记$ {\cal H} $为定义在单位圆盘$ {\Bbb D} $上的调和映射$ f(z) = h(z)+\overline{g(z)} $且满足规范化条件$ f(0) = f_{z}(0)-1 = 0 $的全体,其中$ h(z) $和$ g(z) $在$ {\Bbb D} $上解析,具有展开式

(1.1) $ \begin{align} h(z) = z+\sum\limits_{n = 2}^{\infty}a_{n}z^{n}, g(z) = \sum\limits_{n = 1}^{\infty}b_{n}z^{n}. \end{align} $

记$ {\cal S}_{{\cal H}} $表示$ {\cal H} $中单叶保向调和映射的全体. Lewy^[1]证明了调和映射$ f(z) $在$ {\Bbb D} $内局部单叶当且仅当$ J_{f}(z) = |f_{z}(z)|^{2}-|f_{\overline{z}}(z)|^{2}\neq0 $.记$ {\cal K}_{{\cal H}} $和$ {\cal S}^{*}_{{\cal H}} $分别为单叶保向调和映射,把$ {\Bbb D} $满射到凸域和星形域的函数全体.进一步,记$ {\cal K}_{{\cal H}}^{0} $, $ {\cal S}^{\ast 0}_{{\cal H}} $和$ {\cal S}_{{\cal H}}^{0} $分别表示$ {\cal K}_{{\cal H}} $, $ {\cal S}^{\ast}_{{\cal H}} $和$ {\cal S}_{{\cal H}} $中满足规范化条件$ f_{\bar{z}}(0) = b_{1} = 0 $的函数全体.

Clunie与Sheil-Small^[2]猜想:如果$ f(z)\in{\cal S}_{{\cal H}}^{0} $,则

(1.2) $ \begin{align} |a_{n}|\leq\frac{(2n+1)(n+1)}{6}, |b_{n}|\leq\frac{(2n-1)(n-1)}{6}, n\geq1. \end{align} $

取$ f(z) = \frac{z-\frac{1}{2}z^{2}+\frac{1}{6}z^{3}}{(1-z)^{3}}+\overline{\frac{\frac{1}{2}z^{2}+\frac{1}{6}z^{3}}{(1-z)^{3}}} $时(1.2)式中等号成立.此外,他们在文献[2]中证明了:若$ f(z)\in{\cal K}_{{\cal H}}^{0} $,则

(1.3) $ \begin{align} |a_{n}|\leq\frac{n+1}{2}, |b_{n}|\leq\frac{n-1}{2}, n\geq1. \end{align} $

当$ f(z) = \frac{z-\frac{1}{2}z^{2}}{(1-z)^{2}}+\overline{\frac{-\frac{1}{2}z^{2}}{(1-z)^{2}}}, $ (1.3)式等号成立.

如果定义在$ {\Bbb D} $上的调和映射$ f(z) $且满足$ f(0) = 0 $把每一个$ \{z: |z| = r<1\} $一对一的满射到一个关于原点的星形域上且满足$ \frac{\partial}{\partial\theta}\left(\mbox{arg}\left(f(r{\rm e}^{{\rm i}\theta})\right)\right)>\alpha, $其中$ \theta\in[0, 2\pi) $, $ r\in(0, 1) $且$ \alpha\in[0, 1) $,称$ f(z) $是$ \alpha $阶完全星形的.特别地,若$ \alpha = 0 $,称$ f(z) $为完全星形.

如果定义在$ {\Bbb D} $上的调和映射$ f(z) $把每一个$ \{z: |z| = r<1\} $一对一的满射到一个凸域上且满足$ \frac{\partial}{\partial\theta}\left(\mbox{arg}\frac{\partial}{\partial\theta}\left(f(r{\rm e}^{{\rm i}\theta})\right)\right)>\alpha, $其中$ \theta\in[0, 2\pi) $, $ r\in(0, 1) $且$ \alpha\in[0, 1) $,称$ f(z) $是$ \alpha $阶完全凸的.特别地,若$ \alpha = 0 $,称$ f(z) $为完全凸.

记$ {\cal FK}_{{\cal H}}(\alpha) $和$ {\cal FS}_{{\cal H}}^{*}(\alpha) $分别为$ {\cal K}_{{\cal H}} $中的$ \alpha $阶完全凸函数类和$ {\cal S}_{{\cal H}}^{*} $中的$ \alpha $阶完全星形函数类.对于$ f(z)\in{\cal H} $, Jahangiri^[3]给出了下面结果,内容如下.

定理  A^[3]   设$ f(z) = h(z)+\overline{g(z)} $,其中$ h(z) $和$ g(z) $满足(1.1)式.若

$ \sum\limits_{n = 2}^{\infty} \frac{n-\alpha}{1-\alpha}|a_{n}| +\sum\limits_{n = 1}^{\infty}\frac{n+\alpha}{1-\alpha}|b_{n}|\leq1, $

其中$ \alpha\in[0, 1) $.则$ f(z) $在$ {\Bbb D} $是单叶调和的,且$ f(z)\in{\cal FS}_{{\cal H}}^{*}(\alpha) $.

此外,他们也还在文献[4]中证明了如下结果.

定理 B^[4]  设$ f(z) = h(z)+\overline{g(z)} $,其中$ h(z) $和$ g(z) $满足(1.1)式.若

$ \sum\limits_{n = 2}^{\infty} \frac{n(n-\alpha)}{1-\alpha}|a_{n}| +\sum\limits_{n = 1}^{\infty}\frac{n(n+\alpha)}{1-\alpha}|b_{n}|\leq1, $

其中$ \alpha\in[0, 1) $.则$ f(z) $在$ {\Bbb D} $是单叶调和的,且$ f(z)\in{\cal FK}_{{\cal H}}(\alpha) $.

