Is $\int_{0}^{2\pi} f(x)\, \text{e}^{-i\ell x} \, \text{d}x =0$?

Question

Suppose $f,g \in \mathcal{L}^1(\mathbb{R} / 2 \pi)$ with $f(x)=g(mx),m \in \mathbb{Z}$.

I want to show that

$$ \forall \, \ell \in \mathbb{Z},\,\,\, \text{with} \, \ell \not\equiv 0 \, \text{mod} \, m;\,\,\, \frac{1}{2\pi} \int_{0}^{2\pi} f(x)\, \text{e}^{-i\ell x} \, \text{d}x =0. \tag{*} $$

I tried to use $$f\left(x + \frac{2 \pi}{m} \right) = g \left( mx+2 \pi \right) = g(mx) = f(x)$$ but I am not sure if this helps.

How can I show (\ref{d})?

Answer 1

Without loss of generality, assume that $m>0$.

$\begin{align} \int_{0}^{2\pi} f(x)\, \text{e}^{-ilx} \, \text{d}x&= \sum_{k=1}^m\int_{2\pi (k-1)/m}^{2\pi k/m} f(x)\, \text{e}^{-ilx} \, \text{d}x\\&= \sum_{k=1}^m\int_0^{2\pi/m} f(x+2\pi (k-1)/m)\, \text{e}^{-ilx}\text{e}^{-i2\pi l(k-1)/m} \, \text{d}x \\ &= \sum_{k=1}^m\text{e}^{-i2\pi l(k-1)/m}\int_0^{2\pi/m} f(x)\,\text{e}^{-ilx}\,dx. \end{align}$

But

$$ \sum_{k=1}^m\text{e}^{-i2\pi l(k-1)/m}=0, $$ if $l\not\equiv 0\mod m$.