设$H_k (k=1, 2, \cdots)$表示在单位圆盘$U=\{z:|z|<1\}$内解析函数$f(z)=z+a_{k+1}z^{k+1}+\cdots$的全体构成的类; 以$P_k(\tau)(0\leq\tau<1)$表示$U$内解析并满足条件Re$p(z)>\tau$的所有正实部函数$p(z)=1+p_kz^k+\cdots$的全体; 用$S^*_k{(\tau)}$和$K_k(\tau)$分别表示$H_k$中的$\tau$级星象函数类和$\tau$级凸象函数类.若存在$g(z)\in S^*_k{(\tau)}$, 使得$\frac{zf'(z)}{g(z)}\in P_k(\tau)$则称$f(z)$为$\tau$级近于凸函数, 其全体记为$C_k(\tau)$.

以$\bar{P}_k{(\beta)} (\beta>1)$表示$U$内解析并满足Re$p(z)<\beta$的所有函数$p(z)=1+p_kz^k+\cdots$的全体.若函数$f(z)\in H_k$满足条件$\frac{zf'(z)}{f(z)}\in \bar{P}_k(\beta)$, 则称$f(z)$属于类函数$M_k(\beta)$; 若函数$f(z)\in H_k$满足条件$\frac{(zf'(z))'}{f'(z)}\in \bar{P}_k(\beta)$, 则称$f(z)$属于函数类$N_k(\beta)$; 用$\bar{C}M_k(\beta)$和$\bar{C}N_k(\beta)$分别定义$H_k$中两个子类：

$\begin{array}{l} \bar C{M_k}(\beta ) = \{ f \in {H_k}:\exists g \in {M_k}(\beta ), {\rm{Re}}\frac{{zf'(z)}}{{g(z)}} < \beta, z \in U\}, \\ \bar C{N_k}(\beta ) = \{ f \in {H_k}:\exists g \in {N_k}(\beta ), {\rm{Re}}\frac{{{{(zf'(z))}^\prime }}}{{g'(z)}} < \beta, z \in U\} . \end{array}$

显然$f(z)\in N_k(\beta)\Leftrightarrow zf'(z)\in M_k(\beta), f(z)\in \bar{C}N_k(\beta)\Leftrightarrow zf'(z) \in \bar{C}M_k(\beta).$

文[1-3]中讨论了函数类$\bar{K}_1(\beta)$和$\bar{S}_1^*(\beta)$的性质.

设

$g(z) = z + \sum\limits_{n = 1}^\infty {{b_{k + n}}} {z^{k + n}} \in {H_k}, f(z) = z + \sum\limits_{n = 1}^\infty {{a_{k + n}}} {z^{k + n}} \in {H_k}, $

则用$(f*g)(z)$表示$f(z)$和$g(z)$的Hadamard卷

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

Carlson和Shaffer在文[4]中引进比$D^{\lambda}$更为广泛的线性算子$L(a, c)$.设

$\begin{array}{l} \phi (a, c;z) = \sum\limits_{n = 0}^\infty {\frac{{{{(a)}_n}}}{{{{(c)}_n}}}} {z^{n + 1}}, z \in U, c \ne 0, - 1, - 2, \cdots, \\ L(a, c)f(z) = \phi (a, c;z)*f(z), f(z) \in {H_k}, \end{array}$

(1.1)

其中$(\xi)_n=\frac{\Gamma(\xi+n)}{\Gamma(\xi)}$.由文[4]知$L(a, c)$是$H_k$到自身$H_k$的连续映射, 还容易得到

$\varphi (2(\beta - 1), 1;z) = \frac{z}{{{{(1 - z)}^{2(\beta - 1)}}}}.$

(1.2)

另外, 对$c>a>0, L(a, c)f(z)$有积分表达式

$L(a, c)f(z) = \int_0^1 {{u^{a - 1}}} f(uz)d\eta (a, c - a)(u), $

(1.3)

其中$\eta$是$B$分布$d \eta(a, c-a)(u)=[u^{a-1}(1-u)^{c-a-1} / B(a, c-a)]du. $如果$a\neq0, -1, -2, \cdots$, 则$L(a, c)$存在连续逆映射$L(c, a)$, 从而$L(a, c)$是$H_k$到自身$H_k$的双方单值映射, 显然

