Timeline for Ultraweak topology on B(X): Is the map X\otimes X* -> B(X)* isometric?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 29, 2015 at 19:11 | history | edited | Hannes Thiel | CC BY-SA 3.0 |
added 146 characters in body
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| Mar 6, 2015 at 16:48 | answer | added | Bill Johnson | timeline score: 6 | |
| Mar 2, 2015 at 8:53 | vote | accept | Hannes Thiel | ||
| Feb 28, 2015 at 13:33 | answer | added | Bill Johnson | timeline score: 14 | |
| Feb 26, 2015 at 17:49 | comment | added | Yemon Choi | @MatthewDaws Thanks - I was just going to email you in case you hadn't seen this question! I agree that $B(X,X^{**})$ is the more natural object to look at. | |
| Feb 26, 2015 at 8:53 | comment | added | Matthew Daws | @YemonChoi: I would look at $B(X,X^{**})$ instead of $B(X^*)$, then we're asking if the ball of $B(X,X)$ is weak$^*$-dense in the ball of $B(X,X^{**})$. The PLR shows that this is true for finite-rank operators; and the metric approximation property allows you to reduce from all operators to just the finite-rank ones. If $X$ is reflexive, then there is nothing to prove. What I don't see is how to combine these two rather different viewpoints...! | |
| Feb 26, 2015 at 8:15 | comment | added | weather | Just a wee comment on the sentence on ll. 11, 12.$B(X)$ can only be a dual space if the same is true for $X$. The standard reference for the kind of question you are asking is Grothendieck's Resumé. | |
| Feb 25, 2015 at 20:09 | comment | added | Yemon Choi | This starts to feel like it has something do with the principle of local reflexvity... | |
| Feb 25, 2015 at 20:03 | comment | added | Yemon Choi | [deleted over-hasty comment; will think more] | |
| Feb 25, 2015 at 19:46 | history | asked | Hannes Thiel | CC BY-SA 3.0 |