The Fourth-order Hankel Determinant for Certain Subclasses of Star-like Functions Subordinate to Exponential Function
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摘要: 设
A 表示单位圆盘D={z∈C ∶ |z| < 1}内解析且具有如下形式 f(z)=z+∞∑n=2anzn 的函数族. 文章研究了在单位圆盘D上与指数函数有关的解析函数类Se*: S∗e={f∣zf′(z)f(z)≺ez(f∈A,z∈D)} 的四阶Hankel行列式H4(1), 得到其上界估计.Abstract: LetA be a family of analytic functions with are the form f(z)=z+∞∑n=2anzn on the open unit disk D. A class of analytic functions Se* which are defined on the open unit cicle D and associated with exponential function is introduced, that is S∗e={f∣zf′(z)f(z)≺ez(f∈A,z∈D)}. And the upper bound of the fourth-order Hankel determinant H4(1) for this function class Se* associated with exponential function is given.-
Keywords:
- analytic function /
- Hankel determinant /
- exponential function /
- upper bound
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设$\mathcal{A} $表示单位圆盘D={z∈${\mathbb{C}} $: |z| < 1}内解析且具有如下形式
$$ f(z)=z+\sum\limits_{n=2}^{\infty} a_{n} z^{n} $$ (1) 的函数族.
设$\mathcal{P} $表示单位圆盘D内满足条件p(0)=0及Re p(z)>0的函数p组成的解析函数族.
1992年, MA和MINDA[1]引入某类星像函数类S*(ϕ):
$$ f \in S^{*}(\phi) \Leftrightarrow \frac{z f^{\prime}(z)}{f(z)}\prec\phi(z), $$ 其中ϕ ∈$\mathcal{P} $. 该函数类将单位圆盘映射到关于实轴对称、关于ϕ(0)=1星像且满足ϕ′(0)>0的星像区域.
1996年, SOKÓȽ和STANKIEWICZ[2]引入函数类S*($ \sqrt{1+z}$), 该函数类将单位圆盘映射为伯努利双纽线{w∈${\mathbb{C}} $: |w2-1| < 1}的右半部分星像区域.
2014年, MENDIRATTA等[3]引入函数类S*$(\sqrt{2}- (\sqrt{2}-1) \sqrt{(1-z) /(1+2(\sqrt{2}-1)}))$, 该函数类将单位圆盘映射为伯努利双纽线{w ∈${\mathbb{C}} $: |w2-1| < 1}的左半部分星像区域.
2015年, MENDIRATTA等[4]引入了与指数函数有关的函数类Se*, 该函数类将单位圆盘映射到关于实轴对称、关于1星像的右半平面区域.
另一方面, 对于不同解析函数类的Hankel行列式研究一直是热点问题之一. 1966年, POMMERENKE[5]定义了解析函数f的q阶Hankel行列式Hq(n). 很明显, 当q=2, n=1时, |H2(1)|即是Fekete-Szegö泛函[6-11]. 近年来, 许多学者对各类解析函数的二、三阶Hankel行列式做了大量研究[12-24]. 如: 研究了与指数函数有关的星像函数类的二、三阶Hangkel行列式[12-14]; 研究了有界转动、星像和凸像函数类的三阶Hankel行列式[15]; 研究了近于凸函数类的三阶Hankel行列式[17].
但是, 目前对于与指数函数有关的函数类的Hankel行列式的研究都仅基于二阶和三阶的情形, 而对四阶Hankel行列式的研究还不多见. 基于以上启发, 本文主要研究了与指数函数有关的星像函数类的四阶Hankel行列式H4(1), 得到其上界估计.
1. 预备知识
定义1[25] 设函数f和g在单位圆盘D内解析. 如果存在D内的Schwarz函数ω, 满足: ω(0)=0, |ω(z)| < 1且f(z)=g(ω(z)), 则称f从属于g, 记为f$ \prec $g. 特别地, 如果g在D上是单叶的, 则
$$ f\prec g \quad(z \in D) \Leftrightarrow f(0)=g(0), f(D) \subset g(D) . $$ 设函数f(z)∈$\mathcal{A} $, 若满足条件: Re[zf′(z)/f(z)]>0, 则称f属于星像函数类, 记为f∈S*. 显然, f∈S*将单位圆盘映射到右半平面且星像的区域[1].
