设 $f(z)$与 $g(z)$在单位圆盘 $U=\{z:|z| < 1\}$内解析, 如果存在 $U$内满足 $|\omega(z)|\leq|z|$的解析函数 $\omega(z)$, 使得 $g(z)=f(\omega(z))$, 则称 $g(z)$从属于 $f(z)$, 记作 $g(z)\prec f(z)$.

用 $P(C, D)$ $(-1\leq D < C\leq 1)$表示在单位圆盘 $U$内解析并且满足条件 $p(z)\prec\frac{1+Cz}{1+Dz}$的所有函数 $p(z)=1+\sum\limits_{n=1}^{\infty}p_{n}z^{n}$的全体.显然 $P(1, -1)=P$为熟知的正实部函数类^[1].

设 $S$表示在单位圆盘 $U$内单叶解析函数 $f(z)=z+\sum\limits_{n=2}^{\infty}a_nz^n$构成的函数类.若函数 $f(z)\in S$满足条件 $\frac{zf'(z)}{f(z)}\in P$, 则称 $f(z)$属于星象函数类 $S^*$; 设函数 $f(z)\in S$, 如果存在函数 $g(z) \in S{\rm{*}}$, 使得 $\frac{zf'(z)}{g(z)}\in P$, 则称 $f(z)$属于近于凸函数类 $C$.设 $b$为复数且 $b\neq 0$, $f(z)\in S$, 如果存在 $g(z)\in S^*$使得 $\left\{ {1 + \frac{1}{b}\left( {\frac{{zf'(z)}}{{g(z)}} - 1} \right)} \right\} \in {\rm{ }}P$, 则称 $f(z) $属于复阶近于凸函数类 $C(b)$^[2].设函数 $f(z)\in S$, $\alpha>0$, 如果存在函数 $g(z)\in S^*$, 使得 $\frac{zf'(z)}{f(z)}\left(\frac{f(z)}{g(z)}\right)^\alpha\in P$, 则称 $f(z)$属于Bazilevič函数类 $B_\alpha$.一些学者从不同的角度出发, 研究了一些有趣的Bazilevič函数子族^[3-5].

设 $\alpha > 0$, $\beta\in R$, $f(z)\in S$, 如果存在 $g(z)\in S^*$使得 $\frac{{zf'(z)}}{{f(z)}}{\left( {\frac{{f(z)}}{{g(z)}}} \right)^\alpha }{\left( {\frac{{f(z)}}{z}} \right)^{{\rm{i}}\beta }} \in P, $则称 $f(z)\in B(\alpha, \beta)$^[3].

文献[5]给出了如下 $\alpha$型 $\beta$级Bazilevič函数类 $B_\alpha(\beta)$.

定义1.1^[5]  设 $f(z)\in S$, $\alpha\geq0$, $0\leq\beta < 1$, 若存在 $g(z)\in S^*$, 使得

$ {\rm{Re}}\left\{ {\frac{{zf'(z)}}{{f(z)}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }} \right\} > \beta, $

(1.1)

则称 $f(z)\in B_\alpha(\beta)$, 其中的幂函数取主值.显然 $B_\alpha(0)=B_\alpha$.

本文引入如下Bazilevič函数.

定义1.2  设 $f(z)\in S$, $\alpha > 0$, $-1\leq D < C\leq1$, 如果存在 $g(z)\in S^*$, 使得

$ \frac{{zf'(z)}}{{f(z)}}{\left( {\frac{{f(z)}}{{g(z)}}} \right)^\alpha } \in P(C, D), $

则称 $f(z)\in B_{\alpha}(C, D)$.显然 $B_{\alpha}(1, -1)=B_\alpha$, $B_{\alpha}(1-2\beta, -1)=B_\alpha(\beta)$.

用 $Y$表示圆对称函数类^[6].设 $f(z)\in S$, 若

$ {\rm{log}}\frac{{f(z)}}{z} = 2\sum\limits_{n = 1}^\infty {{b_n}} {z^n}, $

(1.2)

则称 $b_n$为 $f(z)$的对数系数.对数系数的估计在单叶函数的系数估计中有重要作用. Keobe函数 $k(z)=z(1-z)^{-2}$的对数系数为 $b_n=1/n$.对 $b_n$( $n\geq2$)的估计, 现在已经证明

(1) 当 $f(z)\in S^*$时, $b_n\leq \frac{1}{n}$^[7];

(2) 当 $f(z)\in C$时, $b_n\leq A\frac{\mathrm{log}n}{n}$, 其中 $ A $表示一个绝对常数^[8];

(3) 当 $f(z)\in B_\alpha$时, ${b_n} \le {\rm{ }}A(1 + \alpha )\frac{{{\rm{log}}n}}{n}$, 其中 $ A $表示一个绝对常数^[9];

(4) 当 $f(z)\in Y$时, $b_n\leq A\frac{\mathrm{log}n}{n}$, 其中 $ A $表示一个绝对常数^[10];

(5) 当 $f(z)\in B(\alpha, \beta)$时, ${b_n} \le A\frac{{{\rm{log}}n}}{n}$, 其中 $ A $表示一个绝对常数^[11];

(6) 当 $f(z)\in C(b)$时, $b_n\leq A\frac{\mathrm{log}n}{n}$, 其中 $ A $表示一个绝对常数^[12]

不同的地方, $A$表示不同常数.

