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If $ f \in C_0^\infty=\{ g: g\in C^\infty, \lim_{|x|\rightarrow \infty}g(x)=0\}$, then is $f$ uniformly continuous on $\mathbb R$? ($ f : \mathbb R \to \mathbb R $)

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This is true. Hint: You can split $f$ up into a part where you can control it and a different part where it is very small –  Listing Jun 10 '12 at 12:09
    
Doesn't this just follow from the fact that $f$ has compact support, is continuous on that compact set, and hence uniformly continuous on that set (and hence all of $\mathbb{R}$, since $f\equiv0$ outside of the set.) –  Patch Jun 10 '12 at 12:12
    
Doesn't the $0$ in the subscript mean compact support? Then it follows from being continuous on a compact set... –  Seth Jun 10 '12 at 12:13
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People from PDE use $C_0^\infty$ to denote compactly supported functions. People from harmonic analysis (like Rudin) use $C_c$ to denote continuous functions with compact support, while the subscript 0 is used for "vanishing at infinity". –  Siminore Jun 10 '12 at 12:40
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I do not think that this is a duplicate of that question. –  mixedmath Jun 10 '12 at 12:59
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up vote 4 down vote accepted

HINTs

  1. A continuous function on a compact interval is uniformly continuous.
  2. $\lim_{|x| \to \infty} f(x) = 0$ means that $\forall \epsilon...$
  3. Split up the domain to use these two properties.
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