0

Let $\Delta$ denote the unit disc centered at $0$ in the complex plane. I'm interested in proving the following claim.

Assume that $g$ is a function that is bounded and analytic on $\mathbb{C}\setminus \Delta$. Then $\lim_{|z|\to\infty}g(z)$ exists.

It follows from the assumption that there exists some constant $M$ such that $|g(z)|\leq M$. But I am not quite sure how I can make use of this to prove the existence of the above limit.

Any help/hint will be appreciated. Thanks in advance.

neophyte
  • 520

1 Answers1

2

For each $z\in\Delta\setminus\{0\}$, let $f(z)=g\left(\frac1z\right)$. Then $f$ has a singularity at $0$. Since $f$ is bounded, then that singularity has to be a removable singularity, by the Riemann extension theorem. Therefore, the limit $\lim_{z\to0}f(z)$ exists. But $\lim_{|z|\to\infty}g(z)=\lim_{z\to0}f(z)$.

José Carlos Santos
  • 427,504