Suppose \(f(z)\) is entire and \[ \lim_{z\to\infty} {f(z) \over z} = 0 .\]
Show that \(f(z)\) is a constant.
Solution
1. \(f(z)/z\to0\) as \(z\to\infty\) implies \(f\) bounded: for \(|z|>R\), \(|f(z)|\le|z|\).
limit.
2. Cauchy estimate for \(f'\): \(|f'(0)|\le \max_{|z|=R}|f(z)|/R\).
Cauchy.
3. For large \(R\), \(\max_{|z|=R}|f(z)|\le2R\), so \(|f'(0)|\le2\).
1.
4. More generally \(f(z)-f(0)\) satisfies same condition, and \(g(z)=(f(z)-f(0))/z\) entire? Actually consider \(g(z)=(f(z)-f(0))/z\) entire (removable at \(0\)).
\(g\) entire.
5. \(g(z)\to0\) as \(z\to\infty\) (since \(f(z)/z\to0\)), so \(g\) bounded entire, hence constant \(0\) by Liouville.
\(g\) bounded.
6. Hence \(f(z)=f(0)\) constant.
5.
7. Q.E.D.
6.