failure of Hartogs’ theorem in one dimension

It is instructive to see an example where Hartogs’ theoremMathworldPlanetmath fails in one dimensionMathworldPlanetmath. Take U= and let K={0}. The function 1z is holomorphic in UK, but cannot be extended to U.

To understand the example and failure of the theorem it is important to understand the proof ( In the proof, the way we construct an extensionPlanetmathPlanetmath is that we start with a function holomorphic in UK, modify it in a neighbourhood of K to be zero, hence extending as a smooth functionMathworldPlanetmath through K. Then we solve the ¯ operator ( inhomogeneous equation ¯ψ=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 ψ with compact support. This fails in dimension 1. While we always get a solution ψ, the solution can never have compact support. Hence, if we tried the proof with 1z, the new function we obtain in the proof does not agree with 1z 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