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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 can't seem to make any sense of your question while I do know a bit about Fourier analysis. Maybe you should add some additional information about what you exactly want to know.2011-07-27
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    Jonas Teuwen I see your points. I study some kernels. In order to avoid boundary problems I can work on real line, or with periodic data on an interval.Essentially they should be very similar. But I do'nt understand well enough the connection between them. My questions: 1.does everything I wrote make sense. 2. Does the connection between a function and it's periodization similar to the connection between Fourier transform and Fourier coefficients(for example formula for $f(x)$) 3.if this true, what can we say about this connection2011-07-27
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    I still don't understand what you are trying to say. "1." is the only question I understand and I can only answer: "No" to that. I think this is a language issue, what is your native language? Maybe you could ask it in that language so somebody can translate it.2011-07-31
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    I edited the question for clarity. Is it clear now?2011-08-03

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