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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. – Jonas Teuwen Jul 27 '11 at 19:33
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 connection – Katja Jul 27 '11 at 21:42
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. – Jonas Teuwen Jul 31 '11 at 10:29
I edited the question for clarity. Is it clear now? – Katja Aug 3 '11 at 6:58

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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I not sure this is true. $g\chi_{[0,1]}$ should be continuous. For example, let $g(x)=\sum_{j=-\infty}^{\infty}\exp(-\pi\left|x+j\right|)$, then $f(x)=\exp(-\pi\left|x\right|)$. Obviously $f(x)\neq g\chi_{[0,1]}$ – Katja Aug 11 '11 at 10:27
@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. – Florian Aug 13 '11 at 22:54

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