不同区域上广义非线性解析积分算子的单叶性

额尔敦其其格; 李书海

doi:10.6054/j.jscnun.2024013

不同区域上广义非线性解析积分算子的单叶性

额尔敦其其格¹,
李书海^2, ,

1.
呼和浩特民族学院基础教育学院, 呼和浩特 010051
2.
赤峰学院数学与计算机科学学院, 赤峰 024000

基金项目:

国家自然科学基金项目 11561001

内蒙古自治区自然科学基金项目 2019MS01023

内蒙古自治区高等学校科学研究项目 NJZZ19209

详细信息

通讯作者:
李书海, Email: lishms66@163.com

中图分类号: O174.51
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出版历程
- 收稿日期: 2022-06-10
- 网络出版日期: 2024-04-29
- 刊出日期: 2024-02-24

Univalency Conditions of A General Nonlinear Analytic Integral Operator with Different Domains

Eerdunqiqige¹,
LI Shuhai^2, ,

1.
School of Fundamental Education, Hohhot Minzu College, Hohhot 010051, China
2.
School of Mathematics and Computer Science, Chifeng University, Chifeng 024000, China

摘要

摘要:
利用从属关系和单叶函数充分条件, 研究某些广义非线性解析积分算子 $J_{n, \sigma}\left(\mathfrak{A}_n, \mathfrak{B}_n\right)(z)$ , 得到该解析积分算子在不同区域上单叶的充分条件, 推广了相关的非线性解析积分算子单叶的充分性, 并导出有关条形区域的积分算子单叶的充分条件。
- 解析函数 /
- 单叶函数 /
- 非线性 /
- 积分算子 /
- 从属
Abstract:
By using the subordination relationship and the sufficient conditions of univalent functions, some generalized nonlinear analytic integral operators $J_{n, \sigma}\left(\mathfrak{A}_n, \mathfrak{B}_n\right)(z)$ are researched, and the sufficient conditions for the analytic integral operator to be univalent in different regions are obtained. The sufficiency of the relevant nonlinear analytic integral operator to be univalent are generalized. The sufficient conditions for integral operators to be univalent in a strip region are derived.
- analytic function /
- univalent function /
- nonlinear /
- integral operator /
- subordination

HTML全文

设函数 $s(z)$ 和 $t(z)$ 在 $\mathcal{U}=\{z ? \gg: |z|<1\}$ 内解析, 若存在解析函数 $\nu: \nu(0)=0, |\nu(z)|<1(z ? \mathcal{U})$ , 满足条件 $s(z)=t(\nu(z))(z ? \mathcal{U})$ , 则称 $s(z)$ 从属于 $t(z)$ , 记为 $s(z)\prec t(z)$ 。特别地, 由文献[]可知, 设 $t(z)$ 在 $\mathcal{U}$ 内单叶, 则

$s(z) \prec t(z) \Leftrightarrow s(0)=t(0), s(\mathcal{U}) \subset t(\mathcal{U}) 。$

用H表示在 $\mathcal{U}$ 内解析且具有形式

$h(z)=z+\sum\limits_{k=2}^{\infty} a_k z^k$

的函数类；用 $\mathcal{S}$ 表示H中的单叶函数子类；用 $\overline{\mathcal{P}}$ 表示在 $\mathcal{U}$ 内解析且满足条件p(0)=1的函数p(z)组成的类; Re(p(z))表示函数p(z)的实部。

由文献[]可知, 设 $A, B ? \mathbb{C}, |A| \leqslant 1, |B| \leqslant 1$ 和 $p_k(z)=1+\sum_{n=k}^{\infty} p_k z^k ? \overline{\mathcal{P}}$ , 则p_k(z)属于函数类 $\mathcal{H}_k(A, B)$ 当且仅当

$p_k(z) \prec \frac{1+A z}{1+B z} \quad(k ? \mathbb{N}) 。$

由 $\mathcal{H}_k(A, B)$ 易得特殊子类：

$\begin{aligned} & \mathcal{P}(\beta)=\mathcal{H}_1(1-2 \beta, -1)=\{p(z) ? \overline{\mathcal{P}}: \operatorname{Re}(p(z))>\beta \\ & \quad(0 \leqslant \beta<1\}, \\ & \mathcal{H}(\gamma)=\mathcal{H}_1(1-2 \gamma, -1)=\{p(z) ? \overline{\mathcal{P}}: \operatorname{Re}(p(z))<\gamma \\ & \quad(\gamma>1)\}, \end{aligned}$

及文献[3]、[4]中的特殊子类：

$\begin{gathered} L(A, B)=\left\{p(z) ? \mathcal{H}_1(A, B):-1 \leqslant B<A \leqslant 1\right\}, \\ \mathcal{P}(A, B ; C, D)=\{p(z) ? \overline{\mathcal{P}}: p(z) ? L(A, B) \text { 且 } \\ \left.p(z) ? \mathcal{H}_1(C, D)(C<D), z ? \mathcal{U}\right\} 。 \end{gathered}$

