Conjugations Preserving Toeplitz Kernels

Abstract

We study conjugations in $L^2(\mathbb {T})$ and their relation with kernels of Toeplitz operators on $H^2(\mathbb {T})$ space. Such kernels are a generalization of model spaces. We investigate properties of an inequality relation between two unimodular functions defined on the unit circle. This allows us to significantly strengthen previous theorems characterizing all $M_z$-commuting and $M_z$-conjugations leaving invariant model spaces to the Toeplitz kernels setting.

1 Introduction

Let $L^2(\mathbb {T})$ and $H^2(\mathbb {T})$ denote the standard $L^2$ space of functions on the unit circle $\mathbb {T}$ with normalized Lebesgue measure m and the Hardy space, respectively. Recall that functions from $H^2(\mathbb {T})$ naturally extends to holomorphic functions on the unit disc ${\mathbb {D}}$. By $P_+$ we denote the projection from $L^2(\mathbb {T})$ onto $H^2(\mathbb {T})$. For $\varphi \in L^\infty (\mathbb {T})$, a (essentially) bounded element of $L^2(\mathbb {T})$, $M_\varphi$ denotes the multiplication operator on $L^2(\mathbb {T})$, i.e., $M_\varphi f= \varphi f$ for $f\in L^2(\mathbb {T})$. In turn, by $T_\varphi$ we denote the Toeplitz operator on $H^2(\mathbb {T})$, i.e., $T_\varphi f= P_+(\varphi f)$ for $f\in H^2(\mathbb {T})$. Set also $H^\infty (\mathbb {T})=L^\infty (\mathbb {T})\cap H^2(\mathbb {T})$.

In the space $L^2(\mathbb {T})$ there are two natural conjugations (antilinear, isometric involutions): J and $J^\#$. Operation J conjugates the value of the functions (see (2.1)) while $J^\#$ conjugates the Fourier coefficients of the functions (see (2.2)). The behaviour of these two conjugations with respect to $M_z$—the most natural operator on $L^2(\mathbb {T})$—is different. The conjugation $J^\#$ commutes with $M_z$. But, the conjugation J intertwines $M_z$ and $M_{{\bar{z}}}$ (see (2.3)).

Theory of conjugations on the unit circle $\mathbb {T}$ has been intensively studied in the last few years (see e.g. [4, 5, 10, 11]). In [4, Theorems 2.2 and 2.4] and [5, Theorem 4.1], all conjugations commuting (or intertwining) with multiplication operator by the independent variable were described. We recall these results below in (A) and (B). One of the consequences of those results are characterizations of all conjugations leaving invariant model spaces $K^2_\theta := H^2(\mathbb {T})\ominus \theta \,H^2(\mathbb {T})$, where $\theta$ is an inner function (unimodular function in $H^2(\mathbb {T})$), see [4, Theorems 4.2, 4.6] and [5, Theorem 4.2].

The investigation in the following paper goes further in two directions. First, we look at model spaces as a particular case in the broader context of Toeplitz kernels. This approach corresponds to the effort made to understand kernels of Toeplitz operators in recent years (see [6, 7]). In the paper we obtain the characterizations of all $M_z$-commuting conjugations or $M_z$-conjugations (intertwining $M_z$ and $M_{{\bar{z}}}$) leaving a kernel of a given Toeplitz operator with a unimodular symbol invariant.

The second direction leads to stronger characterizations of conjugations than before. This can be seen especially in the $M_z$-commuting conjugations case, where we investigate the problem of existence of such conjugations (see Theorems 5.8 and 6.2).

The importance of conjugations is underlined by its connections to physics, where $\mathcal{PT}\mathcal{}$ symmetries in $L^2(\mathbb {R})$ are considered, see [1,2,3]. For a connection between conjugations on $L^2(\mathbb {T})$ and $L^2(\mathbb {R})$ see [9].

The structure of the paper is as follows. In Sect. 2 we present some natural conjugations in $L^2(\mathbb {T})$ and recall selected results from [4]. In Sect. 3 the basic results on (symmetric) inner functions are shown. The inequality between unimodular functions is introduced and important results for unimodular and inner functions are proved. Main results of the paper are proved in Sects. 5 and 6. In Sect. 5 we obtain characterization of $M_z$-conjugations leaving invariant the kernel of a given Toeplitz operator. In Sect. 6 we describe all $M_z$-commuting conjugations with the same property. Moreover, we illustrate main theorem of Sect. 6 with many examples, which underline the strength of Theorems 5.8, 6.2 with respect to [4, Theorem 4.6]

Throughout the paper we will use symbols $\alpha$, $\beta , \theta$ to denote inner functions and u, v, w to denote unimodular functions. By $\mathbb {N}$, $\mathbb {Z}$, $\mathbb {Z}_+$ we denote natural (starting from 1), integer, non-negative integer numbers.

2 Preliminaries on Conjugations in $L^2(\mathbb {T})$

A bounded, antilinear operator $C :L^2(\mathbb {T})\rightarrow L^2(\mathbb {T})$ is said to be a conjugation if $C^2 = I$ and $\langle C f, C g \rangle = \langle g, f \rangle$ for every f, $g \in L^2(\mathbb {T})$. The following lemma gives a characterization of conjugations ([12, p. 170]).

Lemma 2.1

Let $C:L^2(\mathbb {T})\rightarrow L^2(\mathbb {T})$. Then C is a conjugation if and only if

• (a) C is antilinear,

• (b) $C^2=I$,

• (c) $\Vert Cf\Vert =\Vert f\Vert$ for every $f \in L^2(\mathbb {T})$.

Conjugations here are used to study model spaces and Toeplitz kernels. In the paper we do not distinguish, as in the literature, conjugations differing up to the scalar multiple of modulus one, since the model space and Toeplitz kernel are the same for a given function and its multiples by constant of modulus 1.

In the space $L^2(\mathbb {T})$ there are many conjugation operators. We distinguish two of them, which seems very natural. First of them is J, which is simply a conjugation of the function, i.e.,

$$\begin{aligned} J f = \bar{f}, \quad f \in L^2(\mathbb {T}). \end{aligned}$$

(2.1)

The second natural conjugation is $J^\#$, which derives from conjugating Fourier coefficients of functions in $L^2(\mathbb {T})$. More explicitly,

$$\begin{aligned} J^\# f = f^\#, \quad f \in L^2(\mathbb {T}), \end{aligned}$$

(2.2)

where

$$\begin{aligned} f^\#(z) = \overline{f({\bar{z}})}, \quad z \in \mathbb {T}. \end{aligned}$$

The conjugation $J^\#$ keeps $H^2(\mathbb {T})$ space invariant ($J^\#H^2(\mathbb {T})=H^2(\mathbb {T})$). On the other hand $JH^2(\mathbb {T})=\overline{H^2(\mathbb {T})}$. The behavior of J and $J^\#$ with respect to the multiplication operator $M_z$ is also completely different. More explicitly,

$$\begin{aligned} JM_z=M_{{\bar{z}}}J\qquad \text {and}\qquad J^\#M_z=M_{ z}J^\#. \end{aligned}$$

(2.3)

Now, following [4, 5], we say that a conjugation C on $L^2(\mathbb {T})$ is $M_z$ -conjugation if $CM_z=M_{{\bar{z}}}C$ and C is called $M_z$ -commuting conjugation if $CM_z=M_{ z}C$.

We recall here also some other conjugations on $L^2(\mathbb {T})$. For any inner function $\theta$ there is a conjugation, denoted by $C_\theta$, which plays an important role in the theory of truncated Toeplitz operators on the model space $K^2_\theta :=H^2(\mathbb {T})\ominus \theta H^2(\mathbb {T})$ (see [4, 12, 16]). It is defined by the formula

$$\begin{aligned} C_\theta f= \bar{z}\theta \bar{f}, \quad f \in L^2(\mathbb {T})\end{aligned}$$

(2.4)

and satisfies $C_\theta K^2_\theta =K^2_\theta$. Moreover, for every unimodular function $u \in L^\infty (\mathbb {T})$, the operator

$$\begin{aligned} C_{u,{\mathbb {T}}}:=M_u J =J M_{\bar{u}} \end{aligned}$$

(2.5)

is a conjugation on $L^2(\mathbb {T})$. If in addition u is symmetric (i.e. $u(z)=u({\bar{z}})$ a.e. on ${\mathbb {T}}$), then

$$\begin{aligned} C^\#_{u,{\mathbb {T}}}:=M_u J^\# = J^\#M_{{\bar{u}}} \end{aligned}$$

(2.6)

is also a conjugation on $L^2(\mathbb {T})$. Clearly $C_{\theta } = C_{\theta {\bar{z}},{\mathbb {T}}}$. It is easy to check that $C_{u,{\mathbb {T}}}$ (hence also $C_\theta$) is $M_z$-conjugation and $C^{\#}_{u,{\mathbb {T}}}$ is $M_z$-commuting conjugation.

