Let \( f \in L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n) \), \( g \in L^2(\mathbb{R}^n) \) and \( \mathcal{F} : L^1(\mathbb{R}^n) \to L^\infty(\mathbb{R}^n) \) the Fourier transform . Then
\[ \mathcal{F}(f * g) = (2\pi)^{n/2} \mathcal{F}(f) \cdot \mathcal{F}(g). \]
Remarks
- \( f \in L^1 \) guarantees \( f * g \in L^2 \) (Young’s inequality )
- \( f * g \sim \mathcal{F}^{-1} (\mathcal{F} f \cdot \mathcal{F} g) \)
- \( \mathcal{F}^{-1}(f g) \sim \mathcal{F}^{-1} f * \mathcal{F}^{-1} g \)