近年来, 学者们研究了非线性解析积分算子单叶的充分性及其相关问题。如：PESCAR^[5]在函数类H上引进了非线性积分算子

$F_{\alpha}(z)=\left[\alpha \int_{0}^{z} t^{\alpha-1} h^{\prime}(t) \mathrm{d} t\right]^{\frac{1}{\alpha}} \quad(\alpha ? \mathbb{C}, \operatorname{Re} \alpha>0 ; h ? H),$

给出了该算子的单叶性充分条件; BREAZ等^[6]引进了积分算子 $G_{n, \alpha}(z)$ , 给出该算子的单叶性充分条件, 并推广了文献[5]的结果, 其中

$\begin{gathered} G_{n, \alpha}(z)=\left\{[n(\alpha-1)+1] \int_{0}^{z} \prod\limits_{j=1}^{n}\left(g_{j}(u)\right)^{\alpha-1} \mathrm{~d} u\right\}^{\frac{1}{n(\alpha-1)+1}} \\ \left(\left(\alpha ? \mathbb{C}, \operatorname{Re} \alpha>0, j=1, \cdots, n ; g_{j} ? H\right);\right. \end{gathered}$

BREAZ和TOMA^[7]引进了积分算子 $H_{\gamma_{1}, \gamma_{2}, \cdots, \gamma_{[|\eta|]]}}(z)$ , 给出了该算子单叶性的充分条件, 并进一步推广文献[5]的结果, 其中

$\begin{aligned} & H_{\gamma_{1}, \gamma_{2}, \cdots, \gamma_{[|\eta|]}}(z)=\left\{\eta \beta \int_{0}^{z} u^{\eta \beta-1} \prod\limits_{j=1}^{[|\eta|]}\left(\frac{f_{j}(u)}{u}\right)^{\frac{1}{\gamma j}} \mathrm{~d} u\right\}^{\frac{1}{\eta \beta}} \\ & \quad\left(\gamma_{j}, \beta, \eta ? \mathbb{C}, |\eta| ?[0, 1), j=1, \cdots, [|\eta|]\right. ,\\ & \left.\quad \beta \neq 0 ; f_{j} ? H\right)。 \end{aligned}$

有学者在不同函数类上研究积分算子 $G_{n, \alpha}(z)$ 和 $H_{\gamma_{1}, \gamma_{2}, \cdots, \gamma_{[|\eta|]}}(z)$ 的单叶性充分条件^[8-9], 也有学者围绕着这2个积分算子进行单叶性及其应用研究^[10-19]。已有结果均仅限于在函数的同一个映射区域内讨论了积分算子的单叶性充分条件, 目前还没有发现在不同映射区域内定义的积分算子单叶性的相关研究。

本文研究不同区域上的一类广义非线性解析积分算子 $J_{n, \sigma}\left(\mathfrak{A}_{n}, \mathfrak{B}_{n}\right)(z)$ , 给出某些算子单叶性的充分条件, 推广文献[-]的结果, 并得到文献[]中有关条形区域解析函数 $\mathcal{P}(A, B ; C, D)$ 的非线性积分算子的单叶性充分条件。

1. 预备知识

设 $\lambda_{i}, \theta_{i}, \sigma ? \mathbb{C}, \operatorname{Re} \sigma>0 ; \mu_{i}, \eta_{i} ? \gg , \zeta_{i}, \tau_{i} ? \mathbb{N}(i=\$1, 2, \cdots, n) 。 p_{i}(z), q_{i}(z) ? \overline{\mathcal{P}}$ , 引进广义非线性解析积分算子 $J_{n, \sigma}\left(\mathfrak{A}_{n}, \mathfrak{B}_{n}\right)(z): \overline{\mathcal{P}}^{n} \rightarrow \overline{\mathcal{P}}$

$\begin{align*} & J_{n, \sigma}\left(\mathfrak{A}_{n}, \mathfrak{B}_{n}\right)(z)= \\ & \quad\left\{\sigma \int_{0}^{z} u^{\sigma-1} \exp \left[\int_{0}^{u}\left(\mathfrak{A}_{n}(t)+\mathfrak{B}_{n}(t)\right) \mathrm{d} t\right] \mathrm{d} u\right\}^{\frac{1}{\sigma}}(z ? \mathcal{U}) , \end{align*}$

(1)

其中

$\begin{align*} & \mathfrak{A}_{n}(z)=\frac{1}{z} \sum\limits_{i=1}^{n} \lambda_{i}\left(p_{i}(z)-\mu_{i}\right)^{\tau_{i}} , \\ & \mathfrak{B}_{n}(z)=\frac{1}{z} \sum\limits_{i=1}^{n} \theta_{i}\left(q_{i}(z)-\eta_{i}\right)^{\zeta_{i}}, \end{align*}$

