Let $f: D - K \rightarrow \mathbb{C}$ be holomorphic, where $D$ is a planar domain and $K$ is a compact subset of $D$. Suppose that $f$ extends continuously to all of $D$. On which conditions on $K$ can we conclude that $f$ is actually holomorphic on all of $D$ ?

Note : I ask that $f$ is continuous and not merely bounded on $D$ ; all the references I have found seem concerned with the bounded case. Then the key concept is analytic capacity, and there is a discussion involving the Hausdorff dimension of $K$. But I heard that there are stronger results if we assume continuity instead of just boundedness. Anyway, using Morera's theorem you can see that this holds for say, smooth curves, for which are not removable if you consider only bounded functions.

If someone has a reference, I would be very glad.

  • 1
    $\begingroup$ As far as I recall, also the continuous analytic capacity has been studied extensively. In particular by Tolsa, Melnikov, Vitushkin etc. $\endgroup$
    – J.R.
    Feb 5, 2014 at 20:53
  • $\begingroup$ ah, interesting. thank you, I'll have a look $\endgroup$
    – Albert
    Feb 5, 2014 at 21:09

1 Answer 1


As TooOldForMath said, the key term is continuous analytic capacity, usually denoted $\alpha$ and defined by $$\alpha(K)= \sup_{f} \lim_{z\to\infty}|z (f(z)-f(\infty))|$$ where the supremum is taken over all continuous functions from the Riemann sphere into the closed unit disk that are holomorphic on the complement of $K$. Clearly $\alpha(K)\le \gamma(K)$ where $\gamma$ is the usual analytic capacity. Removability is equivalent to $\alpha(K)=0$. But there is no geometric characterization of such sets. The closest result I know is Tolsa's description of $\alpha$ in terms of Menger curvature of measures supported on $K$, see Theorem B in Bilipschitz maps, analytic capacity, and the Cauchy integral.

In terms of Hausdorff dimension, unqualified continuity does not win you anything: there are non-removable sets of any dimension $\ge 1$. But if you assume Hölder ($C^\alpha$) continuity, the situation greatly simplifies: removable sets are precisely those of $(1+\alpha)$-dimensional Hausdorff measure zero.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .