Let $f:R^n \to R$ be convex. Then there exist signed Radon measures $\mu^{ij}=\mu^{ji}$ such that $$ \int_{R^n} f \frac{\partial^2 \varphi}{\partial x_i \partial x_j} dx= \int_{R^n} \varphi d\mu^{ij} \quad (i,j=1,...,n) $$ for all $\varphi \in C^2_c(R^n)$. Let $[\frac{\partial^2 \varphi}{\partial x_i \partial x_j}]=[D^2 f]$ be the distributional Hessian metrix. The absolutely part w.r.t. n-dim Lebesgue measure $[D^2f]_{ac}$ is a non-negatively definite matrix.

For $C^2$ Riemannian manifolds, since we can take $C_c^2$ functions as test functions, we can define the distributional Hessian metrix just as the the case of $R^n$.

I wonder whether we can define the distributional Hessian metrix in spaces where we have not $C^2_c$ functions. For example, $C^0$ Riemannian manifold, Lipschitz manifold with $L^{\infty}$ Riemannian metric. Metric measure spaces such as $CD(K,N)$ spaces.


In $C^1$ manifolds I guess one could define measure-valued Hessian at least for $C^1$ functions by working in charts and throwing only one derivative on the $C^1$ test function.

In general $CD(K,N)$ spaces I'm not aware of any such notion, but at least in $RCD(K,\infty)$ spaces (=$CD(K,\infty)$ spaces $(X,d,m)$ such that $W^{1,2}(X,d,m)$ is Hilbert - this class is stable w.r.t. mGH convergence) one can define what it means for a function to have Hessian in $L^2$. Of course this is not as general as having the Hessian as a measure, but is something.

The main reason for which it is natural to consider Hessians in $L^2$ for spaces with Ricci curvature bounded from below is in Bochner inequality - which in the appropriate weak sense holds in $RCD(K,\infty)$ spaces - that tells that $$ \Delta\frac{|\nabla f|^2}2\geq |Hf|_{HS}^2+\nabla f\cdot\nabla\Delta f+K|\nabla f|^2 $$ for any 'smooth enough' function $f$. Integrating this inequality we get $$ \int |Hf|^2_{HS} \,dm\leq \int |\Delta f|^2-K|\nabla f|^2\,dm. $$ Since by regularization with the heat flow it is easy to produce functions for which the right hand side of the last expression is finite, we see that in such spaces there must be 'many' functions with Hessian in $L^2$.

While I do understand that mathoverflow is not for self-promotion, these things are very recent and as of now discussed only in a preprint of mine, so allow me to point out the arxiv reference: http://arxiv.org/abs/1407.0809

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  • $\begingroup$ :$C^1$-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. $\endgroup$ – mafan Mar 27 '15 at 12:02
  • $\begingroup$ 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? $\endgroup$ – Nicola Gigli Mar 28 '15 at 18:49
  • $\begingroup$ 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. $\endgroup$ – mafan Mar 29 '15 at 4:24
  • $\begingroup$ 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. $\endgroup$ – Nicola Gigli Mar 29 '15 at 10:04
  • $\begingroup$ 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. $\endgroup$ – mafan Mar 29 '15 at 13:14

For Alexandrov space $A$ (curvature bounded below) the problem is solved by constructing an array of concave functions $f_0,\dots,f_n$ in an open set $U$, where $n=\dim A$. The function suppose to satisfy the following technical condition: for any $p\in U$ the function $$w_p(x)=\min_i \{f_i(x)-f_i(p)\}$$ has maximum at $p$.

Once it is done, it is easy to see that $f_1,\dots,f_n$ form a chart on $U$ and any convex function $h\colon U\to\mathbb R$ if writen in this chart, becomes DC; i.e., difference of two convex functions.

For DC-functions on $\mathbb R^n$ once has a distributional Hessian. Then you can pull it back to the space using standard formulas, as in the smooth case.

So, even for Alexandrov spaces things were not straightforward. Hope it helps.

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