(2)

且幂函数均取主值。

由算子 $J_{n, \sigma}\left(\mathfrak{A}_{n}, \mathfrak{B}_{n}\right)(z)$ 分别推出文献[10-13]中研究的特殊解析积分算子：

$\begin{align*} & J_{\sigma, \lambda}(p)(z)= \\ & \quad\left\{\sigma \int_{0}^{z} u^{\sigma-1} \exp \left[\int_{0}^{u} \frac{\lambda}{t}(p(t)-1) \mathrm{d} t\right] \mathrm{d} u\right\}^{\frac{1}{\sigma}}(p(z) ? \overline{\mathcal{P}}), \end{align*}$

(3)

$J_{n, \theta_{i}}\left(f_{i}, g_{i}\right)(z)= \\ \;\;\;\left\{\sigma \int_{0}^{z}\left[t^{\sigma-1} \prod\limits_{i=1}^{n}\left(\frac{f_{i}(t)}{t}\right)\left(\frac{g_{i}(t)}{t}\right)^{\theta_{i}}\right] \mathrm{d} t\right\}^{\frac{1}{\sigma}}\left(f_{i}, g_{i} ? H\right),$

(4)

$\begin{aligned} & J_{\lambda_i, \theta_i}\left(f_i, g_i\right)(z)= \\ & \quad \int_0^z\left[\prod\limits_{i=1}^n\left(f_i^{\prime}(t)\right)^{\lambda_i}\left(\frac{g_i(z)}{z}\right)^{\theta_i}\right] \mathrm{d} t\left(f_i, g_i ? H\right), \end{aligned}$

(5)

$F_{\lambda_i, \theta_i}(z)=\int_0^z\left[\prod\limits_{i=1}^n\left(f_i^{\prime}(t)\right)^{\lambda_i}\left(\frac{f_i(z)}{z}\right)^{\theta_i}\right] \mathrm{d} t \quad\left(f_i ? H\right) 。$

当 $p_{1}(z), q_{1}(z) ? \overline{\mathcal{P}}$ 时，非线性解析积分算子 $J_{1, \sigma}\left(\mathfrak{A}_{1}, \mathfrak{B}_{1}\right)(z)$ 为文献[14]研究的算子。由式(3)可得到文献[15]的积分算子:

$G_{\alpha_{1}, \cdots, \alpha_{n}, \beta, \sigma}(z)= \\ \;\;\;\left\{\gamma \int_{0}^{z}\left[u^{\gamma-1} \prod\limits_{j=1}^{n}\left(\frac{u}{g_{j}(u)}\right)^{\alpha_{j}}\left(h_{j}^{\prime}(u)\right)^{\beta}\right] \mathrm{d} u\right\}^{\frac{1}{\gamma}}\left(g_{i}, h_{i} ? H\right)。$

下面给出研究积分算子 $J_{n, \sigma}\left(\mathfrak{A}_{n}, \mathfrak{B}_{n}\right)(z)$ 的单叶性所需的引理：

引理1^[20] 设 $p ? L(A, B), -1 \leqslant B<A \leqslant 1$ 和 $|z|=r<1$ , 则

$\frac{1-A r}{1-B r} \leqslant \operatorname{Re}(p(z)) \leqslant|p(z)| \leqslant \frac{1+A r}{1+B r}。$

引理2^[21] 设 $p ? \mathcal{H}_{k}(A, B), k ? \mathbb{N}, -1 \leqslant A<B \leqslant 1$ 和 $|z|=r<1$ , 则

$\frac{1+A r^{k}}{1+B r^{k}} \leqslant \operatorname{Re}(p(z)) \leqslant \frac{1-A r^{k}}{1-B r^{k}}。$

引理3 设 $p ? \mathcal{H}_{1}(A, B), -1 \leqslant A<B \leqslant 1$ 和 $|z|=r<1$ , 则

$\frac{1+A r}{1+B r} \leqslant \operatorname{Re}(p(z)) \leqslant|p(z)| \leqslant \frac{1-A r}{1-B r} 。$

证明根据复变函数的实部和模之间的关系, 有 $\operatorname{Re}(p(z)) \leqslant|p(z)|$ 。取k=1，由引理2易得

$\frac{1+A r}{1+B r} \leqslant \operatorname{Re}(p(z)) \leqslant|p(z)|。$

再由从属关系可知, 若 $p ? \mathcal{H}_{1}(A, B)$ , 则存在解析函数 $\nu: \nu(0)=0, |\nu(z)|<1 \quad(|z|=r<1)$ , 使得

$|p(z)|=\left|\frac{1+A \nu(z)}{1+B \nu(z)}\right| \leqslant \frac{1-A r}{1-B r} 。$