定义2[4] 设Se*表示单位圆盘D={z: |z| < 1} 内满足
$$ \frac{z f^{\prime}(z)}{f(z)}\prec\mathrm{e}^{z} \quad(z \in D, f \in \mathcal{A}) $$ (2) 的解析函数类的全体. 实际上, f∈Se*当且仅当
$$ \left|\log \frac{z f^{\prime}(z)}{f(z)}\right|<1 \quad(z \in D). $$ 定义3[5] 设函数f∈$\mathcal{S} $,
$$ H_{q}(n)=\left|\begin{array}{cccc} a_{n} & a_{n+1} & \cdots & a_{n+q-1} \\ a_{n+1} & a_{n+2} & \cdots & a_{n+q} \\ \vdots & \vdots & & \vdots \\ a_{n+q-1} & a_{n+q} & \cdots & a_{n+2 q-2} \end{array}\right| \quad\left(a_{1}=1\right), $$ 其中, a1=1, n≥1, q≥1. 特别地, 有
$$ H_{4}(1)=\left|\begin{array}{cccc} a_{1} & a_{2} & a_{3} & a_{4} \\ a_{2} & a_{3} & a_{4} & a_{5} \\ a_{3} & a_{4} & a_{5} & a_{6} \\ a_{4} & a_{5} & a_{6} & a_{7} \end{array}\right| \quad(n=1, q=4) . $$ 下面给出本文所需用的引理.
引理1[26] 如果$p(z)=1+\sum\limits_{k=1}^{\infty} c_{k} z^{k} \in \mathcal{P} $, 则存在复数x和z，且|x|≤1, |z|≤1, 使得
$$ 2 c_{2}=c_{1}^{2}+x\left(4-c_{1}^{2}\right) $$ 及
$$ 4 c_{3}=c_{1}^{3}+2 c_{1} x\left(4-c_{1}^{2}\right)-\left(4-c_{1}^{2}\right) c_{1} x^{2}+2\left(4-c_{1}^{2}\right)\left(1-|x|^{2}\right) z. $$ 引理2[27] 如果$p(z)=1+\sum\limits_{k=1}^{\infty} c_{k} z^{k} \in \mathcal{P} $, 则
$$\begin{array}{c} \left|c_{1}^{4}+c_{2}^{2}+2 c_{1} c_{3}-3 c_{1}^{2} c_{2}-c_{4}\right| \leqslant 2,\\\left|c_{1}^{5}+3 c_{1} c_{2}^{2}+3 c_{1}^{2} c_{3}-4 c_{1}^{3} c_{2}-2 c_{1} c_{4}-2 c_{2} c_{3}+c_{5}\right| \leqslant 2, \end{array} \\ \mid c_{1}^{6}+6 c_{1}^{2} c_{2}^{2}+4 c_{1}^{3} c_{3}+2 c_{1} c_{5}+2 c_{2} c_{4}+c_{3}^{2}-c_{2}^{3}-5 c_{1}^{4} c_{2}- \\ \ \ \ \ \ \ \ \ 3 c_{1}^{2} c_{4}-6 c_{1} c_{2} c_{3}-c_{6} \mid \leqslant 2 $$ 及|cn|≤2 (n=1, 2, …).
引理3[28] 如果$p(z)=1+\sum\limits_{k=1}^{\infty} c_{k} z^{k} \in \mathcal{P} $, 那么对0≤μ≤1, 有
$$\begin{array}{c} \left|c_{2}-\frac{c_{1}^{2}}{2}\right| \leqslant 2-\frac{\left|c_{1}\right|^{2}}{2},\\\left|c_{n+k}-\mu c_{n} c_{k}\right|<2,\\\left|c_{n+2 k}-\mu c_{n} c_{k}^{2}\right| \leqslant 2(1+2 \mu) . \end{array} $$ 2. 主要结果
下面给出本文的主要定理.
定理1 若f(z)∈$\mathcal{S} $e*且具有式(1)的形式, 则
$$ \begin{gathered} \left|a_{2}\right| \leqslant 1,\left|a_{3}\right| \leqslant \frac{3}{4},\left|a_{4}\right| \leqslant \frac{31}{72},\left|a_{5}\right| \leqslant \frac{1}{3}, \\ \left|a_{6}\right| \leqslant \frac{17}{60},\left|a_{7}\right| \leqslant \frac{59}{80}. \end{gathered} $$ (3) 证明 设f∈Se*, 则由定义1和式(2), 易知
$$ \frac{z f^{\prime}(z)}{f(z)}=\mathrm{e}^{\omega(z)}, $$ 其中, ω(z)是Schwarz函数, 满足ω(0)=0, |ω(z)| < 1, z∈D. 又
