Suppose \(f: \CC\to \CC\) is entire and \[ \abs{f(z)} \leq \abs{z}^{1\over 2} \quad\text{ when } \abs{z} > 10 .\]
Prove that \(f\) is constant.
Solution
Let \(R> 10\), then by Cauchy: \[ 2\pi \abs{f'(z)} &\leq \oint_{\abs{\xi} = R} { \abs{ f(\xi)} \over \abs{\xi}^2 } \dxi \\ &\leq \oint_{\abs{\xi} = R} R^{-2} \abs{\xi}^{1\over 2} \dxi \\ &= R^{-{3\over 2}} \cdot 2\pi R \\ &\sim R^{-{1\over 2}} \\ &\convergesto{R\to\infty} 0 .\]