# Fourier transform is real if $f$

I want to prove that the Fourier transform $F(\xi)$ of a function $f$ will be a real function when, and only when, $f(x)$ is an even function.

I'm using the following definition of Fourier transform: $F(\xi) = \int_{-\infty}^{\infty} \! e^{-2x\pi i \xi}f(x) \, \mathrm{d}x$. I have problems trying to prove that $f$ is even. Can you gimme a hand? I'm trying to prove that $\int_a^b \! \sin(2\xi x\pi) f(x) \, \mathrm{d}x=0$ implies $f$ even. Would the periodicity of the sin function help?

• Do we have any assumption on $f$, e.g. is it continuous? Commented Jun 21, 2015 at 19:33

$f(x) = f_{odd}(x) + f_{even}(x)$
$e^{-2 x \pi i \xi} = \cos(2 x \pi \xi) - i\sin(2 x \pi \xi)$
$\int e^{-2 x \pi i \xi} f(x) = ...$
$\int \cos(2 x \pi \xi) f_{even}(x) - i \int \sin(2 x \pi \xi) f_{odd}(x)$,
Where the other terms cancel do to (anti)symmetry. If $f_{odd}(x) = 0$, then you are left with only a real result; else you will have an imaginary component in the final integral.
• I'm not sure about this because you're assuming there does not exist $f_{odd}$ nonzero such that $\int f_{odd}(x)\sin(2x\pi\xi)dx = 0$ - i.e., there's no odd functions orthogonal to any frequency of $\sin$. This is sort of a variation on the fundamental lemma of calculus of variations. I think that it's true, but I'm not sure it can go as an assumption. Commented Jun 22, 2015 at 21:16