本文研究 $B_{\alpha}(C, D)$的对数系数, 推广了文献[9]的结果.

为了得到 $B_{\alpha}(C, D)$的对数系数, 需要如下引理

引理2.1^[13]  如果 $p(z)\in P(C, D)$, 其中 $ - 1 \le {\rm{ }}D < C \le 1$, 则 $\frac{1-C}{1-D} < \mathrm{Re}(p(z)) < \frac{1+C}{1+D}$.

引理2.2^[11]  设 $f(z)\in S$, 则 $\mathrm{Re}\frac{zf'(z)}{f(z)}=\frac{\partial}{\partial\theta}\left(\mathrm{arg}\frac{f(z)}{z}\right)+1$.

引理2.3^[11]  设 $f(z)\in S$, $\alpha\in \mathbb{C}$.则对 $z=re^{{\rm i}\theta}$, $0 < r < 1$, 有

$ \frac{\partial }{{\partial \theta }}\left( {{\rm{arg}}{{\left( {\frac{{f(z)}}{z}} \right)}^\alpha }} \right) = \alpha \frac{\partial }{{\partial \theta }}\left( {{\rm{arg}}\frac{{f(z)}}{z}} \right). $

引理2.4^[8]  设 $f(z)\in S$, 则对 $z = r{e^{{\rm{i}}\theta }}$, $\frac{1}{2}\leq r < 1$, 有

(1)$\frac{1}{{2\pi }}\int_0^{2\pi } {{{\left| {\frac{{z{f^\prime }(z)}}{{f(z)}}} \right|}^2}} d\theta \le 1 + \frac{4}{{1 - r}}{\rm{log}}\frac{1}{{1 - \sqrt r }}$

(2)$\frac{1}{{2\pi }}\int_{\frac{1}{2}}^r {{{\int_0^{2\pi } {\left| {\frac{{z{f^\prime }(z)}}{{f(z)}}} \right|} }^2}} d\theta dr \le 1 + 2{\rm{log}}\frac{1}{{1 - r}}$

引理2.5^[14]  设 $g(z)\in S^*$, 则 $\frac{\partial}{\partial\theta}\mathrm{arg}g(z) > 0$且 $\displaystyle\int_{0}^{2\pi}\frac{\partial}{\partial\theta}\mathrm{arg}g(z)=2\pi$.

引理2.6^[15]  设 $f(z)\in S$, 则对 $0 < r < 1$, 有 $\sum\limits_{k=1}^{\infty}k|b_k|^2r^{2k}\leq\mathrm{log}\frac{1}{1-r}$.

引理2.7  设 $f(z)\in B_{\alpha}(C, D)$, $g(z)\in S^*$使得 $\frac{{zf'(z)}}{{f(z)}}{\left( {\frac{{f(z)}}{{g(z)}}} \right)^\alpha } \in {\rm{ }}P(C, D)$.则对 $z=re^{{\rm i}\theta}$, $0\leq r < 1$, 有

(1)${J_1} = \frac{1}{{2\pi }}\left| {\int_0^{2\pi } {{\rm{Re}}} \frac{{zf'(z)}}{{f(z)}}{e^{{\rm{iarg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}d\theta } \right| \le 3;$

(2)$ {J_2} = \frac{1}{{2\pi }}\left| {\int_0^{2\pi } {{\rm{Im}}} \frac{{zf'(z)}}{{f(z)}}{e^{{\rm{iarg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}d\theta } \right| \le 3 + 4\alpha + 8\alpha {\rm{log}}\frac{1}{{1 - r}};$

(3)${J_3} = \frac{1}{{2\pi }}\left| {\int_0^{2\pi } {\frac{{z{f^\prime }(z)}}{{f(z)}}} {e^{2{\rm{iarg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}{e^{{\rm{i}}n\theta }}d\theta } \right| \le 4\alpha {\left( {\frac{1}{{{n^2}}} + \frac{4}{n}{\rm{log}}\frac{1}{{1 - r}}} \right)^{\frac{1}{2}}}{\left( {1 + \frac{4}{{1 - r}}{\rm{log}}\frac{1}{{1 - \sqrt r }}} \right)^{\frac{1}{2}}}.$

