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3
votes
Accepted
The infinity Wasserstein distance $W_\infty$ and the weak topology
First, I interpret your condition that "$X$ has no isolated points" in the following ways: First, every ball $B(x, \varepsilon)$ has non-empty interior. This means, in particular, that we can find arb …
1
vote
Building the Wasserstein space by pushforwards
Perhaps another simple argument that $\mathcal{X}$ is indeed equal to $\mathcal{W}_2(\mathbb{R}^d$):
Starting from the fact that there is a bimeasurable bijection $b : \mathbb{R}^d \rightarrow \mathbb …
3
votes
0
answers
244
views
Metric ($f$-divergence) on space of probability measures that satisfies pythagorean theorem
Let $E$ be a polish space, $\mathcal{P}(E)$ the Borel probability measures on $E$ with the topology of weak convergence and $\mathcal{Q} \subset \mathcal{P}(E)$ a convex and compact set.
First, the m …