Why is $\sup_{x∈[0,1]} {|p'(x)|} ≤ A_d\sup_{x∈[0,1]}{|p(x)|}$ for all polynomials $p$ of degree at most $d$? How can one prove that for any positive integer $d$, there is a constant $A_d &lt; 0$ 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 $d$?
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.
To see a proof that makes this inequality precise, stating that $A_d \leq d^2$, look up Markov's inequality.