8
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ Your definition of $\ell^m_2(\mathbb{N})$ is confusing... $\endgroup$ – David Roberts Jan 27 '15 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 Jan 27 '15 at 13:59
  • $\begingroup$ That's true, I did the modification. Thanks. $\endgroup$ – Goulifet Jan 27 '15 at 14:17
  • 2
    $\begingroup$ 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 Jan 27 '15 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 Jan 28 '15 at 16:15
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.