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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
up vote 5 down vote accepted

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.

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I'd add a complaint/warning that Rudin's "description" of the topology on test functions is needlessly opaque, and does not explain at all why that would be the definition, etc. As L. Schwartz explained nicely c. 1950, this topology is an "inductive limit" of Frechet (so, metric) subspaces (each of which is nowhere dense...!). In effect, Rudin gives a construction of an inductive limit of topological vector spaces, without letting on what's happening, and in effect proves the ind lim properties as lemmas and theorems thereafter. Possibly an anti-Bourbakiste impulse? :) – paul garrett Aug 27 '12 at 16:50
@paulgarrett This is pedagogical impulse – Norbert Aug 27 '12 at 16:55
@Siminore That was elegant – Norbert Aug 27 '12 at 16:56
Rudin's books are all a bit strange; they present very elegant and general theories, but Rudin did not try to offer a perspective towards more general developments of the theories. These are the things you must learn, and do not ask for more! In baby Rudin, the Taylor expansion is contained in a rather useless theorem: no words about approximation, about different forms of the remainder, nothing at all. In Functional Analysis I think that Topological Vector Spaces are a bit overrated, while Banach spaces are confined to a short chapter. – Siminore Aug 27 '12 at 17:03
A good reference concerning all those notions is the book "Topological Vector Spaces, Distributions and Kernels" by F. Tréves. – Ahriman Aug 27 '12 at 17:11

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.

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