Recall from [4, Theorem 2.2 and Theorem 2.4] that

• (A) for any $M_z$-conjugation C, there is a unimodular function u such that $C= C_{u,{\mathbb {T}}}=M_u J =J M_{\bar{u}}$;

• (B) for any $M_z$-commuting conjugation C, there is a unimodular symmetric function u, such that $C= C_{u,{\mathbb {T}}}^\#=M_u J^\# =J^\# M_{\bar{u}}$.

Finally, we mention a well-known lemma, which will be used later in the main section of the article.

Lemma 2.2

Let $C_1$ be an $M_z$-conjugation and $C_2$ be an $M_z$-commuting conjugation. Assume that $u,w \in L^2(\mathbb {T})$ are unimodular. Then

• (a) $M_w C_{u,\mathbb {T}} M_{\overline{w}}$ is an $M_z$-conjugation,

• (b) $J^\# C_2C_1$ is an $M_z$-conjugation.

Proof

Case (a) follows from Lemma 2.1. In order to prove (b) notice that , $C_1 = M_v J$ for some unimodular functions u, $v \in L^2(\mathbb {T})$ (see (A), (B)). Then, by [4, Proposition 2.1],

$$\begin{aligned} (J^\#C_2 C_1)^2 = (J^\#)^2 M_u M_v J (J^\#)^2 M_u M_v J = M_{u\,v }J J M_{\overline{u\,v} } = I. \end{aligned}$$

The remaining two conditions in Lemma 2.1 are clear and hence $J^\# C_2 C_1$ is a conjugation. Finally,

$$\begin{aligned} J^\# C_2 C_1 M_z = J^\# C_2 M_{{\bar{z}}} C_1 = J^\# M_{{\bar{z}}} C_2 C_1 = M_{{\bar{z}}}J^\# C_2 C_1, \end{aligned}$$

which completes the proof. $\square$

3 Preliminaries on Inner Functions

Let $\mu$, $\nu$ be non-negative Borel measures on the $\sigma$-algebra of all Borel sets $\mathfrak {B}$ on the unit circle ${\mathbb {T}}$. We define the conjugation of $\mu$ by the formula

$$\begin{aligned} \mu ^c(\omega )=\mu \big (\{{\bar{z}} : z\in \omega \}\big ), \; \omega \in \mathfrak {B}. \end{aligned}$$

We say that $\mu$ is absolutely continuous with respect to $\nu$ if $\nu (A) = 0$ implies $\mu (A)=0$ for any $A \in \mathfrak {B}$. If this is the case, then we write $\mu \ll \nu$. In addition, there exists Radon-Nikodym derivative $\frac{d \mu }{d \nu }$ (defined a.e. with respect to $\nu$) such that $\mu (A) = \int _A \frac{d \mu }{d \nu } d \nu$, $A \in \mathfrak {B}$. Measures $\mu$ and $\nu$ are said to be mutually singular if there exist disjoint sets A, $B \in \mathfrak {B}$ such that $A \cup B = \mathbb {T}$ and $\mu (A) =\nu (B)=0$. Finally, for two measures $\mu$, $\nu$, we define their infimum by the formula (see [12, p. 89])

$$\begin{aligned} \mu \wedge \nu = \min \Big \{\tfrac{d\mu }{d(\mu +\nu )}, \tfrac{d\nu }{d(\mu +\nu )} \Big \}({\mu +\nu )}. \end{aligned}$$

Then $\mu \wedge \nu$ is a non-negative measure, which is not greater than $\mu$ and $\nu$.

For every $a\in \mathbb {D}\setminus \{0\}$, by $B_a$ we denote a Blaschke factor at $a\in \mathbb {D}$, i.e., $B_a(z) = \frac{\bar{a}}{|a|} \frac{a-z}{1-{\bar{a}} z}$.

With every $\alpha \in H^2(\mathbb {D})$ we associate a function $m_\alpha :\mathbb {D}\rightarrow \mathbb {Z}_+$ such that $m_\alpha (z)$ is the order of zero of $\alpha$ at z. Function $m_\alpha$ generates a Blaschke product

$$\begin{aligned} B_\alpha (z) = z^{m_\alpha (0)}\prod _{a_i \in m_\alpha ^{-1}(\mathbb {N}),\, a_i\ne 0}\Big (B_{a_i}(z) \Big )^{m_\alpha (a_i)}. \end{aligned}$$

We call function $m :\mathbb {D}\rightarrow \mathbb {Z}_+$ a multiplicity function if it satisfies

$$\begin{aligned} \sum _{z \in m^{-1}(\mathbb {N})} (1 - |z|) < \infty . \end{aligned}$$

Notice that for every inner function $\alpha$, the function $m_\alpha$ is a multiplicity function.

Let $\mu$ be a positive measure, which is singular with respect to m (i.e., m and $\mu$ are mutually singular) and let $S_{\mu }$ be defined as

$$\begin{aligned} S_{\mu }(z)= \exp \left( -\int _\mathbb {T}\frac{\xi +z}{\xi -z}\,d\mu (\xi )\right) , \quad z \in \mathbb {D}. \end{aligned}$$

Then $S_\mu \in H^2(\mathbb {T})$. We say that every product of $S_\mu$ and a constant of modulus one is a singular inner function. Notice that $S_\mu$ has no zeros in $\mathbb {D}$.

It turns out that, for every inner function $\alpha$, there exists a unique factorization, i.e., there is a Blaschke product $B_\alpha$ and a non-negative singular measure $\mu _\alpha$ such that $\alpha =\lambda B_\alpha S_{\mu _\alpha }$ for some $|\lambda |=1$ (see [12, Theorem 2.14.]).

Below,we gather simple properties concerning decomposition of inner function and the $\#$ operation.

Lemma 3.1

Let $\alpha =\lambda B_\alpha S_{\mu _\alpha }$ be an inner function. Then

• (a) $m_{\alpha ^\#}(z) = m_{\alpha }({\bar{z}})$ for every $z \in \mathbb {D}$,

• (b) $m_{\alpha \alpha ^\#} = m_{\alpha } + m_{\alpha ^\#}$ and $m_{\alpha \alpha ^\#}$ is symmetric,

• (c) $B_{\alpha \alpha ^\#} = B_\alpha B_\alpha ^\#$,

• (d) $S_{\mu _\alpha }^\# = S_{\mu _\alpha ^c}$,

• (e) $\mu _{\alpha ^\#} = \mu _\alpha ^c$,

• (f) $\mu _{\alpha \alpha ^\#} = \mu _\alpha + \mu _\alpha ^c$,

• (g) $S_{\mu _{\alpha \alpha ^\#}} = S_{\mu _\alpha } S_{\mu _\alpha ^c}$.

Proof

We only give an argument for (d). For any $z\in \mathbb {D}$ we have

$$\begin{aligned} S^\#_{\mu _\alpha }(z)= & {} \overline{S_{\mu _\alpha }(\bar{z})}= \overline{\exp \left( -\int _\mathbb {T}\frac{\xi +\bar{z}}{\xi -\bar{z}}\,d\mu _\alpha (\xi )\right) }\\= & {} \exp \left( -\int _\mathbb {T}\frac{{\bar{\xi }}+z}{{\bar{\xi }}-z}\,d\mu _\alpha (\xi )\right) = \exp \left( -\int _\mathbb {T}\frac{\xi +z}{\xi -z}\,d\mu _\alpha ^c(\xi )\right) = S_{\mu _\alpha ^c}(z). \end{aligned}$$

Thus $S_{\mu _\alpha }^\# = S_{\mu _{\alpha ^c}}$ and we get (d). $\square$

A function $f \in H^2(\mathbb {T})$ is called outer if $\log |f| \in L^1(\mathbb {T})$ and

$$\begin{aligned} f(z) = C\exp \left( \int _\mathbb {T}\frac{\xi +z}{\xi -z} \log |f(\xi )| \text {d}m (\xi )\right) , \,|z| < 1, \end{aligned}$$

for some unimodular constant C. The following lemma allows constructing new outer functions.

Lemma 3.2

([15, p.27]) Let $f \in H^\infty (\mathbb {T})$ and $\Vert f\Vert _\infty < 1$. Then $1+f$ is outer.

