Let $f$ be a continuous function on $\mathbb R^n$ such that $\Delta f \ge 0$ at a point $p$ in the barrier sense. More precisely, for any $\epsilon>0$, there exists a smooth function $f\_{\epsilon}$ which is locally defined on an open set $U$ containing $p$ such that (i) $f(p)=f\_{\epsilon}(p)$, (ii) $f \ge f\_{\epsilon}$ on $U$ and (iii) $\Delta f\_{\epsilon} \ge -\epsilon$ at $p$.


Can we find a sequence of smooth functions $g\_{\epsilon}$ such that $g\_{\epsilon} \to f$ in $C\_{loc}$ and $\Delta g\_{\epsilon} \ge -\epsilon$ at $p$?



Smooth approximation of a subharmonic function in the barrier sense
Withtout loss of generality, we can assume $p=0$ and $f(0)=0$. 


Also, the problem is purely local : we can assume that $f$ and all the functions $f\_\epsilon$ are compactly supported in the unit ball $\text{B}$, which contains also all the neighborhoods $U\_\epsilon$. You can also assume that for all $\epsilon>0$ the estimate $f\leq f\_\epsilon$ is satisfied on $\text{B}$.


Now pick a smooth function $\Phi$ subharmonic on $2 \text{B}$, integrable over $\mathbf{R}^d$ with total mass equal $1$.


Fix $\epsilon$ and take $f\_\epsilon$ given by your statement. Since $f$ and $f\_\epsilon$ vanish outside $\text{B}$, we claim $\Delta \Phi\star (f-f\_\epsilon)\geq 0$ on $\text{B}$, because $\Delta \Phi\geq 0$ on $2\text{B}$. This means in particular that $\Delta(\Phi\star f) \geq \Phi\star \Delta f\_\varepsilon$ is true at the origin. Now replace $\Phi$ by $\Phi\_\delta:=\Phi(x/\delta)/\delta^d$. By standard properties of convolution you have a decreasing family $\delta\_\epsilon\rightarrow 0$ such that $\Phi\_{\delta\_\epsilon} \star \Delta f\_\epsilon(0)\geq-2\epsilon$.


The corresponding sequence $g\_\epsilon:=\Phi\_{\delta\_{\epsilon/2}}\star f$ does the job.


