Distribution with singularities.

Question

I need some help to prove that $f$ defined by $\langle f,\psi\rangle:= \sum_{n=0} ^\infty \psi^{(n)}(n)$ is a distribution which has singularities of infinite order. Here $\psi$ is a test function that belongs to $ \mathcal D(\Bbb R)$.

Thanks.

Answer 1

I will give you the main idea. Then I believe you can deal with the technical details.

You may already suspect that

$$ f=\sum_{n=0}^{\infty} (-1)^n \delta^{(n)}_n = \mathcal D - \lim_{N\rightarrow \infty}\sum_{n=0}^{N} (-1)^n \delta^{(n)}_n $$

where the equality and the limit are taken in the sense of distribution. You can verifie that the distribution $S_N = \sum_{n=0}^{N} (-1)^n \delta^{(n)}_n$ is well defined and also its limit $f$.

Then, since $\delta^{(n)}_x$ is a distribution of order $n$, you have automatically the wanted result.