设$S$表示单位圆盘$E=\{z:|z| < 1\}$内形如$f(z)=z+\sum\limits_{n=2}^{\infty}a_{n}z^{n}$的单叶解析函数类的全体. $ S^*, C$和$K $分别表示通常的星像函数类, 凸函数类和近于凸函数类, 它们都是$S$的子类.

设$f(z)$与$g(z)$在$E$内解析, 若存在$E$内满足$|\varphi(z)|\leq|z|$的解析函数$\varphi(z)$ (不必单叶), 使得$f(z)=g(\varphi(z))$, 则称$f(z)$从属于$g(z)$, 记为$f(z)\prec g(z)$.

Fekete和Szegö于1933年提出函数族$S$上的系数泛函$|a_3-\mu a_2^2|$的精确估计问题, 并得到结果^[1]

$ |a_3-\mu a_2^2|\leq 1+2\exp\left(\frac{-2\mu}{1-\mu}\right), \, 0\leq\mu<1, $

且对任意的$\mu\in[0, 1)$, 等号均能成立.

在文献[2-8]中分别研究了某些星像函数类和近于凸函数类的Fekete-Szegö不等式.本文引进一类$\lambda$ -对数Bazilevic函数, 讨论该函数类的Fekete-Szegö不等式, 并得到对应的极值函数.

定义  设$\lambda\geq 0, \alpha\geq 0, -1\leq B < A\leq1$, 若存在$g(z)\in S^*$, 使得$f(z)\in S$, 且满足条件

$ \left(\frac{zf'(z)}{f(z)}\cdot\left(\frac{f(z)}{g(z)}\right)^\alpha\right)^{1-\lambda}\cdot\left(\frac{(zf'(z))'}{f'(z)}\cdot\left(\frac{f'(z)}{g'(z)}\right)^\alpha\right)^{\lambda}\prec \frac{1+Az}{1+Bz}\quad (z\in E), $

则称$f(z)$为$\lambda$ -对数Bazilevic函数, 这类函数记为$L(\lambda, \alpha, A, B)$, 其中的幂函数取主值.

下面对函数类$L(\lambda, \alpha, A, B)$中建立Fekete-Szegö不等式, 为此需要如下引理.

引理1^[9]  设$\varphi(z)=d_1z+d_2z^2+\cdots$在$E$内解析且满足$|\varphi(z)| < |z|$, 则

$ |d_1|\leq1, \quad |d_2|\leq 1-|d_1|^2. $

引理2^[10]  设$p(z)=1+p_1z+p_2z^2+\cdots$在$E$内解析且对任意$z\in E$, 满足${\rm Re} p(z)>0$, 则

$ |p_k|\leq2\quad(k\geq1), \quad \left|p_2-\frac{p_1^2}{2}\right|\leq 2-\frac{|p_1|^2}{2}. $

定理  设$\lambda\geq0, \alpha>1, -1\leq B < A\leq1$, 若$f(z)=z+\sum\limits_{n=2}^\infty a_nz^n\in L(\lambda, \alpha, A, B)$, 则对任意实数$\mu$, 有

$ \;\;\;\;(2 + \alpha )(1 + 2\lambda )|{a_3} - \mu a_2^2|\\ \le \left\{ {\begin{array}{*{20}{l}} {\alpha (1 + 2\lambda )} & {}\\ { + \frac{{(A - B)|A + B|}}{2} + \frac{{{{[(A-B) + 2\alpha (1 + \lambda )]}^2}[(3 + \alpha )(1 + 3\lambda )-2\mu (2 + \alpha )(1 + 2\lambda )]}}{{2{{(1 + \alpha )}^2}{{(1 + \lambda )}^2}}}, } & {\mu < {\mu _1}, }\\ {(A - B) + \alpha (1 + 2\lambda )} & {}\\ { + \frac{{2{\alpha ^2}{{(1 + \lambda )}^2}(2 - |A + B|)[(3 + \alpha )(1 + 3\lambda )-2\mu (2 + \alpha )(1 + 2\lambda )]}}{{(2 - |A + B|){{(1 + \alpha )}^2}{{(1 + \lambda )}^2} - (A - B)[(3 + \alpha )(1 + 3\lambda )-2\mu (2 + \alpha )(1 + 2\lambda )]}}, } & {{\mu _1} \le \mu \le {\mu _2}, }\\ {(A - B) + \alpha (1 + 2\lambda ), } & {{\mu _2} \le \mu \le {\mu _3}, } \end{array}} \right. $

