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!).

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.