# Fourier coefficients

1. Dec 18, 2013

1. The problem statement, all variables and given/known data
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.

3. 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!

Last edited: Dec 18, 2013
2. Dec 18, 2013

### vanhees71

Note that $f$ is real valued by assumption!

3. Dec 18, 2013

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