用$A$来表示在单位圆$U=\{z\in C :|z|<1\}$内解析,且具有如下形式的泰勒展开式 \begin{equation} f(z)=z+\sum\limits_{n=2}^{\infty}a_{n}z^{n} \label{eq:a1} \tag{1.1} \end{equation} 的函数$f$构成的函数族. $S$表示$A$中的单叶函数子族. 用$S^*(\alpha)$,$K(\alpha)$分别表示为Re$ \{ \frac{zf'(z)}{f(z)}\}$ $>\alpha $和Re$\{ 1+\frac{zf"(z)}{f'(z)}\}>\alpha$ 函数类,其中$0\leq\alpha<1$. 它们都是$S$的子类,分别称为$\alpha$级星像函数类$\alpha$级凸像函数类. 特别地,记$S^*(0)\equiv S^*$, $K(0)\equiv K$. 注意到 $$f(z)\in K(\alpha)\Longleftrightarrow zf'(z)\in S^*(\alpha). $$

设$ f(z)$和$ g(z)$在$ U$ 解析,如果存在$ U$内的解析函数$ w(z)$使$| w(z)|\leq |z|$ 且满足$g(z)\equiv f( w(z))$,则称$ g(z)$从属于$ f(z)$,记作$ g(z)\prec f(z)$或者 $ g\prec f$. 若$ f(z)$在$ U$内单叶, 则$ g(z)\prec f(z)$当且仅当 $ g(0)= f(0)$和$ g(U)\subset f(U)$.

许多作者对$ S^*$,$S^*(\alpha)$ 和 $K(\alpha)$函数类和一些从属关系的充分条件进行了讨论, 得到了很多好的结论,具体可以见文献^{[1, 2, 3, 4, 5, 6, 7]}. 本文利用文献^{[1, 2, 3]} 的引理,对$f(z)$进行讨论, 得到一些$\frac{zf'(z)}{f(z)}$的从属的充分条件和$f(z)$ 星像和凸像的充分条件,推广了文献^{[1, 2, 3, 4]} 的结论.

引理1.1^[1] 设$z\in U$,$\frac{f(z)}{z}eq0 $,$\lambda>1$,若 $f(z)\in A$ 且满足 $$1+\frac{zf"(z)}{f'(z)}\prec \frac{\lambda(1-z)}{\lambda-z} -\frac{(\lambda-1)z}{(1-z)(\lambda-z)},$$ 则 $$\frac{zf'(z)}{f(z)}\prec \frac{\lambda(1-z)}{\lambda-z}.$$

引理1.2^[2] 若 $f(z)\in A$ 且满足 \[ {\rm Re}\bigg\{1+\frac{zf"(z)}{f'(z)}\bigg\}< \left\{{\begin{array}{ll} {\frac{5\lambda-1}{2(\lambda+1)}},~~& (z\in U,1<\lambda\leq 2),\\[3mm] { \frac{\lambda+1}{2(\lambda-1)}},& (z\in U,2<\lambda<3), \end{array}}\right.\] 则 $$\frac{zf'(z)}{f(z)}\prec \frac{\lambda(1-z)}{\lambda-z}.$$ 极值函数为$$f(z)=z (1-\frac{z}{\lambda})^{\lambda-1}.$$

引理1.3^[3] 设$z\in U$,$0<\gamma \leq 1 $,$0<\lambda \leq \frac{\gamma(1+\gamma)}{\sqrt{(1+\gamma)^{2}+\gamma^{2}}}$,若 $f(z)\in A$ 且满足 $$\bigg|f'(z)-(1-\gamma)\frac{f(z)}{z} -\gamma \bigg| < \lambda,$$ 则$f(z)\in S^*(1-\gamma).$

应用引理1.1,我们可以得到定理2.1和定理2.2.

