I am given $f\in L^1 (\mathbb{T})$, and $f(x+\frac{2\pi}{k})=f(x)$ for some natural number $k$.
I want to show that $f$'s Fourier transform gets vanished for $n=rk+d$ where $1\leq d
So here's what I did so far (I write without the normalization factor cause it doesn't make a difference here):
$\hat{f}(n)= \int_{-\pi}^{\pi} f(x) e^{-inx}dx = \int_{-\pi}^{\pi} f(x+\frac{2\pi}{k}) e^{-inx}dx = \int_{\frac{2\pi}{k}-\pi}^{\pi+\frac{2\pi}{k}} f(x) e^{-inx} e^{-i\frac{2\pi}{k} d}dx$
Now here's what I thought if $d$ were odd and $k$ an even integer then we can repeat the above $\frac{k}{2}$ to get:
$\int_{0}^{2\pi} f(x)e^{-inx} e^{-i\pi d} dx = -\int_{0}^{2\pi} f(x) e^{-inx}dx$ Which is zero, but then how would you show show for other cases of $k$ and $d$?
Thanks in advance.