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Let $f\in L^1(\mathbb{R}/2\pi\mathbb{Z})$ and let $F(n)$ denote its Fourier coefficients $$F(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx$$ I want to prove that $f$ is even if and only if $F(n)=F(-n)$ for all $n$.

Suppose $f$ is even. Then I have to prove $\int_{-\pi}^\pi f(x)e^{-inx}dx=\int_{-\pi}^\pi f(x)e^{inx}dx$. This is true because the value on the left-hand side at $a$ is $f(a)e^{-ina}$, while the value of the right-hand side at $-a$ is $f(-a)e^{-ina}=f(a)e^{-ina}$.

What about the converse? Suppose $F(n)=F(-n)$ for all $n$. How can I show that $f(x)=f(-x)$ for all $x$?

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How about making a change of variables and applying an inverse Fourier transform? –  Riemann1337 Oct 26 '13 at 18:34
    
@Riemann1337 Are there more elementary ways than that? (i.e. without making the inverse Fourier transform) –  Paul S. Oct 26 '13 at 18:45
    
Maybe there is a way to argue it by just subtracting the integrals, but I am not sure; using the inverse Fourier transform seems to be the most straightforward way to solve it. –  Riemann1337 Oct 26 '13 at 18:59
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Changing variables, we have $$ F(n)=F(-n)\rightarrow \int_{-\pi}^\pi f(x) e^{inx}dx=\int_{-\pi}^\pi f(x) e^{-inx}dx =\int_{-\pi}^\pi f(-x) e^{inx}dx, $$ so $$ \int_{-\pi}^\pi[f(x)-f(-x)]e^{inx}dx = 0. $$ If $\mathcal{F}$ denotes the Fourier transform, then we have that $$ 0=\mathcal{F}[f(x)-f(-x)]\rightarrow\mathcal{F}[f(x)] = \mathcal{F}[f(-x)]. $$ Applying the inverse Fourier transform should give the result.

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