When can the Fourier transform change order in the inner product of $L^2? $

Question

In $L^2(\mathbb{R}^n)$. Let $(f,g):=\int fg$. If $f\in\mathcal{S}(\mathbb{R}^n)$ and $g\in L^2(\mathbb{R}^n)$. When $(\mathcal{F}^{-1}(f),g)=(f,\mathcal{F}(g))$? This always holds in this case? Thanks.

Answer 1

Yes. It follows from Parseval's identity: $(\mathcal Ff,\mathcal Fg) = (f,g)$ for $f,g\in L^2(\mathbb R^n)$.