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.

  • 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. $\endgroup$ Aug 24 '17 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/… ) $\endgroup$ Aug 25 '17 at 11:52

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].

  • $\begingroup$ Here's the science direct link: sciencedirect.com/science/article/pii/S0166864111003257 $\endgroup$ Aug 25 '17 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 $\ell_2$. $\endgroup$ Aug 25 '17 at 21:40
  • $\begingroup$ @HennoBrandsma: $\mathbb R^n$ is a counterexample to what you claim. $\endgroup$ Aug 25 '17 at 21:58
  • $\begingroup$ @IgorBelegradek everything is meant to be infinite-dimensional of course. $\endgroup$ Aug 25 '17 at 21:59
  • $\begingroup$ @HennoBrandsma: Then you are right: a topological vector space is locally compact iff finite-dimensional. $\endgroup$ Aug 25 '17 at 22:02

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.