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I struggling against the mathematical language used to write the expressions below:

$u \in C^{\infty}_{c}(\rm I\!R^{n})$ -> This says that $u$ belongs to the class of infinitely differentiable funcions on the $\rm I\!R^{n}$, right? What about that $c$ index in $C^{\infty}_{c}$?

$\varphi \in C^{\infty}(\rm I\!R^{n},\rm I\!R)$ -> What does $(\rm I\!R^{n},\rm I\!R)$ tells me about $\varphi$?

Thank you very much.

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    $\begingroup$ The $c$ means that the function has compact support the support of a function being the closure of $\{x\in \mathbb{R}^n,\,f(x)\neq 0\}$ $\endgroup$
    – marwalix
    Oct 27, 2015 at 14:03
  • $\begingroup$ $c$ means that $u$ has compact support, that is $u=0$ out of big enough ball. $(R^n, R)$ tells that $\varphi$ acts from $R^n$ to $R$ $\endgroup$ Oct 27, 2015 at 14:04

1 Answer 1

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The index $c$ means that $\phi$ vanishes outside a compact subset of the $\mathbf R^n$, the so called support $$ \def\s{\operatorname{supp}}\s\def\R{\mathbf R} \phi = \overline{\{x \in \R^n : \phi(x) \ne 0\}} $$ is compact.

$\phi \in C^\infty(\R^n, \R)$ means that $\phi$ is a $C^\infty$ function form $\R^n$ to $\R$, that is $\phi \colon \R^n \to \R$, one can of course also consider functions $\phi \in C^\infty(\R^n, \R^m)$ or with values in some space $X$, written $C^\infty(\R^n, X)$.

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