Timeline for What is vague convergence and what does it accomplish?
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8 events
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| Feb 16, 2020 at 11:47 | comment | added | yada | On $C_c(X)$ one can define the locally convex inductive limit topology $lim$ from the Banach spaces $C_K(X)$ of continuous functions defined on $X$ with support contained in a compact set $K$. $lim$ is finer than the $sup$ norm topology and the dual of $C_c^{lim}$ can be identified with the space of all Radon measures on $X$ (I think this should be your space $\mathcal{M}(X)$). The unbounded Radon measures can be defined as set functions on the $\delta$-ring generated by the compact sets. For more information, see [Bourbaki, Integration] and [Dinculeanu, Vector Measures]. | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Mar 14, 2017 at 4:19 | comment | added | Nate Eldredge | The vague topologies from either $C_0$ or $C_c^{\mathrm{sup}}$ induce the same topology on the set of all subprobability measures (positive measures with total measure $\le 1$), and this topology is compact. That is a major reason for using the vague topology. | |
| Mar 13, 2017 at 14:50 | comment | added | Jochen Wengenroth | $C_c(X)$ is dense in $C^{co}(X)$ and hence both spaces have the same dual which is the space of (complex) measures with compact support. | |
| Mar 13, 2017 at 4:35 | history | edited | Greg Zitelli | CC BY-SA 3.0 |
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| Mar 13, 2017 at 3:11 | comment | added | Greg Zitelli | Right, of course, thanks, their duals will be the same as sets and have the same Banach space topology, but the weak-* topologies from $C_0$ and $C_c^{sup}$ will be different, won't they? | |
| Mar 13, 2017 at 2:07 | comment | added | user95282 | $C_c^{sup}$ is dense in $C_0$, so they have the same dual. For the fourth question: No, there is no problem with thinking this way. The weak convergence in probability is the weak$\ast$ convergence (from $C_b$) in functional analysis. | |
| Mar 13, 2017 at 0:48 | history | asked | Greg Zitelli | CC BY-SA 3.0 |