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Question about measure lemma?

Assuming $1 < p < \infty$, $W^{1,p}(\Omega)$ is reflexive, so bounded sets are weakly precompact by Alaoglu's theorem (the weak-* and weak topologies coincide). Thus $u_j$ has a subsequence convergin …
Nate Eldredge's user avatar
5 votes

Large deviation/concentration inequality for submartingale

This looks like a weak law of large numbers, and in fact a strong law holds: I claim that $\liminf_{t \to \infty} \frac{S_t}{t} \ge \Delta$ almost surely, which implies the desired result. The key is …
Nate Eldredge's user avatar
5 votes

A distribution which is Wasserstein-close to a compactly supported distribution is almost co...

The claimed inequality is not true. The simplest possible counterexample works: let $x,y \in \mathbb{R}^n$ with $|x-y| = \epsilon$, and take $\mu = \delta_x$, $\nu = \delta_y$. Then $W_1(\mu,\nu) = …
Nate Eldredge's user avatar