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S May 24, 2022 at 8:23 history suggested ViktorStein CC BY-SA 4.0
typos fixed
May 24, 2022 at 7:04 review Suggested edits
S May 24, 2022 at 8:23
Apr 28, 2022 at 19:02 vote accept leo monsaingeon
Apr 28, 2022 at 16:51 comment added Tobsn Can not add much to the elaborate answer of @AndréSchlichting below, except for one standard refernce that you may consult in order to find more details, also on André's arguments in the beginning: Bakry et al, Analysis and Geometry of Markov Diffusion Operators
Apr 28, 2022 at 13:22 history edited JHM CC BY-SA 4.0
minor misspelling
Apr 28, 2022 at 13:10 answer added André Schlichting timeline score: 7
Apr 28, 2022 at 4:29 comment added leo monsaingeon @Tobsn: thanks for your input. Allow me a couple questions: 1) I don't see how my above assumption is weaker than yours? The $\lambda$-convexity also leads to contractivity (although I didn't discuss this here, but formally they are also equivalent). And 2) why should this lead to the same spectral gap? Actually this is precisely my question! and I am not aware that for Fokker-Planck the first eigenvalue (0 left aside) is the lowest eigenvalue of the Hessian of the potential. Is this your claim? (I am not saying it isn't, just that I am not aware of it.) Do you have a reference to suggest?
Apr 27, 2022 at 14:19 comment added Tobsn At least on full space, exponential Wasserstein contraction of trajectories i.e. $W(\mu_{t},\mu_{t}')\le e^{-\lambda t}W(\mu_{0},\mu_{0}')$ (which is of course stronger than your assumption above) is actually equivalent to $\lambda$-strong convexity of the potential $V$ and should then also lead to the same spectral gap.
Apr 27, 2022 at 13:12 history edited leo monsaingeon CC BY-SA 4.0
minus sign missing
Apr 27, 2022 at 13:03 history asked leo monsaingeon CC BY-SA 4.0