\(\mathrm {Exercise \ \oplus \ Problem } \ 6 \)
\(\mathrm {Exercise \ \oplus \ Problem } \ 6 \) |
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\( \qquad \)你好，这里是我的个人网站数学分析的每周一题栏目(数学分析每周一题，其中数学分析指的是数学中的分析学, 主要包括微积分，实分析，复分析) \(\qquad \ \)——————Alina Lagrange
Suppose that \(f\) is holomorphic on the open unit disc \(D\), \(f(0)=0\) , if there sxists \(A>0 \) such that \(\Re f(z)\leq A ,\forall z \in D \), prove that \[ |f(z) | \leq \frac{2A |z| }{ 1- |z| }, \ \ \forall z\in D. \]
\(\mathcal{P}roof. \)
Let \[ \zeta = \frac zA -1 , \] \[\tau= \frac{z+1}{z-1} .\] then $$ \tau\circ \zeta \mapsto \frac{ (\frac zA -1 +1 ) }{\frac zA -1 -1}=\frac{ z}{z-2 A}. $$ Here \( f\) maps the unit disc to the special left half-plane \(P =\{ z\in \mathbb C: \Re z \leq A \}\), \( \zeta\) maps \(P \) into the left half-plane. \(\tau\) maps the left half-plane to the unit disc. Using Schwarz's lemma for the composition of the above mapping and \(f\) , we get \[ | \tau \circ\zeta \circ f | \leq |z| \] $$ \frac{| f(z)|}{|f(z)-2 A|} \leq|z|. $$ Hence \[|f(z)| \leq |z| |f(z) -2A|\leq |zf(z)|+2A|z| \] i.e. \[ (1-|z|) | f(z)|\leq 2A|z| . \]