# Smooth approximation of a subharmonic function in the barrier sense

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 $$p$$?

• Are you looking for a functor wich transform a group (sequenced) of smooth functions to the same function? If you do then yes, we can find it. How ... I dono – Izar Urdin Oct 26 '19 at 16:46

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.