One-component inner functions

TL;DR

This paper identifies specific classes of inner functions in $H^$ with connected sublevel sets, highlighting their significance in operator theory.

Contribution

It explicitly characterizes several classes of one-component inner functions with connected level sets in the unit disk.

Findings

Identified classes of inner functions with connected level sets.

Connected level sets occur for specific inner functions.

Highlights importance of one-component inner functions in operator theory.

Abstract

We explicitely unveil several classes of inner functions $u$ in $H^\infty$ with the property that there is $\eta\in ]0,1[$ such that the level set $\Omega_u(\eta):=\{z\in\mathbb D: |u(z)|<\eta\}$ is connected. These so-called one-component inner functions play an important role in operator theory.

Equations

$${\Omega }_u(\eta )=\{z\in \mathbb{D}:|u(z)|<\eta \}$$

$$\underset{z\in \mathbb{D}}{\sup }\left|\frac{1-\overline{u(a)}u(z)}{1-\overline{a}z}\right|\le C\frac{1-|u(a){|}^2}{1-|a{|}^2}.$$

$$S(z)=\exp \left(-\frac{1+z}{1-z}\right),$$

$$D_{\eta }:=\left\{z\in \mathbb{D}:\left|z-\frac{L}{L+1}\right|<\frac{1}{L+1}\right\},L:=\log \frac{1}{\eta },$$

$$\partial H\subseteq \partial {\Omega }_0.$$

$$\mbox{$|f|<\eta$ on $\partial D$}.$$

$$\partial D\stackrel{(\text{ref}hcompo)}{\subseteq }\partial {\Omega }_0\cap \mathbb{D}\subseteq \{z\in \mathbb{D}:|f(z)|=\eta \}.$$

$$\underset{\genfrac{}{}{0ex}{}{\xi \to x}{x\in \partial {\Omega }_0\setminus E}}{\mathrm{limsup}}|u(\xi ){|}^{-1}={\eta }^{-1}.$$

$$\rho (z,w)=\left|\frac{z-w}{1-\overline{z}w}\right|$$

$$D_{\rho }(z_0,r)=\{z\in \mathbb{D}:\rho (z,z_0)<r\}$$

$$G\subseteq \{z\in \mathbb{D}:|B(z)|<\eta \}={\Omega }_B(\eta ),$$

$$|B(z)|=\rho (B(z),B(z_n))\le \rho (z,z_n)<\sigma .$$

$$\rho (z,Z(B))\ge \sigma \text{for every}z\in {\Omega }_0.$$

$$D_{\rho }(z_n,r)=\{z\in \mathbb{D}:\rho (z,z_n)<\epsilon \}$$

$$0<\epsilon <\eta \frac{\delta -\eta }{1-\delta \eta }.$$

$$\delta (B)=\underset{n\in \mathbb{N}}{\inf }(1-|z_n{|}^2)|B^{\prime }(z_n)|\ge \delta ,$$

$$\{z\in \mathbb{D}:|B(z)|<\epsilon \}\subseteq \{z\in \mathbb{D}:\rho (z,Z(B))<\eta \}\subseteq \{z\in \mathbb{D}:|B(z)|<\eta \}.$$

$$G:=\bigcup _n{D}_{\rho }(z_n,\sigma )$$

$$G\subseteq \{z\in \mathbb{D}:|b(z)|<\sigma \}=:\Omega .$$

$$\rho (z,Z(b))\ge \sigma \text{for every}z\in {\Omega }_0.$$

$$0<\eta <\min \{(1-\sqrt{1-{\delta }^2})/\delta ,\sigma \},$$

$$0<\epsilon <\eta \frac{\delta -\eta }{1-\delta \eta }.$$

$$\rho (1-2^{-n},1-2^{-n-1})=\frac{2^{-n}-2^{-n-1}}{2^{-n}+2^{-n-1}+2^{-n}{2}^{-n-1}}=\frac{1}{3+2^{-n}}.$$

$$0<{\eta }_1\le \rho (x_n,x_{n+1})\le {\eta }_2<1.$$

$$\underset{n}{\sup }\frac{1-x_{n+1}}{1-x_n}\le s<1$$

$$\underset{n}{\inf }\rho (x_n,x_{n+1})\ge r>0.$$

$$r_n<r_{n+1},{\theta }_{n+1}<{\theta }_n,$$

$$r_n\nearrow 1,{\theta }_n\searrow 0,$$

$$0<{\eta }_1\le \rho (z_n,z_{n+1})\le {\eta }_2<1.$$

$$x_j:=\frac{3^j-1}{3^j+1}=1-\frac{2}{3^j+1}.$$

Full text

Dedicated to the memory of Vadim Tolokonnikov

18.3.2024

Introduction

Definition 0.1**.**

An inner function $u$ in $H^{\infty}$ is said to be a one-component inner function if there is $\eta\in]0,1[$ such that the level set (also called sublevel set or filled level set) $\Omega_{u}(\eta):=\{z\in{\mathbb{D}}:|u(z)|<\eta\}$ is connected.

