Weierstrass-Casorati Theorem

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $f$ be a holomorphic function defined on the open ball $B \left({a, r}\right) \setminus \left\{{a}\right\}$.

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

Then:

$\forall s < r: f \left({B \left({a, s}\right) \setminus \left\{{a}\right\}}\right)$ is a dense subset of $\C$.



Proof

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

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

$f \left({B \left({0, s}\right) \setminus \left\{{0}\right\}}\right)$ is not a dense subset of $\C$.

Then, by definition of dense subset:

$\exists z_0 \in \C: \exists r_0 > 0: B \left({z_0, r_0}\right) \cap f \left({B \left({0, s}\right) \setminus \left\{{0}\right\}}\right) = \varnothing$

{{explain|This definition of dense still needs to be added to {{ProofWiki}. 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.}}

Hence, the function $\varphi$ defined on $B \left({z_0, r_0}\right)$ by:

$\displaystyle \varphi \left({z}\right) = \frac 1 {f \left({z}\right) - z_0}$

is analytic on $B \left({0, s}\right) \setminus \left\{{0}\right\}$ and bounded near to $0$, because:

$\forall z \in B \left({0, s}\right) \setminus \left\{{0}\right\}: \left|{f \left({z}\right) - z_0}\right| > r_0 \implies \left|{\varphi \left({z}\right)}\right| < \frac 1 {r_0}$

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


Let $\varphi \left({0}\right) \ne 0$.

Then:

$\displaystyle f \left({0}\right) = z_0 + \frac 1 {\varphi \left({0}\right)}$

and the singularity of $f$ was removable.


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

$\displaystyle \varphi \left({z}\right) = \sum_{n \mathop = 1}^{+\infty} a_n z^n$

Then as $\varphi \ne 0$:

$E = \left\{{k \in \N: a_k \ne 0}\right\} \ne \varnothing$

Let $p = \min E$.

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


In each case, the assumption that:

$\exists s < 1: f \left({B \left({0, s}\right) \setminus \left\{{0}\right\}}\right)$ 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.