设区域$D$为复平面$C$上的一个区域,${\cal F}$是区域$D$内的亚纯函数族. 若对于${\cal F}$中任一函数序列均可选出一个子序列在区域$D$上按球距内闭一致收敛,则${\cal F}$在区域$D$内正规.设$f(z)$和$g(z)$是区域$D$内的两个亚纯函数,$a$和$b$是两个复数,若当$f(z)=a$,有$g(z)=b$,则记$f(z)=a\Rightarrow g(z)=b$. 若$f(z)=a\Rightarrow g(z)=b$和$g(z)=b\Rightarrow f(z)=a$,则记$f(z)=a\Leftrightarrow g(z)=b$. 若$f(z)=a\Leftrightarrow g(z)=a$,则称$f(z)$和$g(z)$是区域$D$上IM分担$a$ (参见文献[6]).2000年,庞学诚和Zaclman在文献[1]中证明了如下定理.

定理1.1 设$k$是一个正整数,$b(\neq0)$为有限复数,$M$为正数,${\cal F}$ 是区域$D$ 内的亚纯函数族,其零点重级至少是$k$ . 若对于任意$f\in{\cal F}$,且$\bar{E}_{f}(0)\subset\bar{E}_{f^{(k)}}(b)$,且当$f(z)\in\bar{E}_{f}(0)$时,$0<|f^{(k+1)}(z)|

2010年,吕峰,徐俊峰和仪洪勋^[2]证明了全纯函数族分担一个不恒等于零的全纯函数的正规定则,得到了

定理1.2 设$a(z)(\not\equiv0)$是区域$D$内的全纯函数且零点重级至少为$m$,$k(\geq m+1)$ 是一个正整数,${\cal F}$是区域$D$内的全纯函数族. 若对于任意$f\in{\cal F}$,都有

(1) $f(z),a(z)$没有公共零点;

(2) $f(z)=0\Leftrightarrow f'(z)=a(z)\Rightarrow 0<|f^{(k+1)}(z)|< M$,其中$M$ 为正数. 则${\cal F}$在$D$内正规.

最近,李三华^[3]改进了定理 的结果,证明了

定理1.3 设$k$是一个正整数,$a(z)(\not\equiv0)$是区域$D$内的全纯函数,${\cal F}$是区域$D$内的亚纯函数族,其零点重级至少是$k$. 若对于任意$f\in{\cal F}$,$f(z)=0\Leftrightarrow f^{(k)}(z)=a(z)\Rightarrow 0<|f^{(k+1)}(z)-a'(z)|<|a(z)|$,则${\cal F}$在$D$内正规.

一个自然的问题: 定理$1.3$中的 $a(z)(\not\equiv0)$ 是全纯函数,若将$a(z)$改为不恒为$0,\infty$的亚纯函数,结论是否仍成立？本文研究了这个问题,并证实了定理$1.3$中$a(z)$ 在亚纯函数情况下仍成立.

定理1.4 设$k$为一个正整数,$a(z)(\not\equiv0,\infty)$为区域$D$的亚纯函数,${\cal F}$是区域$D$内的一族亚纯函数,其零点的重级至少为$k$. 若对于任意$f\in{\cal F}$,$f(z)=0\Leftrightarrow f^{(k)}(z)=a(z)\Rightarrow 0<|f^{(k+1)}(z)-a'(z)|<|a(z)|$,则${\cal F}$在$D$内正规.

为了证明定理,我们需要下面的引理.

引理 2.1^[1] 设$k$是一个正整数,${\cal F}$是单位圆盘上的亚纯函数族,其零点重级至少为$k$,且存在$A\geq1$,使得对于任意$f$,在$f$零点处,都有$|f^{(k)}(z)|\leq A$. 假设${\cal F}$在$z_{0}$处不正规,则对任意$0\leq\alpha\leq k$,必存在

a. 点列$z_{n}$,$z_{n}\rightarrow z_{0}$;

b. 函数列$f_{n}\in{\cal F}$;

c. 正数列$\rho_{n}\rightarrow0^+$,使得

$$ g_{n}(\zeta) =\frac{f_{n}(z_{n}+\rho_{n}\zeta)}{\rho^{\alpha}_{n}}\rightarrow g(\zeta) $$ 在复平面$C$按球面距离内闭一致收敛到非常数亚纯函数$g(\zeta)$,其零点重级至少是$k$,级至多为2,并且满足$g^{\sharp}(\zeta)\leq g^{\sharp}(0)=kA+1$.

引理 2.2^[4] 设$k$是一个正整数,$f(z)$是超越亚纯函数,$R(z)(\not\equiv0)$ 是有理函数,且除有限个点外. 若$f(z)$的零点重级至少为$k+2$,那么$f^{(k)}(z)-R(z)$有无穷多个零点.

引理 2.3^[5] 设$k,m$是正整数,$f(z)$是有理函数,其零点重级均至少为$k$. 如果$f^{(k)}(z)=z^{-m}$,那么$f(z)$是常数.

