# failure of Hartogs’ theorem in one dimension

It is instructive to see an example where Hartogs’ theorem^{} fails in one dimension^{}.
Take $U=\u2102$ and let $K=\{0\}.$
The function $\frac{1}{z}$ is holomorphic in $U\setminus K,$ but cannot be extended to $U.$

To understand the example and failure of the theorem it is important to understand the proof (http://planetmath.org/ProofOfHartogsTheorem). In the proof, the way we construct an extension^{} is that we start with a function holomorphic in $U\setminus K,$
modify it in a neighbourhood of $K$ to be zero, hence extending as a smooth function^{} through $K.$ Then we solve the
$\overline{\partial}$ operator (http://planetmath.org/BarPartialOperator) inhomogeneous equation $\overline{\partial}\psi =g$ to “correct” our extension to be holomorphic.
The key point is that $g$ has compact support allowing us to solve the equation and find a $\psi $
with compact support. This fails in dimension 1. While we always get a solution $\psi ,$ the solution can never have compact support. Hence, if we tried the proof with $\frac{1}{z},$ the new function we obtain in the proof does not agree with $\frac{1}{z}$ on any open set and hence is not an extension.

Title | failure of Hartogs’ theorem in one dimension |
---|---|

Canonical name | FailureOfHartogsTheoremInOneDimension |

Date of creation | 2013-03-22 17:46:57 |

Last modified on | 2013-03-22 17:46:57 |

Owner | jirka (4157) |

Last modified by | jirka (4157) |

Numerical id | 4 |

Author | jirka (4157) |

Entry type | Example |

Classification | msc 32H02 |