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Hi,

I was asking myself about some necessary and/or sufficient conditions for a function to be bandlimited (i.e. its Fourier transform is zero t residing out of [-B,B] for some B>0). Of course, if a function is bounded (timelimited), it cannot be bandlimited. But for a non-bounded function, how can we tell if it's bandlimited or not? Or, how can we know if a function is both time- and band- unlimited? Any sufficient/necessary conditions. I'll be glad if you can share with me some known results.

Thanks!

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$x\mapsto 0 \:$ certainly seems like a bounded (timelimited) function that is also bandlimited. $\hspace{.8 in}$ –  Ricky Demer Nov 12 '12 at 8:06
    
a nonzero function cannot be both timelimited and bandlimited, this is a very known result aka Heisenberg's uncertainty –  Ohad Asor Nov 12 '12 at 11:03
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1 Answer

up vote 4 down vote accepted

There is a necessary and sufficient condition for a function $f$ of the real variable to be "band-limited". It is called the Wiener-Paley theorem. $f$ must be a restriction on the real line of an entire function of exponential type. Of course, here one implicitly assumes that $f$ belongs to an appropriate space which permits to interpret it as a "signal". For example, $L^2$ (finite energy), or $L^\infty$ (bounded amplitude), or Schwarz temperate distribution etc.

P.S. This is a mathematical website, and the engineering terminology of the problem may sound strange to some mathematicians. So let me try to translate: We are taking about a function of a real variable for which Fourier transform is defined in some sense. Band limited means that Fourier transform has bounded support, and "time-limited" means that the function itself has bounded support. Of course the function cannot be simultaneously time- and band- limited, unless it is zero. This is a (very crude) form of the "indeterminacy principle", and also follows from the Wiener-Paley theorem.

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Thanks!! I don't know how I forgot about Wiener-Paley. I also didn't know it's engineering terminology, almost all I have is books and internet. Do you recall any other interesting distinguishing property? Essentially, what I'd like to do is to take L_2 and to divide it into 3 subsets: functions with compact support (CS), functions with fourier transform (FT) with CS, functions that neither they nor their FT has CS. I guess requiring those subsets to be Hilbert spaces is too much to ask, not to mention RKHS. –  Ohad Asor Nov 12 '12 at 22:36
    
No, those are not Hilbert spaces, because the $L^2$ limit of functions with bounded support can have unbounded support. And your third set is not even a vector space. –  Alexandre Eremenko Nov 21 '12 at 3:55
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