TITLE: Convergence of test-functions is not induced by any metric.
QUESTION [7 upvotes]: By $\mathcal{D}(\mathbb{R})$ we denote linear space of smooth compactly supported functions. We say that $\{\varphi_n:n\in\mathbb{N}\}\subset\mathcal{D}(\mathbb{R})$ converges to $\varphi\in\mathcal{D}(\mathbb{R})$ if 

for all $k\in\mathbb{Z}_+$ the sequence $\{\varphi_n^{(k)}:n\in\mathbb{N}\}$ uniformly converges to $\varphi^{(k)}$.
there exist a compact $K\subset \mathbb{R}$ such that $\mathrm{supp}(\varphi_n)\subset K$ for all $n\in\mathbb{N}$.

Could you give me a hint to prove the following well known fact.
There is no metric $d$ on $\mathcal{D}(\mathbb{R})$ such that convergence described above is equivalent to convergence in metric space $(\mathcal{D}(\mathbb{R}), d)$.

REPLY [6 votes]: As it is quite easy to prove it directly, I will add another answer.
Let $(\varphi_n)_n$ be a sequence in $\mathcal{D}$ with $\varphi_n(x)=1$ for $|x|\leq n$ and 
$\varphi_n(x)=0$ for $|x|>n+1$. Assume $d$ to be a metric on $\mathcal{D}$ compatible with the topology. Let $B_n$ be the ball around $0$ with radius $1/n$ in this metric. As each $B_n$ is absorbing, there is some $c_n>0$ with $c_n\varphi_n \in B_n$ for each $n$. Hence you have 
$$ c_n\varphi_n \longrightarrow 0 $$
in the metric $d$. By the above definition of the topology, there is a compact set $K\subset\mathbb{R}$ with $\operatorname{supp}c_n\varphi_n \subset K$ for all $n$, which is a contradiction.