其中

$ {\mu _1}{\rm{ }} = {\rm{ }}\frac{{(3 + \alpha )(1 + 3\lambda )[(A-B) + 2\alpha (1 + \lambda )] - (2 - |A + B|){{(1 + \alpha )}^2}{{(1 + \lambda )}^2}}}{{2(2 + \alpha )(1 + 2\lambda )[(A-B) + 2\alpha (1 + \lambda )]}}, \\ {\mu _2}{\rm{ }} = {\rm{ }}\frac{{(3 + \alpha )(1 + 3\lambda )}}{{2(2 + \alpha )(1 + 2\lambda )}}, \\ {\mu _3}{\rm{ }} = {\rm{ }}\frac{{(3 + \alpha )(1 + 3\lambda )[1 + 2\alpha {{(1 + \lambda )}^2}] + {{(1 + \alpha )}^2}{{(1 + \lambda )}^2}}}{{2(2 + \alpha )(1 + 2\lambda )[1 + 2\alpha {{(1 + \lambda )}^2}]}}. $

证  因为$f(z)\in L(\lambda, \alpha, A, B)$, 所以存在$g(z)=z+b_2z^2+b_3z^3+\cdots\in S^*$和$E$内满足条件$|\varphi(z)|\leq|z|$的解析函数$\varphi(z)=d_1z+d_2z^2+\cdots$, 使得

$ \left(\frac{zf'(z)}{f(z)}\cdot\left(\frac{f(z)}{g(z)}\right)^\alpha\right)^{1-\lambda}\cdot\left(\frac{(zf'(z))'}{f'(z)}\cdot\left(\frac{f'(z)}{g'(z)}\right)^\alpha\right)^{\lambda}= \frac{1+A\varphi(z)}{1+B\varphi(z)}. $

将$f(z)=z+\sum\limits_{n=2}^\infty a_nz^n, g(z)$和$\varphi(z)$的幂级数展开式代入上式, 经过一些运算可得

$ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{a_2} = \frac{{(A - B){d_1} + \alpha (1 + \lambda ){b_2}}}{{(1 + \lambda )(1 + \alpha )}}, \\ (2 + \alpha )(1 + 2\lambda ){a_3} = \alpha (1 + 2\lambda ){b_3} + \frac{{\alpha (\alpha - 1)(1 + 3\lambda )}}{{2{{(1 + \alpha )}^2}}}b_2^2 + (A - B){d_2}{\rm{ }}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + \frac{{\alpha (A - B)(3 + \alpha )(1 + 3\lambda )}}{{{{(1 + \alpha )}^2}(1 + \lambda )}}{b_2}{d_1}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; - \frac{1}{2}\left( {({A^2} - {B^2}) - \frac{{{{(A - B)}^2}(3 + \alpha )(1 + 3\lambda )}}{{{{(1 + \alpha )}^2}{{(1 + \lambda )}^2}}}} \right)d_1^2. $

令$x=\frac{(3+\alpha)(1+3\lambda)-2\mu(2+\alpha)(1+2\lambda)}{(1+\alpha)^2(1+\lambda)^2}$, 则由以上两式可得

$ (2 + \alpha )(1 + 2\lambda )({a_3} - \mu a_2^2) = {\rm{ }}\left[{\alpha (1 + 2\lambda ){b_3} + \frac{{{\alpha ^2}{{(1 + \lambda )}^2}x-\alpha (1 + 3\lambda )}}{2}b_2^2} \right]\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + (A - B)\left[{{d_2}-\left( {\frac{{A + B}}{2}-\frac{{(A-B)}}{2}x} \right)d_1^2} \right]\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + \alpha (A - B)(1 + \lambda )x{d_1}{b_2}. $

因为

$ e^{-i\theta}f(e^{i\theta}z)=z+a_2e^{i\theta}z^2+a_3e^{2i\theta}z^3+\cdots $

仍属于$L(\lambda, \alpha, A, B)$, 所以不失一般性, 可以假定$a_3-\mu a_2^2\geq 0$.下面估计${\rm Re} (a_3-\mu a_2^2)$.

