By Kraus D.

A boundary model of Ahlfors' Lemma is verified and used to teach that the classical Schwarz-Carathéodory mirrored image precept for holomorphic services has a in simple terms conformal geometric formula when it comes to Riemannian metrics. This conformally invariant mirrored image precept generalizes clearly to analytic maps among Riemann surfaces and comprises between different effects a characterization of finite Blaschke items because of M. Heins.

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**Additional resources for A boundary version of Ahlfors` lemma, locally complete conformal metrics and conformally invariant reflection principles for analytic maps**

**Sample text**

ROTH AND S. RUSCHEWEYH cλ and Cλ , and let f : → R be an analytic map. 1) z→ξ for every ξ ∈ Γ. 2. (a) Note that λ(f (z)) |f ′ (z)| is the density of the pullback of the metric λ(w) |dw| under the map f , so f ∗ λ(z) |dz| = λ(f (z)) |f ′ (z)| |dz| is a conformal pseudometric on and therefore a well-defined function on . 1, it suffices to assume that λ(w) |dw| is a complete regular conformal metric with curvature bounded below. However, these assumptions cannot be weakened further. For instance, take R = , λ(w) |dw| = |dw| and any function holomorphic in a neighborhood of a point of the unit circle.

Math. 38 (1977), 73–82. [19] D. Minda, The strong form of Ahlfors’ lemma, Rocky Mountain J. Math. 17 (1987), 457–461. [20] D. Minda, A reflection principle for the hyperbolic metric and applications to geometric function theory, Complex Variables Theory Appl. 8 (1987), 129–144. [21] Ch. Pommerenke, Boundary Behaviour of Conformal Maps, Springer-Verlag, Berlin, 1992. [22] O. Roth, A general conformal geometric reflection principle, Trans. Amer. Math. , to appear. [23] H. L. Royden, The Ahlfors–Schwarz Lemma: the case of equality, J.

Finally, π := πR ◦ Ψ is a conformal £ map defined on a neighborhood of such that π(∂ ) = ∂R. 1. 3. We pull the metric λ(w) |dw| back to the unit disk using w = π(u) and get a complete regular conformal metric ν(u) |du| := π ∗ λ(u) |du| = λ(π(u)) |π ′ (u)| |du| . 6 implies lim ν(u) = +∞ . 2) → by g(z) := (π −1 ◦ f )(z). Then λ(f (z)) |f ′ (z)| = ν(g(z)) |g ′ (z)| (z ∈ ) . 1 and therefore assume that f : → R has an analytic continuation across Γ with f (Γ) ⊆ ∂R. 3, g : → also has an analytic extension across Γ with g(Γ) ⊂ ∂ .