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?