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Let $\Omega \subset \mathbb R^n $ be an open subset. The space of smooth functions with compact support in $\Omega$ is then defined as $$C_c^{\infty}(\Omega):=\{ \varphi \in C^{\infty}(\Omega \, \vert \, \mathrm{supp }(\varphi) \subset \Omega \text{ is compact }\} $$ where $\mathrm{supp}(\varphi)=\{x \in \Omega \, \vert \, \varphi(x) \neq 0 \}$.

Very often I read notation $C_0^{\infty}(\Omega)$ which I only started to wonder about now. Why is there a $0$ instead of $c$ for "compact" in the index?

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$C_0(\Omega)$ is the completion of $C_c(\Omega)$ in the sup-norm, i.e., $f\in C_0(\Omega)$ if $f$ is continuous on $\overline{\Omega}\subset\mathbb{R}^n\cup\{\infty\}$ and vanishes at the boundary (including both the finite part and the $\infty$ if $\Omega$ is unbounded).

Similarly, $C_0^k(\Omega)$ is the completion of $C_c^k(\Omega)$ in the $C^k$-uniform norm (so derivative vanishes up to $k$-th order at the boundary), and $C_0^\infty(\Omega)=\bigcap_k C_0^k(\Omega)$.

For example, if $\Omega=(0,1)\subset\mathbb{R}$, then $f(x)=\exp(-x^{-2}(1-x)^{-2})$ belongs to $C_0^\infty(\Omega)$, but not $C^\infty_c(\Omega)$.


There are some who use $C_0(\Omega)$ for what we define as $C_c(\Omega)$, where the subscript $0$ is "justified" as a reminder that the function is 0 outside a compact set (urgh!).

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  • $\begingroup$ Thanks, so $C_0$ are the continuous functions that tend to $0$ as $x$ approaches the boundary, while $C_c$ actually ARE zero outside of a compact subset of $\Omega$ right? $\endgroup$ – Tesla Jun 14 '19 at 17:56
  • $\begingroup$ Yes, that is the idea. $\endgroup$ – user10354138 Jun 14 '19 at 18:02
  • $\begingroup$ Why is it then that for example here on page 15 he defines $C_0^{\infty}$ just as $C_c^{\infty}$? math.ucdavis.edu/~hunter/m218a_09/Lp_and_Sobolev_notes.pdf $\endgroup$ – Tesla Jun 14 '19 at 18:04
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    $\begingroup$ You have to ask the author why he decides to confuse the c and 0. The only justification I hear from people is that $0$ is for them a reminder it vanish outside a compact set, which IMHO is not a very good excuse (these people also tend to use $C_0$ for compactly supported continuous functions, so maybe for them there is no value in considering functions merely vanishing at the boundary.). $\endgroup$ – user10354138 Jun 14 '19 at 18:41
  • $\begingroup$ Ok thanks, just wanted to make sure that the author's definition does not coincide with the one mentioned here $\endgroup$ – Tesla Jun 14 '19 at 18:51

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