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 real-valued if and only if $F(n)=\overline{F(-n)}$ for all $n$.
We have $$\overline{F(-n)}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\overline{f(x)}e^{-inx}dx$$
If $f(x)$ is real, then $f(x)=\overline{f(x)}$, and clearly $F(n)=\overline{F(-n)}$.
If $F(n)=\overline{F(-n)}$, then $$\int_{-\pi}^{\pi}f(x)e^{-inx}dx=\int_{-\pi}^{\pi}\overline{f(x)}e^{-inx}dx$$
How can I conclude that $f$ is real-valued?
Best Answer
Define $g(x):=f(x)-\overline{f(x)}$: we have to show that $g(x)=0$ for all $x$. We have $\int_{-\pi}^\pi g(x)e^{-inx}\mathrm dx=0$ for all $n$. Function of the form $\sum_{j=-N}^Na_je^{ijx}$ are dense in $C[-\pi,\pi]$, hence $$\int_{-\pi}^\pi g(x)h(x)\mathrm dx=0$$ for each $h\in C[-\pi,\pi]$. By an other approximation argument, we get that for each closed subset of $[-\pi,\pi]$, $\int_Fg(x)\mathrm dx=0$.