Weierstrass-Casorati Theorem

Theorem

Let $f$ be a holomorphic function defined on the open ball $\map B {a, r} \setminus \set a$.

Let $f$ have an essential singularity at $a$.

Then:

$\forall s < r: f \sqbrk {\map B {a, s} \setminus \set a}$ is a dense subset of $\C$.

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Proof

Without loss of generality, suppose $a = 0$ and $r = 1$.

Aiming for a contradiction, suppose $\exists s < 1$ such that:

$f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$.

Then, by definition of dense subset:

$\exists z_0 \in \C: \exists r_0 > 0: \map B {z_0, r_0} \cap f \sqbrk {\map B {0, s} \setminus \set 0} = \O$

This article, or a section of it, needs explaining. In particular: This definition of dense still needs to be added to $\mathsf{Pr} \infty \mathsf{fWiki}$. Whichever definition of denseness (either everywhere dense or dense-in-itself) will need to be expanded for the Complex case so as to make it clear that the above definition follows. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

Hence, the function $\varphi$ defined on $\map B {z_0, r_0}$ by:

$\map \varphi z = \dfrac 1 {\map f z - z_0}$

is analytic on $\map B {0, s} \setminus \set 0$ and bounded near to $0$, because:

$\forall z \in \map B {0, s} \setminus \set 0: \cmod {\map f z - z_0} > r_0 \implies \cmod {\map \varphi z} < \dfrac 1 {r_0}$

Therefore, we can extend the domain of $\varphi$ (using the Analytic Continuation Principle).

Let $\map \varphi 0 \ne 0$.

Then:

$\map f 0 = z_0 + \dfrac 1 {\map \varphi 0}$

and the singular point of $f$ was removable.

Otherwise, let the power series of $\varphi$ be written:

$\ds \map \varphi z = \sum_{n \mathop = 1}^{+\infty} a_n z^n$

Then as $\varphi \ne 0$:

$E = \set {k \in \N: a_k \ne 0} \ne \O$

Let $p = \min E$.

Then $0$ is a pole of order $p$ of $f$.

In each case, the assumption that:

$\exists s < 1: f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$ contradicts the fact that $0$ is an essential singularity of $f$.

Hence the result, by Proof by Contradiction.

$\blacksquare$

Also known as

It is also known as the Casorati-Weierstrass Theorem.

Source of Name

This entry was named for Karl Theodor Wilhelm Weierstrass and Felice Casorati.