3

Let $\mathcal{S}$ be the Schwartz space on some Euclidean space with the family of seminorms given by $\{ \lVert \cdot \rVert_n \}_{n \in \mathbb{N}}$ which gives it the standard Frechet topology. By defintion, a set $B \subset \mathcal{S}$ is bounded iff \begin{equation} \sup_{f \in B} \lVert f \rVert_n < \infty \text{ for each } n \in \mathbb{N} \end{equation}

Now, let $\mathcal{S}'$ be the space of tempered distributions. Then, we can give it the strong dual topology following the general definition in the Wikipedia article.

It is well-known that $\mathcal{S}'$ with the strong dual topology has the Heine-Borel property. However, I have difficulty figuring out exactly what it means by "bounded" in $\mathcal{S}'$.

That is,

Could anyone explain precisely what conditions a set $A \subset \mathcal{S}'$ should satisfy to be called "bounded"?

Keith
  • 8,359

1 Answers1

1

A set $A\subset\mathscr{S}'$ is bounded, iff for all bounded set $B\subset \mathscr{S}$, $$ \sup_{T\in A}\left(\sup_{f\in B}|T(f)|\right)<\infty\ . $$ See the brief explanations in

Doubt in understanding Space $\mathscr D(\Omega)$

and for detailed proofs, see my rough set of lecture notes at https://mabdesselam.github.io/MATH7320Fall2017.html

Keith
  • 8,359
Abdelmalek Abdesselam
  • 5,278
  • Thank you. Come to think about it, closedness is quite straightforward as the strong dual topology is a sequential one, and convergence of a sequence is equivalent to that of the weak$^*$ topology. – Keith Aug 22 '24 at 17:00
  • 1
    To get an intuitive grasp of what a bounded set is the best is to use the isomorphism with sequence spaces. A set $A$ in $s'$ the space of sequences which are at most polynomially growing, is bounded iff the "envelope" $\theta_n=\sup_{x\in A}|x_n|$ grows at most polynomially. I don't remember the status of $S'$ being sequential. Perhaps Jochen can give info on that. It's a subtle notion, and I remember seeing an a old article by Dudley about certain distribution spaces not being sequential. I would discourage using sequential criteria for characterizing sets being closed. – Abdelmalek Abdesselam Aug 22 '24 at 17:14
  • Here, the strong dual topology on $\mathcal{S}'$ is the "sequentialization" of the weak$^$ topology. Plus, a sequence (not a general net) in $\mathcal{S}'$ converges in the strong dual topology iff it converges in the weak$^$ topology. – Keith Aug 22 '24 at 20:29
  • I agree that sequential criteria can be misleading and error-prone in many cases, but it seems that $\mathcal{S}'$ is nice enough to simplify most issues. – Keith Aug 22 '24 at 20:31
  • 1
    Unfortunately, the comments to the other statckexchange question do not indicate a reference where one can learn the proofs of all the sequentiality statements. I think the article by Dudley I was trying to remember is "On sequential convergence" in TAMS https://www.jstor.org/stable/1994157 which said that $S'$ is not sequential. However, the MPCPS article "Sequential convergence in locally convex spaces" by Webb ... – Abdelmalek Abdesselam Aug 22 '24 at 21:00
  • 1
    https://doi.org/10.1017/S0305004100042900 says the opposite! That's why I recommend staying away from sequential criteria, especially if all one wants to know is if some set is open or closed. – Abdelmalek Abdesselam Aug 22 '24 at 21:01
  • Hmm...I thought that there are "two different topologies" on $\mathcal{S}'$. One is weak$^$ and the other is strong dual. I am aware that the weak$^$ one is NOT sequential. Perhaps are you talking about it? – Keith Aug 22 '24 at 21:59
  • I will look into the cited papers to figure out what they refer to. Thank you! – Keith Aug 22 '24 at 22:12