Equivalence between Sobolev norm and SobolevSlobodeckij norm for integerorder spaces
This short post explores the equivalence between Sobolev norm and SobolevSlobodeckij norm for $W^{s,p}(\Omega)$ when $s$ is an integer.
Question
A simple argument
My educated guess to the question above is “No”, because when we divide $\overline{\Omega} = \overline{\Omega}_1 \cup \overline{\Omega}_2$ into two nonoverlapping subdomain, for the Sobolev seminorm it is just adding broken norms on the subdomains:
\[u_{1,\Omega}^2 = u_{1,\Omega_1}^2 + u_{1,\Omega_2}^2.\]However for the Slobodeckij norm, it has some cross term like:
\[\int_{\Omega_1} \int_{\Omega_2} \frac{u(x)  u(y)^2}{xy^{n+2}} \,dxdy,\]and apparently,
\[[u]_{1,\Omega}^2 \neq [u]_{1,\Omega_1}^2 + [u]_{1,\Omega_2}^2.\]I am guessing the equivalence relies on the regularities of the $\Omega$, but I could not find any reference on this. Someone on Math.SE gave the reference^{1}.
The reference
The correct behavior of fractional norms in the limit $s \to 1^$ is
\[\lim_{s \to 1^} (1  s)[u]_{s, p}^p = K(p, N)\\nabla u\_p^p,\]where $K(p, N)$ is a constant depending on $p$ and $N$. If $\Omega_1$ and $\Omega_2$ are disjoint, it means that $x  y \geq d > 0$ when $x \in \Omega_1$ and $y \in \Omega_2$, therefore the cross term will tend to zero. As shown in a paper by Bourgain, Brezis, and Mironescu^{2}.
J. Bourgain, H. Brezis, P. Mironescu. Another look at Sobolev spaces, in: J.L. Menaldi, E. Rofman, A. Sulem (Eds.), Optimal Control and Partial Differential Equations, A Volume in Honour of A. Bensoussan’s 60th Birthday, IOS Press, 2001, pp. 439455.

user345872 (https://math.stackexchange.com/users/345872/user345872), Equivalence between Sobolev norm and SobolevSlobodeckij norm for $W^{s,p}(\Omega)$ when $s$ is an integer, URL (version: 20160729): https://math.stackexchange.com/q/1875552 ↩

http://hal.upmc.fr/file/index/docid/747692/filename/another_look_sobolev_spaces_2000.pdf ↩
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