A Detailed Proof of the Riemann Mapping Theorem

This post

Is intended to supply a detailed proofs of the Riemann mapping theorem.

Riemann mapping theorem. Every simply connected region is conformally equivalent to the open unit disc .

Fortunately the proof can be found in many textbooks of complex analysis, but the proof is fairly technical so it can be painful to read. This post can be considered as a painkiller. In this post you will see the proof being filled with many details. However, the writer still encourage the reader to reproduce the proof by their own pen and paper. The writer also hopes that this post can increase the accessibility of this theorem and the proof.

However, there is a bar. We need to assume some background in complex analysis, although they are very basic already. Minimal prerequisite is being able to answer the following questions.

• Contour integration, Cauchy's formula.

• Almost uniform convergence. Let be open and suppose that for all , and uniformly on every compact subset . Does ? What is the uniform limit of ? Informally, we call the phenomenon that a sequence of functions uniformly converges on every compact subset almost uniform convergence. This has nothing to do with almost everywhere in integration theory. In fact, this post does not require background in Lebesgue integration theory.

• Open mapping theorem (complex analysis version).

• Maximum modulus principle and some variants.

• Rouché's theorem. Or even more, the calculus of residues.

Preparation

Despite of the prerequisites, we still need some preparation beforehand.

Simply Connected

Definition 1. Let be a connected topological space. We say is simply connected if every curve is null-homotopic. Let be a closed curve, i.e., it is a continuous map such that . We say is null-homotopic if it is homotopic to a constant map with .

Intuitively, if is simply connected, then contains no "hole". For example, the unit disc is simply connected. However, is not. On the other hand, is still simply connected. Another satisfying result is that every convex and connected open set is simply connected. This is up to a convex combination.

There are a lot of good properties of simply connected region, which will be summarised below.

Proposition 1. For a region (open and connected subset of ), the following conditions are equivalent. Each one can imply other eight.

1. is homeomorphic to the open unit disc .
2. is simply connected.
3. for every path in and , where is the Riemann sphere.
4. is connected.
5. Every can be approximated by polynomials, almost uniformly..
6. For every and every closed path in ,

1. Every has anti-derivative. That is, there exists an such that .
2. If and , then there exists a such that .
3. For such , there also exists a such that .

5~9 are pretty much saying, calculus is fine here and we are not worrying about nightmare counterexamples, to some extent. Most of the implications are not that difficult, but there are some deserve a mention. 4 implying 5 is a consequence of Runge's theorem. In the implication of 7 to 8, one needs to use the fact that is connected. When we have , then we can put from which we obtain . 9 implying 1 is partly a consequence of the Riemann mapping theorem. Indeed, if is the plane then the homeomorphism is easy: is a homeomorphism of onto . But we need the Riemann mapping theorem to give the remaining part, when is a proper subset.

If you know the definition of sheaf, you will realise that is indeed a sheaf. For each open subset , is a ring, even more precisely, a -algebra. The exponential map is a sheaf morphism. However, we now see that it is surjective if and only if is simply connected. I hope this can help you figure out an exercise in algebraic geometry. You know, that celebrated book by Robin Hartshorne.

Since we haven't prove the Riemann mapping theorem, we cannot use the equivalence above yet. However, we can use 9 right away. This gives rise to Koebe's square root trick.

Equicontinuity & Normal Family

Equicontinuity is quite an important concept. You may have seen it in differential equation, harmonic function, maybe just sequence of functions. We will use it to describe a family of functions, where almost uniform convergence can be well established.

Definition 2. Let be a family of functions where is a metric space.

We say that is equicontinuous if, to every , there corresponds a such that whenever , we have for all . In particular, by definition, all functions in are uniformly continuous.

We say that is pointwise bounded if, to every , there corresponds some such that for every .

We say that is uniformly bounded on each compact subset if, to each compact , there corresponds a number such that for all and .

These concepts are talking about "a family of" continuity and boundedness. In our proof of the Riemann mapping theorem, we do not construct the map explicitly, instead, we will use these concepts above to obtain one (which is a limit) that exists. In this post we simply put , a simply connected region and is the natural one.

