1
$\begingroup$

If $f:\mathbb{C}\setminus\{z_0\}\to \mathbb{C}$ is an analytic function with an essential singularity at $z_0$, the Picard's Theorem asserts that in any nbhd of $z_0$, $f$ takes any complex value infinitely often, with perhaps one exception.

Two separate, but related questions, A) and B). Is it possible to have two functions $f,g:\mathbb{C}\setminus\{z_0\}\to \mathbb{C}$, analytic, and with $z_0$ as a essential singularity such that:

A) for any $w\in\mathbb{C}$ and any complex number $\alpha$ there exists $U$ nbhd of $z_0$ such that we have: $$ \{z\in U | f(z)=w\}\cap \{z\in U | g(z)=\alpha w\}=\emptyset $$

B) (suggested by Josh Keneda in the comments) for any $w_1\neq w_2$, there exists $U$ nbhd of $z_0$ such that we have: $$ \{z\in U | f(z)=w_1\}\cap \{z\in U | g(z)=w_2\}=\emptyset $$

Clearly a positive answer for B) will be a also an example for A).

$\endgroup$
3
  • $\begingroup$ Based on the order of your quantifiers, this is impossible. Take $f, g$ with essential singularities at $z_0$. If $w$ is any non-zero value that $f$ takes on in every neighborhood of $z_0$, then for any neighborhood $U$ of $z_0$, we can find $z_1\in U$ with $f(z_1) = w$. Take $\alpha = g(z_1)/w$, and you'll have $z_1 \in \{z\in U | f(z)=w\} \cap \{z \in U | g(z) = \alpha w\}$. $\endgroup$ Sep 15, 2016 at 14:02
  • $\begingroup$ A slightly different question would be whether it's possible to have $f, g$ with essential singularity at $z_0$ such that for all $\alpha, \beta$, there exists a neighborhood $U$ of $z_0$ such that $\{z \in U | f(z) = \alpha $ and $g(z) = \beta\} = \emptyset$. Maybe you meant to ask this? $\endgroup$ Sep 15, 2016 at 14:07
  • $\begingroup$ @JoshKeneda You are right, the order of the last two quantifiers should be reversed. Will make the changes, thank you! $\endgroup$
    – Markus
    Sep 15, 2016 at 16:56

1 Answer 1

1
$\begingroup$

Yaikes, the second time I misread the question, but now I can't delete my answer :)

As Markus points out below $e^{1/z}$ and $ze^{1/z}$ are examples.

$\endgroup$
7
  • 1
    $\begingroup$ Thank you! In my case I have a linear independence, which your example fails. More precisely, if $F:\mathbb{C}\to H$ is an analytic function and $x,y$ are linearly independent vectors, can the analytic functions $<F(\cdot), x>$ and $<F(\cdot), y>$ have the above property? $H$ is a separable Hilbert space here. $\endgroup$
    – Markus
    Sep 16, 2016 at 2:02
  • $\begingroup$ Can't $F = (f,g)$, and $x=(1,0)$ and $y=(0,1)$ with $H$ being $\mathbb{C}^2$? $\endgroup$
    – Jiri Lebl
    Sep 19, 2016 at 18:14
  • $\begingroup$ I think you are right. By the way, I think your original example has to be modified a bit. If $w_1=1$ and $w_2=e^{\beta}$, then the $f,g$ in your example attain these values, respectively, at the same $2\pi i k$ points. $\endgroup$
    – Markus
    Sep 22, 2016 at 17:06
  • $\begingroup$ You are absolutely correct, the example is not correct. $\endgroup$
    – Jiri Lebl
    Sep 22, 2016 at 18:47
  • $\begingroup$ I edited the answer. In fact there is no counterexample except $f=g$. I hope it's correct now. $\endgroup$
    – Jiri Lebl
    Sep 22, 2016 at 19:26

You must log in to answer this question.

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