Let $P_n$ be the <finite-dimensional vector space> of real polynomials of degree at most $n$. For fixed $a>0$,
$$
N(p)=\sup_{x\in[0,a]}|p(x)|
$$
is a <norm>: if it vanishes, the polynomial vanishes on an interval and hence is the zero polynomial. Also
$$
M(p)=N(p)+\sup_{x\in[0,1]}|p'(x)|
$$
is a norm on $P_n$. By <finite-dimensional equivalence of norms>, there is $C>0$, depending only on $n$ and $a$, such that $M(p)\leq C N(p)$. Therefore
$$
\sup_{x\in[0,1]}|p'(x)|\leq C\sup_{x\in[0,a]}|p(x)|.
$$
The <extreme value theorem> supplies $y\in[0,a]$ at which the last supremum is attained, and hence
$$
\boxed{\sup_{x\in[0,1]}|p'(x)|\leq C|p(y)|.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 2 / 12F / c / Solution

Any two norms on a finite-dimensional real or complex vector space are equivalent. Relative to a fixed basis, continuity and positivity of a norm on the compact Euclidean unit sphere give uniform positive lower and finite upper bounds.


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