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Is there a continuous map $S: C^k(M)\to C^{k+1}(M)$ with the following properties?

(1) if $S(f)$ is $C^{k+2}$, then $f$ is $C^{k+1}$,

(2) if $f$ is $C^\infty$, then so is $S(f)$,

(3) $f$ and $S(f)$ are $C^k$-close.

Here $M$ is a manifold, $k\ge 0$, and $C^k(M)$ is equipped with (say) strong $C^k$-topology.

I think (1) fails for the usual convolution smoothing with the $C^1$ kernel, namely, certain functions get smoothed too much. Ideally, the operator should work for all $k$ at once, but the above is what I actually need.

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  • $\begingroup$ 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 $f$. $\endgroup$
    – Anton Petrunin
    Commented Apr 4, 2015 at 2:21
  • $\begingroup$ If the manifold $M$ is non-compact things can get hairy. $\endgroup$
    – Liviu Nicolaescu
    Commented Apr 4, 2015 at 9:22
  • $\begingroup$ @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 $\bar f=\frac{1}{2\pi}\int f$? $\endgroup$
    – Igor Belegradek
    Commented Apr 4, 2015 at 11:59
  • $\begingroup$ @IgorBelegradek, obviously I meant $$\varepsilon\cdot\int\limits_0^x(f(t)-\bar f)\cdot dt.$$ $\endgroup$
    – Anton Petrunin
    Commented Apr 4, 2015 at 15:47

1 Answer 1

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Here is a formal construction, but it will not make you happy:

Fix a $C^\infty$-smoothing $\sigma$. If $f\in [C^{k}\backslash C^{k-1}](M)$ et $$S(f)=\sigma(f)\cdot(1+h_k)$$ Where $(h_k)$ is a fixed sequence of functions such that $h_k\to0$ and $h_k|\Omega\in [C^{k+1}\backslash C^{k}](\Omega)$ for any open set $\Omega\subset M$.

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  • $\begingroup$ This does not work. I need the operator for each $k$ while in your constriction $k$ varies. If we fix $k$, and let $S(f)=\sigma(f)(1+h_i)$ with $h_i\in C^{k+1}-C^k$ and $h_i\to 0$ as $i\to\infty$, then the operator does not have property (2). $\endgroup$
    – Igor Belegradek
    Commented Apr 5, 2015 at 1:21
  • $\begingroup$ @IgorBelegradek corrected. $\endgroup$
    – Anton Petrunin
    Commented Apr 5, 2015 at 3:03
  • $\begingroup$ Still does not work. You cannot really mean $C^k\setminus C^{k-1}$ because this is the empty set. If you mean $C^k\setminus C^{k+1}$, then it is unclear how $S$ is defined on $C^{k+1}$. If you mean to use the formula separately for each $k$, then it is unclear why $S$ is continuous. $\endgroup$
    – Igor Belegradek
    Commented Apr 5, 2015 at 12:33
  • $\begingroup$ @IgorBelegradek, I think you know what I mean, do not you? $\endgroup$
    – Anton Petrunin
    Commented Apr 5, 2015 at 19:33
  • $\begingroup$ I have no idea. Since none of the alternatives works, it puzzles me why you even posted this. $\endgroup$
    – Igor Belegradek
    Commented Apr 5, 2015 at 20:29

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