Bounds of the derivatives of the mollifier function 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!
 A: First we want to study
$$h_\alpha:x\mapsto \frac{\exp\left(-\frac{\alpha}{1-x^2}\right)}{1-x^2}=Xe^{-\alpha X}$$
for $x \in [0,1]$ i.e. $X \in [1,+\infty[$.
The second expression $H_\alpha(X)=Xe^{-\alpha X}$ is positive over $\mathbb R_+$, and zero for $X=0$ and $X\rightarrow +\infty$, so there must be a maximum on which $H_\alpha'(X^*)=0$.
But there is only one zero of the differential, for $X=\frac1\alpha$.
So $h_\alpha$ is bounded by $\frac1{\alpha e}$.
Rewriting:
$$ f^{(n)}(x)=\frac{P_n(x)}{(1-x^2)^{2n}} \exp\left(-\frac{1}{1-x^2}\right)=P_n(x) h_{1/2n}(x)^{2n} $$
we get that:
$$|f^{(n)}|_\infty\leq \left(\frac{2n}e\right)^{2n}\sup_{[-1,1]}|P_n|$$
Now we want to bound $\sup_{[-1,1]}|P_n|$ by $|P_n|_1$ the sum of the absolute values of the coefficients of $P_n$, and use the inductive formula to find a probably give a very rough bound:
$$P_{n+1}(x)=P_n'(x)(1-x^2)^2+4nx(1-x^2) P_n(x)-2xP_n(x)$$
By triangular inequality:
$$ |P_{n+1}|_1\leq|(1-x^2)^2P_n'|_1 + 4n|(1-x^2) P_n|_1 + 2|P_n|_1 $$
Because $P_n$ is of degree less than $3n$:
$$ |P_{n+1}|_1\leq 12n|P_n|_1 + 6n|P_n|_1 + 2|P_n|_1 $$
$$ |P_{n+1}|_1\leq 18(n+1)|P_n|_1 $$
Thus:
$$ |P_n|_1\leq 18^n n! $$
And finally:
$$|f^{(n)}|_\infty \leq \left(\frac{2n\sqrt{18}}e\right)^{2n}n!$$
A: 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.
