Derivation on $ {\cal C}^\infty_0 (\mathbb R)$

Question

It can be proven that

$X \equiv \{f \in {\cal C}^\infty_0(\mathbb R) \ | \ \exists \ g \in {\cal C}^\infty_0(\mathbb R)$ that verifies $ g' = f \}$ is isomorphic to $Y \equiv \{ f \in {\cal C}^\infty_0 (\mathbb R)\ | \ \int_\mathbb R f \ dx = 0\}$

so, in other words

Every derivative of a compactly supported function is compactly supported if and only if it's odd.

Question. Is there an example of a compactly supported function whose derivative is not compactly supported?

Answer 1

Having integral equal to $0$ is not what I would call "being odd": a function $f$ is odd if and only if $f(-x)=-f(x)$ for all $x\in\Bbb R$.

A function with compact support - say, such that it's constant $0$ on $U=(-\infty,a)\cup (b,\infty)$ - will have zero derivative on $U$ (regardless of the overall regularity of the function itself). Therefore, the derivative of a compactly supported differentiable function is compactly supported.