TITLE: Are smooth functions tame?
QUESTION [14 upvotes]: I know the article of Hamilton on the inverse function theorem of Nash and Moser (with the same title) where he proves that $C^\infty(M)$ is a tame Fréchet space, when $M$ is closed or compact with boundary.
However, as far as I see no word is lost on the case of non-compact $M$. In particular, what if $M$ is an open subset of $\mathbb{R}^n$? Of course, we need to endow this with its Fréchet topology (uniform $C^m$ convergence on compact subsets).
I assume that this may not be a tame space, but I couldn't find a reference. At least it has a sequence of seminorms of increasing strength that induce the topology, by taking an exhaustion with compact sets and simultaneously letting the order of differentiation increase.

REPLY [8 votes]: Kofi: "...  what if $M$ is an open subset of $\mathbb R^n$?"
Kofi has already accepted Peter Michor's answer, but anyway I post the following to give a different argument for seeing that $C^\infty(\Omega)$ is not tame. Tameness in the sense of Hamilton implies the interpolation inequalities $$\|x\|_{i+j}\le C_{i,j+k}\|x\|_i^{\frac k{j+k}}\|x\|_{i+j+k}^{\frac j{j+k}}$$ for a taming sequence $\nu_i:x\mapsto\|x\|_i$ of seminorms. This in turn implies existence of a continuous norm on the space. Indeed, every $\nu_i$ is necessarily a norm, since otherwise there would exist a nonzero $x$ with $\|x\|_i=0$ for all $i$, and then the space would not be Hausdorff. Consequently, for $\Omega$ a (nonempty) open set in a Euclidean space, the space $C^\infty(\Omega)$ is not tame.