Any idea or hint to prove this theorem ? The book says the hint is by drawing a picture, but I don't really get it.
If $f\in\mathcal{L}^2([-\pi,\pi])$, extended periodically in $\mathbb{R}$, then $$\int_{x-\pi}^{x+\pi}f(t)dt=\int_{-\pi}^{\pi}f(t)dt$$
Assume $x$ is not a multiple of $2\pi$. Then there exists a unique integer $k$ such that $(2k-1)\pi < x-\pi < (2k+1)\pi < x+\pi < (2k+3)\pi$.
Then $$\int_{x-\pi}^{(2k+1)\pi}{f}=\int_{x-\pi-2k\pi}^{(2k+1)\pi-2k\pi}{f}=\int_{x-(2k+1)\pi}^{\pi}{f},$$ and $x-(2k+1)\pi > -\pi$.
Similarly, $$\int_{(2k+1)\pi}^{x+\pi}{f}=\int_{(2k+1)\pi-(2k+2)\pi}^{x+\pi-(2k+2)\pi}{f}=\int_{-\pi}^{x-(2k+1)\pi}.$$
Therefore, $$\int_{x-\pi}^{x+\pi}{f}=\int_{x-\pi}^{(2k+1)\pi}{f}+\int_{(2k+1)\pi}^{x+\pi}{f}=\int_{x-(2k+1)\pi}^{\pi}{f}+\int_{-\pi}^{x-(2k+1)\pi}{f}=\int_{-\pi}^{\pi}{f}.$$