Integration of periodic function $f \in L^1([0, 2\pi])$

Question

While studying trigonometric series and $L^p$ spaces I was wondering the following:

Let's say we have a $2\pi$-periodic function $f \in L^1([0, 2\pi])$ satisfying $\int_{0}^{2\pi}f(x) \, dx = 0$. Is it true that $\int_{0}^{2\pi}f(nx) \, dx = 0$ for all $n \in \mathbb{N}$?

Answer 1

This is just two changes of variables (assuming your definition of $\mathbb N$ starts at $1$). \begin{align*} \int_0^{2\pi}f(nx) \, dx=\int_0^{2\pi n}f(x)\frac{dx}{n}=\frac{1}{n} \sum_{k=1}^n \int_{2\pi(k-1)}^{2\pi k} f(x) \, dx=\frac{1}{n} \sum_{k=1}^n \int_0^{2\pi} f(x+2\pi(k-1)) \, dx\\ =\frac{1}{n} \sum_{k=1}^n \int_0^{2\pi}f(x) \, dx=\frac1n\cdot n\cdot0=0. \end{align*}