This is an exercise from Functional Analysis By Walter Rudin on page 178 from chapter Test functions and Distribution. I am having trouble arguing for 2nd part.
Question: For $\Omega=(0, \infty)$
$\displaystyle \Lambda(\phi) = \sum_{m=0}^{\infty}{ D^m\phi(1/m)} $
1) $\Lambda$ is a distribution of infinite order.
In small neighborhood around $\frac{1}{m}$ it's restriction would be $\delta^{(m)}$ which has order m, hence $\Lambda$ has to be of infinite order.
2) Here, we need to show that $\Lambda$ can not be extended to $R$. For this we take infinitely differentiable function $\psi$ with compact support and taking value 1 on compact set $K$ around 0.
$\phi(x) = \psi(x) \exp{ax} ; a>1$ Now, this function $\phi$ makes the series given above diverge. But how to argue that $\Lambda(\phi)$ will be the series given above for $\Lambda $ extended to R.