﻿ 涉及导曲线与分担超平面的正规定则
 上海理工大学学报  2022, Vol. 44 Issue (5): 490-496 PDF

Normal criteria relating to derived curves and shared hyperplanes
FAN Chujun, LIU Xiaojun
College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China
Abstract: Based on the theories of value distribution, normal family and the knowledge of advanced algebra, the normal criteria for the holomorphic curves and their derived curves sharing hyperplanes located in t-subgeneral position was studied. Let $\mathcal{F}$ be a family of holomorphic curves of a domain $D \subset \mathbb{C}$ to ${\mathbb{P}^N}(\mathbb{C})$ , ${H_\ell } =\{ {{\boldsymbol{x}} \in {\mathbb{P}^N}(\mathbb{C}):} \left. {\left\langle {{\boldsymbol{x}},{{\boldsymbol{\alpha}} _\ell }} \right\rangle = 0} \right\}$ be hyperplanes in ${\mathbb{P}^N}(\mathbb{C})$ located in t-subgeneral position, where ${{\boldsymbol{\alpha}} _\ell } = {\left( {{a_{\ell 0}},{a_{\ell 1}}, \cdots ,{a_{\ell N}}} \right)^{\text{T}}}$ , $\ell = 1,2, \cdots ,3t + 1$ , ${H_0} = \left\{ {{x_0} = 0} \right\}$ , $t\geqslant N$ . Assume the following conditions hold for every $f \in \mathcal{F}$ : If $f(z) \in {H_\ell }$ , then $\nabla f(z) \in {H_\ell }$ , $\ell = 1,2, \cdots ,3t + 1$ ; If $f(z) \in\displaystyle \bigcup\limits_{\ell = 1}^{3t + 1} {{H_\ell }}$ , then $\dfrac{\left|\langle f(z),{H}_{0}\rangle \right|}{\Vert f(z)\Vert \cdot \Vert {H}_{0}\Vert }\geqslant \delta$ , where $\delta \in \left(0, 1\right)$ is a constant. Then $\mathcal{F}$ is normal on $D$ . For the special case of $N = 3$ and $t = 3,4,5$ , the number of shared hyperplanes can be effectively reduced through this research.
Key words: normal family     holomorphic curves     t-subgeneral position     derived curves     shared hyperplanes
1 问题的提出

2015年，Ye等[1]考虑全纯曲线 $f$ 与其导曲线 $\nabla f(z)$ “强分担”超平面的情形，证明了定理1。

a. $f(z)$ $\nabla f(z)$ $D$ 上“强分担” ${H_j}$ ，其中， $j = 1,2, \cdots ,2N + 1$

b. $f(z) \in \displaystyle \bigcup\limits_{j = 1}^{2N + 1} {{H_j}}$ ，那么， $\dfrac{\left|\langle f(z),{H}_{0}\rangle \right|}{\Vert f(z)\Vert \cdot \Vert {H}_{0}\Vert }\geqslant \delta$ ，其中， ${H_0} = \left\{ {{x_0} = 0} \right\}$ 是一个坐标超平面。

$\mathcal{F}$ $D$ 上正规。

“强分担”超平面 $H$ ，不仅要求 ${f^{ - 1}}(H) = \nabla {f^{ - 1}}(H)$ ，更要求在满足 ${f^{ - 1}}(H) = \nabla {f^{ - 1}}(H)$ 的那些点上有 $f(z) = \nabla f(z)$

2020年，刘晓俊等[2]减弱条件“强分担”并对超平面的第一系数作非零的限制，得到了定理2。

a. $f(z) \in {H_j}$ ，则 $\nabla f(z) \in {H_j}$ ，其中， $j = 1, 2, \cdots ,2N + 1$

b. $f(z) \in \displaystyle \bigcup\limits_{j = 1}^{2N + 1} {{H_j}}$ ，那么， $\dfrac{\left|\langle f(z),{H}_{0}\rangle \right|}{\Vert f(z)\Vert \cdot \Vert {H}_{0}\Vert }\geqslant\delta$ ，其中， ${H_0} = \left\{ {{x_0} = 0} \right\}$ , $0 < \delta < 1$ 是一个常数。

