Suppose \(f\) is entire and, for some integer \(n\ge1\), \[ \lim_{z\to\infty}\frac{f(z)}{z^n}=0. \] Prove that \(f\) is a polynomial of degree at most \(n-1\).
Solution
Write \[ f(z)=\sum_{m=0}^{\infty}a_mz^m \] for the Taylor series of \(f\) at the origin. For \(R>0\), set \[ M(R)=\max_{\abs{\zeta}=R}\abs{f(\zeta)}. \]
1. One has\[ \frac{M(R)}{R^n}\longrightarrow0 \]as \(R\to\infty\).
The hypothesis means that for every \(\varepsilon>0\) there is \(R_0>0\) such that \[ \frac{\abs{f(z)}}{\abs{z}^n}<\varepsilon \] whenever \(\abs{z}\geq R_0\). Hence for every \(R\geq R_0\) and every \(\zeta\) with \(\abs{\zeta}=R\), \[ \frac{\abs{f(\zeta)}}{R^n}<\varepsilon. \] Taking the maximum over the circle gives \[ \frac{M(R)}{R^n}\leq\varepsilon. \]
2. For every integer \(m\geq n\) and every \(R>0\),\[ \abs{a_m} \leq \frac{M(R)}{R^m}. \]
Cauchy’s coefficient formula gives \[ a_m = \frac{1}{2\pi i} \int_{\abs{\zeta}=R} \frac{f(\zeta)}{\zeta^{m+1}} \,d\zeta. \] The circle has length \(2\pi R\), so \[ \abs{a_m} \leq \frac{1}{2\pi}(2\pi R) \frac{M(R)}{R^{m+1}} = \frac{M(R)}{R^m}. \]
3. For every integer \(m\geq n\),\[ a_m=0. \]
Fix \(m\geq n\). For \(R\geq1\), step 2 gives \[ \abs{a_m} \leq \frac{M(R)}{R^n}R^{n-m} \leq \frac{M(R)}{R^n}. \] By step 1, the right-hand side tends to zero as \(R\to\infty\). Thus \(\abs{a_m}=0\).
4. The function \(f\) is a polynomial of degree at most \(n-1\).
Step 3 shows that every Taylor coefficient of degree at least \(n\) vanishes. Therefore \[ f(z)=\sum_{m=0}^{n-1}a_mz^m. \]
5. Q.E.D.
Step 4 is the required conclusion.