4 Division Structure of Unimodular Functions

Let us introduce a new relation, which is a generalization of the classical inequality between two inner functions ($\alpha _1 \leqslant \alpha _2$ if and only if $\alpha _1$ divides $\alpha _2$).

Definition 4.1

Let $v_1$, $v_2 \in L^\infty (\mathbb {T})$ be unimodular functions on $\mathbb {T}$. Then we write

$$\begin{aligned} v_1\leqslant v_2 \text { if and only if } \tfrac{v_2}{v_1}\in H^\infty (\mathbb {T}). \end{aligned}$$

Recall here the basic properties of the relation $\leqslant$ for inner functions (see [12, p. 41]).

Lemma 4.2

Let $\alpha _1=\lambda _1 B_{\alpha _1} S_{\mu _1}$ and $\alpha _2=\lambda _2 B_{\alpha _2} S_{\mu _2}$ be a standard decomposition of inner functions. Then

• (a) $\alpha _1\leqslant \alpha _2$ if and only if $B_{\alpha _1}\leqslant B_{\alpha _2}$ and $S_{\mu _1}\leqslant S_{\mu _2}$,

• (b) $B_{\alpha _1}\leqslant B_{\alpha _2}$ if and only if $m_{\alpha _1}\leqslant m_{\alpha _2}$,

• (c) $S_{\mu _1} \leqslant S_{\mu _2}$ if and only if $\mu _1(\omega ) \leqslant \mu _2(\omega )$ for any $\omega \in \mathfrak {B}$. If any of these conditions is satisfied, then $\mu _1 \ll \mu _2$ and $\frac{d \mu _1}{d \mu _2} \leqslant 1$. Then we write $\mu _1\leqslant \mu _2$.

The inequality relation between two unimodular functions is reflexive, transitive and moreover, since $H^\infty \cap \overline{H^\infty } = \mathbb {C}$,

$$\begin{aligned} v_1 \leqslant v_2 \text { and } v_2 \leqslant v_1 \Longleftrightarrow v_1 = v_2 \,(\text {up to a constant of modulus one}). \end{aligned}$$

(4.1)

It is easy to see that for any unimodular function w the implication

$$\begin{aligned} w\, v_1 \leqslant w\, v_2 \Longrightarrow v_1 \leqslant v_2 \end{aligned}$$

holds. Moreover $v_1 \leqslant v_2$ if and only if $\overline{v_2}\leqslant \overline{v_1}$.

Let us notice that $v \leqslant \alpha$ for any co-analytic function v and analytic function $\alpha$. On the other hand, there are many unimodular functions which are uncomparable with analytic or co-analytic functions. One of such functions is $2\chi _{\mathbb {T}_+}-1$, where $\mathbb {T}_+=\mathbb {C}_+\cap \mathbb {T}$. However, this relation is strongly related with Toeplitz kernels.

Remark 4.3

Let $v_1,v_2\in L^\infty (\mathbb {T})$ be unimodular functions such that $\ker T_{v_2}\not =\{0\}$. Then, by [7, Corollary 7.7],

$$\begin{aligned} v_1\leqslant v_2 \text { if and only if } \ker T_{v_2}\subset \ker T_{v_1}. \end{aligned}$$

Notice that if $v_1\leqslant v_2$ and $v_2 \leqslant v_1$, then $\ker T_{v_1} = \ker T_{v_2}$. From this point of view we will identify two unimodular functions differing up to a constant of modulus one.

Definition 4.4

Let $v \in L^2(\mathbb {T})$ be unimodular. We say that v is self-reflected if $v\leqslant v^\#$ and $v^\# \leqslant v$. In turn, v is called completely non-self-reflected if for any inner function $\alpha \in H^2(\mathbb {T})$

$$\begin{aligned} \alpha \leqslant v, \quad \alpha \leqslant v^\# \Longrightarrow \alpha \text { is constant}. \end{aligned}$$

Lemma 4.5

Let $v_1,v_2\in L^\infty (\mathbb {T})$ be unimodular functions. Then

• (a) $\big (\tfrac{v_2}{v_1}\big )^\# = \tfrac{v_2^\#}{v_1^\#}$,

• (b) $v_1 \leqslant v_2$ if and only if $v_1^\# \leqslant v_2^\#$,

• (c) if $v_1\leqslant v_2$ and $v_2$ is completely non-self-reflected, then $v_1$ is also completely non-self-reflected.

Proof

Notice that (a) is a direct consequence of the definition of $\#$, (b) follows from (a). If $\alpha \in H^2(\mathbb {T})$ is inner, $\alpha \le v_1$, and $\alpha \le v_1^\#$, then $\alpha \le v_2$, $\alpha \le v_2^\#$, which proves (c). $\square$

In the context of inner functions, self-reflectivity and completely non-self-reflectivity can be described in the terms of appropriate properties of multiplicity functions and singular measures.

Proposition 4.6

Let $\alpha =\lambda B_{\alpha } S_{\mu }$ be the canonical decomposition of inner functions; i.e., $B_{\alpha }$ be a Blaschke product, $S_{\mu }$ be a singular inner function. Then the following hold.

• (a) $B_\alpha$ is self-reflected if and only if $m_\alpha (z) =m_\alpha ({\bar{z}})$, $z \in \mathbb {D}$.

• (b) $B_\alpha$ is completely non-self-reflected if and only if $m_\alpha (z)\cdot m_\alpha ({\bar{z}})=0$, $z \in \mathbb {D}$.

• (c) $S_\mu$ is self-reflected if and only if $\mu =\mu ^c$,

• (d) $S_\mu$ is completely non-self-reflected if and only if $\mu \wedge \mu ^c \equiv 0$.

• (e) $\alpha$ is self-reflected if and only if $B_\alpha$ and $S_\mu$ are self-reflected.

• (f) $\alpha$ is completely non-self-reflected if and only if $B_\alpha$ and $S_\mu$ are completely non-self reflected.

Proof

Condition (a)–(c) follows from Lemma 3.1. In order to prove (d), first assume that $S_\mu$ is not completely non-self-reflected. Then there is a positive singular measure $\nu$ such that $S_\nu$ is a divisor of $S_\mu$ and $S_{\mu ^c}$ and hence $\nu \leqslant \mu$ and $\nu \leqslant \mu ^c$. On the other hand $\nu \le \mu \wedge \mu ^c$ and therefore $\mu \wedge \mu ^c \equiv 0$. Now, let us assume that $\mu \wedge \mu ^c\not \equiv 0$. Then $S_{\mu \wedge \mu ^c}$ is a non–constant divisor of $S_\mu$ and $S_{\mu ^c}$. Hence, $S_\mu$ is not completely non-self-reflected. Conditions (e), (f) follow from the uniqueness of decomposition of inner function into Blaschke product and singular function. $\square$

Lemma 4.7

Let $\beta \in H^2({\mathbb {T}})$ be an inner function and let $w \in L^\infty (\mathbb {T})$ be a unimodular function such that $w w^\#$ is an inner function with even multiplicity for each (if any) real zeros. Assume that $w w^\#\leqslant \beta \beta ^\#$. Then there is an inner function $\alpha$ such that

$$\begin{aligned} \alpha \leqslant \beta \text { and } \alpha {\alpha }^\#=ww^\#. \end{aligned}$$

Moreover, if $\beta$ is completely non-self-reflected, then such $\alpha$ is unique.

Proof

The function $v= w w^\#$ is inner and hence $v = B_v S_{\mu _v}$, where $B_v$ is a Blaschke product and $S_{\mu _v}$ is a singular part of v. Notice that v and $B_v$ are self-reflected and as a consequence $S_{\mu _v}$ is also self-reflected. Proposition 4.6c implies that $\mu _v = \mu _v^c$.

