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 Pommerenke (1992) and Krantz (2006).

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

Conversely if ∂U 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 ${\displaystyle \circ }$ γ_r is a Jordan curve. Its length can be estimated using the Cauchy–Schwarz inequality:

${\displaystyle {\displaystyle \ell (f\circ {\gamma }_r)={\int }_{{\gamma }_r}|f^{\mathrm{\prime }}(z)||dz|\le {\left({\int }_{{\gamma }_r}|dz|\right)}^{1/2}\cdot {\left({\int }_{{\gamma }_r}|f^{\mathrm{\prime }}(z){|}^2|dz|\right)}^{1/2}\le (2\pi r{)}^{1/2}\cdot {\left({\int }_{\{\theta :|\zeta +r{e}^{i\theta }|<1\}}|f^{\mathrm{\prime }}(\zeta +r{e}^{i\theta }){|}^{2}rd\theta \right)}^{1/2}.}}$

Hence there is a "length-area estimate":

${\displaystyle {\displaystyle {\int }_0^1\ell (f\circ {\gamma }_r{)}^2\frac{dr}{r}\le 2\pi {\int }_{|z|<1}|f^{\mathrm{\prime }}(z){|}^{2}dxdy=2\pi \cdot \mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}f(D)<\mathrm{\infty }.}}$

The finiteness of the integral on the left hand side implies that there is a sequence r_n decreasing to 0 with ${\displaystyle \ell (f\circ {\gamma }_{r_n})}$ tending to 0. But the length of a curve g(t) for t in (a, b) is given by

${\displaystyle {\displaystyle \ell (g)=\underset{a<t_1<t_2<\cdots <t_k<b}{sup}\sum _{i=1}^{k-1}|g(t_{i+1})-g(t_i)|.}}$

The finiteness of ${\displaystyle \ell (f\circ {\gamma }_{r_n})}$ therefore implies that the curve has limiting points a_n, b_n at its two ends with ${\displaystyle |a_n-b_n|\le \ell (f\circ {\gamma }_{r_n})}$, so this distance, as well that of the curve, tends to 0. These two limit points must lie on ∂U, because f is a homeomorphism between D and U and thus a sequence converging in U has to be the image under f of a sequence converging in D. By assumption there exist a homeomorphism β between the circle ∂D and ∂U. Since β⁻¹ is uniformly continuous, the distance between the two points ξ_n and η_n corresponding to a_n and b_n in ∂U must tend to 0. So eventually the smallest circular arc in ∂D joining ξ_n and η_n is defined. Denote τ_n image of this arc under β. By uniform continuity of β, the diameter of τ_n in ∂U tends to 0. Together τ_n and f ${\displaystyle \circ }$ γ_rn form a simple Jordan curve. Its interior U_n is contained in U by the Jordan curve theorem for ∂U and ∂U_n: to see this, notice that U is the interior of ∂U, as it is bounded, connected and it is both open and closed in the complement of ∂U; so the exterior region of ∂U is unbounded, connected and does not intersect ∂U_n, hence its closure is contained in the closure of the exterior of ∂U_n; taking complements, we get the desired inclusion. The diameter of ∂U_n tends to 0 because the diameters of τ_n and f ${\displaystyle \circ }$ γ_rn tend to 0. Hence the diameter of U_n tend to 0. (For ${\displaystyle {\overline{U}}_n\times {\overline{U}}_n}$ is compact set, hence ${\displaystyle {\overline{U}}_n}$ contains two points u and v such that distance between them is maximal. It is easy to see that u and v must lie in ∂U and the diameters of both U and ∂U equal ${\displaystyle |u-v|}$.)

Now if V_n denotes the intersection of D with the disk |z − ζ| < r_n, then for all sufficiently large n f(V_n) = U_n. Indeed, the arc γ_rn divides D into V_n and complementary region ${\displaystyle V_n^{\prime }}$, so under the conformal homeomorphism f the curve f ${\displaystyle \circ }$ γ_rn divides U into ${\displaystyle f(V_n)}$ and a complementary region ${\displaystyle f(V_n^{\prime })}$; U_n is a connected component of U \ f ${\displaystyle \circ }$ γ_rn, as it is connected and is both open and closed in this set, hence ${\displaystyle U_n}$ equals either ${\displaystyle f(V_n)}$ or ${\displaystyle f(V_n^{\prime })}$. The diameter of ${\displaystyle f(V_n^{\prime })}$ does not decrease with increasing n, for ${\displaystyle n<n^{\prime }}$ implies ${\displaystyle V_n^{\prime }\subset V_{n^{\prime }}^{\prime }}$. Since the diameter of U_n tends to 0 as n goes to infinity, it is eventually less than that of ${\displaystyle f(V_n^{\prime })}$ and then necessarily 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. Arguing as before, its interior V is contained in U and is a connected component of U \ f(X ∪ Y). On the other hand, D \ (X ∪ Y) is the disjoint union of two open sectors W₁ and W₂. Hence, 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

|h(a)| ≤ m^t ⋅ M^{1 − t},

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.