根据Rado-Kneser-Choquet定理知,完全凸调和映射一定是单叶的.事实上, $ \alpha $阶完全凸调和映射也是单叶的.然而,完全星形调和映射却未必单叶.另外,每一个$ \alpha $阶完全凸调和映射是$ \alpha $阶完全星形调和映射,反过来则不一定成立,例如文献[5]中的例题2.6.最近,关于调和映射单叶半径问题和拟共形问题一直是一个热门的研究话题,详细结果请参见文献[6-12].

线性凸组合是构造新函数的一个重要方法.众所周知,解析函数线性凸组合$ tf+(1-t)g $未必单叶,即使$ f $和$ g $是凸的. Abdulhadi等^[13]引入了$ C^{1} $上函数的微分算子$ L = z\frac{\partial}{\partial z}-\overline{z}\frac{\partial}{\partial\overline{z}} $.并指出$ L $满足$ L(af+bg) = aL(f)+bL(g), L(fg) = fL(g)+gL(f), $其中$ a $, $ b $是复常数, $ f $, $ g $是定义在$ {\Bbb C} $上的复函数.在微分算子$ L $下调和性和双调和性保持不变.事实上,很容易验证微分算子$ L^{\epsilon} = z\frac{\partial}{\partial z}-\epsilon\overline{z}\frac{\partial}{\partial\overline{z}} $($ |\epsilon| = 1 $)也是线性的并且保调和性^[14].

调和映射卷积是解析函数卷积的推广形式.单连通区域内两个调和映射的卷积定义为

$ f\ast U = h\ast H+\overline{g\ast G} = z+\sum\limits_{n = 2}^{\infty}a_{n}A_{n}z^{n}+\overline{\sum\limits_{n = 1}^{\infty}b_{n}B_{n}z^{n}}, $

其中

$ f = h+\overline{g} = z+\sum\limits_{n = 2}^{\infty}a_{n}z^{n}+\overline{\sum\limits_{n = 1}^{\infty}b_{n}z^{n}}, U = H+\overline{G} = z+\sum\limits_{n = 2}^{\infty}A_{n}z^{n}+\overline{\sum\limits_{n = 1}^{\infty}B_{n}z^{n}}. $

事实上,两个调和映射的卷积未必保持其原来的性质,例如单叶性、凸性等性质.这使得调和映射的卷积得到了广泛的研究,关于调和映射的更多细节可参见文献[14-17].

在系数满足给定条件下,本文考虑调和映射的一些子类(调和映射微分算子凸组合和调和卷积的微分算子)的$ \alpha $阶完全星形, $ \alpha $阶完全凸的半径问题.

为了帮助证明主要结果,本节需要一些引理.

引理 2.1^[6]  对于$ \alpha\in[0, 1) $和$ t\in[0, 1] $,方程

(2.1) $ \begin{eqnarray} &&2(1-\alpha)(1-r)^{5}-(1-t)[(1+4r+r^{2})(1-r)-\alpha(1-r)^{3}]{}\\ &&-t(1+r)[1-\alpha+(6+2\alpha)r+(1-\alpha)r^{2}] = 0 \end{eqnarray} $

在$ (0, 1) $有唯一解$ r = r(\alpha, t) $.

根据文献[6,定理3.2]和[5,定理4.4] (稍作变形),不难发现下面两个结果.

引理 2.2  对于$ \alpha\in[0, 1) $,方程

(2.2) $ \begin{eqnarray} &&36(1-\alpha)(1-r)^{7}-(13-12\alpha)(1+r)(1-r)^{2}(1+10r+r^{2}){}\\ &&-(1-6\alpha)(1+r)(1-r)^{4}-4(1+57r+302r^{2}+302r^{3}+57r^{4}+r^{5}) = 0 \end{eqnarray} $

在$ (0, 1) $有唯一解$ r = r(\alpha) $.

引理  2.3  对于$ \alpha\in[0, 1) $,方程

(2.3) $ \begin{equation} 4(1-\alpha)(1-r)^{5}-(1-2\alpha)(1-r)^{2}(1+r)-(1+r)(1+10r+r^{2}) = 0 \end{equation} $

在$ (0, 1) $有唯一解$ r = r(\alpha) $.

引理 2.4  对于$ \alpha\in[0, 1) $和$ t\in[\frac{1}{6}, 1] $,方程

(2.4) $ \begin{eqnarray} &&2(1-\alpha)(1-r)^{6}-(1-t)[(1+r)(1-r)(1+10r+r^{2})-\alpha(1+r)(1-r)^{3}]{}\\ &&-t[(1+18r+42r^{2}+18r^{3}+r^{4})-\alpha(1-r)^{2}(1+4r+r^{2})] = 0 \end{eqnarray} $

在$ (0, 1) $有唯一解$ r = r(\alpha, t) $.

证   设

$ \begin{eqnarray*} f(r)& = &2(1-\alpha)(1-r)^{6}-(1-t)[(1+r)(1-r)(1+10r+r^{2})-\alpha(1+r)(1-r)^{3}]\\ &&-t[(1+18r+42r^{2}+18r^{3}+r^{4})-\alpha(1-r)^{2}(1+4r+r^{2})], \end{eqnarray*} $

则

(2.5) $ \begin{equation} f(0) = 1-\alpha>0, f(1) = -80t<0, \end{equation} $

且$ f'(r) = g(r)+th(r) $,其中

$ g(r) = -12(1-\alpha)(1-r)^{5}+4r^{3}+30r^{2}-10-\alpha(1-r)^{2}(4r+2), $

$ h(r) = -8r^{3}-84r^{2}-84r-8+4\alpha(1-r)(1-2r-2r^{2}). $

容易验证在$ r\in(0, 1) $时,有

(2.6) $ \begin{equation} h(r) = -8r^{3}-84r^{2}-84r-8+4\alpha(1-r)(1-2r-2r^{2})<0. \end{equation} $