A famous result of equicontinuity is Arzelà-Ascoli, which says that pointwise boundedness and equicontinuity implies almost uniform convergence.

Theorem 1 (Arzelà-Ascoli) Let be a family of complex functions on a metric space , which is pointwise bounded and equicontinuous. is separable, i.e., it contains a countable dense set. Then every sequence in has then a subsequence that converges uniformly on every compact subset of .

Here is a self-contained proof.

Certainly it is OK to let be a subset of , or their product. We use this in real and complex analysis for this reason. We will need this almost uniform convergence to establish our conformal map. To specify its application in complex analysis, we introduce the concept of normal family.

Definition 3. Suppose , for some region . We call a normal family if every sequence of members of contains a subsequence, which converges uniformly on every compact subset of . The limit function is not required to be in .

We now apply Arzelà-Ascoli to complex analysis.

Theorem 2 (Montel). Suppose is uniformly bounded, then is a normal family.

Proof. We need to show that is "almost" equicontinuous, since uniformly boundedness clearly implies pointwise boundedness, we can apply Arzelà-Ascoli later.

Let be a sequence of compact sets such that (1) and (2) , the interior of . Then for every , there exists a positive number such that where is the disc centred at with radius . If such does not exist, then there exists a point such that whenever , , which is to say, is a boundary point. But this is impossible because lies in the interior of by definition.

For such , we pick such that . Let be the positively oriented circle with centre at and radius , i.e. the boundary of . Recall that the Cauchy formula says We will make use of this. By the formula above, we have Now we make use of our choice of , and . By definition, for (the range of ), we have . Since , we have . Therefore . Bearing this in mind, we see This may looks confusing so we explain it a little more. Since , we must have , therefore whenever , we have . This is where we use the hypothesis of uniformly bounded. we have . The integral of the norm of the integrand , is therefore bounded by . The integral over is therefore bounded by times and the result follows.

What does this inequality imply? For , if we pick , then for every and . That is, for each , the restrictions of the members of to form an equicontinuous family.

Now consider a sequence in . For each , we apply Arzelà-Ascoli theorem to the restriction of to , and it gives us an infinite subset such that converges uniformly on as and . Note we can make sure because if the subsequence converges uniformly within then it converges uniformly within as well. Pick a new sequence where , then we see converges uniformly on every and therefore on every compact subset of . The statement is now proved.

Remarks. We have no idea what the limit is, and this happens in our proof of the Riemann map theorem as well.

The sequence can be constructed explicitly, however. In fact, for every open set in the plane there is a sequence of compact sets such that

• For every compact , there is some such that .
• Every component of contains a component of .

The set is constructed as follows and can be verified to satisfy what we want above. or each , define Then is what we want.

The Schwarz Lemma

Is another important tool for our proof of the Riemann mapping theorem. We need this lemma to establish important inequalities. This lemma as well as its variants show the rigidity of holomorphic maps. We make use of the maximum modulus theorem. For simplicity, let be the Banach space of bounded holomorphic functions on , equipped with supremum norm .

Theorem 3 (Schwarz lemma). Suppose is a holomorphic map in such that and , then on the other hand, if holds for some , or if holds, then for some complex constant such that .

Proof. Since , has a removable singularity at . Hence there exists such that . Fix . For any such that , we have Therefore when , we see for all . Therefore follows. On the other hand, if at some point, the maximum modulus forces to be a constant, say , from which it follows that and .

There are many variances of the Schwarz lemma, and we will be using Schwarz-Pick.

Definition 4. For any , define

This family is a subfamily of Möbius transformation, but we are not paying very much attention to this family right now. We need the fact that such is always a one-to-one mapping which carries (the unit circle) onto and onto and to . This requires another application of the maximum modulus theorem. A direct computation shows that

Theorem 4 (Schwarz-Pick lemma). Suppose , and , . Then

Proof. Consider We see and . What's more important, . By the Schwarz lemma, . On the other hand, we see and therefore In particular, equality holds if and only if for some constant . If this is the case, then The story can go on but we halt here and continue our story of the Riemann mapping theorem.

The Riemann Mapping Theorem

Each determines a direction from the origin, which can be described by Let be a map. We say preserves angles at if exists and is independent of .

