Let $\Omega\subset\mathbb{R}^N$ be a bounded smooth domain. Assume that $u\in W_0^{1,2}(\Omega)$, $u\ge 0$ and $-\Delta u\ge 0$ in the sense of distributions ($u$ is superharmonic).
The standard mollifier process, ensure the existence of a sequence $u_n\in C_0^\infty(\Omega)$, such that $u_n\ge 0$, $u_n$ is superharmonic (for large $n$) in every open set $\omega\subset\subset\Omega$ and $u_n\to u$ in $W^{1,2}(\Omega)$.
My question is: instead of the above regularization, I would like to find one where $u_n\in C_0^\infty(\overline{\Omega})$, $u_n\ge 0$, $-\Delta u_n\ge 0$ in $\Omega$ and $u_n\to u$ in $W^{1,2}(\Omega)$. Is it possible?
One possible way to solve the problem, is to extend $u$ outside $\Omega$, in a nieghbourhood of $\partial \Omega$, such that the extension is superharmonic. This is somehow related with the normal derivative of $u$ (if it exists). When $\Delta u$ is a bounded Radon measure, there is $f\in L^1(\partial \Omega)$, which will be denoted by $\frac{\partial u}{\partial \eta}$, such that $$\int_\Omega\varphi\Delta u=-\int_\Omega \nabla \varphi\nabla u+\int_{\partial\Omega} \varphi\frac{\partial u}{\partial \eta}, \forall \varphi\in C^\infty(\overline{\Omega})\tag{1}, $$
If $U_\delta=\{x\in \mathbb{R}^N\setminus\Omega:\ \operatorname{dist}(x,\partial\Omega)<\delta\}$ for small $\delta>0$ then, one reasonable extension would be $$\overline{u}(x)=\frac{\partial u}{\partial \eta}(\tau_\delta (x))\operatorname{dist}(x,\partial\Omega), \ x\in U_\delta,$$
where $\tau_\delta(x)$ is the unique pointe in $\partial \Omega$ such that $\operatorname{dist}(x,\partial\Omega)=\operatorname{dist}(x,\tau_\delta(x))$. In the classical case, by Hopf's lemma, $\frac{\partial u}{\partial \eta}<0$, so the above extension would be superharmonic, however, I fail to see anything in this generalized sense.
Another approach, which seems intuitive and can work for the general case, is to consider $$\Omega_\delta=\{x\in\Omega: \operatorname{dist}(x,\partial\Omega)\le \delta\}$$
and define $\overline{u}_\delta(x)=u(x)$ in $\Omega\setminus\Omega_\delta$ and $\overline{u}_\delta(x)=\frac{\operatorname{dist}(x,\partial\Omega)}{\delta}u(\tau_\delta (x))$ where $\tau_\delta (x)$ is the unique point in $\partial(\Omega\setminus\Omega_\delta)$, which minimizes $\operatorname{dist}(x,\partial(\Omega\setminus\Omega_\delta)$.
If $-\Delta u_\delta\ge 0$ and $u_\delta\to u$ in $W^{1,2}(\Omega)$ then, we could exnted each $u_\delta$ outside $\Omega$ in the obvius way and regularize it, to obtain the desired result.
