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Jul 1, 2013 at 16:19 vote accept Ryan O'Donnell
Jun 24, 2013 at 18:20 answer added bezirsk timeline score: 3
Jun 24, 2013 at 10:59 comment added Benoît Kloeckner I do not have an answer, but this seems plausible for Lipschitz functions. You could have a look in Federer's book or on functional inequalities where this kind of thing are used (sample keywords: Faber-Krahn, Schwarz symmetrization).
Jun 24, 2013 at 10:34 comment added Ryan O'Donnell @Benoit Kloekcner: Actually, in my application with a little work I can make $u$ piecewise-affine, in which case everything is completely fine. However, I would prefer if I could just leave it Lipschitz and cite something, rather than throw in an extra approximation argument.
Jun 24, 2013 at 10:32 history edited Ryan O'Donnell CC BY-SA 3.0
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May 7, 2013 at 23:29 comment added Ryan O'Donnell @Thomas Richard: no, intuitively one would expect that the boundary of $\{u \geq t\}$ would be $\{u = t\}$. @Benoit Kloekcner: well... suppose it's merely Lipschitz -- am I out of luck?
May 7, 2013 at 16:12 comment added Benoît Kloeckner What kind of regularity do you have on $u$?
May 7, 2013 at 15:13 comment added Thomas Richard Isn't $L_t$ equal to $\{u=t\}$ rather than $\{u\geq t\}$ ?
May 7, 2013 at 14:48 history asked Ryan O'Donnell CC BY-SA 3.0