TITLE: A strange norm in the polynomial space $\mathbb{P}[0,1]$. A contradiction?
QUESTION [6 upvotes]: Let us define a norm $\|f\|=\sup_{k\geq 0}\sup_{t\in [0,1]}|f^{(k)}(t)|$. It is easy to see that in the polynomial space $\mathbb{P}[0,1]$ space, the norm $\|\|$ is well-defined. 
What is the dimension of the completion of $(\mathbb{P}[0,1],\|\|)$ under this norm ? 
It seems to be finite dimensional, because one can show that the unit ball in this new Banach space $X$ is compact as follows: 
We only need to show that every bounded sequence $\{f_n\}$ in $X$ has a convergent subsequence. That is a repeated application of the Ascoli-Arzelà theorem and Cantor's diagonal argument; the boundedness of $\{f^{(k+1)}_n\}$ implies the equicontinuity of $\{ f^{(k)}_n\}$. For details of this argument, we can refer to 
Is there no norm in $C^\infty ([a,b])$? 
Since it is a normed space with Montel-Heine-Borel property, it has to be finite dimensional.  
On the other hand,  $\mathbb{P}[0,1]$ is already infinite dimensional. Then $X$ has to be at least infinite dimensional.
Something must be wrong. 
references: 
Every normed space has a completion?

REPLY [3 votes]: I think the problem is that the unit ball in $X$ is not compact. The Ascoli-Arzelà theorem and Cantor's diagonal argument are not strong enough to yields the convergent subsequence in $X$. Indeed, let us consider $f_n=\frac{x^n}{n!}$. On can show that $\|f_n\|=1$, and $\|f_n-f_m\|=1$ if $n\neq m$.  Therefore, it does not have any convergent subsequence. However, one can find that for any $k$,  it holds that 
$\|f_n\|_{C^k[0,1]}\to 0$ as $n\to\infty$.