$$ \begin{aligned} &\frac{z f^{\prime}(z)}{f(z)}=\frac{z+\sum\limits_{n=2}^{\infty} n a_{n} z^{n}}{z+\sum\limits_{n=2}^{\infty} a_{n} z^{n}}= \\ &\ \ \ \ \left(1+\sum\limits_{n=2}^{\infty} n a_{n} z^{n-1}\right)\left[1-a_{2} z+\left(a_{2}^{2}-a_{3}\right) z^{2}-\left(a_{2}^{3}-2 a_{2} a_{3}+\right.\right. \\ &\ \ \ \ \left.\left.a_{4}\right) z^{3}+\left(a_{2}^{4}-3 a_{2}^{2} a_{3}+2 a_{2} a_{4}-a_{5}\right) z^{4}+\cdots\right]=1+a_{2} z+ \\ &\ \ \ \ \left(2 a_{3}-a_{2}^{2}\right) z^{2}+\left(a_{2}^{3}-3 a_{2} a_{3}+3 a_{4}\right) z^{3}+\left(4 a_{5}-a_{2}^{4}+4 a_{2}^{2} a_{3}-\right. \\ &\ \ \ \ \left.4 a_{2} a_{4}-3 a_{3}^{2}\right) z^{4}+\left(5 a_{6}-5 a_{2} a_{5}+a_{2}^{5}-5 a_{3} a_{4}-5 a_{2}^{3} a_{3}+\right. \\ &\ \ \ \ \left.5 a_{2}^{2} a_{4}+5 a_{2} a_{3}^{2}\right) z^{5}+\left(6 a_{7}-6 a_{2} a_{6}+6 a_{2}^{2} a_{5}-6 a_{3} a_{5}+\right. \\ &\ \ \ \ 18 a_{2} a_{3} a_{4}-a_{2}^{6}-6 a_{2}^{3} a_{4}-3 a_{4}^{2}+2 a_{3}^{3}-9 a_{2}^{2} a_{3}^{2}+6 a_{2}^{4} a_{3}- \\ &\ \ \ \ \left.3 a_{2} a_{4}\right) z^{6}+\cdots. \end{aligned} $$ (4) 令
$$ p(z)=\frac{1+\omega(z)}{1-\omega(z)}=1+c_{1} z+c_{2} z^{2}+\cdots \text {. } $$ 显然有p(z)∈$\mathcal{P} $且
$$ \omega(z)=\frac{p(z)-1}{1+p(z)}=\frac{c_{1} z+c_{2} z^{2}+c_{3} z^{3}+\cdots}{2+c_{1} z+c_{2} z^{2}+c_{3} z^{3}+\cdots}. $$ (5) 由式(5)知,
$$ \begin{aligned} \mathrm{e}^{\omega(z)} &=1+\frac{1}{2} c_{1} z+\left(\frac{c_{2}}{2}-\frac{c_{1}^{2}}{8}\right) z^{2}+\left(\frac{c_{1}^{3}}{48}+\frac{2 c_{3}-c_{1} c_{2}}{4}\right) z^{3}+\\ &\left(\frac{2 c_{4}-c_{1} c_{3}}{4}+\frac{c_{1}^{2} c_{2}}{16}-\frac{c_{2}^{2}}{8}+\frac{c_{1}^{4}}{384}\right) z^{4}+\\ &\left(\frac{2 c_{5}-c_{1} c_{4}-c_{2} c_{3}}{4}+\frac{c_{1}^{2} c_{3}+c_{1} c_{2}^{2}}{16}+\frac{c_{1}^{3} c_{2}}{96}-\frac{19 c_{1}^{5}}{3\,840}\right) z^{5}+\\ &\left(\frac{2 c_{6}-c_{1} c_{5}-c_{2} c_{4}}{4}+\frac{c_{1} c_{2} c_{3}}{8}+\frac{c_{2}^{3}}{48}-\frac{c_{3}^{2}}{8}+\frac{151 c_{1}^{6}}{46\,080}-\frac{19 c_{1}^{4} c_{2}}{768}+\right.\\ &\left.\frac{c_{1}^{2} c_{2}^{2}}{64}+\frac{c_{1}^{2} c_{4}}{16}+\frac{c_{1}^{3} c_{3}}{96}\right) z^{6}+\cdots. \end{aligned} $$ (6) 分别比较式(4)、(6)两边关于z、z2、z3、z4、z5、z6的系数, 可得
$$ \begin{gathered} a_{2}=\frac{c_{1}}{2}, a_{3}=\frac{c_{2}}{4}+\frac{c_{1}^{2}}{16}, a_{4}=\frac{c_{3}}{6}+\frac{c_{1} c_{2}}{24}-\frac{c_{1}^{3}}{288},\\ a_{5}=\frac{c_{4}}{8}+\frac{c_{1} c_{3}}{48}+\frac{c_{1}^{4}}{1\,152}-\frac{c_{1}^{2} c_{2}}{96}, \\ a_{6}=\frac{c_{1} c_{4}}{80}-\frac{c_{2} c_{3}}{120}-\frac{17 c_{1}^{5}}{57\,600}+\frac{11 c_{1}^{3} c_{2}}{2\,880}+\frac{c_{1}^{2} c_{3}}{480}-\frac{c_{1} c_{2}^{2}}{120}+\frac{c_{5}}{10}, \\ \begin{aligned} a_{7}=&\frac{-13 c_{1}^{2} c_{4}}{1\ 920}-\frac{97 c_{1} c_{2} c_{3}}{2\ 880}+\frac{1\,781 c_{1}^{6}}{8\ 294\ 400}+\frac{7 c_{1}^{2} c_{2}^{2}}{11\ 520}-\frac{341 c_{1}^{4} c_{2}}{138\ 240}- \\ &\frac{c_{1} c_{5}}{120}-\frac{c_{2} c_{4}}{96}-\frac{c_{2}^{3}}{576}-\frac{c_{3}^{2}}{144}+\frac{c_{1} c_{3}}{24}+\frac{c_{6}}{12}+\frac{c_{1}^{2} c_{2}}{96}-\frac{c_{1}^{4}}{1\ 152}+ \\ &\frac{211 c_{1}^{3} c_{3}}{34\ 560} . \end{aligned} \end{gathered} $$ (7) 由引理2, 易证|a2|=|c1/2|≤1. 而由引理1, 可得
