TITLE: Prove $|P(0)|\leq 2n+1$
QUESTION [28 upvotes]: Let $P(x)$ be a polynomial with degree $\leq n$ and $|P(x)|\leq\frac{1}{\sqrt{x}}$ for $x\in(0,1]$. Prove that $|P(0)|\leq 2n+1$. The idea should be that if $|P(0)|$ is too large, then the polynomial cannot change values fast enough to avoid intersecting the curve $ \pm 1/\sqrt{x}$, but I don't see how to formalize this.

REPLY [14 votes]: Since $-\frac{1}{\sqrt{x}}\le P(x)\le\frac{1}{\sqrt{x}}$ for $0<x\le 1$, $\;\;\;-1\le \sqrt{x}P(x)\le1\;$ for $0\le x\le1$.
If we let $t=\sqrt{x}$ and $h(t)=tP(t^2)$, then $h$ is a polynomial of degree $2n+1$ with $\big|h(t)\big|\le1$ for $|t|\le1$.
By Bernstein's inequality for polynomials, $\;\;\displaystyle\big|h^{\prime}(t)\big|\le\frac{2n+1}{\sqrt{1-t^2}}\;$ for $|t|<1$;
so $\;\big|P(0)\big|=\big|h^{\prime}(0)\big|\le 2n+1$.

(See (4.2) in http://www.damtp.cam.ac.uk/user/na/people/Alexei/papers/markov.pdf or
Th. 2.8 in http://www.emis.de/journals/HOA/JIA/Volume3_4/156027.pdf)