On the derivative of infinite Blaschke products

Abstract

A well known result of Privalov shows that if $f$ is a function that is analytic in the unit disc $\Delta =\{z\in \mathbb{C} : \vert z\vert <1\} $, then the condition $f'\in H\sp 1$ implies that $f$ has a continuous extension to the closed unit disc. Consequently, if $B$ is an infinite Blaschke product, then $B'\notin H\sp 1$. This has been proved to be sharp in a very strong sense. Indeed, for any given positive and continuous function $\phi $ defined on $[0, 1)$ with $\phi (r)\to\infty $ as $r\to 1$, one can construct an infinite Blaschke product $B$ having the property that

\[ M_1(r,B')\defeq \frac{1}{2\pi } \int_{-\pi }\sp\pi\vert B'(re\sp{it})\vert\,dt=\og\left (\phi (r)\right ) ,\quad\hbox{as $r\to 1$.} \tag{$*$} \]

All examples of Blaschke products constructed so far to prove this result have their zeros located on a ray. Thus it is natural to ask whether an infinite Blaschke product $B$ such that the integral means $M_1(r,B')$ grow very slowly must satisfy a condition \lq\lq close\rq\rq \, to that of having its zeros located on a ray. More generally, we may formulate the following question: Let $B$ be an infinite Blaschke product and let $\{ a_n\} _{n=1}\sp\infty $ be the sequence of its zeros. Do restrictions on the growth of the integral means $M_1(r,B')$ imply some restrictions on the sequence $\{ \Arg (a_n)\} _{n=1}\sp\infty $?

In this paper we prove that the answer to these questions is negative in a very strong sense. Indeed, for any function $\phi $ as above we shall construct two new and quite different classes of examples of infinite Blaschke products $B$ satisfying ($*$) with the property that every point of $\partial \Delta$ is an accumulation point of the sequence of zeros of $B$.