2010
DOI: 10.1007/978-1-4419-6709-1
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Vitushkin’s Conjecture for Removable Sets

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Cited by 31 publications
(17 citation statements)
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“…This result is an important step in the proof of Vitushkin's conjecture (for more details see [35,6]), which states that a compact set with finite one-dimensional Hausdorff measure is removable for bounded analytic functions if and only if it is purely 1-unrectifiable, which means that every 1-rectifiable subset of this set has Hausdorff measure zero. A higher dimensional analogue of Vitushkin's conjecture is proven in [24] but without using a higher dimensional version of Léger's theorem since in the higher dimensional setting there seems to be no connection between the n-dimensional Riesz transform and curvature (cf.…”
Section: Introductionmentioning
confidence: 92%
“…This result is an important step in the proof of Vitushkin's conjecture (for more details see [35,6]), which states that a compact set with finite one-dimensional Hausdorff measure is removable for bounded analytic functions if and only if it is purely 1-unrectifiable, which means that every 1-rectifiable subset of this set has Hausdorff measure zero. A higher dimensional analogue of Vitushkin's conjecture is proven in [24] but without using a higher dimensional version of Léger's theorem since in the higher dimensional setting there seems to be no connection between the n-dimensional Riesz transform and curvature (cf.…”
Section: Introductionmentioning
confidence: 92%
“…In [4] it was also noted that the estimate (3.1) is the best possible one for general compact sets, when the Hausdorff measure H 1 is replaced by the (possibly smaller) Hausdorff content H 1…”
Section: Painlevé Length Estimatesmentioning
confidence: 99%
“…However, since K P has zero one dimensional Hausdorff measure one obtains that F extends conformally to the entire Riemann sphere by Painlevé's theorem (e.g. see [Dud10,Theorem 2.7]). Therefore f : P → Q is the restriction of a Möbius transformation.…”
Section: Standardnessmentioning
confidence: 99%