与条形区域有关的解析函数新子类


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	李书海

	汤获

	马丽娜

	敖恩

与条形区域有关的解析函数新子类

李书海, 汤获, 马丽娜, 敖恩

赤峰学院数学与统计学院内蒙古赤峰 024000

收稿日期: 2014-12-03; 修订日期: 2015-03-13

基金项目: 国家自然科学基金(11561001)、内蒙古自然科学基金(2014MS0101)和内蒙古高等学校科研项目(NJZY240)资助

作者简介: 李书海,lishms66@sina.com;汤获,thth2009@163.com;马丽娜,malina00@163.com;敖恩,aex09@163.com

摘要: 该文引入了条形区域上由Sălăgean算子定义的解析函数新子类,讨论了该类中函数的系数估计和Fekete-Szegö不等式,所得结果推广了一些已知结论.

关键词: 解析函数条形区域 Sălăgean算子从属

A New Subclass of Analytic Functions Associated with the Strip Domains

Li Shuhai, Tang Huo, Ma Lina, Ao En

School of Mathematics and Statistics, Chifeng University, Inner Mongolia Chifeng 024000

Received: 2014-12-03; Revised: 2015-03-13

Abstract: The Sălăgean operator is used here to introduce a new subclass of analytic functions associated with the strip domains. We obtain the bounds of coefficients and Fekete-szegö inequality for functions in this class. The results presented here extend some of the earlier results.

Key words: Analytic functions Strip domain Sălăgean operator Subordination

1 引言

令 ${\cal A}$ 表示单位圆盘 ${\mathbb U}=\{ z\in {\mathbb C}: |z|<1\}$ 内解析且具有如下形式

$\begin{equation}\label{eq:a1} f(z)=z+\sum\limits_{n=2}^{\infty}a_{n}z^{n} \end{equation}$

(1.1)

的函数类, ${\cal S}$ 表示 ${\cal A}$ 中所有的单叶函数子类.

设 $u(z)$ 和 $v(z)$ 在 ${\cal A}$ 中解析,若存在一个 Schwarz 函数 $\omega(z)$ ,在 ${\mathbb U}$ 内满足 $\omega(0)=0$ 和 $|\omega(z)|<1$ , 使得 $u(z)=v(\omega(z))~(z\in {\mathbb U})$ ,则称函数 $u(z)$ 从属于 $v(z)$ , 记作 $u(z)\prec v(z)$ . 另外,若 $v$ 在 ${\mathbb U}$ 内单叶,则有下列关系成立

$u(z)\prec v(z)~~(z\in{\mathbb U})\Longleftrightarrow u(0)=v(0)~~~ \hbox{和}~~~u({\mathbb U})\subset v({\mathbb U}).$

用 ${\cal P}$ 表示在 ${\mathbb U}$ 内解析且具有如下形式

$\begin{equation}\label{eq:a2} p(z)=1+\sum\limits_{n=1}^{\infty}c_{n}z^{n} \end{equation}$

(1.2)

的函数 $p(z)$ 的全体. 若 $\Re(p(z))>0~(z\in {\mathbb U})$ ,则称 $p(z)$ 为 Caratheodory 函数(见文献[1]).

1936 年,Robertson 在文献[]中引入如下的函数子类 $S^*(\alpha)$ 和 $K(\alpha)$

$\Re\left\{\frac{zf'(z)}{f(z)}\right\}>\alpha\quad(f(z)\in {\cal A})$

和

$\Re\left\{1+\frac{zf''(z)}{f'(z)}\right\}>\alpha\quad(f(z)\in {\cal A}).$

特别地,当 $\alpha=0$ 时,上述函数类即为我们熟知的星象函数类 $S^*$ 和凸象函数类 $K$ .

1994 年,Uralegaddi,Ganigi 和 Sarangi 在文献[]中引入如下的函数子类 $M(\beta)$ 和 $N(\beta)$ $(\beta >1)$

$\Re\left\{\frac{zf'(z)}{f(z)}\right\}< \beta\quad(f(z)\in {\cal A})$

和

$\Re\left\{1+\frac{zf''(z)}{f'(z)}\right\}< \beta\quad(f(z)\in {\cal A}).$

定义 1.1 设 $-1\leq B< A\leq 1$ , $C\neq D$ 和 $-1\leq D \leq 1$ . 函数 $p(z)\in P(A,B;C,D)$ 当且仅当 $p(z)$ 满足下列两个从属关系

$\begin{equation}\label{eq:a3} p(z)\prec h_{1}(z)=\frac{1+Az}{1+Bz} \end{equation}$

(1.3)

和

$\begin{equation}\label{eq:a4} p(z)\prec h_{2}(z)=\frac{1+Cz}{1+Dz}. \end{equation}$

(1.4)

在上述函数类 $P(A,B;C,D)$ 中,若令 $A=1-2\alpha~(0\leq\alpha<1)$ , $B=-1$ , $C=1-2\beta~(\beta>1)$ 和 $D=-1$ ,则有

$\begin{equation}\label{eq:a5} p(z)\in P(\alpha,\beta)=P(1-2\alpha,-1;1-2\beta,-1)\Longleftrightarrow\alpha<\Re\{p(z)\}<\beta. \end{equation}$

(1.5)

下面,我们引入如下的解析函数类 $S_{\alpha,\beta}(z):{\mathbb U}\rightarrow {\mathbb C}$ .

定理 1.1 设 $\alpha$ 和 $\beta$ 为实数且 $\alpha<1$ 和 $\beta>1$ ,则定义如下的函数类 $S_{\alpha,\beta}(z)$

$\begin{equation}\label{eq:a6} S_{\alpha,\beta}(z)=1+\frac{\beta-\alpha}{\pi}{\rm i}\log\left(\frac{1-{\rm e}^{\frac{2\pi{\rm i}(1-\alpha)}{\beta-\alpha}}z}{1-z}\right)\quad(z\in {\mathbb U}) \end{equation}$

(1.6)

在 ${\mathbb U}$ 内单叶解析且满足 $S_{\alpha,\beta}(0)=1$ . 此外, $S_{\alpha,\beta}(z)$ 映射 ${\mathbb U}$ 到条形区域 $\omega$ ,这里 $\alpha<\Re\{\omega\}<\beta$ .

