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# 数学代写|复分析代写Complex analysis代考|Lineally convex Hartogs sets

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## 数学代写|复分析代写Complex analysis代考|Lineally convex Hartogs sets

Intuitively, it seems that $E(\emptyset ; f)$ and $E(\operatorname{dom}(f) ; f)$ ought to be lineally convex simultaneously. This is not quite true. We shall note three results in the positive direction, Propositions 9.8.389.8.40 below, and one result in the negative direction, Example 9.8.41. Then we shall establish conditions under which it is true that $f$ is $\mathscr{L}$-closed if and only if $E(\operatorname{dom}(f) ; f)$ is lineally convex (Corollary 9.8.43), as well as conditions which guarantee that $f$ is $\mathscr{L}$-closed if and only if $E(\emptyset ; f)$ is lineally convex (Theorem 9.8.49).

Proposition 9.8.38 If $E(X ; f)$ is lineally convex, then also $X \cup \operatorname{dom}(f)$ and $E(X \cup \operatorname{dom}(f) ; f)$ are lineally convex. In particular, if $E(\emptyset ; f)$ is lineally convex, then so are $\operatorname{dom}(f)$ and $E(\operatorname{dom}(f) ; f)$.

Proof Suppose that $E(X ; f)$ is lineally convex. That $X \cup \operatorname{dom}(f)$ is lineally convex then follows from the easily proved result that the intersection of a lineally convex set and a complex subspace is lineally convex as a subset of the latter. If $E(X ; f)$ is lineally convex, then also $E(X ; f+a)$ is lineally convex for any real number a. Any intersection of lineally convex sets has the same property, so we only need to note that $E(X \cup \operatorname{dom}(f) ; f)$ is equal the intersection of all $E(X ; f-a), a>0$.

Proof Suppose that $E(X ; f)$ is lineally convex. That $X \cup \operatorname{dom}(f)$ is lineally convex then follows from the easily proved result that the intersection of a lineally convex set and a complex subspace is lineally convex as a subset of the latter. If $E(X ; f)$ is lineally convex, then also $E(X ; f+a)$ is lineally convex for any real number a. Any intersection of lineally convex sets has the same property, so we only need to note that $E(X \cup \operatorname{dom}(f) ; f)$ is equal the intersection of all $E(X ; f-a), a>0$.

Proposition 9.8.39 If $f$ is upper semicontinuous and there exists a set $X$ such that $E(X ; f)$ is lineally convex, then $E(\emptyset ; f)$ is lineally convex.

Proof We know from Corollary 9.8 .4 that $E(X ; f)^{\circ}$ is lineally convex if $E(X ; f)$ is lineally convex. Now $E(X ; f)^{\circ}=E(\emptyset ; f)$ if $f$ is upper semicontinuous, hence the result.

However, the semicontinuity of $f$ is not important-it is the fact that the effective domain is open which is relevant. This is shown by the following result.

## 数学代写|复分析代写Complex analysis代考|A necessary differential condition for L -closed functions

It is well known that convex functions as well as plurisubharmonic functions of class $C^2$ can be characterized by differential conditions. Is the same true for $\mathscr{L}$-closed functions? We shall first establish a necessary differential condition.

Proposition 9.8.50 Suppose that $f$ is an $\mathscr{L}$-closed function of class $C^2$ in some open set $\Omega$ of $\mathbf{C}^{1+n} \backslash{0}$. Then
$$\left|\sum\left(f_{z_j z_k}-2 f_{z_j} f_{z_k}\right) b_j b_k\right| \leqslant \sum f_{z_j \bar{z}_k} b_j \bar{b}_k, \quad \text { in } \Omega \text { forall } b \in \mathbf{C}^{1+n} .$$

In particular, if $n=1$ and we define $F(z)=f(1, z), z \in \mathbf{C}$, then
$$\left|F_{z z}-2 F_z^2\right| \leqslant F_{z \bar{z}}$$
Proof Define $g(z)=-\log |\beta \cdot z|$. For every point $a$ where $f(a)$ is finite there is a vector $\beta$ such that $\operatorname{grad} g(a)=\operatorname{grad} f(a)$. Indeed, let us first note that by homogeneity $\sum a_j f_{z_j}(a)=-1 / 2$ for all $a$. If we choose $\beta_j=f_{z_j}(a)$, then $\beta \cdot a=-1 / 2$ and
$$\frac{\partial g}{\partial z_j}(z)=-\frac{1}{2} \frac{\beta_j}{\beta \cdot z}$$
takes the value
$$\frac{\partial g}{\partial z_j}(a)=-\frac{1}{2} \frac{\beta_j}{\beta \cdot a}=\beta_j$$

