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# 数学代写|复分析代写Complex analysis代考|Introduction to the present section

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## 数学代写|复分析代写Complex analysis代考|Introduction to the present section

Lineal convexity is a kind of complex convexity intermediate between usual convexity and pseudoconvexity. More precisely, if $A$ is a convex set which is either open or closed, then $A$ is lineally convex (this is true also in the real category), and if $\Omega$ is a lineally convex open set in $\mathbf{C}^n$, the space of $n$ complex variables, then $\Omega$ is pseudoconvex. Now pseudoconvexity is a local property in the sense that if any boundary point of an open set $\Omega$ has an open neighborhood $\omega$ such that $\Omega \cap \omega$ is pseudoconvex, then $\Omega$ is pseudoconvex; the analogous result holds for convexity. But it is well known that the property of lineal convexity is not a local property in this sense-for easy examples see SubSection 9.4.3. The purpose of this section is to investigate to what extent this is true for sets that are of a special form: the Hartogs domains.
Let us now give the main definition.
Definition 9.4.1 A set $A$ in $\mathbf{C}^n$ is said to be lineally concave if it is a union of hyperplanes. It is called lineally convex if its complement is lineally concave.

A lineally convex set whose boundary is sufficiently smooth satisfies a differential condition. Let $\rho$ be a defining function for $\Omega$ (see Definition 9.4.18), and let $H$ and $L$ denote, respectively, the Hessian and the Levi form at a boundary point a of $\Omega$. Then the differential condition says that
$$|H(s)| \leqslant L(s) \text { for all vectors } s \in T_{\mathbf{C}}(a)$$
where $T_{\mathbf{C}}(a)$ is the complex tangent space at the point $a$. See SubSection 9.4.5 for details. Every lineally convex domain of class $C^2$ satisfies the differential condition-for the converse, see Section 9.6. Here we shall prove that this is so in the special case of Hartogs domains, which we now proceed to define.

## 数学代写|复分析代写Complex analysis代考|Weak lineal convexity

There are several other notions related to lineal convexity:
Definition 9.4.6 An open connected set is called weakly lineally convex if through any boundary point there passes a complex hyperplane which does not intersect the set. An open set is said to be locally weakly lineally convex ${ }^3$ if through every boundary point $a \in \partial \Omega$ there is a complex hyperplane $Y$ passing through a such that a does not belong to the closure of $Y \cap \Omega$.

It is not difficult to prove that local weak lineal convexity implies pseudoconvexity.
For complete Hartogs sets it is very easy to see that weak lineal convexity implies lineal convexity:

Lemma 9.4.7 A complete Hartogs domain which is weakly lineally convex and has a lineally convex base is lineally convex.

Proof Let $\left(z^0, t^0\right) \in \mathbf{C}^n \times \mathbf{C}$ be an arbitrary point in the complement of $\Omega$, a Hartogs domain defined by (9.42). If $R\left(z^0\right)>0$, then the point $\left(z^0, R\left(z^0\right) t^0 /\left|t^0\right|\right)$ belongs to $\partial \Omega$, and if $\Omega$ is weakly lineally convex, there is a hyperplane passing through that point which does not cut $\Omega$. Then the parallel plane through $\left(z^0, t^0\right)$ does not cut $\Omega$ either. If $R\left(z^0\right) \leqslant 0$, then $z^0$ does not belong to the base, and a hyperplane with equation $\zeta \cdot z=\zeta \cdot z^0$ will do, since the base is lineally convex. This proves the lemma.

## 数学代写|复分析代写Complex analysis代考|Introduction to the present section

$$|H(s)| \leqslant L(s) \text { for all vectors } s \in T_{\mathbf{C}}(a)$$

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