设$f(z)$是平面上亚纯函数.用$n(t, f)$表示$f(z)$在$|z|\leq t$上的极点数, 重数按重数算.

记

$ m(r, f)=\frac{1}{2\pi}\int_{0}^{2\pi}\log^{+}|f(r{\rm e}^{{\rm i}\theta})|{\rm d}\theta, $

$ N(r, f)=\int_{0}^{r}\frac{n(t, f)-n(0, f)}{t}{\rm d}t+n(0, f)\log r, $

$ T(r, f)=m(r, f)+N(r, f). $

若$\limsup\limits_{r\to +\infty} \frac{\log T(r, f)} {\log r}=\rho$, 则称$f(z)$为$\rho$级亚纯函数.当$\rho=+\infty$时, 称$f(z)$为无穷级亚纯函数; 当$\rho=+\infty$时, 称$f(z)$为有限级亚纯函数.

用$T(r)\ll U(r)$表示函数$T(r) \leq U(r)$且存在序列$r_n(\to+\infty(n\to+\infty)) $使$T(r_n)= U(r_n)$.这时称$U(r)$为$T(r)$的型函数.

1935年, 熊庆来^[1]首先引进了无穷级亚纯函数的型函数, 并得到了无穷级亚纯函数存在$\rho(r)$级Borel方向.李国平^{[2, p209]}, 庄圻泰^{[3, p171]}用熊庆来的型函数研究了涉及小函数的Borel方向.参照文献[4]单位圆内无穷级代数体函数的结果, 我们认为熊庆来、李国平、庄圻泰的结果可进一步改进.

1959年, Hayman^[5]得到了用亚纯函数的零点与其某阶导数的1值点的幂指量限制它的特征函数的著名不等式(称为Hayman不等式). 1982年, 杨乐^[6]首先证明了存在相应于Hayman不等式的奇异方向(这样的奇异方向称为Hayman方向).之后, 一些文献继续研究Hayman方向, 如顾永兴、龚向宏^[7]得到了有限正级亚纯函数存在涉及小函数的Hayman方向; 张庆彩^[8]用熊庆来的型函数研究了无穷级亚纯函数涉及重值的Hayman方向; Rossi、Fenton^[9]继续研究零级亚纯函数Hayman方向的存在性.同样, 参照文献[4]单位圆内无穷级代数体函数的结果, 无穷级亚纯函数Hayman方向的结果有待进一步改进.再者, 涉及小函数的无穷级亚纯函数Hayman方向也未见研究.

最近, 郑建华^[10]证明了非零级且下级有限与下级无穷两类亚纯函数存在一条涉及小函数为$o(T(r, f))$的T方向. Zheng、Wu^[11]研究了Hayman-T方向的存在性. Wu、Zheng^[12]证明了单位圆内涉及小函数的T点与Hayman-T点的存在性.那么涉及小函数的Hayman-T方向与无穷级亚纯函数涉及小函数的T方向值得研究.

已有平面上有限级亚纯函数涉及小函数的T方向与精确级奇异方向的研究方法与结果不能用于研究无穷级亚纯函数涉及小函数的T方向与精确级奇异方向.

通过建立无穷级亚纯函数涉及小函数的角域内的基本不等式, 我们得到了无穷级亚纯函数存在涉及小函数的精确级Borel方向与Hayman方向, 同时也证明了无穷级亚纯函数存在涉及小函数的T方向与Hayman-T方向.

用$n(r, \theta, \delta, f=a)$表示$f(z)-a$在$\{z:|\arg z-\theta|<\delta, |z|\leq r\}$上的零点数, 重数按重数算.相似于$N(r, f)$, 我们可用$n(r, \theta, \delta, f=a)$写出计数函数$N(r, \theta, \delta, f=a)$.

我们的结果如下:

定理1.1  设$f(z)$是平面上无穷级亚纯函数, 则$f(z)$存在涉及小函数的精确级Borel方向, 即有方向$\arg z=\theta_0(0\leq\theta_0<2\pi)$, 对任意$\delta(>0)$有

$ \limsup\limits_{r\to +\infty} \frac{n(r, \theta_0, \delta, f=g)}{{\rm e}^{\frac{1}{\rho(r)}}\cdot U(r)}>0 $

至多有$2$个例外的$g(z)$, 其中$g(z)\in {\cal G}$, ${\cal G}=$ $\{g(z): T(r, g)=o(U(r))(r \rightarrow+\infty)\}$, $U(r)$是$T(r, f)$的型函数.

注  从收敛指数与小函数的范围, 定理1.1改进了文献[1-3]的结果.

定理1.2  设$f(z)$是平面上无穷级亚纯函数, 则$f(z)$存在涉及小函数的T方向, 即有方向$\arg z=\theta_0(0\leq\theta_0<2\pi)$, 对任意$\delta(>0)$有

$ \limsup\limits_{r\to +\infty} \frac{N(r, \theta_0, \delta, f=g)}{T(r, f)}>0 $

至多有$2$个例外的$g(z)$, 其中$g(z)\in {\cal G}$, ${\cal G}=$ $\{g(z): T(r, g)=o(U(r))(r\to+\infty)\}$, $U(r)$是$T(r, f)$的型函数.

注  若把亚纯函数涉及小函数的T方向的研究分为有限级与无穷级便可得理想结果.

定理1.3  设$f(z)$是平面上无穷级亚纯函数, 则$f(z)$存在涉及小函数的精确级Hayman方向, 即有方向$\arg z=\theta_0(0\leq\theta_0<2\pi)$, 对任意$\delta(>0)$, 正整数$k$与任意有限级亚纯函数$a(z)$, $b(z)$, 当$b(z)-a^{(k)}(z)$不为常数零时有

$\limsup\limits_{r\to +\infty} \frac{n(r, \theta_0, \delta, f=a)+ n(r, \theta_0, \delta, f^{(k)}=b)} {{\rm e}^{\frac{1}{\rho(r)}}\cdot U(r)}>0, $

其中$U(r)=r^{{\rm e}^{\frac{1}{\rho(r)}}}$是$T(r, f)$的型函数.

定理1.4  设$f(z)$是平面上无穷级亚纯函数, 则$f(z)$存在涉及小函数的Hayman-T方向, 即有方向$\arg z=\theta_0(0\leq\theta_0<2\pi)$, 对任意$\delta(>0)$, 正整数$k$与任意有限级亚纯函数$a(z)$, $b(z)$, 当$b(z)-a^{(k)}(z)$不为常数零时有

$\limsup\limits_{r\to +\infty} \frac{N(r, \theta_0, \delta, f=a)+ N(r, \theta_0, \delta, f^{(k)}=b)}{T(r, f)}>0. $

注1  定理1.3与定理1.4的小函数范围是否可取为${\cal G}= \{g(z): T(r, g)=o(U(r))$ $(r\to+\infty)$, $U(r)$是$T(r, f)$的型函数$\}$有待进一步研究.

