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

$ f(z)=z+\sum\limits_{n=2}^{\infty}a_{n}z^{n} $

(1.1)

的函数$f$构成的函数族. $S$表示$A$中的单叶函数子族.用$S^*(\beta)$, $K(\beta)$分别表示为${\rm Re} \{ \frac{zf'(z)}{f(z)}\}>\beta $和${\rm Re} \{ 1+\frac{zf''(z)}{f'(z)}\}>\beta $函数类, 其中$0\leq\beta<1$.它们都是$S$的子类, 分别称为$\beta$级星像函数类和$\beta$级凸像函数类.特别地, 记$S^*(0)\equiv S^*$, $K(0)\equiv K$.注意到$f(z)\in K(\beta)\Longleftrightarrow$ $zf'(z)\in S^*(\beta)$.

函数$f(z)=\sum\limits_{k=1}^{+\infty}a_{k}z^{k}$和$g(z)=\sum\limits_{k=1}^{+\infty}b_{k}z^{k}$的Hadamard乘积或卷积$(f*g)(z)$定义为

$ (f*g)(z)=\sum\limits_{k=1}^{+\infty}a_{k}b_{k}z^{k}. $

设函数$\phi(a, c;z)$定义为

$ \phi(a, c;z):=\sum\limits_{n=0}^{\infty}\frac{(a)_{n}}{(c)_{n}}z^{n+1}~~(c\neq 0, -1, -2, \cdots, z\in U), $

其中$(x)_{n}$定义为

$ (x)_{n}:= \left\{{\begin{array}{ll}{x(x+1)\cdots(x+n-1)}, n\in \mathbf{N}=\{1, 2, \cdots\}, \\ {1 }, \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; n=0. \end{array}}\right. $

Carlson和Shaer^[1]利用了Hadamard乘积定义了线性算子$L(a, c)$, 定义如下:

$ L(a, c)f(z):=\phi(a, c;z)*f(z)= \sum\limits_{n=0}^{\infty}\frac{(a)_{n}}{(c)_{n}}a_{n}z^{n+1}~~(c\neq 0, -1, -2, \cdots, z\in U). $

(1.2)

注意到$L(a, a)f(z)=f(z), ~ L(2, 1)f(z)=zf'(z). $

由式子(1.2), 容易验证

$ z(L(a, c)f(z))'=aL(a+1, c)f(z)-(a-1)L(a, c)f(z). $

(1.3)

许多作者对$ S^*$, $S^*(\beta)$和$K(\beta)$函数类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[2-6].本文结合Hadamard乘积, 利用文[2, 3]的引理, 对算子$L(a, c)f(z)$进行讨论, 得到$L(a, c)f(z)\in S^*(\beta)$和$L(a, c)f(z)\in K(\beta)$的充分条件, 推广了文[2, 3]的结论.

为了定理的证明, 先介绍三个相关的引理.

引理1.1^[2]  若$f(z)\in A$且满足$|f'(z)-1|<\frac{2}{\sqrt{5}}  (z\in U), $则$f(z)\in S^*$.

引理1.2^[2]  若$f(z)\in A$且满足$|\arg f'(z)|<\frac{\pi}{2}\delta~~(z\in U), $则$f(z)\in S^*$, 其中$\delta=0.6165\cdots$是方程$2\tan^{-1}(1-\delta)+(1-2\delta)\pi=0$的唯一解.

引理1.3^[3]  若$f(z)\in A$且满足$|f''(z)|<\frac{1}{\sqrt{5}}=0.4472\cdots~~(z\in U), $则$f(z)\in K$.

应用引理1.1, 可以得到定理2.1.

定理2.1  若$f(z)\in A$且满足

$ \bigg|\bigg(\frac{L(a, c)f(z)}{z}\bigg)^{\frac{1}{1-\beta}} \bigg(\frac{aL(a+1, c)f(z)}{L(a, c)f(z)}-(a-1)-\beta\bigg) -1+\beta \bigg|<\frac{2}{\sqrt{5}}(1-\beta)~~(z\in U), $

其中$0\leq \beta<1$, 则$L(a, c)f(z)\in S^*(\beta)$.

