Fourier transform of $\nabla \circ \operatorname{div}$

Question

Thanks everyone for your attention to this question.

Let $F$ a function from Schwartz space of smooth functions on $\mathbb{R^n}$, which derivatives is rapidly decreasing. So $F$ is locally integrable over $\mathbb{R^n}$ and I can easily compute that:

$$ \mathscr{F}(\operatorname{div} \circ \space \nabla F)(\xi) = \mathscr{F}(\Delta F)(\xi) = -|\xi|^2\mathscr{F}(F). $$

Now I need to calculate the Fourier transform of this composition in reverse order, namely:

$$\mathscr{F}(\nabla \circ \operatorname{div}F).$$

I have a next hypothesis:

$$\mathscr{F}(\nabla \circ \operatorname{div}F) = -\xi\xi^\mathrm{T}\mathscr{F}(F),$$

but my calculations are not entirely accurate, in my opinion, and I am not very sure of the given answer.

Thanks you a lot for your attention!

Answer 1

If you are dealing with Schwartz space then everything should be okay because in Schwartz space, we have the following identities: $$ \mathscr{F}( \text{div} F )= i\xi^T \mathscr{F}(F)$$ for $F: \mathbb{R}^n \rightarrow \mathbb{R}^{n \times m}$ and $$ \mathscr{F}( \nabla G )= i\xi \mathscr{F}(G)$$ for $G: \mathbb{R}^n \rightarrow \mathbb{R}^{1 \times m}$
Thus, $$\mathscr{F}( \text{div} \circ \nabla F)= i\xi^T \mathscr{F}( \nabla F)= - \xi^T \xi F $$ for all $F: \mathbb{R^n} \rightarrow \mathbb{R}^{1 \times m} $
and $$ \mathscr{F}( \nabla \circ \text{div} F)= i\xi \mathscr{F}( \text{div} F)= - \xi \xi^T F $$ for all $F : \mathbb{R}^n \rightarrow \mathbb{R}^n$