Could a functional be defined to be with compact support?

Question

Could a functional, $F:C^\infty\to\mathbb R$ be defined to have the properties of rapidly decreasing and/or with compact support, just like the real-valued functions?

Answer 1

No. In order to understand why, imagine that $F$ had compact support $K \subset C^\infty (X)$. Let $f \in K$. It follows that $F(f) \ne 0$, so by the continuity of $F$ there must exist a whole neighbourhood $U$ of $f$ in $C^\infty (X)$ on which $F \ne 0$, which means that $U \subseteq K$. This means that $f$ has $\overline U$ for a compact neighbourhood. By translating this $\overline U$ (addition is continuous), it follows that every element in $C^\infty (X)$ admits a compact neighbourhood, so $C^\infty (X)$ must be locally-compact. But it is a known result of Weil that a topological vector space is locally-compact if and only if it is finite-dimensional, which $C^\infty (X)$ is clearly not.