Let $\beta =B_\beta S_{\mu _\beta }$ be a decomposition of inner function $\beta$ into Blaschke product and singular function. Then

$$\begin{aligned} \beta \beta ^\#=B_{\beta \beta ^\#}S_{\mu _\beta +\mu _\beta ^c}. \end{aligned}$$

Condition $v=w w^\#\leqslant \beta \beta ^\#$ implies that

$$\begin{aligned} m_v\leqslant m_\beta +m_{\beta ^\#}\quad \text {and }\quad \mu _v \leqslant \mu _\beta +\mu _\beta ^c. \end{aligned}$$

We define the Blaschke product $B_\alpha$ by the multiplicity function $m_\alpha$. We set

$$\begin{aligned} m_\alpha (z) = \left\{ \begin{array}{cc} \Bigl \lceil m_\beta (z) \frac{m_v(z) }{m_\beta (z)+m_{\beta ^\#}(z)} \Bigr \rceil , &{} \text { if Im} \, z \geqslant 0,\\ \\ \Bigl \lfloor m_\beta (z) \frac{m_v(z)}{m_\beta (z)+m_{\beta ^\#}(z)} \Bigr \rfloor , &{} \text { if Im} \, z <0.\\ \end{array} \right. \end{aligned}$$

(Recall the notation: for $x\in \mathbb {R}$ , $\lfloor x\rfloor$ and $\lceil x\rceil$, denote the numbers such that $\lfloor x\rfloor , \lceil x\rceil \in \mathbb {Z}$ and $x-1<\lfloor x\rfloor \leqslant x\leqslant \lceil x\rceil <x+1$.) Then

$$\begin{aligned} m_\alpha (z) \leqslant m_\beta (z), \quad z \in \mathbb {D}, \end{aligned}$$

and thus $m_\alpha$ is a multiplicity function and $B_\alpha \leqslant B_\beta$. For any $t\in (-1,1)$, by Lemma 3.1a and the fact that $m_v(t)$ is even, we have

$$\begin{aligned} m_\alpha (t) = \bigl \lceil \tfrac{m_v(t)}{2} \bigr \rceil = \tfrac{m_v(t)}{2}. \end{aligned}$$

Now, observe that $\lceil x \rceil + \lfloor y \rfloor = x +y$ for any x, $y \in \mathbb {R}$ such that $x+y \in \mathbb {Z}_+$. This fact combined with symmetricity of $m_v$ (see Lemma 3.1b), for any $z\in \mathbb {D}\, \setminus \,\mathbb {R}$, gives us

$$\begin{aligned} m_\alpha (z)+ m_\alpha (\bar{z}) = \Bigl \lceil m_\beta (z) \frac{m_v(z) }{m_\beta (z)\!+\!m_\beta (\bar{z})} \Bigr \rceil + \Bigl \lfloor m_\beta (\bar{z}) \frac{m_v(\bar{z}) }{m_\beta (z)\!+\!m_\beta (\bar{z})} \Bigr \rfloor = m_v(z). \end{aligned}$$

This implies that

$$\begin{aligned} m_\alpha (z) + m_\alpha (\bar{z}) = m_v(z) \text { for any }z \in \mathbb {D}. \end{aligned}$$

Hence $B_{\alpha \alpha ^\#} = B_{v}$.

The inequality $\mu _v\leqslant \mu _\beta +\mu _\beta ^c$ implies that $\mu _v\ll \mu _\beta +\mu _\beta ^c$ and the Radon-Nikodym derivative $\frac{d(\mu _v)}{d(\mu _\beta +\mu _\beta ^c)}$ is not greater than 1 (a.e. on $\mathbb {T}$). Now, we define the singular part of $\alpha$ by the formula

$$\begin{aligned} \mu _\alpha =\tfrac{d( \mu _v)}{d(\mu _\beta +\mu _\beta ^c)}\cdot \mu _\beta . \end{aligned}$$

It is clear that $0 \leqslant \mu _\alpha \leqslant \mu _\beta$ and $\mu _\alpha$ is singular. Therefore $S_{\mu _\alpha } \leqslant S_{\mu _\beta }$. In addition, function v is self-reflected and thus, by Proposition 4.6c, $\mu _v = \mu _v^c$. As a consequence, $\mu _\alpha +\mu _\alpha ^c=\mu _v$ and $S_{\mu _\alpha } S_{\mu _\alpha ^c} = S_{\mu _v}$.

Finally, we set $\alpha = B_\alpha S_{\mu _\alpha }$. It is clear that $\alpha$ is inner and $\alpha \leqslant \beta$. Moreover,

$$\begin{aligned} \alpha \alpha ^\# = B_\alpha B_{\alpha ^\#} S_{\mu _\alpha } S_{\mu _\alpha ^c} = B_v S_{\mu _v} = v =w w^\#, \end{aligned}$$

which gives the first part of the theorem.

Now, let us assume that $\beta =B_\beta S_{\mu _\beta }$ is completely non-self-reflected. Take any inner function $\alpha = B_\alpha S_{\mu _\alpha }$ and a unimodular function w, such that $ww^\# =v=B_v S_{\mu _v}$ is an inner function and

$$\begin{aligned} \alpha \leqslant \beta , \quad \alpha \alpha ^\# =w w^\# \leqslant \beta \beta ^\#. \end{aligned}$$

Then

$$\begin{aligned} m_\alpha (z) +m_\alpha ({\bar{z}}) = m_v(z), \quad z \in \mathbb {D}, \end{aligned}$$

and for any $t\in \mathbb {R}$ we have $2 m_\alpha (t) =m_\alpha (t) + m_\alpha (\bar{t}) = m_v(t)$. On the other hand, non self-reflectivness of $\beta$ implies

$$\begin{aligned} m_\beta (z) \cdot m_\beta ({\bar{z}}) = 0 \text { for any } z \in \mathbb {D}\setminus \mathbb {R}. \end{aligned}$$

Consider two cases for the multiplicity of $z \in \mathbb {D}\, \setminus \, \mathbb {R}$:

• if $m_\beta (z) = 0$, then $m_\alpha (z) \le m_\beta (z)=0$,

• if $m_\beta (z) \ne 0$, then $m_\alpha ({\bar{z}}) \le m_\beta ({\bar{z}}) =0$ and $m_\alpha (z) = m_v(z)$.

Therefore, the Blaschke factor $B_\alpha$ is uniquely determined.

For the uniqueness of $\mu _\alpha$, let

$$\begin{aligned} Z_1 = \Big \{ z \in \mathbb {T}:\tfrac{d \mu _\alpha }{d (\mu _\alpha + \mu _\alpha ^c)}(z) =0\Big \} \text { and } Z_2 = \Big \{ z \in \mathbb {T}:\tfrac{d \mu _\alpha ^c}{d (\mu _\alpha + \mu _\alpha ^c)}(z)=0\Big \}. \end{aligned}$$

By the definition, ${\mu _\alpha }(Z_1)=0$. Notice that the condition $\alpha \leqslant \beta$ and Lemma 4.5c implies that $\alpha$ is also completely non-self-reflected. By Proposition 4.6d, we have $\mu _{\alpha } \wedge \mu _{\alpha }^c=0$. Hence $\mu _\alpha (\mathbb {T}\, \setminus \,(Z_1 \cup Z_2))= 0$. Finally, let $A \subset Z_2$, $A\in \mathfrak {B}$. Then

$$\begin{aligned} \mu _\alpha (A) = \mu _v(A) - \mu _\alpha ^c(A) = \mu _v(A) - \int _A \frac{d \mu _\alpha ^c}{d (\mu _\alpha + \mu _\alpha ^c)} d (\mu _\alpha + \mu _\alpha ^c) = \mu _v(A). \end{aligned}$$

This shows that $\mu _{\alpha }$ is uniquely determined, which ends the proof. $\square$

Remark 4.8

The function $\alpha$ is also unique if $w w^\#$ can be uniquely decomposed. Examples of such functions are $w w^\#= 1$ or $w w^\# = (B_a)^2$ with $a\in (-1,1)$.

Lemma 4.9

Let $\eta \in L^\infty (\mathbb {T})$ be unimodular and $\beta \in H^\infty (\mathbb {T})$ be inner. Then the following conditions are equivalent.

• (a) There is a symmetric unimodular function $w\in L^\infty (\mathbb {T})$ such that

  $$\begin{aligned} {\bar{\beta }} \leqslant \frac{w}{\eta }\leqslant 1. \end{aligned}$$

• (b) There is a symmetric unimodular function $u\in L^\infty (\mathbb {T})$ such that

  $$\begin{aligned} 1 \leqslant u\cdot {\eta }\leqslant \beta . \end{aligned}$$

• (c) $\eta \cdot \eta ^\#$ is an inner function with even multiplicity for each (if any) real zero and $\eta \cdot \eta ^\#\leqslant \beta \cdot \beta ^\#$.

Moreover, if $\beta$ is completely non-self-reflected, then there is exactly one function u fulfilling condition (a) or (b).