$\mathcal{F}$ $D$ 上正规。

a. $f(z) \in {H_\ell }$ ，则 $\nabla f(z) \in {H_\ell }$ ，其中， $\ell = 1,2, \cdots , 3t + 1$

b. $f(z) \in \displaystyle \bigcup\limits_{\ell = 1}^{3t + 1} {{H_\ell }}$ ，则 $\dfrac{\left|\langle f(z),{H}_{0}\rangle \right|}{\Vert f(z)\Vert \cdot \Vert {H}_{0}\Vert }\geqslant\delta$ ，其中， $0 < \delta < 1$ 是一个常数， ${H_0} = \left\{ {{x_0} = 0} \right\}$

$\mathcal{F}$ $D$ 上正规。

$t = N$ ，即 ${H_\ell }$ 处于一般位置时，所需分担的超平面个数为 $3N + 1$

a. $f(z) \in {H_\ell }$ 当且仅当 $\nabla f(z) \in {H_\ell }$ ，其中， $\ell = 1, 2, \cdots ,5$

b. $f(z) \in \displaystyle \bigcup\limits_{\ell = 1}^5 {{H_\ell }}$ ，那么， $\dfrac{\left|\langle f(z),{H}_{0}\rangle \right|}{\Vert f(z)\Vert \cdot \Vert {H}_{0}\Vert }\geqslant\delta$ ，其中， $0 < \delta < 1$ 是一个常数。

$\mathcal{F}$ $D$ 上正规。

a. $f(z) \in {H_\ell }$ 当且仅当 $\nabla f(z) \in {H_\ell }$ ，其中， $\ell = 1, 2, \cdots ,8$

b. $f(z) \in\displaystyle \bigcup\limits_{\ell = 1}^8 {{H_\ell }}$ ，那么， $\dfrac{\left|\langle f(z),{H}_{0}\rangle \right|}{\Vert f(z)\Vert \cdot \Vert {H}_{0}\Vert }\geqslant\delta$ ，其中 $0 < \delta < 1$ 是一个常数。

$\mathcal{F}$ $D$ 上正规。

a. $f(z) \in {H_\ell }$ 当且仅当 $\nabla f(z) \in {H_\ell }$ ，其中， $\ell = 1, 2, \cdots ,2t + 3$

b. $f(z) \in \displaystyle \bigcup\limits_{\ell = 1}^{2t + 3} {{H_\ell }}$ ，则 $\dfrac{\left|\langle f(z),{H}_{0}\rangle \right|}{\Vert f(z)\Vert \cdot \Vert {H}_{0}\Vert }\geqslant\delta$ ，其中， $0 < \delta < 1$ 是一个常数。

2 定义与符号

${\mathbb{P}^N}(\mathbb{C}) = {\mathbb{C}^{N + 1}}\backslash \{ 0\} /\sim$ $N$ 维复射影空间。对任意 ${\boldsymbol{x}} = ({x_0},{x_1}, \cdots ,{x_N})$ , ${\boldsymbol{y}} = ({y_0},{y_1}, \cdots ,{y_N})$ ${\boldsymbol{x}},{\boldsymbol{y}} \in {\mathbb{C}^{N + 1}}\backslash \{ 0\}$ , ${\boldsymbol{x}}\sim{\boldsymbol{y}}$ 当且仅当存在某个 $\lambda \in \mathbb{C}$ ，使得 $({x_0},{x_1}, \cdots ,{x_N}) = \lambda ({y_0},{y_1}, \cdots ,{y_N})$ $({x_0},{x_1}, \cdots ,{x_N})$ 的等价类记作 $[{\boldsymbol{x}}] = [{x_0}: {x_1}: \cdots :{x_N}]$ ，则

 $\begin{split} & {\mathbb{P}^N}(\mathbb{C}) =\\& \left\{ {\left[ {\boldsymbol{x}} \right] = [{x_0}:{x_1}: \cdots :{x_N}]:{\boldsymbol{x}} = ({x_0},{x_1}, \cdots ,{x_N}) \in {\mathbb{C}^{N + 1}}\backslash \{ 0\} } \right\} \end{split}$

