Disclaimer: I am attempting to make progress on the following problem that is part a homework assignment. Thus, I am hoping for hints/suggestions and not necessarily the full solution.
Let the half-ball be $\Omega(r)=\{x\in\mathbb{R}^n: |x|< r, \, x_n>0\}$. Let $u$ be $C^2$ on $\Omega$ and $C^0$ on $\overline{\Omega}$ and harmonic inside $\Omega$: $$(\Delta u)_\Omega=0.$$ We also have that $u(x)=0$ for $x_n=0$. I would like to show that for each $n\ge 1$, there exists $0<c(n)<1$ such that for all $u$ satisfying the above conditions, $$\max_{\overline{\Omega}(r/2)}u\le c(n) \max_{\overline{\Omega}(r)} u$$
Here is my attempt at the solution, but I'm not sure if it's right. Assume by contradiction that the statement is false. That is, there exists $N\ge 1$ such that for all $0<c<1$,
$$\max_{\overline{\Omega}(r/2)}u > c\max_{\overline{\Omega}(r)} u$$
Now, take a sequence $0<c_k\to 1$ as $k\to\infty$. Let $\varepsilon>0$ be arbitrary. Choose $M\ge 1$ such that for all $k\ge M$, $1-\epsilon<c_k<1+\epsilon$. Then we have that for all $k\ge M$,
\begin{aligned}
\max_{\overline{\Omega}(r/2)}u > c_k \max_{\overline{\Omega}(r)} u > (1-\varepsilon)\max_{\overline{\Omega}(r)} u.
\end{aligned}
I would like to pass to the limit $\varepsilon\to 0$ and obtain the clear contradiction
$$\max_{\overline{\Omega}(r/2)}u > \max_{\overline{\Omega}(r)} u.$$
Is it possible to do so?
This approach appears off the right track. As I understand it at a glance, the constant function $f = 1$ seems to present difficulties to the approach: is a harmonic function satisfying that $\text{max}_{\overline{\Omega}_{r/2}} f(x) > c\text{max}_{\overline{\Omega}_r} f(x)$ for all $0 < c < 1$, but not for $1$.
Here's a hint: try to use the mean value property of harmonic functions and the geometry of the domains.