定理2.1 设$z\in U$,$\frac{f(z)}{z}eq0 $,$\lambda>1$,若 $f(z)\in A$ 且满足 $$\bigg\{\frac{1}{1-\alpha} \bigg(\frac{zf'(z)}{f(z)}-\alpha\bigg)+\frac{\frac{z^{2}f"(z)}{f(z)} +\frac{zf'(z)}{f(z)}-[\frac{zf'(z)}{f(z)}]^{2}} {\frac{zf'(z)}{f(z)}-\alpha}\bigg\} \prec \frac{\lambda(1-z)}{\lambda-z} -\frac{(\lambda-1)z}{(1-z)(\lambda-z)},$$ 其中$0\leq\alpha<1$, 则 $$\frac{1}{1-\alpha} \bigg(\frac{zf'(z)}{f(z)}-\alpha\bigg)\prec \frac{\lambda(1-z)}{\lambda-z}.$$

证设$f(z)\in A$,定义 $p(z)$ 如下 \begin{equation}p(z)=\bigg(\frac{f(z)}{z^{\alpha}}\bigg)^{\frac{1}{1-\alpha}} =z+\frac{a_{2}}{1-\alpha}z^{2}+\cdots, \label{eq:b1}\tag{2.1} \end{equation} 其中$0\leq\alpha<1$,则 $p(z)\in A$,且 \begin{equation}\frac{zp'(z)}{p(z)}=\frac{1}{1-\alpha} \bigg(\frac{zf'(z)}{f(z)}-\alpha\bigg). \label{eq:b2}\tag{2.2} \end{equation} 对(2.2)式两边取对数并求导,可得 \begin{equation}1+\frac{zp"(z)}{p'(z)}=\frac{1}{1-\alpha} \bigg(\frac{zf'(z)}{f(z)}-\alpha\bigg)+\frac{\frac{z^{2}f"(z)}{f(z)} +\frac{zf'(z)}{f(z)}-[\frac{zf'(z)}{f(z)}]^{2}} {\frac{zf'(z)}{f(z)}-\alpha}. \label{eq:b3}\tag{2.3} \end{equation} 根据定理2.1的条件可得 $$1+\frac{zp"(z)}{p'(z)}\prec \frac{\lambda(1-z)}{\lambda-z} -\frac{(\lambda-1)z}{(1-z)(\lambda-z)}.$$ 由(2.1)式可得 $$\frac{p(z)}{z}= \bigg(\frac{f(z)}{z}\bigg)^{\frac{1}{1-\alpha}}eq0. $$ 所以由引理1.1可得 $$\frac{zp'(z)}{p(z)}\prec \frac{\lambda(1-z)}{\lambda-z}.$$ 进一步由(2.2)式可得 $$\frac{1}{1-\alpha} \bigg(\frac{zf'(z)}{f(z)}-\alpha\bigg)\prec \frac{\lambda(1-z)}{\lambda-z}.$$

证毕.

注2.1 在定理2.1 中令$\alpha=0$,可以得到引理1.1.

定理2.2 设$z\in U$,$f'(z)eq0 $,$\lambda>1$,若 $f(z)\in A$ 且满足 $$\bigg\{ \frac{1}{1-\alpha} \frac{zf"(z)}{f'(z)}+\frac{\frac{z^{2}f"'(z)}{f'(z)} +\frac{zf"(z)}{f'(z)}- \big[\frac{zf"(z)}{f'(z)}\big]^{2}} {\frac{zf"(z)}{f'(z)}+1-\alpha} \bigg\}\prec \frac{\lambda(1-z)}{\lambda-z} -\frac{(\lambda-1)z}{(1-z)(\lambda-z)} ,$$ 则 $$1+ \frac{1}{1-\alpha} \frac{zf"(z)}{f'(z)} \prec \frac{\lambda(1-z)}{\lambda-z}.$$

