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I have been looking into fractional Sobolev spaces and I see that there are two standard definitions: the one defined via Fourier Transform, and the other via a seminorm. My question is about the latter.

Definition: Let $\Omega$ be an open subset of $\mathbb{R}^N$, $s = m + \sigma$ with $ 0 < \sigma < 1$, $m$ and integer, and $1 \leq p < \infty$. We say that $u \in W^{s, p}(\Omega)$ if $u \in W^{m, p}$ and $$\int_{\Omega} \int_{\Omega} \frac{\vert \partial^{\alpha}u(x) - \partial^{\alpha}u(y)\vert^{p}}{\vert \vert x - y \vert \vert ^{N+\sigma p}} < + \infty \: \forall \, \vert \alpha \vert= m$$

My question is, why does it make sense to define it in this way? What is the importance of the spatial dimension $N$?

Giuseppe Negro
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Kyle W.
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  • Where did you find this definition? I've only ever seen the semi-norm definition for integer-order Sobolev spaces. – Mark Hubenthal Sep 26 '16 at 14:53
  • The book I came across it in was Girault & Raviart Finite Element Methods for Navier Stokes Equations. But it can be found in several other places for example Necas Direct Methods in the Theory of Elliptic Equations – Kyle W. Sep 26 '16 at 14:59
  • I can say that the $N$ is in the exponent because in order for the integral to converge as a weakly singular integral, the decay of the denominator has to be sufficiently large at infinity. In general, if you have an integral operator of the form $T(f)(x) = \int K(x,y)f(y),dy$ where $|K(x,y)| \leq \frac{C}{|x-y|^{N}}$ and satisfies some smoothness-related conditions (see https://en.wikipedia.org/wiki/Singular_integral), then $T$ is bounded from $L^{p}$ to $L^{p}$. – Mark Hubenthal Sep 26 '16 at 15:00
  • This paper seems to provide a nice exposition on these spaces for $s \in (0,1)$, which can be easily extended to the general case you have described above with $s > 0$: http://arxiv.org/pdf/1104.4345.pdf – Mark Hubenthal Sep 26 '16 at 15:05
  • Right, I'd seen that but it didn't give much of an intuition behind the definition. I think your second comment @MarkHubenthal helps quite a bit. – Kyle W. Sep 26 '16 at 15:10
  • I'm sure that some insight can also be gained by taking the fourier transform of $\int_{\Omega} \frac{\partial^{\alpha}u(x) - \partial^{\alpha}u(y)}{|x-y|^{N/p + \sigma}},dy$, since the transform of a convolution is a product of transforms, and smoothness of functions translates into polynomial-like decay of the fourier transform. However, for $p \neq 2$ one has to deal with tempered distributions in order to make sense of the fourier transform, so it can get quite technical. – Mark Hubenthal Sep 26 '16 at 15:23

1 Answers1

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When $\Omega=\mathbb R^N$ and $\sigma \in(0,1)$, one has the following identity: $$ \iint_{\mathbb R^N\times \mathbb R^N} \frac{ |u(x+y)-u(x)|^2}{|y|^{N+2\sigma}}\, dxdy = C_N \int_{\mathbb R^N} |\xi|^{2\sigma} |\hat{u}(\xi)|^2\, d\xi,$$ where $C_N> 0$ is a constant depending on the spatial dimension $N$ only. In particular, if $m\in \mathbb N$, one can write the Sobolev norm for $H^{m+\sigma}(\mathbb R^N)=W^{m+\sigma, 2}(\mathbb R^N)$ in the following form: $$ \| u\|_{H^{k+\sigma}(\mathbb R^N)}^2 = \sum_{|\alpha|\le m} \| \partial^\alpha u \|_{L^2}^2 + \sum_{|\alpha|=m} \iint_{\mathbb R^N\times \mathbb R^N} \frac{ |\partial^\alpha u(x+y)-\partial^\alpha u(x)|^2}{|y|^{N+2\sigma}}\, dxdy.$$ This formulation does not make use of the Fourier transform, which is only available on the free space $\mathbb R^N$, and can therefore be generalized to Sobolev spaces on domains.

Proof (Taken from Bahouri-Chemin-Danchin, proposition 1.37). By Plancherel's identity, one has $$\tag{1} \int_{\mathbb R^N} \frac{|u(x+y)-u(x)|^2}{|y|^{N+2\sigma}}\, dx = C_N \int_{\mathbb R^N} \frac{ |e^{i y\cdot \xi}-1|^2}{|y|^{N+2\sigma}} |\hat{u}(\xi)|^2\, d\xi. $$ The function $$ F(\xi)=\int_{\mathbb R^N} \frac{ |e^{i y\cdot \xi}-1|^2}{|y|^{N+2\sigma}} \, dy$$ is radially symmetric and homogeneous of degree $2\sigma$, so $F(\xi)=C|\xi|^{2\sigma}$. Integrating (1) and using Fubini's theorem one obtains the desired conclusion. $\square$.

EDIT 2025. Reading this answer again after nine years, I do not fully agree with it anymore. Sure, what I said explains the case $p=2, \Omega=\mathbb R^N.$ But in general, the real reason why people defined that norm is more likely to be related to interpolation. These spaces probably arise as interpolation spaces of the $W^{m,p}(\Omega)$ spaces when $m$ is an integer.

Giuseppe Negro
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