引理 2.4^[5] 设$k$,$m(\geq k+1)$是正整数,$\{f_{n}(z)\}$是$\triangle$ 内的亚纯函数列,$\{a_{n}(z)\}$是$\triangle$ 内的全纯函数列,并且一致收敛于1. 如果$f_{n}(z)\neq0,f_{n}^{(k)}(z)\neq\frac{a_{n}(z)}{z^{m}}$,那么$\{f_{n}(z)\}$ 在$\triangle$ 内正规.

引理 2.5^[3] 设$k$是正整数,$\{f_{n}(z)\}$是$\triangle$内的亚纯函数列,其零点重级均至少为$k$. 如果$\{a_{n}(z)\}$是$\triangle$ 内的全纯函数列,并且一致收敛于$a(z)(\neq0)$. 若对于任意的$n$,$f_{n}(z)=0\Leftrightarrow f^{(k)}_{n}(z)=a_{n}(z)\Rightarrow 0<|f^{(k+1)}_{n}(z)-a'_{n}(z)|<|a_{n}(z)|$,则$\{f_{n}(z)\}$在$\triangle$内正规.

证 不妨设$D$为$\triangle$,$z_{0}$为$\triangle$内任意一点. 下证${\cal F}$在$z_{0}$ 处正规. 以下分两种情况讨论.

情形1 当$a(z_{0})\neq\infty$时,则存在$\delta>0$,使得在$\triangle(z_{0},\delta)$内,$a(z)\neq\infty$. 由定理1.3知,${\cal F}$在$z_{0}$处正规.

情形2 当$a(z_{0})=\infty$时,不失一般性,令$a(0)=\infty,a(z)=\frac{\phi(z)}{z^{m}}$,其中$\phi(0)=1$,在$\triangle$内$\phi(z)\neq0,\infty$. 由情形1,${\cal F}$在$\triangle'=\triangle\backslash{0}$内正规. 下证${\cal F}$在$z_{0}=0$处正规. 再分两种情况讨论.

情形2.1 $m\geq k+1$

令${\cal F}_{1}=\{F(z)=f(z)z^{m},f(z)\in{\cal F}\}$. 显然,$f(z),a(z)$没有公共极点. 否则,存在$z_{1}$,使得

$$f(z_{1})=a(z_{1})=\infty, $$ 则$f^{(k)}(z_{1})=a(z_{1})$,根据条件知$f(z_{1})=0$,矛盾. 故$f(0)\not=\infty$,从而$F(0)=0$.

假设${\cal F}_{1}$在$z_{0}=0$处不正规,由引理2.1,存在点列$z_{n}$,$z_{n}\rightarrow0$,函数列$F_{n}(z)\in{\cal F}_{1}$,正数列$\rho_{n}\rightarrow0$,使得$g_{n}(\zeta)=\frac{F_{n}(z_{n}+\rho_{n}\zeta)}{\rho^{k}_{n}}$在复平面$C$上按球距内闭一致收敛于非常数亚纯函数$g(\zeta)$,其零点重级至少为$k$,级至多为2. 且满足

$$g^{\sharp}(\zeta)\leq g^{\sharp}(0)=k(d+1)+1, $$ 其中$d=\max\{|\phi(z)|:|z|\leq1\}$. 下面再分两种情况讨论

情形2.1.1 $\frac{z_{n}}{\rho_{n}}\rightarrow\infty$. 整理得

\begin{eqnarray*}F^{(k)}_{n}(z) &=&\sum^{k}_{j=0}C_{k}^{j}(z^{m})^{(k-j)}f^{(j)}_{n}(z) \\ &=&f^{(k)}_{n}(z)z^{m}+\sum^{k-1}_{j=0}C^{j}_{k}z^{m-k+j}m(m-1)\cdots(m-k+j+1)f^{(j)}_{n}(z). \end{eqnarray*} 于是可得

$\begin{array}{c} g_n^{(k)}(\zeta ) = F_n^{(k)}({z_n} + {\rho _n}\zeta )\\ = f_n^{(k)}({z_n} + {\rho _n}\zeta ){({z_n} + {\rho _n}\zeta )^m} + \sum\limits_{j = 0}^{k - 1} {{c_j}} {({z_n} + {\rho _n}\zeta )^{m - k + j}}f_n^{(j)}({z_n} + {\rho _n}\zeta ), \end{array}$

(3.1)

其中$c_{j}=C^{j}_{k}m(m-1)\cdots(m-k+j+1)$.

