Integration by parts involving $\Delta^\frac{-1}{2}$

Question

Let $f$ and $g$ are well behaved functions. Using the Laplacian and the gradient, is the following equality true?

$$\nabla(f \Delta^\frac{-1}{2}g)= \nabla f \Delta^\frac{-1}{2}g+ f \nabla \Delta^\frac{-1}{2}g=\nabla f \Delta^\frac{-1}{2}g+ fg.$$

I assumed that I can define $\Delta^\frac{-1}{2}g$ as Fourier multiplier.

The other question is: I know that $\int \Delta^\frac{1}{2}f(x) g(x) = \int \Delta^\frac{1}{2}g(x) f(x)$ is passible, is it also passibol for $\Delta^\frac{-1}{2}$? Thanks in advance.

Answer 1

For the sake of simlicity, let $f,g$ be Schwartz functions in $\mathbb R^n$. The first equality $$\nabla (f \Delta^{-1/2}g)= (\Delta^{-1/2}g) \nabla f +f \nabla \Delta^{-1/2}g $$ is true, but it is simply the product rule and has nothing to do with the properties of $\Delta^{-1/2}$. The second inequality $$(\Delta^{-1/2}g) \nabla f +f \nabla \Delta^{-1/2}g = (\Delta^{-1/2}g) \nabla f +f g $$ is certainly not true since the left hand side is a vector while $fg$ is a scalar. For your final question, the Plancherel theorem (assuming $f$ and $g$ are real valued) gives \begin{align*} \int_{\mathbb R^n} f(x)\Delta^{-1/2}g(x) \, dx &= \int_{\mathbb R^n} \hat f (\xi)\mathscr F\{\Delta^{-1/2}g \}(\xi)\, d\xi \\ &=-\int_{\mathbb R^n} \hat f (\xi)\cdot \vert \xi \vert^{-1}\hat g(\xi)\, d\xi\\ &= \int_{\mathbb R^n} \hat g (\xi)\mathscr F\{\Delta^{-1/2}f \}(\xi)\, d\xi \\ &=\int_{\mathbb R^n} g(x)\Delta^{-1/2}f(x) \, dx \end{align*} where $\mathscr F \{f\}=\hat f$ is the Fourier transform of $f$.