证  (1) 由引理2.2和引理2.3可知

$ \begin{array}{l} {J_1} \le \frac{1}{{2\pi }}\left| {{{\int_0^{2\pi } {{e^{{\rm{iarg}}\left( {\frac{{f(z)}}{{g(z)}}} \right)}}} }^\alpha }d\theta } \right| + \frac{1}{{2\pi }}\left| {\int_0^{2\pi } {\frac{\partial }{{\partial \theta }}} \left( {{\rm{arg}}\frac{{f(z)}}{z}} \right){e^{{\rm{iarg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}d\theta } \right|\\ \;\;\;\; = \frac{1}{{2\pi }}\left| {{{\int_0^{2\pi } {{e^{{\rm{iarg}}\left( {\frac{{f(z)}}{{g(z)}}} \right)}}} }^\alpha }d\theta } \right| + \frac{1}{{2\pi }}\left| {\int_0^{2\pi } {\frac{1}{{\alpha {\rm{i}}}}} \frac{\partial }{{\partial \theta }}{e^{{\rm{iarg}}{{\left( {\frac{{f(z)}}{z}} \right)}^\alpha }}}{e^{ - {\rm{iarg}}{{\left( {\frac{{g(z)}}{z}} \right)}^\alpha }}}d\theta } \right| = {J_{11}} + {J_{12}}. \end{array} $

易知 $J_{11}\leq1$, 利用分部积分公式和引理2.5可得

$ \begin{array}{l} {J_{12}} = \frac{1}{{2\pi }}\left| {\int_0^{2\pi } {\frac{\partial }{{\partial \theta }}} {\rm{arg}}\left( {\frac{{g(z)}}{z}} \right){e^{{\rm{iarg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}d\theta } \right|\\ \;\;\;\;\; \le \frac{1}{{2\pi }}\int_0^{2\pi } {\left\{ {\left| {\frac{\partial }{{\partial \theta }}{\rm{arg}}g(z)} \right| + \left| {\frac{\partial }{{\partial \theta }}{\rm{arg}}g(z)} \right|} \right\}} d\theta \le 2. \end{array} $

所以 ${J_1} = \frac{1}{{2\pi }}\left| {\int_0^{2\pi } {{\rm{Re}}} \frac{{zf'(z)}}{{f(z)}}{e^{{\rm{iarg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}d\theta } \right| \le {J_{11}} + {J_{12}} \le 3$.

(2) 记 $\mathrm{Re}\frac{zf'(z)}{f(z)}=u(re^{{\rm i}\theta})$, $\mathrm{Im}\frac{zf'(z)}{f(z)}=v(re^{{\rm i}\theta})$, 由柯西-黎曼条件可知对于 $\frac{1}{2}\leq r < 1$, 有

$ v(r{e^{{\rm{i}}\theta }}) - v(\frac{1}{2}{e^{{\rm{i}}\theta }}) = \int_{\frac{1}{2}}^{\rm{r}} {\frac{{\partial \upsilon (r{e^{i\theta }})}}{{\partial r}}} dr = - \int_{\frac{1}{2}}^\mathit{r} {\frac{1}{r}\frac{{\partial u(r{e^{{\rm{i}}\theta }})}}{{\partial \theta }}dr,} $

则

$ {J_2} \le \frac{1}{{2\pi }}\left| {\int_0^{2\pi } \upsilon (\frac{1}{2}{e^{{\rm{i}}\theta }}){e^{{\rm{iarg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}d\theta } \right| + \frac{1}{{2\pi }}\left| {\int_0^{2\pi } {\int_{\frac{1}{2}}^\mathit{r} {} \frac{1}{r}} \frac{{\partial u(r{e^{{\rm{i}}\theta }})}}{{\partial \theta }}{e^{{\rm{iarg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}drd\theta } \right| = {J_{21}} + {J_{22}}. $

而

$ {J_{21}} \le \frac{1}{{2\pi }}\int_0^{2\pi } {\frac{{\max }}{{\theta \in [{\rm{0}},{\rm{2}}\pi ]}}} \left| {v(\frac{1}{2}{e^{{\rm{i}}\theta }})} \right|d\theta \le \mathop {\max }\limits_{\theta \in [0,2\pi ]} \left| {\frac{{\frac{1}{2}f'(\frac{1}{2}{e^{{\rm{i}}\theta }})}}{{f(\frac{1}{2}{e^{{\rm{i}}\theta }})}}} \right| \le \frac{{1 + \frac{1}{2}}}{{1 - \frac{1}{2}}} = 3. $

由分部积分和引理2.3可知

$ {J_{22}} = \frac{\alpha }{{2\pi }}\left| {\int_{\frac{1}{2}}^r {\int_0^{2\pi } {\frac{1}{r}} } u(r{e^{{\rm{i}}\theta }}){e^{{\rm{iarg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}\left( {\frac{\partial }{{\partial \theta }}{\rm{arg}}\frac{{f(z)}}{z} - \frac{\partial }{{\partial \theta }}{\rm{arg}}\frac{{g(z)}}{z}} \right)d\theta dr} \right|. $