又因为

$$ f^{(j)}_{n}(z) =\Big(\frac{F^{(k)}_{n}(z)}{z^{m}}\Big)^{(j)}= \sum^{j}_{s=0}c_{s}g^{(j-s)}_{n} \Big(\frac{z-z_{n}}{\rho_{n}}\Big)\frac{\rho^{k-j+s}_{n}}{z^{m+s}}, $$ 其中$c_{s}=C^{j}_{s}(-1)^{s}m(m+1)\cdots(m+s-1)$,以及$\lim\limits_{n\rightarrow\infty}\frac{\rho_{n}}{z_{n}+\rho_{n}\zeta}=0$,所以

$f^{(j)}_{n}(z_{n}+\rho_{n}\zeta)(z_{n}+\rho_{n}\zeta)^{m-k+j}=\sum^{j}_{s=0}c_{s}\rho^{k-j+s}_{n}(z_{n}+\rho_{n}\zeta)^{j-k-s}g^{(j-s)}_{n}(\zeta)\rightarrow0.$

(3.2)

由$(3.1)$式和$(3.2)$式可得

$$(z_{n}+\rho_{n}\zeta)^{m}f^{(k)}_{n}(z_{n}+\rho_{n}\zeta)\rightarrow g^{(k)}(\zeta). $$

断言: $g(\zeta)=0\Leftrightarrow g^{(k)}(\zeta)=1\Rightarrow g^{(k+1)}(\zeta)=0$.

事实上,若存在$\zeta_{0}$,使得$g(\zeta_{0})=0$,且$g(\zeta)$不恒为常数. 由Hurwitz 定理,存在$\zeta_{n}$,$\zeta_{n}\rightarrow\zeta_{0}$,使得当$n$ 充分大时,$g_{n}(\zeta_{n})=0$,则$F_{n}(z_{n}+\rho_{n}\zeta_{n})=0$. 又因为$z_{n}+\rho_{n}\zeta_{n}\neq0$ (否则与$\zeta_{n}=-\frac{z_{n}}{\rho_{n}}\rightarrow\infty$矛盾),故$f_{n}(z_{n}+\rho_{n}\zeta_{n})=0$. 根据条件$f^{(k)}_{n}(z_{n}+\rho_{n}\zeta_{n})=a(z_{n}+\rho_{n}\zeta_{n}),$ 则

$$g^{(k)}(\zeta_{0})=\lim\limits_{n\rightarrow\infty}\frac{f^{(k)}_{n}(z_{n}+\rho_{n}\zeta_{n})}{a(z_{n}+\rho_{n}\zeta_{n})}=1. $$

反之,若存在$\zeta_{0}$,使得$g^{(k)}(\zeta_{0})=1$ . 显然,$g(\zeta)\not\equiv1$. 否则,$g(\zeta)=\frac{1}{k!}(\zeta-\zeta_{1})^k$,则

$$g^{\sharp}(0)\leq\left\{ {\begin{array}{ll} {\frac{k}{2}},~~&{|\zeta_{1}|\geq1 },\\[2mm] {1},&{|\zeta_{1}|\leq1},\end{array}} \right. $$ 这与$g^{\sharp}(0)=k(d+1)+1$矛盾.

由于 $(z_{n}+\rho_{n}\zeta)^{m}f^{(k)}_{n}(z_{n}+\rho_{n}\zeta)\rightarrow g^{(k)}(\zeta)$,则$\frac{f^{(k)}_{n}(z_{n}+\rho_{n}\zeta)}{a(z_{n}+\rho_{n}\zeta)}\rightarrow g^{(k)}(\zeta)$. 由Hurwitz 定理,存在$\zeta_{n}$,$\zeta_{n}\rightarrow\zeta_{0}$,使得当$n$充分大时,$f^{(k)}_{n}(z_{n}+\rho_{n}\zeta_{n})=a(z_{n}+\rho_{n}\zeta_{n})$,从而$f_{n}(z_{n}+\rho_{n}\zeta_{n})=0$. 即

$$g^{(k)}(\zeta_{0})=\lim\limits_{n\rightarrow\infty}\frac{(z_{n}+\rho_{n}\zeta_{n})^{m}f_{n}(z_{n}+\rho_{n}\zeta_{n})}{\rho^{k}_{n}} =0. $$

根据条件可得

\begin{eqnarray*}0&<&|\rho_{n}(z_{n}+\rho_{n}\zeta_{n})^{m}f^{(k+1)}_{n}(z_{n}+\rho_{n}\zeta_{n})-\rho_{n}(z_{n}+\rho_{n}\zeta_{n})^{m}a'(z_{n}+\rho_{n}\zeta_{n})|\\&<&|\rho_{n}(z_{n}+\rho_{n}\zeta_{n})^{m}a(z_{n}+\rho_{n}\zeta_{n})|,\end{eqnarray*} 又由于$\lim\limits_{n\rightarrow\infty}|\rho_{n}(z_{n}+\rho_{n}\zeta_{n})^{m}a(z_{n}+\rho_{n}\zeta_{n})|=0$ 及$\lim\limits_{n\rightarrow\infty}\rho_{n}(z_{n}+\rho_{n}\zeta_{n})^{m}a'(z_{n}+\rho_{n}\zeta_{n})=0$,所以 $$\lim\limits_{n\rightarrow\infty}\rho_{n}(z_{n}+\rho_{n}\zeta_{n})^{m}f^{(k+1)}_{n}(z_{n}+\rho_{n}\zeta_{n})=0. $$

