Show that if $u$ is non-negative and harmonic then $|\nabla u(x_0)| \leq \frac{2}{R}u(x_0)$ Suppose $x_0 \in \mathbb{R}^2$ and $u \in C^2(\bar{B}_R(x_0))$ is a non-negative harmonic function. I want to show that $$|\nabla u(x_0)| \leq \frac{2}{R}u(x_0)$$
I know that since $u$ is harmonic it satisfied the mean value property. I'm given two mean value properties (which I earlier proved are equivalent):
$$u(x)=\frac{1}{2\pi r}\int_{B_r(x)}u(y) \, dS(y)$$
and $$u(x)=\frac{1}{\pi r^2}\int_{B_r(x)}u(y) \, dy$$
I think using these is probably the way to get the result, I'm not sure how though.
 A: Rotate your coordinates so that $\nabla u(x_0)$ is in the $e_1$ direction. Note that since $u$ is harmonic, its partial derivative $\partial_1 u$ is too; so the mean value property (ball version) plus the divergence theorem give
$$ \partial_1u(x_0) = \frac{1}{\pi R^2}\int_{B_R} \partial_1u = \frac{1}{\pi R^2}\int_{B_R} {\rm div}(u e_1) = \frac{1}{\pi R^2}\int_{\partial B_R} u e_1 \cdot n.$$
Estimating $|e_1 \cdot n | \le 1$ and applying the mean value property (this time the sphere version) along with the fact $u \ge 0$ we get
$$ |\partial_1 u(x_0)| \le \frac{1}{\pi R^2}\int_{\partial B_R} |u|=\frac{2 \pi R}{\pi R^2}u(x_0).$$
The LHS here is exactly $|\nabla u(x_0)|$ by our choice of coordinates, and the RHS is exactly $\frac 2 R u(x_0)$.
A: It is not hard to prove that if $u \geq 0$ is harmonic on $B(x_0,R)$, then for any $x \in B(x_0,R)$ we have $$u(x) \leq \frac{R+||x-x_0||}{R-||x-x_0||}u(x_0).$$ This you can do by using the mean value property you mentioned above and very simple triangle and reverse triangle inequality of the norm. But then $$u(x)-u(x_0) \leq \left(\frac{R+||x-x_0||}{R-||x-x_0||}-1 \right)u(x_0)=\frac{2||x-x_0||}{R-||x-x_0||}u(x_0).$$ Now, take the norms and divide on both sides by $||x-x_0||$ and let $x \to x_0.$
