Derivative of fractional Laplacian is the fractional Laplacian of the derivative

Question

Is it true that $$\partial_x ((-\Delta)^s u(x)) = ((-\Delta)^s \partial_x u(x))?$$

Answer 1

That is correct. The easiest way to see it is when you define the fractional power of the Laplacian using the Fourier transform. If the Fourier transforms is defined by $$ \hat{f}(\xi)=\int_{\mathbb R^n} f(x)e^{-2\pi i x\cdot\xi}\, dx, $$ then $$ \left(\frac{\partial f}{\partial x_k}\right)^\wedge(\xi)=2\pi i\xi_k \hat{f}(\xi), \quad (-2\pi i x_k f(x))^\wedge(\xi)=\frac{\partial \hat{f}}{\partial \xi_k}(\xi), $$ and for $s>-n/2$ $$ (-\Delta)^s f = ((4\pi^2|\xi|^2)^s\hat{f}(\xi))^\vee. $$ Using these identities one can easily check that the Fourier transform of the left and the right hand side of the expressions in your identity are equal so the identity is true.

Answer 2

Yes, it is true. I assume that you define $(-\Delta)^s$ via a functional calculus. Therefore you have to convince yourself that for two (potentially unbounded!) operators $A,B$ with $AB=BA$ you also have $f(A)B=Bf(A)$ for all suitable functions $f$.