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Let's define the following set of functions:

$$C_c(\mathbb{R}) = \{f \in C(\mathbb{R}) | f(t) = 0, \forall t \text{ s.t. } |t| \geq T, T \geq 0\} $$

Then, can I say that for a general $f \in C_c(\mathbb{R}) $, $f$ is defined in $[-T,T]$ ? Because actually a general function is defined in all $\mathbb{R}$, but it is $0 \text{ } \forall \text{ } t \text{ } \geq T$.

(I need to say that because I want to use the fact that $f$ is defined in a compact set).

ofir_13
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  • While I don't think that it's necessarily wrong to say that $f$ is defined in $[-T, T]$, I think that it would be better to say that $f$ is defined on $\mathbb{R}$ and supported in $[-T, T]$. You can also talk about the restriction of $f$ to $[-T,T]$ (usually denoted by $f \mid_{[-T,T]}$), which is clearly well-defined as $f$ is defined on the whole $\mathbb{R}$. – MSDG Jul 29 '18 at 13:46
  • In either case though, you need to be aware that $f$ and its restriction to $[-T, T]$ are not the same function, strictly speaking. – MSDG Jul 29 '18 at 13:54
  • Because I need to prove that if $f \in C_c(\mathbb{R})$, then $f$ is uniformly continuous. So I was thinking of using the fact that if $f$ is continuous and defined on a compact set, then $f$ is uniformly continuous. – ofir_13 Jul 29 '18 at 13:57
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    In that case, I would probably write something along the lines of: Since $f$ is continuous and vanishes outside of $[-T, T]$, it suffices to prove that its restriction to $[-T,T]$ is uniformly continuous... – MSDG Jul 29 '18 at 14:00
  • If that's the case, then $f$ is uniformly continuous on $[-T, T]$ by Cantor's theorem in analysis. – xbh Jul 29 '18 at 14:04

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