One-component inner functions, the collection of which we denote by $\mathfrak{I}_{c}$, were first studied by B. Cohn [10] in connection with embedding theorems and Carleson-measures. It was shown in [10, p. 355] for instance that arclength on $\{z\in{\mathbb{D}}:|u(z)|=\varepsilon\}$ is such a measure whenever

[TABLE]

is connected and $\eta<\varepsilon<1$.

A thorough study of the class $\mathfrak{I}_{c}$ was given by A.B. Aleksandrov [1] who showed the interesting result that $u\in\mathfrak{I}_{c}$ if and only if there is a constant $C=C(u)$ such that for all $a\in{\mathbb{D}}$

[TABLE]

Many operator-theoretic applications are given in [1, 2, 7, 3]. In our paper here we are interested in explicit examples, which are somewhat lacking in literature. For example, if $S$ is the atomic inner function, which is given by

[TABLE]

then all level sets $\Omega_{S}(\eta)$, $0<\eta<1$ are connected, because these sets coincide with the disks

[TABLE]

which are tangential to the unit circle at $p=1$.

The scheme of our note here is as follows: in section 1 we prove a general result on level sets which will be the key for our approach to the problem of unveiling classes of one-component inner functions. Then in section 2 we first present with elementary geometric/function theoretic methods several examples and then we use Aleksandrov’s criterion to achieve this goal. For instance, we prove that $BS,B\circ S$ and $S\circ B$ are in $\mathfrak{I}_{c}$ whenever $B$ is a finite Blaschke product. Considered are also interpolating Blaschke products. It will further be shown that, under the supremum norm, $\mathfrak{I}_{c}$ is an open subset of the set of all inner functions and multiplicatively closed. In the final section we give counterexamples.

1. Level sets

We first begin with a topological property of the class of general level sets. Although statement (1) is “well-known” (the earliest appearance seems to be in [26, Theorem VIII,  31]), we could nowhere locate a proof. The argument that the result is a simple and direct consequence of the maximum principle is, in our viewpoint, not tenable.

Lemma 1.1**.**

Given a non-constant inner function $u$ in $H^{\infty}$ and $\eta\in\;]0,1[$, let $\Omega:=\Omega_{u}(\eta)=\{z\in{\mathbb{D}}:|u(z)|<\eta\}$ be a level set. Suppose that $\Omega_{0}$ is a component (=maximal connected subset) of $\Omega$. Then

1. (1)

$\Omega_{0}$* is a simply connected domain; that is, ${\mathbb{C}}\setminus\Omega_{0}$ has no bounded components 111A shorter proof can be given by using the advanced definition that a domain $G$ in ${\mathbb{C}}$ is simply connected if every curve in $G$ is contractible in $G$, or equivalently, if for every Jordan curve $J$ in $G$ the interior of $J$ belongs to $G$. That depends though on the Jordan curve theorem..* 2. (2)

$\inf_{\Omega_{0}}|u|=0$.

Proof.

We show that (1) holds for every holomorphic function $f$ in ${\mathbb{D}}$; that is if $\Omega_{0}$ is a component of the level set $\Omega_{f}(\eta)$, $\eta>0$, then it is a simply connected domain 222 This proof, as well as two different ones, including the one mentioned in footnote 1, stem from the forthcoming book manuscript [22] of the second author together with R. Rupp.. Note that each component $\Omega_{0}$ of the open set $\Omega_{f}(\eta)$ is an open subset of ${\mathbb{D}}$. We may assume that $\eta$ is chosen so that $\{z\in{\mathbb{D}}:|f(z)|=\eta\}\not=\emptyset$.

Suppose, to the contrary, that $D$ is a bounded component of ${\mathbb{C}}\setminus\Omega_{0}$. Note that $D$ is closed in ${\mathbb{C}}$. Then, necessarily, $D$ is contained in ${\mathbb{D}}$, because the unique unbounded complementary component of $\Omega_{0}$ contains $\{z\in{\mathbb{C}}:|z|\geq 1\}$. Hence $D$ is a compact subset of ${\mathbb{D}}$. Let $G:=\Omega_{0}^{*}$ be the simply-connected hull of $\Omega_{0}$; that is the union of $\Omega_{0}$ with all bounded complementary components of $\Omega_{0}$. Note that $G$ is open because it coincides with the complement of the unique unbounded complementary component of $\Omega_{0}$. Then, by definition of the simply connected hull, $D\subseteq G$. Now if $H$ is any bounded complementary component of $\Omega_{0}$ then (as it was the case for $D$) $H$ is a compact subset of ${\mathbb{D}}$ and so $\partial H\subseteq{\mathbb{D}}$. Moreover,

[TABLE]

In fact, given $z_{0}\in\partial H$, let $U$ be a disk centered at $z_{0}$. Then $U\cap\Omega_{0}\not=\emptyset$, since otherwise $U\cup H$ would be a connected set strictly bigger than $H$ and contained in the complement of $\Omega_{0}$; a contradiction to the maximality of $H$. Since $z_{0}\in\partial H\subseteq H\subseteq{\mathbb{C}}\setminus\Omega_{0}$, we conclude that $z_{0}\in\partial\Omega_{0}$.