此外,对于$ t\in[\frac{1}{6}, 1] $,得

(2.7) $ \begin{equation} f'(0) = (6\alpha-8)t+6\alpha-22<0, f'(1) = 24-184t<0. \end{equation} $

根据(2.6)式,当$ r\in(0, 1) $,有

$ \begin{eqnarray*} f'(r)\leq g(r)+\frac{1}{6}h(r)& = &-12(1-\alpha)(1-r)^{5}+4r^{3}+30r^{2}-10-\alpha(1-r)^{2}(4r+2)\\ &&+\frac{1}{6}[-8r^{3}-84r^{2}-84r-8+4\alpha(1-r)(1-2r-2r^{2})]\\ &\leq &4r^{3}+30r^{2}-10+\frac{1}{6}[-8r^{3}-84r^{2}-84r-8+4(1-r)]\\ & = &\frac{8}{3}r^{3}+16r^{2}-\frac{44}{3}r-\frac{32}{3}\\ &<&\frac{8}{3}r+16r-\frac{44}{3}r-\frac{32}{3}\\ & = &4r-\frac{32}{3}<0. \end{eqnarray*} $

结合(2.5)式得,存在唯一$ \xi_{1}\in(0, 1) $使得$ f(\xi_{1}) = 0 $.证毕.

根据函数在$ (0, 1) $上的单调性,容易证明下面几个引理中根的唯一性,在此省略其证明过程.

引理 2.5  对于$ \alpha\in[0, 1) $,方程

(2.8) $ \begin{equation} 4(1-\alpha)(1-r)^{6}-(1-2\alpha)(1-r)^{2}(1+4r+r^{2})-(1+26r+66r^{2}+26r^{3}+r^{4}) = 0 \end{equation} $

在$ (0, 1) $有唯一解$ r = r(\alpha) $.

引理 2.6  对于$ \alpha\in[0, 1) $,方程

(2.9) $ \begin{equation} 12(1-\alpha)(1-r)^{6}-(4-5\alpha)(1-r)^{2}(1+4r+r^{2})+\alpha(1-r)^{4}-2(1+26r+66r^{2}+26r^{3}+r^{4}) = 0 \end{equation} $

在$ (0, 1) $有唯一解$ r = r(\alpha) $.

引理 2.7  对于$ \alpha\in[0, 1) $,方程

(2.10) $ \begin{eqnarray} &&12(1-\alpha)(1-r)^{7}-(4-5\alpha)(1+r)(1-r)^{2}(1+10r+r^{2})+\alpha(1+r)(1-r)^{4}{}\\ &&-2(1+57r+302r^{2}+302r^{3}+57r^{4}+r^{5}) = 0 \end{eqnarray} $

在$ (0, 1) $有唯一解$ r = r(\alpha) $.

引理 2.8  对于$ \alpha\in[0, 1) $,方程

(2.11) $ \begin{eqnarray} &&36(1-\alpha)(1-r)^{8}-(13-12\alpha)(1-r)^{2}(1+26r+66r^{2}+26r^{3}+r^{4}){}\\ &&-(1-6\alpha)(1-r)^{4}(1+4r+r^{2}) -4(1+120r+1191r^{2}+2416r^{3}+1191r^{4}+120r^{5}+r^{6}) = 0{}\\ \end{eqnarray} $

在$ (0, 1) $有唯一解$ r = r(\alpha) $.

下面是本文中所用到的幂级数求和.

(3.1) $ \begin{eqnarray} &&\sum\limits_{n = 1}^{\infty}nr^{n-1} = \frac{1}{(1-r)^{2}}, \sum\limits_{n = 1}^{\infty}n^{2}r^{n-1} = \frac{1+r}{(1-r)^{3}}, \sum\limits_{n = 1}^{\infty}n^{3}r^{n-1} = \frac{1+4r+r^{2}}{(1-r)^{4}}, {}\\ &&\sum\limits_{n = 1}^{\infty}n^{4}r^{n-1} = \frac{(1+r)(1+10r+r^{2})}{(1-r)^{5}}, \sum\limits_{n = 1}^{\infty}n^{5}r^{n-1} = \frac{1+26r+66r^{2}+26r^{3}+r^{4}}{(1-r)^{6}}, \\ &&\sum\limits_{n = 1}^{\infty}n^{6}r^{n-1} = \frac{1+57r+302r^{2}+302r^{3}+57r^{4}+r^{5}}{(1-r)^{7}}, {}\\ &&\sum\limits_{n = 1}^{\infty}n^{7}r^{n-1} = \frac{1+120r+1191r^{2}+2416r^{3}+1191r^{4}+120r^{5}+r^{6}}{(1-r)^{8}}.{} \end{eqnarray} $

利用上述幂级数给出下面结果.

定理  3.1   设$ f_{i} = h_{i}+\overline{g_{i}}\in{\cal H} $, $ i = 1, 2, $其中

$ h_{i} = z+\sum\limits_{n = 2}^{\infty}a_{i_{n}}z^{n}, g_{i} = \sum\limits_{n = 2}^{\infty}b_{i_{n}}z^{n} $

且对于$ n\geq2 $

(3.2) $ \begin{equation} |a_{1_{n}}|\leq\frac{n+1}{2}, |b_{1_{n}}|\leq\frac{n-1}{2}, |a_{2_{n}}|\leq\frac{(2n+1)(n+1)}{6}, |b_{2_{n}}|\leq\frac{(2n-1)(n-1)}{6}, \end{equation} $

则$ F^{\epsilon} = (1-t)L^{\epsilon}_{f_{1}}+tL^{\epsilon}_{f_{2}} $,的$ \alpha $阶完全星形半径为$ r_{s} $,且$ r_{s} = r_{s}(\alpha, t) $是方程(2.1)在$ (0, 1) $的唯一解,其中$ t\in[0, 1] $, $ L^{\epsilon}_{f_{i}} = zf_{i_{z}}-\epsilon\overline{z}f_{i_{\overline{z}}} $($ |\epsilon| = 1 $);另外它的单叶半径为$ r_{u} $,其中$ r_{u} $是方程

(3.3) $ \begin{equation} 2(1-r)^{5}-(1-t)(1-r)(1+4r+r^{2})-t(1+r)(1+6r+r^{2}) = 0 \end{equation} $

在$ (0, 1) $的唯一解.上述结果均为最佳.

