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May 16, 2020 at 13:20 vote accept L.F. Cavenaghi
May 16, 2020 at 3:52 answer added Alexandre Eremenko timeline score: 4
May 16, 2020 at 1:57 comment added Deane Yang And the Laplacian of $u$ is defined using weak derivatives as a distribution over the full domain.
May 16, 2020 at 1:10 comment added Deane Yang If $u$ is well-defined as a distribution on $M$, then the answer is yes, if the integral is properly interpreted, namely. as the evaluation of the distribution on the constant function $1$.
May 16, 2020 at 0:15 comment added Nate Eldredge Oh, I didn't see the added assumption of dimension 2.
May 16, 2020 at 0:14 comment added L.F. Cavenaghi @NateEldredge I don't see how your counter example works in dimension greater or equal $2$ since on those it does hold that the integral of the Laplacian of $C^2$ function vanishes.
May 16, 2020 at 0:13 history edited L.F. Cavenaghi CC BY-SA 4.0
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May 16, 2020 at 0:09 comment added Nate Eldredge Now I think my first comment is a counterexample again, with $S$ equal to a single point.
May 16, 2020 at 0:08 history edited L.F. Cavenaghi CC BY-SA 4.0
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May 16, 2020 at 0:02 history edited L.F. Cavenaghi CC BY-SA 4.0
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May 16, 2020 at 0:02 comment added Nate Eldredge If you mean instead something like "almost everywhere equal to a $C^2$ function $v$" then the answer is certainly yes, in the sense of weak derivatives, because then $\Delta u = \Delta v$ almost everywhere.
May 16, 2020 at 0:00 comment added Nate Eldredge You need to be clear what "$C^2$ at almost every point" means. For example if $M$ is $S^1$ realized as $[0,1]$ with endpoints identified, the function $u(x) = x^2$ is $C^2$ except at one point and clearly doesn't satisfy the desired conclusion.
May 15, 2020 at 23:46 history asked L.F. Cavenaghi CC BY-SA 4.0