I have some functions, which are periodic with period 1. Let one of them be $g$.
Function $g$ has the following form $(K\rightarrow\infty)$: $g(x)=\sum_{j=1}^K h\left(\frac{j}{K},x\right) $
Here, $h(x,y)$ is known,has a period of $1$ with respect to both variables.
Now, I can see in plots, but not prove, that the following holds: $g(x)=\sum_{j=-\infty}^\infty \int_{-\infty}^\infty \left(\int_0^1 g(t) e^{-2\pi ist} \;dt\right)e^{2\pi is(x+j)}\;ds $ I.e.: $g$ is the infinite sum of the Fourier transform of its Fourier coefficients. (Actually, I have a proof, but I don't quite trust it.)
So I'am trying to get more understanding why this works for my $g$ and what it means and implies.
Suppose the following integral exists: $f(x)=\int_{-\infty}^\infty \left(\int_0^1 g(t) e^{-2\pi ist}dt\right)e^{2\pi isx} \; ds $
(It seems to exist. Also, $f$ has almost finite support)
It seems, but I can't prove it, that $f$ is the non-periodic analog of $g$. I.e., does the following hold? $g(x)=\sum_{j=-\infty}^{\infty}f(x+j)$