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

2012-04-27T16:18:29Z

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?

Answer by Mark Grant for What is the usual topology of $C^\infty_c(M) $

2012-04-27T16:38:39Z

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

Answer by unknown (google) for What is the usual topology of $C^\infty_c(M) $

2012-04-27T17:28:23Z

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 $M$.

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

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

Answer by Mark Grant for What is the usual topology of $C^\infty_c(M) $

Answer by unknown (google) for What is the usual topology of $C^\infty_c(M) $