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Let $S(\mathbb{N})$ be the space of rapidly decreasing sequences and $S'(\mathbb{N})$ its topological dual, the space of sequences bounded by a polynomial. For $m\in \mathbb{Z}$, we also define $\ell_2^m (\mathbb{N})$ as Hilbert spaces of sequences such that $(u_n (n+1)^m)_{n\in \mathbb{N}} \in \ell_2 (\mathbb{N})$.

It is known than $S(\mathbb{N})$ is the projective limit of the spaces $\ell_2^m (\mathbb{N})$. As such, $S(\mathbb{N})$ is a Frechet space with a nuclear topology. Its dual $S'(\mathbb{N})$ is hence the inductive limit of the spaces $(\ell_2^m)' (\mathbb{N}) = \ell_2^{-m} (\mathbb{N})$. So:

Fact 1: $S'(\mathbb{N})$ has a natural complete nuclear topology defined as a countable inductive limit of Hilbert spaces.

It is also known that any complete nuclear space is isomorphic with the projective limit of a suitable family of Hilbert spaces. See for instance Corollary 3, Section 7.2 of Topological Vector Spaces.

Fact 2: $S'(\mathbb{N})$ has an abstract complete nuclear topology as a projective limit of Hilbert spaces.

Question: Is it possible to describe the topology of $S'(\mathbb{N})$ as a countable projective limit of Hilbert spaces $H_m$, meaning that $S'(\mathbb{N}) =\bigcap_{m\in \mathbb{N}} H_m$, such that the $H_m$ are described as sequence spaces (bigger than $S'(\mathbb{N})$ of course)?

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  • $\begingroup$ Your definition of $\ell^m_2(\mathbb{N})$ is confusing... $\endgroup$
    – David Roberts ♦
    Commented Jan 27, 2015 at 13:31
  • $\begingroup$ I think that the definition should be that $u\in\ell_2^m(\mathbb{N})$ if $(u_n(n+1)^m)_{n\in\mathbb{N}}$ lies in $\ell_2(\mathbb{N})$ (not $\ell_2^m(\mathbb{N})$). $\endgroup$
    – Neil Strickland
    Commented Jan 27, 2015 at 13:59
  • $\begingroup$ That's true, I did the modification. Thanks. $\endgroup$
    – Goulifet
    Commented Jan 27, 2015 at 14:17
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    $\begingroup$ I assume that you are missing an intersection sign in the second bottom line. The answer to your question is yes, by the way---follows from the fact that your spaces are mutually $\alpha$-duals (see Köthe, Garling). $\endgroup$
    – weather
    Commented Jan 27, 2015 at 16:56
  • $\begingroup$ @weather thank you for you answer. If I understand correctly, from Köthe, we can define on $S'(\mathbb{N})$ a family of semi-norms $p_{u}(v) = \sum |u_i||v_i|$ with $u\in S(\mathbb{N})$. I interpret this saying that $S'(\mathbb{N})$ is an intersection of weighted $\ell_1$ spaces (with weights in the space $S(\mathbb{N})$). Also, can I extract a countable family from these weights? This is not clear to me. $\endgroup$
    – Goulifet
    Commented Jan 28, 2015 at 16:15

1 Answer 1

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I can now answer my own question thanks to the following discussion: Which Fréchet spaces have a dual that is a Fréchet space?

  • As explained in the comments, $\mathcal{S}'(\mathbb{N})$ can be defined as the projective limit of a family of semi-norms $(p_u)$ indexed by $u \in \mathcal{S}(\mathbb{N})$.

  • It is however impossible to extract a countable family from the $(p_u)$ due to the much more general fact that the strong dual of a Frechet space $E$ is metrizable if and only if $E$ is normable. Here, $E = \mathcal{S}(\mathbb{N})$ is not normable, so $E' = \mathcal{S}'(\mathbb{N})$ is not metrizable, hence cannot be defined as a projetive limit of a countable family of sequence spaces.

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