Interesting equation in L^1

Question

Consider $L^{1}(T) = \{ f : R \rightarrow C \text{ with period 1 and } \int_{0}^{1} |f (x)| \ dx < \infty\}$.

For $f,g \in L^{1}(T)$ the convolution is given by $(f * g)(x)= \displaystyle\int_{0}^{1}f(x-y) g(y) \ dy$.

The problem is : Solve the equation $f * f = f$ in $L^{1}(T)$.

My best is :

if $f$ is a solution for the equation then for all $n \in Z \ \ \hat{(f *f)} (n) = \hat{f}(n) => (\hat{f}(n))^2 = \hat{f}(n) => \hat{f}(n) = 0 \ or \ 1$.

The Riemann Lebesgue lemma says $\lim_{|n| \rightarrow + \infty} \hat{f}(n) = 0$. then exists a subset of $Z$ ,say $\{ n_1, \dots, n_k \}$ such that the Fourier series of $f$ is

$$ h(x) = e^{2 \pi n_1 x } + \dots + e^{2 \pi n_k x }$$

Now I don't know what to do ...

Answer 1

Your resulting function lies in $L^2(\mathbb{Z})$. Thus the inverse Fourier transform is well defined (it is bijective so $f \in L^2(T)$). So as an element of $L^1$ function $f$ equals to the sum its own Fourier series. Now you just need to solve your equation for finite sum of exponents. But it is easy to calculate the convolution in that case...