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If $M$ is a smooth paracompact manifold, then what is the usual topology of $C^\infty_c(M) $, i.e., the smooth function with compact support?

flag	asked Apr 27 2012 at 16:18 Adterram 299●1●6

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Crossposted to math.SE: math.stackexchange.com/q/137701/264 – Zev Chonoles Apr 27 2012 at 16:29

I don't know if this is usual, but it should be possible to define a metric by $$d(f,g) = \sum_n \frac{1}{2^{n+A(n)}}\sum_{|\alpha|=n}\frac{\left|\sup_K\frac\partial{\partial x^\alpha}(f-g)\right|}{1 + \left|\sup_K\frac\partial{\partial x^\alpha}(f-g)\right|}$$ where $A(n)$ is the number of $\alpha$ s.t. $|\alpha|=n$. The space should be complete in the induced topology. – Todd Leason Apr 27 2012 at 17:46

Added: $K$ has to be taken to include the support of $f,g$. – Todd Leason Apr 27 2012 at 17:50

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Todd: smoothly truncating $e^{-x^2}$ on $\mathbb R$ so as to obtain a sequence of compactly supported functions appropriately should give a Cauchy sequence in that metric which does not converge, no? – Mariano Suárez-Alvarez Apr 27 2012 at 18:38

unknown (google) · Accepted Answer · 2012-04-27 17:28:23Z UTC

Topologizing $C_c^\infty(M)\subseteq C^\infty(M)$ with the subspace topology (where $C^\infty(M)$ has the Whitney topology, generated by the seminorms $\left|\sup_K\frac\partial{\partial x^\alpha}f\right|$), makes it a dense subspace; in particular it is not itself complete. So I wouldn't really call this the "usual topology" on $C_c^\infty(M)$. (it would be sort of like saying the usual topology on $C(M)$ is given by the $L^2$ norm).

To me the usual topology is the inductive limit topology $C_c^\infty(M)=\lim_{K\subseteq M}C_c^\infty(K)$ (which Mariano calls the colimit topology). This topology is not metrizable when $M$ is noncompact (since it's not even first-countable), but is "nicer" in the sense that it gives a well-understood dual space, namely the space of distributions on $M$.

In comparison, the dual space of $C^\infty(M)$ with the Whitney topology is the space of compactly supported distributions on $M$.

Mark Grant · Answer 2 · 2012-04-27 16:38:39Z UTC

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Give $C^\infty(M)$ the Whitney topology, then topologise $C^\infty_c(M)\subseteq C^\infty(M)$ as a subspace.

answered Apr 27 2012 at 16:38

Mark Grant
9,061●1●13●33

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One could also topologise it as the colimit of the $C^\infty(K)$, with $K\subseteq M$ compact, à la Schwartz's space of test functions, depending on what exactly one is trying to achieve. – Mariano Suárez-Alvarez Apr 27 2012 at 16:45

What is the usual topology of $C^\infty_c(M) $

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