证   

$ \begin{eqnarray*} F^{\epsilon} = (1-t)L^{\epsilon}_{f_{1}}+tL^{\epsilon}_{f_{2}} = z+\sum\limits_{n = 2}^{\infty}[(1-t)na_{1_{n}}+tna_{2_{n}}]z^{n}-\epsilon\overline{\sum\limits_{n = 2}^{\infty}[(1-t)nb_{1_{n}}+tnb_{2_{n}}]z^{n}}, \end{eqnarray*} $

其中$ |\epsilon| = 1 $.对于$ r\in(0, 1) $,只需证$ F^{\epsilon }_{r}(z)\in{\cal FS}_{{\cal H}}^{*}(\alpha) $,其中

(3.4) $ \begin{equation} F^{\epsilon}_{r}(z) = \frac{F^{\epsilon}(rz)}{r} = z+\sum\limits_{n = 2}^{\infty}[(1-t)na_{1_{n}}+tna_{2_{n}}]r^{n-1}z^{n}-\epsilon\overline{\sum\limits_{n = 2}^{\infty}[(1-t)nb_{1_{n}}+tnb_{2_{n}}]r^{n-1}z^{n}}. \end{equation} $

考虑

$ S = \sum\limits_{n = 2}^{\infty}\frac{n-\alpha}{1-\alpha}|(1-t)na_{1_{n}}+tna_{2_{n}}|r^{n-1}+\sum\limits_{n = 2}^{\infty}\frac{n+\alpha}{1-\alpha}|(1-t)nb_{1_{n}}+tnb_{2_{n}}|r^{n-1}, $

根据定理A和条件(3.2),只要证明

$ \begin{eqnarray*} &&\sum\limits_{n = 2}^{\infty}\frac{n-\alpha}{1-\alpha}\left[(1-t)\frac{n(n+1)}{2}+t\frac{n(2n+1)(n+1)}{6}\right]r^{n-1}\\ &&+\sum\limits_{n = 2}^{\infty}\frac{n+\alpha}{1-\alpha}\left[(1-t)\frac{n(n-1)}{2}+t\frac{n(2n-1)(n-1)}{6}\right]r^{n-1}\leq1. \end{eqnarray*} $

结合等式(3.1)式,有

$ \begin{eqnarray*} &&2(1-\alpha)(1-r)^{5}-(1-t)[(1+4r+r^{2})(1-r)-\alpha(1-r)^{3}]\\ &&-t(1+r)[1-\alpha+(6+2\alpha)r+(1-\alpha)r^{2}]\geq0. \end{eqnarray*} $

因此对于$ r\leq r_{s} $,有$ F^{\epsilon}_{r}(z)\in{\cal FS}_{{\cal H}}^{*}(\alpha) $,其中$ r_{s} $是方程(2.1)在$ (0, 1) $内的唯一解.当$ \alpha = 0 $, (2.1)式推出(3.3)式.再次运用定理A,得$ F^{\epsilon} $在$ |z|\leq r_{u} = r_{s}(0, t) $单叶调和.

为证最佳性,考虑

$ f_{1_{0}}(z) = z-\sum\limits_{n = 2}^{\infty}\frac{n+1}{2}z^{n}-\overline{\sum\limits_{n = 1}^{\infty}\frac{n-1}{2}z^{n}} = 2z-\frac{z-\frac{1}{2}z^{2}}{(1-z)^{2}}-\overline{\frac{\frac{1}{2}z^{2}}{(1-z)^{2}}}, $

$ \begin{eqnarray*} f_{2_{0}}(z)& = &z-\sum\limits_{n = 2}^{\infty}\frac{(2n+1)(n+1)}{6}z^{n}-\overline{\sum\limits_{n = 1}^{\infty}\frac{(2n-1)(n-1)}{6}z^{n}}\\ & = &2z-\frac{z-\frac{1}{2}z^{2}+\frac{1}{6}z^{3}}{(1-z)^{3}}-\overline{\frac{\frac{1}{2}z^{2}+\frac{1}{6}z^{3}}{(1-z)^{3}}}. \end{eqnarray*} $

取$ \epsilon = 1 $,于是

$ F_{0} = 2z-(1-t)\frac{z}{(1-z)^{3}}-t\frac{z+z^{2}}{(1-z)^{4}}+\overline{(1-t)\frac{z^{2}}{(1-z)^{3}}+t\frac{z^{2}+z^{3}}{(1-z)^{4}}} = H_{0}(z)+\overline{G_{0}(z)}. $

计算给出

$ H'_{0}(r) = 2-(1-t)\frac{1+2r}{(1-r)^{4}}-t\frac{1+5r+2r^{2}}{(1-r)^{5}}, G'_{0}(r) = (1-t)\frac{2r+r^{2}}{(1-r)^{4}}+t\frac{2r+5r^{2}+r^{3}}{(1-r)^{5}}. $

结合(3.3)式,有

$ [H'_{0}(r)-G'_{0}(r)]|_{r = r_{u}} = \frac{2(1-r)^{5}-(1-t)(1-r)(1+4r+r^{2})-t(1+r)(1+6r+r^{2})}{(1-r)^{5}} = 0, $

即$ J_{F_{0}}|_{r = r_{u}} = 0 $.所以,若$ r>r_{u} $,则$ F_{0} $在$ |z|<r $内不单叶.这意味着$ r_{u} $是最佳的.

