Timeline for Is every ideal part of an operator ideal?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 14, 2015 at 22:04 | vote | accept | Chris Ramsey | ||
| Oct 14, 2015 at 22:03 | history | edited | Chris Ramsey | CC BY-SA 3.0 |
Changed compact to finite-rank.
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| Oct 14, 2015 at 21:31 | comment | added | Philip Brooker | Regarding the addition of condition 3., the closure of the finite rank operators is not the compact operators for all Banach spaces; a uniform limit of finite rank operators is usually called an approximable operator. | |
| Oct 14, 2015 at 20:35 | answer | added | Philip Brooker | timeline score: 5 | |
| Oct 14, 2015 at 20:22 | history | edited | Chris Ramsey | CC BY-SA 3.0 |
Further refinements
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| Oct 14, 2015 at 20:05 | comment | added | Philip Brooker | An operator ideal is usually assumed in the literature to be non-trivial in the sense that it contains all the (continuous) finite rank operators (it is probably enough to have a single rank 1 operator and then obtain all finite rank operators using the other axioms). But you also want the second condition of your definition to be $\mathfrak{L} (\mathfrak{W} ,\mathfrak{X}) \mathfrak{I} (\mathfrak{X} ,\mathfrak{Y}) \mathfrak{L}(\mathfrak{Y},\mathfrak{Z}) \subseteq \mathfrak{I} (\mathfrak{W} ,\mathfrak{Z}) $. | |
| Oct 14, 2015 at 19:28 | history | edited | Chris Ramsey | CC BY-SA 3.0 |
Added an edited version of the question.
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| Oct 14, 2015 at 19:18 | comment | added | Chris Ramsey | @ChrisHeunen Yes, it seems you are correct! | |
| Oct 14, 2015 at 18:41 | comment | added | Chris Heunen | Can't you just take $\mathfrak{I}(\mathfrak{X},\mathfrak{X})=J$ and $\mathfrak{I}(\mathfrak{Y},\mathfrak{Z})=\emptyset$ for all $\mathfrak{Y},\mathfrak{Z} \neq \mathfrak{X}$? | |
| Oct 14, 2015 at 18:27 | history | edited | Chris Ramsey | CC BY-SA 3.0 |
Fixed mistake from fixing mistake
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| Oct 14, 2015 at 18:08 | comment | added | Chris Ramsey | Yes, it seems you are correct. I was going on my memory of a talk I went to yesterday. I am certainly no expert on these topics. | |
| Oct 14, 2015 at 18:07 | history | edited | Chris Ramsey | CC BY-SA 3.0 |
Fixed mistake
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| Oct 14, 2015 at 18:00 | comment | added | Bill Johnson | Shouldn't the middle term in (2) be the set of bounded linear operators between the spaces? | |
| Oct 14, 2015 at 17:46 | history | asked | Chris Ramsey | CC BY-SA 3.0 |