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It follows from Whitney extension theorem that for every closed set $ C \subseteq \mathbb{R}^n $ and for every $ k \geq 1 $ there exists a function $ f \in C^k(\mathbb{R}^n) $ such that $ C = \{x : f(x)=0 \} $ and $ D^if(x) =0 $ for every $ x \in C $ and $ i = 1, \ldots , k $.

Is it possible to replace $ k $ with $ \infty $ in the statement above?

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2 Answers 2

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Yes, for example, take an open cover of the complement of $C$ by countably many open balls of radius $r_i$ centered at $v_i$, and use a partition of unity $\{f_i\}_{i∈I}$ subordinate to this cover to glue bump functions $\exp(-(\max(r_i^2-‖x-v_i‖^2,0))^{-2})$ into a globally smooth function $$∑_{i∈I}f_i \exp(-(\max(r_i^2-‖x-v_i‖^2,0))^{-2}).$$

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  • $\begingroup$ Is it smooth on the complement of the boundary though? For instance, if $C$ is a pair of points, isn't $f$ singular on the bisector hyperplane? $\endgroup$
    – Pierre PC
    Commented Mar 30, 2021 at 22:47
  • $\begingroup$ @PierrePC: Yes, somehow I totally forgot to include partitions of unity in the argument, fixed now. $\endgroup$ Commented Mar 30, 2021 at 22:56
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This can be done iteratively. Choose a cover of the complement of $C$ by a countable family of open balls $B_n$. For each of the open balls, consider a smooth function $f_n$ such that $\lbrace f_n>0\rbrace=B_n$. Then for $\varepsilon_n$ going to zero fast enough, $\sum_n\varepsilon_nf_n$ 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 $C$.

I give some details in this answer in the more general case of Hilbert spaces.

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