Aleksandrov maximum principle for semi-convex function 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?
 A: 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. 
