Lagmental Vicfred

Convolution Becomes Multiplication after Fourier Transform by Vicfred

A convolution integral diagonalizes under the Fourier transform. I want the notation, the mechanism, and the failure mode visible at the same time.

Start locally

Fourier analysis expands a function into frequencies. On the circle the coefficients are \(\widehat f(n)\); on \(\mathbf R\) the transform \(\widehat f(\xi)\) is an integral against an oscillatory exponential.

$$ (f*g)(x)=\int_{\mathbf R}f(x-y)g(y)\,dy $$

A reliable calculation names domain and codomain. The notation \(\mathsf{data}\mapsto\mathsf{claim}\) is harmless only after both \(\operatorname{dom}\) and \(\operatorname{cod}\) have been fixed.

$$ \widehat{f*g}(\xi)=\widehat f(\xi)\widehat g(\xi) $$

Compute before generalising

Here is a concrete symbolic test. Reading it from left to right reveals which transformation is reversible and which is only an implication.

$$ \begin{aligned}\widehat{f*g}(\xi)&=\int\!\int f(x-y)g(y)e^{-2\pi ix\xi}\,dy\,dx\\&=\left(\int f(u)e^{-2\pi iu\xi}\,du\right)\left(\int g(y)e^{-2\pi iy\xi}\,dy\right).\end{aligned} $$

The global view

The formula is reusable precisely because it says which pieces are structural and which belong only to the worked example.

$$ \begin{aligned} \mathsf{D}\;&:\quad (f*g)(x)=\int_{\mathbf R}f(x-y)g(y)\,dy,\\[5pt] \mathsf{C}\;&:\quad \widehat{f*g}(\xi)=\widehat f(\xi)\widehat g(\xi). \end{aligned} $$

Edge conditions

Normalization conventions move factors of \(2\pi\) between the transform, inverse transform, derivative rule, and Gaussian formula.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \widehat{f*g}(\xi)=\widehat f(\xi)\widehat g(\xi) \end{gathered}} $$

This is enough machinery for one note: an exact object, a worked case, a structural law, and a clearly marked boundary. Each layer can now be tested independently.

This article was posted on Thu 25 October 2012. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.