# Bounding derivative of a function

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 $a$? –  copper.hat Nov 28 '12 at 22:35
Edited the question, now no typos. –  user51076 Nov 28 '12 at 22:54
How did you find the bound using Bernstein's Theorem? –  dexter04 Nov 28 '12 at 23:00
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. –  user51076 Nov 29 '12 at 6:26
@ dexter04: Based on Bernstein's theorem, $|a'(t)|\leq2f_{max} |a_{max}|$. –  user51076 Nov 29 '12 at 6:29