Let $u$ be a harmonic function on the unit ball in $\mathbb{R}^n$. Show that
$$\sup_{B_{1/2}} \lvert \nabla u \rvert \leq C(n) \sup_{\partial B_1} \lvert u \rvert$$
More generally, show that
$$\sup_{B_{1/2}} \left\lvert \frac{\partial^m u}{\partial x_{i_1} \ldots \partial x_{i_m}} \right\rvert \leq C(n, m) \sup_{\partial B_{1}} \lvert u \rvert$$
for all indicies $i_1, \ldots i_m \in \{1, \ldots, n\}$. Use the solution formula for the Dirichlet problem on the unit ball.
Attempt at solution
The solution formula for the Dirichlet problem on the unit ball is
\begin{equation} u(x) = \frac{1 - \lvert x \rvert^2}{\sigma_{n-1}} \int_{\partial B_1} \lvert x - y \rvert^{-n} u(y) \mathrm{d}y \end{equation}
Taking the gradient yields
\begin{equation} \begin{split} \nabla u(x) &= \nabla \frac{1 - \lvert x \rvert^2}{\sigma_{n-1}} \int_{\partial B_1} \lvert x - y \rvert^{-n} u(y) \mathrm{d}y \\ &= \left( \nabla \frac{1 - \lvert x \rvert^2}{\sigma_{n-1}} \right) \int_{\partial B_1} \lvert x - y \rvert^{-n} u(y) \mathrm{d}y + \frac{1 - \lvert x \rvert^2}{\sigma_{n-1}} \nabla \int_{\partial B_1} \lvert x - y \rvert^{-n} u(y) \mathrm{d}y \\ &= -\frac{\nabla \lvert x \rvert^2}{\sigma_{n-1}} \int_{\partial B_1} \lvert x - y \rvert^{-n} u(y) \mathrm{d}y + \frac{1 - \lvert x \rvert^2}{\sigma_{n-1}} \int_{\partial B_1} \nabla \lvert x - y \rvert^{-n} u(y) \mathrm{d}y \\ &= -\frac{2 \lvert x \rvert \nabla \lvert x \rvert}{\sigma_{n-1}} \int_{\partial B_1} \lvert x - y \rvert^{-n} u(y) \mathrm{d}y + \frac{1 - \lvert x \rvert^2}{\sigma_{n-1}} \int_{\partial B_1} -n \lvert x - y \rvert^{-n - 1} \nabla \lvert x - y \rvert u(y) \mathrm{d}y \\ &= -\frac{2 x \nabla x}{\sigma_{n-1}} \int_{\partial B_1} \lvert x - y \rvert^{-n} u(y) \mathrm{d}y + \frac{1 - \lvert x \rvert^2}{\sigma_{n-1}} \int_{\partial B_1} -n \lvert x - y \rvert^{-n - 2} (x - y) \nabla (x - y) u(y) \mathrm{d}y \\ &= -\frac{2 x}{\sigma_{n-1}} \int_{\partial B_1} \lvert x - y \rvert^{-n} u(y) \mathrm{d}y + \frac{1 - \lvert x \rvert^2}{\sigma_{n-1}} \int_{\partial B_1} -n \lvert x - y \rvert^{-n - 2} (x - y) u(y) \mathrm{d}y \\ &= -\frac{2 x}{\sigma_{n-1}} \int_{\partial B_1} \lvert x - y \rvert^{-n} u(y) \mathrm{d}y + \frac{1 - \lvert x \rvert^2}{\sigma_{n-1}} \int_{\partial B_1} -n \lvert x - y \rvert^{-n} \frac{x - y}{\lvert x - y \rvert^2} u(y) \mathrm{d}y \\ &= -\frac{1}{\sigma_{n-1}} \int_{\partial B_1} 2x \lvert x - y \rvert^{-n} u(y) \mathrm{d}y - \frac{1}{\sigma_{n-1}} \int_{\partial B_1} (1 - \lvert x \rvert^2) n \lvert x - y \rvert^{-n} \frac{x - y}{\lvert x - y \rvert^2} u(y) \mathrm{d}y \\ &= \frac{1}{\sigma_{n-1}} \int_{\partial B_1} \left( -2x - n \frac{x - y}{\lvert x - y \rvert^2} \right) \lvert x - y \rvert^{-n} u(y) \mathrm{d}y \\ &= -\frac{1}{\sigma_{n-1}} \int_{\partial B_1} \left( 2x + n \frac{x - y}{\lvert x - y \rvert^2} \right) \lvert x - y \rvert^{-n} u(y) \mathrm{d}y \\ \end{split} \end{equation}
Hence we have
\begin{equation} \begin{split} \lvert \nabla u(x) \rvert &= \frac{1}{\sigma_{n-1}} \left\lvert \int_{\partial B_1} \left( 2x + n \frac{x - y}{\lvert x - y \rvert^2} \right) \lvert x - y \rvert^{-n} u(y) \mathrm{d}y \right\rvert \\ &\leq \frac{1}{\sigma_{n-1}} \int_{\partial B_1} \left\lvert \left( 2x + n \frac{x - y}{\lvert x - y \rvert^2} \right) \lvert x - y \rvert^{-n} u(y) \mathrm{d}y \right\rvert \\ &\leq \frac{1}{\sigma_{n-1}} \int_{\partial B_1} \left\lvert 2x + n \frac{x - y}{\lvert x - y \rvert^2} \right\rvert \lvert x - y \rvert^{-n} \lvert u(y) \rvert \mathrm{d}y \\ &\leq \frac{1}{\sigma_{n-1}} \int_{\partial B_1} \left( \lvert 2x \rvert + \left\lvert n \frac{x - y}{\lvert x - y \rvert^2} \right\rvert \right) \lvert x - y \rvert^{-n} \lvert u(y) \rvert \mathrm{d}y \\ &\leq \frac{1}{\sigma_{n-1}} \int_{\partial B_1} \left( 2 \lvert x \rvert + \frac{n}{\lvert x - y \rvert} \right) \lvert x - y \rvert^{-n} \lvert u(y) \rvert \mathrm{d}y \\ &\leq \frac{2 \lvert x \rvert}{\sigma_{n-1}} \int_{\partial B_1} \lvert x - y \rvert^{-n} \lvert u(y) \rvert \mathrm{d}y + \frac{n}{\sigma_{n-1}} \int_{\partial B_1} \lvert x - y \rvert^{-n-1} \lvert u(y) \rvert \mathrm{d}y \end{split} \end{equation}
On the other hand
\begin{equation} \begin{split} \lvert u(x) \rvert &= \left\lvert \frac{1 - \lvert x \rvert^2}{\sigma_{n-1}} \int_{\partial B_1} \lvert x - y \rvert^{-n} u(y) \mathrm{d}y \right\rvert \\ &= \frac{1}{\sigma_{n-1}} \lvert 1 - \lvert x \rvert^2 \rvert \left\lvert \int_{\partial B_1} \lvert x - y \rvert^{-n} u(y) \mathrm{d}y \right\rvert \\ &= \frac{1}{\sigma_{n-1}} ( 1 - \lvert x \rvert^2 ) \left\lvert \int_{\partial B_1} \lvert x - y \rvert^{-n} u(y) \mathrm{d}y \right\rvert \end{split} \end{equation}
At this point, I'm stuck. Does anyone have any suggestions as to how to proceed?
If you restrict to $x \in B_{1/2}$ then in the integrals on $\partial B_1$ you have $|y|=1$ and you can find nice bounds for $|x||x-y|^{-n}$ and $|x-y|^{n-1}$.