If $f$ is Schwartz, i.e. $f\in\mathcal{S}\left(\mathbb{R}\right)$, is it true that$f=\mathcal{O}\left(e^{-a|x|^{\epsilon}}\right)$ for some $a,\epsilon>0$ as $|x|\to\infty$?
Functions in the Schwartz class, as well as as their derivatives of all orders, decay faster than any negative power of a monomial. I would expect that the answer to the question is yes, but I am not sure how to start thinking about the positivity or negativity of the statement.
The standard examples of Schwartz functions, such as $e^{-x^2}$, $x^2e^{-x^2}$, and $e^{-x^4}$ all satisfy this bound.
We do know that since $f$ is Schwartz, $f=\mathcal{O}\left(\frac{1}{|x|^m}\right)$, for all $m\in\mathbb{N}$. Additionally, $e^{-a|x|^{\epsilon}}=\mathcal{O}\left(\frac{1}{|x|^m}\right)$ for all natural $m$ and $a, \epsilon > 0$. This means that both $\mathcal{S}$ and $e^{-a|x|^{\epsilon}}$ are contained in $\displaystyle \bigcap_{m=1}^\infty \mathcal{O}\left(\frac{1}{|x|^m}\right)$ (again, for all $a, \epsilon > 0$). This inclusion at least lets us understand where these functions live a little better, but does not answer the initial question. What is an approach that could shed light on whether the statement in the question holds true?
(As a reminder, we say that $f(x)\in \mathcal{O}\left(g(x)\right)$, or equivalently $f(x)= \mathcal{O}\left(g(x)\right)$, if, beyond a sufficiently large value of $x$, $|f|$ is at most a constant times $g$.)
No. For example, if $h(x)$ is a smooth "bump function" with support in $(0,1)$ you could take $$f(x) = \sum_{n=2}^\infty n^{-\log n} h(x-n)$$