 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/5/a/solution

Use the Fourier transform convention

f (ω) = \int_{- \infty}^{\infty} e^{- iω t} f (t) d t, f (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{iω t} f (ω) d ω .

(f * g) (t) = \int_{- \infty}^{\infty} f (τ) g (t - τ) d τ

. For integrable

f, g

, Fubini's theorem and the substitution

s = t - τ

give

f * g (ω) = \int_{R} \int_{R} e^{- iω t} f (τ) g (t - τ) d τ d t = (\int_{R} e^{- iω τ} f (τ) d τ) (\int_{R} e^{- iω s} g (s) d s) .

f * g (ω) = f (ω) g (ω) .

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