Why $C^{\infty}[0,1]$ with natural topology can not be normalized. Consider $C^{\infty}[0,1]$ space, where $f_n \to f$ iff all its derivatives $f_n^{(k)}(x)$ uniformly converge to $f^{(k)}(x)$. The question is prove that this space is metrizable, bot can not be normalized.
My attempt: it's not hard to see that using $\{\sup_{[0,1]} |f^{(k)}(x) - g^{(k)}(x)|\}_{k\ge0}$ we can construct a metric $\rho$, which replicate convergence in it's natural way. So now we need to find a sequence of function $\{f_n\}$ that converges in $\rho$ sense, but there is no norm for which this sequence also converges? The first idea is to consider something easily differentiable $\{\exp(nx)\}_n$. This sequence looks reasonable, but I don't know why this sequence doesn't converge in any norm.
Actually, it looks curious, that there is no appropriate norm to normalize this space.
 A: HINT:
The space $C^{\infty}[0,1]$ cannot be normalized because it does not have any bounded  neighborhoods of $0$. Indeed, every neighborhood of $0$ contains a set of the form $$\{ f \ | \ p_k(f) \le \delta_k, k = 0, \ldots n\}$$ and on such a set every seminorm $p_{m}$ with $m>n$ is unbounded. Indeed, let $f$ a function in $C^n$ but not in $C^{n+1}$, and $f_m$ a sequence in $C^{\infty}$ converging to $f$ in the $C^n$ topology. If we had an inequality $p_{n+1}< C \max_{0\le k \le n} p_k$, then the sequence $f_m$ being  Cauchy in the $C^{n}$ topology, would also be Cauchy in the $C^{n+1}$ topology, so $f_m$ would converge to a function in $C^{n+1}$, contradiction.
$\bf{Added}$ @Jochen's idea: we can write down explicitly a function with first $n$ derivatives small, but $p_{n+1}(f)$ large, for instance $f(x)= \frac{1}{M^{2n+1}} cos M^2 x$, with $M$ large.
A: Another reason why $C^\infty$ is not normable is that it satisfies the Heine-Borel property, that is, every closed and bounded set is compact. (Note that $C^\infty$ is a metric space, as already observed in the question, so it makes sense to speak of "bounded sets").
The proof uses the theorem of Ascoli-Arzelá. Let $f_n\in C^\infty$ be a sequence such that
$$\sup_{n, k}\lVert f_n\rVert_{C^k}< \infty.$$
Then, since the sequence is $C^1$ bounded, it has a $C^0$-converging subsequence. Such subsequence is $C^2$ bounded, thus it has a $C^1$-converging subsequence. Using the diagonal sequence trick (a.k.a. diagonal argument, Cantor argument, etc...) we can produce a subsequence of $f_n$ that is $C^\infty$ convergent. This proves that $C^\infty$ is Heine-Borel.
To conclude we observe that the only normed spaces that are Heine-Borel are the finite-dimensional ones, which is a standard theorem usually attributed to Riesz. Since $C^\infty$ is infinite-dimensional, it cannot be a normable space.