注2  有限级亚纯函数涉及小函数的Hayman-T方向的存在性仍是一个须研究的问题.

用$\Omega(\alpha, \beta)$表示角域$\{z: \alpha\leq\arg z\leq\beta, 0<\beta-\alpha\leq 2\pi\}$.用$n(r, \alpha, \beta, f)$表示$f(z)$在$\Omega(\alpha, \beta)\cap\{|z|\leq r\}$上的极点数, 重数按重数算.用$n(r, \alpha, \beta, f=a)$表示$f(z)-a$在$\Omega(\alpha, \beta)\cap\{|z|\leq r\}$上的零点数.相似于$N(r, f)$, 我们可用$n(r, \alpha, \beta, f)$与$n(r, \alpha, \beta, f=a)$写出计数函数$N(r, \alpha, \beta, f)$与$N(r, \alpha, \beta, f=a)$.记

$ S(r, f)=\frac{1}{\pi}\int\!\!\!\int\limits_{|z|\leq r}\frac{|f'(t{\rm e}^{{\rm i}\theta})|^2} {(1+|f(t{\rm e}^{{\rm i}\theta})|^2)^2}t{\rm d}t{\rm d}\theta, \quad T_{0}(r, f)=\int_0^r \frac{S(t, f)}{t}{\rm d}t, $

$ S(r, \alpha, \beta, f)=\frac{1}{\pi}\int_{0}^{r}\int_{\alpha}^{\beta}\frac{|f'(t{\rm e}^{{\rm i}\theta})|^2} {(1+|f(t{\rm e}^{{\rm i}\theta})|^2)^2}t{\rm d}t{\rm d}\theta, \quad T_{0}(r, \alpha, \beta, f)=\int_0^r \frac{S(t, \alpha, \beta, f)}{t}{\rm d}t. $

因Nevanlinna特征$T(r, f)$与Ahlfors-Shimizu特征$T_{0}(r, f)$仅相差一个有界量^[13], 在本文$T(r, f)$与$T_{0}(r, f)$可交换使用.

设$f(z)$是角域$\Omega(\alpha, \beta)$内的亚纯函数.我们还需下面角域Nevanlinna特征^{[10, 14]}:

$ A_{\alpha, \beta}(r, f)=\frac{\omega}{\pi}\int_{1}^{r} {\Big (}\frac{1}{t^{\omega}}-\frac{t^{\omega}}{r^{2\omega}}{\Big )} (\log^{+}|f(t{\rm e}^{{\rm i}\alpha})|+\log^{+}|f(t{\rm e}^{{\rm i}\beta})|)\frac{{\rm d}t}{t}, $

$ B_{\alpha, \beta}(r, f)=\frac{2\omega}{\pi r^{\omega}}\int_{\alpha}^{\beta} \log^{+}|f(r{\rm e}^{{\rm i}\theta})| \sin \omega(\theta-\alpha){\rm d}\theta, $

$ C_{\alpha, \beta}(r, f)=2\sum\limits_{1<|b_{n}|<r} {\Big (}\frac{1}{|b_{n}|^{\omega}}-\frac{|b_{n}|^{\omega}}{r^{2\omega}}{\Big )} \sin \omega(\theta_{n}-\alpha), $

$ D_{\alpha, \beta}(r, f)=A_{\alpha, \beta}(r, f)+B_{\alpha, \beta}(r, f), $

其中$\omega=\frac{\pi}{\beta-\alpha}$, $b_{n}=|b_{n}|{\rm e}^{{\rm i}\theta_{n}}$是函数$f(z)$在角域$\Omega(\alpha, \beta)$内的极点, 重极点按重数算; 在$C_{\alpha, \beta}(r, f)$的表示式中, 若$f(z)$的极点仅计一次, 我们用记号$\overline{C}_{\alpha, \beta}(r, f)$.记

$S_{\alpha, \beta}(r, f)=A_{\alpha, \beta}(r, f)+B_{\alpha, \beta}(r, f) +C_{\alpha, \beta}(r, f). $

那么由文献[10, 14]有

$ \begin{equation} S_{\alpha, \beta}\bigg(r, \sum\limits_{i=1}^{n}f_{i}\bigg)\leq \sum\limits_{i=1}^{n}S_{\alpha, \beta}(r, f_{i})+3\log n, \quad S_{\alpha, \beta}\bigg(r, \prod\limits_{i=1}^{n}f_{i}\bigg)\leq \sum\limits_{i=1}^{n}S_{\alpha, \beta}(r, f_{i}). \end{equation} $

(2.1)

引理2.1^{[10, 14]}  设$f(z)$是平面上亚纯函数.对任一复数$a\in {\Bbb C}$有

$ \begin{equation} S_{\alpha, \beta}(r, f)=S_{\alpha, \beta}\Big(r, \frac{1}{f-a}\Big)+O(1). \end{equation} $

(2.2)

对任意$q(\geq 3)$个判别复数$a_{i}\in \mathbb{C}$ $(i=1, 2, \cdots, q-1), a_{q}=\infty$有

$ \begin{equation} (q-2)S_{\alpha, \beta}(r, f)\leq \sum\limits_{i=1}^{q}\overline{C}_{\alpha, \beta}\Big(r, \frac{1}{f-a_{i}}\Big)+R_{\alpha, \beta}(r, f), \end{equation} $

(2.3)

其中

$ \begin{eqnarray*} R_{\alpha, \beta}(r, f)&&=&& A_{\alpha, \beta}\Big(r, \frac{f'}{f}\Big)+B_{\alpha, \beta} \Big(r, \frac{f'}{f}\Big) \\ &&&& + \sum\limits_{i=1}^{q-1} A_{\alpha, \beta}\Big(r, \frac{f'}{f-a_{i}}\Big)+ \sum\limits_{i=1}^{q-1} B_{\alpha, \beta}\Big(r, \frac{f'}{f-a_{i}}\Big)+O(1). \end{eqnarray*} $

记

$\dot{S}_{\alpha, \beta}(r, f)=\frac{1}{\pi}\int_{1}^{r}\int_{\alpha}^{\beta} {\Big (}\frac{1}{t^{\omega}}-\frac{t^{\omega}}{r^{2\omega}}{\Big )} \frac{|f'(t{\rm e}^{{\rm i}\theta})|^2} {(1+|f(t{\rm e}^{{\rm i}\theta})|^2)^2} \sin \omega(\theta-\alpha)t{\rm d}t{\rm d}\theta, $

则

$ \begin{equation} S_{\alpha, \beta}(r, f)=\dot{S}_{\alpha, \beta}(r, f)+O(1). \end{equation} $

