2013-12-12: 由偏差定理推出 Koebe $\displaystyle{\frac{1}{4}}$ 掩蔽定理.

证明:  设 $\displaystyle{g(r)=\frac{r}{(1+r)^2}}$, 则 \[ g'(r)=\frac{1}{(1+r)^2}>0. \] 于是 \[ 0\leq r<1\Rightarrow 0\leq g(r)<\frac{1}{4}. \] 由偏差定理, \[ g(|z|)=\frac{|z|}{(1+|z|)^2}\leq |f(z)|,\quad z\in\triangle, \] \[ \left\{w;|w|<\frac{1}{4}\right\}\subset f(\triangle). \]  

\newpage\section{2013-12-12:由偏差定理推出 Koebe$\displaystyle{\frac{1}{4}}$ 掩蔽定理. (\href{}{link})} \label{131212}   证明:  设 $\displaystyle{g(r)=\frac{r}{(1+r)^2}}$, 则[ g'(r)=frac{1}{(1+r)^2}>0. ]于是[ 0leqr<1Rightarrow 0leq g(r)<frac{1}{4}. ] 由偏差定理, [ g(|z|)=frac{|z|}{(1+|z|)^2}leq |f(z)|,quad zintriangle, ][ left{w;|w|<frac{1right}{4}}subset f(triangle). ]