# failure of Hartogs’ theorem in one dimension

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 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 $\bar{\partial}$ operator (http://planetmath.org/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.

Title failure of Hartogs’ theorem in one dimension FailureOfHartogsTheoremInOneDimension 2013-03-22 17:46:57 2013-03-22 17:46:57 jirka (4157) jirka (4157) 4 jirka (4157) Example msc 32H02