Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there some connection between the power-spectrum of a real function $f:\mathbb{R}\to\mathbb{R}$ (that is, its Fourier transform) and the convergence radius of its Taylor expansion around arbitraty $x_0$? Intuitively, I would expect a function with 'limited power at high frequencies' to have 'large convergence radius' around each point, but I could not find such result.

Thanks!

share|improve this question
add comment

2 Answers 2

up vote 2 down vote accepted

A basic phenomenon in the direction of your question is the following. If $\hat{f}(\xi)e^{C |\xi|}$ ($C<+\infty$) is integrable, then $f$ has a holomorphic extension to the strip of width $C$ around the $x$-axis. As a consequence, the Taylor series of $f$ converges on an interval of radius $C$ around each point. One can generalize this to several variables.

share|improve this answer
    
Nice. Can you provide more details on this? Maybe a link or a name for this phenomenon which I can follow? –  Uri Cohen Dec 3 '12 at 12:49
    
Dear Uri Cohen, I have never seen the kind of result you are looking for in a book or paper. It is a very well-known (and elementary) fact that the integrability of $\widehat{f}(\xi)e^{C|\xi|}$ implies the absolute convergence of the integral $F(z):=\frac{1}{2\pi}\int_\R \widehat{f}(\xi)e^{i\xi z}d\xi$ for $|y|<C$, which defines a holomorphic function on the strip that extends $f$. Then complex analysis of one variable tells that the Taylor series converges on intervals of length $2C$. –  Gian Maria Dall'Ara Dec 7 '12 at 13:09
add comment

Another connection is the Wiener-Paley theorem. If the Fourier transform has bounded support then the function is analytic in the whole plane (and has exponential type there). Fourier transform does not have to be integrable in this case, it may exist as a distribution of even a hyperfunction. This fact also has generalization to higher dimensions.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.