${\mathbb{P}^N}(\mathbb{C})$ 上可引入一个自然的度量，即对于点 $[\omega ]$ $[\omega ']$ 之间的距离，采用 ${\mathbb{C}^{N + 1}}$ 中2个圆 $\gamma$ $\gamma '$ 之间的欧几里得距离来表示，其中， $\gamma$ $\gamma '$ 分别代表在球面 ${S^{2N + 1}}$ 上的这2个点（这里取 $\left| \omega \right| = \left| {\omega '} \right| = 1$ ）。简单计算可得

 $\begin{split}& {\rho ^2}\left( {\left[ \omega \right],\left[ {\omega '} \right]} \right) = \mathop {\min }\limits_{\theta ,\theta '} {\left| {\omega {{\rm{e}}^{i\theta }} - \omega '{{\rm{e}}^{i\theta '}}} \right|^2} =\\&\quad \mathop {\min }\limits_{\theta ,\theta '} 2\left\{ {1 - {{\rm{Re}}} \left[ {\left( {\omega ,\omega '} \right){{\rm{e}}^{i\left( {\theta - \theta '} \right)}}} \right]} \right\} = 2\left( {1 - \left| {\left( {\omega ,\omega '} \right)} \right|} \right) \end{split}$

 ${\rho ^2}\left( {\left[ \omega \right],\left[ {\omega '} \right]} \right) = 2\left( {1 - \frac{{\left| {\left( {\omega ,\omega '} \right)} \right|}}{{\left| \omega \right|\left| {\omega '} \right|}}} \right)$

 ${\rm{d}}{s^2} = \frac{{(\omega ,\omega )({\rm{d}}\omega ,{\rm{d}}\omega ) - (\omega ,{\rm{d}}\omega )({\rm{d}}\omega ,\omega )}}{{{{(\omega ,\omega )}^2}}}$

 ${\text{d}}{s^2} = \frac{{{{\left| {{\text{d}}z} \right|}^2}}}{{{{\left( {1 + {{\left| z \right|}^2}} \right)}^2}}}$

$H = \{ {\boldsymbol{x}} \in {\mathbb{P}^N}(\mathbb{C}):\left\langle {{\boldsymbol{x}},{\boldsymbol{\alpha}} } \right\rangle = 0\}$ 是一个超平面，记 $\left\| H \right\| = \left\| {\boldsymbol{\alpha}} \right\| = \mathop {\max }\limits_{0 \leqslant i \leqslant N} \;\left| {{a_i}} \right|$ 。本文仅考虑满足 $\left\| H \right\| = 1$ 的标准化超平面。

 $\left\langle {f(z),H} \right\rangle = \left\langle {\tilde f,{\boldsymbol{\alpha}} } \right\rangle = \sum\limits_{i = 0}^N {{a_i}{f_i}(z)}$

 $\left\| f \right\| = \left\| {\tilde f} \right\| = {\left\{ {\sum\limits_{i = 0}^N {{{\left| {{f_i}(z)} \right|}^2}} } \right\}^{1/2}}$

 ${f^\# }\left( z \right) = \frac{{\left| {f \wedge f'} \right|}}{{{{\left\| f \right\|}^2}}} = \frac{{\sqrt {{{ \displaystyle \sum\limits_{0 \leqslant i < j \leqslant N} {\left| {{f_i}{f_j}^\prime - {f_j}{f_i}^\prime } \right|} }^2}} }}{{ \displaystyle \sum\limits_{i = 0}^N {{{\left| {{f_i}} \right|}^2}} }}$

$f$ $z$ 处的Fubini-Study导数，简记为F-S导数，其中， $\tilde f' = \left( {{{\tilde f}_0}^\prime ,{{\tilde f}_1}^\prime , \cdots ,{{\tilde f}_N}^\prime } \right)$

$N = 1$ 时，则 $f = \dfrac{{{f_1}}}{{{f_0}}} = \left[ {{f_0}:{f_1}} \right]:D \to {\mathbb{P}^1}(\mathbb{C})$ 是亚纯函数，此时，

