Elliptic function - does limit at infinity exist?

1. Mar 26, 2017

binbagsss

1. The problem statement, all variables and given/known data

I am wanting to show that

$lim_{z\to\infty} f(z)=c$ does not exist for $c \in C$, $C$ the complex plane, where $f$ is non-constant periodic meromorphic function. (elliptic)

2. Relevant equations

3. The attempt at a solution

So I want to proove this is not true
.
Beginning with the definition:

For any $\epsilon >0$ can find a $R$ s.t $|z|>R \implies |f(z)-c|<\epsilon$ (1)

Now let $w$ be a period.

Then we also have

$|f(z+mw)-c|<\epsilon$(2), for $m$ some integer

Now I attach the solution:

Questions:

- I thought that (2) would hold automatically, from (1) by periodicty, without additionally needing to have $|z+mw |>R$

- Also this would hold following from (1), if not using periodicity, but simply looking at $|z+mw |>R$, for $z$ large since a shift by some constant is insignificant? Therefore in the solution, I do not understand the need to specify both, by periodicity and include this condition $|z+mw |>R$

- Finally I do not understand the conclusion itself, perhaps due to being unclear on the above points- I understand the goal is to show that if this limit exists it implies that $f$ is holomoprhic (since I then have the theorem that an elliptic function holomorphic is constant, and so proof by contradiction).

I do not understand the conclusion that

$|f(z)-c|=|f(z+mw)-c|<1$ (3) implies the $f(z)$ is a bounded holomorphic function.

My interpretation is that since we can vary $m$ to cover a range of values of $f(z)$ which are bounded this conclusion is made? So $m$ generating alot of points? However isn't this only for large $z$ ? We have only showed bounded for large $z$? (When $|z|>R$ is satisfied)? Or am I barking up the wrong tree as to why we conclude $f(z)$ is constant from (3)?

2. Mar 26, 2017

Staff: Mentor

Your (2) is valid for |z|>R, but that is not really what you want to show.

You want to show that |f(z)-c| < $\epsilon$ for all z, including those not larger than R in magnitude. So you need the reverse direction. Start with those z, then find (using periodicity) a complex value with a magnitude larger than R.

If f(z) does not differ from c by more than 1 for all z, it is bounded. A bounded meromorphic function is constant.

3. Mar 26, 2017

binbagsss

So this is $|z+mw|$ but then how does this show that |f(z)-c| < $\epsilon$ for all z, including those not larger than R in magnitude?

Because we can choose any $m$ integer to make this small?

4. Mar 26, 2017

Staff: Mentor

We can choose an m such that |z+mw|>R, and there we know that |f(z+mw)-c|< ϵ.

Last edited: Mar 27, 2017
5. Mar 27, 2017

binbagsss

but that's large $m$ isn't it? so again only large $z$, not all $z$?

I'm not understanding, what's happened to the $c$?

6. Mar 27, 2017

Staff: Mentor

It is sufficient that some m exists, it does not matter how large it is. z can be anything.
I forgot it. Should be in there of course.

7. Mar 28, 2017

binbagsss

Ok, think I may understand at last, so whilst $|f(z+mw)-c|<\epsilon$ could simply follow from a condition only on $z$ , $|z|>R$ , by periodicity, this is still only considering large $z$.

However if we instead have $|z+mw|>R$, than the periodicity can be used to make |z+mw| large through $m$ and not $z$, and so it holds for small $z$ too?

8. Mar 28, 2017

Staff: Mentor

I don't understand your last post.

Let's make an example, maybe that is easier to understand.

$\epsilon=0.01$, R=1, w=0.5, c=3.
For |z|>1, we know that |f(z)-3|<0.01, or, in words, f(z) is close to 3. We want to show that f(z) is close to 3 everywhere.

What about z=0.2? We don't have direct condition on that, because 0.2<R. But we know f(0.2)=f(0.7)=f(1.2), and |f(1.2)-3|<0.01, i. e. f(1.2) is close to 3. Therefore, |f(0.2)-3|<0.01, which means f(0.2) is close to 3 as well.
z=-0.4+0.2i? Same idea: f(-0.4+0.2i) = f(1.1+0.2i), and |1.1+0.2i|>1. Again we can conclude that |f(-0.4+0.2i)-3|<0.01.

This approach works for all z with |z|<1. Therefore, |f(z)-3|<0.01 for all z.
We can repeat this with every $\epsilon>0$, and conclude that |f(z)-3|<$\epsilon$ for all $\epsilon$. Therefore, f(z)=3 everywhere.

9. Apr 10, 2017

binbagsss

This doesn't need to be $1$ right? The more general definition is |f(z)-c| < $\epsilon$ for $0< \epsilon <1$?

10. Apr 10, 2017

binbagsss

Ahh okay thanks, I think this answers my next question: $m$ is integer so for a fixed $\epsilon$ with the procedure you descried you could only cover a discrete set of points in the complex plane but then by varying $\epsilon$ you can cover continuous points so the whole complex plane? thanks.

11. Apr 10, 2017

Staff: Mentor

It works with every positive real number, yes.
No, you can cover the whole plane with a single value of $\epsilon$.
Check the example in post 8. I picked z=0.2 and z=-0.4+0.2i as examples, but it works for every z.

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