Suppose \(f\) is entire and suppose that for some integer \(n\geq 1\), \[ \lim_{z\to \infty} {f(z) \over z^n} = 0 .\]
Prove that \(f\) is a polynomial of degree at most \(n-1\).
Solution
Goal: If \(f\) is entire and \(\lim_{z\to\infty} \frac{f(z)}{z^n} = 0\) for some integer \(n \geq 1\), prove \(f\) is a polynomial of degree at most \(n-1\).
1. For every \(\varepsilon > 0\) there is \(R_\varepsilon\) such that \(\abs{f(z)} \leq \varepsilon \abs{z}^n\) for all \(\abs{z} \geq R_\varepsilon\).
This is the definition of \(f(z)/z^n \to 0\) as \(z \to \infty\).
2. For each \(k \geq n\), the Taylor coefficient \(\frac{f^{(k)}(0)}{k!} = 0\).
1. Since \(k \geq n\), letting \(R \to \infty\) gives \(\abs{f^{(k)}(0)} \leq 0\) (as \(R^{n-k} \to 0\)), so \(f^{(k)}(0) = 0\). (More precisely, fix \(\varepsilon\) and take the limit \(R\to\infty\): \(\abs{f^{(k)}(0)} \leq \liminf k!\varepsilon R^{n-k} = 0\).)
3. \(f(z) = \sum_{k=0}^{n-1} \frac{f^{(k)}(0)}{k!} z^k\), a polynomial of degree \(\leq n-1\).
The Taylor series of the entire function \(f\) about \(0\) converges to \(f\) everywhere; by 2 all terms with \(k \geq n\) vanish.
4. Q.E.D.
3 proves the claim.