因为

\begin{eqnarray*}g^{(k+1)}_{n}(\zeta)&=&\rho_{n}F^{(k+1)}_{n}(z_{n}+\rho_{n}\zeta)\\&=&\rho_{n}f^{(k+1)}_{n}(z_{n}+\rho_{n}\zeta)(z_{n}+\rho_{n}\zeta)^{m}\\&&+\sum^{k}_{j=0}C_{k+1}^{j}(z_{n}+\rho_{n}\zeta)^{m-(k+1)+j}m(m-1)\cdots(m-k+j)f^{(j)}_{n}(z_{n}+\rho_{n}\zeta),\end{eqnarray*} 又由于 $$f^{(j)}_{n}(z)=\Big(\frac{F^{(k)}_{n}(z)}{z^{m}}\Big)^{(j)}=\sum^{j}_{s=0}c_{s}g^{(j-s)}_{n}\Big(\frac{z-z_{n}}{\rho_{n}}\Big)\frac{\rho^{k-j+s}_{n}}{z^{m+s}}, $$ 其中$c_{s}=C_{j}^{s}(-1)^{s}m(m+1)\cdots(m+s-1)$. 以及$\lim\limits_{n\rightarrow\infty}\frac{\rho_{n}}{z_{n}+\rho_{n}\zeta_{n}}=0$,从而可得 $$\rho_{n}(z_{n}+\rho_{n}\zeta_{n})^{m}f^{(k+1)}_{n}(z_{n}+\rho_{n}\zeta_{n}) \rightarrow g^{(k+1)}(\zeta), $$ 所以 $$g^{(k+1)}(\zeta_{0})=\lim\limits_{n\rightarrow\infty}\rho_{n}(z_{n}+ \rho_{n}\zeta_{n})^{m}f^{(k+1)}_{n}(z_{n}+\rho_{n}\zeta_{n})=0. $$

若$g(\zeta)\neq0$,则$g^{(k)}(\zeta)\neq1$. 由Hayman不等式知,$g(\zeta)$为常数,矛盾.

若存在$\zeta_{0}$,使得$g(\zeta_{0})=0,$ $g^{(k)}(\zeta_{0})=1,$ $g^{(k+1)}(\zeta_{0})=0$. 所以$\zeta_{0}$是$g(\zeta)=0$重级为$k$的零点,并且是$ g^{(k+1)}(\zeta)-1$重级至少为2的零点. 从而由Hurwitz定理,存在两个序列$\zeta_{n_{1}},\zeta_{n_{2}}$,$\zeta_{n_{i}}(i=1,2)\rightarrow\zeta_{0}$,使得当$n$ 充分大时,$f^{(k)}_{n}(z_{n}+\rho_{n}\zeta_{n_{i}})=a(z_{n}+\rho_{n}\zeta_{n_{i}}),$

根据条件$f_{n}(z_{n}+\rho_{n}\zeta_{n_{i}})=0$,则$g_{n}(\zeta_{n_{i}})=0$.

显然,$\zeta_{n_{1}}\neq\zeta_{n_{2}}$. 否则,$\rho_{n}f^{(k+1)}_{n}(z_{n}+\rho_{n}\zeta_{n_{1}})=\rho_{n}a'(z_{n}+\rho_{n}\zeta_{n_{1}})$,矛盾.

所以$\zeta_{0}$ 是$g(\zeta)$重级至少为为$2k$的零点,这与$\zeta_{0}$是$g(\zeta)=0$重级为$k$的零点矛盾.

情形2.1.2 $\frac{z_{n}}{\rho_{n}}\rightarrow\alpha$,其中$\alpha$是有穷复数.

因为

$$ \frac{F_{n}(\rho_{n}\zeta)}{\rho^{k}_{n}} =\frac{F_{n}(\zeta_{n}+\rho_{n}(\zeta-\frac{z_{n}}{\rho_{n}}))}{\rho^{k}_{n}}\rightarrow g(\zeta-\alpha), $$

所以$g(\zeta-\alpha)$的零点重级至少为$k$,并且0是$g(\zeta-\alpha)$的重级至少为$m$的极点.

记$G_{n}(\zeta)=f_{n}(\rho_{n}\zeta)\rho^{m-k}_{n}$,则

$$G_{n}(\zeta)= \frac{F_{n}(\rho_{n}\zeta)}{\rho^{k}_{n}\zeta^{m}}\rightarrow \frac{g(\zeta-\alpha)}{\zeta^{m}}\triangleq G(\zeta), $$ 其中$G(\zeta)$的零点重级至少为$k$. 由于0是$g(\zeta-\alpha)$的重级至少为$m$的极点,所以$G(0)\neq\infty$.

