Uniform (Sobolev-esque) norm involving all derivatives of smooth function I'm trying to construct a norm on the space $\mathcal{D}(\Omega) := \{ f \in C^\infty(\Omega) | f $ is compactly supported on $ \Omega \}$ where $\Omega$ is an open subset of $\mathbb{R}$. I want this norm to include, somehow, the $L^\infty$-norms of all the derivatives of the smooth function to which it is applied. Specifically, I want to be able to encapsulate the statement
"$f_n$$^{(m)} \rightarrow f^{(m)} ($as $n \rightarrow \infty)$ uniformly for all non-negative values of $m$."
as the statement
"$\|f_n - f\| \rightarrow 0$ $(n \rightarrow \infty)$.",
where $\|\cdot\|$ denotes my desired norm.
So far I've considered trying to write it as something along the lines of
\begin{align}\|f\| := \sum\limits_{m \in \mathbb{N}} \frac{1}{m!} \|f^{(m)}\|_\infty, \end{align}
where $\|\cdot\|_\infty$ denotes the usual $L^\infty$-norm. I am very unsure of the validity of this sort of "definition", as I can't see how to prove that this (or a related/similar) series converges. Any help on this would be greatly appreciated! Thanks in advance.
 A: I'm not an expert in functional analysis, but I think this answers your question.
The topology of $\mathcal{D}(\Omega)$ is not induced by a norm. There is a discussion of this in chapter 1 of Rudin's Functional Analysis book. Briefly, a topological vector space whose topology is induced by a norm is locally bounded (proof: the open unit ball is bounded). So we want to show that $\mathcal{D}(\Omega)$ is not locally bounded. There is a result stating that a locally bounded topological vector space satisfying the Heine-Borel property (every closed and bounded set is compact) is finite-dimensional. So it only remains to show the Heine-Borel property for either $\mathcal{D}(\Omega)$ or $C^\infty(\Omega)$.
If you like, I can provide more details from Rudin's book.
A: If $\Omega$ is open then there is no norm on $C^\infty(\Omega)$, it is a classical Montel space.  
In addition, for compact $K\subset\Omega$ we may put $\sup_{x\in K}|f^{(k)}(x)|$ is finite, but $\sup_K\sup_{x\in K}|f^{(k)}(x)|$ may be infinite.