由于$g(z)\in S^*$, 所以存在$E$内具有正实部的解析函数$p(z)=1+p_1z+p_2z^2+\cdots$, 使得$zg'(z)=g(z)p(z)$, 比较系数可得$b_2=p_1, b_3=\frac{1}{2}(p_2+p_1^2)$,

$ {\rm{ }}\;\;\;\;{\rm{Re}}\left[{\alpha (1 + 2\lambda ){b_3} + \frac{{{\alpha ^2}{{(1 + \lambda )}^2}x-\alpha (1 + 3\lambda )}}{2}b_2^2} \right]{\rm{ }}\\ = {\rm{ }}\frac{{\alpha (1 + 2\lambda )}}{2}{\rm{Re}}({p_2} - \frac{{p_1^2}}{2}) + \frac{\alpha }{4}[2\alpha {(1 + \lambda )^2}x + 1]{\rm{Re}}p_1^2{\rm{ }}\\ \le {\rm{ }}\frac{{\alpha (1 + 2\lambda )}}{2}(2 - \frac{{|{p_1}{|^2}}}{2}) + \frac{\alpha }{4}[2\alpha {(1 + \lambda )^2}x + 1]{\rm{Re}}p_1^2{\rm{ }}\\ = {\rm{ }}\alpha (1 + 2\lambda )(1 - {\rho ^2}) + \alpha [2\alpha {(1 + \lambda )^2}x + 1]{\rho ^2}\cos 2\phi, $

其中$b_2=p_1=2\rho e^{i\phi}, 0\leq\rho\leq1$.

$ {\;\;\;(A - B){\rm{Re}}\left[ {{d_2} - \left( {\frac{{A + B}}{2} - \frac{{(A - B)}}{2}x} \right)d_1^2} \right]}\\ { \le (A - B) - \frac{{A - B}}{2}(2 - |A + B|)|{d_1}{|^2} + \frac{{{{(A - B)}^2}}}{2}x{\rm{Re}}d_1^2}\\ { = (A - B) - \frac{{A - B}}{2}(2 - |A + B|){r^2} + \frac{{{{(A - B)}^2}}}{2}x{r^2}\cos 2\theta ,} $

其中$d_1=r e^{i\theta}, 0\leq r\leq1$.所以

$ (2+\alpha)(1+2\lambda){\rm Re}(a_3-\mu a_2^2)\leq \Psi(x), $

其中

$ \Psi (x){\rm{ = }}\alpha (1 + 2\lambda )(1 - {\rho ^2}) + \alpha [2\alpha {(1 + \lambda )^2}x + 1]{\rho ^2}\cos 2\phi + (A - B) - \frac{{A - B}}{2}(2 - |A + B|){r^2}\\ \;\;\;\;\;\;\;\;\; + \frac{{{{(A - B)}^2}}}{2}x{r^2}\cos 2\theta + 2\alpha (A - B)(1 + \lambda )x\rho r\cos (\phi + \theta ). $

(1) 当$\frac{(3+\alpha)(1+3\lambda)[(A-B)+2\alpha(1+\lambda)]-(2-|A+B|)(1+\alpha)^2(1+\lambda)^2}{2(2+\alpha)(1+2\lambda)[(A-B)+2\alpha(1+\lambda)]}\leq \mu \leq \frac{(3+\alpha)(1+3\lambda)}{2(2+\alpha)(1+2\lambda)}$时, $0\leq x\leq \frac{2-|A+B|}{(A-B)+2\alpha(1+\lambda)}$, 有

