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.



Are smooth functions tame?
The idea behind a tame Frechet space is that you have smoothing operators with controlled estimates by which you can counteract loss of derivatives in a nonlinear operator so that a Banach fixed point argument works. You look at a formal version of this. Of course you can have smoothing operators on noncompact manifolds, but the best use for them is if you describe your space as a projective limit (intersection) of Sobolev spaces. See 


* [MR2343536](http://www.ams.org/mathscinet-getitem?mr=2343536); Eichhorn, Jürgen Global analysis on open manifolds. Nova Science Publishers, Inc., New York, 2007. x+644 pp.


for a comprehensive description of situations where you can use the Hamilton-Nash-Moser technique. 


Edit: No, $C^\infty(M)$ with the compact $C^\infty$ topology is not tame. 
Here by tame I mean being a direct summand in a space of rapidly 
falling sequences in a Banach space.


The idea is as follows:
Just take $M=\mathbb R$ with seminorms $\|f\|\_{n,k}=\sup\_{|x|\le n} \max\_{\alpha\le k}|f^{(\alpha)}(x)|$. If it were tame you could find constants $C\_{n,k}$ such that 
$C\_{n,k}\|f\|\_{n,k}$ is rapidly decreasing (even bounded is enough -- then you can tighten the constants). But you always have functions $f$ with $f^{(k)}(n)$ going off faster than the constants you chose. 


$f\mapsto e^f$ is a counter-example to any kind of inverse function theorem on $C^\infty(\mathbb R)$.