Now $\partial H\subseteq\partial\Omega_{0}$ and $\Omega_{0}\subseteq\Omega_{f}(\eta)$ imply that $|f|\leq\eta$ on $\partial H$, and so, by the maximum principle, $|f|\leq\eta$ on $H$. Consequently, $|f|\leq\eta$ on $G$. By the local maximum principle, $|f|<\eta$ on $G$. Since $\partial D\subseteq D\subseteq G$,

[TABLE]

On the other hand,

[TABLE]

Note that the second inclusion follows from the fact that if $|f(z_{0})|<\eta$ for $z_{0}\in\partial\Omega_{0}\cap{\mathbb{D}}$, then $\Omega_{0}$ would no longer be a maximal connected subset of $\Omega_{f}(\eta)$. Hence $|f|=\eta$ on $\partial D$. This is a contradiction to (1.2). Thus $\Omega_{0}$ is a simply connected domain.

(2) If $\overline{\Omega}_{0}\subseteq{\mathbb{D}}$, then, due to $\partial\Omega_{0}\subseteq\{z\in{\mathbb{D}}:|u(z)|=\eta\}$, we obtain from the minimum principle that $u$ must have a zero in $\Omega_{0}$. Now let $E:=\overline{\Omega}_{0}\cap\partial{\mathbb{D}}\not=\emptyset$. In view of achieving a contradiction, suppose that $u$ is bounded away from zero in $\Omega_{0}$. Then $1/|u|$ is subharmonic and bounded in $\Omega_{0}$ and

[TABLE]

Since $u$ is an inner function, $E$ has linear measure zero (by [5, Theorem 4.2]). The maximum principle for subharmonic functions with few exceptional points (here on the set $E$; see [6] or [12]), now implies that $|u|^{-1}\leq\eta^{-1}$ on $\Omega_{0}$. But $|u|<\eta$ on $\Omega$ is a contradiction. We conclude that $\inf_{\Omega_{0}}|u|=0$.

∎

Lemma 1.2**.**

[10]** Let $u$ be an inner function. Then the connectedness of $\Omega_{u}(\eta)$ implies the one of $\Omega_{u}(\eta^{\prime})$ for every $\eta^{\prime}>\eta$.

Proof.

Because $\Omega_{u}(\eta)$ is connected and $\Omega_{u}(\eta)\subseteq\Omega_{u}(\eta^{\prime})$, $\Omega_{u}(\eta)$ is contained in a unique component $U_{1}(\eta^{\prime})$ of $\Omega_{u}(\eta^{\prime})$. Suppose that $U_{0}(\eta^{\prime})$ is a second component of $\Omega_{u}(\eta^{\prime})$. Then $|u|\geq\eta$ on $U_{0}(\eta^{\prime})$, because $U_{0}(\eta^{\prime})$ is disjoint with $U_{1}(\eta^{\prime})$ and hence with $\Omega_{u}(\eta)$. By Lemma 1.1 though, $\inf_{U_{0}(\eta^{\prime})}|u|=0$; a contradiction. Thus $\Omega_{u}(\eta^{\prime})$ is connected. ∎

2. Explicit examples of one-component inner functions

Let

[TABLE]

be the pseudohyperbolic distance of $z$ to $w$ in ${\mathbb{D}}$ and

[TABLE]

the associated disks, $0<r<1$. Here is a first class of examples of functions in $\mathfrak{I}_{c}$. Although the next Proposition must be known (in view of A.B. Aleksandrov’s criterion [1]), see 2.12 below), we include a simple geometric proof for the reader’s convenience.

Proposition 2.1**.**

Let $B$ be a finite Blaschke product. Then $B\in\mathfrak{I}_{c}$.

Proof.

Denote by $z_{1},\dots,z_{N}$ the zeros of $B$, multiplicities included. Let $\eta\in\;]0,1[$ be chosen so close to 1 that $G:=\bigcup_{n=1}^{N}D_{\rho}(z_{n},\eta)$ is connected (for example by choosing $\eta$ so that $z_{j}\in D_{\rho}(z_{1},\eta)$ for all $j$). Now

[TABLE]

because $z\in G$ implies that for some $n$,

[TABLE]

Since $G$ is connected, there is a unique component $\Omega_{1}$ of $\Omega$ containing $G$. In particular, $Z(B)\subseteq G\subseteq\Omega_{1}$. If, in view of achieving a contradiction, we suppose that $\Omega:=\Omega_{B}(\eta)$ is not connected, there is a component $\Omega_{0}$ of $\Omega$ which is disjoint with $\Omega_{1}$, and so with $G$. In particular,

[TABLE]

Since $\overline{\Omega}_{0}\subseteq\overline{\Omega_{B}(\eta)}\subseteq{\mathbb{D}}$, and $|B|=\eta$ on $\partial\Omega_{0}$, we deduce from the minimum principle that $\Omega_{0}$ contains a zero of $B$; a contradiction. ∎

