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I am reading a proof of the Poisson summation formula which states that (with my version of the Fourier transform - I think they sometimes vary by a constant factor) for $f$ a Schwartz function on $\mathbb{R}$ (that is, a smooth function with all derivatives of $f(x)$ of all orders decay faster than any polynomial function as $|x| \to \infty$), the following relationship holds:

$\sum \limits_{n \in \mathbb{Z}}f(n) = \sum \limits_{n \in \mathbb{Z}}\hat{f}(2\pi n) = $

where $\hat{f}$ denotes the Fourier transform. The proof goes as follows: define 2 functions on $\mathbb{T} = \{z: |z| = 1\}$ by $F, G: \mathbb{T} \to \mathbb{C}$ by $F(\theta) = \sum \limits_{n \in \mathbb{Z}}\hat{f}(2\pi n) e^{2 \pi i n \theta}$, and $G(\theta) = \sum \limits_{k \in \mathbb{Z}}f(\theta + k)$. Then take Fourier transforms and show they are equal. After showing also that $F$ and $G$ are both Schwartz functions on $\mathbb{T}$, we apply uniqueness of Fourier series to show that $F=G$. Now uniqueness here relies very much on the fact that both $F$ and $G$ are Schwartz functions on $\mathbb{T}$: but a Schwartz function on $\mathbb{T}$ is simply an element of $C^\infty (\mathbb{T})$, i.e. a smooth function on $\mathbb{T}$.

So, my question is this: how do we know (or show) that $F$ and $G$ are smooth? It is clear both are periodic, so it will suffice to show smoothness within their respective periods I suppose. $f$ is Schwartz on $\mathbb{R}$ which means it is smooth, but we are taking an infinite sum of translations of $f$, so it is not manifestly clear that either $F$ or $G$ will remain smooth. How do we show this? I guess we need to make use of the fact that at large values $f$ is very fast-decaying to show that most terms are "insignificant" in the sums, but whenever I tried to prove the smoothness formally it became messy. Is there a nice trick to showing $F$ and $G$ are smooth on $\mathbb{T}$? I would be very grateful for a proof of the fact. Many thanks in advance.

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up vote 4 down vote accepted

The following result is relevant here: If $f_n : \mathbb{R} \to \mathbb{R} $ is a sequence of continuously differentiable functions, uniformly convergent to $f$ and the sequence $f'_n$ converges uniformly to $g$, then $f$ is differentiable and $f'=g.$

So we can check that $F$ and $G$ are smooth by checking that we can keep on differentiating them, which we do by applying the above theorem, the Weierstrass M-test and the fact that Schwarz functions decay very quickly.

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Thanks Ragib, that was very helpful. – Spyam Jun 9 '12 at 17:06
@Spyam I just noticed that I forgot about $G$ in my original answer, but hopefully you noticed that the exact same method applies. – Ragib Zaman Jun 9 '12 at 17:19

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