distributional Hessian for semiconvex functions on non-smooth manifolds

Question

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.

Answer 1

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

Answer 2

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.