证  设$f(z)\in A$, 定义$p(z)$如下:

$p(z)=\bigg(\frac{L(a, c)f(z)}{z^{\beta}}\bigg)^{\frac{1}{1-\beta}} =z+\frac{a}{c}\cdot\frac{a_{2}}{1-\beta}z^{2}+\cdots, \label{b1}$

(2.1)

则$p(z)\in A$, 且

$ p'(z)=\frac{1}{1-\beta}\bigg(\frac{L(a, c)f(z)}{z}\bigg)^{\frac{1}{1-\beta}} \times \bigg(\frac{z(L(a, c)f(z))'}{L(a, c)f(z)}-\beta\bigg). $

结合式子(1.3) 可以得到

$ p'(z)=\frac{1}{1-\beta}\bigg(\frac{L(a, c)f(z)}{z}\bigg)^{\frac{1}{1-\beta}} \times \bigg(\frac{aL(a+1, c)f(z)}{L(a, c)f(z)}-(a-1)-\beta\bigg). \label{b2}$

(2.2)

由定理的条件可得

$|p'(z)-1|=\frac{1}{1-\beta}\bigg|\bigg(\frac{L(a, c)f(z)}{z}\bigg)^{\frac{1}{1-\beta}} \times \bigg(\frac{aL(a+1, c)f(z)}{L(a, c)f(z)}-(a-1)-\beta\bigg)-1+\beta\bigg|<\frac{2}{\sqrt{5}} .$

应用引理1.1, 可以得到$p(z)\in S^*$.

注意到

$\frac{zp'(z)}{p(z)}=\frac{1}{1-\beta} \bigg(\frac{z(L(a, c)f(z))'}{L(a, c)f(z)}-\beta\bigg), $

所以

${\rm Re} \bigg(\frac{z(L(a, c)f(z))'}{L(a, c)f(z)}\bigg)>\beta(z\in U), $

这意味着$L(a, c)f(z)\in S^*(\beta)$.

在定理2.1中, 令$\beta=\frac{1}{2}$, 可得

推论2.1  若$f(z)\in A$且满足

$ \bigg|\bigg(\frac{L(a, c)f(z)}{z}\bigg)^{2} \bigg(\frac{aL(a+1, c)f(z)}{L(a, c)f(z)}-a+\frac{1}{2}\bigg) -\frac{1}{2} \bigg|<\frac{1}{\sqrt{5}}=0.4472\cdots~~(z\in U), $

则$L(a,c)f(z)\in S^*(\frac{1}{2})$

在定理2.1中, 令$\beta=0$, 可得

推论2.2  若$f(z)\in A$且满足

$ \bigg|\frac{aL(a+1, c)f(z)}{z}-\frac{(a-1)L(a, c)f(z)}{z} -1 \bigg|<\frac{2}{\sqrt{5}}  (z\in U), $

则$L(a, c)f(z)\in S^*$.

在推论2.2中, 令$a=1, c=1$, 可得

推论2.3  若$f(z)\in A$且满足

$ \bigg|\frac{L(2, 1)f(z)}{z} -1 \bigg|<\frac{2}{\sqrt{5}}  (z\in U), $

则$f(z)\in S^*$.

注2.1  在推论2.3中应用$L(2, 1)f(z)=zf'(z)$, 可以得到引理1.1.

注2.2  在定理2.1中令$a=1, c=1$, 可得文[3]中的定理2.1.

应用引理1.2, 可以得到定理2.2.

定理2.2  若$f(z)\in A$且满足

$ \bigg| \arg \bigg(\frac{L(a, c)f(z)}{z}\bigg)+ (1-\beta) \arg \bigg(\frac{aL(a+1, c)f(z)}{L(a, c)f(z)}-(a-1)-\beta\bigg) \bigg|<\frac{\pi}{2}\delta(1-\beta)~~(z\in U), $

其中$0\leq \beta<1$, 则$L(a, c)f(z)\in S^*(\beta)$, 其中$\delta=0.6165\cdots$是方程$2\tan^{-1}(1-\delta)+(1-2\delta)\pi=0$的唯一解.

