Given the holomorphic maximum modulus principle, prove Hopf's lemma

Question

To smooth out my lecture notes, I'm looking for a derivation of Hopf's lemma for harmonic functions $u \colon D \subset \mathbb{R}^2 \to \mathbb{R}$ from the maximum modulus principle (and mean value theorem) for holomorphic functions.

Note: I cannot use Harnack's inequality

Answer 1

Set-up: let $a$ be a boundary point where $u$ attains its maximum, then pick a disk $\Delta$ contained in the domain $D$ such that $ \partial D\cap \partial \Delta=\{a\}$ (this is possible if the boundary is $C^2$ smooth). By rotation and scaling we may assume $\Delta $ is the unit disk and $a=1$. We can also assume the maximum of $u$ is $0$.

Main idea: use a function that perturbs $u$ away from the boundary. One possibility is $$v(z) = \arg\frac{z+i}{1+iz}$$ Indeed, $f(z)=\frac{z+i}{1+iz}$ maps $\Delta$ onto the upper half plane so that the right semi-circle is mapped to the positive real axis. So, $v=0$ on the right semi-circle and $v=\pi$ on the left semi-circle.

For sufficiently small $\epsilon$, the function $u_\epsilon = u+\epsilon v$ satisfies $u_\epsilon\le 0$ on $\partial \Delta$, hence also inside $\Delta$. Therefore, the inward derivative of $u_\epsilon$ at $1$ is non-positive. But the inward derivative of $v$ at $1$ is strictly positive, since $f$ maps $[-1,1]$ onto upper half-circle traced with positive speed. Thus, the derivative of $u$ is strictly negative. $\qquad\Box$

It's possible to rewrite the proof by using the maximum principle for holomorphic functions. It would feature the product of

• $\exp(u+i\tilde u)$, where $\tilde u$ is the harmonic conjugate of $u$.
• $\exp(i \epsilon \log f(z))=f(z)^{i\epsilon}$

I don't think this makes the proof any easier.