The problem I am working on is the following:

PROBLEM: Let $D$ be the unit disc and let $f:D \to \mathbb{C}$ be a conformal mapping such that $f(0) = 0$ and $f'(0) = 1$. Show that there exists some $w$ with $|w| = 1$ such that $w \not \in f(D)$.

THOUGHTS: I am thinking we want to do a proof by contradiction. If the supposition is false, then by the open mapping theorem we know that the image of $f$ contains an open set that contains the entire boundary of the unit disc. Thus, $f$ in some sense expands points outward in all directions from the origin. I know that the derivative at the origin is related to how the function either repels or attracts points near the origin, and so it seems that expanding in all directions past the boundary of the unit disc combined with $f(0) = 0$, $f'(0) = 1$ can somehow lead to a contradiction. Any thoughts?

  • $\begingroup$ Look at $f^{-1}$ and recall Schwarz. $\endgroup$ – Daniel Fischer Aug 15 '17 at 20:19
  • $\begingroup$ But $f$ is only locally invertible? $\endgroup$ – Kyle Liss Aug 15 '17 at 20:55
  • $\begingroup$ Are you sure? In complex analysis, the convention "conformal = biholomorphic" is rather widespread, so unless the problem source explicitly uses the other convention it's not unreasonable to expect that $f$ should be injective. Especially since $$f_n(z) = \frac{1}{n}\bigl(e^{nz} - 1\bigr)$$ is locally conformal, has $f_n(0) = 0$ and $f_n'(0) = 1$, but for large enough $n$, $f_n(D) \supset \partial D$. $\endgroup$ – Daniel Fischer Aug 15 '17 at 21:03
  • $\begingroup$ Ok thanks. I've been using Lang and he defines it as mappings that are locally injective, but I got this problem from a different source so what you say makes sense. $\endgroup$ – Kyle Liss Aug 16 '17 at 0:09

There are two conventions for the meaning of the word conformal.

  1. conformal = biholomorphic, and
  2. conformal = locally biholomorphic.

Here, the first convention is used, since the assertion becomes wrong in the second convention:

$$f_n \colon z \mapsto \frac{1}{n}\bigl(e^{nz}-1\bigr)$$

is, for every $n > 0$, a locally biholomorphic function with $f_n(0) = 0$ and $f_n'(0) = 1$. For large enough $n$, the rectangle $R_n = \{ z \in \mathbb{C} : 0 \leqslant \operatorname{Re} z \leqslant 2\log n/n, \lvert \operatorname{Im} z\rvert \leqslant 2\pi/n\}$ is contained in the unit disk. The map $g_n \colon z \mapsto \frac{1}{n} e^{nz}$ maps $R_n$ onto the annulus $A_n = \bigl\{ w\in \mathbb{C} :\frac{1}{n} \leqslant \lvert w\rvert \leqslant 2\bigr\}$, and $A_n - \frac{1}{n}$ contains the unit circle.

Thus the problem asks to prove that the image of a schlicht function does not contain the entire unit circle. That it can contain all of the unit circle with the exception of a single point is shown by the Koebe functions

$$f_{\alpha} \colon z \mapsto \frac{z}{(1 - \alpha z)^2}$$

with $\lvert \alpha\rvert = 1$. We have $f_{\alpha}(D) = \mathbb{C} \setminus \{ -\overline{\alpha}\cdot t : t \geqslant 1/4\}$.

Now to prove that the image of a schlicht function cannot contain the whole unit circle, we assume that $f$ were a schlicht function with $f(D) \supset \partial D$. Since $f(D)$ is simply connected, it then follows that $f(D) \supset \overline{D}$. By the maximum modulus principle, $g = f^{-1}\lvert_D$ maps the unit disk into itself. Also, $g(0) = 0$ and $g'(0) = 1$. The Schwarz lemma says that then $g = \operatorname{id}_D$, and consequently $f = \operatorname{id}_D$, but that contradicts $f(D) \supset \partial D$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.