乐山师范学院数理学院，四川 乐山

收稿日期：2023年6月29日；录用日期：2023年7月31日；发布日期：2023年8月7日

摘要

陆启铿问题指的是一个域D是否是陆启铿域，陆启铿域是对所有的 $z,w\in D$ ，Bergman核 $K\left(z,w\right)$ 都不等于零的域。本文讨论了一类无界域 $D=\left\{\left(z,u_1,u_2\right)\in ℂ\times \mathbb{B}\times \mathbb{B}:e^{{\lambda }_1{\left|z\right|}^2}{\left|u_1\right|}^2+e^{{\lambda }_2{\left|z\right|}^2}{\left|u_2\right|}^2<1\right\}$ 的陆启铿问题。

关键词

Bergman核，陆启铿问题

Lu Qi-Keng’s Problem on Some Unbounded Domains

Jian Liu, Yuling Zhong, Qiongfang Hu, Yao Yang, Bing Xie, Zhiming Feng^*

School of Mathematics and Physics, Leshan Normal University, Leshan Sichuan

Received: Jun. 29^th, 2023; accepted: Jul. 31^st, 2023; published: Aug. 7^th, 2023

ABSTRACT

The Lu Qi-keng problem refers to whether a domain D is a Lu Qi-keng domain. The Lu Qi-keng domain is a domain where all $z,w\in D$ , and its Bergman kernel $K\left(z,w\right)$ are not equal to zero. In this paper, we investigate the Lu Qi-Keng problem for unbounded domains $D=\left\{\left(z,u_1,u_2\right)\in ℂ\times \mathbb{B}\times \mathbb{B}:e^{{\lambda }_1{\left|z\right|}^2}{\left|u_1\right|}^2+e^{{\lambda }_2{\left|z\right|}^2}{\left|u_2\right|}^2<1\right\}$ .

Keywords:Bergman Kernels, Lu Qi-Keng Problem

Copyright © 2023 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

假定 $D\subset {ℂ}^n$ 中的域，定义D上平方可积解析函数空间

${\mathcal{A}}^2\left(D\right)=\left\{f\in \text{Hol}\left(D\right):{\displaystyle {\int }_D{\left|f\left(z\right)\right|}^2\text{d}V\left(z\right)}<+\infty \right\},$

这里 $z\in D$ ， $\text{Hol}\left(D\right)$ 表示D上解析函数全体， $\text{d}V$ 表示Lebesgue测度。如果 ${\mathcal{A}}^2\left(D\right)\ne 0$ ，记空间 ${\mathcal{A}}^2\left(D\right)$ 的标准正交基为 $\left\{{\phi }_j\left(z\right):1\le j\le \dim {\mathcal{A}}^2\left(D\right)\right\}$ ，称

$K\left(z,w\right)={\displaystyle \underset{j=1}{\overset{\dim {\mathcal{A}}^2\left(D\right)}{\sum }}{\phi }_j\left(z\right)\overline{{\phi }_j\left(w\right)}}$

为域D的Bergman核，Bergman核在中具有再生性，即对任意 $f\in {\mathcal{A}}^2\left(D\right)$ ，有

$f\left(z\right)={\displaystyle {\int }_{D}K}\left(z,w\right)f\left(w\right)\text{d}V\left(w\right)\text{.}$

对于包含原点的Reinhardt域D，D上解析函数有以下幂级数表示：

$f\left(z\right)={\displaystyle {\sum }_{m\ge 0}{c}_m{z}^m},$

这里 $z=\left(z,\cdots ,z_n\right)\in D$ ， $m=\left(m_1,\cdots ,m_n\right)\in {\mathbb{N}}^n$ ， $z^m=z_1^{m_1}\cdots z_n^{m_n}$ 。并且 $\frac{z^m}{\Vert z^m\Vert }$ 是 ${\mathcal{A}}^2\left(D\right)$ 的标准正交基，因此 [1]

$K\left(z,w\right)=\underset{m\in {\mathbb{N}}^n}{{\displaystyle \sum }}\frac{z^m\overline{w^m}}{{\Vert z^m\Vert }^2}\text{.}$

