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Can we say something like monotonicity, growth rate and oscillation of the Fourier transform of a banded function $f$ with support $[0, N]$ $$\mathcal{F}f(\xi) = \int_{0}^N f(x)e^{-ix\xi}dx.$$ Of course generally $\mathcal{F}f(\xi)$ is complex. I guess it is impossible for the real part or imaginary part $\mathcal{F}f(\xi)$ to be monotone globally. But is it possible to be monotone on $[0, cN]$? Or have very small oscillation on $[0, cN]$. Here $c$ is a positive constant.

Thanks,

Jack

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1  
What is a "banded function"? –  drbobmeister Jun 28 '13 at 17:36
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What do you mean by monotone locally? Any C^1 function is monotone on many sufficiently small intervals –  Yemon Choi Jun 28 '13 at 18:15
    
Thanks! Locality is a bad word here. Here I mean the function is monotone on [0, cN] for some universal constant c > 0. –  Jack Lee Jun 28 '13 at 22:05

2 Answers 2

up vote 1 down vote accepted

There is a paper related to your question: MR2072747 Ostrovskii, I. V.; Ulanovskii, A. Non-oscillating Paley-Wiener functions. J. Anal. Math. 92 (2004), 211–232.

The Paley-Wiener theorem completely describes the Fourier transforms of functions with bounded support. Because of the importance of these functions in signal processing, and other applications, there is an enormous literature about them.

On your second question. The answer is no. Take a constant function for example. However, if you require that all Fourier coefficients are zero on $(-M,M)$, one can say something on the sign changes, the most comprehensive account of this phenomenon which I know is MR2038065, and a short exposition is MR2048457. These papers are also avilable from the arxiv and from my web page.

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Thank you very much! I have a further question. If a real-valued function $f\in L^2$ has Fourier coefficients vanishing on $(-M, -1)\cup (1, M)$, where $M>1$, is it guaranteed that $f$ has a sign change in $(0, cM)$? –  Jack Lee Jun 30 '13 at 8:52
    
No. Take a constant function. But if the coefficients are –  Alexandre Eremenko Jun 30 '13 at 15:19
    
Sorry I made a typo. I meant that $f$ has Fourier coefficients supported on $(-M, -1)\cup (1, M)$. Will the corresponding conclusion change? Thanks! –  Jack Lee Jun 30 '13 at 16:50
    
Yes, look at the papers I mentioned. You say "Fourier coefficients", probably meaning "Fourier transform", but this does not matter much. It oscillates if there is a spectral gap. –  Alexandre Eremenko Jun 30 '13 at 17:17
    
Thanks Professor. I have read the paper but the paper proved a lower bound of the sign changes density asymptotically, if I understand correctly. So it does not deal with the case that $\hat f$ is supported on $(-M, -1)\cup (1, M)$. –  Jack Lee Jun 30 '13 at 19:38

By translation and rescaling we can assume that the support is $[-1,1]$. The Paley-Wiener theorem characterizes distributions with support $[-1,1]$. A distribution $u$ is supported in $[-1,1]$ if and only if $\hat u$ is an entire function (holomorphic in $\mathbb C$) such that there exists $M, C$, $$\forall \zeta\in \mathbb C,\quad \vert \hat u(\zeta)\vert\le C(1+\vert\zeta\vert)^M e^{2π\vert\Im \zeta\vert}.\tag{PW} $$ Since it is an iff condition, you can take any entire function $v$ satisfying the estimate that $\hat u$ satisfies in (PW), then its Fourier transform will be supported in $[-1,+1]$: there is no further restrictions.

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