Consider a nonnegative harmonic function $u$ on $B^{+}_{1}(0)= \{x=(x_1,...,x_n)\in B_1(0):x_n>0\}$ satisfying $u=0$ on $\{x_n=0\}\cap B_1(0)$ and $u=1$ on $\{x_n=\frac12\}\cap B_1(0)$. Prove that there exists a constant $C=C(n,\rho)$ such that $$|u|\leq C(n,\rho)\;\;\mathrm{on}\;\; B_{\rho}^+(0),$$where $0<\rho<\frac 12$.
My Idea: Since we only have information on part of the boundary, it's possible that $u$ is not unique. But the boundary condition may yield some compactness on the set of all $u$ which forces it to be bounded. As $u$ is assumed to be nonnegative, Harnack inequality is a reasonable candidate. Also it seems resonable to do some gradient estimates. I do think this problem can be solved by elementary properties of harmonic function, but I just get stuck here.
Any hint or solution is highly appreciated!
Ah, this problem turns out to be a bit artificial. Actually, we just need to notice that every point $x\in B^+_{\frac14}(0)$ is contained in a ball $B_{\frac12}(\overline{x})$, where $\overline{x}\in \{x_n=\frac12\}\cap B_{1}^+(0)$, so we just need to apply Harnack inequality(noticing $u$ is nonnegative): $$u(x)\leq C(n)u(\overline{x})=C(n),$$ where $C(n)$ comes from the Harnack inequalty and only relies on the dimension $n$. Using the same trick, the condition $u(x)=1$ on $\{x_n=\frac12\}$ can even be relaxed to $u(\frac{e_n}{2})=1$ where we just need to apply Harnack inequality twice.