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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 $z_0$?

r123454321
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2 Answers2

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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 $f(z_0)$.

Mike Earnest
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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.

Amitai Yuval
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