I am trying to identify for which $\lambda$ so that I can obtain the Maximum principle, that is, $$\sup_{\overline{\Omega}}u\leq \sup_{\partial \Omega}u$$ for equation $-\triangle u = \lambda u$, where the domain $\Omega$ is open bounded.
I tried to mimic the prove in Evans book. Indeed, by assuming that $\lambda\leq 0$, I have $$-\triangle u \leq 0$$ in the set $V:=\{x,\,\,u(x)>0\}$. Then I can apply the Maximum principle for $u$ in $V$ and obtain that $$\sup_{\overline{V}}u = \sup_{\partial V}u = \sup_{\partial \Omega}u^+\Rightarrow \sup_{\overline{\Omega}}u \leq \sup_{\partial \Omega}u^+$$ However, I can not replace $u^+$ by $u$ above. Any hints will be really welcome!
The obstacle that you encountered is real. Consider one-dimensional situation: for every $\lambda<0$, the function $u(x) = - \cosh (\sqrt{-\lambda} x)$ satisfies $-u''=\lambda u$ and $$\sup_{[-1,1]} u = -1 > -\cosh (\sqrt{-\lambda}) = \max(u(1),u(-1))$$
For eigenfunctions with $\lambda<0$ we have maximum principle for $u^+$, $u^-$, and consequently for $|u|$; but not for $u$ itself.
For eigenfunctions with $\lambda>0$ there is no maximum principle at all.