 ${f^\# }\left( z \right) = \dfrac{{\left| {{f_1}^\prime {f_0} - {f_0}^\prime {f_1}} \right|}}{{{{\left| {{f_0}} \right|}^2} + {{\left| {{f_1}} \right|}^2}}} = \frac{{\left| {{{\left( {\dfrac{{{f_1}}}{{{f_0}}}} \right)}^\prime }} \right|}}{{1 + {{\left| {\dfrac{{{f_1}}}{{{f_0}}}} \right|}^2}}} = \dfrac{{\left| {f'} \right|}}{{1 + {{\left| f \right|}^2}}}$

 ${\rho _f} = \mathop {\overline {\lim } }\limits_{r \to + \infty }\; \frac{{\log \;{T_f}\left( r \right)}}{{\log \;r}} < + \infty$

${H_1},{H_2}, \cdots ,{H_q}$ ${\mathbb{P}^N}(\mathbb{C})$ 中的超平面，则

 ${H_\ell } = \{ {\boldsymbol{x}} \in {\mathbb{P}^N}(\mathbb{C}):\left\langle {{\boldsymbol{x}},{{\boldsymbol{\alpha}} _\ell }} \right\rangle = {a_{\ell 0}}{x_0} + {a_{\ell 1}}{x_1} + \cdots + {a_{\ell N}}{x_N} = 0\}$

 $M \displaystyle \bigcap \left( {\displaystyle \bigcap\limits_{j = {i_0}}^{{i_t}} {H_j}} \right) = \varnothing$

 $\begin{split} {\nabla _\mu }f(z) =& \left[ W\left( {{f_\mu },{f_0}} \right)/d: \cdots :W\left( {{f_\mu },{f_{\mu - 1}}} \right)/d:f_\mu ^2/d:\right.\\&\left.W\left( {{f_\mu },{f_{\mu + 1}}} \right)/d: \cdots :W\left( {{f_\mu },{f_N}} \right)/d \right] \end{split}$

3 主要引理

 ${g_n}(\xi ): = {f_n}\left( {{z_n} + {\rho _n}\xi } \right)$

${\mathbb{C}^m}$ 上内闭一致收敛于从 ${\mathbb{C}^m}$ 映射到 ${\mathbb{P}^N}(\mathbb{C})$ 的非常值全纯映射 $g(\xi )$

1999年，Eremenko证明了一个Picard型定理，即引理3。

 ${G_\ell }\left( \zeta \right) = {a_{\ell 0}} + \sum\limits_{i = 1}^N {{a_{\ell i}}\frac{{{g_i}\left( \zeta \right)}}{{{g_0}\left( \zeta \right)}}} ,\;\;\;\;\ell = 1,2, \cdots ,N + 1$

${G_\ell }\left( \zeta \right) \ne 0$ ${G_\ell }^\prime \left( \zeta \right) \ne 0,{\text{ }}\zeta \in \mathbb{C}$ ，则 $g$ 是线性退化的。

4 主要定理的证明 4.1 定理3的证明

 ${g_n}(\zeta ): = {f_n}\left( {{z_n} + {\rho _n}\zeta } \right)\mathop \Rightarrow \limits_{F - S}^{\zeta \in \mathbb{C}} g\left( \zeta \right)$

4.2 定理6的证明

 ${g_n}(\zeta ): = {f_n}\left( {{z_n} + {\rho _n}\zeta } \right)\mathop \Rightarrow \limits_{F - S}^{\zeta \in \mathbb{C}} g\left( \zeta \right)$ (1)

$\tilde g\left( \zeta \right) = \left( {{g_0},{g_1},{g_2},{g_3}} \right)\left( \zeta \right)$ $g$ 的某个既约表示。由于 ${H_\ell }$ 处于 $t$ 次一般位置， $1 \leqslant \ell \leqslant 2t + 3$ ，则至少存在 $t + 3$ 个超平面的第一系数不为零。不失一般性，假定 ${H_1},{H_2}, \cdots ,{H_{t + 3}}$ 的第一系数均不为零。又由引理4及 $g$ 非常值，存在某个 $i \in \left\{ {1,2, \cdots ,2t + 3} \right\}$ ，不失一般性，假定 $i = 2t + 3$ ，则存在某个 ${\zeta _0} \in \mathbb{C}$ ，使得 $\left\langle {g,{H_{2t + 3}}} \right\rangle \left( {{\zeta _0}} \right) = 0$ ，但 $\left\langle {g,{H_{2t + 3}}} \right\rangle \left( \zeta \right)\not \equiv 0$