$$ \left|a_{3}\right|=\left|\frac{c_{2}}{4}+\frac{c_{1}^{2}}{16}\right|=\left|\frac{3 c_{1}^{2}}{16}+\frac{x\left(4-c_{1}^{2}\right)}{8}\right| $$ 设|x|=t (t∈[0, 1]), c1=c (c ∈[0, 2]), 则利用三角不等式, 得
$$ \left|a_{3}\right| \leqslant \frac{t\left(4-c^{2}\right)}{8}+\frac{3 c^{2}}{16}. $$ 令$F(c, t)=\frac{t\left(4-c^{2}\right)}{8}+\frac{3 c^{2}}{16} $, 对任意的t ∈(0, 1), c ∈(0, 2), 有
$$ \frac{\partial F}{\partial t}=\frac{4-c^{2}}{8}>0, $$ 故F(c, t)在[0, 1]关于t是单调递增函数. 因此, F(c, t)在t=1取得最大值, 即
$$ \max F(c, t)=F(c, 1)=\frac{4-c^{2}}{8}+\frac{3 c^{2}}{16}=\frac{1}{2}+\frac{c^{2}}{16}=G(c). $$ 易证G(c)在c=2处取得最大值, 即|a3|≤G(2)=3/4.
又
$$ \left|a_{4}\right|=\left|\frac{c_{3}}{6}+\frac{c_{1} c_{2}}{24}-\frac{c_{1}^{3}}{288}\right|=\left|\frac{-49 c_{3}}{288}-\frac{c_{1} c_{2}}{48}+\frac{c_{1}^{3}-2 c_{1} c_{2}+c_{3}}{288}\right|. $$ 令c1=c∈[0, 2], 则由引理2可得
$$ \left|a_{4}\right|=\left|\frac{-49 c_{3}}{288}-\frac{c_{1} c_{2}}{48}+\frac{c_{1}^{3}-2 c_{1} c_{2}+c_{3}}{288}\right| \leqslant \frac{25}{72}+\frac{c}{24} \text {. } $$ 令$F(c)=\frac{25}{72}+\frac{c}{24} $, 易证F(c)在c=2处取得最大值, 即|a4|≤F(2)=31/72.
类似地, 因为
$$ \begin{aligned} \left|a_{5}\right|=&\left|\frac{c_{4}}{8}+\frac{c_{1} c_{3}}{48}+\frac{c_{1}^{4}}{1\,152}-\frac{c_{1}^{2} c_{2}}{96}\right|=\\ &\left|\frac{23 c_{1}}{1\,152}\left[c_{3}-\frac{10 c_{1} c_{2}}{23}\right]+\frac{c_{4}}{8}+\frac{c_{1}\left(c_{1}^{3}-2 c_{1} c_{2}+c_{3}\right)}{1\ 152}\right|, \end{aligned} $$ 故对c1=c∈[0, 2], 由引理2和引理3可得
$$ \left|a_{5}\right| \leqslant \frac{1}{4}+\frac{c}{576}+\frac{23 c}{576}. $$ 令$F(c)=\frac{1}{4}+\frac{c}{24} $, 进而可得F(c)在c=2处取得最大值, 即|a5|≤F(2)=1/3.
又
$$ \begin{aligned} \left|a_{6}\right|&=\left|\frac{c_{1} c_{4}}{80}-\frac{c_{2} c_{3}}{120}-\frac{17 c_{1}^{5}}{57\ 600}+\frac{11 c_{1}^{3} c_{2}}{2\ 880}+\frac{c_{1}^{2} c_{3}}{480}-\frac{c_{1} c_{2}^{2}}{120}+\frac{c_{5}}{10}\right|= \\ &\left| \frac{1}{20}\left[c_{5}-\frac{c_{1} c_{4}}{4}\right]+\frac{1}{20}\left[c_{5}-\frac{c_{3} c_{2}}{12}\right]-\frac{c_{3}}{240}\left[c_{2}-\frac{c_{1}^{2}}{2}\right]+ \right. \\ &\left.\frac{17 c_{1}^{3}}{28\ 800}\left[c_{2}-\frac{c_{1}^{2}}{2}\right]-\frac{c_{1} c_{2}}{120}\left(c_{2}-\frac{c_{1}^{2}}{2}\right)-\frac{17 c_{1}^{3} c_{2}}{28\ 800}\right|. \end{aligned} $$ 令c1=c∈[0, 2], 由引理3可得
$$ \left|a_{6}\right| \leqslant \frac{1}{5}+\frac{17 c^{3}\left(2-c^{2} / 2\right)}{28\ 800}+\frac{5}{120}\left(2-\frac{c^{2}}{2}\right)+\frac{17 c^{3}}{14\ 400}. $$ 令
$$ F(c)=\frac{1}{5}+\frac{17 c^{3}\left(2-c^{2} / 2\right)}{28\ 800}+\frac{5}{120}\left(2-\frac{c^{2}}{2}\right)+\frac{17 c^{3}}{14\ 400}, $$ 则
$$ F^{\prime}(c)=\frac{17 c^{2}}{3\ 600}-\frac{17 c^{4}}{11\ 520}-\frac{c}{24}, $$ 因此, c=0是方程F′(c)=0的根. 又因为F″(0) < 0, 所以, F(c)在c=0处取得最大值, 即|a6|≤F(0)=17/60.
