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Oct 22, 2022 at 2:58 comment added Piotr Hajlasz @DeaneYang I posted an answer which gives a much stronger result.
Oct 10, 2022 at 23:09 comment added Deane Yang Ah, yes. I see.
Oct 10, 2022 at 19:46 comment added Alexander Pruss If $C$ is not bounded, $H_v$ might not be defined.
Oct 10, 2022 at 19:24 comment added Deane Yang The assumption that $C$ is bounded is not needed. Even if $C$ is unbounded, its boundary is locally Lipschitz and therefore by Rademacher's theorem, is locally differentiable almost everywhere.
Oct 10, 2022 at 19:06 vote accept Alexander Pruss
Oct 10, 2022 at 18:03 history answered alesia CC BY-SA 4.0