In the proof of the Poisson summation formula, there is a detail which is not clear to me how to resolve.
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a Schwartz-class function.
Let $F(x)=\sum_{n\in\mathbb{Z}}f(x-2\pi n)$. Why does the Fourier series of $F$ converge at any point $x$, i.e. $$F(x)=\sum_{k\in\mathbb{Z}}\hat{F}(k)e^{ikx}?$$
Best Answer
Since $f$ is a Schwartz function, by definition we have that there exists $C>0$ such that $(2\pi+|x|)^2|f(x)|\leq C$ for every $x\in \mathbb{R}$. In particular, for $x\in [0,2\pi]$, we have
$$ \sum_{n\in \mathbb{Z}} |f(x-2\pi n)| \leq \sum_{n\in \mathbb{Z}} \frac{C}{(2\pi +|x-2\pi n|)^2} \leq \sum_{n\in \mathbb{Z}} \frac{C}{(2\pi -|x| +2\pi |n|)^2} \leq \sum_{n\in\mathbb{Z}} \frac{C}{(2\pi |n|)^2}. $$
This implies, by Weierstrass' M-test that the sum defining $F$ is absolutely and uniformly convergent in $\mathbb{R}$ (by periodicity). Since $f'$ is a Schwartz function whenever $f$ is, we can apply the same argument above to the (in principle formal) derivatives of $F$, given by the sum of the derivatives of the $f(x-2\pi n)$, to conclude that $F\in C^\infty(\mathbb{R})$.
Now it's just a matter of integrating by parts to conclude since $\widehat{F^{(m)}}(k)=(ik)^m\hat{F}(k)$, so that, again by Weierstrass, the Fourier series of $F$ converges uniformly in $[0,2\pi]$ to $F$.