由引理2.2和 $\frac{1}{2}\leq r < 1$可知

$ \begin{array}{l} {\mathit{J}_{22}} \le \frac{\alpha }{\pi }\int_{\frac{1}{2}}^r {\int_0^{2\pi } {\left| {{\rm{Re}}\frac{{zf'(z)}}{{f(z)}}} \right|} } \left| {{\rm{Re}}\frac{{zf'(z)}}{{f(z)}} - {\rm{Re}}\frac{{zg'(z)}}{{g(z)}}} \right|d\theta dr\\ \;\;\;\;\; \le 2\alpha \left( {\frac{1}{{2\pi }}{{\int_{\frac{1}{2}}^r {\int_0^{2\pi } {\left| {\frac{{zf'(z)}}{{f(z)}}} \right|} } }^2}d\theta dr + \frac{1}{{2\pi }}\int\limits_{\frac{1}{2}}^r {} \int\limits_0^{2\pi } {\left| {\frac{{zf'(z)}}{{f(z)}}} \right|} \left| {\frac{{zg'(z)}}{{g(z)}}} \right|d\theta dr} \right). \end{array} $

由引理2.4和Schwarz不等式可知

$ \begin{array}{l} {\mathit{J}_{22}} \le 2\alpha \left( {1 + 2{\rm{log}}\frac{1}{{1 - r}}} \right) + 2\alpha {\left( {\frac{1}{{2\pi }}\int_{\frac{1}{2}}^r {} {{\int_0^{2\pi } {\left| {\frac{{z{f^\prime }(z)}}{{f(z)}}} \right|} }^2}d\theta dr\frac{1}{{2\pi }}\int_{\frac{1}{2}}^r {} {{\int_0^{2\pi } {\left| {\frac{{z{g^\prime }(z)}}{{g(z)}}} \right|} }^2}d\theta dr} \right)^{\frac{1}{2}}}\\ \;\;\;\;\; \le 4\alpha \left( {1 + 2{\rm{log}}\frac{1}{{1 - r}}} \right). \end{array} $

所以

$ {J_2} = \frac{1}{{2\pi }}\left| {\int_0^{2\pi } {{\rm{Im}}} \frac{{zf'(z)}}{{f(z)}}{e^{{\rm{iarg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}d\theta } \right| \le {J_{21}} + {J_{21}} \le 3 + 4\alpha \left( {1 + 2{\rm{log}}\frac{1}{{1 - r}}} \right). $

(3) 由(2) 式可知

$ \frac{{zf'(z)}}{{f(z)}} = 1 + z{\left( {{\rm{log}}\frac{{f(z)}}{z}} \right)^\prime } = 1 + \sum\limits_{k = 1}^\infty 2 k{b_k}{z^k}. $

(2.1)

所以

$ \begin{array}{l} \frac{{zf'(z)}}{{f(z)}}{e^{{\rm{i}}n\theta }} = {e^{in\theta }}\left( {1 + \sum\limits_{k = 1}^\infty 2 k{b_k}{z^k}} \right) = {e^{{\rm{i}}n\theta }} + \sum\limits_{k = 1}^\infty 2 k{b_k}{r^k}{e^{{\rm{i}}(n + k)\theta }}\\ \;\;\;\;\;\;\;\;\;\;\;\;\; = \frac{1}{{\rm{i}}}\frac{\partial }{{\partial \theta }}\left( {\frac{{{e^{in\theta }}}}{n} + \sum\limits_{k = 1}^\infty {\frac{{2k{b_k}{r^k}{e^{{\rm{i}}(n + k)\theta }}}}{{n + k}}} } \right) = \frac{1}{{\rm{i}}}\frac{\partial }{{\partial \theta }}F(z). \end{array} $

由引理2.2, 引理2.3和分部积分可得

$ \begin{array}{l} {\mathit{J}_3} = \frac{1}{{2\pi }}\left| {\int_0^{2\pi } {\frac{1}{i}} \frac{\partial }{{\partial \theta }}F(z){e^{2i{\rm{arg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}d\theta } \right| = \frac{\alpha }{\pi }\left| {\int_0^{2\pi } F (z){e^{2i{\rm{arg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}\frac{\partial }{{\partial \theta }}{\rm{arg}}\frac{{f(z)}}{{g(z)}}d\theta } \right|\\ \;\;\;\; = \frac{\alpha }{\pi }\left| {\int_0^{2\pi } F (z){e^{2i{\rm{arg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}\left( {{\rm{Re}}\frac{{zf'(z)}}{{f(z)}} - {\rm{Re}}\frac{{zg'(z)}}{{g(z)}}} \right)d\theta } \right|\\ \;\;\;\; \le \frac{\alpha }{\pi }\left| {\int_0^{2\pi } {\left( {|F(z)||\frac{{zf'(z)}}{{f(z)}}| + |F(z)||\frac{{zg'(z)}}{{g(z)}}|} \right)} d\theta } \right|\\ \;\;\;\; \le 2\alpha {\left( {\frac{1}{{2\pi }}\int_0^{2\pi } | F(z){|^2}d\theta } \right)^{\frac{1}{2}}}{\left( {\frac{1}{{2\pi }}{{\int_0^{2\pi } {\left| {\frac{{zf'(z)}}{{f(z)}}} \right|} }^2}d\theta } \right)^{\frac{1}{2}}}\\ \;\;\;\;\;\;\; + 2\alpha {\left( {\frac{1}{{2\pi }}\int_0^{2\pi } | F(z){|^2}d\theta } \right)^{\frac{1}{2}}}{\left( {\frac{1}{{2\pi }}{{\int_0^{2\pi } {\left| {\frac{{zg'(z)}}{{{\rm{g}}(z)}}} \right|} }^2}d\theta } \right)^{\frac{1}{2}}}. \end{array} $