Bergman核在多复变函数论和复几何中起着重要的作用，虽然 ${ℂ}^n$ 中的有界域都存在Bergman核，但其表达式一般没有显示公式，用显式公式计算Bergman核函数是多复变函数论的一个重要研究方向，用Bergman核的显示表达式，可以研究域的陆启铿问题，所谓陆启铿问题指的是一个域是否是陆启铿域，陆启铿域是对所有的 $z,w\in D$ ，Bergman核 $K\left(z,w\right)$ 都不等于零的域，如果一个域是陆启铿域，则它的表示域存在 [2] 。关于Bergman核的计算，可参考综述文献 [3] [4] ，有界域的陆启铿问题研究可参考综述文献 [5] ，近期这方面的研究可参考 [6] [7] [8] [9] 等。在 [7] 和 [9] 中研究了一类无界域——Fock-Bargmann-Hartogs域

$D_{n,m}=\left\{\left(z,w\right)\in {ℂ}^n\times {ℂ}^m:{\Vert w\Vert }^2<{\text{e}}^{-\mu {\Vert z\Vert }^2}\right\},\mu >0$

的陆启铿问题，本文我们研究另一类无界域(定义见(2.1))的Bergman核的计算，并研究其陆启铿问题。

2. 主要结果

定理2.1. 令 ${\lambda }_1\ge {\lambda }_2>0$ ，

$D=\left\{\left(z,u_1,u_2\right)\in ℂ\times \mathbb{B}\times \mathbb{B}:{\text{e}}^{{\lambda }_1{\left|z\right|}^2}{\left|u_1\right|}^2+{\text{e}}^{{\lambda }_2{\left|z\right|}^2}{\left|u_2\right|}^2<1\right\}$ (2.1)

则域D的Bergman核为

(2.2)

其中

$Z=\left(z,u_1,u_2\right),W=\left(w,v_1,v_2\right)\in D$

${\rho }_1={\text{e}}^{{\lambda }_{1}z\bar{w}}{u}_1\overline{v_1},{\rho }_2={\text{e}}^{{\lambda }_{2}z\bar{w}}{u}_2\overline{v_2}\text{.}$

由定理2.1，我们得到以下结论。

定理2.2. 在定理2.1的假设下，则域D的Bergman核 $K\left(Z,W\right)$ 在 $D\times D$ 上有零点的充要条件是 ${\lambda }_1>2{\lambda }_2>0$ 。

3. 定理2.1的证明

证明. 记

${\mathcal{A}}^2\left(D\right)=\left\{f\in \text{Hol}\left(D\right):{\displaystyle {\int }_D{\left|f\left(Z\right)\right|}^2\text{d}V\left(Z\right)}<+\infty \right\},$

这里 $Z=\left(z,u_1,u_2\right)\in D$ ， $\text{Hol}\left(D\right)$ 表示D上解析函数全体， $\text{d}V$ 表示Lebesgue测度。对于包含原点的Reinhardt域D上的全纯函数f，在D上有幂级数展开式(见 [1] )

$f\left(Z\right)={\displaystyle {\sum }_{m,n\ge 0}{a}_{mn}{z}^m{u}^n},$

这里

$Z=\left(z,u\right)=\left(z,u_1,u_2\right),n=\left(n_1,n_2\right),u^n=u_1^{n_1}{u}_2^{n_2}\text{.}$

容易直接证明 $\left\{\frac{z^m{u}^n}{\Vert z^m{u}^n\Vert }\right\}$ 是Reinhardt域上的一组规范正交系，可以证明它是完备的。因而有

$K\left(Z,W\right)=\underset{m,n\ge 0}{{\displaystyle \sum }}\frac{z^m{u}^n\Vert w^m{v}^n\Vert }{{\Vert z^m{u}^n\Vert }^2}$

其中

$\begin{array}{l}Z=\left(z,u\right)=\left(z,u_1,u_2\right),W=\left(w,v\right)=\left(w,v_1,v_2\right)\in D,\\ {\Vert z^m{u}^n\Vert }^2={\displaystyle {\int }_D{\left|z^m{u}^n\right|}^2\text{d}V}\left(z\right)\text{d}V\left(u_1\right)\text{d}V\left(u_2\right)\text{.}\end{array}$

