We now give some examples of holomorphic functions which fail to have radial limits. First, consider the function holomorphic defined by
where we take the principal branch with . I claim that is bounded on , but does not exist. Observe that, for , , so that
For the second assertion, observe that
we see that oscillates between and infinitely often as .
We now give an example of a holomorphic function which fails to have radial limits on a dense subset of . Let be an enumeration of the rationals. By Weierstrass’ theorem, choosing , we can define a holomorphic function by
I claim that does not exist for , for all . Indeed, we showed above that oscillates between and , hence oscillates between and infinitely often. So
We now give examples of holomorphic functions which have radial limits nowhere. The following results can be found in David G. Cantor’s note A Simple Construction of Analytic Functions without Radial Limits.
We first prove a Tauberian theorem, which we will use to give a wide class of holomorphic functions on without radial limits.
Let be an increasing sequence of positive integers such that
where is the positive root of . Clearly, for all . Observe that, for ,
We can construct such a sequence by induction. Set and . If we choose , then
Suppose we have chosen positive integers and such that
where . Let be the positive solution of . Since , by the archimedean property of the reals, we can choose a sufficiently large positive integer so that
Lemma 1. Suppose is a bounded sequence in satisfying
where is as above. Then .
Proof. Choose so that for all . Then
We need an estimate for the second term. By construction of the sequences and , we have that
Theorem 2. Let be a bounded sequence in not satisfying . If is a sequence of positive integers as above, then defines a holomorphic function on and has radial limits nowhere.
Proof. That is holomorphic in is immediate from the boundedness of and Weierstrass’ theorem. Suppose has a radial limit at . defines a bounded sequence, so by our Tauberian theorem,
But then , which is a contradiction.
Theorem 3. Let be a continuous, strictly positive function, satisfying . Let be a bounded sequence in not satisfying . Then there exists an increasing sequence of integers , such that is holomorphic in , has radial limits nowhere, and .
Proof. We construct the sequence inductively. I claim that we can choose sufficiently large so that
Indeed, for any , the function is continuous and bounded on and . By Weierstrass’ extreme value theorem, attains a maximum at some . By the compactness of , has a convergent subsequence . For very small, we have for all sufficiently large that and for all . Hence,
For the inductive step, suppose we are given . By a completely analogous argument, we can choose sufficiently large so that
and so that satisfy the conditions of the Tauberian theorem above. Hence,
is holomorphic in and has radial limits nowhere. Furthermore,