 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/4/b/solution

Suppose initially that

f, g

are absolutely integrable, so Fubini's theorem justifies exchanging the integrations. The transform of their convolution is

h (ω) = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} f (t) g (z - t) e^{- iω z} d t d z .

s = z - t

, so

z = s + t

and

d z = d s

. The exponential separates:

h (ω) = [\int_{- \infty}^{\infty} f (t) e^{- iω t} d t] [\int_{- \infty}^{\infty} g (s) e^{- iω s} d s] .

f * g (ω) = f (ω) g (ω) .

No extra

2 π

occurs in this direction with the specified normalization.

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