Laplace transform inverse convolution

Question

I'm having a hard time figuring why:

$\mathscr{L}(f*g) =\mathscr{L}(f) \cdot \mathscr{L}(g) \implies \mathscr{L}^{-1}(F \cdot G) = \mathscr{L}^{-1}(F) * \mathscr{L}^{-1}(G)$

I fail to see why that's obvious, is there some way to see it algebraically?

Answer 1

$$\mathscr{L}(\underbrace{f}_{\mathscr{L}^{-1}(F)}*\underbrace{g}_{\mathscr{L}^{-1}(G)}) =\underbrace{\mathscr{L}(f)}_F \cdot \underbrace{\mathscr{L}(g)}_G$$ Now apply Laplace inverse on both sides

Answer 2

Substitute $f = \mathscr{L}^{-1}(F)$ and $g = \mathscr{L}^{-1}(G)$, then apply the inverse transform to both sides of the equality. Hope this helps!