# Continuous extension of a Bounded Holomorphic Function on $\mathbb{C}\setminus K$

Let $f:\mathbb{C}\setminus K\rightarrow\mathbb{D}$ be a holomorphic map, where $K$ is a compact set with empty interior. My question:

Prove or disprove that: $f$ extends continuously on $\mathbb{C}.$

Remark: Observe that if $K$ is discrete then by the Riemann Removable Singularity Theorem we know that infact there is a holomorphic extension.

• In fact, by using increasing good approximations and carving out small measure sets, one can prove your remark even if K is, for instance, the standard cantor set and other such similar things, which means that in such cases no such an $f$ could've existed in the first place.
– Jeff
Jun 30, 2013 at 9:02
• You mean then that f has to be constant as it extends holomorphically? Jun 30, 2013 at 9:05
• So I think I should remove non constant as the question is ill posed. Jun 30, 2013 at 9:08
• Right. Also, I think the answer to your question may be that it you have a contradiction of the same sort if and only if $K$ is a Lebesgue null set. I think you may want to try using Morera's theorem. If $K$ is null, then use Fubini's theorem to see that for almost every size of circle (or triangle or whatever) centered at a given point in $K^c$, the integral around that geometric figure is 0. Then try to use uniform continuity on compacts to deduce that this fact extends via an approximation argument to points in $K$ which are all limit points of $K^c$.
– Jeff
Jun 30, 2013 at 9:08
• When the integral around a circle is 0 for almost all radii, then by continuity it is actually 0 for all radii.
– Jeff
Jun 30, 2013 at 9:10

Here is an example from Which sets are removable for holomorphic functions?

The function $f=z+z^{-1}$ maps the punctured unit disk $\mathbb D$ bijectively onto $\mathbb C\setminus [-2,2]$. The inverse $g=f^{-1}$ is holomorphic in $\mathbb C\setminus [-2,2]$ and is bounded by $1$, but has no holomorphic (or even continuous) extension to $\mathbb C$. Indeed, $g(z)$ approaches both $i$ and $-i$ as $z\to 0$.

A compact set $K$ is removable for bounded holomorphic functions if and only if its analytic capacity is zero. A simple sufficient condition was given by Painlevé: if the $1$-dimensional Hausdorff measure of $K$ is zero, then it's removable for bounded holomorphic functions. That is, every bounded holomorphic function on $\mathbb C\setminus K$ extends to a holomorphic function on $\mathbb C$ (which is necessarily constant by Liouville's theorem).

• Infact I also found this example. Thanks for your answer and for the reference. Jun 30, 2013 at 10:26
• I think I should have been more clear. When I said measure 0 in the cantor set example, I meant 1-dimensionally. (And then it's true, this was a qual question for me.) But by the time I was talking about my attempt at a proof, I meant 2-dimensional, and then I was also mistaken. Of course, since my proof wasn't all that rigorous, it's not that surprising. Thanks for the correction.
– Jeff
Jun 30, 2013 at 16:02

Even for compact subsets of the real line, this is not true. More precisely, suppose that $K \subseteq \mathbb{R}$ is compact. Then $K$ is removable (for bounded holomorphic functions) if and only if $m(K)=0$, where $m$ is the one-dimensional Lebesgue measure.

See e.g. my answer to this question

Hence just take your favorite Cantor set in the real line of positive one-dimensional Lebesgue measure, and it will be non-removable. One can even give an explicit expression for a bounded holomorphic function $f$ on $\mathbb{C}\setminus K$ which is not constant. See the above link.

• Thanks for your answer, but in the above two answers including yours the extension that has been discussed is Holomorphic. Yes that answers my question. But it would be great if someone can provide references for conditions which guarantee the continuous extension. Jul 1, 2013 at 3:27
• It does answer your question : If $K$ is a compact subset of the real line and $f:\mathbb{C} \setminus K$ is holomorphic and continuous on $\mathbb{C}$, then $f$ is in fact holomorphic on $\mathbb{C}$, by Morera's Theorem. Therefore $f$ is constant. Jul 1, 2013 at 12:48
• What you are looking for is called continuous analytic capacity. See my answer to this question : mathoverflow.net/questions/120118/… Jul 1, 2013 at 12:53
• Oh yes, Thanks for pointing it out. And for Continuous analytic capacity I did found a reference, in fact, notes by Garnett. Jul 2, 2013 at 16:06