Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a Schwartz-class function.
Let $F(x)=\sum_{n\in\mathbb{Z}}f(x-2\pi n)$. I understand that because Schwartz-class functions are nice, this series converges absolutely and uniformly for all $x\in\mathbb{R}$.
However, why is $F$ differentiable, and why is its derivative equal to $\sum_{n\in\mathbb{Z}}f'(x-2\pi n)$?
We have $$f(x+h-2\pi n)-f(x-2\pi n)=\int_0^hf'(s+x-2\pi n)\mathrm ds=hf'(h+x-2\pi n)-\int_0^h sf''(s+x-\pi n)\mathrm ds,$$ and since $f$ and its derivatives are Schwartz functions, we can take the sum $n\in\mathbb Z$. Indeed, if $g$ is a Schwartz function, then $|g(x-2\pi n)|\leqslant \frac C{(1+|x-2\pi n|)^2}$ hence for $K$ compact and $n$ such that $\operatorname{diam}(K)\lt 2n\pi$, $\sup_{x\in K}|g(x-2\pi n)|\leqslant{C'}{n^2}$.
This gives $$\frac{F(x+h)-F(x)}h=\sum_{n\in\mathbb Z}f'(x+h-2\pi n)-\sum_{n\in\mathbb Z}\frac 1h\int_0^hsf''(s+x-\pi n)\mathrm ds.$$ The convergence of the first series is uniform on compact sets. For the second one, we use the substitution $s=th$ to get the wanted result.