Proof

Conditions (a) and (b) are equivalent with $u={\bar{w}}$. Assume (b) and let u be a unimodular, symmetric function such that $1\leqslant {u} \cdot {\eta } \leqslant \beta$. This means that $\alpha = u\eta$ is inner and $\alpha \leqslant \beta$. Moreover, u is symmetric and therefore

$$\begin{aligned} \alpha (z)\overline{ \eta (z)} = u(z)=u(\bar{z}) =\alpha (\bar{z})\overline{ \eta (\bar{z})}. \end{aligned}$$

Finally,

$$\begin{aligned} (\eta \cdot \eta ^\#)(z)=\eta (z)\overline{\eta (\bar{z})}=\alpha (z)\overline{\alpha ({\bar{z}})}=(\alpha \cdot \alpha ^\#)(z). \end{aligned}$$

This combined with Lemma 4.5 shows that $\eta \eta ^\# =\alpha \alpha ^\# \leqslant \beta \beta ^\#$.

If $\alpha$ contains any power of Blaschke factor $(B_a)^l$ with $a\in (-1,1)$, $a\ne 0$, $l \in \mathbb {Z}_+$, then $\eta \eta ^\#=\alpha \alpha ^\#$ contains $(B_a)^{2l}$ since $B_a^\#=B_{\bar{a}}=B_a$. The same argument shows that the multiplicity of 0 for $\eta \eta ^\#$ is even and condition (c) holds.

Now, assume (c). By Lemma 4.7, there is an inner function $\alpha$ such that $\alpha \leqslant \beta$ and $\alpha \alpha ^\# = \eta \eta ^\#$. Define $u = {\alpha }{\bar{\eta }}$. Then $1 \leqslant u\cdot {\eta }\leqslant \beta$. Finally,

$$\begin{aligned} u({\bar{z}}) = {\alpha }({\bar{z}}) \overline{\eta ({\bar{z}})}= \frac{\eta (z) \overline{\eta ({\bar{z}})} {\alpha ({\bar{z}})}}{{\eta ( z)}} = \frac{\alpha (z)\overline{\alpha ({\bar{z}})}\alpha ({\bar{z}})}{{\eta ( z)}} = {\alpha ( z)} \overline{\eta ( z)} = u( z) \end{aligned}$$

thus u is symmetric.

For the moreover part, we apply the moreover part of Lemma 4.7 to get the uniqueness of u. $\square$

5 Conjugations and Toeplitz Kernels on $L^2(\mathbb {T})$

Let $\theta \in H^2(\mathbb {T})$ be an inner function. The conjugation $C_{\bar{z}\theta ,{\mathbb {T}}}$ (denoted also by $C_\theta$ in the literature) plays a crucial role in investigating Toeplitz operators on the model space $K^2_\theta$. It keeps $K^2_\theta$ invariant and is the only one $M_z$-conjugation having such property ([5, Theorem 4.3]). Recently kernels of Toeplitz operators as generalizations of model spaces were investigated (see [6, 7]). In the following proposition we will prove that $C_{\bar{z}\theta ,{\mathbb {T}}}$ keeps invariant also $\ker T_v$ for any unimodular function v.

Proposition 5.1

Let $v\in L^\infty (\mathbb {T})$ be unimodular. Then $C_{\overline{v z},{\mathbb {T}}}(\ker T_v)\subset \ker T_v$.

Proof

Let $f\in \ker T_v$. Then $v f\in \overline{z H^2(\mathbb {T})}$ and $C_{\overline{v z},{\mathbb {T}}}f=\overline{v zf}\in H^2(\mathbb {T})$. Moreover, $v\, C_{\overline{v z},{\mathbb {T}}}f=\overline{zf}\in \overline{zH^2(\mathbb {T})}$, which implies that $C_{\overline{v z},{\mathbb {T}}}f\in \ker T_v$. $\square$

Since Toeplitz kernels generalize model spaces, the natural question arises, if the conjugation $C_{\overline{v z},{\mathbb {T}}}$ is also a unique $M_z$-conjugation preserving Toeplitz kernel $\ker T_v$. By [13, Lemma 5] if $\ker T_v$ is non-zero for a given function $v\in L^\infty (\mathbb {T})$, then $\ker T_v= \ker T_{\bar{F}/F}$ for some outer function F. Hence, it is enough to consider Toeplitz operators with unimodular symbols and in this section we will concentrate on such functions. It occurs that the answer to the above question is positive.

Theorem 5.2

Let C be any $M_z$-conjugation on $L^2(\mathbb {T})$ and let $v\in L^\infty (\mathbb {T})$ be a unimodular function such that $\ker T_{ v}\not =\{0\}$. Then the subspace $\ker T_{ v}$ is invariant for C, i.e. $C\big (\ker T_{v}\big )\subset \ker T_{v}$, if and only if $C=\lambda C_{\overline{v z},{\mathbb {T}}}$ with $\lambda \in {\mathbb {C}}$, $|\lambda |=1$.

The above is a special case of the more general Theorem 5.8, which will be proved later. Let us start with mentioning how does the symbol of a Toeplitz operator with non-trivial kernel looks like. In [14, Lemma 3.2] the description of Toeplitz operators on the upper half plane with non-trivial kernel was given. With the similar proof we have an analogue of this result.

Lemma 5.3

Let $v \in L^\infty (\mathbb {T})$ be a unimodular function such that $\ker T_v \ne \{0\}$. Then there exists an inner function $\alpha$ and an outer function F such that

$$\begin{aligned} v = \overline{z\alpha } \frac{\overline{F}}{F}. \end{aligned}$$

(5.1)

It is known that for every nonconstant inner function $\theta$, $\ker T_{\overline{\theta }}=K^2_\theta \ne \{0\}$. Below we show the appropriate factorization of $\overline{\theta }$.

Remark 5.4

Let $\theta$ be a nonconstant inner function. Then, by Lemma 3.2, $k^\theta _0 = 1 - \overline{\theta (0)} \theta$ is an outer function and belongs to $\ker T_{{\bar{\theta }}}$. As a consequence $\overline{\theta } = \overline{z \alpha } \frac{\overline{F}}{F}$, where

$$\begin{aligned} \alpha (z)=\frac{1}{z}\cdot \frac{\theta (z)-\theta (0)}{1-\overline{\theta (0)}\theta (z)}\text { and } F=k^\theta _0. \end{aligned}$$

In the proof of the main theorem of this section we will use the concepts of minimal Toeplitz kernels and maximal vectors for Toeplitz kernels. Following [6], we recall a definition.

Definition 5.5

For a non-zero function $k \in H^2(\mathbb {T})$, let $K_\mathrm{min}(k)$ be the minimal Toeplitz kernel containing k, that is, $K_\mathrm{min}(k) = \ker T_v$ for some $v \in L^\infty (\mathbb {T})$, with $k \in K_\mathrm{min}(k)$, while $\ker T_v \subset \ker T_w$ for every $w \in L^\infty (\mathbb {T})$ such that $k \in \ker T_w$. Let $v\in L^\infty (\mathbb {T})$ be a unimodular function. We say that the function k is a maximal vector for $T_v$ if and only if $\ker T_v = K_\mathrm{min}(k)$.

For further information on minimal kernels and maximal vectors see [6], where, for example, the existence of these objects was proved. Now we are ready to prove the proposition from which we will deduce the main theorem.

Proposition 5.6

Let $v_1,v_2, \eta \in L^\infty (\mathbb {T})$ be unimodular functions such that $\ker T_{v_1}\! \ne \{0\}$ and let C be an $M_z$-conjugation on $L^2(\mathbb {T})$. Then the following conditions are equivalent:

• (a) $C(\eta \ker T_{v_1}) \subset \ker T_{v_2}$,

• (b) $C = C_{\psi ,\mathbb {T}}$ for some unimodular function $\psi \in L^\infty (\mathbb {T})$ such that

  $$\begin{aligned} \eta \overline{v_1z} \leqslant \psi \leqslant \eta \overline{v_2z} \end{aligned}$$

Moreover, there exists $M_z$-conjugation C fulfilling (a) if and only if $v_2\leqslant v_1$.

Proof

(b)$\Rightarrow$(a) Assume that $C = C_{\psi , \mathbb {T}}$ for some unimodular function $\psi \in L^\infty (\mathbb {T})$ such that $\eta \overline{v_1z} \leqslant \psi \leqslant \eta \overline{v_2z}$. Then $v_2 \leqslant v_1$. Let $f \in \ker T_{v_1}$. Then $v_1f\in \overline{zH^2(\mathbb {T})}$ and $\overline{v_1zf} \in H^2(\mathbb {T})$. In addition, from the assumption we know that ${\overline{\eta }} v_1z \psi \in H^\infty$. Hence

$$\begin{aligned} C (\eta f) = \psi \overline{\eta f} = (\psi {\overline{\eta }} v_1 z) \cdot (\overline{v_1zf}) \in H^2(\mathbb {T}). \end{aligned}$$

Moreover, since ${\overline{\eta }} v_2 z \psi \in \overline{H^\infty (\mathbb {T})}$,

$$\begin{aligned} v_2 \cdot C(\eta f) = {\overline{\eta }}v_2 \psi \bar{f} = ({\overline{\eta }} v_2\psi z) \cdot \overline{zf} \in \overline{zH^2(\mathbb {T})}, \end{aligned}$$

which implies that $C(\eta f) \in \ker T_{v_2}$.

