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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)?

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)?

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 an 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 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)?

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)?

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 an 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 projective limit of Hilbert spaces $H_m$, meaning that $S'(\mathbb{N}) =\bigcap_{m\in \mathbb{N}c} H_m$, such that the $H_m$ are described as sequence spaces (bigger than $S'(\mathbb{N})$ of course)?

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 an 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 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)?