Suppose that \(f\) is entire and has polynomial growth in the following sense: \[ \abs{f(z)\over z^n} \leq M \text{ for }\abs{z} \geq R ,\] for some constants \(k\) and \(R\). Show that \(f\) is a polynomial of degree at most \(n\).
Solution
Since \(f\) is entire, it equals its Taylor expansion about \(z_0 = 0\), so for every \(\rho>0\) \[ f(z) = \sum_{k\geq 0} c_k z^k, && c_k = {f^{(k)}(0)\over k! } = {1\over 2\pi i}\int_{\abs{\xi} = \rho} {f(\xi) \over \xi^{k+1}}\dxi .\] For \(\rho\geq R\), a direct estimate yields \[ \abs{c_k} &\leq {1\over 2\pi} \int_{\abs{\xi} = \rho} {\abs{f(\xi)} \over \abs{\xi}^{k+1} }\abs{\dxi}\\ &\leq {1\over 2\pi} \int_{\abs{\xi} = \rho} {M \abs{\xi}^n \over \abs{\xi}^{k+1} }\abs{\dxi} \\ &= {M\over 2\pi} \rho^{n-k-1} \cdot 2\pi \rho \\ &= M\rho^{n-k} ,\] which converges to \(0\) as \(\rho\to \infty\) provided \(k>n\). So \(c_k=0\) for \(k>n\), and \(f\) is a polynomial of degree at most \(n\).