4
$\begingroup$

Let $X$ be a Hausdorff locally convex vector space. Recall (my reference is the book of H. Jarchow, Locally Convex Spaces. B.G. Teubner, 1981) that we say that $X$ is a semi-Montel space if every bounded subset of $X$ is relatively compact (equivalently, every closed and bounded subset of $X$ is compact), and a Montel space if it is semi-Montel and satisfy one (hence all) of the following conditions (equivalent under the semi-Montel hypothesis, see Proposition 11.5.1, pp. 230 of Jarchow's book):

  • $X$ is reflexive;
  • $X$ is barrelled;
  • $X$ is quasi-barrelled.

It is known that the strong dual of a Montel space is also Montel (Jarchow, Proposition 11.5.4, pp. 230-231). In the proof of Theorem 4.11 (5), pp. 39-40 of the book of A. Kriegl and P.W. Michor, The Convenient Setting of Global Analysis (AMS, 1997), it is shown that if $X$ is the strong dual of a Fréchet-Montel space (hence $X$ is a Montel space), then $X$ is a compactly generated topological space (also called a k-space or a Kelley space), i.e. the topology of $X$ is the final topology with respect to the inclusions of compact subsets of $X$. However, it seems to me that the proof of this assertion uses only the fact that $X$ is Montel.

Question: Are Montel spaces compactly generated, or is there a counter-example to this claim?

$\endgroup$
  • 1
    $\begingroup$ Since we are looking at vector spaces here, each question that about a topology has slight variations asking about vector space topologies (i.e. topologies that turn the vector space into a topological vector space), LCTVS topologies etc. So let me ask: What can we say about the TVC or LCTVS topology generated by inclusions of compact sets? $\endgroup$ – Johannes Hahn Oct 15 '14 at 0:26
  • $\begingroup$ @JohannesHahn Well, it seems to me that these inclusions must be extended to the inclusions of the vector subspaces generated by each compact subset, otherwise we cannot guarantee that the final topology will be linear. Since we also want a locally convex topology, it suffices to consider absolutely convex compact subsets (for $X$ semi-Montel, these are the bipolars of bounded subsets of $X$). The picture that seems to emerge is that, for $X$ semi-Montel, the LCTVS topology generated by the inclusions of compact subsets is the bornologification of $X$. $\endgroup$ – Pedro Lauridsen Ribeiro Oct 15 '14 at 2:04
  • $\begingroup$ In view of that, if $X$ is bornological (hence quasi-barrelled) and semi-Montel (hence Montel), I'm willing to bet that $X$ is compactly generated. That would actually be enough for the purposes I have in mind... $\endgroup$ – Pedro Lauridsen Ribeiro Oct 15 '14 at 2:08
  • 1
    $\begingroup$ Kriegl-Michor really mean by $kX$ the finest topology (not necessarily locally convex) making all inclusions of compact subsets continuous. The proof uses the the Banach-Dieudonne theorem for which metrizability is quite essential. $\endgroup$ – Jochen Wengenroth Oct 15 '14 at 11:16
  • 2
    $\begingroup$ A counterexample to the Banach-Dieudonne theorem for non-metrizable spaces was first given by Komura [link.springer.com/article/10.1007%2FBF01361183] $\endgroup$ – Jochen Wengenroth Oct 16 '14 at 10:12
4
$\begingroup$

Komura's example mentioned in the comment is just a big product $\mathbb R^{\mathbb R}$ which is a Montel space and its (strong) dual $X$ is thus also Montel. As Komura showed the finest topology $\tau^f$ which agrees on all compact (=equi-continuous) sets with the weak* (and hence with the strong) topology is not a vector space topology. In particular, it is different from the strong topology. This should show that $X$ is not compactly generated.


The story is quite different if you consider the finest locall convex topology such that all inclusions $K\hookrightarrow X$ ($K$ compact) are continuous. This is indeed the associated bornological locally convex topology.

