Find all entire functions satisfying \[ \abs{f(z)} \leq \abs{z}^{1\over 2} \quad\text{ for } \abs{z} > 10 .\]
Solution
Goal: Find all entire functions with \(\abs{f(z)} \leq \abs{z}^{1/2}\) for \(\abs z > 10\).
1. Write the Taylor expansion \(f(z) = \sum_{n=0}^{\infty} a_n z^n\).
\(f\) is entire, hence has a power series about \(0\) converging on all of \(\CC\).
2. Cauchy’s estimate: \(\abs{a_n} \leq \frac{M(R)}{R^n}\) where \(M(R) := \max_{\abs z = R} \abs{f(z)}\).
Standard Cauchy estimate for the \(n\)-th Taylor coefficient on the circle of radius \(R\).
3. For \(R > 10\) and \(n \geq 1\), \(\abs{a_n} \leq \frac{R^{1/2}}{R^n} = R^{1/2 - n}\).
2 with \(M(R) \leq R^{1/2}\) by hypothesis.
4. \(a_n = 0\) for all \(n \geq 1\).
3 holds for arbitrarily large \(R\), and \(R^{1/2 - n} \to 0\) as \(R \to \infty\) when \(n \geq 1\); hence \(\abs{a_n} = 0\).
5. \(f\) is constant: \(f \equiv a_0\).
1 and 4 leave only the constant term.
6. \(a_0 = 0\).
\(\abs{a_0} = \abs{f(z)} \leq \abs{z}^{1/2}\) for all \(\abs z > 10\) (e.g. along \(z = R > 10\), letting \(R \to \infty\)), forcing \(a_0 = 0\).
7. Q.E.D.
5 and 6 show the only such function is \(f \equiv 0\); conversely \(f \equiv 0\) satisfies the bound, since \(|0| = 0 \le |z|^n\) for every \(z\) and every \(n\).