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 1and 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.

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