$\endgroup$
  • $\begingroup$ That is an interesting counter-example. In this case, $X$ has the Mackey topology and its strong dual $\mathbb{R}^{\mathbb{R}}$ coincides as a vector space with its algebraic dual, hence $X$ is even bornological. Moreover, if one assumes that the continuum hypothesis is true, even $\mathbb{R}^{\mathbb{R}}$ itself is (ultra)bornological. $\endgroup$ – Pedro Lauridsen Ribeiro Oct 16 '14 at 21:17
  • 1
    $\begingroup$ I don't have any references at hand, but I think that products of "moderate cardinality" are bornological, in particular, I believe that $\mathbb R^{\mathbb R}$ is bornological even without assuming the continuum hypothesis. $\endgroup$ – Jochen Wengenroth Oct 17 '14 at 10:10
  • $\begingroup$ I'm actually thinking of the Mackey-Ulam theorem; moreover, Ulam originally showed that $\mathbb R$ does not admit a Ulam measure using the continuum hypothesis. Are there stronger recent results in this respect? $\endgroup$ – Pedro Lauridsen Ribeiro Oct 17 '14 at 15:27
  • 1
    $\begingroup$ The Mackey-Ulam theorem is discussed in Bonet's and Perez-Carreras' book "barrelled locally convex spaces". They write that if a product $\mathbb R^I$ fails to be bornological then the cardinality $d$ of $I$ is "strongly inaccessible". In particular, $a<d$ $\Rightarrow$ $2^a < d$ which is certainly wrong for $c=card(\mathbb R) = 2^{\aleph_0}$. $\endgroup$ – Jochen Wengenroth Oct 18 '14 at 10:29
  • $\begingroup$ Strongly inaccessible cardinals are somewhat beyond my current mathematical knowledge, but of course your explanation makes clear why the continuum hypothesis is not needed regardless of that. Glad I learned something new... Thanks! $\endgroup$ – Pedro Lauridsen Ribeiro Oct 20 '14 at 1:45
5
$\begingroup$

An(other) example of a Montel space which is not compactly generated is $\kern.4mm\mathscr D\kern.4mm(\kern.4mm\mathbb R\kern.4mm)$ . This follows from Theorem 6.1.4(iii) and Proposition 6.2.8(ii) on pages 190 and 195 in

A. Frölicher and A. Kriegl: Linear Spaces and Differentiation Theory, Wiley, Chichester 1988.

$\endgroup$
  • $\begingroup$ Hmm... Come to think of it, this actually could be inferred indirectly from the discussion in the book of Kriegl-Michor (cited in the question) as well: Theorem 4.11 (3), pp. 39-40 states that the Kelley topology of a strict inductive limit of a sequence of Fréchet spaces coincides with its $c^\infty$ topology (i.e. the final topology induced by all smooth curves). If these Fréchet spaces are finite-dimensional and the sequence is strictly increasing, then Proposition 4.26 (iii), pp. 45 entails that the $c^\infty$ (and hence the Kelley) topology is not a vector space topology. $\endgroup$ – Pedro Lauridsen Ribeiro Oct 20 '14 at 1:53
  • $\begingroup$ I do not have the book of Frölicher and Kriegl at hand, but I guess these results are the ones you quoted, right? $\endgroup$ – Pedro Lauridsen Ribeiro Oct 20 '14 at 1:54
  • $\begingroup$ @Pedro Lauridsen Ribeiro: Otherwise YES, but instead Proposition 4.26 (ii), and the spaces should be infinite-dimensional. N.B. Many results in Kriegl and Michor's book are taken almost verbatim from the book of Frölicher anf Kriegl. $\endgroup$ – TaQ Oct 20 '14 at 19:54
  • $\begingroup$ Oops, that was a typo (which unfortunately I can no longer edit out), sorry... Thanks! $\endgroup$ – Pedro Lauridsen Ribeiro Oct 21 '14 at 1:45

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.