Let $\mathcal{O}_M(\mathbb{R}^d)$ be the space of smooth moderately increasing functions

$\{ f \in \mathcal{C}^\infty(\mathbb{R}^d) : \forall \alpha \exists N \text{ such that} \Vert \langle \cdot \rangle^{-N} \partial^\alpha f \Vert_{L^\infty} < \infty \}.$

This space is equipped with the locally convex topology generated by the seminorms $\rho_{\phi, \alpha}(f) : = \Vert \phi \partial^\alpha f \Vert_{L^\infty}$ for $\phi \in \mathcal{S}(\mathbb{R}^d)$ and $\alpha$ a multi-index. Is the space $\mathcal{O}_M(\mathbb{R}^d)$ a compactly generated space?

Various properties of this space and its relation to $\mathcal{S}(\mathbb{R}^d)$ were proved by Schwartz, and Grothendieck proved in his thesis that the space is bornological.

So far, I cannot show this space is compactly generated. Here is what I have found so far:

  1. Locally compact spaces are compactly generated, which is not useful here since any infinite dimensional Hausdorff topological vector space is not locally compact.
  2. If the space was metrizable, then it would be compactly generated. However, I do not know if the uncountable set of seminorms above can be reduced to a countable one giving the same topology.
  3. Bornological spaces can be written as a direct limit of Banach spaces, but a direct limit of compactly generated spaces is not necessarily compactly generated.
  • 2
    $\begingroup$ I doubt very much. Besides mertrizable locally convex spaces, another important class of locally convex spaces which are compactly generated conists of LS- (or DFS- or Silva-) spaces which are countable colimits (in the category of locally convex spaces) of Banach spaces with compact linking maps. That they are compactly generated (equivalently, the locally convex colimit coincides with the colimit in the category of all topological spaces) can be seen as a consequence of the Banach-Dieudonné theorem. $\endgroup$ Feb 24, 2022 at 10:42
  • 1
    $\begingroup$ However, $\mathcal O_M$ is a countable limit of such spaces and, in general topology, being compactly generated is not even stable with respect to finite products. $\endgroup$ Feb 24, 2022 at 10:42
  • $\begingroup$ Thank you for your reply Jochen! I'm tending to agree that $\mathcal{O}_M$ is not compactly generated. Do you know of any properties that show a topological space is not compactly generated? $\endgroup$
    – JMill.
    Feb 25, 2022 at 16:44

1 Answer 1


Not an answer but a simplification of the question. Let $s$ be the space of sequences $x=(x_n)_{n\ge 0}$ in $\mathbb{R}^{\mathbb{N}}$ such that for all $k\in\mathbb{N}$ $$ ||x||_k:=\sup_{n\ge 0}\ (n+1)^{\frac{k}{2}}|x_n|<\infty\ . $$ Give $s$ the locally convex topology defined by the seminorms $||\cdot||_k$, $k\ge 0$. Namely, $s$ is the just the space of sequences of rapid decay.

Let $s_{+}$ be the subset of $s$ made of sequences with nonnegative entries.

Let $s'$ be the space of sequences of temperate (at most polynomial) growth. Let $s'_{+}$ be the subset of $s'$ made of sequences of with nonnegative entries.

Let $\mathcal{M}$ be the spaces of "matrices" $a=(a_{i,j})_{(i,j)\in\mathbb{N}^2}$ such that $$ ||a||_{\omega,\omega'}:=\ \sum_{(i,j)\in\mathbb{N}^2}\omega_i \omega'_j |a_{i,j}| <\infty $$ for all $\omega\in s_{+}$ and all $\omega'\in s'_{+}$. Give $\mathcal{M}$ the locally convex topology defined by the seminorms $||\cdot||_{\omega,\omega'}$.

If I remember correctly, Grothendieck conjectured that $\mathscr{O}_{\rm M}$ is isomorphic to $\mathcal{M}$ and this was proved some thirty years later by Valdivia. Since the question is about the topological vector space structure of $\mathscr{O}_{\rm M}$ only, it might be easier to examine it for $\mathcal{M}$ instead.

  • $\begingroup$ I'm starting to catch on that you like these sequence spaces! :) $\endgroup$ Feb 24, 2022 at 21:47
  • $\begingroup$ I hope I didn't mess up the concrete calculation of the topological tensor product $s'\widehat{\otimes}s$. I wrote the post very quickly. $\endgroup$ Feb 24, 2022 at 22:57
  • $\begingroup$ Thank you for your answer Abdelmalek. I'm a bit confused how the $\omega'$ dependence occurs in the definition of the seminorm $\Vert \cdot \Vert_{\omega,\omega'}$. Is there an $\omega'$ on the right hand side? $\endgroup$
    – JMill.
    Feb 25, 2022 at 16:49
  • $\begingroup$ yes of course. it was a typo. $\endgroup$ Feb 26, 2022 at 16:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .