设${\Bbb C}$表示复平面, ${\cal A}$表示在单位圆盘${\Bbb U}=\{z\in{\Bbb C}:|z|<1\}$内解析且具有如下形式

(1.1) $f(z)=z+\sum\limits_{k=2}^\infty a_kz^k $

的函数类.

1967年, Macgregor^[1]给出了优化的定义.

定义1.1  设函数$f$和$g$在${\Bbb U}$内解析.若存在${\Bbb U}$内的解析函数$\varphi(z)$,使得

$|\varphi(z)|\leq1~~~\textrm{和}~~~f(z)=\varphi(z)g(z)~~(z\in{\Bbb U}), $

则称函数$f$在${\Bbb U}$内优于$g$,记作$f(z)\ll g(z)~~(z\in{\Bbb U}).$

1970年, Roberston^[2]引入了拟从属的概念.

定义1.2  设函数$f$和$g$在${\Bbb U}$内解析.若存在${\Bbb U}$内的解析函数$\varphi(z)$,使得$\frac{f(z)}{\varphi(z)}$在${\Bbb U}$内解析且

$|\varphi(z)|\leq1~~~\textrm{和}~~~|\omega(z)|\leq|z|<1~~(z\in{\Bbb U}), $

满足

$f(z)=\varphi(z)g(\omega(z))~~(z\in{\Bbb U}), $

则称函数$f$在${\Bbb U}$内拟从属于$g$,记作$f(z)\prec_q g(z)~~(z\in{\Bbb U}).$我们注意到,当$\varphi(z)=1$时, $f(z)=g(\omega(z))~~(z\in{\Bbb U})$,此时称函数$f$在${\Bbb U}$内从属于$g$,记作$f(z)\prec g(z)~~(z\in{\Bbb U})$ (参见文献[3]);当$\omega(z)=z$时,拟从属关系即为上述优化关系.因此,从属关系和优化关系是拟从属关系的特殊情形.

1981年, Salagean^[4]引入算子$D^n: {\cal A}\rightarrow{\cal A}$,其定义为(简称Salagean算子)

$D^0f(z)=f(z), $

$D^1f(z)=Df(z)=zf'(z), $

$\vdots $

(1.2) $D^nf(z)=D(D^{n-1}f(z))~~(n\in{\Bbb N}_0=\{0\}\cup\{1, 2, \cdots\}).$

若$f\in{\cal A}$,则由(1.1)式和(1.2)式,我们易得

$D^nf(z)=z+\sum\limits_{k=2}^\infty k^n a_kz^k.$

利用上述Salagean算子和从属关系,我们定义解析函数的两个新子类$L_n^{\gamma}$和$R_n^{b}(a, c)$.

定义1.3  设函数$f\in{\cal A}$,则$f$属于伯努利双纽线左半有界区域内的复$\gamma$阶解析函数类$L_n^{\gamma}$当且仅当

(1.3) $\frac{1}{\gamma}\left\{\frac{D^{n+1}f(z)}{D^{n}f(z)}-1\right\}\prec(\sqrt{2}-1)\left(1- \sqrt{\frac{1-z}{1+2(\sqrt{2}-1)z}}\right), $

这里$n\in{\Bbb N}_0$, $\gamma\in{\Bbb C}\setminus\{0\}$, $z\in {\Bbb U}$.

注1.1  在定义1.3中,若参数$n, ~\gamma$取不同的值,则可得到如下一些函数类:

(1)若取$\gamma=1$,则

$L_n^{1}:=L_n=\left\{f\in {\cal A}:\frac{D^{n+1}f(z)}{D^nf(z)}\prec \sqrt{2}-(\sqrt{2}-1)\sqrt{\frac{1-z}{1+2(\sqrt{2}-1)z}}~~(n\in{\Bbb N}_0)\right\}.$

(2)若取$\gamma=1$和$n=0$,则

$L_0^{1}:=L_0=\left\{f\in {\cal A}:\frac{zf'(z)}{f(z)}\prec\sqrt{2}-(\sqrt{2}-1)\sqrt{\frac{1-z}{1+2(\sqrt{2}-1)z}}~~(z\in {\Bbb U})\right\}^{[5]}, $

蕴含$\frac{zf'(z)}{f(z)}$位于满足$\{w\in{\Bbb C}: {\rm Re} w>0, ~|(w-\sqrt{2})^{2}-1|<1\}$的伯努利双纽线左半有界区域内(如图 1).

(3)若取$\gamma=n=1$,则

$L_1^{1}:=L_1=\left\{f\in {\cal A}:\frac{zf''(z)}{f'(z)}\prec(\sqrt{2}-1)\left(1-\sqrt{\frac{1-z}{1+2(\sqrt{2}-1)z}}\right)~~(z\in {\Bbb U})\right\}, $

蕴含$1+\frac{zf''(z)}{f'(z)}$位于满足$\{w\in{\Bbb C}: {\rm Re} w>0, ~|(w-\sqrt{2})^{2}-1|<1\}$的伯努利双纽线左半有界区域内(如图 2).

定义1.4  设函数$f\in{\cal A}$,则$f$属于伯努利双纽线右半有界区域内复$b$阶解析函数类$R_n^{b}(a, c)$当且仅当

(1.4) $1+\frac{1}{b}\left(\frac{D^{n+1}f(z)}{D^nf(z)}-1\right)\prec\sqrt[a]{c(1+z)}, $

这里$n\in{\Bbb N}_0, a\geq1, b\in{\Bbb C}\setminus\{0\}, c\geq\frac{1}{2}, z\in {\Bbb U}$.

注1.2  在定义1.4中,若参数$n, ~a, ~b, ~c$取不同的值,则也可得如下一些函数类

(1)若取$b=1$,则

$R_n^{1}(a, c):=R_n(a, c)=\left\{f\in {\cal A}:\frac{D^{n+1}f(z)}{D^nf(z)}\prec\sqrt[a]{c(1+z)}~~(n\in{\Bbb N}_0, a\geq1, c\geq\frac{1}{2})\right\}.$

(2)若取$b=1$和$n=0$,则

$R_0^{1}(a, c):=R_0(a, c)=\left\{f\in {\cal A}:\frac{zf'(z)}{f(z)}\prec\sqrt[a]{c(1+z)}~~(a\geq1, c\geq\frac{1}{2})\right\}^{[6]}, $

蕴含$\frac{zf'(z)}{f(z)}$位于满足$\{w\in{\Bbb C}: {\rm Re} w>0, ~~|w^{a}-c|<c\}$的伯努利双纽线右半有界区域内.

(3)若取$b=n=1$,则

$R_1^{1}(a, c):=R_1(a, c)=\left\{f\in {\cal A}:1+\frac{zf''(z)}{f'(z)}\prec\sqrt[a]{c(1+z)}~~(a\geq1, c\geq\frac{1}{2})\right\}, $

蕴含$1+\frac{zf''(z)}{f'(z)}$位于满足$\{w\in{\Bbb C}: {\rm Re} w>0, ~~|w^{a}-c|<c\}$的伯努利双纽线右半有界区域内.

(4)若取$a=2$, $b=c=1$和$n=0$,则

$R_0^{1}(2, 1):=R_0=\left\{f\in {\cal A}:\frac{zf'(z)}{f(z)}\prec\sqrt{1+z}~~(z\in {\Bbb U})\right\}^{[7]}, $

蕴含$\frac{zf'(z)}{f(z)}$位于满足$\{w\in{\Bbb C}: {\rm Re} w>0, ~~|w^{2}-1|<1\}$的伯努利双纽线右半有界区域内(如图 3).

(5)若取$a=2$和$b=c=n=1$,则

$R_1^{1}(2, 1):=R_1=\left\{f\in {\cal A}:1+\frac{zf''(z)}{f'(z)}\prec\sqrt{1+z}~~(z\in {\Bbb U})\right\}, $

蕴含$1+\frac{zf''(z)}{f'(z)}$位于满足$\{w\in{\Bbb C}: {\rm Re} w>0, ~~|w^{2}-1|<1\}$的伯努利双纽线右半有界区域内(如图 4).

1967年, MacGregor^[1]最早开始研究星象函数类的优化问题.随后, 2001年, Altintas等人^{[8, 9]}研究了复阶单(多)叶星象和凸像函数类的优化问题.近年来,许多中外学者对由各种算子定义的不同单(多)叶解析函数类的优化问题作了大量研究,如Goyal等人^{[10, 11]}, Goswami等人^[12], Prajapat等人^[13],汤获等人^[14-16].最近,汤获等人^[17]又讨论了多叶亚纯解析函数类的优化问题.然而,我们发现,很少有人研究伯努利双纽线区域内解析函数类的优化问题.在上述工作的基础上,本文主要研究伯努利双纽线左、右半有界区域内复阶解析函数类$L_n^{\gamma}$和$R_n^{b}(a, c)$的优化问题,所得结果扩充了单复变几何函数论的优化理论.

