Evenness of Fourier coefficients implies even function

Question

Let $f\in L^1(\mathbb{R}/2\pi\mathbb{Z})$ and let $F(n)$ denote its Fourier coefficients $$F(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx$$Assume that the Fourier coefficients determine an $L^1$ function.

How can we prove, without using Fourier transform, that if $F(n)=F(-n)$ for all $n$, then $f(x)=f(-x)$ for all $x$?

Answer 1

Let $g(x) = f(x) - f(-x)$. Then $\hat{g}(n) = \hat{f}(n) - \hat{f}(-n) = 0$ for all $n$. So by the uniqueness theorem $g(x) = 0$ a.e.

I'm not sure what you mean about not using the Fourier transform since the statement itself involves Fourier coefficients, but the usual proof of this uses convolutions and taking limits of them, with not much emphasis on the Fourier transform side.

Answer 2

I hope this satisfies the condition not using Fourier transform: $$ \sum_{n=-N}^NF(n)\,e^{inx}=F(0)+\sum_{n=1}^NF(n)(e^{inx}+e^{-inx})=F(0)+2\sum_{n=1}^NF(n)\cos(n\,x). $$ Since $\cos$ is an even function,...