Integral over product of Fourier transforms

Question

I am struggling showing a property of the Fourier transform. I assume the solution is rather straight forward. Use some properties of the Fourier transform and "calculate". However, I have tried many things and have gotten nowhere. The problem is:

Let $f$ and $g$ be two functions with Fourier transform $\mathcal{F}$ and $\mathcal{G}$ respectively. Prove that the following equality holds,

$$\int_{-\infty}^{\infty}f(t)g(-t)dt = \int_{-\infty}^{\infty}\mathcal{F}(u)\mathcal{G}(u)du.$$

Thanks a lot already.

Best, Sebastian

Answer 1

Here is one possible approach. Recall the convolution of $f$ and $g$ is the function $$ (f\ast g)(x) = \int f(t)g(x-t)\,dt, $$ so the left-hand side is $(f\ast g)(0)$. Recall that for any function $H$, $$ H(0) = \int \hat H(u)\,du, $$ where $\hat H$ is the Fourier transform of $H$. If $H = f\ast g$, what is the Fourier transform of $H$?