One-component inner functions

Dedicated to the memory of Vadim Tolokonnikov.

Abstract

We explicitly 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.

1 Background

Definition 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 Cohn [10] in connection with embedding theorems and Carleson measures. It was shown that [for instance, [10], p. 355] arclength on $\{z\in {\mathbb D}: |u(z)|=\varepsilon \}$ is such a measure whenever

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

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

A thorough study of the class $\mathfrak I_c$ was given by 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 $a\in {\mathbb D}$

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

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

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

then all level sets $\Omega _S(\eta )$, $0<\eta <1$ are connected, because these sets coincide with the disks

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

(1)

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

The scheme of our note here is as follows: in Sect. 2, 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 Sect. 3, we first present several examples with elementary geometric/function theoretic methods 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.

2 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 2

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) $\Omega _0$ is a simply connected domain; that is, ${ \mathbb C}\setminus \Omega _0$ has no bounded components. ¹

• (2) $\it{\inf} _{\Omega _0} |u|=0$.

Proof

We show that item (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.² 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 that, on the contrary, 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 $\{z\in { \mathbb C}: |z|\ge 1\}$. Hence, D is a compact subset of ${\mathbb D}$. Let $G:=\Omega _0^*$ be the simply connected hull of $\Omega _0:$ 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,

$$\begin{aligned} \partial H\subseteq \partial \Omega _0. \end{aligned}$$

(2)

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 $|f|\le \eta$ on $\partial H$, and so, by the maximum principle, $|f|\le \eta$ on H. Consequently, $|f|\le \eta$ on G. By the local maximum principle, $|f|<\eta$ on G. Since $\partial D\subseteq D\subseteq G$,

$$\begin{aligned} |f|<\eta \,\text{ on }\, \partial D. \end{aligned}$$

(3)

On the other hand,

$$\begin{aligned} \partial D \mathop \subseteq \limits _{}^{(2)} \partial \Omega _0\cap {\mathbb D}\subseteq \{z\in {\mathbb D}: |f(z)|=\eta \}. \end{aligned}$$

(4)

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 (3). 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

$$\begin{aligned} \limsup _{\begin{array}{c} \xi \rightarrow x\\ x\in \partial \Omega _0\setminus E \end{array}}|u(\xi )|^{-1}=\eta ^{-1}. \end{aligned}$$

Since u is an inner function, E has linear measure zero (by [5, Theorem 4.2]). The maximum principle for subharmonic functions with a few exceptional points (here on the set E; see [6] or [12]) now implies that $|u|^{-1}\le \eta ^{-1}$ on $\Omega _0.$ But $|u|<\eta$ on $\Omega$ is a contradiction. We conclude that $\inf _{\Omega _0}|u|=0$. $\square$

Lemma 3

[10] Let u be an inner function. Then the connectedness of $\Omega _u(\eta )$ implies the one of $\Omega _u(\eta ')$ for every $\eta '>\eta$.

Proof

Because $\Omega _u(\eta )$ is connected and $\Omega _u(\eta )\subseteq \Omega _u(\eta '),$ $\Omega _u(\eta )$ is contained in a unique component $U_1(\eta ')$ of $\Omega _u(\eta ').$ Suppose that $U_0(\eta ')$ is a second component of $\Omega _u(\eta ').$ Then $|u|\ge \eta$ on $U_0(\eta '),$ because $U_0(\eta ')$ is disjoint with $U_1(\eta ')$ and hence with $\Omega _u(\eta ).$ By Lemma 2 though, $\inf _{U_0(\eta ')} |u|=0;$ a contradiction. Thus $\Omega _u(\eta ')$ is connected. $\square$

3 Explicit examples of one-component inner functions

Let

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

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

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

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 Aleksandrov’s criterion [1]), see Theorem 15 below), we include a simple geometric proof for the reader’s convenience.

Proposition 4

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

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

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

$$\begin{aligned} |B(z)|=\rho (B(z), B(z_n))\le \rho (z,z_n)<\sigma . \end{aligned}$$

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,

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

(5)

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. $\square$

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 $\delta (b):=\inf (1-|z_n|^2)|b'(z_n)|>0$. If b is an interpolating Blaschke product then, for small $\varepsilon$, the pseudohyperbolic disks

$$\begin{aligned} D_\rho (z_n,r)=\{z\in {\mathbb D}: \rho (z,z_n)<\varepsilon \} \end{aligned}$$

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 $\{z\in {\mathbb D}: \rho (z,Z(b))\ge \eta \}$.

