Let $d\in\mathbb N$, $\Omega\subseteq\mathbb R^d$ be bounded and open and $u:\overline\Omega\to\mathbb R$ be continuous. Suppose $M\in\mathbb R$ with $$M>u(x)\;\;\;\text{for all }x\in\Omega\tag1$$ and $x_0\in\partial\Omega$ with $$u(x_0)=M.$$
Is it possible that there is a $\tilde x_0\in\partial\Omega$ with $u(\tilde x_0)>M$?
Remark: This question arose as I was reading Evans proof othe maximum principle, where $\Omega=B_r(0)$ for some $r>0$. Instead of equation (17) therein, it is only clear to me that $\max_{\partial B_r(0)}u\ge u(x)+\varepsilon v(x)$ for all $x\in\partial B_r(0)$.
Best Answer
If $u$ is continuous in $\overline \Omega$ with $u(z) < M$ for all $z \in \Omega$ then $u(z) \le M$ for all $z \in \overline \Omega$.
It is therefore not possible that $u(\tilde x_0)>M$ for some $\tilde x_0\in\partial\Omega$.