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Casorati-Weiestrass theorem for essential singularities

Let $\Omega\subseteq \mathbb{C}$ be open, $ a\in\Omega,\ f\in H(\Omega\backslash \{a\})$ ($f$ analytic on $\Omega\backslash \{a\})$. In the case of an essential singularity: If $ C(a,r)\subseteq\Omega$ ($C(a,r)$ is the disk with origin $a$ and radius $r$), then $f(C(a,r)\backslash \{a\})$ is dense in $\mathbb{C}$.

$\color{green}{\text{(1) What does this theorem actually say? How would you put in descriptive words?}}$

$\color{green}{\text{(2) How would an outline of the proof (in words) look like? What steps should be followed?}}$

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The red, it burns my eyes :( – Ben Millwood Jun 6 '12 at 15:02
it's not a circle, it's a disk. also, you have to suppose it is an essential singularity. 1. it says that near the singularity, $f$ takes a lot of values for example, $e^{1/z}$ for $|z| \leq \epsilon$ (but non zero) can take any value except 0 – Glougloubarbaki Jun 6 '12 at 15:02
Have a look at this:… I think the proof they give is pretty straightforward – M Turgeon Jun 6 '12 at 15:08
Okay, green is better than red, but I wonder why you need colours at all :P – Ben Millwood Jun 6 '12 at 15:10
Thank you very much! – Chris Jun 6 '12 at 15:18
up vote 5 down vote accepted
  1. The theorem says that if $f$ has an essential singularity at $a$, then arbitrarily close to $a$, $f$ takes values arbitrarily close to whatever you like. So the behaviour of $f$ is pretty wild close to $a$ (by comparison to removable singularities, where $f$ is straightforward near $a$, or poles, where $f$ just goes to infinity near $a$).

  2. Suppose $f$ on $D(a,r) \setminus {a}$ doesn't take any values near $b\in \mathbb C$. Then define $g(z) = \frac{1}{f(z)-b}$. Then $g$ is holomorphic and bounded, so it can be extended to $a$. But then $f(z) = \frac{1}{g(z)} + b$ has a pole or removable singularity at $a$. Hence if $f$ does not have a pole or removable singularity at $a$, the above process must fail, i.e. $f$ takes values very close to $b$.

The latter is not a proof as such, but it's a good outline, and the real proof is not much more complex: as a commenter said, see Wikipedia.

It's worth noting that in fact the Casorati-Weierstrass theorem is strengthened by the Big Picard theorem, which states that $f$ doesn't just come close to every value, it in fact takes every value (with at most one exception). The proof of the Picard theorems is unfortunately not so simple.

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Thank you benmachine, got it now! – Chris Jun 6 '12 at 16:01

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