# Prove that $0$ is a removable singularity

Question

Let $D=\{z\in \Bbb{C} : |z|<1 \}, \,f\in H(D)$ and suppose there is a $C>0$ such that $|\frac{f(z)}{z}| \leq \frac{C}{{\sqrt{|z|}}}$ for all $z\in D\setminus 0$. Then the function $g:D\to \Bbb{C}, z \mapsto \frac{f(z)}{z}$ has a removable singularity in $0$.

My attempt

Every isolated singularity is either removable, a pole, or an essential singularity. It's not possible that $0$ is a pole, since then $\lim_{z\to 0} z^m g(z)$ would exist an be nonzero for some $m$, but it's $0$ for all $m\geq 1$. How do I prove that $0$ is not an essential singularity of $g$, hence must be removable?

• If $g$ has an essential singularity at $0$, then so does $\frac{1}{g}$. It is simpler, however, to look what you can deduce about $f$. – Daniel Fischer Jun 5 '15 at 13:01
• What can we deduce about $f$? We have $f(0)=0$ and $|f(z)|/C \leq 1$, so by the Schwarz Lemma, $|f(z)|/C \leq |z|$, hence $g$ is bounded near $0$. Then $0$ must be a removable singularity of $g$. Is that correct? – iwriteonbananas Jun 5 '15 at 13:10
• Well, $f(0) = 0$ is enough. So there is a $h$, holomorphic in a neighbourhood of $0$, with $f(z) = z\cdot h(z)$. – Daniel Fischer Jun 5 '15 at 13:12