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, on . You’d check the end points to find and you’d also find so the maximum is . Also note that takes the open interval to the half open interval .
That is, the image of an open set is NOT an open set, even though is as differentiable as we could hope for.
The situation is different for non-constant analytic functions.
Let be analytic and non constant on some connected open set . Let
Now the function is not identically zero but is analytic and has a zero on ; say it is of order .
Zeros of analytic functions are isolated so one can find some such that has no zero for . So let Now choose .
Then on the circle we have for all then lies on the boundary of the open set . This contradicts that lies in an open set in .
Here is an analytic proof: suppose is not on the boundary. Then we can find where and and then .
This leads to Schwarz’s Lemma: if is analytic on , and for all in the disk, then, for we have .
Here is why: set and note that is analytic on (removable singularity at 0 ). And we have for all in the disk: if which means that for all inside the circle as well. Now let to get the result.
Note also: if at some point in the unit disk; we see that we have an interior maximum which means that is constant, hence where