FailureOfHartogsTheoremInOneDimension 2008-02-06 12:48:59 2008-02-06 12:48:59 Example Hartogs' theorem % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \theoremstyle{theorem} \newtheorem*{thm}{Theorem} \newtheorem*{lemma}{Lemma} \newtheorem*{conj}{Conjecture} \newtheorem*{cor}{Corollary} \newtheorem*{example}{Example} \newtheorem*{prop}{Proposition} \theoremstyle{definition} \newtheorem*{defn}{Definition} \theoremstyle{remark} \newtheorem*{rmk}{Remark} It is instructive to see an example where Hartogs' theorem fails in one dimension. Take $U = {\mathbb C}$ 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 \PMlinkname{the proof}{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 \PMlinkname{$\bar{\partial}$ operator}{BarPartialOperator} inhomogeneous equation $\bar{\partial}\psi = g$ 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.