 ${G_\ell } = {a_{\ell 0}} + {a_{\ell 1}}\frac{{{g_1}}}{{{g_0}}} + {a_{\ell 2}}\frac{{{g_2}}}{{{g_0}}} + {a_{\ell 3}}\frac{{{g_3}}}{{{g_0}}} \ne 0,{\text{ }}\ell = 1,2, \cdots ,t + 3$

 ${k_0}{g_0}(\zeta ) + {k_1}{g_1}(\zeta ) + {k_2}{g_2}(\zeta ) + {k_3}{g_3}(\zeta ) \equiv 0$

 ${g_3}(\zeta ) = k{g_2}(\zeta ) + l{g_1}(\zeta ) + m{g_0}(\zeta )$

 $\frac{{{g_3}(\zeta )}}{{{g_0}(\zeta )}} = k\frac{{{g_2}(\zeta )}}{{{g_0}(\zeta )}} + l\frac{{{g_1}(\zeta )}}{{{g_0}(\zeta )}} + m$

${G_\ell }$ 是全纯的， $\ell = 1,2, \cdots ,t + 3$ 。由Zalcman引理的证明，可得 ${g^\# }\left( \zeta \right) \leqslant 1$ ，则存在常数 $M > 0$ ，使得

 $G_\ell ^\# \left( \zeta \right) \leqslant M$

 ${\boldsymbol{A}} = \left( {\begin{array}{*{20}{c}} {{a_{\sigma (1),0}}}&{{a_{\sigma (2),0}}}&{{a_{\sigma (3),0}}}&{{a_{\sigma (4),0}}} \\ {{a_{\sigma (1),1}}}&{{a_{\sigma (2),1}}}&{{a_{\sigma (3),1}}}&{{a_{\sigma (4),1}}} \\ {{a_{\sigma (1),2}}}&{{a_{\sigma (2),2}}}&{{a_{\sigma (3),2}}}&{{a_{\sigma (4),2}}} \\ {{a_{\sigma (1),3}}}&{{a_{\sigma (2),3}}}&{{a_{\sigma (3),3}}}&{{a_{\sigma (4),3}}} \end{array}} \right)$

${p_1}{B_{\sigma (1)}}{{\rm{e}}^{{A_{\sigma (1)}}\zeta }} + {p_2}{B_{\sigma (2)}}{{\rm{e}}^{{A_{\sigma (2)}}\zeta }} + {p_3}{B_{\sigma (3)}}{{\rm{e}}^{{A_{\sigma (3)}}\zeta }} + {p_4}{B_{\sigma (4)}}{{\rm{e}}^{{A_{\sigma (4)}}\zeta }} = 0$ ，其中， ${p_1},{p_2},{p_3},{p_4}$ 均为常数，因此，

 $\left( {1,\frac{{{g_1}}}{{{g_0}}},\frac{{{g_2}}}{{{g_0}}},\frac{{{g_3}}}{{{g_0}}}} \right){\boldsymbol{A}}\left( {\begin{array}{*{20}{c}} {{p_1}} \\ {{p_2}} \\ {{p_3}} \\ {{p_4}} \end{array}} \right) = 0$

 $\begin{split}& \left( {1,\frac{{{g_1}}}{{{g_0}}},\frac{{{g_2}}}{{{g_0}}},\frac{{{g_3}}}{{{g_0}}}} \right) = \\&\quad\left( {{B_{\sigma (1)}}{{\rm{e}}^{{A_{\sigma (1)}}\zeta }},{B_{\sigma (2)}}{{\rm{e}}^{{A_{\sigma (2)}}\zeta }},{B_{\sigma (3)}}{{\rm{e}}^{{A_{\sigma (3)}}\zeta }},{B_{\sigma (4)}}{{\rm{e}}^{{A_{\sigma (4)}}\zeta }}} \right){{\boldsymbol{A}}^{ - 1}} \end{split}$

