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I am looking for a good way to write down what $H^{-1}(\mathbb{S}^1)$ is. What do I mean by this: Well, certainly one can define this space via charts that map from the sphere into $\mathbb{R}^2$ and then use this definition to give this space some sense, but I have the feeling that this is too abstract in this situation.

For instance, if I want to describe $H^2(\mathbb{S}^1)$ then I can simply notice that this is the domain of the Laplacian, so

$$ H^2(\mathbb{S}^1):=D(-\Delta) = \{f \in L^2(\mathbb{S}^1); \sum_{n=-\infty}^{\infty} |n^2\langle f, e^{-in \cdot} \rangle|^2< \infty \}$$ by the Fourier expansion of the Laplacian, i.e. there is apparently no need for abstract chart definitions in case of $H^2$. So my question is: Is there also a nice characterization of $H^{-1}$ in this case?

Edit: So following Christian Remling's suggestion I assume that the right definition would be something like:

$$\{f \in ?; \sum_{n=-\infty}^{\infty} |(1+|n|)^{-1}\langle f, e^{-in \cdot} \rangle|^{2}< \infty\}.$$

My question would then be: What kind of objects do we allow for $f$?

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    $\begingroup$ I don't think what you wrote down is what one would usually denote by $H^2$ (two derivatives in $L^2$). In any event, the characterization via Fourier coefficients works for negative exponent also. $\endgroup$
    – Christian Remling
    Commented Apr 30, 2016 at 19:49
  • $\begingroup$ @ChristianRemling thank you for pointing out the typo. Unfortunately, I don't really understand. You certainly want $L^2 \subset H^{-1},$ right? So if I would just adapt this notation, then I would still consider $\{f \in L^2; \text{some condition is satisfied}\}$ and there are no tempered distributions on $S^1$. $\endgroup$
    – Lewandowski
    Commented Apr 30, 2016 at 19:53
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    $\begingroup$ There's no reason why Fourier series have to be restricted to ones in $L^2$. $\endgroup$
    – Deane Yang
    Commented Apr 30, 2016 at 20:07
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    $\begingroup$ Is there a reason the standard interpretation of $H^{-1}$ being the dual of $H^1$ doesn't satisfy you? $\endgroup$
    – Johannes Hahn
    Commented Apr 30, 2016 at 20:08
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    $\begingroup$ And why does Johannes Hahn's situation imply the use of charts? You can define $H^{-1}$ as the space of bounded linear functionals on $H^1$. $\endgroup$
    – Deane Yang
    Commented Apr 30, 2016 at 20:31

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I am somewhat confused that, despite saying many true and useful things, no one has said directly that $H^{-1}$ on the circle can be characterized as the set of distributions $\theta$ such that $\sum_{n\in \mathbb Z}(1+n^2)^{-1}\cdot |\theta(x\to e^{inx})|^2<\infty$. (All distributions on the circle are compactly supported, so can be applied to the exponentials, which are smooth.)

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    $\begingroup$ Isn't this what Christian Remling suggested? $\endgroup$
    – Deane Yang
    Commented Apr 30, 2016 at 20:43
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    $\begingroup$ @DeaneYang, probably! And, when I saw that first, I thought "we're done"... But no one else elaborated, and things seemed to go off in a different direction, so I couldn't resist being explicit. $\endgroup$
    – paul garrett
    Commented Apr 30, 2016 at 20:44
  • $\begingroup$ Paul, I agree that your explanation is more explicit and clearer. $\endgroup$
    – Deane Yang
    Commented Apr 30, 2016 at 21:03
  • $\begingroup$ @Deane Yang: I didn't attempt to give a maximally "explicit/clear" explanation since the question seemed too basic for this site, so a helpful (one hopes) hint to the OP seemed appropriate to me. $\endgroup$
    – Christian Remling
    Commented May 1, 2016 at 20:16
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Note first of all that $H^{-1}$ will contain objects that are not functions, such as the Dirac $\delta$ concentrated at a point. (Think that $2\delta_0=\frac{d^2}{dx^2}|x|$.)

The correct definition is that $H^{-1}$ is the completion of $C^\infty(S^1)$ with respect to the norm $\newcommand{\ii}{\boldsymbol{i}}$

$$\Vert f\Vert^2=\sum_{n\in\mathbb{Z}}\bigl(1+n^2\bigr)^{-1} \bigl\vert f_n\bigr\vert^2,\;\;f_n=\int_{S^1} f(\theta)e^{-n\ii\theta} d\theta. $$

Equivalently, $H^{-1}$ consists of distributions $\alpha$ on $S^1$ with the property that there exists $C>0$ such that

$$\big\vert\; \alpha(f)\;\bigl| \leq C\Vert f\Vert_{H^1},\;\;\forall f\in C^\infty(S^1). $$

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  • $\begingroup$ I see, but is there a way to characterize these objects directly? I mean when you do the Fourier characterization on $\mathbb{R}^n$ you get the set of tempered distributions that fulfill a "norm condition". Is there an equivalent class of functionals on $H^{-1}$ which contains the objects I am getting, i.e. can we still make some sense of my Edit by introducing a proper set for $?$? $\endgroup$
    – Lewandowski
    Commented Apr 30, 2016 at 20:28
  • $\begingroup$ I'll add details to the answer. $\endgroup$
    – Liviu Nicolaescu
    Commented Apr 30, 2016 at 20:30

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