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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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  • $\begingroup$ My Dear S. Valera, you look surprisingly like the late great physicist Robert Oppenheimer! What a surprise! Especially considering your year of birth! $\endgroup$ – Robert Lewis May 2 '14 at 4:31
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    $\begingroup$ Ha - nicely noticed. My apologies for any disappointment you may feel to hear that I am but a fan of Mr Oppenheimer and not a reincarnation. $\endgroup$ – S Valera May 2 '14 at 22:04
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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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    $\begingroup$ This speaks to me, thank you. $\endgroup$ – S Valera Apr 20 '14 at 22:54
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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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    $\begingroup$ Brilliant, that ties up a few things nicely. Thanks for your time. $\endgroup$ – S Valera Apr 20 '14 at 22:55
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There's another question that's been marked as a duplicate of this one. The other question asks specifically what the difference is between $\Bbb R$ and $\Bbb C$ here - it seems possibly worthwhile to transplant the answer I gave to the other questions:

In either case $$f(z+h)=f(z)+hf'(z)+\dots,$$where the dots indicate higher order terms that are smaller than the last term if $h$ is small. Now if $f'(z)\ne0$ this shows that $|f|$ cannot have a maximum at $z$, because we can choose $h$ to point in the right direction so $|f(z)+hf'(z)|>|f(z)|$.

But what if $f'(z)=0$? Then $$f(z+h)=f(z)+\frac12h^2f''(z)+\dots.$$In the real case $h^2\ge0$, so if $f''(z)$ has the opposite sign to $f(z)$ then adding the $h^2f''(z)$ makes $|f|$ smaller. But in the complex case $h^2$ can point in any direction; so if we choose the direction for $h^2f''(z)$ to be the same as the direction of $f(z)$ then adding the $h^2f''(z)$ again makes $|f|$ larger.

Hmm, slightly more formally: If $f'(z)\ne0$ then $z$ cannot be a maximum for $|f|$, for very much the same reason in the real and complex case. Suppose that $f'(z)=0$, $f''(z)\ne0$. Now in the real case, if $f(z)>0$ and $f''(z)<0$ then $$|f(z)+\frac12 h^2f''(z)| =|f(z)|-\frac12h^2|f''(z)|<|f(z)|$$ for all small $h\ne0$, regardless of whether $h$ is positive or negative, the only two directions available. But that can't happen in the complex case: If $f''(z)\ne0$ then there exists $\alpha$ so that if $h=re^{i\alpha}$ then $$|f(z)+\frac12h^2f''(z)|=|f(z)|+r^2|f''(z)|>|f(z)|$$ for all small $r>0$.

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