Suppose that \(f\) is entire and \(f\) has sublinear growth in the following sense: \[ \abs{f(z)\over z}\to 0 \text{ as } \abs{z}\to \infty .\] Show that \(f\) must be constant.
Solution: Direct bound
Define \[ g(z) \da \begin{cases} {f(z) - f(0) \over z-0} & z\neq 0 \\ f'(0) & z=0. \end{cases} .\] Note that for \(z\neq 0\), \[ \abs{g(z)} \da \abs{f(z) - f(0)\over z} \leq \abs{f(z) \over z} + \abs{f(0)\over z} \convergesto{\abs{z} \to \infty }0 ,\] where we’ve used the assumption in the last step. The function \(g\) is entire, since \(f(z)-f(0)\) vanishes at \(0\). Given \(\eps>0\), there is \(R_\eps\) with \(\abs{g(z)} < \eps\) for \(\abs{z}\geq R_\eps\). Fix \(z\in\CC\) and let \(R\geq\max(R_\eps,\abs z)\). Then \(\abs{g}<\eps\) on the circle of radius \(R\), so by the maximum modulus principle \(\abs{g(z)} < \eps\). Since \(\eps\) is arbitrary, \(g(z) = 0\) for all \(z\in \CC\), so \(f(z) = f(0)\) is a constant for all \(z\).
Solution: Cauchy bound
Claim: \(f'(z) \equiv 0\). Fix \(z\) and \(\eps>0\). Choose \(R \geq 2\abs z\) so that \(\abs{f(\xi)} \leq \eps \abs{\xi}\) for \(\abs{\xi} \geq R\), and apply Cauchy’s formula: \[ \abs{f'(z)} &= \abs{{1\over 2\pi i } \int_{\abs \xi = R} { f(\xi) \over (\xi - z)^2 }\dxi } \\ &\leq {1\over 2\pi} \int_{\abs \xi = R} { \abs{ f(\xi) } \over \abs{\xi - z}^2 } \abs{\dxi} \\ &\leq {1\over 2\pi} \int_{\abs \xi = R} { \eps \abs{\xi} \over \qty{R - \abs{z}}^2 } \abs{\dxi} \\ &= {1\over 2\pi} \qty{\eps R\over \qty{R-\abs{z}}^2 } \cdot 2\pi R \\ &\leq 4\eps ,\] using \(R-\abs z\geq R/2\). Since \(\eps\) is arbitrary, \(f'(z)=0\); as \(z\) is arbitrary, \(f\) is constant.