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 convex function $w$ (it's equivalent saying that $u$ is semiconvex if exists a $\lambda$ such that the function $z(x)=u(x)+\dfrac{|x|^2}{2\lambda}$ 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 show 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: $$ \sup_{\Omega}u \leq \sup_{\partial\Omega}u+ C ||f||_{L^n(\Gamma^+)}$$

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

I know this result for subsolution $u\in W^{2,n}(\Omega)$, an extension through molltification of the case $u\in C^2(\Omega)$. So I thought that I can deduced the validity of my Aleksandrov maximum principle from its validity for classical subsolution, by mollification or something like this.

Can somebody please help me?