Timeline for Metrization of weak convergence of signed measures
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| Nov 29, 2023 at 0:48 | comment | added | No-one | Moreover, in the specific case of positive Radon measures on a compact $K$ it is true that $|||\cdot|||$ metrizes the weak$^*$ topology. Indeed one can just take $y_0\in C(K)$ with $y_0(x)=1$ for all $x$, and then the convergence in $|||\cdot|||$ implies that the masses of the $u_n$ are bounded. This is what is done in Theorem $3.1$ of the paper jstor.org/stable/25048364 linked by Iian Smith. | |
| Nov 29, 2023 at 0:41 | comment | added | No-one | Of course this is a hand-waving argument just to build some intuition, a proper proof follows from the fact that every weak convergence sequence is (strongly) bounded. | |
| Nov 29, 2023 at 0:22 | comment | added | No-one | Nice answer. It is worth to point out that if $u_j$ is unbounded (with respect to $||\cdot ||$), then it might still be the case that $|||u_n|||\to 0$ but of course $u_j$ doesn't converge weakly. To "see" why this is, notice that $u_j(y_j)\to 0$ for all $y_j$ but not uniformly in $j$. For a generic $y$ not equal to any $y_j$ we can only bound $|u_n(y)|\leq |u_n(y-y_j)|+|u_n(y_j)|\leq ||u_n|| |y_j-y| +|u_n(y_j)|$. So, for a given $n$, if we try to take $j$ so that $||u_n|| |y_j-y|$ is small, it might be the case that $||u_n(y_j)||$ gets large, and $|u_n(y)|$ doesn't go to $0$. | |
| Sep 4, 2013 at 14:33 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Jan 30, 2013 at 19:24 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Jan 30, 2013 at 12:04 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Jan 30, 2013 at 11:20 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Jan 30, 2013 at 11:01 | history | answered | Pietro Majer | CC BY-SA 3.0 |