Let $f\in L^1(R)$ such that $F(f)$ is odd, where $F$ is the Fourier transform. Can I then say that $f$ is odd?


If $F(f)$ is odd, then 


$\int \cos(x\xi) f(x) dx = 0 \:\:\forall \xi\in R$


Can I deduce from it that $f$ is odd?



If the fourier transformed of f is odd, is f odd? 
Write $f^-(x)=f(-x)$ etc. Then $F(f^-)=F(f)^-$. If $F(f)$ is odd then
$$0=F(f)+F(f)^-=F(f+f^-).$$
The only $L^1$ function with zero Fourier transform is $0$ so that $f+f^-=0$,
that is, $f$ is odd.


