Is $\int \hat{g} h = \int g \check{h}$?

Question

Do we have $$\int \hat{g} h = \int g \check{h}?,$$ where $\hat{g}$ is the Fourier transform and $\check{h}$ is the inverse?

Answer 1

With your definitions,$$\begin{align}\int_{\Bbb R^n}\hat{g}(\xi)h(\xi)d\xi&=(2\pi)^{-n/2}\int_{(\Bbb R^n)^2}g(x)e^{-i\xi x}h(\xi)dxd\xi\\&=(2\pi)^{-n}\int_{(\Bbb R^n)^3}g(x)e^{-i\xi(x+y)}\check{h}(y)dxdyd\xi\\&=\int_{(\Bbb R^n)^2}g(x)\delta^{(n)}(x+y)\check{h}(y)dxdy\\&=\int_{\Bbb R^n}g(x)\check{h}(-x)dx.\end{align}$$