$ L(a, c)=L(a, b)L(b, c)=L(b, c)L(a, b), b, c\neq-1, -2, \cdots. $

设$F(a, b;c;z)$为超几何分布函数:

$ F(a, b;c;z)=\sum\limits_{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_n}\frac{z^{n+1}}{k!}, z\in U;a, b, c\neq 0, -1, -2, \cdots, $

显然有

$\frac{z}{{{{(1 - z)}^{2(\beta - 1)}}}} = \varphi (2(\beta - 1), 1;z) = F(2(\beta - 1), 1;1;z).{\rm{ }}$

(1.4)

令

$f_\lambda(z)=\frac{z}{(1-z)^{\lambda+1}}(\lambda>-1).$

Choi等人在文献[5]中引进Choi-Saigo-Srivastava算子$L_{\lambda, \mu}:H_k\rightarrow H_k$,

${L_{\lambda, \mu }}f(z) = ({f_{\lambda, \mu }}*f)(z)(f(z) \in {H_k};\lambda > - 1;\mu > 0), {\rm{ }}$

(1.5)

其中函数$f_{\lambda, \mu}$满足条件

$({f_\lambda }*{f_{\lambda, \mu }})(z) = \frac{z}{{{{(1 - z)}^\mu }}}, (f(z) \in {H_k}, \lambda > - 1;\mu > 0).{\rm{ }}$

(1.6)

特别地, $L_{0, 2}=zf', L_{1, 2}f=f$.从(1.1) 式和(1.5) 式得到^[6]

${L_{\lambda, \mu }}f(z) = L(\mu, \lambda + 1)f(z)(\lambda > - 1;\mu > 0), {\rm{ }}$

(1.7)

由(1.7) 式得到^[6]

$f(z) = L(\lambda + 1, \mu ){L_{\lambda, \mu }}f(z)(\lambda > - 1;\mu > 0).{\rm{ }}$

(1.8)

文[6, 8]中分别用$L_{\lambda, \mu}$算子引进函数类:

定义A^[6] 若函数$f(z)\in H_k$满足条件

${\hbox{Re}} \frac{L(2, 1)L(\mu, \lambda+1)f(z)}{L(\mu, \lambda+1)f(z)}>\tau (\tau<1, z\in U), $

则称$f(z)$属于$K(\lambda, \mu)(\tau)$中.

定义B^[8] 若函数$f(z)\in H_k$满足条件

${\hbox{Re}} \{(1-\alpha)\frac{L(\lambda+1, 1)f(z)}{z}+\alpha \frac {L(2, 1)L(\lambda+1, 1)f(z)}{z}\}>\tau (0\leq \alpha, 0\leq\tau<1, z\in U), $

则称$f(z)$属于$V_{k, \lambda}(\alpha, \tau)$中.

定义C^[8] 设函数$f(z)\in H_k$, 若存在函数$g(z)\in V_{k, \lambda}(\alpha, \tau)$, 使得

${\hbox{Re}} \frac{L(2, 1)L(\lambda+1, 1)f(z)}{L(\lambda+1, 1)g(z)}>\eta (0\leq \eta<1, z\in U), $

则称$f(z)$属于$Q_{k, \lambda}(\alpha, \tau, \eta)$中.

文[6]中利用算子$L_{\lambda, \mu}$研究函数类$K(\lambda, \mu)(\tau)$的性质, 得到类中函数的包含关系、偏差定理, 给出了类中函数的系数不等式和Hadamard乘积性质；作者在文[7]中引进并讨论函数类

$Q_{1, \lambda}\Big(\alpha, (\lambda+\tau)/(\lambda+1), (\lambda+\eta)/(\lambda+1)\Big)$

的性质, 在文[8]中进一步推广文献[7]中函数类, 利用算子理论研究函数类的$Q_{k, \lambda}(\alpha, \tau, \eta)$性质, 得到类中函数的积分表达式, 偏差定理, 给出了类中函数的近于凸半径和卷积性质.

本文中利用算子$L_{\lambda, \mu}$引进如下函数类：

定义1 若函数$f(z)\in H_k$满足条件

${\rm{Re}}\{ (1 - \alpha )\frac{{{L_{\lambda, \mu }}f(z)}}{z} + \alpha ({L_{\lambda, \mu }}f(z))'\} < \beta (\alpha \ge 0, \beta > 1, z \in U), {\rm{ }}$

(1.9)

则称$f(z)$属于$\bar{V}_{k, \lambda, \mu}(\alpha, \beta)$中.