又
$$ \begin{aligned} &\left|a_{7}\right|=\left|\frac{-13 c_{1}^{2} c_{4}}{1\ 920}-\frac{97 c_{1} c_{2} c_{3}}{2\ 880}+\frac{1\ 781 c_{1}^{6}}{8\ 294\ 400}+\frac{7 c_{1}^{2} c_{2}^{2}}{11\ 520}-\frac{341 c_{1}^{4} c_{2}}{138\ 240}-\right.\\ &\left.\frac{c_{1} c_{5}}{120}-\frac{c_{2} c_{4}}{96}-\frac{c_{2}^{3}}{576}-\frac{c_{3}^{2}}{144}+\frac{c_{1} c_{3}}{24}+\frac{c_{6}}{12}+\frac{c_{1}^{2} c_{2}}{96}-\frac{c_{1}^{4}}{1\ 152}+\frac{211 c_{1}^{3} c_{3}}{34\ 560}\right|=\\ &\left|\frac{1\ 781 c_{1}^{4}}{4\ 147\ 200}\left[c_{2}-c_{2}^{2}\right]+\frac{c_{1}^{2}}{576}\left[c_{2}-c_{1}^{2}\right]-\frac{13 c_{1}^{2}\left[c_{4}-\frac{211 c_{1} c_{3}}{234}\right]}{1\ 920}+\right.\\ &\left.\frac{\left[c_{6}-\frac{c_{2} c_{4}}{8}\right]}{12}+\frac{c_{1}\left[c_{5}-\frac{23}{24} c_{2} c_{3}\right]}{120}-\frac{37 c_{1} c_{2}\left[c_{3}-\frac{7 c_{1} c_{2}}{296}\right]}{1440}+\frac{5 c_{1}^{2} c_{2}}{576}-\right.\\ &\left.\frac{c_{2}^{3}}{576}-\frac{c_{3}^{2}}{144}+\frac{c_{1} c_{3}}{24} \right| . \end{aligned} $$ 令c1=c∈[0, 2], 由引理2和引理3可知
$$ \begin{aligned} &\left|a_{7}\right| \leqslant \frac{1}{6}+\frac{c}{60}+\frac{37 c}{360}+\frac{1\ 781 c^{4}\left(2-\frac{c^{2}}{2}\right)}{4\ 147\ 200}+\frac{c^{2}\left(2-\frac{c^{2}}{2}\right)}{576}+\frac{13 c^{2}}{960}+ \\ &\ \ \ \ \ \ \ \ \frac{1}{72}+\frac{1}{36}+\frac{5 c^{2}}{288}+\frac{c}{12} . \end{aligned} $$ 令
$$ \begin{aligned} F(c)=&\frac{1}{6}+\frac{c}{60}+\frac{37 c}{360}+\frac{1\ 781 c^{4}\left(2-\frac{c^{2}}{2}\right) c^{2}\left(2-\frac{c^{2}}{2}\right)}{4\ 147\ 200}+\frac{13 c^{2}}{576}+\frac{5}{960}+ \\ &\frac{1}{72}+\frac{1}{36}+\frac{5 c^{2}}{288}+\frac{c}{12}, \end{aligned} $$ 进而可得F′(c)≥0. 因此, F(c)在c=2处取得最大值, 即|a7|≤F(2)=59/80. 证毕.
定理2 若f∈Se*且具有式(1)的形式, 则
$$ \left|a_{3}-a_{2}^{2}\right| \leqslant \frac{1}{2} $$ (8) 证明 通过式(7)可得
$$ \left|a_{3}-a_{2}^{2}\right|=\left|\frac{c_{2}}{4}-\frac{3 c_{1}^{2}}{16}\right|, $$ 由引理1可知
$$ \left|a_{3}-a_{2}^{2}\right|=\left|\frac{x\left(4-c_{1}^{2}\right)}{8}-\frac{c_{1}^{2}}{16}\right|. $$ 设|x|=t∈[0, 1], c1=c∈[0, 2], 则由三角不等式可得
$$ \left|a_{3}-a_{2}^{2}\right| \leqslant \frac{t\left(4-c^{2}\right)}{8}+\frac{c^{2}}{16}. $$ 令$F(c, t)=\frac{t\left(4-c^{2}\right)}{8}+\frac{c^{2}}{16} $, 则对∀t ∈(0, 1), ∀c ∈(0, 2), 有
$$ \frac{\partial F}{\partial t}=\frac{4-c^{2}}{8}>0 $$ 因此, F(c, t)在[0, 1]关于t是单调递增函数. 故F(c, t)在t=1处取得最大值, 即
$$ \max F(c, t)=F(c, 1)=\frac{\left(4-c^{2}\right)}{8}+\frac{c^{2}}{16}=G(c) . $$ 同理易证G(c)在c=0处取得最大值, 即
$$ \left|a_{3}-a_{2}^{2}\right| \leqslant G(0)=\frac{1}{2}. $$ 证毕.