我们注意到由 (1.6) 式定义的 $S_{\alpha,\beta}(z)$ 可改写为(见文献[4])

$\begin{equation}\label{eq:a7} S_{\alpha,\beta}(z)=1+\sum\limits_{n=1}^{\infty}B_{n}z^{n}, \end{equation}$

(1.7)

这里

$\begin{equation}\label{eq:a8} B_{n}=\frac{\beta-\alpha}{n\pi}{\rm i}\left(1-{\rm e}^{\frac{2n\pi{\rm i}(1-\alpha)}{\beta-\alpha}}\right)\quad(n\in {\mathbb N}). \end{equation}$

(1.8)

由 (1.5) 式和 (1.6) 式,我们有

$\begin{equation}\label{eq:a9} p(z)\in P(\alpha,\beta)\Longleftrightarrow p(z)\prec S_{\alpha,\beta}(z). \end{equation}$

(1.9)

利用文献[5, 6, 7]中的结果,易得

若 $-1< B< A\leq1$ ,则函数 $h_{1}(z)$ 在 ${\mathbb U}$ 内单叶且映射 ${\mathbb U}$ 到圆盘 $\{ w \in \mathbb C:|w-{\sigma _1}| < {r_1}\}$ ,这里

$\begin{equation}\label{eq:a10} \sigma_{1}=\frac{1-AB}{1-B^{2}} \quad \hbox{和} \quad r_{1}=\frac{A-B}{1-B^{2}}. \end{equation}$

(1.10)

利用从属的定义,(1.3) 式等价于

$\begin{equation}\label{eq:a11} \left|p(z)-\frac{1-AB}{1-B^{2}}\right|<\frac{A-B}{1-B^{2}}. \end{equation}$

(1.11)

若 $B=-1$ ,则 (1.3) 式等价于

$\begin{equation}\label{eq:a12} \Re\{ p(z)\}>\frac{1-A}{2}. \end{equation}$

(1.12)

同理,若 $-1

$\begin{equation}\label{eq:a13} \sigma_{2}=\frac{1-CD}{1-D^{2}}\quad \hbox{和} \quad r_{2}=\frac{D-C}{1-D^{2}}. \end{equation}$

(1.13)

利用从属的定义,(1.4) 式等价于

$\begin{equation}\label{eq:a14} \left|p(z)-\frac{1-CD}{1-D^{2}}\right|<\frac{D-C}{1-D^{2}}. \end{equation}$

(1.14)

若 $D=1$ ,则 (1.4) 式等价于

$\begin{equation}\label{eq:a15} \Re\{2-p(z)\}<1+\frac{1-C}{2}. \end{equation}$

(1.15)

综合 (1.10)-(1.15) 式,我们可知函数 $p(z)\in P(A,B;C,D)$ 当且仅当 $p(z)$ 满足下列两个条件

$|p(z)-{\sigma _i}| < {r_i}\quad (i = 1,2;-1 < B < A \le 1;-1 < C < D < 1)$

(1.16)

和

$\begin{equation}\label{eq:a17} \frac{1-A}{2}<\Re\{p(z)\}~(B=-1),~~\Re\{2-p(z)\}<1+\frac{1-C}{2}\quad(D=1), \end{equation}$

(1.17)

其中 $\sigma_{1},r_{1}$ 和 $\sigma_{2},r_{2}$ 分别由 (1.10) 式和 (1.13) 式给出.

在文献 ^[8] 中,Sălăgean 定义算子 $D^{m}f(z): {\cal A}\rightarrow{\cal A}$ 如下

$\begin{eqnarray*} D^{0}f(z)&=&f(z),\\ D^{1}f(z)&=&Df(z)=zf'(z), \end{eqnarray*}$

一般地,有

$\begin{equation}\label{eq:a18} D^{m}f(z)=D(D^{m-1}f(z))=z+\sum_{n=2}^{\infty}n^{m}a_{n}z^{n}~(m\in N_{0}=N\cup\{0\}). \end{equation}$

(1.18)

利用算子 $D^{m}f(z)$ ,我们引入 ${\cal A}$ 中如下的函数子类.

定义 1.2 设 $m \in {N_0},-1 \le B < A \le 1,-1 < C < D \le 1$ . 若 $f(z)\in {\cal A}$ ,则函数 $f(z)\in S_{m}(A,B;C,D)$ 当且仅当 $f(z)$ 满足下列条件

$\begin{equation}\label{eq:a19} \frac{D^{m+1}f(z)}{D^{m}f(z)}\in P(A,B;C,D). \end{equation}$

(1.19)

由定义 1.2,(1.16) 式和 (1.17) 式,我们有

引理 1.1 若 $m\in N_{0}$ 和 $f(z)\in {\cal A}$ ,则函数 $f(z)\in S_{m}(A,B;C,D)$ 当且仅当 $f(z)$ 满足下列两个条件

$\left| {\frac{{{D^{m + 1}}f(z)}}{{{D^m}f(z)}}-{\sigma _i}} \right| < {r_i}\quad (i = 1,2;-1 < B < A \le 1;-1 < C < D < 1)$

(1.20)

和

$\frac{1-A}{2}<\Re\left\{\frac{D^{m+1}f(z)}{D^{m}f(z)}\right\}~(B=-1),~~ \Re\left\{2-\frac{D^{m+1}f(z)}{D^{m}f(z)}\right\}<1+\frac{1-C}{2}\quad(D=1),$

(1.21)

其中 $\sigma_{1},r_{1}$ 和 $\sigma_{2},r_{2}$ 分别由 (1.10) 式和 (1.13) 式给出.

在函数类 $S_{m}(A,B;C,D)$ 中,若取不同的参数 $m,A,B,C$ 和 $D$ ,易得下列已知的函数子类

(1) $S_{0}(1-2\alpha,-1;1-2\beta,-1)=S(\alpha,\beta)~(0\leq \alpha <1,\beta >1)$ (Kuroki 和 Owa^{[4, 9]}; Kwon 等^[10]);

(2) $S_{1}(1-2\alpha,-1;1-2\beta,-1)=K(\alpha,\beta)~(0\leq \alpha <1,\beta >1)$ (Sim 和 Kwon^[11]).

2 预备知识

引理 2.1^[12] 设函数 $p(z)=1+c_{1}z+c_{2}z^{2}+\cdots$ 在 ${\mathbb U}$ 内单叶解析, 又设 $p(z)$ 映射 ${\mathbb U}$ 到一个凸域. 若 $q(z)=1+q_{1}z+q_{2}z^{2}+\cdots$ 在 ${\mathbb U}$ 内解析且满足下列从属关系

$q(z)\prec p(z)~~(z\in {\mathbb U}),$

则有 $| q_{n}| \leq | c_{1}| ~(n=1,2,\cdots).$

利用定义1.1,引理2.1和从属的定义,我们易得下面的引理.

