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