Evans PDE chapter 2 problem 4 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 &gt;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?
Let $u_{\varepsilon} = u + \varepsilon |x|^2, \forall \varepsilon &gt; 0$. Then $\Delta u_{\varepsilon} = 2 \epsilon &gt; 0$ 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 $\max_{\partial U} = \max_{\bar{U}} u$.
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,
$$Tr(H) &lt; 0$$
This implies a contradiction with $u$ is harmonic.