5
$\begingroup$

Let $f$ be holomorphic non-constant on $D=\left\{ 0<\left|z\right|<10\right\}$ . Given that for all $n\in\mathbb{N}$: $\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}$ prove that $f$ has an essential singularity at $0$. Find an example of $f$ satisfying the condition.

My idea was to assume for contradiction the singularity is not essential, which means that either the limit $z\to0$ of $f$ exists or of $1/f$, and from $\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}$ we get that the limit of $f$ exists and is $0$.

I then define $h(z)=f(z)$ for $z\neq0$ and $h(0)=0$, which is holomorphic on the entire disk, and derive a contradiction from there and the uniqueness theorem. But then the suitable holomorphic function is the Gamma function, which we have not really discussed in class and I'm not sure about it's properties.

Any ways of solving this which avoids using the gamma function?

$\endgroup$
1
  • 1
    $\begingroup$ An example of such a function is $f(z)= \exp (-e^{1/z}).$ $\endgroup$
    – zhw.
    Jan 19, 2017 at 18:30

1 Answer 1

3
$\begingroup$

Any ways of solving this which avoids using the gamma function?

Let $g$ be a non-constant entire function with zeros at all integers, and look at

$$f(z) = g(1/z).$$

Then $f$ is holomorphic on $\mathbb{C}\setminus \{0\}$, so surely on the punctured disk $0 < \lvert z\rvert < 10$, and we have

$$\biggl\lvert f\biggl(\frac{1}{n}\biggr)\biggr\rvert = 0 < \frac{1}{n!}$$

for all $n$.

Clearly, $f$ cannot have a pole at $0$, since that would imply $\lvert f(z_n)\rvert \to +\infty$ for all sequences $(z_n)$ in the punctured disk converging to $0$. To exclude a removable singularity at $0$, one shows that if $h \colon \{ z : \lvert z\rvert < 10\} \to \mathbb{C}$ is holomorphic with $\bigl\lvert h\bigl(\frac{1}{n}\bigr)\bigr\rvert \leqslant \frac{1}{n!}$ for all $n\in \mathbb{N}$, then $h \equiv 0$. For that, use the power series expansion

$$h(z) = \sum_{k = 0}^{\infty} a_k z^k$$

and show that if there is an $m$ with $a_m \neq 0$, the inequality $\bigl\lvert h\bigl(\frac{1}{n}\bigr)\bigr\rvert \leqslant \frac{1}{n!}$ cannot hold for large enough $n$.

$\endgroup$
5
  • $\begingroup$ You are right! Now I have an example of a function satisfying the requirements ($e^{2i\pi/z}-1$), but I also realized it doesn't help me all that much arrive at a solution to the problem itself >< $\endgroup$
    – Nescio
    Jan 19, 2017 at 12:57
  • $\begingroup$ I thought $\sin \frac{\pi}{z}$ would be the first one thinks of. But that doesn't matter. What part is still a problem? It seemed as though you knew how to show that $0$ is not a removable singularity. And since a pole is clearly ruled out, that shows the singularity must be essential. Do you need help with showing the singularity cannot be removable? $\endgroup$ Jan 19, 2017 at 13:03
  • $\begingroup$ Yes, this is what I'm missing. I wanted to prove this by showing that a function satisfying the condition stated can't be analytic on the disk, but I'm not certain how to actually go about doing it $\endgroup$
    – Nescio
    Jan 19, 2017 at 13:11
  • $\begingroup$ I've added an outline of how to prove that the singularity can't be removable. $\endgroup$ Jan 19, 2017 at 13:21
  • $\begingroup$ that definitely worked for me. Probably should have been able to figure that out myself >< $\endgroup$
    – Nescio
    Jan 21, 2017 at 9:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .