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I'm looking for a reference to cite regarding the property presented in the title: "Closed and bounded sets of a nuclear Fréchet space are compact"

Thank you in advance for the help!

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Have a look at Proposition 50.2 in

F. Treves: Topological Vectors Spaces, Distributions and Kernels, Academic Press 1995 or Dover 2006

Statement (50.12) in that proposition is precisely what you need.

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does Wikipedia qualify as a reference? for a historical overview, see A pedagogical history of compactness

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    $\begingroup$ Wikipedia is the place where I actually stumbled across the result... the problem is that I doubt wikipedia qualifies as an academic source in an article. Anyway, I prefer not. Thanks anyway! $\endgroup$ Sep 20, 2012 at 13:03
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Maybe not in a single theorem, but you can go for Cor1 in Section 33 and Cor3 in Section 50 in Treves book.

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    $\begingroup$ Thanks for your answer, but I guess Liviu narrowed down a more precise reference, I guess! $\endgroup$ Sep 20, 2012 at 13:04
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    $\begingroup$ and he was a couple of second faster :) $\endgroup$ Sep 20, 2012 at 13:05
  • $\begingroup$ By accident, I had Treves' book right by my side. $\endgroup$ Sep 20, 2012 at 13:12
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Additional references:

[Sch99, §III.7.2, Corollary 2]: every bounded subset of a nuclear space is precompact.

[Pie, §4.4.7]: In each nuclear or dual nuclear locally convex space $E$ all bounded subsets are precompact.

References.

[Pie72]: Albrecht Pietsch, Nuclear Locally Convex Spaces (1972), Ergebnisse der Mathematik und ihrer Grenzgebiete, Springer, Berlin.

[Sch99]: H.H. Schaefer, M.P. Wolff (translator), Topological Vector Spaces, Second Edition (1999), Springer Graduate Texts in Mathematics, Springer, New York.

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Since the garden variety Fréchet nuclear spaces (the Schwartz space, the space of smooth functions on the torus, the space of smooth fuctions with support in a given compact set, etc., Edit: see also Jochen's comment below) are isomorphic as topological vector spaces to $\mathscr{s}$, the space of rapidly decaying sequences, and since the proof of the Montel property for $\mathscr{s}$ is very easy, I might as well explain it here in a self-contained answer.

I will use the convention $\mathbb{N}=\{0,1,2,\ldots\}$ and denote by $\mathscr{s}$ the space of sequences $x=(x_n)_{n\in\mathbb{N}}$ such that $$ ||x||_{\infty,k}:=\sup\limits_{n\ge 0}\ \langle n\rangle^k|x_n|<\infty $$ for all $k\in\mathbb{N}$. Here $\langle n\rangle:=\sqrt{1+n^2}$. The topology is the one given by the collection of seminorms $||\cdot||_{\infty,k}$, $k\ge 0$. An equivalent system of seminorms defining the same locally convex topology is $$ ||x||_{1,k}:=\sum_{n\ge 0}\langle n\rangle^k|x_n|\ . $$ I will denote by $\mathscr{s}_{+}$ the subset of $\mathscr{s}$ made of sequences with only nonnegative entries. Given a subset $A\subset \mathscr{s}$, I will define the envelope of $A$, or ${\rm env}(A)$, as the sequence $z=(z_n)$ given by $$ z_n:=\sup_{x\in A}|x_n|\ . $$ This sequence a priori belongs to $[0,\infty]^{\mathbb{N}}$. Using the $||\cdot||_{\infty,k}$ seminorms, the following is trivial.

Proposition: $A$ is a bounded subset of $\mathscr{s}$ if and only if ${\rm env}(A)\in\mathscr{s}_{+}$.

Conversely, given $\omega\in\mathscr{s}_{+}$, let me define the subset ${\rm box}(\omega)$ made of all sequences $x$ such that, for all $n$, $|x_n|\le \omega_n$.

The wanted Montel property immediately follows from the following result.

Proposition: For any $\omega\in\mathscr{s}_{+}$, we have that ${\rm box}(\omega)$ is a compact subset of $\mathscr{s}$.

Proof: (in the real case, the complex case only needs disks instead of intervals) Let $K:=\prod_{n\ge 0}[-\omega_n,\omega_n]$ with the (metrizable) product topology. Of course, by the countable Tychonov Theorem, $K$ is compact. Let $\tau:K\rightarrow \mathscr{s}$ be the obvious tautological inclusion map. Then the compactness of ${\rm box}(\omega)=\tau(K)$ would follow from the continuity of the map $\tau$. Since the spaces are metrizable, it is enough to show sequential continuity. Now switch to the $||\cdot||_{1,k}$ seminorms for $\mathscr{s}$. The needed sequential continuity is immediate from the discrete Dominated Convergence Theorem (DCT). Indeed, let $(x^{(m)})_{m\ge 0}$ be a sequence in $K$ converging to some element $x\in K$. This means that for all $n$, one has the pointwise convergence $\lim_{m\rightarrow\infty}x_{n}^{(m)}=x_n$. Once mapped inside $\mathscr{s}$, we have, for any $k\ge 0$, $$ ||\tau(x^{(m)})-\tau(x)||_{1,k}=\sum_{n\ge 0}\langle n\rangle^k|x_{n}^{(m)}-x_n| \longrightarrow 0 $$ by the DCT and the use of the dominating function $n\mapsto 2\langle n\rangle^k \omega_n$.

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    $\begingroup$ The Komura-Komura theorem implies that every nuclear Fréchet space is isomorphic to a closed subspace of $s^{\mathbb N}$. Thus knowing that $s$ is Montel implies that every nuclear Fréchet space is Montel -- another example of mathematical overkill. $\endgroup$ Jul 31, 2020 at 15:01
  • $\begingroup$ I suspected there was a result like Komura-Komura, but I didn't want to venture beyond what I knew. I found I think a good reference to this result which solved a conjecture by Grothendieck, Thm 3.18 in www2.math.uni-wuppertal.de/~vogt/vorlesungen/fs.pdf BTW, what is the mathematical overkill you are referring to? I didn't quite get that. $\endgroup$ Jul 31, 2020 at 17:04
  • $\begingroup$ Komura-Komura is much deeper than the fact that nuclear Frechet spaces are Montel. $\endgroup$ Jul 31, 2020 at 19:10

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