引理 2.2 设 $-1\leq B< A\leq 1$ 和 $-1

$|{c_n}| \le \left\{ {\begin{array}{*{20}{l}} {2{\delta _1},\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Re (p(z)) > {\rho _1}(-1 < B < A \le 1,-1 < C < D < 1),}&{}\\ {2{\delta _2},\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Re (p(z)) < {\rho _2}(-1 < B < A \le 1,-1 < C < D < 1),}&{}\\ {2\min \{ {\delta _1},{\delta _2}\} ,\;\;\;\;\;\;\;\;\;\;\;{\rho _1} < \Re (p(z)) < {\rho _2}(-1 < B < A \le 1,-1 < C < D < 1)}&{}\\ \begin{array}{l} 2\min \{ \frac{{1 + A}}{2},\frac{{1-C}}{2}\} ,\frac{{1-A}}{2} < \Re \{ p(z)\} (B =-1),\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Re \{ 2-p(z)\} < 1 + \frac{{1-C}}{2}(D = 1) \end{array}&{} \end{array}} \right.$

(2.1)

其中

$\begin{equation}\label{eq:b2} \delta_{1}=\min \left\{\frac{A-B}{1-B},\frac{D-C}{1+D}\right\}, \end{equation}$

(2.2)

$\begin{equation}\label{eq:b3} \delta_{2}=\min \left\{\frac{A-B}{1+B},\frac{D-C}{1-D}\right\}, \end{equation}$

(2.3)

$\begin{equation}\label{eq:b4} \rho_{1}=\max\left\{\frac{1-A}{1-B},\frac{1+C}{1+D}\right\} \end{equation}$

(2.4)

和

$\begin{equation}\label{eq:b5} \rho_{2}=\min\left\{\frac{1+A}{1+B},\frac{1-C}{1-D}\right\}. \end{equation}$

(2.5)

证设函数 $p(z)=1+c_{1}z+c_{2}z^{2}+\cdots \in P(A,B;C,D)$ ,则由定义1.1 和从属的定义可知

$p(0)= h_{1}(0),\quad p({\mathbb U})\subset h_{1}({\mathbb U})$

和

$p(0)= h_{2}(0),\quad p({\mathbb U})\subset h_{2}({\mathbb U}),$

其中 $h_{1}(z)$ 和 $h_{2}(z)$ 分别由(1.3)式和(1.4)式给出. 因此,我们有

$p(z)= h_{1}(\omega _{1}(z))\quad(\omega _{1}(0)=0,|\omega _{1}(z)|<1)$

和

$p(z)= h_{2}(\omega_{2} (z))\quad(\omega _{2}(0)=0,|\omega _{2}(z)|<1).$

且有

$\begin{equation}\label{eq:b8} |\omega _{1}(z)|=\left|\frac{p(z)-1}{A-Bp(z)}\right|<1,\quad p(z)=u+{\rm i}v \end{equation}$

(2.6)

和

$\begin{equation}\label{eq:b9} |\omega _{2}(z)|=\left|\frac{p(z)-1}{C-Dp(z)}\right|<1,\quad p(z)=u+{\rm i}v. \end{equation}$

(2.7)

由 (2.6) 式和 (2.7) 式,我们不难得到

$\begin{equation}\label{eq:b10} 2u(1-AB)>1-A^{2}+(1-B^{2})(u^{2}+v^{2}) \end{equation}$

(2.8)

和

$\begin{equation}\label{eq:b11} 2u(1-CD)>1-C^{2}+(1-D^{2})(u^{2}+v^{2}). \end{equation}$

(2.9)

因为

$\begin{equation}\label{eq:b12} |p(z)|^{2}\geq [\Re(p(z))]^{2}, \end{equation}$

(2.10)

故由 (2.8) 式,(2.9)式和 (2.10) 式,我们有

$\begin{equation}\label{eq:b13} \frac{1-A}{1-B}< u=\Re(p(z))<\frac{1+A}{1+B} \end{equation}$

(2.11)

和

$\begin{equation}\label{eq:b14} \frac{1+C}{1+D}< u=\Re(p(z))<\frac{1-C}{1-D}. \end{equation}$

(2.12)

下面,我们分四种情形来证明 (2.1) 式.

(i) 若

$\begin{equation}\label{eq:b15} \frac{1-A}{1-B}< u=\Re(p(z)) \end{equation}$

(2.13)

和

$\begin{equation}\label{eq:b16} \frac{1+C}{1+D}< u=\Re(p(z)), \end{equation}$

(2.14)

则由引理2.1,我们有

$\begin{equation}\label{eq:b17} |c_{n}|\leq 2\left(\frac{A-B}{1-B}\right) \end{equation}$

(2.15)

和

$\begin{equation}\label{eq:b18} |c_{n}|\leq 2\left(\frac{D-C}{1+D}\right). \end{equation}$

(2.16)

令

$\begin{equation}\label{eq:b19} \max\left\{\frac{1-A}{1-B},\frac{1+C}{1+D}\right\}=\rho_{1}< u=\Re(p(z)). \end{equation}$

(2.17)

则由 (2.13)-(2.17) 式,得

$|c_{n}| \leq 2\delta_{1}.$

(ii) 若

$\begin{equation}\label{eq:b21} \Re(p(z))< \frac{1+A}{1+B}, \end{equation}$

(2.18)

则由 (2.18) 式,我们有

$\begin{equation}\label{eq:b22} p(z)=\frac{1+A}{1+B}+\left(1-\frac{1+A}{1+B}\right)q(z),\quad(\exists~q(z)=1+q_{1}z+\cdots \in {\cal P}). \end{equation}$

(2.19)

由(2.19)式,可知

$\begin{equation}\label{eq:b23} c_{n}=\left(1-\frac{1+A}{1+B}\right)q_{n}. \end{equation}$

(2.20)

又由文献[]可知 $|q_{n}| \leq 2$ , 则由 (2.20) 式,有

$\begin{equation}\label{eq:b24} |c_{n}|\leq 2\left(\frac{A-B}{1+B}\right). \end{equation}$

(2.21)

另一方面,若

$\begin{equation}\label{eq:b25} \Re(p(z))<\frac{1-C}{1-D}, \end{equation}$

(2.22)

则有

$\begin{equation}\label{eq:b26} |c_{n}|\leq 2\left(\frac{D-C}{1-D}\right). \end{equation}$

(2.23)

利用 (2.18) 式,(2.21) 式,(2.22) 式和 (2.23) 式,易知

若

$\Re(p(z))<\rho_{2}=\min\left\{\frac{1+A}{1+B},\frac{1-C}{1-D}\right\},$

则

$|c_{n}|\leq 2\delta_{2}.$

利用与 (i) 和 (ii) 类似的证明方法,我们易证下面的 (iii) 和 (iv).

(iii) 若 $\rho_{1}<\Re(p(z))<\rho_{2}$ ,则

$|c_{n}|\leq 2\min\{\delta_{1},\delta_{2}\}.$

(iv) 若 $\frac{1-A}{2}<\Re(p(z))~(B=-1)$ 和 $\Re(2-p(z))<1+\frac{1-C}{2}~(D=1)$ ,则

$|c_{n}|\leq 2\min\left\{\frac{1+A}{2},\frac{1-C}{2}\right\}.$

综合 (i)-(iv),(2.1) 式得证.

