TITLE: On the extension of distribution
QUESTION [5 upvotes]: Define a distribution on $(0,+\infty)$ by
$$u(\varphi):=\sum_{k=1}^{\infty} {1 \over {k!}}\partial^k \varphi(1/k)$$
how can I show it cannot be extended to any distribution defined globally on $\mathbb{R}$?

REPLY [2 votes]: For $v\in \mathscr D'(\mathbb R)$ there are $n\in\mathbb N$ and $c>0$ such that
$|v(\varphi)|\le c\sup\lbrace|\varphi^{(k)}(x)|: x\in [0,1], k\le n \rbrace$
for all test functions $\varphi$ with support in $[0,1]$. Construct
$\varphi_n$ with support in a very small interval around $1/n+1$ such that 
$\varphi^{(k)}(1/n+1)=0$ for $k\neq n+1$ and $=2c(n+1)!$ for $k=n$ by multiplying
$(x-1/n+1)^{n+1}$ with a constant times a cut-off function.
This will show that $u$ isn't the restriction of $v$.