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数学物理学报, 2019, 39(1): 156-164 doi:

论文

伯努利双纽线区域内复阶解析函数类的优化问题

汤获,, 李书海,, 牛潇萌,, 马丽娜,

Majorization Problems for the Classes of Analytic Functions of Complex Order on the Domains of Lemniscate of Bernoulli

Tang Huo,, Li Shuhai,, Niu Xiaomeng,, Ma Lina,

通讯作者: 汤获, E-mail: thth2009@163.com

收稿日期: 2018-02-1  

基金资助: 国家自然科学基金.  11561001
国家自然科学基金.  11271045
内蒙古高校青年科技英才支持计划项目.  NJYT-18-A14
内蒙古自然科学基金.  2018MS01026
内蒙古自然科学基金.  2014MS0101
内蒙古自然科学基金.  2017MS0113
内蒙古高校科研基金.  NJZY17300
内蒙古高校科研基金.  NJZY18217
赤峰市自然科学研究课题

Received: 2018-02-1  

Fund supported: the NSFC.  11561001
the NSFC.  11271045
the Program for Young Talents of Science and Technology in Universities of Inner Mongolia Autonomous Region.  NJYT-18-A14
the Natural Science Foundation of Inner Mongolia.  2018MS01026
the Natural Science Foundation of Inner Mongolia.  2014MS0101
the Natural Science Foundation of Inner Mongolia.  2017MS0113
the Higher School Foundation of Inner Mongolia.  NJZY17300
the Higher School Foundation of Inner Mongolia.  NJZY18217
the Natural Science Foundation of Chifeng

作者简介 About authors

李书海,E-mail:lishms66@163.com , E-mail:lishms66@163.com

牛潇萌,E-mail:ndnxm@126.com , E-mail:ndnxm@126.com

马丽娜,E-mail:malina00@163.com , E-mail:malina00@163.com

摘要

该文利用Salagean算子和从属关系分别引入了伯努利双纽线左、右半有界区域内复阶解析函数类LnγRnbac),研究了该函数类的优化问题,并得到了一些有趣的推论.

关键词: 解析函数 ; 伯努利双纽线 ; Salagean算子 ; 从属关系 ; 优化

Abstract

In this paper, we introduce the classes Lnγ and Rnb(a, c) of analytic functions on the left-half and right-half bounded domains of lemniscate of Bernoulli, which are defined by using the Salagean operator and subordination relationship, and investigate majorization problems for functions belonging to these classes. Some interesting corollaries of our main results are also considered.

Keywords: Analytic function ; Lemniscate of Bernoulli ; Salagean operator ; Subordination relationship ; Majorization

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本文引用格式

汤获, 李书海, 牛潇萌, 马丽娜. 伯努利双纽线区域内复阶解析函数类的优化问题. 数学物理学报[J], 2019, 39(1): 156-164 doi:

Tang Huo, Li Shuhai, Niu Xiaomeng, Ma Lina. Majorization Problems for the Classes of Analytic Functions of Complex Order on the Domains of Lemniscate of Bernoulli. Acta Mathematica Scientia[J], 2019, 39(1): 156-164 doi:

1 引言

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

f(z)=z+k=2akzk
(1.1)

的函数类.

1967年, Macgregor[1]给出了优化的定义.

定义1.1  设函数fgU内解析.若存在U内的解析函数φ(z),使得

|φ(z)|1      f(z)=φ(z)g(z)  (zU),

则称函数fU内优于g,记作f(z)g(z)  (zU).

1970年, Roberston[2]引入了拟从属的概念.

定义1.2  设函数fgU内解析.若存在U内的解析函数φ(z),使得f(z)φ(z)U内解析且

|φ(z)|1      |ω(z)||z|<1  (zU),

满足

f(z)=φ(z)g(ω(z))  (zU),

则称函数fU内拟从属于g,记作f(z)qg(z)  (zU).我们注意到,当φ(z)=1时, f(z)=g(ω(z))  (zU),此时称函数fU内从属于g,记作f(z)g(z)  (zU) (参见文献[3]);当ω(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Dn:AA,其定义为(简称Salagean算子)

D0f(z)=f(z),

D1f(z)=Df(z)=zf(z),

Dnf(z)=D(Dn1f(z))  (nN0={0}{1,2,}).
(1.2)

fA,则由(1.1)式和(1.2)式,我们易得

Dnf(z)=z+k=2knakzk.

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LγnRbn(a,c).

定义1.3  设函数fA,则f属于伯努利双纽线左半有界区域内的复γ阶解析函数类Lγn当且仅当

1γ{Dn+1f(z)Dnf(z)1}(21)(11z1+2(21)z),
(1.3)

这里nN0, γC{0}, zU.

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

(1)若取γ=1,则

L1n:=Ln={fA:Dn+1f(z)Dnf(z)2(21)1z1+2(21)z  (nN0)}.

(2)若取γ=1n=0,则

L10:=L0={fA:zf(z)f(z)2(21)1z1+2(21)z  (zU)}[5],

蕴含zf(z)f(z)位于满足{wC:Rew>0, |(w2)21|<1}的伯努利双纽线左半有界区域内(如图 1).

图 1


(3)若取γ=n=1,则

L11:=L1={fA:zf

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).

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

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=1n=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=1n=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).

图 3


(5)若取a=2b=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).

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}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.

2 函数类L_n^{\gamma}R_n^{b}(a, c)的优化问题

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\}|\gamma|\leq\sqrt{2}+1,函数f\in{\cal A}, g\in L_n^{\gamma}.D^{n+1}f(z)\ll D^{n}g(z)(z\in{\Bbb U}, ~n\in{\Bbb N}_0),

\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

\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}
(2.1)

  由于g\in L_n^{\gamma},故由从属关系和(1.3)式,有

\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),
(2.2)

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

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

的解析函数类[18].

根据(2.2)式和(2.3)式,可得

\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|.
(2.4)

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

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

又注意到\varphi\in{\cal P}满足不等式[19]

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

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|\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.

\psi_1(0)=1>0, ~\psi_1(1)=-2<0,故存在r_1,使得当r\in[0, r_1]\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)2c|b|^a\leq(|b|+1)^a.D^{n+1}f(z)\ll D^{n}g(z)(z\in{\Bbb U}, ~n\in{\Bbb N}_0),

\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

[(|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}).
(2.7)

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

\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}),
(2.8)

这里\omega(z)如(2.3)式所述.

利用(2.3)式和(2.8)式,易有

\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|.
(2.9)

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|.

|z|=r|\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.

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.

3 一些推论

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

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

\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

(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.
(3.1)

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

\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}g\in L_1.D^{2}f(z)\ll Dg(z)~~(z\in{\Bbb U}),

\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)c\leq2^{(a-1)}.D^{n+1}f(z)\ll D^{n}g(z)(z\in{\Bbb U}, ~n\in{\Bbb N}_0),

\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

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

在推论3.4中,若分别取n=0n=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)c\leq2^{(a-1)}.zf'(z)\ll g(z)(z\in{\Bbb U}),

\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)c\leq2^{(a-1)}.D^{2}f(z)\ll Dg(z)~~(z\in{\Bbb U}),

\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=2c=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.zf'(z)\ll g(z)~~(z\in{\Bbb U}),

\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

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

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

\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.

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