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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be infinitely differentiable not identically zero. Let $S_f=\{ x : f^{(n)}(x)=0 \ \forall n\in\mathbb{N} \}$. $S_f$ can be non empty e.g. $e^{-1/x^2}$-like examples. In fact, $S_f$ can contain an interval (integrate the previous example). Clearly $S$ is closed. Is there an example of a closed proper subset $K$ of $\mathbb{R}$ such that $K$ is not $S_f$ for any $f$?

We can combine my two examples to make a lot of closed sets, but its not so obvious how to make sure a function with $S_f$ equal to the cantor set.

If such a $K$ exists, is there another way to characterise which sets $S_f$ can be?

J.G.
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user
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2 Answers2

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This can be done iteratively. Choose a cover of the complement of $K$ by a countable family of open balls $_$. For each of the open balls, consider a smooth function $_$ such that $\{_>0\}$=$_$. Then for $_$ going to zero fast enough, $\sum_n __$ and all its derivatives converges uniformly, so the limit exists, is smooth, and the derivatives are the sum are the sum of the derivatives. In particular, they all vanish on $K$

Due to Pierre PC

user
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Given any closed interval $I=[a,b]$, the function: $$ f(x) = \begin{cases} e^{\frac{-1}{(x-a)^2}}, & x < a \\ 0, & a\le x\le b \\ e^{\frac{-1}{(x-b)^2}}, & x > b \end{cases} $$ has $f^{(n)}=0,\forall x\in I, \forall n\in \mathbb{N}$
i.e, $S_f=I$

This work due the fact that all derivaties of $e^{-\frac{1}{(x-p)^2}}$ vanishes at $x=p$.

AMC3F9
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