由引理2.6可知

$ \frac{1}{{2\pi }}\int_0^{2\pi } | F(z){|^2}d\theta = \frac{1}{{{n^2}}} + 4\sum\limits_{k = 1}^\infty {\frac{{{k^2}|{b_k}{|^2}{r^{2k}}}}{{{{(n + k)}^2}}}} \le \frac{1}{{{n^2}}} + \frac{4}{n}\sum\limits_{k = 1}^\infty k |{b_k}{|^2}{r^{2k}} \le \frac{1}{{{n^2}}} + \frac{4}{n}{\rm{log}}\frac{1}{{1 - r}}. $

所以由引理2.4可知

$ {J_3} \le 4\alpha {\left( {\frac{1}{{{n^2}}} + \frac{4}{n}{\rm{log}}\frac{1}{{1 - r}}} \right)^{\frac{1}{2}}}{\left( {1 + \frac{4}{{1 - r}}{\rm{log}}\frac{1}{{1 - \sqrt r }}} \right)^{\frac{1}{2}}}. $

引理2.8  设 $f(z)\in B_{\alpha}(C, D)$( $\alpha > 0$), 则对 $z=re^{{\rm i}\theta}$, $0\leq r < 1$, 有

$ \frac{{1 - |D|r}}{{1 + |C|r}}\frac{{1 - r}}{{1 + r}} \le \left| {{{\left( {\frac{{g(z)}}{{f(z)}}} \right)}^\alpha }} \right| \le \frac{{1 + |D|r}}{{1 - |C|r}}\frac{{1 + r}}{{1 - r}}. $

证  设 $f(z)\in B_{\alpha}(C, D)$, 则存在 $g(z)\in S^*$使得

$ \frac{{zf'(z)}}{{f(z)}}{\left( {\frac{{f(z)}}{{g(z)}}} \right)^\alpha } \prec \frac{{1 + Cz}}{{1 + Dz}} $

由从属关系定义可知, 存在Schwarz函数 $\omega(z)$, 使得

$ \frac{{zf'(z)}}{{f(z)}}{\left( {\frac{{f(z)}}{{g(z)}}} \right)^\alpha } = \frac{{1 + C\omega (z)}}{{1 + D\omega (z)}} $

经简单计算有

$ {\left( {\frac{{g(z)}}{{f(z)}}} \right)^\alpha } = \frac{{1 + D\omega (z)}}{{1 + C\omega (z)}}\frac{{zf'(z)}}{{f(z)}}. $

因为 $|\omega(z)|\leq|z|$, 所以

$ \frac{{1 - |D||z|}}{{1 + |C||z|}}\left| {\frac{{zf'(z)}}{{f(z)}}} \right| \le \left| {{{\left( {\frac{{g(z)}}{{f(z)}}} \right)}^\alpha }} \right| \le \frac{{1 + |D||z|}}{{1 - |C||z|}}\left| {\frac{{zf'(z)}}{{f(z)}}} \right|. $

又因为

$ \frac{{1 - r}}{{1 + r}} \le \left| {\frac{{zf'(z)}}{{f(z)}}} \right| \le \frac{{1 + r}}{{1 - r}}, $

所以

$ \frac{{1 - |D|r}}{{1 + |C|r}}\frac{{1 - r}}{{1 + r}} \le \left| {{{\left( {\frac{{g(z)}}{{f(z)}}} \right)}^\alpha }} \right| \le \frac{{1 + |D|r}}{{1 - |C|r}}\frac{{1 + r}}{{1 - r}}. $

定理3.1  设 $f(z)\in B_{\alpha}(C, D)$, 则对 $n\geq2$,

$ |{b_n}| \le A\frac{1}{n}{\rm{log}}n + B\frac{1}{n} + 16\frac{{1 - C}}{{1 - D}}\frac{{(1 + |D|)n - |D|}}{{(1 - |C|)n + |C|}}, $

其中 $A=32\alpha+8\alpha\left(\frac{1}{4(\mathrm{log}2)^2}+\frac{6}{\mathrm{log}2}+32\right)^{\frac{1}{2}}$, $B=24+16\alpha$.

