removable singularity

Let $U\subset ℂ$ be an open neighbourhood of a point $a\in ℂ$. We say that a function $f:U\backslash \{a\}\to ℂ$ has a removable singularity at $a$, if the complex derivative $f^{\prime }(z)$ exists for all $z\ne a$, and if $f(z)$ is bounded near $a$.

Removable singularities can, as the name suggests, be removed.

Theorem 1

Suppose that $f\mathrm{:}U\mathrm{\backslash }\mathrm{\{}a\mathrm{\}}\mathrm{\to }\mathrm{C}$ has a removable singularity at $a$. Then, $f\mathit{}\mathrm{(}z\mathrm{)}$ can be holomorphically extended to all of $U$, i.e. there exists a holomorphic $g\mathrm{:}U\mathrm{\to }\mathrm{C}$ such that $g\mathit{}\mathrm{(}z\mathrm{)}\mathrm{=}f\mathit{}\mathrm{(}z\mathrm{)}$ for all $z\mathrm{\ne }a$.

Proof. Let $C$ be a circle centered at $a$, oriented counterclockwise, and sufficiently small so that $C$ and its interior are contained in $U$. For $z$ in the interior of $C$, set

$$g(z)=\frac{1}{2\pi i}{\oint }_C\frac{f(\zeta )}{\zeta -z}𝑑\zeta .$$

Since $C$ is a compact set, the defining limit for the derivative

$$\frac{d}{dz}\frac{f(\zeta )}{\zeta -z}=\frac{f(\zeta )}{{(\zeta -z)}^2}$$

converges uniformly for $\zeta \in C$. Thanks to the uniform convergence, the order of the derivative and the integral operations can be interchanged. Hence, we may deduce that $g^{\prime }(z)$ exists for all $z$ in the interior of $C$. Furthermore, by the Cauchy integral formula we have that $f(z)=g(z)$ for all $z\ne a$, and therefore $g(z)$ furnishes us with the desired extension.

Title	removable singularity
Canonical name	RemovableSingularity
Date of creation	2013-03-22 12:56:01
Last modified on	2013-03-22 12:56:01
Owner	rmilson (146)
Last modified by	rmilson (146)
Numerical id	5
Author	rmilson (146)
Entry type	Definition
Classification	msc 30E99
Related topic	EssentialSingularity