(2.4)

引理2.2^[10]  设$f(z)$是平面上亚纯函数, 则

$ C_{\alpha, \beta}(r, f)\leq 4\omega \frac{N(r, \alpha, \beta, f)}{r^{\omega}} +2\omega^{2}\int_{1}^{r} \frac{N(t, \alpha, \beta, f)}{t^{\omega+1}}{\rm d}t, $

$ \dot{S}_{\alpha, \beta}(r, f)\leq 2\omega \frac{T_{0}(r, \alpha, \beta, f)}{r^{\omega}} +\omega^{2}\int_{1}^{r} \frac{T_{0}(t, \alpha, \beta, f)}{t^{\omega+1}}{\rm d}t. $

对任意小$\delta(>0)$有

$ \begin{eqnarray*} \dot{S}_{\alpha, \beta}(r, f)&&\geq&& \omega\sin(\omega\delta) \frac{T_{0}(r, \alpha+\delta, \beta-\delta, f)}{r^{\omega}} -\omega T_{0}(1, \alpha+\delta, \beta-\delta, f)\\ &&&&+ \omega^{2}\sin(\omega\delta)\int_{1}^{r} \frac{T_{0}(t, \alpha+\delta, \beta-\delta, f)}{t^{\omega+1}}{\rm d}t. \end{eqnarray*} $

引理2.3^{[10, 14]}  设$f(z)$是平面上亚纯函数, 记

$m_{\alpha, \beta}\Big(r, \frac{f'}{f}\Big)= \frac{1}{2\pi}\int_{\alpha}^{\beta}\log^{+} \bigg|\frac{f'(r{\rm e}^{{\rm i}\theta})}{f(r{\rm e}^{{\rm i}\theta})}\bigg|{\rm d}\theta. $

则对任意$0<r<R$有

$ B_{\alpha, \beta}\Big(r, \frac{f'}{f}\Big)\leq \frac{4\omega}{r^{\omega}} m_{\alpha, \beta}\Big(r, \frac{f'}{f}\Big), $

$ A_{\alpha, \beta}\Big(r, \frac{f'}{f}\Big)\leq K{\bigg(}\Big(\frac{R}{r}\Big)^{\omega } \int_{1}^{R} \frac{\log T(t, f)}{t^{\omega+1}}{\rm d}t+\log^{+}\frac{r}{R-r} +\log\frac{R}{r}{\bigg)}, $

其中$\omega=\frac{\pi}{\beta-\alpha}$, $K$是与$r$, $R$无关的常数.

引理2.4^[15]  设$f(z)$是$z$平面上亚纯函数, 若$f(0)\neq 0, \infty$, 则对任意$0<r<R$有

$ T(r, f^{(k)})< (k+1)T(r, f)+m\Big(r, \frac{f^{(k)}}{f}\Big), $

$ m\Big(r, \frac{f^{(k)}}{f}\Big)<c_{k}{\Big (}1+\log^{+}\log^{+}\frac{1}{|f(0)|}+\log^{+}\frac{1}{r}+ \log^{+}\frac{1}{R-r}+\log^{+} R+\log^{+} T(R, f){\Big )}, $

其中$k$为正整数, $c_{k}$为仅依于$k$的常数.

下面我们引用无穷级亚纯函数特征函数型函数的孙道椿^[16]结果, 它比熊庆来^[1]无穷级更精确.

引理2.5^[16]  设$T(r)$在$[a, +\infty)$上连续, 且$\limsup\limits_{r\to +\infty}\frac{\log T(r)} {\log r}=+\infty$.则存在连续可微函数$U(r)=r^{{\rm e}^{\frac{1}{\rho(r)}}}$使:

(1)$\rho(r)$单调下降趋于零, $\rho' (r)$单调上升, $\lim\limits_{r\to +\infty}r\rho' (r)\log r\log\log r=0$;

(2)对充分大$r$, 有$T(r)\ll U(r)=r^{{\rm e}^{\frac{1}{\rho(r)}}}$;

(3)$U(r+\frac{r\log r}{\log U(r)\log^2\log U(r)})<(1+o(1))U(r)$.

引理2.6  设$f(z)$, $g_{i}(z)\ (i=1, 2, 3)$是平面上亚纯函数, $g_{i}(z)\ (i=1, 2, 3)$相互判别, 若$\limsup\limits_{r\to +\infty}\frac{\log T(r, f)}{\log r}=+\infty$, $T(r, g_{i})=o(U(r))\ (r\to+\infty)\ (i=1, 2, 3)$, 则对任意小$\delta(>0)$有

$ \begin{eqnarray*} T_{0}(r, \alpha+\delta, \beta-\delta, f) &&\leq&& C\cdot\sum\limits_{i=1}^{3}N(r, \alpha, \beta, f=g_{i})\\ &&&&+ C_{1}\cdot r^{\omega}\cdot\sum\limits_{i=1}^{3}\int_{1}^{r} \frac{N(t, \alpha, \beta, f=g_i)}{t^{\omega+1}}{\rm d}t+o(U(r)), \end{eqnarray*} $

其中$C$, $C_{1}$是仅与$\omega$, $\delta$有关的常数, $U(r)$是$T(r, f)$的型函数, $\omega=\frac{\pi}{\beta-\alpha}$.

证  设

$ \begin{equation} w(z)=\frac{f(z)-g_{1}(z)}{f(z)-g_{2}(z)}\cdot\frac{g_{3}(z)-g_{2}(z)}{g_{3}(z)-g_{1}(z)}, \end{equation} $

(2.5)

那么

$ \begin{equation} f(z)=g_{2}(z)+\frac{g_{2}(z)-g_{1}(z)}{h(z)w(z)-1}, \end{equation} $

(2.6)

其中

$ h(z)=\frac{g_{3}(z)-g_{1}(z)}{g_{3}(z)-g_{2}(z)}. $

由(2.1), (2.6)式与引理2.1的(2.2), (2.3)式得

$ \begin{eqnarray} S_{\alpha, \beta}(r, f)&& \leq &&S_{\alpha, \beta}(r, w)+O \bigg(\sum\limits_{i=1}^{3}S_{\alpha, \beta}(r, g_i)\bigg)+O(1) \\ && \leq&&\sum\limits_{i=1}^{3}C_{\alpha, \beta}(r, f=g_i)+R_{\alpha, \beta}(r, w) +O\bigg(\sum\limits_{i=1}^{3}S_{\alpha, \beta}(r, g_i)\bigg), \end{eqnarray} $

(2.7)

