Consequence of Riemann mapping theorem

Let $\mathbb{H}$ denote the complex upper half plane (not including the real axis). Let $A$ be a bounded subset with $A = \mathbb{H} \cap \overline{A}$ and $\mathbb{H} \backslash A$ simply connected. Why does it follow from the Riemann mapping theorem that there are conformal maps $g: \mathbb{H} \backslash A \rightarrow \mathbb{H}$ with $|g(z)| \rightarrow \infty$ as $z \rightarrow \infty$?

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This condition $A=\mathbb{H}\cap\overline{A}$ is incredibly strange. I assume "bounded subset" means of $\mathbb{H}$ since that is the only set mentioned. But then the condition is just that $A=\overline{A}$, i.e. $A$ is closed. Either I'm missing something or we may as well say "Let $A$ be a bounded closed subset with $\mathbb{H}\setminus A$ simply connected". –  Matt Sep 24 '11 at 23:26
@Matt: I interpreted it that $A$ is bounded in $\mathbb{C}$ and that the closure of $A$ is taken in $\mathbb{C}$, but we'll have to wait to find out the true answer, I guess. –  Jason DeVito Sep 25 '11 at 0:45
@Matt: The notation $\bar A$ must mean the closure in $\mathbb{C}$ (and bounded means bounded in $\mathbb{C}$). Then, $A=\mathbb{H}\cap\bar A$ says that $A$ is closed in $\mathbb{H}$. –  George Lowther Sep 25 '11 at 0:48
I'm not sure why it should even matter for the problem at hand, where the closure is taken....... –  Sam Sep 25 '11 at 1:59

1 Answer

By the Riemann Mapping Theorem you can find a conformal map $f:\mathbb{H}\setminus A \to \mathbb{H}$. Assuming that $A$ is contained in the disk $|z| \le R$, you can extend $f$ by Schwarz reflection meromorphically to $|z| > R$, with a pole or removable singularity at $\infty$. If $f(\infty) = \infty$ for this extension, you are done (set $g = f$), otherwise $g(z) = \frac{1}{f(\infty) - f(z)}$ works.

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