引理 2.3 令 $-1< B< A\leq 1$ 和 $-1

$\begin{equation}\label{eq:b30} p(z)\prec \left\{ \begin{array}{ll} p_{1}(z)=S_{\frac{1-A}{1-B},\frac{1-C}{1-D}}(z),~~&AD\leq BC,\\ p_{2}(z)=S_{\frac{1+C}{1+D},\frac{1+A}{1+B}}(z),&AD\geq BC,\\ p_{3}(z)=S_{\frac{1-A}{1-B},\frac{1+A}{1+B}}(z),&A+C\leq B+D,\\ p_{4}(z)=S_{\frac{1+C}{1+D},\frac{1-C}{1-D}}(z),&A+C\geq B+D,\end{array} \right. \end{equation}$

(2.24)

其中 $S_{\alpha,\beta}(z)$ 定义如 (1.6) 式.

证 (i) 设 $p(z)\in P(A,B;C,D)$ 且 $AD\leq BC$ ,则由 (2.11) 式和 (2.12) 式,我们有

$\begin{eqnarray*}\label{eq:b31} \frac{1-A}{1-B}<\Re\{p(z)\}<\frac{1-C}{1-D}. \end{eqnarray*}$

利用 (1.6) 式,易得

$p(z)\prec p_{1}(z)=S_{\frac{1-A}{1-B},\frac{1-C}{1-D}}(z),\quad AD\leq BC.$

类似于(i),我们易得

(ii) $p(z)\prec p_{2}(z)=S_{\frac{1+C}{1+D},\frac{1+A}{1+B}}(z),\quad AD\geq BC,$

(iii) $p(z)\prec p_{3}(z)=S_{\frac{1-A}{1-B},\frac{1+A}{1+B}}(z),\quad A+C\leq B+D$

和

(iv) $p(z)\prec p_{4}(z)=S_{\frac{1+C}{1+D},\frac{1-C}{1-D}}(z),\quad A+C\geq B+D.$

综上,引理2.3得证.

注 2.1 设 $p_{k}(z)~(k=1,2,3,4)$ 由 (2.24) 式给出且满足 $p_{k}(0)=1$ ,则

(i) 函数 $p_{1}(z)$ 具有如下形式

$p_{1}(z)=1+\sum\limits_{n=1}^{\infty}B_{n,1}z^{n},$

其中

$\begin{equation}\label{eq:b33} B_{n,1}=\frac{\frac{1-C}{1-D}-\frac{1-A}{1-B}}{n\pi}{\rm i}\left(1-{\rm e}^{2n\pi{\rm i}(1-\frac{1-A}{1-B})/(\frac{1-C}{1-D}-\frac{1-A}{1-B})}\right). \end{equation}$

(2.25)

(ii) 函数 $p_{2}(z)$ 具有如下形式

$p_{2}(z)=1+\sum\limits_{n=1}^{\infty}B_{n,2}z^{n},$

其中

$\begin{equation}\label{eq:b35} B_{n,2}=\frac{\frac{1+A}{1+B}-\frac{1+C}{1+D}}{n\pi}{\rm i}\left(1-{\rm e}^{2n\pi{\rm i}(1-\frac{1+C}{1+D})/(\frac{1+A}{1+B}-\frac{1+C}{1+D})}\right). \end{equation}$

(2.26)

(iii) 函数 $p_{3}(z)$ 具有如下形式

$p_{3}(z)=1+\sum\limits_{n=1}^{\infty}B_{n,3}z^{n},$

其中

$\begin{equation}\label{eq:b37} B_{n,3}=\frac{\frac{1+A}{1+B}-\frac{1-A}{1-B}}{n\pi}{\rm i}\left(1-{\rm e}^{2n\pi{\rm i}(1-\frac{1-A}{1-B})/(\frac{1+A}{1+B}-\frac{1-A}{1-B})}\right). \end{equation}$

(2.27)

(iv) 函数 $p_{4}(z)$ 具有如下形式

$p_{4}(z)=1+\sum\limits_{n=1}^{\infty}B_{n,4}z^{n},$

其中

$\begin{equation}\label{eq:b39} B_{n,4}=\frac{\frac{1-C}{1-D}-\frac{1+C}{1+D}}{n\pi}{\rm i}\left(1-{\rm e}^{2n\pi{\rm i}(1-\frac{1+C}{1+D})/(\frac{1-C}{1-D}-\frac{1+C}{1+D})}\right). \end{equation}$

(2.28)

利用引理 2.3,我们易得

引理 2.4 令 $-1< B< A\leq 1$ 和 $-1

$\begin{equation}\label{eq:b40} \frac{D^{m+1}f(z)}{D^{m}f(z)}\prec \left\{ \begin{array}{ll} p_{1}(z),~~&AD\leq BC,\\ p_{2}(z),&AD\geq BC,\\ p_{3}(z),&A+C\leq B+D,\\ p_{4}(z),&A+C\geq B+D, \end{array} \right. \end{equation}$

(2.29)

其中 $p_{k}(z)~(k=1,2,3,4)$ 由 (2.24) 式给出.

引理 2.5^[13] 设函数 $p(z)$ 由 (1.2) 式给出. 若 $p(z)\in {\cal P}$ ,则对任意的复数 $\gamma$ ,有

$|c_{2}-\gamma c_{1}^{2}|\leq 2\max\{1,| 2\gamma-1| \}.$

若取函数

$p(z)= \frac{1+z^{2}}{1-z^{2}},~~p(z)=\frac{1+z}{1-z}.$

则上述结果是精确的.

3 主要结果

定理 3.1 设 $m \in {N_0},-1 \le B < A \le 1,-1 < C < D \le 1$ ,函数 $f(z)$ 由 (1.1) 式给出. 若 $f(z)\in S_{m}(A,B;C,D)$ ,则

$\begin{equation}\label{eq:c1} |a_{n}|\leq\left\{ \begin{array}{ll} \frac{\prod\limits^{n-2}_{k=0}(k+2 \delta_{1})}{(n-1)!n^{m}} ,& \Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)>\rho_{1}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{n-2}_{k=0}(k+2 \delta_{2})}{(n-1)!n^{m}},& \Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<\rho_{2}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{n-2}_{k=0}(k+2 \min \{\delta_{1},\delta_{2}\})}{(n-1)!n^{m}},& \rho_{1}<\Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<\rho_{2}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{n-2}_{k=0}(k+2\min \{\frac{1+A}{2},\frac{1-C}{2}\})}{(n-1)!n^{m}},~& \frac{1-A}{2}<\Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)~(B=-1),\\ & \Re\bigg(2-\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<1+\frac{1-C}{2}~(D=1), \end{array} \right. \end{equation}$

(3.1)

其中 $\delta_{1},\delta_{2},\rho_{1}$ 和 $\rho_{2}$ 分别由 (2.2)式 ,(2.3) 式, (2.4) 式和 (2.5) 式给出.

