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 $n=2$.


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