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Hamilton's paper "The Inverse Function theorem of Nash and Moser" (1982, Bull. Amer. Math. Soc, vol. 7, n. 1, page $137$) proves that $C^{\infty}(M)$ is a tame Fréchet space when $M$ is a compact manifold. It was asked here on MO if this space is tame in the non-compact case, and in the second answer, there was an argument that for a space to be tame, every sufficiently large taming semi-norm would be a norm (which is false in $C^{\infty}(M)$ for the non-compact case), but this is true for $\mathcal{S}(\mathbb{R}^n)$ (the space of functions that are rapidly decreasing, along with its derivatives). So to disprove tameness, a different argument would be necessary.

So the question is, is $\mathcal{S}(\mathbb{R}^n)$ a tame Fréchet space (with the usual Fréchet topology)? Is there a reference?

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Using Fourier series, $C^\infty(S^1)$ is ismorphic to the sequence space $$s=\{(x_k)_{k\in \mathbb N}\in \mathbb C^{\mathbb N}: \sum_{k=1}^\infty k^{2n} |x_k|^2 <\infty \text{ for all $n\in\mathbb N$}\}$$ and this space is isomorphic to $\mathscr S(\mathbb R^d)$. This is very classical, a modern treatment (with much information about structural properties of Fréchet spaces) is in the book Introduction to Functional Analysis of Meise and Vogt.

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    $\begingroup$ As a coda, there is a general rule of thumb: all Fréchet spaces of distribution theory are isomorphic. There are, of course, exceptions but the following holds—if $T$ is a self-adjoint operator on a Hilbert space, then the intersection of the domains of definition of $\{T^n\}$ is, in a natural way, a Fréchet space (Pietsch). If the spectrum of $T$ consists of a sequence of eigenvalues asymptotically like $(|n|^\alpha)$ for some positive $\alpha$, it is isomorphic to $s$. The classical Sturm-Liouville, Laplace and Schrödinger operators generate most spaces of test functions $\endgroup$
    – user131781
    Commented May 10, 2020 at 9:43
  • $\begingroup$ I looked into the book by Meise and Vogt but cannot find any mention of tame Frechet spaces.. Moreover, all Frechet spaces are isomorphic as well-known. So, $\mathcal{S}(\mathbb{R}^n)$ is isomorphic to $C^\infty(M)$ for a non-compact (but locally compact) manifold. However, the OP says $C^\infty(M)$ is NOT tame... $\endgroup$
    – Isaac
    Commented Mar 1 at 18:25
  • $\begingroup$ So...I still find your answer incompete (or unconvincing to me at least..) $\endgroup$
    – Isaac
    Commented Mar 1 at 18:26

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