其中

$ R_{\alpha, \beta}(r, w)=2A_{\alpha, \beta}\Big(r, \frac{w'}{w}\Big)+2B_{\alpha, \beta} \Big(r, \frac{w'}{w}\Big) + A_{\alpha, \beta}\Big(r, \frac{w'}{w-1}\Big)+B_{\alpha, \beta}\Big(r, \frac{w'}{w-1}\Big)+O(1). $

用(2.4), (2.7)式与引理2.2得

$ \begin{eqnarray*} &&&&c\cdot\frac{T_{0}(r, \alpha+\delta, \beta-\delta, f)}{r^{\omega}} \\ && \leq&& \sum\limits_{i=1}^{3}\bigg(4\omega \frac{N(r, \alpha, \beta, f=g_i)}{r^{\omega}} +2\omega^{2}\int_{1 }^{r} \frac{N(t, \alpha, \beta, f=g_i)}{t^{\omega+1}}{\rm d}t\bigg)\\ &&&&+ R_{\alpha, \beta}(r, w) +O\bigg(\sum\limits_{i=1}^{3}\bigg(2\omega \frac{T_{0}(r, \alpha, \beta, g_i)}{r^{\omega}} +\omega^{2}\int_{1}^{r} \frac{T_{0}(t, \alpha, \beta, g_i)}{t^{\omega+1}}{\rm d}t\bigg)\bigg), \end{eqnarray*} $

其中$c=\omega\sin(\omega\delta)$.所以

$ \begin{eqnarray} T_{0}(r, \alpha+\delta, \beta-\delta, f) & \leq& C\cdot\sum\limits_{i=1}^{3} N(r, \alpha, \beta, f=g_i)\\ &&+ C_{1}\cdot r^{\omega}\cdot\sum\limits_{i=1}^{3}\int_{1}^{r} \frac{N(t, \alpha, \beta, f=g_i)}{t^{\omega+1}}{\rm d}t+o(U(r))\\ &&+ r^{\omega}\cdot R_{\alpha, \beta}(r, w)+O\bigg(r^{\omega}\cdot \sum\limits_{i=1}^{3}\int_{1}^{r} \frac{T_{0}(t, \alpha, \beta, g_i)}{t^{\omega+1}}{\rm d}t\bigg), \end{eqnarray} $

(2.8)

其中$C$, $C_{1}$是仅与$\omega$, $\delta$有关的常数.下面估计$r^{\omega} R_{\alpha, \beta}(r, w)$与$r^{\omega}\cdot\int_{1}^{r} \frac{T_{0}(t, \alpha, \beta, g_i)}{t^{\omega+1}}{\rm d}t$.由(2.5)式知

$w(z)=(1+\frac{g_{2}(z)-g_{1}(z)}{f(z)-g_{2}(z)}) \cdot\frac{g_{3}(z)-g_{2}(z)}{g_{3}(z)-g_{1}(z)}. $

那么

$ \begin{equation} T(r, w)\leq T(r, f)+o(U(r))\leq U(r)+o(U(r)). \end{equation} $

(2.9)

设

$R=r+\frac{r\log r}{\log U(r)\log^2\log U(r)}. $

用(2.9)式, 引理2.3, 引理2.4与引理2.5得

$ \begin{eqnarray} r^{\omega} R_{\alpha, \beta}(r, w)&\leq& 2r^{\omega}K\bigg( \Big(\frac{R}{r}\Big)^{\omega } \int_{1}^{R} \frac{\log T(t, w)}{t^{\omega+1}}{\rm d}t+\log^{+}\frac{r}{R-r}+\log\frac{R}{r} \bigg) \\ && +8\omega\cdot m_{\alpha, \beta}\Big(r, \frac{w'}{w}\Big)+4\omega\cdot m_{\alpha, \beta}\Big(r, \frac{w'}{w-1}\Big)+ O(1)\cdot r^{\omega} \\ && + r^{\omega}K\bigg(\Big(\frac{R}{r}\Big)^{\omega } \int_{1}^{R} \frac{\log T(t, w-1)}{t^{\omega+1}}{\rm d}t+\log^{+}\frac{r}{R-r}+\log\frac{R}{r}\bigg) \\ & \leq &3r^{\omega}K \bigg(\frac{2^{\omega }}{\omega}\log T(R, w) + \log\frac{\log U(r)\log^2\log U(r)}{\log r} +\log 2\bigg) \\ && + 8\omega\cdot m\Big(r, \frac{w'}{w}\Big)+4\omega\cdot m\Big(r, \frac{w'}{w-1}\Big)+ O(1)\cdot r^{\omega} \\ & \leq& O\bigg(r^{\omega}\cdot\log T(R, w)+ r^{\omega}\cdot\log\frac{\log U(r)\log^2\log U(r)}{\log r} \bigg) \\ && + O\bigg(r^{\omega}+\log\frac{\log U(r)\log^2\log U(r)}{r\log r}+\log r\bigg) \\ & \leq& O(r^{\omega}\cdot\log U(R)) \\ & =&o(U(r))\ \ (r\to+\infty). \end{eqnarray} $

(2.10)

记

$m(r)=\frac{U(r)}{r^{\omega}}=r^{{\rm e}^{\frac{1}{\rho(r)}}-\omega}. $

由引理2.5(注意: $\rho'(r)<0$)得

$ \begin{equation} r m'(r)= m(r)\bigg(({\rm e}^{\frac{1}{\rho(r)}}-\omega)- \frac{r{\rm e}^{\frac{1}{\rho(r)}}\rho'(r)\log r}{\rho^2(r)}\bigg)>m(r)\cdot ({\rm e}^{\frac{1}{\rho(r)}}-\omega). \end{equation} $

(2.11)

用(2.11)式与L'Hospital法则得

$ \begin{eqnarray} \lim\limits_{r\to +\infty}\frac{r^{\omega}\cdot\int_{1}^{r} \frac{T_{0}(t, \alpha, \beta, g_i)}{t^{\omega+1}}{\rm d}t}{U(r)}&=& \lim\limits_{r\to +\infty}\frac{\int_{1}^{r} \frac{T_{0}(t, \alpha, \beta, g_i)}{t^{\omega+1}}{\rm d}t}{m(r)} = \lim\limits_{r\to +\infty}\frac{T_{0}(r, \alpha, \beta, g_i)}{r^{\omega+1}\cdot m'(r)} \\ & \leq &\lim\limits_{r\to +\infty}\frac{T_{0}(r, \alpha, \beta, g_i)}{r^{\omega}\cdot m(r)({\rm e}^{\frac{1}{\rho(r)}}-\omega)} \\ & =& \lim\limits_{r\to +\infty} \frac{T_{0}(r, \alpha, \beta, g_i)}{U(r)({\rm e}^{\frac{1}{\rho(r)}}-\omega)}=0. \end{eqnarray} $

(2.12)

由(2.8), (2.10)与(2.12)式, 知本引理正确.

