2
$\begingroup$

Suppose $\mathcal{B}$ is a Banach space which is not isomorphic to a Hilbert space. What condition should I pose on $\mathcal{B}$ such that for every closed subspace $\mathcal{M}$, the quotient space $\mathcal{B}/\mathcal{M}$ is isomorphically isometric to some closed subspace of $\mathcal{B}$?

Edit---------------

The comment of M.González looks really good to me. I am considering a revised problem:

Suppose $\mathcal{B}$ is a Banach space which is not isomorphic to a Hilbert space. What condition should I pose on $\mathcal{B}$ such that for every finite dimensional subspace $\mathcal{M}$, the quotient space $\mathcal{B}/\mathcal{M}$ is isomorphically isometric to some closed subspace of $\mathcal{B}$?

Thanks a lot!

Improve this question
$\endgroup$
12
  • 3
    $\begingroup$ This is not quite what you've asked, but just for completeness's sake: If $\mathcal{B}$ has the property that any arbitrary closed subspace $\mathcal{M}$ is complemented, then $\mathcal{B}$ is already isomorphic to a Hilbert space. In other words, you're asking for situations in which $\mathcal{B}/\mathcal{M}$ is isometric to a subspace but that subspace isn't (in general) a complement of $\mathcal{M}$. $\endgroup$
    – Johannes Hahn
    Commented Sep 22, 2020 at 16:38
  • 2
    $\begingroup$ @JohannesHahn Right. That is called Lindenstrauss-Tzafriri theorem. $\endgroup$
    – JohnLee
    Commented Sep 22, 2020 at 16:47
  • 1
    $\begingroup$ Perhaps, you get an idea from non-examples like $\ell^1$ which has every separable Banach space as a quotient. $\endgroup$
    – Jochen Wengenroth
    Commented Sep 22, 2020 at 17:19
  • 8
    $\begingroup$ $\mathcal{B}$ separable containing isometric copies of all separable spaces, like $C[0,1]$. $\endgroup$
    – M.González
    Commented Sep 22, 2020 at 18:45
  • 1
    $\begingroup$ There is some ambiguity in the phrasing of your revised question. When you are asking for $B/M$ to be isometrically isomorphic to some closed subspace of $B$, are you in fact asking for an isometry $B/M \to B$ which is a right inverse to the quotient map $B\to B/M$? That is a much stronger condition, because it implies all closed subspaces of B are complemented $\endgroup$
    – Yemon Choi
    Commented Sep 23, 2020 at 21:02

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .