Solving Real Valued Fourier Coefficients

[DeadOriginal]

Homework Statement

Let $f$ be a $2\pi$ periodic function. Let $\hat{f}(n)$ be the Fourier coefficient of $f$ defined by
$$
\hat{f}(n)=\frac{1}{2\pi}\int_{a}^{b}f(x)e^{-inx}dx.
$$
for $n\in\mathbb{N}$. If $\overline{\hat{f}(n)}=\hat{f}(-n)$ show that $f$ is real valued.

The Attempt at a Solution

Suppose for contradiction that $f$ is not real valued. Then
$$
\overline{\hat{f}(n)}=\frac{1}{2\pi}\overline{\int_{a}^{b}f(x)\cos nxdx}=\frac{1}{2\pi}\left[\overline{\int_{a}^{b}\Re(f(x)\cos nx)dx+i\int_{a}^{b}\Im(f(x)\cos nx)dx}\right]\\
=\frac{1}{2\pi}\left[\int_{a}^{b}\Re(f(x)\cos nx)dx-i\int_{a}^{b}\Im(f(x)\cos nx)dx\right]
$$
and
$$
\hat{f}(-n)=\frac{1}{2\pi}\left[\int_{a}^{b}\Re(f(x)\cos nx)dx+i\int_{a}^{b}\Im(f(x)\cos nx)dx\right]
$$
so $\hat{f}(n)\not=\hat{f}(-n)$ a contradiction. Can anyone verify my proof for me? Thanks!

 

[vanhees71]

Note that f is real valued by assumption!

 

[DeadOriginal]

Woops! This was an if and only if problem and I was having trouble with the converse part. Sorry for the confusion.