Let $\left(a_n\right)_{n\in\mathbb{N}}\subset\mathbb{C}$ such that $$f(z):=\sum_{n=0}^\infty a_nz^{-n}$$ is compact convergent on $B_r(0)\setminus\left\{0\right\}$.

I want to show:

  1. $f$ is holomorphic and $0$ an isolated singularity of $f$
  2. Under which condition $0$ is a removable singularity
  3. $0$ is a essential singularity if and only if the number of $a_n$ with $a_n\ne 0$ is unlimited

Proof: (1) I understand that the hypothesis implies the existence of a $0<\varepsilon <r$ such that $a_nz^{-n}\to f$ uniformly on $K:=\overline{B_\varepsilon (0)\setminus\left\{0\right\}}$. Since $z\mapsto a_nz^{-n}$ is holomorphic on $K$, $f$ is holomorphic on $k$, too. But why if $f$ holomorphic on the entire disc $B_r(0)\setminus\left\{0\right\}$?

(2): By definition, $0$ is removable iff $f$ is bounded on $B_\varepsilon(0)\setminus\left\{0\right\}$ for some $0<\varepsilon <r$. Since $$|f|\le |a_0|+\sum_{n=1}^\infty |a_n|\frac{1}{\left|z\right|^n}$$ $f$ is bounded iff $a_n=0$ for all $n\ge 1$.

(3): By definition, $0$ is essential iff $0$ is not removable and $0$ is not a pole. Let's try to see the first direction: Suppose $0$ is essential. This means, $f$ in unbounded for all $0<\varepsilon <r$. [I think this implies $a_n\ne 0$ for an unlimited number of $n\ge 1$, but I'm not sure - we would not need to use that $f$ is not a pole].

Let's try the second direction: Since $$\left|\sum_{k=1}^na_nz^{-n}\right|\ge \frac{1}{|z|^n}\left\{|a_n|-\sum_{k=1}^{n-1}|a_k||z|^k\right\}$$ we've got $|f(z)|\to\infty $ for $z\to 0$. But this would imply $0$ is a pole - so, something is wrong here.

  • $\begingroup$ What is all the $(a_n)$ are zero? (Are you sure that $0$ is always a singularity?) What about finitely many? (What could you do to make this a product of two things you understand?) What if there are infinitely many that are zero? (Generically, can you transform $z$ to another variable, producing something you better understand?) $\endgroup$ – Eric Towers Jul 5 '14 at 21:50
  • $\begingroup$ Uniform limit of a sequence of holomorphic functions is holomorphic (apply this in a relatively compact domain in punctured disk). $\endgroup$ – Moishe Kohan Jul 6 '14 at 2:00
  • $\begingroup$ You need a relatively compact domain to ensure uniform convergence (this is your hypothesis). Then you conclude that the limit is holomorphic outside of zero. $\endgroup$ – Moishe Kohan Jul 6 '14 at 11:53
  • $\begingroup$ @studiosus I've updated my question and shared my ideas about all (3) sub questions - maybe you can give me some feedback to my questions. $\endgroup$ – 0xbadf00d Jul 6 '14 at 14:39
  • $\begingroup$ What is your definition of "pole"? $\endgroup$ – bryanj Jul 6 '14 at 14:42

The way to approach it might depend on what material (theorem's, etc.) are available at this point. Here are two different approaches.

METHOD #1: Here's one way.

Pick any $a$ in $0 < |z| < r$. There is a small closed disk around $a$ which is also contained in $0 < |z| < r$. The series for $f$ converges uniformly on this small closed (so compact) disk, and so converges uniformly on the small open disk around $a$ of the same radius. By Weierstrass's uniform limit theorem, we know that $f$ is holomorphic in this small open disk around $a$. These small open disks cover the domain $0 < |z| < r$, so $f$ is analytic in the original punctured disk of radius $r$.

It's clear that if all the coefficients $a_n$, $n = 1, 2, 3, \dots$, are zero that $f$ is a constant. In this case $z = 0$ is obviously a removable singularity.

On the other hand, say $z = 0$ is a removable singularity. Then according to your definition, $f$ is bounded in some punctured disk around $0$. Integrate the function $z^n f(z)$ on the circle around $z = 0$ which has radius $\delta$. The circle is compact, so we can interchange the integral and the limit of the sum to get $$ \int_{|z| = \delta} z^n f(z) dz = \int_{|z| = \delta} \frac{z^n a_{n+1}}{z^{n+1}} dz = \int_{|z| = \delta} \frac{a_{n+1}}{z} dz = 2 \pi i a_{n+1} $$ (All the other terms have anti-derivatives, so the integral vanishes.)
But on the other hand, there's the inequality $$ 2 \pi |a_{n+1}| = \Bigg|\int_{|z| = \delta} z^n f(z) dz \Bigg| \le2 \pi \delta^{n+1} M $$ where $M$ bounds $|f(z)|$ near the origin. Taking the limit as $\delta \to 0$ shows that $a_n = 0$ for $n = 1, 2, 3, \dots$. So $f(z) = a_0$, a constant function.
$z = 0$ is removable if and only if $f$ is a constant, which happens if and only if all coefficients except $a_0$ vanish.

