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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 $\Leftrightarrow$ $\mathrm{supp}_{\mathcal F\{f\}}=\mathbb R^+$

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    What does "has no negative frequencies" mean?2010-09-29
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    @Robin With no negative frequencies I mean that the Fourier transform has vanishing components for negative frequencies.2010-09-29
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    Fourier transform? What's that? What would the Fourier transform of $z\mapsto\exp(z)$ be? And what are "negative frequencies"?2010-09-29
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    @Robin: Something proportional to $\delta(\omega - i)$ I guess, although I'm not sure what to do with imaginary frequencies...2010-09-29

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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

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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.

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    Sorry for the confusion, I mean kind of the opposite, I hope my edit clarified that. I mean analytical in the time domain, but vanishing components for negative frequencies in the Fourier-transformed domain2010-09-29
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    The fourier transform is (up to a sign flip and perhaps a constant normalization factor) the same as its inverse, so you can think of either copy of $\mathbb{R}$ as the "frequency space" and the other one as the "time space".2010-09-29
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    @Florian: yes that's just a matter of definition. But my question refers to analyticity in the one domain and a positive-only support in the other, Fourier transformed, domain, while your answer is about analyticity in the latter domain.2010-09-29
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    @Tobias: Since the Fourier transform of $\chi_{[-1,1]}(z)$ is $\sin\omega/\omega$, the Fourier transform of $\sin z/z$ is $\chi_{[-1,1]}(\omega)$. (All of this up to some scaling and normalization, of course, but I'm too lazy to check the exact details.)2010-09-29
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    @Hans Yes. But my question is: Does every analytical function $f(z)$ have a Fourier-transform $f(\omega)$ which vanishes for $\omega<0$ and vice versa? The way I understand Florian's answer it is however a reply to the question "Is the Fourier transform of every function both analytical and vanishing for $\omega<0$?" which he correctly proves wrong. (PS: the product of the normalizations in both domains is $2\pi$)2010-09-29
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    Oh, *now* I get it! Thanks. Is at least the other direction true, i.e. does a support of $\mathbb R^+$ imply an analytic function as Fourier transform or is this only true for compact supports?2010-09-29
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    @Tobias: No. The Heaviside step function has a Fourier transform that is not even a function. Alternatively, $f(x)=1_{[1,\infty)} x^{-2}$ has a Fourier transform that is continuous but not even differentiable.2010-09-29
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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.

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    Thank you, the Paley-Wiener theorem is quite what I was looking for. But do I understand it correctly that this only offers *possible* function supports for which the (inverse) Fourier transform is guaranteed to be analytical, but (as your example of the Gaussian proves) there are exceptions with no limitations on the support?2010-09-29