In which sense the convolution product is an average? [convolution in Fourier transform]

Question

This is probably a silly question. But I'm studying the convolution for Fourier transforms $$ (f*g)(x)=\frac{1}{T} \int_0^T f(y)g(x-y)dy $$ and my professor said that "the product $ f(y)g(x-y) $ is a sort of average of the two functions". Since it's not the first time I hear a product is a kind of average, but I can't convince myself of it, could anyone clarify this a bit? Any other idea on the meaning of convolution will be highly appreciated!

Answer 1

Convolution is a way of gathering like terms. For example, $$ \sum_{k=0}^{\infty}a_k z^k \sum_{n=0}^{\infty}b_n z^n = \sum_{l=0}^{\infty}\left(\sum_{k+n=l}a_kb_n\right)z^l = \sum_{l=0}^{\infty}\left(\sum_{n=0}^{l}a_{l-n}b_{n}\right)z^l. $$ The coefficient of $z^l$ may be also be written as $\sum_{m=0}^{l}a_m b_{l-m}$. Either expression may be considered to be a discrete convolution of sequences $\{ a_n \}_{n=0}^{\infty}$, $\{ b_n \}_{n=0}^{\infty}$.

If you multiply two Fourier integrals, a similar sum appears, but it is an integral sum as opposed to a discrete sum. For example, $$ \int_{-\infty}^{\infty}e^{iux}f(u)du\int_{-\infty}^{\infty}e^{ivx}g(v)dv = \int_{-\infty}^{\infty}e^{iwx}\left(\int_{-\infty}^{\infty}f(w-u)g(u)du\right)dw. $$ So, in this case, convolution takes the form of an integral where the two arguments all sum to $w$: $$ \int_{-\infty}^{\infty}f(w-u)g(u)du=\int_{-\infty}^{\infty}f(u)g(w-u)du. $$ This is the convolution of functions on $\mathbb{R}$.

A similar thing takes place with the Laplace transform, which may be thought of as a continuous power series. Here one gathers all the coefficients of $e^{-st}$ for a fixed $t$: $$ \int_{0}^{\infty}e^{-st}f(t)dt \int_{0}^{\infty}e^{-st'}g(t')dt' = \int_{0}^{\infty}\left(\int_0^t f(t-u)g(u)du\right) e^{-st}dt $$ The convolution that gathers like terms is, in this case, given by $$ \int_0^{\infty} f(t-u)g(u)du $$ I suppose you could see this as some kind of "average," but I don't see it that way. I see this in terms of multiplying sums or integrals and gathering like powers to obtain a new sum or integral of the same type. I'll let you be the judge.

Answer 2

I'm the OP: I found a beautiful explanation. This is actually an average: see here