文献[12, 定理4.1]研究了单位圆内角域内的Hayman不等式, 下面我们给出平面上角域内的Hayman不等式, 其证明方法上属于文献[7, 12], 为了方便阅读, 我们给出详细证明.

引理2.7  设$f(z)$, $\varphi(z)$是平面上非常数的亚纯函数, 若$h(z)=\frac{f^{(k)}(z)}{\varphi(z)}$, $g(z)=\frac{\varphi^{2}(h-1)^{k+2}}{(h')^{k+1}}$不为常数, 则

$ S_{\alpha, \beta}(r, f) \leq 3C_{\alpha, \beta}(r, f=0)+4\overline{C}_{\alpha, \beta}(r, f^{(k)}=\varphi)+ 4C_{\alpha, \beta}\Big(r, \frac{1}{\varphi}\Big)+C_{\alpha, \beta}(r, \varphi)+Q_{\alpha, \beta}(r, f), $

其中

$Q_{\alpha, \beta}(r, f)=3 D_{\alpha, \beta}\Big(r, \frac{h}{f}\Big)+ 3D_{\alpha, \beta}\Big(r, \frac{h'}{f}\Big)+ 3D_{\alpha, \beta}\Big(r, \frac{h'}{h-1}\Big)+ D_{\alpha, \beta}\Big(r, \frac{g'}{g}\Big)+O(1). $

证  用$\frac{1}{f}=\frac{h}{f}-\frac{h-1}{h'}\cdot\frac{h'}{f}$, (2.1)式与(2.2)式得

$ \begin{eqnarray} D_{\alpha, \beta}\Big(r, \frac{1}{f}\Big)& \leq &D_{\alpha, \beta} \Big(r, \frac{h}{f}\Big)+D_{\alpha, \beta}\Big(r, \frac{h'}{f}\Big) +D_{\alpha, \beta}\Big(r, \frac{h-1}{h'}\Big)+O(1)\\ & \leq& D_{\alpha, \beta}\Big(r, \frac{h}{f}\Big)+D_{\alpha, \beta}\Big(r, \frac{h'}{f}\Big) +S_{\alpha, \beta}\Big(r, \frac{h'}{h-1}\Big) - C_{\alpha, \beta}\Big(r, \frac{h-1}{h'}\Big)+O(1)\\ & \leq& D_{\alpha, \beta}\Big(r, \frac{h}{f}\Big)+D_{\alpha, \beta}\Big(r, \frac{h'}{f}\Big) +D_{\alpha, \beta}\Big(r, \frac{h'}{h-1}\Big)+C_{\alpha, \beta}(r, h') \\ &&+ C_{\alpha, \beta}\Big(r, \frac{1}{h-1}\Big)-C_{\alpha, \beta}(r, h-1) -C_{\alpha, \beta}\Big(r, \frac{1}{h'}\Big)+O(1) \\& \leq& D_{\alpha, \beta}\Big(r, \frac{h}{f}\Big)+D_{\alpha, \beta}\Big(r, \frac{h'}{f}\Big) +D_{\alpha, \beta}\Big(r, \frac{h'}{h-1}\Big)+\overline{C}_{\alpha, \beta}(r, h) \\ &&+ \overline{C}_{\alpha, \beta}\Big(r, \frac{1}{h-1}\Big)-C_{\alpha, \beta}^{0}\Big(r, \frac{1}{h'}\Big)+O(1), \end{eqnarray} $

(2.13)

其中$C_{\alpha, \beta}^{0}(r, \frac{1}{h'})$仅算$h'$在$\Omega_{\alpha, \beta}$内的零点, 而不算$h-1$的零点.那么由(2.13)式, 引理2.1与$\overline{C}_{\alpha, \beta}(r, h)\leq \overline{C}_{\alpha, \beta}(r, f)+C_{\alpha, \beta}(r, \frac{1}{\varphi}) $得

$ \begin{eqnarray} S_{\alpha, \beta}(r, f)& \leq& D_{\alpha, \beta}\Big(r, \frac{1}{f}\Big) +C_{\alpha, \beta}\Big(r, \frac{1}{f}\Big)+O(1) \\ & \leq& D_{\alpha, \beta}\Big(r, \frac{h}{f}\Big)+D_{\alpha, \beta}\Big(r, \frac{h'}{f}\Big) +D_{\alpha, \beta}\Big(r, \frac{h'}{h-1}\Big)+\overline{C}_{\alpha, \beta}(r, f) \\ &&+ C_{\alpha, \beta}\Big(r, \frac{1}{\varphi}\Big)+ \overline{C}_{\alpha, \beta}\Big(r, \frac{1}{h-1}\Big)-C_{\alpha, \beta}^{0}\Big(r, \frac{1}{h'}\Big) +C_{\alpha, \beta}\Big(r, \frac{1}{f}\Big)+O(1). \end{eqnarray} $

(2.14)

若用$\overline{C}_{\alpha, \beta}^{(2}(r, f)$表示$f(z)$的多重极点仅算一次, 由(2.14)式得

$ \begin{eqnarray} \overline{C}_{\alpha, \beta}^{(2}(r, f)& \leq& C_{\alpha, \beta}(r, f)-\overline{C}_{\alpha, \beta}(r, f)\leq S_{\alpha, \beta}(r, f)-\overline{C}_{\alpha, \beta}(r, f) \\ & \leq& D_{\alpha, \beta}\Big(r, \frac{h}{f}\Big)+D_{\alpha, \beta}\Big(r, \frac{h'}{f}\Big) +D_{\alpha, \beta}\Big(r, \frac{h'}{h-1}\Big)+ C_{\alpha, \beta}\Big(r, \frac{1}{\varphi}\Big) \\ &&+ \overline{C}_{\alpha, \beta}\Big(r, \frac{1}{h-1}\Big)-C_{\alpha, \beta}^{0}\Big(r, \frac{1}{h'}\Big) +C_{\alpha, \beta}\Big(r, \frac{1}{f}\Big)+O(1). \end{eqnarray} $

(2.15)

若$z_{0}$是$f(z)$的单极点, 且$\varphi(z_{0})\neq 0, \infty$, 由文献[7]知

$ \begin{equation} g(z_{0})\neq 0, \infty;\quad g'(z_{0})=0. \end{equation} $

(2.16)