We now generalize this result to a class of interpolating Blaschke products. Recall that a Blaschke product $b$ with zero set/sequence $\{z_{n}:n\in{\mathbb{N}}\}$ is said to be an interpolating Blaschke product if $\delta(b):=\inf(1-|z_{n}|^{2})|b^{\prime}(z_{n})|>0$. If $b$ is an interpolating Blaschke product  then, for small $\varepsilon$, the pseudohyperbolic disks

[TABLE]

are pairwise disjoint. Moreover, by Hoffman’s Lemma (see below and also [19]), for any $\eta\in]0,1[$, $b$ is bounded away from zero on $\{z\in{\mathbb{D}}:\rho(z,Z(b))\geq\eta\}$.

Theorem 2.2** (Hoffman’s Lemma).**

([18] p. 86, 106 and [13] p. 404, 310). Let $\delta,\eta$ and $\epsilon$ be real numbers, called Hoffman constants, satisfying $0<~{}\delta<~{}1$, $0<\eta<(1-\sqrt{1-\delta^{2}})/\delta$, (that is, $0<\eta<\rho(\delta,\eta)$) and

[TABLE]

If $B$ is any interpolating Blaschke product with zeros $\{z_{n}:n\in{\mathbb{N}}\}$ such that

[TABLE]

then

the pseudohyperbolic disks $D_{\rho}(a,\eta)$ for $a\in Z(B)$ are pairwise disjoint. 2. (2)

The following inclusions hold:

[TABLE]

A large class of interpolating Blaschke products which are one-component inner functions now is given in the following result.

Theorem 2.3**.**

Let $b$ be an interpolating Blaschke product with zero set $\{z_{n}:n\in{\mathbb{N}}\}$. Suppose that for some $\sigma\in\;]0,1[$ the set

[TABLE]

is connected. Then $b$ is a one-component inner function. This holds in particular, if $\rho(z_{n},z_{n+1})<\sigma<1$ for all $n$; for example if $z_{n}=1-2^{-n}$.

Proof.

As in the proof of Proposition 2.1

[TABLE]

Since $G$ is assumed to be connected, there is a unique component $\Omega_{1}$ of $\Omega$ containing $G$. In particular, $Z(b)\subseteq G\subseteq\Omega_{1}$. Now, if we suppose that $\Omega$ is not connected, then there is a component $\Omega_{0}$ of $\Omega$ which is disjoint with $\Omega_{1}$, and so with $G$. In particular,

[TABLE]

Let $\delta:=\delta(b)$,

[TABLE]

[TABLE]

By Lemma 1.1, $\inf_{\Omega_{0}}|b|=0$. Thus, there is $z_{0}\in\Omega_{0}$ be so that $|b(z_{0})|<\varepsilon$. We deduce from Hoffman’s Lemma 2.2 that $\rho(z_{0},Z(b))<\eta<\sigma$. This is a contradiction to (2.2). We conclude that $\Omega$ must be connected. It is clear that the condition $\rho(z_{n},z_{n+1})<\sigma$ for every $n$ implies that $\bigcup_{n}D_{\rho}(z_{n},\sigma)$ is connected. For the rest, just note that

[TABLE]

∎

Corollary 2.4**.**

Let $B$ be a Blaschke product with increasing real zeros $x_{n}$ satisfying

[TABLE]

Then $b\in\mathfrak{I}_{c}$.

Proof.

Just use Theorem 2.3 and the fact that by the Vinogradov-Hayman-Newman theorem, $B$ is interpolating if and only if

[TABLE]

or equivalently

[TABLE]

∎

Using a result of Kam-Fook Tse [25], telling us that a sequence $(z_{n})$ of points contained in a Stolz angle (or cone) $\{z\in{\mathbb{D}}:|1-z|<C(1-|z|)\}$ is interpolating if and only if it is separated (meaning that $\inf_{n\not=m}\rho(z_{n},z_{m})>0$), we obtain:

Corollary 2.5**.**

Let $B$ be a Blaschke product whose zeros $(z_{n})$ are contained in a Stolz angle and are separated. Suppose that $\rho(z_{n},z_{n+1})\leq\eta<1$. Then $B\in\mathfrak{I}_{c}$.

Similarily, using a result by M. Weiss [27, Theorem 3.6] and its refinement in [4, Theorem B], we also obtain the following assertion for sequences that may be tangential at 1 (see also Wortman [28]).

Corollary 2.6**.**

Let $B$ be a Blaschke product whose zeros $z_{n}=r_{n}e^{i\theta_{n}}$ satisfy:

[TABLE]

[TABLE]

[TABLE]

Then $B$ is an interpolating Blaschke product contained in $\mathfrak{I}_{c}$. This holds in paticular if the zeros are located on a convex curve in ${\mathbb{D}}$ with endpoint $1$ and satisfying (2.3).

