Analysis Problem of the Day 64

Today’s problem appeared as Problem 9 on Texas A&M’s August 2021 Complex Analysis Qual:

Problem 9. Let $f$ be a meromorphic function on $\mathbb{C}$ such that $|f(z)| \leq M|z|^n$ for $|z|>R,$ where $z$ is not a pole, for some integer $n$ and $M,R>0.$ Show that $f$ is a rational function.

Solution: Without loss of generality, suppose $0$ is not a pole of $f$ – otherwise, consider $\frac{f(z)}{z^k},$ where $k$ is the order of the pole of $f$ at $0.$ Since we are given the behavior of $f$ in a neighborhood of $\infty,$ we consider $g(z):=f(\frac{1}{z}).$ Then, $g$ is a meromorphic function on $\mathbb{C} \setminus \{0\}$ that is bounded at $\infty.$ In particular, if $w = \frac{1}{z},$ we have that $|g(w)| \leq \frac{M}{|w|^n}$ when $w$ is near $0$ and is not a pole of $g.$ Let $P$ be the set of poles of $g$ excluding the origin. The above estimate implies that $h(w):=g(w) w^{\max(n+1,0)}$ is a meromorphic function on $\mathbb{C} \setminus \{0\}$ that extends continuously to be $0$ on $\mathbb{C} \setminus P.$ But this implies that $|g(w)|<\epsilon$ in a neighborhood of $0 \in \mathbb{C} \setminus P.$ Thus, $0$ cannot be an accumulation point of $P,$ as $\lim_{z \to a} |g(z)|=\infty$ for any pole $a$ of $g.$ In particular, since $g$ is bounded at $\infty$ and is meromorphic, $P$ must be discrete, i.e. $g$ has finitely many poles. Letting $h(w):=\frac{g(w)}{\prod_{k=1}^m (w-a_k)^{n_k}},$ where $a_1,...,a_m$ are the poles of $g$ of orders $n_1,...,n_m,$ respectively, one obtains that $h$ is a bounded holomorphic function and therefore constant by Liouville’s theorem. It follows that $g$ is a rational function, so $f(z) = g(\frac{1}{z})$ is a rational function as well.

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