Is there a nice expression for the Fourier transform of a dot product of two functions?

Question

Suppose we have two vectors $A(x)$ and $B(x)$ on $\mathbb{R}^n$ and we wish to take the Fourier transform of their dot product, that is, we wish to evaluate

\begin{equation} \mathcal{F}\{A(x) \cdot B(x) \}. \end{equation}

My question is is there a nice expression for this in terms of the Fourier transforms of $A(x)$ and $B(x)$?

So far I have determined that if we let $A(x) = (a_1, a_2, \dotsc , a_n) $ and similarly for $B(x)$, this can be written as the sum of the convolutions of individual elements

\begin{equation} \mathcal{F}\{A(x) \cdot B(x) \} = \sum_{j=1}^n \mathcal{F}\{a_j\} \star\mathcal{F}\{b_j\}, \end{equation}

but ideally I would prefer to write this in terms of the FTs of the vectors themselves. I suspect there might not be a nicer expression, but I am very rusty with FTs and convolutions and my googling skills are not turning up anything useful.

(As an aside, the reason I am interested in this is I wish to find the FT of the poisson bracket of two functions which I am writing as

\begin{equation} \{H, \rho\} = \nabla H \cdot \Omega \nabla \rho \end{equation}

so if anyone knows any specific results about FTs of poisson brackets that might also be of help.)

Thanks.

Answer 1

Supposing the dot-product is the Euclidean inner product, you could apply the convolution theorem to each component, i.e. $F(A\cdot B) = F(\sum_i A_i B_i) = \sum_i F(A_i)*F(B_i)$, where $*$ means convolution, $F$ means Fourier transform, and subscripts indicate components of the vectors.

This may not be easier to compute than the original form in some cases.

Answer 2

You could define the dot-convolution of vector valued functions as $$ u\,\dot{\star}\,v := \int u(x-y)\cdot v(y)\,\mathrm d y = \sum_{j=1}^n u_j\star v_j. $$ Then your expression can be written as $$ \mathcal F(u\cdot v) = \mathcal F(u)\,\dot{\star}\,\mathcal F(v). $$ At the end, this is just notations. Some papers even write just $uv$ instead of $u\cdot v$, so that this becomes really like the classical convolution of scalars.