In which sense the convolution product is an average? [convolution in Fourier transform] 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!
 A: 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.
A: I'm the OP: I found a beautiful explanation. This is actually an average: see here
