Define for $u\in C\_c^\infty (\mathbb R^n), 0<s<1$ the integral
$$
I\_s(u) = \int\_{(x,y)\in \mathbb R^{n+n}} \frac{(u(x+y)-u(x))^2}{|y|^{d+2s}} dxdy.
$$
I wish to prove that for some $C=C(s)>1,$
$$
C^{-1}\|u\|\_{\dot H^s} \leq I\_s(u) \leq C\|u\|\_{\dot H^s} .
$$
See [this Wikipedia page](https://en.wikipedia.org/wiki/Sobolev_space#Sobolev%E2%80%93Slobodeckij_spaces) for a reference of this result.


I use the *definition* $\|u\|\_{\dot H^s} = \||\xi|^s\hat u\|\_{L^2}$, where $\hat{}$ denotes Fourier transform.


I try to apply Fouier transform over the integral on $x.$ I end up with an integral of the form
$$
\int \frac{(e^{2\pi i \xi \cdot y}-1)^2}{|y|^{d+2s}} |\hat u(\xi)|^2 dyd\xi,
$$
but apparently that does not provide a factor of $|\xi|^{2s}$ in the integrand. I can easily bound $I\_s(u)$ by $\|u\|\_{\dot H^1},$ but this is a weaker result than what we want.


**How to get the factor $|\xi|^{2s}$ into the integrand?**



Double space integral formulation of homogeneous Sobolev norm
You have done half of the job. We have the absolutely convergent integral which is such that  

$$
f(\xi)=\int\_{\mathbb R^d} \frac{\vert e^{2iπ y\cdot \xi}-1\vert^2}{\vert y\vert^{d+2s}} dy
=c\_{s,d}\vert \xi\vert^{2s},
$$
since for $A\in O(d)$, $f(A\xi)=f(\xi)$ (change of variables $y=Ay'$) and moreover for $\lambda >0$,
$f(\lambda \xi)=\lambda^{2s} f(\xi)$ (change of variables $y=y'/\lambda$).