证  设 $f(z)\in B_{\alpha}(C, D)$, 则存在 $g(z)\in S^*$, 使得 $\frac{{zf'(z)}}{{f(z)}}{\left( {\frac{{f(z)}}{{g(z)}}} \right)^\alpha } \in {\rm{ }}P(C, D).$由引理2.1可知

$ {\rm{Re}}\left\{ {\frac{{zf'(z)}}{{f(z)}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha } - \frac{{1 - C}}{{1 - D}}} \right\} > 0. $

记 $q(z)=\frac{zf'(z)}{f(z)}\left(\frac{f(z)}{g(z)}\right)^\alpha-\frac{1-C}{1-D}$, 则 $\mathrm{Re}q(z) > 0$.由(2.1) 式可知对 $z=re^{{\rm i}\theta}$,

$ 2n{b_n} = \frac{1}{{2\pi {\rm{i}}}}\oint_{|z| = r} {\frac{{zf'(z)}}{{f(z)}}} {z^{ - n - 1}}dz = \frac{1}{{2\pi }}\int_0^{2\pi } {\frac{{zf'(z)}}{{f(z)}}} {r^{ - n}}{e^{ - {\rm{i}}n\theta }}d\theta , $

因此

$ \begin{array}{l} |2n{b_n}{r^n}| = \frac{1}{{2\pi }}\left| {\int_0^{2\pi } {\frac{{zf'(z)}}{{f(z)}}} {e^{ - {\rm{i}}n\theta }}d\theta } \right|\\ \;\;\;\;\;\;\;\;\;\;\;\;\; = \frac{1}{{2\pi }}\left| {{{\int_0^{2\pi } {\left( {\left( {\frac{{g(z)}}{{f(z)}}} \right)} \right.} }^\alpha }q(z){e^{ - {\rm{i}}n\theta }} + {{\left( {\frac{{g(z)}}{{f(z)}}} \right)}^\alpha }\left. {\frac{{1 - C}}{{1 - D}}{e^{ - {\rm{i}}n\theta }}} \right)d\theta } \right|. \end{array} $

由于 $q(z)=2\mathrm{Re}q(z)-\overline{q(z)}$, 所以

$ \begin{array}{l} |2n{b_n}{r^n}| \le \frac{1}{{2\pi }}\left| {{{\int_0^{2\pi } {\left( {\frac{{g(z)}}{{f(z)}}} \right)} }^\alpha }2{\rm{Re}}q(z){e^{ - {\rm{i}}n\theta }}d\theta } \right| + \frac{1}{{2\pi }}\left| {{{\int_0^{2\pi } {\left( {\frac{{g(z)}}{{f(z)}}} \right)} }^\alpha }\overline {q(z)} {e^{ - {\rm{i}}n\theta }}d\theta } \right|\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + \frac{1}{{2\pi }}\frac{{1 - C}}{{1 - D}}\left| {{{\int_0^{2\pi } {\left( {\frac{{g(z)}}{{f(z)}}} \right)} }^\alpha }{e^{ - {\rm{i}}n\theta }}d\theta } \right| = {I_1} + {I_2} + {I_3}. \end{array} $

因为 $\mathrm{Re}q(z)>0$, 所以

$ \begin{array}{l} {I_1} \le \frac{1}{\pi }{\int_0^{2\pi } {\left| {{{\left( {\frac{{g(z)}}{{f(z)}}} \right)}^\alpha }} \right|} ^{}}{\rm{Re}}q(z)d\theta \le \frac{1}{\pi }\left| {{{\int_0^{2\pi } {\left| {{{\left( {\frac{{g(z)}}{{f(z)}}} \right)}^\alpha }} \right|} }^{}}q(z)d\theta } \right|\\ \;\;\; \le \frac{1}{\pi }\left| {\int_0^{2\pi } {\frac{{zf'(z)}}{{f(z)}}} \left| {{{\left( {\frac{{g(z)}}{{f(z)}}} \right)}^\alpha }} \right|{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }d\theta } \right| + \frac{1}{\pi }\frac{{1 - C}}{{1 - D}}\left| {\int_0^{2\pi } {\left| {{{\left( {\frac{{g(z)}}{{f(z)}}} \right)}^\alpha }} \right|} d\theta } \right|\\ \;\;\; \le \frac{1}{\pi }\left| {\int_0^{2\pi } {{\rm{Re}}} \left( {\frac{{zf'(z)}}{{f(z)}}} \right){e^{{\rm{iarg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}d\theta } \right| + \frac{1}{\pi }\left| {\int_0^{2\pi } {{\rm{Im}}} \left( {\frac{{zf'(z)}}{{f(z)}}} \right){e^{{\rm{iarg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}d\theta } \right|\\ \;\;\;\;\;\;\; + \frac{1}{\pi }\frac{{1 - C}}{{1 - D}}\left| {\int_0^{2\pi } {\left| {{{\left( {\frac{{g(z)}}{{f(z)}}} \right)}^\alpha }} \right|} d\theta } \right|. \end{array} $

