On the extension of distribution

Question

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 $\mathbb{R}$?

Answer 1

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 $v$.