Theorem 5

(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

$$\begin{aligned} 0< \varepsilon < \eta \frac{\delta - \eta }{1- \delta \eta }. \end{aligned}$$

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

$$\begin{aligned} \delta (B) = \inf _{n \in {\mathbb {N}}} (1- | z_n |^2) | B'(z_n)| \ge \delta , \end{aligned}$$

then

• (1) The pseudohyperbolic disks $D_\rho (a,\eta )$ for $a\in Z(B)$ are pairwise disjoint.

• (2) The following inclusions hold:

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

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

Theorem 6

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

$$\begin{aligned} G:=\bigcup _n D_\rho (z_n,\sigma ) \end{aligned}$$

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 4

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

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,

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

(6)

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

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

$$\begin{aligned} 0< \varepsilon < \eta \frac{\delta - \eta }{1- \delta \eta }. \end{aligned}$$

By Lemma 2, $\inf _{\Omega _0}|b|=0$. Thus, there is $z_0\in \Omega _0$ be so that $|b(z_0)|<\varepsilon$. We deduce from Hoffman’s Lemma (Theorem 5) that $\rho (z_0,Z(b))<\eta <\sigma$. This is a contradiction to (6). 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

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

$\square$

Corollary 7

Let B be a Blaschke product with increasing real zeros $x_n$ satisfying

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

Then $b\in \mathfrak I_c$.

Proof

Just use Theorem 6 and the fact that by the Vinogradov–Hayman–Newman theorem, B is interpolating if and only if

$$\begin{aligned} \sup _n\frac{1-x_{n+1}}{1-x_n}\le s<1 \end{aligned}$$

or equivalently

$$\begin{aligned} \inf _n\rho (x_n,x_{n+1})\ge r>0. \end{aligned}$$

$\square$

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 the following Corollary

Corollary 8

Let B be a Blaschke product the zeros $(z_n)$ of which are contained in a Stolz angle and are separated. Suppose that $\rho (z_n,z_{n+1})\le \eta <1$. Then $B\in \mathfrak I_c.$

Similarly, using a result by M. Weiss [27, Theorem 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 9

Let B be a Blaschke product the zeros $z_n=r_ne^{i\theta _n}$ of which satisfy:

$$\begin{aligned} r_n<r_{n+1}, ~ \theta _{n+1}< \theta _n, \end{aligned}$$

$$\begin{aligned} r_n\nearrow 1,~ \theta _n\searrow 0, \end{aligned}$$

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

(7)

Then B is an interpolating Blaschke product contained in $\mathfrak I_c$. This holds in particular if the zeros are located on a convex curve in ${\mathbb D}$ with endpoint 1 and satisfying (7).

Other classes of this type can be deduced from [14]. Here are two explicit examples of interpolating Blaschke products in $\mathfrak I_c$ the zeros of which 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 10

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

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

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

$$\begin{aligned} \varphi _{j+1}(z) = (\varphi _ j\circ \varphi )(z)=\frac{z- \frac{\frac{1}{2}+x_j}{1+ \frac{1}{2}x_j}}{1-z \frac{\frac{1}{2}+x_j}{1+ \frac{1}{2}x_j}}. \end{aligned}$$

Since

$$\begin{aligned} \rho (x_j,x_{j+1})=\frac{3^{j+1}-3^j}{3^{j+1}+ 3^j}=\frac{1}{2}, \end{aligned}$$

we deduce from Corollary 7 that the Blaschke product

$$\begin{aligned} B(z):=\prod _{j=1}^\infty \frac{x_j-z}{1-x_j z} \end{aligned}$$

associated with these zeros is in $\mathfrak I_c.$

$\bullet$ Let $\sigma \in \mathrm{Aut}({\mathbb D})$ 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 $\tau '(\xi )<1$. The zeros of the n-th iterate $\tau _n$ of $\tau$ are given by

$$\begin{aligned} z_n=\tau _n^{-1}(0)=(\sigma \circ \varphi ^{-1}_n\circ \sigma ^{-1})(0). \end{aligned}$$

By the grand iteration theorem [23, p.78], since 1 is an attracting fixpoint with $(\varphi ^{-1})'(1)= 1/3<1$, 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 $a=\sigma ^{-1}(0)$,

$$\begin{aligned} \rho (z_n,z_k)= \, &{} \rho \big ( (\varphi ^{-1}_n\circ \sigma ^{-1})(0), (\varphi ^{-1}_k\circ \sigma ^{-1})(0)\big )\\= \, & {} \rho \big (\varphi ^{-1}_{n-k}(a), a\big ) \end{aligned}$$