直接计算得

上面用到

$\begin{array}{l}{\displaystyle {\int }_{{\left|w_1\right|}^2+{\left|w_2\right|}^2<1}{\left|w^n\right|}^2\text{d}V\left(w_1\right)\text{d}V\left(w_2\right)}\\ ={\left(2\pi \right)}^2{\displaystyle {\int }_{\begin{array}{l}{r}_1^2+r_2^2<1\\ r_1\ge 0,r_2\ge 0\end{array}}{r}_1^{2n_1+1}{r}_2^{2n_2+1}\text{d}{r}_1\text{d}{r}_2}={\pi }^2{\displaystyle {\int }_{\begin{array}{l}{x}_1+x_2<1\\ x_1\ge 0,x_2\ge 0\end{array}}{x}_1^{n_1}{x}_2^{n_2}\text{d}{x}_1\text{d}{x}_2}\\ =\frac{{\pi }^2}{n_2+1}{\displaystyle {\int }_0^1{x}_1^{n_1}{\left(1-x_1\right)}^{n_2+1}\text{d}{x}_1}={\pi }^2\frac{n_1!n_2!}{\left(n_1+n_2+2\right)!}\end{array}$

以及

这里用到

所以当 $Z=\left(z,u_1,u_2\right),W=\left(w,v_1,v_2\right)\in D$ 时

$\begin{array}{c}K\left(Z,W\right)=\underset{m,n\ge 0}{{\displaystyle \sum }}\frac{z^m{u}^n\overline{w^m{v}^n}}{{\Vert z^m{u}^n\Vert }^2}\\ =\frac{1}{{\pi }^3}\underset{m,n\ge 0}{{\displaystyle \sum }}\frac{{\left({{\displaystyle \sum }}_{i=1}^2\left(n_i+1\right){\lambda }_i\right)}^{m+1}\left(n_1+n_2+2\right)!}{m!n_1!n_2!}{z}^m{u}^n\overline{w^m{v}^n}\\ =\frac{1}{{\pi }^3}{\displaystyle \underset{n\ge 0}{\sum }\phi }\left(n_1,n_2\right){\text{e}}^{{{\displaystyle \sum }}_{i=1}^2\left(n_i+1\right){\lambda }_{i}z\bar{w}}\frac{\left(n_1+n_2+2\right)!}{n_1!n_2!}{u}^n\overline{v^n}\\ =\frac{{\text{e}}^{{\displaystyle {\sum }_{i=1}^2{\lambda }_{i}z\bar{w}}}}{{\pi }^3}{\displaystyle \underset{n\ge 0}{\sum }\phi }\left(n_1,n_2\right)\frac{\left(n_1+n_2+2\right)!}{n_1!n_2!}{\rho }_1^{n_1}{\rho }_2^{n_2},\end{array}$

这里

${\rho }_1={\text{e}}^{{\lambda }_{1}z\bar{w}}{u}_1\overline{v_1},{\rho }_2={\text{e}}^{{\lambda }_{2}z\bar{w}}{u}_2\overline{v_2},\phi \left(n_1,n_2\right)={\displaystyle {\sum }_{i=1}^2\left(n_i+1\right){\lambda }_i}$ 。

由于

$\begin{array}{l}{\left(1-{\rho }_1-{\rho }_2\right)}^{-\alpha }=\underset{n_1\ge 0,n_2\ge 0}{{\displaystyle \sum }}\frac{\Gamma \left(\alpha +n_1+n_2\right)}{\Gamma \left(\alpha \right)n_1!n_2!}{\rho }_1^{n_1}{\rho }_2^{n_2}\left(\alpha >0\right),\\ {\left(t\frac{\text{d}}{\text{d}t}\right)}^k{t}^n=n^k{t}^n,\end{array}$

