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Aug 31, 2017 at 21:49 comment added Mateusz Kwaśnicki @Rene: (1) Yes, $u(y) > y(x)$ for all $y \ne x$, but in order to sharpen this to $u(y) > u(x) + \varepsilon$ we need a compact set of points $y$: this is why we restrict our attention to $|y - x| \geqslant \rho$. (2) We use compactness to get the improved inequality in the following way: $u(y) - u(x)$ is a positive continuous function of $y$, defined on a compact set $|y - x| \geqslant \rho$. Thus, it attains a (positive) minimal value $m$; that is, $u(y) - u(x) \geqslant m$. Take $\varepsilon=m/2$ to get a sharp inequality. (3) No, the local case is fine: $u_n(0)$ is a local minimum of $u_n$.
Aug 31, 2017 at 21:48 comment added Mateusz Kwaśnicki @Dal: Whoops, I overlooked these comments, sorry.
Aug 31, 2017 at 20:28 comment added user60665 Hello. Thanks for your answer. Before I award the bounty, would you mind addressing the comments from the OP?
Aug 26, 2017 at 10:53 comment added user103450 In the counterexample, the statement seems to be false even in the local case. Am I mistaken?
Aug 26, 2017 at 9:56 comment added user103450 If $x $ is a global minimum doesn't $u(y)>u (x)$ for all $y$? Where is compactness needed? (also, in the original proof?)
Aug 25, 2017 at 20:06 comment added Mateusz Kwaśnicki Compactness is needed to conclude that if $u(y)>u(x)$ for all $y$ such that $|y-x|\geqslant\rho$, then $u(y)>u(x)+\epsilon_\rho$ for these $y$. This is no different from Bressan's notes, where compactness of the sphere $|y-x|=\rho$ is used. I am not sure if there is anything I could add about the counterexample: $u$, $u_n$ and $\phi$ simply do not satisfy the assertion of the lemma (with every occurrence of "local" replaced by "global"). Or maybe I did not understand correctly your question?
Aug 25, 2017 at 18:08 comment added user103450 Thanks for your answer. About the yes part: what role does the compactness condition play? About the no part: I don't understand why it is a counterexample?
Aug 25, 2017 at 9:38 history answered Mateusz Kwaśnicki CC BY-SA 3.0