Aleksandrov maximum principle for semi-convex function

Question

Definition. Let $u:\Omega \rightarrow \mathbb{R} $. A function $u$ is called semiconvex if $u=v+w$ for some $v\in C^{1,1}(\Omega)$ and a convex function $w$.

Note. Saying that $u$ is semiconvex is equivalent to say that there exists a $\lambda$ such that the function $$ z(x)=u(x)+\dfrac{|x|^2}{2\lambda}\text{ is convex}.$$

Consider the elliptic operator of the form $$Lu=a^{ij}D_{ij}u+b^iD_iu$$ and let $L$ be uniformly elliptic.

I want to prove the following statement:

Theorem (Aleksandrov maximum principle): Let $u$ be semiconvex in $\Omega$ and suppose $Lu+f\geq0$ almost everywhere in $\Omega$ for some $f\in L^{n}(\Omega)$. We then have the following estimates: $$ \sup_{\Omega}u \leq \sup_{\partial\Omega}u+ C \Vert f\Vert_{L^n(\Gamma^+)}$$

where $\Gamma^+$ is upper contact set of $u$ (a sub domain of $\Omega$ where the Hessian of $u$ is negative define).

I know that this result holds for subsolutions $u\in W^{2,n}(\Omega)$, as it can be shown by extending the same result for the case $u\in C^2(\Omega)$ through mollification. So I thought that I can deduce the validity of my Aleksandrov maximum principle from its validity for classical subsolution, by mollification or something like this. Could this be true? Can somebody please help me?

Answer 1

The ABP estimate indeed holds in your setting. The key is that the concave envelope of $u$ is in $C^{1,\,1}$, so the area formula is valid for its gradient. Assuming for simplicity that $L = \Delta$, that $\Omega = B_1$ and that $\sup_{\partial B_1} u = 0$, the way I would argue is:

Let $\Gamma$ be the concave envelope (the infimum of linear functions larger than $u$ in $B_1$ and, say, larger than $0$ on $\partial B_2$). Using that $u$ is touched from below by a paraboloid with Hessian $-\lambda^{-1}I$ at every point, one can show (see e.g. the book of Caffarelli-Cabre on fully nonlinear equations) that $\Gamma$ is touched from above by a linear function and from below by a paraboloid of opening $-\lambda^{-1}I$ at every point. In particular, $\Gamma \in C^{1,\,1}$, so we can apply the area formula to $\nabla \Gamma$. By elementary geometry $\Gamma$ is touched from above on the set $K := \{u = \Gamma\}$ by a linear function of slope $p$ for every $p \in B_{\frac{\sup_{B_1}u}{5}},$ so $$(\sup_{B_1}u)^n \leq C(n)|\nabla \Gamma(K)|.$$ Using the area formula and that $D^2u \leq D^2\Gamma \leq 0$ at almost every point in $K$ we conclude that $$(\sup_{B_1}u)^n \leq C(n) \int_{K} |\det D^2\Gamma(x)|\,dx \leq C(n)\int_K |\det D^2u(x)|\,dx.$$ Finally, by the AGM inequality and the equation, the last term is controlled by $C(n)\|f\|_{L^n(K)}^n$, completing the proof.

Remark: The classic example where the semi-convexity (and $W^{2,\,n}$ regularity) and conclusion fail is $u = 1-|x|^{\gamma}$ in $B_1 \subset \mathbb{R}^n,$ with $\gamma \in (0,\,1)$ and $n \geq 2$. Indeed, $u \in W^{2,\,p}$ for $p < \frac{n}{2-\gamma}$ and solves a uniformly elliptic equation of the form $a_{ij}(x)u_{ij} = 0$ away from the origin, but the maximum principle fails.