首先,我们给出并证明函数类$L_n^{\gamma}$的优化性质.

定理2.1  设$\gamma\in{\Bbb C}\setminus\{0\}$且$|\gamma|\leq\sqrt{2}+1$,函数$f\in{\cal A}$, $g\in L_n^{\gamma}$.若$D^{n+1}f(z)\ll D^{n}g(z)$$(z\in{\Bbb U}, ~n\in{\Bbb N}_0), $则

$\left|D^{n+2}f(z)\right|\leq\left|D^{n+1}g(z)\right|~~(|z|\leq r_1), $

其中$r_1=r_1(\gamma)$是如下方程的最小正根

(2.1) $\begin{eqnarray}&&[(2\sqrt{2}-1)|\gamma|^2-4|\gamma|+2(\sqrt{2}+1)]r^5 -[|\gamma|^2+8|\gamma|-8(\sqrt{2}+1)]r^4\\&&+2[(1-2\sqrt{2})|\gamma|^2+4|\gamma|+2(2\sqrt{2}-1)]r^3 +2[2(\sqrt{2}+3)|\gamma|+(1-2\sqrt{2})]r^2\\&&+[(3-2\sqrt{2})|\gamma|^2+4(2+\sqrt{2})|\gamma|-2(5\sqrt{2}+7)]r -[2|\gamma|^2-2(\sqrt{2}+1)|\gamma|\\&&+3+2\sqrt{2}]=0.\end{eqnarray}$

证  由于$g\in L_n^{\gamma}$,故由从属关系和(1.3)式,有

(2.2) $\frac{1}{\gamma}\left\{\frac{D^{n+1}g(z)}{D^{n}g(z)}-1\right\}=(\sqrt{2}-1)\left(1-\sqrt{\frac{1-\omega(z)}{1+2(\sqrt{2}-1)\omega(z)}}\right), $

其中$\omega(z)=c_1z+c_2z^2+\cdots\in{\cal P}$, ${\cal P}$表示在${\Bbb U}$内有界且满足条件

(2.3) $\omega(0)=0 ~~~\textrm{和}~~~|\omega(z)|\leq|z|~~~(z\in{\Bbb U})$

的解析函数类^[18].

根据(2.2)式和(2.3)式,可得

(2.4) $\left|D^{n}g(z)\right|\leq\frac{1}{1-|\gamma|(\sqrt{2}-1)\left(1-\sqrt{\frac{1-|z|}{1+2(\sqrt{2}-1)|z|}}\right)}\left|D^{n+1}g(z)\right|. $

又$D^{n+1}f(z)\ll D^{n}g(z)$,故由定义1.1可知

$D^{n+1}f(z)=\varphi(z)D^{n}g(z).$

对上式两边关于$z$求导,并乘以$z$,可得

(2.5) $D^{n+2}f(z)=z\varphi'(z)D^{n}g(z)+\varphi(z)D^{n+1}g(z).$

又注意到$\varphi\in{\cal P}$满足不等式^[19]

(2.6) $|\varphi'(z)|\leq\frac{1-|\varphi(z)|^2}{1-|z|^2}~~(z\in{\Bbb U}).\$

故将(2.4)式和(2.6)式代入(2.5)式,有

$\left|D^{n+2}f(z)\right|\leq\left(|\varphi(z)|+\frac{|z|(1-|\varphi(z)|^2)}{1-|z|^2}\cdot\frac{1}{1-|\gamma|(\sqrt{2}-1)\left(1-\sqrt{\frac{1-|z|}{1+2(\sqrt{2}-1)|z|}}\right)}\right)\left|D^{n+1}g(z)\right|.$

若取$|z|=r$和$|\varphi(z)|=\rho~~(0\leq\rho\leq1)$,则上式可变为

$\left|D^{n+2}f(z)\right|\leq\Phi_1(r, \rho)\left|D^{n+1}g(z)\right|, $

其中

$\Phi_1(r, \rho)=\frac{r(1-\rho^2)}{(1-r^2)\left[1-|\gamma|(\sqrt{2}-1)\left(1-\sqrt{\frac{1-r}{1+2(\sqrt{2}-1)r}}\right)\right]}+\rho.$