定义2 设函数$f(z)\in H_k$, 若存在函数$g(z)\in \bar{V}_{k, \lambda, \mu}(\alpha, \beta)$, 使得

${\rm{Re}}\frac{{z({L_{\lambda, \mu }}f(z))'}}{{{L_{\lambda, \mu }}g(z)}} < \rho (\rho > 1, z \in U), {\rm{ }}$

(1.10)

则称$f(z)$属于$\bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho)$中.显然

$\begin{array}{l} {{\bar Q}_{k, 0, 2}}(\alpha, \beta, \rho ) = \{ f(z) \in {H_k}:\exists g(z) \in {{\bar V}_{k, 0, 2}}(\alpha, \beta ), {\rm{Re}}\frac{{zf'(z)}}{{g(z)}} < \rho, z \in U\}, \\ {{\bar Q}_{k, 1, 2}}(\alpha, \beta, \rho ) = \{ f(z) \in {H_k}:\exists g(z) \in {{\bar V}_{k, 1, 2}}(\alpha, \beta ), {\rm{Re}}\frac{{{{(zf'(z))}^\prime }}}{{g'(z)}} < \rho, z \in U\} . \end{array}$

本文研究函数类$\bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho)$, 导出类中函数的积分表达式；借助算子理论建立该类中函数的偏差定理, 讨论类中函数的卷积性质和近于凸半径.

利用算子(1.7) 式, 可将(1.8) 是和(1.9) 式写成

${\rm{Re}}\{ (1 - \alpha )\frac{{L(\mu, \lambda + 1)f(z)}}{z} + \alpha \frac{{L(2, 1)L(\lambda + 1, 1)f(z)}}{z}\} < \beta, (\alpha \ge 0, \beta > 1, z \in U)$

(1.11)

和

${\rm{Re}}\frac{{L(2, 1)L(\mu, \lambda + 1)f(z)}}{{L(\mu, \lambda + 1)g(z)}} < \rho, z \in U.$

(1.12)

引理2.1 设$p(z)\in \bar{P}_{k}(\beta) (\beta>1)$, 则存在$X=\{ x:|x|=1\}$上连续的概率测度$\mu(x)$, 使得

$p(z) = \int_{|x| = 1} {\frac{{1 - (2\rho - 1)xz}}{{1 - xz}}} d\mu (x).{\rm{ }}$

(2.1)

证  由$p(z)\in \bar{P}_k(\beta)$可知, Re$\Big(\frac{\beta-p(z)}{\beta-1}\Big)>0$, 利用正实部函数的Herglots表示公式^[9]得

$\frac{\beta-p(z)}{\beta-1}=\int_{|x|=1}\frac{1+xz}{1-xz}d \mu(x).$

由此推出(2.1) 式成立.

利用引理2.1, 结合算子$L_{\lambda, \mu}$的可逆性, 用文[8]中相同的方法不难证明:

定理2.1 设$\alpha>0, \beta>1$, 若$g(z)\in \bar{V}_{k, \lambda}(\alpha, \beta)$, 则存在$X=\{ x:|x|=1\}$上连续的概率密度$\eta(x)$, 使得

$g(z) = L(\lambda + 1, \mu )\{ \frac{1}{{\alpha {z^{\frac{1}{\alpha } - 1}}}}\int_0^z {{t^{\frac{1}{\alpha } - 1}}} [\int_{|x| = 1} {\frac{{1-(2\beta-1)tx}}{{1-tx}}} d\eta (x)]dt\} {\rm{ }}$

(2.2)

或存在$p(z)\in \bar{P}_k(\beta)$, 使得$g(z)=L(\lambda+1, \mu)\Big\{\frac{1}{\alpha z^{\frac{1}{\alpha}-1}}\int_0^z t^{\frac{1}{\alpha}-1}p(t)dt\Big\}.$对于固定的$\lambda, \alpha, \beta$, 函数类$\bar{V}_{k, \lambda, \mu}(\alpha, \beta)$与$X$上左连续的概率测度连续点$\eta(x)$以上述关系构成一一对应.

