Prove or exhibit a counterexample: If \(f(z)\) is an entire function that maps every unbounded sequence to an unbounded sequence, then it is a polynomial.
Solution
1. The statement is true: such an \(f\) is a polynomial.
prove the claim.
2. The hypothesis implies \(|f(z)| \to \infty\) as \(|z| \to \infty\).
2.1. Suppose not; then there is \(M < \infty\) and an unbounded sequence \(z_n\) with \(|f(z_n)| \le M\).
negating “\(|f(z)| \to \infty\)” gives a bounded subsequence of values along an unbounded sequence.
2.2. This contradicts the hypothesis that \(f\) maps every unbounded sequence to an unbounded sequence.
\(\{z_n\}\) is unbounded but \(\{f(z_n)\}\) is bounded.
2.3. Hence \(|f(z)| \to \infty\) as \(|z| \to \infty\).
1 and 2.
3. An entire function with \(|f(z)| \to \infty\) as \(|z| \to \infty\) is a polynomial.
3.1. Consider \(g(w) = f(1/w)\) on \(\CC \setminus \{0\}\).
change of variable.
3.2. \(g\) has a pole (not an essential singularity) at \(w = 0\).
\(|f(z)| \to \infty\) as \(|z| \to \infty\) means \(|g(w)| \to \infty\) as \(w \to 0\), so \(g\) has a pole at \(0\) (by Casorati–Weierstrass, an essential singularity would make \(g\) unbounded in every neighborhood but not tending to \(\infty\)).
3.3. Hence \(g\) has a Laurent expansion \(g(w) = \sum_{n=-N}^{\infty} a_n w^n\) with finitely many negative terms.
a pole has finite order.
3.4. Therefore \(f(z) = g(1/z) = \sum_{n=-N}^{\infty} a_n z^{-n}\) is a polynomial (a finite sum of nonnegative powers of \(z\)).
substitute \(w = 1/z\).
4. Q.E.D.
3.