进一步,得

$ \begin{eqnarray*} \frac{\partial}{\partial\theta}(\mbox{arg}(F_{0}(r{\rm e}^{{\rm i}\theta})))|_{\theta = 0}& = &\frac{rH'_{0}(r)-rG'_{0}(r)}{H_{0}(z)+G_{0}(z)}\\ & = &\frac{2(1-r)^{5}-(1-t)(1-r)(1+4r+r^{2})-t(1+r)(1+6r+r^{2})}{2(1-r)^{5}-(1-t)(1-r)^{3}-t(1+r)(1-r)^{2}}. \end{eqnarray*} $

由(2.1)式得

$ \alpha = \frac{2(1-r)^{5}-(1-t)(1-r)(1+4r+r^{2})-t(1+r)(1+6r+r^{2})}{2(1-r)^{5}-(1-t)(1-r)^{3}-t(1+r)(1-r)^{2}}, $

于是$ \frac{\partial}{\partial\theta}(\mbox{arg}(F_{0}(r{\rm e}^{{\rm i}\theta})))|_{\theta = 0, r = r_{s}} = \alpha. $所以(2.1)式给出的$ r_{s} $是最佳的.证毕.

定理 3.2  在定理3.1的条件下.则$ F^{\epsilon} = (1-t)L^{\epsilon}_{f_{1}}+tL^{\epsilon}_{f_{2}} $ ($ t\in[\frac{1}{6}, 1] $)的$ \alpha $阶完全凸半径为$ r_{c} $,其中$ r_{c} = r(\alpha, t) $是方程(2.4)在$ (0, 1) $的唯一解,该结果最佳.

证   对于$ r\in(0, 1) $,只需$ F^{\epsilon}_{r}(z)\in{\cal FK}_{{\cal H}}(\alpha) $,其中$ F^{\epsilon}_{r}(z) $由方程(3.4)给出.考虑

$ S = \sum\limits_{n = 2}^{\infty}\frac{n^{2}(n-\alpha)}{1-\alpha}|(1-t)a_{1_{n}}+ta_{2_{n}}|r^{n-1}+\sum\limits_{n = 2}^{\infty}\frac{n^{2}(n+\alpha)}{1-\alpha}|(1-t)b_{1_{n}}+tb_{2_{n}}|r^{n-1}. $

根据定理B,需证$ S\leq1 $.即等价证明

$ \begin{eqnarray*} &&\sum\limits_{n = 2}^{\infty}\frac{n^{2}(n-\alpha)}{1-\alpha}\left[(1-t)\frac{n+1}{2}+t\frac{(2n+1)(n+1)}{6}\right]r^{n-1}\\ &&+\sum\limits_{n = 2}^{\infty}\frac{n^{2}(n+\alpha)}{1-\alpha}\left[(1-t)\frac{n-1}{2}+t\frac{(2n-1)(n-1)}{6}\right]r^{n-1}\leq1, \end{eqnarray*} $

结合(3.1)式,得

$ \begin{eqnarray*} &&2(1-\alpha)(1-r)^{6}-(1-t)[(1+r)(1-r)(1+10r+r^{2})-\alpha(1+r)(1-r)^{3}]\\ &&-t[(1+18r+42r^{2}+18r^{3}+r^{4})-\alpha(1-r)^{2}(1+4r+r^{2})]\geq0, \end{eqnarray*} $

因此对于$ r\leq r_{c} $, $ F^{\epsilon}_{r}(z)\in{\cal FK}_{{\cal H}}(\alpha) $,其中$ r_{c} $是方程(2.4)在$ (0, 1) $的唯一解.

下证最佳性,取

$ f_{1_{0}}(z) = z-\sum\limits_{n = 2}^{\infty}\frac{n+1}{2}z^{n}+\overline{\sum\limits_{n = 1}^{\infty}\frac{n-1}{2}z^{n}} = 2z-\frac{z-\frac{1}{2}z^{2}}{(1-z)^{2}}+\overline{\frac{\frac{1}{2}z^{2}}{(1-z)^{2}}}, $

$ \begin{eqnarray*} f_{2_{0}}(z)& = &z-\sum\limits_{n = 2}^{\infty}\frac{(2n+1)(n+1)}{6}z^{n}+\overline{\sum\limits_{n = 1}^{\infty}\frac{(2n-1)(n-1)}{6}z^{n}}\\ & = &2z-\frac{z-\frac{1}{2}z^{2}+\frac{1}{6}z^{3}}{(1-z)^{3}}+\overline{\frac{\frac{1}{2}z^{2}+\frac{1}{6}z^{3}}{(1-z)^{3}}}. \end{eqnarray*} $

取$ \epsilon = 1 $,于是

$ F_{0} = 2z-(1-t)\frac{z}{(1-z)^{3}}-t\frac{z+z^{2}}{(1-z)^{4}}-\overline{(1-t)\frac{z^{2}}{(1-z)^{3}}+t\frac{z^{2}+z^{3}}{(1-z)^{4}}} = H_{0}(z)+\overline{G_{0}(z)}. $

计算给出

$ H'_{0}(r) = 2-(1-t)\frac{1+2r}{(1-r)^{4}}-t\frac{1+5r+2r^{2}}{(1-r)^{5}}, G'_{0}(r) = -(1-t)\frac{2r+r^{2}}{(1-r)^{4}}-t\frac{2r+5r^{2}+r^{3}}{(1-r)^{5}}. $

所以

$ \begin{eqnarray*} &&\frac{\partial}{\partial\theta}(\mbox{arg}\frac{\partial}{\partial\theta}(F_{0}(r{\rm e}^{{\rm i}\theta})))|_{\theta = 0}\\ & = &\frac{H'_{0}(r)+G'_{0}(r)+r(H''_{0}(r)+G''_{0}(r))}{H'_{0}(z)-G'_{0}(z)}\\ & = &\frac{2(1-r)^{6}-(1-t)(1+r)(1-r)(1+10r+r^{2})-t(1+18r+42r^{2}+18r^{3}+r^{4})}{2(1-r)^{6}-(1-t)(1+r)(1-r)^{3}-t(1-r)^{2}(1+4r+r^{2})}. \end{eqnarray*} $