定理2.2 函数$f(z)\in \bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho)$当且仅当存在$X=\{x:|x|=1\}$上左连续的概率测度$\eta(x)$, $\mu(x)$, 使得

$\begin{array}{l} f(z) = L(\lambda + 1, \mu )L(1, 2)\{ [\frac{1}{{\alpha {z^{\frac{1}{\alpha }-1}}}}\int_0^z {{t^{\frac{1}{\alpha }-1}}} (\int_{|x| = 1} {\frac{{1-(2\beta - 1)tx}}{{1 - tx}}} d\eta (x))dt]\\ [\int_{|x| = 1} {\frac{{1-(2\rho-1)xt}}{{1-xt}}} d\eta (x)]\} . \end{array}$

(2.3)

当$\lambda=0, \mu=1$时,

$\begin{array}{l} f(z) = L(1, 2)\{ [\frac{1}{{\alpha {z^{\frac{1}{\alpha }-1}}}}\int_0^z {{t^{\frac{1}{\alpha }-1}}} (\int_{|x| = 1} {\frac{{1-(2\beta - 1)tx}}{{1 - tx}}} d\eta (x))dt]\\ [\int_{|x| = 1} {\frac{{1-(2\rho-1)tx}}{{1-tx}}} d\mu (x)]\} . \end{array}$

(2.4)

对于固定的$\lambda, \alpha, \mu, \rho$, 函数类$\bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho)$与$X$上左连续的概率测度点$\Big\{\Big (\eta(x), \mu(x)\Big)\Big\}$以上述关系式(2.3) 构成一一对应.

证 设$f(z)\in\bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho)$, 则存在$g(z)\in \bar{V}_{k, \lambda, \mu}(\alpha, \beta)$, 使得

${\hbox{Re}} \frac{L(2, 1)L(\mu, \lambda+1)f(z)}{L(\mu, \lambda+1)g(z)}<\rho, z\in U.$

由定理1, 可得

$g(z) = L(\lambda + 1, \mu )\{ \frac{1}{{\alpha {z^{\frac{1}{\alpha } - 1}}}}\int_0^z {{t^{\frac{1}{\alpha } - 1}}} [\int_{|x| = 1} {\frac{{1-(2\beta-1)tx}}{{1-tx}}} d\eta (x)]dt\}, {\rm{ }}$

(2.5)

其中$\eta(x)$为$X$上左连续的概率测度.

由引理2.1得

$\frac{{L(2, 1)L(\mu, \lambda + 1)f(z)}}{{L(\mu, \lambda + 1)g(z)}} = \int_{|x| = 1} {\frac{{1 - (2\rho - 1)xz}}{{1 - xz}}} d\mu (x), $

(2.6)

其中$\mu(x)$为$X$上左连续的概率测度.

由(2.5), (2.6) 两式推出

$\begin{array}{l} L(2, 1)L(\mu, \lambda + 1)f(z) = \{ [\frac{1}{{\alpha {z^{\frac{1}{\alpha }-1}}}}\int_0^z {{t^{\frac{1}{\alpha }-1}}} (\int_{|x| = 1} {\frac{{1-(2\beta - 1)tx}}{{1 - tx}}} d\eta (x))dt]\\ [\int_{|x| = 1} {\frac{{1-(2\rho-1)zx}}{{1-zx}}} d\mu (x)]\} . \end{array}$

利用$L(\mu, \lambda+1)$算子的可逆性, 从上式即得(2.3) 式, 反之亦然；当$\lambda=0$时, (2.3) 式变为(2.4) 式.对于固定的$\lambda, \mu, \alpha, \beta, \rho$, 因$\big\{\big(\eta(x), \mu(x)\big)\big\}$与$\bar{P}_k\times\bar{P}_k$之间构成一一对应, 而$\bar{P}_k\times\bar{P}_k$与函数类$\bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho)$之间也是一一对应, 这表明定理的后一个结论为真.