$ {\Psi (x) \le \alpha [(1 + 2\lambda ) + 2\alpha {{(1 + \lambda )}^2}x] + (A - B) - \frac{{A - B}}{2}[(2 - |A + B|)}\\ {\;\;\;\;\;\;\;\;\;\;\; - (A - B)x\cos 2\theta ]{r^2} + 2\alpha (A - B)(1 + \lambda )xr}\\ {\;\;\;\;\;\;\;\;\; = \alpha [(1 + 2\lambda ) + 2\alpha {{(1 + \lambda )}^2}x] + (A - B) + \frac{{2{\alpha ^2}(A - B){{(1 + \lambda )}^2}{x^2}}}{{(2 - |A + B|) - (A - B)x\cos 2\theta }}}\\ {\;\;\;\;\;\;\;\;\;\;\; - \frac{{A - B}}{2}[(2 - |A + B|) - (A - B)x\cos 2\theta ]{{\left( {r - \frac{{2\alpha (1 + \lambda )x}}{{(2 - |A + B|) - (A - B)x\cos 2\theta }}} \right)}^2}}\\ {\;\;\;\;\;\;\;\;\; \le \alpha [(1 + 2\lambda ) + 2\alpha {{(1 + \lambda )}^2}x] + (A - B) + \frac{{2{\alpha ^2}(A - B){{(1 + \lambda )}^2}{x^2}}}{{(2 - |A + B|) - (A - B)x}}}\\ {\;\;\;\;\;\;\;\;\; = (A - B) + \alpha (1 + 2\lambda ) + \frac{{2{\alpha ^2}{{(1 + \lambda )}^2}(2 - |A + B|)x}}{{(2 - |A + B|) - (A - B)x}}}\\ {\;\;\;\;\;\;\;\;\; = \frac{{2{\alpha ^2}{{(1 + \lambda )}^2}(2 - |A + B|)[(3 + \alpha )(1 + 3\lambda ) - 2\mu (2 + \alpha )(1 + 2\lambda )]}}{{(2 - |A + B|){{(1 + \alpha )}^2}{{(1 + \lambda )}^2} - (A - B)[(3 + \alpha )(1 + 3\lambda ) - 2\mu (2 + \alpha )(1 + 2\lambda )]}}}\\ {\;\;\;\;\;\;\;\;\;\;\;\; + (A - B) + \alpha (1 + 2\lambda ).} $

当$\mu_1\leq\mu\leq \mu_2$时, 不存在对应的极值函数.

(2) 当$\mu\leq\frac{(3+\alpha)(1+3\lambda)[(A-B)+2\alpha(1+\lambda)]-(2-|A+B|)(1+\alpha)^2(1+\lambda)^2}{2(2+\alpha)(1+2\lambda)[(A-B)+2\alpha(1+\lambda)]}$时, $x\geq\frac{2-|A+B|}{(A-B)+2\alpha(1+\lambda)}$, 令$x_0=\frac{2-|A+B|}{(A-B)+2\alpha(1+\lambda)}$, 则由(1) 可得

$ \Psi ({x_0}) \le (A - B) + \alpha (1 + 2\lambda ) + \frac{{2{\alpha ^2}{{(1 + \lambda )}^2}(2 - |A + B|){x_0}}}{{(2 - |A + B|) - (A - B){x_0}}}\\ \;\;\;\;\;\;\;\;\; \le (A - B) + \alpha (1 + 2\lambda ) + \alpha (1 + \lambda )(2 - |A + B|), $

所以

$ \Psi (x) = \Psi ({x_0}) + (x - {x_0})\left( {2{\alpha ^2}{{(1 + \lambda )}^2}{\rho ^2}\cos 2\phi } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\; + \left. {\frac{{{{(A - B)}^2}}}{2}{r^2}\cos 2\theta + 2\alpha (A - B)(1 + \lambda )r\rho \cos (\phi + \theta )} \right)\\ \;\;\;\;\;\;\;\; \le \Psi ({x_0}) + (x - {x_0})\left( {2{\alpha ^2}{{(1 + \lambda )}^2} + \frac{{{{(A - B)}^2}}}{2} + 2\alpha (A - B)(1 + \lambda )} \right)\\ \;\;\;\;\;\;\;\; \le (A - B) + \alpha (1 + 2\lambda ) + \alpha (1 + \lambda )(2 - |A + B|) + \frac{{x - {x_0}}}{2}{[(A-B) + 2\alpha (1 + \lambda )]^2}\\ \;\;\;\;\;\;\;\; = \alpha (1 + 2\lambda ) + \frac{{(A - B)|A + B|}}{2} + \frac{{{{[(A-B) + 2\alpha (1 + \lambda )]}^2}x}}{2}\\ \;\;\;\;\;\;\;\; = \alpha (1 + 2\lambda ) + \frac{{(A - B)|A + B|}}{2}\\ \;\;\;\;\;\;\;\;\;\;\; + \frac{{{{[(A-B) + 2\alpha (1 + \lambda )]}^2}[(3 + \alpha )(1 + 3\lambda )-2\mu (2 + \alpha )(1 + 2\lambda )]}}{{2{{(1 + \alpha )}^2}{{(1 + \lambda )}^2}}}, $

