2
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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 $n=2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.