证  设$p(z)$由式子(2.1) 给出, 则从式子(2.2) 可得

$ | \arg p'(z) | =\bigg|\frac{1}{1-\beta} \arg \bigg(\frac{L(a, c)f(z)}{z}\bigg)+ \arg \bigg(\frac{aL(a+1, c)f(z)}{L(a, c)f(z)}-(a-1)-\beta\bigg) \bigg|\nonumber\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{1}{1-\beta} \bigg|\arg \bigg(\frac{L(a, c)f(z)}{z}\bigg)+ (1-\beta) \arg \bigg(\frac{aL(a+1, c)f(z)}{L(a, c)f(z)}-(a-1)-\beta\bigg) \bigg|.\nonumber $

由引理1.2可以得到, 若满足式子

$\frac{1}{1-\beta} \bigg|\arg \bigg(\frac{L(a, c)f(z)}{z}\bigg)+ (1-\beta) \arg \bigg(\frac{aL(a+1, c)f(z)}{L(a, c)f(z)}-(a-1)-\beta\bigg) \bigg|<\frac{\pi}{2}\delta~~(z\in U), $

则$p(z)\in S^*$.这意味着$L(a, c)f(z)\in S^*(\beta)$.证毕.

在定理2.2中令$\beta=\frac{1}{2}$, 可以得到推论2.4.

推论2.4  若$f(z)\in A$且满足

$\bigg| \arg \bigg(\frac{L(a, c)f(z)}{z}\bigg)+ \frac{1}{2} \arg \bigg(\frac{aL(a+1, c)f(z)}{L(a, c)f(z)}-a+\frac{1}{2}\bigg) \bigg|<\frac{\pi}{4}\delta~~(z\in U), $

则$L(a, c)f(z)\in S^*(\frac{1}{2})$, 其中$\delta=0.6165\cdots$是方程$2\tan^{-1}(1-\delta)+(1-2\delta)\pi=0$的唯一解.

在定理2.2中令$\beta=0$, 可以得到推论2.5.

推论2.5  若$f(z)\in A$且满足

$\bigg| \arg \bigg(\frac{L(a, c)f(z)}{z}\bigg)+ \arg \bigg(\frac{aL(a+1, c)f(z)}{L(a, c)f(z)}-(a-1)\bigg) \bigg|<\frac{\pi}{2}\delta~~(z\in U), $

则$L(a, c)f(z)\in S^*$, 其中$\delta=0.6165\cdots$是方程$2\tan^{-1}(1-\delta)+(1-2\delta)\pi=0$的唯一解.

在推论2.5中令$a=1, c=1$可得推论2.6.

推论2.6  若$f(z)\in A$且满足

$\bigg| \arg \bigg(\frac{L(1, 1)f(z)}{z}\bigg)+ \arg \bigg(\frac{L(2, 1)f(z)}{L(1, 1)f(z)}\bigg) \bigg|<\frac{\pi}{2}\delta~~(z\in U), $

则$L(a, c)f(z)\in S^*$, 其中$\delta=0.6165\cdots$是方程$2\tan^{-1}(1-\delta)+(1-2\delta)\pi=0$的唯一解.

注2.3  在推论2.6中应用$L(1, 1)f(z)=f(z), ~L(2, 1)f(z)=zf'(z)$, 可以得到引理1.2.

注2.4  在定理2.3中令$a=1, ~c=1$可以得到文[3]的定理2.3.

定理2.3  若$f(z)\in A$且满足

$ \;\;\;\;\bigg| \bigg(\frac{[ ~aL(a+1, c)f(z)-(a-1)L(a, c)f(z)~]^{\beta}}{z} \bigg)^{\frac{1}{1-\beta}}\bigg( a(a+1)L(a+2, c)f(z)\\ \;\;\;\;+a (1-\beta-2a)L(a+1, c)f(z)-(a-1)(1-\beta-a)L(a, c)f(z))\bigg) -1+\beta\bigg|\\ <\frac{2}{\sqrt{5}} (1-\beta)~(z\in U), $

其中$0\leq \beta<1$, 则$L(a, c)f(z)\in K(\beta)$.

