Square root of compactly supported C-infinity function

Question

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!

Answer 1

There is an example of a nonnegative $u\in\mathcal{C}^\infty_0(\mathbf{R}^2)$ which does not admit a differentiable square root. Namely, $u = (x^2 + y^2)\varphi$, where $\varphi$ is a bump localised at $0$. The only continuous square roots of $u$ are $\pm\sqrt{x^2 + y^2}\sqrt{\varphi}$, neither of which are differentiable at $0$.

Answer 2

Edit: my previous incorrect answer was: .... "to show that the $n$-th derivative of the square root exists (and vanishes) at a boundary point, use the existence and vanishing of all derivatives up to order $2n$ of the original function, in a Taylor-Maclaurin expansion with remainder."

Edit: But there are already problems at interior points, as comments and examples show.