Convolution Theorem - Proof

Proof

The proof here is shown for a particular normalisation of the Fourier transform. As mentioned above, if the transform is normalised differently, then constant scaling factors will appear in the derivation.

Let f, g belong to L1(Rn). Let be the Fourier transform of and be the Fourier transform of :

where the dot between x and ν indicates the inner product of Rn. Let be the convolution of and

Now notice that

Hence by Fubini's theorem we have that so its Fourier transform is defined by the integral formula

\begin{align} H(\nu) = \mathcal{F}\{h\} &= \int_{\mathbb{R}^n} h(z) e^{-2 \pi i z\cdot\nu}\, dz \\ &= \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} f(x) g(z-x)\, dx\, e^{-2 \pi i z\cdot \nu}\, dz. \end{align}

Observe that and hence by the argument above we may apply Fubini's theorem again (i.e. interchange the order of integration):

Substitute ; then, so:

These two integrals are the definitions of and, so:

QED.

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