Let $X$ and $Y$ be simply connected open regions of $\mathbb{C}$, and let $Z \subset X$ be a Cantor set. Assume we have a homeomorphism $f$ from $X$ to $Y$, which is holomorphic on $X \setminus Z$. Is $f$ necessarily holomorphic on $X$?
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6 Answers
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This belongs to the subject of holomorphic removability. See this Wiki article for more references. In particular, the article implies that any set with Hausdorff dimension smaller than $1$ is holomorphically removable, and if its Hausdroff dimension is greater than $1,$ it is not. If it is equal to $1,$ you remain puzzled.
answered May 16, 2012 at 17:54
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$\begingroup$ Just a note : analytic capacity and those results you mention are for the study of compact sets removable for bounded holomorphic functions $\endgroup$ Commented May 16, 2012 at 18:16
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$\begingroup$ @Malik: good point. The OP did not specify, so s/he can comment on whether this is what is being sought. $\endgroup$ Commented May 16, 2012 at 18:19
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$\begingroup$ Yes, I believe that in the cases I'm interested in one can say that $f$ is bounded. $\endgroup$ Commented May 16, 2012 at 18:45
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1$\begingroup$ I thought the OP's question was about a specific homeomorphism $f:X \to Y$, asking whether the fact of $f$ being holomorphic on $X-Z$ implies that it is holomorphic on all of $X$. Why then should one care that holomorphic removability is inapplicable in the case that the Hausdorff dimension is $>1$? For example, the restriction of any holomorphic map $X \to Y$ to the subset $X-Z$ extends to a holomorphic map $X \to Y$. $\endgroup$ Commented May 17, 2012 at 15:40
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$\begingroup$ As I mention below, the question becomes quite different when asking about "conformal" removability; i.e. removability for conformal homeomorphisms. Of course holomorphic removability in the sense here implies conformal removability, but the converse is far from true. For example, you can easily have conformally removable sets of Hausdorff dimension 2 (but zero Lebesgue measure). $\endgroup$ Commented Feb 11, 2013 at 13:14
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Yes, if $Z$ has Hausdorff 1-dimensional measure $0$. Then for any $\epsilon > 0$ you can cover $Z$ by a finite number of disks the sum of whose circumferences is less than $\epsilon$. The integral of $f$ over the boundary of the union of these disks is bounded by a constant times $\epsilon$. Use this together with Morera's theorem.
answered May 16, 2012 at 18:43
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Removability with respect to homeomorphisms is different from the removability with respect to bounded functions mentioned in another answer.
In particular, it is not necessary to have Hausdorff dimension at most 1. Indeed, any quasicircle is removable with respect to homeomorphisms, as mentioned by Hrant.
For much more complicated sets, see Jeremy Kahn's thesis "Holomorphic Removability of Julia sets": He shows that many Julia sets of quadratic polynomials are in fact removable.
http://arxiv.org/abs/math/9812164
(As above, we consider the set in question to be compact.)
In particular, he discusses the notion of "absolute area zero": A set $K$ has absolute area zero if there is no conformal isomorphism from the complement of $K$ to the complement of some set with positive area. Any such set is removable, and any Cantor set that is well-surrounded has absolute area zero.
On the other hand, as has been noted elsewhere, there are many examples of sets that are not holomorphically removable. The simplest example of a Cantor set would be a Cantor set of positive measure. More interesting examples are provided by Chris Bishop, as cited in Misha's answer.
EDIT: You may also wish to look at the paper "Removability theorems for Sobolev functions and quasiconformal maps" by Peter Jones and Stas Smirnov, which contains a number of sufficient conditions for conformal removability: http://www.unige.ch/~smirnov/papers/hr-j.pdf
Graczyk and Smirnov use these criteria to prove removability of a large class of Julia sets.
answered Jan 31, 2013 at 13:52
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Yes, if the Cantor set has measure zero. This is a consequence of the Measurable Riemann Mapping Theorem which guarantees that the map is quasiconformal, combined with the theorem that if a quasiconformal map is conformal almost everywhere then it is conformal.
answered May 16, 2012 at 18:02
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1$\begingroup$ @Lee and @uncooltoby: Actually, Measurable Riemann Mapping Theorem does not imply quasiconformality of the extension. What you need is that the extension is absolutely continuous on a.e. line parallel to each coordinate axis (ACL). Equivalently, you need the extension to be in $W^{2,1}$ locally. None of this follows from the measure zero assumption on the Cantor set $Z$: Even though $Z$ has measure zero, a set of horizontal lines of positive measure can still hit $Z$. On the other hand, I do not have counter-examples for measure zero Cantor sets. $\endgroup$– MishaCommented Jun 14, 2012 at 12:19
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$\begingroup$ Ah, that's a very good point. $\endgroup$ Commented Jun 14, 2012 at 13:49
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In Theorem 3 of this paper, for every $\alpha>1$, Chris Bishop constructs examples of Cantor sets $E\subset {\mathbb C}$ whose Hausdorff dimension is in the interval $(1, \alpha)$ and homeomorphisms $f: {\mathbb C}\to {\mathbb C}$ which are conformal outside of $E$, but are not conformal on $E$. Furthermore, in his examples, the set $f(E)$ has zero Lebesgue measure.
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Yes, if $H_1(E)=0$. For every $1< t\leq 2$ there are examples of Cantor sets of Hausdorff dimension $t$ which are non-removable. Of course there are also examples of $t$-dimensional sets which are removable (e.g. quasicircles). Complete characterization of removable sets is an interesting and open problem.