从而得到引理3成立。证毕。

引理4^[5] 设 $\alpha ? \mathbb{C}$ 且 $\operatorname{Re} \alpha>0$ 。若 $h ? H$ 满足

$\frac{1-|z|^{2 \operatorname{Re} \alpha}}{\operatorname{Re} \alpha}\left|\frac{z h^{\prime \prime}(z)}{h^{\prime}(z)}\right| \leqslant 1 \quad(z ? \mathcal{U}),$

则算子

$F_{\alpha}(z)=\left[\alpha \int_{0}^{z} t^{\alpha-1} h^{\prime}(t) \mathrm{d} t\right]^{\frac{1}{\alpha}} ? \mathcal{S}。$

2. 主要结果

首先, 给出由式(3)定义的积分算子 $J_{\sigma, \lambda}(p)(z)\$(p ? L(A, B))$ 单叶的充分条件。

定理1 设 $p ? L(A, B)$ 。若 $-1 \leqslant B<A \leqslant 1$ 和 $\sigma ? \mathbb{C}, \lambda ? \mathbb{C}_{+}$ 且

$|\lambda| \leqslant \begin{cases}\frac{\operatorname{Re} \sigma}{(1+|B|)(2+|A+B|)} & (0<\operatorname{Re} \sigma<1), \\ \frac{1}{(1+|B|)(2+|A+B|)} & (\operatorname{Re} \sigma \geqslant 1), \end{cases}$

(6)

则由式(3)定义的算子 $J_{\sigma, \lambda}(p)(z) ? \mathcal{S}$ 。

证明令

$h(z)=\int_{0}^{z} \exp \left(\int_{0}^{u} \frac{\lambda(p(t)-1)}{t} \mathrm{~d} t\right) \mathrm{d} u，$

则易得 $h(0)=h^{\prime}(0)-1=0$ 且

$\begin{equation*} \frac{z h^{\prime \prime}(z)}{h^{\prime}(z)}=\lambda(p(z)-1) \end{equation*}。$

(7)

由引理1和式(7), 可得

$\begin{array}{c} & \frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma}\left|\frac{z h^{\prime \prime}(z)}{h^{\prime}(z)}\right|=\frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma}|\lambda||p(z)-1| \leqslant \\ & \frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma}|\lambda|\left(\frac{1+A|z|}{1+B|z|}+1\right)= \\ & \frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma}|\lambda|\left(\frac{2+(A+B)|z|}{1+B|z|}\right) \leqslant \\ & \frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma}|\lambda|\left(\frac{2+|A+B||z|}{1-|B||z|}\right)(z ? \mathcal{U})。 \end{array}$

(8)

下面对Re σ进行分类讨论:

(ⅰ)若0<Re σ<1, 则由指数函数的单调性可得

$\begin{equation*} 1-|z|^{2\operatorname{Re} \sigma} \leqslant 1-|z|^{2} \leqslant 1-B^{2}|z|^{2} , \end{equation*}$

(9)

其中， $z ? \mathcal{U}, 0 \leqslant B^{2} \leqslant 1$ 。

(ⅱ)若Re σ≥1, 则由指数函数的单调性可得

$\begin{equation*} \frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma} \leqslant 1-|z|^{2} \leqslant 1-B^{2}|z|^{2} , \end{equation*}$

(10)

其中， $z ? \mathcal{U}, 0 \leqslant B^{2} \leqslant 1$ 。由式(6)、(8)~(10), 可得

$\begin{array}{l} \frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma}\left|\frac{z h^{\prime \prime}(z)}{h^{\prime}(z)}\right| \leqslant \\ \;\;\;\;\;\;\;\begin{cases}\frac{|\lambda|(1+|B|)(2+|A+B|)}{\operatorname{Re} \sigma} \leqslant 1 \quad(0<\operatorname{Re} \sigma<1), \\ |\lambda|(1+|B|)(2+|A+B|) \leqslant 1 \quad (\operatorname{Re} \sigma \geqslant 1), \end{cases} \end{array}$

其中 $z ? \mathcal{U}$ 。利用引理4, 推出 $J_{\sigma, \lambda}(p)(z) ? \mathcal{S}。$ 证毕。

其次，给出算子 $J_{\sigma, \lambda}(p)(z)\left(p ? \mathcal{H}_{1}(A, B)\right)$ 单叶的充分条件：

定理2 设 $p ? \mathcal{H}_{1}(A, B)$ 。若－1≤A<B≤1和 $\sigma ? \mathbb{C}, \lambda ? \mathbb{C}_{+}$ 且

$|\lambda| \leqslant \begin{cases}\frac{\operatorname{Re} \sigma}{(1+|B|)(2-(A+B))} & (0<\operatorname{Re} \sigma<1), \\ \frac{1}{(1+|B|)(2-(A+B))} & (\operatorname{Re} \sigma \geqslant 1),\end{cases}$

