# Can the 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? Dec 20, 2014 at 17:17
• @AdamP.Goucher beats me Dec 22, 2014 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. Feb 8, 2017 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. Jul 23, 2017 at 19:05
• (Some people are just angry...) Feb 2, 2018 at 2:30

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. Aug 23, 2013 at 17:19