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Apr 2 at 21:49 vote accept leo monsaingeon
Apr 2 at 19:01 history edited Daniele Tampieri CC BY-SA 4.0
Typo fixes minor additions
Apr 2 at 18:45 answer added rpk timeline score: 5
Nov 24, 2023 at 23:10 history became hot network question
Nov 24, 2023 at 21:34 comment added leo monsaingeon @erz thank you, I was not aware that this had a name. good to know!
Nov 24, 2023 at 21:03 answer added Ayman Moussa timeline score: 6
Nov 24, 2023 at 17:16 comment added erz let me just mention that this property is known as Kadets-Klee or Radon-Riesz property
Nov 24, 2023 at 16:19 comment added Akira Thank you so much for your elaboration!
Nov 24, 2023 at 16:18 comment added leo monsaingeon Let me just give a heuristic reasoning as to why this may hold true: it is well known that lack of strong convergence can only arise form oscillations. But, wild oscillations tend to encode lots of information in the sense of 1/0 bits in information theory. Such bits correspond to high entropy. But since $\limsup E(\rho_n)\leq E(\rho)$ the sequence cannot contain too much information, thus it should not oscillate too fast, and converge better than expected. OK sure, this is a very very rough picture, but I think there's some truth to it and I'm trying to make put this intuition on solid ground
Nov 24, 2023 at 16:12 comment added leo monsaingeon Well, that's the whole point: for some independent reason I'm able to prove that the entropy converges. and I want to improve the convergence, i-e $\rho_n\to\rho$ should hold in a stronger sense than I initially assumed.
Nov 24, 2023 at 16:09 comment added Akira The Boltzmann entropy is only l.s.c. in the weak topology of $L^1 (\Omega)$. Do you have some situations where we have the convergence $\int_\Omega \rho_n(x)\log\rho_n(x) \mathrm d x\to \int_\Omega \rho(x)\log\rho(x) \mathrm d x$?
Nov 24, 2023 at 15:07 history asked leo monsaingeon CC BY-SA 4.0