Let $E$ be a locally compact Hausdorff space and $(\mathcal D(A),A)$ be a closable linear operator on $C_0(E)$. If $(\mathcal D(A),A)$ satisfies the nonnegative maximum principle$^1$, does the same apply to the closure $(\mathcal D(\overline A),\overline A)$?
Let $f\in\mathcal D(A)$ and $x_0\in E$ with $\max(0,f(x))\le f(x_0)$. By definition of $(\mathcal D(\overline A),\overline A)$, there is a $(f_n)_{n\in\mathbb N}\subseteq\mathcal D(A)$ with $$\left\|f_n-f\right\|_\infty\xrightarrow{n\to\infty}0\tag1$$ and $$\left\|Af_n-\overline Af\right\|_\infty\xrightarrow{n\to\infty}0\tag2.$$ Now, it can be shown that there is a $(x_n)_{n\in\mathbb N}\subseteq E$ with $$|f_n(x_n)|=\left\|f_n\right\|_\infty\;\;\;\text{for all }n\in\mathbb N.\tag3$$ Note that $$\tilde f_n:=f_n\operatorname{sgn}f_n(x_n)\in\mathcal D(A)$$ and $$\tilde f_n(x_n)=\left\|f_n\right\|_\infty\ge0$$ for all $n\in\mathbb N$. Moreover, $$\left|\tilde f_n(x_n)-f(x_0)\right|=\left|\left\|f_n\right\|_\infty-\left\|f\right\|_\infty\right|\le\left\|f_n-f\right\|_\infty\xrightarrow{n\to\infty}0\tag4.$$
How can we conclude? Maybe we need to assume the contrary ...
$^1$ If $f\in\mathcal D(A)$ and $x_0\in E$ with $\max(0,f(x))\le f(x_0)$ for all $x\in E$, then $(Af)(x_0)\le0$.