Timeline for Surjectivity of operators on $\ell^\infty$
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Sep 3, 2014 at 15:20 | history | edited | Bill Johnson | CC BY-SA 3.0 |
Fixed spelling in title.
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| Sep 2, 2014 at 18:37 | answer | added | Bill Johnson | timeline score: 11 | |
| Jul 7, 2012 at 23:57 | answer | added | Amir | timeline score: 0 | |
| Jul 7, 2012 at 23:22 | comment | added | Yemon Choi | Now posted at MSE: math.stackexchange.com/questions/168025/… | |
| Jul 6, 2012 at 15:16 | answer | added | Bill Johnson | timeline score: 11 | |
| Jul 5, 2012 at 16:21 | comment | added | Bill Johnson | Here is a much easier question: Does there exist a non surjective bounded linear operator from some Banach space into $\ell_\infty$ that has dense range? I see how to do this but the argument uses something that is not elementary. Is there a simple reason such an operator exists? | |
| Jul 4, 2012 at 19:15 | comment | added | András Bátkai | Even the existence of densely defined closed unbounded operators is nontrivial, see Nagel (ed.): One parameter Semigroups of Positive Operators, page 58. (the section by Lotz on semigroups in Grothendieck spaces). | |
| Jul 4, 2012 at 17:27 | comment | added | András Bátkai | There is a theorem of Lotz stating that there are no strongly continuous semigroups on $l^{\infty}$, meaning that if semigroup is strongly continuous, then the generator is bounded. | |
| Jul 4, 2012 at 16:21 | history | edited | Bill Johnson |
Added tag.
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| Jul 4, 2012 at 0:02 | comment | added | Yemon Choi | Just checking, Bill :-) | |
| Jul 3, 2012 at 23:25 | comment | added | Bill Johnson | Surely, Yemon; for the weak$^*$ topology the problem is trivial. | |
| Jul 3, 2012 at 23:24 | history | edited | Bill Johnson | CC BY-SA 3.0 |
Formatting; punctuation.
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| Jul 3, 2012 at 23:15 | comment | added | Yemon Choi | Just to clarify: I assume that the OP means "dense in the norm topology"? | |
| Jul 3, 2012 at 22:48 | comment | added | Bill Johnson | I guess you mean also that $A$ should be closed and map its domain back into itself. I would have to review semigroup theory (or think more than I care to right now) to see if that is correct. Anyway, how do you get such an $A$? | |
| Jul 3, 2012 at 22:07 | comment | added | Matthias Ludewig | If you take some unbounded densely defined operator A on l infty; does not the corresponding semigroup of operators take values in the domain of A? | |
| Jul 3, 2012 at 21:41 | comment | added | Bill Johnson | Despite the two quick votes to close, I don't find this a trivial question. Am I missing something? | |
| Jul 3, 2012 at 20:26 | history | asked | Amir | CC BY-SA 3.0 |