断言: $\zeta\neq0,G(\zeta)=0\Leftrightarrow G^{(k)}(\zeta)=\frac{1}{\zeta^{m}}\Rightarrow G^{(k+1)}(\zeta)=\frac{-m}{\zeta^{m+1}}$.

事实上,若存在$\zeta_{0}\neq0$,使得$G(\zeta_{0})=0$,且$G(\zeta)\not\equiv0$. 否则,与$g(\zeta-\alpha)\equiv0$矛盾 由Hurwitz定理,存在$\zeta_{n}$,$\zeta_{n}\rightarrow\zeta_{0}$,使得当$n$ 充分大时$G_{n}(\zeta_{n})=0$,则$f_{n}(\rho_{n}\zeta_{n})=0$. 根据条件$f^{(k)}_{n}(\rho_{n}\zeta_{n})=a(\rho_{n}\zeta_{n})$,所以

$$G^{(k)}(\zeta_{0})=\lim \limits_{n\rightarrow\infty}f^{(k)}_{n}(\rho_{n}\zeta_{n})\rho^{m}_{n}=\frac{1}{\zeta^{m}_{0}}. $$

反之,若存在$\zeta_{0}$,使得$G^{(k)}(\zeta_{0})=\frac{1}{\zeta^{m}_{0}}$. 显然,$G(\zeta)\not\equiv\frac{1}{\zeta^{m}}$. 否则,$G(0)=\infty$,这与$G(0)\neq\infty$矛盾. 由于

$$G^{(k)}_{n}(\zeta)-a(\rho_{n}\zeta)\rho^{m}_{n}\rightarrow G^{(k)}(\zeta)-\frac{1}{\zeta^{m}}, $$ 再由Hurwitz 定理,存在$\zeta_{n}$,$\zeta_{n}\rightarrow\zeta_{0}$,使得当$n$ 充分大时$f^{(k)}_{n}(\rho_{n}\zeta_{n})=a(\rho_{n}\zeta_{n})$,根据条件$f_{n}(\rho_{n}\zeta_{n})=0$. 故$G_{n}(\zeta_{n})=0$,从而 $$G(\zeta_{0})=\lim\limits_{n\rightarrow\infty} G_{n}(\zeta_{n})=0. $$ 再根据条件 $$0<|\rho^{m+1}_{n}f^{(k+1)}_{n}(\rho_{n}\zeta_{n})-\rho^{m+1}_{n}a'(\rho_{n}\zeta_{n})|<|\rho^{m+1}_{n}a(\rho_{n}\zeta_{n})|, $$ 又由于$\lim\limits_{n\rightarrow\infty}|\rho^{m+1}_{n}a(\rho_{n}\zeta_{n})|=0$及$\lim\limits_{n\rightarrow\infty}\rho^{m+1}_{n}a'(\rho_{n}\zeta_{n})=0$,所以 $$\lim\limits_{n\rightarrow\infty}\rho^{m+1}_{n}f^{(k+1)_{n}}(\rho_{n}\zeta_{n})=0, $$ $$G^{(k+1)}(\zeta_{0})=\lim\limits_{n\rightarrow\infty}\rho^{m+1}_{n}f^{(k+1)}_{n}(\rho_{n}\zeta_{n})=\lim\limits_{n\rightarrow\infty}\rho^{m+1}_{n}a'(\rho_{n}\zeta_{n})=\frac{-m}{\zeta^{m+1}_{0}}. $$

若$G(\zeta)\neq0$,则$G^{(k)}(\zeta)\neq\frac{1}{\zeta^{m}}$. 由引理2.2,$G(\zeta)$为有理函数,再由引理2.3,$G(\zeta)$ 为常数. 设$G(\zeta)\equiv c$,那么$g(\zeta-\alpha)=c\zeta^{m}$,又因为$g(\zeta)$不恒为常数,故$c\neq0$. 下面再分两种情况

(ⅰ) 若存在$\delta>0$,在$\triangle_{\delta}$内$f_{n}(z)\neq0$,所以$f^{(k)}_{n}(\zeta)\neq\frac{\phi(z)}{z^{m}}$,由引理2.4,$\{f_{n}(\zeta)\}$ 在 $\triangle\delta$内正规,这与${\cal F}$在0处不正规矛盾.

(ⅱ) 若对于任意$\delta>0$,存在$z^{*}_{n}$,$z^{*}_{n}\rightarrow 0$,使得$f_{n}(z^{*}_{n})=0$,其中$z^{*}_{n}$为$f_{n}$ 的模最小的零点. $G_{n}(\zeta)=f_{n}(\rho_{n}\zeta)\rho^{m-k}_{n}\rightarrow c(\neq0)$. 然而$G_{n}(\frac{\zeta^{*}_{n}}{\rho_{n}})=\rho^{m-k}_{n}f_{n}(\zeta^{*}_{n})$,从而可得$\frac{z^{*}_{n}}{\rho_{n}}\rightarrow \infty$.