定理3 若f∈Se*且具有式(1)的形式, 则
$$ \left|a_{2} a_{3}-a_{4}\right| \leqslant \frac{1}{3} $$ (9) 证明 由式(7)可得
$$ \begin{aligned} &\left|a_{2} a_{3}-a_{4}\right|=\left|\frac{c_{1} c_{2}}{8}+\frac{5 c_{1}^{3}}{144}-\frac{c_{3}}{6}+\frac{c_{1} c_{2}}{12}\right|= \\ &\ \ \ \ \ \ \ \ \left|\frac{5\left(c_{1}^{3}-2 c_{1} c_{2}+c_{3}\right)}{144}-\frac{29\left[c_{3}-\frac{22}{29} c_{1} c_{2}\right]}{144}\right| . \end{aligned} $$ 由引理2和引理3可得
$$ \left|a_{2} a_{3}-a_{4}\right| \leqslant \frac{5}{72}+\frac{29}{72}=\frac{17}{36}. $$ 证毕.
定理4 若f∈Se*且具有式(1)的形式, 则
$$ \left|a_{2} a_{4}-a_{3}^{2}\right| \leqslant \frac{1}{4}. $$ (10) 证明 由式(7)可得
$$ \left|a_{2} a_{4}-a_{3}^{2}\right|=\left|\frac{c_{1} c_{3}}{12}-\frac{c_{1}^{2} c_{2}}{48}-\frac{c_{1}^{4}}{288}-\frac{c_{2}^{2}}{16}\right|. $$ 由引理1可得
$$ \begin{aligned} &\left|a_{2} a_{4}-a_{3}^{2}\right|=\left|\frac{c_{1} c_{3}}{12}-\frac{c_{1}^{2} c_{2}}{48}-\frac{c_{1}^{4}}{288}-\frac{c_{2}^{2}}{16}\right|= \\ &\ \ \ \ \left|-\frac{5 c_{1}^{4}}{576}-\frac{x^{2} c_{1}^{2}\left(4-c_{1}^{2}\right)}{48}-\frac{x^{2}\left(4-c_{1}^{2}\right)^{2}}{64}+\frac{c_{1}\left(4-c_{1}^{2}\right)\left(1-|x|^{2}\right) z}{24}\right| . \end{aligned} $$ 设|x|=t∈[0, 1], c1=c∈[0, 2], 则由三角不等式可得
$$ \begin{aligned} &\left|a_{2} a_{4}-a_{3}^{2}\right| \leqslant \frac{t^{2} c^{2}\left(4-c^{2}\right)}{48}+\frac{\left(1-t^{2}\right) c\left(4-c^{2}\right)}{24}+ \\ &\ \ \ \ \ \frac{t^{2}\left(4-c^{2}\right)^{2}}{64}+\frac{5 c^{4}}{576}. \end{aligned} $$ 令
$$ F(c, t)=\frac{t^{2} c^{2}\left(4-c^{2}\right)}{48}+\frac{\left(1-t^{2}\right) c\left(4-c^{2}\right)}{24}+\frac{t^{2}\left(4-c^{2}\right)^{2}}{64}+\frac{5 c^{4}}{576}, $$ 则对∀t∈(0, 1), ∀c∈(0, 2), 有
$$ \frac{\partial F}{\partial t}=\frac{t\left(c^{2}-8 c+12\right)\left(4-c^{2}\right)}{96}>0. $$ 因此, F(c, t)在[0, 1]上关于t是单调递增函数, 从而可得F(c, t)在t=1处取得最大值, 即
$$ \max F(c, t)=F(c, 1)=\frac{c^{2}\left(4-c^{2}\right)}{48}+\frac{\left(4-c^{2}\right)^{2}}{64}+\frac{5 c^{4}}{576}. $$ 设
$$ G(c)=\frac{c^{2}\left(4-c^{2}\right)}{48}+\frac{\left(4-c^{2}\right)^{2}}{64}+\frac{5 c^{4}}{576} , $$ 则
$$ G^{\prime}(c)=\frac{c\left(4-c^{2}\right)}{24}-\frac{c^{3}}{24}-\frac{c\left(4-c^{2}\right)}{16}+\frac{5 c^{3}}{144} . $$ 若G′(c)=0, 则c=0. 又因为G″(0)=-1/12 < 0, 故G(c)在c=0处取得最大值, 即|a2a4-a32|≤G(0)=1/4.证毕.