证根据定义1.2 和从属的定义,我们易知

$\begin{equation}\label{eq:c2} \frac{D^{m+1}f(z)}{D^{m}f(z)}\in h_{1}({\mathbb U}) \end{equation}$

(3.2)

和

$\begin{equation}\label{eq:c3} \frac{D^{m+1}f(z)}{D^{m}f(z)}\in h_{2}({\mathbb U}), \end{equation}$

(3.3)

其中 $h_{1}(z)$ 和 $h_{2}(z)$ 分别由 (1.3) 式和 (1.4) 式给出.

应用 (3.2) 式和 (3.3) 式,我们有

$\frac{D^{m+1}f(z)}{D^{m}f(z)}= p(z)~ (\exists ~p(z)\in P(A,B;C,D))$

或等价于

$\begin{equation}\label{eq:c4} D^{m+1}f(z)=p(z)D^{m}f(z)~ (\exists ~p(z)\in P(A,B;C,D)). \end{equation}$

(3.4)

比较 (3.4) 式两边关于 $z^{n}$ 的系数,可得

$\begin{equation}\label{eq:c5} (n-1)n^{m}a_{n}=(c_{n-1}+c_{n-2}2^{m}a_{2}+\cdots+c_{1}(n-1)^{m}a_{n-1}). \end{equation}$

(3.5)

由 (3.5) 式,可知

$\begin{equation}\label{eq:c6} | n^{m}a_{n}| \leq\frac{1}{n-1}(| c_{n-1}| +| c_{n-2}| 2^{m} | a_{2}| +\cdots+| c_{1}| (n-1)^{m}| a_{n-1}| ). \end{equation}$

(3.6)

利用引理 2.2 和 (3.6) 式,我们得到

$|2^{m}a_{2}|\leq\left\{ \begin{array}{ll} 2\delta_{1},& \Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)>\rho_{1}~(B\neq-1,D\neq 1),\\ 2\delta_{2},& \Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<\rho_{2}~(B\neq-1,D\neq 1),\\ 2\min \{\delta_{1},\delta_{2}\},& \rho_{1}<\Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<\rho_{2}~(B\neq-1,D\neq 1),\\ 2\min \bigg\{\frac{1+A}{2},\frac{1-C}{2}\bigg\},~& \frac{1-A}{2}<\Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)~(B=-1), \\ & \Re\bigg(2-\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<1+\frac{1-C}{2}~(D=1), \end{array} \right.$ $| 3^{m}a_{3}| \leq\left\{ \begin{array}{ll} \frac{\prod\limits^{1}_{k=0}(k+2 \delta_{1})}{2!} ,& \Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)>\rho_{1}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{1}_{k=0}(k+2 \delta_{2})}{2!},& \Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<\rho_{2}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{1}_{k=0}(k+2 \min \{\delta_{1},\delta_{2}\})}{2!},& \rho_{1}<\Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<\rho_{2}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{1}_{k=0}(k+2\min \{\frac{1+A}{2},\frac{1+C}{2}\})}{2!},~& \frac{1-A}{2}<\Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)~(B=-1), \\ & \Re\bigg(2-\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<1+\frac{1-C}{2}~(D=1), \end{array} \right.$

$| 4^{m}a_{4}|\leq\left\{ \begin{array}{ll} \frac{\prod\limits^{2}_{k=0}(k+2 \delta_{1})}{3!} ,& \Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)>\rho_{1}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{2}_{k=0}(k+2 \delta_{2})}{3!},& \Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<\rho_{2}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{2}_{k=0}(k+2 \min \{\delta_{1},\delta_{2}\})}{3!},& \rho_{1}<\Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<\rho_{2}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{2}_{k=0}(k+2\min \{\frac{1+A}{2},\frac{1+C}{2}\})}{3!},~& \frac{1-A}{2}<\Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)~(B=-1), \\ & \Re\bigg(2-\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<1+\frac{1-C}{2}~(D=1) \end{array} \right.$

和

$\begin{equation}\label{eq:c7} | n^{m}a_{n}|\leq\left\{ \begin{array}{ll} \frac{\prod\limits^{n-2}_{k=0}(k+2 \delta_{1})}{(n-1)!} ,& \Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)>\rho_{1}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{n-2}_{k=0}(k+2 \delta_{2})}{(n-1)!},& \Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<\rho_{2}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{n-2}_{k=0}(k+2 \min \{\delta_{1},\delta_{2}\})}{(n-1)!},& \rho_{1}<\Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<\rho_{2}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{n-2}_{k=0}(k+2\min \{\frac{1+A}{2},\frac{1+C}{2}\})}{(n-1)!},~& \frac{1-A}{2}<\Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)~ (B=-1),\\ & \Re\bigg(2-\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<1+\frac{1-C}{2}~(D=1). \end{array} \right. \end{equation}$

(3.7)

下面我们利用数学归纳法来证明. 当 $n=2,3,4$ 时,结论显然成立. 假设 $n=t$ 时, (3.7) 式成立. 我们来分类证明 (3.7) 式.

(i) 若 $\Re(\frac{D^{m+1}f(z)}{D^{m}f(z)})>\rho_{1},$ 则由 (3.6) 式,有

$| t^{m}a_{t}| \leq\frac{2\delta_{1}}{t-1}\sum\limits_{k=2}^{t}(k-1)^m| a_{k-1}| \leq \frac{\prod\limits^{t-2}_{k=0}(k+2\delta_{1})}{(t-1)!}.$

经简单计算,易得

$\begin{eqnarray*} |(t+1)^{m}a_{t+1}|&\leq&\frac{2\delta_{1}}{(t+1)-1}\sum\limits_{k=2}^{t+1}(k-1)^m| a_{k-1}| \\ &=&\frac{2\delta_{1}}{t} \bigg(\sum\limits_{k=2}^{t}(k-1)^m| a_{k-1}| +t^m| a_{t}| \bigg)\\ &\leq&\frac{2\delta_{1}}{t} \bigg(1+\frac{2\delta_{1}}{t-1}\bigg)\sum\limits_{k=2}^{t}(k-1)^m| a_{k-1}| \\ &\leq& \frac{t-1+2\delta_{1}}{t}\frac{\prod\limits^{t-2}_{k=0}(k+2\delta_{1})}{(t-1)!} \\ &=&\frac{\prod\limits^{(t+1)-2}_{k=0}(k+2\delta_{1})}{((t+1)-1)!}, \end{eqnarray*}$

即证 (3.7) 式成立.

