Complex and Symplectic Geometry by Daniele Angella Costantino Medori & Adriano Tomassini

Author:Daniele Angella, Costantino Medori & Adriano Tomassini
Language: eng
Format: epub
Publisher: Springer International Publishing, Cham

Therefore, { f(γ n (z))} is a normal family.

Step 4: Let f lim be any limit point of the sequence { f(γ n (z))}. Since the sequence is non-increasing, it converges, and . Similarly, . By the strong maximum principle, [10], a non-constant holomorphic function on a complex manifold with boundary cannot have local maxima (even non-strict) outside of the boundary. Since does not intersect the boundary of , the function f lim must be constant.

Step 5: Consider now the complement , and suppose it has two distinct points x and y. Let f be a holomorphic function which satisfy f(x) ≠ f(y). Replacing f by an exponent of μf if necessarily, we may assume that | f(x) | < | f(y) |. Since γ fixes Z, which is compact, for any limit f lim of the sequence { f(γ n (z))}, supremum f lim on Z is not equal to infimum of f lim on Z. This is impossible, hence on V, and V is one point.

This finishes the proof of Theorem 1.1. ■

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