Timeline for Is analytic capacity inner regular?
Current License: CC BY-SA 3.0
16 events
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| Aug 14, 2013 at 12:44 | history | edited | Malik Younsi | CC BY-SA 3.0 |
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| Jun 25, 2013 at 3:02 | history | bounty ended | CommunityBot | ||
| Jan 31, 2012 at 21:36 | history | edited | Malik Younsi | CC BY-SA 3.0 |
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| Jan 25, 2012 at 23:09 | comment | added | George Lowther | @Malik: no, I'm not sure that $\gamma(E)=0$ is sufficient anymore. | |
| Jan 25, 2012 at 14:56 | history | edited | Malik Younsi | CC BY-SA 3.0 |
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| Jan 25, 2012 at 14:37 | comment | added | Malik Younsi | @fedja : It is not clear to me at all either. Thank you for the suggestion, I'm going to contact X. Tolsa. Hopefully, I'll be lucky! @George : I don't really see why $\gamma(E)=0$ is a sufficient condition, but I'd be happy to hear more about it. I agree though that for a counterexample, we probably need to think about complicated compact sets. The problem though is that it's very difficult to determine the analytic capacity of even slightly complicated sets! | |
| Jan 25, 2012 at 1:00 | comment | added | George Lowther | ...although, on second thoughts, I'm not quite sure that $\gamma(E)=0$ is really a sufficient condition. | |
| Jan 25, 2012 at 0:53 | comment | added | George Lowther | @Malik: Using the results in the paper mentioned, I think that you get a positive result whenever the set $E=\bigcap_n\overline{\partial K\setminus K_n}$ has zero analytic capacity. $E$ is just the set of points on the boundary of $K$ where $K_n$ are not dense, so it is a nowhere dense closed subset of $\partial K$. For any nowhere dense closed subset of $\partial K$, you can always find $K_n\subseteq\partial K$ with $E=\bigcap_n\overline{\partial K\setminus K_n}$, and $\lim\gamma(K_n)$ is independent of the choice of $K_n$. For a counterexample, you need to find some sufficiently nasty $K,E$. | |
| Jan 24, 2012 at 3:10 | comment | added | fedja | Yes, that's it. It is not clear to me, however, if we can mix that argument with the classical definition somehow and get what you want. We can ask Xavier directly though. If he doesn't know, you are out of luck :). | |
| Jan 23, 2012 at 14:15 | comment | added | Malik Younsi | @Fedja : Thank you for taking the time to comment on this. Are you referring to the quantity $\gamma^{+}$ and the following article from Tolsa on the comparability of $\gamma^{+}$ and analytic capacity ? Tolsa, Xavier "Painlevé's problem and the semiadditivity of analytic capacity." | |
| Jan 23, 2012 at 11:26 | comment | added | fedja | The best I can say is that it is comparable up to a constant factor to a quantity continuous from below in your sense. | |
| Jan 23, 2012 at 1:55 | history | bounty started | Malik Younsi | ||
| Jan 20, 2012 at 20:57 | history | edited | Malik Younsi | CC BY-SA 3.0 |
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| Jan 20, 2012 at 20:56 | comment | added | Malik Younsi | Yes, of course you're right. It is unique in the unbounded component of the complement of $K$. Thank you for the comment, I'll edit accordingly. | |
| Jan 20, 2012 at 17:53 | comment | added | Robert Israel | Slight quibble: the Ahlfors function may be unique in the unbounded component of the complement of $K$, but can't be uniquely determined in any bounded component. | |
| Jan 20, 2012 at 15:54 | history | asked | Malik Younsi | CC BY-SA 3.0 |