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Apr 4, 2015 at 19:45 answer added Anton Petrunin timeline score: 2
Apr 4, 2015 at 15:47 comment added Anton Petrunin @IgorBelegradek, obviously I meant $$\varepsilon\cdot\int\limits_0^x(f(t)-\bar f)\cdot dt.$$
Apr 4, 2015 at 11:59 comment added Igor Belegradek @AntonPetrunin: is your approximation of $f$ a $C^\infty$ function plus a constant? If so, the approximation is also $C^\infty$ regardless of $f$. Also is $\bar f=\frac{1}{2\pi}\int f$?
Apr 4, 2015 at 9:22 comment added Liviu Nicolaescu If the manifold $M$ is non-compact things can get hairy.
Apr 4, 2015 at 2:21 comment added Anton Petrunin For $M=\mathbb S^1$ one can take $C^\infty$-smoothing and add $\varepsilon\cdot\int (f(x)-\bar f)\cdot dx$, where $\bar f$ is the average value of $f$.
Apr 4, 2015 at 1:18 history asked Igor Belegradek CC BY-SA 3.0