引理3.1^[10] 设$p(z)=1+p_kz^k+\cdots\in P_k(0)(z\in U, k\geq1)$, 则对$|z|=r<1$, 有

$\frac{{1 - {r^k}}}{{1 + {r^k}}} \le {\rm{Re}}p(z) \le |p(z)| \le \frac{{1 + {r^k}}}{{1 - {r^k}}}.$

若Re $p(z)<\beta(\beta>1)$, 置$q(z)=\frac{\beta-p(z)}{\beta-1}$, 则Re $q(z)>0$, 且$p(z)=\beta-(\beta-1)q(z)$, 于是由引理3.1推出

引理3.2 设$p(z)=1+p_kz^k+\cdots\in\bar{P}_k(\beta>1, z\in U, k\geq1)$, 则对$|z|=r<1$, 有

$\frac{{1 - (2\beta - 1){r^k}}}{{1 - {r^k}}} \le {\rm{Re}}p(z) \le |p(z)| \le \frac{{1 + (2\beta - 1){r^k}}}{{1 + {r^k}}}.{\rm{ }}$

定理3.1 设$\alpha>0, \beta>1, \rho>1, f(z)\in\bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho)$, 则对$|z|=r<1$, 有

$\begin{array}{l} \frac{{1 - (2\rho - 1){r^k}}}{{\alpha (1 - {r^k})}}\int_0^1 {{t^{\frac{1}{\alpha } - 1}}} \frac{{1 - (2\beta - 1){{(rt)}^k}}}{{1 - {{(rt)}^k}}}dt \le |L(2, 1)L(\mu, \lambda + 1)f(z)|\\ \le \frac{{1 + (2\rho - 1){r^k}}}{{\alpha (1 + {r^k})}}\int_0^1 {{t^{\frac{1}{\alpha } - 1}}} \frac{{1 + (2\beta - 1){{(rt)}^k}}}{{1 + {{(rt)}^k}}}dt. \end{array}$

(3.1)

证 设$f(z)\in \bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho)$, 则存在$g(z)\in\bar{V}_{k, \lambda, \mu}(\alpha, \beta)$, 使得

$ {\hbox{Re}} \frac{L(2, 1)L(\mu, \lambda+1)f(z)}{L(\mu, \lambda+1)g(z)}<\rho, z\in U.$

令

$ \frac{L(2, 1)L(\mu, \lambda+1)f(z)}{L(\mu, \lambda+1)g(z)}=q(z), z\in U, $

则Re$ q(z)<\rho$.

首先证明$|L(\mu, \lambda+1)g(z)|$的偏差性质, 因$g(z)\in \bar{V}_{k, \lambda, \mu}(\alpha, \beta)$存在Re $p(z)<\beta$.由引理3.2, 有

$\begin{array}{l} |L(\mu, \lambda + 1)g(z)|\; \ge \;{\rm{Re}}(L(\mu, \lambda + 1)g(z)) = {\rm{Re}}\{ \frac{1}{{\alpha {z^{\frac{1}{\alpha } - 1}}}}\int_0^1 {{t^{\frac{1}{\alpha } - 1}}} \frac{{1 - (2\beta - 1){{(rt)}^k}}}{{1 - {{(rt)}^k}}}dt\} \\ > \frac{1}{\alpha }\int_0^1 {{t^{\frac{1}{\alpha } - 1}}} \frac{{1 - (2\beta - 1){t^k}}}{{1 - {t^k}}}dt, \end{array}$

(3.2)

$\begin{array}{l} |L(\mu, \lambda + 1)g(z)|\; = \;|\{ \frac{1}{{\alpha {z^{\frac{1}{\alpha } - 1}}}}\int_0^1 {{t^{\frac{1}{\alpha } - 1}}} p(t)dt\} | \le \frac{1}{\alpha }\int_0^1 {{t^{\frac{1}{\alpha } - 1}}} |p(zt)|dt\\ \le \frac{1}{\alpha }\int_0^1 {{t^{\frac{1}{\alpha } - 1}}} \frac{{1 + (2\beta - 1){{(rt)}^k}}}{{1 + {{(rt)}^k}}}dt. \end{array}$

(3.3)

因为

$L(2, 1)L(\mu, \lambda+1)f(z)=q(z)L(\mu, \lambda+1)g(z), z\in U, $

且

$|L(2, 1)L(\mu, \lambda + 1)f(z)| = |q(z)L(\lambda + 1, 1)g(z)|, z \in U, $

(3.4)

