Double space integral formulation of homogeneous Sobolev norm

Question

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 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?

Answer 1

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 $y=y'/\lambda$).