设f(z)在|z|<=1上解析，并且|f(z)|<=1,试证明|f＇(0)|<=1

Question

设f(z)在|z|<=1上解析，并且|f(z)|<=1,试证明|f＇(0)|<=1
详细一点的过程，谢谢

Answer 1

由Cauchy积分公式, f'(0) = 1/(2πi)·∫{|z| = 1} f(z)/z² dz.
故|f'(0)| = 1/(2π)·|∫{|z| = 1} f(z)/z² dz|
≤ 1/(2π)·∫{|z| = 1} |f(z)/z²| |dz|
= 1/(2π)·∫{|z| = 1} |f(z)| |dz|
≤ 1/(2π)·∫{|z| = 1} 1 |dz|
= 1.

Answer 2

修正下：dz趋于0应该改为z趋于0