Other classes of this type can be deduced from [14]. Here are two explicit examples of interpolating Blaschke products in $\mathfrak{I}_{c}$ whose zeros are given by iteration of the zero of a hyperbolic, respectively parabolic automorphism of ${\mathbb{D}}$. These functions appear, for instance, in the context of isometries on the Hardy space $H^{p}$ (see [8]).

Example 2.7**.**

$\bullet$*   Let $\displaystyle\varphi(z)=\frac{z-1/2}{1-(1/2)z}$. Then $\varphi$ is an hyperbolic automorphism with fixed points $\pm 1$. If $\varphi_{j}:=\underbrace{\varphi\circ\cdots\circ\varphi}_{j-{\rm times}}$, then $\varphi_{j}\in{\rm Aut}({\mathbb{D}})$ and vanishes exactly at the point*

[TABLE]

This can readily be seen by using that $x_{j+1}=\varphi^{-1}(x_{j})$ and

[TABLE]

Since

[TABLE]

we deduce from Corollary 2.4 that the Blaschke product

[TABLE]

*associated with these zeros is in $\mathfrak{I}_{c}$.

$\bullet$* Let $\sigma\in{\rm Aut}({\mathbb{D}})$ and $\tau=\sigma\circ\varphi\circ\sigma^{-1}$. Then $\tau$ is also a hyperbolic automorphism fixing the points $\sigma(\pm 1)$, and where $\xi:=\sigma(1)$ is the Denjoy-Wolff point with $\tau^{\prime}(\xi)<1$. The zeros of the $n$-th iterate $\tau_{n}$ of $\tau$ are given by*

[TABLE]

By the grand iteration theorem [23, p.78], since $1$ is an attracting fixpoint with $(\varphi^{-1})^{\prime}(1)=1/3<1$, the sequence $(\varphi^{-1}_{n}(\sigma^{-1}(0)))$ converges nontangentially to $1$. Hence the points $z_{n}$ are located in a cone with cusp at $\xi$. Moreover, if $n>k$ and $a=\sigma^{-1}(0)$,

[TABLE]

Thus, $\rho(z_{n},z_{n+1})=\rho(\varphi(a),a)$ for all $n$ and $\inf_{n\not=k}\rho(z_{n},z_{k})>0$. Now $(z_{n})$ is a Blaschke sequence 333 This also follows form the inequalities $1-|\sigma(\xi_{n})|^{2}=\frac{(1-|a|^{2})(1-|\xi_{n}|^{2})}{|1-\overline{a}\xi_{n}|^{2}}\leq\frac{1+|a|}{1-|a|}(1-|\xi_{n}|^{2})$ and $1-|\psi_{n}(a)|^{2}\leq\frac{1+|a|}{1-|a|}(1-|w_{n}|^{2})$, whenever $(w_{n})$ is a Blaschke sequence and $\psi_{n}(w_{n})=\sigma(a)=0$. ([23, Ex. 6, p. 85]); in fact, use d’Alembert’s quotient criterion and observe that by the Denjoy-Wolff theorem,

[TABLE]

Hence, by Corollary 2.5, $(z_{n})$ is an interpolating sequence (see also [11, p.80]) and the associated Blaschke product  $b=\prod_{n=1}^{\infty}e^{i\theta_{n}}\tau_{n}$ belongs to $\mathfrak{I}_{c}$ (here $\theta_{n}$ is chosen so that the $n$-th Blaschke factor is positive at the origin).

$\bullet$*   Let $\displaystyle\psi(z)=i\;\frac{z-\frac{1+i}{2}}{1-\frac{1-i}{2}\;z}$. Then $\psi$ is a parabolic automorphism with attracting fixed point $1$. The automorphism $\psi$ is conjugated to the translation $w\mapsto w+2$ on the upper half-plane (see figure 1) via the map $M(z)=i(1+z)/(1-z)$ and $\psi_{n}=M^{-1}\circ T_{n}\circ M$. The zeros of the $n$-th iterate $\psi_{n}$ of $\psi$ are given by*

[TABLE]

just use that $z_{n}=(M^{-1}\circ T_{n}^{-1}\circ M)(0)$. These zeros satisfy $\left|z_{n}-\frac{1}{2}\right|=\frac{1}{2}$ and of course also the Blaschke condition $\sum_{n=1}^{\infty}1-|z_{n}|^{2}<\infty$. Moreover,

[TABLE]

*Thus, by, Corollary 2.6, the Blaschke product associated with these zeros is interpolating and belongs to $\mathfrak{I}_{c}$.

Proposition 2.8**.**

Let $B$ be a finite Blaschke product or an interpolating Blaschke product with real zeros clustering at $p=1$. Then $f:=BS\in\mathfrak{I}_{c}$.

