On the Solution of the Dirichlet Problem with Rational Holomorphic ...

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454 P. Ebenfelt and M. Viscardi CMFT Proof of Lemma 3. Consider the function F (w) := 1 ∫ R ′ (z) 2πi R(z) − w dz, ∂Ω w ∈ C\L. Note that F is analytic on C\L since R(∂Ω) = L. Let Z S (g) and P S (g) denote the number of zeros and poles, respectively, of a function g in a domain S (counting multiplicity). The Argument Principle implies (18) F (w) = Z Ω [R(z) − w] − P Ω [R(z) − w] for all w ∈ C\L. Since the right-hand side of equation (18) is always an integer, F must be constant (since F (w) is analytic on C\L and C\L is connected). Also, note that lim w→∞ F (w) = 0, so in fact F ≡ 0. Since R has exactly one (simple) pole in Ω at the point a by definition, equation (18) now yields Z Ω [R(z) − w] = P Ω [R(z) − w] = 1 for all w ∈ C\L. So the equation R(z) = w has exactly one solution in Ω for any w ∈ C\L. Since R −1 (L) is at most a proper real-analytic variety in Ω, we conclude that R is a one-to-one mapping onto R(Ω), and we have (19) P 1 \L ⊆ R(Ω) ⊆ P 1 . We now choose a suitable Möbius transformation ϕ such that ϕ(R(Ω)) does not contain the point ∞, i.e. (20) P 1 \M ⊆ ϕ(R(Ω)) ⊆ C, where M = ϕ(L). Then, since Ω is simply connected and ϕ(R(z)) is one-toone and analytic in Ω, ϕ(R(Ω)) is a simply connected subset of C. It is well known that this is equivalent to saying that M ′ := P 1 \ϕ(R(Ω)) is connected (see [9, p. 202]). Also, note that ϕ(R(Ω)) is an open subset of P 1 by the open mapping theorem (see [9, p. 99]), so M ′ is a closed subset of P 1 . By equation (20), we therefore have ϕ(R(Ω)) = P 1 \M ′ for some closed, connected, possibly empty subset M ′ ⊆ M. Thus, R(Ω) = P 1 \L ′ for some closed, possibly empty subinterval L ′ ⊆ L. If L ′ = ∅, then R(Ω) = P 1 , which is impossible since P 1 is compact while Ω is not. If L ′ = {z 0 } for some z 0 , then R(Ω) = P 1 \{z 0 }, which is also impossible since this would imply that Ω is conformally equivalent to C, violating Liouville’s Theorem. Hence, L ′ = [z 3 , z 4 ] with z 3 < z 4 , as desired. Lemma 4. A Riemann map f : Ω → D is algebraic. Proof of Lemma 4. First, note that by composing R with an appropriate Möbius transformation, we may assume, without loss of generality, that L ′ = [−1, 1]. It is well known that the function φ(w) = 1/2(w + 1/w) maps the unit disk D
5 (2005), No. 2 The Dirichlet Problem with Rational Holomorphic Data 455 conformally and bijectively to the Riemann sphere P 1 minus the segment [−1, 1], so by Lemma 3, the function (21) f(z) := (φ −1 ◦ R)(z) is a biholomorphism from Ω to D. Since φ and R are rational, we conclude that f is algebraic, as desired. Lemma 5. The function f ′ (z) is rational. Proof of Lemma 5. We first establish a relation between the solution to the Dirichlet problem in Theorem 3 and the Bergman kernel K of Ω. This result has already been proven before (see [4, p. 97]), but we include a proof here for completeness. Let h be any function in the Bergman space H 2 (Ω) = L 2 (Ω)∩O(Ω). By Cauchy’s Integral Formula, h(a) = 1 ∫ h(w) 2πi ∂Ω w − a dw = 1 ∫ u(w, w)h(w) dw, 2πi ∂Ω since u(w, w) = 1/(w − a) on ∂Ω. Applying Stokes’ Theorem, and remembering that h is holomorphic in Ω, we have h(a) = − 1 ∫ [ 2i ∂ ] ∂ u(w, w) · h(w) + u(w, w) · 2i 2πi Ω ∂w ∂ w h(w) dA (22) = − 1 ∫ ∂ u(w, w) · h(w) dA. π Ω ∂ w But from the definition of the Bergman kernel, we also have ∫ (23) h(a) = K(a, w) · h(w) dA. Ω Combining equations (22) and (23), we obtain ∫ K(a, w) · h(w) dA = − 1 ∫ ∂ u(w, w) · h(w) dA, Ω π Ω ∂ w i.e. ∫ [ (24) K(a, w) + 1 ] ∂ π ∂ w u(w, w) h(w) dA = 0 for all h ∈ H 2 (Ω). Ω Note that the function K(a, w) + (1/π) · (∂/∂ w)u(w, w) is anti-holomorphic, being the sum of two anti-holomorphic functions. Thus, we may write B(w) := K(a, w) + 1 ∂ u(w, w), π ∂ w where B is holomorphic in Ω. Note that B is also in L 2 (Ω); thus, we have B(w) ∈ H 2 (Ω). Equation (24) now becomes ∫ B(w) · h(w) dA = 0 for all h ∈ H 2 (Ω). Ω
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