Suppose \(f : \mathbb { C } \to \mathbb { C }\) is entire and \(| f ( z ) | \leq | z | ^ { \frac { 1 } { 2 } }\) whenever \(| z | > 1 0\) . Prove that \(f\) is constant.
Erratum: the source asks to prove that \(f\) is the zero function. This is false. The constant function \[ f\equiv1 \] satisfies the printed hypothesis because \[ 1<\sqrt{10}<\abs{z}^{1/2} \] whenever \(\abs{z}>10\). The printed growth bound forces constancy by step <1>3 of the solution, but it does not force the constant to be zero.
Solution
Write the Taylor series of \(f\) at the origin as \[ f(z)=\sum_{n=0}^{\infty}a_nz^n. \]
1. For every \(R>10\) and every integer \(n\geq1\),\[ \abs{a_n}\leq R^{1/2-n}. \]
By Cauchy’s coefficient formula, \[ a_n = \frac{1}{2\pi i} \int_{\abs{\zeta}=R} \frac{f(\zeta)}{\zeta^{n+1}} \,d\zeta. \] Since \(R>10\), the hypothesis gives \[ \abs{f(\zeta)}\leq R^{1/2} \] on the circle \(\abs{\zeta}=R\). Hence \[ \begin{aligned} \abs{a_n} &\leq \frac{1}{2\pi} (2\pi R) \frac{R^{1/2}}{R^{n+1}}\\ &= R^{1/2-n}. \end{aligned} \]
2. For every integer \(n\geq1\),\[ a_n=0. \]
Fix \(n\geq1\). Step 1 holds for every \(R>10\), while \[ R^{1/2-n}\longrightarrow0 \] as \(R\to\infty\). Therefore \[ \abs{a_n}=0. \]
3. The function \(f\) is constant.
By step 2, every coefficient of positive degree in the Taylor expansion of \(f\) vanishes. Thus \[ f(z)=a_0 \] for every \(z\in\CC\).
4. Q.E.D.
Step 3 is the corrected conclusion.