类似于 (i) 的证明,易证

(ii)

$| n^{m}a_{n}|\leq\left\{ \begin{array}{ll} \frac{\prod\limits^{n-2}_{k=0}(k+2 \delta_{2})}{(n-1)!},& \Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<\rho_{2}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{n-2}_{k=0}(k+2 \min \{\delta_{1},\delta_{2}\})}{(n-1)!},& \rho_{1}<\Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<\rho_{2}~(B\neq-1,D\neq 1),\\ \frac{\prod\limits^{n-2}_{k=0}(k+2\min \{\frac{1+A}{2},\frac{1+C}{2}\})}{(n-1)!},~& \frac{1-A}{2}<\Re\bigg(\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)~ (B=-1),\\ & \Re\bigg(2-\frac{D^{m+1}f(z)}{D^{m}f(z)}\bigg)<1+\frac{1-C}{2}~(D=1). \end{array} \right.$

综上,定理 3.1 得证.

定理 3.2 设 $m \in {N_0},-1 < B < A \le 1,-1 < C < D < 1$ , 函数 $f(z)$ 由 (1.1) 式给出. 若 $f(z)\in S_{m}(A,B;C,D)$ ,则

$\begin{equation}\label{eq:c8} |a_{n}|\leq\left\{ \begin{array}{ll} \frac{\prod\limits^{n}_{k=2}(k-2+2((\frac{1-C}{1-D}-\frac{1-A}{1-B})/{\pi})\sin(\pi(1-\frac{1-A}{1-B})/(\frac{1-C}{1-D}-\frac{1-A}{1-B})) )}{(n-1)!n^{m}},&AD\leq BC,\\ \frac{\prod\limits^{n}_{k=2}(k-2+2((\frac{1+A}{1+B}-\frac{1+C}{1+D})/{\pi})\sin(\pi(1-\frac{1+C}{1+D})/(\frac{1+A}{1+B}-\frac{1+C}{1+D})) )}{(n-1)!n^{m}} ,&AD\geq BC,\\ \frac{\prod\limits^{n}_{k=2}(k-2+2((\frac{1+A}{1+B}-\frac{1-A}{1-B})/{\pi})\sin(\pi(1-\frac{1-A}{1-B})/(\frac{1+A}{1+B}-\frac{1-A}{1-B})) )}{(n-1)!n^{m}} ,&A+C\leq B+D,\\ \frac{\prod\limits^{n}_{k=2}(k-2+2((\frac{1-C}{1-D}-\frac{1+C}{1+D})/{\pi})\sin(\pi(1-\frac{1+C}{1+D})/(\frac{1-C}{1-D}-\frac{1+C}{1+D})) )}{(n-1)!n^{m}} ,~&A+C\geq B+D. \end{array} \right. \end{equation}$

(3.8)

证设函数 $p(z)$ 定义如下

$\begin{equation}\label{eq:c9} p(z)= \frac{D^{m+1}f(z)}{D^{m}f(z)}\quad(z\in {\mathbb U}). \end{equation}$

(3.9)

利用引理2.4,我们易知函数

$p(z)$ 满足 (2.29) 式. 再利用引理2.3,我们有

$\begin{equation}\label{eq:c10} p(z)\prec \left\{ \begin{array}{ll} p_{1}(z)=S_{\frac{1-A}{1-B},\frac{1-C}{1-D}}(z),&AD\leq BC,\\ p_{2}(z)=S_{\frac{1+C}{1+D},\frac{1+A}{1+B}}(z),&AD\geq BC,\\ p_{3}(z)=S_{\frac{1-A}{1-B},\frac{1+A}{1+B}}(z),&A+C\leq B+D,\\ p_{4}(z)=S_{\frac{1+C}{1+D},\frac{1-C}{1-D}}(z),~&A+C\geq B+D. \end{array} \right. \end{equation}$

(3.10)

设函数 $p_{k}(z)~(k=1,2,3,4)$ 在 ${\mathbb U}$ 内单叶解析, $p_{k}(0)=1$ ,且具有如下形式

$p_{k}(z)=1+\sum\limits_{n=1}^{\infty}B_{n,k}z^{n}~(k=1,2,3,4),$

其中 $B_{n,1},B_{n,2},B_{n,3}$ 和 $B_{n,4}$ 分别由 (2.25) 式, (2.26) 式,(2.27) 式和 (2.28) 式给出.

若取

$p(z)=1+\sum\limits_{n=1}^{\infty}c_{n}z^{n},$

则由引理2.2和 (3.10) 式,可知

$\begin{equation}\label{eq:c11} |c_{n}|\leq\left\{ \begin{array}{ll} |B_{1,1}|,&AD\leq BC,\\ |B_{1,2}|,&AD\geq BC,\\ |B_{1,3}|,&A+C\leq B+D,\\ |B_{1,4}|,~&A+C\geq B+D, \end{array} \right. \end{equation}$

(3.11)

其中

$\begin{equation}\label{eq:c12} |B_{1,1}|= \bigg(2\bigg(\frac{1-C}{1-D}-\frac{1-A}{1-B}\bigg)\bigg/{\pi}\bigg) \sin\bigg(\pi\bigg(1-\frac{1-A}{1-B}\bigg)\bigg/\bigg(\frac{1-C}{1-D}-\frac{1-A}{1-B}\bigg)\bigg), \end{equation}$

(3.12)

$\begin{equation}\label{eq:c13} |B_{1,2}|= \bigg(2\bigg(\frac{1+A}{1+B}-\frac{1+C}{1+D}\bigg)\bigg/{\pi}\bigg) \sin\bigg(\pi\bigg(1-\frac{1+C}{1+D}\bigg)\bigg/\bigg(\frac{1+A}{1+B}-\frac{1+C}{1+D}\bigg)\bigg), \end{equation}$

(3.13)

$\begin{equation}\label{eq:c14} |B_{1,3}|=\bigg (2\bigg(\frac{1+A}{1+B}-\frac{1-A}{1-B}\bigg)\bigg/{\pi}\bigg) \sin\bigg(\pi\bigg(1-\frac{1-A}{1-B}\bigg)\bigg/\bigg(\frac{1+A}{1+B}-\frac{1-A}{1-B}\bigg)\bigg), \end{equation}$

(3.14)

$\begin{equation}\label{eq:c15} |B_{1,4}|= \bigg(2\bigg(\frac{1-C}{1-D}-\frac{1+C}{1+D}\bigg)\bigg/{\pi}\bigg) \sin\bigg(\pi\bigg(1-\frac{1+C}{1+D}\bigg) \bigg/\bigg(\frac{1-C}{1-D}-\frac{1+C}{1+D}\bigg)\bigg). \end{equation}$

(3.15)

又由 (3.9) 式,可知

$D^{m+1}f(z)=p(z)D^{m}f(z).$

比较上式两边关于 $z^{n}$ 的系数,可得

$\begin{equation}\label{eq:c16} (n-1)n^{m}a_{n}=(c_{n-1}+c_{n-2}2^{m}a_{2}+\cdots+c_{1}(n-1)^{m}a_{n-1}). \end{equation}$

