h (x) = \frac{1}{2 π} \int_{- \infty}^{\infty} f (k) g (k) e^{ik x} d k = \int_{- \infty}^{\infty} f (y) [\frac{1}{2 π} \int_{- \infty}^{\infty} g (k) e^{ik (x - y)} d k] d y = \int_{- \infty}^{\infty} f (y) g (x - y) d y .

Thus multiplication of Fourier transforms corresponds to the convolution $h = f * g$ , which is the convolution theorem in the convention used here.

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14A
Paper 3
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2023
Past exam of the mathematics course of the University of Cambridge
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