Timeline for distributional Hessian for semiconvex functions on non-smooth manifolds
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| Mar 29, 2015 at 17:12 | comment | added | Nicola Gigli | ...things are not (yet) that clear. An alternative is in the use of Bochner inequality. Again, sorry for the self-reference, but if you look at the section `splitting without the Hessian' of my paper on the splitting theorem in RCD(0,N) spaces, you see how to carry on the proof of the splitting theorem without ever (explicitly) mentioning the Hessian of the Busemann function nor the fact that it is affine. Hope it helps. | |
| Mar 29, 2015 at 17:06 | comment | added | Nicola Gigli | ok, then although I don't have a full answer to your question, let me give some hints. First of all, I wouldn't worry too much about the fact that the Sobolev spaces $W^{1,2}(M)$ and $W^{1,2}(M,e^{-\beta})$ do not coincide: it's like in the smooth case, where their intersection is dense in both and their respective 'local' version coincide. Also, I wouldn't insist too much in trying to prove that the Busemann is convex and concave. In the smooth case there is a clear link among 'convexity', 'positivity of the Hessian' and 'contraction of gradient flows' but in RCD spaces... | |
| Mar 29, 2015 at 13:55 | comment | added | mafan | Condider a non-compact non-smooth space without boundary, Alexandrov space $M$ for example, $\Delta$ is the Laplacian w.r.t. Lipshictz function with compact support. I try to prove that $(M,d,e^{-\beta}dvol)$ satisfies $BE(K,\infty)$ for some $K$, hence a $RCD^*(K,\infty)$ space, then deduce the Busemann function $\beta$ is semiconcave. Since $\Delta \beta=-(n-1)$, the Hessian you defined $H[\beta]<0$. $BE(K,\infty)$ hold not so rigorously. But the space is non-compact, $e^{-\beta}$ is not in $W^{1,2}(M)$, the latter doesn't coincide with $W^{1,2}(M,e^{-\beta})$, so no rigorous $BE(K,\infty)$. | |
| Mar 29, 2015 at 13:14 | comment | added | mafan | Sorry, I didn't state the problem completely, add the assumption $\lambda_1=(n-1)^2/4$, the first eigenvalue reaches the sharp upper bound. The Riemannian manifold case is proved by Jiaping Wang and Peter Li, paper "complete manifolds with positive spectrum" I and II. | |
| Mar 29, 2015 at 10:04 | comment | added | Nicola Gigli | I'm not sure I understand: the typical assumption for the splitting is $Ric\geq 0$, not $Ric\geq -(n-1)$. With a negative bound on the Ricci (or the sectional) you can't get that the Hessian is 0, not even in the smooth case. | |
| Mar 29, 2015 at 4:24 | comment | added | mafan | I want to prove a splitting theorem (just like Cheeger-Gromoll) on some non-smooth spaces with curvature $>-(n-1)$, then the Buseman function is surely semiconvex. I have proved the Buseman function has constant Laplacian, but I can't prove it's semiconcave, thus can't prove the function is affine. | |
| Mar 28, 2015 at 18:49 | comment | added | Nicola Gigli | I guess you are right, the first part of my answer does not really make sense. In your generality I don't know any strategy. May I ask from where your question comes from? Maybe you have some additional structure/regularity/rigidity? | |
| Mar 27, 2015 at 12:02 | comment | added | mafan | :$C^1$-Rieman manifold is $C^2$ manifold, so no problem: For a semiconvex function f on a $C^1$ Riemannian manifold, we first use a $C^2$ mollifier to get $C^2$ semiconvex functions $f_{\epsilon}$ to approximate f. For the Hessian of $f_{\epsilon}$, at every point, we can show we have a $n \times n$ matrix A, A-cI is non-negatively definite for some I. Then by approximation, the distributional Hessian matrix have non-negative singular part. But for a $C^0$ Riemannian manifold with curvature bounded below, I don't know how to proof. This is what I am really concerned about. | |
| Mar 26, 2015 at 11:23 | history | answered | Nicola Gigli | CC BY-SA 3.0 |