定理5 若f∈Se*且具有式(1)的形式, 则
$$ \left|a_{2} a_{5}-a_{3} a_{4}\right| \leqslant \frac{73}{144} $$ (11) 证明 若f(z)∈$\mathcal{S} $e*, 则由式(7)可得
$$ \begin{aligned} &\left|a_{2} a_{5}-a_{3} a_{4}\right|=\left|\frac{-c_{1}^{5}}{1\ 536}-\frac{c_{1} c_{4}}{16}+\frac{c_{1} c_{2}^{2}}{96}+\frac{c_{1}^{3} c_{2}}{144}+\frac{c_{2} c_{3}}{24}\right|= \\ &\ \ \ \ \ \ \ \ \left| \frac{c_{1}^{3}\left[c_{2}-\frac{c_{1}^{2}}{2}\right]}{144}+ \frac{c_{2}\left[c_{3}-\frac{c_{1} c_{2}}{2}\right]}{24} -\frac{c_{1}\left[c_{4}-\frac{c_{1}^{2}}{2}\right]}{16}+\frac{13 c_{1}^{5}}{4\ 608}\right| . \end{aligned} $$ 令c1=c∈[0, 2], 由引理3可得
$$ \left|a_{2} a_{5}-a_{3} a_{4}\right| \leqslant \frac{c^{3}\left[2-\frac{c^{2}}{2}\right]}{144}+\frac{1}{6}+\frac{13 c^{5}}{4\ 608}+\frac{c}{8} \text {. } $$ 令
$$ F(c)=\frac{c^{3}\left[2-\frac{c^{2}}{2}\right]}{144}+\frac{1}{6}+\frac{13 c^{5}}{4\ 608}+\frac{c}{8}, $$ 则对∀c∈(0, 2), 有
$$ F^{\prime}(c)=\frac{c^{2}}{24}-\frac{5 c^{4}}{1\ 536}+\frac{1}{8}>0 . $$ 因此, F(c)关于c单调递增, 故F(c)在c=2处取得最大值, 即
$$ \left|a_{2} a_{5}-a_{3} a_{4}\right| \leqslant F(2)=\frac{73}{144}. $$ 证毕.
定理6 若f∈Se*且具有式(1)的形式, 则
$$ \left|a_{5}-a_{2} a_{4}\right| \leqslant \frac{35}{96}. $$ (12) 证明 若f(z)∈$\mathcal{S} $e*, 则由式(7)可得
$$ \begin{aligned} &\left|a_{5}-a_{2} a_{4}\right|=\left|\frac{c_{1}^{4}}{384}-\frac{c_{1} c_{3}}{16}-\frac{c_{1}^{2} c_{2}}{32}+\frac{c_{4}}{8}\right|= \\ &\qquad\left|\frac{c_{1}\left[c_{1}^{3}-2 c_{1} c_{2}+c_{3}\right]}{384}+\frac{5 c_{1}\left[c_{3}-c_{1} c_{2}\right]}{192}-\frac{\left[c_{4}-\frac{35}{48} c_{1} c_{3}\right]}{8}\right|. \end{aligned} $$ 令c1=c∈[0, 2], 由引理2和引理3可得
$$ \left|a_{5}-a_{2} a_{4}\right| \leqslant \frac{c}{192}+\frac{5 c}{96}+\frac{1}{4} . $$ 令$F(c)=\frac{1}{4}+\frac{11 c}{192} $, 则易证F(c)在c=2处取得最大值, 即
$$ \left|a_{5}-a_{2} a_{4}\right| \leqslant F(2)=\frac{35}{96}. $$ 证毕.
