Is integration by parts allowed for the $J^s$ derivative, where $s \in \mathbb R$

Question

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.

Answer 1

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

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