# 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!

• The local extrema of $f^{(n)}(x)$ can be found by computing the critical points of $f^{(n+1)}(x)$. These are the same as the zeroes of $P_{n+1}(x)$. Once you have those, you just evaluate $f^{(n)}(x)$ at them to find your global extrema. This, of course, needs you to find a decent formula for $P_{n+1}(x)$ and its zeros. That could be a good deal more challenging, as finding roots of polynomials is not such a simple task. Playing with wolfram alpha suggests within (-1,1) that there is one root at zero for $n+1$ odd, and no roots when $n+1$ even. I'm not sure of a proof of that, though. Jun 8, 2015 at 1:11
• I modified the formulation a little bit. The above comment is for the initial formulation I used: $f^{(n)}(x)=P_n(x)\cdot f(x)$. Jun 8, 2015 at 4:07
• My comment holds in the new formulation, as well, actually. The inductive formula may be useful for proving the apparent pattern I suggested. Incidentally, you need to replace the $\max$ with a $\sup$: it is not immediately clear that maximum is attained in the open interval. Using $\max |f^{(n)}(x)|$ would also work, but may give you something other than what you were looking for. Jun 8, 2015 at 4:28
• @Pengfei Have you tried $||f^{(n)}||_\infty \le ||\widehat{f^{(n)}}||_1 = ||\xi^n \hat{f}(\xi)||_1 = \int_{-\infty}^\infty |\xi|^n|\hat{f}(\xi)|d\xi$? Jul 21, 2019 at 12:55
• @Pengfei Or just consider your function in the complex plane and use Cauchy bounds choosing radius appropriately. Jul 21, 2019 at 19:01

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:

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

• In fact, numerical calculations appear to show that $|f^{(n)}|_{\infty}=(2ne^{-1}+o(1))^{2n}$. However, it is also interesting to see that your method can not do better than $(4ne^{-1})^{2n}$, since $|P_n(\pm 1)|=2^n$. Jul 22, 2019 at 6:16
• @pre-kidney Yes I have little doubt that this bound is far from the actual maximum by several orders of magnitude!
– FXV
Jul 22, 2019 at 17:54
• It seems that this method cannot be tightened much because it is indeed the case that $\sup_{[-1,1]} P_n \gg |P_n(\pm 1)|$. Oct 11, 2019 at 8:24

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.

• This is a very interesting approach! Oct 14, 2019 at 14:56