No convolution Identity element in $L^1_{per} $ using Fourier series

Question

We have to show that there is no identity element for the ring $ L^1_{per}( ]0,2\pi [) $ specifically using the Fourier coeffcients.

Suppose that : $ \exists e , \: \: e * f = f \: \: \: \: \forall f $

Then $ c_n(f) = \widehat f \widehat e (n) $ and $ \widehat e (x) = 1 $

I know that $ \widehat 1 = 2 \pi \delta_0 $ and this is a distribution hence not in $ L^1 $ but here want don't want to use this result.

Am I allowed to talk about Fourier series for functions in $ L^1_{per}( ]0,2\pi [) $ space ?? I know that there are convergence theorems for continuous periodic functions and also $ L^2 $ functions but what can we say here?

Thanks in advance

Answer 1

I think you'll want to use the Riemann-Lebesgue Lemma for Fourier coefficients. Clearly, $\hat e(n)$ does not vanish as $|n|\to\infty$. Note that the Fourier coefficients of $f$ are perfectly well defined for $f\in L^1$; the issue is the Fourier inversion formula, but that's not needed for the RL Lemma.