用引理2.1得

$ \begin{eqnarray} && D_{\alpha, \beta}\Big(r, \frac{g'}{g}\Big)-D_{\alpha, \beta}\Big(r, \frac{g}{g'}\Big)\\& =&S_{\alpha, \beta}\Big(r, \frac{g'}{g}\Big)-C_{\alpha, \beta}\Big(r, \frac{g'}{g}\Big)+ C_{\alpha, \beta}\Big(r, \frac{g}{g'}\Big)-S_{\alpha, \beta}\Big(r, \frac{g}{g'}\Big)\\& =&C_{\alpha, \beta}(r, g)+C_{\alpha, \beta}\Big(r, \frac{1}{g'}\Big)-C_{\alpha, \beta}(r, g') -C_{\alpha, \beta}\Big(r, \frac{1}{g}\Big)+O(1)\\& =&C_{\alpha, \beta}^{1}\Big(r, \frac{1}{g'}\Big)-\overline{C}_{\alpha, \beta}\Big(r, \frac{1}{g}\Big) -\overline{C}_{\alpha, \beta}(r, g)+O(1), \end{eqnarray} $

(2.17)

其中$C_{\alpha, \beta}^{1}(r, \frac{1}{g'})$表示仅算$g'(z)$的零点, 而不算$g(z)$的零点.用$C_{\alpha, \beta}^{1)}(r, f)$表示仅算$f(z)$的单极点, 那么由(2.16), (2.17)式得

$ \begin{eqnarray} C_{\alpha, \beta}^{1)}(r, f)&\leq& C_{\alpha, \beta}^{1}\Big(r, \frac{1}{g'}\Big)\leq \overline{C}_{\alpha, \beta}\Big(r, \frac{1}{g}\Big)+\overline{C}_{\alpha, \beta}(r, g)+ D_{\alpha, \beta}\Big(r, \frac{g'}{g}\Big)+O(1) \\ & \leq &C_{\alpha, \beta}\Big(r, \frac{1}{\varphi}\Big)+C_{\alpha, \beta}(r, \varphi) +\overline{C}_{\alpha, \beta}^{(2}(r, f)+\overline{C}_{\alpha, \beta}\Big(r, \frac{1}{h-1}\Big)\\ &&+ C_{\alpha, \beta}^{0}\Big(r, \frac{1}{h'}\Big)+D_{\alpha, \beta}\Big(r, \frac{g'}{g}\Big)+O(1). \end{eqnarray} $

(2.18)

这样, 由(2.15), (2.18)式得

$ \begin{eqnarray} \overline{C}_{\alpha, \beta}(r, f)&=& C_{\alpha, \beta}^{^{1)}}(r, f)+ \overline{C}_{\alpha, \beta}^{^{(2}}(r, f) \\& \leq &C_{\alpha, \beta}\Big(r, \frac{1}{\varphi}\Big)+C_{\alpha, \beta}(r, \varphi)+ 2\overline{C}_{\alpha, \beta}^{(2}(r, f)+\overline{C}_{\alpha, \beta}\Big(r, \frac{1}{h-1}\Big) \\ &&+ C_{\alpha, \beta}^{0}\Big(r, \frac{1}{h'}\Big)+D_{\alpha, \beta}\Big(r, \frac{g'}{g}\Big)+O(1) \\ & \leq& 3C_{\alpha, \beta}\Big(r, \frac{1}{\varphi}\Big)+C_{\alpha, \beta}(r, \varphi)+ 3\overline{C}_{\alpha, \beta}\Big(r, \frac{1}{h-1}\Big) \\ &&+ 2D_{\alpha, \beta}\Big(r, \frac{h}{f}\Big)+2D_{\alpha, \beta}\Big(r, \frac{h'}{f}\Big) +2D_{\alpha, \beta}\Big(r, \frac{h'}{h-1}\Big) \\ &&+ 2C_{\alpha, \beta}\Big(r, \frac{1}{f}\Big)+D_{\alpha, \beta}\Big(r, \frac{g'}{g}\Big)+O(1). \end{eqnarray} $

(2.19)

引理2.8  设$f(z)$是平面上无穷级亚纯函数, $a(z)$, $b(z)$是平面上有限级亚纯函数, $b(z)-a^{(k)}(z)$不为常数零, 则对任意小$\delta(>0)$有

$ \begin{eqnarray*} T_{0}(r, \alpha+\delta, \beta-\delta, f) & \leq& C\cdot (N(r, \alpha, \beta, f=a)+ N(r, \alpha, \beta, f^{(k)}=b))\\ &&+ C_{1}\cdot r^{\omega}\cdot\int_{1}^{r} \frac{N(t, \alpha, \beta, f=a)+N(t, \alpha, \beta, f^{(k)}=b)}{t^{\omega+1}}{\rm d}t+o(U(r)), \end{eqnarray*} $

其中$U(r)=r^{{\rm e}^{\frac{1}{\rho(r)}}}$是$T(r, f)$的型函数, $C$, $C_{1}$是仅与$\omega$, $\delta$有关的常数.

证  记$w(z)=f(z)-a(z)$, $\varphi(z)=b(z)-a^{(k)}(z)$, $h(z)=\frac{w^{(k)}(z)}{\varphi(z)}$, $g(z)=\frac{\varphi^{2}(h-1)^{k+2}}{(h')^{k+1}}$.因$w^{(k)}(z)$与$w(z)$有相同的增长级, 所以$h(z)$不为常数, 且增长级为无穷.又若$g(z)$为常数, 那么$\frac{1}{h(z)-1}=c\cdot (\int(\varphi(z))^{\frac{2}{k+1}}{\rm d}z)^{k+1}$与$h(z)$的增长级为无穷矛盾.由引理2.7与(2.1)式得

$ \begin{eqnarray} S_{\alpha, \beta}(r, f)& \leq & 3C_{\alpha, \beta}(r, w=0)+4\overline{C}_{\alpha, \beta}(r, w^{(k)}=\varphi) \\ &&+ 4C_{\alpha, \beta}\Big(r, \frac{1}{\varphi}\Big)+C_{\alpha, \beta}(r, \varphi)+ Q_{\alpha, \beta}(r, w)+S_{\alpha, \beta}(r, a)\\ & \leq &3C_{\alpha, \beta}(r, f=a)+ 4\overline{C}_{\alpha, \beta}(r, f^{(k)}=b) \\ &&+ 4C_{\alpha, \beta}\Big(r, \frac{1}{\varphi}\Big)+C_{\alpha, \beta}(r, \varphi)+ Q_{\alpha, \beta}(r, w)+S_{\alpha, \beta}(r, a), \end{eqnarray} $

(2.20)

其中

$Q_{\alpha, \beta}(r, w)=3 D_{\alpha, \beta}\Big(r, \frac{h}{w}\Big)+ 3D_{\alpha, \beta}\Big(r, \frac{h'}{w}\Big)+ 3D_{\alpha, \beta}\Big(r, \frac{h'}{h-1}\Big)+ D_{\alpha, \beta}\Big(r, \frac{g'}{g}\Big)+O(1).$

