Carathéodory's theorem (conformal mapping)

Proofs of Carathéodory's theorem

The first proof of Carathéodory's theorem presented here is a summary of the short self-contained account in Garnett & Marshall (2005), pp. 14–15; there are related proofs in Pomerenke (1992) and Krantz (2006).

Clearly if f admits an extension to a homeomorphism, then ∂D must be a Jordan curve.

Conversely if ∂D is a Jordan curve, the first step is to prove f extends continuously to the closure of D. In fact this will hold if and only if f is uniformly continuous on D: for this is true if it has a continuous extension to the closure of D; and, if f is uniformly continuous, it is easy to check f has limits on the unit circle and the same inequalities for uniform continuity hold on the closure of D.

Suppose that f is not uniformly continuous. In this case there must be an ε > 0 and a point ζ on the unit circle and sequences z_n, w_n tending to ζ with |f(z_n) − f(w_n)| ≥ 2ε. This is shown below to lead to a contradiction, so that f must be uniformly continuous and hence has a continuous extension to the closure of D.

For 0 < r < 1, let γ_r be the curve given by the arc of the circle | z − ζ | = r lying within D. Then f ∘ γ_r is a Jordan curve. Its length can be estimated using the Cauchy-Schwarz inequality:

Hence there is a "length-area estimate":

The finiteness of the integral on the left hand side implies that there is a sequence r_n decreasing to 0 with ℓ ( f ∘ γ r n ) tending to 0. But the length of a curve g(t) for t in (a, b) is given by

The finiteness of ℓ ( f ∘ γ r n ) therefore implies that the curve has limiting points a_n, b_n at its two ends with |a_n – b_n| ≤ ℓ ( f ∘ γ r n ) , so this difference tends to 0. These limit points must lie on ∂U because f is a homeomorphism between D and U, so a sequence converging in U has to be the image under f of a sequence converging in D. Since ∂U is a Jordan curve, the arc τ_n between a_n and b_n is well-defined. Since ∂U is a homeomorphic image of the circle, the parameters ξ_n and η_n corresponding to a_n and b_n must have |ξ_n − η_n| tending to 0. By uniform continuity it follows that the diameter of τ_n tends to 0. Together τ_n and f ∘ γ_r form a simple Jordan curve. Its interior U_n is contained in U by the Jordan curve theorem for ∂U. The diameter of ∂U_n tends to 0 because the diameters of τ_n and f ∘ γ_r tend to 0. Hence the diameter of U_n tends to 0.

Now if V_n denotes the intersection of D with the disk |z − ζ| < r_n, then f(V_n) = U_n. Indeed, the arc γ r n divides D into V_n and a complementary region; and under the conformal homeomorphism f the curve f ∘ γ r n divides U into U_n and a complementary region. Since f is conformal and preserves the direction of the curve, it will preserve the side of a curve on which a region lies; and therefore f(V_n) = U_n.

So the diameter of f(V_n) tends to 0. On the other hand, passing to subsequences of (z_n) and (w_n) if necessary, it may be assumed that z_n and w_n both lie in V_n. But this gives a contradiction since |f(z_n) − f(w_n)| ≥ ε. So f must be uniformly continuous on U.

Thus f extends continuously to the closure of D. Since f(D) = U, by compactness f carries the closure of D onto the closure of U and hence ∂D onto ∂U. If f is not one-one, there are points u, v on ∂D with u ≠ v and f(u) = f(v). Let X and Y be the radial lines from 0 to u and v. Then f(X ∪ Y) is a Jordan curve so bounds a connected open set V by the Jordan curve theorem. On the other hand, D (X ∪ Y) is the disjoint union of two open sectors W₁ and W₂. For one of them, W₁ say, f(W₁) = V. Let Z be the portion of ∂W₁ on the unit circle, so that Z is a closed arc and f(Z) is a subset of both ∂U and the closure of V. But their intersection is a single point and hence f is constant on Z. By the Schwarz reflection principle, f can be analytically continued by conformal reflection across the circular arc. Since non-constant holomorphic functions have isolated zeros, this forces f to be constant, a contradiction. So f is one-one and hence a homeomorphism on the closure of D.

Two different proofs of Carathéodory's theorem are described in Carathéodory (1954) and Carathéodory (1998). The first proof follows Carathéodory's original method of proof from 1913 using properties of Lebesgue measure on the circle: the continuous extension of the inverse function g of f to ∂U is justified by Fatou's theorem on the boundary behaviour of bounded harmonic functions on the unit disk. The second proof is based on the method of Lindelöf (1914), where a sharpening of the maximum modulus inequality was established for bounded holomorphic functions h defined on a bounded domain V: if a lies in V, then

where 0 ≤ t ≤ 1, M is maximum modulus of h for sequential limits on ∂U and m is the maximum modulus of h for sequential limits on ∂U lying in a sector centred on a subtending an angle 2πt at a.

Discussion

Intuitively, Carathéodory's theorem says that compared to general simply connected open sets in the complex plane C, those bounded by Jordan curves are particularly well-behaved.

Carathéodory's theorem is a basic result in the study of boundary behavior of conformal maps, a classical part of complex analysis. In general it is very difficult to decide whether or not the Riemann map from an open set U to the unit disk D extends continuously to the boundary, and how and why it may fail to do so at certain points.

While having a Jordan curve boundary is sufficient for such an extension to exist, it is by no means necessary . For example, the map

from the upper half-plane H to the open set G that is the complement of the positive real axis is holomorphic and conformal, and it extends to a continuous map from the real line R to the positive real axis R⁺; however, the set G is not bounded by a Jordan curve.