Problem is
Give a direct proof that if $u \in C^{2}(U) \cap C(\overline{U})$ is harmonic within a bounded open set $U$, then $\max_{\overline{U}} u =\max_{\partial U} u$.
What I think is that;
Let $u^{\epsilon} = u+ \epsilon |x|^{2}$, where $\epsilon >0 $. Then $\bigtriangleup u^{\epsilon} = \bigtriangleup u + 2\epsilon |x| = 2\epsilon |x| \geq 0$ since $u$ is harmonic. Equality holds if $x=0$. Suppose $x_{0} \in U$ is maximum point of $u^{\epsilon}(\overline{U})$. Then $\bigtriangleup u^{\epsilon}(x_0) = 2 \epsilon |x_0| \leq 0$.
I want to show that $x_0 \neq 0$, so this is contradiction, but I don't know how to do that. Could anyone have an idea?
If the maximum is attained inside the domain, at point $x_0$, then the Hessian $H$ of $u$ at $x_0$ is symmetric definite negative. You first have that $$Tr(H) = \Delta u$$ and because H is definite negative, $$Tr(H) < 0$$ This implies a contradiction with $u$ is harmonic.