由方程(2.4)得

$ \alpha = \frac{2(1-r)^{6}-(1-t)(1+r)(1-r)(1+10r+r^{2})-t(1+18r+42r^{2}+18r^{3}+r^{4})}{2(1-r)^{6}-(1-t)(1+r)(1-r)^{3}-t(1-r)^{2}(1+4r+r^{2})}, $

于是$ \frac{\partial}{\partial\theta}\left(\mbox{arg}\frac{\partial}{\partial\theta}(F_{0}(r{\rm e}^{{\rm i}\theta}))\right)|_{\theta = 0, r = r_{c}} = \alpha. $这证明了方程(2.4)中给出的$ r_{c} $的最佳性.证毕.

注  取$ t = 0 $或者$ t = 1 $,定理3.1和定理3.2推出文献[14]中的相关结果.

定理 4.1   设$ f = h+\overline{g}\in{\cal H} $满足(1.1)式且有

(4.1) $ \begin{equation} \begin{array}{l} { } |a_{n}|\leq\frac{(n+1)^{2}}{4}, {\quad} n\geq2, \\ { } |b_{n}|\leq\frac{(n-1)^{2}}{4}, {\quad} n\geq1, \end{array} \end{equation} $

则$ L^{\epsilon}_{f} = zf_{z}-\epsilon\overline{z}f_{\overline{z}} $ ($ |\epsilon| = 1 $)在$ |z|<r_{s} $内$ \alpha $阶完全星形,其中$ r_{s} $是方程(2.3)在$ (0, 1) $的唯一解;它的单叶半径为$ r_{u}\approx0.0712 $是方程

(4.2) $ \begin{equation} 4(1-r)^{5}-(1-r)^{2}(1+r)-(1+r)(1+10r+r^{2}) = 0 \end{equation} $

在$ (0, 1) $的解.上述结果均最佳.

证   对于$ r\in(0, 1) $,需证$ L^{\epsilon}_{f _{r}}(z)\in{\cal FS}_{{\cal H}}^{*}(\alpha) $,其中

(4.3) $ \begin{equation} L^{\epsilon}_{f_{r}}(z) = \frac{L^{\epsilon}_{f}(rz)}{r} = z+\sum\limits_{n = 2}^{\infty}na_{n}r^{n-1}z^{n}-\epsilon\overline{\sum\limits_{n = 1}^{\infty}nb_{n}r^{n-1}z^{n}}. \end{equation} $

考虑

(4.4) $ \begin{equation} S = \sum\limits_{n = 2}^{\infty}\frac{n(n-\alpha)}{1-\alpha}|a_{n}|r^{n-1}+\sum\limits_{n = 1}^{\infty}\frac{n(n+\alpha)}{1-\alpha}|b_{n}|r^{n-1}, \end{equation} $

根据定理A,只要证$ S\leq1 $即可.由(4.1)与(4.4)式,即证

$ S\leq\sum\limits_{n = 2}^{\infty}\frac{n(n-\alpha)}{1-\alpha}\frac{(n+1)^{2}}{4}r^{n-1}+\sum\limits_{n = 1}^{\infty}\frac{n(n+\alpha)}{1-\alpha}\frac{(n-1)^{2}}{4}r^{n-1}\leq1, $

结合(3.1)式,有

$ 4(1-\alpha)(1-r)^{5}-(1-2\alpha)(1-r)^{2}(1+r)-(1+r)(1+10r+r^{2})\geq0. $

因此对于$ r\leq r_{s} $, $ L^{\epsilon}_{f_{r}}(z)\in{\cal FS}_{{\cal H}}^{*}(\alpha) $,其中$ r_{s} $是方程(2.3)在$ (0, 1) $唯一解.在方程(2.3)中取$ \alpha = 0 $,有方程(4.2).

取$ \epsilon = 1 $,

$ h_{0}(z) = 2z-\sum\limits_{n = 1}^{\infty}\frac{(n+1)^{2}}{4}z^{n}, g_{0}(z) = -\sum\limits_{n = 2}^{\infty}\frac{(n-1)^{2}}{4}z^{n}. $

于是

$ L_{0}(z) = h_{L_{0}}+\overline{g_{L_{0}}} = 2z-\sum\limits_{n = 1}^{\infty}\frac{n(n+1)^{2}}{4}z^{n}+\overline{\sum\limits_{n = 2}^{\infty}\frac{n(n-1)^{2}}{4}z^{n}}, $

则结合(4.2)式得

$ [h'_{L_{0}}-g'_{L_{0}}]|_{r = r_{u}} = \frac{4(1-r)^{5}-(1-r)^{2}(1+r)-(1+r)(1+10r+r^{2})}{2(1-r)^{5}}|_{r = r_{u}} = 0, $

即$ J_{L_{0}}|_{r = r_{u}} = 0 $.因此, $ r_{u} $最佳.易得

$ \begin{eqnarray*} \frac{\partial}{\partial\theta}\left(\mbox{arg}(L_{0}(r{\rm e}^{{\rm i}\theta}))\right)|_{\theta = 0} & = &\frac{rh'_{L_{0}}(r)-rg'_{L_{0}}(r)}{g_{L_{0}}(r)+h_{L_{0}}(r)}\\ & = &\frac{4(1-r)^{5}-(1-r)^{2}(1+r)-(1+r)(1+10r+r^{2})}{4(1-r)^{5}-2(1-r)^{2}(1+r)}. \end{eqnarray*} $

根据(2.3)式,有

$ \alpha = \frac{4(1-r)^{5}-(1-r)^{2}(1+r)-(1+r)(1+10r+r^{2})}{4(1-r)^{5}-2(1-r)^{2}(1+r)}, $

所以$ \frac{\partial}{\partial\theta}(\mbox{arg}(L_{0}(r{\rm e}^{{\rm i}\theta})))|_{\theta = 0, r = r_{s}} = \alpha. $证毕.

