Let $u\in C^2(B_1(0))\cap C^1(\overline{B_1(0)})$, which satisfies $$\triangle u(x)=f(x)\qquad x\in B_1(0)$$ $$u(x)=0\qquad x\in\partial B_1(0)$$ Prove $$\left|\frac{\partial u}{\partial n}(x)\right|\leq\frac{1}{n}\|f\|_{L^\infty(\partial B_1(0))}\qquad x\in\partial B_1(0)$$
I am learning the maximum principle in elliptic equations, and stuck about the question. I don't see the relations between the $\frac{\partial u}{\partial n}$ and the maximum principle. Can someone give some advice?
The correct inequality is $$\left|\frac{\partial u}{\partial n}(x)\right|\leq\frac{1}{n}\|f\|_{L^\infty( B_1(0))} $$ Let $M=\|f\|_{L^\infty( B_1(0))}$ and consider the function $$v(x) = u(x)+\frac{M}{2n}(|x|^2-1)$$ Since $\Delta v = \Delta u+M\ge 0$, $v$ is subharmonic. By the maximum principle, $v\le 0$ in $B_1(0)$. Since $v\equiv 0$ on the boundary, it follows that $$\frac{\partial v}{\partial n}\ge 0$$ which in terms of $u$ means $$\frac{\partial u}{\partial n} \ge -\frac{M}{n}$$
Repeat the same with $v(x) = u(x) - \frac{M}{2n}(|x|^2-1)$.