# Polynomial boundaries

Suppose polynomial $P_n$ of degree $n$ is such that $|P_n(x)|\le 1$ for $|x|\le 1$. What can you say about the $|P_n'(x)|$ for $|x|\le 1$?

This question is just a generalization of this result: $f(x)=ax^2+bx+c$ where $a, b, c \in R$ and $|f(x)|\leq 1$ on the interval $|x|\leq1$. Prove that $|f'(x)\leq4|$ on the same interval..

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I vaguely recall this: among all polynomials with sup norm less than 1, the Chebyshev polynomials (or some shifted/scaled variant) are the ones with maximum first derivative. So, to get a tight bound, you just have to figure out the bound for derivatives of Chebyshev polynomials.

I’ll try to find a reference, unless someone else finds one first.

Confirmation:
My memory was correct (surprisingly). There is a result known as the Markov-Bernstein inequality. Suppose we use $\|p\|$ to denote $\sup\{|p(x)| : x \in [-1,1] \}$. Then the result says that, for any polynomial $p$ of degree $n$, we have
$$\|p^{(k)}\| \le \|T_n^{(k)}\| \cdot \|p\|$$
where $T_n$ is the Chebyshev polynomial of degree $n$. The bound is tight, and is achieved if $p$ is the Chebyshev polynomial of degree $n$.

Figuring out explicit values for $\|T_n^{(k)}\|$ is straightforward. Putting $n=1$, we see that
$$\|p\| \le 1 \Longrightarrow \|p’\| \le n^2$$