由(3.2), (3.3) 式和引理3.2以及引理2.1, 得到

$\begin{array}{l} \frac{{1 - (2\rho - 1){r^k}}}{{\alpha (1 - {r^k})}}\int_0^1 {{t^{\frac{1}{\alpha } - 1}}} \frac{{1 - (2\beta - 1){{(rt)}^k}}}{{1 - {{(rt)}^k}}}dt \le |q(z) \cdot L(\mu, \lambda + 1)g(z)|\\ \le \frac{{1 + (2\rho - 1){r^k}}}{{\alpha (1 + {r^k})}}\int_0^1 {{t^{\frac{1}{\alpha } - 1}}} \frac{{1 + (2\beta - 1){{(rt)}^k}}}{{1 + {{(rt)}^k}}}dt, \end{array}$

再利用(3.4) 式即得(3.1) 式.证毕

对函数

$f(z) = L(\lambda + 1, \mu )L(1, 2)\{ \frac{{1 - (2\rho - 1){z^k}}}{{\alpha (1 - {z^k}){z^{\frac{1}{\alpha } - 1}}}}\int_0^z {{t^{\frac{1}{\alpha } - 1}}} \frac{{1 - (2\beta - 1){t^k}}}{{1 - {t^k}}}dt\} {\rm{ }}$

(3.5)

分别取$z=re^{i\frac{\pi}{k}}, z=r$时不等式(3.1) 式的等号成立.

引理4.1^[11] 设$\varphi(z), g(z)$在$U$内解析, 满足$\varphi(0)=h(0), \varphi'(0)\neq0, h'(0)\neq0$, 并且对每个适合$|\sigma|=|\tau|=1$的复数$\sigma, \tau$, 有

$\varphi(x)*\frac{1+\tau\sigma z}{1-\sigma z}h(z)\neq0(0<|z|<1).$

设$F(z)$在$U$内解析, 且Re $F(z)>0(0<|z|<1)$, 则

${\hbox{Re}} \Big\{\frac{\varphi*\big(F(z)h(z)\big)}{\varphi*h(z)}\Big\}>0 (0<|z|<1).$

定理4.1 设$\sigma, \tau$满足$|\sigma|=|\tau|=1$的复数,

$f(z)\in\bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho), \phi(z)=z+\sum\limits_{n=k}^{\infty}\alpha_{k+1}z^{k+1}$

在$U$内解析, 且$\varphi(z)*\frac{1+\tau\sigma z}{1-\sigma }z\neq0 (0<|z|<1), $则$f*\phi(z)\in\bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho)$.

证 (i)先证明$g*\phi(z)\in \bar{V}_{k, \lambda, \mu}(\alpha, \beta)$.设

$F(z)=\beta-[(1-\alpha)\frac{L(\mu, \lambda+1)g(z)}{z}+\alpha \frac {L(2, 1)L(\lambda+1, 1)g(z)}{z}], h(z)=z, $

则$F(z)$在$U$内解析, 且Re$F(z)>0(z\in U), \varphi*h(z)=z.$

由于

$\begin{array}{l} \varphi *Fh(z) = \varphi (z)*[\beta z-((1-\alpha )L(\mu, \lambda + 1)g(z) + \alpha L(2, 1)L(\mu, \lambda + 1)g(z))]\\ = \beta z - [(1-\alpha )\varphi *L(\mu, \lambda + 1)g(z) + \alpha \varphi *L(2, 1)L(\mu, \lambda + 1)g(z)]\\ = \beta z - [(1-\alpha )L(\mu, \lambda + 1)(\varphi *g)(z) + \alpha L(2, 1)L(\mu, \lambda + 1)(\varphi *g)(z)]. \end{array}$

(4.1)

利用引理4.1, 得

$\begin{array}{l} {\rm{Re}}\{ \frac{{\varphi *(F(z)h(z))}}{{\varphi *h(z)}}\} = \beta - {\rm{Re}}\{ (1 - \alpha )\frac{{L(\mu, \lambda + 1)(\phi *g)(z)}}{z}\\ + \alpha \frac{{L(2, 1)L(\mu, \lambda + 1)(\varphi *g)(z)}}{z}\} > 0. \end{array}$

(4.2)

即

${\hbox{Re}} \Big\{(1-\alpha)\frac{L(\mu, \lambda+1)(\varphi*g)(z)}{z}+\alpha \frac{L(2, 1)L(\mu, \lambda+1)(\varphi*g)(z)}{z} \Big\}<\beta(z\in U), $

从而$g*\phi(z)\in \bar{V}_{k, \lambda, \mu}(\alpha, \beta)$.

