Define the Fourier transform as $\mathcal{F}f(t)=\int_{\mathbb{R}^n}\exp(-itx)f(x)dx$ which is bijective for $f \in \mathcal{L}^1$.
The convolution theorem states that: $\mathcal{F}(f*g)(t)=\mathcal{F}f(t)\mathcal{F}{g}(t)$.
From knowing this is it possible to prove that inverse fourier transform of a convolution also splits into a product? I.e.
$$ \mathcal{F}^{-1}(f*g)(t) = \mathcal{F}^{-1}{f}(t)\mathcal{F}^{-1}{g}(t). $$ Or would you just have to prove this from the definition?
Since $\mathcal{F}$ is injective, it would suffice to prove that $\mathcal{F}\mathcal{F}^{-1}(f*g)(t)=\mathcal{F}(\mathcal{F}^{-1}{f}(t)\mathcal{F}^{-1}{g}(t))$ but I can't get further than this.
From $$\mathcal{F}^{-1}f(x) = \frac{1}{(2\pi)^n} (\mathcal{F}f)(-x)$$ and $$\mathcal{F}(f*g) = \mathcal{F}f \cdot \mathcal{F}g$$ it follows that $$ \mathcal{F}^{-1}(f*g)(x) = \frac{1}{(2\pi)^n} \mathcal{F}(f*g)(-x) = \frac{1}{(2\pi)^n} (\mathcal{F}f \cdot \mathcal{F}g)(-x) = \frac{1}{(2\pi)^n} \mathcal{F}f(-x) \cdot \mathcal{F}g(-x) \\ = (2\pi)^n \cdot \frac{1}{(2\pi)^n} \mathcal{F}f(-x) \cdot \frac{1}{(2\pi)^n} \mathcal{F}g(-x) = (2\pi)^n \, \mathcal{F}^{-1}f(x) \cdot \mathcal{F}^{-1}g(x). $$ Thus, $$ \mathcal{F}^{-1}(f*g) = (2\pi)^n \, \mathcal{F}^{-1}f \cdot \mathcal{F}^{-1}g. $$