Let $\hat{f}(n)$ be the Fourier coefficients of $f:[0,2\pi]\to \mathbb{C}$ defined as $$\hat{f}(n)=\int_{0}^{2\pi}f(x)e^{-{\rm{i}}nx}\,\mathrm{d}x$$ Note $f$ is Riemann-integrable on $[0,2\pi]$. We are given $\hat{f}(n)=\overline{\hat{f}(-n)}\forall n$, show $f$ is real valued.
What I did was, let $f=u+{\rm{i}} v$ now we want to show $v(x)=0$ for all $x$. We have $\hat{u}(n)+{\rm{i}} \hat{v}(n)=\overline{\hat{u}(n)}-{\rm{i}}\overline{\hat{v}(-n)}$. We also have that $\hat{u}(n)=\overline{\hat{u}(-n)},\hat{v}(n)=\overline{\hat{v}(-n)}$ since $u,v$ are real valued. Now, $\hat v(n)=0$ for all $n$. So we have $$\int_{0}^{2\pi}v(x)e^{-{\rm{i}}nx}\,\mathrm{d}x=0$$ but since any continuous $2\pi$ periodic function can be approximated by trigonometric polynomials, so $$\int_{0}^{2\pi}v(x)g(x)\,\mathrm{d}x=0$$ where $g$ is continuous and $2\pi$ periodic. Since we can approximate $v$ by such a $g$ we also have $$\int_{0}^{2\pi}v^2(x)\,\mathrm{d}x=0$$ But I cannot conclude that $v$ is zero. If $v$ was continuous then I could have concluded that but since continuity is not given I don't see a way out. Can someone help me? Thanks.