Analysis Problem of the Day 19

Today’s problem is Problem 11 from the UCLA Fall 2016 Analysis Qual:

Problem 11: Let $f$ be an entire function such that $z \to f(z) f(\frac{1}{z})$ is bounded on $\mathbb{C} \setminus \{0\}.$

a) If $f(0) \not = 0$ , show that $f$ is constant.

b) If $f(0) = 0,$ show that $f(z) = cz^n$ for some constant $c \in \mathbb{C}$ and natural number $n.$

Solution: a) If $f(0) = a \not = 0,$ then $0<c_1<|f(z)|<c_2$ in a small neighborhood of zero, i.e. $c_1<|f(\frac{1}{z})|<c_2$ in a neighborhood of $\infty.$ Given the problem statement, this implies that $|f(z)|$ is bounded in a neighborhood of $\infty.$ Since $f$ is continuous, it is bounded on any compact set, so $f$ is a bounded entire function and therefore constant by Liouville’s theorem.

b) Since zeroes of holomorphic functions have finite order, let $f(z) = z^n g(z),$ where $g$ is an entire function such that $g(0) \not = 0.$ Then, The statement of the problem is that $z \to z^n g(z) z^{-n} g(\frac{1}{z}) = g(z)g(\frac{1}{z})$ is bounded on $\mathbb{C} \setminus \{0\}.$ By a), $g$ is therefore constant. Thus, $f(z) = cz^n$ for some constant $c \in \mathbb{C}$ and natural number $n.$

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