7
$\begingroup$

An F-space is a completely metrizable topological vector space, i.e. the vector topology is induced by a complete metric. A Fréchet space is, by definition, a locally convex F-space.

It is known that all (infinite dimensional) separable Fréchet spaces are homeomorphic to $l_2$, the space of square summable sequences of real numbers. (See e.g. Anderson, R. D.; Bing, R. H., A complete elementary proof that Hilbert space is homeomorphic to the countable infinite product of lines. Bull. Amer. Math. Soc. 74, 1968, 771–792.)

More precisely any such space, including $l_2$ is homeomorphic to a countable infinite product of copies of real lines.

Can this result be extended to (any) non-locally convex F-spaces?

I am interested in any reference or recent literature on the topology (homeomorphism class) of F-spaces.

More precisely, I am interested in the following examples: $L_p([0,1])$ or $l_p$ with $0<p<1$, which are in fact quasi-Banach spaces. Meaning, the p-sum/integral defines a quasi-norm only, that is $||f+g||_p\leq K(||f||_p+||g||_p)$, holds only for a uniform constant $K>1$. Nevertheless $d(f,g)=||f-g||_p^p$ is a complete metric inducing the vector topology.

This question arose when wondering if $l_p$ has a structure of a Hilbert manifold.

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ The Mazur mappings shows that every $\ell_p$, respectively, $L_p$ is homeomorphic to $\ell_2$, respectively, $L_2$, and hence all are homeomorphic. $\endgroup$
    – Bill Johnson
    Aug 24, 2017 at 23:49
  • $\begingroup$ Thank you for this comment. The extension to $0<p<1$ of Mazur's proof was done by A. Weston in his thesis. The following paper by him on uniform homeomorphisms of $L^p(\mu)$ spaces contains the needed inequalities. ( maths-proceedings.anu.edu.au/CMAProcVol29/… ) $\endgroup$
    – Thibaut Dumont
    Aug 25, 2017 at 11:52

1 Answer 1

Reset to default
6
$\begingroup$

There is a famous linear metric space constructed by R. Cauty [Un espace métrique linéaire qui n'est pas un rétracte absolu, Fund. Math. 146 (1994)] whose completion is a separable $F$-space which is not an AR. I do not have Cauty's paper handy but the latter fact is stated on the first page of Cauty's space enhanced in [Topology Appl. 159 (2012), no. 1, 28–33].

Since $\ell_2$ is an AR, the above $F$-space is not homeomorphic to $\ell_2$.

Incidentally, a separable $F$-space is homeomorphic to $\ell_2$ if and only if it is a non-locally-compact AR, see Corollary 5.2.2 in [Absorbing sets in Infinite-Dimensional Manifolds, T. Banakh, T. Radul, M. Zarichnyi].

Improve this answer
$\endgroup$
5
  • $\begingroup$ Here's the science direct link: sciencedirect.com/science/article/pii/S0166864111003257 $\endgroup$
    – Thibaut Dumont
    Aug 25, 2017 at 11:59
  • $\begingroup$ I think if we add the requirement that the separable, completetely metrisable $F$-space is an AR, we do have that is homeomorphic to $\ell_2$. $\endgroup$
    – Henno Brandsma
    Aug 25, 2017 at 21:40
  • $\begingroup$ @HennoBrandsma: $\mathbb R^n$ is a counterexample to what you claim. $\endgroup$
    – Igor Belegradek
    Aug 25, 2017 at 21:58
  • $\begingroup$ @IgorBelegradek everything is meant to be infinite-dimensional of course. $\endgroup$
    – Henno Brandsma
    Aug 25, 2017 at 21:59
  • $\begingroup$ @HennoBrandsma: Then you are right: a topological vector space is locally compact iff finite-dimensional. $\endgroup$
    – Igor Belegradek
    Aug 25, 2017 at 22:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.