$ onto $\mathcal {H}(b)$. Let $A_{\lambda }$ be an operator on $H^2$ that intertwines $T_{1-{\bar{\lambda }}b}T_{{\overline{F}}_\lambda }$ with the operator $Y^*$, i.e.,

$$\begin{aligned} T_{1-{\bar{\lambda }}b}T_{{\overline{F}}_\lambda }A_\lambda =Y^*T_{1-{\bar{\lambda }}b}T_{{\overline{F}}_\lambda }. \end{aligned}$$

(10)

The operator $A_\lambda$ is given by

$$\begin{aligned} A_\lambda =S^*-F_\lambda (0)^{-1}(S^*F_\lambda \otimes 1). \end{aligned}$$

It follows from the proof of Sarason’s theorem that if one of conditions (i)–(iii) holds true, then the space $\ker (A_\lambda -\bar{z_0})$ is one dimensional and is spanned by the function

$$\begin{aligned} g(z)=\frac{F_{\lambda }(z)}{1-\bar{z_0}z}= F_{\lambda }(z)k_{z_0}(z). \end{aligned}$$

(11)

We also note that condition (iii) in Sarason’s Theorem is equivalent to the fact that the function $k_{z_0}^b$ given by (9) is in $\mathcal {H}(b)$.

Theorem 3

If the assumptions of Sarason’s theorem are satisfied and $\bar{z_0}$ is an eigenvalue of $Y_0^*$, then $\ker (Y_0^*-\bar{z_0})$ is spanned by $k_{z_0}^b$.

Proof

According to the remark at the beginning of this section $\ker (Y_0^*-\bar{z_0})$ is equal to $\ker (Y^*-\bar{z_0})$. Since the operator $T_{1-{\bar{\lambda }}b}T_{{\overline{F}}_\lambda }$ is an isometry of $H^2$ onto $\mathcal {H}(b)$, (11) and (10) imply that the space $\ker (Y^*-\bar{z_0})$ is spanned by

$$\begin{aligned} h= T_{1-{\overline{\lambda }}b}T_{{\overline{F}}_{\lambda }}g= T_{1-{\overline{\lambda }}b}T_{\overline{F}_{\lambda }}(F_{\lambda }k_{z_0}). \end{aligned}$$

We will show that $h= Ck_{z_0}^b$. To this end we use the operator $V_b$ given by (5). We know that for $w\in \mathbb {D}$,

$$\begin{aligned} V_b((1-\overline{b(w)})k_w)=k^b_w \end{aligned}$$

(see [21, III-7], [4, Theorem 20.5]). Since $\mathcal {H}(b)=\mathcal {H}({\bar{\lambda }}b)$, we have

$$\begin{aligned} V_{{\bar{\lambda }}b}((1-\lambda \overline{b(w)})k_{w})(z)&= (1-{\bar{\lambda }}b(z))( 1-\lambda \overline{b(w)}) \int _{\partial \mathbb {D}}\frac{|F_{\lambda }(e^{i\theta })|^2d\theta }{(1-\bar{w}e^{i\theta })(1-ze^{-i\theta })}\\&= (1-{\bar{\lambda }}b(z))T_{{\overline{F}}_{\lambda }}((1-\lambda \overline{b(w)})F_{\lambda }k_w)(z)=k_{w}^b(z). \end{aligned}$$

Let $\{z_n\}$ be a sequence in ${\mathbb {D}}$ converging nontangentially to $z_0$. Then

$$\begin{aligned} T_{1-{\overline{\lambda }}b}T_{\overline{F}_{\lambda }}((1-\lambda \overline{b(z_n)})F_{\lambda }k_{z_n})=k_{z_n}^b. \end{aligned}$$

Observe also that since $\mu _{\lambda }$ is absolutely continuous, $b(z_0)\ne \lambda$ ( [21, VI-7, VI-9]). Moreover, $k_z^b$ tends to $k_{z_0}^b$ in norm as z tends to $z_0$ nontagentially (see [21, VI-4,5], [4, Theorem 21.1]). It then follows that the sequence $\{(1-\lambda \overline{b(z_n)})F_{\lambda }k_{z_n}\}$ converges in $H^2$, which in turn implies compact and pointwise convergence. Hence passing to the limit in the last equality yields

$$\begin{aligned} T_{1-{\overline{\lambda }}b}T_{\overline{F}_{\lambda }}(F_{\lambda }k_{z_0})= Ck_{z_0}^b, \end{aligned}$$

where $C=(1-\lambda \overline{b(z_0)})^{-1}$. $\square$

Our last theorem is an immediate consequence of Theorem 3.

Theorem 4

If $z_1$, $z_2$, ..., $z_s$ are the only eigenvalues of $Y_0$ and each of them is of multiplicity one, then $\mathcal {H}_0(b)$ is spanned by the functions $k_{z_1}^b$, $k_{z_2}^b$, ..., $k_{z_s}^b$.

Finally, we remark that this theorem generalizes results obtained in [5] and in [15] for the case when pairs (b, a) are rational.