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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? –  Robin Chapman Sep 29 '10 at 8:30
    
@Robin With no negative frequencies I mean that the Fourier transform has vanishing components for negative frequencies. –  Tobias Kienzler Sep 29 '10 at 8:34
    
Fourier transform? What's that? What would the Fourier transform of $z\mapsto\exp(z)$ be? And what are "negative frequencies"? –  Robin Chapman Sep 29 '10 at 9:07
    
@Robin: Something proportional to $\delta(\omega - i)$ I guess, although I'm not sure what to do with imaginary frequencies... –  Tobias Kienzler Sep 29 '10 at 9:22

3 Answers 3

up vote 2 down vote accepted

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? –  Tobias Kienzler Sep 29 '10 at 11:59

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 domain –  Tobias Kienzler Sep 29 '10 at 9:18
    
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". –  Florian Sep 29 '10 at 9:30
    
@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. –  Tobias Kienzler Sep 29 '10 at 10:30
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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.) –  Hans Lundmark Sep 29 '10 at 10:53
    
@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 $2\pi$) –  Tobias Kienzler Sep 29 '10 at 11:07

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 $[0,\infty)$

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