The Banach-Dieudonné theorem states that if $X$ is a metrizable locally convex Hausdorff space then the equicontinuous weak-* topology $ew^*$ on $X'$ coincides with the topology of precompact convergence and is therefore a locally convex topology. ($ew^*$ is the final topology on $X'$ coinduced by the inclusions of the equicontinuous sets when equipped with the weak-* topology $w^*$. Note that $ew^*$ is a priori not the locally convex final topology of these inclusions!) If $X$ is complete and thus Fréchet then it also coincides with the topology of compact convergence.

Is this also true in the case that $X$ is a strict inductive limit of a sequence of Banach or Fréchet spaces? In other words, is the equicontinuous weak-* topology on the dual of an LB- or LF-space locally convex or at least linear?

EDIT: I think to have found a counterexample which I try to sketch. Consider the LF-space $\mathcal{D} := C^\infty_c(\mathbb{R})$ of test functions with the locally convex inductive limit topology and its dual $\mathcal{D}'$ - the space of distributions.

  1. Since $\mathcal{D}$ is Montel, the strong dual $\mathcal{D}'_\beta$ is Montel and thus a sequence in $\mathcal{D}'$ is strongly convergent iff it is weakly*-convergent.
  2. It is not hard to see that if $X$ is the strict inductive limit of separable Fréchet spaces then $ew^*$ is a sequential topology and has the same convergent sequences as $w^*$. Thus, $ew^*$ is the sequential coreflection of $w^*$. (The separability of the Fréchet spaces induce separability of $X$ which in turn is used for the (equicontinuous) polars of a neighborhood base of $0$ in $X$ to be metrizable and thus sequential. I can sketch a more detailed proof.) The space $X = \mathcal{D}$ satisfies these assumptions.
  3. Dudley, "Convergence of Sequences of Distributions" (1971) has shown that $\mathcal{D}'_\beta$ is not sequential and that the topology of all sequentially (strongly) open sets (which by 1. and 2. coincides with $ew^*$) is not a vector topology (addition is merely jointly sequentially continuous).

For my applications it is rather of interest, whether for the LB-space $X = C_c(\mathbb{R})$ the $ew^*$-topology on its dual $X'$ (the space of real Radon measures) is a vector topology. We can't use the above proof since point 1. is not satisfied for $X$, i.e. a weakly* convergent sequence in $X'$ needs not be strongly convergent. (Dudley has stated in his paper that $X'_\beta$ is not sequential, but I can't use this fact to check the linearity of the $ew^*$-topology.)

  • $\begingroup$ Would you please remind us of the definition of the equicontinuous weak$^*$ topology? Is it the finest topology on $X'$ which coincides with the weak$^*$ topology on all equicontinuous sets? A corollary of (or at least something closely related to) the Banach-Dieudonne theorem is stated in Köthe's book, page 273: Every precompact set in a metrizable locally convex space is contained in the absolutely convex hull of a sequence converging to $0$. This property also holds in strict LF-spaces $X=\lim X_n$ because every precompact set in $X$ is contained and precompact in some $X_n$. $\endgroup$ May 4 '16 at 8:59
  • $\begingroup$ @JochenWengenroth Yes, your definition of the equicontinuous weak* topology is correct. I somehow doubt that such an extension to LF-spaces is possible. I try to sketch a counterexample in the post above (which uses the Montel property). For me, it still remains open whether we can have such an extension for LB-spaces (without the Montel property). $\endgroup$
    – yada
    May 7 '16 at 5:37
  • $\begingroup$ @yada, did you ever end up figuring this out for the case where $X = C_c (\mathbb{R})$? $\endgroup$ Oct 6 at 22:12
  • 1
    $\begingroup$ @pseudocydonia Yes, I have two independent proofs. One of them is based on the notion of quasi-reflexiveness. Just briefly: for infinite compact $K$ the space $C(K)$ is not quasi-reflexive, whence $C[0,1]$ is not quasi-reflexive. It follows that the direct sum $D := \oplus_{n \in \mathbb{N}} C[0,1]$ is not $B$-complete. Identify $D$ with a closed subspace of $C_c(\mathbb{R})$. Since $B$-completeness is hereditary to closed subspaces, it follows that $C_c(\mathbb{R})$ is not $B$-complete as well. Whence, $ew^*$ on $C_c(\mathbb{R})'$ is not locally convex. $\endgroup$
    – yada
    Oct 17 at 10:39

An interesting paper on spaces for which the $ew^*$ topology coincides with the topology of precompact convergence is S-spaces and the open mapping theorem by Taqdir Husain. He calls such spaces S-spaces, a term that doesn't appear to have become standard.

Proposition 2 of that paper shows that a complete locally convex space is an S-space if and only if the $ew^*$ topology is locally convex. Corollary 2.1 that precedes it states that a complete S-space is always B-complete (i.e. fully complete in the sense of Ptak). Section 9 is on inductive limits, and notes that Grothendieck had already given an example of an LF space that is not B-complete. Unfortunately the sufficient condition he gives for strict inductive limits to be S-spaces can never apply to limits of infinite-dimensional Banach spaces.

  • $\begingroup$ I had already a look on this paper of Husain (and also on his corresponding book "The Open Mapping and Closed Graph Theorems in Topological Vector Spaces") before writing this question on MO. As you mention, the results there do not help for proving local convexity of $ew^*$ on the dual of strict LB-spaces. However, I hope to have shown that $ew^*$ on $C_c(\mathbb{R})'$ is not locally convex and not even linear. $\endgroup$
    – yada
    Nov 25 '16 at 9:32
  • $\begingroup$ Does his book contain any more progress on the conjecture that being a complete S-space is equivalent to hypercompleteness? $\endgroup$ Nov 26 '16 at 0:34
  • $\begingroup$ No, complete $S$-spaces are hypercomplete but it is not known whether the converse holds (1976). I think this and the equivalence to $B$-completeness it is still an open problem. $\endgroup$
    – yada
    Nov 26 '16 at 17:26
  • $\begingroup$ If $X$ is complete and $ew^*$ is locally convex then $X$ is strictly hypercomplete (= a Krein-Smulian space = every $ew^*$ closed convex set is $w^*$-closed), [Wilansky, "Modern Methods in Topological Vector Spaces", Theorem 12-3-11]. Equivalently, a complete $S$-space is strictly hypercomplete. A hypercomplete space need not be strictly hypercomplete [Wilansky, Theorem 12-4-19 and Problems 12-4-203, 12-4-204]. So it remains open whether (i) hypercomplete is equivalent to $B$-complete and (ii) complete $S$-space is equivalent to strictly hypercomplete. $\endgroup$
    – yada
    Dec 30 '16 at 18:53

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.