Conformal mappings preserves angles in a reasonable way. A function is conformal if it is holomorphic and everywhere. We have a theorem describes that, but it is pretty elementary so we are not including the proof in this post.

Theorem 5. Let map a region into the plane. If exists at some and , then preserves angles at . Conversely, if the differential exists and is different from at , and if preserves angles at , then exists and is different from .

There is no confusion about . By differential we mean a linear map such that, writing , we have where as and . To prove this, one can assume that . When the differential exists, one writes We say that two regions and are conformally equivalent if there is a conformal one-to-one mapping of onto . The Riemann mapping theorem states that

Theorem 6 (Riemann mapping theorem). Every proper simply connected region in the plane is conformally equivalent to the open unit disc .

As a famous example, the upper plane is conformally equivalent to by the Cayley transform.

As one may expect, this theorem asserts that the study of a simply connected region can be reduced to to some extent. But a conformal equivalence is not just about homeomorphism. If is a conformal one-to-one mapping, then is also a conformal mapping. In the language of algebra, such a mapping induces a ring isomorphism Therefore, the ring is algebraically the same as . The Riemann mapping theorem also states that, if is a simply connected region, then . From this we can exploit much more information on top of homeomorphism. One can also extend the story to , the Riemann sphere, but that's another story.

The Proof by Arguing A Normal Family

The proof is fairly technical. But it is a good chance to attest to our skill in complex analysis. The bread and butter of this proof is the following set: Our is to prove that there is some such that . Note, once the non-emptiness is proved, since uniformly, we see is a normal family.

Step 1 - Prove Non-emptiness Using Koebe's Square Root Trick

Pick . Then and what is more important, . By 9 of proposition 1, there exists such that , i.e., informally, in . If , then and then . Therefore is one-to-one. On the other hand, if , we still have , and . This is shows that the "square-root" is well-defined here. This is the Koebe's square root trick.

Since is an open mapping, there is an open disc , where , and . But by arguments above we have , and therefore . For this reason, we can put It follows that and therefore . Since is one-to-one, is one-to-one as well and we deduce that , this set is not empty.

Remark. You may have trouble believing that . But if we pick any , we have some such that . We also have but this implies , and therefore . There exists some such that . Hence . It follows that and this is a contradiction.

Since , we have for all and therefore is not a problem either.

Step 2 - Enlarge the Range

If and , and , then there exists a such that .

This step shows that we can "enlarge" the range in some way.

For convenience we use the Möbius transformation Pick . Then and has no zero in . Hence there is some such that Since is one-to-one, another application of Koebe's square root trick shows that is one-to-one. Therefore we have as well. If where , we have (one-to-one). In particular, .

By putting , we have If we put , then the chain rule shows that (Note we used the fact that .) If we can prove that then this step is complete. Note satisfy the condition in Schwarz-Pick lemma and therefore The first equality does not hold because is not of the form for . On the other hand we have Therefore and the this step is complete.

Step 3 - Find the Function with Largest range, Namely the Disc

We take the contraposition of step 2:

Fix . If is an element such that for all , then .

The proof is complete once we have found such a function! To do this, we use the fact that is a normal family. Put By definition of , there is a sequence such that in . By normality of , we pick a subsequence that converges uniformly on compact subsets of . Put the uniform limit to be . It follows that . Since and , cannot be a constant. Since , we must have . But since is open, we are reduced to .

It remains to show that is one-to-one. Fix distinct . Put and , then . Let be a closed disc in centred at with interior denoted by such that

• has no zero point on the boundary of .

We see converges to , uniformly on . They have no zero in since they are one-to-one and have a zero at . By Rouché's theorem, has no zero in either, and in particular . This completes the proof.

Remark. First of all, such a is accessible. This is because zero points of has no limit point in , i.e., they are discrete (when defining , we don't know how many are there yet).

Our choice of enables us to use Rouché's theorem (chances are you didn't get it). Since has no zero on the boundary, we have . When is big enough, we see The second inequality is another application of the maximum modulus theorem. Rouché's theorem applies here naturally as well.

This proof is a reproduction of W. Rudin's Real and Complex Analysis. For a comprehensive further reading, I highly recommend Tao's blog post.