Timeline for Is the space of all borel measures on $\mathbb R^n$ isomorphic to the tensor product of spaces of borel measures on $\mathbb R$?
Current License: CC BY-SA 3.0
21 events
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| Apr 21, 2015 at 15:20 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
Improved Note 2
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| Apr 21, 2015 at 14:57 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
Made it more clear to a hasty reader
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| Apr 21, 2015 at 13:01 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
I give an indication which may completely solve the problem in a restricted case
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| Apr 21, 2015 at 9:45 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
typo(s) formatting
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| Apr 21, 2015 at 9:37 | comment | added | Duchamp Gérard H. E. | .@report I made precise a set of semi-norms which would fit the plan, see the note added in the answer. Thanks again for your comment(s) | |
| Apr 21, 2015 at 9:34 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
Made precise the semi-norms for which the space of measures (not only positive ones) is complete.
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| Apr 21, 2015 at 9:11 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
Return to measure spaces and added a positioning
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| Apr 21, 2015 at 8:24 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
Amended the text w.r.t. unexpected difficulties due to completeness
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| Apr 21, 2015 at 8:17 | comment | added | Duchamp Gérard H. E. | .@report you are right. I tried to repair (using the cone of positive measure and changing the seminorms), but not in general so far. So I amend my answer which becomes only partial. | |
| Apr 20, 2015 at 6:40 | comment | added | report | There are two things that are wrong with the above answer. Firstly it is not true that the space of measures is complete for vague convergence. Secondly, one would require denseness in some norm to finish the argument, not that in vague convergence. I repeat that there is a notion of tensor product for which the result is true, but it is not one in the category of Banach spaces. | |
| Apr 19, 2015 at 18:17 | comment | added | Yemon Choi | Just to clarify for anyone else reading: the tensor product completion Gerard describes above is not in general the same as the projective tensor product of Banach spaces | |
| Apr 16, 2015 at 20:28 | comment | added | Duchamp Gérard H. E. | .@Elesthor As a reference, you can have a look at thm 417C p89 of D.H.Fremlin : Measure theory, Topological Measure Spaces. Once you know the (1) is true (embedding) and that the space of all bm is complete (easy, this is a weak topology), then comes naturally the completed tensor product ! | |
| Apr 14, 2015 at 13:06 | comment | added | Elesthor | That would be very nice of you, thanks! Or a reference if you have. | |
| Apr 10, 2015 at 14:55 | comment | added | Duchamp Gérard H. E. | If needed, we can elaborate a bit. I am at your disposition | |
| Apr 10, 2015 at 12:19 | comment | added | Elesthor | Yes, it was in the sense on completed product. Thank you for the answer. | |
| Apr 10, 2015 at 11:10 | vote | accept | Elesthor | ||
| Apr 9, 2015 at 14:11 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
added 47 characters in body
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| Apr 9, 2015 at 14:04 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
I explained under the form of a route map, why we can consider the tensor product as completed.
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| Apr 8, 2015 at 16:15 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
I put the last formula more explicit in case the asker be a student
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| Apr 8, 2015 at 16:07 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
I added "Hausdorff"
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| Apr 8, 2015 at 16:02 | history | answered | Duchamp Gérard H. E. | CC BY-SA 3.0 |