Fourier series scaling property

Question

This is from Problem 1.5 of Katznelson's book on Harmonic Analysis. The problem is to prove that if $m \nmid n$, $f \in L^1(S^1)$, and $f_{(m)}(x) = f(mx)$, then $\hat{f}(n) = 0$. The fourier coefficient $\hat{f_{(m)}}(n)$ is $$\frac{1}{2\pi} \int_{-\pi}^{\pi} f(mt) e^{-i n t}~dt$$ I tried subsituting $s = mt$ to get $$\frac{1}{2 \pi m} \int_{-m\pi}^{m\pi} f(s)e^{i n/m s}~ds$$ but I still have no idea why that would integrate to $0$.

Answer 1

The considered functions have period $2\pi.$ Their integrals do no change if we replace $[-\pi,\pi]$ with $[0,2\pi].$ Thus $$ {1\over m}\int\limits_{-\pi}^{\pi} f(mt)e^{-i nt}~dt= {1\over m}\int\limits_{0}^{2\pi} f(mt)e^{-i nt}~dt= \int\limits_{0}^{2m\pi} f(s)e^{-ins/m}~ds \\ =\sum_{k=0}^{m-1}\int\limits_{2k\pi}^{2k\pi+2} f(s)e^{-ins/m}~ds=\sum_{k=0}^{m-1}\int\limits_{0}^{2\pi} f(u+2k\pi)e^{-in(u+2k\pi)/m}~du \\ = \sum_{k=0}^{m-1}\int\limits_{0}^{2\pi} f(u)e^{-in(u+2k\pi)/m}~du= \int\limits_{0}^{2\pi} f(u)e^{-inu/m}~du\ \cdot \ \sum_{k=0}^{m-1}e^{-i2kn\pi/m}$$ Let $z=e^{-i2n\pi/m}.$ Then $z\neq 1,$ because $n/m$ is not an integer, and $z^m=1.$ Hence $$\sum_{k=0}^{m-1}e^{-i2kn\pi/m}=\sum_{k=0}^{m-1}z^k={z^m-1\over z-1}=0$$