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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.2016-05-08

2 Answers 2

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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? :)2012-08-27
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    @paulgarrett This is pedagogical impulse2012-08-27
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    @Siminore That was elegant2012-08-27
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    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.2012-08-27
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    A good reference concerning all those notions is the book "Topological Vector Spaces, Distributions and Kernels" by F. Tréves.2012-08-27
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    @Ahriman I was looking in Treves' book, but I was not able to find a proof of the non-metrizability of $\mathcal{D}$. It seems difficult to find in books (except in Rudin's).2012-08-27
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    He doesn't give the proof, but he gives all elements needed to write it. The only thing he doesn't prove is the existence of a translation invariant metric for every metrizable TVS. (this is done in Bourbaki or Schaefer) But once you admit that; you have everything : chapter 13, page 128, remark 13.1 he proves that every LF-space which is not Fréchet is not a Baire space. From this, the previous fact and the fact that every LF-space is complete (theorem 13.1 in Trèves' book), you can deduce that every metrisable LF-space is Fréchet, which is not the case of $\mathcal{D}$.2012-08-27
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    I would add that LF space are complete with respect to the uniform structure induced by the TVS structure. This is why I need a translation invariant metric (which induces the same topology), because such a metric will induce the same uniform structure. Hence, a translation invariant metric on a LF-space is necessarily complete. So we can apply Baire's theorem. (which is valid for complete metric spaces, and not for complete uniform spaces)2012-08-27
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    In my penultimate comment, I was not completely accurate. In fact, Trèves proved that if a LF-space $E$ is a Baire space, then $E= E_n$ for some $n$, where $E_n$ a sequence of definition of $E$. If $\mathcal{D}$ was Baire, by above, it will be equal to some $\mathcal{D}(K)$ for some compact subset $K \subset \Omega$, which is obviously not the case.2012-08-27
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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.