证  设$f(z)\in A$,定义 $p(z)$ 如下 \begin{equation}p(z)=\int_{0}^{z} (f'(t))^{\frac{1}{1-\alpha}} {\rm d}t =z+\frac{a_{2}}{1-\alpha}z^{2}+\cdots,\label{eq:b4}\tag{2.4} \end{equation} 其中$0\leq\alpha<1$,则 $p(z)\in A$,且令 \begin{equation}g(z)=zp'(z)= z \big(f'(z)\big) ^{\frac{1}{1-\alpha}} =z+\frac{2a_{2}}{1-\alpha}z^{2}+\cdots. \label{eq:b7}\tag{2.5} \end{equation} 则 $g(z)\in A$,且 \begin{equation}\frac{zg'(z)}{g(z)}= 1+ \frac{1}{1-\alpha} \frac{zf"(z)}{f'(z)}. \label{eq:b5}\tag{2.6} \end{equation} 对(2.6)式两边取对数并求导,可得 \begin{equation}1+\frac{zg"(z)}{g'(z)}=1+\frac{1}{1-\alpha} \frac{zf"(z)}{f'(z)}+\frac{\frac{z^{2}f"'(z)}{f'(z)}+\frac{zf"(z)} {f'(z)}-\big[\frac{zf"(z)}{f'(z)}\big]^{2}} {\frac{zf"(z)}{f'(z)}+1-\alpha}. \label{eq:b6}\tag{2.7} \end{equation} 根据定理2.2的条件可得 $$1+\frac{zg"(z)}{g'(z)}\prec \frac{\lambda(1-z)}{\lambda-z} -\frac{(\lambda-1)z}{(1-z)(\lambda-z)}. $$ 由(2.5)式可得 $$\frac{g(z)}{z}= \big(f'(z)\big) ^{\frac{1}{1-\alpha}}eq 0. $$ 所以由引理1.1可得 $$\frac{zg'(z)}{g(z)}\prec \frac{\lambda(1-z)}{\lambda-z}.$$ 进一步由(2.6)式可得 $$1+ \frac{1}{1-\alpha} \frac{zf"(z)}{f'(z)} \prec\frac{\lambda(1-z)}{\lambda-z}.$$ 证毕.

在定理2.2 中令$\alpha=0$,可以得到推论2.1.

推论2.1  设$z\in U$,$f'(z)eq0 $,$\lambda>1$, 若 $f(z)\in A$ 且满足 $$1+\frac{z^{2}f"'(z)+2zf"(z)} {zf"(z)+f'(z)} \prec \frac{\lambda(1-z)}{\lambda-z} -\frac{(\lambda-1)z}{(1-z)(\lambda-z)},$$ 则 $$1+ \frac{zf"(z)}{f'(z)} \prec \frac{\lambda(1-z)}{\lambda-z}.$$

应用引理1.2,我们可以得到定理2.3和定理2.4.

定理2.3 若 $f(z)\in A$ 且满足 \begin{eqnarray*} && {\rm Re}\bigg\{\frac{1}{1-\alpha} \bigg(\frac{zf'(z)}{f(z)}-\alpha\bigg)+ \frac{\frac{z^{2}f"(z)}{f(z)}+\frac{zf'(z)}{f(z)}- \big[\frac{zf'(z)}{f(z)}\big]^{2}} {\frac{zf'(z)}{f(z)}-\alpha}\bigg\}\\ &<& \left\{{\begin{array}{ll} {\frac{5\lambda-1}{2(\lambda+1)}},~~& (z\in U,1<\lambda\leq 2),\\[3mm] { \frac{\lambda+1}{2(\lambda-1)}},& (z\in U,2<\lambda<3), \end{array}}\right.\end{eqnarray*} 其中$0\leq\alpha<1$,则 $$\frac{1}{1-\alpha} \bigg(\frac{zf'(z)}{f(z)}-\alpha\bigg)\prec \frac{\lambda(1-z)}{\lambda-z}.$$ 极值函数为 $$f(z)=z (1-\frac{z}{\lambda})^{(\lambda-1)(1-\alpha)}.$$

证明方法和定理2.1的方法类似.

注2.2 在定理2.3 中令$\alpha=0$,可以得到引理1.2.

定理2.4 若 $f(z)\in A$ 且满足 \begin{eqnarray*} &&{\rm Re}\bigg\{ \frac{1}{1-\alpha} \frac{zf"(z)}{f'(z)}+\frac{\frac{z^{2}f"'(z)}{f'(z)} +\frac{zf"(z)}{f'(z)}-\big[\frac{zf"(z)}{f'(z)}\big]^{2}} {\frac{zf"(z)}{f'(z)}+1-\alpha} \bigg\}\\ &<& \left\{{\begin{array}{ll} {\frac{3(\lambda-1)}{2(\lambda+1)}},~~& (z\in U,1<\lambda\leq 2),\\[3mm] { \frac{3-\lambda}{2(\lambda-1)}},& (z\in U,2<\lambda<3), \end{array}}\right.\end{eqnarray*} 其中$0\leq\alpha<1$,则 $$1+ \frac{1}{1-\alpha} \frac{zf"(z)}{f'(z)} \prec \frac{\lambda(1-z)}{\lambda-z}.$$ 极值函数为 $$ f(z) =\int_{0}^{z}(1-\frac{t}{\lambda})^{(\lambda-1)(1-\alpha)}{\rm d}t.$$

证明方法和定理2.2的方法类似.

