Two fun facts
July 31, 2012 1 Comment
Theorem 1 Suppose , and is connected. Then if constant in a neighborhood of some point . Then is constant on all of .
Proof: Since the set is closed, so is it’s preimage. Let denote this preimage. It suffices to show that has no boundary. Then is open, closed, and non-empty. Since is connected, it follows that .
Suppose has a boundary, and let be some point on the boundary. Expand a power series about :
Since is holomorphic at , this power series must have a non-zero radius of convergence so it is valid on some part of the interior of . As we know, the only power series with constant image is the constant series . But now this series is valid in a small ball about . This contradicts the assumption that was on the boundary of , so must be open, proving the theorem.
Theorem 2 Suppose , and is connected. Suppose we have a sequence of points converging to such that . Then is constant on
Proof: By continuity of , . Expanding as a power series about , write
Since we have a sequence of points converging to whose -values all agree with , we see that the derivative at must be
Thus, in a neighborhood of , meaning that the power series: must in fact be constant. Thus, for , meaning that in a neighborhood of .
Now we can apply the previous theorem to conclude that is in fact constant of all of .