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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?

Keith
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    The answer to this question shows that for $g(x) := e^{-x^2}(e^{-x^2} + \sin^2 x),$ $\sqrt{g}$ is not Schwartz. We can bootstrap this to get a counterexample for your case if we can tweak $g$ so that $g - \alpha e^{-x^2}$ is nonnegative. To this end, first note that scaling $g$ doesn't change the main property, so we might as well assume $\alpha = 1$. Then just define $h(x) = g(x/\sqrt 2)$ and $f(x) = h(x) - e^{-x^2}$. – stochasticboy321 Feb 26 '24 at 15:37
  • @stochasticboy321 Thank you for your insightful counterexample. Well... this thwarts my attempt at original problem. – Keith Feb 26 '24 at 15:45
  • @stochasticboy321 Perhaps, could you provide any insight to the original problem as well? – Keith Feb 26 '24 at 15:45

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