The theorem I'm working with is the following:
Let $U \subset \mathbb{C}$ be an open set, $f: U \rightarrow \mathbb{C}$ a holomorphic function, $z_0 \in \mathbb{C}$ an isolated singularity of $f$.
Then the following are equivalent:
- $z_0$ is a removable singularity
- $\lim_{z\to z_0} f(z)$ exists (i.e. $\in \mathbb{C}$)
- $\lim_{z\to z_0}(z-z_0)f(z) = 0$
Now there's a remark in my notes that says:
The implication 3. ⇒ 2. implies also, that there does not exist a holomorphic function $f:\text{B}(0,1)\backslash\{0\} \to \mathbb{C}$ such that there exists a $L \geq 1$ with $$ \frac{1}{L}|z|^{-\frac{1}{2}} \leq |f(z)| \leq L|z|^{-\frac{1}{2}} $$
This is not true in $\mathbb{R}$, for example this $f$ fulfils the inequality above: $$ f\left( x\right) =\begin{cases}\dfrac{1}{\sqrt{x}},x >0\\ -\dfrac{1}{\sqrt{-x}},x <0\end{cases} $$
My question is: I don't see how 3. ⇒ 2. implies the non-existence of such a function in $\mathbb{C}$. Does anyone care to explain this to me?