proof of Casorati-Weierstrass theorem

Assume that $a$ is an essential singularity of $f$. Let $V\subset U$ be a punctured neighborhood of $a$, and let $\lambda \in ℂ$. We have to show that $\lambda$ is a limit point of $f(V)$. Suppose it is not, then there is an $ϵ>0$ such that $|f(z)-\lambda |>ϵ$ for all $z\in V$, and the function

$$g:V\to ℂ,z\mapsto \frac{1}{f(z)-\lambda }$$

is bounded, since for all $z\in V$. According to Riemann’s removable singularity theorem, this implies that $a$ is a removable singularity of $g$, so that $g$ can be extended to a holomorphic function $\overline{g}:V\cup \{a\}\to ℂ$. Now

$$f(z)=\frac{1}{\overline{g}(z)}-\lambda$$

for $z\ne a$, and $a$ is either a removable singularity of $f$ (if $\overline{g}(z)\ne 0$) or a pole of order $n$ (if $\overline{g}$ has a zero of order $n$ at $a$). This contradicts our assumption that $a$ is an essential singularity, which means that $\lambda$ must be a limit point of $f(V)$. The argument holds for all $\lambda \in ℂ$, so $f(V)$ is dense in $ℂ$ for any punctured neighborhood $V$ of $a$.

To prove the converse, assume that $f(V)$ is dense in $ℂ$ for any punctured neighborhood $V$ of $a$. If $a$ is a removable singularity, then $f$ is bounded near $a$, and if $a$ is a pole, $f(z)\to \mathrm{\infty }$ as $z\to a$. Either of these possibilities contradicts the assumption that the image of any punctured neighborhood of $a$ under $f$ is dense in $ℂ$, so $a$ must be an essential singularity of $f$.

Title	proof of Casorati-Weierstrass theorem
Canonical name	ProofOfCasoratiWeierstrassTheorem
Date of creation	2013-03-22 13:32:40
Last modified on	2013-03-22 13:32:40
Owner	pbruin (1001)
Last modified by	pbruin (1001)
Numerical id	4
Author	pbruin (1001)
Entry type	Proof
Classification	msc 30D30
Related topic	PicardsTheorem