注2.3 在定理2.4 中令$\alpha=0$,可以得到文献[2,推论3].

应用引理1.3,我们可以得到定理2.5和定理2.6.

定理2.5 设$z\in U$,$0<\gamma \leq 1 $,$0<\lambda \leq \frac{\gamma(1+\gamma)}{\sqrt{(1+\gamma)^{2}+\gamma^{2}}}$,$0\leq\alpha<1$,若 $f(z)\in A$ 且满足 $$ \bigg|\bigg( \frac{f(z)}{z}\bigg)^{\frac{1}{1-\alpha}} \bigg\{ \frac{zf'(z)}{f(z)}-\alpha-(1-\gamma)(1-\alpha)\bigg\} -\gamma (1-\alpha) \bigg| < \lambda (1-\alpha), $$ 则$f(z)\in S^*(1-\gamma+\alpha\gamma).$

证  设$f(z)\in A$,定义 $p(z)$ 如式(2.1),通过计算和结合定理的条件可得 \begin{eqnarray*} &&\bigg|p'(z)-(1-\gamma)\frac{p(z)}{z} -\gamma \bigg| \\ &= & \frac{1}{1-\alpha} \bigg|\bigg( \frac{f(z)}{z}\bigg)^{\frac{1}{1-\alpha}} \bigg\{ \frac{zf'(z)}{f(z)}-\alpha-(1-\gamma)(1-\alpha)\bigg\} -\gamma (1-\alpha) \bigg| \\ & <& \lambda, \end{eqnarray*} 应用引理1.3可得$p(z)\in S^*(1-\gamma).$ 进一步由式(2.2)可得$f(z)\in S^*(1-\gamma+\alpha\gamma).$

注2.4 在定理2.5 中令$\alpha=0$,可以得到引理1.3.

注2.5  在定理2.5 中令$\gamma=1$,可以得到文献[4,定理2.1].

定理2.6  设$z\in U$,$0<\gamma \leq 1 $,$0<\lambda \leq \frac{\gamma(1+\gamma)}{\sqrt{(1+\gamma)^{2}+\gamma^{2}}}$,$0\leq\alpha<1$,若 $f(z)\in A$ 且满足 $$ \bigg|(f'(z))^{\frac{\alpha}{1-\alpha}}\bigg(\gamma f'(z)+ \frac{1}{1-\alpha}zf"(z)\bigg) -\gamma \bigg| < \lambda, $$ 则$f(z)\in K(1-\gamma+\alpha\gamma).$

证 设$f(z)\in A$,定义 $g(z)$ 如式(2.6),通过计算和结合定理的条件可得 \begin{eqnarray*} \bigg|g'(z)-(1-\gamma)\frac{g(z)}{z} -\gamma \bigg| & =& \bigg|(f'(z))^{\frac{\alpha}{1-\alpha}}\bigg(\gamma f'(z)+\frac{1}{1-\alpha}zf"(z)\bigg) -\gamma \bigg| \\ &<& \lambda.\end{eqnarray*} 应用引理1.3可得$g(z)\in S^*(1-\gamma).$ 进一步由式(2.6)可得 $$ {\rm Re}\bigg( 1+\frac{1}{1-\alpha} \frac{zf"(z)}{f'(z)} \bigg)>(1-\gamma),$$ 即 $$ {\rm Re} \bigg( 1+ \frac{zf"(z)}{f'(z)} \bigg)>(1-\gamma)(1-\alpha)+\alpha.$$ 即$f(z)\in K(1-\gamma+\alpha\gamma).$

注2.6 在定理2.6 中令$\alpha=0$,可以得到文献[3,定理6].

注2.7 在定理2.6 中令$\gamma=1$,可以得到文献[4,定理3.1].