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?
Best Answer
$C_0(\Omega)$ 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!).