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 $B$.