Theory of Functions on Complex Manifolds

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Application of the Ca. In general, global solvability on open sets D S [JfI. From Theorem 1. Hence u satisfies 1. In particular, u is holomorphic outside suppl1 u On the other hand it follows from 1. This contrast becomes even clearer in the following C' theorem of Bartogs, which is 8. Theorem Hartop'theorem.

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Then the compatibility condition 1. Let U be the unbounded component of the complement of supp X. Corollary Non-existence of isolated singularities. Q ft be a connected open set. We only ha. To do this we consider two arbitrary points z, wEB", N f and ihow that z and w belong to the same component of B '" N!

Let X be the complex line which contains 1, and w.

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Then, by Corollary 1. Consequently, z and w belong to the same component of B '" N J. Assume tha. Since, by part i , D "N j is connected, then it follows from Theorem 1. This is a contradiotion, because! In Section 2.

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Here we prove the following result in polydiscs, which also implies local solvability: en 1. Let D c be an open set and let h, E OOO D such tkat the compatibility condition 1. We shall prove inductively that the theorem is true if 1. Assume that it has already been proved for m - 1. Define 1. Domains 01 holomorphy Hartogs' Theorem 1.

In this section we shall examine this phenomenon more closely. Open sets D c 0'" with the property that, roughly speaking, there is no part of the boundary across which every holomorphic function in D can be oontinued holomorphioally are of special interest in the theory of functions of several complex variables. We give a precise definition: on 1. We do not assume that a domain of holomorphy is conneoted. The following definition is closely connected with the notion of a domain of holomorphy: 1.

For every open set D i8 ]'I j-l C,. By Proposition 1.

Let r be the multi-radius of P. Then, for every compac'. For every open set D c cn, the following condition8 are equivalent: i D i8 a domain of holomorphy. Since, by Corollary 1.

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Denote by B". Since all Km are O D -convex, we can then find functionsJ".

E O D such tha. Moreover, we can assume that every 1m is not identically 1 in any of the components of D. Let W be the component of V n D containing U.

Consequently,jis identically o in V, which is a contradiction, becausef is not identically 0 in W. The equivalence of conditions i - iv is proved. First we assume that iii is violated. Then there is a compact set K c: c: D such that KZ is not compact. Now we assume that iii is fulfilled. Then we can choose a sub. J' Since all Kmf are O D -convex, we can find! Hence condition v 22 1. This follows from condition iii in Theorem 1. Let K be a oompact subset of D.

Theory of functions on complex manifolds.

Since by Corollary 1. Let D S is a domain oj kolomorpky. Then D X G Proof. Then, by hypothesis, there is! Hence, by condition v in Theorem 1. Supp08e tkat at lease one oj the following conaitiona is fulJilled: i D is a domain oj kolomorphy. Then F-l G is a domain oj holomorphy. By condition v in Theorem 1. Since one of the conditions i and ii is fulfilled, we only have to consider the case that X is not discrete in D. A bounded open set Dec: A is called an analytic polyhedron if there are holomorphic functions Fv Plurisubharmonic functions 23 for example, condition v in Theorem 1. Plurisubharmonic functions In the next section we introduce the concept of pseudoconvexity. Pseudooonvexity will be defined by means of plurisubharmonic functions, which will be investigated in the present section.

First we recall the notion of a subharmonic function of one complex variable. We give some other equivalent definitions of continuous subharmonic funotions. It is trivial that subharmonicity implies condition i. Since for harmonic functions 1. Then Kc is a non-empty compact subset of the interior of K. Choose a. Let D be an open aet. This theorem is an immediate consequence of the following lemma: 1. We prove 1. Sle Z 1. Plurisubharmonic funotions 1. The set of all continuous plurisubharrnonic functions in D will be denoted by PO D.

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Then by Proposition 1. Then Proof. Let D C on be an open set. Let D c: en be an open set and If E: 0, then we 8et D.


E OOO D. We have e. By differentia. From 1.