Rigorously justifying switching limits in complex analysis

Question

One has, by definition, that if $x_n\rightarrow c\in\mathbb{C}$ then $f(x_n)\rightarrow f(c)$. But how could you rigorously justify, for example, the following statement (cited from p. 139 of Greene & Krantz)?

$$\lim_{|z|\rightarrow\infty}|f(z)| = \infty \Leftrightarrow f(1/z)\text{ has a pole at 0}$$

It's true that this fact serves as a basis for the definition of a pole at infinity, but the book is just beginning the discussion of singularities at infinity and has presented this statement as a fact that should already be apparent with no knowledge of singularities at infinity.

Suppose you wanted to prove right-to-left first. Then $$\lim_{|z|\rightarrow 0}|f(1/z)| = \infty$$ but since the limit of f(1/z) needn't even exist (to say nothing of that it can't be finite) you cannot bring the limit within the absolute value and go from there. Or can you?

This statement appears to be "obvious," but I want to be able to prove it to homework/qual standards.

Answer 1

What you need to use is the limit of composite functions. Hypothesis 2 applies to your case.

In other words if $|z|\rightarrow 0$, $z\ne 0$, then $w=\frac{1} {z}$ has $|w|\rightarrow\infty$, and of course $|w|\ne\infty$ (as $z\ne 0$). So:

$$\lim_{|z|\rightarrow 0}|f(1/z)|=\lim_{|w|\rightarrow\infty}|f(w)|=\infty$$

Answer 2

I read a little bit further on and think that I might have part of an answer (although it would be nice if someone confirmed it just to make sure I'm not reasoning in a circle or something).

You have to think in terms of the topology on the Riemann sphere (which, I should add, had not been introduced by p.139 so I'm wondering if there's a more elementary explanation even if it's uglier). In that topology we have $z_n\rightarrow \infty$ if and only if $|z_n|\rightarrow\infty$ (in particular the concept of -$\infty$ is undefined). So in fact $f(1/z)\rightarrow\infty$ by the definition of a pole and $||$ is trivially continuous at infinity ... so we can switch limits up until we obtain $$\Big| f(1/\big[\lim_{|z|\rightarrow 0}z\big])\Big| = \infty$$ Going anywhere from here, however, seems to rely on that $f$ is continuous at infinity, but that's presumably what we are trying to prove in the first place.

Answer 3

The authors are alluding to the theorem that describes the behavior of a holomorphic function near an essential singularity (Casorati-Weierstrass). So, I doubt you could prove this just by manipulating absolute value signs and limits.

If $f=\sum_0^\infty a_n z^n$ for $z\in \mathbb{C}$, then $f(1/z)=\sum_0^\infty \frac{a_n}{z^n}$ for $z\in\mathbb{C}\backslash 0$. Ignoring constant functions, $f(1/z)$ either has a pole or an essential singularity at $0$.

$f(1/z)$ has a pole at $0$ $\implies$ the latter sum has finitely many terms $\implies$ the former sum has finitely many terms $\implies f(z)$ is a polynomial $\implies f(z)\to\infty$ as $z\to\infty$

$f(1/z)$ has an essential singularity at $0 \implies$ the range of $f(1/z)$ near $0$ is dense in $\mathbb{C}$ $\implies f(1/z)$ has no limit near $0$ $\implies f(z)$ has no limit at $\infty$

"Limit at $\infty$" can be defined easily without reference to the Riemann sphere: $f$ has a finite limit $w$ at $\infty$ if for every $\epsilon$ there is a $\delta$ such that $|f(z)-w|<\epsilon$ for $|z|>\delta$; $f$ goes to $\infty$ at $\infty$ if $|f(z)|>\epsilon$ for $|z|>\delta$. Of course, this is just describing convergence in the topology of the Riemann sphere.