令

$$L_{n}(\zeta)=(z^{*}_{n})^{m-k}f_{n}(z^{*}_{n}\zeta),~~K_{n}(\zeta)=(z^{*}_{n})^{m-k}\phi(z^{*}_{n}\zeta), $$ 则$\{L_{n}(\zeta)\}$为复平面上的亚纯函数列,其零点的重级至少为$k$,$\{K_{n}(\zeta)\}$为亚纯函数列,且在$C/\{0\}$ 内一致收敛于$\frac{1}{\zeta^{m}}$. 再根据条件可得

$L_{n}(z)=0\Leftrightarrow L^{(k)}_{n}(z)=K_{n}(z)\Rightarrow 0<|L^{(k+1)}_{n}(z)-K'_{n}(z)|<|K_{n}(z)|,$

(3.3)

由引理2.5,$\{L_{n}(\zeta)\}$在$C/\{0\}$ 内正规. 因为在$\triangle$内$L_{n}(\zeta)\neq0$,所以$L^{(k)}_{n}(z)\neq K_{n}(z)$,由引理2.4,$\{L_{n}(\zeta)\}$ 在$\triangle$内正规. 故$\{L_{n}(\zeta)\}$在$C$内正规,设$L_{n}(\zeta)\rightarrow L(\zeta)$. 又因为$L_{n}(1)=0$,以及$L_{n}(0)=G_{n}(0)(\frac{z^{*}_{n}}{\rho_{n}})^{m-k}\rightarrow\infty$,所以$L(\zeta)$不是常数函数.

断言: $\zeta\neq0,L(\zeta)=0\Rightarrow L^{(k)}(\zeta)=\frac{1}{\zeta^{m}},L(\zeta)=0\Rightarrow L^{(k+1)}(\zeta)=\frac{-m}{\zeta^{m+1}}$.

事实上,若存在$\zeta_{0}\neq0$,使得$L(\zeta_{0})=0$,且$L(\zeta)\not\equiv0$. 由Hurwitz定理,存在$\zeta_{n}$,$\zeta_{n}\rightarrow\zeta_{0}$,使得当$n$充分大时$L_{n}(\zeta_{n})=0$,根据$(3.3)$式,可得

$$L^{(k)}_{n}(\zeta_{n})=K_{n}(\zeta_{n})=\frac{\phi(z^{*}_{n}\zeta)}{\zeta^{m}_{n}}, $$ 所以 $$L^{(k)}(\zeta_{0})=\lim\limits_{n\rightarrow\infty}L^{(k)}_{n}(\rho_{n}\zeta_{n})=\frac{1}{\zeta^{m}_{0}}. $$

再根据(3.3)式,可得

$$0<|L^{(k+1)}_{n}(\zeta_{n})-K'_{n}(\zeta_{n})|<\Big|\frac{z^{*}_{n}\phi(z^{*}_{n}\zeta)}{\zeta^{m}_{n}}\Big|<|K_{n}(\zeta_{n})|, $$ 所以 $$L^{(k+1)}(\zeta_{0})=\lim\limits_{n\rightarrow\infty} L^{(k+1)}_{n}(\rho_{n}\zeta_{n})=\frac{-m}{\zeta^{m+1}_{0}}. $$ 由于$L_{n}(1)=0,L(1)=0$,由断言知 $$\Big[L^{(k)}(\zeta)-\frac{1}{\zeta^{m}}\Big]\Big|_{\zeta=1}=0,~~\Big[L^{(k)}(\zeta)-\frac{1}{\zeta^{m}}\Big]'\Big|_{\zeta=1}=0. $$ 所以1是$L(\zeta)$的重级为$k$的零点,并且是$L^{(k)}(\zeta)-\frac{1}{\zeta^{m}}$的重级至少为2的零点. 再由Hurwitz定理,存在两个序列$\zeta_{n_{1}},\zeta_{n_{2}}$,$\zeta_{n_{i}}(i=1,2)\rightarrow1$,使得当$n$充分大时,$L^{(k)}_{n}(\zeta_{ni})=K_{n}(\zeta_{ni})$,根据$(3.3)$式可得$L^{(k)}_{n}(\zeta_{ni})=0$,则$f_{n}(z^{*}_{n}\zeta_{ni})=0$,再根据条件$f^{(k)}_{n}(z^{*}_{n}\zeta_{ni})=a(z^{*}_{n}\zeta_{ni})$. 显然,$\zeta_{n_{1}}\neq\zeta_{n_{2}}$. 否则,$z^{*}_{n}f^{(k+1)}_{n}(z^{*}_{n}\zeta_{n1})=z^{*}_{n}a'(z^{*}_{n}\zeta_{n1})$,这与$|f^{(k+1)}_{n}(z^{*}_{n}\zeta_{n1})-a'(z^{*}_{n}\zeta_{n1})|>0$矛盾.

