Suppose that $f\in\mathcal{S}(\mathbb{R})$ where $\mathcal{S}(\mathbb{R})$ is the Schwartz space of distributions. Prove that $$\sum_{n\in\mathbb{Z}}f(x+n)$$ converges absolutely and uniformly.
Edit: I forgot to mention that $x\in K\subset \mathbb{R}$ where $K$ is compact. This fact is mentioned without proof in Stein and Shakarchi's Fourier Analysis so I thought I would ask. Also, the function $$g(x)=\sum_{n\in\mathbb{Z}}f(x+n)$$ is called the periodization of $f$. It is a 1-periodic continuous function satisfying $$g(x)=\sum_{n\in\mathbb{Z}}\hat{f}(n)e^{2\pi inx}$$ where $\hat{f}(n)$ is the Fourier transform of $f$ computed at $n$.
$f\in\mathcal{S}(\mathbb{R})$ so $\forall k \in \mathbb{N}, f^{(k)}(x)=O_{+\infty}(\frac{1}{x^2})$.
Thus $\exists M>0,\forall x \in \mathbb{R},|x|\geq1 \Rightarrow |f(x)|\le \frac{M}{x^2}$.
So $\forall K>0, \forall x \in [-K,K],\forall n \in \mathbb{Z}, |n|>K+1 \Rightarrow |f(x+n)|\le \frac{M}{(x+n)^2}\le \frac{M}{(|n|-K)^2} $. Id est $||f(.+n)|_{[-K,K]}||_\infty=O(\frac{1}{x^2})$.
So it converges normally on all the compacts of $\mathbb{R}$, so it converges uniformly.