show that the formula $\langle S,\phi\rangle= \sum_{n=1}^{+\infty} \phi^{(n)}(1/n)$ defines a distribution over $\mathbb{R^{*}}$

Question

I'm stuck and I can't see the necessity of $\mathbb{R^{*}}$ instead of $\mathbb{R}$

what I noticed so far :

partial sum defines a distribution because for every $\phi \in \mathcal{D}(\mathbb{R})$ (thus for every $\phi \in \mathcal{D}(\mathbb{R^{*}})$ ):

$|\langle S_N,\phi\rangle|\leq\sum_{n=1}^{N} |\phi^{(n)}(1/n)| \leq N\sum_{n=1}^{N}\|\phi^{(n)}\|_{\infty}$

and if $S_N$ converges to a distribution then it must be $S$

any help or hints will be very much appreciated. thanks !

Answer 1

For a test function on $\mathbb{R}^*$, the sum will have finitely many terms since the function is zero on a neighborhood of zero.

If you have a test function in which zero is inside the support, the sum may not converge, eg if $\phi(x)=e^x\psi(x)$, where $\psi$ is smooth compactly supported and $1$ on $(-2,2)$, then each term is $e^{1/n} \rightarrow 1$.