由(2.20)式, 引理2.1与引理2.2得

$ \begin{eqnarray*} && \omega\sin(\omega\delta)\cdot\frac{T_{0}(r, \alpha+\delta, \beta-\delta, f)}{r^{\omega}} \\ & \leq& 12\omega \frac{N(r, \alpha, \beta, f=a)}{r^{\omega}}+6\omega^{2}\int_{1}^{r} \frac{N(t, \alpha, \beta, f=a)}{t^{\omega+1}}{\rm d}t \\ &&+ 16\omega \frac{N(r, \alpha, \beta, f^{(k)}=b)}{r^{\omega}}+8\omega^{2}\int_{1}^{r} \frac{N(t, \alpha, \beta, f^{(k)}=b)}{t^{\omega+1}}{\rm d}\\ &&+ 16\omega \frac{N(r, \alpha, \beta, \varphi=0)}{r^{\omega}}+8\omega^{2}\int_{1}^{r} \frac{N(t, \alpha, \beta, \varphi=0)}{t^{\omega+1}}{\rm d}t \\ &&+ 4\omega \frac{N(r, \alpha, \beta, \varphi)}{r^{\omega}}+2\omega^{2}\int_{1}^{r} \frac{N(t, \alpha, \beta, \varphi)}{t^{\omega+1}}{\rm d}t \\ &&+ 2\omega \frac{T_{0}(r, \alpha, \beta, a)}{r^{\omega}}+\omega^{2}\int_{1}^{r} \frac{T_{0}(t, \alpha, \beta, a)}{t^{\omega+1}}{\rm d}t+ Q_{\alpha, \beta}(r, w). \end{eqnarray*} $

所以

$ \begin{eqnarray} T_{0}(r, \alpha+\delta, \beta-\delta, f) & \leq& C\cdot (N(r, \alpha, \beta, f=a)+ N(r, \alpha, \beta, f^{(k)}=b)) \\ &&+ C_{1}\cdot r^{\omega}\cdot\bigg(\int_{1}^{r} \frac{N(t, \alpha, \beta, f=a)}{t^{\omega+1}}{\rm d}t+\int_{1}^{r} \frac{N(t, \alpha, \beta, f^{(k)}=b)}{t^{\omega+1}}{\rm d}t\bigg) \\ &&+ C_{2}\cdot r^{\omega}\cdot Q_{\alpha, \beta}(r, w)+o(U(r)), \end{eqnarray} $

(2.21)

其中$C$, $C_{i}(i=1, 2)$是仅与$\omega$, $\delta$有关的常数.又因

$ D_{\alpha, \beta}\Big(r, \frac{h}{w}\Big)\leq D_{\alpha, \beta} \Big(r, \frac{w^{(k)}}{w}\Big)+D_{\alpha, \beta}\Big(r, \frac{1}{\varphi}\Big), $

$ D_{\alpha, \beta}\Big(r, \frac{h'}{w}\Big)\leq D_{\alpha, \beta} \Big(r, \frac{\varphi'}{\varphi}\Big)+ 2D_{\alpha, \beta}\Big(r, \frac{1}{\varphi}\Big)+D_{\alpha, \beta}\Big(r, \frac{w^{(k)}}{w}\Big) +D_{\alpha, \beta}\Big(r, \frac{w^{(k+1)}}{w}\Big)+O(1), $

$ D_{\alpha, \beta}\Big(r, \frac{g'}{g}\Big)\leq (k+2)D_{\alpha, \beta}\Big(r, \frac{h'}{h-1}\Big)+ (k+1)D_{\alpha, \beta}\Big(r, \frac{h''}{h'}\Big)+2D_{\alpha, \beta} \Big(r, \frac{\varphi'}{\varphi}\Big)+O(1). $

所以, 通过分别估计$r^{\omega} D_{\alpha, \beta}(r, \frac{w^{(k)}}{w})$, $r^{\omega} D_{\alpha, \beta}(r, \frac{1}{\varphi})$, $r^{\omega} D_{\alpha, \beta}(r, \frac{\varphi'}{\varphi})$, $r^{\omega} D_{\alpha, \beta}(r, \frac{h'}{h-1})$就可估计$r^{\omega}\cdot Q_{\alpha, \beta}(r, w)$.下面估计$r^{\omega} D_{\alpha, \beta}(r, \frac{w^{(k)}}{w})$, $r^{\omega} D_{\alpha, \beta}(r, \frac{h'}{h-1})$, 其它相同.用引理2.3, 引理2.4, 引理2.5并取$R=r+\frac{r\log r}{\log U(r)\log^2\log U(r)}$得

$ \begin{eqnarray} r^{\omega} D_{\alpha, \beta}\Big(r, \frac{w^{(k)}}{w}\Big) &\leq &r^{\omega}{\bigg(} D_{\alpha, \beta}\Big(r, \frac{w^{(k)}}{w^{(k-1)}}\Big) +D_{\alpha, \beta}\Big(r, \frac{w^{(k-1)}}{w^{(k-2)}}\Big)+\cdots+ D_{\alpha, \beta}\Big(r, \frac{w'}{w}\Big){\bigg)} \\& \leq& r^{\omega}K{\bigg(}\Big(\frac{R}{r}\Big)^{\omega } \int_{1}^{R} \frac{\log T(t, w^{(k-1)})}{t^{\omega+1}}{\rm d}t+\log^{+} \frac{r}{R-r}+\log\frac{R}{r}{\bigg)}+\cdots \\ &&+ r^{\omega}K{\bigg(}\Big(\frac{R}{r}\Big)^{\omega } \int_{1}^{R} \frac{\log T(t, w)}{t^{\omega+1}}{\rm d}t+\log^{+} \frac{r}{R-r}+\log\frac{R}{r}{\bigg)} \\ && + 4\omega\cdot m_{\alpha, \beta}\Big(r, \frac{w^{(k)}}{w^{(k-1)}}\Big)+\cdots +4\omega\cdot m_{\alpha, \beta}\Big(r, \frac{w'}{w}\Big) \\& \leq &2^{\omega}K r^{\omega}{\bigg(} \int_{1}^{R} \frac{\log T(t, w^{(k-1)})}{t^{\omega+1}}{\rm d}t+\cdots +\int_{1}^{R} \frac{\log T(t, w)}{t^{\omega+1}}{\rm d}t{\bigg)} \\ &&+4\omega\cdot m\Big(r, \frac{w^{(k)}}{w^{(k-1)}}\Big)+\cdots+4\omega\cdot m \Big(r, \frac{w'}{w}\Big) \\ &&+ k\cdot K\cdot r^{\omega}{\bigg(} \log\frac{\log U(r)\log^2\log U(r)}{\log r} +\log 2{\bigg)} \\& \leq& r^{\omega}\cdot O( \log T(R, w^{(k-1)})+\cdots+\log T(R, w) )\\ &&+ O\bigg(\log R+\log\frac{\log U(r)\log^2\log U(r)}{r\log r}+ r^{\omega} \log\log U(r)\bigg) \\& \leq &r^{\omega}\cdot O(\log T(R, w)+\log\log U(r)) \\& \leq &r^{\omega}\cdot O(\log U(R)) \\ & =&o(U(r))\ \ (r\to+\infty), \end{eqnarray} $

