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?
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? 


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. 


Yes. Using tempered distributions it's immediately obvious, since $f = C \mathcal{F}^3 (\mathcal{F} f)$ for a constant $C$, and $\mathcal{F}$ maps odd distributions into odd distributions. The missing details of the proof are just simple exercises in distributions. Distribution theory is very useful for Fourier transform questions like this, since Fourier inversion works perfectly. 


Look at $L^2$ first: In $L^2$ the FT is diagonalizable. The space of odd functions $\in L^2$ is the direct sum $Eig(\mathcal{F},+i)\oplus Eig(\mathcal{F},i)$ of the eigenspaces of $\mathcal{F}$ with respect to the eigenvalues $+i$ and $i$. Because eigenspace are mapped into themselves, $\mathcal{F}$ maps odd functions to odd functions. In particular this is true for all $f\in L^1\cap L^2$ and by continuity it is true for all $f\in L^1$. (In fact an analogue statement is true for all tempered distributions.) EDIT: Oh, I just saw that you asked for the other direction. Using the same argument you can show that $\mathcal{F}f$ odd $\implies f=\mathcal{F}^3(\mathcal{F}f)$ odd is true for all $f\in L^2$. This time I'm not quite sure if it is possible to extend this from $L^1\cap L^2$ to $L^1$, but maybe the result for $L^1\cap L^2$ is useful for you too. 