则式(3)定义的算子 $J_{\sigma, \lambda}(p)(z) ? \mathcal{S}$ 。

类似定理1的证明方法，易得定理2。

然后, 给出式(1)定义的算子 $J_{n, \sigma}\left(\mathfrak{A}_{n}, \mathfrak{B}_{n}\right)(z)$ 单叶的充分条件：

定理3 设i=1, 2, …, n; $-1<B_{i}<A_{i} \leqslant 1 ;-1<D_{i}<C_{i} \leqslant 1 ; \mu_{i}, \eta_{i} ? \gg;\sigma, \lambda_{i}, \theta_{i} ? \mathbb{C}$ 和 $\operatorname{Re} \sigma>0$ 。若 $p_{i} ? L\left(A_{i}, B_{i}\right), q_{i} ? L\left(C_{i}, D_{i}\right)$ 且

$\left|M_{n}\right| \leqslant\left\{\begin{array}{l} \operatorname{Re} \sigma \quad(0<\operatorname{Re} \sigma<1) , \\ 1 \quad(\operatorname{Re} \sigma \geqslant 1), \end{array}\right.$

(11)

其中

$\begin{aligned} M_{n}= & \sum\limits_{i=1}^{n}\left[\left|\lambda_{i}\right|\left(\frac{1+A_{i}}{1+B_{i}}+\left|\mu_{i}\right|\right)^{\tau_{i}}+\left|\theta_{i}\right|\left(\frac{1+C_{i}}{1+D_{i}}+\left|\eta_{i}\right|\right)^{\zeta_{i}}\right] \\ & \left(\tau_{i}, \zeta_{i} ? \mathbb{N}_{+}\right), \end{aligned}$

则由式(1)定义的算子 $J_{n, \sigma}\left(\mathfrak{A}_{n}, \mathfrak{B}_{n}\right)(z) ? \mathcal{S}$ 。

证明首先假设函数 $h(z)$ ：

$\begin{equation*} h(z)=\int_{0}^{z} \exp \left(\int_{0}^{u}\left(\mathfrak{A}_{n}(t)+\mathfrak{B}_{n}(t)\right) \mathrm{d} t\right) \mathrm{d} u \quad(z ? \mathcal{U}), \end{equation*}$

(12)

其中 $\mathfrak{R}_{n}(z)$ 和 $\mathfrak{B}_{n}(z)$ 由式(2)定义。由式(12)易得h(0)=h′(0)－1=0且

$\begin{equation*} \frac{z h^{\prime \prime}(z)}{h^{\prime}(z)}=z \sum\limits_{i=1}^{n} \lambda_{i}\left(p_{i}(z)-\mu_{i}\right)^{\tau_{i}}+z \sum\limits_{i=1}^{n} \theta_{i}\left(q_{i}(z)-\eta_{i}\right)^{\zeta_{i}} , \end{equation*}$

(13)

其中, $p_{i} ? L\left(A_{i}, B_{i}\right), q_{i} ? L\left(C_{i}, D_{i}\right)(i=1, 2, \cdots, n)$ 。

由引理1和式(13), 可得

$\frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma}\left|\frac{z h^{\prime \prime}(z)}{h^{\prime}(z)}\right|= \\ \;\;\;\;\frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma}\left|z \sum\limits_{i=1}^{n}\left(\lambda_{i}\left(p_{i}(z)-\mu_{i}\right)^{\tau_{i}}+\theta_{i}\left(q_{i}(z)-\eta_{i}\right)^{\zeta_{i}}\right)\right| \leqslant \\ \;\;\;\;\frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma} \sum\limits_{i=1}^{n}\left(\left|\lambda_{i}\right|\left|p_{i}(z)-\mu_{i}\right|^{\tau_{i}}+\left|\theta_{i}\right|\left|q_{i}(z)-\eta_{i}\right|^{\zeta_{i}}\right) \leqslant \\ \;\;\;\;\frac{1-|z|^{2 \operatorname{Re}} \sigma}{\operatorname{Re} \sigma} \sum\limits_{i=1}^{n}\left(\left|\lambda_{i}\right||| p_{i}(z)\left|+\mu_{i}\right|^{\tau_{i}}+\left|\theta_{i}\right||| q_{i}(z)\left|+\eta_{i}\right|^{\zeta \zeta_{i}}\right) \leqslant \\ \;\;\;\;\frac{1-|z|^{2 \operatorname{Re}} \sigma}{\operatorname{Re} \sigma} \sum\limits_{i=1}^{n}\left[\left|\lambda_{i}\right|\left(\frac{1+A_{i}|z|}{1+B_{i}|z|}+\left|\mu_{i}\right|\right)^{\tau_{i}}+\right. \\ \;\;\;\;\;\left.\left|\theta_{i}\right|\left(\frac{1+C_{i}|z|}{1+D_{i}|z|}+\left|\eta_{i}\right|\right)^{\zeta_{i}}\right] 。$

下面对Re σ进行分类讨论:

