2
$\begingroup$

Let $G\subseteq\mathbb{C}$ be a domain and assume $u:G\to\mathbb{R}$ is a harmonic function such that $|u(z)|\leq M$ for all $z\in G$. Show that $|\nabla u(z)|\leq\frac{2M}{r}$ for $0<r<dist(z,\partial G$) and for all $z\in G$. ($\nabla u$ is the gradient of $u$).

Attempt: As we know, there is some harmonic function $v:G\to\mathbb{R}$ such that $f=u+iv$ is holomorphic. I thought it might help to use Cauchy integral formula. For a given $z_0\in G$ and $0<r<dist(z_0,\partial G$) we have:

$f'(z_0)=\frac{1}{2\pi i}\int_{B_r(z_0)}\frac{f(\xi)}{(\xi-z_0)^2} d\xi=\frac{1}{2\pi}\int_0^{2\pi}\frac{f(z_0+re^{it})}{re^{it}}dt=\frac{1}{2\pi r}\int_0^{2\pi}f(z_0+re^{it})(cost-isint)dt$

Alright, now using the fact that $f'(z_0)=u_x(z_0)+iv_x(z_0)$ I tried to compare real parts. What I got is the following:

$u_x(z_0)=\frac{1}{2\pi r}\int_ 0^{2\pi} u(z_0+re^{it})cost+v(z_0+re^{it})sint dt$

I hoped I can get a bound on both $u_x$ and $u_y$ and from there get the required bound on the gradient. However as you can see I got that $u_x$ is dependent on the values of $v$ and this is a function I cannot bound since I don't know anything about it. So I'm stuck. Any ideas?

$\endgroup$
1
$\begingroup$

Suppose $z_0\in G$. let $ |z-z_0| = r < R$ for some $R$ such that $B_R(z_0)\subseteq G.$ Applying Harnack's inequality yields

${R-r \over R+r}u(z_0)\leq u(z_0+re^{it})\leq {R+r \over R-r}u(z_0).$ Then,

$\frac{1}{r}\left({R-r \over R+r}-1\right)u(z_0)\leq \frac{u(z_0+re^{it})-u(z_0)}{r} \leq \frac{1}{r}\left({R+r \over R-r}-1\right)u(z_0).$

Letting $r\to 0$ gives

$\frac{-2}{R}u(z_0)\leq Du_{v=e^{it}}(z_0)\leq \frac{2}{R}u(z_0)\Rightarrow |\nabla u(z_0)|\le \frac{2}{R}|u(z_0)|\le \frac{2}{R}M $.

$\endgroup$
  • $\begingroup$ That's an interesting solution. But can we really use Harnack's inequality here? I thought $u$ needs to be non negative for that. Or does it work for any harmonic functions? $\endgroup$ – Mark Apr 6 '19 at 11:37
  • $\begingroup$ It is ture for nonnegative functions. But the set $\{z:u(z)<0\}\cap G$ is open so you can apply the above to $-u$. $\endgroup$ – Matematleta Apr 6 '19 at 15:30

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.