# 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 $\lambda \in \u2102$.
We have to show that $\lambda $ is a limit point of $f(V)$. Suppose it
is not, then there is an $\u03f5>0$ such that
$|f(z)-\lambda |>\u03f5$ for all $z\in V$, and the function

$$g:V\to \u2102,z\mapsto \frac{1}{f(z)-\lambda}$$ |

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^{} $\overline{g}:V\cup \{a\}\to \u2102$. 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 $\lambda \in \u2102$, so $f(V)$ is dense in $\u2102$ for any punctured neighborhood $V$ of $a$.

To prove the converse, assume that $f(V)$ is dense in $\u2102$ 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 $\u2102$, 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 |