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A record saying that the convolution theorem is trivial since it is identical to the statement that convolution, as a Toeplitz operator, has a Fourier eigenbasis and, therefore, is diagonal in it, has disappeared from Wikipedia.

The convolution theorem states that the convolution of functions, $h$ and $x$, in the time domain, is equivalent to their multiplication in the frequency domain. That is, you convolve them, $h*x$, and take result into frequency domain. Result, $F(h*x)$ must be the same as multiplying their Fourier images, $H$ and $X$, i.e., $F(h*x) = H \cdot X$, where $H$ and $X$ are the Fourier images, $H = F(h)$ and $X = F(x)$.

This was the definition where I used the letters $h$ and $x$ for the functions, instead of conventional $f$ and $g$, because convolution $h*x$ is often represented by $h(x)$, where $h$ is a convolution matrix derived from the first function, $h$, which is applied to the second function, $x$. Being Toeplitz, matrix $h$ has eigenbasis $F$ — the same as apply to vector to take it into Fourier domain (see change of basis to see why base matrix product for base transform). Therefore, matrix $h$, translated into its eigenbasis, happens to be diagonal and $H$. That is, according to the convolution theorem, we must prove that

$$F(h \vec x) = H \vec X$$

But, there is nothing to prove. We just can show that

$$H \vec X = (FhF^{-1})(F\vec x) = F(F^{-1}F)h(F^{-1}F)\vec x = F(h \vec x)$$

or, the other way around

$$F(h \vec x) = F(F^{-1}F)h(F^{-1}F)\vec x = (FhF^{-1})(F\vec x) = H \vec X$$

just to exercise the beautiful algebra of relationships and diagonalization in $F$. We just need to keep in mind that $H = FhF^{-1}$ is diagonal (multiplication operator) because $F$ is the eigenbasis of $h$.

This discussion was classified as nonsense. But why? I find it amazing that Toeplitz (or convolution) has Fourier eigenbasis. Should this precious fact be hidden from the general population? Why should general population appreciate the integral-based proof rather than enjoy this fact? Can I say that Toeplitz operator = Convolution (operator)? Can I say that operator is a (continuous-case) generalization of matrix? Is convolution theorem related with the Fourier eigenbasis? Can it be explained simpler, based on this fact?

Rodrigo de Azevedo
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  • For future reference: It's spelled eigen. No h. – kahen Apr 30 '13 at 19:19
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    Why $H = Fh$ and also = $FhF^{-1}$? – user10024395 Nov 09 '17 at 12:58
  • This is actually exactly what I was looking for when first reading about convolution and Fourier/Laplace transforms - thanks! – MGwynne Mar 24 '18 at 15:06
  • I also found https://math.stackexchange.com/questions/918345/fourier-transform-as-diagonalization-of-convolution?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa useful as well, particularly the linked blog. – MGwynne Mar 24 '18 at 15:33

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