Let \(f\) be entire and suppose that \(\lim_{z \rightarrow \infty} f(z) = \infty\). Show that \(f\) is a polynomial.
Solution
Goal: Prove that if \(f\) is entire and \(\lim_{z \to \infty} f(z) = \infty\), then \(f\) is a polynomial.
1. Define \(g(w) := f\qty(\frac{1}{w})\) for \(w \neq 0\); then \(g\) is holomorphic on \(\CC \setminus \theset{0}\).
Composition of the holomorphic maps \(w \mapsto 1/w\) and \(f\).
2. \(w = 0\) is a pole of \(g\), not an essential singularity.
2.1. \(\lim_{w \to 0} g(w) = \infty\).
As \(w \to 0\), \(1/w \to \infty\) in modulus, and \(\lim_{z \to \infty} f(z) = \infty\) by hypothesis.
2.2. \(0\) is not essential.
If \(0\) were an essential singularity, then by Casorati–Weierstrass the values \(g(w)\) would be dense in \(\CC\) for \(w\) near \(0\), contradicting \(\abs{g(w)} \to \infty\); equivalently \(1/g\) is bounded near \(0\) with limit \(0\), a removable singularity, so \(g\) has a pole.
3. \(g\) has a pole of some order \(m \geq 1\) at \(0\), so its Laurent expansion is \(g(w) = \sum_{k=-m}^{\infty} c_k w^k\).
By 2, the Laurent series about \(0\) has only finitely many negative powers, the most negative being \(-m\).
4. For \(z \neq 0\), \(f(z) = \sum_{k=-m}^{\infty} c_k z^{-k} = \sum_{k=0}^{m} c_{-k} z^k + \sum_{j=1}^{\infty} c_j z^{-j}\).
Substitute \(w = 1/z\) into 3.
5. The tail \(\sum_{j=1}^{\infty} c_j z^{-j}\) vanishes: \(c_j = 0\) for all \(j \geq 1\).
\(f\) is entire, so its Laurent expansion about \(z = 0\) (which is 4, valid for \(z \neq 0\) and hence on a punctured neighborhood of \(0\)) has no negative powers of \(z\); the terms \(c_j z^{-j}\) are exactly the negative powers.
6. \(f\) is a polynomial.
4 and 5 give \(f(z) = \sum_{k=0}^{m} c_{-k} z^k\), a polynomial of degree at most \(m\).
7. Q.E.D.
6 is the claim.