Bounding rapidly decreasing function by Schwartz function Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be continuous, and suppose that it is rapidly decreasing, i.e. for every $n\in\mathbb{N}$ we have
$$\sup\limits_{x\in\mathbb{R}}{|x^nf(x)|}<\infty.$$
Can we always find a Schwartz function $\varphi:\mathbb{R}\rightarrow\mathbb{R}$ such that $|f|\le\varphi$?
In other words, can we bound a function that is "zeroth order Schwartz" by a function nice enough to actually be Schwartz?
I was thinking of explicitly constructing such a $\varphi$ by having it match the values of $|f|$ at its maxima, and smoothly interpolating in between, but it is not clear if it would be easy to prove that such a construction is indeed Schwartz.

Sure with $g(x) = \sup_{y \in [x-1,x+1] } |f(y)|$ then $$g \ast \varphi(x) = \int_{-\infty}^\infty g(y) \varphi(x-y)dy$$ is Schwartz where $$\varphi(x) = 3 \, e^{-1/(1-x^2)} 1_{|x| < 1} \in C^\infty_c$$