由引理2.7可知

$ \begin{array}{l} {I_1} \le 12 + 8\alpha + 16\alpha {\rm{log}}\frac{1}{{1 - r}} + \frac{1}{\pi }\frac{{1 - C}}{{1 - D}}\left| {\int_0^{2\pi } {\left| {{{\left( {\frac{{g(z)}}{{f(z)}}} \right)}^\alpha }} \right|} d\theta } \right|,\\ {I_2} \le \frac{1}{{2\pi }}\left| {{{\int_0^{2\pi } {\overline {\left( {\frac{{zf'(z)}}{{f(z)}}} \right)} } }^{}}\overline {{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }} {{\left( {\frac{{g(z)}}{{f(z)}}} \right)}^\alpha }{e^{ - {\rm{i}}n\theta }}d\theta } \right| + \frac{1}{{2\pi }}\frac{{1 - C}}{{1 - D}}\left| {{{\int_0^{2\pi } {\left( {\frac{{g(z)}}{{f(z)}}} \right)} }^\alpha }{e^{ - {\rm{i}}n\theta }}d\theta } \right|\\ \;\;\;\; = \frac{1}{{2\pi }}\left| {\int_0^{2\pi } {\left( {\frac{{zf'(z)}}{{f(z)}}} \right)} {e^{2{\rm{iarg}}{{\left( {\frac{{f(z)}}{{g(z)}}} \right)}^\alpha }}}{e^{{\rm{i}}n\theta }}d\theta } \right| + \frac{1}{{2\pi }}\frac{{1 - C}}{{1 - D}}\left| {{{\int_0^{2\pi } {\left( {\frac{{g(z)}}{{f(z)}}} \right)} }^\alpha }{e^{ - {\rm{i}}n\theta }}d\theta } \right|. \end{array} $

故由引理2.7可知

$ {I_2} \le 4\alpha {\left( {\frac{1}{{{n^2}}} + \frac{4}{n}{\rm{log}}\frac{1}{{1 - r}}} \right)^{\frac{1}{2}}}{\left( {1 + \frac{4}{{1 - r}}{\rm{log}}\frac{1}{{1 - \sqrt r }}} \right)^{\frac{1}{2}}} + \frac{1}{{2\pi }}\frac{{1 - C}}{{1 - D}}\left| {{{\int_0^{2\pi } {\left( {\frac{{g(z)}}{{f(z)}}} \right)} }^\alpha }{e^{ - {\rm{i}}n\theta }}d\theta } \right|. $

由引理2.8可知

$ \begin{array}{*{20}{l}} {{I_3} \le \frac{1}{{2\pi }}\frac{{1 - C}}{{1 - D}}\int_0^{2\pi } {\left| {{{\left( {\frac{{g(z)}}{{f(z)}}} \right)}^\alpha }} \right|} d\theta \le \frac{{1 - C}}{{1 - D}}\frac{{1 + |D|r}}{{1 - |C|r}}\frac{{1 + r}}{{1 - r}},}\\ {} \end{array} $

所以

$ \begin{array}{l} |{b_n}| \le \frac{1}{n}{r^{ - n}}\left( {6 + 4\alpha + 8\alpha {\rm{log}}\frac{1}{{1 - r}}} \right) + \frac{2}{n}{r^{ - n}}\alpha {\left( {\frac{1}{{{n^2}}} + \frac{4}{n}{\rm{log}}\frac{1}{{1 - r}}} \right)^{\frac{1}{2}}}{\left( {1 + \frac{4}{{1 - r}}{\rm{log}}\frac{1}{{1 - \sqrt r }}} \right)^{\frac{1}{2}}}\\ \;\;\;\;\;\;\;\;\;\; + \frac{2}{n}{r^{ - n}}\frac{{1 - C}}{{1 - D}}\frac{{1 + |D|r}}{{1 - |C|r}}\frac{{1 + r}}{{1 - r}}. \end{array} $