(ii)下面要证明$f*\varphi(z)\in \bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho)$, 设$f(z)\in \bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho)$.

令

$p(z) = \rho - \frac{{L(2, 1)L(\mu, \lambda + 1)f(z)}}{{L(\mu, \lambda + 1)g(z)}}, h(z) = z, $

则$p(z)$在$U$内解析, 且Re$(p(z)>0)(z\in U), \varphi*h(z)=z$.由于

$\phi *L(\mu, \lambda + 1)g(z) \cdot p(z) = \rho \phi *L(\mu, \lambda + 1)g(z) - \phi *L(2, 1)L(\mu, \lambda + 1)f(z), $

(4.3)

注意到

$\begin{array}{l} \phi *L(\mu, \lambda + 1)g(z) = L(\mu, \lambda + 1)(\phi *g)(z), \\ \phi *L(2, 1)L(\mu, \lambda + 1)f(z) = L(2, 1)L(\mu, \lambda + 1)(\phi *f)(z). \end{array}$

由(4.3) 式得到

${\rm{Re}}p(z) = {\rm{Re}}\{ \frac{{L(2, 1)L(\mu, \lambda + 1)(\phi *f)(z)}}{{L(\mu, \lambda + 1)(\phi *g)(z)}}\} < \rho, $

其中由(i)可知$\phi *{\rm{ g(z)}} \in {\overline V _{k,\lambda ,\mu }}(\alpha ,\beta )$.故$f*\varphi(z)\in\bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho)$.

从引理3.2得到函数类$\bar{P}_k(\beta)$与$P_k(\tau)(0\leq\tau<1)$的包含关系.

引理5.1设 $\beta>1, z\in U, k \in N, p(z)=1+p_kz^k+\cdots\in\bar{P}_k(\beta)$, 则对$|z|=r<r_1$, 有

${\rm{Re}}p(z) \ge \frac{{1-(2\beta-1){r^k}}}{{1-{r^k}}} > 0(或p(z) \in {P_k}(0)),$

(5.1)

其中

${r_1} = \mathop {\inf }\limits_{k \ge 1} \{ \frac{1}{{\sqrt[k]{{2\beta - 1}}}}\} .$

(5.2)

令

$P_{\tau, k}=\Big\{p(z)=1+p_kz^k+\cdots:p(z)\text{在}U \text{内解析且}|p(z)-\frac{1}{2\tau}|\leq\frac{1}{2\tau} \Big\} (0\leq\tau<1).$

由文[12]给出$P_{\tau, k}$和$P_k(\tau)$一种关系.

若$q(z)=\frac{1}{p(z)}, p(z)\in P_{\tau, k}$, 则$q(z)\in P_k(\tau)$且

$\begin{array}{l} \frac{{q'(z)}}{{q(z)}} = - \frac{{p'(z)}}{{p(z)}}, \\ |\frac{{p'(z)}}{{p(z)}}| \le \frac{{2k(1 - \tau )|z{|^{k - 1}}}}{{(1 - |z{|^k})(1 + (1 - 2\tau )|z{|^k})}}(z \in U). \end{array}$

(5.3)

引理5.2^[12] 设$p(z)\in P_{\tau, k}(0\leq\tau<1)$, 则

$\Big| \frac{p'(z)}{p(z)}\Big|\leq\frac{2k(1-\tau)|z|^{k-1}}{(1-|z|^{k})(1+(1-2\tau)|z|^k)} (z\in U).$

利用(5.3) 式和引理5.2容易推出

引理5.3 设$q(z)\in P_k(\tau)(0\leq\tau<1)$, 则对$|z|=r<1$, 有

$|\frac{{zq'(z)}}{{q(z)}}| \le \frac{{2k(1 - \tau ){r^k}}}{{(1 - {r^k})(1 + (1 - 2\tau ){r^k})}}(z \in U).$

(5.4)

定理5.1 设$\alpha>0, \beta>1, \rho>1, k\geq2, f(z)\in \bar{Q}_{k, \lambda, \mu}(\alpha, \beta, \rho)$, 则函数$L(\mu, \lambda+1)f(z)$在$|z|<R=\min\{ r_1, r_2\}$内是属于$\bar{C}M_k(\rho)$, $r_1$由(5.2) 式确定, $r_2$为方程

