Why is $\sup_{x∈[0,1]} {|p'(x)|} ≤ A_d\sup_{x∈[0,1]}{|p(x)|}$ for all polynomials $p$ of degree at most $d$?

Question

How can one prove that for any positive integer $d$, there is a constant $A_d < 0$ such that $$ \sup_{x∈[0,1]} {\lvert\, p'(x)\rvert} ≤ A_d\sup_{x∈[0,1]}{\lvert\, p(x)\rvert}, $$ for all polynomials $p$ of degree at most $d$?

Answer 1

This is equivalent of saying that the differential operator is continuous, which is clear, since the vector space of polynomials of degree at most $d$ is finite-dimensional.

Answer 2

Define the functions $f: \mathbb R^{d+1}\to \mathbb R$ and $g: \mathbb R^{d}\to \mathbb R$, as $$ f(a_0,\ldots,a_d)=\sup_{x\in[0,1]}\left|\textstyle\sum_{k=0}^d a_kx^k\right| \quad\text{and}\quad g(a_1,\ldots,a_d)=\sup_{x\in[0,1]}\left|\textstyle\sum_{k=0}^{d-1} ka_kx^{k-1}\right|. $$ It is clear that both $f$ and $g$ are norms in $\mathbb R^{d+1}$ and $\mathbb R^d$, respectively, and hence equivalent to every other norm in those spaces. Thus there exist $c_d,C_d>0$ such that $$ c_d g(a_1,\ldots,a_d)\le \|(a_1,\ldots,a_d)\|_{d}\le \|(a_0,\ldots,a_d)\|_{d+1}\le C_d\, f(a_0,\ldots,a_d), $$ where $\,\|\cdot\|_{d},\,\|\cdot\|_{d+1}\,$ are the standard Euclidean norms in $\mathbb R^d$ and $\mathbb R^{d+1}$, repsectively. Thus $$ \sup_{x\in[0,1]}\lvert p'(x)\rvert=g(a_1,\ldots,a_d)\le \frac{C_d}{c_d}f(a_0,\ldots,a_d)= \sup_{x\in[0,1]}\lvert p(x)\rvert, $$ whenever $p(x)=a_0+\cdots+a_dx^d$.

Answer 3

To see a proof that makes this inequality precise, stating that $A_d \leq d^2$, look up Markov's inequality.