Can the Poisson summation formula break?

Question

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?

Answer 1

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.