How to comprehense the relation between convolution and each Fourier coefficient of f?

Question

I am learning Fourier Analysis by Elias.M.Stein.I don’t understand the motivation of convolution.From my point of view,the convolutions corresponds to the “weighted averages” if we write it in discrete form.On the other hand,we can transform the series to the question of Dirichlet kernel,so my question is how to comprehend the relation between convolution and each Fourier coefficient of f or the relation between the kernel and each coefficient?

Answer 1

Convolution is a way to gather all coefficients of a common power. For example, $$ \sum_{n=0}^{\infty}a_n z^n \sum_{n=0}^{\infty}b_n z^n = \sum_{n=0}^{\infty} \left(\sum_{j+k=n}a_j b_k\right)z^n = \sum_{n=0}^{\infty}\left(\sum_{l=0}^{n}a_l b_{n-l}\right)z^n $$ In this case, $$ (a\star b)(n)=\sum_{l=0}^{n}a_{l}b_{n-l} $$ Other types of objects in math support gather like coefficients. For example, the Laplace transform gives rise to $$ \int_{0}^{\infty}e^{-st}f(t)dt\int_{0}^{\infty}e^{-st}g(t)ds=\int_{0}^{\infty}\left(\int_{0}^{t}f(v)g(t-v)dv\right)e^{-st}dt \\ ( f\star g)(t)=\int_0^t f(v)g(t-v)dv $$ Likewise, with the Fourier transform, $$ \int_{-\infty}^{\infty}f(t)e^{ist}dt\int_{-\infty}^{\infty}g(t)e^{ist}dt =\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(v)g(t-v)dv\right)e^{ist}dt \\ (f\star g)(t)=\int_{-\infty}^{\infty}f(v)g(t-v)dv $$