I am having the following integral:


$$I = \int u\, J^s(\partial\_x \overline{u})- \overline{u}\, J^s(\partial\_x u))dxdy$$


where $J^S= (I-\Delta)^\frac{2}{2}$, $\mathbb{R} \ni s \geq 1$ and $u=u(x,y)$,
$u:\mathbb{R}^2 \to \mathbb{C}$.


My question: Is it allowed to do the integration by parts for the fractional differentiation so I can get zero? That is:


\begin{align}
I &= \int u\, J^s(\partial\_x \overline{u})- \overline{u}\, J^s(\partial\_x u))dxdy\\
&= \int u\, J^s(\partial\_x \overline{u})+ \partial\_x\overline{u}\, J^s( u))dxdy\\
&= \int u\, J^s(\partial\_x \overline{u})-\int u\, J^s(\partial\_x \overline{u})\\
&=0
\end{align}


Are the above calculations right? If so, how to prove it?


Thanks in Advance.



Is integration by parts allowed for the $J^s$ derivative, where $s \in \mathbb R$
Your claim is false, since $J^s$ is self-adjoint, moving $J^s$ from one term to the other does not incur a minus sign. So instead of the two terms cancelling, they actually double.


If $u$ is a Schwartz function (or more generally belonging to a suitable Sobolev space) you can compute by using Plancherel:


You have


$$ \int u J^s (\nabla \bar{u}) = \int \hat{u} (1 + |\xi|^2)^{s/2} (-i) \xi \bar{\hat{u}} = - \int (1 + |\xi|^2)^{s/2} i\xi \hat{u} \bar{\hat{u}} = - \int J^s(\nabla u) \bar{u} $$


