 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ib/paper-4/15d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

For example, it suffices to assume

f, g \in L^{1} (R)

. Indeed the absolute double integral is

\int_{R} \int_{R} ∣ f (t) g (x - t) ∣ d t d x = ∥ f ∥_{1} ∥ g ∥_{1} < \infty.

Thus Fubini's theorem permits interchange in the Fourier transform of the convolution. With

u = x - t

f * g (λ) = \int \int f (t) g (x - t) e^{- iλ x} d t d x = (\int f (t) e^{- iλ t} d t) (\int g (u) e^{- iλ u} d u) .

Hence the convolution theorem in this normalization is

f * g (λ) = \hat{f} (λ) \overset{g}{^} (λ) .

There is no extra

2 π

in this formula; that factor belongs to the inverse transform and Parseval identity for the convention being used.

 New to topics? Read the docs here!