$|f^{(N)}(z)| \leq |z|^{-N}$ in the punctured unit circle, then $0$ is a removable singularity.

Question

Take $f$ analytic in $D_1(0)-\{0\}$ such that: $|f^{(N)}(z)| \leq |z|^{-N}$ then $0$ is a removable singularity.

I know that I can write $f(z) = \sum_{n = - \infty}^{\infty}c_n z^n$

Moreover, I can tell that the function $g(z) = z^N f^{(N)}(z)$ is such that: $|g(z)| \leq 1$. But how do I use this information to deduce that either all $c_n$ such that $n < 0$ are zero or find that $zf(z) \rightarrow 0$ as $z \rightarrow 0$?

Thank you!

Answer 1

You already observed that $g(z)=z^Nf^{(N)}(z)$ is bounded in the unit disc. Then $g$ is already analytic.

Thus, we can write $$g(z)=z^{N}f^{(N)}(z)=\sum_{n=0}^\infty a_nz^{n}.$$ If $$f=\sum_{n=-\infty}^\infty c_nz^n,$$ then $$f^{(N)}(z)=\sum_{n=-\infty}^\infty n(n-1)\cdots(n-N+1)c_nz^{n-N}.$$

Compare the coefficients. We can see that $c_n=0$ for all $n<0$.

Answer 2

By Weierstrass-Casorati, there are only finitely many $-$ve terms (the singularity cannot be essential). Let the most negative power effectively present be $-n$, and you have a contradiction unless $n=0$, as $|f^{(N)}(z)|$ will go like $|z|^{-n-N}$ for $|z|$ small enough.