于是有

$\begin{array}{c}K\left(Z,W\right)=\frac{2{\text{e}}^{{\displaystyle {\sum }_{i=1}^2{\lambda }_{i}z\bar{w}}}}{{\pi }^3}{\displaystyle \underset{n\ge 0}{\sum }\phi }\left(n_1,n_2\right)\frac{\Gamma \left(n_1+n_2+3\right)}{\Gamma \left(3\right)n_1!n_2!}{\rho }_1^{n_1}{\rho }_2^{n_2}\\ ={\frac{2{\text{e}}^{{\displaystyle {\sum }_{i=1}^2{\lambda }_{i}z\bar{w}}}}{{\pi }^3}\phi \left(t_1\frac{\text{d}}{\text{d}{t}_1},t_2\frac{\text{d}}{\text{d}{t}_2}\right)\frac{1}{{\left(1-t_1{\rho }_1-t_2{\rho }_2\right)}^3}|}_{t_1=1,t_2=1}.\end{array}$

记

$\phi \left(n_1,n_2\right)=c_1{n}_1+c_2{n}_2+c_0,$

这里

$c_1={\lambda }_1,c_2={\lambda }_2,c_0={\lambda }_1+{\lambda }_2\text{.}$

根据

$\begin{array}{l}{t}_1\frac{\text{d}}{\text{d}{t}_1}\frac{1}{{\left(1-t_1{\rho }_1-t_2{\rho }_2\right)}^3}=\frac{3{\rho }_1{t}_1}{{\left(1-t_1{\rho }_1-t_2{\rho }_2\right)}^4},\\ t_2\frac{\text{d}}{\text{d}{t}_2}\frac{1}{{\left(1-t_1{\rho }_1-t_2{\rho }_2\right)}^3}=\frac{3{\rho }_2{t}_2}{{\left(1-t_1{\rho }_1-t_2{\rho }_2\right)}^4},\end{array}$

得

$\begin{array}{l}{\phi \left(t_1\frac{\text{d}}{\text{d}{t}_1},t_2\frac{\text{d}}{\text{d}{t}_2}\right)\frac{1}{{\left(1-t_1{\rho }_1-t_2{\rho }_2\right)}^3}|}_{t_1=1,t_2=1}\\ =\frac{3c_1{\rho }_1}{{\left(1-{\rho }_1-{\rho }_2\right)}^4}+\frac{3c_2{\rho }_2}{{\left(1-{\rho }_1-{\rho }_2\right)}^4}+\frac{c_0}{{\left(1-{\rho }_1-{\rho }_2\right)}^3}.\end{array}$

因此

$\begin{array}{c}K\left(Z,W\right)=\frac{2{\text{e}}^{{\displaystyle {\sum }_{i=1}^2{\lambda }_{i}z\bar{w}}}}{{\pi }^3}\left(\frac{3c_1{\rho }_1+3c_2{\rho }_2}{{\left(1-{\rho }_1-{\rho }_2\right)}^4}+\frac{c_0}{{\left(1-{\rho }_1-{\rho }_2\right)}^3}\right)\\ =\frac{2{\text{e}}^{\left({\lambda }_1+{\lambda }_2\right)z\bar{w}}}{{\pi }^3}\frac{\left(2{\lambda }_1-{\lambda }_2\right){\rho }_1+\left(2{\lambda }_2-{\lambda }_1\right){\rho }_2+{\lambda }_1+{\lambda }_2}{{\left(1-{\rho }_1-{\rho }_2\right)}^4}.\end{array}$

证毕

4. 定理2.2的证明

为了研究定理2.1中域D的陆启铿问题，我们先给出以下结论。

定理4.1. 在定理2.1的条件下，以下结论成立：

(i) 如果 $Z,W\in D$ ，那么

$\left|{\rho }_1\right|<1,\left|{\rho }_2\right|<1,\left|{\rho }_1+{\rho }_2\right|<1$ (4.1)

(ii) 如果 ${\theta }_1\in ℝ,{\theta }_2\in ℝ,0\le r_1<1,0\le r_2<1$ 和 $r_1+r_2<1$ ，则存在 $Z,W\in D$ ，使得 ${\rho }_1\left(Z,W\right)=r_1{\text{e}}^{\text{i}{\theta }_1}$ 和 ${\rho }_2\left(Z,W\right)=r_2{\text{e}}^{\text{i}{\theta }_2}$ 。

