prove $(\hat{f}g)^\lor$=$f*g^{\lor}$

Question

I want to prove that $(\hat{f}g)^\lor$=$f*g^{\lor}$ give that $f$,$g$:$\mathbb{R}\to \mathbb{C}$ are integratble functions.

I know that $\hat{f}$ is bounded an the function $g$ is integratble zo the product $\hat{f}.g$ is integratble. This means that the left side of the equation makes sense. Now the function $f$ is intgratble and $\check{g}$ is bounded. This means that $f*\check{g}$ is well defined.

So I tried to fill in the left side of the formula and then with the Fubini positive , I tried to rewrite it to the right side of the formula but it never works out. I think I don't get how you need to fill in with several variables. Is there someone who can prove it?

Answer 1

\begin{align} (\hat f g)^{\vee} (x)&=\int_{\mathbb{R}}\hat f(t) g(t) e^{2\pi ity}dy \\ &= \int_{\mathbb{R}}\int_{\mathbb{R}} f(y) e^{-2\pi ity} g(t) e^{2\pi ity} dtdy \\ &=\int_{\mathbb{R}}f(y)\int_{\mathbb{R}} g(t) e^{2\pi it(x-y)} dtdy \\&=\int_{\mathbb{R}}f(y)\check g(x-y)dy \\&=(\hat fg)^{\vee}(x) \end{align}