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I am teaching a course in basic differential topology, and, following e.g. Milnor, I defined functions of class $C^k$ on subsets of the Euclidean space $\mathbb{R}^n$ as follows.

Let $f\colon X\to \mathbb{R}$ be a function, where $X\subseteq \mathbb{R}^n$. Then $f$ is of class $C^k$ if for every point $x_0\in X$ there exist an open neighbourhood $U$ of $x_0$ in $\mathbb{R}^n$ and a function $F\colon U\to\mathbb{R}$ of class $C^k$ such that $f|_{X\cap U}=F|_{X\cap U}$.

A student in the audience asked the following question:

Is it true that a function $f\colon X\to\mathbb{R}$ is $C^\infty$ if and only if it is $C^k$ for every $k\in\mathbb{N}$? (The ``only if'' statement of course is trivial). Well, I was not able to answer to the question...

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    $\begingroup$ Of course the answer is yes if $X$ is an open subset of the Euclidean space. My question may be reformulated as follows: is it true that if $f\colon X\to\mathbb{R}$ admits a (local) $C^k$ extension $F_k$ for every $k$, then it admits a $C^\infty$ extension? If $F_k\neq F_h$ for every h\neq k$, this does not seem obvious to me. $\endgroup$
    – Roberto Frigerio
    Commented Sep 27, 2018 at 14:05
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    $\begingroup$ I think this will follow from the Whitney extension theorem, after having extended $f$ to $\overline{X}$: en.wikipedia.org/wiki/Whitney_extension_theorem Certainly in one dimension, it's straightforward to deduce the claim from Borel's theorem, and I think a more elaborate version of this argument should work in general. $\endgroup$
    – Christian Remling
    Commented Sep 27, 2018 at 14:46
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    $\begingroup$ Beaten to the punch by Christian Remling. :-) The current Wikipedia page does not explicitly say that the extension can be $C^m$ for $m=\infty$, but the EoM page does. In any case, Whitney's original article does cover the $m=\infty$ case. $\endgroup$
    – Igor Khavkine
    Commented Sep 27, 2018 at 14:53
  • $\begingroup$ You are in Pisa so there are plenty of people around you who would know the answer right away (Alberti, Ambrosio, Magnani...) $\endgroup$
    – Piotr Hajlasz
    Commented Sep 27, 2018 at 15:13

1 Answer 1

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The answer is yes for functions defined on closed sets $X\subset\mathbb{R}^n$. In Section 1.5.5 in [1] we have a necessary and a sufficient condition of the existence of an extension to a $C^m$ function for a finite $m$ and in Section 1.5.6 in [1] we have a necessary and sufficient condition for the existence of an extension to $C^\infty$. It turns out that if the condition for the existence of an extension to $C^m$ is satisfied for all finite $m$, then it is precisely the condition for the existence of an extension to $C^\infty$.

[1] Narasimhan, R. Analysis on real and complex manifolds, North-Holland Mathematical Library, 35. North-Holland Publishing Co., Amsterdam, 1985.

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