# Prove that for every positive integer $d$ there exists $C(d)>0$ such that

for every polynomial $p(x)$ with degree $\leq d$, $\max\limits_{x\in[0,1]}|p'(x)| \leq C(d)\max\limits_{x\in [0,1]} |p(x)|$. There was also a hint given, that says to "use the compactness of a subset in an appropriate finite-dimensional space". However, I am confused by the hint. Should I be using the compactness of $[0, 1]$ in $\mathbb R$, or some other space? Is this perhaps related to functional analysis? In any case, I only want a hint because I want to try to solve this question on my own. Thank you very much!

Let $P_d$ denote the $d+1$-dimensional vector space of real polynomials of degree $\leq d$. The map

$$D:P_d\longrightarrow P_d\qquad p\longmapsto p'$$ is linear and what you are trying to prove says pecisely that $$\|D\|=\sup_{\|p\|\leq 1}\|Dp\|<\infty$$ that is $D$ is bounded (equivalently, continuous) when we equip $P_d$ with the sup norm $\|p\|=\sup_{[0,1]}|p(x)|$.

This relies on the following fundamental result.

Fact: every linear map on a finite-dimensional normed vector space is bounded.

Sketch: one can show first that the unit ball is compact for the $\ell^1$ norm. Then one proves that all norms are equivalent to the $\ell^1$ norm. A fortiori, they're all equivalent and they all have a compact unit ball. Finally, one proves the result easily with the $\ell^1$ norm on the domain. By equivalence, it is bounded with respect to any norm. QED.

• do you perhaps mean the derivative, instead of the derivation? – Aden Dong Apr 16 '13 at 2:12
• and can you expand on your hint a bit more? i'm still pretty clueless. – Aden Dong Apr 16 '13 at 2:13
• Yes, sure. That was French, not English. – Julien Apr 16 '13 at 2:14
• Do you know the unit ball is compact is finite dimension? And, for the record, that's a characterization: it is not compact in infinite dimension. – Julien Apr 16 '13 at 2:15
• i'll think about it, but i think get it. thank you! – Aden Dong Apr 16 '13 at 2:21