# 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 Bernstein'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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Please edit your question for English and math. What is $a(t)$? is this the same as $x(t)$? What sort of estimate do you want? $x(t)$ can be zero at some points where $x'(t)/x(t)=\infty$. –  Alexandre Eremenko Nov 28 '12 at 22:42
Thanks Alexandre, I have corrected the notation and the question. –  Neeks Nov 28 '12 at 22:51
You did not tell me what sort of estimate you want. There is no uniform estimate, of course: $a$ can have complex zeros as close as you wish to the real line, condition $a>0$ does not help, and $a'/a$ can be arbitrarily large at some points. –  Alexandre Eremenko Nov 29 '12 at 2:07
In the case I am dealing with, only real zeros of $a(t)$ are of interest. But you brought out interesting possibility. 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. But polynomials are also entire function and hence band-limited but the bandwidth is too large for it. Please correct me. –  Neeks Nov 29 '12 at 6:24

$a(t)=\cos(t+iϵ)\cos(t−iϵ)$ is a low pass signal and $a^\prime/a$ can be as large as you wish at the point $t=\pi/2,$ if $\epsilon$ is sufficiently small.
1. What "real zeros" are you talking about if one of your conditions is that $a(t)>0$ ? 2. Example which I gave shows that in general there is NO upper bound for $a'/a$ under your conditions. What else are you asking? –  Alexandre Eremenko Nov 29 '12 at 23:38