Prove that $f *e^{2\pi inx} $ is a scalar multiple of $e^{2\pi inx}$ where $f $ is defined in $[0,1] $ and extended to be periodic on $\mathbb R $

Question

Let $f*g= \int_0^1f (x-y)g (y)dy $ be the convolution. Prove that, for a given integer $n $, $f *e^{2\pi inx} $ is a scalar multiple of $e^{2\pi inx}$ where $f $ is defined in $[0,1] $ and extended to be periodic on $\mathbb R $. You may use the fact that if periodic function $f $ and $g $ are integrable on $[0,1] $ then $f*g $ is again a periodic function on $\mathbb R $ that is integrable on $[0,1] $.

I am lost on how to do this. Any help would be greatly appreciated.

Answer 1

We have$$f*e^{2\pi inx}{=\int_{0}^{1}f(u)e^{2\pi in(x-u)}du\\=\int_{0}^{1}f(u)e^{2\pi inx}e^{-2\pi inu}du\\=e^{2\pi inx}\int_{0}^{1}f(u)e^{-2\pi inu}du}$$where $\int_{0}^{1}f(u)e^{-2\pi inu}du$ is constant