partial derivative of Bessel's operator.

Question

Let $J^s = (I- \Delta)^{\frac{s}{2}}$ where $\Delta$ is the Laplacian, and $w(x,y) \in L^2(\mathbb{T}^2)$. During my study in some paper the author stated that

$$\int_{\mathbb{T}^2} J^s (\partial^3_x w) (J^s w)dxdy + \int_{\mathbb{T}^2} J^s (\partial_x \partial_y^2 w)J^s wdxdy=0.$$

I could not reach this result. I used the integration by parts (assuming $J^s$ not intact) by differentiation and integration and obtained

$$\int_{\mathbb{T}^2} J^s (\partial^3_x w) (J^s w)dxdy + \int_{\mathbb{T}^2} J^s (\partial_x \partial_y^2 w)J^s w dxdy=0=\int_{\mathbb{T}^2} J^s (\Delta w) J^s(\partial_xw)dxdy. $$

Any help is appraised.

Answer 1

The trick is by doing the integration by parts three times.

consider

\begin{align} I&=\int_{\mathbb{T}^2} J^s (\partial^3_x w) (J^s w)dxdy + \int_{\mathbb{T}^2} J^s (\partial_x \partial_y^2 w)J^s wdxdy\\ &=-\int_{\mathbb{T}^2} J^s (\partial^2_x w) (J^s \partial_xw)dxdy - \int_{\mathbb{T}^2} J^s ( \partial_y^2 w)J^s \partial_x wdxdy\\ &=\int_{\mathbb{T}^2} J^s (\partial_x w) (J^s \partial^2_x w)dxdy + \int_{\mathbb{T}^2} J^s ( \partial_y w)J^s \partial_y\partial_x wdxdy\\ &=-\int_{\mathbb{T}^2} J^s (w) (J^s \partial^3_x w)dxdy - \int_{\mathbb{T}^2} J^s (w)J^s \partial^2_y\partial_x wdxdy\\ &=-I \end{align}

Hence, $I=0$.