(ⅰ)若0<Re σ<1, 则由指数函数的单调性可得

$1-|z|^{2 \operatorname{Re} \sigma} \leqslant 1-|z|^{2} \leqslant 1,$

其中 $z ? \mathcal{U}。$

(ⅱ)若Re σ≥1, 则由指数函数的单调性可得

$\frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma} \leqslant 1-|z|^{2} \leqslant 1,$

其中 $z ? \mathcal{U}。$

由(ⅰ)和(ⅱ)结合式(12), 可得

$\begin{array}{l} \frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma}\left|\frac{z h^{\prime \prime}(z)}{h^{\prime}(z)}\right| \leqslant \\ \;\;\;\;\left\{\begin{array}{l} \frac{1}{\operatorname{Re} \sigma} \sum\limits_{i=1}^n\left[\left|\lambda_i\right|\left(\frac{1+A_i}{1+B_i}+\left|\mu_i\right|\right)^{\tau_i}+\right. \\ \left.\left|\theta_i\right|\left(\frac{1+C_i}{1+D_i}+\left|\eta_i\right|\right)^{\tau_i}\right] \leqslant 1 \quad(0<\operatorname{Re} \sigma<1), \\ \sum\limits_{i=1}^n\left[\left|\lambda_i\right|\left(\frac{1+A_i}{1+B_i}+\left|\mu_i\right|\right)^{\tau_i}+\left|\theta_i\right|\left(\frac{1+C_i}{1+D_i}+\left|\eta_i\right|\right)^{\tau_i}\right] \leqslant 1 \\ \quad(\operatorname{Re} \sigma \geqslant 1), \end{array}\right. \end{array}$

其中 $z ? \mathcal{U}。$ 应用引理4可得算子 $J_{n, \sigma}\left(\mathfrak{A}_{n}, \mathfrak{B}_{n}\right)(z) ? \mathcal{S}。$ 证毕。

定理4 设 $i=1, 2, \cdots, n ;-1<A_{i}<B_{i} \leqslant 1 ;-1<C_{i}<D_{i} \leqslant 1 ; \mu_{i}, \eta_{i} ? \gg;\sigma, \lambda_{i}, \theta_{i} ? \mathbb{C}$ 和Re σ>0。若 $p_{i} ? \mathcal{H}_{1}\left(A_{i}, B_{i}\right), q_{i} ? \mathcal{H}_{1}\left(C_{i}, D_{i}\right)$ >且

$\left|M_{n}\right| \leqslant\left\{\begin{array}{l} \operatorname{Re} \sigma \quad(0<\operatorname{Re} \sigma<1), \\ 1 \quad(\operatorname{Re} \sigma \geqslant 1), \end{array}\right.$

其中

$\begin{aligned} M_{n}= & \sum\limits_{i=1}^{n}\left[\left|\lambda_{i}\right|\left(\frac{1-A_{i}}{1-B_{i}}+\left|\mu_{i}\right|\right)^{\tau_{i}}+\left|\theta_{i}\right|\left(\frac{1-C_{i}}{1-D_{i}}+\left|\eta_{i}\right|\right)^{\zeta_{i}}\right] \\ & \left(\tau_{i}, \zeta_{i} ? \mathbb{N}_{+} ; \lambda_{i}^{2}+\theta_{i}^{2} \neq 0\right), \end{aligned}$

则由式(1)定义的积分算子 $J_{n, \sigma}\left(\mathfrak{A}_{n}, \mathfrak{B}_{n}\right)(z) ? \mathcal{S}$ 。

利用式(8)和引理4, 如同定理3的证明方法, 易证定理4。

另外，还可得到条形区域的积分算子的单叶性充分条件：

定理5 设 $i=1, 2, \cdots, n ;-1<B_{i}<A_{i} \leqslant 1 ;-1<C_{i}<D_{i} \leqslant 1 ; \mu_{i}, \eta_{i} ? \gg; \sigma, \lambda_{i}, \theta_{i} ? \mathbb{C}$ 和Re σ>0。若 $p_{i} ? L\left(A_{i}, B_{i}\right), q_{i} ? \mathcal{H}_{1}\left(C_{i}, D_{i}\right)$ , 且

$\left|M_{n}\right| \leqslant \begin{cases}\operatorname{Re} \sigma & (0<\operatorname{Re} \sigma<1), \\ 1 & (\operatorname{Re} \sigma \geqslant 1),\end{cases}$

其中

$\begin{aligned} M_{n}= & \sum\limits_{i=1}^{n}\left[\left|\lambda_{i}\right|\left(\frac{1+A_{i}}{1+B_{i}}+\left|\mu_{i}\right|\right)^{\tau_{i}}+\left|\theta_{i}\right|\left(\frac{1-C_{i}}{1-D_{i}}+\left|\eta_{i}\right|\right)^{\zeta_{i}}\right] \\ & \left(\tau_{i}, \zeta_{i} ? \mathbb{N}_{+}\right), \end{aligned}$

