TITLE: norm of differential operator on $P^n[0,1]$
QUESTION [5 upvotes]: Consider the space $P^n[0,1]$ of polynomials of degree $\leq n$ on $[0,1]$, equipped with the sup norm. Now, this is a finite dimensional space, so all linear operators have to be continuous, hence bounded.
My question is: what is the norm of the differential operator $d/dx : P^n[0,1] \rightarrow P^{n-1}[0,1]$? I can't see the relation between the supremum of a polynomial $p(x)=a_0+\dots+a_nx^n$ and the supremum of $p'(x)=a_1 + 2a_2 x +\dots+n a_n x^{n-1}$.

REPLY [2 votes]: The answer is given here for the interval $[-1,1]$ by Markov's inequality for polynomials: The norm is $n^2$ and $||P'||_\infty=n^2||P||_\infty$ for the $n$-th Chebyshev polynomial $T_n$.
For the interval $[0,1]$, we have to use scaling: If $p$ is a polynomial considered on $[0,1]$, then $q(x)=p(\frac{x+1}2)$ is a polynomial considered on $[-1,1]$. Hence $||q'||_{\infty,[-1,1]}\leq n^2||q||_{\infty,[-1,1]}$. Since $q'(x)=\frac12 p'(\frac{x+1}2)$, we find $||p'||_{\infty,[0,1]}\leq 2n^2||p||_{\infty,[0,1]}$ and equality is attained for $t_n(x)=T_n(2x-1)$.