$\mathcal{F}(f)\star\mathcal{F}(g)(\xi)$

Question

Can someone prove that $\mathcal{F}(f)*\mathcal{F}(g)(\xi)= \mathcal{F}(f.g)(\xi)$?

where: $\mathcal{F}f(x)=\int_{\mathbb{R^n}} f(x)e^{-2i\pi x\xi}dx$. Or is there any link? And thank you so much.

Answer 1

The right-hand side is$$\begin{align}\mathcal{F}(f\cdot g)(\xi)&=\int f(x)g(x)\exp(-2\pi ix\xi)dx\\&=\iint f(x)\delta(x-y)g(y)\exp(-2\pi ix\xi)dxdy\\&=\iiint f(x)g(y)\exp(-2\pi ix(\xi-\eta)-2\pi iy\eta)dxdyd\eta\\&=\int\left(\int f(x)\exp(-2\pi ix(\xi-\eta))dx\right)\left(\int g(y)\exp(-2\pi iy\eta)dy\right)d\eta\\&=\int(\mathcal{F}(f))(\xi-\eta)(\mathcal{F}(g))(\eta)d\eta\\&=\mathcal{F}(f)\star\mathcal{F}(g).\end{align}$$