Limit commutes with $\min$.

Question

In my functional analysis course we are studying the weak maximum principle and there is something in a proof that I don't understand. We want to show that, for $u \in C^{2}(\Omega) \cap C^0(\overline{\Omega})$, $-\Delta u \ge 0$ implies that $$\min_{\overline{\Omega}} u = \min_{\partial \Omega} u.$$ First we show the result for $u$ such that $-\Delta u > 0$ in $\Omega$ (this part is not too hard) and then, for $u$ such that $-\Delta u \ge 0$, we set $$v_\varepsilon = u + \varepsilon e^{x_1},$$ so that $-\Delta v_\varepsilon > 0$ in $\Omega$. Therefore, by the first part, we deduce that $$\min_{\overline{\Omega}} v_\varepsilon = \min_{\partial \Omega} v_\varepsilon,$$ for all $\varepsilon.$ After that, my teacher said that we just have to take the limit for $\varepsilon \to 0$ to deduce the result for $u$, but it is not very clear to me that the limit $\varepsilon \to 0$ can commutes with the $\min$. Maybe it will work by using $\liminf$ and $\limsup$ but I am not able to write a rigorous argument.. Any help ?

Answer 1

For all $x\in \overline{\Omega}$,

$$u(x)-v_\epsilon(x) < \epsilon e^{x_1},$$

So the difference of the minima is of order $\epsilon$ ( the auxiliary function must have a positive supremum on every compact subset of $\Omega$)