# Bound on gradient of Harmonic functions

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 $$0) and for all $$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) 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?

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
$$\frac{-2}{R}u(z_0)\leq Du_{v=e^{it}}(z_0)\leq \frac{2}{R}u(z_0)\Rightarrow |\nabla u(z_0)|\le \frac{2}{R}|u(z_0)|\le \frac{2}{R}M$$.
• 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? – Mark Apr 6 '19 at 11:37
• It is ture for nonnegative functions. But the set $\{z:u(z)<0\}\cap G$ is open so you can apply the above to $-u$. – Matematleta Apr 6 '19 at 15:30