Find all entire functions satisfying \[ \abs{f(z)} \leq \abs{z}^{1\over 2} \quad\text{ for } \abs{z} > 10 .\]
Solution
Since \(f\) is entire, take a Laurent expansion at \(z=0\), so \(f(z) = \sum_{k\geq 0} c_k z^k\) where \({2\pi i\over k!} c_k = f^{(k)}(0)\) by Cauchy’s integral formula. Take a Cauchy estimate on a disc of radius \(R>10\): \[ \abs{c_k} &\leq {k!\over 2\pi}\int_{\abs{z} = R} \abs{f(\xi) \over (\xi - 0)^{k+1}}\dxi\\ &\leq {k! \over 2\pi}\int_{\abs z = R}{ \abs{\xi}^{1\over 2} \over \abs{\xi}^{k+1} }\dxi \\ &= {k! \over 2\pi} \cdot {1\over R^{k+{1\over 2}}}\cdot 2\pi R \\ &= \bigo(1/R^{k-{1\over 2}}) .\] So in particular, if \(k\geq 1\) then \(k-{1\over 2}>0\) and \(c_k = 0\). This forces \(f = c_0\) to be constant.