Timeline for Metrization of weak convergence of signed measures
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8 events
| when toggle format | what | by | license | comment | |
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| Nov 9, 2023 at 15:03 | comment | added | Yury Korolev | I've searched forever for a reference for the fact that the closed unit ball in $\mathcal M(\Omega)$ (in the total variation norm) is compact w.r.t. the transportation/Kantorovich-Rubinstein metric (most references only speak of positive measures, cf Dan's reply). The result can be found in Theorem VIII.4.3 in Kantorovich and Akilov. Functional Analysis | |
| Jan 8, 2023 at 12:09 | comment | added | Jochen Wengenroth | 10 years to late, but the extension $d(\omega,o)=1$ need not be a metric. For example, if there are points $x,y\in\Omega$ with $d(x,y)>2$, the extenion does not satisfy the triangle inequality. | |
| Nov 17, 2015 at 14:13 | comment | added | Dirk | As a matter of fact it sticked to the advice and using Kantorovich and Rubinstein for these norms (rather than Wasserstein or Monge), see the paper Imaging with Kantorovich-Rubinstein discrepancy. | |
| Nov 17, 2015 at 14:10 | comment | added | Dirk | Sorry for the late answer: This thing is, that I only want to metrize the notion of weak$*$ convergence not the whole weak* topology. These are in fact two different things. The weak* topology can be metrized on bounded subsets (see Pietro Majers answer). | |
| Mar 11, 2015 at 0:10 | comment | added | Iian Smythe | This is a very late comment, but how does this answer not contradict the fact that the dual space of a Banach space $X$ is weak*-metrizable if and only if $X$ is finite dimensional, as mentioned by Jochen below? It is true that the space of positive measures on $\Omega$ is metrizable, as in Theorem 3.1 of jstor.org/discover/10.2307/25048364. | |
| Feb 6, 2013 at 8:14 | comment | added | Dirk | That is in the direction I was thinking. I just found a similar construction in Gromov's "Metric structures" chapter $3\tfrac{1}{2}$.B: For any metric $d$ defined for measures of the same total mass one may define a metric for all finite measures as follows. The distance between two measures $\mu$ and $\nu$ with total masses $\mu(\Omega) = m$ and $\nu(\Omega)=n$ with $n>m$ define $D(\mu,\nu) = n-m + d(\mu,\tfrac{m}{n}\nu)$ (and the distance to infinite measures is just $\infty$). Apparently, this definition also works for metric measure spaces. | |
| Feb 4, 2013 at 8:33 | vote | accept | Dirk | ||
| Jan 30, 2013 at 22:42 | history | answered | R W | CC BY-SA 3.0 |