The problem is the following : Let $f(x) = x^{k} + \epsilon_{k,n}(x)$,$x \in I := [0,1]$, where $\epsilon_{k,n}(x)$ is a polynomial of degree at most $k$;
Suppose we know that $f \overset{||.||_{\infty,I}}{\longmapsto} x^{k}$, in this way we know that $\epsilon_{k,n} \overset{||.||_{\infty,I}}{\longmapsto} 0$.
I'd like to conclude that all the coefficient of $\epsilon_{k,n} \longmapsto 0$ using the fact that all norm are equivalent on $\mathbb{R}^{n}$.
I tried to identify the polynomial with its coefficients thanks to the vectorial space isomorphism $\mathbb{R}_{k}[x] \cong \mathbb{R}^{k+1}$ which sends $p(x) = a_{n}x^{n}+\cdots +a_{0} \longmapsto \begin{pmatrix} a_{n},\cdots,a_{0}\end{pmatrix}$ and the using the equivalence of $||.||_{2},||.||_{\infty,I}$but I was unable to conlcude.
Any help or hint would be appreciated.
As you said, if your polynomials are of degree bounded, they live in a finite dimensional vector space, so all the norms are equivalent.
Uniform convergence on $[0,1]$ is for the norm $\|.\|_{\infty}$. $\sup_k |a_k|$ is another norm, say $\|.\|_1$. They are equivalent, thus there exists $C >0$ with $\|.\|_1 \leqslant C \|.\|_{\infty}$.
Thus, if $\|P_n\|_{\infty} \to 0$, then $\|P_n\|_1 \to 0$ too.
Be carefull : if the degrees are not bounded, it is not true !