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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?

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    $\begingroup$ 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 $k=0$)... $\endgroup$
    – Martin Hairer
    Commented Jul 16 at 20:42
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    $\begingroup$ 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 $C^{-k}\to C^{-k-1}$. $\endgroup$
    – Pietro Majer
    Commented Jul 16 at 20:55
  • $\begingroup$ @MartinHairer Thank you very much! $\endgroup$
    – CBBAM
    Commented Jul 17 at 1:43
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    $\begingroup$ As $-D^*:(C^{k})^*\to (C^{k+1})^*$... $\endgroup$
    – Pietro Majer
    Commented Jul 17 at 4:15
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    $\begingroup$ @PietroMajer But don't forget that as mentioned above there is no natural convention under which $C^{-k}$ is the dual of $C^k$. $\endgroup$
    – Martin Hairer
    Commented Jul 17 at 5:43

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