4

Does there exist an analytic function $f$, defined in a neighborhood of $\overline{\mathbb{D}}$, such that $f(\overline{\mathbb{D}})=\overline{\mathbb{H}}$ ? where $ \overline{\mathbb{H}} = \{ z \in \mathbb{C} | \ Imz \geq 0\} $ and $\overline{\mathbb{D}} = \{ z \in \mathbb{C} | \ |z| \leq 1\}$.

The first thing that comes to my mind is the linear fractional transformation $T(z)=\frac{i-iz}{1+z}$ (is not analytic in any neighborhood of $ \overline{\mathbb{D}}$, right? since we are not considering the Riemann sphere, or equivalently $f'(-1)=\infty \notin \overline{\mathbb{H}}$). But using this or any other linear fractional transformation you have to map the unit circle onto the real line, hence one point , say $z_0$ on the unit circle gets mapped to $\infty$; therefore $ f(z_0) \notin \overline{\mathbb{H}}$. Is it correct that no linear fractional transformation can do the job ? How can we show that such a map does not exist ?

Any hint or idea is appreciated.

the8thone
  • 4,111

1 Answers1

4

No, there doesn't even exist a continuous function such that this holds: $\overline{\mathbb{D}}$ is compact, and so must be $f(\overline{\mathbb{D}})$.

  • 2
    What if we consider the closure of $\mathbb H$ in the Riemann sphere (which is how I interpreted the question before I read the OP's paragraph)? – Ted Shifrin Dec 16 '13 at 04:44
  • 1
    I think $ \overline{\mathbb{H}}$ in the Riemann sphere is compact , so we are Okay in that case, in other words, The Riemann sphere is the one-point-compactification of the complex plane , and any closed subset of a compact set is compact, i.e. $ \overline{\mathbb{H}}$ in the Riemann sphere is compact. Am I right ? – the8thone Dec 16 '13 at 04:58