要确定$r_1$,我们只需取

$\begin{eqnarray*}r_1&=&\max\left\{r\in[0, 1): \Phi_1(r, \rho)\leq1, \forall~\rho\in[0, 1]\right\}\\&=&\max\left\{r\in[0, 1): \Psi_1(r, \rho)\geq0, \forall~\rho\in[0, 1]\right\}, \end{eqnarray*}$

其中

$\Psi_1(r, \rho)=(1-r^2)\left[1-|\gamma|(\sqrt{2}-1)\left(1-\sqrt{\frac{1-r}{1+2(\sqrt{2}-1)r}}\right)\right]-r(1+\rho).$

显然,当$\rho=1$时, $\Psi_1(r, \rho)$取得最小值,即

$\min\left\{\Psi_1(r, \rho): \rho\in[0, 1]\right\}=\Psi_1(r, 1):=\psi_1(r), $

这里

$\psi_1(r)=(1-r^2)\left[1-|\gamma|(\sqrt{2}-1)\left(1-\sqrt{\frac{1-r}{1+2(\sqrt{2}-1)r}}\right)\right]-2r.$

又$\psi_1(0)=1>0, ~\psi_1(1)=-2<0$,故存在$r_1$,使得当$r\in[0, r_1]$时$\psi_1(r)\geq0$成立,其中$r_1=r_1(\gamma)$为方程(2.1)的最小正根.定理2.1得证.

下面,我们讨论函数类$R_n^{b}(a, c)$的优化性质.

定理2.2  设函数$f\in{\cal A}$, $g\in R_n^{b}(a, c)$且$2c|b|^a\leq(|b|+1)^a$.若$D^{n+1}f(z)\ll D^{n}g(z)$$(z\in{\Bbb U}, ~n\in{\Bbb N}_0), $则

$\left|D^{n+2}f(z)\right|\leq\left|D^{n+1}g(z)\right|~~(|z|\leq r_2), $

其中$r_2=r_2(a, b, c)$是如下方程的最小正根

(2.7) $[(|b|+1)r^2+2r-(|b|+1)]^a=c(1+r)|b|^a(r^2-1)^a~~(a\geq1, b\in{\Bbb C}\setminus\{0\}, c\geq\frac{1}{2}).$

证  因为$g\in R_n^{b}(a, c)$,故由从属关系和(1.4)式,可得

(2.8) $\frac{D^{n+1}g(z)}{D^ng(z)}=1+b\left[\sqrt[a]{c(1+\omega(z))}-1\right]~~(n\in{\Bbb N}_0, a\geq1, b\in{\Bbb C}\setminus\{0\}, c\geq\frac{1}{2}), $

这里$\omega(z)$如(2.3)式所述.

利用(2.3)式和(2.8)式,易有

(2.9) $\left|D^{n}g(z)\right|\leq\frac{1}{1-|b|\left[\sqrt[a]{c(1+|z|)}-1\right]}\left|D^{n+1}g(z)\right|. $

再将(2.6)式和(2.9)式代入(2.5)式,类似于定理2.1的证明,易得

$\left|D^{n+2}f(z)\right|\leq\left(|\varphi(z)|+\frac{|z|(1-|\varphi(z)|^2)}{1-|z|^2}\cdot\frac{1}{1-|b|\left[\sqrt[a]{c(1+|z|)}-1\right]}\right)\left|D^{n+1}g(z)\right|.$

令$|z|=r$和$|\varphi(z)|=\rho~~(0\leq\rho\leq1)$,则上式可写为

$\left|D^{n+2}f(z)\right|\leq\frac{\Phi_2(r, \rho)}{(1-r^2)\left\{1-|b|\left[\sqrt[a]{c(1+r)}-1\right]\right\}}\left|D^{n+1}g(z)\right|, $

其中

$\Phi_2(r, \rho)=-r\rho^2+(1-r^2)\left\{1-|b|\left[\sqrt[a]{c(1+r)}-1\right]\right\}\rho+r.$

当$r\leq r_2$时,函数$\Phi_2(r, \rho)$关于$\rho~(0\leq\rho\leq1)$递增且在$\rho=1$处取得最大值,这里$r_2=r_2(a, b, c)$由(2.7)式给出.于是,有

$\Phi_2(r, \rho)\leq\Phi_2(r, 1):=\Phi_2(r)=(1-r^2)\left\{1-|b|\left[\sqrt[a]{c(1+r)}-1\right]\right\}~~(r\leq r_2), $

即有

$\left|D^{n+2}f(z)\right|\leq\left|D^{n+1}g(z)\right|.$

定理2.2得证.

