5
$\begingroup$

In Stein's Complex Analysis, he presents the following statement and proof of Casorati-Weierstrass:

Suppose $f$ is holomorphic in the punctured disc $D_r(z_0) - \{z_0\}$ and has an essential singularity at $z_0$. Then, the image of $d_r(z_0) - \{z_0\}$ under $f$ is dense in the complex plane.

Proof:

We argue by contradiction. Assume that the range of $f$ is not dense, so that there exists $w \in \mathbb{C}$ and $\delta > 0$ such that

$$|f(z) - w| > \delta$$

for all $z \in D_r(z_0) - \{z_0\}$. We may therefore define a new function on $D_r(z_0) - \{z_0\}$ by

$$g(z) = \frac{1}{f(z) - w}$$

which is holomorphic on the punctured disc and bounded by $1/\delta$. Hence, $g$ has a removable singularity at $z_0$.

Statement 1: If $g(z_0) \neq 0$, then $f(z) - w$ is holomorphic at $z_0$, which contradicts the assumption that $z_0$ is an essential singularity.

Statement 2: In the case that $g(z_0) = 0$, then $f(z) - w$ has a pole at $z_0$ also contradicting the nature of the singularity at $z_0$. The proof is complete.

My Question:

In regards to statement 1, why precisely is this true? I can rewrite $g(z) = \frac{1}{f(z) - w}$ as $f(z) - w = \frac{1}{g(z)}$, and if $g(z_0) \neq 0$, then $f(z_0) - w$ is defined -- does it immediately follow that $f(z_0)$ is holomorphic at $z_0$?

$\endgroup$
5
$\begingroup$

Your reasoning is pretty much correct. Since $f$ is holomorphic on $D_r(z_0)\setminus\{z_0\}$, and $\lim_{z\to z_0}f(z)=\frac{1}{g(z_0)}-w$ exists, it follows the singularity at $z_0$ is removable, contradicting that it is essential. This is one of the many ways to tell if a singularity is removable.

This is what they mean when they say $f$ is holomorphic at $z_0$; they mean the singularity is removable there, so it can be made holomorphic by an appropriate definition of $f(z_0)$.

$\endgroup$
4
$\begingroup$

The fact is that given a holomorphic $f:D_r(z_0)\setminus\{z_0\}\to\mathbb{C}$, it is enough to know that $f$ is bounded, in order to conclude that it can be extended holomorphically to $D_r(z_0).$

To see that, one needs to convince oneself that for any $z\in D_r(z_0)\setminus\{z_0\}$, $$f(z)=\frac{1}{2i\pi}\int_\gamma \frac{f(\zeta)}{\zeta-z}d\zeta,$$where $\gamma$ is the boundary of the disc. This makes the claim obvious, as this expression can clearly be extended holomorphically. To prove this equation, note first that for a given $z$, one can take $\gamma'$ to be another circle around $z_0$, closer to $z_0$ than $z$, and then $$f(z)=\frac{1}{2i\pi}\left(\int_\gamma\frac{f(\zeta)}{\zeta-z}d\zeta-\int_{\gamma'}\frac{f(\zeta)}{\zeta-z}d\zeta\right).$$Taking the circle $\gamma'$ to be smaller and smaller, $f$ being bounded guarantees that the right hand integral vanishes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.