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Apr 13, 2017 at 12:58 history edited CommunityBot
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Jun 25, 2013 at 15:31 vote accept Anton Petrunin
May 30, 2011 at 23:44 answer added fedja timeline score: 5
Apr 14, 2011 at 15:53 answer added Igor Belegradek timeline score: 3
Apr 5, 2011 at 4:22 history bounty ended Anton Petrunin
Apr 3, 2011 at 17:35 history edited Anton Petrunin
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Mar 31, 2011 at 7:31 comment added Syang Chen Alternatively, is it possible to show that(at least for those $\phi$ that you are interested in) $\int_{D} e^{\phi}dx<\infty$? If it is the case, then almost every radius could serve your purpose. As far as I know, $L^p$ integrability of $(\phi)_{+}$ always holds for some $p>0$.
Mar 31, 2011 at 7:30 comment added Syang Chen @Anton, here is another "greedy curve" which will touch the boundary in finite time: let $|\gamma(r)|=r$ such that $\phi (\gamma(r))=$ min$_{|z|=r} \phi (z)$, where $ 0 \leq r < 1 $. But I am not sure if such $\gamma$ exists, and more importantly, if it has finite length.
Mar 29, 2011 at 23:35 comment added Anton Petrunin @Xianghong, this way you at least decrease $\phi$ along $\gamma$, but you might get a curve of infinite length (I do not see why not).
Mar 29, 2011 at 21:14 comment added Syang Chen How about using "greedy algorithm" to choose γ? That is, let γ(0)=0,dγ=−∇ϕ(γ) so that $e^ϕ$ is minimized in each infinitesimal step. Can superharmonicity of ϕ guarantee that γ will touch the boundary eventually, to begin with?
Mar 29, 2011 at 3:23 history bounty started Anton Petrunin
Mar 27, 2011 at 23:32 history edited Anton Petrunin CC BY-SA 2.5
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Mar 17, 2011 at 21:57 history edited Anton Petrunin CC BY-SA 2.5
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Mar 17, 2011 at 20:10 history edited Anton Petrunin CC BY-SA 2.5
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Mar 17, 2011 at 19:24 history edited Anton Petrunin CC BY-SA 2.5
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Mar 17, 2011 at 16:09 history asked Anton Petrunin CC BY-SA 2.5