It's clear that if some finite number of the $a_n$ for $n > 1$ are non zero, that $f$ must have a pole at the origin.
Conversely supposes that $f$ has a pole at the origin. Then according to your definition, $f(z) \to \infty$ as $z \to 0$. In some smaller punctured disk, then, the function $g(z) = 1/f(z)$ has a removable singularity at the origin and so is in fact holomorphic there with $g(0) = 0$. We can then write $g(z) = z^k /h(z)$, where $h$ is analytic and nonzero in a (non-punctured) disk around the origin.
$h(z)$ will have a power series development around $z = 0$, and so we see that $$ f(z) = h(z)/z^k = a_{k}/z^k + \cdots + a_1/z + a_0 + b_1 z + b_2 z^2 + \dots $$ By integrating $f(z)/z^j$ on a circle, for $j = 2, 3, \dots$ - again interchanging integral with the sum of the series - we conclude that all the $b_j$ are zero, because they are zero in the original series representation for $f$ provided in the question.

We have shown that if $z = 0$ is a pole, then only finitely many of the coefficients $a_n$ are non-zero. So the origin is a pole if and only if finitely many of the $a_n$ are non-zero.

The only other possibility is an essential singularity, which must occur if and only if infinitely many coefficients are non-zero.

METHOD #2: Here's another way

Note that this doesn't explicitly use that fact that $f(z)$ converges uniformly on compact subsets contained in your original disk. Instead it relies on knowing a little about how power series work.

Consider $g(z) = \sum_{n=0}^{\infty} a_n z^n = f(1/z)$.
Since $f(z)$ converges on $0 < |z| < r$, see that $g(z)$ converges on $\frac{1}{r} < |z|$.
But, since $g$ is a power series, this means the radius of convergence must be at least $1/r$. So in fact $g(z)$ converges in the whole plane, and so represents a function holomorphic in the whole plane (since power series are holomorpic inside their radius of convergence).

Look at what happens to $g(z)$ as $z \to \infty$. There are three possibilities:
A) $|g(z)|$ is bounded on the whole plane.
B) $g(z) \to \infty$ as $z \to \infty$.
C) Neither A) nor B).

In case A), $g(z)$ is constant by Liouville's theorem, and must be identically equal to $a_0$. Thus all the $a_n$ vanish for $n > 0$.

In case B), $g$ must be a polynomial. If you don't have the theorem that says this, then you can get it a couple of ways:
1) Since $g$ is nonzero outside some bounded disk, all zeros must lie in some fixed compact disk. By the uniqueness/identity theorem there can only be finitely many zeros (else $g$ is identically zero). Divide out by these zeros, take the multiplicative inverse & apply "extended Liouville theorem" to show the quotient is a constant.
2) Alternatively, take the multiplicative inverse of $g$, and subtract of the principle parts of $1/g$ near the zeros of $g$ and get a bounded entire function (which must be identically zero).


Suppose $f$ has a removable singularity at $z = 0$. Then by your definition of removable singularity, $f$ is bounded near zero, so $g$ is bounded on the whole plane. It follows that $g$ is (by Case A) a constant, and so all $a_n$ are zero when $n > 0$.
Conclusion: f has a removable singularity if and only if all $a_n$ are zero for $n > 0$.

Suppose next that $f$ has a pole at $z = 0$. According to your definition, this means that $f(z) \to \infty$ as $z \to 0$. Thus $g(z) \to \infty$ as $z \to \infty$. By Case B), $g(z)$ is a polynomial $g(z) = a_0 + a_1 z + \cdots + a_k z^k$. So finitely many of the $a_k$ are non-zero.
It's easy to show the converse: If finitely many $a_n$ are non-zero then $f$ has a pole.
Conclusion: f has a pole if and only if finitely many $a_n$ are non-zero.

The only remaining possibility is that $f$ has an essential singularity, and this must occur exactly when infinitely many coefficients are non-zero.

  • $\begingroup$ Ugh. Sorry, I think my answer was pretty gross. Maybe some kind person will come along and provide a much better one. $\endgroup$ – bryanj Jul 6 '14 at 21:34

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