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Oct 28, 2022 at 18:11 comment added alesia that's true, the hessians provided by Alexandrov theorem would miss a lot of what's happening. Considering the Hessian as a Radon measure, you'd need to have more information on the discontinuities of u to give a meaning to hess(f)u though
Oct 28, 2022 at 17:35 comment added Piotr Hajlasz @alesia Sure, but the Hessian is defined only as a Peano derivative and the Hessian as the Radon measure is more convenient for someone working with Sobolev spaces.
Oct 28, 2022 at 17:33 comment added alesia In fact, by Alexandrov theorem and the 1-semi-concavity of d^2, d^2 (hence d) has a hessian defined almost everywhere
Oct 28, 2022 at 14:45 comment added Ayman Moussa Thanks Piotr for the precisions ! Well I have a computations which involves the inegral of some quantities like $\text{Hess}(d) u$ for some vector-valued Sobolev function $u$ and I'd like to understand how much I can give it a sense to this expression. Unfortunately, even if $d$ is assumed to be convex, then this seems out of scope because I don't (a priori) have continuity of $u$.
Oct 28, 2022 at 14:30 history edited Piotr Hajlasz CC BY-SA 4.0
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Oct 28, 2022 at 13:33 comment added Piotr Hajlasz @AymanMoussa Asplund's theorem shows that the distance function is semiconcave. I would suggest you to search literature of the semiconcave functions. I am not sure what sort of regularity about $d$ you want to know.
Oct 28, 2022 at 13:25 comment added Ayman Moussa That's indeed a nice trick. This means that $d^2$ can be written as the difference of two convex functions, so that its hessian is the difference of two non-negative measures. However, I don't see how I could say something on $d$ itself knowing this, any clue ?
Oct 28, 2022 at 13:23 history edited Piotr Hajlasz CC BY-SA 4.0
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Oct 28, 2022 at 13:16 history answered Piotr Hajlasz CC BY-SA 4.0