# Convergence of test-functions is not induced by any metric.

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 $(\mathcal{D}(\mathbb{R}), d)$.

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Both answers seem to assume that the metric induces the topology of $\mathcal{D}(\mathbb{R})$. Vobo's answer works if one assumes that $d$ induces some TVS topology on $\mathcal{D}(\mathbb{R})$. But the question as it stands does not require that the metric be compatible with the vector space structure. That would be a sensible requirement - completely arbitrary metrics aren't of much use - so you may want to include it in the question. – Daniel Fischer May 8 at 21:16

It is a consequence of the Baire category theorem. Essentially, $\mathcal{D}$ is of first category in itself and Cauchy sequences converge in $\mathcal{D}$, and this prevents metrizability. You can find a complete discussion in paragraph 6.9 of the book Functional analysis by Walter Rudin.
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.