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Question 1 Let $X$ a separable Banach Space and $Y\subset X$ linear subspace. When can we put a norm on $Y$ in such a way so that $Y$ is a Banach space?

Clearly if $Y$ is closed in the norm topology that's sufficient.

So, I'm coming into this with the perspective of Polish spaces(i.e. completely metrizable separable spaces) where a $G_\delta$ subset of a Polish space is once again Polish and every Polish space is a $G_\delta$ subset of a Polish space.

I was wondering if there is some sort of similar condition about complete norms and how the structure of the space matters.

I know there are linear spaces that can't be Banach spaces, e.g. $c_{0,0}$ the subspace of $c_0$ where cofinitely many elements in each sequence are $0$. In this case, there is a trivial Hamel basis, namely $e_i=(0,\ldots,0, 1, 0, \ldots)$ where $1$ is the $i$th element in the sequence. But this Hamel basis is countable and we know every Hamel basis of an infinite-dimensional Banach space must be uncountable.

We also know that the existence of a complete norm is equivalent to the existence of a complete homogeneous translation-invariant metric $d$, in the sense that

  • $d$ is a complete metric
  • $d(ax,ay)=|a|d(x,y)$
  • $d(x+z,y+z)=d(x,y)$

From here, I believe we can see that $Y$ must at least have a complete metric and so must be a $G_\delta$ subset of $X$.

I haven't been able to come up with a counterexample of $Y$ a complete metric space with linear structure and $Y$ not a Banach Space, so I'm thinking that the $G_\delta$ may be a sufficient condition, so I think this can be restated as

Question 2 If $Y$ a linear space and $d$ a complete metric on $Y$. Then there exists a complete homogeneous translation-invariant metric $d'$ on $Y$.

Kind of related to the questions above, I know every vector space has a norm, just take a Hamel basis $(e_\lambda)$ and $x=\sum a_i e_i$. Now let $||x||=\max{|a_i|}$ or $||x||=\sum |a_i|$. Either of these forms a norm, but this norm isn't necessarily complete, especially in an infinite-dimensional space.

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    $\begingroup$ You haven't asked for there to be any relationship between the norm of $X$ and the new norm of $Y$, so the norm of $X$ is really irrelevant, and we might as well just say "$X$ is a vector space of dimension $\mathfrak{c}$", and ask "Under what conditions does a vector space $Y$ of dimension at most $\mathfrak{c}$ admit a complete norm?" The answer is, iff the dimension of $Y$ is either finite or $\mathfrak{c}$. But I'm not sure that's really what you meant to ask. $\endgroup$
    – Nate Eldredge
    Commented Dec 10, 2015 at 23:19
  • $\begingroup$ By c, do you mean the continuum. $\endgroup$
    – Konrad Wrobel
    Commented Dec 10, 2015 at 23:24
  • $\begingroup$ Yes, $\mathfrak{c} = 2^{\aleph_0}$ is the cardinality of the continuum. If that really is your question, then I can post an answer with details. $\endgroup$
    – Nate Eldredge
    Commented Dec 10, 2015 at 23:24
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    $\begingroup$ No, by "dimension" in my comment I mean Hamel dimension, i.e. cardinality of a Hamel basis. The Hamel dimension of $\ell^p$ is $\mathfrak{c}$; indeed every infinite-dimensional separable Banach space has Hamel dimension $\mathfrak{c}$, so as vector spaces they are all isomorphic. $\endgroup$
    – Nate Eldredge
    Commented Dec 10, 2015 at 23:26
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    $\begingroup$ For a more sensible question, you can ask that the injection map from $Y$ into $X$ with their respective norms is continuous. Thus: what is the condition on a linear subspace of a (separable) Banach space $X$ in order for it to be the range of a linear operator from some Banach space? $\endgroup$
    – Robert Israel
    Commented Dec 10, 2015 at 23:27

1 Answer 1

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The main restriction on putting a complete norm is that Hamel bases in infinite-dimensional separable Banach spaces have cardinality continuum (see Exercise 6.4 on page 191 in Fabian-Habala-Hajek-Montesinos-Pelant-Zizler, Functional Analysis and Infinite-Dimensional Geometry). This gives an answer to your first question: if and only if $Y$ has a Hamel basis of cardinality continuum. (To introduce such norm we use an arbitrary bijection between the Hamel basis of $Y$ and the Hamel bases of an arbitrary separable infinite-dimensional Banach space.)

I think that similar approach can be developed for complete metric spaces.

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  • $\begingroup$ P.S. When I started to print my answer it was not seen that other people are about to post similar answers. $\endgroup$
    – Mikhail Ostrovskii
    Commented Dec 10, 2015 at 23:31
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    $\begingroup$ P.P.S. Actually for complete metric spaces one has to be more inventive than I expected printing my answer: According to Kalton (Studia Math. 116 (1995), no. 2, 167-187) there exist quasi-Banach spaces without basic sequences. Soon this paper will be reprinted with Maurey's comments in the 1st volume of Kalton's selecta. $\endgroup$
    – Mikhail Ostrovskii
    Commented Dec 11, 2015 at 0:15
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    $\begingroup$ If you drop the separability assumption, a similar fact is true: An infinite-dimensional vector space over the reals of dimension $\kappa$ admits a complete norm if and only if $\kappa = \kappa^{\aleph_0}$. (Arthur Kruse, Badly incomplete normed linear spaces. Mathematische Zeitschrift, 1964.) $\endgroup$
    – Goldstern
    Commented Dec 11, 2015 at 1:02

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