则由式(1)定义的算子 $J_{n, \sigma}\left(\mathfrak{A}_{n}, \mathfrak{B}_{n}\right)(z) ? \mathcal{S}$ 。

证明首先假设函数h(z)由式(12)定义, 再由引理1、引理2、引理4和式(13), 可得

$\begin{array}{l} \frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma}\left|\frac{z h^{\prime \prime}(z)}{h^{\prime}(z)}\right| \leqslant \\ \;\;\;\;\frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma} \sum\limits_{i=1}^{n}\left(\left|\lambda_{i}\right||| p_{i}(z)\left|+\mu_{i}\right|^{\tau_{i}}+\left|\theta_{i}\right||| q_{i}(z)\left|+\eta_{i}\right|^{\zeta_{i}}\right) \leqslant \\ \;\;\;\;\frac{1-|z|^{2 \operatorname{Re} \sigma}}{\operatorname{Re} \sigma} \sum\limits_{i=1}^{n}\left[\left|\lambda_{i}\right|\left(\frac{1+A_{i}|z|}{1+B_{i}|z|}+\left|\mu_{i}\right|\right)^{\tau_{i}}+\right. \\ \;\;\;\;\left.\left|\theta_{i}\right|\left(\frac{1-C_{i}|z|}{1-D_{i}|z|}+\left|\eta_{i}\right|\right)^{\zeta_{i}}\right] \end{array}$

接着类似定理3的证明, 易证结论成立, 详细证明过程略。证毕。

3. 若干推论

定理1中取A=1, B=－1, 可得到如下推论, 该推论是文献[10]中的定理2.1：

推论1^[10] 设 $p ? \mathcal{P}$ 。若 $\operatorname{Re} \sigma>0, \sigma ? \mathbb{C}, \lambda ? \mathbb{C}_{+}$ 且

$|\lambda| \leqslant \begin{cases}\frac{\operatorname{Re} \sigma}{4} & (0<\operatorname{Re} \sigma<1), \\ \frac{1}{4} & (\operatorname{Re} \sigma \geqslant 1), \end{cases}$

则由式(3)定义的算子 $J_{\sigma, \lambda}(p)(z) ? \mathcal{S}$ 。

由定理1和定理2可得到文献[4]中函数类非线性积分算子单叶的充分性：

推论2^[4] 设 $p ? \mathcal{P}(A, B ; C, D)$ 。若－1≤B<A≤1;－1≤C<D≤1和 $\sigma ? \mathbb{C}, \lambda ? \mathbb{C}_{+}$ 且

$|\lambda| \leqslant \\ \;\;\;\;\left\{\begin{array}{l} \min \left\{\frac{\operatorname{Re} \sigma}{(1+|B|)(2+|A+B|)}, \frac{\operatorname{Re} \sigma}{(1+|D|)(2-(C+D))}\right\} \\ \quad(0<\operatorname{Re} \sigma<1), \\ \min \left\{\frac{1}{(1+|B|)(2+|A+B|)}, \frac{1}{(1+|D|)(2-(C+D))}\right\} \\ \quad(\operatorname{Re} \sigma \geqslant 1), \end{array}\right.$

则由式(3)定义的算子 $J_{\sigma, \lambda}(p)(z) ? \mathcal{S}$ 。

若定理3中取 $\mathfrak{B}_{n}(z)=0(i=1, 2, \cdots, n)$ , 可得到以下推论：

推论3 设 $i=1, 2, \cdots, n ;-1<B_i<A_i \leqslant 1 ; \mu_i ? \gg; \sigma, \lambda_i ? \mathbb{C} 和 \operatorname{Re} \sigma>0。若 p_i ? L\left(A_i, B_i\right)$ 且

$\left|M_{n}\right| \leqslant\left\{\begin{array}{l} \operatorname{Re} \sigma \quad(0<\operatorname{Re} \sigma<1), \\ 1 \quad(\operatorname{Re} \sigma \geqslant 1), \end{array}\right.$

其中

$M_{n}=\sum\limits_{i=1}^{n}\left|\lambda_{i}\right|\left(\frac{1+A_{i}}{1+B_{i}}+\left|\mu_{i}\right|\right)^{\tau_{i}} \quad\left(\tau_{i} ? \mathbb{N}_{+}\right),$

则由式(1)定义的算子 $J_{n, \sigma}\left(\mathfrak{A}_{n}, 0\right)(z) ? \mathcal{S}$ 。

在定理3中, 令 $\lambda_{i}=\tau_{i}=\zeta_{i}=\mu_{i}=\eta_{i}=1, \mathfrak{A}_{n}(z)=\$\frac{1}{z} \sum\limits_{i=1}^{n}\left(\frac{z f_{i}^{\prime}}{f_{i}}-1\right)+1, \mathcal{B}_{n}(z)=\frac{1}{z} \sum\limits_{i=1}^{n} \theta_{i}\left(\frac{z g_{i}^{\prime}}{g_{i}}-1\right)+1$ , 可得到以下推论：