(2.22)

$ \begin{eqnarray} r^{\omega} D_{\alpha, \beta}\Big(r, \frac{h'}{h-1}\Big) &\leq &r^{\omega}K{\bigg(}\Big(\frac{R}{r}\Big)^{\omega } \int_{1}^{R} \frac{\log T(t, h-1)}{t^{\omega+1}}{\rm d}t +\log^{+}\frac{r}{R-r}+\log\frac{R}{r}{\bigg)} \\& &+ 4\omega\cdot m_{\alpha, \beta}\Big(r, \frac{h'}{h-1}\Big)\\ & \leq& r^{\omega}K{\bigg(}\frac{2^{\omega}}{\omega}\log T(R, h-1) + \log\frac{\log U(r)\log^2\log U(r)}{\log r} +\log 2{\bigg)}\\&&+ 4\omega\cdot m\Big(r, \frac{h'}{h-1}\Big) \\& \leq &r^{\omega}\cdot O(\log T(R, h)+\log U(r)) \\& \leq& r^{\omega}\cdot O(\log U(R)) \\& =&o(U(r))\ \ (r\to+\infty). \end{eqnarray} $

(2.23)

由(2.21), (2.22)与(2.23)式, 知本引理正确.

注  引理2.8可认为是角域内用Ahlfors-Shimizu特征表示的Hayman不等式, 当$f(z)$是有限级时, 是否可建立角域内类似的Hayman不等式是一个困难问题.

定理1.1的证明  由$T(r, f)\ll U(r)$知, 存在方向$\arg z=\theta_0$, 对任意$\delta(>0)$有

$ \limsup\limits_{r\to +\infty} \frac{T_{0}(r, \theta_0-\delta, \theta_0+\delta, f)}{U(r)}>0. $

(3.1)

下证对亚纯函数$g(z)\in{\cal G}$与任意$\varepsilon(>\delta)$必有

$ \limsup\limits_{r\to +\infty} \frac{N(r, \theta_0, \varepsilon, f=g)}{U(r)}>0 $

(3.2)

至多有$2$个例外的$g(z)$.否则, 若有$g_{i}(z)(i=1, 2, 3)$使

$N(r, \theta_0, \delta_{1}, f=g_{i})=o(U(r)). $

记$ m(r)=\frac{U(r)}{r^{\omega}}=r^{{\rm e}^{\frac{1}{\rho(r)}}-\omega}$.用(2.11)式, 仿(2.12)式的证明得$r\to +\infty$时

$r^{\omega}\cdot\int_{1}^{r} \frac{N(r, \theta_0, \varepsilon, f=g_{i})}{t^{\omega+1}}{\rm d}t=o(U(r)). $

(3.3)

用(3.3)式与引理2.6得$r\to +\infty$时

$T_{0}(r, \theta_0, \delta, f)=o(U(r)). $

这与(3.1)式矛盾.故(3.2)式得证.又由引理2.5 (注意: $\rho'(r)<0$)得

$ r U'(r)= U(r)\bigg({\rm e}^{\frac{1}{\rho(r)}}-\frac{r{\rm e}^{\frac{1}{\rho(r)}} \rho'(r)\log r}{\rho^2(r)}\bigg)> U(r)\cdot {\rm e}^{\frac{1}{\rho(r)}}. $

(3.4)

由(3.2)式可知定理1.1正确.否则, 若

$\limsup\limits_{r\to +\infty} \frac{n(r, \theta_0, \varepsilon, f=g)} {{\rm e}^{\frac{1}{\rho(r)}}\cdot U(r)}=0, $

即对任意小$\sigma(>0)$, 存在$r_{0}$, 当$r>r_{0}$时有

$n(r, \theta_0, \varepsilon, f=g)< \sigma\cdot {\rm e}^{\frac{1}{\rho(r)}}\cdot U(r). $

(3.5)

用(3.4), (3.5)式得

$ \begin{eqnarray*} N(r, \theta_0, \varepsilon, f=g)&=&\int_{r_{0}}^{r}\frac{n(t, \theta_0, \varepsilon, f=g)}{t}{\rm d}t \leq\sigma\cdot\int_{r_{0}}^{r} \frac{{\rm e}^{\frac{1}{\rho(t)}}\cdot U(t)}{t}{\rm d}t\\ & \leq& \sigma \cdot\int_{r_{0}}^{r} U'(t){\rm d}t =\sigma \cdot (U(r)-U(r_{0})). \end{eqnarray*} $

因$\sigma(>0)$任意小, 这与(3.2)式矛盾.定理1.1证毕.

定理1.2的证明  由$T(r, f)\ll U(r)$与(3.2)式知本定理正确.

定理1.3的证明  为了完成本定理证明, 下证(3.1)式中的方向$\arg z=\theta_0$具有本定理性质.即对正数$\varepsilon(>\delta)$, 对任意正整数$k$与任意有限级亚纯函数$a(z)$, $b(z)$, 当$b(z)-a^{(k)}(z)$不为常数零时有

$ \limsup\limits_{r\to +\infty} \frac{N(r, \theta_0, \varepsilon, f=a)+ N(r, \theta_0, \varepsilon, f^{(k)}=b)}{U(r)}>0. $

(3.6)

否则, 若有$a(z)$, $b(z)$使

$ N(r, \theta_0, \varepsilon, f=a)+N(r, \theta_0, \varepsilon, f^{(k)}=b)=o(U(r)); $

记$m(r)=\frac{U(r)}{r^{\omega}}=r^{{\rm e}^{\frac{1}{\rho(r)}}-\omega}$, 仿(2.12)式的证明得$r\to+\infty$时

$ r^{\omega}\cdot \int_{1}^{r} \frac{N(t, \theta_0, \varepsilon, f=a)+ N(r, \theta_0, \varepsilon, f^{(k)}=b)}{t^{\omega+1}}{\rm d}t=o(U(r)). $

(3.7)

用(3.7)式与引理2.8得$r\to +\infty$时

$T_{0}(r, \theta_0, \delta, f)=o(U(r)) . $

这与(3.1)式矛盾.故(3.6)式正确.用(3.4), (3.6)式仿定理1.1后面的证明知本定理正确.

定理1.4的证明  由$T(r, f)\ll U(r)$与(3.6)式知本定理正确.