1
$\begingroup$

To avoid trival cases, we assume that $f$ is neither a constant nor a finite Blaschke product.

Two celebrated theorems of Frostman say that $f_a(z)$ is actually a Blaschke product for every $|a|<1$ with the possible exception of a set of logarithmic capacity zero and if $w=f(z)$ is not reducing to a finite Blaschke product or a constant, it will assume infinitely often every value of $|w|<1$ except for a possible set of logarithmic capacity zero. According to above two theorems, there are two exceptional sets associating with each inner function $f(z)$:

${\mathcal E}_1(f)=\{w:|w|<1, f(z)$ assumes $w$ at most finitely often$\}$

and

${\mathcal E}_2(f)=\{w:|w|<1, \frac{f(z)-w}{1-{\overline w}f(z)}$ is not a Blaschke product $\}.$

It is known that the points in ${\mathcal E}_2(f)$ are radial limits of $f$ and the inclusion ${\mathcal E}_1(f)\subseteq{\mathcal E}_2(f)$ always holds. However there maybe also exits points in ${\mathbb D}\setminus{\mathcal E}_2(f)$ are radial limits of $f$. Frostman has constructed the following Blaschke product: $$B(z)=\prod_{k=1}^{\infty}\frac{(1-\frac{1}{k^2})-z}{1-(1-\frac{1}{k^2})z}$$ which has the radial limit $0$ at $z=1$. We denote these points by ${\mathcal E}_3(f)$.

questions:

Question 1. ${\mathcal E}_2(f)\setminus{\mathcal E}_1(f)$ is at most a denumerable set?

Question 2. ${\mathcal E}_3(f)$ is at most a finite set?

Question 3. If $a\in{\mathcal E}_2(f)\setminus{\mathcal E}_1(f)$, then the singular measure of the singular part of $f_a(z)$ is discrete(i.e., it consists entirely of point masses)? Where $f_a(z):=\frac{f(z)-a}{1-{\overline a}f(z)} (|a|<1)$.

$\endgroup$

1 Answer 1

3
$\begingroup$

I think that the answer to all three questions is no. The simplest is 3. Take any inner function $f\neq 1$ with arbitrary non-zero singular measure, and infinitely many zeros. Then $f$ is an not a Blaschke product, thus $0\in E_2(f)\backslash E_1(f)$, and the singular measure of $f=f_0$ is arbitrary.

Negative answer to 2 follows from a result of Otsuka (Proc AMS 1954, 533-535) that there exists an inner function for which every $a$ in the unit disc is a radial limit. Thus $E_3$ is the whole unit disc, while $E_2$ is always of zero capacity.

A counterexample to 1 that I see at this moment is more difficult. It is based on a theorem of Lehto, Ann. Acad. Sci. Fenn. Ser. A. I. (1954). no. 177, which says that if $a$ is a radial limit corresponding to a direct singularity of $f^{-1}$ then $f_a$ is not a Blaschke product. One can construct $f$ with uncountably many direct singularities, but such that $E_1$ will be empty. The construction is a bit too complicated to fit in MO window. And perhaps a simpler example can be found somewhere in Lehto's papers.

$\endgroup$
8
  • $\begingroup$ I believe question 3 should be phrased about singular part of $f_a(z)$ when $a \neq 0$. If we take $f(z) = \exp\left(\frac{z+1}{z-1}\right)$ and $0\neq a \in {\mathcal E}_2(f)\setminus{\mathcal E}_1(f)$, what can we tell about the singular part of $f_a(z)$? $\endgroup$
    – Mambo
    Oct 20, 2017 at 6:30
  • $\begingroup$ If $a\neq 0$ your $f_a$ has no singular part, it is a Blaschke product. $\endgroup$ Oct 20, 2017 at 13:07
  • $\begingroup$ A theorem of Ahern And Clark says if $f$ is inner and $f' \in H^{1/2}$, then $f$ is a Blaschke product. I thought this function was like borderline counterexample. $\endgroup$
    – Mambo
    Oct 20, 2017 at 16:47
  • $\begingroup$ @Eremenko is there a quick way to see why $f_a$ would be Blaschke for every non-zero $a$? $\endgroup$
    – Mambo
    Oct 20, 2017 at 18:04
  • 1
    $\begingroup$ Yes, of course. 0 is not an asymptotic value. Whenever you have a non-trivial singular component, 0 is an asymptotic value. $\endgroup$ Oct 20, 2017 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.