I would like some help with the following problem (Gilbarg/Trudinger, Ex. 2.13):
Let $u$ be harmonic in $\Omega \subset \mathbb R^n$. Use the argument leading to (2.31) to prove the interior gradient bound, $|Du(x_0)|\le \frac{n}{d_0}[\sup_{\Omega} u - u(x_0)], \quad d_0 = \mathrm{dist}(x_0, \partial \Omega)$
The argument mentioned in the exercise is the following: Since $u$ is harmonic, it follows that also $Du$ is harmonic. Hence if $B_R = B_R(x_0)\subset \subset \Omega$ we obtain
$Du(x_0) = \frac{1}{\omega_n R^n} \int_{B_R} Du = \frac{1}{\omega_n R^n} \int_{\partial B_R} u\nu = \frac{n}{R} \frac{1}{n\omega_n R^{n-1}}\int_{\partial B_R} u\nu$
where $\nu$ is the outward pointing unit normal vector (and the integration on the RHS is supposed to happen componentwise). Subtituting $u \mapsto u-u(x_0)$, this shows $Du(x_0) = \frac{n}{R} \frac{1}{n\omega_n R^{n-1}}\int_{\partial B_R} [u-u(x_0)]\nu$ But I don't see how to go on from here...
Of course, if $\sup_{B_R(x_0)} |u-u(x_0)| = \sup_{B_R(x_0)} [u-u(x_0)]$, then there is no problem, since then one can just take norms on both sides. I don't think this needs to be the case in general, however. (integrating a suitable function against the Poisson kernel.)
Your help would be appreciated, thanks!