定理  4.2  在定理4.1的条件下, $ L^{\epsilon}_{f} = zf_{z}-\epsilon\overline{z}f_{\overline{z}} $ ($ |\epsilon| = 1 $)在$ |z|<r_{c} $为$ \alpha $阶完全凸的,其中$ r_{c} $是方程(2.8)在$ (0, 1) $的唯一解,该结果最佳.

证   对于$ r\in(0, 1) $,需要证$ L^{\epsilon }_{f_{r}}(z)\in{\cal FK}_{{\cal H}}(\alpha) $,其中$ L^{\epsilon}_{f_{r}}(z) $满足(4.3)式.考虑

$ S = \sum\limits_{n = 2}^{\infty}\frac{n^{2}(n-\alpha)}{1-\alpha}|a_{n}|r^{n-1}+\sum\limits_{n = 1}^{\infty}\frac{n^{2}(n+\alpha)}{1-\alpha}|b_{n}|r^{n-1}. $

根据定理B,需证$ S\leq1 $.由条件(4.1),即证

$ S\leq\sum\limits_{n = 2}^{\infty}\frac{n^{2}(n-\alpha)}{1-\alpha}\frac{(n+1)^{2}}{4}r^{n-1}+\sum\limits_{n = 1}^{\infty}\frac{n^{2}(n+\alpha)}{1-\alpha}\frac{(n-1)^{2}}{4}r^{n-1}\leq1, $

结合(3.1)式,简单地计算给出

$ 4(1-\alpha)(1-r)^{6}-(1-2\alpha)(1-r)^{2}(1+4r+r^{2})-(1+26r+66r^{2}+26r^{3}+r^{4})\geq0. $

因此对于$ r\leq r_{c} $, $ L^{\epsilon}_{f_{r}}(z)\in{\cal FK}_{{\cal H}}(\alpha) $,其中$ r_{s} $满足(2.8)式.

取

$ h_{0}(z) = 2z-\sum\limits_{n = 1}^{\infty}\frac{(n+1)^{2}}{4}z^{n}, g_{0}(z) = \sum\limits_{n = 2}^{\infty}\frac{(n-1)^{2}}{4}z^{n}, $

则

$ L_{0}(z) = h_{L_{0}}+\overline{g_{L_{0}}} = 2z-\sum\limits_{n = 1}^{\infty}\frac{n(n+1)^{2}}{4}z^{n}-\overline{\sum\limits_{n = 2}^{\infty}\frac{n(n-1)^{2}}{4}z^{n}}, $

且

$ \begin{eqnarray*} &&\frac{\partial}{\partial\theta}\left(\mbox{arg}\frac{\partial}{\partial\theta}\left(L_{0}\left (r{\rm e}^{{\rm i}\theta}\right)\right)\right)\bigg|_{\theta = 0}\\ & = &\frac{h'_{L_{0}}(r)+g'_{L_{0}}(r)+r(h''_{L_{0}}(r)+g''_{L_{0}}(r))}{h'_{L_{0}}(r)-g'_{L_{0}}(r)}\\ & = &\frac{4(1-r)^{6}-(1-r)^{2}(1+4r+r^{2})-(1+26r+66r^{2}+26r^{3}+r^{4})}{4(1-r)^{6}-2(1-r)^{2}(1+4r+r^{2})}. \end{eqnarray*} $

由(2.8)式得

$ \alpha = \frac{4(1-r)^{6}-(1-r)^{2}(1+4r+r^{2})-(1+26r+66r^{2}+26r^{3}+r^{4})}{4(1-r)^{6}-2(1-r)^{2}(1+4r+r^{2})}, $

于是$ \frac{\partial}{\partial\theta}(\mbox{arg}\frac{\partial}{\partial\theta}(L_{0}(r{\rm e}^{{\rm i}\theta})))|_{\theta = 0, r = r_{c}} = \alpha. $最佳性得证.

定理  4.3  设$ f = h+\overline{g}\in{\cal H} $满足(1.1)式且有

$ |a_{n}|\leq\frac{(n+1)^{2}(2n+1)}{12}, {\quad} n\geq2, $

$ |b_{n}|\leq\frac{(n-1)^{2}(2n-1)}{12}, {\quad} n\geq1, $

则$ L^{\epsilon}_{f} = zf_{z}-\epsilon\overline{z}f_{\overline{z}} $ ($ |\epsilon| = 1 $)在$ |z|<r_{s} $内是$ \alpha $阶完全星形的,其中$ r_{s} $是方程(2.9)在$ (0, 1) $内的唯一实根;此外, $ L^{\epsilon}_{f} $的单叶半径$ r_{u}\approx0.0464 $是方程

(4.5) $ \begin{equation} 6(1-r)^{6}-2(1-r)^{2}(1+4r+r^{2})-(1+26r+66r^{2}+26r^{3}+r^{4}) = 0 \end{equation} $

在$ (0, 1) $的根.上述结果均为最佳.

证   过程类似于定理4.1.只需证明

$ \sum\limits_{n = 2}^{\infty}\frac{n(n-\alpha)}{1-\alpha}\frac{(n+1)^{2}(2n+1)}{12}r^{n-1}+\sum\limits_{n = 1}^{\infty}\frac{n(n+\alpha)}{1-\alpha}\frac{(n-1)^{2}(2n-1)}{12}r^{n-1}\leq1, $

结合(3.1)式,计算给出

$ \begin{eqnarray*} &&12(1-\alpha)(1-r)^{6}-(4-5\alpha)(1-r)^{2}(1+4r+r^{2})+\alpha(1-r)^{4}\\ &&-2(1+26r+66r^{2}+26r^{3}+r^{4})\geq0. \end{eqnarray*} $

因此对于$ r\leq r_{s} $, $ L^{\epsilon}_{f_{r}}(z)\in{\cal FS}_{{\cal H}}^{*}(\alpha) $,其中$ r_{s} $是方程(2.9)在$ (0, 1) $内的唯一实根.取$ \alpha = 0 $,则有(4.5)式.

