Supposing I have two integrable functions $f$ and $g$ in $L^2(\mathbb{R})$.
I want to show that:
$$\int_{-\infty}^{+\infty}\hat f(w)\space\overline{\hat g(w)}\ dw = \int_{-\infty}^{+\infty} f(x)\space\overline{g(x)}\ dx$$
Where $\hat f(w)$ and $\hat g(w)$ are the respective Fourier transforms of the function $f(x)$ and $g(x)$.
I don't know where to start exactly. Should I let $f(x)=f_{1}(x)+if_{2}(x)$ and $g(x)=g_{1}(x)+ig_{2}(x)$ and go from there? Starting from the right-hand side of the equation hasn't let me anywhere so I'm guessing I have to compute the Fourier transforms of the respective functions.
We have that
\begin{align*} \int_{-\infty}^\infty \hat{f}(w)\overline{\hat{g}(w)}dw &= \frac{1}{2\pi}\int_{-\infty}^\infty dw\int_{-\infty}^\infty dx\int_{-\infty}^\infty dx' e^{-iw(x-x')}f(x)\overline{g(x')}\\ &= \int_{-\infty}^\infty dx\int_{-\infty}^\infty dx'\underbrace{\frac{1}{2\pi}\int_{-\infty}^\infty dw e^{-iw(x-x')}}_{=\delta(x-x')}f(x)\overline{g(x')}\\ &= \int_{-\infty}^\infty dx f(x)\overline{g(x)} \end{align*}
If this property of the $\delta$ distribution is not known then it wil get a bit more complicated, which guess is what the non-triviality pointed out by copper.hat in the comments refers to.