# Two different notations for $C_c^{\infty}(\Omega)$

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?

$$C_0(\Omega)$$ 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!).

• 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? – Tesla Jun 14 '19 at 17:56
• Yes, that is the idea. – user10354138 Jun 14 '19 at 18:02
• 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 – Tesla Jun 14 '19 at 18:04
• 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.). – user10354138 Jun 14 '19 at 18:41
• Ok thanks, just wanted to make sure that the author's definition does not coincide with the one mentioned here – Tesla Jun 14 '19 at 18:51