[過去ﾛｸﾞ] 多変数関数論4 (1002ﾚｽ)
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489: 01/08(水)11:46 ID:qwVyKE52(2/3) AA×
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489: [sage] 2025/01/08(水) 11:46:17.38 ID:qwVyKE52 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 G={ρ<0}, and ∂G={ρ=0}. http://rio2016.5ch.net/test/read.cgi/math/1724146576/489
名前１３２人目の素数さん 投稿日水 擬凸集合英 は 次元複素空間 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 こういうときはを見るのが定石でして なるほど か 参考

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