Given $s\in (0,1)$ and a measurable function $u:\mathbb{R^n}\to\mathbb{C}$, let us define $$\|u\|\_{\dot H^s(\mathbb{R}^n)}^2:=\iint\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\,dx\,dy$$
and let $\dot H^s(\mathbb{R}^n)$ denote the completion of $C^\infty\_c(\mathbb{R}^n)$ wrt this norm. One can define $H^s(\mathbb{R}^n)$ as the completion of $C^\infty\_c(\mathbb{R}^n)$ wrt the stronger norm $\|u\|\_{L^2}+\|u\|\_{\dot H^s}$, as well.


(1) Clearly, $H^s(\mathbb{R}^n)\subseteq \dot H^s(\mathbb{R}^n)$. Is the inclusion an equality? In light of the open mapping theorem, this is equivalent to ask whether the norms are equivalent on $C^\infty\_c(\mathbb{R}^n)$. If not, does $\dot H^s(\mathbb{R}^n)$ contain the constant functions?


(2) If we define instead $\dot H^s(\mathbb{R}^n)$ as the set of measurable functions $u$ such that $\|u\|\_{\dot H^s}<\infty$, do we get a larger space?



Homogeneous fractional Sobolev spaces
(1) No, not an equality. Look at the characterization in terms of Fourier transforms (\*).


(2) It depends on $s$. For $n=1$ and $\frac12\le s<1$ the completion is a quotient of the semi-Hilbert space defined by the seminorm being finite, with the subspace of constant functions. For all $n$ and $0<s<\frac{n}2$ it is a Hilbert space, and the semi-Hilbert space is the sum of the Hilbert space and the space of constant functions.


(\*) $$\|u\|\_{\dot H^s(\mathbb{R}^n)}^2:=\iint\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\,dx\,dy\\ =\int|h|^{-n-2s}[\int|u(x+h)-u(x)|^2\,dx]\,dh \\ =\int|h|^{-n-2s}[\int|(e^{ih\cdot\xi}-1)\hat u(\xi)|^2\,d\xi]\,dh \\=\int|h|^{-n-2s}[\int|\frac {e^{ih\xi}-1}{|\xi|}|^2|\xi \hat u(\xi)|^2]\,dh \\=\int|\xi \hat u(\xi)|^2[\int|h|^{-n-2s}|\frac {e^{ih\xi}-1}{|\xi|}|^2\,dh]\,d\xi \\=c\int|\xi \hat u(\xi)|^2|\xi|^{2s-2}\,d\xi$$ with $c=\int|e^{ih\_1}-1|^2|h|^{-n-2s}\,dh$


