TITLE: Mollifiers: Integral Bound
QUESTION [5 upvotes]: Desirable is an example such that:
$$\varphi_0\in\mathcal{C}^\infty_0:\quad\int_0^\infty\left|\varphi_0^{(n)}(s)\right|\mathrm{d}s\leq2^n$$
(It should not exist as one would obtain entire elements for generators of semigroups.)
How to prove that they must always exceed the bound?

REPLY [2 votes]: Consider the Fourier transform of $\varphi_0$. We have
$$
\widehat{\varphi_0^{(n)}}(\xi)=C^n\,\xi^n\,\hat{\varphi_0}(\xi),
$$
where $C$ is a constant deppending on the particular definition of the Fourier transform used. Then
$$
|\widehat{\varphi_0^{(n)}}(\xi)|=|C|^n\,|\xi|^n\,|\hat{\varphi_0}(\xi)|\le\int|\varphi_0^{(n)}(x)|\,dx\le2^n\quad\forall\xi\in\mathbb{R},\quad\forall n\in\mathbb{N}.
$$
Then
$$
|\hat{\varphi_0}(\xi)|\le\Bigl(\frac{2\,|C|}{|\xi|}\Bigr)^n\quad\forall\xi\in\mathbb{R},\quad\forall n\in\mathbb{N}.
$$
This implies that $\hat{\varphi_0}$ is supported on $[-2\,|C|,2\,|C|]$, which is not possible since $\varphi_0$ has also compact support.