$(\rho - 1) - 2[m(\rho-1) + k(1-m)]{r^k} - (\rho - 1)(1 - 2m){r^{2k}} = 0$

(5.5)

的最小正跟, 其中

$m = \frac{1}{\alpha }\int_0^1 {{t^{\frac{1}{\alpha } - 1}}} \frac{{1 - (2\beta - 1){r^k}}}{{1 - {r^k}}}dt \in (0, 1).$

(5.6)

证 设$f(z)\in\bar{Q}_{k, \lambda, \rho}(\alpha, \beta, \mu)$, 存在函数$g(z)\in\bar{V}_{k, \lambda, \mu}(\alpha, \beta)$.

下面只要证明$L(\mu, \lambda+1)g(z)$在$|z|<R$内属于$M_k(\rho)$即可.令

$F(z) = \frac{{L(\mu, \lambda + 1)g(z)}}{z}, $

(5.7)

则$F(z)$在$U$内解析.若$g(z)\in\bar{V}_{k, \lambda, \mu}(\alpha, \beta)$, 则利用定理2.1和引理5.1, 存在$ p(z)\in\bar{P}_k(\beta)$, 使得

$\begin{array}{l} {\rm{Re}}F(z) = {\rm{ Re}}[\frac{{L(\mu, \lambda + 1)g(z)}}{z}] = {\rm{ Re}}\{ \frac{1}{{\alpha {z^{\frac{1}{\alpha }}}}}\int_0^1 {{t^{\frac{1}{\alpha } - 1}}} p(t)dt\} \\ > m = \frac{1}{\alpha }\int_0^1 {{t^{\frac{1}{\alpha } - 1}}} \frac{{1 - (2\beta - 1){r^k}}}{{1 - {r^k}}}dt. \end{array}$

(5.8)

利用(5.7) 和(5.8) 式和引理5.3, 得到

$\begin{array}{l} {\rm{Re}}\{ \frac{{z{{(L(\lambda + 1, \mu )g(z))}^\prime }}}{{L(\lambda + 1, \mu )g(z)}}\} = 1 + {\rm{ Re}}\frac{{zF'(z)}}{{F(z)}} \le 1 + |\frac{{zF'(z)}}{{F(z)}}|\\ \le \frac{{(1 - {r^k})(1 + (1 - 2m){r^k}) + 2k(1 - m){r^k}}}{{(1 - {r^k})(1 + (1 - 2m){r^k})}}. \end{array}$

若

$\frac{{(1 - {r^k})(1 + (1 - 2m){r^k}) + 2k(1 - m){r^k}}}{{(1 - {r^k})(1 + (1 - 2m){r^k})}} < \rho, $

(5.9)

则

${\rm{ Re}}\{ \frac{{z{{(L(\lambda + 1, \mu )g(z))}^\prime }}}{{L(\lambda + 1, \mu )g(z)}}\} < \rho, $

(5.8) 式等价于

$\frac{{(\rho - 1) - [m(\rho-1) + k(1-m)]{r^k} - (\rho - 1)(1 - 2m){r^{2k}}}}{{(1 - {r^k})(1 + (1 - 2m){r^k})}} > 0.$

(5.10)

设$\varphi(r)=(\rho-1)-2[m(\rho-1)+k(1-m)]r^k-(\rho-1)(1-2m)r^{2k}$, 则$\varphi(r)$在$[0, 1]$内连续, 且$\varphi(0)=\rho-1>0$, $\varphi(1)=-2k(1-m)<0$, 所以方程(5.5) 的最小正根$r_2$在$(0, 1)$内, 取$R=\min\{r_1, r_2 \}$, 则函数$L(\mu, \lambda+1)g(z)$在$M_k(\rho)$内属于, 由此推出$L(\mu, \lambda+1)f(z)$在$|z|<R$内属于$\bar{C}M_k(\rho)$.

注 在定理(2.1) 至定理(5.1) 中分别取$\lambda=0, \lambda=1$时, 就得到函数类$\bar{Q}_{k, 0, 2}(\alpha, \beta, \rho)$和$\bar{Q}_{k, 1, 2}(\alpha, \beta, \rho)$中函数的相应的结果.