Question on proof of Casorati–Weierstrass theorem

Question

Consider a holomorphic function $f$ with essential singularity in $z_0$. Then $f(B'(z_0,\varepsilon))$ is dense in $\mathbb{C}$ for all $\varepsilon > 0$.

Given proof:

Suppose that there exist $c\in\mathbb{C}$ and $\delta>0$ such that $|f(z)-c|>\delta$ for all $z\in B'(z_0,\varepsilon)$. Then $$ \lim_{z\to z_0}|z-z_0|^{-1}|f(z)-c|=\infty. \quad [1]$$ This shows that $(f(z)-c)/(z-z_0)$ has a pole in $z_0$ $[2]$, of some order $N$. Then $$ \lim_{z\to z_0}|z-z_0|^N|f(z)|\le \lim_{z\to z_0}|z-z_0|^N|f(z)-c|=0.\quad [3]$$ Thus, $z_0$ is not an essential singularity of $f$ $[4]$ - a contradiction .

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[1]: Is my reasoning for correct? - Choose $\varepsilon>0$. For the previous line we now that if $|z-z_0|<\varepsilon$, then $|f(z)-c|>\delta$ for some $c\in\mathbb{C}$ and $\delta>0$. So, $|z-z_0|^{-1}>1/\varepsilon$ and thus $$|z-z_0|^{-1}|f(z)-c|>\delta/\varepsilon \text{ as } z\to z_0.$$

[2]: Same question. - From the previous line, it follows that $\lim_{z\to z_0} |z-z_0||f(z)-c|^{-1} = 0$, or equivalently $\lim_{z\to z_0} (z-z_0)(f(z)-c)^{-1}=0$. Therefore, $\lim_{z\to z_0}(z-z_0)^{-1}(f(z)-c)=\infty$, proving [2].

[3]: Same question.

• inequality: $|z-z_0|^N|f(z)-c+c| \le |z-z_0|^N|f(z)-c|+|z-z_0|^N|c|,$ where to last term tends to zero as $z\to z_0$.

• equality: $z_0$ is a pole of order $N$, thus $(f(z)-c)/(z-z_0)=(z-z_0)^{-N}g(z)$, $g$ holomorphic, $g(z_0)\ne 0$. But then $|z-z_0|^N|f(z)-c|=|z-z_0||g(z)| \to 0\cdot |g(z_0)| = 0$ as $z\to z_0$.

[4]: Same question. From (3) follows that $\lim_{z\to z_0}(z-z_0)^Nf(z)=0$, so $\operatorname{mult}_{z=z_0} f(z) \ge -N$, meaning that multiplicity exists and therefore $z_0$ can't be essential.

Thanks.

Answer 1

[1] The reasoning isn't incorrect, but it only shows that the ratio is $>\delta/\epsilon$ (both of which have fixed values by the supposition). This is not enough to prove it goes to $\infty$. Try it without substituting $\epsilon$ for $z - z_0$.

[2] No, $\lim_{z\to z_0}(z-z_0)^{-1}(f(z)-c)=\infty$, not $0$. This is what was just proved in [1]. And even if true, it would not mean $z_0$ is a pole. Instead [2] follows from an examination of isolated singularities that surely preceeded this proof. There are 3 type:

• removable: $\lim_{z\to z_0} g(z)$ exists (and is finite).
• pole: $\lim_{z\to z_0} (z-z_0)^Ng(z)$ exists for some integer $N > 0$. In this case it must be that $\lim_{z\to z_0} g(z)$ is infinite.
• essential singularity: $\lim_{z\to z_0} (z-z_0)^Ng(z)$ does not exist for any $N$. In this case $\lim_{z\to z_0} g(z)$ does not converge, nor diverge to $\infty$.

Since you know that $\lim_{z\to z_0}\left|\frac {f(z)-c}{z-z_0}\right|=\infty$, the only case possible is a pole.

[3] This is correct.

[4] More simply, this is the condition for $f$ to have a pole at $z_0$, but we are given that $f$ has an essential singularity, not a pole, so this cannot be.