Q:

Extending the Generalised Theorem of Liouville to Negatives: Extending $f$ where $f(z)$ is bounded by $C\left|z\right|^{-\frac{1}{2}}$

I am supposed to show that we can extend a holomorphic function $f(z):\mathbb{D}\setminus \{0\}$ which ist bounded by $\left|f(z)\right| \le C|z|^{-\frac{1}{2}}$ to the entire open unit disk $\mathbb{D}$
I don't really know where to begin.
The well-know (generalized) Theorem of Liouville states that:

An entire function $f$ for which the bound $|f(z)| \le M|z|^n$ holds for all $|z| \ge R_0$ is a polynomial of degree at most $n$.

I have already seen how to use the Theorem for noninteger values of n like $0.5$, but I don't see how we could push the concept to negatives.
Assuming that we can't, more work needs to be done. But what can I use? 
$f$ isn't even defined on the edge $\partial \mathbb{D}$, and since $f$ is obviously not holomorphic on $\mathbb{D}$, I don't see what tools are available.

A:

If $g(z)=zf(z)$, then $\bigl\lvert g(z)\bigr\rvert\leqslant C\sqrt{\lvert z\rvert}$ and therefore $\lim_{z\to0}g(0)=0$. So, by Riemann's theorem on removable singularities, $g$ has a removable singularity at $0$. If we extend $g$ to a holomorphic function $G\colon D\longrightarrow\mathbb C$, then $G(0)=0$ and, if you write$$G(z)=a_1z+a_2z^2+a_3z^3+\cdots,$$then$$f(z)=a_1+a_2z+a_3z^2+\cdots$$

