This question arised to me when trying to prove that the space of infinitely differentiable functions defined in a compact space $K\subset\mathbb{C}$ taking values in $\mathbb{C}$, that is $C^\infty(K)$, is a Banach space. It is well known that $C^n(K)$ (the space of n-times continuously diferentiable functions) is a Banach space with the norm $$||f||:=||f||_\infty+...+||f^{n)}||_\infty$$ When I try to define an analogous norm for $C^\infty(K)$, I have the problem that I dont know how fast is allowed to grow the sequence $$(||f^{n)}||_\infty)_n$$ I've read from several sources that $C^\infty(K)$ is a Frechet space so I'm guessing that there is no restriction about the growth of that sequence. Its that true? Any help would be appreciated.
(Update) [$\approx$ Solved] I've read that the space of analytic functions $C^\omega([0,1])$ is not normable (nor $C^\infty([0,1])$ either then), so my question must have a negative answer: If it were true that $||f^{n)}||_\infty\leq M_n$ for some absolute (i.e. holding for every $f\in C^\omega([0,1])$) sequence $(M_n)_n$ of positive numbers, then we could define $$||f||_\omega:=\sum_{n=0}^\infty2^{-n}\frac{||f^{n)}||_\infty}{M_n}$$ for $f\in C^\omega([0,1])$. It is inmediate to verify that $||\cdot||_\omega$ (would) define a norm over $C^\omega([0,1])$. As this can't happen, there is not such sequence $(M_n)_n$.
[I guess that saying that $C^\omega([0,1])$ is not normable means with respect to the topology "where $f$ and $g$ are close if $f^{n)}$ and $g^{n)}$ are close for every $n\geq0$". That is, with respect to the topology $\tau$ generated by $\{B(\varepsilon,f)\}_{\varepsilon>0,f\in C^\omega([0,1])}$, where $B(\varepsilon,f)=\{g\in C^\omega([0,1]):||g^{n)}-f^{n)}||_\infty<\varepsilon\ \forall n\geq0\}$]
As proving that $C^\omega([0,1])$ is not normable with respect to $\tau$ is the only remaining question, I've changed the tittle. Again, any help would be appreciated :)
Take any exponential on $[0,1]$ for instance $f : x\rightarrow e^{2x}$.
$f^{(n)}(x)=2^nf(x)$, so $\Vert f^{(n)}\Vert=2^ne\rightarrow\infty$