I have been reading up on fractional sobolev spaces (Hitchhikers guide to fractional sobolev spaces https://arxiv.org/pdf/1104.4345.pdf).
Let me denote $H^s(\Omega)=W^{s,2}(\Omega)$ where $\Omega\subset\mathbb{R}^n$ is suitably regular. The norm for $H^{\frac{1}{2}}(\Omega)$ is defined as,
$\|u\|_{H^{\frac{1}{2}(\Omega)}}^2=\int_\Omega |u|^2\ dx+\int_\Omega\int_\Omega \frac{|u(x)-u(y)|^2}{|x-y|^{n+1}}\ dxdy$
The second term is similar to Holder continuity. I want to ask what is the inner product for $H^{\frac{1}{2}}(\Omega)$ Is it.
$\langle u,v\rangle_{H^{\frac{1}{2}}(\Omega)}=\int_\Omega uv\ dx+\int_\Omega\int_\Omega \frac{|u(x)-u(y)||v(x)-v(y)|}{|x-y|^{n+1}}\ dxdy$
I believe in the above inner product the second term on the right is correct. Please correct me if I am wrong. Can someone please explain why this is the case intuitively? Thanks for any help.
Best Answer
Whenever you have a (real) Hilbert space with norm $\|\cdot\|$, and you're not sure what the inner product is, you can just check the polarization identity: $$ \langle u,v\rangle = \frac14\left(\|u+v\|^2-\|u-v\|^2\right).$$ This will give $$\langle u,v\rangle_{H^{\frac{1}{2}}(\Omega)}=\int_\Omega uv\ dx+\int_\Omega\int_\Omega \frac{(u(x)-u(y))\cdot (v(x)-v(y))}{|x-y|^{n+1}}\ dxdy.$$ Another thing, an inner product is supposed to be bilinear in $u,v$; your guess isn't.