${\boldsymbol C} = \left( {\begin{array}{*{20}{c}} {{c_{10}}}&{{c_{11}}}&{{c_{12}}} \\ {{c_{20}}}&{{c_{21}}}&{{c_{22}}} \\ {{c_{30}}}&{{c_{31}}}&{{c_{32}}} \end{array}} \right)$ ，则有 $\left( {1,\dfrac{{{g_1}}}{{{g_0}}},\dfrac{{{g_2}}}{{{g_0}}}} \right) = \left( {{B_{\sigma (1)}}{{\rm{e}}^{{A_{\sigma (1)}}\zeta }},{B_{\sigma (2)}}{{\rm{e}}^{{A_{\sigma (2)}}\zeta }},{B_{\sigma (3)}}{{\rm{e}}^{{A_{\sigma (3)}}\zeta }}} \right){\boldsymbol{C}}$ 。若 $r\left({\boldsymbol C} \right)\geqslant2$ ，则方程 ${\boldsymbol C} \left( {\begin{array}{*{20}{l}} {{x_0}} \\ {{x_1}} \\ {{x_2}} \end{array}} \right) = 0$ 有非零解，因此，存在不全为零的常数 ${q_0},{\text{ }}{q_1},{\text{ }}{q_2}$ ，使得 ${\boldsymbol C} \left( {\begin{array}{*{20}{l}} {{q_0}} \\ {{q_1}} \\ {{q_2}} \end{array}} \right) = 0$ ，则

 $\left( {{B_{\sigma (1)}}{{\rm{e}}^{{A_{\sigma (1)}}\zeta }},{B_{\sigma (2)}}{{\rm{e}}^{{A_{\sigma (2)}}\zeta }},{B_{\sigma (3)}}{{\rm{e}}^{{A_{\sigma (3)}}\zeta }}} \right){\boldsymbol{C}}\left( {\begin{array}{*{20}{l}} {{q_0}} \\ {{q_1}} \\ {{q_2}} \end{array}} \right) = 0$

$\left( {1,\dfrac{{{g_1}}}{{{g_0}}},\dfrac{{{g_2}}}{{{g_0}}}} \right)\left( {\begin{array}{*{20}{l}} {{q_0}} \\ {{q_1}} \\ {{q_2}} \end{array}} \right) = 0$ ，因此， $1,{\text{ }}\dfrac{{{g_1}}}{{{g_0}}},{\text{ }}\dfrac{{{g_2}}}{{{g_0}}}$ 线性相关，矛盾，从而 $r\left( {\boldsymbol{C}} \right) = 3$

${B_{\sigma (1)}}{{\rm{e}}^{{A_{\sigma (1)}}\zeta }},{\text{ }}{B_{\sigma (2)}}{{\rm{e}}^{{A_{\sigma (2)}}\zeta }},{\text{ }}{B_{\sigma (3)}}{{\rm{e}}^{{A_{\sigma (3)}}\zeta }}$ 线性相关，则存在不全为零的常数 ${c_0},{\text{ }}{c_1},{\text{ }}{c_2}$ ，使得

 $\left( {{B_{\sigma (1)}}{{\rm{e}}^{{A_{\sigma (1)}}\zeta }},{B_{\sigma (2)}}{{\rm{e}}^{{A_{\sigma (2)}}\zeta }},{B_{\sigma (3)}}{{\rm{e}}^{{A_{\sigma (3)}}\zeta }}} \right)\left( {\begin{array}{*{20}{c}} {{c_0}} \\ {{c_1}} \\ {{c_2}} \end{array}} \right) = 0$

 $\left( {{B_1}{{\rm{e}}^{{A_1}\zeta }},{B_2}{{\rm{e}}^{{A_2}\zeta }}, \cdots ,{B_{t + 3}}{{\rm{e}}^{{A_{t + 3}}\zeta }}} \right) = \left( {1,\frac{{{g_1}}}{{{g_0}}},\frac{{{g_2}}}{{{g_0}}},\frac{{{g_3}}}{{{g_0}}}} \right){\boldsymbol{B}}$

