Splitting a smooth bump function into a product Given a smooth bump function $\phi$ which vanishes outside of a compact set $K$, can I split it into a product $\phi = \phi_1 \phi_2$ where both $\phi_1$ and $\phi_2$ are smooth bump functions vanishing outside of $K$?
Basically I want to set $\phi_1 = \phi_2 = \sqrt{\phi}$, but taking the square root is not usually smooth. Is there a reason why the square root of a smooth bump function is smooth?
 A: You have to put some mild restrictions on your bump function. For example, if $\phi$ has an isolated zero, then it might not be possible to write $\phi$ as a product of two nonnegative smooth functions. (Think about a function $\phi\colon\mathbb R\to \mathbb R$ that is equal to $x^2$ in a neighborhood of the origin.)
If $\phi>0$ in $\operatorname{Int}K$, where $K = \operatorname{supp}\phi$, then $\sqrt{\phi}$ is smooth.  Clearly it is smooth everywhere in $\operatorname{Int}K$ (because $\phi>0$ there) and everywhere in the complement of $K$ (because $\phi\equiv 0$ there).  If $p\in \partial K$, then there is a sequence of points $p_i\notin K$ such that $p_i\to p$. Because each partial derivative of $\phi$ of any order vanishes at each point $p_i$, by continuity it follows that all partial derivatives of $\phi$ vanish at $p$. From that observation, it follows that $\phi$ vanishes to infinite order at $p$: for every $N$, there is a constant $C_N$ such that
$$
|\phi(p+h)| \le C_N |h|^N.
$$
The same is true for $\sqrt{\phi}$, and from there it's straightforward to prove that all partial derivatives of $\sqrt{\phi}$ at $p$ exist and are equal to zero; thus $\sqrt{\phi}$ is smooth.