证明. (i) 由于

$\begin{array}{c}{\rho }_1={\text{e}}^{{\lambda }_{1}z\bar{w}}{u}_1\overline{v_1}=u_1\overline{v_1}\underset{m=0}{\overset{+\infty }{{\displaystyle \sum }}}\frac{{\left(\sqrt{{\lambda }_1}z\right)}^m}{\sqrt{m!}}\frac{\stackrel{¯}{{\left(\sqrt{{\lambda }_1}w\right)}^m}}{\sqrt{m!}}\\ =\underset{m=0}{\overset{+\infty }{{\displaystyle \sum }}}\left(u_1\frac{{\left(\sqrt{{\lambda }_1}z\right)}^m}{\sqrt{m!}}\right)\left(\stackrel{¯}{v_1\frac{{\left(\sqrt{{\lambda }_1}w\right)}^m}{\sqrt{m!}}}\right)={\displaystyle \underset{m=0}{\overset{+\infty }{\sum }}{\psi }_m^{{\lambda }_1}}\left(z,u_1\right)\overline{{\psi }_m^{{\lambda }_1}\left(w,v_1\right)}\end{array}$

同样得

${\rho }_2={\text{e}}^{{\lambda }_{2}z\bar{w}}{u}_2\overline{v_2}={\displaystyle \underset{m=0}{\overset{+\infty }{\sum }}{\psi }_m^{{\lambda }_2}\left(z,u_2\right)\overline{{\psi }_m^{{\lambda }_2}\left(w,v_2\right)}},$

其中

${\psi }_m^{\lambda }\left(z,u\right)=u\frac{{\left(\sqrt{\lambda }z\right)}^m}{\sqrt{m!}}$

于是由不等式

${\left|{\displaystyle \underset{i=1}{\overset{+\infty }{\sum }}{a}_i\overline{b_i}}\right|}^2\le {\displaystyle \underset{i=1}{\overset{+\infty }{\sum }}{\left|a_i\right|}^2}{\displaystyle \underset{i=1}{\overset{+\infty }{\sum }}{\left|b_i\right|}^2}$

得

$\begin{array}{c}{\left|{\rho }_1+{\rho }_2\right|}^2\le \left({\displaystyle \underset{m=0}{\overset{+\infty }{\sum }}{\psi }_m^{{\lambda }_1}}\left(z,u_1\right)\overline{{\psi }_m^{{\lambda }_1}\left(z,u_1\right)}+{\displaystyle \underset{m=0}{\overset{+\infty }{\sum }}{\psi }_m^{{\lambda }_2}}\left(z,u_2\right)\overline{{\psi }_m^{{\lambda }_2}\left(z,u_2\right)}\right)\\ \times \left({\displaystyle \underset{m=0}{\overset{+\infty }{\sum }}{\psi }_m^{{\lambda }_1}}\left(w,v_1\right)\overline{{\psi }_m^{{\lambda }_1}\left(w,v_1\right)}+{\displaystyle \underset{m=0}{\overset{+\infty }{\sum }}{\psi }_m^{{\lambda }_2}}\left(w,v_2\right)\overline{{\psi }_m^{{\lambda }_2}\left(w,v_2\right)}\right)\\ =\left({\text{e}}^{{\lambda }_1{\left|z\right|}^2}{\left|u_1\right|}^2+{\text{e}}^{{\lambda }_2{\left|z\right|}^2}{\left|u_2\right|}^2\right)\left({\text{e}}^{{\lambda }_1{\left|w\right|}^2}{\left|v_1\right|}^2+{\text{e}}^{{\lambda }_2{\left|w\right|}^2}{\left|v_2\right|}^2\right).\end{array}$

再根据

${\text{e}}^{{\lambda }_1{\left|z\right|}^2}{\left|u_1\right|}^2+{\text{e}}^{{\lambda }_2{\left|z\right|}^2}{\left|u_2\right|}^2<1,{\text{e}}^{{\lambda }_1{\left|w\right|}^2}{\left|v_1\right|}^2+{\text{e}}^{{\lambda }_2{\left|w\right|}^2}{\left|v_2\right|}^2<1$

得

$\left|{\rho }_1+{\rho }_2\right|<1.$

同样得

$\left|{\rho }_1\right|<1,\left|{\rho }_2\right|<1.$

(ii) 令 $Z=\left(z,,u_1,u_2\right)=\left(0,u_1,u_2\right),W=\left(w,v_1,v_2\right)=\left(0,v_1,v_2\right)$ ，则 ${\rho }_1=u_1\overline{v_1},{\rho }_2=u_2\overline{v_2}$ ，并且

