# Maximum modulus principle

First: the last assignment of the year, due Monday: 3.1 2, 6, 14, 20. I hope we can finish 3.1, then do 3.2 before the semester ends.

Now to the new stuff.

Let’s think back to calculus: suppose we wanted to find the maximum of, say, $f(x) = 1-x^2$ on $[-1,1]$. You’d check the end points to find $f(-1) = f(1) =0$ and you’d also find $f'(x) = -2x =0 \text{ when } x = 0$ so the maximum is $(0, f(0)) = (0,1)$. Also note that $f$ takes the open interval $(-1, 1)$ to the half open interval $(0,1]$.

That is, the image of an open set is NOT an open set, even though $f$ is as differentiable as we could hope for.

The situation is different for non-constant analytic functions.

Let $f$ be analytic and non constant on some connected open set $D$. Let $z_0 \in D$

Now the function $f(z) - f(z_0)$ is not identically zero but is analytic and has a zero on $D$; say it is of order $m$.

Zeros of analytic functions are isolated so one can find some $r > 0$ such that $f(z) - f(z_0)$ has no zero for $r >|z-z_0| > 0$. So let $\delta = min\{ |f(z) - f(z_0) | |z-z_0| = r \}$ Now choose $w, |w-f(z_0)| < \delta$.

Then on the circle $|z-z_0| = r$ we have $|(f(z) -w) - (f(z) - f(z_0))| = |w-f(z_0)| |f(z)|$ for all $|z-z_0| < r$ then $f(z_0)$ lies on the boundary of the open set $\{f(z): |z-z_0| < r \}$. This contradicts that $f(z_0)$ lies in an open set in $f(D)$ . Here is an analytic proof: suppose $f(z_0)$ is not on the boundary. Then we can find $\epsilon > 0$ where $|w-f(z_0)| > 0$ and $w = f(z_1)$ and then $|f(z_1) | > |f(z_0) |$.

This leads to Schwarz’s Lemma: if $f$ is analytic on $|z| < 1$, $f(0) = 0$ and $|f(z)| \leq 1$ for all $z$ in the disk, then, for $|z| < 1$ we have $|f(z)| \leq |z|$.

Here is why: set $g(z) =\frac{f(z)}{z}$ and note that $g$ is analytic on $|z| < 1$ (removable singularity at 0 ). And we have for all $z$ in the disk: if $|z| = r, |g(z)| \leq \frac{1}{r}$ which means that $|g(z)| < \frac{1}{r}$ for all $z$ inside the circle $|z| = r$ as well. Now let $r \rightarrow 1$ to get the result.

Note also: if $g(z_0) = 1$ at some point in the unit disk; we see that we have an interior maximum which means that $g$ is constant, hence $g(z) = \frac{f(z)}{z} = \lambda \rightarrow f(z) = \lambda z$ where $|\lambda| = 1$

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