[過去ﾛｸﾞ] 多変数関数論4 (1002ﾚｽ)
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487: 01/08(水)09:40 ID:qwVyKE52(1/3) AAS
お薬増やしておきますね
489: 01/08(水)11:46 ID:qwVyKE52(2/3) AAS
53 名前：１３２人目の素数さん[] 投稿日：2025/01/08(水) 11:37:28.64 ID:Tq8fsyAE
>>51
>擬凸集合（英: pseudoconvex set）は n 次元複素空間 Cn 内のある特殊なタイプの開集合をモデルとして導入された、凸性に似た幾何学的条件で定義される複素多様体上の領域である。
なるほど
こういうときは、en.wikipediaを見るのが定石でして
なるほど、”Every (geometrically) convex set is pseudoconvex.”
C2 (twice continuously differentiable) boundary
Now, G is pseudoconvex iff for every
p∈∂G and w in the complex tangent space at p, that is,
∇ρ(p)w= Σi=1〜n ∂ρ(p)/∂zi wi = 0, we have
?i,j=1〜n ∂2 ρ(p)/∂zj∂¯zj wiw¯j ≧ 0 .
The definition above is analogous to definitions of convexity in Real Analysis.
か・・・
(参考)
en.wikipedia.org/wiki/Pseudoconvexity
Pseudoconvexity
In mathematics, more precisely in the theory of functions of several complex variables, a pseudoconvex set is a special type of open set in the n-dimensional complex space Cn. Pseudoconvex sets are important, as they allow for classification of domains of holomorphy.
Let
G⊂Cn be a domain, that is, an open connected subset. One says that
G is pseudoconvex (or Hartogs pseudoconvex) if there exists a continuous plurisubharmonic function
φ on
G such that the set
{z∈G∣φ(z)<x} is a relatively compact subset of
G for all real numbers x.
In other words, a domain is pseudoconvex if
G has a continuous plurisubharmonic exhaustion function. Every (geometrically) convex set is pseudoconvex.
However, there are pseudoconvex domains which are not geometrically convex.
When G has a C2 (twice continuously differentiable) boundary, this notion is the same as Levi pseudoconvexity, which is easier to work with.
More specifically, with a C2 boundary, it can be shown that
G has a defining function, i.e., that there exists
ρ:Cn→R which is C2 so that
省1
491: 01/08(水)21:01 ID:qwVyKE52(3/3) AAS
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