取$ \epsilon = 1 $,

$ h_{0}(z) = 2z-\sum\limits_{n = 1}^{\infty}\frac{(n+1)^{2}(2n+1)}{12}z^{n}, g_{0}(z) = -\sum\limits_{n = 2}^{\infty}\frac{(n-1)^{2}(2n-1)}{12}z^{n}. $

于是

$ L_{0}(z) = h_{L_{0}}+\overline{g_{L_{0}}} = 2z-\sum\limits_{n = 1}^{\infty}\frac{n(n+1)^{2}(2n+1)}{12}z^{n}+\overline{\sum\limits_{n = 2}^{\infty}\frac{n(n-1)^{2}(2n-1)}{12}z^{n}}. $

结合(4.5)式,有

$ \begin{eqnarray*} [h'_{L_{0}}-g'_{L_{0}}]|_{r = r_{u}} & = &\frac{6(1-r)^{6}-2(1-r)^{2}(1+4r+r^{2})-(1+26r+66r^{2}+26r^{3}+r^{4})}{3(1-r)^{6}}|_{r = r_{u}}\\ & = &0. \end{eqnarray*} $

计算给出

$ \begin{eqnarray*} &&\frac{\partial}{\partial\theta}\left(\mbox{arg}(L_{0}(r{\rm e}^{{\rm i}\theta}))\right)|_{\theta = 0} \\ & = &\frac{12(1-r)^{6}-4(1-r)^{2}(1+4r+r^{2})-2(1+26r+66r^{2}+26r^{3}+r^{4})}{12(1-r)^{6}-5(1-r)^{2}(1+4r+r^{2})-(1-r)^{4}}. \end{eqnarray*} $

此外,根据(2.9)式得

$ \alpha = \frac{12(1-r)^{6}-4(1-r)^{2}(1+4r+r^{2})-2(1+26r+66r^{2}+26r^{3}+r^{4})}{12(1-r)^{6}-5(1-r)^{2}(1+4r+r^{2})-(1-r)^{4}}. $

因此$ \frac{\partial}{\partial\theta}(\mbox{arg}\frac{\partial}{\partial\theta}(L_{0}(r{\rm e}^{{\rm i}\theta})))|_{\theta = 0, r = r_{s}} = \alpha. $说明(2.9)式给出的$ r_{s} $是最佳的.证毕.

定理  4.4  在定理4.3的条件下, $ L^{\epsilon}_{f} = zf_{z}-\epsilon\overline{z}f_{\overline{z}} $ ($ |\epsilon| = 1 $)在$ |z|<r_{c} $内为$ \alpha $阶完全凸,其中$ r_{c} $是方程(2.10)在$ (0, 1) $内的唯一实根.该结果最佳.

证   证明过程类似于定理4.2.因此只要证明最佳性.考虑$ \epsilon = 1 $,

$ h_{0}(z) = 2z-\sum\limits_{n = 1}^{\infty}\frac{(n+1)^{2}(2n+1)}{12}z^{n}, g_{0}(z) = \sum\limits_{n = 2}^{\infty}\frac{(n-1)^{2}(2n-1)}{12}z^{n}. $

证毕.

定理 4.5  设$ f = h+\overline{g}\in{\cal H} $满足(1.1)式且有

$ |a_{n}|\leq\frac{(2n+1)^{2}(n+1)^{2}}{36}, {\quad} n\geq2, $

$ |b_{n}|\leq\frac{(2n-1)^{2}(n-1)^{2}}{36}, {\quad} n\geq1, $

则$ L^{\epsilon}_{f} = zf_{z}-\epsilon\overline{z}f_{\overline{z}} $ ($ |\epsilon| = 1 $)在$ |z|<r_{s} $内是$ \alpha $阶完全星形的,其中$ r_{s} $是方程(2.2)在$ (0, 1) $内的唯一实根;此外$ L_{\epsilon} $的单叶半径$ r_{u}\approx0.0295 $为方程

$ \begin{eqnarray*} &&36(1-r)^{7}-13(1+r)(1-r)^{2}(1+10r+r^{2})-(1+r)(1-r)^{4}\\ &&-4(1+57r+302r^{2}+302r^{3}+57r^{4}+r^{5}) = 0 \end{eqnarray*} $

在$ (0, 1) $的根.上述结果均为最佳.

证   主要证明最佳性.取$ \epsilon = 1 $,

$ h_{0}(z) = 2z-\sum\limits_{n = 1}^{\infty}\frac{(n+1)^{2}(2n+1)^{2}}{36}z^{n}, g_{0}(z) = -\sum\limits_{n = 2}^{\infty}\frac{(n-1)^{2}(2n-1)^{2}}{36}z^{n}. $

经验证该结果达到最佳.

定理 4.6  在定理4.5的条件下, $ L^{\epsilon}_{f}(z) = zf_{z}(z)-\epsilon\overline{z}f_{\overline{z}}(z) $($ |\epsilon| = 1 $)在$ |z|<r_{c} $内属于$ \alpha $阶完全凸,其中$ r_{c} $是方程(2.11)在$ (0, 1) $内的唯一实根.该结果最佳.

证   这里只证最佳性.取$ \epsilon = 1 $,

$ h_{0}(z) = 2z-\sum\limits_{n = 1}^{\infty}\frac{(n+1)^{2}(2n+1)^{2}}{36}z^{n}, g_{0}(z) = \sum\limits_{n = 2}^{\infty}\frac{(n-1)^{2}(2n-1)^{2}}{36}z^{n}. $

经验证该结果达到最佳.