8
$\begingroup$

I apologize if this question is too basic for this site, but I posted it on mathSE and did not get any responses (link can be found here) so I'm crossposting it here.

Let $C^k(\mathbb{R}^n$) be the space of functions with $k$ continuous derivatives, and $H^s(\mathbb{R}^n)$ the Sobolev space $W^{2,s}$. Their continuous dual spaces are commonly denoted as $C^{-k}$ or $H^{-s}$. Similar notations exist for duals to Holder spaces, Besov spaces...etc.

I had always thought the negative index was only notational, but it seems it has a genuine interpretation as "lacking regularity". For example, the space $C^{-k}$ can be seen as "lacking $k$ derivatives". And similarly, if I take the derivative of a $C^0$ function then the result is a distribution belonging to $C^{-1}$. How can one see this?

What justifies the interpretation that a space with index $-k$ "lacks $k$ derivatives"? Why are distributions/dual spaces the right setting for functions that lack some degree of differential regularity?

Improve this question
$\endgroup$
8
  • 4
    $\begingroup$ I would recommend you read through Section 14.3 of hairer.org/notes/RoughPaths.pdf which answers your questions (and more). It's five pages, it's self-contained (except maybe for a couple of definitions), and it uses only the definition of distributions and elementary calculus. Note that $C^{-k}$ is quite different from the dual of $C^k$ (think of $k=0$)... $\endgroup$ Commented Jul 16, 2024 at 20:42
  • 2
    $\begingroup$ I’d say the reason (say in one variable) relies in the fact that, first of all, the distributional derivative takes $C^0(I)$ to its dual, and for the rest, it is defined as $-D^*$, so that from the derivative $C^{k+1}\to C^k$ you still get a derivative $C^{-k}\to C^{-k-1}$. $\endgroup$ Commented Jul 16, 2024 at 20:55
  • $\begingroup$ @MartinHairer Thank you very much! $\endgroup$ Commented Jul 17, 2024 at 1:43
  • 4
    $\begingroup$ As $-D^*:(C^{k})^*\to (C^{k+1})^*$... $\endgroup$ Commented Jul 17, 2024 at 4:15
  • 4
    $\begingroup$ @PietroMajer But don't forget that as mentioned above there is no natural convention under which $C^{-k}$ is the dual of $C^k$. $\endgroup$ Commented Jul 17, 2024 at 5:43

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.