(a)$\Rightarrow$(b) Now, let k be a maximal vector for $\ker T_{v_1}$. Then, by [6, Theorem 2.2], $k = \overline{v_1zp}$ for some outer function $p\in H^2(\mathbb {T})$. From (A) we know that $C = M_\psi J$ for some unimodular function $\psi \in L^\infty (\mathbb {T})$. Then

$$\begin{aligned} {\overline{\eta }}\psi v_1 z p =C(\eta \overline{v_1 z p}) = C(\eta k) \in \ker T_{v_2} \subset H^2(\mathbb {T}). \end{aligned}$$

Since $L^2(\mathbb {T})\ni {\overline{\eta }}\psi v_1 z = \frac{{\overline{\eta }}\psi v_1 z p}{p}$ belongs to Smirnov class, from [8, Theorem 2.11] we deduce that ${\overline{\eta }} \psi v_1 z \in H^2(\mathbb {T})$ and $\eta \overline{ v_1z}\leqslant \psi$. Note that $C_{\overline{v_1z},\mathbb {T}}(k)=p$. Hence, by Proposition 5.1, we get $p\in \ker T_{v_1}$. On the other hand,

$$\begin{aligned} \eta \overline{ \psi v_2 z} p =C_{\overline{v_2z}, \mathbb {T}} (\psi \overline{\eta p}) =C_{\overline{v_2z}, \mathbb {T}}C(\eta p) \in \ker T_{v_2} \subset H^2(\mathbb {T}). \end{aligned}$$

As before $\eta \overline{\psi v_2 z}=\frac{\eta \overline{\psi v_2 z}p}{p}\in L^2(\mathbb {T})$ belongs to Smirnov class which implies that $\eta \overline{ \psi v_2 z} \in H^2(\mathbb {T})$ and $\psi \leqslant \eta \overline{ v_2z}$, which gives the claim. $\square$

Before we prove the theorem, we make a trivial observation using [4, Theorem 2.2].

Remark 5.7

Let $\eta _1$, $\eta _2$, $v_1$, $v_2$ be unimodular functions on $L^2(\mathbb {T})$ and let C be an $M_z$-conjugation. Then $C( \eta _1 \ker T_{v_1}) \subset \eta _2 \ker T_{v_2}$ if and only if $C( \eta _1 \eta _2 \ker T_{v_1}) \subset \ker T_{v_2}$.

Theorem 5.8

Let $v_1$, $v_2$, $\eta _1$, $\eta _2 \in L^\infty (\mathbb {T})$ be unimodular functions, where $\ker T_{v_1} \ne \{0\}$, and let C be an $M_z$-conjugation on $L^2(\mathbb {T})$. Then the following conditions are equivalent.

• (a) $C(\eta _1\ker T_{v_1}) \subset \eta _2\ker T_{v_2}$.

• (b) $C = C_{\psi ,\mathbb {T}}$ for some unimodular function $\psi \in L^\infty (\mathbb {T})$ such that

  $$\begin{aligned} \eta _1\eta _2\overline{ v_1z} \leqslant \psi \leqslant \eta _1 \eta _2\overline{ v_2z} \end{aligned}$$

  (5.2)

Moreover, there exists $M_z$-conjugation C fulfilling (a) if and only if $v_2\leqslant v_1$.

Proof

Apply Proposition 5.6 and Remark 5.7. $\square$

In particular, if $\eta _1=\eta _2=1$, then we obtain the following.

Corollary 5.9

Let $v_1$, $v_2 \in L^\infty (\mathbb {T})$ be unimodular functions such that $\ker T_{v_1} \ne \{0\}$ and let C be an $M_z$-conjugation on $L^2(\mathbb {T})$. If $C(\ker T_{v_1}) \subset \ker T_{v_2}$, then $v_2\leqslant v_1$. Moreover, $C(\ker T_{v_1}) \subset \ker T_{v_2}$ if and only if $C = C_{\psi ,\mathbb {T}}$ for some unimodular function $\psi \in L^\infty (\mathbb {T})$ such that $\overline{v_1 z} \leqslant \psi \leqslant \overline{v_2z}.$

Now, assume that $v\in L^\infty (\mathbb {T})$ is unimodular. Substituting $v_1=v_2=v$ in Theorem 5.8, we get the following.

Corollary 5.10

Let $v,\eta _1,\eta _2 \in L^\infty (\mathbb {T})$ be unimodular functions such that $\ker T_v\! \ne \{0\}$ and let C be an $M_z$-conjugation on $L^2(\mathbb {T})$. Then $C(\eta _1\ker T_v) \subset \eta _2\ker T_v$ if and only if $C = C_{\eta _1 \eta _2 \overline{vz},\mathbb {T}}$.

When we assume that $v_2\leqslant v_1$, then, by (5.2), we have the following two corollaries.

Corollary 5.11

Let $v_1$, $v_2$, $\eta _1, \eta _2$ be unimodular functions in $L^\infty (\mathbb {T})$ such that $\ker T_{v_1} \ne \{0\}$ and $v_2 \le v_1$. Then there always exists an $M_z$-conjugation satisfying $C(\eta _1 \ker T_{v_1}) \subset \eta _2 \ker T_{v_2}$.

Corollary 5.12

Let $v_1, v_2\in \! L^\infty (\mathbb {T})$ be unimodular functions such that $\ker T_{v_1} \ne \{0\}$ and $v_2 \le v_1$. Let C be an $M_z$-conjugation. Then there exists a unimodular function $\eta \in L^\infty (\mathbb {T})$ satisfying $C(\eta \ker T_{v_1}) \subset \ker T_{v_2}$.

6 $M_z$-Commuting Conjugations and Toeplitz Kernels

In this section we are going to prove a counterpart of Theorem 5.8 for $M_z$-commuting conjugations. Before we do this, we state a simple lemma connecting kernel of $T_v$ with $T_{v^\#}$.

Lemma 6.1

Let v be a unimodular function in $L^\infty (\mathbb {T})$. Then $J^\# \ker T_v=\ker T_{v^\#}$.

Proof

This follows from the fact that $f \cdot v^\# = (f^\# \cdot v)^\#$ for every $f \in \ker T_v$. $\square$

The Theorem below relates to [4, Theorem 4.5]. We do not only describe a conjugation leaving invariant wider class of subspaces (Toeplitz kernels instead of model spaces), but we characterize when such conjugation exists. We apply Proposition 5.6 and Lemma 4.9 to obtain a kernel invariance theorem for $M_z$-commuting conjugations.

Theorem 6.2

Let $v_1$, $v_2$, $\eta _1$, $\eta _2 \in L^\infty (\mathbb {T})$ be unimodular functions, let C be an $M_z$-commuting conjugation on $L^2(\mathbb {T})$. Assume that $\ker T_{v_1} \ne \{0\}$. Then the following conditions are equivalent.

• (a) $C(\eta _1 \ker T_{v_1}) \subset \eta _2 \ker T_{v_2}$,

• (b) There is a symmetric, unimodular function $u\in L^\infty (\mathbb {T})$ such that

  $$\begin{aligned} \overline{\eta _1}{\eta _2^\#} \le {\bar{u}} \le \overline{\eta _1}{\eta _2^\#}\ {v_1}\overline{v_2}^{\#} \end{aligned}$$

  (6.1)

  and $C = C^\#_{u, \mathbb {T}}= M_u J^\#$.

Moreover, there exists $M_z$-commuting conjugations C fulfilling (a) if and only if the following three conditions hold:

• (i) $v_2^\#\le v_1$,

• (ii) $\eta _1\, \eta _1^\#\,\overline{\eta _2}\, \overline{\eta _2}^\#$ is an inner function with even multiplicity for each (if any) real zeros,

• (iii) $\eta _1\, \eta _1^\#\,\overline{\eta _2}\, \overline{\eta _2}^\#\leqslant v_1\, v_1^\#\,\overline{v_2}\, \overline{v_2}^\#$.

