Timeline for Self-dual normed spaces which are not Hilbert spaces
Current License: CC BY-SA 3.0
26 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 6 at 14:41 | answer | added | M.González | timeline score: 4 | |
| Mar 16, 2012 at 9:22 | answer | added | Valerio Capraro | timeline score: 11 | |
| Mar 16, 2012 at 7:36 | vote | accept | Uday | ||
| Mar 16, 2012 at 7:17 | comment | added | Denis Serre | See my edits to my answer, given one more example and suggesting infinitely many. | |
| Mar 15, 2012 at 22:32 | comment | added | Bill Johnson | That is a nice problem, Phil. A related problem is to built a space that is isomorphic to a subspace of codimension $n$ but not one of codimension $n-1$. The Kalton-Peck space may be such an example with $n=2$ but I cannot prove it. I don't recall such an example from the G-M theory but I could have forgotten... | |
| Mar 15, 2012 at 22:28 | comment | added | Philip Brooker | Sorry, in my previous comment I meant $n>2$, and $X$ not isomorphic to any previous dual besides the $0$th dual (namely, itself). | |
| Mar 15, 2012 at 22:25 | comment | added | Philip Brooker | @Bill: sorry, my previous comment was ill-thought-out, and I just raced back from having a shower to my computer to try to correct it before anyone noticed! No luck :-) . Anyway, I think I was confused in my head by a related problem that has been on my mind: does there exist a Banach space $X$ and $n>3$ such that $X$ is isomorphic to the $n$th dual of $X$, but to no previous dual? An affirmative answer implies a negative solution to the Schroeder-Bernstein problem for Banach spaces (which Gowers has of course already achieved in general). Such an $X$ would of course be non-reflexive. | |
| Mar 15, 2012 at 22:05 | comment | added | Bill Johnson | @Uday: CW is, IMO, not appropriate for an edited version of this question. | |
| Mar 15, 2012 at 22:04 | comment | added | Bill Johnson | Why can it not be reflexive and HI, Phil? Maybe this is elementary, but I do not see a proof. | |
| Mar 15, 2012 at 21:55 | comment | added | Philip Brooker | It is elementary to show that an indecomposable space that is isomorphic to its dual would necessarily be quasi-reflexive (but not reflexive) and cannot be hereditarily indecomposable. Quasi-reflexive hereditarily indecomposable spaces are known to exist (see the book of Argyros and Todorcevic, Ramsey methods in analysis), but I do not know if anyone has tried and succeeded to construct a quasi-reflexive space that is indecomposable but not hereditarily indecomposable. | |
| Mar 15, 2012 at 21:53 | comment | added | Uday | @Ralph Since the question in your mind is floating around in a half-baked manner. I would prefer(if it is okay for you), you edit the question and mark as CW. | |
| Mar 15, 2012 at 21:40 | comment | added | Ralph | @Yemon: Forgive me, I'm just lazy (i.e. if Uday hadn't posted the question I would had done someday in the future. So I saw the opportunity to get "my question" answered along the way). | |
| Mar 15, 2012 at 21:29 | history | edited | Uday | CC BY-SA 3.0 |
added 7 characters in body
|
| Mar 15, 2012 at 21:24 | comment | added | Bill Johnson | "Indecomposable" means "not isomorphic to the direct sum of two infinite dimensional subspaces". @Ralph: IMO it is reasonable to conjecture that there is no indecomposable space that is even just isomorphic to its dual. | |
| Mar 15, 2012 at 21:19 | comment | added | Ralph | @Andreas: Yes, because the dual of a normed space is complete (at least over $\mathbb{R}$ or $\mathbb{C}$) and hence a self-dual space a implicitely complete. | |
| Mar 15, 2012 at 21:14 | comment | added | Yemon Choi | Voting to close is not a statement "no version of this question is interesting". It is, IMHO, a statement "in its present form we do not want to leave the question open for answers" | |
| Mar 15, 2012 at 21:13 | comment | added | Yemon Choi | Ralph, the question you suggest is indeed research level, but it is somewhat imprecisely defined, and it isn't the question the OP asked. What counts as a "natural" Banach space anyway? Would you say Tsirelson is unnatural? Gowers-Maurey? | |
| Mar 15, 2012 at 21:10 | comment | added | Andreas Blass | @Ralph: Your comment didn't assume completeness of the space; was that intentional? | |
| Mar 15, 2012 at 21:05 | comment | added | Qiaochu Yuan | @Ralph: isn't every Hilbert space of dimension at least $2$ decomposable? | |
| Mar 15, 2012 at 21:01 | answer | added | Denis Serre | timeline score: 24 | |
| Mar 15, 2012 at 20:33 | comment | added | Ralph | @Bill: That's interesting. So "indecomposable normed space that is isometric to its dual $\Rightarrow$ Hilbert space" would be a reasonable conjecture ? | |
| Mar 15, 2012 at 20:14 | comment | added | Bill Johnson | Maybe there exists a indecomposable Banach space that is isometric to its dual. I am pretty sure that it is open whether such a space exists. | |
| Mar 15, 2012 at 19:56 | comment | added | Uday | @David I have got an example. Thanks. @Ralph Are $X \oplus X^{*}$ related to symplectic spaces? | |
| Mar 15, 2012 at 19:49 | comment | added | Ralph | According to Davide's comment: It would in fact be very interesting to have "natural" example for such a space, i.e. one that is not constructed as $X \oplus X^\ast$ (or in a similar way) with a reflexive space $X$. I have googling for such spaces some time ago and couldn't find any. To me, the question (with this additional condition) is reasearch level and shouldn't b closed. | |
| Mar 15, 2012 at 19:37 | comment | added | Davide Giraudo | Look at t.b.'s answer here: math.stackexchange.com/questions/65609/…. | |
| Mar 15, 2012 at 19:33 | history | asked | Uday | CC BY-SA 3.0 |