Proof.

i) Let $B$ be a finite Blaschke product. Chose $\eta\in\;]0,1[$ so close to 1 that the disk $D_{\eta}$ in (0.1), which coincides with the level set $\Omega_{S}(\eta)$, contains all zeros of $B$. Now $D_{\eta}=\Omega_{S}(\eta)\subseteq\Omega_{f}(\eta)$. Now $\Omega_{f}(\eta)$ must be connected, since otherwise there would be a component $\Omega_{0}$ of $\Omega_{f}(\eta)$ disjoint from the component $\Omega_{1}$ containing $D_{\eta}$. But $f$ is bounded away from zero outside $D_{\eta}$; hence $f=BS$ is bounded away from zero on $\Omega_{0}$. This is a contradiction to Lemma 1.1 (2).

ii) If $B$ is an interpolating Blaschke product with zeros $(z_{n})$, then, by Hoffman’s Lemma 2.2, $B$ is bounded away from zero outside $R:=\bigcup D_{\rho}(z_{n},\varepsilon)$ for every $\varepsilon\in\;]0,1[$. Now, if the zeros of $B$ are real, and bigger than $-\sigma$ for some $\sigma\in]0,1[$, this set $R$ is contained in a cone with cusp at 1 and aperture-angle strictly less than $\pi$ (see for instance [21]). Hence $R$ is contained in $D_{\eta}$ for all $\eta$ close to $1$. Thus, as above, we can deduce that $\Omega_{BS}(\eta)$ is connected. ∎

The previous result shows, in particular, that certain non one-component inner functions (for example a thin Blaschke product with positive zeros, see Corollary 3.1), can be multiplied by a one-component inner function into $\mathfrak{I}_{c}$. In particular, $uv\in\mathfrak{I}_{c}$ does not imply that $u$ and $v$ belong to $\mathfrak{I}_{c}$. The reciprocal, though, is true: that is $\mathfrak{I}_{c}$ itself is stable under multiplication, as we are going to show below.

Proposition 2.9**.**

Let $u,v$ be two inner functions in $\mathfrak{I}_{c}$. Then $uv\in\mathfrak{I}_{c}$.

Proof.

Let $\Omega_{u}(\eta)$ and $\Omega_{v}(\eta^{\prime})$ be two connected level sets. Due to monotonicity (Lemma 1.2), and the fact that $\bigcup_{\lambda\in[\lambda_{0},1[}\Omega_{f}(\lambda)={\mathbb{D}}$, we may assume that $\sigma$ satisfies

[TABLE]

and is so close to 1 that $0\in\Omega_{u}(\sigma)\cap\Omega_{v}(\sigma)\not=\emptyset$. Hence $U:=\Omega_{u}(\sigma)\cup\Omega_{v}(\sigma)$ is connected. Now

[TABLE]

If we suppose that $\Omega_{uv}(\sigma)$ is not connected, then there is a component $\Omega_{0}$ of $\Omega_{uv}(\sigma)$ which is disjoint from $U$. In particular, $u$ and $v$ are bounded away from zero on $\Omega_{0}$. This contradicts Lemma 1.1 (2). Hence $\Omega_{uv}(\sigma)$ is connected and so $uv\in\mathfrak{I}_{c}$. ∎

Theorem 2.10**.**

The set of one-component inner functions is open inside the set of all inner functions (with respect to the uniform norm topoplogy).

Proof.

Let $u\in\mathfrak{I}_{c}$. Then, by Lemma 1.2, $\Omega_{u}(\eta)$ is connected for all $\eta\in[\eta_{0},1[$. Choose $0<\varepsilon<\min\{\eta,1-\eta\}$ and let $\Theta$ be an inner function with $||u-\Theta||<\varepsilon$. We claim that $\Theta\in\mathfrak{I}_{c}$, too. To this end we note that

[TABLE]

where $0<\eta-\varepsilon<\eta+\varepsilon<1$. As usual, if we suppose that $\Omega_{\Theta}(\eta+\varepsilon)$ is not connected, then there is a component $\Omega_{0}$ of $\Omega_{\Theta}(\eta+\varepsilon)$ which is disjoint from the connected set $\Omega_{u}(\eta)$, hence disjoint with $\Omega_{\Theta}(\eta-\varepsilon)$. In other words, $|\Theta|\geq\eta-\varepsilon>0$ on $\Omega_{0}$. This contradicts Lemma 1.1 (2). Hence $\Omega_{\Theta}(\eta+\varepsilon)$ is connected and so $\Theta\in\mathfrak{I}_{c}$. ∎

Next we look at right-compositions of $S$ with finite Blaschke products. We first deal with the case where $B(z)=z^{2}$.

Example 2.11**.**

The function $S(z^{2})$ is a one-component inner function.

Proof.

Let $\Omega_{S}(\eta)$ be the $\eta$-level set of $S$. Then, as we have already seen, this is a disk tangent to the unit circle at the point 1. We may choose $0<\eta<1$ so close to $1$ that [math] belongs to $\Omega_{S}(\eta)$. Let $U=\Omega_{S}(\eta)\setminus]-\infty,0]$. Then $U$ is a simply connected slitted disk on which exists a holomorphic square root $q$ of $z$. The image of $U$ under $q$ is a simply connected domain $V$ in the semi-disk $\{z:|z|<1,{\rm Re}\;z>0\}$. Let $V^{*}$ be its reflection along the imginary axis. Then $E:=\overline{V^{*}\cup V}$ is mapped by $z^{2}$ onto the closed disk $\overline{\Omega_{S}(\eta)}$. Then $E\setminus\partial E$ coincides with $\Omega_{S(z^{2})}(\eta)$.

