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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 $p$?

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  • $\begingroup$ 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 $\endgroup$ – Izar Urdin Oct 26 '19 at 16:46
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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.

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