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Mar 17, 2020 at 17:11 comment added Giovanni Febbraro Yes, I have to proof the statement for less regular $u$. The goal is to find the statement as lower as possible regularity of $u$. This can be done through the semiconvex function.
Mar 17, 2020 at 16:45 comment added Giorgio Metafune Still I do not undestand your answer. For $C^2$ functions we agree that Alexandrof maximum principle holds. Then the question is meaningful for less regular $u$ (or $z$, which is the same). Did I understand correctly or convexity palys other roles?
Mar 17, 2020 at 16:18 comment added Giovanni Febbraro No, I don't think.
Mar 17, 2020 at 16:05 answer added Connor Mooney timeline score: 4
Mar 17, 2020 at 15:23 comment added Giorgio Metafune So the question is about the regularity of $u$; for example $u$ could be just a convex function?
Mar 17, 2020 at 14:52 comment added Giovanni Febbraro @GiorgioMetafune I know proofs of Aleksandrov principle when the function $u$ has some useful regularity, as $C^2$ or $W^{2,n}$. In my question all these regularity are not assumed. The only think you know is that $u$ is semiconvex and this implie that $u$ is twice differentiable a.e.. You know some proofs that not use the regularity?
Mar 17, 2020 at 14:41 comment added Giorgio Metafune Sorry, I do not understand. The usual Alexandrov principle holds without any convexity assumption on $u$. What is the question?
S Mar 17, 2020 at 11:37 history suggested Daniele Tampieri CC BY-SA 4.0
Formatting, minor Math Jaxing + typos and grammar
Mar 17, 2020 at 9:52 review Suggested edits
S Mar 17, 2020 at 11:37
Mar 17, 2020 at 8:55 review First posts
Mar 17, 2020 at 9:13
Mar 17, 2020 at 8:52 history asked Giovanni Febbraro CC BY-SA 4.0