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?


  • 3
    $\begingroup$ 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. $\endgroup$ Jun 8, 2015 at 1:11
  • $\begingroup$ I modified the formulation a little bit. The above comment is for the initial formulation I used: $f^{(n)}(x)=P_n(x)\cdot f(x)$. $\endgroup$
    – Pengfei
    Jun 8, 2015 at 4:07
  • 1
    $\begingroup$ 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. $\endgroup$ Jun 8, 2015 at 4:28
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    $\begingroup$ @Pengfei Have you tried $||f^{(n)}||_\infty \le ||\widehat{f^{(n)}}||_1 = ||\xi^n \hat{f}(\xi)||_1 = \int_{-\infty}^\infty |\xi|^n|\hat{f}(\xi)|d\xi$? $\endgroup$ Jul 21, 2019 at 12:55
  • 1
    $\begingroup$ @Pengfei Or just consider your function in the complex plane and use Cauchy bounds choosing radius appropriately. $\endgroup$
    – fedja
    Jul 21, 2019 at 19:01

2 Answers 2


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 $\frac1{\alpha e}$.


$$ f^{(n)}(x)=\frac{P_n(x)}{(1-x^2)^{2n}} \exp\left(-\frac{1}{1-x^2}\right)=P_n(x) h_{1/2n}(x)^{2n} $$

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 $3n$:

$$ |P_{n+1}|_1\leq 12n|P_n|_1 + 6n|P_n|_1 + 2|P_n|_1 $$

$$ |P_{n+1}|_1\leq 18(n+1)|P_n|_1 $$


$$ |P_n|_1\leq 18^n n! $$

And finally:

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

  • $\begingroup$ 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 $|P_n(\pm 1)|=2^n$. $\endgroup$
    – pre-kidney
    Jul 22, 2019 at 6:16
  • $\begingroup$ @pre-kidney Yes I have little doubt that this bound is far from the actual maximum by several orders of magnitude! $\endgroup$
    – FXV
    Jul 22, 2019 at 17:54
  • $\begingroup$ It seems that this method cannot be tightened much because it is indeed the case that $\sup_{[-1,1]} P_n \gg |P_n(\pm 1)|$. $\endgroup$
    – user58955
    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.

  • $\begingroup$ This is a very interesting approach! $\endgroup$
    – Pengfei
    Oct 14, 2019 at 14:56

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