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Let $D$ be an open set in the complex plane and $f(z)$ be a non-constant holomorphic function on D. Then $|f(z)|$ has no local maximum on D. I can follow the proof fine - usually if I don't understand a theorem intuitively beforehand, the proof will offer the insight necessary. Here, however, I can't see the reason for the Maximum Principle to hold - or perhaps I was just too shallow in my grasp of the proof. Does anybody have any shillings of wisdom that they would be willing to offer? Cheers. |
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Think about what the mean value property of analytic functions says: $f(z_0) = \dfrac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) d\theta$, where $f$ is analytic in the disk $B_r(z_0)$. This says that $f$ is equal to the average of the boundary points. How can $|f(z_0)|$ be larger than every single point of which it is the average? Thinking discretely, if $a= \dfrac{a_1 + \cdots + a_n}{n}$, can $a$ be larger than every point in this sum? No, this is not possible. |
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holomorphic means, among other things, that the map is open. This is immediate when $f'(z_0) \neq 0,$ the Inverse Function Theorem says that an open neighborhood of $f(z_0)$ is covered surjectively. Even when $f'(z_0) = 0,$ the surjective part still holds, it is just that the map is locally $k$ to one, where $k$ is the first derivative such that $f^{(k)} (z_0) \neq 0.$ So, there it acts the way $z^3$ acts around the origin, for example. Anyway, your $D$ is open; assume that the modulus takes its maximum at some point $z_0 \in D.$ Well then $f$ maps a neighborhood of $z_0$ onto a neighborhood of $f(z_0),$ including points with larger modulus than $f(z_0)$ |
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