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f is uni-variate polynomial of degree d. I am interested in lower bounding modulus first derivative of f (i.e. |f'|)in interval [a,b] given the promise that in interval [a,b] f' don't have any root.

Like in Markoff upper bound first derivative of polynomial in open interval (a,b) is
|f'(x)|<=(2*M*d*d)/(b-a) where M is upper bound of f in (a,b),but I don't know proof of this theorem.

I know polynomial f i.e. its coefficient,degree and upper bound as well as lower bound of f in interval [a,b], Now I want to lower bound f' in [a,b] in terms of these terms..say like Markoff theorem.

Pls let me know whether this time I able to put my question clearly :)

Thanks Ram

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How about using the Mean Value Theorem, if you know the values of f at some points? – Yemon Choi Feb 13 '11 at 7:16
that would go in recursion of finding min of till d-th derivative of f e.g. min|f'|= min{f'(a),f'(b),min{f'(c)}} where c is root of f" and f"(c)>=0 I need much cleaner approach like "Markoff Theorem" – Ram Feb 13 '11 at 10:32
Ram, you misunderstood me. If you know the minimum value of p, call it $M$, and you know some other value of p, call it $K>M$, then the mean value theorem tells you that there exists u in your interval such that $K-M \leq |f'(u)|(b-a)$. I agree with all the other commenters that your original question may not admit an answer; just because you want such a bound to exist, that is not a reason to expect such a bound to exist. – Yemon Choi Feb 14 '11 at 0:30

I think you need to be a bit more precise. For example, choose any polynomial $g(x)$ that does not vanish on $[a,b]$, let $f'(x)=(x-a+\epsilon)g(x)$, let $F(x)$ be an antiderivative of $f'(x)$, and let $f(x)=F(x)+C$ for some constant $C$. Choosing $\epsilon$ small, you can make $f'(a)=\epsilon g(a)$ small, while choosing an appropriate $C$, you can make $\sup |f(x)|$ or $\inf |f(x)|$ on the interval $[a,b]$ to be anything that you want. So it's really not clear what sort of lower bound you have in mind.

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This is basically repeating what Joe Silverman wrote, in a concrete example. (I do so, as a misunderstanding between asker and answerers seems to persist.)

It might be difficult (or impossible) to get a lower bound similar to the upper bound. Let us fix the interval $[0,1]$.

  1. For $x^2$ the derivative is $0$ in $0$, and this polynomial is thus not admissible by your question.

  2. For $(x+0.001)^2$ the derivative is $0.002$ in $0$, so very small, and this polynomial is admissible by your question.

  3. For $(x+1)^2$ the derivative is $2$ in $0$, so not too small.

So, one would need to distinguish 2 and 3. With this example one could still guess this might be due to the fact that in 2 the polynomial is small at 0. But replacing $(x+0.001)^2$ by 2'. $(x+0.001)^2 + 1$ the minimum is not too small anymore and just looking at rough parameters like maximum and minimum on the interval the polynomials 2' and 3 are not that different. While they are very different with respect to what you are looking for.

Thus, if a bound of the form you are looking for exists, it seems it definitely has to take into account other/finer quantities than maximum/minimum and degree. (I know you do not restrict exclusively to this situation, so this is not a full answer; but is meant to provide some 'bound' on what one can hope for.)

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As already discussed the answer to the question, as stated, is NO. However, one can get close if one is willing to go to a measure theoretic statement. For example, the statement $$ |\{x \in [a,b]:\quad |f'(x)| < \varepsilon\}| \to 0 $$ as $\varepsilon \to 0$, is an easy consequence of continuity. However, one can quantify this. If I remember correctly, it is relatively easy to prove $$ |\{x \in [a,b]:\quad |f'(x)| < \varepsilon\}| \leq C \varepsilon^{1/d} $$ where $C$ is an (universal) constant and $d$ the degree of the polynomial $f$. I think one can give a proof by just rescaling Polya's inequality.

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Okay, here goes:

1. Calculate the coefficients $\langle c_j : j\in d\rangle$ of the degree d-1 polynomial f'.

2. Run quantifier elimination over real closed fields on the formulas

$((\displaystyle\bigwedge_{j\in d} ((a \leq x_j) \land (x_j \leq b))) \implies$ $(((((\displaystyle\sum_{j\in n} 1)\cdot (\displaystyle\sum_{j\in d} (c_j \cdot (\displaystyle\prod_{j\in d} x_j))))+1) \leq 0) \lor (1 \leq ((\displaystyle\sum_{j\in n} 1)\cdot (\displaystyle\sum_{j\in d} (c_j \cdot (\displaystyle\prod_{j\in d} x_j))))))$

until the result is $ \; (0 = 0) \; $.

3. $\frac1n$ is now a lower bound for $|f'|$ in the interval $[a,b]$.

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We have the following inegalities :

$c_1 n (k+1) \leq \sup_{0 \neq f \in \mathcal P_{k,n}} \frac{ \| f' \|_{[-1,1]} }{ \| f \|_{[-1,1]} } \leq c_2 n (k+1)$

Where $c_1 > 0$, $c_2 > 0$ are absolute constants and $\mathcal P_{k,n}$ is the set of all polynomials of degree at most $n$ with real coecients and with at most $k$ ($0 \leq k \leq n$) zeros in the open unit disk.

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Do you know examples of $c_1$ and $c_2$ that work for that? – Ricky Demer Feb 12 '11 at 11:31
We can find them (and proof) in the book : "Polynomials and polynomial inequalities" By Peter B. Borwein,Tamás Erdélyi. – Ilies Zidane Feb 17 '11 at 19:05

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