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 later.
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 spacial dimension $N$?
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. $\square$.