= Solution
For a full <Euclidean lattice> $\Lambda\subseteq\mathbb R^n$, its <dual lattice> is
$$
\Lambda^\vee=\{y:\langle x,y\rangle\in\mathbb Z\text{ for every }x\in\Lambda\}.
$$
For a Schwartz function, the <Poisson summation formula for a Euclidean lattice> states
$$
\sum_{\lambda\in\Lambda}f(\lambda)
=m(\Lambda)^{-1}\sum_{\mu\in\Lambda^\vee}\widehat f(\mu),
\qquad
\widehat f(y)=\int_{\mathbb R^n}f(x)e^{-2\pi i\langle x,y\rangle}\,dx.
$$
To prove it, periodize $f$ over $\Lambda$. The resulting function on $\mathbb R^n/\Lambda$ has Fourier coefficient $m(\Lambda)^{-1}\widehat f(\mu)$ at $\mu\in\Lambda^\vee$. Evaluating its absolutely convergent Fourier series at zero gives the identity. The same proof applies under the usual weaker hypotheses ensuring convergence of both sides.
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