Suppose that $M^2$ is a closed Riemannian manifold and that $u$ is a $C^2(M\setminus S),$ where $S$ is a closed measure set consisting possibly on a enumerable amount of points. Can we still conclude that $\int_M\Delta u =0?$
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2$\begingroup$ 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. $\endgroup$– Nate EldredgeCommented May 16, 2020 at 0:00
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1$\begingroup$ 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. $\endgroup$– Nate EldredgeCommented May 16, 2020 at 0:02
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1$\begingroup$ Now I think my first comment is a counterexample again, with $S$ equal to a single point. $\endgroup$– Nate EldredgeCommented May 16, 2020 at 0:09
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$\begingroup$ @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. $\endgroup$– L.F. CavenaghiCommented May 16, 2020 at 0:14
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1$\begingroup$ 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 $1$. $\endgroup$– Deane YangCommented May 16, 2020 at 1:10
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1 Answer
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You have to say what is the meaning of $\Delta u$, and of $\int\Delta u$. For the integral to have a meaning, $u$ has to be a distribution and $\Delta u$ has to be a (signed, Radon) measure. Such distributions are called $\delta$-subharmonic functions.
Then $\int_M\Delta u=0$ without any assumptions about $C^2$ or $S$. (This is called the "Main Theorem" about compact Riemann surfaces in S. Donaldson, Riemann surfaces, Chap 8, Thm. 5).
But if you understand $\Delta u$ in the naive, classical sense for $C^2$ function, and understand your integral as $\int_{M\backslash S}\Delta u$ then of course this does not have to be $0$.
For example, on the Riemann sphere, let $u(z)=0,\,|z|<1,\; \; u(z)=|z|,\, 1<z<2,\;\; u(z)=2,\, |z|>1$. This function is $C^2$ except on the set of measure zero consisting of two circles $|z|=1$ and $|z|=2$. But $$\left(\int_{|z|<1}+\int_{1<|z|<2}+\int_{|z|>1}\right)u(z)>0.$$
Even simpler example is $u(z)=|z|$ on the Riemann sphere, $\Delta u>0$ when restricted to $0<|z|<\infty$ and $\int_{0<|z|<\infty}u(z)=+\infty$.
answered May 16, 2020 at 3:52