∎

Using Aleksandrov’s criterion (see below), we can extend this by replacing $z^{2}$ with any finite Blaschke product. Recall that the spectrum $\rho(\Theta)$ of an inner function $\Theta$ is the set of all boundary points $\zeta$ for which $\Theta$ does not admit a holomorphic extension; or equivalently, for which $Cl(\Theta,\zeta)=\overline{\mathbb{D}}$, where

[TABLE]

is the cluster set of $\Theta$ at $\zeta$ (see [13, p. 80]).

Theorem 2.12** (Aleksandrov).**

[1, Theorem 1.11 and Remark 2, p. 2915]** Let $\Theta$ be an inner function. The following assertions are equivalent:

1. (1)

$\Theta\in\mathfrak{I}_{c}$. 2. (2)

There is a constant $C>0$ such that for every $\zeta\in\mathbb{T}\setminus\rho(\Theta)$ we have

[TABLE]

and

[TABLE]

Note that, due to this theorem, $\Theta\in\mathfrak{I}_{c}$ necessarily implies that $\rho(\Theta)$ has measure zero.

Proposition 2.13**.**

Let $B$ be a finite Blaschke product. Then $S\circ B\in\mathfrak{I}_{c}$.

Proof.

Let us note first that $\rho(S\circ B)=B^{-1}(\{1\})$. Since the derivative of $B$ on the boundary never vanishes (due to

[TABLE]

$B$ is schlicht in a neighborhood of $1$. The angle conservation law now implies that for $\zeta\in B^{-1}(1)$ the curve $r\mapsto B(r\zeta)$ stays in a Stolz angle at 1 in the image space of $B$. Hence $\liminf_{r\to 1}S(B(r\zeta))=0$ for $\zeta\in\rho(S\circ B)$. Now let us calculate the derivatives:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Hence, for $\zeta\in\mathbb{T}\setminus\rho(S\circ B)$, $|B(\zeta)|=1$ , but $\xi:=B(\zeta)\not=1$, and so, by (2.4),

[TABLE]

∎

Corollary 2.14**.**

Let $S_{\mu}$ be a singular inner function with finite spectrum $\rho(S_{\mu})$. Then $S_{\mu}\in\mathfrak{I}_{c}$.

Proof.

Since $S$ is the universal covering map of ${\mathbb{D}}\setminus\{0\}$, each singular inner function $S_{\mu}$ writes as $S_{\mu}=S\circ v$ for some inner function $v$. Since $\rho(S_{\mu})$ is finite, $v$ necessarily is a finite Blaschke product. (This can also be seen from [15, Proof of Theorem 2.2]). The assertion now follows from Proposition 2.13. ∎

Note that this result also follows in an elementary way from Proposition 2.9 and the fact that every such $S_{\mu}$ is a finite product of powers of the atomic inner function $S$. We now consider left-compositions with finite Blaschke products.

Proposition 2.15**.**

Let $\Theta$ be a one-component inner function. Then each Frostman shift $(a-\Theta)/(1-\overline{a}\Theta)\in\mathfrak{I}_{c}$, too. Here $a\in{\mathbb{D}}$.

Proof.

Let $\tau(z)=(a-z)/(1-\overline{a}z)$. Then $\rho(\tau\circ\Theta)=\rho(\Theta)$. As above,

[TABLE]

for every $\zeta\in\rho(\tau\circ\Theta)$. Now

[TABLE]

from which we easily deduce the first and second derivatives. By using the formulas 2, we obtain

[TABLE]

Hence, the assumption $\Theta\in\mathfrak{I}_{c}$ now yields (via Aleksandrov’s criterion 2.12) that $\sup_{\zeta\in\rho(\tau\circ\Theta)}A(\zeta)<\infty$. Thus $\tau\circ\Theta\in\mathfrak{I}_{c}$.

∎

Corollary 2.16**.**

*Given $a\in{\mathbb{D}}\setminus\{0\}$, the interpolating Blaschke products

$(S-a)/(1-\overline{a}S)$ belong to $\mathfrak{I}_{c}$.*

This also follows from Corollary 2.6 by noticing that the $a$-points of $S$ are located on a disk tangent at $1$ and that the pseudohyperbolic distance between two consecutive ones is constant (see [20]). There it is also shown that the Frostman shift $(S-a)/(1-\overline{a}S)$ is an interpolating Blaschke product.

Corollary 2.17**.**

Let $B$ be a finite Blaschke product and $\Theta\in\mathfrak{I}_{c}$. Then $B\circ\Theta\in\mathfrak{I}_{c}$.

Proof.

This is a combination of Propositions 2.15 and 2.9. ∎

3. Inner functions not belonging to $\mathfrak{I}_{c}$

Here we present a class of Blaschke products that are not one-component inner functions. Recall that a Blaschke product $b$ with zero-sequence $(z_{n})$ is thin if

[TABLE]

It was shown by Tolokonnikov [24, Theorem 2.3] that $b$ is thin if and only if

[TABLE]

Corollary 3.1**.**

Thin Blaschke products are never one-component inner functions.

