Timeline for What are some characterizations of the strong and total variation convergence topologies on measures?
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18 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Sep 21, 2013 at 9:57 | comment | added | user39080 | @user24367 Look at the answer to the linked Stack Exchange question for a counterexample. | |
| Sep 21, 2013 at 1:39 | comment | added | user24367 | would the L to R relation be true if we replace the condition instead by $\int f\mathrm{d}\mu_{n}\to \int f\mathrm{d}\mu$ for every uniformly continuous function? Assume the space is $\mathbb{R}^{N}$ and usual topology. | |
| Sep 20, 2013 at 8:29 | vote | accept | user39080 | ||
| Sep 19, 2013 at 18:29 | answer | added | Pietro Majer | timeline score: 11 | |
| Sep 19, 2013 at 18:04 | comment | added | user39080 | It would be great if you could add some more details or expand your comments into an answer! | |
| Sep 19, 2013 at 18:03 | comment | added | user39080 | I meant the notion from the wiki article. I guess the terminology is confusing if you are thinking in a functional analysis way. | |
| Sep 19, 2013 at 18:01 | comment | added | Pietro Majer | Actually I was thinking to real valued signed measures in general. Maybe I'll add few details below. | |
| Sep 19, 2013 at 17:53 | comment | added | user39080 | I guess you meant to say "probability measures"? That is my interest, and I am glad if L to R holds. | |
| Sep 19, 2013 at 16:48 | comment | added | Pietro Majer | If by "strong convergence" you mean the same as in the wiki article, then I think the implication L to R holds if the sequence μn is also bounded in total variation norm (which is of course true if $\mu_n$ are positive measures). | |
| Sep 19, 2013 at 15:05 | comment | added | Pietro Majer | The condition in 1 does not seem sufficient (wrto the total variation norm of measures) : E.g. if $\lambda$ is the Lebesgue measure on $[0,1]$ and $\mu$ and all $\mu_n$ are a.c. wrto $\lambda$ with densities $g_n$ and $g$ wrto $\lambda$, the condition means $g_n\to g$ weakly in $L^1$, whereas $\mu_n\to\mu$ in norm means $g_n\to g$ in the norm of $L^1$. | |
| Sep 19, 2013 at 14:21 | history | edited | user39080 | CC BY-SA 3.0 |
edited body
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| Sep 19, 2013 at 10:56 | comment | added | user39080 | Not compact, but otherwise you can assume it is a nice space if it helps. | |
| Sep 19, 2013 at 7:48 | comment | added | Pietro Majer | Do you haev any assumption on the base space? That is, it may be any measurable space, or is it a topological space, compact, metric...? | |
| Sep 19, 2013 at 7:36 | history | edited | user39080 | CC BY-SA 3.0 |
added 130 characters in body
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| Sep 19, 2013 at 6:40 | history | edited | user39080 | CC BY-SA 3.0 |
added 11 characters in body; edited tags
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| Sep 14, 2013 at 12:25 | history | edited | user39080 | CC BY-SA 3.0 |
fixed mistake in title
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| Sep 14, 2013 at 8:19 | history | asked | user39080 | CC BY-SA 3.0 |