1
$\begingroup$

Let $U \subset \Bbb R ^n$ be an open subset and let $\mathcal C$ be the space of the smooth functions on $U$ that vanish at infinity, endowed with the seminorms $p_\alpha (f) = \sup \limits _{x \in U} |\partial _\alpha f|$ (or any other equivalent combinations of them), $\alpha$ being a multiindex. Is its dual a well-known space?

I could also make $\mathcal C$ a Banach space by including in the definition the requirement $\sup \limits _\alpha \sup \limits _{x \in U} |\partial _\alpha f| < \infty$. Would this make the dual more convenient to work with?

Or maybe I should just ask: what topology to put on $\mathcal C$ (if any) in order to make it a civilized, well-studied space? I am asking this because its dual would share a number of nice properties with the space of distributions, and might prove to be the correct choice in those contexts where smooth functions with compact support are not an option.

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ I think a better choice would be the space of functions which not only vanish at infinity themselves but also have all their derivatives vanishing at infinity. Then this becomes a certain subspace of the usual $C^\infty$-space of the one-point-compactification of $U$ (which is a manifold in many interesting cases so it makes sense to talk about its $C^\infty$ space) $\endgroup$
    – Johannes Hahn
    Commented Sep 23, 2015 at 21:29
  • $\begingroup$ @JohannesHahn: Indeed, maybe this is what I should look for. Nevertheless, even though the compatification of $U$ might turn out to be a manifold itself, it will not be a Riemannian one, since the restriction of the Euclidean metric does not vanish at infinity on $U$. $\endgroup$
    – Alex M.
    Commented Sep 24, 2015 at 3:11
  • 1
    $\begingroup$ The dual spaces of the first two suggested spaces are just certain spaces of distributions on $U\,$. What is "convenient to work with", depends on what one aims to work for. This does not become clear enough from the question, and so there is no proper answer. $\endgroup$
    – TaQ
    Commented Sep 24, 2015 at 4:17

0

Reset to default

You must log in to answer this question.