If is a simply connected domain and , then there exists such that for all . Such a map is injective, and if then .
Proof Since is simply connected and , we may choose a primitive of the function . By the proposition, is a branch of the logarithm up to a constant. Define . Then for all . Assume that for some . Then , so is injective. Now if , then , so , that is , and thus , yielding a contradiction.
First Step of Riemann Mapping Theorem
Let be a simply connected domain. Let , and define Then is non-empty.
Proof Since , we may choose , . Let be the translated domain , so that . By the above lemma, there exists injective such that for all , and if then . Fix some . Since is injective, it is not constant, the open mapping theorem implies that is open. Thus, there exists such that and . That is, for all , , which is equivalent to for all . Hence so that is an injective, bounded holomorphic function on . And so is injective and vanishes at , and can be bounded into by choosing small enough.
Converse Schwarz lemma
Let be a simply connected domain and suppose . Then there exist an injective such that , and .
Proof We choose . Let and note that and . By the lemma, we can pick injective such that for all . Note that , so . Let , where , so that . Since is injective, is also injective. Since is holomorphic, is holomorphic. Since , we have .
Now let be the following one-sided inverse of : Then , and for all , we have Since , we conclude that , and by differentiating, we have Evaluating at and taking modulus, we have . By the Schwarz lemma, we have . But is not injective since contains both and , so cannot be a rotation, therefore . Hence .
Converse to the Schwarz lemma
This lemma is a sort of converse to the Schwarz lemma, because the Schwarz lemma says that the statement of the lemma is false when .
Proof We will show the contrapositive. Suppose that and . Let , and then choose as in the lemma, and set . Then , , and is injective as a composition of injective functions, so . But as required.
Hurwitz’s Theorem
Hurwitz’s Theorem
Let be a domain, and let with . Suppose that normally on and that for some . Then for all , there is some such that whenever we have .
Proof Since , the uniqueness theorem implies that there is some such that has no zeroes in (if no such existed, we could construct a sequence of zeroes converging to , and would be e identically zero). Let , and . Choose such that implies that for all . This is allowed by the theorem, and is compact.
We now apply Rouche’s theorem to . When , we have for all , So we have $$K_{f_n,\Delta(z_0,r)}=K_{f,\Delta(z_0,r)}\geq1$$$\square$
Remark
This theorem says that can only have a zero if eventually the functions have a zero nearby; that is, a zero cannot just appear in the limit. This is false in the context of real analysis. Consider the functions on the interval . None of the functions has any zeroes, but the limit has a zero at .
Proof Clearly, we have . Since , for all . So taking the limit as , we have for all . Suppose for some . Then is a maximum point of on , so is constant by the strong maximum principle. Since , we must have . Otherwise, for all .
Now we verify the third condition. Suppose , and to a contradiction that for some with . Choose such that . Let and , then normally on , and . By the Hurwitz’s theorem, there exists such that for all . Thus there exists such that Hence . But , so and thus is not injective, a contradiction.
Therefore, is injective, and .
Montel’s Theorem
Lemma
Let be a domain and let . Assume that for every compact there exists an such that for all and . Let Then every sequence in has a subsequence that converges uniformly on .
is bounded: for all , there is a such that for all
is uniformly equicontinuous: for all , there exists a such that implies for all .
Each singleton set is compact, so by assumption, each is bounded. So is bounded. Let , for , let Since is compact, we may choose such that . By assumption, there is such that for all and . Choose such that and . We will show that if , then for all . Fix , Define by . Since , we have , so by the triangle inequality, we have so for all . By construction we have , so we invoke the Schwarz lemma to conclude that for all . Now suppose further that and . Set , and since we have . Thus It follows that This completes the proof of uniform equicontinuity, so by the Arzelà-Ascoli theorem, every sequence in has a subsequence that converges uniformly on .
Montel’s Theorem
Let be a domain and let . Assume that for every compact there exists an such that for all and . Then every sequence in has a normally convergent subsequence.
Proof Let be a sequence, . Clearly each is compact and , since each lies in some , and every compact is a subset of some . By the above lemma, we may choose a subsequence of that converges uniformly on , a subsequence of converges uniformly on , and so on. We then have a family of subsequences that converges uniformly on for all . Now let , a subsequence of . Indeed, this is a subsequence of every eventually after omitting at most finitely many (less than ) terms. Since every compact is contained in some , the sequence converges uniformly on every compact , which is to say that converges normally on .
Part (b), Third Step of Riemann Mapping Theorem
Let be a domain and let . Assume that for every compact there exists an such that for all and . Let Then every sequence in has a subsequence that converges uniformly on .
Proof Given such a simply connected domain , let Let . Choose a sequence such that (this required step 1, that ). By step 3(b), we can find a subsequence that converges normally on to some . Then . Since each is injective, by the corollary, , in particular, . So , therefore is not identically zero. By step 3(a) we have , is injective. Now by step 2, we have . And so is the desired biholomorphic map by the theorem.
Corollary
Every simply connected domain is conformally equivalent to one and only one of the , and .
Proof If contains at least two points, then is conformally equivalent to by the Riemann mapping theorem. If contains only one point, we apply a Möbius transformation that maps this point to . Such a transformation is biholomorphic between and . Finally, if , then .
It remains to show that , and are not conformally equivalent with each other. First of all, and are non-compact, whereas is compact. Secondly, if is a biholomorphic map, then is entire and bounded, and by the |Liouville’s theorem, is constant, which is a contradiction.
Proposition
Every conformal transformation of has the form with .