Let $D$ a bounded coherent space in $2$ or $3$ dimensions.
Let $u: \overline{D} \rightarrow \mathbb{R}$ a function that is continuous at $\overline{D}$ and harmonic at $D$ $\Rightarrow C^2(D)$
Then, both the maximum and minimum of $u$ at $\overline{D}$ are taken at the boundary $\partial{D}$.
$$\max_{(x,y) \in \overline{D}} u(x,y)= \max_{(x,y) \in \partial{D}} u(x,y)$$ $$$$ The proof of this theorem is the following:
Since $u$ I harmonic $\Delta u=0 \Rightarrow \Delta (-u)=0$
It's sufficient to do only the proof for the maximum.
We define the function $v=u(x,y)+\epsilon (x^2+y^2) \in C^0(\overline{D}), C^2(D)$
$\Delta v=4 \epsilon >0$, for $\epsilon >0$
We suppose that $v$ achieves its maximum at $(x_0,y_0) \in D$: $v_x(x_0,y_0)=v_y(x_0,y_0)=0, v_{xx}(x_0,y_0)<0, v_{yy}(x_0,y_0)<0$ $\Rightarrow \Delta v=v_{xx}+y_{yy}<0$
We have a contradiction, so $v$ achieves its maximum at one point at the boundary $(x_0,y_0) \in \partial{D}$.
$u(x,y) \leq v(x,y) \leq v(x_0, y_0)= u(x_0, y_0)+ \epsilon (x_0^2+y_0^2)$
$x^2+y^2<l^2, \forall (x,y) \in \overline{D}$
$\Rightarrow u(x,y) \leq u(x_0,y_0)+\epsilon l^2$
$(x_0,y_0) \in \overline{D} \Rightarrow$ $$\max_{(x,y) \in \overline{D}} u(x,y) \leq \max_{(x,y) \in \partial{D}} u(x,y) + \epsilon l^2$$ Taking the limit $\epsilon \rightarrow 0$ we conclude to the relation : $$\max_{(x,y) \in \overline{D}} u(x,y) = \max_{(x,y) \in \partial{D}} u(x,y)$$
$$$$ I haven't understood how we get from the relation: $\max\limits_{(x,y) \in \overline{D}} u(x,y) \leq \max\limits_{(x,y) \in \partial{D}} u(x,y) + \epsilon l^2$ to the relation: $\max\limits_{(x,y) \in \overline{D}} u(x,y) = \max\limits_{(x,y) \in \partial{D}} u(x,y)$. Taking the limit $\epsilon \rightarrow 0$ we get: $\max\limits_{(x,y) \in \overline{D}} u(x,y) \leq \max\limits_{(x,y) \in \partial{D}} u(x,y)$. But how do we get the equality???
$\max\limits_{(x,y)\in\bar D}u(x,y)\leq\max\limits_{(x,y)\in\partial D}u(x,y)$ says that the maximum value of $u(x,y)$ on $\bar D$ can't be greater than the maximum value of $u(x,y)$ on $\partial D$. Now since $\partial D\subseteq\bar D$ this automatically means that the maximum is attained somewhere on $\partial D$. This is why you can write $\max\limits_{(x,y)\in\bar D}u(x,y)=\max\limits_{(x,y)\in\partial D}u(x,y)$. If you're not yet fully convinced by this intuitive argument, you can prove it by contradiction.