# Nice way to express $H^{-1}(\mathbb{S}^1)$

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 $f$?

• 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. Apr 30, 2016 at 19:49
• @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 $S^1$. Apr 30, 2016 at 19:53
• There's no reason why Fourier series have to be restricted to ones in $L^2$. Apr 30, 2016 at 20:07
• Is there a reason the standard interpretation of $H^{-1}$ being the dual of $H^1$ doesn't satisfy you? Apr 30, 2016 at 20:08
• 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 $H^1$. Apr 30, 2016 at 20:31

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.)

• Isn't this what Christian Remling suggested? Apr 30, 2016 at 20:43
• @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. Apr 30, 2016 at 20:44
• Paul, I agree that your explanation is more explicit and clearer. Apr 30, 2016 at 21:03
• @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. May 1, 2016 at 20:16

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

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

• 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 $?$? Apr 30, 2016 at 20:28
• I'll add details to the answer. Apr 30, 2016 at 20:30