Radial Limits for Holomorphic Functions

Question

Let $U$ be an open disc of center $0$ and radius $R > 0$ in the complex plane, and let $f:U \backslash \{0 \} \rightarrow \mathbb{C}$ be a holomorphic function, such that for some $a \in \mathbb{C}$, we have for any $\theta \in [0,2 \pi)$ \begin{equation} \lim_{r \rightarrow 0} f(r e^{i \theta}) = a, \end{equation} where $r$ is a positive real number. In other terms, all the radial limits of $f$ in $0$ exist and are equal. Does there exist the limit of $f$ in $0$? Clearly, if we suppose that $f$ is bounded in a neighborhood of $0$, then a well known result states that $f$ is holomorphic in $U$, so the answer is yes. The only interesting case is so that of a function $f$ which is unbounded in any neighborhood of $0$. I have no idea of the answer in this case. Thank you very much for your help.

Answer 1

Surprisingly enough, the answer is no!

Let $$ H(z) = \int_0^\infty t^{-t} e^{tz}\,dt. $$ You can show $H$ is entire and that $$|H(x+iy)| \le \frac{1}{|y|- \pi/2} $$ for $|y| > \pi/2$.

In particular, $H(z+i\pi)$ is bounded on every line through $0$. Put $$ F(z) = H(z+i\pi)H(iz+i\pi). $$ Then $F(z) \to 0$ as $z \to \infty$ along lines though $0$. Finally, put $f(z) = F(1/z)$ to get an example for your situation (note that $f$ has an essential singularity at $0$, so the full limit as $z\to 0$ doesn't exist).

See American Mathematical Monthly 114 (2007), David Armitage: "Entire functions that tend to zero on every line" for more detail. If you want, I can fill in the details on how to get the bound on $H$ above.