所以,存在$\zeta_{0}\neq0$,使得

$$G(\zeta_{0})=0,~~G^{(k)}(\zeta_{0})=\frac{1}{\zeta^{m}_{0}},~~G^{(k+1)}(\zeta_{0})=\frac{-m}{\zeta^{m+1}_{0}}. $$ 故$\zeta_{0}$是$G(\zeta)$的重级为$k$的零点,并且是$ G^{(k)}(\zeta)-\frac{1}{\zeta^{m}}$的重级至少为2的零点. 再由Hurwitz定理,存在两个序列$\zeta_{n_{1}},\zeta_{n_{2}}$,$\zeta_{n_{i}}(i=1,2)\rightarrow\zeta_{0}$,使得当$n$充分大时$f^{(k)}_{n}(\rho_{n}\zeta_{ni})=a(\rho_{n}\zeta_{ni})$,再由条件$f_{n}(\rho_{n}\zeta_{ni})=0$,则$G(\zeta_{ni})=0$,同理可得$\zeta_{n_{1}}\neq\zeta_{n_{2}}$.

所以$\zeta_{0}$是$G(\zeta)$重级至少为为$2k$的零点,这与$\zeta_{0}$是$G(\zeta)=0$重级为$k$的零点矛盾.

情形2.2 当$1\leq m\leq k$.

假设${\cal F}$在0处不正规,由引理2.1,存在点列$z_{n}$,$z_{n}\rightarrow0$,函数列$f_{n}(z)\in{\cal F}$,正数列$\rho_{n}\rightarrow0$,使得$g_{n}(\zeta)=\frac{f_{n}(z_{n}+\rho_{n}\zeta)}{\rho^{k-m}_{n}}$ 在复平面$C$上按球距内闭一致收敛于非常数亚纯函数$g(\zeta)$,其零点重级至少为$k$,级至多为2,且满足

$$g^{\sharp}(\zeta)\leq g^{\sharp}(0)=k(d+1)+1, $$ 其中$d=\max\{|a(z)|:|z|\leq1\}$. 以下分两种情况讨论

情形2.2.1 $\frac{z_{n}}{\rho_{n}}\rightarrow\infty.$

令

$$T_{n}(\zeta)=\frac{f_{n}(z_{n}+z_{n}\zeta)}{z^{k-m}_{n}}, ~~S_{n}(\zeta)=z^{m}_{n}a(z_{n}+z_{n}\zeta), $$ 则$\{T_{n}(\zeta)\}$为复平面上的亚纯函数列,其零点的重级至少为$k$,$\{S_{n}(\zeta)\}$为亚纯函数列,且在$\triangle$ 内一致收敛于$\frac{1}{(\zeta+1)^{m}}$. 根据条件可得 $$\begin{array}{rl} T_{n}(z)=0\Leftrightarrow T^{(k)}_{n}(z)=S_{n}(z)\Rightarrow 0<|T^{(k+1)}_{n}(z)-S'_{n}(z)|<|S_{n}(z)| ,\end{array} $$ 由引理2.5,$\{T_{n}(\zeta)\}$在$\triangle$ 内正规. 设$T_{n}(\zeta)\rightarrow T(\zeta)$.

若$T(0)\neq\infty$,则

\begin{eqnarray*} g^{(k-m)}(\zeta) &=&\lim\limits_{n\rightarrow\infty}f^{(k-m)}_{n}(z_{n}+\rho_{n}\zeta) \\ &=&\lim\limits_{n\rightarrow\infty} f^{(k-m)}_{n}(z_{n}+z_{n}(\frac{\rho_{n}}{z_{n}}\zeta)) \\ &=& \lim\limits_{n\rightarrow\infty} T^{(k-m)}_{n} \Big(\frac{\rho_{n}}{z_{n}}\zeta\Big)= T^{(k-m)}(0),\end{eqnarray*} 因此可得$g^{(k-m)}(\zeta)$是常数. 又因为$0\leq k-m\leq k-1$,所以$g(\zeta)$是次数至多为$k-1$ 的多项式,这与$g(\zeta)$的零点重级至少为$k-1$ 矛盾.

若$T(0)=\infty$,则$T_{n}(\frac{\rho_{n}}{z_{n}}\zeta)=z^{m-k}_{n} f_{n}(z_{n}+\rho_{n}\zeta)\rightarrow T(0)=\infty$. 因此

$$ g(\zeta) =\lim\limits_{n\rightarrow\infty}\frac{f_{n}(z_{n}+\rho_{n}\zeta)}{\rho^{k-m}_{n}} =\lim\limits_{n\rightarrow\infty}\Big(\frac{\rho_{n}}{z_{n}}\Big)^{k-m}z^{m-k}_{n}f_{n}(z_{n}+\rho_{n}\zeta)=\infty . $$ 可得$g(\zeta)\equiv\infty$,这与$g(\zeta)$为非常数亚纯函数矛盾.

情形2.2.2 $\frac{z_{n}}{\rho_{n}}\rightarrow\alpha$,其中$\alpha$为有穷复数.

