Fourier transform of product: proof without invoking Fourier inversion theorem Let $f,g$ be Schwartz functions, then $f*g$ and $fg$ are both Schwartz functions. Denote the Fourier tranform of $f$ by $F(f)$.
It is very straightforward to prove $F(f*g)=F(f)F(g)$ from definition and Fubini theorem. However, in most textbooks on Fourier analysis, the other formula $F(fg)=F(f)*F(g)$ is often not proved by direct methods. Instead, they prove the Fourier inversion formula first, and then apply Fourier transform to both sides of $F^{-1}(f*g)=F^{-1}(f)F^{-1}(g)$.
I am wondering whether we can give a proof of $F(fg)=F(f)*F(g)$ without invoking Fourier inversion theorem. In other words, is Fourier inversion theorem essential to this formula? I have tried to prove it directly but find it very difficult. I have little knowledge in mathematical logics, so I don't know whether it is a valid question,  but it really bothers me. Thank you very much if you are willing to help.
 A: Here is a way forward that avoids appeal to the inversion theorem and relies only on the Fubini-Tonelli Theorem (FTT) and the Dominated Convergence Theorem (DCT).  To that end we now proceed.
We denote the Fourier transforms of $f(x)$ and $g(x)$ by $F(k)=\int_{-\infty}^\infty f(x)e^{ikx}\,dx$ and $G(k)=\int_{-\infty}^\infty g(x)e^{ikx}\,dx$, respectively.
Writing the convolution of $F$ and $G$ as $F*G$, we have
$$\begin{align}
(F*G)(k)&= \int_{-\infty}^\infty F(k-k')G(k')\,dk'\\\\
 &=\int_{-\infty}^\infty \left(\int_{-\infty}^\infty f(x)e^{i(k-k')x}\,dx\right)\left(\int_{-\infty}^\infty  g(x')e^{ik'x'}\,dx'\right)\,dk'\\\\
&\overbrace{=}^{\text{FTT}}\lim_{L\to\infty}\int_{-\infty}^\infty f(x)e^{ikx}\int_{-\infty}^\infty g(x')\int_{-L}^L e^{ik'(x'-x)}\,dk'\,dx'\,dx\\\\
&=\lim_{L\to\infty}\int_{-\infty}^\infty f(x)e^{ikx}\int_{\infty}^\infty g(x')\left(\frac{2\sin(L(x'-x))}{(x'-x)}\right)\,dx'\,dx\\\\
&=\lim_{L\to\infty}\int_{-\infty}^\infty f(x)e^{ikx} \int_{\infty}^\infty g(x'/L+x)\frac{2\sin(x')}{x'}\,dx'\,dx \\\\
&\overbrace{=}^{DCT}\int_{-\infty}^\infty f(x)g(x)e^{ikx}\int_{-\infty}^\infty \frac{2\sin(x')}{x'}\,dx'\\\\
&=2\pi \mathscr{F}\{fg\}(k)
\end{align}$$
as was to be shown!
