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?
Best Answer
Let $u_{\varepsilon} = u + \varepsilon |x|^2, \forall \varepsilon > 0$. Then $\Delta u_{\varepsilon} = 2 \epsilon > 0$ since $u$ is harmonic. If $u_{\varepsilon}$ attained a local maximum at an interior point then $\Delta u \le 0$. Thus, $u_{\varepsilon}$ attains its maximum on the boundary and has no maximum at an interior point. If $|x| \le r, \forall x \in U$, it follow that $$ \sup_U u \le \sup_U u_{\varepsilon} \le \sup_{\partial U} u_{\varepsilon} \le \sup_{\partial U} (u + \varepsilon r^2) . $$ Let $\varepsilon \to 0$ we get $\sup_U u \le \sup_{\partial U} u$ and $\max_{\partial U} = \max_{\bar{U}} u$.