Lemma 2.4 at page 8 of these lecture notes on viscosity solutions theory is a classical and frequently used result.
Does the lemma (and its proof) hold true if we replace "local" with "global" throughout?
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7
Yes and no.
Yes: If $\Omega$ is compact, you just need to change $|x-y|=\rho$ to $|x-y|\geqslant\rho$ in lines 3 and 5 of the proof.
No: Take $u(x)=-\cos(x)$ and $\phi(x)=0$ on $\Omega=(-2\pi,2\pi)$. Then $u$ has a strict global minimum on $\Omega$ at $0$, but $u_n(x)=-\cos(x)-x^2/n$ has no global minimum on $\Omega$.
(And $u_n(x)=-\cos(x)-x^{2n}(4\pi^2-x^2)$ does have a global minimum on $\Omega$ somewhere near $\pm 2\pi$).
answered Aug 25, 2017 at 9:38
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$\begingroup$ 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? $\endgroup$– user103450Commented Aug 25, 2017 at 18:08
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$\begingroup$ 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? $\endgroup$ Commented Aug 25, 2017 at 20:06
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$\begingroup$ If $x $ is a global minimum doesn't $u(y)>u (x)$ for all $y$? Where is compactness needed? (also, in the original proof?) $\endgroup$– user103450Commented Aug 26, 2017 at 9:56
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$\begingroup$ In the counterexample, the statement seems to be false even in the local case. Am I mistaken? $\endgroup$– user103450Commented Aug 26, 2017 at 10:53
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$\begingroup$ Hello. Thanks for your answer. Before I award the bounty, would you mind addressing the comments from the OP? $\endgroup$– user60665Commented Aug 31, 2017 at 20:28