Is a function analytical on C iff its Fourier-transform vanishes for negative frequencies? I think Cauchy's integral formula and the Hilbert transform can be used to prove one direction, but is this an equivalence or only an implication?
edit for clarification:
Is a function $f : \mathbb C \to \mathbb C, z\mapsto f(z)$ analytical $\Leftrightarrow$ The Fourier-Transform $\mathcal F\{f\}(\omega) = N \int_{\mathbb R} f(z) e^{i\omega z} dz$ (choose whatever normalization $N$ you like, I prefer symmetry $N=\sqrt{2\pi}$) is zero for all $\omega<0$?
Or shorter: Is the following true? $f$ analytical $\Leftrightarrow$ $\mathrm{supp}_{\mathcal F\{f\}}=\mathbb R^+$
 A: I'm not sure if I got your question right, but as I understand it the answer is no. Let's stick with the fourier transform as an operator $L^2(\mathbb{R}) \to L^2(\mathbb{R})$.
One of the basic facts about the fourier transform is that compactly supported functions transform into analytic functions. So as a counterexample just pick the characteristic function of $[-1,1]$. Its fourier transform is analytical but it certainly has "negative frequencies", namely, it has frequencies almost everywhere in the interval $[-1,1]$.
Edit: To make this more explicit: Let $f := \hat \chi_{[-1,1]}$. Then $f$ is analytic (in fact, we use the definition $\hat u(\xi) = \int e^{i\xi x} f(x) d x$ then $f(\xi)=2 \sin \xi / \xi$ (and $f(0)=2$), which is analytic). By the Fourier inversion theorem, $\hat f(x)=\hat{\hat \chi}_{[-1,1]} (x)=2\pi \chi _{[-1,1]}(-x)$ , whose support is $[-1,1]$. So $f$ is a counterexample to your statement.
A: So, you are only interested in the Fourier transform of $f$
restricted to the real line. Typically the restriction
of such a function won't be in $L^2(\mathbb{R})$ (or any $L^p(\mathbb{R})$
for that matter) and won't have polynomial growth, so won't be a tempered
distribution. I don't know a way in which such a function (e.g. $f(z)=\exp(z)$)
could be said to have a Fourier transform.
But any compactly supported distribution on $\mathbb{R}$ will have
a Fourier transform that is an entire analytic function.
Some other tempered distributions also have this property, notably
the Gaussian $g(z)=\exp(-z^2)$. As Florian points out, that even when
an entire function has a Fourier transform, it need not be supported
on the positive reals; there is absolutely no bias towards positivity
or negativity in its support.
If you are interested in Fourier transforms of analytic functions, you should
look at the Paley-Wiener theorem which translates between properties
of the one and of the other.
A: You may refer to the theorem 19.2 in Walter Rudin's book. Suppose f is a holomorphic function on the upper half plane. If 
$\sup_{0<y<\infty}\int|f(x+iy)|^2dx=c<\infty$. Then f is the inverse Fourier transform of a function support in $[0,\infty)$
