Timeline for Are all linear transformations measurable?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 1, 2014 at 21:36 | vote | accept | Julian Newman | ||
| Aug 1, 2014 at 15:53 | comment | added | Julian Newman | Interesting, thanks - is the proof straightforward? | |
| Aug 1, 2014 at 15:21 | comment | added | Gerald Edgar | Generalization (automatic continuity)... For a homomorphism between complete separable metric groups, if it has the property of Baire, then it is continuous. (In particular, Borel-Borel measurable maps have the property of Baire.) | |
| Aug 1, 2014 at 7:05 | comment | added | blackburne | Baire's category theorem (cover the Banach space by the sets where $|f|\leq n$). By the way, it is not customary for questons of this sort to be answered in MO to the last detsil so this is my final word. | |
| Aug 1, 2014 at 6:51 | comment | added | Julian Newman | I understand that the point of the example is that it is unbounded on any ball - but I do not know what the "standard arguments" are by which one can deduce non-measurability. (I presume you must be using completeness of the Banach space somehow, since on a countable-dimensional normed vector space the same construction gives a map that is unbounded on every ball and yet is measurable.) | |
| Aug 1, 2014 at 6:35 | comment | added | blackburne | This follows by standard arguments from the fact that it is unbounded on any ball. | |
| Aug 1, 2014 at 6:12 | comment | added | Julian Newman | Thank you for your reply. Sorry, I'm probably being a bit dim: how do you prove that the standard example of a discontinuous linear form on an infinite-dimensional Banach space is non-measurable? | |
| Aug 1, 2014 at 6:06 | history | edited | blackburne | CC BY-SA 3.0 |
added 136 characters in body
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| Aug 1, 2014 at 5:55 | history | answered | blackburne | CC BY-SA 3.0 |