8
$\begingroup$

Consider a smooth, connected and complete Riemannian manifold $M$. It is well known that harmonic functions defined on some open subset of $M$ are $C^\infty$.

I'm interested in knowing whether there are quantitative $C^1$ estimates for such harmonic functions which depend only on some bound from below on the curvature (either sectional or Ricci) and an upper bound on the dimension.

To give an example of what I mean, consider the same question with "$C^1$" replaced by "Lipschitz". It is then a classical fact of geometric analysis that Lipschitz estimates follows from a lower bound on the Ricci and an upper bound on the dimension, the proof being an application of the Bochner inequality together with elliptic regularity theory.

Yet, I don't know any result which improves such Lipschitz estimate to $C^1$. I'm interested both in positive statement (e.g. with a lower bound on the sectional there is an estimate) and in negative ones (e.g. a lower bound on the Ricci is not sufficient because along a certain sequence of manifolds the $C^1$ norm of properly defined harmonic functions blows-up).

The question comes from the study of non-smooth spaces arising as (measured)-Gromov-Hausdorff limits of manifolds with prescribed curvature bounds. Basically, what I'm trying to understand is "how smooth" can be functions defined on them, so that it is natural to start looking at harmonic ones. Still, in fact I don't really know how smooth harmonic functions are on Riemannian manifolds, and given that any estimate on limit spaces is (more-or-less) also a quantitative estimate on the smooth ones, before trying to reinvent the wheel I'd like to know if there are known results available in the literature.

Any help is appreciated, thanks.

Improve this question
$\endgroup$
4
  • $\begingroup$ 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 $\endgroup$
    – Otis Chodosh
    Commented Jul 2, 2015 at 15:48
  • $\begingroup$ 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. $\endgroup$
    – Nicola Gigli
    Commented Jul 2, 2015 at 16:58
  • $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Commented Jul 3, 2015 at 19:55
  • $\begingroup$ 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 $\endgroup$
    – Nicola Gigli
    Commented Jul 9, 2015 at 20:38

1 Answer 1

Reset to default
4
$\begingroup$

For the people interested in the solution to this problem, Gigli encouraged Nuñez Zimbrón and De Philippis to look for a solution to this problem, which now can be found here https://arxiv.org/pdf/1909.05220.pdf

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks Raquel! I forgot to update this discussion with its natural answer. $\endgroup$
    – Nicola Gigli
    Commented Aug 29, 2023 at 13:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .