undrestanding Maximum principle with example

Question

I want to know why if a holomorphic function has local maximum or minimum then it is constant,can someone give an example that why this is true?

Answer 1

The things that I am going to say are based on the understanding of this subject. So they may not reflect the true reason for this fact.

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The main reason that this theorem is true is that somehow these functions follow a mean value property that is the Cauchy's integral formula that for a holomorphic function $f$ we have the following property : $$ f(a) = \frac{1}{2\pi i}\int_{|z-a|=r} \frac{f(z)}{z-a}dz.$$

That is if you know that the average of some quantities attains the maximum value of them then all of those quantities should be equal.

I state the proof of this theorem below:

Let $f : U \to \Bbb C$ be a analytic function from the connected open subset, $U$, of $\Bbb C$. if for some $a \in U$ we have $|f(a)| \geq |f(z)|$ in some neighborhood of that point. Then $f$ is constant.

proof: let $C$ be the boundary of a disk centered at $a$ ,subset to $U$, of radius $r$. and suppose $|f(a)|$ is maximum on the this closed disk. Now let $\arg f(z) = \alpha$ then $|f(z)| = e^{-i\alpha}f(z)$ substitute it in the cauchy integral formula , you would get :

$$ |f(a)| = \frac{1}{2\pi i} \int_{|z-a|=r}e^{-i\alpha}\frac{f(z)}{z-a}dz = \frac{1}{2\pi i}\int_{0}^{2\pi} e^{-i\alpha}\frac{f(a+re^{i\theta})}{re^{i\theta}}ire^{i\theta}d\theta\\ = \frac{1}{2\pi}\int_{C} e^{-i\alpha}f(a+re^{i\theta})d \theta.$$ Thus we have :

$$ \Re|f(a)| = \Re \frac{1}{2\pi}\int_{C} e^{-i\alpha}f(a+re^{i\theta})d\theta = \int_{C}\Re (e^{-i\theta} f(a+re^{i\theta})) d\theta \leq \frac{1}{2\pi}\int_{C} |e^{-i\theta} f(a+re^{i\theta})| d\theta \leq \\ \frac{1}{2\pi}\int_{C} |f(a+re^{i\theta})| d\theta \leq \frac{1}{2\pi} \int_{C} |f(a)| d\theta = |f(a)| $$

So we must have the equality $|f(a)| = |f(a+re^{i\theta}|$. and for all $z \in C$ we have: $\Re(e^{-i\theta}f(z)) = |f(z)|=|f(a)|$. So $f(z) = f(a)$ on $C$.

This gives that the function $f$ is constant on an infinite set of numbers with accumulation point. Hence $f$ is constant.