设 $r=1-\frac{1}{n}$, 则

$ \begin{array}{l} |{b_n}| \le (6 + 4\alpha )\frac{1}{n}{\left( {1 - \frac{1}{n}} \right)^{ - n}} + 8\alpha \frac{1}{n}{\left( {1 - \frac{1}{n}} \right)^{ - n}}{\rm{log}}n\\ \;\;\;\;\;\;\;\;\;\; + \frac{2}{n}{\left( {1 - \frac{1}{n}} \right)^{ - n}}\alpha {\left( {\frac{1}{{{n^2}}} + \frac{4}{n}{\rm{log}}n + \frac{4}{n}{\rm{log}}\frac{1}{{1 - \sqrt {1 - \frac{1}{n}} }} + 16{\rm{log}}n{\rm{log}}\frac{1}{{1 - \sqrt {1 - \frac{1}{n}} }}} \right)^{\frac{1}{2}}}\\ \;\;\;\;\;\;\;\;\;\; + \frac{2}{n}{\left( {1 - \frac{1}{n}} \right)^{ - n}}\frac{{1 - C}}{{1 - D}}\frac{{(1 + |D|)n - |D|}}{{(1 - |C|)n + |C|}}(2n - 1). \end{array} $

因为 $(1-\frac{1}{n})^{-n}\leq4$, $\frac{1}{n}\leq\frac{\mathrm{log}n}{n\mathrm{log}2}~(n\geq2)$, $\mathrm{log}\frac{1}{1-\sqrt{1-\frac{1}{n}}}\leq\mathrm{log}2n\leq2\mathrm{log}n$, 所以

$ \begin{array}{l}\begin{split} |b_n|\leq& (6+4\alpha)\frac{1}{n}\cdot4 +8\alpha\frac{1}{n}\cdot4\cdot\mathrm{log}n\nonumber\\ &+\frac{8\alpha}{n}\left(\frac{(\mathrm{log}n)^2}{n^2(\mathrm{log}2)^2}+4\frac{1}{n\mathrm{log}2}(\mathrm{log}n)^2+8\frac{1}{n\mathrm{log}2}(\mathrm{log}n)^2+ 32(\mathrm{log}n)^2\right)^{\frac{1}{2}}\nonumber\\ &+16\frac{1-C}{1-D}\frac{(1+|D|)n-|D|}{(1-|C|)n+|C|}\nonumber\\ =&32\alpha\frac{1}{n}\mathrm{log}n+8\alpha\left(\frac{1}{n^2(\mathrm{log}2)^2} +\frac{12}{n\mathrm{log}2}+32\right)^{\frac{1}{2}} \frac{1}{n}\mathrm{log}n \nonumber\\ &+(24+16\alpha)\frac{1}{n}+16\frac{1-C}{1-D}\frac{(1+|D|)n-|D|}{(1-|C|)n+|C|}\nonumber\\ \leq& A\frac{1}{n}\mathrm{log}n+B\frac{1}{n}+16\frac{1-C}{1-D}\frac{(1+|D|)n-|D|}{(1-|C|)n+|C|}, \end{split} \end{array} $

其中 $A=32\alpha+8\alpha\left(\frac{1}{4(\mathrm{log}2)^2}+\frac{6}{\mathrm{log}2}+32\right)^{\frac{1}{2}}$, $B=24+16\alpha$.

因为当 $C=1, D=-1$时, $B_{\alpha}(1, -1)=B_\alpha$, 可得

推论3.2^[9]  设 $f(z)\in B_\alpha$, 则对 $n\geq2$,

$ |b_{n}|\leq E \left(1+\alpha\right)\frac{1}{n}\mathrm{log}n, $

其中 $E$是绝对常数.

证  由定理3.1可知, 当 $C=1, D=-1$时,

$ \begin{array}{l}\begin{split} |b_{n}|&\leq A\frac{1}{n}\mathrm{log}n+B\frac{1}{n}\leq A\frac{1}{n}\mathrm{log}n+B\frac{1}{n}\frac{\mathrm{log}n}{\mathrm{log}2}= \left(A+B\frac{1}{\mathrm{log}2}\right)\frac{1}{n}\mathrm{log}n\nonumber\\ &=\left(32\alpha+8\alpha\left(\frac{1}{4(\mathrm{log}2)^2} +\frac{6}{\mathrm{log}2}+32\right)^{\frac{1}{2}}+(24+16\alpha)\frac{1}{\mathrm{log}2}\right)\frac{1}{n}\mathrm{log}n\nonumber\\ &=\left\{\frac{24}{\mathrm{log}2}+\left(32+8\left(\frac{1}{4(\mathrm{log}2)^2} +\frac{6}{\mathrm{log}2}+32\right)^{\frac{1}{2}}+\frac{16}{\mathrm{log}2}\right)\alpha\right\}\frac{1}{n}\mathrm{log}n.\nonumber\\ \end{split} \end{array} $

记 $E=32+8\left(\frac{1}{4(\mathrm{log}2)^2}+\frac{6}{\mathrm{log}2}+32\right)^{\frac{1}{2}}+\frac{16}{\mathrm{log}2}$, 则

$ \begin{array}{l}\begin{split} |b_{n}|&\leq E\left(\frac{24}{\mathrm{log}2}\frac{1}{E}+\alpha\right)\frac{1}{n}\mathrm{log}n\leq E \left(1+\alpha\right)\frac{1}{n}\mathrm{log}n. \nonumber\\ \end{split} \end{array} $