In particular, if conditions (i)–(iii) are fulfilled and $v_1 \overline{v_2}^\#$ is completely non-self-reflected, then there is exactly one function u for which (6.1) holds.

Before we start a proof let us make a simple observation based on [4, Theorem 2.4].

Remark 6.3

Let $\eta _1$, $\eta _2$, $v_1$, $v_2$ be unimodular functions on $L^2(\mathbb {T})$. Let C be an $M_z$-commuting conjugation on $L^2(\mathbb {T})$. Then $C( \eta _1 \ker T_{v_1}) \subset \eta _2 \ker T_{v_2}$ if and only if $C( \eta _1 \overline{\eta _2}^\# \ker T_{v_1}) \subset \ker T_{v_2}$.

Proof

For the sake of simplicity, let $\eta =\eta _1\overline{\eta _2}^\#$. Assume condition (a). By Remark 6.3 we get $C(\eta \ker T_{v_1}) \subset \ker T_{v_2}$. Notice that, by Lemma 2.2, the operators ${M_{\eta }} C_{\overline{v_1z},\mathbb {T}} {M_{{\overline{\eta }}}}$ and $J^{\#}C{M_{\eta }} C_{\overline{v_1z},\mathbb {T}} {M_{{\overline{\eta }}}}$ are $M_z$-conjugations.

On the other hand, by Proposition 5.1, the assumption, and Lemma 6.1, we have

$$\begin{aligned} \begin{aligned} J^{\#}C{M_{{\eta }}} C_{\overline{v_1z},\mathbb {T}}{M_{\overline{\eta }}}(\eta \ker T_{v_1})&= J^{\#}C{M_{{\eta }}} C_{\overline{v_1z},\mathbb {T}}(\ker T_{v_1})\\&\subset J^{\#}C(\eta \ker T_{v_1})\subset J^{\#}(\ker T_{v_2})\subset \ker T_{v_2^{\#}}. \end{aligned} \end{aligned}$$

By Proposition 5.6, there is a unimodular function $w\in L^\infty (\mathbb {T})$ such that

$$\begin{aligned} J^{\#}C {M_{{\eta }}}C_{\overline{v_1 z},\mathbb {T}} {M_{\overline{\eta }}}=C_{w,\mathbb {T}}, \quad \eta \overline{v_1 z}\leqslant w\leqslant \eta \overline{{v_2}^\# z}. \end{aligned}$$

In addition, $v_2^\#\leqslant v_1$. Therefore

$$\begin{aligned} C= J^\# C_{w,\mathbb {T}} {M_{\eta }}C_{\overline{v_1 z},\mathbb {T}} {M_{\overline{\eta }}}&= J^\# C_{w,\mathbb {T}} C_{\overline{v_1 z},\mathbb {T}} {M_{\overline{\eta }^2}}\\&= J^\# M_{w v_1 z} M_{\overline{\eta }^2} = J^{\#} M_{w v_1 z{{\bar{\eta }}}^2}. \end{aligned}$$

Thus for a unimodular function $u= \overline{w v_1 z }\eta ^2$ we have

$$\begin{aligned} C= J^\#M_{{\bar{u}}}= C^\#_{u,\mathbb {T}}\quad \text {and}\quad {\bar{\eta }} \leqslant \bar{u}\leqslant {\bar{\eta }} v_1 \overline{v_2}^\#. \end{aligned}$$

Since $C= C^\#_{u,\mathbb {T}}$ is a conjugation thus u is symmetric. Hence (b) is shown.

For the implication (b) $\Rightarrow$ (a), assume that u is symmetric, unimodular function in $L^\infty (\mathbb {T})$ such that $\overline{\eta _1}{\eta _2^\#} \le {\bar{u}} \le \overline{\eta _1}{{\eta _2}^\#}\ {v_1}\overline{v_2}^{\#}$, which is equivalent to the condition $\eta _1\overline{\eta _2}^\#\ \overline{v_1} v_2^{\#} \le u \le \eta _1 \overline{\eta _2}^\#$. After multiplying all inequalities by $\overline{\eta _1} \eta _2^\# v_1$ we get $v_2^\# \le \overline{\eta _1} \eta _2^\# v_1 u \le v_1$. Therefore

$$\begin{aligned} \ker T_{v_1}\subset \ker T_{\overline{\eta _1} \eta _2^\# v_1 u}\subset \ker T_{v_2^\#}. \end{aligned}$$

Moreover, by the definition of $C^\#_{u, \mathbb {T}}$, [4, Proposition 2.3(3)], and Proposition 5.1, we obtain

$$\begin{aligned} M_{\overline{\eta _2}}C^\#_{u, \mathbb {T}}(\eta _1\ker T_{v_1})&=M_{\overline{\eta _2} u}J^\#(\eta _1\ker T_{v_1})\\&=J^\#M_{\eta _1{\overline{\eta _2}^\#} u^\#}(\ker T_{v_1})\\&=J^\# C_{\eta _1\overline{\eta _2}^\#u^\#\overline{z v_1}, \mathbb {T}}C_{\overline{z v_1}, \mathbb {T}}(\ker T_{v_1})\\&\subset J^\# C_{\eta _1\overline{\eta _2}^\#u^\#\overline{zv_1}, \mathbb {T}}(\ker T_{v_1})\\&\subset J^\# C_{\eta _1\overline{\eta _2}^\#\overline{u}\overline{z v_1}, \mathbb {T}}(\ker T_{\overline{\eta _1}\eta _2^\#v_1 u})\\&\subset J^\# \ker T_{\overline{\eta _1}\eta _2^\#v_1 u} \\&\subset J^\# \ker T_{v_2^\#} = \ker T_{v_2}, \end{aligned}$$

which completes the equivalence part. For the moreover part, notice that u must fulfill condition $1 \leqslant {\bar{u}} \eta _1 \overline{\eta _2^\#} \leqslant v_1 \overline{v_2}^\#$. Hence $v_1 \overline{v_2}^\#$ is inner. Now, applying Lemma 4.9 we get the claim. $\square$

Now, we prove that the only conjugation transforming Toeplitz kernel $\ker T_{v_1}$ into $\ker T_{v_2}$ is the $J^\#$ conjugation.

Theorem 6.4

Let $v_1,v_2 \in L^\infty (\mathbb {T})$ be unimodular functions and let C be an $M_z$-commuting conjugation on $L^2(\mathbb {T})$. Assume that $\ker T_{v_1} \ne \{0\}$. Then the following conditions are equivalent:

• (a) $C( \ker T_{v_1}) \subset \ker T_{v_2}$,

• (b) $v^\#_2\leqslant v_1$ and $C=J^\#$ up to the constant multiple.

Proof

Assuming (a) and applying Theorem 6.2 with $\eta _1=\eta _2=1$, we get that there is a symmetric unimodular function $u\in L^\infty (\mathbb {T})$ such that $1\leqslant {\bar{u}}\leqslant v_1\overline{ v_2}^\#$. It means that ${\bar{u}} \in H^\infty (\mathbb {T})$. Since u is symmetric, u have to be a constant. The reverse direction is clear because $J^\# \ker T_{v_1}\subset \ker T_{v^\#_1}\subset \ker T_{v_2}$ since $v_2\leqslant v_1^\#$. $\square$

From Theorem 6.4 we deduce the following.

Corollary 6.5

Let $v \in L^\infty (\mathbb {T})$ be a unimodular function such that $\ker T_{v} \ne \{0\}$. Then the following conditions are equivalent:

• (a) there exists an $M_z$-commuting conjugation C such that $C\ker T_v\subset \ker T_{v}$,

• (b) v is self-reflected.

Moreover, if (b) is fulfilled, then the unique conjugation C satisfying (a) is $C=J^\#$.

Proof

Assuming (a) and applying Theorem 6.4 we have $v^\#\leqslant v$ and $C=J^\#$. This implies that $\ker T_v=\ker T_{v^\#}$. Therefore v is self-reflected. $\square$

7 Examples of $M_z$-Commuting Conjugations

Now, we are going to give examples for theorems from Sect. 6.

Example 7.1

To illustrate Theorem 6.2, in the following example we would like to find all (if any) $M_z$-commuting conjugations C such that

$$\begin{aligned} C(\eta _1 \ker T_{v_1}) \subset \eta _2 \ker T_{v_2}. \end{aligned}$$

(7.1)

According to Remark 6.3 we will always take $\eta _2=1$. In all examples by $\delta _c$ we denote the Dirac delta measure at point $c\in \mathbb {T}$.