 ${\boldsymbol{B}} = \left( {\begin{array}{*{20}{c}} {{a_{10}}}&{{a_{20}}}& \ldots &{{a_{t + 3,0}}} \\ {{a_{11}}}&{{a_{21}}}& \ldots &{{a_{t + 3,1}}} \\ {{a_{12}}}&{{a_{22}}}& \ldots &{{a_{t + 3,2}}} \\ {{a_{13}}}&{{a_{23}}}& \ldots &{{a_{t + 3,3}}} \end{array}} \right)$

$rank\left\{ {\left( {1,\dfrac{{{g_1}}}{{{g_0}}},\dfrac{{{g_2}}}{{{g_0}}},\dfrac{{{g_3}}}{{{g_0}}}} \right)} \right\} = 3$ ，因此， $rank\{ ( {B_1}{{\rm{e}}^{{A_1}\zeta }}, {B_2}{{\rm{e}}^{{A_2}\zeta }}, \cdots ,{B_{t + 3}}{{\rm{e}}^{{A_{t + 3}}\zeta }} )t\} \leqslant 3$

a. ${p_2} \ne 0$ ，则 ${g_2},{g_3}$ 可由 ${g_0},{g_1}$ 线性表出，即存在常数 ${k_1},{k_2},{l_1},{l_2}$ ，有

 $\frac{{{g_2}(\zeta )}}{{{g_0}(\zeta )}} = {k_1}\frac{{{g_1}(\zeta )}}{{{g_0}(\zeta )}} + {l_1} , \frac{{{g_3}(\zeta )}}{{{g_0}(\zeta )}} = {k_2}\frac{{{g_1}(\zeta )}}{{{g_0}(\zeta )}} + {l_2}$ (2)

 $\begin{split}& {G_\ell } = {a_{\ell 0}} + {a_{\ell 1}}\frac{{{g_1}\left( \zeta \right)}}{{{g_0}\left( \zeta \right)}} + {a_{\ell 2}}\frac{{{g_2}\left( \zeta \right)}}{{{g_0}\left( \zeta \right)}} + {a_{\ell 3}}\frac{{{g_3}\left( \zeta \right)}}{{{g_0}\left( \zeta \right)}} = \\&\quad \left( {{a_{\ell 1}} + {a_{\ell 2}}{k_1} + {a_{\ell 3}}{k_2}} \right)\frac{{{g_1}\left( \zeta \right)}}{{{g_0}\left( \zeta \right)}} + {a_{\ell 0}} + {a_{\ell 2}}{l_1} + {a_{\ell 3}}{l_2},\\&\quad\ell = 1,2, \cdots ,2t + 3 \end{split}$

 ${G_{{\ell _t}}}^\prime \ne 0 , \;\; {\left( {\frac{{{g_1}}}{{{g_0}}}} \right)^\prime } \ne 0$ (3)

${H_j}$ 的第一系数均不为零，即 ${a_{j0}} \ne 0$ ，则由断言b可得， $g$ 不取 ${H_j}$ $g\left( \mathbb{C} \right) \subseteq {H_j}$ $j = t + 4,t + 5, \cdots , 2t + 2$ 。由引理4可知， $\tilde g$ 是一个常值映射，矛盾。因此，至少存在2个 ${j_i},{j_k} \in \left\{ {t + 4,t + 5, \cdots ,2t + 2} \right\}, {\text{ }}{j_i} \ne {j_k}$ ，使得 ${a_{{j_i},0}} = {a_{{j_k},0}} = 0$ ，不妨设 ${j_i} = 2t + 1, {\text{ }}{j_k} = 2t + 2$

 $\left\langle {\tilde g,{\alpha _{{k_i}}}} \right\rangle \left( {{\zeta _0}} \right) = {a_{{k_i},1}}{g_1}\left( {{\zeta _0}} \right) + {a_{{k_i},2}}{g_2}\left( {{\zeta _0}} \right) + {a_{{k_i},3}}{g_3}\left( {{\zeta _0}} \right) = 0$

