Let $\Omega_1,\Omega_2\subset\mathbb{R}^n$ be strictly convex bounded domains, possibly with smooth boundary, and such that $\bar\Omega_1\subset\Omega_2$.
Let $u\colon \bar\Omega_1\to \mathbb{R}$ be a convex continuous function such that $u|_{\partial \Omega_1}=0$. Let $\bar u\colon \Omega_2\to \mathbb{R}$ be its extension by 0. Let $v\colon \Omega_2\to \mathbb{R}$ be the lower convex envelope of $\bar u$, i.e., $$v:=\sup\{f\colon\Omega_2\to \mathbb{R}|\,\, f\text{ is convex}, f\leq \bar u\}.$$
Clearly, $v$ is convex.
Question. Is it true that in $\Omega_2\setminus \bar\Omega_1$ $$\det(\frac{\partial^2v}{\partial x_i\partial x_j})=0,$$ where the left hand side is understood in the generalized (Alexandrov) sense as a measure for not necessarily smooth $v$?
