Bounds of the derivatives of the mollifier function

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:

$$|f^{(n)}|_\infty \leq \left(\frac{2n\sqrt{18}}e\right)^{2n}n!$$

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.