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If $U$ is a bounded domain in $\mathbb R^n$ whose boundary is smooth, and $f$ is a smooth function on $U$ whose partial derivatives of all orders have a continuous extensions to $\bar U$. For an arbitrary domain $V \supseteq \bar U$, is there a smooth function $\tilde f$ on $V$ extending $f$?

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    Why doesn't $e^{-1/x}$ on $(0,\infty)$ serve as a counterexample?2012-04-15
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    @William: we can extend $e^{-1/x}$ to $0$ on the left of $0$.2012-04-15
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    Smoothly? How? The derivative develops a singularity to the left of zero, no?2012-04-15
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    If $f(x) = {e^{ - 1/x}}$ for $x>0$ and $f=0$ for $x \leq 0$, $f_ + ^{(n)}(0) = 0 = f_ - ^{(n)}(0)$ for all $n=0,1,2,...$, so it is smooth.2012-04-15
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    Sure, that works.2012-04-15

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