证  设$f(z)\in A$, 定义$p(z)$如下:

$p(z)= \int_{0}^{z} \bigg( (L(a, c)f(t)) ' \bigg)^{\frac{1}{1-\beta}} dt =z+\frac{a}{c}\cdot\frac{a_{2}}{1-\beta}z^{2}+\cdots\label{b3}.$

(2.3)

令

$ g(z)=zp'(z)= z\bigg(\big(L(a, c)f(z)\big) ' \bigg)^{\frac{1}{1-\beta}} =z+\frac{a}{c}\cdot\frac{2a_{2}}{1-\beta}z^{2}+\cdots, $

则$p(z)\in A$, $g(z)\in A$, 且

$g'(z) =\bigg( \big(L(a, c)f(z)\big) ' \bigg)^{\frac{\beta}{1-\beta}} \bigg( \big(L(a, c)f(z)\big) ' +\frac{1}{1-\beta} z \big(L(a, c)f(z)\big) '' \bigg) \nonumber\\ \;\;\;\;\;\;\;=\frac{1}{1-\beta} \bigg( \frac{(z\big(L(a, c)f(z)\big) ')^{\frac{\beta}{1-\beta}}}{z^{\frac{1}{1-\beta}}} \bigg) \bigg( (1-\beta) z\big(L(a, c)f(z)\big) ' + z ^{2} \big(L(a, c)f(z)\big) '' \bigg).\nonumber $

由式子(1.3) 可以得到

$z^{2}(L(a, c)f(z))''=az(L(a+1, c)f(z))'-az(L(a, c)f(z))' \label{b4}$

(2.4)

和

$z(L(a+1, c)f(z))'=(a+1)L(a+2, c)f(z)-aL(a+1, c)f(z). \label{b5}$

(2.5)

结合式子(1.3), (2.4) 和(2.5) 可得

$ g'(z)=\frac{1}{1-\beta} \bigg( \frac{[ ~aL(a+1, c)f(z)-(a-1)L(a, c)f(z)~]^{\beta}}{z} \bigg)^{\frac{1}{1-\beta}}\bigg( a(a+1)L(a+2, c)f(z)\nonumber\\ \;\;\;\;\;\;\;\;\;\;\;+a (1-\beta-2a)L(a+1, c)f(z) -(a-1)(1-\beta-a)L(a, c)f(z))\bigg).\label{b6} $

(2.6)

因此由定理的条件可以得到

$ |g'(z)-1|=\frac{1}{1-\beta}\bigg |\bigg( \frac{[ ~aL(a+1, c)f(z)-(a-1)L(a, c)f(z)~]^{\beta}}{z} \bigg)^{\frac{1}{1-\beta}}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\bigg( a(a+1)L(a+2, c)f(z)+a (1-\beta-2a)L(a+1, c)f(z)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-(a-1)(1-\beta-a)L(a, c)f(z)) \bigg) -1+\beta\bigg |<\frac{2}{\sqrt{5}}, $

所以$g(z)=zp'(z)\in S^{*}$, 这等价于$p(z)\in K$.

注意到

$\frac{zp''(z)}{p'(z)}=\frac{1}{1-\beta}\frac{z(L(a, c)f(z))''}{(L(a, c)f(z))'}, $

可得

${\rm Re}\bigg(1+ \frac{zp''(z)}{p'(z)}\bigg)={\rm Re}\bigg(1+\frac{1}{1-\beta}\frac{z(L(a, c)f(z))''}{(L(a, c)f(z))'}\bigg)>0~(z\in U), $

所以$L(a, c)f(z)\in K(\beta)$.证毕.

在定理2.3中令$\beta=0$, 可以得到推论2.7.

推论2.7  若$f(z)\in A$且满足

$\bigg| \frac{ a(a+1)L(a+2, c)f(z)+a (1-2a) L(a+1, c)f(z) +(a-1)^{2}L(a, c)f(z) }{z} -1\bigg|<\frac{2}{\sqrt{5}}~(z\in U), $

则$L(a, c)f(z)\in K$.

在推论2.7中取$a=1, c=1$, 可得推论2.8.

