设F为单位圆盘△上的一族全纯函数,a和b为2个有限的复数且有b≠a,如果对任意的z∈△且对每个f∈F,若f=α→f′=α,且f=b≥→f′=b,则存在一正整数M且对任意的f∈F,有(1-|z|^2)f^#(z)=(1-|z|^2)|f′(z)|/1+|f(z)|^2≤M.
Let F be a family of holomorphic functions on the unit disk △, and let a and b be two finite complex numbers such thatb≠a. If, for each f∈F and z∈△, f=α→f′=α,and f=b≥→f′=b then there exists a positive number M such that for any f∈F.(1-|z|^2)f^#(z)=(1-|z|^2)|f′(z)|/1+|f(z)|^2≤M.