What makes a norm "appropriate"? Why can't testfunctions be normed appropriately I often hear the term "using an appropriate norm". Then I once read that the $C^\infty_0(\Omega)$ cannot be appropriately normed. Why is that?
Furthermore, when doing some numerical analysis you often use (Sobolev-)spaces which have various norms, why in these cases not use just the $L^2$-norm?
What makes a norm appropriate and why are often different norms necessary? 
Edit: I asked various questions which I believe have the same answer, hope this is ok.
Edit2: For all who are interested, I want to give an answer to my own question, inspired by the comments below and some further thinking about the topic. 
To motivate the norms in Sobolev spaces: To make a Sobolev space a Banach space we need convergence of every Cauchy sequence to an element that also lies in $W^{k,p}$. We demand the elements of $W^{k,p}$ to have a weak derivative that lies in $L^p$ (actually multiple wead derivatives, depending on k). Assume we only used the $L^p$-norm to check for completeness, we would allow a sequence to converge to an element that only lies in $L^p$ but not in $W^{k,p}$. We would have a Cauchy sequence in $W^{k,p}$ that converges to an element in $L^p$ thus we would not have $W^{k,p}$ a Banach space (since it does not converge in the space). To avoid that we need a norm that only allows for sequences (to be Cauchy) that indeed converge to an element in $W^{k,p}$. 
The original $W^{k,p}$-norm allows for that because we need not only our function $f_n \rightarrow f$ in $L^p$ but also the respective weak derivatives. This is only due to the norm. 
Short: The norm tells us, what we need to look at when checking for completeness. And only an appropriate norm covers all demanded requirements for the space. So if we use the wrong norm we converge to an element that does not fulfill the requirements and therefore does not lie in the requested Sobolev space. 
So far the question, what the term appropriate means, this may have been crystal clear to the pros, to me it was not, I felt confused by the seemingly arbitrary choice of norms. 
This leads to the second question: Why cannot the test functions be normed appropriately? Exactly for that reason: whatever norm I chose I cannot make a sequence in $C^\infty_0$ converge to an element of this very space. 
How a topology does the job better is not exactly clear to me, but that is not what the initial question was about. 
Thanks to all people that helped me to shed some light on this. 
 A: You have a notion of convergence in $C_0^\infty(\Omega)$. It's "Convergence on compact sets in $C^k$ for all $k$." (Or maybe you want some uniformity; whatever.) This, with more care than I'm giving here, defines a topology on the space $C_0^\infty(\Omega)$. An "appropriate norm" would mean a norm on this space that induces that topology (that notion of convergence). Theorem: there doesn't exist such a norm. 
Your second question seems to be asking what the point of Sobolev spaces is. You would like to work in a functionally-analytically-nice space to do PDE (Hilbert or Banach), in particular so that you have a working inverse function theorem. One choice is the Holder spaces $C^{r,\alpha}$, $\alpha$ nonzero; these are nice Banach spaces that enjoy some nice regularity properties. Sobolev spaces are another choice: you may think of $W^{p,k}$ as the space of functions that are $L^p$, and whose "first $k$ derivatives are $L^p$". Then an order $\ell$ differential operator maps $W^{p,k} \to W^{p,k-\ell}$ - you've lost $\ell$ derivatives. If you just had $L^2$ functions, what's the codomain of, say, the Laplacian as an operator on $L^2(\Omega)$? Now that you have Sobolev spaces this is fine (take Sobolev index $-2$).
If you're asking "Why can't I just use the $L^2$ norm on the Sobolev spaces?" this is patently absurd: if you did so, your "Sobolev space" would not be complete; its completion would be $L^2(\Omega)$. And since the whole point was to get a nice Hilbert space of functions with some differentiability properties, we've gained nothing by not actually getting a Hilbert space.
