Timeline for Can quotient space be isomorphically isometric to some closed subspace of original one?
Current License: CC BY-SA 4.0
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| Sep 27, 2020 at 16:43 | history | edited | JohnLee | CC BY-SA 4.0 |
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| Sep 24, 2020 at 0:15 | history | edited | JohnLee | CC BY-SA 4.0 |
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| Sep 24, 2020 at 0:12 | comment | added | JohnLee | @LSpice My bad. Thank you for pointing out this. I have revised it. Hope it is clear now. | |
| Sep 24, 2020 at 0:10 | history | edited | JohnLee | CC BY-SA 4.0 |
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| Sep 24, 2020 at 0:06 | comment | added | LSpice | Per @YemonChoi's comment, I think your question should be "What conditions should I impose on $\mathcal B$ such that, for every finite-dimensional subspace $\mathcal M$, …". That is, the condition is chosen before $\mathcal M$. | |
| Sep 24, 2020 at 0:02 | history | edited | JohnLee | CC BY-SA 4.0 |
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| Sep 23, 2020 at 21:28 | comment | added | JohnLee | @YemonChoi In the reviesed one, I suppose M is any finite dimensional subspace of B instead of all of closed subspaces. The isomorphical isometry I wanna construct is between B/M and N in the question. | |
| Sep 23, 2020 at 21:14 | comment | added | Yemon Choi | @G.Rodrigues Sure, I think Jochen mentioned this as a "non-example" to illustrate some constraints | |
| Sep 23, 2020 at 21:10 | comment | added | G. Rodrigues | I am not sure I completely understand the question, but every separable Banach space is a quotient of $\ell_1$ (I know this as the Banach-Mazur theorem) but it is well known that $\ell_1$ does not contain a copy of $c_0$, say because it has the Schur property. This can all be found in a good textbook on Banach space theory. | |
| Sep 23, 2020 at 21:02 | comment | added | Yemon Choi | 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 | |
| Sep 23, 2020 at 20:53 | history | edited | JohnLee | CC BY-SA 4.0 |
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| Sep 22, 2020 at 18:45 | comment | added | M.González | $\mathcal{B}$ separable containing isometric copies of all separable spaces, like $C[0,1]$. | |
| Sep 22, 2020 at 17:19 | comment | added | Jochen Wengenroth | Perhaps, you get an idea from non-examples like $\ell^1$ which has every separable Banach space as a quotient. | |
| Sep 22, 2020 at 16:58 | comment | added | JohnLee | @YemonChoi Yes. | |
| Sep 22, 2020 at 16:57 | comment | added | Yemon Choi | Do you want conditions on $\mathcal B$ that work for all closed subspaces $\mathcal M$? | |
| Sep 22, 2020 at 16:47 | comment | added | JohnLee | @JohannesHahn Right. That is called Lindenstrauss-Tzafriri theorem. | |
| Sep 22, 2020 at 16:40 | history | edited | Johannes Hahn | CC BY-SA 4.0 |
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| Sep 22, 2020 at 16:38 | comment | added | Johannes Hahn | 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 $\mathcal{M}$. | |
| Sep 22, 2020 at 16:16 | history | asked | JohnLee | CC BY-SA 4.0 |