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Consider $a(t)\in\mathbf{L}^{2}(\mathbb{R})$ and $a(t)>0$, is a low pass smooth function with $\hat{a}(f)=0, |f|>f_{max}$. Can we have a upper bound on the following, $\Big|\frac{a'(t)}{a(t)}\Big|$? Using Bernstien's theorem we can upper bound $|a'(t)|$ alone based on $f_{max}$ but how can we upper bound the ratio mentioned here. Any suggestions for it.

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    What is $L(\mathbb{R})$? What is $a$?2012-11-28
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    Edited the question, now no typos.2012-11-28
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    How did you find the bound using Bernstein's Theorem?2012-11-28
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    In the case I am dealing with, only real zeros of $a(t)$ are of interest. Now taking a simple example, $a(t)=1+\mu\sin\omega t$, with $0<\mu<1$ we have,\newline $\Big|\frac{a'(t)}{a(t)}\Big|=\frac{\mu\omega\cos\omega t}{1+\mu\sin\omega t}\leq\frac{\mu\omega}{\sqrt{1-\mu^2}}, \forall t$. So above it can be bounded, though a simple example. Can we have a bound for sum of harmonic sinusoids and generalize it.2012-11-29
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    @ dexter04: Based on Bernstein's theorem, $|a'(t)|\leq2f_{max} |a_{max}|$.2012-11-29

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