# Can Poisson Summation formula break?

The Poisson summation formula states if $f: \mathbb{R} \to \mathbb{R}$ then $\displaystyle \sum_{n \in \mathbb{Z}} f(n) = \sum_{n \in \mathbb{Z}} \hat{f}(n)$ where $$\hat{f}(\xi) = \int_{\mathbb{R}}dx \; e^{-2\pi i x \xi}\;f(x)$$

The fine print is that $f$ needs to be Schwartz class or be a tempered distribution such as the Dirac-delta function

$$\sum_{n \in \mathbb{Z}} \delta_n(x) = \sum_{n \in \mathbb{Z}} e^{2\pi i n x}$$

It then says the Dirac comb is its own Fourier transform.

Are there counterexamples where the left and right sides converge yet these traces do not agree?

Edit As discussed in the comments, it appears in Katznelson's textbook although I don't understand his example very well - convolving a function with the Fejer kernel many times at different scales.

Edit Can anyone fill in details of Lucia's response?

• Why does this question have so many downvotes? – Adam P. Goucher Dec 20 '14 at 17:17
• @AdamP.Goucher beats me – john mangual Dec 22 '14 at 22:29
• @AdamP.Goucher I conjecture the downvotes might've been caused by john's attitude towards the answer he's received. I'm not saying I consider this a valid downvote reason myself though. – Wojowu Feb 8 '17 at 16:08
• @AdamP.Goucher Apparently having an unpopular opinion on the usage of downvotes will lead to lot's of downvotes from users on this site. – 54321user Jul 23 '17 at 19:05
• (Some people are just angry...) – paul garrett Feb 2 '18 at 2:30

## 1 Answer

Exercise 15 of Chapter VI (Section 1) of Katznelson's book "An introduction to Harmonic analysis" gives an example of a continuous $L^1$ function $f$ where both sides of the Poisson summation formula are absolutely convergent, but the formula does not hold.

To add a little to this, the usual proof of Poisson summation begins by periodizing $f$ setting $F(x) = \sum_{n\in {\Bbb Z}} f(x+n)$ and then computing the Fourier series of $F(x)$. The idea in Katznelson's example is that even if $f$ is somewhat nice (e.g. continuous and $L^1$) it need not be the case that $F$ is nice (e.g. it could be discontinuous). In particular Katznelson's example constructs a nice $f$ for which $F$ turns out to be $1$ on all of ${\Bbb R}/{\Bbb Z}$ except for being discontinuous at $0$ where it takes the value $0$. So there is a problem with the Fourier series for $F$ at $0$, and hence with the Poisson summation formula.

• I don't understand why this was downvoted. My answer gave a reference to the literature showing there are counterexamples to the Poisson summation formula, even if $f$ and ${\hat f}$ are somewhat nice. – Lucia Aug 23 '13 at 17:19
• I conjecture that a lot of the time a downvote is a botched upvote (via a mouse slip), so don't lose too much sleep over this. – Igor Rivin Aug 23 '13 at 17:32
• @JohnMangual: Lucia provided an answer for your question. Your not liking it (sorry, "not being crazy about it") is not a reason to downvote. The fact that you googled it in five minutes means that you should have not asked your question, or asked for an answer different from Katznelson (explaining why). Doing what you did is a good way to get people to never look at your questions. – Igor Rivin Aug 23 '13 at 20:32
• I agree with @IgorRivin. (If John was unhappy with the example given by Lucia, and had already seen the example in Katznelson's book, then why not leave a comment while downvoting? – Yemon Choi Aug 23 '13 at 20:44
• @JohnMangual OK, here is a relevant comment: please explain what it is you want displayed or explained in the counterexamples which is not present in the example pointed to by Lucia. If you give people an idea of what you are looking for, they have better chance of giving answers that you are not "not crazy about" – Yemon Choi Aug 23 '13 at 20:47