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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? $g(x)=\sum_{j=-\infty}^{\infty}f(x+j)$

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    I edited the question for clarity. Is it clear now?2011-08-03

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In essence this is just the Fourier inversion formula: If we define the Fourier transform of $u\in L^1(\mathbb{R})$ as $\hat u (\xi)=\int_\mathbb{R} u(x) e^{-2\pi i x\xi} d x$ (note that there are other conventions about where to put factors of $2\pi$ and minus signs) then the inversion formula states that $\hat{\hat u}(x)=u(-x)$ almost everywhere if both $u$ and $\hat u$ are in $L^1(\mathbb{R})$; in other words, the inverse transform is the same as the original transform except for a sign flip. So what you did is to transform $g\chi_{[0,1]}$ and transform it back, so you get $g\chi_{[0,1]}$ again. Thus $f=g\chi_{[0,1]}$.

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    @Katja: Check your calculation; there has to be a mistake in it. Of course if $g\chi_{[0,1]}$ is not continuous then its Fourier transform will not be in $L^1$. The inversion theorem still holds, but the classical definition of Fourier transform must be replaced by a different defnition, for $L^2$-functions for example by approximation by $L^1$-functions.2011-08-13