Proof.

Let $\varepsilon\in\;]0,1[$ be arbitrary close to $1$. Choose $\eta>0$ and $\delta>0$ so close to 1 so that

[TABLE]

By deleting finitely many zeros, say $z_{1},\dots,z_{N}$ of $b$, we obtain a tail $b_{N}$ such that $(1-|z_{n}|^{2})|b_{N}^{\prime}(z_{n})|\geq\delta$ for every $n>N$. Hence, by Theorem 2.2,

[TABLE]

and the disks $D(z_{n},\eta)$ are pairwise disjoint. This implies that the level set $\{z\in{\mathbb{D}}:|b_{N}(z)|<\varepsilon\}$ is not connected. Now choose $r$ so close to 1 that

[TABLE]

for every $z$ with $r\leq|z|<1$. We show that the level set $\{|b|<\varepsilon^{2}\}$ is not connected. In fact, for some $r\leq|z|<1$ we have $|b(z)|<\varepsilon^{2}$, then

[TABLE]

Hence

[TABLE]

Since the disks $D_{\rho}(z_{n},\eta)$ are pairwise disjoint if $n>N$, we are done. ∎

Corollary 3.2**.**

No finite product $B$ of thin interpolating Blaschke products belongs to $\mathfrak{I}_{c}$.

Proof.

Let $\varepsilon\in\;]0,1[$ be arbitrary close to 1. By Corollary 3.1, if $b_{j}$, $(j=1,2)$, are two thin Blaschke products with zero-sequence $(z_{n}^{(j)})_{n}$,

[TABLE]

for suitable $\eta$, the disks $D_{\rho}(z^{(j)}_{n},\eta)$, being pairwise disjoint for $n$ large. Since $\lim_{n}\rho(z_{n}^{(j)},z_{n+1}^{(j)})=1$, we see that a disk $D_{\rho}(z^{(1)}_{n},\eta)$ can meet at most one disk $D_{\rho}(z^{(2)}_{m},\eta)$ for $n$ large. Hence

[TABLE]

where the set on the right hand side obviously is disonnected. The general case works via induction. ∎

Remark**.**

The conditions

[TABLE]

or equivalently

[TABLE]

are not sufficient to guarantee that the interpolating Blaschke product $b$ is a one-component inner function.

Proof.

Take $z_{2n}=1-n^{-n}$ and $z_{2n+1}=1-(n^{-n}+n^{-n})$. Then $(z_{2n})$ and $(z_{2n+1})$ are (thin) interpolating sequences by [16, Corollary 2.4]. Using with $a=n^{-n}$ and $b=2a$ the identity

[TABLE]

we conclude that

[TABLE]

and so the union $(z_{n})$ is an interpolating sequence satisfying (3.3). By Corollary 3.2, the Blaschke product formed with the zero-sequence $(z_{n})$ is not in $\mathfrak{I}_{c}$.

∎

Using the following theorem in [5], we can exclude a much larger class of Blaschke products from being one-component inner functions:

Theorem 3.3** (Berman).**

Let $u$ be an inner function. Then, for every $\varepsilon\in\;]0,1[$, all the components of the level sets $\{z\in{\mathbb{C}}:|u(z)|<\varepsilon\}$ have compact closures in ${\mathbb{D}}$ if and only if $u$ is a Blaschke product and

[TABLE]

In particular this condition is satisfied by finite products of thin Blaschke products (see [17, Proposition 2.2]) as well as by the class of uniform Frostman Blaschke products

[TABLE]

Note that this Frostman condition implies that the associated Blaschke product has radial limits of modulus one everywhere [9, p. 33]. As a byproduct of Theorem 2.3 we therefore obtain

Corollary 3.4**.**

If $b$ is a uniform Frostman Blaschke product with zeros $(z_{n})$ clustering at a single point, then $\limsup_{\rho}(z_{n},z_{n+1})=1$.

Questions 3.5**.**

To conclude, we would like to ask two questions and present three problems*:*

1. (1)

Can every inner function $u$ whose boundary spectrum $\rho(u)$ has measure zero, be multiplied by a one-component inner function into $\mathfrak{I}_{c}$? 2. (2)

Let $S_{\mu}$ be a singular inner function with countable spectrum. Give a characterization of those measures $\mu$ such that $S_{\mu}\in\mathfrak{I}_{c}$. Do the same for singular continuous measures. 3. (3)

In terms of the zeros, give a characterization of those interpolating Blaschke products that belong to $\mathfrak{I}_{c}$. 4. (4)

Does the Blaschke product $B$ with zeros $z_{n}=1-n^{-2}$ belong to $\mathfrak{I}_{c}$?

4. Acknowledgement

We thank Rudolf Rupp and Robert Burckel for their valuable comments concerning Lemma 1.1 (1), the proof of which was orginally developed for the upcoming monograph [22].