Proving Poisson's Formula for Schwartz Functions

Question

Let $\phi \in \mathcal{S}(\mathbb{R})$. Show that $$\sum_{n=-\infty}^{+\infty}\phi(\frac{n}{a}) = a\sum_{n=-\infty}^{+\infty}\hat{\phi}(na)$$

I know the solution will basically be using the Dirac Comb, and we have the following

$$\phi *\Delta_a = \sum_{n=-\infty}^{+\infty}\phi(t-na)$$ and $$\widehat{\phi*\Delta_a} = \hat{\phi} \cdot\hat{\Delta_a} = \hat{\phi} \cdot \frac{1}{a}\Delta_{\frac{1}{a}} = \frac{1}{a}\sum_{n=-\infty}^{+\infty}\hat{\phi} \delta_{\frac{n}{a}} = \frac{1}{a}\sum_{n=-\infty}^{+\infty}\hat{\phi}(\frac{n}{a}) \delta_{\frac{n}{a}}$$ But from these observations, I'm not really sure where to go next.

Answer 1

By the way do you like Fourier series? $f(t)=\sum_n \phi(t-na)$ is smooth and $a$-periodic, it is equal to its Fourier series $\sum_k c_k e^{2i\pi kt/a}$ and (for $a >0$) $c_k=a^{-1}\hat{\phi}(n/a)$. Letting $t=0$ gives the Poisson summation formula.

Answer 2

First, note that $$(\phi * \Delta_a)(0) = \int_\mathbb{R} \widehat{(\phi * \Delta_a)}(0) dt$$

Next, $$(\phi * \Delta_a)(0) = \sum_{n=-\infty}^{+\infty}\phi(na)$$ and $$\int_\mathbb{R} \widehat{(\phi * \Delta_a)}(0) dt = \frac{1}{a}\sum_{n=-\infty}^{+\infty}\hat{\phi}(\frac{n}{a})$$ Thus, $$\sum_{n=-\infty}^{+\infty}\phi(na) = \frac{1}{a}\sum_{n=-\infty}^{+\infty}\hat{\phi}(\frac{n}{a})$$ $$\implies \sum_{n=-\infty}^{+\infty}\phi(\frac{n}{a}) = a\sum_{n=-\infty}^{+\infty}\hat{\phi}(na)$$