The decay conditions of the Poisson summation formula The Poisson summation formula states that for a function $f$ satisfying 
$$
|f(x)|\leq A(1+|x|)^{-n-\delta},~|\hat{f}(\xi)|\leq A(1+|\xi|)^{-n-\delta}
$$
for $\delta>0,$ then the equality 
$$
\sum_{\nu\in\mathbb{Z}^n}f(x+\nu)=\sum_{\nu\in\mathbb{Z}^n}\hat{f}(\nu)e^{2\pi ix\cdot \nu}
$$
holds. Depending on how you interpret the theorem, it can suggest that two ways of making a function periodic are identical. 
My question is why are the boundedness/decay conditions necessary in the theorem? Is it solely to ensure that $f$ and $\hat{f}$ do not blow up to infinity, or is there something deeper? 
Thank you in advance for your help and comments. 
 A: I'm going to talk just about the one-dimensional case for convenience.
There are other versions of the theorem with different hypotheses and sometimes weaker conclusions. Probably the most general is this:
Theorem If $f\in L^1(\Bbb R)$ then $$\sum_{n\in\Bbb Z}f(x+n)\sim
\sum_{j\in\Bbb Z}\hat f(j)e^{2\pi ijx}.$$
Note that the $\sim$ means that I'm not asserting that the RHS converges, just that the RHS is the Fourier series of the LHS.
Proof: Let $F(x)$ equal the LHS. It's easy to see that the series converges in $L^1([0,1])$ and also almost everywhere. Because of the $L^1$ convergence, the Fourier coefficients of $F$ can be calculated by integrating term by term, and it follows that $\hat F(n)=\hat f(n)$. This is exactly what the equation means, given that $\sim$. QED.
The problem with that version is we don't know the RHS converges, in fact it may not; it's just summable to $F$ in various standard ways. And we don't know that the LHS converges at every point; it may not. We need extra conditions to ensure that both sides converge at every point and are equal at every point. The conditions you cite are a convenient example of such a set of conditions.