定理7 若f∈Se*且具有式(1)的形式, 则
$$ \left|a_{5} a_{3}-a_{4}^{2}\right| \leqslant \frac{121}{216}. $$ (13) 证明 若f(z)∈$\mathcal{S} $e*, 则由式(7)可得
$$ \begin{aligned} &\left|a_{5} a_{3}-a_{4}^{2}\right|=\left| \frac{-c_{1}^{4} c_{2}}{6\ 912}+\frac{c_{2} c_{4}}{32}-\frac{5 c_{1} c_{2} c_{3}}{576}+\frac{17 c_{1}^{3} c_{3}}{6\ 912}-\frac{5 c_{1}^{2} c_{2}^{2}}{1\ 152}-\frac{c_{3}^{2}}{36}+\right. \\ &\ \ \ \ \left.\frac{c_{1}^{2} c_{4}}{128}+\frac{7 c_{1}^{6}}{165\ 888}\right|=\left| \frac{c_{3}\left[c_{3}-\frac{5 c_{1} c_{2}}{16}\right] c_{4}\left[c_{2}-\frac{c_{1}^{2}}{2}\right]}{36}+\frac{c_{3}^{2}}{32}- \right.\\ &\ \ \ \ \left.\frac{7 c_{1}^{4}\left[c_{2}-\frac{c_{1}^{2}}{2}\right]}{82\ 944}+\frac{17 c_{1}^{3}\left[c_{3}-\frac{5}{204} c_{1} c_{2}\right]}{6\ 912}+\frac{3 c_{1}^{2}\left[c_{4}-\frac{5 c_{2}^{2}}{27}\right]}{128}-\frac{c_{3}^{2}}{18}\right|. \end{aligned} $$ 令c1=c∈[0, 2], 由引理3可得
$$ \left|a_{5} a_{3}-a_{4}^{2}\right| \leqslant \frac{1}{9}+\frac{2}{9}+\frac{\left[2-\frac{c^{2}}{2}\right]}{16}+\frac{7 c^{4}\left[2-\frac{c^{2}}{2}\right]}{82\ 944}+\frac{17 c^{3}}{3\ 456}+\frac{3 c^{2}}{64} \text {. } $$ 令
$$ F(c)=\frac{1}{9}+\frac{2}{9}+\frac{\left[2-\frac{c^{2}}{2}\right]}{16}+\frac{7 c^{4}\left[2-\frac{c^{2}}{2}\right]}{82\ 944}+\frac{17 c^{3}}{3\ 456}+\frac{3 c^{2}}{64}, $$ 则对∀c∈(0, 2), 有
$$ F^{\prime}(c)=\frac{c}{32}+\frac{17 c^{2}}{1\ 152}+\frac{7 c^{3}}{10\ 368}-\frac{7 c^{5}}{27\ 648}>0 . $$ 因此, F(c)关于c单调递增, 故F(c)在c=2处取得最大值, 即
$$ \left|a_{5} a_{3}-a_{4}^{2}\right| \leqslant F(2)=\frac{121}{216}. $$ 证毕.
定理8 若f∈Se*且具有式(1)的形式, 则
$$ \left|H_{4}(1)\right| \leqslant \frac{52\ 253\ 339}{55\ 987\ 200} \approx 0.933\ 3. $$ (14) 证明 因为
$$ \begin{aligned} H_{4}&(1)=a_{7}\left\{a_{3}\left(a_{2} a_{4}-a_{3}^{2}\right)-a_{4}\left(a_{4}-a_{2} a_{3}\right)+a_{5}\left(a_{3}-a_{2}^{2}\right)\right\}- \\ &a_{6}\left\{a_{3}\left(a_{2} a_{5}-a_{3} a_{4}\right)-a_{4}\left(a_{5}-a_{2} a_{4}\right)+a_{6}\left(a_{3}-a_{2}^{2}\right)\right\}+ \\ &a_{5}\left\{a_{3}\left(a_{3} a_{5}-a_{4}^{2}\right)-a_{5}\left(a_{5}-a_{2} a_{4}\right)+a_{6}\left(a_{4}-a_{2} a_{3}\right)\right\}- \\ &a_{4}\left\{a_{4}\left(a_{3} a_{5}-a_{4}^{2}\right)-a_{5}\left(a_{2} a_{5}-a_{3} a_{4}\right)+a_{6}\left(a_{4}-a_{2} a_{3}\right)\right\}, \end{aligned} $$ 所以, 由三角不等式可得
$$ \begin{aligned} &\left|H_{4}(1)\right| \leqslant\left|a_{7}\right|\left|a_{3}\right|\left|a_{2} a_{4}-a_{3}^{2}\right|+\left|a_{7}\right|\left|a_{4}\right|\left|a_{4}-a_{2} a_{3}\right|+\\ &\ \ \ \ \left|a_{7}\right|\left|a_{5}\right|\left|a_{3}-a_{2}^{2}\right|+\left|a_{6}\right|\left|a_{3}\right|\left|a_{2} a_{5}-a_{3} a_{4}\right|+ \\ &\ \ \ \ \left|a_{6}\right|\left|a_{4}\right|\left|a_{5}-a_{2} a_{4}\right|+\left|a_{6}\right|^{2}\left|a_{3}-a_{2}^{2}\right|+ \\ &\ \ \ \ \left|a_{5}\right|\left|a_{3}\right|\left|a_{3} a_{5}-a_{4}^{2}\right|+\left|a_{5}\right|^{2}\left|a_{5}-a_{2} a_{4}\right|+ \\ &\ \ \ \ \left|a_{5}\right|\left|a_{6}\right|\left|a_{4}-a_{2} a_{3}\right|+\left|a_{4}\right|^{2}\left|a_{3} a_{5}-a_{4}^{2}\right|+ \\ &\ \ \ \ \left|a_{4}\right|\left|a_{5}\right|\left|a_{2} a_{5}-a_{3} a_{4}\right|+\left|a_{4}\right|\left|a_{6}\right|\left|a_{4}-a_{2} a_{3}\right|. \end{aligned} $$ (15) 将式(3)、(8)~(13)代入到式(15), 即得式(14). 证毕.
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