On the set of zero radial limits of bounded analytic functions

Question

Hi,

Let $f$ be a non-identically zero bounded analytic function in the open unit disk $\mathbb{D}$. It is well-known that $f$ has radial limits almost everywhere on the unit circle $\mathbb{T}$. Let $Z_f$ be the set of points in $\mathbb{T}$ where $f$ has zero radial limit :

$$Z_f:= \{e^{i\theta} \in \mathbb{T} : \lim_{r \rightarrow 1}f(re^{i\theta})=0\}.$$

Then it is also well-known that $Z_f$ has measure zero.

My question is the following :

For which sets $E \subseteq \mathbb{T}$ of measure zero does there exist a bounded analytic function $f$ in $\mathbb{D}$ such that $E=Z_f$?

Remark : It is easy to see that every $Z_f$ is a $F_{\sigma \delta}$, so the question could be wether or not every $F_{\sigma \delta}$ of measure zero is $Z_f$ for some $f$.

Answer 1

Lohwater and Piranian proved that for every set $F_\sigma$ of measure zero, there exists a bounded analytic function which has radial limits exactly on the complement of this set. (Ann. Acad. Sci. fenn., 239 (1957) 1-17.)

In the book of Collingwood and Lohwater, Theory of cluster sets (1966) they say that the problem is still not settled completely and that the above result is the best that is known.

I suppose it is still unsolved, but many partial results can be found among the references on Lohwater and Piranian on Mathscinet.

Answer 2

The following classical result of Privalov (1919) gives a partial answer to OP's original question :

Let $E$ be a zero measure subset of $\mathbb{T}$. Then there exists a nonzero bounded analytic function $f$ on $\mathbb{D}$ such that $f(z)$ tends to $0$ as $z$ approaches, in an arbitrary manner (in particular, radially), any point of $E$. If $E$ is, in addition, closed, there exists $f(z)$ analytic on $\mathbb{D}$, continuous for $|z|\leq1$, vanishing on $E$ and only there.

For a reference, and a construction of $f$, see p.276 of

A. Zygmund, Trigonometric series, V. 1, Cambridge, 1959.