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 sequence³ ([23, Ex. 6, p. 85]); in fact, use d’Alembert’s quotient criterion and observe that by the Denjoy–Wolff theorem,

$$\begin{aligned} \frac{1-|z_{n+1}|}{1-|z_n|}= \frac{1-|\tau ^{-1}(z_n)|}{1-|z_n|}\rightarrow (\tau ^{-1})'(\xi )<1. \end{aligned}$$

Hence, by Corollary 8, $(z_n)$ 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 nth 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 Fig. 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

$$\begin{aligned} z_n=\frac{n}{n-i}; \end{aligned}$$

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,

$$\begin{aligned} \rho (z_n,z_{n+1})=\frac{1}{\sqrt{2}}. \end{aligned}$$

Thus, by, Corollary 9, the Blaschke product associated with these zeros is interpolating and belongs to $\mathfrak I_c$.

Fig. 1

The parabolic automorphism

Proposition 11

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 (1), which coincides with the level set $\Omega _S(\eta )$, contains all zeros of B. Note that $D_\eta =\Omega _S(\eta )\subseteq \Omega _f(\eta )$. 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 2 (2).

• (ii) If B is an interpolating Blaschke product with zeros $(z_n)$, then, by Hoffman’s Lemma (Theorem 5), 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 $\pi$ (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.

$\square$

The previous result shows, in particular, that certain non one-component inner functions (for example a thin Blaschke product with positive zeros, see Corollary 21), 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 proceed to show below.

Proposition 12

Let u, v be two inner functions in  $\mathfrak I_c$. Then $uv\in \mathfrak I_c.$

Proof

Let $\Omega _u(\eta )$ and $\Omega _v(\eta ')$ be two connected level sets. Due to monotonicity (Lemma 3), and the fact that $\bigcup _{\lambda \in [\lambda _0,1[} \Omega _f (\lambda )={\mathbb D}$, we may assume that $\sigma$ satisfies

$$\begin{aligned} \max\, \{\eta ,\eta '\}\le \sigma <1 \end{aligned}$$

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

$$\begin{aligned} \Omega _u(\sigma )\cup \Omega _v(\sigma )\subseteq \Omega _{uv}(\sigma ). \end{aligned}$$

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 2 (2). Hence, $\Omega _{uv}(\sigma )$ is connected and so $uv\in \mathfrak I_c.$ $\square$

Theorem 13

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 3, $\Omega _u(\eta )$ 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. Toward this end, we note that

$$\begin{aligned} \Omega _{\Theta }(\eta -\varepsilon )\subseteq \Omega _u(\eta )\subseteq \Omega _\Theta (\eta +\varepsilon ), \end{aligned}$$

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 $\Omega _u(\eta ),$ and hence disjoint with $\Omega _{\Theta }(\eta -\varepsilon ).$ In other words, $|\Theta |\ge \eta -\varepsilon >0$ on $\Omega _0.$ This contradicts Lemma 2 (2). H $\Omega _\Theta (\eta +\varepsilon )$ is connected and so $\Theta \in \mathfrak I_c$. $\square$

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

Example 14

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 (Fig. 2). We may choose $0<\eta <1$ so close to 1 that 0 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 $\{z: |z|<1, \mathrm{Re}\; z>0\}.$ Let $V^*$ be its reflection along the imaginary 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 $\Omega _{S(z^2)}(\eta ).$ $\square$

Fig. 2

The level sets of $S(z^2)$

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

$$\begin{aligned} Cl(\Theta ,\zeta )=\{w\in { \mathbb C}: \exists (z_n)\in {\mathbb D}^{ \mathbb N},~ \lim z_n=\zeta \, \text{ and }\, \lim \Theta (z_n)=w\} \end{aligned}$$

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

Theorem 15

(Aleksandrov) [1, Theorem 11 and Remark 2, p. 2915] Let $\Theta$ be an inner function. The following assertions are equivalent:

• (1) $\Theta \in \mathfrak I_c.$

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

  $$\begin{aligned} i)~~~ |\Theta '' (\zeta )|\le C\; |\Theta ' (\zeta )|^2, \end{aligned}$$

  and

  $$\begin{aligned} ii)~~~ \liminf _{r\rightarrow 1} |\Theta (r\zeta )|<1 \, \text{ for } \text{ all }\, \zeta \in \rho (\Theta ). \end{aligned}$$

Note that, due to the above theorem, $\Theta \in \mathfrak I_c$ necessarily implies that $\rho (\Theta )$ has measure zero.

