Skip to main content
7 events
when toggle format what by license comment
Aug 29, 2023 at 13:53 vote accept Nicola Gigli
Aug 14, 2023 at 16:57 answer added Raquel timeline score: 4
Jul 9, 2015 at 20:38 comment added Nicola Gigli Thanks, I do agree that this line of thoughts should give some results, and in fact it seems likely that if one wants to control the $C^k$ norm of harmonic functions then a control of the first $n(k)$ derivatives of the curvature should suffice. Still, this is not really what I'm looking for, as I'm interested in cases where I only have a bound from below on the curvature (which is sufficient, e.g., to control the Lipschitz constant). Given that "$C^1$" is not that far from "Lipschitz", I hoped something was already known in this direction
Jul 3, 2015 at 19:55 comment added Deane Yang I'm not sure if this works (I'm too lazy to do the tensor calculation), but if you differentiate $\Delta f = 0$ twice, you should get something like $\Delta \nabla^2f + A\nabla^2 f = \nabla\cdot(B\nabla f)$, where $A$ and $B$ depend on the curvature tensor (and maybve its covariant derivative). If so, then a straightforward Moser iteration argument gives an $L^p$ bound on $\nabla^2f$ in terms of the Sobolev constant, curvature (and maybe its covariant derivative). This in turn implies a Holder bound on $\nabla f$, where the constant can also be bounded in terms of geometric invariants.
Jul 2, 2015 at 16:58 comment added Nicola Gigli hello Otis. This looks weaker than what I need, as it gives a bound on the gradient of (the log of) $f$, while I'd like a continuity estimate for it.
Jul 2, 2015 at 15:48 comment added Otis Chodosh Hi Nicola! There is the Cheng--Yau gradient estimate, which only works for positive harmonic functions, I think, but does bound $|\nabla \log f|$, which seems to be what you want. See e.g. arxiv.org/pdf/1106.3560v1.pdf
Jul 2, 2015 at 12:41 history asked Nicola Gigli CC BY-SA 3.0