如果 $r_1$ 和 $r_2$ 至少有一个为零，则结论成立。以下假定 $0<r_1<1,0<r_2<1,r_1+r_2<1$ ，并证明关于 $\left(u_1,u_2,v_1,v_2\right)$ 的不等式组

$\{\begin{array}{l}{u}_1\overline{v_1}=r_1{\text{e}}^{\text{i}{\theta }_1},\\ u_2\overline{v_2}=r_2{\text{e}}^{\text{i}{\theta }_2},\\ {\left|u_1\right|}^2+{\left|u_2\right|}^2<1,\\ {\left|v_1\right|}^2+{\left|v_2\right|}^2<1.\end{array}$ (4.2)

有解。

令 $u_j=\sqrt{x_j}{\text{e}}^{\text{i}{\chi }_j},0<x_j,1\le j\le 2$ ，则上面不等式组(4.2)等价于

$v_j=\frac{r_j}{\sqrt{x_j}}{\text{e}}^{\text{i}\left({\chi }_j-{\theta }_j\right)},0<x_1+x_2<1,\frac{r_1^2}{x_1}+\frac{r_2^2}{x_2}<1.$

进一步化简得

$r_1^2<x_1<1,\frac{r_2^2{x}_1}{x_1-r_1^2}<x_2<1-x_1\text{.}$

于是不等式组(4.2)有解等价于不等式组

$r_1^2<x_1<1,-x_1^2+\left(1+r_1^2-r_2^2\right)x_1-r_1^2>0$

有解。容易验证当 $r_1+r_2<1$ 时， $x_1=\frac{1}{2}\left(1+r_1^2-r_2^2\right)$ 是它的解。证毕

现在给出定理2.2的证明。

定理2.2的证明。(i) 当 ${\lambda }_1={\lambda }_2$ 时，有

$K\left(Z,W\right)=\frac{2{\text{e}}^{2{\lambda }_{1}z\bar{w}}}{{\pi }^3}\frac{{\lambda }_1\left({\rho }_1+{\rho }_2+2\right)}{{\left(1-{\rho }_1-{\rho }_2\right)}^4}\text{.}$

因

$\left|{\rho }_1+{\rho }_2+2\right|\ge 2-\left|{\rho }_1+{\rho }_2\right|\ge 1,$

所以

$K\left(Z,W\right)\ne 0,\forall Z,W\in D\text{.}$

(ii) 当 ${\lambda }_1=2{\lambda }_2$ 时，有

$K\left(Z,W\right)=\frac{2{\text{e}}^{3{\lambda }_{2}z\bar{w}}}{{\pi }^3}\frac{3{\lambda }_2\left({\rho }_1+1\right)}{{\left(1-{\rho }_1-{\rho }_2\right)}^4}\text{.}$

由此得

$K\left(Z,W\right)\ne 0,\forall Z,W\in D\text{.}$

(iii) 当 ${\lambda }_1\ne {\lambda }_2,{\lambda }_1\ne 2{\lambda }_2$ 时，如果 $K\left(Z,W\right)=0$ ，根据(2.2)有

${\rho }_1=\frac{{\lambda }_1-2{\lambda }_2}{2{\lambda }_1-{\lambda }_2}{\rho }_2-\frac{{\lambda }_1+{\lambda }_2}{2{\lambda }_1-{\lambda }_2},{\rho }_1+{\rho }_2=\frac{3\left({\lambda }_1-{\lambda }_2\right)}{2{\lambda }_1-{\lambda }_2}{\rho }_2-\frac{{\lambda }_1+{\lambda }_2}{2{\lambda }_1-{\lambda }_2}\text{.}$ (4.3)

