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Let $f : \mathbb{C}\setminus\{0\} \to \mathbb{C}$ be a holomorphic function satisfying $f(2z) = f(z)$ for all $z \in \mathbb{C}\setminus\{0\}$. Show that $f$ is constant.

Here $f$ is defined as a map $\mathbb{C}\setminus\{0\}\rightarrow \mathbb{C}$. If $0$ is a removable singularity, then it is clear to me that $f(z) = f(0)$ for all $z$. Further, I can rule out the possibility that $0$ is a pole since this would lead to a contradiction on the absolute value of $f$ as I take successive values of $z,z/2,z/4, z/8$, etc. However, I cannot currently rule out the possibility that such an $f$ exists which is non-constant and has an essential singularity at $0$. Can anybody suggest a way please?

Michael Albanese
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user262729
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1 Answers1

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For any $z \in \mathbb{C}\setminus\{0\}$, there is $k \in \mathbb{Z}$ such that $1 \leq |2^kz| \leq 2$ (such a $k$ may not be unique). So, if $A = \{z \in \mathbb{C} \mid 1 \leq |z| \leq 2\}$, $f(\mathbb{C}\setminus\{0\}) = f(A)$ as $f(z) = f(2^kz)$. As $A$ is compact and $f$ is continuous, $f(A)$ is bounded. Therefore, the singularity at the origin is removable by Riemann's theorem.

Now note that $a_n = 2^{-n}$ is a convergent sequence in $\mathbb{C}$ with $f(a_n) = f(a_1)$. By the Identity Theorem, $f$ is a constant function (with value $f(a_1)$).

Michael Albanese
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