On a theorem of Privalov and normal functions

Abstract:

A well known result of Privalov asserts that if $f$ is a function which is analytic in the unit disc $\Delta =\{z\in \mathbb {C} : \vert z\vert <1\}$, then $f$ has a continuous extension to the closed unit disc and its boundary function $f(e^{i\theta })$ is absolutely continuous if and only if $f^{\prime }$ belongs to the Hardy space $H^{1}$. In this paper we prove that this result is sharp in a very strong sense. Indeed, if, as usual, $M_{1}(r, f^{\prime })= \frac {1}{2\pi }\int _{-\pi }^{\pi }\left \vert f^{\prime }(re^{i\theta }) \right \vert d\theta ,$ we prove that for any positive continuous function $\phi$ defined in $(0, 1)$ with $\phi (r)\to \infty$, as $r\to 1$, there exists a function $f$ analytic in $\Delta$ which is not a normal function and with the property that $M_{1}(r, f^{\prime })\leq \phi (r)$, for all $r$ sufficiently close to $1$.