This question has been motivated by original and comments therein.
Let $f$ be a nonnegative Schwartz function on $\mathbb{R}^n$ and let $\alpha \in (0,\infty)$ be fixed. Then, I wonder if \begin{equation} F(x):=\sqrt{ f(x) + \alpha e^{-\lVert x \rVert^2}} \end{equation} is also a Schwartz function.
At least $F$ is well-defined, looks smooth and decays to $0$ as $\lVert x \rVert \to \infty$. Also, by somehow computing partial derivatives of arbitrary order, it seems possible to show that $F$ is indeed Schwartz. However, is there any more elegant way to show that $F$ is a Schwartz function?
Could anyone provide some insight?