I have been trying to solve the following
Let $f$ be continuous in $\overline{B}_{R}$. Suppose $u\in C^{2}(B_{R})\cap C(\overline{B}_R)$ satisfies $\Delta u = f$ in $B_R$. Prove that $|\nabla u(0)|\leq \dfrac{n}{R}\max_{\partial B_{R}}|u|+\dfrac{R}{2}\max_{B_{R}}|f|$.
We have the following hint: in $B_{R}^{+}$, set $$ v(x', x_n)=\dfrac{1}{2}(u(x', x_n)-u(x', -x_n)) $$ and consider the auxiliar function $$ w(x', x_n)=A|x'|^2+Bx_n +Cx_{n}^{2}. $$ Use the comparison principle to estimate $v$ in $B_{R}^{+}$ and then derive a bound for $v_{x_n}(0)$.
Some of conclusions: 1) $v$ is harmonic and $\Delta w= 2A(n-1)+2C$.
2) From (1), I can choose $A, B, C$ such that $\Delta v \geq \Delta w$. But still I can't see any comparison between $v$ and $w$ on the boundary to apply the principle.
Besides that, I am not been able to conclude the exercise at all, even if I suppose that the principle works, I found that $|v|\leq |w|\leq 2nR$ (for $A=1, B=1, c=-n$), but I don't see how to estimate $v_{x_n}(0)$.
Thanks in advance for any help.