Complex Analytic and Differential Geometry (September 2009 by Jean-Pierre Demailly

By Jean-Pierre Demailly

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Apply Th. 24 to ± Re f and ± Im f . It follows that Re f and Im f have pluriharmonic extensions to X, in particular f extends to f ∈ ∞ (X). By density of X A, d′′ f = 0 on X. 26) Corollary. Let A ⊂ Ω (resp. A ⊂ X) be a closed (pluri)polar set. If Ω (resp. X) is connected, then Ω A (resp. X A) is connected. Proof. If Ω A (resp. X A) is a disjoint union Ω1 ∪ Ω2 of non empty open subsets, the function defined by f ≡ 0 on Ω1 , f ≡ 1 on Ω2 would have a harmonic (resp. holomorphic) extension through A, a contradiction.

X A) is connected. Proof. If Ω A (resp. X A) is a disjoint union Ω1 ∪ Ω2 of non empty open subsets, the function defined by f ≡ 0 on Ω1 , f ≡ 1 on Ω2 would have a harmonic (resp. holomorphic) extension through A, a contradiction. § 6. A. Domains of Holomorphy in Cn . Examples Loosely speaking, a domain of holomorphy is an open subset Ω in Cn such that there is no part of ∂Ω across which all functions f ∈ Ç(Ω) can be extended. 1) Definition. Let Ω ⊂ Cn be an open subset. Ω is said to be a domain of holomorphy if for every connected open set U ⊂ Cn which meets ∂Ω and every connected component V of U ∩ Ω there exists f ∈ Ç(Ω) such that f↾V has no holomorphic extension to U .

However, if v is a plurisubharmonic function on X × Cn , the partial minimum function on X defined by u(ζ) = inf z∈Ω v(ζ, z) need not be plurisubharmonic. A simple counterexample in C × C is given by v(ζ, z) = |z|2 + 2 Re(zζ) = |z + ζ|2 − |ζ|2 , u(ζ) = −|ζ|2 . It follows that the image F (Ω) of a pseudoconvex open set Ω by a holomorphic map F need not be pseudoconvex. In fact, if Ω = {(t, ζ, z) ∈ C3 ; log |t| + v(ζ, z) < 0} and if Ω′ ⊂ C2 is the image of Ω by the projection map (t, ζ, z) −→ (t, ζ), then Ω′ = {(t, ζ) ∈ C2 ; log |t| + u(ζ) < 0} is not pseudoconvex.

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