• 1. Let

  $$\begin{aligned} (v_1,v_2, \eta _1,\eta _2) = \Big ( {(B_{\bar{a}})^2\, \overline{S_{\delta _1 }}},\overline{ S_{\delta _1 }},B_a,1\Big ), \end{aligned}$$

  where $a\in \mathbb {D}\setminus \mathbb {R}$. Note that $\dim \ker T_{v_1}=\infty$ by [12, Proposition 5.19] and [7, Theorem 5.1]. In particular, $\ker T_{v_1}\ne \{0\}$. Moreover

  $$\begin{aligned} v_1\overline{v_2}^\#&=(B_{\bar{a}})^2\in H^\infty (\mathbb {T}),\\ \eta _1\, \eta _1^\#\,\overline{\eta _2}\, \overline{\eta _2}^\#&=B_aB_{\bar{a}}\leqslant (B_a)^2(B_{ {\bar{a}}})^2 =v_1\, v_1^\#\,\overline{v_2}\, \overline{v_2}^\# . \end{aligned}$$

  This shows that assumptions (i)–(iii) of Theorem 6.2 are fulfilled. Since $v_1\overline{ v_2}^\#=(B_{\bar{a}})^2$ is completely non-self-reflected, by Theorem 6.2, there exists only one symmetric unimodular function u such that

  $$\begin{aligned} \overline{\eta _1}{\eta _2^\#}=\overline{B_a} \leqslant {\bar{u}} \leqslant \overline{B_{ a}}(B_{\bar{a}})^2 =\overline{\eta _1}{\eta _2^\#}\ {v_1}\overline{v_2}^{\#}. \end{aligned}$$

  On the other hand, we see that only such function is $u=B_a \overline{B_{\bar{a}}}$. Hence the unique $M_z$-commuting conjugation satisfying (7.1) is $C^\#_{B_a\overline{B_{\bar{a}}},\mathbb {T}}$.

• 2. Let

  $$\begin{aligned} (v_1,v_2, \eta _1,\eta _2) = \Big ( B_a B_{\bar{a}}\,\overline{ S_{\delta _1 }},\overline{ S_{\delta _1 }},B_{\bar{a}},1\Big ), \end{aligned}$$

  where $a\in \mathbb {D}\setminus \mathbb {R}$. Then

  $$\begin{aligned} v_1\overline{ v_2}^\#&=B_a B_{\bar{a}}\in H^\infty (\mathbb {T}),\\ \eta _1\, \eta _1^\#\,\overline{\eta _2}\, \overline{\eta _2}^\#&=B_aB_{\bar{a}} \leqslant (B_a)^2 (B_{\bar{a}})^2= v_1\, v_1^\#\,\overline{v_2}\, \overline{v_2}^\#. \end{aligned}$$

  Now (i)–(iii) are also fulfilled. The function $v_1\overline{ v_2}^\#=B_aB_{\bar{a}}$ is not completely non-self-reflected. On the other hand, we see that there are only two symmetric unimodular functions u such that

  $$\begin{aligned} \overline{\eta _1}{\eta _2^\#}=\overline{B_{\bar{a}}} \leqslant {\bar{u}} \leqslant B_a =\overline{\eta _1}{\eta _2^\#}\ {v_1}\overline{v_2}^{\#}. \end{aligned}$$

  Namely, $u\equiv 1$ or $u=\overline{B_a} B_{\bar{a}}$. Hence, the only $M_z$-commuting conjugations fulfilling (7.1) are $J^{\#}$ and $C^{\#}_{\overline{B_a}{B_{\bar{a}}},\mathbb {T}}$.

Example 7.2

Note that if $\eta _1$, $\overline{\eta _2}$ are inner, then condition (ii) of Theorem 6.2 is always fulfilled. On the other hand, for $a\in \mathbb {D}\cap \mathbb {R}$ let

$$\begin{aligned} \eta _1(z)=\left\{ \begin{array}{cc} B_a(z), &{} \text { if } \Im z \geqslant 0, \\ 1, &{} \text { if } \Im z < 0, \end{array} \right. , \text { and } \eta _2 =1. \end{aligned}$$

Then $\eta _1\, \eta _1^\#\,\overline{\eta _2}\, \overline{\eta _2}^\#=B_a$ and (ii) is not valid. Thus, an $M_z$-commuting conjugation fulfilling (7.1) never exists regardless how $v_1, v_2$ are defined.

Example 7.3

Let

$$\begin{aligned} v_1=\overline{S_{s_1\delta _a }S_{t_1\delta _{\bar{a}} }},\quad v_2=\overline{S_{s_2\delta _a }S_{t_2\delta _{\bar{a}} }} \end{aligned}$$

where $a\in \mathbb {T}\,\setminus \,\mathbb {R}$, $0\leqslant s_1, t_1, s_2, t_2 \leqslant 1$ and $s_1^2+t_1^2 >0$. This guarantees that $\ker T_{v_1} \ne \{0\}$. Again, our goal is to find all (if any) $M_z$-commuting conjugations C such that

$$\begin{aligned} C(\eta _1 \ker T_{v_1}) \subset \eta _2 \ker T_{v_2}. \end{aligned}$$

(7.2)

First, observe that

$$\begin{aligned} v_1\overline{ v_2}^\# = {\overline{S_{s_1\delta _a}} S_{t_2\delta _{a}} \overline{S_{t_1\delta _{\bar{a}}}} S_{s_2\delta _{ {\bar{a}}}}}. \end{aligned}$$

(7.3)

Hence, condition (i) is fulfilled if and only if $s_1\leqslant t_2$ and $t_1\leqslant s_2$.

We consider two following cases.

• (a) If $t_1=s_2$ and $t_2>s_1$, then $v_1\overline{ v_2}^\#=S_{(t_2-s_1)\delta _a }$ is completely non-self-reflected. Regardless which $\eta _1, \eta _2$ are considered, there is a unique conjugation C transferring $\eta _1\ker T_{v_1}$ into $\eta _2\ker T_{v_2}$. For example: if $\eta _1=S_{r\delta _{\bar{a}}},\, 0<r<t_2-s_1, \text { and }\eta _2=1$, then (ii) and (iii) are fulfilled and $C=C^\#_{u,\mathbb {R}}$ with symmetric unimodular function $u\in L^\infty (\mathbb {T})$ such that $\overline{S_{r\delta _{\bar{a}}}} \leqslant {\bar{u}}\leqslant \overline{S_{r\delta _{\bar{a}}}} S_{(t_2-s_1)\delta _a }$. Since u is symmetric, $u={S_{r\delta _{\bar{a}}}}\overline{S_{r\delta _{ a}}}$.

• (b) Let $t_2-s_1=s_2-t_1:=s>0$. Then

  $$\begin{aligned} v_1\overline{ v_2}^\#={S_{s\delta _a }S_{s\delta _{\bar{a}} }}. \end{aligned}$$

  Let $\eta _1=S_{s\delta _{ a}}, \eta _2=1$. Then (ii) and (iii) are fulfilled and $C=C^\#_{u,\mathbb {R}}$ with symmetric unimodular function $u\in L^\infty (\mathbb {T})$ such that $\overline{S_{s\delta _{ a}}} \leqslant {\bar{u}}\leqslant S_{s\delta _{\bar{a}} }$. Since u is symmetric, $u={S_{r\delta _{ a}}}\overline{S_{r\delta _{\bar{a}}}}$, $0\leqslant r\leqslant s$. Hence, C is $M_z$-commuting conjugation satisfying (7.2) if and only if $C=C_{u,\mathbb {T}}=M_uJ^\#$ for some $u=S_{r\delta _a }\overline{S_{r\delta _{\bar{a}} }}$, $0\leqslant r\leqslant s$. Note that there are infinitely many such conjugations.

Finally, we give an example for Theorem 6.4.

Example 7.4

Let, as in Example 7.3,

$$\begin{aligned} v_1=\overline{S_{s_1\delta _a }S_{t_1\delta _{\bar{a}} }},\quad v_2=\overline{S_{s_2\delta _a }S_{t_2\delta _{\bar{a}} }} \end{aligned}$$

where $a\in \mathbb {T}\,\setminus \,\mathbb {R}$, $0\leqslant s_1, t_1, s_2, t_2 \leqslant 1$ and $s_1^2+t_1^2 >0$. Consider inclusion

$$\begin{aligned} C( \ker T_{v_1}) \subset \ker T_{v_2}. \end{aligned}$$

(7.4)

By (7.3), there is an $M_z$-commuting conjugation fulfilling (7.4) if and only if $s_1\leqslant t_2$ and $t_1\leqslant s_2$. In each case $C=J^\#$.