## Truth of the Poisson summation formula

The Poisson summation says, roughly, that summing a smooth $L^1$-function of a real variable at integral points is the same as summing its Fourier transform at integral points(after suitable normalization). Here is the wikipedia link.

For many years I have wondered why this formula is true. I have seen more than one proof, I saw the overall outline, and I am sure I could understand each step if I go through them carefully. But still it wouldn't tell me anything about why on earth such a thing should be true.

But this formula is exceedingly important in analytic number theory. For instance, in the book of Iwaniec and Kowalski, it is praised to high heavens. So I wonder what is the rationale of why such a result should be true.

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 Poisson summation is the Fourier inversion formula for the circle in disguise. See mathoverflow.net/questions/89504/… and Darsh Ranjan's answer below. – Phil Isett Feb 26 2012 at 6:48

It is a special case of the trace formula. Both sides are the trace of the same operator.

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A new MO record? A question asked, answered within four minutes or so, and accepted within another five (or less, I can't see the acceptance timestamp). – Harald Hanche-Olsen Feb 7 2010 at 23:33
Yes, I too would imagine it is within less than five minutes. Maybe three, or four. – Feb7 Feb 7 2010 at 23:44
As seen on Wikipedia: "When Γ is the cocompact subgroup Z of the real numbers G=R, the Selberg trace formula is essentially the Poisson summation formula." -- en.wikipedia.org/wiki/Selberg_trace_formula – Konrad Voelkel Feb 8 2010 at 0:36
On the other hand, one mustn't think that Poisson summation comes "for free" via trace fla. The classical proof of Poisson summation uses at one point the standard (but needs a fair bit of justification if one is trying to do everything purely from first principles) proof that (+) a continuous periodic function on R can be written as sum_{m in Z} a_m e^{2 pi i m x}. Poisson summation now follows rather easily. On the other hand, the trace formula for R/Z gives (sthg)=(sthg), but to deduce Poisson summation you have to compute the (something)s and I think you end up having to invoke (+) anyway. – Kevin Buzzard Feb 8 2010 at 10:04
The (something)s in this case are easy to compute. On one side, the orbital integrals are trivial to compute, on the other, one needs to know the spectral decomposition of L^2(R/Z). – MBN Feb 8 2010 at 14:35

In what follows, I'll use the convention $$\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x)e^{-2\pi i x \xi}dx,$$ so that $$f(x) = \int_{-\infty}^{\infty} \hat{f}(\xi)e^{2\pi i x \xi}d\xi.$$

I like the following interpretation of Poisson summation, which also gives a generalization: Consider the Dirac comb distribution $C(x) = \sum_{n\in \mathbb{Z}} \delta(x-n)$. This is a tempered distribution, so it has a Fourier transform. In fact, it is its own Fourier transform. To justify this, I'm going to give a very nonrigorous argument. But if intuition is the main goal, then I think it will help. First, note that $C(x)$ is periodic with period 1. Thus, its "Fourier transform" is actually a Fourier series: its support is in $\mathbb{Z}$. This follows by noting that

\begin{align*}C(x) &= \int_{-\infty}^{\infty} \hat{C}(\xi)e^{2\pi i x \xi}d\xi; \\ C(x) &= \sum_{n\in \mathbb{Z}}a_n e^{2\pi i n x}; \end{align*}

Where the first line is the Fourier inversion formula and the second line is the Fourier series for $C$. It follows by uniqueness that $\hat{C}(\xi) = \sum_{n\in \mathbb{Z}}a_n \delta(\xi - n)$. On the other hand, the (inverse) Fourier transform of $\hat{C}$ is also supported on $\mathbb{Z}$, so $\hat{C}$ is also periodic with period 1. Thus, all the $a_n$ are the same:

$$\hat{C}(\xi) = a\sum_{n\in \mathbb{Z}}\delta(\xi - n),$$ where $a$ is some scalar. It's not hard to see that the scalar has to be 1.

To derive Poisson summation from this, use the convolution theorem: let $f$ be any function. On the one hand, $$(f*C)(x) = \sum_{n\in \mathbb{Z}} f(x+n).$$ On the other hand, we can use the convolution theorem: $$\widehat{(f*C)}(\xi) = \hat{f}(\xi)\hat{C}(\xi) = \hat{f}(\xi)\sum_{n\in \mathbb{Z}} \delta(\xi-n) = \sum_{n\in \mathbb{Z}} \hat{f}(n)\delta(\xi-n).$$The last sum gives the Fourier series of the periodic function $f*C$: $$(f*C)(x) = \sum_{n\in \mathbb{Z}} \hat{f}(n)e^{2\pi i n x}.$$ Plugging in $x=0$ gives the Poisson summation formula, QED. But the result for general $x$ is interesting as well: given a function $f$, you can obtain a periodic function $g(x)$ by (a) adding up $f(x+n)$ over all integers $n$, or (b) taking the Fourier transform of $f$ at integer frequencies and making that the Fourier series of $g$. The result is that (a) and (b) give the same function.

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+1. I always found that distribution proofs are the most beautiful proofs of classical statements and this one is no exception. It lets you do everything as you want to do and you get the answers to technical questions like "Does this function has a fourier transform? May I interchange these limites? Does this series converge?" all for free. – Johannes Hahn Feb 8 2010 at 13:39
One should be careful to note that being supported on ${\mathbb Z}$ is not enough to conclude $C(x) = \sum_n a_n \delta(x)$; for example, take $E(x) = \sum_n \frac{d}{dx} \delta(x - n)$. On the other hand, the property that $e^{2 \pi i \xi} \hat{C}(\xi) = \hat{C}(\xi)$ does imply the desired representation. Similarly, the integer periodicity of $\hat{C}$ follows from how $e^{2 \pi i x} C(x) = C(x)$. Also, instead of convolving, using the fact that $< \hat{f}, C > = <f, \hat{C}>$ is slightly more direct ($C$ and $\hat{C}$ are real and even). – Phil Isett Feb 26 2012 at 6:44