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I have quite a bit of confusion about Poisson's Sum Formula (PSF). With the standard definition of the Fourier Transform (FT),

\begin{align} \hat{f}(\xi) &= \int_{-\infty}^\infty f(x)e^{-2\pi i x \xi}\,dx \\ f(x) &= \int_{-\infty}^\infty \hat{f}(\xi) e^{2\pi i x \xi}\,d\xi \end{align}

Poisson's Sum Formula is,

\begin{align} \sum_{n \in \mathbb{Z}} f(n) = \sum_{n \in \mathbb{Z}} \hat{f}(n) \end{align}

Now, suppose you have an identity you want to prove using PSF, and the identity has the following form \begin{align} \sum_{n \in \mathbb{Z}} g(n) = \sum_{n \in \mathbb{Z}} h(n) \end{align}

for some well-defined $g(n)$, $h(n)$. The thing I don't understand is, if you're just given the identity, it's not possible to make a-priori assumptions about which side is in the "time"-domain and which is in the "frequency"-domain, so to speak. So shouldn't be just as viable to try and show that $h(n)$ is the FT of $g(n)$, as it is to show that $h(n)$ is the Inverse FT of $g(n)$? Shouldn't both be true at the same time?

  • But when you actually go to evaluate the integrals, taking the FT of $g(n)$ yields a different answer than taking the Inverse FT of $g(n)$, due to the sign difference in the integrand.

  • This appears to be further complicated if you pick a different definition of the FT, since you'll get different answers depending on which definition you pick, but the identity should be true independent of the FT.

  • In principle, could you generate an entire family of identities by recursively applying PSF between $g(n)$ and $\hat{g}(n)$, and then between $\hat{g}(n)$ and $\hat{\hat{g}}(n)$, and so on?

  • How do you determine the correct way of proving an identity requiring PSF given there seem to be so many arbitrary elements to a solution?

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    $\begingroup$ If $\sum_{n\in\Bbb Z} g(n)=\sum_{n\in\Bbb Z} h(n)$ then $\sum_{n\in\Bbb Z} g(n)=\sum h_{n\in\Bbb Z}(-n)$. $\endgroup$ – Angina Seng Jun 7 '19 at 6:48
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The Fourier transform is idempotent of period 4, that is, $$\tag{1} \hat{\hat{\hat{\hat{g}}}}(x)=g(x),\quad \forall x\in\mathbb R.$$

Thus, applying the Poisson summation formula multiple times, all we get is this: $$\tag{2} \sum_{n\in\mathbb Z} g(n)=\sum_{n\in\mathbb Z} \hat{g}(n)=\sum_{n\in\mathbb Z}\hat{\hat{g}}(n)=\sum_{n\in\mathbb Z}\hat{\hat{\hat{g}}}(n).$$ However, we also have that $$\tag{3} \hat{\hat{g}}(x)=g(-x), \quad \hat{\hat{\hat{g}}}(\xi)=\hat{g}(-\xi), $$ so the second and the third identity in (2) are obtained from the first by the elementary change of variable $$ \sum_{n\in\mathbb Z}h(n)=\sum_{n\in\mathbb Z}h(-n),$$ as Lord Shark mentions in the comments to the question.

The same thing will happen with different normalizations for the Fourier transform. Changing the normalization in the Fourier transform always amounts to a scaling transformation of the form $$ g(x)\mapsto \alpha g(\beta x), $$ where $\alpha, \beta$ are powers of $2\pi$. So, (1) and (3) are the same in all normalizations.

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