isolated singularity at 0, limit exists

Question

Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be a bounded entire function. Use the isolated singularity $0$ of $z \mapsto f(\frac{1}{z})$ to show that $f(z)$ has a limit, when $|z|$ goes to $\infty$.

How can I tackle this?

Answer 1

Let $M$ denote the upper bound of $|f|$ on $\mathbb{C}$. Consider the function you mentioned, $h(z)=f(\frac{1}{z}):\mathbb{C}\setminus\{0\}\to\mathbb{C}$. Obviously $0$ is an isolated singularity. Observe that $|h(z)|\leq M$ on $\mathbb{C}\setminus\{0\}$, hence $h(z)$ has a removable singularity at $0$. Thus the limit $\displaystyle{\lim_{z\to0}h(z)}$ exists (the holomorphic function that extends $h$ at $0$ is continuous there). Now it is $h(\frac{1}{z})=f(z)$ for all $z\neq 0$. therefore, $\displaystyle{\lim_{z\to\infty}f(z)=\lim_{z\to\infty}h(\frac{1}{z})=\lim_{w\to0}h(w)}$ and the limit you're looking for exists.

P.S.

Judging from the comments, I believe you also want to prove somehow without using Liouville's theorem that $f$ is constant. I will use the maximum modulus principle (hope that's okay!) :

Now $|f|$ is continuous on each compact disk $\overline{D(0,n)}$, hence it attains a maximum value there, say $M_n$. By the maximum modulus principle, $M_n$ is obtained strictly on the circle $C_n=\partial D(0,n)$ and let $\zeta_n$ be a point of $C_n$ such that $|f(\zeta_n)|=M_n$. Since $D(0,n)\subset D(0,n+1)$, obviously we have $M_n\leq M_{n+1}$. But the limit $\displaystyle{\lim_{z\to\infty}f(z)}$ exists, therefore $\displaystyle{\lim_{n\to\infty}M_n=\lim_{n\to\infty}|f(\zeta_n)|=\lim_{z\to\infty}|f(z)|}$. So for all $z\in\mathbb{C}$ it is $|f(z)|\leq\displaystyle{\lim_{w\to\infty}f(w)}.$

Now take the extended function $\bar{h}:\mathbb{C}\to\mathbb{C}$. As proved above, it is $\bar{h}(0)=\displaystyle{\lim_{w\to0}h(w)=\lim_{z\to\infty}f(z)}$. For any $z\in\mathbb{C}\setminus\{0\}$ it is $|\bar{h}(z)|=|f(\frac{1}{z})|\leq|\displaystyle{\lim_{w\to\infty}f(w)|=|\bar{h}(0)|}$. By the maximum modulus principle again, $h$ is constant therefore $f$ is constant.