(3.16)

经简单计算并结合 (3.11) 式,易得

$| n^{m}a_{n}| \leq \frac{| c_{n-1}| +| c_{n-2}| | 2^{m}a_{2}| +\cdots+| c_{1}| | (n-1)^{m}a_{n-1}| }{n-1},$

也即

$| n^{m}a_{n}| \leq\left\{ \begin{array}{ll} \frac{|B_{1,1}|}{n-1}\sum\limits_{k=2}^{n}| (k-1)^{m}a_{k-1}| ,&AD\leq BC,\\ \frac{|B_{1,2}|}{n-1}\sum\limits_{k=2}^{n}| (k-1)^{m}a_{k-1}| ,&AD\geq BC,\\ \frac{|B_{1,3}|}{n-1}\sum\limits_{k=2}^{n}| (k-1)^{m}a_{k-1}| ,&A+C\leq B+D,\\ \frac{|B_{1,4}|}{n-1}\sum\limits_{k=2}^{n}| (k-1)^{m}a_{k-1}| ,~&A+C\geq B+D. \end{array} \right.$

其中 $n=2,3,\cdots,| a_{1}| =1$ ; $|B_{1,1}|$ , $|B_{1,2}|$ , $|B_{1,3}|$ 和 $|B_{1,4}|$ 分别由 (3.12) 式,(3.13) 式,(3.14) 式和 (3.15) 式给出.

要证该定理成立,我们需证

$\begin{equation}\label{eq:c17} Y_{n}\equiv \left\{ \begin{array}{ll} \frac{|B_{1,1}|}{n-1}\sum\limits_{k=2}^{n}| (k-1)^{m}a_{k-1}| \leq \frac{\prod\limits^{n}_{k=2}(k-2+|B_{1,1}|)}{(n-1)!},&AD\leq BC,\\ \frac{|B_{1,2}|}{n-1}\sum\limits_{k=2}^{n}| (k-1)^{m}a_{k-1}| \leq \frac{\prod\limits^{n}_{k=2}(k-2+|B_{1,2}|)}{(n-1)!},&AD\geq BC,\\ \frac{|B_{1,3}|}{n-1}\sum\limits_{k=2}^{n}| (k-1)^{m}a_{k-1}| \leq \frac{\prod\limits^{n}_{k=2}(k-2+|B_{1,3}|)}{(n-1)!},&A+C\leq B+D,\\ \frac{|B_{1,4}|}{n-1}\sum\limits_{k=2}^{n}| (k-1)^{m}a_{k-1}| \leq \frac{\prod\limits^{n}_{k=2}(k-2+|B_{1,4}|)}{(n-1)!},~&A+C\geq B+D \end{array} \right. \end{equation}$

(3.17)

成立.

以下用数学归纳法来证明. 由于

$Y_{2}=\left\{ \begin{array}{ll} |B_{1,1}|,&AD\leq BC,\\ |B_{1,2}|,&AD\geq BC,\\ |B_{1,3}|,&A+C\leq B+D,\\ |B_{1,4}|,~&A+C\geq B+D. \end{array} \right.$

当 $n=2$ 时,结论显然成立. 假设 $n=n_0$ 时结论成立,则经简单计算,易得

$\begin{equation}\label{eq:c18} Y_{n_0+1}= \left\{ \begin{array}{ll} \frac{|B_{1,1}|}{n_0}\bigg(\sum\limits_{k=2}^{n_0}| (k-1)^{m}a_{k-1}| +| n_{0}^{m}a_{n_0}| \bigg),&AD\leq BC,\\ \frac{|B_{1,2}|}{n_0}\bigg(\sum\limits_{k=2}^{n_0}| (k-1)^{m}a_{k-1}| +| n_{0}^{m}a_{n_0}| \bigg),&AD\geq BC,\\ \frac{|B_{1,3}|}{n_0}\bigg(\sum\limits_{k=2}^{n_0}| (k-1)^{m}a_{k-1}| +| n_{0}^{m}a_{n_0}| \bigg),&A+C\leq B+D,\\ \frac{|B_{1,4}|}{n_0}\bigg(\sum\limits_{k=2}^{n_0}| (k-1)^{m}a_{k-1}| +| n_{0}^{m}a_{n_0}| \bigg),~&A+C\geq B+D. \end{array} \right. \end{equation}$

(3.18)

由 (3.18) 式,我们易得

$\begin{equation}\label{eq:c19} Y_{n_0+1}\leq \left\{ \begin{array}{ll} \frac{n_0-1+|B_{1,1}|}{n_0}\frac{|B_{1,1}|}{n_0-1}\sum\limits_{k=2}^{n_0}| (k-1)^{m}a_{k-1}| \leq \prod\limits_{k=2}^{n_0+1}\frac{k-2+|B_{1,1}|}{k-1},&AD\leq BC,\\ \frac{n_0-1+|B_{1,2}|}{n_0}\frac{|B_{1,2}|}{n_0-1}\sum\limits_{k=2}^{n_0}| (k-1)^{m}a_{k-1}| \leq \prod\limits_{k=2}^{n_0+1}\frac{k-2+|B_{1,2}|}{k-1},&AD\geq BC,\\ \frac{n_0-1+|B_{1,3}|}{n_0}\frac{|B_{1,3}|}{n_0-1}\sum\limits_{k=2}^{n_0}| (k-1)^{m}a_{k-1}| \leq \prod\limits_{k=2}^{n_0+1}\frac{k-2+|B_{1,3}|}{k-1},&A+C\leq B+D,\\ \frac{n_0-1+|B_{1,4}|}{n_0}\frac{|B_{1,4}|}{n_0-1}\sum\limits_{k=2}^{n_0}| (k-1)^{m}a_{k-1}| \leq \prod\limits_{k=2}^{n_0+1}\frac{k-2+|B_{1,4}|}{k-1},~&A+C\geq B+D, \end{array} \right. \end{equation}$

(3.19)

即当 $n=n_0+1$ 时,(3.17) 式也成立. 利用数学归纳法,我们得证

$\begin{equation}\label{eq:c20} | n^{m}a_{n}| \leq \left\{ \begin{array}{ll} \frac{1}{(n-1)!}\prod\limits_{k=2}^{n}(k-2+|B_{1,1}|),&AD\leq BC,\\ \frac{1}{(n-1)!}\prod\limits_{k=2}^{n}(k-2+|B_{1,2}|),&AD\geq BC,\\ \frac{1}{(n-1)!}\prod\limits_{k=2}^{n}(k-2+|B_{1,3}|),&A+C\leq B+D,\\ \frac{1}{(n-1)!}\prod\limits_{k=2}^{n}(k-2+|B_{1,4}|),~&A+C\geq B+D, \end{array} \right. \end{equation}$

(3.20)

其中 $|B_{1,1}|,|B_{1,2}|,|B_{1,3}|$ 和 $|B_{1,4}|$ 分别由 (3.12) 式,(3.13) 式,(3.14) 式和 (3.15) 式给出. 定理3.2证毕.

