Proving the equality in weak maximum principle of elliptic problems This one is probably simple, but I just can't prove the result.
Suppose that $\mathop {\max }\limits_{x \in \overline \Omega  } u\left( x \right) \leqslant \mathop {\max }\limits_{x \in \partial \Omega } {u^ + }\left( x \right)$ and $\mathop {\min }\limits_{x \in \overline \Omega  } u\left( x \right) \geqslant \mathop {\min }\limits_{x \in \partial \Omega } {u^ - }\left( x \right)$, where $\overline \Omega  $ is closure of $\Omega $ and ${\partial \Omega }$ is boundary of $\Omega $, ${u^ + } = \max \left\{ {u,0} \right\}$ and  ${u^ - } = \min \left\{ {u,0} \right\}$ (notice that $\left| u \right| = {u^ + } - {u^ - }$).
How to show that $\mathop {\max }\limits_{x \in \overline \Omega  } \left| {u\left( x \right)} \right| = \mathop {\max }\limits_{x \in \partial \Omega } \left| {u\left( x \right)} \right|$?
So far, I've got
$\mathop {\max }\limits_{x \in \overline \Omega  } \left| {u\left( x \right)} \right| = \mathop {\max }\limits_{x \in \overline \Omega  } \left( {{u^ + }\left( x \right) - {u^ - }\left( x \right)} \right) \leqslant \mathop {\max }\limits_{x \in \overline \Omega  } {u^ + }\left( x \right) + \mathop {\max }\limits_{x \in \overline \Omega  } \left( { - {u^ - }\left( x \right)} \right) \leqslant \mathop {\max }\limits_{x \in \overline \Omega  } {u^ + }\left( x \right) - \mathop {\min }\limits_{x \in \overline \Omega  } {u^ - }\left( x \right)$ 
and
$\mathop {\max }\limits_{x \in \partial \Omega } \left| {u\left( x \right)} \right| = \mathop {\max }\limits_{x \in \partial \Omega } \left( {{u^ + }\left( x \right) - {u^ - }\left( x \right)} \right) \leqslant \mathop {\max }\limits_{x \in \partial \Omega } {u^ + }\left( x \right) - \mathop {\min }\limits_{x \in \partial \Omega } {u^ - }\left( x \right)$
Edit: $u \in {C^2}\left( \Omega  \right) \cap C\left( {\overline \Omega  } \right)$, although I don't see how that helps. $Lu=0$, which gives us $\mathop {\max }\limits_{x \in \overline \Omega  } u\left( x \right) \leqslant \mathop {\max }\limits_{x \in \partial \Omega } {u^ + }\left( x \right)$ and $\mathop {\min }\limits_{x \in \overline \Omega  } u\left( x \right) \geqslant \mathop {\min }\limits_{x \in \partial \Omega } {u^ - }\left( x \right)$.
(Renardy, Rogers, An introduction to partial differential equations, p 103)
Edit 2: Come on, this should be super easy, the author didn't even comment on how the equality follows from those two.
 A: $\left. \begin{gathered}
  \mathop {\max }\limits_{x \in \overline \Omega  } u\left( x \right) \leqslant \mathop {\max }\limits_{x \in \partial \Omega } {u^ + }\left( x \right)\mathop  \leqslant \limits^{{u^ + } \leqslant \left| u \right|} \mathop {\max }\limits_{x \in \partial \Omega } \left| {u\left( x \right)} \right| \\
  \mathop {\max }\limits_{x \in \overline \Omega  } \left( { - u\left( x \right)} \right) \leqslant \mathop {\max }\limits_{x \in \partial \Omega }  - {u^ - }\left( x \right)\mathop  \leqslant \limits^{ - {u^ - } \leqslant \left| u \right|} \mathop {\max }\limits_{x \in \partial \Omega } \left| {u\left( x \right)} \right| \\ 
\end{gathered}  \right\} \Rightarrow \mathop {\max }\limits_{x \in \overline \Omega  } \left| {u\left( x \right)} \right| \leqslant \mathop {\max }\limits_{x \in \partial \Omega } \left| {u\left( x \right)} \right|$.
On the other hand, $\partial \Omega  \subseteq \overline \Omega   \Rightarrow \mathop {\max }\limits_{x \in \partial \Omega } \left| {u\left( x \right)} \right| \leqslant \mathop {\max }\limits_{x \in \overline \Omega  } \left| {u\left( x \right)} \right|$