令 ${\rho }_2=r{\text{e}}^{\text{i}\theta }$ ，则

$\begin{array}{l}{\left|{\rho }_1\right|}^2=\frac{{\left({\lambda }_1-2{\lambda }_2\right)}^2{r}^2-2\left({\lambda }_1+{\lambda }_2\right)\left({\lambda }_1-2{\lambda }_2\right)r\cos \left(\theta \right)+{\left({\lambda }_1+{\lambda }_2\right)}^2}{{\left(2{\lambda }_1-{\lambda }_2\right)}^2},\\ {\left|{\rho }_1+{\rho }_2\right|}^2=\frac{9{\left({\lambda }_1-{\lambda }_2\right)}^2{r}^2-6\left({\lambda }_1-{\lambda }_2\right)\left({\lambda }_1+{\lambda }_2\right)r\cos \left(\theta \right)+{\left({\lambda }_1+{\lambda }_2\right)}^2}{{\left(2{\lambda }_1-{\lambda }_2\right)}^2}.\end{array}$

根据

$\left|{\rho }_1\right|<1,\left|{\rho }_2\right|<1,\left|{\rho }_1+{\rho }_2\right|<1,$

得不等式组

$\{\begin{array}{l}0\le r<1,\\ \frac{{\left({\lambda }_1-2{\lambda }_2\right)}^2{r}^2-2\left({\lambda }_1+{\lambda }_2\right)\left({\lambda }_1-2{\lambda }_2\right)r\cos \left(\theta \right)+{\left({\lambda }_1+{\lambda }_2\right)}^2}{{\left(2{\lambda }_1-{\lambda }_2\right)}^2}<1,\\ \frac{9{\left({\lambda }_1-{\lambda }_2\right)}^2{r}^2-6\left({\lambda }_1-{\lambda }_2\right)\left({\lambda }_1+{\lambda }_2\right)r\cos \left(\theta \right)+{\left({\lambda }_1+{\lambda }_2\right)}^2}{{\left(2{\lambda }_1-{\lambda }_2\right)}^2}<1\end{array}$ (4.4)

关于 $\left(r,\theta \right)$ 有解。

记

$\begin{array}{l}f\left(r,\theta \right)={\left({\lambda }_1-2{\lambda }_2\right)}^2{r}^2-2\left({\lambda }_1+{\lambda }_2\right)\left({\lambda }_1-2{\lambda }_2\right)r\cos \left(\theta \right)+{\left({\lambda }_1+{\lambda }_2\right)}^2-{\left(2{\lambda }_1-{\lambda }_2\right)}^2,\\ g\left(r,\theta \right)=9{\left({\lambda }_1-{\lambda }_2\right)}^2{r}^2-6\left({\lambda }_1-{\lambda }_2\right)\left({\lambda }_1+{\lambda }_2\right)r\cos \left(\theta \right)+{\left({\lambda }_1+{\lambda }_2\right)}^2-{\left(2{\lambda }_1-{\lambda }_2\right)}^2.\end{array}$

则 $f=0$ 的解为

$r=\frac{\cos \left(\theta \right)\left({\lambda }_1+{\lambda }_2\right)\pm \sqrt{{\left({\lambda }_1+{\lambda }_2\right)}^2{\left(\cos \left(\theta \right)\right)}^2+3\left({\lambda }_1-2{\lambda }_2\right){\lambda }_1}}{{\lambda }_1-2{\lambda }_2};$

$g=0$ 的解为

$r=\frac{\cos \left(\theta \right)\left({\lambda }_1+{\lambda }_2\right)\pm \sqrt{{\left({\lambda }_1+{\lambda }_2\right)}^2{\left(\cos \left(\theta \right)\right)}^2+3\left({\lambda }_1-2{\lambda }_2\right){\lambda }_1}}{3\left({\lambda }_1-{\lambda }_2\right)}\text{.}$

(iii-1) 当 ${\lambda }_1>2{\lambda }_2$ 时，令 $\theta =0,f=0$ 的解为

$r=-1,\frac{3{\lambda }_1}{{\lambda }_1-2{\lambda }_2}=3+\frac{6{\lambda }_2}{{\lambda }_1-2{\lambda }_2};$

$g=0$ 的解为

$r=-\frac{{\lambda }_1-2{\lambda }_2}{3\left({\lambda }_1-{\lambda }_2\right)},\frac{{\lambda }_1}{{\lambda }_1-{\lambda }_2}=1+\frac{{\lambda }_2}{{\lambda }_1-{\lambda }_2}\text{.}$

