Removable singularity for analytic function on the punctured disk where $|f'(z)| \leq 1/|z|$. Suppose that on the punctured disk $0 < |z| < 1, \ f(z)$ is analytic with its first derivative bounded by $$
  |\,f'(z)| \leq \frac{1}{|z|}.
$$
Show that $f$ has a removable singularity at $z=0$.

I suspect this question may have been answered elsewhere—perhaps someone could point me to an existing answer.
I suspect there may be a clever way to use the Schwarz lemma or perhaps Cauchy's integral formula.
 A: The hypotheses imply $f'(z)$ has at worst a pole of order $1$ at $0.$ So we can write
$$f'(z) = \frac{c}{z} +g(z)$$
for $0<|z|<1,$ where $g$ is analytic in $D(0,1).$ Now $g$ has an antiderivative $G$ in $D(0,1).$ It follows that
$$(f-G)'(z) = \frac{c}{z}, \,\, 0<|z|<1.$$
If $c\ne 0,$ we have discovered an antiderivative of $1/z$ in $\{0<|z|<1\}.$ Is that possible? Insert "no" here: ___.  Thus $c=0,$ leading to the desired conclusion.
A: Let us consider the Laurent series of $f(z)$ and $f'(z)$ in the punctured disk:
$$ f(z) = \sum_{n\in\mathbb{Z}} a_n z^n,\qquad f'(z)=\sum_{n\in\mathbb{Z}}na_n z^{n-1}. $$
If we assume that some $a_n$ with $n<0$ is $\neq 0$, we have a violation of the inequality $\left|f'(z)\right|\leq \frac{1}{|z|}$ for some $z$ sufficiently close to the origin, since in such a case $z f'(z)$ has a singularity at the origin, which is either a pole or an essential singularity. In both cases $zf'(z)$ cannot stay bounded as $z\to 0$. It follows that the only non-zero coefficients are the ones with $n\geq 0$ and $f(z)$ is holomorphic over $|z|<1$. 
A: Note, applying L'Hopital's rule:
$$\lim_{z\to 0} zf(z)=\lim_{z\to0} \frac{f(z)}{1/z}=\lim_{z\to 0} \frac{f'(z)}{-1/z^2}=\lim_{z\to 0} -z^2f'(z)$$
Note that
$$\lim_{z\to 0} |-z^2f'(z)|\le \lim_{z\to0}|z|=0$$
A: Set $\,g(z)=z\,f'(z)$. Clearly,
$$
|g(z)|=|z\,f'(z)|\le 1,
$$
and hence $g$ is bounded in the punctured unit disc $\mathbb D\setminus\{0\}$, and hence, $g$ has removable singularity at $z=0$, and consequently, $g$ extends analytically in the unit disc.
Expand $g$ in the unit disc as $g(z)=\sum_{n=0}^\infty a_nz^n$. Then
$$
f'(z)=\frac{a_0}{z}+\sum_{n=0}^\infty a_{n+1}z^n.
$$
Let $\gamma\subset\mathbb D\setminus\{0\}$ be a closed curve. Then 
$$
0=\int_{\gamma}\big(z\,f(z)\big)'\,dz=\int_{\partial B_r}\big(\,f(z)+z\,f'(z)\big)\,dz=\int_{\partial B_r}\big(\,f(z)+g(z)\big)\,dz=\int_{\partial B_r}f(z)\,dz.
$$
Hence, there exist an analytic function $F$ in $D\setminus\{0\}$, such that $F'=f$. Such an $F$ has a Laurent expansion of the form
$$
F(z)=\sum_{k\in \mathbb Z}b_kz^k
$$
and hence 
$$
f'(z)=F''(z)=\sum_{k\in \mathbb Z}k(k-1)b_kz^{k-2}.
$$
This implies that the coefficient of $z^{-1}$ is the expansion of $f'$ is $0$ and hence $a_0=0$, and thus $f$ is analytic in the disc. 
