2
$\begingroup$

Hi all,

I apologize if this question is too low level for mathoverflow. I'm happy to move it to math.stackexchange if so.

Let $X$ be a closed manifold, let $k$ be a nonnegative integer and let $C^k(X)$ denote the space of $k$-times continuously differentiable functions equipped with the $C^k$ norm.

Is $C^{k+1}(X)$ compactly contained in $C^k(X)$? Does this follow from Arzela-Ascoli?

Thanks.

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ Yes. Please move it to MSE. $\endgroup$
    – timur
    Commented Sep 1, 2012 at 1:23
  • $\begingroup$ Thanks! Will do. Do you have a reference for this? $\endgroup$
    – trex
    Commented Sep 1, 2012 at 2:26

1 Answer 1

Reset to default
5
$\begingroup$

The answer depends on what you mean by a closed manifold and a compact operator. If the manifold is compact without boundary then the spaces involved are Banach spaces and the answer is yes and this is indeed proved using the Arzela-Ascoli theorem, after a standard localisation argument.

If the manifold is without boundary but not compact (think real line), then the spaces are Frechet spaces and the answer depends on what you mean by a compact operator on a locally convex space. If you define these as taking bounded sets into relatively compact ones, then the answer is yes, but if you mean those which take a neighbourhood of zero to a relativeley compact set (the usual definition), then the answer is no.

Improve this answer
$\endgroup$
2
  • 4
    $\begingroup$ I don't think the name "closed manifold" is ambiguous at all, in all the paper and books I read it always meant "compact without boundary". Wikipedia also agrees with me: en.wikipedia.org/wiki/Closed_manifold $\endgroup$
    – YangMills
    Commented Sep 1, 2012 at 18:29
  • $\begingroup$ I certainly meant compact without boundary. $\endgroup$
    – trex
    Commented Sep 2, 2012 at 1:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .