# No Identity for Convolution

Show that there exists no continuous 1-periodic function $$f_0$$ so that $$f\star f_0=f$$ holds for all continuous 1-periodic functions $$f$$. No Lebesgue here. We're dealing with $$L^2$$, the space of continuous Riemann square integrable functions.

Having trouble getting any intuition on convolution integrals. Any one have a hint to point me in the right direction? I'm given as a suggestion to use Riemann-Lebesgue lemma, but I cannot see how it applies.

Suppose $f_0$ exists. Then the convolution identity means that for $x\in [0,1]$ $$\int_0^1f(x-t)f_0(t)dt=f(x)$$ Let's apply this to $f(t)=e^{2i\pi nt}$. We obtain $$\int_0^1 e^{2i\pi n(x-t)}f_0(t)dt=e^{2i\pi nx}$$ That is, for all $n\in\mathbb Z$, $$e^{2i\pi nx}\int_0^1 e^{-2i\pi nt}f_0(t)dt=e^{2i\pi nx}$$ In other words, $$\int_0^1 e^{-2i\pi nt}f_0(t)dt=1$$ Therefore all Fourier coefficients of $f_0$ are equal to 1. This is a contradiction since $f_0$ is assumed to be a continuous function (so its Fourier coefficients must converge to 0 as $|n|\rightarrow +\infty$ by Riemann-Lebesgue).