For every polynomial $p:[0,1] \rightarrow \mathbb{R}$ with $deg(p) \leq n$ we have $\|p'\|_\infty \leq C_n\|p\|_\infty$, for a constant $C_n >0$. We need to prove that, for every $n \in \mathbb{N}$, there exists a constant $C_n >0$ that only depends on $n$, such that for every polynomial function  $p:[0,1] \rightarrow \mathbb{R}$ with $deg(p) \leq n$, we have $\|p'\|_\infty \leq C_n\|p\|_\infty$. 
This exercice was given to me in a functional analysis course. 
I have no idea how to approach this. 
Any hints? 
Any help would be greatly appreciated.
 A: There is a theorem, which says that a linear map between two finite dimensional spaces must be continuous.
The derivative could be seen as a linear operator from $\mathbb{R}_n[x]$ to $\mathbb{R}_{n-1}[x]$, where $\mathbb{R}_n[x]$ means the normed space of polynomials from $[0,1]$ to $\mathbb{R}$ of degree $\le n$, with norm $\|\cdot\|_\infty$.
Here is a hint to prove this theorem:
Since all norms on a finite dimensional space are equivalent, you just need to prove that a linear map between $(\mathbb{R}^n,\|\cdot\|_2)$ and $(\mathbb{R}^m,\|\cdot\|_2)$ is continous. To reach this, you can denote the map by an $m\times n$ matrix $A=(a_{ij})$. Thus,
$$ \|Ax\| = \left(\sum_{i=1}^m \Big|\sum_{j=1}^n a_{ij}x_j \Big|^2 \right)^2, $$
Cauchy's inequality will be useful here.
A: The case $n=0$ is trivial.
For $n>0:$ By contradiction, for $j\in \Bbb N$ suppose $p_j(x)=\sum_{i=0}^nA_{j,i}x^i$ with  $\|p'_j\|>j\|p_j\|$ and  with $\max \{|A_{j,i}|:i\le n\}=1$ for each $j.$
There exists $i_0\le n$ and  a sub-sequence $(p_{j(k)})_{k\in \Bbb N}$  for which  $(A_{j(k),i})_{k\in \Bbb N}$ converges to some $B_i$ for each $i\le n$ and such that $ |A_{j(k),i_0}|=1$ for every $k.$ Then  $(p_{j(k)})_{k\in \Bbb N}$ converges in norm to $p(x)=\sum_{i=0}^nB_ix^i$ and $(p'_{j(k)})_{k\in \Bbb N}$ converges in norm to $p'.$
But $p\ne 0$ because $|B_{i_0}|=1,$ and for all but finitely many $k$ we have  $$0<j(k)\|p\|\le 2j(k)\|p_{j(k)}\|<2\|p'_{j(k)}\|\le 2\sum_{i=0}^n 2i|A_{j(k),i}|\le 4n(n+1).$$
