Let f be analytic in $\mathbb{D}\setminus\{0\}$ and suppose there exists $M>0$
and $m\geq 1$ such that $\left|f^{(m)}(z)\right|\leq \frac{M}{|z|^m}$ for all
$0<|z|<1$. Show that f has a removable singularity at $0$.
My attempt:
Since $f$ is analytic in $\mathbb{D}\setminus\{0\}$ it is holomorphic in in $\mathbb{D}\setminus\{0\}$. Further, we have that$$(\forall z\in\mathbb{D}\setminus\{0\}):|z^m|\left|f^{(m)}(z)\right|\leq M.$$Thus, the function $g(z):=z^mf^{(m)}(z)$ has a removable singularity at $z=0$ by Riemann's singularity theorem and so it can be extended to a function $h(z)$ that is holomorphic on the entire unit disk.
And here, I do not know how to continue, since I was trying to show that $h(z)$ had a zero at $z=0$ but I was not getting very far. So, if anyone has any suggestions on how to move forward, I would appreciate it.
Best Answer
If $f$ has no removable singularity at $0$, then either it has an essential singularity there or a pole. If it has an essential singularity at $0$, then $z^mf^{(m)}(z)$ also has an essential singularity at $0$. But it is also bounded near $0$, and this is impossible, by the Casorati-Weierstrass theorem.
And if it has a pole of order $k$ at $0$, then, near $0$, you have $f(z)=\sum_{n=-k}^\infty a_nz^n$. But then $f^{(m)}$ has a pole of order $k+m$ at $0$, and so $\lim_{z\to0}z^{m+k}f^{(m)}(z)$ exists in $\Bbb C\setminus\{0\}$. So, you cannot have $|f^{(m)}(z)|\leqslant\frac M{|z|^m}$, since this implies that $\lim_{z\to0}z^{m+k}f^{(m)}(z)=0$.