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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 $$ \begin{split} P_0(x) &\equiv 1, \\ P_1(x) &=-2x, \\ \vdots & \quad\vdots\\ P_{n+1}(x) &=P_n'(x)(1-x^2)^2+4nx(1-x^2) P_n(x)-2xP_n(x)\\ \vdots & \quad\vdots\\ \end{split}$$ (here I have borrowed the writing from this Q&A). I am interested in the $L^1$ norm of $f^{(n)}$. Is there a good estimate (upper bound) for $\|f^{(n)}\|_1$? In the referenced question, an estimate for $\|f^{(n)}\|_\infty$ is obtained, but I could not find one for the $L^1$ norm... Thank you!

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    $\begingroup$ (i) What do you mean by "a good estimate (upper bound)"? (ii) If $b_n$ is an upper bound on $\|f^{(n)}\|_\infty$, then obviously $2b_n$ is an upper bound on $\|f^{(n)}\|_1$. However, the upper bound on $\|f^{(n)}\|_\infty$ that you referred to seems far from good, in about any sense. So, what is your hope here? (iii) Is $|\cdot|$ your way to denote $\|\cdot\|$? $\endgroup$
    – Iosif Pinelis
    Commented Sep 1, 2023 at 17:49
  • $\begingroup$ Thank you for your comments. (i) If possible something "close"to the optimal. I was also hoping to get a sense of what the optimal bound is... (ii) perhaps it's not true, but something exponential in $n$... I also wanted to see what bounds were available (iii) let me change the notation $\endgroup$
    – Johnny T.
    Commented Sep 1, 2023 at 18:05
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    $\begingroup$ I think the asymptotic is something like $\|f^{(n)}\|_1\sim\frac cn\,\|f^{(n)}\|_\infty$ for some real $c>0$, and it also seems that $\ln\|f^{(n)}\|_\infty\sim2n\ln n=\ln(n^{2n})$, so that the growth of $\|f^{(n)}\|_1$ seems to be much faster than exponential. But, anyhow, how good do you need the bound to be in your research? $\endgroup$
    – Iosif Pinelis
    Commented Sep 1, 2023 at 18:15
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    $\begingroup$ There is also another construction of the $C^\infty$ compactly supported mollifier using the infinite convolution of characteristic functions of intervals. That one will allow you to control any norm of high derivatives you want rather explicitly and is usually most flexible in this respect for all practical or theoretical purposes. $\endgroup$
    – fedja
    Commented Sep 1, 2023 at 18:22
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    $\begingroup$ A more precise estimate for the sup-norm of the derivatives of $e^{-1/t}$ follows from Cauchy formula for $e^{-1/z}$ on balls of center $t>0$ and radius $t/2$ and it gives $C^n (n!)^2$. The same behaviour for your function follows using Leibnitz rule for the derivatives of a product. The growth cannot be like $C^n n!$ since the function is not analytic. $\endgroup$
    – Giorgio Metafune
    Commented Sep 1, 2023 at 21:19

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