Continuous and periodic

Let $f$ be a continuous real-valued function on $\mathbb{R}$ satisfying

$\left| f(x) \right| \leq \frac{1}{1+x^2} \quad \forall x$

Define a function $F$ on $\mathbb{R}$ by

$F(x) = \sum_{n=-\infty}^{\infty} f \left ( x + n \right )$

Prove that $F$ is continuous and periodic with period $1$ .
Prove that if $G$ is continuous and periodic with period $1$ then
$\int_{0}^{1} F(x)G(x) \, \mathrm{d}x = \int_{-\infty}^{\infty} f(x) G(x) \, \mathrm{d}x$

Solution

We note that
$g\left ( x + 1 \right ) = \sum_{n=-\infty}^{\infty} f \left ( n +x +1 \right ) = \sum_{n'=-\infty}^{\infty} f \left ( x + n \right ) = g(x)$
First of all $G$ is bounded on $[0, 1]$ and $\sum \limits_{n=-N}^{N} f (x + n ) \rightarrow F(x)$ uniformly. Hence,
$\begin{align*} \int_{-\infty}^{\infty} f(x) G(x)\, \mathrm{d}x &= \sum_{n=-\infty}^{\infty} \int_{n}^{n+1} f(x) G(x) \, \mathrm{d}x \\ &= \sum_{n=-\infty}^{\infty} \int_{0}^{1} f(x + n) G(x)\, \mathrm{d}x \\ &= \int_{0}^{1} F(x) G(x) \, \mathrm{d}x \end{align*}$

An interesting function