Proposition 16

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

$$\begin{aligned} z\frac{B'(z)}{B(z)}=\sum _{n=1}^N \frac{1-|a_n|^2}{|a_n-z|^2}, ~~|z|=1, B(a_n)=0,) \end{aligned}$$

(8)

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 $\liminf _{r\rightarrow 1} S(B(r \zeta ))=0$ for $\zeta \in \rho (S\circ B)$. Now let us calculate the following derivatives:

$$\begin{aligned} S'(z)=-S(z) \frac{2}{(1-z)^2}, \end{aligned}$$

$$\begin{aligned} S''(z)= S(z)\left[ \frac{4}{(1-z)^4}- \frac{4}{(1-z)^3}\right] , \end{aligned}$$

$$\begin{aligned} (S\circ B)'= (S'\circ B) B' \end{aligned}$$

$$\begin{aligned} (S\circ B)'' =(S''\circ B) B'^2 + (S'\circ B) B'' \end{aligned}$$

$$\begin{aligned} A:=\frac{(S\circ B)''}{[(S\circ B)']^2} &= {} \frac{S''\circ B}{(S'\circ B)^2}+ \frac{(S'\circ B)}{(S'\circ B)^2}\frac{B''}{B'^2}\\\nonumber &= {} \frac{S''\circ B}{(S'\circ B)^2}+ \frac{1}{S'\circ B}\frac{B''}{B'^2}. \end{aligned}$$

(9)

Hence, for $\zeta \in \mathbb T\setminus \rho (S\circ B)$, $|B(\zeta )|=1$ , but $\xi :=B(\zeta )\not =1$, and so, by (8),

$$\begin{aligned} |A(\zeta )|&\le {} \sup _{\xi \not =1} \frac{|S''(\xi )|}{|S'(\xi )|^2}+ 2\sup _{\xi \not =1}\frac{|1-\xi |^2}{|S(\xi )|}\ C \\&\le {} C' \sup _{\xi \not =1} \frac{|1-\xi |^4}{|1-\xi |^4} +8C\le\infty .\end{aligned}$$

$\square$

Corollary 17

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 $S_\mu$ is written 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]). The assertion now follows from Proposition 16. $\square$

Note that the above result also follows in an elementary way from Proposition 12 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 18

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

Proof

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

$$\begin{aligned} \liminf _{r\rightarrow 1}|\tau \circ \Theta (r\zeta )|<1 \end{aligned}$$

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

$$\begin{aligned} \tau (z)= \frac{1}{\overline{a}}+ \frac{|a|^2-1}{\overline{a}} \frac{1}{1-\overline{a} z}, \end{aligned}$$

from which we easily deduce the first and second derivatives. Using the formulas 9, we obtain

$$\begin{aligned} A:=\left| \frac{(\tau \circ \Theta )''}{[(\tau \circ \Theta )']^2}\right|\le \, {} C\frac{|1-\overline{a} \Theta |^4}{|1-\overline{a}\Theta |^3}+ C' |1-\overline{a} \Theta |^2 \; \frac{|\Theta ''|}{|\Theta '|^2}. \end{aligned}$$

Hence, the assumption $\Theta \in \mathfrak I_c$ now yields (via Aleksandrov’s criterion, Theorem 15) that $\sup _{\zeta \in \rho (\tau \circ \Theta )} A(\zeta )<\infty$. Thus, $\tau \circ \Theta \in \mathfrak I_c.$ $\square$

Corollary 19

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 9 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]). In the cited reference, it is also shown that the Frostman shift $(S-a)/(1-\overline{a} S)$ is an interpolating Blaschke product.