在定理2.1中,若分别取$\gamma=1$, $\gamma-1=n=0$和$\gamma=n=1$,则可得如下推论.

推论3.1  设函数$f\in{\cal A}$和$g\in L_n$.若$D^{n+1}f(z)\ll D^{n}g(z)~~(z\in{\Bbb U}, ~n\in{\Bbb N}_0), $则

$\left|D^{n+2}f(z)\right|\leq\left|D^{n+1}g(z)\right|~~(|z|\leq r_3), $

其中$r_3=r_1(1)$是如下方程的最小正根

(3.1) $(4\sqrt{2}-3)r^5+(8\sqrt{2}-1)r^4+\\2(2\sqrt{2}+7)r^3+2(2\sqrt{2}+7)r^2-(3+8\sqrt{2})r-3=0.$

推论3.2  设函数$f\in{\cal A}$和$g\in L_0$.若$zf'(z)\ll g(z)~~(z\in{\Bbb U}), $则

$\left|f'(z)+zf''(z)\right|\leq\left|g'(z)\right|~~(|z|\leq r_3), $

其中$r_3$由(3.1)式给出.

推论3.3  设函数$f\in{\cal A}$和$g\in L_1$.若$D^{2}f(z)\ll Dg(z)~~(z\in{\Bbb U}), $则

$\left|D^{3}f(z)\right|\leq\left|D^{2}g(z)\right|~~(|z|\leq r_3), $

其中$r_3$由(3.1)式给出.

在定理2.2中,若取$b=1$,则有如下推论3.4.

推论3.4  设函数$f\in{\cal A}$, $g\in R_n(a, c)$且$c\leq2^{(a-1)}$.若$D^{n+1}f(z)\ll D^{n}g(z)$$(z\in{\Bbb U}, ~n\in{\Bbb N}_0), $则

$\left|D^{n+2}f(z)\right|\leq\left|D^{n+1}g(z)\right|~~(|z|\leq r_4), $

其中$r_4=r_2(a, 1, c)$是如下方程的最小正根

(3.2) $2^a(r^2+r-1)^a=c(1+r)(r^2-1)^a~~(a\geq1, c\geq\frac{1}{2}).$

在推论3.4中,若分别取$n=0$和$n=1$,则可得如下推论3.5和推论3.6.

推论3.5  设函数$f\in{\cal A}$, $g\in R_0(a, c)$且$c\leq2^{(a-1)}$.若$zf'(z)\ll g(z)$$(z\in{\Bbb U}), $则

$\left|f'(z)+zf''(z)\right|\leq\left|g'(z)\right|~~(|z|\leq r_4), $

其中$r_4$由(3.2)式给出.

推论3.6  设函数$f\in{\cal A}$, $g\in R_1(a, c)$且$c\leq2^{(a-1)}$.若$D^{2}f(z)\ll Dg(z)~~(z\in{\Bbb U}), $则

$\left|D^{3}f(z)\right|\leq\left|D^{2}g(z)\right|~~(|z|\leq r_4), $

其中$r_4$由(3.2)式给出.

进一步,在推论3.5和推论3.6中,若分别取$a=2$和$c=1$,则可得如下推论3.7和推论3.8.

推论3.7  设函数$f\in{\cal A}$, $g\in R_0$.若$zf'(z)\ll g(z)~~(z\in{\Bbb U}), $则

$\left|f'(z)+zf''(z)\right|\leq\left|g'(z)\right|~~(|z|\leq r_5), $

其中$r_5=r_2(2, 1, 1)$是如下方程的最小正根

(3.3) $r^5-3r^4-10r^3+2r^2+9r-3=0.$

推论3.8  设函数$f\in{\cal A}$, $g\in R_1$.若$D^{2}f(z)\ll Dg(z)$$(z\in{\Bbb U}), $则

$\left|D^{3}f(z)\right|\leq\left|D^{2}g(z)\right|~~(|z|\leq r_5), $

其中$r_5$由(3.3)式给出.