推论2.8  若$f(z)\in A$且满足

$\bigg| \frac{ 2L(3, 1)f(z)- L(2, 1)f(z) }{z} -1\bigg|<\frac{2}{\sqrt{5}} ~(z\in U), $

则$L(a, c)f(z)\in K$.

注2.5  在定理2.3中取$a=1, c=1$, 可得到文[3]中的定理3.1.

定理2.4  若$f(z)\in A$且满足

$ \bigg|\arg \bigg(\frac{[ ~aL(a+1, c)f(z)-(a-1)L(a, c)f(z)~]^{\beta}}{z}\bigg)+(1-\beta)\arg\bigg( a(a+1)L(a+2, c)f(z) \\ +a (1-\beta-2a)L(a+1, c)f(z) -(a-1)(1-\beta-a)L(a, c)f(z))\bigg)\bigg|<\frac{\pi}{2}\delta(1-\beta)~(z\in U), $

其中$0\leq \beta<1$, 则$L(a, c)f(z)\in K(\beta)$, 其中$\delta=0.6165\cdots$是方程$2\tan^{-1}(1-\delta)+(1-2\delta)\pi=0$的唯一解.

证  设$p(z)$由式子(2.3) 给出, 并且设$g(z)=zp'(z)$, 则由式子(2.6) 可得

$g'(z)=\frac{1}{1-\beta} \bigg( \frac{[aL(a+1, c)f(z)-(a-1)L(a, c)f(z)~]^{\beta}}{z} \bigg)^{\frac{1}{1-\beta}}\bigg(a(a+1)L(a+2, c)f(z) \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; +a (1-\beta-2a) L(a+1, c)f(z) -(a-1)(1-\beta-a)L(a, c)f(z))\bigg). $

应用引理1.2, 可以得到

$ |\arg g'(z)|= \bigg| \frac{1}{1-\beta}\arg \bigg( \frac{[ ~aL(a+1, c)f(z)-(a-1)L(a, c)f(z)~]^{\beta}}{z}\bigg)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\arg\bigg(a(a+1)L(a+2, c)f(z) +a (1-\beta-2a) L(a+1, c)f(z)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-(a-1)(1-\beta-a)L(a, c)f(z))\bigg)\bigg|<\frac{\pi}{2}\delta, $

其中$z\in U, $所以$g(z)\in S^{*}$, 这等价于$p(z)\in K$, 即$L(a, c)f(z)\in K(\beta)$.

注2.6  在定理2.4中取$a=1, ~c=1$, 可以得到文[3]中的定理3.3.

定理2.5  若$f(z)\in A$且满足

$\bigg|\frac{aL(a, c)f(z)}{z^{2}}\bigg(\frac{L(a+1, c)f(z)}{L(a, c)f(z)}-1\bigg) \bigg|<\frac{1}{\sqrt{5}}=0.4472\cdots(z\in U), $

则$L(a, c)f(z)\in S^*$.

证  设$f(z)\in A$, 定义

$g(z)= \int_{0}^{z}\frac{L(a, c)f(t) }{t}dt=z+\sum\limits_{n=2}^{\infty}\frac{(a)_{n-1}}{(c)_{n-1}}\frac{a_{n}}{n}z^{n}, $

则经过计算可得

$g''(z)=\frac{L(a, c)f(z)}{z^{2}}\bigg(\frac{z(L(a, c)f(z))'}{L(a, c)f(z)}-1\bigg).$

结合式子(1.3) 可得

$|g''(z)|=\bigg|\frac{aL(a, c)f(z)}{z^{2}}\bigg(\frac{L(a+1, c)f(z)}{L(a, c)f(z)}-1\bigg) \bigg|<\frac{1}{\sqrt{5}}=0.4472\cdots(z\in U).$

应用引理1.3可得$g(z)\in K$.

注意到

${\rm Re} \bigg(1+\frac{zg''(z)}{g'(z)}\bigg) ={\rm Re}\bigg(\frac{zf'(z)}{f(z)}\bigg) >0~(z\in U), $

所以$L(a, c)f(z)\in S^*$.

注2.7  在定理2.5中取$a=1, ~c=1$可以得到文[3]中的定理4.1.