at $z=a$. Then by $\mathscr{L}$-closedness $f(z) \geqslant f(a)+g(z)-g(a)$ for all $z$, for the definition of the $\mathscr{L}$ transformation uses precisely the functions $g$ plus a constant. Take a curve $t \mapsto \gamma(t)$ such that $\gamma(0)=a$ and compare the two functions $\varphi=f^{\circ} \gamma$ and $\psi=f(a)+g^{\circ} \gamma-g(a)$. We have $\varphi(0)=\psi(0)$ and $\varphi^{\prime}(0)=\psi^{\prime}(0)$ and must therefore have $\varphi^{\prime \prime}(0) \geqslant \psi^{\prime \prime}(0)$. We calculate $\varphi^{\prime \prime}$ :
\begin{aligned} & \frac{1}{2} \varphi^{\prime \prime}(t)=\operatorname{Re} \sum f_{z_j z_k}(\gamma(t)) \gamma_j^{\prime}(t) \gamma_k^{\prime}(t) \ & \quad+\sum f_{z_j \bar{z}k}(\gamma(t)) \gamma_j^{\prime}(t) \overline{\gamma_k^{\prime}(t)}+\operatorname{Re} \sum f{z_j}(\gamma(t)) \gamma_j^{\prime \prime}(t) . \end{aligned}
The corresponding formula for $\psi$ simplifies to
$$\frac{1}{2} \psi^{\prime \prime}(t)=\operatorname{Re} \sum g_{z_j z_k}(\gamma(t)) \gamma_j^{\prime}(t) \gamma_k^{\prime}(t)+\operatorname{Re} \sum g_{z_j}(\gamma(t)) \gamma_j^{\prime \prime}(t)$$
since $g_{z_j}$ is holomorphic, i.e., $g_{z_j \bar{z}k}=0$. Also $$g{z_j z_k}(z)=\frac{1}{2} \frac{\beta_j \beta_k}{(\beta \cdot z)^2}=2 g_{z_j}(z) g_{z_k}(z)$$

## 数学代写|复分析代写Complex analysis代考|A necessary differential condition for L -closed functions

$$\left|\sum\left(f_{z_j z_k}-2 f_{z_j} f_{z_k}\right) b_j b_k\right| \leqslant \sum f_{z_j \bar{z}k} b_j \bar{b}_k, \quad \text { in } \Omega \text { forall } b \in \mathbf{C}^{1+n} .$$ 特别是，如果 $n=1$ 我们定义 $F(z)=f(1, z), z \in \mathbf{C}$ ，然后 $$\left|F{z z}-2 F_z^2\right| \leqslant F_{z \bar{z}}$$

$\operatorname{grad} g(a)=\operatorname{grad} f(a)$. 事实上，让我们首先注意到，通过同质性 $\sum a_j f_{z_j}(a)=-1 / 2$ 对全部 $a$. 如果我 们选择 $\beta_j=f_{z_j}(a)$ ，然后 $\beta \cdot a=-1 / 2$ 和
$$\frac{\partial g}{\partial z_j}(z)=-\frac{1}{2} \frac{\beta_j}{\beta \cdot z}$$

$$\frac{\partial g}{\partial z_j}(a)=-\frac{1}{2} \frac{\beta_j}{\beta \cdot a}=\beta_j$$

$$\frac{1}{2} \varphi^{\prime \prime}(t)=\operatorname{Re} \sum f_{z_j z_k}(\gamma(t)) \gamma_j^{\prime}(t) \gamma_k^{\prime}(t) \quad+\sum f_{z_j \bar{z} k}(\gamma(t)) \gamma_j^{\prime}(t) \overline{\gamma_k^{\prime}(t)}+\operatorname{Re} \sum f z_j(\gamma(t)) \gamma_j^{\prime \prime}(t)$$

$$\frac{1}{2} \psi^{\prime \prime}(t)=\operatorname{Re} \sum g_{z_j z_k}(\gamma(t)) \gamma_j^{\prime}(t) \gamma_k^{\prime}(t)+\operatorname{Re} \sum g_{z_j}(\gamma(t)) \gamma_j^{\prime \prime}(t)$$

$$g z_j z_k(z)=\frac{1}{2} \frac{\beta_j \beta_k}{(\beta \cdot z)^2}=2 g_{z_j}(z) g_{z_k}(z)$$

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