# Ngô Quốc Anh

## August 13, 2009

### Two examples of analytic function on a punctured unit disk which has a removable singularity at the origin

Filed under: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh — Ngô Quốc Anh @ 15:54

The following question was proposed in NUS under the QE in AY 2007-2008:

Consider the punctured disk $D=\{ z \in \mathbb C | 0 <|z| <1\}$. Suppose $f : D \to \mathbb C$ is an analytic function such that

$\displaystyle |f''(z)| \leq \frac{2}{|z|^2}$

for all $z \in D$. Is it true that f has a removable singular point at $z=0$?

Proof. Denote the Laurent expansion of $f$ by

$\displaystyle f\left( z \right) = \sum\limits_{n = - \infty }^{ + \infty } {{a_n}{z^n}}$

where

$\displaystyle {a_n} = \frac{1} {{2\pi i}}\int\limits_{\left| z \right| = r < 1} {\frac{{f\left( z \right)dz}} {{{z^{n + 1}}}}}$.

Then from

$\displaystyle f''\left( z\right) =\sum\limits_{n =-\infty }^{+\infty }{\underbrace{\left({n+2}\right)\left({n+1}\right){a_{n+2}}}_{{b_{n}}}{z^{n}}}$

we get

$\displaystyle {b_n} = \frac{1} {{2\pi i}}\int\limits_{\left| z \right| = r < 1} {\frac{{f''\left( z \right)dz}} {{{z^{n + 1}}}}}$.

Thus,

$\displaystyle\left|{{b_{n}}}\right|\leqslant\frac{1}{{2\pi }}\int\limits_{\left| z\right| = r < 1}{\left|{\frac{{f''\left( z\right)}}{{{z^{n+1}}}}}\right|}dz\leqslant\frac{1}{\pi }\int\limits_{\left| z\right| = r < 1}{\left|{\frac{1}{{{z^{n+3}}}}}\right|}dz\leqslant\frac{1}{{\pi{r^{n+3}}}}2\pi r =\frac{2}{{{r^{n+2}}}}$.

When $n \leq -3$, let $r \to 0$ we see that $z=0$ is a removable singularity of $f$ since $b_n=0 (n \leq -3)$ implies $a_n=0 (n \leq -1)$.

Remark. The second derivative can be replaced by an $\mathbb Z\ni m \geq 1$and therefore $\frac{2}{|z|^2}$ should be $\frac{2}{|z|^m}$. This is a question of UCLA QE in Winter 2007.

We also have a similar question proposed in a QE of Indiana University. It says that if $f : D \to \mathbb C$ is an analytic function such that

$\displaystyle |f(z)|\leq\log\frac{1}{|z|}$

for all $z \in D$. Then $f \equiv 0$.

Proof
. As above, one has

$\displaystyle \left|{{b_{n}}}\right| =\left|{\frac{1}{{2\pi i}}\int\limits_{\left| z\right| = r < 1}{\frac{{f\left( z\right)dz}}{{{z^{n+1}}}}}}\right|\leqslant\frac{1}{{{r^{n}}}}\log\frac{1}{r}$.

When $n<0$, letting $r \to 0$ we have $a_n= 0$ for all $n \leq -1$ which implies $z=0$ is a removable singularity of $f$. In other words, $f$ can be extended to an analytic function of the unit disk. Since $\log \frac{1}{|z|} =0$ when $|z|=1$, by the Maximum Modules Principle we obtain $f \equiv 0$.