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Let $U \subset \mathbb{R}^n$ be an open subset of Euclidean space. I feel like there should be a smooth function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ with $f|_U > 0$ and $f|_{\mathbb{R}^n\setminus U} \equiv 0$. Probably one way to show this is to role up your sleeves and try to generalize the proof of paracompactness of manifolds by showing that you can refine to a locally finite cover containing a subset whose union is $U$. After trying to do that for a while, I thought I'd ask if anyone can think of a nicer way to do it. Maybe using a convolution or by applying the standard para compactness result in a clever way. I noticed that distance to the boundary gives a continuous version. Maybe it can be smoothed somehow? I am particularly interested in the case where $U$ is star-shaped, if that's needed (I couldn't imagine why it would be). However, please don't use the result that star-shaped oped sets are diffeomorphic to Euclidean space because that's what I'm trying to prove :)

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up vote 3 down vote accepted

Yes you can use convolution. Take $V_n = \{x : \text{dist}(x,\mathbb{R}\setminus U)>\frac{1}{n} \}$

Than take

$f_n(x) = \frac{1}{2^n}(\chi_V * \Psi_n)(x)$

where $\Psi_n = \Psi(n x)$

and $\Psi$ is this function

And the function you look for should be:

$f(x) = \sum_{n=0}^\infty f_n(x)$

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Fantastic! Thanks, Tom. – Tim kinsella Jan 30 '13 at 9:43
This might be a silly question but what's the best way to verify smoothness of that sum? – Tim kinsella Jan 30 '13 at 18:30
Oh I see. All of the derivatives satisfy the Weierstrass M-test – Tim kinsella Jan 30 '13 at 19:28

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