推论4 设 $i=1, 2, \cdots, n ; f_{i}, g_{i} ? H ;-1<B_{i}<A_{i} \leqslant 1 ;-1<D_{i}<C_{i} \leqslant 1 ; \sigma, \theta_{i} ? \mathbb{C} 和 \operatorname{Re} \sigma>0 。若 \frac{z f_{i}^{\prime}}{f_{i}} ? L\left(A_{i}, B_{i}\right), \frac{z g_{i}^{\prime}}{g_{i}} ? L\left(C_{i}, D_{i}\right)$ 且

$\left|M_{n}\right| \leqslant \begin{cases}\operatorname{Re} \sigma \quad (0<\operatorname{Re} \sigma<1), \\ 1 \quad(\operatorname{Re} \sigma \geqslant 1),\end{cases}$

其中

$M_{n}=\sum\limits_{i=1}^{n}\left[\left(\frac{1+A_{i}}{1+B_{i}}+1\right)+\left|\theta_{i}\right|\left(\frac{1+C_{i}}{1+D_{i}}+1\right)\right],$

则由式(4)定义的算子 $J_{n, \theta_i}\left(f_i, g_i\right)(z) ? \mathcal{S}$ 。

在定理3中，令 $\sigma=1, \tau_{i}=\zeta_{i}=\mu_{i}=\eta_{i}=1, \mathfrak{A}_{n}(z)=\$\frac{1}{z} \sum\limits_{i=1}^{n} \lambda_{i}\left(\frac{z f_{i}^{\prime \prime}}{f_{i}}\right)+1, \mathcal{B}_{n}(z)=\frac{1}{z} \sum\limits_{i=1}^{n} \theta_{i}\left(\frac{z g_{i}^{\prime}}{g_{i}^{\prime}}-1\right)+1$ , 可得到以下推论, 该推论推广了文献[11]的定理3.3：

推论5 设 $i=1, 2, \cdots, n ; f_{i}, g_{i} ? H ;-1<B_{i}<A_{i} \leqslant 1; -1<D_{i}<C_{i} \leqslant 1 ; \sigma, \lambda_{i}, \theta_{i} ? \mathbb{C}$ 和 $\operatorname{Re} \sigma>0$ 。若 $1+\frac{z f_i^{\prime \prime}}{f_i^{\prime}} ?L\left(A_i, B_i\right), \frac{z g_i^{\prime}}{g_i} ? L\left(C_i, D_i\right)$ , 且

$M_{n}=\sum\limits_{i=1}^{n}\left[\left|\lambda_{i}\right|\left(\frac{1+A_{i}}{1+B_{i}}+1\right)+\left|\theta_{i}\right|\left(\frac{1+C_{i}}{1+D_{i}}+1\right)\right] \leqslant 1,$

则由式(5)定义的算子 $J_{\lambda_{i}, \theta_{i}}\left(f_{i}, g_{i}\right)(z) ? \mathcal{S}$ 。

定理4中，令 $\lambda_{i}=\tau_{i}=\zeta_{i}=\mu_{i}=\eta_{i}=1, \mathfrak{A}_{n}(z)=\$\frac{1}{z} \sum\limits_{i=1}^{n}\left(\frac{z f_{i}^{\prime}}{f_{i}}-1\right)+1, \mathcal{B}_{n}(z)=\frac{1}{z} \sum\limits_{i=1}^{n} \theta_{i}\left(\frac{z g_{i}{ }^{\prime}}{g_{i}}-1\right)+1$ , 可得到以下推论:

推论6 设 $i=1, 2, \cdots, n ; f_{i}, g_{i} ? H ;-1<A_{i}<B_{i} \leqslant 1; -1<C_{i}<D_{i} \leqslant 1 ; \sigma, \theta_{i} ? 和 \operatorname{Re} \sigma>0 。若 \frac{z f_{i}^{\prime}}{f_{i}} ? \mathcal{H}_{1}\left(A_{i}\right., \left.B_{i}\right), \frac{z g_{i}^{\prime}}{g_{i}} ? \mathcal{H}_{1}\left(C_{i}, D_{i}\right)$ 且

$\left|M_{n}\right| \leqslant \begin{cases}\operatorname{Re} \sigma & (0<\operatorname{Re} \sigma<1), \\ 1 & (\operatorname{Re} \sigma \geqslant 1),\end{cases}$

其中

$M_{n}=\sum\limits_{i=1}^{n}\left[\left(\frac{1-A_{i}}{1-B_{i}}+1\right)+\left|\theta_{i}\right|\left(\frac{1-C_{i}}{1-D_{i}}+1\right)\right],$

则由式(4)定义的算子 $J_{n, \theta_{i}}\left(f_{i}, g_{i}\right)(z) ? \mathcal{S}$ 。

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