I'm interested in proving the "other way around" from wiki which basically states:
$$\widehat{fg}(\xi)=(\hat{f}*\hat{g})(\xi)$$
Now I've tried some pretty basic things and got almost there, but I think there is a mistake or something I'm not allowed to do, since I don't get the final statement. I would appreciate someone looking through the following.
So basically on the one side we have with the definition of the FT: $$\widehat{fg}(\xi)=\int_{\mathbb{R}^d}e^{-2\pi i x\cdot \xi}f(x)\cdot g(x)dx$$
On the other side I first insert the definition of the convolution: $$(\hat{f}*\hat{g})(\xi)=\int_{\mathbb{R}^d}\hat{f}(\zeta)\hat{g}(\xi-\zeta)d\zeta$$
Next I do the Fourier Transform of the two functions in the integral: $$=\int_{\mathbb{R}^d}[\int_{\mathbb{R}^d}e^{-2\pi i x\cdot \zeta}f(x)dx][\int_{\mathbb{R}^d}e^{-2\pi i x\cdot(\xi-\zeta)}g(x)dx]d\zeta$$
Then I combine the two integrals, since they are over the same variable and same space (but this is not allowed, since $\int_A f(x)dx\int_A g(x)dx\neq\int_A f(x)g(x)dx$ so the error happens here): $$=\int_{\mathbb{R}^d}\int_{\mathbb{R}^d}e^{-2\pi i x\cdot(\zeta+\xi-\zeta)}f(x)g(x)dxd\zeta=\int_{\mathbb{R}^d}\int_{\mathbb{R}^d}e^{-2\pi i x\cdot \xi}f(x)\cdot g(x)dxd\zeta$$
$$=\int_{\mathbb{R}^d}\widehat{fg}d\zeta=\widehat{fg}\int_{\mathbb{R}^d}1d\zeta$$
But as you see the remaining integral over $d\zeta$ ruins everything. I would like to see where the mistake is and understand how the proof goes by using the inversion formula for the FT as it is suggested on wiki.
Found the solution on: Fourier Transform Proof $ \mathcal F(f(x)g(x))=(\frac{1}{2\pi})F(s)*G(s)$
Translated to our notation, this would give:
Starting with the claim $\widehat{f(x)g(x)}=(\hat{f}*\hat{g})(\xi)$ and applying the FT inversion formula on both sides we obtain: $$f(x)g(x)=\int_{\mathbb{R}^d}e^{2\pi i x\cdot\xi}(\hat{f}*\hat{g})(\xi)d\xi$$ Now with the definition of the convolution we get: $$f(x)g(x)=\int_{\mathbb{R}^d}e^{2\pi i x\cdot\xi}\bigg(\int_{\mathbb{R}^d}\hat{f}(\zeta)\hat{g}(\xi-\zeta)d\zeta\bigg)d\xi$$ multiplying the RHS with $1=e^{2\pi i x\cdot \zeta}e^{-2\pi i x\cdot \zeta}$ we have: $$f(x)g(x)=\int_{\mathbb{R}^d}e^{2\pi i x\cdot\xi}\int_{\mathbb{R}^d}\hat{f}(\zeta)\hat{g}(\xi-\zeta)e^{2\pi i x\cdot\zeta}e^{-2\pi i x\cdot \zeta}d\zeta d\xi$$ Because $|\phi(z)e^{az}|=|\phi(z)|$ we can use Fubini on the integrals above, which gives us: $$f(x)g(x)=\int_{\mathbb{R}^d}e^{2\pi i x\cdot\zeta}\hat{f}(\zeta)d\zeta\int_{\mathbb{R}^d}e^{2\pi i x\cdot(\xi-\zeta)}\hat{g}(\xi-\zeta)d\xi$$ from which we finally deduce the true statement: $$f(x)g(x)=f(x)g(x)$$ hence our starting equation $\widehat{fg}=\hat{f}*\hat{g}$ is true and we are done.