5.3 The Riemann Mapping Theorem
The Riemann Mapping Theorem is one of the most profound results in complex analysis; in the case of one dimension, it establishes sufficient conditions for the biholomorphic equivalence between two open subsets of the complex plane.
If there exists a biholomorphism between two regions, then the two regions are said to be conformally equivalent, holomorphically equivalent, or biholomorphically equivalent. As a required intermediate for the proof, we first introduce:
Definition 5.3.1 (Holomorphic Logarithms).
Suppose is holomorphic, where is simply connected. Define the holomorphic logarithm of to be a branch of
for any , where the integral is path-independent is any piecewise curve from to .
Similarly, define the holomorphic powers of to be branches of , where is the holomorphic logarithm.
The path independence of the definition is provided by the simple connectivity of . The result is the heuristic concatenation of several different branches of the complex logarithm, unique up to an additive factor of , where this additive factor is the same throughout.
Theorem 5.3.1 (Riemann Mapping Theorem).
Let (a proper subset, in other words, ) be a simply connected (nonempty) open region. Let be arbitrary. Then there exists a unique biholomorphism such that and .
Proof.
First consider the case for when is a bounded region. In other words, such that .
Define to be the family of all univalent functions (not necessarily surjective) such that . This family is well-defined and nonempty.
To prove this assertion, observe that since , it follows that , , and consequently, . Therefore,
maps to , and it is linear and univalent. This shows that . It is easy to prove that is infinite; any function in the form of for also lies in .
Since is uniformly bounded on , by Montel’s Theorem (Theorem 5.2.3), is a normal family. Let satisfy . Then by Cauchy’s Estimate (Theorem 3.2.2), , on . Hence, we have
where we can assure that is positive since each is univalent at and by Lemma 5.1.1.
If is an accumulation point of , there exists a sequence such that converges to . If is attained as a maximum or that for some , we may let each .
By the normality of , there exists a subsequence such that is locally uniformly convergent in to a function (holomorphy of which is provided by Theorem 4.1.1). By definition, , and define a function sequence with , whose locally uniform limit is . It follows that is a rotation of such that .
Let be arbitrary and different. Choose to satisfy , and let . Since each is univalent in , it follows that each is non-vanishing in and consequently, in . By Theorem 3.3.5, it follows that the locally uniform limit of , or , is either non-vanishing or is identically zero in . The latter is an impossibility since . Hence, has no solutions for . In particular, . By the arbitrariness of and , the univalence of follows.
Additionally, since , in , it follows that . By the Open Mapping Theorem (Theorem 5.1.1), the condition becomes . Since for all and , it follows that .
Lastly, we aim to prove that maps to surjectively. For the sake of contradiction, assume that such that . Consider the unit disk automorphism . Since vanishes when , and since has no solutions in , there exists a holomorphic square root
for . Let , and let
where . Since , it follows that . Let , which is also in . However, since , we have
Additionally, since
it follows that , which is a contradiction of (5.3.1).
Hence, is biholomorphic. To prove the uniqueness of , suppose is an arbitrary biholomorphism such that and . Then, , and by Theorem 3.5.1, for some and . Since , it follows that . Since , and , it follows that . Hence, we have and .
Next, assume that is unbounded. It is easy to show that the boundary contains at least two points. Indeed, if , would be closed because and open by assumption. By Theorem 3.2.9, would either be equal to or , both of which are impossibilities. Additionally, if comprises exactly one point , then in subspace defined by , is clopen (by the same reason as before, open by assumption and closed because is open). It follows that , which is not simply connected.
Suppose and are two distinct points in . Let us apply the linear transformation to , and denote the resulting region by . It follows that . Consider a branch of the holomorphic square root (existent by the holomorphic logarithm from simple connectivity and the fact that ). Trivially, is univalent in .
In addition, we assert that . If not, then such that . By definition, such that and . It would then follow that and . It follows that , which is obtained when . Since and is open, , which is an impossibility.
Fix to be arbitrary. By the Open Mapping Theorem (Theorem 5.1.1), there exists an open neighborhood . It follows that . Therefore, , , and consequently, . Hence, the function maps to a bounded region that lies within the compact disk . Denote by .
It is easy to see that is simply connected. Let . To prove this, it suffices to show that the line integral of any holomorphic function over any closed curve in vanishes. Let be holomorphic, and let be a closed piecewise curve. Then
by Cauchy–Goursat (Theorem 3.1.7), since is simply connected by assumption, is a closed piecewise curve in , and the integrand is holomorphic on . Therefore, is simply connected.
Hence, we may use our previous result and establish a biholomorphism , unique up to a transformation in . Let , which is a biholomorphism from to . Similarly, it is unique up to a transformation in , and the same assertion follows.
Remark.
It is natural that we require ; if there exists a univalent function , then by Liouville’s Theorem (Theorem 3.2.3), would be a constant function.
As we will see in @ sec:multivariatecomplexanalysis, this theorem and many other properties of one-variable holomorphic functions do not extend to functions of several complex variables.