Prove Liouville’s theorem: suppose \(f:\CC\to\CC\) is entire and bounded. Use Cauchy’s formula to prove that \(f'\equiv 0\) and hence \(f\) is constant.
Solution
The main idea: \[ \abs{f'(z)} &\leq {1\over 2\pi }\oint_R {\abs{f(\xi)} \over \abs{\xi}^2 } \dxi\\ &= {1\over 2\pi }\oint_R {\abs{f(\xi)} } R^{-2} \dxi\\ &\leq {1\over 2\pi }\oint_R M R^{-2} \dxi\\ &= {1\over 2\pi} MR^{-2}\cdot 2\pi R \\ &= MR^{-1} \\ &\convergesto{R\to\infty}0 .\] So \(f'\equiv 0\). Since \(\CC\) is connected, \(f'\equiv 0\) implies \(f\) is constant.