Corollary 20

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 12 and 18. $\square$

4 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

$$\begin{aligned} \lim _n\prod _{k\not =n} \rho (z_k,z_n)=\lim _{n\rightarrow 1} (1-|z_n|^2)|b'(z_n)|=1. \end{aligned}$$

It was shown by Tolokonnikov [24, Theorem 3] that b is thin if and only if

$$\begin{aligned} \lim _{|z|\rightarrow 1} (|b(z)|^2+(1-|z|^2)|b'(z)|)=1. \end{aligned}$$

Corollary 21

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

$$\begin{aligned} \varepsilon<\eta ^2\, \text{ and }\, \displaystyle \eta < (1-\sqrt{1-\delta ^2})/ \delta . \end{aligned}$$

By deleting finitely many zeros, say $z_1,\dots , z_N$ of b, we obtain a tail $b_N$ such that $(1-|z_n|^2)|b_N'(z_n)|\ge \delta$ for every $n>N$. Hence, by Theorem 5,

$$\begin{aligned} \{z \in {\mathbb D}: |b_N(z)|< \varepsilon \} \subseteq \{z \in {\mathbb D}: \rho (z, Z(b_N)) < \eta \} \end{aligned}$$

(10)

and the disks ${D_{\rho}}(z_n,\eta )$ 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

$$\begin{aligned} p(z):=\prod _{n=1}^N \rho (z,z_n)\ge \varepsilon \end{aligned}$$

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

$$\begin{aligned} |b_N(z)|= \frac{|b(z)|}{|p(z)|}<\frac{\varepsilon ^2}{\varepsilon }=\varepsilon . \end{aligned}$$

Hence

$$\begin{aligned} \{z: r<|z|<1, |b(z)|<\varepsilon ^2\}\subseteq \{|b_N(z)|<\varepsilon \}\mathop \subseteq \limits _{}^{(3.1)} \bigcup _{n>N} {D_{\rho}}(z_n,\eta ). \end{aligned}$$

Since the disks $D_\rho (z_n,\eta )$ are pairwise disjoint if $n>N$, we are done with the tasks.$\square$

Corollary 22

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 21, if $b_j$, $(j=1,2)$, are two thin Blaschke products with zero-sequence $(z_n^{(j)})_n,$

$$\begin{aligned} \Omega _{b_j}(\varepsilon )\subseteq \bigcup _{n=1}^\infty D_\rho (z^{(j)}_n,\eta ) \end{aligned}$$

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 the most at one disk $D_\rho (z^{(2)}_m,\eta )$ for n large. Hence

$$\begin{aligned} \Omega _{b_1b_2}(\varepsilon ^2)\subseteq \bigcup _{j=1}^2\bigcup _{n=1}^\infty D_\rho (z^{(j)}_n,\eta ), \end{aligned}$$

where the set on the right-hand side obviously is disconnected. The general case works via induction. $\square$

Remark 23

The conditions

$$\begin{aligned} \eta ^*:=\sup _{n\in { \mathbb N}}\rho (z_n, Z(b)\setminus \{z_n\})<1, \end{aligned}$$

(11)

or equivalently

$$\begin{aligned} D(z_n,\eta )\cap \bigcup _{m\not = n}D(z_m, \eta )\not =\emptyset \, \text{ for } \text{ some } \, \eta \in ]0,1[, \end{aligned}$$

(12)

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 $a=n^{-n}$ and $b=2a$, and the identity,

$$\begin{aligned} \rho (1-a,1-b)=\frac{|a-b|}{a+b-ab}, \end{aligned}$$

we conclude that

$$\begin{aligned} \rho (z_{2n}, z_{2n+1})= \frac{n^{-n}}{1-z_{2n} z_{2n+1}}\rightarrow 1/3, \end{aligned}$$

and so the union $(z_n)$ is an interpolating sequence satisfying (12). By Corollary 22, the Blaschke product formed with the zero-sequence $(z_n)$ is not in $\mathfrak I_c.$ $\square$

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

Theorem 24

(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

$$\begin{aligned} \limsup _{r\rightarrow 1} |u(r\xi )|=1\, \textit{ for } \textit{ every }\, \xi \in \mathbb T. \end{aligned}$$

In particular this condition is satisfied by finite products of thin Blaschke products (see [17, Proposition 2]) as well as by the class of uniform Frostman Blaschke products whose zero sequence (z _n ) satisfies

$$\begin{aligned} \sup _{\xi \in \mathbb T} \sum _{n=1}^\infty \frac{1-|z_n|^2}{|\xi -z_n|}<\infty . \end{aligned}$$

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 6, we therefore obtain

Corollary 25

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 25

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

• (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) 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) In terms of the zeros, give a characterization of those interpolating Blaschke products that belong to $\mathfrak I_c$.

• (4) Does the Blaschke product B with zeros $z_n=1-n^{-2}$ belong to $\mathfrak I_c$?