此时不等式 $f<0$ 的解为

$\theta =0,-1<r<3+\frac{6{\lambda }_2}{{\lambda }_1-2{\lambda }_2}.$

不等式 $g<0$ 的解为

$\theta =0,-\frac{{\lambda }_1-2{\lambda }_2}{3\left({\lambda }_1-{\lambda }_2\right)}<r<1+\frac{{\lambda }_2}{{\lambda }_1-{\lambda }_2}\text{.}$

这表明不等式组(4.4)有解

$\theta =0,0\le r<1,$

${\rho }_2=r,{\rho }_1=\frac{{\lambda }_1-2{\lambda }_2}{2{\lambda }_1-{\lambda }_2}r-\frac{{\lambda }_1+{\lambda }_2}{2{\lambda }_1-{\lambda }_2}\text{.}$

当 $0\le r<\min \left\{1,\frac{{\lambda }_1-2{\lambda }_2}{3\left({\lambda }_1-{\lambda }_2\right)}\right\}$ 时，有 $\left|{\rho }_1\right|+\left|{\rho }_2\right|<1$ ，根据定理4.1得 $K\left(Z,W\right)=0$ 在 $D\times D$ 内有解。

(iii-2) 当 ${\lambda }_2<{\lambda }_1<2{\lambda }_2$ 时，如果不等式组(4.4)有解，则 $r=0$ 不可能是(4.4)的解，这是因为

$\frac{{\left({\lambda }_1+{\lambda }_2\right)}^2}{{\left(2{\lambda }_1-{\lambda }_2\right)}^2}>1.$

于是解 $\left(r,\theta \right)$ 满足以下条件

$\begin{array}{l}\max \left\{0,\frac{\cos \left(\theta \right)\left({\lambda }_1+{\lambda }_2\right)+\sqrt{\Delta }}{{\lambda }_1-2{\lambda }_2},\frac{\cos \left(\theta \right)\left({\lambda }_1+{\lambda }_2\right)-\sqrt{\Delta }}{3\left({\lambda }_1-{\lambda }_2\right)}\right\}\\ <r<\min \left\{1,\frac{\cos \left(\theta \right)\left({\lambda }_1+{\lambda }_2\right)-\sqrt{\Delta }}{{\lambda }_1-2{\lambda }_2},\frac{\cos \left(\theta \right)\left({\lambda }_1+{\lambda }_2\right)+\sqrt{\Delta }}{3\left({\lambda }_1-{\lambda }_2\right)}\right\},\end{array}$ (4.5)

这里

$\text{Δ}={\left({\lambda }_1+{\lambda }_2\right)}^2{\left(\cos \left(\theta \right)\right)}^2+3\left({\lambda }_1-2{\lambda }_2\right){\lambda }_1\text{.}$

由(4.5)可知

$\frac{\cos \left(\theta \right)\left({\lambda }_1+{\lambda }_2\right)-\sqrt{\text{Δ}}}{{\lambda }_1-2{\lambda }_2}>0,\frac{\cos \left(\theta \right)\left({\lambda }_1+{\lambda }_2\right)+\sqrt{\text{Δ}}}{3\left({\lambda }_1-{\lambda }_2\right)}>0.$

这等价于

$-\frac{\sqrt{\Delta }}{{\lambda }_1+{\lambda }_2}<\cos \left(\theta \right)<\frac{\sqrt{\Delta }}{{\lambda }_1+{\lambda }_2}.$

进一步化简得

$0<3\left({\lambda }_1-2{\lambda }_2\right){\lambda }_1\text{.}$

这和条件 ${\lambda }_1-2{\lambda }_2<0$ 矛盾。因此在 ${\lambda }_2<{\lambda }_1<2{\lambda }_2$ 时，对一切 $Z,W\in D$ ，Bergman核 $K\left(Z,W\right)\ne 0$ 。证毕

致谢

大学生创新创业训练计划项目(No: S202210649222)。

文章引用

刘 健,钟雨玲,胡琼方,杨 尧,谢 兵,冯志明. 一类无界域的陆启铿问题
Lu Qi-Keng’s Problem on Some Unbounded Domains[J]. 理论数学, 2023, 13(08): 2275-2283. https://doi.org/10.12677/PM.2023.138234

参考文献