定理 3.3 设 $-1< B< A\leq 1$ ,$-1

$\begin{equation}\label{eq:c21} | a_{3}-\mu a_{2}^{2}| \leq \frac{B_{1,k}}{2\cdot3^{m}}\max\{1,| 2\gamma_{k}-1| \}, \end{equation}$

(3.21)

其中

$\gamma_{k}=\frac{1}{2}\left[1-\frac{B_{2,k}}{B_{1,k}}-\left(1-2(\frac{3}{4})^{m}\mu\right)B_{1,k}\right],$ $B_{j,1}=\frac{\frac{1-C}{1-D}-\frac{1-A}{1-B}}{j\pi}{\rm i}\left(1-{\rm e}^{2j\pi{\rm i}(1-\frac{1-A}{1-B})/(\frac{1-C}{1-D}-\frac{1-A}{1-B})}\right)~(j=1,2),$

$B_{j,2}=\frac{\frac{1+A}{1+B}-\frac{1+C}{1+D}}{j\pi}{\rm i}\left(1-{\rm e}^{2j\pi{\rm i}(1-\frac{1+C}{1+D})/(\frac{1+A}{1+B}-\frac{1+C}{1+D})}\right)~(j=1,2),$

$B_{j,3}=\frac{\frac{1+A}{1+B}-\frac{1-A}{1-B}}{j\pi}{\rm i}\left(1-{\rm e}^{2j\pi{\rm i}(1-\frac{1-A}{1-B})/(\frac{1+A}{1+B}-\frac{1-A}{1-B})}\right)~(j=1,2),$

$B_{j,4}=\frac{\frac{1-C}{1-D}-\frac{1+C}{1+D}}{j\pi}{\rm i}\left(1-{\rm e}^{2j\pi{\rm i}(1-\frac{1+C}{1+D})/(\frac{1-C}{1-D}-\frac{1+C}{1+D})}\right)~(j=1,2),$

且此结果是精确的.

证若 $f(z)\in S_{m}(A,B;C,D)$ ,则存在 ${\mathbb U}$ 内的 Schwarz 函数 $\omega(z)$ ,使得

$\begin{equation}\label{eq:c26} \frac{D^{m+1}f(z)}{D^{m}f(z)}=p_{k}(\omega (z))\quad(z\in {\mathbb U}), \end{equation}$

(3.22)

这里 $p_{k}(z)~(k=1,2,3,4)$ 由 (2.24) 式给出.

设函数 $p(z)$ 定义如下

$\begin{equation}\label{eq:c27} p(z)=\frac{D^{m+1}f(z)}{D^{m}f(z)}=1+c_{1}z+c_{2}z^{2}+\cdots, \end{equation}$

(3.23)

则由 (3.22) 式和 (3.23) 式,可知 $p(z)\prec p_{k}(z)$ . 又设

$\begin{equation}\label{eq:c28} q(z)=\frac{1+\omega(z)}{1-\omega(z)}=1+q_{1}z+q_{2}z^{2}+\cdots, \end{equation}$

(3.24)

则 $q(z)$ 在 ${\mathbb U}$ 内解析且为正实部函数. 由 (3.24) 式,易得

$\begin{equation}\label{eq:c29} \omega(z)=\frac{q(z)-1}{q(z)+1}=\frac{1}{2}\left[q_{1}z+ \Big(q_{2}-\frac{q_{1}^{2}}{2}\Big)z^{2}+\cdots\right]. \end{equation}$

(3.25)

再由 (3.25) 式,我们有

$\begin{equation}\label{eq:c30} p_{k}\left(\frac{q(z)-1}{q(z)+1}\right)=1+\frac{1}{2}B_{1,k}q_{1}z +\left[\frac{1}{2}B_{1,k}\Big(q_{2}-\frac{q_{1}^{2}}{2}\Big) +\frac{B_{2,k}q_{1}^{2}}{4}\right]z^{2}+\cdots. \end{equation}$

(3.26)

利用 (3.22) 式,(3.23) 式和 (3.25) 式,可得

$p(z)=p_{k}\left(\frac{q(z)-1}{q(z)+1}\right).$

另一方面,由 (3.26) 式,可知

$\begin{equation}\label{eq:c32} c_{1}=\frac{1}{2}B_{1,k}q_1, \end{equation}$

(3.27)

$\begin{equation}\label{eq:c33} c_{2}=\frac{1}{2}B_{1,k}\Big(q_{2}-\frac{1}{2}q_{1}^{2}\Big)+\frac{1}{4}B_{2,k}q_{1}^{2}. \end{equation}$

(3.28)

接下来,利用 (1.18) 式和 (3.23) 式,我们有

$\begin{equation}\label{eq:c34} 2^{m}a_{2}=c_{1},~~3^{m}a_{3}=\frac{c_{2}+c_1^{2}}{2}. \end{equation}$

(3.29)

综合 (3.27) 式,(3.28) 式和 (3.29) 式,可得

$a_{2}=\frac{B_{1,k}q_{1}}{2\cdot2^{m}},$

$a_{3}=\frac{B_{1,k}q_{2}}{4\cdot3^{m}}+\frac{q_{1}^{2}}{8\cdot3^{m}}(B_{1,k}^{2}-(B_{1,k}-B_{2,k})),$

也即

$a_{3}-\mu a_{2}^{2}= \frac{B_{1,k}}{4\cdot3^{m}}[q_{2}-\gamma_{k} q_{1}^{2}],$

这里

$\gamma_{k}=\frac{1}{2}\left[1-\frac{B_{2,k}}{B_{1,k}}- \Big(1-\frac{2\cdot3^{m}}{(2^{m})^{2}}\mu\Big)B_{1,k}\right].$

最后,应用引理2.5,我们得到

$| a_{3}-\mu a_{2}^{2}| \leq \frac{B_{1,k}}{4\cdot3^{m}}| q_{2}-\gamma_{k} q_{1}^{2}| \leq\frac{B_{1,k}}{2\cdot3^{m}}\max\{1,| 2\gamma_{k}-1| \}.$

若取

$f_{k}(z)=D^{-m}\left\{\int_{0}^{z}\left\{\exp\left(\int_{0}^{\eta}\frac{p_{k}(\xi)-1}{\xi} {\rm d}\xi\right)\right\}{\rm d}\eta\right\},$

则上述结果是精确的,其中 $p_{k}(z)$ 由 (2.24) 式给出. 定理3.3证毕.

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