Given $u \in \mathcal{C}^\infty_0(\mathbb{R}^n)$, $u \geq 0$ everywhere, is $v(x) = \sqrt{u(x)}$ also in $\mathcal{C}^\infty_0$? It is clear that the only problematic points are the boundary of the support, where one must show that all the derivatives vanish.
I would appreciate any help!