Functions holomorphic on a region minus a Cantor set - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:21:35Z http://mathoverflow.net/feeds/question/97138 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97138/functions-holomorphic-on-a-region-minus-a-cantor-set Functions holomorphic on a region minus a Cantor set uncooltoby 2012-05-16T16:51:11Z 2013-02-11T13:10:32Z <p>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 $X$? </p> http://mathoverflow.net/questions/97138/functions-holomorphic-on-a-region-minus-a-cantor-set/97143#97143 Answer by Igor Rivin for Functions holomorphic on a region minus a Cantor set Igor Rivin 2012-05-16T17:54:14Z 2012-05-16T17:54:14Z <p>This belongs to the subject of holomorphic removability. See <a href="http://en.wikipedia.org/wiki/Analytic_capacity#Removable_sets_and_Painlev.C3.A9.27s_problem" rel="nofollow">this Wiki article</a> 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.</p> http://mathoverflow.net/questions/97138/functions-holomorphic-on-a-region-minus-a-cantor-set/97145#97145 Answer by Lee Mosher for Functions holomorphic on a region minus a Cantor set Lee Mosher 2012-05-16T18:02:53Z 2012-05-16T18:02:53Z <p>Yes, if the Cantor set has measure zero. This is a consequence of the <a href="http://en.wikipedia.org/wiki/Measurable_Riemann_mapping_theorem" rel="nofollow">Measurable Riemann Mapping Theorem</a> which guarantees that the map is quasiconformal, combined with the theorem that if a quasiconformal map is conformal almost everywhere then it is conformal.</p> http://mathoverflow.net/questions/97138/functions-holomorphic-on-a-region-minus-a-cantor-set/97150#97150 Answer by Robert Israel for Functions holomorphic on a region minus a Cantor set Robert Israel 2012-05-16T18:43:54Z 2012-05-16T18:43:54Z <p>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.</p> http://mathoverflow.net/questions/97138/functions-holomorphic-on-a-region-minus-a-cantor-set/99574#99574 Answer by Hrant Hakobyan for Functions holomorphic on a region minus a Cantor set Hrant Hakobyan 2012-06-14T09:04:17Z 2012-06-14T10:45:52Z <p>Yes, if $H_1(E)=0$. For every $1&lt; t\leq 2$ 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.</p> http://mathoverflow.net/questions/97138/functions-holomorphic-on-a-region-minus-a-cantor-set/120133#120133 Answer by Misha for Functions holomorphic on a region minus a Cantor set Misha 2013-01-28T18:11:22Z 2013-01-28T18:11:22Z <p>In Theorem 3 of <a href="http://www.acadsci.fi/mathematica/Vol19/bishop.pdf" rel="nofollow">this paper</a>, 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. </p> http://mathoverflow.net/questions/97138/functions-holomorphic-on-a-region-minus-a-cantor-set/120411#120411 Answer by Lasse Rempe-Gillen for Functions holomorphic on a region minus a Cantor set Lasse Rempe-Gillen 2013-01-31T13:52:39Z 2013-02-11T13:10:32Z <p>Removability with respect to homeomorphisms is different from the removability with respect to bounded functions mentioned in another answer.</p> <p>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.</p> <p>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. </p> <p><a href="http://arxiv.org/abs/math/9812164" rel="nofollow">http://arxiv.org/abs/math/9812164</a></p> <p>(As above, we consider the set in question to be compact.)</p> <p>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.</p> <p>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.</p> <p><strong>EDIT:</strong> 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: <a href="http://www.unige.ch/~smirnov/papers/hr-j.pdf" rel="nofollow">http://www.unige.ch/~smirnov/papers/hr-j.pdf</a> </p> <p>Graczyk and Smirnov use these criteria to prove removability of a large class of Julia sets.</p>