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Let us denote by $L^2([0,2\pi])$ the space of all periodic functions that are square integrable. Usually one defines the $H^s$-space for $s>0$ by \begin{align} H^s([0,2\pi]) = \left\{ u \in L^2([0,2\pi]) \, \bigg| \, \int_{(0,2\pi)} \int_{(0,2\pi)} \frac{(u(x)-u(y))^2}{|x-y|^{1+2s}}dx\, dy<\infty\right\}. \end{align}

I've come across several references where also the definition \begin{align} \left\{ u \in L^2([0,2\pi]) \, \bigg| \, \sum_{m\in \mathbb{Z}}(1+m^2)^s |\hat{u}(m)|^2<\infty \right\} \end{align} is used. Here, $\hat{u}$ is the Fourier transform of $u$. Now I'm wondering why both definitions give the same space. I know the equivalent definition via the Fourier transform in the case $H^s(\mathbb{R})$. But somehow, the known techniques do not work if I want to show that both norms are equivalent in the case of periodic functions. Is there an easy trick?

Maybe someone also knows some nice references where fractional Sobolev spaces for periodic functions are treated. I only found references for the case when $s$ is an integer.

cmk
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user406143
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  • In my opinion you should work with mollifiers (https://en.wikipedia.org/wiki/Mollifier): (1) Assume that $u$ satisfies one of the two conditions, then (2) show that also the mollified $u_\epsilon = u*\varphi_\epsilon$ does; (3) Use the characterization on $\mathbb R$ and then (4) let $\epsilon\to 0$ to see that the other condition is satisfied for $u$. For this note that $\hat\varphi(0) = \int\varphi = 1$. – amsmath Sep 05 '19 at 21:10
  • I think your implicit surmise is correct, that it takes some work to show that equivalence from scratch. As @amsmath suggests, one way to not start from scratch is to reduce to the case of the whole line, as black box. But, also, you could just think of how to repeat the argument for the line on the circle, since both are (abelian) Lie groups, and are locally identical. – paul garrett Sep 05 '19 at 22:38

1 Answers1

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If a function is periodic, then we can use Fourier Series instead of the Fourier Transform to represent it. Then the above result is intuitively similar to Parseval's Theorem.

Ninad Munshi
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    I don't see how this helps. I mean, it is clear that $\hat u(m)$ is the $m$-th Fourier coefficient here. Please specify "intuitively similar". – amsmath Sep 05 '19 at 19:52
  • It isn't "clear" that it is the Fourier coefficient since FT and FS are two different things. Using the alternative definition $H^s = {u| \int (1+|\xi|^2)^{s}|\hat{u}(\xi)|^2d\xi < \infty }$, and that the FT of a periodic function will be deltas, it almost looks like the integral becomes a summation, but not really since $\delta^2$ is not a distribution. By intuitively similar I meant that the integral becomes a summation. – Ninad Munshi Sep 05 '19 at 20:26
  • For me, the function wasn't periodic but just defined on $[0,2\pi]$ from the scratch. So for me it was clear that $\hat u(m)$ meant the $m$-th Fourier coefficient. However, I see your point now. But I still don't see how it's supposed to help OP. Maybe it's better to mollify $u|_{[0,2\pi]}$ and use the known characterization on $\mathbb R$. – amsmath Sep 05 '19 at 20:39