关于类${\mathcal U}(λ)$第五系数的一个猜想
On a conjecture for the fifth coefficients for the class ${\mathcal U}(λ)$
摘要：
令$f$是在单位圆${\mathbb D}=\{z:|z|<1\}$中解析的函数，其被归一化使得$f(0)=f'(0)-1=0$，即，类型为$f(z)=z+\sum_{n=2}^{\infty} a_n z^n$。另外，如果是\[ \left| \left(\frac{z}{f(z)}\right)^2 f'(z) -1\right|<λ\quad\quad (z\in{\mathbb D}), \]，则$f$属于类${\mathcal U}(λ)$、$0<λ\le1$。本文证明了当$0.400436\ldots \leλ\le1$满足\[ \frac{f(z)}{z}\prec \frac{1}{(1+z)(1+λz)}, \]（“$\prec$“是通常的从属关系）时，$f$的第五系数模的上界的精确性.
英文摘要：
Let $f$ be function that is analytic in the unit disk ${\mathbb D}=\{z:|z|<1\}$, normalized such that $f(0)=f'(0)-1=0$, i.e., of type $f(z)=z+\sum_{n=2}^{\infty} a_n z^n$. If additionally, \[ \left| \left(\frac{z}{f(z)}\right)^2 f'(z) -1\right|<λ\quad\quad (z\in{\mathbb D}), \] then $f$ belongs to the class ${\mathcal U}(λ)$, $0<λ\le1$. In this paper we prove sharp upper bound of the modulus of the fifth coefficient of $f$ from ${\mathcal U}(λ)$ satisfying \[ \frac{f(z)}{z}\prec \frac{1}{(1+z)(1+λz)}, \] ("$\prec$" is the usual subordination) in the case when $0.400436\ldots \leλ\le1$.
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