3
$\begingroup$

Is it true that for every $m \geq 1$ there is $C = C(m)$ such that if $p(x)$ is an $m$-degree polynomial with $p(x) \in [0,1]$ for every $x \in [0,1]$ then $|p'(x)| \leq C$ for every $x \in [0,1]$?

1 Answers 1

4

Yes it is true. Here is a way to see it: check that $\sup_{x \in [0,1]} |f(x)|$ and $|f(0)|+\sup_{x \in [0,1]} |f'(x)|$ define norms on the space of polynomials of degree at most $m$, and use the equivalence of norms in finite dimension to conclude.

  • 0
    Does this argument give an explicit dependence of $C(m)$ on $m$? – 2017-01-26
  • 0
    @Bill unfortunately not. I don't see any easy way to get that right now, but that's an interesting question. If you were to strengthen the assumption to $|p(z)| \leq 1$ for all complex $z$ with e.g. $|z-\frac{1}{2}| \leq \frac{1}{2}$, the result would follow from the maximum principle – 2017-01-27
  • 0
    but here the bounds $C(m)$ definitely blows up with $m$ – 2017-01-27