当$\mu\leq\mu_1$时, 不存在对应的极值函数.

(3) 当$\frac{(3+\alpha)(1+3\lambda)}{2(2+\alpha)(1+2\lambda)}\leq\mu\leq\frac{(3+\alpha)(1+3\lambda)[1+2\alpha(1+\lambda)^2]+(1+\alpha)^2(1+\lambda)^2}{2(2+\alpha)(1+2\lambda)[1+2\alpha(1+\lambda)^2]}$ 时, $-\frac{1}{1+2\alpha(1+\lambda)^2}\leq x\leq 0$, 有

$ \Psi(0)=\alpha(1+2\lambda)(1-\rho^2)+\alpha\rho^2\cos2\phi+(A-B)-\frac{A-B}{2}(2-|A+B|)r^2\leq(A-B)+\alpha(1+2\lambda). $

令$x_1=-\frac{1}{1+2\alpha(1+\lambda)^2}$, 则

$ \;\;\;\Psi ({x_1}) - [(A-B) + \alpha (1 + 2\lambda )]\\ = - \alpha (1 + 2\lambda ){\rho ^2} + \alpha [2\alpha {(1 + \lambda )^2}{x_1} + 1]{\rho ^2}\cos 2\phi \\ \;\;\; - \frac{{A - B}}{2}[(2-|A + B|)-(A-B){x_1}\cos 2\theta]{r^2} + 2\alpha (A - B)(1 + \lambda ){x_1}\rho r\cos (\phi + \theta )\\ = - \alpha (1 + 2\lambda ){\rho ^2} + \alpha [2\alpha {(1 + \lambda )^2}{x_1} + 1]{\rho ^2}\cos 2\phi + \frac{{2{\alpha ^2}(A - B){{(1 + \lambda )}^2}x_1^2{\rho ^2}{{\cos }^2}(\phi + \theta )}}{{(2 - |A + B|) - (A - B){x_1}\cos 2\theta }}\\ \;\;\; - \frac{{A - B}}{2}[(2-|A + B|)-(A-B){x_1}\cos 2\theta]{\left( {r - \frac{{2\alpha (1 + \lambda ){x_1}\rho \cos (\phi + \theta )}}{{(2 - |A + B|) - (A - B){x_1}\cos 2\theta }}} \right)^2}\\ \le - \alpha (1 + 2\lambda ){\rho ^2} + \alpha [2\alpha {(1 + \lambda )^2}{x_1} + 1]{\rho ^2}\cos 2\phi + \frac{{2{\alpha ^2}(A - B){{(1 + \lambda )}^2}x_1^2{\rho ^2}{{\cos }^2}(\phi + \theta )}}{{(2 - |A + B|) - (A - B){x_1}\cos 2\theta }}\\ = - \alpha {\rho ^2}\left\{ {(1 + 2\lambda ) - [2\alpha {{(1 + \lambda )}^2}{x_1} + 1]\cos 2\phi - \frac{{2\alpha (A - B){{(1 + \lambda )}^2}x_1^2{{\cos }^2}(\phi + \theta )}}{{(2 - |A + B|) - (A - B){x_1}\cos 2\theta }}} \right\}\\ = - \frac{{2\alpha {\rho ^2}}}{{[(2-|A + B|)-(A-B){x_1}\cos 2\theta]{{[1 + 2\alpha (1 + \lambda )]}^2}}}\\ \;\;\; \times \{ [(\lambda + {\sin ^2}\phi ) + \alpha {(1 + \lambda )^2}(1 + 2\lambda )]\\ \;\;\; \cdot [(2-|A + B|)(1 + 2\alpha {(1 + \lambda )^2}) + (A-B)\cos 2\theta] - \alpha (A - B){(1 + \lambda )^2}{\cos ^2}(\phi + \theta )\} . $

