A property of the Fourier transform I need a hand proving this property involving Fourier transforms:
If we have $F,G\in L^2(\mathbb{R})$, and we denote the Fourier transform as $T$, where  $T(F)(\xi)=\int_{\mathbb{R}}F(x)e^{-i\xi x}\;dx$, then the following identity holds:$\:\:\:\:$
$$\int_{\mathbb{R}}T(F)(x)\cdot G(x)\;dx=\int_{\mathbb{R}}F(x)\cdot T(G)(x)\;dx\;. $$
Thanks a lot in advance for any help.
 A: This might looks like a comment.
First I think you understand the Fourier Transform for $L^2(\mathbb{R})$ is only "formally" defined in the formula you wrote. It's actually defined for $L^1(\mathbb{R})$ in that way.
Now the above argument provided by BigM gives you the property of Fourier Transform on $L^1(\mathbb{R})\cap L^2(\mathbb{R})$, the only thing here you need to verify is Fubini's Theorem can be applied when both $f(x),g(x)$ are in $L^1(\mathbb{R})\cap L^2(\mathbb{R}) $.
The subspace $L^1(\mathbb{R})\cap L^2(\mathbb{R})$ is dense in $L^2(\mathbb{R})$. So extend Fourier Transform to $L^2(\mathbb{R})$ by continuity gives the desired result on the whole $L^2(\mathbb{R})$.
A: Hint.plug $T(F)(x)$ in the left hand side of the equality.$\int_{\mathbb{R}}T(f)(x)g(x)dx=\int_{\mathbb{R}}\Big(\int_{\mathbb{R}}f(t)e^{-ixt}dt\Big)g(x)dx=\int_{\mathbb{R}}\int_{\mathbb{R}}f(t)g(x)e^{-ixt}dtdx=\int_{\mathbb{R}}\int_{\mathbb{R}}f(t)g(x)e^{-ixt}dxdt=\int_{\mathbb{R}}f(t)\Big(\int_{\mathbb{R}}g(x)e^{-ixt}dx\Big)dt=\int_{\mathbb{R}}f(t)T(g)(t)dt$ 
Basically its an easy application of Fubini's theorem.
