$L^1$ convolution stays in $L^1$, with norm at most the product of the two $L^1$ norms.Suggest a slogan
If \(f, g \in L^1(\RR^n)\), then the convolution \(f\ast g\) is defined almost everywhere, \(f\ast g\in L^1(\RR^n)\), and \(\norm{f\ast g}_1\le\norm{f}_1\norm{g}_1\).