# Bound of gradient in $0$ of postive harmonic function.

Let $K(0,\rho)\subseteq \mathbb{C}$ be a domain and let $h: K(0,\rho)\to \mathbb{R}$ be a positive harmonic function. Show that $$|\nabla h(0)| \leq \frac{2}{\rho} h(0)$$

I feel like this problem should not be too difficult, but so far my approaches have not been fruitful, so I would appreciate a hint in the right direction. Thanks!

Let $$\omega_n$$ be the area of $$\{ x \in \mathbb{R}^{n-1} : \| x \| = 1 \}$$. Then we have
\begin{align} D_i u(x_0) &= \frac{n}{\omega_n \rho^n}\int_{B_{\rho}(0)} D_iu(y)dy \\ &= \frac{n}{\omega_n \rho^n} \int_{\partial B_{\rho}(0)} u(y) \cdot n(y) dS(y) \end{align}
Now take absolute values on both sides, pull them inside the integral, and use the fact that $$n$$ is a unit normal and you get \begin{align} |D_iu(x_0)| &\leq \frac{n}{\rho}\cdot \frac{1}{\omega_n \rho^{n-1}}\int_{\partial B_{\rho}(0)} |u(y)| dS(y) \\ &= \frac{n}{\rho}\cdot \frac{1}{\omega_n \rho^{n-1}}\int_{\partial B_{\rho}(0)} u(y) dS(y) \end{align} where the second equality comes from the fact that $$u$$ is non-negative. Now simply apply the mean-value property for harmonic functions to obtain the result. Note that your domain is $$\mathbb{C}$$, so $$n=2$$.