TITLE: Bounds of the derivatives of the mollifier function
QUESTION [12 upvotes]: The standard mollifier function is defined as follows 
$$f(x)=\begin{cases} 0 & \text{if } |x| \ge 1\\ \exp \left(-\cfrac{1}{1-x^2}\right) & \text{if } |x|<1.\end{cases}$$
It is well known that $f$ is $C^\infty$, and $f^{(n)}(x)=0$ for $|x| \ge 1$. On the interval $x\in (-1,1)$, the derivative 
$$\displaystyle f^{(n)}(x)=\frac{P_n(x)}{(1-x^2)^{2n}}\cdot f(x)$$
where $P_n$ is a polynomial function of $x$ defined inductively by
$$P_0(x) \equiv 1, \qquad P_1(x)=-2x, \qquad P_{n+1}(x)=P_n'(x)(1-x^2)^2+4nx(1-x^2) P_n(x)-2xP_n(x)$$ 
Note that $\displaystyle \sup_{|x|<1} f^{(n)}(x)<+\infty$, since $f^{(n)}(\pm 1)=0$. So $\displaystyle |f|_n:=\max_{|x|<1} f^{(n)}(x)$ is well defined.
Are there some rough/good estimates on the size $|f|_n$ of the derivatives?
Thanks!

REPLY [4 votes]: As mathworker21 suggested in a comment, 
$$
\|f^{(n)}\|_\infty \leq \|\widehat{f^{(n)}}\|_1 = \|(i\xi)^n\hat f(\xi)\|_1  = \int |\xi|^n |\hat f(\xi)| d\xi.
$$
This paper (https://arxiv.org/abs/1508.04376) shows that $|\hat f(\xi)|$ decays as $|\xi|^{-3/4}e^{-\sqrt{|\xi|}}$. Hence, when $|\xi| \gtrsim (Cn\log n)^2$, we have $|\xi|^n |\hat f(\xi)|\leq |\xi|^{-2}$. On the other hand, $|\hat f(\xi)| \leq \|f\|_1 = 1$. This gives
$$
\int |\xi|^k |\hat f(\xi)| d\xi \lesssim (C n\log n)^{2n+2}.
$$
This improves FXV's bound with an $n!$ factor in it but still seems suboptimal by an $n^2(\log n)^{2n+2}$ factor.