This question is inspired by the MO question Image of $L^1$ under the Fourier transform, but I think it might be much easier so I am posting it here instead.

Let $(X, \|\cdot\|)$ be a separable Banach space, and $E$ a linear subspace.

1. Suppose $E$ is Borel. What can be the complexity of $E$, in the sense of the Borel hierarchy? (For example, as shown here, it cannot be properly $G_\delta$.) Can $E$ have arbitrarily high Borel rank?

Let us say $E$ is Banachable if there is another norm $\|\cdot\|'$ on $E$, stronger than $\|\cdot\|$, under which $E$ is separable Banach. Or equivalently, $E$ is Banachable iff there exists a separable Banach space $Y$ and a continuous injective linear map $T : Y \to X$ whose image is $E$. It is clear that every Banachable subspace is Borel.

2. Is every Borel subspace Banachable?

3. If not, what can be the complexity of a Banachable subspace?

4. Do these answers change if I replace "Banach" by "Hilbert" throughout?

One could also ask the analogous questions for Polish groups.

  • $\begingroup$ I suspect at least some of what you're asking about can be found in my answer to Does there exist a linearly independent and dense subset? and/or the references given in my answer. $\endgroup$ – Dave L. Renfro Jun 5 '14 at 14:15
  • $\begingroup$ @dave: Indeed, it seems to resolve #1 completely. I will have a look at the references you cite. Thanks! $\endgroup$ – Nate Eldredge Jun 5 '14 at 14:22
  • $\begingroup$ Do you know a reference for "every Banachable subspace is Borel"? I was trying to find one to answer this question where the result is quoted with a cryptic remark "by Kuratowski's theorem". Could not find anything connecting Kuratowski to this statement. But found this article which claims that separable Borel subspaces are precisely continuous images of separable Banach spaces, which seems to answer your question 2. No reference given, though... $\endgroup$ – user357151 Mar 13 '17 at 16:50
  • $\begingroup$ @NormalHuman: In general, whenever $X,Y$ are standard Borel spaces and $T : Y \to X$ is Borel and injective, then the image $E = T(Y)$ is Borel in $X$. This is Theorem 15.1 in Kechris Classical Descriptive Set Theory. So it holds in particular when $X,Y$ are separable Banach spaces and $T$ is continuous. $\endgroup$ – Nate Eldredge Mar 14 '17 at 3:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.