复版本严格极大值原理证明中的两处技术疑问
Strict Maximum Principle - Complex Version (Gamelin; p88)
Let $h$ be a bounded complex-valued harmonic function on a domain $D$. If $|h(z)| \le M$ for all $z \in D$, and $|h(z_0)| = M$ for some $z_0 \in D$, then $h(z)$ is constant on $D$.
First I will copy the proof (then explain where I am confused)
Proof
We replace $h(z)$ by $\lambda h(z)$ for an appropriate unimodular constant $\lambda$, and we can assume $h(z_0) = M$. Let $u(z) = \Re h(z)$. Then $u(z)$ is a harmonic function on $D$ that attains its maximum at $z_0$. By the strict maximum principle for real-valued harmonic functions, $u(z) = M$ for all $z \in D$. Since $|h(z)| \le M$ and $\Re h(z) = M$ we must have $\Im h(z) = 0$ for all $z \in D$ hence $h(z)$ is constant.
My Questions
- If I understand correctly, we "redefine" $h$ as $\lambda h(z)$, so essentially we have $h^* = \lambda h(z)$ for an appropriate $\lambda \in \mathbb{C}$. If that's the case, then shouldn't $u(z) = \Re h^*(z)$, and consequently the maximum scaled by $|\lambda|$??
- What if the range of $h$ doesn't include the real-axis, so that we cannot scale $h(z_0)$ to a real value, is that the case?
My Explanation & Answers
Let's work through this step by step to clear up your confusion:
For Question 1
You’re totally right that we’re effectively working with a new function $h^* = \lambda h$, but the critical detail you might have glossed over is that $\lambda$ is a unimodular constant—meaning $|\lambda| = 1$. That’s the linchpin here!
When we pick $\lambda$, we choose it specifically to rotate $h(z_0)$ onto the positive real axis (so $\lambda h(z_0) = M$). Since $|\lambda| = 1$, multiplying by it doesn’t change the magnitude of any value of $h$: $|\lambda h(z)| = |\lambda||h(z)| = 1 \cdot |h(z)| \le M$, so the original bound still holds perfectly for $h^*$.
Then yes, $u(z) = \Re(h^(z))$, and since $h^(z_0) = M$, $u(z_0) = M$. For any other $z \in D$, $u(z) = \Re(\lambda h(z)) \le |\lambda h(z)| = |h(z)| \le M$, so $u$ has an upper bound of $M$ and hits that maximum at $z_0$. Applying the real strict max principle gives $u(z) = M$ everywhere. The "scaling" concern doesn’t apply here because we’re only rotating, not stretching or shrinking the function’s magnitude.
For Question 2
This scenario is impossible—let me explain why! Since $|h(z_0)| = M$, $h(z_0)$ is just some point on the circle of radius $M$ centered at the origin in the complex plane. A unimodular constant $\lambda$ is exactly a complex number on the unit circle, so multiplying $h(z_0)$ by $\lambda$ is just rotating that point around the origin.
No matter where $h(z_0)$ sits on that radius-$M$ circle, we can always find such a $\lambda$ to rotate it to the positive real axis. For example, if $h(z_0) = M e^{i\theta}$ (in polar form), we just take $\lambda = e^{-i\theta}$—this is unimodular, and $\lambda h(z_0) = M e^{i\theta} e^{-i\theta} = M$.
We don’t need the entire range of $h$ to include the real axis—we’re only rotating the entire function so that the specific point $h(z_0)$ lands on the real axis. All other values of $h$ get rotated too, but since we’re only rotating (not scaling), the magnitude bound $|h(z)| \le M$ stays exactly intact for the rotated function.
备注：内容来源于stack exchange，提问作者dp1221
