Let \(f\) be a meromorphic function on the complex plane with the property that \(|f(z)| \leq\) \(M\) for all \(|z|>R\), for some constants \(M>0, R>0\).
Prove that \(f(z)\) is a rational function, i.e., there exist polynomials \(p, q\) so that \(f=\frac{p}{q}\).
Solution
Note that \(f\) must have finitely many poles – either \(z=\infty\) is a pole or a removable singularity, and since poles are isolated, there is some \(R\gg 0\) such that all other poles of \(f\) are in \(\abs{z} \leq R\). The set \(P_f\) of poles is a closed set and \(\bar{\DD_R}\) is compact, so if \(P_f\) is infinite it has an accumulation point, contradicting that poles are isolated.
So enumerate \(P_f\) as \(\ts{p_k}_{k\leq N}\), define \(g(z) \da \prod_{k\leq N}(z-p_k)\), and set \(F(z) \da g(z) f(z)\). Then \(F\) is an entire function, and the claim is that \(F\) is bounded and thus constant by Liouville. Proving the bound: take \(\abs{z} > R\), then \[ \abs{G(z)} &= \abs{f(z)} \abs{g(z)} \\ &\leq M C \abs{z}^N ,\] using that \(g\) is a polynomial of degree \(N\), so \(\abs{g(z)\over z^N}\to 1\) as \(\abs{z}\to \infty\) since \(g\) is monic. So after possibly increasing \(R\), we can choose \(\abs{z}\) large enough so that \(\abs{g(z)\over z^N} < C\) for, say, some constant \(C<2\). In any case, by a common qual question, if \(\abs{G} \in \bigo(\abs{z}^N)\) for \(\abs{z}\) large enough then \(G\) is a polynomial of degree at most \(N\). Then \(f(z) \da G(z)/g(z)\) is a rational function.