断言: $\zeta\neq-\alpha,g(\zeta)=0\Leftrightarrow g^{(k)}(\zeta)=\frac{1}{(\zeta+\alpha)^{m}}\Rightarrow g^{(k+1)}(\zeta)=\frac{-m}{(\zeta+\alpha)^{m+1}}$.

事实上,若存在$\zeta_{0}$,使得$g(\zeta_{0})=0$,且$g(\zeta)$不恒为常数. 由Hurwitz定理,存在$\zeta_{n}$,$\zeta_{n}\rightarrow\zeta_{0}$,使得当$n$充分大时,$g_{n}(\zeta_{n})=0$,则$f_{n}(z_{n}+\rho_{n}\zeta_{n})=0$. 根据条件

$$f^{(k)}_{n}(z_{n}+\rho_{n}\zeta_{n})=a(z_{n}+\rho_{n}\zeta_{n}), $$ 所以 $$g^{(k)}(\zeta_{0})=\lim\limits_{n\rightarrow\infty}\rho^{m}_{n}f^{(k)}_{n}(z_{n}+\rho_{n}\zeta_{n})=\lim\limits_{n\rightarrow\infty}\rho^{m}_{n}a(z_{n}+\rho_{n}\zeta_{n})=\frac{1}{(\zeta+\alpha)^{m}}. $$

反之,若存在$\zeta_{0}$,使得$g^{(k)}(\zeta_{0})=\frac{1}{(\zeta_{0}+\alpha)^{m}}$. 显然,$g(\zeta)\not\equiv\frac{1}{(\zeta+\alpha)^{m}}$. 否则,$-\alpha$为$g^{(k)}(\zeta)$的重级至少为$k+1(>m)$的极点,矛盾.

又由于

$$\rho^{m}_{n}f^{(k)}_{n}(z_{n}+\rho_{n}\zeta)-\rho^{m}_{n}a(z_{n}+\rho_{n}\zeta)\rightarrow g^{(k)}(\zeta)-\frac{1}{(\zeta+\alpha)^{m}}, $$ 则由Hurwitz 定理,存在$\zeta_{n}$,$\zeta_{n}\rightarrow\zeta_{0}$,使得当$n$充分大时$f^{(k)}_{n}(z_{n}+\rho_{n}\zeta_{n})=a(z_{n}+\rho_{n}\zeta_{n})$,根据条件$f_{n}(z_{n}+\rho_{n}\zeta_{n})=0$. 从而$g(\zeta_{0})=\lim\limits_{n\rightarrow\infty} \frac{f_{n}(z_{n}+\rho_{n}\zeta_{n})}{\rho^{k-m}_{n}}=0$. 再根据条件 $$0<|\rho^{m+1}_{n}f^{(k+1)}_{n}(z_{n}+\rho_{n}\zeta_{n})-\rho^{m+1}_{n}a'(z_{n}+\rho_{n}\zeta_{n})|<|\rho^{m+1}_{n}a(z_{n}+\rho_{n}\zeta_{n})|, $$ 又由于$\lim\limits_{n\rightarrow\infty}|\rho^{m+1}_{n}a(z_{n}+\rho_{n}\zeta_{n})|=0$及$\lim\limits_{n\rightarrow\infty}\rho^{m+1}_{n}a'(z_{n}+\rho_{n}\zeta_{n})=\frac{-m}{(\zeta_{0}+\alpha)^{m+1}}$,所以 $$g^{(k+1)}(\zeta_{0})=\lim\limits_{n\rightarrow\infty}\rho^{m+1}_{n}f^{(k+1)}_{n}(z_{n}+\rho_{n}\zeta_{n})=\lim\limits_{n\rightarrow\infty}\rho^{m+1}_{n}a'(z_{n}+\rho_{n}\zeta_{n})=\frac{-m}{(\zeta_{0}+\alpha)^{m+1}}. $$ 余下证明与情形2.1.2的证明过程类似.

所以${\cal F}_{1}$在0处正规. 故存在$\delta>0,\{F_{n}\}$的一子序列(仍记为$\{F_{n}\}$),使得在$\triangle_{\delta}$上,$F_{n}(z)\rightarrow F(z)$. 由于$F_{n}(0)=0$,所以$F(0)=0$,故存在$0<\delta_{1}<\delta$,使得在$\triangle_{\delta_{1}}$内$|F_{n}|\leq1$,所以当$n$充份大时,$|F_{n}|\leq\frac{1}{2}$. 由于在$\triangle_{\delta_{1}}$内,$f_{n}(z)\neq\infty$,则$f_{n}(z)$在$\triangle_{\delta_{1}}$内是全纯函数. 又因为

$$|f_{n}(z)|=\Big|\frac{F_{n}(z)}{z^{m}}\Big|\leq\frac{2^{m-1}}{\delta^{m}_{1}},\ \ |z|=\frac{\delta_{1}}{2}, $$ 再由最大模原理以及Montel定则,故${\cal F}$在0处正规. 从而${\cal F}$在$D$上正规.

即定理1.4成立.证毕.