# Distribution of infinite order

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.

If $\tilde\Lambda$ is an extension of $\Lambda$ and $\phi$ is a compactly supported function which is constantly equal to one on $[-1,1]$, then $\phi\tilde\Lambda$ is a distribution of infinite order, by the same reasoning you used in (1). But $\phi\tilde\Lambda$ has compact support, so that cannot happen.
Hint: If you can find a sequence of test functions $\phi_n$ such that each $\phi_n$ is supported in $(0,\infty)$ and $\phi_n \to \phi$ (in the usual sense), but $\Lambda(\phi_n)$ does not converge, then you will be done. What properties would such $\phi$ have to have?
I don't think there's any special reason to believe that an extension of $\Lambda$ would have to be given by the above series. In fact, in principle $\Lambda$ could have lots of different extensions (an extension could do whatever it wants on $(-\infty,0)$ for instance).