 $\begin{split}& \left( {{a_{{k_i},1}} + {a_{{k_i},2}}{k_1} + {a_{{k_i},3}}{k_2}} \right){g_1}({\zeta _0}) + \left( {{a_{{k_i},2}}{l_1} + {a_{{k_i},3}}{l_2}} \right){g_0}\left( {{\zeta _0}} \right) = 0 , \\& \quad {G_{{k_i}}}\left( {{\zeta _0}} \right) = 0 \end{split}$

${\zeta _0}$ 也是 ${G_{{k_i}}}\left( \zeta \right)$ 的零点。

 ${\left. {{a_{{k_i},1}}{{\left( {\frac{{{f_{n1}}}}{{{f_{n0}}}}} \right)}^\prime } + {a_{{k_i},2}}{{\left( {\frac{{{f_{n2}}}}{{{f_{n0}}}}} \right)}^\prime } + {a_{{k_i},3}}{{\left( {\frac{{{f_{n3}}}}{{{f_{n0}}}}} \right)}^\prime }} \right|_{{z_n} + {\rho _n}{\zeta _n}}} = 0$
 ${\left. {{a_{{k_i},1}}{{\left( {\frac{{{g_{n1}}}}{{{g_{n0}}}}} \right)}^\prime } + {a_{{k_i},2}}{{\left( {\frac{{{g_{n2}}}}{{{g_{n0}}}}} \right)}^\prime } + {a_{{k_i},3}}{{\left( {\frac{{{g_{n3}}}}{{{g_{n0}}}}} \right)}^\prime }} \right|_{{\zeta _n}}} = 0$

$n \to + \infty$ ，则

 ${\left. {{a_{{k_i},1}}{{\left( {\frac{{{g_1}}}{{{g_0}}}} \right)}^\prime } + {a_{{k_i},2}}{{\left( {\frac{{{g_2}}}{{{g_0}}}} \right)}^\prime } + {a_{{k_i},3}}{{\left( {\frac{{{g_3}}}{{{g_0}}}} \right)}^\prime }} \right|_{{\zeta _0}}} = 0$

${G_{{k_i}}}^\prime \left( {{\zeta _0}} \right) = 0$ 。所以， $\left\langle {\tilde g,{\alpha _{{k_i}}}} \right\rangle \left( \zeta \right)$ 的零点都是重级的， ${G_{{k_i}}}\left( \zeta \right)$ 的零点也都是重级的。

 ${\left. {\left( {{a_{{k_i},1}} + {a_{{k_i},2}}{k_1} + {a_{{k_i},3}}{k_2}} \right){{\left( {\frac{{{g_1}}}{{{g_0}}}} \right)}^\prime }} \right|_{{\zeta _0}}} = 0$

${\left( {\dfrac{{{g_1}}}{{{g_0}}}} \right)^\prime }\left( {{\zeta _0}} \right) = 0$ ，矛盾。因此，至少存在1个 ${m_i} \in \left\{ t + 4, \cdots , 2t \right\}$ ，有 ${a_{{m_i},0}} = 0$ 。不失一般性，取 ${m_i} = 2t$

 ${\left. {\left( {{a_{{p_j},1}} + {a_{{p_j},2}}{k_1} + {a_{{p_j},3}}{k_2}} \right){{\left( {\frac{{{g_1}\left( \zeta \right)}}{{{g_0}\left( \zeta \right)}}} \right)}^\prime }} \right|_{{\zeta _0}}} = 0$

b. ${p_2} = 0$ ，则 ${p_0},{p_1}$ 不全为零。

(a) ${p_1} \ne 0$ ，则 ${g_1},{g_3}$ 可由 ${g_0},{g_2}$ 线性表出，类似于a的证明，矛盾；

(b) ${p_0} \ne 0,{\text{ }}{p_1} = 0$ ，而 ${g_0} \ne 0$ ，矛盾。

$N = 3$ $t \geqslant 4$ 时，所需超平面的个数会随着 $t$ 值的增长而增加。目前，仍未能找到明确的上界。

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