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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.

shadow10
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    Pick a $y\in[0;2\pi]$ and a $z\in\mathbb C.$ Define $h:[0;2\pi]\rightarrow\mathbb C$ by $h(y) = z$ and $h(x) = f(x)$ for $x\in[0;2\pi]\setminus{y}$. Let's assume $z\notin\mathbb R.$ Then $h$ is not real. Moreover, since $f$ is Riemann-integrable, $h$ is, too (Riemann-integrability of a function is preserved if we change that function at one point). Now, $h$ has exactly the same Fourier coefficients as $f,$ but $h$ is not real. So the claim from your question ... – jflipp Apr 02 '15 at 13:59
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    ... (i.e. $f$ Riemann-integrable and the relation between Fourier coefficients stated in the question together imply $f$ real-valued) is just not true. Behavior like this is a reason why Fourier analysis is usually done using the Lebesgue integral, with functions defined only almost everywhere. – jflipp Apr 02 '15 at 14:03

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