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?
Best 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.