令

$ M(A - B) = [(\lambda + {\sin ^2}\phi ) + \alpha {(1 + \lambda )^2}(1 + 2\lambda )]\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \times [(2-|A + B|)(1 + 2\alpha {(1 + \lambda )^2}) + (A-B)\cos 2\theta]\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; - \alpha (A - B){(1 + \lambda )^2}{\cos ^2}(\phi + \theta ), $

则$M(A-B)$是$A-B$的一次函数, 且

$ \begin{array}{l} M(0) = (2 - |A + B|)(1 + 2\alpha {(1 + \lambda )^2})[(\lambda + {\sin ^2}\phi ) + \alpha {(1 + \lambda )^2}(1 + 2\lambda )] \ge 0,\\ M(2) = 2[(\lambda + {\sin ^2}\phi ) + \alpha {(1 + \lambda )^2}(1 + 2\lambda )] \cdot [1 + 2\alpha (1 + {\lambda ^2} + \cos 2\theta )]\\ \;\;\;\;\;\;\;\;\;\;\;\; - 2\alpha {(1 + \lambda )^2}{\cos ^2}(\phi + \theta )\\ \;\;\;\;\;\;\;\;\; = 4[(\lambda + {\sin ^2}\phi ) + \alpha {(1 + \lambda )^2}(1 + 2\lambda )] \cdot [\alpha {(1 + \lambda )^2} + {\cos ^2}\theta ]\\ \;\;\;\;\;\;\;\;\;\;\;\; - 2\alpha {(1 + \lambda )^2}({\cos ^2}\phi {\cos ^2}\theta - 2\cos \phi \cos \theta \sin \phi \sin \theta + {\sin ^2}\phi {\sin ^2}\theta )\\ \;\;\;\;\;\;\;\;\; \ge 4\alpha {(1 + \lambda )^2}({\sin ^2}\phi + {\cos ^2}\theta ) - 2\alpha {(1 + \lambda )^2}\\ \;\;\;\;\;\;\;\;\;\;\;\;({\cos ^2}\phi {\cos ^2}\theta - 2\cos \phi \cos \theta \sin \phi \sin \theta + {\sin ^2}\phi {\sin ^2}\theta )\\ \;\;\;\;\;\;\;\;\; \ge 4\alpha {(1 + \lambda )^2}({\sin ^2}\phi {\sin ^2}\theta + {\cos ^2}\phi {\cos ^2}\theta )\\ \;\;\;\;\;\;\;\;\;\;\;\; - 2\alpha {(1 + \lambda )^2}({\cos ^2}\phi {\cos ^2}\theta - 2\cos \phi \cos \theta \sin \phi \sin \theta + {\sin ^2}\phi {\sin ^2}\theta )\\ \;\;\;\;\;\;\;\;\; = 2\alpha {(1 + \lambda )^2}{(\sin \phi \sin \theta + \cos \phi \cos \theta )^2}\\ \;\;\;\;\;\;\;\;\; \ge 0. \end{array} $

故当$0 < A-B < 2$时, $M(A-B)\geq0$, 从而$\Psi(x_1)\leq(A-B)+\alpha(1+2\lambda).$对于$0\leq t\leq1$, 有

$ \Psi (t{x_1}) \le t\Psi ({x_1}) + (1 - t)\Psi (0)\\ \le t[(A-B) + \alpha (1 + 2\lambda )] + (1 - t)[(A-B) + \alpha (1 + 2\lambda )] = (A - B) + \alpha (1 + 2\lambda ). $

当$b_2=0, b_3=1, d_1=0, d_2=1$时等号成立.对应的极值函数为

$ \left\{[(f(z))^\alpha]'\right\}^{1-\lambda}\left\{\left[\left(zf'(z)\right)^\alpha\right]'\right\}^\lambda=\alpha z^{\alpha-1}\left(\frac{1+Az^2}{1+Bz^2}\right)\left(\frac{(1+z^2)^\lambda}{(1-z^2)^{1+\lambda}}\right)^\alpha. $

综上所述, 本定理得证.